A TEXTBOOK OF
STRENGTH OF MATERIALS
[MECHANICS OF SOLIDS]
i
ii
A TEXTBOOK OF
STRENGTH OF MATERIALS
[Mechanics of Solids]
(In S.I. Units)
[For Degree, U.P.S.C. (Engg. Services), GATE and
Other Competitive Examinations]
By
Dr. R.K. BANSAL
B.Sc. Engg. (Mech.), M. Tech., Hons. (I.I.T., Delhi)
Ph.D., M.I.E. (India)
Formerly Professor and Head
Department of Mechanical Engineering,
(University of Delhi)
Delhi College of Engineering, Delhi
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iii
A TEXTBOOK OF STRENGTH OF MATERIALS
Compiled by : Smt. Nirmal Bansal
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v
vi
PREFACE TO THE SIXTH EDITION
The popularity of the fifth edition and reprints of the book A Textbook of Strength
of Materials amongst the students and the teachers of the various Indian universities, has
prompted the bringing out of the sixth edition of the book so soon. The sixth edition has been
thoroughly revised and brought up-to-date. A large number of problems from different B.E.
degree examinations of Indian universities and other examining bodies, such as Institution
of Engineers, U.P.S.C. (Engineering Services) and GATE have been selected and have been
solved at proper places in this edition in S.I. Units.
Four advanced topics of Strength of Materials such as stresses due to rotation in thin and
thick cylinders, bending of curved bars, theories of failure of the material and unsymmetrical
bending and shear centre have been added. These chapters have been written in such a simple
and easy-to-follow language that even an average student can understand easily by self-study.
In the chapter of ‘Columns and Struts’, the advanced articles such as columns with
eccentric load, with initial curvature and beam columns have been included. Also in the chapter
of ‘Principal Stresses and Strains’, strain on an oblique plane and Mohr’s strain circle have
been added.
The notations in this edition have been used up-to-date by the use of sigma and tau for
stresses.
The objective type multiple-choice questions are often asked in the various competitive
examinations. Hence a large number of objective type questions with answers have been added
at the end of the book.
Also a large number of objective type questions which have been asked in most of
competitive examinations such as Engineering Services Examination and GATE with answers
and explanation have been incorporated in this edition.
With these editions, it is hoped that the book will be quite useful for the students of
different branches of Engineering at various Engineering Institutions.
I express my sincere thanks to my colleagues, friends, students and the teachers of
different Indian universities for their valuable suggestions and recommending the book to
their students.
Suggestions for the improvement of this book are most welcome and would be incorporated in the next edition with a view to make the book more useful.
—Author
PREFACE TO THE FIRST EDITION
I am glad to present the book entitled, A Textbook of Strength of Materials to the
engineering students of mechanical, civil, electrical, aeronautical and chemical and also to the
students of A.M.I.E. Examination of Institution of Engineers (India). The course-contents have
been planned in such a way that the general requirements of all engineering students are
fulfilled.
During my long experience of teaching to the engineering students for the past 20 years,
I have observed that the students face difficulty in understanding clearly the basic principles,
fundamental concepts and theory without adequate solved problems along with the text. To
meet this very basic requirement to the students, a large number of the questions taken from
the examinations of the various universities of India and from other professional and competitive
examinations (such as Institution of Engineers, and U.P.S.C. Engineering Service Examinations)
have been solved along with the text, in S.I. units.
This book is written in a simple and easy-to-follow language, so that even an average
students can grasp the subject by self-study. At the end of each chapter, highlights, theoretical
questions and many unsolved numerical problems with answers are given for the students to
solve them.
I am thankful to my colleagues, friends and students who encouraged me to write this
book. I am grateful to Institution of Engineers (India), various universities of India and those
authorities whose work have been consulted and gave me a great help in preparing this book.
I express my appreciation and gratefulness to my publisher, Shri R.K. Gupta (a Mechanical
Engineer) for his most co-operative, painstaking attitude and untiring efforts for bringing out
the book in a short period.
Smt. Nirmal Bansal deserves special credit as she not only provided an ideal atmosphere at home for book writing but also gave inspiration and valuable suggestions.
Though every care has been taken in checking the manuscripts and proof reading, yet
claiming perfection is very difficult. I shall be very grateful to the readers and users of this book
for pointing any mistakes that might have crept in. Suggestions for improvement are most
welcome and would be incorporated in the next edition with a view to make the book more
useful.
—Author
viii
CONTENTS
Chapters
Pages
Chapter 1. Simple Stresses and Strains
1.1.
1.2.
1.3.
1.4.
1.5.
1.6.
1.7.
1.8.
1.9.
1.10.
1.11.
1.12.
1.13.
1.14.
1.15.
1.16.
1.17.
Introduction
Stress
Strain
Types of Stresses
Elasticity and Elastic Limit
Hooke’s Law and Elastic Modulii
Modulus of Elasticity (or Young’s Modulus)
Factor of Safety
Constitutive Relationship between Stress and Strain
Analysis of Bars of Varying Sections
Analysis of Uniformly Tapering Circular Rod
Analysis of Uniformly Tapering Rectangular Bar
Analysis of Bars of Composite Sections
Thermal Stresses
Thermal Stresses in Composite Bars
Elongation of a Bar Due to its Own Weight
Analysis of Bar of Uniform Strength
Highlights
Exercise
1—58
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
Chapter 2. Elastic Constants
2.1.
2.2.
2.3.
2.4.
2.5.
2.6.
2.7.
2.8.
2.9.
2.10.
Introduction
Longitudinal Strain
Lateral Strain
Poisson’s Ratio
Volumetric Strain
Volumetric Strain of a Cylindrical Rod
Bulk Modulus
Expression for Young’s Modulus in Terms of Bulk Modulus
Principle of Complementary Shear Stresses
Stresses on Inclined Sections when the Element is Subjected to Simple
Shear Stresses
2.11. Diagonal Stresses Produced by Simple Shear on a Square Block
2.12. Direct (Tensile and Compressive) Strains of the Diagonals
2.13. Relationship between Modulus of Elasticity and Modulus of Rigidity
Highlights
Exercise
Chapter 3. Principal Stresses and Strains
3.1.
3.2.
3.3.
3.4.
Introduction
Principal Planes and Principal Stresses
Methods of Determining Stresses on Oblique Section
Analytical Method for Determining Stresses on Oblique Section
1
1
2
2
5
6
6
6
6
14
24
27
30
42
44
50
51
53
54
59—84
...
...
...
...
...
...
...
...
...
59
59
59
60
62
68
70
70
73
...
...
...
...
...
...
74
76
77
78
81
82
85—142
...
...
...
...
85
85
85
85
ix
Chapters
3.5.
3.6.
3.7.
Pages
Mohr’s Circle
Strain on an Oblique Plane
Mohr’s Strain Circle
Highlights
Exercise
Chapter 4. Strain Energy and Impact Loading
4.1.
4.2.
4.3.
4.4.
4.5.
4.6.
Introduction
Some Definitions
Expression for Strain Energy Stored in a Body when the Load is Applied
Gradually
Expression for Strain Energy Stored in a Body when the Load is Applied
Suddenly
Expression for Strain Energy Stored in a Body when the Load is Applied
with Impact
Expression for Strain Energy Stored in a Body due to Shear Stress
Highlights
Exercise
Chapter 5. Centre of Gravity and Moment of Inertia
5.1.
5.2.
5.3.
5.4.
5.5.
5.6.
5.7.
5.8.
5.9.
5.10.
5.11.
5.12.
5.13.
5.14.
5.15.
Centre of Gravity
Centroid
Centroid or Centre of Gravity of Simple Plane Figures
Centroid (or Centre of Gravity) of Areas of Plane Figures
by the Method of Moments
Important Points
Area Moment of Inertia
Radius of Gyration
Theorem of the Perpendicular Axis
Theorem of Parallel Axis
Determination of Area Moment of Inertia
Mass Moment of Inertia
Determination of Mass Moment of Inertia
Product of Inertia
Principal Axes
Principal Moments of Inertia
Highlights
Exercise
Chapter 6. Shear Force and Bending Moment
6.1.
6.2.
6.3.
6.4.
6.5.
6.6.
6.7.
6.8.
x
Introduction
Shear Force and Bending Moment Diagrams
Types of Beams
Types of Load
Sign Conventions for Shear Force and Bending Moment
Important Points for Drawing Shear Force and Bending Moment Diagrams
Shear Force and Bending Moment Diagrams for a Cantilever with a
Point Load at the Free End
Shear Force and Bending Moment Diagrams for a Cantilever with a
Uniformly Distributed Load
...
...
...
...
...
123
133
137
137
139
143—170
...
...
143
143
...
143
...
145
...
...
...
...
152
165
166
167
171—236
...
...
...
171
171
171
...
...
...
...
...
...
...
...
...
...
...
...
...
...
171
173
195
196
196
197
198
212
213
219
220
221
229
230
237—294
...
...
...
...
...
...
237
237
237
238
239
240
...
241
...
244
Chapters
6.9.
6.10.
6.11.
6.12.
6.13.
6.14.
6.15.
6.16.
6.17.
6.18.
Pages
Shear Force and Bending Moment Diagrams for a Cantilever
Carrying a Gradually Varying Load
Shear Force and Bending Moment Diagrams for a Simply
Supported Beam with a Point Load at Mid-point
Shear Force and Bending Moment Diagrams for a Simply
Supported Beam with an Eccentric Point Load
Shear Force and Bending Moment Diagrams for a Simply
Supported Beam Carrying a Uniformly Distributed Load
Shear Force and Bending Moment Diagrams for a
Simply Supported Beam Carrying a Uniformly
Varying Load from Zero at Each End to w Per Unit Length at the Centre
Shear Force and B.M. Diagrams for a Simply Supported Beam
Carrying a Uniformly Varying Load from Zero at one End to w Per Unit
Length at the Other End
Shear Force and Bending Moment Diagrams for Over-hanging Beams
S. F. and B. M. Diagrams for Beams Carrying Inclined Load
Shear Force and Bending Moment Diagrams for Beams Subjected
to Couples
Relations between Load, Shear Force and Bending Moment
Highlights
Exercise
Chapter 7. Bending Stresses in Beams
7.1.
7.2.
7.3.
7.4.
7.5.
7.6.
7.7.
7.8.
7.9.
7.10.
7.11.
Introduction
Pure Bending or Simple Bending
Theory of Simple Bending with Assumptions Made
Expression for Bending Stress
Neutral Axis and Moment of Resistance
Bending Stresses in Symmetrical Sections
Section Modulus
Section Modulus for Various Shapes or Beam Sections
Bending Stress in Unsymmetrical Sections
Strength of a Section
Composite Beams (Flitched Beams)
Highlights
Exercise
Chapter 8. Shear Stresses in Beams
8.1.
8.2.
8.3.
Introduction
Shear Stress at a Section
Shear Stress Distribution for Different Sections
Highlights
Exercise
Chapter 9. Direct and Bending Stresses
9.1.
9.2.
9.3.
9.4.
Introduction
Combined Bending and Direct Stresses
Resultant Stress when a Column of Rectangular Section is Subjected to
an Eccentric Load
Resultant Stress when a Column of Rectangular Section is Subjected to a
Load which is Eccentric to both Axes
...
252
...
254
...
256
...
258
...
266
...
...
...
268
272
281
...
...
...
...
286
289
290
291
295—344
...
...
...
...
...
...
...
...
...
...
...
...
...
295
295
296
297
298
300
303
303
315
323
330
340
341
345—380
...
...
...
...
...
345
345
351
376
377
381—412
...
...
381
381
...
382
...
390
xi
Chapters
9.5.
9.6.
9.7.
9.8.
9.9.
Pages
Resultant Stress for Unsymmetrical Columns with Eccentric Loading
Middle Third Rule for Rectangular Sections (i.e., Kernel of Section)
Middle Quarter Rule for Circular Sections (i.e., Kernel of Section)
Kernel of Hollow Circular Section (or Value of Eccentricity
for Hollow Circular Section)
Kernel of Hollow Rectangular Section (or Value of
Eccentricity for Hollow Rectangular Section)
Highlights
Exercise
Chapter 10. Dams and Retaining Walls
10.1. Introduction
10.2. Types of Dams
10.3. Rectangular Dams
10.4. Stresses Across the Section of a Rectangular Dam
10.5. Trapezoidal Dam having Water Face Inclined
10.6. Stability of a Dam
10.7. Retaining Walls
10.8. Rankine’s Theory of Earth Pressure
10.9. Surcharged Retaining Wall
10.10. Chimneys
Highlights
Exercise
Chapter 11. Analysis of Perfect Frames
11.1.
11.2.
11.3.
11.4.
11.5.
Introduction
Types of Frames
Assumptions Made in Finding Out the Forces in a Frame
Reactions of Supports of a Frame
Analysis of a Frame
Highlights
Exercise
Chapter 12. Deflection of Beams
12.1.
12.2.
12.3.
12.4.
Introduction
Deflection and Slope of a Beam Subjected to Uniform Bending Moment
Relation between Slope, Deflection and Radius of Curvature
Deflection of a Simply Supported Beam Carrying a
Point Load at the Centre
12.5. Deflection of a Simply Supported Beam with an Eccentric Point Load
12.6. Deflection of a Simply Supported Beam with a Uniformly Distributed Load
12.7. Macaulay’s Method
12.8. Moment Area Method
12.9. Mohr’s Theorems
12.10. Slope and Deflection of a Simply Supported Beam Carrying a Point Load at
the Centre by Mohr’s Theorem
12.11. Slope and Deflection of a Simply Supported Beam Carrying a
Uniformly Distributed Load by Mohr’s Theorem
Highlights
Exercise
xii
...
...
...
397
402
404
...
405
...
...
...
406
409
410
413—468
...
...
...
...
...
...
...
...
...
...
...
...
413
413
413
417
428
434
447
449
459
462
464
466
469—514
...
...
...
...
...
...
...
469
469
470
470
471
508
508
515—558
...
...
...
515
515
517
...
...
...
...
...
...
519
523
530
535
550
552
...
553
...
...
...
554
555
556
Chapters
Chapter 13. Deflection of Cantilevers
13.1. Introduction
13.2. Deflection of a Cantilever with a Point Load at the Free End by Double
Integration Method
13.3. Deflection of a Cantilever with a Point Load at a Distance ‘a’ from
the Fixed End
13.4. Deflection of a Cantilever with a Uniformly Distributed Load
13.5. Deflection of a Cantilever with a Uniformly Distributed Load for a
Distance ‘a’ from the Fixed End
13.6. Deflection of a Cantilever with a Uniformly Distributed Load for a
Distance ‘a’ from the Free End
13.7. Deflection of a Cantilever with a Gradually Varying Load
13.8. Deflection and Slope of a Cantilever by Moment Area Method
Highlights
Exercise
Chapter 14. Conjugate Beam Method, Propped
Cantilevers and Beams
14.1. Introduction
14.2. Conjugate Beam Method
14.3. Deflection and Slope of a Simply Supported Beam with a Point
Load at the Centre
14.4. Simply Supported Beam Carrying an Eccentric Point Load
14.5. Relation between Actual Beam and Conjugate Beam
14.6. Deflection and Slope of a Cantilever with a Point Load at the Free End
14.7. Propped Cantilevers and Beams
14.8. S.F. and B.M. Diagrams for a Propped Cantilever Carrying a Point Load
at the Centre and Propped at the Free End
14.9. S.F. and B.M. Diagrams for a Propped Cantilever Carrying
a Uniformly Distributed Load and Propped at the Free End
14.10. S.F. and B.M. Diagrams for a Simply Supported Beam with
a Uniformly Distributed Load and Propped at the Centre
14.11. Yielding of a Prop
Highlights
Exercise
Chapter 15. Fixed and Continuous Beams
15.1. Introduction
Pages
559—582
...
559
...
559
...
...
561
562
...
566
...
...
...
...
...
566
572
576
580
581
583—618
...
...
583
583
...
...
...
...
...
583
585
597
597
602
...
603
...
604
...
...
...
...
610
614
615
616
619—678
...
619
15.2. Bending Moment Diagram for Fixed Beams
...
15.3. Slope and Deflection for a Fixed Beam Carrying a Point Load at the Centre ...
620
624
15.4. Slope and Deflection for a Fixed Beam Carrying an Eccentric Point Load
...
628
15.5. Slope and Deflection for a Fixed Beam Carrying a Uniformly Distributed
Load Over the Entire Length
15.6. Fixed End Moments of Fixed Beam Due to Sinking of a Support
15.7. Advantages of Fixed Beams
15.8. Continuous Beams
15.9. Bending Moment Diagram for Continuous Beams
Highlights
Exercise
...
...
...
...
...
...
...
644
654
657
658
658
675
676
xiii
Chapters
Chapter 16. Torsion of Shafts and Springs
16.1. Introduction
16.2. Derivation of Shear Stress Produced in a Circular Shaft Subjected to Torsion
16.3. Maximum Torque Transmitted by a Circular Solid Shaft
16.4. Torque Transmitted by a Hollow Circular Shaft
16.5. Power Transmitted by Shafts
16.6. Expression for Torque in Terms of Polar Moment of Inertia
16.7. Polar Modulus
16.8. Strength of a Shaft and Torsional Rigidity
16.9. Flanged Coupling
16.10. Strength of a Shaft of Varying Sections
16.11. Composite Shaft
16.12. Combined Bending and Torsion
16.13. Expression for Strain Energy Stored in a Body Due to Torsion
16.14. Springs
Highlights
Exercise
Chapter 17. Thin Cylinders and Spheres
Pages
679—746
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
747—788
17.1. Introduction
...
17.2. Thin Cylindrical Vessel Subjected to Internal Pressure
...
17.3. Stresses in a Thin Cylindrical Vessel Subjected to Internal Pressure
...
17.4. Expression for Circumferential Stress (or Hoop Stress)
...
17.5. Expression for Longitudinal Stress
...
17.6. Efficiency of a Joint
...
17.7. Effect of Internal Pressure on the Dimensions of a Thin Cylindrical Shell
...
17.8. A Thin Cylindrical Vessel Subjected to Internal Fluid Pressure and a Torque ...
17.9. Wire Winding of Thin Cylinders
...
17.10. Thin Spherical Shells
...
17.11. Change in Dimensions of a Thin Spherical Shell Due to an Internal Pressure ...
17.12. Rotational Stresses in Thin Cylinders
...
Highlights
...
Exercise
...
Chapter 18. Thick Cylinders and Spheres
18.1.
18.2.
18.3.
18.4.
Introduction
Stresses in a Thick Cylindrical Shell
Stresses in Compound Thick Cylinders
Initial Difference in Radii at the Junction of a Compound Cylinder for
Shrinkage
18.5. Thick Spherical Shells
Highlights
Exercise
Chapter 19. Columns and Struts
19.1. Introduction
19.2. Failure of a Column
19.3. Assumptions Made in the Euler’s Column Theory
xiv
679
679
681
683
684
694
695
695
702
705
713
717
720
728
741
743
747
747
748
748
749
753
757
768
772
777
778
780
783
784
789—816
...
...
...
789
789
797
...
...
...
...
802
808
813
814
817—880
...
...
...
817
817
818
Chapters
Pages
19.4. End Conditions for Long Columns
19.5. Expression for Crippling Load When Both the Ends of the Column are Hinged
19.6. Expression for Crippling Load When One End of the Column is Fixed and
the Other End is Free
19.7. Expression for Crippling Load When Both the Ends of the Column are Fixed
19.8. Expression for Crippling Load When One End of the Column is Fixed and
the Other End is Hinged (or Pinned)
19.9. Effective Length (or Equivalent Length) of a Column
19.10. Limitation of Euler’s Formula
19.11. Rankine’s Formula
19.12. Straight Line Formula
19.13. Johnson’s Parabolic Formula
19.14. Factor of Safety
19.15. Formula by Indian Standard Code (I.S. Code) for Mild Steel
19.16. Columns with Eccentric Load
19.17. Columns with Initial Curvature
19.18. Strut with Lateral Load (or Beam Columns)
Highlights
Exercise
Chapter 20. Riveted Joints
20.1.
20.2.
20.3.
20.4.
20.5.
20.6.
20.7.
20.8.
20.9.
Introduction
Types of Riveted Joints
Chain Riveted Joint
Zig-Zag Riveted Joint
Diamond Riveted Joint
Failure of a Riveted Joint
Strength of a Riveted Joint
Efficiency of a Riveted Joint
Design of a Riveted Joint
Highlights
Exercise
Chapter 21. Welded Joints
21.1.
21.2.
21.3.
21.4.
21.5.
Introduction
Advantages and Disadvantages of Welded Connections
Types of Welded Joints
Analysis of a Compound Weld
Analysis of Unsymmetrical Welded Sections which are Loaded Axially
Highlights
Exercise
Chapter 22. Rotating Discs and Cylinders
22.1.
22.2.
22.3.
22.4.
Introduction
Expression for Stresses in a Rotating Thin Disc
Disc of Uniform Strength
Long Cylinders
Highlights
Exercise
...
...
818
819
...
...
820
822
...
...
...
...
...
...
...
...
...
...
...
...
...
825
827
829
844
856
856
857
857
858
862
867
875
877
881—910
...
...
...
...
...
...
...
...
...
...
...
881
881
882
882
882
886
889
890
902
905
907
911—930
...
...
...
...
...
...
...
911
911
912
916
918
925
927
931—968
...
...
...
...
...
...
931
931
948
952
965
967
xv
Chapters
Pages
Chapter 23. Bending of Curved Bars
969—1016
23.1.
23.2.
23.3.
23.4.
23.5.
Introduction
Assumptions Made in the Derivation of Stresses in a Curved Bar
Expression for Stresses in a Curved Bar
Determination of Factor ‘h2’ for Various Sections
Resultant Stress in a Curved Bar Subjected to Direct Stresses and Bending
Stresses
23.6. Resultant Stress in a Hook
23.7. Stresses in Circular Ring
23.8. Stresses in a Chain Link
Highlights
Exercise
Chapter 24. Theories of Failure
24.1.
24.2.
24.3.
24.4.
24.5.
24.6.
24.7.
24.8.
24.9.
25.1.
25.2.
25.3.
25.4.
25.5.
25.6.
25.7.
Introduction
Properties of Beam Cross-section
Stress in Unsymmetrical Bending
Deflection of Beams in Unsymmetrical Bending
Shear Centre
Determination of Shear Centre for Channel Section
Determination of Shear Centre for I-Section
Highlights
Exercise
Chapter 26. Objective Type Questions
26.1.
26.2.
26.3.
26.4.
Objective Type Questions Generally Asked in Competitive Examinations
Answers of Objective Type Questions
Objective Type Questions from Competitive Examinations
Answers with Explanations
Subject Index
xvi
969
969
969
976
...
...
...
...
...
...
989
990
999
1005
1012
1014
1017—1050
Introduction
Maximum Principal Stress Theory
Maximum Principal Strain Theory
Maximum Shear Stress Theory
Maximum Strain Energy Theory
Maximum Shear Strain Energy Theory
Graphical Representation of Theories for Two Dimensional Stress System
Important Points from Theories of Failures used in Design
Energy of Distortion (or Shear Strain Energy)
Highlights
Exercise
Chapter 25. Unsymmetrical Bending and Shear Centre
...
...
...
...
...
...
...
...
...
...
...
...
...
...
...
1017
1017
1018
1022
1026
1030
1032
1036
1045
1048
1048
1051—1090
...
...
...
...
...
...
...
...
...
1051
1051
1053
1055
1073
1073
1080
1088
1089
1091—1142
...
...
...
...
1091
1118
1119
1127
1143—1144
1
CHAPTER
SIMPLE STRESSES AND
STRAINS
1.1. INTRODUCTION..
When an external force acts on a body, the body tends to undergo some deformation.
Due to cohesion between the molecules, the body resists deformation. This resistance by which
material of the body opposes the deformation is known as strength of material. Within a
certain limit (i.e., in the elastic stage) the resistance offered by the material is proportional to
the deformation brought out on the material by the external force. Also within this limit the
resistance is equal to the external force (or applied load). But beyond the elastic stage, the
resistance offered by the material is less than the applied load. In such a case, the deformation
continues, until failure takes place.
Within elastic stage, the resisting force equals applied load. This resisting force per
unit area is called stress or intensity of stress.
1.2. STRESS..
The force of resistance per unit area, offered by a body against deformation is known as
stress. The external force acting on the body is called the load or force. The load is applied on
the body while the stress is induced in the material of the body. A loaded member remains in
equilibrium when the resistance offered by the member against the deformation and the
applied load are equal.
P
Mathematically stress is written as, σ =
A
where σ = Stress (also called intensity of stress),
P = External force or load, and
A = Cross-sectional area.
1.2.1. Units of Stress. The unit of stress depends upon the unit of load (or force) and
unit of area. In M.K.S. units, the force is expressed in kgf and area in metre square (i.e., m2).
Hence unit of stress becomes as kgf/m2. If area is expressed in centimetre square (i.e., cm2),
the stress is expressed as kgf/cm2.
In the S.I. units, the force is expressed in newtons (written as N) and area is expressed
2
as m . Hence unit of stress becomes as N/m2. The area is also expressed in millimetre square
then unit of force becomes as N/mm2
1 N/m2 = 1 N/(100 cm)2 = 1 N/104 cm2
= 10–4 N/cm2 or 10–6 N/mm2
F∵
GH
1
cm
2
=
1
2
10 mm
1
2
I
JK
STRENGTH OF MATERIALS
∴
1 N/mm2 = 106 N/m2.
Also
1 N/m2 = 1 Pascal = 1 Pa.
The large quantities are represented by kilo, mega, giga and terra. They stand for :
Kilo = 103 and represented by ...... k
Mega = 106 and represented by ...... M
Giga = 109 and represented by ...... G
Terra = 1012 and represented by ...... T.
Thus mega newton means 106 newtons and is represented by MN. The symbol 1 MPa
stands for 1 mega pascal which is equal to 106 pascal (or 106 N/m2).
The small quantities are represented by milli, micro, nano and pico. They are equal to
Milli = 10–3 and represented by ...... m
Micro = 10–6 and represented by ...... µ
Nano = 10–9 and represented by ...... η
Pico = 10–12 and represented by ...... p.
Notes. 1. Newton is a force acting on a mass of one kg and produces an acceleration of 1 m/s2 i.e.,
1 N = 1 (kg) × 1 m /s2.
2. The stress in S.I. units is expressed in N/m2 or N/mm2.
3. The stress 1 N/mm2 = 106 N/m2 = MN/m2. Thus one N/mm2 is equal to one MN/m2.
4. One pascal is written as 1 Pa and is equal to 1 N/m2.
1.3. STRAIN..
When a body is subjected to some external force, there is some change of dimension of
the body. The ratio of change of dimension of the body to the original dimension is known as
strain. Strain is dimensionless.
Strain may be :
1. Tensile strain,
2. Compressive strain,
3. Volumetric strain, and
4. Shear strain.
If there is some increase in length of a body due to external force, then the ratio of
increase of length to the original length of the body is known as tensile strain. But if there is
some decrease in length of the body, then the ratio of decrease of the length of the body to the
original length is known as compressive strain. The ratio of change of volume of the body to
the original volume is known as volumetric strain. The strain produced by shear stress is
known as shear strain.
1.4. TYPES OF STRESSES..
The stress may be normal stress or a shear stress.
Normal stress is the stress which acts in a direction perpendicular to the area. It is
represented by σ (sigma). The normal stress is further divided into tensile stress and compressive
stress.
1.4.1. Tensile Stress. The stress induced in a body, when subjected to two equal and
opposite pulls as shown in Fig. 1.1 (a) as a result of which there is an increase in length, is
known as tensile stress. The ratio of increase in length to the original length is known as tensile
strain. The tensile stress acts normal to the area and it pulls on the area.
2
SIMPLE STRESSES AND STRAINS
Let
P = Pull (or force) acting on the body,
A = Cross-sectional area of the body,
L = Original length of the body,
dL = Increase in length due to pull P acting on the body,
σ = Stress induced in the body, and
e = Strain (i.e., tensile strain).
Fig. 1.1 (a) shows a bar subjected to a tensile force P at its ends. Consider a section x-x,
which divides the bar into two parts. The part left to the section x-x, will be in equilibrium if
P = Resisting force (R). This is shown in Fig. 1.1 (b). Similarly the part right to the section
x-x, will be in equilibrium if P = Resisting force as shown in Fig. 1.1 (c). This resisting force per
unit area is known as stress or intensity of stress.
X
P
P
X
(a)
P
Resisting force (R)
(b)
P
Resisting force (R)
(c)
P
P
R
R
(d)
Fig. 1.1
Resisting force ( R)
Tensile load ( P)
=
Cross-sectional area
A
P
σ=
A
And tensile strain is given by,
∴ Tensile stress = σ =
or
e=
Increase in length
dL
=
.
Original length
L
(∵ P = R)
...(1.1)
...(1.2)
1.4.2. Compressive Stress. The stress induced in a body, when subjected to two equal
and opposite pushes as shown in Fig. 1.2 (a) as a result of which there is a decrease in length
of the body, is known as compressive stress. And the ratio of decrease in length to the original
length is known as compressive strain. The compressive stress acts normal to the area and it
pushes on the area.
Let an axial push P is acting on a body in cross-sectional area A. Due to external push P,
let the original length L of the body decreases by dL.
3
STRENGTH OF MATERIALS
Fig. 1.2
Then compressive stress is given by,
Resisting Force ( R) Push ( P) P

 .
Area ( A)
Area ( A) A
And compressive strain is given by,
σ=
e = Decrease in length  dL .
L
Original length
1.4.3. Shear Stress. The stress induced in a body, when subjected to two equal and
opposite forces which are acting tangentially across the resisting section as shown in Fig. 1.3
as a result of which the body tends to shear off across the section, is known as shear stress. The
corresponding strain is known as shear strain. The shear stress is the stress which acts tangential
to the area. It is represented by τ.
P
P
(a)
P
P
(b)
Fig. 1.3
4
SIMPLE STRESSES AND STRAINS
Consider a rectangular block of height h, length L and width unity. Let the bottom face
AB of the block be fixed to the surface as shown in Fig. 1.4 (a). Let a force P be applied
tangentially along the top face CD of the block. Such a force acting tangentially along a
surface is known as shear force. For the equilibrium of the block, the surface AB will offer a
tangential reaction P equal and opposite to the applied tangential force P.
P
P
D
C
D
C
Resistance
X
X
X
h
A
P
R
Resistance
X
R
X
X
A
B
L
(a)
P
(b)
B
(c)
Fig. 1.4
Consider a section x-x (parallel to the applied force), which divides the block into two
parts. The upper part will be in equilibrium if P = Resistance (R). This is shown in Fig. 1.4 (b).
Similarly the lower part will be in equilibrium if P = Resistance (R) as shown in Fig. 1.4 (c).
This resistance is known as shear resistance. And the shear resistance per unit area is known
as shear stress which is represented by τ.
∴ Shear stress, τ =
=
Shear resistance R
=
A
Shear area
P
L×1
(∵
R = P and A = L × 1) ...(1.3)
Note that shear stress is tangential to the area over which it acts.
As the bottom face of the block is fixed, the face
ABCD will be distorted to ABC1D1 through an angle φ
as a result of force P as shown in Fig. 1.4 (d).
dl
dl
D
C
D1
C1
P
And shear strain (φ) is given by,
or
φ=
Transversal displacement
Distance AD
φ=
DD1 dl
=
AD
h
h
φ
φ
A
...(1.4)
B
L
Fig. 1.4 (d)
1.5. ELASTICITY AND ELASTIC LIMIT..
When an external force acts on a body, the body tends to undergo some deformation. If
the external force is removed and the body comes back to its original shape and size (which
means the deformation disappears completely), the body is known as elastic body. This property,
5
STRENGTH OF MATERIALS
by virtue of which certain materials return back to their original position after the removal of
the external force, is called elasticity.
The body will regain its previous shape and size only when the deformation caused by
the external force, is within a certain limit. Thus there is a limiting value of force up to and
within which, the deformation completely disappears on the removal of the force. The value
of stress corresponding to this limiting force is known as the elastic limit of the material.
If the external force is so large that the stress exceeds the elastic limit, the material
loses to some extent its property of elasticity. If now the force is removed, the material will
not return to its original shape and size and there will be a residual deformation in the material.
1.6. HOOKE’S LAW AND ELASTIC MODULII..
Hooke’s Law states that when a material is loaded within elastic limit, the stress is
proportional to the strain produced by the stress. This means the ratio of the stress to the
corresponding strain is a constant within the elastic limit. This constant is known as Modulus
of Elasticity or Modulus of Rigidity or Elastic Modulii.
1.7. MODULUS OF ELASTICITY (OR YOUNG’S MODULUS)..
The ratio of tensile stress or compressive stress to the corresponding strain is a constant.
This ratio is known as Young’s Modulus or Modulus of Elasticity and is denoted by E.
∴
E=
Tensile stress
Tensile strain
or
Compressive stress
Compressive strain
σ
...(1.5)
e
1.7.1. Modulus of Rigidity or Shear Modulus. The ratio of shear stress to the
corresponding shear strain within the elastic limit, is known as Modulus of Rigidity or Shear
Modulus. This is denoted by C or G or N.
or
E=
Shear stress τ
=
Shear strain φ
Let us define factor of safety also.
∴
C (or G or N) =
...(1.6)
1.8. FACTOR OF SAFETY..
It is defined as the ratio of ultimate tensile stress to the working (or permissible) stress.
Mathematically it is written as
Factor of safety =
Ultimate stress
Permissible stress
...(1.7)
1.9. CONSTITUTIVE RELATIONSHIP BETWEEN STRESS AND STRAIN..
1.9.1. For One-Dimensional Stress System. The relationship between stress and
strain for a unidirectional stress (i.e., for normal stress in one direction only) is given by
Hooke’s law, which states that when a material is loaded within its elastic limit, the normal
stress developed is proportional to the strain produced. This means that the ratio of the normal
6
SIMPLE STRESSES AND STRAINS
stress to the corresponding strain is a constant within the elastic limit. This constant is
represented by E and is known as modulus of elasticity or Young’s modulus of elasticity.
σ
Normal stress
∴
= Constant
or
=E
e
Corresponding strain
where σ = Normal stress, e = Strain and E = Young’s modulus
σ
or
e=
...[1.7 (A)]
E
The above equation gives the stress and strain relation for the normal stress in one
direction.
1.9.2. For Two-Dimensional Stress System. Before knowing the relationship between
stress and strain for two-dimensional stress system, we shall have to define longitudinal
strain, lateral strain, and Poisson’s ratio.
1. Longitudinal strain. When a body is subjected to an axial tensile load, there is an
increase in the length of the body. But at the same time there is a decrease in other dimensions
of the body at right angles to the line of action of the applied load. Thus the body is having
axial deformation and also deformation at right angles to the line of action of the applied load
(i.e., lateral deformation).
The ratio of axial deformation to the original length of the body is known as longitudinal
(or linear) strain. The longitudinal strain is also defined as the deformation of the body per
unit length in the direction of the applied load.
Let
L = Length of the body,
P = Tensile force acting on the body,
δL = Increase in the length of the body in the direction of P.
δL
Then, longitudinal strain =
.
L
2. Lateral strain. The strain at right angles to the direction of applied load is known
as lateral strain. Let a rectangular bar of length L, breadth b and depth d is subjected to an
axial tensile load P as shown in Fig. 1.5. The length of the bar will increase while the breadth
and depth will decrease.
Let
δL = Increase in length,
δb = Decrease in breadth, and
δd = Decrease in depth.
δL
Then longitudinal strain =
...[1.7 (B)]
L
δd
δb
and
lateral strain =
or
...[1.7 (C)]
d
b
b
(d – δd)
d
P
P
(b – δb)
L
L + δL
Fig. 1.5
7
STRENGTH OF MATERIALS
Note. (i) If longitudinal strain is tensile, the lateral strains will be compressive.
(ii) If longitudinal strain is compressive then lateral strains will be tensile.
(iii) Hence every longitudinal strain in the direction of load is accompanied by lateral strains of
the opposite kind in all directions perpendicular to the load.
3. Poisson’s ratio. The ratio of lateral strain to the longitudinal strain is a constant
for a given material, when the material is stressed within the elastic limit. This ratio is called
Poisson’s ratio and it is generally denoted by µ. Hence mathematically,
Lateral strain
...[1.7 (D)]
Longitudinal strain
or
Lateral strain = µ × Longitudinal strain
As lateral strain is opposite in sign to longitudinal strain, hence algebraically, lateral
strain is written as
Lateral strain = – µ × Longitudinal strain
...[1.7 (E)]
4. Relationship between stress and strain. Consider a
σ2
two-dimensional figure ABCD, subjected to two mutually
perpendicular stresses σ1 and σ2.
A
D
Refer to Fig. 1.5 (a).
Let
σ1 = Normal stress in x-direction
σ1
σ1
σ2 = Normal stress in y-direction
Consider the strain produced by σ1.
B
C
The stress σ1 will produce strain in the direction of x and
σ2
also in the direction of y. The strain in the direction of x will be
σ
longitudinal strain and will be equal to 1 whereas the strain
Fig. 1.5 (a)
E
σ
in the direction of y will be lateral strain and will be equal to – µ × 1
E
(∵ Lateral strain. = – µ × longitudinal strain)
Now consider the strain produced by σ2.
The stress σ2 will produce strain in the direction of y and also in the direction of x. The
σ2
strain in the direction of y will be longitudinal strain and will be equal to
whereas the
E
σ
strain in the direction of x will be lateral strain and will be equal to – µ × 2 .
E
Let e1 = Total strain in x-direction
e2 = Total strain in y-direction
Poisson’s ratio, µ =
Now total strain in the direction of x due to stresses σ1 and σ2 =
σ1
σ
−µ 2
E
E
Similarly total strain in the direction of y due to stresses σ1 and σ2 =
∴
8
σ2
σ
−µ 1
E
E
e1 =
σ1
σ
−µ 2
E
E
...[1.7 (F)]
e2 =
σ2
σ
−µ 1
E
E
...[1.7 (G)]
SIMPLE STRESSES AND STRAINS
The above two equations gives the stress and strain relationship for the two-dimensional
stress system. In the above equations, tensile stress is taken to be positive whereas the
compressive stress negative.
1.9.3. For Three-Dimensional Stress System. Fig. 1.5 (b) shows a three-dimensional
body subjected to three orthogonal normal stresses σ1, σ2, σ3 acting in the directions of x, y
and z respectively.
Consider the strains produced by each stress
Y
σ2
separately.
The stress σ1 will produce strain in the direction of
x and also in the directions of y and z. The strain in the
σ1
σ
direction of x will be 1 whereas the strains in the direction
E
σ
σ3
of y and z will be – µ 1 .
X
E
σ
Similarly the stress σ2 will produce strain 2 in
Z
E
Fig. 1.5 (b)
σ
the direction of y and strain of – µ 2 in the direction of x
E
and y each.
σ
σ
Also the stress σ3 will produce strain 3 in the direction of z and strain of – µ × 3 in
E
E
the direction of x and y.
σ
σ
σ
Total strain in the direction of x due to stresses σ1, σ2 and σ3 = 1 − µ 2 − µ 3 .
E
E
E
Similarly total strains in the direction of y due to stresses σ1, σ2 and σ3
σ
σ2
σ
−µ 3 −µ 1
E
E
E
and total strains in the direction of z due to stresses σ1, σ2 and σ3
=
σ3
σ
σ
−µ 1 −µ 2
E
E
E
Let e1, e2 and e3 are total strains in the direction of x, y and z respectively. Then
=
e1 =
σ
σ1
σ
−µ 2 −µ 3
E
E
E
...[1.7 (H)]
e2 =
σ
σ2
σ
−µ 3 −µ 1
E
E
E
...[1.7 (I)]
σ3
σ
σ
...[1.7 (J)]
−µ 1 −µ 2
E
E
E
The above three equations give the stress and strain relationship for the three orthogonal
normal stress system.
Problem 1.1. A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of
20 kN. If the modulus of elasticity of the material of the rod is 2 × 105 N/mm2 ; determine :
(i) the stress,
(ii) the strain, and
(iii) the elongation of the rod.
and
e3 =
9
STRENGTH OF MATERIALS
Sol. Given : Length of the rod,
Diameter of the rod,
L = 150 cm
D = 2.0 cm = 20 mm
π
∴ Area,
A=
(20)2 = 100π mm2
4
Axial pull,
P = 20 kN = 20,000 N
Modulus of elasticity,
E = 2.0 × 105 N/mm2
(i) The stress (σ) is given by equation (1.1) as
P
20000
σ=
=
= 63.662 N/mm2. Ans.
100π
A
(ii) Using equation (1.5), the strain is obtained as
σ
E= .
e
σ
63.662
=
∴ Strain,
e=
= 0.000318. Ans.
E 2 × 10 5
(iii) Elongation is obtained by using equation (1.2) as
dL
e=
.
L
∴ Elongation, dL = e × L
= 0.000318 × 150 = 0.0477 cm. Ans.
Problem 1.2. Find the minimum diameter of a steel wire, which is used to raise a load
of 4000 N if the stress in the rod is not to exceed 95 MN/m2.
Sol. Given : Load, P = 4000 N
Stress,
σ = 95 MN/m2 = 95 × 106 N/m2
(∵ M = Mega = 106)
= 95 N/mm2
(∵ 106 N/m2 = 1 N/mm2)
Let
D = Diameter of wire in mm
∴ Area,
Now
π 2
D
4
Load P
stress =
=
Area A
A=
95 =
∴
4000 4000 × 4
=
π 2
π D2
D
4
or D2 =
4000 × 4
= 53.61
π × 95
D = 7.32 mm. Ans.
Problem 1.3. Find the Young’s Modulus of a brass rod of diameter 25 mm and of
length 250 mm which is subjected to a tensile load of 50 kN when the extension of the rod
is equal to 0.3 mm.
Sol. Given : Dia. of rod, D = 25 mm
∴ Area of rod,
Tensile load,
Extension of rod,
Length of rod,
10
π
(25)2 = 490.87 mm2
4
P = 50 kN = 50 × 1000 = 50,000 N
dL = 0.3 mm
L = 250 mm
A
=
SIMPLE STRESSES AND STRAINS
Stress (σ) is given by equation (1.1), as
P 50,000
σ=
= 101.86 N/mm2.
=
A 490.87
Strain (e) is given by equation (1.2), as
dL 0.3
e=
=
= 0.0012.
250
L
Using equation (1.5), the Young’s Modulus (E) is obtained, as
Stress 101.86 N/mm 2
=
= 84883.33 N/mm2
Strain
0.0012
= 84883.33 × 106 N/m2. Ans.
(∵ 1 N/mm2 = 106 N/m2)
9
2
2
= 84.883 × 10 N/m = 84.883 GN/m . Ans.
(∵ 109 = G)
E=
Problem 1.4. A tensile test was conducted on a mild steel bar. The following data was
obtained from the test :
(i) Diameter of the steel bar
= 3 cm
(ii) Gauge length of the bar
= 20 cm
(iii) Load at elastic limit
= 250 kN
(iv) Extension at a load of 150 kN
= 0.21 mm
(v) Maximum load
= 380 kN
(vi) Total extension
= 60 mm
(vii) Diameter of the rod at the failure
= 2.25 cm.
Determine : (a) the Young’s modulus,
(b) the stress at elastic limit,
(c) the percentage elongation, and (d) the percentage decrease in area.
Sol. Area of the rod,
A=
π 2 π
D =
(3)2 cm2
4
4
= 7.0685 cm2 = 7.0685 × 10–4 m2.
LM∵
MN
cm 2 =
FG 1 mIJ
H 100 K
2
PPO
Q
(a) To find Young’s modulus, first calculate the value of stress and strain within elastic
limit. The load at elastic limit is given but the extension corresponding to the load at elastic
limit is not given. But a load of 150 kN (which is within elastic limit) and corresponding
extension of 0.21 mm are given. Hence these values are used for stress and strain within
elastic limit
∴
and
Stress =
Load
150 × 1000
N/m2
=
Area 7.0685 × 10 −4
(∵ 1 kN = 1000 N)
= 21220.9 × 104 N/m2
Increase in length (or Extension)
Strain =
Original length (or Gauge length)
0.21 mm
=
= 0.00105
20 × 10 mm
∴ Young’s Modulus,
E=
Stress 21220.9 × 10 4
= 20209523 × 104 N/m2
=
0.00105
Strain
11
STRENGTH OF MATERIALS
= 202.095 × 109 N/m2
= 202.095 GN/m2. Ans.
(∵ 109 = Giga = G)
(b) The stress at the elastic limit is given by,
Stress =
250 × 1000
Load at elastic limit
=
Area
7.0685 × 10 −4
4
2
= 35368 × 10 N/m
= 353.68 × 106 N/m2
= 353.68 MN/m2. Ans.
(c) The percentage elongation is obtained as,
Percentage elongation
(∵ 106 = Mega = M)
Total increase in length
× 100
Original length (or Gauge length)
60 mm
=
× 100 = 30%. Ans.
20 × 10 mm
(d) The percentage decrease in area is obtained as,
Percentage decrease in area
=
(Original area − Area at the failure)
× 100
Original area
=
FG π × 3
H4
=
=
F3
GH
2
2
−
π
× 2.25 2
4
π
× 32
4
IJ
K × 100
I
JK
− 2.25 2
(9 − 5.0625)
× 100 =
× 100 = 43.75%. Ans.
2
9
3
Problem 1.5. The safe stress, for a hollow steel column which carries an axial load of
2.1 × 103 kN is 125 MN/m2. If the external diameter of the column is 30 cm, determine the
internal diameter.
Sol. Given :
Safe stress*,
σ = 125 MN/m2 = 125 × 106 N/m2
Axial load,
P = 2.1 × 103 kN = 2.1 × 106 N
External diameter,
D = 30 cm = 0.30 m
Let
d = Internal diameter
∴ Area of cross-section of the column,
A=
Using equation (1.1), σ =
π
π
(D2 – d2) =
(.302 – d2) m2
4
4
P
A
*Safe stress is a stress which is within elastic limit.
12
SIMPLE STRESSES AND STRAINS
2.1 × 10 6
or
125 × 106 =
or
0.09 – d2 = 213.9
∴
π
(.30 2 − d 2 )
4
or (.302 – d2) =
4 × 2.1 × 10 6
π × 125 × 10 6
or 0.09 – 0.02139 = d2
d=
0.09 − 0.02139 = 0.2619 m = 26.19 cm.
Ans.
Problem 1.6. The ultimate stress, for a hollow steel column which carries an axial load
of 1.9 MN is 480 N/mm2. If the external diameter of the column is 200 mm, determine the
internal diameter. Take the factor of safety as 4.
Sol. Given :
Ultimate stress
= 480 N/mm2
Axial load,
P = 1.9 MN = 1.9 × 106 N
(∵ M = 106)
= 1900000 N
External dia.,
D = 200 mm
Factor of safety
= 4
Let
d = Internal diameter in mm
∴ Area of cross-section of the column,
A=
π
π
(D2 – d2) =
(2002 – d2) mm2
4
4
Using equation (1.7), we get
Factor of safety
∴
or
Ultimate stress
Working stress or Permissible stress
480
4=
Working stress
=
480
= 120 N/mm2
4
∴
σ = 120 N/mm2
Now using equation (1.1), we get
Working stress
=
σ=
P
A
or 120 =
1900000 × 4
1900000
=
π
2
2
π
(40000 − d 2 )
(200 − d )
4
1900000 × 4
= 20159.6
π × 120
d2 = 40000 – 20159.6 = 19840.4
d = 140.85 mm. Ans.
40000 – d2 =
or
or
∴
Problem 1.7. A stepped bar shown in Fig. 1.6 is subjected to an
axially applied compressive load of 35 kN. Find the maximum and
minimum stresses produced.
Sol. Given :
Axial load,
P = 35 kN = 35 × 103 N
Dia. of upper part,
D1 = 2 cm = 20 mm
35 kN
2 cm
DIA
3 cm
DIA
Fig. 1.6
13
STRENGTH OF MATERIALS
∴ Area of upper part, A1 =
π
(202) = 100 π mm2
4
π
π
D22 =
(302) = 225 π mm2
4
4
The stress is equal to load divided by area. Hence stress will be maximum where area
is minimum. Hence stress will be maximum in upper part and minimum in lower part.
Area of lower part,
A2 =
∴ Maximum stress
=
Load
35 × 10 3
=
= 111.408 N/mm2. Ans.
A1
100 × π
Minimum stress
=
35 × 10 3
Load
=
= 49.5146 N/mm2. Ans.
225 × π
A2
1.10. ANALYSIS OF BARS OF VARYING SECTIONS..
A bar of different lengths and of different diameters (and hence of different crosssectional areas) is shown in Fig. 1.6 (a). Let this bar is subjected to an axial load P.
Section 3
Section 2
Section 1
P
A1
A2
L1
L2
A3
P
L3
Fig. 1.6 (a)
Though each section is subjected to the same axial load P, yet the stresses, strains and
change in lengths will be different. The total change in length will be obtained by adding the
changes in length of individual section.
Let
P
L1
A1
L2, A2
L3, A3
=
=
=
=
=
Axial load acting on the bar,
Length of section 1,
Cross-sectional area of section 1,
Length and cross-sectional area of section 2,
Length and cross-sectional area of section 3, and
E = Young’s modulus for the bar.
Then stress for the section 1,
Load
P
.
=
Area of section 1 A1
Similarly stresses for the section 2 and section 3 are given as,
σ1 =
P
P
and σ3 =
A2
A3
Using equation (1.5), the strains in different sections are obtained.
σ2 =
∴ Strain of section 1, e1 =
14
σ1
P
=
E
A1 E
FG∵
H
σ1 =
P
A1
IJ
K
SIMPLE STRESSES AND STRAINS
Similarly the strains of section 2 and of section 3 are,
σ2
P
σ
P
=
and e3 = 3 =
.
e2 =
E
A2 E
E
A3 E
Change in length of section 1
But strain in section 1 =
Length of section 1
dL1
or
e1 =
L1
where dL1 = change in length of section 1.
∴ Change in length of section 1, dL1 = e1L1
PL1
=
A1 E
FG∵
H
e1 =
P
A1 E
IJ
K
FG∵
H
e2 =
P
A2 E
IJ
K
F∵
GH
e3 =
P
A3 E
I
JK
Similarly changes in length of section 2 and of section 3 are obtained as :
Change in length of section 2, dL2 = e2 L2
PL2
= A E
2
and change in length of section 3, dL3 = e3 L3
=
PL3
A3 E
∴ Total change in the length of the bar,
dL = dL1 + dL2 + dL3 =
=
LM
N
L
P L1 L2
+
+ 3
E A1 A2
A3
OP
Q
PL3
PL1
PL2
+
+
A1 E A2 E A3 E
...(1.8)
Equation (1.8) is used when the Young’s modulus of different sections is same. If the
Young’s modulus of different sections is different, then total change in length of the bar is
given by,
dL = P
LM L
NE A
1
1
1
+
L3
L2
+
E2 A2 E3 A3
OP
Q
...(1.9)
Problem 1.8. An axial pull of 35000 N is acting on a bar consisting of three lengths as
shown in Fig. 1.6 (b). If the Young’s modulus = 2.1 × 105 N/mm2, determine :
(i) stresses in each section and
(ii) total extension of the bar.
Section 3
Section 1
Section 2
35000 N
35000 N
2 cm DIA
3 cm DIA
5 cm DIA
20 cm
25 cm
22 cm
Fig. 1.6 (b)
15
STRENGTH OF MATERIALS
Sol. Given :
Axial pull,
Length of section 1,
Dia. of section 1,
∴ Area of section 1,
Length of section 2,
Dia. of section 2,
∴ Area of section 2,
Length of section 3,
Dia. of section 3,
P = 35000 N
L1 = 20 cm = 200 mm
D1 = 2 cm = 20 mm
π
(202) = 100 π mm2
4
L2 = 25 cm = 250 mm
D2 = 3 cm = 30 mm
A1 =
π
(302) = 225 π mm2
4
L3 = 22 cm = 220 mm
D3 = 5 cm = 50 mm
A2 =
π
(502) = 625 π mm2
4
Young’s modulus,
E = 2.1 × 105 N/mm2.
(i) Stresses in each section
∴ Area of section 3,
A3 =
Stress in section 1,
σ1 =
Stress in section 2,
σ2 =
P
35000
=
= 49.5146 N/mm2. Ans.
A2 225 × π
Stress in section 3,
σ3 =
P
35000
=
= 17.825 N/mm2. Ans.
A3 625 × π
Axial load
Area of section 1
P
35000
=
=
= 111.408 N/mm2. Ans.
A1 100 π
(ii) Total extension of the bar
Using equation (1.8), we get
Total extension
=
=
=
P
E
FL
GH A
1
+
1
35000
2.1 × 10
5
35000
2.1 × 10 5
L
L2
+ 3
A2 A3
I
JK
FG 200 + 250 + 220 IJ
H 100 π 225 × π 625 × π K
(6.366 + 3.536 + 1.120) = 0.183 mm. Ans.
Problem 1.9. A member formed by connecting a steel bar to an aluminium bar is shown
in Fig. 1.7. Assuming that the bars are prevented from buckling sideways, calculate the
magnitude of force P that will cause the total length of the member to decrease 0.25 mm. The
values of elastic modulus for steel and aluminium are 2.1 × 105 N/mm2 and 7 × 104 N/mm2
respectively.
Sol. Given :
Length of steel bar, L1 = 30 cm = 300 mm
16
SIMPLE STRESSES AND STRAINS
Area of steel bar,
A1 = 5 × 5 = 25 cm2 = 250 mm2
P
Elastic modulus for steel bar,
5 cm × 5 cm
E1 = 2.1 × 105 N/mm2
Steel bar
30 cm
Length of aluminium bar,
L2 = 38 cm = 380 mm
Area of aluminium bar,
10 cm × 10 cm
A2 = 10 × 10 = 100 cm2 = 10000 mm2
Aluminium bar
38 cm
Elastic modulus for aluminium bar,
E2 = 7 × 104 N/mm2
Total decrease in length, dL = 0.25 mm
Fig. 1.7
Let
P = Required force.
As both the bars are made of different materials, hence total change in the lengths of
the bar is given by equation (1.9).
FG L + L IJ
HE A E A K
F
I
300
380
+
0.25 = P G
H 2.1 × 10 × 2500 7 × 10 × 10000 JK
∴
1
dL = P
1
or
2
1
2
2
5
4
= P (5.714 × 10–7 + 5.428 × 10–7) = P × 11.142 × 10–7
∴
P=
0.25
0.25 × 10 7
=
11.142
11.142 × 10 −7
= 2.2437 × 105 = 224.37 kN. Ans.
Problem 1.10. The bar shown in Fig. 1.8 is subjected to a tensile load of 160 kN. If
the stress in the middle portion is limited to 150 N/mm2, determine the diameter of the
middle portion. Find also the length of the middle portion if the total elongation of the bar
is to be 0.2 mm. Young’s modulus is given as equal to 2.1 × 105 N/mm2.
Sol. Given :
Tensile load,
P = 160 kN = 160 × 103 N
Stress in middle portion, σ2 = 150 N/mm2
Total elongation,
dL = 0.2 mm
Total length of the bar,
L = 40 cm = 400 mm
Young’s modulus,
E = 2.1 × 105 N/mm2
Diameter of both end portions, D1 = 6 cm = 60 mm
∴ Area of cross-section of both end portions,
π
A1 =
× 602 = 900 π mm2.
4
160 kN
6 cm DIA
6 cm DIA
160 kN
40 cm
Fig. 1.8
17
STRENGTH OF MATERIALS
Let
D2 = Diameter of the middle portion
L2 = Length of middle portion in mm.
∴ Length of both end portions of the bar,
L1 = (400 – L2) mm
Using equation (1.1), we have
Stress =
Load
.
Area
For the middle portion, we have
P
σ2 =
A2
or
150 =
∴
or
where A2 =
π
D2
4 2
160000
π
D 22
4
D22 =
4 × 160000
= 1358 mm2
π × 150
D2 = 1358 = 36.85 mm = 3.685 cm. Ans.
∴ Area of cross-section of middle portion,
π
A3 =
× 36.85 = 1066 mm2
4
Now using equation (1.8), we get
Total extension,
or
dL =
0.2 =
LM
N
P L1 L2
+
E A1 A2
LM
N
OP
Q
OP
Q
L2
160000 (400 − L 2 )
+
5
900 π
1066
2.1 × 10
[∵ L1 = (400 – L2) and A2 = 1066]
(400 − L2)
L2
0.2 × 2.1 × 10 5
+
=
900 π
1066
160000
1066(400 − L2) + 900 π L 2
0.2625 =
900 π × 1066
0.2625 × 900π × 1066 = 1066 × 400 – 1066 L2 + 900π × L2
791186 = 426400 – 1066 L2 + 2827 L2
791186 – 426400 = L2 (2827 – 1066)
364786 = 1761 L2
or
or
or
or
or
or
∴
L2 =
364786
= 207.14 mm = 20.714 cm.
1761
Ans.
1.10.1. Principle of Superposition. When a number of loads are acting on a body,
the resulting strain, according to principle of superposition, will be the algebraic sum of strains
caused by individual loads.
While using this principle for an elastic body which is subjected to a number of direct
forces (tensile or compressive) at different sections along the length of the body, first the free
body diagram of individual section is drawn. Then the deformation of the each section is
obtained. The total deformation of the body will be then equal to the algebraic sum of
deformations of the individual sections.
18
SIMPLE STRESSES AND STRAINS
Problem 1.11. A brass bar, having cross-sectional area of 1000 mm2, is subjected to
axial forces as shown in Fig. 1.9.
A
B
C
D
80 kN
50 kN
10 kN
20 kN
600 mm
1m
1.20 m
Fig. 1.9
Find the total elongation of the bar. Take E = 1.05 × 105 N/mm2.
Sol. Given :
Area,
A = 1000 mm2
Value of
E = 1.05 × 105 N/mm2
Let
dL = Total elongation of the bar.
The force of 80 kN acting at B is split up into three forces of 50 kN, 20 kN and 10 kN.
Then the part AB of the bar will be subjected to a tensile load of 50 kN, part BC is subjected
to a compressive load of 20 kN and part BD is subjected to a compressive load of 10 kN as
shown in Fig. 1.10.
50 kN
50 kN
A
B
20 kN
20 kN
B
C
10 kN
10 kN
B
D
Fig. 1.10
Part AB. This part is subjected to a tensile load of 50 kN. Hence there will be increase
in length of this part.
∴ Increase in the length of AB
=
=
P1
× L1
AE
50 × 1000
× 600
(∵ P1 = 50,000 N, L1 = 600 mm)
1000 × 1.05 × 10 5
= 0.2857.
Part BC. This part is subjected to a compressive load of 20 kN or 20,000 N. Hence
there will be decrease in length of this part.
∴ Decrease in the length of BC
=
P2
20,000
× L2 =
× 1000
AE
1000 × 1.05 × 10 5
(∵ L2 = 1 m = 1000 mm)
= 0.1904.
19
STRENGTH OF MATERIALS
Part BD. This part is subjected to a compressive load of 10 kN or 10,000 N. Hence
there will be decrease in length of this part.
∴ Decrease in the length of BD
P3
10000
× L3 =
= 2200
AE
1000 × 1.05 × 10 5
(∵ L3 = 1.2 + 1 = 2.2 m or 2200 mm)
= 0.2095.
∴ Total elongation of bar = 0.2857 – 0.1904 – 0.2095
(Taking +ve sign for increase in length and
–ve sign for decrease in length)
= – 0.1142 mm. Ans.
Negative sign shows, that there will be decrease in length of the bar.
=
Problem 1.12. A member ABCD is subjected to point loads P1, P2, P3 and P4 as shown
in Fig. 1.11.
B
C
P1
625 mm
120 cm
2
P2
2500 mm
2
D
A
P3
60 cm
1250 mm
2
P4
90 cm
Fig. 1.11
Calculate the force P2 necessary for equilibrium, if P1 = 45 kN, P3 = 450 kN and
P4 = 130 kN. Determine the total elongation of the member, assuming the modulus of
elasticity to be 2.1 × 105 N/mm2.
Sol. Given :
Part AB :
Part BC :
Part CD :
Area,
Length,
Area,
Length,
Area,
Length,
A1 = 625 mm2 and
L1 = 120 cm = 1200 mm
A2 = 2500 mm2 and
L2 = 60 cm = 600 mm
A3 = 1250 mm2 and
L3 = 90 cm = 900 mm
E = 2.1 × 105 N/mm2.
Value of
Value of P2 necessary for equilibrium
Resolving the forces on the rod along its axis (i.e., equating the forces acting towards
right to those acting towards left), we get
P1 + P3 = P2 + P4
20
SIMPLE STRESSES AND STRAINS
P1 = 45 kN,
P3 = 450 kN and P4 = 130 kN
But
∴
45 + 450 = P2 + 130 or P2 = 495 – 130 = 365 kN
The force of 365 kN acting at B is split into two forces of 45 kN and 320 kN (i.e., 365 –
45 = 320 kN).
The force of 450 kN acting at C is split into two forces of 320 kN and 130 kN (i.e., 450 –
320 = 130 kN) as shown in Fig. 1.12.
From Fig. 1.12, it is clear that part AB is subjected to a tensile load of 45 kN, part BC
is subjected to a compressive load of 320 kN and part CD is subjected to a tensile load 130 kN.
A
45 kN
B
45 kN
320 kN
320 kN
B
C
130 kN
130 kN
C
D
Fig. 1.12
Hence for part AB, there will be increase in length ; for part BC there will be decrease
in length and for part CD there will be increase in length.
∴ Increase in length of AB
P
45000
=
× L1 =
× 1200
(∵ P = 45 kN = 45000 N)
A1 E
625 × 2.1 × 10 5
= 0.4114 mm
Decrease in length of BC
P
320,000
× L2 =
× 600
=
(∵ P = 320 kN = 320000)
A2 E
2500 × 2.1 × 10 5
= 0.3657 mm
Increase in length of CD
P
130,000
=
(∵ P = 130 kN = 130000)
× L3 =
× 900
A3 E
1250 × 2.1 × 10 5
= 0.4457 mm
Total change in the length of member
= 0.4114 – 0.3657 + 0.4457
(Taking +ve sign for increase in length and
–ve sign for decrease in length)
= 0.4914 mm (extension). Ans.
Problem 1.13. A tensile load of 40 kN is acting on a rod of diameter 40 mm and of
length 4 m. A bore of diameter 20 mm is made centrally on the rod. To what length the rod
21
STRENGTH OF MATERIALS
should be bored so that the total extension will increase 30% under the same tensile load. Take
E = 2 × 105 N/mm2.
Sol. Given :
40 kN
40 kN
4m
Fig. 1.12 (a)
Tensile load,
Dia. of rod,
∴ Area of rod,
P = 40 kN = 40,000 N
D = 40 mm
π
A =
(402) = 400π mm2
4
(4 – x)m
xm
d
D
4m
Fig. 1.12 (b)
Length of rod,
Dia. of bore,
L = 4 m = 4 × 1000 = 4000 mm
d = 20 mm
π
∴ Area of bore,
a =
× 202 = 100 π mm2
4
Total extension after bore
= 1.3 × Extension before bore
Value of E
= 2 × 105 N/mm2
Let the rod be bored to a length of x metre or x × 1000 mm. Then length of unbored
portion = (4 – x) m = (4 – x) × 1000 mm. First calculate the extension before the bore is made.
The extension (δL) is given by,
P
40000 × 4000
2
= mm
×L=
δL =
5
π
AE
400π × 2 × 10
Now extension after the bore is made
= 1.3 × Extension before bore
2 2.6
= 1.3 × =
mm
...(i)
π
π
The extension after the bore is made, is also obtained by finding the extensions of the
unbored length and bored length.
For this, find the stresses in the bored and unbored portions.
Stress in unbored portion
Load P 40000 100
N/mm2
=
=
=
=
Area A
400π
π
∴ Extension of unbored portion
Stress
=
× Length of unbored portion
E
22
SIMPLE STRESSES AND STRAINS
=
100
π × 2 × 10
5
× (4 – x) × 1000 =
(4 − x)
mm
2π
Stress in bored portion
=
Load
P
40000
40000
=
=
=
Area ( A − a) (400π − 100π) 300π
∴ Extension of bored portion
=
Stress
× Length of bored portion
E
=
4x
40000
× 1000 x =
mm
5
6π
300π × 2 × 10
∴ Total extension after the bore is made
(4 − x) 4 x
+
2π
6π
Equating equations (i) and (ii),
...(ii)
=
2.6 4 − x 4 x
=
+
2π
6π
π
or
or
4 − x 4x
or
2.6 × 6 = 3 × (4 – x) + 4x
+
2
6
15.6 = 12 – 3x + 4x or 15.6 – 12 = x or 3.6 = x
∴ Rod should be bored upto a length of 3.6 m. Ans.
2.6 =
Problem 1.14. A rigid bar ACDB is hinged at A and supported in a horizontal position
by two identical steel wires as shown in Fig. 1.12 (c). A vertical load of 30 kN is applied at B.
Find the tensile forces T1 and T2 induced in these wires by the vertical load.
E
F
T1
A
T2
C
1m
1m
D
1m
A
B
1m
C
d1
D
d2
B
C¢
D¢
B¢
30 kN
Fig. 1.12 (c)
Fig. 1.12 (d)
Sol. Given :
Rigid bar means a bar which will remain straight.
Two identical steel wires mean the area of cross-sections, lengths and value of E for
both wires is same.
∴
A1 = A2, E1 = E2 and L1 = L2
Load at B = 30 kN = 30,000 N
Fig. 1.12 (c) shows the position of the rigid bar before load is applied at B. Fig. 1.12 (d)
shows the position of the rigid bar after load is applied.
23
STRENGTH OF MATERIALS
T1 = Tension in the first wire
T2 = Tension in the second wire
δ1 = Extension of first wire
δ2 = Extension of second wire
Since the rigid bar remains straight, hence the extensions δ1 and δ2 are given by
δ1
AC 1
=
=
δ2
AD 2
∴
2δ1 = δ2
...(i)
But δ1 is the extension in wire EC
Let
FG T IJ × L
HA K
=
1
Stress in EC × L1
δ1 =
E1
∴
1
1
E1
=
T1 × L1
A1 × E1
T2 × L2
Similarly
δ2 =
A2 × E2
Substituting the values of δ1 and δ2 is equation (i),
2×
or
as :
T1 × L1
T2 × L2
=
A1 × E1
A2 × E2
But A1 = A2, E1 = E2 and L1 = L2. Hence above equation becomes
∴
2T1 = T2
Now taking the moments of all the forces on the rigid bar about A, we get
T1 × 1 + T2 × 2 = 30 × 3
T1 + 2T2 = 90
Substituting the value of T2 from equation (ii), into equation (iii), we get
T1 + 2(2T1) = 90 or 5T1 = 90
90
= 18 kN. Ans.
∴
T1 =
5
From equation (ii),
T2 = 2 × 18 = 36 kN. Ans.
...(ii)
...(iii)
Note. After calculating the values of T1 and T2, the stresses in the two wires can also be obtained
Stress in wire
and
Stress in wire
EC =
Load T1
=
Area
A1
FD =
T2
.
A2
1.11. ANALYSIS OF UNIFORMLY TAPERING CIRCULAR ROD..
A bar uniformly tapering from a diameter D1 at one end to a diameter D2 at the other
end is shown in Fig. 1.13.
Let
P = Axial tensile load on the bar
L = Total length of the bar
E = Young’s modulus.
24
SIMPLE STRESSES AND STRAINS
D1
D2
P
P
dx
x
L
Fig. 1.13
Consider a small element of length dx of the bar at a distance x from the left end. Let
the diameter of the bar be Dx at a distance x from the left end.
Then
Dx = D1 –
FG D
H
1
− D2
L
IJ x
K
D1 − D2
L
Area of cross-section of the bar at a distance x from the left end,
= D1 – kx
where k =
π
π
D2=
(D1 – k.x)2.
4 x
4
Now the stress at a distance x from the left end is given by,
Ax =
σx =
Load
Ax
4P
P
.
=
2
π
2
π
D
k
x
−
(
.
)
1
( D1 − k. x)
4
The strain ex in the small element of length dx is obtained by using equation (1.5).
=
∴
ex =
=
∴
Stress σ x
=
E
E
4P
π ( D1 − k. x) 2
×
1
4P
=
E π E( D1 − k. x) 2
Extension of the small elemental length dx
= Strain. dx = ex . dx
=
4P
π E( D1 − k . x) 2
. dx
...(i)
Total extension of the bar is obtained by integrating the above equation between the
limits 0 and L.
25
STRENGTH OF MATERIALS
∴ Total extension,
dL =
z
L
0
=
4 P . dx
=
π E( D1 − k. x) 2
4P
πE
z
L
0
4P
πE
z
L
0
(D1 – k.x)–2 . dx
( D1 − k . x) −2 × (− k)
. dx [Multiplying and dividing by (– k)]
(− k)
LM (D − k . x) OP 4P LM
MN (− 1) × (− k) PQ = π Ek N ( D
OP
1
1
4P L
−
=
M
π Ek N D − k . L D − k × 0 Q
1
1 O
4P L
−
M
PQ
=
D
−
k
.
L
D
π Ek N
4P
=
πE
−1 L
1
1
0
1
1
1
1
Substituting the value of k =
1
− k . x)
OP
Q
L
0
D1 − D2
in the above equation, we get
L
Total extension,
LM
OP
4P
M F D 1− D I − D1 PP
dL =
D −D I M
F
πE.G
H L JK MN D − GH L JK . L PQ
LM 1
4 PL
1 O
=
−
P
π E . (D − D ) N D − D + D
D Q
LM 1 − 1 OP
4 PL
=
π E . (D − D ) N D
D Q
1
2
1
2
1
1
2
2
1
1
1
2
2
1
1
1
( D − D2 )
4 PL
4 PL
× 1
=
=
...(1.10)
π E . ( D1 − D2 )
D1 D2
π ED1 D2
If the rod is of uniform diameter, then D1 – D2 = D
4 PL
∴ Total extension, dL =
...(1.11)
π E . D2
Problem 1.15. A rod, which tapers uniformly from 40 mm diameter to 20 mm diameter
in a length of 400 mm is subjected to an axial load of 5000 N. If E = 2.1 × 105 N/mm2, find the
extension of the rod.
Sol. Given :
Larger diameter,
D1 = 40 mm
Smaller diameter,
D2 = 20 mm
Length of rod,
L = 400 mm
Axial load,
P = 5000 N
Young’s modulus,
E = 2.1 × 105 N/mm2
Let
dL = Total extension of the rod
26
SIMPLE STRESSES AND STRAINS
Using equation (1.10),
dL =
4 PL
4 × 5000 × 400
=
π E D1 D2
π × 2.1 × 10 5 × 40 × 20
= 0.01515 mm. Ans.
Problem 1.16. Find the modulus of elasticity for a rod, which tapers uniformly from
30 mm to 15 mm diameter in a length of 350 mm. The rod is subjected to an axial load of
5.5 kN and extension of the rod is 0.025 mm.
Sol. Given :
Larger diameter,
D1 = 30 mm
Smaller diameter,
D2 = 15 mm
Length of rod,
Axial load,
Extension,
Using equation (1.10), we get
dL =
or
E=
L = 350 mm
P = 5.5 kN = 5500 N
dL = 0.025 mm
4 PL
π E D1 D2
4 × 5500 × 350
4 PL
=
π
×
30 × 15 × 0.025
π D1 D2 dL
= 217865 N/mm2 or 2.17865 × 105 N/mm2. Ans.
1.12. ANALYSIS OF UNIFORMLY TAPERING RECTANGULAR BAR..
A bar of constant thickness and uniformly tapering in width from one end to the other
end is shown in Fig. 1.14.
X
a
b
P
P
X
x
dx
L
t
Fig. 1.14
27
STRENGTH OF MATERIALS
Let
P = Axial load on the bar
L = Length of bar
a = Width at bigger end
b = Width at smaller end
E = Young’s modulus
t = Thickness of bar
Consider any section X-X at a distance x from the bigger end.
Width of the bar at the section X-X
(a − b) x
=a–
L
= a – kx
where k =
Thickness of bar at section X-X = t
∴ Area of the section X-X
= Width × thickness
= (a – kx)t
∴ Stress on the section X-X
P
Load
=
=
(a − kx)t
Area
Extension of the small elemental length dx
= Strain × Length dx
=
FG∵
H
a−b
L
IJ
K
FG∵ Stress = P IJ
(a − kx)t K
H
Stress
× dx
E
FG P IJ
H (a − kx)t K
=
Strain =
Stress
E
× dx
E
P
=
ax
E(a − kx)t
Total extension of the bar is obtained by integrating the above equation between the
limits 0 and L.
∴ Total extension,
dL =
z
L
0
P
P
dx =
E(a − kx)t
Et
z
L
0
dx
(a − kx)
FG 1IJ = – P [log (a – kL) – log a]
H kK Etk
F a IJ OP
P L
P
log G
M
=
[log a – log (a – kL)] =
Etk N
H a − kL K Q
Etk
LM F
I OP
G a JJ P
FG∵ k = a − bIJ
P
M
log G
=
H
L K
−
−
a
b
a
b
F IJ MM GG a − FG IJ L JJ PP
Et G
H L K N H H L K KQ
=
LM
N
P
. loge (a − kx)
Et
e
e
e
28
OP
Q
L
× −
e
0
e
e
SIMPLE STRESSES AND STRAINS
=
PL
a
log e
.
Et(a − b)
b
...(1.12)
Problem 1.17. A rectangular bar made of steel is 2.8 m long and 15 mm thick. The rod
is subjected to an axial tensile load of 40 kN. The width of the rod varies from 75 mm at one
end to 30 mm at the other. Find the extension of the rod if E = 2 × 105 N/mm2.
Sol. Given :
Length,
L = 2.8 m = 2800 mm
Thickness,
t = 15 mm
Axial load,
P = 40 kN = 40,000 N
Width at bigger end, a = 75 mm
Width at smaller end, b = 30 mm
Value of
E = 2 × 105 N/mm2
Let
dL = Extension of the rod.
Using equation (1.12), we get
PL
a
log e
dL =
Et(a − b)
b
=
40000 × 2800
2 × 10 5 × 15(75 − 30)
log e
FG 75 IJ
H 30 K
= 0.8296 × 0.9163 = 0.76 mm. Ans.
Problem 1.18. The extension in a rectangular steel bar of length 400 mm and thickness
10 mm, is found to be 0.21 mm. The bar tapers uniformly in width from 100 mm to 50 mm. If
E for the bar is 2 × 105 N/mm2, determine the axial load on the bar.
Sol. Given :
Extension,
dL = 0.21 mm
Length,
L = 400 mm
Thickness,
t = 10 mm
Width at bigger end,
a = 100 mm
Width at smaller end,
b = 50 mm
Value of
E = 2 × 105 N/mm2
Let
P = axial load.
Using equation (1.12), we get
dL =
or
0.21 =
FG IJ
H K
PL
a
log e
Et(a − b)
b
P × 400
2 × 10 5 × 10(100 − 50)
log e
FG 100 IJ
H 50 K
= 0.000004 P × 0.6931
∴
P =
0.21
= 75746 N
0.000004 × 0.6931
= 75.746 kN. Ans.
29
STRENGTH OF MATERIALS
1.13. ANALYSIS OF BARS OF COMPOSITE SECTIONS..
A bar, made up of two or more bars of equal lengths but
of different materials rigidly fixed with each other and behaving
as one unit for extension or compression when subjected to an
axial tensile or compressive loads, is called a composite bar.
1
2
L
For the composite bar the following two points are important :
1. The extension or compression in each bar is equal.
Hence deformation per unit length i.e., strain in each bar is
equal.
2. The total external load on the composite bar is equal
P
to the sum of the loads carried by each different material.
Fig. 1.15
Fig. 1.15 shows a composite bar made up of two different
materials.
Let
P = Total load on the composite bar,
L = Length of composite bar and also length of bars of different materials,
A1 = Area of cross-section of bar 1,
A2 = Area of cross-section of bar 2,
E1 = Young’s Modulus of bar 1,
E2 = Young’s Modulus of bar 2,
P1 = Load shared by bar 1,
P2 = Load shared by bar 2,
σ1 = Stress induced in bar 1, and
σ2 = Stress induced in bar 2.
Now the total load on the composite bar is equal to the sum of the load carried by the
two bars.
∴
P = P1 + P2
...(i)
The stress in bar 1,
=
Load carried by bar 1
.
Area of cross-section of bar 1
∴
σ1 =
P1
A1
or
P1 = σ1 A1
...(ii)
Similarly stress in bar 2,
σ2 =
P2
A2
or
P2 = σ2 A2
...(iii)
Substituting the values of P1 and P2 in equation (i), we get
P = σ1A1 + σ2 A2
...(iv)
Since the ends of the two bars are rigidly connected, each bar will change in length by
the same amount. Also the length of each bar is same and hence the ratio of change in length
to the original length (i.e., strain) will be same for each bar.
But strain in bar 1,
=
σ
Stress in bar 1
= 1.
Young’s modulus of bar 1 E1
Similarly strain in bar 2,
=
σ2
.
E2
30
SIMPLE STRESSES AND STRAINS
But strain in bar 1 = Strain in bar 2
σ
σ
∴
= 1 = 2
...(v)
E1 E2
From equations (iv) and (v), the stresses σ1 and σ2 can be determined. By substituting
the values of σ1 and σ2 in equations (ii) and (iii), the load carried by different materials may
be computed.
Modular Ratio. The ratio of
E1
is called the modular ratio of the first material to the
E2
second.
Problem 1.19. A steel rod of 3 cm diameter is enclosed centrally in a hollow copper tube
of external diameter 5 cm and internal diameter of 4 cm. The composite bar is then subjected
to an axial pull of 45000 N. If the length of each bar is equal to 15 cm, determine :
(i) The stresses in the rod and tube, and
(ii) Load carried by each bar.
Take E for steel = 2.1 × 105 N/mm2 and for copper = 1.1 × 105 N/mm2.
Sol. Given :
Dia. of steel rod = 3 cm = 30 mm
∴ Area of steel rod,
π
Copper
As =
(30)2 = 706.86 mm2
tube
4
Steel
rod
15
cm
External dia. of copper tube
= 5 cm = 50 mm
Internal dia. of copper tube
= 4 cm = 40 mm
3 cm
4 cm
∴ Area of copper tube,
π
[502 – 402] mm2 = 706.86 mm2
4
Axial pull on composite bar,
P = 45000 N
Length of each bar,
L = 15 cm
Young’s modulus for steel,
Es = 2.1 × 105 N/mm2
Young’s modulus for copper,
Ec = 1.1 × 105 N/mm2
(i) The stress in the rod and tube
Let
σs = Stress in steel,
Ps = Load carried by steel rod,
σc = Stress in copper, and
Pc = Load carried by copper tube.
Now strain in steel = Strain in copper
Ac =
σs
σ
= c
Es
Ec
or
∴
σs =
5 cm
P = 45000 N
Fig. 1.16
FG∵
H
σ
= strain
E
Es
2.1 × 10 5
× σc =
× σc = 1.909 σc
Ec
1.1 × 10 5
IJ
K
...(i)
31
STRENGTH OF MATERIALS
Load
, ∴ Load = Stress × Area
Area
Load on steel + Load on copper = Total load
σs × As + σc × Ac = P
1.909 σc × 706.86 + σc × 706.86 = 45000
σc (1.909 × 706.86 + 706.86) = 45000
2056.25 σc
=
45000
45000
= 21.88 N/mm2. Ans.
∴
σc =
2056.25
Substituting the value of σc in equation (i), we get
σs = 1.909 × 21.88 N/mm2
= 41.77 N/mm2. Ans.
(ii) Load carried by each bar
As
load = Stress × Area
∴ Load carried by steel rod,
Ps = σs × As
= 41.77 × 706.86 = 29525.5 N. Ans.
Load carried by copper tube,
Pc = 45000 – 29525.5
= 15474.5 N. Ans.
Now
or
or
or
stress =
(∵ Total load = P)
Problem 1.20. A compound tube consists of a steel tube 140 mm internal diameter
and 160 mm external diameter and an outer brass tube 160 mm internal diameter and
180 mm external diameter. The two tubes are of the same length. The compound tube carries
an axial load of 900 kN. Find the stresses and the load carried by each tube and the amount
it shortens. Length of each tube is 140 mm. Take E for steel as 2 × 105 N/mm2 and for brass
as 1 × 105 N/mm2.
Sol. Given :
Internal dia. of steel tube
= 140 mm
External dia. of steel tube
= 160 mm
π
(1602 – 1402) = 4712.4 mm2
∴ Area of steel tube,
As =
4
Internal dia. of brass tube
= 160 mm
External dia. of brass tube
= 180 mm
π
(1802 – 1602) = 5340.7 mm2
∴ Area of brass tube,
Ab =
4
Axial load carried by compound tube,
P = 900 kN = 900 × 1000 = 900000 N
Length of each tube,
L = 140 mm
E for steel,
Es = 2 × 105 N/mm2
E for brass,
Eb = 1 × 105 N/mm2
Let
σs = Stress in steel in N/mm2 and
σb = Stress in brass in N/mm2
32
SIMPLE STRESSES AND STRAINS
Now strain in steel = Strain in brass
σs
Es
σb
Eb
FG∵
H
Strain =
Stress
E
IJ
K
2 × 10 5
Es
× σb =
σb = 2σb
...(i)
Eb
1 × 10 5
Now load on steel + Load on brass = Total load
σs × As + σb × Ab = 900000
(∵ Load = Stress × Area)
2σb×4712.4 + σb × 5340.7 = 900000
(∵ σs = 2σb)
14765.5 σb = 900000
900000
∴
σb =
= 60.95 N/mm2. Ans.
14765.5
Substituting the value of pb in equation (i), we get
σs = 2 × 60.95 = 121.9 N/mm2. Ans.
Load carried by brass tube
= Stress × Area
= σb × Ab = 60.95 × 5340.7 N
= 325515 N = 325.515 kN. Ans.
Load carried by steel tube
= 900 – 325.515 = 574.485 kN. Ans.
Decrease in the length of the compound tube
= Decrease in length of either of the tubes
= Decrease in length of brass tube
= Strain in brass tube × Original length
∴
or
or
or
=
σs =
=
60.95
σb
×L=
× 140 = 0.0853 mm.
Eb
1 × 10 5
Ans.
Problem 1.21. Two vertical rods one of steel and the other of copper are each rigidly
fixed at the top and 50 cm apart. Diameters and lengths of each rod are 2 cm and 4 m respectively.
A cross bar fixed to the rods at the lower ends carries a load of 5000 N such that the cross bar
remains horizontal even after loading. Find the stress in each rod and the position of the load
on the bar. Take E for steel = 2 × 105 N/mm2 and
E for copper = 1 × 105 N/mm2.
Steel
Copper
Sol. Given :
2 cm dia
2 cm dia
Distance between the rods
= 50 cm = 500 mm
Dia. of steel rod
= Dia. of copper rod
= 2 cm = 20 mm
∴ Area of steel rod
= Area of copper rod
=
π
× (20)2 = 100 π mm2
4
400
cm
50 cm
x
Cross bar
5000 N
Fig. 1.17
33
STRENGTH OF MATERIALS
∴ As = Ac = 100 π mm2
Length of each rod
= 4 m = 4000 mm
Total load carried by rods,
P = 5000 N
E for steel,
Es = 2 × 105 N/mm2
E for copper,
Ec = 1 × 105 N/mm2
Let
σs = Stress in steel rod and
σc = Stress in copper rod.
Since the cross bar remains horizontal, the extensions of the steel and copper rods are
equal. Also these rods have the same original length, hence the strains of these rods are
equal.
∴
Strain in steel = Strain in copper
Stress
σc σc
∵ Strain =
=
or
E
Es Ec
FG
H
IJ
K
Es
2 × 10 5
× σc =
× σc = 2 × σc
...(i)
Ec
1 × 10 5
Now load on steel rod + Load on copper rod = Total load
= 5000 N
(∵ Load = Stress × Area)
or
σs × As + σc × Ac = 5000
(∵ As = Ac = 100 mm2)
or
σs × 100π + σc × 100π = 5000
(∵ σs = 2σc)
or
2σc × 100π + σc × 100π = 5000
or
3σc × 100π = 5000
5000
∴
σc =
= 5.305 N/mm2. Ans.
300π
Substituting this value of σc in equation (i), we get
σs = 2 × 5.305 = 10.61 N/mm2. Ans.
Position of the load of 5000 N on cross bar
Let
x = The distance of the 5000 N load from the copper rod (i.e., from
the right hand rod).
Now first calculate the load carried by each rod.
Load carried by steel rod,
(∵ Load = Stress × Area)
Ps = σs × As
= 10.61 × 100π = 3333 N
Load carried by copper rod,
Pc = Total load – Ps
= 5000 – 3333 = 1667 N
Now taking the moments about the copper rod and equating the same, we get
5000 × x = Ps × 50
= 3333 × 50
(∵ Ps = 3333)
3333 × 50
∴
x =
= 33.33 cm. Ans.
5000
Problem 1.22. A load of 2 MN is applied on a short concrete column 500 mm × 500 mm.
The column is reinforced with four steel bars of 10 mm diameter, one in each corner. Find the
∴
34
σs =
SIMPLE STRESSES AND STRAINS
stresses in the concrete and steel bars. Take E for steel as 2.1 × 105 N/mm2 and for concrete as
1.4 × 104 N/mm2.
Sol. Given :
Total load applied,
P = 2 MN = 2 × 106 N
Area of column
= 500 × 500 = 250000 mm2
Steel bars
π
500
2
Area of 4 steel bars, As = 4 × (10) = 314.159 mm2
mm
4
Area of concrete,
Ac = Area of column
– Area of steel bars
= 250000 – 314.159
= 249685.841 mm2
500 mm
E for steel,
Es = 2.1 × 105 N/mm2
Fig. 1.18
E for concrete,
Ec = 1.4 × 104 N/mm2
Let
σs = Stress in steel bar in N/mm2
σc = Stress in concrete in N/mm2
Now
strain in steel = Strain in concrete
σs σc
=
Es Ec
Es
2.1 × 10 5
× σc =
σ c = 15 σ c
Ec
1.4 × 10 4
Now load on steel + Load on concrete = Total load
σs . As + σc . Ac = P
15σc × 314.159 + σc × 249685.841 = 2000000
254398σc = 2000000
∴
or
or
FG∵
H
Strain =
Stress
E
σs =
IJ
K
...(i)
(∵ Load = Stress × Area)
(∵ σs = 15σc)
2000000
= 7.86 N/mm2. Ans.
254398
Substituting this value in equation (i), we get
σs = 15 × 7.86 = 117.92 N/mm2. Ans.
∴
σc =
Problem 1.23. A reinforced short concrete column 250 mm × 250 mm in section is
reinforced with 8 steel bars. The total area of steel bars is 2500 mm2. The column carries a
load of 390 kN. If the modulus of elasticity for steel is 15 times that of concrete, find the
stresses in concrete and steel.
Sol. Given :
Area of concrete column
= 250 × 250 = 62500 mm2
Area of steel bars,
As = 2500 mm2
∴ Area of concrete,
Ac = 62500 – 2500 = 60000 mm2
Total load on column,
P = 390 kN= 390,000 N
E for steel
= 15 × E for concrete
or
Es = 15Ec
Let
σs = Stress in steel in N/mm2
σc = Stress in concrete in N/mm2
35
STRENGTH OF MATERIALS
Now
strain in steel = Strain in concrete
or
σs σc
=
Es Ec
or
σs =
or
or
Es
σ = 15σc
Ec c
FG∵
H
Strain =
Stress
E
IJ
K
...(i)
Also, we know that
Total load = Load on steel + Load on concrete
P = σs . As – σc . Ac
(∵ Load = Stress × Area)
390000 = 15σc × 2500 + σc × 60000
(∵ σs = 15σc)
= 97500σc
390000
= 4.0 N/mm2. Ans.
97500
Substituting this value in equation (i), we get
σs = 15 × 4.0 = 60.0 N/mm2. Ans.
Area of steel required so that column may support a load of 480 kN and maximum
stress in concrete is 4.5 N/mm2.
Let
As* = Area of steel required
Area of concrete
= 62500 – As*
σs* = Stress in steel in N/mm2
σc* = Stress in concrete = 4.5 N/mm2
P* = Total load = 480 kN = 480000 N
We know that
σs* = 15σc*
= 15 × 4.5
(∵ σc* = 4.5)
2
= 67.5 N/mm
Also, we know that
Total load = Load on steel + Load on concrete
or
P* = σs* × As* + σc* × Ac*
or
480000 = 67.5 × As* + 4.5 × (62500 – As*)
= 67.5 × As* + 4.5 × 62500 – 4.5 As*
or
480000 – 4.5 × 62500 = 63 As*
or
198750 = 63 As*
∴
σc =
∴
As* =
198750
= 3154.76 mm2. Ans.
63
Problem 1.24. A steel rod and two copper rods together support a load of 370 kN as
shown in Fig. 1.19. The cross-sectional area of steel rod is 2500 mm2 and of each copper rod
is 1600 mm2. Find the stresses in the rods. Take E for steel = 2 × 105 N/mm2 and for copper
= 1 × 105 N/mm2.
Sol. Given :
Load,
P = 370 kN = 370,000 N
Area of steel rod,
As = 2500 mm2
36
SIMPLE STRESSES AND STRAINS
Area of two copper rods,
Ac = 2 × 1600
370 kN
= 3200 mm2
E for steel,
Es = 2 × 105 N/mm2
E for copper,
Ec = 1 × 105 N/mm2
Length of steel rod,
Ls = 15 + 10
Copper
Steel
Copper
15
cm
= 25 cm = 250 mm
Length of copper rods,
Lc = 15 cm = 150 mm
σs = Stress in steel rod in N/mm2
Let
σc = Stress in copper rods in N/mm2
10 cm
We know that decrease in the length of steel rod
is equal to the decrease in the length of copper rods.
Fig. 1.19
But decrease in length of steel rod
= Strain in steel rod × Length of steel rod
FG
H
IJ
K
Stress
Stress in steel
∵ Strain =
× Lc
E
Es
σs
=
× 250
...(i)
2 × 10 5
Similarly decrease in length of copper rods
= Strain in copper roods × Length of copper rods
=
=
Stress in copper
× Lc
Ec
=
σc
× 150
...(ii)
1 × 10 5
Equating the decrease in length of steel rod to the decrease in the length of copper
rods, we get
σs
2 × 10 5
or
× 250 =
σc
1 × 10 5
σs = σc ×
× 150
2 × 10 5
1 × 10 5
×
150
= 1.2σc
250
...(iii)
Also, we know that
Load on steel + Load on copper = Total load applied
or
σs × As + σc × Ac = P
(∵ Load = Stress × Area)
or
1.2σc × 2500 + σc × 3200 = 370,000
(∵ σs = 1.2σc)
or
6200σc = 370000
370000
= 59.67 N/mm2. Ans.
∴
σc =
6200
Substituting this value in equation (iii), we get
σs = 1.2 × σc
= 1.2 × 59.67 = 71.604 N/mm2. Ans.
37
STRENGTH OF MATERIALS
Problem 1.25. Two brass rods and one steel rod
together support a load as shown in Fig. 1.20. If the stresses
in brass and steel are not to exceed 60 N/mm2 and 120 N/
mm2, find the safe load that can be supported. Take E for
steel = 2 × 105 N/mm2 and for brass = 1 × 105 N/mm2. The
cross-sectional area of steel rod is 1500 mm2 and of each
brass rod is 1000 mm2.
Sol. Given :
Stress in brass,
σb = 60 N/mm2
Stress in steel,
σs = 120 N/mm2
E for steel,
Es = 2 × 105 N/mm2
E for brass,
Area of steel rod,
Area of two brass rods,
Length of steel rod,
Length of brass rods,
Eb = 1 × 105 N/mm2
As = 1500 mm2
Ab = 2 × 1000
= 2000 mm2
Ls = 170 mm
Lb = 100 mm
P
100
mm
Brass
1000
mm2
Steel
1500
mm 2
Brass
1000
mm 2
70 mm
Fig. 1.20
We know that decrease in the length of steel rod should be equal to the decrease in
length of brass rods.
But decrease in length of steel rod
= Strain in steel rod × Length of steel rod
= es × Ls where es is strain in steel
Similarly decrease in length of brass rods
= Strain in brass rods × Length of brass rods
= eb × Lb where eb is strain in brass rod
Equating the decrease in length of steel rod to the decrease in length of brass rods, we get
es Ls = eb × Lb or
Lb 100
es
=
=
Ls 170
eb
(∵ Stress = Strain × E)
...(i)
But
or
stress in steel = Strain in steel × Es
σs = es × Es
Similarly stress in brass is given by,
σb = eb × Eb
Dividing equation (i) by equation (ii), we get
σs
σb
=
...(ii)
es × Es
100 2 × 10 5
=
×
= 1.176
eb × Eb
170 1 × 10 5
Suppose steel is permitted to reach its safe stress of 2 × 105 N/mm2, the corresponding
stress in brass will be
Fσ
GH
38
b
=
σs
2 × 10 5
=
= 1.7 × 10 5 N/mm 2
1.176
1.176
I
JK
SIMPLE STRESSES AND STRAINS
1.7 × 105 N/mm2 which exceeds the safe stress of 1 × 105 N/mm2 for brass. Therefore let brass
be allowed to reach its safe stress of 1 × 105 N/mm2. Then corresponding stress in steel will be
1.176 × 105 N/mm2 which is less than 2 × 105 N/mm2.
∴
Total load = P = Load on steel + Load on copper
= σs × As + σb × Ab
= 1.176 × 105 × 1500 + 1 × 105 × 2000
= 3764 × 105 N or 376.4 × 106 N
= 376.4 MN. Ans.
(∵ M = 106)
Aluminium
σ
σc
σ
= z = a
Ea
Ec
Ez
or
∴
Also
Now
or
Zinc
Copper
Problem 1.26. Three bars made of copper, zinc and aluminium are of equal length and
have cross-section 500, 750 and 1000 square mm respectively. They are rigidly connected at
their ends. If this compound member is subjected to a longitudinal pull of 250 kN, estimate the
proportional of the load carried on each rod and the induced stresses. Take the value of E for
copper = 1.3 × 105 N/mm2, for zinc = 1.0 × 105 N/mm2 and for aluminium = 0.8 × 105 N/mm2.
Sol. Given :
Total load,
P = 250 kN = 250 × 103 N
For copper bar,
Area, Ac = 500 mm2 and Ec = 1.3 × 105 N/mm2
For zinc bar,
Area, Az = 750 mm2 and Ez = 1.0 × 105 N/mm2
For aluminium bar,
Area, Aa = 1000 mm2, and Ea = 0.8 × 105 N/mm2
Let
σc = Stress induced in copper bar,
P = 250 kN
σz = Stress induced in zinc bar,
σa = Stress induced in aluminium bar,
Fig. 1.21
Pc = Load shared by copper rod,
Pz = Load shared by zinc rod,
Pa = Load shared by aluminium rod, and
L = Length of each bar.
Now, we know that the increase in length of each bar should be same, as length of each
bar is equal hence strain in each bar will be same.
∴
Strain in copper = Strain in zinc = Strain in aluminium
Stress in copper
Stress in aluminium
Stress in zinc
or
=
=
Ec
Ea
Ez
σc =
1.3 × 10 5
Ec
× σa =
σa = 1.625σa
Ea
0.8 × 10 5
...(i)
1.0 × 10 5
Ez
× σa =
× σa = 1.25σa
...(ii)
Ea
0.8 × 10 5
total load = Load on copper + Load on zinc + Load on aluminium
250 × 103 = Stress in copper × Ac + Stress in zinc × Az
+ Stress in aluminium × Aa
σz =
39
STRENGTH OF MATERIALS
= σc × Ac + σz × Az + σa × Aa
= 1.625σa × 500 + 1.25σa × 750 + σa × 1000
= 2750σa
(∵ σc = 1.625σa and σz = 1.25σa)
250 × 10 3
= 90.9 N/mm2. Ans.
2750
Substituting the value of σa in equations (i) and (ii), we get
σc = 1.625 × 90.9 = 147.7 N/mm2. Ans.
σz = 1.25 × 90.9 = 113.625 N/mm2. Ans.
Now load shared by copper
= σc × Ac
= 147.7 × 500 = 73850 N. Ans.
Load shared by zinc rod
= σz × Az = 113.625 × 750
= 85218 N. Ans.
Load shared by aluminium rod
= σa × Aa = 90.9 × 1000
= 90900 N. Ans.
∴
and
σa =
Problem 1.27. A steel rod 20 mm in diameter passes centrally through a steel tube of
25 mm internal diameter and 30 mm external diameter. The tube is 800 mm long and is closed
by rigid washers of negligible thickness which are fastened by nuts threaded on the rod. The
nuts are tightened until the compressive load on the tube is 20 kN. Calculate the stresses in the
tube and the rod.
Find the increase in these stresses when one nut is tightened by one-quarter of a turn
relative to the other. There are 4 threads per 10 mm. Take E = 2 × 105 N/mm2.
Sol. Given :
Dia. of rod
= 20 mm
π
(20)2 mm2 = 100π mm2
4
π
Area of tube, At =
(302 – 252) mm2 = 68.75π mm2
4
Length of tube, L = 800 mm
Compressive load on tube, Pt = 20 kN = 20 × 103 N
Value of E = 2 × 105 N/mm2
∴ Area of rod, Ar =
Tube
Rod
Fig. 1.22
When the nuts are tightened, the tube will be compressed and the rod will be elongated.
This means that the tube will be under compression and rod will be under tension. Since no
40
SIMPLE STRESSES AND STRAINS
external forces have been applied, the compressive load on the tube must be equal to the
tensile load on the rod.
Let
σt = Stress in the tube, and
σr = Stress in the rod
Now, Tensile load on the rod = Compressive load on the tube
∴
σr × Ar = σt × At
At
68.75 π
or
σr =
× σt =
× σt = 0.6875σt
...(i)
Ar
100 π
(i) When the compressive load on the tube is 20 kN or 20,000 N.
Then stress in the tube,
Load
20000
=
σt =
Area of tube
68.75 π
= 92.599 N/mm2 (compressive). Ans.
(ii) Substituting this value in equation (i), we get
∴ Stress in the rod, σr = 0.6875 × σt = 0.6875 × 92.599
= 63.66 N/mm2 (tensile). Ans.
(iii) Stresses in the rod and tube, when one nut is tightened by one quarter of a turn.
Let σr* = Stress in the rod and
σt* = Stress in the tube due to tightening of the nut by one-quarter of a turn.
As the stress in the tube is compressive and stress in the rod is tensile hence there will
be decrease in the length of tube but there will be increase in the length of the rod.
∴ Decrease in the length of tube
= Strain × L
Stress in tube
×L
E
σ *t
=
× 800 = 0.004 × σt*
2 × 10 5
Increase in the length of the rod
=
=
Strain =
Stress
E
IJ
K
σ*r
Stress in rod
×L=
×L
E
E
σ *r
=
5
× 800 =
(0.6875 × σ *t ) × 800
2 × 10
2 × 10 5
= 0.00275 × σt*
Axial advancement of the nut = One-quarter of a turn
=
FG∵
H
1
4
(∵
σr = 0.6875σt)
of a turn
But in one turn, the advancement of the nut is
∴ Axial advancement of the nut =
1
4
×
1
4
1
4
th of 10 mm.
× 10 = 0.625 mm
But axial advancement of the nut
= Decrease in length of tube + Increase in the length of rod
41
STRENGTH OF MATERIALS
∴
∴
and
0.625 = 0.004 × σt* + 0.00275σt* = 0.00675 × σt*
0.625
= 92.59 N/mm2. Ans.
σt* =
0.00675
σr* = 0.6875 × 92.59 = 63.65 N/mm2. Ans.
1.14. THERMAL STRESSES..
Thermal stresses are the stresses induced in a body due to change in temperature.
Thermal stresses are set up in a body, when the temperature of the body is raised or lowered
and the body is not allowed to expand or contract freely. But if the body is allowed to expand
or contract freely, no stresses will be set up in the body.
Consider a body which is heated to a certain temperature.
Let
L = Original length of the body,
T = Rise in temperature,
E = Young’s Modulus,
α = Co-efficient of linear expansion,
dL = Extension of rod due to rise of temperature.
If the rod is free to expand, then extension of the rod is given by
...(1.13)
dL = α. T.L.
This is shown in Fig. 1.23 (a) in which AB represents
A
B
B’
the original length and BB′ represents the increase in length
due to temperature rise. Now suppose that an external (a)
compressive load, P is applied at B′ so that the rod is decreased in
dL
L
its length from (L + αTL) to L as shown in Figs. 1.23 (b) and (c).
A
B
B’
Then compressive strain =
Decrease in length
Original length
α.T . L
αTL
≈
=
= α.T
L + α.T . L
L
P
(b)
L
A
B
P
(c)
Stress
L
=E
Strain
Fig. 1.23
∴
Stress = Strain × E = α.T.E
And load or thrust on the rod = Stress × Area = α.T.E × A
If the ends of the body are fixed to rigid supports, so that its expansion is prevented,
then compressive stress and strain will be set up in the rod. These stresses and strains are
known as thermal stresses and thermal strain.
But
∴ Thermal strain,
e=
Extension prevented
Original length
dL
α.T . L
=
= α.T
L
L
And thermal stress, σ = Thermal strain × E
= α.T.E.
Thermal stress is also known as temperature stress.
And thermal strain is also known as temperature strain.
=
42
...(1.14)
...(1.15)
SIMPLE STRESSES AND STRAINS
1.14.1. Stress and Strain when the Supports Yield. If the supports yield by an
amount equal to δ, then the actual expansion
= Expansion due to rise in temperature – δ
= α.T.L – δ.
Actual expansion
(α . T . L − δ)
∴ Actual strain
=
=
Original length
L
And actual stress
= Actual strain × E
(α . T . L − δ)
=
× E.
...(1.16)
L
Problem 1.28. A rod is 2 m long at a temperature of 10°C. Find the expansion of the
rod, when the temperature is raised to 80°C. If this expansion is prevented, find the stress
induced in the material of the rod. Take E = 1.0 × 105 MN/m2 and α = 0.000012 per degree centigrade.
Sol. Given :
Length of rod,
L = 2 m = 200 cm
Initial temperature,
T1 = 10°C
Final temperature,
T2 = 80°C
∴ Rise in temperature,
T = T2 – T1 = 80° – 10° = 70°C
Young’s Modulus,
E = 1.0 × 105 MN/m2
= 1.0 × 105 × 106 N/m2
(∵ M = 106)
11
2
= 1.0 × 10 N/m
Co-efficient of linear expansion, α = 0.000012
(i) The expansion of the rod due to temperature rise is given by equation (1.13).
∴ Expansion of the rod
= α.T.L
= 0.000012 × 70 × 200
= 0.168 cm. Ans.
(ii) The stress in the material of the rod if expansion is prevented is given by equation (1.15).
∴ Thermal stress,
σ=α.T.E
= 0.000012 × 70 × 1.0 × 1011 N/m2
= 84 × 106 N/m2 = 84 N/mm2. Ans. (∵ 106 N/m2 = 1 N/mm2)
Problem 1.29. A steel rod of 3 cm diameter and 5 m long is connected to two grips and
the rod is maintained at a temperature of 95°C. Determine the stress and pull exerted when the
temperature falls to 30°C, if
(i) the ends do not yield, and
(ii) the ends yield by 0.12 cm.
Take E = 2 × 105 MN/m2 and α = 12 × 10 –6/°C.
Sol. Given :
Dia. of the rod,
d = 3 cm = 30 mm
π
× 302 = 225 π mm2
4
Length of the rod,
L = 5 m = 5000 mm
Initial temperature, T1 = 95°C
Final temperature, T2 = 30°C
∴ Area of the rod,
A=
43
STRENGTH OF MATERIALS
∴ Fall in temperature,
Modulus of elasticity,
T = T1 – T2 = 95 – 30 = 65°C
E = 2 × 105 MN/m2
= 2 × 105 × 106 N/m2
= 2 × 1011 N/m2
Co-efficient of linear expansion, α = 12 × 10–6/°C.
(i) When the ends do not yield
The stress is given by equation (1.15).
∴
Stress = α.T.E = 12 × 10–6 × 65 × 2 × 1011 N/m2
= 156 × 106 N/m2 or 156 N/mm2 (tensile). Ans.
Pull in the rod = Stress × Area
= 156 × 225 π = 110269.9 N. Ans.
(ii) When the ends yield by 0.12 cm
∴
δ = 0.12 cm = 1.2 mm
The stress when the ends yield is given by equation (1.16).
(α . T . L − δ)
∴
Stress =
×E
L
(12 × 10 −6 × 65 × 5000 − 1.2)
× 2 × 105 N/mm2
=
5000
(3.9 − 1.2)
=
× 2 × 105 = 108 N/mm2. Ans.
5000
Pull in the rod = Stress × Area
= 108 × 225 π = 76340.7 N. Ans.
1.15. THERMAL STRESSES IN COMPOSITE BARS..
Fig. 1.24 (a) shows a composite bar consisting of two members, a bar of brass and
another of steel. Let the composite bar be heated through some temperature. If the members
are free to expand then no stresses will be induced in the members. But the two members are
rigidly fixed and hence the composite bar as a whole will expand by the same amount. As the
co-efficient of linear expansion of brass is more than that of the steel, the brass will expand
more than the steel. Hence the free expansion of brass will be more than that of the steel. But
both the members are not free to expand, and hence the expansion of the composite bar, as a
whole, will be less than that of the brass, but more than that of the steel. Hence the stress
Brass
Steel
Steel
Brass
Brass
Steel
(a)
(b)
Fig. 1.24
44
(c)
SIMPLE STRESSES AND STRAINS
induced in the brass will be compressive whereas the stress in steel will be tensile as shown
in Fig. 1.24 (c). Hence the load or force on the brass will be compressive whereas on the steel
the load will be tensile.
Let
Ab = Area of cross-section of brass bar
σb = Stress in brass
eb = Strain in brass
αb = Co-efficient of linear expansion for brass
Eb = Young’s modulus for copper
As, σs, es and αs = Corresponding values of area, stress, strain and co-efficient
of linear expansion for steel, and
Es = Young’s modulus for steel.
δ = Actual expansion of the composite bar
Now load on the brass = Stress in brass × Area of brass
= σb × Ab
And load on the steel = σs × As
For the equilibrium of the system, compression in copper should be equal to tension in
the steel
or
Load on the brass = Load on the steel
∴
σb × Ab = σs × As.
Also we know that actual expansion of steel
= Actual expansion of brass
...(i)
But actual expansion of steel
= Free expansion of steel + Expansion due to tensile stress
in steel
= αs . T . L +
σs
.L
Es
And actual expansion of copper
= Free expansion of copper – Contraction due to compressive
stress induced in brass
σb
.L
Eb
Substituting these values in equation (i), we get
= αb . T . L –
αs × T × L +
or
σs
σ
× L = αb × T × L – b × L
Es
Eb
αsT +
σs
σ
= αb × T – b
Es
Eb
where T = Rise of temperature.
Problem 1.30. A steel rod of 20 mm diameter passes centrally through a copper tube of
50 mm external diameter and 40 mm internal diameter. The tube is closed at each end by rigid
plates of negligible thickness. The nuts are tightened lightly home on the projecting parts of the
45
STRENGTH OF MATERIALS
rod. If the temperature of the assembly is raised by 50°C, calculate the stresses developed in
copper and steel. Take E for steel and copper as 200 GN/m2 and 100 GN/m2 and α for steel and
copper as 12 × 10–6 per °C and 18 × 10–6 per °C.
Sol. Given :
Dia. of steel rod
= 20 mm
∴ Area of steel rod, As =
π
× 202 = 100π mm2
4
π
(502 – 402) mm2 = 225π mm2
4
Rise of temperature, T = 50°C
E for steel,
Es = 200 GN/m2
= 200 × 109 N/m2
(∵ G = 109)
= 200 × 103 × 106 N/m2
= 200 × 103 N/mm2
(∵ 106 N/m2 = 1 N/mm2)
E for copper,
Ec = 100 GN/m2 = 100 × 109 N/m2
= 100 × 103 × 106 N/m2 = 100 × 103 N/mm2
α for steel,
αs = 12 × 10–6 per °C
α for copper,
αc = 18 × 10–6 per °C.
As α for copper is more than that of steel, hence the free expansion of copper will be
more than that of steel when there is a rise in temperature. But the ends of the rod and the
tube is fixed to the rigid plates and the nuts are tightened on the projected parts of the rod.
Hence the two members are not free to expand. Hence the tube and the rod will expand by the
same amount. The free expansion of the copper tube will be more than the common expansion,
whereas the free expansion of the steel rod will be less than the common expansion. Hence
the copper tube will be subjected to compressive stress and the steel rod will be subjected to
tensile stress.
Area of copper tube, Ac =
σs = Tensile stress in steel
σc = Compressive stress in copper.
For the equilibrium of the system,
Let
Tensile load on steel = Compressive load on copper
σs . As = σc . Ac
or
σs =
or
=
Ac
× σc
As
225 π
× σc = 2.25σc
100 π
...(i)
We know that the copper tube and steel rod will actually expand by the same amount.
∴
Actual expansion of steel = Actual expansion of copper
...(ii)
But actual expansion of steel
= Free expansion of steel + Expansion due to tensile stress in steel
= αs . T . L +
46
σs
.L
Es
SIMPLE STRESSES AND STRAINS
and actual expansion of copper
= Free expansion of copper – Contraction due to compressive stress in copper
= αc . T . L –
σc
.L
Ec
Substituting these values in equation (ii), we get
αs . T . L +
αs . T +
or
or
σs
σ
. L = αc . T . L – c . L
Es
Ec
12 × 10–6 × 50 +
2.25 σ c
σs
σ
= αc . T – c
Es
Ec
2.25 σ c
200 × 10
3
= 18 × 10–6 × 50 –
σc
100 × 10 3
(∵
σs = 2.25 σc)
σc
= 18 × 10–6 × 50 – 12 × 10–6 × 50
100 × 10 3
200 × 10
1.125 × 10–5 σc + 10–5 σc = 6 × 10–6 × 50
or
3
or
+
2.125 × 10–5 σc = 30 × 10–5
2.125σc = 30
or
or
30
= 14.117 N/mm2. Ans.
2.125
Substituting this value in equation (i), we get
σs = 14.117 × 2.25
= 31.76 N/mm2. Ans.
∴
σc =
Problem 1.31. A steel tube of 30 mm external diameter and 20 mm internal diameter
encloses a copper rod of 15 mm diameter to which it is rigidly joined at each end. If, at a
temperature of 10°C there is no longitudinal stress, calculate the stresses in the rod and tube
when the temperature is raised to 200°C. Take E for steel and copper as 2.1 × 105 N/mm2 and
1 × 10 5 N/mm2 respectively. The value of co-efficient of linear expansion for steel and copper is
given as 11 × 10 –6 per °C and 18 × 10 –6 per °C respectively.
Sol. Given :
Dia. of copper rod
Rise of temperature,
E for steel,
E for copper,
Value of α for steel,
= 15 mm
π
× 152 = 56.25π mm2
Ac =
4
π
As =
(302 – 202) = 125π mm2
4
T = (200 – 10) = 190°C
Es = 2.1 × 105 N/mm2
Ec = 1 × 105 N/mm2
αs = 11 × 10–6 per °C
Value of α for copper,
αc = 18 × 10–6 per °C
∴ Area of copper rod,
Area of steel tube,
47
STRENGTH OF MATERIALS
As the value of α for copper is more than that of steel, hence the copper rod would
expand more than the steel tube if it were free. Since the two are joined together, the copper
will be prevented from expanding its full amount and will be put in compression, the steel
being put in tension.
Let
σs = Stress in steel
σc = Stress in copper.
For equilibrium of the system,
Compressive load on copper = Tensile load on steel
or
σc . Ac = σs . As
125 π
A
∴
σc = σs . s = σs .
= 2.22 × σs
...(i)
56.25 π
Ac
We know that the copper rod and the steel tube will actually expand by the same
amount.
Now actual expansion of steel = Free expansion of steel + Expansion due to tensile stress
σ
= αs . T . L + s . L
Es
and actual expansion of copper = Free expansion of copper
– Contraction due to compressive stress
σ
= αc . T . L – c . L
Ec
But actual expansion of steel = Actual expansion of copper
αs . T . L +
σs
σ
.L = αc . T . L – c . L
Es
Ec
αs . T +
or
11 × 10–6 × 190 +
or
σs
or
2.1 × 10
5
+
σs
Es
= αc . T –
σs
2.1 × 10
2.22 σ s
1 × 10 5
5
σc
Ec
= 18 × 10–6 × 190 –
2.22 σ s
1 × 10 5
(∵
σc = 2.22σs)
= 18 × 10–6 × 190 – 11 × 10–6 × 190
σ s + 2.1 × 2.22 σ s
= 5 × 10–6 × 190
2.1 × 10 5
σs + 4.662σs = 5 × 10–6 × 190 × 2.1 × 105
5.662σs = 199.5
or
or
or
199.5
= 35.235 N/mm2. Ans.
5.662
Substituting this value in equation (i), we get
σc = 2.22 × 35.235 = 78.22 N/mm2. Ans.
∴
σs =
Problem 1.32. A steel tube of 30 mm external diameter and 25 mm internal diameter
encloses a gun metal rod of 20 mm diameter to which it is rigidly joined at each end. The
temperature of the whole assembly is raised to 140°C and the nuts on the rod are then screwed
48
SIMPLE STRESSES AND STRAINS
lightly home on the ends of the tube. Find the intensity of stress in the rod when the common
temperature has fallen to 30°C. The value of E for steel and gun metal is 2.1 × 105 N/mm2 and
1 × 105 N/mm2 respectively. The linear co-efficient of expansion for steel and gun metal
is 12 × 10–6 per °C and 20 × 10–6 per °C.
Sol. Given :
Dia. of gun metal rod
= 20 mm
∴ Area of gun metal rod,
Ag =
π
× 202 = 100π mm2
4
π
(302 – 252) = 68.75π mm2
4
Fall in temperature,
T = 140 – 30 = 110
Value of E for steel,
Es = 2.1 × 105 N/mm2
Value of E for gun metal, Eg = 1 × 105 N/mm2
Value of α for steel,
αs = 12 × 10–6 per °C
Value of α for gun metal, αg = 20 × 10–6 per °C.
As αg is greater than αs, hence the free contraction of the gun metal rod will be more
than that of steel when there is a fall in temperature. But, since the ends of the rods have
been provided with nuts, the two members are not free to contract fully, each of the member
will contract by the same amount. The free contraction of the gun metal rod will be greater
than the common contraction, whereas the free contraction of the steel tube will be less than
the common contraction. Hence the steel tube will be subjected to compressive stress while
the gun metal rod will be subjected to tensile stress.
Let
σs = Stress in steel tube and
σg = Stress in gun metal rod.
For the equilibrium of the system,
Total compressive force in steel = Total tensile force in gun metal
∴
σs . As = σg . Ag
Area of steel tube,
or
As =
σs = σg .
Ag
As
= σg .
100π
68.75π
σs = 1.4545σg
...(i)
We also know that the steel tube and gun metal rod will actually contract by the same
amount.
∴
Actual contraction of steel = Actual contraction of gun metal rod
But actual contraction of steel = Free contraction of steel
+ contraction due to compressive stress in steel
or
= αs . T . L +
σs
.L
Es
Actual contraction of gun metal = Free contraction of gun metal
– expansion due to tensile stress in gun metal
= αg . T . L –
σg
Eg
.L
49
STRENGTH OF MATERIALS
Equating the two values, we get
αs . T . L +
σg
σs
. L = αg . T . L –
.L
Eg
Es
αs . T +
or
12 × 10–6 × 110 +
or
σs
Es
1.4545 σ g
2.1 × 10
5
= αg . T –
σg
Eg
= 20 × 10–6 × 110 –
σg
1 × 10 5
(∵ σs = 1.4545 σg)
σg
1.4545
σg +
= 20 × 10–6 × 110 – 12 × 10–6 × 110
5
1 × 10 5
2.1 × 10
or
1.4545 σ g + 2.1 × σ g
= 8 × 10–6 × 110
2.1 × 10 5
3.5545 σg = 8 × 10–6 × 110 × 2.1 × 105 = 184.8
or
or
184.8
= 51.99 N/mm2. Ans.
3.5545
Substituting this value in equation (i), we get
σs = 1.4545 × 51.99 = 75.62 N/mm2. Ans.
∴
σg =
1.16. ELONGATION OF A BAR DUE TO ITS OWN WEIGHT..
Fig. 1.25 shows a bar AB fixed at end A and hanging freely
under its own weight.
A
Let
L = Length of bar,
A = Area of cross-section,
E = Young’s modulus for the bar material,
w = Weight per unit volume of the bar material.
L
dx
Consider a small strip of thickness dx at a distance x from the
lower end.
x
Weight of the bar for a length of x is given by,
P = Specific weight × Volume of bar upto length x
B
=w×A×x
Fig. 1.25
This means that on the strip, a weight of w × A × x is acting in
the downward direction. Due to this weight, there will be some increase in the length of
element. But length of the element is dx.
Now stress on the element
Weight acting on element
w× A× x
=
=
=w×x
Area of cross-section
A
The above equation shows that stress due to self weight in a bar is not uniform. It
depends on x. The stress increases with the increase of x.
Strain in the element
50
=
Stress w × x
=
E
E
SIMPLE STRESSES AND STRAINS
∴ Elongation of the element
= Strain × Length of element
w× x
× dx
=
E
Total elongation of the bar is obtained by integrating the above equation between limits
zero and L.
∴
δL =
z
L
0
w× x
w
dx =
E
E
LM OP
N Q
w x2
=
E 2
=
WL
2E
L
=
0
z
L
0
x . dx
w L2
×
2
E
...(1.17)
(∵
W = w × L)
..(1.18)
1.17. ANALYSIS OF BAR OF UNIFORM STRENGTH..
In the previous article we have seen that the stress due to self weight of the bar is not
constant but the stress increases with the increase of distance from the lower end. If the self
weight is neglected and a bar of uniform section is subjected to an axial load, then the stress
in the bar would be uniform.
Let us find the shape of the bar of which self weight of the bar is considered and is
having uniform stress on all sections when subjected to an axial P. Such bar is shown in
Fig. 1.26, in which the area of the bar increases from the lower end to the upper end.
Let
A1 = Area of upper end,
A2 = Area of lower end,
w = Weight per unit volume of the bar,
(A + dA)
σ = Uniform stress on the bar.
Consider a strip of length dx at a distance x from
D
C D
dx
the lower end. Let A be the area of the strip at section
A
A
B
AB and (A + dA) be the area at section DC. Consider the
x
A + wAdx
equilibrium of the strip ABCD.
The forces acting on the strip are :
(i) Weight of strip acting downward and equal to
P
w × volume of strip i.e., w × A × dx.
Fig. 1.26
(ii) Force on section AB due to uniform stress (σ)
and is equal to σ × A. This is acting downward.
(iii) Force on section CD due to uniform stress (σ) and is equal to σ (A + dA). This is
acting upwards.
Now, total force acting upwards
= Total force acting downwards
or
σ (A + dA) = σ × A + wAdx
or
σ × A + σdA = σ × A + wAdx
or
σdA = wAdx
dA
w
or
=
dx
σ
A
51
STRENGTH OF MATERIALS
z z
Integrating the above equation, we get
dA
w
w
=
dx or loge A =
x+C
σ
σ
A
where C is the constant of integration.
At
x = 0,
A = A2
Substituting these values in equation (i), we get
w
loge A2 =
×0+C
σ
∴
C = loge A2
Substituting the value of C in equation (i), we get
w
loge A =
x + loge A2
σ
w
A
w
or
loge A – loge A2 =
x or loge
=
.x
σ
A2
σ
...(i)
FG IJ
H K
wx
or
wx
A
=eσ
or
A = A2 e σ
A2
The above equation gives the area at a distance x from lower end.
At
x = L,
A = A1
Substituting these values in equation (ii), we get
A1 = A2 e
wL
σ
...(ii)
...(1.19)
Problem 1.33. A vertical bar fixed at the upper end and of uniform strength carries an
axial tensile load of 600 kN. The bar is 20 m long and having weight per unit volume as
0.00008 N/mm3. If the area of the bar at the lower end is 400 mm2, find the area of the bar at
the upper end.
Sol. Given :
Axial load,
P = 600 kN = 600 × 103 N
Length,
L = 20 m = 20 × 103 mm
Weight per unit volume, w = 0.00008 N/mm3
Area of bar at lower end, A2 = 400 mm2
Let
A1 = Area of bar at upper end.
Now the uniform stress* on the bar,
σ =
P
600 × 10 3
=
= 1500 N/mm2
A2
400
Using equation (1.19), we get
A 1 = A2 e
wL
σ
0.00008 × 20 × 10 3
1500
e
= 400 ×
A1
= e 0.0010667
400
or
*The stress on lower end =
52
= 400 × e0.0010667
P
P
. We want that the stress in the bar should be uniform i.e., equal to
.
A2
A2
SIMPLE STRESSES AND STRAINS
or
loge =
or
2.3 log10
or
log10
A1
= 0.0010667
400
A1
= 0.0010667
400
A1
0.0010667
=
= 0.00046378
2.3
400
A1
= Antilog of 0.00046378 = 1.00107
400
A1 = 400 × 1.00107 = 400.428 mm2. Ans.
∴
∴
HIGHLIGHTS
1.
The resistance per unit area, offered by a body against deformation is known as stress. The
stress is given by
σ=
P
A
where P = External force or load ; A = Cross-sectional area.
2.
Stress is expressed as kgf/m2, kgf/cm2, N/m2 and N/mm2.
3.
1 N/m2 = 10–4 N/cm2 or 10–6 N/mm2.
4.
The ratio of change of dimension of the body to the original dimension is known as strain.
5.
The stress induced in a body, which is subjected to two equal and opposite pulls, is known as
tensile stress.
6.
The stress induced in a body, which is subjected to two equal and opposite pushes, is known as
compressive stress.
7.
Elasticity is the property by virtue of which certain materials return back to their original
position after the removal of the external force.
8.
Hooke’s law states that within elastic limit, the stress is proportional to the strain.
9.
The ratio of tensile stress (or compressive stress) to the corresponding strain is known as Young’s
modulus or modulus of elasticity and is denoted by E.
∴
E=
Tensile or compressive stress
.
Corresponding strain
10.
The ratio of shear stress to the corresponding shear strain within the elastic limit, is known as
modulus of rigidity or shear modulus. It is denoted by C (or G or N).
11.
Total change in the length of a bar of different lengths and of different diameters when subjected to an axial load P, is given by
dL =
LM L + L + L + ....OP
NA A A Q
LL + L + L
=P M
NE A E A E A
P
E
1
2
3
1
2
3
1
1 1
2
2
3
2
3
3
....... when E is same
OP
Q
+ ...
....... when E is different.
53
STRENGTH OF MATERIALS
12.
The total extension of a uniformly tapering circular rod of diameters D1 and D2, when the rod is
subjected to an axial load P is given by
dL =
13.
14.
15.
16.
17.
19.
20.
where L = Total length of the rod.
A composite bar is made up of two or more bars of equal lengths but of different materials rigidly
fixed with each other and behaving as one unit for extension or compression.
In case of a composite bar having equal length : (i) strain in each bar is equal and (ii) total load
on the composite bar is equal to the sum of loads carried by each different materials.
The stresses induced in a body due to change in temperature are known as thermal stresses.
Thermal strain and thermal stress is given by
thermal strain, e = α .T and thermal stress, p = α.T.E
where
α = Co-efficient of linear expansion ,
T = Rise or fall of temperature,
E = Young’s modulus.
Total elongation of a uniformly tapering rectangular bar when subjected to an axial load P is
given by
dL =
18.
4 PL
π ED1D2
a
PL
loge
b
Et(a − b)
where L = Total length of bar ;
t = Thickness of bar
a = Width at bigger end ; b = Width at smaller end
E = Young’s modulus.
In case of a composite bar having two or more bars of different lengths, the extension or compression in each bar will be equal. And the total load will be equal to the sum of the loads carried
by each member.
In case of nut and bolt used on a tube with washers, the tensile load on the bolt is equal to the
compressive load on the tube.
Elongation of a bar due to its own weight is given by
δL =
where
WL
w L2
×
or
2E
2
E
w = Weight per unit volume of the bar material
L = Length of bar.
EXERCISE
(A) Theoretical Questions
1.
Define stress and strain. Write down the S.I. and M.K.S. units of stress and strain.
2.
Explain clearly the different types of stresses and strains.
3.
Define the terms : Elasticity, elastic limit, Young’s modulus and modulus of rigidity.
4.
State Hooke’s law.
5.
Three sections of a bar are having different lengths and different diameters. The bar is subjected to an axial load P. Determine the total change in length of the bar. Take Young’s modulus
of different sections same.
54
SIMPLE STRESSES AND STRAINS
6.
Distinguish between the following, giving due explanation :
(i) Stress and strain,
(ii) Force and stress, and
(iii) Tensile stress and compressive stress.
7. Prove that the total extension of a uniformly tapering rod of diameters D1 and D2, when the rod
is subjected to an axial load P is given by
dL =
8.
9.
10.
11.
12.
13.
14.
4 PL
π ED1D2
where L = Total length of the rod.
Define a composite bar. How will you find the stresses and load carried by each member of a
composite bar ?
Define modular ratio, thermal stresses, thermal strains and Poisson’s ratio.
A rod whose ends are fixed to rigid supports, is heated so that rise in temperature is T °C. Prove
that the thermal strain and thermal stresses set up in the rod are given by,
Thermal strain = α.T and
Thermal stress = α.T.E
where α = Co-efficient of linear expansion.
What is the procedure of finding thermal stresses in a composite bar ?
What do you mean by ‘a bar of uniform strength’ ?
Find an expression for the total elongation of a bar due to its own weight, when the bar is fixed
at its upper end and hanging freely at the lower end.
Find an expression for the total elongation of a uniformly tapering rectangular bar when it is
subjected to an axial load P.
(B) Numerical Problems
1.
2.
3.
4.
5.
6.
A rod 200 cm long and of diameter 3.0 cm is subjected to an axial pull of 30 kN. If the Young’s
modulus of the material of the rod is 2 × 105 N/mm2, determine : (i) stress, (ii) strain and (iii) the
elongation of the rod.
[Ans. (i) 42.44 N/mm2 (ii) 0.000212 (iii) 0.0424 cm]
Find the Young’s modulus of a rod of diameter 30 mm and of length 300 mm which is subjected
to a tensile load of 60 kN and the extension of the rod is equal to 0.4 mm. [Ans. 63.6 GN/m2]
The safe stress, for a hollow steel column which carries an axial load of 2.2 × 103 kN is 120 MN/m2.
If the external diameter of the column is 25 cm, determine the internal diameter.
[Ans. 19.79 cm]
An axial pull of 40000 N is acting on a bar consisting of three sections of length 30 cm, 25 cm and
20 cm and of diameters 2 cm, 4 cm and 5 cm respectively. If the Young’s modulus = 2 × 105 N/mm2,
determine :
(i) stress in each section and
(ii) total extension of the bar.
[Ans. (i) 127.32, 31.8, 20.37 N/mm2, (ii) 0.025 cm]
The ultimate stress for a hollow steel column which carries an
F
6 cm × 6 cm
axial load of 2 MN is 500 N/mm2. If the external diameter of the
Steel bar
column is 250 mm, determine the internal diameter. Take the 20 cm
factor of safety as 4.0.
10 cm × 10 cm
[Ans. – 205.25 mm]
Aluminium bar
30 cm
A member formed by connecting a steel bar to an aluminium
bar is shown in Fig. 1.27. Assuming that the bars are
prevented from buckling sideways, calculate the magnitude of
force P, that will cause the total length of the member to
Fig. 1.27
55
STRENGTH OF MATERIALS
7.
decrease 0.30 mm. The values of elastic modulus for steel and aluminium are 2 × 105N/mm2
and 6.5 × 104 N/mm2 respectively.
[Ans. 406.22 kN]
The bar shown in Fig. 1.28 is subjected to a tensile load of 150 kN. If the stress in the middle
portion is limited to 160 N/mm2, determine the diameter of the middle portion. Find also the
length of the middle portion if the total elongation of the bar is to be 0.25 cm. Young’s modulus
[Ans. 3.45 cm, 29.38 cm]
is given as equal to 2.0 × 105 N/mm2.
150 kN
10 cm DIA
10 cm DIA
150 kN
45 cm
Fig. 1.28
8.
A brass bar, having cross-section area of 900 mm2, is subjected to axial forces as shown in
Fig. 1.29 in which AB = 0.6 m, BC = 0.8 m and CD = 1.0 m.
A
40 kN
B
C
70 kN
D
10 kN
20 kN
Fig. 1.29
Find the total elongation of the bar. Take E = 1 × 105 N/mm2.
9.
[Ans. – 0.111 mm]
A member ABCD is subjected to point loads P1, P2, P3 and P4 as shown in Fig. 1.30. Calculate
the force P3 necessary for equilibrium if P1 = 120 kN, P2 = 220 kN and P4 = 160 kN. Determine
[Ans. 0.55 mm]
also the net change in the length of the member. Take E = 200 GN/m2.
40 mm × 40 mm 25 mm × 25 mm 30 mm × 30 mm
A
B
C
D
P2
P1
0.75 cm
P4
P3
1m
1.2 m
Fig. 1.30
10.
A rod, which tapers uniformly from 5 cm diameter to 3 cm diameter in a length of 50 cm, is
subjected to an axial load of 6000 N. If E = 2 × 105 N/mm2, find the extension of the rod.
[Ans. 0.00127 cm]
11.
Find the modulus of elasticity for a rod, which tapers uniformly from 40 mm to 25 mm
diameter in a length of 400 mm. The rod is subjected to a load of 6 kN and extension of the rod
is 0.04 mm.
[Ans. 76.39 kN/mm2]
12.
A rectangular bar made of steel is 3 m long and 10 mm thick. The rod is subjected to an axial
tensile load of 50 kN. The width of the rod varies from 70 mm at one end to 28 mm at the other.
[Ans. 1.636 mm]
Find the extension of the rod if E = 2 × 105 N/mm2.
56
SIMPLE STRESSES AND STRAINS
13.
The extension in a rectangular steel bar of length 800 mm and of thickness 20 mm, is found to be
0.21 mm. The bar tapers uniformly in width from 80 mm to 40 mm. If E for the bar is 2 × 105
N/mm2, determine the axial tensile load on the bar.
[Ans. 60.6 kN]
14.
A steel rod of 2 cm diameter is enclosed centrally in a hollow copper tube of external diameter
4 cm and internal diameter of 3.5 cm. The composite bar is then subjected to an axial pull of
50000 N. If the length of each bar is equal to 20 cm, determine :
(i) the stress in the rod and tube, and
(ii) load carried by each bar.
Take E for steel = 2 × 105 N/mm2 and for copper = 1 × 105 N/mm2.
[Ans. (i) 54.18 ; 108.36 N/mm2 (ii) 34043.4 N and 15956.6 N]
A load of 1.9 MN is applied on a short concrete column 300 mm ×
200 mm. The column is reinforced with four steel bars of 10 mm
diameter, one in each corner. Find the stresses in the concrete
and steel bars. Take E for steel as 2.1 × 105 N/mm2 and for concrete as 1.4 × 104 N/mm2.
200 mm
Brass
16.
Steel
A mild steel rod of 20 mm diameter and 300 mm long is enclosed centrally inside a hollow copper
tube of external diameter 30 mm and internal diameter of 25 mm. The ends of the tube and rods
are brazed together, and the composite bar is subjected to an axial pull of 40 kN. If E for steel
and copper is 200 GN/m2 and 100 GN/m2 respectively, find the stresses developed in the rod and
tube. Also find the extension of the rod.
[Ans. 94.76 N/mm2, 47.38 N/mm2 and 0.142 mm]
50 kN
Brass
15.
[Ans. 20.13, 301.9 N/mm2]
17.
A reinforced short concrete column 250 mm × 250 mm in section
is reinforced with 8 steel bars. The total area of steel bars is
1608.50 mm2. The column carries a load of 270 kN. If the modulus of elasticity for steel is 18 times that of concrete, find the
stresses in concrete and steel.
100 mm
Fig. 1.31
If the stress in concrete shall not exceed 4 N/mm2, find the area of steel required so that the
column may support a load of 400 kN. [Ans. σc = 3 N/mm2, σs = 54 N/mm2 and As = 2206 mm2]
18.
Two vertical rods one of steel and other of copper are each rigidly fixed at the top and 60 cm
apart. Diameters and length of each rod are 3 cm and 3.5 cm respectively. A cross bar fixed to
the rods at the lower ends carries a load of 6000 N such that the cross bar remains horizontal
even after loading. Find the stress in each rod and the position of the load on the bar. Take E for
steel = 2 × 105 N/mm2 and for copper = 1 × 105 N/mm2.
[Ans. 2.829 and 5.658 N/mm2 ; 39.99 cm]
19.
mm2
and two brass rods each of cross-sectional area of
A steel rod of cross-sectional area 1600
1000 mm2 together support a load of 50 kN as shown in Fig. 1.31.
Find the stresses in the rods. Take E for steel = 2 × 105 N/mm2 and E for brass = 1 × 105 N/mm2.
[Ans. σb = 12.1 N/mm2 and σs = 16.12 N/mm2]
20.
A rod is 3 m long at a temperature of 15°C. Find the expansion of the rod, when the temperature
is raised to 95°C. If this expansion is prevented, find the stress induced in the material of the
rod. Take E = 1 × 105 N/mm2 and α = 0.000012 per degree centigrade.
[Ans. 0.288 cm, 96 N/mm2]
57
STRENGTH OF MATERIALS
21.
A steel rod 5 cm diameter and 6 m long is connected to two grips and the rod is maintained at a
temperature of 100°C. Determine the stress and pull exerted when the temperature falls to
20°C if (i) the ends do not yield, and (ii) the ends yield by 0.15 cm.
Take E = 2 × 105 N/mm2 and α = 12 × 10–6/°C.
[Ans. (i) 192 N/mm2 and 376990 N (ii) 142 N/mm2, 278816.3 N]
22.
A steel rod of 20 mm diameter passes centrally through a copper tube 40 mm external diameter
and 30 mm internal diameter. The tube is closed at each end by rigid plates of negligible thickness.
The nuts are tightened lightly home on the projected parts of the rod. If the temperature of the
assembly is raised by 60°C, calculate the stresses developed in copper and steel. Take E for steel
and copper as 200 GN/m2 and 100 GN/m2 and α for steel and copper as 12 × 10–6 per °C and
[Ans. 16.23, 28.4 N/mm2]
18 × 10–6 per °C.
23.
A vertical bar fixed at the upper end and of uniform strength carries an axial tensile load of 500
kN. The bar is 18 m long and having weight per unit volume as 0.00008 N/mm2. If the area of the
bar at the lower end is 500 mm2, find the area of the bar at the upper end. [Ans. 500.72 mm2]
A straight circular rod tapering from diameter ‘D’ at one end to a diameter ‘d’ at the other end
is subjected to an axial load ‘P’. Obtain an expression for the elongation of the rod.
24.
LMAns.
N
58
δL =
4 PL
πE. D.d
OP
Q
2
ELASTIC CONSTANTS
HAPTER
2.1. INTRODUCTION..
When a body is subjected to an axial tensile load, there is an increase in the length of
the body. But at the same time there is a decrease in other dimensions of the body at right
angles to the line of action of the applied load. Thus the body is having axial deformation and
also deformation at right angles to the line of action of the applied load (i.e., lateral deformation).
This chapter deals with these deformations, Poisson’s ratio, volumetric strains, bulk modulus,
relation between Young’s modulus and modulus of rigidity and relation between Young’s
modulus and bulk modulus.
2.2. LONGITUDINAL STRAIN..
When a body is subjected to an axial tensile or compressive load, there is an axial
deformation in the length of the body. The ratio of axial deformation to the original length of
the body is known as longitudinal (or linear) strain. The longitudinal strain is also defined as
the deformation of the body per unit length in the direction of the applied load.
Let
L = Length of the body,
P = Tensile force acting on the body,
δL = Increase in the length of the body in the direction of P.
δL
.
Then, longitudinal strain =
L
2.3. LATERAL STRAIN..
The strain at right angles to the direction of applied load is known as lateral strain. Let
a rectangular bar of length L, breadth b and depth d is subjected to an axial tensile load P as
shown in Fig. 2.1. The length of the bar will increase while the breadth and depth will
decrease.
Let
δL = Increase in length,
δb = Decrease in breadth, and
δd = Decrease in depth.
δL
L
...(2.1)
δb
δd
or
b
d
...(2.2)
Then longitudinal strain =
and
lateral strain
=
59
STRENGTH OF MATERIALS
b
P
(d – δd)
d
P
(b – δb)
l
l + δl
Fig. 2.1
Note. (i) If longitudinal strain is tensile, the lateral strains will be compressive.
(ii) If longitudinal strain is compressive then lateral strains will be tensile.
(iii) Hence every longitudinal strain in the direction of load is accompanied by lateral strains of
the opposite kind in all directions perpendicular to the load.
2.4. POISSON’S RATIO..
The ratio of lateral strain to the longitudinal strain is a constant for a given material,
when the material is stressed within the elastic limit. This ratio is called Poisson’s ratio
and it is generally denoted by μ. Hence mathematically,
Poisson’s ratio, μ
=
Lateral strain
Longitudinal strain
...(2.3)
Lateral strain = μ × longitudinal strain
As lateral strain is opposite in sign to longitudinal strain, hence algebraically, the lateral
strain is written as
...[2.3 (A)]
Lateral strain = – μ × longitudinal strain
The value of Poisson’s ratio varies from 0.25 to 0.33. For rubber, its value ranges from
0.45 to 0.50.
or
Problem 2.1. Determine the changes in length, breadth and thickness of a steel bar
which is 4 m long, 30 mm wide and 20 mm thick and is subjected to an axial pull of 30 kN in
the direction of its length. Take E = 2 × 105 N/mm2 and Poisson’s ratio = 0.3.
Sol. Given :
Length of the bar,
L = 4 m = 4000 mm
Breadth of the bar,
b = 30 mm
Thickness of the bar,
t = 20 mm
∴ Area of cross-section,
A = b × t = 30 × 20 = 600 mm2
Axial pull,
P = 30 kN = 30000 N
Young’s modulus,
E = 2 × 105 N/mm2
Poisson’s ratio,
μ = 0.3.
Now strain in the direction of load (or longitudinal strain),
60
=
Stress
Load
=
E
Area × E
=
P
30000
=
= 0.00025.
A. E. 600 × 2 × 10 5
FG∵
H
Stress =
Load
Area
IJ
K
ELASTIC CONSTANTS
But longitudinal strain =
δL
.
L
δL
= 0.00025.
L
∴ δL (or change in length) = 0.00025 × L
= 0.00025 × 4000 = 1.0 mm. Ans.
Using equation (2.3),
Lateral strain
=
Poisson’s ratio
Longitudinal strain
Lateral strain
or
0.3 =
0.00025
∴ Lateral strain = 0.3 × 0.00025 = 0.000075.
We know that
δb
δd
δt
Lateral strain =
or
or
b
d
t
∴
δb = b × Lateral strain
= 30 × 0.000075 = 0.00225 mm. Ans.
Similarly,
δt = t × Lateral strain
= 20 × 0.000075 = 0.0015 mm. Ans.
Problem 2.2. Determine the value of Young’s modulus and Poisson’s ratio of a metallic
bar of length 30 cm, breadth 4 cm and depth 4 cm when the bar is subjected to an axial
compressive load of 400 kN. The decrease in length is given as 0.075 cm and increase in breadth
is 0.003 cm.
Sol. Given :
Length, L = 30 cm ; Breadth, b = 4 cm ; and Depth, d = 4 cm.
∴ Area of cross-section,
A = b×d=4×4
= 16 cm2 = 16 × 100 = 1600 mm2
Axial compressive load,
P = 400 kN = 400 × 1000 N
Decrease in length,
δL = 0.075 cm
Increase in breadth,
δb = 0.003 cm
δL 0.075
=
Longitudinal strain
=
= 0.0025
30
L
δb 0.003
Lateral strain
=
=
= 0.00075.
4
b
Using equation (2.3),
Lateral strain
0.00075
=
Poisson’s ratio
=
= 0.3. Ans.
Longitudinal strain
0.0025
Stress
P
Load P
=
∵ Stress =
=
Longitudinal strain
=
E
A× E
Area A
400000
or
0.0025 =
1600 × E
400000
∴
E=
= 1 × 105 N/mm2. Ans.
1600 × 0.0025
∴
FG
H
IJ
K
FG
H
IJ
K
61
STRENGTH OF MATERIALS
2.5. VOLUMETRIC STRAIN..
The ratio of change in volume to the original volume of a body (when the body is subjected
to a single force or a system of forces) is called volumetric strain. It is denoted by ev.
Mathematically, volumetric strain is given by
δV
ev =
V
where δV = Change in volume, and
V = Original volume.
2.5.1. Volumetric Strain of a
Rectangular Bar which is Subjected to an
d
Axial Load P in the Direction of its Length.
Consider a rectangular bar of length L, width b P
P
and depth d which is subjected to an axial load P
b
in the direction of its length as shown in Fig. 2.2.
L
Let
δL = Change in length,
Fig. 2.2
δ b = Change in width, and
δ d = Change in depth.
= L + δL
∴ Final length of the bar
= b + δb
Final width of the bar
Final depth of the bar
= d + δd
Now original volume of the bar, V = L.b.d
Final volume
= (L + δL)(b + δb)(d + δd)
= L.b.d. + bdδL + Lbδd + Ld.δb
(Ignoring products of small quantities)
∴ Change in volume,
δV = Final volume – Original volume
= (Lbd + bdδL + Lbδd + Ldδb) – Lbd
= bdδL + Lbδd + Ldδb
∴ Volumetric strain,
δV
ev =
V
bdδL + Lbδ d + Ldδb
=
Lbd
δL δ d δb
+
+
=
...(2.4)
d
b
L
δL
δd
δb
= Longitudinal strain and
are lateral strains.
But
or
L
d
b
Substituting these values in the above equation, we get
ev = Longitudinal strain + 2 × Lateral strain
...(i)
From equation (2.3A), we have
∴
Lateral strain = – μ × Longitudinal strain.
Substituting the value of lateral strain in equation (i), we get
ev = Longitudinal strain – 2 × μ longitudinal strain
62
ELASTIC CONSTANTS
= Longitudinal strain (1 – 2μ)
δL
(1 – 2μ)
...(2.5)
=
L
Problem 2.3. For the problem 2.1, determine the volumetric strain and final volume of
the given steel bar.
Sol. Given :
The following data is given in problem 2.1 :
L = 4000 mm, b = 30 mm, t or d = 20 mm, μ = 0.3.
Original volume, V = L.b.d = 4000 × 30 × 20 = 2400000 mm3
δL
The value of longitudinal strain i. e.,
in problem 2.1 is calculated
L
δL
= 0.00025
as,
L
Now using equation (2.5), we have
δL
Volumetric strain, ev =
(1 – 2μ)
L
= 0.00025(1 – 2 × 0.3) = 0.0001. Ans.
FG
H
or
∴
∴
IJ
K
δV
= 0.0001
V
δV = 0.0001 × V
= 0.0001 × 2400000 = 240 mm3
Final volume = Original volume + δV
= 2400000 + 240 mm3
= 2400240 mm3. Ans.
FG∵
H
ev =
δV
V
IJ
K
Problem 2.4. A steel bar 300 mm long, 50 mm wide and 40 mm thick is subjected to a
pull of 300 kN in the direction of its length. Determine the change in volume. Take E = 2 × 105
N/mm2 and μ = 0.25.
Sol. Given :
Length,
L = 300 mm
Width,
b = 50 mm
Thickness,
t = 40 mm
Pull,
P = 300 kN = 300 × 103 N
Value of
E = 2 × 105 N/mm2
Value of
μ = 0.25
Original volume,
V = L×b×t
= 300 × 50 × 40 mm3 = 600000 mm3
The longitudinal strain (i.e., the strain in the direction of load) is given by
dL Stress in the direction of load
=
L
E
But stress in the direction of load
=
P
P
=
Area b × t
63
STRENGTH OF MATERIALS
=
dL
150
=
= 0.00075
L
2 × 10 5
∴
or
300 × 10 3
= 150 N/mm2
50 × 40
Now volumetric strain is given by equation (2.5) as
dL
ev =
(1 – 2μ)
L
= 0.00075 (1 – 2 × 0.25) = 0.000375
dV
represents volumetric strain.
Let δV = Change in volume. Then
V
dV
= 0.000375
∴
V
dV = 0.000375 × V
= 0.000375 × 600000 = 225 mm3. Ans.
2.5.2. Volumetric Strain of a
Rectangular Bar Subjected to Three Forces
which are Mutually Perpendicular.
Consider a rectangular block of dimensions
x, y and z subjected to three direct tensile
stresses along three mutually perpendicular
axis as shown in Fig. 2.3.
Then volume of block,
Z
Y
X
X
Y
V = xyz.
Z
Fig. 2.3
Taking logarithm to both sides, we have
log V = log x + log y + log z.
Differentiating the above equation, we get
1
1
1
1
dV = dx + dy + dz
V
x
y
z
dV dx dy dz
=
+
+
V
x
y
z
or
But
dV Change of volume
=
= Volumetric strain
Original volume
V
dx Change of dimension x
=
Original dimension x
x
= Strain in the x-direction = ex
Similarly,
dy
= Strain in y-direction = ey
y
dz
= Strain in z-direction = ez
z
and
64
...(2.6)
ELASTIC CONSTANTS
Substituting these values in equation (2.6), we get
dV
= ex + ey + ez
V
Now, Let σx = Tensile stress in x-x direction,
σy = Tensile stress in y-y direction, and
σz = Tensile stress in z-z direction.
E = Young’s modulus
µ = Poisson’s ratio.
Now σx will produce a tensile strain equal to
σx
in the direction of x, and a compressive
E
µ × σx
in the direction of y and z. Similarly, σy will produce a tensile strain
E
σy
µ × σy
in the direction of y and a compressive strain equal to
in the direction of x
equal to
E
E
σ
and z. Similarly σz will produce a tensile strain equal to z in the direction of z and a compE
µ × σz
ressive strain equal to
in the direction of x and y. Hence σy and σz will produce
E
µ × σy
µ × σz
and
in the direction of x.
compressive strains equal to
E
E
∴ Net tensile strain along x-direction is given by
strain equal to
ex =
Similarly,
and
FG σ
H
Fσ
σ
−µG
=
E
H
ey =
ez
FG
H
IJ
K
σy + σz
σx µ × σy µ × σz σx
−
−
=
−µ
.
E
E
E
E
E
σy
E
−µ
z
x
+ σz
E
x
+ σy
E
IJ
K
IJ
K
Adding all the strains, we get
e x + ey + ez =
=
But
1
2µ
(σ + σy + σz) –
(σx + σy + σz)
E x
E
1
(σ + σy + σz)(1 – 2µ).
E x
ex + ey + ez = Volumetric strain =
dV
.
V
dV
1
(σx + σy + σz)(1 – 2µ)
...(2.7)
=
V
E
Equation (2.7) gives the volumetric strain. In this equation the stresses σx , σy and σz
are all tensile. If any of the stresses is compressive, it may be regarded as negative, and the
dV
above equation will hold good. If the value of
is positive, it represents increase in volume
V
dV
represents a decrease in volume.
whereas the negative value of
V
65
∴
STRENGTH OF MATERIALS
Problem 2.5. A metallic bar 300 mm × 100 mm × 40 mm is subjected to a force of 5 kN
(tensile), 6 kN (tensile) and 4 kN (tensile) along x, y and z directions respectively. Determine
the change in the volume of the block. Take E = 2 ×105 N/mm2 and Poisson’s ratio = 0.25.
Sol. Given :
Dimensions of bar
= 300 mm × 100 mm × 40 mm
∴
x = 300 mm, y = 100 mm and z = 40 mm
∴ Volume,
V = x × y × z = 300 × 100 × 40
= 1200000 mm3
Load in the direction of x
= 5 kN = 5000 N
Load in the direction of y
= 6 kN = 6000 N
Load in the direction of z
= 4 kN = 4000 N
Value of E
= 2 × 105 N/mm2
Poisson’s ratio,
μ = 0.25
4 kN
∴ Stress in the x-direction,
σx =
Load in x-direction
y× z
5 kN
40
mm
5000
=
= 1.25 N/mm2
100 × 40
m
300 mm
6 kN
Similarly the stress in y-direction is given by,
σy =
And stress in z-direction
or
Load in y-direction
x×z
=
6000
= 0.5 N/mm2
300 × 40
=
Load in z-direction
x× y
σz =
4000
300 × 100
= 0.133 N/mm2
Using equation (2.9), we get
dV
1
(σx + σy + σz)(1 – 2μ)
=
V
E
1
(1.25 + 0.5 + 0.113)(1 – 2 × 0.25)
=
2 × 10 5
1.883
=
2 × 10 5 × 2
1.883
∴
dV =
×V
4 × 10 5
1.883
=
× 1200000
4 × 10 5
= 5.649 mm3. Ans.
66
Fig. 2.4
0m
10
ELASTIC CONSTANTS
400 kN
10
0
m
m
Problem 2.6. A metallic bar 250 mm
4 MN
× 100 mm × 50 mm is loaded as shown in
Fig. 2.5.
Find the change in volume. Take
E = 2 × 10 5 N/mm 2 and Poisson’s ratio
50
= 0.25.
mm
Also find the change that should be
250 mm
made in the 4 MN load, in order that there
2 MN
should be no change in the volume of the
Fig. 2.5
bar.
Sol. Given :
Length, x = 250 mm, y = 100 mm and z = 50 mm
∴ Volume, V = xyz = 250 × 100 × 50 = 1250000 mm3
Load in x-direction
= 400 kN = 400000 N (tensile)
Load in y-direction
= 2 MN = 2 × 106 N (tensile)
Load in z-direction
= 4 MN = 4 × 106 N (compressive)
Modulus of elasticity,
E = 2 × 105 N/mm2
Poisson’s ratio,
µ = 0.25.
Now
σx = Stress in x-direction
Load in x-direction
=
Area of cross-section
400000 400000
=
=
= 80 N/mm2 (tension).
y× z
100 × 50
Similarly,
σy =
=
and
σz =
Load in y-direction
x×z
2 × 10 6
= 160 N/mm2
250 × 50
4000000
250 × 100
= 160 N/mm2 (compression).
Using equation (2.7) and taking tensile stresses positive and compressive stresses
negative, we get
or
dV
1
(σx + σy + σz)(1 – 2µ)
=
V
E
dV
1
=
(80 + 160 – 160)(1 – 2 × 0.25)
V
2 × 10 5
80
=
× 0.5 = 0.0002.
2 × 10 5
∴ Change in volume,
dV = 0.0002 × V
= 0.0002 × 1250000
= 250 mm3. Ans.
67
STRENGTH OF MATERIALS
Change in the 4 MN load when there is no change in volume of bar
dV
1
(σx + σy + σz)(1 – 2μ)
Using equation (2.7),
=
V
E
dV
If there is no change in volume, then
=0
V
1
∴
(σ + σy + σz)(1 – 2μ) = 0.
E x
But for most of materials, the value of μ lies between 0.25 and 0.33 and hence the term
(1 – 2μ) is never zero.
∴
σx + σy + σz = 0.
The stresses σx and σy are not to be changed. Only the stress corresponding to the load
4 MN (i.e., stress in z-direction) is to be changed.
∴
σz = – σx – σy = – 80 – 160 = – 240 N/mm2 (compressive)
Load Load
Load
=
But
σz =
or 240 =
Area x × y
250 × 100
∴
Load = 240 × 250 × 100 = 6 × 106 N = 6 MN
But already a compressive load of 4 MN is acting.
∴ Additional load that must be added
= 6 MN – 4 MN = 2 MN (compressive). Ans.
2.6. VOLUMETRIC STRAIN OF A CYLINDRICAL ROD..
Consider a cylindrical rod which is subjected to an axial tensile load P.
Let
d = diameter of the rod
L = length of the rod
Due to tensile load P, there will be an increase in the length of the rod, but the diameter
of the rod will decrease as shown in Fig. 2.6.
L + δL
P
P
d – δd
d
L
Fig. 2.6
∴ Final length
= L + δL
∴ Final diameter
= d – δd
Now original volume of the rod,
π 2
d ×L
4
π
=
(d – δd)2(L + δL)
4
π 2
=
(d + δd2 – 2d × δd)(L + δL)
4
L=
Final volume
68
ELASTIC CONSTANTS
π 2
(d × L + δd2 × L – 2d × L × δd + d2 × δL
4
+ δd2 × δL – 2d × δd × δL)
π 2
(d × L – 2d × L × δd + d2 × δL)
=
4
Neglecting the products and higher powers of two small quantities.
∴ Change in volume, δV = Final volume – Original volume
=
π 2
π 2
(d × L – 2d × L × δd + d2 × δL) –
d ×L
4
4
π 2
(d × δL – 2d × L × δd)
=
4
Change in volume δV
=
∴ Volumetric strain, ev =
V
Original volume
=
π 2
( d × δ L − 2 d × L × δd)
δL 2δd
=
−
= 4
π 2
L
d
d ×L
4
δL
δd
is the strain of length and
is the strain of diameter.
where
d
L
∴ Volumetric strain = Strain in length – Twice the strain of diameter.
...(2.8)
Problem 2.7. A steel rod 5 m long and 30 mm in diameter is subjected to an axial
tensile load of 50 kN. Determine the change in length, diameter and volume of the rod. Take E
= 2 × 105 N/mm2 and Poisson’s ratio = 0.25.
Sol. Given :
Length,
L = 5 m = 5 × 103 mm
Diameter,
d = 30 mm
π 2
π
d ×L=
(30)2 × 5 × 103 = 35.343 × 105
∴ Volume,
V =
4
4
Tensile load,
P = 50 kN = 50 × 103
Value of E
= 2 × 105 N/mm2
Poisson’s ratio,
μ = 0.25
Let
δd = Change in diameter
δL = Change in length
δV = Change in volume
Now
strain of length =
=
=
Stress
E
Load 1
×
Area E
P
π
× d2
4
×
FG∵
H
Stress =
Load
Area
1
1 50 × 10 3
=
×
π
E
2
2 × 10 5
× 30
4
69
IJ
K
STRENGTH OF MATERIALS
=
But
∴
∴
Now
∴
strain of length =
0.4 × 50 × 10 3
= 0.0003536
π × 30 2 × 2 × 10 5
δL
L
δL
= 0.0003536
L
δL = 0.0003536 × 5 × 103
= 1.768 mm. Ans.
Lateral strain
Poisson’s ratio =
Longitudinal strain
Lateral strain = Poisson’s ratio × Longitudinal strain
FG∵
H
= 0.25 × 0.0003536
Longitudinal strain =
δL
L
= 0.0000884
δd
But
Lateral strain =
d
δd
∴
= 0.0000884
d
∴
δd = 0.0000884 × d
= 0.0000884 × 30 = 0.002652 mm
Now using equation (2.8), we get
Volumetric strain,
∴
IJ
K
δV δL 2δd
=
−
V
d
L
= 0.0003536 – 2 × 0.0000884 = 0.0001768
δV = V × 0.0001768
= 35.343 × 105 × 0.0001768
= 624.86 mm3. Ans.
2.7. BULK MODULUS..
When a body is subjected to the mutually perpendicular like and equal direct stresses,
the ratio of direct stress to the corresponding volumetric strain is found to be constant for a
given material when the deformation is within a certain limit. This ratio is known as bulk
modulus and is usually denoted by K. Mathematically bulk modulus is given by
Direct stress
σ
K=
...(2.9)
=
dV
Volumetric strain
V
FG IJ
H K
2.8. EXPRESSION FOR YOUNG’S MODULUS IN TERMS OF BULK MODULUS..
Fig. 2.7 shows a cube A B C D E F G H which is subjected to three mutually perpendicular
tensile stresses of equal intensity.
Let L = Length of cube
dL = Change in length of the cube
70
ELASTIC CONSTANTS
E = Young’s modulus of the material of the cube
σ
E
F
σ = Tensile stress acting on the faces
σ
μ = Poisson’s ratio.
3
A
Then volume of cube, V = L
B
Now let us consider the strain of one of the sides of
σ
σ
H
the cube (say AB) under the action of the three mutually
G
perpendicular stresses. This side will suffer the following
σ
three strains :
D
C
1. Strain of AB due to stresses on the faces AEHD
σ
σ
Fig. 2.7
and BFGC. This strain is tensile and is equal to
.
E
2. Strain of AB due to stresses on the faces AEFB and DHGC. This is compressive
σ
lateral strain and is equal to – μ
.
E
3. Strain of AB due to stresses on the faces ABCD and EFGH. This is also compressive
σ
lateral strain and is equal to – μ
.
E
Hence the total strain of AB is given by
σ
dL σ
σ
σ
=
−μ×
−μ×
=
(1 – 2μ)
E
L
E
E
E
Now original volume of cube, V = L3
If dL is the change in length, then dV is the change in volume.
Differentiating equation (ii), with respect to L,
...(i)
...(ii)
dV = 3L2 × dL
Dividing equation (iii) by equation (ii), we get
...(iii)
dV 3 L2 × dL 3dL
=
=
V
L
L3
Substituting the value of
dL
from equation (i), in the above equation, we get
L
dV 3σ
=
(1 – 2μ)
V
E
From equation (2.9), bulk modulus is given by
σ
σ
=
3σ
dV
(1 − 2μ)
E
V
E
=
3(1 − 2μ)
E = 3K (1 – 2μ)
K=
or
FG IJ
H K
LM∵
N
dV 3σ
=
(1 − 2μ)
V
E
From equation (2.11), the expression for Poisson’s ratio (μ) is obtained as μ =
OP
Q
...(2.10)
...(2.11)
3K − E
.
6K
71
STRENGTH OF MATERIALS
Problem 2.8. For a material, Young’s modulus is given as 1.2 × 105 N/mm2 and
Poisson’s ratio
1
4
. Calculate the Bulk modulus.
Sol. Given : Young’s modulus, E = 1.2 × 105 N/mm2
1
4
Let
K = Bulk modulus
Using equation (2.10),
Poisson’s ratio,
μ =
K =
1.2 × 10 5 1.2 × 10 5
E
=
=
1
2
3(1 − 2μ)
3×
3 1−
2
4
FG
H
IJ
K
2 × 1.2 × 10 5
= 0.8 × 105 N/mm2. Ans.
3
Problem 2.9. A bar of 30 mm diameter is subjected to a pull of 60 kN. The measured
extension on gauge length of 200 mm is 0.1 mm and change in diameter is 0.004 mm. Calculate :
=
(i) Young’s modulus,
(ii) Poisson’s ratio and
(iii) Bulk modulus.
Sol. Given : Dia. of bar, d = 30 mm
π
(30)2 = 225π mm2
∴ Area of bar,
A =
4
Pull,
P = 60 kN = 60 × 1000 N
Gauge length,
L = 200 mm
Extension,
δL = 0.1 mm
Change in dia.,
δd = 0.004 mm
(i) Young’s modulus (E)
Tensile stress,
σ =
Longitudinal strain
=
∴ Young’s modulus,
E =
P 60000
=
= 84.87 N/mm2
A 225π
δL 0.1
=
= 0.0005
L 200
Tensile stress
Longitudinal strain
84.87
= 16.975 × 104 N/mm2
0.0005
= 1.6975 × 105 N/mm2. Ans.
=
(ii) Poisson’s ratio (μ)
Poisson’s ratio is given by equation (2.3) as
Poisson’s ratio
(μ) =
=
72
Lateral strain
Longitudinal strain
FG δd IJ
H dK
0.0005
FG∵
H
Lateral strain =
δL
d
IJ
K
ELASTIC CONSTANTS
=
FG 0.004 IJ
H 30 K = 0.000133 = 0.266.
0.0005
0.0005
Ans.
(iii) Bulk modulus (K)
Using equation (2.10), we get
K =
1.6975 × 10 5
E
=
3(1 − 2μ) 3(1 − 0.266 × 2)
= 1.209 × 105 N/mm2. Ans.
2.9. PRINCIPLE OF COMPLEMENTARY SHEAR STRESSES..
It states that a set of shear stresses across a plane is
t
D
C
always accompanied by a set of balancing shear stresses (i.e.,
of the same intensity) across the plane and normal to it.
t
t
Proof. Fig. 2.8 shows a rectangular block ABCD,
subjected to a set of shear stresses of intensity τ on the faces
AB and CD. Let the thickness of the block normal to the plane
A
B
t
of the paper is unity.
Fig. 2.8
The force acting on face AB
= Stress × Area
= τ × AB × 1 = τ. AB
Similarly force acting on face CD
= τ × CD × 1 = τ.CD
= τ.AB
(∵ CD = AB)
The forces acting on the faces AB and CD are equal and opposite and hence these forces
will form a couple.
The moment of this couple
= Force × Perpendicular distance
= τ. AB × AD
...(i)
If the block is in equilibrium, there must be a restoring couple whose moment must be
equal to the moment given by equation (i). Let the shear stress of intensity τ′ is set up on the
faces AD and CB.
The force acting on face AD = τ′ × AD × 1 = τ′. AD
The force acting on face BC = τ′ × BC × 1 = τ′BC = τ′. AD
(∵ BC = AD)
As the force acting on faces AD and BC are equal and opposite, these forces also forms
a couple.
Moment of this couple = Force × Distance = τ′. AD × AB
...(ii)
For the equilibrium of the block, the moments of couples given by equations (i) and (ii)
should be equal
∴
τ.AB × AD = τ′. AD × AB or τ = τ′.
The above equation proves that a set of shear stresses is always accompanied by a
transverse set of shear stresses of the same intensity.
The stress τ′ is known as complementary shear and the two stresses (τ and τ′) at right
angles together constitute a state of simple shear. The direction of the shear stresses on the
block are either both towards or both away from a corner.
73
STRENGTH OF MATERIALS
In Fig. 2.8, as a result of two couples, formed by the shear forces, the diagonal BD will
be subjected to tension and the diagonal AB will be subjected to compression.
2.10. STRESSES ON INCLINED SECTIONS WHEN THE ELEMENT IS SUBJECTED
TO SIMPLE SHEAR STRESSES
t
D
C
q
t
t
E
A
B
t
C
t
Pn
P
Fig. 2.9 shows a rectangular block ABCD which is in a
state of simple shear and hence subjected to a set of shear
stresses of intensity τ on the faces AB, CD and the faces AD
and CB. Let the thickness of the block normal to the plane of
the paper is unity.
It is required to find the normal and tangential stresses
across an inclined plane CE, which is having inclination θ with
the face CB.
Consider the equilibrium of the triangular piece CEB
of thickness unity. The forces acting on triangular piece CEB
are shown in Fig. 2.10 and they are :
(i) Shear force on face CB,
Q1 = Shear stress × area of face CB
= τ × BC × 1
= τ × BC acting along CB
q
t × BC = Q1
t
P
P
n
(ii) Shear force on face EB,
E
B
Q 2 = Shear stress × area of face EB
t × EB = Q2
= τ × EB × 1 = τ × EB acting along EB
Fig. 2.10
(iii) A force Pn normal to the plane EC
(iv) A force Pt tangential to the plane EC
The force Q1 is acting along the face CB as shown in Fig. 2.11. This force is resolved
into two components, i.e., Q1 cos θ and Q1 sin θ along the plane CE and normal to the plane
CE respectively.
The force Q2 is acting along the face EB. This force is also resolved into two components,
i.e., Q2 sin θ and Q2 cos θ along the plane EC and normal to the plane EC respectively.
For equilibrium, the net force normal to the plane CE
qC
s
should be zero.
co q
Q
Q1
1 s
∴ Pn – Q1 sin θ – Q2 cos θ = 0
in
q
or
Pn = Q1 sin θ + Q2 cos θ
= τ × BC × sin θ + τ × EB × cos θ
q
n
Q1
si
(∵ Q1 = τ × BC and Q2 = τ × EB)
2 (90–q)
Q2
Q
For equilibrium, the net force along the plane CE should E
B
q
be zero.
∴
Pt – Q1 cos θ + Q2 sin θ = 0
or
Pt = Q1 cos θ – Q2 sin θ
(– ve sign is taken due to opposite direction)
Fig. 2.11
= τ × BC × cos θ – τ × EB × sin θ
Q2
co
s
q
74
ELASTIC CONSTANTS
Let
Then
σn = Normal stress on plane CE
σt = Tangential stress on plane CE
Normal force on plane CE
σn =
Area of section CE
Pn
τ × BC × sin θ + τ × EB × cos θ
=
=
CE × 1
CE × 1
BC
EB
= τ×
× sin θ + τ ×
× cos θ
CE
CE
= τ × cos θ × sin θ + τ × sin θ × cos θ
FG∵
H
In triangle EBC,
= 2τ cos θ × sin θ = τ sin 2θ
and
σt =
=
IJ
K
BC
EB
= cos θ and
= sin θ
CE
CE
...(2.12)
Tangential force on plane CE
Area of plane CE
Pt
τ × BC × cos θ − τ × EB × sin θ
=
CE
CE × 1
BC
EB
× cos θ – τ ×
× sin θ
CE
CE
= τ × cos θ × cos θ – τ × sin θ × sin θ
= τ cos2 θ – τ sin2 θ
= τ [cos2 θ – sin2 θ] = τ cos 2θ
...(2.13)
For the planes carrying the maximum normal stress, σn should be maximum. But from
equation (2.12) it is clear that σn will be maximum when sin 2θ = ± 1
= τ×
i.e.,
2θ = ±
π
2
π
which means θ = 45° or – 45°
4
When
θ = 45°, then from equation (2.12), we have
σn = τ sin 90° = τ
D
When
θ = – 45°, then σn = – τ
(Positive sign shows the normal stress is tensile
whereas negative sign shows the normal stress is
compressive.)
When θ = ± 45°, then from equation (2.13), we find
that
A
E
σt = τ cos 2 × 45°
Fig. 2.12
= τ cos 90° = 0
This shows that the planes, which carry the
maximum normal stresses, are having zero shear stresses.
Now from equation 2.13, it is clear that shear stress
will be maximum when cos 2θ = ± 1,
i.e.,
2θ = 0° or 180° or θ = 0° or 90°
or
θ=±
C
45°
45°
B
75
STRENGTH OF MATERIALS
When θ = 0° or 90°, the value of σn from equation (2.12), is zero.
This shows that the planes, which carry the maximum shear stresses, are having zero
normal stresses. These planes are known as planes of simple shear.
Important points. When an element is subjected to a set of shear stresses, then :
(i) The planes of maximum normal stresses are perpendicular to each other.
(ii) The planes of maximum normal stresses are inclined at an angle of 45° to the planes
of pure shear.
(iii) One of the maximum normal stress is tensile while the other maximum normal
stress is compressive.
(iv) The maximum normal stresses are of the same magnitude and are equal to the
intensity of shear stress on the plane of pure shear.
2.11. DIAGONAL STRESSES PRODUCED BY SIMPLE SHEAR ON A SQUARE BLOCK.
Fig. 2.13 shows a square block ABCD of each side equal to ‘a’ and subjected to a set of
shear stresses of intensity τ on the faces AB, CD and faces AD and CB. Let the thickness of
the block normal to the plane of the paper is unity.
D
τ
τ
C
D
D
C
τ
C
θ
45°
τ
τ
A
τ
(a)
B
τ
τ
A
τ
(b)
B
τ
τ
A
τ
B
(c)
Fig. 2.13
The normal stress (σn) on plane AC is given by equation (2.12) as
σn = τ sin 2θ
...(i)
But as shown in Fig. 2.13 (b) the angle made by plane AC with face BC is given by,
AB
a
=
[∵ ABCD is a square of side ‘a’]
a
BC
=1
∴
θ = 45°
Substituting this value of θ in equation (i), we get
σn = τ × sin 2 × 45° = τ × sin 90° = τ
and
σt = τ × cos 2θ = τ × cos 2 × 45°
= τ × cos 90° = 0
Hence on the plane AC, a direct tensile stress of magnitude τ is acting. This tensile
stress is parallel to the diagonal BD. Hence the diagonal BD is subjected to tensile stress of
magnitude τ.
tan θ =
76
ELASTIC CONSTANTS
Similarly it can be proved that on the plane BD, a direct compressive stress of magnitude
τ is acting. This compressive stress is perpendicular to the plane BD or this compressive
stress is along the diagonal AC. Hence the diagonal AC is subjected to compressive stress of
magnitude τ. The pure direct tensile and compressive stresses active on the diagonal planes
AC and BD are called diagonal tensile and diagonal compressive stresses. The stress on the
diagonal plane AC (i.e., along diagonal BD) is tensile whereas on the diagonal plane BD i.e.,
along the diagonal AC is compressive.
Hence the set of shear stresses τ on the faces AB, CD and the faces AD and CB are
equivalent to a compressive stress τ along the diagonal AC and a tensile stress τ along the
diagonal BD.
2.12. DIRECT (TENSILE AND COMPRESSIVE) STRAINS OF THE DIAGONALS..
In Art. 2.11, we have proved that when a square block ABCD of unit thickness is
subjected to a set of shear stresses of intensity q on the faces AB, CD and the faces AD and CB,
the diagonal BD will experience a tensile stress of magnitude q
C
C1
whereas the diagonal AC will experience a compressive stress D1
D
of magnitude q. Due to these stresses the diagonal BD will be
E
elongated whereas the diagonal AC will be shorted. Let us
consider the joint effect of these two stresses on the diagonal
BD.
Due to the tensile stress q along diagonal BD, there will
be a tensile strain in diagonal BD. Due to the compressive stress
q along the diagonal AC, there will be a tensile strain in the
diagonal BD due to lateral strain.*
A
B
Let
µ = Poisson’s ratio
Fig. 2.14
E = Young’s modulus for the material of the block
Now tensile strain in diagonal BD due to tensile stress τ
along BD
Tensile stress along BD τ
=
E
E
Tensile strain in diagonal BD due to compressive stress τ along AC
=
µ×τ
E
∴ Total tensile strain along diagonal BD
=
=
τ µ×τ τ
+
=
(1 + µ)
E
E
E
...(2.14)
Similarly it can be proved that the total strain in the diagonal AC will be compressive
and will be given by
Total compressive strain in diagonal AC
=
τ
(1 + µ) .
E
*Please refer to Art. 2.4, in which it is proved that every strain in the direction of load is accompanied by lateral strain of the opposite kind perpendicular to the direction of load.
77
STRENGTH OF MATERIALS
The total tensile strain in the diagonal BD is equal to half the shear strain. This is
proved as given below :
Due to the shear stresses acting on the faces, the square block ABCD will be deformed
to position ABC1D1 as shown in Fig. 2.14.
Now increase in the length of diagonal BD = BD1 – BD
∴ Tensile strain in the diagonal BD
Increase in length
BD1 − BD
=
...(i)
Original length
BD
From D, draw a perpendicular DE on BD1.
We know that the distortion DD1 is very small and hence angle DBD1 will be very
small. Hence we can take
BD = BE
and
∠CDB = ∠C1D1B = 45°
Now in triangle DD1E, ∠ DD1E = 45°
∴
Length D1E = DD1 cos (DD1E)
=
= DD1 cos 45° =
In triangle
ABD, BD =
=
DD1
2
AB 2 + AD 2
AD 2 + AD 2 =
2 × AD
(∵
AB = AD)
[∵
BD = BE]
Now from equation (i), we have
Tensile strain in diagonal
BD =
BD1 − BD
BD
=
BD1 − BE
BD
=
D1 E
BD
[∵
FG DD IJ
H 2K
FG∵
H
1
=
2 × AD
1
=
=
2× 2
×
D1 E =
DD1
2
BD1 – BE = D1E]
IJ
K
and BD = 2 × AD
DD1 1 DD1
=
AD 2 AD
1
Shear strain*
2
FG∵
H
Shear strain =
DD1
AD
IJ
K
...(2.15)
2.13.RELATIONSHIP BETWEEN MODULUS OF ELASTICITY AND MODULUS OF.
RIGIDITY
We have seen in the last article that when a square block of unit thickness is subjected
to a set of shear stresses of magnitude τ on the faces AB, CD and the faces AD and CB, then
*Please refer to Art. 1.4.3, for shear strain.
78
ELASTIC CONSTANTS
the diagonal strain due to shear stress τ is given by equation (2.14) as
τ
(1 + µ)
E
From equation (2.15) also we have total tensile strain in diagonal BD
Total tensile strain along diagonal BD =
=
1
1 Shear stress
shear strain = ×
2
2
C
FG
H
Shear stress
= modulus of rigidity = C
Shear strain
τ
1
×
C
2
∴ Equating the two tensile strain along diagonal BD, we get
=
(∵
Shear stress = τ)
τ
1
τ
(1 + µ) = ×
2
C
E
or
∴
or
τ
1
(1 + µ) =
2C
E
E = 2C (1 + µ)
C=
E
2 (1 + µ)
(Cancelling τ from both sides)
...(2.16)
...(2.17)
Problem 2.10. Determine the Poisson’s ratio and bulk modulus of a material, for which
Young’s modulus is 1.2 × 105 N/mm2 and modulus of rigidity is 4.8 × 10 4 N/mm2.
Sol. Given :
Young’s modulus,
E = 1.2 × 105 N/mm2
Modulus of rigidity, C = 4.8 × 104 N/mm2
Let the Poisson’s ratio
=µ
Using equation (2.16), we get
E = 2C (1 + µ)
or
1.2 × 105 = 2 × 4.8 × 104 (1 + µ)
or
1.2 × 10 5
= 1.25 or µ = 1.25 – 1.0 = 0.25. Ans.
2 × 4.8 × 10 4
Bulk modulus is given by equation (2.10) as
(1 + µ) =
E
1.2 × 10 5
=
(∵ µ = 0.25)
3 (1 − 2µ) 3(1 − 0.25 × 2)
= 8 × 104 N/mm2. Ans.
Problem 2.11. A bar of cross-section 8 mm × 8 mm is subjected to an axial pull of
7000 N. The lateral dimension of the bar is found to be changed to 7.9985 mm × 7.9985 mm. If
the modulus of rigidity of the material is 0.8 × 105 N/mm2, determine the Poisson’s ratio and
modulus of elasticity.
Sol. Given :
Area of section = 8 × 8 = 64 mm2
Axial pull, P = 7000 N
Lateral dimensions = 7.9985 mm × 7.9985 mm
Volume of C = 0.8 × 105 N/mm2
K =
79
STRENGTH OF MATERIALS
Let μ = Poisson’s ratio and
E = Modulus of elasticity.
Change in lateral dimension
Now lateral strain =
Original lateral dimension
8 − 7.9985 0.0015
=
=
= 0.0001875.
8
8
To find the value of Poisson’s ratio, we must know the value of longitudinal strain. But
in this problem, the length of bar and the axial extension is not given. Hence longitudinal
strain cannot be calculated. But axial stress can be calculated. Then longitudinal, strain will
be equal to axial stress divided by E.
P
7000
σ
=
= 109.375 N/mm2 and longitudinal strain =
∴ Axial stress, σ =
64
Area
E
σ
But lateral strain = μ × longitudinal strain = μ ×
E
μ × 109.375
or
0.0001875 =
(∵ Lateral strain = 0.0001875)
E
E
109.375
=
= 583333.33
∴
μ
0.0001875
or
E = 583333.33μ
...(i)
Using equation (2.17), we get
E
C =
or E = 2C(1 + μ)
2(1 + μ)
(∵ C = 0.8 × 105)
= 2 × 0.8 × 105 (1 + μ)
5
or
583333.33μ = 2 × 0.8 × 10 (1 + μ)
(∵ E = 583333.33μ)
583333.33μ
or
1+μ =
= 3.6458μ
2 × 0.8 × 10 5
∴
1 = 3.6458μ – μ = 2.6458μ
1
= 0.378. Ans.
∴ Poisson’s ratio = μ =
2.6458
Modulus of elasticity (E) is obtained by substituting the value of μ in equation (i).
∴
E = 583333.33μ
583333.33
= 2.2047 × 105 N/mm2. Ans.
∴
E =
2.6458
Problem 2.12. Calculate the modulus of rigidity and bulk modulus of a cylindrical bar
of diameter 30 mm and of length 1.5 m if the longitudinal strain in a bar during a tensile
stress is four times the lateral strain. Find the change in volume, when the bar is subjected to
a hydrostatic pressure of 100 N/mm2. Take E = 1 × 105 N/mm2.
Sol. Given :
Dia. of bar,
d = 30 mm
Length of bar,
L = 1.5 m = 1.5 × 1000 = 1500 mm
π 2
π
∴ Volume of bar,V =
d ×L=
× 30 × 1500
4
4
= 1060287.52 mm3
80
ELASTIC CONSTANTS
Longitudinal strain = 4 × Lateral strain
Hydrostatic pressure, p = 100 N/mm2
1
Lateral strain
∴
= = 0.25
4
Longitudinal strain
Poisson’s ratio,
μ = 0.25
Let
C = Modulus of rigidity
K = Bulk modulus
E = Young’s modulus = 1 × 105 N/mm2
Using equation (2.16), we get
E = 2C (1 + μ)
1 × 105 = 2C(1 + 0.25)
or
or
1 × 105
= 4 × 104 N/mm2. Ans.
2 × 1.25
For bulk modulus, using equation (2.11), we get
E = 3K (1 – 2μ)
1 × 105 = 3K(1 – 2 × 0.25)
∴
or
C =
(∵
μ = 0.25)
1 × 10 5
= 0.667 × 105 N/mm2. Ans.
3 × 0.5
Now using equation (2.9), we get
p
p
=
K =
dV
Volumetric strain
V
2
where p = 100 N/mm
100
∴
0.667 × 105 =
dV
V
∴
K =
FG IJ
H K
FG IJ
H K
dV
V
or
∴
=
100
= 1.5 × 10–3
0.667 × 10 5
dV = V × 1.5 × 10–3 = 1060287.52 × 1.5 × 10–3
= 1590.43 mm3. Ans.
HIGHLIGHTS
1.
Poisson’s ratio is the ratio of lateral strain to longitudinal strain. It is generally denoted by μ.
2.
The tensile longitudinal stress produces compressive lateral strains.
3.
If a load acts in the direction of length of a rectangular bar, then longitudinal strain =
δd
δb
or
d
b
where δl = Change in length,
δb = Change in width,
δd = Change in depth.
δl
and
l
Lateral strain =
81
STRENGTH OF MATERIALS
4.
5.
The ratio of change in volume to original volume is known as volumetric strain.
Volumetric strain (ev) for a rectangular bar subjected to an axial load P, is given by
δl
(1 − 2μ) .
l
6. Volumetric strain for a rectangular bar subjected to three mutually perpendicular stresses is
ev =
given by,
ev =
1
(σ + σy + σz)(1 – 2μ)
E x
where σx, σy and σz are stresses in x, y and z direction respectively.
Principle of complementary shear stresses states that a set of shear stresses across a plane is
always accompanied by a set of balancing shear stresses (i.e., of the same intensity) across the
plane and normal to it.
8. Volumetric strain of a cylindrical rod, subjected to an axial tensile load is given by,
7.
ev = Longitudinal strain – 2 × strain of diameter
δl
δd
−2
.
l
d
9. Bulk modulus K is given by,
=
K=
σ
FG δV IJ .
HVK
10.
The relation between Young’s modulus and bulk modulus is given by,
E = 3K (1 – 2μ).
11. When an element is subjected to simple shear stresses then :
(i) The planes of maximum normal stresses are perpendicular to each other.
(ii) The planes of maximum normal stresses are inclined at an angle of 45° to the plane of pure
shear.
(iii) One of the maximum normal stress is tensile while the other maximum normal stress is
compressive.
(iv) The maximum normal stresses are of the same magnitude and are equal to the shear stress
on the plane of pure shear.
12. The relation between modulus of elasticity and modulus of rigidity is given by
E = 2C (1 + μ) or C =
E
.
2(1 + μ)
EXERCISE
(A) Theoretical Questions
1.
2.
3.
4.
Define and explain the terms : Longitudinal strain, lateral strain and Poisson’s ratio.
Prove that the volumetric strain of a cylindrical rod which is subjected to an axial tensile load is
equal to strain in the length minus twice the strain of diameter.
What is a bulk modulus ? Derive an expression for Young’s modulus in terms of bulk modulus
and Poisson’s ratio.
Define volumetric strain. Prove that the volumetric strain for a rectangular bar subjected to an
axial load P in the direction of its length is given by
δl
ev =
(1 – 2μ)
l
δl
= Longitudinal strain.
where μ = Poisson’s ratio and
l
82
ELASTIC CONSTANTS
5.
6.
7.
8.
9.
10.
(a) Derive an expression for volumetric strain for a rectangular bar which is subjected to three
mutually perpendicular tensile stresses.
(b) A test element is subjected to three mutually perpendicular unequal stresses. Find the change
in volume of the element, if the algebraic sum of these stresses is equal to zero.
Explain briefly the term ‘shear stress’ and ‘complimentary stress’ with proper illustrations.
State the principle of shear stress.
What do you understand by ‘An element in a state of simple shear’ ?
When an element is in a state of simple shear then prove that the planes of maximum normal
stresses are perpendicular to each other and these planes are inclined at an angle of 45° to the
planes of pure shear.
Derive an expression between modulus of elasticity and modulus of rigidity.
(B) Numerical Problems
1.
2.
3.
4.
5.
6.
Determine the changes in length, breadth and thickness of a steel bar which is 5 m long, 40 mm
wide and 30 mm thick and is subjected to an axial pull of 35 kN in the direction of its length.
Take E = 2 × 105 N/mm2 and Poisson’s ratio = 0.32.
[Ans. 0.0729 cm, 0.000186 cm, 0.000139 cm]
For the above problem, determine the volumetric strain and the final volume of the given steel
bar.
[Ans. 0.0000525, 6000317 mm3]
Determine the value of Young’s modulus and Poisson’s ratio of a metallic bar of length 25 cm,
breadth 3 cm and depth 2 cm when the bar is subjected to an axial compressive load of 240 kN.
The decrease in length is given as 0.05 cm and increase in breadth is 0.002.
[Ans. 2 × 105 N/mm2 and 0.33]
A steel bar 320 mm long, 40 mm wide and 30 mm thick is subjected to a pull of 250 kN in the
direction of its length. Determine the change in volume. Take E = 2 × 105 N/mm2 and m = 4.
[Ans. 200 mm3]
A metallic bar 250 mm × 80 mm × 30 mm is subjected to a force of 20 kN (tensile), 30 kN (tensile)
and 15 kN (tensile) along x, y and z directions respectively. Determine the change in the volume
[Ans. 19.62 mm3]
of the block. Take E = 2 × 105 N/mm2 and Poisson’s ratio = 0.25.
A metallic bar 300 mm × 120 mm × 50 mm is loaded as shown in Fig. 2.15.
Find the change in volume. Take E = 2 × 105 N/mm2 and Poisson’s ratio = 0.30.
4.5 MN
500 kN
50
mm
120 mm
300 mm
2.5 kN
Fig. 2.15
Also find the change that should be made in 4.5 MN load, in order that there should be no
change in the volume of the bar.
[Ans. 450 mm2, 4.5 MN]
7.
A steel rod 4 m long and 20 mm diameter is subjected to an axial tensile load of 40 kN. Determine the change in length, diameter and volume of the rod. Take E = 2 × 105 N/mm2 and Poisson’s
ratio = 0.25.
[Ans. 2.5464, 0.05092, 5598 mm3]
83
STRENGTH OF MATERIALS
8.
For a material, Young’s modulus is given as 1.4 × 105 N/mm2 and Poisson’s ratio 0.28. Calculate
the bulk modulus.
[Ans. 1.06 × 105 N/mm2]
9. A bar of 20 mm diameter subjected to a pull of 50 kN. The measured extension on gauge length of
250 mm is 0.12 mm and change in diameter is 0.00375 mm. Calculate :
(i) Young’s modulus
(ii) Poisson’s ratio and
(iii) Bulk modulus.
[Ans. (i) 1.989 × 105 N/mm2, (ii) 0.234, (iii) 1.2465 × 105 N/mm2]
10.
Determine the Poisson’s ratio and bulk modulus of a material, for which Young’s modulus is
[Ans. 0.33, 1.2 × 105 N/mm2]
1.2 × 105 N/mm2 and modulus of rigidity is 4.5 × 104 N/mm2.
11.
A bar of cross-section 10 mm × 10 mm is subjected to an axial pull of 8000 N. The lateral dimension of the bar is found to be changed to 9.9985 mm × 9.9985 mm. If the modulus of rigidity of the
material is 0.8 × 105 N/mm2, determine the Poisson’s ratio and modulus of elasticity.
[Ans. 0.45, 2.4 × 105 N/mm2]
12.
Calculate the modulus of rigidity and bulk modulus of a cylindrical bar of diameter of 25 mm
and of length 1.6 m, if the longitudinal strain in a bar during a tensile test is four times the
lateral strain. Find the change in volume, when the bar is subjected to a hydrostatic
pressure of 100 N/mm2. Take E = 1 × 105 N/mm2.
[Ans. 4 × 104 N/mm2, 0.667 × 105 N/mm2, 1178 mm3]
13.
A bar 30 mm in diameter was subjected to tensile load of 54 kN and the measured extension on
300 mm gauge length was 0.112 mm and change in diameter was 0.00366 mm. Calculate Poisson’s
ratio and values of three modulii.
[Ans. µ = 0.326, E = 204.6 kN/mm2, C = 77.2 kN/mm2, K = 196 kN/mm2]
14.
Derive the relation between E and C. Using the derived relationship, estimate the Young’s modulus
(E) when the modulus of rigidity (C) is 0.80 × 105 N/mm2 and the Poisson’s ratio is 0.3.
[Hint. E = 2C (1 + µ) = 2 × 0.80 × 105 (1 + 0.3) = 2.08 × 105 N/mm2.]
84
3
CHAPTER
PRINCIPAL STRESSES
AND STRAINS
3.1. INTRODUCTION..
In Chapter 2, the concept and definition of stress, strain, types of stresses (i.e., tensile,
compressive and simple shear) and types of strain (i.e., tensile, compressive, shear and
volumetric strains etc.) are discussed. These stresses were acting in a plane, which was at
right angles to the line of action of the force. In many engineering problems both direct (tensile
or compressive stress) and shear stresses are acting at the same time. In such situation the
resultant stress across any section will be neither normal nor tangential to the plane. In this
chapter the stresses, acting on an inclined plane (or oblique section) will be analysed.
3.2. PRINCIPAL PLANES AND PRINCIPAL STRESSES..
The planes, which have no shear stress, are known as principal planes. Hence principal
planes are the planes of zero shear stress. These planes carry only normal stresses.
The normal stresses, acting on a principal plane, are known as principal stresses.
3.3. METHODS FOR DETERMINING STRESSES ON OBLIQUE SECTION..
The stresses on oblique section are determined by the following methods :
1. Analytical method, and
2. Graphical method.
3.4. ANALYTICAL METHOD FOR DETERMINING STRESSES ON OBLIQUE SECTION..
The following two cases will be considered :
1. A member subjected to a direct stress in one plane.
2. The member is subjected to like direct stresses in two mutually perpendicular
directions.
3.4.1. A Member Subjected to a Direct Stress in one Plane. Fig. 3.1 (a) shows a
rectangular member of uniform cross-sectional area A and of unit thickness.
Let P = Axial force acting on the member.
A = Area of cross-section, which is perpendicular to the line of action of the force P.
The stress along x-axis, σ =
P
A
Hence, the member is subjected to a stress along x-axis.
Consider a cross-section EF which is perpendicular to the line of action of the force P.
85
STRENGTH OF MATERIALS
Pt
E
G
E
G
90°
P
P
q
P
(90–q)
F
(90–q)
q
q
P
F
Pn
Fig. 3.1 (a)
Fig. 3.1 (b)
Then area of section,
EF = EF × 1 = A.
The stress on the section EF is given by
σ=
Force
P

Area of EF A
...(i)
The stress on the section EF is entirely normal stress. There is no shear stress (or
tangential stress) on the section EF.
Now consider a section FG at an angle θ with the normal cross-section EF as shown in
Fig. 3.1 (a).
Area of section FG = FG × 1 (member is having unit thickness)
=
EF
1
cos 
=
A
cos 
FGIn  EFG, EF  cos   FG  EF IJ
cos  K
FG
H
( EF × 1 = A)
∴ Stress on the section, FG
=
Force
=
Area of section FG
= σ cos θ
P
FG A IJ
H cos  K

P
cos 
A
FG P  IJ
H A K
...(3.1)
This stress, on the section FG, is parallel to the axis of the member (i.e., this stress is
along x-axis). This stress may be resolved in two components. One component will be normal
to the section FG whereas the second component will be along the section FG (i.e., tangential
to the section FG). The normal stress and tangential stress (i.e., shear stress) on the section
FG are obtained as given below [Refer to Fig. 3.1 (b)].
Let
Pn = The component of the force P, normal to section FG
= P cos θ
Pt = The component of force P, along the surface of the section FG (or tangential
to the surface FG)
= P sin θ
σn = Normal stress across the section FG
σt = Tangential stress (i.e., shear stress) across the section FG.
86
PRINCIPAL STRESSES AND STRAINS
∴ Normal stress and tangential stress across the section FG are obtained as,
Normal stress,
σn =
=
=
Force normal to section FG
Area of section FG
Pn
P cos θ
FG A IJ FG A IJ
H cos θ K H cos θ K
=
(∵ Pn = P cos θ)
P
P
cos θ . cos θ =
cos2 θ
A
A
FG∵
H
= σ cos2 θ
P
=σ
A
IJ
K
...(3.2)
Tangential stress (i.e., shear stress),
σt =
=
Tangential force across section FG
Area of section FG
P sin θ
Pt
FG A IJ = FG A IJ
H cos θ K H cos θ K
(∵ Pt = P sin θ)
=
P
sin θ . cos θ = σ sin θ. cos θ
A
=
σ
× 2 sin θ cos θ
2
=
σ
sin 2θ
2
[Multiplying and dividing by 2]
(∵ 2 sin θ cos θ = sin 2θ)
...(3.3)
From equation (3.2), it is seen that the normal stress (σn) on the section FG will be
maximum, when cos2 θ or cos θ is maximum. And cos θ will be maximum when θ = 0° as
cos 0° = 1. But when θ = 0°, the section FG will coincide with section EF. But the section
EF is normal to the line of action of the loading. This means the plane normal to the axis
of loading will carry the maximum normal stress.
∴ Maximum normal stress,
= σ cos2 θ = σ cos2 0° = σ
...(3.4)
From equation (3.3), it is observed that the tangential stress (i.e., shear stress) across
the section FG will be maximum when sin 2θ is maximum. And sin 2θ will be maximum when
sin 2θ = 1 or 2θ = 90° or 270°
θ = 45° or 135°.
This means the shear stress will be maximum on two planes inclined at 45° and 135° to
the normal section EF as shown in Figs. 3.1 (c) and 3.1 (d).
or
∴ Max. value of shear stress =
σ
σ
σ
sin 2θ =
sin 90° = .
2
2
2
...(3.5)
87
STRENGTH OF MATERIALS
First plane of maximum
shear stress q = 45°
E
E
45°
P
P
135°
P
F
P
F
second plane of maximum
shear stress q = 135°
Fig. 3.1 (c)
Fig. 3.1 (d)
From equations (3.4) and (3.5) it is seen that maximum normal stress is equal to σ
whereas the maximum shear stress is equal to σ/2 or equal to half the value of greatest
normal stress.
Second Method
A Member Subjected to a Direct Stress in one Plane. Fig. 3.2 shows a rectangular
member of uniform cross-sectional area A and of unit thickness. The bar is subjected to a
principal tensile stress σ1 on the faces AD and BC.
D
E
C
θ
Pt
σ1
θ
θ
P1 = σ1 × BC × 1
Pn
A
F
σ1
B
Fig. 3.2
Area of cross-section
= BC × Thickness of bar
= BC × 1
Let the stresses on the oblique plane FC are to be calculated. The plane FC is inclined
at an angle θ with the normal cross-section EF (or BC). This can be done by converting the
stress σ1 acting on face BC into equivalent force. Then this force will be resolved along the
inclined planes FC and perpendicular to FC. (Please note that it is force and not the stress
which is to be resolved).
Tensile stress on face BC = σ1
Now, the tensile force on BC,
P1 = Stress (σ1) × Area of cross-section
(∵ Area = BC × 1)
= σ1 × BC × 1
The above tensile force P1 is also acting on the inclined section FC, in the axial direction
as shown in Fig. 3.2. This force P1 is resolved into two component, i.e., one normal to the plane
FC and other along the plane FC.
Let
Pn = Component of the force P1, normal to the section FC
= P1 cos θ
(∵ P1 = σ1 × BC × 1)
= σ1 × BC × 1 × cos θ
Pt = Component of the force P1, along the section FC
= P1 sin θ
= σ1 × BC × 1 × sin θ
σn = Normal stress on the section FC
88
PRINCIPAL STRESSES AND STRAINS
σt = Shear stress (or tangential stress) across the section FC.
Then normal stress, σn =
Force normal to section FC
Area of section FC
=
Pn
FC × 1
=
σ 1 × BC × cos θ
FC
(∵ bar is of unit thickness)
= σ1 × cos θ × cos θ
= σ1 × cos2 θ
Similarly, tangential (or shear) stress,
σt =
=
FG∵
H
(∵ Pn = σ1 × BC × cos θ)
In triangle FBC,
IJ
K
BC
= cos θ
FC
...(3.5A)
Pt
Force along section FC
=
FC × 1
Area of section FC
σ 1 × BC × 1 × sin θ
FC
FG∵
H
= σ1 × cos θ × sin θ
(∵ Pt = σ1 × BC × 1)
In triangle FBC,
IJ
K
BC
= cos θ
FC
= σ1 × cos θ × sin θ
σ1
=
× 2 × cos θ × sin θ
(Multiplying and dividing by two)
2
σ
...(3.5B) (∵ 2 sin θ cos θ = sin 2θ)
= 1 × sin 2θ
2
From equation (3.5A), it is seen that the normal stress (σn) on the section FC will
be maximum, when cos2 θ or cos θ is maximum. And cos θ will be maximum when θ = 0°
as cos 0° = 1. But when θ = 0°, the section FC will coincide with section EF. But the section
EF is normal to the line of action of the loading. This means the plane normal to the axis
of loading will carry the maximum normal stress.
...(3.5C)
∴ Maximum normal stress = σ1 cos2 θ = σ1 cos2 0° = σ1
From equation (3.5B), it is observed that the tangential stress (i.e., shear stress) across
the section FC will be maximum when sin 2θ is maximum. And sin 2θ will be maximum when
sin 2θ = 1 or 2θ = 90° or 270° or θ = 45° or 135°.
This means the shear stress will be maximum on two planes inclined at 45° and 135° to
the normal section EF or BC as shown in Figs. 3.2 (a) and 3.2 (b).
Second plane of
maximum shear
stress, = 135°
First plane of maximum
shear stress, = 45°
E
45°
P
135°
C
P
C
P
P
135°
F
B
B
(a)
(b)
Fig. 3.2
89
STRENGTH OF MATERIALS
σ1
σ
σ
sin 2θ = 1 sin 90° = 1
...(3.5D)
2
2
2
From equations (3.5C) and (3.5D) it is seen that maximum normal stress is equal to σ1
∴ Max. value of shear stress =
whereas the maximum shear stress is equal to
σ1
or equal to half the value of greatest
2
normal stress.
Note. It is the force which is resolved in two components. The stress is not resolved.
Problem 3.1. A rectangular bar of cross-sectional area 10000 mm2 is subjected to an
axial load of 20 kN. Determine the normal and shear stresses on a section which is inclined at
an angle of 30° with normal cross-section of the bar.
Sol. Given :
Cross-sectional area of the rectangular bar,
A = 10000 mm2
Axial load,
P = 20 kN = 20,000 N
Angle of oblique plane with the normal cross-section of the bar,
θ = 30°
P 20000
Now direct stress
σ=
=
= 2 N/mm2
A 10000
Let
σn = Normal stress on the oblique plane
σt = Shear stress on the oblique plane.
Using equation (3.2) for normal stress, we get
σn = σ cos2 θ
= 2 × cos2 30°
(∵ σ = 2 N/mm2)
(∵ cos 30° = 0.866)
= 2 × 0.8662
= 1.5 N/mm2. Ans.
Using equation (3.3) for shear stress, we get
σ
2
σt =
sin 2θ =
× sin (2 × 30°)
2
2
= 1 × sin 60° = 0.866 N/mm2. Ans.
Problem 3.2. Find the diameter of a circular bar which is subjected to an axial pull of
160 kN, if the maximum allowable shear stress on any section is 65 N/mm2.
Sol. Given :
Axial pull,
P = 160 kN = 160000 N
Maximum shear stress
= 65 N/mm2
Let
D = Diameter of the bar
π 2
∴ Area of the bar
=
D
4
P 160000 640000
∴ Direct stress,
σ=
N/mm2
=
=
2
π 2
A
π
D
D
4
Maximum shear stress is given by equation (3.5).
σ 640000
∴ Maximum shear stress
= =
.
2 2 × πD 2
But maximum shear stress is given as = 65 N/mm2.
Hence equating the two values of maximum shear, we get
640000
∴
65 =
2 × πD 2
90
PRINCIPAL STRESSES AND STRAINS
640000
= 1567
2 × π × 65
∴
D = 39.58 mm. Ans.
Problem 3.3. A rectangular bar of cross-sectional area of 11000 mm2 is subjected to a
tensile load P as shown in Fig. 3.3. The permissible normal and shear stresses on the oblique
plane BC are given as 7 N/mm2 and 3.5 N/mm2 respectively. Determine the safe value of P.
Sol. Given :
C
Area of cross-section, A = 11000 mm2
2
Normal stress,
σn = 7 N/mm
P
P
Shear stress,
σt = 3.5 N/mm2
60°
Angle of oblique plane with the axis of bar = 60°.
B
∴ Angle of oblique plane BC with the normal crosssection of the bar,
Fig. 3.3
θ = 90° – 60° = 30°
Let
P = Safe value of axial pull
σ = Safe stress in the member.
Using equation (3.2),
σn = σ cos2 θ or 7 = σ cos2 30°
= σ (0.866)2.
(∵ cos 30° = 0.866)
7
∴
σ=
= 9.334 N/mm2
0.866 × 0.866
Using equation (3.3),
σ
σt =
sin 2θ
2
σ
σ
σ
or
3.5 =
sin 2 × 30° =
sin 60° =
× 0.866
2
2
2
3.5 × 2
∴
σ=
= 8.083 N/mm2.
0.866
The safe stress is the least of the two, i.e., 8.083 N/mm2.
∴ Safe value of axial pull,
P = Safe stress × Area of cross-section
= 8.083 × 11000 = 88913 N = 88.913 kN. Ans.
Problem 3.4. Two wooden pieces 10 cm × 10 cm
B
in cross-section are glued together along line AB as
shown in Fig. 3.3(a) below. What maximum axial force
P
P
P can be applied if the allowable shearing stress along
30°
AB is 1.2 N/mm2 ?
A
Fig. 3.3 (a)
Sol. Given :
Area of cross-section
= 10 × 10 = 100 cm2
= 100 × 100 mm2 = 10000 mm2
Allowable shear stress,
σt = 1.2 N/mm2
Angle of line AB with the axis of axial force = 30°
∴ Angle of line AB with the normal cross-section,
θ = 90° – 30° = 60°
∴
D2 =
91
STRENGTH OF MATERIALS
Let
or
P = Maximum axial force
σ = Maximum allowable stress in the direction of P.
Using equation (3.3),
σ
σt =
sin 2θ
2
σ
σ
1.2 =
× sin (2 × 60°) =
× sin 120°
2
2
1.2 × 2
2.4
=
∴
σ=
= 2.771 N/mm2
sin 120° 0.866
∴ Maximum axial force,
P = Stress in the direction of P × Area of cross-section
= σ × 10000 = 2.771 × 10000 = 27710 N = 27.71 kN. Ans.
3.4.2. A Member Subjected to like Direct Stresses in two Mutually
Perpendicular Directions. Fig. 3.4 (a) shows a rectangular bar ABCD of uniform crosssectional area A and of unit thickness. The bar is subjected to two direct tensile stresses (or
two-principal tensile stresses) as shown in Fig. 3.4 (a).
σ2
P1 sin θ
D
C
C
θ
σ1
θ
P1
σ1
θ
F
C
P2 cos θ
B
σ2
P1 cos θ
P1 = σ1 × BC × 1
θ Pn
A
F
P1
P2 = σ2 × BF × 1
F
θ
P2
P2 sin θ
Fig. 3.4 (a)
Let FC be the oblique section on which stresses are to be calculated. This can be done
by converting the stresses σ1 (acting on face BC) and σ2 (acting on face AB) into equivalent
forces. Then these forces will be resolved along the inclined plane FC and perpendicular to
FC. Consider the forces acting on wedge FBC.
Let
θ = Angle made by oblique section FC with normal cross-section BC
σ1 = Major tensile stress on face AD and BC
σ2 = Minor tensile stress on face AB and CD
P1 = Tensile force on face BC
P2 = Tensile force on face FB.
The tensile force on face BC,
P1 = σ1 × Area of face BC = σ1 × BC × 1
The tensile force on face FB,
P2 = Stress on FB × Area of FB = σ2 × FB × 1.
92
(∵ Area = BC × 1)
PRINCIPAL STRESSES AND STRAINS
The tensile forces P1 and P2 are also acting on the oblique section FC. The force P1 is
acting in the axial direction, whereas the force P2 is acting downwards as shown in Fig.
3.4 (a). Two forces P1 and P2 each can be resolved into two components i.e., one normal to the
plane FC and other along the plane FC. The components of P1 are P1 cos θ normal to the plane
FC and P1 sin θ along the plane in the upward direction. The components of P2 are P2 sin θ
normal to the plane FC and P2 cos θ along the plane in the downward direction.
Let
Pn = Total force normal to section FC
= Component of force P1 normal to section FC
+ Component of force P2 normal to section FC
= P1 cos θ + P2 sin θ
= σ1 × BC × cos θ + σ2 × BF × sin θ (∵ P1 = σ1 × BC, P2 = σ2 × BF)
Pt = Total force along the section FC
= Component of force P1 along the section FC
+ Component of force P2 along the section FC
(–ve sign is taken due to opposite
= P1 sin θ + (– P2 cos θ)
direction)
= P1 sin θ – P2 cos θ
= σ1 × BC × sin θ – σ2 × BF × cos θ
(Substituting the values P1 and P2)
σn = Normal stress across the section FC
Total force normal to the section FC
=
Area of section FC
=
Pn
σ × BC × cos θ + σ 2 × BF × sin θ
= 1
FC × 1
FC
BC
BF
× cos θ + σ2 ×
× sin θ
FC
FC
= σ1 × cos θ × cos θ + σ2 × sin θ × sin θ
= σ1 ×
FG∵
H
= σ1 cos2 θ + σ2 sin2 θ
= σ1
FG 1 + cos 2θ IJ
H 2 K
*
+ σ2
In triangle FBC,
FG 1 − cos 2θ IJ
H 2 K
IJ
K
BC
BF
= cos θ,
= sin θ
FC
FC
**
[∵ cos2 θ = (1 + cos 2θ)/2 and sin2 θ = (1 – cos 2θ)/2]
σ1 + σ2 σ1 − σ2
+
cos 2θ
2
2
σt = Tangential stress (or shear stress) along section FC
=
=
Total force along the section FC
Area of section FC
* cos 2θ = cos2 θ – sin2 θ
= cos2 θ – (1 – cos2 θ) = 2 cos2 θ – 1
∴ cos2 θ =
(1 + cos 2θ)
2
FG∵
H
Stress =
...(3.6)
Force
Area
IJ
K
** cos 2θ = cos2 θ – sin2 θ
= (1 – sin2 θ) – sin2 θ = 1 – 2 sin2 θ
∴ sin2 θ =
(1 − cos 2θ)
2
93
STRENGTH OF MATERIALS
=
Pt
σ × BC × sin θ − σ 2 × BF × cos θ
= 1
FC
FC × 1
BC
BF
× sin θ – σ2 ×
× cos θ
FC
FC
= σ1 × cos θ × sin θ – σ2 × sin θ × cos θ
= σ1 ×
FG∵
H
In triangle FBC,
IJ
K
BC
BF
= cos θ,
= sin θ
FC
FC
= (σ1 – σ2) cos θ sin θ
(σ − σ 2 )
= 1
× 2 cos θ sin θ
(Multiplying and dividing by 2)
2
(σ − σ 2 )
= 1
sin 2θ
...(3.7)
2
The resultant stress on the section FC will be given as
σR =
σ n2 + σ t2
...(3.8)
Obliquity [Refer to Fig. 3.4 (b)]. The angle made
by the resultant stress with the normal of the oblique
plane, is known as obliquity. It is denoted by φ.
Mathematically,
tan φ =
σt
σn
C
σt
σR
φ
σt
...[3.8 (A)]
A
Maximum shear stress. The shear stress is given
by equation (3.7). The shear stress (σt) will be maximum
when
sin 2θ = 1 or 2θ = 90° or 270°
θ = 45° or 135°
or
D
F
B
σn
Fig. 3.4 (b)
(∵ sin 90° = 1 and also sin 270° = 1)
σ1 − σ2
...(3.9)
2
The planes of maximum shear stress are obtained by making an angle of 45° and 135°
with the plane BC (at any point on the plane BC) in such a way that the planes of maximum
shear stress lie within the material as shown in Fig. 3.4 (c).
And maximum shear stress, (σt)max =
Plane of maximum shear stress
C
C
45°
135°
B
B
Fig. 3.4 (c)
Hence the planes, which are at an angle of 45° or 135° with the normal cross-section
BC [see Fig. 3.4 (c)], carry the maximum shear stresses.
94
PRINCIPAL STRESSES AND STRAINS
Principal planes. Principal planes are the planes on which shear stress is zero. To
locate the position of principal planes, the shear stress given by equation (3.7) should be
equated to zero.
∴ For principal planes,
or
or
∴
σ1 − σ2
sin 2θ
2
sin 2θ
2θ
θ
when θ = 0,
= 0
= 0
= 0 or
= 0 or
σn =
=
=
=
when θ = 90°,
σn =
=
=
=
[∵ (σ1 – σ2) cannot be equal to zero]
180°
90°
σ1 + σ2 σ1 − σ2
+
2
2
σ1 + σ2 σ1 − σ2
+
2
2
σ1 + σ2 σ1 − σ2
+
2
2
σ1
σ1 + σ2 σ1 − σ2
+
2
2
σ1 + σ2 σ1 − σ2
+
2
2
σ1 + σ2 σ1 − σ2
+
2
2
σ2.
cos 2θ
cos 0°
×1
(∵ cos 0° = 1)
cos 2 × 90°
cos 180°
× (– 1)
(∵ cos 180° = – 1)
Note. The relations, given by equations (3.6) to (3.9), also hold good when one or both the stresses
are compressive.
Problem 3.5. The tensile stresses at a point across two mutually perpendicular planes
are 120 N/mm2 and 60 N/mm2. Determine the normal, tangential and resultant stresses on a
plane inclined at 30° to the axis of the minor stress.
Sol. Given :
Major principal stress,
σ1 = 120 N/mm2
Minor principal,
σ2 = 60 N/mm2
Angle of oblique plane with the axis of minor principal stress,
θ = 30°.
Normal stress
The normal stress (σn) is given by equation (3.6),
∴
σn =
=
σ1 + σ2 σ1 − σ2
+
cos 2θ
2
2
120 + 60 120 − 60
+
cos 2 × 30°
2
2
= 90 + 30 cos 60° = 90 + 30 ×
1
2
= 105 N/mm2. Ans.
95
STRENGTH OF MATERIALS
Tangential stress
The tangential (or shear stress) σt is given
by equation (3.7).
120 − 60
sin (2 × 30°)
2
= 30 × sin 60° = 30 × 0.866
= 25.98 N/mm2. Ans.
Resultant stress
The resultant stress (σ R ) is given by
equation (3.8)
=
∴
σR =
=
σ1 = 120 N/mm
2
σ1 − σ2
sin 2θ
2
2
2
σt =
σ1 = 120 N/mm
∴
σ2 = 60 N/mm
Axis of
minor stress
Axis of
major stress
°
30
σ2 = 60 N/mm
2
Fig. 3.5
σ n 2 + σ t 2 = 105 2 + 25.98 2
11025 + 674.96 = 108.16 N/mm2.
Ans.
Problem 3.6. The stresses at a point in a bar are 200 N/mm2 (tensile) and 100 N/mm2
(compressive). Determine the resultant stress in magnitude and direction on a plane inclined
at 60° to the axis of the major stress. Also determine the maximum intensity of shear stress in
the material at the point.
Sol. Given :
Major principal stress,
Minor principal stress,
= 50 + 150 ×
1
2
(∵ cos 60° =
N/mm2.
1
2
)
2
200 N/mm
200 N/mm
2
σ1 = 200 N/mm2
σ2 = – 100 N/mm2
(Minus sign is due to compressive stress)
Angle of the plane, which it makes with the major principal stress = 60°
∴
Angle θ = 90° – 60° = 30°.
Resultant stress in magnitude and direction
First calculate the normal and tangential
stresses.
2
100 N/mm
Using equation (3.6) for normal stress,
σ + σ2 σ1 − σ2
+
cos 2θ
σn = 1
2
2
θ
60°
200 + (− 100) 200 − (− 100)
=
+
Axis of
2
2
major stress
cos (2 × 30°)
(∵ θ = 30°)
200 − 100 200 + 100
2
+
cos 60°
=
100 N/mm
2
2
Fig. 3.6
= 50 + 75 = 125
Using equation (3.7) for tangential stress,
σ − σ2
200 − (− 100)
sin 2θ =
sin (2 × 30°)
σt = 1
2
2
200 + 100
sin 60° = 150 × 0.866 = 129.9 N/mm2.
=
2
96
PRINCIPAL STRESSES AND STRAINS
Using equation (3.8) for resultant stress,
σR =
=
σ n 2 + σ t 2 = 125 2 + 129.9 2
15625 + 16874 = 180.27 N/mm2.
Ans.
The inclination of the resultant stress with the normal of the inclined plane is given by
equation [3.8 (A)] as
σ t 129.9
=
= 1.04
σn
125
φ = tan–1 1.04 = 46° 6′.
tan φ =
∴
Ans.
Maximum shear stress
Maximum shear stress is given by equation (3.9)
σ − σ 2 200 − (− 100) 200 + 100
=
=
= 150 N/mm2. Ans.
∴
(σt)max = 1
2
2
2
Problem 3.7. At a point in a strained material the principal tensile stresses across two
perpendicular planes, are 80 N/mm2 and 40 N/mm2. Determine normal stress, shear stress
and the resultant stress on a plane inclined at 20° with the major principal plane. Determine
also the obliquity. What will be the intensity of stress, which acting alone will produce the
.
2
60 N/mm
C
2
D
Major principal
plane
2 = 80 N/mm
Sol. Given :
Major principal stress,
σ1 = 80 N/mm2
Minor principal stress,
σ2 = 40 N/mm2
The plane CE is inclined at angle 20° with
major principal plane (i.e., plane BC).
∴
θ = 20°
2
1
4
1 = 80 N/mm
same maximum strain if Poisson’s ratio =
E
1
A
B
4
Let σn = Normal stress on inclined plane
2
60 N/mm
CE
Fig. 3.7
σt = Shear stress and
σR = Resultant stress.
Using equation (3.6), we get
σ1 + σ2 σ1 − σ2
80 + 40 80 − 40
+
+
cos 2θ =
cos (2 × 20°)
σn =
2
2
2
2
= 60 + 20 × cos 40° = 75.32 N/mm2. Ans.
The shear stress is given by equation (3.7)
80 − 40
σ − σ2
sin 2θ =
sin (2 × 20°) = 20 sin 40°
∴
σt = 1
2
2
= 12.865 N/mm2. Ans.
The resultant stress is given by equation (3.8)
Poisson’s ratio, µ =
∴
σR =
=
σ n2 + σ t2
75.32 2 + 12.856 2 = 76.4 N/mm2.
Ans.
97
STRENGTH OF MATERIALS
Obliquity (φ) is given by equation [3.8 (A)]
σ t 12.856
=
σn
75.32
12
.856
∴
φ = tan–1
= 9° 41′. Ans.
75.32
Let σ = stress which acting alone will produce the same maximum strain. The maximum
strain will be in the direction of major principal stress.
σ 1 µσ 2
1
−
= (σ 1 − µσ 2 )
∴ Maximum strain
=
E
E
E
1
40
70
=
=
80 −
E
4
E
σ
The strain due to stress
σ=
E
70 σ
Equating the two strains, we get
=
E E
∴
σ = 70 N/mm2. Ans.
Problem 3.8. At a point in a strained material the principal stresses are 100 N/mm2
(tensile) and 60 N/mm2 (compressive). Determine the normal stress, shear stress and resultant
stress on a plane inclined at 50° to the axis of major principal stress. Also determine the
maximum shear stress at the point.
tan φ =
FG
H
IJ
K
Sol. Given :
Major principal stress, σ1 = 100 N/mm2
Minor principal stress, σ2 = – 60 N/mm2 (Negative sign due to compressive stress)
Angle of the inclined plane with the axis of major principal stress = 50°
∴ Angle of the inclined plane with the axis of minor principal stress,
θ = 90 – 50 = 40°.
Normal stress (σn)
Using equation (3.6),
σn =
=
σ1 + σ 2 σ1 − σ 2
+
cos 2θ
2
2
100 + (− 60) 100 − (− 60)
+
cos (2 × 40° )
2
2
100 − 60 100 + 60
+
cos 80°
2
2
= 20 + 80 × cos 80° = 20 + 80 × .1736
= 20 + 13.89 = 33.89 N/mm2. Ans.
=
Shear stress (σt )
Using equation (3.7), σt =
=
98
σ1 − σ 2
sin 2θ
2
100 − ( − 60)
sin (2 × 40°)
2
PRINCIPAL STRESSES AND STRAINS
100 + 60
sin 80° = 80 × 0.9848 = 78.785 N/mm2.
2
Resultant stress (σR)
Using equation on (3.8),
=
σR =
Ans.
σ n2 + σ t2 = 33.89 2 + 78.785 2
= 1148.53 + 6207.07 = 85.765 N/mm2. Ans.
Maximum shear stress
Using equation (3.9),
(σt)max =
σ1 − σ 2 100 − ( − 60)
=
2
2
100 + 60
= 80 N/mm2. Ans.
2
Problem 3.9. At a point in a strained material, the principal stresses are 100 N/mm2
tensile and 40 N/mm2 compressive. Determine the resultant stress in magnitude and direction
on a plane inclined at 60° to the axis of the major principal stress. What is the maximum
intensity of shear stress in the material at the point ?
=
Sol. Given :
The major principal stress, σ1 = 100 N/mm2
The minor principal stress, σ2 = – 40 N/mm2 (Minus sign due to compressive stress)
Inclination of the plane with the axis of major principal stress = 60°
∴ Inclination of the plane with the axis of minor principal stress,
θ = 90 – 60 = 30°.
Resultant stress in magnitude
The resultant stress (σR) is given by equation (3.8) as
σR =
σ n2 + σ t2
where σn = Normal stress and is given by equation (3.6) as
=
σ1 + σ2 σ1 − σ2
+
cos 2θ
2
2
=
100 + ( − 40) 100 − ( − 40)
+
cos (2 × 30°)
2
2
100 − 40 100 + 40
+
cos 60°
2
2
= 30 + 70 × 0.5
= 65 N/mm2
σt = Shear stress and is given by equation (3.7) as
=
and
=
(∵ cos 60° = 0.5)
σ1 − σ 2
100 − ( − 40 )
sin 2θ =
sin (2 × 30°)
2
2
99
STRENGTH OF MATERIALS
=
σR =
100 + 40
sin 60° = 70 × .866 = 60.62 N/mm2
2
65 2 + 60.62 2 = 88.9 N/mm2.
Ans.
Direction of resultant stress
Let the resultant stress is inclined at an angle φ to the normal of the oblique plane.
Then using equation [3.8 (A)].
tan φ =
∴
σ t 60.62
=
σn
65
60.62
= 43°.
65
φ = tan–1
Ans.
Maximum shear stress
Using equation (3.9), (σt)max =
σ1 − σ 2
2
100 − ( − 40 ) 100 + 40
=
= 70 N/mm2. Ans.
2
2
Problem 3.10. A small block is 4 cm long, 3 cm high and 0.5 cm thick. It is subjected to
uniformly distributed tensile forces of resultants 1200 N and 500 N as shown in Fig. 3.7 (a)
below. Compute the normal and shear stresses developed along the diagonal AB.
=
500 N
0.5 cm
B
q
1200 N
1200 N
3 cm
0.5 cm
A
4 cm
500 N
Fig. 3.7(a)
Sol. Given :
Length = 4 cm, height = 3 cm and width = 0.5 cm
Force along x-axis = 1200 N
Force along y-axis = 500 N
Area of cross-section normal to x-axis = 3 × 0.5 = 1.5 cm2
Area of cross-section normal to y-axis = 4 × 0.5 = 2 cm2
100
PRINCIPAL STRESSES AND STRAINS
∴ Stress along x-axis
=
Force along x-axis
Area normal to x-axis
1200
= 800 N/cm2
1.5
σ1 = 800 N/cm2
=
∴
Stress along y-axis, σ2 =
=
Also
∴
Let
Force along y-axis
Area normal to y-axis
500
= 250 N/cm2
2
4
= 1.33
3
θ = tan–1 1.33 = 53.06°
σn = Normal stress on diagonal AB
σt = Shear stress on diagonal AB
tan θ =
Using equation (3.6), σn =
σ1 + σ 2 σ1 − σ 2
+
cos 2θ
2
2
800 + 250 800 − 250
+
cos (2 × 53.06)
2
2
= 525 + 275 × cos 106.12° = 525 + 275 × (– 0.2776)
= 525 – 76.35 = 448.65 N/cm2. Ans.
=
Now using equation (3.7), σt =
σ1 − σ 2
sin 2θ
2
800 − 250
sin (2 × 53.06°)
2
= 275 sin 106.12° = 275 × 0.96 = 264.18 N/cm2.
=
3.4.3. A Member Subjected to a Simple Shear Stress.
Fig. 3.8 shows a rectangular bar ABCD of uniform crosssectional area A and of unit thickness. The bar is subjected to a
simple shear stress (q) across the faces BC and AD. Let FC be
the oblique section on which normal and tangential stresses
are to be calculated.
Let θ = Angle made by oblique section FC with normal
cross-section BC,
Ans.
D
C
Pt
F
A
Pn
B
Fig. 3.8
τ = Shear stress across faces BC and AD.
It has already been proved (Refer to Art. 2.9) that a shear stress is always accompanied
by an equal shear stress at right angles to it. Hence the faces AB and CD will also be subjected
to a shear stress q as shown in Fig. 3.8. Now these stresses will be converted into equivalent
forces. Then these forces will be resolved along the inclined surface and normal to inclined
surface. Consider the forces acting on the wedge FBC of Fig. 3.9.
101
STRENGTH OF MATERIALS
Q1 = Shear force on face BC
D
C Q
1 s
= Shear stress × Area of face BC
in
Q1 cos = τ × BC × 1
Q1 =
(∵ Area of face BC = BC × 1)
× BC × 1
= τ × BC
Q2 =
Q2 = Shear force on face FB
× FB × 1
= τ × Area of FB
A
F B
Q
= τ × FB × 1 = τ . FB
2 co
s
Pn = Total normal force on section FC
Fig. 3.9
Pt = Total tangential force on section FC.
The force Q1 is acting along face CB as shown in Fig. 3.9. This force is resolved into two
components i.e., Q1 cos θ and Q 1 sin θ along the plane CF and normal to the plane
CF respectively.
The force Q2 is acting along the face FB. This force is also resolved into two components
i.e., Q2 sin θ and Q2 cos θ along the plane FC and normal to the plane FC respectively.
∴ Total normal force on section FC,
Pn = Q1 sin θ + Q2 cos θ
= τ × BC × sin θ + τ × FB × cos θ.
(∵ Q1 = τ × BC and Q2 = τ × FB)
And total tangential force on section FC.
(–ve sign is taken due to opposite direction)
Pt = Q2 sin θ – Q1 cos θ.
= τ × FB × sin θ – τ × BC × cos θ
(∵ Q2 = τ . FB and Q1 = τ . BC)
Let
σn = Normal stress on section FC
σt = Tangential stress on section FC
Total normal force on section FC
Then
σn =
Area of section FC
Pn
=
FC × 1
τ . BC . sin θ + τ . FB . cos θ
=
(∵ Area = FC × 1)
FC × 1
BC
FB
=τ.
. sin θ + τ .
. cos θ
FC
FC
= τ . cos θ . sin θ + τ . sin θ . cos θ
BC
FB
∵ In triangle FBC,
= cos θ,
= sin θ
FC
FC
= 2τ cos θ . sin θ
= τ sin 2θ
(∵ 2 sin θ cos θ = sin 2θ) ...(3.10)
Total tangential force on section FC
and
σt-=
Area of section FC
Pt
=
FC × 1
τ × FB × sin θ − τ × BC × cos θ
=
FC × 1
FB
BC
=τ×
× sin θ – τ ×
× cos θ
FC
FC
= τ × sin θ × sin θ – τ × cos θ × cos θ
FG
H
102
2
Q
sin
Let
IJ
K
PRINCIPAL STRESSES AND STRAINS
= τ sin2 θ – τ cos2 θ = – τ [cos2 θ – sin2 θ]
= – τ cos 2θ
(∵ cos2 θ – sin2 θ = cos 2θ) ...(3.11)
–ve sign shows that σt will be acting downwards on the plane CF.
3.4.4. A Member Subjected to Direct Stresses in two Mutually Perpendicular
Directions Accompanied by a Simple Shear Stress. Fig. 3.10 (a) shows a rectangular
bar ABCD of uniform cross-sectional area A and of unit thickness. This bar is subjected to :
2
D
C
D
C Q
1
Pt
1
Q1 cos 1
F
A
Q2 sin Pn
F
B
A
Q2 cos 2
sin
Q1
P1 = 1 × BC × 1
Q2 B
P2 = 2 × FB × 1
(a)
(b )
Fig. 3.10
(i) tensile stress σ1 on the face BC and AD
(ii) tensile stress σ2 on the face AB and CD
(iii) a simple shear stress τ on face BC and AD.
But with reference to Art. 2.9, a simple shear stress is always accompanied by an equal
shear stress at right angles to it. Hence the faces AB and CD will also be subjected to a shear
stress τ as shown in Fig. 3.10 (a).
We want to calculate normal and tangential stresses on oblique section FC, which is
inclined at an angle θ with the normal cross-section BC. The given stresses are converted into
equivalent forces.
The forces acting on the wedge FBC are :
P1 = Tensile force on face BC due to tensile stress σ1
= σ1 × Area of BC
(∵ Area = BC × 1)
= σ1 × BC × 1
= σ1 × BC
P2 = Tensile force on face FB due to tensile stress σ2
= σ2 × Area of FB = σ2 × FB × 1
= σ2 × FB
Q1 = Shear force on face BC due to shear stress τ
= τ × Area of BC
= τ × BC × 1 = τ × BC
Q2 = Shear force on face FB due to shear stress τ
= τ × Area of FB
= τ × FB × 1 = τ × FB.
Resolving the above four forces (i.e., P1, P2, Q1 and Q2) normal to the oblique section
FC, we get
103
STRENGTH OF MATERIALS
Total normal force,
Pn = P1 cos θ + P2 sin θ + Q1 sin θ + Q2 cos θ
Substituting the values of P1, P2, Q1 and Q2, we get
Pn = σ1 . BC . cos θ + σ2 . FB . sin θ + τ . BC . sin θ + τ . FB. cos θ
Similarly, the total tangential force (Pt) is obtained by resolving P1, P2, Q1 and Q2 along
the oblique section FC.
∴ Total tangential force,
Pt = P1 sin θ – P2 cos θ – Q1 cos θ + Q2 sin θ
= σ1 . BC . sin θ – σ2 . FB . cos θ – τ . BC . cos θ + τ . FB . sin θ
(substitute the values of P1, P2, Q1 and Q2)
Now, Let
σn = Normal stress across the section FC, and
σt = Tangential stress across the section FC.
Then normal stress across the section FC,
Pn
Total normal force across section FC
σn =
=
Area of section FC
FC × 1
σ . BC . cos θ + σ 2 . FB . sin θ + τ . BC . sin θ + τ . FB . cos θ
= 1
FC × 1
BC
FB
BC
FB
= σ1 .
. cos θ + σ2 .
. sin θ + τ .
. sin θ + τ .
. cos θ
FC
FC
FC
FC
= σ1 . cos θ . cos θ + σ2 sin θ . sin θ + τ . cos θ . sin θ + τ sin θ . cos θ
FG∵
H
In triangle FBC,
IJ
K
BC
FB
= cos θ and
= sin θ
FC
FC
= σ1 cos2 θ + σ2 sin2 θ + 2τ cos θ sin θ
= σ1
FG 1 + cos 2θ IJ + σ FG 1 − cos 2θ IJ + τ sin 2θ
H 2 K H 2 K
FG∵ cos θ = 1 + cos 2θ , sin θ = 1 − cos 2θ and 2 cos θ sin θ = sin 2θIJ
K
H
2
2
2
2
2
σ1 + σ2 σ1 − σ2
+
cos 2θ + τ sin 2θ
2
2
and tangential stress (i.e., shear stress) across the section FC,
=
...(3.12)
Pt
Total tangential force across section FC
=
Area of section FC
FC × 1
σ1 . BC . sin θ − σ 2 . FB . cos θ − τ . BC . cos θ + τ . FB . sin θ
=
FC × 1
BC
FB
BC
FB
= σ1 .
. sin θ – σ2 .
. cos θ – τ .
. cos θ + τ .
. sin θ
FC
FC
FC
FC
= σ1 . cos θ . sin θ – σ2 . sin θ . cos θ – τ . cos θ . cos θ + τ . sin θ . sin θ
σt =
FG∵
H
In triangle FBC,
= (σ1 – σ2) . cos θ sin θ – τ cos2 θ + τ sin2 θ
=
104
FG σ
H
1
− σ2
2
IJ . 2 cos θ sin θ – τ (cos
K
2
θ – sin2 θ)
IJ
K
BC
FB
= cos θ and
= sin θ
FC
FC
PRINCIPAL STRESSES AND STRAINS
σ1 − σ2
. sin 2θ – τ cos 2θ
(∵ cos2 θ – sin2 θ = cos 2θ) ...(3.13)
2
Position of principal planes. The planes on which shear stress (i.e., tangential stress)
is zero, are known as principal planes. And the stresses acting on principal planes are known
as principal stresses.
The position of principal planes are obtained by equating the tangential stress [given
by equation (3.13)] to zero.
∴ For principal planes,
σt = 0
σ1 − σ2
sin 2θ – τ cos 2θ = 0
or
2
σ1 − σ2
or
sin 2θ = τ cos 2θ
2
sin 2θ
τ
2τ
=
=
or
cos 2θ ( σ1 − σ 2 ) ( σ1 − σ 2 )
2
2τ
or
tan 2θ =
...(3.14)
( σ1 − σ 2 )
But the tangent of any angle in a right angled triangle
N
=
=
Height of right angled triangle
Base of right angled triangle
Height of right angled triangle
2τ
=
Base of right angled triangle
( σ1 − σ 2 )
∴ Height of right angled triangle = 2τ
Base of right angled triangle = (σ1 – σ2).
Now diagonal of the right angled triangle
∴
=±
=
1st Case.
Then
and
2
2
2
( σ1 − σ 2 ) + ( 2τ ) = ± ( σ1 − σ 2 ) + 4 τ
( σ1 − σ 2 ) 2 + 4 τ 2
and –
Diagonal =
2
2
L
(1 – 2)
2
M
Fig. 3.11
( σ1 − σ 2 ) 2 + 4 τ 2
( σ1 − σ 2 ) 2 + 4 τ 2
sin 2θ =
Height
2τ
=
Diagonal
( σ1 − σ 2 ) 2 + 4 τ 2
cos 2θ =
Base
=
Diagonal
(σ 1 − σ 2 )
(σ 1 − σ 2 ) 2 + 4 τ 2
.
The value of major principal stress is obtained by substituting the values of sin 2θ and
cos 2θ in equation (3.12).
∴ Major principal stress
σ1 + σ2 σ1 − σ2
+
=
cos 2θ + τ sin 2θ
2
2
( σ1 − σ 2 )
σ1 + σ 2 σ1 − σ 2
2τ
+
×
+τ×
=
2
2
2
2
( σ − σ )2 + 4τ 2
( σ − σ ) + 4τ
1
2
1
2
105
STRENGTH OF MATERIALS
=
σ1 + σ2 1
+
2
2
=
( σ1 − σ 2 ) 2 + 4 τ 2
σ1 + σ 2
+
2
2 ( σ1 − σ 2 ) 2 + 4 τ 2
=
σ1 + σ 2 1
+
( σ1 − σ 2 ) 2 + 4 τ 2
2
2
σ1 + σ 2
+
=
2
2nd Case.
Diagonal = –
Then
(σ 1 − σ 2 ) 2 + 4 τ 2
FG σ
H
1
− σ2
2
IJ
K
+
2τ 2
(σ 1 − σ 2 ) 2 + 4 τ 2
2
+ τ2
...(3.15)
( σ1 − σ 2 ) 2 + 4 τ 2
sin 2θ =
and
(σ 1 − σ 2 ) 2
cos 2θ =
2τ
− ( σ1 − σ 2 ) 2 + 4 τ 2
( σ1 − σ 2 )
− ( σ1 − σ 2 ) 2 + 4 τ 2
Substituting these values in equation (3.12), we get minor principal stress.
∴ Minor principal stress
=
σ1 + σ2 σ1 − σ2
+
cos 2θ + τ sin 2θ
2
2
=
σ1 + σ 2 σ 1 − σ 2
σ1 − σ 2
2τ
+
×
+τ×
2
2
2
2
− ( σ1 − σ 2 ) + 4 τ
− (σ 1 − σ 2 ) 2 + 4 τ 2
=
σ1 + σ 2
( σ1 − σ 2 ) 2
−
−
2
2 ( σ1 − σ 2 ) 2 + 4 τ 2
=
( σ1 – σ 2 ) 2 + 4 τ 2
σ1 + σ 2
−
2
2 ( σ1 − σ 2 ) 2 + 4 τ 2
=
σ1 + σ 2 1
−
( σ1 − σ 2 ) 2 + 4 τ 2
2
2
σ1 + σ 2
−
=
2
FG (σ
H
1
– σ2 )
2
IJ
K
2τ 2
( σ1 – σ 2 ) 2 + 4 τ 2
2
+ τ2
...(3.16)
Equation (3.15) gives the maximum principal stress whereas equation (3.16) gives
minimum principal stress. These two principal planes are at right angles.
The position of principal planes is obtained by finding two values of θ from equation (3.14).
Fig. 3.11 (a) shows the principal planes in which θ1 and θ2 are the values from equation (3.14).
106
PRINCIPAL STRESSES AND STRAINS
2
D
C
Ma
jor
p
str rinci
pa
es
s
l
1
Principal
planes
1
1
2
2 =
90° + 1
Mi
no
rp
str rinci
es
s pal
A
B
2
Fig. 3.11 (a)
Maximum shear stress. The shear stress is given by equation (3.13). The shear stress
will be maximum or minimum when
d
(σ ) = 0
dθ t
d σ1 − σ 2
sin 2θ − τ cos 2θ = 0
or
2
dθ
σ1 − σ2
or
(cos 2θ) × 2 – τ (– sin 2θ) × 2 = 0
2
(σ1 – σ2) . cos 2θ + 2τ sin 2θ = 0
or
2τ sin 2θ = – (σ1 – σ2) cos 2θ
= (σ2 – σ1) cos 2θ
OP
Q
LM
N
sin 2θ σ 2 − σ1
=
cos 2θ
2τ
σ − σ1
or
tan 2θ = 2
2τ
Equation (3.17) gives condition for maximum or minimum shear
stress.
or
and
sin 2θ = ±
cos 2θ = ±
2
4
+
1) 2
–
2
σ 2 − σ1
( σ 2 − σ1 ) 2 + 4 τ 2
2τ
(σ 2 − σ 1 ) 2 + 4 τ 2
(2 – 1)
Then
σ 2 − σ1
2τ
If tan 2θ =
...(3.17)
2
2
Fig. 3.12
107
STRENGTH OF MATERIALS
Substituting the values of sin 2θ and cos 2θ in equation (3.13), the maximum and
minimum shear stresses are obtained.
∴ Maximum shear stress is given by
(σt)max =
σ1 − σ2
sin 2θ – τ cos 2θ
2
=±
=±
=±
σ1 − σ 2
×
2
( σ 2 − σ1 )
( σ 2 − σ1 ) 2 + 4 τ 2
±τ×
( σ1 − σ 2 ) 2
±
2 ( σ 2 − σ1 ) 2 + 4 τ 2
( σ 2 − σ1 ) 2 + 4 τ 2
2 ( σ 2 − σ1 ) + 4 τ
2
=±
2
2τ 2
( σ 2 − σ1 ) 2 + 4 τ 2
2τ 2
( σ 2 − σ1 ) 2 + 4 τ 2
1
( σ 2 − σ1 ) 2 + 4 τ 2
2
1
( σ 2 − σ1 ) 2 + 4 τ 2
2
1
(σ 1 − σ 2 ) 2 + 4 τ 2
=
...(3.18)
2
The planes on which maximum shear stress is acting, are obtained after finding the
two values of θ from equation (3.17). These two values of θ will differ by 90°.
∴
(σt)max =
The second method of finding the planes of maximum shear stress is to find first principal
planes and principal stresses. Let θ1 is the angle of principal plane with plane BC of Fig. 3.11
(a). Then the planes of maximum shear will be at θ1 + 45° and θ1 + 135° with the plane BC as
shown in Fig. 3.12 (a).
t
t
s2
D
C
5°
f
s o ar
ne she
a
l
.
P ax
m ress
st
(
s1
q1
+4
.
ax
m
s t)
s1
q1 + 135°
A
B
s2
t
t
Fig. 3.12 (a)
Note. The above relations hold good when one or both the stresses are compressive.
108
PRINCIPAL STRESSES AND STRAINS
Problem 3.11. At a point within a body subjected to two mutually perpendicular
directions, the stresses are 80 N/mm2 tensile and 40 N/mm2 tensile. Each of the above stresses
is accompanied by a shear stress of 60 N/mm2. Determine the normal stress, shear stress and
resultant stress on an oblique plane inclined at an angle of 45° with the axis of minor tensile
stress.
Sol. Given :
Major tensile stress,
σ1 = 80 N/mm2
Minor tensile stress,
σ2 = 40 N/mm2
Shear stress,
τ = 60 N/mm2
Angle of oblique plane, with the axis of minor tensile stress,
θ = 45°.
(i) Normal stress (σn )
Using equation (3.12),
σ + σ2 σ1 − σ2
cos 2θ + τ sin 2θ
σn = 1
+
2
2
80 + 40 80 − 40
cos (2 × 45°) + 60 sin (2 × 45°)
+
=
2
2
= 60 + 20 cos 90° + 60 sin 90°
= 60 + 20 × 0 + 60 × 1
(∵ cos 90° = 0)
2
= 60 + 0 + 60 = 120 N/mm . Ans.
2
40 N/mm
2
60 N/mm
2
θ
Axis of minor
tensile stress
2
80 N/mm
80 N/mm
θ
2
60 N/mm
2
40 N/mm
Fig. 3.13
(ii) Shear (or tangential) stress (σt )
Using equation (3.13),
σ1 − σ2
sin 2θ – τ cos 2θ
σt =
2
80 − 40
sin (2 × 45°) – 60 × cos (2 × 45°)
=
2
= 20 × sin 90° – 60 cos 90°
= 20 × 1 – 60 × 0
= 20 N/mm2. Ans.
(iii) Resultant stress (σR )
Using equation,
σR =
σ n2 + σ t2
109
STRENGTH OF MATERIALS
=
120 2 + 20 2 = 14400 + 400
= 14800 = 121.655 N/mm2. Ans.
Problem 3.12. A rectangular block of material is subjected to a tensile stress of 110
N/mm2 on one plane and a tensile stress of 47 N/mm2 on the plane at right angles to the former.
Each of the above stresses is accompanied by a shear stress of 63 N/mm2 and that associated
with the former tensile stress tends to rotate the block anticlockwise. Find :
(i) the direction and magnitude of each of the principal stress and
(ii) magnitude of the greatest shear stress.
Sol. Given :
Major tensile stress, σ1 = 110 N/mm2
Minor tensile stress, σ2 = 47 N/mm2
Shear stress,
τ = 63 N/mm2
(i) Major principal stress is given by equation (3.15).
∴ Major principal stress =
FG σ
H
σ1 + σ2
+
2
1
− σ2
2
IJ
K
47 N/mm
63 N/mm
110 N/mm
2
+ τ2
2
2
q
2
2
110 N/mm
63 N/mm
47 N/mm
2
2
Fig. 3.14
=
110 + 47
+
2
157
+
=
2
FG 63 IJ
H 2K
FG 110 − 47 IJ
H 2 K
2
+ 63 2
2
+ ( 63) 2
= 78.5 + 31.5 2 + 63 2 = 78.5 + 992.25 + 3969
= 78.5 + 70.436 = 148.936 N/mm2. Ans.
Minor principal stress is given by equation (3.16).
∴ Minor principal stress,
110
σ1 + σ 2
−
=
2
FG σ
H
1
− σ2
2
IJ
K
2
+ τ2
PRINCIPAL STRESSES AND STRAINS
=
110 + 47
−
2
FG 110 − 47 IJ
H 2 K
2
+ 63 2 = 78.5 – 70.436
= 8.064 N/mm2. Ans.
The directions of principal stresses are given by equation (3.14).
∴ Using equation (3.14),
tan 2θ =
2τ
2 × 63
=
σ1 − σ 2 110 − 47
2 × 63
= 2.0
63
∴
2θ = tan–1 2.0 = 63° 26′ or 243° 26′
∴
θ = 31° 43′ or 121° 43′. Ans.
(ii)
Magnitude of the greatest shear stress
Greatest shear stress is given by equation (3.18).
Using equation (3.18),
=
(σt)max =
=
1
( σ1 − σ 2 ) 2 + 4 τ 2
2
1
(100 − 47 ) 2 + 4 × 63 2
2
1
1
63 2 + 4 × 63 2 = × 63 × 5
2
2
= 70.436 N/mm2. Ans.
Problem 3.13. Direct stresses of 120 N/mm2 tensile and 90 N/mm2 compression exist on
two perpendicular planes at a certain point in a body. They are also accompanied by shear
stress on the planes. The greatest principal stress at the point due to these is 150 N/mm2.
(a) What must be the magnitude of the shearing stresses on the two planes ?
(b) What will be the maximum shearing stress at the point ?
Sol. Given :
Major tensile stress,
σ1 = 120 N/mm2
Minor compressive stress,
σ2 = – 90 N/mm2
(Minus sign due to compression)
Greatest principal stress
= 150 N/mm2
(a) Let
τ = Shear stress on the two planes.
Using equation (3.15) for greatest principal stress, we get
=
Greatest principal stress =
or
σ1 + σ 2
+
2
150 =
FG σ
H
1
− σ2
2
IJ
K
2
+ τ2
FG 120 − ( − 90) IJ
H 2 K
FG 120 + 90 IJ + τ
H 2 K
120 + ( − 90)
+
2
120 − 90
+
=
2
2
+ τ2
2
2
111
STRENGTH OF MATERIALS
= 15 + 105 2 + τ 2
or
150 – 15 =
105 2 + τ 2
or
135 =
105 2 + τ 2
or
Squaring both sides, we get
1352 = 1052 + τ2
τ2 = 1352 – 1052 = 18225 – 11025 = 7200
∴
τ =
7200 = 84.853 N/mm2. Ans.
(b) Maximum shear stress at the point
Using equation (3.18) for maximum shear stress,
1
( σ1 − σ 2 ) 2 + 4 τ 2
2
1
=
[120 − (− 90)]2 + 4 × 7200
2
1
1
1
210 2 + 28800 =
44100 + 28800 = × 270
=
2
2
2
= 135 N/mm2. Ans.
(σt)max =
(∵ τ2 = 7200)
Problem 3.14. At a certain point in a strained material, the stresses on two planes, at
right angles to each other are 20 N/mm2 and 10 N/mm2 both tensile. They are accompanied by
a shear stress of a magnitude of 10 N/mm2. Find graphically or otherwise, the location of
principal planes and evaluate the principal stresses.
Sol. Given :
10 N/mm
2
2
t = 10 N/mm
Mi
no
rp
str rinci
es
s pal
2
12
1
°4
3¢
Principal
planes
20 N/mm
Ma
31
°4
3¢
jor
p
str rinci
pa
es
s
l
t
2
t = 10 N/mm
10 N/mm
2
Fig. 3.14 (a)
112
PRINCIPAL STRESSES AND STRAINS
Major tensile stress,
σ1 = 20 N/mm2
Minor tensile stress,
σ2 = 10 N/mm2
Shear stress,
τ = 10 N/mm2
Location of principal planes
The location of principal planes is given by equation (3.14).
Using equation (3.14),
2τ
2 × 10
2 × 10
=
=
= 2.0
σ1 − σ 2 20 − 10
10
∴
2θ = tan–1 2.0 = 63° 26′ or 243° 26′
or
θ = 31° 43′ or 121° 43′. Ans.
Magnitude of principal stresses
The major principal stress is given by equation (3.15)
∴ Major principal stress
tan 2θ =
=
σ1 + σ 2
+
2
= 15 +
FG σ
H
1
− σ2
2
IJ
K
2
+ τ2 =
5 2 + 100 = 15 +
20 + 10
+
2
25 + 100 = 15 +
FG 20 − 10 IJ
H 2 K
2
+ 10 2
125 = 15 + 11.18
N/mm2.
= 26.18
Ans.
The minor principal stress is given by equation (3.16).
∴ Minor principal stress
=
σ1 + σ 2
−
2
FG σ − σ IJ
H 2 K
FG 20 − 10 IJ
H 2 K
1
2
+ τ2
2
2
20 + 10
−
+ 10 2
2
= 15 – 11.18 = 3.82 N/mm2. Ans.
=
Problem 3.15. A point in a strained material is subjected to the stresses as shown in
Fig. 3.15.
Locate the principal planes, and evaluate the principal stresses.
2
40 N/mm
60 N/mm
2
60°
2
C
60 N/mm
D
60°
A
B
2
40 N/mm
Fig. 3.15
113
STRENGTH OF MATERIALS
Sol. Given :
The stress on the face BC or AD is not normal. It is inclined at an angle of 60° with face
BC or AD. This stress can be resolved into two components i.e., normal to the face BC (or AD)
and along the face BC (or AD).
∴ Stress normal to the face BC or AD
= 60 × sin 60° = 60 × 0.866 = 51.96 N/mm2
Stress along the face BC or AD
= 60 × cos 60° = 60 × 0.5 = 30 N/mm2
The stress along the face BC or AD is known as shear stress. Hence τ = 30 N/mm2. Due
to complementary shear stress the face AB and CD will also be subjected to shear stress
of 30 N/mm2. Now the stresses acting on the material are shown in Fig. 3.16.
40 N/mm
2
2
2
51.96 N/mm
51.96 N/mm
2
30 N/mm
30 N/mm
2
40 N/mm
2
Fig. 3.16
Major tensile stress,
σ1 = 51.96 N/mm2
Minor tensile stress,
σ2 = 40 N/mm2
Shear stress,
τ = 30 N/mm2
Location of principal planes
Let θ = Angle, which one of the principal planes make with the stress of 40 N/mm2.
The location of the principal planes is given by the equation (3.14).
Using equation (3.14), we get
2τ
2 × 30
=
= 4.999
σ1 − σ 2 51.96 − 40
2θ = tan–1 4.999 = 78° 42′ or 258° 42′
θ = 39° 21′ or 129° 21′. Ans.
tan 2θ =
∴
or
Principal stress
The major principal stress is given by equation (3.15).
∴ Major principal stress
FG σ − σ IJ + τ
H 2 K
51.96 + 40
F 51.96 − 40 IJ
+ G
=
H 2 K
2
σ1 + σ 2
+
=
2
114
2
1
2
2
2
+ 30 2
PRINCIPAL STRESSES AND STRAINS
= 45.98 + 30.6
= 76.58 N/mm2. Ans.
The minor principal stress is given by equation (3.16).
∴ Minor principal stress
=
σ1 + σ2
−
2
FG σ
H
1
FG
H
− σ2
2
IJ
K
2
+ τ2
IJ
K
2
51.96 + 40
51.96 − 40
+ 30 2
−
=
2
2
= 45.98 – 30.6
= 15.38 N/mm2. Ans.
Problem 3.16. The normal stress in two mutually perpendicular directions are 600 N/mm2
and 300 N/mm2 both tensile. The complimentary shear stresses in these directions are of intensity
450 N/mm 2 . Find the normal and tangential stresses on the two planes which are
equally inclined to the planes carrying the normal stresses mentioned above.
Sol. Given :
Major tensile stress,
σ1 = 600 N/mm2
Minor tensile stress,
σ2 = 300 N/mm2
Shear stress,
τ = 450 N/mm2
The normal and tangential stresses are to be calculated on the two planes which are
equally inclined to the planes of major tensile stress and of minor tensile stress. This means
θ = 45° and 135°.
∴
Angle θ = 45° and 135°.
(i) Normal stress (σn) is given by equation (3.12).
σ1 − σ2 σ1 − σ2
+
cos 2θ + τ sin 2θ
2
2
(a) When θ = 45°, the normal stresses (σn) becomes as
∴
σn =
600 + 300 600 − 300
+
cos (2 × 45°) + 450 sin (2 × 45°)
2
2
= 450 + 150 cos 90° + 450 sin 90°
= 450 + 150 × 0 + 450 × 1
(∵ cos 90° = 0 and sin 90° = 1)
= 900 N/mm2. Ans.
(b) When θ = 135°, the normal stress (σn) becomes as
σn =
600 + 300 600 − 300
+
cos (2 × 135) + 450 sin (2 × 135°)
2
2
= 450 + 150 cos (270°) + 450 sin 270°
= 450 + 150 × 0 + 450 × (– 1) (∵ cos 270° = 0 and sin 270° = – 1)
= 450 – 450 = 0. Ans.
(ii) Tangential stress (σt) is given by equation (3.13)
σn =
∴
σt =
σ1 − σ 2
sin 2θ – τ cos 2θ
2
115
STRENGTH OF MATERIALS
(a) When θ = 45°, the tangential stress (σt) becomes as
600 − 300
sin 90° – 450 cos 90°
2
= 150 × 1 – 450 × 0 = 150 N/mm2.
(b) When θ = 135°, the tangential stress (σt) becomes as
σt =
Ans.
600 − 300
sin 270° – 450 cos 270°
2
= 150 × (– 1) – 450 × 0 = – 150 N/mm2. Ans.
Problem 3.17. The intensity of resultant stress on a
2
600 N/cm
plane AB [Fig. 3.16 (a)] at a point in a material under stress
B
is 800 N/cm2 and it is inclined at 30° to the normal to that C
2
plane. The normal component of stress on another plane BC
800 N/cm
2
at right angles to plane AB is 600 N/cm .
30°
Determine the following :
(i) the resultant stress on the plane BC,
(ii) the principal stresses and their directions,
A
(iii) the maximum shear stresses and their planes.
Fig. 3.16 (a)
Sol. Given :
Resultant stress on plane AB
= 800 N/cm2
Angle of inclination of the above stress
= 30°
Normal stress on plane BC
= 600 N/cm2
The resultant stress 800 N/cm2 on plane AB is resolved into normal stress and tangential
stress.
σt =
2
2 = 600 N/cm
2
= 400 N/cm
C
B
2
= 400 N/cm
y
1 = 692.82 N/cm
2
1 = 692.82 N/cm
x
= 400 N/cm
2
D
A
= 400 N/cm
2
2 = 600 N/cm
Fig. 3.16 (b)
The normal stress on plane AB
= 800 × cos 30° = 692.82 N/cm2.
The tangential stress on plane AB
= 800 × sin 30° = 400 N/cm2.
116
2
2
PRINCIPAL STRESSES AND STRAINS
The shear stress on plane AB is, i.e., τAB = 400 N/cm2, then to maintain the equilibrium
on the wedge ABC, another shear stress of the same magnitude, i.e., τBC = 400 N/cm2 must act
on the plane BC. The free body diagram of the element ABCD is shown in Fig. 3.16 (a), showing
normal and shear stresses acting on different faces.
(i) Resultant stress on plane BC
On plane BC, from Fig. 3.16 (a),
σ2 = 600 N/cm2
Shear stress,
τ = 400 N/cm2
∴ Resultant stress on plane BC
=
σ 22 + τ2
=
600 2 + 400 2 = 721 N/cm2.
Ans.
The resultant will be inclined at an angle θ with the horizontal given by,
σ 2 600
=
= 1.5
400
τ
∴
θ = tan–1 1.5 = 56.3°. Ans.
(ii) Principal stresses and their directions
The major principal stress is given by equation (3.15).
∴ Major principal stress
tan θ =
σ1 + σ 2
+
=
2
FG σ
H
1
− σ2
2
FG
H
IJ
K
2
+ τ2
IJ
K
2
692.82 + 600
692.82 − 600
+
+ 400 2
2
2
= 646.41 – 402.68
= 1049.09 N/cm2 (Tensile). Ans.
The minor principal stress is given by equation (3.16)
∴ Minor principal stress
=
σ1 + σ 2
−
=
2
FG σ
H
1
− σ2
2
FG
H
IJ
K
2
+ τ2
IJ
K
2
692.82 + 600
692.82 − 600
−
+ 400 2
2
2
= 646.41 – 402.68
= 243.73 N/cm2 (Tensile). Ans.
The directions of principal stresses are given by equation (3.14), as
2τ
2 × 400
800
=
=
= 8.618
tan 2θ =
( σ1 − σ 2 ) ( 692.82 − 600) 92.82
∴
2θ = tan–1 8.618 = 83.38° or 263.38°
∴
θ = 41.69° or 131.99°. Ans.
(iii) The maximum shear stress and their planes.
The maximum shear stress is given by equation (3.18).
=
117
STRENGTH OF MATERIALS
∴
1
( σ1 − σ 2 ) 2 + 4 τ 2 =
(σt)max =
2
=
FG 692.82 − 600 IJ
H 2 K
FG σ
H
1
− σ2
2
IJ
K
2
+ τ2
2
+ 400 2
= 402.68 N/cm2. Ans.
Problem 3.18. At a certain point in a material under stress the intensity of the resultant
stress on a vertical plane is 1000 N/cm2 inclined at 30° to the normal to that plane and the
stress on a horizontal plane has a normal tensile component of intensity 600 N/cm2 as shown in
Fig. 3.16 (c). Find the magnitude and direction of the resultant stress on the horizontal plane
and the principal stresses.
2
600 N/cm
C
1000 N/cm
B
2
30°
A
Fig. 3.16 (c)
Sol. Given :
Resultant stress on vertical plane AB = 1000 N/cm2
Inclination of the above stress = 30°
Normal stress on horizontal plane BC = 600 N/cm2
The resultant stress on plane AB is resolved into normal and tangential component.
The normal component
= 1000 × cos 30° = 866 N/cm2
Tangential component
= 1000 × sin 30° = 500 N/cm2.
Hence a shear stress of magnitude 500 N/cm2 is acting on plane AB. To maintain the
wedge in equilibrium, another shear stress of the same magnitude but opposite in direction
must act on the plane BC. The free-body diagram of the element ABCD is shown in Fig. 3.16 (d),
showing normal and shear stresses acting on different faces in which :
σ1 = 866 N/cm2,
and
σ2 = 600 N/cm2
τ = 500 N/cm2
(i) Magnitude and direction of resultant stress on horizontal plane BC.
Normal stress on plane BC, σ2 = 600 N/cm2
118
PRINCIPAL STRESSES AND STRAINS
Tangential stress on plane BC, τ = 500 N/cm2
2 = 600 N/cm
2
C
B
2
= 500 N/cm
1 = 866 N/cm
2
1 = 866 N/cm
D
2
A
2
= 500 N/cm
2 = 600 N/cm
2
Fig. 3.16 (d)
∴ Resultant stress
=
σ 22 + τ 2
=
2
600 2 + 500 2 = 781.02 N/cm .
Ans.
The direction of the resultant stress with the horizontal plane BC is given by,
σ 2 600
=
= 1.2
500
τ
θ = tan–1 1.2 = 50.19°.
tan θ =
Ans.
(ii) Principal stresses
The major and minor principal stresses are given by equations (3.15) and (3.16).
∴ Principal stresses
FG σ − σ IJ + τ
H 2 K
866 + 600
F 866 − 600 IJ
± G
=
H 2 K
2
=
σ1 + σ 2
±
2
2
1
2
2
2
+ 500 2
= 733 ± 517.38
= (733 + 517.38) and (733 – 517.38)
= 1250.38 and 215.62 N/cm2.
∴ Major principal stress
= 1250.38 N/cm2. Ans.
∴ Minor principal stress
= 215.62 N/cm2. Ans.
Problem 3.19. At a point in a strained material, on plane BC there are normal and
shear stresses of 560 N/mm2 and 140 N/mm2 respectively. On plane AC, perpendicular to plane
BC, there are normal and shear stresses of 280 N/mm2 and 140 N/mm2 respectively as shown in
Fig. 3.16 (e). Determine the following :
(i) principal stresses and location of the planes on which they act,
(ii) maximum shear stress and the plane on which it acts.
119
STRENGTH OF MATERIALS
A
280 N/mm
140 N/mm
B
2
2
C
140 N/mm
560 N/mm
2
2
Fig. 3.16 (e)
Sol. Given :
On plane AC,
σ1 = – 280 N/mm2
(– ve sign due to compressive stress)
2
τ = 140 N/mm
On plane BC,
σ2 = 560 N/mm2
τ = 140 N/mm2
(i) Principal stresses and location of the planes on which they act.
Principal stress are given by equations (3.15) and (3.16)
∴ Principal stresses
FG σ
H
=
σ1 + σ 2
±
2
=
− 280 + 560
±
2
1
− σ2
2
IJ
K
2
+ τ2
FG − 280 − 560 IJ
H 2 K
2
+ 140 2
= 140 ± 442.7
= 582.7 and (140 – 442.7) N/mm2
= 582.7 and – 302.7 N/mm2
∴ Major principal stress
= 582.7 N/mm2 (Tensile). Ans.
∴ Minor principal stress
= – 302.7 N/mm2. Ans.
The planes on which principal stresses act, are given by equation (3.14) as
tan 2θ =
2τ
2 × 140
280
=
=
= − 0.33
σ1 − σ 2 − 280 − 560 − 840
∴
2θ = tan–1 – 0.33 = –18.26º
– ve sign shows that 2θ is lying in 2nd and 4th quadrant
∴
2θ = (180 – 18.26°) or (360 – 18.26°)
= 161.34° or 341.34°
∴ θ = 80.67° and 170.67°. Ans.
(ii) Maximum shear stress and the plane on which it acts.
Maximum shear stress is given by equation (3.18).
120
PRINCIPAL STRESSES AND STRAINS
∴
(σt)max =
FG σ
H
1
− σ2
2
IJ
K
2
+ τ2
=
FG − 280 − 560 IJ
H 2 K
=
420 2 + 140 2 = 442.7 N/mm2.
2
+ 1402
Ans.
The plane on which maximum shear stress acts is given by equation (3.17) as
tan 2θ =
=
σ 2 − σ1
2τ
560 − ( − 280) 840
=
= 3.0
2 × 140
280
∴
2θ = tan–1 3.0 = 71.56° or 251.56°
∴
θ = 35.78° or 125.78°. Ans.
Problem 3.20. On a mild steel plate, a circle of diameter 50 mm is drawn before the
plate is stressed as shown in Fig. 3.17. Find the lengths of the major and minor axes of an
ellipse formed as a result of the deformation of the circle marked.
20 N/mm
40 N/mm
2
2
D
40 N/mm
C
2
80 N/mm
80 N/mm
2
2
40 N/mm
A
40 N/mm
20 N/mm
2
2
B
2
Fig. 3.17
Take E = 2 × 105 N/mm2 and
1 1
= .
m 4
Sol. Given :
Major tensile stress, σ1 = 80 N/mm2
121
STRENGTH OF MATERIALS
Minor tensile stress, σ2 = 20 N/mm2
Shear stress,
τ = 40 N/mm2
Value of E
= 2 × 105 N/mm2
Major principal stress is given by equation (3.15).
∴ Major principal stress
IJ
K
=
1   2

2
FG 
H
=
80  20

2
FG 80  20 IJ
H 2 K
1
 2
2
2
2
 2
 402
= 50 + 30 2  40 2 = 50 + 50 = 100 N/mm2 (tensile)
Minor principal stress
1   2

=
2
=
80  20

2
FG 
H
1
 2
2
IJ
K
FG 80  20 IJ
H 2 K
2
2
 2
 402 = 50 – 50 = 0.
From Fig. 3.17, it is clear that diagonal BD will be elongated and diagonal AC will be
shortened. Hence the circle will become an ellipse whose major axis will be along BD and
minor axis along AC as shown in Fig. 3.17.The major principal stress acts along BD and
minor principal stress along AC.
∴ Strain along BD
=
=
Major principal stress Minor principal stress

E
mE
100
2  10
5

0
5
2  10  4
FG 1  1 IJ
H m 4K
1
2000
Increase in diameter along BD
=
∴
= Strain along BD × Dia. of hole =
1
 50  0.025 mm
2000
Strain along AC
=
=
Minor principal stress Major principal stress

E
mE
0
2  10
=–
122
5
1
8000

100
4  2  105
(– ve sign shows that there is a decrease in length)
PRINCIPAL STRESSES AND STRAINS
∴
Decrease in length of diameter along AC
1
 50 = 0.00625 mm
8000
∴ The circle will become an ellipse whose major axis will be 50 + 0.025 = 50.025 mm
and minor axis will be
= Strain along AC × Dia. of hole =
50 – 0.00625 = 49.99375 mm.
3.5. MOHR’S CIRCLE..
Mohr’s circle is a graphical method of finding normal, tangential and resultant stresses
on an oblique plane. Mohr’s circle will be drawn for the following cases :
(i) A body subjected to two mutually perpendicular principal tensile stresses of unequal
intensities.
(ii) A body subjected to two mutually perpendicular principal stresses which are
unequal and unlike (i.e., one is tensile and other is compressive).
(iii) A body subjected to two mutually perpendicular principal tensile stresses accompanied by a simple shear stress.
3.5.1. Mohr’s Circle when a Body is Subjected to two Mutually Perpendicular
Principal Tensile Stresses of Unequal Intensities. Consider a rectangular body subjected to two mutually perpendicular principal tensile stresses of unequal intensities. It is
required to find the resultant stress on an oblique plane.
Let
σ1 = Major tensile stress
σ2 = Minor tensile stress, and
θ = Angle made by the oblique plane with the axis of minor tensile stress.
Mohr’s circle is drawn as : (See Fig. 3.18).
E
Take any point A and draw a horizontal
line through A. Take AB = σ1 and AC = σ2 towards
right from A to some suitable scale. With BC as
2θ
diameter describe a circle. Let O is the centre of
θ
B
D
C
O
the circle. Now through O, draw a line OE A
σ2
making an angle 2θ with OB.
σ1
From E, draw ED perpendicular on AB.
Join AE. Then the normal and tangential stresses
on the oblique plane are given by AD and ED
Fig. 3.18
respectively. The resultant stress on the oblique
plane is given by AE.
From Fig. 3.18, we have
Length AD = Normal stress on oblique plane
Length ED = Tangential stress on oblique plane
Length AE = Resultant stress on oblique plane.
1  2
Radius of Mohr’s circle =
2
Angle
φ = obliquity.
Proof. (See Fig. 3.18)
CO = OB = OE = Radius of Mohr’s circle =
1  2
2
123
STRENGTH OF MATERIALS
∴
AO = AC + CO
 1   2 2 2   1   2  1   2


= σ2 +
2
2
2
OD = OE cos 2θ
1  2
cos 2θ
2
AD = AO + OD
1  2 1  2

=
cos 2θ
2
2
= σn or Normal stress
ED = OE sin 2θ
1  2
sin 2θ
=
2
= σt or Tangential stress.
=
∴
and
FG
H
OE 
1  2
2
IJ
K
Important points. (See Fig. 3.18)
(i) Normal stress is along the line ACB. Hence maximum normal stress will be when
point E is at B. And minimum normal stress will be when point E is at C. Hence maximum
normal stress = AB = σ1 and minimum normal stress = AB = σ2.
(ii) Tangential stress (or shear stress) is along a line which is perpendicular to line CB.
Hence maximum shear stress will be when perpendicular to line CB is drawn from point O.
Then maximum shear stress will be equal to the radius of the Mohr’s circle.
1  2
.
∴
(σt)max =
2
(iii) When the point E is at B or at C, the shear stress will be zero.
(iv) The angle φ (which is known as angle of obliquity) will be maximum, when the line
AE is tangent to the Mohr’s circle.
Problem 3.21. Solve problem 3.5 by using Mohr’s circle method.
Sol. The data is given in problem 3.5, is
σ1 = 120 N/mm2 (tensile)
σ2 = 60 N/mm2 (tensile)
θ = 30°.
Scale. Let
1 cm = 10 N/mm2
120
Then
σ1 =
= 12 cm
10
E
60
and
σ2 =
= 6 cm
10
2θ
Mohr’s circle is drawn as : (See Fig. 3.19).
θ
B
A
D
C
O
Take any point A and draw a horizontal
σ2
σ1
line through A. Take AB = σ1 = 12 cm and
AC = σ2 = 6 cm. With BC as diameter (i.e.,
BC = 12 – 6 = 6 cm) describe a circle. Let O is the
Fig. 3.19
centre of the circle. Through O, draw a line OE
124
PRINCIPAL STRESSES AND STRAINS
making an angle 2θ (i.e., 2 × 30 = 60°) with OB. From E, draw ED perpendicular to CB. Join
AE. Measure lengths AD, ED and AE.
By measurements :
Length AD = 10.50 cm
Length ED = 2.60 cm
Length AE = 10.82 cm
Then normal stress
= Length AD × Scale
= 10.50 × 10 = 105 N/mm2. Ans.
Tangential or shear stress
= Length ED × Scale
= 2.60 × 10 = 26 N/mm2. Ans.
Resultant stress
= Length AE × Scale
= 10.82 × 10 = 108.2 N/mm2. Ans.
3.5.2. Mohr’s Circle when a Body is Subjected to two Mutually Perpendicular
Principal Stresses which are Unequal and Unlike (i.e., one is Tensile and other is
Compressive). Consider a rectangular body subjected to two mutually perpendicular principal stresses which are unequal and one of them is tensile and the other is compressive. It is
required to find the resultant stress on an oblique plane.
Let
σ1 = Major principal tensile stress,
σ2 = Minor principal compressive stress, and
θ = Angle made by the oblique plane with the axis
of minor principal stress.
Mohr’s circle is drawn as : (See Fig. 3.20)
E
Take any point A and draw a horizontal line through A
on both sides of A as shown in Fig. 3.20. Take AB = σ1(+) towards
right of A and AC = σ2(–) towards left of A to some suitable
scale. Bisect BC at O. With O as centre and radius equal to CO
2θ
φ
θ
B
or OB, draw a circle. Through O draw a line OE making an C
A
D
O
σ2
σ1
angle 2θ with OB.
ComTensile
From E, draw ED perpendicular to AB. Join AE and CE.
pressive
(+)
(–)
Then normal and shear stress (i.e., tangential stress) on the
oblique plane are given by AD and ED. Length AE represents
the resultant stress on the oblique plane.
Fig. 3.20
∴ From Fig. 3.20, we have
Length AD = Normal stress on oblique plane,
Length ED = Shear stress on oblique plane,
Length AE = Resultant stress on oblique plane, and
Angle φ = Obliquity.
Radius of Mohr’s circle = CO or OB =
1  2
.
2
Proof. (See Fig. 3.20).
CO = OB = OE = Radius of Mohr’s circle
=
1  2
2
125
STRENGTH OF MATERIALS
AO = OC – AC
1  2
   2  2 2  1   2
 2  1

=
2
2
2
∴
AD = AO + OD
= AO + OB cos 2θ
( OD = OE cos 2θ)
1  2 1  2
  2
 OE  Radius = 1

=
cos 2θ
2
2
2
= σn or Normal stress
and
ED = OE sin 2θ
1  2
  2
 OE = 1
sin 2θ
=
2
2
= σt or Tangential (or shear) stress.
Problem 3.22. Solve problem 3.6 by using Mohr’s circle method.
Sol. Given : The data given in problem 3.6, is
σ1 = 200 N/mm2
σ2 = – 100 N/mm2 (compressive)
θ = 30°.
It is required to determine the resultant stress and the maximum shear stress by Mohr’s
circle method. First choose a suitable scale.
Let 1 cm represents 20 N/mm2.
200
= 10 cm
Then
σ1 =
E
20
 100
and
σ2 =
= – 5 cm
20
Mohr’s circle is drawn as given in Fig. 3.21.
60°
f
C
B
A
O
D
Take any point A and draw a horizontal line through
100
200
A on both sides of A. Take AB = σ1 = 10 cm towards right of
A and AC = σ2 = – 5 cm towards left of A. Bisect BC at O.
With O as centre and radius equal to CO or OB, draw a
circle. Through O draw a line OE making an angle 2θ (i.e.,
2 × 30° = 60°) with OB. From E, draw ED perpendicular to
Fig. 3.21
AB. Join AE and CE. Then AE represents the resultant
stress and angle φ represents the obliquity.
By measurement from Fig. 3.21, we have
Length AE = 9.0 cm
Length AD = 6.25 cm and length ED = 6.5 cm
Angle φ = 46°
∴ Resultant stress
= Length AE × Scale
= 9.0 × 20 = 180 N/mm2. Ans.
Angle made by the resultant stress with the normal of the inclined plane = φ = 46°. Ans.
Normal stress = Length AD × 20
= 6.25 × 20 = 125 N/mm2
Shear stress
= Length ED × 20
= 6.5 × 20 = 130 N/mm2.
FG
H
IJ
K
FG
H
126
IJ
K
PRINCIPAL STRESSES AND STRAINS
2q
Maximum shear stress. Shear stress is along a line which is perpendicular to the line
AB. Hence maximum shear stress will be when perpendicular to line AB is drawn from point O.
Then maximum shear stress will be equal to the radius of Mohr’s circle.
∴ Maximum shear stress = Radius of Mohr’s circle
 1   2 200  100

=
= 150 N/mm2. Ans.
2
2
3.5.3. Mohr’s Circle when a Body is Subjected to two Mutually Perpendicular
Principal Tensile Stresses Accompanied by a
s2
Simple Shear Stress. Consider a rectangular body
t
subjected to two mutually perpendicular principal
D
C
tensile stresses of unequal intensities accompanied
Oblique
by a simple shear stress. It is required to find the
q
plane
t
t
resultant stress on an oblique plane as shown in s1
s1
Fig. 3.22.
F
Let σ1 = Major tensile stress
A
B
t
σ2 = Minor tensile stress
τ = Shear stress across face BC and AD
s2
θ = Angle made by the oblique plane
Fig. 3.22
with the plane of major tensile stress.
According to the principle of shear stress, the faces AB and CD will also be subjected to
a shear stress of τ.
Mohr’s circle is drawn as given in Fig. 3.23.
Take any point A and draw a horizontal line through A.
Take AB = σ1 and AC = σ2 towards
right of A to some suitable scale. Draw
H E
perpendiculars at B and C and cut off BF
G
and CG equal to shear stress τ to the same
scale. Bisect BC at O. Now with O as centre
and radius equal to OG or OF draw a circle.
f
D
B
M
Through O, draw a line OE making an
O a
L
A
C
angle of 2θ with OF as shown in Fig. 3.23.
s2
From E, draw ED perpendicular to CB.
Join AE. Then length AE represents the
F
resultant stress on the given oblique plane.
s1
And lengths AD and ED represents the
Fig. 3.23
normal stress and tangential stress
respectively.
Hence from Fig. 3.23, we have
Length AE = Resultant stress on the oblique plane
Length AD = Normal stress on the oblique plane
Length ED = Shear stress on the oblique plane.
Proof. (See Fig. 3.23).
1
1
( CB = σ1 – σ2)
CO = CB = [σ1 – σ2]
2
2
1
AO = AC + CO = σ2 + [σ1 – σ2]
2
127
STRENGTH OF MATERIALS
2 2   1   2  1   2

2
2
AD = AO + OD
1  2
=
+ OE cos (2θ – α)
[ OD = OE cos (2θ – α)]
2
1  2
+ OE [cos 2θ cos α + sin 2θ sin α]
=
2
1  2
=
+ OE cos 2θ cos α + OE sin 2θ sin α
2
1  2
+ OE cos α . cos 2θ + OE sin α . sin 2θ
=
2
1  2
+ OF cos α . cos 2θ + OF sin α . sin 2θ
=
2
( OE = OF = Radius)
1  2
=
+ OB cos 2θ + BF sin 2θ
2
( OF cos α = OB, OF sin α = BF)
1  2
+ CO cos 2θ + τ sin 2θ
( OB = CO, BF = τ)
=
2
  2
1  2 1  2
 CO  1

cos 2θ + τ sin 2θ
=
2
2
2
= σn or Normal stress
ED = OE sin (2θ – α) = OE (sin 2θ cos α – cos 2θ sin α)
= OE sin 2θ cos α – OE cos 2θ sin α
= OE cos α . sin 2θ – OE sin α . cos 2θ
= OE cos α . sin 2θ – OE sin α . cos 2θ
( OE = OF = Radius)
= OB . sin 2θ – BF cos 2θ
( OF cos α = OB, OF sin α = BF)
= CO . sin 2θ – τ cos 2θ
( OB = CO, BF = τ)
1  2
  2
sin 2θ – τ cos 2θ
 CO  1
=
2
2
= σt or Tangential stress.
=
Now
FG
H
IJ
K
FG
H
IJ
K
Maximum and minimum value of normal stress. In Fig. 3.23, the normal stress is
given by AD. Hence the maximum value of AD will be when D coincides with M and minimum
value of AD will be when D coincides with L.
∴ Maximum value of normal stress,
(σn)max = AM = AO + OM
1  2
+ OF
2
1  2
 OB 2  BF 2
=
2
FG
H
=
1   2

=
2
128
FG 
H
1
 2
2
IJ
K
2
(
 2
AO 
IJ
K
1  2
, OM  OF  Radius
2
In triangle OBF, OF = OB 2  BF 2 )
FG
H
OB 
1   2
, BF  
2
IJ
K
PRINCIPAL STRESSES AND STRAINS
Minimum value of normal stress,
(σn)min = AL = AO – LO
  2
– OF
= 1
2
=
1   2

2
FG 
H
1
( LO = OF = Radius)
 2
2
IJ
K
2
 2
(i) For maximum normal stress, the point D coincides with M. But when the point D
coincides with M, the point E also coincides with M. Hence for maximum value of normal
stress,
Angle
2θ = α
( Line OE coincides with line OM)
∴
Also
θ=

2
tan 2θ = tan α =
=
...(i)
FG
H
BF


OB 1   2
2
BF  , OB 
1   2
2
IJ
K
2
.
1   2
(ii) For maximum and minimum normal stresses, the shear stress is zero and hence the
planes, on which maximum and minimum normal stresses act, are known as principal planes
and the stresses are known as principal stresses.
(iii) For minimum normal stress, the point D coincides with point L. But when the point
D coincides with L, the point E also coincides with L. Then
Angle
2θ = π + α
( Line OE coincides with line OL)
 

...(ii)
2 2
From equations (i) and (ii), it is clear that the plane of minimum normal stress is inclined at an angle 90° to the plane of maximum normal stress.
∴
θ=
Maximum value of shear stress. Shear stress is given by ED. Hence maximum value
of ED will be when E coincides with G, and D coincides with O.
∴ Maximum shear stress,
( OH = OF = radius)
(σt)max = OH = OF
=
OB 2  BF 2
=
FG 
H
1
 2
2
IJ
K
2
(
 2
In triangle OBF, OF =
FG
H
OB 
OB 2  BF 2 )
1   2
, BF  
2
IJ
K
Problem 3.23. A point in a strained material is subjected to stresses shown in
Fig. 3.24. Using Mohr’s circle method, determine the normal and tangential stresses across the
oblique plane. Check the answer analytically.
129
STRENGTH OF MATERIALS
2
35 N/mm
2
25 N/mm
Oblique
plane
2
65 N/mm
45°
2
65 N/mm
2
25 N/mm
2
35 N/mm
Fig. 3.24
Sol. Given :
Major principal stress,
σ1 = 65 N/mm2
Minor principal stress,
σ2 = 35 N/mm2
Shear stress,
τ = 25 N/mm2
Angle of oblique plane,
θ = 45°.
Mohr’s circle method
Let
1 cm = 10 N/mm2
Then
σ1 =
65
= 6.5 cm,
10
35
25
= 3.5 cm and τ =
= 2.5 cm
10
10
Mohr’s circle is drawn as given in Fig. 3.25.
σ2 =
G
E
25
B
A
C
L
35
O
2q
D
=9
M
0°
65
25
F
Fig. 3.25
Take any point A and draw a horizontal line through A. Take AB = σ1 = 6.5 cm and
AC = σ2 = 3.5 cm towards right of A. Draw perpendicular at B and C and cut off BF and CG
130
PRINCIPAL STRESSES AND STRAINS
equal to shear stress τ = 2.5 cm. Bisect BC at O. Now with O as centre and radius equal to OF
(or OG) draw a circle. Through O, draw a line OE making an angle of 2θ (i.e., 2 × 45° = 90°)
with OF as shown in Fig. 3.25. From E, draw ED perpendicular to AB produced. Join AE.
Then length AD represents the normal stress and length ED represents the shear stress.
By measurements, length AD = 7.5 cm and
length ED = 1.5 cm.
∴ Normal stress (σn)
= Length AD × Scale = 7.5 × 10 = 75 N/mm2. Ans.
( 1 cm = 10 N/mm2)
And tangential stress (σt) = Length ED × Scale = 1.5 × 10 = 15 N/mm2. Ans.
Analytical Answers
Normal stress (σn) is given by equation (3.12).
∴ Using equation (3.12),
σn =
 1   2 1   2

cos 2θ + τ sin 2θ
2
2
65  35 65  35

cos (2 × 45°) + 25 sin (2 × 45°)
2
2
= 50 + 15 cos 90° + 25 sin 90°
= 50 + 15 × 0 + 25 × 1
( cos 90° = 0, sin 90° = 1)
= 50 + 0 + 25 = 75 N/mm2. Ans.
Tangential stress is given by equation (3.13)
∴ Using equation (3.13),
=
σt =
1   2
sin 2θ – τ cos 2θ
2
65  35
sin (2 × 45) – 25 cos (2 × 45)
2
= 15 sin 90° – 25 cos 90° = 15 × 1 – 25 × 0 = 15 – 0
= 15 N/mm2. Ans.
Problem 3.24. At a certain point in a strained material, the intensities of stresses on
two planes at right angles to each other are 20 N/mm2 and 10 N/mm2 both tensile. They are
accompanied by a shear stress of magnitude 10 N/mm2. Find graphically or otherwise, the
location of principal planes and evaluate the principal stresses.
Sol. Given :
Major tensile stress,
σ1 = 20 N/mm2
G
Minor tensile stress,
σ2 = 10 N/mm2
Shear stress,
τ = 10 N/mm2
This problem may be solved analytically or graphically. Here we shall solve it graphically (i.e., by Mohr’s
B
M
circle method).
A
L
C
O
2q
2
10
Scale. Take 1 cm = 2 N/mm
10
20
20
10
Then σ1 =
= 10 cm, σ2 =
= 5 cm
2
2
F
10
= 5 cm.
and
τ=
2
Fig. 3.26
Mohr’s circle is drawn as given in Fig. 3.26.
=
131
STRENGTH OF MATERIALS
Take any point A and draw a horizontal line through A. Take AB = σ1 = 10 cm and
AC = σ2 = 5 cm towards right side of A. Draw perpendiculars at B and C and cut off BF = CG
= τ = 5 cm. Bisect BC at O. Now with O as centre and radius equal to OG (or OF), draw a circle
cutting the horizontal line through A, at L and M as shown in Fig. 3.26. Then AM and AL
represent the major principal and minor principal stresses.
By measurements, we have
Length AM = 13.1 cm and Length AL = 1.91 cm
∠FOB (or 2θ) = 63.7°.
∴ Major principal stress
= Length AM × Scale
= 13.1 × 2 N/mm2
( 1 cm = 2 N/mm2)
= 26.2 N/mm2. Ans.
Minor principal stress
= Length AL × Scale
= 1.91 × 2 = 3.82 N/mm2. Ans.
Location of principal planes
2θ = 63.7°
63.7
= 31.85°. Ans.
∴
θ=
2
The second principal plane is given by
θ + 90° or 31.85° + 90° or 121.85°. Ans.
Problem 3.25. An elemental cube is subjected to tensile stresses of 30 N/mm2 and
10 N/mm2 acting on two mutually perpendicular planes and a shear stress of 10 N/mm2 on
these planes. Draw the Mohr’s circle of stresses and hence or otherwise determine the magnitudes
and directions of principal stresses and also the greatest shear stress.
Sol. Given :
Major tensile stress,
σ1 = 30 N/mm2
Minor tensile stress,
σ2 = 10 N/mm2
Shear stress,
τ = 10 N/mm2
Scale. Take
1 cm = 2 N/mm2
30
Then
σ1 =
= 15 cm
2
10
10
σ2 =
= 5 cm
and
τ=
= 5 cm
2
2
Mohr’s circle of stresses is drawn as given in
H
Fig. 3.27.
G
Max.
Take any point A and draw a horizontal line
shear
stress
through A.
10
Take AB = σ1 = 15 cm and AC = σ2 = 5 cm
B
towards right side of A. Draw perpendiculars at B and
M
A
L
C
O 2q
C and cut off BF = CG = τ = 5 cm. Bisect BC at O. Now
10
10
with O as centre and radius equal to OG (or OF), draw
30
a circle cutting the horizontal line through A at L and
M as shown in Fig. 3.27. Then AM and AL represents
F
the major and minor principal stresses respectively.
And OH represents the maximum shear stress.
Fig. 3.27
132
PRINCIPAL STRESSES AND STRAINS
By measurements, we have
Length AM = 17.1 cm
Length AL = 2.93 cm
Length OH = Radius of Mohr’s circle
= 7.05 cm
∠FOB (or 2θ) = 45°.
∴ Major principal stress
= Length AM × Scale
= 17.1 × 2
( 1 cm = 2 N/mm2)
2
= 34.2 N/mm . Ans.
Minor principal stress = Length AL × Scale
= 2.93 × 2
( 1 cm = 2 N/mm2)
2
= 5.86 N/mm . Ans.
∠FOB or 2θ = 45°
45
= 22.5°. Ans.
∴
θ=
2
The second principal plane is given by θ + 90°.
∴ Second principal plane
= 22.5 + 90 = 112.5°. Ans.
The greatest shear stress
= Length OH × Scale
= 7.05 × 20 = 14.1 N/mm2. Ans.
3.6. STRAIN ON AN OBLIQUE PLANE..
To determine the strain on an oblique plane due to stresses σx, σy and τ, we will first
consider strains due to each stress separately. After that we will combine the result.
σx). A rectangular bar ABCD is
3.6.1. Strain on an Oblique Plane due to Stress (σ
subjected to a stress σx. Due to this stress there is an increase in length ‘dx’ as shown in
Fig. 3.28. It is required to find the strain on the oblique plane AC (i.e., strain in the diagonal
AC). Final position of the bar is shown by AB′C′D.
D
C
C¢
q
q
F
sx
sx
q
A
B
x
B¢
dx
Fig. 3.28
Let dx = Increase in length x due to stress σx
ex = Strain in x-direction
=
∴
But
dx
x
dx = x × ex
dx = BB′ = CC′ = x × ex
133
STRENGTH OF MATERIALS
Let e is the strain in diagonal AC.
e=
∴
AC   AC
AC
But AC′ = AF + FC′ = AC + FC′
( AF = AC and FC′ = CC′ cos θ)
=
( AC  CC  cos )  AC
AC
=
CC  cos 
AC
=
( x  ex )  cos 
AC
=
AC cos   ex  cos 
= ex cos2 θ
AC
(
CC′ = dx = x × ex)
3.6.2. Strain on an Oblique Plane due to
σy). Refer to Fig. 3.29.
Stress (σ
Let dy = Increase in length ‘y’ due to stress σy
ey = Strain in y-direction
e = Strain in diagonal AC
Now
CC′ = DD′ = dy
But
ey = strain in y-direction
D¢
dy = y × ey
CC′ = DD′ = dy = y × ey
Now e = strain in diagonal AC due to stress σy
AC   AC
AC
C¢
F
dy
C
D
(90–q)
y
But
A
B
sy
Fig. 3.29
AC′ = AF + FC′
= AC + FC′
=
But
∴
(
AF = AC)
( AC  FC  )  AC FC 

AC
AC
FC′ = CC′ cos (90 – θ)
= CC′ sin θ
= (y × ey) sin θ
= (AC × sin θ × ey × sin θ)
(
CC′ = dy = y × ey)
( y = AC sin θ)
AC  sin 2   e y
e = FC  =
AC
AC
= ey × sin2 θ
134
(90–q)
q
∴
=
...(3.19)
sy
dy
=
y
or
Also x = AC cos θ
...(3.20)
PRINCIPAL STRESSES AND STRAINS
3.6.3. Strain on an Oblique Plane due to Shear
Stress τ. Refer to Fig. 3.30.
Let φ = shear strain due to shear stress τ
t
D
C
F
CC 
=
CB
f
CC′ = φ × CB
= φ × AC sin θ ( CB = AC sin θ)
We know that the strain in diagonal AC is given by
∴
e=
C¢
q
D¢
AC   AC
AC
But
q
A
B
t
Fig. 3.30
AC′ = AF + FC′
= AC + FC′
( AC  FC  )  AC
AC
FC

=
But
AC
=
FC′ = CC′ × cos θ
= (φ × AC sin θ) × cos θ
( CC′ = φ × AC sin θ)
  AC  sin   cos 
= φ × sin θ × cos θ
AC


...(3.21)
=  2 sin   cos  =  sin 2
2
2
3.6.4. Strain on an Oblique Plane due to Stresses σx, σ y and τ . The total strain on
an oblique plane will be obtained by adding the R.H.S. of equations (3.19), (3.20) and (3.21).
Let e = Total strain in diagonal AC due to stresses σx, σy, and τ
=
= Strain due to σx + strain due to σy + strain due to τ
= ex cos2 θ + ey sin2 θ +
= ex

sin 2θ
2
(1  cos 2)
(1  cos 2) 
 ey
 sin 2
2
2
2
1  cos 2
1  cos 2 

2
, sin2  
 cos  

2
2


=
ex  e y
2

ex  e y
2
 cos 2 

 sin 2
2
...(3.22)
3.6.5. Maximum or Minimum Value of Strain on Oblique Plane. The strain ‘e’ on
an oblique plane given by equation (3.22) will be maximum or minimum if de = 0
d
or
LM
MN
FG
H
IJ
K
OP
PQ
d ex  e y  ex  e y cos 2   sin 2 = 0
d
2
2
2
135
STRENGTH OF MATERIALS
0
FG e
H
x
 ey
2
IJ ( sin 2)  2   (cos 2)  2 = 0
2
K
+

(e x  e y )
2θ = tan 1
∴
y) 2
tan 2θ =
e
or
f
–
sin 2

=
cos 2
(e x  e y )
x
or
f
– (ex – ey) sin 2θ + φ × cos 2θ = 0
±Ö
(e
or
2
or
2q
(ex – ey)

(ex  e y )
Fig. 3.30 (a)

θ = 1 tan 1
( ex  e y )
2
or
From Fig. 3.30 (a), we have
( ex  e y )
cos 2θ = 
(ex  e y ) 2   2
sin 2θ = 
and

(ex  e y ) 2   2
Hence the diagonal strain ‘e’ will be maximum or minimum when the value of
tan 2θ =

. By substituting the +ve values of cos 2θ and sin 2θ in equation (3.22), we
(e x  e y )
get maximum value of strain.
∴
emax =
=
=
ex  e y
2
ex  e y
2
ex  e y
2



ex  e y
2

( ex  e y )
2
(ex  e y )  
2



2

(ex  e y ) 2   2
(ex  e y ) 2   2
2 ( ex  e y ) 2   2
ex  e y
1

( ex  e y )2  2 
2
2
( ex  e y )2
2

 
2
2
...(3.23)
Similarly by substituting the (–ve) values of cos 2θ and sin 2θ in equation (3.22), we get
the minimum value of strain.
∴
emin =
=
ex  e y
2
ex  ey
2

ex  e y
2

( ) ( e x  e y )
(ex  e y ) 2   2



2
( )
( ex  e y ) 2   2
Equations (3.23) and (3.24) give the values of principal strains.
136
2
2
ex  ey
 ex  ey 

1

 
(ex  ey )2  2 
   
2
2
2
 2 
...(3.24)
PRINCIPAL STRESSES AND STRAINS
3.7. MOHR'S STRAIN CIRCLE..
In case of Mohr’s strain circle (which is similar to Mohr’s stress circle) the linear strains ex
1 
and ey are taken along horizontal axis and half of shear strain    is taken along vertical axis.
2 

If we compare equations (3.12) and (3.22), we find that σ1 = ex, σ2 = ey and τ = (Half of
2
shear strain). Also by comparing equations (3.15), (3.16) with (3.23), (3.24) respectively, we

get σ1 = ex, σ2 = ey and τ = . Hence for drawing Mohr’s strain circle, the linear strains ex, ey
2
1 
are taken along horizontal axis whereas half of shear strain    along vertical axis.
2 
Suppose the linear strains (ex and ey) and shear strain φ are given. Construct Mohr’s
strain circle and find the values of principal strains (e1 and e2) and position of principal plane.
The followings are the steps: (Refer to Fig. 3.31)
(i) Take any point A. Draw horizontal and vertical lines through A.
(ii) On horizontal line, take AB = ex and AC = ey towards right side of A if they are
positive.
(iii) Draw perpendiculars at B and C and cut off BF = CG = half of shear strain i.e., φ/2.
(iv) Join points G and F cutting the
horizontal line at O. Now with
O as centre and radius equal to
OG (or OF), draw a circle cutting
the horizontal line through A at
L and M as shown in Fig. 3.31.
Then AM and AL represent the
major and minor principal
strains e1 and e2 respectively.
And OH represents the half of
maximum shear strain. Measure
angle MOF. Half of angle MOF
gives the position of principal
Fig. 3.31
plane. The second principal
plane will be at an angle of
90 + θ, where θ = Half of angle MOF.
HIGHLIGHTS
1.
2.
3.
4.
The planes, which have no shear stress, are known as principal planes.
The stresses, acting on principal planes, are known as principal stresses.
Analytical and graphical methods are used for finding the stresses on an oblique section.
When a member is subjected to a direct stress (σ) in one plane, then the stresses on an oblique
plane (which is inclined at an angle θ with the normal cross-section) are given by :
Normal stress,
σn = σ cos2 θ

Tangential stress,
σt =
sin 2θ
2
137
STRENGTH OF MATERIALS
Max. normal stress
5.
6.
=σ

Max. shear stress
= .
2
When a member is subjected to two like direct stresses in two mutually perpendicular directions, then the stresses on an oblique plane inclined at an angle θ with the axis of the minor
stress (or with the plane of major stress) are given by:
Normal stress,
σn =
1   2 1   2

cos 2θ.
2
2
Tangential stress,
σt =
 1  2
sin 2θ
2
Resultant stress,
σR =
 n2   t 2 .
The angle made by the resultant stress with the normal of the oblique plane, is known as obliquity.
It is denoted by φ. Mathematically, tan φ =
7.
When a member is subjected to a simple shear stress (τ), then the stresses on an oblique plane
are given as:
Normal stress,
σn = τ sin 2θ
Tangential stress,
8.
t
.
n
σt = – τ cos 2θ.
When a member is subjected to two direct stresses (σ 1 , σ 2 ) in two mutually
perpendicular directions accompanied by a simple shear stress (τ), then the stresses, on an
oblique plane inclined at an angle θ with the axis of minor stress, are given by:
Normal stress,
σn =
1   2  1   2

cos 2θ + τ sin 2θ.
2
2
Tangential stress,
σt =
 1  2
sin 2θ – τ cos 2θ
2
(a) Position of principal planes is given by tan 2θ =
2
 1  2
(b) Major principal stress
=
1  2

2
FG 
H
1
 2
2
IJ
K
2
=
1  2

2
FG 
H
1
 2
2
IJ
K
2
(c) Minor principal stress
(d) Maximum shear stress
=
1
( 1  2 )2  4 2
2
 2
 2
 2  1
.
2
9. Mohr’s circle of stresses is a graphical method of finding normal, tangential and resultant stresses
on an oblique plane.
10. Maximum shear stress by Mohr’s circle method, is equal to the radius of the Mohr’s circle.
11. The planes of maximum and minimum normal stresses are at an angle of 90° to each other.
(e) Condition for maximum shear stress, tan 2θ =
138
PRINCIPAL STRESSES AND STRAINS
EXERCISE
(A) Theoretical Questions
1. Define the terms : Principal planes and principal stresses.
2. A rectangular bar is subjected to a direct stress (σ) in one plane only. Prove that the normal and
shear stresses on an oblique plane are given by

σn = σ cos2 θ and σt =
sin 2θ
2
where θ = Angle made by oblique plane with the normal cross-section of the bar,
σn = Normal stress, and
σt = Tangential or shear stress.
3. A rectangular bar is subjected to two direct stresses (σ1 and σ2) in two mutually perpendicular
directions. Prove that the normal stress (σn) and shear stress (σt) on an oblique plane which is
inclined at an angle θ with the axis of minor stress are given by
1   2 1   2
cos 2θ

2
2
  2
and
σt = 1
sin 2θ.
2
4. Define the term ‘obliquity’ and write how it is determined.
σn =
5. Derive an expression for the stresses on an oblique plane of a rectangular body, when the body
is subjected to a simple shear stress.
6. A rectangular body is subjected to direct stresses in two mutually perpendicular directions
accompanied by a shear stress. Prove that the normal stress and shear stress on an oblique
plane inclined at an angle θ with the plane of major direct stress, are given by
1   2 1   2

cos 2θ + τ sin 2θ
2
2
  2
sin 2θ – τ cos 2θ.
and
σt = 1
2
7. Derive an expression for the major and minor principal stresses on an oblique plane, when the
body is subjected to direct stresses in two mutually perpendicular directions accompanied by a
shear stress.
σn =
8. Write a note on Mohr’s circle of stresses.
9. A body is subjected to direct stresses in two mutually perpendicular directions accompanied by a
simple shear stress. Draw the Mohr’s circle of stresses and explain how you will obtain the principal stresses and principal planes.
10.
A body is subjected to direct stresses in two mutually perpendicular directions. How will you
determine graphically the resultant stress on an oblique plane when :
(i) the stresses are unequal and unlike, and
(ii) the stresses are unequal and like.
(B) Numerical Problems
1. A rectangular bar of cross-sectional area 12000 mm2 is subjected to an axial load of 360 N/mm2.
Determine the normal and shear stresses on a section which is inclined at an angle of 30° with
the normal cross-section of the bar.
[Ans. 9.25 N/mm2, 1.3 N/mm2]
2. Find the diameter of a circular bar which is subjected to an axial pull of 150 kN, if the maximum
[Ans. 3.989 cm]
allowable shear stress on any section is 60 N/mm2.
139
STRENGTH OF MATERIALS
3. A rectangular bar of cross-sectional area 10000 mm2 is subjected to a tensile load P as shown
in Fig. 3.32. The permissible normal and shear stresses on the oblique plane BC are given
as 8 N/mm2 and 4 N/mm2 respectively. Determine the safe value of P.
[Ans. 92.378 kN]
C
P
P
30°
B
Fig. 3.32
4. The principal tensile stresses at a point across two mutually perpendicular planes are
100 N/mm2 and 50 N/mm2. Determine the normal, tangential and resultant stresses on a plane
inclined at 30° to the axis of the minor principal stress.[Ans. 0.875 N/mm2, 21.65 N/mm2, 90.138]
5. The principal stresses at a point in a bar are 160 N/mm2 (tensile) and 80 N/mm2 (compressive).
Determine the resultant stress in magnitude and direction on a plane inclined at 60° to the axis
of the major principal stress. Also determine the maximum intensity of shear stress in the material
at the point.
[Ans. 144.22 N/mm2, φ = 46° 5.7, 120 N/mm2]
6. At a point in a strained material, the principal stresses are 140 N/mm2 (tensile) and 60 N/mm2
(compressive). Determine the resultant stress in magnitude and direction on a plane inclined at
45° to the axis of the major principal stress. What is the maximum intensity of shear stress in
the material at the point ?
[Ans. 107.7 N/mm2 , φ = 61° 11.9, 100 N/mm2]
7. At a point within a body subjected to two mutually perpendicular directions, the stresses are
100 N/mm2 (tensile) and 75 N/mm2 (tensile). Each of the above stresses, is accompanied by a
shear stress of 75 N/mm2. Determine the normal, shear and resultant stresses on an oblique
plane inclined at an angle of 45° with the axis of minor tensile stress.
[Ans. 150, 25, 152.07 N/mm2]
8. For the problem 7, determine : (i) the direction and magnitude of each of the principal stress
and (ii) magnitude of the greatest shear stress.
[Ans. 154.057, – 4.057 N/mm2, θ = 35°, 468′ and 125° 46.8′ N/mm2]
9. Direct stresses of 160 N/mm2 tensile and 120 N/mm2 compressive exist on two perpendicular
planes at a certain point in a body. They are also accompanied by shear stresses on the planes.
The greatest principal stress at the point due to these is 200 N/mm2.
(i) What must be the magnitude of the shearing stresses on the two planes?
(ii) What will be the maximum shearing stress at the point?
[Ans. (i) 113.137 N/mm2 (ii) 180 N/mm2]
10.
At a certain point in a strained material, the stresses on the two planes at right angles to each
other are 40 N/mm2 and 20 N/mm2 both tensile. They are accompanied by a shear stress of
magnitude 20 N/mm2. Find graphically or otherwise, the location of principal planes and evaluate
the principal stresses.
[Ans. θ = 31° 43′, 121° 43′ and 52.36, 7.64 N/mm2]
11.
Solve problem 4, by graphical method.
12.
Solve problem 5, by graphical method.
13.
Solve problem 4, using Mohr’s circle of stresses.
14.
Solve problem 5, using Mohr’s circle of stresses.
140
PRINCIPAL STRESSES AND STRAINS
15.
A point in a strained material is subjected to stresses shown in Fig. 3.33.
Using Mohr’s circle method, determine the normal and tangential stresses across the oblique
plane. Check the answer analytically.
[Ans. 105 N/mm2, 15 N/mm2]
2
50 N/mm
2
40 N/mm
Oblique
plane
2
45°
80 N/mm
2
80 N/mm
2
40 N/mm
2
50 N/mm
Fig. 3.33
16.
An elemental cube is subjected to tensile stresses of 60 N/mm2 and 20 N/mm2 acting on two
mutually perpendicular planes and a shear stress of 20 N/mm2 on these planes. Draw the Mohr’s
circle of stresses and hence or otherwise determine the magnitudes and directions of principal
stresses and also the greatest shear stress.
[Ans. 68.214, 11.72 N/mm2, θ = 25.5° and 112.5°, 28.28 N/mm2]
17.
A strained material is subjected to two dimensional stresses. Prove that the sum of the normal
components of stresses on any two mutually perpendicular planes is constant.
[Hint. Normal stresses on a plane inclined at θ with major principal plane is given by
1   2 1  2

cos 2θ
2
2
Normal stress on a plane inclined at (θ + 90°) is given by
σn =
18.
σn* =
1   2 1  2

cos [2(θ + 90°)]
2
2
=
1   2 1  2

cos (180° + 2θ)
2
2
=
1   2 1  2

cos 2θ
2
2
...(i)
...(ii)
Adding (i) and (ii), σn + σn* = σ1 + σ2 = constant].
At a point in a two-dimensional system, the normal stress on two mutually perpendicular planes
are σ1 and σ2 (both alike) and shear stress is τ. Show that one of the principal stresses is zero if
τ=
1   2 .
[Hint. Principal stresses =
1   2

2
FG 
H
1
 2
2
=
1   2

2
FG 
H
1
 2
2
IJ
K
IJ
K
2
 2
2
 1 2
141
STRENGTH OF MATERIALS
=
19.
1  2

2
FG 
H
1
 2
2
IJ
K
2

1  2 1  2

2
2
= σ1 + σ2 and zero].
A rectangular block of material is subjected to a tensile stress of 100 N/mm2 on one plane and a
tensile stress of 50 N/mm2 on a plane at right angles, together with shear stresses of 60 N/mm2
on the faces. Find :
(i) the direction of principal planes,
(ii) the magnitude of principal stresses and
(iii) magnitude of the greatest shear stress.
[Ans. (i) 33° 41′ or 123° 41′ (ii) 140 N/mm2 and 10 N/mm2 tensiles (iii) 65 N/mm2]
142
4
CHAPTER
STRAIN ENERGY AND
IMPACT LOADING
4.1. INTRODUCTION..
Whenever a body is strained, the energy is absorbed in the body. The energy, which is
absorbed in the body due to straining effect is known as strain energy. The straining effect
may be due to gradually applied load or suddenly applied load or load with impact. Hence the
strain energy will be stored in the body when the load is applied gradually or suddenly or
with an impact. The strain energy stored in the body is equal to the work done by the applied
load in stretching the body.
4.2. SOME DEFINITIONS..
Before deriving the expressions for the strain energy stored in a body due to gradually
applied load or suddenly applied load or load with an impact, the following terms will be defined:
1. Resilience
2. Proof resilience, and
3. Modulus of resilience.
4.2.1. Resilience. The total strain energy stored in a body is commonly known as
resilience. Whenever the straining force is removed from the strained body, the body is capable of doing work. Hence the resilience is also defined as the capacity of a strained body for
doing work on the removal of the straining force.
4.2.2. Proof Resilience. The maximum strain energy, stored in a body, is known as
proof resilience. The strain energy stored in the body will be maximum when the body is
stressed upto elastic limit. Hence the proof resilience is the quantity of strain energy stored
in a body when strained upto elastic limit.
4.2.3. Modulus of Resilience. It is defined as the proof resilience of a material per
unit volume. It is an important property of a material. Mathematically,
Proof resilience
Modulus of resilience =
.
Volume of the body
4.3. EXPRESSION FOR STRAIN ENERGY STORED IN A BODY WHEN THE LOAD
IS APPLIED GRADUALLY.
In Art. 4.1, we have mentioned that the strain energy stored in a body is equal to the
work done by the applied load in stretching the body.
Fig. 4.1 shows load extension diagram of a body under tensile test upto elastic limit.
The tensile load P increases gradually from zero to the value of P and the extension of the
body increases from zero to the value of x.
143
STRENGTH OF MATERIALS
K
M
P
Load
The load P performs work in stretching the body.
This work will be stored in the body as strain energy
which is recoverable after the load P is removed.
Let
P = Gradually applied load,
x = Extension of the body,
A = Cross-sectional area,
L = Length of the body,
V = Volume of the body,
E = Young’s modulus,
U = Strain energy stored in the body, and
σ = Stress induced in the body.
Now work done by the load = Area of load extension curve (Shaded area in Fig. 4.1)
= Area of triangle ONM
1
= × P × x.
2
But
load, P = Stress × Area = σ × A
and extension,
FG∵
H
x = Strain × Length
Strain =
N
O
x
Extension
Fig. 4.1
...(i)
IJ
K
FG∵ Strain = Stress IJ
H
E K
Extension
∴ Extension = Strain × L
Length
Stress
×L
E
σ
=
× L.
E
Substituting the values of P and x in equation (i), we get
=
...(4.1)
1
σ
1 σ2
×σ×A×
×L=
×A×L
2
E
2 E
σ2
×V
(∵ Volume V = A × L)
=
2E
But the work done by the load in stretching the body is equal to the strain energy
stored in the body.
∴ Energy stored in the body,
Work done by the load
=
σ2
× V.
...(4.2)
2E
Proof resilience. The maximum energy stored in the body without permanent deformation (i.e., upto elastic limit) is known as proof resilience. Hence if in equation (4.2), the
stress σ is taken at the elastic limit, we will get proof resilience.
U=
σ *2
× Volume
2E
where σ* = Stress at the elastic limit.
Modulus of resilience = Strain energy per unit volume
∴ Proof resilience
=
σ2
×V
Total strain energy
σ2
=
= 2E
=
Volume
V
2E
144
...(4.3)
...(4.4)
STRAIN ENERGY AND IMPACT LOADING
4.4. EXPRESSION FOR STRAIN ENERGY STORED IN A BODY WHEN THE LOAD
IS APPLIED SUDDENLY
When the load is applied suddenly to a body, the load is constant throughout the process of the deformation of the body.
Consider a bar subjected to a sudden load.
Let
P = Load applied suddenly,
L = Length of the bar,
A = Area of the cross-section,
V = Volume of the bar = A × L,
E = Young’s modulus,
x = Extension of the bar,
σ = Stress induced by the suddenly applied load, and
U = Strain energy stored.
As the load is applied suddenly, the load P is constant when the extension of the bar
takes place.
∴ Work done by the load = Load × Extension = P × x.
The maximum strain energy stored (i.e., energy stored upto elastic limit) in a body is
given by
σ2
× Volume of the body
2E
σ2
× A × L.
(∵ Volume = A × L)
=
2E
Equating the strain energy stored in the body to the work done, we get
σ
σ2
σ
∵ From equation (4.1), x = × L
×A×L=P×x=P×
× L.
E
2E
E
σ×L
Cancelling
from both sides, we get
E
σ× A
P
= P or σ = 2 ×
.
...(4.5)
2
A
From the above equation it is clear that the maximum stress induced due to suddenly
applied load is twice the stress induced when the same load is applied gradually.
After obtaining the value of stress (σ ), the values of extension (x) and the strain energy
stored in the body may be calculated easily.
Problem 4.1. A tensile load of 60 kN is gradually applied to a circular bar of 4 cm
diameter and 5 m long. If the value of E = 2.0 × 105 N/mm2, determine :
(i) stretch in the rod,
(ii) stress in the rod,
(iii) strain energy absorbed by the rod.
Sol. Given :
Gradually applied load,
P = 60 kN = 60 × 1000 N
Dia. of rod,
d = 4 cm = 40 mm
π
× 402 = 400 π mm2
∴ Area,
A=
4
Length of rod,
L = 5 m = 500 cm = 5000 mm
U=
LM
N
OP
Q
145
STRENGTH OF MATERIALS
A × L = 400 π × 5000 = 2 × 106 π mm3
2 × 105 N/mm2.
stretch or extension in the rod,
stress in the rod, and
strain energy absorbed by the rod.
Load
P
60000
Now
stress, σ =
=
=
= 47.746 N/mm2. Ans.
Area
400π
A
The stretch or extension is given by equation (4.1),
σ
47.746
×L=
× 5000 = 1.19 mm. Ans.
x =
E
2 × 10 5
The strain energy absorbed by the rod is given by equation (4.2),
∴ Volume of rod, V=
Young’s modulus, E =
Let
x =
σ =
U =
σ2
47.476 2
×V=
× 2 × 106 π = 35810 N-mm = 35.81 N-m. Ans.
2E
2 × 2 × 10 5
Problem 4.2. If in problem 4.1, the tensile load of 60 kN is applied suddenly determine:
(i) maximum instantaneous stress induced,
(ii) instantaneous elongation in the rod, and
(iii) strain energy absorbed in the rod.
Sol. Given :
The data given in problem 4.1 is d = 40 mm, Area = 400 π mm2, L = 5000 mm, Volume
= 2 × 106 π mm3, E = 2 × 105 N/mm2 and suddenly applied load, P = 60000 N.
(i) Maximum instantaneous stress induced
Using equation (4.5),
P
60000
σ=2×
=2×
= 95.493 N/mm2. Ans.
400π
A
(ii) Instantaneous elongation in the rod
Let x = Instantaneous elongation
σ
95.493
Then
x=
×L=
× 5000
[see equation (4.1)]
E
2 × 10 5
= 2.38 mm. Ans.
(iii) Strain energy is given by,
U =
95.4932
σ2
×V=
× 2 × 106 π = 143238 N-mm
2 × 2 × 10 5
2E
= 143.238 N-m. Ans.
Problem 4.3. Calculate instantaneous stress produced in a bar 10 cm2 in area and 3 m
long by the sudden application of a tensile load of unknown magnitude, if the extension of the
bar due to suddenly applied load is 1.5 mm. Also determine the suddenly applied load. Take
E = 2 × 105 N/mm2.
Sol. Given :
Area of bar,
A = 10 cm2 = 1000 mm2
Length of bar,
L = 3 m = 3000 mm
Extension due to suddenly applied load,
x = 1.5 mm
U=
146
STRAIN ENERGY AND IMPACT LOADING
E = 2 × 105 N/mm2.
σ = Instantaneous stress due to sudden load, and
P = Suddenly applied load.
The extension x is given by equation (4.1),
σ
σ
x =
× L or 1.5 =
× 3000
E
2 × 10 5
Young’s modulus,
Let
∴
σ =
1.5 × 2 × 10 5
= 100 N/mm2. Ans.
3000
Suddenly applied load
The instantaneous stress produced by a sudden load is given by equation (4.5) as
P
P
σ = 2×
or 100 = 2 ×
A
1000
1000 × 100
∴
P =
= 50000 N = 50 kN. Ans.
2
Problem 4.4. A steel rod is 2 m long and 50 mm in diameter. An axial pull of 100 kN is
suddenly applied to the rod. Calculate the instantaneous stress induced and also the instantaneous elongation produced in the rod. Take E = 200 GN/m2.
Sol. Given :
Length,
L = 2 m = 2 × 1000 = 2000 mm
Diameter,
d = 50 mm
π
∴ Area,
A =
× 502 = 625 π mm2
4
Suddenly applied load,
P = 100 kN = 100 × 1000 N
Value of
E = 200 GN/m2 = 200 × 109 N/m2
(∵ G = Giga = 109)
200 × 10 9
N/mm2 (∵ 1 m = 1000 mm ∴ m2 = 106 mm2)
10 6
= 200 × 103 N/mm2
Using equation (4.5) for suddenly applied load,
P
100 × 1000
σ = 2×
=2×
N/mm2 = 101.86 N/mm2. Ans.
A
625 π
Let
dL = Elongation
P
101.86
Then
dL =
×L=
× 2000 = 1.0186 mm. Ans.
E
200 × 10 3
Problem 4.5. A uniform metal bar has a cross-sectional area of 700 mm2 and a length
of 1.5 m. If the stress at the elastic limit is 160 N/mm2, what will be its proof resilience ?
Determine also the maximum value of an applied load, which may be suddenly applied without
exceeding the elastic limit. Calculate the value of the gradually applied load which will produce the same extension as that produced by the suddenly applied load above.
Take E = 2 × 105 N/mm2.
Sol. Given :
Area,
A = 700 mm2
Length,
L = 1.5 m = 1500 mm
∴ Volume of bar,
V = A × L = 700 × 1500 = 1050000 mm2
=
147
STRENGTH OF MATERIALS
Stress at elastic limit, σ* = 160 N/mm2
Young’s modulus,
E = 2 × 105 N/mm2
(i) Proof resilience is given by equation (4.3), as
σ *2
160 2
× Volume =
× 1050000
2E
2 × 2 × 10 5
= 67200 N-mm = 67.2 N-m. Ans.
(ii) Let
P = Maximum value of suddenly applied load, and
P1 = Gradually applied load.
Using equation (4.5) for suddenly applied load,
P
(change σ to σ*)
σ* = 2 ×
A
σ*× A
160 × 700
∴
P =
=
= 56000 N = 56 kN. Ans.
2
2
For gradually applied load,
P1
σ* =
A
or
P1 = σ* × A = 160 × 700 = 112000 N = 112 kN. Ans.
Problem 4.6. A tension bar 5 m long is made up of two parts, 3 metre of its length has
a cross-sectional area of 10 cm2 while the remaining 2 metre has a cross-sectional area of
20 cm2. An axial load of 80 kN is gradually applied. Find the total strain energy produced in
the bar and compare this value with that obtained in a uniform bar of the same length and
having the same volume when under the same load. Take E = 2 × 105 N/mm2.
Sol. Given :
Total length of bar,
L = 5 m = 5000 mm
Length of 1st part,
L1 = 3 m = 3000 mm
Area of 1st part,
A1 = 10 cm2 = 10 × 100 mm2 = 1000 mm2
∴ Volume of 1st part,
V1 = A1 × L1 = 1000 × 3000 = 3 × 106 mm3
Length of 2nd part,
L2 = 2 m = 2000 mm
Area of 2nd part,
A2 = 20 cm2 = 20 × 100 mm2 = 2000 mm2
∴ Volume of 2nd part,V2 = 2000 × 2000 = 4 × 106 mm3
Axial gradual load, P = 80 kN = 80 × 1000 = 80000 N
Proof resilience
=
80 kN
10 cm
2
20 cm
3 cm
2 cm
Fig. 4.2
148
2
80 kN
STRAIN ENERGY AND IMPACT LOADING
E = 2 × 105 N/mm2
80000
P
Stress in 1st part,
σ1 =
=
= 80 N/mm2
1000
A1
P
80000
=
Stress in 2nd part,
σ2 =
= 40 N/mm2
A2
2000
Strain energy in 1st part,
Young’s modulus,
U1 =
80 2
σ 21
× V1 =
× 3 × 106 = 48000 N-mm = 48 N-m
2 × 2 × 10 5
2E
Strain energy in 2nd part,
σ 22
40 2
× V2 =
× 4000000 = 16000 N-mm = 16 N-m
2E
2 × 2 × 10 5
∴ Total strain energy produced in the bar,
U = U1 + U2 = 48 + 16 = 64 N-m. Ans.
Strain energy stored in a uniform bar
Volume of uniform bar,V = V1 + V2 = 3000000 + 4000000 = 7000000 mm2
Length of uniform bar, L = 5 m = 5000 mm
Let
A = Area of uniform bar
Then
V = A × L or 7000000 = A × 5000
7000000
∴
A =
= 1400 mm2
5000
P 80000
=
Stress in uniform bar, σ =
= 57.143 N/mm2
5000
A
∴ Strain energy stored in the uniform bar,
U2 =
σ2
57.1432
×V =
× 7000000
2E
2 × 2 × 10 5
= 57143 N-mm = 57.143 N-m
64
Strain energy in the given bar
=
= 1.12. Ans.
∴
57.143
Strain energy in the uniform bar
Problem 4.7. A bar of uniform cross-section ‘A’ and length ‘L’ hangs vertically, subjected to its own weight. Prove that the strain energy stored within the bar is given by
U =
A × ρ2 × L3
6E
where E = Modulus of Elasticity,
ρ = Weight per unit volume of the bar.
Sol. Given :
A = Cross-sectional area,
L = Length of bar,
E = Modulus of Elasticity,
ρ = Weight per unit volume.
Consider an element at a distance ‘x’ from the lower end of the bar as shown in
Fig. 4.2 (a). Let ‘dx’ be the thickness of the element.
U=
149
STRENGTH OF MATERIALS
The section X – X will be acted upon by the weight of the
bar of length x.
Let
Wx = Weight of the bar of length x
= (Volume of the bar of length x)
× Weight of unit volume
= (A × x) × ρ = ρAx
As a result of this weight, the portion dx will experience a
small elongation dδ. Then
Elongation in dx
Strain in portion dx =
Length of dx
dδ
=
dx
Stress in portion dx =
=
Also
E=
=
dx
X
L
X
x
Weight acting on section X-X
Area of section
ρ× A× x
=ρ× x
A
Fig. 4.2 (a)
Stress
Strain
ρ × x ρ × x × dx
=
dδ
dδ
dx
FG IJ
H K
ρ × x × dx
E
Now the strain energy stored in portion dx is given by,
dU = Average Weight × Elongation of dx
∴
dδ =
FG 1 × W IJ × dδ
H2 K
F 1 I ρ × x × dx
= G × ρAxJ ×
H2 K
E
=
x
(∵ Wx = ρAx)
1
dx
× ρ2 Ax 2 ×
2
E
Total strain energy stored within the bar due to its own weight W is obtained by integrating the above equation from 0 to L.
=
∴
U=
=
z
z
L
dU
0
L
0
=
150
1
dx
× ρ 2 Ax 2 ×
2
E
1 ρ2 × A
×
E
2
z
L
0
x 2 dx
STRAIN ENERGY AND IMPACT LOADING
LM
MN
1 ρ2 × A
x3
= 2× E × 3
=
OP
PQ
L
0
1 ρ2 × A L3
×
×
E
2
3
A × ρ 2 × L3
. Ans.
6E
Problem 4.8. The maximum stress produced by a pull in a bar of length 1 m is
150 N/mm2. The area of cross-sections and length are shown in Fig. 4.3. Calculate the
strain energy stored in the bar if E = 2 × 105 N/mm2.
=
P
2
A
475 mm
200 mm
100 mm
B
C
50 mm
P
2
2
200 mm
D
475 mm
Fig. 4.3
Sol. Given :
Length of bar,
L = 1 m = 1000 mm
Max. stress,
σ = 150 N/mm2
Part AB : Length,
L1 = 475 mm
Area,
A1 = 200 mm2
Part BC : Length,
L2 = 50 mm
Area,
A2 = 100 mm2
Part CD : Length,
L3 = 475 mm
Area,
A3 = 200 mm2
Value of
E = 2 × 105 N/mm2
In this problem, maximum stress is given. Axial pull P is not known. But stress is
equal to load/area. As load (or axial pull) for the bar is same, hence stress will be maximum,
when area will be minimum. Part BC is having less area and hence stress in part BC will be
maximum. As parts AB and CD are having same areas, hence stresses in them will be equal.
Let
σ2 = Stress in part BC = 150 N/mm2
σ1 = Stress in part AB or in part CD
Now
load = Stress × Area
or
load = σ1 × A1 = σ2 × A2
σ A
150 × 100
∴
σ1 = 2 2 =
= 75 N/mm2
A1
200
Now strain energy stored in part AB,
U1 =
σ12
× V1
2E
...(i)
151
STRENGTH OF MATERIALS
where
V1 = Volume of part AB
= A1 × L1 = 200 × 475
= 95000 mm3
Substituting this value in equation (i), we get
U1 =
=
σ12
× 95000
2E
75 2
2 × 2 × 105
× 95000
= 1335.938 N-mm
Strain energy stored in part BC,
U2 =
=
=
σ 22
× V2
2E
150 2
2 × 2 × 105
150 2
2 × 2 × 105
× A 2 × L2
(∵ V2 = A2 × L2)
× 100 × 50 = 281.25 N-mm
Energy stored in part CD,
σ 32
× V3 = 1335.938 N-mm (∵ V3 = V1, σ3 = σ1 ∴ U3 = U1)
2E
∴ Total strain energy stored,
U = U1 + U2 + U3 = 1335.938 + 281.25 + 1335.938 N-mm
= 2953.126 N-mm. Ans.
U3 =
4.5. EXPRESSION FOR STRAIN ENERGY STORED
IN A BODY WHEN THE LOAD IS APPLIED WITH
IMPACT
The load dropped from a certain height before the
load commences to stretch the bar is a case of a load applied
with impact. Consider a vertical rod fixed at the upper end
and having a collar at the lower end as shown in Fig. 4.4.
Let the load be dropped from a height on the collar. Due to
this impact load, there will be some extension in the rod.
Let P = Load dropped (i.e., load applied with impact)
L = Length of the rod,
A = Cross-sectional area of the rod,
V = Volume of rod = A × L,
h = Height through which load is dropped,
δL = Extension of the rod due to load P,
E = Modulus of elasticity of the material of rod,
σ = Stress induced in the rod due to impact load.
152
Vertical
rod
Load
L
h
Collar
δL
Fig. 4.4
STRAIN ENERGY AND IMPACT LOADING
i.e.,
The strain in the bar is given by,
Stress
Strain =
E
δL σ
=
L
E
σ
∴
δL =
×L
E
Work done by the load = Load × Distance moved
= P(h + δL)
The strain energy stored by the rod,
...(4.6)
...(i)
σ2
σ2
×V =
× AL
2E
2E
Equating the work done by the load to the strain energy stored, we get
U=
σ2
. AL
2E
σ
σ2
P h+ .L =
. AL
2E
E
FG
H
or
or
or
or
P(h + δL) =
FG∵
H
IJ
K
Ph + P .
...(ii)
δL =
IJ
K
σ
.L
E
σ2
σ
.L=
. AL
2E
E
σ
σ2
. AL – P .
. L – Ph = 0
E
2E
Multiplying by 2E to both sides, we get
AL
σ
2E
2E
.L×
– Ph.
=0
σ2 – P .
E
A. L
AL
2P
2PEh
σ2 –
.σ–
= 0.
A
A. L
The above equation is a quadratic equation in ‘σ ’,
∴
σ=
2P
±
A
FG 2P IJ
H AK
2
+ 4.
F
GH
2 PEh
A. L
2×1
2
2
P
±
A
=
P
+
A
=
P P
P P
2 PEh A 2
2 AEh
+
1+
× 2 =
+
1+
A A
A. L
A A
P.L
P
=
P
2 AEh
1+ 1+
A
P.L
F
GH
P
4P
8 . PEh
+
=
±
4 A2 4 . A . L A
FG P IJ
H AK
=
FG P IJ
H AK
2
+
+
2
I
JK
2 PEh
A. L
2 PEh
A. L
I
JK
−b ± b − 4 ac
roots =
2a
(Neglecting – ve root)
...(4.7)
After knowing the value of ‘σ ’, the strain energy can be obtained.
153
STRENGTH OF MATERIALS
Important Conclusions
(i) If δL is very small in comparison with h.
The work done by load = P. h
Equating the work done by the load to the strain energy stored in the rod, we get
P.h=
σ2
. AL
2E
2EPh
2E . P . h
and σ =
...(4.8)
A. L
A. L
(ii) In equation (4.7), if h = 0, we get
P
P
2P
(1 + 1 + 0 ) =
(1 + 1) =
σ=
A
A
A
which is the case of suddenly applied load.
Once the stress σ is known, the corresponding instantaneous extension (δL) and the
strain energy (U) can be obtained.
Problem 4.9. A weight of 10 kN falls by 30 mm on a collar rigidly attached to a vertical
bar 4 m long and 1000 mm2 in section. Find the instantaneous expansion of the bar. Take
E = 210 GPa. Derive the formula you use.
∴
Sol. Given :
Falling weight,
Falling height,
Length of bar,
Area of bar,
Value of
σ2 =
P
h
L
A
E
=
=
=
=
=
10 kN = 10,000 N
30 mm
4 m = 4000 mm
1000 mm2
210 GPa = 210 × 109 N/m2
(∵ G = Giga = 109 and Pa = Pascal = 1 N/m2)
210 × 10 9 N
(∵ 1 m = 1000 mm and m2 = 106 mm2)
10 6 mm 2
= 210 × 103 N/mm2 = 2.1 × 105 N/mm2
Let
dL = Instantaneous elongation due to falling weight
σ = Instantaneous stress produced due to falling weight
Using equation (4.7), we get
=
σ =
=
F
GH
P
2 EAh
1+ 1+
A
P×L
F
GG
H
I
JK
2 × 2.1 × 10 5 × 1000 × 30
10000
1+ 1+
10000 × 4000
1000
e
j
e
= 10 1 + 1 + 315 = 10 1 + 316
Now
∴
154
j
I
JJ
K
= 10 × 18.77 = 187.7 N/mm2
δL σ
Stress
σ
E =
or
=
=
δL
L
E
Strain
L
σ
187.7 × 4000
δL =
×L=
= 3.575 mm. Ans.
E
2.1 × 10 5
FG IJ
H K
STRAIN ENERGY AND IMPACT LOADING
Problem 4.10. A load of 100 N falls through a height of 2 cm onto a collar rigidly
attached to the lower end of a vertical bar 1.5 m long and of 1.5 cm2 cross-sectional area. The
upper end of the vertical bar is fixed.
Determine :
(i) maximum instantaneous stress induced in the vertical bar,
(ii) maximum instantaneous elongation, and
(iii) strain energy stored in the vertical rod.
Take
E = 2 × 105 N/mm2.
Sol. Given :
Impact load,
P = 100 N
Height through which load falls,
h = 2 cm = 20 mm
Length of bar,
L = 1.5 m = 1500 mm
Area of bar,
A = 1.5 cm2 = 1.5 × 100 mm2 = 150 mm2
∴ Volume,
V = A × L = 150 × 1500 = 225000 mm3
Modulus of elasticity,E = 2 × 105 N/mm2
Let
σ = Maximum instantaneous stress induced in the vertical bar,
δL = Maximum elongation, and
U = Strain energy stored.
(i)Using equation (4.7),
σ =
=
F
GH
P
2 AEh
1+ 1+
A
P.L
I = 100 F
JK 150 GG 1 +
H
1+
2 × 150 × 2 × 10 5 × 20
100 × 1500
I
JJ
K
100
(1 + 1 + 8000 ) = 60.23 N/mm2. Ans.
150
(ii)Using equation (4.6),
60.23 × 1500
σ
×L=
= 0.452 mm. Ans.
E
2 × 10 5
(iii)Strain energy is given by,
δL =
σ2
60.232
×V=
× 225000 = 2045 N-mm
2E
2 × 2 × 10 5
= 2.045 N-m. Ans.
Problem 4.11. The maximum instantaneous extension, produced by an unknown falling weight through a height of 4 cm in a vertical bar of length 3 m and of cross-sectional area
5 cm2, is 2.1 mm.
Determine :
(i) the instantaneous stress induced in the vertical bar, and
(ii) the value of unknown weight. Take E = 2 × 105 N/mm2.
Sol. Given :
Instantaneous extension, δL = 2.1 mm
Length of bar,
L = 3 m = 3000 mm
Area of bar,
A = 5 cm2 = 500 mm2
∴ Volume of bar,
V = 500 × 3000 = 1500000 mm3
U =
155
STRENGTH OF MATERIALS
Height through which weight falls, h = 4 cm = 40 mm
Modulus of elasticity, E = 2 × 105 N/mm2
Let
σ = Instantaneous stress produced, and
P = Unknown weight.
We know
E =
∴ Instantaneous stress
Stress
Strain
or Stress = E × Strain
= E × Instantaneous strain = E ×
δL
L
2.1
N/mm2 = 140 N/mm2. Ans.
3000
Equating the work done by the falling weight to the strain energy stored, we get
= 2 × 105 ×
P(h + δL) =
σ2
×V
2E
140 2
× 1500000 = 73500
2 × 2 × 10 5
73500
P =
= 1745.8 N. Ans.
42.1
or
P(40 + 2.1) =
or
Note. The value of P can also be obtained by using equation (4.7).
Problem 4.12. An unknown weight falls through a height of 10 mm on a collar rigidly
attached to the lower end of a vertical bar 500 cm long and 600 mm2 in section. If the maximum
extension of the rod is to be 2 mm, what is the corresponding stress and magnitude of the
unknown weight ? Take E = 2.0 × 105 N/mm2.
Sol. Given :
Height through which the weight falls, h = 10 mm
Length of the bar,
L = 500 cm = 5000 mm
Area of the bar,
A = 600 mm2
Maximum extension, δL = 2 mm
Young’s modulus,
E = 2.0 × 105 N/mm2
Let
σ = Instantaneous stress produced in the bar, and
P = Weight falling on the collar.
Stress
We know
E =
Strain
δL
δL
∵ Strain =
∴
Stress = E × Strain = E ×
L
L
Substituting the known values, we get
2
= 80 N/mm2. Ans.
σ = 2.0 × 105 ×
5000
Value of weight falling on the collar
Using equation (4.7),
FG
H
σ =
156
F
GH
P
2A . E . h
1+ 1+
A
P.L
I
JK
IJ
K
STRAIN ENERGY AND IMPACT LOADING
or
or
or
80 =
F
GG
H
P
2 × 600 × 2.0 × 10 5 × 10
1+ 1+
600
P × 5000
I
JJ
K
48000
480000
= 1 + 1+
P
P
48000
480000
–1 =
1+
P
P
Squaring both sides,
FG 48000 IJ
H P K
2
2 × 48000
480000
= 1+
P
P
2304000000 96000 480000
or
(cancelling 1 from both sides)
−
=
P
P
P2
576000
2304000000 480000 96000
=
or
=
+
P
P
P
P2
1
2304000000
cancelling from both sides
or
= 576000
P
P
2304000000
or
P =
= 4000 N = 4 kN. Ans.
576000
Problem 4.13. A bar 12 mm diameter gets stretched by 3 mm under a steady load of
8000 N. What stress would be produced in the same bar by a weight of 800 N, which falls
vertically through a distance of 8 cm onto a rigid collar attached at its end ? The bar is initially
unstressed. Take E = 2.0 × 105 N/mm2.
+ 1−
FG
H
Sol. Given :
Dia. of bar,
∵ Area of bar,
Increase in length,
Steady load,
Falling weight,
Vertical distance,
Young’s modulus,
Let
d = 12 mm
π
(12)2 = 113.1 mm2
A =
4
δL = 3 mm
W = 8000 N
P = 800 N
h = 8 cm = 80 mm
E = 2.0 × 105 N/mm2
L = Length of the bar, and
σ = Stress produced by the falling weight.
With steady load
Stress
E=
=
Strain
We know
or
2.0 × 105
IJ
K
FG Steady load IJ
H Area K
FG 8000 IJ
H 1131. K = 8000 × L
=
FG 3 IJ 1131. 3
H LK
δL
L
157
STRENGTH OF MATERIALS
2.0 × 105 × 1131
. ×3
= 8482.5 mm
8000
Now using equation (4.7), we get
∴
L =
σ =
=
F
GH
P
2 AEh
1+ 1+
A
PL
F
GG
H
I
JK
. × 2.0 × 105 × 80
800
2 × 1131
1+ 1+
.
1131
8.0 × 8482.5
I
JJ N/mm
K
2
= 7.0734(1 + 1 + 533.33 ) = 7.0734 × 24.1155
= 170.578 N/mm2. Ans.
Problem 4.14. A rod 12.5 mm in diameter is stretched 3.2 mm under a steady load
of 10 kN. What stress would be produced in the bar by a weight of 700 N, falling through
75 mm before commencing to stretch, the rod being initially unstressed ? The value of E may
be taken as 2.1 × 105 N/mm2.
Sol. Given :
Dia. of rod,
d = 12.5 mm
∴ Area of rod,
A =
Increase in length,
Steady load,
Falling load,
Falling height,
Young’s modulus,
Let
We know
δL
W
P
h
E
L
σ
=
=
=
=
=
=
=
E =
=
2.1 × 105 =
or
=
158
π
× 12.52 = 122.72 mm2
4
3.2 mm
10 kN = 10,000 N
700 N
75 mm
2.1 × 105 N/mm2
Length of the rod,
Stress produced by the falling weight.
Stress
Strain
FG Steady load IJ
H Area K
FG δL IJ
H LK
FG 10,000 IJ
H 122.72 K
FG 3.2 IJ
H LK
FG 10,000 IJ × FG L IJ
H 122.72 K H 3.2 K
STRAIN ENERGY AND IMPACT LOADING
∴
2.1 × 10 5 × 122.72 × 3.2
= 8246.7 mm
10,000
L =
Now using equation (4.7), we get
F
GH
P
2 AEh
1+ 1+
A
P ×L
σ =
=
F
GG
H
I
JK
2 × 122.72 × 21
700
. × 105 × 75
1+ 1+
700 × 8246.7
122.72
I
JJ
K
= 153.74 N/mm2. Ans.
Problem 4.15. A vertical round steel rod 1.82 metre long is securely held at its upper
end. A weight can slide freely on the rod and its fall is arrested by a stop provided at the lower
end of the rod. When the weight falls from a height of 30 mm above the stop the maximum
stress reached in the rod is estimated to be 157 N/mm2. Determine the stress in the rod if the
load had been applied gradually and also the minimum stress if the load had fallen from a
height of 47.5 mm.
Take E = 2.1 × 105 N/mm2.
Sol. Given :
Length of rod,
L = 1.82 m = 1.82 × 1000 = 1820 mm
Height through which load falls,
h = 30 mm
Maximum stress induced in the rod,
σ = 157 N/mm2
Modulus of elasticity,
E = 2.1 × 105 N/mm2
Let
σ1 = Stress induced in the rod if the load is applied gradually and
σ2 = Maximum stress if the load had fallen from a height of 47.5 mm.
Strain energy stored in the rod when load falls through a height of 30 mm,
U =
σ2
157
× Volume =
×V
2E
. × 105
2 × 21
= 0.05868 × V N-m
The extension of the rod is given by equation (4.6),
δL =
=
σ
×L
E
157
21
. × 105
× 1820 = 1.36 mm
∴ Total distance through which load falls
= h + δL = 30 + 1.36 = 31.36 mm
∴ Work done by the falling load = Load × Total distance
= P × 31.36
Equating the work done by the falling load to the strain energy stored, we get
P × 31.36 = 0.05868 × V
159
STRENGTH OF MATERIALS
P 0.05868
=
= 0.001871
31.36
V
or
P
= 0.001871
A. L
or
or
(∵ V = A.L)
P
= 0.001871 × L = 0.001871 × 1820 = 3.4
A
1st Case. If the load had been applied gradually, the stress induced is given by,
Load P
=
Area A
= 3.4 N/mm2. Ans.
2nd Case. If the load had fallen from a height of 47.5 mm.
Let
σ2 = Maximum stress induced.
Using equation (4.7), we get
σ1 =
LM
MN
L
= 3.4 M1 +
MN
= 3.4 e1 +
σ2 =
P
2 AEh
1+ 1+
A
P×L
1+
OP
PQ
[Here σ = σ2]
2 × 21
. × 105 × 47.5
3.4 × 1820
1 + 3219.24
j
OP
PQ
FG∵ P = 3.4, h = 47.5IJ
H A
K
N/mm2.
= 196.64
Ans.
Problem 4.16. A vertical compound tie member fixed rigidly at its upper end, consists
of a steel rod 2.5 m long and 200 mm in diameter, placed within an equally long brass tube
21 mm in internal diameter and 30 mm external diameter. The rod and the tube are fixed
together at the ends. The compound member is then suddenly loaded in tension by a weight of
10 kN falling through a height of 3 mm on to a flange fixed to its lower end. Calculate the
maximum stresses in steel and brass. Assume Es = 2 × 105 N/mm2 and Eb = 1.0 × 105 N/mm2.
Sol. Given :
30
21
Length of steel rod, L = 2.5 m = 2500 mm
Dia. of steel rod, ds = 20 mm
∴ Area of steel rod,
As =
Internal dia. of brass tube
External dia. of brass tube
∴ Area of brass tube,
Length of brass tube,
160
=
=
=
π
× 202
4
100 π mm2
21 mm
30 mm
π
(302 – 212)
4
= 114.75 π mm2
= 250 cm
= 2500 mm
Ab =
Brass
tube
P = 10 kN
Steel
rod
2.5 m
20
3 mm
Flange
Fig. 4.5
STRAIN ENERGY AND IMPACT LOADING
Weight,
P = 10 kN = 10,000 N
Height through which weight falls,
h = 3 mm
Young’s modulus for steel, Es = 2 × 105 N/mm2
Young’s modulus for brass, Eb = 1.0 × 105 N/mm2
Let
σs = Stress in steel tube, and
σb = Stress in brass tube.
As both the ends are fixed together,
Strain in steel rod = Strain in brass tube
i.e.,
σs
σ
= b
E s Eb
or
σs =
LM∵ Strain = Stress OP
E Q
N
σb
2 × 10 5
× Es = σ b ×
= 2 × σb
Eb
1 × 10 5
...(i)
Now volume of steel rod,
Vs = Area × Length
= As × L = 100 π × 2500
= 250000 π mm3
Volume of brass tube,
Vb = Ab × L = 114.75 π × 2500 = 286875 π mm3
∴ Strain energy stored in steel rod,
Us =
(2 × σ b ) 2
σ 2s
× 250000 π
× Vs =
2 Es
2 × 2 × 10 5
(∵ σs = 2σb)
2
= 7.854 σ b .
and strain energy stored in brass tube,
σ 2b
σ b2
× Vb =
× 286875 π
2 Eb
2 × 1 × 10 5
= 4.506 σb2.
Total strain energy stored in the compound bar;
U = Us + Ub
= 7.854 σb2 + 4.506 σb2
= 12.36 σb2.
Work done by the falling weight = Weight (h + δL)
= 10000 (3 + δL)
As both the ends are fixed,
The strain in steel rod = Strain in brass rod
Ub =
But strain in brass rod
or
∴
=
σb
Eb
σb
δL
=
L 1 × 106
σb
× 2500
δL =
1 × 106
= 0.025 σb.
...(ii)
...(iii)
FG∵ Strain = Stress IJ
H
E K
(∵ L = 2500 mm)
161
STRENGTH OF MATERIALS
Substituting this value of δL in equation (iii), we get
Work done by falling weight = 10000 (3.0 + 0.025 σb)
...(iv)
Now equating the work done by the falling weight to the total strain energy stored
[i.e., equating equations (iv) and (ii)], we get
2
10000 [ 3 + 0.025 σb] = 12.36 σ b
30000 + 250 σb = 12.36 σ 2b
or
12.36 σ 2b – 250 σb – 30000 = 0
or
σ b2 −
or
250
30000
σb −
= 0
12.36
12.36
σ 2b – 20.226 σb – 2427.18 = 0
or
The above equation is a quadratic equation.
∴
σb =
20.226 ± 20.226 2 + 4 × 242718
.
2
20.226 ± 409.09 + 9708.72
=
2
20.226 ± 100.587
=
2
20.226 + 100.587
=
(Neglecting – ve root)
2
= 60.4 N/mm2. Ans.
From equation (i), we get σs = 2 × σb = 2 × 60.4
= 120.8 N/mm2. Ans.
Problem 4.17. A vertical bar 4 metre long and of 2000 mm2 cross-sectional area is
fixed at the upper end and has a collar at the lower end. Determine the maximum stress induced when a weight of :
(i) 3000 N falls through a height of 20 cm on the collar,
(ii) 30 kN falls through a height of 2 cm on the collar.
Take E = 2.0 × 105 N/mm2.
Sol. Given :
Length of bar,
L = 4 m = 4000 mm
Area of bar,
A = 2000 mm2
∴ Volume of bar,
V = A × L = 2000 × 4000 = 8000,000 mm3
1st Case. Falling weight, P1 = 3000 N
Height,
h1 = 20 cm = 200 mm
Let
σ1 = Maximum stress induced.
In this case the falling weight is small as compared to second case. The small weight
will produce a small extension of the bar. Hence the extension in the bar will be negligible as
compared to the height of 20 cm through which the weight falls.
∴ Using equation (4.8), we get
σ=
162
2EPh
A. L
STRAIN ENERGY AND IMPACT LOADING
or
σ1 =
2 EP1h1
A. L
or
σ1 =
2 × 2 × 105 × 3000 × 200
2000 × 4000
(∵ A = 2000 mm2, L = 4000 mm)
= 173.2 N/mm2. Ans.
2nd Case. Falling weight, P2 = 30 kN = 30000 N
Height,
h2 = 2 cm = 20 mm
Let
σ2 = Maximum stress induced.
In this case falling weight is having a large value. Hence the extension produced by a
large weight will be large. Moreover the height through which this weight falls is 2 cm only.
Hence the extension in the bar, in comparison to the height through which weight falls, is not
negligible.
∴ Using equation (4.7), we get
LM
MN
P L
M1 +
A MN
P
2 AEh
σ = A 1 + 1 + P. L
σ2 =
or
=
1+
2
OP
PQ
2 AEh2
P2 . L
LM
MN
OP
PQ
2 × 2000 × 2 × 10 5 × 20
30000
1+ 1+
30000 × 4000
2000
OP
PQ
= 15 (1 + 11.590) = 188.85 N/mm2. Ans.
Problem 4.18. A crane-chain whose sectional area is 6.25 cm2 carries a load of 10 kN.
As it is being lowered at a uniform rate of 40 m per minute, the chain gets jammed suddenly,
at which time the length of the chain unwound is 10 m. Estimate the stress induced in the
chain due to the sudden stoppage. Neglect the weight of the chain. Take E = 2.1 × 105 N/mm2.
Sol. Given :
Area,
Load,
A = 6.25 cm2 = 625 mm2
W = 10 kN = 10,000 N
Velocity,
V = 40 m/min =
Length of chain unwound
∴
L =
Value of
E =
Let
σ =
K.E. of the crane
=
40
2
m/s =
m/s
60
3
= 10 m = 10 × 1000 mm
10,000 mm
2.1 × 105 N/mm2
Stress induced in the chain due to the sudden stoppage.
FG IJ
H K
1
1 W
mV 2 =
×V2
2
2 g
163
STRENGTH OF MATERIALS
=
IJ FG IJ
K H K
FG
H
1 10000
2
×
2 9.81
3
2
N-m = 226.5 N-m
= 226.5 × 1000 N-mm = 226500 N-mm
...(i)
When the chain gets jammed suddenly, the whole of the K.E. of the crane is absorbed
in the chain. But the energy stored or absorbed in the chain
=
=
σ2
×A×L
2E
σ2
2 × 2.1 × 10 5
× 625 × 10,000 N-mm
...(ii)
Now K.E. of crane = Energy stored in the chain
or
226500 =
∴
σ2 =
∴
σ=
σ2
2 × 2.1 × 10 5
× 625 × 10,000
226500 × 2 × 2.1 × 10 5
625 × 10,000
226500 × 2 × 2.1 × 10 5
625 × 10,000
= 123.37 N/mm2. Ans.
Problem 4.19. A cage weighing 60 kN is attached to the end of a steel wire rope. It is
lowered down a mine shaft with a constant velocity of 1 m/s. What is the maximum stress
produced in the rope when its supporting drum is suddenly jammed ? The free length of the
rope at the moment of jamming is 15 m, its net cross-sectional area is 25 cm 2 and E
= 2 × 105 N/mm2. The self-weight of the wire rope may be neglected.
Sol. Given :
Weight,
Velocity,
Free length,
Area,
Value of
K.E. of the cage
W = 60 kN = 60,000 N
V = 1 m/s
L = 15 m = 15,000 mm
A = 25 cm2 = 25 × 100 mm2
E = 2 × 105 N/mm2
FG IJ × V
H K
1 F 60,000 I
J × 1 N-m = 30000
= ×G
N-m
9.81
2 H 9.81 K
=
1
1 W
mV 2 =
2
2 g
2
2
30000 × 1000
N-mm
...(i)
9.81
This energy is to be absorbed (or stored) by the rope.
Let σ = Maximum stress produced in the rope when its supporting drum is suddenly
jammed.
=
164
STRAIN ENERGY AND IMPACT LOADING
But the maximum energy stored
σ2
σ2
×A×L=
× 2500 × 15000 N-mm
=
2E
2 × 2 × 105
But K.E. of the cage = Energy stored in the rope
...(ii)
30000 × 1000
σ2
=
× 2500 × 15000
9.81
2 × 2 × 10 5
or
σ2 =
or
σ =
30000 × 1000 × 2 × 2 × 105
9.81 × 2500 × 15000
30000 × 1000 × 2 × 2 × 10 5
= 180.61 N/mm2. Ans.
9.81 × 2500 × 15000
4.6. EXPRESSION FOR STRAIN ENERGY STORED IN A BODY DUE TO SHEAR
STRESS
Fig 4.6 shows a rectangular block of length l, height
h and breadth b, fixed at the bottom face AB. Let a shear
force P is applied on the top face CD and hence the top
face moves a distance equal to CC1.
Let
τ = Shear stress produced,
φ = Shear strain, and
C = Modulus of rigidity.
Now shear stress,
Shear force
τ=
Area
(∵ Area of top face = l × b)
P
=
l×b
∴
P=τ×l×b
CC1
and shear strain, φ =
CB
∴
CC1 = CB. φ
D
D1
C
C1
P
h
f
A
f
l
B
Fig. 4.6
If the shear force P is applied gradually, then average load will be equal to
P
.
2
∴ Work done by gradually applied shear force
= Average load × Distance
P
1
× CC1 = (τ × l × b). (CB. φ)
=
2
2
(∵ P = τ × l × b and CC1 = CB. φ)
1
= . τ × l × b × h. φ
(∵ CB = h)
2
1
1
τ
× Volume of block
= .τ×φ×l×b×h= .τ×
2
2
C
Shear stress
φ = Shear strain =
C
2
1
τ
= ×
×V
(∵ V = l × b × h)
2
C
FG
H
IJ
K
165
STRENGTH OF MATERIALS
But the work done is equal to the strain energy stored.
τ2
×V
...(4.9)
2C
Problem 4.20. The shear stress in a material at a point is given as 50 N/mm2. Determine the local strain energy per unit volume stored in the material due to shear stress. Take
C = 8 × 104 N/mm2.
Sol. Given :
Shear stress,
τ = 50 N/mm2
Modulus of rigidity, C = 8 × 104 N/mm2
Using equation (4.9),
∴ Strain energy stored =
τ2
50 2
× Volume =
× Volume
2C
2 × 8 × 10 4
= 0.015625 × Volume
∴ Strain energy per unit volume
0.015625 × Volume
= 0.015625 N/mm2. Ans.
=
Volume
Strain energy
=
HIGHLIGHTS
1.
2.
3.
The energy stored in a body due to straining effect is known as strain energy.
Resilience is the total strain energy stored in a body. Resilience is also defined as the capacity of
a strained body for doing work on the removal of the straining force.
The maximum strain energy stored in a body is known as proof resilience. The proof resilience is
given by,
σ2
× Volume
2E
where σ = Stress at the elastic limit.
The proof resilience of a body per unit volume is known as modulus of resilience.
The maximum stress induced in a body is given by
P
....... if the load P is applied gradually
σ=
A
P
..... if the load P is applied suddenly
=2
A
Proof resilience =
4.
5.
=
where
6.
7.
F
GH
2 AEh
P
1+ 1+
A
P.L
I ..... if the load P is applied with impact
JK
A = Cross-sectional area of the body,
h = Height through which load falls,
E = Modulus of rigidity,
L = Length of the body.
The maximum stress induced in a body due to suddenly applied load is twice the stress induced
when the same load is applied gradually.
If the extension produced in a rod due to impact load is very small in comparison with the height
through which the load falls, then the maximum stress induced in body is given by
σ=
166
2E . P . h
A. L
STRAIN ENERGY AND IMPACT LOADING
where
8.
9.
P = Impact load,
h = Height through which load falls.
To find the expression for the stress induced in a body either by suddenly applied load or by an
impact load, the strain energy stored in a body is equated to the work done by the load.
The energy stored in a body due to shear stress (τ) is given by
U=
where
τ2
×V
2C
V = Volume of the body, and
C = Modulus of rigidity.
EXERCISE
(A) Theoretical Questions
1.
2.
3.
4.
5.
6.
Define the following terms :
(i) Resilience
(ii) Strain energy
(iii) Impact loading, and
(iv) Spring.
Define resilience, proof resilience and modulus of resilience.
Find an expression for the strain energy stored in a body when
(i) the load is applied gradually
(ii) the load is applied suddenly, and
(iii) the load is applied with an impact.
Prove that the maximum stress induced in a body due to suddenly applied load is twice the
stress induced when the same load is applied gradually.
Derive an expression for the stress induced in a body due to suddenly applied load and hence
find the value of extension produced in the body.
Prove that the maximum strain energy stored in a body is given by,
σ2
× Volume
2E
where σ = Stress at the elastic limit.
Explain the terms : Gradually applied load, suddenly applied load, and load applied with an
impact.
Prove that the stress induced in a body when the load is applied with impact is given by,
U=
7.
8.
σ=
F
GH
P
2 AEh
1+ 1+
A
P .L
I
JK
where
9.
P = Load applied with impact,
A = Cross-sectional area of the body,
h = Height through which load falls,
L = Length of the body, and
E = Modulus of elasticity.
If the extension produced in a rod due to impact load is very small in comparison with the height
through which the load falls, prove that stress induced in the body will be given by
2EPh
.
A. L
Prove that the strain energy stored in a body due to shear stress is given by,
σ=
10.
U=
where
τ2
×V
2C
τ = Shear stress,
C = Modulus of rigidity, and
V = Volume of the body.
167
STRENGTH OF MATERIALS
11.
Explain the following terms : (i) Proof stress, (ii) Proof resilience, and (iii) Modulus of resilience.
(B) Numerical Problems
1.
A tensile load of 50 kN is gradually applied to a circular bar of 5 cm diameter and 4 m long. If the
value of E = 2.0 × 105 N/mm2, determine : (i) stretch in the rod, (ii) stress in the rod, and
(iii) strain energy absorbed by the rod. [Ans. (i) 0.0509 cm, (ii) 25.465 N/mm2 (iii) 12.73 N-m]
2. If in question 1, the tensile load of 50 kN is applied suddenly, determine : (i) maximum instantaneous stress induced, (ii) instantaneous elongation in the rod, and (iii) strain energy absorbed
in the rod.
[Ans. (i) 50.93 N/mm2, (ii) 0.1018 cm (iii) 50.93 N-m]
3. Calculate instantaneous stress produced in a bar 10 cm2 in area and 4 m long by the sudden
application of a tensile load of unknown magnitude, if the extension of the bar due to suddenly
applied load is 1.35 mm. Also determine the suddenly applied load. Take E = 2 × 105 N/mm2.
[Ans. 67.5 N/mm2, 33.75 kN]
2
4. A uniform metal bar has a cross-sectional area of 6 cm and a length of 1.4 m. If the stress at the
elastic limit is 1.5 tonne/cm2, find the proof resilience of the bar. Determine also the maximum
value of an applied load, which may be suddenly applied without exceeding the elastic limit.
Calculate the value of the gradually applied load which will produce the same extension as that
produced by the suddenly applied load above. Take E = 2000 tonnes/cm2.
5. A tension bar 6 m long is made up of two parts, 4 metre of its length has a cross-sectional area
12.5 cm2 while the remaining 2 m length has a cross-sectional area of 25 cm2. An axial load of
5 tonnes is gradually applied. Find the total strain energy produce in the bar and compare this
value with that obtained in a uniform bar of the same length and having the same volume when
0
[Ans. 242 kgf/cm, 1.054]
under the same load. Take E = 2 × 106 kgf/cm2.
6. A load of 200 N falls through a height of 2.5 cm on to a collar rigidly attached to the lower end of
a vertical bar 2 m long and of 3 cm2 cross-sectional area. The upper end of the vertical bar is
fixed. Determine : (i) maximum instantaneous stress induced in the vertical bar, (ii) maximum
instantaneous elongation, and (iii) strain energy stored in the vertical rod. Take E
= 2 × 106 kgf/cm2.
[Ans. (i) 58.4 N/mm2 (ii) 0.0584 cm (iii) 511.5 N-m]
7. The maximum instantaneous elongation, produced by an unknown falling weight through a
height of 4 cm in a vertical bar of length 5 m and of cross-sectional area 5 cm2, is 1.80 mm.
Determine : (i) the instantaneous stress induced in the vertical bar and (ii) the values of
[Ans. (i) 72 N/mm2 and (ii) 775.1 N]
unknown weight. Take E = 2 × 106 kgf/cm2.
8. An unknown weight falls through a height of 20 mm on a collar rigidly attached to the lower end
of a vertical bar 5 m long and 800 mm2 in cross-section. If the maximum extension of the rod is
to be 2.5 mm, what is the corresponding stress and magnitude of the unknown weight ? Take
E = 2.0 × 106 kgf/cm2.
[Ans. 1000 kgf/cm2, 444.44 kgf]
9. A bar 1.5 cm diameter gets stretched by 2.5 mm under a steady load of 100 kgf. What stress
would be produced in the same bar by a weight of 120 kgf, which falls vertically through a
distance 5 cm on to a rigid collar attached at its end ? The bar is initially unstressed. Take
[Ans. 1309.44 kgf/cm2]
E = 2.0 × 106 kgf/cm2.
10. A vertical round steel rod 2 m long is securely held at its upper end. A weight can slide freely on
the rod and its fall is arrested by a stop provided at the lower end of the rod. When the weight
falls from a height of 2.5 cm above the stop, the maximum stress reached in the rod is estimated
to be 1450 kgf/cm2. Determine the stress in the rod if the load had been applied gradually and
also the maximum stress if the load had fallen from a height of 4.5. Take E = 2.0 × 106 kgf/cm2.
[Ans. 39.743 kgf/cm2, 193.42 kgf/cm2]
11. A vertical compound tie member fixed rigidly at its upper end, consists of a steel rod 3 m long
and 20 mm diameter, placed within an equally long brass tube 20 mm internal diameter and
20 mm external diameter. The rod and the tube are fixed together at the ends. The compound
168
STRAIN ENERGY AND IMPACT LOADING
member is then suddenly loaded in tension by a weight of 1200 kgf falling through a height of
5 mm on to a flange fixed to its lower end. Calculate the maximum stresses in steel and brass.
Assume Es = 2 × 106 kgf/cm2 and Eb = 1.0 × 106 kgf/cm2. [Ans. 1173.5 kgf/cm2, 586.76 kgf/cm2]
12.
A circular rod 5 cm in diameter and 3 metre long hangs vertically and has a collar securely
attached to the lower end. Find the maximum stress induced : (i) when a weight of 250 kgf falls
through 15 cm on the collar, (ii) when a weight of 2500 kgf falls 1.5 cm on the collar. Take
E = 2.1 × 106 kgf/cm2.
[Ans. (i) 1635 kgf/cm2, (ii) 1767 kgf/cm2]
13.
The shear stress in a material at a point is given as 45 N/mm2. Determine the local strain
energy per unit volume stored in the material due to shear stress. Take C = 8 × 104 N/mm2.
[Ans. 0.01265 N/mm2]
169
.
5
CHAPTER
CENTRE OF GRAVITY AND
MOMENT OF INERTIA
5.1. CENTRE OF GRAVITY..
Centre of gravity of a body is the point through which the whole weight of the body
acts. A body is having only one centre of gravity for all positions of the body. It is represented
by C.G. or simply G.
5.2. CENTROID..
The point at which the total area of a plane figure (like rectangle, square, triangle,
quadrilateral, circle etc.) is assumed to be concentrated, is known as the centroid of that area.
The centroid is also represented by C.G. or simply G. The centroid and centre of gravity are at
the same point.
5.3. CENTROID OR CENTRE OF GRAVITY OF SIMPLE PLANE FIGURES..
(i) The centre of gravity (C.G.) of a uniform rod lies at its middle point.
(ii) The centre of gravity of a triangle lies at the point where the three medians* of the
triangle meet.
(iii) The centre of gravity of a rectangle or of a parallelogram is at the point, where its
diagonal meet each other. It is also the point of intersection of the lines joining the
middle points of the opposite sides.
(iv) The centre of gravity of a circle is at its centre.
5.4.CENTROID (OR CENTRE OF GRAVITY) OF AREAS OF PLANE FIGURES BY
THE METHOD OF MOMENTS
Fig. 5.1 shows a plane figure of total area A whose centre of Y
Area a3
gravity is to be determined. Let this area A is composed of a number
Area a2
Area a4
Area a1
of small areas a1, a2, a3, a4, ...... etc.
∴
A = a1 + a2 + a3 + a4 + ...
G
Let x1 = The distance of the C.G. of the area a1 from axis OY
x1
x2 = The distance of the C.G. of the area a2 from axis OY
x2
x3 = The distance of the C.G. of the area a3 from axis OY
x3
x4 = The distance of the C.G. of the area a4 from axis OY
x4
and so on.
O
X
The moments of all small areas about the axis OY
x
= a1x1 + a2x2 + a3x3 + a4x4 + ...
...(i)
Fig. 5.1
*The line connecting the vertex and the middle point of the
opposite side of a triangle is known as median of the triangle.
171
STRENGTH OF MATERIALS
Let G is the centre of gravity of the total area A whose distance from the axis OY is x .
...(ii)
Then moment of total area about OY = A x
The moments of all small areas about the axis OY must be equal to the moment of total
area about the same axis. Hence equating equations (i) and (ii), we get
a1x1 + a2x2 + a3x3 + a4x4 + ... = A x
a x + a2 x2 + a3 x3 + a4 x4 + ...
x = 1 1
or
...(5.1)
A
where A = a1 + a2 + a3 + a4 ...
If we take the moments of the small areas about the axis OX and also the moment of
total area about the axis OX, we will get
a1 y1 + a2 y2 + a3 y3 + a4 y4 + ...
...(5.2)
A
y = The distance of G from axis OX
where
y1 = The distance of C.G. of the area a1 from axis OX
y2, y3, y4 = The distance of C.G. of area a2, a3, a4 from axis OX respectively.
5.4.1. Centre of Gravity of Areas of Plane Figures by Integration Method. The
equations (5.1) and (5.2) can be written as
y=
x=
ai xi
ai
and
y=
ai yi
ai
where
i = 1, 2, 3, 4, .....
xi = Distance of C.G. of area ai from axis OY and
yi = Distance of C.G. of area ai from axis OX.
The value of i depends upon the number of small areas. If the small areas are large in
number (mathematically speaking infinite in number), then the summations in the above
equations can be replaced by integration. Let the small areas are represented by dA instead
of ‘a’, then the above equations are written as :
z x * dA
z dA
z y * dA
y=
z dA
x=
and
...(5.2 A)
...(5.2 B)
∫ x* dA = Σxiai
∫ dA = Σai
∫ y*dA = Σyiai
Also
x* = Distance of C.G. of area dA from axis OY
y* = Distance of C.G. of area dA from axis OX.
5.4.2. Centroid (or Centre of Gravity) of a Line. The centre of gravity of a line
which may be straight or curve, is obtained by dividing the given line, into a large number of
small lengths as shown in Fig. 5.1 (a).
The centre of gravity is obtained by replacing dA by dL in equations (5.2 A) and (5.2 B).
where
z
Then these equations become x = x * dL
dL
172
z
...(5.2 C)
CENTRE OF GRAVITY AND MOMENT OF INERTIA
Y
B
L
dL
x*
y*
A
O
and
y=
X
z y * dL
z dL
Fig. 5.1 (a)
...(5.2 D)
where x* = Distance of C.G. of length dL from y-axis, and
y* = Distance of C.G. of length dL from x-axis.
If the lines are straight, then the above equations are written as :
and
x=
L1 x1 + L2 x2 + L3 x3 + .......
L1 + L2 + L3 + .......
...(5.2 E)
y=
L1 y1 + L2 y2 + L3 y3 + .......
L1 + L2 + L3 + .......
...(5.2 F)
5.5. IMPORTANT POINTS..
(i) The axis, about which moments of areas are taken, is known as axis of reference. In
the above article, axis OX and OY are called axis of reference.
(ii) The axis of reference, of plane figures, is generally taken as the lowest line of the
figure for determining y , and left line of the figure for calculating x .
(iii) If the given section is symmetrical about X-X axis or Y-Y axis, then the C.G. of the
section will lie on the axis of symmetry.
5.5.1. Centre of Gravity of Composite Bodies. The centre of gravity of composite
bodies or sections like T-section, I-section, L-sections etc. are obtained by splitting them into
rectangular components. Then equations (5.1) and (5.2) are used.
Problem 5.1. Find the centre of gravity of the T-section shown in Fig. 5.2 (a).
Sol. The given T-section is split up into two rectangles ABCD and EFGH as shown in
Fig. 5.2 (b). The given T-section is symmetrical about Y-Y axis. Hence the C.G. of the section
will lie on this axis. The lowest line of the figure is line GF. Hence the moments of the areas
are taken about this line GF, which is the axis of reference in this case.
Let
y = The distance of the C.G. of the T-section from the bottom line GF
(which is axis of reference)
a1 = Area of rectangle ABCD = 12 × 3 = 36 cm2
3
y1 = Distance of C.G. of area a1 from bottom line GF = 10 +
= 11.5 cm
2
173
STRENGTH OF MATERIALS
a2 = Area of rectangle EFGH = 10 × 3 = 30 cm2
y2 = Distance of C.G. of area a2 from bottom line GF =
12 cm
10
= 5 cm.
2
12 cm
Y
A
B
1
3 cm
D
3 cm
H
E
C
10 cm
10 cm
2
G
3 cm
Fig. 5.2 (a)
3 cm
Y
F
Fig. 5.2 (b)
Using equation (5.2), we have
a y + a2 y2 a1 y1 + a2 y2
y= 1 1
=
(∵ A = a1 + a2)
A
a1 + a2
36 × 11.5 + 30 × 5 414 + 150
=
= 8.545 cm. Ans.
=
36 + 30
66
Problem 5.2. Find the centre of gravity of the I-section shown in Fig. 5.3 (a).
Sol. The I-section is split up into three rectangles ABCD, EFGH and JKLM as shown
in Fig. 5.3 (b). The given I-section is symmetrical about Y-Y axis. Hence the C.G. of the section
will lie on this axis. The lowest line of the figure line is ML. Hence the moment of areas are
taken about this line, which is the axis of reference.
10 cm
Y
A
10 cm
B
2 cm
1
2 cm
D E
F C
15 cm
15 cm
2
2 cm
2 cm
J
H
G
K
2 cm
3
2 cm
M
20 cm
L
20 cm
(a)
(b)
Fig. 5.3
Let
y = Distance of the C.G. of the I-section from the bottom line ML
a1 = Area of rectangle ABCD = 10 × 2 = 20 cm2
y1 = Distance of C.G. of rectangle ABCD from bottom line ML = 2 + 15 +
174
2
= 18 cm
2
CENTRE OF GRAVITY AND MOMENT OF INERTIA
a2 = Area of rectangle EFGH = 15 × 2 = 30 cm2
y2 = Distance of C.G. of rectangle EFGH from bottom line ML
15
= 2 + 7.5 = 9.5 cm
=2+
2
a3 = Area of rectangle JKLM = 20 × 2 = 40 cm2
y3 = Distance of C.G. of rectangle JKLM from bottom line ML =
a1 y1 + a2 y2 + a3 y3
A
a1 y1 + a2 y2 + a3 y3
a1 + a2 + a3
20 × 18 + 30 × 9.5 + 40 × 1
20 + 30 + 40
360 + 285 + 40 685
=
90
90
7.611 cm. Ans.
2
= 1.0 cm
2
Now using equation (5.2), we have y =
=
=
=
=
(∵ A = a1 + a2 + a3)
Problem 5.3. Find the centre of gravity of the L-section shown in Fig. 5.4.
Sol. The given L-section is not symmetrical about any
2 cm
section. Hence in this case, there will be two axis of references.
A
B
The lowest line of the figure (i.e., line GF) will be taken as axis
of reference for calculating y . And the left line of the L-section
(i.e., line AG) will be taken as axis of reference for calculating
10 cm
1
x.
12 cm
The given L-section is split up into two rectangles ABCD
and DEFG, as shown in Fig. 5.4.
D
C
To Find y
2
Let y = Distance of the C.G. of the L-section from bottom
G
line GF
8 cm
a1 = Area of rectangle ABCD = 10 × 2 = 20 cm2
Fig. 5.4
y1 = Distance of C.G. of rectangle ABCD from bottom
line GF
10
=2+
= 2 + 5 = 7 cm
2
a2 = Area of rectangle DEFG = 8 × 2 = 16 cm2
y2 = Distance of C.G. of rectangle DEFG from bottom line GF
2
= = 1.0 cm.
2
Using equation (5.2), we have
a y + a2 y2
where A = a1 + a2
y= 1 1
A
a1 y1 + a2 y2 20 × 7 + 16 × 1 140 + 16
=
=
=
20 + 16
36
a1 + a2
=
E
2 cm
F
156 13
=
= 4.33 cm.
36
3
175
STRENGTH OF MATERIALS
To Find x
Let x = Distance of the C.G. of the L-section from left line AG
x1 = Distance of the rectangle ABCD from left line AG
2
=
= 1.0 cm
2
x2 = Distance of the rectangle DEFG from left line AG
8
=
= 4.0 cm.
2
Using equation (5.1), we get
a x  a2 x2
where A = a1 + a2
x 1 1
A
a x  a2 x2 20  1  16  4

= 1 1
( a1 = 20 and a2 = 16)
a1  a2
20  16
20  64 84 7

 = 2.33 cm.
=
36
36 3
Hence the C.G. of the L-section is at a distance of 4.33 cm from the bottom line GF and
2.33 cm from the left line AG. Ans.
Problem 5.4. Using the analytical method, determine the centre of gravity of the plane
uniform lamina shown in Fig. 5.5.
(U.P. Tech. University, 2001-2002)
Sol. Let y be the distance between C.G. of the lamina and the bottom line AB.
Area 1
Area 2
a1 = 10 × 5 = 50 cm2
5
y1 =
= 2.5 cm
2


× r2 =
× 2.52 = 9.82 cm2
2
2
5
y2 =
= 2.5 cm
2
C
55
= 12.5 cm2
2
5
y3 = 5 + = 6.67 cm.
3
Using the relation,
a y  a2 y2  a3 y3
y 1 1
a1  a2  a3
a3 =
2.5
cm
5 cm
3
a2 =
Area 3
2.5
cm
2.5
cm
2
D
A
1
10 cm
12.5 cm
Fig. 5.5
232.9
50  2.5  9.82  2.5  12.5  6.67
cm =
= 3.22 cm.
72.32
50  9.82  12.5
Similarly, let x be the distance between C.G. of the lamina and the left line CD.
Area 1
a1 = 50 cm2
10
x1 = 2.5 +
= 7.5 cm
2
=
176
5 cm
5 cm
B
CENTRE OF GRAVITY AND MOMENT OF INERTIA
Area 2
a2 = 9.82 cm2
4r
4 × 2.5
x2 = 2.5 –
= 2.5 –
cm = 1.44 cm
3π
3π
Area 3
a3 = 12.5 cm2
x3 = 2.5 + 5 + 2.5 = 10 cm.
Now using the relation,
x=
a1 x1 + a2 x2 + a3 x3 50 × 7.5 + 9.82 × 1.44 + 12.5 × 10
=
cm
50 + 9.82 + 12.5
a1 + a2 + a3
514.14
= 7.11 cm.
72.32
Hence the C.G. of the uniform lamina is at a distance of 3.22 cm from the bottom line
AB and 7.11 cm from the left line CD. Ans.
Problem 5.5. From a rectangular lamina ABCD
10 cm
10 cm × 12 cm a rectangular hole of 3 cm × 4 cm is cut as
A
B
shown in Fig. 5.6.
Find the C.G. of the remaining lamina.
Sol. The section shown in Fig. 5.6, is having a cut
12
hole. The centre of gravity of a section with a cut hole is
cm
1
3
1
determined by considering the main section first as a comcm cm cm
E F
plete one, and then subtracting the area of the cut-out hole,
4 cm
i.e., by taking the area of the cut-out hole as negative.
H
G
Let y is the distance between the C.G. of the section
with a cut hole from the bottom line DC.
2 cm
a1 = Area of rectangle ABCD = 10 × 12 = 120 cm2
D
C
y1 = Distance of C.G. of the rectangle ABCD from bottom line DC
Fig. 5.6
12
= 6 cm
=
2
a2 = Area of cut-out hole, i.e., rectangle EFGH,
= 4 × 3 = 12 cm2
y2 = Distance of C.G. of cut-out hole from bottom line DC
4
= 2 + = 2 + 2 = 4 cm.
2
Now using equation (5.2 ) and taking the area (a2) of the cut-out hole as negative, we get
=
FG a y
H
IJ
K
− a2 y2
*
where A = a1 – a2
A
a1 y1 − a2 y2
=
(– ve sign is taken due to cut-out hole)
a1 − a2
y=
1 1
a1 y1 + a2 y2
but for cut-hole area a2 is taken – ve. Hence
a1 + a2
a y − a2 y2 .
y= 1 1
a1 − a2
*y =
177
STRENGTH OF MATERIALS
=
120 × 6 − 12 × 4 720 − 48
=
= 6.22 cm.
120 − 12
108
To Find x
Let x = Distance between the C.G. of the section with a cut hole from the left line AD
x1 = Distance of the C.G. of the rectangle ABCD from the left line AD
10
= 5 cm
=
2
x2 = Distance of the C.G. of the cut-out hole from the left line AD
3
= 5 + 1 + = 7.5 cm.
2
Using equation (5.1) and taking area (a2) of the cut hole as negative, we get
a x − a2 x2
x= 1 1
(∵ A = a1 – a2)
a1 − a2
120 × 5 − 12 × 7.5 600 − 90 510
=
=
=
= 4.72 cm.
108
108
120 − 12
Hence the C.G. of the section with a cut hole will be at a distance of 6.22 cm from
bottom line DC and 4.72 cm from the line AD. Ans.
Y
150 mm
100 mm
100
mm
75 mm
Fig. 5.6 (a)
Y
100 mm
100 mm
2
75 mm
3
1
Yc
75 mm
X
Xc
200 mm
Fig. 5.6 (b)
178
X
200 mm
150 mm
Problem 5.5 (A). Determine the co-ordinates
XC and YC of the centre of a 100 mm diameter circular
hole cut in a thin plate so that this point will be the
centroid of the remaining shaded area shown in Fig.
5.6 (a).
(U.P. Tech. University, 2001-2002)
Sol. The given shaded area is equal to area of a
thin rectangular plate of size 200 mm × 150 mm minus
the area of a triangle of length 100 mm and height 75 mm
minus the area of circular hole of dia. 100 mm as shown
in Fig. 5.6 (b).
Let
A1 = Area of rectangular plate
= 200 × 150 = 30000 mm2
A2 = Area of triangle
100 × 75
= 3750 mm2
=
2
A3 = Area of hole
π
=
(1002) = 2500π mm2
4
The centre of hole is the centroid of the shaded
area. Hence XC and YC is the co-ordinates of the centre of the hole and also the co-ordinates of the centroid
of the shaded area.
For area A1,
200
150
= 100 mm, y1 =
= 75 mm
x1 =
2
2
CENTRE OF GRAVITY AND MOMENT OF INERTIA
For area A2,
x2 = 100 +
2
× 100 = 166.67 mm
3
2
× 75 = 125 mm
3
x3 = XC and y3 = YC
For area A3,
Now using equation (5.1) and taking areas A2 and A3 as negative, we get
A x − A2 x 2 − A3 x 3 30000 × 100 − 3750 × 166.67 − 2500 π × X C
x = XC = 1 1
=
(30000 − 3750 − 2500 π)
A1 − A2 − A3
or
XC (30000 – 3750 – 2500π) = 30000 × 100 – 3750 × 166.67 – 2500π × XC
XC (30000 – 3750) – 2500π × XC = 30000 × 100 – 3750 × 166.67 – 2500π × XC
or
XC (30000 – 3750) = 30000 × 100 – 3750 × 166.67
(Cancelling 2500 × π × XC on both sides)
∴
26250 XC = 3000000 – 625012.5 = 2374987.5
2374987.5
∴
XC =
= 90.47 mm. Ans.
26250
A y − A2 y2 − A3 y3
Similarly,
y = YC = 1 1
A1 − A2 − A3
30000 × 75 − 3750 × 125 − 2500π × YC
=
(30000 − 3750 − 2500π)
or
YC (30000 – 3750 – 2500π) = 30000 × 75 – 3750 × 125 – 2500π × YC
or
YC (30000 – 3750) = 30000 × 75 – 3750 × 125
(Cancelling 2500π × YC on both sides)
or
26250YC = 30000 × 75 – 3750 × 125
= 225000 – 468750 = 1781250
1781250
= 67.85 mm. Ans.
∴
YC =
26250
Problem 5.5 (B). A semi-circular area is
removed from the trapezoid as shown in
Y
3
Fig. 5.6 (c). Determine the centroid of the remaining area.
(U.P. Tech. University, 2000-2001)
Sol. The given shaded area is equal to the
1
area of a thin rectangular plate of size 100 mm ×
(150 + 100) mm plus the area of the triangle of
2
length 250 mm and of height (150 – 100) = 50 mm
X
100 mm
minus the area of semi-circular area of diameter
150 mm
100 mm as shown in Fig. 5.6 (c).
Let
A1 = Area of rectangular plate
Fig. 5.6 (c)
= 100 × 250 = 25000 mm2
150 mm
100 mm
y2 = 75 +
πr 2 π × 50 2
= 1250π mm2
=
2
2
250 × 50
A3 = Area of the triangle =
= 6250 mm2
2
A2 = Area of semi-circle =
179
STRENGTH OF MATERIALS
250
= 125 mm
2
100
y1 = Distance of C.G. of area A1 from x-axis =
= 50 mm
2
100
x2 = Distance of C.G. of area A2 from y-axis = 150 +
= 200 mm
2
4 r 4 × 50 200
=
=
y2 = Distance of C.G. of area A2 from x-axis =
3π
3π
3π
2 500
x3 = Distance of C.G. of area A3 from y-axis = 250 × =
3
3
50 350
=
y3 = Distance of C.G. of area A3 from x-axis = 100 +
mm
3
3
x , y = Distance of C.G. of the shaded area from y and x-axis.
Now using equation (5.1) and taking area A2 as negative, we get
x1 = Distance of C.G. of area A1 from y-axis =
x=
x=
A1 x1 − A2 x2 + A3 x3
A1 − A2 + A3
25000 × 125 − 1250π × 200 + 6250 ×
500
3
25000 − 1250π + 6250
3125000 − 785398 + 1041666
=
= 123.75 mm. Ans.
27323
Similarly,
200
350
+ 6250 ×
25000 × 50 − 1250π ×
A1 y1 − A2 y2 + A3 y3
3π
3
y=
=
A1 − A2 + A3
27323
1250000 − 83333 + 729166
=
= 69.38 mm. Ans.
27323
∴ Centroid of the given section = ( x , y ) = (123.75 mm, 69.38 mm).
5.5.2. Problems of Finding Centroid or Centre of Gravity of Areas by
Integration Method
Problem 5.6. Determine the co-ordinates of the C.G. of the area OAB shown in Fig. 5.7,
if the curve OB represents the equation of a parabola, given by
y = kx2
in which
OA = 6 units
and
AB = 4 units.
Sol. The equation of parabola is y = kx2
...(i)
First determine the value of constant k. The point B is lying on the curve and having
co-ordinates
x = 6 and y = 4
Substituting these values in equation (i), we get
4 = k × 62 = 36 k
4
1
=
∴
k=
36 9
180
CENTRE OF GRAVITY AND MOMENT OF INERTIA
or
Substituting the value of k in equation (i), we get
1
...(ii)
y = x2
9
2
x = 9y
...(iii)
x =3 y
Consider a strip of height y and width dx as shown in
Fig. 5.7. The area dA of the strip is given by
dA = y × dx
y
The co-ordinates of the C.G. of this area dA are x and
2
∴ Distance of C.G. of area dA from y-axis = x
y
and distance of C.G. of area dA from x-axis =
2
y
∴
x* = x and y* =
2
Let x = Distance of C.G. of total area OAB from axis OY
Y
2
B
y = kx
4.0
or
y
y/2
O
dx
A
X
6
Fig. 5.7
y = Distance of C.G. of total area OAB from axis OX.
Using equation (5.2 A), we get
x=
But y =
∴
z
x∗ dA
z
dA
=
z
6
0
x × y dx
z
y dx
0
x2
from equation (ii).
9
6
1
x2
x×
× dx
9
x= 0
= 9
6 x2
1
dx
9
0 9
z
=
z
z
z
6
0
6
0
(∵ dA = ydx, x* = x)
6
x3
x2
Lx O
dx M 4 P
N Q
=
dx L x O
MN 3 PQ
4 6
0
3 6
z
z
6
0
6
x 3 dx
x 2 dx
0
1
× 64
4
=
1
× 63
3
0
1 3
= × × 6 = 4.5. Ans.
4 1
Using equation (5.2 B), we get
y=
z
z
y∗ dA
dA
where y* = Distance of C.G. of area dA from x-axis
y
=
(here)
2
dA = ydx
181
STRENGTH OF MATERIALS
∴
z
y∗ dA =
z
y
× dA =
2
z
z
1
=
2
6
0
1
=
2
6
0
z
6
0
1
y dx =
2
2
Also
0
Fx I
GH 9 JK
6
0
y dx =
6
0
z
2
z
6
0
y2
dx
2
LM OP
N Q
1 1 x5
x dx = ×
0
2 81 5
6
4
LM OP
N Q
x2
1 x3
dx =
9
9 3
5
F∵
GH
dx
5
1 1 6
6
×
×
=
2 81 5
810
z z z
z
z
dA =
z
6
2
x4
1 1
dx = ×
81
2 81
5
=
y
× y dx =
2
6
=
0
y=
x2
9
I
JK
6
0
1 63 63
×
=
9 3 27
6
27 6 5
∴
= 810
=
×
y=
810 6 3
63
dA
27
1
36 6
= . Ans.
=
× 62 =
30
30 5
Problem 5.7. Determine the co-ordinates of the C.G.
y* dA
y=x
Y
2
x
and the
of the shaded area between the parabola y =
4
straight line y = x as shown in Fig. 5.8.
Sol. The equations of parabola and straight line are
2
x
y=—
4
A
D
y
x2
y=
...(i)
4
y2
y=x
...(ii)
O
dx
x
The point A is lying on the straight line as well as
on the given parabola. Hence both the above equations
Fig. 5.8
holds good for point A. Let the co-ordinates of point A are
x, y.
Substituting the value of y from equation (ii) in equation (i), we get
y1
X
x2
x2
or 4 =
=x
4
x
Substituting the value of x = 4, in equation (ii),
y=4
Hence the co-ordinates of point A are 4, 4.
Now divide the shaded area into a large number of small areas each of height y and
width dx as shown in Fig. 5.8. Then area dA of the strip is given by
dA = ydx = (y1 – y2) dx
...(iii)
where y1 = Co-ordinate of point D which lies on the straight line OA
y2 = Co-ordinate of the point E which lies on the parabola OA.
x=
182
CENTRE OF GRAVITY AND MOMENT OF INERTIA
The horizontal co-ordinates of the points D and E are same.
The values of y1 and y2 can be obtained in terms of x from equations (ii) and (i),
x2
4
Substituting these values in equation (iii),
y1 = x and y2 =
F
GH
dA = x −
I dx
JK
x2
4
...(iv)
The distance of the C.G. for the area dA from y-axis is given by,
x* = x
And the distance of the C.G. of the area dA from x-axis is given by,
y1 − y2
y
y* = y2 +
= y2 +
(∵ y = y1 – y2)
2
2
2 y2 + y1 − y2 y1 + y2
=
=
2
2
2
x
x+
x2
4
∵ y1 = x and y2 =
=
4
2
=
F
GH
x2
1
x+
2
4
F
GH
I
JK
I
JK
...(v)
Now let x = Distance of C.G. of shaded area of Fig. 5.8 from y-axis
y = Distance of C.G. of shaded area of Fig. 5.8 from x-axis.
Now using equation (5.2 A),
z x* dA where x* = x
z dA
F x I dx
dA = G x −
H 4 JK
F x I dx
x* dA = xG x −
H 4 JK
F x I dx = LM x − x OP
= Gx −
H 4 JK N 3 4 × 4 Q
x=
2
∴
z
z
z
2
4
(∵ x varies from 0 to 4)
0
4
2
3
3
0
and
[See equation (iv)]
4
4
0
43
44
64
−
=
=
– 16
3
4×4
3
64 − 48 16
=
=
3
3
2
4
x
dA =
x−
dx
0
4
z z FGH
Lx
=M
N2
I
JK
OP
Q
4
42
43
x3
=
−
−
3×4 0
2
3×4
16 16 48 − 32 16
=
−
=
=
2
3
6
6
2
...(vi)
183
STRENGTH OF MATERIALS
∴
x=
Now using equation (5.2 B),
y=
where
y* =
LM
N
x2
1
x+
2
4
F
GH
z
16
16
6
= 3 =
×
= 2. Ans.
16
3
16
dA
6
y* dA
dA
x* dA
z
z
OP
Q
[From equation (v)]
I dx
JK
x IF
x I
1F
y* dA =
x+
x−
dx
G
J
G
2H
4 KH
4 JK
1 F
1Lx
x I
x O
x −
dx = M
−
=
P
G
J
2 H
16 K
2 N 3 5 × 16 Q
1 L4
4 O 1 L 64 64 O
−
−
= M
P=
2 N 3 5 × 16 Q 2 MN 3
5 PQ
64 L 1 1 O
F 5 − 3 IJ
= 32 G
−
=
H 15 K
2 MN 3 5 PQ
dA = x −
∴
z
z
z
x2
4
2
4
0
z
4
2
2
4
0
3
and
z
= 32 ×
dA =
5
4
0
5
2
64
=
15 15
16
6
z
3
[From equation (vi)]
64
64
6
8
∴
y=
= 15 =
×
= . Ans.
16
15
16
5
dA
6
5.5.3. Problems of Finding Centroid or Centre of Gravity of Line-Segment by
Integration Method
Problem 5.8. Determine the centre of gravity of a
Y
quadrant AB of the arc of a circle of radius R as shown in
Fig. 5.9 (a).
B
Sol. The centre of gravity of the line AB, which is an
arc of a circle radius R, is obtained by dividing the curved
dy
line AB into a large number of elements of length dL as
dL
R
shown in Fig. 5.9 (a).
The equation of curve AB is the equation of circle of
y*
dx
radius R.
∴ The equation of curve AB is given by
A
O
X
x*
x2 + y2 = R2
R
Differentiating the above equation,
Fig. 5.9 (a)
2x dx + 2y dy = 0
[∵ R is constant]
184
z
y* dA
CENTRE OF GRAVITY AND MOMENT OF INERTIA
or
2y dy = – 2x dx
− 2 x dx − x dx
=
...(i)
or
dy =
2y
y
Consider an element of length dL as shown in Fig. 5.9 (a). The C.G. of the length dL is
at a distance x* from y-axis and y* from x-axis.
z
Now using equation (5.2 D) for y , we get
y=
z
y* dL
...(ii)
dL
Let us express dL in terms of dx and dy.
But
dx 2 + dy 2
dL =
=
dx 2 +
FG − x dx IJ
H y K
=
dx 2 +
x2
dx 2
2
y
= dx
1+
= dx
R2
y2
FG∵
H
2
From (i), dy =
− x dx
y
IJ
K
y2 + x 2
y2
x2
= dx
y2
(∵ x2 + y2 = R2)
R
. dx.
y
Substituting the value of dL in equation (ii),
R
R
y×
dx
y* × . dx
y
y
y=
=
dL
dL
=
z
=
z
z
z
z
R dx
=
dL
z
z
z
R
R
dx
0
dL
LM OP
NQ
(∵ y* = y)
R
R x
=
FG 2πR IJ
H 4K
0
(∵ ∫ dL is total length of arc of one quadrant of a circle)
R × R 2R
=
=
. Ans.
2 πR
π
4
Similarly, the value of x can be calculated. Due to symmetry this value will also be
equal to
∴
2R
.
π
x=y=
2R
. Ans.
π
185
STRENGTH OF MATERIALS
2nd Method
Here
Now
Y
dL = R dθ
y* = R sin θ
x* = R cos θ
y=
z
z
z
y* dL
=
π/2
dL
π/2
0
z
0
LM
N
R sin θ dθ
2
π/2
R dθ
0
R 2 − cos θ
LO
RMθP
NQ
=
OP
Q
π/2
x*
( R sin θ) × ( R dθ)
z
0
dL
π/2
=
z
B
2
z
π/2
0
R
z
y*
dq
R dθ
R
=
R
R
q
sin θ dθ
π/2
0
O
A
x*
X
R
dθ
Fig. 5.9 (b)
LM FG π IJ − cos 0OP
N H 2K
Q
LM π − 0OP
N2 Q
π/2
− R cos
=
0
0
− R[0 − 1] 2R
=
=
. Ans.
π
π
2
Similarly,
x=
z
z
x* dL
=
dL
L O
R Msin θP
N Q
=
LO
R MθP
NQ
( R cos θ) × ( R dθ)
z
0
π/2
0
π/2
0
π/2
0
z
π/2
=
R dθ
=
R2
z
π/2
0
R
z
cos θ dθ
π/2
0
dθ
R[sin 90° − sin 0° ]
R
2R
=
. Ans.
=
π
π
π
−0
2
2
FG
H
IJ
K
FG IJ
H K
Problem 5.9. Determine the centre of gravity of the area of the circular sector OAB of
radius R and central angle α as shown in Fig. 5.10.
Sol. The given area is symmetrical about x-axis. Hence
Y
B
the C.G. of the area will lie on x-axis. This means y = 0. To
D
find x , the moment of small areas are to be taken along
R
G
dq
y-axis. Divide the area OAB into a large number of triangular
C
elements each of altitude R and base Rdθ as shown in
q
Fig. 5.10. Such triangular element is shown by OCD in which
a
O
X
altitude OC = R and base CD = Rdθ. The area dA of this trianx*
gular element is given by,
R
OC × CD R × Rdθ
=
dA =
2
2
A
R 2 dθ
Fig.
5.10
=
2
186
CENTRE OF GRAVITY AND MOMENT OF INERTIA
The C.G. of this triangular element is at G
2
2
where
OG = × OC = × R
3
3
The distance of C.G. of area dA from y-axis is given by,
2
x* = OG × cos θ = R × cos θ
3
Now using equation (5.2 A),
x=
z
z
2
x* dA
=
dA
z
α /2
0
FG 2 R cos θIJ FG R dθ IJ
H3
KH 2 K
2
2
z
α /2
0
R2
dθ
2
LMsin θOP
cos θ dθ
2R N
Q
=
3
R
LMθOP
dθ
2
NQ
α
F I
sin G J
H 2 K 4R F α I
2R
=
3
FG α JI = 3a sin GH 2 JK . Ans.
H 2K
R3
= 3
z
α/2
α/2
0
2
z
α/2
0
0
α/2
0
The area OAB is symmetrical about the x-axis, hence
y = 0. Ans.
For a semi-circle, α = π = 180°, hence
FG IJ
H K
FG IJ
H K
4R
π
sin
3α
2
4R
4R
180
sin
=
=
. Ans.
3×π
2
3π
x=
Problem 5.10. Determine the centre of gravity of a semi-circle of radius R as shown in
Fig. 5.10 (a).
Sol. This problem can also be solved by the
Y
method given in problem 5.9. The following other
2
2
2
methods can also be used. Due to symmetry, x = 0.
x +y =R
The area AOB is symmetrical about the Y-axis, hence
x = 0. The value of y is obtained by taking the moments of small areas and total area about x-axis.
y
1. Considering the strip parallel to Y-axis
C.G.
Area of strip, dA = y. dx
y/2
The distance of the C.G. of the area dA from
B
A X
O
y
x-axis is equal to
dx
2
Moment of area dA about x-axis
x
y
= dA.
Fig. 5.10 (a)
2
187
STRENGTH OF MATERIALS
y
. dA
2
y
=
. ydx
2
=
(∵ dA = y.dx)
y2
. dx
2
Moment of total area A about x-axis is obtained by integrating the above equation.
∴ Moment of total area A about x-axis
=
=
z
z
y2
. dx
2
y2
dx
(∵ x varies from – R to R)
−R 2
But equation of semi-circle is
x2 + y2 = R2
or
y2 = R2 – x2
2
Substituting this value of y in the above equation, we get
Moment of total area A about x-axis
=
=
z
R
R
−R
( R2 − x 2 )
dx
2
LM
OP
N
Q
1 LF
R I R
(− R) UO
− S R (− R) −
= MG R . R −
VP
J
2 MNH
3 K T
3 WPQ
R I R
1 LF
( − R ) UO
− S− R −
= MG R −
VP
J
2 MNH
3 K T
3 WPQ
F
F 2R
1 L 2R
R I O 1 L 2R
= M
− G−
− G− R +
= M
P
J
2 MN 3
3 K PQ 2 MN 3
H
H 3
1 L 2R
2R O 1 4 R
2R
+
= ×
=
= M
P
2N 3
3 Q 2
3
3
1 2
x3
R .x−
=
2
3
3
3
2
3
3
3
3
−R
3
2
3
R
3
3
3
3
3
3
I OP
JK PQ
3
...(i)
Let y = Distance of C.G. of the total area of semi-circle from x-axis.
The total area of semi-circle is also equal to
∴ Moment of this total area about x-axis
πR 2
2
πR 2
2
Equating the two values given by equations (i) and (ii), we get
= y×
y×
188
πR 2 2 R 3
=
2
3
...(ii)
CENTRE OF GRAVITY AND MOMENT OF INERTIA
2 R3
2
4R
. Ans.
×
=
2
3
3π
πR
4R
. Ans.
Hence the location of C.G. of semi-circle is 0,
3π
2. Considering the strip parallel to X-axis
Area of strip, dA = 2x . dy
The distance of the C.G. of this area from x-axis is y
∴ Moment of this area about x-axis
= y. dA
= y. 2xdy
= 2xy dy
...(i)
2
2
2
But, we know x + y = R
∴
x2 = R2 – y2
B
∴
or
y=
2
R −y
x =
FG
H
IJ
K
Y
C.G.
x
dy
x
y
O
A
X
Fig. 5.10 (b)
2
Substituting the above value of x in equation (i), we get
Moment of area dA about x-axis,
= 2 R 2 − y 2 . y . dy
Moment of total area A about x-axis will be obtained by integrating the above equation
from 0 to R.
∴ Moment of area A about x-axis
=
z
R
0
=–
2 R 2 − y 2 . y dy
z
R
0
2
R −y
2
(∵ y varies from 0 to R)
L (R
. (− 2 y) dy = − M
N
2
− y2 )3/ 2
3/ 2
2
2 R3
[0 – R2] =
3
3
Also the moment of total area A about x-axis = A × y
=–
where
OP
Q
R
0
...(ii)
πR 2
2
y = Distance of C.G. of area A from x-axis
A = Total area of semi-circle =
πR 2
×y
2
Equating the two values given by equations (ii) and (iii),
∴ Moment of total area A about x-axis =
...(iii)
πR 2
2 R3
×y=
2
3
or
y=
2 R3
2
4R
.
×
=
2
3
3π
πR
Ans.
189
STRENGTH OF MATERIALS
Problem 5.11. To determine the centre of
gravity of the area shown in Fig. 5.10 (c) given by
x2
+
2
y2
B
= 1.
a
b2
Sol. Consider a small strip of thickness dx
parallel to y-axis at a distance of x from the
y-axis.
2
2
y
2 = 1
2 +
b
a
x
b
C.G.
Area of the strip, dA = y.dx
y
y/2
y
The C.G. of area dA is at a distance from
2
x-axis.
Moment of the area dA about x-axis
O
A
a
x
dx
Fig. 5.10 (c)
y
. dA
2
y
. ydx
(∵ dA = y.dx)
=
2
y2
=
. dx
2
∴ Moment of the total area about x-axis
=
z
y2
. dx
0 2
Let us substitute the value of y2 in terms of x.
=
or
x2
(∵ x varies from 0 to a) ...(i)
y2 = 1
a 2 b2
x 2 a2 − x 2
y2
=
=
1
–
a2
a2
b2
b2
y2 = 2 (a2 – x2)
a
Substituting the value of y2 in equation (i), we get
Moment of total area about x-axis
The given equation is
or
a
=
+
1
2
z
a
b2
0
a2
(a 2 − x 2 ) dx =
LM
N
OP
Q
b2
2a 2
...(ii)
LMa x − x OP
3Q
N
2
b2
a3
b2
2a3 ab2
3
a
−
=
×
=
=
3
3
3
2a 2
2a 2
The total area A of the given figure is given by
A=
From equation (ii),
y=
Now equation (iv) is, A =
190
zL z
dA =
MN ba
z
a
0
2
2
y . dx
(a 2 − x 2 )
OP
Q
1/2
=
3 a
0
...(iii)
...(iv)
b 2
(a – x2)1/2
a
b 2
(a − x 2 ) 1/ 2 . dx
a
...(v)
CENTRE OF GRAVITY AND MOMENT OF INERTIA
b
a
=
=
LM
N
z
a
0
2 1/ 2
2
(a − x )
O
. dx P
Q
*
LM OP
N Q
F∵
GH
b πa 2
=
a 4
z
π . ab
4
*
a
a 2 − x 2 dx =
0
πa 4
4
I
JK
...(vi)
Let y = the distance of C.G. of the total area A from x-axis.
Then moment of total area A about x-axis
=A× y
πab
=
. y
...(vii)
4
The equations (iii) and (vii) give the moment of total area about x-axis. Hence equating
these equations, we get
πab
ab2
.y=
4
3
ab2
4
4b
∴
. Ans.
=
y=
.
3 πab 3π
To find x , take the moment of small area dA about y-axis.
The C.G. of area dA is at a distance of x from y-axis.
∴ Moment of area dA about y-axis = x.dA
= x.y.dx
Moment of total area A about y-axis is obtained by integration
Now moment of total area A about y-axis
=
=
z
z
a
x.y.dx
0
a
0
=
b
a
z
x.
a
0
b 2
(a − x 2 ) 1/ 2 . dx
a
x . (a 2 − x 2 ) 1/ 2 dx =
LM
N
2
2 3/ 2
OP
Q
a
b
a
LM∵
N
z
a
0
(∵ x varies from 0 to a)
y=
b 2
(a − x 2 ) 1/2 from equation (v)
a
(− 2)
. x(a 2 − x 2 ) 1/2 . dx
(− 2)
−b
ba 2
[0 – a3] =
3a
3
0
Also the moment of total area A about y-axis
=A× x
=
(∵ dA = ydx)
b
(a − x )
3/ 2
− 2a
=
OP
Q
...(viii)
...(ix)
where x = Distance of C.G. of total area A from y-axis.
Equating the two values given by equations (viii) and (ix),
A× x=
ba 2
3
*Please refer some standard Textbooks of Mathematics.
z
a
0
LM 1 x a − x + 1 a
2
N2
1 F π I πa
= a G J=
.
2 H 2K
4
a2 − x 2 dx =
2
2
2
4
2
sin −1
x
a
OP = LM0 + 1 a
Q N 2
a
0
2
sin −1 (1)
OP
Q
191
STRENGTH OF MATERIALS
ba 2
ba 2
=
3 A 3 × πab
4
4a
=
. Ans.
3π
The co-ordinates of the C.G. of given area are
4a
4b
x=
and y =
.
3π
3π
5.5.4. Centroid of Volume. Centroid
of volume is the point at which the total volume
of a body is assumed to be concentrated. The
volume is having three dimensions i.e., length,
width and thickness. Hence volume is measured in [length]3. The centroid [i.e., or centre
of gravity] of a volume is obtained by dividing
the given volume into a large number of small
volumes as shown in Fig. 5.10 (d). Similar
method was used for finding the centroid of an
area in which case the given area was divided X
into large number of small areas. The centroid
of the volume is hence obtained by replacing
dA by dv in equations (5.2A) and (5.2B).
∴
LM∵
N
x=
z
z
z
z
z
z
A=
πab
see equation (vi)
4
dv
Z
OP
Q
V
C.G.
Z*
z
O
x*
y*
Y
x
y
Fig. 5.10 (d)
Then these equations become as
x=
and
y=
x* dv
...(5.3 A)
dv
y* dv
...(5.3 B)
dv
As volume is having three dimensions, hence third equation is written as
z=
where
z* dv
...(5.3 C)
dv
x* = Distance of C.G. of small volume dv from y-z plane (i.e., from axis oy)
y* = Distance of C.G. of small volume dv from x-z plane (i.e., from axis ox)
z* = Distance of C.G. of small volume dv from x-y plane
and x , y , z = Location of centroid of total volume.
Note. If a body has a plane of symmetry, the centre of gravity lies in that plane. If it has two
planes of symmetry, the line of intersection of the two planes gives the position of centre of gravity. If
it has three planes of symmetry, the point of intersection of the three planes gives the position of centre
of gravity.
192
CENTRE OF GRAVITY AND MOMENT OF INERTIA
Problem 5.12. A right circular cone of radius R at the base and of height h is placed as
shown in Fig. 5.10 (e). Find the location of the centroid of the volume of the cone.
Sol. Given :
Radius of cone = R
Height of cone = h
Y
h
x
dx
R
r
O
X
Z
Fig. 5.10 (e)
In Fig. 5.10 (e), the axis of the cone is along x-axis. The centroid will be at the x-axis.
Hence, y = 0 and z = 0.
To find x , consider a small volume dv. For this, take a thin circular plate at a distance
x from O. Let the thickness of the plate is dx as shown in figure and radius of the plate is r.
The centroid of the plate is at a distance ‘x’ from O. Hence x* = x.
Now volume of the thin plate,
dv = πr2 × dx
...(i)
Let us find the value of r in terms of x.
From similar triangles, we get
or
R h
=
r
x
R× x
r=
h
Substituting the value of r in equation (i), we get
dv = π
FG R × x IJ
H h K
2
dx
z z
z F zI
z GH JK
z FGH IJK
...(ii)
Now x is given by equation (5.3A) as
x=
x* dv
dv
x.π
=
π
=
x dv
[∵ Here x* = x]
dv
R× x
h
2
R× x
h
2
. dx
dx
LM∵
MN
dv = π
FG R × x IJ
H h K
2
dx from equation (ii)
193
OP
PQ
STRENGTH OF MATERIALS
z
z
π × R2 h 3
x dx
2
0
= h 2
R h 2
x dx
π× 2
h 0
[∵ Limits of integration are w.r.t. x. And x varies from 0 to h]
LM F x I OP
M GH 4 JK PP
= M
MM x3 PP
N Q
4
h
=
3
3h
. Ans.
4
0
Problem 5.13. A hemisphere of radius R is placed as shown in Fig. 5.10 (f). The axis of
symmetry is along z-axis. Find the centroid of the hemisphere.
Sol. The hemisphere is placed as shown in
Z
Fig. 5.10 (f). The axis of symmetry is taken as Z-axis.
The centroid will be at the Z-axis. Hence x = 0 and
y = 0.
Radius of hemisphere = R.
To find z , consider a small volume dv of the
hemisphere. For this, take a thin circular plate at a
height z and thickness dz. Let ‘y’ is the radius of this
plate.
Then
dv = Area of section × thickness
= πy2 × dz
...(i)
(∵ Area of any section for sphere
or hemisphere = πr2, Here r = y)
dz
y
z
R
O
Y
X
The centre of gravity of the small volume is at
a distance z from O.
Fig. 5.10 (f )
Let us now, find the value of y in terms of z.
From Fig. 5.10 (f), we have
R 2 = z2 + y2
y2 = R2 – z2
or
Substituting the value of y2 in equation (i), we get
dv = π[R2 – z2] × dz
...(ii)
As in this case, the axis of symmetry is Z-axis. Hence x and y are zero. The distance of
the centroid from x-y plane is given by equation (5.3C) as
z=
z
z∗ dv
z
dv
where z* = Distance of centroid of the small volume dv from x-y plane.
= z [In the present case]
194
∴
z=
=
z
z
z
z
z
z
z dv
dv
z × π( R 2 − z 2 ) dz
R
=
CENTRE OF GRAVITY AND MOMENT OF INERTIA
0
π( R 2 − z 2 ) dz
π( R 2 z − z 3 ) dz
R
0
[∵ From equation (ii), dv = π(R2 – z2) × dz]
π( R 2 − z 2 ) dz
[The limits of integration are according to dz. Here z varies from 0 to R]
L R z − z OP
πM
N 2 4Q
=
L
z O
π MR z −
P
3Q
N
2 2
2
4
3
R
0
R
0
LM R × R − R OP F R I
2
4 Q GH 4 JK
3
=N
=
=
LM R × R − R OP 2 R 8 R.
3
3 Q
N
2
2
2
4
3
4
Ans.
3
5.6. AREA MOMENT OF INERTIA..
Y
Lamina of
Consider a thick lamina of area A as shown in Fig. 5.11.
area A
Let x = Distance of the C.G. of area A from the axis OY.
y = Distance of the C.G. of area A from the axis OX.
C.G.
Then moment of area about the axis OY
= Area × perpendicular distance of C.G. of area from
axis OY
y
x
= Ax
...(5.3D)
Equation (5.3D) is known as first moment of area about
O
X
the axis OY. This first moment of area is used to determine the
Fig. 5.11
centre of gravity of the area.
If the moment of area given by equation (5.3D) is again multiplied by the perpendicular
distance between the C.G. of the area and axis OY (i.e., distance x), then the quantity
(Ax). x = Ax2 is known as moment of the moment of area or second moment of area or area
moment of inertia about the axis OY. This second moment of area is used in the study of
mechanics of fluids and mechanics of solids.
Similarly, the moment of area (or first moment of area) about the axis OX = Ay.
And second moment of area (or area moment of inertia) about the axis OX = (Ay) . y = Ay2.
If, instead of area, the mass (m) of the body is taken into consideration then the second
moment is known as second moment of mass. This second moment of mass is also known as
mass moment of inertia.
Hence moment of inertia when mass is taken into consideration about the axis OY
= mx2 and about the axis OX = my2.
Hence the product of the area (or mass) and the square of the distance of the centre of
gravity of the area (or mass) from an axis is known as moment of inertia of the area (or mass)
about that axis. Moment of inertia is represented by I. Hence moment of inertia about the
axis OX is represented by Ixx whereas about the axis OY by Iyy.
195
STRENGTH OF MATERIALS
The product of the area (or mass) and the square of the distance of the centre of gravity
of the area (or mass) from an axis perpendicular to the plane of the area is known as polar
moment of inertia and is represented by J.
Consider a plane area which is split up into small areas a1, a2, a3, ... etc. Let the C.G. of
the small areas from a given axis be at a distance of r1, r2, r3, ... etc. as shown in Fig. 5.12.
Then the moment of inertia of the plane area about the given axis is given by
I = a1r12 + a2r22 + a3r32 + ...
...(5.4)
or
I = Σar2.
...(5.5)
5.7. RADIUS OF GYRATION..
Radius of gyration of a body (or a given lamina) about an axis
is a distance such that its square multiplied by the area gives moment of inertia of the area about the given axis.
For Fig. 5.12, the moment of inertia about the given axis is
given by equation (5.4) as
I = a1r12 + a2r22 + a3r32 + ...
...(i)
Let the whole mass (or area) of the body is concentrated at a
distance k from the axis of reference, then the moment of inertia of
the whole area about the given axis will be equal to Ak2.
If Ak2 = I, then k is known as radius of gyration about the
given axis.
∴
k=
I
.
A
Given
axis
Area a2 Area a3
Area a1
r1
r2
r3
Fig. 5.12
...(5.6)
5.8. THEOREM OF THE PERPENDICULAR AXIS..
Theorem of the perpendicular axis states that if IXX and IYY be the moment of inertia of
a plane section about two mutually perpendicular axis X-X and Y-Y in the plane of the section,
then the moment of inertia of the section IZZ about the axis Z-Z, perpendicular to the plane
and passing through the intersection of X-X and Y-Y is given by
IZZ = IXX + IYY.
The moment of inertia IZZ is also known as polar moment of inertia.
Proof. A plane section of area A and lying in plane xZ
y is shown in Fig. 5.13. Let OX and OY be the two mutually
perpendicular axes, and OZ be the perpendicular axis.
x
O
Consider a small area dA.
X
r
y
Let
x = Distance of dA from the axis OY
y = Distance of dA from axis OX
dA
Plane
r = Distance of dA from axis OZ
Y
section
Then r2 = x2 + y2.
of area A
Now moment of inertia of dA about x-axis
Fig. 5.13
= dA × (Distance of dA from x-axis)2
= dA × y2.
196
CENTRE OF GRAVITY AND MOMENT OF INERTIA
∴ Moment of inertia of total area A about x-axis,
IXX = ΣdAy2.
Similarly, moment of inertia of total area A about y-axis, IYY = ΣdAx2
and moment of inertia of total area A about z-axis,
IZZ = ΣdAr2
= ΣdA [x2 + y2]
(∵ r2 = x2 + y2)
2
2
= ΣdA x + ΣdA y
= IYY + IXX
or
IZZ = IXX + IYY.
...(5.7)
The above equation shows that the moment of inertia of an area about an axis at origin
normal to x, y plane is the sum of moments of inertia about the corresponding x and y-axis.
In equation (5.7), IZZ is known as Polar Moment of Inertia.
5.9. THEOREM OF PARALLEL AXIS..
Plane
area A
It states that if the moment of inertia of a plane area about
y
an axis in the plane of area through the C.G. of the plane area be
G
represented by IG, then the moment of the inertia of the given plane
X
X
area about a parallel axis AB in the plane of area at a distance h
from the C.G. of the area is given by
h
2
IAB = IG + Ah .
where IAB = Moment of inertia of the given area about AB
A
B
IG = Moment of inertia of the given area about C.G.
Fig. 5.14
A = Area of the section
h = Distance between the C.G. of the section and the axis AB.
Proof. A lamina of plane area A is shown in Fig. 5.14.
Let X-X = The axis in the plane of area A and passing through the C.G. of the area.
AB = The axis in the plane of area A and parallel to axis X-X.
h = Distance between AB and X-X.
Consider a strip parallel to X-X axis at a distance y from the X-X axis.
Let the area of the strip = dA
Moment of inertia of area dA about X-X axis = dAy2.
∴ Moment of inertia of the total area about X-X axis,
IXX or IG = ΣdAy2
...(i)
Moment of inertia of the area dA about AB
= dA(h + y)2
= dA[h2 + y2 + 2hy].
∴ Moment of inertia of the total area A about AB,
IAB = ΣdA[h2 + y2 + 2hy]
= ΣdAh2 + ΣdAy2 + ΣdA 2hy.
As h or h2 is constant and hence they can be taken outside the summation sign. Hence
the above equation becomes
IAB = h2ΣdA + ΣdAy2 + 2hΣdAy.
But ΣdA = A. Also from equation (i), ΣdAy2 = IG. Substituting these values in the above
equation, we get
IAB = h2. A + IG + 2h ΣdAy.
...(ii)
197
STRENGTH OF MATERIALS
But dA . y represents the moment of area of strip about X-X axis. And ΣdAy represents
the moments of the total area about X-X axis. But the moments of the total area about X-X
axis is equal to the product of total area (A) and the distance of the C.G. of the total area from
X-X axis. As the distance of the C.G. of the total area from X-X axis is zero, hence ΣdAy will be
equal to zero.
Substituting this value in equation (ii), we get
IAB = h2. A + IG + 0
...(5.8)
or
IAB = IG + Ah2
Thus if the moment of inertia of an area with respect to an axis in the plane of area
(and passing through the C.G. of the area) is known, the moment of inertia with respect to
any parallel axis in the plane may be determined by using the above equation.
5.10. DETERMINATION OF AREA MOMENT OF INERTIA..
The area moment of inertia of the following sections will be determined by the method
of integration :
1. Moment of inertia of a rectangular section,
2. Moment of inertia of a circular section,
3. Moment of inertia of a triangular section,
4. Moment of inertia of a uniform thin rod.
5.10.1. Moment of Inertia of a Rectangular Section
1st Case. Moment of inertia of the rectangular section about the X-X axis passing through the C.G. of the section.
Fig. 5.15 shows a rectangular section ABCD having width = b and depth = d. Let X-X is
the horizontal axis passing through the C.G. of the rectangular section. We want to determine
the moment of inertia of the rectangular section about X-X axis. The moment of inertia of the
given section about X-X axis is represent by IXX.
Consider a rectangular elementary strip of thickness dy
b
A
B
at a distance y from the X-X axis as shown in Fig. 5.15.
Area of the strip = b . dy.
d
dy
2
Moment of inertia of the area of the strip about X-X axis =
y
Area of strip × y2
d
X
X
= (b . dy) × y2 = by2dy.
d
Moment of inertia of the whole section will be obtained by
2
d
d
integrating the above equation between the limits –
to .
2
2
D
C
∴
IXX =
z
d/2
by2dy = b
−d/2
z
Fig. 5.15
d/2
−d/2
y2dy
(∵ b is constant and can be taken outside the integral sign)
L y O = b LMFG d IJ − FG − d IJ OP
=b M P
N 3 Q 3 MNH 2 K H 2 K PQ
F d I OP = b LM d + d OP
b Ld
= M
− G−
3 MN 8 H 8 JK PQ 3 N 8
8 Q
3
d/2
3
3
− d /2
3
198
3
3
3
CENTRE OF GRAVITY AND MOMENT OF INERTIA
b 2d 3 bd 3
.
=
.
...(5.9)
3 8
12
Similarly, the moment of inertia of the rectangular section about Y-Y axis passing
through the C.G. of the section is given by
=
db3
.
12
Refer to Fig. 5.15 (a)
Area of strip, dA = d × dx
M.O.I. of strip above Y-Y axis = dA × x2
= (d × dx) × x2
= d × x2 × dx
IYY =
∴
IYY
z
...(5.10)
A
(∵ dA = d . dx)
Lx O
=
d × × dx = d M P
N3Q
d LF b I
F bI O
= MG J − G − J P
H 2 K PQ
3 MNH 2 K
d Lb
b O d b
db
+
= .
=
= M
.
P
3N8
8Q 3 4
12
b
Y
B
d
3 b/ 2
b/ 2
x2
−b/ 2
−b/ 2
3
3
3
3
3
3
D
C
x
b
2
b
2
dx
Y
Fig. 5.15 (a)
2nd Case. Moment of inertia of the rectangular section about a line passing
through the base.
Fig. 5.16 shows a rectangular section ABCD having width = b
b
and depth = d. We want to find the moment of inertia of the rectanguA
B
lar section about the line CD, which is the base of the rectangular
section.
Consider a rectangular elementary strip of thickness dy at a
distance y from the line CD as shown in Fig. 5.16.
d
Area of strip = b . dy.
dy
Moment of inertia of the area of strip about the line CD
y
2
= Area of strip . y
D
C
= b . dy . y2 = by2 dy.
Fig. 5.16
Moment of inertia of the whole section about the line CD is
obtained by integrating the above equation between the limits 0 to d.
∴ Moment of inertia of the whole section about the line CD.
=
z
d
0
=b
by2dy = b
LM y OP
N3Q
z
d
y2dy
0
3 d
=
0
bd 3
.
3
...(5.11)
3rd Case. Moment of inertia of a hollow rectangular section.
Fig. 5.17 shows a hollow rectangular section in which ABCD is the main section and
EFGH is the cut-out section.
199
STRENGTH OF MATERIALS
The moment of inertia of the main section ABCD about X-X axis is given by
equation (5.11),
bd 3
=
12
where b = Width of main section
d = Depth.
The moment of inertia of the cut-out section EFGH about
X-X axis
b
A
B
E
d
F
b1
X
d1
H
b1d13
12
where b1 = Width of the cut-out section, and
G
D
=
X
C
Fig. 5.17
d1 = Depth of the cut-out section.
Then moment of inertia of hollow rectangular section about X-X axis,
IXX = Moment of inertia of rectangle ABCD about X-X axis – Moment of inertia
of rectangle EFGH about X-X axis
bd 3 b1d13
.
−
12
12
5.10.2. Moment of Inertia of a Circular Section.
Fig. 5.18 shows a circular section of radius R with O as centre.
Consider an elementary circular ring of radius ‘r’ and thickness
‘dr’. Area of circular ring
=
Y
dr
R
r
X
O
X
= 2πr. dr.
Y
In this case first find the moment of inertia of the circular
section about an axis passing through O and perpendicular to the
Fig. 5.18
plane of the paper. This moment of inertia is also known as polar
moment of inertia. Let this axis be Z-Z. (Axis Z-Z is not shown in Fig. 5.18). Then from the
theorem of perpendicular axis, the moment of inertia about X-X axis or Y-Y axis is obtained.
Moment of inertia of the circular ring about an axis passing through O and perpendicular to the plane of the paper
= (Area of ring) × (radius of ring from O)2
= (2πr . dr) . r2
= 2πr3dr
...(i)
Moment of inertia of the whole circular section is obtained by integrating equation (i)
between the limit 0 to R.
∴ Moment of inertia of the whole section about an axis passing through O and perpendicular to the plane of paper is given as
IZZ =
z
R
2πr3 dr = 2π
0
R
r3dr
0
Lr O
= 2π M P
N4Q
4
200
z
R
= 2π
0
R 4 πR 4
=
.
4
2
CENTRE OF GRAVITY AND MOMENT OF INERTIA
D
2
where D = Diameter of the circular section
But
R=
∴
or
IZZ =
FG IJ
H K
π
D
×
2
2
4
=
πD 4
32
...(5.12)
πD 4
32
But from the theorem of perpendicular axis given by equation (5.7), we have IZZ = IXX + IYY.
But due to symmetry,
IXX = IYY
Polar moment of inertia =
∴
IXX = IYY =
I ZZ
2
πD 4 1 πD 4
× =
32
2
64
Moment of Inertia of a hollow circular section
Fig. 5.19 shows a hollow circular section.
Let D = Diameter of outer circle, and
d = Diameter of cut-out circle.
Then from equation (5.13), the moment of inertia of the
π
outer circle about X-X axis =
D4.
64
And moment of inertia of the cut-out circle about X-X axis
=
...(5.13)
Y
D
d
X
O
X
Y
π 4
d.
Fig. 5.19
64
∴ Moment of inertia of the hollow circular section, about X-X axis,
IXX = Moment of inertia of outer circle – moment of inertia of cut-out circle
=
π
π 4
π
D4 –
d =
[D4 – d4]
64
64
64
π
Similarly,
IYY =
[D4 – d4].
64
5.10.3. Moment of Inertia of a Triangular Section
1st Case. Moment of inertia of a triangular section about its base.
Fig. 5.20 shows a triangular section ABC of base
A
width = b and height = h. Consider a small strip of thickness
dy at a distance y from the vertex A.
y
Area of the strip, = DE . dy
...(i)
dy
D
h
The distance DE in terms of y, b and h is obtained from
two similar triangles ADE and ABC as
DE y
B
=
b
BC h
y
Fig. 5.20
∴
DE = BC .
h
=
E
C
201
STRENGTH OF MATERIALS
b. y
h
Substituting this value of DE in equation (i), we get
(∵ BC = b)
=
by
. dy.
h
Distance of the strip from the base = (h – y)
∴ Moment of inertia of the strip about the base
= Area of strip × (Distance of strip from base)2
Area of strip
=
by
by
. dy . (h – y)2 =
(h – y)2 . dy.
h
h
The moment of inertia of the whole triangular section about the base (IBC) is obtained
by integrating the above equation between the limits 0 to h.
=
∴
IBC =
z
h
0
by
[ h − y]2 dy
h
z
z
b h
y(h – y)2 dy
h 0
(∵ b and h are constants and can be taken outside the integral sign)
=
=
b
h
h
y (h2 + y2 – 2hy) dy =
0
b
h
z
h
0
(yh2 + y3 – 2hy2) dy
LM
OP
N
Q
2h . h O b L h
b Lh . h
h
+
−
= M
P= M
4
3 Q hN 2
hN 2
b
L 6 + 3 − 8 OP = bh . 1
=
.h M
h
12
N 12 Q
=
b y 2 h 2 y 4 2hy 3
+
−
2
4
3
h
2
4
2
h
0
4
3
4
+
h 4 2h 4
−
4
3
OP
Q
3
bh3
12
2nd Case. Moment of inertia of the triangular section about
through the C.G. and parallel to the base.
Consider a triangular section of base = b and height = h
as shown in Fig. 5.21. Let X-X is the axis passing through the
C.G. of the triangular section and parallel to the base.
The distance between the C.G. of the triangular section
h
h
X
and base AB = .
3
Now from the theorem of parallel axis, given by equaB
tion (5.8), we have
=
...(5.14)
an axis passing
A
C.G
X
h
3
C
b
Fig. 5.21
202
CENTRE OF GRAVITY AND MOMENT OF INERTIA
Moment of inertia about
BC = Moment of inertia about C.G. + Area × (Distance between X-X and BC)2
FG h IJ
H 3K
F hI
=I –A×G J
H 3K
bh
F b × h IJ . FG h IJ
=
−G
12 H 2 K H 3 K
2
or
IBC = IG + A ×
2
∴
IG
BC
3
2
=
bh 3 bh 3 bh 3 (3 − 2)
−
=
12
18
36
=
bh3
36
Problem 5.13 (A). Determine the moment of inertia of the section about an axis passing through the base
BC of a triangular section shown in Fig. 5.21 (a).
(U.P. Tech. University, 2002-2003)
Sol. Given :
Base, b = 100 mm ; height, h = 90 mm.
Moment of inertia of a triangular section about an
axis passing through the base is given by equation
(5.14) as
IBC =
F∵
GH
I BC =
bh3
b× h
and Area =
12
2
I
JK
...(5.15)
A
90 mm
B
C
100 mm
Fig. 5.21 (a)
bh3
12
100 × 90 3
= 6.075 × 106 mm4. Ans.
12
Y
5.10.4. Moment of Inertia of a Uniform Thin Rod.
Consider a uniform thin rod AB of length L as shown in Fig. 5.22.
A
Let m = Mass per unit length of rod, and
M = Total mass of the rod
=m×L
...(i)
Suppose it is required to find the moment of inertia of
the rod about the axis Y-Y. Consider a strip of length dx at a
distance x from the axis Y-Y.
Mass of the strip
= Length of strip × Mass per unit length
= dx . m
or
m . dx.
Moment of inertia of the strip about Y-Y axis
= Mass of strip × x2
= (m . dx) . x2
= mx2dx.
=
x
dx
B
L
Fig. 5.22
203
STRENGTH OF MATERIALS
Moment of inertia of the whole rod (IYY) will be obtained by integrating the above
equation between the limits 0 to L.
∴
IYY =
z
L
0
mx 2 dx = m
Lx O
=m M P
N3Q
L
3
=
0
z
L
x 2 dx
(∵ m is constant)
0
mL3
3
mL . L2 ML2
=
[∵ m . L = M from equation (i)]
3
3
5.10.5. Moment of Inertia of Area Under a Curve of given Equation. Fig. 5.22 (a)
shows an area under a curve whose equation is parabolic and is given by
=
x = ky2
in which
y = b when x = a
Suppose it is required to find the moment of
inertia of this area about y-axis. Consider a strip of
thickness dx at a distance x from y-axis.
The area of strip, dA = y dx
...(i)
Let us substitute the value of y in terms of x
in the above equation. The equation of curve is
...(ii)
x = ky2
First find the value of k.
When y = b, x = a. Hence above equation
becomes
a = kb2
or
k=
y
2
x = ky
b
y
O
x
a
x
dx
5.22 (a)
a
b2
Substituting the value of ‘k’ in equation (ii), we get
x=
a
b2
. y2 or y2 =
F b xI
y=G
H a JK
2
or
1/ 2
=
b
b2 x
a
x
a
Substituting this value of y in equation (i), we get
dA =
b
a
. x . dx
The moment of inertia of elemental area (dA) about y-axis
= x2. dA = x2 .
204
b
a
. x dx
...(iii)
CENTRE OF GRAVITY AND MOMENT OF INERTIA
∴ Moment of inertia of the total area about y-axis is obtained by integrating the
above equation between the limits 0 to a.
(∵ x varies from 0 to a)
∴
Iyy =
=
z
a
b
x2 .
a
0
. x . dx =
LM x OP
a N 7/2 Q
7/2
b
a
b
a
.
z
a
x 5 / 2 . dx
0
2
2 b
.
. a7 / 2 =
ba2. Ans.
7
7
a
=
0
To find the moment of inertia of the given area about x-axis, the element shown in
Fig. 5.22 (a) can be considered to be a rectangle of thickness dx. The moment of inertia of this
element about x-axis is equal to the moment of inertia of the rectangle about its base.
∴ Moment of inertia of the element about x-axis
LM
N
OP
Q
bd 3
dx . y 3
∵ it is
where b = dx and d = y
3
3
The moment of inertia of the given area about x-axis is obtained by integrating the
above equation between the limits 0 to a.
=
∴
IXX =
z
LM
zN
a
0
dx . y 3
=
3
a
=
=
=
0
b
b
. x
a
3
3
3a 3 / 2
z
b3
3 . a3 / 2
a
0
z
OP
Q
a
0
3
y3
. dx
3
dx
. x 3 / 2 dx =
b
3
3a 3 / 2
LM x OP
N 5/2 Q
5/ 2 a
LM∵
N
b
y=
a
x from equation (iii)
OP
Q
0
2
2 3
2
. . a5 / 2 =
b .a=
ab3. Ans.
5
15
15
Problem 5.14. Fig. 5.23 shows a T-section of dimensions
10 × 10 × 2 cm. Determine the moment of inertia of the section
about the horizontal and vertical axes, passing through the centre of gravity of the section. Also find the polar moment of inertia
of the given T-section.
Sol. First of all, find the location of centre of gravity of
the given T-section. The given section is symmetrical about the
axis Y-Y and hence the C.G. of the section will lie on Y-Y axis.
The given section is split up into two rectangles ABCD and EFGH
for calculating the C.G. of the section.
10 cm
A
B
1
D
2 cm
H
E
C
10 cm
2
G
2 cm
F
Let
y = Distance of the C.G. of the section from the
Fig. 5.23
bottom line GF
a1 = Area of rectangle ABCD = 10 × 2 = 20 cm2
y1 = Distance of C.G. of the area a1 from the bottom line GF = 8 + 1 = 9 cm
a2 = Area of rectangle EFGH = 8 × 2 = 16 cm2
8
y2 = Distance of C.G. of rectangle EFGH from the bottom line GF =
= 4 cm
2
205
STRENGTH OF MATERIALS
a1 y1 + a2 y2 20 × 9 + 16 × 4 180 + 64
244
=
=
=
= 6.777 cm.
36
20 + 16
36
a1 + a2
Hence the C.G. of the given section lies at a distance of 6.777 cm from GF. Now find the
moment of inertia of the T-section.
Now, Let IG1 = Moment of inertia of rectangle (1) about the horizontal axis and passing
through its C.G.
IG2 = Moment of inertia of rectangle (2) about the horizontal axis and passing
through the C.G. of the rectangle (2)
h1 = The distance between the C.G. of the given section and the C.G. of the
rectangle (1)
= y1 – y = 9.0 – 6.777 = 2.223 cm
h2 = The distance between the C.G. of the given section and the C.G. of the
rectangle (2)
= y – y2 = 6.777 – 4.0 = 2.777 cm.
Using the relation,
y =
10 × 23
= 6.667 cm4
12
2 × 83
IG2 =
= 85.333 cm4.
12
From the theorem of parallel axes, the moment of inertia of the rectangle (1) about the
horizontal axis passing through the C.G. of the given section
Now
IG1 =
= IG1 + a1h12 = 6.667 + 20 × (2.223)2
= 6.667 + 98.834 = 105.501 cm4.
Similarly, the moment of inertia of the rectangle (2) about the horizontal axis passing
through the C.G. of the given section
= IG2 + a2h22 = 85.333 + 16 × (2.777)2
= 85.333 + 123.387 = 208.72 cm4.
∴ The moment of inertia of the given section about the horizontal axis passing through
the C.G. of the given section is,
Ixx = 105.501 + 208.72 = 314.221 cm4. Ans.
The moment of inertia of the given section about the vertical axis passing through the
C.G. of the given section is,
2 × 10 3 8 × 2 3
+
Iyy =
12
12
= 166.67 + 5.33 = 172 cm4. Ans.
Now the polar moment of inertia (Izz) is obtained from equation (5.7) as
Izz = Ixx + Iyy
= 314.221 + 172 = 486.221 cm4. Ans.
Problem 5.15. Find the moment of inertia of the section
shown in Fig. 5.24 about the centroidal axis X-X perpendicular
to the web.
206
10 cm
Y
A
B
2 cm
1
D H
E C
10 cm
2
2 cm
J
G
F
K
Y
20 cm
L
2 cm
3
M
Fig. 5.24
CENTRE OF GRAVITY AND MOMENT OF INERTIA
Sol. First of all find the location of centre of gravity of the given figure. The given
section is symmetrical about the axis Y-Y and hence the C.G. of the section will lie on Y- Y
axis. The given section is split up into three rectangles ABCD, EFGH and JKLM. The centre
of gravity of the section is obtained by using
y =
a1 y1 + a2 y2 + a3 y3
–
a1 + a2 + a3
...(i)
where y = Distance of the C.G. of the section from the bottom line ML
a1 = Area of rectangle ABCD = 10 × 2 = 20 cm2
y1 = Distance of the C.G. of the rectangle ABCD from the bottom line ML
2
= 2 + 10 + = 12 + 1 = 13 cm
2
a2 = Area of rectangle EFGH = 10 × 2 = 20 cm2
y2 = Distance of the C.G. of rectangle EFGH from the bottom line ML
10
= 2 + 5 = 7 cm
=2+
2
a3 = Area of rectangle JKLM = 20 × 2 = 40 cm2
y3 = Distance of the C.G. of rectangle JKLM from the bottom line ML
2
= = 1.0 cm.
2
Substituting the above values in equation (i), we get
20 × 13 + 20 × 7 + 40 × 1
y =
20 + 20 + 40
260 + 140 + 40 440
=
=
= 5.50 cm.
80
80
The C.G. of the given section lies at a distance of 5.50 cm from the bottom line ML. We
want to find the moment of inertia of the given section about a horizontal axis passing through
the C.G. of the given section.
Let IG1 = Moment of inertia of rectangle (1) about the horizontal axis passing through
its C.G.
IG2 = Moment of inertia of rectangle (2) about the horizontal axis passing through
the C.G. of rectangle (2)
IG3 = Moment of inertia of rectangle (3) about the horizontal axis passing through
the C.G. of rectangle (3)
h1 = The distance between the C.G. of the rectangle (1) and the C.G. of the given
section
= y1 – y = 13.0 – 5.50 = 7.50 cm
h2 = The distance between the C.G. of rectangle (2) and the C.G. of the given
section
= y2 – y = 7.0 – 5.50 = 1.50 cm
h3 = The distance between the C.G. of the rectangle (3) and the C.G. of the given
section
207
STRENGTH OF MATERIALS
= y – y3 = 5.50 – 1.0 = 4.5 cm
10 × 2 3
= 6.667 cm4
12
2 × 10 3
IG2 =
= 166.667 cm4
12
20 × 2 3
IG3 =
= 13.333 cm4.
12
From the theorem of parallel axes, the moment of inertia of the rectangle (1) about the
horizontal axis passing through the C.G. of the given section
Now
IG1 =
= IG1 + a1h12 = 6.667 + 20 × (7.5)2
= 6.667 + 1125 = 1131.667 cm4.
Similarly, the moment of inertia of the rectangle (2) about the horizontal axis passing
through the C.G. of the given section
= IG2 + a2h22 = 166.667 + 20 × 1.52
= 166.667 + 45 = 211.667 cm4.
And moment of inertia of the rectangle (3) about the horizontal axis, passing through
the C.G. of the given section
= IG3 + a3h32 = 13.333 + 40 × 4.52
= 13.333 + 810 = 823.333 cm4
Now moment of inertia of the given section about the horizontal axis, passing through
the C.G. of the given section
= Sum of the moment of inertia of the rectangles (1), (2) and (3) about
the horizontal axis, passing through the C.G. of the given section
= 1131.667 + 211.667 + 823.333 = 2166.667 cm4. Ans.
Problem 5.15(A). Determine the polar moment of inertia of I-section shown in
Fig. 5.24 (a). (All dimensions are in mm).
(U.P. Tech. University, 2001-2002)
Sol. Let us first find the location of C.G. of the given
80
section. It is symmetrical about the vertical axis, hence C.G.
lies on this section.
12
1
Now,
A1 = Area of first rectangle
= 80 × 12 = 960 mm2
A 2 = Area of second rectangle
12
150
128
[(150 – 12 – 10) × 12]
2
2
= 128 × 12 = 1536 mm
A3 = Area of third rectangle
3
10
= 120 × 10 = 1200 mm2
120
y1 = Distance of C.G. of area A1
Fig. 5.24 (a)
from bottom line
12
= 150 –
= 144 mm
2
y2 = Distance of C.G. of area A2 from bottom line
128
= 74 mm
= 10 +
2
208
CENTRE OF GRAVITY AND MOMENT OF INERTIA
y3 = Distance of C.G. of area A3 from
bottom line =
10
= 5 cm.
2
y = Distance of C.G. of the given section
from bottom line.
The C.G. of the section is obtained by using,
y =
A1 y1 + A2 y2 + A3 y3
A1 + A2 + A3
960 × 144 + 1536 × 74 + 1200 × 5
960 + 1536 + 1200
138240 + 113664 + 6000
257904
=
=
3696
3696
= 69.779 ~
69.78
cm
−
Location of centroidal axis is shown in Fig. 5.24 (b).
(i) Moment of inertia of the given section about X-X
M.O.I. of the rectangle ¬ about centroid axis X-X is given by,
=
IXX1
= ( IG1 )X + A1 × h12 where h1 = (y1 – y )
Y
80 × 12 3
=
+ 960(144 – 69.78)2 = 5.3 × 106 mm4
12
M.O.I. of rectangle ­ about centroid axis X-X is given by,
and
IXX2
= ( IG2 )X + A2 × h22 where h2 = (y2 – y )
IXX3
12 × 128 3
=
+ 1536 × (74 – 69.78)2
12
= 2.12 × 106 mm4
= ( IG3 )X + A3 × h32 where h3 = (y3 – y )
120 × 10 3
=
+ 1200 × (5 – 69.78)2 = 5.04 × 106 mm4
12
∴ IXX = IXX1 + IXX2 + IXX3
= 5.3 × 106 + 2.12 × 106 + 5.04 × 106 mm4
= 12.46 × 106 mm4
(ii) Moment of inertia of the given section about Y-Y
80
X
X
69.78 mm
Y
Fig. 5.24 (b)
12 × 80 3
= 521 × 103 mm4 = 0.521 × 106 mm4
12
128 × 12 3
IYY2= ( IG2 )Y =
= 18.432 × 103 mm4 = 0.018432 × 106 mm4
12
10 × 120 3
IYY3= ( IG3 )Y =
= 1.44 × 106 mm4
12
IYY = IYY1 + IYY2 + IYY3
= 0.521 × 106 + 0.018432 × 106 + 1.44 × 106 mm4 = 1.979 × 106 mm4
IYY1= ( IG1 )Y =
209
STRENGTH OF MATERIALS
∴ Polar moment of inertia (IZZ) is given by,
IZZ = IXX + IYY
= 12.46 × 106 + 1.979 × 106 mm4
= 14.439 × 106 mm4. Ans.
C
R
Problem 5.16. Find the moment of inertia of the area shown
shaded in Fig 5.25, about edge AB.
Sol. Given :
Radius of semi-circle, R = 10 cm
Width of rectangle,
b = 20 cm
A
Depth of rectangle,
d = 25 cm
Moment of inertia of the shaded portion about AB
= M.O.I. of rectangle ABCD about AB
– M.O.I. of semi-circle on DC about AB
M.O.I. of rectangle ABCD about AB
=
bd 3
3
= m
c
10
D
Semi
circle
25 cm
B
20 cm
Fig. 5.25
[see equation (5.11)]
20 × 253
= 104167
3
M.O.I. of semi-circle about DC
=
=
1
× [M.O.I. of a circle of radius 10 cm about a diameter]
2
=
π 4
1
1 π
×
d = ×
× 204 = 3.925 cm4
2
64
2 64
LM
N
OP
Q
Distance of C.G. of semi-circle from DC
=
4 r 4 × 10
=
= 4.24 cm
3π
3π
πr 2 π × 10 2
=
= 157.1 cm2
2
2
M.O.I. of semi-circle about a line through its C.G. parallel to CD
= M.O.I. of semi-circle about CD – Area × [Distance of C.G. of semi-circle from DC]2
= 3925 – 157.1 × 4.242
= 3925 – 2824.28 = 1100.72 cm4
Distance of C.G. of semi-circle from AB
= 25 – 4.24 = 20.76 cm
M.O.I. of semi-circle about AB = 1100.72 + 157.1 × 20.762
= 1100.72 + 67706.58 = 68807.30 cm4
∴ M.O.I. of shaded portion about AB
= 104167 – 68807.30 = 35359.7 cm4. Ans.
Area of semi-circle, A =
210
CENTRE OF GRAVITY AND MOMENT OF INERTIA
Problem 5.16 (A). Find the moments of inertia about the centroidal XX and YY axes of
the section shown in Fig. 5.25 (a).
(U.P. Tech. University, 2002-2003)
Sol. First find the location of the C.G. of the given figure:
Y
Let
a1 = Area of complete rectangle
B/2
=B×D
a2 = Area of removed rectangle portion
2
B D BD
×
=
=
2 2
4
D
B
D
x1 =
, y1 =
and
1
D/2
2
2
3B
B 1 B
+
=
x2 =
,
2 2 2
4
X
B
D 1 D
3D
Fig. 5.25 (a)
=
+
y2 =
2 2 2
4
where (x1, y1) and (x2, y2) are the co-ordinates of the C.G. of the complete rectangle and cut out
rectangle respectively. Area a2 is negative.
B B × D 3B
×
BD × −
a1 x1 − a2 y2
2
4
4
=
=
Now
x
3
a1 − a2
BD
4
5
B2 × D
3 2
−
B D
BD 2
5
16
2
16
=
=
×D
=
3
3
12
BD
BD
4
4
D BD 3 D
−
×
BD ×
a y − a2 y2
2
4
4
=
Similarly,
y = 1 1
3
a1 − a2
BD
4
5
BD 2
3
−
BD 2
BD 2
5
16
2
16
=
=
=
×D
3
3
12
BD
BD
4
4
Now draw the centroidal axes XX and YY as shown in Fig. 5.25 (b).
Let IXX1 = M.O.I. of complete rectangle ¬ about centroidal axis X-X
= M.O.I. of complete rectangle ¬ about horizontal axis passing through its C.G.
+ Area of complete rectangle ¬
× Distance between X-X axis and horizontal axis passing through the C.G. of
FG
H
FG
H
IJ
K
IJ
K
rectangle ¬ (By theorem of parallel axis)
BD 3
+ (B × D) [y1 – y ]2
12
2
BD 3
D 5D
+ BD
−
=
12
2
12
[∵ IXX1 = IG1xx + A1h12]
=
BD 3
=
12
LM
OP
N
Q
LDO
+ BD M P
N 12 Q
2
LM∵
N
y1 =
D
5D
,y=
2
2
211
OP
Q
STRENGTH OF MATERIALS
BD 3 BD 3
13
BD 3
+
=
12
144 144
= (IG2X) + A2 × h22
Y
=
Similarly,
IXX2
FG IJ
H K
B
D
×
2
2
=
12
D/12
+
LM
N
LM
N
BD 3 BD 3 D 5 D
+
−
192
4
4
12
LM∵
N
=
B/2
3
BD
× [y2 – y ]2
4
BD
∵ A2 =
, h2 = ( y2 − y)
4
=
Y
B/2
y2 =
2
=
X
OP
Q
y = 5D/12
X
1
D/2
X
5B/12
2
Y
x
3D
5D
,y=
2
12
FG IJ
H K
BD 3 BD
4D
+
×
192
4
12
OP
Q
D/2
OP
Q
Fig. 5.25 (b)
BD 3 16 BD 3
+
192 4 × 144
BD 3 BD 3 3 BD 3 + 16 BD 3 19 BD 3
+
=
=
192
36
576
576
IXX = M.O.I. of given section about centroidal axis X-X
= IXX1 – IXX2
=
Now
13 BD 3 19 BD 3 52 BD 3 − 19 BD 3 33 BD 3
−
=
=
= 0.0573 BD3. Ans.
144
576
576
576
Similarly, the M.O.I. of the given section about centroidal axis Y-Y is given by
IYY = IYY1 – IYY2
where
IYY1 = M.O.I. of rectangle ¬ about centroidal axis Y-Y
= IG1y + A1 × [x1 – x ]2
=
=
LM
N
DB3
B 5B
+ BD ×
−
12
2 12
IYY2 = IG2y + A2[x2 – x ]2
and
FG IJ
H K
D
B
×
2
2
=
12
∴
IYY =
3
+
LM
N
OP
Q
2
=
BD 3 B 5 B
−
4
4
12
DB 2 BD × B 2
13
DB3
+
=
12
144
144
OP
Q
2
=
DB3 DB3 19 DB3
+
=
192
36
576
13
19 DB 3
33
DB3 −
=
DB 3 = 0.0573 DB3. Ans.
144
576
576
5.11. MASS MOMENT OF INERTIA..
Consider a body of mass M as shown in Fig. 5.26.
Let x = Distance of the centre of gravity of mass M from axis OY
y = Distance of the centre of gravity of mass M from axis OX
Then moment of the mass about the axis OY = M . x
212
CENTRE OF GRAVITY AND MOMENT OF INERTIA
The above equation is known as first moment of
Y
Body of
mass about the axis OY.
Mass M
If the moment of mass given by the above equation is again multiplied by the perpendicular distance
between the C.G. of the mass and axis OY, then the
Centroid
quantity (M . x) . x = M . x2 is known as second moment
of mass about the axis OY. This second moment of the
mass (i.e., quantity M . x2) is known as mass moment
y
of inertia about the axis OY.
Similarly, the second moment of mass or mass
O
moment of inertia about the axis OX
x
2
= (M.y) . y = M.y
Fig. 5.26
Hence the product of the mass and the square of
the distance of the centre of gravity of the mass from an
axis is known as the mass moment of inertia about that
axis. Mass moment of inertia is represented by Im.
Given axis
Hence mass moment of inertia about the axis OX is
Mass m3
Mass m2
represented by (Im)xx whereas about the axis OY by
(Im)yy.
Mass m1
Consider a body which is split up into small
masses m1, m2, m3 ...... etc. Let the C.G. of the small
areas from a given axis be at a distance of r1, r2, r3
...... etc. as shown in Fig. 5.27. Then mass moment of
r1
inertia of the body about the given axis is given by
r2
Im = m1r12 + m2r22 + m3r32 + ......
= Σmr2
r3
If small masses are large in number then the
summation in the above equation can be replaced by
Fig. 5.27
integration. Let the small masses are replaced by dm
instead of ‘m’, then the above equation can be written as
Im =
z
r2 dm
X
...(5.16)
5.12. DETERMINATION OF MASS MOMENT OF INERTIA..
The mass moment of inertia of the following bodies will be determined by the method of
integration :
1. Mass moment of inertia of a rectangular plate,
2. Mass moment of inertia of a circular plate,
3. Mass moment of inertia of a hollow circular cylinder.
5.12.1. Mass Moment of Inertia of a Rectangular Plate
(a) Mass moment of inertia of a rectangular plate about X-X axis passing through
the C.G. of the plate.
Fig. 5.28 shows a rectangular plate of width b, depth ‘d’ and uniform thickness ‘t’. Consider a small element of width ‘b’ at a distance ‘y’ from X-X axis as shown in Fig. 5.29.
Here X-X axis is the horizontal line passing through the C.G. of the plate.
Area of the element
= b × dy
213
STRENGTH OF MATERIALS
Y
B
A
dy
d/2
y
X
d/2
d X
d
X
C.G.
d/2
X
d/2
D
Y
b
C
b
t
Fig. 5.28
Fig. 5.29
∴ Mass of the element = Density × Volume of element
= ρ × [Area × thickness of element]
= ρ × [b × dy × t]
[∵ ρ = Density and t = thickness]
= ρbt dy
Mass moment of inertia of the element about X-X axis
= Mass of element × y2
= (ρbt dy) × y2 = ρbt y2 dy
Mass moment of inertia of the plate will be obtained by integrating the above equation
d
d
between the limits – to .
2
2
∴
(Im)xx =
z
d/2
− d/2
ρbt y2 dy = ρ bt
z
d/2
− d/2
y2 dy
[∵ ρ, b, t are constant and can be taken outside the integral sign]
L y O = ρbt LMFG d IJ − FG − d IJ OP
= ρbt M P
H 2 K PQ
3 MNH 2 K
N3Q
F d I OP = ρbt LM d + d OP = ρbt × 2d
ρbt L d
− G−
=
M
3 N 8
8 Q
3
8
3 MN 8 H 8 JK PQ
3
d/2
3
3
− d/2
3
=
3
ρbt 3
bd 3
d =ρ×t
12
12
3
3
3
...(5.17)
bd 3
is the moment of inertia of the area of the rectangular section about X-X axis.
12
This moment of inertia of the area is represented by Ixx.
∴
(Im)xx = ρ × t × Ixx
...(5.18)
where (Im)xx = Mass moment of inertia of the plate about X-X axis passing through C.G. of the
plate.
Ixx = Moment of inertia of the area of the plate about X-X axis.
But
214
CENTRE OF GRAVITY AND MOMENT OF INERTIA
Again from equation (5.17), we have
(Im)xx = ρ × t ×
bd 3
12
d2
12
= ρb × d × t ×
d2
12
(∵ M = Mass of the plate = ρ × Volume of the plate = ρ × [b × d × t])
1
=
Md2
...(5.19)
12
Similarly, the mass moment of inertia of the rectangular plate about Y-Y axis passing
through the C.G. of the plate is given by
1
(Im)yy =
Mb2.
...(5.20)
12
(b) Mass moment of inertia of the rectangular
b
A
plate about a line passing through the base.
B
Fig. 5.30 shows a rectangular plate ABCD, having
width = b, depth = d and uniform thickness = t. We want to
find the mass moment of inertia of the rectangular plate
about the line CD, which is the base of the plate. Consider a
d
rectangular elementary strip of width b, thickness t and
depth ‘dy’ at a distance y from the line CD as shown in
dy
Fig. 5.30.
dA = b . dy
Area of strip,
y
Volume of strip
= dA × t = b . dy . t = b . t . dy
D
C
Mass of the strip, dm = Density × Volume of strip
Fig. 5.30
= ρ(b . t . dy) = ρ . b . t . dy
Mass moment of inertia of the strip about the line CD
= Mass of strip . y2
= dm . y2 = y2 . dm
Mass moment of inertia of the whole rectangular plate about the line CD is obtained by
integrating the above equation between the limits 0 to d.
∴ Mass moment of inertia of the rectangular plate about the line CD
=M×
=
z
d
0
y2 . dm =
=ρ.b.t
z
d
0
z
y2 . (ρ . b . t . dy)
0
y2 dy
Ly O
=ρ.b.t. M P
N3Q
3
M . d2
=
3
d
d
=ρ.b.t.
0
[∵ dm = ρ . b . t . dy]
[∵ ρ, b and t are constant]
d3
d2
=ρ.b.t.d.
3
3
...(5.21)
[∵ ρ . b . t . d = Mass of rectangular plate = M]
215
STRENGTH OF MATERIALS
(c) Mass moment of inertia of a hollow rectangular plate.
Fig. 5.31 shows a hollow rectangular plate in which
b
B
ABCD is the main plate and EFGH is the cut-out section.
A
The mass moment of inertia of the main plate ABCD
about X-X is given by equation
E
F
d
1
2
b1
Md
=
12
X
The mass moment of inertia of the cut-out section X
d
1
EFGH about X-X axis
H
G
1
md12
=
12
where M = Mass of main plate ABCD
D
C
=ρ.b.d.t
Fig.
5.31
m = Mass of the cut-out section EFGH
= ρ . b1 . d1 . t
Then mass moment of inertia of hollow rectangular plate about X-X axis is given by
1
1
Md 2 −
md12 .
...(5.22)
(Im)xx =
12
12
5.12.2. Mass Moment of Inertia of a Circular Plate
Fig. 5.32 shows a circular plate of radius R and thickness t with O as centre. Consider an
elementary circular ring of radius ‘r’ and width dr as shown
Y
in Fig. 5.32 (a).
Area of ring, dA = 2πr . dr
Volume of ring
= Area of ring × t = dA . t
dr
R
= 2πr . dr . t
r
Mass of ring, dm = Density × Volume of ring
= ρ(2πr.dr.t)
X
O
X
In this case first find the mass moment of inertia about
an axis passing through O and perpendicular to the plane of
the paper i.e., about axis Z-Z.
∴ Mass moment of inertia of the circular ring about
axis Z-Z
Y
Fig. 5.32
= (Mass of ring) × (radius of ring)2
= dm × r2 = (ρ . 2πr dr . t) × r2 = ρ . t . 2πr3 dr
The mass moment of inertia of the whole circular plate will be obtained by integrating the
above equation between the limits 0 to R.
∴ Mass moment of inertia of circular plate about Z-Z axis is given by
(Im)zz =
z
R
0
ρ . t . 2πr3 dr = 2π . ρ . t
Lr O
= 2πρ . t M P
N4Q
216
4
R
0
z
R
0
r3 dr
CENTRE OF GRAVITY AND MOMENT OF INERTIA
= 2π . ρ . t .
R4
R4
=π.ρ.t.
4
2
Y
Now mass of circular plate,
M = ρ × Volume of plate
= ρ × πR2 × t
[Volume of plate = Area × t = πR2 × t]
Substituting this value in above equation, we get
t
R 2 MR 2
=
...(5.23)
2
2
But from the theorem of perpendicular axis given by equation (5.7), we have
Izz = Ixx + Iyy
or
(Im)zz = (Im)xx + (Im)yy
And due to symmetry, we have (Im)xx = (Im)yy
∴
(Im)xx = (Im)yy = (Im)zz /2
∴
R
X
O
(Im)zz = ρ × πR2 × t ×
F MR I
GH 2 JK
2
=
MR 2
4
2=
Fig. 5.32 (a)
...(5.24)
5.12.3. Mass Moment of Inertia of a Hollow Circular Cylinder
Let R0 = Outer radius of the cylinder
Ri = Inner radius of the cylinder
L = Length of the cylinder
M = Mass of cylinder
= Density × Volume of cylinder
...(i)
= ρ × π[R02 – Ri2] × L
dm = Mass of a circular ring of radius ‘r’ width ‘dr’ and length L [Refer to Fig. 5.32]
= Density × Volume of ring = ρ × Area of ring × L
= ρ × 2πr dr × L
Now mass moment of inertia of the circular ring about Z-Z axis
= Mass of ring × (radius)2
= (ρ × 2πrdr × L) × r2
The mass moment of inertia of the hollow circular cylinder will be obtained by integrating the above equation between the limits Ri to R0.
∴ Mass moment of inertia of the hollow circular cylinder about Z-Z axis is given by,
(Im)zz=
z
R0
Ri
(ρ × 2πr dr . L) r2
z
Lr O
= ρ × 2π × L
dr = ρ × 2π × L M P
N4Q
L R − R OP
= ρ × 2π × L × M
MN 4 PQ
R0
Ri
4
R0
r3
4
0
Ri
4
i
217
STRENGTH OF MATERIALS
LR
= ρ × 2π × L × M
MN
2
0
OP
PQ
− Ri 2
[R02 + Ri2]
4
[∵ R04 – Ri4 = (R02 – Ri2)(R02 + Ri2)]
= ρ × π[R02 – Ri2] × L ×
( R0 2 + Ri 2 )
2
M ( R0 2 + Ri 2 )
[∵ ρ × π × (R02 – Ri2) = M]
2
( Im ) zz M ( R0 2 + Ri 2 )
=
Now
(Im)xx = (Im)yy =
2
4
5.12.4. Mass Moment of Inertia of a Right Circular Cone of base Radius R,
Height H and Mass M about its Axis
Let
R = Radius of the base of the cone,
H = Height of the cone,
M = Mass of the cone
1
O
= Density × Volume of cone = ρ × πR2 × H
3
Consider an elemental plate of thickness dy and of radius x
y
a
at a distance y from the vertex (as shown in Fig. 5.32 (b)).
x
R
x R
H
∴ x=
×y
We have, tan α = =
H
y H
dy
Mass of the elemental plate,
dm = ρ × Volume
B
= ρ × (πx2 × dy)
=
L R ×y
= ρ × Mπ
N H
2
2
O
× dyP
Q
LM∵
N
OP
Q
R
A
R× y
Fig. 5.32 (b)
n=
2
H
The mass moment of inertia of the circular elemental plate about the axis of the cone (here
axis of the cone is Z-Z axis of the circular elemental plate) is given by equation (5.23) as
(Im)zz =
=
Mass of plate × radius 2
2
(dm) × r 2 dm × x 2
=
2
2
LM
N
= ρ×
=
=
OP
Q
(∵ r = x)
x2
πR 2 y 2
dy
×
×
2
H2
ρ × πR 2 y 2
H2
LM R y OP × 1
NH Q 2
2 2
× dy ×
ρ × πR 4 × y 4
2
LM∵
N
OP
H
Q
LM∵ x = Ry OP
HQ
N
dm = ρ ×
πRy
2
dy
dy
2H 4
Now the total mass moment of inertia of the circular cone will be obtained by integrating
the above equation between the limits 0 to H.
218
CENTRE OF GRAVITY AND MOMENT OF INERTIA
∴
(Im)zz =
=
But mass of cone,
∴
z
H
ρ πR 4 × y 4
dy =
ρπR 4
0
2H 4
ρπR 4
H 5 ρπR 4 × H
×
=
5
2×5
2H 4
2H 4
LM y OP
N5Q
5
×
H
0
ρπR 2 × H
3
ρπR 2 × H R 2 × 3
(Im)zz =
×
3
10
3
3
=M×
R2 =
MR2
10
10
M=
...(5.25)
5.13. PRODUCT OF INERTIA..
Fig. 5.33 shows a body of area A. Consider a small
area dA. The moment of this area about x-axis is
y . dA. Now the moment of y . dA about y-axis is xy dA.
Then xy dA is known as the product of inertia of area dA
with respect to x-axis and y-axis. The integral
z
Y
dA
xy dA is
known as the product of inertia of area A with respect to x
and y axes. This product of inertia is represented by Ixy.
...(5.26)
∴
Ixy = ∫ xy dA
Hence the product of inertia of the plane area is obtained if an elemental area is multiplied by the product of
its co-ordinates and is integrated for entire area.
The product of inertia (Ixy) can also be written mathematically as
Ixy = Σxi yi Ai = x1y1A1 + x2 y2 A2 + ......
where xi yi = co-ordinates of the C.G. of area Ai.
Note. (i) The product of inertia may be positive, negative or zero depending upon distance x and y which could be
positive, negative or zero.
(ii) If area is symmetrical with respect to one or both
of the axes, the product of inertia will be zero as shown in
Fig. 5.34. The total area A is symmetrical about y-axis. The
small area dA which is symmetrical about y-axis has co-ordinates (x, y) and (– x, y). The corresponding product of inertia
for small area are xydA and – xydA respectively. Hence product of inertia for total area becomes zero.
(iii) The product of inertia with respect to centroidal
axis will also be zero.
x
y
O
X
Fig. 5.33
...(5.26A)
Y
dA
dA
x
x
X
Fig. 5.34
Problem 5.17. Fig. 5.35 (a) shows a plane area. Determine the product moment of inertia of the given area. All dimensions are in mm.
Sol. Divide the given area into two parts. The first part is a rectangle and second part is
a right angled triangle. Take x-axis and y-axis as shown in Fig. 5.35 (b). The areas and location
of their C.G. are as follows :
219
STRENGTH OF MATERIALS
Y
40
90
2
C.G 1
90
C.G 2
(20, 45)
(50, 30)
1
O
70
40
30
X
70
(a)
(b)
Fig. 5.35
Area of rectangle, A1 = 90 × 40 = 3600 mm2.
The co-ordinates of C.G. of rectangle À are : x1 = 20 mm, y1 = 45 mm.
90 × 30
Area of triangle,
A2 =
= 1350 mm2.
2
The co-ordinates of C.G. of triangle Á are :
1
1
x2 = 40 +
× 30 = 40 + 10 = 50 mm ; y2 =
× 90 = 30 mm.
3
3
The product of inertia of given area is given by equation (5.26A) as
Ixy = x1y1A1 + x2y2A2
= A1x1y1 + A2x2y2
= 3600 × 20 × 45 + 1350 × 50 × 30
= 3240000 + 2025000 = 5265000 mm4. Ans.
5.14. PRINCIPAL AXES..
The principal axes are the axes about which the product of inertia is zero.
The product of inertia (Ixy) of plane area A with respect to x and y axes is given by equation
(5.26) as
Ixy =
z
z
xy dA
z
But the moment of inertia of plane area A about x-axis [Ixx] or about y-axis [Iyy] is given by
Ixx =
y2 dA
and
Iyy =
x2 dA
The moment of inertia is always positive but product of inertia may be positive (if both
x and y are positive), may be negative (if one co-ordinate is positive and other is negative) or may
be zero (if any co-ordinate is zero).
Fig. 5.36 (a) shows a body of area A. Consider a small area dA. The product of inertia of
the total area A with respect to x and y-axes is given as
Ixy =
220
z
xy dA
...(i)
CENTRE OF GRAVITY AND MOMENT OF INERTIA
y
x1
Total area A
dA
dA
x
y′
y
O
x
x′
O
y1
y′
[Here x′ is + ve, but y ′ is –ve]
(a)
(b)
Fig. 5.36
Let now the axes are rotated anticlockwise by 90° as shown in Fig. 5.36 (b) keeping the
total area A in the same position. Let x1 and y1 are the new axes. The co-ordinates of the same
small area dA with respect to new axes are x′ and y′.
Hence the product of inertia of the total area A with respect to new axes x1 and y1 becomes
as
I x1 y1 =
z
x ′y ′ dA
z
( y)(− x) dA = −
...(ii)
Now let us find the relation between old and new co-ordinates. From Figs. 5.36 (a) and
5.36 (b), we get
x = – y′ and y = x′
or
y′ = – x and x′ = y
Substituting the values of x′ and y′ in equation (ii), we get
I x1 y1 =
z
xy dA = – Ixy
FH∵
z
xy dA = I xy
IK
The above result shows that by rotating the axes through 90°, the product of inertia has
become negative. This means that the product of inertia which was positive previously has now
become negative by rotating the axes through 90°. Hence product of inertia has changed its sign.
It is also possible that by rotating the axes through certain angle, the product of inertia will
become zero. The new axes about which product of inertia is zero, are known as principal axes.
Note. (i) The product of inertia is zero about principal axes.
(ii) As the product of inertia is zero about symmetrical axis, hence symmetrical axis is the principal axis of inertia for the area.
(iii) The product of inertia depends upon the orientation of the axes.
5.15. PRINCIPAL MOMENTS OF INERTIA..
Fig. 5.37 (a) shows a body of area A with respect to old axes (x, y) and new axes (x1, y1). The
new axes x1 and y1 have been rotated through an angle θ in anticlockwise direction. Consider a
small area dA. The co-ordinates of the small area with respect to old axes is (x, y) whereas with
respect to new axes, the co-ordinates are x′ and y′. The new co-ordinates (x′, y′) are expressed in
terms of old co-ordinates (x, y) and angle θ as [Refer to Figs. 5.37 (b) and 5.37 (c)]
221
STRENGTH OF MATERIALS
y
Total area A
y1
Small area dA
x
θ
x1
y′
x′
y
θ
θ
x
O
(a)
y
ys
dA
(x, y)
in θ
y
in θ
)
nθ
sθ
y co
ys
i
(x s
Y1
y
θ
x
θ
os
yc
x′
x co s
θ
in θ
ys
O
X1
y′
sθ
x co
O
X1
θ
x
(x sin θ) θ
x
(b)
(c)
Fig. 5.37
and
x′ = y sin θ + x cos θ
...(i)
y′ = y cos θ – x sin θ
...(ii)
The moment of inertia and product of inertia of area A with respect to old axes are
Ixx =
z
z
z
z
z
z
y2 dA, Iyy =
z
x2 dA and Ixy =
z
z
xy dA.
z
...(5.27)
Also the moment of inertia and product of inertia of area A with respect to new axes will be
I x1 x1 =
(y′)2 dA, I y1 y1 =
(x′)2 dA and
I x1 y1 =
x′y′ dA
Let us substitute the values of x′, y′ from equation (i) and (ii) in the above equations,
we get
I x1x1 =
=
=
=
222
(y′)2 dA
(y cos θ – x sin θ)2 dA
[∵ y′ = y cos θ – x sin θ]
(y2 cos2 θ + x2 sin2 θ – 2xy cos θ sin θ) dA
y2 cos2 θ dA +
z
x2 sin2 θ dA –
z
2xy cos θ sin θ dA
= cos2 θ
z
CENTRE OF GRAVITY AND MOMENT OF INERTIA
y2 dA + sin2 θ
z
x2 dA – 2 cos θ sin θ
z
xy dA
(∵ After rotation, the angle θ is constant and hence
cos2 θ, sin2 θ and 2 cos θ sin θ are constant)
= (cos2 θ)Ixx + (sin2 θ)Iyy – (2 cos θ sin θ)Ixy
...(5.27A)
Similarly
I y1 y1 =
=
=
=
z
z
z
z
FH∵
z
y 2 dA = I xx ,
z
x 2 dA = I yy and
z
xy dA = I xy
IK
(x′)2 dA
(y sin θ + x cos θ)2 dA
[∵ x′ = y sin θ + x cos θ]
(y2 sin2 θ + x2 cos2 θ + 2xy sin θ cos θ) dA
y2 sin2 θ dA +
= sin2 θ
z
z
x2 cos2 θ dA +
y2 dA + cos2 θ
z
z
2xy sin θ cos θ dA
x2 dA + 2 sin θ cos θ
z
xy dA
(∵ θ is constant and hence sin θ and cos θ are constants)
...(5.27B)
= sin2 θ . Ixx + cos2 θ Iyy + 2 sin θ cos θ Ixy
FH∵
z
y 2 dA = I xx ,
Adding equations (5.27A) and (5.27B), we get
z
x 2 dA = I yy and
z
xy dA = I xy
IK
I x1 x1 + I y1 y1 = Ixx [sin2 θ + cos2 θ] + Iyy [sin2 θ + cos2 θ]
+ 2 sin θ cos θ Ixy – 2 sin θ cos θ Ixy
= Ixx + Iyy
[∵ sin2 θ + cos2 θ = 1] ...(5.27C)
Equation (5.27C) shows that sum of moments of inertia about old axes (x, y) and new axes
(x1, y1) are same. Hence the sum of moments of inertia of area A is independent of orientation of
axes. Now let us find the value of I x x − I y y .
1 1
1 1
Subtracting equation (5.27B) from equation (5.27A), we get
I x1 x1 − I y1 y1 = cos2 θ Ixx + sin2 θ Iyy – 2 cos θ sin θ Ixy
– [sin2 θ Ixx + cos2 θ Iyy + 2 cos θ sin θ Ixy]
= Ixx
θ–
θ) + Iyy
θ – cos2 θ) – 4 cos θ sin θ . Ixy
= Ixx (cos2 θ – sin2 θ) – Iyy (cos2 θ – sin2 θ) – 4 cos θ sin θ Ixy
= (Ixx – Iyy) (cos2 θ – sin2 θ) – 2 × 2 cos θ sin θ × Ixy
= (Ixx – Iyy) cos 2θ – 2Ixy sin2 θ
...(5.27D)
cos
1
2
1
+
θ
− cos 2θ
∵ cos 2 θ =
, sin 2 θ =
2
2
(cos2
sin2
(sin2
FG
H
∴ cos2 θ – sin2 θ = cos 2θ and 2 sin θ cos θ = sin 2θ
Now let us find the values of I x1 x1 and I y1 y1 in terms of Ixx, Iyy and θ.
Adding equations (5.27C) and (5.27D), we get
2 I x1 x1 = [Ixx + Iyy] + [(Ixx – Iyy) cos 2θ – 2Ixy sin 2θ]
223
I
K
STRENGTH OF MATERIALS
( I xx + I yy )
I x1x1 =
or
+
( I xx − I yy )
cos 2θ – Ixy sin 2θ
...(5.27E)
2
2
To find the values of I y1 y1 , subtract equation (5.27D) from (5.27C). Now subtracting
equation (5.27D) from equation (5.27C), we get
2 I y1 y1 = (Ixx + Iyy) – [(Ixx – Iyy) cos 2θ – 2Ixy sin 2θ]
( I xx + I yy )
I y1 y1 =
∴
2
Product of Inertia about New Axes
−
( I xx − I yy )
2
cos 2θ + Ixy sin 2θ
...(5.27F)
Let us now find the value of I x1 y1 in terms of Ixy and angle θ.
I x1 y1 =
We know that
z
z
z
z
( x ′)( y ′) dA
Substituting the values of x′ and y′, we get
I x1 y1 =
I x1 y1 =
or
=
(y sin θ + x cos θ)(y cos θ – x sin θ) dA
(∵ x′ = y sin θ + x cos θ and y′ = y cos θ – x sin θ)
(y2 sin θ cos θ – xy sin2 θ + xy cos2 θ – x2 cos θ sin θ) dA
y2 sin θ cos θ dA –
= sin θ cos θ
I x1 y1 =
=
z
2 sin θ cos θ
2
z
xy sin2 θ dA +
y2 dA – sin2
z
z
z
xy cos2 θ dA –
xy dA + cos2 θ
=
=
I yy
I xx
sin 2θ + Ixy (cos2 θ – sin2 θ) –
sin θ
2
2
2
( I xx − I yy )
2
xy dA – cos θ sin θ
y2 dA – sin2 θ Ixy + cos2 θ Ixy –
sin 2θ
sin 2θ
. Ixx + Ixy (cos2 θ – sin2 θ) –
Iyy
2
2
( I xx − I yy )
x2 cos θ sin θ dA
z
x2 dA
(∵ θ is constant and hence sin θ, cos θ are constants)
FH∵
=
z
z
z
2 cos θ sin θ
2
FH∵
y 2 dA = I xx ,
z
x2 dA
z
xy dA = I xy
IK
z
x 2 dA = I yy
IK
sin 2θ + Ixy (cos2 θ – sin2 θ)
sin 2θ + Ixy cos 2θ
...(5.27G)
(∵ cos2 θ – sin2 θ = cos 2θ)
Direction of Principal Axes
We have already defined the principal axes. Principal axes are the axes about which the
product of inertia is zero. Now the new axes (x1, y1) will become principal axes if the product of
inertia given by equation (5.27G) is zero (i.e., I x1 y1 = 0).
224
CENTRE OF GRAVITY AND MOMENT OF INERTIA
∴ For principal axes, I x1 y1 = 0
or
( I xx − I yy )
2
sin 2θ + Ixy cos 2θ = 0
( I xx − I yy )
or
2
sin 2θ = – Ixy cos 2θ
or
− 2 I xy
2 I xy
sin 2θ
=
=
cos 2θ I xx − I yy I yy − I xx
or
tan 2θ =
2I xy
...(5.27H)
I yy − I xx
The above equation will give the two values of 2θ or θ. These two values of θ will differ by
90°. By substituting the values of θ in equations (5.27E) and (5.27F), the values of principal
moments of inertia ( I x1 x1 and I y1 y1 ) can be obtained. If from equation (5.27H), the values of
sin 2θ and cos 2θ in terms of Ixy, Ixx and Iyy are substituted in equation (5.27E), we get
I xx + I yy
±
( I xx − I yy ) 2
+ I xy2 .
2
2
These are the values of principal moment of inertia.
Problem 5.18. For the section shown in Fig. 5.38 (a) determine :
(i) Moment of inertia about its centroid along (x, y) axis.
(ii) Moment of inertia about new axes which is turned through an angle of 30° anticlockwise to the old axis.
(iii) Principal moments of inertia about its centroid.
All dimensions are in cm.
Sol. Given :
Fig. 5.38 (a) shows the given section. It is symmetrical about x-axis. The C.G. of the
section lies at O (origin of the axes). To find moment of inertia of the given section, it is divided
into three rectangles as shown in Fig. 5.38 (b). First the moment of inertia of each rectangle
about its centroid is calculated. Then by using parallel axis theorem, the moment of inertia of the
given section about its centroid is obtained.
I x1 x1 =
y
10
30
40
20
C.G.
10
O
1
30
40
40
30
C.G.2
2
x
C.G. 3 10
30
30
3
10
30
10
(a)
(b)
Fig. 5.38
225
STRENGTH OF MATERIALS
(a) Consider rectangle (1)
The C.G. of rectangle (1) is at a distance of 20 cm from x-axis and at a distance of 25 cm
from y-axis.
(Ixx)1 = (IG)1x + A1(k1x)2
...(1)
where (Ixx)1 = M.O.I. of rectangle (1) about x-axis passing through the centroid of the given figure
of the given section.
(IG)1x = M.O.I. of rectangle (1) about an axis passing through C.G. of rectangle (1) and
parallel to
x-axis =
bd 3
12
10 × 303
(Here b = 10 and d = 30)
12
= 2.25 × 104 cm4
A1 = Area of rectangle (1) = 10 × 30 = 300 cm2
(k1x) = Distance of C.G. of rectangle (1) from x-axis
= 20
Substituting the above values in equation (1), we get
(Ixx)1 = 2.25 × 104 + 300 × 202
= 2.25 × 104 + 12 × 104
= 14.25 × 104 cm4
...(A)
Similarly, the M.O.I. of rectangle (1) about y-axis passing through the centroid of the
given figure is given by,
(Iyy)1 = (IG)1y + A1(k1y)2
=
where
bd 3 30 × 10 3
= 0.25 × 104 cm4
=
12
12
(k1y) = Distance of C.G. of rectangle (1) from y-axis = 25
(∵ A1 = 300)
∴
(Iyy)1 = 0.25 × 104 + 300 × 25
4
4
= 0.25 × 10 + 18.75 × 10
= 19 × 104 cm4
...(B)
(b) Consider rectangle (2)
The C.G. of this rectangle coincides with the C.G. of the given section. Hence
(IG)1y =
(Ixx)2 =
bd 3 60 × 10 3
= 0.5 × 104 cm4
=
12
12
...(C)
10 × 60 3
= 18 × 104 cm4
...(D)
12
(c) Consider rectangle (3)
The C.G. of rectangle (3) is at a distance of 20 cm from x-axis and at a distance of 25 cm
from y-axis. Hence k3x = 20 cm and k3y = 25 cm.
Now
(Ixx)3 = (IG)3x + A3(k3x)2
and
(Iyy)2 =
=
226
10 × 30 3
+ (10 × 30)(20)2 = 2.25 × 104 + 12 × 104 = 14.25 × 104 cm4
12
CENTRE OF GRAVITY AND MOMENT OF INERTIA
(Iyy)3 = (IG)3y + A3(k3 y)2
and
30 × 10 3
+ 300 × 252 = 0.25 × 104 + 18.75 × 104 = 19 × 104 cm4.
12
(i) Moment of inertia of complete section about its centroid
Ixx = (Ixx)1 + (Ixx)2 + (Ixx)3
= 14.25 × 104 + 0.5 × 104 + 14.25 × 104 cm4
= 29 × 104 cm4. Ans.
and
Iyy = (Iyy)1 + (Iyy)2 + (Iyy)3
= 19 × 104 + 18 × 104 + 19 × 104
= 56 × 104 cm4. Ans.
(ii) Moment of inertia of complete section
y
about new axes which is turned through an angle
C.G.1
of 30° anticlockwise.
20
1
Here θ = 30°.
Let us first calculate the product of inertia of
2
C.G.2
x
whole area about old axes x, y.
20
3
(a) Consider rectangle (1)
25
2
C.G.3
A1 = 10 × 30 = 300 cm
The C.G. of rectangle (1) is at a distance of 20 cm
25
above x-axis and at a distance of 25 cm from y-axis.
Hence co-ordinates of this C.G. are
Fig. 5.38 (c)
x1 = – 25 cm and y1 = 20 cm.
(b) For rectangle (2)
A2 = 10 × 60 = 600 cm2. The C.G. of rectangle (2) lies on the origin (O). Hence x2 = 0 and
y2 = 0.
(c) For rectangle (3)
A3 = 10 × 30 = 300 cm2
The C.G. of rectangle (3) is at a distance of 20 cm below x-axis and at a distance of 25 cm
from y-axis.
Hence co-ordinate of this C.G. are : x3 = 25 cm and y3 = (– 20 cm).
The product of inertia (Ixy) of the whole figure is given by equation (5.26A) as
Ixy = A1x1y1 + A2x2y2 + A3x3y3
= 300 × (– 25) × 20 + 600 × 0 × 0 + 300 × 25 × (– 20)
= – 15 × 104 + 0 + (– 15 × 104)
y
y1
= – 30 × 104 cm4
x1
Now the moment of inertia of the complete section
q
about the new axes (x1, y1) can be obtained from equations
(5.27E) and (5.27F) as
q
=
I x1x1 =
( I xx + I yy )
+
( I xx − I yy )
x
cos 2θ – Ixy sin 2θ
2
2
where Ixx = 29 × 104 cm4, Iyy = 56 × 104 cm4,
Ixy = – 30 × 104 cm4 and θ = 30°
Fig. 5.38 (d)
227
STRENGTH OF MATERIALS
∴
29 × 10 4 + 56 × 10 4
29 × 10 4 − 56 × 10 4
+
cos 60° – (– 30 × 104) sin 60°
2
2
1
= 42.5 × 104 – 13.5 × 10 × + 30 × 104 × 0.866
2
= 35.75 × 104 + 26 × 104 = 61.75 × 104 cm4. Ans.
I x1 x1 =
I y1 y1 =
and
I xx + I yy
2
−
I xx − I yy
2
cos θ + Ixy sin 2θ
29 × 10 4 + 56 × 10 4 29 × 10 4 − 56 × 10 4
cos 60° + (– 30 × 104) sin 60°
−
2
2
= 42.5 × 104 + 6.75 × 104 – 26 × 104 = 23.25 × 104 cm4. Ans.
=
(iii) Principal moments of inertia about the centroid
The principal moments of inertia are the moments
of inertia about the principal axes.
The direction of principal axes is given by equation (5.27H) as
tan 2θ =
=
=
2I xy
y
y1
57.12°
32.88°
I yy − I xx
x
2 × (− 30 × 10 4 )
56 × 10 4 − 29 × 10 4
− 60 × 10 4
x1
= – 2.222
Fig. 5.38 (e)
27 × 10 4
As 2θ is negative, hence it lies in 2nd and 4th quadrant.
∴
2θ = tan–1 (– 2.222)
= – 65.77° and 114.23°
or
θ = – 32.88° and 57.12°
The +ve angle is taken anti-clock and – ve angle is taken clockwise to the existing axes
x and y. The principal axes are shown as x1 and y1 in Fig. 5.38 (e). The moment of inertia along
these axes is the principal moment of inertia. Hence by substituting θ = – 32.88° and 57.12°, in
equations (5.27E) and (5.27F), we get principal moment of inertia.
∴
FI I
GH JK
max.
=
x1 x1
min.
I xx + I yy
2
+
I xx − I yy
2
cos 2θ – Ixy sin 2θ
29 × 10 4 + 56 × 10 4 29 × 10 4 − 56 × 10 4
+
2
2
× cos (– 2 × 32.88) – (– 30 × 104) sin (– 2 × 32.88)
[∵ θ = – 32.88°]
= 42.5 × 104 – 13.5 × 104 × 0.41 + 30 × 104 × (– 0.912)
= 42.5 × 104 – 5.535 × 104 – 27.36 × 104
= 9.605 × 104 cm4
=
228
CENTRE OF GRAVITY AND MOMENT OF INERTIA
and
FI I
GH JK
max.
=
y1 y1
min.
I xx + I yy
2
−
I xx − I yy
2
cos 2θ + Iyy sin 2θ
= 42.5 × 104 + 5.535 × 104 + 27.36 × 104 = 75.395 × 104 cm4
Hence principal moment of inertia are
Imax. = 75.395 × 104 cm4. Ans.
Imin. = 9.605 × 104 cm4. Ans.
Alternate Method
The principal moments of inertia can also be obtained by
Imax. =
min.
=
I xx + I yy
2
±
( I xx − I yy ) 2
2
29 × 10 4 + 56 × 10 4
±
2
= 42.5 × 104 ±
+ I xy 2
(29 × 10 4 − 56 × 10 4 ) 2
+ (− 30 × 10 4 ) 2
2
(− 13.5 × 10 4 ) 2 + (− 30 × 10 4 ) 2
= 42.5 × 104 ± 104 × 32.89
= (42.5 + 32.89) × 104 and (42.5 – 32.89) × 104
= 75.39 × 104 and 9.61 × 104 cm4
∴
Imax. = 75.39 × 104 cm4 and Imin. = 9.61 × 104 cm4
Now Imax. and Imin. are the required principal moment of inertia. Ans.
HIGHLIGHTS
1. The point, through which the whole weight of the body acts, is known as centre of gravity.
2. The point, at which the total area of a plane figure is assumed to be concentrated, is known as
centroid of that area. The centroid and centre of gravity are at the same point.
3. The centre of gravity of a uniform rod lies at its middle point.
4. The C.G. of a triangle lies at a point where the three medians of a triangle meet.
5. The C.G. of a parallelogram or a rectangle is at a point where its diagonal meet each other.
6. The C.G. of a circle lies at its centre.
7. The C.G. of a body consisting of different areas is given by
x=
a1x1 + a2 x2 + a3 x3 + ......
a1 + a2 + a3 + ......
and
y=
a1 y1 + a2 y2 + a3 y3 + ......
a1 + a2 + a3 + ......
where x and y = Co-ordinates of the C.G. of the body from axis of reference
a1, a2, a3, ...... = Different areas of the sections of the body
= Distances of the C.G. of the areas a1, a2, a3, ...... from Y-axis
x1, x2, x3, ......
= Distances of the C.G. of the areas a1, a2, a3, ...... from X-axis.
y1, y2, y3, ......
8. If a given section is symmetrical about X-X axis or Y- Y axis, the C.G. of the section will lie on the
axis symmetry.
9. The moment of inertia of an area (or mass) about an axis is the product of area (or mass) and
square of the distance of the C.G. of the area (or mass) from that axis. It is represented by I.
229
STRENGTH OF MATERIALS
10. Radius of gyration of a body (or a given lamina) is the distance from an axis of reference where the
whole mass (or area) of the given body is assumed to be concentrated so as not to after the
moment of inertia about the given axis. It is represented by k. Mathematically, k =
I
.
A
11. According to theorem of perpendicular axis IZZ = IXX + IYY where IXX and IYY = Moment of inertia of
a plane section about two mutually perpendicular axes X-X and Y-Y in the plane of the section,
IZZ = Moment of inertia of the section perpendicular to the plane and passing through the intersection of X-X and Y-Y axes.
12. According to the theorem of parallel axis IAB = IG + Ah2 , where
IG = Moment of inertia of a given area about an axis passing through C.G. of the area
IAB = Moment of inertia of the given area about an axis AB, which is parallel to the axis passing
through G
h = Distance between the axis passing through G and axis AB
A = Area of the section.
13. Moment of inertia of a rectangular section :
(i) about an horizontal axis passing through C.G. =
(ii) about an horizontal axis passing through base =
bd3
12
bd3
.
3
πD4
.
64
Moment of inertia of a triangular section :
14. Moment of inertia of a circular section =
15.
(i) about the base =
bh3
12
(ii) about an axis passing through C.G. and parallel to the base =
bh3
.
36
where b = Base width, and h = Height of the triangle.
16. The C.G. of an area by integration method is given by
x=
where
z x* dA
z dA
and
z x* dL
z dL
and
y=
z y* dA
z dA
x* = Distance of C.G. of area dA from y-axis
y* = Distance of C.G. of area dA from x-axis.
17. The C.G. of a straight or curved line is given by
x=
y=
z y* dL .
z dL
EXERCISE
(A) Theoretical Questions
1. Define centre of gravity and centroid.
2. Derive an expression for the centre of gravity of a plane area using method of moments.
230
CENTRE OF GRAVITY AND MOMENT OF INERTIA
3. What do you understand by axes of reference ?
4. Define the terms : moment of inertia and radius of gyration.
5. State the theorem of perpendicular axis. How will you prove this theorem ?
6. State and prove the theorem of parallel axis.
7. Find an expression for the moment of inertia of a rectangular section :
(i) about an horizontal axis passing through the C.G. of the rectangular section, and
(ii) about an horizontal axis passing through the base of the rectangular section.
8. Prove that the moment of inertia of a circular section about an horizontal axis (in the plane of
the circular section) and passing through the C.G. of the section is given by
πD4
.
64
9. Prove that moment of inertia of a triangular section about the base of the section
bh3
12
b = Base of triangular section, and
=
where
h = Height of triangular section.
10.
Derive an expression for the moment of inertia of a triangular section about an axis passing
through the C.G. of the section and parallel to the base.
11.
Show that IO = IG + Ah2, where IG is the moment of inertia of a lamina about an axis through its
centroid and lying in its plane and h is the distance from the centroid to a parallel axis in the
same plane about which its moment of inertia is IO, A being the area of the lamina.
12.
State and prove the parallel axes theorem on moment of inertia for a plane area.
13.
Prove that the moment of area of any plane figure about a line passing through its centroid is
zero.
14.
Show that the product of inertia of an area about two mutually perpendicular axis is zero, if the
area is symmetrical about one of these axis.
(U.P. Tech. University, 2002-2003)
15.
Determine an expression for mass moment of inertia of hollow steel cylinder of mass M, outer
radius R0, inner radius Ri and length L about its axis. The hole in the cylinder is concentric.
16.
(U.P. Tech. University, 2002-2003)
Derive an expression for mass moment of inertia of a right circular cone of base radius R, height
H and mass M about its axis.
(U.P. Tech. University, 2001-2002)
(B) Numerical Problems
1. Find the centre of gravity of the T-section shown in Fig. 5.39.
[Ans. 8.272 cm]
12 cm
2 cm
12 cm
2 cm
Fig. 5.39
231
STRENGTH OF MATERIALS
2. Find the centre of gravity of the I-section shown in Fig. 5.40.
[Ans. 6.44 cm]
[Hint. a1 = 8 × 2 = 16 cm2, a2 = 12 × 2 = 24 cm2,
a3 = 16 × 2 = 32 ; y1 = 2 + 12 + 1 = 15,
y2 = 2 + 6 = 8, y3 = 1
∴
a y + a2 y2 + a3 y3
y= 1 1
a1 + a2 + a3
=
8 cm
2 cm
1
12 cm 2
2
cm
16 × 15 + 24 × 8 + 32 × 1
16 + 24 + 32
3
240 + 192 + 32 464
=
= 6.44 cm .]
72
72
3. (a) Find the centre of gravity of the L-section shown in Fig. 5.41.
=
2 cm
16 cm
Fig. 5.40
[Ans. x = 1.857, y = 3.857]
2 cm
9 cm
10 cm
2 cm
6 cm
Fig. 5.41
(b) Find the moment of inertia of ISA 100 × 75 × 6 about the centroidal XX and YY axis, shown in
Fig. 5.41 (a).
(U.P. Tech. University, 2001-2002)
6
1
100
69
2
75 mm
Fig. 5.41 (a)
[Hint. Locate first x and first y
a1 = 100 × 6 = 600 mm2, x1 = 3 mm, y1 = 50
69
= 40.5
a2 = 69 × 6 = 414 mm2, x2 = 6 +
2
y2 = 3 mm
a x + a2 x2 600 × 3 + 414 × 40.5
= 18.31 mm
∴ x = 1 1
=
a1 + a2
600 + 414
a1 y1 + a2 y2 600 × 50 + 414 × 3
=
= 30.81 mm
y =
a1 + a2
600 + 414
232
6
CENTRE OF GRAVITY AND MOMENT OF INERTIA
Now find the moment of inertia about centroidal X-X axis :
IXX1 = ( IG1 )x + a1h12
=
6 × 1003
6 × 1003
+ 600 × (y1 – y )2 =
+ 600(50 – 30.81)2
12
12
= 720.95 × 103 mm4.
IXX2 = ( IG2 )x + a2h22 =
69 × 63
+ 414(y2 – y )2
12
69 × 63
+ 414(3 – 30.81) = 321.428 × 103
12
= IXX1 + IXX2 = 720.95 × 103 + 321.428 × 103 = 1042.378 × 103 mm4.
=
∴
IXX
To find M.O.I. about centroidal axis Y-Y
IYY1 = ( IG1 )y + a1(x1 – x )2 =
100 × 63
+ 600(3 – 18.31)2 = 142.437 × 103 mm4
12
6 × 693
+ 414(40.5 – 18.31)2 = 368.1 × 103 mm4
12
= IYY1 + IYY2 = (142.437 + 368.1) × 103 mm4 = 510.537 × 103 mm4. ]
IYY2 = ( IG2 )y + a2(x2 – x )2 =
∴
IYY
4. From a rectangular lamina ABCD 10 cm × 14 cm a rectangular hole of 3 cm × 5 cm is cut as shown
in Fig. 5.42. Find the centre of gravity of the remaining lamina.
[Ans. x = 4.7 cm, y = 6.444 cm]
10 cm
14 cm
3
1
1
cm cm cm
5 cm
2 cm
Fig. 5.42
5. For the T-section shown in Fig. 5.39, determine the moment of inertia of the section about the
horizontal and vertical axes, passing through the centre of gravity of the section.
[Ans. 567.38 cm4, 294.67 cm4]
6. For the I-section shown in Fig. 5.40, find the moment of inertia about the centroidal axis X-X
perpendicular to the web.
[Ans. 2481.76 cm4]
7. Locate the C.G. of the area shown in Fig. 5.43 with respect to co-ordinate axes. All dimensions are
in mm.
[Hint. a1 = 10 × 30 = 300 mm2, x1 = 5 mm, y1 = 15 mm,
a2 = 40 × 10 = 400 mm2, x2 = 10 + 20 = 30 mm,
y2 = 5 mm
a3 = 10 × 20 = 200 mm2, x3 = 5 mm,
y3 = – 10 mm
233
STRENGTH OF MATERIALS
a4 = 10 × 10 = 100 mm2, x4 = 45 mm,
Y 10
y4 = 10 + 5 = 15 mm
x=
a1x1 + a2 x2 + a3 x3 + a4 x4
(a1 + a2 + a3 + a4 )
1500 + 12000 + 1000 + 4500
=
1000
= 1.5 + 12 + 1 + 4.5 = 19 mm.
10
20
1
10
4
50
10
2
O
a y + a2 y2 + a3 y3 + a4 y4
y= 1 1
(a1 + a2 + a3 + a4 )
X
40
3
4500 + 2000 − 2000 + 1500
Fig. 5.43
1000
= 4.5 + 2 – 2 + 1.5 = 6 mm. ]
8. A thin homogeneous wire is bent into a triangular shape ABC such that AB = 240 mm, BC =
260 mm and AC = 100 mm. Locate the C.G. of the wire with respect to co-ordinate axes. Angle at
A is right angle.
=
[Hint. First determine angles α and β. Use sine rule
Y
BC
AC
AB .
=
=
sin 90° sin α sin β
a
b
B
240
AB
α = 22.62°. Also sin β =
× sin 90° =
260
BC
β = 67.38°
D
m
∴
90°
0m
∴
0
24
10
∴
AC × sin 90° 100
=
sin α =
BC
260
A
mm
C
260 mm
Fig. 5.44
Using equation 5.2 (E) and 5.2 (F)
x=
L1x1 + L2 x2 + L3 x3
, where L1 = AB = 240,
( L1 + L2 + L3 )
x1 = distance of C.G. of AB from y-axis
240
× cos α = 120 × cos 22.62° = 110.77 mm
2
L2 = BC = 260 mm, x2 = Distance of C.G. of BC from y-axis = 130
L3 = AC = 100 mm, x3 = Distance of C.G. of AC from y-axis
100
cos β = 240 cos α + 50 cos β
= BD +
2
= 240 × cos 22.62° + 50 cos 67.38° = 240.77
=
∴
x=
240 × 110.77 + 260 × 130 + 100 × 240.77
= 140.77 mm.
240 + 260 + 100
y=
L1 y1 + L2 y2 + L3 y3 , where y = 240 sin α = 120 × sin 22.62° = 46.154
1
2
L1 + L2 + L3
100
sin β = 50 sin 67.38° = 48.154
2
240 × 46.154 + 260 × 0 + 100 × 46.154
y=
600
= 26.154 mm.]
y2 = 0, y3 =
∴
234
X
CENTRE OF GRAVITY AND MOMENT OF INERTIA
9. Determine the C.G. of the uniform plane lamina shown in Fig. 5.45. All dimensions are in cm.
[Hint. The figure is symmetrical about y-y axis.
y=
∴
where
Y
a1 y1 + a2 y2 + a3 y3 + a4 y4
a1 + a2 + a3 + a4
a1 = 40 × 30 = 1200 cm2, y1 =
a2 = 30 × 20 = 600 cm2, y2 = 30 +
a3 = –
a4 = –
∴
y=
=
4
30
= 15 cm
2
30
= 45 cm
2
90°
2
20
10
π × 102
4 r 4 × 10 40
=
=
= – 50π, y3 =
2
3π
3π
3π
10
10
1
30
10 170
20 × 10
=
= – 100 cm2, y4 = 60 –
3
3
2
3
40
170
− 100 ×
3π
3
1200 + 600 − 50 π − 100
1200 × 15 + 600 × 45 − 50 π ×
10
10
O
20
X
10
Fig. 5.45
18000 + 27000 − 666.7 − 5666.7
1700 − 50 π
38666.6
= 25.06 cm from origin O]
1542.92
10. From a circular plate of diameter 100 mm a circular part of diameter 50 mm is cut as shown in
Fig. 5.46. Find the centroid of the remainder.
(U.P. Tech. University, 2002-2003)
=
y
100 mm
x′
x
y′
50 mm
Fig. 5.46
[Hint. Fig. 5.46 is symmetrical about x-axis. Hence centroid lies on x-axis.
∴
y = 0.6. The value of x is given by x =
a1x1 + a2 x2
a1 − a2
π
100
× 1002 = 7853.98 mm2, x1 =
= 50 mm
4
2
π
× 502 = – 1963.5 mm2, x2 = 100 – 25 = 75 mm
a2 = –
4
7853.98 × 50 − 1963.5 × 75
x=
= 41.67 mm
∴
7853.98 − 1963.5
Hence centroid is at (41.67 mm, 0)]
But
a1 =
FG
H
IJ
K
235
6
CHAPTER
SHEAR FORCE AND
BENDING MOMENT
6.1. INTRODUCTION..
The algebraic sum of the vertical forces at any section of a beam to the right or left of the
section is known as shear force. It is briefly written as S.F. The algebraic sum of the moments of
all the forces acting to the right or left of the section is known as bending moment. It is written
as B.M. In this chapter, the shear force and bending moment diagrams for different types of
beams (i.e., cantilevers, simply supported, fixed, overhanging etc.) for different types of loads
(i.e., point load, uniformly distributed loads, varying loads etc.) acting on the beams, will be
considered.
6.2. SHEAR FORCE AND BENDING MOMENT DIAGRAMS..
A shear force diagram is one which shows the variation of the shear force along the length
of the beam. And a bending moment diagram is one which shows the variation of the bending
moment along the length of the beam.
Before drawing the shear force and bending moment diagrams, we must know the different
types of beams and different types of load acting on the beams.
6.3. TYPES OF BEAMS..
The following are the important types of beams :
1. Cantilever beam,
2. Simply supported beam,
3. Overhanging beam,
4. Fixed beams, and
5. Continuous beam.
6.3.1. Cantilever Beam. A beam which is fixed at one end and free at the other end, is
known as cantilever beam. Such beam is shown in Fig. 6.1.
Fig. 6.1
Fig. 6.2
6.3.2. Simply Supported Beam. A beam supported or resting freely on the supports at
its both ends, is known as simply supported beam. Such beam is shown in Fig. 6.2.
237
STRENGTH OF MATERIALS
6.3.3. Overhanging Beam. If the end portion of a beam is extended beyond the support,
such beam is known as overhanging beam. Overhanging beam is shown in Fig. 6.3.
Simply supported
portion
Overhanging
portion
Support
Fig. 6.3
Fig. 6.4
6.3.4. Fixed Beams. A beam whose both ends are fixed or built-in walls, is known as fixed
beam. Such beam is shown in Fig. 6.4. A fixed beam is also known as a built-in or encastred beam.
6.3.5. Continuous Beam. A beam which is provided more than two supports as shown in Fig. 6.5, is
known as continuous beam.
6.4. TYPES OF LOAD..
Fig. 6.5
A beam is normally horizontal and the loads acting on the beams are generally vertical.
The following are the important types of load acting on a beam :
1. Concentrated or point load,
2. Uniformly distributed load, and
3. Uniformly varying load.
6.4.1. Concentrated or Point Load. A concentrated load is one which is considered to
act at a point, although in practice it must really be distributed over a small area. In Fig. 6.6, W
shows the point load.
W
Fig. 6.6
w N/m
Fig. 6.7
6.4.2. Uniformly Distributed Load. A uniformly distributed load is one which is spread
over a beam in such a manner that rate of loading w is uniform along the length (i.e., each unit
length is loaded to the same rate) as shown in Fig. 6.7. The rate of loading is expressed as w
N/m run. Uniformly distributed load is, represented by u.d.l.
For solving the numerical problems, the total
uniformly distributed load is converted into a point load,
acting at the centre of uniformly distributed load.
6.4.3. Uniformly Varying Load. A uniformly varying load is one which is spread over a beam in such a manner that rate of loading varies from point to point along the
beam as shown in Fig. 6.8 in which load is zero at one end
Fig. 6.8
and increases uniformly to the other end. Such load is known
as triangular load.
238
SHEAR FORCE AND BENDING MOMENT
For solving numerical problems the total load is equal to the area of the triangle and this
total load is assumed to be acting at the C.G. of the triangle i.e., at a distance of 23 rd of total
length of beam from left end.
6.5. SIGN CONVENTIONS FOR SHEAR FORCE AND BENDING MOMENT..
(i) Shear force. Fig. 6.9 shows a simply supported beam AB, carrying a load of 1000 N
at its middle point. The reactions at the supports will be equal to 500 N. Hence RA = RB =
500 N.
Now imagine the beam to be divided into two portions by the section X-X. The resultant of
the load and reaction to the left of X-X is 500 N vertically upwards. (Note in this case, there is no
load to the left of X-X). And the resultant of the load and reaction to the right of X-X is (1000 ↓ –
500 ↑ = 500 ↓ N) 500 N downwards. The resultant force acting on any one of the parts normal to
the axis of the beam is called the shear force at the section X-X. Here the shear force at the
section X-X is 500 N.
The shear force at a section will be considered positive when the resultant of the forces to
the left to the section is upwards, or to the right of the section is downwards. Similarly the shear
force at a section will be considered negative if the resultant of the forces to the left of the section
is downwards, or to the right of the section is upwards. Here the resultant force to the left of the
section is upwards and hence the shear force will be positive.
X
A
RA
1000 N
C
RB
X
X
B
Convexity
500 N
X
500 N
Fig. 6.9
Convexity
Concavity
1000 N
(a) Positive B.M.
Concavity
(b) Negative B.M.
Fig. 6.10
(ii) Bending moment. The bending moment at a section is considered positive if the bending
moment at that section is such that it tends to bend the beam to a curvature having concavity at
the top as shown in Fig. 6.10 (a). Similarly the bending moment (B.M.) at a section is considered
negative if the bending moment at that section is such that it tends to bend the beam to a
curvature having convexity at the top as shown in Fig. 6.10 (b). The positive B.M. is often called
sagging moment and negative B.M. as hogging
1000 N
X
moment.
A
B
C
Consider the simply supported beam AB,
carrying a load of 1000 N at its middle point. Reactions
1m
RA and RB are equal and are having magnitude 500 N
X
as shown in Fig. 6.11. Imagine the beam to be divided
2m
2m
into
two
portions
by
the
section
RB = 500 N
RA = 500 N
X-X. Let the section X-X is at a distance of 1 m from A.
Fig. 6.11
239
STRENGTH OF MATERIALS
The moments of all the forces (i.e., load and reaction) to the left of X-X at the section
X-X is RA × 1 = 500 × 1 = 500 Nm (clockwise). Also the moments of all the forces (i.e., load and
reaction) to the right of X-X at the section X-X is RB × 3 (anti-clockwise) – 1000 × 1 (clockwise)
= 500 × 3 Nm – 1000 × 1 Nm = 1500 – 1000 = 500 Nm (anti-clockwise).
Hence the tendency of the bending moment at X-X is to bend the beam so as to produce
concavity at the top as shown in Fig. 6.12.
X
X
X
X
Fig. 6.12
Fig. 6.13
The bending moment at a section is the algebraic sum of the moments of forces and
reactions acting on one side of the section. Hence bending moment at the section X-X is 500 Nm.
The bending moment will be considered positive when the moment of the forces and reaction
on the left portion is clockwise, and on the right portion anti-clockwise. In Fig. 6.12, the bending
moment at the section X-X is positive.
Similarly the bending moment will be considered negative when the moment of the forces
and reactions on the left portion is anti-clockwise, and on the right portion clockwise as shown in
Fig. 6.13. In Fig. 6.13, the bending moment at the section X-X is negative.
6.6. IMPORTANT POINTS FOR DRAWING SHEAR FORCE AND BENDING MOMENT
DIAGRAMS
In Art. 6.2, it is mentioned that the shear force diagram is one which shows the variation
of the shear force along the length of the beam. And a bending moment diagram is one which
show the variation of the bending moment along the length of beam. In these diagrams, the
shear force or bending moment are represented by ordinates whereas the length of the beam
represents abscissa.
The following are the important points for drawing shear force and bending moment
diagrams :
1. Consider the left or the right portion of the section.
2. Add the forces (including reaction) normal to the beam on one of the portion. If right
portion of the section is chosen, a force on the right portion acting downwards is positive while a
force acting upwards is negative.
If the left portion of the section is chosen, a force on the left portion acting upwards is
positive while a force acting downwards is negative.
3. The positive values of shear force and bending moments are plotted above the base line,
and negative values below the base line.
4. The shear force diagram will increase or decrease suddenly i.e., by a vertical straight
line at a section where there is a vertical point load.
5. The shear force between any two vertical loads will be constant and hence the shear
force diagram between two vertical loads will be horizontal.
240
SHEAR FORCE AND BENDING MOMENT
6. The bending moment at the two supports of a simply supported beam and at the free
end of a cantilever will be zero.
6.7.SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR A CANTILEVER WITH
A POINT LOAD AT THE FREE END
Fig. 6.14 shows a cantilever AB of length L fixed at A and free at B and carrying a point
load W at the free end B.
W
x
X
( a)
A
B
L
(b)
+
W
A
W
B
S.F. diagram
Base line
Base line
A
B
–
(c) W × L
C
W×x
B.M. diagram
Fig. 6.14
Let
Fx = Shear force at X, and
Mx = Bending moment at X.
Take a section X at a distance x from the free end. Consider the right portion of the
section.
The shear force at this section is equal to the resultant force acting on the right portion at
the given section. But the resultant force acting on the right portion at the section X is W and
acting in the downward direction. But a force on the right portion acting downwards is considered
positive. Hence shear force at X is positive.
∴
Fx = + W
The shear force will be constant at all sections of the cantilever between A and B as there
is no other load between A and B. The shear force diagram is shown in Fig. 6.14 (b).
Bending Moment Diagram
The bending moment at the section X is given by
Mx = – W × x
...(i)
(Bending moment will be negative as for the right portion of the section, the moment of W
at X is clockwise. Also the bending of cantilever will take place in such a manner that convexity
will be at the top of the beam).
241
STRENGTH OF MATERIALS
From equation (i), it is clear that B.M. at any section is proportional to the distance of the
section from the free end.
At x = 0 i.e., at B, B.M. = 0
At x = L i.e., at A, B.M. = W × L
Hence B.M. follows the straight line law. The B.M. diagram is shown in Fig. 6.14 (c). At
point A, take AC = W × L in the downward direction. Join point B to C.
The shear force and bending moment diagrams for several concentrated loads acting on a
cantilever, will be drawn in the similar manner.
Problem 6.1. A cantilever beam of length 2 m carries the point loads as shown in
Fig. 6.15. Draw the shear force and B.M. diagrams for the cantilever beam.
Sol. Given :
Refer to Fig. 6.15.
300 N
500 N
800 N
C
D
B
A
(a)
0.5 m
0.7 m
I
J
H
(b)
0.8 m
300 N
G
1600 N
F
+
500 N
E
800 N
A
B
A
B
C
Base line
D
Base line
C
–
(c)
2350 Nm
D
C¢
640 Nm
B¢
1550 Nm
A¢
2350 Nm
Fig. 6.15
Shear Force Diagram
The shear force at D is + 800 N. This shear force remains constant between D and C.
At C, due to point load, the shear force becomes (800 + 500) = 1300 N. Between C and B, the shear
force remains 1300 N. At B again, the shear force becomes (1300 + 300) = 1600 N. The shear
force between B and A remains constant and equal to 1600 N. Hence the shear force at different
points will be as follows :
242
SHEAR FORCE AND BENDING MOMENT
S.F. at D,
FD = + 800 N
S.F. at C,
FC = + 800 + 500 = + 1300 N
S.F. at B,
FB = + 800 + 500 + 300 = 1600 N
S.F. at A,
FA = + 1600 N.
The shear force, diagram is shown in Fig. 6.15 (b) which is drawn as :
Draw a horizontal line AD as base line. On the base line mark the points B and C below
the point loads. Take the ordinate DE = 800 N in the upward direction. Draw a line EF parallel
to AD. The point F is vertically above C. Take vertical line FG = 500 N. Through G, draw a
horizontal line GH in which point H is vertically above B. Draw vertical line HI = 300 N.
From I, draw a horizontal line IJ. The point J is vertically above A. This completes the shear
force diagram.
Bending Moment Diagram
The bending moment at D is zero :
(i) The bending moment at any section between C and D at a distance x and D is given by,
Mx = – 800 × x which follows a straight line law.
At C, the value of x = 0.8 m.
∴ B.M. at C,
MC = – 800 × 0.8 = – 640 Nm.
(ii) The B.M. at any section between B and C at a distance x from D is given by
(At C, x = 0.8 and at B, x = 0.8 + 0.7 = 1.5 m. Hence here x varies from 0.8 to 1.5).
Mx = – 800 x – 500 (x – 0.8)
...(i)
Bending moment between B and C also varies by a straight line law.
B.M. at B is obtained by substituting x = 1.5 m in equation (i),
∴
MB = – 800 × 1.5 – 500 (1.5 – 0.8)
= – 1200 – 350 = – 1550 Nm.
(iii) The B.M. at any section between A and B at a distance x from D is given by
(At B, x = 1.5 and at A, x = 2.0 m. Hence here x varies from 1.5 m to 2.0 m)
Mx = – 800 x – 500 (x – 0.8) – 300 (x – 1.5)
...(ii)
Bending moment between A and B varies by a straight line law.
B.M. at A is obtained by substituting x = 2.0 m in equation (ii),
∴
MA = – 800 × 2 – 500 (2 – 0.8) – 300 (2 – 1.5)
= – 800 × 2 – 500 × 1.2 – 300 × 0.5
= – 1600 – 600 – 150 = – 2350 Nm.
Hence the bending moments at different points will be as given below :
MD = 0
MC = – 640 Nm
MB = – 1550 Nm
and
MA = – 2350 Nm.
The bending moment diagram is shown in Fig. 6.15 (c) which is drawn as.
Draw a horizontal line AD as a base line and mark the points B and C on this line. Take
vertical lines CC′ = 640 Nm, BB′ = 1550 Nm and AA′ = 2350 Nm in the downward direction. Join
points D, C′, B′ and A′ by straight lines. This completes the bending moment diagram.
243
STRENGTH OF MATERIALS
6.8. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR A CANTILEVER
WITH A UNIFORMLY DISTRIBUTED LOAD
Fig. 6.16 shows a cantilever of length L fixed at A and carrying a uniformly distributed
load of w per unit length over the entire length of the cantilever.
w Per unit length
x
X
( a)
B
A
L
C
(b) w L
+
w.x
A
B
S.F. diagram
Base line
Base line
A
B
w. x 2
2
–
(c) w × L2
2
A′
B.M. diagram
Fig. 6.16
Take a section X at a distance of x from the free end B.
Let
Fx = Shear force at X, and
Mx = Bending moment at X.
Here we have considered the right portion of the section. The shear force at the section X
will be equal to the resultant force acting on the right portion of the section. But the resultant
force on the right portion = w × Length of right portion = w.x.
This resultant force is acting downwards. But the resultant force on the right portion
acting downwards is considered positive. Hence shear force at X is positive.
∴
Fx = + w.x
The above equation shows that the shear force follows a straight line law.
At B, x = 0 and hence
Fx = 0
At A, x = L and hence
Fx = w.L
The shear force diagram is shown in Fig. 6.16 (b).
Bending Moment Diagram
It is mentioned in Art. 6.4.3 that the uniformly distributed load over a section is converted
into point load acting at the C.G. of the section.
The bending moment at the section X is given by
Mx = – (Total load on right portion)
× Distance of C.G. of right portion from X
= – (w . x) .
244
x
x
x2
=–w.x. =–w.
2
2
2
...(i)
SHEAR FORCE AND BENDING MOMENT
(The bending moment will be negative as for the right portion of the section, the moment
of the load at x is clockwise. Also the bending of cantilever will take place in such a manner that
convexity will be at the top of the cantilever).
From equation (i), it is clear that B.M. at any section is proportional to the square of the
distance of the section from the free end. This follows a parabolic law.
At B, x = 0 hence
Mx = 0
L2
.
2
The bending moment diagram is shown in Fig. 6.16 (c).
Problem 6.2. A cantilever of length 2.0 m carries a uniformly distributed load of 1 kN/m
run over a length of 1.5 m from the free end. Draw the shear force and bending moment diagrams
for the cantilever.
Sol. Given :
U.D.L.,
w = 1 kN/m run
Refer to Fig. 6.17.
Mx = – w .
At A, x = L hence
1 kN/m Run
B
C
A
( a)
1.5 m
2.0 m
D 1.5 kN
E
(b) 1.5 kN
+
A
C
A
C
–
(c) 1.875
S.F. diagram
B
1.125
C′
A′
B
Straight line
Parabolic
B.M. diagram
Fig. 6.17
Shear Force Diagram
Consider any section between C and B a distance of x from the free end B. The shear force
at the section is given by
(+ve sign is due to downward force on right portion of the section)
Fx = w.x
= 1.0 × x
(∵ w = 1.0 kN/m run)
At B, x = 0 hence
Fx = 0
At C, x = 1.5 hence Fx = 1.0 × 1.5 = 1.5 kN.
The shear force follows a straight line law between C and B. As between A and C there is
no load, the shear force will remain constant. Hence shear force between A and C will be represented
by a horizontal line.
245
STRENGTH OF MATERIALS
The shear force diagram is shown in Fig. 6.17 (b) in which
FB = 0, FC = 1.5 kN and FA = FC = 1.5 kN.
Bending Moment Diagram
(i) The bending moment at any section between C and B at a distance x from the free end
B is given by
Mx = – (w.x.).
F
GH
I
JK
x2
x2
x
= − 1.
=−
2
2
2
...(i)
(The bending moment will be negative as for the right portion of the section the moment
of load at x is clockwise).
At B, x = 0 hence
MB = −
02
=0
2
1.5 2
= – 1.125 Nm
2
From equation (i) it is clear that the bending moment varies according to parabolic law
between C and B.
(ii) The bending moment at any section between A and C at a distance x from the free end
B is obtained as :
(here x varies from 1.5 m to 2.0 m)
Total load due to U.D.L.= w × 1.5 = 1.5 kN.
At C, x = 1.5 hence MC = −
1.5
= 0.75 m from the free end B or at a distance of
2
(x – 0.75) from any section between A and C.
∴ Moment of this load at any section between A and C at a distance x from free end
= (Load due to U.D.L.) × (x – 0.75)
...(ii)
∴
Mx = – 1.5 × (x – 0.75)
(– ve sign is due to clockwise moment for right portion)
From equation (ii) it is clear that the bending moment follows straight line law between A
and C.
At C, x = 1.5 m hence
MC = – 1.5 (1.5 – 0.75) = – 1.125 Nm
At A, x = 2.0 m hence
MA = – 1.5 (2 – 0.75) = – 1.875 Nm.
Now the bending moment diagram is drawn as shown in Fig. 6.17 (c). In this diagram line
CC′ = 1.125 Nm and AA′ = 1.875 Nm. The points B and C′ are on a parabolic curve whereas the
points A′ and C′ are joined by a straight line.
Problem 6.3. A cantilever of length 2.0 m carries a uniformly distributed load of
2 kN/m length over the whole length and a point load of 3 kN at the free end. Draw the S.F. and
B.M. diagrams for the cantilever.
Sol. Given :
Length,
L = 2.0 m
U.D.L.,
w = 2 kN/m length
Point load at free end = 3 kN
This load is acting at a distance of
246
SHEAR FORCE AND BENDING MOMENT
Refer to Fig. 6.18.
2 kN/m
(a)
3 kN
A
B
2m
D
(b) 7 kN
C
+
3 kN
A
S.F. diagram
B
Base line
A
(c) 10 kNm
Base line
B
–
B.M. diagram
A¢
Fig. 6.18
Shear Force Diagram
The shear force at B = 3 kN
Consider any section at a distance x from the free end B. The shear force at the section is
given by,
Fx = 3.0 + w.x
(+ve sign is due to downward force on
right portion of the section)
= 3.0 + 2 × x
(∵ w = 2 kN/m)
The above equation shows that shear force follows a straight line law.
At B, x = 0 hence
FB = 3.0 kN
At A, x = 2 m hence FA = 3 + 2 × 2 = 7 kN.
The shear force diagram is shown in Fig. 6.18 (b) in which FB = BC = 3 kN and FA = AD
= 7 kN. The points C and D are joined by a straight line.
Bending Moment Diagram
The bending moment at any section at a distance x from the free end B is given by,
FG
H
F 2x
= − G 3x +
H 2
Mx = − 3 x + wx .
2
x
2
I
JK
IJ
K
(∵
w = 2 kN/m)
...(i)
= – (3x + x2)
(The bending moment will be negative as for the right portion of the section, the moment
of loads at x is clockwise).
247
STRENGTH OF MATERIALS
Equation (i) shows that the B.M. varies according to the parabolic law. From equation
(i), we have
At B, x = 0 hence
MB = – (3 × 0 + 02) = 0
At A, x = 2 m hence
MA = – (3 × 2 + 22) = – 10 kN/m
Now the bending moment diagram is drawn as shown in Fig. 6.18 (c). In this diagram,
AA′ = 10 kNm and points A′ and B are joined by a parabolic curve.
Problem 6.4. A cantilever of length 2 m carries a uniformly distributed load of 1.5 kN/m
run over the whole length and a point load of 2 kN at a distance of 0.5 m from the free end. Draw
the S.F. and B.M. diagrams for the cantilever.
Sol. Given :
Length,
L=2m
U.D.L.,
w = 1.5 kN/m run
Point load,
W = 2 kN
Distance of point load from free end = 0.5 m
Refer to Fig. 6.19.
2 kN
1.5 kN/m
C
A
B
(a)
1.5 m
0.5 m
2m
F
E
(b) 5.0
2.0
+
D
0.75
A
S.F. diagram
C
B
Base line
Base line
A
–
B
C
C¢
0.1875
(c) 6.0
A¢
B.M. diagram
Fig. 6.19
Shear Force Diagram
(i) Consider any section between C and B at a distance x from the free end. The shear force
at the section is given by,
(+ve sign is due to downward
Fx = + w.x
force on right portion)
= 1.5 × x
...(i)
248
SHEAR FORCE AND BENDING MOMENT
In equation (i), x varies from 0 to 0.5. The equation (i) shows that shear force varies by a
straight line law between B and C.
At B, x = 0 hence
FB = 1.5 × 0 = 0
At C, x = 0.5 hence
FC = 1.5 × 0.5 = 0.75 kN
(ii) Now consider any section between A and C at a distance x from free end B. The shear
force at the section is given by
(+ve sign is due to downward force
Fx = + w.x + 2 kN
on right portion of the section)
= 1.5x + 2
...(ii)
In equation (ii), x varies from 0.5 to 2.0. The equation (ii) also shows that shear force
varies by a straight line law between A and C.
At C, x = 0.5 hence
FC = 1.5 × 0.5 + 2 = 2.75 kN
At A, x = 2.0 hence
FA = 1.5 × 2.0 + 2 = 5.0 kN
Now draw the shear force diagram as shown in Fig. 6.19 (b) in which CD = 0.75 kN, DE
= 2.0 kN or CE = 2.75 kN and AF = 5.0 kN. The point B is joined to point D by a straight line
whereas the point E is also joined to point F by a straight line.
Bending Moment Diagram
(i) The bending moment at any section between C and B at a distance x from the free end
B is given by
Mx = – (w.x) .
x
2
= – (1.5 × x).
x
2
(∵
w = 1.5 kN/m)
= – 0.75x2
...(iii)
(The bending moment will be negative as for the right portion of the section the moment at
the section is clockwise).
In equation (iii), x varies from 0 to 0.5. Equation (iii) shows that B.M. varies between
C and B by a parabolic law.
At B, x = 0 hence
MB = – 0.75 × 0 = 0
At C, x = 0.5 hence MC = – 0.75 × 0.52 = – 0.1875 kNm.
(ii) The bending moment at any section between A and C at a distance x from the free end
B is given by
x
x
– 2(x – 0.5) = – (1.5 × x) . – 2(x – 0.5)
2
2
(∵ w = 1.5 kN/m)
2
= – 0.75 x – 2(x – 0.5)
...(iv)
In equation (iv), x varies from 0.5 to 2.0. Equation (iv) shows that B.M. varies by a parabolic
law between A and C.
At C, x = 0.5 hence MC = – 0.75 × 0.52 – 2(0.5 – 0.5) = – 0.1875 kN/m
At A, x = 2.0 hence MA = – 0.75 × 22 – 2(2.0 – 0.5) kNm = – 3.0 – 3.0 = – 6.0 kNm
Now the bending moment diagram is drawn as shown in Fig. 6.19 (c). In this diagram line
CC′ = 0.1875 and AA′ = 6.0. The points A′, C′ and B are on parabolic curves.
Mx = – (w.x.) .
249
STRENGTH OF MATERIALS
Problem 6.5. A cantilever 1.5 m long is loaded with a uniformly distributed load of
2 kN/m run over a length of 1.25 m from the free end. It also carries a point load of 3 kN at a
distance of 0.25 m from the free end. Draw the shear force and bending moment diagrams of
the cantilever.
Sol. Given :
Length,
L = 1.5 m
U.D.L.,
w = 2 kN/m
Point load,
W = 3 kN
Refer to Fig. 6.20.
3 kN
2 kN/m
A
C
D
(a)
0.25 m
1m
0.25 m
B
1.25 m
1.5 m
H
5.5
3.5
G
F
+
(b) 5.5
3.0
E
0.5
A
D
A
D
C
S.F. diagram
Base line
(c)
5.94
B
C
C′
–
B
Base line
0.0625
Parabolic
D′
A′
4.563
Straight line
B.M. diagram
Fig. 6.20
Shear Force Diagram
The shear force at B is zero.
The shear force increases to 2 × 0.25 = 0.5 kN by a straight line at C. Due to point load of
3 kN, the shear force suddenly increases to 0.5 + 3 = 3.5 kN at C.
The shear force further increases to 3.5 + 2 × 1 = 5.5 kN by a straight line at D. The shear
force remains constant between A and D as there is no load between A and D.
Now the shear force diagram is drawn as shown in Fig. 6.20 (b). In this diagram line CE
= 0.5 kN, CF = 3.5 kN, DG = 5.5 kN and AH = 5.5 kN. The point B is joined to E by a straight
line. The point F is also joined to G by a straight line. Line GH is horizontal.
250
SHEAR FORCE AND BENDING MOMENT
Bending Moment Diagram
B.M. at B = 0
B.M. at D = – 2 × 0.25 ×
0.25
= – 0.0625 kNm
2
B.M. at D = – 2 × 1.25 ×
1.25
– (3 × 1) = – 4.563 kNm
2
B.M. at A = – 2 × 1.25 ×
FG 1.25 + 0.25IJ – 3 × (1 + 0.25) = – 5.94 kNm.
K
H 2
The bending moment between B and C and between C and D varies by a parabolic law.
But B.M. between A and D varies by a straight line law.
Now the bending moment diagram is drawn as shown in Fig. 6.20 (c). In this diagram line
CC′ = 0.0625, DD′ = 4.563 and AA′ = 5.9. The points B, C′ and D′ are on parabolic curve whereas
points A′ and D′ are joined by a straight line.
Problem 6.6. A cantilever of length 5.0 m is loaded as shown in Fig. 6.21. Draw the
S.F. and B.M. diagrams for the cantilever.
Sol. The shear force at B is 2.5 kN and remains constant between B and C.
The shear force increases by a straight line law to 2.5 + 2 × 1 = 4.5 kN at D. The shear
force remains constant between D and E. At point E, the shear force suddenly increases to 4.5
+ 3 = 7.5 kN due to point load at E. Again the shear force remains constant between A and E. Now
the shear force diagram is drawn as shown in Fig. 6.21 (b).
3 kN
1 kN/m
E
A
D
2.5 kN
C
B
(a)
1m
1.5 m
0.5 m
2m
5m
K
J
7.5
4.5
(b) 7.5
I
A
A
4.5
+
H
D
–
(c) 22.5
F
C
B
C
B
2.5
E
D
S.F. diagram
E
2.5
G
Base line
Base line
D′
8.25 kNm
C′
1.25 kNm
Parabolic
E′
A′
Straight lines
15 kNm
B.M. diagram
Fig. 6.21
251
STRENGTH OF MATERIALS
Bending Moment Diagram
B.M. at B = 0
B.M. at C = – 2.5 × 0.5 = – 1.25 kNm
B.M. at D = – 2.5 × 2.5 – 2 × 1 × 1 = – 8.25 kNm
B.M. at E = – 2.5 × 4 – 2 × 1 × (1.5 + 1.0) = – 10 – 5 = – 15 kNm
B.M. at A = – 2.5 × 5 – 2 × 1 × (1 + 1.5 + 1.0) – 3 × 1
= – 12.5 – 7.0 – 3 = – 22.5 kNm.
Now the bending moment diagram is drawn as shown in Fig. 6.21 (c). In this diagram,
the B.M. varies according to parabolic law between points C and D only. Between other points
B.M. varies according to straight line law.
6.9.
SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR A CANTILEVER
CARRYING A GRADUALLY VARYING LOAD
Fig. 6.22 shows a cantilever of length L fixed at A and carrying a gradually varying load
from zero at the free end to w per unit length at the fixed end.
w
w.x
L
C
(a)
A
B
X
x
L
Load diagram
Parabolic curve
w×L
(b) 2
2
w.x
2L
+
A
S.F. diagram
B
C
Base line
Base line
C
A
B
3
(c) w × L2
w. x
6L
–
6
Cubic curve
B.M. diagram
Fig. 6.22
Take a section X at a distance x from the free end B.
Let
Fx = Shear force at the section X, and
Mx = Bending moment at the section X.
Let us first find the rate of loading at the section X. The rate of loading is zero at B and is
w per metre run at A. This means that rate of loading for a length L is w per unit length. Hence
w
rate of loading for a length of x will be
× x per unit length. This is shown in Fig. 6.22 (a) by
L
w. w
.
CX, which is also known as load diagram. Hence CX =
L
252
SHEAR FORCE AND BENDING MOMENT
The shear force and the section X at a distance x from free end is given by,
Fx = Total load on the cantilever for a length x from the free end B
= Area of triangle BCX
XB . XC
=
2
w. x 2
=
2L
=
x
FG w. x IJ
H LK
FG∵
H
2
XB = x, XC =
w. x
L
IJ
K
...(i)
Equation (i) shows that the S.F. varies according to the parabolic law.
w × 02
=0
2L
w. L2 w. L
At A, x = L hence
FA =
=
2L
2
The bending moment at the section X at a distance x from the free end B is given by,
Mx = – (Total load for a length x) × Distance of the load from X
= – (Area of triangle BCX) × Distance of C.G. of the triangle from X
At B, x = 0 hence
FB =
=−
F wx I × x = − wx
GH 2L JK 3 6 L
2
3
...(ii)
Equation (ii) shows that the B.M. varies according to the cubic law.
w×0
At B, x = 0 hence
MB = −
=0
6L
w. L3
w. L2
=−
.
At A, x = L hence
MA = −
6L
6
Problem 6.7. A cantilever of length 4 m carries a gradually varying load, zero at the
free end to 2 kN/m at the fixed end. Draw the S.F. and B.M. diagrams for the cantilever.
Sol. Given :
Length,
L=4m
w = 2 kN/m
Load at fixed end,
Shear Force Diagram
The shear force is zero at B. The shear force at C will be equal to the area of load diagram
ABC.
4×2
∴
Shear force at C =
= 4 kN
2
The shear force between A and B varies according to parabolic law.
Bending Moment Diagram
The B.M. at B is zero. The bending moment at A is equal to −
w. L2
.
6
w. L2
2 × 42
= – 5.33 kNm.
=−
6
6
The B.M. between A and B varies according to cubic law.
∴
MA = −
253
STRENGTH OF MATERIALS
C
2 kN/m
(a)
B
A
4m
Load diagram
D
(b)
4 kN
+
B
A
S.F. diagram
A
B
–
(c)
5.33
A¢
B.M. diagram
Fig. 6.23
6.10. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR A SIMPLY
SUPPORTED BEAM WITH A POINT LOAD AT MID-POINT
Fig. 6.24 shows a beam AB of length L simply supported at the ends A and B and carrying
a point load W at its middle point C.
W
as the load is acting at the middle point
The reactions at the support will be equal to
2
W
.
of the beam. Hence RA = RB =
2
Take a section X at a distance x from the end A between A and C.
Let
Fx = Shear force at X,
and
Mx = Bending moment at X.
Here we have considered the left portion of the section. The shear force at X will be equal
to the resultant force acting on the left portion of the section. But the resultant force on the left
W
portion is
acting upwards. But according to the sign convention, the resultant force on the
2
left portion acting upwards is considered positive. Hence shear force at X is positive and its
W
.
magnitude is
2
W
∴
Fx = +
2
W
Hence the shear force between A and C is constant and equal to +
.
2
254
SHEAR FORCE AND BENDING MOMENT
W
x
(a)
C
B
L/2
RA = W
—
2
(b)
X
A
W
RB = —
2
L
W
2
+
Base line
A
B
C
W
2
–
S.F. diagram
C¢
(c)
+
A
W×L
4
C
B.M. diagram
B
Base line
Fig. 6.24
Now consider any section between C and B at distance x from end A. The resultant force
on the left portion will be
FG W − W IJ = − W .
H2 K 2
This force will also remain constant between C and B. Hence shear force between C and B
W
.
is equal to −
2
W
W
At the section C the shear force changes from +
to −
.
2
2
The shear force diagram is shown in Fig. 6.24 (b).
Bending Moment Diagram
(i) The bending moment at any section between A and C at a distance of x from the end A,
is given by
W
.x
...(i)
2
(B.M. will be positive as for the left portion of the section, the moment of all forces at X is
clockwise. Moreover, the bending of beam takes place in such a manner that concavity is at the
top of the beam).
Mx = RA.x
At A, x = 0 hence
At C, x =
L
hence
2
or
Mx = +
W
×0=0
2
W L W×L
MC =
.
× =
2
2
4
MA =
255
STRENGTH OF MATERIALS
From equation (i), it is clear that B.M. varies according to straight line law between
W×L
A and C. B.M. is zero at A and it increases to
at C.
4
(ii) The bending moment at any section between C and B at a distance x from the end A, is
given by
FG
H
IJ
K
L
W
L WL 2 x
=
. x − Wx + W × =
−
2
2
2
2
2
L
WL W L W × L
At C, x =
hence MC =
−
× =
2
2
2
2
4
WL W
At B, x = L hence MB =
× L = 0.
−
2
2
WL
Hence bending moment at C is
and it decreases to zero at B. Now the B.M. diagram
4
can be completed as shown in Fig. 6.24 (c).
Mx = RA.x – W × x −
Note. The bending moment is maximum at the middle point C, where the shear force changes its
sign.
6.11. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR A SIMPLY
SUPPORTED BEAM WITH AN ECCENTRIC POINT LOAD
Fig. 6.25 shows a beam AB of length L, simply supported at the ends A and B and carrying
a point load W at C at a distance of ‘a’ from the end A.
W
x
X
( a)
a
RA
(b)
C
B
A
W×b
L
b
RB
L
+
B
C
A
W×a
L
–
S.F. diagram
W×a×b
L
(c)
+
A
B.M. diagram
C
B
Fig. 6.25
Let
RA = Reaction at the support A, and
RB = Reaction at the support B.
First calculate the reactions, by taking moments about A or about B.
Taking moments of the forces on the beam about A, we get
RB × L = W × a
W. a
∴
RB =
L
256
SHEAR FORCE AND BENDING MOMENT
W. a
L
a
L−a
W ×b
= W 1−
=W
=
(∵ L – a = b)
L
L
L
Consider a section X at a distance x from the end A between A and C.
The shear force Fx at the section is given by,
W. b
...(i)
Fx = + RA = +
L
(The shear force will be positive as the resultant force on the left portion of the section is
acting upwards).
W. b
.
The shear force between A and C is constant and equal to
L
Now consider any section between C and B at a distance x from the end A. The resultant
force on the left portion will be RA – W
W. b
L−b
W. a
b− L
or
=−W
=−
(∵ L – b = a)
− W = W.
L
L
L
L
W. a
The shear force between C and B is constant and equal to −
. At the section C, the
L
W. a
W. b
to −
. The shear force diagram is shown in Fig. 6.25 (b).
shear force changes from
L
L
and
RA = W – RB = W –
IJ
K
FG
H
FG
H
FG
H
IJ
K
IJ
K
FG
H
IJ
K
Bending Moment Diagram
The bending moment at any section between A and C at a distance x from the end A, is
given by
W. b
.x
(Plus sign due to sagging)
Mx = RA × x = +
L
W. b
×0=0
At A, x = 0 hence MA =
L
W . a. b
W. b
At C, x = a hence MC =
.a=
L
L
W . a. b
Hence the B.M. increases from zero at A to
at C by a straight line law. The B.M. is
L
W . a. b
at C to zero at B following a straight line law.
zero at B. Hence B.M. will decrease from
L
The B.M. diagram is drawn in Fig. 6.25 (c).
From the shear force and bending moment diagrams, it is clear that the B.M. is maximum
at C where the S.F. changes its sign.
Problem 6.8. A simply supported beam of length 6 m, carries point load of 3 kN and 6 kN
at distances of 2 m and 4 m from the left end. Draw the shear force and bending moment
diagrams for the beam.
Sol. First calculate the reactions RA and RB.
Taking moments of the force about A, we get
RB × 6 = 3 × 2 + 6 × 4 = 30
30
= 5 kN
∴
RB =
6
∴
RA = Total load on beam – RB = (3 + 6) – 5 = 4 kN
257
STRENGTH OF MATERIALS
A
3 kN
6 kN
C
D
2m
(a)
2m
4m
5 kN
6m
4 kN
(b)
B
+
4 kN
3 kN
1 kN
A
B
1 kN
C
D
Base line
–
5 kN
S.F. diagram
5 kN
(c)
8 kNm
A
+
C
10 kNm
D
B.M. diagram
B
Base line
Fig. 6.26
Shear Force Diagram
Shear force at A,
FA = + RA = + 4 kN
Shear force between A and C is constant and equal to + 4 kN
Shear force at C,
FC = + 4 – 3.0 = + 1 kN
Shear force between C and D is constant and equal to + 1 kN.
Shear force at D, FD = + 1 – 6 = – 5 kN
The shear force between D and B is constant and equal to – 5 kN.
Shear force at B,
FB = – 5 kN
The shear force diagram is drawn as shown in Fig. 6.26 (b).
Bending Moment Diagram
B.M. at A,
MA = 0
B.M. at C,
MC = RA × 2 = 4 × 2 = + 8 kNm
B.M. at D,
MD = RA × 4 – 3 × 2 = 4 × 4 – 3 × 2 = + 10 kNm
B.M. at B,
MB = 0
The bending moment diagram is drawn as shown in Fig. 6.26 (c).
6.12. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR A SIMPLY
SUPPORTED BEAM CARRYING A UNIFORMLY DISTRIBUTED LOAD
Fig. 6.27 shows a beam AB of length L simply supported at the ends A and B and
carrying a uniformly distributed load of w per unit length over the entire length. The reactions
at the supports will be equal and their magnitude will be half the total load on the entire
length.
258
SHEAR FORCE AND BENDING MOMENT
x
w/Unit length
X
B
A
( a)
C
RA
(b)
RB
L
Base line
w.L
2
+
A
Fx
S.F. diagram
–
L/2
(c)
w.L
2
L/2
Mx
A
B
C
w.L
8
2
C
B.M. diagram
B
Base line
Fig. 6.27
Let
RA = Reaction at A, and
RB = Reaction at B
w. L
∴
RA = RB =
2
Consider any section X at a distance x from the left end A. The shear force at the section
(i.e., Fx) is given by,
w. L
Fx = + RA – w . x = +
–w.x
...(i)
2
From equation (i), it is clear that the shear force varies according to straight line law.
The values of shear force at different points are :
w. L
w. L w . 0
At A, x = 0 hence FA = +
−
=+
2
2
2
w. L
w. L
At B, x = L hence FB = +
− w. L = −
2
2
L
L
w. L
− w. = 0
At C, x =
hence FC = +
2
2
2
The shear force diagram is drawn as shown in Fig. 6.27 (b).
The bending moment at the section X at a distance x from left end A is given by,
x
Mx = + RA . x – w . x .
2
w. L
w. L
w. x2
∵ RA =
=
...(ii)
.x−
2
2
2
From equation (ii), it is clear that B.M. varies according to parabolic law.
FG
H
IJ
K
259
STRENGTH OF MATERIALS
The values of B.M. at different points are :
w. L
w. 0
At A, x = 0 hence
MA =
.0 −
=0
2
2
w. L
w
At B, x = L hence
MB =
.L−
. L2 = 0
2
2
2
L
w . L2 w . L2
w . L2
w. L L w L
−
=+
hence MC =
=
.
At C, x =
. − .
2
4
8
8
2
2 2
2
FG IJ
H K
w . L2
at the
8
middle point of the beam and from this value the B.M. decreases to zero at B according to the
parabolic law.
Now the B.M. diagram is drawn as shown in Fig. 6.27 (c).
Problem 6.9. Draw the shear force and bending moment diagram for a simply supported
beam of length 9 m and carrying a uniformly distributed load of 10 kN/m for a distance of 6 m
from the left end. Also calculate the maximum B.M. on the section.
Sol. First calculate reactions RA and RB.
Thus, the B.M. increases according to parabolic law from zero at A to +
10 kN/m
C
A
( a)
B
6m
RA
RB
20 kN
9m
40 kN
(b)
Base line
40
+
A
C
D
B
–
S.F. diagram
20
Parabolic
Parabolic
Straight
line
(c)
+
A
80
D
B.M. diagram
60
C
B
Base line
Fig. 6.28
Taking moments of the forces about A, we get
6
RB × 9 = 10 × 6 × = 180
2
180
∴
RB =
= 20 kN
9
∴
RA = Total load on beam – RB = 10 × 6 – 20 = 40 kN.
260
SHEAR FORCE AND BENDING MOMENT
Shear Force Diagram
Consider any section at a distance x from A between A and C. The shear force at the
section is given by,
...(i)
Fx = + RA – 10 x = + 40 – 10 x
Equation (i) shows that shear force varies by a straight line law between A and C.
At A, x = 0 hence
FA = + 40 – 0 = 40 kN
At C, x = 6 m hence
FC = + 40 – 10 × 6 = – 20 kN
The shear force at A is + 40 kN and at C is – 20 kN. Also shear force between A and C
varies by a straight line. This means that somewhere between A and C, the shear force is zero.
Let the S.F. is zero at x metre from A. Then substituting the value of S.F. (i.e., Fx) equal to zero
in equation (i), we get
0 = 40 – 10x
40
=4m
∴
x=
10
Hence shear force is zero at a distance 4 m from A.
The shear force is constant between C and B. This equal to – 20 kN.
Now the shear force diagram is drawn as shown in Fig. 6.28 (b). In the shear force diagram,
distance AD = 4 m. The point D is at a distance 4 m from A.
B.M. Diagram
The B.M. at any section between A and C at a distance x from A is given by,
x
Mx = RA × x – 10 . x . = 40x – 5x2
2
Equation (ii) shows that B.M. varies according to parabolic law between A and C.
At A, x = 0 hence
MA = 40 × 0 – 5 × 0 = 0
At C, x = 6 m hence
MC = 40 × 6 – 5 × 62 = 240 – 180 = + 60 kNm
At D, x = 4 m hence
MD = 40 × 4 – 5 × 42 = 160 – 80 = + 80 kNm
The bending moment between C and B varies according to linear law.
B.M. at B is zero whereas at C is 60 kNm.
The bending moment diagram is drawn as shown in Fig. 6.28 (c).
...(ii)
Maximum Bending Moment
The B.M. is maximum at a point where shear force changes sign. This means that the
point where shear force becomes zero from positive value to the negative or vice-versa, the B.M.
at that point will be maximum. From the shear force diagram, we know that at point D, the
shear force is zero after changing its sign. Hence B.M. is maximum at point D. But the B.M. at
D is + 80 kNm.
∴
Max. B.M. = + 80 kN. Ans.
Problem 6.10. Draw the shear force and B.M. diagrams for a simply supported beam of
length 8 m and carrying a uniformly distributed load of 10 kN/m for a distance of 4 m as shown
in Fig. 6.29.
Sol. First calculate the reactions RA and RB.
Taking moments of the forces about A, we get
FG
H
RB × 8 = 10 × 4 × 1 +
IJ
K
4
= 120
2
261
STRENGTH OF MATERIALS
10 kN/m
A
(a)
D
C
1m
4m
B
3m
RA
RB
15 kN
8m
25 kN
25
(b)
Base line
+
25
D
A
C
E
S.F. diagram
(c)
B
–
15
15
Parabolic
Straight
line
Parabolic
Straight
line
+
56.25
45
E
D
25
A
C
B
B.M. diagram
Fig. 6.29
120
= 15 kN
8
= Total load on beam – RB
= 10 × 4 – 15 = 25 kN
∴
RB =
∴
RA
Shear Force Diagram
The shear force at A is + 25 kN. The shear force remains constant between A and C and
equal to + 25 kN. The shear force at B is – 15 kN. The shear force remains constant between
B and D and equal to – 15 kN. The shear force at any section between C and D at a distance x
from A is given by,
Fx = + 25 – 10(x – 1)
...(i)
At C, x = 1 hence
FC = + 25 – 10(1 – 1) = + 25 kN
At D, x = 5 hence
FD = + 25 – 10(5 – 1) = – 15 kN
The shear force at C is + 25 kN and at D is – 15 kN. Also shear force between C and D
varies by a straight line law. This means that somewhere between C and D, the shear force is
zero. Let the S.F. be zero at x metre from A. Then substituting the value of S.F. (i.e., Fx) equal
to zero in equation (i), we get
0 = 25 – 10(x – 1)
or
0 = 25 – 10x + 10
or
10x = 35
35
∴
x=
= 3.5 m
10
Hence the shear force is zero at a distance 3.5 m from A.
Hence the distance AE = 3.5 m in the shear force diagram shown in Fig. 6.29 (b).
262
SHEAR FORCE AND BENDING MOMENT
B.M. Diagram
B.M. at A is zero
B.M. at B is also zero
B.M. at C = RA × 1 = 25 × 1 = 25 kNm
The B.M. at any section between C and D at a distance x from A is given by,
( x − 1)
Mx = RA . x – 10(x – 1) .
= 25 × x – 5(x – 1)2
...(ii)
2
At C, x = 1 hence
MC = 25 × 1 – 5(1 – 1)2 = 25 kNm
At D, x = 5 hence
MD = 25 × 5 – 5(5 – 1)2 = 125 – 80 = 45 kNm
At E, x = 3.5 hence
ME = 25 × 3.5 – 5(3.5 – 1)2 = 87.5 – 31.25 = 56.25 kNm
B.M. will increase from 0 at A to 25 kNm at C by a straight line law. Between C and D the
B.M. varies according to parabolic law as is clear from equation (ii). Between C and D, the B.M.
will be maximum at E. From D to B the B.M. will decrease from 45 kNm at D to zero at B
according to straight line law.
Problem 6.11. Draw the S.F. and B.M. diagrams of a simply supported beam of length
7 m carrying uniformly distributed loads as shown in Fig. 6.30.
10 kN/m
5 kN/m
C
A
D
3m
(a)
2m
B
2m
RA = 25
(b) 25
RB = 15
+
E
C
5
D
A
B
–
15
S.F. diagram
31.25
(c)
30
A
B.M. diagram
E
20
D
C
B
Fig. 6.30
Sol. First calculate the reactions RA and RB.
Taking moments of all forces about A, we get
RB × 7 = 10 × 3 ×
∴
RB =
FG
H
IJ
K
2
3
= 45 + 60 = 105
+5×2× 3+2+
2
2
105
= 15 kN
7
263
STRENGTH OF MATERIALS
RA = Total load on beam – RB
= (10 × 3 + 5 × 2) – 15 = 40 – 15 = 25 kN
and
S.F. Diagram
The shear force at A is + 25 kN
The shear force at C = RA – 3 × 10 = + 25 – 30 = – 5 kN
The shear force varies between A and C by a straight line law.
The shear force between C and D is constant and equal to – 5 kN
The shear force at B is – 15 kN
The shear force between D and B varies by a straight line law.
The shear force diagram is drawn as shown in Fig. 6.30 (b).
The shear force is zero at point E between A and C. Let us find the location of E from A. Let
the point E be at a distance x from A.
The shear force at E = RA – 10 × x = 25 – 10x
But shear force at E = 0
∴
25 – 10x = 0
or
10x = 25
25
= 2.5 m
or
x=
10
B.M. Diagram
B.M. at A is zero
B.M. at B is zero
B.M. at C,
MC = RA × 3 – 10 × 3 ×
3
= 25 × 3 – 45 = 75 – 45 = 30 kNm
2
At E, x = 2.5 and hence
2.5
= 25 × 2.5 – 5 × 6.25
2
= 62.5 – 31.25 = 31.25 kNm
3
B.M. at D,
MD = 25(3 + 2) – 10 × 3 ×
+ 2 = 125 – 105 = 20 kNm
2
The B.M. between AC and between BD varies according to parabolic law. But B.M. between
C and D varies according to straight line law. Now the bending moment diagram is drawn as
shown in Fig. 6.30 (c).
Problem 6.12. A simply supported beam of length 10 m, carries the uniformly distributed load and two point loads as shown in Fig. 6.31. Draw the S.F. and B.M. diagram for the
beam. Also calculate the maximum bending moment.
Sol. First calculate the reactions RA and RB.
Taking moments of all forces about A, we get
4
RB × 10 = 50 × 2 + 10 × 4 × 2 +
+ 40(2 + 4)
2
= 100 + 160 + 240 = 500
500
∴
RB =
= 50 kN
10
and
RA = Total load on beam – RB
= (50 + 10 × 4 + 40) – 50 = 130 – 50 = 80 kN
B.M. at E,
ME = RA × 2.5 – 10 × 2.5 ×
FG
H
FG
H
264
IJ
K
IJ
K
SHEAR FORCE AND BENDING MOMENT
50 kN
D
C
A
(a)
B
4m
10 m
2m
R A = 80
(b)
40 kN
10 kN/m
4m
R B = 50
80
50
+
30
A
E
D
C
B
10
S.F. diagram
(c)
160
A
C
+
205
200
E
D
–
50
B
B.M. diagram
Fig. 6.31
S.F. Diagram
The S.F. at A,
FA = RA = + 80 kN
The S.F. will remain constant between A and C and equal to + 80 kN
The S.F. just on R.H.S. of C = RA – 50 = 80 – 50 = 30 kN
The S.F. just on L.H.S. of D = RA – 50 – 10 × 4 = 80 – 50 – 40 = – 10 kN
The S.F. between C and D varies according to straight line law.
The S.F. just on R.H.S. of D = RA – 50 – 10 × 4 – 40 = 80 – 50 – 40 – 40 = – 50 kN
The S.F. at B = – 50 kN
The S.F. remains constant between D and B and equal to – 50 kN
The shear force diagram is drawn as shown in Fig. 6.31 (b).
The shear force is zero at point E between C and D.
Let the distance of E from point A is x.
Now shear force at E = RA – 50 – 10 × (x – 2)
= 80 – 50 – 10x + 20 = 50 – 10x
But shear force at E = 0
50
∴
50 – 10x = 0 or x =
=5m
10
B.M. Diagram
B.M. at A is zero
B.M. at B is zero
265
STRENGTH OF MATERIALS
B.M. at C,
MC = RA × 2 = 80 × 2 = 160 kNm
4
B.M. at D,
MD = RA × 6 – 50 × 4 – 10 × 4 ×
2
= 80 × 6 – 200 – 80 = 480 – 200 – 80 = 200 kNm
At E, x = 5 m and hence B.M. at E,
5−2
ME = FA × 5 – 50(5 – 2) – 10 × (5 – 2) ×
2
= 80 × 5 – 50 × 3 – 10 × 3 × 32 = 400 – 150 – 45 = 205 kNm
The B.M. between C and D varies according to parabolic law reaching a maximum value
at E. The B.M. between A and C and also between B and D varies according to linear law. The
B.M. diagram is shown in Fig. 6.31 (c).
FG
H
IJ
K
Maximum B.M.
The maximum B.M. is at E, where S.F. becomes zero after changing its sign.
∴
Max. B.M. = ME = 205 kNm. Ans.
6.13. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR A SIMPLY
SUPPORTED BEAM CARRYING A UNIFORMLY VARYING LOAD FROM ZERO
AT EACH END TO w PER UNIT LENGTH AT THE CENTRE
Fig. 6.32 shows a beam of length L simply supported at the ends A and B and carrying a
uniformly varying load from zero at each end to w per unit length at the centre. The reactions at
the supports will be equal and their magnitude will be half the total load on the entire length, as
the load is symmetrical on the beam.
But
total load on the beam = Area of load diagram ABO
AB × CO L × w
w. L
=
or
=
2
2
2
∴
RA = RB = Half the total load
FG
H
IJ
K
1 w. L
w. L
=
2
2
4
Consider any section X between A and C at a distance x from end A.
The rate of loading at X
= Vertical distance XD in load diagram
=
LM
x
CO
XD
=
FG L IJ . w MM∵ FG L IJ = x
H 2 K MN H 2 K
∴
OP
x × CO 2 x . w
XD =
FG L IJ = L PP
PQ
H 2K
2w
.x
L
Now load on the length AX of the beam = Area of load diagram AXD
=
x . XD
=
=
2
w
= . x2
L
266
x.
2w
.x
L
2
FG∵
H
XD =
IJ
K
2w
.x
L
SHEAR FORCE AND BENDING MOMENT
O
D
w/m
A
B
X
x
(a)
C
Load diagram
L/2
L
(b )
w.L
4
+
Fx
C
A
B
S.F. diagram
w.L
4
2
w.L
(c )
+
A
B.M. diagram
B
C
Fig. 6.32
This load is acting at a distance of
x
from X
3
Now S.F. at X is given by,
Fx = RA – load on the length AX
=
w. L w 2
− x
4
L
FG∵
H
RA =
IJ
K
w. L
...(i)
4
Equation (i) shows that shear force varies according to parabolic law between A and C.
At A, x = 0 hence
At C, x =
L
hence
2
Fx =
w. L w
w. L
− .0 =
4
4
L
FC =
w. L w L
−
4
L 2
FG IJ
H K
2
=
w . L wL
−
=0
4
4
w. L
4
The shear force is shown in Fig. 6.32 (b). The shear force between A and C and also
between C and B is parabolic.
The shear force at B = – RB = −
267
STRENGTH OF MATERIALS
B.M. Diagram
The bending moment is zero at A and B.
The B.M. at X is given by,
Mx = RA . x – Load of length AX .
x
3
w. L
x w. L
w
w
. x − x2 . =
.x−
. x3
4
3
4
wL
L
Equation (ii) shows that B.M. between A and C varies according to cubic law.
At A, x = 0 hence
Mx = 0
=
At C, x =
FG IJ
H K
L
w. L L
w
L
hence MC =
. −
.
2
4
2 3L 2
...(ii)
3
w . L2 wL2 3w . L2 − wL2 wL2
−
=
=
8
24
24
12
The maximum B.M. occurs at the centre of the beam, where S.F. becomes zero after
changing its sign.
=
wL2
12
wL
But total load on the beam,
W=
2
wL L w . L
. =
∴
Max. B.M. =
.
2 6
6
∴ Max. B.M. is at C,
MC =
6.14. SHEAR FORCE AND B.M. DIAGRAMS FOR A SIMPLY SUPPORTED BEAM
CARRYING A UNIFORMLY VARYING LOAD FROM ZERO AT ONE END TO w
PER UNIT LENGTH AT THE OTHER END
Fig. 6.33 shows a beam AB of length L simply supported at the ends A and B and carrying
a uniformly varying load from zero at end A to w per unit length at B. First calculate the
reactions RA and RB.
Taking moments about A, we get
RB × L =
∴
and
RB =
FG w. L IJ . 2 L
H 2K 3
w. L
3
LMTotal load FG = w. L IJ is acting 2 L from AOP
H 2K
3
N
Q
w. L w. L w. L
−
=
2
3
6
Consider any section X at a distance x from end A. The shear force at X is given by,
w. L w. x x
Fx = RA – load on length AX =
−
.
6
L 2
AX . CX x . w . x
Load on AX =
=
2
2. L
RA = Total load on beam – RB =
FG
H
wL wx 2
−
6
2L
Equation (i) shows that S.F. varies according to parabolic law.
=
268
IJ
K
...(i)
SHEAR FORCE AND BENDING MOMENT
O
w.x
L
C
w
A
B
x
(a)
X
Load diagram
RA
L
+
w .L/6
(b)
RB
B
C
A
L /√3
(c)
–
w.L
3
S.F . diagram
+
A
B
C
B.M . diagram
Fig. 6.33
w. L
w
w. L
−
×0=
6
2L
6
w . L w . L2 w . L w . L
w . L − 3w . L
2w . L
w. L
=−
=−
At B, x = L, hence, FB =
−
=
−
=
6
6
3
6
2L
6
2
w. L
w. L
The shear force is +
at A and it decreases to −
at B according to parabolic law.
6
3
Somewhere between A and B, the S.F. must be zero. Let the S.F. be zero to a distance x from A.
Equating the S.F. to zero in equation (i), we get
At A, x = 0 hence,
FA =
wL wx 2
−
6
2L
w
.
L
2 L L2
x2 =
×
=
6
w
3
L
x=
= 0.577 L
3
or
0=
or
∴
wx 2 w . L
=
2L
6
B.M. Diagram
The B.M. is zero at A and B.
The B.M. at the section X at a distance x from the end A is given by,
Mx = RA.x – Load on length AX .
=
x
3
FG∵
H
Load on AX is acting at
x
from X
3
w. L
wx 2 x wL
wx 3
.x−
. =
.x−
6
2L 3
6
6L
269
IJ
K
STRENGTH OF MATERIALS
Equation (ii) shows the B.M. varies between A and B according to cubic law.
Max. B.M. occurs at a point where S.F. becomes zero after changing its sign.
L
L
from A. Hence substituting x =
in equation (ii), we
That point is at a distance of
3
3
get maximum B.M.
w. L L
w
Max. B.M. =
−
.
.
6
3 6L
∴
=
wL2
6 3
−
wL2
18 3
=
FG L IJ
H 3K
3
3w . L2 − wL2
18 3
=
wL2
.
9 3
Problem 6.13. A simply supported beam of length 5 m carries a uniformly increasing
load of 800 N/m run at one end to 1600 N/m run at the other end. Draw the S.F. and B.M.
diagrams for the beam. Also calculate the position and magnitude of maximum bending moment.
Sol. The loading on the beam is shown in Fig. 6.34. The load may be assumed to be
consisting of a uniformly distributed load of 800 N/m over the entire span and a gradually
varying load of zero at A to 800 N/m at B.
Then load on beam due to uniformly distributed load of 800 N/m = 800 × 5 = 4000 N
1
Load on beam due to triangular loading = × 800 × 5 = 2000 N
2
DE = 160 x
G
X1 E
F
1600 N
D
800 N
(a)
A
x
RA
(b)
2666.67
B
C
X
5m
–
S.F. diagram
3333.33
3761.5 Nm
+
B.M. diagram
Fig. 6.34
270
RB
+
2.637
(c)
Load diagram
SHEAR FORCE AND BENDING MOMENT
Now calculate the reactions RA and RB.
Taking the moments about A, we get
2
5
RB × 5 = 4000 × + 2000 ×
of 5
5
2
∴
RB = 2000 + 1333.33 = 3333.33 N
RA = Total load on beam – RB
= (4000 + 2000) – 3333.33 = 2666.67 N
Consider any section X-X at a distance x from A.
Rate of loading at the section X-X
= Length CE = CD + DE
FG
H
and
= 800 +
IJ
K
x
× 800 = 800 + 160x
5
Total load on length AX
= Area of load diagram ACDEF
= Area of rectangle + Area of Δ DEF
160 x × x
= 800x + 80x2
2
Now the S.F. at the section X-X is given by,
= 800 × x +
F x = RA – load on length AX
...(i)
= 2666.67 – (800x + 80x2) = 2666.67 – 800x – 80x2
Equation (i) shows that shear force varies between A and B according to parabolic law.
At A, x = 0 hence Fx = 2666.67 – 800 × 0 – 80 × 0 = + 2666.67 N
At B, x = 5 hence Fx = 2666.67 – 800 × 5 – 80 × 52
= 2666.67 – 4000 – 2000 = – 3333.33 N
Let us find the position of zero shear. Equating the S.F. equal to zero in equation (i), we get
0 = 2666.67 – 800x – 80x2
or
2666.67
= 0
or
x2 + 10x – 33.33 = 0
80
The above equation is a quadratic equation. Its solution is given by,
x2 + 10x −
− 10 ± 10 2 + 4 × 33.33 − 10 ± 233.33
=
2
2
− 10 + 15.274
=
(Neglecting – ve root)
2
= 2.637 m
x =
B.M. Diagram
The B.M. at the section X-X is given by
x 1
x
− . x . 160 x .
2 2
3
80 3
x
= 2666.67x – 400x2 –
3
Mx = RA × x – 800 × x ×
...(ii)
271
STRENGTH OF MATERIALS
Equation (ii) shows that B.M. between A and B varies according to cubic law.
At A, x = 0,
Mx = 0
At B, x = 5,
MB = 0
Maximum B.M. occurs where S.F. is zero. But S.F. is zero at a distance of 2.637 m
from A. Hence maximum B.M. is obtained by substituting x = 2.637 m in equation (ii).
80
∴
Max. B.M. = 2666.67 × 2.637 – 400 × 2.6372 –
× 2.6373 = 3761.5 Nm. Ans.
3
6.15. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR OVER-HANGING
BEAMS
If the end portion of a beam is extended beyond the support, such beam is known as
overhanging beam. In case of overhanging beams, the B.M. is positive between the two supports,
whereas the B.M. is negative for the over-hanging portion. Hence at some point, the B.M. is zero
after changing its sign from positive to negative or vice versa. That point is known as the point
of contraflexure or point of inflexion.
6.15.1. Point of Contraflexure. It is the point where the B.M. is zero after changing its
sign from positive to negative or vice versa.
Problem 6.14. Draw the shear force and bending moment diagrams for the over-hanging
beam carrying uniformly distributed load of 2 kN/m over the entire length as shown in Fig. 6.35.
Also locate the point of contraflexure.
Sol. First calculate the reactions RA and RB
Taking moments of all forces about A, we get
6
RB × 4 = 2 × 6 × = 36
(∵ Total load on beam = 2 × 6 = 12 kN. This
2
load is acting at a distance 3 m from A)
2 kN/m
A
C
B
(a)
4m
2m
RB = 9
RA = 3
4
3
+
3
(b)
4
D
A
–
1.5
+
B
C
5
S.F. diagram
5
(c)
+
A
2.25
D
B
E
4.0
B.M. diagram
Fig. 6.35
272
–
C
SHEAR FORCE AND BENDING MOMENT
∴
and
36
= 9 kN
4
= Total load – RB = 2 × 6 – 9 = 3 kN
RB =
RA
Shear Force Diagram
Shear force at A
= + RA = + 3 kN
(i) The shear force at any section between A and B at a distance x from A is given by,
FA = RA – 2x
(∵ RA = 3)
= 3 – 2x
...(i)
At A, x = 0 hence
FA = 3 kN
At B, x = 4 hence
FB = 3 – 2 × 4 = – 5 kN
The shear force varies according to straight line law between A and B. At A, the shear
force is positive whereas at B, the shear force is negative. Between A and B somewhere S.F. is
zero. The point, where S.F. is zero, is obtained by substituting Fx = 0 in equation (i).
3
∴
0 = 3 – 2x or x = = 1.5 m
2
Hence S.F. is zero at a distance of 1.5 m from A (or S.F. is zero at point D).
(ii) The S.F. at any section between B and C at a distance x from A is given by,
F x = + RA – 4 × 2 + RB – (x – 4) × 2 = 3 – 8 + 9 – 2(x – 4)
= 4 – 2(x – 4)
...(ii)
At B, x = 4 m hence FB = 4 – 2(4 – 4) = + 4 kN
At C, x = 6 m hence FC = 4 – 2(6 – 4) = 0
Between B and C also S.F. varies by a straight line law. At B, S.F. is + 4 kN and at C, S.F.
is zero.
The S.F. diagram is shown in Fig. 6.35 (b).
Bending Moment Diagram
The B.M. at A is zero.
(i)The B.M. at any section between A and B at a distance x is given by,
Mx = RA × x – 2 × x ×
x
2
= 3x – x2
At A, x = 0 hence
MA = 0
At B, x = 4 hence
MB = 3 × 4 – 42 = – 4 kNm
Max. B.M. occurs at D, where S.F. is zero after changing its sign.
At D, x = 1.5 hence MD = 3 × 1.5 – 1.5 = 4.5 – 2.25 = 2.25 kNm
The B.M. between A and B varies according to parabolic law.
(ii)The B.M. at any section between B and C at a distance x is given by,
x
Mx = RA × x – 2 × x × + RB × (x – 4)
2
= 3x – x2 + 9(x – 4)
At B, x = 4 hence
MB = 3 × 4 – 42 + 9(4 – 4) = 4 kNm
At C, x = 6 hence
MC = 3 × 6 – 62 + 9(6 – 4) = 18 – 36 + 18 = 0
The B.M. diagram is shown in Fig. 6.35 (c).
...(iii)
...(iv)
273
STRENGTH OF MATERIALS
Point of Contraflexure
This point will be between A and B where B.M. is zero after changing its sign. But B.M. at
any section at a distance x from A between A and B is given by equation (iii) as
Mx = 3x – x2
Equation Mx to zero for point of contraflexure, we get
0 = 3x – x2 = x(3 – x)
or
3–x=0
(∵ x cannot be zero as B.M. is not
changing sign at this point)
∴
x=3
Hence point of contraflexure will be at a distance of 3 m from A.
Problem 6.15. Draw the S.F. and B.M. diagrams for the overhanging beam carrying
uniformly distributed load of 2 kN/m over the entire length and a point load of 2 kN as shown
in Fig. 6.36. Locate the point of contraflexure.
Sol. First calculate the reactions RA and RB.
Taking moments of all forces about A, we get
RB × 4 = 2 × 6 × 3 + 2 × 6 = 36 + 12 = 48
48
∴
RB =
= 12 kN
4
and
RA = Total load – RB = (2 × 6 + 2) – 12 = 2 kN
2 kN
2 kN/m
A
B
(a)
C
4m
2m
RA
RB
6
(b)
+
2
+
2
D
B
A
–
1m
C
6
S.F. diagram
(c)
+
A
1.0
–
2m
B.M. diagram
Fig. 6.36
274
B
E
D
C
SHEAR FORCE AND BENDING MOMENT
S.F. Diagram
S.F. at A = + RA = + 2 kN
(i)The S.F. at any section between A and B at a distance x from A is given by,
F x = + RA – 2 × x
= 2 – 2x
...(i)
At A, x = 0 hence FA = 2 – 2 × 0 = 2 kN
At B, x = 4 hence FA = 2 – 2 × 4 = – 6 kN
The S.F. between A and B varies according to straight line law. At A, S.F. is positive and
at B, S.F. is negative. Hence between A and B, S.F. is zero. The point of zero S.F. is obtained by
substituting Fx = 0 in equation (i).
2
∴
0 = 2 – 2x or x = = 1 m
2
The S.F. is zero at point D. Hence distance of D from A is 1 m.
(ii)The S.F. at any section between B and C at a distance x from A is given by,
F x = + RA – 2 × 4 + RB – 2(x – 4)
= 2 – 8 + 12 – 2(x – 4) = 6 – 2(x – 4)
...(ii)
At B, x = 4 hence FB = 6 – 2(4 – 4) = + 6 kN
At C, x = 6 hence FC = 6 – 2(6 – 4) = 6 – 4 = 2 kN
The S.F. diagram is drawn as shown in Fig. 6.36 (b).
B.M. Diagram
B.M. at A is zero
(i) B.M. at any section between A and B at a distance x from A is given by,
x
= 2x – x2
...(iii)
2
The above equation shows that the B.M. between A and B varies according to parabolic
Mx = RA × x – 2 × x ×
law.
At A, x = 0 hence MA = 0
At B, x = 4 hence MB = 2 × 4 – 42 = – 8 kNm
Max. B.M. is at D where S.F. is zero after changing sign
At D, x = 1 hence MD = 2 × 1 – 12 = 1 kNm
The B.M. at C is zero. The B.M. also varies between B and C according to parabolic law.
Now the B.M. diagram is drawn as shown in Fig. 6.36 (c).
Point of Contraflexure
This point is at E between A and B, where B.M. is zero after changing its sign. The
distance of E from A is obtained by putting Mx = 0 in equation (iii).
∴
0 = 2x – x2 = x(2 – x)
2–x = 0
and
x = 2 m. Ans.
Problem 6.16. A beam of length 12 m is simply supported at two supports which are 8 m
apart, with an overhang of 2 m on each side as shown in Fig. 6.37. The beam carries a concentrated load of 1000 N at each end. Draw S.F. and B.M. diagrams.
275
STRENGTH OF MATERIALS
Sol. As the loading on the beam is symmetrical. Hence reactions RA and RB will be equal
and their magnitude will be half of the total load.
(1000 + 1000)
= 1000 N
∴
RA = RB =
2
1000 N
1000 N
A
( a)
B
C
D
2m
8m
2m
RB = 1000 N
RA = 1000 N
+
C
(b )
1000 N
C
(c )
A
S.F. diagram
–
A
2000 Nm
–
1000 N
B
D
B
D
2000 Nm
B.M. diagram
Fig. 6.37
S.F. at C
S.F. remains constant (i.e.,
S.F. at A
S.F. remains constant (i.e.,
= – 1000 N
= – 1000 N) between C and A
= – 1000 + RA = – 1000 + 1000 = 0
= 0) between A and B
S.F. at B
= 0 + 1000 = + 1000 N
S.F. remains constant (i.e., = 1000 N) between B and D
S.F. diagrams is drawn as shown in Fig. 6.37 (b).
B.M. Diagram
B.M. at C = 0
B.M. at A = – 1000 × 2 = – 2000 Nm
(– ve sign is due to hogging moment)
B.M. between C and A varies according to straight line law.
The B.M. at any section in AB at a distance of x from C is given by,
Mx = – 1000 × x + RA(x – 2)
= – 1000 × x + 1000(x – 2) = – 2000 Nm
Hence B.M. between A and B is constant and equal to – 2000 Nm.
B.M. at D = 0.
∴ B.M. diagram is shown in Fig. 6.37 (c).
Note. In this particular case, the S.F. is zero between AB and B.M. is constant. Hence length AB
is subjected to only constant B.M. The length between A and B is absolutely free from shear force.
276
SHEAR FORCE AND BENDING MOMENT
Problem 6.17. Draw the S.F. and B.M. diagrams for the beam which is loaded as shown
in Fig. 6.38. Determine the points of contraflexure within the span AB.
Sol. First calculate the reactions RA and RB.
Taking moments about A, we have
RB × 8 + 800 × 3 = 2000 × 5 + 1000(8 + 2)
or
8RB + 2400 = 10000 + 10000
20000 − 2400 17600
∴
RB =
=
= 2200 N
8
8
and
RA = Total load – RB = 3800 – 2200 = 1600
800 N
2000 N
C
A
1000 N
B
D
E
(a)
5m
8m
3m
2m
RA = 1600
(b)
+
C
800
A
–
800
RB = 2200
D
+
B
–
1200
1000
E
S.F. diagram
A
C
(c)
–
O1
+
1600
D
O2
B
E
–
2000
2400
B.M. diagram
Fig. 6.38
S.F. Diagram
S.F at C
= – 800 N
S.F. between C and A remains – 800 N
S.F. at A
= – 800 + RA = – 800 + 1600 = + 800 N
S.F. between A and D remains + 800 N
S.F. at D
= + 800 – 2000 = – 1200 N
S.F. between D and B remains – 1200 N
S.F. at B
= – 1200 + RB = – 1200 + 2200 = + 1000 N
S.F. between B and E remains + 1000 N
S.F. diagram is shown in Fig. 6.38.
277
STRENGTH OF MATERIALS
B.M. Diagram
B.M. at C
B.M. at A
B.M. at D
=0
= – 800 × 3 = – 2400 Nm
= – 800 × (3 + 5) + RA × 5
= – 800 × 8 + 1600 × 5
= – 6400 + 8000 = + 1600 Nm
B.M. at B
= – 1000 × 2 = – 2000 Nm
B.M. at E
=0
The B.M. diagram is drawn as shown in Fig. 6.38 (c).
Points of Contraflexure
There will be two points of contraflexure O1 and O2, where B.M. becomes zero after changing
its sign. Point O1 lies between A and D, whereas the point O2 lies between D and B.
(i) Let the point O1 is x metre from A.
Then B.M. at O1
= – 800(3 + x) + RA × x = – 800(3 + x) + 1600x
= – 2400 – 800x + 1600x = – 2400 + 800x
But B.M. at O1 is zero
∴
0 = – 2400 + 800x
or
x=
2400
= 3 m. Ans.
800
(ii) Let the point O be x metre from B.
Then B.M. at O2
= 1000(x + 2) – RB × x = 1000x + 2000 – 2200 × x = 2000 – 1200x
But B.M. at O2
=0
∴
0 = 2000 – 1200x
2000 5
= = 1.67 m from B. Ans.
1200 3
Problem 6.18. A horizontal beam 10 m long is carrying a uniformly distributed load of
1 kN/m. The beam is supported on two supports 6 m apart. Find the position of the supports, so
that B.M. on the beam is as small as possible. Also draw the S.F. and B.M. diagrams.
Sol. The beam CD is 10 m long. Let the two supports 6 m apart are at A and B.
Let
x = Distance of support A from C in metre
∴
x=
Then distance of support B from end D
= 10 – (6 + x) = (4 – x) m
First calculate the reactions RA and RB.
Taking moments about A, we get
1×x×
x
(10 − x)
+ RB × 6 = (10 – x) × 1 ×
2
2
(10 − x) 2
x2
or x2 + 12RB = (10 × x)2 = 100 + x2 – 20x
+ 6 RB =
2
2
12RB = 100 + x2 – 20x – x2 = 100 – 20x
or
or
∴
278
RB =
100 − 20 x 4(25 − 5 x)
1
5
=
= (25 – 5x) = (5 – x)
12
12
3
3
SHEAR FORCE AND BENDING MOMENT
RA = Total load – RB
5
30 − 25 + 5 x
= 10 × 1 – (5 – x) =
3
3
5 + 5x 5
= (1 + x)
=
3
3
In the present case of overhanging beam, the maximum negative B.M. will be at either of
the two supports and the maximum positive B.M. will be in the span AB. If the B.M. on the beam
is as small as possible, then the length of the overhanging portion should be so adjusted that the
maximum negative B.M. at the support is equal to the maximum positive B.M. in the span AB.
The B.M. will be maximum in the span AB at a point where S.F. is zero.
Let B.M. is maximum (or S.F. is zero) at a section in AB at a distance of y m from C.
and
1 kN/m
D
C
A
x
B
6m
RA
(4 – x)
RB
10 m
Fig. 6.39
or
But S.F. at this section = y × 1 – RA
∴
y × 1 – RA = 0
5
y×1–
(1 + x) = 0
3
5
∴
y = (1 + x)
3
Now B.M. at the support A
=–1×x×
and
...(i)
x
x3
=−
5
2
...(ii)
B.M. at a distance y from C
y
+ RA × (y – x)
2
=–1×y×
y2 5
+ (1 + x)(y – x)
2
3
2
1 5
5
5
−
(1 + x) + (1 + x)
(1 + x) − x
2 3
3
3
5
5 (1 + x) 5
(1 + x) −
+ (1 + x) − x
3
3×2
3
5
− 5 − 5 x + 10 + 10 x − 6 x
(1 + x)
3
6
5− x
5
5
=
(1 + x)
(– x2 + 4x + 5)
6
18
3
=−
=
=
=
=
LM
N
LM
N
LM
N
LM
N
LM
N
OP
Q
OP
Q
OP
Q
OP
Q
LM∵ R = 5 (1 + x)OP
3
N
Q
OP LM∵ y = 5 (1 + x)OP
3
Q
Q N
A
...(iii)
279
STRENGTH OF MATERIALS
For the condition that the B.M. shall be as small as possible, the hogging moment at the
support A and the maximum sagging moment in the span AB should be numerically equal.
∴ Equating equations (ii) and (iii) and ignoring the – ve sign of B.M. at A, we get
5
x2
(– x2 + 4x + 5) =
18
2
2
– 5x + 20x + 25 = 9x2
or
14x2 – 20x – 25 = 0
The above equation is a quadratic equation. Hence its solution is given by
∴
or
x =
20 ± 20 2 + 4 × 14 × 25 20 ± 400 + 1400 20 ± 42.42
=
=
2 × 14
28
28
20 ± 42.42
28
= 2.23 m
Substituting this value of x in equation (i), we get
=
(Neglecting – ve value)
5
5 × 3.23
= 5.38 m
(1 + 2.23) =
3
3
Now the values of reactions RA and RB are obtained as :
y =
RA =
and
5
5
(1 + x) = (1 + 2.23) = 5.38 kN
3
3
5
5
(5 − x) = (5 – 2.23) = 4.62 kN
3
3
Now the S.F. and B.M. diagrams can be drawn as shown in Fig. 6.40.
RB =
S.F. Diagram
S.F. at C
= 0
S.F. just on L.H.S. of
A = – 1 × 2.23 = – 2.23 kN
Shear force varies between C and A by a straight line law.
S.F. just on L.H.S. of
A = – 2.23 + RA
= – 2.23 + 5.38 = + 3.15 kN
S.F. just on L.H.S. of
B = + 3.15 – 1 × 6 = – 2.85 kN
Shear force between A and B varies by a straight line law.
S.F. just on R.H.S. of
B = – 2.85 + RB
= – 2.85 + 4.62 = + 1.17 kN
S.F. at
D = 1.17 – 1 × 1.77 = 0
S.F. between B and D varies by a straight line law.
S.F. diagram is drawn as shown in Fig. 6.40 (c).
B.M. Diagram
B.M. at C
B.M. at A
280
= 0
= – 1 × 2.23 ×
2.23
= – 2.49 kNm
2
SHEAR FORCE AND BENDING MOMENT
1 kN/m
C
D
A
2.23
(a)
RA = 5.38 N
3.15
+
–
1.77
+
E
–
2.23 kN
1.77
RB = 4.62 N
10 m
A
(b ) C
B
6m
D
B
2.85
y = 5.38 m
S.F. diagram
2.49
+
C
(c )
A
–
B
E
2.49
kNm
1.06
D
–
B.M. diagram
Fig. 6.40
B.M. at E (i.e., at a distance y = 5.38 m from point C)
= – 1 × 5.38 ×
5.38
+ RA × (5.38 – 2.23)
2
5.38 2
+ 5.38 × 3.15 = 2.49 kNm
2
1.77
= – 1.06 kNm
B.M. at B
= – 1 × 1.77 ×
2
The B.M. between C and A ; between A and B ; and between B and D varies according to
parabolic law. B.M. diagram is shown in Fig. 6.40 (c).
=−
6.16. S.F. AND B.M. DIAGRAMS FOR BEAMS CARRYING INCLINED LOAD..
The shear force is defined as the algebraic sum of the vertical forces at any section of a
beam to the right or left of the section. But when a beam carries inclined loads, then these
inclined loads are resolved into their vertical and horizontal components. The vertical components only will cause shear force and bending moments.
281
STRENGTH OF MATERIALS
The horizontal components of the inclined loads will introduce axial force or thrust in the
beam. The variation of axial force for all sections of the beam can be shown by a diagram known
as thrust diagram or axial force diagram.
In most of the cases, one end of the beam is hinged and the other end is supported on
rollers. The roller support cannot provide any horizontal reaction. Hence only the hinged end
will provide the horizontal reaction.
Problem 6.19. A horizontal beam AB of length 4 m is hinged at A and supported on
rollers at B. The beam carries inclined loads of 100 N, 200 N and 300 N inclined at 60°, 45° and
30° to the horizontal as shown in Fig. 6.41. Draw the S.F., B.M. and thrust diagrams for the
beam.
Sol. First of all, resolve the inclined loads into their vertical and horizontal components.
The inclined load at C is having horizontal component
= 100 × cos 60° = 100 × 0.5 = 50 N,
whereas the vertical component= 100 × sin 60° = 100 × 0.866 = 86.6 N
100 N
200 N
60°
A
300 N
45°
C
30°
D
B
E
(a)
1m
1m
86.6 N
HA =
451.8 N
1m
141.4 N
50 N
A
1m
150 N
141.4 N
259.8 N
(b )
C
D
RA = 173.15 N
(c )
Load diagram
86.55
A
C
141.4
RB = 204.85
E
D
( d)
B
54.85
150
S.F. diagram
–
204.85
259.7
+
173.15 Nm
A
C
D
50 N
451.2 N
(e )
A
+
401.2
C
204.8
B.M. diagram
E
B
E
B
141.4
259.8
D
Thrust diagram
Fig. 6.41
282
B
86.6
+
173.15
E
SHEAR FORCE AND BENDING MOMENT
Similarly the inclined load at D is having horizontal component
= 200 × cos 45° = 141.4 N,
whereas the vertical component = 200 × sin 45° = 141.4 N
The inclined load at E is having horizontal component
= 300 × cos 30° = 300 × 0.866 = 259.8 N,
whereas the vertical component = 300 × sin 30° = 150 N
The horizontal and vertical components of all inclined loads are shown in Fig. 6.41 (b).
As beam is supported on rollers at B, hence roller support at B will not provide any
horizontal reaction. The horizontal reaction will be only provided by hinged end A.
Let
HA = Horizontal reaction at A
= Sum of all horizontal components of inclined loads
= 50 + 141.4 + 259.8
(All horizontal components are acting in the same direction)
= 451.20 N
To find the reactions RA and RB, take the moments of all forces about A,
∴
RB × 4 = 86.6 × 1 + 141.4 × 2 + 150 × 3 = 819.4
819.4
or
RB =
= 204.85 N
4
∴
RA = Total vertical load – RB
= (86.6 + 141.4 + 150) – 204.85 = 173.15 N
S.F. Diagram
The S.F. is due to vertical loads (including vertical reactions) only
S.F. at A
= + RA = + 173.15 N
S.F. remains constant between A and C and equal to 173.15 N
S.F. suddenly changes at C due to point load and S.F. at C
= 173.15 – 86.6 = 86.55 N
S.F. remains constant between C and D and is equal to 86.55 N
S.F. at D
= 86.55 – 141.40 = – 54.85 N
The S.F. remains constant between E and D and is equal to – 54.85 N
The S.F. at B
= – 54.85 – 150.00 = – 204.85 N
The S.F. diagram is shown in Fig. 6.41 (c).
B.M. Diagram
The B.M. is only due to vertical loads (including vertical reactions) only
The B.M. at A
=0
B.M. at C
= RA × 1 = 173.15 × 1 = 173.15 Nm
B.M. at D
= RA × 2 – 86.6 × 1
= 173.15 × 2 – 86.6 = 259.7 Nm
B.M. at E
= RA × 3 – 86.6 × 2 – 141.4 × 1
= 173.15 × 3 – 86.6 × 2 – 141.4 = + 204.85 Nm
B.M. at B
=0
The B.M. diagram is shown in Fig. 6.41 (d).
283
STRENGTH OF MATERIALS
Thrust Diagram or Axial Force Diagram
The thrust diagram is due to horizontal components including horizontal reaction.
Axial force at A
= + HA = 451.20 N
The axial force remains constant between A and C and is equal to 451.20 N
Axial force at C
= HA – 50 = 451.20 – 50 = 401.2 N
Axial force remains constant between C and D and is equal to 401.2 N
Axial force at D
= 401.2 – 141.40 = 259.8 N
Axial force remains constant between D and E and is equal to 259.8 N
Axial force at E
= 259.8 – 259.8 = 0
Axial force between E and B is zero.
Thrust diagram or axial force diagram is shown in Fig. 6.41 (e).
Problem 6.20. A horizontal beam AB of length 8 m is hinged at A and placed on rollers
at B. The beam carries three inclined point loads as shown in Fig. 6.42. Draw the S.F., B.M.
and axial force diagrams of the beam.
Sol. First resolve the inclined loads into their vertical and horizontal components.
Vertical component of force at C
= 4 sin 30° = 4 × 0.5 = 2 kN
Horizontal component of force at C
= 4 × cos 30° = 4 × 0.866 = 3.464 kN →
Vertical component of force at D
= 8 × sin 60° = 8 × 0.866 = 6.928 kN
Horizontal component of force at D
= 8 × cos 60° = 8 × 0.5 = 4 kN ←
Vertical component of force at E
= 6 × sin 45° = 6 × 0.707 = 4.242 kN
Horizontal component of force at E
= 6 × cos 45° = 6 × 0.707 = 4.242 kN ←
The horizontal and vertical components of all inclined loads are shown in Fig. 6.42 (b).
The horizontal reaction will be provided by the hinged end A.
∴ Horizontal reaction at A,
HA = – 3.464 + 4 + 4.242 = 4.778 kN
To find vertical reactions RA and RB, take the moments of all forces about A.
∴
RB × 8 = 2 × 2 + 6.928 × 4 + 4.242 × 6 = 57.164
57.164
∴
RB =
= 7.1455 kN
8
Now
RA = Total vertical loads – RB
= (2 + 6.928 + 4.242) – 7.1455 = 6.0245 kN
S.F. Diagram
S.F. is due to vertical loads
S.F. at A
= + RA = + 6.0245 kN
284
SHEAR FORCE AND BENDING MOMENT
8 kN
4 kN
60°
30°
A
(a)
C
2m
6.928
3.464
2m
4.242
4 kN
A
C
B
E
2m
2 kN
HA
45°
D
2m
(b)
6 kN
4.242
D
E
B
RB
RA
+
6.0245
4.0245
(c)
6.928
E
A
C
D
B
2.9035
7.1455
S.F. diagram
4.242
20.098
12.049
(d)
A
C
+
B.M. diagram
3.464
(e)
D
14.291
E
B
4.0
8.248
4.778
4.242
A
C
D
Axial force diagram
E
B
Fig. 6.42
S.F. remains 6.0245 kN between A and C
S.F. at C
= + 6.0245 – 2 = + 4.0245 kN
S.F. remains 4.0245 kN between C and D
S.F. at D
= + 4.0245 – 6.928 = – 2.9035 kN
S.F. remains – 2.9035 kN between D and E
S.F. at E
= – 2.9035 – 4.242 = – 7.1455 kN
S.F. remains constant between E and B and equal to – 7.1455
S.F. diagram is shown in Fig. 6.42 (c).
285
STRENGTH OF MATERIALS
B.M. Diagram
B.M. is only due to vertical loads
B.M. at A
=0
B.M. at C
= RA × 2 = 6.0245 × 2 = 12.049 kNm
B.M. at D
= 6.0245 × 4 – 2 × 2 = 20.098 kNm
B.M. at E
= 6.0245 × 6 – 2 × 4 – 6.928 × 2 = 14.291 kNm
B.M at B
=0
B.M. diagram is shown in Fig. 6.42 (d).
Axial Force Diagram
Axial force is due to horizontal components including horizontal reaction.
Axial force at A
= + HA = + 4.778 kN
Axial force remains 4.778 kN between A and C
Axial force at C
= + 4.778 + 3.464 = + 8.242
Axial force remains 8.242 kN between C and D
Axial force at D
= 8.242 – 4.0 = + 4.242
Axial force remains 4.242 kN between D and E
Axial force at E
= + 4.242 – 4.242 = 0
Axial force remains zero between E and B
Axial force diagram is shown in Fig. 6.42 (e).
6.17. SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR BEAMS SUBJECTED
TO COUPLES
When a beam is subjected to a couple at a section, only the bending moment at the section
of the couple changes suddenly in magnitude equal to that of the couple. But the S.F. does not
change at the section of the couple as there is no change in load due to couple at the section. But
while calculating the reactions, the magnitude of the couple is taken into account.
The sudden change in B.M. at the section of the couple can also be obtained by calculating
B.M. separately with the help of both the reactions.
Problem 6.21. A simply supported beam AB of length 6 m is hinged at A and B. It is
subjected to a clockwise couple of 24 kNm at a distance of 2 m from the left end A. Draw the
S.F. and B.M. diagrams.
Sol. Fig. 6.43 (a) shows the simply supported beam AB, hinged at A and B. The clockwise
couple at C will try to lift the beam up at the support A, and to depress the beam down at the
support B. Hence the reaction at A will be downwards and at B the reaction will be upwards as
shown in Fig. 6.43 (b).
To find reactions of RA and RB, take the moments about A.
(∵ Moment due to RB is anti-clockwise and moment
∴
RB × 6 – 24 = 0
at C is clockwise)
24
= 4 kN ↑
∴
RB =
6
Since there is no external vertical load on the beam, therefore the reaction at A will be the
same, as that of B, but in opposite direction.
(∵ Load on beam = 0)
∴
RA = Load on beam – RB
= – RB = – 4 kN.
286
SHEAR FORCE AND BENDING MOMENT
S.F. Diagram
S.F. at A
= RA = – 4 kN
The S.F. remains constant (i.e., equal to – 4 kN) between A and B.
The S.F. diagram is shown in Fig. 6.43 (c).
C
A
24 kNm
B
(a)
2m
6m
RA
C
A
4m
24 kNm
B
(b)
2m
4m
A
(c)
Base line
C
RB
B
–
4 kN
4 kN
S.F. diagram
16
+
(d)
A
–
C
8 kNm
B.M. diagram
B
Fig. 6.43
B.M. Diagram
B.M. at A
=0
B.M. just on the L.H.S. of C
= RA × 2 = – 4 × 2 = – 8 kNm
B.M. just on the R.H.S. of C
= RB × 4 = 4 × 4 = + 16 kNm
(B.M. just on the R.H.S. of C can also be calculated as the sum of moments due to RA and
moment due to couple. But moment due to RA is anti-clockwise whereas due to couple is clockwise. Hence net B.M. on R.H.S. of C = – 8 + 24 = + 16 kNm).
There is a sudden change in B.M. at C due to couple.
B.M. at B
=0
B.M. diagram is shown in Fig. 6.43 (d).
Problem 6.22. A beam 10 m long and simply supported at each end, has a uniformly
distributed load of 1000 N/m extending from the left end upto the centre of the beam. There is
also an anti-clockwise couple of 15 kNm at a distance of 2.5 m from the right end. Draw the S.F.
and B.M. diagrams.
Sol. The reaction at A will be upwards. To find whether the reaction at B is upwards or
downwards, take the moments about A.
The following are the moments at A :
(i) Moment due to U.D.L. = 1000 × 5 ×
5
= 12500 Nm (clockwise)
2
287
STRENGTH OF MATERIALS
(ii) Moment of couple
∴ Net moment
= 15000 Nm (Anti-clockwise)
= 15000 – 12500
= 2500 Nm (Anti-clockwise)
This moment must be balanced by the moments due to reaction at B. Hence the moment
about A due to reaction at B should be equal to 2500 Nm (clockwise). This is only possible when
RB is acting downwards. This is shown in Fig. 6.44 (b).
15000 N/m
1000 N/m
C
B
A
D
( a)
5m
2.5 m
2.5 m
RB
C
(b)
D
A
RA
(c)
+
5250 N
250 N
250 N
A
C
S.F. diagram
D
B
Straight line
Parabolic
(d)
+
A
B.M. diagram
13750
C
14375
D
625 –
B
Fig. 6.44
∴
RB × 10 = 2500
∴
RB =
and
288
2500
= 250 N
10
RA = Total load on beam + RB
(Here RB is +ve as acting downwards)
= 1000 × 5 + 250 = 5250 N.
SHEAR FORCE AND BENDING MOMENT
S.F. Diagram
S.F. at A
= + RA = 5250 N
S.F. at C
= 5250 – 5 × 1000 = + 250 N
S.F. between A and C varies according to straight line law.
S.F between C and B remains constant and equal to + 250 N
S.F. diagram is shown in Fig. 6.44 (c).
B.M. Diagram
B.M at A
=0
5
B.M. at C
= RA × 5 – 1000 × 5 ×
2
= 5250 × 5 – 12500 = 13750 Nm
B.M. just on the left hand side of D
5
+ 2.5
= 5250 × 7.5 – 1000 × 5 ×
2
= 39375 – 25000 = 14375 Nm
B.M. just on the right hand side of D
= – RB × 2.5 = – 250 × 2.5 = – 625 Nm
B.M. at B
=0
The B.M. diagram is shown in Fig. 6.44 (d).
FG
H
IJ
K
6.18. RELATIONS BETWEEN LOAD, SHEAR FORCE AND BENDING MOMENT..
Fig. 6.45 shows a beam carrying a uniformly distributed load of w per unit length. Consider
the equilibrium of the portion of the beam between sections 1-1 and 2-2. This portion is at a
distance of x from left support and is of length dx.
1
2
w/m run
x
A
B
1
M
dx
2
M + dM
F
F + dF
Fig. 6.45
Let
F = Shear force at the section 1-1,
F + dF = Shear force at the section 2-2,
M = Bending moment at the section 1-1,
M + dM = Bending moment at the section 2-2.
The forces and moments acting on the length ‘dx’ of the beam are :
(i) The force F acting vertically up at the section 1-1.
(ii) The force F + dF acting vertically downwards at the section 2-2.
(iii) The load w × dx acting downwards.
(iv) The moments M and (M + dM) acting at section 1-1 and section 2-2 respectively.
289
STRENGTH OF MATERIALS
The portion of the beam of length dx is in equilibrium. Hence resolving the forces acting
on this part vertically, we get
F – w.dx – (F + dF) = 0
dF
= – w.
dx
The above equation shows that the rate of change of shear force is equal to the rate of
loading.
Taking the moments of the forces and couples about the section 2-2, we get
dx
M – w.dx .
+ F.dx = M + dM
2
w (dx) 2
+ F.dx = dM
or
−
2
Neglecting the higher powers of small quantities, we get
F.dx = dM
dM
dM
or
F=
or
= F.
dx
dx
The above equation shows that the rate of change of bending moment is equal to the shear
force at the section.
or
– dF = w.dx
or
HIGHLIGHTS
1. Shear force at a section is the resultant vertical force to the right or left of the section.
2. The diagram which shows the variation of the shear force along the length of a beam, is known as
shear force diagram.
3. Bending moment at a section is algebraic sum of the moments of all the forces acting to the left
or right of the section.
4. The diagram which shows the variation of the bending moment along the length of a beam, is
known as bending moment diagram.
5. A beam which is fixed at one end and free at the other is known as cantilever beam. But a beam
supported or resting freely on the supports at its both ends, is known as simply supported beam.
6. If the end portion of a beam is extended beyond the support, such beam is known as overhanging
beam.
7. A load acting at a point, is known as concentrated load or a point load.
8. If a left portion of a section is considered, then S.F will be positive at the section if the resultant
of the vertical forces (including reactions) to the left of the section is upwards. But if the resultant
is acting downwards then S.F. at the section will be negative.
9. If a right portion of a section is considered, the S.F. will be positive at the section if the resultant
of the vertical forces to the right of the section is downwards. But if the resultant is acting
upwards then S.F. at the section will be negative.
10. If a left portion of a section is considered, the B.M. will be positive at the section if the moment
of all vertical forces and of reaction, at the section is clockwise. But if the resultant moment at
the section is anti-clockwise, then B.M. will be negative.
11. If a right portion of a section is considered, the B.M. will be positive at the section if the resultant
moment at the section is anti-clockwise. But if the resultant moment at the section is clockwise,
then B.M. will be positive.
12. The shear force changes suddenly at a section where there is a vertical point load.
290
SHEAR FORCE AND BENDING MOMENT
13.
14.
15.
16.
17.
18.
The shear force between any two vertical loads remains constant.
Shear force for a uniformly distributed load varies according to a straight line law whereas B.M.
varies according to a parabolic curve.
B.M. is maximum at a section where S.F. is zero after changing its sign.
The point where B.M. is zero after changing its sign, is known as point of contraflexure or point of
inflexion.
When an inclined load is acting on a beam, then inclined load is resolved into two components.
Vertical component will cause S.F. and B.M. whereas horizontal component will cause axial
thrust in the beam.
When a beam is subjected to a couple at a section, then B.M. changes suddenly at the section but
S.F. remains unaltered at the section.
EXERCISE
(A) Theoretical Questions
1. Define and explain the following terms :
Shear force, bending moment, shear force diagram and bending moment diagram.
2. What are the different types of beams ? Differentiate between a cantilever and a simply supported beam.
3. What are the different types of loads acting on a beam ? Differentiate between a point load and
a uniformly distributed load.
4. What are the sign conventions for shear force and bending moment in general ?
5. Draw the S.F. and B.M. diagrams for a cantilever of length L carrying a point load W at the free
end.
6. Draw the S.F. and B.M. diagrams for a cantilever of length L carrying a uniformly distributed
load of w per m length over its entire length.
7. Draw the S.F. and B.M. diagrams for a cantilever of length L carrying a gradually varying load
from zero at the free end to w per unit length at the fixed end.
8. Draw the S.F. and B.M. diagrams for a simply supported beam of length L carrying a point load
W at its middle point.
9. Draw the S.F. and B.M. diagrams for a simply supported beam carrying a uniformly distributed
load of w per unit length over the entire span. Also calculate the maximum B.M.
10. Draw the S.F. and B.M. diagrams for a simply supported beam carrying a uniformly varying load
from zero at each end to w per unit length at the centre.
11. What do you mean by point of contraflexure ? Is the point of contraflexure and point of inflexion
different ?
12. How many points of contraflexure you will have for simply supported beam overhanging at one
end only ?
13. How will you draw the S.F. and B.M. diagrams for a beam which is subjected to inclined loads ?
14. What do you mean by thrust diagram ?
15. Draw the S.F. and B.M. diagrams for a simply supported beam of length L which is subjected to
a clockwise couple μ at the centre of the beam.
(B) Numerical Problems
1. A cantilever beam of length 2 m carries a point load of 1 kN at its free end, and another load of 2
kN at a distance of 1 m from the free end. Draw the S.F. and B.M. diagrams for the cantilever.
[Ans. Fmax = + 3 kN ; Mmax = – 4 kNm]
291
STRENGTH OF MATERIALS
2. A cantilever beam of length 4 m carries point loads of 1 kN, 2 kN and 3 kN at 1, 2 and 4 m from
the fixed end. Draw the shear force and B.M. diagrams for the cantilever.
[Ans. Fmax = + 6 kN ; Mmax = – 17 kNm]
3. A cantilever of length 2 m carries a uniformly distributed load of 3 kN/m run over a length of 1 m
from the fixed end. Draw the S.F. and B.M. diagrams. [Ans. Fmax = + 3 kN ; Mmax = – 1.5 kNm]
4. A cantilever of length 5 m carries a uniformly distributed load of 2 kN/m length over the whole
length and a point load of 4 kN at the free end. Draw the S.F. and B.M diagrams for the cantilever.
[Ans. Fmax = + 14 kN ; Mmax = – 45 kNm]
5. A cantilever of length 4 m carries a uniformly distributed load of 1 kN/m run over the whole
length and a point load of 2 kN at a distance of 1 m from the free end. Draw the S.F. and B.M.
[Ans. Fmax = + 14 kN ; Mmax = – 14 kNm]
diagrams for the cantilever.
6. A cantilever 2 m long is loaded with a uniformly distributed load of 2 kN/m run over a length of
1 m from the free end. It also carries a point load of 4 kN at a distance of 0.5 m from the free end.
[Ans. Fmax = + 6 kN ; Mmax = – 9 kNm]
Draw the shear force and B.M. diagrams.
7. A cantilever of length 6 m carries two point loads of 2 kN and 3 kN at a distance of 1 m and 6 m
from the fixed end respectively. In addition to this the beam also carries a uniformly distributed
load of 1 kN/m over a length of 2 m at a distance of 3 m from the fixed end. Draw the S.F. and B.M.
[Ans. Fmax = + 7 kN ; Mmax = – 28 kNm]
diagrams.
8. A cantilever of length 6 m carries a gradually varying load, zero at the free end to 2 kN/m at the
fixed end. Draw the S.F. and B.M. diagrams for the cantilever.
[Ans. Fmax = + 6 kN ; Mmax = – 12 kNm]
9. A simply supported beam of length 8 m carries point loads of 4 kN and 6 kN at a distance of
2 m and 4 m from the left end. Draw the S.F. and B.M. diagrams for the beam.
[Ans. Mmax = + 20 kNm]
10. A simply supported beam of length 10 m carries point loads of 30 kN and 50 kN at a distance of
3 m and 7 m from the left end. Draw the S.F. and B.M diagrams for the beam.
[Ans. Mmax = + 132 kNm]
11. A simply supported beam of length 8 m carries point loads of 4 kN, 10 kN and 7 kN at a distance
of 1.5 m, 2.5 m and 2 m respectively from left end A. Draw the S.F. and B.M. diagrams for the
[Ans. Mmax = + 90 kNm]
simply supported beam.
12. A simply supported beam is carrying a uniformly distributed load of 2 kN/m over a length of 3 m
from the right end. The length of the beam is 6 m. Draw the S.F. and B.M. diagrams for the beam
[Ans. Mmax = + 5.06 kNm]
and also calculate the maximum B.M. on the section.
13. A beam of length 6 m is simply supported at the ends and carries a uniformly distributed load of
1.5 kN/m run and three concentrated loads of 1 kN, 2 kN and 3 kN acting at a distance of
1.5 m, 3 m and 4.5 m respectively from left end. Draw the S.F. and B.M. diagrams and determine the maximum bending moment.
[Ans. 12.75 kNm]
14. A beam of length 10 m is simply supported and carries point loads of 5 kN each at a distance of
3 m and 7 m from left support and also a uniformly distributed load of 1 kN/m between the point
[Ans. Mmax = + 23 kNm]
loads. Draw S.F. and B.M. diagrams for the beam.
15. A beam of length 6 m is simply supported at its ends. It is loaded with a gradually varying load
of 750 N/m from left hand support to 1500 N/m to the right hand support. Construct the S.F. and
B.M. diagrams and find the amount and position of the maximum B.M. over the beam.
[Ans. Mmax = 5077.5 Nm at 3.165 m from left hand support]
16. A simply supported beam of length 8 m rests on supports 6 m apart, the right hand end is
overhanging by 2 m. The beam carries a uniformly distributed load of 1500 N/m over the entire
length. Draw S.F. and B.M. diagrams and find the point of contraflexure, if any.
[Ans. Mmax = 5.33 kNm ; 5.33 from left hand support]
292
SHEAR FORCE AND BENDING MOMENT
17.
A simply supported beam of length 8 m rests on supports 5 m apart, the right hand end is
overhanging by 2 m and the left hand end is overhanging by 1 m. The beam carries a uniformly
distributed load of 5 kN/m over the entire length. It also carries two point loads of 4 kN and 6 kN
at each end of the beam. The load of 4 kN is at the extreme left of the beams, whereas the load of
6 kN is at the extreme right of the beam. Draw S.F. and B.M. diagrams for the beam and find the
points of contraflexure.
[Ans. 1.405 m and 4.955 from the extreme left of the beam]
18. A beam is loaded as shown in Fig. 6.46. Draw the S.F. and B.M. diagrams and find :
(i) maximum S.F.
(ii) maximum B.M.
(iii) point of inflexion.
50 kN
50 kN
2m
2m
40 kN
2.33 m
40 kN
2m
2m
Fig. 6.46
19.
[Ans. 50 kN ; 100 kN ; none]
A beam is loaded as shown in Fig. 6.47. Find the reactions at A and B. Also draw the S.F., B.M.
and thrust diagrams.
2 kN
A
45°
1m
1 kN
C
D
1.5 m
45°
1.5 m
3 kN
E
B
30°
2m
Fig. 6.47
20.
[Ans. RA= 2.09 kN ; RB = 1.53 kN ; HA = – 1.893 kN]
A simply supported beam of length 5 m, carries a uniformly distributed load of 100 N/m extending
from the left end to a point 2 m away. There is also a clockwise couple of 1500 Nm applied at the
centre of the beam. Draw the S.F. and B.M. diagrams for the beam and find the maximum
bending moment.
[Ans. 845 Nm at a distance of 1.3 m from left end]
293
7
CHAPTER
BENDING STRESSES IN
BEAMS
7.1. INTRODUCTION..
When some external load acts on a beam, the shear force and bending moments are set up
at all sections of the beam. Due to the shear force and bending moment, the beam undergoes
certain deformation. The material of the beam will offer resistance or stresses against these
deformations. These stresses with certain assumptions can be calculated. The stresses introduced
by bending moment are known as bending stresses. In this chapter, the theory of pure bending,
expression for bending stresses, bending stress in symmetrical and unsymmetrical sections,
strength of a beam and composite beams will be discussed.
7.2. PURE BENDING OR SIMPLE BENDING..
If a length of a beam is subjected to a constant bending moment and no shear force
(i.e., zero shear force), then the stresses will be set up in that length of the beam due to B.M. only
and that length of the beam is said to be in pure bending or simple bending. The stresses set up
in that length of beam are known as bending stresses.
W
W
B
A
(a)
C
D
a
L
a
RB = W
RA = W
+
C
(b )
A
C
D
S.F. diagram
–
W
W
B
B
A
D
–
(c )
wxa
B.M. diagram
wxa
Fig. 7.1
A beam simply supported at A and B and overhanging by same length at each support
is shown in Fig. 7.1. A point load W is applied at each end of the overhanging portion. The
295
STRENGTH OF MATERIALS
S.F. and B.M. for the beam are drawn as shown in Fig. 7.1 (b) and Fig. 7.1 (c) respectively.
From these diagrams, it is clear that there is no shear force between A and B but the B.M.
between A and B is constant.
This means that between A and B, the beam is subjected to a constant bending moment
only. This condition of the beam between A and B is known as pure bending or simple bending.
7.3. THEORY OF SIMPLE BENDING WITH ASSUMPTIONS MADE..
Before discussing the theory of simple bending, let us see the assumptions made in the
theory of simple bending. The following are the important assumptions :
1. The material of the beam is homogeneous* and isotropic**.
2. The value of Young’s modulus of elasticity is the same in tension and compression.
3. The transverse sections which were plane before bending, remain plane after bending
also.
4. The beam is initially straight and all longitudinal filaments bend into circular arcs
with a common centre of curvature.
5. The radius of curvature is large compared with the dimensions of the cross-section.
6. Each layer of the beam is free to expand or contract, independently of the layer, above or
below it.
Theory of Simple Bending
Fig. 7.2 (a) shows a part of a beam subjected to simple bending. Consider a small length δx
of this part of beam. Consider two sections AB and CD which are normal to the axis of the beam
N – N. Due to the action of the bending moment, the part of length δx will be deformed as shown
in Fig. 7.2 (b). From this figure, it is clear that all the layers of the beam, which were originally
of the same length, do not remain of the same length any more.
The top layer such as AC has deformed to the shape A′C′. This layer has been shortened in
its length. The bottom layer BD has deformed to the shape B′D′. This layer has been elongated.
From the Fig. 7.2 (b), it is clear that some of the layers have been shortened while some of them
are elongated. At a level between the top and bottom of the beam, there will be a layer which is
neither shortened nor elongated. This layer is known as neutral layer or neutral
M
M
A
C
A′
N
N
N′
C′
N′
Axis of beam
B
δx
B′
D
(a) Before bending
D′
(b) After bending
Fig. 7.2
*Homogeneous means the material is of the same kind throughout.
** Isotropic means that the elastic properties in all directions are equal.
296
BENDING STRESSES IN BEAMS
surface. This layer in Fig. 7.2 (b) is shown by N′ – N′ and in Fig. 7.2 (a) by N – N. The line of
intersection of the neutral layer on a cross-section of a beam is known as neutral axis (written as
N.A.).
The layers above N – N (or N′ – N′) have been shortened and those below, have been
elongated. Due to the decrease in lengths of the layers above N – N, these layers will be subjected
to compressive stresses. Due to the increase in the lengths of layers below N – N, these layers
will be subjected to tensile stresses.
We also see that the top layer has been shortened maximum. As we proceed towards the
layer N – N, the decrease in length of the layers decreases. At the layer N – N, there is no change
in length. This means the compressive stress will be maximum at the top layer. Similarly the
increase in length will be maximum at the bottom layer. As we proceed from bottom layer
towards the layer N – N, the increase in length of layers decreases. Hence the amount by which
a layer increases or decreases in length, depends upon the position of the layer with respect to
N – N. This theory of bending is known as theory of simple bending.
7.4. EXPRESSION FOR BENDING STRESS..
Fig. 7.3 (a) shows a small length δx of a beam subjected to a simple bending. Due to the
action of bending, the part of length δx will be deformed as shown in Fig. 7.3 (b). Let A′B′ and C′D′
meet at O.
Let R = Radius of neutral layer N′N′
θ = Angle subtended at O by A′B′ and C′D′ produced.
O
q
R
A
C
N
N
A′
E
B
y
dx
(a)
F
D
C′
N′
E′
N′
F′
y
B′
D′
(b)
(c) Stress Diagram
Fig. 7.3
7.4.1. Strain Variation Along the Depth of Beam. Consider a layer EF at a distance
y below the neutral layer NN. After bending this layer will be elongated to E′F′.
Original length of layer
EF = δx.
Also length of neutral layer NN = δx.
After bending, the length of neutral layer N′N′ will remain unchanged. But length of
layer E′F′ will increase. Hence
N′N′ = NN = δx.
297
STRENGTH OF MATERIALS
Now from Fig. 7.3 (b),
N′N′ = R × θ
and
E′F′ = (R + y) × θ
(∵ Radius of E′F′ = R + y)
But
N′N′ = NN = δx.
Hence
δx = R × θ
∴ Increase in the length of the layer EF
= E′F′ – EF = (R + y) θ – R × θ
(∵ EF = δx = R × θ)
= y×θ
∴ Strain in the layer EF
Increase in length
=
Original length
y×θ y×θ
=
=
(∵ EF = δx = R × θ)
EF
R×θ
y
=
R
As R is constant, hence the strain in a layer is proportional to its distance from the neutral axis.
The above equation shows the variation of strain along the depth of the beam. The variation of
strain is linear.
7.4.2. Stress Variation
Let
σ = Stress in the layer EF
E = Young’s modulus of the beam
Stress in the layer EF
Then
E =
Strain in the layer EF
σ
y
=
∵ Strain in EF =
y
R
R
y E
∴
σ = E×
=
×y
...(7.1)
R R
Since E and R are constant, therefore stress in any layer is directly proportional to the
distance of the layer from the neutral layer. The equation (7.1) shows the variation of stress
along the depth of the beam. The variation of stress is linear.
In the above case, all layers below the neutral layer are subjected to tensile stresses whereas
the layers above neutral layer are subjected to compressive stresses. The Fig. 7.3 (c) shows the
stress distribution.
Equation (7.1) can also be written as
σ E
...(7.2)
=
y R
FG IJ
H K
FG
H
IJ
K
7.5. NEUTRAL AXIS AND MOMENT OF RESISTANCE..
The neutral axis of any transverse section of a beam is defined as the line of intersection
of the neutral layer with the transverse section. It is written as N.A.
In Art. 7.4, we have seen that if a section of a beam is subjected to pure sagging moment,
then the stresses will be compressive at any point above the neutral axis and tensile below the
298
BENDING STRESSES IN BEAMS
neutral axis. There is no stress at the neutral axis. The stress at a
distance y from the neutral axis is given by equation (7.1) as
E
× y.
σ=
R
Fig. 7.4 shows the cross-section of a beam. Let N.A. be the
neutral axis of the section. Consider a small layer at a distance y
from the neutral axis. Let dA = Area of the layer.
Now the force on the layer
= Stress on layer × Area of layer
= σ × dA
dy
y
N
A
Fig. 7.4
IJ
K
FG
H
E
E
×y
× y × dA
...(i) ∵ σ =
R
R
Total force on the beam section is obtained by integrating the above equation.
∴ Total force on the beam section
=
z
E
× y × dA
R
E
y × dA
=
(∵ E and R is constant)
R
But for pure bending, there is no force on the section of the beam (or force is zero).
=
∴
or
E
R
z
z
z
y × dA = 0
y × dA
=0
FG as E cannot be zeroIJ
K
H R
Now y × dA represents the moment of area dA about neutral axis. Hence ∫ y × dA represents the moment of entire area of the section about neutral axis. But we know that moment of
any area about an axis passing through its centroid, is also equal to zero. Hence neutral axis
coincides with the centroidal axis. Thus the centroidal axis of a section gives the position of
neutral axis.
7.5.1. Moment of Resistance. Due to pure bending, the layers above the N.A. are subjected to compressive stresses whereas the layers below the N.A. are subjected to tensile stresses.
Due to these stresses, the forces will be acting on the layers. These forces will have moment
about the N.A. The total moment of these forces about the N.A. for a section is known as moment
of resistance of that section.
The force on the layer at a distance y from neutral axis in Fig. 7.4 is given by equation (i), as
E
× y × dA
R
Moment of this force about N.A.
= Force on layer × y
Force on layer
=
E
× y × dA × y
R
E
× y2 × dA
=
R
=
299
STRENGTH OF MATERIALS
z
z
Total moment of the forces on the section of the beam (or moment of resistance)
E
E
y 2 × dA
× y2 × dA =
R
R
Let M = External moment applied on the beam section. For equilibrium the moment of
resistance offered by the section should be equal to the external bending moment.
=
z
E
y 2 × dA .
R
But the expression ∫ y2 × dA represents the moment of inertia of the area of the section
about the neutral axis. Let this moment of inertia be I.
∴
M=
E
× I or
R
But from equation (7.2), we have
∴
M=
M E
=
I
R
...(7.3)
σ E
=
y R
M σ E
= =
∴
...(7.4)
I
y R
Equation (7.4) is known as bending equation.
In equation (7.4), the different quantities are expressed in consistent units as given
below :
M is expressed in N mm ; I in mm4
σ is expressed in N/mm2 ; y in mm
and E is expressed in N/mm2 ; R in mm.
7.5.2. Condition of Simple Bending. Equation (7.4) is applicable to a member which is
subjected to a constant bending moment and the member is absolutely free from shear force. But
in actual practice, a member is subjected to such loading that the B.M. varies from section to
section and also the shear force is not zero. But shear force is zero at a section where bending
moment is maximum. Hence the condition of simple bending may be assumed to be satisfied at
such a section. Hence the stresses produced due to maximum bending moment, are obtained
from equation (7.4) as the shear forces at these sections are generally zero. Hence the theory and
equations discussed in the above articles are quite sufficient and give results which enables the
engineers to design beams and structures and calculate their stresses and strains with a reasonable
degree of approximation where B.M. is maximum.
7.6. BENDING STRESSES IN SYMMETRICAL SECTIONS..
The neutral axis (N.A.) of a symmetrical section (such as circular, rectangular or square)
lies at a distance of d/2 from the outermost layer of the section where d is the diameter (for a
circular section) or depth (for a rectangular or a square section). There is no stress at the neutral
axis. But the stress at a point is directly proportional to its distance from the neutral axis. The
maximum stress takes place at the outermost layer. For a simply supported beam, there is a
compressive stress above the neutral axis and a tensile stress below it. If we plot these stresses,
we will get a figure as shown in Fig. 7.5.
300
BENDING STRESSES IN BEAMS
sc
d/2
N
A
Stress distribution
across a section
d
st
Fig. 7.5
Problem 7.1. A steel plate of width 120 mm and of thickness 20 mm is bent into a
circular arc of radius 10 m. Determine the maximum stress induced and the bending moment
which will produce the maximum stress. Take E = 2 × 105 N/mm2.
Sol. Given :
Width of plate,
b = 120 mm
Thickness of plate,
t = 20 mm
bt 3 120 × 20 3
= 8 × 104 mm4
=
12
12
Radius of curvature,
R = 10 m = 10 × 103 mm
Young’s modulus,
E = 2 × 105 N/mm2
Let
σmax = Maximum stress induced, and
M = Bending moment.
σ E
Using equation (7.2),
=
y R
E
∴
σ=
×y
...(i)
R
Equation (i) gives the stress at a distance y from N.A.
Stress will be maximum, when y is maximum. But y will be maximum at the top layer or
bottom layer.
t 20
∴
ymax = =
= 10 mm.
2
2
Now equation (i) can be written as
E
σmax =
× ymax
R
2 × 10 5
=
× 10 = 200 N/mm2. Ans.
10 × 10 3
From equation (7.4), we have
M E
=
I
R
E
2 × 10 5
∴
M=
×I=
× 8 × 104
R
10 × 10 3
= 16 × 105 N mm = 1.6 kNm. Ans.
∴ Moment of inertia,
I=
301
STRENGTH OF MATERIALS
Problem 7.2. Calculate the maximum stress* induced in a cast iron pipe of external
diameter 40 mm, of internal diameter 20 mm and of length 4 metre when the pipe is supported
at its ends and carries a point load of 80 N at its centre.
Sol. Given :
External dia.,
D = 40 mm
Internal dia.,
d = 20 mm
Length,
L = 4 m = 4 × 1000 = 4000 mm
Point load,
W = 80 N
In case of simply supported beam carrying a point load at the centre, the maximum
bending moment is at the centre of the beam.
40 mm
20 mm
80 N
4m
(b) Area of cross-section
(a)
Fig. 7.6
W×L
4
80 × 4000
∴ Maximum B.M.
=
= 8 × 104 Nmm
4
∴
M = 8 × 104 Nmm
Fig. 7.6 (b) shows the cross-section of the pipe.
Moment of inertia of hollow pipe,
π
I=
[D4 – d4]
64
π
π
=
[404 – 204] =
[2560000 – 160000]
64
64
= 117809.7 mm4
Now using equation (7.4),
M σ
=
...(i)
I
y
when y is maximum, stress will be maximum. But y is maximum at the top layer from the N.A.
D 40
=
∴
ymax =
= 20 mm
2
2
And maximum B.M.
=
*The bending stress will be maximum at the section where B.M. is maximum. This is because
M σ
=
I
y
302
or σ =
M
× y.
I
BENDING STRESSES IN BEAMS
Equation (i) can be written as
M σ max
=
I
ymax
M
∴
σmax =
× ymax
I
8 × 10 4 × 20
=
= 13.58 N/mm2. Ans.
117809.7
7.7. SECTION MODULUS..
Section modulus is defined as the ratio of moment of inertia of a section about the neutral
axis to the distance of the outermost layer from the neutral axis. It is denoted by the symbol Z.
Hence mathematically section modulus is given by,
1
Z=
...(7.5)
ymax
where
I = M.O.I. about neutral axis
and
ymax = Distance of the outermost layer from the neutral axis.
From equation (7.4), we have
M σ
=
I
y
The stress σ will be maximum, when y is maximum. Hence above equation can be
written as
M σ max
=
I
ymax
∴
M = σmax .
I
ymax
I
=Z
ymax
...(7.6)
∴
M = σmax . Z
In the above equation, M is the maximum bending moment (or moment of resistance
offered by the section). Hence moment of resistance offered the section is maximum when section
modulus Z is maximum. Hence section modulus represent the strength of the section.
But
7.8. SECTION MODULUS FOR VARIOUS SHAPES OR BEAM SECTIONS..
1. Rectangular Section
Moment of inertia of a rectangular section about an axis
through its C.G. (or through N.A.) is given by,
bd 3
I=
12
Distance of outermost layer from N.A. is given by,
d
ymax =
2
∴ Section modulus is given by,
b
d/2
N
A
d
Fig. 7.7
303
STRENGTH OF MATERIALS
Z=
I
ymax
bd 3
=
12 ×
2. Hollow Rectangular Section
Here
∴
FG d IJ
H 2K
=
bd 3 2 bd 2
× =
12
6
d
BD 3 bd 2
−
12
12
1
=
[BD3 – bd3]
12
D
ymax=
2
I
Z =
ymax
...(7.7)
I =
B
b
D
D/2
d
N
1
[ BD 3 − bd 3 ]
12
=
D
2
1
=
[BD3 – bd3]
6D
FG IJ
H K
A
Fig. 7.8
...(7.8)
3. Circular Section
For a circular section,
I=
∴
Z=
π 4
d
d and ymax =
64
2
I
ymax
π 4
d
π 3
64
=
=
d
d
32
2
FG IJ
H K
...(7.9)
4. Hollow Circular Section
Here
and
ymax
∴
π
[D4 – d4]
64
D
=
2
I=
Z=
=
I
ymax
D/2
D
π
[ D4 − d4 ]
64
=
D
2
N
d
FG IJ
H K
π
[D4 – d4]
32D
...(7.10)
Fig. 7.9
Problem 7.3. A cantilever of length 2 metre fails when a load of 2 kN is applied at the free
end. If the section of the beam is 40 mm × 60 mm, find the stress at the failure.
Sol. Given :
Length, L = 2 m = 2 × 103 mm
Load,
W = 2 kN = 2000 N
304
BENDING STRESSES IN BEAMS
Section of beam is 40 mm × 60 mm.
∴ Width of beam,
b = 40 mm
Depth of beam,
d = 60 mm
2 kN
40 mm
60
mm
2m
Fig. 7.10
Fig. 7.10 (a)
Fig. 7.10 (a) shows the section of the beam.
Section modulus of a rectangular section is given by equation (7.7).
bd 2 40 × 60 2
= 24000 mm3
=
6
6
Maximum bending moment for a cantilever shown in Fig. 7.10 is at the fixed end.
∴
M = W × L = 2000 × 2 × 103 = 4 × 106 Nmm
Let
σmax = Stress at the failure
Using equation (7.6), we get
M = σmax . Z
∴
Z=
M 4 × 10 6
=
= 166.67 N/mm2. Ans.
24000
Z
Problem 7.4. A rectangular beam 200 mm deep and 300 mm wide is simply supported
over a span of 8 m. What uniformly distributed load per metre the beam may carry, if the
bending stress is not to exceed 120 N/mm2.
w/m length
Sol. Given :
Depth of beam,
d = 200 mm
Width of beam,
b = 300 mm
L
w.L
w.L
Length of beam,
L=8m
2
2
Max. bending stress,
Fig. 7.11
σmax = 120 N/mm2
Let w = Uniformly distributed load per
metre length over the beam
300 mm
(Fig. 7.11 (a) shows the section of the beam.)
Section modulus for a rectangular section is given by equa200
mm
tion (7.7).
∴
∴
σmax =
Z=
bd 2
300 × 200 2
=
= 2000000 mm3
6
6
Fig. 7.11 (a)
305
STRENGTH OF MATERIALS
Max. B.M. for a simply supported beam carrying uniformly distributed load as shown in
Fig. 7.11 is at the centre of the beam. It is given by
w × L2
w × 82
=
(∵ L = 8 m)
8
8
= 8w Nm = 8w × 1000 Nmm
= 8000w Nmm
(∵ 1 m = 1000 mm)
Now using equation (7.6), we get
M = σmax. . Z
or
8000w = 120 × 2000000
120 × 2000000
= 30 × 1000 N/m = 30 kN/m. Ans.
∴
w =
8000
Problem 7.5. A rectangular beam 300 mm deep is simply supported over a span of
4 metres. Determine the uniformly distributed load per metre which the beam may carry,
if the bending stress should not exceed 120 N/mm2. Take I = 8 × 106 mm4.
M =
Sol. Given :
Depth,
d = 300 mm
Span,
L = 4m
Max. bending stress, σmax = 120 N/mm2
Moment of inertia,
I = 8 × 106 mm4
Let,
w = U.D.L. per metre length over
the beam in N/m.
The bending stress will be maximum, where
bending moment is maximum. For a simply
supported beam carrying U.D.L., the bending
moment is maximum at the centre of the beam
[i.e., at point C of Fig. 7.11 (b)]
∴
Max. B.M. = 2w × 2 – 2w × 1
= 4w – 2w
w/m length
A
B
C
2m
2m
4m
2w
2w
Fig. 7.11 (b)
F Also M = w × L
GH
8
2
= 2w Nm
= 2w × 1000 Nmm
M = 2000w Nmm
Now using equation (7.6), we get
M = σmax × Z
or
where
8 × 10
I
=
150
ymax
Hence above equation (i) becomes as
Z =
6
FG∵
H
or
306
I
JK
w × 4 2 16w
=
= 2w
8
8
...(i)
ymax =
8 × 10 6
150
120 × 8 × 10 6
w =
= 3200 N/m. Ans.
2000 × 150
2000w = 120 ×
=
d 300
=
= 150 mm
2
2
IJ
K
BENDING STRESSES IN BEAMS
Problem 7.6. A square beam 20 mm × 20 mm in section and 2 m long is supported at the
ends. The beam fails when a point load of 400 N is applied at the centre of the beam. What
uniformly distributed load per metre length will break a cantilever of the same material 40 mm
wide, 60 mm deep and 3 m long ?
400 N
Sol. Given :
Depth of beam,
d = 20 mm
2m
Width of beam,
b = 20 mm
Length of beam,
L = 2 m
Fig. 7.12
Point load,
W = 400 N
In this problem, the maximum stress for the simply supported beam is to be calculated
first. As the material of the cantilever is same as that of simply supported beam, hence maximum stress for the cantilever will also be same as that of simply supported beam.
Fig. 7.12 (a) shows the section of beam.
20 mm
The section modulus for the rectangular section of simply supported beam is given by equation (7.7).
bd 2 20 × 20 2 4000
mm3
=
=
6
6
3
Max. B.M. for a simply supported beam carrying a point load
at the centre (as shown in Fig. 7.12) is given by,
w × L 400 × 2
=
= 200 Nm
M=
4
4
= 200 × 1000 = 200000 Nmm
Let σmax = Max. stress induced
Now using equation (7.6), we get
M = σmax . Z
4000
or
200000 = σmax ×
3
200000 × 3
∴
σmax =
= 150 N/mm2
4000
Now let us consider the cantilever as shown in Fig. 7.13.
Let w = Uniformly distributed load per m run.
Maximum stress will be same as in case of simply
supported beam.
∴
σmax = 150 N/mm2
Width of cantilever, b = 40 mm
Depth of cantilever, d = 60 mm
Length of cantilever, L = 3 m
Fig. 7.13 (a) shows the section of cantilever beam.
∴
Z=
Section modulus of rectangular section of cantilever =
∴
Z=
40 × 60 2
= 24000 mm3
6
bd 2
6
20
mm
Fig. 7.12 (a)
wN/m RUN
3m
Fig. 7.13
40 mm
60
mm
Fig. 7.13 (a)
307
STRENGTH OF MATERIALS
Maximum B.M. for a cantilever
wL2 w × 32
= 4.5w Nm = 4.5 × 1000w Nmm
=
2
2
∴
M = 4.5 × 1000w Nmm
Now using equation (7.6), we get
M = σmax . Z
or
4.5 × 1000w = 150 × 24000
150 × 24000
∴
w =
= 800 N/m. Ans.
4.5 × 1000
Problem 7.7. A beam is simply supported and carries a uniformly distributed load of
40 kN/m run over the whole span. The section of the beam is rectangular having depth as
500 mm. If the maximum stress in the material of the beam is 120 N/mm2 and moment of
inertia of the section is 7 × 108 mm4, find the span of the beam.
Sol. Given :
U.D.L.,
w = 40 kN/m = 40 × 1000 N/m
Depth,
d = 500 mm
Max. stress,
σmax = 120 N/mm2
M.O.I. of section,
I = 7 × 108 mm4
Let
L = Span of simply supported beam.
Section modulus of the section is given by equation (7.5), as
I
Z =
ymax
d 500
=
where
ymax =
= 250 mm
2
2
7 × 10 8
= 28 × 105 mm3
∴
Z =
250
The maximum B.M. for a simply supported beam, carrying a U.D.L. over the whole span
w . L2
.
is at the centre of the beam and is equal to
8
w . L2 40000 × L2
=
∴
M =
8
8
= 5000L2 Nm = 5000L2 × 1000 Nmm
Now using equation (7.6), we get
M = σmax . Z
or
5000 × 1000 × L2 = 120 × 28 × 105
=
120 × 28 × 10 5
= 2.4 × 28
5000 × 1000
L = 2.4 × 28 = 8.197 m say 8.20 m. Ans.
L2 =
or
∴
Problem 7.8. A timber beam of rectangular section is to support a load of 20 kN uniformly
distributed over a span of 3.6 m when beam is simply supported. If the depth of section is to be
twice the breadth, and the stress in the timber is not to exceed 7 N/mm2, find the dimensions of
the cross-section.
308
BENDING STRESSES IN BEAMS
How would you modify the cross-section of the beam, if it carries a concentrated load of
20 kN placed at the centre with the same ratio of breadth to depth ?
Sol. Given :
Total load,
W = 20 kN = 20 × 1000 N
Span,
L = 3.6 m
Max. stress,
σmax = 7 N/mm2
Let
b = Breadth of beam in mm
Then depth,
d = 2b mm
Section modulus of rectangular beam =
bd 2
6
b × (2b) 2 2b3
mm3
=
6
6
Maximum B.M., when the simply supported beam carries a U.D.L. over the entire span,
WL
wL2
or
.
is at the centre of the beam and is equal to
8
8
WL 20000 × 3.6
=
∴
M =
= 9000 Nm
8
8
= 9000 × 1000 Nmm
Now using equation (7.6), we get
M = σmax . Z
∴
or
2b 3
3
3 × 9000 × 1000
= 1.92857 × 106
7×2
(1.92857 × 106)1/3
124.47 mm say 124.5 mm. Ans.
2b = 2 × 124.5 = 249 mm. Ans.
9000 × 1000 = 7 ×
b3 =
or
∴
and
L =
b =
=
d =
Dimension of the section when the beam carries a point load at the centre.
W×L
B.M. is maximum at the centre and it is equal to
when the beam carries a point
4
load at the centre.
W × L 20000 × 3.6
=
= 18000 Nm
∴
M =
4
4
= 18000 × 1000 Nmm
σmax = 7 N/mm2
and
or
2b 3
3
Using equation (7.6), we get
M = σmax.Z
Z =
18000 × 1000 = 7 ×
(∵ In this case also d = 2b)
2b 2
3
309
STRENGTH OF MATERIALS
3 × 18000 × 1000
= 3.85714 × 106
7×2
∴
b = (3.85714 × 106)1/3 = 156.82 mm. Ans.
and
d = 2 × 156.82 = 313.64 mm. Ans.
Problem 7.9. A timber beam of rectangular section of length 8 m is simply supported.
The beam carries a U.D.L. of 12 kN/m run over the entire length and a point load of 10 kN at
3 metre from the left support. If the depth is two times the width and the stress in the timber is
not to exceed 8 N/mm2, find the suitable dimensions of the section.
Sol. Given :
Length,
L = 8m
U.D.L.,
w = 12 kN/m = 12000 N/m
Point load,
W = 10 kN = 10000 N
Depth of beam
= 2 × Width of beam
∴
d = 2b
Stress,
σmax = 8 N/mm2
First calculate the section where B.M. is maximum. Where B.M. is maximum, the shear
force will be zero. Now the equations of pure bending can be used. For doing this, calculate the
reactions RA and RB as shown in Fig. 7.14.
∴
b3 =
10 kN
A
12 kN/m
C
B
3m
8m
RA
RB
Fig. 7.14
Taking moments about A, we get
RB × 8 = 12000 × 8 × 4 + 10000 × 3
12000 × 32 + 30000
∴
RB =
= 51750 N
8
∴
RA = Total load – RB
= (12000 × 8 + 10000) – 51750 = 54250 N
Now
S.F. at A = + RA = + 54250 N
S.F. just L.H.S. at C
= 54250 – 12000 × 3 = + 18250 N
S.F. just R.H.S. of C
= 18250 – 10000 = 8250 N
S.F. at B
= – RB = – 51750 N
The S.F. is changing sign between section CB and hence at some section in C and B the
S.F. will be zero.
Let S.F. is zero at x metre from B.
Equating the S.F. at this section to zero, we have
12000 × x – RB = 0
or
12000 × x – 51750 = 0
51750
∴
x =
= 4.3125 m
12000
310
BENDING STRESSES IN BEAMS
∴ Maximum B.M. will occur at 4.3125 m from B.
∴ Maximum B.M.
= M = RB × 4.3125 – 12000 × 4.3125 ×
4.3125
2
= 51750 × 4.3125 – 111585.9375
= 111585.9375 Nm = 111585.9375 × 1000 Nmm
Section modulus for rectangular beam is given by,
bd 2 b × (2b) 2 2b3
=
=
6
6
3
Now using equation (7.6), we get
M = σmax . Z
Z =
2b 3
3
3
×
111585
.9375 × 1000
∴
b3 =
= 20.9223 × 106
16
∴
b = (20.9223 × 106)1/3 = 275.5 mm. Ans.
and
d = 2 × 275.5 = 551 mm. Ans.
Problem 7.10. A rolled steel joist of I section has the dimensions as shown in Fig. 7.15.
This beam of I section carries a u.d.l. of 40 kN/m run on a span of 10 m, calculate the maximum
stress produced due to bending.
Sol. Given :
u.d.l.,
w = 40 kN/m = 40000 N/m
200 mm
Span,
L = 10 m
20 mm
Moment of inertia about the neutral axis
or
111585.9375 × 1000 = 8 ×
200 × 400 3 (200 − 10) × 360 3
−
12
12
= 1066666666 – 738720000
= 327946666 mm4
Maximum B.M. is given by,
=
or
w × L2 40000 × 10 2
=
M =
8
8
= 500000 Nm
= 500000 × 1000 Nmm
= 5 × 108 Nmm
Now using the relation,
M σ
=
I
y
M
∴
σ =
×y
I
M
5 × 10 8
× ymax =
× 200
σmax =
I
327946666
= 304.92 N/mm2. Ans.
360 mm
N
A
400 mm
10 mm
20 mm
Fig. 7.15
(∵ ymax = 200 mm)
311
STRENGTH OF MATERIALS
Problem 7.11. An I-section shown in Fig. 7.16, is simply supported over a span of 12 m.
If the maximum permissible bending stress is 80 N/mm2, what concentrated load can be carried
at a distance of 4 m from one support ?
Sol. Given :
Bending stress, σmax = 80 N/mm2
Let
W = Concentrated load carried at a distance of 4 m from support
B in Newton
To find the maximum bending moment (which will be
100 mm
at point C where concentrated load is acting), first calculate
11.5 mm
the reactions RA and RB.
Taking moments about point A, we get
7.5 mm
RB × 12 = W × 8
8W 2
= W
∴
RB =
12 3
W
2
N
A
225 mm
and
RA = W – RB = W – W =
3
3
8
W
× 8 = W Nm
B.M. at point C
= RA × 8 =
3
3
But B.M. at C is maximum
11.5 mm
∴ Maximum B.M.,
8
8
(a)
Mmax =
W Nm = W × 1000 Nmm
3
3
W
8000
A
C
B
=
W Nmm
3
Now find the moment of inertia of the given I-sec4m
tion about the N.A.
12 m
100 × 225 3 (100 − 7.5) × (225 − 2 × 11.5) 3
−
12
12
92.5 × (202) 3
= 94921875 –
12
= 94921875 – 63535227.55 = 31386647.45 mm4
Now using the relation,
M σ
=
I
y
M σ max
=
or
I
ymax
225
where
ymax =
= 112.5 mm.
2
Now substituting the known values, we get
∴ I=
FG 8000 W IJ
H 3 K
31386647.45
or
=
(b)
Fig. 7.16
80
112.5
W =
312
w
3
80
3
× 31386647.45 ×
= 8369.77 N. Ans.
112.5
8000
2w
3
BENDING STRESSES IN BEAMS
Problem 7.12. Two circular beams where one is solid of diameter D and other is a
hollow of outer dia. Do and inner dia. Di, are of the same length, same material and of same
weight. Find the ratio of section modulus of these circular beams.
Sol. Given :
Dia. of solid beam
=D
Dias. of hollow beam
= Do and Di
Let
L = Length of each beam (same length)
W = Weight of each beam (same weight)
ρ = Density of the material of each shaft (same material)
Now weight of solid beam
= ρ × g × Area of section × L
π 2
=ρ×g×
D ×L
4
Weight of hollow beam
= ρ × g × Area of section × L
π
[D02 – Di2] × L
=ρ×g×
4
But the weights are same
π
π
[D02 – Di2] × L
∴
ρ × g × D2 × L = ρ × g ×
4
4
or
D2 = D02 – Di2
...(i)
Now section modulus of solid section,
π
D3
[See equation (7.9)]
Z =
32
And section modulus of hollow section,
π
[D04 – Di4]
[See equation (7.10)]
Z1 =
32 D0
=
∴
π
[D02 + Di2] [D02 – Di2]
32 D0
Section modulus of solid section
Section modulus of hollow section
=
=
=
=
π 3
D
32
π
[ D0 2 + Di 2 ] [ D0 2 − Di 2 ]
32 D0
D 3 × D0
[ D0 2 + Di 2 ] [ D0 2 − Di 2 ]
=
D × D0 × D 2
[ D0 2 + Di 2 ] [ D0 2 − Di 2 ]
D × D0 × [ D0 2 − Di 2 ]
[ D0 + Di ] ( D0 − Di )
2
2
2
2
D × D0
( D0 2 + Di 2 )
Also from equation (i),
D2 = D02 – Di2 or Di2 = D02 – D2
[∵ D2 = D02 – Di2 from equation (i)]
...(ii)
313
STRENGTH OF MATERIALS
Substituting the value of Di2 in equation (ii), we get
D × D0
D × D0
Section modulus of solid shaft
=
=
2
2
2
Section modulus of hollow shaft D0 + D0 − D
(2 D0 2 − D 2 )
2
2
Section modulus of hollow shaft 2 D0 2 − D 2 = 2 D0 − D
=
D × D0 D × D0
Section modulus of solid shaft
D × D0
D0
D
−
=2
. Ans.
D
D0
Problem 7.13. A water main of 500 mm internal diameter and 20 mm thick is running
full. The water main is of cast iron and is supported at two points 10 m apart. Find the maximum
stress in the metal. The cast iron and water weigh 72000 N/m3 and 10000 N/m3 respectively.
or
Sol. Given :
Internal dia.,
Di = 500 mm = 0.5 m
Thickness of pipe,
t = 20 mm
∴ Outer dia.,
D0 = Di + 2 × t = 500 + 2 × 20 = 540 mm = 0.54 m
Weight density of cast iron
= 72000 N/m3
Weight density of water
= 10000 N/m3
π
π
Internal area of pipe
=
D2=
× 0.52 = 0.1960 m2
4 i
4
This is also equal to the area of water section.
∴ Area of water section = 0.196 m2
π
π
Outer area of pipe
=
D 2=
× 0.542 m2
4 0
4
20
mm
(a)
500 mm
540 mm
20
mm
(b)
Fig. 7.17
π
π
D 2–
D2
4 0
4 i
π
π
= [D02 – Di2] = [0.542 – 0.52] = 0.0327 m2
4
4
Moment of inertia of the pipe section about neutral axis,
π
π
[D04 – Di4] =
[5404 – 5004] = 1.105 × 109 mm4
I=
64
64
Let us now find the weight of pipe and weight of water for one metre length.
∴ Area of pipe section =
314
BENDING STRESSES IN BEAMS
Weight of the pipe for one metre run
= Weight density of cast iron × Volume of pipe
= 72000 × [Area of pipe section × Length]
= 72000 × 0.0327 × 1
(∵ Length = 1 m)
= 2354 N
Weight of the water for one metre run
= Weight density of water × Volume of water
= 10000 × (Area of water section × Length)
= 10000 × 0.196 × 1 = 1960 N
∴ Total weight on the pipe for one metre run
= 2354 + 1960 = 4314 N
Hence the above weight is the U.D.L. (uniformly distributed load) on the pipe. The
maximum bending moment due to U.D.L. is w × L2/8, where w = Rate of U.D.L. = 4314 N per
metre run.
∴ Maximum bending moment due to U.D.L.,
w × L2
4314 × 10 2
=
(∵ L = 10 m)
8
8
3
= 53925 Nm = 53925 × 10 N mm
M σ
= .
Now using
I
y
M
∴
σ =
×y
I
The stress will be maximum, when y is maximum. But maximum value of
M =
D0 540
=
= 270 mm.
2
2
= 270 mm
y =
∴
ymax
M
53925 × 10 3
× ymax =
× 270
I
1.105 × 10 9
= 13.18 N/mm2. Ans.
∴ Maximum stress, σmax =
7.9. BENDING STRESS IN UNSYMMETRICAL SECTIONS..
In case of symmetrical sections, the neutral axis passes through the geometrical centre of
the section. But in case of unsymmetrical sections such as L, T sections, the neutral axis does
not pass through the geometrical centre of the section. Hence the value of y for the topmost layer
or bottom layer of the section from neutral axis will not be same. For finding the bending stress
in the beam, the bigger value of y is used. As the neutral axis passes through the centre of
gravity of the section, hence in unsymmetrical sections, first the centre of gravity is calculated
in the manner as explained in chapter 5.
Problem 7.14. A cast iron bracket subject to bending has the cross-section of I-form
with unequal flanges. The dimensions of the section are shown in Fig. 7.18. Find the position of
the neutral axis and moment of inertia of the section about the neutral axis. If the maximum
bending moment on the section is 40 MN mm, determine the maximum bending stress. What is
the nature of the stress ?
315
STRENGTH OF MATERIALS
Sol. Given :
Max. B.M., M = 40 MN mm = 40 × 106 Nmm
Let us first calculate the C.G. of the section. Let y is the distance of the C.G. from the
bottom face. The section is symmetrical about y-axis and hence y is only to be calculated. Then,
A1 y1 + A2 y2 + A3 y3
( A1 + A2 + A3 )
Area of bottom flange = 130 × 50 = 6500 mm2
Distance of C.G. of A1 from bottom face
50
= 25 mm
2
Area of web = 200 × 50 = 10000 mm2
Distance of C.G. of A2 from bottom face
200
= 150 mm
50 +
2
Area of top flange = 200 × 50 = 10000 mm2
Distance of C.G. of A3 from bottom face
50
= 275 mm.
50 + 200 +
2
y =
where
A1 =
y1 =
=
A2 =
y2 =
=
A3 =
y3 =
=
200 mm
50 mm
Bracket
50 mm
200
mm
50 mm
130 mm
Fig. 7.18
6500 × 25 + 10000 × 150 + 10000 × 275
6500 + 10000 + 10000
162500 + 1500000 + 2750000
=
26500
4412500
=
= 166.51 mm
26500
Hence neutral axis is at a distance of 166.51 mm from the bottom face. Ans.
∴
316
y =
BENDING STRESSES IN BEAMS
Moment of inertia of the section about the N.A.
I = I 1 + I 2 + I3
where I1 = M.O.I. of bottom flange about N.A.
= M.O.I. of bottom flange about an axis passing through its C.G.
+ A1 × (Distance of its C.G. from N.A.)2
130 × 50 3
+ 6500 × (166.51 – 25)2
12
= 1354166.67 + 130163020 = 131517186.6 mm4
=
σt
200 mm
3
133.49
mm
N
200
mm
2
A
50 mm
166.51
mm
1
130 mm
σc = 23.377
Fig. 7.19
Similarly
I2 = M.O.I. of web about N.A.
50 × 200 3
+ A2 . (166.51 – y2)2
12
50 × 200 3
=
+ 10000 (166.51 – 150)2
12
= 33333333.33 + 272580.1
= 33605913.43 mm4
I3 = M.O.I. of top flange about N.A.
=
and
200 × 50 3
+ A3 . (y3 – 166.51)2
12
200 × 50 3
=
+ 10000 × (275 – 166.51)2
12
= 2083333.33 + 117700801 = 119784134.3 mm4
∴
I = I1 + I2 + I3 = 131517186.6 + 33605913.43 + 119784134.3
= 284907234.9 mm4. Ans.
Now distance of C.G. from the upper top fibre
= 300 – y = 300 – 166.51 = 133.49 mm
=
317
STRENGTH OF MATERIALS
and the distance of C.G. from the bottom fibre
= y = 166.51 mm
Hence we shall take the value of y = 166.51 mm for maximum bending stress.
Now using the equation
M σ
=
I
y
M
40 × 10 6
∴
σ=
×y=
× 166.51 = 23.377 N/mm2
I
284907234.9
∴ Maximum bending stress
= 23.377 N/mm2. Ans.
This stress will be compressive. In case of cantilevers, upper layer is subjected to tensile
stress, whereas the lower layer is subjected to compressive stress.
Problem 7.15. A cast iron beam is of I-section as shown in Fig. 7.20. The beam is simply
supported on a span of 5 metres. If the tensile stress is not to exceed 20 N/mm2, find the safe
uniformly load which the beam can carry. Find also the maximum compressive stress.
80 mm
20 mm
3
169.34 mm
200 mm
2
20 mm
N
A
90.66 mm
40 mm
1
160 mm
Fig. 7.20
Sol. Given :
Length of beam,
L=5m
Maximum tensile stress, σt = 20 N/mm2
First calculate the C.G. of the section. Let y is the distance of the C.G. from the bottom
face. As the section is symmetrical about y-axis, hence y is only to be calculated.
Now
y=
A1 y1 + A2 y2 + A3 y3
( A1 + A2 + A3 )
(160 × 40) .
=
318
FG
H
IJ
K
FG
H
20
200
40
+ (200 × 20) 40 +
+ (80 × 20) . 40 + 200 +
2
2
2
160 × 40 + 200 × 20 + 80 × 20
IJ
K
BENDING STRESSES IN BEAMS
128000 + 560000 + 400000 1088000
=
= 90.66 mm
6400 + 4000 + 1600
12000
N.A. lies at a distance of 90.66 mm from the bottom face or 260 – 90.66 = 169.34 mm from
the top face.
Now moment of inertia of the section about N-axis is given by,
I = I 1 + I2 + I3
where I1 = M.O.I. of bottom flange about N.A.
= M.O.I. of bottom flange about its C.G. + A1
× (Distance of its C.G. from N.A.)2
=
160 × 40 3
+ 160 × 40 × (90.66 – 20)2
12
= 853333.33 + 31954147.84 = 32807481.17 mm4
I2 = M.O.I. of web about N.A.
= M.O.I. of web about its C.G. + A2
× (Distance of its C.G. from N.A.)2
=
20 × 200 3
+ 200 × 20 × (140 – 90.66)2
12
= 13333333.33 + 9737742.4 = 23071075.73 mm4
I3 = M.O.I. of top flange about N.A.
= M.O.I. of top flange about its C.G. + A3
× (Distance of its C.G. from N.A.)2
=
80 × 20 3
+ 80 × 20 × (250 – 90.66)2
12
= 53333.33 + 40622776.96 = 40676110.29 mm4
∴
I = 32807481.17 + 23071075.75 + 40676110.29 = 96554667.21 mm4.
For a simply supported beam, the tensile stress will be at the extreme bottom fibre and
compressive stress will be at the extreme top fibre.
Here maximum tensile stress = 20 N/mm2
Hence for the maximum tensile stress,
y = 90.66 mm
[i.e., y is the distance of the extreme bottom fibre (where the tensile stress is maximum) from the
N.A.]
=
Using the relation,
M σ
=
I
y
σ
×I
y
20
× 96554667.21
(∵ σ = σt = 20 N/mm2)
=
90.66
= 21300389.85 Nmm
...(i)
Let w = Uniformly distributed load in N/m on the simply supported beam.
∴
M=
The maximum B.M. is at the centre and equal to
wL2
8
319
STRENGTH OF MATERIALS
w × 25 × 1000
w × 52
Nm =
Nmm = 3125 w Nmm
8
8
Equating the two values of M, given by equations (i) and (ii), we get
3125w = 21300389.85
21300389.85
∴
w =
= 6816.125 N/m. Ans.
3125
Maximum Compressive Stress
Distance of extreme top fibre from N.A.,
yc = 169.34 mm
M = 21300389.85
I = 96554667.21
Let
σc = Max. compressive stress
∴
M =
Using the relation,
M σ
=
I
y
…(ii)
M
×y
I
M
21300389.85
or
σc =
× yc =
× 169.34 = 37.357 N/mm2. Ans.
96554667.21
I
Problem 7.16. A cast iron beam is of T-section as shown in Fig. 7.21. The beam is
simply supported on a span of 8 m. The beam carries a uniformly distributed load of 1.5 kN/m
length on the entire span. Determine the maximum tensile and maximum compressive stresses.
∴
σ =
100 mm
20 mm
1
32.23
mm
N
N
80 mm
2
100
mm
67.77 mm
20 mm
Fig. 7.21
Sol. Given :
Length,
L=8m
U.D.L.,
w = 1.5 kN/m = 1500 N/m
To find the position of the N.A., the C.G. of the section is to be calculated first. The C.G.
will be lying on the y-y axis.
Let
320
y = Distance of the C.G. of the section from the bottom
BENDING STRESSES IN BEAMS
FG
H
IJ
K
80
20
(100 × 20) × 80 +
+ 80 × 20 ×
A1 y1 + A2 y2
2
2
∴
y =
=
A1 + A2
(100 × 2) + (80 × 20)
180000 + 64000 244000
=
=
= 67.77 mm
2000 + 1600
3600
∴ N.A. lies at a distance of 67.77 mm from the bottom face or 100 – 67.77 = 32.23 mm
from the top face.
Now moment of inertia of the section about N.A. is given by,
I = I1 + I2
where
I1 = M.O.I. of top flange about N.A.
= M.O.I. of top flange about its C.G. + A1 × (Distance of its C.G. from N.A.)2
100 × 20 3
+ (100 × 20) × (32.23 – 10)2
12
= 66666.7 + 988345.8 = 1055012.5 mm4
I2 = M.O.I. of web about N.A.
= M.O.I. of web about its C.G. + A2 × (Distance of its C.G. from N.A.)2
=
20 × 80 3
+ (80 × 20) × (67.77 – 40)2
12
= 853333.3 + 1233876.6 = 2087209.9 mm4
I = I1 + I2 = 1055012.5 + 2087209.9 = 3142222.4 mm4.
For a simply supported beam, the maximum tensile stress will be at the extreme bottom
fibre and maximum compressive stress will be at the extreme top fibre.
Maximum B.M. is given by,
=
w × L2 1500 × 8 2
=
= 12000 Nm
8
8
= 12000 × 1000 = 12000000 Nmm
M=
Now using the relation
M
M σ
or σ =
×y
=
I
I
y
(i) For maximum tensile stress,
y = Distance of extreme bottom fibre from N.A. = 67.77 mm
12000000
∴
σ=
× 67.77 = 258.81 N/mm2. Ans.
3142222.4
(ii) For maximum compressive stress,
y = Distance of extreme top fibre from N.A. = 32.23 mm
M
12000000
∴
σ=
×y=
× 32.23 = 123.08 N/mm2. Ans.
3142222.4
I
Problem 7.17. A simply supported beam of length 3 m carries a point load of 12 kN at a
distance of 2 m from left support. The cross-section of the beam is shown in Fig. 7.22 (b).
Determine the maximum tensile and compressive stress at X-X.
Sol. Given :
Point load,
W = 12 kN = 12000 N
First find the B.M. at X-X. And to do this first calculate reactions RA and RB.
321
STRENGTH OF MATERIALS
X
12 kN
25
50
25
25
mm
75
mm
150
mm
50
mm
A
B
1.5 m
1m
100 mm
3m
RA = 4 kN
(a)
RB = 8 kN
(b)
Fig. 7.22
Taking moments about A, we get
RB × 3 = 12 × 2
12 × 2
∴
RB =
= 8 kN and RA = W – RB = 12 – 8 = 4 kN
3
B.M. at X-X
= RA × 1.5 = 4 × 1.5 = 6 kNm
= 6 × 1000 Nm = 6000 × 1000 Nmm
= 6000,000 Nmm
∴
M = 6000,000 Nmm
Now find the position of N.A. of the section of the beam. This can be obtained if we know
the position of C.G. of the section.
y = Distance of the C.G. of the section from the bottom edge
Let
A1 y1 − A2 y2
(Negative sign is due to cut out part)
=
A1 − A2
FG
H
(150 × 100) × 75 − (75 × 50) × 50 +
=
75
2
IJ
K
150 × 100 − 75 × 50
1125000 − 328125 796875
=
=
= 70.83 mm
15000 − 3750
11250
Hence N.A. will lie at a distance of 70.83 mm from the bottom edge or 150 – 70.83 = 79.17
mm from the top edge as shown in Fig. 7.23.
Now the moment of inertia of the section about N.A. is given by,
I = I 1 – I2
where I1 = M.O.I. of outer rectangle about N.A.
= M.O.I. of rectangle 100 × 150 about its C.G. + A1
× (Distance of its C.G. from N.A.)2
100 × 150 3
+ 100 × 150 × (75 – 70.83)2
12
= 28125000 + 260833.5 = 28385833.5 mm4
=
322
BENDING STRESSES IN BEAMS
I2 = M.O.I. of cut out rectangular part about N.A.
= M.O.I. of cut out part about its C.G. + A2
× (Distance of its C.G. from N.A.)2
=
50 × 75
12
3
+ 50 × 75
FG
H
IJ
K
100 mm
25
50
mm
25
25
mm
2
75
− 70.83
2
75
150
= 1757812.5 + 1042083.375
mm
mm
N
A
= 2799895.875 mm4
∴
I = I1 – I2 = 28385833.5 – 2799895.875
50
mm
= 25585937.63 mm4
The bottom edge of the section will be subjected to
Fig. 7.23
tensile stress whereas the top edge will be subjected to
compressive stress. The top edge is at 79.17 mm from
N.A. whereas bottom edge is 70.83 mm from N.A.
Now using the relation,
M σ
=
I
y
M
∴
σ=
×y
I
(i) For maximum tensile stress, y = 70.83 mm
∴ Maximum tensile stress,
6000000
σ=
× 70.83 = 16.60 N/mm2. Ans.
25585937.63
(ii) For maximum compressive stress,
y = 79.17 mm.
∴ Maximum compressive stress,
M
6000000
σ=
×y=
× 79.17 = 18.56 N/mm2. Ans.
25585937.63
I
× 50 +
79.17
mm
70.83
mm
7.10. STRENGTH OF A SECTION..
The strength of a section means the moment of resistance offered by the section and
moment of resistance is given by,
σ
M σ
I
or M = × I = σ × Z where Z =
∵
=
M=σ×Z
I
y
y
y
where M = Moment of resistance
σ = Bending stress, and
Z = Section modulus.
For a given value of allowable stress, the moment of resistance depends upon the section
modulus. The section modulus, therefore, represents the strength of the section. Greater the
value of section modulus, stronger will be the section.
The bending stress at any point in any beam section is proportional to its distance from
the neutral axis. Hence the maximum tensile and compressive stresses in a beam section are
FG
H
IJ
K
323
STRENGTH OF MATERIALS
proportional to the distances of the most distant tensile and compressive fibres from the neutral
axis. Hence for the purposes of economy and weight reduction the material should be concentrated
as much as possible at the greatest distance from the neutral axis. This idea is put into practice,
by providing beams of I-section, where the flanges alone with-stand almost all the bending stress.
We know the relation :
M σ
M
M
M
=
=
or σ =
×y=
I
y
I
Z
I
FG IJ
H yK
where
Z = Section modulus.
For a given cross-section, the maximum stress to which the section is subjected due to a
given bending moment depends upon the section modulus of the section. If the section modulus is
small, then the stress will be more. There are some cases where an increase in the sectional area
does not result in a decrease in stress. It may so happen that in some cases a slight increase in
the area may result in a decrease in section modulus which result in an increase of stress to
resist the same bending moment.
Problem 7.18. Three beams have the same length, same allowable bending stress and
the same bending moment. The cross-section of the beams are a square, rectangle with depth
twice the width and a circle. Find the ratios of weights of the circular and the rectangular
beams with respect to square beams.
Sol. Given :
Fig. 7.24 shows a square, a rectangular and a circular section.
x
x
b
d
(b )
(c)
2b
(a)
Fig. 7.24
Let
x = Side of a square beam
b = Width of rectangular beam
∴
2b = Depth of the rectangular beam
d = Diameter of a circular section.
The moment of resistance of a beam is given by,
M=σ×Z
where Z = Section modulus.
As all the three beams have the same allowable bending stress (σ), and same bending
moment (M), therefore the section modulus (Z) of the three beams must be equal.
324
BENDING STRESSES IN BEAMS
Section modulus of a square beam
bd 3
I
x . x3 2
= 12 =
=
×
y
d
12
x
2
FG JI
H K
(∵ b = d = x)
x2
6
Section modulus of a rectangular beam
=
bd 3
b × (2b) 3
12
= 12 =
d
2b
2
2
FG JI
H K
(∵ d = 2b)
b × 8b 3
2
2
×
= b3
12
2b 3
Section modulus of a circular beam
=
πd 4
4
3
64 = πd × 2 = πd
=
d
d
64
32
2
Equating the section modulus of a square beam with that of a rectangular beam, we get
or
x3 2 3
= b
6
3
3x3
x3
=
∴
b3 =
= 0.25x3
6×2 4
b = (0.25)1/3 x = 0.63x
...(i)
Equating the section modulus of a square beam with that of a circular beam, we get
x 3 πd 3
=
6
32
1/3
32 x 3
32
∴
d3 =
or d =
. x = 1.1927x
6π
6π
The weights of the beams are proportional to their cross-sectional areas. Hence
Weight of rectangular beam Area of rectangular beam
=
Weight of square beam
Area of square beam
b × 2b
0.63 x × 2 × 0.63 x
=
=
x×x
x×x
= 0.7938. Ans.
Weight of circular beam Area of circular beam
=
Weight of square beam
Area of square beam
FG IJ
H K
and
πd 2
2
x × (1.1927 x) 2
4 = πd
=
=
4x2
x2
4 x2
= 1.1172. Ans.
(∵ d = 1.1927x)
325
STRENGTH OF MATERIALS
Problem 7.19. A beam is of square section of the side ‘a’. If the permissible bending
stress is ‘σ’, find the moment of resistance when the beam section is placed such that (i) two
sides are horizontal, (ii) one diagonal is vertical. Find also the ratio of the moments of the
resistance of the section in the two positions.
Sol. Given :
Bending stress = σ
1st Case
Fig. 7.25 (a) shows the square beam section when two sides are horizontal.
a
A
A
B
a
a/2
a/√2
a
N
A
D
D
a/√2
a/√2
C
(a )
B
C
(b)
Fig. 7.25
Let M1 = Moment of resistance of the square beam when two sides are horizontal.
Moment of resistance is given by,
M=σ×Z
∴
M1 = σ × Z1
where Z1 = Section modulus
=
I
ymax
∴
M1 = σ ×
...(i)
a × a3
a 4 2 a3
= 12 =
× =
12 a
6
a/2
a3
. Ans.
6
...(ii)
2nd Case
Fig. 7.25 (b) shows the square beam section when one diagonal is vertical.
Let
M2 = Moment of resistance of the beam in this position
∴
M2 = σ × Z2
where Z2 = Section modulus for the section shown in Fig. 7.25 (b).
=
I2
ymax
F∵
GH
326
=
2×
bh3
12
a
2
FG IJ
H K
M.O.I. of a triangle about its base =
I
JK
bh3
. There are two triangles
12
BENDING STRESSES IN BEAMS
FG a IJ
H 2K
FG a IJ
H 2K
2
× 2a
12
=
a3
=
3
FG∵
H
Here base = b = 2 a and h =
IJ
2K
a
6× 2
a3
∴
M2 = σ ×
. Ans.
6× 2
Ratio of moment of resistance of the section in two positions
Fσ × a I
GH 6 JK
3
=
M1
=
M2
Fσ × a I =
GH 6 × 2 JK
2 = 1.414. Ans.
3
Problem 7.20. Prove that the moment of a resistance of a beam of square section, with
its diagonal in the plane of bending is increased by flatting top and bottom corners as shown in
8a
Fig. 7.26 and that the moment of resistance is a maximum when y =
. Find the percentage
9
increase in moment of resistance also.
Sol. Given :
Fig. 7.26 (a) shows a square section with diagonal AC vertical. Let the portions AEF and
CGH be cut off.
Let I1 = M.O.I. of the square ABCD about diagonal B.D.
Z1 = Section modulus of square ABCD
M1 = Moment of resistance of the square ABCD
I2 = M.O.I. of the new section with cut off portion (i.e., M.O.I. of DEFBHG about
diagonal BD)
Z2 = Section modulus of new section
M2 = Moment of resistance of the new section.
In Fig. 7.26 (a), diagonal AC = 2a
(2a – 2y) = 2(a – y)
A
E
A
F
y
N D
B
F
K
L
a
D
y
2a
G
E
G
H
B A
y
H
C
C
(a )
(b)
Fig. 7.26
327
STRENGTH OF MATERIALS
∴ Diagonal
DB = AC = 2a.
Now moment of inertia of the square ABCD about N.A. (i.e., diagonal BD) is given by
∴
I1 = M.O.I. of two triangles ABD and BCD about their base BD
=2×
=
bh3
2a × a 3
=2×
12
12
a4
3
(Here b = 2a and h = a)
Fa I
GH 3 JK
=
3
Section modulus,
Z1 =
I1
ymax
(Here ymax = a)
a
a4 1 1 3
a
× =
3
a 3
Moment of resistance is given by,
M= σ × Z
1
...(i)
∴
M1 = σ × Z1 = σ × a3 = σ × 0.3333a3
3
Now the M.O.I. of the new section with cut off portion (i.e., M.O.I. of DEFBHG) about the
diagonal BD is given by [Refer to Fig. 7.26 (b)].
I2 = M.O.I. of four triangles (i.e., triangles DEK, FLB, DGH and HLB) plus
M.O.I. of rectangle EFHG about N.A. (i.e., diagonal BD)
=
=
4 × bh3 EF × EG 3
4 × y × y 3 2(a − y) × (2 y) 3
=
+
+
12
12
12
12
(∵ Here b = y, h = y, EF = 2(a – y) and EG = 2y)
y4 4
y 4 4 ay 3 4 y 4 4
+ (a – y) × y3 =
+
−
= ay3 – y4
3
3
3
3
3
3
and section modulus of new section is given by,
4 3
ay − y 4
I2
3
Z2 =
=
ymax
y
4
ay2 – y3
=
3
Now moment of resistance of the new section is given by,
4 2
ay − y 3
M2 = σ × Z2 = σ ×
3
=
LM
N
(∵ Here ymax = y)
OP
Q
The moment of the resistance of the new section will be maximum, if
...(ii)
dM2
= 0.
dy
Hence differentiating equation (ii) w.r.t. y and equating it to zero, we get
LM FG
IJ OP = 0
KQ
N H
I
F4
σ G a × 2y − 3y J = 0
K
H3
d
4 2
σ
ay − y 3
dy
3
2
or
328
(∵ σ and a are constants)
BENDING STRESSES IN BEAMS
or
∴
or
4
a × 2y – 3y2 = 0
3
8
a×y
3y2 =
3
8 a × y 8a
=
y=
3 3× y
9
(∵ σ cannot be zero)
...(iii)
Substituting this value of y in equation (ii), we get
(M2)max = σ ×
LM 4 × a × FG 8a IJ − FG 8a IJ OP = σ × L 4 × 64 a
MN 3 × 81
MN 3 H 9 K H 9 K QP
2
3
3
−
512 3
a
729
OP
Q
= σ × [1.0535a3 – 0.7023a3] = σ × 0.3512 a3
...(iv)
But from equation (i), M1 = σ × 0.3333 a3
∴ M2 is more than M1. And from equation (iii), it is clear that M2 is maximum when
8a
. Ans.
9
Now increase in moment of resistance
= (M2)max – M1 = σ × 0.3512 a3 – σ × 0.3333 a2
= σ × 0.0179 a3
Percentage increase in moment of resistance
Increase in moment of resistance
=
× 100
Original moment of resistance
y=
σ × 0.0719 × a3
× 100 = 5.37%. Ans.
σ × 0.3333 × a 3
Problem 7.21. Prove that the ratio of depth to width of the strongest beam that can be
cut from a circular log of diameter d is 1.414. Hence calculate the depth and width of the
strongest beam that can be cut of a cylindrical log of wood whose diameter is 300 mm.
Sol. Given :
Dia. of log = d
b
D
C
Let ABCD be the strongest rectangular section which
can be cut out of the cylindrical log.
Let
b = Width of strongest section.
d
h
d = Depth of strongest section.
d
Now section modulus of the rectangular section
=
F bh I
G J
I H 12 K bh
Z= =
=
y
GHF h2 IJK 6
3
or
2
...(i)
A
B
Fig. 7.27
In the above equation, b and h are variable.
From ∆BCD,
b2 + h2 = d2
h2 = d2 – b2
329
STRENGTH OF MATERIALS
Substituting the value of h2 in equation (i), we get
b 2
1
[d – b2] = [bd2 – b3]
...(ii)
6
6
In the above equation, d is constant and hence only variable is b.
Now for the beam to be strongest, the section modulus should be maximum (or Z should
be maximum).
For maximum value of Z,
Z=
LM
N
d bd 2 − b3
6
db
or
or
dZ
=0
db
OP = 0
Q
or
d 2 − 3b2
=0
6
d2 – 3b2 = 0
or
d2 = 3b2
But from triangle BCD,
d2 = b2 + h2
Substituting the value of d2 in equation (iii), we get
b2 + h2 = 3b2 or h2 = 2b2
2 ×b
or
h=
or
h
= 2 = 1.414. Ans.
b
...(iii)
...(iv)
Numerical Part
Given,
d = 300 mm
But for equation (iii), d2 = 3b2 or 3b2 = d2 = 3002 = 90000
90000
= 30000
3
b = (30000)1/2 = 173.2 mm. Ans.
b2 =
or
∴
From equation (iv),
h=
2 × b = 1.414 × 173.2 = 249.95 mm. Ans.
7.11. COMPOSITE BEAMS (FLITCHED BEAMS)..
A beam made up of two or more different materials assumed to be rigidly connected together
and behaving like a single piece is known as a composite beam or a wooden flitched beam.
Fig. 7.27 (a) shows a wooden beam (or timber beam) reinforced by steel plates. This arrangement
is known as composite beam or a flitched beam. The strain at the common surfaces will be same
for both materials. Also the total moment of resistance will be equal to the sum of the moments
of individual sections.
When such a beam is subjected to bending, the bending stresses and hence strains due to
bending stresses at a point are proportional to the distance of the point from the common neutral
axis. Consider the composite beam as shown in Fig. 7.27 (a) and let at a distance y from the N.A.,
the stresses in steel and wood are f1 and f2 respectively.
330
BENDING STRESSES IN BEAMS
Let E1 = Young’s modulus of steel plate
Steel
Wooden
Steel
plate
piece
I1 = Moment of inertia of steel about N.A.
plate
M1 = Moment of resistance of steel
E2 = Young’s modulus of wood
I2 = M.O.I. of wood about N.A.
y
1
2
M2 = Moment of resistance of wood.
Strain in steel at a distance y from N.A.
d
Stress σ 1
=
=
N
A
E
E1
(∵ Stress in steel = σ1)
Strain in wood at a distance y from N.A.
σ2
=
E2
But strain at the common surface is same
b
t
t
σ1
σ
∴
= 2
E1
E2
Fig. 7.27 (a)
...(7.11)
E1
× σ2
or
σ1 =
E2
= m × σ2
...(i)
E1
where m =
and is known as modular ratio between steel and wood.
E2
M σ
= , we get
Using the relation
I
y
σ
M=
×I
y
Hence moment of resistance of steel and wood are given by,
σ
σ
M1 = 1 × I1
and
M2 = 2 × I2
y
y
∴ Total moment of resistance of the composite section,
M = M1 + M2
σ
σ
= 1 × I1 + 2 × I2
y
y
mσ 2 × I1 σ 2
+
× I2
=
(∵ σ1 = mσ2 from equation i)
y
y
σ
= 2 [mI1 + I2]
...(7.12)
y
In equation (7.12) I2 + mI1 can be treated as equivalent moment of inertia of the crosssection, as if all made of material 2 (i.e., wood) which will give the same amount of resistance as
the composite beam. Let this be denoted by I.
...(7.13)
∴
I = mI1 + I2
σ2
×I
...(7.14)
Then
M=
y
331
STRENGTH OF MATERIALS
The equivalent section is produced by using I = I2 + mI1. This can be done by multiplying
the dimensions of the material 1 in the direction parallel to the N.A. by m. The equivalent figure
can be used for finding the position of N.A. and equivalent moment of inertia.
Problem 7.22. A flitched beam consists of a wooden joist 10 cm wide and 20 cm deep
strengthened by two steel plates 10 mm thick and 20 cm deep as shown in Fig. 7.28. If the
maximum stress in the wooden joist is 7 N/mm2, find the corresponding maximum stress
attained in steel. Find also the moment of resistance of the composite section. Take Young’s
modulus for steel = 2 × 105 N/mm2 and for wood = 1 × 104 N/mm2.
Sol. Given :
Let suffix 1 represent steel and suffix 2 repre1 cm
1 cm
10 cm
sent wooden joist.
Width of wooden joist,
b2 = 10 cm
Depth of wooden joist,
d2 = 20 cm
Width of one steel plate, b1 = 1 cm
Depth of one steel plate, d1 = 20 cm
20 cm
Number of steel plates
=2
Max. stress in wood,
σ2 = 7 N/mm2
N
A
E for steel,
E1 = 2 × 105 N/mm2
E for wood,
E2 = 1 × 104 N/mm2
Now M.O.I. of wooden joist about N.A.,
b2 d2 3
10 × 20 3
=
12
12
= 6666.66 cm4
= 6666.66 × 104 mm4
M.O.I. of two steel plates about N.A.,
I2 =
Steel
plate
Wooden
joist
Steel
plate
Fig. 7.28
2 × b1d13
2 × 1 × 20 3
I1 =
=
12
12
= 1333.33 cm4 = 1333.33 × 104 mm4.
Now modular ratio between steel and wood is given by,
E1
2 × 10 5
=
= 20
E2
1 × 10 4
The equivalent moment of inertia (I) is given by equation (7.13).
∴
I = mI1 + I2
= 20 × 1333.33 × 104 + 6666.66 × 104
= 104(26666.6 + 6666.66) = 104 × 33333.2
Moment of resistance of the composite section is given by equation (7.14).
σ2
×I
∴
M=
y
m=
7 × 10 4 × 33333.2
=
(∵ y = 10 cm = 10 × 10 mm)
10 × 10
= 233332.4 × 102 N mm = 23333.24 Nm. Ans.
332
BENDING STRESSES IN BEAMS
Maximum Stress in Steel
Let
σ1 = Max. stress in steel.
Now using equation, we get
σ1
σ2
=
E1
E2
E1
∴
σ1 =
× σ2
E2
FG∵
H
= 20 × 7
= 140 N/mm2. Ans.
E1
= m = 20 and σ 2 = 7 N / mm 2
E2
IJ
K
2nd Method
Total moment of resistance is equal to the sum of moment of resistance of individual
member.
∴
M = M1 + M2
...(i)
FG∵
H
σ1
× I1
y
140
=
× 1333.33 × 104
100
= 18666620 Nmm = 18666.620 Nm
σ
M2 = 2 × I2
y
where M1 =
and
(∵
M σ
=
I
y
IJ
K
y = 10 × 10 = 100 mm)
7
× 6666.66 × 104 Nmm
100
= 46666.62 Nmm = 4666.662 Nm
∴
M = M1 + M2 = 18666.620 + 4666.662
= 23333.282 Nm. Ans.
3rd Method
The equivalent moment of inertia (I) is obtained by producing equivalent section.
(a) The equivalent wooden section is obtained by multiplying the dimension of steel plate
=
F
H
in the direction parallel to the N.A. by the modular ratio between steel and wood i.e., by
multiplying by
I
JK
Es
2 × 10 5
= 20 . But the width of one steel plate parallel to N.A. is 1 cm.
=
Ew
1 × 10 4
Hence equivalent wooden width for this steel
plate will be 20 × 1 = 20 cm. This is shown in
Fig. 7.29.
∴ Equivalent M.O.I. is given by,
∴
bd 3
I=
12
20 cm
10 cm
20 cm
20 cm
A
N
Fig. 7.29. Equivalent wooden section
333
STRENGTH OF MATERIALS
(20 + 10 + 20) × 20 3
12
= 33333.33 cm4 = 33333.33 × 104 mm4
∴ Total moment of resistance
= Moment of resistance of the equivalent wooden section
=
=
σ
×I
y
=
Stress in wood
×I
y
7
× 33333.33 × 104 = 23333333.33 Nmm
100
= 23333.333 Nm. Ans.
(b) The equivalent steel section is obtained by multiplying the dimensions of wooden joist
=
F
H
in the direction parallel to N.A. by the modular ratio between wood and steel i.e., by multiplying
by
IJ
K
1 × 10 4
1
Ew
=
=
.
5
20
Es
2 × 10
But the width of wooden joist parallel to N.A. is 10 cm. Hence
1
equivalent steel width will be 10 ×
= 0.5 cm. This is shown in
20
Fig. 7.30.
Hence equivalent M.O.I. is given by
I=
bd 3
12
(1 + 0.5 + 1) × 20 3
12
= 1666.66 cm4
= 1666.66 × 104 mm4
=
∴
M=
20 cm
N
A
σ
×I
y
140
× 1666.76 × 104
100
(Here σ is the stress in steel
and = 140 N/mm2)
= 23333240 Nmm
= 23333.240 Nm. Ans.
=
0.5 cm
1 cm 1 cm
Fig. 7.30. Equivalent
steel section
Note. The width of the single wooden beam for the total moment of resistance of 23333.33 Nm
should be 20 + 10 + 20 = 50 cm as shown in Fig. 7.29. But the width of flitched beam for the same moment
of resistance is only 1 + 10 + 1 = 12 cm as shown in Fig. 7.28. Hence flitched beams require less space.
334
BENDING STRESSES IN BEAMS
Problem 7.23. A timber beam 100 mm wide and 200 mm deep is to be reinforced by
bolting on two steel flitches each 150 mm by 12.5 mm in section. Calculate the moment of
resistance in the following cases : (i) flitches attached symmetrically at the top and bottom ;
(ii) flitches attached symmetrically at the sides. Allowable stress in timber is 6 N/mm2. What is
the maximum stress in the steel in each case ? Take Es = 2 × 105 N/mm2 and Et = 1 × 104 N/mm2.
Sol. Given :
1st Case. Flitches attached symmetrically at the
150 mm
top and bottom.
Steel
12.5 mm
plate
(See Fig. 7.31).
Let suffix 1 represents steel and suffix 2 represents timber.
Timber
200 mm
Width of steel,
b1 = 150 mm
Depth of steel,
d1 = 12.5 mm
N
A
Width of timber,
b2 = 150 mm
Depth of timber,
d2 = 200 mm
Number of steel plates
= 2
Max. stress in timber, σ2 = 6 N/mm2
Steel
12.5 mm
E for steel,
E1 = Es = 2 × 105 N/mm2
plate
E for timber,
E2 = Et = 1 × 104 N/mm2
Distance of extreme fibre of timber from N.A.,
Fig. 7.31
y2 = 100 mm
Distance of extreme fibre of steel from N.A.,
y1 = 100 + 12.5 = 112.5 mm.
Let
σ1* = Max. stress in steel
σ1 = Stress in steel at a distance of 100 mm from N.A.
Now we know that strain at the common surface is same. The strain at a common distance
of 100 mm from N.A. is steel and wood would be same. Hence using equation (7.11), we get
σ1
σ2
=
E1
E2
E1
2 × 10 5
× σ2 =
× 6 = 120 N/mm2.
E2
1 × 10 4
But σ1 is the stress in steel at a distance of 100 mm from N.A. Maximum stress in steel
would be at a distance of 112.5 mm from N.A. As bending stresses are proportional to the distance
from N.A.
σ1 *
σ1
Hence
=
112.5
100
112.5
112.5
∴
σ1* =
× σ1 =
× 120 = 135 N/mm2. Ans.
100
100
Now moment of resistance of steel is given by
σ1 *
× I1 (where σ1* is the maximum stress in steel)
M1 =
y1
135
=
× I1
112.5
∴
σ1 =
335
STRENGTH OF MATERIALS
where
I1 = M.O.I. of two steel plates about N.A.
= 2 × [M.O.I. one steel plate about its C.G. + Area of one steel plate
× (Distance between its C.G. and N.A.)2]
LM b d + b d × F 100 + d I OP
J
GH
2 K PQ
MN 12
L 150 × 12.5 + 150 × 12.5 × FG 100 + 12.5 IJ OP
=2× M
H
2 K PQ
MN 12
=2×
3
1 1
2
1 1
1
3
∴
Similarly,
2
= 2 × [24414.06 + 21166992.18]
= 42382812.48 mm4
135
M1 =
× 42382812.48
112.5
= 50859374.96 Nmm = 50859.375 Nm
σ2
× I2
M2 =
y2
6
150 × 200 3
×
100
12
= 6000000 Nmm = 6000 Nm
∴ Total moment of resistance is given by,
M = M1 + M2
= 50859.375 + 6000 = 56859.375 Nm. Ans.
2nd Case. Flitches attached symmetrically at the sides (See Fig. 7.32)
Here distance of the extreme fibre of steel from N.A.
150
=
= 75 mm.
2
150 mm
In the first case we have seen that stress in steel at
12.5 mm 12.5 mm
a distance of 100 mm from N.A. is 120 N/mm2.
Hence the stress in steel at a distance of 75 mm
from N.A. is given by,
120
σ1** =
× 75
N
100
150 mm
(∵ Stress are proportional
to the distance from N.A.)
= 90 N/mm2
∴ Maximum stress in steel
Fig. 7.32
= σ1** = 90 N/mm2. Ans.
Total moment of resistance is given by,
M = M1 + M2
where M1 = Moment of resistance of two steel plates
σ1 * *
× I1
=
ymax
(Here σ1** = Maximum stress in steel = 90 N/mm2)
=
336
A
200 mm
BENDING STRESSES IN BEAMS
90
× I1
75
I1 = M.O.I. of two steel plates about N.A.
(ymax = 75 mm)
=
=2×
12.5 × 150 3
= 7031250 mm4
12
90
× 7031250 Nmm = 8437500 Nmm = 8437.5 Nm.
75
Similarly, M2 = Moment of resistance of timber section
σ2
=
× I2
y2
∴
M1 =
F
GH
I
JK
150 × 200 8
6
150 × 200 3
∵ I2 =
×
12
100
12
= 6000000 Nmm = 6000 Nm
∴ Total moment of resistance,
M = M1 + M2
= 8437.5 + 6000 = 14437.5 Nm. Ans.
Problem 7.24. Two rectangular plates, one of steel and the other of brass each 40 mm
wide and 10 mm deep are placed together to form a beam 40 mm wide and 20 mm deep, on two
supports 1 m apart, the brass plate being on the top of the steel plate. Determine the maximum
load, which can be applied at the centre of the beam, if the plates are :
(i) separate and can bend independently,
(ii) firmly secured throughout their length.
Maximum allowable stress in steel = 112.5 N/mm2 and in brass = 75 N/mm2. Take
Es = 2 × 105 N/mm2 and Eb = 8 × 104 N/mm2.
Sol. Given :
Width of plates,
b = 40 mm
40 mm
Depth of plates,
d = 10 mm
Brass
Span,
L=1m
10 mm
Stress in steel,
σs = 112.5 N/mm2
Stress in brass,
σb = 75 N/mm2
Value of E for steel, Es = 2 × 105 N/mm2 10 mm
Steel
Value of E for brass, Eb = 8 × 104 N/mm2.
1st Case. The plates are separate and can
Fig. 7.33
bend independently.
Since the two materials bend independently, each will have its own neutral axis. It will be
assumed that the radius of curvature R is the same for both the plates.
E
σ
Using the relation
=
R
y
E× y
or
R=
σ
E × yb
Es × ys
or
R=
= b
σs
σb
σs
E × ys
or
= s
σb
Eb × yb
=
337
STRENGTH OF MATERIALS
But ys = yb as the two plates are having their own N.A. The distance of the extreme
fibre of brass from its own N.A. is 5 mm. Also the distance of extreme fibre of steel from its
N.A. = 5 mm.
2 × 10 5
σs
E
= s =
= 2.5
σb
Eb
8 × 10 4
∴
i.e.,
Now the allowable stress in steel is 112.5 N/mm2
σs = 112.5 N/mm2.
Then maximum stress in brass will be,
112.5
σs
=
= 45 N/mm2
2.5
2.5
This is less than the allowable stress of 75 N/mm2.
σb =
Note. If maximum stress in brass is taken as 75 N/mm2. Then the stress in steel will be σs = 2.5
× σb = 2.5 × 187.5 N/mm2. This stress is more than the allowable stress in steel.
The total moment of resistance is given by,
M = Ms + Mb
where Ms = Moment of resistance of steel plate.
=
σs
× Is
ys
F
GH
40 × 10 3
112.5
40 × 10 3
∵ I s = M.O.I. of steel plate =
×
12
5.0
12
= 75000 Nmm = 75 Nm
Mb = Moment of resistance of brass plate
=
and
=
I
JK
σb
× Ib
yb
45
40 × 10 3
×
= 30000 Nmm = 30 Nm
5.0
12
∴
M = Ms + Mb = 75 + 30
= 105 Nm
...(i)
Let W = Maximum load applied at the centre in N to a simply supported beam.
Then maximum bending moment will be at the centre of the beam. And it is equal to,
=
W×L
W × 1.0
=
Nm
4
4
Equation (i) and (ii), we get
M=
...(ii)
W
= 105 or W = 4 × 105 = 420 N. Ans.
4
2nd Case. The plates are firmly secured throughout their length.
In this case, the two plates act as a single unit and thus will have a single N.A. Let us
convert the composite section into an equivalent brass section as shown in Fig. 7.34 (b).
The equivalent brass section is obtained by multiplying the dimensions of steel plate in
the direction parallel to the N.A. by the modular ratio between steel and brass (i.e., by multiplying
338
BENDING STRESSES IN BEAMS
by
I
JK
Es
= 2.5 . But the width of steel plate parallel to N.A. is 40 mm. Hence equivalent brass
Eb
width for the steel plate will be 40 × 2.5 = 100 mm. This is shown in Fig. 7.34.
40 mm
40 mm
10 mm
Brass
Steel
10 mm
2
N
7.86 mm
10 mm
A
10 mm
1
40 × 2.5 = 100 mm
( a) Composite beam
(b) Equivalent brass section
Fig. 7.34
Let
y = Distance between C.G. of the equivalent brass section and
bottom face.
A1 y1 + A2 y2
=
A1 + A2
100 × 10 × 5 + 40 × 10 × (10 + 5)
=
100 × 10 + 40 × 10
11000
5000 + 6000
=
=
= 7.86 mm.
1400
1000 + 400
Hence N.A. of the equivalent brass section is at a distance of 7.86 mm from the bottom
face.
Now the moment of inertia of the equivalent brass section about N.A. is given as
I = [M.O.I. of rectangle 100 × 10 about its C.G.
+ Area of rectangle 100 × 10 × (Distance between its C.G. and N.A.)2]
+ [M.O.I. of rectangle 40 × 10 about its C.G. + Area of rectangle 40 × 10
× (Distance between its C.G. and N.A.)2]
=
LM 100 × 10
N 12
3
OP
Q
+ 100 × 10 × (7.86 − 5) 2 +
40 × 10 3
+ 40 × 10 × [5 + (10 – 7.86)]2
12
= 8333.33 + 8179.6 + 3333.33 + 20391.84
= 40238.1 mm4.
Distance of upper extreme fibre from N.A.
= 20 – 7.86 = 12.14 mm
Distance of lower extreme fibre from N.A.
= 7.86 mm
Now allowable stress in brass is given 75 N/mm2. As the upper plate is of brass.
Hence the upper extreme fibre will have a stress of 75 N/mm2. Then the lowermost fibre
75
will have the stress =
× 7.86 = 48.56 N/mm2. In Fig. 7.34 (b), the lowermost fibre is also
12.14
of brass. Hence the actual stress in the lowermost fibre of steel will be
= 48.56 × 2.5 = 121.4 N/mm2.
339
STRENGTH OF MATERIALS
But the safe stress in steel is given as 112.5 N/mm2. Hence the brass cannot be fully
stressed.
If we take maximum stress in steel at the bottom to be 112.5 N/mm2, then the
corresponding stress in brass at the bottom fibre will be
112.5
= 45 N/mm2.
2.5
∴
σs = 112.5 N/mm2 and σb = 45 N/mm2.
Now using the relation,
σ
M
=
y
I
σ
or
M =
×I
y
45
=
× 40238.1 = 230370.8 Nmm
7.86
= 230.3708 Nm
...(iii)
The maximum bending moment at the centre of a simply supported beam, carrying a
point load W at the centre is given by,
W×L
W × 1.0
M =
=
...(iv)
4
4
Equating (iii) and (iv), we get
W
= 230.3708
4
∴
W = 4 × 230.3708 = 921.48 N. Ans.
HIGHLIGHTS
1. The stresses produced due to constant bending moment (with zero shear force) are known as
bending stresses.
2. The bending equation is given by,
M σ E
= =
I
y R
where
M = Bending moment
σ = Bending stress
I = Moment of inertia about N.A.
y = Distance of the fibre from N.A.
R = Radius of curvature
E = Young’s modulus of beam.
3. The bending stress in any layer is directly proportional to the distance of the layer from the
neutral axis (N.A.).
4. The bending stress on the neutral axis is zero.
5. The neutral axis of a symmetrical section (such as circular, rectangular or square) lies at a
distance of
340
d
from the outermost layer of the section where d is the depth of the section.
2
BENDING STRESSES IN BEAMS
6. If the top layer of the section is subjected to compressive stress then the bottom layer of the
section will be subjected to tensile stress.
7. The ratio of moment of inertia of a section about the neutral axis to the distance of the outermost
layer from the neutral axis is known as section modulus. It is denoted by Z.
I
ymax
Section modulus for various sections are given as :
∴
8.
Z=
bd2
6
1
(BD3 – bd3)
=
6D
...For rectangular section
Z=
...For a hollow rectangular section
πd3
...For a circular section
32
π
=
[D4 – d4]
...For a hollow circular section.
32D
For finding bending stresses in unsymmetrical section, first their C.G. is to be obtained. This
gives the position of N.A. The bigger value of y is to be used in bending equation.
The moment of resistance offered by the section is known as the strength of the section.
A beam made up of two or more different materials assumed to be rigidly connected together and
behaving like a single unit, is known as a composite beam or flitched beam.
The strain at the common surface of a composite beam is same.
=
9.
10.
11.
12.
σ1
σ
= 2.
E1
E2
∴
E1
is known as modular ratio of first material to the second material.
E2
13.
The ratio of
14.
Total moment of resistance of a composite beam is the sum of the moment of resistance of
individual section.
EXERCISE
(A) Theoretical Questions
1. Define the terms : bending stress in a beam, neutral axis and section modulus.
2. What do you mean by ‘simple bending’ or ‘pure bending’ ? What are the assumptions made in the
theory of simple bending ?
3. Derive an expression for bending stress at a layer in a beam.
4. What do you understand by neutral axis and moment of resistance ?
5. Prove that relation,
where
M σ E
= =
I
y R
M = Bending moment,
I = M.O.I.
σ = Bending stress,
E = Young’s modulus,
y = Distance from N.A.
and
R = Radius of curvature.
341
STRENGTH OF MATERIALS
6. What do you mean by section modulus ? Find an expression for section modulus for a rectangular,
circular and hollow circular sections.
7. How would you find the bending stress in unsymmetrical section ?
8. What is the meaning of ‘Strength of a section’ ?
9. Define and explain the terms : modular ratio, flitched beams and equivalent section.
10. What is the procedure of finding bending stresses in case of flitched beams when it is of
(i) a symmetrical section and (ii) an unsymmetrical section ?
11. Explain the terms : Neutral axis, section modulus, and moment of resistance.
12. Show that for a beam subjected to pure bending, neutral axis coincides with the centroid of the
cross-section.
13. Prove that the bending stress in any fibre is proportional to the distance of that fibre from
neutral layer in a beam.
(B) Numerical Problems
1. A steel plate of width 60 mm and of thickness 10 mm is bent into a circular arc of radius 10 m.
Determine the maximum stress induced and the bending moment which will produce the maximum stress. Take E = 2 × 105 N/mm2.
[Ans. 100 N/mm2 ; 100 Nm]
2. A cast iron pipe of external diameter 60 mm, internal diameter of 40 mm, and of length 5 m is
supported at its ends. Calculate the maximum bending stress induced in the pipe if it carries a
point load of 100 N at its centre.
[Ans. 7.34 N/mm2]
3. A rectangular beam 300 mm deep is simply supported over a span of 4 m. What uniformly
distributed load per metre, the beam may carry if the bending stress is not to exceed 120 N/mm2 ?
Take I = 8 × 106 mm4.
[Ans. 3.2 kN/m]
4. A cast iron cantilever of length 1.5 metre fails when a point load W is applied at the free end. If
the section of the beam is 40 mm × 60 mm and the stress at the failure is 120 N/mm2, find the
point load applied.
[Ans. 1.92 kN]
5. A cast iron beam 20 mm × 20 mm in section and 100 cm long is simply supported at the ends. It
carries a point load W at the centre. The maximum stress induced is 120 N/mm2. What uniformly
distributed load will break a cantilever of the same material 50 mm wide, 100 mm deep and 2 m
long ?
[Ans. 5 kN per m run]
6. A timber beam is 120 mm wide and 200 mm deep and is used on a span of 4 metres. The beam
carries a uniformly distributed load of 2.8 kN/m run over the entire length. Find the maximum
bending stress induced.
[Ans. 7 N/mm2]
7. A timber cantilever 200 mm wide and 300 mm deep is 3 m long. It is loaded with a U.D.L. of
3 kN/m over the entire length. A point load of 2.7 kN is placed at the free end of the cantilever.
Find the maximum bending stress produced.
[Ans. 7.2 N/mm2]
8. A timber beam is freely supported on supports 6 m apart. It carries a uniformly distributed load
of 12 kN/m run and a point load of 9 kN at 3.5 m from the right support. Design a suitable section
of the beam making depth twice the width, if the stress in timber is not to exceed 8 N/mm2.
[Ans. 230 mm × 460 mm]
9. A beam of an I-section shown in Fig. 7.35 is simply supported over a span of 4 metres.
Determine the load that the beam can carry per metre length, if the allowable stress in the
[Ans. 2.5 kN/m run]
beam is 30.82 N/mm2.
342
BENDING STRESSES IN BEAMS
60 mm
20 mm
100 mm
20 mm
100 mm
Fig. 7.35
10.
A beam is of T-section as shown in Fig. 7.36. The beam is simply supported over a span of 4 m and
carries a uniformly distributed load of 1.7 kN/m run over the entire span. Determine the maximum tensile and maximum compressive stress.
[Ans. 8 N/mm2 and 4.8 N/mm2]
150 mm
50 mm
150 mm
50 mm
Fig. 7.36
11.
A simply supported beam of length 4 m carries a point load of 16 kN at a distance of 3 m from left
support. The cross-section of the beam is shown in Fig. 7.37. Determine the maximum tensile
and compressive stress at a section which is at a distance of 2.25 m from the left support.
[Ans. 24.9 N/mm2 ; 27.84 N/mm2]
343
STRENGTH OF MATERIALS
25
50 mm
25
25
mm
75
mm
150
mm
50
mm
100 mm
Fig. 7.37
x3
where ‘σ ’ is the
6
permissible stress in bending, x is the side of the square beam and beam is placed such that its
two sides are horizontal.
Find the moment of resistance of the above beam, if it is placed such that its one diagonal is
12. Prove that the moment of resistance of a beam of square section is equal to σ ×
13.
vertical, the permissible bending stress is same (i.e., equal to ‘σ’).
[Ans. x3 × σ /6 × 2 ]
14. A flitched beam consists of a wooden joist 150 mm wide and 300 mm deep strengthened by a
steel plate 12 mm thick and 300 mm deep on either side of the joist. If the maximum stress in the
wooden joist is 7 N/mm2, find the corresponding maximum stress attained in steel. Find also the
moment of resistance of the composite section. Take E for steel = 2 × 105 N/mm2 and for
[Ans. 140 N/mm2, 66150 Nm]
wood = 1 × 104 N/mm2.
15. A timber beam 60 mm wide by 80 mm deep is to be reinforced by bolting on two steel flitches,
each 60 mm by 5 mm in section. Find the moment of resistance in the following cases : (i) flitches
attached symmetrically at top and bottom ; (ii) flitches attached symmetrically at the sides.
Allowable timber stress is 8 N/mm2. What is the maximum stress in the steel in each case ? Take
E for steel = 2.1 × 105 N/mm2 and for timber = 1.4 × 104 N/mm2.
[Ans. (i) 3768 Nm, σs = 135 N/mm2 (ii) 1052 Nm, σs = 90 N/mm2]
16. Two rectangular plates, one of steel and the other of brass each 37.5 mm by 10 are placed to
either to from a beam 37.5 mm wide by 20 mm deep, on two supports 75 cm apart, the
brass component being on top of the steel component. Determine the maximum central load if
the plates are (i) separate and can bend independently, (ii) firmly secured throughout their
length. Permissible stresses for brass and steel are 70 N/mm2 and 100 N/mm2. Take
[Ans. (i) 472.2 N (ii) 1043.5 N]
Eb = 0.875 × 105 N/mm2 and Es = 2.1 × 105 N/mm2.
17. A timber beam 150 mm wide and 100 mm deep is to be reinforced by two steel flitches each
150 mm × 10 mm in section. Calculate the ratio of the moments of the resistance in the twomentioned cases : (i) Flitches attached symmetrically on the sides (ii) Flitches attached at
top and bottom.
[Ans. 0.31]
344
8
SHEAR STRESSES
IN BEAMS
CHAPTER
8.1. INTRODUCTION..
In the last chapter, we have seen that when a part of a beam is subjected to a constant
bending moment and zero shear force, then there will be only bending stresses in the beam. The
shear stress will be zero as shear stress is equal to shear force divided by the area. As shear force
is zero, the shear stress will also be zero.
But in actual practice, a beam is subjected to a bending moment which varies from section
to section. Also the shear force acting on the beam is not zero. It also varies from section to
section. Due to these shear forces, the beam will be subjected to shear stresses. These shear
stresses will be acting across transverse sections of the beam. These transverse shear stresses
will produce a complimentary horizontal shear stresses, which will be acting on longitudinal
layers of the beam. Hence beam will also be subjected to shear stresses. In this chapter, the
distribution of the shear stress across the various sections (such as Rectangular section,
Circular section, I-section, T-sections etc.) will be determined.
8.2. SHEAR STRESS AT A SECTION..
Fig 8.1 (a) shows a simply supported beam carrying a uniformly distributed load. For a
uniformly distributed load, the shear force and bending moment will vary along the length of the
beam. Consider two sections AB and CD of this beam at a distance dx apart.
A
C
(σ
B
D
C
(σ +dσ )
H
τ
y1
A
(M+dM)
E
N
y1
dx
B
G
dA
×
σ
σ
A
(M)
dA
dx
( a)
N
σ)
+d
F
A
y
b
D
(c )
( b)
Area, A = Area of EFGH
Fig. 8.1
345
STRENGTH OF MATERIALS
Let at the section AB,
F = Shear force
M = Bending moment
and at section CD, F + dF = Shear force
M + dM = Bending moment
I = Moment of inertia of the section about the neutral axis.
Let it is required to find the shear stress on the section AB at a distance y1 from the
neutral axis. Fig. 8.1 (c) shows the cross-section of the beam. On the cross-section of the beam,
let EF be a line at a distance y1 from the neutral axis. Now consider the part of the beam above
the level EF and between the sections AB and CD. This part of the beam may be taken to consists
of an infinite number of elemental cylinders each of area dA and length dx. Consider one such
elemental cylinder at a distance y from the neutral axis.
∴
dA = Area of elemental cylinder
dx = Length of elemental cylinder
y = Distance of elemental cylinder from neutral axis
Let σ = Intensity of bending stress* on the end of the elemental cylinder on the
section AB
σ + dσ = Intensity of bending stress on the end of the elemental cylinder on the section
CD.
The bending stress at distance y from the neutral axis is given by equation (7.6) as
M σ
=
I
y
M
×y
I
For a given beam, the bending stress is a function of bending moment and the distance y
from neutral axis. Let us find the bending stress on the end of the elemental cylinder at the
section AB and also at the section CD.
∴ Bending stress on the end of elemental cylinder on the section AB, (where bending
moment is M) will be
∴
σ=
M
×y
I
Similarly, bending stress on the end of elemental cylinder on the section CD, (where
bending moment is M + dM) will be
σ=
( M + dM )
×y
I
(∵ On section CD, B.M. = M + dM and bending stress = σ + dσ)
Now let us find the forces on the two ends of the elemental cylinder.
Force on the end of the elemental cylinder on the section AB
= Stress × Area of elemental cylinder
= σ × dA
σ + dσ =
*Bending stresses are acting normal to the cross-section.
346
SHEAR STRESSES IN BEAMS
M
× y × dA
I
Similarly, force on the end of the elemental cylinder on the section CD
= (σ + dσ) dA
=
=
FG∵ σ = M × yIJ
K
H
I
LM∵ σ + dσ = (M + dM) × yOP
I
N
Q
( M + dM )
× y × dA
I
At the two ends of the elemental cylinder, the forces are different. They are acting along
the same line but are in opposite direction. Hence there will be unbalanced force on the elemental
cylinder.
∴ Net unbalanced force on the elemental cylinder
=
( M + dM )
M
× y × dA –
× y × dA
I
I
dM
× y × dA
...(i)
I
The total unbalanced force above the level EF and between the two sections AB and CD
may be found out by considering all the elemental cylinders between the sections AB and CD and
above the level EF (i.e., by integrating the above equation (i)).
∴ Total unbalanced force
=
where
z
z
dM
dM
y × dA
× y × dA =
I
I
dM
=
×A× y
I
A = Area of the section above the level EF (or above y1)
= Area of EFGH as shown in Fig. 8.1 (c)
=
(∵
z y × dA = A × y)
y = Distance of the C.G. of the area A from the neutral axis.
Due to the total unbalanced force acting on the part of the beam above the level EF and
between the sections AB and CD as shown in Fig. 8.2 (a), the beam may fail due to shear. Hence
in order the above part may not fail by shear, the horizontal section of the beam at the level EF
must offer a shear resistance. This shear resistance at least must be equal to total unbalanced
force to avoid failure due to shear.
Area A
C
A
G
H
dM × A × y
I
E
N
A
dx
B
D
(a )
y1
y
F
A
N
b
(b)
Fig. 8.2
347
STRENGTH OF MATERIALS
∴ Shear resistance (or shear force) at the level EF
= Total unbalanced force
dM
×A× y
I
Let
τ = Intensity of horizontal shear at the level EF
b = Width of beam at the level EF
∴ Area on which τ is acting
= b × dx
∴Shear force due to τ
= Shear stress × Shear area
= τ × b × dx
Equating the two values of shear force given by equations (ii) and (iii), we get
...(ii)
=
...(iii)
dM
×A× y
I
dM
A× y
×
τ =
I
b × dx
τ × b × dx =
∴
=
dM Ay
.
dx I × b
= F×
Ay
I×b
FG∵ dM = F = Shear forceIJ
K
H dx
...(8.1)
The shear stress given by equation (8.1) is the horizontal shear stress at the distance y1
from the neutral axis. But by the principal of complementary shear, the horizontal shear stress
is accompanied by a vertical shear stress τ of the same quantity.
Sometimes A × y is also expressed as the moment of area A about the neutral axis.
Note. In equation (8.1), b is the actual width at the level EF (Though here b is same at all
levels, in many cases b may not be same at all levels) and I is the total moment of inertia of the
section about N.A.
Problem 8.1. A wooden beam 100 mm wide and 150 mm deep is simply supported over
a span of 4 metres. If shear force at a section of the beam is 4500 N, find the shear stress at a
distance of 25 mm above the N.A.
Sol. Given :
50
C.G.
Width,
b = 100 mm
mm
75 mm
Depth,
d = 150 mm
y
25
Shear force,
F = 4500 N
y1
mm
Let τ = Shear stress at a distance of 25 mm above
A
N
the neutral axis.
150 mm
Using equation (8.1), we get
Ay
...(i)
I .b
A = Area of the beam above y1
= 100 × 50 = 5000 mm2
(Shaded area of Fig. 8.2)
τ = F.
where
348
100 mm
Fig. 8.3
SHEAR STRESSES IN BEAMS
y = Distance of the C.G. of the area A from neutral axis
50
= 50 mm
2
I = M.O.I. of the total section
= 25 +
=
bd 3
12
100 × 150 3
= 28125000 mm4
12
b = Actual width of section at a distance y1 from N.A. = 100 m
Substituting these values in the above equation (i), we get
=
τ =
4500 × 5000 × 50
= 0.4 N/mm2. Ans.
28125000 × 100
Problem 8.2. A beam of cross-section of an isosceles triangle is subjected to a shear
force of 30 kN at a section where base width = 150 mm and height = 450 mm. Determine :
(i) horizontal shear stress at the neutral axis,
(ii) the distance from the top of the beam where shear stress is maximum, and
(iii) value of maximum shear stress.
Sol. Given :
B
Shear force at the section, F = 30 kN = 30,000 N
Base width, CD = 150 mm
300 mm
Height, h = 450 mm.
C.G.
(i) Horizontal shear stress at the neutral axis
The neutral axis of the triangle is at a distance of
h
from
3
100 mm
y
N
A
2h
from the apex B. Hence distance of neutral axis from B
3
C
2 × 450
Fig. 8.3 (a)
= 300 mm as shown in Fig. 8.3 (a). The width of the
will be
3
section at neutral axis is obtained from similar triangles BCD and BNA as
base or
or
NA 300
=
CD 450
300
300
NA =
× CD =
× 150 = 100 mm.
450
450
The shear stress at any section is given by equation (8.1) as
A× y
I×b
τ = Shear stress at the section
F = Shear force = 30,000 N
A = Area above the axis at which shear stress is to be obtained
[i.e., shaded area of Fig. 8.3 (a)]
τ=F×
where
D
...(i)
349
STRENGTH OF MATERIALS
NA × 300
100 × 300
=
= 15000 mm2
2
2
y = Distance of the C.G. of the area A from neutral axis
=
1
× 300 = 100 mm
3
I = M.O.I. of the total section about neutral axis
=
F
GH
I
JK
B × h3
Base width × Height 3
where B = Base Width of Triangle
∵
36
36
=
150 × 450 3
mm4
36
b = Actual width of the section at which shear stress is to be obtained
= NA = 100 mm.
Substituting these values in equation (i), we get
=
τ = 30,000 ×
FG
H
15000 × 100
150 × 450
36
IJ
K
3
N/mm2
× 100
= 1.185 N/mm2. Ans.
(ii) The distance from the top of the beam where shear stress is maximum
Let the shear stress is maximum at the section EF
B
at a distance x from the top of the beam as shown in
x 2x/3
Fig. 8.3 (b). The distance EF is obtained from similar
triangles BEF and BCD as
2h
EF
CD
x
450
x
x
x
∴
EF =
× CD =
×150 = .
450
450
3
The shear stress at the section EF is given by equation (8.1) as
3
=
τ = F×
A× y
I×b
F
E
y1
N
A
(h/3)
C
...(ii)
Fig. 8.3 (b)
EF × x
x
x
= ×
2
3
2
x2
6
= Distance of C.G. of the Area A from neutral axis
=
y
=
350
2h 2 x
2 × 450 2 x
−
=
=
−
3
3
3
3
FG 300 − 2x IJ
H
3K
D
150 mm
where F = 30,000 N
A = Area of section above EF i.e., Area of shaded triangle BEF
=
y 450 mm
FG∵ EF = x IJ
H
3K
SHEAR STRESSES IN BEAMS
I = M.O.I. of ΔBCD about neutral axis
=
150 × 450 3
mm4
36
b = Width of section EF =
x
.
3
Substituting these values in equation (ii), we get
F x I × FG 300 − 2x IJ
GH 6 JK H
3K
F 2x IJ
= 0.0000395x G 300 −
τ=
H
3K
F 150 × 450 I × x
GH 36 JK 3
F
2x I
= 0.0000395 G 300 x −
3 JK
H
30,000 ×
2
3
2
For maximum shear stress
...(iii)
dτ
=0
dx
2
4x
× 2x = 0 or 300 =
3
3
300 × 3
or
x=
= 225 mm. Ans.
4
Hence, shear stress is maximum at a distance of 225 mm from the top of the beam.
(iii) Value of Maximum Shear Stress
The value of maximum shear stress will be obtained by substituting x = 225 mm in
equation (iii).
or
300 –
∴ Maximum shear stress
FG
H
= 0.0000395 300 × 225 −
2
× 225 2
3
IJ
K
= 1.333 N/mm2. Ans.
8.3. SHEAR STRESS DISTRIBUTION FOR DIFFERENT SECTIONS.
The following are the important sections over which the shear stress distribution is to be
obtained :
1. Rectangular Section,
2. Circular Section,
3. I-Section,
4. T-Sections, and
5. Miscellaneous Sections.
8.3.1. Rectangular Section. Fig. 8.4 shows a rectangular section of a beam of width b
and depth d. Let F is the shear force acting at the section. Consider a level EF at a distance y
from the neutral axis.
The shear stress at this level is given by equation (8.1) as
Ay
τ=F.
b× I
where A = Area of the section above y (i.e., shaded area ABFE)
d
=
−y ×b
2
FG
H
IJ
K
351
STRENGTH OF MATERIALS
B
A
(d/2 – y)
d/2
E
q
F
y
d
qmax
A
N
C
D
(b )
b
(a )
Fig. 8.4
y = Distance of the C.G. of area A from neutral axis
=y+
FG
H
IJ
K
FG
H
d y
1 d
y d
1
d
=
y+
−y =y+ − = +
4 2
2 4
2
2
2 2
IJ
K
b = Actual width of the section at the level EF
I = M.O.I. of the whole section about N.A.
Substituting these values in the above equation, we get
F.
τ=
=
FG d − yIJ × b × 1 FG y + d IJ
H 2 K 2 H 2K
F
GH
b× I
F d2
− y2
2I 4
I
JK
...(8.2)
From equation (8.2), we see that τ increases as y decreases. Also the variation of τ with
respect to y is a parabola. Fig. 8.4 (b) shows the variation of shear stress across the section.
At the top edge, y =
d
and hence
2
τ=
At the neutral axis,
FG IJ OP = F
H K PQ 2I
2
×0=0
y = 0 and hence
F
GH
I
JK
=
Fd 2
=
8I
Fd
bd 3
8×
12
=
12 F
F
−
= 1.5
8
bd
bd
τ=
352
LM
MN
F d2
d
−
2I 4
2
F d2
F d2
−0 =
×
2I 4
2I
4
F∵ I = bd I
GH
12 JK
3
...(i)
SHEAR STRESSES IN BEAMS
Now average shear stress, τavg =
Shear force
F
=
.
Area of section b × d
Substituting the above value in equation (i), we get
τ = 1.5 × τavg
...(8.3)
Equation (8.3) gives the shear stress at the neutral axis where y = 0. This stress is also
the maximum shear stress.
∴
τmax = 1.5τavg
...(8.4)
From equation (8.1), τ =
Ay
. In this equation the value of A y can also be calculated as
Ib
given below :
A y = Moment of shaded area of Fig. 8.4 (a) about N.A.
Consider a strip of thickness dy at a distance y from N.A. Let dA is the area of this strip.
Then
dA = Area of strip = b × dy
Moment of the area dA about N.A.
= dA. y or y × dA
= y × bdy
(∵ dA = b × dy)
The moment of the shaded area about N.A. is obtained by integrating the above equation
d
between the limits y to .
2
∴ Moment of shaded area about N.A.
=
z
d /2
y
=b
z
y × b × dy
d /2
y × dy
y
L y OP
=b M
N2Q
2
d/2
=
y
(as b is constant)
b
2
LM FG d IJ
MN H 2 K
2
− y2
OP = b L d
PQ 2 MN 4
2
− y2
OP
Q
But moment of shaded area about N.A. is also equal to A y
∴
Ay =
b
2
LM d
N4
2
− y2
OP
Q
Substituting the value of A y in equation (8.1), we get
F×
τ=
F
GH
b d2
− y2
2 4
I×b
I
JK
=
F
GH
F d2
− y2
2I 4
I
JK
This equation is same as equation (8.2).
Problem 8.3. A rectangular beam 100 mm wide and 250 mm deep is subjected to a
maximum shear force of 50 kN. Determine :
(i) Average shear stress,
(ii) Maximum shear stress, and
(iii) Shear stress at a distance of 25 mm above the neutral axis.
353
STRENGTH OF MATERIALS
Sol. Given :
100 mm
Width,
b = 100 mm
Depth,
d = 250 mm
Maximum shear force, F = 50 kN = 50,000 N.
(i) Average shear stress is given by,
N
A
50,000
F
=
τavg =
Area
b× d
125
50,000
mm
=
= 2 N/mm2. Ans.
100 × 250
(ii) Maximum shear stress is given by equation (8.4)
τmax = 1.5 × τavg
∴
Fig. 8.4 (c)
= 1.5 × 2 = 3 N/mm2. Ans.
(iii) The shear stress at a distance y from N.A. is given by equation (8.2).
∴
τ=
=
F
GH
F d2
− y2
2I 4
F
GH
I
JK
50000 250 2
− 252
2I
4
FG
H
I
JK
250
mm
(∵ y = 25 mm)
IJ
K
50000 62500
50000 × 12
− 625 =
× 15000 N/mm2
4
bd 3
2 × 100 × 250 3
2×
12
= 2.88 N/mm2. Ans.
Alternate Method [See Fig. 8.4 (d)]
The shear stress at a distance 25 mm from neutral
axis is given by equation (8.1) as
100 mm
A× y
τ=F×
C.G.
I×b
100 125
mm mm
where
F = 50,000 N
A = Area of beam above 25 mm (i.e., shaded area
y
25 mm
in Fig. 8.4 (d))
A
2
250
= 100 × 100 = 10000 mm
mm
y = Distance of the C.G. of the area A from
neutral axis
100
= 75 mm
= 25 +
2
Fig. 8.4 (d)
I = M.O.I. of total section about neutral axis
=
bd 3 100 × 250 3
=
12
12
b = Actual width of the section at a distance 25 mm from neutral axis = 100.
Substituting these values in equation (8.1), we get
10000 × 75
τ = 50,000 ×
100 × 250 3
× 100
12
=
F
GH
354
I
JK
SHEAR STRESSES IN BEAMS
=
50000 × 10000 × 75 × 12
= 2.88 N/mm2. Ans.
100 × 250 3 × 100
Problem 8.4. A timber beam of rectangular section is simply supported at the ends and
carries a point load at the centre of the beam. The maximum bending stress is 12 N/mm2 and
maximum shearing stress is 1 N/mm2, find the ratio of the span to the depth.
Sol. Given :
Maximum bending stress,
σmax = 12 N/mm2
W
Maximum shear stress,
τmax = 1 N/mm2.
A
B
Let
b = Width of the beam,
d = Depth of the beam,
L
L = Span of the beam,
RA = W
RB = W
2
2
W = Point load at the centre.
Maximum shear force,
and
maximum B.M.,
Now average shear stress,
or
Fig. 8.5
W
F =
2
W×L
M =
4
τavg =
FG W IJ
Shear force
H 2K
=
Area
Maximum shear stress is given by equation (8.4)
∴
τmax = 1.5 × τavg
W
1 = 1.5 ×
2bd
b×d
=
W
.
2bd
FG∵ τ
H
max
= 1, τ avg =
W
2
4
=
=
bd
1.5
3
or
W
2bd
IJ
K
...(i)
Now using bending equation
M
σ
=
I
y
or σ =
M
×y
I
∴ Maximum bending stress,
M
× ymax
I
W×L d
×
4
2
=
bd 3
12
σmax =
W. L
12 W . L. d
×
= 1.5
3
bd 2
8
bd
W. L
12 = 1.5
bd 2
W L
= 1.5
.
bd d
FG∵ y
H
max
=
d
2
IJ
K
=
or
(∵ σmax = 12)
355
STRENGTH OF MATERIALS
= 1.5 ×
∴
=2×
LM∵ W = 4 from equation (i)OP
N bd 3
Q
4
L
×
3
d
L
d
L 12
=
= 6. Ans.
d
2
Problem 8.5. A simply supported wooden beam of span 1.3 m having a cross-section
150 mm wide by 250 mm deep carries a point load W at the centre. The permissible stress are
7 N/mm2 in bending and 1 N/mm2 in shearing. Calculate the safe load W.
Sol. Given :
W
Span,
L = 1.30 mm
Width,
b = 150 mm
Depth,
d = 250 mm
W
W
1.3 m
Bending stress,
σ = 7 N/mm2
2
2
Shearing stress,
τ = 1 N/mm2
∴
Maximum B.M.,
W×L
W
M=
=
× 1.3
4
2
Fig. 8.6
Nm
W
× 1.3 × 1000 Nmm = 325 W Nmm
4
W
Maximum S.F.
=
N.
2
(i) Value of W for bending stress consideration
Using bending equation
M
σ
=
I
y
where M = 325 W Nmm
=
bd 3
150 × 250 3
=
= 195312500 mm4
12
12
σ = 7 N/mm2
d
205
y=
=
= 125.
2
2
Substituting these values in the above equation (i), we get
325W
7
=
195312500
125
7 × 195312500
∴
W=
= 33653.8 N.
325 × 125
(ii) Value of W for shear stress consideration
Average shear stress,
I=
and
FG IJ
H K
W
W
Shear force
2
=
=
τavg =
2
×
150
× 250
Area
b× d
356
...(i)
SHEAR STRESSES IN BEAMS
Maximum shear stress is given by equation (8.4)
3
× τavg
2
= 1 N/mm2
∴
τmax =
But
τmax
∴
1=
3
W
×
2
2 × 150 × 250
2 × 2 × 150 × 250
= 50000 N.
3
Hence, the safe load is minimum of the two values (i.e., 33653.8 and 50000 N) of W. Hence
safe load is 33653.8 N. Ans.
or
W=
8.3.2. Circular Section. Fig. 8.7 shows a
circular section of a beam. Let R is the radius of
the circular section of F is the shear force acting
on the section. Consider a level EF at a distance y
from the neutral axis.
The shear stress at this level is given by
equation (8.1) as
τ=
where
F× A× y
I×b
...(i)
E
R
N
B
O
τ
F
y
τmax
A
(a )
(b )
Fig. 8.7
A y = Moment of the shaded area about
the neutral axis (N.A.)
I = Moment of inertia of the whole circular section
b = Width of the beam at the level EF.
Consider a strip of thickness dy at a distance y from N.A. Let dA is the area of strip.
Then
dA = b × dy = EF × dy
(∵ b = EF)
= 2 × EB × dy
(∵ EF = 2 × EB)
R 2 − y 2 × dy
=2×
(∵ In rt. angled triangle OEB, side EB =
R2 − y2 )
Moment of this area dA about N.A.
= y × dA
=y×2
R 2 − y 2 × dy
(∵ dA = 2 R 2 − y 2 dy)
R 2 − y 2 dy.
= 2y
Moment of the whole shaded area about the N.A. is obtained by integrating the above
equation between the limits y and R
∴
Ay =
z
R
2y
y
=–
z
R
y
R 2 − y 2 dy
(– 2y)
R 2 − y 2 dy.
357
STRENGTH OF MATERIALS
Now (– 2y) is the differential of (R2 – y2). Hence, the integration of the above equation
becomes as
Ay = –
LM ( R
N
2
− y2 )3/ 2
3/ 2
OP
Q
R
y
2
[(R2 – R2)3/2 – (R2 – y2)3/2]
3
2
2
= – [0 – (R2 – y2)3/2] = (R2 – y2)3/2
3
3
Substituting the value of A y in equation (i), we get
= –
F×
τ =
But
2 2
( R − y2 )3/ 2
3
I×b
b = EF = 2 × EB = 2 ×
R2 − y2
Substituting this value of b in the above equation, we get
2
F ( R2 − y2 )3/ 2
F
3
τ =
=
(R2 – y2)
2
2
EI
I×2 R − y
...(8.5)
Equation (8.5) shows that shear stress distribution across a circular section is parabolic.
Also it is clear from this equation that with the increase of y, the shear stress decreases. At
y = R, the shear stress, τ = 0. Hence, shear stress will be maximum when y = 0 i.e., at the
neutral axis.
∴ At y = 0 i.e., at the neutral axis, the shear stress is maximum and is given by
F 2
R
τmax =
3I
π
π
But
I=
D4 =
× (2 R)4
(∵ D = 2R)
64
64
π 4
R
=
4
F
4
F × R2
∴
τmax =
= ×
...(8.6)
π 4
3
πR 2
3× R
4
But average shear stress,
F
Shear force
τavg =
=
Area of circular section
πR 2
Hence equation (8.6) becomes as,
4
× τavg
...(8.7)
3
Problem 8.6. A circular beam of 100 mm diameter is subjected to a shear force of 5 kN.
Calculate :
(i) Average shear stress,
(ii) Maximum shear stress, and
(iii) Shear stress at a distance of 40 mm from N.A.
τmax =
358
SHEAR STRESSES IN BEAMS
Sol. Given :
Diameter,
D = 100 mm
100
= 50 mm
2
Shear force,
F = 5 kN = 5000 N.
(i) Average shear stress is given by,
∴ Radius,
R=
τavg =
Shear force
Area of circular section
5000
= 0.6366 N/mm2. Ans.
π(50) 2
(ii) Maximum shear stress for a circular section is given by equation (8.7).
=
4
× τavg
3
4
= × 0.6366 = 0.8488 N/mm2. Ans.
3
(iii) The shear stress at a distance 40 mm from N.A. is given by equation (8.5).
∴
τmax =
∴
τ=
F
(R2 – y2)
3I
FG
H
π
5000
∵ y = 40 and I =
100 4
(502 – 402)
π
64
4
× 100
3×
64
5000 × 64
=
(2500 – 1600)
3 × π × 100000000
= 0.3055 N/mm2. Ans.
=
8.3.3. I-Section
Fig. 8.8 shows the I-section of a beam.
B
Flange
d
2
Web
d
N
τmax
D
A
b
d
2
(a )
(b )
Fig. 8.8
359
IJ
K
STRENGTH OF MATERIALS
Let
B = Overall width of the section,
D = Overall depth of the section,
b = Thickness of the web, and
d = Depth of web.
The shear stress at a distance y from the N.A., is given by equation (8.1) as
Ay
.
I×b
In this case the shear stress distribution in the web and shear stress distribution in the
flange are to be calculated separately. Let us first calculate the shear stress distribution in the
flange.
(i) Shear stress distribution in the flange
Consider a section at a distance y from N.A. in the
flange as shown in Fig. 8.8 (c).
Width of the section = B
τ=F×
D
Shaded area of flange, A = B FG − yIJ
H2 K
D/2
y
Distance of the C.G. of the shaded area from neutral
axis is given as
FG
H
IJ
K
1 D
−y
2 2
D y
−
=y+
4 2
D y
1 D
+ =
=
+y
4 2
2 2
Hence shear stress in the flange becomes,
F × Ay
τ=
I×B
D
1 D
F×B
−y ×
+y
2
2 2
=
I×B
y =y+
FG
H
FG
H
IJ
K
IJ
K
FG
H
LMFG D IJ − y OP
MNH 2 K PQ
I
F FD
−y J
=
G
2I H 2
K
=
d/2
N
A
b
Fig. 8.8 (c)
(∵ Here width = B)
IJ
K
2
F
2I
2
2
2
...(8.8)
Hence, the variation of shear stress (τ) with respect to y in the flange is parabolic. It is also
clear from equation (8.8) that with the increase of y, shear stress decreases.
(a) For the upper edge of the flange,
D
y=
2
Hence shear stress,
360
LM
MN
FG IJ OP = 0.
H K PQ
F D2
D
−
τ=
2I 4
2
2
SHEAR STRESSES IN BEAMS
(b) For the lower edge of the flange,
Hence
y=
d
2
τ=
F D2
d
−
2I 4
2
LM
MN
FG IJ OP = F FG D
H K PQ 2I H 4
2
2
−
F
(D2 – d2)
8I
(ii) Shear stress distribution in the web
Consider a section at a distance y in the web from the
N.A. as shown in Fig. 8.9.
Width of the section = b.
Here A y is made up of two parts i.e., moment of the
flange area about N.A. plus moment of the shaded area of the
web about the N.A.
∴ A y = Moment of the flange area about N.A.
+ moment of the shaded area of
web about N.A.
1 d
d
D d
1 D d
×
+
+b
−y ×
+y
=B
−
2
2 2
2 2
2 2 2
d2
4
I
JK
=
FG
H
IJ
K
FG
H
IJ
K
F
GH
I
JK
FG
H
IJ
K
FG
H
D/2
d/2
y
N
A
b
IJ
K
b d2
B
− y2
(D2 – d2) +
2 4
8
Hence the shear stress in the web becomes as
=
...(8.9)
Fig. 8.9
LM
MN
F
GH
I OP
JK PQ
B
b d2
F × Ay
F
(D2 − d2 ) +
− y2
=
×
...(8.10)
8
2 4
I×b
I×b
From equation (8.10), it is clear that variation of τ with respect to y is parabolic. Also with
the increase of y, τ decreases.
At the neutral axis, y = 0 and hence shear stress is maximum.
τ=
∴
LM
OP
N
Q
F L B( D − d ) bd O
=
M 8 + 8 PQ
I×b N
τmax =
F
B
b d2
( D2 − d 2 ) + ×
2
4
I×b 8
2
2
2
...(8.11)
At the junction of top of the web and bottom of flange,
d
y=
2
Hence shear stress is given by,
τ=
=
LM
MN
F
GH
FG
H
F
B
b d2
d
−
(D2 − d 2 ) +
2 4
2
I×b 8
F × B × ( D2 − d 2 )
8I × b
IJ
K
2
I OP
JK P
Q
...(8.12)
361
STRENGTH OF MATERIALS
The shear stress distribution for the web and flange is shown in Fig. 8.8 (b). The shear
stress at the junction of the flange and the web changes abruptly. Equation (8.9) gives the stress
at the junction of the flange and the web when stress distribution is considered in the flange. But
equation (8.12) gives the stress at the junction when stress distribution is considered in the web.
From these two equations it is clear that the stress at the junction changes abruptly
F
B
F
(D2 – d2) to
×
(D2 – d2).
from
8I
b
8I
150 mm
310 mm
20 mm
350 mm
Problem 8.7. An I-section beam 350 mm × 150 mm
has a web thickness of 10 mm and a flange thickness of
20 mm. If the shear force acting on the section is 40 kN,
find the maximum shear stress developed in the I-section.
Sol. Given :
Overall depth,
D = 350 mm
Overall width,
B = 150 mm
Web thickness,
b = 10 mm
Flange thickness,
= 20 mm
∴ Depth of web, d = 350 – (2 × 20) = 310 mm
Shear force on the section, F = 40 kN = 40,000 N.
Moment of inertia of the section about neutral axis,
N
150 × 350 3 140 × 310 3
I=
mm4
−
12
12
= 535937500 – 347561666.6
= 188375833.4 mm4.
Maximum shear stress is given by equation (8.11)
∴
τmax =
=
LM
N
F
B( D 2 − d 2 ) bd 2
+
8
8
I×b
LM
N
A
10 mm
20 mm
Fig. 8.10
OP
Q
40000
150(350 2 − 310 2 ) 10 × 310 2
+
188375833.4 × 10
8
8
= 0.000021234
LM 150 (122500 − 96100) + 120125OP
N8
Q
OP
Q
= 13.06 N/mm2. Ans.
Alternate Method
The maximum shear stress developed in the I-section will be at the neutral axis. This
shear stress is given by,
τmax =
where
F× A× y
I×b
F = 40,000 N
A × y = Moment of the area above the neutral axis about the neutral axis
= Area of flange × Distance of C.G. of the area of flange from neutral axis + Area
of web above neutral axis × Distance of the C.G. of this area from neutral axis
362
SHEAR STRESSES IN BEAMS
= (150 × 20) ×
FG 310 + 20 IJ + FG 310 × 10IJ × FG 310 × 1IJ
H 2 2 K H 2 K H 2 2K
= 3000 × 165 + 1550 × 77.5
= 495000 + 120125 = 615125 mm3
I = Moment of inertia of the whole section about neutral axis
= 188375833.4 mm4 (Already Calculated)
b = Width of the web at neutral axis
= 10 mm
∴ τmax =
40,000 × 615125
= 13.06 N/mm2. Ans.
188375833.4 × 10
Problem 8.8. For problem 8.7, sketch the shear stress distribution across the section.
Also calculate the total shear force carried by the web.
Sol. Given :
From problem 8.7, we have
B = 150 mm ; D = 350 mm
d = 310 mm ; b = 10 mm
I = 188.375 × 106 mm4
F = 40000 N ;
τmax = 13.06 N/mm2.
Shear stress distribution in the flange
The shear stress at the upper edge of the flange is zero.
Actually shear stress distribution in the flange is given by equation (8.8) as
τ=
F
2I
FD
GH 4
2
− y2
I
JK
...(i)
150 mm
10.51
0.707
310 mm
350 mm
20 mm
13.06
N
A
10 mm
0.707
20 mm
(b )
( a)
Fig. 8.11
363
STRENGTH OF MATERIALS
For the upper edge of the flange,
∴
y =
D
2
τ =
F
2I
F D − FG D IJ I = F F D
GH 4 H 2 K JK 2I GH 4
2
2
2
−
D2
4
I
JK
=0
For the lower edge of the upper flange (i.e.,) at the joint of web and flange,
d
2
∴ Substituting this value in equation (i), we get
y =
τ =
LM
MN
FG IJ OP = F F D
H K PQ 2I GH 4
F D2
d
−
2I 4
2
2
2
−
d2
4
I
JK
F
40000
(D2 – d2) =
(3502 – 3102)
8I
8 × 188.375 × 10 6
= 0.7007 N/mm2.
=
Shear stress distribution in the web
The shear stress is maximum at N.A. and it is given by,
τmax = 13.06 N/mm2
(calculated in problem 8.7)
The shear stress at the junction of web and flange is given by equation (8.12) as
F×B
τ =
(D2 – d2)
8I × b
40000 × 150
(3502 – 3102) = 10.51 N/mm2
8 × 188.375 × 10 6 × 10
(The shear stress at the junction can also be
obtained as equal to
B
150
× 0.7007 =
× 0.7007 = 10.51 N/mm2)
10
b
Now shear stress distribution which is symmetrical about N.A., can be plotted as shown
in Fig. 8.11 (b). The shear stress for web and flange are parabolic. The shear stress at the
junction suddenly changes from 0.707 to 10.51 N/mm2.
=
Total Shear force carried by the web
Total shear force carried by the web will be equal to the total shear force carried by the
I-section minus the total shear force carried by the two flanges.
∴ Total shear force carried by the web
= Total shear force carried by I-section minus two times the shear force
carried by one flange
= 40,000 – 2 × Shear force carried by one flange
...(ii)
To find the shear force carried by one flange, first calculate the shear stress in the flange
at a distance ‘y’ from neutral axis. Now consider an elemental strip of flange of thickness ‘dy’.
Then area of strip will be width of the flange × thickness of strip i.e., dA = 150 × dy. Now the
shear force carried by the elemental strip
364
SHEAR STRESSES IN BEAMS
= Shear stress at a distance y in the flange × Area of the strip
= τ × 150 × dy
Total shear force carried by the flange will be obtained by integrating the above equation
310
350
to
(i.e., from 155 to 175).
from
2
2
∴ Total shear force carried by one flange
=
z
175
155
τ × 150 × dy
...(iii)
The value of ‘τ’ (i.e., shear stress) in the flange at a distance y from neutral axis is given
by
τ=
where
F× A× y
I×b
F = 40,000
A y = Moment of area of the flange above y, about neutral axis
[i.e., shaded area of Fig. 8.8 (c) on page 360]
FG D − yIJ × 1 FG D + yIJ
H2 K 2H2 K
F 350 − yIJ × 1 FG 350 + yIJ
= 150 G
H 2 K 2H 2 K
= B
(∵ Here B = 150, D = 350)
1
(175 + y)
2
= 75 (1752 – y2) = 75 (30625 – y2)
I = Moment of inertia of the whole section about neutral axis
(Already calculated)
= 188.375 × 106 mm4
b = Width of flange
= 150 mm.
Substituting the above values, we get
= 150 (175 – y) ×
40,000 × 75 (30625 − y 2 )
= 0.000106 (30625 – y2)
188.375 × 10 6 × 150
Substituting this value of τ in equation (iii), we get
Total shear force carried by one flange
∴
τ=
=
z
175
0.000106(30625 – y2) × 150 × dy
155
= 0.000106 × 150
z
175
(30625 – y2) dy
155
L
y O
P
= 0.0159 M30625 y −
3 Q
N
1
L
= 0.0159 M30625 (175 − 155) − (175
3
N
3 175
155
3
− 1553 )
OP
Q
365
STRENGTH OF MATERIALS
LM
N
= 0.0159 612500 −
1
(5359375 − 3723875)
3
= 0.0159 612500 − 545166.66
OP
Q
= 1070.61 N
Substituting this value in equation (ii), we get
Total shear force carried by web
= 40,000 – 2 × 1070.61
= 37858.78 N = 37.858 kN. Ans.
8.3.4. T-Section. The shear stress distribution over a T-section is obtained in the same
manner as over an I-section. But in this case the position of neutral axis (i.e., position of C.G.) is
to be obtained first, as the section is not symmetrical about x-x axis. The shear stress distribution diagram will also not be symmetrical.
Problem 8.9. The shear force acting on a section of a beam is 50 kN. The section of the
beam is of T-shaped of dimensions 100 mm × 100 mm × 20 mm as shown in Fig. 8.12. The
moment of inertia about the horizontal neutral axis is 314.221 × 104 mm4. Calculate the shear
stress at the neutral axis and at the junction of the web and the flange.
Y
35.35
100 mm
20
mm
1
7.07
35.285
A
N
80 mm
100 mm
32.22
2
67.78
20 mm
(a )
Y
(b )
Fig. 8.12
Sol. Given :
Shear force,
F = 50 kN = 50000 N
Moment of inertia about N.A.,
I = 314.221 × 104 mm4.
First calculate the position of neutral axis. This can be obtained if we know the position of
C.G. of given T-section. The given section is symmetrical about the axis Y-Y and hence the C.G.
of the section will lie on Y-Y axis.
Let
y* = Distance of the C.G. of the section from the top of the flange.
Then
366
y* =
A1 y1 + A2 y2
( A1 + A2 )
SHEAR STRESSES IN BEAMS
FG
H
(100 × 20) × 10 + (20 × 80) × 20 +
=
80
2
(100 × 10) + (10 × 90)
IJ
K
20000 + 96000
= 32.22.
2000 + 1600
Hence, neutral axis will be at a distance of 32.22 mm from the top of the flange as shown
in Fig. 8.12 (a).
Shear stress distribution in the flange
Now the shear stress at the top edge of the flange, and bottom of the web is zero.
Shear stress in the flange just at the junction of the flange and web is given by,
=
τ=
F × Ay
I×b
A = 100 × 20 = 2000 mm2
where
y = Distance of C.G. of the area of flange from N.A.
20
= 22.22 mm
2
b = Width of flange = 100 mm
= 32.22 –
∴
τ=
50000 × 2000 × 22.22
314.221 × 10 4 × 100
= 7.07 N/mm2.
Shear stress distribution in the web
The shear stress in the web just at the junction of the web and flange will suddenly
100
= 35.35 N/mm2. The shear stress will be maximum at
increase from 7.07 N/mm2 to 7.07 ×
20
N.A. Hence shear stress at the N.A. is given by
τ=
F × Ay
I×b
where A y = Moment of the above N.A. about N.A.
= Moment of area of flange about N.A. + Moment of area of web about N.A.
= 20 × 100 × (32.22 – 10) + 20 × (32.22 – 10) ×
22.22
2
= 44440 + 4937.28 = 49377.284 mm2
b = 20 mm
∴
τ=
50000 × 49377.284
= 39.285 N/mm2
314.221 × 10 4 × 20
Now the shear stress distribution diagram can be drawn as shown in Fig. 8.12 (b).
8.3.5. Miscellaneous Sections. The shear stress distribution over miscellaneous sections is obtained in the same manner as over a T-section. Here also the position of neutral axis is
obtained first.
367
STRENGTH OF MATERIALS
Problem 8.10. The shear force acting on a beam at an I-section with unequal flanges is
50 kN. The section is shown in Fig. 8.13. The moment of inertia of the section about N.A. is
2.849 × 104. Calculate the shear stress at the N.A. and also draw the shear stress distribution
over the depth of the section.
0.952
200 mm
50
mm
3
133.49
3.808
83.49
200 mm
4.4196
N
A
116.51
50 mm
166.51
2
3.22
50 mm
1
1.239
130 mm
Fig. 8.13
Sol. Given :
Shear force,
F = 50 kN = 50,000 N
Moment of inertia about N.A.,
I = 2.849 × 108 mm4.
Let us first calculate the position of N.A. This is obtained if we know the position of the
C.G. of the given I-section. Let y* is the distance of the C.G. from the bottom face. Then
y* =
where
368
A1 y1 + A2 y2 + A3 y3
( A1 + A2 + A3 )
A1 = Area of bottom flange
= 130 × 50 = 6500 mm2
A2 = Area of web = 200 × 50 = 10000 mm2
A3 = Area of top flange = 200 × 50 = 10000 mm2
y1 = Distance of C.G. of A1 from bottom face
50
= 25 mm
=
2
y2 = Distance of C.G. of A2 from bottom face
200
= 50 +
= 150 mm
2
y3 = Distance of C.G. of A3 from bottom face
50
= 50 + 200 +
= 275 mm
2
SHEAR STRESSES IN BEAMS
∴
y* =
6500 × 25 × 10000 × 150 + 10000 × 275
= 166.51 mm
6500 + 10000 + 10000
Hence N.A. is at a distance of 166.51 mm from the bottom face (or 300 – 166.51 = 133.49 mm
from upper top fibre).
Shear stress distribution
(i) Shear stress at the extreme edges of the flanges is zero.
(ii) The shear stress in the upper flange just at the junction of upper flange and web is
given by,
τ=
where
F × Ay
I×b
A y = Moment of the area of the upper flange about N.A.
= Area of upper flange × Distance of the C.G. of upper flange from N.A.
= (200 × 50) × (133.49 – 25) = 1084900
b = Width of upper flange = 200 mm
50000 × 1084900
= 0.9520 N/mm2.
2.849 × 108 × 200
(iii) The shear stress in the web just at the junction of the web and upper flange will
200
suddenly increase from 0.952 to 0.952 ×
= 3.808 N/mm2.
50
(iv) The shear stress will be maximum at the N.A. This is given by
∴
τ=
τmax =
where
F × Ay
I×b
A y = Moment of total area (about N.A.) about N.A.
= Moment of area of upper flange about N.A. + Moment of area of web about N.A.
= 200 × 50 × (133.49 – 25) + (133.49 – 50) × 50 ×
(133.49 − 50)
2
= 1084900 + 174264.5 = 1259164.5
b = 50 mm
and
∴
τmax =
50000 × 1259164.5
= 4.4196 N/mm2.
2.849 × 10 8 × 50
(v) The shear stress in the lower flange just at the junction of the lower flange and the web
is given by
τ=
where
F × Ay
I×b
A y = Moment of the area of the lower flange about N.A.
= 130 × 50 × (166.51 – 25) = 918125
b = Width of lower flange = 130 mm
∴
τ=
50000 × 918125
2.849 × 10 8 × 130
= 1.239 N/mm2.
369
STRENGTH OF MATERIALS
(vi) The shear stress in the web just at the junction of the web and lower flange will
1.239 × 130
= 3.22 N/mm2.
suddenly increase from 1.239 to
50
The shear stress diagram is shown in Fig. 8.13 (b).
Problem 8.11. The shear force acting on a beam at a section is F. The section of the
beam is triangular base b and of an altitude h. The beam is placed with its base horizontal. Find
the maximum shear stress and the shear stress at the N.A.
Sol. Given :
Base = b
Altitude = h
The N.A. of the triangle ABC will lie at the C.G. of the triangle. But the C.G. of the
2h
triangle will be at a distance of
from the top.
3
∴ Neutral axis will be at a distance of
2h
from the top.
3
C
x
2h/3
h
2
F
E
τmax
y
h
N
A
A
B
b
(a)
(b)
Fig. 8.14
Consider a level EF at a distance y from the N.A. The shear stress at this level is given by,
τ=
where
F × Ay
I×b
...(i)
A y = Moment of the shaded area about the neutral axis
= Area of triangle CEF × Distance of C.G. of triangle CEF from N.A.
FG 1 × EF × xIJ × FG 2h − 2x IJ
K H 3 3K
H2
F 1 bx × xIJ × 2 (h – x)
=G ×
H2 h K 3
=
(∵ As triangles CEF and CAB are similar.
Hence
370
=
1 bx 2
2
×
× (h – x)
2
3
h
=
1 bx 2
×
× (h – x)
3
h
EF
x
=
AB h
or
EF =
x
x×b
× AB =
h
h
IJ
K
SHEAR STRESSES IN BEAMS
I = Moment of inertia of the whole triangular section CAB about N.A.
b = Actual width at the level EF
x×b
.
h
Substituting these values in equation (i), we get
= EF =
F×
τ=
1 bx 2
.
. (h − x)
1 F . x (h − x)
3 h
= .
x×b
3
I
I×
h
F
(xh – x2)
...(ii)
3I
From equation (ii), it is clear that variation of τ with respect to x is parabolic. At the top,
x = 0 and hence τ is also zero. At the bottom x = h and τ is also zero.
=
At the N.A., x =
2h
, and hence the shear at the N.A. becomes as,
3
LM
MN
F L 2h
M
3I N 3
=
bh3
36
2
τ =
×
27
But
FG IJ OP
H K PQ
4 h O F (6 h
−
P= ×
9 Q 3I
F 2h
2h
×h−
3I 3
3
τ =
2
2
2
2
F 2h 2
2
Fh 2
− 4h2 )
=
×
=
×
3I
9
27
I
9
I=
∴
=
Fh 2
F bh I
GH 36 JK
3
2
36 × Fh 2
×
27
bh 3
=
8 F
.
3 bh
...(8.13)
Maximum shear stress
The shear stress of any depth x from the top is given by equation (ii). The maximum shear
stress will be obtained by differentiating equation (ii) with respect to x and equating to zero.
∴
or
or
or
LM
N
OP
Q
d F
( xh − x 2 ) = 0
dx 3 I
F
(h – 2x) = 0
3I
h – 2x = 0
(∵ F and I are constants and cannot be zero)
h
2
Now substituting this value of x in equation (ii), we get
x=
τmax =
LM
MN
FG
H
F h
h
×h−
3I 2
2
IJ
K
2
OP
PQ
371
STRENGTH OF MATERIALS
=
LM
N
F h2 h2
−
3I 2
4
Fh 3
bh 3
12 ×
36
36
Fh 2
=
×
12
bh 3
OP = F × h
Q 3I 4
2
=
Fh 2
12 I
F∵ I = bh I
GH
12 JK
3
=
3F
bh
Now draw the shear stress diagram as shown in Fig. 8.14 (b).
=
...(8.14)
Note. In the above case, the shear stress is not maximum at the N.A., but it is maximum at a
depth of h/2 from the top. In all other cases, the shear stress was maximum at the N.A.
Problem 8.12. A beam of triangular cross-section is subjected to a shear force of 50 kN.
The base width of the section is 250 mm and height 200 mm. The beam is placed with its base
horizontal. Find the maximum shear stress and the shear stress at the N.A.
Sol. Given :
Shear force,
F = 50 kN = 50000 N
Base width,
b = 250 mm
Height,
h = 200 mm
Maximum shear stress is given by equation (8.14).
3F
3 × 50000
=
= 3 N/mm2. Ans.
bh
250 × 200
Shear stress at N.A. is given by equation (8.13).
∴
τmax =
∴
τ =
8F
8 × 50000
=
= 2.67 N/mm2. Ans.
3bh
3 × 250 × 200
Problem 8.13. A beam of square section is used as a beam with one diagonal horizontal.
The beam is subjected to a shear force F, at a section. Find the maximum shear in the crosssection of the beam and draw the shear distribution diagram for the section.
Sol. Given :
Fig. 8.15 shows a square beam ABCD, having diagonal AC horizontal.
B
F
E
x
b
2
y
N
C
b
3b
8
τmax
A
τmax
3b
8
D
(a )
(b )
Fig. 8.15
372
SHEAR STRESSES IN BEAMS
Let b = Length of diagonal AC. This is also length of diagonal BD.
The N.A. of the beam shown in Fig. 8.15 (a), passes through diagonal AC.
Consider a level EF at a distance y from the N.A. The shear stress at this level is given by,
τ=
where
F × Ay
I×b
...(i)
A y = Moment of the shaded area about N.A.
= Area of triangle BEF × Distance of C.G. of triangle BEF from N.A.
FG 1 × EF × xIJ FG b − 2 xIJ
H2
K H2 3 K
I Fb 2 I
F1
= G × 2 x × xJ G − xJ
K H2 3 K
H2
F b 2x IJ
=x G −
H2 3 K
=
FG∵ EF = x , ∴ EF = x × b = 2xIJ
(b / 2)
H CA (b/ 2)
K
2
I = Moment of inertia of the whole section about N.A.
b×
3
b4
48
12
b = Actual width at the level EF = 2x
Substituting these values in equation (i), we get
=2×
and
FG bIJ
H 2K
=
FG b − 2x IJ
H 2 3 K = F × 24 x L 3b − 4 x O
MN 6 PQ
F b I × 2x
b
GH 48 JK
F × x2
τ=
=
4F
b4
At the top, x = 0 hence τ = 0
At the N.A., x =
4
4
x (3b – 4x)
FG
H
b
4 F b 3b − 4 × b
hence τ = 2 .
2
2
2
b
4F b
2F
= 4 . .b= 2
2
b
b
...(ii)
IJ
K
Maximum shear stress
Maximum shear stress will be obtained by differentiating equation (ii) with respect to x
and equating to zero.
∴
or
LM
N
OP
Q
d 4F
(3bx − 4 x 2 ) = 0
dx b4
4F
b4
(3b – 8x) = 0
373
STRENGTH OF MATERIALS
or
or
FG∵ 4 F cannot be zeroIJ
K
H b
3b – 8x = 0
4
3b
8
Substituting this value of x in equation (ii), we get maximum shear stress.
x=
∴
τmax =
4F
b
4
×
3b
8
FG 3b − 4 × 3bIJ = 4 F × 3b × 3b = 9 × F
H
8
2
4 b
8K
b
4
2
.
The shear stress distribution is shown in Fig. 8.15 (b).
Problem 8.14. Fig. 8.16 shows a section, which is subjected to a shear force of 100 kN.
Determine the shear stresses at A, B, C and D. Sketch the shear stress distribution also.
A
25 mm
5.165
B
25 mm
150 mm
C
50
22.25
mm
N
A
D
35.46
25
mm
25 mm
125 mm
(a )
(b )
Fig. 8.16
Sol. Given :
Shear forced, F = 100 kN = 100000 N.
The neutral axis will be at a distance of
125
= 62.5 mm from the top, as the given section
2
is symmetrical about X-X and Y-Y axis.
Moment of inertia of the given section about N.A. is given by,
I = M.O.I. of rectangle 125 × 150 about N.A.
–M.O.I. two semi-circle (or one circular hole) about N.A.
125 × 150 3
π
× 1004 mm4 = 3.025 × 107 mm4
−
12
64
The shear stress is given by,
=
τ=
F × Ay
I×b
At A,
A y = 0 and hence τ = 0
At B,
A y = Moment of area (125 × 25) about N.A.
374
SHEAR STRESSES IN BEAMS
FG
H
IJ
K
FG
H
25
25
∵ A = 125 × 25 and y = 50 +
2
2
5
3
= 125 × 25 × 62.5 = 1.953 × 10 mm
b = 125 mm
= (125 × 25) × 50 +
∴
τ=
At C,
100000 × 1.9531 × 10 5
= 5.165 N/mm2. Ans.
3.025 × 107 × 125
A y = Moment of area above an horizontal line passing through C.C.,
about N.A.
= Moment of area of rectangle 125 × 50 about N.A.
– Moment of area of circular portion between C and B about N.A.
FG
H
= (125 × 50) × 25 +
= 3.125 × 105 –
z
z
50
25
= 3.125 × 105 –
50
25
= 3.125 × 105 +
IJ
K
50
−
2
z
y = 50
y = 25
2x . dy.y
FH∵ x =
2 × 2500 − y 2 × y × dy
R2 − y2
LM (2500 − y ) OP
N 3/ 2 Q
2 3 / 2 60
25
A
B¢ dy
B
= 125 – 2 ×
R 2 − 25 2
25
25
50 2 − 25 2
= 125 – 86.6
= 38.4 mm
100000 × 258374
N
E
C
C
x
R
50
= 125 – 2 ×
D
y
A
D
25
mm
3.025 × 107 × 38.4
= 22.25 N/mm2. Ans.
IK
− 2500 − y 2 (– 2y) × dy
2
= 3.125 × 105 + [(2500 – 502)3/2 – (2500 – 252)3/2]
3
2
= 3.125 × 105 + [0 – (2500 – 625)3/2]
3
2
= 3.125 × 105 + (– 81189)
3
= 3.125 × 105 – 54126 = 258374 mm3
b = Width of beam at C (i.e., length C-C)
125
= Width of complete section – 2
A
× Width of circular portion at C
25
(i.e., length EC)
∴ τ=
IJ
K
Fig. 8.17
375
STRENGTH OF MATERIALS
At D,
A y = Moment of area above N.A., about N.A.
= Moment of area of rectangle 125 × 75 about N.A.
– Moment of area of circular portion between D and B about N.A.
75
−
2
= 125 × 75 ×
= 351500 –
z
z
50
0
50
= 351500 –
= 351500 +
= 351500 +
0
z
y = 50
y = 50
2.x.dy.y
2 × 2500 − y 2 × y × dy
FH∵ x =
2500 − y 2
IK
− 2500 − y 2 (– 2y)dy
LM (2500 − y ) OP
N 3/ 2 Q
2 3 / 2 50
0
2
[(2500 – 502)3/2 – (2500 – 0)3/2]
3
2
[0 – 125000]
3
= 351500 – 83333.33 mm3
= 268166.67 mm3
b = Width of beam at D (i.e., length D-D)
= 25 mm
= 351500 +
100000 × 268166.67
= 35.46 N/mm2. Ans.
3.025 × 107 × 25
The variation of shear stress is shown in Fig. 8.16 (b).
∴
τ=
HIGHLIGHTS
1. The stresses produced in a beam, which is subjected to shear force is known as shear stresses.
2. The shear stress at a fibre in a section of a beam is given by,
F × Ay
I×b
F = Shear force acting at the given section.
A = Area of the section above the fibre.
τ=
where
y = Distance of the C.G. of the area A from the N.A.
I = Moment of inertia of whole section about N.A.
b = Actual width at the fibre.
3. The shear stress distribution across a rectangular section is parabolic and is given by,
τ=
where
F
2I
Fd
GH 4
2
− y2
I
JK
d = Depth of the beam
y = Distance of the fibre from N.A.
4. The maximum shear stress is at the N.A. for a rectangular section and is given by,
τmax = 1.5 τavg.
376
SHEAR STRESSES IN BEAMS
5. The shear stress distribution across a circular section is parabolic and is given by,
F
(R2 – y2).
3I
6. The shear stress is maximum at the N.A. for a circular section and is given by,
τ=
4
× τavg .
3
7. The shear stress distribution in I-section is parabolic. But at the junction of web and flange, the
shear stress changes abruptly. The shear stress at the junction of the flange and the web changes
τmax =
F
B
F
×
(D2 – d2) to
(D2 – d2) abruptly,
8I
b 8I
where
D = Overall depth of the section,
d = Depth of web,
b = Thickness of web,
B = Overall width of the section.
8. The shear stress distribution for unsymmetrical sections is obtained after calculating the position
of N.A.
9. In case of triangular section, the shear stress is not maximum at the N.A. The shear stress is
maximum at a height of h/2.
10. The shear stress distribution diagram for a composite section, should be drawn by calculating
the shear stress at important points.
from
EXERCISE
(A) Theoretical Questions
1. What do you mean by shear stresses in beams ?
2. Prove that the shear stress at any point (or in a fibre) in the cross-section of a beam which is
subjected to a shear force F, is given by
Ay
b× I
where A = Area of the section above the fibre,
τ=F×
y = Distance of the C.G. of the area A from N.A.,
b = Actual width at the fibre, and
I = Moment of inertia of the section about N.A.
3. Show that for a rectangular section of the maximum shear stress is 1.5 times the average stress.
4. Prove that the shear stress distribution in a rectangular section of a beam which is subjected to
a shear force F is given by
τ=
F
2I
Fd
GH 4
2
I
JK
− y2 .
5. Prove that the maximum shear stress in a circular section of a beam is 4/3 times the average
shear stress.
6. Derive an expression for the shear stress at any point in a circular section of a beam, which is
subjected to a shear force F.
7. How will you draw the shear stress distribution diagram for composite section ?
8. How will you prove that the shear stress changes abruptly at the junction of the flange and the
web of an I-section ?
377
STRENGTH OF MATERIALS
9. The shear stress is not maximum at the N.A. in case of a triangular section. Prove this statement.
10. Prove that the maximum shear stress in a triangular section of a beam is given by
τmax =
3F
bh
where b = Base width, and
h = Height.
11. Show that the ratio of maximum shear stress to mean shear stress in a rectangular cross-section
is equal to 1.50 when it is subjected to a transverse shear force F. Plot the variation of shear
stress across the section.
12. Sketch the distribution of shear stress across the depth of the beams of the following crosssections :
(i) T-section, and
(ii) Square section with diagonal vertical.
(B) Numerical Problems
1. A rectangular beam 100 mm wide and 150 mm deep is subjected to a shear force of 30 kN.
Determine : (i) average shear stress and (ii) maximum shear stress. [Ans. 2 N/mm2 ; 3 N/mm2]
2. A rectangular beam 100 mm wide is subjected to a maximum shear force of 100 kN. Find the
[Ans. 250 mm]
depth of the beam if the maximum shear stress is 6 N/mm2.
3. A timber beam of rectangular section is simply supported at the ends and carries a point load at
the centre of the beam. The length of the beam is 6 m and depth of beam is 1 m. Determine the
maximum bending stress and the maximum shear stress.
[Ans. 12 N/mm2 ; 1 N/mm2]
4. A timber beam 100 mm wide and 150 mm deep supports a uniformly distributed load of intensity
w kN/m length over a span of 2 m.
If the safe stresses are 28 N/mm2 in bending and 2 N/mm2 in shear, calculate the safe intensity
of the load which can be supported by the beam.
[Ans. 20 kN/m]
5. A circular beam of 105 mm diameter is subjected to a shear force of 5 kN. Calculate : (i) average
shear stress, and (ii) maximum shear stress. Also sketch the variation of the shear stress along
the depth of the beam.
[Ans. (i) 0.577 N/mm2 (ii) 0.769 N/mm2]
6. The maximum shear stress in a beam of circular section of diameter 150 mm, is 5.28 N/mm2.
Find the shear force to which the beam is subjected.
[Ans. 70 kN]
7. A beam of I-section is having overall depth as 500 mm and overall width as 190 mm. The
thickness of flanges is 25 mm whereas the thickness of the web is 15 mm. The moment of inertia
about N.A. is given as 6.45 × 108 mm4. If the section carries a shear force of 40 kN, calculate the
maximum shear stress. Also sketch the shear stress distribution across the section.
[Ans. 62.33 N/mm2]
8. An I-section has flanges of width b and the overall depth is 2b. The flanges and web are of uniform
thickness t. Find the ratio of the maximum shear stress to the average shear stress.
[Ans. 2.25]
9. An I-section has the following dimensions :
flanges : 150 mm × 20 mm
web
: 30 mm × 10 mm
The maximum shear stress developed in the beam is 16.8 N/mm2. Find the shear force to which
the beam is subjected.
[Ans. 50 kN]
10. A 12 cm by 5 cm I-section is subjected to a shearing force of 10 kN. Calculate the shear stress at
the neutral axis and at the top of the web. What percentage of shearing force is carried by the
web ? Given I = 220 × 104 mm4, area = 9.4 × 102 mm2, web thickness = 3.5 mm and flange thickness
= 5.5 mm.
[Ans. 27.2 N/mm2 ; 20.1 N/mm2 ; 9.5 kN. i.e., 95% of the total]
378
SHEAR STRESSES IN BEAMS
11.
12.
13.
The shear force acting on a section of a beam is 100 kN. The section of the beam is of T-shaped of
dimensions 200 mm × 250 mm × 50 mm. The flange thickness and web thickness are 50 mm.
Moment of inertia about the horizontal neutral axis is 1.134 × 108 mm4. Find the shear stress at
the neutral axis and at the junction of the web and the flange.
[Ans. 11.64 N/mm2 ; 2.76 N/mm2 and 11.04 N/mm2]
A beam is of T-section, flange 12 cm by 1 cm, web 10 cm by 1 cm. What percentage of the shearing
force at any section is carried by the web ?
[Ans. 93.5%]
For the section shown in Fig. 8.18, determine the average shearing stresses at A, B, C and D for
a shearing force of 20 kN. Draw also the shear stress distribution across the section.
[Ans. 0 ; 6.47 N/mm2 ; 27.7 N/mm2 ; 44.4. N/mm2]
50 mm
A
A
10
mm B
10
mm
B
C
10
mm
20
20
m
m
m
D
D
60 mm
C
m
10
mm
Fig. 8.18
14.
A rectangular beam is simply supported at the ends and carries a point load at the centre. Prove
that the ratio of span to depth
Maximum bending stress
.
2 × Maximum shear stress
W = Point load at centre,
b = width, and d = Depth.
=
[Hint. Let
Max. Shear force =
W
WL
, Max. bending moment =
.
4
2
FG
H
F
GH
WL
M
4
∴ Max. bending stress =
=
Z
bd 2
6
Max. shear stress =
=
∴
IJ
K = WL × 6
I 4 bd
JK
2
=
3 WL
2 bd 2
3
× Average shear stress
2
3 Max. shear force 3 W
1
3
W
×
= ×
×
= ×
Area of section
b×d
2
2
2
4 b×d
OP
L
Max. bending stress
= .P
=
P
3 W I
2 × Max. shear stress F
GH 2 × 4 b × d JK d PPQ
FG 3 WL IJ
H 2 bd K
2
379
9
CHAPTER
DIRECT AND BENDING
STRESSES
9.1. INTRODUCTION..
Direct stress alone is produced in a body when it is subjected to an axial tensile or
compressive load. And bending stress is produced in the body, when it is subjected to a bending
moment. But if a body is subjected to axial loads and also bending moments, then both the
stresses (i.e., direct and bending stresses) will be produced in the body. In this chapter, we shall
study the important cases of the members subjected to direct and bending stresses. Both these
stresses act normal to a cross-section, hence the two stresses may be algebraically added into a
single resultant stress.
9.2. COMBINED BENDING AND DIRECT STRESSES..
Consider the case of a column* subjected by a compressive load P
acting along the axis of the column as shown in Fig. 9.1. This load will
cause a direct compressive stress whose intensity will be uniform across the
cross-section of the column.
Let
σ 0 = Intensity of the stress
A = Area of cross-section
P = Load acting on the column.
Then stress,
P
Load P
=
Area A
P
Now consider the case of a column subjected by a compressive load P
Fig. 9.1
whose line of action is at a distance of ‘e’ from the axis of the column as
shown in Fig. 9.2 (a). Here ‘e’ is known as eccentricity of the load. The eccentric load shown in
Fig. 9.2 (a) will cause direct stress and bending stress. This is proved as discussed below :
1. In Fig. 9.2 (b), we have applied, along the axis of the column, two equal and opposite
forces P. Thus three forces are acting now on the column. One of the forces is shown in Fig.
9.2 (c) and the other two forces are shown in Fig. 9.2 (d).
2. The force shown in Fig. 9.2 (c) is acting along the axis of the column and hence this
force will produce a direct stress.
3. The forces shown in Fig. 9.2 (d) will form a couple, whose moment will be P × e. This
couple will produce a bending stress.
σ0 =
* Column is a vertical member subjected to a compressive load.
381
STRENGTH OF MATERIALS
Hence an eccentric* load will produce a direct stress as well as a bending stress. By
adding these two stresses algebraically, a single resultant stress can be obtained.
e
P
P
P
P
P
(a )
e
P
P
(b )
(c )
(d)
Fig. 9.2
9.3. RESULTANT STRESS WHEN A COLUMN OF RECTANGULAR SECTION IS SUBJECTED TO AN ECCENTRIC LOAD
A column of rectangular section subjected to an eccentric load is shown in Fig. 9.3. Let the
load is eccentric with respect to the axis Y-Y as shown in Fig. 9.3 (b). It is mentioned in Art. 9.2
that an eccentric load causes direct stress as well as bending stress. Let us calculate these
stresses.
Let
P = Eccentric load on column
e = Eccentricity of the load
σ0 = Direct stress
σb = Bending stress
b = Width of column
d = Depth of column
∴ Area of column section, A = b × d
Now moment due to eccentric load P is given by,
M = Load × eccentricity
=P×e
The direct stress (σ 0) is given by,
Load ( P) P
=
...(i)
Area
A
This stress is uniform along the cross-section of the column.
The bending stress σb due to moment at any point of the column section at a distance y
from the neutral axis Y-Y is given by
σ0 =
σ
M
= b
I
±y
* Eccentric load is a load whose line of action does not coincide with the axis of the column. The
accentricity of the load may be about one of the axis, or about both the axis.
382
DIRECT AND BENDING STRESSES
M
×y
...(ii)
I
where I = Moment of inertia of the column section about
∴
σb = ±
e
P
d . b3
12
Substituting the value of I in equation (ii), we get
the neutral axis Y-Y =
12 M
M
×y
× y =±
3
d . b3
d.b
12
The bending stress depends upon the value of y from
the axis Y-Y.
The bending stress at the extreme is obtained by
b
substituting y = in the above equation.
2
12 M b
6M
× =±
∴
σb = ±
3
2
d.b
d . b2
σb = ±
=±
6P×e
(∵ M = P × e)
d . b2
Elevation
(a)
D
Y
C
Position
of load P
d
X
X
A
b
Y
Plan
(b )
B
6P×e
6P×e
=±
d.b.b
A×b
(∵ Area = b × d = A) σmin
The resultant stress at any point will be the algebraic
σmax
sum of direct stress and bending stress.
If y is taken positive on the same side of Y-Y as the
(c )
load, then bending stress will be of the same type as the
Fig. 9.3
direct stress. Here direct stress is compressive and hence
bending stress will also be compressive towards the right of
the axis Y-Y. Similarly bending stress will be tensile towards the left of the axis Y-Y. Taking
compressive stress as positive and tensile stress as negative we can find the maximum and
minimum stress at the extremities of the section. The stress will be maximum along layer BC
and minimum along layer AD.
Let σmax = Maximum stress (i.e., stress along BC)
σmin = Minimum stress (i.e., stress along AD)
Then σmax = Direct stress + Bending stress
= σ0 + σb
=±
=
FG
H
(Here bending stress is +ve)
IJ
K
6×e
P
1+
A
b
σmin = Direct stress – Bending stress
= σ0 – σb
=
and
P 6 P.e
+
A
A.b
...(9.1)
383
STRENGTH OF MATERIALS
FG
H
IJ
K
P 6 P.e
6×e
P
−
1−
=
...(9.2)
A
A.b
A
b
These stresses are shown in Fig. 9.3 (c). The resultant stress along the width of the
column will vary by a straight line law.
If in equation (9.2), σmin is negative then the stress along the layer AD will be tensile. If
σmin is zero then there will be no tensile stress along the width of the column. If σmin is positive
then there will be only compressive stress along the width of the column.
=
Problem 9.1. A rectangular column of width 200 mm
and of thickness 150 mm carries a point load of 240 kN at an
eccentricity of 10 mm as shown in Fig. 9.4 (i). Determine the
maximum and minimum stresses on the section.
Sol. Given :
Width,
b = 200 mm
Thickness,
d = 150 mm
∴ Area,
A = b×d
= 200 × 150 = 30000 mm2
Eccentric load,
P = 240 kN
= 240000 N
Eccentricity,
e = 10 mm
Let
σmax = Maximum stress, and
σmin = Minimum stress.
(i) Using equation (9.1), we get
σmax
FG
H
6×e
P
1+
=
A
b
FG
H
IJ
K
σmin
FG
H
P
6×e
1−
=
A
b
FG
H
IJ
K
200 mm
Y
e
150
mm
Y
Fig. 9.4 (i)
IJ
K
240000
6 × 10
1+
30000
200
= 8(1 + 0.3) = 10.4 N/m2. Ans.
(ii) Using equation (9.2), we get
=
240 kN
10 mm
200 mm
smin
= 5.6
smax
= 10.4
Fig. 9.4 (ii)
IJ
K
6 × 10
240000
1−
= 8(1 – 0.3) = 5.6 N/mm2. Ans.
200
30000
These stresses are shown in Fig. 9.4 (ii).
=
Problem 9.2. If in Problem 9.1, the minimum stress on the section is given zero then
find the eccentricity of the point load of 240 kN acting on the rectangular column. Also calculate
the corresponding maximum stress on the section.
Sol. Given :
The data from Problem 9.1 is :
b = 200 mm, d = 150 mm, P = 240000 N, A = 30000 mm2
384
DIRECT AND BENDING STRESSES
Minimum stress,
σmin = 0
Let
e = Eccentricity
Using equation (9.2), we get
σmin =
or
or
0 =
FG
H
P
6×e
1−
A
b
FG
H
IJ
K
240000
6×e
1−
30000
200
200 mm
max = 16
IJ
K
6×e
6×e
= 0 or 1 =
200
200
Fig. 9.5
200
∴
e =
= 33.33 mm. Ans.
6
Corresponding maximum stress is obtained by using equation (9.1).
1–
∴
σmax =
FG
H
6×e
P
1+
A
b
F
GG
GH
IJ
K
I
JJ
JK
6 × 200
240000
1+
=
= 8(1 + 1) = 16 N/mm2 Ans.
6
30000
200
The stresses are shown in Fig. 9.5.
Problem 9.3. If in Problem 9.1, the eccentricity is given 50 mm instead of 10 mm then
find the maximum and minimum stresses on the section. Also plot these stresses along the
width of the section.
Sol. Given :
200 mm
The data from Problem 9.1 is :
2
4 N/mm
b = 200 mm
d = 150 mm
P = 240000 N
A = 30000 mm2
2
20 N/mm
Eccentricity,
e = 50 mm
(i) Maximum stress (σmax) is given by
Fig. 9.6
equation (9.1) as
σmax =
=
FG
H
P
6×e
1+
A
b
FG
H
IJ
K
IJ
K
240000
6 × 50
1+
= 8(1 + 1.5) = 20 N/mm2. Ans.
30000
200
(ii) Minimum stress (σmin) is given by equation (9.2) as
σmin =
FG
H
P
6×e
1−
A
b
IJ
K
385
STRENGTH OF MATERIALS
FG
H
IJ
K
240000
6 × 50
1−
= 8(1 – 1.5) = – 4 N/mm2. Ans.
30000
200
Negative sign means tensile stress.
The stresses are plotted as shown in Fig. 9.6.
=
Note. From the above three problems, we have
(i) The minimum stress is zero when e =
Problem 9.2.
200
mm or
6
b
mm (as b = 200). This is clear from
6
(ii) The minimum stress is +ve (i.e., compressive) when e <
which e = 10 mm which is less than
200
(i.e., 33.33).
6
(iii) The minimum stress is –ve (i.e., tensile) when e >
e = 50 mm which is more than
b
. This is clear from Problem 9.1 in
6
b
. This is clear from Problem 9.3 in which
6
200
(i.e., 33.33).
6
Problem 9.4. The line of thrust, in a compression testing specimen 15 mm diameter, is
parallel to the axis of the specimen but is displaced from it. Calculate the distance of the line of
thrust from the axis when the maximum stress is 20% greater
e P
than the mean stress on a normal section.
Sol. Given :
Diameter,
d = 15 mm
π
× 15 2
∴ Area,
A=
4
= 176.714 mm2
σmax = 20% greater than mean
120
× mean stress
=
100
= 1.2 × mean stress.
Let
P = Compressive load on specimen
e = Eccentricity
Load
P
Mean stress =
=
N/mm2
Area
176.714
15 mm
We know that moment,
e
M= P×e
Now bending stress is given by
M σb
=
I
y
M
×y
∴
σb =
I
∴ Maximum bending stress will be when y = ±
d
.
2
smin
smax
Hence maximum bending stress is given by,
σb =
386
FG IJ
H K
d
M
× ±
I
2
Fig. 9.7
DIRECT AND BENDING STRESSES
=±
M d
×
2
I
FG∵
H
d
M
×
=± π
2
4
d
64
=±
=±
I=
π 4
d
64
IJ
K
32 M
πd 3
32 P × e
(∵ M = P × e)
πd 3
Direct stress due to load is given by,
P
P
=
A 176.714
∴ Maximum stress = Direct stress × Bending stress
= σ0 + σb
σ0 =
32 P × e
P
+
176.714
πd 3
= 1.2 × Mean stress
σmax =
or
But
σmax
= 1.2 ×
P
176.714
...(i)
...(ii)
FG∵
H
(given)
Mean stress =
P
176.714
IJ
K
Equating equations (i) and (ii), we get
P
32 P × e
P
+
= 1.2 ×
3
176.714
176.714
πd
32 × P × e
or
πd
3
32 × e
or
πd
∴
3
=
0.2 P
1.2 P
P
−
=
176.714 176.714
176.714
=
0.2
176.714
e=
(Cancelling P from both sides)
0.2 × π × d 3
0.2 × π × 153
=
= 0.375 mm. Ans.
32 × 176.714
32 × 176.714
Problem 9.5. A hollow rectangular column of external depth 1 m and external width
0.8 m is 10 cm thick. Calculate the maximum and minimum stress in the section of the column
if a vertical load of 200 kN is acting with an eccentricity of 15 cm as shown in Fig. 9.8.
Sol. Given :
External width,
B = 0.8 m = 800 mm
External depth,
D = 1.0 m = 1000 mm
Thickness of walls,
t = 10 cm = 100 mm
Inner width,
b = B – 2 × 100
= 800 – 200 = 600 mm
Inner depth,
d =D–2×t
= 1000 – 2 × 100 = 800 mm
387
STRENGTH OF MATERIALS
∴ Area,
A = B×D–b×d
= 800 × 1000 – 600 × 800
= 800000 – 480000
= 320000 mm2
M.O.I. about Y-Y axis is given by,
1000 × 800 3 800 × 600 3
−
12
12
= 42.66 × 109 – 14.4 × 109
= 28.26 × 109 mm4
Eccentric load, P = 200 kN = 200,000 N
Eccentricity, e = 15 cm = 150 mm
We know that the moment,
M = P×e
= 200,000 × 150
= 3000000 Nmm
The bending stress is given by,
I =
σb
M
=
I
y
M
×y
y
Maximum bending stress will be when
y = ± 400
∴
σb =
M
× (± 400)
I
30000000
× 400
= ±
28.26 × 109
= ± 0.4246 N/mm2
Direct stress is given by,
∴
σb =
σ0 =
∴ Maximum stress
Minimum stress
=
=
=
=
=
P 200000
=
A 320000
0.625 N/mm2
σ 0 + σ b = 0.625 + 0.4246
1.0496 N/mm2 (Compressive).
σ 0 – σ b = 0.625 – 0.4246
0.2004 N/mm2 (Compressive).
Fig. 9.8
Ans.
Ans.
Problem 9.6. A short column of external diameter 40 cm and internal diameter 20 cm
carries an eccentric load of 80 kN. Find the greatest eccentricity which the load can have
without producing tension on the cross-section.
Sol. Given :
External dia.,
D = 40 cm = 400 mm
Internal dia.,
d = 20 cm = 200 mm
388
DIRECT AND BENDING STRESSES
∴ Area of cross-section,
π
(D2 – d2)
A=
4
π
=
(4002 – 2002) = 30000 × π mm2
4
Moment of inertia
π
(D4 – d4)
64
π
=
(4004 – 2004) = 3.75 × 108 × π mm4
64
P = 80 kN = 80000 N
e = Eccentricity when there is no tension.
e
P
I=
Eccentric load,
Let
P
80000
=
Now direct stress,σ0 =
A 30000 × π
We know that moment,
M = P × e = 80000 × e
Now bending stress (σb) is given by
200 mm
400 mm
...(i)
e
M σb
=
I
y
M×y
I
The bending stress will be maximum when
∴
σb =
D
400
=±
= ± 200 mm
2
2
∴ Maximum bending stress is given by,
smax
y = ±
M × (± 200)
M × 200
=±
σb =
I
I
= ±
80000 × e × 200
3.75 × 10 8 × π
Fig. 9.9
...(ii)
Now minimum stress is given by,
σmin = σ0 – σb
=
80000
80000 × e × 200
−
30000 × π
3.75 × 10 8 × π
There will be no tension if σmin = 0
∴ For no tension, we have
0=
or
80000
80000 × e × 200
−
30000 × π
3.75 × 10 8 × π
80000 × e × 200
80000
=
30000 × π
3.75 × 10 8 × π
389
STRENGTH OF MATERIALS
or
e =
3.75 × 10 8 × π × 80000
= 62.5 mm. Ans.
30000 × π × 80000 × 200
Problem 9.7. If in the Problem 9.6, the eccentricity of the point load is given as 150 mm,
then calculate the maximum and minimum stress in the section.
Sol. Given :
The data from Problem 9.6 is :
400 mm
D = 400 mm, d = 200 mm
P = 80000 N, A = 30000 × π mm2 s
min
Moment of inertia,I = 3.75 × 108 × π mm4
Eccentricity,
e = 150 mm
P
80000
=
A 30000 × π
= 0.8488 N/mm2
We know that moment,
M = P × e = 80000 × 150
= 12000000 Nmm
Maximum bending stress is given by,
smax
Now direct stress,σ0 =
Fig. 9.10
12000000 × (± 200)
M × ymax
=
(∵ ymax = ± 200 mm)
3.75 × 10 8 × π
I
= ± 2.037 N/mm2
∴ Maximum stress = σ0 + σb
= 0.8488 + 2.037 = 2.8858 N/mm2 (Compressive). Ans.
Minimum stress
= σ0 – σb
= 0.8488 – 2.037 = – 1.1882 N/mm2 (Tensile). Ans.
The stress distribution across the width is shown in Fig. 9.10.
σb =
9.4. RESULTANT STRESS WHEN A COLUMN OF RECTANGULAR SECTION IS
SUBJECTED TO A LOAD WHICH IS ECCENTRIC TO BOTH AXES
A column of rectangular section ABCD, subjected to a load which is eccentric to both axes,
is shown in Fig. 9.11.
Let P = Eccentric load on column
C
Y
D
e x = Eccentricity of load about X-X axis
Load point
e y = Eccentricity of load about Y-Y axis
b
d
σ0
σbx
σby
Mx
My
390
=
=
=
=
=
=
=
=
=
Width of column
Depth of column
Direct stress
Bending stress due to eccentricity ex
Bending stress due to eccentricity ey
Moment of load about X-X axis
P × ex
Moment of load about Y-Y axis
P × ey
By
X
P
ex
d
X
O
A
Y Plan
b
Fig. 9.11
B
DIRECT AND BENDING STRESSES
Ixx = Moment of inertia about X-X axis
bd 3
12
Iyy = Moment of inertia about Y-Y axis
=
db3
12
Now the eccentric load is equivalent to a central load P, together with a bending moment
P × ey about Y-Y and a bending moment P × ex about X-X.
(i) The direct stress (σ 0) is given by,
=
P
A
(ii) The bending stress due to eccentricity ey is given by,
σ0 =
σ by =
My × x
I yy
=
P × ey × x
I yy
...(i)
(∵ My = B × ey) ...(ii)
b
b
to +
2
2
(iii) The bending stress due to eccentricity ex is given by,
In the above equation x varies from –
σ bx =
M x × y P × ex × y
=
I xx
I xx
d
d
to +
2
2
The resultant stress at any point on the section
= σ0 ± σby ± σbx
In the above equation, y varies from –
=
P M y × x Mx . y
±
±
A
I yy
I xx
...(9.3)
(i) At the point C, the co-ordinates x and y are positive hence the resultant stress will be
maximum.
(ii) At the point A, the co-ordinates x and y are negative and hence the resultant stress will
be minimum.
(iii) At the point B, x is +ve and y is –ve and hence resultant stress
=
P M y . x Mx . y
+
−
I yy
I xx
A
(iv) At the point D, x is –ve and y is +ve and hence resultant stress
=
P M y . x Mx . y
−
+
.
A
I yy
I xx
Problem 9.8. A short column of rectangular cross-section 80 mm by 60 mm carries a
load of 40 kN at a point 20 mm from the longer side and 35 mm from the shorter side. Determine
the maximum compressive and tensile stresses in the section.
391
STRENGTH OF MATERIALS
Sol. Given :
Width,
b = 80 mm
D
Depth,
d = 60 mm
∴ Area,
A = 80 × 60 = 4800 mm2
Point load,
P = 40 kN = 40000 N
60 mm
Eccentricity of load about X-X axis,
X
e x = 10 mm
Eccentricity of load about Y-Y axis,
e y = 5 mm
A
Moment of load about X-X axis,
Mx = P × ex = 40000 × 10
= 400000 Nmm
Moment of load about Y-Y axis,
My = P × ey = 40000 × 5 = 200000 Nmm
Moment of inertia about X-X axis,
Y
80 mm
Load
point
20 mm
Shorter
side
30 mm
35 mm
5
10
C
X
O
B
Y
40 mm
Fig. 9.12
1
× 80 × 60 3 = 1440000 mm4
12
1
× 60 × 80 3 = 2560000 mm4
Similarly,
Iyy =
12
(i)The maximum compressive stress will be at point C where x and y are positive. The
value of x = 40 mm and y = 30 mm at C.
Hence maximum compressive stress is given by equation (9.3)
Ixx =
=
P M y × x Mx × y
+
+
A
I yy
I xx
(Taking +ve sign)
40000 200000 × 40 400000 × 30
+
+
4800
2560000
1440000
= 8.33 + 3.125 + 8.33 = 19.785 N/mm2. Ans.
(ii)The maximum tensile stress will be at point A where x = – 40 mm and y = – 30 mm.
Hence using equation (9.3), we get
=
Resultant stress at A =
P M y × x Mx × y
+
+
I yy
I xx
A
40000 200000 × 40 400000 × 30
−
−
4800
2560000
1440000
= 8.33 – 3.125 – 8.33 = – 3.125 N/mm2. Ans.
=
Problem 9.9. A column is rectangular in cross-section of 300 mm × 400 mm in dimensions. The column carries an eccentric point load of 360 kN on one diagonal at a distance of
quarter diagonal length from a corner. Calculate the stresses at all four corners. Draw stress
distribution diagrams for any two adjacent sides.
Sol. Given :
Width,
Depth,
392
b = 300 mm
d = 400 mm
DIRECT AND BENDING STRESSES
∴ Area,
A = b × d = 300 × 400
= 12 × 104 mm2
Eccentric load,
P = 360 kN = 360000 N
The eccentric load is acting at point E, where
distance EC = one quarter of diagonal AC.
Now diagonal
∴
AC =
D
Y
4
3
3
4
cos θ =
and sin θ =
5
5
1
OE = EC = of AC
4
1
= × 500 = 125 mm
4
C
Load
point
ey
75
100 ex
300 2 + 400 2 = 500
In ΔCAB, tan θ =
Also
300 mm
400
mm
O
X
B
A
Fig. 9.12 (a)
4
= 100 mm
5
3
ey = OF = OE cos θ = 125 × = 75 mm
5
Moment of load about x-x axis,
Mx = P × ex = 360000 × 100 = 36000000 Nmm
Moment of load about y-y axis,
My = P × ey = 360000 × 75
= 27000000 Nmm
And
ex = EF = OE sin θ = 125 ×
1
× 300 × 4003 = 16 × 108 mm4
12
1
Iyy =
× 400 × 3003 = 9 × 108 mm4
12
The resultant stress at any point is given by equation (9.3) as
Also
Resultant stress
Ixx =
=
P M y × x Mx × y
+
+
A
I yy
I xx
(i) Resultant stress at point C
At point C, x = 150 mm and y = 200 m
∴ Resultant stress at C
=
P M y × 150 M x × 200
+
+
I yy
I xx
A
27000000 × 150 36000000 × 200
+
12 × 10
9 × 10 8
16 × 10 8
= 3 + 4.5 + 4.5 N/mm2
= 12 N/mm2 (compressive). Ans.
=
360000
4
+
393
STRENGTH OF MATERIALS
(ii) Resultant stress at point B
At point B, x = 150 mm and y = – 200 mm
Resultant stress at B
=
P M y × 150 M x × (− 200)
+
+
A
I yy
16 × 10 8
=
360,000 27000000 × 150 36000000 × 200
−
+
12 × 10 4
9 × 10 8
16 × 10 8
= 3 + 4.5 – 4.5
= 3 N/mm2 (compressive). Ans.
(iii) Resultant stress at point A
At point A, x = – 150 mm and y = – 200 mm
∴ Resultant stress at point A
=
=
P M y × (− 150) M x × (− 200)
+
+
A
I yy
I xx
360000
12 × 10
4
−
27000000 × 150
9 × 10
8
−
36000000 × 200
16 × 10 8
= 3 – 4.5 – 4.5
= – 6 N/mm2 (Tensile). Ans.
(iv) Resultant stress at point D
At point D, x = – 150 mm and y = 200 mm
∴ Resultant stress at point D
=
P M y (− 150) M x × 200
+
+
A
I yy
I xx
360000
=
12 × 10
4
−
27000000 × 150
9 × 10
8
+
36000000 × 200
16 × 10 8
= 3 – 4.5 + 4.5
= 3 N/mm2 (compressive). Ans.
Stress distribution for AB and BC (i.e., two adjacent sides)
Fig. 9.12 (b) shows the stress distribution along two adjacent sides (i.e., AB and BC).
At point A, resultant stress is 6 N/mm2 (tensile) whereas at point B, the resultant stress is
3 N/mm2 (compressive). Take AE = 6 N/mm2 and BF = 3 N/mm2. Join E to F.
For side BC, the resultant stress at B is 3 N/mm2 (compressive) whereas at point C the
resultant stress is 12 N/mm2 (compressive).
Take BH = 3 N/mm2 (compressive) and CG = 12 N/mm2 (compressive).
394
DIRECT AND BENDING STRESSES
2
12 N/mm
D
C
A
B
E
6 N/mm
C
G
H
B
2
3 N/mm
2
B
A
2
3 N/mm
F
Fig. 9.12 (b)
Problem 9.10. A masonry pier of 3 m ×
4 m supports a vertical load of 80 kN as shown in
Fig. 9.13.
D
(a) Find the stresses developed at each corner of the pier.
(b) What additional load should be placed
1.5 m
at the centre of the pier, so that there is no tension
anywhere in the pier section ?
(c) What are the stresses at the corners with 3 m X
the additional load in the centre ?
Sol. Given :
Width
b = 4m
A
Depth,
d = 3m
2
∴ Area,
A = 4 × 3 = 12 mm
Point load,
P = 80 kN
Eccentricity of load about X-X axis,
e x = 0.5 m
Eccentricity of load about Y-Y axis,
e y = 1.0 m
Moment of load about X-X axis,
Mx = P × ex = 80 × 0.5 = 40 kNm
Similarly,
My = P × ey = 80 × 1.0 = 80 kNm
Moment of inertia about X-X axis,
1
× 4 × 33 = 9 mm4
Ixx =
12
Y
2m
4m
C
Load point
1m
0.5 m
O
Y
X
B
Fig. 9.13
395
STRENGTH OF MATERIALS
1
× 3 × 43 = 16 m4.
12
(a) Stresses developed at each corner of the pier
The resultant stress at any point is given by equation (9.3).
P M y × x Mx × y
+
+
...(i)
Hence resultant stress =
I yy
I xx
A
(i)At point A, x = – 2.0 m and y = – 1.5 m. Hence resultant stress at A (i.e., σA) is obtained
by substituting these values in the above equation (i).
80 80 × (− 2.0) 40 × (− 1.5)
∴
σA =
+
+
12
16
9
= 6.66 – 10 – 6.66
= – 10 kN/m2 (Tensile). Ans.
(ii)A point B, x = 2.0 m and y = – 1.5 m. Hence resultant stress at B (i.e., σB) is obtained by
substituting these values in equation (i).
80 80 × 2.0 40 × (− 1.5)
∴
σB =
+
+
12
16
9
= 6.66 + 10 – 6.66
= 10 kN/m2 (Compressive). Ans.
(iii)At point C, x = 2.0 m and y = 1.5 m. Hence resultant stress at C (i.e., σC) is given by,
80 80 × 2.0 40 × 1.5
+
+
∴
σC =
12
16
9
= 6.66 + 10 + 6.66
= 23.33 kN/m2 (Compressive). Ans.
(iv)At point D, x = – 2.0 m and y = 1.5 m. Hence resultant stress at D (i.e., σD) is given by,
80 80 × (− 2.0) 40 × 1.5
+
+
∴
σD =
12
16
9
= 6.66 – 10 + 6.66
= + 3.33 kN/m2 (Compressive). Ans.
(b) Additional load at the centre of the pier, so that there is no tension anywhere in the
pier section.
Let W =Additional load (in kN) placed at the centre for no tension anywhere in the pier
section.
The above load is compressive and will cause a compressive stress
W W
=
=
kN/m2
(∵ A = 12 m2)
A 12
As this load is placed at the centre, it will produce a uniform compressive stress across
the section of the pier. But we know that there is tensile stress at point A having magnitude
= 10 kN/m 2. Hence the compressive stress due to load W should be equal to tensile
stress at A.
W
= 10
∴
12
or
W = 10 × 12 = 120 kN. Ans.
(c) Stresses at the corners with the additional load at the centre
Similarly,
Iyy =
Stress due to additional load =
396
W 120
=
= 10 kN/m2 (Compressive)
12
A
DIRECT AND BENDING STRESSES
This stress is uniform across the cross-section of the pier. Hence to find the stresses at the
corners with this additional load, we must add the stress 10 kN/m2 in each value of the stresses
already existing in the corners.
∴ Stress at A, σA = – 10 + 10 = 0. Ans.
Similarly,
σB = 10 + 10 = 20 kN/m2. Ans.
σC = 23.33 + 10 = 33.33 kN/m2. Ans.
and
σD = 3.33 + 10 = 13.33 kN/m2. Ans.
9.5. RESULTANT STRESS FOR UNSYMMETRICAL COLUMNS WITH ECCENTRIC
LOADING
In case of unsymmetrical columns which are subjected to eccentric loading, first the centre
of gravity (i.e., C.G.) of the unsymmetrical section is determined. Then the moment of inertia of
the section about the axis passing through the C.G. is calculated. After that the distances between
the corners of the section and its C.G. is obtained. By using the values of the moment of inertia
and distances of the corner from the C.G. of the section, the stresses on the corners are then
determined.
Problem 9.11. A short column has a square section 300 mm × 300 mm with a square
hole of 150 mm × 150 mm as shown in Fig. 9.14. It carries an eccentric load of 1800 kN,
located as shown in the figure. Determine the maximum compressive and tensile stress
across the section.
Y
D
75 mm
75 mm
150 mm
C
50 mm
H
G
Load
point
158.34
mm
30 mm
150 mm
300 mm
e
X
X
41.66
E
141.66
mm
F
100 mm
A
75 mm
75 mm
B
300 mm
Y
Fig. 9.14
Sol. Given :
Dimension of column
Dimension of hole
= 300 mm × 300 mm
= 150 mm × 150 mm
397
STRENGTH OF MATERIALS
∴ Area of section,
A = 300 × 300 – 150 × 150
= 90000 – 22500
= 67500 mm2
Point load,
P = 1800 kN = 1800000 N
The point load is acting on Y-Y axis. The given section is also symmetrical about
Y-Y axis. But it is unsymmetrical to X-X axis. Let us first find the position of X-X axis. For
this, find the distance of C.G. from the bottom line AB. Let y is the distance of the C.G. of
the section from the bottom line AB.
y=
Then
A1 y1 + A2 y2
( A1 + A2 )
where A1 = Area of outer square = 300 × 300 = 90000 mm2
y1 = Distance of C.G. of outer square from line AB = 150 mm
A2 = Area of square hole = 150 × 150 = 22500 mm2 = – 22500 mm2
(–ve sign due to cut out portion)
y2 = Distance of C.G. of square hole from line AB = 100 +
∴
y=
150
= 175 mm
2
90000 × 150 − 22500 × 175
(90000 − 22500)
13500000 − 3937500
= 141.66 mm
67500
∴ The axis X-X lies at a distance 141.66 mm from line AB or at a distance of 300 –
141.66 = 158.34 mm from line CD.
The load is unsymmetrical to X-X axis.
Hence eccentricity, e = 158.34 – (50 + 30) = 78.34 mm
∴ Moment about X-X axis,
M = P × e = 1800000 × 78.34
= 14101200 Nmm
Now let us calculate the moment of inertia of the section about X-X axis.
Let I1 = M.O.I. of outer square ABCD about X-X axis.
= M.O.I. of ABCD about an axis parallel to X-X and passing through
its C.G. + Area of ABCD (Distance of C.G. of ABCD from X-X axis)2
=
300 × 300 3
+ 300 × 300 × (158.34 – 150)2
12
= 675000000 + 6260004 = 681260004 mm4
I2 = M.O.I. of square hole about X-X axis
= M.O.I. of hole about its C.G. + Area of hole (Distance of C.G. of
hole from X-X )2
=
150 × 150 3
= 150 × 150(175 – 141.66)4
12
= 42187500 + 25010001 = 67197501 mm4
=
398
DIRECT AND BENDING STRESSES
∴ Net moment of inertia of the section about X-X axis is given by
I = I1 – I 2
= 681260004 – 67197501 = 614062503 mm4
Now direct stress is given by,
P 1800000
=
= 26.66 N/mm2
σ0 =
67500
A
This stress is uniform across the section.
Bending stress is given by,
M σb
=
I
y
M×y
or
σb =
...(i)
I
The maximum value of y from X-X axis is 158.34 mm. This is the distance of the line CD
from X-X axis. As load is acting above the X-X axis, hence the bending stress will be compressive
on the edge CD. This stress is obtained by substituting y = 158.34 mm in equation (i).
∴ Bending stress at the edge CD due to moment
M × 158.34 141012000 × 158.34
=
614062503
614062503
= 36.36 N/mm2 (Compressive).
Bending stress at the edge AB will be tensile. The distance of AB from X-X is 141.66 mm.
Bending stress at the edge AB due to moment will be obtained by substituting y = –
141.66 in equation (i).
∴ Bending stress at the edge AB due to moment
=
M × 141.66
141012000 × 141.66
=–
(Tensile)
614062503
I
= – 32.529 N/mm2
∴ Resultant stress at the edge CD
= σ0 + σb
= 26.66 + 36.66 = 63.32 N/mm2 (Compressive). Ans.
and resultant stress at the edge AB
= 26.66 – 32.529 = – 5.869 (Tensile). Ans.
Problem 9.12. A short column has a rectangular section 160 mm × 200 mm with a
circular hole of 80 mm diameter as shown in Fig. 9.15. It carries an eccentric load of 100 kN, at
a point as shown in the figure. Determine the stresses at the four corners of the section.
Sol. Given :
Width,
B = 160 mm
Depth,
D = 200 mm
Area of rectangular ABCD,
A1 = 160 × 200 = 32000 mm2
Dia. of hole, d = 80 mm
π
∵ Area of hole, A2 = × 802 = 5026.5 mm2
4
399
=–
STRENGTH OF MATERIALS
Y
83.72 mm
76.27 mm
C
D
ey
Load
point
100 mm
ex
50 mm
200 mm
X
X
60 mm
80 mm
A
B
160 mm
Y
Fig. 9.15
∴ Area of section,
Eccentric load,
A = A1 – A2 = 32000 – 5026.5
= 26973.5 mm2
P = 100 kN = 100 × 103 N
The given section is symmetrical about X-X axis. But it is unsymmetrical to Y-Y axis.
Let us first find the position of Y-Y axis. For this find the distance of the C.G. of the section
from the reference line AD. Let x is the distance of the C.G. of the section from the reference line AD.
Then
x=
A1 x1 + A2 x2
( A1 + A2 )
where A1 = Area of rectangle ABCD = 32000 mm2
x1 = Distance of C.G. of rectangle ABCD from reference line AD = 80 mm
(–ve sign due to cut out portion)
A2 = Area of hole = – 5026.5 mm2
x2 = Distance of C.G. of hole from line AD = 60 mm
∴
x=
32000 × 80 − 5026.5 × 60
(32000 − 5026.5)
2560000 − 301590
= 83.73 mm.
26973.5
Hence the axis Y-Y will lie at a distance of 83.73 mm from the line AD or at a distance of
160 – 83.73 = 76.27 mm from line BC as shown in Fig. 9.15.
The load is unsymmetrical to X-X axis as well as Y-Y axis.
Eccentricity of load about X-X axis,
e x = 50 mm
=
400
DIRECT AND BENDING STRESSES
Eccentricity of load about Y-Y axis,
e y = 83.73 – 60 = 23.73 mm
Moment of eccentric load about X-X axis,
Mx = P × ex
= 100 × 103 × 50 = 5 × 106 Nmm
Moment of eccentric load about Y-Y axis,
My = P × ey
= 100 × 103 × 23.73
= 2.373 × 106 Nmm
Now find the moment of inertia of the section about X-X axis and Y-Y axis.
...(i)
...(ii)
I xx1 = M.O.I. of rectangle ABCD about X-X axis
Let
= M.O.I. of rectangle about its C.G. + Area of rectangle
(Distance of C.G. of ABCD from X-X axis)2
160 × 200 3
+ 160 × 200 (0)
12
= 1.066 × 108 mm4
=
I xx2 = M.O.I. of the hole about X-X axis
π
× 804 = 2.01 × 106 mm4
64
The moment of inertia of the section about X-X is given by
∴
Ixx = I xx1 – I xx2
=
= 1.066 × 108 – 2.01 × 106
= 104.59 × 106 mm4
...(iii)
Similarly, Iyy = I yy1 − I yy2
...(iv)
where I yy1 = M.O.I. of ABCD about Y-Y axis
= M.O.I. of ABCD about its C.G. + A1 (Distance of C.G. of ABCD from Y-Y)2
200 × 160 3
+ 200 × 160 (83.73 – 80)2
12
= 6.826 × 107 + 4.45 × 105
= 687.05 × 105 mm4
=
and
I yy2 = M.O.I. of hole about Y-Y axis
= M.O.I. of hole about its C.G. + A2 (Distance of its C.G. from Y-Y)2
π
=
× 804 + 5026.5 (83.73 – 60)2
64
= 2.01 × 106 + 2.83 × 106
= 4.84 × 106
Hence substituting these values in equation (iv), we get
Iyy = 687.05 × 105 – 4.84 × 106
= 63.865 × 106 mm4
401
STRENGTH OF MATERIALS
The resultant stress at any point is obtained from equation (9.3).
∴ Resultant stress
=
P M y . x Mx . y
±
±
A
I yy
I xx
The values of x and y are taken to be positive on the same side of X-X and Y-Y as the load.
Here O is the origin. Hence at point D, x and y are positive. At point B, x and y are both negative.
At point C, x is negative whereas y is positive.
At point A, x is positive whereas y is negative.
(i) At point A, x = 83.73 mm and y = – 100 mm. Hence resultant stress at A,
σA =
=
P M y × 83.73 M x × (− 100)
+
+
A
I yy
I xx
100000 2.373 × 10 6 × 83.73 5 × 10 6 × 100
+
−
26973.5
63.865 × 10 6
104.56 × 10 6
= 3.707 + 3.111 – 4.781
= 2.037 N/mm2. Ans.
(ii)At point B, x = – 76.27 and y = – 100 mm. Hence resultant stress at B,
σB =
=
P M y × (− 76.27) M x × (− 100)
+
+
I yy
I xx
A
100000 2.373 × 10 6 × 76.27 5 × 10 6 × 100
−
−
26973.5
104.56 × 10 6
63.865 × 10 6
= 3.707 – 2.833 – 4.781
= – 3.907 N/mm2. Ans.
(iii)At point C, x = – 76.27 and y = 100 mm. Hence resultant stress at C,
P M y × (− 76.27) M x × 100
+
+
σC =
I xy
I xx
A
= 3.707 – 2.833 + 4.781
= 5.655 N/mm2. Ans.
(iv)At point D, x = 83.73 and y = 100 mm. Hence resultant stress at D,
P M y (83.73) M x × 100
+
+
σD =
A
I yy
I xx
= 3.707 + 3.111 + 4.781
= 11.599 N/mm2. Ans.
9.6. MIDDLE THIRD RULE FOR RECTANGULAR SECTIONS [i.e., KERNEL OF
SECTION]
The cement concrete columns are weak in tension. Hence the load must be applied on
these columns in such a way that there is no tensile stress anywhere in the section. But when an
eccentric load is acting on a column, it produces direct stress as well as bending stress. The
resultant stress at any point in the section is the algebraic sum of the direct stress and bending
stress.
402
DIRECT AND BENDING STRESSES
Consider a rectangular section of width ‘b’
and depth ‘d’ as shown in Fig. 9.16. Let this section
is subjected to a load which is eccentric to the axis
Y-Y.
Let
P = Eccentric load acting on the
column
e = Eccentricity of the load
A = Area of the section.
Then from equation (9.2), we have the minimum stress as
FG
H
Y
b
d/6
D
X
C
A
d
3
d
X
B
b/6
b/6
b/3
IJ
K
Y
6×e
P
Fig. 9.16
1−
...(i)
σmin =
A
b
If σmin is –ve, then stress will be tensile. But if σmin is zero (or positive) then there will be
no tensile stress along the width of the column.
Hence for no tensile stress along the width of the column,
σmin ≥ 0
6×e
P
6×e
1−
1−
or
≥0
or
≥0
b
A
b
6×e
b
or
≥e
or
1≥
6
b
b
or
e≤
...(9.4)
6
b
The above result shows that the eccentricity ‘e’ must be less than or equal to . Hence the
6
b
greatest eccentricity of the load is
from the axis Y-Y. Hence if the load is applied at any
6
b
distance less than from the axis, on any side of the axis Y-Y, the stresses are wholly compressive.
6
Hence the range within which the load can be applied so as not to produce any tensile stress, is
within the middle third of the base.
Similarly, if the load had been eccentric with respect to the axis X-X, the condition that
tensile stress will not occur is when the eccentricity of the load with respect to this axis X-X does
d
not exceed . Hence the range within which the load may be applied is within the middle third
6
of the depth.
If it is possible that the load is likely to be eccentric about both the axis X-X and Y-Y, the
condition that tensile stress will not occur is when the load is applied anywhere within the
b
d
rhombus ABCD whose diagonals are AC =
and BD =
as shown in Fig. 9.16. This figure
3
3
ABCD within which the load may be applied anywhere so as not to produce tensile stress in any
part of the entire rectangular section, is called the Core or Kernel of the section. Hence the
kernel of the section is the area within which the line of action of the eccentric load P must cut
FG
H
IJ
K
FG
H
IJ
K
403
STRENGTH OF MATERIALS
the cross-section if the stress is not to become tensile in any part of the entire rectangular
section.
Note. (i) If direct stress (σ0) is equal to bending stress (σb), then the tensile stress will be zero.
(ii) If the direct stress (σ0) is more than bending stress (σb), then the stress throughout the section
will be compressive.
(iii) If the direct stress (σ0) is less than bending stress (σb), then there will be tensile stress.
(iv) Hence for no tensile stress, σ0 ≥ σb.
9.7. MIDDLE QUARTER RULE FOR CIRCULAR SECTIONS [i.e., KERNEL OF
SECTION]
Consider a circular section of diameter ‘d’ as shown
in Fig. 9.17. Let this section is subjected to a load which is
eccentric to the axis Y-Y.
Let
P = Eccentric load
e = Eccentricity of the load
π
A = Area of the section = d 2
4
Now direct stress,
Y
O
X
X
d
4 d
8
4P
P
P
=
=
π
A
πd 2
d2
4
Moment,
M=P×e
Bending stress (σ b) is given by,
σ0 =
d
Y
Fig. 9.17
M σb
M×y
=
or σ b =
I
y
I
Maximum bending stress will be when
d
2
∴ Maximum bending stress is given by,
y=±
d
2 = ± 32 × P × e
π 4
πd 3
d
64
FG IJ
H K
P×e×
M
d
× ±
=±
σb =
2
I
Now minimum stress is given by,
σmin = σ0 – σb
=
For no tensile stress,
4P
or
πd 2
−
4P
πd
σmin ≥ 0
32 P × e
πd 3
2
−
≥ 0 or
8×e
1–
≥ 0 or
d
or
404
32 P × e
πd 3
4P
πd
1≥
2
FG 1 − 8d IJ ≥ 0
H dK
8×e
d
or e ≤
d
8
...(9.5)
DIRECT AND BENDING STRESSES
d
. It means
8
d
that the load can be eccentric, on any side of the centre of the circle, by an amount equal to .
8
Thus, if the line of action of the load is within a circle of diameter equal to one-fourth of the main
circle as shown in Fig. 9.17, then the stress will be compressive throughout the circular section.
The above result shows that the eccentricity ‘e’ must be less than or equal to
9.8. KERNEL OF HOLLOW CIRCULAR SECTION (OR VALUE OF ECCENTRICITY FOR
HOLLOW CIRCULAR SECTION)
Let
D0
Di
P
e
A
=
=
=
=
=
External diameter, and
Internal diameter
Eccentric load
Eccentricity of the load
Area of section
π
[D02 – Di2]
4
M = Moment due to eccentric load P = P × e
Z = Section modulus
I
=
ymax
=
π
[ D0 4 − Di 4 ]
64
=
D0
2
π
= 32 D [D04 – Di4]
0
Now direct stress (σ 0) is given by
P
σ0 =
A
The direct stress is compressive and uniform throughout the section.
Bending stress (σ b) is given by
FG∵
H
FG IJ
H K
∴
M σb
=
I
y
M
M
σb =
× y=
I
I
y
FG IJ
H K
ymax =
D0
2
IJ
K
...(i)
FG
H
IJ
K
M
I
∵
=Z
...(ii)
Z
y
The bending stress may be tensile or compressive. The resultant stress at any point is the
algebraic sum of direct stress and bending stress. There will be no tensile stress at any point if
the bending stress is less than or equal to direct stress at that point.
=
405
STRENGTH OF MATERIALS
or
Hence for no tensile stress,
Bending stress ≤ Direct stress
σ b ≤ σ0
Substituting the values of σ 0 and σ b from equations (i) and (ii), we get
M P
≤
Z
A
P×e P
≤
Z
A
1
e
≤
Z A
Z
e≤
A
or
or
(∵ M = P × e)
(cancelling P from both sides)
...(9.6)
π
[ D0 4 − Di 4 ]
32 D0
≤
π
[ D0 2 − Di 2 ]
4
≤
4 π ( D0 2 + Di 2 )( D0 2 − Di 2 )
32πD0
( D0 2 − Di 2 )
≤
1
(D02 + Di2)
8 D0
...(9.7)
The above result shows that the eccentricity ‘e’ must be less than or equal to
(D02 + Di2)/(8D0). It means that the load can be eccentric, on any side of the centre of the circle,
by an amount equal to (D02 + Di2)/(8D0). Thus, if the line of action of the load is within a circle
of diameter equal to (D02 + Di2)/(4D0), then the stress will be compressive throughout.
∴ Diameter of kernel =
D0 2 + Di 2
.
4 D0
9.9. KERNEL OF HOLLOW RECTANGULAR SECTION (OR VALUE OF ECCENTRICITY FOR HOLLOW RECTANGULAR SECTION)
Refer to Fig. 9.17 (a).
Let B = Outer width of rectangular section
D = Outer depth
b = Internal width
d = Internal depth
A = Area of section
=B×D–b×d
BD 3 bd 3
−
12
12
D
=
2
Y
B
b
D
X
X
d
Ixx =
ymax
406
Y
Fig. 9.17 (a)
DIRECT AND BENDING STRESSES
∴
Zxx =
I xx
ymax
F BD
GH 12
=
3
Similarly,
Zyy =
I yy
xmax
−
bd 3
12
I
JK
D/2
=
BD 3 − bd 3
6D
DB3 db3
−
3
3
12 = DB − db
= 12
6B
B/ 2
For no tensile stress at any section, the value of e is given by equation (9.6).
∴
e≤
Z
A
or ex ≤
ex
LM (BD − bd ) OP
6D
Q≤
≤ N
ey
F DB − db I
GH 6 B JK
=
3
or
and ey ≤
Z yy
A
3
( BD − bd)
3
and
Z xx
A
( BD 3 − bd 3 )
6 D( BD − bd)
...(9.8)
3
( BD − bd)
=
DB3 − db3
6 B( BD − bd)
...[9.8 (A)]
It means that the load can be eccentric on either side of the geometrical axis by an amount
equal to
DB3 − db3
( BD 3 − bd 3 )
and
along x-axis and y-axis respectively.
6 B( BD − bd)
6 D( BD − bd)
Problem 9.13. Draw neat sketches of kernel of the following cross-sections :
(i) Rectangular section 200 mm × 300 mm
(ii) Hollow circular cylinder with external dia = 300 mm and thickness = 50 mm
(iii) Square with 400 cm2 Area.
Sol. Given :
200 mm
(i) Rectangular Section
Y
B = 200 mm
D = 300 mm
Value of ‘e’ for no tensile stress along width is given by
equation (9.4) as
D
300 mm
B 200
e≤
≤
≤ 33.33 cm
6
6
Hence take OA = OC = 33.33 cm
The value of ‘e’ for no tensile stress along the depth is
given by,
D 300
≤
≤ 100 cm
3
3
Hence take OD = OB = 100 cm
e≤
X
A
O
C
X
B
Y
Fig. 9.18
407
STRENGTH OF MATERIALS
Now join A to B, B to C, C to D and D to A. The figure ABCD represents the kernel of the
given rectangular section as shown in Fig. 9.18.
(ii) Kernel for Hollow Circular Section
Given :
External dia.,
D0= 300 m
108.32
mm
Thickness,
t = 50 mm
∴ Internal dia.,
Di = D0 – 2 × t
= 300 – 2 × 50 = 200 mm
For hollow cylindrical section, for no tensile stress,
the value of e is given by equation (9.7) as
e≤
1
= (D02 + Di2)
8 D0
O
1
(3002 + 2002)
≤
8 × 300
1
(90000 + 40000)
2400
130000
≤ 54.16 mm
≤
2400
Taking O as centre and radius equal to 54.16 mm
(or dia. = 2 × 54.16 = 108.32 mm) draw a circle. This
circle is the kernel of the hollow circular section of external dia. = 300 mm and internal dia. = 200 mm, as shown
in Fig. 9.19.
≤
Di = 200 mm
Do = 300 mm
Fig. 9.19
20 cm
(iii) Kernel for Square Section
Given :
Area = 400 cm2
∴ One side of square = 400 = 20 cm
For no tensile stress, the value of ‘e’ for the square
section is given by equation (9.4) as
20
Side
[B or D] ≤
≤ 3.33 cm
6
6
Hence take OA = OC = OB = OD = 3.33 cm
Join ABCDA as shown in Fig. 9.20. Then ABCD is
the kernel of given square section.
e≤
D
20/3 cm
A
O
20 cm
C
B
20/3 cm
Fig. 9.20
Problem 9.14. Draw neat sketch of kernel of a
hollow rectangular section of outer cross-section 300 mm × 200 mm and inner cross-section 150
mm × 100 mm.
Sol. Given :
Outer rectangular section, B = 300 mm, D = 200 mm
Inner rectangular section, b = 150 mm and d = 100 mm.
For no tensile stress the value of ‘e’ along x-axis and along y-axis are given by equations
(9.8) and (9.8 A) respectively.
408
DIRECT AND BENDING STRESSES
Using equation (9.8), we get
ex ≤
( BD 3 − bd 3 )
6 D( BD − bd)
(300 × 200 3 − 150 × 100 3 )
100 3 × (2400 − 150)
=
6 × 200(300 × 200 − 150 × 100)
12 × 100 3 × (6 − 1.5)
2250
=
= 41.67 mm
12 × 4.5
Hence take OA = OC = 41.67 mm in Fig. 9.21
Using equation (9.8 A), we get
=
ey ≤
( DB3 − db3 )
6 B( BD − bd)
=
(200 × 300 3 − 100 × 150 3 )
6 × 300(300 × 200 − 150 × 100)
=
100 3 (5400 − 337.5)
6 × 3 × 10 3 (6 − 1.5)
Y
300 mm
150 mm
D
200 mm
62.5 mm
5062.5
=
= 62.5 mm
18 × 4.5
Hence take OD = OB = 62.5 mm in
Fig. 9.21.
Now join A to B, B to C, C to D and D
to A. The figure ABCD represents the kernel of
the given hollow rectangular section.
X
A
O
41.67
C
B
100
mm
X
Y
Fig. 9.21
HIGHLIGHTS
1. The axial load produces direct stress (σ 0).
2. Eccentric load produces direct stress as well as bending stress (σ b).
3. The maximum and minimum stress at any point in a section which is subjected to a load which
is eccentric to Y-Y axis is given by,
σmax = Direct stress + Bending stress
=
and
FG
H
IJ
K
...For a rectangular section
FG
H
IJ
K
...For a rectangular section
P
6×e
1+
A
b
σmin = Direct stress – Bending stress
=
6×e
P
1−
A
b
where P = Eccentric load
A = Area of section
e = Eccentricity
b = Width of the section.
4. If σ0 = σb, the tensile stress will be zero across the section.
5. If σ0 > σb, there will be no tensile stress across the section.
409
STRENGTH OF MATERIALS
6. If σ 0 < σ b, there will be tensile stress across the section.
7. The resultant stress at any point when a symmetrical column section is subjected to a load which
is eccentric to both the axis, is given by,
σ=
P M y × x Mx . y
±
±
A
I yy
I xx
where
P = Eccentric load
A = Area of the section
My = Moment of load about Y-Y axis
Iyy = Moment of inertia about Y-Y axis
Mx and Ixx = Moment and moment of inertia about X-X axis respectively.
The values of x and y are positive on the same side on which load is acting.
8. For unsymmetrical sections, subjected to eccentric load, first of all the C.G. of the section is
determined. Then moment of inertia of the section about an axis passing through the C.G. is
obtained. After that stresses are obtained.
9. For a rectangular section, there will be no tensile stress if the load is on either axis within the
middle third of the section.
10. For a circular section of diameter ‘d’, there will be no tensile stress if the load lies in a circle of
d
with centre O of the main circular section. This is known as ‘middle quarter rule for
4
circular sections’.
11. For no tensile stress, the value of eccentricity e is given by
diameter
e≤
d
8
...For circular section
1
≤ 8 D (D02 + Di2)
0
...For hollow circular section with D0 as
external dia. and Di as internal dia.
b
d
and
≤
6
6
≤
...For rectangular section
One side of square
6
ex ≤
( BD3 − bd3 )
6 D( BD − bd)
ey ≤
( DB3 − db3 )
.
6 B( BD − bd)
...For square section
...For hollow rectangular section with B and D as outer
width and depth and b and d as inner width and depth
EXERCISE
(A) Theoretical Questions
1. What do you mean by direct stress and bending stress ?
2. Prove that an eccentric load causes a direct stress as well as bending stress.
3. Find an expression for the maximum and minimum stresses when a rectangular column is
subjected to a load which is eccentric to Y-Y axis.
410
DIRECT AND BENDING STRESSES
4. Prove that for rectangular section subjected to eccentric load, the maximum and minimum stresses
are given by :
σmax =
FG
H
P
6e
1+
A
b
IJ
K
and σmin =
FG
H
P
6e
1−
A
b
IJ
K
where P = Eccentric load,
A = Area of the system,
b = Width of section,
and e = Eccentricity.
5. How will you find the maximum and minimum stresses at the base of a symmetrical column,
when it is subjected to load which is eccentric to both axis ?
6. Find and expression for the maximum and minimum stresses at the base of an unsymmetrical
column which is subjected to an eccentric load.
7. What do you mean by the following terms :
(i) Middle third rule for rectangular sections, and
(ii) Middle quarter rule for circular sections.
8. Prove that for no tension at the base of a short column :
(i) of rectangular section, the line of action of the load should be within the middle third, and
(ii) of circular section, the line of action of the load should be within the middle quarter.
9. Draw a neat sketches of Kernel of the following cross-sections :
(i) Rectangular 200 mm × 300 mm
(ii) Hollow circular cylinder with external dia. = 300 mm, thickness = 50 mm.
(iii) Square with 400 cm2 area.
(B) Numerical Problems
1. A rectangular column of width 120 mm and of thickness 100 mm carries a point load of 120 kN
at an eccentricity of 10 mm. Determine the maximum and minimum stresses at the base of the
column.
[Ans. 15 N/mm2, 5 N/mm2]
2. If in the above problem, the minimum stress at the base of the section is given as zero then find
the eccentricity of the point load of 120 kN acting on the rectangular section. Also calculate the
corresponding maximum stress on the section.
[Ans. 20 mm, 20 N/mm2]
3. If in Q. 1, the eccentricity is given as 30 mm, then find the maximum and minimum stress on the
section. Also plot these stress along the width of the section.
[Ans. – 5 N/mm2, 25 N/mm2]
4. In a tension specimen 13 mm in a diameter the line of pull is parallel to the axis of the specimen
but is displaced from it. Determine the distance of the line of pull from the axis, when the
maximum stress is 15% greater than the mean stress on a section normal to the axis.
[Ans. 0.25 mm]
5. A hollow rectangular column is having external and internal dimensions as 120 cm deep × 80 cm
wide and 90 cm deep × 50 cm wide respectively. A vertical load of 200 kN is transmitted in the
vertical plane bisecting 120 cm side and at an eccentricity of 10 cm from the geometric axis of the
section. Calculate the maximum and minimum stresses in the section.
[Ans. 0.61 N/mm2 and 0.17 N/mm2]
6. A short column of diameter 40 cm carries an eccentric load of 80 kN. Find the greatest eccentricity
which the load can have without producing tension on the cross-section.
[Ans. 5 cm]
7. A short column of external diameter 50 cm and internal diameter 30 cm carries an eccentric load
of 100 kN. Find the greatest eccentricity which the load can have without producing tension on
the cross-section.
[Ans. 8.5 cm]
411
STRENGTH OF MATERIALS
8. A hollow circular column of 25 cm external and 20 cm internal diameter respectively carries an
axial load of 200 kN. It also carries a load of 100 kN on a bracket whose line of action is 20 cm from
the axis of the column. Determine the maximum and minimum stress at the base section.
[Ans. 39 N/mm2 (comp.), 5.13 N/mm2 (tension)]
9. A column section 30 cm external diameter and 15 cm internal diameter supports an axial load of
2.6 MN and an eccentric load of PN at an eccentricity of 40 cm. If the compressive and tensile
stresses are not to exceed 140 N/mm2 and 60 N/mm2 respectively, find the magnitude of load P.
[Ans. 766.8 kN]
10. A rectangular pier of 1.5 m × 1.0 m is subjected to a compressive load of 450 kN as shown in
Fig. 9.18. Find the stresses on all four corners of the pier.
[Ans. σA = 0.45 N/mm2, σB = + 0.15 N/mm2, σC = 1.05 N/mm2, σD = 0.45 N/mm2]
Y
D
C
0.25 m
Load
point
0.25 m
1.0 m
X
X
B
A
1.5 m
Y
Fig. 9.22
412
10
CHAPTER
DAMS AND RETAINING
WALLS
10.1. INTRODUCTION..
A large quantity of water is required for irrigation and power generation throughout the
year. A dam is constructed to store the water. A retaining wall is constructed to retain the earth
in hilly areas. The water stored in a dam, exerts pressure force on the face of the dam in contact
with water. Similarly, the earth, retained by a retaining wall, exerts pressure on the retaining
wall. In this chapter, we shall study the different types of dams, stresses across the section of a
dam, stability of a dam and minimum bottom width required for a dam section.
10.2. TYPES OF DAMS..
There are many types of dams, but the following types of dams are more important :
1. Rectangular dams
2. Trapezoidal dams having
(a) Water face vertical, and
(b) Water face inclined.
A trapezoidal dam as compared to rectangular dam is economical and easier to construct.
Hence these days trapezoidal dams are mostly constructed.
10.3. RECTANGULAR DAMS..
Fig. 10.1 shows a rectangular dam having water on one of its sides.
Let h = Height of water
F = Force exerted by water on the side of the dam
W = Weight of dam per metre length of dam
H = Height of dam
b = Width of dam
w0 = Weight density of dam.
Consider one metre length of the dam.
The forces acting on the dam are
(i) The force F due to water in contact with the side of the dam.
The force F* is given by
F = wA h
*The derivation for F can be seen in any standard book of Fluid Mechanics.
413
STRENGTH OF MATERIALS
= w × (h × 1) ×
=
FG∵
H
h
2
A = h × 1 and h =
h
2
IJ
K
w × h2
.
2
b
Free surface of water
DAM
G
H
h
F
O
h/3
W
C
e
D
B
N
M
R
x
Fig. 10.1
The force F will be acting horizontally at a height of
h
above the base as shown in
3
Fig. 10.1.
(ii) The weight W of the dam. The weight of the dam is given by
W = Weight density of dam × Volume of dam
[∵ Length of dam = 1 m]
= w0 × [Area of dam] × 1
= w0 × b × H
The weight W will be acting downwards through the C.G. of the dam as shown in
Fig. 10.1.
These are only two forces acting on the dam. The resultant force may be determined by
the method of parallelogram of forces as shown in Fig. 10.1. The force F is produced to intersect
the line of action of the W at O. Take OC = F and OB = W to some scale. Complete the rectangle
OBDC. Then the diagonal OD will represent the resultant R to the same scale.
∴ Resultant
...(10.1)
F2 + W2
And the angle made by the resultant with vertical is given by
BD F
=
...(10.2)
tan θ =
OB W
10.3.1. The Horizontal Distance between the Line of Action of W and the Point
through which the Resultant Cuts the Base. In Fig. 10.1 the diagonal OD represents the
414
R=
DAMS AND RETAINING WALLS
16 m
Then
20 m
resultant of F and W. Let the diagonal OD is extend so that it cuts the base of the dam at point
M. Also extend the line OB so that it cuts the base at point N. Then the distance MN is the
horizontal distance between the line of action of W and the point through the resultant cuts
the base.
Let
x = Distance MN
The distance x is obtained from similar triangles OBD and ONM as given below
NM BD
=
i.e.,
ON OB
x
F
=
or
(∵ Distance ON = h/3, BD = OC = F and OB = W)
(h/3) W
F h
×
...(10.3)
∴
x=
W 3
The distance x can also be calculated by taking moments of all forces (here the forces F
and W) about the point M.
h
=W×x
∴
F×
3
F h
× .
∴
x=
W 3
Problem 10.1. A masonry dam of rectangular
10 m
section, 20 m high and 10 m wide, has water upto a
height of 16 m on its one side. Find :
(i) Pressure force due to water on one metre
length of the dam,
(ii) Position of centre of pressure, and
(iii) The point at which the resultant cuts the
base.
F
F
Take the weight density of masonry = 19.62
kN/m3, and of water = 9.81 kN/m3.
R
Sol.
Given :
Height of dam,
H = 20 m
W
Width of dam,
b = 10 m
x
Height of water,
h = 16 m
Fig. 10.2
Weight density of masonry,
w0 = 19.62 kN/m3 = 19620 N/m3
For water,
w = 9.81 kN/m3 = 9.81 × 1000 N/m3
(i) Pressure force due to water on one metre length of dam
Let
F = Pressure force due to water
F = wA h
h
2
(∵ w for water = 9.81 kN/m3 = 9.81 × 1000 N/m3)
16
= 1255680 N. Ans.
= 9.81 × 1000 × (16 × 1) ×
2
= 9.81 × 1000 × (h × 1) ×
415
STRENGTH OF MATERIALS
(ii) Position of centre of pressure
The point, at which the force F is acting, is known as centre of pressure. The force F is
16
h
i. e.,
= 5.33 m above the base.
acting horizontally at a height of
3
3
∴ Position of centre of pressure from base
= 5.33 m. Ans.
(iii) The point at which the resultant cuts the base
Let
x = Horizontal distance between the line of action of W and the point through
which the resultant cuts the base
W = Weight of dam per metre length of dam
= Weight density of masonry × (Area of dam) × 1
= w0 × b × H × 1 = 19620 × 10 × 20 × 1 = 3924000 N
Using equation (10.3),
F h
1255680 16
×
× =
= 1.706 m. Ans.
x=
3924000
3
W 3
Problem 10.2. A masonry dam of rectangular cross-section 10 m high and 5 m wide has
water upto the top on its one side. If the weight density of masonry is 21.582 kN/m3. Find :
(i) Pressure force due to water per metre length of the dam
(ii) Resultant force and the point at which it cuts the base of the dam.
Sol. Given :
5m
Height of dam,
H = 10 m
Width of dam,
b=5m
Height of water,
h = 10 m
Weight density of masonry
w0 = 21.582 kN/m3
= 21582 kN/m3.
F
F
(i) Pressure force due to water is given by
10
F = wA h = 9.81 × 1000 × (10 × 1) ×
10/3
2
= 490500 N. Ans.
N
M
(ii) Resultant force is given by equation (10.1).
W
R
IJ
K
10 m
FG
H
∴
R=
F2 + W2
...(i)
x
Fig. 10.3
where W = Weight of masonry dam
= Weight density of masonry × Area of dam × 1
= w0 × b × H × 1 = 21582 × (10 × 5) × 1 = 1079100 N.
Substituting the values of F and W in equation (i), we get
R=
490500 2 + 1079100 2
= 1185048 N = 1.185 MN. Ans.
The point at which the resultant cuts the base
Let x = Horizontal distance between the line of action of W and the point through which
the resultant cuts the base.
416
DAMS AND RETAINING WALLS
Using equation (10.3),
x* =
F h
490500
10
×
× =
= 1.51 m. Ans.
1079100
3
W 3
10.4. STRESSES ACROSS THE SECTION OF A RECTANGULAR DAM..
Fig. 10.4 shows a rectangular dam of height H and width b.
The dam is having water upto a depth of h. The forces acting on dam are
(i) The force F due to water at a height of
b
h
above the base of the dam,
3
(ii) The weight W of the dam at the C.G.
of the dam.
The resultant force R is cutting the base
H
G
of the dam at the point M as shown in Fig. 10.4.
h
Let x = The horizontal distance between
F
O
F
the line of action of W and the
h/3
point through which the resultW
ant (R) cuts the base (i.e., disR
B
A
M
N
tance MN in Fig. 10.4). This disb
x
tance is given by equation (10.3).
2
d
F h
×
=
Fig. 10.4
W 3
d = The distance between A and the
point M, where the resultant R cuts the base
= Distance AM = AN + NM
b F h
×
= +
(∵ Distance AN = Half the width of dam)
2 W 3
b
The resultant R meets the base of the dam at point
M. This resultant force R acting at M may be resolved into
vertical and horizontal components. The vertical component
will be equal to W whereas the horizontal component will be
equal to F as shown in Fig. 10.4 (a). The vertical component
W acting at point M on the base of the dam is an eccentric
load as it is not acting at the middle of the base. The point
F
N in Fig. 10.4 for a rectangular dam is the middle point of
the base.
R W
But an eccentric load produces direct stress and
W
bending stress as mentioned in chapter 9.
∴ Eccentricity of the vertical component W is equal
A
N F M B
b
x
to distance NM which is equal to x in this case.
2
Fig. 10.4 (a)
*x can also be obtained by taking moments of all forces (i.e., force F and W) about the point M.
∴ F×
10
F 10
10
490500
×
= W × x or x =
=
×
= 1.51 m.
W
3
3
3
1079100
417
STRENGTH OF MATERIALS
∴ Eccentricity,
e = Distance x (or Distance NM)
...(10.4)
b
...(10.5)
= AM – AN = d –
2
Due to the eccentricity, there will be a moment on the base of the dam. This moment will
cause some bending stresses at the base section of the dam.
Now the moment on the base section
= W × Eccentricity
Axis
b
= W.e
∴ Moment,
M = W.e
M σb
=
I
y
We know that
...(i)
1m
where M = Moment
I = Moment of inertia
1 × b3
Base section
(See Fig. 10.5)
of dam
12
3
Fig. 10.5
b
=
12
σ b = Bending stress at a distance y from the centre of gravity of the base section
y = Distance between the C.G. of the base section and extreme edge of the base (which is
b
equal to ± in this case).
2
Substituting the values in equation (i), we get
=
W. e
3
(b /12)
=
σb
(± b/2)
b 12
6W . e
× 3 =±
2 b
b2
The bending stress across base at point B (see Fig. 10.4)
6 W. e
.
=
b2
And the bending stress across base at point A
6 W. e
=–
.
b2
But the direct stress on the base section due to direct load is given by
Weight of dam
W
W
σ0 =
=
=
.
Area of base
b×1
b
∴ Total stress across the base at B
∴
σb = ± W . e
σmax = σ 0 + σb =
FG
H
W
6. e
W 6 W. e
1+
+
=
b
b
b
b2
and total stress across the base at A,
σmin = σ 0 + Bending stress at point A
W 6 W. e
−
=
b
b2
418
IJ
K
...(10.6)
DAMS AND RETAINING WALLS
IJ
K
FG
H
W
6. e
1−
...(10.7)
b
b
If the value of σmin is negative, this means that at the point A the stress is tensile.
Problem 10.3. For the Problem 10.1, find the maximum and minimum stress intensities
at the base of the dam.
Sol. The data given for Problem 10.1 is
H = 20 m, b = 10 m, h = 16 m, w0 = 19620 N/m3
The force calculated are
F = 1255680 N, W = 3924000 N
And distance,
x = 1.706 m.
From equation (10.4), we know
Eccentricity,
e = Distance x
= 1.706 m
(∵ x = 1.706 m)
Maximum stress at the base of the dam (i.e., σmax)
Using equation (10.6), we have
6. e
6 × 1.706
W
3924000
1+
1+
=
σmax =
b
10
b
10
= 392400 (1 + 1.0236)
= 794060.64 N/m2 = 0.794 N/mm2 (compressive). Ans.
Minimum stress at the base of the dam (i.e., σmin)
Using equation (10.7), we get
W
6 × 1.706
6. e
3924000
σ min =
=
1−
1−
b
10
10
b
= 392400 (1 – 1.0236) = – 9260.64 N/m2
= 0.00926 N/mm2 (Tensile). Ans.
Problem 10.4. For the Problem 10.2, find the maximum and minimum stress intensities
at the base of the dam.
Sol. The data given for Problem 10.2 is
H = h = 10 m, b = 5 m and w0 = 21582 N/m3
Calculated values are
F = 490500 N, W = 1079100 N, x = 1.51 m.
From equation (10.4), we know
Eccetricity,
e = x = 1.51 m.
Maximum stress at the base of the dam (i.e., σ max)
Using equation (10.7), we have
6. e
W
6 × 1.51
1079100
1+
1+
=
= 215820 (1 + 1.812)
σmax =
b
b
5
5
= 606885.84 N/m2 (compressive). Ans.
Minimum stress at the base of the dam (i.e., σ min)
Using equation (10.7), we have
6. e
W
6 × 1.51
1079100
1−
1−
=
σmin =
b
b
5
5
= 215820 (2 – 1.812) = – 175245.84 N/m2
= 175245 N/m2 (tensile). Ans.
=
FG
H
IJ
K
IJ
K
FG
H
FG
H
IJ
K
FG
H
IJ
K
FG
H
IJ
K
FG
H
IJ
K
FG
H
IJ
K
FG
H
IJ
K
419
STRENGTH OF MATERIALS
10.4.1. Trapezoidal Dam having Water Face Vertical. Fig. 10.6 shows a trapezoidal
dam having water face vertical. Consider one metre length of the dam.
a
C
D
G
H
h
F
F
O
h/3
W
M
A
N
x
B
R
d
b
Fig. 10.6
Let H = Height of dam
h = Height of water,
a = Top width of dam,
b = Bottom width of dam,
w0 = Weight density of dam masonry,
w = Weight density of water = ρ × g = 1000 × 9.81 N/m3
= 9.81 kN/m3 = 9810 N/m3
F = Force exerted by water
W = Weight of dam per metre length of dam.
Now the forces acting on the dam are
(i)
F = Force exerted by water
h
h2
=w×
.
2
2
The force F will be acting horizontally at a height of h/3 above the base.
(ii) W = Weight of dam per metre length of dam
= Weight density of dam × (Area of cross-section) × 1
a+b
1
= w0 ×
× H × 1 [∵ Area = (Sum of parallel sides) × Height]
2
2
(a + b)
= w0 ×
×H
2
The weight W will be acting downwards through the C.G. of the dam.
(i) The distance of the C.G. of the trapezoidal section (shown in Fig. 10.6) from the vertical
face AC is obtained by splitting the dam section into a rectangle and a triangle, taking the
moments of their areas about line AC and equating the same with the moment of the total area
of the trapezoidal section about the line AC.
= w × A × h = w × (h × 1) ×
FG
H
420
IJ
K
DAMS AND RETAINING WALLS
i.e.,
or
Area of rectangle × Distance of C.G. of rectangle from AC + Area of triangle
× Distance of C.G. of triangle from AC = Total area of trapezoidal × Distance AN
FG
H
IJ FG
K H
IJ
K
b−a
a+b
a (b − a) × H
a+
=
+
× H × AN
2
2
3
2
From the above equation distance AN can be calculated.
(ii) The distance AN can also be calculated by using the relation given below.
(a × H) ×
a 2 + ab + b2
...(10.8)
3(a + b)
Now let x* = Horizontal distance between the line of action of weight of dam and the
point where the resultant cuts the base
= Distance MN and it is given by equation (10.3)
F h
×
=
W 3
d = Distance between A and the point M where the resultant cuts the base (i.e.,
distance AM)
= AN + NM
...(10.9)
The distances AN and NM can be calculated and hence the distance ‘d’ will be known.
AN =
Now the eccentricity, e = d – half the base width of the dam
b
=d–
2
Then the total stress across the base of the dam at point B,
6. e
W
1+
σ max =
...(10.10)
b
b
and the total stress, acros the base at A,
W
6. e
1−
...(10.11)
σ min =
b
b
Problem 10.5. A trapezoidal masonry dam is of 18 m height. The dam is having water
upto a depth of 15 m on its vertical side. The top and bottom width of the dam are 4 m and 8 m
respectively. The weight density of the masonry is given as 19.62 kN/m3. Determine :
(i) The resultant force on the dam per metre length.
(ii) The point where the resultant cuts the base, and
(iii) The maximum and minimum stress intensities at the base.
Sol. Given :
Height of dam,
H = 18 m
Depth of water,
h = 15 m
Top width of dam,
a =4m
Bottom width of dam, b = 8 m
Weight density of masonry,
w0 = 19.62 kN/m3 = 19620 N/m3
FG
H
FG
H
IJ
K
IJ
K
* The distance x can also be calculated by taking moments of all forces about the point M.
∴ F×
h
=W×x
3
∴ x=
h
F
×
W 3
421
STRENGTH OF MATERIALS
(i) Resultant force on dam
Let us find first the force F and weight of the
4m
D
C
dam.
Force F = w × A × h
= 9810 × (h × 1) ×
15 m
15
= 1103625 N
2
h
And it is acting at a distance of i.e.,
3
= 9810 × 15 ×
18 m
h
2
F
O
F
5m
15
= 5.0 m above the base.
3
Now weight of dam is given by
W = Weight density of masonry
× Area of dam × 1
A
N
M
W
B
R
x
d
F a + b IJ × H × 1
= w × GH
2 K
F 4 + 8 IJ × 18 × 1 N
= 19620 × GH
2 K
8m
Fig. 10.7
0
= 19620 × 6 × 18 = 2118960 N.
The distance of the line of action of W from the line AC is obtained by splitting the dam into
rectangle and triangle, taking the moments of their areas about the line AC and equating to the
moment of the area of the trapezoidal about the line AC.
or
or
FG
H
IJ FG
K H
IJ
K
1
4+8
4 × 18
× 4+ ×4 =
× 18 × AN
3
2
2
144 + 36 [5.33] = 108 × AN
144 + 36 × 5.33
∴
AN =
= 3.11 m.
108
AN can also be calculated as given by equation (10.8)
4 × 18 × 2 +
a 2 + ab + b2
42 + 4 × 8 + 82
=
(∵ a = 4 m and b = 8 m)
3(4 + 8)
3(a + b)
16 + 32 + 64 112
=
=
= 3.11 m.
36
36
The resultant force R is given by
∴
AN =
F 2 + W 2 = 1103625 2 + 2118960 2
= 238925.5 N = 2.389 MN. Ans.
(ii) The point where the resultant cuts the base
Let x = The horizontal distance between the line of action of W and the point at which
the resultant cuts the base.
Using equation (10.3), we get
F h
1103625 15
×
× =
x=
= 2.604 m
2118960
3
W 3
R=
422
DAMS AND RETAINING WALLS
The distance x can also be calculated by taking moments of all forces about the point M.
∴
F×5=W×x
F
1103625
×5=
× 5 = 2.604 m
∴
x=
2118960
W
From Fig. 10.6, the distance AM = d.
∴
d = AN + NM
= 3.11 + x = 3.11 + 2.604 = 5.714 m
b
Now eccentricity,
e=d–
2
8
= 5.714 – = 5.714 – 4.0 = 1.714 m.
2
(iii) The maximum and minimum stress intensities
Let
σmax = Maximum stress, and
σmin = Minimum stress
Using equation (10.10), we get
LM
N
OP
Q
LM
N
OP
Q
6. e
W
2118960
6 × 1.714
1+
1+
=
b
b
8
8
= 264870 (1 + 1.2855) = 605360 N/m2. Ans.
Using equation (10.11), we get
σmax =
LM
N
OP
Q
LM
N
OP
Q
FG IJ
H K
4/3 m
6×e
W
2118960
6 × 1.714
1−
1−
=
b
b
8
8
= 264870 (1 – 1.2855) = – 75620 N/m2. Ans.
– ve sign shows that stress is tensile.
Problem 10.6. A masonry trapezoidal dam 4 m high, 1 m wide at its top and 3 m width
at its bottom retains water on its vertical face. Determine the maximum and minimum stresses
at the base (i) when the reservoir is full, and (ii) when the reservoir is empty. Take the weight
density of masonry as 19.62 kN/m3.
Sol. Given :
1m
C
Height of dam,
H = 4m
Top width of dam,
a = 1m
Bottom width of dam, b = 3 m
Depth of water,
h = 4m
Weight density of masonry,
4m
w0 = 19.62 kN/m3 = 19620 N/m3
Consider one metre length of dam.
F
F
(i) When the reservoir is full of water
The force exerted by water on the vertical face
of the dam per metre length is given by,
M
N
A
B
4
x
W
F = w × A × h = 9810 × (4 × 1) ×
R
2
d
(∵ w = 9810 N/m3 for water)
3m
= 78480 N
σmin =
Fig. 10.8
423
STRENGTH OF MATERIALS
The weight of dam per metre length is given by
W = Weight density of masonry dam × Area of trapezoidal × 1
FG a × b IJ × H
H 2 K
F 1 + 3 IJ × 4 = 156960 N.
= 19620 × GH
2 K
= w0 ×
Now let us find the position of the C.G. of the dam section. This is done by splitting the
trapezoidal into rectangle and triangle, taking the moments of their areas about the line AC and
equating to the moment of the area of the trapezoidal about the line AC.
∴
or
FG 4 × 1 × 1IJ + LM 4 × 2 × FG 1 + 1 × 2IJ OP = FG 1 + 3 IJ × 4 × AN
H
H 3 KQ H 2 K
2K N 2
2 + 4 × 1.67 = 8 × AN
2 + 6.68 8.68
=
= 1.08 m
8
8
AN can also be calculated from equation (10.8), as
∴
AN =
AN =
=
a 2 + ab + b2
3(a + b)
12 + 1 × 3 + 32 1 + 3 + 9 13
=
=
= 1.08 m.
3(1 + 3)
12
12
The horizontal distance x, between the line of action of W and the point at which the
resultant cuts the base, is obtained by using equation (10.3),
∴
x=
F h
×
W 3
78480
4
× = 0.67 m
156960 3
∴ Horizontal distance AM from Fig. 10.7 is given by
d = AN + x
= 1.08 + 0.67 = 1.75 m
=
∴ Eccentricity,
b
2
e=d–
3
= 1.75 – 1.50 = 0.25 m.
2
Now let σ max = Maximum stress at the base of the dam, and
σ min = Minimum stress.
Using equation (10.10), we get
= 1.75 –
σ max =
=
424
FG
H
W
6e
1+
b
b
FG
H
IJ
K
IJ
K
156960
6 × 0.25
= 78480 N/m2. Ans.
1+
3
3
DAMS AND RETAINING WALLS
Using equation (10.11), we get
W
6. e
1−
σmin =
b
b
156960
6 × 0.25
1−
=
= 26163 N/m2. Ans.
3
3
(ii) When the reservoir is empty
1m
When the reservoir is empty, the force F exerted by water
C
will be zero as there is no water retained by the dam. Hence only
the weight of the dam will be acting as shown in Fig. 10.9.
The weight of dam, W = 156960 N as before. The position of
the C.G. of the dam will also remain the same.
∴ Distance AN = 1.08 m as before.
Now the resultant force on the dam is equal to the weight 4 m
of the dam, as force F is zero. Hence the horizontal distance at the
base of dam between A and the point at which the resultant (i.e.,
force W in this case) cuts the base is equal to distance AN.
W
∴
d = AN = 1.08 m.
A
N
As W is not acting at the middle of the base, this load is an
3m
eccentric load.
d
b
Now eccentricity, e = d –
2
Fig. 10.9
3
= 1.08 – = 1.08 – 1.5 = – 0.42 m.
2
(Minus sign only indicates that stress at A will be more than at B).
Now using equation (10.10), we get
FG
H
FG
H
FG
H
IJ
K
IJ
K
IJ
K
W
6. e
1+
b
b
6 × 0.42
156960
1+
=
3
3
= 96265 N/m3. Ans.
Using equation (10.11), we get
W
6. e
1−
σmin =
b
b
156960
6 × 0.42
1−
=
3
3
= 8367.93 N/m2. Ans.
σmax =
FG
H
FG
H
FG
H
IJ
K
IJ
K
B
(Numerically e = 0.42)
IJ
K
Problem 10.7. A masonry dam, trapezoidal in cross-section, 4 m high, 1 m wide at its
top and 3 m wide at its bottom, retains water on its vertical face to a maximum height of 3.5 m
from its base. Determine the maximum and minimum stresses at the base (i) When the reservoir
is empty, and (ii) When the reservoir is full. Take the unit weight of masonry as 19.62 kN/m3.
Sol. Given :
H = 4 m, a = 1 m, b = 3 m
h = 3.5 m, w0 = 19.62 kN/m3 = 19620 N/m3
425
STRENGTH OF MATERIALS
a = 1m
h = 3.5 m
H = 4m
F
F
W
h/3
A
R
M
N
B
x
d
b = 3m
Fig. 10.9 (a)
Weight density of water, w = 9810 N/m3
Consider 1 m length of the dam.
The force F exerted by water is given by
F = w × A × h = 9810 × (h × 1) ×
= 9810 × (3.5 × 1) ×
h
2
3.5
= 9.81 × 6125 N = 60086 N
2
h 3.5
=
= 1.167 m above the ground.
3
3
The weight W of the dam per metre length is given by
This force acts at a height of
FG a + bIJ × H × 1
H 2 K
F 1 + 3 IJ × 4 × 1 = 156960 N
= 19620 × GH
2 K
W = w0 ×
The distance of C.G. of the dam section from point A [i.e., distance AN of Fig. 10.9 (a)] is
given by equation (10.8) as
a 2 + ab + b2 12 + 1 × 3 + 32
=
3(a + b)
3(1 + 3)
1+ 3 + 9
13
=
=
= 1.08 m
12
12
The horizontal distance ‘x’, between the line of action of W and the point at which the
resultant cuts the base, is obtained by using equation (10.3) as
F h
60086
3.5
×
× =
= 0.446 m
x=
156960
3
W 3
(i) Maximum and Minimum stresses at the base when the reservoir is empty:
When the reservoir is empty (i.e., there is no water), the only force acting on the dam will
be its own weight i.e., 156960 N. The position of C.G. of the dam section will remain same. Hence
AN =
426
DAMS AND RETAINING WALLS
distance AN = 1.08 m. Also the resultant force here will be W only. The distance of the point
where resultant cuts the base from A will be
d = AN = 1.08 m
Hence eccentricity ‘e’ is given by
e=d–
b
2
3
= – 0.42 m
2
(Minus sign only indicates that stress at A will be more than that of at B)
The stresses are given by equation (10.10) as
= 1.08 –
∴ Maximum stress
and
Minimum stress
FG
H
IJ
K
LM
N
OP
Q
=
6×e
156960
6 × (− 0.42)
W
1±
1±
=
b
3
3
b
=
6 × 0.42
156960
1±
= 52320 (1 ± 0.84)
3
3
FG
H
IJ
K
= σmax
= 52320 (1 + 0.84) = 52320 (1.84) N/m2
= 96268.8 N/m2. Ans.
= σmin
= 52320 (1 – 0.84) = 52320 (0.16)
= 8371.2 N/m2. Ans.
(ii) Maximum and minimum stresses when reservoir is full:
In this case, two forces i.e., F and W are acting on the dam. The resultant (R) of these two
forces cuts the base at the point M. The distance AM is given by,
d = AM = AN + x
= 1.08 + 0.446 = 1.526 m
Now eccentricity is obtained as
b
3
= 1.526 – = 1.526 – 1.5 = 0.026 m
2
2
∴ Maximum stress is given by,
e=d–
σ max =
FG
H
W
6×e
1+
b
b
FG
H
IJ
K
IJ
K
156960
6 × 0.026
1+
= 52320 (1 + 0.052)
3
3
= 55040.64 N/m2. Ans.
=
and
σ min =
FG
H
6×e
W
1−
b
b
FG
H
IJ
K
IJ
K
156960
6 × 0.026
1−
= 52320 (1 – 0.052)
3
3
= 52320 × 0.948 = 49599.36 N/m2. Ans.
=
427
STRENGTH OF MATERIALS
10.5. TRAPEZOIDAL DAM HAVING WATER FACE INCLINED..
Fig. 10.10 shows a trapezoidal dam section having its water face inclined.
a
C
D
E
F
Fy
h
H
G
F
Fx
O
F
A
N
M
B
x
R
W
d
b
Fig. 10.10
Let H = Height of dam,
h = Height of water,
a = Top width of dam,
b = Bottom width of dam,
w0 = Weight density of dam masonry,
w = Weight density of water = 9810 N/m3
θ = Inclination of face AC with vertical,
F = Force exerted by water on face AC,
Fx = Component of F in x-direction = F cos θ,
Fy = Component of F in vertically downward direction = F sin θ,
W = Weight of dam per metre length of dam,
= w0 ×
FG a + b IJ × H
H 2 K
L = Length of sloping side AE which is subjected to water pressure.
Consider one metre length of the dam.
Now in triangle AEF,
AF h
=
cos θ =
AE L
h
∴
L=
cos θ
428
...(i)
DAMS AND RETAINING WALLS
The force acting on the dam are
(i) The force exerted by water on face AE is given by,
F = w.A. h
A = Area of face AE
= AE × 1
(∵ Length of dam perpendicular to plane of paper = 1 m)
where
FG∵
H
h
cos θ
h
h =
2
=
∴
F=w×
AE = L =
h
cos θ
IJ
K
h w × h2
h
× =
2 2 cos θ
cos θ
h
The force F acts perpendicular to the face AE as shown in Fig. 10.10 at a height above
3
the base.
Now,
Fx = F × cos θ
=
w × h2
× cos θ
2 cos θ
w × h2
2
= Force exerted by water on vertical face AF
Fy = F sin θ
F∵
GH
F=
w × h2
2 cos θ
I
JK
F∵
GH
F=
w × h2
2 cos θ
I
JK
EF
AF
IJ
K
=
and
w × h2
=
× sin θ
2 cos θ
=
w × h2
× tan θ
2
=
EF
w × h2
×
AF
2
=
EF
w × h2
×
h
2
=w×
FG
H
In Δ AEF , tan θ =
(∵ AF = h)
h × EF
2
= w × Area of triangle AEF
FG
H
Area of Δ AEF =
EF × h
2
IJ
K
= w × Area of triangle AEF × 1
= Weight of water in the wedge AEF.
Hence the force F, acting on inclined face AE is equivalent to force Fx acting on the
vertical face AF and force Fy which is equal to the weight of water in the wedge AEF.
429
STRENGTH OF MATERIALS
The force Fx acts at a height
h
above the base whereas the force Fy acts through the C.G.
3
of the triangle AEF.
(ii) Weight of dam per metre length of the dam and it is given by
a+b
× H × w0.
W=
2
The weight W will be acting through the C.G. of the trapezoidal section of the dam. The
distance of the C.G. of the trapezoidal section shown in Fig. 10.10 from the point A is obtained by
splitting the dam section into triangles and rectangle, taking the moments of their areas about
the point A and equating the same with the moment of the total area of the trapezoidal section
about the point A. By doing so the distance AN will be known.
(iii) The force R, which is the resultant of the forces F and W, cuts the base of the dam at
point M. The distance AM can be calculated by taking moments of all forces (i.e., forces Fx, Fy
and W) about the point M. But the distance AM = d.
b
Now the eccentricity, e = d – .
2
Then the total stress across the base of the dam at point B,
6. e
V
1+
σmax =
...(10.12)
b
b
and the total stress across the base of the dam at point A,
V
6. e
1−
...(10.13)
σmin =
b
b
where V = Sum of the vertical forces acting on the dam
= Fy + W.
Problem 10.8. A masonry dam of trapezoidal section is 10 m high. It has top width
of 1 m and bottom width 7 m. The face exposed to water has a slope of 1 horizontal to 10
vertical. Calculate the maximum and minimum stresses on the base, when the water level
coincides with the top of the dam. Take weight density of masonry as 19.62 kN/m3.
Sol. Given :
Height of dam,
H = 10 m
Top width of dam,
a=1m
Bottom width of dam, b = 7 m
Slope of face exposed to water = 1 hor. to 10 vertical
∴ Length of EC in Fig. 10.11 = 1 m
Depth of water,
h = 10 m
Weight density of masonry, w0 = 19.62 kN/m3 = 19620 N/m3
Consider one metre length of the dam.
Let the weight of dam (W) cut the base at N whereas the resultant R cuts the base at M.
The force F due to water acting on the face AC is resolved into two components Fx and Fy
as shown in Fig. 10.11.
But
Fx = Force due to water on vertical face AE
FG
H
IJ
K
FG
H
FG
H
IJ
K
IJ
K
=w×A× h
430
DAMS AND RETAINING WALLS
1m
E
10 m
Fy
C
G
F
Fx
O
F
10/3 m
N
A
1m
M
B
R
W
d
7m
Fig. 10.11
= 9810 × (10 × 1) ×
10
2
(∵ Area, A = AE × 1)
= 490500 N
10
The force Fx will act at a height of
m above the base of the dam.
3
Fy = Weight of water in wedge AEC
= w × Area of AEC × 1
(∵ Length of dam = 1 m)
10 × 1
× 1 = 49050 N.
= 9810 ×
2
The force Fy will act downward through the C.G. of the triangle AEC i.e., at a distance
1
3
× 1=
1
3
m from AE.
Weight of dam,
W = w0 ×
FG a + b IJ × H = 19620 × FG 1 + 7 IJ × 10 = 784800 N.
H 2 K
H 2 K
The weight W will be acting through the C.G. of the dam.
The position of C.G. of the dam (i.e., distance AN) is obtained by splitting the trapezoidal
into triangles and rectangle, taking the moments of their areas about A and equating to the
moment of area of the trapezoidal about the point A.
5
10 × 5
10 × 1 2
(a + b)
+ (10 × 1 × 1.5) +
× 2+
×
=
× H × AN
∴
2
3
2
2
3
1+ 7
or
3.33 + 15 + 91.67 =
× 10 × AN = 40 × AN
2
110
∴
AN =
= 2.75 m
40
The resultant force R cuts the base at M. To find the distance of M from A (i.e., distance
AM), take the moments of all forces about the point M.
FG
H
IJ
K
FG
H
IJ
K
FG
H
IJ
K
431
STRENGTH OF MATERIALS
∴
or
or
10
– Fy × (AM – 0.33) – W × NM = 0
3
10
490500 ×
– 49050 × (AM – 0.33) – 784800 × (AM – AN) = 0
3
(∵ NM = AM – AN)
4905000
– 490500 AM + 16350 – 784800 AM + 784800 × 2.75 = 0
3
(∵ AN = 2.75)
Fx ×
490500
+ 16350 + 784800 × 2.75 = 784800 AM + 49050 AM
3
or
3809550 = 833850 AM
3809550
= 4.568
∴
AM =
833850
or
d = 4.568
(∵ AM = d)
b
Now the eccentricity,
e=d–
2
7
= 4.568 – = 4.568 – 3.5 = 1.068 m.
2
Maximum and Minimum stresses on the base
Let
σ max = Maximum stress on the base,
σ min = Minimum stress on the base.
Using equation (10.12), we get
V
6.e
1+
σ max =
b
b
where V = Total vertical forces on the dam
= W + Fy = 784800 + 49050 = 833850 N
6 + 1.068
833850
1+
∴
σ max =
7
7
= 228167 N/m2. Ans.
Using equation (10.13), we get
V
6.e
1−
σmin =
b
b
6 × 1.068
833850
1−
=
6
7
= 10077.8 N/m2. Ans.
Problem 10.9. A masonry dam of trapezoidal section is 10 m high. It has top width
of 1 m and bottom width 6 m. The face exposed to water has slope of 1 horizontal to 10 vertical.
Calculate the maximum and minimum stresses on the base when the water level coincides with the top of the dam. Take weight density of masonry as 22.563 kN/m3.
Sol. Given :
Height of dam,
H = 10 m
Height of water,
h = 10 m
Top width of dam,
a=1m
Bottom width of dam,
b=6m
or
IJ
K
FG
H
IJ
K
FG
H
FG
H
432
FG
H
IJ
K
IJ
K
DAMS AND RETAINING WALLS
Slope of the face AC which is exposed to
water = 1 horizontal to 10 vertical.
∴
EC = 1 m
(∵ AE = 10 m)
Weight density of masonry,
w0 = 22.563 kN/m3
= 22563 N/m3
Consider one metre length of dam.
Now the force F due to water acting on the face
AC is resolved into two components Fx and Fy as shown
in Fig. 10.12.
Force,
Fx = Force due to water acting on
vertical face AE
E
1m
C
Fy
G
10 m
F
Fx
O
F
10/3 m
W
=w×A× h
= 9810 × (10 × 1) ×
and
A
10
2
FG∵
H
h=
10
2
R
N
M
B
d
IJ
K
4m
1m 1m
6m
Fig. 10.12
= 490500 N
Fy = Weight of water in the wedge AEC
= w × Area of triangle AEC × 1
EC × AE
=w×
×1
2
1 × 10
= 9810 ×
× 1 = 49050 N
2
Weight of dam,
Force,
W = w0 ×
FG a + b IJ × H = 22563 × FG 1 + 6 IJ × 10
H 2 K
H 2 K
= 789705 N.
The position of the C.G. of the dam (i.e., distance AN ) is obtained by splitting the trapezoidal
into triangles and rectangle, taking the moments of their areas about A, and equating to the
moment of the area of the trapezoidal about point A.
1
10 × 4
1
10 × 1
2
a+b
× + 10 × 1 × 1 +
+
∴
× 1+ 1+ × 4 =
× H × AN
2
2
3
2
2
3
10 (1 + 6)
=
× 10 × AN
or
3.33 + 15 + 20 ×
3
2
85 = 35 × AN
85 17
∴
AN =
=
= 2.43 m.
35
7
Now let the resultant R of forces F and W cut the base at M.
Taking the moments of all forces (i.e., force Fx, Fy and W ) about the point M, we get
1
10
= W × NM + Fy × AM − × 1
Fx ×
3
3
1
10
490500 ×
= 789705 × (AM – AN) + 49050 AM −
3
3
FG
H
IJ
K
FG
H
IJ
K
FG
H
FG
H
IJ FG
K H
IJ
K
FG
H
IJ
K
IJ
K
433
STRENGTH OF MATERIALS
49050
4905000
= 789705 × AM – 789705 × AN + 49050 × AM –
3
3
17
17 49050
∵ AN =
−
= AM × (789705 + 49050) – 789705 ×
7
7
3
49050
= AM × 838755 – 1917855 –
3
4905000
49050
∴
AM × 838755 =
+ 1917855 +
= 3569205
3
3
3569205
∴
AM =
= 4.255 m
838755
b
∴ Eccentricity,
e = AM –
(∵ AM = d)
2
6
= 4.255 – = 4.255 – 3.0 = 1.255 m.
2
Maximum stress on the base
Using equation (10.12), we get
FG
H
IJ
K
FG
H
V
6.e
1+
b
b
where V = Total vertical forces on the dam
= W + Fy = 789705 + 49050 = 838755 N
σmax =
IJ
K
FG
H
IJ
K
6 + 1.255
838755
1+
= 315232 N/m2. Ans.
6
6
Using equation (10.13), we get
∴
σmax =
FG
H
IJ
K
V
6.e
1−
b
b
6 + 1.255
838755
=
= 35647 N/m2. Ans.
1−
6
6
σmin =
FG
H
IJ
K
10.6. STABILITY OF A DAM..
A dam should be stable under all conditions. But the dam may fail :
1. by sliding on the soil on which it rests,
2. by overturning,
3. due to tensile stresses developed, and
4. due to excessive compressive stresses.
10.6.1. Condition to Prevent the Sliding of the Dam. Fig. 10.13 shows a dam of
trapezoidal section of height H and having water upto a depth of h. The forces acting on the dam
are :
h
above the base.
3
(ii) Weight of the dam W acting vertically downwards through the C.G. of the dam.
The resultant R of the forces F and W is passing through the point M. The dam will be in
equilibrium if a force R* equal to R is applied at the point M in the opposite direction of R. Here
R* is the reaction of the dam. The reaction R* can be resolved into two components. The vertical
(i) Force due to water pressure F acting horizontally at a height of
434
DAMS AND RETAINING WALLS
component of R* will be equal to W whereas the horizontal component will be equal to frictional
force at the base of the dam.
H
G
h
F
O
F
h/3
R
A
W
B
M
N
R
x
d*
Fractional
force = mW
Fig. 10.13
Let μ = Co-efficient of friction between the base of the dam and the soil.
Then maximum force of friction is given by,
Fmax = μ × W
...(10.14)
If the force of friction i.e., Fmax is more than the force due to water pressure (i.e., F), the
dam will be safe against sliding.
10.6.2. Condition to Prevent the Overturning of the Dam. If the resultant R of the
weight W of the dam and the horizontal F due to water pressure, strikes the base within its
width i.e., the point M lies within the base AB of Fig. 10.13, there will be no overturning of the
dam. This is proved as :
For the dam shown in Fig. 10.13, taking moments about M.
Moment due to horizontal force F about point M
h
=F×
...(i)
3
Moment due to weigth W about point M
=W×x
...(ii)
The moment, due to horizontal force F, tends to overturn the dam about the point B ;
whereas the moment due to weight W tends to restore the dam. If the moment due to weight W
is more than the moment due to force F, there will be no overturning of the dam. For the
equilibrium of the dam, the two moments should be equal.
h
∴
F×
=W×x
...(iii)
3
Since overturning can take place about point B, hence restoring moment about the
point B
= W × NB
435
STRENGTH OF MATERIALS
But overturning moment due to force F about point B
h
h
∵ from equation (iii) F × = W × x
=W×x
=F×
3
3
There will be no overturning about point B, if restoring moment about B is more than the
overturning moment about B i.e.,
if
W × NB > W × x
or
NB > x
> NM
(∵ x = NM)
This means that there will be no overturning of the dam if point M lies between N and B
or between A and B.
10.6.3. Condition to Avoid Tension in the Masonry of the Dam at its Base. The
masonry of the dam is weak in tension and hence the tension in the masonry of the dam should
be avoided. The maximum and minimum stresses across the base of the dam are given by
equations (10.10) and (10.11). The maximum stress is always compressive but the minimum
FG
H
IJ
K
FG
H
stress given by equation (10.11) will be tensile if the term 1 −
case, there will be no tensile stress at the base of dam
6.e
if
1–
≥0
b
or
b–6.e≥0
or
6.e≤b
or
or
IJ
K
6.e
is negative. In the limiting
b
b≥6.e
b
e≤
6
...(10.15)
where e = Eccentricity and b = Base width of dam.
b
on the either side of
6
the middle point of the base section. Hence the resultant must lie within the middle third of the
base width, in order to avoid tension. Refer to Fig. 10.13.
If d* = Maximum distance between A and the point through which resultant force R
meets the base.
b
...(i)
Then
e = d* –
2
But to avoid tension at the base of the dam, maximum value of eccentricity,
b
e≤
...(ii)
6
From equations (i) and (ii), we have
b b
d* – ≤
2 6
b b b + 3b 4b 2
≤
≤ b
∴
d* ≤ + ≤
...(10.16)
6 2
b
5 3
Hence if the maximum distance between A and the point through which resultant force R
meets the base (i.e., distance d*) is equal to or less than two third of the base width, there will be
no tension at the base of dam.
10.6.4. Condition to Avoid the Excessive Compressive Stresses at the Base of
the Dam. The maximum and minimum stresses across the base of the dam are given by
This means that the eccentricity of the resultant can be equal to
436
DAMS AND RETAINING WALLS
equations (10.10) and (10.11). The condition to avoid the excessive compressive stresses in the
masonry of the dam is that the pmax i.e., maximum stress in the masonry should be less than the
permissible stress in the masonry.
Problem 10.10. A trapezoidal masonry dam having 4 m top width, 8 m bottom width
and 12 m high, is retaining water upto a height of 10 m as shown in Fig. 10.14. The density of
masonry is 2000 kg/m3 and coefficient of friction between the dam and soil is 0.55. The allowable compressive stress is 343350 N/m2. Check the stability of dam.
Sol. Given :
Top width of dam,
a=4m
Bottom width of dam,
b=8m
Height of dam,
H = 12 m
Depth of water,
h = 10 m
Density of masonry,
ρ0 = 2000 kg/m3
∴ Weight density of masonry,
w0 = 2000 × 9.81 N/m3
Co-efficient of friction,
μ = 0.55
4m
G
12 m
10 m
F
F
O
10/3 m
M
N
A
W
B
R
x
d
8m
Fig. 10.14
Allowable compressive stress
= 343350 N/m2
Consider on metre length of dam.
The horizontal force F exerted by water on the vertical side of dam is given by
F=w×A× h
= 1000 × 9.81 × (10 × 1) ×
10
2
= 490500 N
437
STRENGTH OF MATERIALS
10
m above the base.
3
Weight of the dam per metre length is given by,
w = Weight density of masonry × Area of trapezoidal section × 1
The force F will be acting at a height of
= w0 ×
FG a + bIJ × H × 1 = 2000 × 9.81 × FG 4 + 8 IJ × 12 × 1
H 2 K
H 2 K
= 1412640 N.
The weight W will be acting at the C.G. of the dam. The C.G. of the dam is obtained by
splitting the trapezoidal section into rectangle and triangle, taking the moments of their areas
about the point A and equating to the moment of the area of the trapezoidal about A.
∴ 4 × 12 × 2 +
IJ
K
IJ FG
K H
16
= 72 × AN
3
96 + 128 = 72 × AN
or
96 + 24 ×
or
96 + 128
224
=
= 3.11 m.
72
72
Taking the moments of the forces acting on the dam about the point M.
10
=W×x
F×
3
F 10
490500
10
x=
×
=
= 1.157 m
×
1412640
3
3
W
∴ Distance
AM = AN + x
= 3.11 + 1.157 = 4.267 m.
(i) Check for the tension in the masonry of the dam
Now from equation (10.15), we have
∴
or
FG
H
4+8
4 × 12
4
× 4+
=
× 12 × AN
2
2
3
AN =
2
2
× b ≤ × 8.0
3
3
≤ 5.33 m.
d* ≤
As the distance AM is less than d* or
2
× b (i.e., 5.33 m), the dam is safe against the
3
tension in its masonry at the base. Ans.
(ii) Check for overturning
The resultant is passing through the base AB of the dam and hence there will be no
overturning.
(iii) Check for sliding of the dam
From equation (10.14), the maximum force of friction is obtained as,
Fmax = μ × W
= 0.55 × 1412640 = 776952 N.
Since force of friction is more than the horizontal force due to water (i.e., F = 490500), the
dam is safe against sliding.
438
DAMS AND RETAINING WALLS
(iv) Check for excessive compressive stress at the base of the dam
From equation (10.10), the maximum stress at the base of the dam is given by
σmax =
FG
H
6e
W
1+
b
b
b
b
= AM –
2
2
8
= 4.267 – = 0.267 m
2
W = 1412640 N
IJ
K
where e = d –
(∵ d = AM)
IJ
K
FG
H
6 × 0.267
1412640
1+
= 211940 N/m2.
8
8
Since the maximum stress is less than the allowable stress, hence the masonry of the dam
is safe against excessive compressive stress. Ans.
Problem 10.11. A trapezoidal masonry dam having top width 1 m and height 8 m, is
retaining water upto a height of 7.5 m. The water face of the dam is vertical. The density of
masonry is 2240 kg/m3 and co-efficient of friction between the dam and soil is 0.6. Find the
minimum bottom width of the dam required.
Sol. Given :
1m
Top width,
a = 1.0 m
Height of dam,
H = 8.0 m
Depth of water,
h = 7.5 m
Density of masonry, ρ0 = 2240 kg/m3
∴ Weight density of masonry,
G
8m
7.5 m
w0 = ρ0 × g = 2240 × 9.81 N/m3
F
F
O
Co-efficient of friction, μ = 0.60
Let b = Width of dam at the base. Con2.5 m
sider one metre run of the dam. Horizontal force
A
N
M
B
F exerted by water is given by,
∴
σmax =
R
W
F=w×A× h
= 1000 × 9.81 × (7.5 × 1) ×
7.5
2
(∵ w for water = ρ × g
= 9810 N/m3)
= 275906.25 N.
The weight of dam per metre run is given by,
x
d
b
Fig. 10.15
FG a + bIJ × H × 1
H 2 K
F 1 + bIJ × 8 × 1 = 87897.6 (b + 1) N.
= 2240 × 9.81 GH
2 K
W = w0
The weight W will be acting through the C.G. of the dam. The distance of the C.G. of the
dam from A is given by equation (10.8).
439
STRENGTH OF MATERIALS
∴
AN =
a 2 + ab + b2
3(a + b)
12 + 1 × b + b2 1 + b + b2
=
.
3(1 + b)
3(1 + b)
The horizontal distance x, between the line of action of W and the point at which the
resultant force R cuts the base, is obtained by using equation (10.3).
F h
×
∴
x=
W 3
275906.25
7.5
7.847
×
=
=
.
87897.6 (b + 1)
3
(b + 1)
∴ Distance
d = AM = AN + NM = AN + x
=
=
1 + b + b2 7.847
+
3(1 + b)
(b + 1)
(i) There will be no tension in the dam at the base if d ≤
Hence for the limiting case d =
or
or
or
...(i)
2
b
3
2
b
3
2
1 + b + b2 7.847
+
= b
[Substituting the value of d from equation (i)]
3
3(1 + b)
b+1
1 + b + b2 + 3 × 7.487 = 2b(b + 1)
1 + b + b2 + 22.461 = 2b2 + 2b
b2 + b – 23.461 = 0.
The above equation is a quadratic equation. Its solution is
− 1 ± 12 + 4 × 1 × 23.461
− 1 ± 9.7387
=
2
2
− 1 + 9.7387
=
(Neglecting negative value)
2
= 4.37 m
...(ii)
(ii) There will be no sliding of the dam if
μW > F
or
0.6 × 87897.6 (b + 1) > 275906.25
275906.25
or
(b + 1) >
> 5.23
0.6 × 87897.6
or
b > 4.23
...(iii)
Hence the minimum bottom width of the dam, so that there is no tension at the base of the
dam and also there is no sliding of the dam, should be greater of the two values given by equations
(ii) and (iii).
∴ Minimum bottom width = 4.37 m. Ans.
Problem 10.12. A masonry gravity dam is vertical at the water face and has a height of
8.5 m above its base. It is 1.2 m wide at the top. It retains water upto a height of 8 m above the
base. The density of masonry is 2300 kg/m3. Determine the minimum bottom width required
b=
440
DAMS AND RETAINING WALLS
to satisfy “no tension” condition in the section and also to ensure that there is no sliding at the
base. The co-efficient of friction between the dam and foundation is 0.5.
1.2 m
8.5 m
8m
F
O
h
3
F
R
W
A
N
B
M
x
d
b
Fig. 10.15 (a)
Sol. Given :
Height of dam,
H = 8.5 m
Width at top,
a = 1.2 m
Depth of water,
h = 8m
Density of masonry, ρ0 = 2300 kg/m3
∴ Weight density of masonry,
w0 = ρ0 × g = 2300 × 9.81 N/m3
Weight density of water,
w = 1000 × 9.81 N/m3 = 9810 N/m3
Co-efficient of friction, μ = 0.5
Let b = Bottom width at the base
Consider 1 m length of the dam
The force F exerted by water is given by
F = w × A × h = 9810 × (h × 1) ×
= 9810 × (8 × 1) ×
h
2
8
2
= 313920 N
The weight of dam is given by
...(i)
FG a + b IJ × H × 1
H 2 K
F 1.2 + bIJ × 8.5 × 1
= (2300 × 9.81) × GH
2 K
W = w0 ×
= 95892.75 (1.2 + b)
...(ii)
441
STRENGTH OF MATERIALS
The horizontal distance x, between the line of action of W and the point M at which the
resultant force R cuts the base, is given by equation (10.3) as
F h
×
x =
W 3
313920
8
26.19
=
× =
95892.75 (1.2 + b) 3
3(1.2 + b)
The weight W will be acting through the C.G. of the dam. The horizontal distance of the
C.G. of the dam section from point A is given by equation (10.8) as
AN =
a 2 + ab + b2
3(a + b)
1.2 2 + 1.2b + b2
1.44 + 1.2b + b2
=
3(1.2 + b)
3(1.2 + b)
d = AM = AN + NM = AN + x
=
Now distance,
=
1.44 + 1.2b + b2
26.19
+
3(1.2 + b)
3(1.2 + b)
=
1.44 + 1.2b + b2 + 26.19
3(1.2 + b)
27.63 + 1.2b + b2
3(1.2 + b)
(a) Width at the base for no tension at the base
There will be no tension in the dam at the base if
2
d≤ b
3
2
27.63 + 1.2b + b
2
≤ b
3(1.2 + b)
3
27.63 + 1.2b + b2 ≤ 2b(1.2 + b)
≤ 2.4b + 2b2
0 ≤ 2.4b + 2b2 – 27.63 – 1.2b – b2
0 ≤ b2 + 1.2b – 27.63
2
b + 1.2b – 27.63 ≥ 0
For limiting case, b2 + 1.2b – 27.63 = 0.
The above equation is a quadratic equation. Hence its roots are given by
=
or
or
or
or
or
− 1.2 ± 1.2 2 + 4 × 1 × 27.63
− 1.2 ± 10.58
=
2
2×1
= 4.69 m
(Neglecting –ve root which is not possible)
Hence there will be no tension at the base, if width b is more than 4.69 m.
(b) Width of dam for no sliding of dam at the base
There will be no sliding of the dam at the base if μW ≥ F.
Substituting the values of W from equation (ii) and of F from equation (i), we get
μ × 95892.75 (1.2 + b) ≥ 313920
0.5 × 95892.75 (1.2 + b) ≥ 313920
b =
or
...(iii)
442
DAMS AND RETAINING WALLS
313920
≥ 6.54
0.5 × 95892.75
or
b ≥ (6.54 – 1.2) ≥ 5.34 m.
(c) Width of dam for no tension condition and also for no sliding at the base
For no tension, we have
b ≥ 4.69 m
For no sliding at the base
b ≥ 5.34 m
To satisfy both the conditions, b ≥ 5.34 m
∴ Minimum bottom width
= 5.34 m. Ans.
Problem 10.13. A masonry dam of trapezoidal section is 12 m high with a top width of
2 m. The water face has a better of 1 in 12. Find the minimum bottom width necessary so that
tensile stresses are not induced on the base section. Assume density of masonry = 2300 kg/m3,
that of water = 1000 kg/m3 and no free board.
(1.2 + b) ≥
or
Sol. Given :
Height of dam,
H = 12 m
Top width,
a=2m
Slope of water face
= 1 in 12
CD CD
1
=
=
or
tan θ =
12 AD 12
∴ Length
CD = 1 m
Density of masonry,ρ0 = 2300 kg/m3
∴ Weight of density of masonry,
w0 = 2300 × 9.81 N/m3
Density of water,
ρ = 1000 kg/m3
∴ Weight density of water,
w = 1000 × 9.81 N/m3.
No free board means the depth of water is equal
to the height of dam.
∴ Depth of water, h = 12 m
Consider one metre length of the dam.
The forces acting on the dam are :
The force F due to water acting on the face AC is
perpendicular to the face AC. This force F is resolved
into two components Fx and Fy as shown in Fig. 10.16.
(i) Force Fx = Force due to water acting on
vertical face AD
1m
D
Fy
12 m
2m
C
G
F
Fx
O
F
q
12/3 m
R
W
A
N
M
x
d
=w×A× h
= 1000 × 9.81 × (12 × 1) ×
B
b
12
2
Fig. 10.16
= 72000 × 9.81 N.
12
= 4 m above the base of the dam i.e., from point A.
3
Fy = Weight of water in the wedge ADC
= w × Area of triangle ADC × 1
12 × 1
× 1 = 6000 × 9.81 N.
= 1000 × 9.81 ×
2
443
The force Fx acts at a height of
(ii) The force
STRENGTH OF MATERIALS
1
m from line AD.
3
(iii) Weight of dam, W = w0 × Area of trapezoidal × 1
The line of action of Fy is at a distance of
FG a + b IJ × H × 1
H 2 K
F 2 + bIJ × 12 × 1
= 2300 × 9.81 × GH
2 K
= 2300 × 9.81 ×
= 13800 × 9.81(2 + b) N.
The weight of the dam (W ) is acting at the C.G. of the dam. The position of C.G. of the dam
(i.e., distance AN in Fig. 10.16) is obtained by splitting the trapezoidal into triangles and rectangle,
taking the moments of their areas about A and equating to the moment of area of the trapezoidal
about the point A.
∴
or
or
FG 12 × 1IJ × 2 + 12 × 2 × (1 + 1) + 12 × (b − 3) × LM3 + 1 × (b − 3)OP = FG a + bIJ × H × AN
H 2 K 3
2
N 3
Q H 2 K
F b − 3 IJ = FG 2 + bIJ × 12 × AN
4 + 48 + 6(b – 3) × G 3 +
H 3 K H 2 K
F b + 6 IJ = 6(2 + b) × AN
52 + 6(b – 3) G
H 3 K
156 + 6(b − 3)(b + 6)
.
3 × 6 × (2 + b)
The resultant R cuts the base at M. To find the distance of M from A (i.e., distance AM or
d), take the moments of all forces about the point M.
or
AN =
∴
FG
H
Fx × 4 – Fy × d −
IJ
K
1
–W×x=0
3
FG
H
72000 × 9.81 × 4 – 6000 × 9.81 × d −
or
But
IJ
K
1
– 13800 × 9.81(2 + b) × x = 0
3
x = d – AN
...(i)
156 + 6(b − 3)(b + 6)
=d–
.
18(2 + b)
Substituting the value of x in equation (i), we get
FG
H
9.81 × 72000 × 4 – 6000 × 9.81 d −
IJ
K
1
– 13800 × 9.81(2 + b)
3
LM
N
× d−
To avoid the tension at the base of the dam, the distance d ≤
∴ Taking the limiting value, we get
d=
444
2
b.
3
OP
Q
156 + 6(b − 3)(b + 6)
= 0.
18(2 + b)
2
b.
3
DAMS AND RETAINING WALLS
Substituting this value of d in above equation, we get
2b 1
−
– 13800 × 9.81(2 + b)
9.81 × 288000 – 6000 × 9.81
3 3
2b 156 + 6(b − 3)(b + 6)
−
×
= 0.
3
18(2 + b)
FG
H
or 288000 – 2000(2b – 1) – 13800(2 + b) ×
or
or
or
or
or
or
IJ
K
LM
N
OP
Q
LM 2b × 6 × (2 + b) − 156 − 6(b − 3)(b + 6) OP = 0
18(2 + b)
N
Q
13800
[12b(2 + b) – 156 – 6(b – 3)(b + 6)]
18
138
2880 = 20(2b – 1) –
[24b + 12b2 – 156 – 6(b2 + 3b – 18)]
18
23
[6b2 + 6b – 48]
2880 – 40b + 20 –
3
2880 – 40b + 20 – 23 [2b2 + 2b – 16]
2880 – 40b + 20 – 46b2 – 46b + 36.8
– 46b2 – 86b + 3268
46b2 + 86b – 3268
The above equation is a quadratic equation. Hence its solution is
288000 – 2000(2b + 1) –
= 0
= 0
= 0
=
=
=
=
0
0
0
0.
− 86 ± 86 2 + 4 × 46 × 3268
− 86 ± 780.19
=
92
2 × 46
− 86 + 780.12
=
(Neglecting –ve roots)
92
= 7.545 m. Ans.
Problem 10.14. A mass concrete dam shown in Fig. 10.17 (a) has a trapezoidal crosssection. The height above the foundation is 61.5 m and its water face is vertical. The width at
the top is 4.5 m.
Calculate the necessary minimum width of the dam at its bottom, to ensure that no
tension shall be developed when water is stored upto
4.5
60 metres. Draw the pressure diagram at the base of
m
D
C
the dam, for this condition, and indicate the maximum pressure developed.
Take density of concrete as 2,400 kg/m3 and that
of water as 1,000 kg/m3.
Sol. Given :
Height of dam,
H = 61.5 m
Top width of dam,
a = 4.5 m
Height of water,
h = 60 m
Density of concrete, ρ0 = 3400 kg/m3
Weight density of concrete,
w0 = 2400 × 9.81 N/m3
Density of water,
ρ = 1000 kg/m3
A
B
∴ Weight density of water,
b
w = 1000 × 9.81 Nm3
Fig. 10.17 (a)
60 m
61.5 m
b=
445
STRENGTH OF MATERIALS
Let b = Minimum width of the dam at its bottom in metres,
F = Total water pressure on the dam per metre length,
x = Horizontal distance between the C.G. of the dam section and point M.
wh 2
2
1000 × 9.81(60 2 )
=
= 17658000 N
...(i)
2
We know that the weight of dam per metre length,
(a × b)
W = w0 ×
×H×1
2
4.5 + b
= 2400 × 9.81 ×
× 61.5 N
2
= 723978(4.5 + b) N
...(ii)
Now let us find out the position of the C.G. of the dam section. We know that the distance
AN from equation (10.8) is given by
Using the relation,
F=
a 2 + ab + b2 (4.5) 2 + 4.5b + b2
=
3(a + b)
3(4.5 + b)
2
20.25 + 4.5b + b
=
3(4.5 + b)
AN =
Now from equation (10.3)
F h
×
x=
W 3
17658000
60
488
×
=
=
723978(4.5 + b)
3
(4.5 + b)
∴ Horizontal distance AM
4.5
m
D
20.25 + 4.5b + b2
3(4.5 + b)
488
+
4.5 + b
2
20.25 + 4.5b + b + 1464
=
3(4.5 + b)
1484.25 + 4.5b + b2
=
3(4.5 + b)
There will be no tension in the dam at
the base if
2
d≤ b
3
2b
Hence for the limiting case d =
3
1484.25 + 4.5b + b2 2b
=
or
3(4.5 + b)
3
or
1484.25 + 4.5b + b2 = 2b(4.5 + b) = 9b + 2b2
or
b3 + 4.5b – 1484.25 = 0.
C
446
G
60 m
F
F
O
20 m
61.5 m
d = AN + x =
A
N
W
M
x
B
R
b
max
Fig. 10.17 (b)
DAMS AND RETAINING WALLS
Solving this equation, as a quadratic equation for b,
− 4.5 ± (4.5) 2 + 4 × 1484.25
2
− 4.5 + 77.05
=
= 36.725 m. Ans.
2
Pressure diagram at the base of the dam
Let σmax = Maximum stress across the base at B.
Substituting the value of b in equation (ii)
W = 723978(4.5 + b) = 723978(4.5 + 36.275)
= 29520252 N
Using equation (10.10),
W
6e
1+
σmax =
b
b
But from equation (10.15),
b
e= .
6
2W 2 × 29520252 N
=
= 1627479 N/m2
∴
σmax =
b
26.275
= 1.6275 MN/m2 and σmin = 0.
The pressure diagram at the base of the dam is shown in Fig. 10.17(b). Ans.
∴
b=
FG
H
IJ
K
10.7. RETAINING WALLS..
The walls which are used for retaining the soil or earth, are known as retaining walls.
The earth, retained by a retaining wall, exerts pressure on the retaining wall in the same way as
water exerts pressure on the dam. A number of theories have been evolved to determine the
pressure exerted by the soil or earth on the retaining wall. One of the theories is Rankine’s
theory of earth pressure. Before discussing Rankine’s theory, let us define the angle or repose
and study the equilibrium of a body on an inclined plane.
10.7.1. Angle of Repose. It is defined as the maximum inclination of a plane at which a
body remains in equilibrium over the inclined plane by the assistance of friction only. The earth
particles lack in cohesion and have a definite angle of repose. And angle of repose* is equal to
angle of friction (φ). Angle of friction is the angle made by the resultant of the normal reaction
and limiting force of friction with the normal reaction.
10.7.2. Equilibrium of a Body on an Inclined Plane. If the inclination of the inclined
plane is less than the angle of repose, the body will be in equilibrium entirely by friction only.
But if the inclination of the plane is greater than the angle of repose, the body will be in equilibrium
only with the assistance of an external force. Let an external horizontal force P is applied on a
body, which is placed on an inclined plane having inclination greater than angle of repose, to
keep the body in equilibrium. There are two cases :
(i) The body may be on the point of moving down the plane, and
(ii) The body may be on the point of moving the plane.
1st Case. The body is on the point of moving down the plane.
Let W = Weight of the body
P = Horizontal force applied on the body in order to prevent the body from moving
down the plane
447
STRENGTH OF MATERIALS
θ = Angle of inclination of the plane
φ = Angle of limiting friction i.e., angle made by the resultant of normal reaction and
limiting force of friction with the normal reaction as shown in Fig. 10.18 (b).
R ′ = Resultant of normal reaction and limiting force of friction.
The forces acting on the body are shown in Fig. 10.18 (a). The body is in equilibrium
under the action of three forces W, P and R ′. Applying Lami’s theorem* to the forces acting on
the body, we get
P
W
=
sine of angle between R′ and P
sine of angle between W and R′
P
W
or
=
sin (θ + 90 − φ)
sin (90 − θ + 90 + φ)
P
W
or
=
sin (90 + θ − φ)
sin [180 − (θ − φ)]
W
N
re orm
ac a
ti o l
n
R
E
φ
90° – θ
Fo
fric rce o
ti o f
n
P
90°
μR
φ
R′
θ
–φ
90
–θ
90
θ
R′
No
rm
a
l
A
(b )
(a )
Fig. 10.18. Body moving down.
or
W sin [180 − (θ − φ)]
W sin (θ − φ)
=
sin [90 + (θ − φ)]
cos (θ − φ)
= W tan (θ – φ)
...(10.17)
2nd Case. The body is on the point of moving up the plane.
Let W = Weight of the body,
P = Horizontal force applied on the body in order to prevent the body from moving up
the plane,
θ = Angle of inclination,
φ = Angle of limiting friction i.e., angle made by the resultant (R′) of normal
reaction and limiting force of friction with the normal reaction as shown in
Fig. 10.19 (b),
R ′ = Resultant of normal reaction and limiting force of friction.
P=
*Please refer to some standard book of Engineering Mechanics.
448
DAMS AND RETAINING WALLS
The forces acting on the body are shown in Fig. 10.19 (a). The body is in equilibrium
under the action of three forces W, P and R′.
Applying Lami’s theorem to the forces acting on the body, we get
W
P
=
sine of angle between R′ and P
sine of angle between W and R′
P
W
or
=
sin (90 − φ + 90 − θ) sin (θ + 90 + φ)
P
W
or
=
sin [90 + (θ + φ)]
sin [180 − (θ + φ)]
W sin [180 − (θ + φ)]
or
P=
sin [90 + (θ + φ)]
W sin (θ + φ)
=
cos (θ + φ)
= W tan (θ + φ).
...(10.18)
F
fri orce
ct
io of
n
W
90 – φ
90 – θ
R′
φ
P
θ
φ
90°
R′
or
m
al
al
rm tion
o
N ac
re
N
θ
( b)
(a )
Fig. 10.19. Body moving up.
10.8. RANKINE’S THEORY OF EARTH PRESSURE..
Rankine’s theory of earth pressure is used to determine the pressure exerted by the earth
or soil on the retaining wall. This theory is based on the following assumptions :
1. The earth or soil retained by a retaining wall is cohesionless.
2. Frictional resistance between the retaining wall and the retained material (i.e., earth
or soil) is neglected.
3. The failure of the retained material takes place along a plane, known as rupture plane.
Fig. 10.20 shows a trapezoidal retaining wall ABCD retaining the earth upto a height h on
the vertical face AD. Let the earth surface is horizontal and it is in level with the top of the
retaining wall. Let AE is the rupture plane which means if the wall AD is removed the wedge AED
of earth will move down along the plane AE. Let P is the horizontal force offered by the retaining
wall, to keep the wedge AED in equilibrium. Let w is the weight density of the earth or soil.
449
STRENGTH OF MATERIALS
Consider one metre length of the retaining wall.
The forces acting on the wedge AED of the retained material are :
(i) Weight of wedge AED,
W = Weight density of earth × Area of AED × 1
AB × ED
×1
2
h × h cot θ
=w×
2
=w×
FG∵
H
tan θ =
IJ
K
AD
AD
= h × cot θ
or ED =
tan θ
ED
w × h 2 × cot θ
.
2
(ii) The horizontal force P exerted by the retaining wall on the wedge.
=
Earth
surface
h cot q
E
D
C
q
Rupture
plane
W
h
R
mR
f
P
R
q
A
B
Fig. 10.20
(iii) The resultant reaction R′ at the plane AE. The reaction R′ is the resultant of normal
reaction R and force of friction μ R. The resultant reaction R′ makes an angle φ with the normal
of the plane AE.
(iv) The frictional resistance along the contact face AD is neglected.
These forces are similar as shown in Fig. 10.18 (a). The wedge AED is in equilibrium
under the action of three forces P, W and R′. The value of horizontal force P is given by equation
(10.17) as
P = W tan (θ – φ)
...(i)
But here
W = Weight of wedge AED
wh 2
cot θ
2
Substituting the value of W in equation (i), we get
=
P=
450
wh 2
cot θ . tan(θ – φ)
2
...(ii)
DAMS AND RETAINING WALLS
In the above equation the angle θ is the angle of the rupture plane. The earth is having
maximum tendency to slip along rupture plane. Hence the supposing force P should be maxidP
= 0.
mum. But P will be maximum if
dθ
Hence differentiating equation (ii), w.r.t. θ, we get
LM
N
OP
Q
dp
d wh2
cot θ . tan (θ − φ) = 0
=
dθ
2
dθ
or
or
wh 2
{cot θ sec2 (θ – φ) – cosec2 θ tan (θ – φ)} = 0
2
cot θ sec2 (θ – φ) – cosec2 θ tan (θ – φ) = 0
Let tan θ = t and tan (θ – φ) = t1.
Equation (iii) becomes as
FG IJ × t = 0
H K
F t + 1I × t = 0
−G
H t JK
1
1
(1 + t12 ) − 1 + 2
t
t
or
or
or
or
or
or
or
or
or
or
1 + t12
t
1
2
2
RS∵
T
cot θ =
...(iii)
1
1
1
= and cosec 2 θ = 1 + cot 2 θ = 1 + 2
tan θ t
t
1
t(1 + t12) – (t2 + 1) × t1 = 0
t + tt12 – t1t2 – t1 = 0
t – t1t2 + tt12 – t1 = 0
t[1 – t1t] – t1[1 – tt1] = 0
(1 – t1t)(t – t1) = 0
Either (1 – t1t) = 0 or (t – t1) = 0
∴ tt1 = 1 or t = t1
If t = t1, then θ = tan (θ – φ).
This is not possible
∴
tt1 = 1
tan θ tan (θ – φ) = 1
1
tan θ =
tan (θ − φ)
= cot (θ – φ) = tan [90 – (θ – φ)]
∴
θ = 90 – (θ – φ)
θ + θ – φ = 90°
2θ – φ = 90°
90 + φ
φ
= 45° +
∴
θ=
2
2
φ
Thus, the plane of rupture is inclined at 45° +
with the horizontal.
2
Substituting the value of θ in equation (ii), we get
FG
H
P=
IJ
K
wh 2
wh 2 tan (θ − φ)
cot θ . tan (θ – φ) =
tan θ
2
2
451
UV
W
STRENGTH OF MATERIALS
wh 2
=
2
wh 2
=
2
wh 2
=
2
wh 2
=
2
=
wh
2
2
FG
H
IJ
K
φ
−φ
2
φ
tan 45° +
2
φ
tan 45° −
2
φ
tan 45° +
2
φ
tan 45° − tan
2 ×
φ
1 + tan 45° tan
2
φ
1 − tan
1 − tan
2 ×
φ
1 + tan
1 + tan
2
tan 45° +
F
GG
GH
F
GG
GH
FG
H
FG
H
FG
H
IJ
K
IJ
K
IJ
K
I
JJ
JK
F 1 − tan φ I
GG
2J
GH 1 + tan 2φ JJK
I
JJ
JK
F
GG
GH
wh 2
2
R| cos φ − sin φ U|
S| 2φ 2φ V|
T cos 2 + sin 2 W
R| cos φ + sin φ − 2 cos φ sin φ U|
2
2
2
S| 2φ
φ
φ
φV
cos
+ sin
+ 2 cos sin |
2
2
2W
T 2
φ
φ
R| 1 − 2 cos sin U|
2
2
S|
φ
φV
1 + 2 cos sin |
2
2W
T
LM 1 − sin φ OP
N 1 + sin φ Q
2
=
wh
2
2
wh 2
=
2
=
wh 2
2
F 1 − tan 45° tan φ I
GG
2J
φ J
GH tan 45° + tan 2 JK
φI
2J
(∵
φJ
J
2K
R| F sin φ I U|
|| GG 1 − 2φ JJ ||
|S GH cos 2 JK |V
|| FG sin 2φ IJ ||
|| GG 1 + cos φ JJ ||
2K W
TH
2
2
2
2
θ = 45° +
φ
2
IJ
K
tan 45° = 1)
2
2
=
FG∵
H
wh 2
...(10.19)
2
But P is the horizontal force exerted by the retaining wall on the wedge. The wedge of the
earth will also exert the same horizontal force on the retaining wall. Hence equation (10.19) gives
also horizontal force exerted by the earth on the retaining wall.
∴
P=
The horizontal force P acts at a height of
452
h
above the base.
3
DAMS AND RETAINING WALLS
Pressure intensity at the bottom. If we assume a linear variation of the pressure
intensity varying from zero at the top to the maximum value p at the bottom, then we have
p× h
.
P=
2
But from equation (10.19),
LM
OP
N
Q
p× h
wh L 1 − sin φ O
=
MN 1 + sin φ PQ
2
2
L 1 − sin φ OP
p = wh M
N 1 + sin φ Q
wh 2 1 − sin φ
.
1 + sin φ
2
Equating the two values of P, we get
P=
∴
or
2
...(10.20)
Problem 10.15. A masonry retaining wall of trapezoidal section is 6 metre high and
retains earth which is level upto the top. The width at the top is 1 m and the exposed face is
vertical. Find the minimum width of the wall at the bottom in order the tension may not be
induced at the base. The density of masonry and earth is 2300 and 1600 kg/m3 respectively. The
angle of repose of the soil is 30°.
Sol. Given :
Height of wall,
h =6m
Width at the top,
a =1m
Density of masonry,
ρ0 = 2300 kg/m2
∴ Weight density of masonry
w0 = ρ0 × g = 2300 × 9.81 N/m3
Density of earth,
ρ = 1600 kg/cm3
∴ Weight density of earth, w = ρ × g = 1600 × 9.81 N/m3
Angle of repose, φ = 30°
Let b = Minimum width at the bottom.
Earth surface
G
6m
P
P
O
2m
A
N
M
x
W
B
R
d
b
Fig. 10.21
453
STRENGTH OF MATERIALS
Consider one metre length of the retaining wall.
The thrust of earth on the vertical face is given by equation (10.19),
P=
FG
H
1 − sin φ
1
wh2
1
+ sin φ
2
IJ
K
FG
H
IJ
K
1 − sin 30°
1
× 1600 × 9.81 × 62
1 + sin 30°
2
1 − 0.5
800 × 9.81 × 36 × 0.5
=
= 800 × 9.81 × 36
1 + 0.5
1.5
= 94176 N.
=
FG
H
The thrust P will be acting at a height of
IJ
K
6
= 2 m above the base. Weight of 1 m length
3
of trapezoidal wall,
W = Weight density of masonry × Area of trapezoidal × 1
a+b
×h×1
= 2300 × 9.81 ×
2
1+ b
= 2300 × 9.81 ×
× 6 = 67689 (1 + b) N.
2
The weight W will be acting through the C.G. of the trapezoidal section. The distance of
the C.G. of the trapezoidal from the point A is obtained by using equation (10.8).
FG
H
FG
H
∴
IJ
K
IJ
K
a 2 + ab + b2
AN =
3(a + b)
12 + 1 × b + b2 1 + b + b2
=
3(1 + b)
3(1 + b)
The horizontal distance x, between the line of action of W and the point at which the
resultant force R cuts the base, is given by equation (10.3).
P h
×
∴
x=
(∵ Here P = F)
W 3
94176
6 2.782
× =
=
67689(1 + b) 3 (1 + b)
Hence in Fig. 10.21, d = AN + x
=
1 + b + b2
2.782
+
3(1 + b)
(1 + b)
2
1 + b + b + 3 × 2.782 1 + b + b2 + 8.346
b2 + b + 9.346
=
=
=
3(1 + b)
3(1 + b)
3(1 + b)
If the tension at the base is just avoided,
2
d= b
3
b2 + b + 9.346
2
= b
3(1 + b)
3
b2 + b + 9.346 = 2b(1 + b) = 2b + 2b2
b2 + b – 9.346 = 0
=
or
or
or
454
DAMS AND RETAINING WALLS
The above equation is quadratic equation. Its solution is given by
− 1 ± 6.195
− 1 ± 12 + 4 × 1 × 9.346
=
2
2
− 1 + 6.195
=
(Neglecting – ve value)
2
= 2.597 m. Ans.
Problem 10.16. A masonry retaining wall of trapezoidal section is 10 m high and retains
earth which is level upto the top. The width at the top is 2 m and at the bottom 8 m and the
exposed face is vertical. Find the maximum and minimum intensities of normal stress at the
base.
Take : Density of earth = 1600 kg/m3,
Density of masonry = 2400 kg/m3,
Angle of repose of earth = 30°.
Sol. Given :
Height of wall,
h = 10 m
Width of wall at top, a = 2 m
Width at the bottom, b = 8 m
Density of earth,
ρ = 1600 kg/m3
∴ Weight density of earth,
w = ρ × g = 1600 × 9.81 N/m3
Density of masonry, ρ0 = 2400 kg/cm3
∴ Weight density of masonry,
w0 = ρ0 × g = 2400 × 9.81 N/m3
Angle of repose,
φ = 30°
Consider 1 m length of the wall.
b=
10 m
2m
G
P
O
10/3 m
A
W
N
M
B
R
x
d
8m
Fig. 10.22
Thrust of earth on the vertical face of the wall is given by equation (10.19),
P=
FG
H
1 − sin φ
1
wh2
1 + sin φ
2
IJ
K
455
STRENGTH OF MATERIALS
FG
H
IJ
K
1 − sin 30°
1
× 1600 × 9.81 × 102
1 + sin 30°
2
1 − 0.5
= 800 × 9.81 × 100
1 + 0.5
0.5
80000 × 9.81
= 80000 × 9.81 ×
=
N.
1.5
3
10
The thrust P will be acting at a height of
m above the ground. Weight of 1 m length of
3
trapezoidal wall.
W = Weight density of masonry × Volume of wall
= 2400 × 9.81 × [Area of cross-section of trapezoidal wall] × 1
=
FG
H
IJ
K
FG 8 + 2 IJ × 10 × 1 LM∵
H 2 K
N
= 120000 × 9.81 N.
= 2400 × 9.81 ×
Area =
FG 8 + 2 IJ × 10 m OP
H 2 K
Q
2
The weight W will be acting through the C.G. of the trapezoidal section. The distance of the
C.G. of the trapezoidal section from the point A is obtained by using equation (10.8).
∴
AN =
a 2 + ab + b2
3(a + b)
84
2 2 + 2 × 8 + 8 2 4 + 16 + 64
=
=
= 2.8 m.
30
3(2 + 8)
30
The horizontal distance x between the line of action of W and the point at which the
resultant force R cuts the base, is given by equation (10.3).
P h
×
∴
x=
(∵ Here P = F )
W 3
80000 × 9.81
10
×
=
= 0.74 m
3 × 120000 × 9.81 3
Hence in Fig. 10.22, d = AN + x
= 2.8 + 0.74 = 3.54 m
b
∴ Eccentricity,
e=d–
2
8
= 3.54 – = 3.54 – 4.0 = – 0.46 m
2
(Minus sign only indicates that stress at A will be more than at B).
The maximum and minimum stresses at the base are given by equations (10.10) and
(10.11).
∴ Stresses (σmax and σmin)
6.e
W
1±
=
b
b
6 × 0.46
120000 × 9.81
1±
=
= 147150 (1 ± 0.345)
8
8
= 147150 × 1.345 and 147150 × (1 – 0.345)
= 197916.75 N/m2 and 96383.25 N/m2
∴
σ max = 197916.75 N/m2 and is acting at A. Ans.
σ min = 96383.25 N/m2 and is acting at B. Ans.
=
FG
H
456
IJ
K
FG
H
IJ
K
DAMS AND RETAINING WALLS
Problem 10.17. A masonry retainting wall of trapezoidal section is 1.5 m wide at the top,
3.5 m wide at the base and 6 m high. The face of the wall retaining earth is vertical and the earth
level is upto the top of the wall. The density of the earth is 1600 kg/m3 for the top 3 m and 1800
kg/m3 below this level. The density of masonry is 2300 kg/m3. Find the total lateral pressure on
the retaining wall per m run and maximum and minimum normal pressure intensities at the
base. Take the angle of repose = 30° for both types of earth.
Sol. Given :
Width at the top,
a = 1.5 m
Width at the bottom,
b = 3.5 m
Height of the wall,
h=6m
Density of upper earth, ρ1 = 1600 kg/m3
∴ Weight density of upper earth,
w1 = 1600 × 9.81 N/m3
Depth of upper earth, h1 = 3 m
Density of lower earth, ρ2 = 1800 kg/m3
∴ Weight density of lower earth,
w2 = 1800 × 9.81 N/m3
Depth of lower earth,
h2 = 3 m
Density of mesonry,
ρ0 = 2300 kg/m3
∴ Weight density of masonry,
w0 = 2300 × 9.81 N/m3
Angle of repose for both earth,
φ = 30°.
Total lateral pressure on the retaining wall per m run
1.5 m
Earth surface
A
w1 =
1600 kgf
m
3
3m
A
P1
1800 kgf
m
3
W
3m
w2 =
C
(a )
O
P
6m
B
D
B
R
N
M
x
3.5 m
C
E
F
(b )
Pressure diagram
Fig. 10.23
The pressure diagram on the retaining wall is shown in Fig. 10.23 (b)
Let
P = Total lateral pressure force
P1 = Pressure force due to upper earth.
P2 = Pressure force due to lower earth.
457
STRENGTH OF MATERIALS
The pressure intensity at a depth h is given by equation (10.20) as
LM 1 − sin φ OP
N 1 + sin φ Q
Pressure intensity at B,
F 1 − sin 30° IJ = 1600 × 9.81 × 3 FG 1 − 0.5 IJ
p =w h G
H 1 + sin 30° K
H 1 + 0.5 K
p = wh
∴
B
1 1
0.5
= 1600 × 9.81 N/m2.
1.5
This is represented by length BD in pressure diagram.
∴ Length BD = pB = 1600 × 9.81 N/m3
Similarly pressure intensity at C,
= 4800 × 9.81 ×
pC = pB + w2h2
LM 1 − sin φ OP
N 1 + sin φ Q
= 1600 × 9.81 + 1800 × 9.81 × 3 ×
FG 1 − 0.5 IJ
H 1 + 0.5 K
= 1600 × 9.81 + 1800 × 9.81 N.
This is represented by length CF in pressure diagram
∴
CF = 1600 × 9.81 + 1800 × 9.81 = 3400 × 9.81 N/m2
But
CE = BD = 1600 × 9.81
∴
EF = CF – CE
= (1600 + 1800) × 9.81 – 1600 × 9.81 = 1800 × 9.81 N/m2
∴ Pressure force due to upper earth,
P1 = Area of triangle ABD
1
1
= × AB × BD = × 3 × 1600 × 9.81 = 23544 N
2
2
1
This force acts at a height of × 3 = 1 m above B or at a height of (3 + 1) = 4 m above point C.
3
Pressure force due to lower earth,
P2 = Area of BDFC =
1
[BD + CF] × BC
2
1
[1600 + 3400] × 9.81 × 3.0 = 73575 N.
2
This force acts at a height from C
3
= [Area of rectangle CEDB ×
2
+ Area of triangle EFD × 1] ÷ Total area
3 1800 × 9.81 × 3
×1
1600 × 9.81 × 3 × +
2
2
=
1800 × 9.81 × 3
1600 × 9.81 × 3 +
2
9.81 × 7200 + 2700 × 9.81 9900
=
=
= 1.32 m from C.
9.81 × 4800 + 2700 × 9.81 7500
=
458
DAMS AND RETAINING WALLS
∴ Total pressure force,
P = P1 + P2 = 23544 + 73575 = 97119 N. Ans.
Maximum and minimum normal stresses at base
Weight of retaining wall per m run,
W = Weight density of masonry ×
= 2300 × 9.81 ×
FG a + bIJ × h × 1
H 2 K
FG 1.5 + 3.5IJ × 6 × 1 = 338445 N.
H 2 K
The weight W will be acting at the C.G. of the retaining wall. The distance of the C.G. of
the retaining wall from point C is given by equation (10.8) as,
1.52 + 1.5 × 3.5 + 3.52
a 2 + ab + b2
=
= 1.32 m
3 (1.5 + 3.5)
3(a + b)
Let x = Distance between the line of action of W and the resultant of W and P at the base.
Taking moments of W, P1 and P2 about the point M, we get
P1 × 4 + P2 × 1.32 = W × x
P1 × 4 + P2 × 1.32
∴
x=
W
23544 × 4 + 73575 × 1.32
94176 + 97119
=
=
= 0.565 m
338445
338445
∴ Distance
CM = CN + x = 1.32 + 0.565 = 1885 m
b
3.5
= 0.135 m.
∴ Eccentricity,
e = CM – = 1.885 –
2
2
Now using equations (10.10) and (10.11), we get
CN =
and
σ min
FG
H
IJ
K
W
6.e
1+
b
b
338445
6 × 0.135
1+
=
= 119073.78 N/m2. Ans.
3.5
3.5
6×e
W
1−
=
b
b
338445
6 × 0.135
1−
=
= 74320.56 N/m2. Ans.
3.5
3.5
σ max =
FG
H
FG
H
FG
H
IJ
K
IJ
K
IJ
K
10.9. SURCHARGED RETAINING WALL.
Fig. 10.24 shows a retaining wall of height h and retaining earth which is surcharged at
an angle α with the horizontal. Then the total earth pressure exerted on the retaining wall is
given by,
P* =
wh 2
cos α − cos 2 α − cos 2 φ
cos α .
2
cos α + cos 2 α + cos 2 φ
...(10.21)
where φ = Angle of repose.
*The proof of this expression may be seen in some standard book of theory of structure.
459
STRENGTH OF MATERIALS
h
3
above the base of the retaining wall and parallel to the
free surface of the earth.
The pressure P is resolved into two components
i.e., horizontal and vertical components.
The horizontal component, PH = P cos α and acts
h
at a height of above base.
3
The vertical component, Pv = P sin α and acts
along DA.
Earth
surface
The total earth pressure P acts at a height of
a
C
Pv
P
PH
D
h
a
h/3
B
A
Fig. 10.24
Problem 10.18. A masonry retaining wall of trapezoidal section 2 m wide at its top, 3 m
wide at its bottom is 8 m high. It is retaining a soil on its vertical side at a surcharge of 20°. The
soil has a density of 2000 kg/m3 and has an angle of repose of 45°. Find the total pressure on the
wall per metre length and the point, where the resultant cuts the base.
Also find maximum and minimum intensities of stress at the base. Take density of the
masonry as 2400 kg/m3.
Sol. Given :
2m
Top width,
a =2m
20°
C
D
Base width,
b =3m
Height of wall,
h =8m
Angle of surcharge, α = 20°
Density of soil,
ρ = 2000 kg/m3
8m
∴ Specific weight of soil,
3
w = 2000 × 9.81 N/m
Angle of repose,
φ = 45°
Density of masonry, ρ0 = 2400 kg/m3
A
∴ Weight density of masonry,
B
w0 = ρ0 × g
3m
= 2400 × 9.81 N/m3.
Fig. 10.25
Total pressure on the wall per metre length
Let
P = Total pressure on the wall per metre length.
Using equation (10.21),
P =
cos α − cos 2 α − cos 2 φ
wh 2
cos α
2
cos α + cos 2 α + cos 2 φ
P =
cos 20° − cos 2 20° − cos 2 45°
2000 × 9.81(8) 2
cos 20° ×
N
2
cos 20° + cos 2 20° + cos 2 45°
= 64000 × 9.81 × 0.9397 ×
= 627840 N. Ans.
460
0.9397 − 0.9397 2 − 0.70712
0.9397 + 0.9397 2 + 0.70712
N
DAMS AND RETAINING WALLS
2m
The point, where the resultant cuts the base
C
D
Let the resultant cut the base at M as shown in
Fig. 10.26.
Let x = Horizontal distance between the c.g. of the
vertical load of wall and M (i.e., NM).
Pv
We know that the horizontal component, of the
8
m
pressure,
PH
PH = 627840 cos 20° N
= 627840 × 0.9397 N = 114090.3 N
and vertical component of the pressure,
PV = 627840 sin 20° N
A
N
M B
R
= 627840 × 0.3420 N = 41535.54 N
3m
Weight of dam
Fig. 10.26
a+b
= w0 ×
×h
2
2+3
× 8 = 470880 N
= 2400 × 9.81 ×
2
∴ Total load acting vertically down,
W = 470880 + 41535.54 = 512415.54 N.
First of all, let us find out the position of c.g. of the vertical load. Taking moments of the
vertical loads about A and equating the same,
8×1 7
×
W × AN = PV × 0 + 2400 × 9.81 × 2 × 8 × 1 + 2400 × 9.81 ×
2
3
∴
512415.54 AN = 60,8000 × 9.81
60,8000 × 9.81
or
AN =
= 1.164 m
512415.54
Now using the relation,
PH h
× with usual notations.
x =
3
W
114090.3 8
× = 0.594 m
=
512415.54 3
or
Distance NM = x = 0.594 m.
∴ Horizontal distance between A and the point M, where the resultant cuts the base,
d = AN + NM = 1.164 + 0.594 m = 1.758 m. Ans.
Maximum and minimum intensities of stress at the base
Let
σ max = Maximum intensity of stress at the base.
σ min = Minimum intensity of stress at the base.
We know that the eccentricity of the resultant,
b
3
e = d – = 1.758 – = 0.258 m
2
2
FG
H
IJ
K
FG
H
Using the relation, σ max =
IJ
K
FG
H
6.e
W
1+
b
b
IJ
K
461
STRENGTH OF MATERIALS
=
Now using the relation,
FG
H
IJ
K
IJ
K
FG
H
IJ
K
6 × 0.258
512415.54
1+
N/m2 = 259082.1 N/m3
3
3
FG
H
6.e
W
1−
b
b
512415.54
6 × 0.258
1−
=
N/m2 = 82659.06 N/m2. Ans.
3
3
σ min =
Chimney
Wind
10.10 CHIMNEYS.
Chimneys are tall structures subjected to horizontal
wind pressure. The base of the chimneys are subjected to
bending moment due to horizontal wind force. This bending
moment at the base produces bending stresses. The base of
the chimney is also subjected to direct stresses due to self
weight of the chimney. Hence at the base of the chimney, the
bending stress and direct stress are acting. The direct stress
σ0 is given by,
Weight of chimney
W
=
σ0 =
Area of section at the base
A
The bending stress (σb) is obtained from
h
W
M σb
=
I
y
M
M
M
× y=
=
...(10.22)
I
( I / y) Z
where M = Bending moment due to horizontal wind force
and
Fig. 10.27. Chimney subjected
Z = Modulus of section.
to wind force.
The wind force (F ) acting in the horizontal direction
on the surface of chimney is given by,
F=K×p×A
...[10.22 (A)]
where
K = co-efficient of wind resistance, which depends upon the shape of the area exposed to
wind.
= 1 for rectangular and square chimneys
2
= for circular chimney
3
p = intensity of wind pressure
A = projected area of the surface exposed to wind.
= D × h for circular chimney
= b × h for rectangular or square chimney
b = width of chimney exposed to wind
h = height of chimney.
h
The wind force F will be acting at .
2
462
or
σb =
DAMS AND RETAINING WALLS
h
.
2
Hence bending moment (M) at the base of chimney is given by,
h
.
M=F×
2
Problem 10.19. Determine the maximum and minimum stresses at the base of an hollow
circular chimney of height 20 m with external diameter 4 m and internal diameter 2 m. The
chimney is subjected to a horizontal wind pressure of intensity 1 kN/m2. The specific weight of
the material of chimney is 22 kN/m3.
Sol. Given :
Height, H = 20 m ; External dia, D = 4 m ; Internal dia, d = 2 m.
Horizontal wind pressure, p = 1 kN/m2
Specific weight,
w = 22 kN/m3
Let us first find the weight of the chimney and horizontal wind force (F).
Weight (W ) of the chimney is given by,
W = ρ × g × Volume of chimney
= Weight density × Volume of chimney
= w × [Area of cross-section] × height
The moment of F at the base of the chimney will be F ×
= 22 ×
LM π (D
N4
2
OP
Q
− d 2 ) × h kN
π 2
(4 – 22) × 20 kN = 4146.9 kN
4
∴ Direct stress at the base of the chimney,
W
where A = Area of cross-section
σ0 =
A
4146.9
4146.9
=
= 440 kN/m2
=
π 2
π
3
2
(4 − 2 )
4
Now let us find the wind force (F). This force is given by equation [10.22(A)].
∴
F=K×p×A
2
where K = as the section is circular
3
A = projected area of the surface exposed to wind
=D×h
where D = External dia = 4 m
= 4 × 20 = 80 m2
p = horizontal wind pressure = 1 kN/m2
2
160
∴
F = × 1 × 80 =
= 53.33 kN
3
3
The bending moment (M ) at the base,
h
20
= 53.33 ×
= 533.3 kNm
M=F×
2
2
The bending stress (σ b) is given by equation (10.22) as
I
M
where Z = section modulus =
σb =
y
Z
= 22 ×
463
STRENGTH OF MATERIALS
π
D
(D4 – d4), y =
64
2
π
4
I=
[44 – 24] = 11.78 m4 and y = = 2 m.
64
2
I
11.78
Z=
=
= 5.89 m3
y
2
533.3
σb =
= 90.54 kN/m2
5.89
Now the maximum and minimum stresses at the base are given by,
σmax = σ 0 + σ b = 440 + 90.54 = 530.54 kN/m2 (comp)
σ min = σ 0 – σ b = 440 – 90.54 = 349.46 kN/m2 (comp). Ans.
I=
∴
∴
∴
∴
and
HIGHLIGHTS
1. A dam is constructed to store water whereas a retaining wall is constructed to retain the earth.
2. Trapezoidal dams, as compared to rectangular dams, are economical and easier to construct.
3. Thrust due to water on the vertical side of a dam is given by
wh2
2
where w = Weight density of water = 1000 × 9.81 N/m3
h = Depth of water.
The horizontal distance between the line of action of W and the point through which the resultant
cuts the base is given by
F=
4.
h
F
×
W 3
where
F = Force exerted by water,
h = Depth of water.
5. The eccentricity is given by,
x=
W = Weight of dam and
b
2
where d = The distance between A and the point where the resultant R cuts the base
e=d–
b F×h
+
2 W ×3
F×h
= AN +
W ×3
and
b = Base width of the dam.
6. The position of the C.G. of the dam from the point A is given by,
=
AD =
a2 + ab + b2
3(a + b)
a = Top width of the dam, and
b = Bottom width of the dam.
=
where
464
b
2
... For a rectangular dam
....For a trapezoidal dam
...For a rectangular dam
...For a trapezoidal dam
DAMS AND RETAINING WALLS
7. The maximum and minimum stresses at the base of a dam having water face vertical are given by,
FG
IJ
H
K
W F
6 . eI
=
G 1 − b JK
b H
σ max =
σ min
and
6.e
W
1+
b
b
where W = Weight of the dam
= w0 × b × H × 1
= w0 ×
...For a rectangular dam
FG a + b IJ × H × 1
H 2 K
...For a trapezoidal dam
b = Bottom width of dam, and
e = Eccentricity.
8. If the reservoir is empty, then the only force acting on the dam is the weight of the dam.
9. In case of a trapezoidal dam, if water face is inclined, then the force due to water acting on the
inclined face is resolved into two components. The components in the x-direction and y-directions
are given by,
Fx = Force exerted by water on the vertical face
and
Fy = Weight of the water included with the vertical face and inclined face.
10. The maximum and minimum stresses induced at the base of a trapezoidal dam having water face
inclined are given by,
FG
IJ
H
K
W F
6 . eI
G 1 − b JK
=
b H
σ max =
and
σ min
W
6.e
1+
b
b
where V = Sum of the forces acting on the dam = Fy + W.
where W = Weight of dam
and
Fy = Weight of water included with the vertical face and inclined face.
11. If the force of friction is more than the force due to water pressure, there will be no sliding of the
dam. But force of friction is equal to µ × W, where µ is the co-efficient of friction between the base
of the dam and the soil and W = weight of dam.
12. There will be no overturning of the dam if the resultant of water pressure and weight of dam
strikes the base within its width.
13. There will be no tension in the masonry of the dam at its base if
2
b
3
where e = Eccentricity,
b = Base width,
and
d = Distance between the point A and the point through which resultant force meets the
base.
14. The pressure exerted by earth on the retaining wall is given by Rankine’s theory of earth pressure.
According to this theory the pressure exerted by earth, which is level upto the top, on the retaining
wall is given by
where P =
h=
w=
and
φ=
e≤
b
6
P=
wh 1 − sin φ
2 1 + sin φ
d ≤
or
LM
N
OP
Q
Pressure exerted by earth on retaining wall,
Height of retaining wall,
Weight density of earth retained by the wall,
Angle of repose.
465
STRENGTH OF MATERIALS
15.
Angle of repose is the maximum inclination of a plane at which a body remains in equilibrium over
the inclined plane by the assistance of friction only. The earth particles lack in cohesion and have
a definite angle of repose.
EXERCISE
(A) Theoretical Questions
1. What is the difference between a dam and a retaining wall ?
2. Describe the different types of dams. Why a trapezoidal dam is mostly used these days ?
3. A masonry dam of rectangular section of height H and bottom width b retains water upto a depth
of h. How will you find the point at which the resultant cuts the base. Take the weight density of
masonry as w0.
4. Prove that the horizontal distance between the line of action of the weight of the dam and the point
through the resultant cuts the base of a rectangular dam is given by
h
F
×
W 3
where F = Force exerted by water
W = Weight of dam,
and h = Depth of water.
x=
F
h
× .
W 3
6. Find an expression for the stresses developed at the base of a rectangular dam which retains water
upto a given depth.
7. Prove that the maximum and minimum stresses at the base of a rectangular dam are given by
5. Prove that the eccentricity in case of a rectangular dam is given by e =
8.
FG
H
FG
H
IJ
K
IJ
K
W
6.e
W
6.e
1−
and
σ min =
1+
b
b
b
b
where W = Weight of the dam,
b = Width of dam at the base, and
e = Eccentricity.
Prove that in case of a trapezoidal dam having water face vertical, the distance between A and the
point through resultant passes at the base is given by
σ max =
d=
a2 + ab + b2
F
h
+
×
3(a + b)
W 3
where a = Top width of dam,
b = Bottom width of dam,
F = Force exerted by water,
W = Weight of dam, and
a = Depth of water.
9. A trapezoidal dam is having one of the face vertical. If the reservoir is empty, how will you find the
stresses at the base of the dam.
10. Find an expression for the stresses induced at the base of a trapezoidal dam having water face
inclined.
11. What do you mean by stability of a dam ? What are the different conditions under which a dam is
going to fail ?
12. Prove the statement that the resultant (of the water pressure force and weight of the dam) must lie
within middle third of the base width, in order to avoid tension in the masonry of the dam at the
base.
466
DAMS AND RETAINING WALLS
13. How will you find the minimum bottom width of a dam, if the dam is safe against sliding,
overturning and tensile stress at the base.
14. Define the terms : Retaining wall, dam and angle of repose.
15. A dam of weight W is placed on an inclined plane, having inclination more than the angle of
repose. Prove that the minimum horizontal force applied on the body to keep it in equilibrium
when the body is on the point of moving down the plane is given by
P = W tan (θ – φ)
where θ = Angle of inclination of plane and
φ = Angle of repose.
16. If in the baove question, the body is on the point of moving up the plane then prove that minimum
horizontal force is given by P = W tan (θ + φ).
17. What are the assumptions made in Rankine’s theory of earth pressure ? How is this theory is used
to determine the pressure exerted by the earth on the retaining wall ?
18. What do you mean by plane of rupture ? Prove that the pressure exerted by the earth on the
retaining wall when earth is level upto the top is given by
P=
LM
N
wh2 1 − sin φ
2
1 + sin φ
OP
Q
where W = Weight density of retained earth by the wall,
h = Height of the retaining wall, and
φ = Angle of repose.
19. Defined angle of repose.
20. Distinguish between active and passive earth pressure. Draw the active earth pressure diagram
against a smooth vertical back retaining wall, and hence explain the intensity of pressure at any
depth Z, the centre of pressure and the total pressure.
(B) Numerical Problems
1. A masonry dam of rectangular section, 16 m high and 8 m wide, has water upto a height of 15 m
on its one side. Find :
(i) Pressure force due to water on one metre length of the dam,
(ii) Position of centre of pressure, and
(iii) The point at which the resultant cuts the base. Take density of masonry = 2000 kg/m3.
[Ans. 1103625 N, 5 m, 2.197 m]
2. A masonry dam of rectangular cross-section 12 m high and 5 m wide has water upto the top on its
one side. If the density of masonry is 2300 kg/m3, find : (i) Pressure force due to water per metre
length of dam (ii) Resultant force and the point at which it cuts the base of the dam.
[Ans. (i) 706320 N (ii) 1.527 MN, 2.087 m]
3. For the question 1, find the maximum and minimum stress intensities at the base of the dam.
[Ans. 831181.68 and 203341.68 N/m2]
4. For the question 2, find the maximum and minimum stress intensities at the base of the dam.
[Ans. 948833 and 407321 N/m2]
5. A trapezoidal masonry dam is of 20 m height. The dam is having water upto a depth of 16 m on its
vertical side. The top and bottom width of the dam are 3 m and 9 m respectively. The density of the
masonry is given as 2000 kg/m2. Determine :
(i) the resultant force on the dam per metre length,
(ii) the point where the resultant cuts the base, and
(iii) the maximum and minimum stress intensities at the base.
[Ans. 2.1667 MN ; 6.094 m ; 539.6 kN/m2, – 16382]
467
STRENGTH OF MATERIALS
6. A masonry trapezoidal dam 5 m high, 1 m wide at its top and 3 m wide at its bottom remains
water on its vertical face. Determine the maximum and minimum stresses at the base :
(i) when the reservoir is full and
(ii) when the reservoir is empty. Take the density of masonry as 2000 kg/m3.
[Ans. (i) 14954, – 1621.33 (ii) 12266 ; 1066.66]
7. A masonry dam of trapezoidal section is 12 m high. It has top width of 1 m and bottom width 6 m.
The face exposed to water has a slope of 1 horizontal to 12 vertical. Calculate the maximum and
minimum stresses on the base, when the water level coincides with the top of the dam. Take
density of masonry as 2000 kg/m3.
[Ans. 228965.4, 65334.6 N/m2]
8. A trapezoidal masonry dam having 4.5 m top width, 9.5 m bottom width and 15 m high, is
retaining water upto a height of 12 m. The density of masonry is 2000 kg/m2 and co-efficient of
friction between the dam and soil is 0.6. The allowable stress is 392400 N/m2. Check the stability
of the dam.
[Ans. Dam is safe]
9. A trapezoidal masonry dam having top width 2 m and height 10 m, is retaining water upto a
height of 9 m. The water face of the dam is vertical. The density of masonry is 2200 kg/m3 and coefficient of friction between the dam and soil is 0.6. Find the minimum bottom width of the dam
required.
[Ans. 4.46 m]
10. A masonry retaining wall of trapezoidal section is 8 m high and retains earth which is level upto
the top. The width at the top is 1.5 m and exposed face is vertical. Find the minimum width of the
wall at the bottom in order the tension may not be induced at the base. Masonry and earth has
densities 2300 kg/m3 and 1600 kg/m3 respectively. The angle of repose of the soil is 30.
[Ans. 3.45 m]
11. A masonry retaining wall of trapezoidal section is 12 m high and retains earth which is level upto
the top. The width at the top is 3 m and at the bottom 6 m and exposed face is vertical. Find the
maximum and minimum intensities of normal stress at the base. Take density of earth = 1600
kg/m3 and density of masonry = 2300 kg/m3 and angle of repose of earth = 30°.
[Ans. 318138.3, 87985.9 N/m2]
468
11
CHAPTER
ANALYSIS OF PERFECT
FRAMES
11.1. INTRODUCTION..
A structure made up of several bars (or members) riveted or welded together is known as
frame. If the frame is composed of such members which are just sufficient to keep the frame in
equilibrium, when the frame is supporting an external load, then the frame is known as perfect
frame. Though in actual practice the members are welded or riveted together at their joints, yet
for calculation purposes the joints are assumed to be hinged or pin-joined. In this chapter, we
shall discuss how to determine the forces in the members of a perfect frame, when it is subject to
some external load.
11.2. TYPES OF FRAMES..
The different types of frames are :
(i) Perfect frame, and
C
(ii) Imperfect frame.
Imperfect frame may be a deficient frame or a redundant
frame.
11.2.1. Perfect Frame. The frame which is composed of
such members, which are just sufficient to keep the frame in
equilibrium, when the frame is supporting an external load, is
A
B
known as perfect frame. The simplest perfect frame is a triangle
as shown in Fig. 11.1 which consists three members and three
Fig. 11.1
joints. The three members are : AB, BC and AC whereas the three
joints are A, B and C. This frame can be easily analysed by the condition of equilibrium.
Let the two members CD and BD and a joint D are added to the triangular frame ABC.
Now, we get a frame ABCD as shown in Fig. 11.2 (a). This frame can also be analysed by the
conditions of equilibrium. This frame is also known as perfect frame.
C
D
A
B
C
D
A
(a)
B
E
(b)
Fig. 11.2
469
STRENGTH OF MATERIALS
Suppose we add a set of two members and a joint again, we get a perfect frame as shown in
Fig. 11.2 (b). Hence for a perfect frame, the number of joints and number of members are given
by,
n = 2j – 3
where n = Number of members, and
j = Number of joints.
11.2.2. Imperfect Frame. A frame in which number of members and number of joints
are not given by
n = 2j – 3
is known, an imperfect frame. This means that number of members in an imperfect frame will
be either more or less than (2j – 3).
(i) If the number of members in a frame are less than (2j – 3), then the frame is known
as deficient frame.
(ii) If the number of members in a frame are more than (2j – 3), then the frame is known
as redundant frame.
11.3. ASSUMPTIONS MADE IN FINDING OUT THE FORCES IN A FRAME..
The assumptions made in finding out the forces in a frame are :
(i) The frame is a perfect frame
(ii) The frame carries load at the joints
(iii) All the members are pin-joined.
11.4. REACTIONS OF SUPPORTS OF A FRAME..
The frames are generally supported
(i) on roller support or
(ii) on a hinged support.
If the frame is supported on a roller support, then the line of action of the reaction will
be at right angles to the roller base as shown in Figs. 11.3 and 11.4.
C
E
D
A
B
Hinged
support
F
G
RA
Roller
support
RB
Roller
base
Fig. 11.3
If the frame is supported on a hinged support, then the line of action of the reaction will
depend upon the load system on the frame.
470
ANALYSIS OF PERFECT FRAMES
The reactions at the supports of a frame are determined by the conditions of equilibrium. The external load on the frame and the reactions at the supports must form a system of
equilibrium.
C
D
B
A
E
VA
RB
Roller base
Fig. 11.4
11.5. ANALYSIS OF A FRAME..
Analysis of a frame consists of :
(i) Determinations of the reactions at the supports and
(ii) Determination of the forces in the members of the frame.
The reactions are determined by the condition that the applied load system and the
induced reactions at the supports form a system in equilibrium.
The forces in the members of the frame are determined by the condition that every joint
should be in equilibrium and so, the forces acting at every joint should form a system in
equilibrium.
A frame is analysed by the following methods :
(i) Method of joints,
(ii) Method of sections, and
(iii) Graphical method.
11.5.1. Method of Joints. In this method, after determining the reactions at the supports, the equilibrium of every joint is considered. This means the sum of all the vertical forces
as well as the horizontal forces acting on a joint is equated to zero. The joint should be selected
in such a way that at any time there are only two members, in which the forces are unknown.
The force in the member will be compressive if the member pushes the joint to which it is
connected whereas the force in the member will be tensile if the member pulls the joint to
which it is connected.
Problem 11.1. Find the forces in the members AB, AC and BC of the truss shown in
Fig. 11.5.
20 kN
Sol. First determine the reactions RB and RC. The
A
line of action of load of 20 kN acting at A is vertical. This
load is at a distance of AB × cos 60° from the point B. Now
let us find the distance AB.
The triangle ABC is a right-angled triangle with an60°
30°
B
C
gle BAC = 90°. Hence AB will be equal to BC × cos 60°.
∴
AB = 5 × cos 60° = 5 × 21 = 2.5 m
Now the distance of line of action of 20 kN from B is
AB × cos 60° or 2.5 ×
1
2
5m
RB
Fig. 11.5
RC
= 1.25 m.
471
STRENGTH OF MATERIALS
and
Taking the moments about B, we get
RC × 5 = 20 × 1.25 = 25
25
= 5 kN
∴
RC =
5
RB = Total load – RC = 20 – 5 = 15 kN
Now let us consider the equilibrium of the various joints.
Joint B
Let
A
1
F1
B
60°
F2
C
2
R = 15 kN
B
F1 = Force in member AB
Fig. 11.6
F2 = Force in member BC
Let the force F1 is acting towards the joint B and the force F2 is acting away* from the joint
B as shown in Fig. 11.6. (The reaction RB is acting vertically up. The force F2 is horizontal. The
reaction RB will be balanced by the vertical component of F1. The vertical component of F1 must
act downwards to balance RB. Hence F1 must act towards the joint B so that its vertical component
is downward. Now the horizontal component of F1 is towards the joint B. Hence force F2 must act
away from the joint to balance the horizontal component of F1).
Resolving the forces acting on the joint B, vertically
F1 sin 60° = 15
15
15
=
∴
F1 =
= 17.32 kN (Compressive)
sin 60° 0.866
As F1 is pushing the joint B, hence this force will be compressive. Now resolving the forces
horizontally, we get
1
F2 = F1 cos 60° = 17.32 × = 8.66 kN (tensile)
2
As F2 is pulling the joint B, hence this force will be tensile.
Joint C
Let
F3 = Force in the member AC
F2 = Force in the member BC
The force F2 has already been calculated in magnitude and
A
3
direction. We have seen that force F2 is tensile and hence it will
F3
pull the joint C. Hence it must act away from the joint C as shown
in Fig. 11.7.
30°
C
Resolving forces vertically, we get
B
2
F2
F3 sin 30° = 5 kN
5
RC = 5 kN
∴
F3 =
= 10 kN (Compressive)
sin 30°
Fig. 11.7
As the force F3 is pushing the joint C, hence it will be
compressive. Ans.
Problem 11.2. A truss of span 7.5 m carries a point load of 1 kN at joint D as shown in
Fig. 11.8. Find the reactions and forces in the members of the truss.
*The direction of F2 can also be taken towards the joint B. Actually when we consider the equilibrium of the joint B, if the magnitude of F1 and F2 comes out to be positive then the assumed direction
of F1 and F2 are correct. But if any one of them is having a negative magnitude then the assumed
direction of that force is wrong. Correct direction then will be the reverse of the assumed direction.
472
ANALYSIS OF PERFECT FRAMES
Sol. Let us first determine the reactions RA and RB
Taking moments about A, we get RB × 7.5 = 5 × 1
5
2
= = 0.667 kN
∴
RB =
7.5 3
C
1
4
3
A
30°
2
60°
30°
D
5m
B
5
1 kN
7.5 m
RA
RB
Fig. 11.8
∴
RA = Total load – RB
∴
= 1 – 0.667 = 0.333 kN
Now consider the equilibrium of the various joints.
Joint A
Let
F1 = Force in member AC
F2 = Force in member AD.
Let the force F1 is acting towards the joint A and F2 is acting
C
1
away from the joint A as shown in Fig. 11.9.
F1
Resolving the forces vertically, we get
30°
F1 sin 30° = RA
A
D
F2 2
RA
0.333
=
or
F1 =
sin 30°
0.5
RA = .333 kN
= 0.666 kN (Compressive)
Resolving the forces horizontally, we get
Fig. 11.9
F2 = F1 × cos 30°
= 0.666 × 0.866 = 0.5767 kN (Tensile)
C
1
Joint B
F4
Let
F4 = Force in member BC
30°
F5 = Force in member BD
B
D
F5
5
Let the direction of F4 and F5 are assumed as shown in
Fig. 11.10.
RB = 0.667
Resolving the forces vertically, we get
F4 sin 30° = RB = 0.667
Fig. 11.10
0.667
or
F4 =
= 1.334 kN (Compressive)
sin 30°
Resolving the forces horizontally, we get
F5 = F4 cos 30° = 1.334 × 0.866 = 1.155 kN (Tensile)
473
STRENGTH OF MATERIALS
Joint D
Let F3 = Force in member CD. The forces F2 and F5 have been already calculated in
magnitude and direction. The forces F2 and F5 are tensile and hence they will be pulling the joint
D as shown in Fig. 11.11. Let the direction* of F3 is assumed as shown in Fig. 11.11.
Resolving the forces vertically, we get
C
F3 sin 60° = 1
3
1
1
=
∴
F3 =
F3
sin 60° 0.866
= 1.1547 kN (Tensile)
60° D
Hence the forces in the members are :
F5 5 B
F1 = 0.666 kN (Compressive)
A
2 F2
F2 = 0.5767 kN (Tensile)
1 kN
F3 = 1.1547 kN (Tensile)
Fig. 11.11
F4 = 1.334 kN (Compressive)
F5 = 1.155 kN (Tensile). Ans.
Problem 11.3. A truss of span 5 m is loaded as shown in Fig. 11.12. Find the reactions
and forces in the members of the truss.
Sol. Let us first determine the reactions RA and RB.
10 kN
D
7
12 kN
1
E
6
3
5
60°
A
60°
2
30°
60°
C
B
4
5m
RA
RB
Fig. 11.12
Triangle ABD is a right-angled triangle having angle ADB = 90°.
∴
AD = AB cos 60° = 5 × 0.5 = 2.5 m
The distance of the line of action of the vertical load 10 kN from point A will be AD cos
60°
or
2.5 × 0.5 = 1.25 m.
From triangle ACD, we have AC = AD = 2.5 m
∴
BC = 5 – 2.5 = 2.5 m
In right-angled triangle CEB, we have
BE = BC cos 30° = 2.5 ×
3
2
*The horizontal force F5 is more than F2. Hence the horizontal component of F3 must be in the
direction of F2. This is only possible if F3 is acting away from D.
474
ANALYSIS OF PERFECT FRAMES
The distance of the line of action of the vertical load of 12 kN from point B will be
BE × cos 30°
or
BE ×
F
GH
I
JK
3
3
3
= 2.5 ×
×
= 1.875 m.
2
2
2
∴ The distance of the line of action of the load of 12 kN from point A will be (5 – 1.875) =
3.125 m.
Now taking the moments about A, we get
RB × 5 = 10 × 1.25 + 12 × 3.125 = 50
50
= 10 kN
∴
RB =
5
∴
RA = Total load – RB = (10 + 12) – 10 = 12 kN
Now consider the equilibrium of the various joints.
Joint A
Let
F1 = Force in member AD, and
D
F2 = Force in member AC
1
Let the directions of F1 and F2 are assumed as shown in
Fig. 11.13.
F1
Resolving the forces vertically,
F1 sin 60° = 12
60° F2
A
C
12
2
∴
F1 =
sin 60°
RA = 12 kN
= 13.856 kN (Compressive)
Fig. 11.13
Resolving the forces horizontally,
F2 = F1 cos 60° = 13.856 × 0.5
= 6.928 kN (Tensile)
Now consider the joint B.
Joint B
Let
F3 = Force in member BE, and
F4 = Force in member BC
E
3
Let the directions of F3 and F4 are assumed as shown in
F3
Fig. 11.14.
Resolving the forces vertically, we get
30°
B
C
F3 sin 30° = 10
4
F4
10
∴
F3 =
= 20 kN
(Compressive)
RB = 10 kN
sin 30
Now resolving the forces horizontally, we get
Fig. 11.14
F4 = F3 cos 30° = 20 × 0.866
= 17.32 kN (Tensile)
Now consider the joint C.
Joint C
Let
F5 = Force in member CE
F6 = Force in member CD
475
STRENGTH OF MATERIALS
Let the directions of F5 and F6 are assumed as shown
in Fig. 11.15.
The forces F2 and F4 are already known in magnitude
and directions. They are tensile and hence will be pulling
the joint C as shown in Fig. 11.15.
Resolving forces vertically, we get
F6 sin 60° + F5 sin 60° = 0
or
F6 = – F5
...(i)
Resolving forces, horizontally, we get
F2 – F6 cos 60° = F4 – F5 cos 60°
F6
F
= 17.32 – 5
2
2
− F6 + F5
or
= 17.32 – 6.928 = 10.392
2
or
– F6 + F5 = 10.392 × 2 = 20.784
or
F5 + F5 = 20.784
20.784
= 10.392 kN
∴
F5 =
2
and
F6 = – F5 = – 10.392 kN
The magnitude of F6 is –ve, hence the assumed
direction of F6 is wrong. The correct direction F6 will be as
shown in Fig. 11.15 (a).
∴
F5 = 10.392
(Compressive)
and
F6 = 10.392
(Tensile)
Now consider the joint E.
or
D
E
6
60°
A
2
60°
C
F2
5
F5
F6
F4
4
B
Fig. 11.15
6.928 –
(∵ – F6 = F5)
D
E
F5
F6
60°
60°
A
F2
F4
B
Joint E
Let
F7 = Force in member ED
Fig. 11.15 (a)
Let F7 is acting as shown in Fig. 11.16.
The forces F3 and F5 are known in magnitude and
12 kN
directions. They are compressive hence they will be push7
D
ing the joint E as shown in Fig. 11.16.
60°
F7
Resolving the forces along BED, we get
E
3
F7 + 12 cos 60° = F3
30°
F5
or
F7 = F3 – 12 × 0.5
F3
5
= 20 – 6 = 14 kN
30°
60°
B
C
A
(Compressive)
As F7 is positive hence the assumed direction of
Fig. 11.16
F7 is correct. Ans.
Problem 11.4. A truss of span 9 m is loaded as shown in Fig. 11.17. Find the reactions
and forces in the member of the truss.
Sol. Let us first calculate the reactions RA and RB.
Taking moments about A, we get
RB × 9 = 9 × 3 + 12 × 6 = 27 + 72 = 99
C
476
ANALYSIS OF PERFECT FRAMES
C
D
E
F
4m
A
G
H
9 kN
3m
12 kN
3m
RA
B
9m
3m
RB
Fig. 11.17
99
= 11 kN
9
and
RA = Total load – RB = (9 + 12) – 11 = 10 kN
In this problem, there are some members in which force is zero.
These members are obtained directly as given below :
“If three forces act at a joint and two of them are along the same straight line, then for the
equilibrium of the joint, the third force should be equal to zero.”
1. Three forces are acting at the point A (i.e., RA, FAC and FAG), two of which (i.e., RA,
RAC) are along the same straight line. Hence the third force (i.e., RAG) is zero.
2. Similarly, three forces are acting at the joint B (i.e., RB,
C
FBF and FBH), two of which (i.e., RB and FBH) are along the same
straight line. Hence the third force FBH should be zero.
3. At the joint E also, three forces (i.e., FED, FEF and FEH) are
acting, two of which (i.e., FED and FEF) are along the same straight
A
line. Hence the third force FEH must be zero.
G
Now the equilibrium of various joints can be considered.
Joint A [See Fig. 11.17 (a)]
FAG = Force in member AG = 0
RA = 10 kN
FAC = Force in member AC
Fig. 11.17 (a)
= RA = 10 kN (Compressive)
FCD
C
D
Now consider the equilibrium of joint C.
Joint C [See Fig. 11.17 (b)]
θ
∴
Let
RB =
FCG
FCD = Force in member CD
FCG = Force in member CG
4m
10 kN
FAC = 10 kN (Compressive)
Let the directions of FCG and FCD are assumed as shown in Fig.
11.17 (b).
Resolving the forces vertically, we get
FCG cos θ = 10
A
G
3m
Fig. 11.17 (b)
477
STRENGTH OF MATERIALS
∴
FCG =
10
cos θ
AG 4
=
(∵ CG = 3 2 + 4 2 = 5)
CG 5
5
10
∴
FCG =
= 10 × = 12.5 kN (Tensile)
4
(4/5)
Resolving forces horizontally, we get
FCD = FCG sin θ
3
= 12.5 × = 7.5 kN (Compressive)
5
Now consider the equilibrium of joint G.
Joint G
D
The force in member CG is 12.5 kN (Tensile).
C
Hence at the joint G, this force will be pulling the joint G
12.5 kN
as shown in Fig. 11.17 (c).
FGD
Resolving the forces vertically, we get
12.5 cos θ + FGD = 9
∴
FGD = 9 – 12.5 cos θ
A
B
G FGH
4
4
9 kN
∵ sin θ =
= 9 – 12.5 ×
5
5
Fig. 11.17 (c)
= 9 – 10 = – 1 kN.
As the magnitude of FGD is negative, hence its assumed
D
direction is wrong. The correct direction will be as shown in
C
Fig. 11.17 (d).
Then,
FGD = 1 kN (Compressive)
12.5 kN
FGD
Resolving the forces horizontally, we get
12.5 sin θ = FGH
θ
3
3
A
B
G
FGH
or
FGH = 12.5 ×
∵ cos θ =
5
5
9 kN
= 7.5 kN (Tensile)
Fig. 11.17 (d)
Now consider the equilibrium of joint D.
Joint D
The forces in the members CD and GD have been already calculated. They are 7.5 kN and
1 kN respectively. Both are compressive.
Let
FDH = Force in member DH, and
E
C 7.5 kN
D
FDE = Force in member DE
θ
Resolving the forces vertically, we get
FDH cos θ = 1 kN
1 kN
1
1
=
∴
FDH =
H
cos θ (4/5)
G
5
= = 1.25 kN (Tensile)
Fig. 11.17 (e)
4
But
478
cos θ =
FG
H
IJ
K
FG
H
IJ
K
ANALYSIS OF PERFECT FRAMES
Resolving the forces horizontally, we get
7.5 + FDH sin θ = FDE
FG
H
3
3
∵ sin θ =
5
5
= 7.5 + 0.75 = 8.25 kN (Compressive)
Now consider the equilibrium of joint E.
Joint E
As shown in Fig. 11.17 (f), at joint E three forces are
D 8.25 kN
F
E 8.25 kN
acting. The forces i.e., FDE and FEF are in the same straight
line.
Hence force FEH must be zero.
Force in EF, i.e.,
FEF = FDE
= 8.25 kN (Compressive)
Now consider the joint H.
H
Joint H
Fig. 11.17 (f )
It is already shown that forces in the members EH and
BH are zero.
E
Also the forces in the member GH is 7.5 kN tensile
F
D
and in the member DH is 1.25 kN tensile.
1.25 kN
Let FHF is the force in the member HF.
Resolving forces vertically, we get
1.25 cos θ + FHF cos θ = 12
θ θ
or
FDE = 7.5 + 1.25 ×
FG
H
4
4
4
∵ cos θ =
+ FHF × = 12
5
5
5
1.0 + 0.8 FHF = 12
12 − 1.0 11
=
= 13.75
or
FHF =
0.8
0.8
(Tensile)
Now consider the joint B.
Joint B
See Fig. 11.17 (h).
or
1.25 ×
IJ
K
G
7.5 kN
B
H
12 kN
Fig. 11.17 (g)
F
11 kN
The force in member BF
= 11 kN (Compressive)
Now the forces in each member are known.
They are shown in Fig. 11.18. Also these forces are shown
in a tabular form.
B
H
RB = 11 kN
Fig. 11.17 (h)
479
IJ
K
STRENGTH OF MATERIALS
7.5 kN
8.25 kN
D
θ
8.25 kN
E
F
θ
1 kN
kN
kN
5
1.2
.5
12
10 kN
θ
7.5 kN
A
N
θ
13
.75
k
C
θ
B
H
G
9 kN
3m
RA = 0 kN
11 kN
12 kN
3m
3m
RB = 11 kN
Fig. 11.18
Member
Force in member
AC
AG
CG
CD
DG
DE
DH
GH
EH
EF
HB
HF
BF
10 kN (Comp.)
0
12.5 kN (Tens.)
7.5 kN (Comp.)
10 kN (Comp.)
8.25 kN (Comp.)
1.25 kN (Tens.)
7.5 kN (Tens.)
0
8.25 kN (Comp.)
0
13.75 kN (Tens.)
11 kN (Comp.)
Problem 11.5. A plane truss is loaded and supported as shown in Fig. 11.19. Determine
the nature and magnitude of the forces in the members 1, 2 and 3.
Sol. First calculate the reactions RA and RB
Taking moments about A, we get
RB × 4 = 1 × 1000
1000
= 250 N
∴
RB =
4
∴
RA = 1000 – 250 = 750 N
From figure, we know that
CH 2.25
=
= 0.75
tan θ =
AH
4
3
AH
=
cos θ =
∵ AC = 3 2 + 2.25 2 = 3.75
CH 3.75
= 0.8
CH 2.25
=
= 0.6
and
sin θ =
AC 3.75
480
FH
IK
ANALYSIS OF PERFECT FRAMES
C
G
1000 N
1
2.25 m
D
A
2
θ
E
1m
B
3
θ
F
1m
H
1m
1m
4m
RA
RB
Fig. 11.19
Consider the equilibrium of joint A.
Joint A [See Fig. 11.19 (a)]
D
Resolving the forces vertically,
FAD sin θ = 750
750 750
θ
=
∴
FAD =
A
sin θ 0.6
E
= 1250 N (Compressive)
RA = 750 kN
Resolving the forces horizontally, we get
Fig. 11.19 (a)
FAE = FAD cos θ = 1250 × 0.8
= 1000 N (Tensile)
Now consider joint E.
Joint E
Three forces, i.e., FAE, FEF and FED are acting at the joint E. Two of the forces, i.e., FAE and
FEF are in the same straight line. Hence the third force, i.e., FED should be zero
and
FEF = FAE = 1000 N (Tensile)
Now consider the equilibrium of joint D.
Joint D
Let
F1 = Force in member DG
FDF = Force in member DF.
1000 N
Let us assume their directions as shown in Fig. 11.19 (b).
1
G
The forces in the member AD and DE are 1250 N
D
(Compressive) and 0 respectively.
N
50
Resolving forces vertically, we get
12
1250 sin θ + F1 sin θ + FDF sin θ = 1000
θ
θ
A
E
F
or
1250 × 0.6 + F1 × 0.6 + FDF × 0.6 = 1000
Fig. 11.19 (b)
(∵ sin θ = 0.6)
or
or
1000
= 1666.66
0.6
F1 + FDF = 1666.66 – 1250 = 416.66
1250 + F1 + FDF =
...(i)
481
STRENGTH OF MATERIALS
or
Resolving the forces horizontally, we get
1250 cos θ + F1 cos θ = FDF × cos θ
1250 + F1 = FDF or F1 – FDF = – 1250
Adding (i) and (ii), we get
2F1 = 416.66 – 1250 = – 833.34
...(ii)
833.34
= – 416.67 N
2
1000 N
Substituting the value of F1 in equation (i), we get
7N
6.6
41
– 416.67 + FDF = 416.66
D
or
FDF = 416.66 + 416.67
83
3.3
N
= 833.33 N (Comp.)
3N
50
2
1
The magnitude of F1 is negative. Hence its assumed
direction is wrong. The correct direction of F1 is shown in
A
E
F
Fig. 11.19 (c).
Fig. 11.19 (c)
∴
F1= 416.67 N (Compressive). Ans.
To find the forces F2 and F3, consider the joint F.
Joint F
The forces in the members DF and EF are already
G
known. They are :
2
FDF = 833.33 N (Compressive)
D
83
FEF = 1000 N (Tensile).
3.3
3N
These forces are acting at the joint F as shown in
1
Fig. 11.19 (d).
θ
E
F
H
Let
F2 = Force in member FG, and
3
1000 N
F3 = Force in member FH
Fig. 11.19 (d)
Resolving forces vertically, we get
833.33 sin θ = F2
or
F2 = 833.33 × 0.6
(∵ sin θ = 0.6)
= 499.998 N ~
500
N
(Tensile).
Ans.
−
Resolving forces horizontally, we get
F3 + 833.33 cos θ = 1000
or
F3 = 1000 – 833.33 × 0.8
(∵ cos θ = 0.8)
= 333.336 N (Tensile). Ans.
11.5.1.1 Method of Joints Applied to Cantilever Trusses. In case of cantilever trusses,
it is not necessary to determine the support reactions. The forces in the members of cantilever
truss can be obtained by starting the calculations from the free end of the cantilever.
Problem 11.6. Determine the forces in all the members of a cantilever truss shown in
Fig. 11.20.
Sol. Here the calculations can be started from end C. Hence consider the equilibrium of
the joint C.
Joint C
Let
FCD = Force in member CD, and
FCA = Force in member CA.
∴
482
F1 = –
ANALYSIS OF PERFECT FRAMES
Their assumed directions are shown in Fig. 11.20.
Resolving the force vertically, we get
FCD × sin 60° = 1000
∴
FCD =
60°
A
1000 N
Fig. 11.20
D
B
30° 30°
.7
54
N
1154.7 cos 30°
= 1154.7 N
cos 30°
A
Resolving the forces horizontally, we get
FBD = FAD sin 30° + FDC sin 30°
= 1154.7 × 0.5 + 1154.7 × 0.5 = 1154.7 N (Tensile)
Now the forces are shown in a tabular form below :
∴
C
Fig. 11.20 (a)
Member
Force in the member
AC
CD
AD
577.35 N
1154.7 N
1154.7 N
Compressive
Tensile
Compressive
BD
1154.7 N
Tensile
Problem 11.7. Determine the forces in all the members of a cantilever truss shown in Fig. 11.21.
Sol. Start the calculations from joint C.
From triangle ACE, we have
AS 3
=
tan θ =
AC 4
EC =
2
3 +4
2
Nature of force
1000 N
1000 N
2m
2m
A
B
3m
C
D
=5
AC 4
= = 0.8
CE 5
AE 3
= = 0.6.
sin θ =
CE 5
cos θ =
C
4m
(Compressive)
Also
60°
11
FAD =
D
1000
1000
=
= 1154.7 N (Tensile)
sin 60° 0.866
Resolving the forces horizontally, we get
FCA = FCD × cos 60°
= 1154.7 × 0.5
= 577.35 N (Compressive)
Now consider the equilibrium of the joint D.
Joint D
[See Fig. 11.20 (a)]
The force FCD = 1154.7 N (tensile) is already calculated.
Let
FAD = Force in member AD, and
FBD = Force in member BD
Their assumed directions are shown in Fig. 11.20 (a).
Resolving the forces vertically, we get
FAD cos 30° = 1154.7 cos 30°
∴
B
E
Fig. 11.21
483
STRENGTH OF MATERIALS
Joint C
The direction of forces at the joint C are shown in Fig. 11.21.
Resolving the forces vertically, we get
FCD sin θ = 1000
1000 1000
=
∴
FCD =
= 1666.66 N (Compressive)
sin θ
0.6
Resolving the forces horizontally, we get
FCB = FCD × cos θ = 1666.66 × 0.8 = 1333.33 N (Tensile)
Now consider the equilibrium of joint B.
Joint B
Resolving vertically, we get
FBD = 1000 N (Compressive)
FBA = FCB = 1333.33 (Tensile)
Now consider the joint D.
Joint D
The forces in member CD and BD have already been
A
C
calculated. They are 1666.66 N and 1000 N respectively as shown
θ
θ
in Fig. 11.21 (a).
1000 N
N
Let
FDA = Force in member DA, and
66
6.
6
θ
FDE = Force in member DE
16
D
θ
Resolving forces vertically, we get
1000 + 1666.66 sin θ = FAD sin θ + FED sin θ
or
1000 + 1666.66 × 0.6 = FAD × 0.6 + FED × 0.6
θ
1000
E
+ 1666.66 = 3333.32
...(i)
or
FAD + FED =
0.6
Fig. 11.21 (a)
Resolving forces horizontally, we get
1666.66 cos θ + FAD cos θ = FED cos θ
or
1666.66 + FAD = FED or FED – FAD = 1666.66
...(ii)
Adding equations (i) and (ii), we get
2FED = 3333.32 + 1666.66 = 4999.98
4999.98
= 2499.99 ~
∴
FED =
− 2500 N (Compressive)
2
Substituting this value in equation (i), we get
FAD + 2500 = 3333.32
∴
FAD = 3333.32 – 2500 = 833.32 N (Tensile)
Now the forces are shown in a tabular form below :
484
Member
Force in the member
AB
BC
CD
DE
AD
BD
1333.33 N
1333.33 N
1666.66 N
2500 N
833.32 N
1000 N
Nature of force
Tensile
Tensile
Compressive
Compressive
Tensile
Compressive
ANALYSIS OF PERFECT FRAMES
11.5.1.2 Method of Joints Applied to Trusses Carrying Horizontal Loads. If a
truss carries horizontal loads (with or without vertical loads), hinged at one end and supported on
rollers at the other end, then the support reaction at the roller supported end will be normal,
whereas the support reactions at the hinged end will consists of : (i) horizontal reaction and (ii)
vertical reaction.
The horizontal reaction will be obtained by adding algebraically all the horizontal loads ;
whereas the vertical reaction will be obtained by subtracting the roller support reaction from
the total vertical loads. Now the forces in the members of the truss can be determined.
Problem 11.8. Determine the forces in the truss shown in Fig. 11.22 which carries a
horizontal load of 12 kN and a vertical load of 18 kN.
Sol. The truss is supported on rollers at B and hence the reaction at B must be normal
to the roller base i.e., the reaction at B, in this case, should be vertical.
At the end A, the truss is hinged and hence the support reactions at the hinged end A
will consists of a horizontal reaction HA and a vertical reaction RA.
D
12 kN
1.5 m
HA
A
θ
θ
C
B
Roller
18 kN
2m
2m
RA
RB
Fig. 11.22
Taking moments of all forces at A, we get
RB × 4 = 18 × 2 + 12 × 1.5 = 36 + 18 = 54
54
= 13.5 kN (↑)
4
∴
RA = Total vertical load – RB = 18 – 13.5 = 4.5 kN (↑)
HA = Sum of all horizontal loads = 12 kN (← )
Now the forces in the members can be calculated.
∴
and
In triangle BCD,
∴
RB =
BD =
cos θ =
BC 2 + CD 2 = 2 2 + 1.5 2 = 2.5 m
BC
2
=
= 0.8
BD 2.5
CD 1.5
=
= 0.6
BD 2.5
Let us first consider the equilibrium of joint A.
sin θ =
485
STRENGTH OF MATERIALS
Joint A
The reactions RA and HA are known in magnitude and
direction. Let the directions of the forces in the members AC and
AD are as shown in Fig. 11.22 (a).
Resolving the forces vertically, we get
FAD sin θ = RA
D
FAD
A
RA
4.5
=
or
FAD =
= 7.5 kN
(Compressive)
sin θ 0.6
Resolving the forces horizontally, we get
FAC = HA + FAD cos θ
= 12 + 7.5 × 0.8 = 18 kN (Tensile)
Now consider the joint C.
Joint C
At the joint C, the force O in member CA and vertical load
18 kN are known in magnitude and directions. For equilibrium
of the joint C.
FBC = FCA = 18 kN (Tensile)
FCD = 18 kN (Tensile)
Now consider the joint B.
Joint B
At the joint B, RB and force FBC are known in magnitude and
direction.
Let FBD is the force in member BD.
Resolving the forces vertically, we get
FBD × sin θ = RB
RB
13.5
=
∴
FBD =
0.6
sin θ
= 22.5 kN (Compressive)
Now the forces are shown in a tabular form below :
Member
Force in the member
HA = 12 kN
C
FAC
RA = 4.5 kN
Fig. 11.22 (a)
D
A
18 kN
C
B
18 kN
Fig. 11.22 (b)
D
θ
C
18 kN
13.5 kN
B
RB
Fig. 11.22 (c)
Nature of force
AC
AD
CD
CB
18 kN
7.5 kN
18 kN
18 kN
Tensile
Compressive
Tensile
Tensile
BD
2.5 kN
Compressive
Problem 11.9. Determine the forces in the truss shown in Fig. 11.23 which is subjected
to horizontal and vertical loads. Mention the nature of forces in each case.
486
ANALYSIS OF PERFECT FRAMES
C
D
8 kN
E
θ
1.5 m
HA
A
θ
θ
θ
G
3 kN
4m
θ
θ
θ
F
B
Roller
6 kN
4m
4m
RA
RB
Fig. 11.23
Sol. The truss is supported on rollers at B and hence RB will be vertical. The truss is
hinged at A and hence the support reactions at A will consists of a horizontal reaction HA and
a vertical reaction RA.
C
Taking moment about A, we get
RB × 12 = 8 × 1.5 + 3 × 4 + 6 × 8
= 72
1.5 m
72
= 6 kN (↑)
∴
RB =
12
and
RA = Total vertical loads – RB
A
C*
F
2m
= (3 + 6) – 6
4m
= 3 kN (↑)
and
HA = Sum of all horizontal loads
Fig. 11.23 (a)
= 8 kN (← )
In the triangle ACC*, AC =
AC* 2 + CC * 2 = 2 2 + 1.5 2 = 2.5
AC*
CC* 1.5
2
=
=
= 0.8 and sin θ =
= 0.6
AC 2.5
AC 2.5
Now the forces in the members can be calculated. Consider the joint A.
Joint A
The reactions RA and HA are known in magnitude and
FCA
direction. Let the directions of the forces FCA and FFA are as
shown in Fig. 11.23 (b).
θ
A
Resolving the forces vertically, we get
HA = 8 kN
FFA
FCA × sin θ = 3 kN
∴
cos θ =
3 kN
C
E
3
3
=
= 5 kN
(Compressive)
∴
FCA =
RA
sin θ 0.6
Resolving the forces horizontally, we get
Fig. 11.23 (b)
FFA = FCA cos θ + HA
= 5 × 0.8 + 8 = 12 kN (Tensile)
Now consider joint C.
Joint C
The force FCA is known in magnitude and direction. The assumed directions of the forces
FCD and FCF are shown in Fig. 11.23 (c).
487
STRENGTH OF MATERIALS
F
D
F
Resolving forces vertically, we get
FCD
FCA sin θ = FCF sin θ
C
θ
∴
FCA = FCF = 5 kN (∵ FCA = 5 kN)
D
θ
5
∴
FCF = 5 kN (Tensile)
FCF
=
A
FC
Resolving forces horizontally, we get
FCD = FCA cos θ + FCF cos θ
A
F
= 5 × 0.8 + 5 × 0.8 = 8 kN
Fig. 11.23 (c)
(Compressive)
Now consider the joint F.
Joint F
The forces FFA and FFC are known in magnitude and directions. The assumed directions of
the forces FDF and FGF are shown in Fig. 11.23 (d).
Resolving the forces vertically, we get
C
D
5 × sin θ + FDF sin θ = 3
FCF = 5
5 sin θ + 3
or
FDF = –
sin θ
3
3
=–5+
=–5+
=–5+5=0
A
G
F
F
=
12
FGF
sin θ
0.6
FA
Resolving the forces horizontally, we get
3 kN
12 + 5 cos θ = FGF + FDF cos θ
Fig. 11.23 (d)
or
12 + 5 × 0.8 = FGF + 0 or 12 + 4 = FGF
∴
FGF = 12 + 4 = 16 kN (Tensile)
Now consider the joint D.
Joint D
The forces FDC and FFD are known in magnitude and
direction. The assumed directions of FDG and FDE are shown
in Fig. 11.23 (e).
Resolving vertically, we get
FDG sin θ = FDF × sin θ = 0
∴
FDG = 0
Resolving forces horizontally, we get
FDE = FCD = 8 kN
∴
FDE = 8 kN (Compressive)
Now consider the joint G.
Joint G
The forces FDG and FFG are known in magnitude and
direction. The assumed directions of FGE and FGB are shown
in Fig. 11.23 (f).
Resolving the forces vertically, we get
FGE sin θ = FDG sin θ + 6 = 6
6
6
=
FGE =
= 10 kN (Tensile)
sin θ 0.6
or
488
FDE
8 kN D
θ
θ
C
E
FDG
F
G
Fig. 11.23 (e)
E
D
FGE
θ
F
16 kN
θ
G
FGB
6 kN
Fig. 11.23 (f)
B
ANALYSIS OF PERFECT FRAMES
Resolving forces horizontally, we get
FGB = 16 – FGE cos θ
= 16 – 10 × 0.8 = 8 kN (Tensile)
Now consider the joint E.
Joint E
The forces FGE and FDE are known in magnitude and directions. Let FBE is acting in a
direction as shown in Fig. 11.23 (g).
Resolving forces vertically, we get
D
FGE sin θ = FBE sin θ
∴
FBE = FGE = 10
(Compressive)
∴
FBE = 10 kN
G
If we have calculated the forces in member BE and BG,
considering joint B, we would have got the same results.
Now the forces in each member are shown in Fig. 11.23 (h).
C
5 kN
8 kN
D
5 kN
8 kN
E
8 kN
10 kN
8 kN
E
FBE
B
Fig. 11.23 (g)
8 kN
10 kN
10 kN
1.5 m
8 kN
HA
12 kN
16 kN
F
A
G
3 kN
3 kN
4m
8 kN
B
6 kN
4m
RA
6 kN
4m
RB
Fig. 11.23 (h)
11.5.1.3. Method of Joints Applied to Trusses Carrying Inclined Loads. If a truss
carries inclined loads, hinged at one end and supported on rollers at the other end, then the
support reaction at the roller supported end will be normal, whereas the support reactions at the
hinged end will consists of :
(i) Horizontal reaction and
(ii) Vertical reaction.
The inclined loads are resolved into horizontal and vertical components.
The horizontal reaction will be obtained by adding algebraically all the horizontal components of the inclined loads ; whereas the vertical reaction will be obtained by subtracting the
roller support reaction from the total vertical components of the inclined loads. Now the forces
in the members of truss can be determined.
Problem 11.10. Determine the forces in the truss shown in Fig. 11.24 which is subjected
to inclined loads.
Sol. The truss is supported on roller at B and hence RB will be vertical.
The truss is hinged at A and hence the support reactions at A will consists of a horizontal reaction HA and a vertical reaction RA.
Now length
AC = 4 × cos 30 = 4 × 0.866 = 3.464 m
and
length
AD = 2 × AC = 2 × 3.464 = 6.928 m
489
STRENGTH OF MATERIALS
1 kN
D
2 kN
C
1 kN
60°
A
G
60°
30°
60°
60°
30°
F
E
HA
B
1 kN
4m
4m
RA
4m
RB
Fig. 11.24
Now taking moments about A, we get
RB × 12 = 2 × AC + 1 × AD + 1 × AE
= 2 × 3.464 + 1 × 6.928 + 1 × 4 = 17.856
17.856
= 1.49 kN
∴
RB =
12
Total vertical components of inclined loads
= (1 + 2 + 1) × sin 60°
= 4 × 0.866 = 3.464 kN
Total horizontal components of inclined loads
= (1 + 2 + 1) cos 60° = 4 × 0.5 = 2 kN
Now RA = Vertical components of inclined loads + 1.0 – RB
= 4.464 – 1.49 = 2.974 kN (↑)
and
HA = Sum of all horizontal components = 2 kN
Now the forces in the members can be calculated.
Consider the equilibrium of joint A.
Joint A
1 kN
Let
FAE = Force in member AE
FAC
and
FAC = Force in member AC
Their directions are assumed as shown in Fig. 11.24 (a).
60°
30°
Resolving the forces vertically, we get
H
A
FAE
2 kN
FAC × sin 30° + 1 × sin 60° = 2.974
2.974 kN
or
FAC × 0.5 + 0.866 = 2.974
RA
2.974 − 0.866
∴
FAC =
Fig. 11.24 (a)
0.5
= 4.216 kN (Compressive)
Resolving the forces horizontally, we get
FAE = 2 + FAC cos 30° – 1 × cos 60°
= 2 + 4.216 × 0.866 – 0.5 = 5.15 kN (Tensile)
490
C
E
ANALYSIS OF PERFECT FRAMES
Now consider the joint C.
Joint C
From Fig. 11.24 (b), we have
FCD = FAC = 4.216
and
FCE = 2 kN
Now consider joint E.
2 kN
D
FCD
C
(Compressive)
(Compressive)
16
4.2
kN
FCE
A
Joint E [See Fig. 11.24 (c)]
Resolving forces vertically, we get
1 + 2 × sin 60° = FED × sin 60°
1
or
FED = 2 +
= 3.155
sin 60°
(Tensile)
Resolving forces horizontally, we get
5.15 – 2 × cos 60° – FED cos 60° – FEF = 0
1
1
A
or
5.15 – 2 × – 3.15 × – FEF = 0
2
2
FEF = 5.15 – 1 – 1.57 = 2.58 kN
(Tensile)
At the joint G, two forces, i.e., FBG and FDG are in the
same straight line and hence the third force, i.e., FGF should be zero.
∴
FGF = 0
Now consider the joint F.
Joint F [See Fig. 11.24 (d)]
Resolving forces vertically, we get
FDF × sin 60° = 0
∴
FDF = 0
E
Resolving horizontally, we get
FFB = FEF = 2.58 kN
∴
F FB
=
2.58
kN (Compressive)
Now consider the joint B.
Joint B
Resolving vertically, we get
FBG × sin 30° = 1.49
1.49
= 2.98 kN
(Compressive)
∴
FBG =
0.5
Joint G
FGD = FBG = 2.98 kN
(Compressive)
E
Fig. 11.24 (b)
C
D
2 kN
FED
60°
60°
E
5.15 kN
F
FEF
1 kN
Fig. 11.24 (c)
D
G
60°
2.58 kN
F
B
2.58 kN
Fig. 11.24 (d)
G
30°
F
B
2.58 kN
RB = 1.49 kN
Fig. 11.24 (e)
491
STRENGTH OF MATERIALS
The forces are shown in a tabular form as
Member
Force in the member
AC
AE
CE
CD
ED
EF
DF
DG
GB
FB
FG
Nature of force
4.216 kN
5.15 kN
2 kN
4.216 kN
3.155 kN
2.58 kN
0
2.98 kN
2.98 kN
2.58 kN
0
Compressive
Tensile
Compressive
Compressive
Tensile
Tensile
Nil
Compressive
Compressive
Compressive
Nil
11.5.2. Method of Sections
When the forces in a few members of a truss are to be determined, then the method of
section is mostly used. This method is very quick as it does not involve the solution of other
joints of the truss.
In this method, a section line is passed through the members, in which forces are to be
determined as shown in Fig. 11.25. The section line should be drawn in such a way that it does
not cut more than three members in which the forces are unknown. The part of the truss, on
any one side of the section line, is treated as a free body in equilibrium under the action of
external forces on that part and forces in the members cut by the section line. The unknown
forces in the members are then determined by using equations of equilibrium as
ΣFx = 0, ΣFy = 0 and ΣM = 0.
2 kN
1
1
C
2 kN
2 kN
D
1
C
2 kN
E
2 kN
2 kN
E
D
A
B A
F
3 kN
B
G
1
(a) Given Truss
F
3 kN
G
1
3 kN
(b) Left Part
1
3 kN
(c) Right Part
Fig. 11.25
If the magnitude of the forces, in the members cut by a section line, is positive then the
assumed direction is correct. If magnitude of a force is negative, then reverse the direction of
that force.
Problem 11.11. Find the forces in the members AB and AC of the truss shown in Fig.
11.26 using method of section.
(U.P. Tech. University, 2002-2003)
492
ANALYSIS OF PERFECT FRAMES
∴
and
20 × 1.25
= 5 kN
RC =
5
RB = 20 – 5 = 15 kN
20 kN
m
A
1
2.5
Sol. First determine the reaction RB and RC.
The distance of line of action of 20 kN from point B is
1
AB × cos 60° or 2.5 × = 1.25 m
2
Taking moments about point B, we get
RC × 5 = 20 × 1.25
90°
30°
60°
B
C
1
5m
RB
RC
Fig. 11.26
F
BA
2.5
m
Now draw a section line (1-1), cutting the members
AB and BC in which forces are to be determined. Now conA
sider the equilibrium of the left part of the truss. This part is
shown in Fig. 11.27.
1
90°
Let the directions of FBA and FBC are assumed as shown
in Fig. 11.27.
30°
60°
Now taking the moments of all the forces acting on the
B
C
FBC
1
left part about point C, we get
5m
15 × 5 + (FBA × AC)* = 0
15 kN
(∵ The perpendicular distance between the line of
Fig. 11.27
action of FBA and point C is equal to AC)
or
75 + FBA × 5 × cos 30° = 0
(∵ AC = BC × cos 30°)
− 75
= – 17.32 kN
or
FBA =
5 × cos 30°
The negative sign shows that FBA is acting in the opposite direction (i.e., towards point B).
Hence force FBA will be a compressive force.
∴
FBA = 17.32 kN (Compressive). Ans.
Again taking the moments of all the forces acting on the left part about point A, we get
15 × Perpendicular distance between the line of action of
15 kN and point C = FBC × Perpendicular distance between FBC and point A
15 × 2.5 × cos 60° = FBC × 2.5 × sin 60°
15 × 2.5 × cos 60° 15 × 0.5
=
∴
FBC =
2.5 × sin 60°
0.866
= 8.66 kN (Tensile). Ans.
These forces are same as obtained in Problem 11.1.
Problem 11.12. A truss of span 5 m is loaded as shown in Fig. 11.28. Find the reactions
and forces in the members marked 4, 5 and 7 using method of section.
*The moment of the force FBA about point C, is also taken by resolving the force FBA into vertical
and horizontal components at point B. The moment of the horizontal component about C is zero, whereas
the moment of vertical component will be (FBA × sin 60°) × 5 = FBA × 5 × sin 60° or FBA × 5 × cos 30°.
(∵ sin 60° = cos 30°)
493
STRENGTH OF MATERIALS
10 kN
D
1
m
7
2.5
Sol. Let us first determine the reactions RA and RB.
Triangle ABD is a right-angled triangle having angle
ADB = 90°
∴
AD = AB cos 60° = 5 × 0.5 = 2.5 m
The distance of line of action the vertical load 10 kN
from point A will be AD cos 60° or 2.5 × 0.5 = 1.25 m.
From triangle ACD, we have
AC = AD = 2.5 m
∴
BC = 5 – 2.5 = 2.5 m
In right-angled triangle CEB, we have
E
60°
A
3
BE = BC cos 30° = 2.5 ×
2
∴ The distance of line of action of vertical load 12 kN
F
GH
5
60°
30°
C
B
4
1
5m
RA
3
from point B will be BE cos 30° or BE ×
2
= 2.5 ×
12 kN
RB
Fig. 11.28
I
JK
3
3
×
= 1.875 m
2
2
∴ The distance of the line of action of the load of 12 kN from point A will be
(5 – 1.875) = 3.125 m
Now taking the moments about A, we get
RB × 5 = 10 × 1.25 + 12 × 3.125 = 50
50
= 10 kN and RA = (10 + 12) – 10 = 12 kN
5
Now draw a section line (1-1), cutting the members 4,
D 1
5 and 7 in which forces are to be determined. Consider the
12 kN
F7
equilibrium of the right part of the truss (because it is smaller
7
√3
than the left part).
2.5 × —
E
2
This part is shown in Fig. 11.29. Let F4, F5 and F7 are
F5
the forces in members 4, 5 and 7. Let their directions are
5
30°
assumed as shown in Fig. 11.29.
F4
C
4
Now taking the moments of all the forces acting on
1
the right part about point E, we get
2.5 m
RB × BE cos 30° = F4 × (BE × sin 30°)
∴
RB =
F
GH
10 × 2.5 ×
or
or
I
JK
3
3
3
×
= F4 × 2.5 ×
× 0.5
2
2
2
10 ×
∴
494
RB = 10 kN
3
= F4 × 0.5
2
F4 = 10 ×
3
1
×
= 17.32 kN (Tensile).
2
0.5
Fig. 11.29
B
ANALYSIS OF PERFECT FRAMES
or
Now taking the moments of all the forces about point B acting on the right part, we get
12 × BE cos 30° + F5 × BE = 0
12 × cos 30° + F5 = 0
∴
F5 = – 12 × cos 30° = – 10.392 kN
– ve sign indicates that F5 is compressive.
∴
F5 = 10.392 kN (Compressive). Ans.
Now taking the moments about point C of all the forces acting on the right parts, we get
12 × (2.5 – BE cos 30°) = F7 × CE + RB × BC
F
GH
12 × 2.5 − 2.5 ×
or
3
3
×
2
2
I =F
JK
7
× 2.5 × sin 30° + 10 × 2.5
12 × (2.5 – 1.875) = F7 × 1.25 + 25 or 7.5 = 1.25F7 + 25
or
7.5 − 25
= – 14 kN
1.25
Negative sign shows that F7 is compressive.
∴
F7 = 14 kN (Compressive). Ans.
These forces are same as obtained in Problem 11.3.
Problem 11.13. A truss of span 9 m is loaded as shown in Fig. 11.30. Find the reactions
and forces in the members marked 1, 2 and 3.
Sol. Let us first calculate the reactions RA and RB.
Taking moments about A, we get
RB × 9 = 9 × 3 + 12 × 6 = 27 + 72 = 99
F7 =
or
99
= 11 kN
9
RA = (9 + 12) – 11 = 10 kN
∴
RB =
and
1
C
D
E
F
2
4m
A
3
G
H
9 kN
RA
3m
B
12 kN
3m
RB
3m
9m
Fig. 11.30
Now draw a section line (1-1), cutting the members 1, 2 and 3 in which forces are to be
determined. Consider the equilibrium of the left part of the truss (because it is smaller than
495
STRENGTH OF MATERIALS
the right part). This part is shown in Fig. 11.30 (a). Let F1,
F2 and F3 are the forces members 1, 2 and 3 respectively.
Let their directions are assumed as shown in Fig. 11.30 (a).
Taking moments of all the forces acting on the left
part about point D, we get
10 × 3 = F3 × 4
10 × 3
∴
F3 =
4
= 7.5 kN (Tensile). Ans.
Now taking the moments of all the forces acting on
the left part about point G, we get
10 × 3 + F1 × 4 = 0
− 30
∴
F1 =
= – 7.5 kN
4
Negative sign shows that force F1 is compressive.
∴
F1 = 7.5 kN (Compressive). Ans.
Now taking the moments about the point C, we get
F2 × 3 – 9 × 3 + F3 × 4 = 0
or
F2 × 3 – 27 + 7.5 × 4 = 0
1
F1
C
1
D
2
4m
F2
3
A
G
10 kN
F3
9 kN
1
3m
Fig. 11.30 (a)
(∵ F3 = 7.5)
27 − 7.5 × 4 − 3
=
= – 1.0 kN
3
3
Negative sign shows that force F2 is compressive.
∴
F2 = 1.0 kN (Compressive). Ans.
Problem 11.14. For the pin-joined truss shown in Fig. 11.31, find the forces in the
members marked 1, 2 and 3 with the single load of 80 kN as shown.
Sol. First calculate reactions RA and RB.
F2 =
or
H
0.5 m
D
3
2
4.5 m
4m
A
B
1
C
E
4m
4m
80 kN
4m
4m
RB
RA
Fig. 11.31
Taking moments about A, RB × 16 = 80 × 12
80 × 12
= 60 kN
∴
RB =
16
∴
RA = Total vertical load – RB = 80 – 60 = 20 kN
496
ANALYSIS OF PERFECT FRAMES
Let us now find the forces in the members 1, 2 and 3
by the method of section. Take a section Y-Y passing through
the members 1, 2 and 3. Now consider the equilibrium of
left portion shown in Fig. 11.31 (a).
Let F1, F2 and F3 are the forces in the members 1, 2
and 3 respectively. Their assumed directions are also
shown in Fig. 11.31 (a).
Taking moments of all forces (here RA, F1, F2 and
F3) about point D, we get
RA × 4 = F1 × 4.5
∴
F3
D
y
H
3
0.5 m
F2
4.5 m
2
A
RA × 4 20 × 4
=
F1 =
4.5
4.5
= 17.78 kN (Tensile). Ans.
E
20 kN
4m
RA
F1
y
1
C
4m
Fig. 11.31 (a)
Now taking the moments about C, we get
...(i)
RA × 8 = F3 × Perpendicular distance between F3 and point C
To find the perpendicular distance between the line of action of F3 and point C, first find
angle CDH
DE 4.5
=
tan θ =
EC 4.0
4.5
0.5
∴
θ = tan–1
= 48.37° and tan α =
4.0
4.0
0
.
5
∴
α = tan–1
= 7.125°
4.0
∴
∠CDH = θ + α = 48.37 + 7.125 = 55.495
L
F3
From triangle DEC, we know that
D
2
2
CD = 4.5 + 4 = 6.02 m
Now from C, draw a perpendicular CL on the line of
action of F3 as shown in Fig. 11.31 (b).
∴ From right-angled triangle CDL,
θ
α
CL
θ
CD
E
C
A
∴
CL = CD sin (α + θ)
= 6.02 × sin (55.495)
Fig. 11.31 (b)
= 4.96 m
Substituting the value of OL (i.e., perpendicular distance between F3 and C) in equation
(i), we get
RA × 8 = F3 × 4.96
RA × 8 20 × 8
=
= 32.26 kN (Compressive). Ans.
∴
F3 =
4.96
4.96
To find the force F2, resolve the forces (i.e., RA, F3, F2 and F1) vertically. Hence, we get
RA – F3 sin α + F2 sin θ = 0
or
20 – 32.26 × sin (7.125) + F2 × sin (48.37) = 0
sin (α + θ) =
497
STRENGTH OF MATERIALS
or
20 – 4 + F2 × 0.7474 = 0
16
= 21.4 kN (Compressive). Ans.
or
F2 =
0.7474
11.5.2.1. Method of Section, Cutting more than Three Members. In method of section, in general a section should cut only three members, since only three unknowns can be
determined from three equations of equilibrium. However, there are special cases where we may
cut more than three members. It is illustrated in the following example. A section line can cut
four members if the axes of the three of them intersect in one point, thus making it possible to
determine the axial force in the fourth member by taking moments about the point of intersection
of the axes of the three members.
Problem 11.15. For the frame shown in
O
20 kN
Fig. 11.32 find the forces in the members BD, BG, GA,
90°
AC and AB of the bottom bay only. State their nature.
2m
Sol. Let us first find the reactions. The frame carries horizontal loads. As the frame is supported on rollF
H
E
ers at B, hence the reaction RB will be vertical.
20 kN
90°
At the point A, the frame is hinged and hence the
2m
support reactions at A will consist of a horizontal reaction HA and a vertical reaction RA.
20 kN
G
C
Taking moments of all forces about A, we get
D
1
90°
RB × 4 = 20 × 2 + 20 × 4 + 20 × 6 = 240
2m
240
2
= 60 kN (↑)
∴
RB =
2
4
B
Now
RA = Total vertical loads – RB
HA A
1 Roller
= 0 – 60 = – 60 kN
2m
2m
– ve sign means, RA is acting downwards.
RA
RB
∴
RA = 60 kN (↓)
Fig. 11.32
and
HA = Sum of all horizontal loads
= (20 + 20 + 20) = 60 kN (← )
2m
Now draw a section line (1-1), cutting the memC
G
D
bers BD, BG and BA in which forces are to be determined.
1
Consider the equilibrium of the right part. This part is
90°
shown in Fig. 11.32 (a). Let FBD, FBG and FBA are the
2m
F BD
forces in the members BD, BG and BA respectively. Let
F BC
θ
their directions are assumed as shown in Fig. 11.32 (a).
B
In this particular case, all the three forces are meetA
F AB
ing at one point B. Hence by cutting these members by
60 kN
1
section line (1-1), we may not get the results.
RB
Let us draw a section line (2-2), cutting four memFig. 11.32 (a)
bers AC, CG, GD and BD in which forces are to be determined. The axes of three members, i.e., AC, CG and GD are intersecting at point C. And hence
taking moments about point C, we can find force in member BD.
Similarly the axes of BD, GD and CG are meeting at point D. And hence taking moments
about point D, we can find the force in member AC.
498
ANALYSIS OF PERFECT FRAMES
The section line (2-2), cutting the four members,
is shown in Fig. 11.32 (b). Let the forces in the members
are FCA, FCG, FDG and FBD. Let their directions are assumed as shown in Fig. 11.32 (b). Consider the equilibrium of part above the section line (2-2). Taking the
moments of all the forces (acting on the upper part) about
point C, we get
20 × 2 + 20 × 4 + FBD × 4 = 0
FBD =
or
FCA =
FBG =
2m
FCG
C
FDG
G
D
20 kN
FBD
90°
FCA
2
2
A
B
4m
Fig. 11.32 (b)
D
G
45°
60 − 30
30
=
cos 45° 1/ 2
2 ×
FBD
= 30
FBG
45°
A
B
FBA
2 (Compressive)
Resolving horizontally, we get
FBA = FBG sin 45°
RB = 60 kN
1
2
Fig. 11.32 (c)
= 30 kN (Tensile)
Now consider joint A.
Joint A
At the joint A ; RA, HA, FAC and FAB are known in magnitude and direction.
Resolving horizontally, we get
60 = 30 + FAG × cos 45°
60 − 30
30
=
∴
FAG =
cos 45°
1
2
FG IJ
H K
= 30 ×
20 kN
90°
120
= 30 kN (Tensile). Ans.
4
= 30 ×
= 30 ×
F
H
E
Now consider joint B.
Joint B
Resolving forces vertically, we get
FBG cos 45° + FBD = RB
or
FBG × cos 45° + 30 = 60
or
20 kN
2m
− 40 − 80
4
− 120
=
= – 30 kN
4
– ve sign means the force FBD is compressive.
∴
FBD = 30 kN (Compressive). Ans.
Now taking moments of all forces (acting on the
upper part) about point D, we get
FCA × 4 = 20 × 2 + 20 × 4 = 120
∴
O
2 (Tensile)
C
G
30 kN
FAG
45°
HA = 60
A
30 kN
RA = 60
Fig. 11.32 (d)
499
B
STRENGTH OF MATERIALS
Now the forces are shown in a tabular form below :
Member
Force in the member
BD
30 kN
BG
30 ×
AB
AC
30 kN
30 kN
AG
30 ×
Nature of force
Compressive
2 kN
Compressive
Tensile
Tensile
2
Tensile
Problem 11.16. A truss of 12 m span is loaded as shown in Fig. 11.33. Determine the
forces in the members DG, DF and EF, using method of section.
Sol. The truss is supported on rollers at B and hence RB will be vertical. The truss is
hinged at A and hence the support sections at A will consists of a horizontal section HA and a
vertical section RA.
In triangle AEC,
AC = AE × cos 30°
= 4 × 0.866 = 3.464 m
Now length
AD = 2 × AC = 2 × 3.464
= 6.928 m
1 kN
D
1
2 kN
C
1 kN
A
G
30°
60°
60°
HA
1 kN
4m
30°
F
E
B
1
4m
RA
4m
RB
Fig. 11.33
Now taking the moments about A, we get
RB × 12 = 2 × AC + 1 × AD + 1 × AE
= 2 × 3.464 + 1 × 6.928 + 1 × 4 = 17.856
17.856
= 1.49 kN
∴
RB =
12
Now draw the section line (1-1), passing through members DG, DF and EF in which the
forces are to be determined. Consider the equilibrium of the right part of the truss. This part is
shown in Fig. 11.33 (a). Let FDG, FFD and FEF are the forces in members DG, FD and EF
respectively. Let their directions are assumed as shown in Fig. 11.33 (a). Taking moments of
all forces acting on right part about point F, we get
RB × 4 + FDG × FG = 0
or
1.49 × 4 + FDG × (4 × sin 30°) = 0
(∵ FG = 4 × sin 30°)
500
ANALYSIS OF PERFECT FRAMES
− 1.49 × 4
1
= – 2.98 kN
4 × sin 30°
D
– ve sign shows that the force FDG is compressive.
FDG
∴
FDG = 2.98 kN (Compressive). Ans.
G
FFD
Now taking the moments about point D, we get
30°
RB × BD cos 30° = FFE × BD × sin 30°
B
E
F
1.49 kN
FFE
or
RB × cos 30° = FFE × sin 30°
1
4m
1.49 × cos 30° 1.49 × 0.866
RB
=
∴
FFE =
sin 30°
0.5
Fig. 11.33 (a)
= 2.58 kN (Tensile). Ans.
Now taking the moments of all forces acting on the right part about B, we get
FFD × ⊥ distance between FFD and B = 0
∴
FFD = 0. Ans.
(∵ ⊥ distance between FFD and B is not zero)
11.5.3. Graphical Method
The force in a perfect frame can also be determined by a graphical method. The analytical
methods (such as method of joints and method of sections) give absolutely correct results, but
sometimes it is not possible to get the results from analytical methods. Then a graphical method
can be used conveniently to get the results. The graphical method also provides reasonable accurate results.
The naming of the various members of a frame are done according to Bow’s notations. According to this notation of force is designated
A
by two capital letters which are written on either side of the line of
B
action of the force. A force with letters A and B on either side of the line
of action is shown in Fig. 11.34. This force will be called AB.
Force AB
The following steps are necessary for obtaining a graphical soluFig. 11.34
tion of a frame.
(i) Making a space diagram
(ii) Constructing a vector diagram
(iii) Preparing a force table.
1. Making a space diagram. The given truss or frame is drawn accurately according
to some linear scale. The loads and support reactions in magnitude and directions are also
shown on the frame. Then the various members of the frame are named according to Bow’s
notation. Fig. 11.35 (a) shows a given truss and the forces in the members AB, BC and AC are
to be determined. Fig. 11.35 (b) shows the space diagram to same linear scale. The member AB
is named as PS and so on.
2. Constructing a vector diagram. Fig. 11.35 (c) shows a vector diagram, which is
drawn as given below :
(i) Take any point p and draw pq parallel to PQ vertically downwards. Cut pq = 4 kN to
same scale.
(ii) Now from q draw qr parallel to QR vertically upwards and cut qr = 2 kN to the same
scale.
or
FDG =
501
STRENGTH OF MATERIALS
4 kN
4 kN
B
B
P
A
45°
45°
C
A
2 kN
Q
S
C
R
4m
2 kN
p
2 kN
s
r
2 kN
q
(a) Given Diagram
(b) Space Diagram
(c) Vector Diagram
Fig. 11.35
(iii) From r draw rp parallel to RP vertically upwards and cut rp = 2 kN to the same scale.
(iv) Now from p, draw a line ps parallel to PS and from r, draw a line rs parallel to RS,
meeting the first line at s. This is vector diagram for joint (A). Similarly the vector diagrams
for joint (B) and (C) can be drawn.
3. Preparing a force table. The magnitude of a force in a member is known by the
length of the vector diagram for the corresponding member, i.e., the length ps of the vector
diagram will give the magnitude of force in the member PS of the frame.
Nature of the force (i.e., tensile or compressive) is determined according to the following
procedure :
(i) In the space diagram, consider any joint. Move round that joint in a clockwise direction. Note the order of two capital letters by which the members are named. For example, the
members at the joint (A) in space diagram Fig. 11.35 (b) are named as PS, SR and RP.
(ii) Now consider the vector diagram. Move on the vector diagram in the order of the
letters (i.e., ps, sr and rp).
(iii) Now mark the arrows on the members of the space diagram of that joint (here
joint A).
(iv) Similarly, all the joints can be considered and arrows can be marked.
(v) If the arrow is pointing towards the joint, then the force in the member will be
compressive whereas if the arrow is away from the joint, then the force in the member will be
tensile.
Problem 11.17. Find the forces in the members AB, AC and BC of the truss shown in
Fig. 11.36.
Sol. First determine the reactions RB and RC.
1
= 2.5 m
2
Distance of line of action of 20 kN from point B
From Fig. 11.36 (a), AB = BC × cos 60° = 5 ×
= AB cos 60° = 2.5 ×
502
1
= 1.25 m
2
ANALYSIS OF PERFECT FRAMES
p
20 kN
20 kN
A
A
90°
Q
P
R
30°
60°
B
C
B
5m
RB
B
C
S
15 kN
RC
(a) Given Diagram
s
q
15 kN
(b) Space Diagram
r
(c) Vector Diagram
Fig. 11.36
Now taking moments about B, we get
RC × 5 = 20 × 1.25 = 25
25
= 5 kN and RB = 20 – 5 = 15 kN
5
Now draw the space diagram for the truss alongwith load of 20 kN and the reactions RB
and RC equal to 15 kN and 5 kN respectively as shown in Fig. 11.36 (b). Name the members
AB, AC and BC according to Bow’s notations as PR, QR and RS respectively. Now construct
the vector diagram as shown in Fig. 11.36 (c) and as explained below :
(i) Take any point p and draw a vertical line pq downward equal to 20 kN to some
suitable scale. From q draw a vertical line qs upward equal to 5 kN to the same scale to represent the reaction at C. Then sp will represent the reaction RB to the scale.
(ii) Now draw the vector diagram for the joint (B). From p, draw a line pr parallel to PR
and from s draw a line sr parallel to SR, meeting the first line at r. Now prs is the vector
diagram for the joint (B). Now mark the arrows on the joint B. The arrow in member PR will be
towards joint B, whereas the arrow in the member RS will be away from the joint B as shown
in Fig. 11.36 (b).
(iii) Similarly draw the vector diagrams for joint A and C. Mark the arrows on these
joints in space diagram.
Now measure the various sides of the vector diagram. The forces are obtained by multiplying the scale factor. The forces in the members are given in a tabular form as :
∴
RC =
Member
Force in member
According to given truss
According to
Bow’s notation
AB
AC
BC
PR
QR
RS
17.3 kN
10.0 kN
8.7 kN
Nature of force
Compressive
Compressive
Tensile
Problem 11.18. A truss of span 7.5 m carries a point load of 1000 N at joint D as shown
in Fig. 11.37. Find the reactions and forces in the member of the truss.
503
STRENGTH OF MATERIALS
Sol. First determine the reactions RA and RB.
C
C
P
30°
A
5m
R
30°
D
B
A
7.5 m
RA
D
T
1000 N
q
P
Q
S
s
B
p
1000 N
RB
333 N
(a) Given Diagram
667 N
(b) Space Diagram
r
t
(c) Vector Diagram
Fig. 11.37
Taking moments about A, we get
RB × 7.5 = 5 × 1000
5000
= 667 N and RA = 1000 – 667 = 333 N.
7.5
Now draw the space diagram for the truss alongwith load of 1000 N and reactions RA and
RB equal to 333 N and 667 N respectively as shown in Fig. 11.37 (b). Name the members AC, CB,
AD, CD and DB according to Bow’s notations as PR, PQ, RT, QR and QS respectively. Now
construct the vector diagram as shown in Fig. 11.37 (c) and as explained below :
(i) Take any point s and draw a vertical line st downward equal to load 1000 N to some
suitable scale. From t draw a vertical line tp upward equal to 333 N to the same scale to
represent the reaction at A. The ps will represent the reaction RB to the scale.
(ii) Now draw the vector diagram for the joint A. From p, draw a line pr parallel to PR
and from t draw a line tr parallel to RT, meeting the first line at r. Now prt is the vector
diagram for the joint A. Now mark the arrows on the joint A. The arrow in the member PR will
be towards the joint A, whereas the arrow in the member RT will be away from the joint A as
shown in Fig. 11.37 (b).
(iii) Similarly draw the vector diagrams for the joint C, B and D. Mark the arrows on
these joints as shown in Fig. 11.37 (b).
Now measure the various sides of the vector diagrams. The forces in the members are
obtained by multiplying the scale factor to the corresponding sides of the vector diagram. The
forces in members are given in a tabular form as :
∴
RB =
Member
Force in member
Nature of force
According to given truss
According to
Bow’s notation
AC
PR
666 N
Compressive
AD
RT
576.7 N
Tensile
504
CB
PQ
1333 N
Compressive
CD
QR
1155 N
Tensile
DB
QS
1555 N
Tensile
ANALYSIS OF PERFECT FRAMES
Problem 11.19. Determine the forces in all the members of a cantilever truss shown in
Fig. 11.38.
Sol. In this case the vector diagram can be drawn without knowing the reactions. First of
all draw the space diagram for the truss along with loads of 1000 N of joints B and C. Name the
members AB, BC, CD, DE, AD and BD according to Bow’s notation as PT, QS, SR, RV, VT and
ST respectively. Now construct the vector diagram as shown in Fig. 11.38 (c) and as explained
below :
(i) The vector diagram will be started from joint C where forces in two members are
unknown. Take any point q and draw a vertical line qr downward equal to load 1000 N to some
suitable scale. From r, draw a line rs parallel to RS and from q draw a line qs parallel to QS,
meeting the first line at s. Now qrs is the vector diagram for the joint C. Now mark the arrows
on the joint C. The arrow in the member RS will be towards the joint C, whereas the arrow in
the member SQ will be away from the joint C as shown in Fig. 11.38 (b).
1000 N
1000 N
2m
A
B
1000 N
R1
2m
1000 N
P
Q
B
C
C
R
p
t
A
T
3m
S
V
D
R2
E
(a) Given Figure
v
q
D
s
r
E
(b) Space Diagram
Fig. 11.38
(c) Vector Diagram
(ii) Now draw the vector diagram for the joints B and D similarly.
Mark the arrows on these joints as shown in Fig. 11.38 (b).
Now measure the various sides of the vector diagram. The forces in the members are given
in a tabular form as :
Member
Force in member
According to given truss
According to
Bow’s notation
AB
BC
CD
DE
AD
BD
PT
QS
SR
RV
VT
ST
1333 N
1333 N
1666 N
2500 N
833 N
1000 N
Nature of force
Tensile
Tensile
Compressive
Compressive
Tensile
Compressive
From the vector diagram, the reactions R1 and R2 at A and E can be determined in magnitude
and directions.
Reaction R2 = rv = 2500 N. This will be towards point E.
505
STRENGTH OF MATERIALS
Reaction R1 = vp = 2000 N. This will be away from the point A as shown in Fig. 11.38 (b).
The reaction R1 is parallel to vp.
Problem 11.20. Determine the support reacE
tions and nature and magnitude of forces in the mem200 kN
2m
bers of truss shown in Fig. 11.39.
C
θ
(U.P. Tech. University, 2001–2002)
A
θ
θ
Sol. Let us start from joint A where forces in
2m
two members are unknown.
(90 – θ)
Joint A
D
B
4m
4m
In triangle ABC,
BC 2
Fig. 11.39
=
tan θ =
CA 4
4
4
AC
200 kN
∴
cos θ =
=
=
2
2
AB
20
F
2 +4
AC
sin θ =
and
C
2
20
Refer to Fig. 11.39 (a). The forces are shown at joint A. Resolving
forces vertically, we get
FAB sin θ = 200
200
200
200 × 20
=
=
= 447.2 kN. Ans.
sin θ 2 / 20
2
Resolving forces horizontally, we get
FAC = FAB cos θ
4
= 400 kN (Tensile).
= (100 × 20 ) ×
20
Joint B
Refer to Fig. 11.39(b)
∠ABC = 90 – θ
Resolving forces vertically,
FBC = FAB cos (90 – θ)
= FAB sin θ
2
= (100 × 20 ) ×
20
2
∵ FAB = 100 × 20 and sin θ =
20
∴
FBC = 200 kN (Tensile)
Resolving forces horizontally, we get
FBD = FAB sin (90 – θ) = FAB cos θ
4
= 400 kN (Comp.)
= (100 × 20 ) ×
20
Joint C
Refer to Fig. 11.39(c)
Resolving forces horizontally,
FCE cos θ + FCD cos θ = FAC
∴
FAB
B
Fig. 11.39 (a)
FAB =
FG
H
506
A
θ
IJ
K
C
(90–θ)
FBC
D
FAB
FBD B
Fig. 11.39 (b)
A
ANALYSIS OF PERFECT FRAMES
FCE ×
or
∴
4
20
+ FCD ×
FCE + FCD = 400 ×
4
20
E
= 400
20
4
…(i)
FCE – FCD =
200
=
sin θ
= 100 ×
Adding equations (i) and (ii),
A
FBC
FCD
D
B
FCE sin θ – FCD sin θ – FBC = 0
(FCE – FCD) sin θ = FBC = 200
∴
C FAC
θ
= 100 × 20
Resolving forces vertically, we get
or
FCE
θ
200
2
20
20
FG
H
Fig. 11.39 (c)
(∵ FBC = 200 kN)
FG∵
H
IJ
K
2
sin θ =
20
IJ
K
…(ii)
2FCE = 200 ×
or
20
FCE = 100 × 20 (Tensile)
Substituting this value in equation (i), we get
FCD = 100 ×
20 – 100 ×
20 = 0
To find the support reactions, consider joint D and E.
Joint D
The force
FBD = 400 kN
whereas FDC = 0. Hence at joint D, there will be only horizontal reaction RDH , which will balance force FBD.
∴
RDH = FBD = 400 kN.
Joint E
At joint E, the force FEC = 100 × 20 kN. To balance this
force, there will be horizontal reaction and vertical reaction at E.
Let
REV = Vertical component of reaction at E
REH = Horizontal component of reaction at E
Resolving forces horizontally, we get
4
REH = FEC cos θ = (100 × 20 ) ×
= 400 kN. Ans.
20
Resolving forces vertically, we get
REV = FEC sin θ = (100 × 20 ) ×
2
= 200 kN. Ans.
20
Now the nature and magnitude of forces in the members are:
AB → 447.2 kN (Compressive)
BC → 200 kN (Tensile)
AC → 400 kN (Tensile)
BD → 400 kN (Compressive)
CD → 0
CE → 447.2 kN (Tensile).
C
D
FDC = 0
B
FBD = 400 kN
RDH
Fig. 11.39 (d)
REV
E
REH
F
θ
FEC = θ
Fig. 11.39 (e)
507
C
STRENGTH OF MATERIALS
HIGHLIGHTS
1. The relation between number of joints (j) and number of members (n) in a perfect frame is given
by n = 2j – 3.
2. Deficient frame is a frame in which number of members are less than (2j – 3) whereas a redundant
frame is a frame in which number of members are more than (2j – 3).
3. The reaction on a roller support is at right angles to the roller base :
4. The forces in the members of a frame are determined by :
(i) Method of joints
(ii) Method of sections, and
(iii) Graphical method.
5. The force in a member will be compressive if the member pushes the joint to which it is connected
whereas the force in the member will be tensile if the member pulls the joint to which it is
connected.
6. While determining forces in a member by method of joints, the joint should be selected in such a
way that at any time there are only two members, in which the forces are unknown.
7. If three forces act at a joint and two of them are along the same straight line then third force would
be zero.
8. If a truss (or frame) carries horizontal loads, then the support reaction at the hinged end will
consists of (i) horizontal reaction and (ii) vertical reaction.
9. If a truss carries inclined loads, then the support reaction at the hinged end will consists of : (i)
horizontal reaction and (ii) vertical reaction. They will be given as :
Horizontal reaction = Horizontal components of inclined loads
Vertical reaction = Total vertical components of inclined loads – Roller support reaction.
10. Method of section is mostly used, when the forces in a few members of a truss are to be determined.
11. The following steps are necessary for obtaining a graphical solution of a frame :
(i) Making a space diagram,
(ii) Constructing a vector diagram, and
(iii) Preparing a force table.
12. The various members of a frame are named according to Bow’s notation.
EXERCISE
A. Theoretical Questions
1. Define and explain the terms : Perfect frame, imperfect frame, deficient frame and a redundant
frame.
(U.P. Tech. University, 2002–2003)
2. (a) What is a frame ? State the difference between a perfect frame and an imperfect frame.
(b) What are the assumptions made in finding out the forces in a frame ?
3. What are the different methods of analysing (or finding out the forces) a perfect frame ? Which one
is used where and why ?
4. How will you find the forces in the members of a truss by method of joints when
(i) the truss is supported on rollers at one end and hinged at other end and carries vertical loads.
(ii) the truss is acting as a cantilever and carries vertical loads.
(iii) the truss is supported on rollers at one end and hinged at other end and carries horizontal and
vertical loads.
508
ANALYSIS OF PERFECT FRAMES
5.
6.
7.
8.
9.
10.
(iv) the truss is supported on rollers at one end and hinged at other end and carries inclined loads.
(a) What is the advantage of method of section over method of joints ? How will you use method of
section in finding forces in the members of a truss ?
(b) Explain with simple sketches the terms (i) method of sections and (ii) method of joints, as
applied to trusses.
How will you find the forces in the members of a joint by graphical method ? What are the
advantages or disadvantages of graphical method over method of joints and method of section ?
What is the procedure of drawing a vector diagram for a frame ? How will you find out (i) magnitude of a force, and (ii) nature of a force from the vector diagram ?
How will you find the reactions of a cantilever by graphical method ?
What are the assumptions made in the analysis of a simple truss.
Explain what you understand by perfect frame, deficient frame and redundant frame.
B. Numerical Problems
1. Find the forces in the members AB, AC and BC of the truss shown in Fig. 11.40.
[Ans. AB = 4.33 kN (Comp.)
5 kN
AC = 2.5 kN (Comp.)
A
BC = 2.165 kN (Tens.)]
30°
60°
C
B
5m
Fig. 11.40
2. A truss of span 7.5 m carries a point load of 500 N at joint D as shown in Fig. 11.41. Find the
reactions and forces in the members of the truss.
[Ans. RA = 166.5 N
C
RB = 333.5 N
1
4
=
333
N (Comp.)
F
1
3
F2 = 288.5 N (Tens.)
2
60°
A
B
30°
30°
F3 = 577.5 N (Tens.)
D
5m
F4 = 667 N (Comp.)
500 N
F5 = 577.5 N (Tens.)]
7.5 m
Fig. 11.41
3. A truss of span 7.5 m is loaded as shown in Fig. 11.42. Find the reactions and forces in the
members of the truss.
[Ans. AD = 3.464 kN (Comp.)
2.5 kN
AC = 1.732 kN (Tens.)
D
CD = 2.598 kN (Tens.)
3 kN
CE = 2.598 kN (Comp.)
E
DE = 3.50 kN (Comp.)
BE = 5 kN (Comp.)
60°
60°
60°
30°
BC
= 4.33 kN (Tens.)]
A
B
C
7.5 m
Fig. 11.42
509
STRENGTH OF MATERIALS
4. A truss is shown in Fig. 11.43. Find the forces in
all the members of the truss and indicate it is in
tension or compression.
(U.P. Tech. University 2000–2001)
10 kN
15 kN
C
B
A
20 kN
F
60°
60°
E
D
10 kN
RA
3m
3m
RE
Fig. 11.43
[Hint. In the problem, length of members are not given. Assume AD = DE = 3 m and ∠DAC = ∠DEC
= 60 as from figure it appears that AD = DE and ∠DAC = ∠DEC
MA = 0, 10 × 3 + 15 × 3 + 20 × 6 – 6 × RE = 0,
30 + 45 + 120
= 32.5 kN
6
and
RA = 10 + 15 + 20 + 10 – RE = 55 – 32.5 = 22.5
Start from joint B where forces in two members are unknown
Joint B
FBA = 10 kN (Comp.)
FBC = 0
or
RE =
10 kN
B
C
A
Joint A
ΣV = 0
22.5 – 10 – FAC sin 60° = 0
C
B
10
∴ FAC =
60°
A
D
12.5
= 14.43 kN
sin 60°
RA = 22.5
ΣH = 0, FAD = FAC cos 60° = 7.215
Joint D
10
A
7.215
D
10 kN
510
7.215
E
ANALYSIS OF PERFECT FRAMES
Joint F
FFE = 20 kN (Comp.)
FFC = 0
20
C
F
20
E
C
ΣV = 0, 32.5 – 20 – FCE sin 60° = 0
Joint E
12.5
= FCE = 14.43 kN
sin 60°
ΣH = 0, FED = FCE cos 60° = 7.215 kN]
20 kN
60°
E
D
RE = 32.5
5. Determine the forces in the various members of the truss shown in Fig. 11.44.
[Ans. AB = 1200 N (Comp.)
400 N
BC = 800 N (Comp.)
CD = 800 N (Comp.)
400 N
400 N
C
DE = 1200 N (Comp.)
200 N
200 N
B
EF = 600 N (Tens.)
D
AF = 600 N (Tens.)
30°
30°
BF = DF = 400 N (Comp.)
A
E
F
FC = 400 N (Tens.)]
10 m
Fig. 11.44
6. A plane truss is loaded and supported as shown in Fig. 11.45. Determine the nature and magnitude of forces in the members 1, 2 and 3.
[Ans. F1 = 833.34 N (Comp.)
C
F2 = 1000 N (Tens.)
F3 = 666.66 (Tens.)]
G
2000 N
1
4.5 m
D
A
2
θ
E
2m
3
θ
F
2m
B
H
2m
2m
8m
Fig. 11.45
511
STRENGTH OF MATERIALS
7. Determine the forces in all the members of a cantilever truss shown in Fig. 11.46.
[Ans. AC = 1154.7 N (Comp.)
B
D
CD = 2309.4 N (Tens.)
AD = 2309.4 N (Comp.)
BD = 2309.4 N (Tens.)]
60°
A
60°
C
5m
2000 N
Fig. 11.46
8. A cantilever truss is loaded as shown in Fig. 11.47. Find the force in member AB.
[Ans. AB = 15 kN (Tens.)]
5 kN
6m
B
A
5 kN
6m
C
θ
4m
F
D
E
3m
6m
Fig. 11.47
9. Find the axial forces in all the members of the truss shown
in Fig. 11.48.
[Hint. Start from joint B
First find angles θ and α
ED 3 1
= =
∴ θ = tan–1 0.5 = 26.56°
EB 6 2
EA 3
= = 1 ∴ α = tan–1 1.0 = 45°
tan α =
ED 3
tan θ =
8000
Joint B
FBA
A
θ
FBC
C
512
B
12000 N
E
3m
α
A
8000 N
3m
θ
B
3m
C
D
Fig. 11.48
ΣFy = 0, FBC sin θ = 8000
8000
8000
=
= 17891 N (Comp.)
sin θ sin 26.56°
ΣFx = 0 = FBC cos θ
= 17891 × cos 26.56° = 16002 N (Tensile)
∴ FBC =
ANALYSIS OF PERFECT FRAMES
Joint C
A
B
θ
C FBC = 17891
ΣFx = 0, FBC cos θ = FCD cos θ
∴ FCD = FBC = 17891 N (Comp.)
ΣFy = 0, FCA – FBC sin θ + FCD sin θ = 0
∴ FCA = 0 (∵ FBC = FCD)
D
ΣFy = 0, FAD cos α = 12000
Joint A
12000
= 16970
sin 45°
E FAE
A
B
ΣFx = 0, FAD = cos α – FAE + FBA = 0
α
FBA = 16002
or FAD cos 45° – FAE + 16002 = 0
16970 cos 45° – FAE + 16002 = 0
FAD
C
D
∴ FAE = 16002 + 16970 cos 45°
= 16002 + 11999 = 28001 N (Tens.)]
10. Determine the forces in the truss shown in Fig. 11.49 which carries a horizontal load of 16 kN and
a vertical load of 24 kN.
[Ans. AC = 24 kN (Tens.)
AD = 10 kN (Comp.)
D
CD = 24 kN (Tens.)
16 kN
CB = 24 kN (Tens.)
1.5 m
BD = 30 kN (Comp.)]
12000 N
A
∴ FAD =
θ
θ
B
C
24 kN
2m
2m
Fig. 11.49
11. Find the forces in the member AB and AC of the truss shown in Fig. 11.40 of question 1, using
method of sections.
[Ans. AB = 4.33 kN (Comp.)
AC = 2.5 kN (Comp.)]
12. Find the forces in the members marked 1, 3, 5 of truss shown in Fig. 11.41 of question 2, using
method of sections.
[Ans. F1 = 333 N (Comp.)
F3 = 577.5 N (Tens.)
F5 = 577.5 N (Tens.)]
13. Find the forces in the members DE, CE and CB of the truss, shown in Fig. 11.42 of question 3,
using method of sections.
[Ans. DE = 3.5 kN (Comp.)
CE = 2.598 kN (Comp.)
BC = 4.33 kN (Tens.)]
14. Using method of section, determine the forces in the members CD, FD and FE of the truss shown
in Fig. 11.43 of question 5.
[Ans. CD = 800 N (Comp.)
FD = 400 N (Comp.)
FE = 600 N (Tens.)]
513
STRENGTH OF MATERIALS
15. Using method of section, determine the forces in the members CD, ED and EF of the truss shown
in Fig. 11.50.
[Ans. CD = 4.216 kN (Comp.)
ED = 3.155 kN (Tens.)
1 kN
EF = 2.58 kN (Tens.)]
D
2 kN
C
1 kN
A
30°
G
60°
60°
60°
60°
30°
F
E
B
1 kN
Fig. 11.50
16. Find the forces in the members AB, AC and BC of the truss shown in Fig. 11.40 of question 1, using
graphical method.
17. Using graphical method, determine the magnitude and nature of the forces in the members of the
truss shown in Fig. 11.41 of question 2.
18. Determine the forces in all the members of a cantilever truss shown in Fig. 11.46 of question 7,
using graphical method. Also determine the sections of the cantilever.
19. A cantilever truss is loaded and supported as shown in Fig. 11.51. Find the value of load P which
would produce an axial force of magnitude 3 kN in the member AC using method of section.
(U.P. Tech. University, 2002–2003)
P
3m
3m
E
A
C
2m
B
D
F
1.5 m
3m
Fig. 11.51
[Hint. Force in member AC, FAC = 3 kN
Now pass a section ➀-➀ as shown in Fig. 11.51 (a).
P
C
1
FAC
E
A
FCD
FDF
D
F
1
3m
Fig. 11.51 (a)
Take moments about point D.
ΣMD = 0 ; FAC × 2 – P × 1.5 = 0
∴ 3 × 2 – P × 1.5 = 0
or
6 = 1.5P or P = 4 kN.]
514
But
FAC = 3 kN
12
DEFLECTION OF BEAMS
CHAPTER
12.1. INTRODUCTION..
If a beam carries uniformly distributed load or a point load, the beam is deflected from
its original position. In this chapter, we shall study the amount by which a beam is deflected
from its position. Due to the loads acting on the beam, it will
B
be subjected to bending moment. The radius of curvature of A
C
M E
the deflected beam is given by the equation
= . The raI
R
(a) Beam position before loading
I×E
dius of curvature will be constant if R =
= constant.
W
M
B
A
The term (I × E)/M will be constant, if the beam is subjected
to a constant bending moment M. This means that a beam for
which, when loaded, the value of (E × I)/M is constant, will
C′
bend in a circular arc.
(b) Beam position after loading
Fig. 12.1 (a) shows the beam position before any load is
Fig. 12.1
applied on the beam whereas Fig. 12.1 (b) shows the beam
position after loading.
12.2. DEFLECTION AND SLOPE OF A BEAM SUBJECTED TO UNIFORM BENDING
MOMENT
A beam AB of length L is subjected to a uniform bending moment M as shown in Fig. 12.1 (c). As the beam is subjected to a constant bending moment, hence it will bend into
a circular arc. The initial position of the beam is shown by
ACB, whereas the deflected position is shown by AC′B.
Let R = Radius of curvature of the deflected beam,
y = Deflection of the beam at the centre (i.e., distance CC′),
I = Moment of inertia of the beam section,
E = Young’s modulus for the beam material, and
θ = Slope of the beam at the end A (i.e., the angle
made by the tangent at A with the beam AB).
For a practical beam the deflection y is a
small quantity.
D
O
R
(90 – )
A
R
C
B
y
C
L
2
L
2
Fig. 12.1 (c)
515
STRENGTH OF MATERIALS
Hence tan θ = θ where θ is in radians. Hence θ becomes the slope as slope is
dy
= tan θ = θ.
dx
L
Now
AC = BC =
2
Also from the geometry of a circle, we know that
AC × CB = DC × CC′
L L
×
= (2R – y) × y
(∵ DC = DC′ – CC′ = 2R – y)
2 2
L2
= 2Ry – y2
or
4
For a practical beam, the deflection y is a small quantity. Hence the square of a small
quantity will be negligible. Hence neglecting y2 in the above equation, we get
or
L2
= 2Ry
4
L2
∴
y=
8R
But from bending equation, we have
M E
=
I
R
E×I
R=
M
Substituting the value of R in equation (i), we get
y=
or
...(i)
...(ii)
L2
EI
8×
M
ML2
...(12.1)
8 EI
Equation (12.1) gives the central deflection of a beam which bends in a circular arc.
Value of Slope (θ)
From triangle AOB, we know that
y=
FG IJ
H K
L
AC
L
2
=
sin θ =
=
AO
R
2R
Since the angle θ is very small, hence sin θ = θ (in radians)
L
∴
θ=
2R
EI
L
∵ R=
from equation (ii)
=
EI
M
2×
M
M×L
=
...(12.2)
2 EI
Equation (12.2) gives the slope of the deflected beam at A or at B.
FG
H
516
IJ
K
DEFLECTION OF BEAMS
12.3. RELATION BETWEEN SLOPE, DEFLECTION AND RADIUS OF CURVATURE..
Let the curve AB represents the deflection of a beam as shown in Fig. 12.2 (a). Consider
a small portion PQ of this beam. Let the tangents at P and Q make angle ψ and ψ + dψ with
x-axis. Normal at P and Q will meet at C such that
PC = QC = R
Y
C
Y
dψ
C
B
R
dψ
ψ
R
Q
ds
d
(ψ +
ψ)
P
ψ
dx
A
A
ψ
(ψ + dψ)
B
dy
ψ + dψ
ψ
O
X
(9
0–
Q
90–(ψ+dψ)
P
ψ)
(ψ+dψ)
O
(a )
X
(b )
Fig. 12.2
or
and
The point C is known as centre of curvature of the curve PQ.
Let the length of PQ is equal to ds.
From Fig. 12.2 (b), we see that
Angle PCQ = dψ
∴
PQ = ds = R.dψ
ds
R=
dψ
But if x and y be the co-ordinates of P, then
dy
tan ψ =
dx
dy
sin ψ =
ds
dx
cos ψ =
ds
Now equation (i) can be written as
...(i)
...(ii)
FG ds IJ FG 1 IJ
ds H dx K H cos ψ K
R=
=
=
dψ F dψ I
GH dx JK FGH ddxψ IJK
or
R=
sec ψ
FG dψ IJ
H dx K
...(iii)
517
STRENGTH OF MATERIALS
Differentiating equation (ii) w.r.t. x, we get
sec2 ψ .
dψ d 2 y
=
dx dx 2
F d yI
G J
dψ H dx K
=
2
2
or
sec 2 ψ
dx
Substituting this value of
R=
dψ
in equation (iii), we get
dx
sec ψ
F dyI
GG
J
dx J
GG sec ψ JJ
H
K
=
sec ψ . sec 2 ψ
2
2
d2 y
dx 2
=
sec 3 ψ
F d yI
GH dx JK
2
2
2
Taking the reciprocal to both sides, we get
d2 y
d2 y
2
1
dx 2
= dx3 =
R sec ψ (sec 2 ψ ) 3 / 2
d2 y
dx 2
=
(1 + tan 2 ψ ) 3 / 2
For a practical beam, the slope tan ψ at any point is a small quantity. Hence tan2 ψ can
be neglected.
1 d2 y
=
R dx 2
From the bending equation, we have
M E
=
I
R
1 M
=
R EI
Equating equations (iv) and (v), we get
∴
or
M d2 y
=
EI dx 2
d2 y
∴
M = EI
dx 2
Differentiating the above equation w.r.t. x, we get
But
∴
518
dM
d3 y
= EI
dx
dx 3
dM
= F shear force (See page 288)
dx
d3 y
F = EI
dx 3
...(iv)
...(v)
...(12.3)
...(12.4)
DEFLECTION OF BEAMS
Differentiating equation (12.4) w.r.t. x, we get
But
dF
d4 y
= EI
dx
dx 4
dF
= w the rate of loading
dx
∴
w = EI
d4 y
...(12.5)
dx 4
Hence, the relation between curvature, slope, deflection etc. at a section is given by :
Deflection
=y
dy
dx
Slope
=
Bending moment
= EI
Shearing force
= EI
d2 y
dx 2
d3 y
dx 3
d4 y
.
dx 4
Units. In the above equations, E is taken in N/mm2
I is taken in mm4,
y is taken in mm,
M is taken in Nm and
x is taken in m.
The rate of loading
= EI
12.3.1. Methods of Determining Slope and Deflection at a Section in a Loaded
Beam. The followings are the important methods for finding the slope and deflection at a
section in a loaded beam :
(i) Double integration method
(ii) Moment area method, and
(iii) Macaulay’s method
Incase of double integration method, the equation used is
M = EI
d2 y
dx
2
or
d2 y
dx
2
=
M
EI
First integration of the above equation gives the value of
dy
or slope. The second intedx
gration gives the value of y or deflection.
The first two methods are used for a single load whereas the third method is used for
several loads.
12.4. DEFLECTION OF A SIMPLY SUPPORTED BEAM CARRYING A POINT LOAD
AT THE CENTRE
A simply supported beam AB of length L and carrying a point load W at the centre is
shown in Fig. 12.3.
As the load is symmetrically applied the reactions RA and RB will be equal. Also the
maximum deflection will be at the centre.
519
STRENGTH OF MATERIALS
W
L
2
L
2
C
A
B
yc
X
x
L
W
2
W
2
Fig. 12.3
W
2
Consider a section X at a distance x from A. The bending moment at this section is
given by,
Mx = RA × x
W
×x
(Plus sign is as B.M. for left portion at X
=
2
is clockwise)
But B.M. at any section is also given by equation (12.3) as
Now
RA = RB =
M = EI
d2 y
dx 2
Equating the two values of B.M., we get
EI
d2 y
2
dx
On integration, we get
=
W
×x
2
...(i)
dy W x 2
+ C1
...(ii)
=
×
dx 2
2
where C1 is the constant of integration. And its value is obtained from boundary conditions.
EI
FG IJ
H K
L
dy
, slope
= 0 (As the maximum deflection is at the
2
dx
centre, hence slope at the centre will be zero). Substituting this boundary condition in equation (ii), we get
The boundary condition is that at x =
FG IJ
H K
W
L
×
0=
4
2
or
2
+ C1
WL2
16
Substituting the value of C1 in equation (ii), we get
C1 = –
dy Wx 2 WL2
...(iii)
=
−
4
16
dx
The above equation is known the slope equation. We can find the slope at any point on
the beam by substituting the values of x. Slope is maximum at A. At A, x = 0 and hence slope
at A will be obtained by substituting x = 0 in equation (iii).
EI
520
DEFLECTION OF BEAMS
∴
EI
FG dy IJ
H dx K
=
at A
W
WL2
×0−
4
16
LMFG dy IJ
MNH dx K
2
or
is the slope at A and is represented by θ A
at A
OP
PQ
WL
16
WL2
∴
θA = –
16 EI
The slope at point B will be equal to θA, since the load is symmetrically applied.
EI × θA = –
WL2
16 EI
Equation (12.6) gives the slope in radians.
∴
θB = θA = –
...(12.6)
Deflection at any point
Deflection at any point is obtained by integrating the slope equation (iii). Hence integrating equation (iii), we get
W x 3 WL2
.
−
x + C2
4 3
16
where C2 is another constant of integration. At A, x = 0 and the deflection (y) is zero.
Hence substituting these values in equation (iv), we get
EI × 0 = 0 – 0 + C2
or
C2 = 0
Substituting the value of C2 in equation (iv), we get
EI × y =
...(iv)
Wx 3 WL2 . x
...(v)
−
12
16
The above equation is known as the deflection equation. We can find the deflection at
any point on the beam by substituting the values of x. The deflection is maximum at centre
L
L
point C, where x = . Let yc represents the deflection at C. Then substituting x =
and y = yc
2
2
in equation (v), we get
EI × y =
FG IJ
H K
W L
EI × yc =
12 2
3
−
FG IJ
H K
WL2
L
×
16
2
WL3 WL3 WL3 − 3WL3
−
=
96
32
96
3
3
2WL
WL
=–
=−
96
48
3
WL
∴
yc = –
48 EI
(Negative sign shows that deflection is downwards)
=
∴ Downward deflection, yc =
WL3
48 EI
...(12.7)
521
STRENGTH OF MATERIALS
Problem 12.1. A beam 6 m long, simply supported at its ends, is carrying a point load
of 50 kN at its centre. The moment of inertia of the beam (i.e. I) is given as equal to
78 × 106 mm4. If E for the material of the beam = 2.1 × 105 N/mm2, calculate : (i) deflection at
the centre of the beam and (ii) slope at the supports.
Sol. Given :
Length,
L = 6 m = 6 × 1000 = 6000 mm
Point load,
W = 50 kN = 50,000 N
M.O.I.,
I = 78 × 106 mm4
Value of
E = 2.1 × 105 N/mm2
Let
yc = Deflection at the centre and
θA = Slope at the support.
(i) Using equation (12.7) for the deflection at the centre, we get
WL3
48 EI
50000 × 6000 3
=
48 × 2.1 × 10 5 × 78 × 10 6
= 13.736 mm. Ans.
(ii) Using equation (12.6) for the slope at the supports, we get
yc =
θB = θA = –
=
=
WL2
16 EI
WL2
16 EI
50000 × 6000 2
16 × 2.1 × 10 5 × 78 × 10 6
= 0.06868 radians
(Numerically)
radians
FG
H
IJ
K
180
180
∵ 1 radian =
degree
degree
π
π
= 3.935°. Ans.
Problem 12.2. A beam 4 metre long, simply supported at its ends, carries a point load W
at its centre. If the slope at the ends of the beam is not to exceed 1°, find the deflection at the
centre of the beam.
Sol. Given :
Length,
L = 4 m = 4000 mm
Point load at centre
=W
= 0.06868 ×
1× π
= 0.01745 radians
180
Let
yc = Deflection at the centre
Using equation (12.6), for the slope at the supports, we get
Slope at the ends,
θA = θB = 1° =
WL2
16 EI
WL2
0.01745 =
16 EI
θA =
or
522
(Numerically)
...(i)
DEFLECTION OF BEAMS
Now using equation (12.7), we get
yc =
=
WL3
48 EI
F∵ WL = WL × L I
GH 48 EI 16 EI 3 JK
LM∵ WL = 0.01745 from equation (i)OP
N 16 EI
Q
3
WL2
L
×
16 EI 3
4000
3
= 23.26 mm. Ans.
= 0.01745 ×
2
2
Problem 12.3. A beam 3 m long, simply supported at its ends, is carrying a point
load W at the centre. If the slope at the ends of the beam should not exceed 1°, find the deflection at the centre of the beam.
Sol. Given :
Length, L = 3 m = 3 × 1000 = 3000 mm
Point load at centre
Slope at the ends,
=W
θA = θB = 1°
1× π
= 0.01745 radians
180
Let yc = Deflection at the centre
=
Using equation (12.6), we get
θA =
WL2
16 EI
or 0.01745 =
WL2
16 EI
...(i)
Now using equation (12.7), we get
yc =
WL2
L
WL3
=
×
48 EI 16 EI 3
L
= 0.01745 ×
3
3000
3
= 17.45 mm. Ans.
= 0.01745 ×
F∵ WL = 0.01745I
GH 16 EI
JK
2
(∵ L = 3000 mm)
12.5. DEFLECTION OF A SIMPLY SUPPORTED BEAM WITH AN ECCENTRIC
POINT LOAD
A simply supported beam AB of length L and carrying a point load W at a distance a
from support A and at a distance b from support B is shown in Fig. 12.4.
The reactions at A and B can be calculated by taking moments about A.
We find that reaction at A is given by
RA =
W ×b
L
and RB =
W×a
L
523
STRENGTH OF MATERIALS
W
x
C
A
B
X
b
a
W.b
L
L
W.a
L
Fig. 12.4
(a) Now consider a section X at a distance x from A in length AC. The bending moment
at this section is given by,
Mx = RA × x
W ×b
×x
L
But B.M. at any section is also given by equation (12.3) as
=
M = EI
(Plus sign due to sagging)
d2 y
dx 2
Equating the two values of B.M., we get
W ×b
×x
L
dx
Integrating the above equation, we get
EI =
d2 y
2
=
dy W × b x 2
+ C1
=
×
dx
L
2
where C1 is the constant of integration.
Integrating the equation (i), we get
EI
...(i)
W . b x3
+ C1.x + C2
...(ii)
.
2L
3
where C2 is another constant of integration. The values of C1 and C2 are obtained from boundary conditions.
(i) At A, x = 0 and deflection y = 0
Substituting these values in equation (ii), we get
0 = 0 + 0 + C2
∴
C2 = 0
Substituting the value of C2 in equation (ii), we get
EI.y =
EI.y =
W.b
. x3 + C1 . x
6L
...(iii)
dy
= θ C . (Note that value of θC is unknown).
dx
The value of C1 is obtained by substituting these values in equation (i). Hence, we get
(ii) At C, x = a and slope
EI. θC =
524
W . b a2
+ C1
.
L
2
DEFLECTION OF BEAMS
W . b . a2
2L
Substituting the value of C1 in equations (i) and (iii), we get
∴
C1 = EI × θC –
EI
dy W . b
W . b . a2
=
. x2 + EI × θC –
2L
dx
2L
EIy =
F
GH
...(iv)
I
JK
W .b
W . b . a2
x
. x3 + EI . θ C −
6L
2L
...(v)
Equation (iv) gives the slope whereas equation (v) gives the deflection at any point in
section AC. But the value of θC is unknown.
(b) Now consider a section X at a distance x from A in length CB as shown in Fig. 12.5.
Here x varies from a to L. The B.M. at this section is given by,
Mx = RA.x – W(x – a)
W .b
. x – W(x – a)
=
L
W
A
C
B
x
x
a
W.b
L R
A
b
L
RB
W.a
L
Fig. 12.5
But B.M. at this section is also given by equation (12.3) as
d2 y
dx 2
Equating the two values of B.M., we get
M = EI
d2 y
W .b
. x – W(x – a)
L
dx
Integrating the above equation, we get
EI
2
=
dy W . b x 2 W ( x − a) 2
+ C3
=
.
−
dx
L
2
2
where C3 is the constant of integration.
Integrating the equation (vi) again, we get
EI
...(vi)
W . b x 3 W ( x − a) 3
−
+ C3x + C4
...(vii)
.
.
2L
3
2
3
where C4 is another constant of integration. The values of C3 and C4 are obtained from boundary conditions.
(i) At B, x = L and y = 0. Substituting these values in equation (vii), we get
EI.y =
0=
W . b L3 W ( L − a) 3
−
+ C3 × L + C4
.
.
2L
3
2
3
525
STRENGTH OF MATERIALS
W . b . L2 W . b3
+ C3.L + C4
−
6
6
Wb3 W . b . L2
– C3 × L
−
C4 =
6
6
=
∴
(∵ L – a = b)
...(viii)
dy
= θC. (The value of θC is unknown).
dx
The value of C3 is obtained by substituting these values in equation (vi).
Hence, we get from equation (vi)
(ii) At C, x = a and slope
EI.θC =
=
W . b . a2 W
(a – a)2 + C3
−
2L
2
W . b . a2
– 0 + C3
2L
W . b . a2
2L
Substituting the value of C3 in equation (viii), we get
∴
C3 = EI.θC –
C4 =
=
=
=
=
=
=
=
=
=
FG
H
IJ
K
F
GH
...(ix)
I
JK
W . b3 W . b . L2
W . b . a2
−
− EI . θ C −
L
6
6
2L
W . b3 W . b . L2
W . b . a2
−
− EI. θ C . L +
6
6
2
W .b 2
W . b . a2
(b − L2 ) − EI. θ C . L +
6
2
2
W .b 2
W. b . a
– EI.L.θC
(b − L2 ) +
6
2
W .b 2
[b – L2 + 3a2] – EI.L.θC
6
W .b 2
[b – (a + b)2 + 3a2] – EI.L.θC
6
W .b 2
[b – a2 – b2 – 2ab + 3a2] – EI.L.θC
6
W .b
[2a2 – 2ab] – EI.L.θC
6
W .b
× 2a(a – b) – EI.L.θC
6
W. ab
[a – b] – EI.L.θC
3
(∵ L = a + b)
dy
at any point in CB is obtained by substituting the value of C3 in
dx
equation (vi). Hence, we get from equation (vi),
The slope i. e.,
EI
526
dy W . b 2 W
W . b . a2
=
.x −
(x – a)2 + EI.θC −
dx
2L
2
2L
...(x)
DEFLECTION OF BEAMS
The deflection (i.e., y) at any point in CB is obtained by substituting the values of C3 and
C4 in equation (vii). Hence, we get from equation (vii),
EI .
F
GH
I
JK
W .b 3 W
W . b . a2
.x −
x
(x – a)3 + EI. θ C −
6L
6
2L
W . ab
(a – b) – EI . L . θC
...(xi)
3
The deflection at the point C is obtained by substituting x = a in the above equation. Let
yC = the deflection at C. Hence, we get
+
F
GH
I
JK
W . b . a2
W . b . a3 W
3
EI
.
θ
−
a
EI.yC =
(a – a) +
−
C
2L
6L
6
+
LM
N
1 LW . b . a
=
M
EI N 6 L
or
yC =
W .a.b
(a – b) – EI.L.θC
3
W . b . a3
1 W . b . a3
W .a.b
− 0 + EI . a . θ C −
(a − b) − EI . L . θ C
+
EI
6L
2L
3
3
−
W . a . b3 W . a . b
+
(a − b) + EI.a.θC – EI.L.θC]
2L
3
OP
Q
...(A)
The deflection at the point C can also be obtained by substituting x = a in equation (v).
Hence, we get
EI.yC =
yC =
or
F
GH
W . b . a2
W . b . a3
+ EI. θ C −
6L
2L
LM
N
I .a
JK
W . b . a3
1 W . b . a3
+ EI. θ C . a −
EI
6L
2L
OP
Q
...(B)
Equating the two values of yC given by equations (A) and (B), we get
LM
N
OP
Q
LM
N
W . b . a3
1 W . b . a3
1 W . b . a3 W . b . a3 W . a . b
+ EI. θ C . a −
−
+
=
(a – b)
EI
6L
2L
EI
6L
2L
3
+ EI . a . θ C − EI . L . θ C
W .a.b
(a – b) – EI.L.θC
3
W .a.b
or
EI.L.θC =
(a – b)
3
W .a.b
or
θC =
(a – b)
...(12.8)
3 EI . L
The above equation gives the value of θC (i.e., slope at point C). Substituting this value
of θC in equation (iv), we get the slope at any point in AC. Hence, we get from equation (iv),
or
0=
EI
W . b . a2
W .a.b
dy W . b 2
(a − b) −
=
. x + EI ×
2L
3 EI . L
2L
dx
=
W . b . a2
W .b 2 W .a.b
.x +
(a − b) −
2L
3L
2L
527
STRENGTH OF MATERIALS
W .b
[3x2 + 2a(a – b) – 3a2]
6L
W .b
=
[3x2 – 2ab – a2]
6L
=
...(C)
As the length AC is more than length CB, hence maximum slope will be at the support
dy
at A will be equal to θA.
A, where x = 0. Let the slope at A is represented by θA. Hence
dx
Substituting x = 0 in equation (C), we get
EI
FG dy IJ
H dx K
=
at A
W .b
[3 × 0 – 2ab – a2]
6L
W .b
(– 2ab – a2)
6L
− W . a. b
or
θA =
(a + 2b)
...(12.9)
6 EI. L
[Negative sign with the slope means that tangent at the point A makes an angle in the
anti-clockwise or negative direction].
or
EI.θA =
Value of Maximum Deflection
Since ‘a’ is more than ‘b’ hence maximum deflection will be in length AC. The deflection
at any point in length AC is given by equation (v) as
EI.y =
=
F
Ix
GH
JK
L W . a . b (a − b) − W . b . a OP . x
+ M EI .
2L Q
N 3EI. L
LM∵ θ = W . a . b (a − b) from Eq. (12.8)OP
3 EI. L
N
Q
L W . a . b (a − b) − W . b . a OP . x
+M
2L Q
N 3L
W . b . a2
W .b 3
x + EI . θ C −
6L
2L
W .b 3
x
6L
2
C
=
W .b 3
x
6L
2
W .b 3
[x + 2a (a – b)x – 3a2 . x]
6L
W .b 3
[x + 2a2x – 2abx – 3a2x]
=
6L
W .b 3
W .b 3
[x – a2x – 2abx] =
[x – x(a2 + 2ab)]
=
6L
6L
W .b 3
[x – x(a2 + 2ab)]
or
y=
6 EIL
dy
=0
The deflection will be maximum if
dx
dy W . b
=
[3x2 – (a2 + 2ab)]
But
dx 6 EIL
dy
=0
For maximum deflection,
dx
528
=
...(D)
DEFLECTION OF BEAMS
∴
W .b
[3x2 – (a2 + 2ab)] = 0
6 EIL
FG∵ W . b cannot be zeroIJ
K
H 6 EIL
3x2 – (a2 + 2ab) = 0
or
x2 =
or
a 2 + 2ab
3
La
x= M
N
2
OP
Q
1/ 2
+ 2 ab
3
Substituting this value of x in equation (D), we get maximum deflection.
∴
LMF a + 2abI
MNGH 3 JK
W . b L (a + 2ab)
=
M 3× 3
6 EIL N
ymax =
W .b
6 EIL
2
2
3/ 2
−
3/ 2
−
LM
N
Fa
GH
2
+ 2ab
3
1/2
(a2 + 2ab) 3 / 2
3
1
1
W .b
−
. (a 2 + 2ab) 3 / 2
6 EIL
3× 3
3
(1 − 3)
W .b 2
(a + 2ab) 3 / 2
=
6 EIL
3 3
W .b
2
(a + 2ab) 3 / 2
=–
9 3 EI. L
Negative sign means the deflection is in downward direction.
W .b
∴ Downward,
ymax =
(a2 + 2ab)3/2
9 3 EI. L
=
I
JK
OP
Q
(a 2 + 2ab)
OP
Q
OP
PQ
...(12.10)
Deflection Under the Point Load
Let
yC = Deflection under the point load
The deflection at any point in length AC is given by equation (D), as
W .b 3
[x – x(a2 + 2ab)]
yC =
6 EIL
The deflection under the point load will be obtained by substituting x = a in the above
equation.
W .b
[a3 – a(a2 + 2ab)]
∴
yC =
6 EIL
W .b
[a3 – a3 – 2a2b]
=
6 EIL
W .b
Wa 2 b2
=
× (– 2a2b) = –
6 EIL
3 EIL
Negative sign means the deflection is downward.
∴ Downward,
yC =
Wa 2 b2
3 EIL
...(12.11)
Note. The above method for finding the slope and deflection is very laborious. There is a simple
method of finding the slope and deflection at any point in a beam. This method is known as Macaulay’s
method which will be discussed later on.
529
STRENGTH OF MATERIALS
Problem 12.4. Determine : (i) slope at the left support, (ii) deflection under the load and
(iii) maximum deflection of a simply supported beam of length 5 m, which is carrying a point
load of 5 kN at a distance of 3 m from the left end. Take E = 2 × 105 N/mm2 and I = 1 × 108 mm4.
Sol. Given :
Length,
L = 5 m = 5000 mm
Point load,
W = 5 kN = 5000 N
Distance between point load and left end,
a = 3 m = 3000 mm
∴
b = L – a = 5 – 3 = 2 m = 2000 mm
Value of
E = 2 × 105 N/mm2
M.O.I.,
I = 1 × 108 mm4
Let
θA = Slope at the left support,
yC = Deflection under the load, and
ymax = Maximum deflection.
(i) Using equation (12.9), we get
W .a.b
(a + 2b)
θA = –
6 EI. L
5000 × 3000 × 2000
=–
× (3000 + 2 × 2000)
(radians)
6 × 2 × 10 5 × 10 8 × 5000
= – 0.00035 radians. Ans.
Negative sign means that the angle made by tangent at A is anti-clockwise.
(ii) The deflection under the load is given by equation (12.11), as
yC =
=
Wa 2 . b2
3 EIL
5000 × 3000 2 × 2000 2
= 0.6 mm. Ans.
3 × 2 × 10 5 × 10 8 × 5000
(iii) The maximum deflection is given by equation (12.10), as
W. b
ymax =
(a2 + 2ab)3/2
9 3 EI. L
5000 × 2000
=
(30002 + 2 × 3000 × 2000)3/2
9 × 3 × 2 × 10 5 × 10 8 × 5000
1
=
(9000000 + 12000000)3/2
9 × 3 × 10 10
= 0.6173 mm. Ans.
12.6. DEFLECTION OF A SIMPLY SUPPORTED BEAM WITH A UNIFORMLY DISTRIBUTED LOAD
A simply supported beam AB of length L and carrying a uniformly distributed load of w
per unit length over the entire length is shown in Fig. 12.6. The reactions at A and B will be
w× L
equal. Also the maximum deflection will be at the centre. Each vertical reaction =
.
2
530
DEFLECTION OF BEAMS
ω/Unit length
x
A
B
x
C
L
RA = ω × L
2
RB = ω × L
2
Fig. 12.6
w× L
2
Consider a section X at a distance x from A. The bending moment at this section is given
∴
RA = RB =
by,
x
w. L
w . x2
=
.x−
2
2
2
But B.M. at any section is also given by equation (12.3), as
Mx = RA × x – w × x ×
d2 y
dx 2
Equating the two values of B.M., we get
M = EI
w. L
w . x2
x
−
2
2
dx 2
Integrating the above equation, we get
EI
d2 y
=
dy w . L x 2 w x 3
=
.
− .
+ C1
dx
2
2
2 3
where C1 is a constant of integration.
Integrating the above equation again, we get
EI
...(i)
w . L x3 w x4
− .
+ C1x + C2
...(ii)
.
4
3
6 4
where C2 is another constant of integration. Thus two constants of integration (i.e., C1 and C2)
are obtained from boundary conditions. The boundary conditions are :
(i) at x = 0, y = 0 and
(ii) at x = L, y = 0
Substituting first boundary condition i.e., x = 0, y = 0 in equation (ii), we get
0 = 0 – 0 + 0 + C2 or C2 = 0
Substituting the second boundary condition i.e., at x = L, y = 0 in equation (ii), we get
EI.y =
w . L L3 w L4
− .
+ C1 . L
.
4
3
6 4
w . L4 w . L4
+ C1.L
−
=
12
24
wL3 wL3
wL3
+
=−
C1 = –
12
24
24
Substituting the value of C1 in equations (i) and (ii), we get
0=
or
EI
dy w . L 2 w 3 wL3
=
.x − x −
dx
4
6
24
(C2 is already zero)
...(iii)
531
STRENGTH OF MATERIALS
and
EI.y =
or
EIy =
F
GH
I
JK
w. L 3 w
wL3
x −
. x4 + −
x+0
12
24
24
(∵ C2 = 0)
w. L 3 w
wL3
x −
. x4 −
x
12
24
24
...(iv)
FG
H
Equation (iii) is known as slope equation. We can find the slope i. e., the value of
dy
dx
IJ
K
at any point on the beam by substituting the different values of x in this equation.
Equation (iv) is known as deflection equation. We can find the deflection (i.e., the value
of y) at any point on the beam by substituting the different values of x in this equation.
Slope at the Supports
Let
θA = Slope at support A. This is equal to
θB = Slop at support B =
and
FG dy IJ
H dx K
FG dy IJ
H dx K
at A
at B
dy
= θA.
dx
Substituting these values in equation (iii), we get
At A, x = 0 and
wL
w
wL3
×0− ×0−
4
6
24
3
2
wL
WL
=
(∵ w . L = W = Total load)
=−
24
24
WL2
...(12.12)
∴
θA = –
24 EI
(Negative sign means that tangent at A makes an angle with AB in the anti-clockwise
direction)
EI.θA =
By symmetry,
θB = –
WL2
24 EI
...(12.13)
Maximum Deflection
L
. Let
2
L
and x =
in
2
The maximum deflection is at the centre of the beam i.e., at point C, where x =
yC = deflection at C which is also maximum deflection. Substituting y = yC
equation (iv), we get
EI.yC =
FG IJ
H K
w. L L
.
12
2
3
−
FG IJ
H K
w
L
.
24 2
4
−
FG IJ
H K
wL3
L
−
24
2
5w. L4
w. L4 wL4 wL4
−
−
=−
96
384
48
384
5 wL4
5 W . L3
yC = –
.
.
=−
EI
384 EI
384
=
∴
532
(∵ w.L = W = Total load)
DEFLECTION OF BEAMS
Negative sign indicates that deflection is downwards.
∴ Downward deflection,
5 WL3
.
...(12.14)
384 EI
Problem 12.5. A beam of uniform rectangular section 200 mm wide and 300 mm deep
is simply supported at its ends. It carries a uniformly distributed load of 9 kN/m run over the
entire span of 5 m. If the value of E for the beam material is 1 × 104 N/mm2, find :
(i) the slope at the supports and
(ii) maximum deflection.
Sol. Given :
Width,
b = 200 mm
Depth,
d = 300 mm
yC =
bd 3 200 × 300 3
= 4.5 × 108 mm4
=
12
12
U.d.l.,
w = 9 kN/m = 9000 N/m
Span,
L = 5 m = 5000 mm
∴ Total load,
W = w . L* = 9000 × 5 = 45000 N
Value of
E = 1 × 104 N/mm2
Let
θA = Slope at the support
yC = Maximum deflection.
(i) Using equation (12.12), we get
M.O.I.,
and
I=
θA = –
=–
W . L2
24 EI
45000 × 5000 2
24 × 1 × 10 4 × 4.5 × 10 8
= 0.0104 radians. Ans.
(ii) Using equation (12.14), we get
radians
5 W . L3
.
EI
384
5
45000 × 5000 3
×
=
384 1 × 10 4 × 4.5 × 10 8
= 16.27 mm. Ans.
Problem 12.6. A beam of length 5 m and of uniform rectangular section is simply
supported at its ends. It carries a uniformly distributed load of 9 kN/m run over the entire
length. Calculate the width and depth of the beam if permissible bending stress is 7 N/mm2 and
central deflection is not to exceed 1 cm.
Take E for beam material = 1 × 104 N/mm2.
Sol. Given :
Length,
L = 5 m = 5000 mm
U.d.l.,
w = 9 kN/m
yC =
*Here L should be taken in metre. Hence for calculating total load, L must be in metre and in
other calculations L is taken in mm.
533
STRENGTH OF MATERIALS
∴ Total load,
Bending stress,
Central deflection,
Value of
Let
and
W = w.L = 9 × 5 = 45 kN = 45000 N
f = 7 N/mm2
yC = 1 cm = 10 mm
E = 1 × 104 N/mm2
b = Width of beam is mm
d = Depth of beam in mm
bd 3
12
Using equation (12.14), we get
∴ M.O.I.,
I =
5 W . L3
.
384
EI
5
45000 × 5000 3
×
10 =
384
bd 3
1 × 10 4 ×
12
yC =
F I
GH JK
or
5
45000 × 5000 3 × 12
×
384
1 × 10 4 × 10
= 878.906 × 107 mm4
...(i)
The maximum bending moment for a simply supported beam carrying a uniformly distributed load is given by,
bd3 =
or
w . L2 W . L
(∵ W = w.L = Total load)
=
8
8
45000 × 5
45000 × 5
Nm =
× 1000 Nmm
=
8
8
= 28125000 Nmm
Now using the bending equation as
M f
=
I
y
d
28125000
7
∵ Here y =
=
2
d
bd 3
2
12
28125000 × 12 14
=
d
bd 3
28125000
× 12
bd2 =
= 24107142.85 mm3
...(ii)
14
Dividing equation (i) by equation (ii), we get
M=
or
or
or
F I
GH JK
FG
H
FG IJ
H K
838.906 × 107
= 364.58 mm. Ans.
24107142.85
Substituting this value of ‘d’ in equation (ii), we get
b × (364.58)2 = 24107142.85
d=
∴
534
b=
24107142.85
364.58 2
= 181.36 mm. Ans.
IJ
K
DEFLECTION OF BEAMS
Problem 12.7. A beam of length 5 m and of uniform rectangular section is supported at
its ends and carries uniformly distributed load over the entire length. Calculate the depth of
the section if the maximum permissible bending stress is 8 N/mm2 and central deflection is not
to exceed 10 mm.
Take the value of E = 1.2 × 104 N/mm2.
Sol. Given :
Length,
L = 5 m = 5000 mm
Bending stress,
f = 8 N/mm2
Central deflection, yC = 10 mm
Value of
E = 1.2 × 104 N/mm2
Let
W = Total load
and
d = Depth of beam
The maximum bending moment for a simply supported beam carrying a uniformly distributed load is given by,
w. L2 W . L
=
8
8
Now using the bending equation,
M f
=
I
y
f × I 8× I
=
M =
y
(d / 2)
16I
∴
M =
d
Equating the two values of B.M., we get
W . L 16 I
=
8
d
16 × 8 I 128 I
=
W =
L×d
L×d
Now using equation (12.14), we get
M=
or
or
5
×
384
5
×
10 =
384
yC =
or
=
or
WL3
EI
128 I
L3
×
L × d EI
(∵ W = w.L)
...(i)
FG∵ y = d IJ
H
2K
...(ii)
...(iii)
FG∵ y
H
C
= 10 mm and W =
128 I
L×d
IJ
K
5
128 × L2
×
384
d×E
5
128 × L2
5
128 × 5000 2
=
×
×
d=
384
10 × E
384 10 × 1.2 × 10 4
= 347.2 mm = 34.72 cm. Ans.
12.7. MACAULAY’S METHOD..
The procedure of finding slope and deflection for a simply supported beam with an
eccentric point load as mentioned in Art. 12.5, is a very laborious. There is a convenient method
for determining the deflections of the beam subjected to point loads.
535
STRENGTH OF MATERIALS
This method was devised by Mr. M.H. Macaulay and is known as Macaulay’s method.
This method mainly consists in the special manner in which the bending moment at any section is expressed and in the manner in which the integrations are carried out.
12.7.1. Deflection of a Simply Supported Beam with an Eccentric Point Load.
A simply supported beam AB of length L and carrying a point load W at a distance ‘a’ from left
support and at a distance ‘b’ from right support is shown in Fig. 12.7. The reactions at A and
B are given by,
W. b
W. a
and RB =
RA =
L
L
W
a
b
C
A
B
L
RA = W.b
L
RB = W.a
L
Fig. 12.7
The bending moment at any section between A and C at a distance x from A is given by,
W. b
×x
L
The above equation of B.M. holds good for the values of x between 0 and ‘a’. The B.M. at
any section between C and B at a distance x from A is given by,
Mx = RA.x – W × (x – a)
W. b
. x – W(x – a)
=
L
The above equation of B.M. holds good for all values of x between x = a and x = b.
Mx = RA × x =
The B.M. for all sections of the beam can be expressed in a single equation written as
W. b
x – W (x – a)
...(i)
L
Stop at the dotted line for any point in section AC. But for any point in section CB, add
the expression beyond the dotted line also.
Mx =
The B.M. at any section is also given by equation (12.3) as
M = EI
d2 y
...(ii)
dx 2
Hence equating (i) and (ii), we get
– W(x – a)
dx 2
Integrating the above equation, we get
EI
d2 y
EI
536
=
W. b
.x
L
dy W . b x 2
=
+ C1
dx
L 2
–
W ( x − a) 2
2
...(iii)
...(iv)
DEFLECTION OF BEAMS
where C1 is a constant of integration. This constant of integration should be written after the
first term. Also the brackets are to be integrated as a whole. Hence the integration of (x – a) will
be
( x − a) 2
x2
and not
− ax .
2
2
Integrating equation (iv) once again, we get
W ( x − a) 3
W. b x3
+ C1x + C2 −
...(v)
.
2L 3
2
3
where C2 is another constant of integration. This constant is written after C1x. The integration
EIy =
of (x – a)2 will be
FG x − a IJ
H 3 K
3
. This type of integration is justified as the constant of integrations
C1 and C2 are valid for all values of x.
The values of C1 and C2 are obtained from boundary conditions. The two boundary
conditions are :
(i) At x = 0, y = 0 and
(ii) At x = L, y = 0
(i) At A, x = 0 and y = 0. Substituting these values in equation (v) upto dotted line only,
we get
0 = 0 + 0 + C2
∴
C2 = 0
(ii) At B, x = L and y = 0. Substituting these values in equation (v), we get
0=
W . b L3
W ( L − a) 3
+ C1 × L + 0 –
.
2L 3
2
3
(∵ C2 = 0. Here complete Eq. (v) is to be taken)
W . b . L2
W b3
+ C1 × L –
(∵ L – a = b)
2 3
6
W
W . b. L2
W. b 2
∴
C1 × L =
. b3 –
(L – b2)
=−
6
6
6
W. b 2
(L – b2)
...(vi)
∴
C1 = –
6L
Substituting the value of C1 in equation (iv), we get
2
dy W . b x 2
W. b 2
− W ( x − a)
=
+ −
L − b2
EI
6L
dx
L 2
2
=
LM
N
jOPQ
e
2
W. b . x2 W. b 2
− W ( x − a)
L − b2
...(vii)
−
2L
6L
2
Equation (vii) gives the slope at any point in the beam. Slope is maximum at A or B. To
find the slope at A, substitute x = 0 in the above equation upto dotted line as point A lies in AC.
dy
W .b
Wb 2
∵
at A = θ A
×0−
(L – b2)
∴
EI.θA =
dx
2L
6L
Wb 2
=–
(L – b2)
6L
Wb
(L2 – b2)
(as given before)
∴
θA = –
6 EIL
=
e
j
IJ
K
FG
H
537
STRENGTH OF MATERIALS
Substituting the values of C1 and C2 in equation (v), we get
LM
N
OP
Q
W
Wb 2
W .b 3
(x – a)3
...(viii)
.x + −
( L − b2 ) x + 0 –
6
6L
6L
Equation (viii) gives the deflection at any point in the beam. To find the deflection yc
under the load, substitute x = a in equation (viii) and consider the equation upto dotted line
(as point C lies in AC). Hence, we get
W .b
W .b
W .b
. a3 –
(L2 – b2)a =
. a (a2 – L2 + b2)
EIyc =
6L
6L
6L
W .a.b 2
=–
(L – a2 – b2)
6L
W .a.b
=–
[(a + b)2 – a2 – b2]
(∵ L = a + b)
6L
W .a.b 2
=–
[a + b2 + 2ab – a2 – b2]
6L
W .a.b
Wa 2 . b2
[2ab] = –
=–
6L
3L
2
2
Wa . b
∴
yc = –
...(same as before)
3 EIL
EIy =
Note. While using Macaulay’s Method, the section x is to be taken in the last portion of the beam.
Problem 12.8. A beam of length 6 m is simply supported at its ends and carries a point
load of 40 kN at a distance of 4 m from the left support. Find the deflection under the load and
maximum deflection. Also calculate the point at which maximum deflection takes place. Given
M.O.I. of beam = 7.33 × 107 mm4 and E = 2 × 105 N/mm2.
Sol. Given :
Length,
L = 6 m = 6000 mm
Point load,
W = 40 kN = 40,000 N
Distance of point load from left support, a = 4 m = 4000 mm
∴
b = L – a = 6 – 4 = 2 m = 2000 mm
Let
yc = Deflection under the load
ymax = Maximum deflection
Using equation
yc = –
∴
yc = –
W . a 2 . b2
3 EIL
40000 × 4000 2 × 2000 2
3 × 2 × 10 5 × 7.33 × 10 7 × 6000
= – 9.7 mm. Ans.
Problem 12.9. A beam of length 6 m is simply supported at its ends and carries two
point loads of 48 kN and 40 kN at a distance of 1 m and 3 m respectively from the left support.
Find :
(i) deflection under each load,
(ii) maximum deflection, and
(iii) the point at which maximum deflection occurs.
Given E = 2 × 105 N/mm2 and I = 85 × 106 mm4.
538
DEFLECTION OF BEAMS
Sol. Given :
I = 85 × 105 mm4 ; E = 2 × 105 N/mm2
First calculate the reactions RA and RB.
Taking moments about A, we get
RB × 6 = 48 × 1 + 40 × 3 = 168
168
∴
RB =
= 28 kN
6
∴
RA = Total load – RB = (48 + 40) – 28 = 60 kN
A
48 kN
40 kN
C
D
B
1m
3m
6m
RA
RB
Fig. 12.8
Consider the section X in the last part of the beam (i.e., in length DB) at a distance x
from the left support A. The B.M. at this section is given by,
EI
d2 y
dx
2
= RA.x
– 48(x – 1)
= 60x – 48(x – 1) Integrating the above equation, we get
– 40(x – 3)
– 40(x – 3)
dy 60 x 2
( x − 1) 2 ( x − 3) 2
+ C1 – 48
–
40
=
dx
2
2
2
= 30x2 + C1 – 24(x – 1)2 – 20(x – 3)2
Integrating the above equation again, we get
EI
30 x 3
+ C1x + C2
3
...(i)
− 24( x − 1) 3 − 20( x − 3) 3
3
3
20
= 10x3 + C1x + C2 – 8(x – 1)3 −
(x – 3)3
...(ii)
3
To find the values of C1 and C2, use two boundary conditions. The boundary conditions are:
(i) at x = 0, y = 0, and
(ii) at x = 6 m, y = 0.
(i) Substituting the first boundary condition i.e., at x = 0, y = 0 in equation (ii) and
considering the equation upto first dotted line (as x = 0 lies in the first part of the beam), we get
0 = 0 + 0 + C2 ∴ C2 = 0
(ii) Substituting the second boundary condition i.e., at x = 6 m, y = 0 in equation (ii) and
considering the complete equation (as x = 6 lies in the last part of the beam), we get
20
0 = 10 × 63 + C1 × 6 + 0 – 8(6 – 1)3 –
(6 – 3)3
(∵ C2 = 0)
3
20
or
0 = 2160 + 6C1 – 8 × 53 –
× 33
3
= 2160 + 6C1 – 1000 – 180 = 980 + 6C1
EIy =
539
STRENGTH OF MATERIALS
− 980
= – 163.33
6
Now substituting the values of C1 and C2 in equation (ii), we get
20
(x – 3)3
...(iii)
EIy = 10x3 – 163.33x – 8(x – 1)3 –
3
(i) (a) Deflection under first load i.e., at point C. This is obtained by substituting x = 1 in
equation (iii) upto the first dotted line (as the point C lies in the first part of the beam). Hence,
we get
EI. yc = 10 × 13 – 163.33 × 1
= 10 – 163.33 = – 153.33 kNm3
= – 153.33 × 103 Nm3
= – 153.33 × 103 × 109 Nmm3
= – 153.33 × 1012 Nmm3
∴
C1 =
− 153.33 × 10 12
− 153.33 × 10 12
=
mm
EI
2 × 10 5 × 85 × 10 6
= – 9.019 mm. Ans.
(Negative sign shows that deflection is downwards).
(b) Deflection under second load i.e. at point D. This is obtained by substituting x = 3 m
in equation (iii) upto the second dotted line (as the point D lies in the second part of the beam).
Hence, we get
EI.yD = 10 × 33 – 163.33 × 3 – 8(3 – 1)3
= 270 – 489.99 – 64 = – 283.99 kNm3
= – 283.99 × 1012 Nmm3
∴
yc =
− 283.99 × 10 12
= – 16.7 mm. Ans.
2 × 10 5 × 85 × 10 6
(ii) Maximum Deflection. The deflection is likely to be maximum at a section between C
dy
and D. For maximum deflection,
should be zero. Hence equate the equation (i) equal to
dx
zero upto the second dotted line.
∴
30x2 + C1 – 24(x – 1)2 = 0
or
30x2 – 163.33 – 24(x2 + 1 – 2x) = 0
(∵ C1 = – 163.33)
2
or
6x + 48x – 187.33 = 0
The above equation is a quadratic equation. Hence its solution is
∴
yD =
x=
− 48 ± 48 2 + 4 × 6 × 187.33
= 2.87 m.
2×6
(Neglecting – ve root)
Now substituting x = 2.87 m in equation (iii) upto the second dotted line, we get maximum deflection as
EIymax = 10 × 2.873 – 163.33 × 2.87 – 8(2.87 – 1)3
= 236.39 – 468.75 – 52.31
= 284.67 kNm3 = – 284.67 × 1012 Nmm3
∴
540
ymax =
− 284.67 × 10 12
= – 16.745 mm. Ans.
2 × 10 5 × 85 × 10 6
DEFLECTION OF BEAMS
Problem 12.10. A beam of length 8 m is simply supported at its ends. It carries a uniformly distributed load of 40 kN/m as shown in Fig. 12.9. Determine the deflection of the beam
at its mid-point and also the position of maximum deflection and maximum deflection. Take E
= 2 × 105 N/mm2 and I = 4.3 × 108 mm4.
40 kN/m
C
A
D
1m
B
4m
3m
8m
RA
RB
Fig. 12.9
Sol. Given :
Length,
L=8m
U.d.l.,
W = 40 kN/m
Value of
E = 2 × 105 N/mm2
Value of
I = 4.3 × 108 mm4
First calculate the reactions RA and RB.
Taking moments about A, we get
FG
H
RB × 8 = 40 × 4 × 1 +
IJ
K
4
= 480 kN
2
480
= 60 kN
8
∴
RA = Total load – RB = 40 × 4 – 60 = 100 kN
In order to obtain the general expression for the bending moment at a distance x from
the left end A, which will apply for all values of x, it is necessary to extend the uniformly
distributed load upto the support B, compensating with an equal upward load of 40 kN/m over
the span DB as shown in Fig. 12.10. Now Macaulay’s method can be applied.
∴
RB =
40 kN/m
A
B
C
D
1m
4m
40 kN/m
3m
8m
RA
RB
Fig. 12.10
The B.M. at any section at a distance x from end A is given by,
EI
or
d2 y
dx
2
= RA.x – 40(x – 1) ×
d2 y
( x − 1)
2
= 100x – 20 (x – 1)2 dx 2
Integrating the above equation, we get
EI
EI
dy 100 x 2
+ C1 =
dx
2
–
+ 40 × (x – 5) ×
( x − 5)
2
+ 20 (x – 5)2
20( x − 1) 3
3
+ 20
( x − 5) 3
3
...(i)
541
STRENGTH OF MATERIALS
Integrating again, we get
20 ( x − 5) 4
20 (n − 1) 4 +
3
3
3
4
3
5
5
x
= 50
+ C1x + C2 – (x – 1)4 + (x – 5)4
...(ii)
3
3
3
where C1 and C2 are constants of integration. Their values are obtained from boundary conditions
which are :
(i) at x = 0, y = 0 and
(ii) at x = 8 m, y = 0
(i) Substituting x = 0 and y = 0 in equation (ii) upto first dotted line (as x = 0 lies in the first
part AC of the beam), we get
0 = 0 + C1 × 0 + C2
∴ C2 = 0
(ii) Substituting x = 8 and y = 0 in complete equation (ii) (as point x = 8 lies in the last part
DB of the beam), we get
50
5
5
0=
× 83 + C1 × 8 + 0 – (8 – 1)4 + (8 – 5)4
(∵ C2 = 0)
3
3
3
= 8533.33 + 8C1 – 4001.66 + 135
or
8C1 = – 4666.67
− 4666.67
= – 583.33
or
C1 =
8
Substituting the value of C1 and C2 in equation (ii), we get
50 3
5
5
x – 583.33x – (x – 1)4 + (x – 4)4
...(iii)
EIy =
3
3
3
(a) Deflection at the centre
By substituting x = 4 m in equation (iii) upto second dotted line, we get the deflection at the
centre. [The point x = 4 lies in the second part (i.e., CD) of the beam].
50
5
∴
EI.y =
× 43 – 583.33 × 4 – (4 – 1)4
3
3
= 1066.66 – 2333.32 – 135 = – 1401.66 kNm3
= – 1401.66 × 1000 Nm3
= – 1401.66 × 1000 × 109 Nmm3
= – 1401.66 × 1012 Nmm3
EIy = 50
x3
+ C1x + C2 3
−
− 1401.66 × 10 12
− 1401.66 × 10 12
=
EI
2 × 10 5 × 4.5 × 10 8
= – 16.29 mm downward. Ans.
(b) Position of maximum deflection
The maximum deflection is likely to lie between C and D. For maximum deflection the
dy
slope
should be zero. Hence equating the slope given by equation (i) upto second dotted line to
dx
zero, we get
∴
y=
20
x2
+ C1 –
(x – 1)3
3
2
0 = 50x2 – 583.33 – 6.667(x – 1)3
The above equation is solved by trial and error method.
0 = 100
542
...(iv)
DEFLECTION OF BEAMS
Let x = 1, then R.H.S. of equation (iv)
= 50 – 583.33 – 6.667 × 0 = – 533.33
Let x = 2, then R.H.S. = 50 × 4 – 583.33 – 6.667 × 1 = – 390.00
Let x = 3, then R.H.S. = 50 × 9 – 583.33 – 6.667 × 8 = – 136.69
Let x = 4, then R.H.S. = 50 × 16 – 583.33 – 6.667 × 27 = + 36.58
In equation (iv), when x = 3 then R.H.S. is negative but when x = 4 then R.H.S. is positive.
Hence exact value of x lies between 3 and 4.
Let x = 3.82, then R.H.S.
= 50 × 3.82 – 583.33 – 6.667 (3.82 – 1)3
= 729.63 – 583.33 – 149.51 = – 3.22
Let x = 3.83, then R.H.S.
= 50 × 3.832 – 583.33 = 6.667 (3.83 – 1)3
= 733.445 – 583.33 – 151.1 = – 0.99
The R.H.S. is approximately zero in comparison to the three terms (i.e., 733.445, 583.33
and 151.1).
∴ Value of x = 3.83. Ans.
Hence maximum deflection will be at a distance of 3.83 m from support A.
(c) Maximum deflection
Substituting x = 3.83 m in equation (iii) upto second dotted line, we get the maximum
deflection [the point x = 3.83 lies in the second part i.e., CD of the beam.]
50
5
∴
EI.ymax =
× 3.833 – 583.33 × 3.83 – (3.83 – 1)4
3
3
= 936.36 – 2234.15 – 106.9 = – 1404.69 kNm3
= – 1404.69 × 1012 Nmm3
∴
ymax =
− 1404.69 × 10 12
= – 16.33 mm. Ans.
2 × 10 5 × 4.3 × 10 8
Problem 12.11. An overhanging beam ABC is loaded as shown in Fig. 12.11. Find the
slopes over each support and at the right end. Find also the maximum upward deflection between the supports and the deflection at the right end.
Take E = 2 × 105 N/mm2 and I = 5 × 108 mm4.
10 kN
A
B
6m
C
3m
RA
RB
Fig. 12.11
Sol. Given :
Point load,
W = 10 kN
Value of
E = 2 × 105 N/mm2
Value of
I = 5 × 108 mm4
First calculate the reaction RA and RB.
Taking moments about A, we get
RB × 6 = 10 × 9
543
STRENGTH OF MATERIALS
10 × 9
= 15 kN
6
∴
RA = Total load – RB = 10 – 15 = – 5 kN
Hence the reaction RA will be in the downward direction. Hence Fig. 12.11 will be modified
as shown in Fig. 12.12. Now write down an expression for the B.M. in the last section of the beam.
∴
RB =
10 kN
A
B
6m
C
3m
RA = 5 kN
RB = 15 kN
Fig. 12.12
The B.M. at any section at a distance x from the support A is given by,
EI
d2 y
dx 2
= – RA × x + RB × (x – 6)
= – 5x + 15(x – 6)
Integrating the above equation, we get
dy − 5 x 2
+ C1 =
dx
2
Integrating again, we get
EI
+
15 ( x − 6) 2
2
(∵ RA = 5)
...(i)
15 ( x − 6) 3
5 x3
+ C1x + C2 +
2
3
2 3
5 3
5
= – x + C1x + C2 + (x – 6)3
...(ii)
6
2
where C1 and C2 are constant of integration. Their values are obtained from boundary conditions
which are :
(i) at x = 0, y = 0 and
(ii) at x = 6 m, y = 0.
(i) Substituting x = 0 and y = 0 in equation (ii) upto dotted line (as x = 0 lies in the first part
AB of the beam), we get
0 = 0 + C1 × 0 + C2
∴ C2 = 0
(ii) Substituting x = 6 m and y = 0 in equation (ii) upto dotted line (as x = 6 lies in the first
part AB of the beam), we get
−5
× 63 + C1 × 6 + 0 = – 5 × 36 + 6C1
(∵ C2 = 0)
0=
6
+ 5 × 36
∴
C1 =
= 30
6
Substituting the values of C1 and C2 in equations (i) and (ii), we get
dy
5
15
= – x2 + 30 +
(x – 6)2
...(iii)
EI
dx
2
2
5
5
and
EIy = – x3 + 30x + (x – 6)3
...(iv)
6
2
EI.y = –
544
DEFLECTION OF BEAMS
(a) Slope over the support A
By substituting x = 0 in equation (iii) upto dotted line, we get the slope at support A (the
point x = 0 lies in the first part AB of the beam).
5
dy
∴
EI.θA = – × 0 + 30 = 30 kNm2 = 30 × 1000 Nm2 ∵
at A = θ A
2
dx
= 30 × 1000 × 106 Nmm2 = 30 × 109 Nmm2
∴
FG
H
IJ
K
30 × 10 9
30 × 10 9
=
E×I
2 × 10 5 × 5 × 10 8
= 0.0003 radians. Ans.
θA =
(b) Slope at the support B
By substituting x = 6 m in equation (iii) upto dotted line, we get the slope at support B (the
point x = 6 lies in the first part AB of the beam).
5
× 62 + 30 = – 90 + 30
2
= – 60 kNm2 = – 60 × 109 Nmm2
EI.θB = –
∴
FG∵ dy
H dx
at B = θ B
IJ
K
− 60 × 10 9
− 60 × 10 9
=
θB =
E×I
2 × 10 5 × 5 × 10 8
= – 0.0006 radians. Ans.
(c) Slope at the right end i.e., at C
By substituting x = 9 m in equation (iii), we get the slope at C. In this case, complete
equation is to be taken as point x = 9 m lies in the last part of the beam.
dy
5
15
∵
at C = θ C
(9 – 6)2
∴
EI.θC = – × 92 + 30 +
dx
2
2
= – 202.5 + 30 + 67.5 = – 105 kNm2
= – 105 × 109 Nmm2
FG
H
∴
IJ
K
− 105 × 10 9
− 105 × 10 9
=
E× I
2 × 10 5 × 5 × 10 8
= – 0.00105 radians. Ans.
θC =
(d) Maximum upward deflection between the supports
dy
should be zero. Hence equating the
dx
slope given by equation (iii) to be zero upto dotted line, we get
5
0 = – x2 + 30 = – 5x2 + 60
2
For maximum deflection between the supports,
or
5x2 = 60
or x =
60
= 12 = 3.464 m
5
Now substituting x = 3.464 m in equation (iv) upto dotted line, we get the maximum
deflection as
5
EIymax = – × 3.4643 + 30 × 3.464
6
545
STRENGTH OF MATERIALS
= – 34.638 + 103.92 = 69.282 kNm3
= 69.282 × 1000 × 109 Nmm3 = 69.282 × 1012 mm3
69.282 × 10 12
2 × 10 5 × 5 × 10 8
= 0.6928 mm (upward). Ans.
∴
ymax =
(e) Deflection at the right end i.e., at point C
By substituting x = 9 m in equation (iv), we get the deflection at point C. Here complete
equation is to be taken as point x = 9 m lies in the last part of the beam.
5
5
∴
EI yC = – × 93 + 30 × 9 + (9 – 6)3
6
2
= – 607.5 + 270 + 67.5
= – 270 kNm3 = – 270 × 1012 Nmm3
− 270 × 10 12
2 × 10 5 × 5 × 10 8
= – 2.7 mm (downwards). Ans.
Problem 12.12. A beam ABC of length 9 m has one support of the left end and the other
support at a distance of 6 m from the left end. The beam carries a point load of 1 kN at right end
and also carries a uniformly distributed load of 4 kN/m over a length of 3 m as shown in
Fig. 12.13. Determine the slope and deflection at point C.
Take E = 2 × 105 N/mm2 and I = 5 × 108 mm4.
Sol. Given :
Point load,
W = 12 kN
U.d.l.,
w = 4 kN/m
Value of
E = 2 × 105 N/mm2
Value of
I = 5 × 108 mm4
First calculate the reactions RA and RB.
∴
yC =
12 kN
4 kN/m
D
B
A
C
3m
6m
3m
RA
RB
Fig. 12.13
Taking moments about A, we get
FG
H
IJ
K
3
+ 12 × 9
2
= 54 + 108 = 162
RB × 6 = 4 × 3 × 3 +
∴
and
546
162
= 27 kN (↑)
6
RA = Total load – RB = 24 – 27 = – 3 kN (↓)
RB =
DEFLECTION OF BEAMS
Negative sign shows that RA will be acting downwards. In order to obtain general expression for the bending moment at a distance x from the left end A, which will apply for all values of
x, it is necessary to extend the uniformly distributed load upto point C, compensating with an
equal upward load of 4 kN/m over the span BC as shown in Fig. 12.14. Now Macaulay’s method
can be applied.
12 kN
4 kN/m
B
D
C
A
4 kN/m
3m
3m
3m
RB = 27 kN
RA = 3 kN
Fig. 12.14
The B.M. at any section at a distance x from the support A is given by,
( x − 3) ( x − 6)
= – RA × x – 4(x – 3)
+ RB(x – 6) + 4(x – 6)
2
2
dx
2
2
= – 3x – 2(x – 3) + 27(x – 6) + 2(x – 6)
Integrating the above equation, we get
EI
d2 y
2
dy
3x2
+ C1 =−
dx
2
Integrating again, we get
EI
–
2 ( x − 3) 3
3
+
2 (x − 6) 3
27 ( x − 6) 2
+
3
2
...(i)
3 x3
2 ( x − 3) 4 27 ( x − 6) 3 2 ( x − 6) 4
+ C1x + C2 –
+
+
2 3
3
4
2
3
3
4
3
4
9
1
x
( x − 3) ( x − 6) 3 + (x – 6)4 ..(ii)
or
EI.y = –
+ C1x + C2 –
+ 2
6
2
6
where C1 and C2 are constant of integration. Their values are obtained from boundary conditions which are :
(i) at x = 0, y = 0 and
(ii) at x = 6 m, y = 0.
(i) Substituting the x = 0 and y = 0 in equation (ii) upto first dotted line (as x = 0 lies in
the first part AD of the beam), we get
0 = 0 + C1 × 0 + C2
∴ C2 = 0
(ii) Substituting x = 6 and y = 0 in equation (ii) upto second dotted line (as x = 6 lies in
the second part DB of the beam), we get
EI.y = –
63
(6 − 3) 4
+ C1 × 6 + 0 –
2
6
= – 108 + 6C1 – 13.5 = – 121.5 + 6C1
121.5
C1 =
= 20.25
6
Substituting the values of C1 and C2 in equations (i) and (ii), we get
3
dy
27
2
2 ( x − 3) 3 2 3
= − x2 + 20.25 –
EI
+ 2 (x – 6) + 3 (x – 6) ..(iii)
2
dx
3
|
|
1
9
1
x3
EIy = –
+ 20.2 × x | – (x – 3)4 | + (x – 6)3 + (x – 6)4
|
|
6
2
6
2
...(iv)
0=–
or
and
547
STRENGTH OF MATERIALS
(a) Slope at the point C
By substituting x = 9 m in equation (iii), we get the slope at C. Here complete equation is to
be taken as point x = 9 m lies in the last part of the beam.
3
2
27
2
(9 – 6)2 + (9 – 6)3
∴
EI.θC = – × 92 + 20.25 – (9 – 3)3 +
2
3
2
3
dy
∵
at C = θ C
dx
= – 121.5 + 20.25 – 144 + 121.5 + 18 = – 105.75 kNm2
= – 105.75 × 103 × 106 Nmm2 = – 105.75 × 109 Nmm2
FG
H
∴
θC = –
105.75 × 10 9
2 × 10 5 × 5 × 10 8
IJ
K
= – 0.0010575 radians. Ans.
(b) Deflection at the point C
By substituting x = 9 m in complete equation (iv), we get the deflection at C.
1
9
1
93
+ 20.25 × 9 – (9 – 3)4 + (9 – 6)3 + (9 – 6)4
6
2
6
2
= – 364.5 + 182.25 – 216 + 121.5 + 13.5
= – 263.25 kNm3 = – 263.25 × 1012 Nmm3
∴
EI × yC = –
263.25 × 10 12
= – 2.6325 mm. Ans.
2 × 10 5 × 5 × 10 8
Problem 12.13. A horizontal beam AB is simply supported at A and B, 6 m apart. The
beam is subjected to a clockwise couple of 300 kNm at a distance of 4 m from the left end as
shown in Fig. 12.15. If E = 2 × 105 N/mm2 and I = 2 × 108 mm4, determine :
(i) deflection at the point where couple is acting and
(ii) the maximum deflection.
∴
yC = –
C
A
B
300 kNm
4m
6m
RA
RB
Fig. 12.15
Sol. Given :
Length,
L=6m
Couple
= 300 kNm
Value of
E = 2 × 105 N/mm2
Value of
I = 2 × 108 mm4
First calculate the reactions RA and RB.
Taking moments about A, we get
RB × 6 = 300
300
= 50 kN (↑)
∴
RB =
6
and
RA= Total load – RB = 0 – 50 kN
= – 50 kN
548
(∵ There is no load on beam)
DEFLECTION OF BEAMS
Negative sign shows that RA is acting downwards as shown in Fig. 12.16.
C
A
B
300 kNm
2m
4m
RB = 50 kN
RA = 50 kN
Fig. 12.16
The B.M. at any section at a distance x from A, is given by
EI
d2 y
dx 2
= – 50x + 300
= – 50x + 300(x – 4)0
Integrating the above equation, we get
dy
− 50 x 2
+ C1 =−
dx
2
Integrating again, we get
EI
+ 300(x – 4)
...(i)
50 x 3
300 ( x − 4) 2
×
+ C1x + C2 +
2
3
2
25 3
=–
x + C1x + C2 + 150(x – 4)2
...(ii)
3
where C1 and C2 are constants of integration. Their values are obtained from boundary conditions
which are :
(i) at x = 0, y = 0 and
(ii) at x = 6 m and y = 0.
(i) Substituting x = 0 and y = 0 in equation (ii) upto dotted line, we get
0 = 0 + C1 × 0 + C2 ∴ C2 = 0
(ii) Substituting x = 6 m and y = 0 in complete equation (ii), we get
25
× 63 + C1 × 6 + 0 + 150(6 – 4)2
0=–
3
= – 1800 + 6C1 + 600
1800 − 600
= 200
∴
C1 =
6
Substituting the values of C1 and C2 in equation (ii), we get
25 3
x + 200x + 150(x – 4)2
(∵ C2 = 0) ...(iii)
EIy = –
3
(i) Deflection at C (i.e., yC)
By substituting x = 4 in equation (iii) upto dotted line, we get the deflection at C.
25
× 43 + 200 × 4
∴
EI yC = –
3
= – 533.33 + 800 = + 266.67 kNm3
= 266.67 × 1012 Nmm3
EIy = –
∴
yC =
266.67 × 10 12
2 × 10 5 × 2 × 10 8
= 6.66 mm upwards. Ans.
549
STRENGTH OF MATERIALS
(ii) Maximum deflection
First find the point where maximum deflection takes place. The maximum deflection is
dy
likely to occur in the larger segment AC of the beam. For maximum deflection
should be zero.
dx
Hence equating the slope given by equation (i) upto dotted line to zero, we get
50 2
–
x + 200 = 0
(∵ C1 = 200)
2
2
or
– 25x + 200 = 0
or
200
=2×
25
x=
2 in equation (iii) upto dotted line, we get the maximum
Now substituting x = 2 ×
deflection.
∴
∴
25
× (2 × 2 )3 + 200(2 × 2 )
3
= – 188.56 + 565.68
= 377.12 kNm3 = 377.12 × 1012 Nmm3
EI × ymax = –
ymax =
377.12 × 10 12
2 × 10 5 × 2 × 10 8
12.8. MOMENT AREA METHOD..
Fig. 12.17 shows a beam AB carrying some type of loading, and hence subjected to bending moment as shown in
Fig. 12.17 (a). Let the beam bent into
AQ1P1B as shown in Fig. 12.17 (b).
Due to the load acting on the beam.
Let A be a point of zero slope and zero
deflection.
Consider an element PQ of small
length dx at a distance x from B. The
corresponding points on the deflected
beam are P1Q1 as shown in Fig. 12.17 (b).
Let R = Radius of curvature of deflected part P1Q1
dθ = Angle subtended by the
arc P1Q1 at the centre O
M = Bending moment between
P and Q
P1C = Tangent at point P1
Q1D = Tangent at point Q1.
The tangent at P1 and Q1 are cutting the vertical line through B at points
C and D. The angle between the normals
at P1 and Q1 will be equal to the angle
550
2 m
= 9.428 mm upwards. Ans.
Area = M.dx
(a)
M
P
B
Q
dx
x
A
B.M. Diagram
L
A
D
Q1
P1
y
dy
(b)
dθ
C
B
R
dθ
O
Fig. 12.17
DEFLECTION OF BEAMS
between the tangents at P1 and Q1. Hence the angle between the lines CP1 and DQ1 will be equal
to dθ.
For the deflected part P1Q1 of the beam, we have
P1Q1 = R.dθ
But
P1Q1 ≈ dx
∴
dx = R.dθ
dx
...(i)
∴
dθ =
R
But for a loaded beam, we have
M E
EI
=
or R =
I
R
M
Substituting the values of R in equation (i), we get
dx
M dx
dθ =
=
...(ii)
EI
EI
M
Since the slope at point A is assumed zero, hence total slope at B is obtained by integrating
the above equation between the limits 0 and L.
FG IJ
H K
∴
θ=
z
L
0
M . dx
1
=
EI
EI
z
L
M . dx
0
But M.dx represents the area of B. M. diagram of length dx. Hence
z
L
0
M . dx represents the
area of B. M. diagram between A and B.
1
∴
θ=
[Area of B. M. diagram between A and B]
EI
But
θ = slope at B = θB
∴ Slope at B,
Area of B. M . diagram between A and B
θB =
...(12.15)
EI
If the slope at A is not zero then, we have
“Total change of slope between B and A is equal to the erea of B. M. diagram between B
and A divided by the flexural rigidity EI”
Area of B.M. between A and B
or
θ B – θA =
...(12.16)
EI
Now the deflection, due to bending of the portion P1Q1 is given by
dy = x.dθ
Substituting the value of dθ from equation (ii), we get
M . dx
dy = x .
...(iii)
EI
Since deflection at A is assumed to be zero, hence the total deflection at B is obtained by
integrating the above equation between the limits zero and L.
z
y=
L
z
xM . dx
1 L
=
xM . dx
0
EI
EI 0
But x × M.dx represents the moment of area of the B.M. diagram of length dx about
point B.
551
∴
STRENGTH OF MATERIALS
Hence
z
L
0
xM . dx represents the moment of area of the B.M. diagram between B and A
about B. This is equal to the total area of B.M. diagram between B and A multiplied by the
distance of the C.G. of the B.M. diagram area from B.
1
Ax
× A×x =
∴
y=
...(12.17)
EI
EI
where A = Area of B.M. diagram between A and B
x = Distance of C.G. of the area A from B.
12.9. MOHR’S THEOREMS..
The results given by equation (12.15) for slope and (12.17) for deflection are known as
Mohr’s theorems. They are state as :
I. The change of slope between any two points is equal to the net area of the B.M.
diagram between these points divided by EI.
II. The total deflection between any two points is equal to the moment of the area of
B.M. diagram between the two points about the last point (i.e., B) divided by EI.
The Mohr’s theorems is conveniently used for following cases :
1. Problems on cantilevers (zero slope at fixed end).
2. Simply supported beams carrying symmetrical loading (zero slope at the centre).
3. Beams fixed at both ends (zero slope at each end).
The B.M. diagram is a parabola for uniC
formly distributed loads. The following prop- D
erties of area and centroids or parabola are
given as :
x2
G2
Let
BC = d
Area A
d
AB = b
In Fig. 12.18, ABC is a parabola and
x
G
ABCD is a surrounding rectangle.
Let A1 = Area of ABC
B
x 1 = Distance of C.G. of A1 from AD A
b
A2 = Area of ACD
=
Distance
of
C.G.
of
A
from
AD
x2
2
Fig. 12.18
G1 = C.G. of area A1
G2 = C.G. of area A2.
Then
A1 = Area of parabola ABC
2
= bd
3
A2 = Area ACD = Area ABCD – Area ABC
2
1
= b × d – bd = bd
3
3
5
x1 = b
8
1
x2 = b.
4
552
1
1
1
DEFLECTION OF BEAMS
12.10. SLOPE AND DEFLECTION OF A SIMPLY SUPPORTED BEAM CARRYING A
POINT LOAD AT THE CENTRE BY MOHR’S THEOREM
Fig. 12.19 (a) shows a simply supported AB of length L and carrying a point load W at the
centre of the beam i.e., at point C. The B.M. diagram is shown in Fig. 12.19 (b). This is a case of
symmetrical loading, hence slope is zero at the centre i.e., at point C.
But the deflection is maximum at the centre.
L
2
A
W
C
B
(a)
W
2
W
2
L
D
(b)
WL
4
A
B
C
2×L
3 2
B.M. Diagram
L
2
Fig. 12.19
Now using Mohr’s theorem for slope, we get
Area of B.M. diagram between A and C
Slope at
A=
EI
But area of B.M. diagram between A and C
= Area of triangle ACD
1 L WL WL2
× ×
=
2 2
4
16
2
WL
∴ Slope at A or θA
=
EI
Now using Mohr’s theorem for deflection, we get from equation (12.17) as
Ax
y=
EI
where A = Area of B.M. Diagram between A and C
=
WL2
16
x = Distance of C.G. of area A from A
2 L L
= × =
3 2 3
WL2 L
×
3
3 = WL .
∴
y = 16
EI
48 EI
=
553
STRENGTH OF MATERIALS
12.11. SLOPE AND DEFLECTION OF A SIMPLY SUPPORTED BEAM CARRYING A
UNIFORMLY DISTRIBUTED LOAD BY MOHR’S THEOREM
Fig. 12.20 (a) shows a simply supported beam AB of length L and carrying a uniformly
distributed load of w/unit length over the entire span. The B.M. diagram is shown in Fig. 12.20
(b). This is a case of symmetrical loading, hence slope is zero at the centre i.e., at point C.
L
w/Unit length
A
B
w.L
2
(a)
w.L
2
5×L
8 2
D
2
w.L
8
(b)
A
C
L
2
B.M. Diagram
B
Fig. 12.20
(i) Now using Mohr’s theorem for slope, we get
Area of B.M. diagram between A and C
Slope at
A=
EI
But area of B.M. diagram between A and C
= Area of parabola ACD
2
= × AC × CD
3
2 L wL2 w . L3
= × ×
=
3 2
8
24
3
w. L
∴ Slope at
A=
24 EI
(ii) Now using Mohr’s theorem for deflection, we get from equation (12.17) as
Ax
y=
EI
where A = Area of B.M. diagram between A and C
w . L3
24
x = Distance of C.G. of area A from A
and
5
5 L 5L
= × AC = × =
8
8 2 16
w. L3 5 L
×
4
16 = 5 w. L .
∴
y = 24
384 EI
EI
554
=
DEFLECTION OF BEAMS
HIGHLIGHTS
1. The relation between curvature, slope, deflection etc. at a section is given by :
Deflection = y
Slope =
dy
dx
B.M. = EI
S.F. = EI
w = EI
d2 y
dx2
d3 y
dx3
d4 y
dx4
dy
= tan θ = θ .
dx
2. Slope at the supports of a simply supported beam carrying a point load at the centre is given by :
As deflection is very small, hence slope is also given by
WL2
16 EI
where
W = Point load at the centre,
L = Length of beam
E = Young’s modulus,
I = M.O.I.
The deflection at the centre of a simply supported beam carrying a point load at the centre is
θ A = θB = –
3.
given by yC = –
WL3
.
48 EI
4. The slope and deflection of a simply supported beam, carrying a uniformly distributed load of
w/unit length over the entire span, are given by,
WL2
5 WL2
and yC =
.
24 EI
384 EI
5. Macaulay’s method is used in finding slopes and deflections at any point of a beam. In this
method :
(i) Brackets are to be integrated as a whole.
(ii) Constants of integrations are written after the first term.
(iii) The section, for which B.M. equation is to be written, should be taken in the last part of the
beam.
θA = θB =
dy
is zero.
dx
7. The slope at point B if slope of A is zero by moment-area method is given by,
6. For maximum deflection, the slope
Area of B. M. diagram between A and B
.
EI
8. The deflection by moment area method is given by
θB =
Ax
EI
A = Area of B.M. diagram between A and B
y=
where
x = Distance of C.G. of area from B.
555
STRENGTH OF MATERIALS
EXERCISE
(A) Theoretical Questions
1. Derive an expression for the slope and deflection of a beam subjected to uniform bending moment.
2. Prove that the relation that M = EI
d2 y
dx2
where M = Bending moment, E = Young’s modulus, I = M.O.I.
3. Find an expression for the slope at the supports of a simply supported beam, carrying a point load
at the centre.
4. Prove that the deflection at the centre of a simply supported beam, carrying a point load at the
centre, is given by yC =
WL3
48 EI
where W = Point load, L = Length of beam.
5. Find an expression for the slope and deflection of a simply supported beam, carrying a point load
W at a distance ‘a’ from left support and at a distance ‘b’ from right support where a > b.
6. Prove that the slope and deflection of a simply supported beam of length L and carrying a uniformly distributed load of w per unit length over the entire length are given by
Slope at supports = –
WL2
, and
24 EI
Deflection at centre =
5 WL3
384 EI
where W = Total load = w × L.
7. What is a Macaulay’s method ? Where is it used ? Find an expression for deflection at any section
of a simply supported beam with an eccentric point load, using Macaulay’s method.
8. What is moment-area method ? Where is it conveniently used ? Find the slope and deflection of a
simply supported beam carrying a (i) point load at the centre and (ii) uniformly distributed load
over the entire length using moment-area method.
(B) Numerical Problems
1.
2.
3.
4.
5.
6.
A wooden beam 4 m long, simply supported at its ends, is carrying a point load of 7.25 kN at
its centre. The cross-section of the beam is 140 mm wide and 240 mm deep. If E for the beam =
6 × 103 N/mm2, find the deflection at the centre.
[Ans. 10 mm]
A beam 5 m long, simply supported at its ends, carries a point load W at its centre. If the slope at
the ends of the beam is not to exceed 1°, find the deflection at the centre of the beam.
[Ans. 29.08 mm]
Determine : (i) slope at the left support, (ii) deflection under the load and (iii) maximum deflection
of a simply supported beam of length 10 m, which is carrying a point load of 10 kN at a distance
6 m from the left end.
[Ans. 0.00028 rad., 0.96 mm and 0.985 mm]
Take E = 2 × 105 N/mm2 and I = 1 × 108 mm4.
A beam of uniform rectangular section 100 mm wide and 240 mm deep is simply supported at its
ends. It carries a uniformly distributed load of 9.125 kN/m run over the entire span of 4 m. Find
[Ans. 6.01 mm]
the deflection at the centre if E = 1.1 × 104 N/mm2.
A beam of length 4.8 m and of uniform rectangular section is simply supported at its ends. It
carries a uniformly distributed load of 9.375 kN/m run over the entire length. Calculate the
width and depth of the beam if permissible bending stress is 7 N/mm2 and maximum deflection
is not to exceed 0.95 cm.
[Ans. b = 240 mm and d = 336.8 mm]
Take E for beam material = 1.05 × 104 N/mm2.
Solve problem 3, using Macaulay’s method.
556
DEFLECTION OF BEAMS
7. A beam of length 10 m is simply supported at its ends and carries two point loads of 100 kN and
60 kN at a distance of 2 m and 5 m respectively from the left support. Calculate the deflections
under each load. Find also the maximum deflection.
Take I = 18 × 108 mm4 and E = 2 × 105 N/mm2.
[Ans. (i) – 4.35 mm (ii) – 6.76 mm (iii) ymax = – 6.78 mm]
8. A beam of length 20 m is simply supported at its ends and carries two point loads of 4 kN and
10 kN at a distance of 8 m and 12 m from left end respectively. Calculate : (i) deflection under
each load (ii) maximum deflection.
Take E = 2 × 106 N/mm2 and I = 1 × 109 mm4.
[Ans. (i) 10.3 mm and 10.6 downwards, (ii) 11 mm]
9. A beam of length 6 m is simply supported at its ends. It carries a uniformly distributed load of 10
kN/m as shown in Fig. 12.21. Determine the deflection of the beam at its mid-point and also the
position and the maximum deflection.
Take EI = 4.5 × 108 N/mm2.
[Ans. – 2.578 mm, x = 2.9 m, ymax = – 2.582 mm]
10 kN/m
1m
3m
2m
6m
Fig. 12.21
10. A beam ABC of length 12 metre has one support at the left end and other support at a distance of
8 m from the left end. The beam carries a point load of 12 kN at the right end as shown in
Fig. 12.22. Find the slopes over each support and at the right end. Find also the deflection at the
right end.
Take E = 2 × 105 N/mm2 and I = 5 × 108 mm4.
[Ans. θA = 6.00364, θB = – 0.00128, θC = – 0.00224, yC = – 7.68 mm]
12 kN
A
B
8m
C
4m
Fig. 12.22
11. An overhanging beam ABC is loaded as shown in Fig. 12.23. Determine the deflection of the beam
at point C.
[Ans. yc = – 4.16 mm]
Take E = 2 × 105 N/mm2 and I = 5 × 108 mm4.
8 kN
2 kN/m
B
A
4m
4m
C
4m
Fig. 12.23
12. A beam of span 8 m and of uniform flexural rigidity EI = 40 MN-m2, is simply supported at its
ends. It carries a uniformly distributed load of 15 kN/m run over the entire span. It is also
subjected to a clockwise moment of 160 kNm at a distance of 3 m from the left support. Calculate
the slope of the beam at the point of application of the moment.
[Ans. 0.0061 rad.]
557
13
CHAPTER
DEFLECTION OF
CANTILEVERS
13.1. INTRODUCTION..
Cantilever is a beam whose one end is fixed and other end is free. In this chapter we
shall discuss the methods of finding slope and deflection for the cantilevers when they are
subjected to various types of loading. The important methods are (i) Double integration method
(ii) Macaulay’s method and (iii) Moment-area-method. These methods have also been used for
finding deflections and slope of the simply supported beams.
13.2.DEFLECTION OF A CANTILEVER WITH A POINT LOAD AT THE FREE END
BY DOUBLE INTEGRATION METHOD
A cantilever AB of length L fixed at the point A and free at the point B and carrying a
point load at the free end B is shown in Fig. 13.1. AB shows the position of cantilever before
any load is applied whereas AB′ shows the position of the cantilever after loading.
L
x
L–x
x
A
B
y
yB
B
Fig. 13.1
Consider a section X, at a distance x from the fixed end A. The B.M. at this section is
given by,
Mx = – W (L – x)
(Minus sign due to hogging)
But B.M. at any section is also given by equation (12.3) as
M = EI
d2 y
dx 2
Equating the two values of B.M., we get
d2 y
= – W (L – x) = – WL + W.x
dx 2
Integrating the above equation, we get
EI
EI
dy
Wx 2
= – WLx +
+ C1
dx
2
...(i)
559
STRENGTH OF MATERIALS
Integrating again, we get
x2 W x3
+ C1x + C2
...(ii)
+
2
2 3
where C1 and C2 are constant of integrations. Their values are obtained from boundary condidy
tions, which are : (i) at x = 0, y = 0 (ii) x = 0,
=0
dx
[At the fixed end, deflection and slopes are zero]
(i) By substituting x = 0, y = 0 in equation (ii), we get
0 = 0 + 0 + 0 + C2 ∴ C2 = 0
dy
= 0 in equation (i), we get
(ii) By substituting x = 0,
dx
0 = 0 + 0 + C1 ∴ C1 = 0
Substituting the value of C1 in equation (i), we get
EIy = – WL
EI
dy
Wx 2
= – WLx +
dx
2
F
GH
= – W Lx −
x2
2
I
JK
...(iii)
Equation (iii) is known as slope equation. We can find the slope at any point on the
cantilever by substituting the value of x. The slope and deflection are maximum at the free
end. These can be determined by substituting x = L in these equations.
Substituting the values of C1 and C2 in equation (ii), we get
EIy = – WL
=–W
x 2 Wx 3
+
2
6
F Lx
GH 2
2
−
x3
6
(∵ C1 = 0, C2 = 0)
I
JK
...(iv)
Equation (iv) is known as deflection equation.
Let
FG dy IJ at B = θ
H dx K
θB = slope at the free end B i.e.,
B and
yB = Deflection at the free end B
(a) Substituting θB for
dy
and x = L in equation (iii), we get
dx
F
GH
EI.θB = – W L . L −
L2
2
I =–W. L
JK
2
2
WL2
...(13.1)
2 EI
Negative sign shows that tangent at B makes an angle in the anti-clockwise direction
with AB
∴
θB = –
∴
θB =
560
WL2
2 EI
...(13.1A)
DEFLECTION OF CANTILEVERS
(b) Substituting yB for y and x = L in equation (iv), we get
F
GH
EI.yB = – W L .
∴
yB = –
L2 L3
−
2
6
I =–W FL
GH 2
JK
3
−
L3
6
I =–W. L
JK
3
3
WL3
3 EI
...(13.2)
(Negative sign shows that deflection is downwards)
∴ Downward deflection, yB =
WL3
3 EI
...(13.2 A)
13.3. DEFLECTION OF A CANTILEVER WITH A POINT LOAD AT A DISTANCE ‘a’
FROM THE FIXED END
A cantilever AB of length L fixed at point A and free at point B and carrying a point load
W at a distance ‘a’ from the fixed end A, is shown in Fig. 13.2.
L
W
a
L–a
C
A
B
yc
yB
C
B
Fig. 13.2
Let
θC = Slope at point C i.e.,
FG dy IJ at C
H dx K
yC = Deflection at point C
yB = Deflection at point B
The portion AC of the cantilever may be taken as similar to a cantilever in Art. 13.1
(i.e., load at the free end).
Wa 2
[In equation (13.1 A) change L to a]
2 EI
Wa 3
[In equation (13.2 A) change L to a]
and
yC =
3 EI
The beam will bend only between A and C, but from C to B it will remain straight since
B.M. between C and B is zero.
Since the portion CB of the cantilever is straight, therefore
Slope at C = slope at B
∴
or
θC = +
θ C = θB =
Wa 2
2 EI
Now from Fig. 13.2, we have
yB = yC + θC(L – a)
=
Wa 3 Wa 2
(L – a)
+
3 EI
2 EI
...(13.3)
F∵
GH
θC =
Wa 2
2 EI
I
JK
...(13.4)
561
STRENGTH OF MATERIALS
Problem 13.1. A cantilever of length 3 m is carrying a point load of 25 kN at the free
end. If the moment of inertia of the beam = 108 mm4 and value of E = 2.1 × 105 N/mm2, find
(i) slope of the cantilever at the free end and (ii) deflection at the free end.
Sol. Given :
Length,
L = 3 m = 3000 mm
Point load,
W = 25 kN = 25000 N
M.O.I.,
I = 108 mm4
Value of E = 2.1 × 105 N/mm2
(i) Slope at the free end is given by equation (13.1 A).
2
∴
θB =
25000 × 3000
WL2
=
= 0.005357 rad. Ans.
2 × 2.1 × 10 5 × 10 8
2 EI
(ii) Deflection at the free end is given by equation (13.2 A),
25000 × 3000 3
WL3
=
= 10.71 mm. Ans.
3 EI 3 × 2.1 × 10 5 × 10 8
Problem 13.2. A cantilever of length 3 m is carrying a point load of 50 kN at a distance
of 2 m from the fixed end. If I = 108 mm4 and E = 2 × 105 N/mm2, find (i) slope at the free end
and (ii) deflection at the free end.
Sol. Given :
Length,
L = 3 m = 3000 mm
Point load,
W = 50 kN = 50000 N
Distance between the load and the fixed end,
a = 2 m = 2000 mm
M.O.I.,
I = 108 mm4
Value of E = 2 × 105 N/mm2
(i) Slope at the free end is given by equation (13.3) as
yB =
Wa 2
50000 × 2000 2
=
= 0.005 rad. Ans.
2 EI
2 × 2 × 10 5 × 10 8
(ii) Deflection at the free end is given by equation (13.4) as
θB =
Wa 3 Wa 2
+
(L – a)
3 EI
2 EI
50000 × 2000 3
50000 × 2000 2
+
(3000 – 2000)
=
3 × 2 × 10 5 × 10 8 2 × 2 × 10 5 × 10 8
= 6.67 + 5.0 = 11.67 mm. Ans.
yB =
13.4. DEFLECTION OF A CANTILEVER WITH A UNIFORMLY DISTRIBUTED LOAD..
A cantilever AB of length L fixed at the point A and free at the point B and carrying a
uniformly distributed load of w per unit length over the whole length, is shown in Fig. 13.3.
Consider a section X, at a distance x from the fixed end A. The B.M. at this section is
given by,
( L − x)
Mx = – w (L – x) .
(Minus sign due to hogging)
2
562
DEFLECTION OF CANTILEVERS
L
x
(L – x)
w/Unit length
X
A
B
yB
B
Fig. 13.3
But B.M. at any section is also given by equation (12.3) as
M = EI
d2 y
dx 2
Equating the two values of B.M., we get
d2 y
w
(L – x)2
2
dx
Integrating the above equation, we get
EI
2
=–
dy
w ( L − x) 3
=–
(– 1) + C1
dx
2
3
w
=
(L – x)3 + C1
6
Integrating again, we get
EI
...(i)
w ( L − x) 4
.
(– 1) + C1x + C2
6
4
w
=–
(L – x)4 + C1x + C2
...(ii)
24
where C1 and C2 are constant of integrations. Their values are obtained from boundary condidy
tions, which are : (i) at x = 0, y = 0 and (ii) at x = 0,
= 0 (as the deflection and slope at fixed
dx
end A are zero).
EIy =
(i) By substituting x = 0, y = 0 in equation (ii), we get
0=–
∴
C2 =
w
wL4
(L – 0)4 + C1 × 0 + C2 = –
+ C2
24
24
wL4
24
(ii) By substituting x = 0 and
0=
∴
dy
= 0 in equation (i), we get
dx
w
wL3
(L – 0)3 + C1 =
+ C1
6
6
C1 = –
wL3
6
563
STRENGTH OF MATERIALS
Substituting the values of C1 and C2 in equations (i) and (ii), we get
dy
w
wL3
=
(L – x)3 –
...(iii)
dx
6
6
w
wL3
wL4
(L – x)4 –
...(iv)
x+
and
EIy = –
24
6
24
Equation (iii) is known as slope equation and equation (iv) as deflection equation. From
these equations the slope and deflection can be obtained at any sections. To find the slope and
deflection at point B, the value of x = L is substituted in these equations.
EI
Let
θB = Slope at the free end B i.e.,
FG dy IJ at B
H dx K
yB = Deflection at the free end B.
From equation (iii), we get slope at B as
w
wL3
wL3
(L – L)3 –
=–
6
6
6
wL3
WL2
∴
θB = –
=–
(∵ W = Total load = w.L) ...(13.5)
6 EI
6 EI
From equation (iv), we get the deflection at B as
EI.θB =
w
wL3
wL4
(L – L)4 –
×L+
24
6
24
wL4 wL4
3
wL4
+
=–
=–
wL4 = –
24
6
24
8
4
3
wL
WL
∴
yB = –
=–
8 EI
8 EI
∴ Downward deflection at B,
EI.yB = –
(∵ W = w.L)
wL4 WL3
=
...(13.6)
8 EI 8 EI
Problem 13.3. A cantilever of length 2.5 m carries a uniformly distributed load of
16.4 kN per metre length over the entire length. If the moment of inertia of the beam = 7.95
× 107 mm4 and value of E = 2 × 105 N/mm2, determine the deflection at the free end.
Sol. Given :
Length,
L = 2.5 m = 2500 mm
U.d.l.,
w = 16.4 kN/m
∴ Total load,
W = w × L = 16.4 × 2.5 = 41 kN = 41000 N
Value of
I = 7.95 × 107 mm4
Value of
E = 2 × 105 N/mm2
Let
yB = Deflection at the free end,
Using equation (13.6), we get
yB =
WL3
41000 × 2500 3
=
8 EI
8 × 2 × 10 5 × 7.95 × 10 7
= 5.036 mm. Ans.
Problem 13.4. A cantilever of length 3 m carries a uniformly distributed load over the
entire length. If the deflection at the free end is 40 mm, find the slope at the free end.
yB =
564
DEFLECTION OF CANTILEVERS
Sol. Given :
Length,
Deflection at free end,
Let
Using equation (13.6),
L = 3 m = 3000 mm
yB = 40 mm
θB = slope at the free end
we get
WL3
8 EI
WL2 × L WL2 × 3000
=
40 =
8 EI
8 EI
40 × 8
WL2
=
3000
EI
Slope at the free end is given by equation (13.5),
yB =
or
or
∴
θB = –
...(i)
40 × 8 1
WL2
WL2 1
×
=–
× =–
3000 6
6 EI
EI
6
LM∵
N
From equation (i),
WL2 40 × 8
=
EI
3000
OP
Q
= 0.01777 rad. Ans.
Problem 13.4 (A). A cantilever 120 mm wide and 200 mm deep is 2.5 m long. What is
the uniformly distributed load which the beam can carry in order to produce a deflection of
5 mm at the free end ? Take E = 200 GN/m2.
Sol. Given :
Width,
b = 120 mm
Depth,
d = 200 mm
Length,
L = 2.5 m = 2.5 × 1000 = 2500 mm
Deflection at free end, yB = 5 mm
Value of E = 200 GN/m2 = 200 × 109 N/m2
(∵ G = Giga = 109)
200 × 10 9 N
(1000) 2 mm 2
= 2 × 105 N/mm2
=
[∵ 1 m2 = (1000 mm)2]
bd 3 120 × 200 3
=
= 8 × 107 mm4
12
12
Let
w = uniformly distributed load per m length in N
W = Total load
=w×L
(Here L is in metre)
= w × 2.5 = 2.5 × w N
Using equation (13.6), we get
Moment of inertia,
I=
yB =
or
5=
WL3
8 EI
2.5 w × 2500 3
8 × 2 × 10 5 × 8 × 10 7
565
STRENGTH OF MATERIALS
5 × 8 × 2 × 10 5 × 8 × 10 7
= 16384 N/m
2.5 × 2500 3
= 16.384 kN/m. Ans.
or
w=
13.5. DEFLECTION OF A CANTILEVER WITH A UNIFORMLY DISTRIBUTED LOAD
FOR A DISTANCE ‘a’ FROM THE FIXED END
A cantilever AB of length L fixed at the point A and free at the point B and carrying
a uniformly distributed load of w/m length for a distance ‘a’ from the fixed end, is shown
in Fig. 13.4.
The beam will bend only between A and C, but from C to B it will remain straight since
B.M. between C and B is zero. The deflected shape of the cantilever is shown by AC′B′ in which
portion C′B′ is straight.
θC = Slope at C, i.e.,
Let
FG dy IJ at C
H dx K
yC = Deflection at point C, and
yB = Deflection at point B.
L
a
A
L–a
w/m Length
C
B
yC C
c
yB
B
Fig. 13.4
The portion AC of the cantilever may be taken as similar to a cantilever in Art. 13.4.
w . a3
[In equation (13.5) put L = a]
6 EI
w . a4
[In equation (13.6) put L = a]
yC =
8 EI
Since the portion C′B′ of the cantilever is straight, therefore slope at C = slope at B
∴
and
θC =
θC = θB =
or
wa 3
6 EI
...(13.7)
Now from Fig. 13.4, we have
yB = yC + θC (L – a)
wa 4 w . a3
+
(L – a)
...(13.8)
8 EI
6 EI
13.6. DEFLECTION OF A CANTILEVER WITH A UNIFORMLY DISTRIBUTED LOAD
FOR A DISTANCE ‘a’ FROM THE FREE END
A cantilever AB of length L fixed at the point A and free at the point B and carrying
a uniformly distributed load of w/m length for a distance ‘a’ from the free end is shown in
Fig. 13.5 (a).
=
566
DEFLECTION OF CANTILEVERS
The slope and deflection at the point B is determined by considering :
(i) the whole cantilever AB loaded with a uniformly distributed load of w per unit length
as shown in Fig. 13.5 (b).
(ii) a part of cantilever from A to C of length (L – a) loaded with an upward uniformly
distributed load of w per unit length as shown in Fig. 13.5 (c).
L
(L – a)
A
C
a
w/Unit Length
A
C
w/Unit Length
(a)
(b)
(c)
A
B
B
B
C
w/Unit Length
Fig. 13.5
and
Then slope at B = Slope due to downward uniform load over the whole length
– slope due to upward uniform load from A to C
deflection at B = Deflection due to downward uniform load over the whole length
– deflection due to upward uniform load from A to C.
(a) Now slope at B due to downward uniformly distributed load over the whole length
wL3
6 EI
(b) Slope at B or at C due to upward uniformly distributed load over the length (L – a)
=
w( L − a) 3
6 EI
Hence net slope at B is given by,
=
wL3 w( L − a) 3
...(13.9)
−
6 EI
6 EI
The downward deflection of point B due to downward distributed load over the whole
length AB
θB =
=
wL4
8 EI
567
STRENGTH OF MATERIALS
The upward deflection of point B due to upward uniformly distributed load acting on
the portion AC = upward deflection of C + slope at C × CB
w( L − a) 4 w . ( L − a) 3
+
×a
8 EI
6 EI
∴ Net downward deflection of the free end B is given by
=
yB =
LM
N
wL4
w( L − a) 4 w( L − a) 3
−
+
×a
8 EI
8 EI
6 EI
(∵ CB = a)
OP
Q
...(13.10)
Problem 13.5. Determine the slope and deflection of the free end of a cantilever of length
3 m which is carrying a uniformly distributed load of 10 kN/m over a length of 2 m from the
fixed end.
Take I = 108 mm4 and E = 2 × 105 N/mm2.
Sol. Given :
Length,
L = 3 m = 3000 mm
10000
N/mm = 10 N/mm
1000
Length of u.d.l. from fixed end, a = 2 m = 2000 mm.
Value of
I = 108 mm4
Value of
E = 2 × 105 N/mm2
U.d.l.,
w = 10 kN/m = 10000 N/m =
θB = Slope of the free end and
yB = Deflection at the free end.
(i) Using equation (13.7), we have
Let
wa 3
10 × 2000 3
=
= 0.00066. Ans.
6 EI
6 × 2 × 10 5 × 10 8
(ii) Using equation (13.8), we get
θB =
yB =
wa 4 w . a3
+
(L – a)
8 EI
6 EI
10 × 2000 4
+
10 × 2000 3
× (3000 – 2000)
8 × 2 × 10 × 10
6 × 2 × 10 5 × 10 8
= 1 + 0.67 = 1.67 mm. Ans.
Problem 13.6. A cantilever of length 3 m carries a uniformly distributed load of
10 kN/m over a length of 2 m from the free end. If I = 108 mm4 and E = 2 × 105 N/mm2 ; find
: (i) slope at the free end, and (ii) deflection at the free end.
Sol. Given :
Length,
L = 3 m = 3000 mm
=
U.d.l.,
5
8
w = 10 kN/m = 10000 N/m =
Length of u.d.l. from free end, a = 2 m = 2000 mm
Value of
I = 108 mm4
Value of
E = 2 × 105 N/mm2
568
10000
N/mm = 10 N/mm
1000
DEFLECTION OF CANTILEVERS
θB = Slope at the free end i.e.,
Let
FG dy IJ at B and
H dx K
yB = Deflection at the free end.
(i) Using equation (13.9), we get
θB =
wL3 w( L − a) 3
−
6 EI
6 EI
10 × 3000 3
10(3000 − 2000) 3
6 × 2 × 10 × 10
6 × 2 × 10 5 × 10 8
= 0.00225 – 0.000083 = 0.002167 rad. Ans.
(ii) Using equation (13.10), we get
=
yB =
LM
N
5
8
−
wL4
w( L − a) 4 w( L − a) 3
−
+
×a
8 EI
8 EI
6 EI
LM
N
OP
Q
OP
Q
10 × 3000 4
10(3000 − 2000) 4 10(3000 − 2000) 3
× 2000
+
−
8 × 2 × 10 5 × 10 8
8 × 2 × 10 5 × 10 8
6 × 2 × 10 5 × 10 8
= 5.0625 – [0.0625 + 0.1667] = 4.8333 mm. Ans.
Problem 13.7. A cantilever of length 3 m carries two point loads of 2 kN at the free end
and 4 kN at a distance of 1 m from the free end. Find the deflection at the free end.
Take E = 2 × 105 N/mm2 and I = 108 mm4.
Sol. Given :
Length,
L = 3 m = 3000 mm
Load at free end, W1 = 2 kN = 2000 N
Load at a distance one m from free end,
W2 = 4 kN = 4000 N
Distance AC,
a = 2 m = 2000 mm
Value of
E = 2 × 105 N/mm2
Value of
I = 108 mm4
Let
y1 = Deflection at the free end due to load 2 kN alone
y2 = Deflection at the free end due to load 4 kN alone.
=
2 kN
4 kN
C
A
B
1m
3m
Fig. 13.6
Downward deflection due to load 2 kN alone at the free end is given by equation (13.2 A)
as
WL3
2000 × 3000 3
=
= 0.9 mm.
3 EI
3 × 2 × 10 5 × 10 8
Downward deflection at the free end due to load 4 kN (i.e., 4000 N) alone at a distance
2 m from fixed end is given by (13.4) as
569
y1 =
STRENGTH OF MATERIALS
Wa 3 Wa 2
+
(L – a)
3 EI
2 EI
4000 × 2000 3
4000 × 2000 2
+
(3000 – 2000)
=
3 × 2 × 10 5 × 10 8 2 × 2 × 10 5 × 10 8
= 0.54 + 0.40 = 0.94 mm
∴ Total deflection at the free end
= y1 + y2
= 0.9 + 0.94 = 1.84 mm. Ans.
Problem 13.8. A cantilever of length 2 m carries a uniformly distributed load of
2.5 kN/m run for a length of 1.25 m from the fixed end and a point load of 1 kN at the free
end. Find the deflection at the free end if the section is rectangular 12 cm wide and 24 cm
deep and E = 1 × 104 N/mm2.
Sol. Given :
Length,
L = 2 m = 2000 mm
U.d.l.,
w = 2.5 kN/m = 2.5 × 1000 N/m
2.5 × 1000
=
N/mm = 2.5 N/mm
1000
Point load at free end, W = 1 kN = 1000 N
Distance AC,
a = 1.25 m = 1250 mm
Width,
b = 12 cm
Depth,
d = 24 cm
y2 =
bd 3 12 × 24 3
=
12
12
= 13824 cm4 = 13824 × 104 mm4 = 1.3824 × 108 mm4
E = 1 × 104 N/mm2
y1 = Deflection at the free end due to point load 1 kN alone
y2 = Deflection at the free end due to u.d.l. on length AC.
Value of
I=
Value of
Let
2.0 m
1.25 m
2.5 kN/m
A
1 kN
0.75 m
C
B
Fig. 13.7
(i) Now the downward deflection at the free end due to point load of 1 kN (or 1000 N) at
the free end is given by equation (13.2 A) as
WL3
1000 × 2000 3
=
= 1.929 mm.
3 EI
3 × 10 4 × 1.3824 × 10 8
(ii) The downward deflection at the free end due to uniformly distributed load of
2.5 N/mm on a length of 1.25 m (or 1250 mm) is given by equation (13.8) as
y1 =
y2 =
570
wa 4 w . a3
+
(L – a)
8 EI
6 EI
DEFLECTION OF CANTILEVERS
2.5 × 1250 4
2.5 × 1250 3
+
(2000 – 1250)
8 × 10 4 × 1.3824 × 10 8 6 × 10 4 × 1.3824 × 10 8
= 0.5519 + 0.4415 = 0.9934
∴ Total deflection at the free end due to point load and u.d.l.
= y1 + y2 = 1.929 + 0.9934 = 2.9224 mm. Ans.
Problem 13.9. A cantilever of length 2 m carries a uniformly distributed load 2 kN/m
over a length of 1 m from the free end, and a point load of 1 kN at the free end. Find the slope
and deflection at the free end if E = 2.1 × 105 N/mm2 and I = 6.667 × 107 mm4.
Sol. Given : (See Fig. 13.8)
Length,
L = 2 m = 2000 mm
2 × 1000
U.d.l.,
w = 2 kN/m =
N/mm = 2 N/mm
1000
Length BC,
a = 1 m = 1000 mm
Point load,
W = 1 kN = 1000 N
Value of
E = 2.1 × 105 N/mm2
Value of
I = 6.667 × 107 mm4.
=
1 kN
A
2 kN/m
C
B
1m
2m
Fig. 13.8
(i) Slope at the free end
Let
θ1 = Slope at the free end due to point load of 1 kN i.e., 1000 N
θ2 = Slope at the free end due to u.d.l. on length BC.
The slope at the free end due to a point load of 1000 N at B is given by equation (13.1 A) as
θ1 =
WL2
2 EI
(∵ θB = θ1 here)
1000 × 2000 2
= 0.0001428 rad.
2 × 2.1 × 10 5 × 6.667 × 107
The slope at the free end due to u.d.l. of 2 kN/m over a length of 1 m from the free end is
given by equation (13.9) as
=
θ2 =
wL3 w( L − a) 3
−
6 EI
6 EI
(∵ θB = θ2 here)
2 × 2000 3
2 × (2000 − 1000) 3
−
6 × 2.1 × 10 5 × 6.667 × 107 6 × 2.1 × 10 5 × 6.667 × 107
= 0.0001904 – 0.000238 = 0.0001666 rad.
=
∴ Total slope at the free end
= θ1 + θ2 = 0.0001428 + 0.0001666 = 0.0003094 rad. Ans.
571
STRENGTH OF MATERIALS
(ii) Deflection at the free end
Let
y1 = Deflection at the free end due to point load of 1000 N
y2 = Deflection at the free end due to u.d.l. on length BC.
The deflection at the free end due to point load of 1000 N is given by equation (13.2 A) as
y1 =
WL3
3 EI
(∵ Here y1 = yB)
1000 × 2000 3
= 0.1904 mm.
3 × 2.1 × 10 5 × 6.667 × 107
The deflection at the free end due to u.d.l. of 2 N/mm over a length of 1 m from the free
end is given by equation (13.10) as
=
y2 =
LM
N
wL4
w( L − a) 4 w( L − a) 3
−
+
×a
8 EI
8 EI
6 EI
LM
N
OP
Q
2 × 2000 4
2(2000 − 1000) 4
=
5
7 –
8 × 2.1 × 10 × 6.667 × 10
8 × 2.1 × 10 5 × 6.667 × 107
+
2(2000 − 1000) 3 × 1000
6 × 2.1 × 10 5 × 6.667 × 107
= 0.2857 – [0.01785 + 0.0238] = 0.244 mm
∴ Total deflection at the free end
= y1 + y2 = 0.1904 + 0.244 = 0.4344 mm. Ans.
OP
Q
13.7. DEFLECTION OF A CANTILEVER WITH A GRADUALLY VARYING LOAD..
A cantilever AB of length L fixed at the point A and free at the point B and carrying a
gradually varying load from 0 at B to w per unit run at the fixed end A, is shown in Fig. 13.9.
L
x
L–x
C
w
A
B
X
yB
Fig. 13.9
Consider a section X at a distance x from the fixed end A.
w
w
(L – x) per unit run. Hence vertical height XC =
(L – x).
L
L
Hence the B.M. at this section is given by
Mx = – (Load on length Bx) × (Distance of C.G. of the load on BX from section X)
= – (Area of ∆BXC) × (Distance of C.G. of area BXC from X)
(Minus sign is due to hogging)
572
The load at X will be
DEFLECTION OF CANTILEVERS
FG BX . XC IJ × FG 1 of length BX IJ
K
H 2 K H3
( L − x) w
L1
O w (L – x) .
.
=–
(L – x) × M ( L − x)P = –
2
L
3
N
Q 6L
=–
3
But B.M. at any section is also given by equation (12.3) as
M = EI
d2 y
dx 2
Equating the two values of B.M., we get
d2 y
w
(L – x)3
6
L
dx
Integrating the above equation, we get
EI
2
=–
dy
w ( L − x) 4
=–
(– 1) + C1
dx
6L
4
w
=
(L – x)4 + C1
24 L
Integrating again, we get
EI
...(i)
w ( L − x) 5
(– 1) + C1x + C2
24 L
5
w
=–
(L – x)5 + C1x + C2
...(ii)
120 L
where C1 and C2 are constant of integrations. Their values are obtained from boundary conditions, which are :
dy
(i) at x = 0, y = 0 and
(ii) at x = 0,
= 0.
dx
(i) By substituting x = 0 and y = 0 in equation (ii), we get
EIy =
w
wL4
(L – 0)5 + C1 × 0 + C2 or C2 =
.
120 L
120
dy
(ii) By substituting x = 0 and
= 0 in equation (i), we get
dx
w
0=
(L – 0)4 + C1
24 L
wL4
wL3
∴
C1 = –
=–
24 L
24
Substituting the values of C1 and C2 in equations (i) and (ii), we get
0=–
dy
w
wL3
=
(L – x)4 –
...(iii)
dx
24 L
24
w
wL3
wL4
(L – x)5 –
x+
...(iv)
and
EIy = –
120 L
24
120
Equation (iii) is known as slope equation and equation (iv) as deflection equation. The
slope and deflection at the free end (i.e., point B) can be obtained by substituting x = L in these
equations.
EI
Let
θB = Slope at the free end B i.e.,
yB = Deflection at the free end B.
FG dy IJ at B and
H dx K
573
STRENGTH OF MATERIALS
dy
= θB in equation (iii), we get
dx
(a) Substituting x = L and
EI θB =
w
wL3
wL3
(L – L)4 –
=–
24 L
24
24
wL3
radians.
24 EI
(b) Substituting x = L and y = yB in equation (iv), we get
∴
θB = –
EI yB = –
w
wL3
wL4
(L – L)5 –
.L+
120 L
24
120
=0–
wL4 wL4
5wL4 + wL4
wL4
=–
=–
+
120
24
120
30
wL4
30 EI
∴ Downward deflection of B is given by
∴
...(13.11)
yB = –
(Minus sign means downward deflection)
wL4
...(13.12)
30 EI
Problem 13.10. A cantilever of length 4 m carries a uniformly varying load of zero
intensity at the free end, and 50 kN/m at the fixed end.
If E = 2.0 × 105 N/mm2 and I = 108 mm4, find the slope and deflection at the free end.
Sol. Given :
Length,
L = 4 m = 4000 mm
yB =
50 × 1000
= 50 N/mm
1000
Value of
E = 2 × 105 N/mm2
Value of
I = 108 mm4
Let
θB = Slope at the free end and
yB = Deflection at the free end.
(i) Using equation (13.11), we get
Load at fixed end,
w = 50 kN/m =
wL3
50 × (4000) 3
=–
= 0.00667 rad. Ans.
24 EI
24 × 2 × 10 5 × 10 8
(ii) Using equation (13.12), we get
θB = –
wL4
50 × (4000) 4
=
= 21.33 mm. Ans.
30 EI
30 × 2 × 10 5 × 10 8
Problem 13.11. A cantilever of length 2 m carries a uniformly varying load of 25 kN/m
at the free end to 75 kN/m at the fixed end. If E = 1 × 105 N/mm2 and I = 108 mm4, determine
the slope and deflection of the cantilever at the free end.
Sol. Given :
Length,
L = 2 m = 2000 mm
25 × 1000
Load at the free end
= 25 kN/m =
= 25 N/mm
1000
yB =
574
DEFLECTION OF CANTILEVERS
75 kN/m
50 kN/m
Load at fixed end
= 75 kN/m = 75 N/mm
Value of
E = 1 × 105 N/mm2
Value of
I = 108 mm4.
The load acting on the cantilever is shown in Fig. 13.10. This load is equivalent to a
uniformly distributed load of 25 kN/m (or 25 N/mm) over the entire length and a triangular load
of zero intensity at free end and (75 – 25 = 50 kN/m or 50 N/mm) 50 N/mm at the fixed end.
25 kN/m
B
A
2m
Fig. 13.10
(i) Slope at the free end
Let θ1 = Slope at free end due to u.d.l. of 25 N/mm
θ2 = Slope at free end due to triangular load of intensity 50 N/mm at fixed end.
The slope at the free end due to u.d.l. of 25 N/mm (i.e., w = 25 N/mm) is given by equation (13.5) as
θ1 =
wL3
6 EI
25 × 2000 3
(Here θ1 = θB, and w = 25)
= 0.0033 rad.
6 × 1 × 10 5 × 10 8
The slope at the free end due to triangular load of intensity of 50 N/mm (i.e. w =
50 N/mm) is given by equation (13.11) as
=
θ2 =
=
wL3
24 EI
50 × 2000 3
(Here w = 50 N/mm)
24 × 1 × 10 5 × 10 8
= 0.00167 rad.
∴ Total slope at the free end
= θ1 + θ2 = 0.0033 + 0.00167 = 0.00497. Ans.
(ii) Deflection at the free end
Let y1 = Deflection at the free end due to u.d.l. of 25 N/mm
y2 = Deflection at the free end due to triangular load.
Using equation (13.11), we get deflection at the free end due to u.d.l.
wL4
25 × 2000 4
=
= 5 mm
8 EI
8 × 1 × 10 5 × 10 8
Using equation (13.12), we get deflection at the free end to uniformly varying load of
zero at the free end and 50 N/mm at the fixed end.
∴
y1 =
∴
y2 =
wL4
50 × 2000 4
=
= 2.67 mm
30 EI
30 × 1 × 10 5 × 10 8
575
STRENGTH OF MATERIALS
∴ Total deflection at the free end
= y1 + y2 = 5 + 2.67 = 7.67 mm. Ans.
13.8. DEFLECTION AND SLOPE OF A CANTILEVER BY MOMENT AREA METHOD..
The moment area method is discussed in Art. 12.8, where this method was applied to a
simply supported beam. Let us apply this method to a cantilever. According to this method the
change of slope between any two points is equal to the net area of the B.M. diagram between
these two points divided by EI. If one of the points is having zero slope, then we can obtain the
slope at the other point.
Similarly if the deflection at a point A is zero, then the deflection at the point B according to this method is given by
Ax
y=
EI
where A = Area of B.M. diagram between A and B, and
x = Distance of C.G. of the area A from B.
13.8.1. Cantilever Carrying a Point Load at the Free end. Fig. 13.11 (a) shows a
cantilever of length L fixed at end A and free at the end B. It carries a point load W at B.
W
A
B
(a)
L
x = 2L
3
A
B
W.L.
(b)
B.M. Diagram
C
Fig. 13.11
The B.M. will be zero at B and will be W.L at A. The variation of B.M. between A and B
is linear as shown in Fig. 13.11 (b).
At the fixed end A, the slope and deflection are zero.
Let
θB = Slope at B i.e.,
FG dy IJ at B and
H dx K
yB = Deflection at B
Then according to moment area method,
Area of B.M. Diagram between A and B
θB =
EI
1
× AB × AC
2
=
(Area of triangle ABC)
EI
1
× L×W.L
WL2
2
=
=
EI
2 EI
576
DEFLECTION OF CANTILEVERS
and
yB =
Ax
EI
...(i)
where A = Area of B.M. diagram between A and B =
W . L2
2
x = Distance of C.G. of area of B.M. diagram from B =
2L
3
W . L2 2 L
×
3
2
3 = W.L .
∴
yB =
EI
3 EI
13.8.2. Cantilever Carrying a Uniformly Distributed load. Fig. 13.12 (a) shows a
cantilever of length L fixed at end A and free at the end B. It carries a uniformly distributed
load of w/unit length over the entire length.
w/Unit Length
A
(a)
x = 3L
4
A
(b)
B
B
C.G
2
w.L
2
B.M. Diagram
C
Fig. 13.12
w . L2
at A. The variation of B.M. between A and
2
B is parabolic as shown in Fig. 13.12 (b). At the fixed end A, the slope and deflection are zero.
The B.M. will be zero at B and will be
1
w . L2 w . L3
=
.L.
3
2
6
and the distance of the C.G. of the B.M. diagram from B,
3L
x =
4
Let
θB = Slope at B, i.e., dy and at B
dx
yB = Deflection at B.
Then according to moment area method,
Area of B.M. diagram (ABC),
A=
FG IJ
H K
Area of B. M. diagram wL3
=
EI
6 EI
Ax
w . L3 3 L
w . L4
×
yB =
=
=
.
EI
6 EI
4
8 EI
θB =
and
13.8.3. Cantilever Carrying a Uniformly Distributed Load upto a Length ‘a’
from the Fixed end. Fig. 13.13 (a) shows a cantilever of length L fixed at end A and free at
the end B. It carries a uniformly distributed load of w/unit length over a length ‘a’ from the
fixed end.
577
STRENGTH OF MATERIALS
w/Unit length
A
B
C
(a )
a
(L – a)
L
x = 3a
4
A
(b )
(L – a)
C
B
C.G.
2
w.a
2
D
Fig. 13.13
w . a2
. The variation of B.M.
2
between C and A will be parabolic as shown in Fig. 13.13 (b). At the fixed end the slope and
deflection are zero.
The B.M. will be zero at B and C. But B.M. at A will be
1
w . a2 w . a3
.a.
=
3
2
6
and the distance of the C.G. of B.M. diagram from B,
3a
x = (L – a) +
4
Area of B.M. diagram
A=
Let
θB = Slope at B i.e.,
FG dy IJ at B and
H dx K
yB = Deflection at B.
Then according to moment area method,
A w . a3
=
EI
6 EI
w . a4
Ax
w . a3
3a
w . a3
( L − a) +
× ( L − a) +
and
yB =
=
=
.
EI
6 EI
8 EI
6 EI
4
Problem 13.12. A cantilever of length 2 m carries a point load of 20 kN at the free end
and another load of 20 kN at its centre. If E = 105 N/mm2 and I = 108 mm4 for the cantilever
then determine by moment area method, the slope and deflection of the cantilever at the free
end.
Sol. Given :
Length,
L=2m
Load at free end,
W1 = 20 kN = 20000 N
Load at centre,
W2 = 20 kN = 20000 N
Value of
E = 105 N/mm2
Value of
I = 108 mm4
First draw the B.M. diagram,
B.M. at B
=0
B.M. at C
= – 20 × 1 = – 20 kNm = – 20 × 103 × 103 Nmm
θB =
578
LM
N
OP
Q
DEFLECTION OF CANTILEVERS
B.M. at A
= – 20 × 1 – 20 × 2 = – 60 kNm = – 60 × 103 × 103 Nmm
B.M. diagram is shown in Fig. 13.14 (b).
20kN
20kN
( a)
A
C
B
1m
A
( b)
1m
C
F
60
kNm
2
3
B
1
20 kNm
D
B.M. Diagram
E
Fig. 13.14
To find the area of B.M. diagram, divide the Fig. 13.14 (b) into two triangles and one
rectangle.
1
1
Now
area A1 = × CD × BC = × 20 × 1
2
2
(∵ m2 = 106 mm2)
= 10 kNm2 = 10 × 103 × 106 Nmm2
10
2
= 10 Nmm
Similarly area A2 = CD × AC = 20 × 1 = 20 kNm2
1
1
and
area A3 = × FD × EF = × 1 × 40 = 20 kNm2
2
2
∴ Total area of B.M. diagram,
A = A1 + A2 + A3 = 10 + 20 + 20 = 50 kNm2
= 50 × 103 × 106 Nmm2
(∵ m2 = 106 mm2)
Slope and deflection at the fixed end is zero.
Let
θB = Slope at the free end B.
Then according to the moment area method,
Area of B.M. diagram
θB =
EI
3
50 × 10 × 10 6
=
= 0.005 radians. Ans.
10 5 × 10 8
Let
yB = Deflection at the free end B.
Then according to moment area method,
Ax
EI
Now let us find x or A x .
Then total moment of the bending moment diagram about B is given by
yB =
...(i)
A . x = A1 x1 + A2 x2 + A3 x3
= 10 ×
FG 2 × 1IJ + 20 × FG 1 + 1IJ + 20 × FG 1 + 2 × 1IJ
H3 K
H 2K
H 3 K
579
STRENGTH OF MATERIALS
20
100
+ 30 +
= 70 kNm3
3
3
= 70 × 103 × 109 Nmm3
= 7 × 1013 Nmm3
Substituting this value in equation (i), we get
=
yB =
(∵ m3 = 109 mm3)
7 × 10 13
= 7 mm. Ans.
10 5 × 10 8
HIGHLIGHTS
1.
The slope i.e.,
dy
or θ of a cantilever at the free end is given by,
dx
θB =
WL2
2 EI
θB = θC =
θB =
Wa2
2 EI
w . L2
6 EI
θB = θC =
2.
when the point load is at the free end
wa3
6 EI
when the point load is at a distance of ‘a’ from the fixed end
when it carries a uniformly distributed load over the whole length.
when it carries a uniformly distributed load over a length
‘ a’ from the fixed end.
θB =
w . L3 w . ( L − a)3
−
6 EI
6 EI
θB =
w . L3
24 EI
when it carries a uniformly distributed load over a
distance ‘ a’ from the free end
when it carries a gradually varying load from zero at the free end to
w / m run at fixed end.
where W = Point load,
w = Uniformly distributed load,
L = Length of beam,
I = Moment of inertia, and
E = Young’s modulus.
The deflection i.e., y of a cantilever of length L, at the free end is given by,
580
yB =
WL3
3 EI
yB =
Wa3 Wa2
+
(L – a)
3 EI
2 EI
yB =
wL4
8 EI
yB =
wa4 w . a3
(L – a)
+
8 EI
6 EI
when the point load is at the free end
when the point load is at a distance of ‘ a’ from the
fixed end
when it carries a uniformly distributed load over the whole length
when it carries a uniformly distributed load over a
length ‘ a’ from the fixed end.
DEFLECTION OF CANTILEVERS
3.
LM
MN
yB =
wL4
w( L − a)4 w( L − a)3
−
+
×a
8 EI
8 EI
6 EI
yB =
wL4
30 EI
OP
PQ
when it carries a uniformly distributed
load over a length ‘ a’ from the free end
when it carries a gradually varying load from zero at the free end to
w / m run at the fixed end.
For a cantilever, at the fixed end slope and deflections are zero. Hence moment area method can
be easily applied for finding slope and deflections of cantilevers. The slope (θB) and deflection (yB)
at the free end is given by,
Area of B. M. diagram between free end and fixed end
EI
Ax
and
yB =
EI
where A = Area of B.M. Diagram and
θB =
4.
x = Distance of C.G. of B.M. diagram from free end.
Area of B.M. diagram sometimes is found easily by splitting the combined areas into triangles
and rectangles.
EXERCISE
(A) Theoretical Questions
1.
2.
3.
4.
What is a cantilever ? What are the different methods of finding of slope and deflection of a
cantilever ?
Derive an expression for the slope and deflection of a cantilever of length L, carrying a point load
W at the free end by double integration method.
Solve question 2, by moment area method.
Prove that the slope and deflection of a cantilever carrying uniformly distributed load over the
whole length are given by,
wL3
wL4
and yB =
6 EI
8 EI
where w = Uniformly distributed load and
EI = Flexural rigidity.
Find the expression for the slope and deflection of a cantilever of length L which carries a uniformly distributed load over a length ‘a’ from the fixed end by
θB =
5.
6.
(i) Double integration method and
(ii) Moment area method.
Prove that the slope and deflection of a cantilever length L, which carries a gradually varying
load from zero at the free end to w/m run at the fixed end are given by :
θB =
wL3
24 EI
and yB =
wL4
30 EI
where EI = Flexural rigidity.
(B) Numerical Problems
1.
A cantilever of length 2 m carries a point load of 30 kN at the free end. If moment of inertia
I = 108 mm4 and value of E = 2 × 105 N/mm2, then find :
(i) slope of the cantilever at the free end and
(ii) deflection at the free end.
[Ans. (i) 0.003 rad., (ii) 4 mm]
581
STRENGTH OF MATERIALS
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
A cantilever of length 3 m carries a point load of 60 kN at a distance of 2 m from the fixed end. If
E = 2 × 105 and I = 108, find :
(i) slope at the free end and
(ii) deflection at the free end.
[Ans. 0.006 rad., 14 mm]
A cantilever of length 30 m carries a uniformly distributed load of 24 kN/m length over the entire
length. If moment of inertia of the beam = 108 mm3 and value of E = 2 × 105 N/mm2, determine
the slope and deflection at the free end.
[Ans. 0.0054 rad. ; 12.15 mm]
A cantilever of length 3 m carries a uniformly distributed load over the entire length. If the slope
at the free end is 0.01777 radians, find the deflection at the free end.
[Ans. 39.99 mm]
Determine the slope and deflection at the free end of a cantilever of length 4 m which is carrying
a uniformly distributed load of 12 kN/m over a length of 3 m from the fixed end. Take EI = 2 ×
[Ans. 0.0027 rad., 8.775 mm]
1013 N/mm2.
A cantilever of length 3 m carries a uniformly distributed load of 15 kN/m over a length of 2 m
from the free end. If I = 108 mm4 and E = 2 × 105 N/mm2, find :
(i) slope at the free end and
(ii) deflection at the free end.
[Ans. 0.00326 rad., 7.25 mm]
A cantilever of length 2 m carries a load of 20 kN at the free end and 30 kN at a distance 1 m
from the end. Find the slope and deflection at the free end. Take E = 2.0 × 105 N/mm2 and
I = 1.5 × 108 mm4.
[Ans. 0.00183 rad., 2.6 mm]
Determine the deflection at the free end of a cantilever which is 2 m long and carries a point load
of 9 kN at the free end and a uniformly distributed load of 8 kN/m over a length of 1 m from the
fixed end.
[Ans. 6.54 mm]
Take I = 2.25 × 107 mm3 and E = 2.2 × 105 N/mm2.
A cantilever of length 2 m carries a uniformly varying load of zero intensity at the free end, and
45 kN/m at the fixed end. If E = 2 × 105 N/mm2 and I = 108 mm4, find the slope and deflection of
the free end.
[Ans. 0.00075 rad., ; 1.2 mm]
A cantilever of length 2 m carries a point load of 30 kN at the free end and another load of 30 kN
at its centre. If EI = 1013 N/mm2 for the cantilever then determine by moment area method, the
slope and deflection at the free end of cantilever.
[Ans. 0.0075 rad. ; 10.50 mm]
L
from the fixed end.
2
Determine the slope and deflection at the free end using area moment method.
A cantilever of length ‘L’ carries a U.D.L. of w per unit for a length of
LM
N
L
. Hence slope at free end,
2
Hint. See Article 13.8.3 on page 577. Here a =
3
θB =
and deflection at free end,
w×a
=
6 EI
w×
FG L IJ
H 2K
6 EI
3
=
w × L3
.
48 EI
w × a3
w × a4
yB =
(L – a) +
6 EI
8 EI
FG L IJ
F LI
w×G J
H
LI
2K F
=
GH L − 2 JK + 8HEI2 K = w48×EIL × L2 + w128× EIL
6 EI
FG 4 + 3 IJ
w× L
wL
w× L F 1
=
=
+
G + 1 IJ = wL
EI H 128 × 3 K
96 EI
128 EI
EI H 96 128 K
WL
7
7 w× L O
×
=
=
PQ
EI
384
384 EI
3
4
w×
3
4
4
582
4
4
4
4
4
14
CHAPTER
CONJUGATE BEAM METHOD,
PROPPED CANTILEVERS
AND BEAMS
14.1. INTRODUCTION..
The slopes and deflections of beams and cantilevers may be obtained from various methods like double integration method, moment area method, Macaulay’s method, etc. But these
methods become laborious, when applied to beams whose flexural rigidity (i.e., the product of
E and I is known as flexural rigidity) is not uniform throughout the length of the beam. The
slopes and deflections of such beams can be easily obtained by conjugate beam method.
14.2. CONJUGATE BEAM METHOD..
Before describing the conjugate beam method, let us first define conjugate beam.
Conjugate beam is an imaginary beam of length equal to that of the original beam but
M
for which the load diagram is the
diagram* (i.e., the load at any point on the conjugate
EI
beam is equal to B.M. at that point divided by EI).
The slopes and deflection at any section of a beam by conjugate beam method is given
by :
1. The slope at any section of the given beam is equal to the shear force at the corresponding section of the conjugate beam.
2. The deflection at any section for the given beam is equal to the bending moment at
the corresponding section of the conjugate beam.
Hence before applying the conjugate beam method, conjugate beam is constructed. The
load on the conjugate beam at any point is equal to the B.M. at that point divided by EI. Hence
the loading on the conjugate beam is known. Then the shear force at any point on the conjugate beam gives the slope at the corresponding point of actual beam. And the B.M. at any
point on the conjugate beam gives the deflection at the corresponding point of the actual beam.
14.3. DEFLECTION AND SLOPE OF A SIMPLY SUPPORTED BEAM WITH A POINT
LOAD AT THE CENTRE
Fig. 14.1 (a) shows a simply supported beam AB of length L carrying a point load W at
the centre C. The B.M. at A and B is zero and at the centre B.M. will be W . L/4 . The B.M.
varies according to straight line law. The B.M. diagram is shown in Fig. 14.1 (b). Now the
conjugate beam AB can be constructed. The load on the conjugate beam will be obtained by
*M/EI diagram is a diagram which shows the variation of M/EI over the length of the beam.
583
STRENGTH OF MATERIALS
dividing the B.M. at that point by EI. The shape of the loading on the conjugate beam will be
same as of B.M. diagram. The ordinate of loading on conjugate beam will be equal to
FG
H
IJ
K
W. L
W×L
M
4
W. L
=
=
. Hence ordinate at the centre will be
as shown in Fig. 14.1 (c). The
EI
EI
4 EI
4 EI
load diagram for conjugate beam is shown in Fig. 14.1 (c).
W
C
A
(a)
B
L/2
RA = W
2
RB = W
2
L
D′
W.L
4
(b)
A
C′
B
B.M. Diagram
D*
Load
Diagram
W.L
4 EI
A
(c)
B
C*
2
Conjugate Beam
R*A = WL
16 EI
2
R*B = WL
16 EI
Fig. 14.1
RA* = Reaction at A for conjugate beam
RB* = Reaction at B for conjugate beam
Total load on the conjugate beam [See Fig. 14.1 (c)]
= Area of the load diagram
1
1
WL
= × AB × C*D* = × L ×
2
2
4 EI
2
WL
=
8 EI
Reaction at each support for the conjugate beam will be half of the total load
Let
∴
Let
RA* = RB* =
WL2
1 WL2
×
=
2 8 EI 16 EI
θA = Slope at A for the given beam i.e.,
FG dy IJ at A
H dx K
yc = deflection at C for the given beam.
Then according to conjugate beam method,
θA = Shear force at A for the conjugate beam
584
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
= RA*
(∵ S.F. at A for conjugate beam = RA*)
WL2
16 EI
yC = B.M. at C for the conjugate beam
[See Fig. 14.1 (c)]
L
– Load corresponding to AC*D*
= RA* ×
2
× Distance of C.G. of AC*D* from C
=
and
=
FG
H
IJ FG
K H
1 L WL
1 L
WL2 L
. −
× ×
×
×
16 EI 2
2 2 4 EI
3 2
IJ
K
WL3
WL3
3WL3 − WL3
−
=
32 EI 96 EI
96 EI
3
WL
=
.
48 EI
=
14.4. SIMPLY SUPPORTED BEAM CARRYING AN ECCENTRIC POINT LOAD..
Fig. 14.2 (a) shows a beam AB of length L, simply supported at A and B and carrying a
point load of W at a distance ‘a’ from the end A. The reactions at A and B are given by
W .b
W.a
RA =
and RB =
L
L
W ×b
W .b. a W . a.b
The B.M. will be zero at A and B. At C, the B.M. will be RA × a =
×a=
=
.
L
L
L
Now the B.M. diagram can be drawn as shown in Fig. 14.2 (b).
W
C
A
a
B
b
(a)
L
RA = W.b
L
RB = W.a
L
D′
W.a.b
L
( b)
A
B
C′
B.M. Diagram
D*
Load Diagram
W.a.b
E.I.L
( c)
A
B
C*
R*A
a
Conjugate beam
b
R*B
Fig. 14.2
585
STRENGTH OF MATERIALS
Construct now the conjugate beam. The load at any point on the conjugate beam will be
equal to B.M. at that point divided by EI. Fig. 14.2 (c) shows the conjugate beam with the
loads. The vertical load on conjugate beam at C* will be
M W .a.b W .a.b
=
=
.
EI
L × EI
EI . L
Let RA* = Reaction at A for conjugate beam
RB* = Reaction at B for conjugate beam
Taking moments about A of the conjugate beam, we get
RB* . L = Load AC*D* × Distance of C.G., of AC*D* from A + Load BC*D*
× Distance of C.G. of BC*D* from A
FG 1 × AC * × C * D *IJ × FG 2 × AC *IJ + FG 1 . BC*. C * D *IJ × FG AC * + 1 × BC *IJ
H2
K H3
K H2
K H
K
3
F 1 W . a . bIJ × FG 2 × aIJ + FG 1 × b × W . a . bIJ × FG a × bIJ
= G ×a×
H2
K H2
3K
EIL K H 3
EI . L K H
bI
W . a . b Wab F
+
a+ J
=
G
3 EIL
2 EIL H
3K
=
3
2
W a 3 b Wa 2 b2 W . a . b3
+
+
3 EIL
2 EIL
6 EIL
W a.b
=
[2a2 + 3ab + b2]
6 EIL
W a.b 2
(a + b2 + 2ab + a2 + ab)
=
6 EIL
W .a.b
=
[(a + b)2 + a(a + b)]
6 EIL
W ab
[L2 + a . L]
(∵ a + b = L)
=
6 EIL
Wa . ( L − a)
Wa
=
[L2 – a2]
. L( L + a) =
6 EIL
6 EI
W.a
∴
RB* =
(L2 – a2)
6 EIL
Similarly the reaction at A can be obtained as
Wb
RA* =
(L2 – b2)
(Substitute b for a)
6 EIL
dy
Let θA = Slope at A for the given beam i.e.,
at A
dx
yC = Deflection at C for the given beam.
Then according to conjugate beam method,
θA = Shear force at A for the conjugate beam
= RA*
(∵ S.F. at A for conjugate beam = RA*)
W .b
=
(L2 – b2)
6 EIL
yC = B.M. at C for conjugate beam
[See Fig. 14.2 (c)]
= RA* . a – Load AC*D* × Distance of C.G. of AC*D* from C*
=
FG IJ
H K
and
586
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
FG
H
IJ FG IJ
K H K
W .b
W .a.b
a
1
(L2 – b2) . a –
×
×a×
6 EIL
EIL
2
3
W . a3b
W .a.b 2
=
( L − b2 ) −
6 EIL
6 EIL
W .a.b 2
2
2
=
[L – b – a ].
6 EIL
Problem 14.1. A simply supported beam of length 5 m carries a point load of 5 kN at a
distance of 3 m from the left end. If E = 2 × 105 N/mm2 and I = 108 mm4, determine the slope at
the left support and deflection under the point load using conjugate beam method.
Sol. Given :
Length,
L=5m
Point load,
W = 5 kN
Distance AC,
a =3m
Distance BC,
b =5–3=2m
Value of
E = 2 × 105 N/mm2 = 2 × 105 × 106 N/m2
= 2 × 105 × 103 kN/m2 = 2 × 108 kN/m2
Value of
I = 1 × 108 mm4 = 10–4 m4
Let
RA = Reaction at A
and
RB = Reaction at B.
Taking moments about A, we get
RB × 5 = 5 × 3
5×3
= 3 kN
∴
RB =
5
and
RA = Total load – RB = 5 – 3 = 2 kN
The
B.M. at A = 0
B.M. at B = 0
B.M. at C = RA × 3 = 2 × 3 = 6 kNm.
Now B.M. diagram is drawn as shown in Fig. 14.3 (b).
Now construct the conjugate beam as shown in Fig. 14.3 (c). The vertical load at C* on
conjugate beam
B.M. at C 6 kNm
=
=
EI
EI
Now calculate the reaction at A* and B* for conjugate beam
Let
RA* = Reaction at A* for conjugate beam
RB* = Reaction at B* for conjugate beam.
Taking moments about A*, we get
RB* × 5 = Load on A*C*D* × distance of C.G. of A*C*D* from A*
+ Load on B*C*D* × Distance of C.G. of B*C*D* from A*
=
=
FG 1 × 3 × 6 IJ × FG 2 × 3IJ + FG 1 × 2 × 6 IJ × FG 3 + 1 × 2IJ
H 2 EI K H 3 K H 2 EI K H 3 K
18
6
11
8
22 40
+
×
=
+
=
3
EI EI
EI EI EI
40 1
8
RB* =
× =
EI 5 EI
=
∴
587
STRENGTH OF MATERIALS
5 kN
C
A
(a )
B
3m
5m
RA
RB
D
6 kNm
(b )
A
B.M. Diagram
B
C
D*
6
EI
(c )
B*
A*
C*
3m
2m
Conjugate Beam
R*A
R*B
Fig. 14.3
∴
Let
RA* = Total load (i.e., load A*B*D*) – RB*
=
FG 1 × 5 × 6 IJ − 8
H 2 EI K EI
=
15
8
7
−
=
EI EI EI
θA = Slope at A for the given beam i.e.,
FG dy IJ at A
H dx K
yC = Deflection at C for the given beam
Then according to conjugate beam method,
θA = Shear force at A* for conjugate beam = RA*
7
7
=
(∵ E = 2 × 108 kN/m2 and I = 10–4 m4)
=
EI
2 × 10 8 × 10 −4
= 0.00035 radians. Ans.
yC = B.M. at C* for conjugate beam
= RA* × 3 – Load A*C*D* × Distance of C.G. of A*C*D* from C*
FG
H
IJ FG
K H
IJ
K
7
1
6
1
×
×3
×3−
×3×
2
EI
EI
3
21
9
12
−
−
=
EI EI EI
12
6
6 × 1000
=
=
mm = 0.6 mm. Ans.
8
−4
4 m =
10000
2 × 10 × 10
10
=
588
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
Problem 14.2. A simply supported beam of length 4 m carries a point load of 3 kN at a
distance of 1 m from each end. If E = 2 × 105 N/mm2 and I = 108 mm4 for the beam, then using
conjugate beam method determine : (i) slope at each end and under each load ( i i ) d e f l e c t i o n
under each load and at the centre.
Sol. Given :
Length,
L=4m
Value of
E = 2 × 105 N/mm2 = 2 × 105 × 106 N/m2
= 2 × 105 × 103 kN/m2 = 2 × 108 kN/m2
I = 108 mm4 =
10 8
m4 = 10–4 m4.
10 12
As the load on the beam is symmetrical as shown in Fig. 14.4 (a), the reactions RA and
RB will be equal to 3 kN.
Value of
Now B.M. at A and B are zero.
3 kN
3 kN
C
D
A
1m
(a)
2m
B
1m
4m
RB = 3 kN
RA = 3 kN
(b )
A
E
F
3 kN
3 kN
C
D
E*
J*
3
E.I
(c )
B
F*
3
E.I
A*
B
C*
1m
R*A
H*
D*
1m
Conjugate beam
R*B
Fig. 14.4
B.M. at C = RA × 1 = 3 × 1 = 3 kNm
B.M. at D = RB × 1 = 3 × 1 = 3 kNm
Now B.M. diagram can be drawn as shown in Fig. 14.4 (b).
Now by dividing the B.M. at any section by EI, we can construct the conjugate beam as
shown in Fig. 14.4 (c). The loading are shown on the conjugate beam.
Let
RA* = Reaction at A* for the conjugate beam and
RB* = Reaction at B* for conjugate beam
589
STRENGTH OF MATERIALS
The loading on the conjugate beam is symmetrical
∴
RA* = RB* = Half of total load on conjugate beam
1
[Area of trapezoidal A*B*F*E*]
=
2
1 ( E * F * + A * B*)
× E*C*
=
2
2
4.5
1 (2 + 4)
3
=
×
=
EI
EI
2
2
(i) Slope at each end and under each load
LM
N
LM
N
Let
OP
Q
OP
Q
θA = Slope at A for the given beam i.e.,
FG dy IJ at A
H dx K
θB = Slope at B for the given beam
θC = Slope at C for the given beam and
θD = Slope at D for the given beam
Then according to conjugate beam method,
θA = Shear force at A* for conjugate beam = RA*
4.5
4.5
=
= 0.000225 rad. Ans.
=
EI 2 × 10 8 × 10 −4
4.5
= 0.000225 rad. Ans.
θB = RB* =
EI
θC = Shear force at C* for conjugate beam
= RA* – Total load A*C*D*
4.5 1
3
3
=
− × 1×
=
EI 2
EI EI
3
=
= 0.00015 rad. Ans.
2 × 10 8 × 10 −4
Similarly,
θD = 0.00015 rad. Ans.
(By symmetry)
(ii) Deflection under each load
Due to symmetry, the deflection under each load will be equal
Let
yC = Deflection at C for the given beam and
yD = Deflection at D for the given beam.
Now according to conjugate beam method,
yC = B.M. at C* for conjugate beam
= RA* × 1.0 – (Load A*C*E*) × Distance of C.G. of A*C*E* from C*
1
3
1
4.5
×
=
× 1−
× 1×
2
3
EI
EI
4.5 0.5 4.0
=
−
=
EI EI EI
4
4 × 1000
=
mm
m=
2 × 10 8 × 10 −4
2 × 10 4
= 0.2 mm. Ans.
Also
yD = 0.2 mm.
FG
H
590
IJ
K
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
Deflection at the centre of the beam
= B.M. at the centre of the conjugate beam
= RA* × 2.0 – Load A*C*E*
× Distance of C.G. of A*C*E* from the centre of beam
– Load C*H*J*E*
× Distance of C.G. of C*H*J*E* from the centre of beam
FG
H
IJ FG
K H
IJ FG
K H
IJ
K
3
1
3
1
1
4.5
× 1+
− 1×
×
× 2.0 −
× 1×
2
3
2
EI
EI
EI
9
2
1.5 6.5
−
−
=
=
EI EI EI EI
6.5
6.5 × 1000
m=
=
mm
8
−4
2 × 10 × 10
2 × 10 4
= 0.325 mm. Ans.
Problem 14.3. A simply supported beam AB of span 4 m carries a point of 100 kN
at its centre C. The value of I for the left half is 1 × 108 mm4 and for the right half portion I is
2 × 108 mm4. Find the slopes at the two supports and deflection under the load.
Take E = 200 GN/m2.
Sol. Given :
Length,
L=4m
Length
AC = Length BC = 2 m
Point load,
W = 100 kN
Moment of inertia for AC
=
I = 1 × 108 mm4 =
10 8
10 12
m4 = 10–4 m4
Moment of inertia for BC
= 2 × 108 mm4
= 2 × 10–4 m4 = 2I
(∵ 10–4 = I)
2
9
2
Value of
E = 200 GN/m = 200 × 10 N/m
= 200 × 106 kN/m2.
The reactions at A and B will be equal, as point load is acting at the centre.
100
= 50 kN
∴
RA = RB =
2
Now B.M. at A and B are zero.
B.M. at C
= RA × 2 = 50 × 2 = 100 kNm
Now B.M. can be drawn as shown in Fig. 14.5 (b).
Now we can construct the conjugate beam by dividing B.M. at any section by the product of E and M.O.I.
The conjugate beam is shown in Fig. 14.5 (c). The loading are shown on the conjugate
beam. The loading on the length A*C* will be A*C*D* whereas the loading on length B*C*
will be B*C*E*.
100
B.M. at C
=
The ordinate C*D*
=
E × M.O.I. for AC EI
591
STRENGTH OF MATERIALS
B.M. at C
100
50
=
=
Product of E and M.O.I. for BC
E × 2 I EI
RA* = Reaction at A* for conjugate beam
RB* = Reaction at B* for conjugate beam
The ordinate C*E*
Let
=
100 kN
C
B
A
(a )
2m
2m
RB = 50 kN
RA = 50 kN
D
(b )
100 kNm
B
A
C
B.M. Diagram
D*
100
EI
(c )
A*
E*
100 50
=
2EI EI
B*
C*
2m
2m
R* A
Conjugate beam
R* B
Fig. 14.5
First calculate RA* and RB*
Taking moments of all forces about A*, we get
RB* × 4 = Load A*C*D* × Distance of C.G. of A*C*D* from A
+ Load B*C*E* × Distance of C.G. of B*C*E* from A*
=
FG 1 × 2 × 100 IJ × FG 2 × 2IJ + FG 1 × 2 × 50 IJ × FG 2 + 1 × 2IJ
K H 2 EI K H 3 K
H2
EI K H 3
400 400 800
+
=
3 EI 3 EI 3 EI
200
RB* =
3EI
RA* = Total load on conjugate beam – RB*
=
∴
and
592
=
FG 1 × 2 × 100 + 1 × 2 × 50 IJ − 200
H2
EI
EI K 3 EI
2
=
150 200 250
.
−
=
3 EI 3 EI
EI
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
(i) Slopes at the supports
Let
θB
and
FG dy IJ at A for the given beam
H dx K
F dy I
= Slope at B i.e., G J at B for the given beam
H dx K
θA = Slope at A i.e.,
Then according to the conjugate beam method,
θA = Shear force at A* for conjugate beam = RA*
250
=
3EI
250
=
= 0.004166 rad. Ans.
3 × 200 × 10 6 × 10 4
θB = Shear force at B* for conjugate beam = RB*
200
200
=
= 0.003333 rad. Ans.
=
3 EI 3 × 200 × 10 6 × 10 −4
(ii) Deflection under the load
Let
yC = Deflection at C for the given beam.
Then according to the conjugate beam method,
yC = B.M. at point C* of the conjugate beam
= RA* × 2 – (Load A*C*D*) × Distance of C.G. of A*C*D* from C*
FG
H
IJ FG
K H
IJ
K
100
1
1
250
×
×2
×2−
×2×
EI
3
2
3 EI
500 200 100
−
=
=
3 EI 3 EI
EI
100
=
m
200 × 10 6 × 10 −4
1
1
× 1000 = 5 mm. Ans.
m=
=
200
200
Problem 14.4. A beam ABCD is simply supported at its ends A and D over a span of
30 metres. It is made up of three portions AB, BC and CD each 10 m in length. The moments of
inertia of the section of these portion are I, 3I and 2I respectively, where I = 2 × 1010 mm4. The
beam carries a point load of 150 kN at B and a point load of 300 kN at C. Neglecting the weight
of the beam calculate the slopes and deflections at A, B, C and D. Take E = 2 × 102 kN/mm2.
Sol. Given :
Length,
L = 30 m
Length AB = Length BC = Length CD = 10 m
=
M.O.I. of AB,
M.O.I. of BC,
M.O.I. of CD,
Point load at
Point load at
Value of
I = 2 × 1010 mm4 =
2 × 10 10
10 12
m4 = 2 × 10–2 m4.
3I = 6 × 10–2 m4
2I = 4 × 10–2 m4
B = 150 kN
C = 300 kN
E = 2 × 102 kN/mm2 = 2 × 102 × 106 kN/m2 = 2 × 108 kN/m2
593
STRENGTH OF MATERIALS
150 kN
I
A
300 kN
3I
B
C
2I
D
(a )
10 m
RA = 200 kN
10 m
10 m
RB = 250 kN
F
E
(b )
2500 kNm
2000 kNm
A
B
D
C
B.M. Diagram
E*
(c )
2000
EI
K*
A*
2000
3 EI
2500
3 EI
F*
H*
J*
B*
10 m
R*4
2500
2 EI
D*
C*
10 m
Conjugate beam
10 m
R*D
Fig. 14.6
To find reactions RA and RD, take moments about A.
∴
RD × 30 = 150 × 10 + 300 × 20 = 7500
7500
= 250 kN
∴
RD =
30
∴
RA = Total load – RD
= (150 + 300) – 250 = 200 kN.
Now draw B.M. diagram
B.M. at A and D = 0
B.M. at B = RA × 10 = 200 × 10 = 2000 kNm
B.M. at C = RD × 10 = 250 × 10 = 2500 kNm
B.M. diagram is shown in Fig. 14.6 (b).
Now construct the conjugate beam as shown in Fig. 14.6 (c) by dividing B.M. at any
section by their product of E and I. For the portion AB corresponding conjugate beam is A*B*C*,
for the portion BC corresponding conjugate beam is B*C*H*K* and for the portion CD the
corresponding conjugate beam is C*D*F*. The loading are shown in Fig. 14.6 (c).
2000
2000
The ordinates
B*E* =
, B*K* =
3EI
EI
2500
2500
C*F* =
, C*H* =
2EI
3EI
594
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
2500 2000 500
−
=
3 EI
3 EI
3 EI
Let
RA* = Reaction at A* for conjugate beam
RD* = Reaction at D* for conjugate beam.
To find RA* and RD*, take the moments of all loads acting on the conjugate beam about
A*, we get
RD* × 30 = ( 21 × A*B* × B*E*) × ( 23 × A*B*) + (B*C* × B*K*)
H*J* =
× (10 +
10
2
) + ( 21 × K*J* × H*J*) × (10 + 10 ×
+
=
( 21
× C*F* × C*D*) × (20 + 10 ×
2
3
1
3
)
)
FG 1 × 10 × 2000 IJ × FG 2 × 10IJ + FG 10 × 2000 IJ × (15)
H2
K H 3EI K
EI K H 3
500 I F 50 I F 1 2500
F1
+ G × 10 ×
J × G J + G × × 10IJK × FGH 703 IJK
H2
3 EI K H 3 K H 2 2 EI
200000 300000 125000 437500
+
+
+
3 EI
3 EI
9 EI
3 EI
600000 + 900000 + 125000 + 1312500 2937500
=
=
9 EI
9 EI
2937500 293750
=
∴
RD* =
9 EI × 30
27 EI
∴
RA* = Total load on conjugate beam – RD*
1
2000
2000 1
500 1
2500
× 10 ×
+ 10 ×
+ × 10 ×
+ × 10 ×
=
EI
2
3 EI
2
3 EI 2
2 EI
293750
–
27EI
10000 20000 2500 6250
293750
=
−
+
+
+
EI
3 EI
3 EI
EI
27 EI
30000 + 20000 + 2500 + 18750
293750
=
−
3 EI
27 EI
347500
71250 293750 641250 − 293750
=
=
−
=
27EI
3 EI
27 EI
27 EI
(i) Slopes at A, B, C and D
According to conjugate beam method
(a) Slope at A for the given beam
= S.F. at A* for conjugate beam
347500
347500
=
∴
θA = RA* =
27 EI
27 × 2 × 10 8 × 2 × 10 −2
= 0.003218 rad. Ans.
(b) Slope at B for the given beam
= S.F. at B* for conjugate beam
= RA* – Load A*B*E*
2000
347500 1
=
− × 10 ×
27 EI
2
EI
347500 10000 347500 − 270000
=
−
=
27 EI
27 EI
EI
595
=
FG
H
FG
H
FG
H
IJ
K
IJ
K
IJ
K
STRENGTH OF MATERIALS
77500
77500
=
27 EI
27 × 2 × 10 8 × 2 × 10 −2
= 0.0007176 radians. Ans.
(c) Slope at C for the given beam
= S.F. at C* for conjugate beam
= RD* – Load D*C*F*
293750 1
2500
− × 10 ×
=
27 EI
2
2 EI
125000
293750 6250 293750 − 27 × 6250
=
=
−
=
27EI
27 EI
27 EI
EI
125000
=
= 0.001157 rad. Ans.
27 × 2 × 10 8 × 2 × 10 −2
(d) Slope at D for the given beam
= S.F. at D* for conjugate beam
293750
= RD* =
27EI
293750
=
= 0.00272 rad. Ans.
27 × 2 × 10 8 × 2 × 10 −2
(ii) Deflection at A, B, C and D
(a) Deflection at A for the given beam
= B.M. at A* for the conjugate beam
= 0. Ans.
(b) Deflection at B for the given beam
= B.M. at B* for the conjugate beam
= RA* × 10 – Load A*B*E* × Distance of C.G. of A*B*E* from B*
=
FG
H
IJ
K
347500
1
2000
10
×
× 10 −
× 10 ×
27 EI
2
EI
3
3475000 100000
=
−
27 EI
3 EI
3475000 − 900000 2575000
=
=
27 EI
27 EI
2575000
=
= 0.02384 m
27 × 2 × 10 8 × 2 × 10 −2
= 23.84 mm. Ans.
(c) Deflection at C for the given beam
= B.M. at C* for the conjugate beam
= RD* × 10 – Load D*C*F* × Distance of C.G. of D*C*F* from C*
293750
1
2500 10
=
× 10 − × 10 ×
×
27 EI
2
2 EI
3
2937500 62500 2937500 − 62500 × 9
=
−
=
27 EI
3 EI
27 EI
=
596
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
2375000
2375000
=
= 0.02199 m
27 × 2 × 10 8 × 2 × 10 −2
27EI
= 21.99 mm. Ans.
(d) Deflection of D for the given beam = 0. Ans.
=
14.5. RELATION BETWEEN ACTUAL BEAM AND CONJUGATE BEAM..
The relations between an actual beam and the corresponding conjugate beam for different end conditions are given in Table 14.1.
TABLE 14.1
S. No.
Actual beam
Conjugate beam
1.
Simply supported or roller supported end
(Deflection = 0 but slope exists)
Simply supported end B.M. = 0 but S.F. exists
2.
Free end (slope and deflection exist)
Fixed end (S.F. and B.M. exist)
3.
Fixed end (slope and deflection are zero)
Free end (S.F. and B.M. are zero)
4.
Slope at any section
S.F. at the corresponding section
5.
Deflection at any section
B.M. at the corresponding section
6.
Given system of loading
The loading diagram is M/EI diagram
7.
B.M. diagram positive (sagging)
M/EI load diagram is positive (i.e., loading
is downward)
8.
B.M. diagram negative (hogging)
M/EI load diagram is negative (i.e., loading is
upward)
14.6.DEFLECTION AND SLOPE OF A CANTILEVER WITH A POINT LOAD AT THE
FREE END
Fig. 14.7 (a) shows a cantilever AB of length L and carrying a point load W at the free
end B. The B.M. is zero at the free end B and B.M. at A is equal to W.L. The B.M. diagram is
shown in Fig. 14.7 (b). The conjugate beam can be drawn by dividing the B.M. at any section
by EI. Fig. 14.7 (c) shows the conjugate beam A*B* (free at A* and fixed at B*). The loading on
the conjugate beam will be negative (i.e., upwards) as B.M. for cantilever is negative. The
loading on conjugate beam is shown in Fig. 14.7 (c).
dy
Let
θB = Slope at B i.e.,
at B for the given cantilever and
dx
yB = Deflection at B for the given cantilever.
Then according to the conjugate beam method,
θB = S.F. at B* for the conjugate beam
= Load B*A*C*
1
W.L
1
W . L2
=
= × A*B* × A*C* = × L ×
2
2
EI
2 EI
and
yB = B.M. at B* for the conjugate beam
= Load B*A*C* × Distance of C.G. of B*A*C* from B*
FG IJ
H K
FG 1 . L . WL IJ × FG 2 × LIJ = WL .
H 2 EI K H 3 K 3EI
3
=
597
STRENGTH OF MATERIALS
W
A
(a )
B
L
Given beam
A
(b )
B
–
W.L
B.M. Diagram
C
A*
(c )
B*
Fixed end
W.L
E.I
Conjugate beam
C*
Fig. 14.7
Problem 14.5. A cantilever of length 3 m carries a point load of 10 kN at a distance of
2 m from the fixed end. If E = 2 × 105 N/mm2 and I = 108 mm4, find the slope and deflection at
the free end using conjugate beam method.
Sol. Given :
Length,
Point load,
Distance,
Value of
Value of
L=3m
W = 10 kN
AC = 2 m
E = 2 × 105 N/mm2
= 2 × 105 × 106 N/m2 = 2 × 108 kN/m2
I = 108 mm4
= 108 ×
1
10 12
m4 = 10–4 m4
B.M. at B = 0
B.M. at C = 0
B.M. at A = – 10 × 2 = – 20 kNm
Now B.M. can be drawn as shown in Fig. 14.8 (b). Now construct conjugate beam A*B*
(free at A* and fixed at B*) by dividing the B.M. at any section by EI, as shown in Fig. 14.8 (c).
The loading on the conjugate beam will be negative (i.e., acting upwards) as B.M. is negative.
Let
θB = Slope at the free end for the given cantilever i.e.,
yB = Deflection at B for the given cantilever.
Then according to the conjugate beam method,
θB = S.F. at B* for conjugate beam
= Load A*C*D* =
598
1
2
× A*C* × A*D*
FG dy IJ at B and
H dx K
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
20 20
1
×2×
=
2
EI EI
20
=
= 0.001 rad. Ans.
2 × 10 8 × 10 −4
=
10 kN
1m
C
A
(a )
B
3m
Given beam
C
A
B
(b )
20 kNm
B.M. Diagram
D
C*
A*
(c )
20
E.I
B*
Conjugate beam
D*
Fig. 14.8
and
yB = B.M. at B* for the conjugate beam
= (Load A*C*D*) × Distance of C.G. of A*C*D* from B*
=
FG 1 × 2 × 20 IJ × FG 1 + 2 × 2IJ
H 2 EI K H 3 K
20 7
20
7
× =
×
8
−4
EI 3 2 × 10 × 10
3
= 0.00233 m = 2.33 mm. Ans.
Problem 14.6. A cantilever beam AB of length 2 m is carrying a point load 10 kN at B.
The moment of inertia for the right half of the cantilever is 108 mm4 whereas that for the left
half is 2 × 108 mm4. If E = 2 × 108 kN/m2, find the slope and deflection at the free end of the
cantilever.
Sol. Given :
Length,
L=2m
Point load,
W = 10 kN
Length,
AC = length BC = 1 m
M.O.I. of length BC, I = 108 mm4 = 10–4 m4
M.O.I. of length AC
= 2 × 108 mm4 = 2 × 10–4 m4 = 2I
8
Value of E = 2 × 10 kN/m2
B.M. at B = 0
=
599
STRENGTH OF MATERIALS
B.M. at C = – 10 × 1 = – 10 kNm
B.M. at A = – 10 × 2 = – 20 kNm.
10 kN
2I
(a)
C
I
A
B
C
A
B
10 kNm
20 kNm
D
(b )
B.M. Diagram
E
5
EI
A*
20 10 H*
(c) 2 EI = EI
C*
10
F* E.I
B*
D*
Conjugate beam
E*
Fig. 14.9
Now B.M. diagram can be drawn as shown in Fig. 14.9 (b). Now construct conjugate
beam A*B* (free at A* and fixed at B*) by dividing the B.M. at any section by their EI factor.
The loading diagram will be as shown in Fig. 14.9 (c) in which,
B.M. at A
20
10
=
=
A*E* =
E × (M. O.I. of AC) E × 2 I EI
B.M. at C
10
5
=
=
C*E* =
E × (M. O.I. of AC) E × 2 I EI
B.M. at C
10
10
C*D* =
=
=
E × (M. O.I. of BC) E × I EI
dy
Let
θB = Slope at B i.e.,
at B for the given cantilever
dx
yB = Deflection at B for the given cantilever.
Then according to conjugate beam method,
θB = S.F. at B* for conjugate beam
= Load A*C*F*E* + Load B*C*D*
FG IJ
H K
=
1
2
(A*E* + C*F*) × A*C* +
FG
H
IJ
K
1
2
B*C* × C*D*
1 10
5
1
10
× 1+ × 1×
+
2 EI EI
2
EI
15
10
25
=
+
=
2 EI 2 EI 2 EI
25
=
= 0.000625 rad. Ans.
2 × 2 × 10 8 × 10 −4
=
600
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
yB = B.M. at B* for the conjugate beam
= Load A*C*F*H* × Distance of its C.G. from B*
and
FG
H
= 1×
FG
H
IJ
K
+ Load H*E*F* × Distance of its C.G. from B*
+ Load A*C*D* × Distance of its C.G. from B*
5
1
5
× 1.5 +
× 1×
EI
EI
2
IJ × FG 1 + 2 × 1IJ + FG 1 × 1 × 10 IJ × (
K H 3 K H 2 EI K
2
3
× 1)
7.5
25
10
45 + 25 + 20
+
+
=
6 EI
EI 6 EI 3 EI
90
15
15
=
=
=
m
6 EI EI 2 × 10 8 × 10 −4
= 0.00075 m = 0.75 mm. Ans.
Problem 14.7. A cantilever of length 3 m carries a uniformly distributed load of
80 kN/m over the entire length. If E = 2 × 108 kN/m2 and I = 108 mm4, find the slope and
deflection at the free end using conjugate beam method.
Sol. Given :
Length,
L=3m
=
U.d.l.,
Value of
Value of
B.M. at
w = 80 kN/m
E = 2 × 108 kN/m2
I = 108 mm4 =
10 8
10 12
m4 = 10–4 m4
B=0
3
L
= – 80 × 3 × = – 360 kNm
2
2
The variation of B.M. between A and B is parabolic as shown in Fig. 14.10 (b).
Now construct conjugate beam A*B* (free at A* and fixed at B*) by dividing the B.M. at
any section by EI. The loading diagram will be as shown in Fig. 14.10 (c).
B.M. at
A = – (w.L) .
θB = Slope at B for the given cantilever and
yB = Deflection at B for the given cantilever.
Then according to conjugate beam method,
θB = S.F. at B* for conjugate beam
= Load B*A*C* or Area of B*A*C*
Let
1
of the rectangle containing parabola
3
1
= × (A*B* × A*C*)
3
360
1
= ×3×
3
EI
360
360
=
=
EI
2 × 10 8 × 10 −4
= 0.008 rad. Ans.
=
601
STRENGTH OF MATERIALS
80 kN/m
A
B
( a)
3m
B
A
360 kNm
(b )
Parabolic
B.M. Diagram
C
3L
4
A*
B*
Parabolic
(c )
360
E.I
Conjugate beam
C*
Fig. 14.10
and
yB = B.M. at B* for conjugate beam
= Load A*C*B* × Distance of its C.G. from B*
=
=
FG 1 × 3 × 360 IJ × 3L = 360 × 3 × 3 = 810
H3
EI K
4
4
EI
EI
810
2 × 10 8 × 10 −4
= 0.0405 m = 40.5 mm. Ans.
14.7. PROPPED CANTILEVERS AND BEAMS..
When a cantilever or a beam carries some load, maximum deflection occurs at the free
end in case of cantilever and at the middle point in case of simply supported beam. The deflection
can be reduced by providing vertical support at these points or at any suitable point. Propped
cantilevers means cantilevers supported on a vertical support at a suitable point. The vertical
support is known as prop. The props which does not yield under the loads is known as rigid.
The prop (or support) which is of the same height as the original position of the (unloaded)
cantilever or beam, does not allow any deflection at the point of support (or prop) when the
cantilever or beam is loaded. The prop exerts an upward force on the cantilever or beam. As
the deflection at the point of prop is zero, hence the upward force of the prop is such a magnitude
as to give an upward deflection at the point of prop equal to the deflection (at the point of prop)
due to the load on the beam when there is no prop.
Hence the reaction of the prop (or the upward force of the prop) is calculated by equating the downward deflection due to load at the point of prop to the upward deflection due to
prop reaction.
602
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
14.8.S.F. AND B.M. DIAGRAMS FOR A PROPPED CANTILEVER CARRYING A
POINT LOAD AT THE CENTRE AND PROPPED AT THE FREE END
Fig. 14.11 (a) shows a cantilever AB of length L fixed at A and supported on a prop at B
carrying a point load W at the centre.
W
C
A
B
( a)
L/2
L
P=
5W
16
11 W
16
B
C
A
5
W
16
5
W
16
( b)
S.F. Diagram
E
( c)
A
3W
16
5WL
32
3L
11
C
O
B.M. Diagram
B
D
Fig. 14.11
Let
P = Reaction at the rigid prop.
To find the reaction P at the prop*, the downward deflection due to W at the point of
prop should be equal to the upward deflection due to prop reaction at B.
Now we know that downward deflection at point B due to load W
W
=
FG L IJ
H 2K
3 EI
3
FG L IJ
H 2 K × FG L IJ
H 2K
2 EI
2
W
+
(See equation 13.4 )
WL3
WL3
+
24 EI 16 EI
2WL3 + 3WL3 5WL3
=
=
48 EI
48 EI
The upward deflection at the point B due to prop reaction P alone
=
=
PL3
3 EI
...(i)
...(ii)
*Never calculate P by equating the clockwise moment due to the load W to the anticlockwise
moment due to P at the fixed end, as at the fixed end there exist a fixing moment.
603
STRENGTH OF MATERIALS
Equating equations (i) and (ii), we get
PL3 5WL3
=
3 EI 48 EI
5
P=
W
16
∴
(i) S.F. Diagrams
S.F. at
...(14.1)
B=–P
5W
=–
16
(Minus sign due to right upwards)
The S.F. will remain constant between B and C and equal to (–)
S.F. at
C=–
5W
16
5W
11W
+W =+
16
16
11W
between C and A.
16
The S.F. diagram is shown in Fig. 14.11 (b).
(ii) B.M. Diagram
B.M. at
B=0
5W L 5WL
× =
B.M. at
C=
16
2
32
5W
W.L
B.M. at
A=
×L−
16
2
5WL − 8WL
3WL
=
=−
16
16
The B.M. diagram is shown in Fig. 14.11 (c). As the B.M. is changing sign between C
and A, hence there will be a point of contraflexure between C and A. To find its location,
equate the B.M. between A and C to zero.
The B.M. at any section between C and A at a distance x from B
5W
L
× x−W x−
=
16
2
Equating the above B.M. to zero, we get
The S.F. will remain +
FG
H
FG
H
or
or
or
IJ
K
IJ
K
5W
L
.x−W x−
=0
16
2
5x
L
−x+
=0
16
2
11
L
–
x=−
16
2
16 L
8L
=
x=
11 × 2 11
Hence the point of contraflexure will be at a distance 8L/11 from B or 3L/11 from A.
14.9.S.F. AND B.M. DIAGRAMS FOR A PROPPED CANTILEVER CARRYING A
UNIFORMLY DISTRIBUTED LOAD AND PROPPED AT THE FREE END
Fig. 14.12 (a) shows a cantilever AB of length L fixed at A and propped at B, carrying a
uniformly distributed load of w/unit length over its entire length.
604
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
x
w/Unit Length
A
B
(a)
X
L
P=
(b )
5wL
8
3L
8
C
A
S.F. Diagram
3WL
8
B
3wL
8
E
A
(c )
9wL2
128
O
C
2
w.L
8
B
3L
4
B.M. Diagram
Fig. 14.12
Let P = Reaction at the prop.
To find the reaction P at the prop, the downward deflection due to uniformly distributed
load at B should be equated to the upward deflection due to prop reaction at B.
We know that downward deflection at point B due to u.d.l.
wL4
...(i) (See equation 13.6 )
8 EI
The upward deflection at point B due to prop reaction P alone
=
PL3
3 EI
Equating equations (i) and (ii), we get
=
∴
PL3 w . L4
=
3 EI
8 EI
3
P=
w.L
8
...(ii)
...(14.2)
(i) S.F. diagram
S.F. at
B=–P
(Minus sign due to right upwards)
3
= – wL
8
The S.F. at any section at a distance x from B is given by
3
...(iii)
Fx = – w . L + w . x
8
The S.F. varies by a straight line law between A and B. S.F. at A is obtained by substituting x = L in the above equation.
605
STRENGTH OF MATERIALS
3
5
wL + w . L = + w . L
8
8
To find the point, at which S.F. is zero, equation (iii) should be equated to zero.
3wL
3L
+ wx or x =
∴
0=–
8
8
Hence the S.F. is zero at a distance 3L/8 from B. The point of zero shear is shown by C.
The S.F. diagram is shown in Fig. 14.12 (b).
∴
FA = –
(ii) B.M. diagram
B.M. at
B=0
B.M. at any section at a distance x from B is given by,
3
x
Mx = w . L . x – w . x
...(iv)
8
2
At A, x = L and hence B.M. at A is given by,
3
L
MA = w . L . L – w . L .
8
2
3
1
2
2
= w. L − w. L
8
2
(3 − 4) w . L2
w . L2
=
=−
8
8
3L
The S.F. is zero at x =
, hence B.M. at the point of zero shear is obtained by substitut8
3L
in equation (iv). Hence B.M. at C is given by
ing x =
8
3
3L
3L 3L
− w.
.
MC = w . L .
8
8
8 8×2
9w L2 9w . L2
9
−
=
w . L2
64
128
128
The B.M. diagram is shown in Fig. 14.12 (c).
=
or
or
(iii) Point of contraflexure
Putting Mx = 0 in equation (iv), we get
3
x
0= w.L.x–w.x.
8
2
3
x
0 = L−
8
2
3 × 2L 3L
=
x =
.
8
4
(Cancelling w.x)
(iv) Deflection
The B.M. at any section at a distance x from B is given by equation (iv).
∴
Mx =
wx 2
3
w. L. x −
8
2
But B.M. at any section is also equal to EI
606
d2 y
dx 2
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
3
wx 2
.
.
−
w
L
x
2
dx 2 8
Integrating the above equation, we get
∴
EI
d2 y
=
dy 3w . L . x 2 w x 3
=
− .
+ C1
dx
8×2
2 3
3
w
. w . L . x2 −
=
. x3 + C1
16
6
Integrating again, we get
EI
EIy =
...(v)
x3 w x4
3
w. L.
− .
+ C1x + C2
16
3
6 4
w . L . x3
w
. x4 + C1x + C2
...(vi)
−
16
24
where C1 and C2 are constants of integration. At the fixed end the slope and deflection are
zero. At the end B, deflection is zero. Hence at B, x = 0 and y = 0.
Substituting x = 0 and y = 0 in equation (vi), we get
0 = C2
Substituting x = L and y = 0 in equation (vi), we get
=
w . L . L3
w
. L4 + C1 . L + 0
−
16
24
wL3 w. L3
+ C1
−
=
16
24
wL3 wL3 2wL3 − 3wL3
wL3
C1 =
.
−
=
=−
24
16
48
48
Substituting the values of C1 and C2 in equation (vi), we get
0=
or
w . L . x3
w
wL3
.x
−
. x4 −
16
24
48
The above equation gives the deflection at any section of the cantilever.
EIy =
The deflection at the centre of the cantilever is obtained by substituting x =
(∵ C2 = 0)
...(vii)
L
in equa2
tion (vii). If yC is the deflection at the centre then, we have
EI . yC =
∴
FG IJ
H K
wL
L
×
16
2
3
−
FG IJ
H K
w L
24 2
4
−
wL3 L
.
48 2
=
wL4
wL4
wL4
−
−
16 × 8 24 × 16
96
=
3wL4 − wL4 − 4wL4
2wL4
wL4
=−
=–
24 × 16
24 × 16
192
yC = –
∴ Downward deflection,
yC =
wL4
(Negative sign means that deflection is downwards)
192 EI
wL4
192 EI
...(14.3)
607
STRENGTH OF MATERIALS
(v) Maximum deflection
Maximum deflection takes place where
dy
is zero. Differentiating equation (vii) w.r.t.
dx
x, we get
EI
Putting,
dy 3wL 2 4w . x 3 wL3
=
x −
−
dx
16
24
48
3w . L 2 w . x 3 w . L3
=
x −
−
16
6
48
dy
= 0, we get
dx
wx 3 w . L3
3
w . L . x2 −
−
16
6
48
2
3
0 = 9w . L . x – 8w . x – wL3.
The above equation is solved by trial and error. Hence we get
x = 0.422L
Substituting this value in equation (vii), we get maximum deflection.
0 =
∴
...(14.4)
wL
w
wL3
× (.422L)3 –
(0.422L)4 –
× (0.422L)
16
24
48
= – 0.005415wL4
EIymax =
0.005415w. L4
EI
∴ Maximum downward deflection
0.005415
=
w . L4
...(14.5)
EI
Problem 14.8. A cantilever of length 6 m carries a point load of 48 kN at its centre. The
cantilever is propped rigidly at the free end. Determine the reaction at the rigid prop.
Sol. Given :
Length,
L=6m
Point load,
W = 48 kN
Let
P = Reaction at the rigid prop
Using equation (14.1), we get
5
×W
P=
16
5
=
× 48 = 15 kN. Ans.
16
Problem 14.9. A cantilever of length 4 m carries a uniformly distributed load of
1 kN/m run over the whole length. The cantilever is propped rigidly at the free end. If the value
of E = 2 × 105 N/mm2 and I of the cantilever = 108 mm4, then determine :
(i) Reaction at the rigid prop,
(ii) The deflection at the centre of the cantilever,
(iii) Magnitude and position of maximum deflection.
Sol. Given :
Length,
L=4m
∴
608
ymax =
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
U.d.l.
Value of
Value of
w = 1 kN/m run
E = 2 × 105 N/mm2 = 2 × 105 × 106 N/m2
= 2 × 1011 N/m2
I = 108 mm4 = 108 × 10–12 m4 = 10–4 m4
(i) Reaction at the rigid prop
Let
P = Reaction at the rigid prop
Using equation (14.2), we get
3
P= ×w.L
8
3
= × 1 × 4 = 1.5 kN. Ans.
8
(ii) The deflection at the centre of the cantilever
Let
yC = Deflection at the centre of cantilever
Using equation (14.3), we get
yC =
=
wL4
192 EI
1000 × 4 4
m
192 × 2 × 10 11 × 10 −4
256
2 1000
m= ×
mm
=
4
3 10 4
384 × 10
= 0.0667 mm. Ans.
(∵ w = 1 kN = 1000 N)
(iii) Magnitude and position of maximum deflection
The position of the maximum deflection is given by equation (14.4).
∴
x = 0.422 × L
= 0.422 × 4 = 1.688 m.
Hence maximum deflection will be at a distance 1.688 m from the free end of the cantilever.
Maximum deflection is given by equation (14.5)
0.005415w . L4
EI
0.005415 × 1000 × 4 4
m
=
(∵ w = 1 kN = 1000 N)
2 × 10 11 × 10 −4
0.005415 × 1000 × 256 × 1000
mm
=
2 × 107
= 0.0693 mm. Ans.
Problem 14.10. A cantilever ABC is fixed at A and rigidly propped at C and is loaded
as shown in Fig. 14.13. Find the reaction at C.
Sol. Given :
Length,
L=6m
U.d.l.,
w = 1 kN/m
Loaded length,
L1 = 4 m
Let
P = Reaction at the prop C.
609
∴
ymax =
STRENGTH OF MATERIALS
1 kN/m
A
C
B
4m
2m
P
Fig. 14.13
To find the reaction P at the prop, the downward deflection due to uniformly distributed
load on the AB at point C should be equated to the upward deflection due to prop reaction at C.
We know that downward deflection at point C due to u.d.l. on length AB is given by,
wL14 wL13
(L – L1)
+
8 EI
6 EI
1 × 44 1 × 43
32
64
(6 − 4) =
=
+
+
EI 3 EI
8 EI
6 EI
96 + 64 160
=
=
3 EI
3 EI
The upward deflection at point C due to prop reaction P alone
y=
PL3 P × 6 3 72 P
=
=
3 EI
3 EI
EI
Since both the deflections given by equations (i) and (ii) should be equal.
160 72 P
∴
=
3 EI
EI
160
P=
= 0.741 kN. Ans.
3 × 72
=
or
...(i)
...(ii)
14.10. S.F. AND B.M. DIAGRAMS FOR A SIMPLY SUPPORTED BEAM WITH A UNIFORMLY DISTRIBUTED LOAD AND PROPPED AT THE CENTRE
Fig. 14.14 (a) shows a simply supported beam AB of length L propped at its centre C
and carrying a uniformly distributed load of w/unit length over its entire span.
Let P = Reaction of the prop at C
To find the reaction P at the prop, the downward deflection at C due to uniformly distributed load should be equated to the upward deflection at C due to prop reaction.
The downward deflection at the centre of a simply supported beam due to uniformly
distributed load is given by,
5wL4
...(i)
384 EI
The upward deflection of the beam at C due to prop reaction P alone is given by,
yC =
PL3
48 EI
Equating equations (i) and (ii), we get
yC =
PL3
5wL4
=
48 EI 384 EI
610
...(ii)
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
5wL4
48 EI
×
384 EI
L3
5W
5
= .w. L =
8
8
or
P=
x
(a)
(∵ W = w . L)
w/Unit Length
A
B
C
X
L/2
L/2
P
RA
RB
5 wL
16
3 wL
16
3 wL
16
5 wL
16
(b)
S.F. Diagram
2
2
9wL
512
9wL
512
(c)
wL2
32
B.M. Diagram
Fig. 14.14
Now reactions RA and RB can be calculated. Due to symmetry, the reactions RA and RB
would be equal.
But
RA + RB + P = Total load on beam
=w.L=W
∴
RA + RA +
or
or
5W
=W
8
3W
1
5W
1 3W
RA =
= ×
=
W−
16
2
8
2
8
3W
RA = RB =
.
16
FG
H
IJ
K
FG∵
H
RB = RA and P =
5W
8
IJ
K
(i) S.F. Diagram
3W
16
The S.F. at any section X at a distance x from A is given by,
3W
Fx =
− wx
16
S.F. at
A = RA =
...(i)
611
STRENGTH OF MATERIALS
L
and hence S.F. at C will be,
2
3W wL
−
FC =
16
2
3W W
−
=
16
2
3W − 8W
5W
=−
=
16
16
at C,
x=
(∵ W = w . L)
3W
5W
at A to –
at C.
16
16
5W
3W
Similarly for the span CB, the S.F. will change uniformly from +
at C to –
at B.
16
16
Let at a distance x from A in the span AC, the S.F. is zero. Equating S.F. as zero in
equation (i), we get
Hence for the span AC, the S.F. changes uniformly from +
3W
–w×x
16
3w . L
–w.x
=
16
3L
=
−x
16
3L
x=
16
0=
∴
(∵ W = w . L)
3L
3L
from A. Also S.F. will be zero at a distance
from
16
16
B due to symmetry. Now the S.F. diagram can be drawn as shown in Fig. 14.14 (b).
Hence S.F. is zero at a distance
(ii) B.M. Diagram
B.M. at A is zero and also at B is zero.
B.M. at any section X at a distance x from A is given by,
Mx = RA . x – w . x .
=
x
2
FG∵
H
3wL
w . x2
.x−
16
2
The B.M. at C will be obtained by substituting x =
∴
MC =
3wL L
. −
16 2
2
=
FG L IJ
H 2K
3W
3w . L
or
16
16
IJ
K
...(ii)
L
in the above equation.
2
2
2
3w . L
w. L
3wL2 − 4wL2
−
=
32
8
32
=–
612
w.
RA =
wL2
32
2
...(14.6)
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
Now the B.M. will be maximum where S.F. is zero after changing its sign. But S.F. is
3L
from A.
zero after changing its sign at a distance x =
16
Hence by substituting x =
3L
in equation (ii), we get maximum B.M.
16
∴ Max. B.M.
=
3wL 3 L w 3 L
.
− .
16 16 2 16
=
9wL2
9wL2
18wL2 − 9wL2
−
=
256
2 × 256
2 × 256
FG IJ
H K
2
9wL2
.
512
To find the position of point of contraflexure, the B.M. must be equated to zero. Hence
substituting Mx = 0, in equation (ii), we get
3wL
w
.x−
0=
. x2
16
2
3L x
−
=
(Cancelling w . x from both sides)
16 2
3L
3L
or
x=
×2=
8
16
Now the B.M. diagram can be drawn as shown in Fig. 14.14 (c).
Problem 14.11. A uniform girder of length 8 m is subjected to a total load of 20 kN
uniformly distributed over the entire length. The girder is freely supported at its ends. Calculate the B.M. and the deflection at the centre.
If a prop is introduced at the centre of the beam so as to nullify this deflection, find the
net B.M. at the centre.
Sol. Given :
Length,
L=8m
Total load,
W = 20 kN
W 20
∴ U.d.l.,
w=
=
= 2.5 kN/m.
8
L
(i) The deflection at the centre of a simply supported beam carrying a uniformly distributed load is given by (without prop)
=
5 × 2.5 × 8 4 400
5wL4
=
. Ans.
=
3EI
384 EI
384 EI
where EI = Stiffness of the girder.
(ii) The B.M. at the centre of a simply supported beam due to uniformly distributed load
only (i.e., without prop) is given by
y=
wL2 2.5 × 8 2
= 20 kNm. Ans.
=
8
8
(iii) Net B.M. at the centre when a prop is introduced at the centre
Let MC = Net B.M. at centre when a prop is provided.
Now using equation (14.6), we get
M=
MC = –
wL2
2.5 × 8 2
= – 5 kNm. Ans.
=−
32
32
613
STRENGTH OF MATERIALS
14.11. YIELDING OF A PROP..
In case of a rigid prop the downward deflection due to load is equal to the upward
deflection due to prop reaction. But if the prop sinks down by some amount say δ, then downward deflection due to load is equal to the upward deflection due to prop reaction plus the
amount by which the prop sinks down.
If
y1 = Downward deflection of beam at the point of prop due to load,
y2 = Upward deflection of the beam due to prop reaction, and
δ = Amount by which the prop sinks down
Then
y1 = y2 + δ
...(14.7)
Problem 14.12. A cantilever of length L carries a uniformly distributed load w per unit
length over the whole length. The free end of the cantilever is supported on a prop. If the prop
sinks by δ, find the prop reaction.
Sol. Given :
Length
=L
U.d.l.
=w
Sinking of prop
=δ
The downward deflection (y1) of the free end of cantilever due to uniformly distributed
load is equal to
wL4
.
8 EI
The upward deflection (y2) of the free end due to prop reaction P will be equal to
PL3
.
3 EI
Now using equation (14.7), we get
y1 = y2 + δ
wL4 PL3
=
+δ
8 EI 3 EI
PL3 wL4
=
−δ
3 EI 8 EI
or
F
GH
I
JK
3EI wL4
− δ . Ans.
L3 8EI
Problem 14.13. A simply supported beam of span 10 m carries a uniformly distributed
load of 1152 N per unit length. The beam is propped at the middle of the span. Find the amount,
by which the prop should yield, in order to make all the three reactions equal.
Take E = 2 × 105 N/mm2 and I for beam = 108 mm4.
Sol. Given :
Span,
L = 10 m
U.d.l.,
w = 1152 N/m
Value of
E = 2 × 105 N/mm2 = 2 × 105 × 106 N/m2
= 2 × 1011 N/m2
Value of
I = 108 mm4 = 108 × 10–12 m4 = 10–4 m4
Total load on beam, W = w . L = 1152 × 10 = 11520 N
If all the three reactions (i.e., RA, RB and P) are equal, then each reaction will be one
third of the total load on the beam.
or
P=
614
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
1152 N/m
A
B
C
5m
5m
P
RA
RB
Fig. 14.15
W 11520
=
= 3840 N.
3
3
Let δ = Amount by which the prop should yield if all the three reactions are equal.
Now the downward deflection of the beam at the centre due to uniformly distributed
load alone is given by,
∴
RA = RB = P =
5wL4
5
1152 × 10 4
=
×
m
384 2 × 10 11 × 10 −4
384 EI
7.5
7.5
m=
× 103 mm = 7.5 mm.
=
10 3
10 3
The upward deflection due to prop reaction at the point of prop is given by,
y1 =
y2 =
PL3
3840 × 10 3
=
m
48 EI
48 × 2 × 10 11 × 10 −4
40
m=
(∵ P = 3840 N)
40 × 10 3
mm = 4 mm
10 4
10 4
Now using equation (14.7), we get
y 1 = y2 + δ
δ = y1 – y2 = 7.5 – 4.0 = 3.5 mm. Ans.
=
or
HIGHLIGHTS
1.
2.
3.
4.
5.
6.
7.
8.
9.
The conjugate beam method is used to find the slope and deflections of such beams whose flexural
rigidity (i.e., EI) is not uniform throughout of its length.
Conjugate beam is an imaginary beam of length equal to that of original beam but for which load
diagram is M/EI diagram.
The load on conjugate beam at any point is equal to the B.M. at that point divided by EI.
The slope at any section of the given beam = S.F. at the corresponding section of the conjugate
beam.
The deflection at any point of the given beam = B.M. at the corresponding point of the conjugate
beam.
Propped cantilevers means cantilevers supported on a vertical supported at a suitable point.
The rigid prop does not allow any deflection at the point of prop.
The reaction of the prop (or the upward force of the prop) is calculated by equating the downward
deflection due to load at the point of prop to the upward deflection due to prop reaction.
For a cantilever carrying a uniformly distributed load over the entire span and propped rigidly
at the free end, we have
(i) Prop reaction,
P=
3
w.L
8
615
STRENGTH OF MATERIALS
(ii) B.M. at fixed end,
M=
(iii) Point of contraflexure,
w. L2
8
3L
x=
4
wL4
192 EI
0.005415wL4
(v) Maximum deflection, ymax =
EI
where
w = Uniformly distributed load,
x = Distance from free end.
For a simply supported beam, carrying a uniformly distributed load over the entire span and
propped at the centre, we have
(iv) Deflection at the centre, yC =
10.
(i) Prop reaction,
P=
5
W
8
(ii) Support reactions, RA = RB =
(iii) B.M. at centre, M = –
3W
16
wL2
32
3L
8
W = Total load on beam
= w.L
w = Uniformly distributed load on beam
x = Distance from the support.
(iv) Point of contraflexure, x =
where
EXERCISE
(A) Theoretical Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
Define and explain the terms : Conjugate beam, conjugate beam method, flexural rigidity and
propped beam.
What is the use of conjugate beam method over other methods ?
How will you use conjugate beam method for finding slope and deflection at any section of a
given beam ?
Find the slope and deflection of a simply supported beam carrying a point load at the centre,
using conjugate beam method.
A cantilever carries a point load at the free end. Determine the deflection at the free end, using
conjugate beam method.
What is the relation between an actual beam and the corresponding conjugate beam for different
end conditions ?
What do you mean by propped cantilevers and beams ? What is the use of propping the beam ?
How will you find the reaction at the prop ?
A cantilever of length L, carries a uniformly distributed load of w/m run over the entire length .
It is rigidly propped at the free end. Prove that :
3
w . L and
8
WL4
.
(ii) Deflection at the centre =
192 EI
(i) Prop reaction =
616
CONJUGATE BEAM METHOD, PROPPED CANTILEVERS AND BEAMS
10.
A simply supported beam of length L, carries a uniformly distributed load of w/m run over the
entire span. The beam is rigidly propped at the centre. Determine :
(i) Prop reactions,
(ii) Support reactions,
(iii) B.M. at the centre, and
(iv) Point of contraflexure, if any.
(B) Numerical Problems
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
A beam 6 m long, simply supported at its ends, is carrying a point load at 50 kN at its centre. The
moment of inertia of the beam is 76 × 106 mm4. If E = 2.1 × 105 N/mm2, determine the slope at the
supports and deflection at the centre of the beam using conjugate beam method.
[Ans. 3.935 and 13.736 mm]
A simply supported beam of length 10 m, carries a point load of 10 kN at a distance 6 m from the
left support. If E = 2 × 105 N/mm3 and I = 1 × 108 mm4, determine the slope at the left support and
deflection under the point load using conjugate beam method.
[Ans. 6.00028 rad. and 0.96 mm]
A beam of length 6 m is simply supported at its ends and carries two point loads of 48 kN and
40 kN at a distance of 1 m and 3 m respectively from the left support. Find the deflection under
each load. Take E = 2 × 105 N/mm2 and I = 85 × 106 mm4. Use conjugate beam method.
[Ans. 9.019 mm and 16.7 mm]
A beam AB of span L is simply supported at A and B and carries a point load W at the centre C of
the span. The moment of inertia of the beam section is I for the left half and 2I for the right half.
Calculate the slope at each end and deflection at the centre.
LMAns. θ
MN
A
=
WL8
5WL2
WL2
, θB =
and yC =
96 EI
24 EI
68 EI
OP
PQ
A cantilever of length 3 m is carrying a point load of 25 kN at the free end. If I = 108 mm4 and
E = 2.1 × 105 N/mm3, then determine : (i) slope of the cantilever at the free end and (ii) deflection
at the free end using conjugate beam method.
[Ans. 0.005357 rad. and 10.71 mm]
A cantilever of length 3 m is carrying a point load of 50 kN at a distance of 2 m from the fixed end.
If I = 108 mm4 and E = 2 × 105 N/mm2, find (i) slope at the free end, and (ii) deflection at the free
end using conjugate beam method.
[Ans. 0.005 rad. and 11.67 mm]
A cantilever of length 5 m carries a point load of 24 kN at its centre. The cantilever is propped
rigidly at the free end. Determine the reaction at the rigid prop.
[Ans. 7.5 kN]
A cantilever of length 4 m carries a uniformly distributed load of 2 kN/m run over the whole
length. The cantilever is propped rigidly at the free end. If E = 1 × 105 N/mm2 and I = 108 mm4,
then determine :
(i) reaction at the rigid prop
(ii) the deflection at the centre of the cantilever, and
(iii) magnitude and position of maximum deflection.
[Ans. (i) 3 kN (ii) 0.0667 mm (iii) x = 1. 688 m, ymax = 0.0693 mm]
A simply supported beam of length 8 m carries a uniformly distributed load of 1 kN/m run over
the entire length. The beam is rigidly propped at the centre. Determine : (i) reaction at the prop
(ii) reactions at the supports (iii) net B.M. at the centre and (iv) positions of points of contraflexures.
[Ans. (i) 5 kN (ii) 1.5 kN (iii) – 2.0 kNm (iv) 3 m from each support]
A cantilever of length 10 m carries a uniformly distributed load of 800 N/m length over the
whole length. The free end of the cantilever is supported on a prop. The prop sinks by 5 mm. If
[Ans. 2750 N]
E = 3 × 105 N/mm2 and I = 108 mm4, then find the prop reaction.
617
15
CHAPTER
FIXED AND CONTINUOUS
BEAMS
15.1. INTRODUCTION
A beam whose both ends are fixed is known as a fixed beam. Fixed beam is also called a
built-in or encaster beam. In case of a fixed beam both its ends are rigidly fixed and the slope
and deflection at the fixed ends are zero. But the fixed ends are subjected to end moments.
Hence end moments are not zero in case of a fixed beam.
Fig. 15.1
In case of simply supported beam, the deflection is zero at the ends. But the slope is not
zero at the ends as shown in Fig. 15.1 (a).
In case of fixed beam, the deflection and slope are zero at the fixed ends as shown in
Fig. 15.1 (b). The slope will be zero at the ends if the deflection curve is horizontal at the ends.
To bring the slope back to zero (i.e., to make the deflection curve horizontal at the fixed ends),
the end moments MA and MB will be acting in which MA will be acting anti-clockwise and MB
will be acting clockwise as shown in Fig. 15.1 (b).
A beam which is supported on more than two supports is known as continuous beam.
This chapter deals with the fixed beams and continuous beam. In case of fixed beams the B.M.
diagram, slope and deflection for various types of loading such as point loads, uniformly distributed load and combination of point load and u.d.l., are discussed. In case of continuous
beam, Clapeyron’s equation of three moments and application of this equation to the continuous beam of simply supported ends and fixed ends are explained.
619
STRENGTH OF MATERIALS
15.2. BENDING MOMENT DIAGRAM FOR FIXED BEAMS
Fig. 15.1 (c) shows a fixed beam AB of length L subjected to two loads W and 2W at a
L
distance of
from each ends.
4
W
2W
L/4
L/2
L/4
A
B
MA
MB
L
RB
RA
Fig. 15.1 (c)
Let
RA = Reaction at A
RB = Reaction at B
MA = Fixed end moment at A
MB = Fixed end moment at B
The above four quantities i.e., RA, RB, MA and MB are unknown.
The values of RA, RB, MA and MB are calculated by analysing the given beam in the
following two stages :
(i) A simply supported beam subjected to given vertical loads as shown in Fig. 15.2.
Consider the beam AB as simply supported.
Let
RA* = Reaction at A due to vertical loads
RB* = Reaction at B due to vertical loads.
Taking moments about A, we get
2W
W
L/4
L/2
D
C
L/4
B
A
(a)
RA* = 5W
4
F
RB* = 7W
4
E
+
(b)
A
C
D
B.M. diagram considering beam as simply supported
Fig. 15.2
L
3L
+ 2W ×
4
4
W 6W
7W
RB* =
=
+
4
4
4
RA* = Total load – RB*
RB* × L = W ×
∴
and
620
B
FIXED AND CONTINUOUS BEAMS
= 3W –
7W
5W
=
4
4
B.M. at A = 0,
B.M. at B = 0.
5W
L
5WL
×
=
4
16
4
7WL
7W L
B.M. at D =
=
×
16
4
4
Now B.M. diagram can be drawn as shown in Fig. 15.2 (b). In this case, B.M. at any
point is a sagging (+ve) moment.
(ii) A simply supported beam subjected to end moments only (without given loading) as
shown in Fig. 15.3.
Let
MA = Fixed end moment at A
MB = Fixed end moment at B
R = Reaction* at each end due to these moments.
As the vertical loads acting on the beam are not symmetrical (they are W at distance
L/4 from A and 2W at a distance L/4 from B), the fixed end moments will be different.
Suppose MB is more than MA and reaction R at B is acting upwards. Then reaction R at
A will be acting downwards as there is no other load on the beam. (ΣFy = 0). Taking moments
about A, we get clockwise moment at A = Anti-clockwise moment at A.
MB = MA + R.L
B.M. at C =
∴
R=
MB − M A
L
...(A)
A
B
MB
MA
( a)
R
R
A
(b )
MA
B
–
MB
H
J
B.M. diagram due to end moments
Fig. 15.3
As MB has been assumed more than MA, the R.H.S. of equation (A) will be positive. This
means the magnitude of reaction R at B is positive. This also means that the direction of
reaction R at B is according to our assumption. Hence the reaction R will be upwards at B and
downwards at A as shown in Fig. 15.3 (a). The B.M. diagram for this condition is shown in
Fig. 15.3 (b). In this case, B.M. at any point is a hogging (–ve) moment.
Since the directions of the two bending moments given by Fig. 15.2 (b) and Fig. 15.3 (b)
are opposite to each other, therefore their resultant effect may be obtained by drawing the two
moments on the same side of the base AB, as shown in Fig. 15.4.
*The reaction at each end will be equal. There is no vertical load on the beam hence reaction at
A + reaction at B = 0. Or reaction at A = – reaction at B.
621
STRENGTH OF MATERIALS
F
E
J
+
H
–
MB
–
MA
A
C
D
B
Resultant B.M. diagram
Fig. 15.4
Now the final reactions RA and RB are given by
RA = RA* – R
and
RB = RB* + R
In the above two equations, RA* and RB* are already calculated. They are : RA* = 5W/4
and RB* = 7W/4. But the value of R is in terms of MB and MA. It is given by R = (MB – MA)/L.
Hence to find the value of R, we must calculate the value of MB and MA first.
To find the values of MA and MB
Let
Mx = B.M. at any section at a distance x from A due to vertical loads
Mx′ = B.M. at any section at a distance x from A due to end moments.
The resultant B.M. at any section at a distance x from A
= Mx – Mx′
(Mx is +ve but Mx′ is –ve)
But B.M. at any section is also equal to EI
d2 y
dx 2
d2 y
= Mx – Mx′
dx 2
Integrating the above equation for the entire length, we get
∴
EI
EI
LM dy OP
N dx Q
L
=
0
z
L
0
M x dx −
z
L
0
...(i)
M x ′ dx
dy
represents the slope. And slope at the fixed ends i.e., at A and B are zero. The
dx
above equation can be written as
But
EI
LMFG dy at
NH dx
IJ FG dy at
K H dx
x=L −
z
z
z
L
=
EI [0 – 0] =
or
Now
z
0=
0
M x . dx −
L
0
M x . dx −
L
0
M x . dx −
IJ OP
KQ
x=0
z
z
z
L
0
M x ′ dx
L
0
M x ′ dx
L
0
M x ′ dx
L
0
M x . dx represents the area of B.M. diagram due to vertical loads and
represents the area of B.M. diagram due to end moments.
622
z
...(ii)
L
0
M x ′. dx
FIXED AND CONTINUOUS BEAMS
Let
a = Area of B.M. diagram due to vertical loads
a′ = Area of B.M. diagram due to end moments.
z
z
Then
L
0
L
and
0
M x . dx = a
M x ′. dx = a′
Substituting these values in equation (ii), we get
0 = a – a′
or
a = a′
...(15.1)
The above equation shows that area of B.M. diagram due to vertical loads is equal to the
area of B.M. diagram due to end moments.
Again consider the equation (i)
d2 y
= Mx – Mx′
dx 2
Multiplying the above equation by x, we get
EI
d2 y
= x.Mx – x.Mx′
dx 2
Integrating for the whole length of the beam i.e., from 0 to L, we get
EI.x.
z
L
0
EI. x
z
L
d2 y
dx
2
d2 y
. dx =
z
z
L
0
L
x. M x . dx −
L
0
x. M x ′ dx
L
x. M x ′ . dx
...(iii)
0
0
dx 2
In the above equation, Mx.dx represents the area of B.M. diagram due to vertical loads
at a distance x from the end A. And the term (x.Mx.dx) represents the moment of area of B.M.
EI
0
x.
. dx =
diagram about the end A. Hence
z
x. M x . dx −
z
z
L
0
x. M x . dx represents the moment of the total area of B.M.
diagram due to vertical loads about A, and it is equal to total area of B.M. diagram due to
vertical loads multiplied by the distance of C.G. of area from A,
z
∴
L
0
x. M x dx = ax
where x = Distance of the C.G. of B.M. diagram due to vertical loads.
z
Similarly,
L
0
x. M x ′ . dx = a′ x ′
where x′ = Distance of the C.G. of B.M. diagram due to end moments.
Substituting the above values in equation (iii), we get
EI
z
L
0
or
x.
d2 y
dx 2
. dx = ax − a′ x ′
L dy − yOP
EI M x
N dx Q
L
= ax − a′ x ′
0
LM∵
MN
FG
H
IJ FG
K H
I
JK
d2 y
d 2 y dy
dy
dy
d
x
−y = x
x
+
−
=
dx
dx
dx
dx 2
dx 2 dx
623
OP
PQ
STRENGTH OF MATERIALS
EI
or
or
LMFG xdy − yIJ
MNH dx K
FG
H
− x
at x = L
IJ
K
dy
−y
dx
at x = 0
OP ax − a′ x ′
PQ =
EI [(LθB – yB) – (0 × θA – yA) ] = ax − a′ x ′ .
Since slope and deflection at A and B are zero, hence θA, θB, yA and yB are zero.
0 = ax − a′ x ′
∴
ax = a′ x ′
or
...(15.2)
But from equation (15.1), we have
a = a′
∴
...(15.3)
x = x′
Hence the distance of C.G. of B.M. diagram due to vertical loads from A is equal to the
distance of C.G. of B.M. diagram due to end moments from A.
Now by using equations (15.1) and (15.3) the unknowns MA and MB can be calculated.
This also means that MA and MB can be calculated by
(i) equating the area of B.M. diagram due to vertical loads to the area of B.M. diagram
due to end moments.
(ii) equating the distance of C.G. of B.M. diagram due to vertical loads to the distance of
C.G. of B.M. diagram due to end moments. The distance of C.G. must be taken from the same
end in both cases.
15.3. SLOPE AND DEFLECTION FOR A FIXED BEAM CARRYING A POINT LOAD
AT THE CENTRE
Fig. 15.5 (a) shows a fixed beam AB of length L, carrying a point load W at the centre C
of the beam.
W
A
( a)
B
C
MB
MA
RB
RA
D
+
E
(b )
W.L H
4
–
MA
C
A
RA = W
2
(c )
F
–
B.M. diagram
MB = W.L
8
B
+
B
A
C
–
S.F. diagram
Fig. 15.5
624
W.L
8
RB =
W
2
FIXED AND CONTINUOUS BEAMS
MA = Fixed end moment at A
MB = Fixed end moment at B
RA = Reaction at A
RB = Reaction at B.
The above four are unknown i.e., RA, RB, MA and MB are unknown.
Let
(i) B.M. Diagram
Due to symmetry, the end moments MA and MB will be equal. Hence the B.M. diagram
due to end moments will be a rectangle as shown in Fig. 15.5 (b) by AEFB. Here the magnitude
of MA and MB are unknown. The bending moment diagram for a simply supported beam carrying
a point load at the centre will be a triangle with the maximum B.M. at the centre equal to
W. L
. The B.M. diagram for this case is shown in Fig. 15.5 (b) by a triangle ADB in which
4
W. L
.
CD =
4
Now according to equation (15.1), area of B.M. diagram due to vertical loads should be
equal to the area of B.M. diagram due to end moments.
∴ Equating the areas of the two bending moment diagrams, we get
Area of triangle ADC = Area of rectangle AEFB
or
or
1
× AB × CD = AB × AE
2
1
W. L
×L×
= L × MA
2
4
W. L
∴
MA =
8
W. L
Also
MB = MA =
8
Now the B.M. diagram can be drawn as shown in Fig. 15.5 (b).
...(15.4)
(ii) S.F. Diagram
Equating the clockwise moments and anti-clockwise moments about A, we get
RB × L + MA = MB + W .
But
∴
or
L
2
MA = MB
RB × L = W .
L
2
W
2
W
Due to symmetry,
RA =
2
Now the S.F. diagram can be drawn as shown in Fig. 15.5 (c).
RB =
There will be two points of contraflexure at a distance of
L
from the ends.
4
625
STRENGTH OF MATERIALS
(iii) Slope and Deflection
The B.M. at any section between AC at a distance x from A is given by,
EI
d2 y
= R A × x – MA
dx 2
FG∵ M
H
W. x W. L
−
2
8
Integrating the above equation, we get
=
A
=
W. L
W
, RA =
8
2
IJ
K
dy
W x2 W. L
=
.
−
. x + C1
dx
2 2
8
where C1 is a constant of integration.
EI
dy
= 0. Hence C1 = 0
dx
Therefore, the above equation becomes as
At x = 0,
dy
Wx 2 W . L
=
x
−
dx
4
8
Equation (i) gives the slope of the beam at any point :
EI
...(i)
Integrating equation (i) again, we get
EIy =
W x3 WL x2
.
−
.
+ C2
4 3
8
2
where C2 is another constant of integration. At x = 0, y = 0. Hence C2 = 0.
Therefore, the above equation becomes as
W . x 3 W . Lx 2
...(ii)
−
12
16
Equation (ii) gives the deflection of the beam at any point. The deflection is maximum
L
L
in equation (ii), we get
at the centre of the beam, where x = . Hence substituting x =
2
2
EIy =
EIymax
FG IJ
H K
W L
=
12 2
=
or
ymax =
3
−
FG IJ
H K
W. L L
.
16
2
2
WL3 WL3
2WL2 − 3WL3
WL3
=
=–
−
96
64
192
192
− WL3
192 EI
Minus sign means that the deflection is downwards.
∴
626
Downward deflection, ymax =
WL3
192 EI
...(15.5)
FIXED AND CONTINUOUS BEAMS
Note. The deflection at the centre of a simply supported beam carrying a point load W at the
centre is
WL3
. Hence the deflection of the simply supported beam is four times the deflection of the
48 EI
fixed beam.
Or in other words, the deflection of a fixed beam is one fourth times the deflection of the simply
supported beam. Hence when fixed beams are used, the deflection will be less.
Problem 15.1. A fixed beam AB, 6 m long, is carrying a point load of 50 kN at its centre.
The moment of inertia of the beam is 78 × 106 mm4 and value of E for beam material
is 2.1 × 105 N/mm2. Determine :
(i) Fixed end moments at A and B, and
(ii) Deflection under the load.
Sol.
Given :
Length,
L = 6 m = 6000 mm
Point load, W = 50 kN = 50000 N
M.O.I.,
I = 78 × 106 mm4
Value of
E = 2.1 × 105 N/mm2
Let
MA = Fixed end moment at A,
MB = Fixed end moment at B,
ymax = Deflection under the central point load.
Using equation (15.4), we get
MA = MB =
=
W. L
8
50 × 6
= 37.5 kNm.
8
Ans.
Using equation (15.5), we get
ymax =
=
WL3
192 EI
50000 × 6000 3
192 × 2.1 × 10 5 × 78 × 10 6
= 3.434 mm. Ans.
Alternate Method
Fig. 15.5A(b) shows the simply supported beam, which is having Max. B.M. at the centre equal to RA* × 3 = 25 × 3 = 75 kNm. Fig. 15.5A (c) shows the B.M. diagram for simply
supported beam.
Fig. 15.5A(d) shows the fixed beam with end moments only. Due to symmetry end moments are equal. Hence MA = MB. Fig. 15.5A (e) shows the B.M. diagram due to end moments
only. This diagram is a rectangle.
627
STRENGTH OF MATERIALS
50 kN
(a) A
B
50 kN
B
A
(b )
RA* = 25 kN
RB* = 25 kN
75 kNm
+
(c )
+
B.M. diagram for simply supported beam
MB
MA
(d)
MB
MA
(e )
–
B.M. diagram due to end moments only
Fig. 15.5A
i.e.,
or
Equating the areas of two B.M. diagrams, we get
Area of B.M. diagram for simply supported beam
= Area of B.M. diagram due to end moments.
75 × 6
= MA × 6
2
75
MA =
= 37.5 kNm
2
But
MA = MB
∴
MA = MB = 37.5 kNm. Ans.
15.4. SLOPE AND DEFLECTION FOR A FIXED BEAM CARRYING AN ECCENTRIC
POINT LOAD
Fig. 15.6 (a) shows a fixed beam AB of length L, carrying a point load W at C at a
distance of ‘a’ from A and at a distance of ‘b’ from B. The fixed end moments MA and MB and
also reactions at A and B i.e., RA and RB are shown in the same figure.
628
FIXED AND CONTINUOUS BEAMS
(i) B.M. Diagram
As the load is not acting symmetrically, therefore MA and MB will be different. In this
case MB will be more than MA as the load is nearer to point B. The B.M. diagram due to end
moments will be trapezium as shown in Fig. 15.6 (b) by AEFB. Here the length AE (i.e., MA)
and BF (i.e., MB) are unknown.
The B.M diagram for a simply supported beam carrying an eccentric point load will be
W . a. b
. The B.M. diagram for this
triangle with maximum B.M. under the point load equal to
L
W . a. b.
.
case is shown in Fig. 15.6 (b) by a triangle ADB in which CD =
L
W
A
B
C
MA
MB
(a)
RA
RB
D
F
+
E
(b )
–
MA
–
W.a.b
L
A
MB
C
B
+
RA
B
(c )
C
A
–
RB
Fig. 15.6
or
Equating the areas of the two bending moment diagrams, we get
Area of trapezium AEFB = Area of triangle ADB
1
1
( AE + BF ). AB = × AB × CD
2
2
1
1
W . a. b
( M A + M B ). L = × L
L
2
2
W . a. b
MA + MB =
L
Now using equation (15.3),
...(i)
x = x′
or
Distance of C.G. of B.M. diagram due to vertical loads from A = Distance of C.G. of
B.M. diagram due to end moments from A.
629
STRENGTH OF MATERIALS
A1 x1 + A2 x2
( A1 + A2 )
x′ =
Now
[See Fig. 15.7 (a)]
L 1
2L
+ . L (M B − M A ) ×
2 2
3
=
1
M A . L + . L (MB − M A )
2
L
L
M A . + (M B − M A ) .
2
3 = 3M A . L + 2M B . L − 2M A . L
=
1
3(2 M A + M B − M A )
M A + (MB − M A )
2
M A . L + 2 M B .L
(M A + 2M B ) . L
=
=
3 (M A + MB )
3 (M A + M B )
( M A . L) .
F
2
E
( a)
MB
1
MA
B
A
D
W.a.b
L
3
(b )
4
A
B
Fig. 15.7
and
x =
A3 x3 + A4 x4
( A3 + A4 )
[See Fig. 15.7 (b)]
FG 1 × a × CDIJ × 2a + 1 . b . CD × FG a + b IJ
H2
K 3 2
H 3K
=
FG
H
1
1
. a . CD + . b . CD
2
2
2a 2
b
+b a+
3
3
=
a+b
IJ
K
FG Cancelling CD IJ
H
2 K
2a 2 + 3ab + b2
2a 2 + 2ab + ab + b2
=
3 (a + b)
3 (a + b)
2a (a + b) + b (a + b)
(2a + b) (a + b)
=
=
3 (a + b)
3 (a + b)
=
=
630
2a + b
a + (a + b)
a+ L
=
=
3
3
3
(∵ a + b = L)
FIXED AND CONTINUOUS BEAMS
x′ = x
But
∴
or
(M A + 2M B ) . L
a+ L
=
3 (M A + M B )
3
MA + 2MB =
(a + L) ( M A + MB )
L
LM
N
OP
Q
W . a. b
(a + L) W . a. b
∵ M A + MB =
from equation (i)
.
L
L
L
W . a. b
= (a + L) .
...(ii)
L2
Subtracting equation (i) from equation (ii), we get
=
MB = (a + L) .
W . a. b
2
L
−
W . a. b
L
FG
IJ
H
K
W . a .b
W . a.b F a + L − L I
=
=
G
J
K L
L
L H
=
W . a.b a + L
−1
L
L
2
2
...(iii)
Substituting the value of MB in equation (i), we get
MA +
W . a2 . b
2
L
=
W . a.b
L
W . a . b Wa 2 b
–
L
L2
W . a.b
W . a. b. b
=
(L – a) =
(∵ L – a = b)
2
L
L2
W . a . b2
...(iv)
=
L2
Now MA and MB are known and hence bending moment diagram can be drawn. From
equations (iii) and (iv), it is clear that if a > b than MB > MA.
∴
MA =
(ii) S. F. Diagram
Equating the clockwise moments and anticlockwise about A,
RB × L + MA = MB + W . a
(MB − M A ) + W . a
∴
RB =
L
(M A − M B ) + W . b
Similarly
RA =
L
By substituting the values of MA and MB from equations (iii) and (iv), in the above
equations, we shall get RA and RB. Now S.F. can be drawn as shown in Fig. 15.6 (c).
(iii) Slope and Deflection
The B.M. at any section between AC at a distance x is given by
EI
d2 y
= RA × x – MA
dx 2
631
STRENGTH OF MATERIALS
Substituting the value of RA in the above equation, we get
EI
d2 y
dx
2
=
LM (M
N
=
(M A − MB )
W .b
. x − MA
x+
L
L
A
OP
Q
− MB ) + W . b
. x – MA
L
LM
N
W .b
x
. x − M A + (M B − M A )
L
L
Substituting the values of MA and MB, we get
=
EI
d2 y
dx
2
LM
MN
OP
Q
F
GH
=
W . a . b2
W . a 2 . b W . a . b2
W .b
.x−
+
−
L
L2
L2
L2
=
W . b . x W . a . b2
W . a 2 . b W . a . b2
−
−
−
L
L2
L2
L2
=
W . b . x W . a . b2 W . a . b
x
−
−
a−b .
2
2
L
L
L
L
=
W. b . x W. a . b
x W . a . b2
−
−
.
−
a
b
L
L
L2
L2
=
W. b
W . a . b2
2x – a(a – b) x] –
[L
L3
L2
=
F
GH
b
b
W. b
I x OP
JK L PQ
Ix
JK L
g
g
( L2 − a 2 + ab) x −
W . a . b2
L3
L2
But
L=a+b
∴
L2 = (a + b)2 = a2 + b2 + 2ab.
Substituting the value of L2 in the above equation, we get
∴
EI
d2 y
dx 2
=
=
=
W. b
(a2 + b2 + 2ab – a2 + ab)x –
L3
W. b
L3
(b2 + 3ab) x −
W . b2
(b + 3a) x −
L3
Integrating the above equation, we get
W . a . b2
L2
W . a . b2
L2
W . a . b2
L2
dy W . b2
x 2 W . a . b2
=
(
b
+
3
a
)
.
−
x + C1
dx
2
L3
L2
where C1 is a constant of integration.
EI
At x = 0,
∴
632
dy
= 0. Hence C1 = 0.
dx
EI
dy
W . b2
W . a . b2
(b + 3a) . x 2 −
.x
=
3
dx
2L
L2
...(v)
FIXED AND CONTINUOUS BEAMS
Integrating again, we get
x 3 W . a . b2 x 2
.
−
+ C2
3
2
2 L2
L2
where C2 is another constant of integration. At x = 0, y = 0. Hence C2 = 0.
EIy =
W . b2
(b + 3a) .
W . b2
(b + 3a) x 3 −
W . a. b2
. x2
...(vi)
6 L3
2 L2
The deflection under the load is obtained by substituting x = a in the above equation.
Let yc is the deflection under the load, then
∴
EIy =
EIyc =
=
W . b2
6 L3
W . b2 . a 3
6 L3
=–
=–
=–
=–
∴
(b + 3a) . a3 –
yc = –
W . a. b2
2 L2
. a2
.(b + 3a − 3 L)
W . a 3 . b2
6 L3
W . a 3 . b2
6 L3
W . a 3 . b2
6 L3
(3L – 3a – b)
[3(L – a) – b]
(3b – b)
(∵ L – a = b)
W . a 3 . b3
3 L3
W . a 3 . b3
...(15.6)
3 EIL3
Maximum deflection
Since a > b, hence maximum deflection will take place between A and C. For maximum
dy
dy
deflection,
should be zero. Hence substituting
= 0 in equation (v), we get
dx
dx
0=
W . a . b2
2
L
x=
W . b2
2 L3
W . b2
2 L3
(b + 3a). x 2 –
W . a . b2
L2
.x
(b + 3a) x 2
2aL
W . a . b2
2 L3
×
=
...(15.7)
2
2
(b + 3a)
L
W . b (b + 3a)
Substituting this value of x in equation (vi), we get maximum deflection. If ymax represents the maximum deflection, then
∴
x=
EI.ymax =
FG
H
W . b2
2aL
(b + 3a)
3
b + 3a
6L
IJ
K
3
−
W . a. b2
2 L2
×
FG 2aL IJ
H b + 3a K
2
633
STRENGTH OF MATERIALS
F 2aL IJ LM(b + 3a) . 2aL − 3aLOP
.G
=
(b + 3a)
H b + 3a K N
6L
Q
Wb F 2aL I
.G
=−
J . aL
6 L H b + 3a K
2
W . b2
3
2
2
3
=−
Wb2
4 a 2 L2
2 Wa 3 b2
aL
=
−
.
.
.
3 (b + 3a) 2
6 L3 (b + 3a) 2
2
Wa 3 b2
.
...(15.8)
.
3 EI (b + 3a) 2
Problem 15.2. A fixed beam AB of length 3 m carries a point load of 45 kN at a distance
of 2 m from A. If the flexural rigidity (i.e., EI) of the beam is 1 × 10 4 kNm2, determine :
(i) Fixed end moments at A and B,
(ii) Deflection under the load,
(iii) Maximum deflection, and
(iv) Position of maximum deflection.
Sol. Given :
Length,
L=3m
Point load,
W = 45 kN
Flexural rigidity, EI = 1 × 104 kNm2
Distance of load from A,
a=2m
∴ Distance of load from B,
b=1m
Let
MA and MB = Fixed end moments,
yc = Deflection under the load
ymax = Maximum deflection and
x = Distance of maximum deflection from A.
(i) The fixed end moments at A and B are given by
∴
ymax = −
MA =
and
MB =
W . a . b2
2
L
W . a2 . b
=
32
= 10 kNm. Ans.
45 × 22 × 1
= 20 kNm.
32
L2
(ii) Deflection under load is given by equation (15.6) as
yc = –
=
45 × 2 × 12
W . a 3 . b3
45 × 2 3 × 13
= – 0.000444 m
3 × 1 × 10 4 × 33
3 EIL3
= – 0.444 mm. Ans.
–ve sign means the deflection is downwards.
(iii) Maximum deflection is given by equation (15.8) as
ymax = –
634
=–
Ans.
2
Wa 3 . b2
×
3 EI (b + 3a) 2
FIXED AND CONTINUOUS BEAMS
2
=–
45 × 23 × 12
=
2
=−
16 × 45
3 × 1 × 10
3 × 10 4 × 49
(1 + 3 × 2)
= – 0.00049 m = – 0.49 m. Ans.
(iv) The distance of maximum deflection from point A is given by equation (15.7) as
x=
4
.
2a . L
(b + 3a)
2×2×3
12
=
= 1.714 m. Ans.
1+ 3 × 2
7
Alternate Method
Fig. 15.7A (b) shows the simply supported beam with vertical load of 45 kN at a distance
2 m from A.
The reactions RA* and RB* due to vertical load will be :
3RB* = 45 × 2 or RB* = 90/3 = 30 kN and RA* = 45 – 30 = 15 kN.
Fig. 15.7A (c) shows the B.M. diagram with max. B.M.at C and equal to RA* × 2 = 15 ×
2 = 30 kNm.
Fig. 15.7A (d) shows the fixed beam with end moments and reactions. As the vertical
load is not acting symmetrically, therefore MA and MB will be different. In this case MB will be
more than MA, as load is nearer to point B. The B.M. diagram is shown in Fig. 15.7A(e)
(i) Fixed end moments at A and B. To find the value of MA and MB, equate the areas of
two B.M. diagrams.
∴ Area of B.M. diagram due to vertical loads
= Area of B.M. diagram due to end moments
∴
A1 + A2 = A3 + A4 where A1 =
30 × 2
30 × 1
= 30, A2 =
= 15
2
2
A3 = 3MA, A4 =
or
or
(MB − M A ) × 3
2
= 1.5 (MB – MA)
30 + 15 = 3MA + 1.5MB – 1.5 MA
45 = 1.5MA + 1.5MB
45
= MA + MB
or
MA + MB = 30
...(i)
1.5
Now equating the distance of C.G. of B.M. diagram due to vertical load to the distance
of C.G. of B.M. diagram due to end moments from the some end (i.e., from end A)
or
x = x′
A × x3 + A4 × x4
A1 x1 + A2 x2
= 3
A3 + A4
A1 + A2
or
or
30 ×
or
or
FG
H
1
4
+ 15 × 2 +
3
3
30 + 15
IJ
K
3
+ 1.5 ( M B − M A ) × 2
2
3 M A + 1.5 M B − 1.5 M A
3M A ×
=
40 + 35
4.5 M A + 3 M B − 3 M A
1.5 M A + 3 M B
=
=
45
1.5 M A + 1.5 M B
1.5 M A + 1.5 M B
635
STRENGTH OF MATERIALS
or
or
or
M A + 2MB
1.5 ( M A + 2 M B )
75
5
=
or
=
M A + MB
45
3
1.5 ( M A + M B )
5MA + 5MB = 3MA + 6MB
MA = MB
Solving equations (i) and (ii), we get
MA = 10 kNm and MB = 20 kNm. Ans.
...(ii)
45 kN
(a)
A
B
C
2m
3m
45 kN
R B*
RA*
2m
(b)
3m
RA* = 15 kN
RB* = 30 kN
30 kNm
1
(c)
2
C
2m
1m
3m
B.M. diagram for simply supported beam with vertical loads
MA
MB
(d)
R
R
3m
MA
3
4
(MB–MA)
(e)
MB
B.M. diagram due to end moments only
Fig. 15.7A
Let us now find the reaction R due to end moments only. As the end moments are different,
hence there will be reaction at A and B. Both the reactions will be equal and opposite in direction,
as there is no vertical load, when we consider end moments only. As MB is more, the reaction R
will be upwards at B and downwards at A as shown in Fig. 15.7A (d).
636
FIXED AND CONTINUOUS BEAMS
Taking the moments about A for Fig. 15.7A(d), we get clockwise moment at A = Anticlockwise moments at A
MB = MA + R × 3
MB − M A
20 − 10
10
∴
R=
=
=
kN
3
3
3
Now the total reaction at A and B will be,
10
35
RA = RA* – R = 15 –
=
kN
3
3
10
100
and
RB = RB* + R = 30 +
=
kN
3
3
Now, consider the fixed beam as shown in Fig. 15.7B.
The B.M. at any section between AC at a distance x from A is given by RA × x – MA
MA = 10 kNm
45 kN
A
C
B
2m
3m
x
RA = 35
3
Fig. 15.7B
or
EI
d2 y
dx 2
= RA × x − M A
=
35
× x − 10
3
Integrating, we get
EI
at x = 0,
dy 35 x 2
=
×
− 10 x + C1
dx
3
2
dy
=0
dx
∴
∴ C1 = 0
EI
dy
35 2
x − 10 x
=
dx
6
...(iii)
Integrating again, we get
EI × y =
at x = 0, y = 0,
∴
35 x 3 10 x 2
×
−
+ C2
6
3
2
∴ C2 = 0
EI × y =
(ii) Deflection under the load
From equation (iv), we have
y=
LM
N
35 3
x − 5x2
18
1 35 3
x − 5x2
EI 18
...(iv)
OP
Q
637
STRENGTH OF MATERIALS
To find the deflection under the load, substitute x = 2 m in the above equation.
∴
y=
LM
N
1
=
or
or
or
OP
Q
LM 35 × 8 − 20OP
N 18
Q
1 35
× 23 − 5 × 22
EI 18
1 × 10 4
= – 0.000444 m = – 0.444 mm. Ans.
(– ve sign means the deflection is downwards).
(iii) Maximum deflection
dy
= 0.
Deflection (y) will be maximum when
dx
dy
Hence substituting the value of
= 0 in equation (iii), we get
dx
35 2
x − 10 x
0=
6
0 = 35x2 – 60x
0 = x (35x – 60)
This means that either x = 0 or 35x – 60 = 0 for maximum deflection.
But x cannot be zero, because when x = 0, y = 0.
∴
35x – 60 = 0
(∵ EI = 1 × 104)
60 12
=
= 1.714 m
35
7
Substituting x = 1.714 m in equation (iv), we get maximum deflection.
x=
∴
or
EIymax =
ymax =
35
(1714
.
) 3 − 5(1714
.
)2
18
LM
N
1 35
(1714
.
) 3 − 5(1714
.
)2
EI 18
1
OP
Q
9.79 − 14.69
1 × 10 4
= 0.00049 m = 0.49 mm. Ans.
(iv) Position of maximum deflection
The maximum deflection will be at a distance of 1.714 m (i.e., x = 1.714 m) from end A.
Ans.
Problem 15.3. A fixed beam AB of length 6 m carries point loads of 160 kN and 120 kN
at a distance of 2 m and 4 m from the left end A. Find the fixed end moments and the reactions
at the supports. Draw B.M. and S.F. diagrams.
Sol. Given :
Length
=6m
Load at C, WC = 160 kN
Load at D, WD = 120 kN
Distance
AC = 2 m
Distance
AD = 4 m
=
638
FIXED AND CONTINUOUS BEAMS
For the sake of convenience, let us first calculate the fixed end moments due to loads at C
and D and then add up the moments.
(i) Fixed end moments due to load at C.
For the load at C, a = 2 m and b = 4 m
∴
WC . a . b2
M A1 =
L2
160 × 2 × 4 2
=
62
WC . a 2 . b
M B1 =
=
= 142.22 kNm
160 × 2 2 × 4
62
L2
(ii) Fixed end moments due to load at D.
Similarly for the load at D, a = 4 m and b = 2 m
∴
WD . a . b2
M A2 =
L2
120 × 4 × 2 2
=
62
WD . a 2 . b
M B2 =
and
L2
=
160 kN
(a)
= 71.11 kNm
= 53.33 kNm
160 × 4 2 × 2
62
= 106.66 kNm
120 kN
A
B
MA
MB
RA
RB
E
F
+
(b)
–
195.55
A
149.63
(c)
293.34
266.66
C
D
–
177.77
B
+
10.37
120
–
130.37
Fig. 15.8
∴ Total fixing moment at A,
MA = M A1 + M A2 = 142.22 + 53.33
= 195.55 kNm. Ans.
639
STRENGTH OF MATERIALS
and total fixing moment at B,
MB = M B1 + M B2 = 71.11 + 106.66
= 177.77 kNm. Ans.
B.M. diagram due to vertical loads
Consider the beam AB as simply supported. Let RA* and RB* are the reactions at A and B
due to simply supported beam. Taking moments about A, we get
RB* × 6 = 160 × 2 + 120 × 4
= 320 + 480 = 800
800
∴
RB* =
= 133.33 kN
6
and
RA* = Total load – RB* = (160 + 120) – 133.33
= 146.67 kN
B.M. at A = 0
B.M. at C = RA* × 2 = 146.67 × 2 = 293.34 kNm
B.M. at D = RB* × 2 = 133.33 × 2 = 266.66 kNm
B.M. at B = 0.
Now the B.M. diagram due to vertical loads can be drawn as shown in Fig. 15.8 (b).
In the same figure the B.M. diagram due to fixed end moments is also shown.
S.F. Diagram
Let
RA = Resultant reaction at A due to fixed end moments and vertical loads
RB = Resultant reaction at B.
Equating the clockwise moments and anti-clockwise moments about A, we get
RB × 6 + MA = 160 × 2 + 120 × 4 + MB
or
RB × 6 + 195.55 = 320 + 480 + 177.77
800 + 177.77 − 195.55
∴
RB =
= 130.37 kN
6
and
RA = Total load – RB
= (160 + 120) – 130.37 = 149.63 kN
S.F. at A = RA = 149.63 kN
S.F. at C = 149.63 – 160 = – 10.37 kN
S.F. at D = – 10.37 – 120 = – 130.37 kN
S.F. at B = – 130.37 kN
Now S.F. diagram can be drawn as shown in Fig. 15.8 (c).
Alternate Method
Fig. 15.8A (b) shows the simply supported beam with vertical loads.
Let RA* and RB* are the reactions at A and B due to vertical loads. Taking moments about
A, we get
RB* × 6 = 160 × 2 + 120 × 4 = 320 + 480 = 800
800 400
∴
R B* =
= 133.33 kN
=
6
3
and
RA* = Total load – RB*
= (160 + 120) – 133.33 = 146.67 kN
B.M. at A = 0
B.M. at C = RA* × 2 = 146.67 × 2 = 293.34 kNm
B.M. at D = RB* × 2 = 133.33 × 2 = 266.66 kNm
640
FIXED AND CONTINUOUS BEAMS
Now the B.M. diagram due to vertical loads can be drawn as shown in Fig. 15.8A(c)
Fig. 15.8A (d) shows the fixed beam with end moments only. As the load 160 kN is nearer
to end A, hence MA will be more than MB. The B.M. diagram due to end moments is shown in Fig.
15.8A(e).
To find the values of MA and MB, equate the areas of two B.M. diagrams.
∴
Area of B.M. diagram due to vertical loads
= Area of B.M. diagram due to end moments
160 kN
120 kN
MB
MA
A
B
C
2m
(a)
D
4m
RA
6m
160
RB
120
B
A
(b)
R A*
R B*
26.66
E
3
G
293.33
(c)
1
F
266.67
2
4
D
C
A
B
2m
2m
2m
B.M. diagram for simply supported beam with vertical loads
MA
(d)
MB
B
A
(e)
5
MB
MA
6
(MA–MB)
B.M. diagram due to end moments only
Fig. 15.8A
641
STRENGTH OF MATERIALS
∴
A1 + A2 + A3 + A4 = A5 + A6
...(i)
AC × CE 2 × 293.33
= 293.33
=
2
2
A2 = CD × DF = 2 × 266.67 = 533.34
where A1 =
GF × GE 2 × 26.66
=
= 66.66
2
2
DB × DF 2 × 266.67
=
= 266.67
A4 =
2
2
6 × (M A − M B )
A5 = MB × 6 = 6MB, A6 =
= 3 (MA – MB) = 3MA – 3MB
2
Substituting these values in equation (i), we get
293.33 + 533.34 + 66.66 + 266.67 = 6MB + 3MA – 3MB
or
1119.98 = 3MB + 3MA = 3 (MB + MA)
1119.98
∴
MB + MA =
= 373.33
...(ii)
3
To get the other equation between MA and MB , equate the distance of C.G. of B.M. diagram
due to vertical loads to the distance of C.G. of B.M. diagram due to end moments from end A.
or
x = x′
A3 =
A1 x1 + A2 x2 + A3 x3 + A4 x4
A x + A6 x6
= 5 5
A1 + A2 + A3 + A4
A5 + A6
or
or
or
or
2.95 =
or
or
or
or
or
FG
H
IJ
K
FG
H
IJ
K
2
2
4
+ 533.34 × 3 + 26.66 × 2 +
+ 266.66 × 4 +
3
3
3
293.33 + 533.34 + 26.66 + 266.66
1
6 M B × 3 + 3 (M A − 3M B ) × × 6
3
=
6 M B + 3M A − 3M B
3 (6 M B + 2 M A − 2 M B )
391.1 + 1600 + 70.91 + 1245.35
=
3 (MB + M A )
1119.98
293.33 ×
4 MB + 2M A
MB + M A
2.95MB + 2.95MA = 4MB + 2MA
2.95MA – 2MA = 4MB – 2.95MB
0.95MA = 1.05MB
1.05
M B = 1.1MB
0.95
Substituting this value of MA in equation (ii), we get
MB + 1.1 MB = 373.33
MA =
373.33
= 177.77 kNm. Ans.
2.1
From equation (iii), MA = 1.1 × 177.77 = 195.55 kNm. Ans.
MB =
642
...(iii)
FIXED AND CONTINUOUS BEAMS
Combined B.M. Diagram
MA = 195.55 kNm and MB = 177.77 kNm. Now the combined B.M. diagram can be drawn
as shown in Fig. 15.8 (b).
To draw the S.F. diagram, let us first find the values of resultant reactions due to vertical
loads and fixed end moments RA and RB. Refer to Fig. 15.8A(a). Taking moments about A, we get
clockwise moments at
A = Anti-clockwise moments at A
∴
160 × 2 + 120 × 4 + MB = MA + RB × 6
or
320 + 480 + MB = MA + 6RB
or
800 + 177.77 = 195.55 + 6RB
800 + 177.77 − 195.55
= 130.37 kN
6
RA = Total load – RB = (160 + 120) – 130.37 = 149.63 kN
∴
RB =
and
S.F. Diagram
S.F. at A = RA = 149.63 kN
S.F. at C = 149.63 – 160 = – 10.37 kN
S.F. at D = – 10.37 – 120 = – 130.37 kN
S.F. at B = – 130.37 kN
Now S.F. diagram can be drawn as shown in Fig. 15.8(c).
Problem 15.4. A fixed beam of length 6 m carries two point loads of 30 kN each at a
distance of 2 m from both ends. Determine the fixed end moments and draw the B.M. diagram.
Sol. Given :
Length,
L=6m
Point load at C, W1 = 30 kN
Point load at D, W2 = 30 kN
Distance
AC = 2 m
Distance
AD = 4 m
The fixing moment at A due to loads at C and D is given by
MA = Fixing moment due to load at C + Fixing moment due to load at D
=
W1a1b12
2
L
+
30 × 2 × 4 2
W2 a2 . b2 2
L2
30 × 4 × 2 2
80 40
+
= 40 kNm.
3
3
6
6
Since the beam and loading is symmetrical, therefore fixing moments at A and B should be
=
2
+
2
=
equal.
∴
MB = MA = 40 kNm. Ans.
To draw the B.M. diagram due to vertical loads, consider the beam AB as simply supported. The reactions at the simply supported beam will be equal to 30 kN each.
B.M. at A and B = 0
B.M. at C = 30 × 2 = 60 kNm
B.M. at D = 30 × 2 = 60 kNm.
Now the B.M. diagram due to vertical loads and due to end moments can be drawn as
shown in Fig. 15.9 (b).
643
STRENGTH OF MATERIALS
A
30 kN
30 kN
C
D
B
(a)
+
–
40 kNm
A
60 kNm
60 kNm
C
D
–
40 kNm
B
(b)
Fig. 15.9
15.5. SLOPE AND DEFLECTION FOR A FIXED BEAM CARRYING A UNIFORMLY
DISTRIBUTED LOAD OVER THE ENTIRE LENGTH
Fig. 15.10 (a) shows a fixed beam of length L, carrying uniformly distributed load of
w/unit length over the entire length.
Let
MA = Fixed end moment at A
MB = Fixed end moment at B
RA = Reaction at A
RB = Reaction at B.
(i) B.M. Diagram
Since the loading on the beam is symmetrical, hence MA = MB. The B.M. diagram due to
end moments will be a rectangle as shown in Fig. 15.10 (b) by AEFB. The magnitude of MA or
MB is unknown.
The B.M. diagram for a simply supported beam carrying a uniformly distributed load
will be parabola whose central ordinate will be w.L2/8. The B.M. diagram for this case is
w. L2
.
8
Equating the areas of the two bending moment diagrams, we get
Area of rectangle AEFB = Area of parabola ADB
2
AB × AE =
× [AB × CD]
3
2
w. L2
w. L2
×L×
or
MA =
L × MA =
3
8
12
2
w. L
∴
MB = MA =
12
Now the B.M. diagram can be drawn as shown in Fig. 15.10 (b).
shown in Fig. 15.10 (b) by parabola ADB in which CD =
644
...(15.9)
FIXED AND CONTINUOUS BEAMS
w/unit length
A
MA
(a)
+
E
B.M. Diagram
+
(c)
A
–
w.L
8
A
RA
F
2
–
MA
RB
D
RA
(b)
B
MB
C
MB
C
B
C
B
S.F. Diagram
–
RB
Fig. 15.10
(ii) S.F. Diagram
Equating the clockwise moments and anti-clockwise moments about A, we get
L
RB × L + MA = w.L. + MB
2
But
MA = MB
L
w. L
∴
RB × L = w.L.
or
RB =
2
2
Due to symmetry,
w. L
...(15.10)
RA = RB =
2
Now the S.F. diagram can be drawn as shown in Fig. 15.10 (c).
(iii) Slope and deflection
The B.M. at any section at a distance x from A is given by,
EI
d2 y
dx
2
= RA × x – MA – w.x
x
2
w. L
wL2 wx 2
=
−
.x−
2
12
2
wL. x wx 2 wL2
−
−
2
2
12
Integrating the above equation, we get
F∵
GH
RA =
w. L
wL2
, MA =
2
12
=
I
JK
...(i)
dy w. L x 2 w x 3 wL2
. x + C1
=
.
−
−
dx
2
2
2 3
12
wL 2 w 3 wL2
=
.x − x −
. x + C1
4
6
12
where C1 is a constant of integration.
EI
645
STRENGTH OF MATERIALS
dy
= 0. Hence C1 = 0.
dx
Therefore, the above equation becomes as
At x = 0,
dy w. L 2 w 3 wL2
=
.x − . x −
.x
dx
4
6
12
Integrating the above equation, we get
EI
...(ii)
wL x 3 w x 4 wL2 x 2
.
− .
−
.
+ C2
4
3
6 4
12
2
wL 3 w
wL2 2
=
.x −
. x4 −
. x − C2
12
24
24
where C2 is another constant of integration.
At x = 0, y = 0. Hence C2 = 0.
Therefore, the above equation becomes as
EIy =
EIy =
wL 3 w
wL2 2
.x −
. x4 −
.x
12
24
24
...(iii)
The deflection at the centre is obtained by substituting x =
the deflection at the centre is yc
∴
EIyc =
FG IJ
H K
wL L
.
12
2
3
−
FG IJ
H K
w L
24 2
4
−
wL2
24
FG L IJ
H 2K
L
in the above equation. Let
2
2
w. L4 wL4 wL4
wL4
−
−
=−
96
384
96
384
4
wL
∴
yc = –
384 EI
Minus sign means that the deflection is downwards.
=
...(15.11)
Note. The deflection at the centre of a simply supported beam carrying a uniformly distributed
load over the entire length is 5/384, wL4. This means that the central deflection for the fixed beam is
one-fifth of the central deflection of the simply supported beam.
(iv) Points of contraflexures
For the points of contraflexures, B.M. given by equation (i) should be zero. Hence equating equation (i) to zero, we get
wLx wx 2 wL2
−
−
2
2
12
wL2
L2
0 = wLx – wx2 –
= Lx – x2 –
6
6
0=
or
or
L2
=0
6
Solving the above quadratic equation, we get
x2 – Lx +
+L±
x=
646
L2 −
2
4 × L2
6
L2
3
L±
=
2
=
L
L
±
.
2 2 3
FIXED AND CONTINUOUS BEAMS
As L/2 represents the centre of the beam. Hence the two points of contraflexures occur
at a distance of L/2 3 from the centre of the beam.
Problem 15.5. A fixed beam of length 5 m carries a uniformly distributed load
of 9 kN/m run over the entire span. If I = 4.5 × 10–4 m4 and E = 1 × 107 kN/m2, find the
fixing moments at the ends and the deflection at the centre.
Sol. Given :
Length,
L=5m
U.d.l.
w = 9 kN/m
Value of
I = 4.5 × 10–4 m4
Value of
E = 1 × 107 kN/m2.
(i) The fixing moments at the ends is given by equation (15.9) as
w. L2 9 × 5 2
= 18.75 kNm. Ans.
=
12
12
(ii) The deflection at the centre is given by equation (15.11) as
MA = MB =
wL4
9 × 54
=–
384 EI
384 × 1 × 107 × 4.5 × 10 −4
≡ 0.003254 m = – 3.254 mm. Ans.
Problem 15.6. Find the fixing moments and support reactions of a fixed beam AB of
length 6 m, carrying a uniformly distributed load of 4 kN/m over the left half of the span.
Sol. Given :
Length, L = 6 m
U.d.l., w = 4 kN/m
(i) B.M. diagram due to end moments
Let
MA = Fixing moment at A
MB = Fixing moment at B.
The value of MA will be more than MB as load due u.d.l. is nearer to point A.
The B.M. diagram due to end moments will be trapezium as shown in Fig. 15.11 (b) by
AEFB.
The area of B.M. diagram due to end moments is given by,
yc = –
a′ =
and
1
2
(MA + MB) × 6 = 3(MA + MB)
...(i)
(ii) B.M. diagram due to vertical loads
Now draw the B.M. diagram due to u.d.l. for a simply supported beam.
Let
RA* = Reaction at A for a simply supported beam
RB* = Reaction at B for a simply supported beam.
Taking moments about A for a simply supported beam, we get
RB* × 6 = 4 × 3 × 1.5 = 18
18
= 3 kN
∴
RB* =
6
RA* = Total load – RB*
= 4 × 3 – 3 = 9 kN
The B.M. at A and B are zero.
647
STRENGTH OF MATERIALS
4 kN/m
MA
A
C
MB
RB
(a)
RA
B
E
MA
F
H
MB
A
B.M. diagram for fixing moments
(b)
D
B
9 kNm
A
(c)
B
C
dx
B.M. diagram for vertical loads
Fig. 15.11
B.M. at C = RB* × 3 = 3 × 3 = 9 kNm.
The B.M. diagram from A to C will be parabolic and from C to A the B.M. diagram will
follow a straight line law as shown in Fig. 15.11 (c).
The area of the B.M. due to vertical loads is given by
a = Area of parabola ACD + Area of triangle BCD
= Area of parabola ACD +
1
2
×9×3
...(ii)
To find the area of the parabola ACD, consider a strip of length ‘dx’ at a distance x from
A in portion AC.
The B.M. at a distance x from A is given by
x
(∵ RA* = 9)
Mx = RA* × x – 4 × x . = 9x – 2x2
2
Area of B.M. diagram of length dx
= Mx.dx = (9x – 2x2).dx
Total area of parabola from A to C is obtained by integrating the above equation between
the limits of 0 and 3.
∴ Area of parabola ACD
=
z
3
0
(9 x − 2 x 2 ). dx
L 9x
=M
N2
2
2x3
−
3
OP
Q
3
=
0
9 × 3 2 2 × 33
= 40.5 – 18 = 22.5
−
2
3
Substituting this value in equation (ii), we get
1
a = 22.5 + × 9 × 3 = 36.0
2
648
...(iii)
FIXED AND CONTINUOUS BEAMS
or
Equating the two areas given by equations (i) and (iii), we get
3(MA + MB) = 36.0
MA + MB = 12.0
Now moment of B.M. diagram due to vertical loads about A is given by
ax =
=
z
z
z
3
0
3
0
=
...(iv)
3
0
x. M x . dx + Area of triangle BCD
x(9 x − 2 x 2 ) . dx +
FG
H
× Distance of C.G. of BCD from A
IJ
K
1
1
×9×3× 3+ ×3
2
3
(9 x 2 − 2 x 3 ) . dx + 54
L 9x
=M
N3
3
2x4
−
4
OP
Q
3
LM
N
+ 54 = 3 × 33 −
0
OP
Q
1
× 3 4 + 54
2
= (81 – 40.5) + 54 = 94.5
...(v)
Moment of B.M. diagram due to end moments about A is given by [see Fig. 15.11 (b)].
a′ x ′ = Area ABFH × Distance of C.G. of ABFH from A
+ Area HFE × Distance of its C.G. from A
L 1
1
+ × L × (MA – MB) × × L
= (MB × L) ×
2 2
3
6 1
6
= MB × 6 × + × 6 × (MA – MB) ×
2 2
3
= 18MB + 6MA – 6MB
= 6MA + 12MB = 6(MA + 2MB)
...(vi)
But
∴
or
ax = a ′ x ′
94.5 = 6(MA + 2MB)
94.5
= 15.75
MA + 2MB =
6
Subtracting equation (iv) from (vii), we get
MB = 15.75 – 12.0 = 3.75 kNm. Ans.
Substituting this value in equation (iv), we get
MA = 12 – 3.75 = 8.25 kNm. Ans.
...(vii)
Support reactions
Let
or
and
RA = Resultant reaction at A
RB = Resultant reaction at B.
Equating the anti-clockwise moments and clockwise moments about A,
RB × 6 + MA = 4 × 3 × 1.5 + MB
RB × 6 + 8.25 = 18 + 3.75 = 21.75
21.75 − 8.25 13.50
∴
RB =
= 2.25 kN. Ans.
=
6
6
RA = Total load – RB
= 4 × 3 – 2.25 = 9.75 kN. Ans.
649
STRENGTH OF MATERIALS
Second Method for Problem 15.6
Macaulay’s method can be used and directly the fixing moments and end reactions can
be calculated. This method is used where the areas of B.M. diagrams cannot be determined
conveniently.
4 kN/m
MA
A
MB
C
B
4 kN/m
RB
RA
Fig. 15.12
For this method it is necessary that u.d.l. should be extended upto B and then compensated for upward u.d.l. for length BC as shown in Fig. 15.12.
The B.M. at any section at a distance x from A is given by
EI
d2 y
dx 2
= RA.x – MA – w × x ×
= RA.x – MA –
4 × x2
2
= RA . x − M A − 2 x 2
x
( x − 3)
+ w × (x – 3) ×
2
2
+
4( x − 3) 2
2
+ 2( x − 3) 2
...(A)
Integrating, we get
EI
dy
2x3
2( x − 3) 3
x2
= RA.
− MA. x −
+ C1 +
dx
2
3
3
...(i)
dy
= 0.
dx
Substituting this value in the above equation upto dotted line, we get
C1 = 0.
Therefore equation (i) becomes as
when x = 0,
2x3
2( x − 3) 3
x2
dy
= RA .
− MA. x −
+
2
3
3
dx
Integrating again, we get
EI
EI y =
RA x 3 M A . x 2 2 x 4
2 ( x − 3) 4
.
−
−
+ C2 +
2
3
2
3 4
3
4
...(ii)
...(iii)
when x = 0, y = 0.
Substituting this value upto dotted line, we get
C2 = 0
Therefore equation (iii) becomes as
EI y =
when x = 6, y = 0.
650
RA . x 3 M A . x 2
1 4
1
−
−
x + ( x − 3) 4
6
2
6
6
...(iv)
FIXED AND CONTINUOUS BEAMS
Substituting this value in equation (iv) [Here complete equation is taken], we get
RA × 6 3 M A × 6 2
1
1
−
−
× 64 +
× (6 – 3)4
6
6
2
6
= 36RA – 18MA – 216 + 13.5
202.50 = 36RA – 18MA
101.25 = 18RA – 9MA
dy
At x = 6 m,
= 0.
dx
Substituting these values in the complete equation (ii), we get
0=
or
...(v)
62
2
2
− M A × 6 − × 6 3 + (6 – 3)3
2
3
3
= 18RA – MA × 6 – 144 + 18
126 = 18RA – 6MA
...(vi)
Subtracting equation (v) from equation (vi), we get
126 – 101.25 = 9MA – 6MA
or
24.75 = 3MA
24.75
∴
MA =
= 8.25 kNm. Ans.
3
Substituting this value in equation (vi), we get
126 = 18RA – 6 × 8.25
126 + 6 × 8.25
= 9.75 kN. Ans.
∴
RA =
18
Now
RB = Total load – RA
= 4 × 3 – 9.75 = 2.25 kN. Ans.
To find the value of MB, we must equate the clockwise moments and anti-clockwise
moments about B. Hence
Clockwise moments about B = Anti-clockwise moments about B.
MB + RA × 6 = MA + 4 × 3 × (4.5)
or
MB + 9.75 × 6 = 8.25 + 54
(∵ RA = 9.75 and MA = 8.25)
or
MB + 58.50 = 62.25
∴
MB = 62.25 – 58.50 = 3.75 kNm. Ans.
Problem 15.7. A fixed beam of length 20 m, carries a uniformly distributed load of
8 kN/m on the left hand half together with a 120 kN load at 15 m from the left hand end.
Find the end reactions and fixing moments and magnitude and the position of the maximum
deflection. Take E = 2 × 108 kN/m3 and I = 4 × 108 mm4.
Sol. Given :
Length,
L = 20 m
U.d.l.,
w = 8 kN/m
Point load,
W = 120 kN
Value of
E = 2 × 108 kN/m3
Value of
I = 4 × 108 mm4 = 4 × 10–4 m4
Lengths,
AC = 10 m, AD = 15 m
Fig. 15.13 shows the loading on the fixed beam.
651
0 = RA ×
STRENGTH OF MATERIALS
120 kN
MA
A
MB
8 kN/m
C
B
D
RB
RA
Fig. 15.13
Let
RA and RB = End reactions at A and B
MA and MB = Fixing moments at A and B
Let us apply Macaulay’s method for this case. Hence it is necessary that the u.d.l. should
be extended upto B and then compensated for upward u.d.l. for length BC as shown in Fig. 15.14.
The B.M. at any section at a distance x from A is given by,
EI
d2 y
dx
2
= RA.x – MA – w × x ×
FG x IJ
H 2K
− 120( x − 15) + w
× (x – 10) ×
= RA × x – MA – 8 ×
x2
2
− 120( x − 15) +
FG x − 10 IJ
H 2 K
8 × ( x − 10) 2
2
= RA.x – MA – 4 x 2 − 120( x − 15) + 4( x − 10) 2
Integrating the above equation, we get
EI
120( x − 15) 2
dy
x2
x3
= RA .
− MA. x − 4 .
+ C1 −
2
3
2
dx
+
4( x − 10) 3
3
...(i)
dy
= 0. Substituting this value in the above equation upto first dotted line, we get
dx
C1 = 0. Therefore, equation (i) becomes as
when x = 0,
120 kN
8 kN/m
MA
MB
D
C
A
10 m
B
8 kN/m
15 m
RA
20 m
RB
Fig. 15.14
EI
652
4
dy RA 2
=
. x − M A . x − x 3 − 60( x − 15) 2
2
3
dx
+
4
(x – 10)3
3
...(ii)
FIXED AND CONTINUOUS BEAMS
Integrating again, we get
RA . x 3 M A . x 2
4x4
60( x − 15) 3
4 ( x − 10) 4
−
−
+ C2 −
+
...(iii)
6
2
3×4
3
3
4
when x = 0, y = 0. Substituting this value in the above equation upto first dotted line, we get
C2 = 0. Therefore equation (iii) becomes as
EIy =
1
RA . x 3 M A . x 2 x 4
−
−
− 20( x − 15) 3 + (x – 10)4
3
6
2
3
when x = 20, y = 0. Substituting these values in complete equation (iv), we get
EIy =
...(iv)
1
RA × 20 3 M A × 20 2 20 4
−
−
– 20(20 – 15)3 + (20 – 10)4
3
6
2
3
2
4
M
20
20
125 1 10
=
RA − A −
−
+ ×
(Dividing by 202)
6
2
3
20
3 400
M
20
400 12.5 25
RA − A −
−
+
=
6
2
3
3
2
20 RA − 3 M A − 800 − 37.5 + 50
=
6
20RA – 3MA = 800 + 37.5 – 50 = 787.5
...(v)
dy
= 0. Substituting these values in complete equation (ii), we get
At x = 20,
dx
RA
4
4
0=
× 202 – MA × 20 – × 203 – 60(20 – 15)2 + (20 – 10)3
3
2
3
4 1000
4 × 400
− 3 × 25 + ×
= 10RA – MA –
(Dividing by 20)
3
20
3
200
1600
= 10RA – MA –
− 75 +
3
3
1600
200 1400
+ 75 −
=
10RA – MA =
+ 75
3
3
3
10RA – MA = 541.66
20RA – 2MA = 1083.32
(Multiplying by 2 both sides) ...(vi)
Subtracting equation (v) from equation (vi), we get
MA = 1083.32 – 787.50 = 295.82 kNm. Ans.
Substituting this values of MA in equation (vi), we get
20RA – 2 × 295.82 = 1083.32
1083.32 + 2 × 295.82
∴
RA =
20
= 83.748 kN. Ans.
Now
RB = Total load on beam – RA
= (10 × 8 + 120) – 83.748 = 116.252 kN. Ans.
Equating the clockwise moment and anti-clockwise moment about B, we get
MB + RA × 20 = MA + 120 × 5 + 8 × 10 × 15
MB + 83.748 × 20 = 295.82 + 600 + 1200
MB = 2095.82 – 83.748 × 20 = 420.86 kNm. Ans.
0=
or
or
or
or
or
or
653
STRENGTH OF MATERIALS
Maximum deflection and position of maximum deflection
Since the point load is more than the toal distributed load and acts at an equal distance
from the nearest end, hence maximum deflection will be in the portion AD. For maximum
dy
dy
should be zero. Substituting the value of
= 0 in equation (ii) [the term
deflection,
dx
dx
2
– 60(x – 15) in equation (ii) should be ignored as this term is for the portion DB], we get
RA × x 2
4
4
− M A . x − x 3 + (x – 10)3
2
3
3
83.748
4 x 3 4( x − 10) 3
=
× x 2 − 295.82 x −
+
2
3
3
4
4
3
= 41.874x2 – 295.82x – x + [x3 – 1000 – 3x × 10(x – 10)]
3
3
4000
4
4
− × 3x × 10 × x + × 3x × 10 × 10
= 41.874x2 – 295.82x –
3
3
3
4000
= 41.874x2 – 295.82x –
– 40x2 + 400x
3
= 1.874x2 + 104.18x – 1333.33
This is a quadratic equation. Hence its solution is
0=
− 104.18 ± 104.18 2 + 4 × 1.874 × 1333.33
2 × 1.874
− 104.81 + 144.387
=
(Neglecting –ve root)
2 × 1.874
= 10.727 m. Ans.
Hence maximum deflection occurs at a distance of 10.27 m from A. Maximum deflection
is obtained by substituting x = 10.727 m in equation (iv) [neglecting the term – 20(x – 15)3]
x=
∴
∴
RA . x 3 M A . x 2 x 4 1
−
−
+ (x – 10)4
6
2
3
3
3
1
83.748 × 10.727
295.82 × 10.727 2 10.727 4
=
+
(10.727 – 10)4
−
−
3
6
2
3
= 17228.9 – 17019.8 – 4413.6 + 0.09
= – 4204.5
− 4204.5
− 4204.5
ymax =
=
EI
2 × 10 8 × 4 × 10 −4
= 0.05255 m = 52.56 mm. Ans.
EIymax =
15.6. FIXED END MOMENTS OF FIXED BEAM DUE TO SINKING OF A SUPPORT
If the ends of a fixed beam are not at the same level, then the support which is at a
lower level is known as sinking support. Fig. 15.15 (a) shows a fixed beam AB of length L
whose ends A and B are fixed at different levels. The end A is at a higher level than the end B.
The beam carries no load. Hence rate of loading on the beam is zero.
Let
δ = Difference of level between the ends
MA = Fixing moment at the end A
654
FIXED AND CONTINUOUS BEAMS
MB = Fixing moment at the end B
RA = Normal reaction at A and
RB = Normal reaction at B.
In this particular case, MA is a negative (hogging) and MB is a positive moment.
Numerically MA and MB are equal.
We know that
d4 y
= Rate of loading
dx 4
=0
EI
Integrating, we get EI
d3 y
dx 3
= C1
...(i)
where C1 is a constant of integration. And EI
equal to RA. Hence EI
d3 y
dx 3
A
d3 y
dx 3
represents the shear force. At x = 0, S.F. is
(at x = 0) is RA.
MA
RB
( a)
RA
δ
MB
B
MB
(b )
MA
B.M. diagram
Fig. 15.15
Substituting this value in equation (i), we get
RA = C1
∴ Equation (i) becomes as
EI
d3 y
dx 3
Integrating again, we get
EI
d2 y
dx 2
= RA
= R A . x + C2
where C2 is another constant of integration. And EI
equal to – MA.
Hence at x = 0, EI
d2 y
dx 2
...(ii)
d2 y
dx 2
represents the B.M. at x = 0, B.M. is
= – MA.
655
STRENGTH OF MATERIALS
or
d2 y
= – MA in equation (ii), we get
dx 2
– MA = RA × 0 + C2
C2 = – MA
Substituting C2 in equation (ii), we get
Substituting x = 0 and EI
d2 y
= RA.x – MA
dx 2
Integrating the above equation again, we get
EI
EI
...(A)
x2
dy
= RA .
– MA.x + C3
dx
2
where C3 is another constant of integration. At x = 0,
dy
= 0.
dx
Hence
C3 = 0.
Therefore the above equation becomes as
dy RA . x 2
=
− MA. x
dx
2
Integrating again, we get
EI
...(iii)
RA x 3
x2
.
− MA .
+ C4
2
3
2
where C4 is a constant of integration. At x = 0, y = 0. Hence C4 = 0. Therefore the above
equation becomes as
RA 3 M A 2
.x −
.x
...(iv)
EIy =
6
2
At x = L, y = – δ. Hence the above equation becomes as
R
M
– EI.δ = A . L3 − A . L2
...(v)
6
2
dy
= 0. Substituting these values in equation (iii), we get
At x = L,
dx
RA 2
0=
.L – MA.L
2
2M A
or
RA =
...(vi)
L
Substituting the value of RA in equation (v), we get
2M A
M
– EI.δ =
. L3 − A . L2
6L
2
M A . L2 M A . L2 2 M A . L2 − 3 M A . L2
=
−
=
3
2
6
1
=–
M .L2
6 A
6 EIδ
∴
MA =
L2
Now the B.M. at any section at a distance x from A is given by equation (A) as
EIy =
EI
656
d2 y
dx 2
= RA.x – MA
FIXED AND CONTINUOUS BEAMS
FG∵ R
H
FG∵ M
H
2M A
.x – MA
L
2 6 EIδ
6 EIδ
×
.x−
=
L
L2
L2
12 EIδ
6 EIδ
=
.x−
3
L
L2
=
At x = L, EI
becomes as
d2 y
dx 2
MB =
=
represents B.M. at B i.e., EI
12 EIδ
L3
12 EIδ
2
L
×L−
−
L
Hence numerically MA = MB =
dx 2
=
A
=
2M A
L
6 EIδ
L2
IJ
K
IJ
K
= MB. Hence the above equation
6 EIδ
6 EIδ
2
d2 y
A
L2
=
6 EIδ
L2
6 EIδ
. This means that if the ends of a fixed beam are at
L2
different levels (or one end sinks down by an amount δ with respect to other end), the fixing
moment at each end is equal. At the higher end, this moment is a hogging moment and at the
lower end this moment is a sagging moment. The B.M. diagram is shown in Fig. 15.15 (b).
Problem 15.8. A fixed beam AB of length 3 m is having moment of inertia
I = 3 × 106 mm4. The support B sinks down by 3 mm. If E = 2 × 105 N/mm2, find the
fixing moments.
Sol. Given :
Length, L = 3 m = 3000 mm
Value of I = 3 × 106 mm4
Value of E = 2 × 105 N/mm2
The amount by which the support B sinks down,
δ = 3 mm.
The fixing moments at the ends is given by,
6 EIδ
MA = MB =
L2
6 × 2 × 10 5 × 3 × 10 6 × 3
=
3000 2
5
= 12 × 10 Nmm = 12 × 103 Nm = 12 kNm. Ans.
The fixing moment at A will be a hogging moment whereas at B it will be a sagging
moment.
15.7. ADVANTAGES OF FIXED BEAMS
The following are the advantages of a fixed beam over a simply supported beam :
(i) For the same loading, the maximum deflection of a fixed beam is less than that of a
simply supported beam.
(ii) For the same loading, the fixed beam is subjected to a lesser maximum bending
moment.
(iii) The slope at both ends of a fixed beam is zero.
(iv) The beam is more stable and stronger.
657
STRENGTH OF MATERIALS
15.8. CONTINUOUS BEAMS
Continuous beam is a beam which is supported on more than two supports. Fig. 15.16
shows such a beam, which is subjected to some external loading (here a uniformly distributed
load). The deflection curve for the beam is shown by dotted line. The deflection curve is having
convexity upwards over the intermediate supports, and concavity upwards over the mid of the
span. Hence there will be hogging moments (i.e., negative) over the intermediate supports and
sagging moments (i.e., positive) over the mid of the span. The end supports of a simply supported continuous beam will not be subjected to any bending moment. But the end support of
fixed continuous beam will be subjected to fixing moments. If the moments over the intermediate supports are known, then the B.M. diagram can be drawn.
w/Unit length
A
E
B
C
D
Deflection curve
Fig. 15.16
Fig. 15.16 shows a simply supported continuous beam. In this figure the end supports
at A and E will not be subjected to any bending moment. Hence in this case MA = ME = 0.
Fig. 15.16 (a) shows a continuous beam with fixed ends at A and E. Here the end supports
at A and E will be subjected to fixing moments. Hence MA and ME will not be zero.
E
A
B
C
D
Fig. 15.16 (a)
15.9. BENDING MOMENT DIAGRAM FOR CONTINUOUS BEAMS
In Art. 15.8 it is mentioned that if the moments over the intermediate supports of a
continuous beam are known, then the B.M. diagram can be drawn easily. The moments over
the intermediate supports are determined by using Clapeyron’s theorem of three moments
which states that :
If BC and CD are any two consecutive span of a continuous beam subjected to an external
loading, then the moments MB, MC and MD at the supports B, C and D are given by,
6 a1 x1 6 a2 x2
+
MB.L1 + 2MC(L1 + L2) + MD.L2 =
...(15.12)
L1
L2
where L1 = Length of span BC
L2 = Length of span CD
a1 = Area of B.M. diagram due to vertical loads on span BC
a2 = Area of B.M. diagram due to vertical loads on span CD
658
FIXED AND CONTINUOUS BEAMS
x1 = Distance of C.G. of the B.M. diagram due to vertical loads on BC from B
x2 = Distance of C.G. of the B.M. diagram due to vertical loads on CD from D.
Equation (15.12) is known as the equation of three moments or Clapeyron’s equation.
15.9.1. Derivation of Clapeyron’s Equation of three Moments. Fig. 15.17 shows the
length BCD (two consecutive spans) of a continuous beam which is shown in Fig. 15.16. Let MB,
MC and MD are the support moments at B, C and D respectively.
C
B
D
(a)
–x
2
–x
1
Mx
(b)
B
D
dx
–
x1′
C
B.M. diagram due to vertical loads
K
–
x2′
J
(c)
L
Mx′
MB
dx
B
Mc
C
B.M. diagram due to support moments
MD
D
MC
(d)
MB
MD
B
C
Resultant B.M. Diagram
D
Fig. 15.17
Let
L1 = Length of span BC
L2 = Length of span CD
a1 = Area of B.M. diagram due to vertical loads on span BC
a2 = Area of B.M. diagram due to vertical loads on span CD
a1′ = Area of B.M. diagram due to support moments MB and MC
a2′ = Area of B.M. diagram due to support moments MC and MD
x1 = Distance of C.G. of B.M. diagram due to vertical loads on BC
x2 = Distance of C.G. of B.M. diagram due to vertical loads on CD
x1 ′ = Distance of C.G. of B.M. diagram due to support moments on BC
x2 ′ = Distance of C.G. of B.M. diagram due to support moments on CD.
659
STRENGTH OF MATERIALS
Fig. 15.17 (b) and (c) show the B.M. diagrams due to vertical loads and due to supports
moments respectively.
(i) Consider the span BC
Let
Mx = B.M. due to vertical loads at a distance x from B (sagging)
Mx′ = B.M. due to support moments at a distance x from B (hogging)
∴ Net B.M. at a distance x from B is given by,
d2 y
= Mx – Mx′
dx 2
Multiplying by x to both sides, we get
EI
d2 y
= x.Mx – x.M′x
dx 2
Integrating from zero to L1, we get
EI.x.
z
L1
0
or
EI . x.
LM
N
EI x
d2 y
dx
2
. dx =
dy
−y
dx
OP
Q
L1
z
L1
0
x. M x . dx −
z
L1
0
x. M x ′ . dx
= a1 x1 − a1 ′ x1 ′
...(i)
0
(∵ Mx.dx = Area of B.M. diagram of length dx. And x.Mx.dx
= Moment of area of B.M. diagram of length dx about B.
Hence
z
L1
0
x. M x . dx = a1 x1 . And so on)
Substituting the limits in L.H.S. of equation (i), we have
EI
LMR|L F dy I
MNS|T GH dx JK
U|
V|
W
R| FG dy IJ
S| H dx K
T
− yC − 0 ×
1
at C
= a1 x1 − a1 ′ x1 ′
− yB
at B
U|OP
V|P
WQ
EI[(L1.θC – yC) – (0 – yB)] = a1 x1 − a1 ′ x1 ′ .
or
LM∵ FG dy IJ
MN H dx K
= θC
at C
OP
PQ
But deflection at B and C are zero. Hence yB = 0 and yC = 0. Hence above equation becomes
as
[EI.L1.θC = a1 x1 − a1 ′ x1 ′
But
a1′ = Area of B.M. diagram due to supports moments
= Area of trapezium BCKJ
=
and
1
2
(MB + MC) × L1
x1 ′ = Distance of C.G. of area BCKJ from B
2 L1
L1 1
+ × ( MC − M B ). L1 ×
2
2
3
=
1
M B . L1 + ( MC − M B ). L1
2
3 M B L1 + 2 L1 ( MC − M B )
L
L
M B . 1 + ( MC − M B ) × 1
6
2
3 =
=
2 M B + MC − M B
M B + ( MC − M B ). 21
2
M B . L1.
660
...(ii)
FIXED AND CONTINUOUS BEAMS
L1
[3 M B + 2 MC − 2 M B ]
M B + 2 MC
L
= 3
=
× 1
M B + MC
M B + MC
3
F
GH
I
JK
Substituting the values of a1 and x1 ′ in equation (ii), we get
EI.L1.θC = a1 x1 −
= a1 x1 −
F
GH
I
JK
M B + 2 MC
L
1
( M B + MC ). L1 ×
× 1
2
3
M B + MC
L12
(MB + 2MC)
6
6 a1 x1
– L1(MB + 2MC)
...(iii)
L1
(ii) Consider the span CD
Similarly considering the span CD and taking D as origin and x positive to the left, it can
be shown that
6 a2 x2
– L2(MD + 2MC)
6EI.(– θC) =
L2
[In the above case the slope at C (i.e., θC) will have opposite sign than that given by
equation (iii). The reason is that the direction of x from B for the span BC, and from D for span CD
are in the opposite direction].
Hence the above equation becomes as
6 a2 x2
∴
– 6EIθc =
– L2(MD + 2MC)
...(iv)
L2
Adding equations (iii) and (iv), we get
6 a1 x1
6 a2 x2
0=
– L1(MB + 2MC) +
– L2(MD + 2MC)
L1
L2
6 a1 x1
6 a2 x2
+
– L1MB – 2L1MC – L2MD – 2L2MC
=
L1
L2
6 a1 x1
6 a2 x2
or
L1.MB + L2MD + 2MC (L1 + L2) =
+
L1
L2
6 a1 x1
6 a2 x2
or
MBL1 + 2MC(L1 + L2) + MDL2 =
+
L1
L2
15.9.2. Application of Clapeyron’s equation of Three Moments to Continuous
Beam with Simply Supported ends. The fixing moments on the ends of a simply supported
beam is zero. The continuous beam with simply supported ends may carry uniformly distributed
load or point loads as given in the following problems:
Problem 15.9. A continuous beam ABC covers two consecutive span AB and BC of lengths
4 m and 6 m, carrying uniformly distributed loads of 6 kN/m and 10 kN/m respectively. If the
ends A and C are simply supported, find the support moments at A, B and C. Draw also B.M.
and S.F. diagrams.
Sol.
Given :
Length AB,
L1 = 4 m
Length BC,
L2 = 6 m
or
6EI.θC =
661
STRENGTH OF MATERIALS
U.d.l. on AB,
w1 = 6 kN/m
U.d.l. on BC,
w2 = 10 kN/m
Since the ends A and C are simply supported, the support moments at A and C will be zero.
∴
MA = MC = 0
To find the support moment at B (i.e., MB), Clapeyron’s equation of three moments should
be applied. Hence, we get
6 a1 x1
6 a2 x2
MA.L1 + 2MB(L1 + L2) + MC .L2 =
+
L1
L2
6a1 x2 6 a2 x2
or
0 × 4 + 2MB(4 + 6) + 0 × L2 =
+
4
6
3a1 x1
or
20MB =
+ a2 x2
...(i)
2
10 kN/m
6 kN/m
A
C
B
(a)
RB
RA
RC
45
31.8
(b )
12
A
B
B.M. diagram
C
35.30
(c )
4.05 +
19.95
24.7
S.F. diagram
Fig. 15.18
The B.M. diagram on a simply supported beam carrying u.d.l. is a parabola having an
2
2
altitude of wL . And area of B.M. diagram = × Span × Altitude. The distance of C.G. of this
3
8
Span
area from one end =
.
2
Now
a1 = Area of B.M. diagram due to u.d.l. on AB
2
2
wL2
× AB × Altitude = × AB × 1 1
3
3
8
2
2
6×4
= ×4×
= 32
3
8
=
662
FIXED AND CONTINUOUS BEAMS
L1 4
= =2m
2
2
a2 = Area of B.M. diagram due to u.d.l. on BC
x1 =
w L 2 2
2
10 × 6 2
× BC × 2 2 = × 6 ×
= 180
3
8
3
8
L
6
and
x2 = 2 = = 3 m.
2
2
Substituting these values in equation (i), we get
3 × 32 × 2
20MB =
+ 180 × 3
2
= 96 + 540 = 636
636
∴
MB =
= 31.8 kNm.
20
Now B.M. diagram due to supports moments is drawn as shown in Fig. 15.18 (b) in which
MA = 0, MC = 0 and MB = 31.8 kNm.
The B.M. diagram due to vertical loads (here u.d.l.) on span AB and span BC are also
shown by parabolas of altitudes
=
w1 L12
w L 2
6 × 42
10 × 6 2
=
= 12 kNm and 2 2 =
= 45 kNm
8
8
8
8
respectively in Fig. 15.18 (b).
S.F. Diagram
First calculate the reactions RA, RB and RC at A, B and C respectively. For the span AB,
taking moments about B, we get
4
(The support B has moment MB)
RA × 4 – 6 × 4 × = M B
2
= – 31.8
(∵ MB = 31.8. Negative sign is taken as the moment at B is hogging)
or
4RA – 48 = – 31.8
− 31.8 + 48
= 4.05 kN.
or
RA =
4
Similarly for the span BC, taking moments about B, we get
6
RC × 6 – 6 × 10 × = MB = – 31.8
2
or
6RC – 180 = – 31.8
180 − 31.8
= 24.7 kN.
or
RC =
6
Now
RB = Total load on ABC – (RA + RC)
= (6 × 4 + 10 × 6) – (4.05 + 24.7) = 55.25 kN.
Now complete the S.F. diagram as shown in Fig. 15.18 (c).
Problem 15.10. A continuous beam ABCD of length 15 m rests on four supports covering
3 equal spans and carries a uniformly distributed load of 1.5 kN/m length. Calculate the moments
and reactions at the supports. Draw the S.F. and B.M. diagrams also.
663
STRENGTH OF MATERIALS
Sol.
Given :
Length AB,
L1 = 5 m
Length BC,
L2 = 5 m
Length CD,
L3 = 5 m
U.d.l.,
w1 = w2 = w3 = 1.5 kN/m.
Since ends A and D are simply supported, the support moments at A and D will be zero.
∴
MA = 0 and MD = 0
From symmetry
MB = MC
To find the support moments at B and D, Clapeyron’s equation of three moments is applied
for ABC and for BCD.
1.5 kN/m
(a)
A
RA
B
C
RB
RC
4.6875
4.6875
D
RD
4.6875
(b )
A
B
C
B.M. diagram due to vertical loads
3.75 kNm
D
3.75 kNm
(c )
A
B
C
B.M. diagram due to support moments
D
A
B
C
Resultant B.M. diagram
D
( d)
4.5
3.75
3.0
B
(e )
A
D
C
3.75
B
4.5
S.F. diagram
Fig. 15.19
664
3.0
FIXED AND CONTINUOUS BEAMS
For ABC, we get
6 a1 x1 6 a2 x2
+
L1
L2
6 a1 x1 6 a2 x2
0 × 5 + 2MB(5 + 5) + MC × 5 =
+
5
5
6
20MB + 5MC = ( a1 x1 + a2 x2 )
5
Now
a1 = Area of B.M. Diagram due to u.d.l. on AB when AB
is considered as simply supported beam
2
= × AB × Altitude of parabola
3
w1 L1
2
2
1.5 × 5 2
= ×5×
= 15.625
= ×5×
3
3
8
8
L1
5
= = 2.5 m
x2 =
2
2
Due to symmetry a2 = a1 = 15.625 and x2 = x1 = 2.5
MAL1 + 2MB(L1 + L2) + MC . L2 =
or
or
or
or
...(i)
Substituting these values in equation (i), we get
6
20MB + 5MC = (15.625 × 2.5 + 15.625 × 2.5)
5
6
= × 2 × 15.625 × 2.5 = 93.750
5
(∵ MB = MC due to symmetry)
20MB + 5MB = 93.750
93.750
= 3.75 kNm
MB =
25
∴
MB = MC = 3.75 kNm. Ans.
Now the B.M. diagram due to supports moments is drawn as shown in Fig. 15.19 (c), in
which
MA = 0, MD = 0, MB = MC = 3.75 kNm.
The B.M. diagram due to vertical loads (here u.d.l.) on span AB, BC and CD (considering
wL2
1.5 × 5 2
= 4.6875
each span as simply supported) are shown by parabolas of altitudes 1 1 =
8
8
kNm each in Fig. 15.19 (b). Resultant B.M. diagram is shown in Fig. 15.19 (d).
Support Reactions
Let RA, RB, RC and RD are the support reactions at A, B, C and D respectively.
Due to symmetry,
RA = RD
RB = RC
For the span AB, taking moments about B, we get
5
MB = RA × 5 – 1.5 × 5 ×
2
or
– 3.75 = RA × 5 – 18.75
(∵ MB = – 3.75)
or
5RA = 18.75 – 3.75 = 15
15
= 3.0 kN. Ans.
∴
RA =
5
665
STRENGTH OF MATERIALS
∴ Due to symmetry, RD = RA = 3.0 kN. Ans.
Now
RA + RB + RC + RD = Total load on ABCD
or
RA + RB + RB + RA = 1.5 × 15
(∵ RC = RB, RD = RA)
or
2(RA + RB) = 22.5
22.5
or
RA + RB =
= 11.25
2
or
RB = 11.25 – RA = 11.25 – 3.00 = 8.25
(∵ RA = 3.0)
∴
RB = RC = 8.25 kN. Ans.
Now the S.F. diagram can be drawn as shown in Fig. 15.19 (e).
Problem 15.11. A continuous beam ABCD, simply supported at A, B, C and D is loaded
as shown in Fig. 15.20 (a). Find the moments over the beam and draw B.M. and S.F. diagrams.
Sol.
Given :
Length AB,
L1 = 6 m
Length BC,
L2 = 5 m
Length CD,
L3 = 4 m
Point load in BD,
W1 = 9 kN
Point load in BC,
W2 = 8 kN
U.d.l. on CD,
w = 3 kN/m.
(i) B.M. diagram due to vertical loads taking each span as simply supported
Consider beam AB as simply supported
W ×a×b
9×2×4
B.M. at point load at E = 1
=
(∵ Here a = 2 m, b = 4 m)
6
L1
= 12 kNm
Similarly B.M. at F, considering beam BC as simply supported
8×2×3
W .a.b
=
(∵ Here a = 2, b = 3 and L2 = 5)
= 2
5
L2
= 9.6 kNm
The B.M. at the centre of a simply supported beam CD, carrying u.d.l.
w × L3 2
3 × 42
=
= 6 kNm.
8
8
Now the B.M. diagram due to vertical loads taking each span as simply supported can be
drawn as shown in Fig. 15.20 (b).
(ii) B.M. diagram due to support moments
Let MA, MB, MC and MD are the supports moments at A, B, C and D respectively. But the
end supports of a simply supported beam are not subjected to any bending moment. Hence the
support moments at A and D will be zero.
∴ MA = 0 and MD = 0
To find the support moments at B and C, Clapeyron’s equation of three moments is applied
for ABC and for BCD.
(a) For spans AB and BC from equation of three moments, we have
6 a1 x1 6 a2 x2
MA.L1 + 2MB (L1 + L2) + MC . L2 =
+
L1
L2
=
666
FIXED AND CONTINUOUS BEAMS
6 a1 x1 6 a2 x2
+
6
5
6
22 M B + 5 M C = a1 x1 + a2 x2
5
0 + 2MB (6 + 5) + MC × 5 =
or
or
9 kN
8 kN
E
F
...(i)
3 kN/m
A
( a)
RB
RA
12.0
E
RD
RC
9.6
(b )
A
D
C
B
6.0
B
F
C
B.M. diagram due to loads taking each span
as simply supported
6.84
(c)
A
D
4.48
B
C
D
B.M. diagram due to support moments
(d)
A
(e)
E
B
F
Resultant B.M. diagram
4.14
D
7.12
5.27
4.86
C
2.73
4.88
S.F. diagram
Fig. 15.20
Now
a1 x1 = Moment of area of B.M. diagram due to vertical load on AB
when AB is considered as simply supported beam about point A.
1
2×2 1
1
= × 2 × 12 ×
+ × 4 × 12 × 2 + × 4
2
3
2
3
= 16 + 80 = 96
FG
H
IJ
K
a2 x2 = Moment of area of B.M. diagram due to vertical load on BC
when BC is considered as simply supported beam about point C
1
1
1
2
= × 3 × 9.6 × × 3 + × 2 × 9.6 × 3 + × 2
2
3
2
3
= 28.8 + 35.2 = 64.0
FG
H
IJ
K
667
STRENGTH OF MATERIALS
Substituting these values in equation (i), we get
6
22MB + 5MC = 96 + × 64
5
= 172.8
(b) For spans BC and CD from equation of three moments, we have
MB . L2 + 2MC(L2 + L3) + MD . L3 =
...(ii)
6 a2 x2 6 a3 x3
+
L2
L3
6 a2 x2 6 a3 x3
+
(∵ MD = 0)
5
4
6
6
...(iii)
5MB + 18MC = a2 x2 + a3 x3
5
5
a2 x2 = Moment of area of B.M. diagram due to vertical load on BC when
or
MB × 5 + 2MC(5 + 4) + 0 =
or
where
BC is considered as simply supported beam, about point B
1
2
1
1
= × 2 × 9.6 × × 2 + × 3 × 9.6 × 2 + × 3
2
3
2
3
= 12.8 + 43.2 = 56.0
FG
H
IJ
K
a3 x3 = Moment of area of B.M. diagram due to u.d.l. on CD, when CD is
and
considered as simply supported beam, about point D
=
FG 2 × Base × AltitudeIJ × Base
K 2
H3
2
4
× 4 × 6 × = 32
3
2
Substituting these values in equation (iii), we get
6
6
...(iv)
5MB + 18MC = × 56 + × 32 = 115.2
5
4
Solving equations (ii) and (iv), we get
MB = 6.84 kNm and MC = 4.48 kNm.
Now the B.M. diagram due to supports moments is drawn as shown in Fig. 15.20 (c), in
=
which
MA = 0, MB = 6.84, MC = 4.48 and MD = 0.
The B.M. diagram due to supports moments will be negative. Resultant B.M. diagram is
shown in Fig. 15.20 (d).
(iii) Support Reactions
Let RA, RB, RC and RD are the support reactions at A, B, C and D respectively,
For the span AB, taking moments about B, we get
MB = RA × 6 – 9 × 4
or
– 6.84 = 6RB – 36
(∵ MB = – 6.84)
36 − 6.84
= 4.86 kN. Ans.
or
RB =
6
For the span CD, taking moments about C, we get
MC = RD × 4 – 3 × 4 ×
668
4
2
FIXED AND CONTINUOUS BEAMS
or
or
(∵ MC = – 4.48)
– 4.48 = 4RD – 24
24 − 4.48
= 4.88 kN. Ans.
∴
RD =
4
Now taking moments about C for ABC, we get
MC = RA × (6 + 5) – 9 (5 + 4) + RB × 5 – 8 × 3
– 4.48 = 4.86 × 11 – 9 × 9 + RB × 5 – 24
(∵ MC = – 4.48, RA = 4.86)
∴
5RB = 81 + 24 – 4.86 × 11 – 4.48 = 47.06
47.06
= 9.41 kN. Ans.
5
Now
RC = Total load on ABCD – (RA + RB + RD)
= (9 + 8 + 4 × 3) – (4.86 + 9.41 + 4.88)
= 9.85 kN. Ans.
Now complete the S.F. diagram as shown in Fig. 15.20 (e).
15.9.3. Clapeyron’s Equation of Three Moments Applied to Continuous Beam
with Fixed end Supports. We have seen in Art. 15.9.2 that fixing moments on the ends of a
simply supported continuous beam are zero. But in case of a continuous beam fixed at its one or
both ends, there will be fixing moments at the ends, which are fixed. To analyse the continuous
beam which is fixed at the ends by the equation of three moments an imaginary support of zero
span is introduced. The fixing moment at this imaginary support is always equal to zero.
∴
RB =
A
C
A1
A
C
B
B
Zero
Span
(a) Continuous beam fixed at A
(b) Continuous beam with zero span
Fig. 15.21
If the beam is fixed at the left end A, than an imaginary zero span is introduced to the left
of A as shown in Fig. 15.21 (b). But if the beam is fixed at the right end, then an imaginary zero
span is introduced to the right end support. After this Clapeyron’s equation of three moments is
applied.
Problem 15.12. A continuous beam ABC of uniform section, with span AB and BC as
4 m each, is fixed at A and simply supported at B and C. The beam is carrying a uniformly
distributed load of 6 kN/m run throughout its length. Find the support moments and the reactions. Also draw the bending moment and S.F. diagrams.
Sol.
Given :
Length AB,
L1 = 4 m
Length BC,
L2 = 4 m
U.d.l.,
w = 6 kN/m.
(i) B.M. diagram due to u.d.l. taking each span as simply supported
Consider beam AB as simply supported. The B.M. at the centre of the span AB
=
w . L12
6 × 42
=
= 12 kNm
8
8
669
STRENGTH OF MATERIALS
Similarly B.M. at the centre of span BC, considering beam BC as simply supported
w . L2
6 × 42
=
= 12 kNm
8
8
The B.M. diagram due to u.d.l. taking each span as simply supported is drawn in
Fig. 15.22 (c).
(ii) B.M. diagram due to support moments
As beam is fixed at A, therefore introduce an imaginary zero span AA1 to the left of A as
shown in Fig. 15.22 (b). The support moment at A1 is zero.
Let
M0 = Support moment at A1 and is zero
MA = Support moment at A
MB = Support moment at B
MC = Support moment at C.
The extreme end C is simply supported hence MC = 0. To find MA and MB theorem of three
moments is used.
=
6 kN/m
A
( a)
C
B
6 kN/m
(b )
A
A′
C
B
Zero
span
RA
RB
RC
12.0
(c)
10.28
12.0
6.86
A
B
C
14.57
11.14
C
(d)
A
B
9.43
12.86
Fig. 15.22
Applying the theorem of three moments for the spans A1A and AB, we have
M0 × 0 + 2MA(0 + L1) + MBL1 =
670
6 a0 x0 6 a1 x1
+
L0
L1
FIXED AND CONTINUOUS BEAMS
6 a1 x1
4
3
or
8MA + 4MB =
...(i)
a x
2 1 1
where
a1 x1 = Moment of area of B.M. diagram due to u.d.l. on AB when AB is
considered as simply supported beam about point B
L
2
× Base × Altitude × 1
=
3
2
2
4
= × 4 × 12 × = 64.
3
2
Substituting this value in equation (i), we get
3
8MA + 4MB = × 64 = 96
2
or
2MA + MB = 24
...(ii)
Now applying the theorem of three moments for the spans AB and BC, we get
6 a1 x1 6 a2 x2
MA . L1 + 2MB (L1 + L2) + MC .L2 =
+
L1
L2
6 a1 x1 6 a2 x2
+
or
MA × 4 + 2MB(4 + 4) + 0 =
(∵ MC = 0)
4
4
3
3
or
4MA + 16MB = a1 x1 +
...(iii)
a x
2
2 2 2
where
a1 x1 = Moment of area of B.M. diagram due to u.d.l. on AB when
or
0 + 2MA (0 + 4) + MB × 4 = 0 +
FG
H
IJ
K
AB is considered as simply supported beam about C
2
4
= × 4 × 12 × = 64
3
2
a2 x2 = Moment of area of B.M. diagram due to u.d.l. on BC when
or
or
or
BC is considered as simply supported beam about C
2
4
= × 4 × 12 × = 64.
3
2
Substituting these values in equation (iii), we get
3
3
4MA + 16MB = × 64 + × 64 = 192
2
2
MA + 4MB = 48
Multiplying the above equation by 2, we get
2MA + 8MB = 96
Subtracting equation (ii) from equation (iv), we get
7MB = 96 – 24 = 72
72
MB =
= 10.28 kNm. Ans.
7
Substituting this value in equation (ii), we get
2MA + 10.28 = 24
24 × 10.28
MA =
= 6.86 kNm. Ans.
2
...(iv)
671
STRENGTH OF MATERIALS
Now B.M. diagram due to support moments is drawn as shown in Fig. 15.22 (c) in which
MA = 6.86, MB = 10.28, and MC = 0. The B.M. due to supports moments will be negative. Resultant
B.M. diagram is also shown in Fig. 15.22 (c).
(iii) Support Reactions
Let RA, RB and RC are the support reactions at A, B and C respectively.
For the span BC, taking moments about B, we get
4
MB = RC × 4 – 6 × 4 ×
2
or
– 10.28 = 4RC – 48
(∵ MB is negative)
48 − 10.28
= 9.43 kN. Ans.
or
RC =
4
For the span AB, taking moments about B, we get
4
MB = MA + RA × 4 – 6 × 4 ×
2
or
– 10.28 = – 6.86 + 4RA – 48
(∵ MB and MA are negative)
48 + 6.86 − 10.28
= 11.14 kN. Ans.
or
RA =
4
and
RB = Total load – (RA + RC)
= 6 × 8 – (11.14 + 9.43) = 27.43 kN. Ans.
Now complete the S.F. diagram as shown in Fig. 15.22 (d).
Problem 15.13. A continuous beam ABC of uniform section, with span AB and BC as
6 m each, is fixed at A and C and supported at B as shown in Fig. 15.23 (a). Find the support
moments and the reactions. Draw the S.F. and B.M. diagrams of the beam.
Sol.
Given :
Length AB,
L1 = 6 m
Length BC,
L2 = 6 m
U.d.l. in AB,
w = 2 kN/m
Point load in BC, W = 12 kN.
(i) B.M. diagram due to vertical loads taking each span as simply supported
Consider beam AB as simply supported. The B.M. at the centre of AB
w L12
2 × 62
=
= 9 kNm.
8
8
Consider beam BC as simply supported. The B.M. at the centre of BC
W × L2
12 × 6
=
=
= 18 kNm
4
4
The B.M. diagram due to vertical loads is drawn as shown in Fig. 15.23 (c).
(ii) B.M. diagram due to support moments
As beam is fixed at A and C, therefore introduce an imaginary zero span AA1 and CC1 to the
left of A and to the right of C respectively as shown in Fig. 15.23 (b). The support moments at A1
and C1 are zero.
Let M0 = Support moment at A1 and C1 and it is zero
MA = Fixing moment at A
MB = Support moment at B
=
672
FIXED AND CONTINUOUS BEAMS
MC = Fixing moment at C.
To find MA, MB and MC, theorem of three moments is used.
(a) Applying the theorem of three moments for the spans A1A and AB, we get
M0 × 0 + 2MA (0 + L1) + MB. L1=
6 a0 x0 6 a1 x1
+
L0
L1
12 kN
2 kN/m
A
B
D
C
B
12 kN
D
C
(a)
(b )
A1
2 kN/m
A
Zero
span
(c )
RA
9.0
5.25
A
RC
RB
B.M. diagram
5.625
18
7.5
D
B
5.625
( d)
A
S.F. diagram
C1
Zero
span
9.75
C
C
B
6.375
D
6.375
Fig. 15.23
or
or
where
or
6 a1 x1
6
12MA + 6MB = a1 x1
0 + 2MA(0 + 6) + MB × 6 = 0 +
...(i)
a1 x1 = Moment of area of B.M. diagram due to u.d.l. on AB when it is
considered as simply supported beam about B.
L1
2
= × Base × Altitude ×
3
2
2
6
= × 6 × 9 × = 108.
3
2
Substituting this value in equation (i), we get
12MA + 6MB = 108
2MA + MB = 18
(b) Applying the theorem of three moments for the span AB and BC, we get
6 a1 x1 6 a2 x2
+
MA . L1 + 2MB(L1 + L2) + MC.L2 =
L1
L2
...(ii)
673
STRENGTH OF MATERIALS
or
MA × 6 + 2MB(6 + 6) + MC × 6 =
6 a1 x1 6 a2 x2
+
6
L1
6MA + 24MB + 6MC = a1 x1 + a2 x2
or
...(iii)
2
6
× 6 × 9 × = 108
3
2
a2 x2 = Moment of area of B.M. diagram due to point load on BC when it is
where a1 x1 =
or
considered as simply supported beam about C
1
= × 6 × 18 × 3 = 162
2
Substituting these values in equation (iii), we get
6MA + 24MB + 6MC = 108 + 162 = 270
MA + 4MB + MC = 45
(c) Now applying the theorem of three moments for the span BC and CC1, we get
6 a2 x2 6 a0 x0
+
L2
L0
6 a2 x2
+0
MB × 6 + 2MC(6 + 0) + 0 =
6
6MB + 12MC = a2 x2
...(iv)
MB.L2 + 2MC(L2 + 0) + M0 × 0 =
or
or
where
...(v)
a2 x2 = Moment of area of B.M. diagram due to point load on BC when it is considered
as simply supported beam about B
1
= × 6 × 18 × 3 = 162.
2
Substituting this value in equation (v), we get
6MB + 12MC = 162
...(vi)
or
MB + 2MC = 27
Solving equations (ii), (iv) and (vi), we get
MA = 5.25 kNm,
MB = 7.5 kNm
and
MC = 9.75 kNm.
Now B.M. diagram due to support moments is drawn as shown in Fig. 15.23 (c). The B.M.
due to support moments is negative.
(iii) Support reactions
Let RA, RB and RC are the support reactions at A, B and C respectively.
For the span AB, taking moments about B, we get
MB = RA × 6 – 6 × 2 × 3 + MA
or
– 7.5 = RA × 6 – 36 – 5.25
(∵ MB and MA are negative)
36 + 5.25 − 7.5
= 5.625 kN. Ans.
or
RA =
6
For the span BC, taking moments about B, we get
MB = RC × 6 – 12 × 3 + MC
or
– 7.5 = RC × 6 – 36 – 9.75
(∵ MB and MC are negative)
36 + 9.75 − 7.5
or
RC =
= 6.375 kN. Ans.
6
674
FIXED AND CONTINUOUS BEAMS
Now
RB = Total load – (RA + RC)
= (6 × 2 + 12) – (5.625 + 6.375) = 12 kN. Ans.
The S.F. is shown in Fig. 15.23 (d).
HIGHLIGHTS
1. A beam whose both ends are fixed is known as fixed beam. And a beam which is supported on
more than two supports is known as a continuous beam.
2. In case of a fixed beam :
(i) a = a′
(ii) ax = a′ x ′
x = x′
and
Or
(i) The area of B.M. diagram due to vertical loads is equal to the area of B.M. diagram due to end
moments.
(ii) Distance of C.G. of B.M. diagram due to vertical loads is equal to the distance of C.G. of B.M.
diagram due to end moments from the same point.
3. The deflection at the centre of a fixed beam carrying a point load at the centre is given by
yc =
WL3
192 EI
where
W = Point load,
L = Length of beam.
4. The deflection at the centre of a fixed beam carrying a point load at the centre is one-fourth of the
deflection of a simply supported beam.
5. The deflection of a fixed beam with an eccentric load, under the point load is given by,
Wa3b3
.
3 EIL3
6. (a) For a fixed beam carrying uniformly distributed load over the whole length :
yc =
End moments =
W × L2
12
w . L4
384 EI
(b) The deflection at the centre of a fixed beam carrying uniformly distributed load over the whole
span is one-fifth of the deflection of a simply supported beam.
7. The end moments of a fixed beam due to sinking of a support is given by
Max. deflection =
MA = MB
6 EIδ
L2
where δ = Sinking of one support with respect to the other. At the higher end this moment is –ve
whereas at the lower end it is positive.
8. Clapeyron’s theorem of three moments for a continuous beam ABC is given by,
6 a1x1 6 a2 x2
+
L1
L2
where a1 = Area of B.M. diagram due to vertical loads on span AB
a2 = Area of B.M. diagram due to vertical loads on span BC
MA.L1 + 2MB(L1 + L2) + MC × L2 =
x1 = Distance of C.G. of B.M. diagram due to vertical loads on AB from A
x2 = Distance of C.G. of B.M. diagram due to vertical loads on BC from point C.
675
STRENGTH OF MATERIALS
9. To apply the theorem of three moments to a fixed continuous beam, an imaginary support of zero
span is introduced.
EXERCISE
(A) Theoretical Questions
1. What do you mean by a fixed beam and a continuous beam ?
2. Prove that for a fixed beam :
(i) Area of B.M. diagram due to vertical loads is equal to the area of B.M. diagram due to end
moments.
(ii) Distance of C.G. of B.M. diagram due to vertical loads is equal to the distance of C.G. of B.M.
diagram due to end moment from the same point.
3. Find an expression for the deflection for a fixed beam carrying a point load at the centre. Also
obtain the value of maximum deflection.
4. Prove that the deflection at the centre of a fixed beam is one-fourth the deflection of a simply
supported beam of the same length, when they carry a point load W at the centre.
5. Draw the S.F. and B.M. diagrams for a fixed beam, carrying an eccentric load.
6. Prove that the deflection at the centre of a fixed beam, carrying a uniformly distributed load is
given by
yc =
wL4
384 EI
Determine the position of points of contraflexures also.
7. Derive an expression for the fixing moments, when one of the supports of a fixed beam sinks down
by δ from its original position.
8. What are advantages and disadvantages of a fixed beam over a simply supported beam ?
9. What is the Clapeyron’s theorem of three moments ? Derive an expression for Clapeyron’s theorem of three moments.
10. How will you apply Clapeyron’s theorem of three moments to a
(i) continuous beam with simply supported ends
(ii) continuous beam with fixed end supports ?
(B) Numerical Problems
1. A fixed beam AB, 5 m long, carries a point load of 48 kN at its centre. The moment of inertia of the
beam is 5 × 107 mm4 and value of E for the beam material is 2 × 105 N/mm2. Determine :
(i) Fixed end moments at A and B, and
(ii) Deflection under the load.
[Ans. (i) MA = MB = 30 kNm, (ii) 3.125 mm]
2. A fixed beam of length 5 m carries a point load of 20 kN at a distance of 2 m from A. Determine the
fixed end moments and deflection under the load, if the flexural rigidity of the beam is
[Ans. MA = 14.4 kNm, MB = 9.6 kNm, yC = 1.15 mm]
1 × 104 kNm2.
3. A fixed beam of length 6 m carries point loads of 20 kN and 15 kN at distances 2 m and 4 m from
the left end A. Find the fixed end moments and the reactions at the supports. Draw B.M. and S.F.
diagrams.
[Ans. MA = 24.44 kNm, MB = 22.22 kNm, RA = 18.70 kN, RB = 16.30 kN]
4. A fixed beam of length 3 m carries two point loads of 30 kN each at a distance of 1 m from
both the ends. Determine the fixing moments and draw the B.M. diagram.
[Ans. MA = MB = 20 kNm]
676
FIXED AND CONTINUOUS BEAMS
5. A fixed beam AB of length 6 m carries a uniformly distributed load of 3 kN/m over the left half of
the span together with a point load of 4 kN at a distance of 4.5 m from the left end. Determine the
fixing end moments and the support reactions.
[Ans. MA = 7.3 kNm, MB = 6.2 kNm, RA = 7.93 kN, RB = 5.07 kN]
6. A fixed beam AB of length 6 m is having moment of intertia I = 5 × 106 mm4. The support B sinks
[Ans. MA = MB = 1000 Nm ]
down by 6 mm. If E = 2 × 105 N/mm2 find the fixing moments.
7. A continuous beam ABC of length 10 m rests on three supports A, B and C at the same level in
which span AB = 6 m and span BC = 4 m. In span AB , there is a point load of 3 kN at a distance
of 2 m from the end A, whereas in the span BC, there is a uniformly distributed load of 1 kN/m run
over the whole length. Determine the support moments and support reactions. Draw S.F. and
B.M. diagrams also.
[Ans. (i) MA = MC = 0, MB = 2.4 kNm,
(ii) RA = 1.6 kN, RB = 4 kN, RC = 1.4 kN]
8. A continuous beam consists of three successive span of 8 m, 10 m and 6 m and carries loads of
6 kN/m, 4 kN/m and 8 kN/m respectively on the spans. Determine the bending moments and
reactions at the supports.
[Ans. (i) MA = MD = 0, MC = 32.2 kNm, MB = 40.16 kNm,
(ii) RA = 18.98 kN, RB = 49.82 kN, RC = 48.57 kN, RD = 18.63 kN]
9. A continuous beam ABC consists of two consecutive spans AB and BC of length 8 m and 6 m
respectively. The beam carries a uniformly distributed load of 1 kN/m throughout its length. The
end A is fixed and the end C is simply supported. Find the support moments and the reactions.
Also draw the S.F. and B.M. diagrams.
[Ans. (i) MA = 5.75 kNm, MB = 4.5 kNm, MC = 0,
(ii) RA = 4.15 kN, RB = 7.6 kN, RC = 2.25 kN]
10. Draw the S.F. and B.M. diagram of a continuous beam ABC of length 10 m which is fixed at A and
is supported on B and C. The beam carries a uniformly distributed load of 2 kN/m length over the
entire length. The spans AB and BC are equal to 5 m each.
[Ans. (i) MA = 3.57 kNm, MB = 5.357 kNm, MC = 0,
(ii) RA = 5.357 kN, RB = 8.571 kN, RC = 6.071 kN]
677
16
CHAPTER
TORSION OF SHAFTS
AND SPRINGS
16.1. INTRODUCTION
A shaft is said to be in torsion, when equal and opposite torques are applied at the two
ends of the shaft. The torque is equal to the product of the force applied (tangentially to the
ends of a shaft) and radius of the shaft. Due to the application of the torques at the two ends,
the shaft is subjected to a twisting moment. This causes the shear stresses and shear strains
in the material of the shaft.
16.2. DERIVATION OF SHEAR STRESS PRODUCED IN A CIRCULAR SHAFT
SUBJECTED TO TORSION
When a circular shaft is subjected to torsion, shear stresses are set up in the material of
the shaft. To determine the magnitude of shear stress at any point on the shaft, consider a
shaft fixed at one end AA and free at the end BB as shown in Fig. 16.1. Let CD is any line on
the outer surface of the shaft. Now let the shaft is subjected to a torque T at the end BB as
shown in Fig. 16.2. As a result of this torque T, the shaft at the end BB will rotate clockwise
and every cross-section of the shaft will be subjected to shear stresses. The point D will shift to
D′ and hence line CD will be deflected to CD′ as shown in Fig. 16.2 (a). The line OD will be
shifted to OD′ as shown in Fig. 16.2 (b).
B
A
DD
D
C
O
B
A
L
Fig. 16.1. Shaft fixed at one end AA before torque T is applied.
Let
R = Radius of shaft
L = Length of shaft
T = Torque applied at the end BB
τ = Shear stress induced at the surface of the shaft due to torque T
C = Modulus of rigidity of the material of the shaft
φ = ∠DCD′ also equal to shear strain
679
STRENGTH OF MATERIALS
θ = ∠DOD′ and is also called angle of twist.
T
B
A
D′
D′
C
φ
D
D
A
T
θ
O
B
L
(a)
(b)
Fig. 16.2. Shaft fixed at AA and subjected to torque T at BB.
Now distortion at the outer surface due to torque T
= DD′
∴ Shear strain at outer surface
= Distortion per unit length
Distortion at the outer surface DD′
=
=
Length of shaft
L
DD′
=
= tan φ
CD
_ φ)
=φ
(if φ is very small then tan φ ~
∴ Shear strain at outer surface,
DD′
...(i)
φ=
L
Now from Fig. 16.2 (b).
Arc
DD′ = OD × θ = Rθ
(∵ OD = R = Radius of shaft)
Substituting the value of DD′ in equation (i), we get
Shear strain at outer surface
R×θ
φ=
...(ii)
L
Now the modulus of rigidity (C) of the material of the shaft is given as
Shear stress induced
Shear stress at the outer surface
C=
=
Shear strain produced
Shear strain at outer surface
Rθ
τ
∵ From equation (ii), shear strain =
=
L