TM
Strength of
Materials
R.K. Kaushik
Distributed by:
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Strength of Materials
Dr. R.K. Kaushik
Professor
Department of Mechanical Engineering
Ganga Institute of Technology & Management
Kablana, Dist. Jhajjar (Haryana), India
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©Copyright 2019 I.K. International Pvt. Ltd., New Delhi-110002.
This book may not be duplicated in any way without the express written consent of the publisher,
except in the form of brief excerpts or quotations for the purposes of review. The information
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ISBN: 978-93-89307-31-3
EISBN: 978-93-89447-89-7
Edition: 2019
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Dedicated to my wife dear Saroj Gaur Kaushik
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Preface
Though in my 48 years experience, 28 years is in MNC ABB (earlier Taylor), but for 20
years I taught various subjects to Higher Diploma (equivalent to B.Tech, Mech), B.Tech
subjects. Strength of Materials, Material Science and Machine Design (Mech) have been
my favourite subjects.
This book covers the syllabuses of most universities of B.Tech as well as Diploma in
Mechanical Engineering. Enough solved examples and problems for exercise are given in
the book.
I wish to express my special thanks to my wife Saroj Gaur Kaushik who assisted me a
lot during preparation of the book.
I request students and teachers to point out mistakes, if any to give me opportunity to
correct the same in next edition.
R.K. Kaushik
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Contents
Preface
vii
1.
Introduction
1
2.
Stress and Strains
Stress
Deformation of a Body Due to Self Weight
Extension of Tapered Rectangular Strip
Bar of Uniform Strength
Exercise
4
4
11
17
20
26
3.
Temperature Stress and Strain
Composite Tube or Bar
Thermal Stresses in a Bar of Tapering Section
Exercise
29
30
32
38
4.
Elastic Constants
Relation between Modulus of Elasticity and Modulus of Rigidity
Relation between Modulus of Elasticity and Bulk Modulus
Exercise
40
43
44
45
5.
Principal Stresses and Strains
Stresses on a Oblique Section
Material Subjected to Two Perpendicular Stresses
Material Subjected to Shear Stresses
Material Subjected to Direct and Shear Stresses
Graphical Method (Mohr’s Circle of Stresses)
Graphical (Mohr’s) Method
Exercise
47
47
50
54
55
59
69
77
6.
Shearing Force and Bending Moment
Shearing Force
Bending Moment
79
79
80
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Contents
Shearing Force and Bending Moment Diagrams
Cantilevers
Exercise
82
89
94
7.
Centre of Gravity
Centroid for Plane Figures
Centre of Gravity for Solid Bodies
Centroid of Different Sections
Exercise
97
97
99
101
106
8.
Moment of Inertia
Rectangular Section
Radius of Gyration
Theorem of Parallel Axis
Theorem of the Perpendicular Axis
Moment of Inertia of a Hollow Rectangular Section
Moment of Inertia of a Circular Section
Moment of Inertia of a Triangle
Moment of Inertia of Semi-circular Lamina about its Centroidal Axis
Product of Inertia of Rectangle
Exercise
111
111
112
113
113
114
114
115
116
122
122
9.
Bending of Beams
Relationship Between Curvature and Strain
Moment of Resistance
Modulus of Section
Beams of Uniform Strength
Composite Beams or Flitched Beams
Combined Bending and Direct Stresses
Conditions for No Tension in the Section
Exercise
125
125
126
128
136
138
142
146
151
10.
Shear Stresses in Beams
(A) Shear Stress Distribution for Beam of Rectangular Section
(B) Shear Stress Distribution of a Solid Circular Section
(C) Shear Stress Distribution in an I Section
Exercise
157
159
160
162
170
11.
Torsion
Assumptions
Torsional Moment of Resistance
Twist of the Shaft
Composite Shaft
Twisting Beyond the Limit of Proportionality
175
175
176
182
185
185
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Contents •
xi
Torsion of a Tapering Shaft
Thin Circular Tube Subjected to Torsion
Exercise
187
190
193
12.
Thin Cylindrical and Spherical Shells
(i) Hoop Stress or Circumferential Stress
(ii) Longitudinal Stress
Change in Volume
Wire-Bound Thin Cylindrical Shells
Exercise
196
196
197
198
205
212
13.
Thick Cylinders and Spheres
Solid Circular Shaft Subjected to External Pressure
Thick Spherical Shells
Exercise
214
219
229
231
14.
Deflection of Beams
Relation between Slope, Deflection and Radius of Curvature
(A) Double Integration Method
(B) Macaulay’s Method
(C) Moment Area Method
(D) Conjugate Beam Method
(E) Superposition Method
(F) Strain Energy Method
Props
Exercise
234
234
235
251
259
266
277
278
285
286
15.
Strain Energy, Impact Loading and Deflection Due to Bending
(A) Strain Energy Stored in a Body When the Load is Gradually Applied
(B) Suddenly Applied Load
(C) Strain Energy Stored in a Body, When the Load is Applied with Impact
Strain Energy in Pure Shearing
Strain Energy in Torsion
Strain Energy Due to Bending
Maxwell’s Reciprocal Theorem
Betti’s Theorem of Reciprocal Deflections
Exercise
291
291
292
294
305
306
307
311
312
318
16.
Theories of Elastic Failure
1. Maximum Principal Stress Theory (Rankine’s Theory)
2. Maximum Shear Stress Theory (Guest or Teresa’s Theory)
3. Maximum Principal Strain Theory (St. Venant’s Theory)
320
320
321
322
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Contents
4. Maximum Strain Energy Theory (Haigh’s Theory)
5. Maximum Shear Strain Energy Theory (Von Mises and Henkey’s Theory)
Exercise
323
324
339
17.
Combined Stresses (Direct, Bending, Torsion)
Combined Bending and Twisting
Exercise
342
343
355
18.
Fixed Beams
Fixed Beam with a Point Load at the Centre
Fixed Beam with Uniformly Distributed Load Over the Span
By Area Moment Method
Deflection for a Fixed Beam with Concentrated Load Anywhere
on the Span
Exercise
357
357
359
361
19.
Continuous Beams
Clapeyron’s Equation of Three Moments
Exercise
377
377
387
20.
Springs
Wahl’s Correction Factor
Close-coiled Helical Spring with Axial Couple
Springs in Series and Parallel
Open-coiled Helical Spring
Composite Action of Axial Load and Couple
Flat Spiral Spring
Leaf, Laminated or Carriage Springs or Semi-elliptic Spring
Quarter-Elliptic Leaf Spring
Exercise
388
390
391
396
402
404
405
408
410
421
21.
Columns and Struts
(a) Both Ends Hinged
(b) Column with One End Fixed and the Other Free
(c) Both Ends Fixed
(d) One End Fixed, Other End Hinged
Limitation for the Use of Euler’s Theory
Rankine’s Formula
Columns Subjected to Eccentric Loading (Secant Formula)
Perry’s Formula for Eccentrically Loaded Column
Exercise
424
426
427
428
429
437
438
446
449
452
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362
375
Contents •
xiii
22.
Bending of Curved Bars
Beams with Large Radius of Curvature (or Small Curvature)
Beam with Small Initial Radius of Curvature (or Large Curvature)
Rectangular Section
Circular Section
Triangular Section
Trapezoidal Section
Stress in a Chain Link
Exercise
456
456
459
464
464
465
466
484
488
23.
Unsymmetrical Bending
Determination of Principal Axes and Principal Moments of Inertia
Beam with Unsymmetrical Bending Moment
Momental Ellipse
Deflection of Beams due to Unsymmetrical Bending
Method for Finding Bending Stream is Unsymmetrical Bending
Shear Centre for Channel Section
Exercise
491
492
493
495
496
497
508
516
24.
Rotating Discs and Cylinders
Rotating Ring
Rotating Disc of Constant Thickness
Solid Disc
Hollow Disc
Rotating Long Cylinder
Solid Cylinder
Disc of Uniform Strength
Exercise
520
520
524
526
527
542
545
554
557
25.
Frameworks
Bow's Notation for Graphical Solution
The Method of Sections
Exercise
559
560
580
587
26.
Dams
Rectangular Dams
Trapezoidal Dams with Water Face Vertical
Rule of Middle Third
Trapezoidal Dams with Water Face Inclined
Exercise
591
591
596
599
603
608
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•
Contents
27.
Riveted Joints
Rivet
Types of Rivet Heads
Material of Rivets
Types of Riveted Joints
Some Definitions Related to Riveted Joints
Failure of a Riveted Joint
Efficiency of a Riveted Joint
Diamond Riveting
Exercise
610
610
610
612
612
614
615
617
621
625
28.
Welding Joints
Classification of Welding
Advantages and Disadvantages of Welded Joints over Riveted Joints
Unsymmetrical Welded Section
Fillet Welds Under Bending Moment
Eccentrically Loaded Welded Joint
Exercise
626
626
627
632
636
643
647
29.
Mechanical Testing of Materials
1. Tensile Test
2. Compression Test
3. Shear Tests
4. Modulus of Rupture
5. Impact Test
6. Cupping Test
7. Modulus of Rigidity of Rubber
8. Torsion of a Round Bar
9. Verification of Macaulay’s Method for Beam Deflection
10. Verification of Maxwell’s Reciprocal Theorem
11. Hardness Testing Machines
12. Experiment: Close-coiled Helical Springs
13. Experiment: Value of the Modulus of Rigidity C of
a Close-coiled Helical Spring
14. Experiment: Young’s Modulus of Elasticity of a Material Using Simply
Supported Beam
15. Fatigue Testing
Exercise
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649
649
653
657
659
661
664
665
667
669
670
671
681
683
685
686
688
Contents •
30.
Miscellaneous Solved Problems: Stresses and Strains
Complex Stress and Strain
Shear Force and Bending Moment Diagram
Bending Stresses
Shear Stresses
Torsion
Springs
Application of Castigliano’s Theorem
Moment of Inertia
Strain Energy
Theories of Failure
Rotating Discs and Cylinder
Bending of Curved Bars
Fixed Beams
Columns and Struts
Combined Loading
Index
xv
689
697
698
699
702
703
706
710
714
716
719
726
727
728
730
732
737
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C HAPTER
1
INTRODUCTION
In our life we come across many steel structures such as bridges, machine tools, etc., it is impossible
to construct these without thorough knowledge of strength of materials. As we know if the load
reaches a maximum stress beyond permissible limits, the structure is going to fail. Hence, it is very
important to study the strength of materials before their application in practical field. During one
study we will come across fundamental quantities such as length, stress and time.
In this book S.I. units are used everywhere for the convenience of students. Earlier units such
as F.P.S., C.G.S and M.K.S. have been discarded.
Following dimensions are used for various quantities:
Force: N (Newton = kg m/s2 )
Pressure: Pa (Pascal = N/m2 )
Length: metre or mm
Stress: Pa (1 N/m2 )
Remember 1 GN/m2 = 1 kN/mm2 , 1 MPa = 1 N/mm2 , G = 109 called giga.
Mass density = kg/m3
Before we plunge into details of strength of materials it is better to understand the following
abbreviations:
T (tera)
1012
G (giga)
109
M (mega) 106
k (kilo)
103
Some important formulae:
sin2 A + cos2 A = 1
sec2 A = 1 + tan2 A
cosec2 A = 1 + cot2 A
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Strength of Materials
The law of sines: In any triangle
b
c
a
=
=
sin A sin B sinC
If a, b, and c are the sides and A, B and C their opposite angles.
sin (A + B) = sin A cos B + cos A sin B
sin (A − B) = sin A cos B − cos A sin B
cos (A + B) = cos A + cos B − sin A sin B
tan (A + B) =
tan A + tan B
1 − tan A tan B
tan (A − B) =
tan A − tan B
1 + tan A tan B
1 − cos 2A
2
1 + cos 2A
cos2 A =
, sin 2A = 2 sin A cos A
2
sin2 A =
2 cos A sin B = sin (A + B) − sin (A − B), cos 2A = cos2 A − sin2 A
If
ax2 + bx + c = 0
Then,
√
−b ± b2 − 4ac
x=
2a
dxn
= nxn−1
dx
dc
= a where c is a constant.
dx
d (u.v)
dv
du
= u. + v.
dx
dx
dx
d (sin x)
d (cos x)
= cos x,
= − sin x
dx
dx
d (tan x)
d (cot x)
= sec2 x,
= −cosec2 x
dx
dx
d
d (sec x)
= sec x tan x; (cosec x) = −cosec x · cot x
dx
dx
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Introduction
xn dx =
(ax + b)n dx =
xn+1
+ c,
n+1
c is a constant
(ax + b)n+1
(n + 1) × a
logxe = 2.3(log x)
(a − bx)−n+1
−n
(a − bx) dx =
(−n + 1)(−b)
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3
C HAPTER
2
STRESS AND STRAINS
Stress
Stress can be classified broadly in three types as described below:
l
dl
1. Tensile stress: It is illustrated in Fig. 2.1
where a tensile load W is applied to a uniform rod fixed at one end.
Tensile stress, σ =
W
Figure 2.1
W
W
=
Cross-sectional area of rod
A
unit is N/mm2 or MN/m2 , σ (Greek letter sigma).
2. Compressive stress: As shown in Fig. 2.2
when load W tends to compress a rod
of cross-section area A, then compressive
W
stress = .
A
W
Figure 2.2
3. Shear stress: If two plates are joined
together with rivet as shown in Fig. 2.3.
The stress in rivet is known as shear stress,
it is denoted by τ (Greek letter tau), shear
F
stress in rivet, τ = .
A
F
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F
Figure 2.3
Stress and Strains
•
5
Behaviour of an elastic limit when subjected to a varying load.
Strain: It is defined as the ratio of change in length to original length. It is denoted by Greek
letter ε (epsilon).
Thus strain,
Change in length dl
=
ε=
Original length
l
It is illustrated in Fig. 2.1.
Hooke’s law: It states that stress is proportional to strain within elastic limit.
∴
Stress ∝ Strain
or
Stress
= a constant = E
Strain
Modulus of elasticity: E is called modulus of elasticity or Young’s modulus.
Following Table gives value of E for some important materials.
E(N/mm2 ) approx.
E(GN/m2 or kN/mm2 ) approximately
Steel
200 × 109
200
Cast iron
115 × 109
115
Wrought iron
175 × 109
175
Brass
85 × 109
85
Aluminium alloys
70 × 109
70
Copper
120 × 109
120
Timber
10 × 109
10
Spheroidal C.I.
175 × 109
175
Rubber
0.8 × 106
−
Material
Stress-Strain Curve: Figure 2.4 shows a graph for plain carbon steel during a tensile test.
D
F
C
σ
B
C'
A
(Stress)
ε (Strain)
Figure 2.4
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Strength of Materials
It may be noted that point A is the limit of proportionality and B is the elastic limit. Between point
A and B it is a curve thus not linear relationship. Therefore, actually E is constant within the limit
of proportionality, though in Hooke’s law, we had mentioned, within ‘elastic limit’, because A and
B are very close to each other. Let us name the important points on the graph:
A: Limit of proportionality
B: Elastic limit: It may be noted that on removal of load up to elastic limits, specimen comes
back to its original dimension.
C: Higher yield point: This is the point where yielding of the material begins.
C : Lower yield point: The stress associated with the lower yield point is known as yield strength.
D: Maximum stress: Here the stress is maximum because due to plastic behaviour of the material, area of cross section is very low.
E: Point of fracture: At this point ‘waisting occurs’ as shown in Fig. 2.5.
Cone
Waisting
Cap
Figure 2.5
If the material is loaded beyond the elastic limit and then load is removed, a permanent extension
remains, called permanent set.
Proof Stress: For engineering purposes it is desirable to know the stress to which a highly
ductile material such as aluminium can be loaded safely before a permanent extension takes place.
This stress is known as the proof stress or offset stress and is defined as the stress at which
a specified permanent extension has taken place in the tensile test. Proof stress is found from the
stress-strain curve as given in Fig. 2.6. The extension specified is usually 0.1, 0.2 or 0.5 per cent of
gauge length.
Stress
0.1%
Proof
stress
0
0.001
Strain
Figure 2.6
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Stress and Strains
•
7
The proof stress here is found on the basis of 0.1 per cent strain.
Procedure: Draw a line parallel to the initial slope of the curve. The stress at the point where this
line cuts the curve is the 0.1% proo f stress. The 0.2 per cent proof stress is also found in the same
manner.
Note: Though we define Hooke’s law to be taken without elastic limit, but strictly speaking it is
applicable up to the point of proportionality B in Fig. 2.4.
Brittle Materials: Fig. 2.6 shows that the stress-strain graph for brittle materials such as cast iron.
The metal is almost elastic and up to fracture but does not obey Hooke’s law. A material such as
this which has little plasticity or ductility and does not neck down before fracture is termed ‘brittle’.
The modulus of elasticity for cast iron is not constant but depends on the portion of the curve from
which it is calculated.
Fracture
Stress
Strain
Figure 2.7 Cast iron in tension
The following table is very useful for mechanical properties:
Material
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
Copper annealed
Copper hard
Aluminium soft
Aluminium hard
Black mild steel
Bright mild steel
Structural steel
Cast iron
Spheroidal
Graphite
Cast iron (annealed)
Stainless steel
Percentage
elongation
Yield stress
MN/m2
0.1% proof stress
MN/m2
Ultimate tensile stress
MN/m2
60
4
35
5
25–26
14–17
20
−
−
−
−
−
60
320
30
140
220
400
90
150
220–250
−
430–500
−
230
−
–
280–340
600
−
60
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Strength of Materials
stress
strain
Let F, L, A, dl be the force, length, area of cross section, and extension or contraction
Now E =
respectively,
then E =
F.L
A.dl
E XAMPLE 2.1: A bar of mild steel has an overall length of 2.1 m. The diameter up to 700 mm
length is 56 mm, the diameter of the remaining 1.4 m is 35 mm. Calculate the extension of the bar
due to a tensile load of 55 kN.
E = 200 GN/m2 .
S OLUTION :
∴
Remember 1 GN/m2 = 1 kN/mm2
E = 200 kN/mm2
Fl
F.l
We know E =
∴ dl =
Adl
A.E
∴
Therefore, for portion of 700 mm,
the extension dl1 =
55000 × 700 × 7 × 4
5
=
mm = 0.0178 mm
200000 × 22 × 56 × 56 64
Now dl2 for 1400 mm length of 35 mm dia,
55000 × 1400 × 7 × 4
= 0.4 mm
200000 × 22 × 35 × 35
Total extension = dl1 + dl2 = 0.0178 + 0.4 = 0.4178 mm
dl2 =
Compound bars: When two or more materials (members) are rigidly fixed together so that they
share the same load and extend or compress by same amount, the two members form compound
bar. Let us say that in Fig. 2.7 we have to find stress in each material and amount of compression.
P
Material A
of ES
l
Material B
of EB
Let the outer tube of material A has outside
dia as d1 and inside dia as d2 and inner tube of
material B has outside dia as d3 and inside dia
as d4 . Both ends are joined rigidly to make compound bar of length l.
Figure 2.8
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•
9
Now the basic method is very simple as we know that load P is shared in some proportion by
materials is A and B. That means
σa Aa + σb Ab = P
(i)
Also, since the tubes are rigidly connected so they will have same strain as length l is common.
stress
σ
=
ε
strain
σa σb
ε=
=
Ea Eb
E=
∴
since strain ε is equal.
(ii)
Since the modulus of material A and B is known, so equations (i) and (ii) can be solved to find σa
and σb . Now for compression,
σa
σ
x
or b
strain = , pick any strain
l
Ea
Eb
σa
Then x =
× l = Compression
Ea
E XAMPLE 2.2: A column is made up of a steel tube, 70 mm inside diameter filled with concrete.
If the maximum stress in the concrete is not to exceed 21 N/mm2 and the column is to carry a
compressive load of 200 kN, calculate the minimum outside diameter of the tube. For concrete
E = 20 kN/mm2 and for steel, E = 200 kN/mm2 .
S OLUTION :
Let suffixes c and s denote the concrete and steel, respectively
Ac =
π × 702
= 3850 mm2
4
As =
π (d 2 − 702 )
mm2
4
where d in mm = outside diameter of the tube. Since the steel and concrete are of equal length of
the compression of both are the same, strains are equal, then working throughout in kN and mm, we
know, strain in concrete = strain in steel
σs σc
σ
=
because
=E
Es Ec
ε
Es
200
× 21 = 210 N/mm2
σs =
× σc =
Ec
20
Total load, P = σc Ac + σs As
P = σc Ac + σs As
∴
200000 = 21 × 3850 + 210 ×
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π (d 2 − 702 )
4
10
•
Strength of Materials
Hence,
or
d 2 = 5630 mm2
d = 75 mm
E XAMPLE 2.3: A steel bar of 20 mm diameter and 400 mm long is placed concentrically inside
a gunmetal tube (Fig. 2.9). The tube has inside diameter 22 mm and thickness 4 mm. The length of
the tube exceeds the length of the steel bar by 0.12 mm. Rigid plates are placed on the compound
assembly. Find: a) the load which will just make tube and bar of same length and b) the stresses in
the steel and gunmetal when a load of 50 kN is applied. E for steel = 213 GN/m2 , E for gunmetal =
100 GN/m2 .
S OLUTION :
P
0.12 mm
Area of gunmetal tube, Ag =
π
(0.032 − 0.0222 )
4
= 0.000327 m2
Area of steel bar As =
π
(0.02)2 = 0.0003142 m2
4
Figure 2.9
a) For tube to compress 0.12 mm:
0.12
= 0.0003, Let σ1 be the stress in the tube
400
σ1
σ1
= 0.0003
= 0.0003,
Eg
100
strain =
∴
σ1 = 0.0003 × 100 = 0.03 GN/m2 = 30000 kN/m2
Hence, load = 30000 × 0.000327 = 9.81 kN
b) Load available to compress bar and tube as a compound bar is given by, let σ2 be the additional stress produced in the gunmetal tube due to this load and σs be the corresponding stress
in the steel bar, then
Load on compound bar = 50 − 9.81 = 40.19 kN
P = σ2 Ag + σs As
40.19 = σ2 × 0.000327 + σ3 × 0.0003142
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(i)
Stress and Strains
•
11
Also
σ2 σs
= ,
Eg Es
σ2 =
∴
100
σs
2100
(ii)
From Eqns. (i) and (ii)
σ2 = 40, 600 kN/m2 = 40.6 MN/m2
σs = 85300 kN/m2 = 85.3 MN/m2
Final stress in gunmetal = σ1 + σ2
= 40600 + 30000 = 70, 600 kN/m2
= 70.6 MN/m2
Deformation of a Body Due to Self Weight
B
Let us consider a bar AB which is hanging freely under its
own weight (see Fig. 2.10)
Let w = specific weight of the bar material
Now consider a small section dx at a distance x from A.
Weight of the bar for a length = w× volume
= wAx (A is cross section of the bar)
dx
l
x
A
Figure 2.10
Now elongation of the elementry length dx due to weight of the bar for length x, (wAx)
=
pl
(wAx) dx wx.dx
=
=
A.E
A.E
E
l
Total elongation =
wxdx w
=
E
E
0
∴
elongation, dl =
x.dx
0
=
l
2 l
w x
E 2
0
wl 2
2E
Because total weight of bar, W = wA.l.
wAl.l
Now elongation dl can be written as
2AE
Wl
Hence, dl =
2AE
@seismicisolation
@seismicisolation
12
•
Strength of Materials
This result also proves that the extension due to own weight is half if same weight is applied at
the end (of course neglecting extension due to self weight).
E XAMPLE 2.4: A steel bar ABC 18 m long is having cross-sectional area 4 mm2 weighs 22.5 N
(Refer Fig. 2.11). If modulus of elasticity of wire is 210 GN/m2 , find the deflections at C and B.
Deflection at C due to self weight of wire AC = dlc
A
9m
dlc =
B
22.5 × 18000
Wl
=
= 0.241 mm
2AE
2 × 4 × 210000
Deflection at B:
Now deflection at B is due to two reasons: i) due to self
weight of AB and ii) due to weight of BC.
9m
C
dlB =
W /2 × l/2 W /2 × l/2
+
2AE
A.E
dlB =
W /2 × l/2
AE
Figure 2.11
∴
=
1
+1
2
22.5 × 9000
(1.5) = 0.181 mm
2 × 4 × 210000
Sometimes a machine member is a acted upon by a number of forces, some acting at outer edges
while some are acting inside the body. In such cases in order to find out the total extension or
contraction, the principle of superposition is applied. This has been very well made clear by the
following examples:
E XAMPLE 2.5: A steel bar ABC of 400 mm length and 20 mm diameter is subjected to a point
load as shown in Fig. 2.12. Determine the total change in the length of bar. Take E = 200 GPa.
A
B
60 kN
C
20 kN
200 mm
200 mm
Figure 2.12
@seismicisolation
@seismicisolation
40 kN
Stress and Strains
•
13
S OLUTION :
For simplification split it into two parts as under:
A
C
40 kN
40 kN
400 mm
A
B
C
20 kN
20 kN
δ=
200 mm
A=
Pl
AE
π
(20)2 = 314 mm2
4
δAC =
40 × 103 × 400
= 0.255 mm
314 × 200000
δAB =
20 × 103 × 200
= 0.064 mm
314 × 200000
δ = 0.255 + 0.064 = 0.319 mm Ans.
Total
E XAMPLE 2.6: A copper rod ABCD of 800 mm2 cross-sectional area and 7.5 m long is subjected
to forces as shown in Fig. 2.13. Find the total elongation of the bar. Take E = 100 GPa
3.5 m
2.5 m
1.5 m
B
A
C
30 kN
40 kN
D
50 kN
20 kN
3.5 m
1.5 m
2.5 m
S OLUTION :
Splitting into three figures as shown below:
A
D
40 kN
40 kN
7.5 m
C
B
20 kN
20 kN
1.5 m
B
D
10 kN
10 kN
4m
Figure 2.13
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14
•
Strength of Materials
S OLUTION :
δ=
δAD =
pl
;
AE
40 × 103 × 7500
800 × 100 × 10000
= 3.75 mm
δBC =
20 × 103 × 1500
800 × 100 × 1000
= 0.375 mm
δBD =
10 × 103 × 4000
800 × 100 × 1000
= 0.50 mm
∴ Total extention = δAD + δBC + δBD = 3.75 + 0.375 + 0.50
= 4.6 mm Ans.
Extension of Tapering Rod (Circular):
l
p
d1
d2
x
dx
Figure 2.14
Let the dia. of elementary strip x from dia d1 be d .
∴
let
d = d1 −
d1 − d2
l
d1 − d2
=k
l
∴ d = d1 − kx
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x
p
Stress and Strains
Hence, cross-sectional area at distance x from larger end A =
∴
π d 2 π
= (d1 − kx)2
4
4
σ =
Extension of elementary length
4P
Total extention of the bar = δ =
πE
∴
15
P
4P
=
A π (d1 − kx)2
σ
4P
Strain = ε =
=
E
π E(d1 − kx)2
Stress at this section,
∴
•
l
0
dx = ε dx =
4P dx
π E(d1 − kx)2
l
4P (dl − kx)−1
dx
=
(d1 − kx)2 π E −1 × −k o
=
l
4P
1
π Ek d1 − kx 0
=
4P
π Ek
=
4P
1
1
−
π E(d1 − d2 ) d1 − d1 + d2 d1
=
4Pl
π E(d1 − d2 )
=
d1 − d2
4Pl
·
π E(d1 − d2 ) d1 d2
δ=
1
1
−
d1 − kl d1
but k =
d1 − d2
l
1
1
−
d2 d1
4Pl
π Ed1 d2
If both the diameters are equal to d.
Then
δ=
4Pl
π Ed 2
E XAMPLE 2.7: A round steel rod of different cross-sections is loaded as shown in Fig. 2.15. Find
the maximum stress induced in the rod and its deformations. Take E = 210 GPa.
@seismicisolation
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16
•
Strength of Materials
A
S OLUTION :
Let AB be part I, BC part II and CD part III
A
75 mm φ
0.9 m
π
(75)2 = 4415.6 mm2
4
π
A2 = (45)2 = 1589.6 mm2
4
π
A3 = (452 − 302 ) = 883.1 mm2
4
A1 =
B
B
120 kN
45 mm φ
2.2 m
60 kN
C
C
D
D
To simplify the force 120 kN acting at B − B.
0.9 m
30 mm φ
20 kN
Figure 2.15
80 kN
A
0.9 m
Tensile stress in Part I
A
80 kN
B
40 kN
B B
80 kN
C
σ1 =
B
2.2 m 20 kN
C
C
C
0.9 m
40 kN
D
D
∴
Compressive stress in Part II
PBC
40000
= 25.16 MPa
σ2 =
=
A2
1589.6
Tensile stress in Part III
PCD 20000
σ3 =
=
= 22.65 MPa
A3
883.1
20 kN
Max stress induced is in BC = 25.16 MPa.
PAB
80000
= 18.1 MPa
=
A1
4415.6
Ans
80 × 103 × 0.9 × 1000
δ1 =
P1 l1
=
A1 E
δ2 =
40 × 103 × 2.2 × 1000
P2 l2
=
= 0.264 mm (Contr.)
A2 E
1589.6 × 210000
δ3 =
P3 l3
20 × 103 × 0.9 × 1000
=
= 0.097 mm (Ext)
A3 E
883.1 × 210000
4415.6 × 210000
= 0.0776 mm (Ext)
Deformation of the rod = δ1 − δ2 + δ3 = 0.0776 − 0.264 + 0.097
= −0.0894 mm
@seismicisolation
@seismicisolation
Contraction
Ans
Stress and Strains
Extension of Tapered Rectangular Strip
x
P
a
P b
x
x
dx
t
Figure 2.16
Consider any section x − x distant x from the bigger end
Width of the section = t
∴
∴
Area of the section =
Extension of an elemental length dx =
∴
P
t(a − kx)
Pdx
t(a − kx)E
Total extension of the rod = δ =
P
tE
l
0
dx
a − kx
P 1
· − loge [(a − kx)]l0
tE k
P
= − [loge (a − kl) − loge a]
tE
a
P
=
tkE a − k
=−
But
a−b
l
P.l
a
δ=
log e
Et(a − b)
b
k=
@seismicisolation
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•
17
18
•
Strength of Materials
E XAMPLE 2.8: A straight bar of steel rectangular in section is 3 m long and of thickness of 12 mm.
The width of rod varies uniformly from 110 mm or one end to 35 mm at the other end. If the rod is
subjected to an axial load (tensile) of 25 kN, find the extension of the rod. Take E = 200000 N/mm2 .
Extension of the rod, δ =
a
Pl
log e
Et(a − b)
b
P = 25000 N, l = 3000 mm, t = 12 mm
a = 110 mm, ab = 35 mm & E = 200000 N/mm2
∴
δ=
=
25000 × 3000
110
loge
5
35
2 × 10 × 12(110 − 35)
25000 × 3000
× 1.1452
2 × 105 × 12 × 75
= 0.477 mm
Ans
E XAMPLE 2.9: A rigid bar AB is attached to two vertical rods as shown in Fig. 2.17 is horizontal
before the load is applied. Determine the vertical movement of P if it is of magnitude 60 kN.
Steel
Aluminium
For aluminium
L=3m
A = 500 mm2
E = 75 GPa
C
A
3.5 m
2.5 m
60 kN
Figure 2.17
@seismicisolation
@seismicisolation
For steel
B L=4m
A = 300 mm2
E = 210 GPa
Stress and Strains
S OLUTION :
For Al, ∑ MB = 0, 6PAl = 2.5 × 60
2.5 × 60
= 25 kN = 25000 N
6
25000 × 3000
PL
=
= 2 mm
δAl =
A.E
500 × 75000
∴
PA =
For steel ∑ MA = 0 gives
Pst = 3.5 × 60
3.5 × 60
= 35 kN
6
35000 × 4 × 1000
=
= 2.33 mm
300 × 200000
Pst =
σST
A
C
B
2 mm
C1
B1
2.33 mm
Y
A1
C2
B2
Figure 2.18
Now from similar triangles A1 C1 C2 and A1 , B1 , B2
B1 B2
Y
=
;
A1C1 A1 B1
∴
Y
2.33 − 2
=
3.5
6
Y = 0.1925 mm
Now vertical movement of
P = CC2
= CC1 +Y
= 2 + 0.1925 = 2.1995 mm
@seismicisolation
@seismicisolation
Ans
•
19
20
•
Strength of Materials
Bar of Uniform Strength
As we have seen earlier that the stress due to self weight is not constant. It increases with the
increase of distance from the lower end.
We wish to find the shape of the bar of which the self weight is considered and is having uniform
stress on all sections when subjected to an axial P. Figure 2.19 shows such a bar of uniform stress
in which the area increases from the lower end to the upper end.
Area A1
σ(A+dA)
Area A1
A
dA
L
dx
x
A
σA+wAdx
A2
Area A2
P
(a)
(b)
Figure 2.19
Let L be the length of bar, having area A1 , and area A2 be cross-sectional areas of the bar at top
and bottom, respectively.
Let w be the specific weight of the bar material (1.2. weight per unit volume of the bar).
The forces acting on the elementary stripe are:
i) Weight of the strip acting downward and is equal to w× volume of strip.
ii) Force on section AB due to uniform stress is equal to σ × A. This is acting downward. A is
area of elementary stripe.
iii) Force on section CD due to uniform (σ ) is equal to σ (A + dA). This is acting upwards.
Total force acting upwards = Total force acting downwards
σ (A + dA) = σ × A + wA dx
σA + σ dA = σ A + wA.dx
or
dA w
= dx
A
σ
@seismicisolation
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Stress and Strains
•
21
Integrating,
w x
x
σ 0
A
wx
ln
=
A2
σ
wx
A = A2 .e
σ
[ln A]AA2 =
Therefore, cross-sectional area at upper side
A1 = A2 e
wL
σ
In order to calculate the extension of the bar, consider extension of the elementary stripe. Let du be
the extension of small length dx.
du σ
=
dx E
σ
or du = dx
E
Then strain =
Integrating
σ
u=
E
L
dx
(because σ is constant in the bar)
0
∴
u=
σL
E
E XAMPLE 2.10: A vertical bar fixed at the upper end and of uniform strength carries an axial
tensile load of 800 kN. The bar is 22 m long and having weight per unit volume as 0.000075 N/mm3 .
If the area of the bar at the lower end is 450 mm2 , find the area of the bar at the upper end.
S OLUTION :
P = 800 kN = 800000 N
L = 22 m = 22000 mm
w = 0.000075 N/m3
A2 = 450 mm2
Let the area at upper end be A1
uniform stress on the bar;
σ=
P
800000
= 1777 N/mm2
=
A2
450
@seismicisolation
@seismicisolation
•
22
Strength of Materials
Using equation,
wL
σ
0.000075 × 22000
−4
1777.8
= 450e9.28×10
A1 = 450 e
A1 = A2 e
A1 = 450.4 mm2
Ans
E XAMPLE 2.11: A steel rod of 25 mm dia passes centrally through a copper tube of 30 mm inside
diameter and 40 mm outside diameter. Copper tube is 850 mm long and is closed by rigid washers of
negligible thickness, which are fastened by nut threaded on the rod as shown in Fig. 2.20. The nuts
are tightened till the load on the assembly is 20 kN. Calculate: i) the initial stresses on the copper
tube and steel rod and ii) also calculate increase in the stresses, when one nut is tightened by onequarter of a turn relative to the other. Take pitch of the thread as 1.5 mm. E for copper = 100 GPa,
E for steel = 100 GPa
S OLUTION :
Steel rod
Washer
on each side
Copper tube
Figure 2.20
Let σs = Stress in steel rod
σc = Stress in copper rod
i)
π
π
(Ds )2 = (25)2 = 156.25π mm2
4
4
π 2
π
Ac = (D − d 2 ) = (402 − 302 ) = 175π mm2
4
4
As =
Tensile rod on steel = Compressive load on copper tube
σs =
Ac
175π
× σc =
× σc
As
156.25π
@seismicisolation
@seismicisolation
Stress and Strains
∴
•
23
σs = 1.12 σc
P = 20000 N
20000 = σs As + σc Ac
20000 = 156.25π × 1.12 σc + σc × 175π
20000 = 350πσc
σc =
20000
= 18.2 MPa
350π
Ans.
σs = 1.12σc , σs = 1.12 × 18.2 = 20.38 MPa Ans.
σs1 = Increase in stress in steel rod
ii)
σc1 = Increase in the stress in the copper rod
Increase in the length of steel rod,
δ ls =
σs1 l 1.12σc1 × 850
= 4.76 × 10−3 σc1
=
Es
200, 000
Decrease in the length of copper rod
δ lc =
σc1 l σc1 × 850
=
= 8.5 × 10−3 σc1
Ec
100000
Since the nut is tightened by 1/4 of the turn then its axial advancement
1
× pitch
4
1
= × 1.5 = 0.375 mm
4
=
Since the axial advancement of the nut is equal to the decrease in the length of the tube plus
increase in the length of the rod, therefore
0.374 = 4.76 × 10−3 σc1 + 8.5 × 10−3 σc1 = 13.26 × 10−3 σc1
0.374 × 103
= 28.2 MPa (Compressive) Ans.
13.26
σs1 = 1.12 × σc1 = 1.12 × 28.2 = 31.584 MPa (Tensile)
σc1 =
Ans.
E XAMPLE 2.12: Three long parallel wires equal in length are supporting a rigid bar connected at
their bottoms as shown in Fig. 2.21. If the cross-sectional area of each wire is 100 mm2 , calculate
the stresses in each wire. Take EB = 100 GPa, ES = 200 GPa.
@seismicisolation
@seismicisolation
24
•
Strength of Materials
Steel
Brass
Brass
10 kN
Figure 2.21
S OLUTION :
σs × 100 + 100σb + 100σb
(i)
100σs + 200σb = 10000
σs + 2σb = 100 N/mm2
(ii)
σs
σb
=
3
200 × 10
100 × 103
∴
σs = 2σb
(iii)
substituting for σs in (ii)
2σb + 2σb = 100;
σb =
100
= 25 MPa Ans.
4
σs = 2 × 25 = 50 MPa Ans.
E XAMPLE 2.13: Two steel rods and one copper rod each of 20 mm diameter together support a
load of 50 kN as shown in Fig. 2.22. Find the stress in each rod. Take Es = 200 GPa, Eb = 100 GPa
50 kN
Copper
2m
Brass
Brass
Figure 2.22
@seismicisolation
@seismicisolation
1.5 m
Stress and Strains
•
25
π
(20)2 = 314 mm2
4
Total area of steel A s + 314 × 2 = 628 mm2
Ac = As =
σs As + σc Ac = 50000
628σs + 314σc = 50000
2σs + σc = 159.24
(i)
σs ls σc lc
=
;
Es
Ec
σs × 2000 σc × 1500
=
200000
100000
σs = 1.5σc
(ii)
Substituting for σs from Eqn. (ii) in Eqn. (i)
2 × 1.5σc + σc = 159.24
∴
σc = 39.81 MPa Ans.
σs = 1.5 × 39.81 = 59.7 MPa Ans.
E XAMPLE 2.14: A uniform bar ABCD has built-in ends A&D. It is subjected to two point loads
P1 and P2 equal to 80 kN and 40 kN at B and C as shown in Fig. 2.23. Find values of reactions at
A and D.
A
B
C
P1
500 mm
D
P2
1000 mm
500 mm
S OLUTION :
A
B
RA
RA
500 mm
RB
RB
1000 mm
RC
RC
500 mm
Figure 2.23
@seismicisolation
@seismicisolation
26
•
Strength of Materials
RA + RB = P1 = 80 kN
(i)
RC − RB = P2 = 40 kN
(ii)
Adding Eqns. (i) and (ii)
RA + RC = 120 kN
RA × 500 RB × 1000 RC × 500
−
−
A×E
A×E
A×E
δ = RA − 2RB − RC = 0
0=δ =
substituting,
(120 − RC ) − 2(RC − 40) − RC = 0
200
= 50 kN Ans.
4
RA + RC = 120;
solving
∴
RC =
RA = 120 − 50 = 70 kN
Ans.
Note: RA and RC are reactions at A and D respectively.
Exercise
2.1 A bar of mild steel has an overall length of 2100 mm. The diameter of 700 mm length is 56
mm. The diameter of the remaining 1400 mm is 35 mm. Calculate the extension of the bar
due to a tensile load of 55 kN. E = 200 GN/m2 .
[Ans Extension = 0.478 mm]
2.2 The block of weight W hangs from the point A, the bars AB and AC are pinned to the support
at B and C. The cross-sectional area for AB is 800 mm2 and for AC is 400 mm2 . Neglecting
the weight of the bars, determine the maximum safe value of W , if the stress in AB is limited
to 110 MPa and in AC to 120 MPa (Figure 2.24).
B
C
40°
A
60°
W
[Ans
Figure 2.24
81.7 kN]
2.3 A rectangular bar of 2 m length and 12.5 mm thickness uniformly tapers from 100 mm at
one end to 20 mm at the other. If the bar is subjected to a tensile force of 25 kN, find its
deformation. Take E as 200 GPa.
[Ans 0.4 mm]
@seismicisolation
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Stress and Strains
•
27
2.4 A brass bar having cross-sectional area of 900 mm2 is subjected to axial forces as shown in
Fig. 2.25 in which AB = 0.6 m, BC = 0.8 m and CD = 1.0 m.
A
40 kN
B
C
D
50 kN
20 kN
10 kN
Figure 2.25
Find the total elongation of the bar. Take E = 100 GPa.
[Ans. 0.11mm]
2.5 A 1200 mm long composite rod consists of a steel tube of 50 mm external diameter and
40 mm internal diameter. A copper rod of 30 mm dia is placed coaxially into the steel
tube. The assembly is held between to rigid plates and is subjected to an axial compressive
force of 200 kN. Find the stress induced in each material and the contraction produced. Take
Es = 200 GPa, Ec = 100 GPa.
[Ans σs = 188.63 MPa,
σc = 94.31 MPa
Contraction = 1.132 mm]
2.6 A 10 m long and 10 mm thick flat steel bar tapers from 60 mm at one end to 20 mm at the
other. Determine the change in length of the bar when a tensile force P = 12 kN is acting
along its axis. E = 200 GN/m2
[Ans 1.648 mm]
2.7 Two elastic rods, A and B, of equal length hang vertically 0.6 m apart and support a rigid bar
horizontally. The bar remains horizontal when a vertical load of 60 kN is applied to the bar
0.2 m from A. If the stress in A is 100 MN/m2 , find the stress in B and the cross-sectional
areas of the two rods. EA = 200 GN/m2 , EB = 130 GN/m2 .
[Ans 65 MN/m2 , 400 mm2 , 307.5 mm2 ]
2.8 If the modulus of elasticity of steel is twice that of the modulus of elasticity of brass, calculate
the outside diameter of a brass tube that should sheath a steel bar of diameter 50 mm so that
the steel and brass equally share an axial compressive load.
[Ans 86.6 mm]
2.9 A tubular steel tie-rod has an outside diameter of 50 mm, an inside diameter of 40 mm, and is
subjected to a tensile load of 88000 N. Taking E = 200 GN/m2 , calculate the extension over
a length of 2.7 m. If corrosion now reduces the effectual outside diameter of the tube to 49
mm and increases the effectual inside diameter to 41 mm, calculate the increase in extension,
over the length of 2.7 m, due to corrosion.
[Ans 1.68 mm, 0.42 mm]
@seismicisolation
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28
•
Strength of Materials
2.10 Figure 2.26 shows a rigid bar ABC hinged at A and suspended at two points B and C by two
bars BD and CE, made of aluminium and steel, respectively. The bar carries a load of 20 kN
midway between B and C. The cross-sectional area of aluminium bar BD is 3 mm2 and that
of steel bar CE is 2 mm2 . Determine the loads taken by the two bars BD and CE.
D
E
B
C
1000 mm
500 mm
1000 mm
1000 mm
20 kN
[Ans
Figure 2.26
@seismicisolation
@seismicisolation
Pa = 3.481 kN, Ps = 13.26 kN]
C HAPTER
3
TEMPERATURE STRESS AND STRAIN
When the temperature of a body is raised (specially, metallic bodies), the dimensions of the body
tend to increase, if cooled the dimensions of the bodies tend to decrease. If we prevent this increase
or decrease in dimensions due to rise or fall in temperature, the thermal stresses are bound to occur.
Let a steel rod having linear coefficient of expansion (α ), shown in Fig. 3.1 be clamped or fixed
at both ends. Now if its temperature is raised by t ◦ c
L
Figure 3.1
Then prevented expansion = α l t
Now since this expansion has been prevented so thermal stress of compressive nature will be
induced.
α lt
strain =
= αt
l
stress
E=
, so stress will be E α t.
strain
Remember that thermal strain is =
Prevented expansion or contraction
Original length
E XAMPLE 3.1: A steel spacer of length 400 mm at 20◦ C has its temperature increased to 80◦ C. If
the spacer has a diameter of 56 mm, calculate the compressive load in the spacer in newtons, if it is
assumed that:
a) the expansion is completely prevented
b) the spacer has a length of 400.1 mm at 80◦ C,
Take E = 200 GN/m2 , α = 11 × 10−6 /C◦ ,
a) For completely restricted expansion:
E = 200 GN/m2 = 200000 N/mm2 , t = 80 − 20 = 60◦ C
@seismicisolation
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30
•
Strength of Materials
Temperature stress = E α t
= 200000 × 11 × 10−6 × 60
= 132 N/mm2
Load = stress × area
π
= 132 × (56)2 = 32495.2 N
4
b) For partially restricted expansion:
Ans.
Natural expansion = l α t = 400 × 11 × 10−6 × 60
= 0.264 mm
Permitted expansion = 400.1 − 400 = 0.1 mm
Compression effect = 0.264 − 0.1 = 0.164 mm
x 0.164
= 4.1 × 10−4
Compressive strain = =
l
400
= 0.00041 mm
stress
E=
strain
∴ stress = E × strain
= 200000 × 0.00041 = 82 N/mm2
Load = stress × area
π
= 82 × (56)2 = 201864 N Ans.
4
It is obvious from the above answers that the stress is more when expansion is completly
prevented.
Composite Tube or Bar
Imagine we have a copper tube fitted with a steel rod inside. And this assembly is brazed at both the
ends. Now if its temperature is raised by t ◦ C we will analyse what happens:
xc
αslt
A
xs
A
αclt
l
Copper
Figure 3.2
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Steel
Temperature Stress and Strain
•
31
Let αs and αc be the coefficient of linear expansion for steel and copper, respectively. When
we heat, steel will expand (if allowed) less than copper as shown in Fig. 3.2. Since copper expands
more than steel and both materials are bound together so copper will pull steel upward and steel
will push the copper downward. Let line A − A be where both settle eventually
xc + xs = αc lt − αs lt
xc xs
+ = αc t − αs t = (αc − αs ) t
l
l
εc + εs = (αc − αt )
(i)
Also push on copper = pull on steel
σc Ac = σs As
σc = Thermal stress in copper
σs = Thermal stress in steel
As = Cross-sectional area of steel
Ac = Cross-sectional area of copper
(ii)
Solving equations (i) and (ii) σs and σc can be found out.
E XAMPLE 3.2: A copper flat measuring 60 mm × 40 mm is brazed to another steel flat 60 mm ×
50 mm as shown in Fig. 3.3. If this composite flat is heated through 140◦ C, determine:
i) the stress produced in each of the bar
ii) shear force which tends to rupture the brazing
iii) shear force.
Take
αc = 18.5 × 10−6 /◦ C
αs = 12 × 106 /◦ C
Ec = 110 GN/m2
Es = 220 GN/m2
length of each flat = 450 mm
40 mm
Copper
50 mm
Steel
450 mm
Figure 3.3
S OLUTION :
i) Stress produced in each of the bar.
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•
32
Strength of Materials
We shall solve this problem by the method derived from the last article.
εs + εc = (αc − αs ) t
αs
αc
+
= (13.5 − 12) + 10−6 × 140
220000 110000
αs
αc
+
= 910 × 10−6
220000 110000
Pull on steel = Push on copper
(i)
αs × As = αc Ac
αs × 60 × 50 = αc 60 × 40
or
αs = 0.8αc
(ii)
Substituting for σs in Eqn. (i)
αc
0.8αc
+
= 910 × 10−6
220000 110000
Multiplying the whole equation by 220000, we get
∴ 2.8σc = 200.2
0.8σc + 2σc = 200.2;
or σc = 71.5 MPa Ans.
σc = 0.8σc = 0.8 × 71.5 = 57.2 MPa
Ans.
ii) Shear force:
Shear force = σs As = σc Ac = 71.5 × 60 × 40
= 171600 N = 171.6 kN Ans.
iii)
171600
Shear force
=
Shear area
450 × 60
= 6.36 MN/m2
Shear stress =
Thermal Stresses in a Bar of Tapering Section
Consider a tapered round bar whose temperature is raised by Δt.
A
B
d2
d1
A
L
Figure 3.4
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Temperature Stress and Strain
•
33
As the temperature increases the bar will expand. If the tapered bar is fixed at both ends, the
stress will develop in tapered bar.
Since we know δ L = L α Δ t
(i)
and also
δL =
4 PL
π Ed1 d2
(ii)
where P is the load required to bring the deformed bar to the original.
From Eqns. (i) and (ii)
π E d1 d2 α Δt
4 PL
∴ P=
π Ed1 d2
4
P
π Ed1 d2 α Δt
=
Maximum stress, σ t = π
π
d22
4 × d22
4
4
E. α . Δt.d1
d1
(σt )max =
= E α Δt
d2
d2
L α Δt =
∴
∴
When bar is of uniform cross section d1 = d2
σt = E α Δt
E XAMPLE 3.3: A steel rod of 320 mm2 cross-sectional area and a coaxial copper tube of 800 mm2
cross-sectional area are rigidly bounded together at their ends. An axial compressive load of 40 kN
is applied to the composite bar, and the temperature is then raised by 100◦ C.
Determine the stresses in the copper and steel after heating as stated above. The modulii of elasticity for steel and copper are 200 GN/m2 and 100 GN/m2 and the coefficients of linear expansion
are 12 × 10−6 /◦ C and 16 × 10−6 /◦ C, respectively.
S OLUTION :
as = 320 mm2
40 kN
ac = 800 mm2
t = 100◦ C
Steel
Es = 200 GN/m2
Copper
Ec = 100 GN/m2
αs = 12 × 10−6 /◦ C
Figure 3.5
αc = 16 × 10−6 /◦ C
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34
•
Strength of Materials
S OLUTION :
Due to 40 kN load, let the stresses developed be σs and σc .
σs as + σc ac = 40000
320 σs + 800 σc = 40000
σs σc
=
Es
Ec
∴
(i)
σs
σ
= c
200 100
σs = 2σc
(ii)
Substituting Eqn. (i) from Eqn. (ii)
2 × 320σc + 800σc = 40000
∴ σc = 27.78 MPa(Compressive)
Hence,
σs = 27.78 × 2 = 55.56 N/mm2 (Compressive)
Now due to temperature:
εs + εc = (αc − αs )t
Let σs & σc be the stresses due to thermal effect.
σs σc
+
= (16 − 12)100 × 10−6
Es
Ec
σs
σc
+
= 400 × 10−6
200000 100000
σs
+ σc = 400 × 10−6
2
Also 320σs = 800 σc
∴
σs + 2σc = 80
(iii)
∴
σs = 2.5σc
(iv)
Substituting Eqn. (iv) in Eqn. (iii),
2.5σc + 2σc = 80
∴
σc = 17.78 N/mm2
σs = 2.5σc = 2.5 × 17.78 = 44.45 N/m2
σc =
44.45
= 17.78 N/mm2
2.5
(Compressive)
(Tensile)
(Compressive)
Net stresses:
σs = −55.56 + 44.45 = 11.11 N/mm2
(Compressive)
Ans.
σc = −17.78 − 27.78 = 45.56 N/mm2
(Compressive)
Ans.
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Temperature Stress and Strain
•
35
E XAMPLE 3.4: A composite bar shown in Fig 3.6 is rigidly attached to the end supports. The
temperature of the composite system is raised by 65◦ C. Find out the stresses in three portions of the
bar: a) if the supports are rigid and b) the supports yield by 0.5 mm.
Es = 200 GPa;
Ea = 90 GPa;
αs = 12 × 10−6 /◦ C;
40 mm dia
Ec = 100 GPa
αa = 20 × 10−6 /◦ C;
90 mm dia
Copper
Aluminium
300 mm
450 mm
αc = 16 × 10−6 /◦ C
40 mm dia
Steel
300 mm
Figure 3.6
S OLUTION :
Free elongation of copper section = αctlc
= 16 × 10−6 × 65 × 300 = 0.312 mm
Free elongation of aluminium = 20 × 10−6 × 65 × 450 = 0.585 mm
Free elongation of steel = 12 × 10−6 × 65 × 300 = 0.234 mm
Total free extension of composite bar
= 0.312 + 0.585 + 0.234 = 1.131 mm
(i)
Because the extension is prevented by the rigid supports, therefore compressive stresses will set up
in the bar.
Let F be the compressive force in the bar in N.
∴ Stress are
F
4F
=
= 7.962 × 10−4 F N/mm2
Ac π (40)2
P
4F
σa =
=
= 1.573 × 10−4 F N/mm2
Aa π (90)2
P
4F
σs =
=
= 7.962 × 10−4 F N/mm2
As π (40)2
σc =
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36
•
Strength of Materials
Strains are:
σc 7.962 × 10−4 F
=
= 7.962 × 10−9 F
Ec
100000
σa 1.573 × 10−4 F
= 1.75 × 10−9 F
εa =
=
Ea
90000
σs 7.962 × 10−4 F
εs =
=
= 3.981 × 10−9 F
Es
200000
εc =
Extensions
δ lc = εc × lc = 7.962 × 10−9 F × 300 = 2.389 × 10−6 F
δ la = 1.75 × 10−9 F × 450 = 0.7875 × 10−6 F
δ ls = 3.981 × 10−9 F × 300 = 1.194 × 10−6 F
Total extension = (2.389 × 10−6 + 0.7875 × 10−6 + 1.194 × 10−6 )F
= 4.3705 × 10−6 F
Also, δ lc + δ la + δ ls = 1.131 mm from Eqn. (i)
∴ 4.3705 × 10−6 F = 1.131
∴ F = 258780 N = 258 kN
4 × 258000
= 205.4 N/mm2 = 205.4 MPa
π × 1600
4 × 258000
= 40.6 N/mm2 = 40.6 MPa
σa =
π × 8100
4 × 258000
σs =
= 205.4 MPa
π × 1600
σc =
b) When the supports yield by 0.5 mm, then
δ la + δ lc + δ ls = 1.131 − 0.5 = 0.631 mm
43.05 × 10−6 F = 0.631
F = 144.4 kN
σs = σc = 7.962 × 10−4 × 144400 = 115 N/mm2 = 115 MPa
σa = 1.573 × 10−4 × 144400 = 22.7 N/mm2 = 22.7 MPa
E XAMPLE 3.5: A compound bar is made up by connecting a steel member and a copper member
rigidly fixed at their ends as shown in Fig. 3.7.
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Temperature Stress and Strain
1 mm dia
copper bar
37
Es = 200 GN/m2 ,
15 dia
αs = 12 × 10−6 /◦ C
20 dia
Steel
200 mm
•
Ec = 100 GN/m2 ,
200 mm
αc = 16 × 10−6 /◦ C
Figure 3.7
If the composite bar’s temperature is raised by 70◦ C, determine the stresses in copper end two
areas of steel.
S OLUTION :
Let σs be the stress in 15 mm dia of steel & σs is the stress in 20 mm dia of steel σc is the
stress in copper bar of dia 15 mm. Total tension in the steel bar is the same as the compression in
the copper bar.
π
π
π
σc × (15)2 = σs × (15)2 = σs × (20)2
4
4
4
225σc = 225σs = 400σs
∴ σc = σs = 1.78σs
If l in the actual length of the compound bar due to rise in temperature, then
σc
×l
Ec
σs
σ
× 200 + s × 200
For steel rod l − l(1 + αst) =
Es
Es
l(1 + αct) − l =
For copper rod
Adding the two eqs.,
σc
σs
σ
× l + × 200 + s × 200
Ec
Es
Es
1.786s
1.786s
× 400 +
× 200
l(αct − αst) =
100000
200000
σs
+
× 200
200000
400 × 10−6 (16 × 70 − 12 × 70) = 7.12 × 10−3 σs + 1.78 × 10−3 σs
l(1 + αct) − l(1 + αst) =
+ σs × 1 × 10−3
0.112 = 9.9σs × 10−3
∴
∴
σs = 11.31 N/mm2 = 11.31 MPa
σc = σs = 1.78 × σs
σc = σs = 1.78 × 11.31 = 20.1318 MPa
(Compressive in copper & tensile in steel)
@seismicisolation
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(i)
(ii)
38
•
Strength of Materials
Exercise
3.1 A steel bar of 100 mm diameter is rigidly clamped at both ends so that all axial extension
is prevented. A hole of 40 mm diameter is drilled out at one third of the length. If the bar is
raised in temperature by 30◦ C above that of the clamps, calculate the maximum axial stress
in the bar. E = 210 GN/m2 , α = 0.000012/◦ C
[Ans 84.7 MN/m2 ]
3.2 Two steel bars are connected together so as to form a rod of total length 625 mm. One of
mild steel is 225 mm long and 25 mm diameter. If the bar is heated to 50◦ C above 50◦ C room
temperature calculate the stress in each part of the rod. For stainless steel, E = 175 GN/m2 ,
α = 18 × 10−6 /◦ C. For mild steel E = 200 GN/m2 , α = 12 × 10−6 /◦ C
[Ans Mild steel, 44.7 MN/m2 , stainless steel 195 MN/m2 ]
3.3 A steel bar of 50 mm diameter is placed between two stops, with an end clearance of 0.05
mm. The temperature of the bar is raised by 60◦ C and the stops are found to have been forced
apart a distance of 0.05 mm. Calculate the maximum stress in the bar if its total length is 250
mm and there is a hole of 25 mm diameter drilled along the length for a distance of 100 mm.
E = 200 kN/m2 , α = 12 × 10−6 /◦ C.
[Ans 75.3 MN/m2 ]
3.4 A compound tube is formed by a stainless steel outer tube of 50 mm outside diameter and
47 mm inside diameter, together with a concentric mild steel inner tube of wall thickness 6
mm. The radial clearance between inner and outer tubes is 2 mm. The two tubes are welded
together at their ends, the compound tube being free to expand when heated. Calculate the
stress in each tube due to a rise of 50◦ C. For stainless, E = 175 kN/m2 , α = 18 × 10−6 /◦ C.
For mild steel, E = 200 kN/m2 , α = 12 × 10−6 /◦ C.
[Ans Mild steel, 13.4 MN/m2 tensile; stainless steel, 41 MN/m2 comp]
3.5 A stainless steel bar of 25 mm diameter is placed inside and concentric with a mild steel tube
of 30 mm inside diameter and 50 mm outside diameter. The tube and bar are welded together
at the ends but are otherwise free to expand. Calculate the stress in each part of the compound
bar so formed due to a temperature rise of 25◦ C. If the length of the compound bar is 250 mm
what is the extension? For stainless steel E = 170 kN/m2 , α = 18 × 10−6 /◦ C. For mild
steel E = 196 kN/m2 , α = 12 × 10−6 /◦ C.
[Ans Steel bar, −19.05 MN/m2 , tube 7.42 MN/m2 ; 0.0845 mm]
3.6 A compound bar is made up of steel plate 50 mm wide, 10 mm thick clad on both sides
by copper plates each 50 mm wide, 5 mm thick. The plates are bonded together along their
length and at room temperature the original length is 1.2 m. Find the load taken by each plate
and the increase in length when the temperature rises to 100◦ C. For steel, E = 200 kN/m2 ,
α = 12 × 10−6 /◦ C. For copper E = 110 kN/m2 , α = 18 × 10−8 /◦ C.
[Ans Steel +21.28 kN, copper −10.64 kN; 1.41 mm]
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Temperature Stress and Strain
•
39
3.7 In the arrangement shown in Fig 3.8, the steel bar is 30 mm in diameter and the bronze
bar 50 mm in diameter. The bars are of equal length and just fit between the end fixings at
room temperature. The distance between the fixings cannot change. Calculate the stresses
produced in the steel and bronze by a temperature rise of 100◦ C. For steel, E = 200 GN/m2 ,
α = 12 × 10−6 /◦ C. For bronze E = 110 kN/m2 , α = 18 × 10−6 /◦ C.
Bronze
Steel
Figure 3.8
[Ans
−363 MN2 /m2 , bronze, −130.5 MN/m2 , steel]
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C HAPTER
4
ELASTIC CONSTANTS
Whenever a tensile load is applied axially to a bar, there will be a longitudinal or linear strain. But,
naturally there will be compressive strain on lateral side (sometimes called secondary strain). The
ratio of lateral strain to longitudinal strain is known as Poisson’s ratio.
or, Poisson’s ratio =
Lateral strain
= Constant
Linear strain
This is denoted by Greek letter μ (mu).
Material
Poisson’s ratio (μ )
Steel
Cast iron
Copper
Brass
Aluminium
Concrete
Rubber
Stainless steel
Wrought iron
Bronze
0.25 to 0.33
0.23 to 0.27
0.31 to 0.34
0.32 to 0.42
0.32 to 0.38
0.08 to 0.18
0.45 to 0.50
0.305
0.278
0.350
E XAMPLE 4.1: A metal bar 60 mm × 60 mm in section is subjected to an axial compressive load
of 600 kN. If the contraction of a 200 mm gauge length was found to be 0.65 mm and the increase in
thickness as 0.05 mm, find the values of Young’s modulus and Poisson’s ratio for the bar material.
P.L
600000 × 200
=
= 0.65
A.E
60 × 60 × E
600000 × 200
= 51282 N/mm2
∴ E=
0.65 × 60 × 60
= 51.3 GN/m2 approx. Ans.
0.65
Linear strain =
= 0.00325
200
0.05
= 0.000833
Lateral strain =
60
δl =
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Elastic Constants
•
41
lateral 0.000833
=
= 0.256
linear
0.00325
∴ μ = 0.256 Ans.
Change in volume
Volumetric strain =
Original volume
δv
εv =
v
μ = Poisson’s ratio =
σy
σz
σx
σy
σx
σz
Figure 4.1
Referring to Fig. 4.1.
new length of side x = x(1 + εx )
new length of side y = y(1 + εy )
and new length of side z = z(1 + εz )
∴ new volume = xyz (1 + εx )(1 + εy )(1 + εz )
= xyz (1 + εx + εy + εz )
(Neglecting products of strains)
xyz (1 + εx + εy + εz ) − xyz
∴ εv =
xyz
i.e., volumetric strain = sum of perpendicular strains
∴
εv = εx + εy + εz
Alternately, v = xyz
∴
log v = log x + log y + log z
Differentiating w.r.t. x
∴
1 dv 1 1 dy 1 dz
·
= + ·
+ ·
v dx x y dx z dx
Multiplying Eqn. (i) by dx
∴
dv dx dy dz
=
+
+
v
x
y
z
@seismicisolation
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(i)
42
•
Strength of Materials
εv = εx + εy + εz
or
E XAMPLE 4.2: A rectangular bar 600 m long and 120 mm × 60 mm in cross section is subjected
to forces as shown in Fig. 4.2. What is the change in the volume of bar? Take E = 200 GPa, μ = 0.3.
320 kN
180 N
150 kN
150 kN
60 mm
180 kN
320 kN
120 mm
600 mm
Figure 4.2
Original volume = l × b × t = 600 × 120 × 60 = 4320000 mm3
Stress in x-x direction
σx =
Px
150000
= 20.83 N/mm2 (Tension)
=
Ax 120 × 60
σy =
180000
= 5 N/mm2 (Tension)
600 × 60
σz =
320000
= 4.44 N/mm2 (Compressive)
600 × 120
Now resultant strain in each direction
σx μσy μσz
20.83
0.25 × 5 0.25 × 4.44
20.69
−
+
=
−
+
=
E
E
E
200000 200000
200000
200000
σy μσx μσz
5
0.25 × 20.83 0.25 × 4.44
0.9025
εy = + −
+
=
−
+
=
E
E
E
200000
200000
200000
200000
4.44
0.25 × 20.83 0.25 × 5
10.9
σz μσx μσy
−
=−
−
−
=
εz = + −
E
E
E
200000
200000
200000
200000
20.69
0.9025
10.9
δv
=
+
−
Volumetric strain =
v
200000 200000 200000
10.69
= 5.345 × 10−5
=
200000
εx = +
∴
δv = v × 5.345 × 10−5 = 4320000 × 0.5345 × 10−5
= 2309.04 mm3 .
Ans.
@seismicisolation
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Elastic Constants
•
43
Bulk Modulus: When a body is subjected to three mutually perpendicular stresses of equal intensity, the ratio of direct stress to the corresponding volumetric strain is known as bulk modulus.
It is denoted by K.
Direct stress
σ
K=
=
δv
Volumetric strain
v
Direct stress occurs, for example, as hydrostatic pressure.
Principle of shear stress
If we apply shear stress in clockwise direction,
then automatically shear stress in anticlockwise
will set as complementary.
τ′
τ
τ
τ′
Figure 4.3
Relation between Modulus of Elasticity and Modulus of Rigidity
Modulus of rigidity =
Shear stress
Shear strain
τ'
Consider a cube
τ
AD2 + AD2 = BD2
∴ 2AD2 = BD2
√
BD = AD 2
τ
D
D1
C1 C
φ
τ'
Before
distortion
P
D2
φ
A
After
distortion
B
Figure 4.4
Strain of BD =
BD1 − BD D1 D2 DD1 cos 45 φ
√
=
=
=
BD
BD
2
AD 2
Linear strain of the diagonal BD,
φ
τ
=
2
2C
τ = shear stress
C = modulus of rigidity
=
(i)
We know that the effect of this stress will cause tensile stress on the diagonal BD and compressive
stress on the diagonal AC. Therefore, tensile strain on the diagonal BD due to tensile stress on the
diagonal BD is given by
τ
BD =
(ii)
E
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44
•
Strength of Materials
And the tensile strain on the diagonal BD due to compressive stress on the
diagonal AC = μ ×
Z
E
(iii)
The combined effect of the above two stresses on the
diagonal BD =
τ
τ
τ
+ μ = (1 + μ )
E
E
E
(iv)
Equating Eqns. (i) & (iv),
τ
τ
= (1 + μ ) or
2C E
E
2C
=
1
1+μ
or
1
1
= (1 + μ )
2C E
C=
E
2(1 + μ )
C = Shear modulus or modulus of rigidity
Relation between Modulus of Elasticity and Bulk Modulus
Let us take a cube of each side l and subject it to direct stress (hydrostatic pressure) on the faces of
the cube.
Initial volume of cube v = l 3
By differentiation,
∴
∴
dv = 3l 2 dL
dv
Volumetric strain, εv =
v
3l 2 dl
l2
dl
= 3εl
εv = 3
l
εv =
∴
Now linear strain
dl
of any side of cube is,
l
dl
σ
σ
σ
= −μ −μ
l
E
E
E
σ
εL = (1 − 2μ )
E
3σ
εv =
(1 − 2μ )
E
σ
K=
εv
εl =
∴
∴
@seismicisolation
@seismicisolation
(i)
(ii)
Elastic Constants
∴
εv =
•
45
σ
K
Substituting in Eqn. (i)
∴
σ
3σ
=
(1 − 2μ )
K
E
∴
E = 3K(1 − 2μ )
Relation between E, C & K
Since C =
E
; E = 2C(1 + μ )
2(1 + μ )
∴
E
−1 = μ
2C
(i)
Also E = 3K(1 − 2μ )
E
− 1 = −2μ
3K
or
μ=
1
E
−
2 6K
(ii)
Equating Eqns. (i) & (ii)
1
E
E
−1 = −
2C
2 6K
1
1
3
1
+
= +1 =
E
2C 6K
2
2
1
1
E
+
=3
C 3K
E=
9CK
C + 3K
Exercise
4.1 A bar of 50 mm diameter and 300 mm length is made of a material having E = 200 GPa and
μ = 0.3, calculate its modulus of rigidity. Also determine the change in the volume of the bar
when subjected to a hydrostatic stress of 100 MPa.
[Ans C = 76.92 GPa, 353 mm3 ]
4.2 A square piece of steel 150 mm long by 25 mm square is subjected to a compressive load
of 100 kN. Find the change in length of the piece if all lateral strain is presented by the
application of uniform lateral external pressure of suitable intensity. E = 200 GPa, μ = 0.25.
[Ans 0.1 mm]
4.3 The gauge length marked on a steel rod of diameter of 10 mm is 60 mm. When this rod is
subjected to tension test, the gauge length increases to 80 mm. The rod yields at 40 kN and
maximum load applied is 80 kN after which the rod breaks at 45 kN. Determine: a) percentage
elongation, b) yield strength, c) ultimate strength and d) breaking strength.
[Ans 33.3%, 509.3 MPa, 1018.6 MPa, 572.9 MPa]
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46
•
Strength of Materials
4.4 A bar of cross-section 10 × 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 0.43, 229 GPa]
4.5 A 50 mm diameter steel bar is subjected to a tensile load of 100 kN. The extension over
its 300 mm length was found to be 0.08 mm and change of its diameter was 0.0035 mm.
Determine the modulus of rigidity of bar.
[Ans 75.6 GPa]
4.6 A rectangular block is subjected to stresses 25 MPa (tensile), 20 MPa (comp) and 30 MPa
(tensile) in the directions x, y and z, respectively. Determine the strains in three directions
and value of the bulk and the rigidity modulii. E = 200 GPa, μ = 0.28.
[Ans 0.083, 0.0001, 0.00015 K = 151.5 GPa,C = 78.125 GPa]
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5
C HAPTER
PRINCIPAL STRESSES AND STRAINS
While designing a component of a machine, the essential feature in mechanical engineering and
design is based on the knowledge of principal stresses at the critical section of a component. While
designing a crankshaft in an engine which converts the translatory motion of the piston into rotary
m, there are complex stresses induced, the study of the same is very important from design purpose
of view.
Principal Planes: In any strained material, there are three planes mutually perpendicular to each
other which carry direct stresses only and no shear stress. Out of these three one plane carries the
maximum stress, the other minimum stress and the third plane carries intermediate stress (which is
not much required in design calculations).
Principal Stress: The magnitude of direct stress, across a principal plane, is known as principal
stress. The determination of principal planes and principal stress is an important consideration in
the design of a machine component or structures.
For determination of stresses of an oblique section of a strained body two methods are normally
used i.e., (i) analytical methods (ii) graphical method employing Mohr’s circle.
While deriving equations we shall take tensile stresses and strains as positive and compressive
stresses and strain an negative.
The shear stress which tends to rotate the element in clockwise direction is taken as positive and
the one tensing to rotate in anticlockwise direction as negative.
There are two methods used
(i) Analytical Method (ii) Graphical Method
Stresses on a Oblique Section
Figure 5.1(a) shows a piece of material subjected to a tensile force P. If the cross-sectional area is
P
a, the tensile stress on the cross-section, σ = .
a
C
C
θ
P
A
θ
P
A
B
(a)
θ
T
B
(b)
Figure 5.1
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N
P
48
•
Strength of Materials
The part ACB is in equilibrium under the forces acting upon it, so that the resultant force on
CB is equal and opposite to the applied force P. This can be resolved into normal and tangential
components, N and T , producing direct and shear stresses σθ and τθ respectively. The area of the
oblique section CB is a sec θ , so that
N
P cos θ
=
= σ cos2 θ
a sec θ
a sec θ
T
P sin θ
σ
τθ =
=
= σ sin θ cos θ = sin 2θ
a sec θ
a sec θ
2
σθ =
(5.1)
(5.2)
σθ
C
α
σr
zθ
B
Figure 5.2
The resultant stress σθ is given by σθ = σθ2 + τθ2 since each of the stresses acts on the same
τθ
area, and α = tan−1
. Where in the simple case considered above α = θ . σθ has a maximum
σθ
value σ when θ = 0 and τθ has a maximum value σ /2 when θ = 45◦ . Thus, any material whose
ultimate shear stress is less than half the ultimate stress in tension or compression, will fail due to
shear when subjected to a tensile or compressive load.
It is usual to work in terms of the applied and induced stresses rather than forces and to assume
the material to be of unit thickness.
E XAMPLE 5.1: A steel bar is subjected to a tensile stress of 45 MPa. What will be the values of
normal and shear stresses across a section, which makes an angle of 60◦ with the tensile stress.
S OLUTION :
C
30º
P
P
60º
A
B
Figure 5.3
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Principal Stresses and Strains
•
49
σθ = σ cos2 θ
= 45 cos2 30 = 40 × 0.7499
= 33.75 MPa Ans.
σ
45
sin 2 × 30 = 22.5 × 0.866
τθ = sin 2θ =
2
2
= 19.485 MPa Ans.
E XAMPLE 5.2: Two wooden rods of cross section 150 mm × 110 mm are joined together along a
line AB as shown in Fig. 5.4. Find the maximum force which can be applied if the shear stress along
AB is 2 MPa.
C
P
P
70°
150 mm
B
110 mm
Figure 5.4
S OLUTION :
θ = 90◦ − 70◦ = 30◦
Cross-sectional area = 150 × 110 = 16500 mm2
Let σ be the safe stress in the rods.
σ
sin 2θ
2
σ
2.5 = sin 2 × 30◦ = σ × 0.433
2
2.5
= 5.77 MPa Ans.
Safe stress σ =
0.433
τθ =
∴
E XAMPLE 5.3: A tension member is formed by joining two wooden rectangular rods 250 mm ×
120 mm in cross section as shown in Fig. 5.5. Determine the safe value of force F which can be
applied if allowable normal and shear stresses in the joint are 0.7 MPa and 1.4 MPa respectively.
C
P
25°
65°
D
P
250 mm
120 mm
Figure 5.5
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50
•
Strength of Materials
S OLUTION :
90◦ − 65◦ = 25◦
Let σ be the safe stress in N/mm2 .
Area of cross section = 250 × 120 = 30, 000 mm2
Normal stress, σθ = σ cos2 θ
σ=
Shear stress τθ =
∴
0.7 = σ cos2 25 = σ × 0.821
0.7
= 0.853 MPa
0.821
σ
σ
sin 2θ , 1.4 = sin 2 × 25 = 0.383σ
2
2
τθ =
1.4
= 3.655 MPa
0.383
We can see that 0.853 MPa is least of two. Hence, safe stress = 0.853 MPa
∴
Safe load = 0.853 × 30000 = 25590 MPa
= 25.59 kN Ans.
Material Subjected to Two Perpendicular Stresses
Figure 5.6 shows an element of unit thickness which is subjected to perpendicular tensile stresses σx
and σy . The wedge ABC is in equilibrium under the forces acting upon it, so that, resolving forces
are normal to AC,
σθ × AC = σx × AB cos θ + σy × BC sin θ
∴
σθ = σx cos2 θ + σy sin2 θ
σx + σy σx − σy
=
+
cos 2θ
2
2
(5.3)
The maximum value is σx or σy , whichever is greater and the minimum value is σx or σy whichever
is smaller.
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51
σy
A
6θ
θ
σx
σx
zθ
B
C
σy
Figure 5.6
Resolving forces parallel to AC,
τθ × AC = σx × AB sin θ − σy × BC cos θ
∴ τθ = (σx − σy ) sin θ cos θ
σx − σy
=
sin 2θ
2
(5.4)
The maximum value is σx − σy when θ = 45◦ .
If σx = σy , τθ = 0 for all values of θ .
E XAMPLE 5.4: In a strained structure, at a point an element is subjected to two mutually perpendicular tensile stresses of 250 MPa and 130 MPa. Determine the intensities of normal, shear and
resultant stresses on a plane inclined at 25◦ with the axis of minor tensile stress.
S OLUTION :
Let tensile stress along x − x axis = 250 MPa and along y − y axis = 130 MPa
σx + σy σx − σy
+
cos 2θ
2
2
250 + 130 250 − 130
=
+
cos 2 × 25
2
2
250
= 190 + 60 × 0.643 = 190 + 38.57
= 228.57 MPa Ans.
σx − σy
250 − 130
sin 2θ =
sin 2θ × 25
Shear stress τθ =
2
2
= 60 sin 50 = 45.96 MPa Ans.
Resultant stress σr = σθ2 + τθ2 = 228.572 + 45.962
√
= 52244.2 + 2112.3
= 233.14 MPa Ans.
130
Normal stress, σθ =
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25°
250 MPa
130 Mpa
Figure 5.7
52
•
Strength of Materials
E XAMPLE 5.5: At a certain point of a machine component, the mutually perpendicular stresses
are 200 MPa and 75 MPa, both tensile. Determine the normal, shear, resultant stresses on a plane
inclined at an angle of 60◦ with the axis of major tensile stress. Also find the magnitude of the
maximum shear stress in the component.
S OLUTION :
75
30°
200
200 MPa
60°
75 MPa
Figure 5.8
σx + σy σx − σy
+
cos 2θ
2
2
200 + 75 200 − 75
+
cos 2 × 30
=
2
2
1
= 137.5 + 62.5 ×
2
Normal stress, σθ =
= 168.75 MPa Ans.
Shear stress, τθ =
σx − σy
sin 2 × 30
2
τθ = 62.5 × 0.866 = 54.13 MPa Ans.
Resultant stress = σθ2 + τθ2
= 168.752 + 54.132 = 177.22 MPa Ans.
τmax , Shear stress =
σx − σy
sin 2θ
2
τmax will be when 2θ = 90◦
∴
τmax = ±
or
θ = 45◦
σx − σy
200 − 75
=±
= ±62.5 MPa Ans.
2
2
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53
E XAMPLE 5.6: In a structural element, the stresses at a point are 150 MPa (tensile) and 80 MPa
(compressive). Find the magnitude of the normal and shear stresses on a plane inclined at 55◦ with
tensile stress. Also find the direction of the resultant stress and the magnitude of the maximum shear
stress.
S OLUTION :
80
C
B
35°
150
150 MPa
55°
D
A
80 MPa
Figure 5.9
Normal stress, σθ =
=
σx + σy σx − σy
+
cos 2θ
2
2
150 + (−80) 150 − (−80)
+
cos 2 × 35
2
2
= 35 + 115 × 0.342 = 74.33 MPa
Shear stress, τθ =
=
Ans.
σx − σy
sin 2θ
2
150 − (−80)
sin 2 × 35 = 115 × 0.939
2
= 107.98 MPa Ans.
Resultant stress, σr = σθ2 + τθ2
=
74.332 + 107.982
= 131.1 MPa
τmax = ±
Ans.
σx − σy 150 − (−80)
=
= 115 MPa Ans.
2
2
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•
Strength of Materials
Material Subjected to Shear Stresses
When a material is subjected to put shear stress on one plane, an equal shear stress is induced on
the perpendicular plane to present rotation of the element, such a state is shown in Fig. 5.10.
τ
A
θ
σθ
B
τ
τθ
τ
τ
C
Figure 5.10
Resolving forces normal to AC,
σθ × AC = τ × AB sin θ + τ × BC cos θ
∴ σθ = τ cos θ sin θ + τ sin θ cos θ
σθ = τ sin 2θ
(5.5)
The maximum value is τ when θ = 45◦
Resolving forces parallel to AC,
τθ × AC = −τ × AB cos θ + τ BC sin θ
∴
τθ = −τ cos2 θ + τ sin2 θ
= −τ cos 2θ
(5.6)
The maximum value is τ when θ = 0 or 90◦ .
Note: Shear stress induces numerically equal tensile and compressive stresses on planes at 45◦ to
the planes of the shear stress as shown in Fig. 5.11.
τ
τ
τ
τ
τ
τ
τ
τ
Figure 5.11
E XAMPLE 5.7: A steel block is fixed in a vice so that about half of the height is in the vice.
Determine the normal stress and shear stress at a oblique section inclined at 30◦ to the vertical as
shown in Fig. 5.12. When 25 MPa shear stress is applied.
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Principal Stresses and Strains
•
55
25 MPa
A
30°
Vice
B
25 MPa
Figure 5.12
Normal stress, σθ = τ sin 2θ = 25 sin 2 × 30◦
= 21.65 MPa Ans.
Shear stress, τθ = −τ cos 2θ
= −25 cos 2 × 30◦ = −12.5 MPa
Ans.
Material Subjected to Direct and Shear Stresses
This is the most general case of stresses in two dimensions, since any system of stresses in two
dimensions can be reduced to this form. Refer Fig. 5.13.
σy
τ
A
τ
θ
σx
σx
τ
B
C
τ
σy
Figure 5.13
From the combinations of previous formulae,
σx + σy σx − σy
+
cos 2θ + τ sin 2θ
2
2
σx − σy
τθ =
sin 2θ − τ cos 2θ
2
d σθ
=0
For σθ to be maximum or minimum
dθ
σθ =
∴
− (σx − σy ) sin 2θ + 2τ cos 2θ = 0
2τ
tan 2θ =
σx − σy
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(5.7)
(5.8)
(5.9)
56
•
Strength of Materials
σy
τ
y)2
+
4τ 2
σ2
τ
(σ
x
–
σ
2τ
σx
σ1
τ
σx
θ
σ1
σ2
τ
2θ
σy
σx – σ y
Figure 5.15
Figure 5.14
2τ
Therefore, from Fig. 5.14, sin 2θ = (σx − σy )2 + 4τ 2
σx − σy
and cos 2θ = (σx − σy )2 + 4τ 2
(5.10)
(5.11)
For maximum and minimum values of σθ , substituting the value of sin 2θ and cos 2θ in Eqn. (5.7),
σx + σy σx − σy
σx − σy
2τ
+τ +
·
2
2
2
2
(σx − σy ) + 4τ
(σx − σy )2 + 4τ 2
1
=
(σx + σy ) ± (σx − σy )2 + 4τ 2
2
σθ =
(5.12)
Equation (5.12) gives principal stresses σ1 and σ2 .
It is a very important and useful equation. The planes upon which these stresses are mutually
perpendicular, one of these stresses acts on one plane and the other stress acts on the perpendicular
plane. The association between the stresses and planes is usually obvious by considering the equilibrium of the wedge ABC but when in doubt, a unique rotation for θ can be obtained from either of
Equations (5.13) or (5.14), derived as follows: (Refer Fig. 5.10).
Since there is no shear stress on a principal plane, then if AC is such a plane, the only stress
acting on it is the principal stress, σ .
Hence, resolving forces horizontally,
σ × AC cos θ = σx × AB + τ × BC
∴ σ = σx + τ tan θ
σ − σx = τ tan θ
(5.13)
σ × AC sin θ = σy × BC + τ × AB
∴ σ = σy + τ cot θ
(5.14)
Resolving vertically,
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57
Now substituting the values of sin 2θ and cos 2θ from Eqns. (5.10) and (5.11) into expression
for τθ Eqn. (5.8) gives τθ = 0 on these planes. Such planes are referred to as principal planes and
the direct stresses acting on them are the principal stresses.
Thus, at any point in a stressed material, there are always mutually perpendicular planes upon
which the stresses are wholly tensile or compressive and these are respectively the greatest and least
stresses at that point.
Such an arrangement is shows in Fig. 5.13 where σ1 and σ2 are the principal stresses and θ is
the angle given by Eqn. (5.9). Since this system of stress is identical with that shown in Fig. 5.6, it
follows from Eqn. (5.4) that the maximum shear stress in the body is given by:
σ1 − σ2
acting on planes at 45◦ to the principal.
2
1 1
2
2
Thus, τmax =
(σx + σy ) + (σx − σy ) + 4τ
2 2
1
−
(σx + σy ) − (σx − σy )2 + 4τ 2
2
1
∴ τmax =
(σx − σy )2 + 4τ 2
2
τmax =
(5.15)
In case of one direct stress only,
σ=
1
σx + σx2 + 4τ 2
2
2τ
σx
1
τmax =
σx2 + 4τ 2
2
(5.16)
tan 2θ =
(5.17)
E XAMPLE 5.8: In a complex stress system, an element has σx = 30 MPa compressive, σy = 50 MPa
tensile and τ = 15 MPa. Determine the principal stresses at the point.
S OLUTION :
1
2
2
Principal stresses σ1 σ2 =
(σx + σy ) ± (σx − σy ) + 4τ
2
1
2
2
σ1 or σ2 =
(−30 + 50) ± (−30 − 50) + 4 × 15
2
1
20 ± (−80)2 + 4 × 225
=
2
=
1
{20 ± 85.44}
2
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58
•
Strength of Materials
σ1 = 52.72 MPa (Tensile)
σ2 = −32.72 MPa (Compressive)
tan 2θ =
=
2τ
σx − σy
2 × 15
30
3
=
= − = −0.375
−30 − 50 −80
8
tan 2θ = −0.375
∴
or
2θ = −20.55
θ = 10.28◦
79.72◦
or
Ans.
E XAMPLE 5.9: A component has a plane element which is subjected to 120 MPa tensile accompanied by a shear stress of 25 MPa. Find: i) the normal and shear stresses on a plane inclined at an
angle of 25◦ with the tensile stress and ii) the maximum shear stress on the plane.
S OLUTION :
θ with vertical axis = 90 − 25 = 65◦
25
120
65°
120
25°
25
Figure 5.16
σx + σy σx − σy
+
cos 2θ + 2 sin 2θ
2
2
120 + 0 120 − 0
=
+
cos 2 × 65 + 25 sin 2 × 65
2
2
= 60 + 60 × (−0.643) + 25 × 0.766
= 40.57 MPa
σx − σy
sin 2θ − τ cos 2θ = 60 × 0.766 − 25 × (−0.643)
τθ =
2
= 62.07 MPa Ans.
1
τmax =
(σx − σy )2 + 4τ 2
2
1
1
=
1202 + 4 × 252 =
1202 + 4 × 625
2
2
= 65 MPa Ans.
σθ =
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•
59
E XAMPLE 5.10: An element in a stress material has tensile stress of 600 MN/m2 and a compressive
stress of 300 MN/m2 acting on two mutually perpendicular planes and equal shear stresses of
120 MN/m2 on these planes. Determine principal stresses and position of the principal planes. Also
find maximum shear stress.
S OLUTION :
1
2
2
σ1 & σ2 =
(σx + σy ) ± (σx − σy ) + 4τ
2
1
2
2
600 + (−300) ± (600 − (−300)) + 4 × 120
=
2
=
1
300 ± 9002 + 4 × 14400
2
=
1
{300 ± 931.45}
2
= 615.7 MPa (Tensile) or
− 315.7 MPa (Compressive)
Ans.
Position of principal planes θ1 , θ2
tan 2θ =
2τ
2 × 120
=
σx − σy 600 − (−300)
tan 2θ =
240
= 0.2666
900
∴
2θ = 14.93
∴
θ1 = 7.46◦ , θ2 = 187.46◦
Ans.
Max. stress (shear)
τmax =
σ1 − σ2 615.7 − (−315.7)
=
2
2
= 465.7 MPa Ans.
Graphical Method (Mohr’s Circle of Stresses)
This method is easier than analytical method. Here, we will adopt a sign correction that: i) all the
angles in the anticlockwise direction to the x − x axis are taken as negative and for the angles in
the clockwise direction an positive; ii) the tensile stresses are drawn towards right from origin and
compressive towards left and iii) the measurements above x − x axis and right of y − y axis are
positive and those below x − x axis and left to y − y axis are taken as negative.
Since this method involves drawing so small errors are bound to occur.
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•
Strength of Materials
E XAMPLE 5.11: At a point in a stressed body the stresses are shown in Fig. 5.17. Determine the
principal and maximum shearing stresses with the help of analytical method and Mohr’s circle
method. Also find principal planes.
S OLUTION :
30 MPa
τ 25 MPa
60 MPa
60 MPa
τ
τ = 25 MPa
τ
30 MPa
Figure 5.17
Analytical Method:
1
2
2
Principal stress σ1 or σ2 =
(σx + σy ) ± (σx − σy ) + 4τ
2
1
(60 + 30) ± (60 − 30)2 + 4(25)2
=
2
=
1
90 ± 302 + 2500
2
σ1 = 74.15 MPa
σ2 = 15.85 MPa
Maximum shear stress
τmax =
1
2
(σx − σy )2 + 4τ 2
Substituting resultant values τmax = 29.15 MPa
Principal planes:
tan 2θ =
50
2τ
2 × 25
=
=
σx − σy 60 − 30 30
= 1.667
∴
2θ = 59◦
θ1 = 29.5◦ , θ2 = 119.5◦ with the plane AB of the element.
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61
Mohr’s Circle Method
+τ
τmax
0
σ2
τ
–τ
σy
By measuerment and correction
σ1 = 75 MPa tensile
σ2 = 17 MPa tensile
τmax = 29 MPa
σx
2θp = 60°, θp1 = 30°
θp2 = 120°
Anticlockwise
σ1
Figure 5.18
E XAMPLE 5.12: A rectangular block of material is subjected to a tensile stress 100 MPa on a plane
and a tensile stress of 40 MPa at right angles to the former together with a shear stress of 60 MPa
on the same plane. Determine: a) analytically and b) by Mohr’s circle method i) the direction of
principal planes ii) magnitude of principal stress iii) magnitude of the maximum shear stress and
the corresponding plane.
S OLUTION :
40 MPa
60 MPa
100 MPa
100 MPa
60 MPa
40 MPa
Analytical solution:
Figure 5.19
Principal stresses:
1
2
2
(σx + σy ) ± (σx − σy ) + 4τ
σ1 & σ2 =
2
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62
•
Strength of Materials
1
2
2
(100 + 40) ± (100 − 40) + 4 × 60
=
2
√
1
1
=
140 ± 3600 + 14400 = {140 ± 134.16}
2
2
i) & ii) = 137.08 MPa (tensile), 2.22 MPa (Tensile) Ans.
1
1
2τ
2 × 60
θ = tan−1
= tan−1
2
σx − σy 2
100 − 40
1
1
= tan−1 2 = × 63.43 = 31.7◦ , 121.7◦ Ans.
2
2
1
(σx − σy )2 + 4τ 2
iii) τmax =
2
1
134.16
(100 − 40)2 + 4 × 602 =
=
= 67.08 MPa Ans.
2
2
60 MPa = τ
C
0
Principal
stress σ1 = 3 MPa
B
E 20 = 63° A
60 MPa = τ
τmax = 66 MPa
Mohr’s circle solution:
D
σy = 40 MPa
2θ = 63°
θ = 315°
σx = 100 MPa
Principal stress σ2 = 135 MPa
Figure 5.20
E XAMPLE 5.13: Various stresses acting on a block are shown in Fig. 5.21. Find analytically and
using Mohr’s circle method, the principal and maximum shearing stresses, and the angles of the
principal planes.
40 MPa
τ
τ
30 MPa
30 MPa
τ
τ = 20 MPa
40 MPa
Figure 5.21
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Principal Stresses and Strains
S OLUTION :
Analytical Method
1
2
2
(σx + σy ) ± (σx − σy ) + 4τ
Principal stresses σ1 or σ2 =
2
1
2
2
(−30−40)± (−30 − 40) +4(20)
=
2
σ1 = −4.59 MPa (Compressive)
σ2 = −65.41 MPa (Compressive)
1
Maximum shear stress τmax =
(σx − σy )2 + 4τ 2
2
By substituting max shear stress = 30.41 MPa
tan 2θ =
60
2τ
60
=
=6
=
σx − σy −30 − (−40) 10
2θ p = 80.53 θ1 = 40.27, θ2 = 130.27 Ans.
Principal stresses: σ1 = −4.1 MPa, σ2 = −65 MPa
Ans.
Max shear stress: τmax = 30.5 MPa Ans.
Angles of the principal planes
θ1 = 40.5◦
and
θ2 = 130.5◦ (anticlockwise) Ans.
+τaxis
τmax
σy = 40
σx = −30
–σaxis
B
C
A
τ
σ1
σ2
–τaxis
Figure 5.22
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+σaxis
0
•
63
64
•
Strength of Materials
E XAMPLE 5.14: On a machine bracket the stresses on two mutually perpendicular planes at a point
are 350 MPa tensile and 280 MPa tensile. The shear stress across these planes is 160 MPa.
Determine graphically and analytically the magnitude and directions of principal stresses and
maximum shear stress.
S OLUTION :
1
2
2
σ1 or σ2 =
(σx + σy ) ± (σx − σy ) + 4τ
2
1
=
(350 + 280) ± (350 − 280)2 + 4 × 1602
2
=
1
630 ± 702 + 102400
2
=
1
{630 ± 327.6}
2
σ1 = 478.8 MPa (Tensile) Ans.
σ2 = 151.2 MPa (Tensile). Ans.
For direction of principal stresses θ1 , θ2 :
tan 2θ =
2τ
2 × 160
320
=
=
= 4.57
σx − σy 350 − 280
70
∴
2θ = 77.66
∴
θ1 = 38.83◦
Maximum shear stress, τmax =
=
θ2 = 128.83◦
σ1 − σ2
2
478.8 − 151.2
2
= 163.8 MPa
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Principal Stresses and Strains
•
65
Graphical Method
E
σ2 = 151
MPa
Ans.
0
T
τ
τmax
M N
S
2θ
L
2θ = 76° & 256° Ans.
P
X
τ
F
σy = 280 MPa
σx = 350 MPa
σ1 = 478 MPa Ans
Figure 5.23
E XAMPLE 5.15: In a structure the principal stresses at a point are 80 MN/m2 (tensile) and 30 MN/m2
(tensile). Find the normal, tangential stresses and the resultant stress and its obliquity on a plane at
25◦ with the major principal plane.
S OLUTION :
Analytical Method:
σx = 80 MN/m2 (Tensile)
σy = 30 MN/m2 (Tensile)
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•
Strength of Materials
30 N/mm2
σx = 80 N/mm2
80 N/mm2
25°
30 N/mm2
Figure 5.24
Normal stress = σθ =
σθ =
Normal stress,
σx + σy σx − σy
+
cos 2θ
2
2
80 + 30 80 − 30
+
cos(2 × 25)
2
2
σθ = 55 + 25 × 0.6428
= 55 + 16.07
= 71.07 N/mm2 (Tensile). Ans.
Shear stress or tangential stress,
τθ =
=
σx − σy
sin 2θ
2
80 − 30
sin(2 × 25)
2
= 25 × 0.766
= 19.15 N/mm2
Hence, resultant stress
=
Ans.
σ 2 + τθ2 = σr
71.072 + 19.152
√
= 5050.9 + 366.7
σr =
∴
σr = 73.6 N/mm2
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Principal Stresses and Strains
•
67
Angle of obliquity,
tan φ =
=
τθ
σθ
18.15
71.17
= 0.269
∴
φ = 15.06◦
Ans.
Graphical (Mohr’s) Method:
D
2θ = 50°
φ
O
σy =
A
C
E
B
X
30 N/mm2
бx = 80 N/mm2
Figure 5.25
plot OA = σy = 30 N/mm2
scale 10 N/mm2 = 1 cm
and OB = σx = 80 N/mm2
Bisect AB to locate centre C.
Draw CD at angle 2θ (2 × 25◦ ) = 50◦
Now by measurement
Normal stress, σθ = OE = 71 N/mm2 Ans.
Shear stress τθ = DE = 19.8 N/mm2 Ans.
Resultant stress σr = 72.8 N/mm2 Ans.
Angle of obliquity, φ = 15◦ Ans.
Note: Slight difference is due to small mistakes in drawing Mohr’s circle.
E XAMPLE 5.16: At a point in a machine member the principal stresses are 120 MN/m2 (tensile)
and 70 MN/m2 (compressive). Determine the normal stress and the shear stress on a plane inclined
at 50◦ to the axis of major principal stress. Also determine the maximum shear stress at the point.
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•
Strength of Materials
S OLUTION :
σx = +120 N/mm2
σy = −70 N/mm2
(−ve sign for compression)
θ = 55◦
Analytical Method:
Normal stress,
σθ =
σx + σy σx − σy
+
cos 2θ
2
2
70 MN/m2
120 MN/m2
50º
70 MN/m2
Figure 5.26
σθ =
120 − 70 120 − (−70)
+
cos(2 × 50◦ )
2
2
= 25 + 95 × (−0.174)
= 25 − 16.53
= 8.47 N/mm2
(Tensile)
Ans.
Shear stress or tangential stress,
τθ =
=
σx − σy
sin 2θ
2
120 − (−70)
sin(2 × 50)
2
= 95 × 0.985
= 93.575 N/mm2
Ans.
Maximum shear stress = τmax
=
σx − σy
2
=
120 − (−70)
2
= 95 N/mm2
Ans.
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120 MN/m2
Principal Stresses and Strains
•
69
M L
P
τθ = 92 N/mm2
C
100˚
B
O
τmax = 95.5 N/mm2
A
N C
σx = 120 N/mm2
σy = 70 N/mm2
Tensile
Compressive
Figure 5.27
Graphical (Mohr’s) Method
Plot
OA = 120 N/mm2
2
OB = 70 N/mm
(Tensile)
(Compressive)
Bisect BA to get centre C and draw a circle as shown. Draw CM at an angle 2θ i.e. 100◦ . Drop perpendicular MN.
On measurement, we find τθ = 92 N/mm2 (OP)
τθ = 92 N/mm2 (OP)
Max shear stress,
Normal stress,
τmax = 95.5 N/mm2 (CL)
σθ = 8 N/mm2 (ON)
E XAMPLE 5.17: On a steel bridge, there is a point which is subjected to perpendicular stresses of
60 MN/m2 tensile and 40 MN/m2 tensile. Compute the normal, tangential stresses and resultant
stress with its angle of obliquity on a plane making an angle of 35◦ with the axis of the second
stress.
S OLUTION :
σx = 60 MN/m2 (Tensile),
θ = 90◦ − 35◦ = 55◦
σy = 40 MN/m2 (Tensile)
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•
Strength of Materials
Analytical Method:
σx + σy σx − σy
+
cos 2θ
2
2
60 + 40 60 − 40
=
+
cos(2 × 55)
2
2
= 50 + 10 × (−0.342)
Normal stress = σθ =
2
= 50 − 3.42 = 46.6 N/mm
40
55°
Ans.
60
60
40
Figure 5.28
Shear or tangential stress,
σx − σy 60 − 40
=
2
2
τθ = 10 N/mm2 Ans.
τθ =
Resultant stress,
σθ2 + τθ2
= 46.62 + 102
√
= 2171.56 + 100
σr =
∴
= 47.67 N/mm2 Ans.
τ
tan φ = θ
σθ
10
= 0.2146
=
46.6
φ (obliquity) = 12.11◦ Ans.
Graphical Method:
Plot OA = 40 N/mm2 and OB = 60 N/mm2
Bisect AB and draw circle as shown.
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71
σθ = 46.2 N/mm2
M
O
φ
A
110°
=2θ
Q C
B
τθ = 9 N/mm2
40 N/mm2
σx = 60 N/mm2
Figure 5.29
On measurement:
σθ = OQ = 46.2 N/mm2
τθ = MQ = 9 N/mm2
σr = 47 N/mm2
φ = 10.8◦ Ans.
E XAMPLE 5.18: At a point in a member of a machine, the stresses on two mutually perpendicular
planes are 40 N/mm2 tensile and 20 N/mm2 tensile. The shear stress across the plane is 10 N/mm2 .
Find the magnitude and direction of the resultant stress on a plane making an angle of 35◦ with the
plane of the first stress. Find also the normal and tangential stresses on the planes.
S OLUTION :
Analytical Method:
σx = 40 N/mm2
σy = 20 N/mm2
τxy = 10 N/mm2
20 N/mm2
10 N/mm2
40 N/mm2
40 N/mm2
35°
10 N/mm2
20 N/mm2
Figure 5.30
σx + σy σx − σy
+
cos 2θ + τxy sin 2θ
2
2
40 + 20 40 − 20
=
+
cos(2 × 35◦ ) + 10 sin(2 × 35◦ )
2
2
Normal stress, σθ =
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•
Strength of Materials
= 30 + 10 × 0.342 + 10 × 0.9396
= 30 + 3.42 + 9.4
= 42.82 N/mm2
Ans.
Shear stress on tangential stress:
σx + σy
sin 2θ − τxy cos 2θ
2
40 − 20
τθ =
sin (2 × 35◦ ) − 10 cos (2 × 35◦ )
2
= 10 × 0.94 − 10 × 0.342
= 9.4 − 3.42
= 5.98 N/mm2
τθ =
Resultant stress:
(42.82)2 + (5.98)2
√
= 1833.55 + 35.76
σr =
= 43.23 N/mm2
5.98
42.88
tan φ = 0.139
Angle of obliquity = 7.9◦ Ans.
tan φ =
∴
τθ
σθ
Ans.
tan φ =
∴
Plot OA = 40 N/mm2 and OB = 20 N/mm2
Drop perpendicular LA and MB each 10 N/mm2 as shown join ML to get C. Draw Mohr’s circle
taking MC as radius and C as centre. Draw CP at 70◦ to CL.
P
M
O
φ = 20°
20 N/mm2
C
B
2θ = 70°
N A
10 N/mm2
L
40 N/mm2
Figure 5.31
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By measurement:
σθ = ON = 40 N/mm2 tensile Ans.
σr = OP = 41.5 N/mm2
τ = PN = 11 N/mm2
φ = 9◦ Ans.
Ans.
Ans.
Difference is due to errors in plotting the diagram.
Principal Strains: The principal strains are the strains in the direction of the principal stresses. If
the principal stresses on an element are σx and σy (Refer Fig. 5.32).
σy
σx
σx
σy
Figure 5.32
Then strain in direction of σx due to σx =
σx
E
and strain in direction of σx due to σy = −μ
σy
E
σy
σx
−μ
(5.18)
E
E
σy
σx
Similarly, resultant stain in y direction, εy =
−μ
(5.19)
E
E
If these strains are measured and it is required to find the corresponding stresses, then multiplying
Eqn. (5.19) by μ and adding to Eqn. (5.18),
∴
resultant stain in x direction, εx =
σx
(1 − μ 2 )
E
E
σx =
(εx + μεy )
1 − μ2
E
σy =
(εy + μεx )
1 − μ2
εx + μεy =
∴
(5.20)
(5.21)
Strains on an Oblique Section: Figure 5.33 shows an element ABCD which is subjected to pure
strains εx and εy , the distorted shape relative to the point F being shown dotted. The line FG,
inclined at θ to AB moves to the position FG, the displacements of G in the x and y directions being
dx and dy respectively.
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•
Strength of Materials
εy
D
C
dr
εx
G
εx
E
r
θ
A
F
dθ
x
θ
dy
y dx
B
εy
Figure 5.33
Since the movement of G is very small, the distance GG may be regarded as the change in the
dr
length of FG, so that the strain on FG, εθ =
r
r2 = x2 + y2
∴
2r dr = 2x dx + 2y dy
∴
dr dx x
=
r
x r
2
+
dy y
y r
2
εθ = εx cos2 θ + εy sin2 θ
=
εx + εy εx + εy
+
cos 2θ
2
2
Similarly, it can be shown that the strain on FE is
(5.22)
εx + εy εx − εy
−
cos 2θ
2
2
y
x
x dy − y dx
sec2 θ d θ =
x2
dy y dx y
=
· −
·
y x
x x
= (εy − εx ) tan θ
∴ d θ = (εy − εx ) sin θ cos θ
εx − εy
=−
sin 2θ
2
tan θ =
∴
The negative sign indicates that when εx > εy , as assumed in Fig. 5.33, the angle θ decreases.
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A similar analysis on the triangle AFE shows that FE also rotates through the same angle as
FG, so that the total change in the right angle EFG is 2 d θ . This is the shear strain in direction
inclined at angle θ to the faces of ABCD.
i.e., φθ = (εx − εy ) sin 2θ
(5.23)
Equations (5.22) and (5.23) are similar to Eqns. (5.3) and (5.4) and they can therefore be represented
by a similar graphical construction. Figure 5.34 shows Mohr’s strain circle, in which OA represents
εx , OB represents εy and the angle AQC is 2θ . Then OD and CD represents respectively, εθ and
φθ /2.
εθ
C
2θ
B
Q
0
εx+εy
2
φθ
2
D A
εx−εy
2
εy
εx
Figure 5.34
These days for measurements electric resistance strain ganges are used very extensively.
Ellipse of Stress:
In order to draw an ellipse of stress, the procedure is given below with steps of construction:
1. Take O as centre and draw two circles σx and σy respectively.
2. Through O draw OM perpendicular to the FN as shown, to cut the inner circle in P and after
circle in M.
3. Through M draw a line MR perpendicular to OX and through P, draw line PQ perpendicular
to OY to cut the line MR in Q. Join OQ.
Refer diagram: OR = σx cos θ , QR = σy sin θ
OQ = OR2 + QR2
= σx2 cos2 θ + σy2 sin2 θ = Resultant stress σr
∴
tan α =
QR
OR
tan α =
σy sin θ
σy
=
tan θ
σx cos θ
σx
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•
Strength of Materials
The point Q may be drawn for different values of θ . The locus of Q will give an ellipse.
y
σy
F
σx
σx
θ
σy
M
F
o P
x
θ
β
N
N
θ
α
R
x
Q
y
Figure 5.35
E XAMPLE 5.19: A point in a strained component is subjected to two mutually perpendicular tensile
stresses of 270 MPa and 150 MPa. Determine the resultant stress on a plane which makes 35◦ with
the major principal stress by ellipse of stress.
Resultant stress σr
= OQ = 303.5 MPa
α = 40.5◦
Y
M
F
150 MPa
θ=
°
55
Q
θ = 55°
X
α = 40.5°
o
R
N
Y
270 MPa
Figure 5.36
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Principal Stresses and Strains
•
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Exercise
5.1 The state of stress of a machine component is shown in Fig. 5.37. Calculate the principal
stresses and the maximum shear stress.
100 N/cm2
300 N/cm2
500 N/cm2
Figure 5.37
[Ans σ1 = 600 N/cm2 inclined at 18◦ 26
to stress of 500 N/cm2 σ2 = −400 N/cm2
perpendicular to σ1 maximum shear stress
τmax = ±500 N/cm2 . It is inclined to σ1 at
+45◦ and converges towards σ1 ]
5.2 A bar of uniform cross section 20 × 30 mm is subjected to an axial pull of 54 kN, applied at
each end of the bar. Determine the normal and shear components of a force acting on a plane
inclined at 20◦ to the line of action of the load and also find normal and shear stresses acting
on the same plane.
[Ans Pn = 18.47 kN, Ps = 50.77 kN
σθ = 10.53 MPa, τθ = 28.83 MPa]
5.3 At a certain point in a strained material, the principal stresses are 100 MN/m2 and 40 MN/m2 ,
both tensile. Using and explaining Mohr’s circle of stress, find the normal, tangential and
resultant stresses across a point in the plane of 48◦ to major principal plane.
[Ans σθ = 66.86 MPa, τθ = 29.84 MPa
σr = 73.22 MPa]
5.4 At a certain point in a structural member the value of σx = 45 MPa, σy = 75 MPa and τ =
45 MPa. Calculate the principal stresses and locate exactly the planes on which (Fig. 5.38)
these exist. Represent the stress system by means of Mohr’s circle. Also find the maximum
shearing stresses and the planes on which they act.
τ
σx
σy
τ
τ
τ
σy
Figure 5.38
σx
[Ans σ1 = 90 MPa, σ2 = −60 MPa
θ1 , θ2 = 18◦ 26 , 108◦ 20
τmax = 75 MPa, θ3 , θ4 = 63◦ , 26◦ , 153◦ 20 ]
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•
Strength of Materials
5.5 Direct stresses of 120 MPa (tensile) and 90 MPa (compressive) exist on two perpendicular
planes at a certain point in a body. They are also accompanied by shear stress on the planes.
The greater principal stress at the point due to these is 150 MPa. Determine: i) the shear stress
on these planes and ii) also find the maximum shear stress at the point.
[Ans τ = 84.85 N/mm2 , τmax = 135 N/mm2 ]
5.6 A point in a load carrying member is subjected to the following stress condition:
σx = −120 MN/m2 , σy = 180 MPa, τxy = 80 MPa counterclockwise. Then
a) Draw the initial stress element
b) Draw complete Mohr’s circle
c) Draw complete principal stress element
d) Draw the complete shear stress element
180 MPa
τ = 80 MPa CCW
120 MPa
120 MPa
180 MPa
Figure 5.39
[Ans b) σr = 170 MPa, α = 28.07◦
θ = 75.96◦ ccw, θ = 59.04◦ cw
σ1 = 200 MPa, σ2 = 140 MPa]
5.7 Prove that in a two-dimensional stress system, if εx and εy are the strains in two particular
directions, the normal stresses in these directions are given by
σx =
E
E
(εx + μεy ), σy =
(εy + μεx )
2
1+μ
1 − μ2
By means of strain gauges, the strains in two perpendicular directions at 30◦ to the directions
of the principal stresses are found to be 425 × 10−6 and 86 × 10−6 (both tensile) respectively,
both tensile. Determine the normal stresses in the direction of the measured strains, and also
the principal stresses. E = 200 GN/m2 , μ = 0.3.
[Ans 98.8 and 86 MPa, 125.2 and 19.6 MPa]
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6
C HAPTER
SHEARING FORCE AND BENDING MOMENT
Shearing Force
Consider a horizontal beam AB (Fig. 6.1) which is simply supported at each end and which carries
a single load W at the centre. Such a load which is assumed to act at a point, is called a concentrated
or point load. Since the beam is in equilibrium, the reactions RA and RB of the supports at A and B,
respectively, will each be W /2 units.
A
W x− L
2
L
2
X
B
x
W
2
X L−x
W
2
L
RB = W
2
RA = W
2
Figure 6.1
Now let us consider any section XX of the beam, distance x from the left-hand end A. It will
be noticed that two forces are acting on the part of the beam to the left of the section XX. These
forces are:
i) The reaction RA = W /2 acting vertically upwards
ii) The concentrated load W acting vertically downwards
There is, therefore a resultant force of
W−
W
W
=
2
2
tending to shear the left hand part of the beam downwards across X − X. This resultant force is
called the shearing force of the beam at the section X − X.
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•
Strength of Materials
For equilibrium, the resultant of the forces acting on the part of the beam to the right of X − X
must be equal in magnitude, but part of the beam to the left of X − X. As a matter of fact, it will be
noticed that the only force acting on the right-hand side of section X − X of the beam is the reaction
RB = W /2 tending to shear this part of the beam upwards across. We may, therefore conclude that:
‘The shearing force at a particular section of a beam is the resultant force, of all the forces acting
on either side of that section.’
Bending Moment
Now let us think again of the part of beam to the left of the section X − X (Fig. 6.1).
Taking moments about X − X, we get
W
x (clockwise)
2
L
(anticlockwise)
Moment due to load W = W x −
2
Moment due to reaction RA =
There is therefore, a resultant moment at X − X,
L
W
= x −W x −
2
2
=
W
(L − x) (clockwise)
2
Acting upon the beam in a clockwise direction, and the left-hand part of the beam will tend to rotate
about XX in the direction of this resultant moment. The resultant moment is called the bending
moment at the section X − X.
For equilibrium, the resultant moment of all the forces acting on the part of the beam to the
right of X − X must be equal in magnitude, but opposite in direction to the resultant moment of all
the forces acting on the part of the beam to the right of X − X. It will be noticed that the only force
acting on this part of the beam in the reaction RB = W /2 acting upwards at the end B, its distance
being (L − x) from X − X.
Now taking moments about X − X, we get:
Moment due to reaction RB =
W
(L − x) (anticlockwise).
2
Hence, the right-hand part of the beam will tend to rotate about X − X in the direction of this
moment.
It should therefore be emphasised that there can only be one value for the bending moment (and
also for the shearing force) at a particular section of a beam regardless of which side of the section
is considered.
From the foregoing discussion we may then conclude that:
‘The bending moment at a particular section of a beam is the resultant moment about a section
of all the forces acting on either side of the section.’
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•
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Sign convention for the shearing force and bending moment:
i) Shear Force
There are several conventions in use for mathematical sign with the magnitude of a shearing
force and with the magnitude of a bending moment, but the system given below appears to
be most popular.
The shearing force (SF) which tends to slide the left-hand side of the section of the beam
upwards relative to the right-hand side will be considered positive, and that which tends to
slide the left-hand side of the section of the beam downwards relative to the right-hand side
will be considered negative (see Fig. 6.2).
Positive SF
(left up, right down)
Negative SF
(left down, right up)
Figure 6.2
Hence, the rule is:
Left up, right down is positive
Left down, right up is negative
ii) Bending Moment:
The bending moment (BM) which tends to cause the beam concave upwards, sometimes
called sagging, will be considered positive and that which tends to cause the beam concave
downwards, sometimes called hogging, will be considers negative (see Fig. 6.3). Notice that
this diagram also gives the sign convention for cantilever beams.
W
Positive BM (sagging)
Negative BM (hogging)
Figure 6.3
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Shearing Force and Bending Moment Diagrams
Diagrams indicating the value of the shearing force and the bending moment at any section along
a loaded beam or cantilever are called Shear Force Diagram and Bending Moment Diagram. These
diagrams are drawn immediately underneath the loading diagram to same horizontal scale. In each
case positive values are plotted upwards and negative values downwards and, in addition, the principal values should be shown on both diagrams. Also scale for drawing (for example, 5 kN = 10 mm
to suit the size of paper) and for Bending Moment Diagram (for example, 15 kNm = 10 mm to suit
the size of paper) should be shown beneath each diagram.
Method: In case of beams always start from the left hand end and for cantilevers from the free end.
E XAMPLE 6.1: A beam ABCDE, 6 m long is simply supported at A and E. It carries concentrated loads as shown in Fig. 6.4. Draw Shear Force Diagram (SFD) and Bending Moment Diagram
(BMD) showing principal values.
S OLUTION :
200 N
A
400 N
B
1.5 m
300 N
C
1.5 m
D
2 mm
E
1m
Figure 6.4
6m
(a) Loading Diagram
400 N
400 N
200 N
200 N
o
o
(b) S.F.D.
200 N
500 N
500 N
900 Nm
600 Nm
500 Nm
(c) B.M.D.
o
o
Figure 6.5
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•
83
Calculations:
Beam Reactions
Taking moments about A, and working in N and m, we get
RE × 6 = (200 × 1.5) + (400 × 3) + (300 × 5)
= 300 + 1200 + 1500 = 3000
∴
RE =
3000
= 500 N
6
RA + RE = Sum of loads
Now,
RA = 200 + 400 + 300 − 500 = 400 N
SFD Calculations:
Starting from left point A,
SF at A = 0 rising to 400 N
SF at B = 400 falls to 400 − 200 = 200 N
SF at C = 200 N falls to + 200 − 400 = −200 N
SF at D = −200 N falls to − 200 − 300 = −500 N
SF at E = −500 N rises to − 500 + 500 = 0
BMD Calculations:
BM at A = 0
BM at B = +(400 × 1.5) = +600 Nm
BM at C = +(400 × 3) − (200 × 1.5) = 1200 − 300
= +900 Nm
BM at D = +(400 × 5) − (200 × 3.5) − (400 × 2)
= +2000 + 700 − 800 = +500 Nm
BM at E = +(400 × 6) − (200 × 4.5) − (400 × 3)
= −(300 × 1) = +2400 − 900 − 1200 − 300 = 0
Note: Scales for drawing SFD and BMD should be chosen such that it suits the size of the paper
and place allotted for diagrams.
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•
Strength of Materials
E XAMPLE 6.2: An overhang beam (refer Fig. 6.6) is loaded as shown. Draw to seek the shearing
force and bending moment diagram and mark all the important points.
S OLUTION :
Reactions: Taking moments about A and working in kN and m units
RB × 3 + ×0.3 = (10 × 0.9) + (15 × 2.1) + (10 × 3.6)
RB × 3 = 9 + 31.5 + 36 − 1.5 = 75
75
= 25 kN
3
RA + RB = Sum of loads
RB =
Now
RA + 25 = 5 + 10 + 15 + 10 = 40 kN
∴
RA = 40 − 25 = 15 kN
10 kN
5 kN
A
C
0.3 m
15 kN
D
E
0.9 m 3 m
10 kN
F
B
1.2 m
0.9 m
0.6 m
Figure 6.6
(a) Loading Diagram
10 kN
10 kN
o
10 kN
o
10 kN
o
o
(b) SF Diagram
Values in kN
15 kN
15 kN
7.5 kNm
7.5 kNm
o
o
1.5 kNm
BM Diagram
Values in kNm
6 kNm
Figure 6.7
SFD Calculations:
At C = 0 falls to − 5 kN
At A = −5 rising to − 5 + 15 = +10 kN
At D = +10 kN falls to + 10 − 10 = 0
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Shearing Force and Bending Moment
•
85
At E = 0 falls, to 0 − 15 = −15 kN
At B = −15 kN rises to − 15 + 25 = +10 kN
At F = +10 falls to + 10 − 10 = 0
BMD Calculations:
At C = 0
At A = −5 × 0.3 = −1.5 kNm
At D = (−5 × 1.2) + (15 × 0.9) = +7.5 kNm
At E = (−5 × 2.4) + (15 × 2.1) − (10 × 1.2) = +7.5 kNm
At B = (−5 × 3.3) + (15 × 3) − (10 × 2.1) − (15 × 0.9) = −6 kNm
At F = (−5 × 3.9) + (15 × 3.6) − (10 × 2.7) − (15 × 1.5) + (25 × 0.6) = 0
The above values are then plotted to scale (for both SFD & BMD) and indicating beneath diagram
beams carrying uniformly distributed loads.
E XAMPLE 6.3: A 20 m long beam is simply supported between B and F. It carries uniformly
distributed loads of 200 N/m along overhang AB and 100 N/m run over the length EF as shown in
Fig. (6.8(a)). It also carries concentrated loads (refer Fig. 6.8(b)). Draw Shear Force and Bending
Moment diagrams for the loaded beam and mark all important values.
600 N
200 N/m
A
800 N
C
B
D
E
2m P 2m
400 N
100 N/m
2m
4m
1100 N
4m
1100 N
500 N
4m
2m
4m
500 N
300
SFD
Values in N m
Figure 6.8(a)
o
o
800 N
F
300
700
1100 N
4800 Nm
3600
2800
BMD
Values in N m
2000
o
0
400
1600 Nm
Figure 6.8(b)
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86
•
Strength of Materials
Beam Reactions:
Taking moments about F and working in N and m units, we get:
RB × 16 = (200 × 4 × 18) + (600 × 12) + (800 × 8)
+ (400 × 4) + (100 × 4 × 2)
= 14400 + 7200 + 6400 + 1600 + 800 = 30400
30400
= 1900 N
16
RB + RF = Sum of loads
∴
Now,
RB =
1900 + RF = (200 × 4) + 600 + 800 + 400 + (100 × 4)
∴
RF = 3000 − 1900 = 1100 N
SFD Calculations:
At A = 0
For uniformly distributed load, let us take mid-point P and Q so that we know exactly whether
it will be a straight line, concave or convex.
At P = −200 × 2 = −400 N
At B = −200 × 4 = −800 N
At C = (−200 × 4) + 1900 = 1100 N falls to 1100 − 600 = 500 N
At D = 500 N falls to 500 − 800 = −300 N
At E = −300 N falls to − 300 − 400 = −700 N
At Q = −700 − 100 × 2 = −900 N
At F = −900 N − 100 × 2 = −1100 N rises to − 1100 + 1100 = 0
BMD Calculations:
At A = 0
At P = −200 × 2 × 2 = −400 Nm
At B = −200 × 4 × 2 = −1600 Nm
At C = −200 × 4(2 + 4) + 1900 × 4 = 19200 + 7600 = +2800 Nm
At D = −200 × 4(2 + 8) + 1900 × 8 − 600 × 4 = +4800 Nm
At E = −200 × 4(2 + 12) + 1900 × 12 − 600 × 8 − 800 × 4 = +3600 Nm
At Q = −200 × 4(2 + 14) + 1900 × 14 − 600 × 10 − 800 × 6
− 400 × 2 − 100 × 2 × 1 = +2000 Nm
At F = 0
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Shearing Force and Bending Moment
•
87
E XAMPLE 6.4: For the loaded beam in Fig. 6.9, draw BM and SF diagrams and show important
points.
8 kN
0.5 m
A
B
C
5m
3m
Figure 6.9(a)
This beam (6.9b) is equivalent to shown in (6.9a)
8 kN
A
C
RA = 3.5 kN
4 kNm
5m
B
RB = 4.5 kN
3m
Figure 6.9(b)
Reactions:
Taking moments about B,
RA × 8 = 8 × 3 + 4 = 28
∴
RA =
28
= 3.5 kN
8
RB = 8 − 3.5 = 4.5 kN
3.5 kN
3.5 kN
o
o
4.5 kN
4.5 kN
17.5 kNm
13.5 kNm
o
o
Figure 6.9(c)
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88
•
Strength of Materials
SFD Calculations:
At A = 0 rises to + 3.5 kN
At C = 3.5 to 3.5 − 8 = −4.5 kN
At B = −4.5 rises to − 4.5 + 4.5 = 0
BMD Calculations:
At A = 0
At C = +3.5 × 5 = 17.5 kNm
Also at C = 17.5 − 4 = 13.5 kNm
Remember: i) where shear force is zero or change sign, bending moment is maximum; ii) where
B.M. is zero that point on Bending Moment Diagram is known as point of contraflexure or point of
inflexion.
Relation between intensity of loading, shearing force and bending moment:
Consider a short length, dx of a beam Fig. 6.10, carrying a uniformly distributed load w per
unit length. Over this length, let the shear force change from F to F + dF and the bending moment
change from M to M + dM.
Equating vertical force on the element,
F + wdx = F + dF
∴w=
dF
dx
(i)
Taking moments about the right-hand end of the element,
w/unit length
F + dF
M
M + dM
dx +
F
Figure 6.10
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Shearing Force and Bending Moment
M + F.dx + w.dx
•
89
dx
= M + dM
2
ignoring the second order
of small quantities
F.dx = dM
∴
F=
dM
dw
(ii)
Therefore, intensity of loading is the rate of change of shearing force and shearing force is the rate
of change of bending moment. This latter relation shows that the maximum bending moment occurs
where the shearing force is zero. Combining Eqns. (i) and (ii)
d2M
dF
.
=
dx
dx2
w=
Cantilevers
E XAMPLE 6.5: Draw SF and BM diagrams for the cantilever loaded as shown in Fig. (6.11) and
show important values on both diagrams.
25 kN
15 kN
C
15 kN/m
D
E
B
A
F
1.5 m
1.5 m
2m
Figure 6.11
1m
1m
o
o
15 kN
SFD
15 kN
45 kN
70 kN
70 kN
15 kNm
37.5 kNm
BMD
75 kNm
m
.5
47
kN
142.5 kNm
2
Figure 6.12
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90
•
Strength of Materials
SF Calculations:
SF at B = 0 falling to − 15 kN
SF at E = −15 kN
SF at D = −15 kN − 15 × 2 = −45 kN
SF at C = −45 kN falls to − 45 − 25 = −70 kN
SF at A = −70 kN
BM Calculations:
BM at = 0
BM at E = −15 × 1 = −15 kNm
BM at D = −15 × 3 − 15 × 2 × 1 = −75 kNm
BM at C = −15 × 4.5 − 15 × 2 × 2.5 = −142.5 kNm
BM at A = −15 × 6 − 15 × 2 × 4 − 25 × 1.5 = −247.5 kNm
Take extra mid-point F on u.d.l. to find the nature of curve
BM at F = −15 × 2 − 15 × 1 ×
1
= −37.5 kNm
2
E XAMPLE 6.6: Figure 6.10 shows a partial variable loaded cantilever. Draw SF and BM diagrams.
D
X
4 kN/m
A
B
2m
E
F
X
x
C
3m
Figure 6.13(a)
Consider a section X − X between B and C at a distance x from the free end C.
ΔBCD is similar to ΔCEF For Rate of loading at X − X
FC
EF
x
EF
=
;
=
DB
BC
4
3
4
EF = x kN/m
3
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Shearing Force and Bending Moment
•
91
Now,
Shear force at X − X = Total load from C to F
=
2x2
1 4x
× ×x =
kN
2
3
3
(i)
For SF at C, put x = 0 in Eqn. (i) ∴ SF at C = 0
SF at B, put x = −3 in Eqn. (i)
∴
SF at B =
2 × 32
= −6 kN
3
SF at A = −6 kN
BM Calculations:
BM at X − X = −
2x2 x −2x3
× =
kN/m
3
3
9
Equation (ii) is that of a cubic parabola
BM at C = 0
BM at B, put x = 3 in Eq. (ii) =
−2(3)3
= −6 kNm
9
3
BM at A = Total load on BC × 2 +
3
1
=−
× 4 × 3 × (2 + 1)
2
= −18 kNm
Between A and B, BMD will be a straight line.
o
o
SFD
6 kN
o
o
BMD
Cubic parabola
6 kNm
18 kNm
Figure 6.13(b)
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(ii)
92
•
Strength of Materials
E XAMPLE 6.7: In a gradually varying loaded beam as shown in Fig. 6.14, the span of simply
supported beam is 6 m. Load at B is 1.6 kN/m run and gradually increases to 4 kN/m run at the
other end.
Determine the position and amount of maximum bending moment. Also draw the shear force
and bending moment diagram.
4 kN/m
1.6 kN/m
A
B
6m
Figure 6.14
Let us split the beam in two parts:
i) a rectangular of length 6 m and height 1.6 kN/m run.
ii) a triangular zero at B and 4 − 1.6 = 2.4 kN/m at A.
Let RA and RB be reactions due to part (i)
RA = RB =
1.6 × 6
= 4.8 kN
2
Let RA and RB be reactions due to part (ii)
Taking moments about B,
RA × 6 =
2.4 × 6 2
× ×6
2
3
∴
RA = 4.8 kN
RA + RB = Total triangular beam (part ii)
4.8 + RB =
2.4 × 6
2
∴
RB = 2.4 kN
Net reactions:
RA = RA + R = 4.8 + 4.8 = 9.6 kN
RB = RB + RB = 4.8 + 2.4 = 7.2 kN
SFD Calculations:
S.F. at A = 0 rises to 9.6 kN
S.F. at B = 7.2 rises to 7.2 − 7.2 = 0
@seismicisolation
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Shearing Force and Bending Moment
BMD Calculations:
Obviously, BM at A = 0
BM at B = 0
Maximum bending moment occurs where SF changes sign.
Let P be the point where maximum bending moment occurs and S.F. changes sign.
Let x be the distance between P and B.
As, SF is zero, therefore,
0 = −7.2 + 1.6x +
2.4x
1
×x×
6
2
0 = −7.2 + 1.6x + 0.2x2
0.2x2 + 1.6x − 7.2 = 0
x2 + 8x − 36 = 0
√
−8 ± 64 + 4 × 36 −8 ± 14.42
x=
=
2
2
= 3.21 m.
Maximum BM = +7.2 × 3.21 − 1.6 × 3.21 ×
3.21 2.4 × 3.21 3.21
−
×
2
6
3
− 23.112 − 8.24 − 1.37 = 13.5 kNm
Ans.
4 kN/m
1.6 kN/m
B
A
6m
RB = 7.2 kN
RA = 9.6 kN
9.6
o
SFD
P
o
7.2 kN
13.5 kNm
BMD
o
o
Figure 6.15
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•
93
94
•
Strength of Materials
Exercise
6.1 Draw and SF and BM diagrams for the loaded cantilever as shown is in Fig. (6.16) and state
BM at D.
40 kN
2m
30 kN
1m
20 kN
10 kN/m
1m
[Ans
Figure 6.16
(310 kNm)]
6.2 A beam AB of length 8 m is simply supported at A and B. It carries a load which varies from 2
kN/m at A to 5 kN at B. Draw SF and BM diagrams. State what is maximum bending moment
and its position. Show all principal values on SF and BM on diagrams
[Ans
(28.1 kNm) at 4.3m from A]
6.3 A beam AB is hinged at A and simply supported at B. A rigid bracket is attached to the
middle point C of the beam is loaded as shown in Fig. 6.17. Draw SF and BM diagrams
showing important values. What is the maximum BM and its position.
0.8 m
Hinge
1000 N
B
C
A
2m
2m
[Ans
Figure 6.17
Max. BM 1400 Nm at 2 m from A]
6.4 For the beam loaded as shown in Fig. 6.18, at 2 m from A draw SF and BM diagrams. Indicate
maximum BM and its position.
C
A
B
30 kNm
6m
4m
[Ans
Figure 6.18
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18 Nm, 6 m from A]
Shearing Force and Bending Moment
•
95
6.5 Draw the SF and BM diagram for beam shown in Fig. 6.19. Locate the point of contraflexure.
1 kN/m
6 kN
B
30°
A
2m
C
3 kN
E
D
2m
4m
1m
[Ans 0.8 m from D]
Figure 6.19
6.6 For the loaded beam shown in Fig. 6.20. Draw SF and BM diagrams and indicate all principal
values on both diagrams. Also find maximum bending moment and its position.
4 kN
8 kN
2 kN/m
1 kNm
6 kN
C
D
A
5m
5m
E
5m
B
F
5m
5m
20 m
[Ans
Figure 6.20
65 kNm 5 m from F]
6.7 Draw SF and BM diagrams for a cantilever loaded as shown in Fig. 6.21. Show all the principal values on both diagrams.
20 kN
20 kN/m
C
A
10 kN
D
E
B
1m
1m
2m
1m
Figure 6.21
@seismicisolation
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[Ans
Max BM = 190 kNm]
•
96
Strength of Materials
6.8 Draw SF and BM diagrams for loaded cantilever as shown in Fig. 6.22 and indicate position
and value of maximum BM.
D
4 kN/m
A
B
2m
C
3m
[Ans
Figure 6.22
18 kNm at A]
6.9 A simply supported beam of 5 m span carries a triangular load of total 30 kN. Draw SF and
BM diagrams and indicate the maximum value of BM and its position.
[Ans 25 kNm at centre]
6.10 Draw the SF and BM diagrams for the cantilever beam shown in Fig. 6.23. The vertical load
of 2 kN at C is partly supported by a force of 3 kN at the prop B. State: a) the reaction at the
built in end; b) the maximum BM and its position and c) where the BM is zero.
700 mm
250 mm
A
2 kN
C
B
3 kN
Figure 6.23
[Ans a) 1 kN; b) 0.5 kNm; c) At B downward]
6.11 Draw SF and BM diagrams for the beam shown in Fig. 6.24. Determine the points of contraflexure. What is the maximum BM.
100 N
100 N/m
C
A
1m
D
B
6m
Figure 6.24
2m
[Ans
@seismicisolation
@seismicisolation
7.79 m and 3.38 m from free end D]
C HAPTER
7
CENTRE OF GRAVITY
Centroid: This is the point at which the whole area of a plane figure is assumed to be concentrated.
It is applicable for plane geometrical figures such as rectangle, square, triangle, etc.
Centre of Gravity: The centre of gravity of a body is the point in the body through which the whole
weight of the body acts.
Centroid for Plane Figures
a) A circle:
G
π d2
Area =
4
Centroid is at centre O
d
2
Figure 7.1
A
b) A triangular plane figure:
1
Area = × b × h
2
G h
h
3
B
C
b
Figure 7.2
c) Square:
Area = A2
G
A
centroid at from each side.
2
A
A
Figure 7.3
Y
d) Rectangle: Area = A × L
L
A
X̄ = , Ȳ =
2
2
A
A
2
L
2
L
Figure 7.4
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@seismicisolation
X
•
Strength of Materials
e) Semicircle:
π r2
2
4r
from X − X axis
Ȳ =
3π
Area =
G
r
Y
X
X
Figure 7.5
f) Trapezium:
a
1
Area = c(a + b)
2
a + 2b c
X̄ =
a+b 3
b
X
c
Figure 7.6
Y
g) Qudrant of Circle:
1
Area = π r2
4
4r
Ȳ = X̄ =
3π
G
X
X
o
X
r
Figure 7.7
h) Sector of a Circle:
O
r2 θ
Area =
2
Y
θ is in radius
θ
r
G
2r sin α
Ȳ =
3 α
Figure 7.8
In denominator α is to be taken into radians.
l
i) Circular Segment:
Y
Area = A
Length of area = l
Angle in degrees = α
Radius = r
Height of segment = h
G
h
C
r
98
α
1
C = 2 h(2r − h) A = [rl −C(r − h)]
2
@seismicisolation
@seismicisolation
Figure 7.9
Centre of Gravity
j) Ellipse:
Area = π ab
G
b
a
Figure 7.10
Centroid is where major axis and minor axis intersect. Ȳ = b
k) Parabola:
x
4
Area = xy
3
3x
X̄ =
5
y
x
Figure 7.11a
For one half of parabola:
Area =
2xy
3
X̄ =
3x
5
and Ȳ =
3y
8
G
Y
X
x
Figure 7.11b
l) Semicircular arc:
Ȳ =
2r
π
y
G
r
Figure 7.12
Centre of Gravity for Solid Bodies
a) Cylinder:
G
h
2
Volume = π r h
h
ȳ =
r
2
x̄ = 0 (since cylinder is symmetrical about y axis). Figure 7.13
2
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y
h
•
99
100
•
Strength of Materials
b) Rod:
π d2
V=
·l
4
l
ȳ = 0 (because of symmetry)
x̄ =
2
y
x
x
l
y
Figure 7.14
c) Hemisphere:
v=
1 4 3
× πr
2 3
y
3r
3r
from the base so ȳ =
8
8
x̄ = 0 (due to symmetry)
C.G. is at
G
3r
8
x
r
Figure 7.15
d) Sphere:
4
V = π r3 for solid sphere
3
4
= π (r13 − r23 ) for hollow sphere
3
r2
r1
C.G. is at centre.
Figure 7.16
e) Qudrent of a sphere:
3
x̄ = r,
8
Y
3
ȳ = r
8
G
3
8
r
3
8
X
r
Figure 7.17
f) Solid cone:
1
V = π r2 h
3
h
from the base
Ȳ =
4
G
Figure 7.18
@seismicisolation
@seismicisolation
Y
x
Centre of Gravity
g) Hollow cone:
h
Ȳ =
3
•
101
from the base
G
Y
Figure 7.19
Centroid of Different Sections
E XAMPLE 7.1: Determine the centroid of 120 mm × 180 mm × 30 mm T section.
120 mm
30
1
180
2
A
y
B
30 mm
Figure 7.20
S OLUTION :
In such cases divide the section into two different smaller sections, e.g., 1 and 2. Since it is
symmetrical about Ȳ axis so X̄ = 0. a1 = 120 × 30 = 3600 mm2 y1 = 165, a2 = 150 × 30 =
150
= 75 mm
4500 mm2 , y2 =
2
A, total area = 3600 + 4500 = 8100 mm2
Aȳ = a1 y1 + a2 y2
a1 y1 + a2 y2
A
3600 × 165 + 4500 × 75
= 115 mm
=
8100
E XAMPLE 7.2: Determine centroid of plane Fig. 7.21.
S OLUTION :
∴
ȳ =
C
25 mm
1
300 mm
x G
y
25 mm
2
A
B
200 mm
Figure 7.21
@seismicisolation
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from AB Ans
102
•
Strength of Materials
Divide the sections into two rectangular strips 1 and 2 ., Now
a1 = 275 × 25 = 6875, y1 = 162.5, x1 = 12.5, a2 = 200 × 25 = 5000 mm2 , x2 = 100 mm, y2 =
12.5 mm A = a1 + a2 = 6875 + 5000 = 11875 mm2
a1 y1 + a2 y2
A
6875 × 162.5 + 5000 × 12.5
Ȳ =
11875
1117187.5 + 62500
=
11875
Ȳ =
= 99.34 mm
X̄ =
(from AB)
6875 × 12.5 + 5000 × 100
11875
= 49.34 mm
Ans
E XAMPLE 7.3: Determine the centroid of plane Fig. 7.22.
S OLUTION :
Let us divide the section into simple rectangles 1, 2 and 3.
x̄ = 0 because of symmetry
120 mm
1
30 mm
2
a1 = 200 × 30 = 6000 mm2
y1 = 15 + 250 + 40 = 305 mm
250 mm
G
30 mm
y
a2 = 250 × 30 = 7500 mm2 ,
250
y2 =
+ 40 = 165 mm
2
3
A
350 mm
40
= 20 mm
Figure 7.22
2
A = a1 + a2 + a3 = 6000 + 7500 + 14000 = 27500 mm2
a1 y1 + a2 y2 + a3 6000 × 305 + 7500 × 165 + 14000 × 20
ȳ =
=
A
27500
ȳ = 121.73 mm from AB Ans
a3 = 350 × 40 = 14000 mm2 ,
y3 =
E XAMPLE 7.4: Find the centroid of section shown in Fig. 7.23
y1
140 mm
3
60 mm
1
2
x
200 mm
y
Figure 7.23
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40 mm
B
60 mm
55 mm
x
Centre of Gravity
•
103
S OLUTION :
55
π 552
= 1187.31 mm2 y1 =
= 27.5 mm
4×2
2
4r
4 × 27.5
x1 = 27.5 −
= 27.5 −
x1 = 15.82 mm
3π
3 × 3.14
a2 = 200 × 55 = 11000 mm2 , y2 = 27.5 mm, x2 = 100 + 27.5 = 127.5 mm
60 × 60
60
= 1800 mm2 , y3 =
+ 55 = 75 mm, x3 = 30 + 140 + 27.5, x3 = 197.5 mm
a3 =
2
3
A = a1 + a2 + a3 = 1187.31 + 11000 + 1800 = 13987.31
a1 x1 + a2 x2 + a3 x3 1187.31 × 15.82 + 11000 × 127.5 + 1800 × 197.5
x̄ =
=
A
13987.31
= 126.98 mm
a1 y1 + a2 y2 + a3 y3 1187.31 × 27.5 + 11000 × 27.5 + 1800 × 75
ȳ =
=
A
13987.31
ȳ = 33.613 mm Ans
a1 =
E XAMPLE 7.5: Determine the centroid of the shaded area shown in Fig. 7.24.
C
75 mm
75 mm
D
180 mm
m
0m
10
A
150 mm
B
150 mm
Figure 7.24
The figure is symmetrical about y-axis so x = 0. Take whole trapezium as 1 including whole
and take hole as 2 alone.
180
{(75 + 75) + (150 + 150)} = 60 (150 + 300) = 27000 mm2
2
180
300 + 2 × 150
y1 =
×
= 80 mm
3
300 + 150
a1 =
π r2 3.14 × 1002
=
= 15700 mm2
2
2
4r
4 × 100
y2 =
=
= 63.694 mm
3π
2 × 3.14
a2 =
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104
•
Strength of Materials
In this case
A = a1 − a2 = 27500 − 15700 = 11800 mm2
In this case
ȳ =
=
a1 y1 − a2 y2
A
27500 × 80 − 15700 × 63.694
= 101.695 mm from AB
11800
Ans.
E XAMPLE 7.6: A frustum of a solid right circular cone has from bottom an axial hole of 60 cm
diameter as shown in Fig. 7.25. Find the centre of gravity of the body.
Figure 7.25
S OLUTION :
The frustum is symmetrical about vertical axis. Therefore, its centre of gravity lies on this axis.
A cross-sectional view is shown in Fig. 7.26 of the frustum.
Due to similar triangle: i) Cone OAB.
a
x
0
=
1.7 0.8
a = 1.6
a
1.8 + a
0.8 m
=
1.7
0.8
C
D
x = 3.4 m
(1.8 + a) 0.8 = 1.7a
1.8 m
1.44 + 0.8a = 1.7a
1.44 = 0.9a
A
1.44
B
0.6
∴ a=
= 1.6 m
left AB be the
1.7 m
0.9
reference
Figure 7.26
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Centre of Gravity
•
105
Taking AB as reference,
i) OAB
π 2
π 1.7 2
× 3.4 = 2.5711 m3
R H=
3
3
2
3.4
= 0.085 m
y1 =
4
ii) Right circular cone OCD
π
π
V2 = r2 h = (0.4)2 × 1.6 = 0.268 m3
3
3
1.6
y2 =
+ 1.8 = 2.2 m
4
iii) Circular hole
π
1.8
= 0.9, V3 = (0.6)2 × 1.8
y3 =
2
4
3
= 0.5087 m ;
V1 =
V = 2.5711 − 0.268 − 5087 = 1.7944 m3
V1 y1 +V2 y2 −V3 y3
Ȳ =
V
2.5711 × 0.85 + 0.268 × 2.2 − 0.5087 × 0.9
=
1.7944
2.185 − 0.5896 − 0.45783
=
1.7944
= 0.634 m from AB Ans
Total volume,
E XAMPLE 7.7: A hemisphere and a cone have their bases joined together, the two being of same
size of dia 300 mm. If the height of cone is 500 mm, find the centre of gravity from the bottom of
hemisphere (Refer Fig. 7.27).
D
500 mm
1
A
B
2
C
300 mm
Figure 7.27
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150 mm
106
•
Strength of Materials
Let cone be 1 and hemisphere be 2.
Let C.G. be calculated from bottom of hemisphere C.
1. Cone: Volume,
π 2
π
R H = (150)2 × 500 = 11775000 mm3
3
3
500
y1 =
+ 150 = 275 mm
4
V1 =
2. Hemisphere: Volume,
V2 =
1 4 3 2
× π R = × 3.14 × (150)3 = 7064929, mm3
2 3
3
y2 = 150 −
3 × 150
= 93.75 mm,
8
Total volume = 11775000 + 7064929 = 18839929
ȳ =
V1 y1 +V2 y2
V
=
11775000 × 275 + 7064929 × 93.75
18839929
=
3238125000 + 662337094
18893329
= 207.03 mm
from C Ans
Exercise
7.1 A circular hole of 60 mm diameter is cut out from a circular disc of 120 mm diameter as
shown in Fig. 7.28. Find the centre of gravity of the section from A.
120 mm
A
B
60 mm
[Ans: 50 mm from A]
Figure 7.28
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Centre of Gravity
•
107
7.2 Find the centre of gravity from AB of channel section shown in Fig. 7.29.
100
20
300 mm
20 mm
20
[Ans: 27.39 mm from AB]
Figure 7.29
7.3 Find the centroid of area of figure shown in Fig. 7.30.
A
100 mm
B
20 mm
G
H
180 mm
C
D
20 mm
F
120 mm
E
Figure 7.30
[Ans: x̄ = 106.74 from DE
ȳ = 95.26 from FE]
7.4 Determine the centroid of T section as shown in Fig. 7.31.
160 mm
10 mm
150 mm
10 mm
[Ans: ȳ = 43.71 mm]
Figure 7.31
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108
•
Strength of Materials
7.5 Determine the centroid of figure shown below:
100 mm
30
15
70
20 mm
20
50 mm
[Ans: 44.2 mm from top surface]
Figure 7.32
7.6 A cylinder with a hemispherical cavity and a conical cap is shown in Fig. 7.33. All dimensions
are in mm. Determine the centre of gravity of composite volume if the cylinder is made of
steel and conical cap is made of aluminium. Assume density of steel = 7870 kg/m3 and
density of aluminium = 2770 kg/m3 .
A
420 mm
C
500 mm
B
600 mm
[Ans:
Figure 7.33
371.13 mm from AB]
7.7 A square hole is punched out of a circular lamina as shown in Fig. 7.34. The diagonal of the
square which is punched out is equal to the radius of circle. Find the centroid of remaining
laminar.
Y
X
X
a
Y
[Ans: x̄ = 0.095 a from y axis]
Figure 7.34
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Centre of Gravity
•
109
7.8 Find the centroid of the cross-sectional area of a Z section as shown in Fig. 7.35.
Q
10 cm
B
2.5 cm
E
2.5 cm
20 cm
5 cm
A
B
20 cm
Figure 7.35
[Ans: x̄ = 13.557 cm from GE
ȳ = 7.69 cm from AB]
7.9 Locate the centroid of a shaded area in Fig. 7.36.
Y
R2 = 6 cm
R2
R1 = 2 cm
R1
X
3 cm
3 cm
[Ans: 2.417 cm, 3.03 cm]
Figure 7.36
7.10 Determine the centroid of the cross-sectional area of an I section shown in Fig. 7.37.
Y
20 cm
5 cm
15 cm
[Ans:
X
30 cm
Y
5 cm
X
Figure 7.37
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ȳ = 10.96 cm]
110
•
Strength of Materials
7.11 A hemisphere of radius 40 mm is cut out from a right circular cylinder of diameter 80 mm
and height 160 mm as shown in Fig. 7.38. Find the centre of gravity of the body from the
base AB.
160 mm
A
B
80 mm
[Ans: 77.2 mm]
Figure 7.38
7.12 Determine the centre of gravity of the component shown in Fig. 7.39.
4 cm dia
2.5 cm
A
6 cm dia
3.5 cm
B
[Ans 1.517 cm from bottom]
Figure 7.39
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C HAPTER
8
MOMENT OF INERTIA
Measurement using moment of inertia with respect to an axis is an important property, which gives
a measure of resistance to bending in the case of thin plates or plane areas.
Y
x
dA
y
X
0
Figure 8.1
Let dA be a small elementary area of plane figure shown in Fig. 8.1.
Now elementary moment of inertia of dA about X axis will be dA.y × y = dA y2 .
Then total moment of inertia of whole area will be:
IXX = ∑ dA y2
Similarly, IYY = ∑ dA x2
Rectangular Section
D
Moment of inertia (second moment of area) of
elementary strip dy = b.dy.y2
Y
dy
C
h
2
Y
X
h
h
2
A
Y
b
B
Figure 8.2
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X
112
•
Strength of Materials
+h/2
by2 dy
Total moment of inertia =
−h/2
=b
+h/2
y3
3
−h/2
⎡
⎢
= b⎣
h
2
3
3
−
− h2
3
3
⎤
⎥
⎦
bh3
12
hb3
=
12
=
Similarly,
IYY
Radius of Gyration
The second moment of area has dimensions of length to the fourth power, it can be expressed as the
product of area ‘A’ and length ‘k’ squared.
Thus I = Ak2
k is known as radius of gyration.
Product of Inertia:
Let x and y be the distances from y-axis and x-axis of an elemental area dA.
Y
x
dA
y
X
Figure 8.3
The moment of this area about x-axis is dA.y. And the moment of dAy about y-axis is xydA.
This term xydA is called the product of inertia of area dA with respect to x-axis and y-axis. And
xydA is called the product inertia of the entire area, denoted as IXY .
IXY =
xydA
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Moment of Inertia
•
113
Theorem of Parallel Axis
It states that if the moment of inertia of a plane over an axis in the same plane through the centre
of gravity of the area be represented as IG , then the moment of inertia of a given plane area about
parallel axis AB in the plane of area at a distance h from C.G. of the area is given by
Plane Area A
IAB = IG + Ah2
G
x
x
h
A
B
Figure 8.4
Theorem of the Perpendicular Axis
It states that if IXX and IYY are the moment of inertia of a plane section about two mutually perpendicular axis X − X and Y −Y in the planes of the section, then the moment of inertia of the section
IZZ perpendicular to the plane and passing through the intersection of X − X and Y −Y is given by:
Z
IZZ = IXX + IYY
x
0
X
y
dA
Y
Figure 8.5
Moment of inertia of the rectangular section about a line passing through the base
b
A
bd 3
IGG =
12
Applying the parallel axis theorem,
IDC = IGG + Ah2
2
d
bd 3
bd 3 bd 3
+ bd
+
IDC =
=
12
2
12
4
Y
G
X
d
D
bd 3
IDC =
3
Y
Figure 8.6
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B
X
C
114
•
Strength of Materials
b
Moment of Inertia of a Hollow Rectangular Section
b1
IXX =
bd 3 b1 d13
−
12
12
d
d1
X
X
Figure 8.7
Moment of Inertia of a Circular Section
Consider an elementary circular ring of radius r
and thickness dr.
Area of circular ring = 2π rdr
Let us first find the moment of inertia about an
axis passing through O (polar axis), perpendicular to the plane of the paper.
Figure 8.8
Moment of inertia of the circular ring about an axis passing through O (polar axis),
= (Area of ring) × (Radius of ring)2
= (2π rdr) × r2 = 2π r3 dr
R
2π r3 dr
Total moment of inertia =
0
=
=
R
2π r4
4
π d4
32
=
0
π R4
2
=
where d = 2r
π
d
2
4
2
(i)
Now according to the perpendicular axis theorem,
IZZ = IXX + IYY
In circular section,
IXX = IYY
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(ii)
•
Moment of Inertia
115
Substituting from (i) & (ii)
π d4
= 2IXX
32
π d4
Also IYY =
64
∴
π d4
64
IXX =
where, d = diameter = 2r
The moment of inertia of a hollow circular section
Y
Z
π D4 π d 4
Polar moment of inertia =
−
32
32
π 4
= (D − d 4 )
32
π D4 π d 4
−
IYY = IXX =
64
64
4
πd
(D4 − d 4 )
=
64
X
d
X
D
Z
Y
Figure 8.9
Moment of Inertia of a Triangle
A
a) About its base:
Consider an elemental strip at y from A, of
thickness dy.
y
dy
Area of strip = DE · dy
∴
∴
B
y
DE
=
BC
h
h
C.G.
h
3
b
X
Now triangles ADE and ABC are similar.
E
D
Figure 8.10
y by
∴ DE = BC · =
h
h
by
Area of stripe =
· dy
h
Therefore, moment of inertia of the stripe about the base BC,
= Area of stripe × (h − y)2 =
by
· dy(h − y)2
h
Moment of inertial of whole triangle about the base BC,
IBC =
h
by
0
h
(h − y)2 dy =
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b
h
h
0
y(h − y)2 dy
X
C
116
•
Strength of Materials
b h 2
y(h + y2 − 2hy) dy
h 0
b h 2
=
(yh + y3 − 2hy2 ) dy
h 0
h
b y2 h2 y4 2hy3
+ −
=
h
2
4
3 0
4
4
4
b h
h
2h
=
+ −
h 2
4
3
=
Hence,
IBC =
bh3
12
b) About an axis passing through C.G. and parallel to the base:
Using parallel axis theorem,
IBC = IG + A
2
h
3
Ah2 bh3 bh h2
=
−
×
9
12
2
9
bh3 bh3
=
−
12
18
bh3
IGG =
36
or IG = IBC −
Hence,
Moment of Inertia of Semi-circular Lamina about its Centroidal Axis
IG = 0.00686d 4
X
R
G
X
4R
3π
d
Figure 8.11a
Moment of inertia of a quarter of a circle:
IXX = 0.00343d 4
X
G
R
d
2
Figure 8.11b
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X
4R
3π
Moment of Inertia
•
117
E XAMPLE 8.1: Find the IXX for the figure shown:
S OLUTION :
100 mm
10 mm
15 mm
130 mm
X
X
α=
100–15 = 42.5 mm
2
α
10 mm
100 mm
Figure 8.12
In this case we can find IX by subtracting IX of hollow figure in to IX of whole figure (100 ×
130 mm).
100 × 1303
42.5 × 1103
Required IX =
−
×2
12
12
= 18308333.3 − 9427916.7
= 8880416.6 mm4
Ans
E XAMPLE 8.2: Find the second moment of area and the radius of gyration about axis XX passing
through centroid shown in figure. (The section is symmetrical about Y -axis.)
S OLUTION :
80
20 mm
I
20
2
X
C
X
120 mm
Y
3
A
30 mm
B
160 mm
Figure 8.13
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118
•
Strength of Materials
S OLUTION :
To find centre of gravity, divide the section in the three simple sections.
a1 = 80 × 20 = 1600 mm2 ,
y1 = 160 mm
a2 = 120 × 20 = 2400 mm ,
y2 = 90 mm
a3 = 160 × 30 = 4800 mm2 ,
y3 = 15 mm
2
A = a1 + a2 + a3 = 1600 + 2400 + 4800 = 8800 mm2
1600 × 160 + 2400 × 90 + 4800 × 15
ȳ =
8800
= 61.82 mm, Now using the parallel axis theorem,
80 × 203
20 × 1203
+ 1600 (160 − 61.82)2 +
IXX =
12
12
160 × 303
+ 4800 (15 − 61.82)2
+ 2400 (90 − 61.82) +
12
= [53333.3 + 15422899.8 + 2880000 + 67632
+ 360000 + 10522139.5]
2
= 29306004.6 mm4
Now I = Ak
Ans
2
where k is radius of gyration
I
29306004.6
=
k=
A
8800
= 57.71 mm Ans
∴
E XAMPLE 8.3: Find the IX and IY of the Fig. 8.14 shown below:
5 mm
80 mm
X
Let us take x & y from AB & AC.
Y
C
x
G
X
y
1
2
A
5 mm
B
45 mm
Y
Figure 8.14
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Moment of Inertia
•
119
Let us divide the area in to two, i.e., 1 & 2
a1 = 80 × 5 = 400
a2 = 40 × 5 = 200
x1 = 2.5, y1 = 40
x2 = 25, y2 = 2.5
Total area = 400 + 200 = 600 mm2
400 × 2.5 + 200 × 2.5
= 10 mm
x̄ =
600
400 × 40 + 200 × 2.5
ȳ =
= 27.5 mm
600
Using parallel axis theorem, IX = IG + Ah2
5 × 803
40 × 53
IX =
+ 400 (40 − 27.5)2 +
+ 200 (2.5 − 27.5)2
12
12
= [213333.3 + 62500 + 416.67 + 125000]
= 401249.97 = 401250 mm4 Ans
80 × 53
5 × 403
+ 400 (10 − 2.5)2 +
+ 200 (10 − 25)2
IY =
12
12
= [833.33 + 22500 + 26666.7 + 45000]
= 950000 mm4
Ans
E XAMPLE 8.4: Determine the moment of inertia of a plate with a circular hole (as shown in Fig.
8.15) about X-axis passing through its centroid.
Y
1
G
X
2
50
mm
y
40 mm
80 mm
Y
Figure 8.15
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X
180 mm
120
•
Strength of Materials
Let total area including hole is 1 and hole alone is 2
a1 = 180 × 80 = 14400,
a2 =
π 2
50 = 1962.5,
4
y1 = 90
y2 = 40
A = 14400 − 1962.5 = 12437.5 mm2
ȳ =
14400 × 90 − 1962.5 × 40
= 97.89 mm
12437.5
80 × 1803
I¯X1 =
= 38880000 mm4
12
Moment of inertia of the area a1 about the centroidal x-axis of the shaded figure, G.
IX = (I¯X )1 + a1 d 2
= 38880000 + 14400 (90 − 97.89)2
IX = 38880000 + 896430.24
= 39776430 mm4
π
Moment of inertia of the area a2 about its own G, I¯X2 = (50)4
64
I¯X 2 = 306640.6
Moment of inertia of the area a2 about the centroidal axis
(Ix )2 = (I¯X )2 + a2 d 2
= 306640.6 + 1962.5(97.89 − 40)2
= 306640.6 + 6576832.2
= 6883472.8
Moment of inertia of composite (shaded) figure,
Ix = I X − (I X )2 = 39776430 − 6883472.8
= 32892957.2 mm4
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Ans
Moment of Inertia
E XAMPLE 8.5: Determine IX and IY of the cross section of iron beam shown in Fig. 8.16.
Y
12 cm
A
5 cm
X
G
G′
X
B
15 cm
Y
Figure 8.16
IX = IX of rectangular 12 × 15 − IX of circle of radius 5 cm
12 × 15 π
=
− (10)4
12
64
= 3375 − 490.87 = 2884.13 cm4 Ans
Similarly,
Iy = Iy of rectangle − Iy of semicircular portions
Iy of rectangle =
15 × 123
= 2160 cm4
12
For the semicircle part ACB,
IAB =
1 π
× (10)4 = 245.3 cm4
2 64
4r
4×5
=
= 2.12 cm
3π
3π
1
1
Area A = π r2 = π × 52 = 39.27 cm2
2
2
Using IAB = IGG + Ah2
Distance of G of semicircle from AB =
IGG = 245.3 − 39.27 (2.12)2 = 68.8 cm4
Again from theorem, IYY = IGG + Ah21
h1 = 6 − 2.12 = 3.88 cm
∴
IYY = 68.8 + 39.27 × 3.882 = 659.98 cm4 = 660 cm4 app.
Because there are two semicircular cutouts,
∴
IYY for both = 2 × 660 = 1320
IY for section = 2160 − 1320 = 840
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Ans
•
121
122
•
Strength of Materials
Product of Inertia of Rectangle
Y
b
Consider on elementary area dA = dx.dy
dy
dx
Product of inertia of the element = xydA = xy.dx.dy
x
y
For whole rectangle (b × h), the product of
inertia,
h
b h
=
X
0
Figure 8.17
b
xydxdy =
0
2 b
x
=
2
y2
×
2
0
0
h
=
0
xdx ×
b
y. dy
0
b2 h2 b2 h2
×
=
2
2
4
Note: If a section in symmetrical about an axis, its product of inertia about that axis is zero.
E XAMPLE 8.6: Find the product of inertia of the hatched area about xy shown in Fig 8.18.
60 mm
15 mm
55 mm
X
X
With respect to x and y-axes, product of inertia
b2 h2
of rectangle in
4
Now, product of inertia of the hatched area =
Product of inertia of rectangle
60 × 55 − Product of inertia of rectangle 45 × 40
602 × 552 452 × 402
−
= 2722500 − 810000
4
4
= 1912500 mm4 Ans
15 mm
=
Figure 8.18
Exercise
8.1 Determine the moment of inertia about a target X − X of semicircular of dia 4 cm as shown
in Fig. 8.19.
Y
X
X
o
4 cm
[Ans: 10.15 cm4 ]
Figure 8.19
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Moment of Inertia
•
123
8.2 Find the moment of inertia of the section shown in Fig. 8.20, about the centroidal axis X − X
perpendicular to web.
10 cm
2 cm
2 cm
10 cm
2 cm
20 cm
[Ans: 2666.67 cm4 ]
Figure 8.20
8.3 Find the moment of the area shown in Fig. 8.21 about edge AB.
R=10
cm
25 cm
A
B
20 mm
[Ans: 35359.7 cm4 ]
Figure 8.21
8.4 Calculate the second moment of area of an equal sized trapezium as shown in Fig. 8.22 about
the centroidal axis parallel to the base.
C
40 mm
D
30 mm
A
80 mm
B
[Ans: 130909.1 cm4 ]
Figure 8.22
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•
124
Strength of Materials
8.5 Calculate second moment of area for the Fig. 8.23 about the centroidal axis parallel to the
AB.
100 mm
30 mm
15 mm
70 mm
20 mm
A
B
50 mm
[Ans: 802.885 × 104 mm4 ]
Figure 8.23
8.6 Find the second moment of area of Fig. 8.24 about centroidal axis.
12 cm
5 cm
5 cm
20 cm
5 cm
24 cm
[Ans: 70086 cm4 ]
Figure 8.24
8.7 Determine moment of inertia IXX and IYY of section shown in Fig. 8.25.
4 cm
2 cm
G
3
8 cm
3 cm
2 cm
[Ans: 448 cm4 , 58 cm4 ]
4 cm
Figure 8.25
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C HAPTER
9
BENDING OF BEAMS
The shearing forces and bending moments at the various sections of a loaded beam give rise to
shearing and bending stresses in the beam and, by making a number of assumptions, the relationship
existing between the stresses, the strains, the bending moment, the curvature as well as the elasticity
of the beam will be established in the following sections of this chapter.
If a beam ABCD of uniform section is subjected to bending moment M the beam will bend
with radius R, subtending an angle θ at sector P OQ . After bending the beam, has taken shape as
A BC D and centroidal axis PQ as P Q . Now upper longitudinal layer is subjected to compression and has become A B . The lowermost layer DC has become DC and is in tension. Since
at top it is compression and at bottom it is tension, so obviously at centroidal axis, there will
be no stress and is called neutral axis. Here P Q is neutral axis and layers of fibres here suffer
no stress or strain due to bending. Consequently, its original length PQ remain, unchanged, i.e.,
PQ = P Q .
Before we derive any relationship between stresses, we have to make a number of assumptions
which are given below:
i)
ii)
iii)
iv)
v)
The material obeys Hooke’s law and is within elastic limit.
The beam is initially straight and unstressed.
The material is homogeneous and isotropic.
The beam bends in the form of a circular arc.
The value of modulus of elasticity E for the material of the beam has the same value in
compression as in tension.
vi) Plane transverse sections of the beam before bending remain plane after bending.
Relationship Between Curvature and Strain
After application of bending moment M, the radius of curvature R of the bent beam is measured
from the centre O to the neutral surface. If θ is the angle is radius subtended by the arc A B at
centre O, then since the neutral surface remains unchanged in length then,
Line PQ = Arc P Q
P Q = Rθ
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126
•
Strength of Materials
Now let us consider the bottom layer of the beam distance y from the neutral surface.
Length of DC before bending = PQ = P Q = Rθ
Length of DC after bending = DC = (R + y)θ
∴
Extension of DC = (R + y)θ − Rθ
= yθ
yθ
Extension
=
Original length Rθ
y
ε=
R
Strain in DC, ε =
∴
Also, strain in DC, ε =
σ
E
∴
y σ
E
σ
= ;
=
E
R y
R
(i)
Since, Young’s modulus E is constant for the material of the beam, and R is also constant for
the particular curvature considered, the stress σ varies across the depth of the beam from zero at the
neutral axis XX to a maximum value at the outer layers of fibres. Since, we are only considering
sections that are symmetrical about the neutral axis XX, the distance y from the neutral axis XX
to the outer layers of fibres is equal to half the overall depth of the beam. The stress distribution is
drawn in Fig. 9.1 where compressive stresses are plotted to the left of axis Y −Y , and tensile stresses
to the right.
Figure 9.1
Moment of Resistance
The moment of resistance of a beam is the resultant moment about the neutral axis of the internal
tensile and compressive forces resisting the applied bending moment M.
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Bending of Beams
•
127
Referring again to Fig. 9.1, let us consider an elemental strip of the cross section of the beam at
a distance x from the neutral axis XX and of thickness dx.
If σ1 is the stress at the strip then, since the stresses at the various layers of the beam section are
proportional to their distances from the neutral axis XX.
σ1 x
= ;
σ
y
x
σ1 = σ
y
∴
σ is the stress at y.
x
Force (F) on strip = Stress × Area = σ × b dx
y
Moment of this force about the neutral axis XX
= Force × Distance from the neutral axis XX
x
σ
= σ × b dx x = bx2 dx
y
y
E 2
bx dx
R
The total moment of resistance is the sum of the moments of all the elemental stripes above and
below the neutral axis (NA), XX, i.e., from x = y = d/2 to x = y = −d/2 and is given by:
From Eqn. (i), =
+d/2
M=
−d/2
M=
i.e., M =
The quantity
E 2
bx dx
R
E bx
R 3
∴
E
M=
R
+d/2
bx2 dx
−d/2
3 +d/2
−d/2
E bd 3
×
R
12
bd 3
is called the second moment of area of the section about NA and is denoted by I.
12
∴
E
×I
R
M E
=
I
R
M=
(ii)
Combining Eqn. (i) and (ii)
M E
σ
= =
I
R Y
This is known as the simple bending equation.
Note: If we had taken any other section, for example, circular, the result would have been same.
Only I has to be determined for a particular section.
@seismicisolation
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128
•
Strength of Materials
Modulus of Section
M σ
=
I
y
I
or M = σ
y
Since y represents the distance of the outer layer of fibres from NA, the ratio of the second moment
of area (moment of inertia) I to y is called modulus of section and is denoted by τ .
Thus, Z =
I
y
∴
M = Zσ
(iii)
Equation (iii) is very important since it enables us to determine readily the permissible or maximum
bending moment on a beam.
Since the beam of Fig. 9.1 is of rectangular section.
∴
bd 3
12
d
2
bd 3 2 bd 2
I
× =
Modulus of section, Z = =
Y
12
d
6
I=
y=
and
Note: EI is known as flexural rigidity.
E XAMPLE 9.1: A mild steel bar of 35 mm × 15 mm section and 1.2 m length is simply supported
at its ends with the 35 mm edge horizontal. If a load of 100 N is applied at the centre of the bar,
determine the maximum stress, produced in the material due to bending.
S OLUTION :
100 N
0.6 m
15 mm
A
35 mm
50 N
B
1.2 m
50 N
Figure 9.2
bd 3 35 × 153
=
= 9843.75 mm4
12
12
100
Reactions, RA = RB =
= 50 N
2
I=
M = 50 × 0.6 = 30 Nm = 30000 Nmm
@seismicisolation
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Bending of Beams
•
129
Maximum stress will occur at 7.5 mm from NA.
y = 7.5 mm
Now,
M
M σ
=
∴ σ= y
I
y
I
30000
σ=
× 7.5 = 22.86 N/mm2 = 22.86 MN/m2
9843.75
E XAMPLE 9.2: A tubular frame member of 85 mm external diameter is subjected to a bending
moment of 600 Nm. If the stress set up in the material due to bending is 50 MN/m2 , find the
internal diameter of the member.
S OLUTION :
Let D and d be the external and internal diameters of the tubular frame.
Then,
π (D4 − d 4 )
,
σ = 50 MN/m2 = 50 N/mm2
64
M σ
85
= ;
Here y =
= 42.5
I
y
2
I=
M = 600 Nm = 600000 Nmm,
I=
π 2 (854 − d 4 ) π (52200625 − d 4 )
=
64
64
Substituting
600000 × 64
50
=
π (52200625 − d 4 ) 42.5
π (52200625 − d 4 ) × 50 = 600000 × 64 × 42.5
52200625 − d 4 =
600000 × 64 × 42.5
π × 50
= 10385454.5
or d 4 = 52200625 − 10385454.5
∴
∴
d 4 = 41815170.5
d = 80.41 mm. Ans
E XAMPLE 9.3: A beam, the cross-section of which is shown in Fig. 9.3, acts as a cantilever, which
projects 1.8 m from the wall. The cantilever carries a load of 5 kN at the free end. Calculate the
maximum bending stress.
@seismicisolation
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•
130
Strength of Materials
75 mm
10 mm
10 mm
85 mm
10 mm
Figure 9.3
S OLUTION :
Since the beam cross section is symmetrical about X and Y axis, therefore I can be obtained as
follows:
5 kN
75 × (85 + 20)3 (75 − 10)(85)3
−
12
12
75 × 1157625 65 × 614125
−
=
12
12
I=
1.8 m
Figure 9.4
= 7235156.25 − 3326510.42
I = 3908645.83 mm4
M = 5 × 1000 × 1.8 × 1000 = 9000000 mm4
y = 10 +
Now,
85
= 52.5
2
M σ
=
I
y
∴
σ=
M
y
I
σ=
9000000
× 52.5 = 120.89 N/mm2
3908645.83
or σ = 120.89 MN/m2 . Ans
E XAMPLE 9.4: A spring steel strip, 25 mm wide and 1.5 mm thick, is bent to an arc of a circle
of 2 m radius. Calculate the bending moment necessary and the maximum stress set up. E for steel
= 200 GN/m2 .
S OLUTION :
b = 25 mm;
∴
I=
d = 1.5 mm
bd 3 25 × 1.53
7.03
=
= 7.03 mm4 = 12 m4
12
12
10
@seismicisolation
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Bending of Beams
Distance from N.A. to extreme layer of fibres } = y =
•
131
1.5
d
=
mm
2
2
0.75
m
103
EI
M E
=
OR M =
I
R
R
9
m4
200 × 10 × 7.03 N
×
M=
m
2 × 1012
m2
y = 0.75 mm =
Now
∴
Bending moment = 0.703 Nm.
For maximum stress
Ey 200 × 109 × 0.75 N
m
=
σ=
×
R
2 × 103
m2 m
σ = 75 × 106 N/m2 = 75 MN/m2 .
E XAMPLE 9.5: The beam cross section shown in Fig. 9.5 is subjected to a bending moment of
12 kNm. Calculate the maximum stress and its nature indicating its place.
110 mm
C
D
20 mm
20 mm
100 mm
25 mm
A
B
180 mm
Figure 9.5
110 mm
Y
C
20 mm
y
D
20 mm
1
G
2
100 mm
3
A
B
Y
180 mm
Figure 9.6
@seismicisolation
@seismicisolation
25 mm
132
•
Strength of Materials
We see that the section is symmetrical about Y −Y axis. Let the centre of gravity G be at Y from
AB. In order to find G, let us divide the section into three simple rectangles, i.e., 1, 2 & 3.
a1 = 110 × 20 = 2200 mm2 ; y1 = 135, a2 = 100 × 20 = 2000 mm2 ;
y2 = 75; a3 = 180 × 25 = 4500 mm2 ; y3 = 12.5 mm
Total area, A = 2200 + 2000 + 4500 = 8700 mm2
y=
∴
2200 × 135 + 2000 × 75 + 4500 × 12.5 503250
=
8700
8700
y = 57.84 mm
Now to find I of the whole section, using parallel axis theorem:
110 × 203
2
+ 2200 (135 − 57.84)
I=
12
20 × 1003
2
+ 2000 (75 − 57.84)
+
12
180 × 253
2
+
+ 4500 (12.5 − 57.84)
12
= [73333.3 + 13098064.3] + [1666666.7 + 588931.2]
+ [234375 + 15054595.2]
∴
I = 30715965.7 mm4
M σ
=
I
y
yt = 57.84, yc = (100 + 25 + 20) − 57.84 = 87.16 mm
t for tension, c for compression
M
y
I
12000000
× 57.84 = 22.596 N/mm2
σt =
30715965.7
σ=
= 22.6 MN/m2 (tensile at AB)
σc =
12000000
× 87.16 = 34.05 N/mm2
30715965.7
= 34 MN/m2 (compressive at CD)
Hence, maximum stress is 34 MN/m2 at CD of compressive nature. Ans
@seismicisolation
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Bending of Beams
•
133
E XAMPLE 9.6: The horizontal beam of channel cross section as shown in Fig. 9.7 is 2.8 m long
and is simply supported at the ends. Determine the maximum uniformly distributed load it can carry
if the tensile and compressive stresses must not exceed 35 and 52 N/mm2 , respectively. Neglect the
weight of channel.
2
1
y1
G
110 mm
20
20 mm
3
y
y2
20 mm
110 mm
Figure 9.7
In order to find position of G, let us divide the channel in three simple rectangles. Being symmetrical about Y axis, we have to only find y.
a1 = 90 × 20 = 1800; y1 = 65 mm; a2 = 1800 mm2 ,
y2 = 65 mm, a3 = 150 × 20 = 3000 mm2 ; y3 = 10 mm.
Total area, A = 1800 + 1800 + 3000 = 6600 mm2
y=
1800 × 65 + 1800 × 65 + 3000 × 10
= 40 mm
6600
y = Y2 = 40 mm, y1 = 110 − 40 = 70 mm
20 × 903
2
I=
+ 1800 (65 − 40)
12
150 × 203
2
+ 3000 (10 − 40)
×2+
12
= [1215000 + 1125000] × 2 + [100000 + 2700000]
= 4680000 + 2800000 = 7480000 mm4
σ=
M
y
I
M = 1.4w ×
1.4
2
= 0.98w Nm
w/m
2.8 m
1.4 w
1.4 w
Figure 9.8
@seismicisolation
@seismicisolation
134
•
Strength of Materials
Since y1 > y2 , so stress will be greater at y1 of
52 =
0.98w × 1000
× 70 . . . (i)
7480000
∴
w = 5669.97 N/m
Now taking lower side, i.e., y2 = 40
35 =
0.98w × 1000
× 40 . . . (ii)
7480000
∴
w = 6678.6 N/m
Obviously, if we choose 6.678 kN/m uniformly distributed load then stress in equation (i) will
exceed 52 N/mm2 which is not permissible.
Hence, safe uniformly distributed load (w) = 5.67 kN/m Ans.
E XAMPLE 9.7: A rectangular beam is to be cut from a circular log of wood of diameter D. Find
the dimensions of the strongest section in bending.
S OLUTION :
M σ
=
I
y
or M = σ
I
= σZ
y
Let the strongest section cut out of the
circular log of diameter D be of width b and
depth d.
The section modulus for it is:
Z=
Also b2 + d 2 = D2
d
D
bd 2
6
or
b
Figure 9.9
d 2 = D2 − b2
Z=
b (D2 − b2 ) b D2 − b3
=
6
6
For strongest section Z should be maximum. Therefore,
dZ
= 0.
db
1 2
(D − 3b2 ) = 0
6
D2 − 3b2 = 0
b2 =
But
3b2 = D2
or
D2
3
or
D
b= √
3
d 2 = D2 − b2
= D2 −
@seismicisolation
@seismicisolation
D2 2D2
=
3
3
Bending of Beams
For strongest section d = D
2
3
•
135
Ans
E XAMPLE 9.8: A horizontal cantilever 3.2 m long is having rectangular cross section 65 mm wide
throughout its length. Its depth varies from 55 mm at free end to 200 mm at fixed end. A load of
6 kN at free end is applied, determine the position of maximum stress induced in the section. Also
find the value of maximum bending stress.
S OLUTION :
6 kN
X
65 mm
65 mm
x
200 mm
55 mm
200 mm
X
3.2 m
Fixed end
55 mm
Free end
Figure 9.10
Let us consider X − X section at a distance of x from the free end.
dx, depth at section X − X = 55 +
Sectional modulus Z =
145x 176000 + 145x
=
3200
3200
1
bdx2 65
= (176000 + 145x)2 ×
6
6
32002
Z = (176000 + 145x) × 1.058 × 10−6
σ=
=
M
6000x
=
Z
(176000 + 145x)2 × 1.058 × 10−6
6000 × 0.945 × 106 x
5670 × 106 x
=
2
(176000 + 145x)
(176000 + 145x)2
For maximum stress
dσ
d
=
dx
dx
=
5670 × 106 x
(176000 + 145x)2
5670 × 106 (176000 + 145x)2 + 2(176000 + 145x)145 × 5670 × 106 x
(176000 + 145x)4
@seismicisolation
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136
•
Strength of Materials
176000 + 145x = 290x
x = 1213.8 mm
The maximum stress will be at 1213.8 mm from the free end.
Ans
For maximum stress:
σ=
(Substituting for x) =
6000x
(176000 + 145x)2 × 1.058 × 10−6
6000 × 1213.8 × 0.945 × 106 6882246 × 106
=
(176000 + 145 × 1213.8)2
1.24 × 1011
= 55.5 N/mm2 = 55.5 MN/m2
Ans
Beams of Uniform Strength
In simply supported beams, carrying a uniformly distributed load, the maximum bending moment
will occur at its centre. As we go near the supports, the bending moment reduces untill it become
zero. Therefore, it is in interest to save the material near supports. Lot of material can be saved by
designing beams of uniform strength. Naturally, we will have to reduce the section gradually untill
supports. The section of a beam of uniform strength may be varied in the following ways:
a) By keeping the width uniform and varying depth
b) By keeping the depth uniform and varying width
c) By varying both width and depth.
Generally, uniform strength is maintained by keeping the width uniform and varying the depth.
E XAMPLE 9.9: A simply supported beam of 2.6 m span has a constant width of 120 mm throughout its length with varying depth of 160 mm at the centre to minimum at the ends as shown in
Figure 9.11. The beam is carrying a concentrated load P at its centre.
P
0.5 m X
A
C
dx
X
B
2.6 m
Figure 9.11
Find the minimum depth of the beam at a section 0.5 m from the left hand support, such that
the maximum bending stress at this section is equal to that at the mid-span of the beam.
@seismicisolation
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Bending of Beams
•
137
S OLUTION :
Width = 120 mm, span = 2.6 m = 2600 mm
depth at centre dc = 160 mm.
Let depth at 0.5 m from A be dx at X and σx be the bending stress at X.
P
Reaction RA = RB =
2
P
Bending moment at C, Mc = × 1300 = 650 P
2
P
and bending moment at X, Mx = × 500 = 250 P
2
Section modulus at centre of beam,
Zc =
b.dc2 120 (160)2
=
= 512000 mm3
6
6
Zx =
b.dx2 120 dx2
=
= 20 dx2 mm3
6
6
Bending moment at C (Mc ),
650P = σc × Zc = σc × 512000
σc =
650P
512000
(i)
Similarly, for stress at X,
Bending moment at X,
250P = σx × Zx = σx × 20 dx2
250P 12.5P
=
20 dx2
dx2
σy =
for uniform strength,
σc = σx
Equating (i) and (ii)
12.5P
650P
=
512000
dx2
dx2 =
∴
12.5 × 512000
= 9846
650
dx = 99.23 mm
Ans
@seismicisolation
@seismicisolation
(ii)
138
•
Strength of Materials
Composite Beams or Flitched Beams
A composite beam is one which consists of two or more materials rigidly fixed together throughout
their length. Generally, wooden beams are reinforced with steel plates to make them stronger. The
reinforcing material should have higher modulus of elasticity then the material of the beam. Some
examples are given in Figure 9.12.
Composite beam
Figure 9.12
If M1 and M2 are the parts of the applied bending moment M carried by the two materials, then
M1 + M2 = M
(i)
Here the radius of curvature of the two materials will be same as they are fixed together,
R1 = R2
E1 I1 E2 I2
=
M1
M2
or
M1 E1 I1
=
M2 E1 I2
(ii)
M1 and M2 can be determined from Eqns. (i) and (ii) and the stresses in the two materials are then
given by
M1
M2
σ1 =
and σ2 =
Z1
Z2
Alternately, the composite section may be replaced by an equivalent homogeneous section. Thus,
the section shown in Fig. 9.13(a) is equivalent to the section in Fig. 9.13(b) composed entirely of
E2
= n.
material (1) or to the section in Fig. 9.13(c) composed entirely of material (2). Where
E1
(2)
(1)
(2)
b
b
(a)
nb
(b)
b
n
b
(c)
Figure 9.13
In this method first I of the equivalent section is obtained. Then stresses may be obtained from
M
σ = y, these being the stresses which would exist in the homogeneous section. In section (b), the
I
@seismicisolation
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Bending of Beams
•
139
actual stresses in material (2) would be n times those in the equivalent section and in section (c), the
actual stresses in material (1) would be 1/n times those in the equivalent section.
In this analysis, it has been assumed that the compound sections are all symmetrical about the
plane of bending, otherwise twisting of the section would occur.
E XAMPLE 9.10: A flitched beam consists of a wooden joist 110 mm wide and 220 mm deep
reinforced by two steel plates 12 mm thick and 220 mm deep as shown in Fig. 9.14. If the maximum
stress in wooden joist is 10 N/mm2 , determine the corresponding stress induced in steel. Also
find the moment of resistance of composite section. E for steel = 200 GN/m2 and E for wood
= 12 GN/m2 .
S OLUTION :
Modular ratio m =
200
Es
=
= 16.7
Ew
12
220 mm
Maximum stress in wood = 10 N/mm2
Corresponding maximum stress in steel,
2
σs = mσw = 16.7 × 10 = 167 N/mm
12 mm
Ans
110 mm
12 mm
Figure 9.14
Total moment of resistance = Moment of resistance of wood
+ Moment of resistance by steel
=
σw Iw σs Is
+
y
y
Moment of inertia of wood = Iw =
110 × 2203
= 97606666.7 mm4
12
12 × 2203
= 21296000 mm4
12
σw
10
× Iw =
× 97606666.7
Moment of resistance by wood = Mw =
y
110
Moment of inertia of steel = Is = 2 ×
= 8873333.3 Nmm
Moment of resistance by steel, Ms =
=
σs
× Is
y
167
× 21296000 = 32331200 Nmm
110
Total moment of resistance = Mw + Ms = 8873333.3 + 32331200
= 41204533.3 Nmm
= 41.2 × 106 Nmm
@seismicisolation
@seismicisolation
Ans
140
•
Strength of Materials
Alternative method
Equivalent width in terms of steel
30.59 mm
bw
110
=
= 6.59 mm
m
16.7
bs =
220 mm
bd 3 30.59 × 2203
=
12
12
Moment of inertia, I =
12
6.59
Equivalent steel section
12
4
= 27143526.7 mm
Figure 9.15
Moment of resistance of equivalent steel section
M=
σs
167
I=
× 27143526.7 = 41208809 Nmm
y
110
= 41.2 × 106 Nmm
Ans
Note: Same as before.
Another alternative method
Equivalent width of steel in terms of wood,
510.8 mm
bw = mbs = 16.7 × 12 = 200.4 mm
Moment of inertia I =
220
mm
bd 3 510.8 × 2203
=
12
12
= 453249867 mm4
200.4 mm 110 mm 200.4 mm
Equivalent wooden section
Figure 9.16
Moment of resistance of equivalent wooden section,
M=
10 × 453249867
σw
×I =
= 41.2 × 106 Nmm Ans
y
110
Note: Same as before.
E XAMPLE 9.11: A wooden beam 120 mm wide and 180 mm deep is to reinforced by two plates
120 mm × 10 mm and 100 × 5 mm. The thicker plate is secured to the top and thinner one to the
bottom surface as shown in Fig. 9.17. The permissible stress in steel is 140 MPa and the value
of modular ratio is 18. Calculate moment of resistance of strengthened section. Also determine
maximum stress in wood (timber).
@seismicisolation
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Bending of Beams
•
141
1
10 mm
N
10 mm
A
180
mm
G
180 mm 6.7 mm
3
2
5 mm
B
A
120 mm
Flitched beam
y = 114.46 mm
5 mm
Equivalent steel section
Figure 9.17
Let us find the equivalent steel section of flitched beam.
Equivalent width of timber in steel =
120
bt
=
= 6.7 mm
m
18
To find position of neutral axis, first find the centre of gravity of equivalent steel section from AB. For
that we split equivalent steel section in three simple rectangles 1, 2 & 3. a1 = 120 × 10 = 1200 mm2 ;
y1 = 190 mm; a2 = 180 × 6.7 = 1206 mm2 ; y2 = 95 mm; a3 = 120 × 5 = 600 mm2 ; y3 = 2.5 mm.
Total area = 1200 + 1206 + 600 = 3006 mm2
1200 × 190 + 1206 × 95 + 600 × 2.5
= 114.46 mm
3006
120 × 103
2
+ 1200(190 − 114.46)
I=
12
6.7 × 1803
2
+
+ 1206(95 − 114.46)
12
120 × 53
2
+
+ 600(2.5 − 114.46)
12
Y (from AB) =
= [10000 + 3423775] + [3256200 + 456702]
+ [1250 + 7521025] = 14668952 mm4
= 14.67 × 106 mm4
M = σs ×
I
ymax
= 140 ×
;
y from AB = 114.46 is max
14668952
= 17942104 Nmm
114.46
= 17.94 kNm
Ans
@seismicisolation
@seismicisolation
142
•
Strength of Materials
Maximum stress in timber will exist at the top of bottom flange, so ytmax = 114.6 − 5 = 109.6 mm
σtmax =
M
17.94 × 106
× 109.6
×Ytmax =
I
14.67 × 106
= 134.03 N/mm2
Actual σtmax =
134.03
= 7.45 N/mm2 = 7.45 MPa. Ans
18
Combined Bending and Direct Stresses
P
In Fig. 9.18, a short column is shown with eccentric
loading. In this case there will be two stresses (i) direct
P.e
P
× distance of load
stress and (ii) bending stress i.e.,
A
I
from the neutral axis.
e
e
Figure 9.18
A combination of bending and direct stresses may occur in a variety of circumstances but in every
case, the stresses due to bending moment and direct load may be calculated separately and the
results combined to give the resultant stresses. Thus σ = σd ± σb where σd and σb are the direct
and bending stresses. The shape of the resultant stress distribution diagram will depend on whether
σb is greater or less than σd . Figure 9.18 shows a bar which is subjected to an axial load P and a
bending moment due to load being eccentric. Figures 9.19(a), (b) and (c) show the possible forms
of the resultant stress distribution.
P
a
M
P
<
Z
a
(a)
M
Z
P
a
M
Z
M
P
=
Z
a
(b)
Figure 9.19
@seismicisolation
@seismicisolation
P
a
M
Z
M
P
>
Z
a
(c)
Bending of Beams
•
143
E XAMPLE 9.12: A steel flat 180 mm wide and 30 mm thick is subjected to a pull of 180 kN,
which is off the geometrical axis by 4 mm in the plane which bisects the thickness. Determine the
maximum and minimum stresses induced in the section.
S OLUTION :
4 mm
P
30 mm
180 mm
28.89
MPa
37.77
MPa
Figure 9.20
P = 180000 N
A = 180 × 30 = 5400 mm2 ,
P = 180000 N
M = P.e = 180000 × 4 = 720000 Nmm
Z=
bd 2 30 × 1802
=
= 162000 mm3
6
6
σd =
P 180000
=
= 33.33 N/mm2
A
5400
σb =
M 720000
=
= 4.44 N/mm2
Z
162000
σmax = σd + σb = 33.33 + 4.44 = 37.77 N/mm2 = 37.77 MPa
σmin = σd − σb = 33.33 − 4.44 = 28.89 N/mm2 = 28.89 MPa
E XAMPLE 9.13: A mild steel T section as shown in Fig. 9.21 carries a load of 100 kN in the central
plane bisecting the web at 45 mm from the base. Determine the maximum and minimum stresses
induced in the section.
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•
Strength of Materials
180 mm
A
10 mm
1
B
y
G
45 mm
120 mm
2
10 mm
Figure 9.21
Let us first find the G of section from AB,
a1 = 180 × 10 = 1800; y1 = 5 mm;
a2 = 110 × 10 = 1100 mm2 ; y2 = 65 mm
Total area = 1800 + 1100 = 2900 mm2
1800 × 5 + 1100 × 65
= 27.76 mm = yt ;
2900
180 × 103
2
I=
+ 1800(27.76 − 5)
12
10 × 1103
2
+
+ 1100(27.76 − 65)
12
y=
yc = 92.24
= [15000 + 932432] + [1109167 + 1525499] = 3582098 mm4
Eccentricity of loading, e = 45 − 27.76 = 17.24 mm
M = P.e = 100000 × 17.24 = 1724000 Nmm
M.yc 1724000 × 92.24
=
= 44.39 MPa
I
3582098
1724000 × 27.76
M.yt
=
= 13.36 MPa
σt =
I
3582098
P 100000
= 34.48 MPa
σd = =
A
2900
σmax = σd + σc
σmax = 34.48 + 44.39 = 78.87 MPa Ans
σmin = σd − σt
σc =
σmin = 34.48 − 13.36 = 21.12 MPa Ans
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•
145
Modulus of rupture:
If a metallic beam, simply supported at its ends, is loaded by a gradually increasing transverse
central load till rupture or breaking takes place, the actual stresses at the outer layers at rupture
σ
M
= , since the condition
or breaking, are not the ones calculated by the bending equation:
I
Y
of elasticity assumed in this formula has ceased to exist. However, the stress calculated from the
equation:
M
M
y = = σ or say σr , is known as transverse rupture stress or modulus or rupture. Since, this
I
Z
test is generally applied to different qualities of cast iron or timber beams, the modulus of rupture
σr so calculated is called a guide to compare the strength of various qualities of these materials.
This test is conducted on brittle materials.
E XAMPLE 9.14: A 150 mm × 150 mm pine wood beam was supported at its ends on a 4.5 m span
and loaded as shown in Fig. 9.22. The beam failed when a 8 kN load was placed at 1.5 m from each
end. Find the modulus of rupture.
8 kN
1.5 m
A
8 kN
1.5 m
C
1.5 m
B
D
RA
RB
Figure 9.22
Beam is of square section, side = 150 mm = b
Section modulus =
=
b3
6
1503
= 56.25 × 10 mm4
6
Reactions RA = RB = 8 kN
Portion CD of the beam is under pure bending,
MC = MD = 8 × 1.5 = 12 kNm
So maximum, M = 12 × 106 Nmm
Modulus of rupture, σm =
=
M
Z
12 × 106
56.25 × 104
= 21.33 N/mm2
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Ans
146
•
Strength of Materials
Note: The theoretical value of the fracture stress obtained by flexure formula using ultimate bending
moment is known as modulus of rupture.
Conditions for no Tension in the Section
P
M
> , then as per
As we have seen in eccentric loading when load is of compressive nature, if
Z
A
stress distribution diagram, stress changes sign, being somewhere tensile while compressive stress
at other places. Now in masonry work, it is not desirable to have tensile stresses to avoid failure
of structure. This limits eccentricity e to a certain value which we shall investigate for rectangular
section and circular section.
1. Rectangular section: If a rectangular section ABCD is loaded at a point distant e along X − X
axis and off the Y − Y axis, as shown in Fig. 9.23. Naturally, bending will take place about Y − Y
axis.
Here Compressive
load P
B
A
a
Core of kernel
of the section
b
3
b
e
D
C
d
d
3
Figure 9.23
bd 3
and Area, A = bd.
12
2
= AkYY
Here IYY =
Also IYY
∴
2 =
Equating the above two AkYY
bd 3
12
2
=
bd kYY
bd 3
12
∴
2
kYY
=
d2
12
(i)
Now, for no reverse stress,
σd ≥ σb ≥
M
Z
P×e×d
P P×e×d
, we know for symmetrical section y = yt = yc = d/2.
≥
≥
A
2I
2 A k2
or e ≤
2k2
d
where k is radius of gyration.
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(ii)
Bending of Beams
•
147
2k2
.
Thus, for no tension in the section, the eccertricity must not exceed
d
Equation (ii) becomes with the help of (i)
e≤
2d 2
d
≤
d × 12 6
Therefore, the load should not be placed at a distance more than d/6 on either side of centroid on
X − X axis to avoid any reverse stress occurs.
d d
d
And therefore the limit of eccertricity (e) = + = .
6 6
3
Hence, the stress will be of the same sign throughout the section if the load line is within the
middle third of the section. This is known as middle third rule for rectangular section.
Likewise, if the load is placed on Y − Y axis, off the X-axis, middle on YY -axis which is b/3
is safe zone. If we join the four points of middle, third distances on XX and YY axis are joined, a
rhombus of diamond shape is obtained as shown in Fig. 9.24. This portion is known as the core or
kernel of the section. The reverse stress will not occur in any part of the entire rectangular section,
if the load is placed anywhere inside he rhombus.
2. Circular section
In this case IXX = IYY
∴
k2 =
π d4
= Ak2
64
y
π d 4 /64 d 2
=
16
π d 2 /4
2k2
e≤
d
d
2 × d2
=
e≤
d × 16 8
X d
X
We know that for no tension,
d
4
∴
e≤
d
8
y
Figure 9.24
d
d
= .
8
4
Hence for load must fall within the middle fourth of the section.
Hence, diameter of core or kernel = 2e = 2 ×
E XAMPLE 9.15: A short column of hollow cylindrical section 300 mm outside diameter and
150 mm inside diameter carries a vertical load of 500 kN at 80 mm away from the axis of the
column. Determine maximum and minimum stresses and state their nature.
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•
Strength of Materials
S OLUTION :
Area, A =
π 2
π
(D − d 2 ) = (3002 − 1502 )
4
4
80 mm
500 kN
= 52987.5 mm2
P
500000
Direct stress, σd = =
= 9.44 N/mm2
A 52987.5
P×e
Bending stress, σ =
Z
Z=
σb =
π /64(3004 − 1504 )
I
=
= 2483789 mm3
y
300/2
500000 × 80
= 16.1 N/mm2
2483789
150 mm
300 mm
Figure 9.25
Maximum stress = 9.44 + 16.1 = 25.54 N/mm2 = 25.54 MPa (Compressive) Ans
Minimum stress = 9.44 − 16.1 = −6.66 N/mm2 = 6.66 MPa (Tensile) Ans
E XAMPLE 9.16: A masonry column of 3.5 m × 4.3 m supports a load of 50 kN as shown in
Fig. 9.26, i) determine the stresses developed at each corner of the column; ii) in order that there is
no tension anywhere in the section of column, what additional load should be applied at the centre
of column? iii) what are the stresses at the corners with the additional load in the centre?
Here the load is applied
Y
A
1m
B
0.5m
X
3.5m
D
4.3m
Y
C
Figure 9.26
S OLUTION :
i) Area, A = 4.3 × 3.5 = 15.05 m2
Load, P = 50 kN
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X
Bending of Beams
•
149
4.3 × 3.53
= 15.363 m4
12
3.5 × 4.33
= 23.2 m4
Moment of inertia about Y −Y , IY =
12
Moment of inertia about X − X, IX =
eX = 1 m
and
eY = 0.5 m
∴
Moment Mx = P × 0.5 = 50 × 0.5 = 25 kNm
My = P × 1 = 50 × 1 = 50 kNm
4.3
= 2.15 m
2
3.5
= 1.75 m
Distance between XX axis and corners A & D =
2
Distance between YY axis and corners A & B =
σA =
P Mx × y My × x
50
25 × 1.75 50 × 2.15
−
=
+
+
−
A
IX
IYY
15.05
15.363
23.2
= 3.32 + 2.848 − 4.63 = 1.538 kN/m2
σB =
P Mx × y My × x
50
25 × 1.75 50 × 2.15
+
=
+
+
+
A
IX
IY
15.05
15.363
23.2
= 3.32 + 2.848 + 4.63 = 10.798 kN/m2
σC =
Ans
P Mx × y My × x
50
25 × 1.75 50 × 2.15
+
=
−
−
+
A
IX
IY
15.5
15.363
23.2
= 3.32 − 2.848 + 4.63 = 5.102 kN/m2
σD =
Ans
Ans
25 × 1.75 50 × 2.15
P Mx × y My × x
50
−
−
−
−
=
A
IX
IY
15.5
15.363
23.2
= 3.32 − 2.848 − 4.63 = −4.158 kN/m2 = 4.158 kN/m2 (Tensile) Ans
ii) Compressive stress due to additional load P ,
=
P
P
=
kN/m2
A
15.05
If there is to be no tension in the column section, the compressive stress due to P kN should
be equal to the tensile stress, i.e., 4.158 kN/m2 .
∴
P
= 4.158 Hence P = 62.58 kN
15.05
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Ans
150
•
Strength of Materials
iii) Stresses at the corners with the additional load in the centre:
Stress due to additional load =
P 62.58
=
= 4.158 kN/m2 = 4.16 kN/m2 app.
A
15.05
σa = Stress at A = 1.538 + 4.16 = 5.698 kN/m2
Ans
2
σb = Stress at B = 10.798 + 4.16 = 14.958 kN/m
2
σc = Stress at C = 5.102 + 4.16 = 9.262 kN/m
Ans
Ans
σd = Stress at D = −4.158 − 4.158 = 0 Ans
E XAMPLE 9.17: Determine the maximum force that the screw on c-clamp can exert on the wooden
block in Fig. 9.27, if the allowable stress at section A − A is not to exceed 83 MN/m2 .
250 mm
A
A
50 mm
12.5 mm
Cross section A-A
Figure 9.27
S OLUTION :
Eccentricity of force, e = 250 mm
Permissible stress, σt = 83 MN/m2 = 83 N/mm2
To find P exerted by the screw of c-clamp,
Area of cross section at A − A, a = 50 × 12.5 = 625 mm2
Moment of inertia about N.A. at A − A, I =
∴
bh3 12.5 × 503
=
12
12
I = 13.0208 × 104 mm4
Bending moment due to P = P × 250 Nmm
P
P
(i) Direct stress due to P, σd = =
= 0.0016 P
A 625
M
P × 250
(ii) Bending stress, σb = y =
× 25 = 0.048 P
I
13.0208 × 104
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•
Bending of Beams
151
Total stress = σd + σb = 0.0016P + 0.048P = 0.0496P
Total stress must not exceed allowable stress 83 N/mm2
∴
0.0496P = 83
∴
P = 1673.38N = 1.673 kN
Ans
Exercise
9.1 A light wooden bridge is supported by six parallel timber beams, each 300 mm deep, and
200 mm wide. Each beam may be considered simply supported over a span on 4.5 m. If the
allowable stress in the timber is 5.60 MN/m2 , calculate the greatest uniformly distributed
load the bridge can support.
[Ans
179.2 kN]
9.2 The crane beam shown in Fig. 9.28 is made up of two 25 mm thick steel plates each 400
mm deep at the middle section. Calculate the maximum allowable central load W if the
span is 4.5 m and the ultimate tensile stress of the steel is 370 MN/m2 . Allow a factor of
safety of 16.
[Ans
73 kN]
W
Figure 9.28
9.3 Calculate the maximum bending moment which may be applied to the cast iron section shown
in Fig. 9.29, if the permissible tensile stress of the material is 280 MN/m2 and a factor of
safety of 10 is to be used.
100 mm
10 mm
120 mm
10 mm
10 mm
100 mm
Figure 9.29
9.4 The bar of section shown in Fig. 9.30 is simply supported over a span of 1 m and carries a
central load of 20 kN. Find the maximum bending moment and the maximum bending stress
in the material. [I = 77.44 × 104 mm4 , 5 kNm, 194 MN/m2 ]
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•
Strength of Materials
20 mm
rad
60 mm
50 mm
Figure 9.30
9.5 A cast iron beam is in the shape of T section as shown in Fig. 9.31. 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 beam span. Determine the maximum and minimum tensile and compressive
stresses.
100 mm
20 mm
100 mm
20 mm
[Ans
Figure 9.31
123.1 MN/m2 ]
9.6 An I section beam as shown in Fig. 9.32, if the ratio of maximum compressive stress to
tensile stress induced is 4:3, determine the dimension b assuming that the section is subjected
to sagging moment. Show also the stress variation diagram across the section.
Compressive
side
b
300 mm
20 mm
260 mm
200 mm
20 mm
Tensile
side
Figure 9.32
[Ans
b = 112.5 mm]
9.7 A composite beam is made by placing two steel plates, 12 mm thick and 240 mm deep,
one each on both sides of a wooden section 90 mm wide and 240 mm deep. Determine
the moment of resistance of the beam. Take Es /Ew = 15, the stress in the wood not to
exceed 7 MPa.
[Ans
@seismicisolation
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30.24 Nm]
Bending of Beams
•
153
9.8 An uniformly tapering vertical post of height 10 m has a diameter of 150 mm at the top. A
horizontal pull of 250 N is applied at the top of the post. Find the maximum bending stress
for the post and state where it occurs.
[Ans 1.12 N/mm2 , 5 m from top]
9.9 A timber beam 150 m wide is reinforced by a steel plate 100 mm wide and 10 mm deep
attached at the lower face of the timber beam. Calculate the moment of resistance of the
beam, if the allowable stresses in timber and steel are 6 MPa and 60 MPa respectively. Take
Es = 16 Et .
[Ans 9.45 kNm]
9.10 Prove that the moment of resistance of a beam of square section with diagonal in the plane of
bending is increased by flattening the top and bottom corners, as shown in Fig. 9.33, and that
8
the moment of resistance in a maximum when, y = Y .
9
Y
y
Figure 9.33
9.11 A tie bar 75 mm wide and 25 mm thick sustains an axial load of 100 kN. What depth of metal
may safely be removed from one of the narrow sides in order that the maximum stress over
the reduced width may not exceed 100 MN/m2 .
[Ans 12.05 mm]
9.12 A horizontal cantilever 3 m long is of rectangular cross section 60 mm wide throughout its
length, the depth varying uniformly form 60 mm at the free end to 180 mm at the fixed
end. A load of 4 kN acts at the free end. Determine the position of the most highly stressed
section, and find the value of the maximum bending stress induced. Neglect the weight of the
cantilever itself.
[Ans 41.6 MN/m2 ]
9.13 A steel tube of 40 mm outside diameter and 30 mm inside diameter is used as a simply
supported beam on a span of 1 m end. It is found that the maximum safe load it can carry at
midpoint is 1.2 kN.
Four of these tubes are placed parallel to another and firmly fixed together to form in effect
a simple beam, the centres of the tubes forming a square of 40 mm side with one pair of
centres vertically over the other pair. Find the maximum central load which this beam can
carry if the maximum stress is not to exceed that of the single tube above.
[Ans 8.64 kN]
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154
•
Strength of Materials
9.14 A short column has the cross section as shown in Fig. 9.34. An axial compressive force of
150 kN is applied at point F. Calculate the stress at each corner.
20 mm
F
30 mm
200 mm
[Ans σa = −7.5 N/mm2 (Compressive)
σb = 23.2 N/mm2 (Tensile)
σc = −9.8 N/mm2 (Compressive)
100 mm
σd = 8.2 N/mm2 (Tensile)]
Figure 9.34
9.15 A flitched beam consists of a wooden joist 120 mm wide and 200 mm deep, strengthened
by a steel plates 10 mm thick and 180 mm deep on either side of the joist at centre. If the
stresses in wood and steel are not to exceed 7.5 MN/m2 and 127.5 MN/m2 , find the moment
of resistance of section of the beam. Take Es = 20 Ew .
[Ans
19.436 kNm]
9.16 A steel bar 120 mm in diameter is completely encased in an aluminium tube of 180 mm outer
diameter and 120 mm inner diameter so as to make a composite beam. The composite beam
is subjected to a bending moment of 15 kNm. Determine the maximum stress due to bending
in each material. Take Es = 3Eal .
[Ans
σs = 37.56 MN/m2 , σal = 18.78 MN/m2 ]
9.17 A cantilever of uniform strength and rectangular cross section and has the depth twice the
width throughout. The beam is of 1.2 m long and to be loaded with a uniformly distributed
load of 16 kN/m length. Allowable bending stress = 20 N/mm2 . Calculate the dimensions of
the beam at the middle and at the fixed end.
[Ans
120 mm at middle, 190.5 mm at fixed end]
9.18 Calculate the maximum tensile and compressive stresses induced in a box girded carrying
a uniformly distributed load of 16 N/m on a simply supported length of 8 m. The box section is made of ISA 75 × 75 × 5 mm thick angles welded to four 360 mm × 6 mm structured steel plates as shown in Fig. 9.35. For each angle section area ‘A’= 7.27 × 102 mm2 ,
IXX IYY = 38.7 × 104 mm4 and the distance of centre of gravity from the outer face of each
leg of angle Cxx = Cyy = 20.2 mm.
[Ans
σmax = 89.168 MPa both in tension and compression]
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Bending of Beams
6 mm
•
155
16 kN/m
360 mm
Angle 75 × 75 × 5
8m
5
75
75
6 mm
360 mm
Figure 9.35
9.19 A tapering chimney of hollow circular section is 30 mm high. Its external diameters at the
top end base are 1.5 m and 2 m respectively as shown in Fig. 9.36. There is wind pressure
of 30 kN/m2 from right to left. If the weight of the chimney is 50,000 kN and the internal
diameter at the base is 1 m, determine the maximum and minimum stresses at the base and
show their variation.
1.5 m
Wind force
30 m
[Ans σmax = 18.552 N/m2 ,
1m
σmin = −5.62 N/m2 (Compressive)]
2.5 m
Figure 9.36
9.20 A steel pipe of external diameter 1224 mm and internal dia 1200 mm is running full if the
bending stress is not to exceed 56 N/mm2 , find the greatest span on which the pipe may be
freely supported. Steel and water weight 76800 N/m3 and 10000 N/m3 respectively.
[Ans
20.36 m]
9.21 A 4 m long horizontal beam is simply supported at the ends in a rectangular cross section 40
mm wide and 80 mm deep. It is loaded with centrally concentrated load of P. Find the value
of P which produces a bending stress of 20 MPa at a point 1 m from the left support and 20
mm below the top surface of beam. What is the maximum bending tensile stress in the beam
and where does it occur?
[Ans
3413 N, 80 MPa at the bottom of beam at midspan]
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156
•
Strength of Materials
9.22 A beam of I section with flange t mm thick 3 t mm wide and web 2t deep ×t thick (making
overall depth of beam section as 4t) is simply supported at ends. The beam carries uniformly
distributed load over whole of its length in addition to a central concentrated load equal to
the total uniformly distributed load. If the length of the beam is 3 m, find the relationship
between t and uniformly distributed load for a maximum bending stresses of 120 MN/m2 .
[Ans w = 261 × 106t 3 where w in N/m and t in m]
9.23 A rectangular beam, simply supported over a span of 4 m, is carrying a uniformly distributed
load of 50 kN/m. Find the dimensions of the beam, its depth if the beam section is 2.5 times
its width. Take maximum bending stress as 60 MPa.
[Ans 125 mm, 300 mm]
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10
C HAPTER
SHEAR STRESSES IN BEAMS
Normally, beams are designed for bending stresses and then checked for shear stresses, as in several
situations arise in design in which mode of failure is likely to be shear rather than bending. For
example, wooden beams which are weak in shear along the planes parallel to the grain of the wood,
and thin webbed beams, where if the web is excessively thin, it would not have sufficient stiffness
and stability to hold its shape and it would fail due to shearing stress.
Therefore, it is necessary to study shear stress distribution as detailed under.
On application of shear force, the shear stress on the cross section tends to slide the transverse
elements of the beam and the complimentary shear stress of elements act on it (for example, wooden
beams when tested to destruction). Though the mean shear stress is equal to shear force divided by
cross-sectional area but the shear stress in fact is not uniform in the cross section. It is zero on top
and bottom of section.
In order to derive an expression for shear stress at any point in the cross section of beam, let us
consider any normal section AB of a beam where bending moment and shear force be M and F act
respectively. Now consider another section CD at a distance δ x from AB, where bending moment
and shear force are M + δ M and F + δ F, respectively
M
δy
A
y2
N
P
y E
M+δM
C
Q
F
y1
y
G
A
B
D
δx
Longitudinal section
y1
b
Cross section
Figure 10.1
Now consider any elementary stripe PQ of thickness δ y at a distance Y from neutral axis.
M
y
I
M +δM
y
Bending stress at Q =
I
Bending stress at P =
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158
•
Strength of Materials
where I is the moment of inertia of cross section about neutral axis
∴ Area at P or Q = bδy
M
Total force on face P = · y × b.δy
I
M +δM
× y × b.δy
Total force on face Q =
I
Hence, unbalanced force PQ,
=
M
M +δM
× y.b.δ y − · y × b · δy
I
I
∴ Total unbalanced force on portion AEFC,
y2
=
y1
δM
δM
.ybδ y =
I
I
y2
b.y.δ y
(i)
y1
This force is balanced by the shear force across EF
shear force across EF = τ × Area of EF
where τ is the shear stress,
Shear force at EF = τ × b.δx
(ii)
Equating (i) and (ii),
δM
I
y2
b.y.δy = Zb.δx
y1
δM 1
×
τ=
δ x Ib
We know that
y2
b.y.δy
y1
δM
= F, shear force at the section AB.
δx
y2
b.y.δy = Moment of area of part on force AB. (transverse)
y1
between A and E about neutral axis
Now if A= Area of the part on face AB (transverse) between A and E.
y = distance of its centre of gravity from neutral axis
@seismicisolation
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(iii)
Shear Stresses in Beams
Then,
y2
b.y.δy = Ay
y1
δM
and
Substituting the values of
δx
y2
by · δy in Eqn. (iii),
y1
we get,
τ =F·
1
Ay
Ib
or
τ=
FAy
, where I is moment of inertia of the whole section.
Ib
This is an important relation.
Now we shall discuss shear stress distribution across few important cross sections:
(A) Shear Stress Distribution for Beam of Rectangular Section
Take a section ABCD as shown in Fig. 10.2.
b
A
d/2
B
E
y
τ
F
y
A
N
D
Parabola
Section of beam
C
d
Max shear
stress
τmax
Shear stress
distribution diagram
Figure 10.2
y = distance of centre of gravity of shaded area from neutral axis
d
−y b
A=
2
1 d
−y
y = y+
2 2
1 d
=
+y
2 2
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•
159
160
•
Strength of Materials
Moment of inertia of the shaded area about neutral axis
bd 3
12
F.Ay
Shear stress, τ =
I.b
I=
Substituting for the values of A & y, we have
1 d
d
−y b
+y
2
2 2
τ=
I.b
2
b d
2
−y
F×
2 4
=
I.b
2
F d
2
−y
∴τ =
2I 4
F.
(i)
The shear stress distribution shown in Fig. 10.2 is parabolic.
d
When, y = , τ = 0 (by substituting in Eqn. (i))
2
At N.A, y = 0 so τ will be maximum
Fd 2
F d2
−0 =
Hence, τmax =
2I 4
8I
bd 3
12
2
Fd
3 F
= .
We get, τmax =
3
2
bd
bd
8×
12
F
= average shear stress at the section, and is denoted by τav
But,
bd
Substituting for I =
3
∴ τmax = τav
2
Therefore, maximum shear stress for a rectangular section is 1.5 times the average shear stress.
(B) Shear Stress Distribution of a Solid Circular Section
Let us consider a solid circular section of diameter D, radius R and Shear force F at that shaded
section.
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Shear Stresses in Beams
•
161
b
Parabolic
B
A
δy
y
y
τ
D
τmax
R
Shear stress
distribution diagram
Beam section
Figure 10.3
Consider any elementary strip of thickness δy over the layer AB at a distance y from NA.
AB = b = 2 R2 − y2
∴ Area of shaded strip = b.dy = 2 R2 − y2 × dy
Moment of this area about neutral axis
R
Ay =
R
2
2
2y R − y dy = by dy
y
y
Because
b = 2 R2 − y2
b2 = 4 R2 − y2
Differentiating both sides,
2bdb = 4(−2y)dy
1
or ydy = − b.db
4
Substituting in Eqn. (i),
R Ay =
y
1
b − bdb
4
When y = y, b = 0 and when y = R, b = 0
∴
1
Ay = −
4
0
1 −b3
b3
b db = −
=
4 3
12
2
b
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(i)
162
•
Strength of Materials
Now shear stress at any section,
τ=
∴
FAy
Ib
b3
2
12 = Fb
Shear stress at section AB =
I ×b
12I
2
2
F4(R − y )
=
12I
F (R2 − y2 )
=
3
I
F×
(ii)
we find from Eqn. (ii) that when y = R, τ = 0 (at the highest point of the section) and at y = 0, i.e.,
at neutral axis, shear stress is maximum.
2
D
F
4
2
∴ τmax (at y = 0) = R2 = F
π 4 = 3 τav
3I
3× D
64
F
Because average, τ = π
D2
4
Hence, for solid circular beam section,
4
τmax = τav = 1.33 τav
3
(C) Shear Stress Distribution in an I Section
F (D2–d2)
8I
B
y
y
d
b
A
N
Beam section
Parabolic
BF (D2–d2)
8Ib
D
Parabolic
Shear stress distribution diagram
Figure 10.4
i) Shear stress distribution in the flange
D
A = Shaded area = B
−y
2
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F [B(D2-d2)+bd2]
8Ib
Shear Stresses in Beams
•
163
y = Centroidal distance of this area from neutral axis
1 D
1 D
1 D
−y =
− y + 2y =
+y
= y+
2 2
2 2
2 2
1 D
D
∴ Ay = B
−y ×
+y
2
2 2
FAy
Ib
Substituting values for A, y & b,
we know, τ =
1 D
D
−y ×
+y
F ×B
F D2
2
2 2
2
=
−y
τ=
IB
2I 4
d
At the outer surface of the flange, where y = ; τ = 0
2
d
At the inner surface of the flange, where y =
2
2
d2
F 2
F D
−
=
τ=
D − d2
2I 4
4
8I
Obviously shear stress distribution in the flange is parabolic curve.
ii) Shear stress distribution in the web:
Width of the section in the web at a distance y from neutral axis = b
Ay = Moment of flange area about neutral axis + Moment of web area about neutral axis
d 1 D−d
d
1 d
D−d
×
+
+b
−y × y+
−y
=B
2
2 2
2
2
2 2
D−d
d D
d
d y
=B
×
+
+b
−y
+
2
2 4
2
4 2
b d2
B 2
2
2
−y
= (D − d ) +
8
2 4
Substituting this value of A y in Eqn., τ =
FAy
Ib
F B 2
b d2
2
2
(D − d ) +
−y
τ=
Ib 8
2 4
F
=
B(D2 − d 2 ) + b(d 2 − 4y2 )
8Ib
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(i)
164
•
Strength of Materials
Therefore, in the web, the shear stress τ varies with respect to y follows a parabolic curve.
Also, τ increases as y decreases. At y = 0, τ is maximum.
Substituting in Eqn. (i),
τmax =
F
B(D2 − d 2 ) + b d 2
8Ib
On the junction of the web and flange, y =
d
,
2
Substituting in Eqn. (i)
F
d2
2
2
2
B(D − d ) − b(d − )
τ=
8Ib
4
=
FB 2
(D − d 2 )
8Ib
Of course in web also shear stress distribution is parabolic. It may be noted that abrupt change
in width of flange at bottom surface of the flange, the numerical value of shearing stress
F
B F
B
suddenly changes from (D2 − d 2 ) to × (D2 − d 2 ) in the proportion (i.e., increases).
8I
b 8I
b
E XAMPLE 10.1: An R.S.J. is of I section of overall height 200 mm and flange width 125 mm. The
web thickness is 7 mm and the flange thickness 11 mm. The standard taper on the flange may be
neglected and all corners may be assumed sharp. The beam is subjected to transverse loads acting
parallel to the web, and at one section the shear force is 100 kN. Determine the maximum vertical
shearing stress in the web at this section.
Also determine the shear stress at top and bottom of the flange.
S OLUTION :
125 mm
11 mm
7
X
X
200 mm
Shear force, F = 100 kN, Working in metre & N,
0.125 × 0.23 (0.125 − 0.007) × (0.2 − 0.022)3
−
Ixx =
12
12
= 8.83 × 10−5 − 5.546 × 10−5
11 mm
= 2.784 × 10−5 m4
Figure 10.5
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Shear Stresses in Beams
•
165
Maximum stress will be in the web at neutral axis
F.A.y
Ib
100000(0.125 × .011 × 0.945 + 0.007 × 0.089 × 0.0445)
τmax =
2.784 × 10−5
= 80.78 MN/m2 Ans
Now,
τ=
Shear stress at the top and bottom of the web,
τ=
100000 × 0.125 × .011 × 0.0945
= 66.6 MN/m2
2.784 × 10−5
Ans
E XAMPLE 10.2: A cast iron beam with flange and web section is 250 mm deep overall. The top
flange is 125 mm × 50 mm deep, the bottom flange 200 mm × 50 mm deep and the web is 40 mm
thick. If the transverse shearing force is 140 kN. Calculate the consequent shear stress is the web at
the top and bottom junctions with flanges and maximum shear stress and sketch a diagram showing
the variation of shear stresses over the depth of the beam.
S OLUTION :
Working in m and N throughout;
14.93 MN/m2
1
0.125 m
0.05 m
18.7 MN/m2
0.04 m
N
G
Zmax
2
y
0.15 m
A
3
0.05 m
17.1 MN/m2
A
0.20 m
Shear stress distribution
diagram
Figure 10.6
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B
166
•
Strength of Materials
For y from AB,
a1 = 0.125 × 0.05 = 6.25 × 10−3 ; y1 = 0.225 m
a2 = 0.04 × 0.15 = 6 × 10−3 ; y2 = 0.125 m
a3 = 0.20 × 0.05 = 0.01; y3 = 0.025 m
A = a1 + a2 + a3 = 6.25 × 10−3 + 6 × 10−3 + 0.01
= 0.02225 m2
6.25×10−3 ×0.225 + 6×10−3 ×0.125 + 0.01×0.025
0.02225
= 0.1083 m
y=
0.125 × 0.053
2
+ 0.125 × 0.05 × (0.1083 − 0.225)
I=
12
0.04 × 0.153
2
+ 0.15 × 0.04 × (0.1083 − 0.125)
+
12
0.2 × 0.053
2
+ 0.2 × 0.05 × (0.1083 − 0.025) = 0.000171 m4
+
12
Shear stress at top of web, τ =
τ=
FAy
Ib
140000 × 0.125 × 0.05 × 0.1167
= 14.93 MN/m2
0.000171 × 0.04
Ans
Shear stress at bottom of web,
τ=
140000 × 0.2 × 0.05 × 0.0833
= 17.1 MN/m2
0.000171 × 0.04
Ans
Max stress shear stress,
τmax =
140000 × [0.125 × 0.05(0.1167) + 0.04 × 0.0917 × 0.04585
= 18.5 MN/m2
0.000171 × 0.04
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Ans
Shear Stresses in Beams
•
167
Hint:
0.1167
0.0917 m
N
0.1417 m
A
0.04585
Figure 10.7
E XAMPLE 10.3: Figure 10.8 shows the section of a Tee-beam made of a uniform material, which is
subjected to a shear force of 200 kN and a bending moment of 25 kNm. Calculate (a) the maximum
bending stress and (b) the maximum shear giving sketches to show the form of stress distribution in
each case.
S OLUTION :
67.16 MN/m2
150 mm
A
1
B
25 mm
y
0.05625
150 mm
2
25 mm
65.3 MN/m2
= τmax
0.11875
σmax = 137.5 MN/m2
Figure 10.8
Let us first find position of C.G. from AB,
a1 = 0.150 × 0.025 = 3.75 × 10−3 m2 ; y1 = 0.0125 m; a2 = 0.150 × 0.025 = 3.75 × 10−3 m2
y2 = 0.1 m; A = a1 + a2 = 0.0075
3.75 × 10−3 × 0.0125 + 3.75 × 10−3 × 0.1
= 0.05625 m
0.0075
0.15 × 0.0253
Ixx =
+ 3.75 × 10−3 (0.05625 − 0.0125)2
12
0.025 × 0.153
−3
2
+ 3.75 × 10 (0.1 − 0.05625)
+
12
y=
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168
•
Strength of Materials
= 2158 × 10−8 m4
M
25000
ymax =
× 0.11875 = 137.5 MN/m2 Ans
I
2158 × 10−8
M
25000
σmin = ymin =
× 0.05625 = 65.16 MN/m2 Ans
I
2158 × 10−8
σmax =
τ=
F.Ay
200000 × [0.15 × .025 × 0.04375 + 0.025 × 0.03125 × 0.0156]
; τmax =
Ib
2158 × 10−8 × 0.025
= 65.3 MN/m2
Ans
E XAMPLE 10.4: A cantilever of I Section 200 mm × 100 mm has rectangular flanges 10 mm thick
web and 7.5 mm thick. It carries a uniformly distributed load. Determine the length of the cantilever
if the maximum bending stress is three times the maximum shearing stress. What is the ratio of the
stresses halfway along the length of cantilever?
100 mm
w N/m
10 mm
l
7.5 mm
X
200 mm
(b)
X
10 mm
100 mm
(a)
Figure 10.9
S OLUTION :
0.1 × 0.23 0.0925 × 0.183
−
= 2.17 × 10−5 m4
Ixx =
12
12
Maximum BM =
wl 2
M
wl 2 × 0.1
= 2304.15 wl 2
, σmax = y =
2
I
2 × 2.17 × 10−5
FAy
, Shear force(max) = wl
Ib
wl × (0.1 × 0.01 × 0.095 + 0.090 × .0075 × 0.045)
τmax =
2.17 × 10−5 × 0.0075
τmax is at neutral axis & τ =
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•
169
wl(9.5 × 10−5 + 3.0375 × 10−5 ) wl × 10−5 (12.54)
=
0.016275 × 10−5
0.016275 × 10−5
(i)
Shear Stresses in Beams
=
= 770.5 wl
Since σmax is three times τmax ;
∴
w l = 2304.15
w l2
3 × 770.5
l = 1 m Ans
∴
σmax = 2304.15 w(1)2 = 2304.15 w
τmax = 770.5 w × 1 = 770.5 w
Ratio of stresses half-way:
0.0125w
w(0.5)2
0.125 w
= 0.125 w ∴ σ =
× 0.1 =
2
I
I
−5
0.5w × 10 (12.54)
w
= 10−5 × 836
F = w × 0.5 = 0.5w From Eq(i) τ =
I × 0.0675
I
0.0125w/I
σ
= 1.5 Ans
Required ratios = =
τ
836 × 10−5 w/I
M=
E XAMPLE 10.5: Find the maximum shear stress in a hollow circular section of 100 mm external
diameter and 75 mm internal when subjected to a total shearing force of 160 kN.
S OLUTION :
FAy
Ib
b = 2(R − r)
π
Shaded area = (R2 − r2 )
2
π R2 4R π r2 4r
−
·
·
2 3π
y = 2 π3π
(R2 − r2 )
2
(R3 − r3 )
y=
3π (R2 − r2 )
τ=
y
b
Figure 10.10
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(i)
170
•
Strength of Materials
Substituting in Eqn. (i)
π
F × (R2 − r2 )
4(R3 − r3 )
2
τ=
×
I × 2(R − r)
3π (R2 − r2 )
=
Substituting in Eqn. (ii);
∴
F R3 − r3
.
3I R − r
(ii)
I=
π
(0.14 − 0.0754 ) = 0.00000336 m4
64
R=
0.1
0.075
= 0.05; r =
= 0.375
2
2
τ=
160000
0.053 − 0.03753
×
3 × 0.00000336
0.05 − 0.0375
= 91.7 MN/m2
Ans
Exercise
10.1 Plot the distribution of shear stress over the section shown in Fig. 10.11, which is subjected
to a shearing force of 300 kN, giving essential values.
300 mm
25 mm
300 mm
25 mm
Figure 10.11
[Ans
τmax = 51.5 MN/m2 ]
10.2 A transverse shear force F and a bending moment FL are applied to a uniform beam having
the symmetrical cross section shown in Fig. 10.12. If the ratio of the transverse shear stress
at the natural axis to the maximum direct stress due to bending is not to be less then 0.5,
determine the maximum permissible value of l in terms of a.
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•
171
0.8a
4a
2a
[Ans
Figure 10.12
l = 9.15a]
10.3 A rolled steel section is shown in Fig. 10.13, it is subjected to a vertical force of 20 kN.
Determine shear stress at points A, B and C of the section.
A
30 mm
B
R = 30 mm
60 mm
dia
C
30 mm
100 mm
Figure 10.13
[Ans
τA = 0, τB = 1.96 N/mm2 , τc = 5.88 N/mm2 ]
10.4 An extruded aluminium alloy section is of shape and dimensions as per Fig. 10.14. If a
vertical force on the section is 4 kN. Find the dimension d, if the average shear stress in the
section is 8 MN/m2 . What is the shear stress at neutral axis?
30 mm
d/2
45°
d
60 mm
Figure 10.14
[Ans 20 mm; τ at neutral axis = 11.4 MN/m2 ]
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172
•
Strength of Materials
10.5 A T-section beam symmetrical about a vertical axis, is made with a top flange 100 mm wide
and 14 mm thick to which a vertical web plate 150 mm deep and 10 mm wide is welded.
At a certain point, the total shearing force is 40 kN. Calculate the percentage shear carried
by the vertical web and the shearing force per metre run in the welded section.
[Ans
94.2%; 308 kN]
10.6 An I Section shown in Fig. 10.15 is used as a beam. Determine the percentage of shear force
resisted by web if the beam is subjected to a shear force F.
150 mm
20 mm
20 mm
250 mm
20 mm
Figure 10.15
[Ans
92.53%]
10.7 A beam of triangular section having base width 200 mm and height of 300 mm is subjected
to a shear force 3 kN. Determine the value of maximum shear stress and sketch the shear
stress distribution diagram along the depth of beam.
[Ans
3
Hint: τmax = τmean × ; 150 kN/m2 ]
2
10.8 A beam of channel section as shown in Fig. 10.16, is subjected at a vertical section, a shear
force of 50 kN. Draw shear stress distribution diagram. Find the ratio of maximum and mean
shear stresses.
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•
173
[Ans
2.2]
Shear Stresses in Beams
60 mm
15 mm
120 mm
15 mm
15 mm
60 mm
Figure 10.16
10.9 Find the maximum shear stress induced by a force of 4 kN is the vertical section of a hollow
beam of a square section, if the outside width is 100 mm and the thickness is 20 mm.
[Ans
1.35 MN/m2 ]
10.10 The beams of cross section (shown in Fig. 10.17) has shear force of 5.3 kN. Determine shear
stress at points A and B.
325 mm
12 mm
B
A
200 mm
100 mm
100 mm
100 mm
12 mm
Figure 10.17
[Ans τA = 287 kPa; τB = 396 kPa]
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174
•
Strength of Materials
10.11 The I section shown in Fig. 10.18 is used as a simply supported beam so that maximum
shear stress developed in the beam is 16.8 N/mm2 . Determine the shear force to which the
beams is subjected.
150 mm
20 mm
10 mm
30 mm
20 mm
150 mm
Figure 10.18
[Ans 50 kN]
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11
C HAPTER
TORSION
When a cylindrical shaft is subjected to equal and opposite couples at the end, shear stresses develop
in the shaft whether it is rotating or in equilibrium. Our objective is to derive an equation connecting
torque, modulus of rigidity, length, polar moment of inertia, length, shear strain, shear stress, radius
and length. For that we have to make certain important assumptions before going ahead.
Assumptions
i)
ii)
iii)
iv)
v)
vi)
vii)
The material is homogeneous and isotropic.
Twist along the shaft is uniform.
The shear stress is directly proportional to the radial distance from the axis of shaft.
Transverse planes of the shaft remain plane.
The modulus of rigidity is same throughout the material.
All radii remain straight.
The distortion along the shaft is uniform throughout.
Now consider a circular shaft of outer radius R, fixed at one end. A torque T is applied at the
force end as shown in Fig. 11.1.
T
φ
B'
B θ
A
0
Radius = R
Diameter = D
l
Figure 11.1
Let us take a straight line AB, which after application of torque T in clockwise direction takes
the new position as AB so that shear strain φ as shown in Fig. 11.1 is set up. At the same time angle
θ (in radians) is made at the free end as ∠BOB .
Now shear strain = φ
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176
•
Strength of Materials
∴ BB = l φ ,
Also BB = Rθ
Rθ
Therefore, l φ = Rθ , or φ =
l
τ
Shear stress at outer radius
=
Also we know shear strain φ =
Modulus of rigidity
C
(i)
(ii)
Equating Eqns. (i) and (ii)
Rθ
τ
= ,
l
C
We get
Cθ
τ
=
l
R
(iii)
Torsional Moment of Resistance
Take an elementary ring at radius r of thickness
dr. Now as per assumption shear stress τ at r is
proportional to shear stress τ at radius R.
∴
τ
τ
=
R
r
∴
τ =
τ
.r
R
Figure 11.2
Shear force acting on elementary ring = τ 2π rdr
τ
Substituting for τ , shear force acting on elementary ring = 2π r2 dr
R
τ
Moment of this shear force at centre O = 2π r3 dr
R
R
Total moment of resistance, T =
τ
2πτ
2π r3 dr =
R
R
0
R
r3 dr
0
=
2πτ r4
R
4
R
0
2πτ R4 πτ R4
×
=
∴ T=
R
4
2R
D 4
4
4
π 2
πR
πD
=
=
Now
2
2
32
4
πD
= J, Polar moment of inertia
32
IZZ
[As we know from perpendicular axis theorem, IXX + IYY = IZZ
is known as polar moment of inertia]
@seismicisolation
@seismicisolation
∴
π D4 π D4
π D4
+
=
64
64
32
Torsion
Hence, T =
τJ
;
R
or
•
τ
T
=
J
R
177
(iv)
Combining Eqns. (iii) and (iv), we get torsion equation:
τ
Cθ
T
= =
J
R
l
Power Transmitted: If T is torque and N is rotational speed in revolutions per minute, then,
Power,
P=
T × 2π N
kW, if T is in Nm.
60 × 1000
T
= CJ , CJ is known as torsional rigidity,
Torsional Rigidity: The stiffness or torsional rigidity =
θr
which can also be defined as the ratio of the torsion to angle of twist (in radians) per unit length of
the shaft.
E XAMPLE 11.1: A circular shaft of 60 mm diameter transmits torque from one shaft to another.
Find the safe torque, which the shaft can transmit, if the shear stress is not to exceed 50 MPa.
S OLUTION :
T
τ
=
J
R
∴
Safe torque =
or,
T=
J
π d4
π d3
τ=
τ=
τ
R
32.d/2
16
π (60)3
× 50 = 2119500 Nmm
16
= 2.1195 kNm
Ans
E XAMPLE 11.2: A solid circular shaft is to transmit a torque of 15 kNm. If the permissible shear
stress is 55 MPa, find the diameter of the shaft.
S OLUTION :
τ
π d3
J
T
=
or, T = τ =
τ
J
R
R
16
T = 15000 Nm,
τ = 55 MPa, d =?
Substituting,
π d3
× 55 × 106
16
15000 × 16
∴
d3 =
= 1.39 × 10−3
π × 55 × 106
or, d = 0.112 m = 111.6 mm Ans
15000 =
@seismicisolation
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178
•
Strength of Materials
E XAMPLE 11.3: Torque is required to transmit through a hollow shaft of external diameter 100
mm and internal diameter 55 mm. If the permissible shear stress in 55 MPa, what is the safe torque
that can be transmitted?
S OLUTION :
π (D4 − d 4 )
32
τ
π (D4 − d 4 )
T = J × ×2 =
× 55
D
16 × D
π (1004 − 554 )
π (100000000 − 9150625) × 55
=
× 55 =
16 × 100
16 × 100
= 9806054 Nmm = 9.8 kNm Ans
J=
E XAMPLE 11.4: A solid steel shaft 60 mm diameter and 800 mm long transmits 35 kW at 200
r.p.m. Calculate: i) the maximum shear stress produced; ii) the angle of twist in degrees and iii) the
shear stress at a radius of 25 mm. C = 80 GN/m2
S OLUTION :
P=
T=
i)
T
τ
=
J
r
2π NT
60
or, T =
P × 60
2π N
35000 × 60
= 1671.97 Nm
2π × 200
∴
τ=
1671.97 × 1000 × 30 × 32
T
r=
J
π (60)4
= 39.44 N/mm2 = 39.44 MPa Ans
60
= 30 mm
2
Cθ
1671.97 × 1000 × 800 × 32
T.l
T
=
∴ θ=
=
= 0.01315 radians
J
l
C.J
80000 × π (60)4
180
× 0.01315 = 0.754◦ Ans
=
π
39.44 × 25
τ
τ1 39.44 τ1
= ;
=
∴ τ1 =
= 32.87 N/mm2
r
r1 30
25
30
= 32.87 MPa Ans
Because maximum stress will be at radius
ii)
iii)
E XAMPLE 11.5: Compare the weights of hollow and solid shaft with equal lengths to transmit a
2
given torque for the same maximum shear stress if the inside diameter is of the outside.
3
@seismicisolation
@seismicisolation
Torsion
S OLUTION :
Whollow
=
Wsolid
π /4
D2o −
2
Do
3
•
179
2
× l × density
π /4 D4 × l × density
Do = Outside diameter of hollow shaft
D = Diameter of solid shaft
4
D2o − D2o 5D2
o
9
=
=
D2
9D2
Now, Thollow = Tsolid ;
τ ×π
Thollow =
D4o −
2
Do
3
(i)
4
×2
τπ
=
32 Do
65 4
D
81 o
16Do
= 0.1575D3o τ
τ
π
Tsolid = × (D)4 × 2 = 0.19625D3 τ
D 32
Equating, Tsolid = Thollow
0.19625D3 τ = 0.1575D3o τ
D3o
= 1.246
D3
Substituting in Equation (i) =
∴
Do = 1.076D
Whollow 5(1.0760D)2
=
= 0.643 Ans
Wsolid
9D2
E XAMPLE 11.6: A hollow marine propeller shaft turning at 110 rev/min is required to propel
a vessel at 47 km/h for the expenditure of 6.4 MW, the efficiency of the propeller being 68 per
2
cent. The diameter ratio of the shaft is to be and the direct stress due to thrust is not to exceed
3
8 MN/m2 . Calculate: (a) the shaft diameter and (b) the maximum shearing stress due to the torque.
S OLUTION :
Output power =
47 × 103
× P Watt
3600
where P is the propulsive force in N.
47 × 103
P = 0.68 × 6.4 × 106
3600
∴ P = 334 kN
π
∴ 334 × 103 = (D2 − d 2 ) × 8 × 106
4
∴
@seismicisolation
@seismicisolation
180
•
Strength of Materials
where D and d are outside and inside diameters, respectively
2
π 5 2
× D × 8 × 106 ; because d = D
4 9
3
D = 0.3093 m and d = 0.2062 m Ans
334 × 103 =
∴
6.4 × 106 × 10
= 556000 Nm
2π × 110
J
π
556000 = τ = τ × (D4 − d 4 )
R
16
π 0.30934 − 0.20624
=τ×
16
0.3093
T=
∴
Solving, we get
τ = 128 MN/m2
∴
Ans
E XAMPLE 11.7: A solid shaft 220 mm diameter has the same cross-sectional area of that of a
hollow shaft of the same material with inside diameter of 140 mm. Find the ratio of the power
transmitted by the two shafts at the same speed.
S OLUTION :
Dh = External dia of the hollow shaft
dn = 140 mm, the internal dia of the hollow shaft
Because the two shafts have the same area,
π 2
π
(D − 1402 ) = × 2202
4 h
4
2
Dh − 19600 = 48400
∴
D2h = 48400 + 19600 = 68000
Dh = 260.77 mm
Ratio of power transmitted by the two shafts,
Phollow Thollow Zhollow
=
=
Psolid
Tsolid
Zsolid
4
4
π (Dh − dn ) π (260.774 − 1404 )
Zhollow =
=
= 3190915.75 mm3
16Dh
16 × 260.77
π (220)3
Zsolid =
= 2089670 mm3
16
Phollow 3190915.75
= 1.527 Ans
=
Psolid
2089670
@seismicisolation
@seismicisolation
Torsion
•
181
E XAMPLE 11.8: A hollow shaft is to transmit 320 kW at 100 r.p.m. If the shear stress is not to
exceed 65 N/mm2 and the internal diameter is 0.5 of the external diameter, determine the external
and internal diameters, assuming the maximum torque is 1.5 times the mean torque.
S OLUTION :
Dh = External diameter of the hollow shaft
Di = 0.5 Dh ,
P = 320 kW
2 π N Tmean 2 × 100 × Tmean
=
60
60
320000 × 60 × 1000
= 30573.2 Nm
Tmean =
2π × 100
P=
Tmax = 1.5 × 30573.2 = 45859.8 Nm
= 45859.8 × 1000 = 45859800 Nmm
Zn =
π (D4h − D4i ) π (D4h − (0.5Dh )4 )
=
16Dh
16Dh
=
π × 0.9375D4h
= 0.184D3h
16Dh
Tmax = τ · Zh = 65 × 0.184D3h
45859800 = 65 × 0.184D3h
∴
∴
Dh = (3834431.4)1/3 = 156.52 mm
Internal diameter, Di = 156.52 × 0.5 = 78.26 mm
E XAMPLE 11.9: A shaft shown in Fig. 11.3
rotates at 220 r.p.m. with 35 kW and 20 kW
taken off at A and B, respectively and 50 kW
applied at C. Find the maximum shear stress
developed in the shaft and the angle of twist
(degrees) of the gear A relative to gear C. Take
C = 85 GN/m2 .
Ans
Ans
B
A
85 mm dia
50 mm dia
4.5 m
2m
Figure 11.3
S OLUTION :
Shaft between B and C: Power of the shaft = 50 kW, let the torque in this part be Tbc
2π NT
2π × 220Tbc
; 50000 =
60
60
Tbc = 2171.4 Nm = 2171400 Nmm
Power =
∴
@seismicisolation
@seismicisolation
C
182
•
Strength of Materials
Let σs be the maximum shear stress in this part (between B and C) of the shaft.
τs
16Tbc
π d3
= Tbc
or τs =
16
π d3
16 × 2171400
34742400
τs =
=
= 18 N/mm2
1928352.5
π 853
Shaft between B and A,
Power of the shaft = 50 − 20 = 30 kW
2π × 220 × Tab
For Tab , 30000 =
60
∴ Tab = 1302.84 Nm
Let τs be the maximum shear stress in this part of the shaft,
π d3
T × 16
= Tab ∴ τ = ab 3
16
πd
16 × 1302.84 × 1000
∴ τ =
= 53.11 N/mm2
π (50)3
τs ×
Ans
Therefore, maximum shear stress occurs at the 50 mm diameter shaft.
Twist of the Shaft
Let θbc be the twist of the shaft BC,
l Tbc
2000
2171400 × 32
.
=
×
C J
85000
π (50)4
= 0.0833 radians
θbc =
Now let θab be the twist of the shaft AB,
l Tab
4500
1302.84 × 1000 × 32
×
=
×
C
J
85000
π (85)4
= 0.013466 radians
θab =
Hence, angle of twist of A with respect to C
= θbc + θab = 0.0833 + 0.013466 = 0.096766 radians
0.096766 × 180
=
= 5.55◦ = 5◦ 33 Ans
π
E XAMPLE 11.10: A maximum shear stress of 160 MN/m2 is induced in a hollow shaft of 120
mm and 70 mm external and internal diameters, respectively. What maximum shear stress will be
developed in a solid shaft of the same weight, material and length, subjected to the same torque?
@seismicisolation
@seismicisolation
Torsion
•
183
S OLUTION :
RH =
120
= 60 mm,
2
rH =
70
= 35 mm
2
Let τ be the maximum shear stress in the hollow shaft and τ1 be the maximum shear stress in
the solid shaft
TH =
τ
π
τ1 × π Rs
4
× (R4H − rH
); Ts =
; Rs is the radius of solid shaft.
RH 2
2
Since the torque is same,
τ
π 4
τ1 × π Rs
4
=
×
R − rH
RH 2 H
2
(i)
putting the values of τ , RH and rH in Eqn. (i)
π
160 π 4
× (60 − 354 ) = τ1 × R3s
60 2
2
2.67(12960000 − 1500625) = τ1 R3s
30596531 = τ1 R3s
(ii)
Because the weight, length and material is same,
Cross-sectional area of hollow shaft = Cross-sectional area of solid shaft
π (RH − rH ) = π Rs
2
2
2
602 − 352 = R2s ∴ Rs =
∴ Rs = 48.73 mm
√
3600 − 1225
Substituting in Eqn. (ii),
30596531 = τ1 (48.73)3
∴
τ1 = 264.41 N/mm2
264.41
Maximum shear stress in solid shaft
=
= 1.65 Ans
Maximum shear stress in hollow shaft
160
Hence, hollow shaft is 1.65 stronger than solid shaft.
E XAMPLE 11.11: The stepped steel shaft shown in Fig. 11.4 is subjected to a torque T at free end
and a Torque (1.7T ) in the opposite direction at the junction of the two sizes.
@seismicisolation
@seismicisolation
184
•
Strength of Materials
1.7 T
60 mm dia
T
120 mm dia
C
A
B
1.3 mm
2 mm
Figure 11.4
What is the total angle of twist at the free end, if maximum shear stress in the shaft is limited to
80 MPa? Take C = 85 GN/m2 .
S OLUTION :
Now the torques at B and C are in opposite directions, therefore the effect of these two torques
will be studied independently (sum of the two twists, one in clockwise direction and the other in
anticlockwise direction).
To find value of T at C: It must be noted that torque in AB will induce more stress in BC because
of smaller diameter between B and C. Hence, let us first calculate the torque in BC because it is less
stressed than permissible in AB.
T=
π
π
× 80 × 603 = 3391200 Nmm
τ (DBC )3 =
16
16
Polar moment of inertia,
π
(DAB )4 =
32
π
JBC = (DBC )4 =
32
JAB =
π
(120)4 = 20347200 mm4
32
π
(60)4 = 1271700 mm4
32
For angle of twist due to T at C
T.l
T lAB lBC
+
=
J.C C JAB JBC
3391200
1300
2000
=
+
85000
20347200 1271700
θ=
= 39.9 6.39 × 10−5 + 157.27 × 10−5
= 39.9 × 163.66 × 10−5 = 0.0653 radians
Now angle of twist at C due to torque 1.7T at B,
θ=
T lAB
1.7T × 1300
=
×
C JAB 85000 × 20347200
Substituting for
θ=
1.7 × 3391200 × 1300
= 0.00433 radians
85000 × 20347200
@seismicisolation
@seismicisolation
Torsion
•
185
Hence, angle of twist = 0.0653 − 0.00433 = 0.06097 radians
0.06097 × 180
=
= 3.495◦ Ans
π
Composite Shaft
In a composite shaft, there are two or more shafts of different materials fixed together. The applied
torque is shared by each material.
1
Total torque T is shared as torque T1 by material
number 1 and T2 by material number 2.
∴
2
T1 + T2 = T
(i)
Since the angle of twist of the two shafts is
same,
θ1 = θ2
T2
T1
=
C1 J1 CJ2
Figure 11.5
or
T1 C1 J1
=
T1 C2 J2
(ii)
Solving Eqns. (i) and (ii) T1 and T2 can be found out and the stresses in the two materials are
given by:
T1
T2
τ1 = ; τ2 =
Z1
Z2
Twisting Beyond the Limit of Proportionality
If the torque applied to a shaft is sufficient to cause yielding in the material, the relation between the
shear stress and the angle of twist is assumed to be similar to that between the direct stress and angle
of bending for an overstrained beam. Thus, the stress is proportional to the radians up to the limit
of proportionality, after which it remains constant over the remainder of the section of the shaft.
dx
r
R
x
τ
Let us consider a shaft section of radius R
Fig. (11.6), which is subjected to a torque sufficient to cause yielding to a radius r; let the shear
stress at the limit of proportionality be τ .
For the plastic part, then,
π
T = τ Z = τ × r3
2
For the plastic part, the torque on an elementary
ring of radius x and thickness dx is z×2π x×dx×x.
Figure 11.6
@seismicisolation
@seismicisolation
186
•
Strength of Materials
R
Total torque on plastic part =
r
τ × 2π x2 dx
2
= τ × π (R3 − r3 )
3
The total torque carried by the shaft is then the sum of torques carried by the elastic and plastic
parts.
Note: T. J is called torsional rigidity.
Polar Modulus:
ZP =
J
= Polar modulus
R
E XAMPLE 11.12: A composite shaft consists of a steel rod of 70 mm diameter surrounded by a
closely fitting tube of brass. Find the outside diameter of the tube so that when a torque of 1200
Nm is applied to the composite shaft, it will be shared equally by the two materials. C for steel =
85 GN/m2 , C for brass = 45 GN/m2 . Find also the maximum shear stress in each material and
common angle of twist in a length of 3.8 m.
S OLUTION :
Total torque, T = Ts + Tb = Ts + Ts (∵ Ts = Tb given)
T
1200000
=
= 600000 Nmm
2
2
Tb = Ts = 600000 Nmm
Cθ
C.θ .J
T
=
OR, T =
J
l
l
Cs θs Js
For steel, Ts =
ls
Cb θb Jb
For brass, Tb =
lb
Cs θs Js Cb θb Jb
since Ts = Tb ;
∴
=
ls
lb
Ts =
As we know ls = lb
∴
Cs θs Js = Cb θb Jb
But in composite shaft, the angle of twist in each shaft is same.
∴ θs = θb
OR, Cs Js = Cb Jb
π
π
85000 × (70)4 = 45000 × [D4 − 704 ]
32
32
where D is the outside diameter of brass tube
2.04 × 1012 = 45000 D4 − 24010000
@seismicisolation
@seismicisolation
Torsion
D4 − 24010000 = 4.533 × 109
D4 = 45333333 + 24010000
D4 = 69343333,
OR,
D = 91.25 mm
Ans
For maximum shear stresses:
τ=
T ×R
;
J
For steel, τs
Ts × d/2
Js
70
600000 ×
2 = 600000 × 16 = 8.913 N/mm2 Ans
τs =
π
π (70)3
(70)4
32
D
D
Tb ×
600000 ×
300000D
2
2
τb =
= π
=
Jb
0.1067(D4 − 24010000)
(D4 − 704 )
32
300000 × 91.25
273750000
=
=
= 5.66 N/mm2
4
4835819.3
0.1067 × (91.25 − 24010000)
Common angle of twist,
π
4
Cs × θs × Js 85000 × θs × 32 (70)
=
= 52699580.6 θs
Ts =
ls
3800
substituting for Ts ,
600000 = 52699580.6θs
∴ θs = 0.0114 radians
0.0114 × 180
=
= 0.6535◦
π
Ans
Torsion of a Tapering Shaft
B
x
A
R1
R
x
dx
l
Figure 11.7
@seismicisolation
@seismicisolation
R2
Ans
•
187
188
•
Strength of Materials
Let the tapered shaft in Fig. 11.7 be subjected to a torque T .
τ1 = Maximum shear stress at A
τ2 = Maximum shear stress at B
The shear stress at a distance x and section X = τ ,
π
π
π
Now, T = τ1 × R31 = τ2 × R32 = τ × × R3
2
2
2
or, τ1 R31 = τ2 R32 = τ R3
Angle of twist of small length dx,
R is the radius at section x.
T dx
2T dx
T dx
=
π 4 =
CJ
C× 2R
Cπ R4
R2 − R1
R = R1 +
× x = R1 + kx
l
dθ =
Also,
where k =
(i)
R2 − R1
, k is constant for this shaft.
l
dθ =
2T
dx
·
Cπ (R1 + kx)4
(ii)
∴ Total angle of twist for length l of the shaft = θ d θ
l
1
dx
2T 1
2T
=
.
=− .
Cπ (R1 + kx)4
Cπ 3k (R1 + kx)3 0
0
2T 1
1
1
=− .
−
Cπ 3k (R1 + kl)3 R31
l
Now,
k=
R2 − R1
l
or, kl = R2 − R1 ; OR
∴
R2 = R1 + kl
2 T
1
1
1
2 T
1
−
−
.
.
=
3k Cπ R32 R31
3k Cπ R31 R32
3
R2 − R31
2T
l
θ=
3Cπ R2 − R1
R31 R32
2 T l R21 + R1 R2 + R22
θ= .
3 Cπ
R31 R32
θ =−
In case of a shaft of uniform radius R,
R1 = R2 = R
@seismicisolation
@seismicisolation
(iii)
Torsion
•
189
2 T l 3R2
2 T l R2 + R2 + R2
=
θ= .
.
3 Cπ
3 Cπ R6
R6
2 Tl
3
Tl
θ= .
×
=
,
3 Cπ R4 CJ
which is same as before.
E XAMPLE 11.13: A 1.2 m long shaft tapers uniformly from a diameter of 100 mm to a diameter
of 140 mm. If the shaft transmits a torque of 18 kNm, find:
i) Angle of twist
ii) Maximum shear stress developed. Take C = 85 GN/m2
S OLUTION :
R1
R2
l
Figure 11.8
140
100
= 70 mm, R1 =
= 50 mm
2
2
T = 18000 Nm, l = 1.2 m, Now working in N and m;
R2 =
Angle of twist, θ :
2 T l R21 + R1 R2 + R22
; working in N and m
θ= .
3 Cπ
R31 R32
2 18000 × 1.2 0.052 + 0.05 × 0.07 + 0.072
= ×
3
85 × 109 π
0.053 × 0.073
−3
+ 3.5 × 10−3 + 4.9 × 10−3
−8 2.5 × 10
= 5.3953 × 10
1.25 × 10−4 × 3.43 × 10−4
10.9 × 10−3
−8
= 5.3953 × 10
= 0.01372 radian
4.2875 × 10−8
@seismicisolation
@seismicisolation
190
•
Strength of Materials
=
0.01372 × 180
= 0.786◦
π
Ans
Maximum shear stress developed:
Tmax = τmax ×
π
× D31
16
Because maximum shear stress occurs at the smallest diameter.
π
(0.1)3
16
τmax = 91.72 MN/m2 Ans
18 × 103 = τmax ×
∴
Thin Circular Tube Subjected to Torsion
Consider a thin circular tube of external diameter D and thickness t. Whereas t is very small
compared to diameter D.
t
J = Area of the section × Square of radius
D 2 π D3t
= π Dt
=
2
4
J
π D3t × 2
=τ×
R
4×D
2
τπ D t
T=
2
T =τ×
∴
D
Thin circular tube
(i)
Figure 11.9
and
4T l
T.l
T.l.4
=
=
3
C.J C.π D t
π D3tC
4T l
θ=
π D3tC
θ=
∴
Strength weight ratio =
T
W
W = π Dtlw
T=
∴
Hence,
(ii)
where w is the weight density of the material.
τπ D2t
2
T
τπ D2t
τD
=
=
W
2π Dtlw 2lw
τD
T
=
W
2lw
@seismicisolation
@seismicisolation
(iii)
Torsion
•
191
E XAMPLE 11.14: A thin steel tube of 90 mm diameter is 3 mm thick. If the allowable shear stress
is 75 MN/m2 and C = 82 GN/m2 , find:
i) Safe twisting moment that can be applied to the tube;
ii) The twist in a length of 550 mm.
S OLUTION :
D = 90 mm = 0.09 m
t = 3 mm = 0.003 m
i) Safe twisting moment T :
τπ D2t 75 × 106 × (0.09)2 (0.003)
=
2
2
= 911.25 Nm Ans
T=
ii) Angular twist θ :
4 × 911.25 × 0.55
4T l
=
3
π D tC π (0.09)3 (0.003) × 82 × 109
= 0.00356 radians
0.00356 × 180
=
= 0.204◦ Ans
π
θ=
E XAMPLE 11.15: A shaft LMN of 550 mm length and 45 mm external diameter is having a hole
for a part of its length LM, of a 22 mm diameter and for the remaining length MN having a hole of
32 mm diameter. If the shear stress is not to exceed 80 N/mm2 , find the maximum power, the shaft
can transmit at a speed of 250 r.p.m.
If the angle of twist in the length of 22 mm diameter hole is equal to that in the 32 mm diameter
hole, find the length of the shaft that has been bored to 22 mm and the length of the shaft that has
been bored to 32 mm diameter.
4
do − di4
π
T=
τ
16
do
22 mm
32 mm
N
L
M
l2
l1
500 mm
Figure 11.10
@seismicisolation
@seismicisolation
45 mm
192
•
Strength of Materials
For shaft LM,
4
π
45 − 224
= 0.00436 × 80 [4100625 − 234256]
τ
16
45
= 1348589.5 Nmm
T1 =
For shaft MN,
4
4100625 − 1048576
45 − 324
π
π
=
× 80
τ
T2 =
16
45
16
45
= 1064826 Nmm
So safe torque is minimum torque 1064826 Nmm or 1064.826 kNm.
2π × 250 × 1064.826
2π NT
=
= 27862.95
60
60
P = 27.86 kW Ans
P=
Now θ =
T.l
C.J
T l1
C.J1
T l2
Angle of twist for MN, θ2 =
C.J2
Angle of twist for
LM, θ1 =
But θ1 = θ2
T l1
T l2
l1
l2
=
∴
=
CJ1 CJ2
J1 J2
π
π
J1 = [454 − 224 ] = [4100625 − 234256]
32
32
= 379387.46 mm4
π
π
J2 = [454 − 324 ] = [4100625 − 1048576]
32
32
= 299482.3 mm4
l1
l2
Substituting in Eqn. (i)
=
J1 J2
or,
l1 J1 379387.46
= 1.267
=
=
l2 J2
299482.3
l1 = 1.267 l2
= 1.267(500 − l1 ) ∵ l1 + l2 = 500 mm
∴
@seismicisolation
@seismicisolation
(i)
Torsion
•
193
l1 = 633.5 − 1.267 l1
or
l1 + 1.267 l1 = 633.5
633.5
= 279.4 mm Ans
2.267
32, l2 = 500 − 279.4 = 220.6 mm Ans
2.267 l1 = 633.5
of dia
∴
l1 =
Exercise
11.1 A shaft is transmitting 100 kW at 160 r.p.m. Find a suitable diameter for the shaft if the maximum torque transmitted exceeds the mean by 25%. Take allowable shear stress as 70 N/mm2 .
[Ans 80 mm]
11.2 A hollow shaft is to transmit 337.5 kW at 100 r.p.m. If the shear stress is not to exceed
65 N/mm2 and the internal diameter is 0.6 of the external diameter, find the external and
internal diameters assuming that the maximum torque is 1.3 times the mean.
[Ans 155.2 mm, 93.12 mm]
11.3 A hollow shaft of external diameter 120 mm transmits 300 kW power at 200 r.p.m. Determine
the maximum internal diameter if the maximum stress in the shaft is not to exceed 60 N/mm2 .
[Ans 88.5 mm]
11.4 A solid cylindrical shaft is to transmit 300 kW power at 100 r.p.m.
a) If the shear stress is not to exceed 80 N/mm2 , find its diameter.
b) What percent saving in weight would be obtained if this shaft is replaced by a hollow
one, whose internal diameter equals to 0.6 of the external diameter, the length, the
material and maximum shear stress being the same.
[Ans a) 122 mm b) 29.55%]
11.5 A hollow shaft, having an inside diameter 50% of its outer diameter transmits 600 kW at
150 r.p.m. Determine the external diameter of the shaft if the shear stress is not to exceed
65 N/mm2 and the twist in a length of 3 m should not exceed 1.4 degrees. Assume maximum
torque = 1.2 mean torque and C = 100 GMPa.
[Ans 157.0 mm]
11.6 A solid shaft of 150 mm diameter is to be replaced by a hollow shaft of the same material with
internal diameter equal to 60% of external diameter. Find the saving in material, if maximum
allowable shear stress is the same for both the shafts.
[Ans 30.9%]
11.7 A hollow shaft has an external diameter of 0.3 m and internal diameter of 0.15 m. Compare
its strength with that of a solid shaft of the same weight per unit length.
[Ans Hollow 44%, stronger than solid]
@seismicisolation
@seismicisolation
194
•
Strength of Materials
11.8 Two shafts of the same material and same length are subjected to same torque. If the first
shaft is of a solid circular section and second shaft is of hollow circular section whose internal
diameter is 2/3 of the external diameter and the maximum shear stress developed in each is
the same, compare the weights of two shafts.
Wh
[Ans
= 0.643]
Ws
11.9 A shaft running at 180 r.p.m. has to transmit 97.5 kW. The shaft must not be stressed beyond
70 N/mm2 and also it must not twist more than 1 degree in a length of 3 m. What diameter
would you recommend? C = 90 GN/m2 .
[Ans d = 87.82 mm]
11.10 A mild steel tube 80 mm external diameter has a wall thickness of 5 mm. Find the torque
necessary to induce a maximum shear stress of 80 N/mm2 in a length of 1.8 m of the tube
and the amount of twist in degrees. C = 90 GN/m2
[Ans T = 3.328 kNm, θ = 2.292◦ ]
11.11 A solid circular shaft tapers uniformly from D at one end to d at the other end. It D = 1.2d,
determine the percentage error committed, if the angle of twist for a given length is calculated
using the mean diameter.
[Ans 2.727%]
11.12 A compound shaft is formed by surrounding a 50 mm diameter solid brass shaft by a steel
tube having wall thickness of 5 mm. The metals are securely attached to each other at their
ends. Determine the increase in torque carrying capacity of the composite shaft over that of
brass shaft alone as above. Assuming C for brass = 50 GMPa; C for steel = 120 GMPa. The
working stress for brass and steel may be taken as 75 N/mm2 and 120 N/mm2 respectively.
[Ans 98.7%]
11.13 A composite shaft consists of a steel rod of 60 mm diameter surrounded by a close fitting
tube of brass. Find the outside diameter of the brass tube, when a torque of 1 kNm is applied
on the composite shaft end shared equally by the two materials. C for steel as 84 GPa and C
for brass as 42 GPa. Also find the common angle of twist in 4 m.
[Ans 79 mm, 1.07◦ ]
11.14 A solid shaft 3.6 m long and 75 mm in diameter is fixed at both its ends. A twisting moment
of 22.8 kNm is applied at a distance of 15 mm from one end. Calculate i) twisting moment
shared by each portion; ii) the angle of twist on both sides of the place of application of the
twisting moment, and iii) the maximum shear stress at each side. C = 90 GN/m2 .
[Ans 13.3 kNm, 9.5 kNm, 0.0715 radian, 114.7 MN/m2 , 160 MN/m2 ]
11.15 A steel shaft 1 m long, 30 mm diameter is rigidly fixed at the ends. A torque of 600 Nm
is applied at a distance of 250 mm from one end. Calculate: i) Fixing couples at the ends,
ii) maximum shearing stress and iii) angle of twist at the point of application of torque.
C = 82 GN/m2
[Ans 150 Nm, 450 Nm, 84.8 MN/m2 ; 0.01725 radian]
11.16 Design a suitable diameter for a circular shaft required to transmit 117.6 kW at 100 r.p.m.
The shear stress in the shaft is not to exceed 70 MPa and the maximum torque exceeds the
mean by 40%. Also calculate the angle of twist in a length of 2 m. Take C = 90 GPa.
[Ans 78 mm; 2.336◦ ]
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Torsion
•
195
11.17 A hole equal to half the diameter of C shaft is drilled in a shaft throughout its length. By what
percentage are its weight and torsional strength reduced?
[Ans Weight = 25%; Strength = 6.3%]
11.18 A steel bar 19 mm diameter is encased in a closely fitting brass tube of 32 mm external
diameter, securely fixed together at the ends. The compound bar is subjected to a torque of
520 Nm and the angle of twist measured on a gange length of 250 mm is found to be 1.8◦ . If
Csteel = 80 GN/m2 and calculate C for the brass.
Find also the maximum shearing stresses in the two materials and the proportions of the
total strain energy taken up by each part.
[Ans 34.6 GN/m2 ; 95.5 MN/m2 , 69.5 MN/m2 ;
1
0.248 : 0.752; Hint: Strain Energy = T θ ]
2
11.19 A hollow circular shaft 20 mm thick transmits 300 kW power at 200 r.p.m. Determine the
external diameter of the shaft if the shear strain due to torsion is not to exceed 0.00086. Take
C = 80 GPa.
[Ans 107.5 mm]
11.20 A shaft is required to transmit a power of 300 kW running at a speed of 120 r.p.m. If the shear
strength of the shaft material is 70 N/mm2 , design a hollow shaft with inner diameter equal
to 0.75 times the outer diameter.
[Ans Do = 136 mm Di = 102 mm]
11.21 A stepped shaft of 2 m length consists of three lengths of diameter of 90 mm, 70 mm and
50 mm in sequence. If the angle of twist is the same for each section, compute the length of
each section and the total twist. The maximum shear stress in the entire shaft is not to exceed
50 N/mm2 . Take C = 80 GPa.
[Ans l1 = 1368.2 mm, l2 = 501.7 mm, l3 = 130.3 mm]
11.22 Prove that a hollow shaft is always stronger than a solid shaft of the same material, weight
and length when subjected to simple torque.
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C HAPTER
12
THIN CYLINDRICAL AND SPHERICAL SHELLS
It is very important to study thin cylindrical and spherical shells as we generally come across in
industries these containing fluids for example, boilers, tanks, compressed air receivers and in chemical industry particularly. The thickness of a thin shell is less than 1/10th to 1/15th of its diameter.
Types of stresses in thin shells: Due to the pressure of internal fluid, the following types of stresses
develop in thin cylinders:
i) Hoop or circumferential stress
ii) Longitudinal stress
iii) Radial stress, but it is negligible as compared to hoop stresses or longitudinal stresses, so not
considered.
i) Hoop Stress or Circumferential Stress
Thickness 't'
d X
P
X
(b)
l
(a)
't'
(c)
Figure 12.1
If pressure p succeeds in bursting the cylinder into two halves as shown in Fig. 12.1(c) [one
half is not shown], it is evident that resisting area is tl + tl = 2tl. Let σh be the hoop stress or
circumferential stress then,
Bursting force = p d l
(i)
[Note: d is the internal dia and dl is the projected area where pressure p acts to split it into two
halves.]
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Thin Cylindrical and Spherical Shells
Resisting force = σh × 2tl
197
p
d = dia
(ii)
Equating Eqns. (i) and (ii),
•
Figure 12.2
σh × 2tl = pdl
pd
∴ hoop stress, σh =
2t
(iii)
ii) Longitudinal Stress
As both ends of cylinder shown in Fig. 12.1(a) are closed with plates by rivets or welding, the force
acting is on σh both ends.
π
So bursting force = p d 2
4
(iv)
Let σl be the longitudinal stress developed in thin shell, then,
Resisting force = σl × resisting area
d
= σl × 2π t
2
(v)
Equating Eqns. (iv) and (v),
d
π
σl × 2π t = p d 2
2
4
pd
So longitudinal stress, σl =
4t
If joint efficiency of longitudinal joint of the shell is ηl then Eqn. (iii) becomes:
Hoop or circumferential stress, σh =
ρd
2t ηl
And if joint efficiency of circumferential joint in given as ηc then Eqn. (v) becomes:
Longitudinal stress, σl =
ρd
4t ηc
Sometimes only one common efficiency of joint is given on set η , then,
σh =
ρd
2t η
and σl =
ρd
4t η
Change in the dimensions of a thin cylindrical shell:
Let μ be the Poisson’s ratio of the material of shell.
pd
pd
and longitudinal stress σl =
As we know, hoop stress σh =
2t
4t
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(vi)
198
•
Strength of Materials
σ
σ
Now, circumferential strain = h − μ l = εl
E
E
Since the circumference is proportional to the diameter, so that the diameter strain is equal to
circumferential strain.
δd
∴ diameter strain,
=ε
d
Using the above two equations, increase in diameter can be calculated.
δh
δl δl
Again longitudinal strain
=
−μ
= ε2 from which the increase in length can be deterl
E
E
mined.
Change in Volume
Volume of the shell, v =
π 2
d l
4
Taking log of both sides of above equation
log v = log
π
+ 2 log d + log l
4
Differentiating,
δv
δd δl
=2
+
v
d
l
= 2ε1 + ε2
From this equation the increase in volume can be found out.
It may be noted that in deriving above formulae, the following, assumptions have been made:
i) The radial stresses in the cylinder wall are negligible.
ii) The stresses are uniformly distributed throughout the wall of the cylinder.
iii) There is no longitudinal stays in the cylinder.
iv) Material of the cylinder obeys Hooke’s law and does not exceed elastic limit.
v) Young’s modulus (E) is constant in material at all points.
E XAMPLE 12.1: A cylindrical shell of 1.5 m diameter is made up of 20 mm thick plates. Determine the hoop and longitudinal stresses in the plates, if the shell is subjected to an internal pressure
of 3 MPa. Take efficiency of the joints as 75%.
S OLUTION :
Hoop stress,
Longitudinal stress,
pd
2t η
3 × 1500 × 100
= 150 MN/m2 Ans
=
2 × 20 × 75
pd
σl =
4t η
3 × 1500 × 100
= 75 MN/m2 Ans
=
4 × 20 × 75
σh =
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Thin Cylindrical and Spherical Shells
•
199
E XAMPLE 12.2: A thin cylinder containing a fluid is of internal diameter 50 mm and of 5 mm
thick sheet. If tensile stress in the material is not to exceed 38 MPa, find the maximum pressure to
which a fluid can be subjected to.
S OLUTION :
Let p be the maximum pressure which can be allowed in the cylinder.
Now,
pd
p × 50
σ=
=
=5p
2t
2×5
Now we know hoop stress is of tensile nature and maximum permissible tensile stress is
38 MPa.
∴
38 = 5 p
38
= 7.6 MN/m2
Hence, maximum pressure, p =
5
Ans
E XAMPLE 12.3: The internal pressure in a thin cylindrical shell of diameter 450 mm is 6 MPa.
Determine the minimum thickness of the shell, if allowable tensile strength in the sheet material is
350 MPa and efficiency of the joints is 70 %. Take factor of safety as 4.
S OLUTION :
pd
2t η
6 × 450 × 100
350 =
2 × t × 70
6 × 450 × 100
= 5.51 mm
∴ t=
2 × 70 × 350
σh =
Considering factor of safety as 4,
Required thickness = 5.51 × 4
= 22.04 mm
Ans
E XAMPLE 12.4: A cylinder, 2.8 m long and 800 mm in diameter is subjected to an internal
pressure of 2.5 N/mm2 . Determine the minimum thickness of the metal if the stress in the material
of the shell is not to exceed 70 N/mm2 .
Calculate:
(i) the change in diameter
(ii) the change in length
(iii) the change in volume
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200
•
Strength of Materials
Given E = 200 GN/m2 and μ = 0.25
S OLUTION :
∴
σh =
pd
2t
σh =
2.5 × 800
,
2×t
but σh is given as 70 N/mm2
70 =
2.5 × 800
2×t
∴
σh =
2.5 × 800
= 70.03 MN/m2
2 × 14.28
σl =
2.5 × 800
pd
=
= 35.015 MN/m2
4t
4 × 14.28
Now, diametral strain, ε1 =
t=
2.5 × 800
= 14.28 mm
2 × 70
Ans
σ
σh
−μ l
E
E
δd
1
= ( σh − μ σl )
d
E
=
1
(70.03 − 0.25 × 35.015)
200000
= 3.064 × 10−4
∴
Change in diameter, δ d = 3.064 × 10−4 × 800 = 0.24512 mm (increase)
Longitudinal strain,
ε2 =
=
ε2 =
=
Ans
σh
σl
−μ
E
E
1
(σl − μσh )
E
δl
1
= (σl − μσh )
l
E
1
(35.015 − 0.25 × 70.03)
200000
= 0.0000874
∴
Change in length, δ l = 0.0000874 × 2800 = 0.24472 mm (increase)
δV
= 2ε1 + ε2
V
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Ans
Thin Cylindrical and Spherical Shells
•
201
= 2 × 3.064 × 10−4 + 8.75375 × 10−5
= 0.0003064 + 0.000874 = 0.0003939375
∴
Change in volume, δ V = 0.0003938 ×V
π 2
π
d × l = (800)2 × 2800
4
4
= 1406720000 mm3
δ V = 0.0003938 × 1406720000
Volume, V =
∴
= 553966.3 mm3
(Increase)
Ans
E XAMPLE 12.5: A boiler is made of 12 mm thick plates and is subjected to a steam pressure
of 2 N/mm2 . The efficiencies of the longitudinal and circumferential joints are 70% and 40%,
respectively. If permissible tensile stress in boiler plates is 110 N/mm2 , determine the maximum
allowable diameter.
S OLUTION :
σh =
pd
2 × d × 100
= 0.119d
=
2t ηl
2 × 12 × 70
Allowable σh = 110 N/mm2
∴
or,
110 = 0.119d
d = 924.37 mm
Again,
σl =
pd
2 × d × 100
=
= 0.1042d
4t ηc
4 × 12 × 40
Permissible tensile stress is 110 N/mm2
∴ 110 = 0.1042 d
Hence, d = 1055.6 mm
For safety minimum of the two diameters is the required diameter
Hence, d = 924 mm
Ans
E XAMPLE 12.6: A cylindrical water tank 6.5 m in diameter with its axis vertical is made from
steel plates 3 mm thick. Determine the maximum height to which the tank may be filled if the hoop
stress is permissible upto 65 MPa.
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202
•
Strength of Materials
S OLUTION :
p × 6.5
pd
or, 65000000 =
2t
2 × 0.003
(Note: working in N and m) ∴ p = 60000 N/m2
σh =
We know p = wh; w for water = 10 × 103 = 10000 N/m3 [Taking g = 10 m/s2 appr.]
60000
60000 = 10000 × h ∴ height of water =
=6m
10000
Hence, h = 6 metre.
Ans
E XAMPLE 12.7: A thin cylinder of diameter 500 mm is made of 6 mm thick plate. The efficiencies
of the longitudinal and circumferential joints are 70 and 40%, respectively. Determine the largest
permissible pressure, if the tensile stress of the plate is limited to 100 MPa.
pd
2t ηl
p × 500 × 100
100 =
2 × 6 × 70
∴ p = 1.68 MPa
pd
σl =
2t ηc
p × 500 × 100
100 =
2 × 6 × 40
∴ p = 0.96 MPa
σh =
Hence, permissible pressure, i.e., p = 0.96 MPa Ans
Thin Spherical Shells:
π 2
d
4
Bursting force = projected area × pressure
π
(i)
= d2 × p
4
Projected area =
Resisting force: let σ be the stress developed in
the material of spherical shell.
Resisting force = σ × π dt
t
d
(ii)
p
Figure 12.3
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Thin Cylindrical and Spherical Shells
•
203
Equating (i) and (ii)
π 2
d p
4
pd
σ=
4t
σ π dt =
∴
If η is the efficiency of the joint, then
σ =
pd
4t η
Ans
E XAMPLE 12.8: A spherical gas vessel of 1.4 m diameter is subjected to a pressure of 2.2 MPa.
Determine the stress induced in the plate of vessel, if thickness of the vessel plate is 6 mm.
S OLUTION :
pd
4t
2.2 × 1400
= 128.3 MPa
=
4×6
σ=
Ans
E XAMPLE 12.9: A spherical shell of 1.5 m diameter is subjected to an internal pressure of 3 MPa.
Determine the minimum thickness of the plates required, if the permissible stress induced in the
material of the shell is not to exceed 90 MPa. Take efficiency of joint as 75%.
S OLUTION :
pd
4t η
3 × 1500 × 100
90 =
4 × t × 75
3 × 1500 × 100
= 16.67 mm
∴ t=
90 × 4 × 75
σ=
Ans
Change in diameter and volume of a thin spherical shell due to an internal pressure:
In case of stress everywhere in spherical shell is same try σ . Then,
σ μσ
−
E
E
σ
= (1 − μ )
E
pd
(1 − μ )
=
4tE
Circumferential strain =
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204
•
Strength of Materials
Because circumference is proportional to the diameter, the diametral strain is given by
pd
δd
=
(1 − μ )
d
4tE
If V is the volume of the spherical shell,
4
4
V = π r3 = π
3
3
V=
so,
3
d
π
= d3
2
6
π d3
6
Now, because the volumetric strain in a sphere is 3 times the linear strain,
3pd
δV
δd
=3
=
(1 − μ )
V
d
4tE
From the above equation change in volume, δ V can be calculated.
E XAMPLE 12.10: Calculate the increase in diameter and volume of a spherical shell 1200 mm
in diameter and 12 mm thick when it is subjected to an internal pressure of 2 N/mm2 . Take E =
200 GPa, μ = 0.3.
S OLUTION :
pd
δd
=
(1 − μ )
d
4tE
Change in diameter,
δd =
=
pd 2
(1 − μ )
4tE
2 × (1200)2
(1 − 0.3)
4 × 12 × 200000
= 0.21 mm increase
δV
3δ d
3pd
=
=
(1 − μ )
V
d
4tE
π
π
Volume, V = d 3 = (1200)3
6
6
= 904320000 mm3
Change in volume,
δV =
V × 3pd
(1 − μ )
4tE
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Thin Cylindrical and Spherical Shells
=
•
205
904320000 × 3 × 2 (1200)
(1 − 0.3)
4 × 12 × 200000
= 678240 mm3 (increase)
Ans
Wire-Bound Thin Cylindrical Shells
In order to strengthen a cylindrical shell against bursting in longitudinal section due to hoop or
circumferential stress, a wire is bound tightly around the outer diameter of the shell. This wire will
be under tension and is closely wound around the shell as shown in Fig. 12.4. This increases the
strength of the thin cylinder to withstand high internal pressure without excessive increase in wall
thickness. By winding wire, an initial tensile stress σw in wire is produced. This wire is of high
tensile material.
The bursting force due to internal pressure per mm length is equal to the resisting force due to
pipe section plus wire section. The circumferential stress in the pipe is also equal to the strain in the
steel wire.
Wire
Cylinder
Wire wound
around a pipe
t
1R
d
σwσc
σcσw
l
Figure 12.4
E XAMPLE 12.11: A cast iron pipe of 350 mm internal diameter and 15 mm thick is wound closely
with a single layer of circular steel wire of 6 mm diameter under tension of 70 MPa. Determine the
initial compressive stress in the pipe section. Also find stresses set up in the pipe and steel wire,
when water under pressure of 7 MPa is admitted into the pipe. E for cast iron = 100 GPa, E for
wire steel = 200 GPa and Poisson’s ratio = 0.3.
S OLUTION :
Initial compressive stress in the pipe section:
Before admission of pressurised water, pipe is subjected to compression due to tension in the
wire. Let us consider one mm length of the pipe.
In 6 mm there are two sections (one at bottom and the other at top). Therefore 1 mm will have
2
= 0.333 wire sections.
6
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206
•
Strength of Materials
Initial compressive force in the wire, before the pressurised water is admitted into the pipe,
π
(6)2 × 70 = 658.7 N
= 0.333 ×
4
Hence, initial compressive stress in the pipe section,
σc =
658.7
= 21.96 N/mm2
2 × 15
Stresses set up in the pipe and steel wire:
Let
σ p = Stress in the pipe section in N/mm2 and
σw = Stress in the steel wire in N/mm2
Bursting force per mm length of the pipe, when water under pressure is admitted
= p × d × l = 7 × 350 × 1 = 2450 N
Total resisting force = Resisting force in pipe + Resisting force in wire
π
= [2σPtl] + σw × 0.333 × (6)2
4
= [2σP × 15 × 1] + [σw × 9.411]
= 30 σP + 9.411 σw
(i)
(ii)
Now the bursting force in pipe = total resisting force
Therefore, equating Eqns. (i) and (ii),
2450 = 30 σP + 9.411 σw
(iii)
Circumferential strain in the pipe,
σp
1
0.3 × 7 × 350
pd
1
=
σp −
− μ
×
E
4t
Ec
Ec
4 × 15
σ p − 12.25
=
Ec
σw
Strain in the steel wire =
Es
=
(iv)
(v)
Since circumferential strain in the pipe is equal to the strain in the steel wire, so equating
Eqns. (iv) & (v)
σ p − 12.25 σw
=
Ec
Es
Es
σw =
× (σ p − 12.25)
Ec
200000
× (σ p − 12.25) = 2(σ p − 12.25)
σw =
100000
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(vi)
Thin Cylindrical and Spherical Shells
•
207
Substituting the value of σw in Eqn. (iii),
2450 = 30 σ p + 9.411 × 2(σ p − 12.25)
2450 = 30 σ p + = 18.822 σ p − 230.57
2450 = 48.822 σ p − 230.57
∴
σ p = 54.9 N/mm2
σw = 2 (σ p − 12.25)
{from Eqn. (vi)
σw = 2 (54.9 − 12.25) = 85.3 N/mm2
Hence, final stress in the pipe section
= 54.9 − 21.96 = 32.94 N/mm2
Final stress in steel wire = 70 + 85.3 = 155.3 N/mm2
Ans
Ans
E XAMPLE 12.12: A brass tube of internal diameter 160 mm and external diameter 180 mm, is
wound closely with a steel wire of diameter 2.5 mm at an initial tension of 120 N/mm2 . If the tube is
Es
= 2, μ
subjected to an internal pressure of 3 N/mm2 , find the stresses in the brass and the steel,
Eb
for brass = 0.25.
S OLUTION :
Let us consider 10 mm length (l) of the wire bound pipe, number of turns of wire in 10 mm
length
10
=
=4
2.5
Axial
plane
180 mm dia
10 mm
Figure 12.5
An axial plane will cut the wire at eight sections
π
Whose area = 8 × (2.5)2 = 39.2 mm2
4
Before internal pressure,
Total tensile force in the wire
= 120 × 39.2 = 4704 N
180 − 160
Thickness of tube =
= 10 mm = t
2
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208
•
Strength of Materials
Area of brass section by the axial plane
= 2 × l × t = 2 × 10 × 10 = 200 mm2
If the compressive stress in the tube is σbc , the total compressive resistance in the tube
= σbc × 200 N
Equating this to the tensile force in the wire, we get
200 σbc = 4704
∴
σbc = 23.52 N/mm2 (Compressive)
Therefore, before the tube is subjected to internal pressure, the compressive stress in it is
23.52 N/mm2 .
Let the hoop stress in the tube and wire induced by the internal pressure be σb and σs ,
respectively.
Bursting force in the tube = pdl = 3 × 160 × 10 = 4800 N
Resistance offered by wire = σs × 39.2 N
Resistance offered by tube = σb × 200 N
Total resisting force = bursting force
39.2σs + 200σb = 4800
Hoop strain in the wire =
(i)
σs
Es
pd
Hoop stress in the tube is σb which is not equal to
due to force exerted by the wire. The
2t
pd
.
longitudinal stress in the tube =
4t
σb
pd
−μ
Eb
4tEb
0.25 × 3 × 160
σ
= b−
Eb
4 × 10 × Eb
3
σb
=
−
Eb Eb
Hoop strain in the tube =
Strain in wire and tube is equal
∴
σs σb 3
= −
Es Eb Eb
Es
σs =
(σb −3)
Eb
σs = 2 (σb −3)
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(ii)
Thin Cylindrical and Spherical Shells
•
209
Solving Eqns. (i) and (ii)
Substituting for σs from Eqn. (ii) in Eqn. (i);
39.2 × 2 (σb −3) + 200σb = 4800
78.4 (σb −3) +200σ b = 4800
78.4σ b −235.2 + 200σ b = 4800
278.4σ b = 5035.2
σb = 18.086 N/mm2 (Tensile)
From Eqn. (ii) & substituting for σb ,
σs = 2 (σb − 3)
σs = 2 (18.06 − 3)
= 30.12 N/mm2 tensile
The initial stresses in the wire and tube are 120 N/mm2 (tensile) and 23.52 N/mm2 (compressive),
respectively,
Therefore, the final stresses are:
Stress in the wire = 120 + 30.12 = 150.12 N/mm2 (tensile)
Ans
2
Stress in the tube = 23.52 − 14.1 = 9.42 N/mm (compression)
Ans
E XAMPLE 12.13: A cast iron pipe of 220 mm inside diameter and 250 mm external diameter
is closely wound with a layer of 6 mm diameter steel wire with initial tensile stress of 40 MPa.
Find the stresses developed in the pipe and the steel wire when water is admitted into the pipe at a
pressure of 4 MPa. Esteel = 200 GPa, Ecast iron = 100 GPa, Poisson’s ratio = 0.3.
S OLUTION :
Equating the circumferential strains of the wire and tube,
Initial stresses
Equivalent wire thickness,tw =
Initial compressive hoop stress in the tube, σ =
σ=
π ·d π ×6
=
= 4.712 mm
4
4
tw
σw
t
4.712
× 40 = 12.57 N/mm2
15
Stresses due to fluid pressure alone:
When water is admitted at pressure p, let the stresses be σ tensile (hoop) in the tube and σw
tensile in the wire due to pressure alone.
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210
•
Strength of Materials
For equilibrium,
σ .2t + σw 2tw = pd
where d is internal diameter of pipe
or
σ ×2 × 15 + σ w ×2 × 4.712 = 4 × 220
or
30σ + 9.424σw = 880
σ = −0.314σw + 29.33
∴
(i)
4 × 220
(σ − μσl )
pd
σ
=
= w where σl =
Eci
Es
4t
4 × 15
2
σl = 14.67 N/mm
Es
(σ − 0.3 × 14.67)
σw
∵
=
=2
Eci
2Eci
Eci
2σ − 8.802 = σw
σ = 0.5 σw + 4.401
Equating Eqns. (i) and (ii)
−0.314 σw + 29.33 = 0.5 σ w + 4.401
−0.814 σw = − 24.93
σw = 30.63 N/mm2
Substituting value of σw is Eqn. (ii)
σ = 0.5 × 30.63 + 4.401 = 24.12 N/mm2
Final stresses:
In the pipe = 24.12 − 12.57 = 11.55 N/mm2
2
In the wire = 30.63 + 40 = 70.63 N/mm
Cylinder with hemispherical ends
Hoop stress in the cylindrical part:
σh c =
pd
2tc
σlc =
pd
4tc
Longitudinal stress in the cylindrical part
@seismicisolation
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Ans
Ans
(ii)
Thin Cylindrical and Spherical Shells
•
211
tc
d
ts
p
l
tc = Thickness of cylinder
ts = Thickness of hemisphere
Figure 12.6
Hoop stress in the hemispherical part
σh s =
pd
4ts
Circumferential strain in the cylindrical part
= εc =
σhc μσlc
pd
−
=
(2 − μ )
E
E
4tc E
Circumferential strain in hemispherical part
= εh =
σhs μσhs
pd
−
=
(1 − μ )
E
E
4tc E
For the condition of no distortion on the junction,
εc = εh
pd
(2 − μ ) =
4tc E
ts
=
tc
pd
(1 − μ )
4ts E
1−μ
2−μ
E XAMPLE 12.14: A thin cylindrical shell of diameter 1.6 m is provided with hemispherical ends.
If it is subjected to an internal pressure of 3 N/mm2 , find the thickness of the cylindrical and
hemispherical parts for a permissible stress of 120 N/mm2 .
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212
•
Strength of Materials
S OLUTION :
Cylindrical part:
pd
= 120
2tc
∴
tc =
3 × 1600
= 20 mm
2 × 120
Ans
pd
= 120
4ts
∴
ts =
3 × 1600
= 10 mm
4 × 120
Ans
Hemispherical part:
Exercise
12.1 A thin cylindrical shell of diameter 1200 mm is subjected to a fluid pressure of 4 MPa. What
should be the thickness of the wall if the maximum stress is not to exceed 100 MPa. Hence,
what will be the change in volume per metre length. E = 200 GPa; μ = 0.3.
[Ans δ v = 1.074 × 106 mm3 ]
12.2 A cylinder of internal diameter 2.5 m and of thickness 50 mm contains a gas. If the tensile
stress in the material is not to exceed 80 N/mm2 , determine the internal pressure of gas.
[Ans 3.2 N/mm2 ]
12.3 A thin cylinder of internal diameter 1.25 m contains a fluid at an internal pressure of 2 N/mm2 .
Determine the maximum thickness of the cylinder if :
i) The longitudinal stress is not to exceed 30 N/mm2 .
ii) The hoop stress is not to exceed 45 N/mm2 .
[Ans 28 mm]
12.4 A cylindrical vessel of 3 m diameter is used for processing rubber and is 10 m long. If the steel
plates have the thickness of 24 mm, and vessel operates at 800 kPa internal pressure. Determine: i) the change in length, ii) change in diameter and iii) change in volume. E = 200 GPa,
μ = 0.3.
[Ans δ l = 0.492 mm, δ d = 0.62 mm
δ v = 32 × 106 mm3 ]
12.5 In a thin cylinder, the hoop strain is 3.5 times the longitudinal strain. How much is the Poisson’s ratio?
[Ans μ = 0.25]
12.6 In a thin cylinder the longitudinal stress is 60 MPa. E = 200 GPa and μ = 0.25. How much
is the volumetric strain.
δV
[Ans
= 1.2 × 10−3 ]
V
12.7 A steel pipe 900 mm diameter has to carry water under a head of 200 m. If the allowable
tensile stress is 90 MPa, determine the minimum value of thickness of pipe.
[Ans 9.8 mm]
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Thin Cylindrical and Spherical Shells
•
213
12.8 A spherical shell of internal diameter 0.9 m and of thickness 10 mm is subjected to an internal
pressure of 1.4 N/mm2 . Determine the increase in volume and increase in diameter. Take
1
E = 200 GPa, μ = .
3
[Ans δ V = 12028.5 mm3 , δ d = 0.0954 mm]
12.9 A boiler shell is to be made of 20 mm thick plate having a permissible tensile stress of
125 N/mm2 . If the efficiencies of the longitudinal and circumferential joints are 80% and
30%, respectively, determine:
i) The maximum permissible diameter of the shell for an internal pressure of 2.5 N/mm2 ,
and
ii) Permissible intensity of internal pressure when the shell diameter is 1.6 m.
12.10
12.11
12.12
12.13
12.14
12.15
[Ans i) 1200 mm ii) 1.875 N/mm2 ]
A thin spherical shell of internal diameter 1.5 m and of thickness 8 mm is subjected to an
internal pressure of 1.5 N/mm2 . Determine the increase in diameter and increase in volume.
Take E = 2 × 105 N/mm2 and μ = 0.3.
[Ans 0.369 mm, 1304 × 103 mm3 ]
Find the increase in volume of a spherical shell of 900 mm diameter and 10 mm thick when
subjected to an internal pressure of 1.5 N/mm2 . Take E = 205 GPa, μ = 0.3.
[Ans 131900.1 mm3 ]
A copper tube is closely wound with a steel wire of 1 mm diameter. If the internal and external
diameters of the tube are 42 mm and 45 mm, find the required tension in the wire so that an
internal pressure of 2000 kPa produces a circumferential tensile stress of 8000 kPa in the
tube. Take Es = 1.65 Ec .
[Ans 10.5 N]
A cast iron thin cylindrical pipe of internal diameter 200 mm and 15 mm thick is closely
wound by a simple layer of steel wire of 3mm diameter under a tension of 50 N/mm2 . Find
the stressed set up in the pipe when the pipe is empty. Also determine the stresses set up in
the pipe and steel wire, when water is admitted in the pipe under a pressure of 5 N/mm2
ECI = 100 GPa, Es = 200 GPa and μ = 03.
[Ans 7.89 N/mm2 ; σp = 1865 N/mm2 ; σw = 93.08 N/mm2 ]
A hollow cylindrical shell of diameter 1.5 m, thickness 10 mm and length 3 m is subjected to
an internal pressure of 4 N/mm2 . Find the change in volume. Take E = 200 GPa, μ = 0.3.
[Ans 15.1 × 106 mm3 ]
An 1800 mm internal diameter cylindrical shell is provided with hemispherical ends, and
subjected to internal fluid pressure of 5 N/mm2 . Determine the thickness of cylindrical and
hemispherical portions if the maximum allowable stress s 90 N/mm2 .
[Ans 50 mm, 25 mm]
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C HAPTER
13
THICK CYLINDERS AND SPHERES
We shall now consider thick cylinders for which the thickness of the wall is not small as compared
with the diameter of the cylinder. The thickness of a thick cylinder is more than 1/10 to 1/20 of the
diameter. When the cylinder is subjected to internal pressure p, it will expand, resulting in increased
length and diameter, i.e., axial and circumferential stresses develop in the walls of the cylinder and
both these stresses are tensile stresses which in case of thick cylinders are analysed as per ‘Lame’s
theory’.
Assumptions made in Lame’s theory:
i) The material is homogeneous and isotropic.
ii) Expansion or contraction of all fibres are unstrained by adjacent fibres.
iii) Increase in length is uniform throughout, irrespective of radius and thus the axial strain is the
same at any radius of the cylinder.
iv) The material is stressed within elastic limit.
v) Plane sections, perpendicular to the longitudinal axis of the cylinder, remain plane even after
applications of pressures.
Let r1 and r2 be the external and internal radii of a thick cylinder of length l subjected to radial
pressure. A cross section of a thick cylinder is shown in Fig. 13.1.
dx
dx
r1
px
x
x
x
r2
x
x
px+dx
Figure 13.1
Consider an elementary ring of the shell at radius x and thickness dx. Let px and (px + d px) be
the intensities of radial pressure at the inner and outer sides of the elementary ring
Bursting force, P = px × 2x × l − (px + dx).2(x + dx) × l
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Thick Cylinders and Spheres
•
215
The resisting force = σx × 2dx × l
Equating,
σx 2dx l = px 2x l − (px + d px ).2(x + dx) × l
Neglecting small quantities of the second order,
σx = −px − x
d px
d
= − (xpx )
dx
dx
(1)
Another relation between px and σx will now be obtained from the consideration that longitudinal
strain at any point in the section of the shell is constant. At any point in the section of the elementary
ring, the principal stresses are:
i) The radial pressure px
ii) The hoop stress σx and
iii) a longitudinal tensile stress σl equal to
p r22
as explained under:
r12 − r22
σl
r1
p
r2
σl
Figure 13.2
p × π r22 = σl × π (r12 − r22 )
∴
σl = p
r22
r12 − r22
The longitudinal strain at any point in the ring of radius x and thickness dx is, obviously equal to
σl μσx μ px
−
+
E
E
E
This is independent of x.
∴
σ l μσx μ px
−
+
= a constant
E
E
E
∴ (σx − px ) = a constant since σl , μ and E are constants.
∴ (σx − px ) = 2a, where a is a constant
@seismicisolation
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(2)
216
•
Strength of Materials
From the relations (1) and (2), we can solve for σx and px . Substituting (px + 2a) for σx in (1),
we get
d px
dx
d px
2dx
∴
=−
(px + a)
x
2dx
d px
=−
or
px + a
x
(px + 2a) = −px − x
Integrating,
loge (px + a) = −2 loge x + loge b
where loge b is a constant of integration.
∴
b
x2
b
px = 2 − a
x
b
σx = 2 + a
x
px + a =
and from (2),
(3)
(4)
Equations (3) and (4) are Leme’s formulae or equations for the radial pressure and hoop stress at
any specified point in the section of a thick cylindrical shell.
The constants a and b can be calculated for given conditions.
E XAMPLE 13.1: A steel pipe of 350 mm internal diameter and 90 mm thickness carries water at a
pressure of 10 N/mm2 . Determine the maximum and minimum intensities of hoop stress across the
section. Show the radical pressure distribution and hoop stress distribution, across the section.
S OLUTION :
350
= 175 mm, thickness = 90 mm
2
350 + 2 × 90
= 265 mm
r1 =
2
r2 =
r2
r1
Figure 13.3
px at radius 175 mm = 10
Using Lame’s equation,
b
−a
x2
b
− a OR
10 =
1752
b
− 10
∴ a=
30625
px =
10 =
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@seismicisolation
b
−a
30625
(i)
Thick Cylinders and Spheres
•
217
px at radius 265 mm = 0
b
−a
2652
b
a=
70225
0=
(ii)
Equating Eqns. (i) and (ii),
b
b
− 10 =
30625
70225
b
b
−
= 10
30625 70225
3.2653 × 10−5 b − 1.42399 × 10−5 b = 10
1.84131 × 10−5 b = 10
∴ b = 543091
Substituting b in Eqn. (ii),
a=
543091
= 7.73
70225
Now, hoop stress at 175 mm radius,
b
b
+a =
+ 7.73
2
x
1752
543091
σx =
+ 7.73 = 25.46 N/mm2
30625
σx =
At 265 mm,
543091
+ 7.73
2652
543091
+ 7.73
=
70225
σx =
= 7.81 N/mm2
mm
175
265
mm
Ans
p =10 N/mm2
7.81 N/mm2
sx=25.46 N/mm2
Radial pressure and
hoop stress distribution
Figure 13.4
@seismicisolation
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Ans
218
•
Strength of Materials
E XAMPLE 13.2: A thick metallic cylindrical shell has 160 mm internal diameter. It is to withstand
an internal pressure of 12 N/mm2 . Determine the necessary thickness of the shell, if the permissible
tensile stress in the section is 25 N/mm2 .
160
= 80 mm.
2
b
b
−a
12 = 2 − a; 12 =
6400
80
b
− 12
a=
6400
r2 =
(i)
From Lame’s equation the permissible stress (σx ),
b
+a
r22
b
25 = 2 + a or,
r2
σx =
a = 25 −
b
6400
Equating Eqns. (i) and (ii)
b
b
− 12 = 25 −
6400
6400
b
b
b
+
= 25 + 12;
= 37
6400 6400
3200
∴ b = 118400
Substituting the value of b in Eqn. (i)
a=
118400
− 12 = 6.5
6400
We know internal pressure at outer radius is zero,
∴
b
118400
−a =
− 6.5
2
r1
r12
118400
118400
or r12 =
= 18215.4
6.5 =
2
6.5
r1
√
r1 = 18215.4 = 134.96 mm
0=
So required thickness = 134.96 − 80 = 54.96 mm
Ans
@seismicisolation
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(ii)
Thick Cylinders and Spheres
•
219
Solid Circular Shaft Subjected to External Pressure
p
Let the radius of solid shaft be r, and external pressure applied on it be p. Applying Lame’s
equations,
px =
b
− a,
x2
σx =
b
+a
x2
p
Figure 13.5
It is evident from the equation of radial pressure that px is infinite at radius = 0. Therefore, the value
of constant b is zero, because it is not possible for pressure to be infinite at radius. Therefore, we get
px = −a = p, Hence, pressure p is constant everywhere and its value is equal to external pressure p.
Also, σx = 0 + a , where a = −p
Therefore, σx = −p
The minus sign shows that σx is compressive. Hence, the intensity of pressure is constant everywhere and is compressive, its value is equal to the external pressure p.
Compound cylinders: A compound cylinder consists of two concentric cylinders as shown in
Fig. 13.5, the outer cylinder being shrunk onto the inner cylinder so that the later is initially compression before the application of internal pressure. The final stresses are then the resultants of those
due to pre-stressing and those due to the internal pressure.
If the radius of common surface is r0 and the pressure at this surface before the application of
the internal pressure is p0 , then the initial stresses are determined by considering the two cylinders
separately, the boundary conditions for the outer cylinder being px = p0 when x = r0 and px = 0
when r = r1 and for the inner cylinder, px = p0 when r = r0 and pr = 0 when r = r2 . The stresses
due to internal pressure are obtained by considering the cylinder to be homogeneous with pr = p at
r = r2 and pr = 0 at r = r1 .
The various stresses are then combined algebrically as shown in Fig. 13.7, from which it is
clear that the maximum resultant stress is less
than that for a homogeneous cylinder of the same
cross section with same internal pressure.
r1
r2
r0
Figure 13.6
px
σc
r2
r0
Initial stress
r1 Stresses due to po
Resultant stresses
Figure 13.7
@seismicisolation
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220
•
Strength of Materials
E XAMPLE 13.3: A compound tube is composed of a tube 240 mm internal diameter and 21
mm thick shrunk on a tube of 240 mm external diameter and 23 mm thick. The radial pressure
at the junction in 10 N/mm2 . The compound tube is subjected to an internal fluid pressure of
80 N/mm2 . Find the variation of hoop stress over the well of the compound tube.
S OLUTION :
Outer tube: px =
b1
− a1
x2
and σx =
b1
+ a1
x2
r1 =
r0
r2
r1
240
240
+ 21 = 141 mm r0 =
= 120 mm
2
2
r2 = 120 − 23 = 97 mm
x = 141 mm, px = 10 N/mm2
b1
− a1 = 0
(141)2
b1
b1
− a1 = 0 ∴ a1 =
19881
19881
b1
− a1 = 10
Again
(120)2
b1
b1
− a1 = 10 ∴ a1 =
− 10
14400
14400
At
Figure 13.8
Equating Eqns. (i) and (ii),
b1
b1
=
− 10
19881 14400
b1
b1
−
= 10; 6.944 × 10−5 b1 − 5.0299 × 10−5 b1 = 10
14400 19881
1.9141 × 10−5 b1 = 10
b1 = 522439
From Eqn. (i)
∴
a1 =
522439
= 26.28
19881
Hoop stresses for the outer tube,
522439
+ 26.28 = 52.56 N/mm2 (Tensile)
19881
522439
+ 26.28 = 62.56 N/mm2 (Tensile)
σ120 =
14400
b2
Inner tube : px = 2 −a2
x
b2
σx = 2 + a2
x
σ141 =
@seismicisolation
@seismicisolation
(i)
(ii)
Thick Cylinders and Spheres
px = 10 N/mm2
b2
b2
∴
− a2 = 10 ∴ a2 =
− 10
14400
14400
At x = 97 mm, px = 0
b2
b2
− a2 = 0
− a2 = 0 ∴
∴
2
9409
97
•
221
At x = 120,
(iii)
(iv)
Substituting for a2 from Eqn. (iii) in Eqn. (iv),
b2
b
b2
b2
−
− 10 = 0,
−
+ 10 = 0
9409
14400
9409 14400
1.063 × 10−4 b2 − 0.694 × 10−4 b2 = −10
0.369 × 10−4 b2 = −10
∴
b2 =
−10
= −271003
0.369 × 10−4
Putting value of b2 in Eqn. (iii)
a2 =
−271003
− 10 = −28.82
14400
∴
a2 = −28.82
Hence, the hoop stresses for the inner tube are calculated as follows:
−271003
+ (−28.82)
14400
= −47.64 N/mm2 (Compressive)
−271003
σ97 =
+ (−28.82)
972
= −57.62 N/mm2
σ120 =
Stresses due to internal fluid pressure alone:
B
−A
x2
B
σx = 2 +A
x
px =
At x = 97 mm, px = 80 N/mm2
80 =
B
B
−A
− A, 80 =
2
9409
97
(v)
At x = 141 mm; px = 0
0=
B
−A
1412
∴
A=
@seismicisolation
@seismicisolation
B
19881
(vi)
222
•
Strength of Materials
Substituting for A from Eqn. (vi) into Eqn. (v)
B
B
−
9409 19881
80 = 1.0628 × 10−4 B − 0.503 × 10−4 B
80 =
80 = 0.5598 × 10−4 B,
∴
B = 1429082
Substituting for B in Eqn. (vi),
A=
1429082
= 71.88
19881
Hence, the hoop stresses due to fluid pressure alone are given by,
1429082
+ 71.88 = 223.76 N/mm2 (Tensile)
972
1429082
σ120 =
+ 71.88 = 171.12 N/mm2 (Tensile)
1202
1429082
σ141 =
+ 71.88 = 143.76 N/mm2 (Tensile)
1412
σ97 =
Now due to the combined effect of shrinking tube (outer) on the inner tube and internal fluid pressure, the final hoop stresses are as follows:
Outer tube:
σ120 = 62.56 + 71.88 = 134.44 N/mm2
Ans
σ141 = 52.56 + 143.76 = 196.32 N/mm (Tensile) Ans
2
Inner tube:
σ120 = −47.64 + 171.12 = 123.48 N/mm2 (Tensile) Ans
σ97 = −57.62 + 223.76 = 166.14 N/mm2 (Tensile) Ans
Shrinkage Allowance: While making a compound cylinder, shrinkage allowance is kept, i.e., if
it is necessary that the inner diameter of the outer cylinder to be slightly smaller than the outer
diameter of the inner cylinder. Normally, in order to fit, outer cylinder is heated till it slides over the
inner cylinder. Naturally, on cooling outer cylinder will shrink and exert the required pressure at the
common surface. Interface pressure is p0 (at common surface).
Outer cylinder: If E1 and μ1 are the modulus of material and Poisson’s ratio of material of outer
cylinder, respectively. At outer surface of the outer cylinder, let the hoop stress be σc1 and radial
stress being p0 = σr
Then circumferential strain at outer cylinder:
σc1 μ1 p0
+
E1
E1
2r0
Increase in diameter = 2r0 εc1 =
(σc1 + μ1 p0 )
E1
εc1 =
@seismicisolation
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Thick Cylinders and Spheres
•
223
Inner cylinder: Let E2 and μ2 be the modulus of elasticity and Poisson’s ratio of the inner cylinder,
respectively. At the outer surface let the hoop stress be σc2 and radial stress be p0 = σr
σc2 μ2 p0
−
E2
E2
2r0
Decrease in diameter: 2r0 εc2 =
(σc2 − μ2 p0 )
E2
Therefore, shrinkage allowance = Increase in diameter of outer cylinder
+ Decrease in diameter of inner cylinder
2r0
2r0
=
(σc1 + μ1 p0 ) +
(σc2 − μ2 p0 )
E1
E2
σc1 + μ1 p0 σc2 − μ2 p0
= 2r0
+
E1
E2
Circumferential strain: εc2 =
If the two cylinders are made of same material, then E and μ will be common, so the above expression becomes,
Shrinkage allowance =
2r0
(σc1 + σc2 )
E
E XAMPLE 13.4: A steel tube of 120 mm inside diameter and 170 mm outside diameter is to
have an outer tube of same material having 200 mm outside diameter shrunk on it, the shrinkage
allowance being such that the radial pressure between the two tube is to be 25 N/mm2 . Take E =
200 GPa, μ = 0.25, calculate:
i)
ii)
iii)
iv)
v)
Hoop stress at the inner surface of outer tube
Increase in internal diameter of outer tube
Hoop stress at the outer surface of inner tube
Reduction in external diameter of inner tube
Shrinkage allowance
S OLUTION :
R1 = 100 mm R0 =
∴
170
= 85 mm
2
R0 = 85 mm
σr = p0 = 25
b
b
−a
25 = 2 − a, 25 =
7225
85
b
b
b
−a ∴ a =
=
p = σr = 0 =
2
2
10000
100
100
Substituting the value of a in Eqn. (i)
b
b
1
1
25 =
−
=b
−
7225 10000
7225 10000
@seismicisolation
@seismicisolation
(i)
(ii)
224
•
Strength of Materials
25 = b (0.0001384 − 0.0001)
25
∴ b=
= 651042
3.84 × 10−5
From Eqn. (ii)
a=
651042
= 65.1
10000
b
651042
+ 65.1 = 155.21 N/mm2 (Tensile) Ans
+a =
2
7225
85
i)
σc =
ii)
155.21 0.25 × 25 155.21 + 6.25
σc μσr
+
=
+
=
E
E
200000
200000
200000
= 8.073 × 10−4
Increase in diameter of outer tube = 8.073 × 10−4 × 170
= 0.1372 mm. Ans
iii) R0 = 85 mm
σr = p0 = 25 =
b
b
− a; 25 =
−a
2
7225
85
(iii)
120
= 60 mm
2
b
6
σr = 0 = 2 − a a =
3600
60
when R2 =
(iv)
Putting value of a from Eqn. (iv) in Eqn. (iii),
b
1
1
b
−
=b
−
7225 3600
7225 3600
25 = −0001394b ∴ b = −179340
179340
= −49.82
∴ a=−
3600
R0 = 85 mm
b
−179340
σ c1 = 2 + a =
− 49.82 = −74.64 N/mm2 (Compressive)
7225
85
25 =
iv) Circumferential strain of outer diameter of inner tube:
74.64
0.25 × 25
σc1 μ p0
+
=−
+
E
E
200000
200000
1
(−74.64 + 6.25) = −3.4195 × 10−4 (Compressive)
=
200000
εc =
@seismicisolation
@seismicisolation
Ans
Thick Cylinders and Spheres
∴
•
225
Reduction in external diameter = 3.4195 × 10−4 × 170
= 581.3 × 10−4
= 0.05813 mm
Shrinkage allowance = 0.1372 + 0.0581 = 0.1953 mm
Ans
E XAMPLE 13.5: A compound thick cylinder is formed by shrinking a tube of external diameter
320 mm over another tube of internal diameter 140 mm. After shrinking, the diameter at the junction
of the tubes is found to be 240 mm and radial compression as 30 N/mm2 . Find the original difference
in radii at the junction. Take E = 200 GPa.
S OLUTION :
320
140
= 160 mm; Inner radius, r2 =
= 70 mm
2
2
240
radius at junction, r0 =
= 120 mm
2
b2
b2
30 =
− a2 =
− a2
1202
14400
b2
− a2
30 =
14400
r1 =
(i)
Also
0=
b2
− a2
702
∴
a2 =
b2
4900
Substituting the value of a2 from Eqn. (ii) in Eqn. (i)
b2
b2
−
; 30 = b2 6.944 × 10−5 − 2.041 × 10−4
14400 4900
30
30
=−
b2 =
(.00006944 − 0.0002041)
1.3466 × 10−4
b2 = −222783
30 =
Substituting b2 value in Eqn. (i),
−222783
− a2
14400
a2 = −15.47 − 30 = −45.47
30 =
Considering the outer cylinder when r1 = 160 mm,
px = 30 N/mm2
b
30 = 2 − a
x
@seismicisolation
@seismicisolation
(ii)
226
•
Strength of Materials
b2
b1
−a
− a1 =
2
14400
(120)
b1
− 30
a=
14400
30 =
(iii)
Similarly,
b1
b1
− a1 =
− a1
2
25600
160
b1
a2 =
25600
0=
(iv)
Substituting, value of a2 in Eqn. (iii)
b1
b1
b1
b1
=
− 30 OR, 30 =
−
25600 14400
14400 25600
30 = b1 (6.944 × 10−5 − 3.906 × 10−5 )
30
987492
= 987492, a2 =
= 38.57
b1 =
−5
25600
3.038 × 10
b1
σc1 = 2 + a1
x
987492
=
+ 38.57 = 88.24 N/mm2
1412
b2
σc2 = 2 + a2
x
−222783
=
+ (−45.47)
1202
= −15.47 − 45.47
= −60.94 N/mm2 (Compressive)
1 2r0
(σc1 + σc2 )
Difference in radii at junction =
2 E
r0
(σc1 + σc2 )
E
120
=
{88.24 + (−60.94)}
200000
= 0.01638 mm Ans
=
E XAMPLE 13.6: A compound tube is made of a tube 270 mm internal diameter 30 mm thick
shrunk on a tube of 220 mm internal diameter. At junction radial pressure is 10 N/mm2 . The compound tube is subjected to an internal pressure of 70 N/mm2 . Find the variation of hoop stress over
the wall of the compound tube.
@seismicisolation
@seismicisolation
Thick Cylinders and Spheres
270 + 60
= 165 mm
2
220
= 110 mm
r2 =
2
270
r0 =
= 135 mm
2
b
b
−a
Outer tube : p = 10 =
− a; 10 =
2
18225
135
b
− 10
a=
18225
b
b
0=
−a =
−a
27225
1652
b
a=
27225
•
227
r1 =
(i)
(ii)
Substituting for a from Eqn. (ii) into Eqn. (i)
b
b
1
1
=
− 10 or, 10 = b
−
27225 18225
18225 27225
10 = b (5.487 × 10−5 − 3.673 × 10−5 )
∴
b=
10
1.814 × 10−5
b = 551268
551268
a=
= 20.25
27225
b
σx = 2 + a
x
551268
At radius = 165, σc =
+ 20.25 = 40.50 N/mm2
1652
551268
At radius = 135, σc =
+ 20.25 = 50.50 N/mm2
1352
Ans
Ans
Inner tube
At radius 135 mm, σr = pr = 10 =
b
−a
18225
b
−a
At radius 110 mm, σr =
1102
b
−a
0=
12100
b
a=
12100
10 =
@seismicisolation
@seismicisolation
b
−a
1352
(i)
(ii)
228
•
Strength of Materials
Substituting for a from Eqn. (ii) into Eqn. (i),
b
b
−
18225 12100
10 = b (5.487 × 10−5 − 8.264 × 10−5 )
10
b=−
2.777 × 10−5
b = −360101
360101
a=−
= −29.76
12100
10 =
or
At radius r0 = 135 mm
σc =
=
−360101
+ (−29.76)
1352
−360101
− 29.76
18225
= −49.52 N/mm2 (Compressive) Ans
At radius r2 = 110
−360101
− 29.76
1102
= −59.52 N/mm2 (Compressive) Ans
σc =
Stresses due to internal fluid pressure:
At radius r2 = 110
p110 = σ110 =
b
−a
1102
b
−a
12100
b
a=
− 70
12100
70 =
(i)
At radius r1 = 165
σ165 = p165 = 0 =
∴
a=
b
−a
1652
b
27225
@seismicisolation
@seismicisolation
(ii)
Thick Cylinders and Spheres
•
229
Substituting the value of a from Eqn. (ii) into Eqn. (i),
b
b
=
− 70
27225 12100
1
1
70 = b
−
12100 27225
70 = b(8.264 × 10−5 − 3.673 × 10−5 )
70 = b(4.591 × 10−5 )
∴ b = 1524722
1524722
= 56
and a =
27225
At radius r3 = 165 mm:
1524722
+ 56
1652
= 112 N/mm2 (Tensile) Ans
σc =
At radius r0 = 135 mm:
1524722
+ 56
1352
= 139.66 N/mm2 (Tensile) Ans
σc =
At radius r2 =110:
1524722
+ 56
1102
= 182 N/mm2 (Tensile) Ans
σc =
Inner tube
Radii
σc due to shrinkage
σc due to inner fluid pressure N/mm2
Resultant σc N/mm2
N/mm2
Outer tube
180 mm
135 mm
165 mm
135 mm
−59.22
182
122.78
−49.52
139.66
90.14
40.50
112
152.5
50.50
139.66
190.16
Thick Spherical Shells
The thick spherical shells are used to store fluids under high pressure. Equations for hoop stress and
radial stresses are obtained in a similar manner as in thick cylindrical shells, by applying Lame’s
theory. The equations for circumferential (hoop) stress and radial stresses are given below:
b
+a
r3
2b
σr = 3 − a
r
σc =
@seismicisolation
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230
•
Strength of Materials
E XAMPLE 13.7: A spherical shell of 150 mm internal diameter has an internal pressure of 40
MN/m2 . If the permissible tensile stress in 85 MN/m2 . Determine thickness of the shell.
150
= 75 mm = 0.075 m
2
(p) = 40 MN/m2
ri =
Internal pressure
Permissible stress
(σh ) = 85 MN/m2
2b
σr = 3 − a
r
b
σh = 3 + a
r
At r = 0.075 m, the radial stress = 40 MN/m2
2b
− a = 40
(0.075)3
2b
− a = 40
4.2187 × 10−4
4740.8b − a = 40
(i)
At r = 0.075 m, σh or σc = 85 MN/m2
b
+ a = 85
(0.075)3
b
+ a = 85
4.2187 × 10−4
2370.4b + a = 85
a = 85 − 2370.4b
Substituting the value of a in Eqn. (i)
4740.8b − (85 − 2370.4b) = 40
4740.8b − 85 + 2370.4b = 40
7111.2b = 125 ∴ b = 0.01758
from Eqn. (ii) a = 85 − 2370.4 × 0.01758
= 43.33
Let the external radius be r1 , σr = 0
2b
2b
− a or, a = 3
3
r
r
2b 2 × 0.01758
3
r =
=
= 8.114 × 10−4
a
43.33
0=
@seismicisolation
@seismicisolation
(ii)
Thick Cylinders and Spheres
∴
•
231
r = 0.0933 m = 93.3 mm
thickness = 0.0933 − 0.075 = 0.0183 m
= 18.3 mm Ans
Principal and shear stresses: Because there is no torque acting on the cylinder, so σc , σr and σl
are the principal stresses. σr is compressive and σc and σl are tensile.
σc is maximum and σr is minimum,
We know, τmax (maximum shear stress) =
τmax =
or
=
=
τmax =
σmax − σmin
2
σc − (−σr )
(∴ σr is compressive)
2
σc + σr
2
1
b
b
+a + 2 +a
2
r2
r
b
r2
Changes in cylinder dimensions:
Circumferential strain =
1
[σc − μσr − μσl ]
E
Change in diameter at any radius r of the cylinder is given by
δd =
Change in length, δ l =
2r
[σc − μσr − μσl ]
E
1
[σl − μσr − μσc ]
E
Exercise
13.1 A thick pipe of steel is of 450 mm diameter and of thickness 100 mm. It is subjected to an
internal fluid pressure of 60 bar. Find the hoop stress at internal and outer surfaces.
[Ans σc = 15.6 MN/m2 and 9.6 MN/m2 at inner and outer surface respectively]
13.2 A cylinder 200 mm outside diameter and 100 mm inside diameter has an internal fluid pressure of 20 MN/m2 . Determine the longitudinal stress, assuming it to be uniform.
[Ans 6.67 MN/m2 ]
13.3 Calculate the thickness of the metal necessary for a steel cylindrical shell of internal diameter
0.15 m to withstand an internal pressure of 50 MN/m2 , the permissible tensile stress is not to
exceed 150 MN/m2 .
[Ans 31 mm]
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232
•
Strength of Materials
13.4 A thick walled closed-end cylinder is made of an aluminium alloy having E = 72 GPa and
μ = 0.33, has inside diameter of 200 mm and outside diameter of 800 mm. The cylinder
is subjected to internal fluid pressure of 150 MPa. Determine the circumferential stresses at
0.1m radius and 0.4 m radius
[Ans 170 MN/m2 and 20 MN/m2 ]
13.5 Calculate the ratio of thickness to internal diameter for a tube subjected to internal pressure
when the pressure is 5/8 of the value of the maximum permissible circumferential stress.
Find the increase in internal diameter of such a tube 100 mm internal diameter when the
internal pressure is 80 MN/m2 . Also find the change in wall thickness.
t
= 0.54, δ d = 0.07306 mm, decrease in wall thickness = 0.01438 mm]
[Ans
d
13.6 A hub is shrunk onto a solid shaft and the final dimensions are:
Diameter at the junction = 150 mm
External diameter of hub = 200 mm
Find the initial difference in the diameters of the hub and the shaft to produce a radial contact
pressure of 30 MPa. E = 200 GPa.
[Ans 0.1028 mm]
2
13.7 An external pressure of 10 MN/m is applied to a thick cylinder of internal diameter 150 mm
and external diameter 300 mm. If the maximum hoop stress permitted on the inside wall is
35 MN/m2 . Calculate:
i) The maximum internal pressure that can be applied
ii) The change in outside diameter if the cylinder has the closed ends, E = 210 GPa,
μ = 0.3
[Ans p = 37 MN/m2 , δ d = 0.03423 mm]
13.8 A compound cylinder formed by shrinking one tube onto another is subjected to an internal
pressure of 60 N/m2 . Before the fluid is admitted, the internal and external diameters of the
compound cylinder are 120 mm and 200 mm and the diameter at the junction is 160 mm. If,
after shrinking on, the radial pressure at the common surface is 8 N/mm2 . Calculate the final
stress set up by the section
Ans
[Ans See table below:]
Hoop stress
Inner tube
Outer tube
N/mm2
x = 60 mm
x = 80 mm
x = 80 mm
x = 100 mm
Initially
Due to fluid pressure
Resultant
−36.58
+127.5
+90.92
−28.58
+86.48
+57.9
+36.44
+86.48
+122.92
+28.44
+67.5
+95.94
‘+’ for tension
‘-’ for compression
@seismicisolation
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Thick Cylinders and Spheres
•
233
13.9 A compound thick cylinder is formed by shrinking a tube of external diameter 300 mm over
another tube of internal diameter 150 mm. After shrinking, the diameter at the junction of
the tubes is found to be 250 mm and radial compression at 28 N/mm2 . Find the original
difference in radii at the junction. Take E for the cylinder metal as 200 GPa.
[Ans 0.13 mm]
13.10 A spherical shell of 80 mm internal diameter has to withstand an internal fluid pressure of
30 MN/m2 . Find the thickness of the shell if permissible stress in 80 MN/m2 .
[Ans 7.7 mm]
@seismicisolation
@seismicisolation
C HAPTER
14
DEFLECTION OF BEAMS
Relation between Slope, Deflection and Radius of Curvature
To find out relation between slope, deflection and radius of curvature, consider Fig. 13.1 where a
small portion AB of a beam bent into an arc as shown in the figure.
C = Centre of arc into which the beam has been bent
α = Angle which the tangent at A makes with x − x axis
ds = length of beam AB
α + d α = Angle which the tangent at B makes with x − x axis.
Y
C
da
A
a
0
B
R
ds
dy
D
dx
a+da
X
Figure 14.1
It is clear from the geometry of the figure that
or
∠ACB = d α and ds = Rd α
ds
dx
∴ R=
=
(ds = dx, being very small)
dα
dα
dα
1
=
R
dx
(i)
dy
dy
or α = , because tan θ = θ for very small angles.
Now, tan α =
dx
dx
Differentiating the above equation with respect to x, we have
d2y
dα
= 2
dx
dx
1
d2y
= 2
R dx
[because
1
dα
= as shown in Eqn. (i)]
dx
R
@seismicisolation
@seismicisolation
(ii)
Deflection of Beams
•
235
EI
1
M
M E
= or, M =
or =
I
R
R
R EI
1
Substituting in Eqn. (ii) for ,
R
From bending equation,
M
d2y
= 2
EI
dx
∴
M = EI
d2y
dx2
(iii)
Equation (iii) is called the curvature-moment relation or the differential equation of flexure.
The term EI is known as flexural rigidity.
Deflection of beams: It can be found by the following methods:
(A)
(B)
(C)
(D)
(E)
(F)
Double integration method
Macaulay’s method
Moment area method
Conjugate beam method
Superposition method
Strain energy method.
(A) Double Integration Method
Cantilevers
1. Cantilever with a Concentrated Load at its Free End
W
l
X
x
B
A
yB
Figure 14.2
Consider section X at a distance of x from free end.
Mx = −W x
EI
d2y
dx2
(due to hogging minus sign is taken)
= −W x
(i)
Integrating Eqn. (i)
E
W x2
dy
=−
+C1
dx
2
C1 is the constant of integration, we know that when x = l,
(ii)
dy
=0
dx
Eqn. (ii),
O=−
W l2
+C1
2
or,
@seismicisolation
@seismicisolation
C1 =
W l2
2
dy
is slope . Substituting in
dx
236
•
Strength of Materials
Now Eqn. (ii) becomes,
EI
dy
W x2 W l 2
=−
+
dx
2
2
(iii)
This is the required equation for the slope at any point on cantilever. The maximum slope occurs
dy
as slope i for
at the free end. For maximum slope, substituting x = 0 in Eqn. (iii). (Let us denote
dx
very small angles).
EI
or
W l2
2
W l2
iB =
EI
iB =
Integrating the Eqn. (iii) once again,
EIy = −
W x3 W l 2 x
+
+C2
6
2
(iv)
C2 is constant for integration. When x = 0, y = 0, substituting these values of x and y in the
above equation,
W l3 W l3
W l3
+
+C2 =
+C2
6
2
3
W l3
C2 = −
3
O=−
∴
Substituting for C2 in Eqn. (iv),
EIy = −
W x3 W l 2 x W l 3
+
−
6
2
2
From Eqn. (v), deflection y can be calculated at any point of cantilever.
For maximum deflection, x = 0
W l3
3
W l3
yB = −
3EI
W l3
=
3EI
EIyB = −
or
(minus sign is dropped because it is due to sign convention only).
@seismicisolation
@seismicisolation
(v)
Deflection of Beams
•
237
2. Cantilever with a Concentrated Load not at the Free End
W
l
l1
C
yC
A
B
C'
yB
B'
Figure 14.3
It should be noted that portion AC will bend as AC ; while the portion CB remains straight.
W l12
Slope, ic =
2EI
Since portion C B is straight,
W l12
iB = iC =
2EI
and from previous article,
y=
W l13
3EI
From the geometry of the figure, we find out that
yB = yC = ic (l − l1 )
=
W l13 W l12
+
(l − l1 )
3EI 2EI
3. Cantilever with Uniformly Distributed Load
l
A
w/m
x
x
x
B
iB
B'
yB
Figure 14.4
wx2
2
2
d y
wx2
EI 2 = −
2
dx
Mx = −
Integrating the above equation
EI
dy
wx3
=−
+C1
dx
3
@seismicisolation
@seismicisolation
(i)
238
•
Strength of Materials
When x = l, then slope
dy
= 0, substituting in the above equation;
dx
O=−
wl 3
+C1
6
or C1 =
wl 3
6
So, Eqn. (i) because,
EI
wx3 wl 3
dy
=−
+
dx
6
6
(ii)
wl 3
This is the equation for slope at any point on cantilever. Maximum slope at free end B, iB =
6EI
Integrating Eqn. (ii), we get,
EIy = −
wx4 wl 3 x
+
+C2
24
6
(iii)
When x = l, y = 0 substituting in Eqn. (iii)
We get
O=
wl 4 wl 4
+
+C2
24
6
∵ C2 =
wl 4
8
Substituting in Eqn. (iii)
EIy = −
wx4 wl 3 x wl 4
+
+
24
6
8
This equation can give deflection at any point on cantilever. Maximum deflection is at free and B,
putting x = 0.
wl 4
8EI
wl 4
=
8EI
yB = −
(Minus sign is removed. It depends on sign convention) or yB =
W l3
; where W = wl.
8EI
4. Cantilever Partially Loaded with Uniformly Distributed Load
l1
A
l
C B
iB yc
C'
B'
Figure 14.5
@seismicisolation
@seismicisolation
yB
Deflection of Beams
•
239
Portion AC of the cantilever will bend into AC , while the portion CB will remain straight and
will take new position as C as shown in Fig. 14.5, Cantilever with uniformly distributed load (partially) will have slope iB if the tangent is drawn from point B .
ic =
wl13
6EI
(from previous article)
Because the portion CB of the cantilever is straight,
∴
iB = ic =
yc =
wll3
6EI
wll4
8EI
(from previous article)
From the geometry of the figure, we have
yB = yc + ic (l − l1 ) =
wl 3
wl14
+ 1 (l − l1 )
8EI
6EI
4
3
l
l
7wl 4
w
w
l
+
× =
8EI 2
6EI 2
2
384EI
5. Cantilever Loaded from the Free End
Cor. if l1 = l/2 then yB =
A
w/unit length
B
C
l1
l
(a)
B
A
C
l1
l
(b)
Figure 14.5a,b
In each case redraw the figure with uniformly distributed load throughout the length l. And draw
an upward distributed load (w/unit length from fixed end to point C for a length of (l − l1 ).
Now obtain slope and reflections due to newly loaded beam as per Article 3 and Article 4. Then
the slope at B is equal to the slope due to the total load minus the slope due to superimposed load
upwards.
Now, the deflection at B is equal to the deflection due to the total load minus the deflection due
to superimposed load.
@seismicisolation
@seismicisolation
240
•
Strength of Materials
wl 3
w(l − l1 )3
−
6EI
6EI
4 wl
w(l − l1 )4 w(l − l1 )3
yB =
−
+
l1
8EI
8EI
6EI
iB =
and
The slope and deflection at A due to superimposed uniformly distributed load from A to C is
obtained by substituting (l − l1 ) for l
6. Cantilever with a Gradually Varying Load
X
x
w
per unit
length
B
A
iB
l
yB
Figure 14.6
1 wx
x
wx3
Mx = − ×
×x× = −
2
l
3
6l
d2y
wx3
EI 2 = −
6l
dx
Integrating,
EI
when x = l, then
wx4
dy
=−
+C1
dx
24l
(i)
dy
= 0, substituting in the equation above,
dx
wl 4
+C1 OR,
24l
dy
wx
wl 3
EI
=−
+
dx
24l
24
O= −
∴
C1 = +
wl 3
24
(ii)
This equation will give slope at any distance x from free end integrating Eqn. (ii) again,
EI iB =
wl 3
24
or,
iB =
wl 3
radian
24EI
Integrating Eqn. (ii) again,
EIy = −
w
wl 3
+
+ C2
120l
24
@seismicisolation
@seismicisolation
(iii)
Deflection of Beams
•
241
when x = l, y = 0, substituting in Eqn. (iii)
wl 4
wl 4 wl 4
+
+ C2 , or C2 = −
120
24
30
wx5 wl 3 x wl 4
EIy = −
+
−
120l
24
30
O=−
∴
(iv)
This is required equation for deflection at any distance x from free end. For maximum deflection,
x=0
∴
wl 4
30 EI
wl 4
=
30 EI
y=−
(minus sign means downward deflection)
E XAMPLE 14 A-1: A cantilever beam is 130 mm wide, 165 mm deep and 2 m long. Determine
the slope and deflection at the free end of the beam, if it is loaded with a concentrated load of 25 kN
at the free end. Take E = 200 Gpa
S OLUTION :
I=
130 × 1653
= 48664687.5 mm4
12
W l2
iB =
2EI
iB =
25 kN
A
B
(as derived before)
25000 (2000)2
2 × 200000 × 48664687.5
2m
= 0.00514 radian
Ans
Figure 14.6a
Deflection at free end B,
yB =
W l3
3EI
yB =
25000 (2000)3
= 6.85 mm
3 × 200000 × 48664687.5
(as derived before)
Ans
E XAMPLE 14 A-2: A cantilever of length 2.5 m is carrying a point load of 60 kN at a distance of
1.8 m from the fixed end. If I = 9 × 107 mm4 and E = 200 GPa, find: (i) slope at the free end and
(ii) deflection at the free end.
@seismicisolation
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242
•
Strength of Materials
S OLUTION :
60 kN
W l2
θB = 1
2EI
60000 × (1800)2
θC = θB =
2 × 200000 × 9 × 107
= 0.0054 radian
C
A
B
1.8 m
2.5 m
Figure 14.7
W (l1 )3 W (l1 )2
+
(l − l1 )
3EI
2EI
60000 (1800)3
60000 (1800)3
=
+
(2.5 − 1.8)
3 × 200000 × 9 × 107 2 × 200000 × 9 × 107
= 6.48 + 6.804 = 13.284 mm Ans
yB =
E XAMPLE 14 A-3: A cantilever of length 2.8 m carries a uniformly distributed load of 10 kN/m
over a length of 2 m from the free end. If I = 8 × 109 mm4 and E = 200 GPa; determine: i) the slope
at the free end, and (ii) deflection at the free end.
S OLUTION :
In such cases assume uniformly distributed load
is extended up to fixed end A. And now assume
10 kN/m is acting upwards from A to C as show
Fig. 14.9.
Now using the results of derivations earlier
in this section, we get
θB =
A
2m
2.8 m
(l1)
(l)
Figure 14.8
10 kN/m
B
A
wl 3 w(l − l1 )3
−
6EI
6EI
10 kN/m
B
C
C
10 kN/m
(l)
Figure 14.9
10 × (2800)3
10 (2800 − 800)3
θB =
−
6 × 200000 × 8 × 109 6 × 200000 × 8 × 109
= 0.0000229 − 0.00000833
= 0.00001457 radian
Ans
@seismicisolation
@seismicisolation
Note here
10000
= 10 N/mm
1000
Deflection of Beams
•
243
wl 4
w(l − l1 )3 l1
−
yB =
8EI
6EI
(ii)
10 (2800 − 800)3 800
10 (2800)4
−
=
8 × 200000 × 8 × 109
6 × 200000 × 8 × 109
= 0.04802 − [0.00667]
= 0.04135 mm
Ans
E XAMPLE 14 A-4: A cantilever of length 3.5 m carries two point loads of 1.5 kN at the free end
and 3 kN at a distance of 1.2 m from the free end. Find the deflection at free end. E = 200 GPa,
I = 109 mm4 .
S OLUTION :
3 kN
Deflection at free end due to 1.5 kN,
yB1
l1 = 2.5 m
wl 3
=
3EI
=
C
A
1500 × (3500)3
3 × 200000 × 109
1.5 kN
B
1m
3.5 m
Figure 14.10
= 0.1072 mm
Deflection due to 3 kN at free end,
yB2 =
=
wl13 W l12 (l − l1 )
+
3EI
2EI
3000 × (2500)2 (1000)
3000 (2500)3
+
3 × 200000 × 109
2 × 200000 × 109
= 0.078125 + 0.046875
= 0.125 mm
Total deflection = yB1 + yB2 = 6.1072 + 0.125 = 0.2322 mm
Ans
E XAMPLE 14 A-5: A cantilever of length 2.5 m carries a uniformly distributed load of 2 kN/m run
for a length of 1.4 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: 120 mm wide 250 mm deep. E = 10000 N/mm2
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244
•
Strength of Materials
S OLUTION :
I=
120 × 2503
= 156250000 mm4
12
2 kN/m
A
1 kN
B
C
1.1 m
1.4 m = l1
2.5 m
Figure 14.11
Deflection y1 due to 1 kN,
y1 =
W l3
1000 (2500)3
=
= 3.33 mm
3EI
3 × 10000 × 156250000
y2 due to uniformly distributed load of 2 kN/m at free end
y2 =
=
wl14
wl
+
(l − l1 )
8EI 6EI
2 (1400)3 (2500 − 1400)
2 (1400)4
+
8 × 10000 × 156250000 6 × 10000 × 156250000
= 0.6147 + 0.6439 = 1.2586 mm
Total deflection yB ,
yB = 3.33 + 1.2586 = 4.5886 mm Ans
E XAMPLE 14 A-6: A cantilever of length 3.5 m carries a uniformly varying load of zero intensity
at the free end and 20 kN at fixed end. E = 200 GPa, I = 0.9 × 108 mm4 . Find the slope and
deflection at the free end.
20 kN/m
B
A
3.5 m
Figure 14.12
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Deflection of Beams
•
245
S OLUTION :
θB =
wl 3
24EI
(derived before)
20 (3500)3
= 0.001985 radian
24 × 200000 × 0.9 × 108
wl 4
(derived before)
yB =
30EI
20 (3500)4
=
= 5.56 mm Ans
30 × 200000 × 0.9 × 108
=
1. Simply Supported Beam with a Central Point Load
X
W
x
A
C
X
l/2
W/2
yc
B
l/2
W/2
Figure 14.13
W
Reactions: RA = RB =
2
Consider a section X at x from point A,
Wx
2
d2y W x
EI 2 =
2
dx
Mx =
(i)
Integrating Eqn. (i)
EI
dy W x2
= × +C1
dx
2
2
(ii)
l dy
= 0, substituting
When x = ,
2 dx
O=
∴
2
W
l
1
×
× +C1 ;
2
2
2
C1 = −
W l2
16
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=
W l2
+C1 = 0
16
246
•
Strength of Materials
Substituting for C1 in Eqn. (ii),
EI
This is required equation for slope
dy W x2 W l 2
=
−
dx
4
16
(iii)
dy
(in radians). Integrating Eqn. (iii) again,
dx
W x3 W l 2 K
−
+C2
12
16
EIy =
(iv)
When x = 0; y = 0, substituting in Eqn. (iv),
∴
0 = 0 − 0 +C2
C2 = 0
Now Eqn. (iv) becomes,
EIy =
W x3 W l 2 x
−
12
16
This is the required equation for deflection.
l
For maximum deflection at midpoint where point load W is acting, substituting x = ; we get,
2
3
W
EIy =
=
12
y=
−
W l2
l
2
16
W l3 W l3
−
96
32
y=−
So
l
2
W l3
48EI
(minus sign shows downward deflection)
W l3
48EI
2. Simply Supported Beam with Uniform Distributed Load
x
X
w unit of length
A
wl
RA = 2
C
yc
l
Figure 14.14
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B
wl
RB = 2
Deflection of Beams
•
247
wl
Reactions = RA = RB = , c is the centre.
2
Consider a section X at a distance x from A,
x2
2
wlx wx2
=
−
2
2
d 2 y wlx wx2
EI 2 =
−
2
2
dx
Mx = RA x − w
Integrating
EI
dy wlx2 wx3
=
−
+C1
dx
4
6
(i)
l dy
= o, substituting in Eqn. (i)
When x = ,
2 dx
o=
wl 3 wl 3
−
+C1
16
48
∴
C1 = −
wl 3
24
Equation (i) becomes
EI
dy wlx2 wx3 wl 3
=
−
−
dx
4
6
24
(ii)
This is required equation for slope at any point.
Integrating Eqn. (ii),
EIy =
wlx3 wl 4 wl 3 x
−
−
+C2
12
24
24
when x = 0, y = 0
∴
0 = 0 − 0 − 0 +C2
∴
EIy =
wlx3
12
−
wx4
24
−
C2 = 0
wl 3 x
24
(iii)
This is the required equation for deflection at one point. For maximum deflection yc at centre,
x = l/2, substituting in Eqn. (iii),
EIy =
5wl 4
wl 4 wl 4 wl 4
−
−
=−
96 384 48
384
∴ Maximum deflection at mid span
5wl 4
384EI
5wl 4
yc =
384EI
yc = −
or
(minus sign indicates downward deflection)
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248
•
Strength of Materials
If we substitute wl = W , then
yc = −
5W l 3
384EI
Deflection due to impact: If a mass M drops through a height h produces a maximum instantaneous
displacement y and let a load W gradually applied produces the same deflection y. Then the work
done in the two cases is the same,
1
Mg (h + y) = wy
2
Y = kw
∴
where k depends on the nature of the beam and the position of load
∴
1
Mg (h + kw) = kw2
2
This is a quadratic from which W , the equivalent static load, can be obtained.
Then, y = kw
3. Simply Supported Beam with Point Load Eccentric
Consider the beam shown in Fig. 14.15.
W
In this case point where slope is zero is not
defined because of eccentric load.
C
A
Wb
RA =
;
l
Wa
RB =
l
a
B
b
l
Wb
l
Wa
l
Figure 14.15
Let as consider portion AC, any point at a distance x from A,
W bx
l
2
dy
W bx
EI 2 =
l
dx
M=
Integrating,
EI
dy W bx2
=
+C1
dx
2l
Integrating again,
EIy =
W bx3
+C1 x +C2
6l
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(i)
Deflection of Beams
•
249
At A, x = 0, y = 0, substituting in the above equation,
0 = 0 + 0 +C2
∴
EIy =
W bx3
6l
∴
C2 = 0
+C1 x
(ii)
Now consider portion CB, at any point at a distance x from A,
W bx
−W (x − a)
l
d 2 y W bx
EI 2 =
−W (x − a)
l
dx
Mx =
Integrating,
ET
dy wbx2 W (x − a)2
=
−
+C3
dx
2l
2
(iii)
EIy
W bx3
(x − a)3
−W
+C3 x +C4
6l
6
(iv)
Integrating again,
W bx2
+C1
2l
W bx2 W (x − a)2
And slope at C from Eqn. (iii) =
−
+C3
2l
2
Therefore, equating the above two equations,
Now slope at C from Eqn. (i) =
W bx2 W (x − a)2
W bx2
+C1 =
−
+C3
2l
2l
2
When x = a
Equation (v) becomes,
W bx2
W bx2
+C1 =
+C3
2l
2l
or C1 = C3
Also deflection at C form Eqn. (ii) = deflection at C from Eqn. (iv)
Equating,
W bx3
W bx3 W (x − a)3
+C1 x =
−
+C3 x +C4
6l
6l
6
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(v)
250
•
Strength of Materials
At C, x = a, C1 = C3 , substituting x = a,
W ba3
W ba3 W (a − a)3
+C1 a =
−
+C3 a +C4
6l
6l
6
Since C1 = C3
∴
At B, x = l, y = 0
From Eqn. (iv),
C4 = 0
0=
(l − a)3
W bl 3
−W
+C3 l
6l
6
∵
C4 = 0
W bl 3 W b3
−
+C3 l = 0
6l
6
or,
∴
C3 = −
W bl 3 W b3
+
6l
6l
Wb 2
l − b2
6l
From Eqn. (ii), if x ≤ a, then
Since C1 = C3
∴
C1 = −
W bx3
W b(l 2 − b2 )3 x
−
2l
6l
W bx3
w(x − a)3
Wb 2
EIy =
−
−
(l − b2 )x
6l
6
6l
EIy =
Also,
This is for x ≥ a,
∴ yc =
(from Eqn. iv)
Wb 2
a + b2 − l 2 if x = a
6EIl
Note: Thus from the above two equations, deflection at any point can be determined, provided right
equation is used, depending upon value of x. If it is less or more than a.
For maximum deflection we can work as follows:
Let a > b,
dy
W bx2 W b 2
From Eqn. (i) EI
=
−
(l − b2 )
dx
2l
6l
dy
=0
We know that at maximum deflection, slope,
dx
0=
Solving,
x=
W bx2
Wb 2
−
(l − b2 )
2l
6l
l 2 − b2
3
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Deflection of Beams
Thus, maximum deflection,
Wb
EIy =
6l
ymax =
l 2 − b2
3
3
−
Wb 2
(l − b2 )
6l
•
251
l 2 − b2
3
W b (l 2 − b2 )3/2
√
9 3 EIl
Distance of point of maximum deflection from the centre =
l
l
l
The maximum value of which can be = √ − =
3 2 13
Therefore it is to be noted that maximum deflection lies within
l 2 − b2 l
−
3
2
l
of the centre.
13
(B) Macaulay’s Method
As we have seen in the preceding article that we have to make separate expression for different
sections in double integration method. This method is okay for simple cases, but becomes very
tedious and complicated for complex problems. To make calculations easier Macaulay’s method is
very convenient. In this method (x − a) is expressed in brackets and while solving the problems it
is treated in a special manner which will be explained through problems. Let us take previous case
again (Refer Fig. 14.16).
W
A
a
Wb
Wa
RA =
; RB =
l
l
B
b
k
l
Wa
l
Wb
l
Figure 14.16.
Consider a section X at a distance x from A,
Mx = EI
d 2 y W bx =
−W [x − a] ;
l
dx2
The left of this line is for AC and the whole for CB.
Integrating,
EI
X
C
W [x − a] 2
dy W bx2
=
+C1 −
dx
2l
2
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252
•
Strength of Materials
Again integrating,
EIy =
W [x − a]3
W bx3
+C1 x +C2 −
2l
6
[x − a]2
is integrated in special manner in, Macaulay’s method.
Note: Now
2
At A, x = 0, y = 0, The equation for portion AC,
W bx3
+C1 x +C2
2l
0 = 0 + 0 +C2
∴
EIy =
C2 = 0
At B, x = l, y = 0. Now as indicated before, the whole equation is for CB.
∴
∴
0=
W b3
W bl 3
+C1 l −
2l
6
C1 = −
or,
0=
W bl 2
W b3
+C1 l −
2
6
Wb 2
(l − b2 )
6l
Therefore, the slope and deflection are given by equations:
[x − a]2
dy W bx2 W b 2
=
−
(l − b2 )−W
dx
2l
6l
2
[x − a]3
W bx3 W b 2 2 −
(l −b )x−W
EIy =
6l
6l
6
EI
Deflection under load,
Wab 2 2 2 W ba3 W b 2 2 −
l −b a = −
l −b −a
6l
6l
6l
Wab (a + b)2 −b2 −a2
=−
6l
Wab 2 2
a +b +2ab − b2 −a2
EIyc = −
6l
EIyc =
yc = −
Wa2 b2
3l EI
(minus sign indicates downward deflection)
For maximum deflection we have to proceed as done already in double integration method.
E XAMPLE 14 B-1: A simple supported beam has a span of 15 m and carries two point loads of 4
kN and 9 kN at 6 m and 10 m, respectively from one end. Find the deflection under each load and
maximum deflection. E = 200 GPa, I = 400 × 106 mm4 .
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Deflection of Beams
•
253
S OLUTION :
4 kN
9 kN
A
D
6m
RA = 5.4 kN
5m
4m
B
7.6 kN
10 m
15 m
Figure 14.17
Taking moments about B, 15RA = 4 × 9 + 9 × 5 = 81
∴
RA =
5.4) = 7.6 kN; Using Macaulay’s method,
Mx = EI
81
= 5.4 kN RB = (4 + 9 −
15
d2y
= 5.4x − 4 (x − 6) − 9 (x − 10)
dx2
x2
(x − 6)2
(x − 10)2
dy
= 5.4 −4
−9
+C1
dx
2
2
2
EI
EIy = 5.4
(i)
x3 4 (x − 6)3 9 (x − 10)3
−
−
+C1 x +C2
6
6
6
When x = 0, y = 0
0 = 0 − 0 − 0 + 0 +C2
∴
C2 = 0,
When x = 15, y = 0
0=
5.4 (15)3 4 (15 − 9)3 9 (15 − 10)3
−
−
+C1 x
6
6
6
0 = 3037.5 − 486 − 187.5 + 15C1
∴
C1 = −157.6;
Working in N and m, I =
∴
EIy =
400 × 106
(1000)4
5.4x3 4 (x − 6)3 9 (x − 10)3
−
−
−157.6x
6
6
6
(ii)
= 4 × 10−4 m4
E = 2 × 1011 N/m2
Since we have been working in kN, so will multiply by 1000 the equation of y. For deflection at C,
put x = 6 m in Eqn. (ii)
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254
•
Strength of Materials
1 5.4 × 63
yc =
− 0 − 0 − 157.6 × 6
EI
6
=
1000
2 × 1011 × 4 × 10−4
=−
[194.6 − 945.6]
751.2 × 1000
= −0.00939 m = −9.39 mm
2 × 1011 × 4 × 10−4
Ans
Minus sign means downward deflection.
∴ deflection at c, yc = 9.39 mm Ans
Deflection at D: Putting x = 10 m
1000
5.4 × 103 4 (10 − 6)3
yD =
−
−0 − 157.6 × 10
6
6
2 × 1011 ×4 × 10−4
= 1.25 × 10−5 [900 − 42.67 − 1576]
=−
1.25 × 718.67
= −0.00898 m = 8.98 mm
105
Ans
dy
For maximum deflection
=0
dx
Using Eqn. (i)
0 = 2.7x2 − 2 x2 + 36 − 12x − 4.5 x2 + 100 − 20x − 157.6
0 = 3.8x2 − 114x + 679.6
0 = x2 − 30x + 178.84
√
+30 ± 900 − 715.36 30 ± 13.6
=
x=
2
2
= 8.2 m
Ans
Maximum deflection will occur at 8.2 m.
Substituting x = 8.2 in Eqn. (ii)
EIy =
5.4
4(8.2 − 6)3
(8.2)3 −
− 0 − 157.6 × 8.2
6
2
EIy = 496.23 − 7.09 − 1292.32
= −803.18 or
y=−
803.18 × 1000
2 × 1011 × 4 × 10−4
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Deflection of Beams
•
255
y = 0.010 m
y = 10 mm
or
Ans
Note: If term is in bracket is −ve, then, it is regarded as zero in Macaulay’s method, for example,
(8.2 − 10)3 in Eqn. (ii) will be regarded as zero as shown above.
E XAMPLE 14 B-2: A horizontal beam AB a freely supported at A and B, 9 m apart and carries a
uniformly distributed load of 12 kN/m run. A clockwise moment of 120 kNm is applied to the beam
at a point C, 3.5 m from the left hand support A. Calculate the slope and defection of the beam at C.
If flexural rigidity EI = 40 × 103 kNm2 .
S OLUTION :
120 kN/m 12 kN/m
A
3.5 m
X
x
40.7 kN
B
C
67.33 kN
9m
Figure 14.18
Taking moments about A,
9
+ 120
RB × 9 = 12 × 9 ×
2
= 606 kN
606
= 67.33 kN
∴ RB =
9
RA = (12 × 9) − 67.33 = 40.7 kN
Using Macaulay’s method, taking A as origin and considering section X at a distance of x metres
from A,
d2y
x
Mx = EI 2 = 40.7x − 12x × + 120 (x − 3.5)◦
2
dx
Note: (x − 3.5)◦ is taken when couple is applied.
Integrating,
x2
x3
dy
= 40.7 − 12 + 120 (x − 3.5) +C1
dx
2
6
dy
= 20.35x2 − 2x3 + 120 (x − 3.5) +C1
EI
dx
EI
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(i)
256
•
Strength of Materials
Integrating again,
x3 2x4 120 (x − 3.5)2
−
+
+C1 x +C2
3
4
2
EIy = 6.78x3 − 0.5x4 + 60 (x − 3.5)2 +C1 x +C2
EIy = 20.35
(ii)
Now when x = 0, y = 0
0 = 0 − 0 + 0 + 0 +C2
∴
C2 = 0
Eqn. (ii) becomes,
EIy = 6.78x3 − 0.5x4 + 60 (x − 3.5)2 +C1 x
(iii)
when x = 9, y = 0, substituting Eqn. (iii),
∴
0 = 6.78 (9)3 − 0.5 × 94 + 60 (9 − 3.5)2 + 9C1
0 = 4942.62 − 3280.5 + 1815 + 9C1
C1 = −386.35
So Eqn. (ii) and Eqn. (iii) become,
EI
For slope
dy
= 20.35x2 − 2x3 + 120 (x − 3.5) − 386.35
dx
EIy = 6.78x3 − 0.5x4 + 60 (x − 3.5)2 − 386.35x
dy
at C, substituting x = 3.5 in Eqn. (i)
dx
EI
dy
= 20.35 (3.5)2 − 2 (3.5)3 + 120 (3.5 − 3.5) − 386.39
dx
= 249.3 − 85.75 + 0 − 386.35
1
[−222.45]
=
40 × 103
= −0.00556 radian
= 0.00556 radian. Ans
For deflection at C, using Eqn. (v) and substituting x = 3.5;
EIy = 6.78 (3.5)3 − 0.5 (3.5)4 + 60 (3.5 − 3.5)2 − 386.35 × 3.5
= 290.7 − 75 + 0 − 1352.225
1
(−1136.5)
y=
40 × 103
= −0.0284 m
(minus sign means deflection is downward)
= 28.4 mm Ans
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(iv)
(v)
Deflection of Beams
•
257
Beams of composite section (Flitched Beams):
When a beam is made up of two or more materials joined together such that it behaves like a
single piece and the deflection of each material at a certain point is same.
It is possible to find slope and deflection of beam of composite section by algebraically adding
the flexural rigidities (EI) of different materials.
∑ EI = E1 I1 + E2 I2
If should be noted that the moment of inertia of the composite section is to be found out about the
centre of gravity of the section.
E XAMPLE 14 B-3: A flitched beam of space 7 m consists of a timber section 170 mm wide and
250 mm deep. Two plates of 170 mm and 15 mm thick are fixed at the top and bottom of the
timber section. The composite beam is subjected to a point load of 90 kN at the middle of the beam.
Determine the deflection of the beam under the load. Take E for steel as 200 GPa and E for timber
as 10 GPa, respectively.
S OLUTION :
In this case centre of gravity of steel plates and
of timber is at the centre of the whole section
because of symmetry.
15 mm
250 mm
∴
EItimber = 10000
15 mm
170 × 2503
12
= 2.21 × 1012 N mm2
170 mm
Figure 14.19
170 (15)3
Esteel = 200000 2 ×
+ 2 (170 × 15) × (132.5)2
12
= 200000 [95625 + 89536875]
= 17.9 × 1012
Total flexural rigidity of the composite section about its centre of gravity,
ΣEI = 2.21 × 1012 + 17.9 × 1012
= 20.11 × 1012
Deflection at the centre of the beam,
yc =
W l3
90000 × (7 × 1000)3
=
48ΣEI
48 × 20.11 × 1012
= 31.98 mm
Ans
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258
•
Strength of Materials
E XAMPLE 14 B-4: A composite beam consists of two timber joists 130 mm wide and 300 mm
deep with a steel plate 240 mm deep and 15 mm thick fixed symmetrically between the timber joists.
The beam carries a uniformly distributed load of 4 kN/m and is simply supported over a span of 5
metres. Find the slopes at supports and deflection of the beam at its centre. Take
Es = 200 GPa
and Et = 10 GPa.
S OLUTION :
It is clear from the figure that the centre of gravity of the beam section coincides with the centre
of gravity of the steel plate.
Therefore, flexural rigidity for the timber joists,
15 mm
300 mm
240 mm
130 mm
130 mm
(a)
4 kN/m
A
B
5m
(b)
Figure 14.20
130 × 2803
EIt = 10 × 1000 2 ×
12
= 4.756 × 1012 Nmm2
15 × 2403
EIs = 200 × 1000
12
= 3.456 × 1012 Nmm2
Total flexural rigidity of the composite beam about its centre of gravity,
ΣEI = 4.756 × 1012 + 3.456 × 1012
= 8.212 × 1012 Nmm2
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Deflection of Beams
•
259
Slope at the supports,
Because of symmetry θB = θA =
wl 3
24ΣEI
3
4 × 5 × 103
=
24 × 8.212 × 1012
∵
5 × 1000
= 5 N/mm
5 kN/m
1000
= 2.537 × 10−3 = 0.00254 radian
Deflection of the beam at its centre,
Yc =
=
5wl 4
384ΣEI
5 × 4 (5 × 1000)4
384 × 8.212 × 1012
= 3.96 mm
Ans
(C) Moment Area Method
This method is based on the application of two theorems known as Mohr’s theorem. This method
is partially convenient in case of beams acted upon with concentrated loads, in which case bending
moment area consists of triangles and rectangles. However, this method may be conveniently used
in certain cases of distributed load where the position of the centroid of the bending moment area is
known.
Theorem I: The change in slope between any two points on a beams is equal to the net area of the
bending moment diagram between these points, divided by the flexural rigidity (EI) of the beam.
Let AB [Fig. 14.21(a)] be part of C beam, which has been deflected to the position A B and let
Fig. 14.21(b) represent the corresponding part of the BM diagram.
x2
x1
A
y1
A'
B
θ1
y2
B'
(a)
M
dx
(b)
Figure 14.21
@seismicisolation
@seismicisolation
θ2
260
•
Strength of Materials
If the slopes at A and B, distances x1 and x2 from origin are θ1 and θ2 , respectively,
We know EI
d2y
=M
dx2
∴ slope of beam at any point
θ=
dy
=
dx
=
M
dx
EI
1
EI
M.dx,
as E and I are constants.
1
× area of BM diagram between points A and B.
Then θ2 − θ1 =
EI
If θ1 = 0, than θ2 the actual slope of the beam at B.
M
d2y
=
2
EI
dx
Multiplying both sides by x,
x
d2y
M
=
x
2
EI
dx
Integrating by parts between the limits of x, and x2 ,
x2
x2
dy
1
x −y =
M.x.dx
dx
EI
x1
x1
=
1
x
EI
moment of area of BM diagram about origin.
dy
By a suitable choice of origin, x can usually be made zero at both limits and y can be made zero
dx
at one limit leaving the value of y at other limit to represent the required deflection.
In should be noted that moment-area method is usually convenient to use only when a point of
zero-slope is known.
For ready reference following table gives the value of area A, distance from origin x̄, and BM
diagrams.
@seismicisolation
@seismicisolation
Deflection of Beams
BM Diagram
x
Area
Remarks
b
h
1 bh
2
2 b
3
Concetrated Load for
centilever
1 bh
3
3 b
4
Uniformly distributed
load on centilever
1 bh
4
4 b
5
Uniformly varying
load on a centilever
2 bh
3
5 b
8
Uniformly distributed
load on single
supported beam
x
b
h
Square parabola
x
b
h
Cubic parabola
x
h
x
b
Figure 14.22
Let us take some standard cases:
(a) Cantilever with concentrated end load; Fig. 14.23:
θb − θa =
1
1
× W l.l
EI 2
W
A
θb
W l2
θb =
since θa = 0
2EI
Taking the origin at B,
1
1
2
dy
− y]l =
× W l.l × l
dx
EI 2
3
wl 3
[0 − 0] − [0 − yb ] = yb =
3EI
[x
wl
yb
G
2l
3
Figure 14.23
@seismicisolation
@seismicisolation
B
BM Diagram
•
261
262
•
Strength of Materials
(b) Cantilever with uniformly distributed load; Fig. 14.24:
1
1 wl 2
θb − θa =
×
.l
EI 3 2
A
wl 3
i.e. θb =
since θa = 0
6EI
w/unit length
B
θb
yb
l
Taking the origin at B,
l
1 1 wl 2 3
dy
×
l× l
x −y =
dx
EI 3 2
4
6
i.e.
(0 − 0) − (0 − yb ) = yb =
wl2
2
wl 4
8EI
3 l
4
BM Diagram
Figure 14.24
(c) Cantilever with end couple; Fig. 14.25:
B
A
1
θb − θa = ×Ml
EI
Ml
since θa = 0
θb =
EI
Taking the origin at B,
l
l
1
dy
×Ml×
x −y =
dx
EI
2
0
(0 − 0) − (0 − yb ) = yb =
θb
M
yb
l
M
l/2
BM Diagram
Figure 14.25
Ml 2
2EI
(d) Simply supported beam with central concentrated load Fig. 14.26
1 1 wl l
.
θb − θc = ×
elastic curve
W
EI 2 4 2
B y
A
2
C
Wl
c
since θc = 0
θB =
qb W/2
W/2
16EI
l
Taking the origin at B and considering the
part BC only,
Wl
l/2
4
dy
1 1 Wl l 2 l
x −y
=
× .
× ×
2 l
dx
EI 2 4 2 3 2
0
3 2
BM Diagram
W l3
(0 − yc ) − (0 − 0) =
48EI
Figure 14.26
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Deflection of Beams
•
263
(e) Simply supported beam with uniformly distributed load Fig. 14.27:
1
2 wl 2 l
× .
.
EI 3 8 2
θb = wl 3 since θc = 0.
w/unit length
B
yc
θb − θc =
∴
A
l
w l3
since θc = 0
24EI
wl2
Taking the origin at B and considering BC
8
only,
5 l
l/2
8 2
3
dy
1 2 wl
l 5 l
BM Diagram
x −y
=
× .
× × .
dx
EI 3 8
2 8 2
0
Figure 14.27
4
5 wl
.
yc
(0 − yc ) − (0 − 0) =
384 EI
5wl 4
=
384 EI
dy
in radians.
Note: θ is in radians and is slope
dx
Alternately Mohr’s theorem related to the moment-area method can be written as follows.
i.e θb =
Theorem I
Change in slope between A and B
=
Net area of bending moment diagram A and B
EI
Theorem II
Deflection of B with respect to A
=
Moment of area of bending moment diagram between A and B
EI
E XAMPLE 14 C-1: A cantilever of length 3 m carries a concentrated load of 50 kN at the free end
and another 40 kN at 1 m from fixed end. Calculate the slope and deflection at the free end using
moment-area method. E = 200 GPa and I = 107 mm4 .
40 kN
50 kN
C
A
1m
B
2m
Figure 14.28
@seismicisolation
@seismicisolation
264
•
Strength of Materials
Bending moments:
MB = 0, MC = −50 × 2 = −100 kNm
MA = −50 × 3−40 × 1 = −190 kNm
A
B
2
3
1
100 kNm
Bending moment diagram
190 kNm
Figure 14.29
Let us divide the bending moment diagram in three parts as shown in Fig. 14.29.
1
A1 = 100 × 2 × = 100 kNm
2
A2 = 100 × 1 = 100 kNm
1
A3 = (190 − 100) × 1× = 45 kNm
2
Now slope at B will be
A1 + A2 + A3
A
=
EI
EI
100 + 100 + 45 245
=
=
EI
EI
θC =
245 × 106
= 0.0001225 radians
200 × 103 ×107
180
= 0.0001225 ×
= 0.007◦
π
=
Deflection at B =
Ax̄
;
EI
Ans
x̄ is measured from free end B,
2
2
100× ×2 + 100 × 2.5 + 45× 2 + ×1
3
3
=
EI
133.33 + 250 + 120 503.33
=
=
EI
EI
=
503.33 × 109
= 0.25 mm
200 × 103 ×107
@seismicisolation
@seismicisolation
Ans
Deflection of Beams
•
265
E XAMPLE 14 C-2: A beam 7 m long is simply supported at its ends and carries point loads of
25 kN each at points 1.7 m from the ends. Calculate by moment-area method: i) the maximum slope
and ii) deflection under each load. Take EI = 60000 kNm2
25 kN
Due to symmetry
RA = RB = 25 kN
MC = 25 × 1.7 = 45.9 kNm
A
25 kN 1.7 m
Slope at the M is zero because of maximum
deflection at M (the midpoint).
Slope at A or B is maximum, slope at B with
respect to M
=
Area of bending moment diagram between M and B
EI
25 kN
M
C
D
B
1.7 m 25 kN
7m
45.9 kNm
1.7 m 1.8 m
M
1.8 m
45.9 kNm
1.7 m
Figure 14.30
1
45.9 × 1.7× +45.9 × 1.7 39.015 + 78.03
2
=
θB =
EI
EI
117.045
= 0.001951 radian
=
60000
180
= 0.001951×
= 0.112◦ Ans
π
Moment of area bending moment diagram from M to B about B
Deflection at M = ym =
EI
1.8
45.9
2
1.8 × 45.9 ×
+ 1.7 +
× 1.7 × 1.7 ×
2
2
3
=
EI
214.812 + 44.217
=
60000
= 0.00432 m
= 4.32 mm (maximum deflection)
Deflection under load 25 kN,
= ym − deflection of D with respect to M;
Deflection of C with respect to M
Moment of bending moment diagram area between M and D about D
EI
1.8
45.9 × 1.8×
2 = 0.00124 m
=
60000
= 1.24 mm
=
@seismicisolation
@seismicisolation
266
•
Strength of Materials
Hence, deflection under load 25 kN at D
= 4.32 − 1.24 = 3.08 mm
Ans
(D) Conjugate Beam Method
The basis of this method depends upon the modification of moment-area method. The momentarea method is convenient for the beams of constant flexure rigidity, but conjugate beam method is
particularly useful for the beams of different flexure rigidity. In fact conjugate beam is an imaginary
M
beam of length equal to that of the original beam but for which the load diagram is
diagram.
EI
M
M
This
diagram shows the variation of
over the length of the beam. In other words, the load at
EI
EI
any section on the conjugate beam is equal to the bending moment at that point divided by flexural
rigidity (EI). In nutshell, 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.
Therefore, before applying the conjugate beam method, conjugate beam is constructed. The
load on the conjugate beam at any section is equal to the bending moment at that section divided by
EI. Hence, the loading on the conjugate beam is known.
The shear force at any point on the conjugate beam gives the slope at the corresponding point
of actual beam. And the bending moment at any section on the conjugate beam gives the deflection
at the corresponding point of the actual beam.
Simply supported beam with a concentrated load at the centre
A simply supported beam AB of length l carrying a point load W at the centre C is shown in
W
Fig. 14.31.
C
A
B
l/2
(a)
l
RA = W
2
(b)
A
(c)
A
R'A = Wl2
16 EI
D'
Wl
4
C'
D''
Wl
EI
Loaded
c
Conjugate Beam
Figure 14.31
@seismicisolation
@seismicisolation
RB = W
2
Bending moment diagram
B
Load diagram
B
R'B = Wl2
16 EI
Deflection of Beams
•
267
(i) The bending moment diagram is drawn in usual way. For constructing a conjugate beam
diagram, the load on the conjugate beam is obtained by dividing the bending moment at that
point by EI. The shape of the loading on the conjugate beam is same as of bending moment
diagram. It should be remembered that bending moment diagram (modified) as shown in Fig.
14.31 (c) is the loading diagram known as conjugate beam.
ii) Bending moment diagram from loaded conjugate diagram gives the deflection at any section.
iii) Shear force diagram drawn on the basis of conjugate diagram provides the slope at the desired
section.
Now, referring to Fig. 14.31, in case of simply supported beam with a central point load, because
total load on the conjugate beam = Area of the load diagram
Wl
W l2
1
=
= ×l×
2
4EI
8 EI
∴ Reaction RA = RB =
dy
Slope at A
dx
W l2
8 EI
= θA
And deflection at C = yc
According to conjugate beam method,
θA = Shear force at A for the conjugate beam
= RA =
W l2
,
16EI
because shear force at A for conjugate beam = RA
yc = Bending moment at C for the conjugate beam
1 l
Wl
1 l
W l2 l
· −
× ×
×
×
=
16EI 2
2 2 4EI
3 2
=
W l2
3W l 3 −W l 3
W l3
−
=
32EI 96EI
96EI
=
W l3
48EI
@seismicisolation
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268
•
Strength of Materials
Simply supported beam carrying an eccentric concentrated load
W
C
B
A
a
(a)
RA = Wb
l
(b)
A
(c)
A'
b
l
D
Wab
l
BM Diagram C1
Load diagram
D1
B
Wab
EIl
R'A
Conjugate beam C2
a
RB = Wa
l
b
B'
R'B
Figure 14.32
After constructing bending moment diagram, the conjugate beam shown in Fig. 14.32(C) is
Wab
.
drawn. The vertical load on conjugate beam will be
EIl
To find RA and RB ; taking moments about A of the conjugate beam, we get
RB × l
1
Wab 2
1
Wab
b
= ×a×
× ×a+ ×b×
× a+
2
EIl
3
2
EIl
3
b
Wa3 b Wab2
+
a+
=
3EI l 2EI l
3
Wa3 b Wa2 b2 Wab3
+
+
3EI l
2EI l
6EI l
Wab
=
+ 2a2 +3ab + b2
6EI l
Wab 2 2
a +b +2ab + ab + a2
=
6EI l
Wab =
(a + b)2 + a(a + b)
6EI l
Wab 2
l +a l
=
6EI l
Wa(l − a)
. l (l + a)
=
6EI l
Wa 2 2
(l −a )
=
6EI
=
@seismicisolation
@seismicisolation
Deflection of Beams
Similarly,
RA =
•
269
Wb 2 2
(l −b )
6EI l
dy
at A
dx
yc = deflection at C for the given beam
Using conjugate beam method,
If θA = slope at A or
θA = Shear force at A for the conjugate beam
= RA
=
(because shear forece at A for conjugate beam = RA )
Wb 2 2
(l −b )
6EI l
yc = Bending moment at C for conjugate beam
1
Wab
a
×a×
×
= RA ×a−
2
EI l
3
Wa3 b
Wb 2 2
(l −b ).a −
6EI l
6EI l
Wab 2 2 2
=
l −b −a
6EI l
=
E XAMPLE 14 D-1: Using conjugate beam method, find the slope at the support points and deflection under the given load for the beam shown in Fig. 14.33(a). Take E = 200 GPa and I = 4 ×
10−5 m4 .
S OLUTION :
8 kN
2.5 I
A
RA = 3 kN
I
D
C
4.8 m
6m
3.6 m
D is the midpoint of beam AB.
Figure 14.33(a)
Taking moments about B,
RA × 9.6 = 8 × 3.6
RA = 3 kN
RB = 8 − 3 = 5 kN
@seismicisolation
@seismicisolation
B
RB = 5 kN
270
•
Strength of Materials
(b)
18 kNm
14.4 kNm
A
B
BMD for actual beam
A'
(c)
5.76
EI
7.2
EI
18
EI
B'
R'A
R'B
Loaded conjugate beam
Figure 14.33 (b, c)
Bending moment diagram for actual beam:
MC = 3 × 6 = 18 kNm
MD = 3 × 4.8 = 14.4 kNm
Conjugate beam (Fig. 14.33C)
7.2
1
=
E (2.5I)
EI
18
1
=
Load intensity just to the right of C = 18 ×
EI
EI
1
Load intensity at D = RA × 4.8 ×
E (2.5I)
Load intensity just to the left of C = 18 ×
=
5.76
3 × 4.8
=
2.5 EI
EI
Reactions RA and RB . Taking moments about B,
7.2
1
6
18
1 3.6
RA × 8 =
×6× ×
+ 3.6 +
× 3.6 ×
EI
2
3
EI
2 3
120.96
38.88
=
+
EI
EI
=
or
159.84
EI
RA = 19.98/EI
1
7.2
1 18
RA + RB =
×
×6 +
×
× 3.6
2
EI
2 EI
@seismicisolation
@seismicisolation
Deflection of Beams
•
21.6
32.4
+
EI
EI
54
=
EI
19.98
54
+ RB =
EI
EI
34.02
∴ RB =
EI
=
Now
Slope at A, θA = shear force at A = RA
=
19.98 × 1000
19.98
=
EI
200 × 109 × 4 × 10−5
[Multiplying by 1000 to convert into newtons]
= 0.002497 radian
= 0.143◦
Slope at B, θB = shear force at B = RB
34.02
EI
34.02 × 1000
=
200 × 109 × 4 × 10−5
=
= 0.00425 radian
= 0.244◦
Deflection under the given load,
Yc = Bending moment at C of the conjugate beam
=
6
19.98 × 6 7.2 × 6
−
×
EI
EI × 2
3
119.88 43.2
−
EI
EI
76.68
=
EI
76.68 × 1000
=
200 × 109 × 4 × 10−5
=
= 0.0096 m
= 9.6 mm
Ans
@seismicisolation
@seismicisolation
271
272
•
Strength of Materials
5.76
1
4.8
19.98 × 4.8
−
× 4.8 × ×
EI
EI
2
3
22.12
95.9
−
=
EI
EI
73.78 × 1000
=
200 × 109 × 4 × 10−5
Similarly,
yD =
= 0.00922
= 9.22 mm
Ans
E XAMPLE 14 D-2: Find the deflection at centre of the beam shown in the Fig. 14.34. Take E =
210 GPa and I = 280 × 10−5 m4
S OLUTION :
280 kN
E
C
A
RA = 140 kN
(a)
I
2I
1.8 m
3.6 m
D
B
I
RB = 140 kN
1.8 m
(a) Actually loaded beam
504 kNm
252 kNm
(b)
252 kNm
(b) BMD for actual beam
P
126
EI
Q
A′
C′
1.8 m
T
126
EI
E′
1.8 m
126
EI
S
252
EI
126
EI
(c)
RA′
R
D′
1.8 m
Loaded conjugate beam
Figure 14.34
@seismicisolation
@seismicisolation
B′
1.8 m
RB′ =
Deflection of Beams
•
273
RB × 7.2 m = 280 × 3.6
RB = 140 kN
RA + RB = 280
∴
RA = 280 − 140 = 140 kN
ME = 140 × 3.6 = 504 kNm
MC = 140 × 1.8 = 252 kNm
MD = 140 × 5.4 − 280 × 1.8
= 756 − 504 = 252 kNm
Diagram (b) is drawn accordingly
Conjugate Beam:
Load at E =
504
2EI
=
252
EI
Load just to the left of C =
252
= PC EI
Load just to the right of D = T D
Load just to the right of C =
(Due to symmetry)
252 126
=
2EI
EI
Load just to the right of D = SD
Therefore, PQ = QC =
and T S = SE =
126
EI
126
EI
Because load on the conjugate beam is symmetrical, so the support reactions are equal each
being equal to one-half of the total load on the conjugate beam.
@seismicisolation
@seismicisolation
274
•
Strength of Materials
RA = RB = Area of Δ A PC + area of trapezium C QRD
1
252
1 126 252
= ×1.8×
+
+
×1.8
2
EI
2 EI EI
=
226.8 340.2
+
EI
EI
=
567
EI
Deflection of mid-point E of the actual beam:
yD = Bending moment at E of the conjugate beam
1.8
yD = RA × 3.6 − area A PC × 1.8+
3
1.8 126
1 1.8
126
×1.8×
+
×1.8× +
−
EI
2
EI
2 3
567 × 3.6 252
1
1.8
204.12 68.04
=
−
×1.8× 1.8+
−
+
EI
EI
2
3
EI
EI
=
2041.2 544.32 204.12 68.04
−
−
−
EI
EI
EI
EI
=
1224.52
EI
=
1224.52 × 1000
210 × 109 ×280 × 10−5
= 0.00208 m
= 2.08 mm Ans
E XAMPLE 14 D-3: Find the deflection at the free end of cantilever shown in Fig. 14.35. Take
E = 200 GPa, I = 1.2 × 10−4 m4
Bending moment along cantilever:
MB = 0
MC = 12 × 1.75 = 21 kNm
MA = 12 × 3.5 + 22 × 1.75
= 42 + 38.5
= 80.5 kNm
@seismicisolation
@seismicisolation
Deflection of Beams
22 kN
A
(a)
Figure 14.35
12 kN
C
B
1.75 m
1.75 m
21 kNm
BMD for actual beam
80.5 kNm
(b)
BMD for actual beam
A′
C′
80.5
EI
B′
21
EI
59.5
EI
Loaded conjugate beam
(c)
Figure 14.35(a)
Deflection at B,
yB = Bending moment at B of conjugate beam
1.75
21
× 1.75 ×
+1.75
=
EI
2
59.5
1 2 × 1.75
+
× 1.75 ×
+ 1.75
EI
2
3
+
1 2
21
× 1.75 × × ×1.75
EI
2 3
=
96.47 106.3 21.44
+
+
EI
EI
EI
=
224.21
EI
@seismicisolation
@seismicisolation
•
275
276
•
Strength of Materials
=
224.21 × 1000
200 × 109 ×1.2 × 10−4
= 0.00934 m
= 9.34 mm
Ans
E XAMPLE 14 D-4: For the cantilever shown in Fig. 14.36 (a), find slope and deflection at free
end. Take E = 200 GPa, I = 1.8 × 10−5 m4 .
S OLUTION :
3 kN/m
A
B
2m
(a)
6 kNm
Bending moment diagram for actual beam
1.333
A′
6
EI
B′
G
Loaded conjugate beam
Figure 14.36
MA =
3×2×2
= 6 kNm
2
Variation of bending moment will be parabolic.
Slope at B
= θB = Shear force at B of conjugate beam
1 1.333
6
×2× =
=
3EI
3
EI
1.333 × 1000
=
= 0.00037 radian
200 × 109 × 1.8 × 10−5
0.00037 × 180
= 0.0212◦ Ans
=
π
@seismicisolation
@seismicisolation
Deflection of Beams
•
277
Centroid of the loaded conjugate beam = 34 × 2 = 1.5 from B
Deflection at B = yB = Bending moment at B of conjugate beam
1
2
6
×3× × 1.5 =
3EI
3
EI
2 × 1000
=
= 0.00055 m
200 × 109 ×1.8 × 10−5
= 0.55 mm Ans
=
(E) Superposition Method
The resultant deflection at a point in a beam, in this method is obtained by adding the deflections
at this point due to individual load on the beam. As long as stresses are within the elastic limit, the
deflections at a point due to each individual load are superimposed or each other algebraically to
give the resultant deflection due to the forces acting on the beam.
E XAMPLE 14 E-1: Determine the deflection at centre C and point D, one meter from left hand
support, for the simply supported beam shown in Fig. 14.37, EI = 1500 kN/m2
S OLUTION :
D
A
1m
12 kN
C
3 kN/m
B
2m
2m
Figure 14.37
Let us split the problem in two parts as shown in Fig. 14.38(a) and (b)
12 kN
C
A
B
D
1m
2m
2m
Figure 14.38a
D
A
3 kN/m
B
C
2m
2m
1m
Figure 14.38b
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•
Strength of Materials
i) Deflection due to concentrated load alone,
yc =
12 × (4)3
W l3
=
= 0.011 m = 11 mm
48EI
48 × 1500
For point D, using general equation,
W x3 W l 2 x
−
12EI 16EI
12(1)3
12(4)2 × 1
=
−
12 × 1500 16 × 1500
= 0.000667 − 0.008
= 0.00733 = 7.33 mm (discarding minus sign)
yD =
ii) Deflection due to uniformly distributes load alone :
5wl 4
384EI
5 × 3(4)4
=
384 × 1500
= 0.00667 m = 6.67 mm
yc =
To find deflection at D due to general equation
1 wlx3 wx4 wl 3 x
yD =
−
−
EI 12
24
24
Substitution for x = 1,
1
3 × 4(1)3
3(1)4
3(4)3 1
−
−
1500
12
24
24
1
=
[1 − 0.125 − 8]
1500
= 0.00475 m (ignoring minus sign)
= 4.75 mm
YD =
Total deflection at C = 11 + 6.67 = 17.67 mm Ans
Total deflection at D = 7.33 + 4.75 = 12.08 mm Ans
(F) Strain Energy Method
Strain energy due to bending: Fig. 14.39 shows an element of a beam subjected to a bending
moment which varies as shown from M at one end to M + dM at the other. The length of the
element in dx, the mean radius, of curvature in R and the change of slope between the ends in d φ .
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•
279
dx
dφ
M+dM
M
R
dφ
Figure 14.39
Then the work done in bending moment
=
1
× mean bending moment × angle of bending
2
(assuming that the moment is gradually applied)
This work is stored in the element as strain energy.
M + α dM
Therefore, strain energy =
× d φ where < α < 1
2
1
Md φ (to the first order of small quantities)
2
dx dx M
M
E
dφ =
=
∵
=
R
EI
I
R
i.e. du =
But
(i)
Substituting for d φ in Eqn. (i)
∴
∴
du =
M2
dx
2EI
l
total strain energy =
M2
dx
2EI
(ii)
0
where l is the length. When total strain energy is required M is terms of a to be substituted in the
expression (ii)
Castigliano’s Theorem: If a structure is subjected to a number of external loads (or couples) the
partial derivation of the total strain energy with respect to any load or couple provides the deflection
in the deflection of that load (couple).
Figure 14.40 shows an elastic body which is subjected to forces W1 , W2, W3 , etc. At each load
point, the deflection can be resolved into components in the direction of, and perpendicular to, the
line of action of the force at that point.
Let x1 be the deflection at A in the direction of W1 ,
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•
Strength of Materials
W3
W2
B
C
A
D
W4
W1
Figure 14.40
x2 be the deflection at B in the direction of W2
x3 be the deflection at C in the direction of W3 etc.
1
1
1
U = W1 x1 + W2 x2 + W3 x3 + .....
2
2
2
or 2U = W1 x1 + W2 x2 + W3 x3 + .....
Work done,
Differentiating with respect to W1 , remembering that the deflection at each point is a function of all
the loads.
∂U
∂ x1
∂ x2
∂ k2
= x1 + W1
+ W2
+ W3
+...
(iii)
2
∂ w1
∂ w1
∂ w1
∂ w2
If W increases gradually to W1 + σ W1 , the other forces remaining constant. This change will make
x1 increase to x1 + δ x1 , x2 to x2 + δ x2 , etc.
Then additional work done,
1
δ W1 ) δ x1 + W2 dx2 +W3 dx3 + ....
2
1
δU
δ x1
δ x2
δ x3
= W1
+ dx1 +W2
+ W3
δ W1
δ W1 2
δ W1
δ W1
δ U = (W1 +
∴
(iv)
Subtracting Eqn. (iv) from Eqn. (iii)
x1 =
∂U
∂ W1
(v)
That is, the movement of A in the direction of W1 is equal to the partial derivative of the total strain
energy of the system with respect to W1
Similarly,
x2 =
∂U
;
∂ W2
x3 =
∂U
etc.
∂ W3
Castigliano theorem applies equally well to couples and to mixtures of forces and couples. The
rotation at any point is the partial derivative of the total strain energy with respect to the couple at
that point.
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•
281
Application of Castigliano’s theorem to deflection (including curved bars).
From Eqn. (i),
l
U =
M2
dx
2EI
0
If it is required to find the deflection δ at a point at which there is a load P in the direction of
the required deflection, then
1
∂U
=
δ=
∂P
2EI
1
=
2EI
l
0
l
0
∂ (M 2 )
dx assuming I is constant
∂P
1
∂M
dx =
2M
∂P
EI
l
M
0
∂M
dx
∂P
Note: If there is no force P at the required point and in required direction, much force can be
applied and when an expression for the deflection is obtained, this force is then made zero.
For the rotation in the direction of a couple C,
1
φ =
EI
l
M
0
∂M
dx
∂C
E XAMPLE 14 F-1: A simply supported beam of span l carries a concentrated load W at a distance
a and b respectively from the two ends A and B. Obtain an expression for the deflection under the
load.
S OLUTION :
W
a
b
A
x
C
dx
Wb
l
l
Figure 14.41
Wb
l
Wb
x
At any point between A and C, M =
l
∴ UAC (strain energy between A and C)
Reaction at A =
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B
Wa
l
282
•
Strength of Materials
1
=
2EI
1 Wb
x
l
2
dx =
0
W 2 b2 a3
6EI l 2
Similarly, strain energy between B and C,
UBC =
W 2 a3 b2
6EI l 2
W 2 a2 b2
(a + b)
6EI l 2
(Because total strain energy = UAC + UBC )
Therefore, total strain energy =
∴
Total strain energy =
W 2 a2 b2
6EI l
1
Also work down = W δ , which is equal to total strain energy.
2
Hence,
1
W 2 a2 b2
Wδ =
2
6EI l
W a2 b2
δ=
3EI l
Note: The sign (+ve or −ve) of M does not matter, because M is squared subsequently.
E XAMPLE 14 F-2: Derive an expression for the maximum deflection (using Castigliano’s theorem) of a cantilever of length l carries a uniformly distributed load w/unit length over it entire span,
when a point load P is applies at it free end.
S OLUTION :
P
w unit length
l
x
Figure 14.42
Consider a section ‘x’ from p;
Bending moment due to P, M = Px
Bending moment due to uniformly distributed load, Me =
∴
Mx = Px +
wx2
2
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wx2
2
Deflection of Beams
•
283
Strain energy stored in the beam in given as
e
M 2 dx
2EI
U =
0
As per Castigliano’s theorem,
∂U
∂⎡
P
δ=
l
∂⎣
0
=
∂P
l
=
0
But
⎤
M 2 dx ⎦
2EI
2M ∂ M
.
.dx
2EI ∂ P
∂M
=x
∂P
Therefore,
δ=
l
M
xdx
EI
0
1
=
EI
l wx2
Px +
2
x.dx
(subsituting for M)
0
1
=
EI
l Px2 +
wx3
2
dx
0
l
wx4
1 Px3
+
=
EI 3
8 0
3
1 Pl
wl 4
=
+
Ans
EI 3
8
when P = 0, then
δ=
wl 4
8EI
E XAMPLE 14 F-3: A steel spring ABC of the dimensions shown in Fig. 4.43 in fully clamped at
A. If a vertical force of 20 N in applied at C, find the vertical deflection of this point E = 200 GPa.
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•
Strength of Materials
S OLUTION :
For BC, taking origin at C
P = 20 N
100 mm
dx
B
C
M = 20x
∴
x
θ
0.10
1
UBC =
2EI
(20x)2 dx
dθ
0
60 mm R
0.2
=
J
3EI
2 mm
A
20 mm
Figure 14.43
For AB, taking the origin at B,
M = 20(0.10 + 0.06 sin θ )
dx = 0.06 d θ
∴
1
UAB =
2EI
π
20(0.10 + 0.06 sin θ )2 × 0.06 d θ
0
=
12
EI
π
(0.01 + 0.012 sin θ + 0.0036 sin2 θ )d θ
0
12
[0.01 × 0.024 + 0.0018π ]
=
EI
0.733
=
EI
Total strain energy
0.733
0.7996
0.2
+
=
3EI
EI
EI
1
Work done = × 20 × δ
2
=
Equating Eqns. (i) and (ii)
1
× 20 × δ =
2
0.7996
0.02 × 0.0023
200 × 109 ×
12
δ = 0.03 m or 30 mm Ans
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(i)
(ii)
Deflection of Beams
•
285
Props
Props are sometimes needed in the loaded beams to counterbalance the deflection partially or completely. There are generally three types of props: i) Right props in which prop do not deflect
at all (ii) Sinking props are those props which on loading sink into ground by some amount.
iii) Elastic props are those which get compressed due to the elastic property of their material. On
removal of load, these regain their original position.
E XAMPLE 14.1: A simply supported beam carrying a uniformly distributed load over the whole
span is propped to its with the help of a rigid prop. Calculate the load carried by the prop and each
of the supports.
S OLUTION :
The downward deflection of point C due to uniformly distributes load.
w/unit length
B
C
A
l
P
Figure 14.44
δ=
5wl 4
384 EI
(i)
Pl 3
48EI
(ii)
The upward deflection due to prop,
δp=
To make the downward deflection at C, disappear equation (i) should be equal to (ii) therefore
equating
5wl 4
Pl 3
=
48EI
384EI
5wl
∴ P=
8
Also since RA = RB
1
RA = RB = (wl − P)
2
5
1
= (wl − wl)
2
8
3
= wl. Ans
16
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•
Strength of Materials
E XAMPLE 14.2: Solve example 14.1, assuming the prop to be elastic having the stiffness ‘S’ Find
the Prop’s load and reactions.
S OLUTION :
P
Downward deflection of the prop due to its elasticity =
S
Therefore,
Pl 3
P
5wl 4
−
=
384EI 48EI
S
P=
5wl 4 s
8 [48EI + Sl 3 ]
Ans
Then
RA = RB =
1
(wl − P)
2
5Sl 3
1
= wl
2
8(48EI + Sl 3 )
3
Sl + 128EI
3
Ans
wl
=
16
Sl 3 + 48EI
Exercise
14.1 A cantilever beam of length 2 m carries a load of 50 kN at the free end. Find the maximum
slope and deflection of the beam. Take E = 2 × 105 N/mm2 , I = 2 × 108 mm4 .
[Ans: 0.0025 radian, 3.33 mm]
14.2 A cantilever beam of length 3 m having a rectangular cross section. Determine the slope and
deflection at the force and of cantilever and under the load also when (i) point load of 20 kN
acts at a distance of 1.5 m from fixed end and (ii) point load of 20 kN acts at a distance of 2
m from fixed end. Take EI = 8 × 1012 Nmm2
[Ans: 0.0028 radian, 0.878 mm downward, 0.005 radian, 11.67 mm, 6.67 mm]
14.3 A cantilever with a span of 4 m carries a point load at its free end. If the maximum slope is
1.5 degrees, calculate the deflection at the free end.
[Ans: 69.81 mm]
14.4 A cantilever beam with a span of 3 m carries a point load 30 kN at a distance of 2 m from the
fixed end. Determine the slope and deflection at the free end and at the point where load is
applied. I = 11924 cm4 , E = 200 GPa.
[Ans: 0.00256 radian, 3.35 mm]
14.5 A cantilever 3 m long is loaded with a uniformly distributed load of 15 kN/m over a length of
2 m from the fixed end. Determine the slope and deflection at the free end of the cantilever.
Take E = 210 GPa and I = 0.000095 m4 .
[Ans: 0.001 radian, 2.5 mm]
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14.6 A cantilever beam of 3 m span is 15 cm wide and 25 cm deep. If carries a uniformly distributed load of 20 kN/m over its whole span and 25 kN load at the free end. Calculate the
maximum slope and deflection. E = 210 GN/m2 .
[Ans: 0.004937 radian, 10.42 mm]
14.7 A cantilever of uniform section has a length ‘l’. It is propped at the free end and carries a
point load W at a distance ‘a’ from the fixed end.
(a) If the prop holds the free end at the level of the fixed end, find the prop reaction.
(b) If now the prop is removed what will be the deflection at the free end.
Wa2
Wa2
(3l − a);
(3l − a)]
[Ans:
3
6EI
2l
14.8 A cantilever beam is loaded by a moment M1 at the free end as well as with uniformly
distributed load of w kN/m over half of its length at free end. Prove that the deflection at the
free end in given by expression:
M1 L2
7wl 4
+
2EI
384EI
14.9 A 2 m simply supported beam having cross section 150 mm × 500 mm carries a point load
of 20 kN at a distance of 0.5 m from the left end. Find the slope at the two ends, deflection
under the load and the maximum deflection. E = 2 × 104 mm2
[Ans : 8.02 × 10−3 degree
5.7 × 10−3 degree
0.06 mm
0.071 mm]
14.10 A wooden beam is 100 mm wide by 200 mm deep carries a load of 1.5 kN/m run over a
simply supported span of 4 m. Take E = 200 GPa. Find the deflection at the centre of the
span.
[Ans: 8.33 mm]
14.11 A cast iron pipe 150 mm outer diameter and 120 mm inner diameter rests on simple supports
1m apart and is subjected to a central load. Determine the load necessary to cause a maximum
deflection of 1.3 mm E = 105 N/mm2 .
[Ans: 91.5 kN]
14.12 A horizontal cantilever AB, 1.5 m long is subjected to uniformly distributed to load 1.5
kN/mm to a length of 1 m from the fixed end A and a point load of 500 N at its free
end B. The cantilever is of rectangular section of width 75 mm and depth 150 mm and its
E = 1.25 × 105 N/mm2 . Determine the deflection at the free end B.
[Ans: 0.146 mm]
14.13 Two point loads of 5 kN and 15 kN are acting on a 5 m simple beam AB, at 1 m and 2 m
respectively from the left end. Determine: a) slopes at the two ends, b) deflections under each
load c) position and magnitude of maximum deflection. E = 90 GPa. I = 18 × 106 mm4 .
[Ans:
y5 = 17.1 mm, y15 = 26.85 mm,
Maximum deflection = 27.49 mm at 2.327 m from A.]
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θA = 1.06◦ , θB = 0.884◦
288
•
Strength of Materials
14.14 A uniform simply supported beam AB is of 6 m span. At a distance of 4 m from left hand
support it carries a load of 10 kN on a bracket as shown in Fig. 14.45. Determine deflection
at the point C. Take E = 200 kN/m2 and I = 2 × 107 mm4 .
10 kN
0.5 m
B
A
C
4m
2m
[Ans:
Figure 14.45
7.8 mm]
14.15 A cantilever 150 mm wide, 200 mm deep and of 2 m span carries a uniformly varying load
of 50 mm at the fixed and as shown in Fig. 14.46. Determine the slope and deflection at the
free end. Take E = 100 GPa.
150 kN/m
50 kN/m
A
B
2m
[Ans:
Figure 14.46
0.01rad, 15.3 mm]
14.16 A cantilever 2 m long consists of a rectangular timber joint 150 mm × 240 mm deep. Two
steel plates 150 mm × 10 mm thick are fixed at the top and bottom faces of the timber joist as
shown in Fig. 14.47.
150 mm
10 mm
240 mm
10 mm
Figure 14.47
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•
289
Determine the slope and deflection of the cantilever at its free end, when it is carrying a
uniformly distributed load of 10 kN/m. Take Es = 200 GPa and Et = 10 GPa.
[Ans: 0.0012 rad, 1.8 mm]
14.17 A horizontal beam AB is freely supported at A and B, 8 m apart and carries a uniformly
distributed load of 15 kN/m run. A clockwise moment of 160 kNm is applied to the beam at
a point C, 3 m from the left hand support A. Calculate the slope and deflection of the beam at
C, if EI = 40 × 103 kNm2 .
[Ans: 0.0061 rad, 23.5 mm]
14.18 A simply supported beam of 2 m span carries a point load of 20 kN at its mid span. Determine the maximum slope and deflection of the beam. Take EI = 500 × 109 Nmm2 . Solve the
problem by conjugate beam method.
[Ans: 0.01 rad, 6.67 mm]
14.19 A simply supported beam is loaded as shown is Fig. 14.48. The beam has moment of inertia
of its section as 2I for the middle left of its length, whereas it is only I at the quarter length on
each end. Use conjugate beam method to find deflection at points C and E. Take E = 200 GPa
and I = 4 × 10−6 m4 .
200 kN
A
C
1.5 m
1.5 m
0.48 mm, 0.67 mm]
B
D
E
1.5 m
[Ans:
1.5 m
Figure 14.48
14.20 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 end 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.
Take E = 200 GPa and I = 85 × 106 mm4
Use Macaulay’s method.
[Ans: 9.019 mm, 16.7 mm, 16.745 mm, 2.87 m from left support]
14.21 A cantilever beam as shown in Fig. 14.49 is loaded by 500 N at the free end. Determine the
value of W so that deflection at the free end is 10 mm. Take E = 15 GN/m2 , I = 10×106 mm4
W
1.5 m
500 N
1.5 m
Figure 14.49
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[Ans:
774.2 N]
290
•
Strength of Materials
14.22 Using moment-area method, find the deflection at the centre of a simply supported beam
carrying w N/m. The span of beam in l.
5wl 4
Ans : yc =
384 EI
14.23 A horizontal cantilever 2 m long has its free and attached to a vertical tie-rod 3 m long and
300 mm2 cross-section area, which is initially unstrained. If I of the cantilever cross-sectional
is 6.5 × 106 m4 , determine the load taken by the tie-rod and the deflection of the cantilever
when a uniformly distributed load of 30 kN/m is placed on the outer 1 m of the cantilever.
Assume E for both cantilever and tie-rod to be 200 GPa.
[Ans: 18.76 kN, 0.938 mm]
14.24 The loads acting on a simply supported beam are shown in Fig. 14.50. Use Macaulay’s
method to determine deflection at points C, D and E.
26 kN
C
E
26 kN
D
50 kN/m
A
B
1.5 m
1.5 m
1.5 m
1.5 m
Figure 14.50
[Ans: 0.54 mm, 0.54 mm, 0.756 mm]
14.25 Use the method of superposition to find the slope at end A and the deflection at centre point
D for the beam AB loaded as shown in Fig. 14.51.
W
W
W
2.5Wa2 19 wa3
;
Ans : θA =
EI
6 EI
C
D
E
B
A
a
a
a
a
Figure 14.51
14.26 Determine the deflections at the ends and centre of the beam shown in Fig. 14.52. Take
EI = 5 × 106 N/m2 . Use superposition method.
40 kN
3m
20 kN
10 kN/m
40 kN
E
A
C
2m
B
6m
D
2m
Figure 14.52
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[Ans: yc = 63.3 mm,
yD = 63.3 mm, yE = 20.25 mm]
C HAPTER
15
STRAIN ENERGY, IMPACT LOADING AND
DEFLECTION DUE TO BENDING
Whenever an elastic body is subjected to a force, it is stretched or compressed and on removal
of that force, elastic body comes back to its original form. This happens because of strain energy
or resilience stored in the body during compression or tension. For example, bow and arrow, the
moment stretched bow is released, its tensile strain energy is taken by arrow which moves forward
with great speed. This energy, which is absorbed in a body, when strained within elastic limit is
known as strain energy. Strain energy is always capable of doing some work. The amount of strain
energy in a body is found out by the principle of work. Therefore,
Strain energy = Work done.
Resilience: Many times total strain energy stored in a body is termed resilience. It is also defined as
the capacity of a strained body for doing work when the strained body springs back to its original
shape.
Proof Resilience: Maximum strained energy which can be stored in a body is termed proof resilience.
Modulus of Resilience: This is the proof resilience per unit volume of a material. This is an important
property of the material.
(A) Strain Energy Stored in a Body When the Load is Gradually Applied
It an axial load P is gradually applied to a bar and producer extension x, then the work done or strain
energy is represented by the area under load-extension diagram
P
U = Strain energy
Load
x
Extension
Figure 15.1
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292
•
Strength of Materials
1
P.x, and P = stress × area = σ × a
2
σl
and x =
E
σl
1
∴ U = σ ·a×
2
E
U=
=
The term
1 σ2
× Volume
2 2E
σ2
× Volume represents strain energy.
2E
(B) Suddenly Applied Load
Sometimes if load P is suddenly applied, for
example, load P is suddenly placed on the collar
in Fig. 15.2, causing deflection x. (It should be
noted that in previous case of gradually applied
load, it was assumed that load is increased to P
from zero value gradually.)
l
P
Now work done = P.x
But strain energy, U =
σ2
σ2
× Volume =
× Al
2E
2E
where A is the area of cross section of the rod l
is the length of rod supporting collar.
Figure 15.2
Now because the energy stored is equal to the work done, therefore
σ2
σ
× A × l = P.x = P. l
2E
E
P
or σ = 2 ×
A
It proves that stress induced in this case is twice the stress induced when the load is applied
gradually. After stress σ is found out, the corresponding instantaneous deformation x and the strain
energy are obtained as usual.
E XAMPLE 15.1: On a steel rod 3 m long, an axial pull of 25 kN is suddenly applied. If diameter
of rod is 40 mm, calculate the strain energy which is absorbed in the rod. Take E = 200 GPa.
S OLUTION :
Cross-sectional area =
π
(40)2 = 1256 mm2
4
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•
293
P
A
2 × 25000
= 39.81 N/mm2
=
1256
For suddenly applied load, σ = 2 ×
Volume of the rod, V = length × Area
= 3000 × 1256 = 3768000 mm3
Hence, strain energy,
U=
=
σ2
× Volume
2E
(39.81)2
× 3768000
2 × 200000
= 14929 Nmm = 14.929 Nm
Ans
E XAMPLE 15.2: Find the maximum instantaneous stress and work done at maximum elongation,
when an axial load of 80 kN is suddenly applied to the steel rod of 3.5 m length having a diameter
of 35 mm. Also calculate the maximum instantaneous force in the rod. Take E = 200 Gpa.
S OLUTION :
π 2 π 2
d =
35 = 961.625 mm2 .
4
4
P
σmax (instantaneous stress) = 2 ×
A
80000
= 166.38 N/mm2 Ans
= 2×
961.625
A=
Work done at maximum elongation,
σmax × l
E
166.38 × 3500
= 2.91 mm
=
200000
Work done = P.x
= 80000 × 2.91
= 232800 Nmm
= 232.8 Nm Ans
x=
@seismicisolation
@seismicisolation
294
•
Strength of Materials
Maximum instantaneous force developed in the rod
= σmax × area
= 166.38 × 961.625
= 159995.17 N
= 159.995 kN Ans
(C) Strain Energy Stored in a Body, When the Load is Applied with Impact
Load
P
As we see in Fig. 15.3, if load P is suddenly
released to fall through a height h, if will produce extension x in the rod.
∴
Work done = load × distance moved
= P(h + x)
(i)
And strain energy stored =
h
σ2
× Al
2E
(ii)
Where A and l are area of rod and length of rod,
respectively.
Collar
Figure 15.3
Equating Eqns. (i) and (ii),
σ2
× Al = P(h + x)
2E
σ = P h+ l
E
where σ is the instantaneous stress induced in the rod due to impact.
Pσ l
σ2
× Al = Ph +
E
2E
Al
Pl
∴ σ2
−σ
− Ph = 0
2E
E
E
Simplifying by multiplying both sides by
,
Al
σ2
PEh
P
−σ
−
=0
2
A
Al
@seismicisolation
@seismicisolation
Strain Energy, Impact Loading and Deflection Due to Bending
•
295
Solving this quadratic equation,
PEh
P 2
1
+ 4×
A
2
Al
σ=
1
2×
2
P
2EAh
1± 1+
σ=
A
Pl
P
±
A
∴
After finding instantaneous stress, the corresponding deformation and strain energy can be found
out as usual,
2P
If h = 0, σ =
that is the case of a suddenly applied load.
A
In case instantaneous extension x is very small as compared to h, then
Work done = Ph
σ2
× Al = Ph
2E
2EPh
σ2 =
Al
2EPh
σ=
Al
But, of course, this will give approximate instantaneous stress.
E XAMPLE 15.3: A steel bar 4 m long and 3000 mm2 in area hangs vertically and is secured to
a fixed collar at its lower end. If a weight of 20 kN falls on the collar from a height of 12 mm,
determine the stress developed in the bar. What will be the instantaneous strain energy stored in the
bar? Take E = 200 GPa.
S OLUTION :
We know in this case,
P
σ=
A
2EAh
1+
Pl
2 × 200000 × 3000 × 12
20000
1+
3000
20000 × 4000
√
= 6.67 1 + 180
√
= 6.67 181
=
= 89.73 N/mm2
Ans
@seismicisolation
@seismicisolation
296
•
Strength of Materials
Strain energy stored in the bar
U=
∴
σ2
× Volume
2E
(89.73)2
× 3000 × 4000
2 × 200000
= 241544.2 Nmm
U=
= 241.54 Nm
Ans
E XAMPLE 15.4: A steel bar 15 mm diameter gets stretched by 0.8 mm under a steady load of
8 kN. What stress would be produced in the bar by a weight 600 N, if this weight falls through 100
mm before striking the collar rigidly fixed to the lower end? Take E = 200 GPa.
First we must find the length of bar.
E=
Pl
Ax
π
EAx
, A = (15)2 = 176.625
P
4
200000 × 176.625 × 0.8
= 3532.5 mm
l=
8000
∴
l=
Now this is the case of impact loading,
∴
σ=
=
P
A
1+
1+
600
176.625
1+
2AEh
P l
2 × 176.625 × 200000 × 100
600 × 3532.5
√
= 3.397(1 + 1 + 3333.3)
= 3.397(1 + 57.74)
= 199.55 N/mm2
Ans
E XAMPLE 15.5: An unknown weight falls through 15 mm on a collar rigidly fixed to the lower
end of a vertical bar of 3.5 m length and 750 mm2 area in section. If the extension is 2.5 mm, what
is the corresponding stress and the value of the unknown weight? E = 200 GPa.
Maximum stress = E × Maximum strain
2.5
= 200000 ×
3500
= 142.86 N/mm2
If W is the unknown weight, then
Loss of potential energy = Strain energy stored in the rod
W (h + x) =
σ2
× Volume
2E
@seismicisolation
@seismicisolation
Strain Energy, Impact Loading and Deflection Due to Bending
•
297
(142.86)2
× 750 × 3500
2 × 200000
17.5W = 133934.1
133934.1
W=
17.5
= 7653.3
= 7.653 kN Ans
W (15 + 2.5) =
E XAMPLE 15.6: A 3.5 m long bar of 20 mm in diameter hangs vertically and has a collar attached
at the lower end. Determine the maximum stress induced when a weight of 1000 N falls on the
collar from a height of 40 mm.
If the bar is turned down to half the diameter along half of its length, what will be the value of
the maximum stress? E = 200 GPa.
S OLUTION :
π
(20)2 = 314 mm2
4
P
2AEh
σ1 1 =
1+ 1+
A
Pl
2 × 314 × 200000 × 40
1000
1+ 1+
=
314
1000 × 3500
√
= 3.185[1 + 1 + 1435.4]
= 3.185[1 + 37.89]
A=
= 123.86 N/mm2
Now the reduced diameter =
Ans
20
= 10 mm, A1 = 314 mm2
2
A2 =
1750 mm
A1
π
(10)2 = 78.5 mm2 , l1 = l2 = 1750 mm
4
P = Falling load = 1000 N
Pe = Equivalent gradually applied load which
produces the same maximum stress and extension
as is caused by the falling load, P.
A2
1750 mm
Figure 15.4
@seismicisolation
@seismicisolation
298
•
Strength of Materials
Now the total extension,
x = x1 + x2
=
Pe l1 Pe le
+
A1 E A2 E
=
1
Pe l1 1
+
E A1 A2
=
1
Pe × 1750 1
+
200000 314 78.5
∴
l1 = l2
x = 0.00875 Pe [0.00318 + 0.01274]
x = 0.00875 Pe [0.01592]
or
x = 0.0001393 Pe
Also,
1
P h + x = Pe × x
2
1
1000 (40 + 0.0001393 Pe ) = Pe × 0.0001393 Pe
2
40000 + 0.1393 Pe = 0.00006965 Pe2
0.00006965 Pe2 − 0.1393 Pe − 40000 = 0
Pe2 − 2000 Pe − 574300072
√
+2000 ± 4000000 + 4 × 574300072
Pe =
2
√
2000 ± 4000000 + 2297200288
=
2
2000 + 47971
= 24985.5
=
2
Maximum stress is in smaller section,
σ2 =
24985.5
= 318.3 N/mm2
78.5
Ans
E XAMPLE 15.7: A steel rod is 2.5 m long and 60 mm in diameter. An axial pull of 130 kN is
suddenly applied to the rod. Calculate the instantaneous stress induced and also the instantaneous
elongation to the rod. Take E = 200 GN/m2 .
@seismicisolation
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Strain Energy, Impact Loading and Deflection Due to Bending
•
299
S OLUTION :
π
(60)2 = 2826 mm2 , l = 2500 mm
4
P = 130000 N, E = 200000 N/mm2
P 2 × 130000
σ = 2× =
= 92 N/mm2 Ans
A
2826
σl
E=
x
σ l 92 × 2500
Instantaneous elongation = x =
=
E
200, 000
= 1.15 mm Ans
Area =
E XAMPLE 15.8: A tension bar 6 m long is 3.5 m long having cross-sectional area of 1200 mm2
while remaining 2.5 m long is having 2500 mm2 cross-sectional area, An axial load of 120 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 under the same load.
E = 200 GN/m2 .
S OLUTION :
1200 mm2
2
1
120 kN
2500 mm2
3.5 m
2.5 m
Figure 15.5
120000
= 100 N/mm2
1200
120000
= 48 N/mm2
σ2 =
2500
σ1 =
Strain energy in portion 1,
U1 =
σ2
× Volume
2E
@seismicisolation
@seismicisolation
120 kN
300
•
Strength of Materials
(100)2
× 1200 × 3500
2 × 200000
= 105000 Nmm
= 105 Nm
=
Strain energy in portion 2,
σ2
× Volume
2E
(48)2
=
× 2500 × 2500
2 × 200000
= 36000 Nmm
U2 =
= 36 Nm
Total energy produced in the bar = 105 + 36
U1 = 141 Nm
Total volume of the first bar, v = v1 + v2
v = 1200 × 3500 + 2500 × 2500
= 4200000 + 6250000
= 10450000 mm3
Length of uniform bar = 6000 mm
v = A×l
v 10450000
=
= 1741.7 mm2
l
6000
120000
Stress in the uniform bar =
= 68.9 N/mm2
1741.7
σ2
Strain energy in uniform bar =
× Volume
2E
(68.9)2
U2 =
× 10450000
2 × 200000
= 124021 Nmm
Area of uniform bar, A =
= 124.02 Nm
∴
141
Strain energy in the given bar, U1
=
Strain energy in the uniform bar, U2 124.02
= 1.137 Ans
@seismicisolation
@seismicisolation
Strain Energy, Impact Loading and Deflection Due to Bending
•
301
E XAMPLE 15.9: A bar of uniform cross section a and length l hangs vertically, subjected to its
own weight. Prove that the strain energy stored in bar is given by
U=
A × ρ2 × l
6E
where ρ is weight per unit volume.
S OLUTION :
ρ = Weight per unit volume.
Consider section X − X which is acted upon by
the weight of the bar of length X.
Wx = (A × x) × ρ = ρ Ax
dx
x
x
Due to this weight the portion dx will have a
small elongation d δ . Then
l
elongation in dx
dx
dδ
=
dx
ρ Ax
Stress in portion δ x =
A
= ρ ×x
Strain in portion dx =
x
Figure 15.6
Since,
∴
Stress
Strain
ρx
ρ xdx
= =
dδ
dδ
dx
ρ × x × dx
dδ =
E
E=
Strain energy stored in portion dx is given by
dU = Average weight × elongation of dx
1
W x dδ
=
2
1
ρ × x × dx
× ρ Ax ×
=
2
E
1 2 2 dx
= ρ Ax ×
2
E
Total strain energy stored within the bar due to its own weight W is obtained by integrating from
0 to l.
@seismicisolation
@seismicisolation
302
•
Strength of Materials
l
U=
dU
0
l
=
1 2 2 dx
ρ Ax
2
E
0
=
ρ 2A
2E
l
x2 dx
0
l3
1
× ρ 2A ×
2E
3
2
3
Aρ l
=
6E
=
E XAMPLE 15.10: The maximum stress produced by a pull in a bar of length 1.2 m is 180 N/mm2 .
Figure 15.7 shows the cross section and length. Determine the strain energy stored in the bar if
E = 200 GPa.
300 mm2
P
A
150 mm2
B
0.4 mm
300 mm2
D
C
0.4 mm
P
0.4 mm
Figure 15.7
S OLUTION :
Stress will be maximum where cross-sectional area is minimum. Therefore, maximum stress,
σ = 180 N/mm2 in area 150 mm2 .
Now strain energy stored in part AB will be found out σBC × 150 = σAB × 300, because the load
is equal in every portion.
180 × 150
= 90 N/mm2
300
σ AB2
× Volume
UCD = UAB =
2E
902
=
× 300 × 400
2 × 200000
= 2430 Nmm
= 2.430 Nm
σAB =
@seismicisolation
@seismicisolation
Strain Energy, Impact Loading and Deflection Due to Bending
•
303
2
σbc
× Volume
2E
902
× 150 × 400
=
2 × 200000
= 1215 Nmm
= 1.215 Nm
UBC =
Total strain Energy, U = UBC +UAB +UCD
U = 1.215 + 2 × 2.43
= 1.215 + 4.86
= 6.075 Nm Ans
(∵
UAB = UCD )
E XAMPLE 15.11: A vertical compound tie member fixed rigidly at its upper end, consists of a
steel rod 3 m long and 25 mm in diameter, placed within an equally long brass tube 27 mm is
internal diameter and 35 mm external diameter. The rod and the tube are fixed together at the ends.
The compound bar is then suddenly loaded in tension by a weight of 15 kN falling through a height
of 5 mm on to a flange fixed to its lower end. Determine maximum stresses in steel and brass. Take
Es = 200 GPa, Eb = 1.0 × 105 N/mm2 .
S OLUTION :
π
(25)2 = 490.625 mm2
4
π
π
Ab = (352 − 272 ) = (1225 − 729)
4
4
2
= 389.36 mm
Length = 3 mm = 3000 mm
P = 15 kN = 15000 N
35 mm
27 mm
As =
Steel rod
Brass
tube
3m
25 mm
5 mm
Flange
Figure 15.8
Since both the ends are fixed together,
Strain in steel rod = Strain in brass tube
σs σb
=
Es Eb
20000
σ
σs = b × Es = σb ×
Eb
100000
σs = 2σb
@seismicisolation
@seismicisolation
(i)
304
•
Strength of Materials
Volume of steel rod,
Vs = Area × length
= 490.625 × 3000
= 1471875 mm3
Volume of brass tube = 389.36 × 3000
= 1168080 mm3
Strain energy in steel rod,
σs2
× Volume
2Es
(2σb )2
=
× 1471875
2 × 200000
= 14.72 σb2
Us =
Strain energy in brass tube,
σb2
× Volume
2Eb
σ 2 × 1168080
= b
2 × 100000
= 5.840 σb2
Ub =
Total strain energy stored in compound bar,
U = Us +Ub
= 14.72σb2 + 5.840 σb2
= 20.56 σb2
(ii)
Work done by falling load = P(h + x)
= 15000 (3 + x)
σ
Strain in brass rod = b
Eb
x
σb
or
=
l
1 × 105
σ × 3000
x= b
= 0.03σb
1 × 105
Substituting this value of x in Eqn. (iii),
15000 (3 + 0.03σb )
@seismicisolation
@seismicisolation
(iii)
Strain Energy, Impact Loading and Deflection Due to Bending
•
Now equating the work done by the falling weight to the total strain energy.
15000 (3 + 0.03σb ) = 20.56 σb2
45000 + 450σb = 20.56
(from Eqn. (ii))
σb2
20.56 σb2 − 450σb = 45000 = 0
σb2 − 21.89σb − 2188.7 = 0
This is a quadratic equation.
∴
√
21.89 ± 21.892 + 4 × 2188.7
σb =
2
√
21.89 ± 479.17 + 8754.8
=
2
21.89 + 96.1
=
2
= 59 N/mm2 Ans
σs = 2σb
= 2 × 59 = 118 N/mm2
Ans
Strain Energy in Pure Shearing
Let a rectangular block (refer Fig. 15.9) of material be subjected to shearing forces F acting on
the opposite faces as shown.
1
Now, the work done = F ×CC
2
1
= F ×CB × φ
2
MM [∵
= tan φ = φ for small value of φ ]
MN
D′
D
F
φ
A
C′
C
φ
F
B
Figure 15.9
Now F = τ × DC where τ is the shearing stress
τ
(φ = εs = shear strain)
Also φ =
C
Taking unit depth normal to diagram, we get
Strain energy = work done
τ
τ2
1
× DC ×CB
Work done = τ × DC ×CB × =
2
C 2E
Now DC ×CB is the volume of rectangular block, since it has unit depth normal to DCBA.
τ2
× Volume of block.
∴ Strain energy =
2E
@seismicisolation
@seismicisolation
305
306
•
Strength of Materials
Strain Energy in Torsion
a) Solid shaft
T
l
φ
θ
R
l
Figure 15.10
Let a solid shaft be subjected to a torque T , which produces a twist θ in the length of shaft.
Work done =
But
∴
∴
Cθ
τ
T
=
=
J
l
R
T=
Work done =
=
Now J =
∴
1
T.θ , which is stored as strain energy.
2
Work done =
τ
.J
R
and
θ=
τL
CR
1 τJ τL
×
×
2 R CR
1 τ2 J × l
× × 2
2 C
R
π R4
2
1 τ 2 π R4 × l
× ×
2 C
2R2
1 τ2
× × π R2 l
4 C
τ2
× Volume ∵
or Strain energy =
4C
=
Volume = π R2 l
b) Strain Energy for a Hollow Shaft
1
Work done = T θ
2
Work done =
and
θ=
τl
CR
Jl
τ2
×
2C R2
@seismicisolation
@seismicisolation
and T =
τ
J (as with solid shaft)
R
Strain Energy, Impact Loading and Deflection Due to Bending
π 4 4
R −r
2
τ 2 π l R4 − r4
∴ Work done =
×
2C
2R2
2
τ 2 π l R + r2 R2 − r2
=
×
2C
2R2
2
R + r2
τ2
×
Thus, strain energy, V =
×
Volume
∵
4C
R2
•
307
J=
π l R2 − r2 = Volume
Strain Energy Due to Bending
As we have already proved in chapter of Deflection of Beams that strain energy stored in beam
l 2
M dx
.
in given by U =
0 2EI
In case M is constant over the length l, then
U=
M2l
2EI
Strain Energy and Deflection Due to Bending
If y is the deflection under the load W , then
1
U = Wy
2
2U
or y =
W
Strain Energy of a Simply Supported Beam having One Eccentric Load
X
W
b
a
A
C
x
RA=Wb
l
l
X
B
RB=Wa
l
Figure 15.11
Refer Fig. 14.37, we have already proved that total strain energy = U = UAC +UCB =
@seismicisolation
@seismicisolation
w2 a2 b2
6EIl
308
•
Strength of Materials
Strain Energy of a Simply Supported Beam having Uniformly Distributed Load on
Whole Span
X
A
RA= wl
2
w/unit length
B
RB= wl
2
x
X
Figure 15.12
Considering a section X − X, at x from A,
wx2
wl
x−
2
2
l 2
M
U=
dx
0 2EI
2
l
wlx wx2
1
dx
−
=
2EI 0
2
2
w2 l 2 2
l x + x4 − 2lx3 dx
=
8EI 0
3
l
w2 l 2 x
x5 2lx4
=
+ −
8EI 3
5
4
Mx =
∴
Strain energy,
0
=
Hence,
U=
w2
l5
8EI
3
+
l5
5
−
l5
4
w2 l 5
240EI
Strain Energy and Deflection in a Cantilever having Point Load at Free End
W
Y
d
A
l
X
x
Figure 15.13
@seismicisolation
@seismicisolation
B
b
Strain Energy, Impact Loading and Deflection Due to Bending
Mx = −W x
l
Strain energy,
U=
M2
dx
2EI
0
l
=
W 2 x2
dx
2EI
0
W2
=
2EI
l
x2 .dx
0
=
Hence,
U=
W2
2EI
x3
3
l
0
W 2l3
6EI
If y at free end is the deflection (maximum)
1
Then work done = W Ymax
2
Equating, work done and strain energy,
1
W 2l3
W ymax =
2
6EI
W l3
(as before)
ymax =
3EI
Now if, σ = Maximum bending stress
Wl
Mmax
6W l
=
=
2
Z
bd 2
bd
6
σ bd 2
W=
6l
σ=
∴
Putting value of W in the expression for U,
We have,
σ 2 b2 d 4 l 3
bd 3
36l 2 × 6E ×
12
σ2
=
× bdl
18E
σ2
× Volume of the beam
U=
18E
U=
Hence,
@seismicisolation
@seismicisolation
•
309
310
•
Strength of Materials
Strain Energy of a Simply Supported Beam with Point Load at Centre
W
X
d
A
x
C
X
l
W/2
B
l/2
b
W/2
Figure 15.14
At x from A,
Mx =
W
x
2
Strain energy for AC,
=
l/2 2 2
W x
4
0
×
1
dx
2EI
l
W2
W2
=
x2 dx =
8EI
=
8EI
0
2
3
W l
x3
3
l/2
o
192EI
Because of symmetry,
For whole beam, strain energy U = 2 ×
W 2l3
Therefore, U =
96EI
Now,
∴
W 2l3
192EI
σ=
W l/4
Mmax
3W l
= 2 =
Z
bd /6
2bd 2
σ=
3W l
2bd 2
or W =
2σ bd 2
3l
Substituting in total strain energy (U) formula deriver above, we have
U=
=
2σ bd 2
3l
2
×
l3
96EI
4σ 2 b2 d 4
l3
×
2
96EI
9l
@seismicisolation
@seismicisolation
Strain Energy, Impact Loading and Deflection Due to Bending
=
4σ 2 b2 d 4
×
9l 2
•
311
l3
96E ×
bd 3
12
σ2
(lbd)
18E
σ2
U=
× Volume of the beam
18E
=
Maxwell’s Reciprocal Theorem
It states that “the work done by the first system of loads due to displacements caused by a second
system of loads equals the work done by the second system of loads due to displacements caused
by the first system of loads”.
Or in simple words: “In any beam or truss the deflection at any point M due to load W at any
other point C is the same as the deflection at C due to the same load W at D.” For example, the
central deflection (at M) of a simply supported beam carrying an offset load, (See Fig. 15.15) is the
same as the deflection at C if the load were moved to the centre M.
W
W
A
C
M
y
B
A
C
B
M
y
(b)
(a)
Figure 15.15
Another example: If a cantilever carries a concentrated load not at the free end (Fig. 15.16), the
deflection at C due to the load at B is same as the deflection at B if the load were moved to C.
W
B
(a)
W
B
y
C
C
y
(b)
Figure 15.16
Proof: Let an elastic body be subjected to forces
Wa and Wb at points A and B respectively (see
Fig. 15.17).
B
A
Wa
Figure 15.17
@seismicisolation
@seismicisolation
Wb
312
•
Strength of Materials
Let δaa be the deflection at A in the direction of Wa due to Wa .
Let δab be the deflection at A in the direction of Wa due to Wb .
Let δbb be the deflection at B in the direction of Wb due to Wb .
Let δba be the deflection at B in the direction of Wb due to Wa .
1
Let Wa be applied first then work done = Wa δaa , assuming the load is gradually applied.
2
1
1
If Wb is now applied, the additional work done = Wb δbb + Wa δab , the whole of Wa moving
2
2
through the additional distance δab .
1
1
Thus, the total work done = Wa δaa + Wb δbb +Wa δab .
2
2
If the loads are removed and then reapplied in reverse order (i.e., Wb is applied first), it will be
1
1
seen, by analogy, that the total work done = Wa δaa + Wb δbb +Wb δba .
2
2
Irrespective of the order in which the loads are applied, the body will assume the same strained
position and have the work done by the loads will be the same, so that Wa δab = Wb δba .
In the case where Wa = Wb , δab = δba , i.e., the deflection at A due to a load at B is the same as
the deflection at B if the load is applied at A, the deflections being the movements of the points A
and B in the directions shown in Fig. 15.17.
Betti’s Theorem of Reciprocal Deflections
It states “In an elastic system, the external work done by a force F acting at P during the deflections caused by another force at Q is equal to the external work done by the force at Q during the
deflections caused by the force P.”
Mathematically,
Fp δ pq = Q p δqp
E XAMPLE 15.12: A thin proving ring of radius r is subjected to a diametral tensile load W . Determine (a) the increase in diameter in the direction of W , (b) the decrease in diameter perpendicular
to the direction of W .
S OLUTION :
W
P
A
Q
Mo
r
dθ θ
B
W
(a)
(b)
Figure 15.18
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Strain Energy, Impact Loading and Deflection Due to Bending
•
313
It is sufficient to consider one quarter of the ring AB (Fig. 15.18b) fixed at the midpoint of the
W
at the free end. Due to continuity of the ring at A, however
sides and subjected to a force P =
2
the tangent there must remain horizontal and hence a moment M0 must be applied for this purpose,
the moment corresponding to the bending moment in the actual ring at that point.
The increase in the vertical diameter of the ring will be twice the deflection of A relative to B
in the quadrant and the decrease in horizontal diameter will be twice the horizontal movement of A
relative to B. For this purpose, it is necessary to add a horizontal force Q of zero magnitude.
Taking the origin at A,
M = M0 − Pr sin θ − Qr (1 − cos θ )
∂M
=1
∂ M0
∂M
= −r sin θ
∂P
∂M
= −r(1 − cos θ )
∂Q
and dx = r d θ
There is no rotation in the direction of M0 ,
∴
1
∂U
=
∂ M0 EI
π /2
i.e.,
0
from which M0 =
2Pr
π
l
M.
0
∂M
dx = 0
∂ M0
{M0 − Pr sin θ } × 1 × r d θ = 0
∂U
2 l ∂M
M
=
dx
δv = 2
∂P
EI 0
∂P
2 π /2 2pr
− Pr sin θ × {−r sin θ } × r d θ
=
EI 0
π
2Pr3 π 2
W r3
−
=
= 0.1484
EI
4 π
EI
2 l ∂M
∂U
M
=
dx
∂ Q EI 0 ∂ Q
2 π /2 2Pr
=
− Pr sin θ × {−r (1 − cos θ )} × r d θ
EI 0
π
2Pr3 2 1
=
−
EI
π 2
δh = 2
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•
Strength of Materials
= 0.137
W r3
EI
Note: Since Q = 0, the term involving Q may be omitted from all the integrals but it cannot be
∂M
will be obtained.
omitted from the original expression for M, otherwise no value for
∂Q
E XAMPLE 15.13: Find the strain energy stored by the structure shown in Fig. 15.19 and hence
compute the vertical deflection of the end A. Assume that the section of the member is uniform.
S OLUTION :
a
B
At any section in AB distant x from A, the bending moment is given by
x
y
M = Wx
∴ Strain energy stored by AB
l
M 2 dx
2EI
a 2 2
W x .dx
=
2EI
0
W 2 a2
UAB =
6EI
UAB =
C
Figure 15.19
Any section is BC distant y from B the bending moment. is given by,
M = Wa
∴ Strain energy stored by BC
UBC =
l 2
M dx
0
2EI
=
l 2 2
W a dy
0
2EI
=
W 2 a2 l
2EI
∴ Total strain energy stored
U = UAB + UBC
=
W 2 a2 W 2 a2 l
W 2 a2
+
=
(a + 3l) Ans
6EI
2EI
6EI
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A
W
Strain Energy, Impact Loading and Deflection Due to Bending
•
315
Let δ be the vertical deflection at the end A, equating the work done to the strain energy stored,
we have,
1
W 2 a2
Wδ =
(a + 3l)
2
6EI
Wa2 (a + 3l)
∴ δ=
Ans
EI
Bending Under Impact Loads: If a weight W drops from a height h at the point C on simply
supported beam as shown is Fig. 15.20, then in order to find instantaneous deflection under the load
is sometimes required.
W
A
h
C
y
B
Figure 15.20
Let We be the equivalent weight which when gradually applied at C will produce the same
deflection y. As we know in both cases the deflection is same, then the strain energy in respective
cases should also be same. And strain energy of the beam is equal to the work done by respective
weights in both cases.
1
Therefore, W (h + y) = We y
(i)
2
Depending upon type of beam (end conditions), the deflection y can be found out in terms of We
using methods described in chapter ‘Deflection of Beams’. The deflection is directly proportional
to the load applied on beam.
So,
or
y ∝ We
y = kWe
(ii)
where k is a constant which depends upon the type of beam and its end conditions.
Substituting for y from Eqn. (ii) in Eqn. (i)
1
W (h + y) = We y
2
1
W (h + kWe ) = We2 k
2
(iii)
Equation (iii) is a quadratic equation and its solution will give the value of We . Then deflection
y can be calculated from Eqn. (ii)
E XAMPLE 15.14: A concentrated load of 12 kN applied to a simply supported beam at midpoint
of the beam, produces a deflection of 7 mm and a maximum bending stress of 25 MN/m2 . Calculate
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•
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the maximum value of the instantaneous stress produced when a weight of 6 kN is allowed to fall
through a height of 20 mm on the beam at middle of the span.
S OLUTION :
Let We be the static load equivalent to the given impact or falling load. Since 12 kN static load
produces a deflection of 7 mm, then We will produce a deflection y so that
7
×We = 0.583We mm = 0.000583We m
y=
12
1
Now using equation W (h + y) = We y
2
1
We × 0.000583 We
2
0.12 + 0.003498 W = 0.0002915 We2
6(0.020 + 0.000583 We ) =
0.0002915 We2 − 0.003498 W − 0.12 = 0
We2 − 12 W − 411.664 = 0
√
+12 ± 144 + 1646.66
We =
2
+12 ± 42.32
=
2
= 27.16 kN
Now because a static load of 12 kN produces a maximum bending stress of 25 MN/m2 , in the
impact case the maximum bending stress produced for which static load is 27.16 kN, will be
25 × 27.16
= 56.58 MN/m2 Ans
σmax =
12
Load factor: In impact loading on a collar we have derived the instantaneous stress,
σ=
P
A
1+
1+
2hAE
Pl
Static deflection δ l due to load P is given by
δl =
Pl
AE
∴
σi =
p
A
1+
1+
2h
δl
[σ is replaced by σi for instantaneous stress]
OR
σi = σ
σi
= 1+
σ
1+
1+
1+
2h
δl
2h
δl
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The dimensionless ratio
•
317
σi
is usually called load factor and is denoted by n.
σ
∴
n = 1+
1+
2h
δl
E XAMPLE 15.15: A 1.2 m long beam rectangular in section 40 mm wide × 50 mm deep is supported on rigid supports at its ends. If it is struck at the centre by a 15 kg mass falling through a
height of 70 mm, find:
i) The instantaneous stress developed;
ii) The instantaneous strain energy is the beam.
Take E = 200 GN/m2
15 kg
h = 70 mm
40 mm
50 mm
Beam
A
B
y
Beam Cross-section
1.2 m
Figure 15.21
S OLUTION :
I=
40 × 503
bd 3
=
= 416666.7 mm4
12
12
Static deflection due to W
W l3
48EI
(15 × 9.81) (1.2)3 (1000)3
=
48 × 200000 × 416666.7
= 0.0636 mm
δl =
√
2h
2 × 70
= 1+
= 1 + 2202.25
δl
0.0636
n = 46.93 (where n is load factor)
n=
1+
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•
Strength of Materials
i) Instantaneous stress developed, σi :
Static stress,
∴
M Wl
15 × 9.81 × 1200
=
=
1
Z
4Z
4 × × 40 × 502
6
= 2.65 N/mm2
σ=
σi = nσ = 46.93 × 2.65 = 124.36 MN/m2
Ans
ii) Strain energy stored:
Strain energy,
1
We δ li
2
1
U = (nW ) × (nδ l) [∵ We = nW, δ li = nδ l]
2
n2
(46.93)2
= W.δ l =
× 15 × 9.81 × 0.0636
2
2
= 10306 Nmm
= 10.31 Nm or J Ans
U=
Exercise
15.1 A steel rod at 100 mm in diameter is 5 m long. Find the maximum instantaneous stress
induced when a pull of 200 kN is suddenly applied to it. Find also the instantaneous elongation. Take E = 200 GN/m2 .
[Ans 1.273 mm]
15.2 Compare the strain energies of the two bars A and B of the same material and subjected to
the same axial tensile loads. Bar A is of 50 mm diameter throughout while the bar B, though
of the same total length as A, has diameter of 25 mm over the middle third of its length, the
remainder being of 50 mm diameter. Also compare their proof resilience in simple tension.
1
[Ans
, 8]
2
15.3 Two shafts A and B are of same length and made of steel and bronze, respectively. Both the
shafts are subjected to equal torques. Find the ratio of the shaft diameters so that (i) strain
energy stored per unit volume (ii) total strain energy stored, by each is the same. Take C for
steel as 80 GPa and for bronze as 50 GPa.
[Ans 0.924, 0.889]
15.4 A steel rod of cross-sectional area 1000 mm2 and 2 m long has a collar at its lower end,
while its upper end is fixed. A weight of 200 N falling from height h and striking the collar
produces an instantaneous maximum stress of 50 MPa in the rod. Assuming 5 per cent energy
loss during impact, determine the value of h. Take E = 200 GPa for steel.
15.5 A cantilever of uniform section with length l and flexural rigidity EI carries a uniformly
distributed load of w per unit length from its mid-span to free-end. Use Castiglieno’s theorem
to find deflection at the free end of the cantilever.
[Ans W l 4 /128EI]
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15.6 A hollow shaft subjected to a pure torque attains a maximum shearing stress τ . Given that the
strain energy stored per unit volume is τ 2 /3C, calculate the ratio of shaft diameters. If such a
shaft is required to transmit 3700 kW at 110 r.p.m. with uniform torque and the energy stored
2
is 20 kJ/m3 , determine the actual diameters. Take C = 80 GN/m√
.
3 : 1, 298.2 mm, 172 mm]
[Ans
15.7 A weight of 2 kN falls from 24 mm on to a collar fixed to a steel bar which is 14 mm in
diameter and 5.5 m long. Determine the maximum stress induced in the bar. E = 205 GPa.
[Ans 166 MPa]
15.8 A steel bar of constant section, second moment of area I is bent as shown in Fig. 15.22 and
fixed at one end. Find the horizontal and vertical deflections at the free end.
P
a
l
Ans
Figure 15.22
a Pal 2
Pa2
(l + );
EI
3 2EI
15.9 Fig. 15.23 shows a flat ring made of steel 25 mm wide by 6 mm thick, loaded with a central
load of 600 N. Calculate the maximum bending moment in the ring. Sketch the bending
moment diagram and find the position of the point of inflexion.
Taking E as 200 GN/m2 , find the vertical deflection produced by the load of 600 N.
600 N
150 mm
75 mm
Mean radius
150 mm
6 mm
600 N
Figure 15.23
[Ans 38.7 Nm (at load point);
129 mm from load point, 8.9 mm]
15.10 A simply supported rectangular beam has a width of 120 mm and depth of 240 mm covering
a span of 3 m. A load of 12 kN is dropped at the midspan of the beam from a height of
12 mm. Find the maximum instantaneous deflection and the stress induced in the beam.
E = 200 GN/m2 .
[Ans 3.95 mm, 126 N/mm2 ]
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C HAPTER
16
THEORIES OF ELASTIC FAILURE
A material is regarded as failed if it is loaded beyond the elastic limit and permanent deformation occurs when the particles of a material separate from each other (as in use of brittle material)
accompanied by considerable plastic deformation.
It is easy to guess failure of material when the material is subjected to simple stress followed
by an axial loading. But when the material is subjected to complex stresses followed by biaxial or
triaxial loading then it is difficult to predict the failure of material.
For complex systems, we shall discuss five important theories of failure. In these five theories,
the complex state has been related to the elastic limit in simple tension or compression.
As we have studied that in any complex loading system, three principal stresses exist: σ1 , σ2
and σ3 such that σ1 > σ2 > σ3 . σ1 is the maximum principal stress, σ2 is the intermediate principal
stress end σ3 is the minimum principal stress.
In 2-D stress system we will discuss the following five theories of failure:
1.
2.
3.
4.
5.
Maximum Principal Stress Theory (Rankine’s Theory)
Maximum Shear Stress Theory (Guest’s or Tresca’s Theory)
Maximum Principal Strain Theory (St. Venant’s Theory)
Maximum Strain Energy Theory (Haigh’s Theory)
Maximum Shear Strain Energy Theory (Von Mises’s Theory):
Now we shall discuss these theories in details.
1. Maximum Principal Stress Theory (Rankine’s Theory)
According to this theory, failure will occur when the maximum principal tensile stress (σ1 ) in the
complex system reaches the value of the maximum stress at the elastic limit (σet ) in simple tension
or the minimum principal stress (that is, the maximum principal compressive stress) reaches the
elastic limit stress (σec ) in simple compression, i.e.
σ1 = σet (in simple tension) = σo = σu (yield stress is simple tension or compression)
σ3 = σec (in simple compression)
σ3 means numerical value of σ3 .
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321
Graphical representation is shown in Fig. 16.1.
This theory is found to have good results when applied to brittle materials such as cast iron.
σu
σu
and σ2 = ±
Note: Working stress, σ1 = ±
Factor of Safety
F.O.S
where σy , σx , τ direct and shear stresses on given planes is the complex system, σ1 = maximum
principal stress
σ2
σο
Square
(−) σο
(+) σο
σ1
(−) σο
Figure 16.1
2. Maximum Shear Stress Theory (Guest or Teresa’s Theory)
According to this theory the failure occurs when the maximum shear stress τmax reaches the value
of the maximum shear stress in simple tension at the elastic limit, i.e.,
τmax =
or
σ1 − σ3 σet
σu
=
=
2
2
2
in simple tension
σ1 − σ3 = σet = σu
While designing σet is replaced by the safe stress.
This theory does not give accurate results for the state of stress of pure shear in which the
maximum amount of shear in developed.
This theory is preferred in case of ductile materials such as mild steel. In 3-D stress system,
σ1 − σ3
τmax =
2
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•
Strength of Materials
σ2
B
+
σ1− σ2 = − σο
A
σ2 = σο
σο
σ1 = σο
45˚
(−) σο
C
O
D
+ σο
σ1
σ1− σ2 = σο
σ2 = − σο
F
E (−)σο
σ2 = − σο
Figure 16.2 Graphical representation of maximum shear stress theory
3. Maximum Principal Strain Theory (St. Venant’s Theory)
Failure occurs when the greatest principal strain reaches the strain at the elastic limit in a simple test.
According to this theory, failure of material occurs when the maximum strain in the complex stress
system equals the value of maximum strain at yield point in simple tension or compression test.
εcomplex = εsimple
1
σ0
(σ1 − μσ2 ) = ±
E
E
σ1 − μσ2 = ±σ0 (for tensile test)
Also σ2 − μσ1 = ±σ0 (for compressive test)
Except for brittle material, this theory does not match with experimental results, and so finds
little general support.
σ2
A
σο
σ2 − μσ1 = σο
B
Rhomboid
σ1 − μσ2 = σο
σο
(−) σο
C
σ1 − μσ2 = − σο
D
σ2 − μσ1 = − σο
σ1
Shear
diagonal
(−) σο
Figure 16.3 Graphical representation of maximum principal strain theory
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•
323
By plotting the above equation a rhomboid is obtained.
4. Maximum Strain Energy Theory (Haigh’s Theory)
Failure occurs when the energy stored per unit volume in a strained material reaches the strain
energy per unit volume at the elastic limit in a simple tension test, i.e., the maximum energy a body
can store without permanent deformation in a fixed quantity, irrespective of the manner in which it
is strained.
σ2
1 2
σ1 + σ22 − 2μσ1 σ2 = 0
2E
2E
σ12 + σ22 − 2μσ1 σ2 = σ02
2 2
σ1
σ2
σ1
σ2
·
=1
+
− 2μ
σ0
σ0
σ0
σ0
or
or
σ0
σ0
This is the equation of an ellipse with semi-major and semi-minor axes √
; √
respec1−μ
1+μ
tively each at 45◦ . This theory gives good results for ductile materials to the coordinate axes as
shown graphically in Fig. 16.4.
σ2
Ellipse
C
B
D
A
E
F
Figure 16.4
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σ1
Shear
diagonal
324
•
Strength of Materials
5. Maximum Shear Strain Energy Theory (Von Mises and Henkey’s Theory)
This is also known as distortion energy theory. Failure occurs when the shear strain energy per unit
volume in a strained material reaches the shear strain energy per unit volume at the elastic limit in a
simple tension test, this is similar to the preceding theory but it is assumed that the volumetric strain
energy plays no part in producing elastic failures.
(a) For 3D stress system:
σ2
1 (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 = 0
12C
6C
or
(σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 = 2σ 20
(b) For 2D stress system, putting σ3 = 0, we get
σ12 + σ22 − σ1 σ2 = σ02
σ12 + σ22 − σ1 σ2
1
2 = σ0
σ2
45°
45°
(–) σo
σo
(–) σo
Figure 16.5
This theory is good for ductile materials.
This is the best theory of the above five theories.
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σ1
Theories of Elastic Failure
•
325
When σt = σc and εt = εc .
σ2
(– , +)
I
II
Maximum
strain energy
theory
(+ , +)
Maximum shear
strain theory
Maximum shear
stress theory
45°
45°
σ1
(+, –)
III
(– , –)
Maximum principal
strain theory
Maximum principal
stress theory
IV
Figure 16.6 Graphical representation of various theories of failure on the same diagrams
E XAMPLE 16.1: The load on a bolt is an axial pull of 15 kN together with transverse shear force
of 7 kN. Estimate the diameter of the bolt. Using all theories of failure. σut = 280 N/mm2 , factor
of safety = 4. Poisson’s ratio (μ ) = 0.3
S OLUTION :
280
= 70 N/mm2 = σ0
Allowable simple tensile stress =
4
Let d = diameter of bolt (core dia).
15000 60000
=
=σ
Then, normal stress is = π
π d2
d2
4
7000 28000
and shear stress (τ ) = π
=
π d2
d2
4
Therefore, principle stresses are
σx
σ1 , σ2 =
±
2
σ 2
x
2
+ τ2
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and σ3 = 0
326
•
Strength of Materials
σ1 σ2 =
σ1 =
60000
±
2π d 2
60000
2π d 2
2
+
2
30000
1 √
+ 2 900000000 + 784000000
2
πd
πd
=
1
30000
+ 2 × 41036.6
π d2
πd
=
71036.6
π d2
σ2 =
28000
π d2
30000 41036.6
−
π d2
π d2
=−
110.36
π d2
σ3 = 0
1. Applying Rankine Theory (Principal Stress Theory)
Maximum principal stress in bolt
σ1 =
σ 1
+
2 2
σ 2 + 4τ 2
60000
π d2
2
60000 1
+
2
2π d 2
=
30000
1 √
+
3600000000 + 3136000000
π d2
2π d 2
=
1
30000
+
× 82073.14
π d2
2π d 2
σ1 =
+4
28000
π d2
2
=
71036.57
π d2
Maximum stream in simple tension = 70 N/mm2 = σ0
Now equating σ1 = σ0
71036.57
= 70
π d2
∴ d 2 = 323.2
or d = 17.977 = 18 mm
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•
327
2. Maximum Shear Stress Theory (Guest on Tresca’s Theory)
τmax =
1
2
1
=
2
Now
σ 2 + 4τ 2
60000
π d2
2
28000
+4
π d2
2
=
1 √
3600000000 + 3136000000
2π d 2
=
1
× 82073.14
2π d 2
=
41036.6
π d2
τ=
10
σ2 − σ1 σ
= =
= 35 N/mm2
2
2
2
41036.6
= 35
π d2
∴
d2 =
∴
41036.6
= 373.4
35 × π
d = 19.32 mm
3. Maximum Strain Energy Theory (Von Mises & Hencky Theory) or Distortion Theory
σ12 + σ22 − 2μσ1 σ2 = σ02
71036.6
π d2
2
11036
11036 2
71036.6
= 702
+ −
+
2
×
0.3
π d2
π d2
π d2
5046198540 121793296 4703759506
+
+
= 4900
π 2d4
π 2d4
π 2d4
9871751392
= 4900
π 2d4
1001232443
= d4
4900
d 4 = 204333
d = 21.26 mm
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•
Strength of Materials
4. Maximum Strain Energy Theory (Haigh Theory)
(σ1 − σ2 )2 + σ22 = σ02
71036.6 11036
+
π d2
π d2
82072
π d2
1
π 2d4
2
+
2
11036
= 702
+ −
π d2
121793296
= 4900
π 2d4
(6735813184 + 121793296) = 4900
6747992480
= d4
π 2 × 4900
d 4 = 139675
d = 19.33 mm
Maximum diameter d is 21.26 mm
So the answer is 21.26 mm
E XAMPLE 16.2: A subject is subjected to a maximum torque of 12 kNm and a maximum bending
moment of 8.5 kNm at a particular section. If the allowable equivalent stress in simple tension is
170 MN/m2 , find the diameter of the shaft according to the maximum shear strain theory.
Maximum torque, T = 12 kNm
Maximum bending moment, M = 8.5 kNm
Allowable equivalent stress in simple tension,
σt = 170 MN/m2
∴
M=
π 3
d σb
32
σb =
M × 32
π d3
τ=
and
16T
π d3
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T =τ×
π 3
d
16
Theories of Elastic Failure
•
329
Principal stresses are given by:
σ
σ1 , σ2 = b ±
2
σ 2
b
2
+ τ2
1
σb ± σb2 + 4τ 2
2
⎡
⎤
1 ⎣ 32M
32M 2
32T 2 ⎦
=
±
+
2 π d3
π d3
π d3
=
16 M+
π d3
16 σ1 = 3 M +
πd
=
M2 + T 2
M2 + T 2
σ2 = 0
and σ3 =
16 M−
π d3
M2 + T 2
According to the maximum shear stress theory,
σt = σ1 − σ3 =
=
d3 =
16 M+
π d3
16 M2 + T 2 − 3 M −
πd
M2 + T 2
32
[M 2 + T 2 ]
π d3
32 × 103
×
π × 160 × 106
8.52 + 122
d 3 = 6.37 × 10−5 × 14.705
d 3 = 0.000937
d = 97.8 mm
∴ d = 0.0978 m
Ans
E XAMPLE 16.3: A steel shaft is subjected to an end thrust of 90 MPa and the maximum shearing
stress on the surface arising from torsion is 68 MPa. The yield point of the material in simple tension
was found to be 300 MPa. Calculate the factor of safety of the shaft according to the following
theories:
i) Maximum shear stress theory
ii) Maximum distortion energy theory
(AMIE Summer 2000)
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•
Strength of Materials
S OLUTION :
Given σ1 =? σ2 = 0, σ3 = −90 MN/m2
τmax = 60 MN/m2 ; σl t = 300 MN/m2
i) Maximum shear stress theory
σ1 − σ2
= 60
2
or σ − (−90) = 120 or σ1 = 30 MN/m2
Also σ1 − σ3 = σt
τmax =
30 − (−90) = σt ∴ σt = 120 MN/m2
300
σet
F. O. S. =
=
= 2.5 Ans
σt
120
ii) Maximum distortion energy theory
σt2 = σl2 + σ32 − σ3 σ1
= 302 + (−90)2 − (−90)(30)
= 11700
∴
σt = 108.17 MN/m2
σet
300
F.O.S. =
=
= 2.77 Ans
σt
108.17
E XAMPLE 16.4: In a steel member, at a point the major principal stress is 200 MN/m2 and the
minor principal stress is compressive. If the tensile yield point of the steel is 250 MN/m2 , find
the value of the minor principal stress at which yielding will commence, according to each of the
following criteria of failure:
(i) Maximum shearing stress,
(ii) Maximum total strain energy, and
(iii) Maximum shear strain energy
Take, Poisson’s ratio = 0.3
S OLUTION :
σ1 = 200 MN/m2
Yield point stress = σe = 250 MN/m2
Minor principal stress, σ2:
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331
i) Maximum shearing stress criterion:
σ1 − σ2 = σe
σ2 = σ1 − σe = 200 − 250 = −50 MN/m2
σ2 = 50 MN/m2
(Compressive)
Ans
ii) Maximum total strain energy criterion:
σ12 + σ22 − 2μσ1 σ2 = σe2
(or σ02 )
(200)2 + σ22 − 2 × 0.3 × 200 × σ2 = 2502
40000 + σ22 − 120 σ2 = 62500
σ22 − 120σ − 62500 + 40000 = 0
σ22 − 1206 − 22500 = 0
√
+120 − 14400 + 90000
σ2 =
2
√
+120 − 104400
σ2 =
2
=
(only −ve sign is taken as σ2 is to be compressive)
+120 − 323
2
= 101.5 MN/m2
(Compressive)
Ans
iii) Maximum shear strain energy criterion:
(σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 = 2σe2
putting σ3 = 0, we get
(σ1 − σ2 )2 + σ22 + σ12 = 2σe2
σ12 + σ22 − 2σ1 σ2 + σ22 + σ12 = 2σe2
σ12 + σ22 − σ1 σ2 = σ22
(200)2 + σ22 − 200 σ2 = (250)2
σ22 + 40000 − 62500 − 200σ2 = 0
σ22 − 200σ2 − 22500 = 0
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332
•
Strength of Materials
√
+200 − 40000 + 90000
σ2 =
2
= −80.27 MN/m2
(Compressive)
Ans
E XAMPLE 16.5: At a point two of the principal stresses are 150 N/mm2 and 100 N/mm2 .
Determine the safe range of the third principal stress at the point by five different theories. Take
E = 200 GPa and μ = 0.25, failure stress in tension test to be 220 N/mm2 . Failure stress in tension
and compression is the same.
S OLUTION :
a) Maximum principal stress theory
σ1 = σy = 220 N/mm2
σ3 = σy
(σy = yield stress)
2
= −220 N/mm
Hence, the range is −220 ≤ σ ≤ 220
b) Maximum strain theory
σ1 − 0.3 (150 + 100) = σy = 220
Also,
σ1 = 220 + 75 = 295 N/mm2
σ3 − 0.3 (150 + 100) = σ y = −220
σ3 = −220 + 75 = −145 N/mm2
Hence, the range is −145 ≤ σ ≤ 220 N/mm2
c) Maximum strain energy theory
σ 2 + (100)2 + (150)2 − 2 × 0.3 (150σ + 100σ + 150 × 100) = 2202
σ 2 + 10000 + 22500 − 150σ − 9000 = 48400
σ 2 − 150σ − 24800 = 0
∴
√
150 ± 22500 + 99600
σ=
2
150 ± 349.4
σ=
2
σ1 = 249.7, σ2 = −99.7
Hence, the range is −99.7 ≤ σ ≤ 249.7.
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333
d) Maximum shear stress theory
σ1 − 100 = σy = 210
σ1 = 220 + 100 = 320 N/mm2
Also, 150 − σ3 = 220
σ3 = 150 − 220 = −70 N/mm2
Hence, the range is −70 ≤ σ ≤ 320 N/mm2
e) Maximum distortion theory or shear strain theory
(σ − 100)2 + (σ − 150)2 + (150 − 100)2 = 2 (220)2
σ 2 + 10000 − 200σ + σ 2 + 22500 − 300σ + 2500 = 96800
2σ 2 − 500σ − 61800 = 0
σ 2 − 250σ − 30900 = 0
√
250 ± 62500 + 123600
σ=
2
+250 ± 431.4
σ=
2
σ1 = 340.7, σ3 − 181.4
Hence, the range is −181.4 ≤ σ ≤ 340.7 N/mm2 .
E XAMPLE 16.6: Principal stresses at a point in an elastic material are 120 MPa tensile, 60 MPa
tensile and 20 MPa compressive. Determine the factor of safety against the failure based on various
theories. The elastic limit in simple tension is 240 MPa and Poisson’s ratio 0.3.
S OLUTION :
i) Maximum principal stress theory
Failure takes place when the maximum principal stress reaches the value of maximum stress
at that limit.
Thus, maximum principal stress, σ = 120 MPa
240
=2
So factor of safety =
120
ii) Maximum shear stress theory
σ = 120 − (−20) = 140 MPa
240
= 1.71
Factor of safety =
140
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334
•
Strength of Materials
iii) Maximum principal strain theory
σ1 − μσ2 − μσ3 = 120 − 0.3 × 60 − 0.3(−20) = 108 MPa
Factor of safety =
240
= 2.03
108
iv) Maximum strain energy theory
σ 2 = σ12 + σ22 + σ32 − 2 μ (σ1 σ2 + σ2 σ3 + σ3 σ1 )
= 1202 + 602 + (−20)2 − 2 × 0.3 [120 × 60 + 60 (−20) + (−20 × 120)]
= 14400 + 3600 + 400 − 3600
= 14800
∴
σ = 121.6
Factor of safety =
240
= 1.97
121.6
v) Maximum shear strain energy theory
2 σ 2 = (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2
= (120 − 60)2 + [60 − (−20)]2 + [−20 − 120]2
= 3600 + 6400 + 19600
= 29600
σ2 =
∴
29600
= 14800
2
σ = 121.65 MPa
Factor of safety =
240
= 1.97
121.65
E XAMPLE 16.7: A solid shaft transmits 1000 kW at 300 r.p.m. Maximum torque is 2 times the
mean. The shaft is subjects to a bending moment, which is 1.5 times the mean torque. The shaft is
of ductile material for which the permissible tensile and shear stresses are 120 MPa and 60 MPa
respectively. Determine the shaft diameter using suitable theory of failure. Give justification of the
theory used.
[AMI.E. Sec B Winter 94]
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•
335
S OLUTION :
Principal stresses:
σb =
Principal stresses =
32M
;
π d3
1
σb ±
2
τ=
16T
π d3
σb2 + 4τ 2
32
1 32M
± 3 M2 + T 2
3
2 πd
πd
16 M ± M2 + T 2
=
3
πd
16 σ1 = 3 M+ M 2 + T 2 , σ2 = 0
πd
=
and σ3 =
16 M−
π d3
M2 + T 2
Best theory is the distortion energy theory, which gives
σ12 + σ32 − σ1 σ3 = σt2
or
16
π d3
2 M+
M2 + T 2
2
+ M−
M2 + T 2
2
×
16 2
16 2 2
2
2
− M −M +T
= σt2
π d3
π d3
16 2 2
2
2
2
2M
= σt2
or
+
2M
+
2T
+
T
π d3
16 2 2
2
4M
= σt2
or
+
3T
π d3
32
σt = 3
πd
3
M2 + T 2
4
2π NTmean
= 1000 × 103
60
2π × 300 × Tmean
= 1000 × 103
60
∴
Tmean =
@seismicisolation
@seismicisolation
100 × 103 100
=
kNm
π
π
(i)
336
•
Strength of Materials
T = Max. torque =
200
kNm
π
M = Maximum bending moment =
150
kNm.
π
Putting in Eqn. (i), we get
150 2
200 2 3
× 103
+
π
π
4
32 × 103
3
3
d =
(150)2 + (200)2
6
4
120 × 10 π
32
3
=
1502 + (200)2
4
120 π 2 × 100
3
−3
= d = 6.191 × 10
d = 0.1836 m.
= 183.6 mm Ans
32
120 × 10 =
π d3
6
Justification of using this theory: The theories used for ductile materials are: (i) Maximum shear
stress theory and (ii) distortion energy theory. Out of these theories, distortion energy theory is
best for ductile materials, as the experimental results for these materials fit very well in this theory.
Maximum shear stress theory gives results on safer side, that is, a little more material is used than
the actual required, as given by distortion energy theory.
Maximum shear stress theory
π 3
d σt
32
200 2
150 2
3
or 10
+
π
π
π 3
or = d × 120 × 106
32
200 2
150 2
32 ×
+
π
π
3
d =
× 103
6
π × 120 × 10
d 3 = 6.755 × 10−3
= 0.183 m
∴ d = 189 mm.
M2 + T 2 =
Because σt = 2τ (given), the equivalent torque criterion gives same results.
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•
337
E XAMPLE 16.8: A mild steel shaft 100 mm diameter is subjected to a maximum torque of 15 kNm
and a maximum bending moment of 10 kNm at a particular section. Find the factor of safety according to the maximum shear stress theory if the elastic limit in simple tension is 240 MN/m2 .
[UPTU 2006–2007]
S OLUTION :
Tmax = 15 kNm = 15 × 106 Nmm
Mmax = 10 kNm = 10 × 106 Nmm
τ=
=
16T
π d3
16 × 15 × 106
π × [100]3
= 76.36 N/mm2 .
σb =
10 × 105
M
y=
I
I
y
=
10 × 106
Z
10 × 106
= π
(100)3
32
107 × 32
= 101.81 N/mm2
π × 106
σb
σb 2
σ1 , σ2 =
+ τ2
±
2
2
=
= 50.90 ±
(50.90)2 + (76.36)2
= 50.90 ± 91.76
= 142.66 or
τmax =
=
− 40.86 N/mm2
1
(σmax − σmin )
2
1
(142.66 + 40.86)
2
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338
•
Strength of Materials
= 91.76 N/mm2
τmax =
240 × 106
2 × 106
= 120 N/mm2
Now,
∴
91.76 =
Factor of safety =
120
Factor of Safety
120
91.76
= 1.3 Ans
E XAMPLE 16.9: Find the diameter of a shaft according to the maximum shear stress theory if the
shaft is subjected to a maximum torque of 15 kNm and a maximum bending moment of 10 kNm at
a particular section. Take allowable equivalent stress in simple tension as 200 MN/m2 .
M=
π 3
d σb
32
σb =
32M
π d3
and T =
π 3
d τ
16
16T
π d3
1
2
2
σ=
σb ± σb + 4τ
2
⎡
⎤
32 2
32T ⎦
1 ⎣ 32
M±
+
=
2 π d3
π d3
π d3
τ=
16 M±
π d3
16 M+
σ1 =
π d3
=
∴
M2 + T 2
M2 + T 2
According to the maximum shear stress theory,
σ1 − σ3 =
32
π d3
M2 + T 2
d3 =
32
πσt
M2 + T 2
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Theories of Elastic Failure
=
32 × 103
π × 200 × 106
•
339
102 + 152
= 5.095 × 10−5 (18.03)
d = 0.0947 m
= 94.7 mm
Ans
Exercise
16.1 A steel shaft is subjected to two principal stresses of +40 MN/m2 and −90 MN/m2 . If the
elastic limit in simple tension or compression is 360 MN/m2 , find factor of safety according
to each maximum principal stress, maximum shear stress and shear strain theories.
[Ans 4,2.77 and 3.12]
16.2 A bar of circular section is subjected to an axial tensile load of 10 kN and a transverse
sheer force of 5 kN. Find the diameter of the bar if the allowable stress in simple tension
is 100 MN/m2 using each of the following theories of failure:
i) Maximum principal stress theory
ii) Maximum shear stress theory
iii) Maximum Strain energy theory if μ = 0.25
[Ans i) 12.35 mm, ii) 13 mm and iii) 12.8 mm]
16.3 Determine the allowable shear stress for a circular solid shaft transmitting torque according
to each of the following theories of failure:
a) Maximum principal stress theory
b) Maximum principal strain theory
c) Maximum shear stress theory
d) Maximum strain energy theory
e) Distortion energy theory (shear strain energy theory)
Allowable tensile stress in simple tension for the material is 160 MN/m2 .
[Ans 160, 123.1, 80, 99.2 and 92.4 MN/m2 ]
16.4 A circular shaft of steel is subjected to combined bending and torsion, the bending moment
being 20 kNm and torque 10 kNm. If safe equivalent stress in simple tension is 200 MN/m2
and μ = 0.25, find suitable diameter of the shaft based on the following theories.
a) Maximum principal stress theory
b) Maximum shear stress theory
c) Maximum shear strain energy theory
[Ans
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a) 102.56 mm; b) 104 mm; c) 103.5 mm]
340
•
Strength of Materials
16.5 An axial pull of 20 kN along with a shear force of 15 kN is applied to a circular bar of 20 mm
diameter. The elastic limit of the bar material is 230 MPa and the Poisson’s ratio, μ = 0.3.
Determine the factor of safety against failure based on:
a) maximum shear stress theory
b) maximum strain energy theory
c) maximum principal strain energy theory
d) maximum shear strain energy theory
[Ans
2; 2.3; 2.37; 2.2]
16.6 A thin cylindrical shell of diameter 200 mm and wall thickness t is subjected to material
pressure of 3 N/mm2 . Yield strength of the material is 280 N/mm2 . Taking factor of safety
of 3 and using the maximum shear stress theory, determine the wall thickness t of the cylinder.
[Ans 3.45 mm minimum]
16.7 The load on a bolt consists of an axial thrust of 8 kN together with a transverse shear load of
4 kN. Calculate the diameter of the bolt using strain energy theory. Take factor of safety as 3.
σy = 285 MN/m2 , μ = 0.3.
[Ans 11.73 mm]
16.8 In a steel drum that is subjected to axial compressive force, the internal pressure is 10 N/mm2 .
The maximum circumferential stress is 100 MPa and the longitudinal stress is 30 MPa. Find
the equivalent tensile stress in a simple tensile test according to each of the five theories of
failure. Take Poisson’s ratio, μ = 0.3.
[Ans 96.43 MPa]
16.9 A mild steel shaft of 40 mm diameter when subjected to a pure torsion ceases to be elastic
when torque reaches 2 kNm. A similar shaft is subjected to a torque of 1.2 kNm and a bending
moment M kNm. If the maximum strain energy is the criterion for the elastic failure, find the
value of M. Take Poisson’s ratio = 0.28.
[Ans M = 1.28 kNm]
16.10 A shaft of 100 mm diameter is subjected to a bending moment of 5 kNm. Find the value of
the maximum torque which can be applied to the shaft for each of the following conditions:
a) maximum direct stress not to exceed 120 N/mm2 ; b) the maximum shearing stress not
to exceed 60 N/mm2 ; c) maximum shear strain energy per unit volume not to exceed that
induced by simple shear stress of 80 N/mm2 .
[Ans a) 17.85 kNm; b) 10.65 kNm; c) 14.6 kNm]
16.11 A cast iron cylinder has outside and inside diameters of 200 mm and 125 mm. If the ultimate
tensile strength of the cast iron is 150 MN/m2 , find according to each of the following theories, the internal pressure which would cause rupture: a) Maximum principal stress theory;
b) Maximum strain theory and c) Maximum strain energy theory. Assume no longitudinal
stress in the cylinder. μ = 0.25
[Ans a) 65.7 N/mm2 ; b) 59.2 N/mm2 ; c) 55.2 N/mm2 ]
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341
16.12 A hollow mild steel shaft having 60 mm internal diameter and 120 mm as external diameter
is subjected to a torque of 10 kNm and a bending moment of 3 kNm. Calculate the direct
stress which acting alone would produce the same
i) Maximum strain energy
ii) Maximum shear strain energy
iii) Maximum shear stress
[Ans
(i) −22.39 N/mm2 ; (ii) 56.78 N/mm2 and (iii) 64.642 N/ mm2 ]
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C HAPTER
17
COMBINED STRESSES (DIRECT, BENDING,
TORSION)
In engineering there are not always direct stresses, but combination of direct and bending or direct
and torsion or combination of three is involved. Here, in this chapter we shall study such problems.
E XAMPLE 17.1: A bending moment of 500 Nm and torque of 350 Nm is applied to a shaft of
diameter 80 mm. Find: i) the maximum normal stress on a section perpendicular to the axis; b) the
maximum shear stress on a section perpendicular to the axis; c) the principal stresses and d) the
maximum shear stress:
S OLUTION :
a) Bending stress, σb =
σb =
b) τ =
My
I
500 × 0.080 × 64
2 × π (0.080)4
= 9952229.3 N/m2 = 9.95 MN/m2
T.r 350 × 0.040 × 32
=
= 3483280 N/m2 = 3.48 MN/m2
J
π × (0.08)4
Ans
c) Principal stresses:
σb 1
σb2 + 4τ 2
±
2 2
9.95 1
±
=
(9.95)2 + 4(3.48)2
2
2
1√
= 4.975 ±
99 + 48.44
2
σ1,2 =
= 4.975 ± 6.07
= 11.045 N/mm2 , −1.095 N/mm2
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Ans
Ans
Combined Stresses (Direct, Bending, Torsion)
•
343
d) Maximum shear stress:
σ1 − σ2
2
1
= (11.045 + 1.095)
2
= 6.07 N/mm2 Ans
τmax =
E XAMPLE 17.2: A cast iron column, hollow circular section, has a projecting bracket carrying a
load of 120 kN. The load line is off axis of column by 0.3 m. The external diameter of the column
is 0.35 m and thickness of metal 30 mm. Determine the extreme stresses in the section.
S OLUTION :
π
× (0.352 − 0.292 ) = 0.03 m2
4
π
I=
(0.354 −0.294 ) = 0.0003891 m4
64
M = 120000 × 0.3 = 36000 Nm
A=
Direct stress,
120000
= 4000000 N/m2
0.03
= 4 MN/m2 (Compressive)
σd =
Bending stress,
σb =
36000 × 0.35
My
=
= 16.19 MN/m2 (Tensile) also (Compressive)
I
2 × 0.0003891
Extreme stresses:
σ1 = −4 + 16.19 = +12.19 MN/m2
(Tensile) Ans
σ2 = −4 − 16.19 = −20.19 MN/m2
= 20.19 MN/m2
(Compressive) Ans
Combined Bending and Twisting
2τ
σb
A common application of combined stresses is that of a shaft subjected to bending and twisting
and it is often convenient to express the resulting direct and shear stresses directly in terms of the
applied moment and torque.
We had already proved in the case of one direct stress only tan 2θ =
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344
•
Strength of Materials
d
σx
T
τ
τ
σx
M
(b)
(a)
Figure 17.1
If the bending moment is M and the torque is T (Fig. 17.1(a)) then the stresses acting on an
element on the upper surface are shown in plan view (Fig. 17.1(b)) (those on the lower surface are
the same, except that σx is compressive).
M
σx = π
d3
32
and
T
τ= π
d3
16
assuming solid shaft.
Thus, the maximum principal stress, σ1 is given by
σ1 =
i.e.,
1
σx + σx2 + 4τ 2
2
⎫
⎧
⎛
⎛
⎞2
⎞2
⎪
⎪
⎬
⎨
1
M
⎜ T ⎟
⎜ M ⎟
=
π 3 ⎪ + ⎝ π 3 ⎠ + 4 ⎝ π 3 ⎠
2⎪
⎩ d ⎭
d
d
32
32
16
π 3
d τmax = M 2 + T 2
16
π 2
d τmax is evidently the equivalent torque, i.e., that torque which, acting alone,
16
would produce the same maximum shear stress as M and T acting together.
The quantity
i.e., Te =
M2 + T 2
E XAMPLE 17.3: A shaft of 100 mm diameter is subjected to at a certain section: a) a bending
moment of 8 kNm and b) a twisting moment of 11 kNm. Compute the maximum direct stress
induced in the section, indicating the position of plane on which it acts. Determine the stress, which
acting alone, will produce the same maximum i) strain and ii) strain energy. μ = 0.3.
S OLUTION :
d = 100 mm; M = 8 kNm = 8 × 106 Nmm; T = 9 kNm = 9 × 106 Nmm.
@seismicisolation
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Combined Stresses (Direct, Bending, Torsion)
Shear stress, τ =
16 T
16 × 11 × 106
=
= 56 N/mm2
3
πd
π (100)3
Bending stress, σb =
32 M 32 × 8 × 106
=
= 81.53 N/mm2
3
π d3
π (100)
tan 2 θ =
∴
2τ
11
= 1.375
=
σb
8
2 θ = 53.97 = 53◦ 58 ; θ1 = 26.5◦ 29
or
with the axis of the shaft.
θ2 = 116◦ 59 with the axis of the shaft.
σb
σb 2
σ1 =
+ τ2
+
2
4
81.53
81.53 2
=
+ 562
+
2
4
√
= 40.765 + 1662 + 3136
= 110 N/mm2
σb
σb 2
σ2 =
+ τ2
−
2
4
81.53 2
81.53
−
=
+ 562
2
4
√
= 40.765 − 1162 + 3136
= −24.79 N/mm2
i) For maximum strain, we have
σ = E ε = σ1 − μσ2
= 110 − 0.3 × (−24.79)
= 117.44 N/mm2
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63◦ 1
•
345
346
•
Strength of Materials
ii) For maximum strain energy, we have
σ 2 = σ12 + σ22 − (μ × σ1 σ2 )
= (110)2 + (−24.79)2 − {0.3 × 110 × (−24.79)}
= 12100 + 614.5 + 818;
√
σ = 135325
σ = 116.33 N/mm2 . Ans
E XAMPLE 17.4: A ship’s heavy duty propeller shaft is of 0.6 m diameter and supports a propeller
weighing 170 kN. The propeller overhangs the supporting bracket by 2.5 m as shown in Fig. 17.2.
The speed of the ship is 50 km/h. The engine develops 30 MW and the propeller receives it at
350 r.p.m. Assuming a propulsive efficiency of 72%, determine the maximum principal stresses and
shear stress in the shaft.
S OLUTION :
T = Torque on the shaft
P = Axial thrust
2π × 350 × T
= 30 × 106
60
30 × 106 × 60
T=
= 818926.3 Nm
2π × 350
Let
P
2.5 m
170 kN
Figure 17.2
Bending moment in shaft, Mmax = 170 × 2.5 = 425 kNm
Work done by propeller = Axial thrust × Velocity
P × 50 × 103
3600
30 × 106 × 0.72 × 3600
P=
= 1555200 N = 1555.2 kN
50 × 103
30 × 106 × 0.72 =
P −1555.2 × 103 × 4
=
= −5.50 N/mm2
A
π (0.6)2
32 × 425 × 1000
Bending stress, σb =
π (0.6)3
Direct stress, σd =
= 20.05 N/mm2
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Combined Stresses (Direct, Bending, Torsion)
•
347
Resultant maximum direct stress = −σd − σb = −5.50 − 20.05 = −25.55 N/mm2
16T
16 × 818926.3
=
= 19.32 N/mm2
3
3
π
(0.6)
πd
Principal stresses,
Shear stress, τ =
1
1
σ1,2 = (σd + σb ) ±
(σd + σb )2 + 4τ 2
2
2
1
1
= (−25.55) ±
(−25.55)2 + 4 (19.32)2
2
2
1√
= −12.775 ±
652.8 + 1493.05
2
1√
= −12.775 ±
2145.85
2
= −12.755 ± 23.16
σ1 = 10.385 N/mm2
2
σ2 = −35.93 N/mm
Ans
Ans
Maximum shear stress
1 2
σ + 4 τ2
2
1
=
(25.55)2 + 4 (19.32)2
2
1√
=
2145.85
2
= 23.16 N/mm2 Ans
=
E XAMPLE 17.5: A solid shaft of diameter d is subjected to a bending moment M = 15 kNm and
a twisting moment T = 25 kNm. What is the minimum diameter of the shaft if the shear stress in
the shaft is not to exceed 160 N/mm2 and the direct stress is not to exceed 200 N/mm2 .
S OLUTION :
Bending moment = M = 15 kNm = 15 × 106 Nmm
Torque T = 25 kNm = 25 × 106 Nmm
Maximum permissible shear stress, τ = 160 N/mm2
Maximum permissible bending stress, σ = 200 N/mm2
@seismicisolation
@seismicisolation
348
•
Strength of Materials
Minimum diameter of the shaft
√
π d3
Equivalent torque, Te = M 2 + T 2 = τmax
16
√
√
π d3
M2 + T 2
M 2 + T 2 16
3
or d =
×
=
∴
16
τmax
τmax
π
2 2
15 × 106 + 25 × 106
16
3
d =
×
160
π
d = 97.53 mm
√
M + M2 + T 2
π d3
Similarly, equivalent bending moment = Me =
= σmax
2
32
√
√
M+ M 2 + T 2 32
π d 3 M+ M 2 + T 2
=
×
∴
∴ d3 =
32
2 × σmax
σmax × 2
π
6+
6 2 + 25 × 106 2
15
×
10
15
×
10
d3 =
× 32
2 × 200 × π
d = 95.77 mm
Naturally, between these two diameters the larger value is answer. Hence, d = 97.53 mm
Ans.
E XAMPLE 17.6: A circular shaft, transmitting 60 kW power at 150 rpm is supported in bearings
that are 4.5 m apart. At 1.7 m from one bearing, it carries a pulley which exerts a transverse load
of 18 kN on the shaft. Determine the suitable diameter of the shaft if: a) the maximum direct stress
is not to exceed 120 N/mm2 , b) the maximum shear stress is not to exceed 70 N/mm2 and c) the
stress which acting alone would produce the maximum strain energy is not to exceed 130 N/mm2 .
Take μ = 0.3
S OLUTION :
60P
60 × 60000
=
= 382.66 Nm = 382660 Nmm
2π N
2π × 150
Wab 18000 × 1.7 (4 − 1.7) 18000 × 1.7 × 2.3
=
=
= 17595 Nm
M=
l
4
4
= 17595000 Nmm
T=
i) Maximum principal stress criterion
16 σ = 3 M + M2 + T 2
πd
16 17595000 + 175950002 + 3826602
d3 =
π × 120
d 3 = 0.0425 [17595000 + 17599161]
d= 116.8 mm
@seismicisolation
@seismicisolation
Combined Stresses (Direct, Bending, Torsion)
•
349
ii) Maximum shear stress criterion
16 2
M +T2
π × d3
16 d3 =
175950002 + 3826602
π × 70
τmax =
= 0.073 × 1.7599 × 107 = 1284727
∴
d = 108.71 mm
iii) Maximum strain energy criterion,
σ=
16
4M 2 +
π d3 2
1
+1
μ
T2
1
μ
1
+1
2
16
0.3
2
d3 =
4
(17595000)
+
(382660)2
1
π × 140 0.3
= 0.0364 1.23834 × 1015 + 2.6 (1.4643 × 1011 )
= 0.0364 1.23834 × 1015 + 3.808 × 1011
= 0.0364 1.23834 × 1015 + 0.0003808 × 1015
= 1281280
d = 108.61
Hence, maximum diameter is chosen for safety,
∴
d = 117 mm
Ans
E XAMPLE 17.7: A solid circular shaft while transmitting power is subjected to a twisting moment
of 10 kNm and a bending moment of 6 kNm at a particular location. If the shaft diameter is 80 mm,
compute principal stresses and maximum shearing stress.
S OLUTION :
Diameter of shaft, d = 80 mm
Torque, T = 10 kNm = 10 × 106 Nmm
Bending moment, M = 6 kNm = 6 × 106 Nmm
@seismicisolation
@seismicisolation
350
•
Strength of Materials
Maximum principal stress
16 2 +T2
M
+
M
π d3
!
16
6
6 2 + 10 × 106 2
6
×
10
σmax =
+
6
×
10
π 803
= 9.95 × 10−6 6 × 166 + 36 × 1012 + 100 × 1012
= 9.95 × 10−6 6 × 106 + 11.66 × 106
σmax =
= 174.83 MPa (tensile) Ans
16 6
6
6
×
10
σmin =
−
11.66
×
10
π 803
= 9.95 × 10−6 −5.66 × 106
= −56.32 MPa
= 56.32 MPa
(Compressive)
Ans
Maximum shearing stress:
16 2
2
M
+
T
π d3
16
6 2 + 10 × 106 2
=
6
×
10
π 803
= 9.95 × 10−6 (36 × 1012 + 100 × 1012 )
τmax =
= 9.95 × 10−6 × 11.66 × 106
= 116.02 MPa
Ans
E XAMPLE 17.8: A propeller shaft 180 mm external diameter and 90 mm internal diameter transmits 1000 kW power at 110 r.p.m. It is also subjected to a bending moment of 10 kNm and an end
thrust of 130 kN. Find: (i) the principal stresses and (ii) the stress which acting alone would produce
the same maximum strain. Take μ = 0.3
S OLUTION :
do = 180 mm di = 90 mm P = 1000 kW, N = 110 r.p.m,
M = 10 kNm = 10 × 106 Nmm,
A=
End Thrust, E = 120 kN = 120 × 103 N
π
(1802 − 902 ) = 19075.5 mm2
4
@seismicisolation
@seismicisolation
Combined Stresses (Direct, Bending, Torsion)
Z=
π
32
do4 − di4
do
π
32
=
π
Torsional sectional modulus =
16
1804 − 904
180
1804 − 904
180
= 536498.4 mm4
= 1072996.8 mm4
130000
= 6.82 N/mm2
19075.5
10 × 106
= 18.64 N/mm2
σb =
536498.4
σd o =
Maximum compressive direct stress:
σc = 6.82 + 18.64 = 25.46 N/mm2
60P
60 × 1000
T=
=
= 86.86 kNm
2π N
2π × 110
= 86.86 × 106 Nmm
86.86 × 106
= 80.95 N/mm2
Maximum shear stress = τ =
1072996.8
Principal stresses:
σc
σ1 =
+
2
σ 2
c
25.46
+
=
2
2
+ τ2
25.46
2
2
+ 80.952
√
= 12.73 + 671.5
= 12.73 + 81.94
= 94.67 N/mm2
σ2 = 12.73 − 81.94
(Tensile) Ans
= −69.21 N/mm2
= 69.21 N/mm2
(Compressive) Ans
Max shear stress:
τmax =
=
25.46
2
2
+ 80.952
√
6715
= 81.94 N/mm2
@seismicisolation
@seismicisolation
Ans
•
351
352
•
Strength of Materials
Stress to give maximum strain
σ = σ1 + μ σ2
= 94.67 − 0.3 (−69.21)
= 94.67 + 20.76
= 115.43 N/mm2
Ans
E XAMPLE 17.9: The maximum normal stress and the maximum shear stress, analysed for a
shaft of 150 mm diameter under combined bending and torsion were found to be 120 MN/m2
and 80 MN/m2 , respectively. Find the bending moment and torque to which the shaft is subjected.
If the maximum shear stress is limited to 100 MN/m2 , find by how much the torque can be
increased if the bending moment is kept constant?
(AMIE Summer 1995)
S OLUTION :
d = 150 mm = 0.15 m
Maximum normal stress, σmax = 120 MN/m2
Maximum shear stress, σmax = 80 MN/m2
For combined bending and torsion,
16 2 +T2
M
+
M
π d3
16 2
τmax = 3
M +T2
πd
σmax =
(i)
(ii)
Substituting the values in Eqns. (i) and (ii), we get
16
2 +T2
M
+
M
π (0.15)2
16
2 +T2
80 × 106 =
M
π (0.15)2
80 × 106 × π (0.15)3
M2 + T 2 =
16
2
2
M + T = 53014.4
120 × 106 =
∴
Putting this value in Eqn. (iii), we get
16
[M + 53014.4]
π (0.15)3
120 × 106 × π (0.15)3
M + 53014.4 =
16
M + 53014.4 = 79521.4
M= 26507 Nm Ans
120 × 106 =
@seismicisolation
@seismicisolation
(iii)
(iv)
(v)
Combined Stresses (Direct, Bending, Torsion)
•
353
Substituting for M in Eqn. (v), we get
(26507)2 + T 2 = 53014.4
(26507.2)2 + T 2 = (53014.4)2
70262 × 104 + T 2 = 281052.6 × 104
T 2 = 104 (281052.6 − 70262)
= (210790.6)104
T = 459 × 102 = 45900 Nm
Ans
Now,
τmax = 100 MN/m2 = 100 × 106 N/m2
Since the bending moment is kept constant,
M = 26507 Nm
(already found out)
Substituting in Eqn. (ii), we have,
16
100 × 10 =
π (0.15)3
6
!
(26507)2 + T 2
100 × 106 × π × (0.15)3
16
2
2
26507 + T = 66267.9
265072 + T 2 =
Squaring both sides,
265072 + T 2 = 4.39 × 109
7.02 × 108 + T 2 = 4.39 × 109
T 2 = 108 (36.88)
∴
∴
T = 104 × 6.07
= 60700 Nm Ans
The increased torque = (60700 − 45900)
= 14800 Nm Ans
E XAMPLE 17.10: A shaft rotating at 220 r.p.m. is required to transmit 20 kW. It is simply supported between bearings 2.8 m apart. It carries at its centre a flywheel of mass 120 kg. Determine
the suitable diameter of the shaft if neither the maximum shear stress nor the tensile stress increases
above 50 MN/m2 and 70 MN/m2 respectively.
@seismicisolation
@seismicisolation
354
•
Strength of Materials
S OLUTION :
Torque, T =
25000 × 60
= 1085.7 Nm
2π × 220
Maximum bending moment =
Wl
120 × 9.81 × 2.8
=
= 824.04 Nm
4
4
Equivalent torque,
Te =
M2 + T 2 =
(824.04)2 + (1085.7)2
√
Te = 679041.9 + 1178744.5
= 1363 Nm
Equivalent bending moment,
1
M + M2 + T 2
2
1
2
2
824.04 + (824.04) + (1363)
=
2
√
1
=
824.04 + 679041.9 + 1857769
2
1
= (824.04 + 1592.74)
2
= 1208.39 Nm
Me =
If maximum shear stress is not to exceed 40 MN/m2 ,
Te = 1363 Nm
π 3
d τmax
Also Te =
16
π
1363 = d 3 × 50 × 106
16
1363 × 16
d3 =
= 0.0001389
π × 50 × 106
∴ d = 0.0518 m = 51.8 mm
@seismicisolation
@seismicisolation
Combined Stresses (Direct, Bending, Torsion)
•
355
If maximum tensile stress or principal stress is not to exceed 70 MN/m2 .
Me = 1208.39 Nm
π
Also Me = d 3 σmax
32
π
∴ 1208.39 = d 3 ×70 × 106
32
1208.39 × 32
3
d =
= 0.000176
π × 70 × 106
∴ d = 0.056 m
= 56 mm Ans
Hence, suitable diameter out of 51.8 and 56 mm is 56 mm. Ans
Exercise
17.1 A shaft 100 mm diameter carries a flywheel of mass 2 tonne at the centre of a simply supported span of 2.4 m. The centre of gravity of flywheel is 2.5 mm from the axis of rotation. If
the shaft transmits 750 kW at 300 r.p.m. estimate the maximum stress at the centre of gravity
(a) when centre of gravity is vertically below the axis and b) when it is vertically above the
axis.
[Ans a) 10.6 N/mm2 Tensile b) 7.33 N/mm2 Compressive]
17.2 Find the principal stresses in a propeller shaft and the inclination to the axis of the planes over
which they act. Given that: (a) the thrust results in an axis compressive stress of 15 MN/m2
and (b) the engine torque results in a shear stress at the shaft surface of 60 MN/m2 .
[Ans σmax = 52.97 MPa Tensile at 41◦ 21 , σmin = 67.97 MPa Compressive at 31◦ 26 ]
17.3 A rolled steel of I section having flanges 100 mm × 20 mm and web 260 mm × 10 mm is used
on a short column to carry a load of 100 kN. The load acts eccentrically 50 mm to the left
of axis YY passing through the centre of web and 60 mm above the axis XX passing through
the centroid of the section. Determine the maximum and minimum stress intensities induced
in the section.
[Ans 99.32 N/mm2 (Compressive), 69 N/mm2 (Tensile).]
17.4 A shaft of dia 100 mm is subjected to tensile load 157 kN, bending moment 6.28 Nm
and torque 9.42 kNm simultaneously. Calculate maximum tensile, compressive and shearing
stresses produced in the shaft material.
[Ans 106 MPa, 74.8 MPa, 63.8 MPa]
17.5 A horizontal beam, 3 m long, 50 mm wide, 75 mm deep, is simply supported at its left and
pin-jointed on its right end. It is subjected to a vertical downward concentrated load of 0.5
kN at its midspan and a horizontal thrust of 5 kN acting from left end to the right side, 20
mm below the longitudinal axis. Determine the stress at the extreme fibre at the midspan.
[Ans 6.67 N/mm2 compressive (top fibre) 5.67 N/mm2 tensile (bottom fibre)]
17.6 A solid shaft 127 mm diameter transmits 630 kW at 300 r.p.m. It is also subjected to a
bending moment of 9.1 kNm and an end thrust. If the maximum principal stress is limited to
77 N/mm2 , find the end thrust.
[Ans 31.223 kN]
@seismicisolation
@seismicisolation
356
•
Strength of Materials
17.7 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 located in the first
quadrant. Determine the maximum compressive and tensile stresses in the section.
[Ans 19.785 N/mm2 , −3.125 N/mm2 ]
17.8 A hollow rectangular column is having external and internal dimensions as 2.4 m × 1.8 m and
1.2 m × 1.2 m respectively. Calculate the safe load, that can be placed at an eccentricity of
0.5 m on a plane bisecting the longer side, if the maximum compressive stress is not to exceed
5 MN/m2 .
[Ans 720 kN]
17.9 A screw clamp is tightened on a workpiece exerting clamping force as 11 kN. Screw C
clamp is shown in Fig. 17.3. Find the maximum tensile and compressive stresses in the
material at section A − B due to bending and direct loading. Area of section = 480 mm2 ,
Ixx = 6.4 × 104 mm4 .
Workpiece
15 mm
90 mm
Centroid
40 mm
B
Section AB
A
Figure 17.3
[Ans
314.4 MN/m2 at top face, −414.4 MN/m2 (Compressive)]
@seismicisolation
@seismicisolation
18
C HAPTER
FIXED BEAMS
Fixed beams and continuous beams are also called indeterminate beams because reactions of these
cannot be determined from equations of static equilibrium, [ΣFx = 0, ΣFy = 0 and ΣM = 0]. Fixed
beams are also called built-in or encastre beams.
Fixed Beam with a Point Load at the Centre
A
W
C
x
B
x
MA
l2
l
D
Wl
4
+
A
A
C
MA
B
B
−
MB
E
E
Free bending moment
diagram
F
+
−
−−
A
Fixed bending moment diagram
F
B
Figure 18.1
Because of symmetry fixing moment at A and B are equal
∴
MA = MB
@seismicisolation
@seismicisolation
358
•
Strength of Materials
W
Wl
. (Because of symmetry, RA = RB = ; So maximum bending
Maximum bending moment is
4
2
Wl
W l
× =
)
moment =
2 2
4
Area of triangle ADB = Area of rectangle AEFB
1
Wl
×l×
= MA × l
2
4
Wl
∴ MA =
8
Wl
Due to symmetry, MA = MB =
8
Consider a point X, x from A,
Mx = EI
d2y
= RA · x − MA
dx2
d2y
= RA · x − MA
dx2
d2y W x W l
−
EI 2 =
2
8
dx
EI
Integrating,
EI
dy
W x2
W lx
=
−
+C1
dx
4
8
C1 is the constant of integration
dy
= 0,
When x = 0,
dx
Therefore, C1 = 0
The equation becomes,
EI
W x2
Wl
dy
=
−
x
dx
4
8
(i)
Equation (i) given the slope at any point.
Integrating Eqn. (i) again,
EI y =
W x2
W lx2
−
+ C2
12
16
where C2 is constant of integration,
When x = 0, y = 0 ∴ C2 = 0
Therefore,
EI y =
W x3
W lx2
−
12
16
@seismicisolation
@seismicisolation
(ii)
Fixed Beams
•
359
Equation (ii) deflection given at any point
For deflection at centre, substitute x = /2 in Eqn. (ii). Note that deflection at centre in this case
will be maximum.
W
EI ymax =
12
=
W l3
W l3
−
96
64
=−
∴
3
Wl l 2
l
−
2
16 2
ymax =
W l3
192
(minus sign for downward deflection)
W l3
192EI
It is interesting to compare this deflection with similar case of simply supported beam having
wl 3
which is four times more than beam with fixed ends.
load at centre which is
48EI
This shows that fixed beams are more stronger and stiffer.
Fixed Beam with Uniformly Distributed Load Over the Span
x
w/m
X
A
M
RA A
D
wl2
12 A
MB
X
RB
l
E
+
C
−
−
wl
.
2
Also due to symmetry, fixing moments,
MA = MB .
Maximum bending moment for simply supported beam,
Due to symmetry RA = RB =
B
wl2
8
B
Mmax =
Figure 18.2
Consider a point at X, x from A, using Macaulay’s method,
Mx = RA x − MA x0 −
EI
wx2
2
wx2
d2y
0
=
R
x
−
M
x
−
A
A
2
dx2
Integrating,
EI
dy wl x2
wx3
=
− MA x −
+C1
dx
2 2
6
C1 is the constant of integration.
@seismicisolation
@seismicisolation
wl 2
8
360
•
Strength of Materials
At x = 0,
dy
=0
dx
∴
C1 = 0
∴
EI
wx3
dy wlx2
=
− MA x −
dx
4
6
(i)
From Eqn. (i) slope at any point can be computed.
Integrating Eqn. (i) again,
EIy =
At x = 0, y = 0
∴
w l x3
x2 w x4
− MA −
+C2
12
2
24
C2 = 0
EIy =
w l x3 w x4 MA x2
−
−
12
24
2
(ii)
When x = l, y = 0
0=
∴
MA
l2 w l4 w l4
=
−
2
12
24
MA
l2 w l4
=
2
24
∴
Because of symmetry MA = MB =
w l 4 w l 4 MA l 2
−
−
12
24
2
MA =
w l2
12
w l2
12
Substituting for MA in Eqn. (ii)
w l x3 w x4 w l 4
−
−
12
24
24
1 w l x3 w x4 w l 2 x2
−
−
y=
EI 12
24
24
EIy =
∴
This equation will give deflection at any point.
l
For maximum deflection which will be at centre, i.e., x = , substituting for x in Eqn. (iii)
2
1
wl 4
wl 4
wl 4
ymax =
−
−
EI 12(2)3 24(2)4 24(2)2
1 wl 4 wl 4 wl 4
ymax =
−
−
EI 96
384
96
@seismicisolation
@seismicisolation
(iii)
Fixed Beams
∴
Hence,
ymax = −
ymax =
wl 4
384 EI
•
361
(minus for downward deflection)
wl 4
384 EI
By Area Moment Method
For fixed beam ΣA = 0 and ΣA x̄ = 0
2 w l2
×
×l +M×l = 0
3
8
∴
M=−
w l2
12
wm
C
l
w
−wl2
8
Free Moment Diagram
A1
M = −wl
2
Fixed Moment Diagram
A2
12
+
−
−
Total BM Diagram
wl2
24
−wl2
12
Figure 18.3
For maximum deflection at midpoint C,
1
[A1 x̄1 + A2 x̄2 ]
EI
2 wl 2 l
3 l
Ml l
1
×
×
+
×
=
EI
3 8
2
8 2
2
4
1 3wl 4 wl 4
−
=
EI 384
96
yc = ymax =
=−
wl 4
384 EI
ymax = yc =
wl 4
384 EI
Therefore, we find that the central deflection for a fixed beam having uniformly distributed load ‘w’
is one-fifth of the central deflection of the simply supported beam.
@seismicisolation
@seismicisolation
•
362
Strength of Materials
Deflection for a Fixed Beam with Concentrated Load Anywhere on the Span
a
A
MA
W
b
x
x
C
MB
x
l
RA
Wb
l
Wab
So free bending moment at C =
l
For free beam reaction at A =
B
RB
Wab
l
+
Free B.M.D
−
Ma
−
Mb
+
−
Figure 18.4
At x from A, (using Macaulay’s method), bending moment,
Mx = RA x − MA −W [x − a]
EI
d2y
= RA x − MA −W [x − a]
dx2
Integrating,
EI
x2
W
dy
= RA − MA x − [x − a]2 +C1
dx
2
2
Integrating again,
EIy = RA
When x = 0, y = 0,
x3
x2 W
− MA − [x − a]3 +C1 x +C2
6
2
6
dy
=0
dx
∴
C1 = C2 = 0
∴
1
dy
=
dx EI
and
1
y=
EI
x2
W
RA − MA x − [x − a]2
2
2
x3 MA x2 W
RA −
− [x − a]3
6
2
6
@seismicisolation
@seismicisolation
(i)
(ii)
Fixed Beams
•
363
dy
=0
dx
Substituting in Eqns. (i) and (ii)
At x = l, y = 0,
W
l2
− MA l − b2
2
2
3
2
l
MA l
W
0 = RA −
− b3
6
2
6
0 = RA
and
Simplifying the above two equations
0 = RA l 2 − 2MA l −W b2
and
0 = RA l 3 − 3MA l 2 −W b3
Solving further,
RA =
W b2 (l + 2a)
l3
(iii)
Wab2
l2
(iv)
and
MA =
For static equilibrium,
RA +RB −W = 0
(v)
RA l −W (l − a) − MA +M B = 0
(vi)
and
From Eqns. (iv) and (vi),
Wa2 b
MB =
l2
For deflection under load, etc., using Eqn. (ii), putting x = a
a3
a2
1
RA − MA
yc =
EI
6
2
1 W b2 (l + 2a)a3 Wa3 b2
=
−
EI
6l 3
2l 2
Simplifying,
yc = −
or yc =
Wa3 b3
3E I l
Wa3 b3
3E I l
@seismicisolation
@seismicisolation
364
•
Strength of Materials
By Moment Area Methods:
W
a
A
b
B
l
MA
MB
A1
Wab
l
A
+
+
Free bending moment diagram
+
B
2a
3
l
2
(a + 3b
)
A2
−
MA
MB
Fixed bending moment diagram
l
3
Figure 18.5
We know A1 + A2 = 0
MA +M B
1 W.a.b
·
·l −
×l = 0
2
l
2
Wab
or MA +M B =
l
Now
A1 x̄1 + A2 x̄2 = 0
1 Wab
2a
1 Wab
b
·a ·
+
·
b
a+
2 l
3
2
l
3
l
l
l
− MB · l ·
− (MA − MB )
× =0
2
2
3
3
l2
l2
l2
Wa b Wa2 b2 Wab3
+
+
− MB · − MA · + MB ·
=0
or
3l
2l
6l
2
6
6
or
or
2Wa3 b + 3Wa2 b2 +Wab3 − 2MB l 3 − MA l 3 = 0
1
2Wa3 b + 3Wa2 b2 +Wab3
l3
Wab
= 3 2a2 +3ab + b2
l
MA +2M b =
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(i)
Fixed Beams
•
Wab 2
(a
+
b)
+a
(a
+
b)
l3
Wab
or MA + 2Mb = 3 (a + b) (a + b + a)
l
Wab
= 3 × l × (l + a)
l
Wab
or MA + 2Mb = 2 (l + a)
l
365
=
(ii)
By solving Eqns. (i) and (ii)
Wab2
l2
Wa2 b
MB = 2
l
MA =
E XAMPLE 18.1: A fixed beam of 8 m span is loaded with point loads of 160 kN at a distance
of 2.5 m from each support. Draw the bending moment and shear force diagrams. Also find the
maximum deflection.
Take E = 200 GN/m2 and I = 12 × 109 mm4 .
S OLUTION :
160 kN
160 kN
For simply supported beam
C
A
RA = RB = 160 kN
MA = 160 × 2.5
= 400 kNm
D
3
2.5 m
MA
8m
400 kNm
B
2.5 m
MB
400 kNm
Free bending moment diagram
+
A
B
−
MA
MB = MA = 275 kNm
Fixed bending moment diagram
+
125 kNm
−
−
275 kNm
160 kN
160 kN
SF Diagram
Figure 18.6
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366
•
Strength of Materials
By equating the areas of free and fixed bending moment diagram
400
MA × 8 = (8 + 3)
2
2200
kNm = 275 kNm
∴ MA =
8
Bending moment at centre = 400 − 275 = 125 kNm
Points of Contraflexure:
Bending moment (actual) at any section in AC at a distant x from A,
M = Free moment − Fixed moment
= 160x − 275
At point of contraflexure M should be zero
∴
160x − 275 = 0
275
= 1.72 m
or x =
160
from either end due to symmetry.
The bending moment at any section between A and D at a distance of x from A is given by,
EI
d2y
= 160x − 275 − 160 (x − 2.5)
dx2
Integrating,
EI
dy
=0 ∴
dx
Integrating again,
When x = 0,
dy
x2
160
= 160 − 275x −
(x − 2.5)2 +C1
dx
2
2
C1 = 0
EIy
160x3
x2 160
− 275 −
(x − 2.5)3 +C2
6
2
6
When x = 0, y = 0, ∴ C2 = 0
For maximum deflection which will occur at the centre in this case, putting x = 4 m
160 (4)3
(4)2 160
− 275
−
(4 − 2.5)3
6
2
6
= 1706.7 − 2200 − 90
= −5833
5833
5833
=−
ymax = −
EI
200 × 106 × 12 × 109 × 10−12
= −0.00243 m
= −2.43 mm
= −2.43 mm Ans
EIymax =
minus sign indicates downward deflection.
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Fixed Beams
•
367
E XAMPLE 18.2: A fixed beam of 5 m span carries point loads of 120 kN and 85 kN as shown in
Fig. 18.7. Find the following:
1. Fixing moments at the ends
2. Reactions at the supports
Also draw bending moment and shear force diagrams.
S OLUTION :
85 kN
120 kN
C
A
D
2m
1.5 m
MA
B
MB
1.5 m
5m
164.25 kNm
143.25 kNm
+
Free bending moment diagram
−
Fixed bending moment diagram
100.275 kNm
115 kNm
+
−
−
7.55
Final bending moment diagram
92.5 kN
7.55
112.45 kN
SFD
Figure 18.7
For free reactions:
Taking moments about B,
RA × 5 = 120 × 3.5 + 85 × 1.5
RA = 109.5
RB = (120 + 85) − 109.5
= 95.5 kN
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•
Strength of Materials
MC = 109.5 × 1.5 = 164.25
MD = 109.5 × 3.5 − 120 × 2 = 143.25 kNm
Fixing moments:
w a b2
l2
120 × 1.5 × 3.52 85 × 3.5 × 1.52
=
+
52
52
= 88.2 + 26.8
= 115 kNm Ans
MA = ∑
w a2 b
l2
85 × 3.52 × 1.5 120 × 1.52 × 3.5
=
+
52
52
= 62.475 + 37.8
= 100.275 kNm Ans
MB = ∑
Reaction (R) at each support due to end moments alone,
115 − 100.275
5
= 2.945
R=
Since MA > MB ,
The reaction R at A is upward and at B is downward,
Final reaction at A
= R f A = 109.5 + 2.945 = 112.45 kN
R f B = 95.5 − 2.945 = 92.5 kN Ans
Ans
E XAMPLE 18.3: A fixed beam of 6.5 m span is subjected to a concentrated couple of 250 kNm
applied at a section in from the left hand. Find the end moments by Macaulay’s method. Also draw
bending moment and shear force diagrams.
S OLUTION :
MA
250 kNm
B
A
C
4m
Figure 18.8
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2.5 m
Fixed Beams
•
369
Let us assume directions of fixing moment and reaction as shown in Fig. 18.8. If after calculations, these turn out to be negative, then our assumed direction will be wrong.
The bending at any section at a distance x from end A is given by
d2y
= RA x+ MA + 250
dx2
Now to facilitate application of Macaulay’s method is arranged as given below:
EI
d2y
= RA x+ MA + 250 (x − 4)0
dx2
(i)
x2
dy
= RA + MA x + 250 (x − 4) +C1
dx
2
(ii)
EI
Integrating,
EI
dy
= 0, ∴ C1 = 0
dx
Integrating again, we here
when x = 0,
EIy = RA
x3
x2 250
+ MA +
(x − 4)2 +C2
6
2
2
(iii)
when x = 0, y = 0, ∴
C2 = 0
dy
=0
when x = 6.5 m,
dx
Substituting these values in Eqn. (ii),
(6.5)2
+ 6.5 MA + 250 (6.5 − 4)
2
0 = 21.1 RA + 6.5 MA + 625
MA + 3.25 RA = −96.15
0 = RA
(iv)
At B, the deflection is zero
∴ When x = 6.5 m, y = 0
Substituting these values in Eqn. (iii), we get
(6.5)3
(6.5)2
+ MA ×
+ 125 (65.4)2
6.5
2
0 = 42.25 RA + 21.12 MA + 781.25
42.25 RA + 21.12 MA = −781.25
0 = RA ×
Solving Eqns. (iv) and (v),
42.25 RA + 21.11 (−96.15 − 3.25 RA ) = −781.25
42.25 RA − 2030 − 68.61 RA = −781.25
−26.36 RA = 1249
or RA = −47.38 kN
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(v)
370
•
Strength of Materials
−ve sign shows our assumed direction was wrong.
Substituting the value of RA in Eqn. (iv),
3.25 (−47.28) + MA = −96.15
∴ MA = 57.85 kNm
Ans
Now for fixed beam, bending moment diagram and shear force diagrams are shown below.
MA = 57.85 kNm
250 kNm
A
4.0 m
47.38 kN
B
.
C
2.5 m
47.38 kN
Fixed beam
118.5 kNm
57.85 kNm
+
+
Bending moment diagram
−
131.7 kNm
47.38 kN
47.38 kN
SFD
Figure 18.9
Bending moment calculations:
MA = 57.83 (sagging)
just before C, Mc = 57.83 − 47.38 × 4 = −131.7 kNm (hogging)
just after C, Mc = −131.7 + 250 = 118.5 (sagging)
MB = 57.83 − 47.38 × 6.5 + 250 = 0
E XAMPLE 18.4: A built-up beam of span AB, 5 m carries a concentrated load of 70 kN at 1.3 m
and 130 kN at 3.6 m from A.
Determine the fixing end moments and draw shear force and bending moment diagrams.
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Fixed Beams
S OLUTION :
70 kN
A
MA
130 kN
D
C
B
1.3 m
2.3 m
1.4 m
MB
3.6 m
Figure 18.10
By principle of superposition, fixing end moments are:
wab2
wab2
MA =
+
l2
l2
70
130
=
70 × 1.3 × 3.72 130 × 3.6 × 1.42
+
52
52
= 49.83 + 36.69
= 86.52 kNm Ans
2 2 wa b
wa b
MB =
+
l2
l2
70
130
=
70 × 1.32 × 3.7 130 × 3.62 × 1.4
+
52
52
= 17.51 + 94.35
= 111.86 kNm
Ans
For reactions at supports:
Taking moments about B,
5 RA = 70 × 3.7 + 130 × 1.4
5 RA = 259 + 182
441
=
5
= 88.2 kN
RA + RB = 70 + 130 = 200 kN
∴ RB = 200 − 88.2
= 111.8 kN
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•
371
372
•
Strength of Materials
70 kN
A
C
B
D
1.3 m
86.52
130 kN
2.3 m
RA
114.66 kNm
1.4 m
156.52
111.86
RB
112.86 kNm
86.52 kNm
83.13 kN
BMD
13.13 kN
SFD
116.87 kN
Figure 18.11
Bending moment calculations:
Bending moment at A and bending moment at B = 0
Mc , bending moment at C = 88.2 × 1.3
= 114.66 kNm
MD , bending moment at D = 111.8 × 1.4
= 156.52 kNm
Support reactions:
Taking moments about B,
RA × 5 + 111.86 = 70 × 3.7 + 130 × 1.4 + 86.52
= 527.52
RA = 83.13 kN
RA + RB = 70 + 130 = 200
RB = (70 + 130) − 83.13 = 116.87 kN
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•
373
Shear force calculations:
Shear force at A = 0 rising to 83.13 kN
Shear force at C = 83.13 falling to 83.13 − 70 = 13.13 kN
Shear force at D = 13.13 falling to 13.13 − 130 = 116.87 kN
Shear force at B = 116.87 rising to 116.87 − 116.87 = 0
E XAMPLE 18.5: Find reactions and fixing moments at fixed ends and draw shear force and bending moment diagrams for the fixed beam shown in Fig. 18.12. Use Macaulay’s method.
2 kN
2 kN
1.5 kN/m
B
A
1m
2m
C
1m
D
Figure 18.12
For solving the problem extend uniformly distributed load up to D and to cancel the extended
load (between C and D), let there be an upward uniformly distributed load from C to D. Now
2 kN
3 kN
1.5 kN/m
D
B
A
C
MA
MD
1.5 kN/m
RA
RD
1m
3.66
2m
1m
Shear force diagram
1.66 kN
E
x'
1.34 kN
1.52 kN
4.34 kN
0.6 kNm
0.92 kNm
Bending moment diagram
3.06 kNm
3.42 kNm
Figure 18.13
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374
•
Strength of Materials
EI
d2y
1.5(x − 1)2 1.5(x − 3)2
◦
+
− 3(x − 3)
=
R
x
−
M
x
−
2(x
−
1)
−
A
A
2
2
dx2
Integrating twice, we get
dy RA x2
1.5(x − 1)3 1.5(x − 3)3 3
=
− MA x − (x − 1)2 −
+
− (x − 3)2 + A
dx
2
6
6
2
3
2
3
4
4
MA x
(x − 1)
(x − 1)
(x − 3)
(x − 3)2
RA x
−
−
−
+
−
+C1 x +C2
EIy =
6
2
3
16
16
2
EI
dy
= 0 at x = 0 and at x = 4
dx
We have C1 = 0
We know that,
∴
0 = 8 RA − 4 MA − 9 −
27 1 3
+ −
4
4 2
8 RA − 4 MA = 17
(i)
y = 0 at x = 0 giving C2 = 0
and y = 0 at x = 4 giving
0=
64RA
1
1
81
− 8 MA − 9 −
+
−
6
16 16 2
Which comes to:
64RA − 48 MA = 87
Solving Eqns. (i) & (ii),
RA = 3.66 kN Ans
MA = 3.06 kNm Ans
RB = 2 + 3 + 3 − RA = 8 − 3.66 = 4.34 kN Ans
MB = −3.06 + 3.66 × 1 = +0.6 kNm Ans
MC = −3.06 + 3.66 × 3 − 2 × 2 − 3 × 1 = 0.92 kNm
MD = −3.06 + 3.66 × 4 − 2 × 3 − 3 × 2 − 3 × 1
= −3.42 kNm
Bending moment where symmetry fixing is zero (ME ): Due to similar triangles
x
2 − x
=
1.66
1.34
On solving this equation, we have,
@seismicisolation
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(ii)
•
Fixed Beams
375
As shear force is zero, x = 1.11 m
∴
x
2
= −3.06 + 3.66 × 2.11 − 2 × 1.11 − 0.75(1.11)2
= 1.52 kNm Ans
ME = −MA + RA (1 + x ) − 2 × x = 1.5x ×
Exercise
18.1 A fixed beam of 6 m span is loaded with point loads of 150 kN at a distance 2 m from each
support. Draw bending moment and shear force diagrams. Find also the maximum deflection.
Take E = 2 × 108 kN/m2 and I = 8 × 108 mm4 .
[Ans Fixing moments: 200 kNm, 200 kNm, Ymax = 1.56 mm]
18.2 A fixed beam of 6 m span carries point loads of 100 kN and 75 kN as shown in Fig. 18.14.
Find:
i) Fixing moments at the ends
ii) Reactions at the supports
100 kN
C
75 kN
D
A
B
2m
2m
2m
Figure 18.14
18.3
18.4
18.5
18.6
Draw bending moment and shear force diagrams.
[Ans MA = 122.22 kNm, MB = 111.11 kNm, RA = 93.52 kN, RB = 83.33 kN]
A fixed-ended beam of span 9 m carries a uniformly distributed load of 15 kN/m and two
equally concentrated loads, of 200 kN at the 3 m from left hand and another 3 m from right
hand. Find the fixing moments and the deflection at the centre. EI = 210 MN/m2 .
[Ans 392.75 kNm, yc = 6.835 mm]
A horizontal I beam, rigidly built in at the ends, and 7.5 m long carries a total uniformly
distributed load of 100 kN as well as a concentrated load of 40 kN. If the bending stress is
limited to 75 MN/m2 and the deflection must not exceed 2.5 mm, find the depth of section
required. E = 200 GN/m2 .
[Ans 594 mm]
A horizontal beam built-in each at end has a clear span of 4.5 m and carries loads of 50 kN
at 1.5 m and 70 kN at 2.5 m from left hand and claculate the fixing moment and the position
and amount of the maximum bending moment.
[Ans 67.5 kNm, 60.5 kNm, 67.5 kNm at left hand end]
A built-in beam of 5 m span carries a uniformly distributed load of 30 kN/m extending from
one support to the centre of the span. If the moment of inertia of the section is 200 × 10−6 m4 ,
Calculate: (a) the end fixing moments; (b) the end reactions and (c) the position and magnitudes of the maximum deflection. Sketch the shear force and bending moment diagrams.
E = 200 GN/m2 .
[Ans 19.53 kNm; 43 kNm; 14.06 kNm; 60.94 kN, 3.15 mm at 2.8 m from unloaded end]
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•
Strength of Materials
18.7 A fixed beam of 6 m span carries uniformly distributed unloaded of 10 kN/m for a distance 4
m from the left hand end. Find the fixing moments at the supports.
[Ans 26.66 kNm; 17.77 kNm]
18.8 A fixed beam of 6 m span supports two point loads of 300 kN each at 2 m from each end.
Find the fixing moments at the ends and draw the shear force and bending moment diagrams.
Also find the central deflection. Take I = 9 × 108 mm4 and E = 200 GPa.
[Ans 400 kNm; yc = 2.78 mm]
18.9 A fixed beam of 6 m span carries a uniformly distributed load of 20 kN/m run over the right
half and 30 kN/m over the right half and a concentrated load of 40 kN at the centre of the
span. Calculate the fixed end moments. Assume uniform flexural rigidity.
[Ans 93.375 kNm; 110.625 kNm]
18.10 A 250 mm × 112.5 mm steel beam, I = 47.6 × 10−6 m4 , is used as a horizontal beam with
fixed ends and a clear span of 3 m. Calculate from the first principles the load which can be
applied at one-third span if the bending stress is limited to 120 MN/m2
[Ans 103 kN]
18.11 A built-in beam of span 10 m carries a uniformly distributed load of 15 kN/m on the entire
span along with two point loads each of 200 kN at a distance of 3 m and 7 m from left end.
Find the fixing moments and the deflection at the centre. Take EI = 210 MN/m2
[Ans 545 kNm; 8.29 mm]
18.12 A girder of span 5 m is fixed at each end and carries two point loads of 90 kN and 120 kN
placed at 1 m and 2 m from the left support respectively. Find the reactions and fixing
moments at the ends. E = 200 GPa, I = 8 × 107 mm4 . Also find the deflections at the load
points.
[Ans 158.36 kN; 51.64 kN, 144 kNm, 72 kNm y under 90 kN = 2.84 mm
y under 120 kN = 5.741 mm]
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19
C HAPTER
CONTINUOUS BEAMS
When a beam rest on more than two supports, it is called continuous beams. Their applications are
in long bridges, buildings, etc. The continuous beam is statically indeterminate and can be analysed
by various methods, for example, by the theorem of three moments or the Clapeyron’s theorem
method.
Clapeyron’s Equation of Three Moments
A
C
B
l1
l2
x2
x1
a1
(a)
A
x
D
(b)
MA
A x′
(c)
MA
A
MxC.G.
dx
Mx′
dx
a2
C.G.
a′1
Bending moment diagram
due to vertical loads
C
B
E
a2′
MB
F
MC
B
C
Bending moment diagram
due to support moments
MB1
MC
B
Resultant bending moment diagram
C
Figure 19.1
The beam ABC may be extended both ways but we are considering only portion ABC.
Span AB: Now consider the span AB and let Mx be the bending moment due to vertical loads at a
distance x as shown. Mx is positive, being sagging moment, and Mx is the bending moment due to
support moments at a distance x from support A. This is negative because of hogging effect.
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378
•
Strength of Materials
Hence, net bending moment at x is given by,
EI
d2y
= Mx − Mx
dx2
Integrating from zero to l1 after multiplying by x (both sides)
EIx
l
EI
0
d2y
= xMx − xMx
dx2
d2y
x 2 dx =
dx
l1
x Mx dx −
0
l1
xMx dx
0
l1
dy
EI x − y = a1 x1 − a1 x1
dx
0
(i)
Whereas a1 = area of BMD due to vertical loads
Also Mx .dx = Area of bending moment diagram of length dx and, x Mx dx = moment of area of
bending moment diagram of length dx about A.
l1
x Mx dx = a1 x1
Hence,
and similarly
0
l1
xMx dx = a1 x1
0
Equation (i) becomes
EI
dy
dy
l1
− yB − 0 ×
− yA
dx atB
dx atA
= a1 x1 − a1 x1
EI [(l1 θB − yB ) − (0 − yA )] = a1 x1 − a1 x1
dy
Because
at B = θB . And deflections at A and B are zero, therefore yA = 0 and yB = 0. So the
dx
above equation becomes:
@seismicisolation
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Continuous Beams
EI θB = a1 x1 − a1 x1
•
379
(ii)
a1 = area of bending moment diagram due to supports moments
= Area of trapezium ABED
=
1
(MA + MB ) × l1
2
x = distance of centre of gravity of area ABED from A
l1 1
2l1
+ (MB − MA )l1 ×
2 2
3
1
MA l1 + (MB − MA ) l1
2
MA l1 .
=
l1
l1
+ (MB − MA ) ×
2
3
1
MA + (MB − MA )
2
MA .
=
3MA l1 + 2l1 (MB − MA )
6
=
2MA + MB − MA
2
l1
[3MA + 2MB − 2MA ]
= 3
MB + MC
MA + 2MB l1
=
MA + MB 3
Substituting the values of a1 and x1 in Eqn. (ii), we have
1
MA + 2MB l1
EIl1 θB = a1 x1 − (MA + MB )l1 ×
2
MA + MB 3
= a1 x1 −
6EIl 1 θB =
l12
(MA + 2MB )
6
6a1 x1
− l1 (MA +2M B )
l1
(iii)
Span BC: Likewise considering the span BC with origin C and x positive to the left, it can be shown
that
6a2 x2
6EI (−θB ) =
− l2 (MC +2M B )
l2
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•
Strength of Materials
Slope at B in span BC will be obviously of opposite sign, i.e., negative because of opposite directions.
Therefore, the above equation becomes as
−6EI θ B =
6a2 x2
− l2 (MC +2M B )
l2
(iv)
If we add Eqn. (iii) and (iv), then
0=
6a1 x1
6a2 x2
−l1 (MA +2M B ) +
−l 2 (MC +2M B )
l1
l2
0=
6a1 x1 6a2 x2
+
−l 1 MA +2l 1 MB −l 2 MC +2l 2 MB
l1
l2
= l1 MA +l 2 MC +2M B (l1 +l 2 ) =
MA l1 +2M B (l1 +l 2 ) +MC l2 =
6a1 x1 6a2 x2
+
l1
l2
6a1 x1 6a2 x2
+
l1
l2
This is known as Clapeyron’s equation of three moments.
E XAMPLE 19.1: A continuous beam ABC is shown in Fig. 19.2, find the support moments at A, B
and C if A and C are simply supported. Also draw bending moment and shear force diagrams.
12 kN/m
9 kN/m
A
5m
B
C
7m
Figure 19.2
S OLUTION :
Because the ends A and C are simply supported, the support moments at A and C will be zero.
∴
MA = MC = 0 Ans.
Using Clapeyron’s equation of three moments to find support moment at B, we get
MA l1 + 2MB (l1 + l2 ) + MC l2 =
6a1 x1 6a2 x2
+
l1
l2
0 + 2MB (5 + 7) + 0 =
6a1 x1 6a2 x2
+
5
7
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(i)
Continuous Beams
For maximum bending moment between A and B,
RA = RB =
9×5
2
=
45
= 22.5 kN
2
MABmax = 22.5 × 2.5 − 22.5 ×
2.5 2.5
×
2
4
= 56.25 − 17.6
= 38.65 kNm
For maximum bending moment between B and C,
RB = RC =
12 × 7
2
=
84
= 42 kN
2
7
3.5
MBCmax = 42× −12 × 3.5×
2
2
= 147 − 73.5
= 73.5 kNm
12 kN/m
9 kN/m
A
C
B
5m
7m
59.27
73.5 kNm
A
38.65
Bending moment diagram
C
B
50.47
18.15 kN
Shear force diagram
26.85 kN
53.53 kN
Figure 19.3
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•
381
382
•
Strength of Materials
Further calculations for shear force diagram:
18.15 − 9 × 5 = 26.85 kN,
50.47 = 77.32 − 26.85
The bending moment diagram on a simply supported beam carrying uniformly distributed load
2
will have area = × span × altitude.
3
So, now
2
a1 = ×5×38.65 = 129.48
3
5
x1 = =2.5 m
2
2
a2 = ×7×73.5=344.7
3
7
x2 = = 3.5 m
2
Substituting these values in Eqn. (i)
2M B (5 + 7) =
6 × 129.48 × 2.5 6 × 344.7 × 3.5
+
5
7
24 MB = 388.44 + 1034.1
MB = 59.27 kNm
Ans
Shear force diagram calculations:
For span AB, taking moments about B,
5
RA ×5 − 12 × 5× = MB = −59.27 (MB is −ve because t is hogging)
2
5RA = −59.27+150
∴
RA = 18.15 kN
Similarly for span BC, taking moments about B,
RC × 7 − 12 × 7 ×
7
= −59.27
2
∴ RC = 33.53 kN
RB = Total load on ABC − (RA + RC )
= (9 × 5 + 12 × 7) − (18.5 + 33.53)
= 129 − 51.68
= 77.32 kN
@seismicisolation
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Continuous Beams
•
383
E XAMPLE 19.2: A continuous beam ABCD of length 12 m resting on four supports covering
3 equal spans carries a uniformly distributed load of 2 kN/m length. Calculate the moments and
reactions at the supports. Draw the shear force and bending moment diagrams.
S OLUTION :
2 kN/m
A
B
4m
RA
RB
4m
RC
RD
4 kNm
4 kNm
B
A
D
C
4m
4 kNm
C
3.201
kNm
D
3.201
kNm
A
B
C
A
B
C
D
Bending moment diagram
due to vertical loads
Bending moment diagram
due to support moments
Resultant bending moment diagram
4 kN
3.2 kN
A
4.8 kN
B
C
4 kN
4.8 kN
D
SF diagram
3.2 kN
Figure 19.4
Beam being simply supported, the moments at A and D are zero.
MA = 0
By symmetry
and MD = 0
MB = Mc
For finding the support moments at B and D, Clapeyron’s equation of three moments is applied for
ABC and BCD. For ABC:
MA l1 + 2MB (l1 + l2 ) + MC .l2 =
6a1 x1 6a2 x2
+
l1
l2
6a1 x1 6a2 x2
+
4
4
6a1 x1 6a2 x2
16MB + 4MC =
+
l1
l2
0 × 4 + 2MB (4 + 4) + MC × 4 =
16MB + 4MC =
@seismicisolation
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6
(a1 x1 + a2 x2 )
4
(i)
384
•
Strength of Materials
a1 = Area of bending moment diagrams due to uniformly distributed load on AB.
2
× AB × altitude of parabola
3
(4 × 2)
2
Altitude of parabola =
× 2 − 2 × 2 × = 8 − 4 = 4 kNm
2
2
= Maximum bending moment of free bending moment diagram.
=
∴
2
× 4 × 4 = 10.67
3
4
x2 = = 2 m
2
a1 =
Owing to symmetry, a2 = a1 = 10.67
and x2 = x1 = 2 m
Putting the values is Eqn. (i),
16M B + 4MC =
6
(10.67 × 2 + 10.67 × 2)
4
Now MB = MC due to symmetry
16MB + 4MB = 1.5 (21.34 + 21.34)
20MB = 64.02
MB = 3.201 kNm = MC
Ans
Now,
MA = 0, MD = 0, MB = MC = 3.201 kNm
In bending moment diagram considering AB, BC and CD as simply supported, the parabolas have
altitudes 4 kNm each.
For support reactions:
Due to symmetry RA = RD
RB = Rc
For the span AB, taking moments about B, we get
MB = RA × 4 − 2 × 4 ×
4
2
Since MB in hogging moment so giving it negative sign,
−3.201 = 4RA − 16
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Continuous Beams
•
385
4RA = 16 − 3.201
∴
RA = 3.2 kN
Ans
Due to symmetry, RD = RA = 3.2 kN
Ans
Now
RA + RB + RC + RD = Total load on ABCD
= 2 × 12
2(RA + RB ) = 24
RA + RB = 12
∴
RB = 12 − RA
= 12 − 3.2 = 8.8 kN
RB = RC = 8.8 kN
Ans
Now bending moment and shear force diagrams are drawn as shown in Fig. 19.4.
For shear force at B = 3.2 − 4 × 2 = −4.8 kN
8.8 − 4.8 = 4 kN
and
For shear force at C = 4 − 4 × 2 = −4 kN
and
8.8 − 4 = 4.8 kN
E XAMPLE 19.3: A continuous beam ABC consists of two spans AB of length 4 m and BC of
length 3 m. The span AB carries a concentrated load of 100 kN at its midpoint. The span BC carries
a point load of 120 kN at 1 m from C. Find: (a) moments at the supports and (b) reactions at the
supports.
(Poona University)
S OLUTION :
100 kN
A
120 kN
B
C
2m
3m
4m
1m
80 kNm
100 kNm
O
x1= 2
G
x2= 1.33
O
O Free bending moment
diagram
O
65.66 kNm
Figure 19.5
@seismicisolation
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386
•
Strength of Materials
WP
= 100 kNm
4
1
a) Area of bending moment diagram for span AB = a1 = × 100 × 4 = 200 kNm2
2
4
Due to symmetry x1 = = 2
2
1
Area of bending moment diagram for span BC = a2 = × 80 × 3 = 120 kNm2
2
The bending moment under the load 100 kN =
x2 =
l +b 3+1
=
= 1.33 m
3
3
Let MA , MB & MC be the support moments at A, B and C, respectively.
MA l1 +2M B (l 1 +l 2 ) + MC l2 =
6a1 x1 6a2 x2
+
l1
l2
Being simply supported MA = MC = 0, we have
200×2 120×1.33
+
2MB (4 + 3) = 6
4
3
14MB = 6 (100 + 53.2)
MB =
6×153.2
14
= 65.66 kNm Ans
b) Considering the span AB simply supported and taking moments about B,
RA × 4 − 100 × 2 + 65.66 = 6
200 − 65.66
4
RA = 35.58 kN Ans
RA =
∴
Similarly, taking span BC as simply supported and taking moments about B, we get
RC × 3 − 120 × 2 + 65.66 = 0
RC =
∴
Because
240 − 65.66
3
RC = 58.12 kN Ans
RA + RB +RC = 100 + 120
∴
RB = 220 − 35.58 − 58.12
= 126.3 kN Ans
@seismicisolation
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Continuous Beams
•
387
Exercise
19.1 A continuous beam ABC 8 m long consists of two spans AB = 3 m and BC = 5 m. The span
AB carries load of 50 kN/m while the span BC carries a load of 10 kN/m. Find:
(i) Support moment at B
(ii) Reactions at the supports
[Ans MB = 179.7 kNm, RA = 48.4 kN, RB = 192.5 kN RC = 59.1 kN]
19.2 A continuous beam ABCD having three equal spans of length l each. It carries a uniformly
distributed load w/unit length over its entire length. It is freely supported on all supports,
which are at the same level.
Draw the bending moment and shear force diagrams for this beam.
Ans
MB = MC = −
wl 2
2wl
11wl
, MA = 0, RD =
= RA , RB =
= RC
10
5
10
2 2 2
wl at l from A
25
5
wl 2
or D respectively. Maximum bending moment in span BC =
40
The maximum bending moment in span AB or CD =
19.3 A continuous beam 12 m long supported over spans AB = BC = CD = 4 m carries a uniformly
distributed load of 3 kN/m run over span AB, a concentrated load of 4 kN at a distance of 1 m
from point B on support BC and a load of 3 kN at the centre of the span CD. Find: i) support
moments and ii) support reactions
[Ans MA = 0, MB = −4.05 kNm, MC = −1.05 kNm
RA = 4.99 kN, RB = 10.76 kN, RC = 2.01 kN, RD = 1.24 kN]
19.4 A continuous beam ABC 8 m long consists of two spans AB = 3 m and BC = 5 m. The span
AB carries a load of 50 kN/m while the BC carries a load of 10 kN/m Find:
i) Bending moment at B
ii) Reactions at the supports
[Ans MB = 179.7 kNm, RA = 48.4 kN, RB = 192.5 kN, RC = 59.1 kN]
@seismicisolation
@seismicisolation
C HAPTER
20
SPRINGS
Springs are of various types for serving different purposes. Some types are listed below:
1. Helical springs:
a) Close-coiled helical springs
b) Open-coiled helical springs
c) Torsion helical springs
d) Compression helical springs
2.
3.
4.
5.
Flat springs
Belleville springs
Leaf or laminated springs
Torsion springs
Springs are usually made of high carbon steel (0.7 to 1.0% carbon) or medium carbon alloy
steels. Sometimes phosphor–bronze or 18 Cr/ 8 Ni stainless steel are used for better corrosion resistance.
These are used for railway carriages, cars, motorcycles, etc., for absorbing the shocks. Whereas
sometimes they find use in brakes or clutches or spring balances, etc.
Close-coiled helical spring with Axial load: If helix angle is 10◦ or less, the helical spring is called
close-coiled helical spring. Fig. 20.1 shows a close-coiled spring under an axial load w.
D = mean coil diameter
d = wire diameter
n = number of coils
l = length of wire π Dn
δ = axial deflection.
Figure 20.1
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Springs
•
389
If may be noted that it is assumed to be subjected to torsion only. The effects of bending and
direct shear are being negligible.
D
The torque on the wire of all sections = W ×
2
T
τ
As we know =
J
R
T.R
∴ Maximum shear stress, τ =
J
τ=
W × D/2 × d/2
π 4
d
32
=
8W D 16W R
=
π d3
π d3
(i)
The twist of one end of the wire relative to the other end is given by
or
∴
θ=
T.l
CJ
θ=
W D/2 × π Dn
π
C × d4
32
=
Cθ
T
=
J
l
16W D2 n
Cd 4
Since the axial movement of the free end = θ ×
D
2
i.e., δ =
8W D3 n
Cd 4
(ii)
Also it is possible to derive this formula this way:
Work done by the load = strain energy in the wire
1
1
Wδ = Tθ
2
2
1
T ·l
= T×
2
C·J
=
T 2l
2CJ
@seismicisolation
@seismicisolation
∵
T
Cθ
=
J
l
390
•
Strength of Materials
Substituting for T and J,
WD 2
×l
T 2l
2
=
π
2CJ
2C d 4
32
4W 2 D3 n
=
Cd 4
1
2
As this is equal to
1
W δ so equating,
2
4W 2 D3 n
1
Wδ =
2
Cd 4
8W D3 n
δ=
same as proved in Eqn. (i).
Cd 4
In terms of mean radius R which is D/2,
8W (2R)3 n
Cd 4
64W R3 n
=
Cd 4
δ=
Wahl’s Correction Factor
While deriving the above equations, the effect of curvature of spring and direct shear stress is
neglected.
8W D 16W R
=
is modified to include these effects by introducing a factor K called
π d3
π d3
Wahl’s correction factor, so that
16 W R
τ =
K
π d3
Equation τ =
where K is found experimentally and is given by
K=
4S − 1 0.615
+
4S − 4
S
where
S=
D
= Spring index.
d
Total strain energy of the spring:
U=
1
Tθ
2
@seismicisolation
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Springs
•
391
τ Cθ
T
= =
J
r
l
π c
τl
1 τ 32 d
×
U=
2 d/2
d/2C
Using
=
π τ 2d2l
16 C
1 τ2 π
= . × d2l
4 C
4
τ2
× volume of the spring wire
4C
Stiffness of the spring is defined as force per unit deflection,
So strain energy, U =
Stiffness
=k=
W
W
Cd 4
=
=
δ
64 R3 n
64W R3 n
4
Cd
Close-coiled Helical Spring with Axial Couple
Figure 20.2 shows a close-coiled helical spring under an axial couple M.
The bending moment on the wire at all sections = M
M
∴ Maximum bending stress, σ =
Z
M
σ= π
d3
32
32 M
=
π d3
M
M
M
M
Figure 20.2
Owing to couple M, the radius of curvature of the coils changes from R to R and the number of
coils changes fromn to n . 1
1
Then, M = EI
approximately
−
R
R
The length of wire, l = 2π Rn = 2π R n
∴
∴
2π n
1
1
2π n
=
and
=
R
l
R
l
2π (n − n)
M = EI ×
l
@seismicisolation
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392
•
Strength of Materials
2π (n − n) is the angle of twist of the spring, φ ,
EI φ
∴ M=
l
Ml
M × π Dn
or φ =
=
π
EI
E d4
64
64MDn
∴ φ=
E d4
Strain energy stored is the spring,
M2l
U=
2EI
2σ I
M
σ
=
OR M =
and
I
d/2
d
2
2σ I
l
U=
×
d
2EI
−4σ 2 I
l
×
2E
d 2
π 4
d
4σ 2
l
64
×
U=
2
2E
d
π σ 2d2l
σ 2 π d2l
=
=
32 E
8E
4
U=
or
U=
Stiffness for bending, k =
(i)
σ2
× volume of the spring wire
8E
M
φ
M
× Ed 4
64 MDn
Ed 4
k=
64 Dn
k=
E XAMPLE 20.1: A closely-coiled helical spring is to carry a load of 750 N. Its mean coil diameter
is to be 10 times that of the wire diameter. Calculate mean coil diameter and wire diameter if the
maximum shear stress in the material of the spring is to be 100 MN/m2 .
S OLUTION :
W = 750 N
D = 10d
τ = 100 MN/m2
@seismicisolation
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Springs
•
393
We know,
τ=
100 × 106 =
16W R
π d3
16 × 750 ×
10d
2
π d3
16 × 750 × 5
=
π d2
16 × 750 × 5
d2 =
π × 100 × 106
= 0.000191
d = 0.0138
= 13.8 mm
∴
Mean coil dia,
D = 10 × 13.8 = 138 mm
Ans
E XAMPLE 20.2: A helical spring is made of 13 mm diameter steel wire wound on a 140 mm
diameter mandrel. If there are 12 active coils, what is spring constant? Take C = 85 GN/m2 . What
force must be applied to the spring to elongate 30 mm?
S OLUTION :
Diameter of steel wire, d = 13 mm = 0.013 m
Diameter of mandrel, D = 140 mm = 0.14 m
Number of active coils, m = 12
Modules of rigidity C = 85 GN/m2
Elongation of the spring,
δ = 30 mm = 0.03 m
Spring constant = stiffness of spring k
W
δ
Cd 4
0.14
k=
= 0.07
, R=
2
64R3 n
85 × 109 × (0.013)4
=
64(0.07)3 × 12
= 9215.9 N/m Ans
W
= 9215.9
δ
W
= 9215.9
0.03
W = 276.5 N Ans
k=
Now,
or
@seismicisolation
@seismicisolation
394
•
Strength of Materials
E XAMPLE 20.3: A close-coiled helical spring has mean diameter of 75 mm and spring constant
of 80 kN/m. It has 8 coils. What is the suitable diameter of the spring wire if maximum shear stress
is not to exceed 250 MN/m2 ? Module of rigidity of the spring wire material is 80 GN/m2 .
What is the maximum axial load the spring can carry?
(AMIE Summer, 2000)
S OLUTION :
R=
D 75
=
= 37.5 mm = 0.0375 m
2
2
Spring constant, k = 80 kN/m
Active coils = 8
Maximum shear stress = 250 MN/m2
Modulus of rigidity, C = 80 GN/m2
π
Now, T = τ × d 3
16
W × 0.0375 = (250 × 106 ) ×
π
(d)3
16
(i)
Also we know w = k.δ
W = 80000 × δ
∴
(ii)
64W R3 4 64W × (0.0375)3 × 8
=
Cd 4
80 × 109 × d 4
W
= 3.375 × 10−13 4
d
δ=
Putting value of δ in Eqn. (ii)
W = 80000 × 3.375 × 10−13
d 4 = 80000 × 3.375 × 10−13
d = 0.0128 m
= 12.8 mm Ans
W
d4
For maximum axial load the spring can carry, W :
From Eqn. (i),
π
(0.0128)3
16
W = 2744 N Ans
W × 0.0375 = (250 × 106 )
E XAMPLE 20.4: A helical compression spring has a coil diameter of 75 mm and must carry
a maximum load of 900 N with a compression of 100 mm. Calculate the diameter of the wire and
number of free turns required. Allow a maximum shearing stress of 400 MN/m2 and
C = 80 GN/m2 .
@seismicisolation
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Springs
•
395
S OLUTION :
D = 0.075 m
W = 900 N
τ = 400 MN/m2
C = 80 GN/m2 ,
D =?, n =?
δ = 100 mm = 0.1 m
We know,
τ=
8W D
π d3
d3 =
8W D
π .τ
=
8 × 900 × 0.075
π × 400 × 106
= 0.42972 × 10−6
∴
d = 10−2 (0.7546)
= 0.007546 m
= 7.55 mm
Ans
Also,
δ=
8W D3 n
Cd 4
h=
8Cd 4
8W D3
=
0.1 × 80 × 109 × 0.007554
8 × 900 × 0.0753
= 8.56 Ans
E XAMPLE 20.5: A close-coiled helical spring is to have a stiffness of 80 kN/m and to exert a
force of 2.7 kN. If the mean diameter of the coils is to be 75 mm and the maximum stress is not to
exceed 250 MN/m2 , calculate the required number of coils and the diameter of the steel rod from
which the spring should be made.
The modulus of rigidity of the material is 80 GN/m2 .
@seismicisolation
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396
•
Strength of Materials
S OLUTION :
δ=
8W D34
Cd 4
n=
δ Cd 4
8W D3
Since stiffness = 80 kN/m
So for a force of 2.7 kN, deflection
2.7
80
= 0.03375
=
Also,
τ=
8W D
π d3
d3 =
8W D
πτ
=
8 × 2700 × 0.075
π × 250 × 106
d 3 = 10−2 (1.273)
∴
d = 0.01273 m
= 12.73 mm
Now,
n=
=
8Cd 4
8W D3
0.03375 × 80 × 109 × (0.01273)4
8 × 2700× (0.075)3
= 7.81
= 8 Nos
Ans
Springs in Series and Parallel
For springs in series, the load in the same for both springs (if two springs are attached in series).
So, net deflation δ = δ1 + δ2 .
@seismicisolation
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Springs
•
397
W W
W
+
=
where k1 and k2 are the stiffness of the respective springs. And k is the
k1 k2
k
equivalent stiffness.
1
1
k1 k2
1
= +
or k =
∴
k k1 k2
k1 +k2
or δ =
If two springs are of stiffness k1 and k2 and are joined together in parallel, they have the common
deflection δ and the load is the mm of load taken by each
W1 W2
W
=
=
Thus, δ =
k
k1
k2
Since W = W1 +W2 or kδ = k1 δ + k2 δ
or k = k1 + k2
where W1 and W2 are loads shared by each spring and k in the effective stiffness.
Concentric (Cluster) Springs: When close coiled springs are placed inside the other, concentric
(cluster) spring in formed.
Do
do
di
Di
Figure 20.3
In case if the material and the free length of outside spring and inside springs are the same, then
8W i Di 8W o Do
=
maximum shear stress is equal, τ =
π di
π do
Since deflection is same, so
8W 3 D3 n 8W 3 D3 n
δ = i 4i = o 4o
Cd i
Cd o
Suffix I is for inside spring and o for outside spring.
In cluster springs, the solid length is the same,
ni di = no do
E XAMPLE 20.6: A composite spring has two close-coiled springs connected is series; one spring
has coils of a mean diameter of 30 mm and wire diameter 3 mm. Find the wire diameter of the other
spring, if it has 16 coils of mean diameter 42 mm. The stiffness of the composite spring is 1.8 kN/m.
Determine the maximum load that can be carried by the composite spring and the corresponding
extension of the maximum stream in 260 MN/m2 . Modulus of rigidity of the material of spring is
80 GN/m2 .
@seismicisolation
@seismicisolation
398
•
Strength of Materials
S OLUTION :
W W
W
= +
k
k1 k2
where k is stiffness.
1
1
1
= +
k k1 k2
Stiffness,
k1 =
=
Cd 41
80 × 109 × (0.003)4
=
8 × 0.033 × 12
8D31 n1
6.48
2.592 × 10−3
= 2500 N/m
k2 =
80 × 109 ×d 42
8 × (0.042)3 ×16
= 8.436 × 1012 d24
1
1
1
=
+
k 2500 8.436 × 1012 d24
8.436 × 1012 d24 +2500
1
=
1800
21090 × 1012 d24
21090 × 1012 d24 = 15184.8 d24 + 4500000
2.109 × 1016 d24 = 4500000
d24 =
4500000
2.109 × 1016
=
45 × 105
2.109 × 1016
= 21.34 × 10−11
∴
dz = 3.82 mm
Ans
For maximum load we will take spring with minimum wire diameter
τ=
8W G
π d13
W=
τπ d13
8D1
@seismicisolation
@seismicisolation
Springs
=
399
260 × 106 × π × (0.003)3
8 × 0.03
= 91.85 N
Total compression =
•
Ans
91.85
W
=
= 0.051 m
k
1800
= 51 mm
Ans
E XAMPLE 20.7: In a compound helical spring the inner spring is arranged within and concentric
with the outer one but is 10 mm shorter. The outer spring has 10 coils of mean diameter 25 mm and
the diameter of the wire is 3 mm. Find the stiffness of the inner spring if an axial load of 150 N
causes the outer one to compress 20 mm. If the radial clearance between the two springs is to be 1.6
mm, find the diameter of the wire of the inner spring when it has 8 coils C = 80 GN/m2 for both
springs.
n0 = 10 coils
D0 = 25 mm
d0 = 3 mm
0.010 m
8W D3 n
Cd 4
δ Cd 4
W=
8D3 n
0.02 × 80 × 109 × (0.003)4
W=
8 (0.025)3 × 10
= 103.68 N
δ=
0.025
Figure 20.4
Load shared by inner spring = 150 − 103.68 − 46.32 N
Stiffness =
46.32
W
=
= 4.62 kN/m
δ
0.01
Ans
Diameter for inner spring
= (0.025 − 0.0016 × 2 − 0.003) − d
= 0.0188 − d
δ Cd 4
8D3 n
0.01 × 86 × 109 × d 4
46.32 =
8(0.0188 − d)3 × 8
W=
Solving the above equation,
d = 2.03 mm Ans
@seismicisolation
@seismicisolation
400
•
Strength of Materials
E XAMPLE 20.8: Determine amount of compression and maximum shear stress produced when a
load of 2200 N is dropped axially on a close-coiled helical spring from a height of 250 mm. The
spring has 24 coils having mean diameter of 200 mm and wire diameter of 30 mm. Modulus of
rigidity, C = 80 GPa.
S OLUTION :
Mean diameter of coil D = 200 mm
Diameter of wire, d = 30 mm
Number of coils, n = 24
Height of fall, h = 250 mm
Falling load, w = 2200 N
Modulus of rigidity,
C = 80 GPa = 80000 N/mm2
For compression δ ,
Let We = Equivalent gradually applied load which shall produce the same deflection as by the
falling load 2200 N.
Work done by the falling load = work done by We
1
W (h + δ ) = We δ
2
Now
δ=
∴
We =
8We D3 n
Cd 4
δ Cd 4
8D3 n
2200(250 + δ ) =
1 δ Cd 4
·δ
×
2 8D3 n
2200(250 + δ ) =
1 80000(30)4 2
δ
2 8(200)3 24
2200(250 + δ ) = 21.1δ 2
550000 + 2200δ = 21.1δ 2
21.1δ 2 − 2200δ − 550000 = 0
δ 2 − 104.3δ − 26066.4 = 0
@seismicisolation
@seismicisolation
Springs
•
401
√
104.3 ± 104.32 + 4 × 26066.4
δ=
2
√
104.3 ± 10878.5 + 104265.6
=
2
104.3 + 339.3
=
2
= 221.8 mm
Ans
For maximum stress:
We =
=
δ Cd 4
8D3 n
221.8 × 80000(30)4
8(200)3 × 24
= 9357.2 N
τ=
=
8We D
π d3
8 × 9357.2 × 200
π (30)3
= 176.6 N/mm2
Ans
E XAMPLE 20.9: A close-coiled helical spring of steel wire of 7 mm diameter and having 12
complete turns is subjected to an axial couple M. The mean coil diameter is 85 mm. If the maximum
bending stress in spring wire is not to exceed 250 MN/m2 , determine: i) the magnitude of axial
couple M and ii) the angle through which one end of the spring is turned relative to the other end.
Take modulus of elasticity for steel as 200 GN/m2
S OLUTION :
Number of complete turns = 12
Diameter of steel wire = 7 mm
Mean diameter of coil = 85 mm
Maximum bending stress
σb = 250 MN/m2
Esteel = 200 GN/m2
@seismicisolation
@seismicisolation
402
•
Strength of Materials
i) Axial couple:
32M
π d3
32M
250 × 106 =
π (0.007)3
σb =
∴
250 × 106 × π × (0.007)3
32
= 8.41 Nm Ans
M=
ii)
64MDn
Ed 4
64 × 8.41 × 0.085 × 12
=
200 × 109 (0.007)4
= 1.1433 radians
180
= 1.1433 ×
π
= 65.54◦ Ans
φ=
Open-Coiled Helical Spring
In open-coiled helical spring, helix angle is more than 10◦ . Let the helix angle of the wire be α and
mean radius of the coil be R. Then the length of wire,
l = 2π Rn sec α
n = no. of coils
l
α
2πR
Y
X
α
X
Y
W
α
WR
Figure 20.5
@seismicisolation
@seismicisolation
Springs
•
403
(a) Axial load
Component in plane X − X
= W R sin α , bending the wire
Component in plane Y −Y
= W R cos α , twisting the wire
For axial deflection:
Equating the external work to the strain energy in torsion and bending
We get,
1
(W R sin α )2 l (W R cos α )2 l
Wδ =
+
2
2EI
2CJ
2
Sin α cos2 α
+
∴ δ = W R2 l
EI
CJ
2π R
2π W R3 n sin2 α cos2 α
+
Because l =
=
cos α
EI
CJ
cos α
64W R3 n 2 sin2 α cos2 α
+
δ= 4
E
C
d cos α
8W D3 n 2 sin2 α cos2 α
δ= 4
+
E
C
d cos α
(i)
(ii)
Strain Energy:
1
U = Wδ
2
U=
4W 2 D3 n 2 sin2 α cos2 α
+
E
C
d 4 cos α
(b) Axial couple:
Y
Xα
X
M
<
Y
M
Figure 20.6
@seismicisolation
@seismicisolation
<
<
M
α
(iii)
404
•
Strength of Materials
Figure 20.6 shows part of the spring, subjected to an axial couple M which is regarded positive if
it increases the curvature of wire, i.e., winding up the spring. This moment may be resolved into
components perpendicular and parallel to the wire.
Component in plane X − X = M cos α , bending the wire
Component in plane Y −Y = M sin α , twisting the wire
Equating the work done by the couple to the strain energy in the wire,
1
(M cos α )2 l (Msinα )2 l
Mφ =
+
2
2EI
2CJ
2
cos α sin2 α
+
∴ φ = Ml
EI
CJ
M π Dn 32 2 cos2 α sin2 α
+
φ=
cos α π d 4
E
C
32MDn 2 cos2 α sin2 α
= 4
+
E
C
d cos α
(iv)
(v)
Shear stress, τ :
The torque component W R cos α causes the shear stress
τ=
=
∴
τ=
W R cos α d
π 4 ×2
d
32
16W R cos α
π d3
8W D cos α
π d3
Bending stress , σb
The moment component W R sin α causes the bending stress
σb =
W R sin α d
π 4 ×2
d
64
=
32W R sin α
π d3
=
16W D sin α
π d3
(vi)
Composite Action of Axial Load and Couple
When an axial load is applied to an open-coiled spring the spring winds up as well as extends. And
similarly, when a winding-up couple is applied, the spring extends as well as twists. These movements and the formulae already derived can be obtained by the application of Castigliano’s theorem.
@seismicisolation
@seismicisolation
Springs
•
405
If the load W end couple M are applied simultaneously, total movement in plane
X − X = −W R sin α + M cos α , bending the wire and total moment in plane Y − Y = W R cos α +
M sin α , twisting the wire
∴
1
1
1
1
(−W R sin α + M cos α )2
+ (W R cos α + M sin α )2
2
EI 2
CJ
1 W 2 R2 sin2 α −W RM sin 2α + M 2 cos2 α W 2 R2 cos2 α −W RM sin 2α + M 2 sin2 α
+
=
2
EI
CJ
1 2W R2 sin2 α − RM sin 2α 2W R2 cos2 α − RM sin 2α
∂U
=
+
δ=
∂W
2
EI
CJ
2
sin α cos2 α
sin 2α 1
1
= W R2 l
+
+ MRl
−
(vii)
EI
CJ
2
CJ EI
l −W R sin 2α + 2M cos2 α W R sin 2α + 2M sin2 α
∂U
=
+
φ=
∂M 2
EI
CJ
2
sin 2α 1
1
cos α sin2 α
+
+W Rl
−
.
(viii)
= Ml
EI
CJ
2
CJ EI
U=
Additional terms obtained in Eqns. (vii) end (viii) represent the extension due to M and angle
of twist due to W , respectively.
If α → 0 these formulae reduce to those obtained for close-coiled springs.
Note that Eqns. (vii) and (viii) can be used if only load or only couple is applied or both.
Flat Spiral Spring
It is made of uniform thin metallic stripe wound into a spiral is one plane. The inner end is fixed to
a winding spindle C.
R
x
b
t
B
H
A l
y
D
V
C
E
Flat spiral spring
Figure 20.7
Refer Fig. 20.7, let a torque T be applied to the spindle C for winding. And let H and V be the
horizontal and vertical reactions at D. Consider an element AB of length ol. Coordinates of element
AB are xy as shown as Fig. 20.7.
@seismicisolation
@seismicisolation
406
•
Strength of Materials
AB = (V x − Hy)
Bending moment on the element
(i)
If d θ is the change in angle between the tangents at A and B and if the length of the element is
dl
dl, then curvature of this element is r =
dθ
dl
r
M E
but
=
I
R
1
M
=
where R is the radius of curvature.
R EI
M.dl
∴ dθ =
EI
Vx −H y
dl
Substituting from Eqn. (i), d θ =
EI
The change in angle between targets at the extreme points of spring is:
or
dθ =
θ=
=
Vx − Hy
dl
EI
V
EI
xdl −
H
EI
ydl
Now xld is the moment of the whole length of the spring about the direction of Y whereas winding
spindle being the centroid of the spring, we have
xdl = moment of the profile of the spring about Y-axis
= lR
where l in the total length of spring.
And ydl = the moment of the whole spring length about the axis joining the centres of the winding
spindle and the pin D. So we have
ydl = 0
V
H
V IR
∴ θ=
lR −
×0 =
EI
EI
EI
But V R is equal to the torque T applied to the spindle
θ=
TP
EI
@seismicisolation
@seismicisolation
(ii)
Springs
•
407
The greatest bending moment occurs in the spring at E, which is
M = V × 2R = 2T
If σ is the bending stress at E, then
σ =
M
Z
where Z is the section modulus for the area of cross section of the spring about N.A. perpendicular
to the plane of spring.
∴
2T × 6
12T
= 2
2
bt
bt
σ bt 2
or T =
12
Maximum bending stress,
σ=
(iii)
The energy stored in winding the spring is
1
Tθ
2
1
TR
T 2l
= ×T ×
=
2
EI
2EI
2
1
σ bt 2
σ bt 2
1
12
×
=
×
× 3
=
12
2EI
12
2E
bt
U=
σ2
× bt l
24E
σ2
=
× volume of spring
24E
=
(iv)
From the geometry of the circle into which the plates are initially formed.
y
l
2
l
2
2R-y
Radius of circle = R
@seismicisolation
@seismicisolation
408
•
Strength of Materials
2
l
y (2R − y) =
2
or y =
l2
8R
Leaf, Laminated or Carriage Springs or Semi-elliptic Spring
Laminated or leaf springs are made up of a number of stripes, made of 0.9% carbon (approximately)
steel, these are also called carriage springs. The leaf spring is designed so that maximum stress is
the same in all plates at all sections giving maximum utilisation of material. The arrangement of
the spring is shown in Fig. 20.8, each plate free to slide relative to the adjacent plates as the spring
deflects. The ends of each plate are tapered to provide a uniform change in effective breadth between
the centre and the ends and if the plates are cut along their centre lines and placed side by side, they
form a diamond-shaped plate.
W
W
2
x
W
nb
x
Figure 20.8
Let
b = breadth of plate
t = thickness of each plate
n = number of plates
@seismicisolation
@seismicisolation
2
Springs
•
409
Then effective width of plate at centre = nb
Consider, section at a distance x from one end,
x
−W
M
2
σ=
= Z
t2
x
× nb ×
l/2
6
−3W l
=
2nbt 3
(i)
Note that this expression is independent of x. That means σ is constant at all sections.
At any section M is proportional to x and I is proportional to x, so that M is proportional to I.
E
M
= ; R is a constant. Thus, if the spring is to become flat when loaded, the plates
Now since
I
R
must initially be bent in the arc of a circle.
The load which causes the plates to become flat is called the proof load.
E
d2y M
−W x/2
−3W l
=
=
=
2
3
I
dx
nbt 3
t
x
× nb ×
l/2
12
Integrating Eqn. (ii),
E
l
When x = ,
2
dy
=0
dx
∴
A=
dy
3W lx
+A
=−
dx
nbt 3
3W l l
·
nbt 3 2
3W l
EIy = − 3
nbt
∴
x2 l
− x +B
2 2
When x = 0, y = 0 ∴ B = 0
The maximum deflection occurs at the centre
l
where x =
2
i.e., ymax =
3W l 3
8E n bt 3
Strain energy,
l
M2
dx
2EI
U=
0
l/2
=2
0
3W l W
· × dx
2Enbt 3 2
@seismicisolation
@seismicisolation
(from Eqn. ii)
(ii)
410
•
Strength of Materials
U=
The strain energy
3 W 2l3
16 Enbt 3
3 W 2l2
may be written as
16 Enbt 3
3W l 2 nbtl
U=
12E
2nbt 2
σ 2 nbtl
×
6E
2
σ2
U=
× volume
6E
=
∴
Quarter-Elliptic Leaf Spring
Figure 20.9 shows a quarter-elliptic leaf spring. In this case we substitute l = 2l and W = 2W in
formulae of stress σ and deflection ymax of elliptical leaf spring discussed above.
W
l
3 2W × 2l
6W l
×
=
2
nbt 3
nbt 3
3 2W × (2l)3
ymax = ×
8
Enbt 3
3
6W l
=
Enbt 3
σ=
t
Figure 20.9
E XAMPLE 20.10: Find the mean radius of an open-coiled spring having helix angle 30◦ to
given vertical displacement of 25 mm and an angular rotation of the loaded and of 1.25◦ under
an axial load of 40 N. The material available is steel rod of 6 mm diameter. E = 200 GN/m2 ,
C = 80 GN/m2 .
S OLUTION :
δ = WR l
2
So
∴
sin2 α cos2 α
+
EI
CJ
200
E
=
= 2.5
C
80
E = 2.5 C
J = 2I
J
EI = 2.5 C ×
2
EI = 1.25 CJ
@seismicisolation
@seismicisolation
(i)
Springs
•
411
Substituting in Eqn. (i)
sin2 30◦ cos2 30◦
+
0.025 = W R l
1.25 CJ
CJ
2
W R l 0.25
+ 0.75
=
CJ
1.25
2
0.025 =
W R2 l
(0.95)
CJ
or,
W R2 l
= 0.0263
CJ
Also
φ = Ml
cos2 α sin2 α
+
EI
CJ
(i)
sin 2α 1
1
+W Rl
−
2
CJ EI
Since, M = 0
So
sin 2α 1
1
φ = W Rl
−
2
CJ EI
1.25 × π W Rl sin 60 1
1
=
−
180
2
CJ 1.25 CJ
1
0.433 W Rl
1−
0.0218 =
CJ
1.25
0.0218 = 0.0866
W Rl
CJ
W Rl
= 0.252
CJ
(ii)
From Eqns. (i) and (ii)
0.252R = 0.263
∴
R = 0.1044 m
= 104.4 mm
Ans
E XAMPLE 20.11: An open-coiled helical spring is made of 10 mm diameter steel wire, the coils
having 13 complete turns and a mean diameter of 102 mm, the helix angle being 14◦ . Calculate the
deflection under 250 N and intensities of direct and shearing stresses induced in the section wire.
@seismicisolation
@seismicisolation
412
•
Strength of Materials
If 250 N axial load is replaced by an axial torque of 9 Nm, calculate the angle of rotation of the
coil and the axial deflection. E = 210 GPa, C = 85 GPa.
S OLUTION :
δ = WR l
2
sin2 α cos2 α
+
EI
CJ
sin 2α 1
1
+ MRl
−
2
CJ EI
M = 0, W = 250 N, l = 2π Rn sec α
l + 2π ×
∴
102
1
1
× 13 ×
= 4163.64 ×
2
cos 14
cos 14
l = 4292.4 mm
δ = 250 (51)2 × 4292.4
cos2 14
sin2 14
+
3
210 × 10 × I 85 × 103 J
J = 2I
I=
∴
π (10)4
= 490.6 mm4
64
J = 2 × 490.6 = 981.2 mm4
0.0585
09415
δ = 2791133100
+
210000 × 490.6 85000 × 981.2
= 2791133100 5.68 × 10−10 + 1.13 × 10−8
= 2791133100 5.68 × 10−10 + 113 × 10−10
= 33.12 mm
Ans
Stresses:
Bending stress, σ =
=
32W R sin α
π d3
32 × 250 × 51 × sin 14
π (10)3
= 31.43 N/mm2
@seismicisolation
@seismicisolation
Ans
Springs
Shear stress, τ =
=
16W R cos α
π d3
16 × 250 × 51 × cos 14
π (10)3
= 63.04 N/mm2
When only torque of 9 Nm in applied:
φ = Ml
cos2 α sin2 α
+
EI
CJ
sin 2α
+W Rl
2
1
1
−
CJ EI
W = 0, M = 9 Nm = 9000 Nmm
0.0585
0.9415
+
φ = 9000 × 4292.4
210000 × 490.6 85000 × 981.2
φ = 38631600 9.14 × 10−3 + 7.014 × 10−10
+0
d = 0.380 radian
= 0.38 ×
180
π
= 21.8◦
Ans
For axial deflection;
δ = WR l
2
sin2 α cos2 α
+
EI
CJ
sin 2α 1
1
+ MRL
−
2
CJ EI
W = 0, M = 9 Nm = 9000 Nmm
sin 2α 1
1
δ = MRl
−
2
CJ EI
1
1
sin 2 × 14◦
−
= 9000 × 51 × 4292.4
2
85000 × 981.2 210000 × 490.6
= 4624791594 12 × 10−9 − 9.7 × 10−9
= 10.6 mm
Ans
@seismicisolation
@seismicisolation
•
413
414
•
Strength of Materials
E XAMPLE 20.12: An open-coiled helical spring consists of 14 coils each of mean diameter 68
mm. If the spring wire of diameter 8 mm and helix angle 28◦ , determine:
(i) The load required to elongate the spring by 23 mm and the bending and shear stresses caused
by that load;
(ii) The axial twist that would cause a bending stress of 60 MN/m2 in the coils.
Take E = 200 GN/m2 and C = 85 GN/m2
S OLUTION :
Wire diameter = 8 mm
Coil mean diameter = 68 mm
No. of turns = 14
Helix angle = 28◦
E = 200 GN/m2
C = 85 GN/m2
δ = W R2 l
sin2 α cos2 α
+
EI
CJ
l = 2π Rn sec α ;
I=
π
(8)4 = 200.96 mm4 ,
64
J = 2 × 200.96
= 401.92 mm4
1
68
× 14
= 3385 mm
2
cos 28
cos2 28
sin2 28
2
+
23 = W (34) × 3385
200000 × 200.96 85000 × 401.92
5.878 × 10−6 = W 5.484 × 10−9 + 22.8 × 10−9
l = 2π ×
W=
5.878 × 10−6
= 207.8 N Ans
28.284 × 10−9
Bending stress,σb =
=
16W D sin α
π d3
16 × 207.8 × 68 × sin 28
π (8)3
= 66 MN/m2
@seismicisolation
@seismicisolation
Ans
Springs
Shear stress, τ =
=
•
415
8W D cos α
π (8)3
8 × 207.8 × 68 cos 28
π (8)3
= 62.08 MN/m2
Ans
(ii) Axial twist:
T = Axial twist required to cause bending stress of 60 MN/m2
Component of axial torque causing bending = T cos α
Now,
σb =
32M 32T cos α
=
π d3
π d3
60 =
32T cos α
π d3
T =
60 π d 3
32 cos α
=
60 × π × 83
32 cos 28
= 3413.8 Nm
= 3.1414 Nm
Ans
E XAMPLE 20.13: A flat spring is 6 mm wide, 0.3 mm thick and 3.5 m long. Calculate the torque,
the work stored and the number of turns required to wind up the spring whereas the maximum
bending stress does not exceed 520 N/mm2 , E = 200 GN/m2 .
S OLUTION :
b = 6 mm,
t = 0.3 mm,
l = 3.5 m = 3500 mm
σ = 520 N/mm2 , E = 200 GN/m2 = 200000 N/mm2
σ bt 2
12
520 × 6 × (0.3)2
= 23.4 Nmm
=
12
= 0.0234 Nm Ans
Torque, T =
@seismicisolation
@seismicisolation
416
•
Strength of Materials
Work stored = Strain energy
σ2
× volume of spring
24E
5202
U=
× 6 × (0.3) × 3500
24 × 200000
=
= 354.9 Nmm
= 0.3549 Nm
Ans
Now
θ=
Tl
,
EI
I=
6 × (0.3)3
12
=
23.4 × 3500 × 12
200000 × 6 × (0.3)3
= 30.33 radians
=
30.33
2π
= 4.8 turns
Ans
E XAMPLE 20.14: A flat spiral spring is having 10 mm broad and 0.6 mm thick steel stripe. The
length of the steel stripe is 7 m. The end at the greatest radius is attached to a fixed point and
the other end to a spindle. Find: (a) the maximum turning moment which can be applied to the
spindle if the stress in the stripe is not to exceed 560 MN/m2 ; (b) the number of turns required to
be given to the spindle and (c) the energy stored in the spring. Take E = 200 GN/m2 .
S OLUTION :
b = 10 mm = 0.01 m
t = 0.6 mm = 0.0006 m
l=7m
E = 200 GN/m2
@seismicisolation
@seismicisolation
Springs
•
417
a) We know,
T=
=
σ bt 2
12
560 × 106 × 0.01 × (0.0006)2
12
= 0.168 Nm
b)
Ans
0.01 × (.0006)3
= 1.8 × 10−13
12
0.168 × 7
=
200 × 109 × 1.8 × 10−13
θ=
Tl
,
EI
I=
= 32.67 radians
=
32.67
2π
= 5.2 turns
Ans
σ2
× volume of the spring
24E
2
560 × 106
× (0.01)(0.0006) × 7
=
24 × 200 × 109
Strain energy, U =
=
313600 × 1012 × 0.01 × 0.0006 × 7
24 × 200 × 109
= 2.74 Nm
Ans
E XAMPLE 20.15: A laminated steel spring, simply supported at the ends and centrally loaded
with a span of 0.75 m is repaired to carry a load of 7.5 kN and the central deflection is not to exceed
50 mm. The bending stress must not be greater than 400 MN/m2 . Plates are available in multiples
of 4 mm for width.
Determine the suitable value for the thickness, width and number of plates and the radius to
which the plates should be formed.
Assume the width to be twelve times the thickness. E = 200 GN/m2 .
@seismicisolation
@seismicisolation
418
•
Strength of Materials
S OLUTION :
We know
3W l 3
8Enbt 3
W = 7.5 kN
l = 0.75 m
b = 12 t
y=
E = 200 GN/m2
Substituting in the formula,
3 × 7.5 × 103 ×0.753
8 × 200 × 109 ×n × 12t × t 3
4
nt = 9.89 × 10−9
0.05 =
Which gives
(i)
Also,
3W l
2nbt 2
3 × 7.5 × 0.75
400 × 106 =
2 × n × 12t × t 2
σ=
nt 3 = 1.758 × 10−6
or,
(ii)
From Eqns. (i) & (ii)
nt 4
9.89 × 10−9
=
nt 3 1.758 × 10−6
∴ t= 0.00562 m Ans
The nearest suitable value is 6 mm.
b = 12 t (given)
b = 12 × 6 = 72 mm
(thickness is rounded to 6)
Now nt 3 = 1.758 × 10−6 from Eqn. (ii)
∴
n=
1.758 × 10−6
(0.006)3
= 8.13
Hence, 9 plates are required
@seismicisolation
@seismicisolation
Ans.
Springs
•
419
It must be noted that higher value to round off figure is taken so that laminated steel spring is
safer.
3 × 7.5 × 103 × 0.753
3W l
Actual deflection under load =
∴
y
=
8 × 200 × 109 × 9 × 0.072 × 0.0063
8En bt 3
= 0.0424 m
= 42.4 mm
Now
y=
Ans
l2
8R
0.0424 =
0.752
8R
∴
0.752
8 × 0.0424
R=
R = 1.66 m
Ans
E XAMPLE 20.16: A carriage spring has 12 plates each 65 mm wide by 6 mm thick and the
longest plate is 0.8 m long. The greatest bending stress is not to exceed 185 MN/m2 and the central
deflection when the spring is fully loaded is not to exceed 20 mm. Estimate the magnitude of the
greatest central load that can be applied to the spring. Take E = 200 GN/m2 .
S OLUTION :
n = 12 plates
b = 65 mm
t = 6 mm
l = 0.8 m
δ = 185 MN/m2
Maximum deflection (central) = 20 mm.
ymax =
3W l 3
8Enbt 3
W=
ymax ×8Enbt 3
3W l 3
W=
0.02 × 8 × 200 × 109 × 12 × 0.065 × (0.006)3
3 × (0.8)3
= 3510 N
= 3.51 kN
Ans
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•
Strength of Materials
E XAMPLE 20.17: A leaf spring is required to satisfy the following specifications: l = 0.75 m,
w = 5 kN, b = 75 mm, maximum stress = 210 MN/m2 , maximum deflection = 25 mm and
E = 200 GN/m2 . Find the number of leaves and their thickness. If the leaves become straight
when this load in applied, find its radius of curvature.
S OLUTION :
We know
σ=
3W l
2nbt 2
nt 2 =
3W l
2bσ
=
3 × 5000 × 0.75
2 × (0.075) × 210 × 106
nt 2 =3.571 × 10−4
(i)
Also,
3W l 3
8Enbt 3
3 × 5000 × (0.75)3
nt 3 =
8 × 200 × 109 × (0.075) × 0.025
ymax =
nt 3 = 2.10 9 × 10−6
From Eqns. (i) and (ii)
nt 3 2.1093 × 10−6
=
nt 2
3.571 × 10−4
t = 5.906 × 10−3 m
= 6 mm (say) Ans
Using Eqn. (i)
n=
3.571 × 10−4
(5.906 × 10−3 )2
3.571 × 10−4
(5.906)2 × 10−6
= 10 Nos Ans
n=
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(ii)
Springs
•
421
We know,
3 × 5000 × 0.75
M 3W l
=
=
3
I
nbt
10 × 0.075 × (0.006)3
= 69444444000
M E
=
I
R
E.I
∴ R=
M
=
200 × 109
69444444000
= 2.8 m Ans
Exercise
20.1 A close-coiled helical spring is to have a stiffness of 70 kN/m and to exert a force of 2.25 kN.
If the mean diameter of the coils is to be 90 mm, and the working stress 230 MN/m2 , find the
required number of coils and the diameter of the steel rod from which the spring should be
made. Take the modulus of rigidity as 80 GN/m2 .
[Ans 6.58; 13.55 mm]
20.2 A close-coiled helical spring made of round steel wire is required just to fit a rod 30 mm
diameter and to carry an axial load of 120 N without causing the deflection to exceed 20 mm.
The maximum allowable shearing stress is 200 MN/m2 , and modulus of rigidity of steel is
80 GN/m2 . Find the diameter of wire, the mean diameter of coil and the number of turns.
[Ans 3.72 mm; 33.72 mm; 8.52]
20.3 A close-coiled helical spring of circular section having a mean coil diameter of 60 mm a
subjected to an axial load of 80 N applied at the end of spring producing a shear stress of
100 N/mm2 and a deflection of 50 mm. Find the diameter, the number of coils, the length of
the spring wire and the strain energy stored in the spring.
[Ans 4.96 mm; 17.5; 2 joules]
20.4 A close-coiled helical spring has a mean diameter of the coils 12 times the wire diameter. It is
to be designed to absorb 300 J of energy with an extension of 150 mm. The maximum shear
stress is not to exceed 140 MN/mm2 . Determine the mean diameter of the coil, diameter of
the wire and number of turns. Also find the load with which an extension of 250 mm could
be produced in the spring. Take modulus of rigidity, C = 80 GN/m2 .
[Ans 30 mm; 360 mm, 1333.4 N]
20.5 Determine the maximum shear stress and the amount of compression produced when a mass
of 200 kg is dropped axially on a close-coiled helical spring from a height of 250 mm.
The spring has 20 coils each of mean diameter 200 mm and the wire diameter is 25 mm,
C = 84 GN/m2 .
[Ans 0.287 m, 239.6 kN/m2 ]
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•
Strength of Materials
20.6 Determine the mass of closely-coiled helical spring which would absorb the energy of a truck
of mass 10 tonnes while moving with a velocity of 1 m/s. If the spring is compressed by the
impact. Working bending stress is 300 N/mm2 ; C = 80 GN/m2 and specific gravity of the
spring material = 7.8
20.7 Two springs A and B are connected is series Spring A consists of 12 coils of 6 mm steel
wire and of 30 mm outside diameter. Spring B consists of 18 coils of 8 mm steel wire and
40 mm outside diameter. What is spring constant (stiffness) for the common system? What
is the longest force that can be applied to these springs without exceeding a shear stress of
360 MPa? C = 82 GN/m2 .
[Ans 19.44 N/mm, 1017.8 N]
20.8 One helical spring is placed inside the coil of the other helical spring, having same material,
same number of coils and free axial lengths. The two springs are compressed by a load of
1.5 kN. The mean coil diameter and wire diameter of the outer spring and inner spring are
80 mm, 60 mm and 10 mm, 7 mm, respectively. Calculate the load taken and the maximum
shear stress in each spring.
[Ans 543.47 N; 956.53 N; 194.86 N/mm2 ; 242.09 N/mm2 ]
20.9 An open-coiled helical spring is made of 9.525 mm diameter steel wire, the coils having 14
complete turns and a mean diameter of 101.6 mm, the helix angle being 15◦ . Calculate the
deflections under 225 N and intensity of direct and shearing stresses induced in the section
of wire.
If 225 N axial load is replaced by an axial torque of 8.5 Nm, calculate the angle of rotation
of the coil and the axial deflection. E = 210 GPa, C = 84 GPa.
[Ans 39.03 mm; 49.87 N/mm2 , 67.4 N/mm2 , 27◦ , 14.7 mm]
20.10 An open-coiled helical spring has 10 coils of a 12 mm diameter steel wire with a mean
diameter of 150 mm. The helix angle of the coils is 32◦ . Find the axial extension produced
by a load of 250 N. Any formulas used must be proved from fundamentals. E = 210 GN/m2 ,
C = 70 GN/m2 .
[Ans 49.7 mm]
◦
20.11 Find the mean radius of an open-coiled spring having helix angle of 30 to give a vertical
displacement of 25 mm and an angular rotation of the loaded end of 1.25◦ under an axial load
of 40 N. The material available is steel rod of 6 mm diameter. E = 200 GPa, C = 80 GPa.
[Ans R = 0.1044 m]
20.12 An open-coiled helical spring made of 10 mm diameter rod has six free coils 100 mm mean
diameter. The ends of the spring are fastened to two discs kept 0.75 m apart, which is the free
length of the spring. Calculate the force or the discs, acting along the axis of the spring, when
one disc is rotated through 10◦ to coil the spring. E = 200 GPa, C = 80 GPa.
[Ans 11.5 N (Compressive)]
20.13 An open-coiled spring consists of 10 coils, each of mean diameter 50 mm, the wire forming
the coil being 6 mm is diameter. Each coil makes an angle of 30◦ with the plane perpendicular
to the axis of the spring.
(i) Determine the load repaired to elongate the spring by 20 mm and the bending and shear
stresses caused by that load.
(ii) Calculate the axial twist that would cause a bending stress of 60 MN/m2 .
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Springs
20.14
20.15
20.16
20.17
20.18
20.19
•
423
Take E = 200 GN/m2 , C = 82 GN/m2 .
[Ans 192.7 N; 113.6 N/mm2 , 98.37 N/mm2 , ii) 1.469 Nm]
An open coiled helical spring is subjected to a combined axial load and torque. The spring
has mean coil diameter of 120 mm, number of coils 10, wire diameter 12 mm and helix angle
of 30◦ . Find the value of axial load and torque which would extend the spring by 5 mm with
no rotation of coils, indicating if the torque tends to wind or unwind the spring. Find also the
maximum normal and shear stresses in the spring. Take E = 210 GPa, C = 84 GPa.
[Ans −0.355 Nm (trying to unwind), 58.667 N, 16.37 N/mm2 , 10.325 N/mm2 ]
A carriage spring 600 mm long is constructed out of 60 mm wide steel plates. Find the
thickness of leaves and the number of plates required, if the spring is subjected to a maximum
load of 2680 N with a maximum central deflection of 12 mm and a maximum bending stress
of 160 N/mm2 . Take E = 200 GPa.
[Ans 6 mm; 7]
A semi-elliptic leaf spring 800 mm long has 12 leaves. The cross section of leaves is 60 mm ×
6 mm. For a central deflection of 15 mm and a bending stress not greater than 200 N/mm2 ,
find the value of the central load. E = 200 GPa.
[Ans 3.24 kN]
A steel carriage spring is 762 mm in span and carries a central load of 50 kN. The stress is to
be limited to 188 MPa in plates which are 76.5 mm wide and 6.35 mm thick. What will be
the stress in any plate and deflection of the spring at the centre? Also determine how much a
plate will overhang the one immediately below it and to what radius of curvature should each
plate be curve? Take E = 210 GPa.
[Ans 185.3 MPa; 20.17 mm; 38.1 mm; R = 3.6 m]
A 5 mm wide and 0.3 mm thick flat spiral spring is 2 m long. The maximum stress is 600 MPa
at the greatest bending moment. Determine the torque the work stored and number of turns
to wind up the spring. E = 208 GPa.
[Ans T = 22.5 Nmm, 3.82 turns, 270 Nmm]
A flat spiral spring is 5 mm wide, 0.25 mm thick and 3 m long. Assuming maximum stress of
1000 MN/m2 to occur at the point of greatest bending moment, calculate:
i) the torque
ii) the work that can be stored in the spring
iii) the number of turns required to wind up the spring
Take E = 200 GPa.
[Ans 0.026 Nm, 0.781 Nm, 9.533 turns]
20.20 A quarter elliptic spring has a clear span of 800 mm and carries a load of 10 kN at the free
end. The bending stress in stripe and deflection at free end are not to exceed 320 MPa and
80 mm, respectively. Find the number of plates if the width of a plate is 8 times the thickness.
Take E = 210 GN/m2 .
3 W l3
3 Wl
[Hint: δ =
]
; σ=
3
8 nbt E
2 nbt 2
[Ans t = 12.2 mm; 10]
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C HAPTER
21
COLUMNS AND STRUTS
A strut usually means a compression member which is long in comparison with its cross-sectional
area. If the strut is vertical, then it is known as column, pillar or stanchion. For very long column or
strut, the effect of direct compression is negligible and they fail due to buckling load (or critical load
or crippling load). Practically, many machine parts and members of structure behave as columns
or struts. For example, connecting rod of an internal combustion engine is a strut with both ends
hinged.
Since, both the ends of a strut or column need be considered while analysing them. For columns
and struts combination of end conditions are very important and are considered during derivation of
formulae related to the same. The possible end conditions are:
a)
b)
c)
d)
Both ends hinged
Both ends fixed
One end is fixed and the other hinged
One end is fixed and the other free.
It may be noted that if the member of the structure is not vertical and one or both of its ends are
hinged or pin joined, the member is known as strut. As told before, connecting rods, piston rods,
etc., are struts.
Very long columns: Let us first consider the case of very long columns for which crippling load is
loaded so that the effect of direct stress is negligible compared to bending stress. Long columns
are those whose slenderness ratio is more than 120 or whose length is more than 30 times the least
diameter. Long columns are analysed using Euler’s analysis. Leonhard Euler was Swiss mathematician, who was the first to analyse long columns mathematically. Following assumptions are made
during derivation of Euler’s formulae:
1.
2.
3.
4.
5.
The columns are made up of homogeneous material.
The columns are initially straight.
The cross section of column is constant throughout.
The columns carry perfectly axial loads.
Plane cross sections of column normal to centre line remain plane during buckling and the
effect of shear stress is neglected.
6. Self-weight of columns is neglected.
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425
7. The column is long compared to lateral dimensions.
8. The stresses do not exceed the limit of proportionality.
9. Longitudinal fibres of the column are free to expand or contract independently without any
effect on adjoining fibres.
10. Shortening of the columns due to direct compression is negligible.
Now we will derive formulae for different end conditions (as stated before) using Euler’s
method. Before this let us first define important terms:
Slenderness Ratio: It is the ratio of unsupported length of the column to the minimum radius of
le
is the slenderness ration where le is the
gyration of the cross-sectional ends of the column.
k
equivalent length.
Both ends
hinged
(a)
One end fixed
other free
(b)
Both ends
fixed
(c)
l
le= √2
le=l
le=2l
le=l/2
Equivalent Length (le ): It is the effective length of the column as shown in Fig. 21.1.
One end fixed
other hinged
(d)
Figure 21.1
Buckling factor =
le
, quad where k is the radius of gyration
Minimum k
Buckling load (or critical load or crippling load):
The maximum limiting load at which the column tends to have lateral displacement or tends to
buckle. The buckling takes place about the axis having minimum radius of gyration.
It may be noted that columns are of the three types:
i) Short columns: l = 8 times diameters or slenderness ratio is less than 32.
ii) Medium columns: l = 8 to 30 times diameter or slenderness ratio is 30 to 100.
iii) Long columns: l = more than 30 times diameter or slenderness ratio is more than 100.
Safe load =
Buckling load
Factor of safety
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•
Strength of Materials
(a) Both Ends Hinged
P
The column AB shown is Fig. 21.1 is hinged at
both ends and let P be the critical load. Consider a section of column at a distance x from A,
where the deflection is y. Moments due to critical
load P,
o B
M = −Py
∴
EI
dx2
X
x
d2y
EI 2 = −Py
dx
d2y
y
l
oA
Figure 21.2
+ Py = 0
d 2 y Py
+
=0
dx2 EI
Solving the above differential equation,
y = C1 cos
x
P
EI
+C2 sin
To determine constants C1 and C2 , when x = 0, y = 0
∴
x
P
EI
C1 = 0
Now applying end condition at B. Here, x = l and y = 0; 0 = C2 sin
If C2 is zero, then column would not bend at all therefore, sin l
P
= 0, π , 2π , 3π , etc.
For which l
EI
Taking the least significant value,
P
EI
P
l
EI
=0
P
=π
EI
π 2 EI
π 2 EI
P= 2
or = 2
l
le
l=
or
Note is this case equivalent length le = l.
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(i)
Columns and Struts
•
427
(b) Column with One End Fixed and the Other Free
In this case, refer to Fig. 21.3, consider point x,
P
e
B
M = P(e − y)
d2y
= P(e − y)
dx2
d 2 y P(e − y)
=
EI
dx2
d2y
P
P
+ y=
e
EI
dx2 EI
EI
or
y
l
X
x
A
Figure 21.3
Solution of this differential equation is
y = e +C1 cos x
P
EI
+C2 sin x
P
EI
When x = 0, y = 0
∴
Now Eqn. (i) becomes
y = e − e cos x
or,
At A,
or C1 = −e
0 = e +C1 ;
P
EI
+C2 sin x
P
EI
dy
P
P
P
P
= +e
sin x
+C2
cos x
dx
EI
EI
EI
EI
dy
= 0 and x = 0
dx
∴
Now since P cannot be zero
Equation (i) becomes
∴
0 = C2
P
EI
C2 = 0
y = e − e cos x
@seismicisolation
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P
EI
(ii)
428
•
Strength of Materials
At B, x = l, y = e
∴
Since, e is not zero,
∴
l = e − e cos l
P
EI
p
=0
cos EI
P
π 3 π 5π
= , ,
or EI
2 2 2
Taking the least value,
etc
P
π
=
EI
2
π 2 EI
Hence P =
42
∴
Note: In this case equivalent length le = 2l.
(c) Both Ends Fixed
P
Since both end are fixed, so by symmetry end
moments are equal (say M).
At section x from end A,
EI
M
d2y
= −(Py − M)
dx2
d2y
P
M
or
+ y=
2
EI
EI
dx
B
l/4
y
l/2
l
x
l/4
A
M
Figure 21.4
Solution of the above differential equation is
P
P
M
+C2 sin x
y = +C1 cos x
P
EI
EI
P
P
P
P
dx
= −C1
sin x
+C2
cos x
dy
EI
EI
EI
EI
At A, x = 0 and
dx
=0
dy
∴
C2
P
=0
EI
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Columns and Struts
•
429
Since, P cannot be zero, therefore C2 = 0
Also at A, x = 0 and y = 0
M
+C1 = 0
P
M
P
M M
P
y = − cos x
P
P
EI
C1 = −
∴
But at B, x = l, y = 0
P
M M
− cos x
=0
P
P
EI
M
P
M
cos x
=
∴
P
EI
P
P
=1
Or cos l
EI
P
= 0, 2π , 4π , . . .
l
EI
Taking least significant value, we get
P=
4π 2 EI
l2
(iii)
Note: In this case equivalent length le = l/2.
(d) One End Fixed, Other End Hinged
Let M be the fixing moment at A and for equilibrium F is the horizontal force as shown in
Fig. 21.5.
At x from A,
d2y
EI 2 = −Py + F(l − x)
dx
d 2 y Py
F
+
=
(l − x)
2
EI
EI
dx
P
B
F
y
l
x
A
M
Figure 21.5
@seismicisolation
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•
Strength of Materials
Solution of the above differential equation is
P
P
F
x +C2 Sin
x + (l − x)
y = C1 Cos
EI
EI
P
At A, y = 0 when x = 0
Fl
P
Fl
P
P
F
∴ y = − cos
x +C2 sin
x + (l − x)
P
EI
EI
P
P
P
P
P
F
dy Fl
=
sin
x +C2
cos
x −
or
dx
P EI
EI
EI
EI
P
C1 = −
∴
dy
=0
dx
Also, at A, x = 0 and
∴
C2
F
P
− =0
EI P
F
EI
∴ C2 = ×
P
P
F EI
F
Fl
P
P
x +
sin
x + (l − x)
∴ y = − cos
P
EI
P P
EI
P
At B, x = l and y = 0
Fl
0 = − cos
P
F EI
P
P
l +
sin
l
EI
P
P
EI
Dividing equation stated above by cos
or
P
l ,
EI
P
Fl F EI
tan
l
0=− +
P
P
P
EI
P
P
Fl
P
tan
l =
×
EI
P
F EI
P
P
l =
×l
tan
EI
EI
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Columns and Struts
•
431
From mathematical tables, we find only two values for which tan θ = θ as in this case. Either
θ = 0 or θ = 4.493 radians. Since θ = 0 is not admissible as then, P will be zero which is not
possible. Therefore, we adopt θ = 4.493 radians.
∴
or
l
P
= 4.493
EI
4.4932 EI
P=
l2
2π 2 EI
P=
l2
(iv)
Note: In this case equivalent length le = √l .
2
Alternative methods: Equations (ii), (iii) and (iv) can be obtained by putting equivalent length le
in Eqn. (i) in place of actual length l. For example, for case (b): le = 2l (Refer Fig. 21.1)
Substituting in Eqn. (i) for equivalent length,
P=
π 2 EI
l2
After substituting equivalent length 2l in place of l 2 to find critical load for one end fixed and the
other free.
π 2 EI π 2 EI
=
P=
(2l)2
4l 2
which is same as before. Similarly, Eqn. (iii) and Eqn. (iv) can be derived by substituting in place
l
l
of l; and √ .
2
2
For reference equivalent lengths are given again in the following table:
Ends conditions
Both ends hinged
One end fixed and the other free
Both ends fixed
One end fixed and the other hinged
Equivalent length
le = l
le = 2l
le = l/2
l
le = √
2
E XAMPLE 21.1: A steel column is of length 7.5 m and diameter 550 mm with both ends hinged.
Determine the crippling load by Euler’s formula. Take E = 210 GPa.
S OLUTION :
Actual length of the column l = 7.5 m = 7500 nmm
Diameter of column, d = 550 mm
Young’s modulus, E = 210 GPa = 210000 N/mm2
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432
•
Strength of Materials
Least moment of inertia of the column section,
π d4
64
π
(550)4 = 4489525391 mm4
=
64
I=
Since, the column is hinged at both ends,
∴ Equivalent length = l
π 2 EI
l2
2
π × 210000 × 4498525391
=
75002
8
= 1.65 × 10 N Ans
P=
E XAMPLE 21.2: Compare the crippling load of solid circular column of diameter 250 mm and
hollow circular column of same cross-sectional area and thickness 35 mm. Materials and lengths
are same. Assume both are hinged at both ends.
S OLUTION :
π
Sectional area of solid column = (250)2
4
Thickness of the hollow circular column = 35 mm
Let the external diameter of hollow column = D mm
∴
Internal diameter of hollow circular column
= D · 2t = D − 2 × 35 = (D − 70) mm
∴
Sectional area of hollow circular section
π 2
=
D − (D − 70)2
4
Since, sectional areas of both are same,
π
π 2
∴
D − (D − 70)2 = (250)2
4
4
D2 − (D2 + 4900 − 140 D) = 62500
−4900 + 140 D = 62500
140 D = 62500 + 4900
D = 481.43 mm,
Inner dia = 481.43 − 70
= 411.43 mm
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Columns and Struts
•
433
Least moment of inertia:
π
481.434 − 411.434
64
π
5.372 × 1010 − 2.865 × 1010
=
64
IH =
= 0.123 × 1010 mm4
IS =
π
(250)4
64
= 191650391 mm4
π 2 E IH
2
PH
IH
0.123 × 1010
= 2l
=
=
PS
IS
191650391
π E IS
2
l
= 6.42 Ans
E XAMPLE 21.3: A steel tube 4.5 m long, 40 mm internal diameter and 6 mm thick is used as a
column: the crippling load if:
i) both ends are hinged,
ii) both ends are built-in, and
iii) one end is built-in and other is free. Take E = 200 GPa
S OLUTION :
i) External diameter = 40 mm + 6 × 2 = 52 mm
Internal diameter = 40 mm
π
524 − 404
64
π
(7311616 − 2560000)
I=
64
= 233126.16 mm4
I=
i)
π 2 EI
for both ends hinged
l2
π 2 × 200000 × 233126.16
=
45002
= 22701.5 N Ans
P=
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•
Strength of Materials
ii) For both ends built-in (fixed):
In this case effective length,
l
2
4500
= 2250 mm
=
2
π2 E I
P= 2
le
le =
π 2 × 200000 × 233126.16
22502
= 90806 N Ans
=
iii) One end is built-in and other end free:
Equivalent length
le = 2l
le = 2 × 4500 = 9000 mm
π2 E I
le2
2
π × 200000 × 233126.16
=
(9000)2
= 5675.4 N Ans
P=
E XAMPLE 21.4: Determine the limiting length of a column with both ends hinged and having a
cross-section 70 mm × 120 mm so that the critical stress is 300 N/mm2 . Take E = 200 GPa.
Column section = 70 mm × 120 mm
Critical stress = 300 N/mm2
E = 200000 N/mm2
Cross-sectional area = 70 × 120 = 8400 mm2
Critical load = critical stress × sectional area
= 300 × 8400 = 2520 kN
Imin =
120 × 703
= 3430000 m4
12
Using Euler’s formula,
π 2 EI
l2
π 2 × 200000 × 3430000
2520000 =
l2
P=
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Columns and Struts
•
435
π 2 × 200000 × 3430000
2520000
= 2684002
l = 1638.3 mm
l2 =
Limiting length of column =1.638 m Ans
E XAMPLE 21.5: A column having a T section with a flange 150 mm × 15 mm and web
150 mm × 15 mm is 3.5 m long. Assuming the column to be hinged at both ends, find the buckling
load by using Euler’s formula. Take E = 200 GPa.
S OLUTION :
1
150 mm
A
B
15 mm
Y
X
X
G
2
150 mm
15 mm
Y
Figure 21.6
Let us first find y form AB
a1 = 150 × 15 = 2250 mm2 , y1 =
a2 = 150 × 15 = 2250 mm2 y2 =
15
= 7.5 mm
2
150
+ 15 = 90 mm
2
A = a1 + a2 = 2250 + 2250 = 4500
ȳ =
=
2250 × 7.5 + 2250 × 90
4500
16875 + 202500
4500
= 48.75 mm.
IXX =
150 × 153
15 × 1503
+ 2250 (48.75 − 7.5)2 +
+ 2250 (48.75 − 90)2
12
12
@seismicisolation
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436
•
Strength of Materials
= [42187.5 + 3828.5] + [4218750 + 1701.6]
= [46016] + [4220452]
= 4266468 mm4
IYY =
15 × 1503 150 × 153
+
12
12
= 4218750 + 42187.5
= 4260937.5 mm4
IYY < IXX , so we will take IYY in the formula of Euler.
π 2 EI
Buckling load = P = 2 (for both ends hinged)
l
=
π 2 × 200000 × 4260937.5
(3500)2
= 685896 N
= 685.9 kN
Ans
E XAMPLE 21.6: Determine the Euler’s crippling load for a strut, square in section and 4.5 m long
with one end hinged and the other end free. The same bar is found to deflect 4 mm at the centre
when a load of 90 N is placed at midspan with the bar simply supported at the ends.
S OLUTION :
Length of strut = 4.5 m = 4500 mm
δ=
W l3
48EI
4=
90 × 45003
48EI
EI =
90 × 45003
48 × 4
= 4.27 × 1010 N mm2
Equivalent length for one end hinged and the other free, le = 2l = 2 × 4500 = 9000 mm
P=
=
π 2 EI
le2
π 2 × 4.27 × 1010
(9000)2
@seismicisolation
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Columns and Struts
= 5197.6 N
= 5.197 kN
•
437
Ans
Limitation for the Use of Euler’s Theory
In case of short column, Euler’s theory may give a critical load which is greater than that required
to produce failure due to direct compression. The limiting case occurs when
Pc = σc A
where σc is the compression stress at the yield point and A is the cross-sectional area.
π 2 EI
l2
2
nπ E Ak2
=
l2
P=n
But
where n is a constant depending on the end fixing conditions and k is radius of gyration (least) of
the cross section
∴
n
π 2 E A k2
= σc
l2
l
nπ 2 E
=
or
k
σc
The quantity l/k as we know is called the slenderness ratio and its value at this point is called
validity limit for Euler’s theory.
l
If the slenderness ration is small, the crippling stress will be high. But for the column material,
k
the crippling stress cannot be greater than the crushing stress. Hence, when the slenderness ratio is
less than a certain limit, Euler’s formula gives a value of crippling stress greater than the crushing
l
stress which is wrong. In the limiting case, we can find the value of for which crippling stress is
k
equal to crushing stress.
For example, for a mild steel column with both ends hinged.
Crushing stress = 335 N/mm2
Young’s modulus, E = 200 GN/m2
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•
Strength of Materials
Equating the crippling stress to the crushing stress corresponding to the minimum value of
slenderness ratio, we have
Crippling stress = Crushing stress
Crippling stress =
=
Crippling load P
=
Area
A
π 2E × A
2
le
A
k
π 2E
= 2
le
k
∴
π 2E
(For a column with both ends hinged
2 = 335
le
k
2
π 2 × 200000
l
=
k
335
= 5886
l
= 76.7, say 77
∴
k
le = l)
Therefore, if the slenderness ratio is less than 80 for mild steel column (with both ends hinged),
then Euler’s formula is not valid.
Rankine’s Formula
As we have discussed earlier that Euler’s formula gives correct results for only long columns, which
fail mainly due to buckling. Short columns fail mainly due to directly crushing whereas medium
columns which are neither long nor short fail by both buckling and direct crushing. Rankine derived
an empirical formula based on practical experiments for determining the crippling load which is
applicable to all columns irrespective of whether they are short or long.
Let PC = Crushing load
PR = Actual crippling load for column
PE = Crippling load per Euler’s formula
The Rankine hypothesis is
1
1
1
=
+
PR PC PE
or
1
PE + PC
=
PR PC × PE
@seismicisolation
@seismicisolation
Columns and Struts
PC × PE
PE + PC
PC
or PR =
PC
1+
PE
•
439
or PR =
Substituting for PC = σC A and for PE =
PR =
or PR =
∴
PR =
where a =
σC A
σC × A
1+ 2
π EI
le2
π 2 EI
in Eqn. (i)
le2
σc A
σc × A × le2
1+ 2
π × E × Ak2
σc A
2
le
1+a
k
σc
;
π 2E
(i)
(This is known as Rankine formula)
(Also known as Rankine-Gordon formula)
a is Rankine constant
σc and a both are constants in Rankine formula. The following table shows the values of σc and a
for different materials.
Material
σc (MN/m2 )
Mild steel
320
Cast iron
550
Wrought iron
250
Timber
50
Aluminium
120
a=
σc
for both ends hinged
π 2E
1
7500
1
1600
1
9000
1
750
1
5000
E XAMPLE 21.7: A rolled steel joist ISMB 300 is to be used as a column of 3.5 m length with
both ends fixed. Find the safe load on the column taking factor of safety as 4, σc = 330 N/mm2 ,
1
. Properties of the column section:
a=
7500
Area = 5626 mm2 , Ixx = 8.603 × 107 mm4 , Iyy = 4.539 × 107 mm4
@seismicisolation
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440
•
Strength of Materials
S OLUTION :
Since, IYY is less than IXX therefore, the column will tend to buckle about y-y axis
Least moment of inertia of column section,
I = 4.539 × 107 mm4
∴
I = Ak2
k=
k=
I
A
4.539 × 107
5626
= 89.82 mm
Crippling load as given by Rankine formula,
Pcr =
For both fixed ends, le =
σc A
2
le
1+a
k
l
3500
=
= 1750 mm
2
2
Pcr =
=
330 × 5626
1
1750 2
1+
7500 89.82
1856580
1856580
=
1 + 0.0506
1.0506
= 1767161.6 N
Allowing factor of safety 4,
Safe load =
1767161.6
4
= 441790 N
= 441.79 kN Ans
E XAMPLE 21.8: A hollow cast iron pipe of length 4 m is used as a column. The external diameter
is 60 mm and internal diameter is 35 mm. Determine the crippling load if both of its ends are fixed.
1
.
Take σc = 540 N/mm2 and a =
1600
@seismicisolation
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Columns and Struts
S OLUTION :
D, external diameter = 60 mm
d, internal diameter = 35 mm
π 2
(D − d 2 )
4
π
= (662 − 352 )
4
Area =
= 1864.375 mm2
Moment of inertia,
π
(604 − 354 )
64
π
= (12960000 − 1500625)
64
I=
= 562225.6 mm4
I
k=
A
562225.6
=
= 17.36 mm
1864.375
Crushing stress σc = 540 N/mm2
1
Rankine constant, a =
1600
Using Rankine formula,
PR =
For both ends fixed le =
σc A
2
le
1+a
k
4000
l
=
= 2000 mm
2
2
540 × 1864.375
P=
2000 2
1
1+
1600 17.36
=
1006762.5
1 + 8.295
= 550293.8
= 550.29 kN Ans
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•
441
442
•
Strength of Materials
E XAMPLE 21.9: A hollow cylindrical cast iron column is 4.5 m long with both ends fixed. Determine the minimum diameter of the column if it has to carry a safe load of 350 kN with a factor of
safety of 4. Take the internal diameter as 0.7 times the external diameter. Take σc = 540 N/mm2 and
1
in Rankine’s formula.
a=
1600
S OLUTION :
Length of column = 4.5 m = 4500 mm
Equivalent length le for both ends fixed =
∴
4500
l
=
2
2
le = 2250 mm
Safe load = 350 kN
If external diameter = D
Then internal diameter = 0.7D
Crushing stress σc = 540 N/mm2
a=
1
1600
for cast iron
Taking factor of safety into consideration
Crippling load
Area,
P = 540 × 4 = 2160 kN
π 2
A=
D − (0.7 D)2
4
π 2
=
D − 0.49 D2
4
= 0.1275 π D2
π 4
I=
D − (0.7D)4
64
π 4
=
D − 0.2401 D4
64
= 0.0119 π D4
But,
I = A × k2
I
k=
A
0.0119 π D4
=
0.1275 π D2
= 0.3055 D
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Columns and Struts
•
443
Now, using Rankine’s formula,
P=
2160000 =
2160000 =
σc A
2
le
1+a
k
540 × 0.1275 π D2
2
1
2250
1+
1600 0.3055D
216.189D2
33901.8
1+
D2
2160000
D2
= 2
216.189
D + 33901.8
D2
2160000D2 + 7.323 × 1010 = 216.189 D4
D4 − 9991.25D2 − 338731388 = 0
√
9991.25 + 99825076 + 1354925552
D =
2
2
D2 =
9991.25 + 38141.2
2
D2 = 155.13 mm
Ans
Internal diameter,
d = 155.13 × 0.7
= 108.59 mm Ans
E XAMPLE 21.10: A column for a crane gantery consists of two ISMB 400 × 140 mm connected
by two plates 10 mm thick as shown in Fig. 21.7. The length of column is 8 m and both ends are
hinged. Taking a factor of safety of 6, calculate the safe load for the column using Rankine’s formula
1
. For ISMB 400 × 140 : Iyy = 422.11 cm4 area
with failure stress σc = 320 MN/m2 and a =
7500
= 78.46 cm2 and Ixx = 20458.4 cm4 .
@seismicisolation
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444
•
Strength of Materials
Y
320 mm
10 mm
400 mm
X
X
10 mm
480 mm
Y
Figure 21.7
S OLUTION :
Iyy = 2 (IC.G + A x̄2 ) +
1
(48)3 × 1
12
= 2 422.11 + 78.46 × (16)2 +
48 × 48 × 48
12
= 59240 cm4
Ixx = 2 2058.4 +
48
(1)3 + (20.5)2 × 1 × 48
12
= 91124.8 cm4
Area of cross section = 78.46 × 2 + 48 × 1 × 2
= 252.92 cm2
Taking the least moment of inertia
k=
59240
258.92
= 15.3 cm
le = l = 8 m = 800 cm
le
800
=
= 52.29
k
15.3
@seismicisolation
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Columns and Struts
•
445
252.92
320
×
× 106
A σc
100
×
100
6
P=
2 =
1
le
(52.29)2
1+
1+a
7500
k
P=
1348907
= 988500 N
1.3646
= 988.5 kN
Ans
The following formulae are in addition to above, but are rarely used, and are mentioned for their
academic importance only.
1. Johnson’s Parabolic formula
P = σc A 1 − b
le
k
2 For mild steel, σc = 300 MPa and b = 0.00003 for pinned ends.
2. Straight line formula
P
l
= σc − Z
A
k
where Z is an empirical constant and where the range of application is strictly defined.
Some of the approximate empirical formulae used in practical designing are given below:
i) Stress at critical load for cast iron
P
= 23.8 − 0.6
A
l
N/mm2
k
ii) Stress at critical load for structural steel
P
= 367.5 − 2
A
l
N/mm2
k
Gordon’s formula
P=
σc A
2
l
1+a
b
where b is the least diameter or the width of the strut and a is the Gordon constant and is given by
σc Ab2
a =
.
EI π 2
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•
Strength of Materials
Columns Subjected to Eccentric Loading (Secant Formula)
(A) Euler’s Formula:
AB is a column of length l subjected to an eccentric load P at eccentricity e. Here the top of
column is free and bottom is fixed.
a
P
e
B
y
l
X
x
A
Figure 21.8
The bending moment at the section x from fixed end is given by
d2y
= P(a + e − y)
dx2
P(a + e)
d2y
P
+ y=
EI
dx2 EI
EI
∴
This is differentiation equation and the solution is given by
Y = C1 cos x
P
+C2 sin x
EI
P
+ (a + e)
EI
The slope at any section is given by,
dy
= −C1
dx
At A, x = 0 and y = 0 and also
P
sin x
EI
P
+C2
EI
P
cos x
EI
dy
=0
dx
∴
0 = C1 + (a + e)
P
and 0 = C2
EI
∴ C2 = 0 and C1 = −(a + e)
@seismicisolation
@seismicisolation
P
EI
Columns and Struts
At B, x = l, y = a
•
447
P
+ (a + e)
EI
P
∴ a = (a + e) 1 − cos l
EI
P
∴ (a + e) cos l
=e
EI
P
a + e = e sec l
EI
∴
a = −(a + e) cos l
The maximum bending moment for the column occurs at A and is equal to P (a + e)
P
∴ Maximum bending moment = M = P.e sec l
EI
Therefore, the maximum compressive stress for the column section at A,
P
P.e sec l
P
EI
σmax = σd + σb = +
(i)
A
Z
Z is section modulus.
If both the ends are hinged Eqn. (i) can be modified as
P
le
P.e sec
P
2 EI
σmax = σd + σb = +
(ii)
A
Z
because for a column with one end fixed and the other free, equivalent length le = 2l (as one end is
fixed and the other is free).
In general, for any end condition, formula can be written as
P
le
Pe sec
P
2 EI
σmax = +
(iii)
A
Z
where le is equivalent length depending upon the end condition and is given in the earlier table
before for ready reference.
It may be noted that in the
case of short columns (with no buckling) maximum moment is P.e
le
P
which is increased to Pe sec
in case of long columns.
2 EI
Formula of Eqn. (iii) is also known as Secant formula for eccentric loads.
@seismicisolation
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448
•
Strength of Materials
E XAMPLE 21.11: A circular hollow column having external diameter of 250 mm and internal
diameter of 170 mm is having a length of 4.2 m with both ends fixed. The column is eccentrically
loaded by a load of 220 kN with eccentricity 30 mm. Determine the extreme stresses on the column
section. Also find the maximum eccentricity in order that there may be no tension anywhere in the
section. Take E = 94 GN/m2
S OLUTION :
π
(0.252 − 0.172 )
4
= 0.026376 m2
Area =
Moment of inertia of the section,
π
(0.254 − 0.174 )
64
= 1.507 × 10−4 m4
I=
Equivalent length (both ends fixed) le =
l
2
le =
4.2
= 2.1 m
2
Maximum bending moment
le
M = P.e. sec
2
P
EI
le
P
To calculate the angle
2 EI
√
2.1
220000
=
1.05
0.0155
2
94 × 109 × 1.507 × 10−4
= 0.131 radians
0.131 × 180
= 7.51◦
=
π
1
sec 7.51◦ =
cos 7.51
1
= 1.0087
=
0.9914
∴
Maximum bending moment,
Mmax = P · e · ×1.0087
= 220 × (0.03) × 1.0087
= 6.66 kNm
@seismicisolation
@seismicisolation
Columns and Struts
•
449
Maximum compressive stress,
P M
+ ; P.e = 220 × 0.03 = 6.66 kNm
A Z
6.66 × (0.125)
200
0.25
+
=
Because y =
0.026376
2
1.507 × 10−4
= 7582.6 + 5524.2
σmax =
and
Z=
I
y
= 13106.8 kN/m2
= 13.11 MN/m2
Ans
For no tension corresponding to the maximum eccentricity:
P M
=
A
Z
P
le
P · e sec
P
2 EI
=
A
Z
220 × e × 1.0087 × 0.125
220
=
0.026376
1.507 × 10−4
1.507 × 10−4
e=
0.026376 × 1.0087 × 0.125
= 0.0453 m
= 45.3 mm Ans
Perry’s Formula for Eccentrically Loaded Column
Prof. Perry has given a formula to determine the safe load of an eccentrically loaded column. This
is known as Perry’s formula. Before deriving this formula which is limited, following assumptions
are made:
i) The formula gives the average value of stress in a column at failure.
ii) The formula is applicable only when the column is subjected to both axial as well as eccentric
loading.
l
yc
iii) Prof. Robertson also performed experiments and found that when the value of 2 = 0.003 ,
k
k
Perry’s formula is valid for initial curvature as well as eccentric loading.
( yc = distance of extreme fibre from neutral axis.)
In a column of effective length le subjected to a load P at an eccentricity of e, them bending
moment = P · e
σd = stream due to direct load
P
So σd =
A
@seismicisolation
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450
•
Strength of Materials
σmax = Maximum permissible stress
le = Effective or equivalent length of the column.
σb = Maximum compressive stress due to bending moment
M Myc
=
=
Z
Ak2 P
P · e sec l2e
EI
=
· yc
Ak2
π
P
P · e · yc
=
sec
2
2 PEuler
Ak
where PEuler =
π 2 EI
le2
P P · e · yc
π
σmax = +
sec
2
A
2
Ak
∴
Substituting
P
= σd
A
σmax = σd
As per Prof. Perry,
sec
∴
∴
π
2
π
e yc
1 + 2 sec
2
k
P
PEuler
P
PEuler
1.2PEuler
P
=
PEuler PEuler − P
approximately
PEuler
Let σEuler =
A
1.2 PEuler
π
P
1.2 σEuler
sec
=
=
2 PEuler PEuler − P σEuler − σd
σmax = σd 1 +
eyc 1.2 PEuler
·
k2 PEuler − P
σmax = σd 1 +
eyc 1.2 σEuler
·
k2 σEuler − σd
σmax
−1 =
σd
σmax
σEuler − σd
−1
=
σd
σEuler
σmax
σ
=
−1
1− d
σd
σEuler
eyc 1.2 σEuler
·
k2 σEuler − σd
1.2 e · yc
k2
1.2 e · yc
k2
This is known as Prof Perry’s formula.
@seismicisolation
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Columns and Struts
•
451
E XAMPLE 21.12: A hollow mild steel column has an external diameter 35 cm and internal diameter 32 cm and is 4.5 m long. It is subjected to a vertical load P acting at an eccentricity of 5.5 cm
when both the ends are fixed. The maximum compressive stress is limited to 3800 N/cm2 . Find the
maximum value of P it can carry. Use Perry’s formula and take E = 2 × 107 N/cm2 .
S OLUTION :
l
450
= 225 cm.
Effective length le = =
2
2
Cross-sectional area,
π 2
35 − 322
4
= 157.785 cm2
A=
Eccentricity of vertical load, e = 5.5 cm.
Maximum compressive stream, σmax = 3800 N/cm2
Young’s modulus, E = 2 × 107 N/cm2
Moment of inertia,
π 4
35 − 324
I=
64
= 22178.6 cm4
I = Ak2
k2 =
∴
k2 =
I
A
22178.6
157.785
= 140.56 cm2
PEuler =
π 2 EI π 2 × 2 × 107 × 22178.6
=
= 86388987.5 N
le2
2252
σEuler =
86388987
157.785
= 547510.8 N/cm2
yc =
D 35
=
= 17.5 cm
2
2
Applying Perry’s formula,
σmax
σ
1.2 eyc
−1
1− d
=
σd
σEuler
k2
σd
1.2 × 5.5 × 17.5
3800
−1 1−
=
σd
547510.8
140.56
@seismicisolation
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452
•
Strength of Materials
3800 − σd
σd
547510.8 − σd
547510.8
= 0.822
(3800 − σd ) (547510.8 − σd ) = 0.822 × σd × 547510.8
2080541040 − 547510.8 σd − 3800 σd + σd2 = 450 σd
σd2 − 551760.8σd + 2080541040 = 0
√
551760.8 ± 3.044 × 1011 − 8.32 × 109
σd =
2
551760.8 − 544132.3
σd =
2
= 3814.25 N/cm2
Hence, maximum value of P = 3814.25 × Area
= 3814.25 × 157.785
= 601831 N
= 601.8 kN
Ans
Exercise
21.1 A built-up beam shown in Fig. 21.9 is simply supported at its ends. Calculate its length, given
that when it is subjected to a load of 40 kN/m length, it deflects by 10 mm. Find out the safe
load, if this beam is used as a column with both ends fixed. Assume a factor of safety as 4.
Use Euler’s formula, take E = 210 GN/m2 .
30 cm
5 cm
2 cm
100
cm
5 cm
30
cm
Figure 21.9
[Ans
@seismicisolation
@seismicisolation
l = 14.15 m, 2332.5 kN]
Columns and Struts
•
453
21.2 A hollow cylindrical cast iron column is 4 m long with both ends fixed. Determine the minimum diameter of the column, if it has to carry a safe load of 250 kN with a factor of safety of
1
in Rankine’s
5. Take the internal diameter as 0.3 times the external diameter. Take a =
1600
formula and σc = 550 N/mm2 .
[Ans 108.8 mm]
21.3 A straight cylindrical bar has 16 mm diameter and is 1.2 m long. It is freely supported at its
two ends in a horizontal position and loaded at the centre with a concentrated load of 90 N.
The central deflection is found to be 5 mm.
If placed vertical and loaded along its axis, what load would cause it to buckle? What is
the ratio of the maximum stresses in the two cases?
[Ans 4.444 kN, 3.04]
21.4 Calculate the diameter of the piston rod from the following data:
Diameter of engine cylinder = 350 mm
Maximum effective stress pressure in the cylinder = 700 kN/m2
Distance from piston to cross-head centre = 1.5 m
Factor of safety = 4.
σc = 330 MN/m2
1
a=
for both ends fixed.
30, 000
[Ans 41.9 mm]
21.5 A strut 3 m long is constructed of steel tube 75 mm outside diameter and 3 mm thick. The
ends are pin-jointed, but the end load of 50 kN is applied eccentrically through a line parallel
to and 2.5 mm away from the axis of the strut, which is initially straight. Find the deflection
and the maximum stress at the centre of length, E = 200 GN/m2 .
[Ans 3.38 mm, 98.7 MN/m2 ]
21.6 A column having a section as shown in Fig. 21.10 is 3 m long. If the column is hinged at both
ends, find the crippling load by using Euler’s formula. E = 200 GPa.
Y
120 mm
A
B
16 mm
y = 54.1 mm
X
X
G
150 mm
16 mm
Y
Figure 21.10
[Ans 5.16.5 kN]
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•
Strength of Materials
21.7 A hollow column of external and internal diameters 150 mm and 110 mm, respectively and of
length 2.5 m is hinged at top and bottom ends. A load of 80 kN is applied at an eccentricity of
50 mm from the geometrical axis of the column. Determine the maximum stress in the column
section and also calculate the maximum eccentricity so that no tensile stress develops in the
column. Take E = 200 GPa.
[Ans 26.78 N/mm2 , 28.81 mm]
21.8 Find (Euler’s critical load) for a hollow cylindrical cast iron column 200 mm external diameter and 25 mm thick, 5 m long. The column is hinged at both ends. Take E = 80 GPa.
For what length the critical load by Euler’s and Rankine’s formula be equal? Take
1
.
σc = 600 N/mm2 and a =
1600
[Ans 1692.8 kN, 5.3 m]
21.9 A circular rod of diameter 40 mm and length 800 mm is subjected to an axial compressive
load of 100 kN at an eccentric distance of e from the axis of the rod, Assuming the lateral
deflection at the midpoint of the rod to be 0.8 mm, find the distance of eccentricity, e and the
maximum stress in the rod. Take E = 200 GPa.
[Ans 0.59 mm, 92.33 N/mm2 ]
21.10 A 2 m long column has a circular cross section of 60 mm diameter. One of the ends of the
column is fixed in direction end position and the end is free. Taking a factor of safety as 3,
calculate the safe load using:
i) Rankine’s formula. Take yield stress = 550 N/mm2 and a =
ii) Euler’s formula, E for material of column = 130 GPa
1
for pinned ends.
1600
[Ans 11.4 kN, 170 kN]
21.11 A hollow cast iron column with fixed ends supports an axial load of 1 MN. If the column is
4.5 m long and has an external diameter of 250 mm, find the thickness of metal required. Use
1
the Rankine’s formula, taking a constant of
for pinned ends and a working stress of 80
6400
2
MN/m .
[Ans 28 mm]
21.12 A pin-ended square cross-section column of length 3 m is subjected to a compressive stress
of 10 MPa. Using a factor of 2.5, find the cross section if the column is to safely support
(a) a 100 kN load and (b) a 200 kN load. Take E = 15 GPa.
[Ans 117 × 117 mm cross-section]
21.13 A 1.2 m long column has a circular cross section of 5 mm diameter. One of the ends of
the column is fixed in the direction and position and other end is free. Taking factor of safety
1
as 3, calculate the safe load using (i) Rankine’s formula with σC = 560 N/mm2 and a =
1600
for pinned ends and (ii) Euler’s formula with E for C.I = 120 GPa.
[Ans i) 9.9 kN, ii) 13.45 kN]
21.14 A straight bar, 1 m long, 25 mm wide and 25 mm thick is bent in the shape of a bow, the
elastic deflection at the middle being 50 mm. Its ends are joined by a bowstring. Assuming
the modulus of elasticity, E = 200 GPa, determine the force in bow string and the maximum
stress in the bar.
[Ans P = 0.645 N, σmax = 125 N/mm2 ]
@seismicisolation
@seismicisolation
Columns and Struts
•
455
21.15 An I-joist ISMB 250 @ 37.3 kg/m has an effective length of 5 m. It is used as a stanchion
with two plates 250 mm × 10 mm welded to it sides, as shown in Fig. 21.11. Compare the
load carrying capacity. What will be its load capacity if one plate is attached to each flange?
For ISMB 250 @ 37.5 kg/m, a = 47.55 cm2 , Ixx = 5137.6 mm4 , Iyy = 334.5 × 104 mm4
10 mm
10 mm
250 mm
125 mm
Figure 21.11
[Ans
@seismicisolation
@seismicisolation
814.25 kN, 876.2 kN]
22
C HAPTER
BENDING OF CURVED BARS
M σ
E
So far we have dealt with the bending of straight bars using equation
= = . But for bending
I
Y
R
curved bars like crane hooks, chain links, rings, etc., simple bending equation cannot be used.
Actually curved bars mean the bars with small or large initial curvature.
We have to make certain assumptions while deriving formulas for curved bars, these are stated
as follows:
1. Each layer of the beam is free to expand or contract independently of the layer above or
below it.
2. The bar material obeys Hooke’s law and is loaded within elastic limit.
3. The longitudinal fibres of the bar, parallel to centroidal axis exert no pressure on each other.
In other words, the distance between longitudinal fibre from centroidal axis is same before
and after bending.
4. The transverse sections which are plane before bending, remain plane after bending.
5. The value of Young’s modulus E is the same in tension and compression.
6. Radial strain is neglected.
Beams with Large Radius of Curvature (or Small Curvature)
Consider a curved bar as shown in Fig. 22.1(a) having Ri as its initial radius of curvature and R1 as
radius of curvature after application of end moments, M.
y
C
N
D
A
Ri
θ+δθ
(b)
o
D'
A'
Ri
M
θ
(a)
y
C'
N'
Figure 22.1
Let
Ri = Initial radius of curvature of the bar
R1 = Final radius of curvature
@seismicisolation
@seismicisolation
o
M
Bending of Curved Bars
•
457
θ = Initial angle subtended at the centre of the bar
θ + δ θ = Final angle after bending.
Change in length
After bending strain, ε =
Original length
C D −CD
CD
(R +y)(θ + δ θ ) − (Ri +y)θ
= 1
(Ri −y)θ
=
ε=
R1 (θ + δ θ ) + yδ θ − Ri θ
(Ri +y)θ
(i)
As shown in Fig. 21.1(b), NA = Ri θ and N A = R1 (θ + δ θ ) and, N A = NA
As we know the length of neutral axis NA will remain same, so NA = N A Ri θ = R1 (θ + δ θ ) = R1 θ + R1 δ θ
Ri δ θ = θ (Ri −R1 )
(R −R1 )
δθ
= i
θ
R1
(ii)
Substituting for Ri θ in Eqn. (i)
ε=
=
Now put the value of
R1 (θ + δ θ ) + yδ θ − R1 (θ + δ θ )
(Ri − y)θ
yδ θ
(Ri − y)θ
(iii)
δθ
from Eqn. (ii) in Eqn. (iii)
θ
ε=
=
Ri −R1
y
×
Ri + y
R1
y(Ri −R1 )
R1 Ri
Since, y is very small compared Ri and R1 , so negligible
1
1
−
Hence, ε = y
R1 Ri
1
σ
1
−
=y
E
R1 Ri
@seismicisolation
@seismicisolation
458
•
Strength of Materials
Therefore,
Hence,
1
1
−
R1 Ri
M
1
1
σ
=
=E
−
y
I
R1 Ri
σ
=E
y
E XAMPLE 22.1: A round steel bar of 55 mm in diameter is bent into a circular arc of 3.8 m radius
and the subtended angle is 85◦ . A couple is applied at each end so that the slope is changed to 90◦
at one end relative to other. If E is 200 GN/m2 , determine the maximum stress due to couple.
S OLUTION :
Diameter of bar = 55 mm
Initial radius of curvature = 3.8 m = 3800 mm
Initial angle, θ = 85◦
Final angle (θ + δ θ ) = 90◦
E = 200 GN/m2 = 200 × 103 N/mm2
55
Distance between outermost layer of bar and neutral axis =
= 27.5 mm and δ θ = 90 − 85 = 5◦
2
R1 = Final radius of arc
Now
δθ
(Ri − R1 )
=
θ
R1
3800 − R1
5
=
85
R1
5R1 = 323000 − 85R1
90R1 = 323000
323000
= 3588.8 mm
90
1
1
σ = E ·y
−
R1 Ri
R1 =
Therefore,
σmax = 200000 × 27.5
1
1
−
3588.8 3800
= 5500000 2.79 × 10−4 − 2.63 × 10−4
= 5500000 0.16 × 10−4
= 88 N/mm2
Ans
@seismicisolation
@seismicisolation
Bending of Curved Bars
•
459
Beam with Small Initial Radius of Curvature (or Large Curvature)
If the radius of curvature is more than 5 times the depth of the beam, a simple flexure formula can be
used. But analysis for beams having small initial radius of curvature was first dealt with by Winkler
and later by Andrews and Pearson. The following derivation is known as Winkler-Bach theory:
The beam ABCD is having an initial radius R. EF is its centroidal axis and NA is neutral axis
on NA, there will be no change in length when moment M is applied. After application of moment
M, A and B are new positions of points A and B such that A O B subtends an angle φ . Here, M
is considered positive which tends to increase the curvature of the beam. EF is centroidal axis. Let
PQ be any fibre at a distance of y from centroidal axis. From Fig. 22.2 it is understood that after
moment is applied ABCD takes the form A B C D .
Note: Same assumptions as stated in beam with small curvature are applicable here.
A
A'
C'
M
G
H
y
F
E
E'
N
C
D' A F'
C'
φ
o'
B
R'
B'
H'
M
θ
R
o
Figure 22.2
Now strain in fibre GH,
G H − GH
GH
(R + y) φ
(R + y) φ − (R + y) θ
=
−1
=
(R + y) θ
(R + y) θ
ε=
It can be written as
Strain in fibre EF = ε =
R +y
φ
= (1 + ε )
θ
R+y
(i)
E F − EF
R φ
R φ − Rθ
=
−1
=
EF
Rθ
Rθ
or
R
φ = 1+ε θ
R
@seismicisolation
@seismicisolation
(ii)
460
•
Strength of Materials
From Eqns. (i) and (ii)
(1 + ε )
R+y
R
=
1
+
ε
R + y
R
R R + y
(1 + ε ) = 1 + ε ×
R
R+y
R + y/R ε = 1+ε
−1
R + y/R
1 + y/R ε = 1+ε
−1
1 + y/R
or
Adding and subtracting ε on right-hand side,
ε = (1 + ε )
1 + y/R −1−ε +ε
1 + y/R
1 + y/R − 1(1 + ε ) + ε 1 + y/R
1 + y/R −1 +ε
= (1 + ε )
1 + y/R
1 + y/R − 1 − y/R
+ε
= (1 + ε )
1 + y/R
⎤
⎡ 1
1
⎢ y R − R ⎥
⎥
= ε + (1 + ε ) ⎢
⎣ 1 + y/R ⎦
= (1 + ε )
1
1
−
= ε + (1 + ε )
R
R
σ = E ε = E ε + (1 + ε )
y
1 + y/R
1
y
1
+
1 + y/R R R
(iii)
Total force on cross section C A or D B in the circumferential direction,
σ dA y = E ε dA (∵ σ = ε E)
y
1
1
=E
ε dA + (1 + ε )
+
dA
1 + y/R R R
1
1
y
F = E ε A + E(1 + ε )
+
dA
R R
1 + y/R
F=
@seismicisolation
@seismicisolation
(iv)
Bending of Curved Bars
•
461
The moment of resistance about an axis through the centroid,
M=
σ dA · y = E
ε y dA +
=E
ε y dA
(1 + ε )
y2
1 + y/R
1
1
−
R R
dA
Since y = 0 at the centroidal axis end
y dA = 0
∴
and
1
1
+
M = E(1 + ε )
R R
y2
dA =
1 + y/R
y2
dA
1 + y/R
(v)
yR + y2 − yR
dA
1 + y/R
yR (1 + y/R) − yR
yR
dA = yR −
dA
1 + y/R
1 + y/R
y
y
dA = R ydA − R
dA
y−
=R
1 + y/R
1 + y/R
=
= 0−R
yR
dA = −R
1 + y/R
let
Now let us find out the value of
As we assumed
yR
dA
1 + y/R
y2 dA
= Ah2
(1 + y/R)
y
dA
(1 + y/R)
y2
dA = Ah2
(1 + y/R)
Now,
y2
2
y + − y /R
y
R
dA =
dA
(1 + y/R)
(1 + y/R)
y y2
−
y
1
+
R
R
dA
=
(1 + y/R)
y2
1
=
dA
y− ×
R (1 + y/R)
@seismicisolation
@seismicisolation
(vi)
462
•
Strength of Materials
1
y2
×
dA
R (1 + y/R)
1
y2
= 0−
y dA = 0
dA
∵
R (1 + y/R)
1
y2
2
= − Ah
dA = Ah2
∵
R
(1 + y/R)
=
y dA −
Now Eqn. (iv) becomes,
1
1 h2
F = EA ε − (1 − ε )
−
R R R
As net force F = 0
∴
ε = (1 + ε )
1
1
−
R R
h2
R
Putting this value of ε in Eqn. (iv)
or
1
1 h2
y
1
1
σ = E (1 + ε )
−
−
+ (1 + ε )
R R R
1 + y R R
1
1 1 2
yR
−
h +
= E(1 + ε )
R R R
1 + y/R
y2
1
1
dA
−
Also we know, M = E(1 + ε )
R R
(1 + y/R)
1
1
∴ M = 1−ε
Ah2
−
R R
1
M
1
= 2
E 1 + ε
−
R R
Ah
M 1 2
yR2
σ = 2. h +
R+y
Ah R
M
R2 y
=
1+ 2
AR
h R +Y
(vii)
At inside of the centroidal axis, y is negative thus,
σ=
M
R2 y
1− 2
AR
h R−y
@seismicisolation
@seismicisolation
(viii)
Bending of Curved Bars
•
463
The position of neutral axis can be found out because we know σ = 0 at NA.
M
R2 y
1+ 2
=0
∴ σ=
AR
h R+y
yR2 = −Rh2 − yh2
or
or
y=
−Rh2
R2 + h2
(ix)
Note:
1. If bending moment M tends to decrease the curvature, then Eqn. (vi) will be compressive
stress and Eqn. (vii) will be tensile stress.
2. The bending moment is positive when it decreases the radius of curvature and negative when
it increases the radius of curvature.
Therefore, σ should be positive if it is tensile in nature.
3. y is positive if it is measured away from the centre of curvature and negative when measured
towards the centre of curvature.
Crane Hooks:
(Inside)
(Outside)
W
M
R2 y1
σx1 = +
1− 2
A AR
h (R − y1 )
W
M
R2 y2
σx2 = −
1+ 2
A AR
h (R + y2 )
Centre of
curvature
X2
G
X1 O
y2
W
y1
x
R
Figure 22.3
Values of h2 for Various Sections: h is known as link radius.
h2 =
we know,
1
A
y2
dx
1 + y/R
R
y2
dx
A R+y
2
R
R dA
=
ydA− RdA +
A
R+y
2
R dA
R
0 − RA +
=
A
R+y
=
∴
h2 =
R2
A
dA
− R2
R+y
@seismicisolation
@seismicisolation
464
•
Strength of Materials
Y
Rectangular Section
A rectangular section with centre of curvature 0
lying on Y -Y axis is shown in Fig. 22.4. Consider
an elementary stripe of width b and depth dy at
a distance y from the centroidal axis.
Area of the strip, dA = b dy
Area of section, A = b d
R3
h =
bd
+d/2
2
−d/2
dA
dy
y
X
b
b.dy
− R2
R+y
R3
R
Axis of curvature
O
Centre of curvature
+d/2
[loge (R + y)]−d/2 − R2
d
2R + d
R3
2
loge
− R2
h =
d
2R − d
2R + d
2
2 R
or h = R
loge
−1
d
2R − d
=
X d
G
Figure 22.4 Rectangular section
b
Circular Section
dy
π
A = d2
4
Considering a stripe of width b and the depth dy
at a distance y from the centroidal layer as shown
in Fig. 22.5
2
d
b=2
− y2
2
Area of strip, dA = b.dy
2
d
=2
− y2 · dy
2
R3
h =
bd
+d/2
2
−d/2
8R3
=
π d2
d2
R
O
Axis of curvature
Centre of curvature
Figure 22.5 Circular section
d 2
2×
− y2
2
.dy − R2
R+y
d
+d/2
−d/2
y
2
− y2
4
dy − R2
R+y
@seismicisolation
@seismicisolation
Bending of Curved Bars
•
465
Expanding binomial expression and then integrating, we have
h2 =
d2
d4
+...
+
16 128 R2
Triangular Section
b'
R+y = a
dy = da
Width of elementary strip b =
R2 −a
b
d
G
b
Area of elementary strip,
dy
d
R1
R
R2
R3
=
A
R1
O
dA
−R2
R+y
R2 b da
R1
a
Centre of curvature
Figure 22.6 Triangular section
−R2
da
R2 −a
b. −R2
d
a
R1
⎡
⎤
R2
R2
3
R ⎣ R2 b da b
−
=
da⎦ −R2
A
d
a d
R1
R1
3
R R2 b
R2 b
loge − (R2 −R1 ) −R2
=
A
d
R d
3
R
2h b
3R − 2d
=
d+
loge
−b −R2
A
3 d
3R − d
R3
=
A
3
Axis of curvature
Now,
R3
h =
A
d
R2
dA = b .dy = b .da
2
y
R2
Because,
d
3
d
R1 = R−
3
R2 − R1 = d
R2 = R+
@seismicisolation
@seismicisolation
466
•
Strength of Materials
3R + 2d
b
2R3
(3d + 2d)
loge
− b −R2
bd
3d
3R − d
1
Because A = bd
2
3
3R + 2d
2R 3R + 2d
2
loge
−1 −R2
h =
d
3d
3R − d
∴
or
h2 =
Trapezoidal Section
Figure 22.7 shows a trapezoidal section. Consider an elementary strip of width b and depth dy at
distance y from centroidal axis.
r = R + y ∴ dr = 0 + dy OR dr = dy
dy
b1 +b2
y
A=
×d
2
R2
Area of stripe, dA = b × dy
b1 +b2
(R2 −r)
where
b = b2 +
r
R1
d
b1 −b2
, (R2 −r) dy
dA = b2 +
d
b1 −b2
(R2 −r) dr {∴
= b2 +
d
Now
b2
d
b
G
b1
R
Axis of curvature
O
dy = dr}
Centre of curvature
Figure 22.7
R3
dA
−R2
A
(R + y)
b1 −b2
(R2 −r) dr
b
+
2
R3
d
=
−R2
A
(R + y)
h2 =
R3 b2 + (b1 −b2 ) (R2 −r) dr
−R2
A
r
R3
b2 (b1 −b2 ) ×R2 (b1 −b2 ) ×r
=
+
−
dr − R2
A
r
d ×r
d ×r
⎡
⎤
R2
3
R ⎣
b1 −b2
(b1 −b2 ) R2 ⎦ 2
b2 +
×R2 (log e r)RR21 −
(r)R1 −R
=
A
d
d
R1
3
R
b1 −b2
R2
b1 −b2
=
b2 +
×R2 log e r
(R2 −R1 ) −R2
−
A
d
R1
d
R3
(b1 −b2 )
R2
2
Hence, h =
b2 +
×R2 loge
− (b1 −b2 ) −R2
A
d
R1
= [∴ R2 −R1 = d]
h2 =
@seismicisolation
@seismicisolation
Bending of Curved Bars
T-Section
•
467
b2
Let its initial radius of curvature be r
Then, r = R + y
∴
dy
t2
dr = 0 + dy
R3
h =
A
2
dA
−R2
R+y
y
G
t1
Let T section be made of two rectangles b1 × t1 R3
Axis of
and b2 ×t2 .
curvature
Therefore,
⎡
⎤
R2
R2
R3 ⎣ b1 ×dr
b2 ×dr ⎦ 2
2
−R
h =
+
A
r
r
R1
R1
R2
R2
R3
= × b1 loge
+b2 loge
−R2
A
R1
R2
where, A = b1t1 +b2t2
r
b1
R2
R
O
R1
Centre of
curvature
Figure 22.8
I-Section
t3
b3
r = R+y
dr = 0 + dy
dr = dy
dy
R3
1
dA − R2
h =
A
R+y
R3 dA
=
−R2
A
r
2
y
G
b2
r
R4
R3
t1
b1
R2 R1
R
O
Axis of
curvature
Centre of
curvature
Figure 22.9
I section is made up of three rectangles: b1t1 , b2t2 and b3t3
⎡
⎤
R
R
R
3 2 dA 3 dA 4 dA
R
⎣
⎦ −R2
Now, h2 =
+
+
A
r
r
r
R1
R2
R3
⎡
⎤
R
R
3
2
R4
R3 ⎣ b1 dr
b2 dr
b3 dr ⎦ 2
=
+
+
−R
A
r
r
r
R1
R2
R3
3
R2
R3
R4
l
+ b2 loge
+ b3 loge
− R2
=
b1 loge
A
R1
R2
R3
@seismicisolation
@seismicisolation
t2
468
•
Strength of Materials
Total area of cross section of the beam is
A = b1 t1 + b2 t2 + b3 t3
E XAMPLE 22.2: A curved bar of square section 85 mm sides and mean radius of curvature 125
mm is initially unstressed if a bending moment of 800 Nm is applied to straighten it then determine
the stresses that develop at the inner and the outer faces of the bar.
S OLUTION :
85
d = 85 mm
Area of cross section, A = 852 = 7225 mm2
Radius of curvature R = 125 mm
Applied bending moment = −800 Nm
85 mm
Axis of
curvature
125 mm
= −800 × 103 Nmm
(negative sign is because curvature is being
decreased)
R3
2R + d
2
Now, h =
loge
− R2
d
2R − d
1253
2 × 125 + 85
=
loge
− 1252
85
2 × 125 − 85
Figure 22.10
325
− 15625
165
= 22978 × −15625
= 22978 loge
= 643.4 mm2
M
y
R2
Now σ =
1+ 2 ×
AR
R+y
h
For outer surface,
85
= 42.5 mm
2
800 × 103
1252
42.5
σ =−
1+
×
7225 × 125
643.4 125 + 42.5
y=
= −0.886 [1 + 6.16]
= −6.34 MN/m2
Stress at inner surface:
85
= 42.5 mm
2
M
y
R2
σ=
1− 2 ×
AR
R−y
h
y=
@seismicisolation
@seismicisolation
(Compressive) Ans
Bending of Curved Bars
=−
•
469
1252
42.5
800 × 103
1−
×
7225 × 125
643.4 125 − 42.5
= −0.886 [1 − 12.51]
= +10.19 MN/m2 (Tensile) Ans
E XAMPLE 22.3: A beam of circular section of diameter 18 mm has its centre line curved to a
radius of 44 mm. Find the maximum stresses in the beam when subjected to 4.5 kNmm moment.
S OLUTION :
Diameter,
d = 18 mm
Radius of curvature,
R = 44 mm
Bending moment = 4.5 kNmm = 4500 Nmm
π
Area = (18)2 = 254.34 mm2
4
Distance between centre line and extreme fibre,
18
= 9 mm
2
d2
d2
h2 =
+
16 128R2
182
182
=
+
16 128(44)2
y=
= 20.25 + 0.00131
= 20.251 mm2
Maximum stress at bottom surface
R2 y
M
1+ 2
σ=
AR
h (R + y)
4500
442 × 9
=
1+
254.34 × 44
20.251(44 + 9)
= 0.4021 (1 + 16.23)
= 6.93 MN/m2 (Tensile) Ans
Maximum stress at top surface,
M
R2 y
σ=
1− 2
AR
h (R − y)
@seismicisolation
@seismicisolation
470
•
Strength of Materials
=
4500
(1 − 11.31)
254.34 × 44
= 0.4021 (−10.31)
= −4.14 MN/m2
(Compressive)
Ans
E XAMPLE 22.4: A curved bar with a mean radius of curvature of 140 mm is initially unstressed
and is 70 mm wide by 86 mm deep rectangular section in the plane of bending. To straighten it a
bending moment of 2.4 kNm is applied.
Determine the position of neutral axis and also the maximum bending stresses.
S OLUTION :
b = 70 mm; d = 86 mm
R = 140 mm; M = 2.4 kNm = −2.4 × 106 Nmm
(negative sign is taken as bending moment tends to decrease curvature)
Area of cross-section, A = bd = 70 × 86 = 6020 mm2
2R + d
R
loge
−1
h2 = R2
d
2R − d
2 × 140 + 86
140
= 1402
loge
−1
86
2 × 140 − 86
= 1402 [1.628 loge 1.89 − 1]
= 19600 [0.0363]
= 711.48 mm2
Position of neutral axis:
y=−
=
Rh2
R2 + h2
−140 × 711.48
(140)2 + 711.48
=−
99607.2
20311.5
= −4.9 mm
Ans
(−ve sign means the neutral axis is on compression zone (i.e., concave side) of the centroidal
axis) maximum bending stresses:
@seismicisolation
@seismicisolation
Bending of Curved Bars
•
471
86
= 43 mm
2
R2
M
y
1− 2 ·
σ max =
AR
h R−y
2.4 × 106
1402
43
=−
1−
×
6020 × 140
711.48 140 − 43
Inner side of curve concave side y =
= −2.848 (1 − 12.21)
= +31.93 MN/m2
(Tensile) Ans
Outer side (convex side) of curve:
R2 y
H
1+ 2
σ =−
AR
h R+y
2.4 × 106
1402
43
=−
1+
×
6020 × 140
711.48 140 + 43
= −2.848 (1 + 6.47)
= −21.27 MN/m2 (Compressive) Ans
E XAMPLE 22.5: Figure 22.11 shows a frame of rectangular cross section subjected to a load of
3 kN. Find: i) the resultant stresses at points X and Y and ii) position of neutral axis.
3 kN
130 mm
X
Y
52 mm
3 kN
20
G
50 mm
Figure 22.11
@seismicisolation
@seismicisolation
472
•
Strength of Materials
S OLUTION :
Area cross section at X-Y
A = 20 × 50 = 1000 mm2 = 1 × 10−3 m2
(m) Bending moment = −3 × 103 (130 + 50) × 10−3 = −540 Nm
(Negative sign because moment tends to decrease the curvature.)
(i) Resultant stresses at points X and Y :
3 × 103
= 3 MN/m2 (Tensile)
1 × 10−3
2R + d
2
2 R
h =R
loge
−1
d
2R − d
Direct stress =
R = 52 mm = 0.052 m, d = 50 mm = 0.05 m
0.052
2 × 0.052 + 0.05
h2 = (0.052)2
loge
−1
0.05
2 × 0.052 − 0.05
0.154
−3
= 2.7 × 10
1.04 loge
−1
0.054
= 2.7 × 10−3 [1.04 × 1.048 − 1]
= 2.43 × 10−4 m2
Bending stress due to bending moment at Y : y =
0.05
= 0.025
2
M
R2 y
σby =
1− 2 .
AR
h R−y
(0.052)2
0.025
−540
1−
×
=
1 × 103 × 0.052
2.43 × 10−4 0.052 − 0.025
= −10.385 [1 − 10.303]
= +96.61 MN/m2
(Tensile)
Bending stress due to bending moment at X : y =
Ans
0.05
= 0.025
2
M
y
R2
1+ 2 ·
AR
h R+y
(0.052)2
0.025
−540
1
+
×
=
3
−4
0.052 + 0.025
1 × 10 × 0.052
2.43 × 10
= −10.385[1 + 3.61]
σbx =
= −47.87 MN/m2
(Compressive)
@seismicisolation
@seismicisolation
Ans
Bending of Curved Bars
•
473
Resultant stress at point Y ,
σY = 3 + 96.61 = 99.61 MN/m2
(Tensile) Ans
Resultant stress at point X,
σX = 3 − 47.87 = 44.87 MN/m2
(Compressive)
Ans
(ii) Position of neutral axis:
Y =−
Rh2
R2 + h2
=−
0.052 × 2.43 × 10−4
(0.052)2 + 2.43 × 10−4
=−
0.12636 × 10−4
27.04 × 10−4 + 2.43 × 10−4
= −0.00429 m
= −4.29 mm
Negative sign means neutral axis is at 4.29 mm away from centroidal axis towards the centre
of curvature. Ans
E XAMPLE 22.6: Figure 22.12 shows a circular ring of rectangular section, with a slit and subjected to load P.
Calculate the magnitude of the force P if the maximum stress along the section 1-2 is not to
exceed 250 MN/m2 .
55
mm
P
G
85
2
G
1
m 10
m
R
1
1
m 55
m
200
A
Figure 22.12
Area at section 1 − 2, A = 0.055 × 0.085 = 4.675 × 10−3 m2
Allowable stress, σ = 250 MN/m2
@seismicisolation
@seismicisolation
474
•
Strength of Materials
Bending moment
M = P(0.155) = 0.155 PNm
(M is taken +ve because it tends to increase the curvature)
Magnitude of the force P:
P
P
=
= 213.9 PN/m2
A 4.675 × 10−3
R
2R + d
loge
−1
h2 = R2
d
2R − d
Direct stress =
R = 0.155 m, d = 0.085 m
2 × 0.155 + 0.085
2 0.155
2
loge
−1
h = (0.155)
0.085
2 × 0.155 − 0.085
0.395
−1
= 0.024 1.8235 loge
0.225
= 0.024 [1.0262 − 1]
= 6.28 × 10−4 m2
Resultant stress at point 2, y =
0.085
= 0.0425
2
M
y
R2
P
σmax =
1− 2 ·
+ (Compressive)
AR
A
h R−y
0.1552
P
0.155P
0.0425
1
−
250 × 106 =
+
×
−3
−4
0.155 − 0.0425
4.675 × 10 × 0.155
6.28 × 10
4.675 × 10−3
250 × 106 = 213.9P [1 − 14.45] + 0.214 × 103 P
= −2876.9P − 214P
= 3090.9P
∴
P=
250 × 106
3090.9
= 80882.6 N
= 80.88 kN Ans
E XAMPLE 22.7: A crane hook is shown in Fig 22.13, lifting a load of 160 kN. Determine the
maximum compressive and tensile stresses in the critical section A-B of the crane hook.
@seismicisolation
@seismicisolation
Bending of Curved Bars
•
475
B
O
G
b1 = 130
A
b2 = 40
d = 175
d2
d1
10
0
Section along AB
All dimensions are in mm
160 kN
Figure 22.13
S OLUTION :
b1 = 130 mm = 0.13 m
b2 = 40 mm = 0.04 m
d = 175 mm = 0.175 m
d b1 + 2b2
for locating centre of gravity, d1 =
3 b1 + b2
Centre of gravity may be found out by splitting the section as usual or use the above formula.
0.175 0.13 + 2 × 0.04
d1 =
3
0.13 + 0.04
= 0.072 m
d2 = 0.175 − 0.072 = 0.103 m,
A=
0.175
(0.13 + 0.04)
2
= 0.014875 m2 = 0.0149
R = 0.1 + 0.072 m = 0.172 mm, R1 = 0.1 m, R2 = 0.1 + 0.175 = 0.275 m
R2
R3
b1 − b2
2
b2 +
× R2 loge
h =
− (b1 − b2 ) − R2
A
d
R1
(0.13 − 0.04)
0.275
0.1723
2
0.04 +
× 0.275 loge
− (0.13 − 0.04)
h =
0.0149
0.175
0.1
− 0.1722
= 0.3415 [{0.04 + 0.514 × 0.275} loge 2.75 − (0.09)] − 0.0296
= 0.3415 [0.18135 × 1.012 − 0.09] − 0.0296
= 0.00234 m2
Bending moment = −160 × 103 × 0.172 = −27520 Nm
(Negative sign is taken because bending moment is tending to decrease the curvature.)
@seismicisolation
@seismicisolation
476
•
Strength of Materials
160 × 103
× 10−6 = 10.74 MN/m2 (Tensile)
0.0149
Bending stress at A,
R2
M
y
1+ 2 ×
σbA =
AR
h R+y
27520
0.1722
0.103
=−
1+
×
0.0149 × 0.172
0.00234 0.172 + 0.103
Direct stress, σbA =
= −10738255 [1 + 4.735] ×10−6
= −61.85 MN/m2
(Compressive) Ans.
Bending stress at B,
M
y
R2
σbB =
1+ 2 ×
AR
h R−y
27520
0.1722
0.172
=−
1+
×
× 10−6
0.0149 × 0.172
0.00234 0.172 − 0.072
= −10738255 [1 − 9.1] ×10−6
= 86.98 MN/m2
(Tensile) Ans
E XAMPLE 22.8: When a curved beam of trapezoidal section of bottom width 35 mm, top width
24 mm and height 44 mm is subjected to pure bending moment of +650 Nm. The bottom width is
towards the centre of curvature. The radius of curvature is 54 mm and beam is curved in a plane
parallel to depth. Determine i) location of neutral axis and ii) Maximum and minimum stresses.
Also plot the variation of stresses across the section.
S OLUTION :
55.53
24
d2= 23.36 mm
Centroidal
Axis
N
d1= 20.64 mm
G
44 mm
A
35 mm
R = 54 mm
15
10
5
42.58
33.21
22.26
A 9.27
N
6.37 5
25.55
10
49.66
15
85.47
Stresses are in MPa
Figure 22.14
@seismicisolation
@seismicisolation
Bending of Curved Bars
•
b1 = 35 mm, b2 = 24 mm, d = 44 mm
M = +650 Nm = 650000 Nmm; R = 54 mm
b1 + 2b2 d
35 + 2 × 24
44
d1 =
=
×
= 1.407 × 14.67
b1 + b2 3
35 + 24
3
= 20.64 mm
R1 = R − d1 = 54 − 20.64 = 33.36 mm
R2 = R + d2 = 54 + 23.36 = 77.36 mm
Area,
44
= 1298 mm2
A = (35 + 24)
2
R3
b1 − b2
R2
2
h =
b2 +
× R2 loge
− (b1 − b2 ) − R2
A
d
R1
35 − 24
77.36
543
24 +
× 77.36 loge
− (35 − 24) − 542
=
1298
44
33.36
= 121.31 [{24 + 0.25 × 77.36} 0.842 − 11] − 2916
= 121.31 [{43.34} 0.842 − 11] − 2916
= 176.47 mm2
i) Location of neutral axis:
Let y = distance of neutral axis from centroidal axis.
The location of neutral axis is given by,
y=−
=−
R × h2
54 × 176.47
=− 2
2
2
R h
54 + 176.47
9.529.4
= −3.08 mm
3092.47
Ans
−ve sign means the neutral axis is below the centroidal axis at a distance of 3.08 mm.
ii) Stresses:
y
M
R2
σ=
1− 2
RA
R−y
h
For max stress, y = 20.64, h2 = 176.77 mm2
M = +650 Nm = 650000 Nmm
R = 54 mm, A = 1298 mm2
650000
542
20.64
=
1−
54 × 1298
176.47 54 − 20.64
@seismicisolation
@seismicisolation
477
478
•
Strength of Materials
= 9.27 [1 − 10.22]
= −85.47 MN/m2
= 85.47 MN/m2 (Compressive) Ans
Minimum stress occur at y = 23.36 mm
542
23.36
650000
1+
54 × 1298
176.47 54 + 23.36
= 9.27 [1 + 4.99]
σmin =
= 55.53 MN/m2 (Tensile) Ans
iii) Plotting the stresses across the section
General formula:
R2
y
M
1+ 2
R×A
R+y
h
650000
542
y
=
1+
54 × 1298
176.47 54 + y
y
= 9.27 1 + 16.53
54 + y
σ=
At y = 23.36, σ = +55.53 MN/m2
At y = 15 mm, σ = 9.27 1 + 16.53
15
= 42.58 MN/m2
54 + 15
10
At y = 10 mm, σ = 9.27 1 + 16.53
= 33.21 MN/m2
54 + 10
5
At y = 5 mm, σ = 9.27 1 + 16.53
= 22.26 MN/m2
54 + 5
0
= 9.27 MN/m2
At y = 0 mm, σ = 9.27 1 + 16.53
54 + 0
−5
At y = −5 mm, σ = 9.27 1 + 16.53
= −6.37 MN/m2
54 − 5
−10
= −25.55 MN/m2
At y = −10 mm, σ = 9.27 1 + 16.53
54 − 10
−15
= −49.66 MN/m2
At y = −15 mm, σ = 9.27 1 + 16.53
54 − 15
y = −20.64 mm σ = −85.47 MN/mm2 (already calculated)
@seismicisolation
@seismicisolation
Bending of Curved Bars
•
479
E XAMPLE 22.9: Fig. 22.15 shows an open ring having T -section. Determine the stresses at the
points X and Y . If the ring in subjected to a load of 160 kN.
160 kN
t2=220 mm
A
R
b1 = 160 mm X
b2 = 35 mm
y2
5
27
Y
y1
R1
B t = 32 mm
1
R2
R3
160 kN
Figure 22.15
Area of T section = b1 t1 + b2 t2 = A
A = 160 × 32 + 35 × 220
A = 5120 + 7700
A = 12820 mm2 = 0.01282 m2
Let position of centre of gravity from AB = y1
y1 =
160 × 32 × 16 + 35 × 220 × 142
= 91.7 mm = (0.0917 m)
12820
y2 = (220 + 32) − 91.7 = 160.3 mm = (0.1603 m)
R3
R2
R3
2
h =
+ b2 loge
b1 loge
− R2
A
R1
R2
R1 = 275 mm = 0.275 m
R2 = 275 + 32 = 307 mm = (0.307 m)
R3 = 275 + 32 + 220 = 527 mm = (0.527 m)
R = 275 + 91.7 = 366.7 mm = (0.3667 m)
(0.3667)2
0.307
0.527
2
h =
0.160 loge
+ 0.035 loge
− (0.3667)2
0.01282
0.275
0.307
= 3.846 [0.0181 + 0.0019] − 0.1344
= 0.00829 m2
@seismicisolation
@seismicisolation
480
•
Strength of Materials
160000 × 10−6
= 12.48 MN/m2 (Compressive)
0.01282
Bending moment = M = +P × R
(+ve sign is taken because bending moment is tending to increase the curvature)
Bending stress at X,
M
R2
y2
(σb )X =
1+ 2
AR
R + y2
h
(0.3667)2
0.1603
160000 × 0.3667
1+
=
0.01282 × 0.3667
0.00829
(0.3667 − 0.1603)
Direct stress =
= 12480499.2 [1 + 12.6]
= 169.7 MN/m2 tensile
M
R2
y1
(σb )Y =
1+ 2
AR
R − y1
h
0.0917
160000 × 0.3667
(0.3667)2
=
1−
0.01282 × 0.3667
0.00829
(0.3667 − 0.0917)
= 12480499.2 [1 − 5.41]
= −55.04 MN/m2 (Compressive)
Hence, resultant stress at x = σd + (σb )X = −12.48 + 169 = 156.52 MN/m2 (Tensile) Ans
resultant stress at y = σd + (σb )Y = −12.48 − 55.04 = 67.52 MN/m2 (Compressive)
Stresses in circular ring:
P
R
M
R
θ
d
d1
d2
O
P
Figure 22.16
In such case there will be two stresses:
P
× sin θ
(i) Direct stress, σ0 =
2A
@seismicisolation
@seismicisolation
θ
Pcos θ
P
2
Ans
Bending of Curved Bars
(ii) Bending stress due to bending moment M,
where
M = bending moment = P × R
M
R2
y
Bending stress, σb =
1+ 2
A×R
R+y
h
1 sin θ
−
π
2
•
481
Bending stress will be maximum at the outer edge and inner edge of the cross section
Where y = +d2 for extreme outer edge from centre line
and y = −d1 for extreme inner edge from centre line
Resultant stress, σ = σ0 + σb
E XAMPLE 22.10: A ring of mean radius of curvature 100 mm is subjected to a pull of 4 kN.
The line of section of load passes through the centre of ring. Determine the maximum tensile and
compressive stresses in the material of the ring if the diameter of cross section of the ring is 20 mm.
S OLUTION :
R = 100 mm
Pull , P = 4 kN, diameter of cross section = 20 mm.
π
Area of cross section, A = (20)2 = 314 mm2
4
h2 =
=
1 d4
d2
+
16 128 R2
1
204
202
+
×
16 128 1002
= 25 + 0.125
= 25.125 mm2
(i) direct stress, σ0 =
P
x sin θ
2A
Direct stress at θ = 0
σ0 =
Bending moment at θ = 0
P
× sin θ = 0
2A
1 1
M = P×R
− sin θ
π 2
1 1
− sin 0
= P×R×
π 2
=
4000 × 100
= 127388.5 Nmm
π
@seismicisolation
@seismicisolation
482
•
Strength of Materials
Bending stress at θ = 0, at outer edge:
R2
M
y
1+ 2 ×
σb o =
A×R
R+y
h
127388.5
1002
10
=
1+
×
314 × 100
25.12 100 + 10
20
+ sign because y =
= 10 mm at outer edge
2
= 4.06 [1 + 36.19]
= 146.93 N/mm2 tensile
20
Bending stress at the innermost of the cross section when θ = 0 and y = − = −10 mm.
2
[−ve sign because y is towards centre of curvature.]
1002
10
(σb )i = 4.06 1 −
×
25.12 100 − 10
= 4.06 [1 − 44.23]
= −175.51 N/mm2 (Compressive)
∴
Ans
Resultant stress at the outermost edge,
(σr )0 = σ0 + (σb0 )
= 0 + 146.93 = 146.93 (Tensile)
Ans
And resultant stress if the innermost edge
(σr )i = σ0 + (σbi )
= 0 − 175.51
= −175.51 = 175.51 N/mm2 (Compressive) Ans
(ii) Stresses when θ = 90◦ =
π
2
P
p
× sin 90◦ =
2A
2A
4000
×1
=
4 × 314
Direct stress, σ0 =
= 3.185 N/mm2 (Tensile)
@seismicisolation
@seismicisolation
Bending of Curved Bars
Bending moment
1 1
− × sin 90◦
π 2
1 1
= 4000 × 100
−
π 2
M = P×R
= 400000 (0.3185 − 0.5)
= −72600 Nmm
The bending stress at outermost edge
where y = 10 mm
R2
M
y
1+ 2 ×
(σb )o =
AR
R+y
h
1002
10
−72600
1+
×
=
314 × 100
25.12
100 + 25.12
= −2.312 [1 + 31.82]
= −73.57 N/mm2 (Compressive)
The bending stress at innermost edge
where y = −10 mm
R2
M
y
1− 2 ×
A×R
R −y
h
1002
10
−72600
1−
×
=
314 × 100
25.12
100 − 25.12
(σb )i =
= −2.312 [1 − 53.16]
= +120.6 N/mm2 (Tensile)
Resultant stress at outer most edge,
(σr )o = σo + (σb )o
= 3.185 − 73.57
= −70.385 (Compressive) Ans
@seismicisolation
@seismicisolation
•
483
484
•
Strength of Materials
Resultant stress at innermost edge,
(σr )i = σo + (σb )i
= 3.185 + 120.6
= 123.78 N/mm2
Hence maximum tensile stress = 146.93 N/mm2
Maximum compressive stress = 175.51 N/mm2
Ans
Ans
Ans
Stress in a Chain Link
P
A
A
R
Direct stress σo =
X
P
sin θ
2A
Bending moment, M in curved portion
P × R 2R + l
=
− sin θ
2
2R − l
θ
B
l
Bending moment, M in straight portion
P × R 2R + l
−1
=
2
2R + l
P
Figure 22.17
E XAMPLE 22.11: On a simple chain link as shown in Fig. 22.18, a pull of 30 kN is applied. The
radius of curvature of the semicircular ends is 43 mm. The circular cross section of the link has
diameter of 46 mm. If the length of straight portion is 48 mm, determine the stresses in curved
portion and straight portion of link.
@seismicisolation
@seismicisolation
Bending of Curved Bars
•
485
S OLUTION :
30 kN
P = 30 kN = 30000 N
R = 45 mm
θ
45 mm
Cross-sectional diameter d = 46 mm
l = 43 mm
l = 43 mm
Area of the cross-section =
π
(46)2
4
= 1661.1 mm2
h2 =
=
30 kN
462
1
464
+
× 4
16 128 45
= 132.25 + 8.5 × 10−3
Figure 22.18
= 132.258 mm2
(a) Stress in curved portion:
P
× sin θ
2A
R2
M
y
1+ 2 ×
Bending stress, σb =
A×R
R+y
h
Direct stress, σo =
(Sign of y depends on its position)
Where M is the bending moment in curved portion,
P × R 2R + l
M=
− sin θ
2
2R + l
When θ = 0
1
d4
d2
+
× 4
16 128 R
P × R 2R + l
=
− sin 0
2
πR + l
P × R 2R + l
(as sin 0 = 0)
=
2
πR + l
30000 × 45 2 × 45 + 43
=
2
π × 45 + 43
@seismicisolation
@seismicisolation
486
•
Strength of Materials
= 675000 [0.722]
= 487350 Nmm
46
487350
452
23
At outer fibre, y =
= 23, (σb )o =
1+
2
1661.1 × 45
132.258 45 + 23
= 6.52[1 + 5.18]
= 40.29 N/mm2 (Tensile)
∴
Resultant stream at the extreme outer edge,
Since, θ = 0, σ0 =
So
P
× sin 0 = 0
A×R
(σr )0 = r0 + (σb )0
= 0 + 40.29 = 40.29 N/mm2 (Tensile) Ans
Bending stress at the extreme inner edge where θ = 0 and y = r = −23 mm
23
(45)2
(σb )i = 6.52 1−
×
132.26 45.23
= 6.52[1 − 16]
= −97.8 N/mm2 (Compressive) Ans
Resultant stress at the extreme inner edge,
( σ r )i = σ 0 + ( σ b )i
= 0 − 97.8 N/m2
= −97.8 N/mm2 (Compressive) Ans
(b) Stresses at θ = 900 :
P
× sin θ
2A
30000
=
× sin 90◦
2 × 1661.1
Direct stress σ0 =
= 9.03 N/mm2
P × R 2R + l
M=
−1
2
πR + l
@seismicisolation
@seismicisolation
Bending of Curved Bars
30000 × 45 2 × 45 + 43
−1
=
2
π × 45 + 43
133
= 675000
−1
1843
= −187886.6 Nmm
Bending stress at outer edge y = 23 mm
(σb )0 =
452
23
−187886.6
1+
×
1661.1 × 45
132.26
45 + 23
= −2.513(1 + 5.18)
= −15.53 N/mm2 (Compressive)
Resultant stress = σ0 + (σb0 )
= 9.03 − 15.53
= −6.5 N/mm2 (Compressive) Ans
(c) Stresses in straight portion:
σ0 = direct stress
=
30000
P
=
= 9.03 N/mm2
2A 2 × 1661.1
σb = bending stress
M
×y where M is bending moment
I
P × R 2R + l
σb =
−1
2
2R + l
=
= −187886.6 Nmm
I=
(Refer (b))
π d4
64
π (46)4
= 219675.2 mm4
64
−187886.6
M
×23 at outer edge, y = 23 mm
(σb )0 = ×y =
I
219675.2
=
= −19.67 N/mm2 (Compressive)
@seismicisolation
@seismicisolation
•
487
488
•
Strength of Materials
Resultant stress at outer edge,
= σ0 + (σb )0
= 9.03 − 19.67
= −10.64 N/mm2 (Compresive) Ans
(σb )i = bending stress at inner edge of straight
position where y = −23 mm
(σb )i =
−187886.6
× (−23)
219675.6
= 19.67 N/mm2 (Tensile)
Resultant stress at inner edge = σ0 + (σb )i
= 9.03 + 19.67
= 25.7 N/mm2 (Tensile) Ans
Exercise
22.1 Determine the ratio of maximum and minimum value of the stresses for a curved bar of
rectangular section in pure bending. Radius of curvature is 80 mm and depth of the beam is
60 mm. Locate neutral axis also.
[Ans 1.682,4 mm]
22.2 A crane hook whose horizontal cross section is trapezoidal, 50 mm wide at the inside and
25 mm wide at the outside, thickness 50 mm carries a vertical load of 9810 N whose line of
action is 38 mm from the inside edge of this section. The centre of curvature is 50 mm from
the inside edge. Calculate the maximum tensile stress and compressive stress set up.
[Ans 49.89 N/mm2 (Tensile), 30.92 N/mm2 (Compressive)]
22.3 A central horizontal section of a hook is a symmetrical trapezium of inner width = 67.5 mm
and outer width = 22.5 mm. The depth of section is 90 mm. The load line passes through the
centre of curvature. The radius of the hook is 52.5 mm. Determine the maximum compressive
and tensile stresses in the section of the hook.
[Ans 83.7 MN/m2 (Tensile), 43.2 MN/m2 (Compression)]
22.4 A round bar of steel 38 mm diameter is bent into a curve of mean radius 31.7 mm. If a bending
moment of 4.6 Nm tending to increase the curvature acts on the bar find the maximum tensile
and compressive stresses.
[Ans 0.564 MN/m2 (Tensile), 1.605 MN/m2 (Compressive)]
22.5 A ring of mean diameter 250 mm is made of round steel of 20 mm diameter. A pull of 4 kN
acts on the ring. Calculate the greatest tensile and compressive stresses.
[Ans 33.6 MN/m2 (Tensile), 30.9 MN/m2 (Compressive)]
22.6 The links of a chain are made of 12.5 mm diameter round steel and have semicircular ends,
the mean radius of which is 37.5 mm. The ends are connected by straight pieces 37.5 mm
@seismicisolation
@seismicisolation
Bending of Curved Bars
•
489
long. Find the tensile and compressive stresses at various points of the link under a load of
10 kN.
[Ans Stress in straight position: 310.2 MPa (Tensile) 228.8 MPa (Compressive). At
junction of semicircle and straight position: 349.6 MPa (Tensile) 200.6 MPa (Compressive).
Curved portion: 633.44 MPa (Tensile), 810 MPa (Compressive))]
22.7 A crane hook whose horizontal cross section is trapezoidal 50 mm wide at the inner side,
25 mm wide at the outer side and 50 mm thick carries a load of 10 kN whose line of action is
60 mm from the inner edge of the section. The centre of curvature is 50 mm from the inside
edge. Determine the maximum tensile and compressive stresses set up in the material.
[Ans 67.68 MPa (Tensile), 43.52 MPa (Compressive)]
22.8 Figure 22.19 shows a frame subjected to a load of 2.4 kN. Find (i) The resultant stress at
points 1&2; (ii) Position of neutral axis.
2.4 kN
1
2
[Ans 91.72 MN/m2 (Tensile)
39.84 MN/m2 (Compressive)
4.35 mm]
120
48
2.4 kN
Dimensions are in mm.
18
48
Figure 22.19
22.9 A circular ring made of a steel bar of 20 mm diameter is subjected to a pull of 15 kN along its
vertical diameter. The outer diameter of the ring is 140 mm. Determine the maximum tensile
and compressive stresses in the ring.
[Ans 321.3 MPa (Tensile), 413.5 MPa (Compressive)]
22.10 A curved beam of T section as shown is Fig. 22.20 is subjected to pure bending moment of
900 Nm which try to decrease the curvature. Find the position of neutral axis and the bending
stresses at the outer most and inner most fibres.
20 mm
C. G.
60 mm
[Ans σouter = −29.45 N/mm2 (Compressive), σinner = 17.54 N/mm2 (Tensile) −1.562
mm]
R
20 mm
80 mm
Figure 22.20
300 mm
Axis of curvature
@seismicisolation
@seismicisolation
490
•
Strength of Materials
22.11 A circular ring of 60 mm wide and 90 mm deep rectangular cross section is subjected to an
axial vertical diameter load W . Calculate the magnitude of W , if the maximum stress at the
section cut by horizontal diametral plane does not exceed 225 MN/m2 .
[Ans 83.19 kN]
22.12 A chain link made of round bar of 12 mm diameter has a straight length of 60 mm and mean
radius of curvature of 40 mm. The ring is subjected to an axial pull of 1.5 kN. Determine the
maximum stress of the intrados of the link.
[Ans 149.7 MN/m2 ]
@seismicisolation
@seismicisolation
23
C HAPTER
UNSYMMETRICAL BENDING
Unsymmetrical bending results if the section is unsymmetrical but the load line is inclined to both
the principal axes or if the section itself is unsymmetrical, for example, angle section of channel
section (with vertical web), and the load line is along any centroidal axis.
For studying unsymmetrical bending let us revise the product of inertia for an angle section as
discussed below.
In order to determine
the product of inertia of the angle section shown in Fig. 23.1, we proceed
with the formula xy da: The quantity xy da is called product of inertia.
Let us find the product of inertia with respect to X and Y axes.
Y
15 mm
A1
45 mm
A2
15 mm
45 mm
X
Figure 23.1
Using parallel axis theorem,
Rect. 1.
IXY = IXY
45
+ A1 (7.5)
2
A1 = 45 × 15 = 675 mm2
A2 = (45 − 15) × 15 = 450 mm2
@seismicisolation
@seismicisolation
and
xy da = 0
492
•
Strength of Materials
IXY = 0 + 675 × 7.5 × 22.5 (because
xy da = 0)
= 113906.2 mm4
Rect. 2.
IXY = IXY
+ A2
30
15 +
2
15
2
= 0 + 450 × 30 × 7.5
= 101250 mm4
Product of inertia of the total area,
= 113906.2 + 101250 = 215156.2 mm4
In unsymmetrical bending the principal axes and principal moments of inertia are important to be
found before proceeding for further calculations.
To understand principal axis, it may be noted that if a figure has an axis of symmetry, that axis
is a principal axis. During unsymmetrical bending, the principal axes has to be found as follows.
Determination of Principal Axes and Principal Moments of Inertia
Figure 23.2 shows a figure of axis A in which the principal axes through centroid G are UU and VV .
Axes XX and YY are another pair of perpendicular axes passing through G making an angle α with
principal axes. Axes XX and YY are through centre of gravity of section which is unsymmetrical
or on where inclined load is applied. After application of load on unsymmetrical bending new axes
UV and UV are inclined at α with XX and YY and about which the bending takes place.
Let us consider any figure where da in positive quadrant has co-ordinates u and v relative to VV
and UU and x and y relative to YY and XX respectively.
V
If IXX , IYY and IXY are given about XX and
Y − Y axes, d is required to find the principal
moments of inertia and direction of the principal axes.
Y
x da
u
y V
X
U
α
X
G
u = x cos α + y sin α
v = y cos α + x sin α
U
Y
V
Figure 23.2
IVV =
2
v da =
y2 cos2 α − 2xy sin α cos α + x2 sin2 α da
= IXX cos2 α − IXY sin 2α + IYY sin2 α
@seismicisolation
@seismicisolation
(i)
Unsymmetrical Bending
IVV =
IUV
u2 da =
493
x2 cos2 α + 2xy sin α cos α + y2 sin2 α da
= IXX sin2 α + IXY sin 2α + IYY cos2 α
xy(cos2 α − sin2 α ) + (y2 − x2 sin α cos α ) da
= uvda =
= IXY cos 2α + (IXX − IYY )
∴
•
tan 2α =
(ii)
sin 2α
=0
2
2IXY
IYY − IXX
(iii)
Equation (iii) gives two values of 2α differing by π , i.e., two values of α differing by π /2.
Given the principal moments of inertia, it is required to find the moments of inertia about XX
and YY
x = u cos α − v sin α
IXX =
IYY
y = v cos α + u sin α
v2 cos2 α + uv sin 2α + u2 sin2 α da
y2 da =
= IUV cos2 α + IVV sin2 α (because IUV = 0)
= x2 da =
u2 cos2 α − uv sin 2α + v2 sin2 α da
(iv)
= IVV cos2 α + IUU sin2 α
(v)
∴
(vi)
IXX + IYY = IUU + IVV (= J)
Beam with Unsymmetrical Bending Moment
Figure 23.3 shows the cross section of a beam in which the applied bending moment act in the plane
YY , inclined at an angle α to the principal axis VV .
Let the applied bending moment is M.
Y
Component in plane VV ,
V
N
= M cos α
u
M
u da
v
U
α
Component in plane UU
= M sin α
X
A
U
Y
Figure 23.3
@seismicisolation
@seismicisolation
X
β
G
V
494
•
Strength of Materials
Therefore, stress on element da, of coordinates u and v relative to the principal axis,
M cos α
M sin α
×v+
×u
IVV
IVV
v cos α u sin α
σ =M
+
IUU
IVV
=
(i)
Equation (i) applies to all points in the cross section provided that the appropriate signs are
given to the coordinates u and v.
For points on the neutral axis, σ = 0
i.e.,
u cos α
u sin α
=−
IUU
IVV
IUU
or v = −
tan α xu
IVV
(ii)
IUU
tan α .
IVV
If β is the inclination of this axis to UU, than
This is a straight line of slope −
−1
β = − tan
IUU
tan α
IVV
(iii)
The most highly stressed point is that which is the farthest from the neutral axis at an angle of β as
shown in Fig. 23.4. The stress at this point is then obtained by substituting the appropriate values of
u and v in Eqn. (i).
To Find I UU and I VV using Mohr’s Circle
As we determine principal stresses σ1 and σ2
using Mohr’s circle, it is also possible to find IUU
and IVV , as explained in Fig. 23.3A. Diagram is
self-explanatory. C is the centre of AB. And as a
result OA = IVV and OB = IUU . It is understood
that perpendicular DG and EF are equal to IXY .
Hence, if IXX , IYY and IXY are known, Mohr’s
circle can help to determine IUU and IVV .
G
IYY
IXY
IVV
O
C
A
D α
E
2α
IXY
F
IXX
IUU
Figure 23.3A
@seismicisolation
@seismicisolation
Unsymmetrical Bending
•
495
Radius of the circle = IXX cosec 2α
=
IXX − IYY
2
2
2
+ IXY
Momental Ellipse
Figure 23.3B
1
1
and Gb =
in Fig. 23.3B and draw an ellipse with Ga and Gb as the major
kUU
kVV
1
IUU
1
. Then Gc =
and minor semi-axes kUU =
and Gd =
.
A
kXX
kYY
Let Ga =
Proof. We know from the equation of the ellipse,
u2
1
kUU
2 + v2
1
2 = 1
kVV
But we know u = Gc cos α
and v = Gc sin α
Therefore,
2
2
cos2 α + Gc2 kVV
sin2 α = 1
Gc2 kUU
∴
2
2
kUU
cos2 α + A kVV
sin2 α =
@seismicisolation
@seismicisolation
A
Gc2
496
•
Strength of Materials
We have already proved that,
IXX = IUU cos2 α + IVV sin2 α
A
Then, IUU cos2 α + IVV sin2 α =
= IXX
Gc2
1
Hence, Gc =
kXX
For figures such as circle, square, equilateral triangle or other regular polygon having more than
two axes of symmetry, the momental ellipse becomes a circle and therefore the moment of inertia
about any axis through G is the same as about the principal axes.
Deflection of Beams due to Unsymmetrical Bending
Consider Fig. 23.4 where components of load W are shown acting at G. The load W is resolved in to
i) W sin α along UG
ii) W cos α along V G
Y
V
U
N
G
Let
X
δu
δ
δu = deflection due to W sin α as shown
δv = deflection due to W cos α as shown
α
β δv
X
β
A
U
Y
V
Figure 23.4
Then depending upon the end conditions of the beam, values of δu and δv are given by:
δu =
K (W sin α ) l 3
EIVV
(i)
δv =
K (W cos α ) l 3
EIUU
(ii)
1
K = A constant depending on end conditions of the beam. For example, K =
for a beam
48
simply supported beam with a point load at centre.
l = length of the beam
Resultant deflection δ =
(δ u)2 + (δ v)2
@seismicisolation
@seismicisolation
Unsymmetrical Bending
•
497
Hence,
Kl 3
δ=
E
or δ =
W sin α 2
W cos α 2
+
IVV
IUU
sin2 α cos2 α
+ 2
2
IVV
IUU
KW l 3
E
(iii)
The inclination β of the deflection δ , with the line GV is given by:
tan β =
δu IUU
=
tan α
δv
IVV
(iv)
Method for Finding Bending Stream is Unsymmetrical Bending
i) Find centre of gravity of the given section.
ii) Determine IXX , IYY and IXY of the given section.
2IXY
. If the value of α is positive, then
IYY − IXX
the principal axis UU will be in the anticlockwise direction with X axis. And the location of
VV axis will be at right angle to UU axis.
iv) Find the values of IUU and IVV . If IXX = IYY , then value of IUU will be obtained from
equation:
iii) Calculate the value of α from relation tan 2α =
IUU = IXX cos α 2 − IXY sin 2α + IYY sin2 α
v) Find M and its components along principal axes GU and GV .
vi) Find the resultant bending stress using relation:
δb =
(M sin α ) × u (M cos α ) ×v
+
IVV
IUU
=M
u sin α u cos α
+
IVV
IUU
E XAMPLE 23.1: An unequal angle of dimensions 120 mm by 75 m and 12 mm thick is shown in
Figure 23.5. Determine.
i) Position of the principal axes end
ii) Magnitude of principal moment of inertia for the given angle.
@seismicisolation
@seismicisolation
498
•
Strength of Materials
S OLUTION :
Y
75 mm
A
C
1
15 mm
y
x
X
120 mm
X
C.G.
2
Y
B
15 mm
Figure 23.5
a1 = 75 × 15 = 1125 mm2 ; y1 = 7.5, x1 =
a2 = 105 × 15 = 1575 mm2 y2 =
75
= 37.5 mm
2
105
15
+ 15 = 67.5; x2 =
= 7.5 mm
2
2
Total area = a1 + a2 = 1125 + 1575 = 2700 mm2
ȳ =
=
x̄ =
=
IXX =
1125 × 7.5 + 1575 × 67.5
1575
8437.5 + 106312.5
= 72.86 mm
1575
1125 × 37.5 + 1575 × 7.5
1575
42187.5 + 11812.5
= 34.29 mm
1575
75 × 153
15 × 1053
+ 1125 × (65.36)2 +
+ 1575 (5.36)2
12
12
∵
(72.86 − 7.5 = 65.36 and
72.86 − 67.5 = 5.36)
@seismicisolation
@seismicisolation
Unsymmetrical Bending
•
= [21093.75 + 4805.92] + [1447031.25 + 45249.12]
= 1518180 mm4
IYY =
105 × 153
15 × 753
+ 1125 (34.29 − 37.5)2 +
+ 1575 (34.29 − 7.5)2
12
12
= [527343.75 + 11592.1] + [29531.25 + 1130384]
= 1698851.1 mm4
IXY = A1 h1 k1 + A2 h2 k2
h1 = (72.86 − 7.5) = 65.36
k1 = (34.29 − 37.5) = +3.21
h2 = (72.86 − 67.5) = 5.36
It is below X − X & hence − ve = −5.36
k2 = (34.29 − 7.5) = 26.79 = It is towards left of Y −Y. Hence, − ve = −26.79
IXY = 1125 × 65.36 × 3.21 × +1575 × (−5.36) × (−26.79)
= 236031.3 + 226161.2
= 462192.5 mm4
i) Position of principal axes
tan 2α =
2IXY
IYY − IXX
=
2 × 462192.5
1658851.1 − 1518180
=
924385
140671
= 6.571
2α = 81.35
∴
α = 40.67◦
Axis UU will be 40.67◦ anticlockwise.
@seismicisolation
@seismicisolation
499
500
•
Strength of Materials
ii) Value of principal of moment of inertia
IUU = IXX cos2 α − IXY sin 2α + IYY sin2 α
= 1518180 cos2 40.67 − 462192.5 sin 2 × 40.67 + 1658851.7 sin2 40.67
= 873386.7 − 456923 + 704538.6
= 1121002.3 mm4
IVV = IXX + IYY − IUU
= 1518180 + 1658851.1 − 1121002.3
= 2056028.8 mm4
E XAMPLE 23.2: A 80 mm × 80 mm angle as shown in Fig. 23.6 is used as a freely supported
beam with one leg vertical. IXX = IYY = 0.8736 × 10−6 m4 . When a bending moment is applied
in the vertical plane YY , the mid-section of the beam deflects in the direction AA at 30◦ 15 to the
vertical. Calculate the second moments of area of the section about its principal axes. Find also the
bending stress at the corner B, if the bending moment is 2 kNm.
B
Y
Y
A
10 mm
V
U
30˚15'
80 mm
α = 45˚
23.4 mm
X
X
X
β
X
10 mm
80 mm
Y
U
A
Y
V
Figure 23.7
Figure 23.6
Because the angle has an axis of symmetry, axis (UU in Fig. 23.7) is a principal axis and
α = 45◦ .
From equations:
IXX + IYY = IUU + IVV and
the inclination of the resultant deflection to VV
= tan
−1
δUU
δVV
−1
= tan
IUU + IVV = IXX + IYY = 2 × 0.8736 × 10−6 m4
@seismicisolation
@seismicisolation
IUU
tan α
IVV
Unsymmetrical Bending
= 1.7472 × 10−6 m4
IUU
IUU
tan α =
(∵ tan α = 1)
tan β =
IVV
IVV
IUU
∴
= tan 45◦ + 30◦ 15 = 3.79
IVV
•
501
(i)
(ii)
From Eqns. (i) and (ii)
IUU = 1.382 × 10−6 m4
IVV = 0.365 × 10−6 m4
and
By drawing or calculation, u = 23.5 mm, v = 56.56 mm
v cos α u sin α
∴ σ =M
+
IUU
IVV
◦
0.0235 sin 45◦
3 0.05656 cos 45
N/m2
+
= 2 × 10
1.382 × 10−6
0.365 × 10−6
= 149 MN/m2
Ans
E XAMPLE 23.3: A beam of T section having flange 120 mm × 20 mm and web 150 mm × 10 mm
is 2.5 m in length and is simply supported at its ends. It carries a load of 3.5 kN inclined at 20◦ to
the vertical and passing through the centroid of the section. If E = 200 GN/m2 , calculate:
i)
ii)
iii)
iv)
Maximum tensile stress
Maximum compressive stress
Deflection due to the load
Position of neutral axis.
S OLUTION :
Y,V
20˚ 3.5 kN
A
Wv
B
120 mm
1
3.5 kN
20˚ 70˚
Wu
20 mm
42.7 = y
G
X,U
2
150 mm
127.3
C
X,U
D
10 mm
Y,V
Figure 23.8
@seismicisolation
@seismicisolation
1.25 m
1.25 m
502
•
Strength of Materials
a1 = 120 × 20 = 2400 mm2 , y1 = 10 mm
a2 = 150 × 10 = 1500 mm2 , y2 =
150
+ 20 = 95 mm
2
A = 2400 + 1500 = 3900
ȳ =
2400 × 10 + 1500 × 95
3900
= 42.7 mm
(150 + 20) − 42.7 = 127.3 mm
Because the section is symmetrical about the vertical axis, therefore the principal axes pass through
the centroid G and are along UU and VV axis shown.
IXX = IUU =
10 × 1503
120 × 203
+ 2400 (42.7 − 10)2 +
+ 1500 (42.7 − 95)2
12
12
= [80000 + 2566296] + [2812500 + 4102935]
= 9561731 mm4 = 9.562 × 10−6 m4
IYY = IVV =
20 × 1203 150 × 103
+
12
12
= [2880000 + 12500]
= 2892500 mm4 = 2.892 × 10−6 m4
Components of W
Wu = W sin 20 = 3.5 sin 20◦
= 1.197 kN
Wv = W cos 20 = 3.5 cos 20◦
= 3.29 kN
Bending moments:
Mu =
1.197 × 2.5
Wu × l
=
4
4
= 0.748 kNm
@seismicisolation
@seismicisolation
Unsymmetrical Bending
Mv =
•
503
3.29 × 2.5
Wv × l
=
4
4
= 2.056 kNm
Mu will cause maximum compressive stresses at B and D and maximum tensile stress at A and C.
Mv will cause maximum compressive stresses at A and B and maximum tensile stress at C and D.
i) Maximum tensile stress:
Maximum tensile stress at C,
Mu × (5 × 10−3 ) Mv × (127.3 × 10−3 )
+
IVV
IUU
σc =
0.748 × 5 × 10−3
2.056 × 127.3 × 10−3
−3
×
10
+
× 10−3 MN/m2
9.562 × 10−3
2.892 × 10−6
=
= 1.293 + 27.37
= 28.663 MN/m2
Ans
ii) Maximum compressive stress:
Maximum compressive stress at B,
σB =
Mu × (60 × 10−3 ) Mv × (42.7 × 10−3 )
+
IVV
IUU
0.748 × (60 × 10−3 )
2.056 × 42.7 × 10−3
× 10−3 +
× 10−3 MN/m2
−6
2.892 × 10
9.562 × 10−6
=
= 15.52 + 9.181
= 24.7 MN/m2
Ans
iii) Deflection due to load,
δ=
Here K =
∴
KW l 3
E
sin2 α cos2 α
+
2
IUU
IVV
1
for a beam simply supported and carrying a load at the centre.
48
KW l 3
δ=
EIVV
δ=
sin α
2
IUU
IVV
2
+ cos2 α
3.5 × 103 × (2.5)3
48 × 200 × 109 × 9.562 × 10−6
sin2 20◦ ×
@seismicisolation
@seismicisolation
9.562 × 10−6
+ cos2 20◦
2.892 × 10−6
504
•
Strength of Materials
√
= 0.000596 ± 0.117 × 10.93 + 0.883
= 0.000596 × 1.47
= 8.7 × 10−4 m
= 0.876 mm Ans
iv) Position of neutral axis:
tan β =
IUU
tan α
IVV
9.562 × 10−6
tan 20◦
2.892 × 10−6
= 1.20
=
β = 50.19◦
Ans
E XAMPLE 23.4: The section of a beam is rectangular 100 mm deep and 75 mm wide. It is subjected to a bending moment of 10 kNm. The loading plane is inclined at 45◦ to the Y axis of the
section. Locate the neutral axis of section. Also find the maximum bending stress anywhere in
section.
[A.M.I.E Summer 2000]
S OLUTION :
Figure 23.9 shows a rectangular section with the given dimensions. This section is symmetrical
about XX and YY axis, therefore, XX and YY are the principal axes UU and VV .
1
× 75 × 1003 = 6250000 mm4
12
1
× 100 × 753 = 3515625 mm4
=
12
IUU = IXX =
IVV = IYY
Bending moment M = 10 kNm = 10 × 103 kNmm
Component of bending moment,
M1 = M sin 45◦ = 10 × 103 × 0.707 = 7070 kNm
M2 = M cos 45◦ = 10 × 103 × 0.707 = 7070 kNm
Position of neutral axis (NA)
Let β be the angle which the neutral axis makes with the principal axis UU.
IUU
tan α
IVV
6250, 000
=
tan 45◦ = 1.7777
3515625
tan β =
@seismicisolation
@seismicisolation
Unsymmetrical Bending
∴
β = tan−1 1.7777
= 60.64◦ = 60◦ 38 •
505
Ans
Maximum bending stress anywhere in section.
E XAMPLE 23.5: The section of a beam is rectangular 100 mm deep and 75 mm wide. It is subjected to a bending moment of 10 kNm as shown is Fig. 23.9. The loading plane is inclined at 45◦
to the y-axis. Find the maximum stress in the section.
S OLUTION :
On examining Fig. 23.9 we see that the maximum bending stress in the section may occur either
at the point B or D.
Let us determine the coordinates x, y for the point B
x=
75
= +37.5 mm;
2
y=−
100
= −50 mm
2
y is minus because it is downward.
Then u = x cos α + y sin α , α = 45◦
= 37.5 × cos 45◦ − 50 × sin 45◦
= 37.5 × 0.707 − 50 × 0.707
= 26.5125 − 35.35
= −8.8375 mm
v = y cos θ − x sin θ
= −50 × 0.707 × −37.5 × 0.707
= −35.35 − 26.5125 = −61.8625 mm
σB =
Y,V
M = 10 kNm
A
M2
D
E 8'
45˚
60˚3
β=
XU
XU
M1
G
A
N
B
=
100 mm
M1 u M2 v
+
IVV
IUU
7070 × (−8.8375) 7070 − (−61.8625)
+
3515625
6250000
= −0.0177 − 0.0699
= −0.0876 kN/mm2
= −87.6 N/mm2 (Tensile)
Ans
Maximum bending stress at the point D will be 87.6 kN/mm2 (Compressive).
Ans
Y,V
75 mm
Figure 23.9
@seismicisolation
@seismicisolation
506
•
Strength of Materials
E XAMPLE 23.6: Figure 23.10 shows an unequal angle section, for which IXX = 0.8 × 10−6 m4
and IYY = 0.382 × 10−6 m4 . Find the moment of inertia about the principal axes UU and VV , given
1◦
that the angle between the axes UU and X − X is 28 .
2
If the angle with the 80 mm leg vertical, is used as a beam, freely supported on a span of 2 m
carrying a vertical load of 2 kN at the centre, find: (a) the maximum stress at the point A and (b)
the direction and magnitude of the maximum deflection. Neglect the weight of the beam and take
E = 200 GN/m2
S OLUTION :
V
V
10 mm
Y
80 mm
G
28.5˚
X
X
X
26.4 mm
60 mm
U
Y
V
X
δ VV
δ UU
10 mm
U
U
Y
U
β
δ
(i)
Y
16.4 mm
V
Figure 23.11
Figure 23.10
IUU + IVV = IXX + IYY
= (0.8 + 0.382) × 10−6 = 1.132 × 10−6 m4
As we know
IXX = IUU cos2 α + IVV sin2 α
and
∴
IYY = IVV cos2 α + IUU sin2 α
IXX − IYY = (IUU − IVV ) cos2 α
IUU − IYY =
=
IXX − IYY
cos2 α
(0.8 − 0.382) × 10−6
cos 57◦
= 0.2072 × 10−6 m4
@seismicisolation
@seismicisolation
(i)
Unsymmetrical Bending
•
507
(a) Maximum bending moment
2 × 103 × 2
Wl
=
4
4
= 1000 Nm
=
By drawing or calculation, u = 20 mm, v = 50 mm
(See Fig. 23.11)
∴
v cos α u sin α
σ =M
+
IUU
IVV
0.05 cos 28.5◦ 0.02 × sin 28.5◦
N/m2
+
= 1000
0.9748 × 10−6
0.2072 × 10−6
= 91.2 MN/m2
Ans
(b)
W l3
δ=
48E
cos α
IVV
2
2 × 103 × 23
=
48 × 200 × 109
sin α
+
IVV
2
cos 28.5◦
0.9748 × 10−6
= 0.004125 m = 4.125 mm
2
sin 28.5◦
+
0.2072 × 10−6
2
Ans
Now
β = tan
= tan
−1
−1
IUU
tan α
IVV
0.9748 × 10−6
tan 28.5◦
0.2072 × 10−6
= 68◦ 48 ∴
Angle to vertical = 68◦ 48 − 28◦ 30 = 40◦ 18 . Ans
Shear Centre
In case of symmetrical section such as rectangular and I sections, the applied shear force F is
balanced by the set of shear forces summed over the rectangular section as over the flanges and the
web of I section and shear centre coincides with the centriod of the section. If the applied load is not
@seismicisolation
@seismicisolation
508
•
Strength of Materials
placed at the shear centre, the sections twists about this point and this point is also known as centre
of twist. Therefore, the shear centre of a section can be defined as a point about which the applied
shear force in balanced by the set of shear forces obtained by summing the shear stresses over the
section.
The shear centre discussed above is a point in or outside a section through which the shear force
applied produces no torsion or twist of the member.
Shear Centre for Channel Section
Figure 23.12 shows a channel section. Let e be the distance of the shear centre from the web.
B
dA
F1 =
t1
X
F.AȲ
dA
It1
A = t1 x
F1
F
τ dA =
dA = t1 dx
t1
x
d
o
X
C.G.
S.C.
e
h
2
F b
h
t1 x · t1 dx
F1 =
It1 o
2
Ft1 h 2
b
=
4I
2
h
1
I = t1 h3 + 2Bt1
12
2
ȳ =
h
F1
t1
B
Figure 23.12
(i)
1 t1 h
1
= Bt1 h2 1 +
2
6 Bt1
Substituting value of I in Eqn. (i)
F1 =
=
Ft1 h B2
1
l t1 h
4 × Bt1 h2 1 + ·
2
6 Bt1
=
Fb
1 t1 h
2h 1 +
6 t1 B
Fb
1 ht1
2h 1 +
6 Bt1
Taking moments about 0,
F × e = F1 ×
h
h
+ h1 × = F1 h
2
2
@seismicisolation
@seismicisolation
(ii)
Unsymmetrical Bending
•
509
Substituting for F1 from Eqn. (ii),
FB h
1 t1 h
2h 1 + ·
6 t1 B
1
B
2
e=
1 Aw
1+
6 Af
F.e =
∴
Aw = area of web
A f = area of flange.
E XAMPLE 23.7: Find the position of shear centre for a channel section 50 mm × 50 mm × 5 mm.
47.5 mm
F1
F
5 mm
50 × 503 45 × 403
−
12
12
= 520833.3 − 240000
I=
22.5 mm
e
N
A
S.C.
45 mm
= 280833.3 mm4
5 mm
5 mm
F1
50 mm
Figure 23.13
Horizontal maximum shear stress in the flange,
F.AȲ
I.b
F (50 − 2.5) (25 − 2.5) × 5
=
I ×5
F
= 1068.75
I
Zmax =
Shear force in flange:
F
× 1068.75 × (50 − 2.5) × 2.5
I
F
= 126914
I
F1 =
@seismicisolation
@seismicisolation
510
•
Strength of Materials
Clockwise moment due to F1 ,
= F1 × (50 − 5)
= 126914 × 45 ×
= 5711130
F
I
F
I
Equating this to anticlockwise moment, F × e,
F × e = 5711130 ×
F
I
5711130
280833.3
= 20.34 mm
e=
Ans
Alternatively,
1
B
2
e=
1 Aw
1+ .
6 Af
1
× 47.5
2
=
1
45 × 5
1+ ×
6
47.5 × 5
23.75
=
1 × 0.158
= 20.51 mm
Ans
Shear centre for unequal I section
F
b2
b1
F2
F1
x
dx
h
X
F3
O
t1
F
e
X
S.C.
t2
F2
F1
b2
Figure 23.14
@seismicisolation
@seismicisolation
b1
t1
Unsymmetrical Bending
Shear stress is any layer, τ =
FAȲ
It
I = IXX
t3
= 2 (b1 + b2 ) 1 + (b1 + b2 )t1 ×
12
2 h
h3
+ t2 .
2
12
For shear force F1
dA = t1 x dx
Aȳ = t1 x
b1
F1 =
h
2
τ dA =
0
b1
=
0
F
× h.t1 .x dx
2IXX
Fht1 x2
=
2IXX 2
=
F.x.t1 h
. .t1 dx
IXX t1 2
b1
0
Fht1 b21
4IXX
Also likewise,
F2 =
Fht1 b22
4IXX
Taking moment of the shear forces about the centre of the web 0, we get.
F2 h = F1 h + F.e (F3 = F
(F2 − F1 ) h = F.e
Fh2t1 2
b2 − b21 = F.e
4IXX
t1 h2 b22 − b21
or e =
4IXX
@seismicisolation
@seismicisolation
for equilibrium)
•
511
512
•
Strength of Materials
E XAMPLE 23.8: Determine the position of the shear centre of a section of a beam shown in
Fig. 23.15.
70 mm
55 mm
35 mm
15 mm
X
X
250 mm
35 mm
70
55
Figure 23.15
t1 = 35 mm
b1 = 55 mm
b2 = 70 mm
h = 250 − 35 = 215 mm
IXX = 2
15 × 1803
125 × 253
+ 125 × 35 × 107.52 +
12
12
= 2 [446614.6 + 50558.6] + 7290000
= 2 [497173.2] + 7290000
= 8284346.4 mm4
We know,
∴
t1 h2 b22 − b21
e=
4IXX
35 × 2152 702 − 552
e=
4 × 8284346.4
1617875(1875)
33137385.6
= 91.54 mm Ans
=
@seismicisolation
@seismicisolation
Unsymmetrical Bending
•
E XAMPLE 23.9: Determine the position of shear centre for the Fig. 23.16.
60 mm
F
F5
F4
3 mm
F3
N
S.C.
3
e
F1
F2
25 mm
35 mm
A
35 mm
25 mm
60 mm
Figure 23.16
F1 , F2 , F3 , F4 , F5 = shear forces in different positions.
It may be noted that F3 is shear force in upward direction.
FAȳ
We know that shear stress, τ =
shear force F1 will be equal to F5
IXX × b
60 mm
dx
x
N
35
mm
C
A
B
dy 25 mm
60 mm
y
3 mm
A
Figure 23.17
Consider an element of length y, width 3 mm and thickness dy.
A = area of element = 3 × y = 3y
y
ȳ = distance of centre of gravity of area A from NA = 35 + ; width of element = 3 mm
2
IXX = moment of inertia of whole section about x axis.
y
F × 3y × 35 +
2 = F y 35 + y
∴ Z=
IXX × 3
IXX
2
dA = area of thickness dy = 3dy
@seismicisolation
@seismicisolation
513
514
•
Strength of Materials
Shear force F1 (= F5 ) = τ dA
25
=
0
y
y 35 +
× 3dy
IXX
2
3F
=
IXX
F
y2
dy
35y +
2
25
0
25
3F 35y2 y3
=
+
IXX
2
6 0
3F
(25)3
2
=
17.5 (25) +
IXX
6
∴
=
3F
[10937.5 + 2604.2]
IXX
=
3F
(13541.7)
IXX
F1 = F5 = 40625.1
F
IXX
Shear force F2 (= F4 ) in top and bottom flanges.
Shear stress,
τ=
FAȳ
IXX × b
Note that Aȳ is made up of two parts here.
i) Moment of flange area upto length x about neutral axis.
ii) Moment of shaded area of vertical projection about neutral axis.
dx
x
25
35 mm
A
N
Figure 23.18
@seismicisolation
@seismicisolation
Unsymmetrical Bending
∴
Aȳ = (3x) × (35 + 25) + (3 × 25) × 35 +
25
2
= 180x + 3562.5
b = 3 mm
τ=
∴
F× (180x + 3562.5)
IXX ×3
Area of thickness dx, dA = 3dx ∴ Shear force, F4 (= F2 ) = τ .dA
60
=
0
=
F (180x + 3562.5)
× 3dx
IXX × 3
60
F(180x + 3562.5)
IXX
0
=
180x2
+ 3562.5x
2
F
IXX
=
=
∴
F
IXX
dx
60
0
180 (60)2
+ 3562.5 (60)
2
F
[324000 + 213750]
IXX
F4 = F =
F
IXX
[537750]
Taking the moments of all forces about the centre of web, we get
F × e = F4 × 60 + F2 × 60 + (F5 + F1 ) × 50
= 120 F4 + (F5 + F1 ) × 50
(∵
F4 = F2 & F5 = F1 )
= 120F4 + 100F1
= 120 ×
=
or
e=
F
IXX
F
IXX
F
(537750) + 100 × 40625.1
IXX
[64530000 + 4062510]
68592510
IXX
@seismicisolation
@seismicisolation
•
515
516
•
Strength of Materials
IXX = Moment of inertia of the given section about x − x
60 × 33
3 × 1203
2
+2
+ 60 × 3 × 60
=
12
12
+2
3 × 253
+ 3 × 25 × (60 − 12.5)2
12
= 432000 + 2 [135 + 648000] + 2 [3906.25 + 169218.7]
= 432000 + 1296270 + 173124.95
= 1901394.95
∴
e=
68592510
= 36 mm
1901394.95
Ans
Exercise
A
23.1 A steel beam 50 mm × 25 mm in cross section is supported over a span of 1 m with one 50
mm face inclined at 30◦ to the vertical. A load of 600 N acts vertically at the centre of the
span. Neglecting the weight of the beam, and assuming that the ordinary beam theory applies,
find: (a) the maximum stress in the beam due to bending and (b) the magnitude and direction
of the maximum deflection. E = 200 GN/m2 .
[Ans 26.9 MN/m2 , 0.523 mm at 36◦ 35 to vertical]
23.2 A 8 cm × 8 cm angle as shown in Fig. 23.19 is used as a freely supported beam with one leg
vertical IXX = IYY = 0.8736 × 10−6 m4 . When a bending moment is applied in the vertical
plane YY , the mid section of the beam deflects in the direction AA at 30◦ 15 of the vertical.
Calculate the second moments of the area of the section about its principal axis. Also find the
bending stress at the corner B, if the bending moment is 2 kNm.
1 cm
B
8 cm
30˚15'
Y
X
X
1 cm
A
8 cm
Y
Figure 23.19
[Ans 1.382 × 10−6 m4 ; 0.365 × 10−6 m4 ; 149 MPa]
@seismicisolation
@seismicisolation
Unsymmetrical Bending
•
517
23.3 Determine the centroidal principal moments of an equal angle 30 mm × 30 mm × 10 mm.
[Ans
84166.67 mm4 , 12166.67 mm4 ]
23.4 The tension flange of a girder of I section is 240 mm × 40 mm whereas the compressive
flange is 120 mm × 20 mm. The web is 300 mm deep and 20 mm thick. If the girder is used
as a simply supported beam of 8 m span, determine the load per metre run if the allowable
stress is 90 MPa in compression and 30 MPa in tension.
Hint: This question is not of unsymmetrical bending. [Ans
9,277 kN/m]
23.5 A 1200 mm long cantilever of I section 50×30 mm having horizontal flange each of thickness
2.5 mm vertical web of thickness 2 mm is subjected to a load of 40 N at its free end. The load
is inclined at an angle of 15◦ to the vertical.
Determine the resultant bending stress at top corners of the upper flange.
[Ans
4.917 N/mm2 (Compressive); 28.12 N/mm2 (Tensile)]
23.6 A I section beam shown is Fig. 23.20, 2.4 metre is used as cantilever beam to support the load
of 200 N at the free end which makes 30◦ with the vertical. Determine the resulting bending
stresses at corners A and B.
200 N
30°
Y
A
B
2.5 mm
45 mm X
X
2 mm
C
2.5 mm
30 mm
Y
D
[Ans
Figure 23.20
11985 × 105 N/m2 Tensile at A,
8799 × 105 N/m2 Tensile at B]
23.7 An equal angle 100 mm × 100 mm × 20 mm has its one horizontal leg on top. As a cantilever
of 1.5 m span, it carries, at it free end, a point load of 6 kN inclined at an angle of 30◦ to the
vertical and its line of action passing through the centroid of the section (Fig. 23.21).
Determine stresses at points A, B and C. Also locate neutral axis.
@seismicisolation
@seismicisolation
518
•
Strength of Materials
B
C
20 mm
100 mm
100 mm
20
mm
A
Figure 23.21
[Ans
σA = 193.63 N/mm2 comp, σB = 296.183 N/mm2 tension,
σC = 126.61 MPa compression.]
23.8 Determine the position of shear centre for a channel having dimensions: flanges 120 mm ×
20 mm and web 160 mm × 10 mm.
[Ans e = 50.84 mm]
23.9 Locate the shear centre of section shown is Fig. 23.22.
t1
b1
t2
t3
X
X h
t2
b1
b2
t1
Ans
Figure 23.22
e=
h2 b22t2
b1 b2t1
h2 2b21
−
+
+
IXX
2
3
4IXX
23.10 Find the shear centre of the section shown in Fig. 23.23.
10 mm
25 mm
10 kN
210 mm
e
10 mm
25 mm
110 mm
Figure 23.23
[Ans
@seismicisolation
@seismicisolation
e = 44.7 mm to the left of web corner]
Unsymmetrical Bending
•
23.11 Determine the position of shear centre for the unequal I section shown in Fig. 23.24.
60 mm
80 mm
40 mm
220 mm
40 mm
60 mm
80 mm
Figure 23.24
[Ans
e = 9.08 mm]
23.12 Find the shear centre of an arc of a circle as shown in Fig. 23.25.
F
θ
θ
e
R
t
Figure 23.25
Ans
@seismicisolation
@seismicisolation
e=
2R(sin θ − θ cos θ )
(θ − sin θ cos θ )
519
C HAPTER
24
ROTATING DISCS AND CYLINDERS
ω
Fc
Rotating Ring
Consider a thin ring which is rotating about its
centre of gravity at an angular velocity of ‘ω ’
radians per second. Fig. 24.1 shows the ring of
thickness ‘t’. Let r be its mean radius and ρ the
density of the material of the ring.
dθ
θ
r
σc
t
σc
Figure 24.1
Let us consider an element of the ring subtending an angle d θ at θ from horizontal axis. Centrifugal force on the element per unit length = mrw2
= [ρ (rd θ )t]rw2
Total vertical force per unit length,
π
=
ρ r2 d θ t ω 2 sin θ
0
= ρr t ω
2
π
2
sin θ d θ
0
= ρ r t ω (− cos θ )π0
2
2
= 2ρ r2t ω 2
If σc is the hoop stress induced in the ring.
Then for equilibrium,
σc (2t)1 = 2ρ r2t ω 2
@seismicisolation
@seismicisolation
Rotating Discs and Cylinders
2
V
σc = ρ r ω = ρ r
r
∴
2
2
[ω =
2
•
521
V
]
r
σc = ρ r 2
where V is the linear velocity.
E XAMPLE 24.1: A steel ring of cross-sectional area 900 mm2 rotating about its axis passing
through its centre of gravity at an angular speed of 2500 r.p.m. The mean diameter of the ring is
275 mm. Determine the hoop stress produced in the material of the ring. Mass density of the steel
is 7800 kg/m3 .
If the rotational speed is increased by 25%, what will be then the increase in the centrifugal
stress?
S OLUTION :
σc = ρω 2 r2 = 7800
2π × 2500
60
2
×
0.275
2
2
= 7800 × 68469.4 × 0.0189
= 10093758.9
= 10.09 MN/m2
Ans
Increased speed = 1.25 × 2500 = 3125 r.p.m.
2π × 3125
Stress at increased speed = 7800
60
2
0.275
×
2
2
= 7800 × 106983.5 × 0.0189
= 15771507.6
= 15.7 MN/m2
% increase in stress =
15.7 − 10.09
× 100
10.09
= 55.6 % Ans
E XAMPLE 24.2: A flywheel with a moment of inertia of 350 kgm2 rotates at 275 r.p.m. If the
maximum stress is not to exceed 8 MPa, find the thickness of the rim. Take the width of the rim as
160 mm and mass density of the material 7600 kg/m2 . Neglect the effect of inertia of spokes.
S OLUTION :
σc = 8 N/mm2 = 8 MN/m2 ,
ρ = 7600 kg/m3
@seismicisolation
@seismicisolation
522
•
Strength of Materials
Width = 0.16 m, let t be the thickness of rim.
ω=
2π × 275
= 28.78 radians/s
60
At the outer radius, stress will be maximum,
σc = ρ .ro2 .ω 2
8 × 106 = 7600 × ro2 (28.78)2
ro2 = 1.271
ro = 1.127 m, say 1 m (as mean radius)
Let us assume radius of gyration be 0.9 m
Moment of inertia = mk2
350 = (7600 × 2π × 1 × 0.16 × t)0.92
t = 0.0566 m
t = 56.6 mm
Ans
E XAMPLE 24.3: A composite ring is made of two materials. The outer ring is of steel and the
inner ring is of copper. The diameter of the common surface is 850 mm. Each ring has a width of
35 mm and a thickness of 25 mm in radial direction. The ring rotates at 1850 rpm. Find the stresses
set up in the steel and copper. Take Es = 2 Ec . Density of steel = 7600 kg/m3 k, Density of copper
is 9000 kg/m3 .
S OLUTION :
Rc
d = 850 mm
r = 425 mm = 0.425 m
t = 25 mm = 0.025 m
2π × 1850
ω=
60
= 193.63 rad/s
Rs
Copper
Steel
Figure 24.2
850 25
+
= 437.5 mm
2
2
850 25
Rc =
−
= 412.5 mm
2
2
Rs =
Let p be the pressure due to shrinkage at the common surface when disc is not rotating.
@seismicisolation
@seismicisolation
Rotating Discs and Cylinders
Then hoop stress due to shrinkage in the steel ring =
Hoop stress in copper ring =
•
523
p × 850
pd
=
= 17 p (Tensile)
2t
2 × 25
pd
p × 800
=
= 17 p (Compressive)
2t
2 × 25
Hoop stress due to rotation,
In steel
σc = ρ r2 ω 2
= 7600 × (0.4375)2 × 193.632
= 54.54 MN/m2 (Tensile)
In copper
σc = ρ r2 ω 2
= 9000 (0.4125)2 (193.63)2
= 57.42 MN/m2 (Tensile)
As the strain in steel = strain in copper
17p + 54.54 −17p + 57.42
=
Es
Ec
17p + 54.54 −17p + 57.42
=
2Ec
Ec
17p + 54.54 = 2(−17p + 57.42)
17p + 54.54 = −34p + 114.82
51p = 60.28
p = 1.18 MN/m2
Total stress in steel = 20 × 1.18 + 54.54
= 23.6 + 54.54
= 78.14 MPa Ans
Total stress in copper = −20 × 1.18 + 57.42
= −23.6 + 57.42
= 33.82 MPa Ans
E XAMPLE 24.4: A wheel 850 mm in diameter has a thin rim. If density is 7500 kg/m3 and E = 200
GPa, calculate:
(i) How many revolutions per minute may it make without the hoop stress exceeding 150 MN/m2 ?
(ii) Change in diameter
Neglect the effect of spokes.
@seismicisolation
@seismicisolation
524
•
Strength of Materials
S OLUTION :
r=
850
= 425 mm = 0.425 m
2
ρ = 7500 kg/m3
E = 200 GN/m2
σc = 150 MN/m2
(i)
σc = ρω 2 r2
2π N 2
150 × 106 = 7500
0.4252
60
N2 = 10107304.5
N = 3179.2 r.p.m.
Ans
(ii)
σc
E
δ d 150 × 106
=
d
200 × 109
150 × 106
δd =
× 0.85
200 × 109
= 0.00064 m
= 0.64 mm Ans
Circumferential strain εc =
or
Rotating Disc of Constant Thickness
In engineering, the design of steam and gas turbines and other discs require analysis of stresses and
deformation, as they have to rotate at high speeds. The constant thickness discs are often called flat
discs. It is important to study radial and hoop stresses developed in them. Interestingly the theory
related to these is quite similar to thick cylinders where we used Lame’s theory.
ρr2ω2δrδθ
Derivations of equations for hoop and radial stresses:
Figure 24.3 shows an element of rotating disc,
extending angle θ at the centre.
Let σr and σc be the radial and hoop stresses
respectively developed in the rotating disc. Both
stresses are tensile in nature.
(σr+δσr) (r+δr)δθ
σcδr
σrrδθ
dθ
Figure 24.3
@seismicisolation
@seismicisolation
σc δr
Rotating Discs and Cylinders
•
525
If at a radius r from axis of rotation σr and σc are developed when disc rotates uniformly with
speed ω . Then if u is the radial shift, the stress-strain equations are:
du
= σr − μσc
dr
u
E · = σc − μσr
r
E·
Obtaining from E ·
(i)
(ii)
du
Eqn. (ii) and equating to Eqn. (i), we get
dr
(σc − σr ) (1 + μ ) + r d σc /dr − μ r(d σr /dr) = 0
(iii)
If the density of the material of disc is ρ , then for the element shown in Fig. 24.3, the centrifugal
force
= (ρ r δ θ .δ r) rω 2
= ρ r2 ω 2 δ r.δ θ
for unit thickness
In the radial direction equilibrium equation is
1
2σc dr sin δ θ + σr rδ θ − (σr + δ σr )(r + δ r)δ θ = ρ r2 ω 2 δ r δ θ
2
Dividing the whole equation by δ rδ θ ,
1
2σc dr sin δ θ σ rδ θ (σ + δ σ )(r + δ r)δ θ
r
r
r
2
+
−
= ρ r2 ω 2
δrδθ
δrδθ
δrδθ
δθ
(r + δ r)
dr sin 2
r
+ σ r − ( σ r + δ σr )
= σc ·
= ρ r2 ω 2
δθ
δr
δr
δr
2
δθ
r
r
(rδ δ r)
dr sin 2
− σ2 ·
− σr −
− δ σr = ρ r 2 ω 2
= σc
·
+ σ2 ·
δθ
δθ
δr
δr
δr
2
Neglecting δ σr being too small
And taking limits,
= σc − σr − r
d σr
= ρ r2 ω 2
dr
Substitute for σc − σr from Eqn. (iv) in Eqn. (iii)
d σr
d σc
d σr
2 2
r
) + ρr ω
− μr
=0
(1 + μ ) + r
dr
dr
dr
@seismicisolation
@seismicisolation
(iv)
526
•
Strength of Materials
Rearranging,
d σc d σr
+
= − ρ r ω 2 (1 + μ )
dr
dr
Integrating,
σc + σr = −(ρ r2 ω 2 /2) (1 + μ ) + 2A
(v)
2A is constant of integration
Subtracting Eqn. (iv)
2
d σr
2ω
= − ρr
(3 + μ ) + 2A
2σr + r
dr
2
1 d(σr r2 )
ρ r2 ω 2 (3 + μ )
=−
+ 2A
r dr
2
or,
Integrating,
σr r2 = −(ρ r4 ω 2 /8) (3 + μ ) + Ar2 − B
or σr = A −
From (v), σc = A +
B
− (3 + μ ) + (ρ r2 ω 2 /8)
r2
(vi)
B
− (1 + 3μ ) + (ρ r2 ω 2 /8)
r2
(vii)
Solid Disc
The stress cannot be infinite at the centre in a solid disc
∴
σr = 0 = A −
B
0
or B = 0
If R is the outside radius, then at the outer surface, σR = 0
3+μ
3+μ
ρω 2 R2 or A =
ρ R2 ω 2
8
8
3+μ
therefore σr =
ρω 2 R2 − r2
8
ρω 2 (3 + μ )R2 − (1 + 3μ )r2
and σc =
8
∴
0 = A−
(viii)
(ix)
At the centre, r = 0, then,
σr = σc =
3+μ
ρω 2 R2
8
(maximum value of stress)
@seismicisolation
@seismicisolation
(x)
Rotating Discs and Cylinders
•
527
And the outer surface,
σc =
1−μ
ρω 2 R2
4
and σr = 0
3 + 0.3
If μ = 0.3, σr = σc =
ρω 2 R2 = 0.413 ρω 2 R2 at the centre (maximum)
8
and at the end surface,
1 − 0.3
ρω 2 R2 = 0.175ω 2 R2
4
0.175
σc =
σc (max) = 0.424σc (max)
0.413
σc =
σe
Stress
σr
Radius
Figure 24.4
Hollow Disc
In this case the boundary conditions are
From Eqn. (vii),
At r = ri ,
σr = 0
At r = r0 ,
σr = 0
2
B
2ω
σc = A + 2 − (3 + μ ) ρ r
8
r
At inner radius,
0 = A−
B (3 + μ ) 2 2
−
ρ ri ω
8
r2
0 = A−
B (3 + μ )
−
ρω 2 ro2
8
r2
and outer radius,
@seismicisolation
@seismicisolation
(xi)
528
•
Strength of Materials
Solving above two equations
B=
A=
and
3+μ
8
3+μ
8
ρω 2 .ri2 . ro2
ρω 2 ri2 + ro2
If we now put the value of A and B in Eqns. (vi) and (vii), we get the values of σr and σc respectively
as given below:
σr =
and
σc =
3+μ
8
3+μ
8
ρω 2 ri2 + ro2 −
ri2 ro2
− r2
r2
(xii)
ρω 2 ri2 + ro2 −
ri2 ro2 1 + 3μ 2
−
r
3+μ
r2
(xiii)
Let the value of r at which σr is maximum be R1 . Now differentiating Eqn. (xii) and equating the
same to zero, we get,
d σr
=
dr
∴
3+μ
8
ρ w2
2ri2 ro2
=0
R31
R41 = ri2 ro2
√
R1 = ri r0
σr is maximum at the value of radius which is the geometric mean of the inner and outer radius of
the disc
√
Substituting r = ri r0 in Eqn. (xii),
σr(max) =
=
r2 r2
3+μ
ρω 2 ri2 + ro2 − i o − ri r0
8
ri r0
3+μ
ρω 2 (ro − ri )2
8
It is clear from Eqn. (xiii) that σc is maximum when r is minimum, i.e., when r = ri
∴
σc(max) =
=
r2 r2 1 + 3μ
3+μ
ρω 2 ri2 + ro2 − i 2o −
ri
8
3+μ
ri
3+μ
1−μ 2
ρω 2 ro2 +
.r
8
3+μ i
@seismicisolation
@seismicisolation
(xiv)
Rotating Discs and Cylinders
•
529
The stress distribution, i.e., σr and σc as given by Eqns. (xii) and (xiii) are shown in figure below.
ro
σc
σr
ri
Figure 24.5
E XAMPLE 24.5: A disc of uniform thickness and 650 mm diameter rotates at 1850 r.p.m. Find
the maximum stress developed in the disc. If a hole of 95 mm diameter is made at the centre of
the disc, find the maximum values of radial and hoop stresses. Density of the material of the disc
= 7800 kg/m3 and μ = 0.3.
S OLUTION :
R=
650
= 0.325 m,
2
ρ = 7800 kg/m3
ω=
and
2π × 1850
= 193.63 rad/s
60
μ = 0.3
Maximum radial stress and hoop stress are at the centre are equal.
3+μ
ρω 2 r2
8
3 + 0.3
× 7800 (193.63)2 (0.325)2
σr = σc =
8
σr = σc =
= 0.4125 × 7800 × 37492.6 × 0.105625
= 12741801.5 N/m2
= 12.74 MN/m2
With hole of radius ri =
r0 = 0.325 m
0.095
= 0.0475 m
2
Maximum radial stress is at
i.e.,
Ans
ri2 ro2 radius
√
0.0475 × 0.325 = 0.124 m
@seismicisolation
@seismicisolation
530
•
Strength of Materials
3+μ
ρω 2 (ro + ri )2
8
3 + 0.3
=
× 7800(193.63)2 (0.325 − 0.0475)2
8
σr =
= 0.4125 × 7800(37492.6)(0.2775)2
= 9289452 N/m2
= 9.28 MN/m2
Ans
Maximum hoop stress is at the inner radius,
σc =
=
3+μ
1−μ 2
ρω 2 ro2 +
r
8
3+μ i
3 + 0.3
1 + 0.3
× 7800 (193.63)2 0.3252 −
(0.0475)2
8
3 + 0.3
= 0.4125 × 292442099.8 [0.106 + 0.212 × 0.00226]
= 120632366 (0.1065)
= 12847347
= 12.85 MPa Ans
E XAMPLE 24.6: A thin steel disc of uniform thickness and of 240 mm diameter with a central hole
of 50 mm diameter rotates at 9500 r.p.m. calculate the maximum principal stress and the maximum
shear stress in the disc. ρ = 7000 kg/m3 , μ = 0.3.
S OLUTION :
B
3+μ
We know, radical stress, σr = A − 2 −
ρω 2 r2
8
r
B
1 + 3μ
and hoop stress, σc = A + 2 −
ρω 2 r2
8
r
9500 × 2π
= 994.3 rad/s
60
B
3 + 0.3
7000 × 994.32 r2
σr = A − 2 −
8
r
ω=
B
− 2854676315 r2
r2
B
1 + 3 × 0.3
σc = A + 2 −
× 7000 × 994.32 r2
8
r
= A−
= A+
B
− 1643601515 r2
r2
@seismicisolation
@seismicisolation
Rotating Discs and Cylinders
•
531
Now σr = 0 when r = 0.025 m and also when r = 0.120 m
0 = A−
B
− 2854676315 (0.025)2
(0.025)2
B
− 1784173
0.000625
B
0 = A−
− 2854676315 (0.12)2
(0.120)2
0 = A−
Also
= A−
B
− 41107339
0.0144
(i)
(ii)
From Eqn. (ii)
A=
B
+ 41107339
0.0144
Equating (i) & (ii),
B
B
+ 1784173 =
+ 41107339
0.000625
0.0144
B
B
−
= 41107339 − 1784173
0.000625 0.0144
1600B − 69.44B = 39323166
∴
Using (iii)
∴
∴
∴
39323166
= 25692
1530.56
25692
+ 41107339
A=
0.0144
B=
= 1784166.7 + 41107339 = 42891506
B
1 + 3μ
σc(max) = A + 2 −
ρω 2 r2 [∵ r = 0.025 m]
8
r
σc(max) = 42891506 + 42506080 − 1027251
= 84370335
= 84.37 MPa Ans
σmax
2
84.37
=
2
Max shear stress τmax =
= 42.185 MPa Ans
@seismicisolation
@seismicisolation
(iii)
532
•
Strength of Materials
E XAMPLE 24.7: Calculate to the largest value of radial and hoop stress for a rotating disc of
internal diameter 180 mm and external diameter 360 mm. The disc is rotating at 1700 r.p.m.
ρ = 7200 kg/m3
S OLUTION :
and
μ = 0.3
B
3+μ
σr = A − 2 −
ρω 2 r2
8
r
B
1 + 3μ
and σc = A + 2 −
ρω 2 r2
8
r
ω=
Now
also at
∴
2π × 1700
= 177.93 rad/s
60
σr = 0 when r = 0.09 m
r = 0.180 m
3 + 0.3
0 = A−
−
× 7200 × 177.932 × 0.092
2
8
(0.09)
B
0 = A − 123.4B − 761622.6
A = 123.4B + 761623
B
3 + 0.3
Also 0 = A −
−
× 7200 × 177.932 × 0.0182
8
(0.18)2
(i)
0 = A − 30.86B − 3046490
A = 30.86B + 3046490
Equating (i) and (ii)
123.4B + 761623 = 30.86B + 3046490
92.54B = 2284867
B = 24690
A = 30.86 × 24690 + 3046490
= 3808423
∴
24690
3+μ
−
ρω 2 r2
8
r2
24690
1 + 3μ
σc = 3808423 + 2 −
ρω 2 r2
8
r
σr = 3808423 −
@seismicisolation
@seismicisolation
(ii)
Rotating Discs and Cylinders
σr is maximum when
•
533
r2 = ri ro
= 0.09 × 0.18
= 0.0162
∴
σr( max) = 3808423 −
24690
− 1523245
0.0162
= 3808423 − 1524074 − 1523245
= 761104 N/m2
= 0.76 MN/m2
Ans
σc in maximum when,
r = 0.09 m
∴
σc max = 3808423 +
24690
(0.09)2
−
1 + 3 × 0.3
× 7200 × 177.932 × (0.09)2
8
= 3808423 + 3048148 − 438510
= 6418061
= 6.42 MPa Ans
E XAMPLE 24.8: A solid disc of uniform thickness and having a diameter of 500 mm rotates at
7800 r.p.m. Determine the radial and the hoop stresses at radii of 0, 100 mm, 150 mm and 250 mm.
ρ = 7500 kg/m3 , μ = 0.25.
S OLUTION :
2π × 7800
= 816.4 rad/s; R = 0.25 m
60
3+μ
σr =
ρω 2 R2 − r2
8
3 + 0.25
=
ρω 2 R2 − r2
8
= 0.406 × 7800 × 816.42 0.252 − r2
= 2110700575 0.0625 − r2
= 2110.7 0.0625 − r2 MPa
ω=
At r = 0
@seismicisolation
@seismicisolation
534
•
Strength of Materials
σr = 2110.7 × 0.0625 = 131.92 MPa
At r = 100 mm = 0.1 m
σr = 2110.7 0.0625 − 0.12
= 2110.7 (0.0625 − 0.01)
= 110.81 MPa
At r = 150 mm = 0.15 m
σr = 2110.7 0.0625 − 0.152
= 2110.7 (0.04)
= 84.43 MPa
At r = 250 mm = 0.25 m
σr = 2110.7 0.0625 − 0.252
=0
Now
ω2 (3 + μ ) R2 − (1 + 3μ ) r2
8
7800 × 816.42 3.25 × 0.252 − (1 + 3 × 0.3) r2
=
8
= 649846236 0.2031 − 1.9r2
σc = ρ
= 649.85[0.2031 − 1.9r2 ]
At r = 0
σθ = 649.85 [0.2031 − 0]
= 131.98 MPa
At r = 150 mm = 0.15 m
σθ = 649.85 0.2031 − 1.9 × 0.152
= 649.85 [0.16035]
= 104.20 MPa
At r = 250 mm = 0.25 m
@seismicisolation
@seismicisolation
Rotating Discs and Cylinders
•
535
σθ = 649.85 0.2031 − 1.9 × 0.252
= 649.85 [0.08435]
= 54.81 MPa
130
σθ
110
Stress
(MPa)
σr
90
0
100 mm 150 mm 250 mm
r
Figure 24.6 Stress distribution curves
E XAMPLE 24.9: A thin disc of uniform thickness is of 900 mm outer diameter and 60 mm inner
diameter. It rotates at 3500 r.p.m. Determine the radial and hoop stresses at radii 0, 25 mm, 50 mm,
100 mm, 150 mm, 200 mm, 300 mm and 450 mm. ρ = 7800 kg/m3 . μ = 0.25
S OLUTION :
Ri =
60
= 30 mm
2
= 0.03 m
Ro =
900
= 450 mm
2
= 0.45 m,
ρ = 7800 kg/m3
2π × 3500
= 366.33 rad/s
60
r2 r2
3+μ
σr =
ρω 2 ri2 + r2o − 1 2o − r2 , σro = 0
8
r
3 + 0.25
.032 × 0.452
7800 (366.33)2 (0.03)2 + (.45)2 −
σ0.025 =
− 0.0252
8
0.0252
ω=
= 425238863 [0.0009 + 0.2025 − 0.00292 − 0.000625]
= 425238863 × 0.199855
= 84986113
= 84.99 MPa
@seismicisolation
@seismicisolation
(i)
536
•
Strength of Materials
From Eqn. (i)
σr = 425238863 0.2034 −
σr0.05 = 425238863 0.2034 −
0.0001822
− r2
r2
0.0001822
2
(0.05)
− (0.05)2
= 425238863 [0.2034 − 0.073 − 0.0025]
= 425238863 [0.1279]
= 54388051
= 54.39 MPa
σr0.1 = 425238863 0.2034 −
0.0001822
− 0.12
0.12
= 425238863 [0.2034 − 0.01822 − 0.01]
= 425238863 × 0.17518
= 74493344
= 74.49 MPa
σr0.15 = 425238863 0.2034 −
0.0001822
− 0.152
0.152
= 425238863 [0.2034 − 0.0081 − 0.0225]
= 425238863 [0.1728]
= 73.48 MPa
σr0.2 = 425238863 0.2034 −
0.0001822
− 0.22
0.22
= 425238863 [0.2034 − 0.00455 − 0.04]
= 425238863 [0.15885]
= 67.55 MPa
σr0.3 = 425238863 0.2034 −
0.0001822
− 0.32
0.32
= 425238863 [0.2034 − 0.002 − 0.09]
= 425238863 [0.1114]
= 47.37 MPa
@seismicisolation
@seismicisolation
Rotating Discs and Cylinders
σr0.45 = 425238863 0.2034 −
•
537
0.0001822
− 0.452
0.452
= 425238863 [0.2034 − 0.0008997 − 0.2025]
= 425238863 [0.0000003]
Now,
= 0 approx.
r2 r2 1 + 3μ 2
3+μ
σc =
ρ w2 ri2 + ro2 + i 2o −
r
8
3+μ
r
3 + 0.25
(0.03)2 (0.45)2 1 − 3 × 0.25 2
=
−
× 7800 × (366.33)2 (0.03)2 + (0.45)2 +
r
8
3 + 0.25
r2
= 425238863 0.0009 + 0.2025 +
0.0001822
− 0.077r2
r2
σco = 0
σc0.025 = 425238863 0.0009 + 0.2025+
0.0001822
−0.077 (0.025)2
0.0252
= 425238863 [0.0009 + 0.2025 + 0.29125 − 0.0000481]
= 425238863 [0.4946]
= 210.32 MPa
σc0.05 = 425238863 0.0009 + 0.2025+
0.0001822
−0.077 (0.05)2
0.052
= 425238863 [0.0009 + 0.2025 + 0.07288 − 0.0001925]
= 425238863 [0.2761]
= 117.41 MPa
σc0.1 = 425238863 0.0009 + 0.2025+
0.0001822
−0.077 (0.1)2
0.12
= 425238863 [0.0009 + 0.2025 + 0.01822 − 0.00077]
= 425238863 [0.22085]
= 93.91 MPa
σc0.15 = 425238863 0.0009 + 0.2025+
0.0001822
−0.077 (0.15)2
0.0225
= 425238863 [0.0009 + 0.2025 + 0.0081 − 0.00173]
@seismicisolation
@seismicisolation
538
•
Strength of Materials
= 425238863 [0.20977]
= 89.20 MPa
σc0.2 = 425238863 0.0009 + 0.2025+
0.0001822
−0.077 (0.2)2
2
0.2
= 425238863 [0.0009 + 0.2025 + 0.00455 − 0.00308]
= 425238863 [0.20487]
= 87.11 MPa
σc0.30 = 425238863 0.0009 + 0.2025+
0.0001822
−0.077 (0.3)2
2
0.3
= 425238863 [0.0009 + 0.2025 + 0.00202 − 0.00693]
= 425238863 [0.19849]
= 84.4 MPa
σc0.45 = 425238863 0.0009 + 0.2025+
0.0001822
2
(0.45)
−0.077 (0.45)2
= 425238863 [0.0009 + 0.2025 + 0.000899 − 0.0156]
= 425238863 [0.188699]
= 80.24 MPa
200
100
σc
80
σr
60
Stress
40
20
0
25
50
100
200
300
450
r (mm)
Figure 24.7
E XAMPLE 24.10: Calculate the largest value of radial and hoop stresses for a rotating disc of internal dia 160 mm and external diameter 340 mm. The disc is rotating at 1600 r.p.m.
ρ = 7100 kg/m3 , μ = 0.3.
@seismicisolation
@seismicisolation
Rotating Discs and Cylinders
•
539
B
3+μ
σr = A − 2 −
ρω 2 r2
8
r
B
1 + 3μ
σc = A + 2 −
ρω 2 r2
8
r
S OLUTION :
ω=
2π × 1600
= 167.47 rad/s
60
σr = 0 when r = 0.08 m
3.3
B
7100× (167.47)2 (0.08)2
−
o = A−
8
(0.08)2
o = A − 156.25 B − 525698
Also, σr = 0
when
(i)
r = 0.170 m
3.3
B
−
×7100 (167.47)2 (0.170)2
o = A−
8
0.1702
= A − 34.6 B − 2373855
From Eqns. (i) and (ii)
A = 156.25B + 525698
Also
∴
A = 34.6B + 2373855
156.25B + 525698 = 34.6B + 2373855
121.65B = 1848157
B = 15192
A = 156.25 × 15192 + 525698
A = 2899448
15192 3.3
−
× 7100× (167.47)2 r2
8
r2
15192
σr = 2899448 − 2 −47292906r2
r
σr = 2899448 −
σ is maximum when
r2 = ri ro
= 0.080 × 0.170
@seismicisolation
@seismicisolation
(ii)
540
•
Strength of Materials
= 0.0136 m2
15192
= 2899448 −
σr
− 47292906(0.0136)
(max)
0.0136
= 2899448 − 1117059 − 643183.5
= 1139205.5
= 1.14 MPa Ans
σc is max when r = 0.08
σc(max) = 2899448 −
15192
− 47292926 (0.08)2
r2
= 2899448 − 2373750 − 302675
= 223023
= 0.22 MPa Ans
E XAMPLE 24.11: Determine the largest values of radial and hoop stresses for a rotating disc of
internal diameter 160 mm and external diameter 340 mm. The disc is rotating at 1650 r.p.m. For the
disc material, density = 7100 kg/m3 , μ = 0.3.
S OLUTION :
B
3+μ
σr = A − 2 −
ρω 2 r2
8
r
B
1 + 3μ
σc = A + 2 −
ρω 2 r2
8
r
ω=
2π × 1650
= 172.7 rad/s
60
Now σr = 0
when
when r = 80 mm = 0.08 m
r = 170 mm = 0.17 m
3.3
B
−
7100 × (172.7)2 (0.08)2
o = A−
8
(0.08)2
= A − 156.25B − 559045
(i)
σr = 0 when r = 0.17 m
B
3.3
∴ o = A−
7100 × (172.7)2 (0.17)2
−
8
(0.17)2
Also
= A − 34.6 B − 2524439
@seismicisolation
@seismicisolation
(ii)
Rotating Discs and Cylinders
From Eqn. (i), A = 156.25 B + 559045
From Eqn. (ii), A = 34.6 B + 2524439
Equating,
156.25 B + 559045 = 34.6 B + 2524439
121.65 B = 1965394
B = 16156
A = 156.25 × 16156 + 559045
= 30832420
∴
16156
3.3
7100 × 172.72 r2
σr = 30832420 − 2 −
8
r
16156
− 87350818 r2
r2
16156 1.9
× 172.72 r2
σc = 30832420 + 2 −
8
r
16156
= 30832420 + 2 − 21171 r2
r
= 30832420 −
and
Now σr is maximum when r2 = ri ro
= 0.08 × 0.17
= 0.0136
σr(max) = 30832420 −
16156
− 87356818 × 0.0136
0.0136
= 30832420 − 1187941 − 1187971
= 28456508
= 28.46 MPa
Ans
σc is maximum when r = 0.08 m
σc(max) = 30832420 +
16156
− 21171(0.08)2
0.082
= 30832420 + 2524375 − 135.4944
= 30832420
= 30.83 MPa
@seismicisolation
@seismicisolation
•
541
542
•
Strength of Materials
Rotating Long Cylinder
Centrifugal force
σr+δσr
σz
δθ
2
r0
ri
σc
dθ
2
w
δr
σc
L
σz
σr
r
δθ
ω
Figure 24.8
Let the density of the cylinder is ρ and length L, rotating about its axis with an angular velocity ω .
and, σr = radial stress produced at radius r
σr + δ σc = radial stress produced at radius (r + δ r)
σc = hoop stress at radius r assumed constant over δ r.
σz = Axial stress constant over cylinder cross section.
u = radial shift displacement at radius r
u + δ u = radial shift (displacement) at radius (r + δ r)
2π (r + u) − 2π r u
Circumferential strain, εc =
=
2π r
r
u + δu
Applying Hooke’s law,
δr
Unstrained
εc =
u
Strained
u
1
[σc − μ (σr + σz )] =
E
r
Eu = r [σc − μ (σr + σz )]
∴
r
∴
Figure 24.9
1
[σc − μ (σr + σz )]
E
From Fig. 24.9,
radial shift at radius r is u, whereas of radius
(r + δr ), it is (u + δu )
δu
∴ Radial strain =
δr
@seismicisolation
@seismicisolation
Rotating Discs and Cylinders
Also, radial strain =
•
543
1
[σr − μ (σc + σz )]
E
By equating,
∴
E
1
u
= [σr − μ (σc + σz )]
r
E
δr → 0
when
E
du
= σ r − μ (σ c + σ z )
dr
(i)
Differentiating Eqn. (i), we get
E
d σc
dy
d σr d σz
=r
+ σ c − μ (σ r + σ z ) − μ r
+
dr
dr
dr
dr
= σ r − μ (σ c + σ z )
∴
(σc − σr ) (1 + μ ) + r
d σc
−μ
dr
from Eqn. (i)
d σr d σz
+
=0
dr
dr
(ii)
Now the centrifugal force on the element (Fig. 24.6) is
Centrifugal force = ρ (rδθ .δr .L) ω 2 r
= ρω 2 r2 Lδθ δr
Now equating forces on the element in the direction of centrifugal force, we have
ρ w2 r2 δθ δr L + (σr + δ σr ) [(r + δ r) δθ L]
= σr .rδθ L + 2σc .δr .
δθ
δ
δ
L (Because sin θ = θ )
2
2
2
After simplifying,
σc − σr = ρ w2 r2 + r
d σr
dr
(iii)
Substituting value of (σc − σr ) from Eqn. (iii) to Eqn. (ii), we get
d σr d σz
d σr
d σc
(1 + μ ) + r
−μ
+
=0
ρ w2 r2 + r.
dr
dr
dr
dr
ρ w2 r2 (1 + μ ) + r
d σr
d σr
σc
d σr
d σz
+ μr
+ r − rμ
− rμ
=0
dr
dr
dr
dr
dr
d σc d σr
d σz
+
= −ρ w2 r (1 + μ ) + μ
dr
dr
dr
1
[σz − μ (σc + σr )]
E
∴ E εz = σz − μ (σc + σr ) = a constant say C
∴ σz = C + μ (σc + σr )
Now,
εz =
@seismicisolation
@seismicisolation
(iv)
544
•
Strength of Materials
or
d σc d σr
d σz
=μ
+
dr
dr
dr
(v)
Substituting this in Eqn. (iv)
d σc d σz
d σc d σr
+
= −ρ ω 2 r (1 + μ ) + μ 2
+
dr
dr
dr
dr
∴
ρ ω 2r
−ρ ω 2 r (1 + μ )
d σc d σr
=
−
+
=
dr
dr
1−μ
1 − μ2
Integrating, we get
σc + σr =
ρ ω 2 r2
+ 2A
2 (1 − μ )
(vi)
where 2A is a constant of integration.
From Eqns. (iii) and (vi),
2σr =
d σr
ρ ω 2 r2
+ 2A − ρ ω 2 r2 − r
2(1 − μ )
dr
3 − 2μ
d σr
= −ρω 2 r2
+ 2A
2 σr + r
dr
2 (1 − μ )
1 d 3 − 2μ
+ 2A
σr .r2 = −ρω 2 r2
r dr
2 (1 − μ )
d σr r 2
3 − 2μ
= −ρω 2 r3
+ 2Ar
or
dr
2 (1 − μ )
Integrating,
σr .r2 = −
ρω 2 r4
+ Ar2 − B
8
Where −B is a constant of integration.
σr = A −
B
3 − 2μ
−
ρ ω 2 r2
2
8(1 − μ )
r
Substituting this value of σr in Eqn. (vi)
σc = −
B
3 − 2μ
ρ ω 2 r2
+ 2A − A + 2 −
ρω 2 r2
2 (1 + μ )
8 (1 − μ )
r
= A+
3 − 2μ
ρ ω 2 r2
B
1−
−
4
r2 2 (1 − μ )
@seismicisolation
@seismicisolation
(vii)
Rotating Discs and Cylinders
•
545
or
σc = A +
B
1 + 2μ
−
ρ ω 2 r2
8 (1 − μ )
r2
(viii)
Now to evaluate σz , from Eqn. (v)
d σc d σr
d σz
=μ
+
dr
dr
dr
2B
1 + 2μ
2B
3 + 2μ
2
2
ρω r + − 3 −
ρω r
=μ − 3 −
4 (1 − μ )
4 (1 − μ )
r
r
μ
ρ ω 2r
=−
1−μ
Integrating,
σz = −
μ ρ ω 2 r2
+C
1−μ
2
where C is a constant of integration.
To determine C, consider the equilibrium of cylinder in the axial direction
ro
σz (2π rdr) = 0
ri
or
ro
σz rdr = 0 or
ri
−
μ
1−μ 8
ρω 2
ro4 − ri4 +
c
2
ro ri
μ ρω 2 r3
+Cr = 0
−
(1 − μ ) 2
ro2 − ri2 = 0
c=
μ ρω 2 2 2 ro + ri
1−μ 4
μ ρω 2 r2
μ ρω 2 2 2 .
+
ro − ri
1−μ
2
1−μ 4
μ
ρω 2 ri2 + ro2 − 2r2
or σz =
4 (1 − μ )
∴
σz =
(ix)
Solid Cylinder
If we consider Eqns. (vii) and (viii) carefully, we find that the constant B becomes zero because
stresses become infinite at r = 0
Now for solid cylinder boundary conditions is r = ro ; σr = 0
@seismicisolation
@seismicisolation
546
•
Strength of Materials
Therefore, Eqns. σr = A −
Becomes
B
3 − 2μ
−
ρω 2 r2
2
8 (1 − μ )
r
3 − 2μ
ρω 2 ro2
8 (1 − μ )
3 − 2μ
or, A =
ρω 2 ro2
8 (1 − μ )
0 = A−
[since constant
B = 0]
Substituting this value of A in Eqns. (vii) and (viii),
3 − 2μ
σr =
ρω 2 ro2 − r2
8 (1 − μ )
ρω 2 σc =
(3 − 2μ ) ro2 − (1 + 2μ ) r2
and
8 (1 − μ )
(x)
(xi)
Now for σz , putting ri = 0 in Eqn. (viii), maximum values of σr , σc and σz when r = 0 at the centre.
3 − 2μ
σr(max) =
ρω 2 ro2
8 (1 − μ )
3 − 2μ
σc(max) =
ρω 2 ro2
8 (1 − μ )
i.e.,
σr(max) = σc(max)
μ
and σz(max) = −
ρω 2 ro2 (Compressive)
4(1 − μ )
Also σz = 0 when 2r2 = ro2
√
or when r = ro 2
σc
σz
σr
ro
ro
√2
Figure 24.10 Stress distribution of σr , σc and σz in a rotating cylinder
@seismicisolation
@seismicisolation
Rotating Discs and Cylinders
•
547
Hollow cylinder: In case of hollow cylinder constants A and B can be determined from Eqns. (vii)
and (viii) taking boundary conditions, σr = 0 when r = ri and r = ro . From Eqn. (vii)
and
0 = A−
3 − 2μ
B
−
ρω 2 ri2
2
8 (1 − μ )
ri
0 = A−
B
3 − 2μ
−
ρω 2 ro2
2
ro 8 (1 − μ )
Solving above two equations,
B=
σr =
∴
and σc =
3 − 2μ
ρω 2 ri2 ro2
8 (1 − μ )
and,
A=
3 − 2μ
ρω 2 ri2 + ro2
8 (1 − μ )
r2 r2
3 − 2μ
ρω 2 ri2 + ro2 − i 2 o − r2
8 (1 − μ )
r
(i)
r2 r2 1 + 2μ 2
3 − 2μ
ρω 2 ri2 + ro2 − i 2 o −
r
8 (1 − μ )
3 − 2μ
r
(ii)
Also it was derived earlier that
σz =
Now σr is maximum where
μ
ρω 2 ri2 + ro2 − 2r2
4 (1 − μ )
d σr
=0
dr
∴
i.e., r =
σr(max) =
(iii)
√
ri ro
3 − 2μ
ρω 2 (ro − ri )2
8 (1 − μ )
(iv)
Also σc is maximum where r is minimum, i.e., r = ri
∴
σc(max) =
σz is maximum at r = ri and r = ro
∴
σz(max) =
when
μ
ρω 2 ro2 − ri2
4(1 − μ )
(Tensile)
r = ri
and when r = r0
σz(max) =
∴
1 − 2μ ri2
3 − 2μ
ρω 2 ro2 1 +
.
4 (1 − μ )
3 − 2μ ro2
μ
ρω 2 ro2 − ri2
4(1 − μ )
Also σz = 0
ri2 + ro2
r=
2
when
2r2 = ri2 + ro2
@seismicisolation
@seismicisolation
(Compressive)
(v)
548
•
Strength of Materials
σc
σz
σr
ri
ro
Figure 24.11 Variation of stresses in rotating hollow cylinder
E XAMPLE 24.12: A shaft of 250 mm diameter is rotating at 3200 r.p.m. Determine the maximum
radial and hoop stresses in the shaft. For the shaft material take ρ = 7800 kg/m3 and μ = 0.3.
S OLUTION :
The shaft may be taken as a solid cylinder.
∴
σr = A −
ω=
Now,
∴
3 + 2μ
ρω 2 r2
8 (1 + μ )
3200 × 2π
= 334.9 rad/s
60
σr = 0 when r = 125 mm = 0.125 m
0 = A−
3 − 2 × 0.3
× 7800 × 334.92 × (0.125)2
8 (1 + 0.3)
0 = A−
2.4
× 13669257
10.4
A = 3154444
σr = 3154444 −
2.4
× 7800 × 334.92 × r2
10.4
σr = 3154444 − 201884418r2
Now σr and σc are maximum at the centre of the cylinder (shaft), where r = 0
σr(max) = σc(max) = 3154444
= 3.15 MPa Ans
@seismicisolation
@seismicisolation
Rotating Discs and Cylinders
•
549
E XAMPLE 24.13: Also determine radial stress and hoop stress at radius 20 mm, 40 mm, 100 mm
and 250 mm and show stress distribution for the example 24.11.
σr = A −
As already found
3 − 2μ
ρω 2 r2
8 (1 − μ )
A = 3154444
σr = 3154444 −
2.4
× 7800 × 334.92 r2
10.4
σr = 3154444 − 201884418 r2
μ ρω 2 2 2 σc =
ro − r
1−μ 4
0.3 7800 × 334.92 ×
0.1252 − r2
=
0.7
4
σc = 93732051 0.015625 − r2
At r = 0.02 m
Using Eqns. (i) and (ii)
σr = 3154444 − 201884418(0.02)2
= 3073690
= 3.07 MPa Ans
σc = 93732051(0.015625 − 0.022 )
= 93732051(0.015225)
= 1427070
= 1.4 MPa Ans
At r = 0.04 m
σr = 3154444 − 201884418(0.04)2
= 3154444 − 323015
= 2831429
= 2.8 MPa Ans
σc = 93732051(0.015625 − 0.042 )
= 93732051 × 0.014025
= 1314592
= 1.31 MPa Ans
@seismicisolation
@seismicisolation
(i)
(ii)
550
•
Strength of Materials
Now σr and σc are maximum at the centre of the cylinder (shaft) where r = 0
σr(max) = σc(max) = 3154444
= 3.15 MPa Ans
E XAMPLE 24.14: For example 24.11, determine the radial and hoop stresses at radii 20 mm,
40 mm, 100 mm and 125 mm and show stress distribution in a diagram.
S OLUTION :
σr = A −
3 − 2μ
ρω 2 r2
8 (1 − μ )
A = 31544444
Now,
σr = 3154444 −
(already calculated)
2.4
× 7800 × 334.92 r2
10.4
= 3154444 − 201884418 r2
σc = A −
1 + 2μ
ρω 2 r2
8 (1 − μ )
= 3154444 −
1.6
× 7800 × 334.92 r2
5.6
= 3154444 − 185861845 r2
Using Eqns. (i) and (ii)
At
r = 20 mm = 0.02 m
σr = 3154444 − 201884418 (0.02)2
= 3073690
= 3.07 MPa
Ans
σc = 3154444 − 185861845 (0.02)2
= 3154444 − 74345
= 3080099
At
(i)
= 3.08 MPa Ans
r = 40 mm = 0.04 m
σr = 3154444 − 201884418 r2
= 3154444 − 201884418 (0.04)2
= 3154444 − 323015
@seismicisolation
@seismicisolation
(ii)
Rotating Discs and Cylinders
= 2831429
= 2.8 MPa
Ans
σc = 3154444 − 185861845 r2
= 3154444 − 185861845 (0.04)2
= 3154444 − 297379
= 2857065
= 2.86 MPa Ans
At r = 100 mm = 0.1 m
σr = 3154444 − 201884418 (0.1)2
= 3154444 − 2018844
= 2952599
= 1.13 MPa
Ans
σc = 3154444 − 185861845 r2
= 3154444 − 185861845 (0.1)2
= 3154444 − 1858618
= 1295825
= 1.29 MPa
At r = 125 mm = 0.125 m
Ans
σr = 3154444 − 201884418 (0.125)2
=0
σc
Figure 24.12
@seismicisolation
@seismicisolation
σr
•
551
552
•
Strength of Materials
E XAMPLE 24.15: A steel cylinder of 350 internal diameter and 700 mm external diameter is
rotating at 2200 r.p.m. Calculate the maximum value of radial, circumferential and longitudinal
stresses. Also determine the maximum shear stress in the cylinder. Take ρ = 7700 kg/m3 and
μ = 0.3
S OLUTION :
σr = A −
B
3 − 2μ
−
ρ ω 2 r2
r2 8 (1 − μ )
B
1 + 2μ
−
ρ ω 2 r2
2
8 (1 − μ )
r
μ
σz =
− ρω 2 (ri2 + r02 − 2r2 )
4 (1 − μ )
σc = A +
2π × 2200
= 230.3 rad/s
60
B 3 × 2 × 0.3
σr = A − 2 −
× 7700 × 230.32 r2
8 (1 − 0.3)
r
ω=
B 2.4
× 7700 × 230.32 r2
−
r2 5.6
B
= A − 2 − 175 × 106 r2
r
B
σc = A + 2 − 116.83 × 106 r2
r
0.3
σz =
× 7700 × 230.32 0.1752 + 0.352 − 2r2
4 (1 − 0.3)
= 43.76 × 106 0.153 − 2r2
σr = A −
Similarly,
(i)
(ii)
(iii)
In order to find out constants A and B,
From Eqn. (i) σr = 0 when r = 0.175 m and r = 0.35 m
B
− 175 × 106 (0.175)2
0 = A−
(0.175)2
0 = A − 32.65 B − 5359375
Also
0 = A−
B
2
(0.35)
− 175 × 106 (0.35)2
0 = A − 8.16 B − 21.4 × 106
32.65 B + 5359375 = 8.16 B + 21.4 × 106
24.49 B = 21400000 − 5359375
B = 654987
@seismicisolation
@seismicisolation
(iv)
Rotating Discs and Cylinders
Putting the value of B in Eqn. (iv)
0 = A − 32.65 × 654987 − 5359375
A = 21385325 − 5359375
= 16025950
∴
Equations (i) and (ii) become,
654987
− 175 × 106 r2
r2
654987
σc = 16025950 −
− 11683 × 106 r2
r2
σr = 16025950 −
Now σr is maximum when
r2 = ri ro = (0.175) (0.35)
= 0.06125
∴
σrmax = 16025950 −
654987
− 1.75 × 106 × 0.06125
0.06125
= 16025950 − 10693665 − 107187
= 5225098
= 5.22 MPa
Ans
σc is maximum at r = 0.175 m
σc(max) = 16025950 +
654987
(0.175)2
− 116.83 × 106 × (0.175)2
= 16025950 + 21387331 − 3577918
= 33835363
= 33.83 MPa Ans
σz is maximum r = ri = 0.175 m (Tensile)
and r = ro = 0.35 (Compressive)
From Eqn. (iii),
σzmax = ± 43.76 × 106 0.153 − 2r2
= ± 43.76 × 106 0.153 − 2 (0.175)2
= ± 43.76 × 106 (0.153 − 0.06125)
= ± 4.01 MPa Ans
@seismicisolation
@seismicisolation
•
553
554
•
Strength of Materials
For shear stress,
σc = 33.83 MPa,
σr = 0, σz = 4.01 MPa
σmax − σmin
∴
2
= 16.92 MPa Ans
τmax =
Zmax =
33.83 − 0
2
Disc of Uniform Strength
As we have seen that radial and hoop stresses in a rotating disc of uniform thickness keep on
changing with the change in radius. Since hoop stress is maximum at the centre and decreases
towards the periphery. And even the radial stress is zero at the periphery. Therefore, it is possible to
design a disc of uniform strength and vary the thickness to make it economical.
Let us consider a flat rotating disc of uniform strength (refer it Figure 24.13).
Centrifugal force
t+dt
dr
r
Element
dθ
2
A
σ
B
σ
D
C
σ
dθ
2
t
dθ
t0
Axis of rotation
(a)
o
(b)
Figure 24.13
Figure (a) shows the elevation of half such a disc and (b) shows the free body diagram of an element
ABCD of the disc which subtends an angle d θ at centre o. Let the uniform stress in the radial and
circumferential directions be σ as shown in figure (b)
Volume of the element = r.d θ .t.dr
Centrifugal force acting on the element ABCD due to rotation
= ρ .r.d θ .t.dr.ω 2 r
= ρ d θ .t.dr.ω 2 r2
Radial force on the face DC = r.d θ .t.σ
Radial force on the face AB = (r + dr)d θ .(t + dt)σ
Circumferential force on faces BC and DA = t.dr.σ
Resolving all the forces along the radial direction and considering equilibrium, we get
dθ
ρ .d θ .t.dr.ω 2 .r2 + (r + dr)d θ (t + dt)σ = rd θ .t.σ + 2tdr sin
.σ
2
@seismicisolation
@seismicisolation
Rotating Discs and Cylinders
•
555
dθ
dθ
=
because d θ is very small.
2
2
Cancelling d θ on both sides, the above equation is simplied as follows:
It may be noted that sin
p.t.dr.ω 2 r2 + r.dt.σ = 0
dt
σ = −ρω 2 r dr
t
or
Integrating both sides, we get
loge t =
ρω 2 r2
+ loge
2σ
(loge A is a constant of integration)
loge
t
ρω 2 r2
=−
A
2σ
−ρω 2 r2
t
= e 2σ
A
At r = 0, t = to
to = A
∴
(Note this point)
ρ ω 2 r2
2σ
Thickness at any radius, t = t0 e
−
E XAMPLE 24.16: The rotor disc of a turbine is of 900 mm diameter at the blade ring and is
fixed to a 76 mm diameter shaft. If the minimum thickness of the disc is to be 10 mm, find the
thickness at the shaft for a uniform stress of 200 MPa at 8000 r.p.m. Density ρ of the disc material is
7800 kg/m3 .
S OLUTION :
−ρω 2 r2
We know t = t0 e 2σ
At
At
r = 0.45 m,
r = 0.038 m,
−ρω 2 (0.45)2
2σ
10 = t0 e
−ρω 2 (0.038)2
2σ
t = t0 e
Dividing Eqn. (ii) by (i)
−ρω 2 (0.038)2
ρω 2 (0.2025 − 0.00144) /2σ
2σ
=
8e
t = 10 ×
−ρω 2 (0.45)2
t0 e
2σ
t0 e
@seismicisolation
@seismicisolation
(i)
(ii)
•
556
Strength of Materials
Where
2π × 8000 2
0.20106
ρω 2 × 0.20106
= 7800 ×
×
2σ
60
2 × 200 × 106
= 2.749
∴
t = 8e−2.749
= 8 × 1.5627
= 125 mm
Ans
E XAMPLE 24.17: A grinding wheel is 320 mm diameter with the bore at the centre 30 mm diameter. If the thickness of the wheel at the outer radius is 25 mm, what should be the thickness at the
bore diameter for a uniform allowable stress of 12 MN/m2 at 3000 r. p. m. Take density of the wheel
material as 2750 kg/m3 .
S OLUTION :
r1 = 0.16 m, t1 = 0.025 m.
r2 = 0.015 m, t2 =?
3000 × 2π
= 314 rad/s
60
σ = 12 × 106 N/m2 and ρ = 2750 kg/m3
ω=
Now,
−ρω 2 r2
t = t0 e 2σ
ρω 2 r2 2750 × (314 × 0.16)2
=
= 0.2892
2σ
2 × 12 × 106
0.025 = t0 e−0.2892
t0 =
0.025
e−0.2892
m
0.025
m
1
e0.2892
0.025
m
t0 =
0.749
t0 = 0.0333 m
t0 =
= 33.3 mm
Again,
−2750×(314×0.0125)2
2×12×106
t2 = 33.3e
t2 = 33.3e
−1765.2
106
@seismicisolation
@seismicisolation
Rotating Discs and Cylinders
•
557
= 33.3e−0.001765
= 33.3 ×
1
e0.001765
1
1.00177
= 33.24 mm Ans
= 33.3 ×
Exercise
24.1 A flywheel rim with a mean diameter of 6 m rotates at a speed such that the hoop stress in
the material is 10 MPa. The density of the material of the rim is 7000 kg/m3 . Determine the
speed ignoring the effect of arms.
[Ans 120.3 r.p.m.]
24.2 Calculate the hoop stress in a thin rim, 0.6 m mean diameter revolving about its axis at
800 r.p.m. Steel weighs 7700 kg/m3 .
[Ans 47.7 N/m2 .]
24.3 Determine the greatest values of radial and hoop stresses for a rotating disc in which the outer
and inner radii are 0.3 m and 0.15 m. The angular speed is 150 rad/s. Take μ = 0.309 and
ρ = 7700 kg/m3 .
[Ans 1.6 MN/m2 ; 13.6 MN/m2 ]
24.4 A steel cylinder of 300 mm internal and 600 mm external diameters is rotating at 2000 r.p.m.
Calculate the maximum value of radial, circumferential and longitudinal stresses. Also find
the maximum shear stress is the cylinder. Take ρ = 7800 kg/m3 and μ = 0.3
[Ans σr(max) = 3.3 MPa, σcmax = 27.48 MPa, τmax = 13.74 MPa]
24.5 A disc of uniform thickness having inner and outer diameter 100 mm and 400 mm respectively is rotating at 5000 r.p.m. about its axis. Take ρ = 7800 kg/m3 , and μ = 0.28.
Determine the stress distribution along the radius of disc
[Ans σr(0.15) = 0; σr(0.1 m) = 19.74 MPa σr(0.15) = 13.64 MPa
σr(0.2) = 0 σr(max) = 19.74 MPa σc0.05 m = 71.09 MPa; σc0.1 m = 41.11 MPa σc0.15 = 30.09 MPa
σc(0.2) = 19.78 MPa; σc(max) = 71.09 MPa, τmax = 35.54 MPa]
24.6 A disc having inner and outer diameters 150 mm and 300 mm respectively is rotating at an
angular speed of 150 rad/s. Calculate the greatest values of radial and circumferential stresses.
Take ρ = 7700 kg/m3 and μ = 0.304.
[Ans 16 MN/m2 , 13.6 MN/m2 ]
24.7 A stress turbine rotor running at 3500 r.p.m. is to be designed so that the radial and circumferential stresses are to be same and constant throughout and equal to 80 MPa. If the
axial thickness at the centre is 15 mm, what is the thickness at a radius of 500 mm? Density
= 7800 kg/m3 .
[Ans
@seismicisolation
@seismicisolation
2.92 mm]
558
•
Strength of Materials
24.8 A disc of turbine rotor is 0.5 m diameter. At the blade ring its thickness is 55 mm. It is keyed
to a shaft of 50 mm diameter. If the uniform stress in the rotor disc is limited to 200 MN/m2
at 9000 r.p.m. Determine the thickness of the disc at the start. Take ρ = 7700 kg/m3 .
[Ans 158.43 mm]
24.9 A solid cylinder, 250 mm diameter is rotating at 1500 r.p.m. Determine the maximum hoop
stress and the radial stress produced in the cylinder if the density of material is 7800 kg/m3
and μ = 0.28. Also determine the places where these stresses are maximum. What is the
radial stress at the outer radius?
[Ans
σc = 1.273 MPa at the centre, σr = 1.273 MPa at the centre, zero]
@seismicisolation
@seismicisolation
C HAPTER
25
FRAMEWORKS
A framework is an assembly of bars connected by hinged or pinned joints and intended to carry
loads at the joints only. Each hinge joint is assumed to rotate freely without friction, hence all the
bars in the frame exert direct forces only and are therefore in tension or compression. A tensile load
is taken as positive and a member carrying tension is called a tie. A compressive load is negative
and a member is compression is called a strut. The bars are usually assumed to be light compared
with the applied loads. In practice the joints of a framework may be riveted on welded but the
direct forces are often calculated assuming pin joints. This assumption gives values of tension or
compression which are on the safe side.
Frames are sometimes classified as 1. Perfect frame and 2. Imperfect frame.
Perfect frame: In a perfect frame, this is made up of members sufficient to keep it in equilibrium
without any change in its frame. The simplest perfect frame is a triangle which has three members.
The number of members, in a perfect frame, may also be expressed by the relation:
W
A
n = (2 j − 3)
n = no. of members
j = no. of joints
C
B
Figure 25.1
Deficient frame: This is an imperfect frame which does not satisfy the equation n = (2 j − 3). In this
case members are more or less than required number n.
Redundant frame: This is also an imperfect frame in which the number of members is more than
(2 j − 3).
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560
•
Strength of Materials
To solve the frames to determine if the members are in tension or in compression, there are three
methods, named
i) Graphical method (Bow’s notation)
ii) Analytical method or method of joints
iii) Method of sections
Bow’s Notation for Graphical Solution
E XAMPLE 25.1: The framework shown in Fig. 25.2 is composed of equilateral triangles. Determine the nature and magnitude of the forces in each member of the framework.
800 N
400 N
B
3
4
C
A
F
E
1
5
G
2
D
R1 = 700 N
R2 = 500 N
Figure 25.2
S OLUTION :
Taking moments about the extreme left hand pin-joint to find R2 :
Clockwise moments = Anticlockwise moments
1
3
800
span + 400
span = CD(span)
4
4
800 1200
+
= CD
4
4
∴ CD = 200 N + 300 N = 500 N
Now, downward forces = Upward forces
800 + 400 = R1 + 500
∴
R1 = 1200 − 500 = 700 N
Note: It may be noted that reaction R2 is called CD, reading clockwise around this joint.
Framework
Remember if we read member 1 − 3 around joint A, it will be read clockwise as AE (i.e., A to E
near joint A and EA near joint 3). In vector diagram we use small letters.
Now we proceed to draw the vector diagram to a suitable scale (for example, 1 kN = 1.5 cm).
The vector diagram is shown in Fig. 25.3.
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Frameworks
•
561
The sequence used in constructing the diagram is given below:
The equilibrium of external loading is a vertical line of loads:
ab = 800 units downward
bc = 400 units downward
cd = 500 units upward
da = 700 units upward.
a
g
e
f
d
b
c
Figure 25.3 Vector diagram
Proceeding to the extreme left hand pin-joint, a line from a is drawn parallel to member AE, and
a line from d is drawn parallel to member ED. The point e is the intersection of these lines.
The same procedure is applied to the extreme right-hand pin-joint permitting the positioning of
point g.
The next joint chosen is the pin-joint at which 800 N load in acting. Starting from member EA
and proceeding clockwise round the joint, vectors ea and ab already appear on the vector diagram.
From b a line is drawn parallel of FE, this positions the point f . The closing line g f is obtained
by considering the forces acting at the pin-joint under the load of 400 N. This makes the diagram
complete. And it should be observed that the diagram is a set of closed figures, there being no lines
which do not form a part of a polygon.
We can now return to the force diagram and place arrows near the pin-joints. For instance, for
the extreme left-hand pin-joint, the arrow on member AE points towards the joint, since the sense
of e to d on the vector diagram is ‘horizontally to the right’. Arrows at the other ends of AE and ED
are in opposing direction to those near the extreme left-hand pin-joint.
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•
Strength of Materials
B
A
F
E
C
G
D
Figure 25.4
The same procedure is applied to the remaining joints, producing arrows shown in Fig. 25.4.
We are now in position to state the nature of the forces in the various members, since if arrows point
together or pointing towards each other, the member is in tension while if they point apart (away)
the member is in compression. The magnitude of forces could have been obtained by scaling the
diagram, but the vectors of ef and f g are relatively small, and inaccuracy in the graphical work
could cause significant errors. Now db represents 100 N hence e f = f g = 100 N × (cosec 60◦ ) =
115.5 N. It would be advisable in many cases to obtain results by scaling the diagram. The solution
is completed by using trigonometry to obtain length of vectors and by tabulating the answers to a
degree of accuracy of three significant figures.
Answers:
Member
Load(N)
Nature
AE
808
C
BF
346
C
CG
577
C
DG
289
T
DE
404
T
EF
115
C
FG
115
T
C = Compression, T = Tension
E XAMPLE 25.2: The framed structure shown in Figure 25.5 is maintained in the position by a
force P. Determine:
(a) the magnitude of the force P;
(b) the reaction R at the hinge in magnitude and direction;
(c) the magnitude and nature of the force in each member.
400 N
C
P
B
D
G
F
H
1m
R
Hinge
2m
A
400 N
2m
E
Figure 25.5
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@seismicisolation
400 N
Frameworks
•
563
Taking moments about the hinge:
Clockwise moments = Anticlockwise moments
400 × 2 + 400 × 3 + 400 × 4 = P × 1
800 + 1200 + 1600 = P
∴
P = 3600 N
The vector diagram is shown in Fig. 25.6 (a) and the structure with arrows on the members is
shown in Fig. 25.6 (b).
g
c
b
d
e
h
a
f
Figure 25.6a
B
n=
ctio
Rea
400 N
C
P
R
D
G
H
F
A
E
400 N
400 N
Figure 25.6b
Answers:
P = 3600 N
Hinge reaction, R = 3790 at 18◦ 26 Member
Load(N)
Nature
BF
1700
C
CG
1200
T
DH
566
T
with horizontal
EH
400
C
C = Compression, T = Tension
@seismicisolation
@seismicisolation
AF
2400
C
FG
1700
T
GH
1130
C
564
•
Strength of Materials
E XAMPLE 25.3: The wall crane shown in Fig. 25.7 is loaded by a vertical force of 10 kN at the
joint ABD.
Determine graphically the magnitude and nature of the force in each member and the vertical
support reactions at the pin-joints ADC and BCD.
1 kN
A
B
D
30°
60°
C
RCA
RBC
Figure 25.7
S OLUTION :
Let DC be 5 units, then joint ABD is a right angle, it can be worked out A = 4.33 and
BD = 2.5 units
P
s
nit
3u
3
.
4
30°
2.5 units
60°
M
5 units O
N
Figure 25.8
In
ONP,
∴
ON
= cos 60◦
NP
ON
1
=
2.5
2
2.5
= 1.25 units
ON =
2
Taking moments about joint DBC (Fig. 25.7)
RCA × 5 = 1 × 1.5
1.25
RCA =
= 0.25 kN
5
RBC = 1 − 0.25 = 0.75 kN
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Frameworks
•
565
Now we proceed to draw vector diagram:
a
c
d
b
Figure 25.9 Vector diagram
1 kN
A
B
D
C
0.25k N
0.75 kN
Figure 25.10
Note: According to Bow’s notation, force 1 kN will be read as ABD (reading clockwise around
joint).
After measurement from vector diagram, results are tabulated below:
Member
Load (kN)
Nature
AD
0.5
C
BD
0.867
C
CD
0.433
T
Reactions : RCA = 0.25 kN
RBC = 0.75 kN
E XAMPLE 25.4: The wall crane shown in Fig. 25.11(a) is loaded by a vertical force of 10 kN
at joint ABD. The bar AD is horizontal and the framework is pinned to the vertical wall at joints
ADC and BC. Determine the magnitude and nature of the forces in the framework. What are the
magnitude and direction at the top wall joint?
Note: The arrowheads on members and reactions are shown in Fig. 25.12(a) as answer are not
in question.
@seismicisolation
@seismicisolation
566
•
Strength of Materials
A
45
45°
D
45
10 kN
C
C
d
a
B
30°
b
(a)
joint ABD
(b)
Figure 25.11
S OLUTION :
It may be noted that vertical line of the wall is not a load or bar of the frame. Space C separates
the reaction BC at the lower wall joint and the reaction CA at the upper wall joint.
The reaction CA is unknown in magnitude and direction. The reaction of the wall at the lower
joint is equal and opposite to the force in bar BC.
Joint ABD Fig. 25.11(b)
Draw ab vertically downwards from a to represent the 10 kN load AB. Draw bd parallel to bar
BD, and ad parallel to bar DA.
The intersection d of bd and ad complete the force diagram for the joint.
The directions of forces at the joint are fixed by the known downward vector ab. Following the
vectors in order round triangle abd determines the direction bd (towards the joint) and the direction
da (away from the joint). These directions are shown in Fig. 25.11(b).
Joint DBC (Fig. 25.12 a):
The known force at this joint is db equal and opposite to bd.
From b draw force bc parallel to BC.
From d draw force dc parallel to CD.
Thus joint c is determined.
The force directions at the joint are fixed by the vector arrow from d to e as shown.
Joint ADC (Fig. 25.12(b)):
The known force is ad in AD, equal and opposite to da, already found. Join ca. This, then
represents the force in CA, the reaction at the joint. It acts away from the joint. The polygon abdc
in complete force diagram for the framework. By construction, the forces are found as follows:
Reaction at top well joint =7.8 kN (CA) at 20◦ .
Reaction at bottom = 14.6 as shown in Fig. 25.11(a)
Member
Load (kN)
Nature
BC
14.6
C
CD
3.8
C
@seismicisolation
@seismicisolation
DB
14.14
C
DA
10
T
•
Frameworks
567
c
c
d
a
Joint DBC
a
d
b
(b)
Joint ADC
b
(a)
Figure 25.12
E XAMPLE 25.5: The Warren girder shown in Fig. 25.13(a) is loaded vertically by forces of 20
and 30 kN at the lower panel pin-joints. It is simply supported at the joints ABG and AED.
Determine the magnitude and nature of forces in each bar and the magnitudes of the support
reactions.
Note: Arrowheads shown in Fig. 25.13(a) are not in question but meant as answers.
S OLUTION :
A
E
A
G
F
D
b
A
B
C
20 kN
30 kN
b
(a)
b
e
g
a
c
f
a
g
Joint ABG
(b)
a
g
c
f
Joint GBCF
(c)
Figure 25.13
@seismicisolation
@seismicisolation
d
(d)
568
•
Strength of Materials
There are more than two unknown forces at every joint; hence we cannot begin the force diagram
immediately. It is necessary first to calculate the vertical reactions at the supports by taking moments
about each support in turn.
let
L = vertical reaction DA
R = vertical reaction AB
Assume each horizontal bar of the frame to be of length 2 units.
Taking moments about joint AED,
4 × R = (30 × 3) + (20 × 1)
R = 27.5 kN
Taking moments about joint ABG,
4 × L = (30 × 1) + (20 × 3)
L = 22.5 kN
Check: L + R = 22.5 + 27.5 = 50 kN = total load. Joint ABG (Fig. 25.13(b))
Draw ab vertically upwards to represent the 27.5 kN reaction AB.
Draw bg through b parallel to BG.
Draw ga through a parallel to GA.
The force directions at the joints are fixed by the upward vector ab. The direction of the force in
BG is away from the joint. The direction of the force GA is towards the joint. The direction arrows
are now added to Fig. 25.13(a).
Joint GBCF (Fig 25.13(c)):
Draw bc to represent the downward load BC of 30 kN. Draw f g through g parallel to FG. Draw
e f through e parallel to CF.
The intersection of f g and e f given point f and complete the diagram for the joint.
The directions of the forces at the joint are fixed by the known direction bc of the load BC.
Tranversing the polygon be f g in order determines the force directions.
Joint CDEF (Fig 25.13 d):
Draw cd vertically downward from c to represent the known load CD of 20 kN.
Draw de parallel to DE.
Draw e f parallel to EF.
If correctly drawn, the point e lies on the intersection of de and e f , and also on the horizontal line
ae, parallel to AE.
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Frameworks
•
569
The directions of the forces at the joint are fixed by the known direction of cd. The directions
are found by following the arrows round the polygon cde f in order.
Results:
Reaction AB = 27.5 kN
Reaction DA = 22.5 kN
Member
Load (kN)
Nature
BG
39
T
GA
27.5
C
CF
25
T
FG
3.5
T
DE
3.2
T
EF
3.5
C
EA
22.5
C
T for tension and C for compression.
E XAMPLE 25.6: Figure 25.14 shows a truss of total span 30 m. Determine the magnitude and
nature of forces in all members of the framework.
P
P
P C
K
P/2
B
A
D
E
M
L
N
O
2m
2m
R1 = 3P
2m
P
F
P
J
2m
P
Q
R
S
2m
R2 = 3P
6P
= 3P
2
b
p
c
r
j,s
R,g
n,o
k
d
i
e
f
m
Vector diagram
Figure 25.14
@seismicisolation
@seismicisolation
H
2m
Framework
Due to symmetry reaction R1 = R2 =
P/2
G
g
570
•
Strength of Materials
Answers:
Member
Magnitude
Nature
T for Tension C for Compression
BJ, GS
JI, IS
JK, RS
CK, RF
KL, QR
LI, IQ
LM, PQ
DM, EP
MN, OP
NI, IO
NO
6.73 P
6.25 P
1.00 P
6.73 P
1.60 P
1.00 P
1.50 P
5.40 P
1.95 P
4.75 P
0
C
T
C
C
T
T
C
C
T
T
−
Method of joints: In this method, every joint is considered separately and conditions of equilibrium
are applied. Of course reactions in some cases are found out beforehand.
E XAMPLE 25.7: The framework shown in Fig. 25.15 has a span of 6 m and a load of 10 kN is
applied at top. Find the magnitude of nature of forces in the members AB, AC and BC.
S OLUTION :
10 kN
A
60°
B
RB
30°
D
6m
Figure 25.15
For taking moments we will need BD, so
BD
= cos 60
AB
∴ BD = 0.56 AB
In a right angled triangle BAC
cos 60◦ =
AB AB
=
BC
5
AB = 6 cos 60◦
@seismicisolation
@seismicisolation
C
RC
Frameworks
∴ AB = 6 ×
571
F1
1
2
60°
=3m
Now
•
F2
7.5 kN
BD = 0.5 AB
Figure 25.16.
∴ BD = 0.5 × 3
= 1.5 m
Now taking moments about B,
10 × 1.5 = RC × 6 = 0
∴ RC =
15
= 2.5 kN
6
RB = 10 − 2.5
and
= 7.5 kN
Joint B:
Let force F1 act towards the joint B and F2 acting away from the joint B.
Resolving the force at B, vertically,
F1 sin 60 = 7.5
7.5
sin 60
7.5
=
0.866
F1 =
F1 = 8.66 kN
As F1 is pushing the joint B, therefore this force will make member BA compressive. Now
resolving horizontally,
F1 cos 60 = F2
1
8.66 × = F2
2
∴ F2 = 4.33 kN
(member BC will be in tension)
@seismicisolation
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572
•
Strength of Materials
Joint C:
F2
F3
30°
F2
is already known equal to 4.33 kN.
Now resolving vertically, F3 sin 30 = 2.5
2.5
F3 =
0.5
= 5 kN
2.5 kN
Figure 25.17
(It will make the member AC in compression).
Ans: Results :
RB = 7.5 kN RC = 2.5 kN
Member
Force (kN)
Nature
BC
4.33
T
AB
8.66
C
AC
5
C
E XAMPLE 25.8: A truss as shown in Fig. 25.18 in loaded. Determine the magnitude and nature
of force in each member.
S OLUTION :
2000 N
4000 N
C
B
A
60°
60°
60°
60°
E
4m
RA = 2500 N
D
4m
RD = 3500 N
Figure 25.18
Taking moments about D,
RA × B = 2000 × 6 + 4000 × 2
= 12000 + 8000
20000
= 2500 N
RA =
8
@seismicisolation
@seismicisolation
Frameworks
RA + RD = 2000 + 4000
= 6000
2500 + RD = 6000
RD = 6000 − 2500
= 3500 N
Joint A: Resolving vertically
B
F2 sin 60 = 2500
F2
2500
sin 60
2500
=
0.866
F2 =
60°
A
E
F1
2500 N
Figure 25.19
= 2887 N
Resolving horizontally
F2 cos 60 = F1
∴ F1 = 2887 × 0.5
= 1443.5 N (∴
member AE will be in tension)
Joint D
Resolving vertically
C
F4 sin 60 = 3500
3500
F4 =
0.866
= 4041.6 N
∴
60°
E
Member CD will be under compression
F4
D
F3
3500 N
Figure 25.20
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•
573
574
•
Strength of Materials
Resolving horizontally,
F3 = F4 cos 60◦
F3 = 4041.6 × 0.5
= 2020.8 N
∴ Member ED will be under tension.
Joint B:
We already know from study of joint A that F2 = 2887 N
2000 N
Resolving vertically
2000 + F6 sin 60 = 2887 sin 60
0.866F6 = 2500 − 2000
500
F6 =
0.866
= 577 N
F5
B
60°
887 N
2
C
60°
F6
A
E
Figure 25.21
Resolving horizontally,
2887 cos 60 + 577 cos 60◦ = F5
(∵
F6 = 577 N)
1443.5 + 288.5 = F5
F5 = 1732 N
Therefore, BC will be under compression.
Joint C:
4000 N
We have already found out found in
CD = 4042 N and force in BC = 1732 N
B
1732 N
E
Resolving vertically,
4000 = F6 sin 60 + 4042 sin 60
4000 = 0.866 F6 + 3500
F6 =
4000 − 3500
0.866
= 577.4 N
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C
60°
F6
60°
4042 N
A
Frameworks
•
575
Results: RA = 2500, RB = 3500 N
Member
Force N
Nature
AB
2887 N
Comp
BC
1732 N
Comp
CD
4041.6
Comp
BE
577
Tension
CE
577 N
Comp
AE
1443.5
Tension
ED
2021
Tension
E XAMPLE 25.9: For the truss shown in Fig. 25.22, determine the force in all the members in
magnitude and nature.
5 kN
F
G
H
J
K
5m
θ
A
C
3.75 m
RA = 7.5 kN
D
3.75 m
E
3.75 m
B
3.75 m
RB = 7.5 kN
5 kN
5 kN
Figure 25.22
S OLUTION :
Due to symmetry of truss and loading,
RA = RB =
5+5+5
= 7.5 kN
2
√
In right angled triangle ACF, CF = 3.752 + 52
CF =
√
14 + 25
CF = 6.245 m
∴
sin θ =
5
= 0.8
6.245
θ = 53.13◦
cos θ =
3.75
= 0.6
6.245
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576
•
Strength of Materials
Joint A:
Resolving horizontally
F1
F2 = 0
A
F2
Resolving vertically
F1 = 7.5 kN
RA = 7.5 kN
∴ Member AF is under compression.
Figure 25.23
Joint F:
Resolving horizontally,
F
F4 = F3 cos θ
= F3 × 0.6
F4
θ
F3
F1 = 7.5 kN
Figure 25.24
Resolving vertically,
7.5 = F3 sin θ
7.5 = F3 × 0.8
7.5
F3 =
= 9.375 kN (Member FC is under tension.)
0.8
F4 = 0.6 F3
F4 = 0.6 × 9.375
= 5.625 kN
∴ Member FG under compression.
Joint C:
As we found out earlier F2 = 0
Resolving horizontally,
F5
F3 = 9.375
F6 = 9.375 cos θ
= 9.375 × 0.6
= 5.625 kN
F6
θ
F2 = 0
C
5 kN
∴
Member CD is under tension.
Figure 25.25
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•
577
Resolving vertically,
F5 + 9.375 sin θ = 5
F5 + 9.375 × 0.8 = 5
F5 = 5 − 7.5 = −2.5 kN
The negative sign shows that the choice in assuming direction of F5 was wrong. Therefore, the
assumed direction of F5 is member CG need to be reversed. Hence, CG is under compression.
Joint G:
Resolving horizontally,
5.625 + F7 cos θ = F8
5.625 + F7 × 0.6 = F8
Resolving vertically,
G
FA = 5.625 kN
2.5 = F7 sin θ
2.5 = F7 × 0.8
2.5
= 3.125 kN
F7 =
0.8
∴
∴
F7
F5 = 2.5 kN
Figure 25.26
GD is under tension
∴
F8
θ
5.625 + 3.125 × 0.6 = F8
∴ F8 = 5.625 + 1.875 = 7.5 kN
GH is under compression
Joint H:
Resolving horizontally,
5 kN
7.5 = F10
∴
F10
member HJ is under compression.
Resolving vertically,
5 = F9
∴
F8 = 7.5 kN
H
F9
F5 = 2.5 kN
member HD is under compression.
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Figure 25.27
578
•
Strength of Materials
Since the the truss is symmetrical, there is no need to consider remaining joints.
Results: RA = RB = 7.5 kN.
Member
Force (kN)
Nature
AF, BK
7.5
C
AC, BE
0
-
CF, EK
9.375
T
FG, JK
5.625
C
GC, JE
2.5
C
CD, DE
5.625
T
DG, DJ
3.125
T
GH, HJ
7.5
C
HD
5
C
E XAMPLE 25.10: A cantilever truss is shown in Fig. 25.28. Determine the magnitude and nature
of the force in each member.
50 kN
F
60°
A
37.5 kN
50 kN
E
60°
60°
B
12.5 m
D
60°
37.5 kN
C
12.5 m
Figure 25.28
Joint A:
Resolving horizontally,
F1
F2 = F1 cos 60
F2 = 0.5 F1
60
A
F2
37.5 kN
Figure 25.29
Resolving vertically,
37.5 = F1 sin 60
37.5 = F1 × 0.866
F1 =
37.5
= 43.3 kN
0.866
∴
F2 = 0.5 × 43.3 = 21.65 kN
Therefore, the member AF is under tension and the member AB is under compression.
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Frameworks
Joint F:
Resolving horizontally,
50 kN
60°
F
60° F4
B
F3
F
1=
43
.3
F4 = 43.3 cos 60 + F3 cos 60◦
F4 = 21.65 + 0.5F3
A
Figure 25.30
Resolving vertically,
F1 sin 60◦ + 50 = F3 sin 60
43.3 × 0.866 + 50 = 0.866F3
37.5 + 50 = 0.866 F3
F3 =
87.5
0.866
= 101.04 kN
Therefore, member FB is under compression
F4 = 21.65 + 0.5 × 101.04
= 21.65 + 50.52
= 72.17 kN
Member FE is under tension
Joint B:
E
F5
Resolving horizontally,
21.65 + 101.04 cos 60 + F5 cos 60 = F6
21.65 + 50.52 + 0.5F5 = F6
101.04
60°
60°
C
A
F6
21.65 kN B
Resolving vertically,
37.5 + 101.04 sin 60 = F5 × sin 60
37.5 + 101.04 × 0.866 = 0.866F5
37.5 + 87.5 = 0.866F5
125
= 144.34 kN
F5 =
0.866
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37.5 kN
Figure 25.31
•
579
580
•
Strength of Materials
Member BF is under tension
Now,
21.65 + 50.52 + 0.5 × 144.34
= F6
F6 = 144.34 kN
Therefore, member BC is under compression.
Joint E:
Resolving horizontally,
50 kN
72.17 + 144.34 cos 60 + F7 cos 60◦ = F8
72.17 + 72.17 + 0.5F7 = F8
72.17 kN
F
60°
144.34 kN
Resolving vertically,
E
F8
60°
◦
D
F7
50 + 144.34 sin 60 = F7 sin 60
50 + 144.34 × 0.866 = 0.866F7
Figure 25.32
175 = 0.866F7
F7 =
175
= 202.08 kN
0.866
72.17 + 72.17 + 0.5 × 202.08 = F8
F8 = 245.38 N
Therefore member EC is under compression.
And member ED is under tension
Answer:
Member
Force
Nature
AF
43.3
T
AB
21.65
C
BF
101.04
C
FE
72.17
T
BE
144.34
T
BC
144.34
C
CE
202.08
C
CD
245.38
T
The Method of Sections
The general conditions of equilibrium can be developed into a method of finding the magnitudes
of forces in frame structures; it is particularly useful when one or few of the members is being
considered. The method simply consists of finding reactions, and there making a section cut the
member under consideration, finally applying the general condition of equilibrium:
Leftward forces = Rightward forces
Upward forces = Downward forces
Clockwise moments = Anticlockwise moments.
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581
In certain cases, a simple solution is effected by arranging for the section to cut the member
under consideration and two other members. Moments are taken about the intersection of the other
two members. We will now solve the following example by this method.
E XAMPLE 25.11: Figure 25.33 shows a loaded framework. Determine the magnitude of the force
in the horizontal member CF, and state whether this member is in tension or compression.
2000 N
4000 N
C
F
B
E
R1
D
G
4m A
R2
Figure 25.33
We will first find R1 by taking moments about the right handed support.
Clockwise moments = Anticlockwise moments
R1 × 4 m = 4000 N × 1 m + 2000 N × 3 m
= 400 Nm + 6000 Nm
= 10000 Nm
10000
= 2500 N, ∴ R2 = 2000 + 4000 − 2500
∴ R1 =
4
= 3500 N
Although it is not completely necessary in the solution, let us find R2 to check the value of R1 ,
Moments about the left hand support:
Clockwise moments = Anticlockwise moments
2000 N × 1 m = 4000 N × 3 m = R2 × 4 m
2000 Nm + 1200 Nm = R2 × 4 m
1400
= 3500 N.
∴ R2 =
4m
Upward forces = Downward forces
2500 × 3500 N = 2000 N + 4000 N
6000 N = 6000 N
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582
•
Strength of Materials
The reactions therefore appears to be correct. We now make a section, if we can, to cut the
members under consideration (the force in that member being designates as P) and two other members. This gives us Fig. 25.34. It does not matter, for the moment, in which directions we assume
the force P to be acting.
2000 N
Section
line
P
O
2500 N
Figure 25.34
We now consider the loadings on the portion cut off by the section. We will take the left-hand
side but it does not matter which. The two other members intersect at 0.
Taking moments about 0:
Clockwise moments = Anticlockwise moments
P × 1.732 m + 2500 N × 2 m = 2000 N × 1 m
P × 1.732 m = 2000 Nm − 5000 Nm
= −3000 Nm
−3000 Nm
∴ P=
1.732 Nm
= −1732 N
The negative result tells us the direction we assumed P to be acting was incorrect. P in fact
pushes on the joint BCFE. The arrow on the other end of members CF pushes on the opposing joint
CDCF. The arrows on members CF therefore point away from each other, and members CF is in
compression.
Answer: Force in member CF is a compressive force of 1732 N (Note : Had the value of P
been positive it would have indicated that the direction in which the force P was assumed to act was
correct.)
The question has now been solved, but to show the validity of the method, let us repeat the solution with another section, with the members to the right of this section line as shown in Fig. 25.34.
This time we will let P push on the joint.
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Frameworks
P
•
583
4000 N
O
3500 N
Figure 25.35
Taking moments about 0:
Clockwise moments = Anticlockwise moments
P × 1.732 m + 4000 N × 1 m = 3500 N × 2 m
P × 1.732 m = 7000 Nm − 4000 Nm = 3000 Nm
3000 Nm
∴ P=
= 1732 N
1.732 m
P is positive, hence the assumption for the direction was correct. The arrow on the other end of
member CF points in the opposite direction. The arrows point away from each other. Member CF
is therefore in compression. Hence, CF is loaded is compression by a force of 1730 N, as before.
In introducing this method, it was suggested that the section cuts the member under consideration and two other members, moments then being taken about the point of intersection of the other
two members. This can not apply on every section and in cases it may be necessary to adopt aspects
of the general conditions of equilibrium other than moments. To demonstrate this we will follow
another question.
E XAMPLE 25.12: The framework in Fig. 25.36 is an equilibrium triangle, and the member AD is
kept horizontal by a force P. Determine, either by calculation or graphically:
a) the magnitude of the force P
b) the direction and magnitude of the higher reaction R, and
c) the force in member AD, stating it is in tension or in compression.
P
B
C
D
R
A
500 N
Figure 25.36
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584
•
Strength of Materials
S OLUTION :
Moments about hinge, taking the length of side of the equilateral triangle as L:
P × L = 500 N × L
∴ P = 500 N
Leftward forces = Rightward forces
Let the hinge reaction = R at θ .
Horizontal component of P = Horizontal component of R
P cos 30◦ = R cos θ
500 × 0.866 = R cos θ = 433 N
Upward forces = Downward forces
Vertical component of P+ vertical component of R = 500 N
P sin 30◦ + R sin θ = 500 N
500 N × 0.5 = R sin θ = 500 N
∴
R sin θ = 500 N − 250 N = 250 N
R sin θ
250
=
R cos θ
433
or
tan θ = 0.5774
∴
θ = 30◦
R sin θ = 250 N
∴
R=
=
250 N
250
=
sin θ
sin 30
250
0.5
= 500 N
∴
R = 500 N at 30◦
We, now take a section to cut the member under consideration and at least one other member.
For simplicity, the section is an indicated in Fig. 25.37.
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Frameworks
•
585
Section line
Q
P
500 N
Figure 25.37
Although we only require the value of the force in the member AD, it is necessary to include one
other member. Further more, we cannot take moments, so we must use other aspects of the general
conditions of equilibrium.
Upward forces = Downward forces
Q sin 60 = 500
Q × 0.866 = 500
Q=
500
= 577.4 N
0.866
Leftward forces = Rightward forces
P = Q cos 66◦
= 577.4 × 0.5
= 288.7 N
A positive value of P shows that the assumption made for its direction is correct. The arrow
on the other end of member AD points in the opposite direction The arrows point away from each
other, hence member AD is in compression.
Answers:
a) P = 500 N
b) R = 500 N at 30◦
c) Force in D = 289 N compressive.
E XAMPLE 25.13: The framework shown in Fig. 25.38 carries a load of 19.6 kN at the lower
middle point. Find the forces in bars 1, 2, 3, and 4 and state whether the state whether the bars are
in tension or compression.
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586
•
Strength of Materials
A
1
4
2
60º
30º
60º
30º
3
19.6 kN
A
Figure 25.38
S OLUTION :
√
3 m, bar 4 = 1 m and the height
Let the horizontal
members
each
of
length
2
m,
then
bar
2
=
√
3
of the frame is
m. The weight of 19.6 kN is acting as shown. From symmetry the reaction
2
19.6
R1 = R2 =
= 9.8 kN
2
A
22.63 kN
19
.6
11.32 kN
19.6 kN
60º
30º
30º
5.66 kN
19.6 kN
9.8 kN
11.32 kN
kN
60º
5.66 kN
9.8 kN
A
Figure 25.39
Get the frame cut by an imaginary plane AA, shown in Fig. 25.38 and let X,Y, Z be the internal
forces in the cut bars 1, 2, 3 respectively. If we assume that the cut bars are all in tension then the
forces X,Y, Z will act as shown in Fig. 25.40, i.e. forces X and Y are pulling away from the top
right-hand joint and force Z is pulling away from the lower right hand support.
Q cos 60
1
X
2
4
Q
Y
3
Z
Q sin 60
Z
R = 9.8 kN
R = 9.8 kN
(a)
(b)
Figure 25.40
Consider the equilibrium of the frame section to the right of the cutting plane AA. To find force
Y resolve external and internal forces on the frame section, thus in the vertical direction:
Y sin 30◦ = R = 9.8
Y = +19.6 kN
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•
587
Y is positive therefore the assumed direction is correct and bar 2 is in tension.
To find force X we may take moments about the joint carrying weight 19.6 kN, then eliminating
the forces Y and Z.
√
3
+R×2 = 0
X
2
9.8 × 4
X =− √
= −22.63 kN
3
X is negative (therefore the direction of the force must be reversed) end bar 1 is in compression.
To find force Z, resolve forces horizontally for the frame section, thus:
X +Y cos 30◦ + Z = 0
−22.63 + 19.6 × 0.866 + Z = 0
Z = +5.66 kN
Z is positive, therefore bar 3 is in tension.
To find the force in bar 4, we may consider the forces acting at the lower right hand support
joint. There are three forces R, Z and the force in bar 4 say Q. Resolving forces horizontally,
Q × cos 60◦ = Z = 5.66
Q = 11.32 kN
The forces in the remaining bars can be found from symmetry and the complete solution is
shown in Fig. 25.39. In practice the correct directions of the forces in the bars can often be obtained
by inspection.
Exercise
25.1 Fig 25.41 shows a simple roof truss. A wind load normal to the longer sloping side is assumed
to be equivalent to a 10 kN load at each pin joint. The reaction at the right hand joint may
be taken vertical. Determine the reactions and the nature and magnitude of the force in each
member.
10 kN
B
A
10 kN
E
60º
30º
D
[Ans Members: AE, 10 comp; BE, 0; DE
−5 Comp. Reactions CD: 8.66, DA, 13.25 at
41◦ to and above horizontal.
Note:- −ve sign denotes compression otherwise member is in tension.]
C
Figure 25.41
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588
•
Strength of Materials
25.2 Figure 25.42 shows a loaded framework, the triangle being equilateral. Find by the method
of sections, the magnitude of the force in member FG stating whether this is tensile on compressive.
1000 N
3000 N
C
B
D
F
E
G
60º
60º
60º
60º
A
R1
R2
[Ans
Figure 25.42
577 N in compression]
25.3 Determine by the method of sections, the magnitude and nature of forces in members AB, BC
and CD only, of the pin-jointed framework shown in Fig. 25.43. (All sloping members lie at
45◦ to the horizontal).
R1
1 kN
A
B
C
D
2 kN
[Ans AB = 5 kN (Compressive)
BC = 0
CD = 5 kN (Tensile)]
5 kN
R2
Figure 25.43
25.4 In the warren girder shown in Fig. 25.44 all bars are of equal length and the loads are vertical.
Find the magnitude and nature of the forces in the members A, B and C. Both reactions are
vertical.
20 kN
60°
80 kN
60°
60°
B
C
60°
60°
60° A 60°
60° 60°
[Ans
Figure 25.44
@seismicisolation
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A: 46.2, B: −23.1 C: −69.3 kN]
Frameworks
•
589
25.5 For the framework shown in Fig. 25.45, find analytically the magnitude and nature of the
forces in all the members of the frame and state the reactions at the supports.
F
B
A
J
45°
C
H
E
45°
45°
X
Y
D
G
100 kN
[Ans AB − 66.7; DA = vertical
AJ = Horizontal AJ = 0, BC, 94.3;
BE = HG = FH = 33.3 HJ = EF = −47.1,
EG, 66.7, AF, −33.3, AC = −66.7 kN;
reaction at X = 66.7 kN vertical,
reaction at Y = 33.3 kN vertical.]
Figure 25.45
25.6 Determine the magnitude and nature of the forces in the framework shown in Fig. 25.46 by
any method and state magnitude and direction of the reactions at the supports.
[Ans AD, 11.54; BD, −5.77; DC, −11.54;
CA, 11.54 kN Reaction at X = 11.54 kN
horizontal; Reaction at Y = 15.27 at 48◦ 8 to
vertical.]
A
X
C
D
B
Y
10 kN
Figure 25.46
25.7 For the framework shown in Fig. 25.47, find by any method magnitude and nature of force in
each member. Also find magnitude and direction at hinge X, and force in cable at Y .
Cable
Y
A
30°
30°
E
90°
C
D
B
10 kN
[Ans AE, 17.32, EB, −20, BD, −17.32;
CD = 5; DE, −10 kN
Reaction at X = 13.23 kN
at 40◦ 54 to vertical.
Reaction at Y = pull in cable
= 8.66 kN]
30°
X
Figure 25.47
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590
•
Strength of Materials
25.8 For the framework shown in Fig. 25.48, find by any method, the force and its nature of each
member. Also find reactions with directions.
Note: 10 tonnes should be converted into kN.
A
10 tonne
C
75°
B
[Ans AC = 98.1, BC = −170,
CD = −25.4, AD = 94.8 kN
Reaction at X = 94.8 kN
parallel to bar AD
Reaction at Y = 178 kN
at 76◦ 12 to horizontal]
D
45°
60°
45°
X
Y
Figure 25.48
25.9 Find graphically the force and its nature in each member of the framework (Fig. 25.49). Also
find reaction at X and Y with direction.
A
5m
1.5 m
E
D
C
X
30°
5m
B
20 kN
30°
Figure 25.49
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Y
[Ans AD = AE = −33.3 kN,
DC = BE = 29.6,
DE = 33.3 kN
Reaction at X & Y
are vertical = 10 kN]
C HAPTER
26
DAMS
Throughout the year large quantity of water is required for irrigation and power generation. In order
to meet this requirement, construction of dams is important to store the water. A retaining wall is
constructed to retain the earth in hilly areas to construct a dam. There are many types of dams for
example, rectangular dams, trapezoidal dams and so on. Trapezoidal dams are more popular because
of being economical and easier to construct.
Rectangular Dams
Figure 26.1 shows a rectangular dam. For study we will consider a unit length of this dam which is
retaining water on one of its vertical sides.
b
Water
Dam
G
h
O
F
h/
3
N
H
C
θ
M
K
A J
W
R
Water pressure
diagram
x
Figure 26.1
w = Specific weight of water
h = Height of water
F = Force exerted by water on the side of the dam
H = Height of dam
W = Weight of dam per metre length of dam
b = Width of dam
wd = Weight density of dam masonry
Weight of dam per unit length,
W = wd b.H
591
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592
•
Strength of Materials
This weight acts through centre of gravity, G of the dam.
Intensity of water pressure will be zero at water surface and will increase linearly to wh at the
bottom surface.
wh
The average intensity of water pressure on the face of the dam =
2
The force F is given by
F = wA h
= w (h × 1) ×
=
h
2
h=
h
2
w × h2
2
h
This force F will act horizontally at a height of above the base (called centre of pressure).
3
Thus, there are only two forces acting on the dam as shown in Fig. 26.1. The resultant force
R may be found out by parallelogram of forces as shown in figure. Take OC = F and ON = W to
some suitable scale. On completing rectangle, diagonal OD will represent the resultant force R.
√
∴ Resultant R = F 2 +W 2
F
MN
=
tan θ =
ON
W
so θ can be calculated.
The diagonal OM represents the resultant of F and W . If we extend the diagonal OM so that it
cuts the base of the dam at point K. And if line of weight W is extended to meet the base at J,
Let
x = distance JK
The distance x is obtained from similar triangles ONM and OJK as explained below:
∴
NM
JK
=
h/ 3
ON
x
F
=
h/ 3 W
F h
x= ×
W 3
E XAMPLE 26.1: A masonry dam of rectangular section, 22 m high and 12 m wide has water up to
a height of 18 m on its one side. Determine:
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 the weight density of masonary = 20 kN/m3 and weight density of water = 9.81 kN/m3 .
@seismicisolation
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Dams
•
593
S OLUTION :
(i) F = pressure force due to water on one metre length of dam
F = wAh
h
2
18
= 9.81 × 1000 × (18 × 1) ×
2
= 9.81 × 1000 × (h × 1) ×
= 1589220 N
∴
h
A = h×1 & h =
2
Ans
(ii) Position of centre of pressure:
The force F is acting horizontally at a height of h/3 above the base.
∴
18
3
= 6 m Ans
Position of centre of pressure =
(iii) Let x be horizontal distance between the line of action of W and the point through which the
resultant cuts the base (refer Fig. 26.1)
W = weight of dam per length of dam
= weight density of masonry × Area of dam × 1
= Wd × b × H × 1
= 20000 × 12 × 22 × 1
= 5280000 N
We know,
F h
×
W 3
1589220 18
=
×
5280000
3
= 1.806 m Ans
x=
E XAMPLE 26.2: A concrete dam of rectangular section 18 m high and 7 m wide contain water up
to a height of 14 m. Determine:
(a) Total pressure per metre length of the dam.
(b) Point where the resultant cuts the base and
(c) Maximum and minimum intensities of stress at the base. Assume weight of water and concrete as 10000 N/m3 and 25000 N/m3 respectively.
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594
•
Strength of Materials
S OLUTION :
Total pressure per metre length of the dam:
F=
=
wh2
2
10000 (14)2
= 980000 = 980 kN
2
Ans
7m
Water
18 m
14 m
O
F
14 m
3
A
J
K
W
B
R
σmin
σmax
Figure 26.2
Point where the resultant cuts the base:
Let the resultant R cut the base at K as shown in Fig. 26.2.
The weight of the concrete per metre length,
W = ρ ×b×H
= 25 × 7 × 18
= 3150 kN
Horizontal distance between the centre of gravity of the dam section and the point where the resultant cuts the base (i.e., distance JK),
F h
×
W 3
14
980
×
=
3150
3
= 1.452 m Ans
x=
Maximum and minimum stress at the base:
eccentricity of the resultant
e = x = 1.452 m
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Dams
•
595
b
F h
b
×
d = AJ + JK = + x = +
2
2
W 3
and the eccentricity of the resultant,
e=d−
b
2
[d = AK]
It is understood that as a result of eccentricity e, some moment will come into play causing some
bending stress at the base section of the dams, let this moment be M,
M = weight of the dam × eccentricity
= W.e
For unit length, moment of inertia of the base section about its centre of gravity,
l × b3 1 × b3
=
12
12
b3
=
12
I=
Let y be the distance between the centre of gravity of the base section and extreme fibre of the
b
base
and σb be the bending stress in the fibre at a distance y from the centre of gravity of the
2
base section.
M σ
Also,
=
I
y
M.y
∴ σb =
I
b
W.e ×
2
=
b3
12
σ We
= 2
b
Now direct stress at the base
Weight of dam
Width of dam
W
=
b
σd =
Now the stress across the base at B will be maximum and at A stress will be minimum.
W 6We W
6e
∴ σmax = σd + σb = + 2 =
1+
b
b
b
b
@seismicisolation
@seismicisolation
596
•
Strength of Materials
W 6We W
σmin = σd − σb = − 2 =
b
b
b
6e
1−
b
In this example,
∴
e = x = 1.452 m
W
6e
σmax =
1+
b
b
3150
6 × 1.452
=
1+
7
7
= 450 (1 + 1.245)
= 1010.25 kN/m2 (Compressive) Ans
W
6e
σmin =
1−
b
b
3150
6 × 1.452
=
1−
7
7
= 450 (1 − 1.245)
= −110.25 kN/m2
= 110.25 kN/m2 (Tensile) Ans
Trapezoidal Dams with Water Face Vertical
Figure 26.3 shows trapezoidal dam having its water face vertical.
a
Water
h
G
L
W
H
M
N
h/
3
A
K
J
B
R
W
b
σmin
Figure 26.3
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σmax
Dams
•
597
Let a be top width of the dam and b be bottom width of the dam.
Weight of the dam per unit length,
W = w×
(a + b)
×H
2
wh2
2
Horizontal distance between the centre of gravity of the dam and the point where the resultant
R cuts the base,
F h
x= ×
W 3
Total pressure on a unit length of the trapezoidal dam = F =
The distance between the toe of the dam A and the point where the resultant R cuts the base
which is distance AK,
F h
+
d = AJ + JK = AJ +
W 3
AJ may be found either by splitting the dam section into a rectangle and a triangle or
a2 + b2 + ab
AJ =
.
3(a + b)
Now taking the moments about A and equating the same with the moment of the dam section
about A,
∴
Eccentricity,
e = d − AJ
The stress at A will be minimum and at B will be maximum, such that:
6e
W
1+
σmax =
b
b
W
6e
and σmin =
1−
b
b
E XAMPLE 26.3: A concrete dam of trapezoid section having water on vertical face is 17 m high.
The base of the dam is 8.5 m and top 4 m wide. Determine:
(a) resultant thrust on the base per water length of dam;
(b) point, where the resultant thrust cuts the base; and
(c) maximum and minimum stresses across the base.
Take weight of the concrete as 25 kN/m3 and water level upto top of the dam.
@seismicisolation
@seismicisolation
598
•
Strength of Materials
S OLUTION :
Water
4m
D
17 m
F
C
L
J
A
K
B
R
W
8.5 m
σmin
σmax
Figure 26.4
(a) Resultant thrust on the base per metre length:
wh2
2
9.81 × 172
F=
2
F = 1417.5 kN
F=
Weight of concrete per metre length
a+b
H
W =ρ
2
4 + 8.5
× 17
= 25
2
= 2656.25 kN
√
Resultant thrust per metre length = R = F 2 +W 2
√
= 1417.52 + 2656.252 = 3010.8 kN Ans
(b) Point, where the resultant cuts the base:
Let the resultant cut the base at K (refer Fig. 26.4).
To find the centre of gravity of the dam section, taking moments of the area about and
equating the same,
17 × 4.5
4
4.5
4.5
(17 × 4) +
AJ = 17 × 4 ×
+ 17 ×
(4 +
2
2
2
3
(68 + 38.25) AJ = [136] + [210.375]
@seismicisolation
@seismicisolation
Dams
•
599
346.375
106.25
= 3.26 m
AJ =
Now
F h
×
W 3
17
1417.5
×
=
2656.25
3
= 3.06 m
JK = x =
∴
horizontal distance AK,
d = AJ + x
= 3.26 + 3.06
= 6.32 m Ans
(c) Maximum and minimum stresses:
Eccentricity,
e=d−
b
2
= 6.32 −
∴
8.5
2
= 2.07 m
6e
W
1+
σmax =
b
b
2656.25
6 × 2.07
=
1+
8.5
8.5
= 312.5 (1 + 1.461)
= 769.06 kN/m2
W
6e
σmin =
1−
b
b
(Tensile) Ans
= 312.5(1 − 1.461)
= −144.06 kN/m2
(Compressive)
Ans
Rule of Middle Third
As far as possible, there should not be any tension in the base of a dam wall as masonry work is
weak in tension.
We have seen that at A (which is called the heel of the wall), the direct compressive stress is
W/b and the bending stress is tensile as determined below:
@seismicisolation
@seismicisolation
600
•
Strength of Materials
Modulus of section of the base AB,
Z=
1
b2
× 1 × b2 =
6
6
Maximum bending stress on the section
σb =
M 6We
= 2
Z
b
This is tensile at A and compressive at B (refer Fig. 26.4)
Now there will be no tension at A, if
6We
W
> 2
b
b
b
e<
6
If K lies at a distance not greater than b/6 from J. In other words, if the resultant of F and W
cuts the base width within middle third, there would be no tension anywhere in the base. This is
called the rule of the middle third.
b
6
b b
x + JK − =
2 6
If e =
or
or Ak = x + JK =
2b
3
Note: e is the distance between the point where the resultant R cuts the base and the centre point
of the dam.
E XAMPLE 26.4: 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 4 m and 9 m respectively.
The weight density of the masonry is given as 20 kN/m3 . Determine:
a) the resultant force on the dam per metre length;
b) the point where the resultant cuts the base; and
c) the maximum and minimum stress intensities cut the base.
S OLUTION :
Height of dam, H = 20 m
Depth of water, h = 16 m
Top width of dam, a = 4 m
Bottom width of dam, b = 9 m
Weight density of masonry, wd = 20 kN/m3 = 20000 N/m3 .
@seismicisolation
@seismicisolation
Dams
C
16 m = h
a
20 m = H
M
N
h
3
A
601
D
G
L
F
•
J
K
B
R
W
b
Figure 26.5
a) Resultant force on dam
Force, F = w A h
4
2
16
= 9810 × (16 × 1) ×
2
= 1255680 N
= 9810 × (h × 1) ×
16
h
= 5.33 m above the base.
It is acting at a distance of , i.e.,
3
3
Weight of dam is given by
w = weight density of masonry × area of dam × 1
a+b
= wd ×
×H ×1
2
4+9
= 20000 ×
× 20 × 1
2
= 2600000 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.
A × 20 × 2 +
4 × 20
4
4+9
4+
=
× 20 AJ
2
3
2
160 + 213.3 = 130 AJ
AJ = 2.87 m
@seismicisolation
@seismicisolation
602
•
Strength of Materials
Resultant force R is given by
F 2 +W 2
= 12556802 + 26000002
R=
= 2887340
= 2.887 MN
Ans
b) The point where the resultant cuts the base.
x = JK = the horizontal distance between the line of action of W and the point at which the
resultant cuts the base
We know,
F h
×
ω 3
1255680 16
×
=
2600000
3
= 2.57 m
x=
The distance
AK = AJ + JK = d
= 2.87 + 2.57
Eccentricity,
= 5.44 m
b
e=d−
2
= 5.44 −
= 0.94 m
9
2
Ans
c) The maximum and minimum stress intensities
W
6e
σmax =
1+
b
b
2600000
6 × 0.94
=
1+
9
9
= 288889 [1 + 0.627]
= 470022 N/m2 Ans
6e
W
1−
σmin =
b
b
2600000
6 × 0.94
=
1−
9
9
@seismicisolation
@seismicisolation
Dams
= 288889 [1 − 0.627]
= 107756 N/m2
Ans
Trapezoidal Dams with Water Face Inclined
E
h F
a
D
I
C
H
G
θ
h
3
θ
J
K
B
A
R
W
b
Figure 26.6
Let us consider a trapezoidal dam with water face inclined.
Let
a = Top width of the dam
b = Bottom width of the dam
H = Height of the dam
wd = Specific weight of the dam masonry
h = Height of water retained by the dam
w = Specific weight of the water
θ = Inclination of the water face with the vertical.
AI = l
Length of sloping side AI, which is subjected to water surface.
h
= cos θ
l
h
∴ l=
cos θ
The weight of the dam per unit length,
W = ωd ×
(a + b)
×H
2
Total pressure on a unit length of the dam,
F=
wh
whl
×l =
2
2
@seismicisolation
@seismicisolation
•
603
604
•
Strength of Materials
Refer Fig. 26.6, the water pressure F will act at a height of h/3 from the bottom of the dam.
Horizontal component at this water pressure F,
FH = F cos θ =
whl h wh2
× =
2
l
2
And vertical component of this water pressure,
Fr = F sin θ =
whl EI
×
2
l
w
× EI × h
2
= Weight of the wedge AEI of water.
=
w2 h2
on the
It is evident that such a dam may be taken to have a horizontal pressure equal to
2
imaginary vertical face AE. The weight of wedge AFE of water may be considered a part of the
weight of dam when determining the centre of gravity of the dam section.
The distance between the centre of gravity of the dam section and the point, where the resultant
R cuts the base is given by the relation,
x=
∴
F h
×
W 3
(x = JK)
Total stress across the base at B,
σmax =
W
b
6e
1+
b
σmax =
W
b
6e
1−
b
And total stress across the base at A,
E XAMPLE 26.5: A masonry dam of trapezoidal section is 12 m high. The top width of dam is
1.5 m and bottom width is 8 m. The face exposed to water has a slope of 1 horizontal to 10 vertical
as shown in Fig. 26.7.
Calculate the maximum and minimum stresses on the base, when the water level coincides with
the top of dam. Take weight of masonry as 21 kN/m3 and that of water as 10 kN/m3 .
S OLUTION :
H = 10 m
a = 1.5 m
@seismicisolation
@seismicisolation
Dams
b=8m
h = 12 m
wd = 21 kN/m3
w = 10 kN/m3
1.5 m
C
θ
E
D
12 m
F
θ
J
K
W
R
B
A
1.2 m
1.5 m
S
8m
Figure 26.7
10
1
θ = 84.29◦
12
tan θ =
AS
12
AS =
tan θ
12
AS =
10
= 1.2 m
tan θ =
∴
Total water pressure F per meter length of dam,
wh2
2
10 × (12)2
=
2
= 720 kN
F=
@seismicisolation
@seismicisolation
•
605
606
•
Strength of Materials
Weight of the dam per metre length (including wedge AEC of water)
(a + b)
h
+ wd ×
H
W = w×
2
2
(1.5 + 8)
12
+ 21 ×
× 12
= 10 ×
2
2
= 60 + 1197
= 1257 kN
In order to find out the centre of gravity of the dam section (including wedge AEC of water).
Taking moments about A and equating the same,
12 2
12 1
×
+ 21 ×
×
W × AJ = 12 ×
2
3
2
3
1.5
+ 21 × 12 × 1.2 +
2
12 × (8 − 1.2 − 1.5) (8 − 1.2 − 1.5)
+ 21 ×
×
+ (1.2 + 1.5)
2
3
1257 × AJ = 24 + 84 + 491.4 + (21 × 31.8 × 4.47)
= 599.4 + 2985
AJ = 2.85 m
F h
JK = x = ×
W 3
720
12
=
×
= 2.291 m
1257
3
∴
AK = d = AJ + x
= 2.85 + 2.291 = 5.141 m
b
and eccentricity, e = d −
2
8
= 5.141 − = 1.141 m
2
W
6e
1257
6 × 1.141
∴ σmax =
1+
=
1+
b
b
8
8
= 157.125 (1.856) = 291.6 kPa Ans
6e
1257
6 × 1.141
W
1−
=
1−
σmin =
b
b
8
8
= 157.125 × 0.144 = 22.63 kPa Ans
Horizontal distance
@seismicisolation
@seismicisolation
Dams
•
607
Stability of dam depends upon several factors and conditions as described below.
Condition A: Resultant thrust cuts the base within the middle third of the base width to avoid
tensile stress at the base. When bending stress is greater than the direct stress, tension is developed
at the base of the dam. To avoid the tension at the base of the dam, bending stress must be equal to
or less than the direct stress.
σb = bending stress and d = direct stress.
σb ≤ σd
6We ω
6We
or
≤
when σb = 2
b
b2
b
6e
or
≤1
b
b
or e ≤
6
b
Obviously, it means eccentricity must not be greater than at either side from the centre of the
6
base width. Hence, the resultant thrust R must interact with the base within the middle-third of the
base width to avoid tension at the base, which is known as middle-third-rule.
Let
d = AJ + JK
4
F
×
= AJ +
W
3
Eccentricity, e
F
A
J
K
W
b
= d − AJ (as before)
R
Figure 26.8
Condition B: To avoid crushing of masonry at the base of the dam (or wall) σmax should not be
more than the permissible or safe stress of the dam material.
Condition C: The total horizontal thrust F causes sliding of the dam. To counteract sliding of
the dam, the maximum μ w is set up at the base of the dam, where μ is the coefficient of friction
between the masonry of the dam (or wall) and the soil on which it rests.
Hence, condition to prevent sliding of the dam (or wall) at the base, μ w > F is to be satisfied.
In other words,
μw
>1
F
μw
is taken at least 1.5.
For designing purposes
F
Condition D: A dam (or wall) can be overturned tilted about the toe B as shown in Fig. 26.9
as the water almost F causes an overturning moment. To maintain the equilibrium condition, the
weight of dam W provides a restoring moment about B.
@seismicisolation
@seismicisolation
608
•
Strength of Materials
h
H
F
h
3
G
J
K
A
B
W
b
R
Figure 26.9
h
3
Restoring moment abut B = W × JB
Overturning moment about B = F ×
Condition to prevent overturning about the toe B is:
Restoring moment > Overturning moment
h
3
h
or W (b − AJ) > F
3
JK
F
F h
=
or (b − AJ) > .
W 3
h/3 W
or b > AJ + JK
∴
W × JB > F.
∴ AB > AK as shown in Fig. 26.9.
This condition will satisfy if K lies between AB.
For no tension condition, we have such that point E must lie within the middle third of the
base width AB. So, if the a dam is checked against no tension condition at the base, it will be
automatically checked for overturning of the dam (or wall).
Exercise
26.1 A rectangular masonry dam 6 m high and 3 m wide has water level up to its top. Find: (i)
total pressure per metre length of the dam; (ii) point at which the resultant cuts the base and
(iii) maximum and minimum intersection of stresses at the bottom of the dam. Assume the
weight of water and masonry as 10 kN/m3 and 20 kN/m3 respectively.
[Ans 180 kN, 1.0 m, 360 kPa, −120 kPa Tensile]
26.2 A masonry dam 12 m traperzoidal in section has top width 1 m and bottom width 7.2 m. The
face exposed to water has a slope of 1 horizontal to 10 vertical. Check the stability of the
@seismicisolation
@seismicisolation
Dams
•
609
dam, when the water level rises 10 m high. The coefficient of friction between the bottom of
the dam and the soil is 0.6. Take the weight of masonry as 22 kN/m3 .
[Ans Safe against tension, safe against or returning, safe against sliding]
26.3 A masonry dam of trapezoidal section is 6 m high, 2 m wide at the top and 5 m wide at the
bottom. The water face of the dam has a slope 1 in 6. Calculate the maximum and minimum
stresses at the base of the dam, when the height of water is same as the height of dam. Take
unit weight of masonry and water as 20 kN/m3 and 9.81 kN/m3 respectively.
[Ans 100133 N/m2 ; 79638.9 N/m2 ]
26.4 A masonry dam of trapezoidal section is 10 m high and retains water up to the top. The width
at the top is 3 m and at the bottom 8 m. Water face has a slope of 1 in 10. Find the maximum
and minimum intersities of stress at the base. Density of water = 10000 N/m3 and density of
masonary = 24000 N/m3 .
[Ans σmax = 231600 N/m2 Comp, σmin = 110900 N/m2 Comp]
26.5 A concrete dam of trapezoidal section has a vertical face on the water side. Its height is 5 m,
top width 2 m and bottom width 3 m. Determine the maximum and minimum stresses at
its base when the reservoir is full. Take weight of water 9810 N/m3 and weight of masonry
20000 N/m3 .
[Ans 178335 N/m2 , −11667 N/m2 Tensile]
@seismicisolation
@seismicisolation
C HAPTER
27
RIVETED JOINTS
Riveted joints are permanent joints used to fasten one plate with another to counteract tensile forces,
shear forces, etc. These are used in pressure vessels such as boiler, at structure works such bridge
truss, roof truss, etc. In ship building also they are extensively used. We shall now study and investigate the strength of riveted joints.
Rivet
A rivet is a cylindrical solid bar with a load integral to it. The middle portion of the rivet is called
‘body’ or ‘shank’ and the bottom portion is called tail. According to IS: 2998-1982 (Reaffirmed
1992), the material of a rivet must have tensile strength not less than 40 N/mm2 and it should not
crack when flattened. Elongations should not be less than 25 per cent. In cold condition or when
heated to 650◦ C and quenched, the shank shall be bent on itself through 180◦ C without cracking.
The rivet when hot must flatten without cracking to a diameter 2.5 times the diameter of shank.
Head
Tail
Body or shank
Figure 27.1 Rivet
These are manufactured either by cold heating or by hot forging. In case if rivets are cold forged,
they shall be properly heat-treated to remove stresses.
Types of Rivet Heads
According to Indian standard specifications, the rivet heads are available into many types.
Rivet heads for general purposes. According to IS: 2155 − 1982 (Reaffirmed 1996) some types
are shown in Fig. 27.2 (having diameter below 12 mm).
@seismicisolation
@seismicisolation
Riveted Joints
d
1.6 d
0.7 d
0.7 d
d
d
1.6 d
Pan Head
Snap Head
2d
0.25 d
1.516 d
0.5 d
d
d
2.25 d
Flat Head
Mush Room Head
2d
90°
d
Flat Countersunk Head 90°
Figure 27.2
@seismicisolation
@seismicisolation
0.5 d
•
611
612
•
Strength of Materials
Material of Rivets
It should be tough and ductile. They are usually made of steel (low carbon). Sometimes brass, aluminium or copper is used for special applications. The rivets for general purposes are manufactured
from steel conforming to the following Indian standards:
(a) IS: 1148–1982 (Reaffirmed 1992): Specification for hot rolled rivet bars upto 40 mm
diameter for structural purposes.
(b) IS: 1149:1982 (Reaffirmed 1992): Specification for high tensile steel rivet bars for structural
purposes.
For boiler work rivets are made conforming to IS: 1990–1973 (Reaffirmed 1992).
Types of Riveted Joints
1. Lap joint
2. Butt joint.
1. Lap Joints
10°
d
M = 1.5 d
p
Single riveted lap joint
Figure 27.3
@seismicisolation
@seismicisolation
Riveted Joints
•
613
d
d
p
F
F
pb
Double rivets lap joint with chain riveting
M = 1.5 d
d
F
F
p
pb
Double rivets lap joint with jig-jag riveting
Figure 27.4
2 Butt Joints: In this type of joint, the plate to be joined and placed end to end as shown in
Fig. 27.5. The covering plates are called cover straps.
@seismicisolation
@seismicisolation
614
•
Strength of Materials
M=1.5 d
F
p
F
Single riveted butt joint with two cover plates with chain riveting
M = 1.5 d
Section A-A
F
p
p/2
F
A
A
Double riveted butt joint with two cover plates with zig-zag riveting
Figure 27.5
Some Definitions Related to Riveted Joints
1. Pitch of rivets: It is the distance from the centre of any rivet to the centre of next rivet in the
same row. It is denoted by p.
2. Margin: It is the distance between the edge of the plate and centre of the nearest hole. It is
denoted by M. In all types of riveted joint, margin (M) is taken as equal or greater than 1.5 d
where d is the diameter of hole.
3. Back pitch: Back pitch is the distance between any two consecutive rows of rivets in the same
plate. It is denoted by pb (Fig. 27.4).
4. Diagonal pitch: It is the diagonal distance between two adjacent rivets in two consecutive
rows in the same plate. It is denoted by pd (Fig. 27.6).
pb
pd
Figure 27.6
@seismicisolation
@seismicisolation
p
Riveted Joints
•
615
Failure of a Riveted Joint
Possible ways of failure of riveted joint are described below.
1. Tearing of the plate at an edge:
A joint may fail due to tearing of the plate at an edge as shown in Figs. 27.7 and 27.8.
This can be prevented by keeping the margin M = 1.5d, d is the diameter of hole.
M
F
d
F
F
p
p-d
F
d
Tearing of the plate at an edge
Tearing of the plate across the rows of rivets
Figure 27.7
Figure 27.8
2. Tearing of the plate across a row of rivets:
Because of tensile stresses in the main plates, the main plate or cover plates may tear off
across a row of rivets as shown in Fig. 27.8. For such cases, only one pitch length is considered.
The resistance offered by the plates against tearing is known as tearing resistance or tearing strength of the plate.
Let
p = pitch of the rivets.
d = diameter of rivets hole
t = thickness of the plate
σt = permissible tensile stress for the plate material
Tearing area per pitch length,
At = (p − d)t
Therefore, tearing resistance or pull required to tear off the plate per pitch length,
Ft = At σt = (p − d)t σt
If the tearing resistance Ft is greater then the applied load F, then this type of failure will
not occur.
3. Shearing of the rivets:
The rivets are being exerted by tensile forces and thus inducing shearing stresses in the rivets
as is evident from Fig. 27.4 and Fig. 27.5 and if not properly designed, can be sheared off as
shown in Fig. 27.9.
@seismicisolation
@seismicisolation
616
•
Strength of Materials
F
F
Shearing off a rivet in a lap joint
F
F
Shearing off a rivet in a single cover butt joint
F
F
Shearing off a rivet in a double cover butt joint
Figure 27.9
Let d be the diameter of the rivet hole.
τ = safe permissible shear stress for the rivet material, and
π
n = number of rivets per pitch length. Shearing area is As = × d 2 (in single shear)
4
π
2
As = 2 × × d (in double shear)
4
Therefore, shearing resistance or pull required to shear off the rivet per pitch length,
π
× d 2 × τ (in single shear)
4
π
Fs = n × 2 × × d 2 × τ (in double shear)
4
Fs = n ×
4. Crushing of the plate or rivets:
Instead of shearing, sometimes, the rivets are crushed as shown in Fig. 27.10. This results in
oval shape of rivet hole and hence the joint becomes loose. Resisting area against crushing in
the projected area of the hole or rivet.
The resistance offered by a rivet to be crushed is known as crushing resistance or crushing
strength of the rivet.
F
F
Crushins of a rivet
Figure 27.10
@seismicisolation
@seismicisolation
Riveted Joints
•
617
d = diameter of the rivet hole
t = thickness of the plate
σc = safe permissible crushing stress for the rivet or plate material
n = number of rivets per pitch length under crushing.
Crushing area = projected area per rivet
Ac = dt
Total crushing area = n.d.t
Therefore, crushing resistance or pull required to crush the rivet per pitch length,
Fc = n · d · t · σc
If Fc is greater than applied load F then the joint will fail due to crushing.
Efficiency of a Riveted Joint
It is given by
Least of Ft , Fs and Fc
p × t × σt
p = pitch of the rivets
t = thickness of plate
σt = Allowable tensile stress of the plate material.
η=
E XAMPLE 27.1: Determine the efficiency of the following rivets joints:
(i) Simple riveted lap joint of 8 mm plates with 24 mm diameter rivets having a pitch of 55 mm.
(ii) Double riveted lap joint of 8 mm plates with 24 mm diameter rivets having a pitch of 70 mm.
Assume
Permissible tensile stress in plate = 130 GPa
Permissible shearing stress in rivets = 100 MPa
Permissible crushing stress in rivets = 185 MPa.
S OLUTION :
t = 8 mm, d = 24 mm, σt = 130 MPa = 130 N/mm2 , τ = 100 MPa = 100 N/mm2 ,
σc = 185 MPa = 185 N/mm2
(i) Efficiency of the first joint:
Pitch = 55 mm
Let us find the tearing resistance of plate, shearing and crushing resistance of the rivets.
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618
•
Strength of Materials
a) Tearing resistance of the rivet:
Ft = (p − d)t σt
= (55 − 24) × 8 × 130
= 32240 N
b) Shearing resistance of the rivet:
Because the joint is a single lap joint, the strength of one rivet in single stress is taken. The
shearing resistance of one rivet,
π 2
d ×τ
4
π
= (24)2 × 100
4
= 45216 N
Fs =
c) Crushing resistance of the rivet:
Again as the joint in a single riveted, therefore strength of one rivet is taken. The crushing
resistance of one rivet,
Fc = d × t × σc = 24 × 8 × 185 = 35520 N.
Hence strength of joint = least of Ft , Fs and Fc
= 32240 N
Now strength of unriveted or solid plate,
F = p × t × σt = 55 × 8 × 130
= 57200 N
∴ Efficiency of joint,
η=
=
Least of
Ft , Fs and Fc
F
32240
= 0.563 or 56.3 %
57200
Ans
(ii) Efficiency of the second joint:
As we know pitch = 70 mm
a) Tearing resistance of the plate,
The tearing resistance of the plate per pitch length,
Ft = (p − d) × t × σt = (70 − 24) × 8 × 120
= 44160 N
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Riveted Joints
•
619
b) Shearing resistance of the rivets:
Because the joint is double riveted lap joint, strength of two rivets in single layer is
taken. The shearing resistance of the rivets,
π
× d2 × Z
4
π
= 2 × (24)2 × 100
4
= 90432 N
Fs = n ×
c) Crushing resistance of the rivet
Because the joint is double riveted, strength of two rivets must be taken. The crushing
resistance of rivets,
Fc = n × d × t × σc
= 2 × 24 × 8 × 185
= 71040 N
Therefore strength of the joint,
= Least of Ft , Fs and Fc
= 44160 N
Now the strength of the unriveted or solid plate,
F = p × t × σt
= 70 × 8 × 130
= 72800 N
∴
Efficiency of the joint,
least of Ft , Fs , and Fc
F
44160
=
72800
= 0.606 or 60.6% Ans
η=
E XAMPLE 27.2: A double riveted double cover butt joint in plates 25 mm thick is made with
30 mm diameter rivets at 110 mm pitch. The allowable stresses are:
σt = 135 MPa, τ = 95 MPa, σc = 140 MPa.
Find the efficiency of joint, taking the strength of the rivet in double shear as twice that of single
shear.
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•
Strength of Materials
S OLUTION :
t = 25 mm, d = 30 mm, p = 110 mm, σt = 135 MPa, τ = 95 MPa, σc = 140 MPa.
Let us first find the tearing resistance of the plate, shearing resistance and crushing resistance of the
rivet.
(i) Tearing resistance of the plate:
Tearing resistance of the plate per pitch length,
Ft = (p − d) × t × σt = (110 − 30) × 25 × 135
= 270000 N
(ii) Shearing resistance of the rivets
Because the joint is double riveted butt joint, the strength of two rivets in double shear is
taken. Shearing resistance of the rivets,
π
× d2 × 7
4
π
= 2 × 2 × × (30)2 × 95
4
= 268470 N
Fs = n × 2 ×
(iii) Crushing resistance of the rivets
Because the joint is double riveted, the strength of two rivets is taken. Crushing resistance of
the rivets,
Fc = n × d × t × σc
= 2 × 30 × 25 × 140
= 210000 N
Therefore strength of the joint = Least of Ft , Fs and Fc
= 210000 N
Strength of the unriveted or solid plate,
F = P × t × σt
= 110 × 25 × 135
= 371250 N
Therefore efficiency of the joint
Least of Ft , Fs and Fc
F
210000
=
371250
= 0.566 or 56.6% Ans
=
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Riveted Joints
•
621
Note: When thickness of the plate is greater than 8 mm, Unwin’s empirical formula to determine
the rivet hole is quite useful:
√
d=6 t
Diamond Riveting
In many structural constructions, for example, bridge truss members, it becomes necessary to provide a large number of rivets to connect a member to the gusset plate. If F is the force to be transF
. Obviously the
mitted, the number of rivets necessary will be given by n =
strength of one rivet
strength of the rivet will be taken least of Ft , Fs and Fc .
In the case of tension members of a roof truss, the number of rivets required for the joint is
small and the section is usually weakened by one rivet hole only. But in bridge truss, the number of
rivets required is large and the section will depend upon how these rivets are arranged. For example,
suppose a bridge diagonal is a flat of thickness t requiring 9 rivets to connect it by a butt joint to
a gusset plate, we may arrange the rivets in three rows of three each as diamond riveting shown in
Figure 27.11. Such type of arrangement is also known as Lozenge joint.
F
1
2
3
4
1
2
3
4
Figure 27.11
For the joint shown in Fig 27.11, assuming the section 1-1 through the first rivet hole to be the
weakest, the width of flat will be given by,
F = σt (b − d)t
where σt is tensile stress
b is the width of flat and
d is the rivet hole.
In diamond riveting section 1-1 which is weakened by one rivet hole only is usually the weakest,
sections 2-2, 3-3 and 4-4 successively getting stronger. If σt is the permissible tensile stress for the
plate, the pull required to tear the plate at section 1-1 is given by,
F1 = σt (b − d)t
To fail at section 2–2 by tearing of the plate, the rivet in front must first give way. The pull required
for the failure of the plate at section 2–2 is given by,
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622
•
Strength of Materials
F2 = σt (b − 2d)t + strength of one rivet in front. Similarly, for failure at section 3–3 by tearing
of the plate, the three rivets in front must give way and therefore the pull required for tearing at
section 3–3 is given by,
F3 = σt (b − 3d)t + strength of three rivets in front.
Similarly,
F4 = σt (b − 3d)t+ strength of six rivets in front, for failure at section 4-4.
Out of forces F1 , F2 , F3 and F4 , usually F1 is the least, which means that the weakest section is
1-1. The efficiency of such a joint is the greatest, and is given by
η=
F1 (b − d)t σt
b−d
=
=
F
b t σt
b
Such a joint thus provides the most economical arrangement of rivets.
E XAMPLE 27.3: Two lengths of a mild steel tie rod, having width 200 mm and thickeness 12.5
mm are to be connected by means of a butt joint with double cover straps. Design the riveted joint,
if the allowable stresses are 100 N/mm2 in tension, 70 N/mm2 in shear and 180 N/mm2 in crushing.
Also determine the efficiency of joint.
S OLUTION :
b = 200 mm;
t = 12.5 mm;
σt = 100 N/mm ;
2
τ = 70 N/mm2 ,
σc = 180 N/mm2 .
Diameter of rivet (d)
For t ≥ 8 mm the diameter of rivet hole
√
is d = 6 t (using Unwin’s empirical formula)
√
d = 6 12.5 = 21.21 mm.
The standard diameter of rivet hole in 21.5 mm and the corresponding diameter of rivet is
20 mm.
∴ d = 21.5 mm
Number of rivets:
The pull acting on the joint is
Ft = (b − d)t σt
= (200 − 21.5) × 12.5 × 100
= 223125 N
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Riveted Joints
The shearing strength of one rivet is
π
Fs = 1.75 d 2 × τ
4
π
= 1.75 × (21.5)2 × 70
4
= 44451 N
Note: Instead of 2, value of 1.75 is taken per Indian standard for safety.
The crushing strength of one rivet is
Fc = d × t × σc
= 21.5 × 12.5 × 180
= 48375 N
The number of rivets for joint is
Ft
minimum of ps MPa
223125
=
44451
= 5 rivets
n=
Number of rows:
Five rivets are arranged as shown in Fig. 27.12.
1
2
3
1
2
3
Figure 27.12
Thickness of butt cover:
t1 = 0.75t
= 0.75 × 12.5 = 9.375
= 9.4 mm
Pitch of rivet (p)
p = 3d + 5
= 3 × 21.5 + 5
= 69.5 mm.
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•
623
624
•
Strength of Materials
Spacing of rows of rivets (Pb ) and margin (M)
Pb = 3 d = 3 × 21.5
= 64.5 mm
M = 1.5 d
= 1.5 × 12.5
= 32.25 mm.
Efficiency of joint (η ):
The tearing resistance of joint along section 1–1 is
Ft1 = (b − d)t σt .
= (200 − 21.5) × 12.5 × 100
= 223125 N
The tearing resistance of joint along 2–2 as
Ft2 = (b − 2d) × t × σt + strength of one rivet in front
= (b − 2d) × t × σt + Fs
= (200 − 2 × 21.5) × 12.5 × 100 + 44451
= 240701 N
The tearing resistance of joint along section 3–3 is,
Ft3 = (b − 2d) × t × σ t + strength of three rivets in front.
= (b − 2d) × t × σ t + 3Fs
= (200 − 2 × 21.5) × 12.5 × 100 + 3 × 44451
= 196250 + 133353
= 329603 N
The shearing resistance of five rivets is
n × Fs = 5 × 44451
= 222255 N
Crushing resistance of five rivets is
n × Fc = 5 × 48375
= 241875 N
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Riveted Joints
•
625
Or efficiency of joint η
Minimum of Ft1 , Ft2 , Ft3 , nFs , nFc
Strength of solid plate
223125
=
260 × 12.5 × 100
= 0.8925 or 89.25% Ans
=
Exercise
27.1 Find the efficiency of the following joints:
a) Single riveted lap joint of 6 mm plates with 20 mm diameter rivets having a pitch of
50 mm.
b) Double riveted lap joint of 6 mm plates with 20 mm diameter rivets having a pitch of
60 mm.
Assume:
Permissible tensile stress in plate = 100 MPa
Permissible shearing stress in rivets = 80 MPa
Permissible crushing stress in rivets = 160 MPa
[Ans 60%, 66.6%]
27.2 Two mild steel bars of 350 mm width and 15 mm thickness are to be connected by means of
Lozenge joint with two straps. The tensile load acting on the joint is 15 × 104 N. Design and
sketch the joint, if the permissible stresses are: 80 MPa in tension, 120 MPa in crushing and
60 MPa in shear.
[Ans Total number of rivets = 9; Thickness of butt straps = 12 mm, efficiency of
joint = 93%, Pitch of rivets = 72 mm margin at the ends = 36 mm]
27.3 A double riveted double cover butt joint with plates 20 mm thick made with 20 mm diameter rivets at 110 mm pitch, the permissible stresses are σt = 120 MPa, τ = 100 MPa and
σc = 150 MPa. Find the efficiency of joint taking strength of rivet in double shear is twice
than that of single shear.
[Ans 56.82%]
27.4 A double riveted butt joint, in which the pitch of the rivets in outer rows is twice that in the
inner rows, connects two 16 mm thick plates with two cover plates each 12 mm thick. The
diameter of rivets is 22 mm. Determine the pitches of the rivets in the two rows if the working
stresses are not to exceed in the following limits: Tensile stress in plates = 100 MPa, shear
stress in rivets = 75 MPa and bearing stress in rivets and plates = 150 MPa. Make a fully
dimensioned sketch of the joint by showing at least two views.
[Ans 107 mm, 53.5 mm]
27.5 A diamond riveted double strap butt joint for two plates each 400 mm wide and 20 mm thick.
The covers are 15 mm thick each while the rivets used are of 25 mm diameter. Determine the
strength and efficiency of the joint. The maximum allowable stresses in tearing, shearing and
bearing (crushing) are 120 MPa, 100 MPa and 180 MPa respectively.
[Ans 900 kN, 93.75%]
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C HAPTER
28
WELDED JOINTS
Welding is a process of joining metals in which parent metals are fused together to form a single
piece. The difference between welding and soldering or brazing is that in later filler material is used
whose melting temperature is lower than that of the workpieces. Welding is used wherever strength
is required and soldering and brazing are primarily employed as simple joints to take up only light
loads. In forge welding pressure is used to join parent metals whereas if two parts are joined without
any pressure but with a separate weld material, the process is known as fusion welding. The joints
so made are called welded joints.
Classification of Welding
Broadly various welding processes are classified as given below:
Gas Welding
i)
ii)
iii)
iv)
Air-acetylene welding
Oxy acetylene welding
Oxy-hydrogen welding
Pressure gas welding
Arc Welding
i)
ii)
iii)
iv)
v)
vi)
vii)
viii)
ix)
Carbon arc welding
Flux coated arc welding
Shielded metal arc welding
Submerged arc welding
Tungsten Inert Gas (TIG) welding
Metal Inert Gas (MIG) welding
Plasma Arc welding
Electroslag welding
Stud arc welding
Resistance Welding
i)
ii)
iii)
iv)
Spot welding
Seam welding
Projection welding
Flash butt welding
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•
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v) High frequency resistance welding
vi) Percussion welding
vii) Resistance butt welding.
Solid State Welding
i) Cold welding welding
ii) Explosion welding
iii) Forge welding
iv) Roll welding
v) Ultrasonic welding
vi) Hot pressure welding
vii) Friction welding
viii) Diffusion welding.
Thermo-chemical Welding Processes
i) Thermit welding
ii) Atomic hydrogen welding
Radiant Energy Welding Processes
i) Electron beam welding
ii) Laser beam welding
Advantages and Disadvantages of Welded Joints over Riveted Joints
Advantages:
i) The welded joints provide maximum efficiency (sometimes upto 100%) which is not possible
in riveted joints.
ii) Changes, alterations and additions can be easily made in the welded structures.
iii) The welded structures are usually lighter than riveted structures.
iv) As the welded structure is smooth in appearance, so it is better in looks.
v) Sometimes certain structures are difficult to be riveted, for example, circular steel pipe, but
they can be easily welded.
vi) Welding provides rigid joints. This is in line with the modern trend of providing rigid frames.
vii) The process of welding is less time-consuming compared to riveting.
viii) It is possible to weld any part of structure at any point. But riveting requires enough room.
Disadvantages:
i) Due to uneven heating and cooling, undesirable stresses develop in structures.
ii) Welding requires highly skilled labour and supervision.
iii) As in welding, there is no provision of expansion and contraction is provided, therefore there
is a possibility of cracks developing in it.
iv) The inspection of welding work is more difficult than riveting work.
Types of welds: Welds may be classified mainly into two types:
i) Fillet weld: This type of weld is used when the members to be connected overlap each other
as shown in Fig. 28.1.
The section of the fillet weld for design purposes will be taken as an isosceles right-angle
triangle. The length of either of the equal sides of the triangle is called size of the weld.
@seismicisolation
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628
•
Strength of Materials
let
fil
fillet
Figure 28.1
t = Throat thickness
Reinforcement
C
A
45°
D
s
B
s
s = Leg as size of weld
Figure 28.2 Enlarged view of circled portion
For determining the strength of the fillet joint, it is assumed that the section of fillet as a right
angled triangle as discussed before. BD is hypotenuse making equal angles with other two sides CB
and AB (refer Fig. 28.2). The length of each side is called leg or size of the weld. Perpendicular
distance BD is known as throat thickness. For calculation and design purposes, the minimum area
of the weld is obtained at the throat BD. This minimum area is the product of the throat thickness
and length of weld.
It is evident from Fig. 28.2 that,
Throat thickness,
t = 5 × sin 45◦
= 0.707 s
As described, minimum area of the weld as
Throat area,
A = Throat thickness × Length of weld
= t × p = 0.707 s × l
In order to determine tensile strength F,
F = Throat area × Permissible tensile stress
= 0.707 s × l × σt (for, single fillet weld as shown in Fig. 28.2)
Tensile strength of the welded joint for double fillet weld shown in Fig. 28.1,
F = 2 × 0.707 s × l × σt
Many times weld is weaker than the plate owing to slag and blow holes. That is why weld is
given a reinforcement shown is Fig. 28.1 to make weld stronger.
ii) Butt weld: Figures 28.3 and 28.6 shows butt welds.
@seismicisolation
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Welded Joints
•
629
Reinforcement
a)
Reinforcement
Single V - butt weld
Figure 28.3
b)
Double V - butt weld
Figure 28.4
c)
Single U - butt weld
Figure 28.5
d)
Double U - butt weld
Figure 28.6
E XAMPLE 28.1: A plate 90 mm wide and 12 mm thick is to be welded to another plate by means
of double parallel fillets. The plates are subjected to a load of 90 kN. Find the length of the weld if
the allowable shear stress in the weld does not exceed 65 MPa.
S OLUTION :
Width = 90 mm
Thickness = 12 mm
Load = 90 kN = 90 × 103 N
Zmax = 65 MPa = 65 N/mm2
Note: Size of weld is generally equal to thickness of plate.
∴
s = 12 mm
@seismicisolation
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630
•
Strength of Materials
So maximum load (F) which the plates can carry for double parallel fillet joint weld is given by
the following equation:
90000 = 2 × 0.707 × 5 × l × τ
90000 = 1.414 × 12 × l × 65
90000
l=
1.414 × 12 × 65
= 81.6 mm, say 82 mm
Ans
E XAMPLE 28.2: A tie bar 120 mm × 15 mm thick is to be welded to another plate as per
Fig. 28.7. Find the minimum overlap required if 8 mm fillet welds are used. σt =120 N/mm2 ,
τ = 80 N/mm2 .
120 mm
x
Figure 28.7
Total length of weld, taking tensile load,
= 120 × 2
= 240 mm
Maximum tensile force in the tie bar = 120 × 15 × 125
= 225000 N
(i)
Throat thickness, t = 0.707 × s
Total shear load taken by joint,
= 2x × 0.707 × 15 × 80
= 1697 x
(ii)
Total tensile load taken by weld joint,
= 120 × 0.707 × 15 × σt
= 120 × 0.707 × 15 × 125
= 159075 N
Adding loads (ii) and (iii) and equating with total 225000 N
@seismicisolation
@seismicisolation
(iii)
Welded Joints
1697x + 159075 = 225000
225000 − 159075
x=
1697
65925
=
1697
= 38.85, say 40 mm
•
631
Ans
E XAMPLE 28.3: A steel plate 120 mm wide 15 mm thick is required to join with another plate by
a single lap weld and parallel fillet as shown in Fig. 28.8. The maximum tensile and shear stresses
are 100 N/mm2 and 60 N/mm2 respectively. Find the length of each parallel fillet if the joint is
subjected to a total load of 180 kN.
S OLUTION :
l2
F
120 mm
l1
F
l2
Figure 28.8
Width of plate = 120 mm
Thickness of plate = 15 mm
Maximum tensile stress = 100 N/mm2
Maximum shear stress = 60 N/mm2
Now, length of single fillet parallel lab weld,
l1 = width of plate
= 120 mm
l2 = length of each parallel fillet weld.
Let F1 be the load carried by single fillet lap weld and F2 be the load carried b parallel fillet
weld
Total load carried by the plates,
= 180 kN = 180000 N
F1 = 0.707 × l1 × t × σt
= 0.707 × 120 × 15 × 100
= 127260 N
@seismicisolation
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632
•
Strength of Materials
F2 = 2 × 0.707 × l2 × t × τ
= 2 × 0.707 × l2 × 15 × 60
= 1272.6 l2
F1 + F2 = 180000
127260 + 1272.6 l2 = 180000 − 127260
180000 − 127260
l2 =
1272.6
= 41.4 mm, say 42 mm Ans
Unsymmetrical Welded Section
l1
F1
a
G
F
b
F2
l2
Figure 28.9
For equilibrium ∑ H = 0 and ∑ M = 0
F = F1 + F2
F1 × a = F2 × b
F1 = 0.707 s × L1 × σ
F1 = 0.707 s × L2 × σ
We know
Solving eqns.
F1 L1 b
=
=
F2 L2 a
F ×b
F1 =
a+b
F ×a
F2 =
a+b
Axial load in terms of shear stress:
Axial load, F = Shear stress × Area of weld
= τ × area of well
But area of weld = (Total length of weld) × Throat thickness
= (l1 + l2 ) × 0.707 × s
Hence, F = τ × (l1 + l2 ) 0.707 s
@seismicisolation
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(i)
(ii)
(iii)
Welded Joints
•
633
E XAMPLE 28.4: A 220 mm × 150 mm × 15 mm angle carries a load of 250 kN. The angle is
welded to a steel plate by fillet welds or shown in Fig. 28.10(a). Find the lengths of the weld at the
top and bottom. Permissible shear stress is 100 N/mm2 .
l1
15 mm
220 mm
148.98 mm
G
71.02 mm
15 mm
l2
150 mm
To find C. G. of angle
Figure 28.10a
15 mm
2
220 mm
G
A
150 mm
1
B
15 mm
Figure 28.10b
a1 = 150 × 15 = 2250 mm2 ,
a2 = 205 × 15 = 3075 mm ,
2
y1 = 7.5 mm
y1 = 117.5 mm
2
A = 2250 + 3075 = 5325 mm .
Let bottom line AB be reference line,
2250 × 7.5 + 3075 × 117.5
5325
= 71.02 mm
ȳ =
Thickness of angle = 15 mm
Size of weld = 15 mm
Allowable shear stress = 100 N/mm2
Distance of the top edge of the angle section from neutral axis = 148.98 mm
Distance of the bottom edge of the angle section from neutral axis = 71.02 mm
l1 = length of weld at the top
l2 = length of weld at the bottom
@seismicisolation
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634
•
Strength of Materials
Now , F = τ (l1 + l2 ) × 0.707 × t
250000 = 100(l1 + l2 ) × 0.707 × 15
250000
100 × 0.707 × 15
Total length, l = 235.74 mm
(l1 + l2 ) =
let
l1 + l2 = l
b×l
a+b
71.02 × 235.74
=
148.98 + 71.02
16742.3
=
220
= 76.1 mm Ans
l1 =
Now l = l1 + l2
l2 = 235.74 − 76.1
= 159.64 mm
Ans
E XAMPLE 28.5: A welded lap joint is to be provided to connect two tie bars 200 mm × 15 mm a
shown in Fig. 28.11. The working stress in the bar is 180 N/mm2 . Investigate the design of the size
of the fillet welds be 14 mm. Safe stress for the weld may be taken as 110 N/mm2 .
60 mm
60 mm
60 mm
90 mm
120 mm
90 mm
F
F
Figure 28.11
S OLUTION :
Safe load in the tie bar = 200 × 15 × 180
= 540000 N
Total length of weld = 2 × 60 + 4 602 + 902
√
= 120 + 4 × 11700
@seismicisolation
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Welded Joints
•
635
= 120 + 4 × 108.17
= 552.7 mm
Strength of weld = 0.707 × 14 × 552.7 × 110
= 601768.7 N
Because the strength of the weld is greater than even the maximum tension in the tie bar, the
design is safe.
T
Fillet welds under torsion
A. Circular fillet weld:
Figure 28.12 shows a circular shaft welded
to a plate. Let this be subjected to torque T . In a
horizontal plane, the shear stress in the weld,
t
Figure 28.12
T × d/2
J
2
d
and J = π td
2
τ=
because
τ
T
=
J
R
T × d/2
2T
2 =
π
td 2
d
π td
2
t
Throat thickness = √
2
√
2T 2
Maximum shear stress, τmax =
π td 2
2.83T
=
π td 2
Therefore, τ =
∴
(i)
z
B. Long adjacent fillet welds:
A vertical plate is welded to a horizontal
plate by two identical fillet welds as shown in
Fig. 28.13.
Let torque T be applied on vertical plate as
shown.
Now t = leg length of weld
l = length of the joint.
T
b
t
t
Figure 28.13
@seismicisolation
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l
636
•
Strength of Materials
Torque T will induce shear stress varying from zero at the axis and maximum at the plate ends.
T × l/2
3T
= 2.
Induced shear stress, τ =
1 3
tl
2 × tl
12
The maximum value of the shear stress in the throat of weld is given by
√
3T 2 4.24T
τmax =
=
(ii)
tl 2
tl 2
Fillet Welds Under Bending Moment
Let a circular rod connected to a rigid plate
by a fillet weld.
d = Diameter of rod
M = Bending moment acting on the rod
s = Size or leg of the weld
t = Throat thickness
Z = Section modulus of the weld section
=
d
M
s
t
s
Figure 28.14
π td 2
4
Therefore, bending stress,
M
Z
M
=
π td 2
4
4M
=
π td 2
σb =
The maximum bending stress occurs on the throat of the weld which is inclined at 45◦ to the
horizontal plane,
∴
length of throat t = s sin 45◦
= 0.707 s
4m
σ b(max) =
π × 0.707 s × d 2
5.66 m
=
π sd 2
For ready reference the table is given below showing polar moment of inertia J and section
modulus Z for different type of weld.
@seismicisolation
@seismicisolation
Welded Joints
•
Note: s = Size of weld
Type of weld
b
Polar moment of inertia, J
Section modulus, Z
sl 3b2 + l 2
8.49
sbl
√
2
sl b2 + 3l 2
8.49
sb2
4.242
l
b
l
l
y
G
b
s
(b + l)4 − 6b2 l 2
16.97 (l + b)
4lb + b2
at bottom
8.49
2
b (4lb + b)
at top
s
8.49 (2l + b)
s
x
l2
2l+b
b2
y=
2(l+b)
x=
s
b2
√ bl +
3
2
s (b + l)3
8.49
b
l
l
G
b
s
√
2
(b + 2l)3 l 2 (b + l)2
−
12
b + 2l
s
b2
√ lb +
6
2
x
x=
l2
2(l+b)
d
π sd 3
5.66
s
@seismicisolation
@seismicisolation
π sd 2
5.66
637
638
•
Strength of Materials
E XAMPLE 28.6: Two steel shafts of 55 mm diameter are connected by means of a flange coupling.
The flanges are welded on the shaft ends. Determine the sign of the welds required on the surface
of each shaft, both on the inner and outer faces of the flange, to transmit full torque capacity of the
shafts. Assume the permissible shear stress in the shaft in 70 MPa and that in the weld is 100 MPa.
S OLUTION :
d = 55 mm,
τshaft = 70 MPa, τweld = 100 MPa
Maximum torque, Ts transmitted is given by
π d3
× τshaft
16
π
Ts =
× 553 × 70
16
= 2285576.6 Nmm
Ts =
∴
(i)
Since the welds are provided at two places on each flange, the torque capacity, Tw of the welds,
is given by,
Tw = 2 ×
π sd 2 τweld
2.83
(as proved earlier)
where s is the size of weld
2 × π s × 552 × 100
2.83
= 671272.1 s Nmm
Tw =
(ii)
Equatity Eqns. (i) and (ii)
2285576.6 = 671272.1 s
2285576.6
∴ s=
671272.1
= 3.4 mm Ans
E XAMPLE 28.7: A circular shaft of dia 60 mm and length 220 mm is welded to a flat plate. A
force of 15 kN is acting at 220 mm long shaft at the free end. The shaft is horizontal and the size of
weld is 22 mm. Determine the maximum normal stress in the weld.
@seismicisolation
@seismicisolation
Welded Joints
S OLUTION :
d = 60 mm, F = 15 kN, l = 220 mm,
s, size of weld = 22 mm.
22
Throat thickness, t = √ = 15.5 mm
2
Throat area, A = π d
= π × 60 × 15.5
= 2920.2 mm2
F
Transverse shear stress =
A
15000
=
2920.2
= 5.14 N/mm2
Bending moment = F × l
= 15000 × 220
= 3300000 Nmm
For a circular filled weld section modulus Z is given by
π sd 2
5.66
π × 22 × 602
or Z =
5.66
= 43937.8 mm3
M
Bending stress, σb =
Z
3300000
=
43937.8
= 75.1 N/mm2
Z=
Resultant normal stress, σmax =
τ 2 + σb2
= 5.142 + 75.12
√
= 26.4 + 5640
= 75.27 N/mm2
@seismicisolation
@seismicisolation
Ans
•
639
640
•
Strength of Materials
E XAMPLE 28.8: A rectangular cross-section bar is welded to support by means of fillet welds as
shown in Fig 28.15. Determine the size of the welds, if the permissible shear stress in the weld is
limited to 90 MPa.
30 kN
160
450 mm
120
Figure 28.15
S OLUTION :
F = 30, kN = 30 × 103 N, τmax = 90 MPa, l = 120, b = 160 mm, L = 450 mm.
The throat area for a rectangular fillet weld,
A = t(2b + 2l)
= 0.707 s (2b + 2l)
= 0.707 s (2 × 160 × 2 × 120)
= 0.707 s (320 + 240)
= 396 s
F
30 × 103
=
A
396 s
75.75
=
s
Direct shear stream, τ =
b2
lb +
3
1602
= 0.707 s 120 × 160 +
3
= 0.707 s (19200 + 8533.3)
5
From table, Z section modulus = √
2
Z = 19607.5 s mm3
Bending stress,
M
Z
M = P×L
σb =
= 30000 × 450
= 13500000
@seismicisolation
@seismicisolation
Welded Joints
•
641
13500000
19607.5 s
688.5
N/mm2
=
s
∴
σb =
1
2
(σb )2 + 4τ 2
Maximum shear stress (τmax ),
90 =
1
=
2
688.5
s
2
75.75
+4
s2
2
1√
474632.25 + 22952
2s
352.5
90 =
s
352.5
or s =
90
= 3.92 mm
=
= Say 4 mm
Ans
E XAMPLE 28.9: The bracket shown in Fig. 28.16 is designed to carry a dead weight of 15 kN.
What sizes of fillets are required at the top and bottom of the bracket? Assume that the forces act
through A and B.
25 mm
Fillet weld
A
75 mm
B
15 kN
50 mm
Figure 28.16
The welds are produced through shielded arc welding process with a permissible strength of
150 N/mm2 .
S OLUTION :
Shearing stress =
Shearing force
Area of weld group
@seismicisolation
@seismicisolation
642
•
Strength of Materials
Let t be the throat thickness and s be the size of weld.
t = 0.707 s
Area of weld group,
A = 2 × 25 t
= 50 t
15000
τ=
50t
300
=
N/mm2
t
Therefore,
(i)
To determine bending stress σ , moment of inertia of weld group is important to calculate.
Ixx = 2
25t 3 25t × 752
+
12
4
Because t is a small quantities as compared to other quantities like
∴
25t × (75)2
2
= 70312.5t mm4
M = 15000 × 50
= 750000 Nmm
M 75
σ=
×
Ixx
2
75
750000
×
=
70312.5t
2
400
=
t
Ixx =
Bending moment,
∴
The maximum stress (shearing) on the plane of throat,
τmax =
σ 2
=
25t 3
can be neglected.
12
2
200
t
+ τ2
2
+
300
t
1√
40000 + 90000
t
360.6
N/mm2
=
t
=
@seismicisolation
@seismicisolation
2
(ii)
Welded Joints
•
643
As stated in the problem, permissible stress is 150 N/mm2
∴
360.6
t
360.6
or t =
150
= 2.4 mm
2.4
= 3.39
∴ s=
0.707
Say 3.4 mm Ans
150 =
Eccentrically Loaded Welded Joint
F
B
R
b
A
θ
θ
τ1
G
τ2
a
e
dA
A
B
r
R
G
Figure 28.17
Let us choose an elemental area dA at a distance r from G and assume that stress acting on dA
is τ , so that
τ
τ2
=
r
R
r
or τ = τ2
(i)
R
The force acting on elemental area dA,
dF = τ dA
The moment due to dF about G
dM = τ r dA
From Eqns. (i) and (ii)
dM =
τ2 r2 dA
R
@seismicisolation
@seismicisolation
(ii)
644
•
Strength of Materials
Integrating,
dM = M = F.e
τ2
r2 dA
R
τ2
F.e R
Fe =
JG ∴ τ2 =
R
JG
or F.e =
JG is the polar moment of inertia of a fillet weld about G. Now τ , and τ2 act at an angle θ .
The resultant stress, τA = τ12 + τ22 + 2 τ1 τ2 cos θ
To determine JG , the following procedure is adopted.
i) Find the C.G. of weld group with reference to inside edges of vertical and horizontal fillets.
ii) Determine the second moment of inertia IXX1 and IYY 1 respectively with respect to horizontal
and vertical axes passing through C.G. And then transfer them to axes of C.G. of weld group.
iii) Similarly find Ixx2 and IYY 2 and then transfer them to horizontal and vertical axes passing
through C.G. of weld group.
iv) In order to find JG, add the above four moment of inertia:
JG = I XX1 + I YY 1 + I XX2 + I YY 2
, etc. neglect terms like a t 2 , etc.
Remember since t is too small while computing IXX1
, IYY
1
E XAMPLE 28.10: A bracket is welded to its support as shown in Fig. 28.18. All welds are fillet welds of equal thickness. Determine the fillet size if the permissible stress in the welds is
80 N/mm2 .
[U.P.S.C. Engg. Services 1979]
50 mm
60 mm
1
s
E
Y
θ
A
θ
τ2
τ1
150
3
OG
D
13.33
mm
B
C
2
16.7 mm
Figure 28.18
@seismicisolation
@seismicisolation
X
25 kN
Welded Joints
•
645
To find C.G:
A1 = 60 s, A2 = 60 s, A3 = 150 s
As the weld group is symmetrical about horizontal axis, C.G. will be on OX-axis, let its distance
from DE be x.
A1 (30) + A2 (30) + A3 (0)
A1 + A2 + A3
60 s × 30 + 60 s × 30
x=
60 s + 60 s + 150 s
2 × 60 × 30
=
270
= 13.33 mm
x=
To find polar moment of inertia of weld group about C.G.
Let Ixx and Iyy be the second moments of inertia of plane rectangular about horizontal and
and I be the moments of
vertical axes passing through C.G. of respective rectangles. And Ixx
yy
inertia about horizontal and vertical axes passing through C.G. of weld group, G. The suffixes 1, 2,
3 are for areas 1, 2 and 3. Since s is much smaller, it can be neglected. Also value of s is not known
initially.
60 s3
= 5 s2
12
60 s3
= 5 s3
=
12
t (150)3
= 281.25 × 103 s
=
12
= 5 s2 + 60 s (75)2
Ixx1 =
Ixx2
Ixx3
Ixx
1
= 5 s2 + 337.5 × 103 s
Ixx
= 5 s2 + 60 s (75)2
2
= 5 s2 + 337.5 × 103 s
= Ixx3 = 281.25 × 103 s
Ixx
3
s (60)3
= 18 × 103 s
12
s (60)3
=
= 18 × 103 s
12
150 (s)3
=
= 12.5 s3
12
= 18 × 103 s + 60 s (16.7)2
Iyy
=
1
Iyy
2
Iyy3
Iyy
1
= 34.7 × 103 s
@seismicisolation
@seismicisolation
646
•
Strength of Materials
= 18 × 103 t + 60 s (16.7)2
Iyy
2
= 34.7 × 103 s
Iyy
= 12.5 s3 + 120 s (13.33)2
3
= 12.5 s2 + 21.32 × 103 s
The polar moment of inertia of the weld group about C.G.
JG = Ixx
+ Ixx
+ Ixx
+ Iyy
+ Iyy
+ Iyy
1
2
3
1
2
3
= 5 s3 + 337.5 × 103 s + 5 s3 + 337.5 × 103 s + 281.25 × 103 s + 34.7 × 103 s
+ 34.7 × 103 s + 12.5 s3 + 21.32 × 103 s
= 22.5 s3 + 1047 × 103 s mm4
Because s3 is a small quantity, s3 may be neglected.
Hence, JG = 1047 × 103 t mm4 .
The secondary shearing stress at any point on the fillet at a distance R from G is given by
τ =
FeR
JG
This stress will act perpendicular to R, its value will be higher if R is larger R at A and C is
greatest. From triangle ABG,
RA =
(AB)2 + (BG)2
=
(75)2 + (46.7)2
= 88.35 mm
e = 50 + BG
= 50 + 46.7
= 96.7 mm
25000 × 96.7 × 88.35
1047 × 103 s
204
=
MPa
s
τ2 =
For resultant shearing stress, τ1 and τ2 are to be added vectorially,
τs =
τs =
τ12 + τ22 + 2 τ1 τ2 cos θ
1
s
(132.3)2 + (204)2 + 2 × 132.3 × 204 × 0.53
@seismicisolation
@seismicisolation
Welded Joints
•
647
1√
17503.3 + 41616 + 28608.55
s
296.2
MPa
=
s
=
Permissible stress in the weld is 80 MPa or 80 N/mm2 .
296.2
s
296.2
s=
80
= 3.70 mm
80 =
Say 4 mm
Ans
Exercise
28.1 A steel plate 100 mm wide and 12.5 mm thick is welded by a parallel fillet welds to another
plate. The plates are subjected to a load of 60 kN. Find the length of the weld if the maximum
shear stress is not to exceed 56 N/mm2 .
[Ans 60.6 mm]
28.2 A 12 mm thick bracket plate is connected to column flange as shown as Fig. 28.19 and
transmits a load of 180 kN. Design a suitable weld. Permissible stresses in the weld are:
in bending = 15.45 kN/cm2 , in shear = 10.05 kN/cm2 .
180 kN
150 mm
150 mm
600
mm
[Ans
Figure 28.19
Size of weld = 3.68 mm]
28.3 A 100 mm wide and 12.5 mm thick plate is welded to another plate by two parallel fillet
welds. A load of 50 kN acts along the axis of this plate. Find the length of each fillet if the
permissible shear stress is 55 MPa.
[Ans 38.5 mm]
28.4 A plate is attached to a frame by two side fillet welds as shown in Fig. 28.20. Determine the
size of the welds to resist vertical load of 50000 N. The permissible stress is 10250 N/cm2 .
@seismicisolation
@seismicisolation
648
•
Strength of Materials
50 kN
D
100 mm
E
100
mm
B
150 mm
A
[Ans Size of weld DE = 12.3 mm, size of
weld AB = 11 mm]
Figure 28.20
28.5 A rectangular cross-section bar is welded to a support by means of fillet welds as shown in
Fig. 28.21. Determine the size of welds, if the permissible shear stress in the weld is limited
to 75 MPa.
25 kN
150
500 mm
100 mm
[Ans
Figure 28.21
@seismicisolation
@seismicisolation
5.32 mm]
C HAPTER
29
MECHANICAL TESTING OF MATERIALS
1. Tensile Test
a) To investigate for a tensile specimen, the relationship between gauge length and percentage
elongation.
b) To find percentage reduction in area after fracture.
Apparatus: Denison Universal testing machine (Fig. 29.2) or equivalent machine, mild steel specimen (dimensions shown in Fig. 29.1) centre punch, hammer, rule, divider, vernier calliper.
26 mm dia
Gauge length = 5.65 √A
A = cross-sectional area of specimen
20 Rad
20
mm
14 mm
70 mm
Gauge
length
90 mm
55 mm
240 ± 1 mm
Figure 29.1
Procedure:
1. With specimen aid of divider and punch mark two points 70 mm apart (gauge length) with
centre punch, as shown in Fig. 29.1.
2. The specimen should be in annealed condition
3. Mount the specimen symmetrically. Fix Lindley extensometer on gauge length tightly.
4. Start apply load after bringing the load printer to zero. And take 5 − 6 readings of load and
extension.
@seismicisolation
@seismicisolation
650
•
Strength of Materials
5. Before yield point remove the extensometer after taking reading of extension. This is very
important otherwise extensometer will get damaged.
6. Now if automatic graph is fitted, keep an eye on it and see how two yield points on mild steel
appears.
7. Apply load until fracture takes place. It will be observed that before breaking, necking shape
takes place.
8. Observe the fractured piece after taking them out from U.T.M. It will be cup and cone shaped
fracture.
Results
Place the broken pieces together so that no gap appears between two pieces. Now measure the
gauge length which will obviously be more than before let original gauge length is equal to L1 and
after breaking and putting the two pieces to make them straight without any gap and measure the
new gauge length. Let new gauge length after fracture = l2 . Then,
l2 − l1
× 100 = % elongation
l1
Maximum load indicated by pointer on dial
B. Ultimate tensile strength =
Original area of cross section
C. Percentage reduction in area:
A.
Let initial area = A1
Now measure the diameter of neck where after necking fracture has taken place and calculate
this area. Let it be A2 . Now,
Percentage reduction in area =
A1 − A2
× 100
A1
Conclusion: Note down carefully the sources of error and draw sketch of portion of both broken
pieces (near portion of fracture)
Cup
Cone
@seismicisolation
@seismicisolation
Mechanical Testing of Materials
•
651
Plot stress-strain curve, if automatic graph is not available on U.T.M. and mark all important
points.
E
B
A
C
F
D
Figure 29.2 Universal testing machine
@seismicisolation
@seismicisolation
652
•
Strength of Materials
Figure 29.3 Use of Lindley extensometer
Brittle material
Ductile material
Figure 29.4 Fractured ends of specimens.
Note: For brittle material graph will be almost straight line without showing yield point.
For future reference the following table is quite useful, showing various properties:
@seismicisolation
@seismicisolation
Mechanical Testing of Materials
•
653
Table 29.1 Typical Mechanical Properties of Some Metals & Alloys
Metal
or
Alloy
Lead
Aluminium
Duralumin
Magnesium
with 10% Al
Copper
70 − 30 brass
cartridge
metal
Phosphorbronze 95%
Cu, 5% Sn
Wrought
iron
Mild
steel
Nickel-chromiummolybdenum steel
2.5% Ni, 0.75%
Cr, 0.5% Mo
Stainless
steel
Cast iron (grey)
Condition
Soft sheet
Wrought
and annealed
Extruded
and heattreated
Cast and
heat-treated
Wrought and
annealed
Annealed
sheet, deep
drawn
Hard
rolled
Hot-rolled
bar
Hot rolled
sheet
Hardened
and
tempered
0.1% Proof
stress MN/m2
−
Ultimate
tensile strength
MN/m2
17.7
% elongation
on 70 mm
gauge length
64
Brinell
hardness
number
4
Fatigue
limit
MN/m2
2.78
−
58.7
60
15
30.9
278
432
15
115
170
116
247
2
80
61.8
46.3
216
60
42
66.4
84.9,
371
324,
463
67.5,
19.5
62,
132
114,
151
649
711
5.5
188
185
201
0.5% proof
stress 232
309
30
100
185
309
28
100
185
1310
1540
10
444
432
185
−
463
309
30
0.0
170
250
263
137
Softened
cast
Fatigue Limit: The fatigue limit of a metal is the maximum range of stress which the test piece will
endure without failure after an infinite number of reversals usually 107 cycles about a mean stress
of zero.
2. Compression Test
To investigate the behaviour of the following materials when subjected to compressive loading:
i) cast iron, ii) mild steel, iii) aluminium, iv) copper and v) concrete
@seismicisolation
@seismicisolation
654
•
Strength of Materials
Apparatus: Denison universal testing machine or equivalent machine with attachments for compression tests, a cylindrical specimen of suitable length / diameter ratio of each of the above materials, micrometer, dial test-indicator with supporting stand.
The compression testing attachment consists of upper and lower hardened steel compression
plates.
The upper plate is located in a recess in the lower wedge-box of the machine and is retained by a
headed peg which forms a baynot fastening. The lower plate is located in a reverse on the transverse
beam of the machine.
Procedure: Measure the length and diameter of the specimen and check that the ends are machined
perfectly flat and perpendicular to the longitudinal axis.
Locate the cast-iron specimen vertically on the lower compression plate and lower the straining
wedge-box until the upper compression plate just touches the specimen.
Position the dial test indicator so that it is bearing on the top of the lower wedge-box and zero
the pointer.
Gradually, apply the axial compressive load to the specimen and record corresponding readings
of contractor, as given by the dial test indicator reading, and load until fracture occurs.
Examine and sketch the fractured specimen and note particularly the approximate angular position of the plane of fracture.
Test the specimens of the other materials using the same procedure and if, after a sufficient
period of testing, failure seems unlikely to occur for any specimen, discontinue the test. Plot graphs
for each test, using the readings recorded, of load against contraction.
∴
σθ = σ cos θ ·
AB
= σ cos2 θ .
AC
Resolving parallel to plane AC,
τθ AC = σ sin θ · AB
AB
∴ τθ = σ sin θ
= σ sin θ cos θ
AC
Due to friction between the surfaces the resistance to sliding will be increased by an amount
μσθ where μ may be regarded as the coefficient of friction.
Let τ be the ultimate shear stress in cast-iron. Then if fracture is to occur,
τθ = τ + μσθ
∴ τ = τθ − μσθ
= σ sin θ cos θ − μσ cos2 θ
σ
μσ
(cos 2θ + 1)
= sin 2θ −
2
2
@seismicisolation
@seismicisolation
Mechanical Testing of Materials
•
655
For a maximum value of τ ,
dτ
=0
dθ
σ cos 2θ + μσ sin 2θ = 0
∴
tan 2θ = −
1
μ
For cast-iron μ is assumed to be 0.364
∴
1
0.364
= −2.747
2θ = 110◦
θ = 55◦
tan 2θ = −
That is, fracture occurs on a plane making an angle of 55◦ with the ends of the specimen. Typical
forms of specimens after testing.
Copper
Aluminium
Mild steal
Cast iron
Concrete
Figure 29.5
Theory: Compression tests are normally carried out on specimens of circular cross section which,
to eliminate bending, are normally such that the length /diameter ratio is less than 2. It is extremely
important that the compressive load should be axial, and for this reason the ends of the specimen
must be perfectly flat and perpendicular to the axis of the specimen. Even if the applied load is
absolutely axial and the surfaces are in perfect contact the stress distribution is not uniform over the
cross sections of the specimen, due to friction, and for this reason the specimens tend to become
barrel shaped.
@seismicisolation
@seismicisolation
656
•
Strength of Materials
The behaviour of the specimens when subjected to loading depends largely on the particular
material. For ductile materials their behaviour is similar to that when they are subjected to tensile
loading. During the elastic stage the load is proportional to the contraction produced, and after
the yield point, which is not so pronounced as when tested in tension, the deformation is plastic.
Fracture of ductile materials is seldom obtained, and the compressed cylinder ultimately becomes a
flat disc.
Brittle materials when tested do not exhibit a yield point, and the plastic deformation is much
less noticeable than for ductile materials. Since the shear resistance is respectively small, failure
occurs due to shear on internal planes, which make an angle of approximately 55◦ , with the crosssection of the specimen. Theoretically, the plane on which the maximum shear stress occurs is at
45◦ , but due to friction between the sliding surfaces this angle is always somewhat greater.
For example, consider a cast-iron specimen subjected to a compressive stress σ . Let the direct
and shear components on a plane make the equilibrium of the portion ABC and assume unit thickness.
σ
σθ
C
σ
θ
σ cosθ
τθ
A
θ
B
σ sin
σ
Figure 29.6
Resolving perpendicular to plane AC.
σθ AC = σ cos θ · AB
Results:
i) Cast-iron
Original diameter of specimen, d = mm
Original length of specimen, l = mm
Load W (kN)
Contraction (mm)
@seismicisolation
@seismicisolation
θ
Mechanical Testing of Materials
•
657
Record similar results for each different materials.
Conclusions: Draw up separate conclusions for each specimen with regard to the following:
i) Whether fracture occured or not,
ii) Nature of fracture, if any.
Compare the graphs of load against contraction obtained for each material.
What do you infer from the tests on the cast-iron and concrete specimens with regard to the
ratio between their ultimate shear and ultimate compressive strengths?
3. Shear Tests
Object: To determine experimentally, the ultimate shear strength, in single and double shear of the
following materials: i) mild steel, ii) cast-iron, iii) copper, iv) brass and v) aluminium.
Apparatus: Denison universal testing machine together with attachments for single and double
shear tests, round specimens of the above materials, micrometre, etc.
The shear attachments are shown in Fig. 29.7. The shear die engages with the lower straining
wedge-box, which is provided with a spigot and headed pegs to form a bayonet fastening. The shear
box, which locates the specimen horizontally, rests on the transverse beam of the testing machine
which locates the specimen horizontally, rests on the transverse beam of the testing machine.
Figure 29.7
Procedure: Attach the shear die to the wedge-box, and suitably position the shear box on the transverse beam.
Check the diameters of the specimens using a micrometre.
Insert a single shear specimen of mild steel in the hole in one side of the shear box so that it
projects two-thirds of the distance across the gap. Apply the load gradually to the specimens by
means of the shear die, and record the ultimate (maximum) load required before fracture occurs.
Test a second mild steel specimen in single shear as before, and record the average value of the
ultimate load.
Repeat the procedure for the single shear specimens of cast-iron, copper, brass and aluminium
and in each case record the average value required for fracture in double shear.
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Insert a specimen of mild steel symmetrically through both holes in the shear box. Apply the
load to the specimen and record the average ultimate load required for fracture in double shear.
Test a second mild steel specimen in double shear an before and record the average value of the
ultimate load.
Repeat the procedure for the double shear specimens of cast-iron, copper, brass and aluminum,
and record for each material the average value of the ultimate load.
Examine the form of the fractured specimen for each test.
Results:
Average ultimate load kN
Material
Diameter of specimen (mm) Cross-sectional area mm2 Single shear Double shear
Mild steel
Cast iron
Copper
Brass
Aluminium
Calculations:
Ultimate shear strength (N/mm2 )
Material
Single shear
Double shear
Mild steel
Cart iron
Copper
Brass
Aluminium
Conclusions:
Compare for each material the values obtained for the ultimate shear stress in single and double
shear and account for any discrepancies.
Compare also the experimental values with those normally accepted for the particular materials.
Deduce from the form of the facture obtained, what effects other than pure shear will have
influenced the results given.
Compare the values of ultimate shear strengths of the materials with ultimate tensile strength of
respective material.
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Table 29.2 Mechanical properties of metals
Material
Young’s modulus GN/m2
Tensile strength MN/m2
% Elongation
0.2% C steel
0.4% C steel
0.8% C steel
18/8 Stainless steel
Grey cast iron
White cast iron
Brass (70 Cu 30 Zn)
Bronze 95 Cu 5 Sn
210
210
210
210
210
210
110
110
350
600
800
700
250
−
460
600
30
20
8
65
−
−
50
30
4. Modulus of Rupture
Object: To determine experimentally, by commercial transverse tests on stand cast-iron specimens
i) The transverse breaking load
ii) The deflection at fracture
iii) Modulus of rupture.
Apparatus: Denison universal machine or equivalent machine with transverse test attachments,
standard cast-iron specimens, dial test-indicator with supporting stand, steel rule.
The transverse test attachment are a shown in Fig. 29.8 and consist of a loading knee together
with two supporting dogs.
Figure 29.8
The loading knee is located in a recess in the lower wedge-box of the machine and is retained
by a headed peg which forms a bayonet fastening. The support dogs are located in recesses on the
transverse beam of the machine to give the required span.
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Procedure: Measure the diameter and length of the specimens and check that they conform to the
standard requirements.
Locate the transverse attachments in the testing machine ensuring that the support dogs are at
the required distance apart.
Place a specimen symmetrically on the supports and then lower the loading knee until it just
touches the surface of the specimen.
Position the dial test indicator so that it is bearing on the underside of the lower wedge-box and
zero the pointer.
Apply the load gradually to the specimen, through the loading knee, and record corresponding values of load and dial test indicator reading until fracture occurs. Repeat this procedure for
several specimens, in each case recording corresponding readings of load, and the dial test-indicator
reading until fracture occurs.
Note in particular, for each specimen, the breaking load and maximum deflection.
Plot graphs of load against deflection from the readings recorded. Determine the modulus of
rupture using average breaking load by substitution in the equation given in the theory.
Theory: The transverse test is universally recognized as the standard test for cast-iron. The strength
of the material is indicated by the maximum central load the simply supported standard bar carries,
and the maximum deflection at the point of application of the load gives and approximate measure
of the toughness of the material.
The transverse strength of cast-iron is commonly expressed by a figure known as either modulus
of rupture or transverse rupture stress, this being the maximum stress that would here existed if
the material conformed to the theory of bending. However, this is not so, since the material is
stressed beyond the elastic limit. However, the modulus, of rupture is normally accepted as being a
convenient way of expressing the results of a transverse test.
Consider a circular specimen of diameter d simply supported over a span L. Let the maximum
load required for fracture be W .
From the theory of bending, modulus of rupture or transverse rupture stress,
σ=
and
My
I
where
Simply supported span, L =
WL
4
d
π d4
and I =
2
64
WL
d/2
8W L
4
σ=
=
4
πd
π d3
64
y=
Results:
Diameter of specimen, d =
M=
mm
mm
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Specimen Nos
Load (kN)
Deflection (mm)
Draw up similar tables for the other specimens.
Calculations:
Average breaking load =
Modulus of rupture, σ =
kN
kN/mm2
σ=
8W L
π d3
Conclusion:
Compare the values found for the breaking load and maximum deflection with the minimum
standard requirements for the particular grade of cast-iron.
Comment on the form of the graphs obtained of load against deflection.
5. Impact Test
Object: To investigate the effect upon the Izod impact value for an alloy steel.
Apparatus: Avevy Izod impact testing machine or equivalent machine, selection of about eight
rounds or square Izod specimens being hardened with manufacturer’s instructions, but one only
being correctly tempered. Of the remainder, one is left in the dead hard condition, and the others
are tempered at various temperature, differing from the prescribed temperature, quenched in various
media, e.g., air, oil, water. The dimensions of the specimen are given below:
28 mm
28 mm
28 mm 28 mm
Hammer Strikes here
22
10 mm
45˚
130 mm
2 mm
10 mm
Izod specimen
Hammer Strikes
here 60 mm
10 mm
square
2 mm deep, 1 mm Radius Notch
30 mm
40 mm
No
Charpy specimen
Figure 29.9
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Here specimen’s dimensions for charpy test are also shown.
The machine consists of a rigid construction as shown in Fig. 29.10 which can carry Izod test
and Charpy test both.
The specimens are held in gripping dies mounted in a vice on the machine base, the dies being
clamped by means of a screw and spanner. A position gauge is used to ensure proper mounting of
specimen.
(a) Impact testing machine
(b) Izod specimen and striker
(c) Charpy support block.
Figure 29.10
A trigger retains the pendulum in the raised position while the specimen is set up in the vice.
The scale is an inverted quadrant graduated in joules.
Procedure: Before beginning of the experiment, the pendulum is released to see that the pointer
comes to zero. Because there in no specimen mounted, so the energy consumed is nil.
Now fix the pendulum at the required height and hold the hammer (pendulum) in position with
the help of ratchet or another mechanism.
After fixing Izod specimen in the vice, release the hammer, so that in breaking the specimen,
some energy will be spent and the pointer will give a reading in joules. This energy is the indicator
of toughness of material. More the energy spent, more is strong the specimen in toughness.
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Striker
5˚
22 mm
10˚
1
mm
4
rad
2 to 3 mm radius
specimen
45˚
Gripping dies
Figure 29.11 Izod test
After breaking the specimen, the hammer will swing backwards which is either stopped by a
handle actuating a mechanism or if mechanism is not working, it should be carefully caught by the
hand and lowered onto the rest.
Now remove the broken piece (both) from gripping dies and raise the hammer to the right height
held by a mechanism provided with a trigger.
Repeat the test in respect of each specimen and record the Izod impact value in each case.
Compare the impact value obtained for each specimen with that obtained for the specimen
heat-treated in accordance with manufacturer’s instructions.
Results:
Write the exact composition of the steel used, the recommended heat-treatment and the minimum Izod value specified by the manufacturers for this condition.
Specimen No
Hardening
temp(◦ C)
Tempering
temperature (◦ C)
1
2
3
4
5
6
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Quenching
media
Izod impact
value (J)
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Conclusion:
From the tabulated results deduce the effects of variation of tempering temperature and punching media upon the impact value of the meterial. Compare the relative severities of air, oil and water
punching and deduce evidence to support your comparison from results.
6. Cupping Test
This test is used to determine the ductility and drawing properties of sheet metal.
Pressure
ring
d
20 mm dia
Material
under test
Crack
Die
Figure 29.12 In Erichsen test, the material
The sheet is kept between the die and pressure ring as shown in Fig. 29.12. The force on the
ball is applied gradually over a steel (hardened) ball of 20 mm diameter. The sheet is then checked
if after how much depth (d) the crack develops in specimen sheet.
This test is very useful to test ductility of sheets, drawn sheets or deep-drawn sheets.
Figure 29.13 Erichsen test machine
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7. Modulus of Rigidity of Rubber
To determine the relationship between shear stress and sheet strain for rubber, and also to find
modulus of rigidity of the material, a wall mounted rubber specimen fitted with dial-indicator and
hanger for weights is needed.
Set the dial test indicator to zero with the hanger in position. Now apply a load to the hanger.
The rubber block in new under sheet (refer Fig. 29.14) and the loading plate will move downwards
relative to the fixed plate, the relative displacement can be read from the dial indicator.
Figure 29.14
This experiment should be repeated with increasing loads and record the varied displacement of
the loading plate in each case. Unload and note the corresponding displacements with the decreasing
loads.
All results (readings) should be tabulated. Calculate the shear stress and corresponding shear
strain and from the graph, determine the modulus of rigidity of rubber.
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Theory (Refer Fig. 29.15)
P = load on hanger
l = length of rubber block
h = width of rubber block
t = thickness of rubber block
d = vertical displacement of loading plate
h
P
Area under stress = l × t
P
∴ Shear stress =
A
P
=
l ×t
l
P
Figure 29.15
Neglecting any displacement due to bending, shear strain, φ =
d
h
Results:
Displacement
Load P(N)
Load increasing d1
Load decreasing d2
Calculations:
Load P(N)
Mean displacement
(d + d2 )
mm
d= 1
2
t
Shear stress
P
σ = (N/mm2 )
lt
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Shear strain
d
φ = (rad)
h
Mechanical Testing of Materials
B
Shear
stress
N/mm2
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•
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From the graph (Fig. 29.16) plotted of shear
stress against shear stain the modulus of rigidity
C for
Shear stress
Shear strain
= Slope of graph
AB
or C =
N/mm2
OA
rubber =
A
Shear strain φ
While performing the experiment be careful in taking reading, from dial indicator.
Quote the value obtained for modulus of rigidity and list probable causes of inaccuracy in the
result.
8. Torsion of a Round Bar
Using the torsion apparatus shown in Fig. 29.16 we have to determine: (i) the relationship between
the angle of twist and the torque applied for a round bar and (ii) the modulus of rigidity of the
material of the bar.
Effective length of specimen
Pointer
Graduated circular
dial for angle
of twist
Light pulleys
set up to
form a couple
Pointer
Scale
x
F
Figure 29.16
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We shall require metal bar, load-hangers, steel rule and weights in addition to torsion-testing
apparatus.
Clamp the specimen securely at its ends, and adjust the position of the upper protractor scale to
give a convenient gauge length.
Pass the draw cords over the pulleys and attach load-hangers to the free-ends. Zero the protractor
scales with the hangers in position.
Attach equal loads to the hangers, thus applying a couple to the bar, and record the angle of
twist over the gauge length by observing the difference of the protractor-scale readings.
Repeat this procedure for increasing values of load, and in each case, record the corresponding
angle of twist.
Measure the diameter of the bar, the diameter of the pulley, and the gauge length.
Calculate the torque applied for each load and compile a table of readings of load, torque and
angle of twist.
Draw a graph against angle of twist and by selecting two points on the graph, find the law
relating the two variables in the form T = K θ . Using the slope of the graph, also determine the
modulus of rigidity.
The two forces F constitute a couple Fx, i.e., applied torque, T = Fx (Nmm)
Using Formula,
∴
F
Cθ
T
=
J
l
Tl
N/mm2
C=
θ
x/2
x/2
F
where J =
Figure 29.17
π d4
mm4
32
C = modulus of rigidity (N/mm4 )
θ = angle of twist (radians)
l = effective length of specimen (mm)
∴
C=
T l
× (N/mm2 )
θ J
l
Therefore C = slope of graph × N/mm2
J
Results and calculations:
Diameter of bar, d =
mm
∴
J=
π d4
=
32
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Effective length of rod, l =
669
mm
Effective diameter of pulley, x =
mm
Angle of twist
Load F(N)
•
Angle of twist
π
(rad)
θ =θ×
180
θ (degrees)
T = Fx (Nmm)
After the experiment, compare value C obtained with normally accepted for the material of the
bar. If there is some error, state the sources of error.
9. Verification of Macaulay’s Method for Beam Deflection
To verify Macaulay’s method, we need a beam deflection apparatus, steel beam, two dial test indicators and stands, micrometer, steel rule, two hangers, and weights.
w1
w2
C
A
a
x
RA
D
X
b
RB
l
Figure 29.18
Set the beam on apparatus. Take convenient points C and D. Make sure that beam is simply
supported and are free at A and B. Calculate reactions RA and RB after loading W1 and W2 at distance
a and b from A respectively. Take point X at x mm from point A such that both loads are covered by
x (loads W1 & W2 )
Now measure the vertical deflection at C and D with dial indicators, so that to verify Macaulay’s
method we can compare these deflections with theoretical values. Now,
EI
Mx = RA x −W1 [x − a] −W2 [x − b]
(i)
d2y
= RA x −W1 [x − a] −W2 [x − b]
dx2
(ii)
As we know in Macaulay’s method the bracketed terms are integrated as a whole.
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Integrating Eqn. (ii)
EI
x2 W1
W2
dy
= RA −
[x − a]2 −
[x − b]2 +C1
dx
2
2
2
EIy = RA
(iii)
W2
x3 W1
−
[x − a]3 −
[x − b]2 +C1 x +C2
6
6
2
(iv)
When x = 0, y = 0
Substituting in Eqn. (iv)
C2 = 0
Now, C1 can be found out by substituting x = l, because at B, deflection is zero so finally we
have,
W1
W2
x3
−
[x − a]3 −
[x − b]3 + C1 x
EI y = RA
6
6
6
Since C1 is known, so find out deflection at C and D. If breadth of beam b = mm and
bd 3
depth = mm, then I =
. E is already known for the material of the beam.
12
While substituting to find yc and yD , remember any term in bracket becomes –ve, then it is to
be ignored.
Finally, to compare the deflection use following table:
Experimental deflections
At C
At D
Theoretical values
or Calculated values using Macaulay’s method
At C
At D
If results are somewhat differing, state the sources of error.
10. Verification of Maxwell’s Reciprocal Theorem
The reciprocal theorem asserts that the deflection at any point A due to load at any point B is equal
to the deflection at point B when the same load is placed at A.
Let us use a small cantilever apparatus. We will need in addition hanger, weights and a dial
indicator.
First place hanger at B and apply load on
the hanger. Using dial indicator, find the deflection at A. Let it be YBA . Now remove hanger and
load. And place the same load at A and observe
the deflection at B with dial indicator. Let this
deflection be YAB .
W
A
B
l
L
Figure 29.19
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Now you will notice that
YBA = YAB
Hence, Maxwell’s theorem is verified. If there is some error, find out the sources of error. May
be due to vibration because of some compressor or some other machine, dial indicator is not giving
correct readings.
So try again at some stable place.
11. Hardness Testing Machines
The most common hardness testing machines are:
i)
ii)
iii)
iv)
Brinell Hardness Testing Machine
Rockwell Hardness Testing Machine
Vickers hardness Testing Machine
Shore Scleroscope
(i) Brinell Hardness Testing Machine
BN (Brinell number) P is the load (kgf). An alternative and more precise equation than given is
BN =
P
√
πD
(D − D2 − d 2 )
2
(i)
d = diameter of ball impression and D = diameter of ball.
It is apparent that the harder the metal, the smaller the indentation diameter, and consequently
the higher the Brinell number. For thin sheets, however the load of 300 kgf is too heavy, assuming
the relation of indentation depth is D2 is constant. The ball and load are therefore variable and the
load is applied slowly for 30 seconds. So here also some Eqn. (i) is applicable.
When the load is varied in standard machines, the variation increases by 500 kgf each. After
application of the load to the softer metals for 30 s minimum, the pressure is released and the test
piece removed by lowering the table or platform. The impression diameter is measured by a low
power microscope with a graduated scale in the eyepiece. The middle of the indentation should be
more than 2.5× diameter from the testpiece edge and the thickness more than 7× indentation depth.
The indentation must not be visible on the other side of the thin testpiece.
Uses of the Brinell test: Following are the main purposes for which the test is useful.
(a) As a test of hardness: It is possible to find the amount of hardness induced by cold working.
(b) To explore machineability. Brinell number gives a guide to the machining possibilities. If To
explore BN a material is more 280–320, machining becomes difficult. But if Brinell number
is below 120–100 the material may be too soft and may tear under the cutting edge.
(c) To estimate the tensile strength (of steel):
For annealed (normal steel): tensile strength (N/mm2 ) = 3.6 × B.N
For hardened and tempered steel: tensile strength (N/mm2 ) = 3.2 × B.N
(d) The shape of ball impression gives an indication of the work-hardening capacity.
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Description:
P
(a) Metal being tested.
(b) Piling up.
(c) Sinking.
(W&T. Avery Ltd.
Brinell test.
Figure 29.20
In this machine a small hardened and tempered chromium steel or tungsten carbide ball of 10 mm
diameter is forced by pressure into the metal to be tested. The test-piece is placed on the platform
of the testing machine either universal or specially designed, the platform being elevated by screw
to make contact with the ball. This achieved, an oil pump applies increasing pressure, which is read
from a manometer or pressure gauge fixed to the machine. A regulated balance is also fitted, the
centre pin of which rests on the balls in the cylinders. These act as a piston and carry weights,
rendering pressure variable with hardness. The load is known in advance as is usually 3000 kgf,
under which load the weights float.
The pressure is applied for 10 seconds and then released, the machined and fine emery polished
test piece being removed and taken to a microscope whose inner lens has a millimetre scale divided
into tenths, so that the diameter of the indentation made by the ball can be measured with accuracy.
Non-ferrous metals are mostly subjected to a load of 500 kgf for 30 seconds. The Brinell number,
corresponding to the hardness of the metal is obtained from the equation:
BN =
P
Spherical area of impression (mm2 )
Brinell hardener values
Material
Brinell Number
Soft brass
60
Mild steel
130
235
Annealed chisel steel
415
White cast-iron
Nitride surface
750
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(ii) Rockwell Hardness Testing Machine
Figure 29.21 shows a Rockwell hardness testing machine like the Vickers machine, this
employs a diamond but one conical in form, though both Vickers and Rockwell machines balls can
be used if desired. The ball in these instances is 1.588 mm diameter. However, the modern tendency
is to use the diamond cone (angle 120◦ ) and though machines are graduated in two scales, C and B,
most values are given on the C scale for the harder metals. There are other scales, namely A, D, E
and F, each representing a different combination of indenter and load. A recent machine of British
manufacture based on the same principles in the Avery hardness tester, which is an improvement on
the Rockwell as regards the means of measurement and in certain other respects.
The hardness numeral is obtained by measuring the depth of the impressions superimposed
on one another. The first impression made is produced by a another. The first impression made is
produced by a load of 10 kgf, termed the preliminary, which penetrates the superficial layers of the
steel or other metal. The minor load is applied and is indicated on a small subsidiary dial at a point
marked ‘set’. The primary pointer is then set at zero by revolving the inclined edge of the did until it
shows zero. The load is now raised to 150 kgf (140 kgf is added to the initial load 10 kgf) by button
pressure. The 140 kgf load is then removed by a side lever and the hardness number is shown on
the instrument dial, which is marked off in graduations from 0 to 100, representing a single rotation
of the pointer and a penetrator movement limited to 0.02 mm. In the Rockwell testing machine the
diamond scale load limit is 10 to 140 kgf.
The test takes only about 10 seconds which means that it is almost invisible to the naked eye, no
particular harm is done to the finish of the work. The surface concerned needs no other preparation
than the elimination of oxide. On the other hand, the machine’s scale is not nearly so open, so that
errors may occur in some instances from misreading of the scale rebound, while should there be any
modification in the form of the penetrator face, this may also impair the accuracy of the readings.
To ensure that the maximum accuracy is obtained, therefore certain precautions are essential.
The machine should be so positioned that vibration has no serious disturbing effect on the dial group
readings. When the minor load is being used, the large scale needles on the dial range must be ‘set’
within + or −5 scale divisions. It may happen that as the platform of the machine is raised, the
Figure 29.21 Rockwell testing machine
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needle stops, though it does not do so on ‘set’. It is then necessary to adjust the lifting screw again,
the dial being rotated so that ‘set’ comes under the needle.
Assuming the machine is of orthodox pattern, the dashpot speed in giving if the major load
should allow the crack to finish its traverse in 5 seconds when no test piece is in the machine, the
machine being set-up to give 100 kgf load. Similarly, the surface testing machine is set for seconds
to give a major lord of 30 kgf.
Calibrating the Rockwell Hardness Testing Machine
The makers of the Rockwell machine suggest that the user should check it each time against the test
blocks to ensure that the penetrator is not damaged and the machine is not out of calibration. The
practice they advise is to check at the high, low and middle ranges of the given scale. Thus, to check
the full C scale, the machine should be checked at C 63, 45 and 25. If only a single range or two
ranges are employed, blocks should fall within ±5 hardness numbers on the C scale or any scale
using diamond penetrator and within ±10 numbers on the B scale on induced, any scale using ball
penetrators.
(iii) The Vicker’s Diamond Test Machine
Because of the limitations of the Brinell test, it was found desirable to replace ball by a pyramidal
from of diamond. When the Brinell machine is put to work on soft metal, the surface of the metal is
evaluated by plastic flow about its initial height an shown in Fig. 29.22. This means that indentation
diameter d is greater than it should be, so that the hardness numeral is lower than the true value.
Deformation of the half by extremely hard metals in another source of error. The square pyramidal
diamond in studied under a microscope and to ensure that one corner just makes contact with a fixed
knife-edge that can be seen through microscope eyepiece.
d'
Figure 29.22
The basic principle underlying the diamond pyramid test (also called Vicker’s hardness test)
is similar to that of the Brinell hardness number, designated as HV or sometime also as DPN
representing diamond pyramid number is determind as the ratio of the impressed load in kg to
the area of the impression in square mm made by a pyramid with an accurately ground square base.
The indenter is a diamond in the form of a square based pyramid with an included angles between
opposite faces equal to 136◦ (Refer Fig. 29.23. Vicker’s hardness testing machine and impression
as seen through microscope (vertical lines are edges of microscope measuring shutters).
On examing Fig. 29.24 it reveals that there is a geometric similarity for all values of load divided
by areas ratio. Apex angle is 136◦ is obtained by keeping the value of d/D = 0.375.
Load (kg)
should remain the same
It is quite clear that the hardness value equal to
Impressed area (mm2 )
under conditions, for a specimen of uniform hardness irrespective of the impressed load on the
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(Vickers-Armstrong, Ltd.
-Impression as seen thorugh
Microscope (Vertical lines
are edges of Microscope
Measuring Shutters).
(Vickers-Armstrong, Ltd.
Figure 29.23
x = 0.375D
Diamond pyramid
D
x
d
d2
1
Test specimen
Figure 29.24
indenting pyramid. The Vicker’s hardness number is defined as the load per unit surface of contact,
i.e., impressed, in kg per square mm, and can be calculated from the average diagonal as follows:
2P sin θ2
d2
p
= 1.854 2
d
HV =
θ = 136◦
d1 + d2
where, d = length of average diagonal
in mm
2
since
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θ = 136◦ apex angle
P = impressed load in kg.
The main advantages in this form of indentor are:
i) The impressions irrespective of size are geometrically similar.
ii) The deformation of the diamond is practically nil owing to its enormous hardness value.
iii) The surface of impression is extremely well defined and being square can be measured with
great accuracy across the diagonal corners.
iv) The hardness numbers found on homogeneous materials are unaffected by variation of load.
(iv) Shore Scleroscope
This is shown in Fig. 29.25. The share scleroscope employs the principle of the drop and rebound
of diamond-tipped hammer inside a glass tube. The hammer which is raised to the top of the tube
by vacuum, is allowed to fall freely from a fixed height onto the test specimen. The height of the
resulting rebound of the hammer is read at the graduated scale and indicates the degree of hardness
of the material. The scale is graduated in units, which are obtained by dividing the average rebound
from quenched pure high carbon steel into 100 part and then extended proportionally for metals
having exceptional hardness. A lens and pointer are provided for accurate scale reading. These
are principally used in connections with the soft materials. There is one important advantage of
scleroscope that it can be taken out from laboratory and test may be carried out in field on heavy
machinery.
Figure 29.25
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677
Table I Approximate Equivalent Hardness Number Conversion Table
Rockwell
Hardness
C Scale No.
Vicker’s Diamond
Pyramid hardness No.
Brinell Hardness No 10 mm ball,
3000 kg load standard ball
Rockwell
Hardness
B Scale No
Shore Scleroscope
No
68
940
–
–
97
67
900
–
–
95
66
865
–
–
92
65
832
–
–
91
64
800
–
–
88
63
772
–
–
87
62
746
–
–
85
61
720
–
–
83
60
697
–
–
81
59
674
–
–
80
58
653
–
–
78
57
633
–
–
76
56
613
–
–
75
55
595
–
–
74
54
577
–
–
72
53
560
–
–
71
52
544
500
–
69
51
528
487
–
68
50
513
475
–
67
49
498
464
–
66
48
484
451
–
64
47
471
442
–
63
46
458
432
–
62
45
446
421
–
60
44
434
409
–
58
43
423
400
–
57
42
412
390
–
56
41
402
381
–
55
40
392
371
–
54
39
382
362
–
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52
(Continued )
678
•
Strength of Materials
Table I (Continued)
Rockwell
Hardness
C Scale No.
Vicker’s Diamond
Pyramid hardness No.
Brinell Hardness No 10 mm ball,
3000 kg load standard ball
Rockwell
Hardness
B Scale No
Shore Scleroscope
No
38
372
353
–
51
37
363
344
–
50
36
354
336
(109)
49
35
345
327
(108.5)
48
34
336
319
(108.0)
47
33
327
311
(107.5)
46
32
318
301
(107.0)
44
31
310
294
(106.0)
43
30
302
286
(105.5)
42
29
294
279
(104.5)
41
28
286
271
(104.0)
41
27
279
264
(103.0)
40
26
272
258
(102.5)
38
25
266
253
(101.5)
38
24
260
247
(101.0)
37
23
254
243
100.0
36
22
248
237
99.0
35
21
243
231
98.5
35
20
238
226
97.8
34
(18)
230
219
96.7
33
(16)
222
212
95.5
32
(14)
213
203
93.9
31
(12)
204
194
92.3
29
(10)
196
187
90.7
28
(8)
188
179
89.7
27
(6)
180
171
87.1
26
(4)
173
165
85.5
25
(2)
166
158
83.5
24
(0)
160
152
81.7
24
Note: The Brinell hardness No. can be used to estimate the approximate tensile strength (N/mm2 ) of steels from the
relationship: T S = 3.4 HB
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Table II Comparative Hardness Scaler for steels (un herdened) and non-ferrous Alloys
Rockwell B scale 1/16” ball
Penetrator 100 kg load
Brinell scale 10 mm
standard Ball 3000 kg load
100
240
99
234
98
228
97
222
96
216
95
210
94
205
93
200
92
195
91
190
90
185
89
180
88
176
87
172
86
169
85
165
84
162
83
159
82
156
81
153
80
150
79
147
78
144
77
141
76
139
75
137
74
135
73
132
72
130
71
127
70
125
(Continued )
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•
Strength of Materials
Table II (Continued)
Rockwell B scale 1/16” ball
Penetrator 100 kg load
Brinell scale 10 mm
standard Ball 3000 kg load
69
123
68
121
67
119
66
117
65
116
64
114
63
112
62
110
61
108
60
107
59
106
Some Other Hardness Tests
Knoop Hardness Number: The Knoop hardness test is applied to extremely thin metals, plated
surfaces, exceptionally hard and brittle materials, very shallow carburized or nitrided surfaces, or
whenever the applied load must be kept below 3600 g. The knoop indentor is a diamond ground to
an elongated pyamidal form and it produces an inclination having long and short diagonals with a
ratio of approximately 7 to 1. The longitudinal angle of the indentor is 172 degrees 30 minutes and
tranverse angle 130 degrees. The tukon tester in which the Knoop indentor is used is fully automatic
under electronic control. The Knoop hardness number equals load in kilogram divided by the projected area of indentation in square millimeters. The indentation number corresponding to the long
diagonal and for a given load, may be determined from a table computed for a theoretically perfect
indentor. The load, which may be varied from 25 to 3600 grams is applied for a definite period and
always normal to the surface tested. Lapped place surfaces free from scratches are required.
Monotron Hardness Number: With this apparatus a standard steel drill is caused to make a definite
number of revolutions, while it is pressed with standard force against the specimen to be tested. The
hardness is automatically recorded on a diagram on which deed soft material given a horizontal
line, while a material as hard as the drill itself given a vertical line, intermediate hardness being
represented by the corresponding angle between 0 to 90◦ .
Turner’s Sclerometer: In making this test a weighted diamond point is drawn, once forward and
once backward, over the smooth surface of the material to be tested. The hardness number is the
weight required to produce a standard scratch.
Mohr’s Hardness Scale: Hardness in general is determined by what is known as Mohr’s scale,
a standard for hardness which is mainly applied to non-metallic elements and minerals. In this
hardness scale there are ten degrees or steps each designated by a mineral, the difference in the
series will scratch any of the preceding members.
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681
This scale is as follows:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Talc
Gypsum
Calcite
Fluor spar
Apatite
Orthoclase
Quartz
Topaz
Sapphire or corundum
Diamond.
These minerals, arbitrarily selected as standards, are successively harder, from talc, the softest
of all minerals to diamond, the hardest. This scale, which is now universally used for non-metallic
minerals, is however, not applied to metals.
12. Experiment
1) Close-coiled Helical Springs
Object: To determine experimentally, for close-coiled helical springs, the relationship between
deflection and i) applied load; ii) number of free coils; iii) mean coil diameter and iv) wire diameter.
And to compare these with those indicated by the theory.
Apparatus:
Spring deflection apparature as shown in Fig. 29.26 close coils springs, steel rule, calipers, micrometer, hanger and weights.
Figure 29.26
Procedure:
i) Set up any spring in the deflection apparatus and apply
initial load zero the vernier scale. Apply increasing
loads to the spring in suitable increments and in each
case record the corresponding deflection produced as
indicated by the vernier scale.
Plot using the readings, a graph of deflection against
load.
ii) Select a suitable number of springs having the same
wire end mean coil diameters but a different number
of ‘free’ coils. Place each spring, in turn, in the deflection apparatus and apply the same load to each. Record
corresponding readings and deflection and number of
‘free’ coils for each spring. Plot a graph of deflection
against number of ‘free’ coils for each spring.
iii) Select a suitable number of springs which have the
same wire diameter and number of ‘free’ coils but
which differ in their mean coil diameter.
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•
Strength of Materials
Apply the same load to each spring, in turn, and record corresponding readings of mean coil
diameter and deffection produced.
Plot a logarithmic graph, using the readings which have the same mean coils diameter and
number of coil but differ in their wire diameter.
Apply the same load to each spring, in turn, and record corresponding readings of wire diameter
and deflection produced. Plot a logarithmic graph, using the readings obtained, of deflection against
wire diameter. Determine from the graphs plotted the four required relationships.
Theory: In a close-coiled helical spring the helix angle is very small, and it may be assumed that
each coil is in a plane perpendicular to the spring axis.
When a load is applied to a close-coiled spring stream are set up due to torsion, direct share
and bending. The stresses due to direct share and bending are very small and may be nested in
comparison with those set up due to torsion.
Consider a spring of mean coil diameter D, wire diameter, d, and number of ‘free’ coils n,
objected to an axial load W which produces deflection S (see Fig. 29.27).
The work done by the load in deflecting spring, that is,
W
T
1 D
δ = θ = W. θ
2
2
2
2
wire
diameter
d
where T = applied torque = W.D
2
θ = angle of twist of wire in radius. Therefore deflection,
D/2
δ=
Dθ
2
From the torsion equation,
Cθ
T
=
J
l
W
Figure 29.27
∴
θ=
T.l
C.J
where l = length of wire = π Dn
J = polar second moment of area of cross section of wire,
π d4
=
32
WD
π Dn
∴ θ = 2π
C d4
32
θ=
16W D2 n
Cd 4
δ=
8w D3 n
C d4
Therefore deflection
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683
Results:
(i)
(ii)
(iii)
(iv)
Load (N)
Deflection δ (mm)
Number of free coils n
Deflection δ (mm)
Mean coil Diameter D (mm)
Wire diameter d (mm)
Deflection δ
Deflection δ (mm)
Calculations: From the graphs of deflection against load and deflection against number of ‘free’
coils show that
δ α W and δ α n
By determining the slopes of the logarithmic graphs show that
δ α D3
and
1
δα 4
d
Conclusions:
Comment on the form of the graphs obtained and compare the relationship found with those indicated by the theoretical equation for the deflection.
13. Experiment
To determine, experimentally, the value of the modulus of rigidity of C material of a close-coiled
helical spring.
Apparatus: Spring deflection apparatus as used (shown in Fig. 29.25) in previous experiment. Close
coiled spring, steel rules, calipers, micrometer, load hanger and weights.
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•
Strength of Materials
Procedure: Measure the wire diameter, mean coil diameter and count the number of ‘from’ coils.
Set up the spring in the deflection apparatus and apply a suitable initial load to open the coils,
and zero the vernier scale. Apply increasing loads to the spring in suitable increments and in each
case record the corresponding deflection. Using the slope of this graph, in conjunction with spring
data, determine the modulus of rigidity of the material of the spring.
Theory: As proved for the previous experiment, when an axial load, W, is applied to a close-coiled
spring the deflection, δ is given by,
δ=
8W D3 n
Cd 4
D = mean coil diameter
d = diameter of wire
n = number of ‘free’ coils
Therefore, c =
8W D3 n
δ d4
=
8D3 n
d4
=
8D3 n
× slope of graph
d4
W
δ
Results:
Wire diameter, d =
mm
Mean coil diameter, D =
mm
Number of ‘free’ coils, n =
Spring load W (N)
Spring Deflection δ (mm)
Calculations:
Slope of graph =
W
=
δ
Modulus of rigidity =
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N/mm2
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•
685
Conclusions:
Comment on the form of the graph obtained and compare the value found for modulus of rigidity
with that normally accepted for the material.
14. Experiment
To find Young’s modulus of elasticity of a material using simply supported beam.
Apparatus: A simply supported beam, hanger, weights, dial test indicator with its stand.
Dial test indicator, steel rule
d
l
Weights
Apparatus
Figure 29.28
Theory: If a central load W is applied at the centre of a simply supported beam, then deflection of
the centre is given by relation:
W l3
48EI
W l3
E=
48δ I
δ=
or
Procedure:
1. Measure breadth and depth of steel beam (it can be any other material also), let it be b and d
respectively.
2. At the centre of beam fix a book as shown in diagram (see Fig. 29.27). Also measure length
l of the beam between supports.
3. Place the dial test indicator on the top of the hanger (placed at the centre of beam). Make the
dial read zero.
4. Place the hanger whose weight is known on the hanger and take the reading of deflection with
the help of dial test indicator.
5. Remove the weight and repeat the experiment two times and note deflection.
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•
Strength of Materials
Readings:
Weight (N) +
Weight of Hanger
δ (mm)
1.
W
δ1
2.
3.
W
W
δ2
δ3
Sr. No.
Average
deflection
δ1 + δ2 + δ3
3
=δ
Calculations:
I=
bd 3
= . . . . . . mm4
12
Average deflection from above table,
δ = . . . . . . mm
L = . . .. . . mm
E=
Now
W l3
(N/mm2 )
48δ I
Conclusion: If the value of E found from the experiment are different from standard value of the
material, then find out the sources of error, e.g., length (l) measured is wrong or hook (weight) is
not at exact centre or dial indicator is sticky etc.
Note: This experiment can also be used to find deflection at the centre using moment area method.
Deflection at centre C,
=
Moment of area bending moment diagram between left hand support and centre of span
EI
Of course, material is to be taken whose modulus of elasticity, E is already known.
δ=
W l3
48EI
Then the deflection by experiment can be compared by theoretical value.
15. Fatigue Testing
It has been observed during that if a component is subjected to repeated stresses, it fails stresses
below the yield point stresses. Particularly, this kind of failure happens if stresses are due to reserved
bending or reversed torque. Such kind of failure is known as endurance or fatigue limit.
Experiment: To find the endurance limit of a metal specimen.
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687
Theory: In order to study the effect of fatigue of a material, a rotating fine standard specimen as
shown in Fig. 29.29 is rotated in a fatigue testing machine while the specimen is loaded in bending.
As the specimen rotates, the bending stream at the upper fibres varies from maximum tensile to
maximum compressive as shown in the Fig. 29.30. In other words, the specimen is subjected to a
completely reversed stress cycle.
7.675 mm
R = 251 mm
875 mm
Figure 29.29 Standard specimen
Standard specimen: In the market, variety of fatigue testing instruments are available. Basically, it
should consist of some method to produce alternating loads on the specimen. Also a counter should
be incorporated in it which can show number of cycles before the specimen fails. Figure 29.30
shows a rotary bending fatigue machine.
Bearing
Bearing
Motor
COUNTER
Shaft from motor
fixed specimen
by fastening system
W
Figure 29.30
Procedure:
1.
2.
3.
4.
5.
6.
7.
8.
9.
The standard specimen is fitted in the bearings as shown in Fig. 29.30.
Care should be taken that specimen in concentric with bearing within 0.04 mm.
Apply suitable load by adjusting the arrangement of supporting the load.
Set the revolution counter to zero.
Start the motor of the machine. It is understood from the set up shown that the upper layer
is under tension and when the same layer comes to bottom, it is under compression, thus
stresses are completely reversed.
Keep motor running until the specimen fails due to fatigue count the number of revolutions
from the counter.
Test is repeated with increased load on a new specimen and the procedure is repeated.
For every test calculate the stress (σ ) applied.
Plot a curve between σ and log N. Where N denotes cycles before failure.
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•
Strength of Materials
Diameter of the
specimen d (mm)
Load
W (N)
Number of cycles
(N)
Stress (σ )
in N/mm2
Observation:
The experiment is rotated at least four times with different loads on say stresses induced at the
centre of the specimen in case of material of one kind. while testing new specimens.
Endurance limit (106 cycles) = . . .. . .
It may be noted that the term endurance limit is used for reversed bending only while for other
types of loading, the term endurance strength may be used when referring to the fatigue strength
of the material. It may be defined as the maximum stress which can be applied to the machine part
working under actual conditions. Endurance limit or fatigue limit (σc ) a the maximum value of
the completely reversed bending stress which a polished standard specimen can withstand without
failure, for infinite number of cycles (usually 107 cycles).
Exercise
29.1
29.2
29.3
29.4
29.5
29.6
29.7
29.8
29.9
Discuss the term ‘fatigue’ and ‘fatigue stress’.
What is the necessity of a bend test?
Explain how would you determine the modulus of rupture of a timber specimen.
Draw stress-strain cure of tensile test on mild steel specimen and show important points on it.
Explain the procedure of fatigue testing.
State the names of tests which can be carried out on Universal Testing Machine.
Describe briefly how a torsion test on a mild steel specimen is performed.
What is ‘necking’ in tensile test and why it occurs?
Explain each of following property:
(i) Malleability
(ii) Ductility
(iii) Hardness
(iv) Elasticity
(v) Plasticity
(vi) Brittleness.
29.10 How do you find the yield point in tensile test on mild steel?
29.11 Giving dimensions, draw sketch of a standard tensile test specimen of mild steel.
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30
C HAPTER
Miscellaneous Solved Problems:
Stresses and Strains
E XAMPLE 30.1: A tapered rod shown in Fig. 30.1 carries a tensile load of 18 kN at the free end.
Calculate the total extension of the rod E = 200 GPa. Neglect self-weight.
65 mm dia
A
480 mm
B
35 mm dia
480 mm
C
12 mm dia
18 kN
Figure 30.1
S OLUTION :
For portion BC,
Extension δ =
4Pl
π Ed2 d1
689
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•
Strength of Materials
Extension δBC =
4 × 18000 × 480
π × 200000 × 35 × 12
= 0.131 mm
For portion AB,
Extension δAB =
4 × 18000 × 480
π × 200000 × 65 × 35
= 0.0242 mm
Total extenstoin = δBC + δAB
= 0.131 + 0.242
= 0.373 mm
Ans
E XAMPLE 30.2: A bar of length 25 cm has varying cross section. It carries a load of 15 kN. Find
x2
cm2 where x is the distance from one end
the extension if the cross section is given by 6 +
100
in cm. Take E = 200 GPa. Neglect weight of the bar.
S OLUTION :
Stress at distance x from the given end,
=
=
Load
=
Area
15000
N/m2
x2
10−4
6+
100
1.5 × 108
x2
6+
100
If da is the extension of an element of length dx at distance x from the given end, then strain at the
section,
=
da σ
=
dx E
da
0.075
1.5 × 108
=
=
2
dx
600 + x2
x
× 200 × 109
6+
100
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Miscellaneous Solved Problems: Stresses and Strains
da =
∴
a=
•
691
0.075
dx (if x is in cm, a will also be in cm)
600 + x2
25
0.075
0
600 + x2
dx
25 k
0.075
−1
Total extension, a = √
∴
tan √
600
600
25
0.075
tan−1 √
∴ a=
24.475
600
dx
x
1
= tan−1
2
2
a
a
x +a
= 3.062 × 10−3 tan−1 1.02
= 3.062 × 10−3 × 45.57◦
= 3.062 × 10−3 × 0.795
= 2.434 × 10−3 cm
= 0.02434 mm
[∴
degrees has been changed to radians]
Ans
E XAMPLE 30.3: A rigid beam is loaded as shown in Fig. 30.2. Determine the stresses in the steel
wire and in the aluminium member. What is the reaction at the hinge A? Take E for aluminium as
60 GPa and E for steel as 200 GPa.
Steel wire
Area = 0.6 cm2
Hinge B
20 cm
100 cm
Al bar Area = 50 cm2
Hinge A
40 cm
100 cm
Rigid support
40 cm
50 kN
Figure 30.2
S OLUTION :
For static equilibrium sum of vertical forces should be equal to zero. Let the forces in wire and
steel wire be Ps and Pa respectively as shown in Fig. 30.3.
Then ΣV = 0
Pa + 50 = Ps + R
And sum of moments about the hinge A should be zero.
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(i)
692
•
Strength of Materials
Pa
Ps
Hinge A
40 cm
100 cm
R
40 cm
50 kN
Figure 30.3
Then,
ΣM=0
Pa × 40 + Ps × 140 = 50 × 100
140 50 × 100
Pa + Ps ×
=
40
40
∴ Pa + 3.5 Ps = 125
(ii)
Pa and Ps are in kN.
If we consider the deformations of the aluminium bar and the wire. The deformed condition of
the system is shown in Fig. 30.4.
Pa
40 cm
100 cm
40 cm
δa
δs
R
Ps
Figure 30.4
From Fig. 30.4, we get
δs
δa
=
140
40
140
δa
∴ δs =
40
∴
or, δ s = 3.5 δ a
Pa la
Ps ls
= 3.5
As Es
Aa Ea
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Miscellaneous Solved Problems: Stresses and Strains
•
693
Substituting,
20
Pa ×
× 103
Ps × 1 × 103
100
= 3.5
0.6
50
× 200 × 109
× 60 × 109
100 × 100
100 × 100
∴
Pa = 35.7 Ps
(iii)
From Eqns. (i), (ii) and (iii), the values of Ps , Pa and R can be found out.
From Eqns. (ii) and (iii), we get
35.7 Ps + 3.5 Ps = 125
39.2 Ps = 125
∴
Ps =
125
39
= 3.205 kN
Hence, Pa = 35.7 × 3.205
= 114.42 kN
Stress in bar =
Pa
114.42 × 103
= 22.88 MN/m2
=
50
Aa
100 × 100
Stress in wire =
=
Ans
Ps
As
3.205 × 103
60
100 × 100
= 0.534 MN/m2
Ans
From the Eqn. (i)
Pa + 50 = Ps + R
114.42 + 50 = 3.205 + R
∴
R = 161.215 kN Ans
E XAMPLE 30.4: A rigid steel plate is supported by three vertical posts of 2 m height. But the
middle post is 0.5 mm less as shown in Fig. 30.5. The cross-sectional area of each post is 180 mm ×
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•
Strength of Materials
P
0.5 mm
2.0 m
A
B
C
Figure 30.5
180 mm. Determine the safe value of the load P if the permissible stress for concrete in compression
is 16 N/mm2 . Take E for concrete as 12 kN/mm2
S OLUTION :
Allowable stress in concrete, σ = 16 N/mm2 . Due to the action of load P, there will be less
strain or less stress in the middle post B. Therefore, stress in the outer posts A and C = 16 N/mm2 .
Now,
16
σ
× 2000 −
× 2000 = 0.5
Econ
Econ
32000 − 2000 σ = 12000 × 0.5 = 6000
2000 σ = 26000
∴
σ = 13 N/mm2
Safe load,
P = 2 × 180 × 180 + 180 × 180 × 13
= 1036800 + 421200
= 1458000 N
= 1458 kN Ans
E XAMPLE 30.5: A composite bar of steel and copper is shown in Fig. 30.6. Determine the compressive force developed in the bars after the rise in temperature by 90◦ C. Also find the change of
length in the copper bar. The area of cross section of copper bar is 650 mm2 and that of steel bar is
1100 mm2
Ecu = 105 GPa,
Es = 200 GPa
αcu = 18 × 10−6 /◦ C,
σs = 11 × 10−6 /◦ C
Clearance between support and composite bar is 0.6 mm.
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Miscellaneous Solved Problems: Stresses and Strains
0.6 mm Copper
Steel
0.5 m
0.5 m
Figure 30.6
S OLUTION :
Free expansion in bars due to temperature rise,
= 18 × 10−6 × 90 × 500 + 11 × 10−6 × 90 × 500
= 0.81 + 0.495
= 1.305 mm
Temperature rise = 90◦ C
Clearance = 0.6 mm
Contraction in length of bars = 1.305 − 0.6
= 0.705
Compressive force in the copper bar,
0.705 =
500
P
500
P
×
+
×
650 105000 1100 200000
P = 73445 N
= 73.445 kN (Compressive)
Ans
Change in length of the copper bar,
= 18 × 10−6 × 90 × 500 −
73445
500
×
650
105000
= 0.81 − 0.538
= 0.272 mm
Ans
@seismicisolation
@seismicisolation
•
695
696
•
Strength of Materials
E XAMPLE 30.6: A rod having constant cross section is tightly clamped at both ends as shown in
Fig. 30.7. If the temperature is raised by 65◦ C, determine the stress in the rod.
Es = 200 GPa, Ec = 100 GPa
αs = 10 × 10−6 /◦ C, αc = 15 × 10−6 /◦ C
Copper
Steel
1.6 m
0.9 m
Figure 30.7
S OLUTION :
For free extensions,
δcu = αcn lcu × t
= 15 × 10−6 × 900 × 65
= 0.8775 mm
δs = αs × ls × t
= 10 × 10−6 × 1600 × 65
= 10.4 mm
Total extension = 0.8775 + 1.04
= 1.9175 mm
If P is the compressive force in bar
Acu = As = A
P lcu
ls
δ=
+
A Ecu Es
1600
900
1.9175 = σ
+
100000 200000
1.9175 = σ 9 × 10−3 + 8 × 10−3
1.9175 = 17 × 10−3 σ
∴
σ = 112.79 MN/mm2
@seismicisolation
@seismicisolation
Ans
Miscellaneous Solved Problems: Stresses and Strains
•
697
Complex Stress and Strain
E XAMPLE 30.7: A point in a strained component has the following stresses as shown in Fig. 30.8.
Determine:
a) Principal stresses and principal planes
b) Maximum shear stress with its orientation.
140 MPa
70 MPa
80 MPa
P
Q
80 MPa
70 MPa
R
S
140 MPa
Figure 30.8
S OLUTION :
σx = −80 MPa, σy = 140 MPa, τ = 70 MPa
Principal stress = σ1,2 =
σx + σy
±
2
−80 + 140
±
2
√
= 30 ± 6700
=
σx + σy 2
+ τ2
2
(−80 + 140)2
+ 702
2
= 30 ± 81.85
= −51.85
= 51.85 MPa
(Compressive)
Location of principal plane,
tan 2θ =
=
2τ
σx − σy
140
2 × 70
=−
−80 − 140
320
= −0.636
tan(180 − 2θ ) = −(−0.636)
tan(180 − 2θ ) = 0.636
@seismicisolation
@seismicisolation
Ans
698
•
Strength of Materials
180 − 2θ = 32.46◦
2θ = 147.54
θ1 = 73.77◦
Ans
θ2 = θ1 + 90 = 73.77 + 90◦ = 163.77 Ans
σ1 − σ2
2
111.85 − (−51.85)
=
2
= 81.85 MPa Ans
Maximum shear stress, τmax =
Location of planes having maximum shear stress,
θ3 = 73.77 + 45 = 118.77◦
Ans
θ4 = 163.77 + 45 = 208.77◦
Ans
Shear Force and Bending Moment Diagram
E XAMPLE 30.8: Draw shear force and bending diagrams for the beam shown in Fig. 30.9 and
indicate the values at important points.
6 kN
B
A
2m
7 kN
C
2m
E
D
2m
1m
Figure 30.9
Taking moments about A,
6 × 4 + 7 × 7 = 14 + RD × 6
24 + 49 = 14 + 6RD ∴ RD = 9.8 kN
RA = 6 + 7 − RD = 13 − 9.8 = 3.2 kN
∴ RA = 3.2 kN
SF Calculations
SF at A = 0 rising to 3.2 kN
SF at B = 3.2 kN
SF at C = 3.2 kN falling to 3.2 − 6 = −2.8 kN
SF at D = −2.8 kN rising to − 2.8 + 2.8 = 7.0 kN
SF at E = 7 kN finally to zero
@seismicisolation
@seismicisolation
Miscellaneous Solved Problems: Stresses and Strains
•
699
BM Calculations
at A = 0
at B = 3.2 × 2 = 6.4 at left
6.4 − 14 = 7.6 kNm at right
BM at C = 3.2 × 4 − 14 = −1.2 kNm
BM at D = 3.2 × 6 − 14 − 6 × 2 = −6.8
BM at E = 3.2 × 7 − 14 − 6 × 3 + 9.8 × 1 = 0.2 0
BM
BM
6 kN
B
A
3.8 kN
7 kN
C
D
E
9.8 kN
2m
2m
2m
(a)
1m
7 kN
3.2
o
o
2.8 kNm
(b)
2.8 kNm
SFD
6.4 kNm
o
o
1.2 kNm
7.6 kNm
(c)
6.8 kNm
BMD
Figure 30.10
Bending Stresses
E XAMPLE 30.9: A timber beam 160 mm wide and 200 mm deep is reinforced by bolting on two
steel flitches each 160 mm × 12 mm in section. Find the moment of resistance when (a) flitches are
attached symmetrically at the sides and (b) the flitches are attached symmetrically at the top and
bottom. Permissible stress in timber is 8 N/mm2 . What is the maximum stress in steel in each case?
Take Es = 20Et
@seismicisolation
@seismicisolation
700
•
Strength of Materials
S OLUTION :
160
mm
160
mm
200 mm
12 mm
160 mm
Figure 30.11
a) Let stress in timber σt = 8 N/mm2 at 100 mm from N.A.
Stress in timber at 80 mm from N.A,
80
× 8 = 6.4 MPa
100
Maximum stress in steel, σs = mσw = 20 × 6.4
σt1 =
= 128 MPa
∴
Moment of inertia, It =
=
bd 3
13
160 × 200 3
= 106.7 × 106 mm4
12
Is = 2
td 3 2 × 12 × 1603
=
= 8.2 × 106 mm4
12
12
Moment of resistance = Ms + Mt
=
σs
σt
× Is +
× It
y
y
=
128
8
× 8.20 × 106 +
× 106.7 × 106
100
80
= 10.5 × 106 + 10.67 × 106
= 21.17 × 106 mm
(b) If the stress in timber, σt = 8 MPa
Then stress in steel σs = 20 × 8 = 160 MPa
@seismicisolation
@seismicisolation
Ans
Miscellaneous Solved Problems: Stresses and Strains
160 mm
12 mm
200 mm
12 mm
160 mm
Figure 30.12
For maximum stress in steel,
160
σs
=
112 100
σs = 179.2 MPa
Moment of Inertia:
MOI of timber, It =
MOI of steel, Is ,
160 × 2003
= 106.7 × 106 mm4
12
=2
12
160 × 123
+ (160 × 12)(100 + )2
12
2
= 2 [276480 + 21573]
= 43699200 mm4
= 43.7 × 106 mm4
Moment of resistance = Mt + Ms
=
σt
σs
× It + × Is
yt
ys
=
8
179.2
× 106.7 × 106 +
× 43.7 × 106
100
112
= 8.536 × 106 + 69.92 × 106
= 78.456 × 106 Nmm
Ans
@seismicisolation
@seismicisolation
•
701
702
•
Strength of Materials
Shear Stresses
E XAMPLE 30.10: For Fig. 30.13, evaluate the maximum shear stress if it is subjected to a shear
force of 120 kN.
A
B
110 mm
25 mm
D
14.88 N/mm2
T
K
x
40.56 N/mm2
M P
25 mm
L
90 mm
C
Figure 30.13
S OLUTION :
For this section it is obvious that maximum shear stress will develop at the centre, i.e., at M
and P
90 × 1103 π (50)4
−
12
64
= 9.98 × 106 − 0.307 × 106
I=
= 9.673 × 106 mm4
Now, Aȳ = Moment of area above N.A.
FAȳ
Shear stress, τ =
bI
110 − 50
Area A between KT and BA = 90 ×
2
90 × 30 = 2700 mm2
110 30
−
= 40 mm
2
2
120000 × 2700 × 40
At T, τ =
90 × 9.673 × 106
C.G. of this area from N.A; ȳ =
= 14.88 N/mm2
Aȳ = Moment of area above N.A.
= (Moment of area of rectangle 90 × 55 above N.A.) − (Moment of area of circular portion
between T.P. about N.A.)
@seismicisolation
@seismicisolation
Miscellaneous Solved Problems: Stresses and Strains
Aȳ = 90 × 55 ×
= 136125 −
= 136125 −
55
−
2
25
25
0
703
25
2xdy.y
0
2×
0
•
625 − y2 × y × dy ∵
x=
252 − y2
− 625 − y2 (−2y).dy
25
= 136125 +
0
625 − y2 (−2y).dy
625 − y2
= 136125 +
3/2
3/2 25
2
(625 − 252 )3/2 − (625 − 02 )3/2
3
2
= 136125 + −6251.5
3
2
= 136125 − × 15625
3
= 136125 +
= 136125 − 10416.6
= 125708.4 mm3
Now, τmax =
F.A.ȳ
bI
=
120000 × 125708.4
[90 − (2 × 25)] × 9.673 × 106
=
120000 × 125708.4
40 × 9.673 × 106
= 38.63 N/mm2
Ans
Torsion
E XAMPLE 30.11: A steel bar of 2.4 cm diameter is subjected to a torque of 310 Nm produces an
angle of 1.4◦ in a length of 22 cm. The same bar when subjected to tension elongates 0.012 cm in
length of 16 cm under a load of 75 kN. Determine the Poisson’s ratio for the material.
@seismicisolation
@seismicisolation
704
•
Strength of Materials
S OLUTION :
Bar diameter, d = 2.4 cm
Length, l = 22 cm
Torque, T = 310 Nm
Angle of twist θ = 1.4◦ =
1.4 × π
180
= 0.0244 radians
J=
π d 4 π (2.4)4
=
32
32
= 3.255 cm4 = 3.255 × 10−8 m4
We know,
T
Cθ
=
J
l
T.l
C=
Jθ
=
310 × 22
100 × 3.255 × 10−8 × 0.0244
= 8.587 × 109 N/m2
= 85.87 GPa
Cross-sectional area A =
(i)
π × 2.42
4
= 4.5216 cm2
= 4.5216 × 10−4 m2
E=
=
P.l
A dl
75000 × 16 × 100
4.5216 × 10−4 × 0.012 × 100
= 2.211 × 1011
= 221 GPa
@seismicisolation
@seismicisolation
(ii)
Miscellaneous Solved Problems: Stresses and Strains
•
705
E
2(1 + μ )
Substituting for C and E from Eqns. (i) and (ii),
We know, C =
85.87 =
221
;
2(1 + μ )
1+μ =
221
= 1.287
85.87 × 2
1 + μ = 1.287
∴
μ = 0.287 Ans
E XAMPLE 30.12: A shaft of diameter D is subjected to a combined bending moment M and
torsion T . The allowable stress in shear is 72% of the allowable stress in tension. If the factor of
safety for failure due to shear and failure due to maximum tensile stress are the same, determine the
ratio of M/T.
S OLUTION :
Let σb be the stress in bending and τs the maximum shear stress due to torsion.
M M(D/2) 32 M
=
=
I
π D3
(π D4 )
64
τ
T
=
R
J
τs
16T
=
(D/2)
π D4
)
(
32
16T
τs =
π D3
σb + σb2 + 4τs2
The maximum tensile stress is σmax =
2
32M
16T
32M 2
+4
+
π D3
π D3
π D3
=
2
16
=
(M + M 2 + T 2 )
π D3
σb =
And
or
But
τmax = 0.72σmax
@seismicisolation
@seismicisolation
706
•
Strength of Materials
∴
16T
16
= 0.72 ×
(M +
3
πD
π D3
M2 + T 2)
T = 0.72 M + 0.72 M 2 + T 2
T − 0.72M = 0.72 M 2 + T 2
or
Squaring both sides
T 2 + 0.52M 2 − 1.44T M = 0.52M 2 + 0.52T 2
T 2 − 0.52T 2 = 1.44T M
0.48T 2 = 1.44T M
M
= 0.333 Ans
T
or
Springs
E XAMPLE 30.13: Calculate the thickness and number of leaves of a laminated spring which is
required to support a central load of 3 kN on a span of 1.0 m, if the maximum stress is limited to
235 MN/m2 and central deflection to 82 mm. Breadth of each leaf can be assumed to be 110 mm.
Take E for spring material = 200 GN/m2
S OLUTION :
W = 3 × 103 N span, l = 1.0 m
Maximum stress σmax = 235 × 106 N/m2
E = 200 × 109 N/m2 , b = 110 mm = 110 × 10−3 m
Wl
3
Because, σmax = ×
2
nbt 2
235 × 106 =
3000 × 1
3
×
2 n × (110 × 10−3 )t 2
where t is thickness of a leaf.
nt 2 =
3000 × 1
3
1
×
×
2 110 × 10−3 235 × 106
nt 2 = 1.741 × 10−4
@seismicisolation
@seismicisolation
(i)
Miscellaneous Solved Problems: Stresses and Strains
3
Now central deflection is yc =
8
82 × 10−3 =
nt 3 =
W l3
nEbt 3
•
707
3
3000 × 13
×
8 n(200 × 109 )(110 × 10−3 )t 3
3000
3
×
8 200 × 109 × 82 × 10−3 × 110 × 10−3
nt 3 = 6.236 × 10−7
(ii)
Dividing (ii) by (i)
t=
∴
6.236 × 10−7
1.741 × 10−4
t = 3.582 × 10−3 m = 3.582 mm = 3.6 mm
Ans
Substituting value of t in Eqn. (i),
n=
1.741 × 10−4
= 13.56
(3.582 × 10−3 )2
The nearest whole number is 14 Ans
E XAMPLE 30.14: An open-coiled helical spring made of 12 mm diameter rod has seven free
coils each of 110 mm mean diameter. The ends of the spring are fastened to two discs kept 0.8 m
apart, which is the free length of the spring. Calculate the force on the discs, acting along the axis
of the spring, when one disc is rotated through 12◦ to coil. Take the spring E = 200 GN/m2 and
C = 80 GN/m2 .
S OLUTION :
0.8
= 0.114 m
7
0.114
tan α =
= 0.33
0.11π
Pitch of coils =
∴
∴
α = 18.26◦
l = 2π Rn sec α
= 2π × 0.055 × 7 ×
1
cos 18.26◦
= 2π × 0.055 × 7 × 0.9496
= 2.29 m
@seismicisolation
@seismicisolation
708
•
Strength of Materials
200
E
=
= 2.5C
C
80
And J = 2I ∴
We know
EI = 1.25 CJ
sin 2α 1
1
sin2 α cos2 α
+
+ MRl
−
=0
EI
CJ
2
CJ EI
2
sin2 ×18.26 1
1
sin 18.26 cos2 18.26
+
+M×
−
=0
W × 0.055
1.25CJ
CJ
2
CJ 1.25CJ
0.3132
1
0.5951
W × 0.055
+ 0.94962 + M ×
1−
=0
1.25
2
1.25
δ = W R3 l
or
or
0.055 W {0.0784 + 0.9} + M × 0.298 × 0.2 = 0
4.312 × 10−3 W + 0.0495 W + 0.0595 M = 0
0.0538 W = −0.0596 M
0.0538
W
0.0596
or
M=−
or
M = −0.903 W
Now,
φ = Ml
= 12 ×
or
cos2 α sin2 α
+
EI
CJ
sin2 α
+W Rl
2
1
1
−
CJ EI
π
180
2
Wl
cos 18.26
2
−0.903
+ sin 18.26
CJ
1.25
1
sin 2 × 18.26
1−
= 0.2093
+0.055 ×
2
2.5
W × 2.29
× [−0.903 {0.721 + 0.0982}
π
80 × 109 ×
× 0.0124
32
+ {0.055 × 0.0298 × 0.6}] = 0.2093
0.122 W × [−0.7397 + 0.009834] = 0.2093
− 0.122 W × 0.073 = 0.2093
W = −23.5 N (Compressive)
Ans
@seismicisolation
@seismicisolation
Miscellaneous Solved Problems: Stresses and Strains
•
709
E XAMPLE 30.15: A composite spring has two close-coiled helical steel springs in series; each
spring has a mean diameter of 8 times the diameter of its wire. One spring has 20 coils and wire
diameter of 2.5 mm. Find the diameter of the wire of the other spring, if it has 15 coils and the
stiffness of the composite spring is 1.5 kN/m.
Find the greatest axial load that can be applied to the spring and the corresponding extension
for a maximum shearing stress of 310 MN/m2 C = 80 GN/m2 .
S OLUTION :
Figure 30.14 shows the arrangement of the two springs, which are subjected to an axial load W .
d1 = 0.0025 m,
∴
S1 =
=
Total extension, δ =
∴
Effective stiffness =
W
W W
+
S1 S2
1.5 =
=
∴
D1 = 0.02 m
Cd14
W
=
d1 8D31 n1
80 × 109 × 0.00254
= 2.44 kN/m
8 × 0.02 × 20
W W
+
S1 S2
S1 S2
S1 + S2
2.44 S2
2.44 + S2
3.66 + 1.5 S2 = 2.44 S2
S2 =
∴
3.66
= 3.894 kN/m
0.94
2.57 × 103 =
80 × 109 × d24
8 × (8d2 )3 × 15
from which, d2 = 0.001975 m
8W D 64W
Also, τ =
=
Since D = 4d
π d3
π d2
The maximum shear stress therefore occurs in spring (2)
310 × 106 =
δ=
64W
;
π × 0.0019752
∴
W = 59.33 N
W
59.33
= 0.0395 m
=
S
1.5 × 103
@seismicisolation
@seismicisolation
Ans
Ans
710
•
Strength of Materials
W
1
2
W
Figure 30.14
Application of Castigliano’s Theorem
E XAMPLE 30.16: The ring shown in Fig. 30.15 is made of a flat steel strip 20 mm × 3 mm and is
shaped in the form of a circle of mean diameter 0.2 m. The ends at B are cut square and not joined. A
pull P is applied along the diameter CD which is at right angles to the diameter AB. If the maximum
tensile stress due to P is 125 MN/m2 , find the increase in the opening at B due to P. E = 200 GN/m2 .
3 mm
P
P
m
0m
20
A
C
C
dθ dθ
B
A
θ
Q
θ
B
D
P
Figure 30.15
Figure 30.16
S OLUTION :
The increase in opening at B is twice the movement of B relation to A in the half ring shown in
Fig. 30.16. Since there is no force at B in the direction of the required deflection, a force Q, of zero
magnitude, must be applied there.
@seismicisolation
@seismicisolation
Miscellaneous Solved Problems: Stresses and Strains
•
711
The required deflection is given by
δ =2
2
∂ UAB
=
∂Q
EI
M
∂M
dx
∂Q
For BC, taking the origin at B,
M = Qr(1 − cos θ )
∂M
= r(1 − cos θ ) and dx = rd θ
∂Q
For AC, taking the origin at A,
M = Qr(1 + cos θ ) + Pr cos θ
∂M
= r(1 + cos θ ) and dx = rd θ
∂Q
π /2
2
∴ δ=
Qr(1 − cos θ ) × r(1 − cos θ ) × rd θ
EI 0
+
π /2
0
2Pr3
=
EI
=
{Qr(1 + cos θ ) + Pr cos θ } × r(1 + cos θ ) × rd θ
π /2
0
cos θ (1 + cos θ )d θ · · · since Q = 0
2Pr3
π
(1 + )
EI
4
The maximum stress occurs at A, where M = Pr,
∴
Pr = σ Z
= 125 × 106 ×
∴
∴
0.02 × 0.0032
6
= 3.75 Nm
3.75
P=
= 37.5 N
0.10
2 × 37.5 × 0.103
π
1+
δ=
3
4
0.02 × 0.003
200 × 109 ×
12
= 0.0149 m or 14.9 mm Ans
E XAMPLE 30.17: Obtain the deflection under a concentrated load of 65 kN applied to a simply
supported beam has shown in Fig. 30.17. Use Castigliano’s theorem. EI = 2.4 MN/m2 .
@seismicisolation
@seismicisolation
712
•
Strength of Materials
65 kN
X
B
A
1m
3m
x
X
Figure 30.17
For purpose of calculation let us replace load 65 kN at C by load W .
RA × 4 = W × 1
W
RA =
4
Let there be a section X − X at x m beyond load W . Then,
Mx = RA × x −W × (x − 3)
Wx
−W (x − 3)
=
4
∂M
x
= − (x − 3)
∂W
4
M ∂M
·
· dx
EI ∂ W
1 3 Wx x
1 4 Wx
x
=
× dx +
−W (x − 3) ×
− (x − 3) dx
EI 0 4
4
EI 3
4
4
3
4
2
W
W
x
− (x − 3) dx
=
x2 dx +
16EI 0
EI 3 4
3
W 4 x − 4x + 12 2
W
2
x dx +
dx
=
16EI 0
EI 3
4
Again δ =
=
W
16EI
3
0
x2 dx +
9W
16EI
4
3
x2 − 8x + 16 dx
3 3
9W
x
W
+
[12.33 − 28 + 16]
=
16EI 3 0 16EI
0.75W
9
(1 + 0.33) =
, Now substituting for W & EI
16EI
EI
0.7 × 65 × 103
δ=
= 0.0203 m = 20.3 mm Ans
2.4 × 106
=
E XAMPLE 30.18: A steel bar of constant section moment of area I is bent as shown in Fig. 30.18
and fixed at one end. Find the horizontal and vertical deflections at the free end.
@seismicisolation
@seismicisolation
Miscellaneous Solved Problems: Stresses and Strains
P
C
a
l
A
B
Figure 30.18
Horizontal δ :
∂M
=x
M = Px,
∂P
1 a 2
δbc =
Px dx
EI 0
=
δab =
=
Pa3
3EI
1
EI
l
Pa2 dx
0
Q
Pa2 l
P
EI
Q is introduced as zero force.
Figure 30.19
Total horizontal deflection = δbc + δab
=
Pa2 Pa2 l
+
3EI
EI
=
Pa2 a
+l
EI 3
Ans
Vertical deflection δ ,
BC : it is zero
AB :
M = Pa + Qx,
δab =
1
EI
δab =
1
EI
=
l
0
dM
=x
dQ
(Pa + Qx) x · dx
l
Pal 2
2EI
Pa xdx
0
Ans
@seismicisolation
@seismicisolation
•
713
714
•
Strength of Materials
Moment of Inertia
E XAMPLE 30.19: Derive expression for moment of inertia of a cirle about X − X axis passing
through centroid.
A
B
dy
C
y
o
X
θ
θ
X
R
Figure 30.20
S OLUTION :
Consider an elementary strip of thickness dy as shown in Fig. 30.20. This strip is at a distance
y from X − X axis.
We have: BC = OC cos θ = R cos θ
∴
Length of strip = AC = 2BC
= 2R cos θ
Area of strip = 2R cos θ × dy
But y = r sin θ
Differentiating both sides, dy = R cos θ d θ
∴
Area of strip = 2R cos θ × R cos θ d θ
= 2R2 cos2 θ · d θ
Moment of inertia of strip about X-axis,
= 2R2 cos2 θ d θ y2
= 2R2 cos2 θ × R2 sin2 θ
= 2R4 sin2 θ cos2 θ d θ
The moment of inertia of the whole circle about X-X axis is
IXX =
+π /2
−π /2
2R4 sin2 θ cos2 θ d θ
@seismicisolation
@seismicisolation
Miscellaneous Solved Problems: Stresses and Strains
Now sin 2θ = 2 sin θ cos θ
4
IXX = R
∴
sin2 2θ
= 2 sin2 θ cos2 θ
2
+π /2
sin2 2θ
−π /2
2
dθ
+π /2
1 − cos 4θ
1 − cos 4θ
dθ
= sin2 2θ
∵
=R
4
2
−π /2
R4
π sin 2π
π sin(−2π )
=
−
− − −
4
2
4
2
4
4
=
R4 π π
+
4 2 2
Because sin 2π = sin(−2π ) = 0
=
Now radius R =
π R4
4
D
2
4
D
π
2
=
4
∴
IXX =
π D4
64
Ans
or
Alternate Method
Y
dr Z
X
r
X
R
Z
Y
Figure 30.21
Consider an elementary ring at radius r and of thickness dr.
Area of elementary ring = 2π r · dr
@seismicisolation
@seismicisolation
•
715
716
•
Strength of Materials
Moment of inertia of this elementary ring about central axis, i.e., Z-Z axis. = 2π r dr · r2 =
2π r3 dr
Total moment of inertia about Z axis
R
=
0
= 2π
2π r3 dr
r4
4
R
0
2π R4 π R4
=
=
4
2
(i)
Substituting R = D/2 in Eqn. (i)
4
D
2
=
2
π D4
=
32
π
IZZ
IZZ
Also IXX + IYY = IZZ
Since IXX = IYY being symmetrical area
2IXX =
π D4
32
or IXX =
π D4
64
Ans
Strain Energy
E XAMPLE 30.20: A bar of 120 cm in length is subjected to an axial pull, such that the maximum
stress is equal to 160 MN/m2 . Its area of cross section is 2.5 cm2 over a length of 120 cm except for
the middle 6 cm length, where the area of cross section is 1.2 cm2 . If E = 200 GN/m2 , calculate
the strain energy stored in the bar.
S OLUTION :
2
1.2 cm2
1
P
2.5 cm2
2
P
6 cm
120 cm
Figure 30.22
2 and central portion of length as 1.
Portions on left and right are marked as @seismicisolation
@seismicisolation
Miscellaneous Solved Problems: Stresses and Strains
•
717
Maximum stress will be at lowest are, i.e., 1.2 cm2
σ1 = 160 MN/m2
Because σ1 A1 = σ2 A2
160 × 1.2 × 10−4 = σ2 × 2.5 × 10−4
∴
σ2 = 76.8 MN/m2
Strain energy stored in the bar,
σ12 A1 l1 σ22 A2 l2
+
2E
2E
2 2
160 × 106 × 1.2 × 10−4 × 6 × 10−2
76.8 × 106 × 2.5 × 10−4 × 114 × 10−2
=
+
2 × 200 × 109
2 × 200 × 109
U=
2 = 120 − 6 = 114 cm]
[∵ Total portion of = 0.4608 + 4.2025
= 4.6633 Nm or J
Ans
E XAMPLE 30.21: A vertical composite tie bar rigidity fixed at the upper end consists of a steel
rod of 15 mm diameter enclosed in a brass tube of 15 mm internal diameter and 25 mm external
diameter, each being 1.8 m long. Both are fixed together at the ends. The tie bar is suddenly loaded
by a weight of 7.5 kN falling through a distance of 5 mm. Determine the maximum stresses in the
steel rod and the brass tube. Take Es = 200 GPa & Eb = 100 GPa.
S OLUTION :
Refer Fig. 30.22.
π 2
15 = 176.625 mm2
4
π
Ab = (252 − 152 ) = 314 mm2
4
As =
Let x = Extension in mm
σs =
Es x
l
and σb =
@seismicisolation
@seismicisolation
Eb x
l
718
•
Strength of Materials
Steel rod
Brass tube
1.8 m
7.5 kN
5 mm
15 mm
25 mm
Figure 30.23
Strain energy of the bar,
σ2
σs2
As l + b Ab l
2Es
2Eb
2
2
E2
E x
= 2s
As l + 2 b Ab l
l · 2Es
l · 2Eb
Eb x2
Es x2
=
As +
Ab
2l
2l
x2
= (Es As + Eb Ab )
2l
x2
=
(200000 × 176.625 + 100000 × 314)
2 × 1800
x2
(35325000 + 31400000)
=
3600
= 18534.7x2 Nmm
=
(i)
Potential energy lost by weight,
= W (h + x) = 7500(5 + x) Nmm
Equating Eqns. (i) and (ii)
18534.7x2 = 7500(5 + x)
18534.7x2 − 7500x − 37500 = 0
x2 − 0.405x − 2.02 = 0
x=
+(0.405) ±
@seismicisolation
@seismicisolation
(−0.4052 − 4 × 1 × (−2.02))
2
(ii)
Miscellaneous Solved Problems: Stresses and Strains
•
719
√
+0.405 ± 0.164 + 8.08
=
2
=
+0.405 ± 2.87
2
= 1.6375 mm
σs =
Es x 200000 × 1.6375
=
= 181.91 MPa
l
1800
σb =
Eb x 100000 × 1.6375
=
= 90.97 MPa
l
1800
Ans
Ans
Theories of Failure
E XAMPLE 30.22: A bolt has an axial pull of 15 kN together with a transverse shear force of 8 kN.
Find the diameter of the bolt required according to:
i)
ii)
iii)
iv)
v)
Maximum principal stress theory
Maximum shear stress theory
Maximum principal strain theory
Maximum strain energy theory
Maximum distortion energy theory.
Take allowable tensile stress at elastic limit = 120 MPa and Poisson’s ratio = 0.3
S OLUTION :
Pt = 15 kN, Ps = 8 kN, σt at elastic limit = 120 MPa, μ = 0.3
Let d be the diameter of bolt in mm. Then cross-sectional area of the bolt, A =
= 0.785d 2 mm2
Axial tensile stress,
σ1 =
Pt
19.1
15
= 2 kN/mm2
=
A
0.785d 2
d
and transverse shear stress,
τ=
8
Ps
10.19
=
=
kN/mm2
A
0.785d 2
d2
(i) Maximum principal stress theory:
σ1 + σ2 1
2
2
+
σt1 =
(σ1 − σ2 ) + 4τ
2
2
σ1 1
2
2
(σ1 ) + 4τ
+
=
2
2
@seismicisolation
@seismicisolation
π d2
4
720
•
Strength of Materials
Because σ2 = 0
⎡
2
=
19.1 1 ⎣
+
2d 2 2
=
1 √
9.55
+ 2
364.81 + 415.34
2
d
2d
=
9.55 13.97 23.52
+ 2 =
kN/mm2
d2
d
d2
=
23520
N/mm2
d2
19.1
d2
+4
10.19
d2
2
According to the maximum principal stress theory,
σt at elastic limit
2
23520 120
=
2
d2
d = 19.8 mm Ans
τmax =
or
or
(ii) According to the maximum shear stress theory,
1
2
2
τmax =
(σ1 − σ2 ) + 4τ
2
1
2
2
(σ1 ) + 4τ
=
2
Because σ2 = 0
⎡
⎤
10.19 2 ⎦
19.1 2
1⎣
+4
=
2
d2
d2
1
= 2
2d
19.12 + 4(10.19)2
=
1 √
364.81
+
415.34
2d 2
=
13.97
kN/mm2
d2
=
13970
N/mm2
d2
@seismicisolation
@seismicisolation
⎤
⎦
Miscellaneous Solved Problems: Stresses and Strains
According to the maximum shear stress theory,
σt at elastic limit
2
13970 120
=
2
d2
∴ d = 15.26 mm Ans
τmax =
(iii) According to the maximum principal strain theory,
σ1 1
+
σt1 =
(σ )2 + 4τ 2
Maximum
2
2
=
and minimum
σ1 1
−
σt2 =
2
2
23520
d2
( σ1
⎡
=
19.1 1 ⎣
−
2d 2 2
=
1
9.55
− 2
2
d
2d
[already calculated in (i)]
)2 + 4τ 2
19.1
d2
2
+4
10.19
d2
2
(19.1)2 + 4(10.19)2
1 √
9.55
−
364.81
+
415.34
d2
2d 2
9.55 13.97
= 2 −
2
d
4420
4.42
= − 2 kN/mm2 = − 2 N/mm2
d
d
=
We know that according to the maximum principal strain theory,
σt1 μσ2 σt at elastic limit
−
=
E
E
E
or σt1 − μσ2 = σt at elastic limit
∵
or
d = 14.38 mm
23520 (−4420) × 0.3
−
= 120
d2
d2
23520 1326
+ 2 = 120
d2
d
24846
= 120
d2
Ans
@seismicisolation
@seismicisolation
⎤
⎦
•
721
722
•
Strength of Materials
(iv) According to the maximum strain energy theory,
(σt1 )2 + (σt2 )2 − 2μσt1 σt2 = (σt at elastic limit)2
23520 2
−4420 2
23520
−4420
= 1202
+
− 2 × 0.3 ×
×
d2
d2
d2
d2
553190400 19536400 62375040
+
+
= 14400
d4
d4
d4
635101840
= 14400
d4
∴ d = 14.49 mm
Ans
(v) According to the maximum distortion theory,
(σt1 )2 + (σt2 )2 − 2σt1 σt2 = (σt at elastic limit)2
23520 2
−4420 2
23520
−4420
= 1202
+
−
2
×
×
d2
d2
d2
d2
553190400 19536400 207916800
+
+
= 14400
d4
d4
d4
780643600
= 14400
d4
∴ d = 15.26 mm
Ans
E XAMPLE 30.23: A compound tube is made of a tube 300 mm internal diameter of 30 mm
thick shrunk on a tube of 250 mm internal diameter. The interface radial pressure at the junction
is 12 N/mm2 . The compound tube is subjected to an internal pressure of 75 N/m2 . Determine the
variation of hoop stress over the wall of the compound tube.
S OLUTION :
r2 = 125
r0 = 150
r1
=1
80
mm
Figure 30.24
@seismicisolation
@seismicisolation
Miscellaneous Solved Problems: Stresses and Strains
•
723
Interface radial pressure p0 = 12 N/mm2
Stress due to shrinkage:
Outer tube:
At radius r0 = 150
σr = 12 =
b
−a
1502
(i)
σr = 0 =
b
−a
1802
(ii)
At radius r1 = 180 mm
Subtracting Eqn. (ii) from Eqn. (i)
12 =
b
b
−
1502 1802
12 =
b
b
−
22500 32400
12 =
b(32400 − 22500)
22500 × 32400
{∴
a is cancelled}
12 = 1.358 × 10−5 b
∴ b = 883652.4
For A, from Eqn. (ii)
a=
b
1802
a=
883652.4
32400
= 27.27
At radius r1 = 180 mm,
σc =
=
b
+a
1802
883652.4
+ 27.27
32400
= 54.54 N/mm2
At radius r0 = 150 mm,
σc =
b
+a
1502
@seismicisolation
@seismicisolation
(tensile)
724
•
Strength of Materials
=
883652.4
+ 27.27
22500
= 66.54 N/mm2
(Tensile)
Inner tube:
At radius r0 = 150 mm
σr = 12 =
b
−a
1502
(iii)
σr = 0 =
b
−a
1252
(iv)
At radius r2 = 125 mm
Subtracting Eqn. (iv) from Eqn. (iii),
12 =
=
12 =
b
b
−
2
150
1252
b
b
−
22500 15625
b(15625 − 22500)
22500 × 15625
12 = −
6875b
351562500
∴ b = −613636.4
From Eqn. (iv)
−613636.4
−a
15625
a = −39.27
0=
And at radius r0 = 150,
σc =
=
b
+a
1502
−613636.4
− 39.27
22500
= −66.54 N/mm2
(Compressive)
At radius r2 = 125 mm,
σc =
−613636.4
− 39.27
15625
= −78.54 N/mm2
(Compressive)
@seismicisolation
@seismicisolation
Miscellaneous Solved Problems: Stresses and Strains
•
725
Stresses due to internal fluid pressure:
At radius r2 = 125
σr = 75 =
b
−a
1252
(v)
σr = 0 =
b
−a
1802
(vi)
At radius r1 = 180
Subtracting Eqn. (vi) from Eqn. (v),
b
b
−
1252 1802
b
b
−
75 =
15625 32400
b(32400 − 15625)
75 =
15625 × 32400
75
∴ b=
3.3136 × 10−5
75 =
b = 2263399.3
Using Eqn. (vi)
a=
2263399.3
= 69.86
32400
∴ a = 69.86
At radius r1 = 180 mm
b
+a
1802
2263399.3
+ 69.86
=
32400
σc =
= 139.72 N/mm2
(Tensile)
At radius r0 = 150 mm,
σc =
2263399.3
+ 69.86
1502
= 170.45 N/mm2
(Tensile)
At radius r2 = 125 mm,
σc =
2263399.3
+ 69.86
1252
= 214.72 N/mm2
@seismicisolation
@seismicisolation
(Tensile)
726
•
Strength of Materials
+ for tensile and − for compressive
Inner tube
Radii
125 mm 150 mm
σc due to shrinkage (N/mm2 )
−78.54
−66.54
σc due to inner pressure (N/mm2 ) +214.72 +170.45
RESULTANT σc N/mm2
+136.18 +103.91
Outer tube
180 mm 150 mm
+54.54
+66.54
+139.72 +170.45
+194.26 +236.99
Rotating Discs and Cylinder
E XAMPLE 30.24: A disc of uniform thickness and 625 mm diameter rotates at 2000 r.p.m. Determine the maximum stress developed in the disc. If a hole of 100 mm diameter is made at the centre
of the disc. Find the maximum values of radial and hoop stresses.
Density of the material = 7700 kg/m3 and μ = 0.27
S OLUTION :
625
2π × 2000
R=
= 372.5, w =
= 209.33 rad/s, ρ = 7700 kg/m3 and μ = 0.27.
2
60
Maximum radial stress and hoop stress are at the centre and equal.
3+μ 2 2
ρw r
8
3 + 0.27
=
× 7700 × (209.33)2 (0.3125)2
8
= 13.4682597 N/m2
σr = σc =
= 13.47 MN/m2
100
= 50 mm = 0.05 m r0 = 0.3125 m
2 √
Maximum radial stress is at ri2 · ro2 radius, i.e. 0.052 × 0.31252 = 0.015625 m
With the hole of radius ri =
3+μ 2
ρ w (r0 + ri )2
8
3 + 0.27
× 7700 × 209.332 (0.3125 + 0.05)
=
8
σr =
= 137914979 × 0.3625
= 49994179.9 N/m2
= 49.99 MN/m2
Ans
Maximum hoop stress is at the inner radius,
3+μ 2 2 1−μ 2
σc =
ρ w r0 +
r
8
3+μ i
@seismicisolation
@seismicisolation
Miscellaneous Solved Problems: Stresses and Strains
•
727
1 − 0.27
3 + 0.27
2
2
2
× 7700 × 209.33 0.3125 +
× 0.05
=
8
3 + 0.27
= 137914979 0.0977 + 5.58 × 10−4
= 13551250 N/m2 = 13.55 MN/m2
Ans
Bending of Curved Bars
E XAMPLE 30.25: A curved bar of rectangular section of 25 mm width and 42 mm depth and mean
radius of curvature of 65 mm is initially unstressed. If a bending moment of 480 Nm is applied to
the bar which tends to straighten it, determine the stresses at the inner and outer surfaces. Also find
the position of neutral axis.
S OLUTION :
2R + d
R3
− R2
h = ln
d
2R − d
2 × 65 + 42
623
ln
− 652
=
42
2 × 65 − 42
172
= 6538.7ln
− 4225
88
2
= 6538.7 × 0.67 − 4225
= 155.93 mm3
Bending moment is negative as it tends to straighten the bar. y is negative at inside and positive
at outside.
Stress at outside force
M
y
R2
1+ 2 ·
AR
h R+y
652
21
−480 × 1000
1+
×
=
65 × (25 × 42)
155.93 65 + 21
569
= −7.033 1 +
86
= −53.56 MPa (Compressive)
σ0 =
@seismicisolation
@seismicisolation
728
•
Strength of Materials
Stress at inside force
R2
M
y
1− 2 ·
AR
h R−y
652
21
= −7.033 1 −
×
155.93 65 − 21
569
= −7.033 1 −
44
σi =
= 83.92 MPa (Tensile)
Position of neutral axis
y=
−Rh2
R2 + h2
=
−65 × 155.93
652 + 155.93
=
−10135.45
4069.07
= −2.491 mm
Fixed Beams
E XAMPLE 30.26: A fixed beam of 7 m span is loaded with concentrated loads of 200 kN at distance 2.5 m from each support. Draw the BM and SF diagrams. Find also the maximum deflection.
Take E = 200 GPa and I = 9 × 108 mm4
S OLUTION :
By equating the areas of free and fixed BM diagrams,
1
MA × 7 = (7 + 2) × 500
2
MA = 321.43 kNm
MB = 321.43 kNm
Position of contraflexure:
BM (actual) at any section in AC at a distance x from A is given by
M = Free moment − fixed moment
= 200x − 321.43
@seismicisolation
@seismicisolation
Miscellaneous Solved Problems: Stresses and Strains
200 kN
(a) Fixed Beam
200 kN
C
A
2m
2.5 m
MA
B
D
2.5 m
7m
fixing moments MA = MB due to symmetry
RA = RB = 200 kN
MB
BMD + C = 200 × 2.5 = 500 kNm
500 kNm
500 kNm
+
(b) Free BMD
A
(c) Fixed BMD
B
321.43
kNm
321.43
kNm
200 kN
200 kN
C
A
MA
2.5 m
B
D
2m
2.5 m
7m
MB
178.57 kNm
(d) Final BMD
321.43 kNm
500−321.43 = 178.57
200 kN
(e) SFD
200 kN
Figure 30.25
For point of contra-flexure equate M = 0
200x − 32143 = 0
x = 1.607 m from either end
@seismicisolation
@seismicisolation
•
729
730
•
Strength of Materials
Slope and Deflection: Take BM at x between A and D, then
M = EI
d2y
= 200x − 321.57 − 200(x − 2)
dx2
Integrating,
EI
x2
dy
200
= 200 − 321.57x +C1 −
(x − 2)2
dx
2
2
dy
When x = 0,
= 0 ∴ C1 = 0
dx
Integrating again,
EIy = 200
x2 200
x3
− 321.57 −
(x − 2)3 +C2
6
2
6
When x = 0, y = 0 C2 = 0
Hence,
EIy = 33.33x3 − 160.785x2 − 33.33(x − 2)3
To get maximum deflection which occurs at the centre. Putting x =
7
= 3.5 m
2
EIymax = 33.33(3.5)3 − 160.785(3.5)2 − 33.33(3.5 − 2)3
= 1429 − 1969.6 − 112.49
= −653.1
653.1
ymax = −
m
2 × 108 × 9 × 108 × 10−12
Note: Since we are working in kN so to convert 200 × 109 N/m2 into kN/m2 , divide it by 103 to
make it 2 × 108
∴ ymax = −3.63 × 10−3 m
= −3.63 mm Ans
Columns and Struts
E XAMPLE 30.27: A T -section 150 mm × 120 mm × 20 mm is used as a column of 4.5 m long
with hinged at both ends. Determine the crippling load. Take E = 200 GPa.
S OLUTION :
E = 200 GPa
@seismicisolation
@seismicisolation
Miscellaneous Solved Problems: Stresses and Strains
= 200 kN/mm2
= 200000 N/mm2
1 and 2
Let AB be the reference line and divide the area into two simple sections a1 = 100 × 20 = 2000 mm2
100
= 50 mm
y1 =
20
a2 = 150 × 20 = 3000 mm2
20
y2 = 120 −
2
= 110 mm, Total Area, A = 2000 + 3000 = 5000 mm2
2000 × 50 + 3000 × 110
ȳ =
5000
= 86 mm
150 mm
Y
20 mm
2
X
X
G
1
120 mm
20 mm
A
B
Y
Figure 30.26
IXX =
20 × (100)3
150 × (20)3
+ 2000(36)2 +
+ 3000 × (24)2
12
12
= 4.26 × 106 + 1.83 × 106 = 6.09 × 106 mm4
100 × (20)3 20 × (150)3
+
12
12
= 66666.7 + 5625000
IYY =
= 5691666.7 mm4
@seismicisolation
@seismicisolation
•
731
732
•
Strength of Materials
Now because 5691666.7 (IYY ) is less than 6090000 mm4 (IXX ). So we shall take IYY as the value
of I in formula (Euler’s). Also as the column is hinged at its both end, so length of columns,
Le = l = 4500 mm
Now, Euler’s crippling load,
π 2 EI
Le2
π 2 × 200000 × 5691666.7
=
45002
= 554247 N
= 554.25 kN Ans
ρE =
Combined Loading
E XAMPLE 30.28: For the inclined cantilever shown in Fig. 30.26, find the dimensions of the
rectangular section of the beam if the permissible maximum direct stress is 120 MN/m2 . Check for
the maximum shear stress which in this case should not exceed the value of 75 MN/m2 . Neglect the
additional bending moment due to deflection of end of cantilever. Take value of P = 130 kN.
S OLUTION :
Refer to Figs. 30.27 and 30.28.
√ Load P is resolved along the beam axis and perpendicular to it.
3
The resolved component
P produces bending and shear force, while the component 0.5 P
2
produces direct tension.
A
B 1m
b
√3 P
2
P
P
2
30°
P Se
at Action
B
Figure 30.27
@seismicisolation
@seismicisolation
2b
Figure 30.28
Miscellaneous Solved Problems: Stresses and Strains
•
733
Depth of beam = 2b (given).
P
M
y+
I
A
√
3
P×1
P
= 2
×b+
3
b × 2b
b(2b)
12
Maximum tensile stress in section =
120 × 106 =
=
0.433 × 130000 × 1 130000
+
0.667b3
2b2
0.433 × 130000 + 130000 × 0.3335b
0.667b3
80040000b3 = 56290 + 43355b
80040000b3 − 43355b − 56290 = 0
By trial b = 0.101 m
∴ b = 101 mm
depth = 2b = 202 mm
Ans
Check for shear stress:
0.101 × 0.2023
= 6.94 × 10−5 m4
12
0.101
130000 × 0.866 × 0.101 × 0.101 ×
2
=
0.101 × 6.94 × 10−5
= 8273.98 N = 8.27 kN which is quite safe Ans
τmax =
FAȳ
,
bI
I=
E XAMPLE 30.29: A shaft is subjected to torque T and a load F as shown in Fig. 30.29. Determine
the maximum shear stress on a cross section of the shaft. Diameter of shaft is 5 cm, torque T = 1.2
kNm and force F = 22 kN.
S OLUTION :
In Fig. 30.30, point A has maximum shear stress because both the shearing stresses are directed
downward.
Resultant shear stress at A = τ1 + τ2
Resultant shear stress at B = τ1 − τ2
T
4F
∴ Maximum stress = π
+
d 3 3A
16
4F
∵ τr =
3A
@seismicisolation
@seismicisolation
τr = modulus of rupture
734
•
Strength of Materials
A
T
τ1
τ1
τ1
τ1
A
2 cm
Torsional shear
stress
A
8 cm
Figure 30.29
∴ τmax =
B
τ2
τ 2 τ2
Direct shear
stress
Figure 30.30
16 × 1000 × 1.2
4 × 22000
+
π
π × (0.05)3
3 × (0.05)2
4
= 48917197.4 + 14946921.4
= 63864118.8
= 63.86 MN/m2
Ans
E XAMPLE 30.30: A shaft of 12 cm diameter is subjected to a bending moment of 1200 Nm and
a torque of 600 Nm. Find (a) the maximum normal stress on a section perpendicular to the axis (b)
the maximum shear stress on a section perpendicular to the axis and (c) the principal stresses.
S OLUTION :
(a) The maximum normal stress on a section perpendicular to the axis.
Let σb be the maximum normal stress
Now,
M
y
I
1200 × 64 0.12
since
×
=
2
π (0.12)4
σb =
π
(0.12)4 m4
64
0.12
= 7077140.835 N/mm2 , y =
2
= 7.08 MN/m2 Ans
I=
(b) The maximum shear stress on a section perpendicular to the axis.
Let τ be the maximum shear stress
We know,
T ·r
J
600 × 0.06
= π
(0.12)4
32
τ=
@seismicisolation
@seismicisolation
Miscellaneous Solved Problems: Stresses and Strains
= 1769285.2 N/mm2
= 1.77 MN/mm2
Ans
(c)
σb 2
1
σ1,2 = σ1,2 − σb ±
+ τ2
2
2
7.08 2
1
= × 7.08 ±
+ 1.772
2
2
√
= 3.54 ± 12.53 + 3.133
= 3.54 ± 3.96
= 7.5 MPa, −0.42 MPa
@seismicisolation
@seismicisolation
Ans
•
735
Index
A
Arc welding 626
Area moment method 361
B
Beams
of composite section 257
of uniform strength 136
with large radius of curvature 456
with unsymmetrical bending moment 493
deflection of 496
Bending moment 80
Bending stream 497
Bending under impact loads 315
Betti’s theorem of reciprocal deflections 312
Bow’s notation for graphical solution 560
Brinell hardener values 672
Brinell hardness testing machine 671
Brinell test 671
Buckling factor 425
Buckling load 425
Bulk modulus 43, 44
Butt joints 613
C
Cantilever with a,
concentrated load at its free end 235
concentrated load not at the free end 237
loaded from the free end 239
partially loaded with uniformly distributed
load 238
uniformly distributed load 237
gradually varying load 240
of concentrated end load 261
end couple 262
uniformly distributed load 262
Carriage springs 408
Centre of gravity 97
for solid bodies 99
Centroid 97
of different sections 101
Circumferential stress 196
Clapeyron’s equation of three moments 377
Close-coiled helical spring with,
axial couple 391
axial load 388
Columns and struts 424
Combined bending 142
Combined bending and twisting 343
Combined stresses 342
Composite action of axial load and couple 404
Composite beams 138
Composite shaft 185
Compound cylinders 219
Compression test 653
Compressive stress 4
Concentric (cluster) springs 397
Conjugate beam method 266
Continuous beams 377
Crippling load 425
Critical load 425
Cupping test 664
Curvature and strain, relationship 125
Curved bars, bending 456
Cylinders 520
@seismicisolation
@seismicisolation
738
• Index
D
Dams 591
Deficient frame 559
Deflection 234
due to bending 307
due to impact 248
Diamond riveting 621
Direct stresses 142
Disc of uniform strength 554
Distortion theory 327
Double integration method 235, 251
Hardness testing machines 671
Hollow cylinder 547
Hollow disc 527
Hoop stress 196
E
Eccentrically loaded welded joint 643
Elastic constants 40
Equivalent hardness number conversion table 677
Equivalent length 425
Erichsen test 664
Euler’s formula 446
Euler’s theory 437
L
Laminated spring 408
Lap joints 612
Large curvature 459
Leaf spring 408
Limit of proportionality 185
Lindley extensometer 652
Longitudinal stress 197
Longitudinal tensile stress 215
I
Impact test 661
J
Johnson’s parabolic formula 445
K
Knoop hardness number 680
F
Fatigue limit 653
Fillet welds,
under bending moment 636
under torsion 635
Fixed beam,
deflection for 362
uniformly distributed load over the span 359
with a point load at the centre 357
Flat spiral spring 405
Flexural rigidity 235
Flitched beams 138, 257
Fractured ends of specimens 652
Frameworks 559
G
Gas welding 626
Gordon’s formula 445
Graphical (Mohr’s) method 69
Guest or Teresa’s theory 321, 327
H
Haigh’s theory 323
Hardness scaler for steels 679
M
Macaulay’s method 251, 669
Material subjected to
direct and shear stresses 55
shear stresses 54
two perpendicular stresses 50
Materials, mechanical testing of 649
Maximum principal strain theory 322
Maximum principal stress theory 320
Maximum shear strain energy theory 324
Maximum shear stress theory 321, 327
Maximum strain energy theory 323, 327
Maxwell’s reciprocal theorem 311, 670
Metals and alloys, mechanical properties of 653
Metals, mechanical properties of 653, 659
Method of sections 580
Middle third rule for rectangular section 147
Modulus of elasticity 43, 44, 45
Modulus of resilience 291
Modulus of rigidity 43, 665
Modulus of rupture 659
Modulus of section 128
@seismicisolation
@seismicisolation
Index •
Mohr’s circle 494
Mohr’s circle of stresses 59
Mohr’s hardness scale 680
Mohr’s theorem 263
Moment area method 259
Moment of inertia 111
Moment of inertia of a,
circular section 114
hollow rectangular section 114
triangle 115
semi-circular lamina 116
rectangular section 113
Moment of resistance 126
Momental ellipse 495
Monotron hardness number 680
N
Non-ferrous alloys 679
O
Offset stress 6
Open-coiled helical spring 402
Over riveted joints 627
P
Perfect frame 559
Permanent set 6
Perry’s formula 449
Principal stress 47, 231
Principal strains 47
Product of inertia of rectangle 122
Proof resilience 291
Proof stress 6
Props 285
Q
Quarter-elliptic leaf spring 410
R
Radiant energy welding processes 627
Radius of curvature 234
Radius of gyration 112
Rankine’s formula 438
Rankine’s theory 320
Rectangular dams 591
Rectangular section 111
Resilience 291
Resistance welding 626
Rivet 610
efficiency of a 617
failure of a 615
material of 612
Rivet heads, types of 610
Riveted joints 610, 612
Rockwell hardness testing machine 673
Rotating disc of constant thickness 524
Rotating discs 520
Rotating long cylinder 542
Rotating ring 520
Rule of middle third 599
S
Secant formula 446
Semi-elliptic spring 408
Shear centre 507
for channel section 508
for unequal I section 510
Shear force diagram calculations 382
Shear stress 4, 231
in beams 157
principle of 43
Shear stress distribution,
for beam of rectangular section 159
in an I section 162
Shear tests 657
Shearing force 79
and bending moment diagrams 82
Shore scleroscope 676
Shrinkage allowance 222
Simply supported beam with,
a central point load 245
a concentrated load at the centre 266
a uniform distributed load 246
point load eccentric 248
with central concentrated load 262
Slenderness ratio 425
Slope 234
Small curvature 456
Small initial radius of curvature 459
@seismicisolation
@seismicisolation
739
740
• Index
Solid cylinder 545
Solid disc 526
Solid state welding 627
Springs 388
in series and parallel 396
St. Venant’s theory 322
Straight line formula 445
Strain 29
Strain energy method 278
Strain energy 291
due to bending 307
in pure shearing 305, 307
in torsion 306
Stress 4
at critical load for cast iron 445
at critical load for structural steel 445
on a oblique section 47
Stress-strain curve 5
Suddenly applied load 292
Superposition method 277
Thick spherical shells 229
Thin circular tube subjected to torsion 190
Torsion 175
of a round bar 667
of a tapering shaft 187
Torsional moment of resistance 176
Trapezoidal dams,
with water face inclined 603
with water face vertical 596
Turner’s sclerometer 680
U
Universal testing machine 651
Unsymmetrical bending 491
Unsymmetrical welded section 632
V
Vicker’s diamond test machine 674
Von mises and Henkey’s theory 324, 327
T
Temperature stress 29
Tensile stress 4
Theorem of parallel axis 113
Theorem of the perpendicular axis 113
Theories of elastic failure 320
Thermal stresses in a bar of tapering section 32
Thermo-chemical welding processes 627
Thick cylinders and spheres 214
W
Wahl’s correction factor 390
Welded joints 627
Welded joints 626
Welding,
classification of 626
Wire-bound thin cylindrical shells 205
Y
Young’s modulus 5
@seismicisolation
@seismicisolation
Strength ofofMaterials
Strength
Materials
978-93-89307-19-1
978-93-89307-31-3
Kaushik
Strength ofof Materials
Strength
Materials Kaushik
This book comprehensively covers all the major topics involving application of concepts of
strength
materials which covers
a mechanical
engineer
encounters
day. of
Structural
This
bookofcomprehensively
all the major
topics
involving every
application
conceptsand
of
machine
elements
which
come
under
the
purview
of
this
subject
and
covered
in
the
book and
are
strength of materials which a mechanical engineer encounters day to day. Structural
beams ofelements
all kinds, which
thin and
thickunder
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and spherical
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machine
come
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are:
and cylinders,
springs,
dams,
trusses
so on. Solid
parameters
covered
beams
of all kinds;
thinframes,
and thick
cylinders
andand
spherical
cells;mechanics
columns and
struts; bars;
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in the
book aresprings;
all typesframes;
of stresses
and
strains,
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moments ofcovered
inertia,
and
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dams;
trusses
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Solid mechanics
centre
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Various
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the book
are allelastic
the types
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coveredcentre
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dedicated chapter
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will
aid
inertia,
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elasticA constants,
and so on.
Various theories
of of
elastic
failure
have
students
in understanding
strength
of materials
is determined
in laboratory
been
covered
in a focusedhow
chapter.
A dedicated
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on mechanical
testing conditions.
of materials will
aid students in understanding how strength of materials is determined in laboratory conditions.
Learning Aids:
Aids:
ŸLearning
At the end
of the chapters, a large number (more than 300) of questions have been given.
•
At
the
end of the chapters a large number (more than 300) of questions have been
Ÿ The text is supplemented with ample line diagrams/pictures so that students can
given.
understand the concepts easily.
• The text is supplemented with line diagrams/pictures profusely so that the students can
Ÿ The topics
are discussed
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sequence of gradually increasing level of difficulty, making it
understand
the concepts
easier
for
students
to
follow
the
after building
appropriate
background
strongly. so that
• The topics are discussed intopics
a sequence
of gradually
increasing
level of difficulty,
Ÿ Relatively
difficult
like principal
stresses,
theory
failure, bending
of curved
bars,
it is easier
fortopics
the students
to follow
the topics
afterofbuilding
appropriate
background
thick strongly.
cylinders, strain energy, continuous beams, fixed beams, etc., have been treated with
• Relatively
difficult
topics
like is
principal
stresses,
theory of failure, bending of curved
special
care to see
that the
material
accessible
to all students.
bars, thick cylinders, strain energy, continuous beams, fixed beams, etc., have been
withand
special
care tothis
see book
that the
is accessible
to all
With its treated
coverage
approach,
willmaterial
be immensely
useful
to students.
students of B. Tech
(Mechanical Engineering) as well as that of A.M.I.E. (India). It will also be useful for competitive
With
its coverage
and approach,
this book
will be immensely useful to the students of B. Tech
examinations
like GATE,
IES and Civil
Services.
(Mechanical Engineering) as well as that of A.M.I.E. (India). It will also be useful for competitive
examinations
like[ GATE,
and Civil
Services.
R. K. Kaushik
D.M.E.,IES
Gr.I.E
(I), M.Tech
(Mech.), M.I.E, Chartered Engineer, Ph.D.] is
Professor, Department of Mechanical Engineering, Ganga Institute of Technology &
R. K. Kaushik
[ D.M.E.,
Gr.I.EHe
(I),was
M.Tech
(Mech.),
Chartered
Engineer,atPh.D.]
is
Management,
Kablana
- Jhajjar.
formerly
head ofM.I.E,
department
(Mechanical)
Haryana
Professor,
Department
of
Mechanical
Engineering,
Ganga
Institute
of
Technology
&
Institute of Technology. Dr Kaushik has overall 47 years of experience, out of which 28 years
Management,
Kablana
Jhajjar.
He
was
formerly
head
of
department
(Mechanical)
at
Haryana
were in different departments of ABB Ltd., earlier known as Taylor Instrument Ltd (both MNCs),
Institute
Technology.
Dr Kaushik
47 in
years
of experience,
which
years
where
heofmostly
discharged
the rolehas
of aoverall
manager
its R&D
cell. He hasout
19of
years
of 28
teaching
were
in
different
departments
of
ABB
Ltd.,
earlier
known
as
Taylor
Instrument
Ltd
(both
MNCs),
experience, teaching subjects like Strength of Materials, Mechanical Design and Engineering
where he mostly
the roleengineering)
of a managerstudents.
in its R&D
hasworked
19 years
teaching
Mechanics
to B. discharged
Tech (mechanical
Hecell.
hasHe
also
asoflecturer
in
experience, teaching subjects like Strength of Materials, Mechanical Design and Engineering
Kenya and Botswana for 12 years, teaching at the degree level. He has also been to
Mechanics to B. Tech (mechanical engineering) students. He has also worked as lecturer in
Switzerland in 2007 briefly on a promotion assignment for an India-based company.
Kenya and Botswana for 12 years, teaching at the degree level. He has also been to
Switzerland in 2007 briefly on a promotion assignment for an India-based company.
Strength of
Materials
Strength of
R.K. Kaushik
Materials
R.K. Kaushik
Distributed by:
38
89
9 3
99 778899 3
30
07
7311931
@seismicisolation
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TM
TM