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SOLUTIONS MANUAL FOR
MECHANICAL
DESIGN
OF MACHINE
COMPONENTS
SECOND EDITION: SI VERSION
by
ANSEL C. UGURAL
SOLUTIONS MANUAL FOR
MECHANICAL DESIGN
OF MACHINE
COMPONENTS
SECOND EDITION: SI VERSION
by
ANSEL C. UGURAL
Boca Raton London New York
CRC Press is an imprint of the
Taylor & Francis Group, an informa business
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CONTENTS
Part I
BASICS
Chapter 1
INTRODUCTION
1
Chapter 2
MATERIALS
16
Chapter 3
STRESS AND STRAIN
24
Chapter 4
DEFLECTION AND IMPACT
48
Chapter 5
ENERGY METHODS AND STABILITY
68
Part II
FAILURE PREVENTION
Chapter 6
STATIC FAILURE CRITERIA AND RELIABILITY
100
Chapter 7
FATIGUE FAILURE CRITERIA
117
Chapter 8
SURFACE FAILURE
135
Part III
APPLICATIONS
Chapter 9
SHAFTS AND ASSOCIATED PARTS
145
Chapter 10
BEARINGS AND LUBRICATION
164
Chapter 11
SPUR GEARS
176
Chapter 12
HELICAL, BEVEL, AND WORM GEARS
194
Chapter 13
BELTS, CHAINS, CLUTCHES, AND BRAKES
208
Chapter 14
MECHANICAL SPRINGS
225
Chapter 15
POWER SCREWS, FASTENERS, AND CONNECTIONS
240
Chapter 16
MISCELLANEOUS MACHINE COMPONENTS
261
Chapter 17
FINITE ELEMENT ANALYSIS IN DESIGN
278
Chapter 18
CASE STUDIES IN MACHINE DESIGN
308
vi
NOTES TO THE INSTRUCTOR
The Solutions Manual to accompany the text MECHANICAL DESIGN of Machine Components
supplements the study of machine design developed in the book. The main objective of the manual is to
provide efficient solutions for problems in design and analysis of variously loaded mechanical components.
In addition, this manual can serve to guide the instructor in the assignment of problems, in grading these
problems, and in preparing lecture materials as well as examination questions. Every effort has been made
to have a solutions manual that cuts through the clutter and is self –explanatory as possible thus reducing
the work on the instructor. It is written and class tested by the author.
As indicated in its preface, the text is designed for the junior-senior courses in machine or
mechanical design. However, because of the number of optional sections which have been included,
MECHANICAL DESIGN of Machine Components may also be used to teach an upper level course. In
order to accommodate courses of varying emphases, considerably more material has been presented in the
book than can be covered effectively in a single three-credit-hour course. Machine/mechanical design is one
of the student’s first courses in professional engineering, as distinct from basic science and mathematics.
There is never enough time to discuss all of the required material in details.
To assist the instructor in making up a schedule that will best fit his classes, major topics that will
probably be covered in every machine design course and secondary topics which may be selected to
complement this core to form courses of various emphases are indicated in the following Sample
Assignment Schedule. The major topics should be covered in some depth. The secondary topics, because
of time limitations and/or treatment on other courses, are suggested for brief coverage. We note that the
topics which may be used with more advanced students are marked with asterisks in the textbook.
The problems in the sample schedule have been listed according to the portions of material they
illustrate. Instructor will easily find additional problems in the text to amplify a particular subject in
discussing a problem assigned for homework. Answers to selected problems are given at the end of the text.
Space limitations preclude our including solutions to open-ended web problems. Since the integrated
approach used in this text differs from that used in other texts, the instructor is advised to read its preface,
where the author has outlined his general philosophy. A brief description of the topics covered in each
chapter throughout the text is given in the following. It is hoped that this material will help the instructor in
organizing his course to best fit the needs of, his students.
Ansel C. Ugural
Holmdel, N.J.
vii
DESCRIPTION OF THE MATERIAL CONTAINED IN
“MECHANICAL DESIGN of Machine Components”
Chapter 1 attempts to present the basic concepts and an overview of the subject. Sections 1.1 through 1.8
discuss the scope of treatment, machine and mechanical design, problem formulation, factor of safety, and
units. The load analysis is normally the critical step in designing any machine or structural member (Secs.
1.8 through 1.9). The determination of loads is encountered repeatedly in subsequent chapters. Case studies
provide a number of machine or component projects throughout the book. These show that the members
must function in combination to produce a useful device. Section 1.10 review the work, energy, and power.
The foregoing basic considerations need to be understood in order to appreciate the loading applied to a
member. The last two sections emphasize the fact that stress and strain are concepts of great importance to a
comprehension of design analysis.
Chapter 2 reviews the general properties of materials and some processes to improve the strength
of metals. Sections 2.3 through 2.14 introduce stress-strain relationships, material behavior under various
loads, modulus of resilience and toughness, and hardness, selecting materials. Since students have
previously taken materials courses, little time can be justified in covering this chapter. Much of the material
included in Chapters 3 through 5 is also a review for students. Of particular significance are the Mohr’s
circle representation of state of stress, a clear understanding of the three-dimensional aspects of stress,
influence of impact force on stress and deformation within a component, applications of Castigliano’s
theorem, energy of distortion, and Euler’s formula. Stress concentration is introduced in here, but little
applications made of it until studying fatigue (Chap.7).
The first section of Chapter 6 attempts to provide an overview of the broad subject of “failure”,
against which all machine/mechanical elements must be designed. The discipline of fracture mechanics is
introduced in Secs. 6.2 through 6.4. Yield and fracture criteria for static failure are discussed in Secs. 6.4
through 6.12. The last 3 sections deal with the method of reliability prediction in design. Chapter 7 is
devoted to the fatigue and behavior of materials under repeated loadings. The emphasis is on the Goodman
failure criterion. Surface failure is discussed in Chapter 8. Sections 8.1 through 8.3 briefly review the
corrosion and friction. Following these the surface wear is discussed. Sections 8.6 through 8.10 deal with
the surfaces contact stresses and the surface fatigue failure and its prevention. The background provided
here is directly applied to representative common machine elements in later chapters.
Sections 9.1 through 9.4 of Chapter 9 treat the stresses and design of shafts under static loads.
Emphasis is on design of shafts for fluctuating loading (Secs. 9.6 and 9.7). The last 5 sections introduce
common parts associated with shafting. Chapter 10 introduces the lubrication as well as both journal and
roller bearings. As pointed out in Sec. 8.9, rolling element bearings provide interesting applications of
contact stress and fatigue. Much of the material covered in Secs. 11.1 through 11.7 of Chapter 11 introduce
nomenclature, tooth systems, and fundamentals of general gearing. Gear trains and spur gear force analysis
are taken up in Secs. 11.6 and 11.7. The remaining sections concern with gear design, material, and
manufacture. Non-spur gearing is considered in Chapter 12. Spur gears are merely a special case of helical
gears (Secs. 12.2 through 12.5) having zero helix angle. Sections 12.6 through 12.8 deal with bevel gears.
Worm gears are fundamentally different from other gears, but have much in common with power screws to
be taken up in Chap. 15.
viii
Chapter 13 is devoted to the design of belts, chains, clutches, and brakes. Only a few different
analyses are needed, with surface forms effecting the equations more than the functions of these devices.
Belts, clutches, and brakes are machine elements depending upon friction for their function. Design of
various springs is considered in Chapter 14. The emphasis is on helical coil springs (Secs. 14.3 through
14.9) that provide good illustrations of the static load analysis and torsional fatigue loading. Leaf springs
(Sec. 14.11) illustrate primarily bending fatigue loading. Chapter 15 attempts to present screws and
connections. Of particular importance is the load analysis of power screws and a clear understanding of the
fatigue stresses in threaded fasteners. There are alternatives to threaded fasteners and riveted or welded
joints. Modern adhesives (Secs. 15.17 and 15.18) can change traditional preferred choices.
It is important to assign at least portions of the analysis and design of miscellaneous mechanical
members treated in Chapter 16. Sections 16.3 through 16.7 concern with thick-walled cylinders, press or
shrink fits, and disk flywheels. The remaining sections concerns with the bending of curved frames, plate
and shells-like machine and structural components, and pressure vessels. Buckling of thin-walled cylinders
and spheres is also briefly discussed. Chapter 17 represents an addition to the material traditionally
covered in “Machine/Mechanical Design” textbooks. It attempts to provide an introduction to the finite
element analysis in design. Some practical case studies illustrate solutions of problems involving structural
assemblies, deflection of beams, and stress concentration factors in plates. Finally, case studies in
preliminary design of the entire crane with winch and a high-speed cutting machine are introduced in
Chapter 18.
ix
SAMPLE ASSIGNMENT SCHEDULE
MACHINE/MECHANICAL DESIGN (3 credits.)
Prerequisites:
A. C. Ugural, MECHANICAL DESIGN of Machine Components, 2nd SI Version, CRC
Press (T & F Group).
Courses on Mechanics of Materials and Engineering Materials.
WEEK
TOPICS
TEXT:
SECTIONS
PROBLEMS
1
Introduction
Materials*
1.1 to 1.12
2.1 to 2.5,2.8 to 2.11
1.6,1.17,1.26,1.34
2.7,2.11,2.15,2.20
2
A Review of Stress Analysis*
3.1 to 3.14, 4.1 to 4.9
5.1 to 5.12
3.12,3.34,3.45,4.24,4.37
5.20,5.26,5.30,5.54, 5.73
3
Static Failure Criteria & Reliability
6.1 to 6.15
6.7,6.14,6.25,6.29,6.40
4
Fatigue Criteria and Surface Failure
7.1 to 7.15, 8.1 to 8.10
7.6,7.17,7.28,7.32,7.34
8.3,8.8, 8.14, 8.21
5
EXAM # 1
Shafts and Associated Parts
- - - - 9.1 to 9.12
- - - - 9.7,9.14,9.19,9.24,9.28
6
Lubrication and Bearings
10.1 to 10.16
10.4,10.10,10.16,10.31,10.36
7
Spur Gears
11.1 to 11.12
11.15,11.18,11.22,11.33,11.38
8
Helical, Bevel, and Worm Gears*
12.1 to 12.10
12.10,12.12,12.18,12.21,12.32
9
EXAM # 2
Belt and Chain Drives
- - - - 13.1 to 13.6
- - - - 13.1,13.8,13.9, 13.10,13.13
10
Clutches and Brakes
13.8 to 13.15
13.21,13.27,13.34,13.39,13.46
11
Mechanical Springs
14.1 to 14.11
14.8,14.12,14.22,14.25,14.35
12
EXAM # 3
Power Screws and Fasteners
- - - - 15.1 to 15.12
- - - - 15.2,15.6,15.14,15.19,15.30
13
Connections*
15.13 to 15.16
15.34,15.40,15.48,15.54
14
Miscellaneous Machine
Components*
16.1 to 16.5, 16.8
to 16.11
16.6,16.17,16.25,16.38
15
FEA in Design and
Case Studies in Machine Design*
FINAL EXAM
17.1 to 17.3,17.5,17.7
18.1 and 18.2
- - - - -
17.2,17.4,17.8,17.12
18.2,18.4,18.6,18.10
- - - - -
* Secondary topics. The remaining, major topics constitute the “main stream” of the machine design
course.
x
Section I
BASICS
CHAPTER 1
INTRODUCTION
SOLUTION (1.1)
Free Body: Angle Bracket (Fig. S1.1)
(a)

(b)


M
B
 0;
F ( 0 .5 )  6 8 ( 0 .2 5 )  2 8 .8 ( 0 .5 )  0 ,
Fx  0 :
R B x  2 8 .8  F  0 ,
Fy  0 :
R B y  6 8  2 1 .6  0 ,
F  6 2 .8 k N 
R Bx  3 4 k N 
R B y  8 9 .6 k N 
Thus
RB 
(3 4 )  (8 9 .6 )
2
2
 9 5 .8 k N
and
  ta n
1
34
8 9 .6
 2 0 .8
o
F
A
0.5
34
21.6 kN
68 kN
B
89.6
R Bx
Figure S1.1
28,8 kN
B
R By
C

RB
0.25
0.25
SOLUTION (1.2)
Free Body: Beam ADE (Fig. S1.2)

M
A
 0:
 W (3 .7 5 a )  F B D ( 2 a )  0 ,
E
D
W
FBD
A
2a
1.75a
R Ay
a
Fx  0 :
R Ax  0
Fy  0 :
 R Ay  FBD  W  0 ,
Free-Body: Entire structure (Fig. S1.2)

V
M
A
 0:
R C (3 a )  W (3 .7 5 a )  0 ,
R C  1 .2 5 W 
F
O
0.875 kN


R A y  0 .8 7 5W 
R Ax
A
F B D  1 .8 7 5 W 
M”
Free Body: Part AO (Fig. S1.2)
Figure S1.2
V  0 .8 7 5W 
F  0
M  0 .8 7 5 a W
1
SOLUTION (1.3)
29 kN/m
C
D
32 kN  m
A
0.6 m
B
E
RA
1,2 m


M
A
 0:
Fy  0 :
R B  1 1 .0 2 k N 
R A  4 5 .8 2 k N 
RB
2.4 m
2.4 m
Segment CD
29 kN/m
M
C
M
0.6 m D V
D

D
( 2 9 )( 0 .6 )  5 .2 2 k N  m
2
1
2
V D  1 7 .4 k N
D
Segment CE
3 4 .8 k N
C
VE
M
0.6 m
M
1.2 m A
E
1.2 m E
 3 4 .8 (1 .8 )  4 5 .8 2 (1 .2 )  7 .6 5 6 k N  m
E
V E  1 1 .0 2 k N
45 . 82
SOLUTION (1.4)

(a)
8 kN
1m
m
M
B
 0:
2m
0 .8 R C ( 6 )  0 .6 R C ( 2 )  2 4 ( 4 )  0
 R C  26 . 667
10 kN
kN
R Cx  16 kN , R Cy  21 . 334 kN
C
3m D
RC
B 2m
R Bx
4
3
R By
Then


2m
A
F x  0 : R Bx  16 kN
F y  0 : R By  12 . 66 kN
( b ) Segment CD
8
M
4 M
16
C
3m D V
D
21.334
D
 21 . 334 ( 3 )  12 (1 . 5 )  6 ( 2 )  34 kN  m
F D  16 kN
D
FD
V D  21 . 334  18  3 . 334
2
kN
SOLUTION (1.5)
3m
(a)
A
2

M
A
 0:
RCx 
3
2
RCy
C
RCy
RCx
4
3
R Cy
R By
RCx
FD
40 kN
C
 0:
4 0 (5 )  4 R C y  0
M
RCy  5 0 k N 
R Cx  75 kN 
Then
R By  1 0 k N  ,
R Bx  7 5 k N 
D
0.75
VD
B
1
m
4m
(b)
M
2m
3
R Bx B

D
F D  75 ( 53 )  50 ( 54 )  85 kN
1.0
V D  75 ( 54 )  50 ( 53 )  30 kN
75
C
5
50 4 3
M
D
 7 5 (1)  5 0 ( 0 .7 5 )  3 7 .5 k N  m
SOLUTION (1.6)
(a)
Free body entire connection
B
P A
T

M
C
0 :
0,5 m
T  0 .7 R A
0.2 m
Segment AB
A
AB 
F AB
B

0.15 m
M
( 0 .5 )  ( 0 .1 5 )
2
B
0 :

18 
 0 .5 2 2 m
18 ( 0 . 15 )  R A ( 0 . 5 )  0
T  3 .7 8 k N  m
and
Fx  0 :
2
R A  5 .4 k N
0.5 m
RA
(b)
R A ( 0 .7 )  T  0
C
RA
18 kN
0.15 m
0 .5
0 .5 2 2
FAB  0,
F A B  1 8 .7 2 9 k N
SOLUTION (1.7)
120 mm
120 mm
y
A
R A  2 kN
B
RB  2 kN
50 mm
4 kN


70 mm
D
z
Free Body: Entire Crankshaft (Fig. S1.7a)
( a ) From symmetry: R A  R B
T
Fz  0 : R A  R B  2 k N
M
x
 0 :  4 ( 0 .0 5 )  T  0 ,
T  0 .2 k N  m  2 0 0 N  m
x
C
(a)
(CONT.)
3
1.7 (CONT.)
M
y
( b ) Cross Section at D (Fig. S1.7b)
D
T
Vz
Vz  2 kN
(b)
T  200 N  m
M
Figure S 1.7
y
 2 ( 0 .0 7 )  0 .1 4 k N  m
 140 N  m
SOLUTION (1.8)
30 kN
Free-Body Diagram, Beam AB
B
1.8 m
4
7
C
FC D
1.2 m

Fx  0 :

Fy  0 : R A 

M
A

 0:
7
65
FC D  6 0  0 ,
4
65
FC D  3 0  0 ,
 6 0 (1 .8 ) 
M
60 kN
A
F C D  6 9 .1 1 k N
7
65
R A  6 4 .3 k N 
FC D ( 3 )  M
A
 0,
 72 kN  m
1.8 m
A
M
A
RA
SOLUTION (1.9)
B 1.2 m C
129.6 kN
0.9 m 0.9 m 1.2 m
2
A
Free body entire frame
2.4 m
1
1
2
D

M
A
0 :
 129 . 6 ( 0 . 9 ) 
R D y  4 8 .6 k N ,
1
2
R Dy (1 . 2 )  R Dy ( 3 )  0
R D x  2 4 .3 k N
R Dy
R Dy R D
B
R By
R B x 1.2 m
D
C
Free body BCD
2.4 m


24.3
Fx  0 :
R B x  2 4 .3 k N
Fy 0 :
R B y  4 8 .6 k N
RB 
2 4 .3  4 6 .6
2
48.6
4
2
 5 2 .6 k N
SOLUTION (1.10)
y
4 kN 3 kN
R Ay
A
0.3 m
R Az
T
R Ey
5 kN
z
C
E
1m
D
R Ez
1m
2 kN
x
B
0.5 m
0.5 m
0.3 m
(a)

M
x




3 ( 0 .1 5 )  4 ( 0 .1 5 )  T  5 ( 0 .1 5 )  2 ( 0 .1 5 )  0
T  0 . 6 kN  m
or
(b)
 0:
M
z
 0 : ( 4  3 )( 1 )  R Ey ( 2 . 5 )  0 , R Ey   2 . 8 kN
M
y
 0 :  R Ez ( 2 . 5 )  ( 5  2 )( 3 )  0 , R Ez  8 . 4 kN
F y  0 : R Ay  4  3  R Ey  0 , R Ay   4 . 2 kN
F z  0 : R Az  R Ez  5  2  0 , R Az   1 . 4 kN
Thus
RA 
4 .2
RE 
2 .8
2
 1 .4
2
 4 . 427
2
 8 .4
2
 8 . 854 kN
kN
SOLUTION (1.11)
(a) Free-body Diagrams, Arm BC and shaft AB
V
z
T
C
y
B
x
M
T
V
A
T
Figure S1.11
M
V
M
V
B
T
(b) At C:
V  2 kN
T  50 N  m
At end B of arm BC:
V  2 kN
T  50 N  m
M  200 N  m
At end B of shaft AB:
V  2 kN
At A:
V  2 kN
T  200 N  m
T  200 N  m
M  50 N  m
M  300 N  m
5
SOLUTION (1.12)
Free Body: Entire Pipe
200 N
0.15 m
D
y
36 N  m
C
Ry
T
Rx
A
0.2 m
Rz
Mz
My
z
0.3 m
B
x
Reactional forces at point A:



Fx  0 :
Rx  0
Fy  0 :
R y  200  0,
Fz  0 :
Rz  0
R y  200 N
Moments about point A:



M
x
 0:
T  2 0 0 ( 0 .1 5 )  0 ,
M
y
 0:
M
y
M
z
 0:
M
z
T  30 N  m
 0
 2 0 0 ( 0 .3 )  3 6  0 ,
M
z
 96 N  m
The reactions act in the directions shown on the free-body diagram.
SOLUTION (1.13)
Free Body : Entire Pipe
0.15 m
D
y
Rz
z
36 N  m
C
WCD
T
Mz
0.075 m
Ry
Rx
200 N
WAB
0.2 m
WBC
A
My
0.15 m
0.3 m
B
x
(CONT.)
6
1.13 (CONT.)
We have 1 lb/ft=14.5939 N/m (Table A.2).
Thus, for 3 in. or 75-mm pipe (Table A.4): 14.5939(7.58)=110.62 N/m
Total weights of each part acting at midlength are:
W A B  1 1 0 .6 2 ( 0 .3)  3 3 .2 N
W B C  1 1 0 .6 2 ( 0 .2 )  2 2 .1 N
W C D  1 1 0 .6 2 ( 0 .1 5 )  1 6 .6 N
Reactional forces at point A:



Fx  0 :
Fy  0 :
Rx  0
R y  W AB  W BC  W CD  2 0 0  0,
Fz  0 :
R y  2 7 1 .9 N
Rz  0
Moments about point A:



M
x
 0:
M
y
 0:
M
z
 0:
T  W C D ( 0 .0 7 5 )  1 0 ( 0 .1 5 )  0 ,
T  2 .7 4 5 N  m
M
M
z
y
 0
 ( 2 0 0  W C D  W B C )( 0 .3 )  W A B ( 0 .1 5 )  3 6  0
M
z
 1 1 2 .6 N  m .
SOLUTION (1.14)
1.6 kN
RCy
(a)
R Ay
D
R Ax A
C
Dy
0.5 m
RCx
1m
Dx
Bx
E
By
D
1.6 kN
0.15 m
Dx
1.6 kN
By
Dy
Free body pulley B


B
Bx
0.25 m
B
150 mm
F x  0 : B x  1 . 6 kN 
F y  0 : B y  1 . 6 kN 
Free body CED



M
D
 0 : R Cx ( 0 . 4 )  1 . 6 ( 0 . 15 )  0 ,
F x  0 :  D x  0 .6  1 .6  0 ,
F y  0:
R Cy  D
R Cx  0 . 6 kN 
D x  1 kN 
y
Free body ADB


M
A
 0 : D y ( 0 . 5 )  B y (1 . 5 )  0 , D y  4 . 8 kN , R Cy  4 . 8 kN 
F x  0 :  R Ay  D y  B y  0 ,
R Ay  3 . 2 kN 
(CONT.)
7
1.14 (CONT.)

(b)
M
F x  0 : R Ax  D x  B x  0 ,
V G 0.6 m B
G
FG
G
M
1.6
G
R Ax  0 . 6 kN 
 1 . 6 ( 0 . 6 )  960
N  m , V G  1 . 6 kN
F G  1 . 6 kN
1.6
SOLUTION (1.15)
Free body entire rod

M

 0 : R Dy ( 0 . 25 )  300 ( 0 . 1 ),
x
M
z

 0:
R Dy  120
N 
 2 0 0 ( 0 .3 5 )  R B y ( 0 .2 5 )  ( 3 0 0  1 2 0 ) ( 0 .2 )  0
C
R By  136
N 
Free body ABE
y
B
A
200 N
Vy
100
175

M
z

E
 0:  M
z
x
E
136 N

M
z
 200 ( 0 . 275 )  136 ( 0 . 175 )  0 , M
z
z
 31 . 2 N  m
F y  0 : V y  200  136  64 N
SOLUTION (1.16)
Free body entire rod:
 M  0:
 M  
R D y ( 0 .2 5 )  4 0 0 ( 0 .1)  0 ,
x
z
R B y ( 0 .2 5 )  ( 4 0 0  1 6 0 ) ( 0 .2 )  0 ,
0:
C
Segment ABE
A
y
B
Vy
0.175 m
192 N
z
x
E
M
z
At point E:
M
z
  192 ( 0 . 175 )   33 . 6 N  m
V y  192
RDy  160 N 
N
8
R By  1 9 2 N 
SOLUTION (1.17)
Side view
Top view
F2
50 mm
F2
B
C
D
Td
50 mm
100 mm
Figure (c)
Figure (a)
F1
150 N  m
A
50 mm



Fig. (b):
Fig. (a):
Fig. (c):
F1
Figure (b)
M
A
 0 : F 1 ( 0 . 05 )  150  0 , F 1  3 kN
M
B
 0 : F 1 ( 0 . 1 )  F 2 ( 0 . 05 )  0 , F 2  6 kN
M
D
 0:
F 2 ( 0 .0 5 )  T d  0 ,
T d  0 .3
kN  m
SOLUTION (1.18)
18 kN
13.5 kN
1m
C
3m
B
A
Ry
6m
Rx
3
4
4m
R
Free body-entire frame

M
A
 0:
R y (1 0 )  1 3 .5 ( 6 )  1 8 ( 4 )  0 ,
Free body-member BC

M
C
 0:
R x (3)  R y ( 4 )  0
and
Rx 
4
3
(1 5 .3)  2 0 .4 k N
Thus
FBC  R 
( 2 0 .4 )  (1 5 .3 )
2
2
 2 5 .5 k N
9
R y  1 5 .3 k N
SOLUTION (1.19)
P
Free body-member AB
B
R Ax
A
E 40
o

a


 0:
o
o
 R E  1.1 5 6 P
RE
R Ay
A
R E ( 4 a )  P c o s 4 0 ( a )  P s in 4 0 ( 6 a )  0
2a
4a
M
F x  0:
R Ax  P co s 4 0
 0 .7 6 6 P 
F y  0:
R A y  1.1 5 6 P  P s in 4 0
o
o
 0,
R A y  0 .5 1 3 P 
Free body-member CD
RE
RCx
C 2a
E
RD
RCy


M
C

D
4a
 0:
30
M
D
 0:
 R C y  0 .7 7 1 P 
o
R D s in 3 0 ( 6 a )  R E ( 2 a )  0 ,
o
F x  0:
RCx  R D co s 3 0
R E ( 4 a )  RCy ( 6 a )  0
o
R D  0 .7 7 1 P
 0 .6 6 8 P 
SOLUTION(1.20)
( a ) Power= P = ( p A ) ( L ) ( n /6 0 )
 (1 .2 )( 2 1 0 0 )( 0 .0 6 )( 1 56 00 0 )  3 .7 8 k W
Power required 
P
e

3 .7 8
0 .9
 4 .2 k W
( b ) Use Eq.(1.15),
T 
9549 kW
n

9 5 4 9 ( 4 .2 )
1500
 2 6 .7 4 N  m
SOLUTION (1.21)
a=1.5 m, b=0.55 m, c=0.625 m, L=2.7 m, V=29 m/s, W=14.4 kN, kW=14
( a ) From Eq. (1.15), the drag force equals,
Fd 
1000 kW
V

1 0 0 0 (1 4 )
29
 4 8 2 .8 N
See: Fig. P1.21:

Fx  0 :
F d  4 8 2 .8 N
It follows that

M
A
 0:
 W a  Fd c  R f L  0
or
 1 4 4 0 0 (1 .5 )  ( 4 8 2 .8 )( 0 .6 2 5 )  R f ( 2 .7 )  0
(CONT.)
10
1.21 (CONT.)
Solving,
R f  7 .8 8 8 k N
and

Fy  0 :
R r  1 4 .4  7 .8 8 8  0
or
R r  6 .5 1 2 k N
(b)
We have V  0 ,
Fd  0 ,
F  0.
See Fig. P1.21:

M
A
Rf 
a
L
 0:
Wa  Rf L  0
Thus
So,

W 
1 .5
2 .7
(1 4 .4 )  8 k N
F y  0 gives
R r  W  R f  1 4 .4  8  6 .4 k N
SOLUTION (1.22)
Refer to Solution of Prob. 1.21.
a=1.5 m, b=0.55 m, c=0.625 m, L=2.7 m, V=29 m/s, W=14.4 kN, kW=14
Now we have
W t  1 4 .4  5 .4  1 9 .8 k N
( a ) From Eq. (1.15), the drag force equals,
Fd 
1000 kW

V
1 0 0 0 (1 4 )
29
 4 8 2 .8 N
See: Fig. 1.21 (with W  W t ):

M
A
 0:
 W t a  Fd c  R f L  0
  1 9 8 0 0 (1 .5 )  ( 4 8 2 .8 )( 0 .6 2 5 )  R f ( 2 .7 )  0
from which
R f  1 0 .8 8 8 k N
and

Fy  0 :
R r  1 9 .8  1 0 .8 8 8  0
R r  8 .9 1 2 k N
( b ) V  0,
Fd  0 ,

M
A
Then
Rf 
a
L
So,

F  0 , as before,
 0:
W 
 Wta  R f L  0
1 .5
2 .7
(1 9 .8 )  1 1 k N
F y  0 gives
R r  W t  R f  1 9 .8  1 1  8 .8 k N
11
SOLUTION (1.23)
( a ) Free-Body Diagram: Gears (Fig. S1.23).
Applying Eq. (1.15):
T AC 
9550 P
n
9550 (35 )

 6 6 8 .5 N  m
500
Therefore,
TA
F 
rA

 5 .3 4 8 k N
6 6 8 .5
0 .1 2 5
( b ) T D E  F rD  5, 3 4 8 ( 0 .0 7 5 )  4 0 1 .1 N  m
TDE
D
rD
T AC
A
F
rA
F
Figure S1.23
SOLUTION (1.24)
n1  1 8 0 0 rp m ,
n 2  4 2 5 rp m ,
kW  20
From Eq. (1.15), we obtain T = 9549 kW/n. Thus
For input shaft
T 
9549 ( 23)
1800
 122 N  m
For output shaft
T 
9549 ( 20 )
425
 4 4 9 .4 N  m
Equation (1.14) gives
e 
20
23
 100  87 %
SOLUTION (1.25)
N=150, F=2.25 kN, s=62.5 mm, e=88%
( a ) Referring to Eq. (1.12):
power output  F s ( 6N0 )  2 2 5 0  0 .0 6 2 5 ( 16500 )
 3 5 1 .6 W
( b ) Using Eq. (1.14), power transmitted by the shaft:
power input  3 5 1 .6 ( 0 .8 8 )  3 9 9 .5 W
SOLUTION (1.26)
Equation (1.10) becomes
Ek 
1
2
I (

2
max
2
min
(1)
)
Here, mass moment inertia with 5 percent added:
I  ( 1 . 05 )

32
( d o  d i )  l
4
4
(Table A.5)
(CONT.)
12
1.26 (CONT.)

 1 . 05
( 0 .4
32
4
 1 . 299 kg  m
 0 . 3 )( 0 . 1 )( 7 , 200 )
4
2
 m a x  1 2 0 0 ( 610 )  2 0 r p s  1 2 5 .7 ra d s
 m in  1 1 0 0 ( 610 )  1 8 .3 r p s  1 1 5 ra d s
Equation (1) is therefore
Ek 
1
2
( 1 . 299 )( 125 . 7
 115
2
2
)
 1, 6 7 3 J
SOLUTION (1.27)
Final length of the wire:
L AC ' 
( 2 )  (1 .2 6 )
2
2
 2 .3 6 3 8 m
Initial length of the wire is
L AC 
( 2 )  (1 2 .5 )
2
2
 2 .3 5 8 5 m
Hence, Eq. (1.20):
 AC 
L AC '  L AC
L AC

2 .3 6 3 8  2 .3 5 8 5
2 .3 5 8 5
 0 .0 0 2 2 5  2 2 5 0 μ
SOLUTION (1.28)
(a) c 
2 ( r  r ) 2 r
2 r
( c ) i 
0 .3
150
( c ) o 
(b) r 
r
r

μ
 2000
μ
 800
0 .2
250
 r o   ri

r o  ri
0 .3  0 .2
250  150
 1000
μ
SOLUTION (1.29)
LO B  d ,
( a )  OB 
L AB  L BC  d
0 .0 0 1 2 d
d
 1200 μ
( b ) L AB '  L CB '  d
 AB   BC 
( c ) C A B  ta n
1
2  1 .4 1 4 2 1 d
2
 (1 . 0012 d )
1 .4 1 5 0 6  1 .4 1 4 2 1
1 .4 1 4 2 1
 1 .0 0d1 2 d  
2

1
2
 1 . 41506 d
 601 μ
4 5 .0 3 4 4
o
Increase in angle C A B is 4 5 .0 3 4 4  4 5  0 .0 3 4 4 .
Thus
o
  0 .0 3 4 4  18 0   6 0 0 μ
13
SOLUTION (1.30)
(a) x 
0 .8  0 .5
250

μ
 1200
y
 0 .4  0
200

 2000 μ
( b ) L ' AD  L AD   x L AD  L AD (1   x )
 250 (1 . 0012 )  250 . 3 mm
SOLUTION (1.31)
 L AB  800 (10
6
)150  120 (10
6
 L AD  1000 (10
3
) 200  200 (10
) mm
3
) mm
We have
L BD  L AB  L AD
2
2
2
2 L BD  L BD  2 L AB  L AB  2 L AD  L AD
or
L AB
 L BD 

LBD
 150
250
 L AB 
L AD
LBD
 L AD
(120 ) 
200
250
( 200 ) 10
(1)
3
 0 . 232
mm
SOLUTION (1.32)
AC  BD 
300
2
 300
2
 424 . 26 mm
B ' D '  4 2 4 .2 6  0 .5  4 2 3 .7 6 m m ,
A ' C '  4 2 4 .2 6  0 .2  4 2 4 .4 6 m m
Geometry: A ' B '  A ' D '
x  
C'
B'

D'

A'
xy

y

A ' D '  AD
AD
  4 2 3 .7 6  2  4 2 4 .4 6  2 

 
 
2
2


 
 
1
2
 300
 363 μ
300

2
  

2
 2 tan
 1 423 . 76 2
424 . 46 2
 1651
μ
SOLUTION (1.33)
0.5 m
A'
0.00064 m
A
F
1m
B
C
B
 800  10
6
( 0 .8 )
 0 .0 0 0 6 4 m
B'
x
We have
 AD   AD L AD
0.005 m
0.33 m
C'
(CONT.)
14
1.33 (CONT.)
From triangles A ' A F and C ' C F :
0 .0 0 0 6 4
x

0 .0 0 5
1 .5  x
,
x  0 .1 7 m
From triangles B ' B F and C ' C F :
0 .3 3
 01.0.3035 ,  B  0 .0 0 1 2 4 m    B E (contraction)

B
Therefore
 BE 
  BE
LBE

0 .0 0 1 2 4
1
  1 2 4 0 (1 0
6
)
  1, 2 4 0 
SOLUTION (1.34)
(a) x 

xy
0 . 006
50
 120
μ

y
  1000  200   800

 0 . 004
25
  160
μ
μ
( b ) L ' A B  L A B (1   y )  2 5 (1  0 .0 0 0 1 6 )  2 4 .9 9 6 m m
L ' A D  L A D (1   x )  5 0 (1  0 .0 0 0 1 2 )  5 0 .0 0 6 m m
End of Chapter 1
15
CHAPTER 2
MATERIALS
SOLUTION (2.1)
A0 

4
(1 2 .5 )  1 2 2 .7 m m ,
2
We have  a 
2
 1500 μ ,
0 .3
200
Af 
t 

0 .0 0 6
1 2 .5
(1 2 .5  0 .0 0 6 )  1 2 2 .6 m m
2
4
 480 μ
Thus
S
p

E 
S
p
a
3
1 8 (1 0 )

P
A0

 1 4 6 .8 M P a
1 2 2 .7
6
1 4 6 .7 (1 0 )
1 5 0 0 (1 0
6
 
 9 7 .8 G P a ,
)
t
a
 0 .3 2
Also
% e lo n g a tio n 
0 .3
200
(1 0 0 )  0 .1 5
in area 
% reduction
122 . 7  122 . 6
122 . 7
(100 )  0 . 082
SOLUTION (2.2)
Normal stress is
 

P
A
 2 8 6 .8 M P a
2200

( 3 .1 2 5 )
2
4
This is below the yield strength of 350 MPa (Table B.1).
We have
  L  576.50 0  0 .0 0 1 3 3 9  1 3 3 9 μ
Hence
E 



6
2 8 6 .8 (1 0 )
1 3 3 9 (1 0
6
 2 1 4 .2 G P a
)
SOLUTION (2.3)
The cross-sectional area: A  w o t o  1 2 .7 ( 6 .1)  7 7 .4 7 m m
( a ) Axial strain and axial stress are
 a  0 6.038.54 1  0 .0 0 1 3 2 4  1 3 2 4 μ

a

Because 
a
 S y (See Table B.1), Hooke's Law is valid.
P
A

2 1 ,5 0 0
7 7 .4 7 (1 0
6
)
 2 7 7 .5 M P a
( b ) Modulus of elasticity,
E 

a
a

6
2 7 7 .5 (1 0 )
1 3 2 4 (1 0
6
)
 2 0 9 .6 G P a
( c ) Decrease in the width and thickness
 w   w o  0 .3(1 2 .7 )  3 .8 1 m m
 t   t o  0 .3( 6 .1)  1 .8 3 m m
16
2
2
SOLUTION (2.4)
Assume Hooke's Law applies. We have
 t   15.5   3 0 0 μ
a  
t
 

300
0 .3 4
 822 μ
Thus,
  E  a  (1 0 5  1 0 )(8 2 2  1 0
9
9
)  9 2 .6 1 M P a
  S y , our assumption is valid.
Since
So
P   A  (9 2 .6 1)(  4 )(5 )  1 .8 1 8 k N
2
SOLUTION (2.5)
We obtain
L AC  L BD 


 L AC
x


 L BD
y
(a) E 
L AC
L BD

x

x
y
(b)  
x
(c) G 


15
 15
2

21 . 17  21 . 21
21 . 21
  1886

21 . 22  21 . 21
21 . 21
 471 μ
100 ( 10
6
 1886 ( 10
471
1886
53
2 ( 1  0 . 25 )
)
6
2
 21 . 21 mm
μ
 53 GPa
)
 0 .2 5
 21 . 2 GPa
SOLUTION (2.6)
Use generalized Hooke’s law:
x   y  z 
1 2 
E
( x  
y
For a constant triaxial state of stress:
x   y  z  ,  x  
y
Then, Eq. (1) becomes  
1  2  0
or
1 2 
E
 
 z)
(1)
 z  
 . Since  and 
must have identical signs:
1
2
SOLUTION (2.7)
We have 
x

3
4 5 0 (1 0 )
50 (75 )
 120
M Pa
(CONT.)
17
2.7 (CONT.)
(a) x 
 
(b) E 
x
x
(c) z  
6
1 2 0 (1 0 )


2 0 0 0 (1 0
6
 500 μ
0 .0 2 5
50
 60 G Pa
)
6
1 2 0 (1 0 )
  0 .2 5
x
E
6
 a   5 0 0 (1 0
(d) G 
 
y
 0 .2 5
500
2000


 2000 μ,
0 .5
250
9
6 0 (1 0 )
 500 μ
9
6 0 (1 0 )
) 7 5   3 7 .5 (1 0
3
a '  7 5  0 .0 3 7 5  7 4 .9 6 2 5 m m
) mm;
 24 G Pa
2 (1  0 .2 5 )
SOLUTION (2.8)
We have
 y  z  0

x

3
25 ( 10 )
20  10 ( 10
6
)
 125
MPa
Thus
y  0 
1
E
[
  (  x   z )]
(1)
z  0 
1
E
[ z   (  x   y )]
(2)
x 
y
[ x   ( 
1
E
  z )]
y
(3)
Equations (1) and (2) become
 y   z   x
 z  
Adding :  ( 
(1’)
  x
y
(2’)
  z )  2   x (1   ) . Then, Eq. (3):
2
y
x 

2
1   2 
1 
x
E
Substituting the data:

x

6
1  0 . 3  0 . 18 125 ( 10 )
9
0 .7
70 ( 10 )
 1327
μ
SOLUTION (2.9)
Hooke's Law. We have 
x 


10



9
x
E
10
6
7 2 1 0
z  
E
6
7 2 1 0
y  


x
E

9
x
E
10
6
7 2 1 0
9
y
 0 and
y


z
E
[(8 0 )  0  0 .3 (1 4 0 )]  0 .0 0 0 5 2 8  5 2 8 μ



y
E

z
E
[  0 .3 (8 0 )  0  0 .3 (1 4 0 )]   9 1 7 μ


E
y


z
E
[  0 .3 (8 0 )  0  1 4 0 ]  1 6 1 1 μ
(CONT.)
18
2.9 (CONT.)
(a)
 L AB   x a ,
Change in length
6
 L A B  (5 2 8  1 0
(b)
)(3 2 0 )  0 .1 6 9 m m
Change in thickness
 t   yt  (917  10
(c)
6
)(1 5 )   0 .0 1 4 m m
Change in volume,
e  x 
  z  5 2 8  9 1 7  1 6 1 1  1 .2 2 2
y
 V  e V o  1 .2 2 2 (3 2 0  3 2 0  1 5 )  1 .8 7 7 m m
3
SOLUTION (2.10)
By assumptions, rubber in triaxial stress:

x

z

  p,
y
 
 
F
d
2
4
4F
d
2

Stains are  x   z  0 . Hooke's law gives
x  0 
1
E
[
  (
x
y
y
  z )]
or
0  p 

4F

Solving,
p 
x
2
 d (1   )
4 F
z
Q.E.D.
2
 d (1   )
Substitute the data:
p 
3
4 ( 0 .5 )(1 0  1 0 )
2
 ( 6 2 .5 ) (1  0 .5 )
 3 .2 6 M P a (C)
SOLUTION (2.11)
Hooke’s law gives
90 MPa
y 10 mm
150 MPa

x

1
E
(
x

y

1
E
(
y

z
 
x
100 mm

E
(
 
y
) 
 x ) 
x

y
10
6
100 ( 10
10
9
6
9
1 0 0 (1 0 )
)  
)
(150 
(90 
( 1 3 ) 10
100 ( 10
6
9
)
 L  1800  (100 )  180 μm
 a   1400  ( 50 )   70 μm
 b   200  (10 )   2 μm
and
mm ,
a '  49 . 993
19
mm ,
b '  9 . 9998
150
3
)  1800
μ
)  1400 μ
(150  90 )   200 μ
Thus
L '  100 . 018
90
3
mm
SOLUTION (2.12)
We have
x  
y
 z  p
Gen. Hooke’s law:

x
 
y
 
z
 
(1  2 )  
p
E
120 ( 10
100 ( 10
6
9
) 1
) 3
  400 μ
Thus
 L   400  (100 )   40 μm
 a   400  ( 50 )   20 μm
 b   400  (10 )   4 μm
and
L '  99 . 96 mm ,
a '  49 . 98 mm ,
b '  9 . 996
SOLUTION (2.13)
We have

x

Vo 
(a) x  

r 
3
  p . The volume is
z
4
3
(1 2 5 )  8 .1 8 1(1 0 ) m m
3
[   (   )]  
1
E
6
1 6 8 (1 0 )

4
3
y
9
7 0 (1 0 )

E
6
3
(1  2 )
(1  0 .5 )   1 2 0 0 μ
Change in diameter,
 d   x d   1 2 0 0 (1 0
6
) 2 5 0   0 .3 m m
Decrease in circumference:
 (  d )   0 .3    0 .9 4 2 5 m m
( b )  V  e V o  (1  2 )  xV o
 ( 0 .5 )(  1 2 0 0  1 0
6
)(8 .1 8 1  1 0 )   4 9 0 9 m m
6
SOLUTION (2.14)
From Fig.2.3b and Eq.2.20:
U
t

S y  Su
2

f

250  440
2
( 0 . 27 )  93 MPa
We have L f  50  50 ( 0 . 27 )  63 . 5 mm
Using Eq.(2.1): % e lo n g a tio n 
6 3 .5  5 0
50
(1 0 0 )  2 7 %
20
3
mm
SOLUTION (2.15)
Table B.1:
S
 260
y
MPa , E  70 GPa
We have
V  AL 
For U

U
r
U
app
S
2
y
2E

( 0 . 005 ) ( 3 )  58 . 9  10
2
4
( 260  10

6
2
)
9
2 ( 70  10
)
 482 . 9 kJ
m
6
m
3
3
 U r V  482 . 9  10 ( 58 . 9  10
3
6
)  28 . 44 J
 9 J :
app
n 
2 8 .4 4
9
 3.1 6
SOLUTION (2.16)
( a ) ASTM-A242. E  2 0 0 G P a and 
U
o

Sy
2
2E
6
( 3 4 5 1 0 )

2
9
2 ( 2 0 0 1 0 )
y
 298
 345 M Pa
kJ
m
3
( b ) Stainless (302). E  1 9 0 G P a and S y  5 2 0 M P a
U
o

Sy
2
2E
6
( 5 2 0 1 0 )

2
9
2 ( 2 0 0 1 0 )
 712
kJ
m
3
SOLUTION (2.17)
( a ) Aluminum 2014-T6. E  7 2 G P a and 
Uo 
Sy
2
2E

6
( 4 1 0 1 0 )
y
 410 M Pa
2
9
2 ( 7 2 1 0 )
 1167
kJ
m
3
( b ) Annealed yellow brass. E  1 0 5 G P a and S y  1 0 5 M P a
Uo 
Sy
2
2E
6
(1 0 5  1 0 )

9
2
2 (1 0 5  1 0 )
 5 2 .5
kJ
m
3
SOLUTION (2.18)
Referring to Fig. P2.18: E 
(a) Uo 
Sy
2
2E

6
(1 9 2 .5  1 0 )
9
2 ( 4 2 1 0 )
2
6
1 0 5 (1 0 )
0 .0 0 2 5
 42 G Pa,
 4 4 1 .5 k J m
S y  1 9 2 .5 M P a
3
( b ) Total area under    diagram:
U
 2 6 2 .5 (1 0 )( 0 .1 7 6 )  4 6 .2 M J m
6
t
21
3
SOLUTION (2.19)
( a ) V  50  50  1, 500  3 . 75 (10 ) mm
6
Thus
nU 
or
S
S
V
2E
 [
y
2
y
1
2 EnU
V
]2
9
 [ 2  200 10
r

S
2
y
6
( 253  10

2E
)
2
9
2 ( 200  10
)
1
 1 . 5  400
3 . 75 ( 10
(b) U
3
3
] 2  253
)
 160
MPa
kPa
SOLUTION (2.20)
S y  250
Table B.1:
E  200
M Pa,
GPa
We have
U  nU
U
r
S

app
2
y
6
( 2 5 0 1 0 )

2E
 5 (1 7 )  8 5 N  m
2
 1 5 6 .2 5
9
2 ( 2 0 0 1 0 )
kN  m m
3
Therefore
V 
U
Ur

85
3
1 5 6 .2 5 (1 0 )
V  AL :
Also
3
 0 .5 4 4 (1 0
0 . 544 (10
3
) 

) m
3
2
d ( 2 .4 )
4
or
d  0 .0 1 7 m = 1 7 m m
SOLUTION (2.21)
Refer to Fig. P2.21. We have
E 
(a) Uo 
Sy
2

2E
6
1 9 0 (1 0 )
0 .0 0 1
3
( 2 4 5 1 0 )
2
9
2 (1 9 0  1 0 )
 190 G Pa,
 158
S y  245 M Pa
kJ
m
3
( b ) Total area under    diagram:
 3 5 0  1 0 ( 0 .2 8 )  9 8
3
U
t
MJ
m
3
SOLUTION (2.22)
V  ( 0 .0 5 )( 0 .0 5 )(1 .2 )  0 .0 0 3 m
( a ) n U 
S
3
and S y  S p
2
p
2E
V,
S
2
p

nU ( 2 E )
V
Substituting the data given,
(CONT.)
22
2.22 (CONT.)
Sp 
2
9
1 .8 (1 5 0 )( 2  2 1 0  1 0 )
0 .0 0 3
 3 7 .8  1 0
15
or
S p  1 9 4 .4 M P a
(b) Uo 
Sp
2

2E
3 7 .8  1 0
15
9
2 ( 2 1 0 1 0 )
 90
kJ
m
3
SOLUTION (2.23)
Applying Eq. (2.22), we find
S u  3 .4 5 H
B
M P a  3 .4 5 (1 4 9 )  5 1 4 M P a
Equation (2.24):
S y  3 .6 2 (1 4 9 )  2 0 7  3 3 2 .4 M P a
SOLUTION (2.24)
Using Eq. (2.22),
S u  3 .4 5 H
B
M P a  3 .4 5 (1 7 9 )  6 1 8 M P a
Formula (2.24):
S y  3 .6 2 (1 7 9 )  2 0 7  4 4 1 M P a
SOLUTION (2.25)
Formula (2.22),
S u  3 .4 5 H
B
M P a  3 .4 5 (1 5 6 )  5 3 8 M P a
Equation (2.24):
S y  3 .6 2 (1 5 6 )  2 0 7  3 7 8 M P a
SOLUTION (2.26)
Equation (2.22) gives
S u  3 .4 5 H
B
M P a  3 .4 5 ( 2 9 3)  1 0 1 0 .9 M P a
Formula (2.24):
S y  3 .6 2 ( 2 9 3 )  2 0 7  8 5 3 .7 M P a
End of Chapter 2
23
CHAPTER 3
STRESS AND STRAIN
SOLUTION (3.1)
(a) 
s
3
4 5 (1 0 )

 ( 2 5 1 0
3
)
 9 1 .6 7
4
3
(b) h 
 ( 2 5 .7 8  1 0
(c) t 
 ( 2 1 .1 2  1 0
(d) b 
2
4 5 (1 0 )
3
3
)(1 0 .9 4  1 0
3
3
)(1 2 .5  1 0
3
 5 0 .7 9 M P a
)
4 5 (1 0 )
M Pa
 5 4 .2 6
)
3
4 5 (1 0 )

[( 5 0  1 0
3
2
)  ( 2 5 .7 8  1 0
M Pa
 3 1 .2 2 M P a
3
2
) ]
4
SOLUTION (3.2)
A   [( a 
a
9
16
Thus
a
4
)  a ]
2
2

2
Pn
Sy
a
9
16
,
a
2

2

16 Pn
9 S y
all

a 
,
S
y
n
A 
,

P

all
Pn
Sy
Pn
4
3
 Sy
Substituting the data given:
a 
1 . 2 ( 10
4
3
6
)( 2 . 2 )
6
 ( 280  10 )
 73 mm  a min
SOLUTION (3.3)
Free body-Member ACD
R Ay
A

R Ax
 0 :  45 sin 15 ( 0 . 8 ) 
o
A

FBC
0.4 m
3

0.1 m
0.4 m
Fx  0 :
3
5
F BC  45 sin 15
 R A x  5 .8 2 5

D
o
RA 
Resultant is
F y  0 : 45 cos 15
(5 .8 2 5 )  ( 6 6 .7 6 3 )
2

We therefore have

n
RA 2
 d
2
4
,
2
d  (
o
 6 6 .9 9
2 R An

1
)2
Hence
3
dA  [
2 ( 6 6 .7 6 3  1 0 )( 2 )
dB  [
2 ( 2 9 .1 2  1 0 )( 2 )
6
 (1 9 6  1 0 )
3
6
 (1 9 6  1 0 )
1
] 2  0 .0 2 0 8 m = 2 0 .8 m m
1
] 2  0 .0 1 3 7 5 m = 1 3 .7 5 m m
24
F BC ( 0 . 3 )
kN
o
 R Ax  0
kN 
 R Ay 
 R A y  6 6 .7 6 3
45 kN
3
5
F BC ( 0 . 1)  0
4
5
 F B C  R B  2 9 .1 2
4
C
15
M
kN 
kN
4
5
F BC  0
SOLUTION (3.4)
R Ay
5 kN
B 1m 1m
C
0.75 m
R Ax A 1 m
1.5 m



E
3
A
 0:
5( 3 ) 
RE
RA 
3
5
R E ( 0 .7 5 ) 
F x  0 : R Ax  7 . 82 kN 
F y  0 : R Ay  5 . 43
7 .8 2  5 .4 3
2
2
kN 
 9 .5 2 k N
Free body-beam ABC
7.82 A
1m
(b) 
A

  1 . 018
8 ,145
3
8 ( 10
)
3
9 .5 2 (1 0 ) 2
 ( 0 .0 2 5 )
2
M
C
 0 : F BD  8 . 145
RCy
FBD
 
BD

2m
5.43
(a) 
RCx
C
B
MPa
 9 .6 9 7 M P a
4
SOLUTION (3.5)
RA
A


L cos 
L ta n 

RB
B
 0:
A
F y  0:
R B  FBC 
R A  F AC 
P
s in 
A AC 
 sin 
L AC 
L
cos 
A BC 
P
P
 tan 
L BC  L
Total weight, W   ( A A C L A C  A B C L B C ) :
W 
 PL

( sin  1cos  
cos 
sin 
) 
 PL

2
1  cos 
sin  cos 
Therefore
dW
d
P L


2
[
(s in  c o s  )
2
The foregoing reduces to
2
or
  5 4 .7
2
2
s in  c o s  ( 2 c o s  )(  s in  )  (1  c o s  )(  s in   c o s  )
3 cos   1
P
ta n 
We have
C
P
L
M
or
cos  
1
3
o
25
4
5
R E  13 . 04 kN
or
4
D
M
] 0
kN ( C )
RE (2)  0
SOLUTION (3.6)
Refer to Table 3.1 (Case C)
a  b  0 .0 9  0 .0 4 5  4 .0 5  1 0
t
3
m
2
45 mm
t
 
T
;
30  10 
6
2abt
90 mm
1 .2  1 0
2 ( 4 .0 5  1 0
Solving,
t  4 .9 4 m m
We have
  1 .5  0 .0 2 6 ra d and  a ll   L  0 .0 2 6 0 .8  0 .0 3 3 ra d m .
o
Hence
 
( a  b ) tT
2
2

2
2t a b G
( a  b )T
2
2
2 ta b G
0 .1 3 5 (1 .2  1 0 )
3
0 .0 3 
2 ( 0 .0 4 5 ) ( 0 .0 9 ) ( 2 8  1 0 ) t
2
2
9
Solving: t  5 .3 4 m m
Use
t a ll  5 .5 m m
SOLUTION (3.7)
Refer to Solution of Prob. 3.6. We now have a=b.
 
T
2
;
30  10 
6
2a t
70 mm
t
Solving:
70 mm
Similarly,
 
T
3
;
0 .0 3 3 
a Gt
Solving:
Use
1 .2  1 0
3
( 0 .0 7 ) ( 2 8  1 0 ) t
3
9
t  3 .8 m m
t a ll  4 .1 m m
26
1 .2  1 0
3
2
2 ( 0 .0 7 ) t
t  4 .1 m m
3
3
)t
SOLUTION (3.8)
Refer to Case C, Table 3.1.
We now have a=b=31.25 mm,  a ll  1 2 2 .4 5  4 .9  0 .0 8 6 ra d m .
o
T
 a ll 
2
105 10
;

6
2a t
T
2 ( 3 1 .2 5  1 0
3
) ( 4 .7  1 0
2
3
,
T  9 6 3 .8 7 N  m
,
T  3 5 4 .0 2 N  m
)
Likewise,
 a ll 
T
3
0 .0 8 6 
;
a tG
T
( 3 1 .2 5  1 0
3
) ( 4 .7  1 0
3
3
) ( 2 8 .7  1 0 )
9
T a ll  3 5 4 .0 2 N  m
Thus,
SOLUTION (3.9)
We have a  b  ( ro  ri ) 2  1 1 .7 m m Case E, Table 3.1).
From Eq. (3.11),
 m ax 
Tr
J

T (1 2 .4 )(1 0
4
3
4
)
(  2 )(1 2 .4  1 1 )(1 0
 8 7 7 (1 0 ) T
3
12
)
From Table 3.1:
 m ax 
T
2 abt

 8 3 0 .5 (1 0 ) T
3
T
2
2  (1 1 .7 ) (1 .4 )(1 0
9
)
Therefore, error in the thin-wall estimate:
8 7 7  8 3 0 .5
877
 0 .0 5 3  5 .3 %
SOLUTION (3.10)
( a ) Maximum moment occurs at midlenth B.
Thus
60 kN

a ll
d 
V(,kN)
A
3
30 kN
30 mm
14 mm
B

14 mm
30
E
D

32 M
m ax
d
3
3
32 M

m ax
a ll
32 ( 435 )
 2 6 .1 m m
6
 ( 2 5 0 1 0 )
( b ) Maximum shear is at D and E.
Hence, from Table 3.2:
x
-30
M, (N  m)
435
210
Mc
I
or
d
30 kN

 a ll 
4 V m ax
3
A
d 
1 6 V m ax
4 V m ax
3 d2 4


1 6 V m ax
3 d
2
or
2
x

Use d a ll  2 6 .1 m m
27
3   a ll
3
2
1 6 ( 3 0 1 0 )
6
3  (1 5 0  1 0 )
 1 8 .4 3 m m
SOLUTION (3.11)
12 kN
24 kN
21 kN
2m
2m
B
3m
4m
RB
D
2m
4m
A
C
RD
RA
RC
Reactions


M
D
M
F
 0 : R B  21 kN
F y  0 : R D  15 kN
 0 : R A  14 kN
C
 0 : R C  7 kN
y
21 kN
A
C
2m
14
V
(kN)
4m
7
14
+
x
_
7
M
( kN  m )
28
+
x
SOLUTION (3.12)
4
2.7
1m
1m 1m
A


B
3.13
V
(kN)
x
3.57
_
 0:
A
Fy  0 :
a ll

M
( kN  m )
R B  3 .5 7 k N  m
R A  3 .1 3 k N  m
m ax c
:
I
1 2 .5 (1 0 ) 
6
3
3 .5 6 (1 0 )( h 2 )
( 0 .0 5 )( h
3
12 )

3
3 .5 6  1 0  6
2
0 .0 5 ( h )

+
max

3 V
2 A

3
3 . 57
2 0 . 05 ( 0 . 185 )
 578 . 9  700 OK
x
SOLUTION (3.13)
b  h
2
2
 d ,
2
h
2
 d
2
 b
S 
2
We obtain
dS
db
Thus

b 
2
2
d
6

d
and h  d
3
b
2
 0:
,
h  0 .1 8 5 m
3.56
3.13
M

0.43
+
M
b
2

2
d
3
2
3
28
bh
6
2

b
6
(d
2
 b ) 
2
bd
6
2

3
b
6
SOLUTION (3.14)
c 
I 
200 ( 25 )( 162 . 5 )  150 ( 15 )( 75 )
 135 . 35 mm
200 ( 25 )  150 ( 15 )
1
12
 200  25 ( 27 . 15 )
3
( 200 )( 25 )
 135 . 35  15 ( 1352. 35 )
2
 13 . 24  10
2
We have at N.A.:
Q m a x  1 3 5 .3 5 (1 5 )( 1 3 52.3 5 )  1 3 7 .4  1 0

VQ

max
6
3
22 ( 10 )( 137 . 4  10

Ib
13 . 24 ( 10
6
)
3
3
mm
 15 . 22 MPa
)( 0 . 015 )
SOLUTION (3.15)
I 
1
12
[ 200 ( 300 )
 100 ( 200 ) ]  383 . 3 (10
3
3
3
Q  A y  200  50 (100  25 )  1 . 25 (10
*
q 
VQ
I
F (2)  q  s,
,

2F
s
6
) m
) m
4
3
VQ
I
Thus
V a ll 
3
2 (1 5  1 0 )( 3 8 3 .3  1 0

2 FI
sQ
0 .1 (1 .2 5  1 0
3
6
)
)
 92 kN
SOLUTION (3.16)
V max  2 . 4 kN
M
max
Design is based on 
all
 12 MPa :

all

M
max
c

1
2
) 
6
: 12 (10
I
wL
2
 1 . 44 kN  m
3
1 . 44 ( 10 ) b
2b
4
3

2 ,160
b
3
, b  56 . 5 mm
Check
 all  0 . 18 MPa :

max

3
3 V max
2
A
:
2 . 4 ( 10 )
3
2 2 ( 0 . 0565 ) 2
 564 kPa  810
kPa
OK .
SOLUTION (3.17)
V max 
1
2
( 50  6 )  150
Design based on 
S min 
M

all
max
kN ,
 170

all
max

1
8
( 50 )( 6 )
2
 225
kN  m
MPa :
3
225 ( 10 )
170 ( 10
M
6
Table A.7: S 4 6 0  8 1 .
)
 1 . 324 (10
6
) mm
3
(smallest possible, others will work)
Shear stress S 460  81 : A web  d ( t w )  457 (11 . 7 )  5 . 347 (10 ) mm
3

max

V max
A web

3
150 ( 10 )
5 . 347 ( 10
3
)
 28 . 1 MPa
29
 100
MPa
OK .
2
6
mm
4
SOLUTION (3.18)
M
 (w0 x
x
M
S 

x
bh
6
;
all
2 L )( x 3 )  w 0 x
2
2
w0x


6L
6L
(1)
all
bh 1
h  h1
At x=L:
3
3
w0x

6

3
6L
(2)
all
Divide Eq.(1) by (2):
2
h

2
h1
3
x
L
h  h 1 ( Lx )
or
3
3
2
SOLUTION (3.19)
RA  RB 
1
2
wL,
M
x

wLx 
1
2
bh
6
Equation (3.27), at a distance x :
At x 
L
2
2
bh 1
:

6
w
2
2
(
h
Divide Eq.(1) by Eq.(2):
or h  2 h 1
x
L
 ( Lx )
SOLUTION (3.20)
4

L
2
)
2

2

w
2

( Lx  x )
2
Lx  x
L
2
2
1

(1)
all
2
y
Es
n 
(20)200=40(103)
3
( 4  1 0 )( 3 2 5 )
 2 .8 9  1 0

126  10 
6
20 M
s
( 0 .1 6 2 5 )
2 .8 9  1 0
3
3

12
9
3
( 3 .8  1 0 )( 3 0 0 )
mm
6
;
 1 4 1 .6
w
M
3
12
 7 .3 5  1 0 
w
 M
;
 20
Ew
It 
12.5 mm
I
bh
6
4
300 mm
nM y

2
z

x
a ll
(2)
12.5 mm
s

all
200 mm

M
S 
2
wx ,
1
L
4
h1
2
1
2
4
My
I
 2 .8 9  1 0

M
w
( 0 .1 5 )
2 .8 9  1 0
3
kN  m
 112 kN  m
s
Stress in steel governs:
M  112 kN  m
SOLUTION (3.21)
Es
n 
I 
Ew
1
12
M 
 20
[( 7 5  4 0 t )( 2 2 5 )] 
3
1
8
wL 
2
[ 7 5 (1 0 0 )]  ( 6 4 .9 4 1  3 7 .9 6 9 t )  1 0
3
1
12
[3 5  1 0 ( 2 .5 )(1 2 )]  2 7 .3 4
3
1
8
kN  m
2
Therefore, we write

s

w

nM y
I
;
135  10
6

3
2 0 ( 2 7 .3 4  1 0 )( 0 .1 1 2 5 )
( 6 4 .9 4 1  3 7 .9 6 9 t )  1 0
6
,
t  1 0 .2 9 m m
Similarly

My
I
;
8  10
6

3
( 2 7 .3 4  1 0 )( 0 .1 1 2 5 )
( 6 4 .9 4 1  3 7 .9 6 9 t )  1 0
Stress in steel governs: t  1 0 .2 9 m m
30
6
,
t  8 .4 1 5 m m
6
m
4
3
m
4
SOLUTION (3.22)
Transformed Section of Brass
y
Es
n 
It  [
89.1 mm
C1
z
3
 A d ]1  [
2
bh
36
3
bh
12
 [ 3 6 ( 2 0 0 )(1 2 0 ) 
3
1
9.1 mm
43.4 mm
y  55.9 mm
 2
Eb
 A d ]2
2
200
2
 1 2 0 ( 9 .1 ) ]1
2
 [ 1 2 (1 0 0 )( 2 5 )  1 0 0 ( 2 5 )( 4 3.4 ) ] 2
3
1
C2
 15 . 43  10
1100mm
6
mm
2
4
200 mm
Thus
120 (10
6
3
) 
M ( 55 . 9  10
) 
2 M ( 89 . 1  10
)
6
15 . 43  10
M  33 . 12 kN  m
,
Similarly,
140 (10
6
15 . 43  10
3
)
6
M  12 . 12 kN  m
,
(governs)
SOLUTION (3.23)
n 
Es
Eb
 2
I t  I b  nI
s
 [
30 ( 50 )
3
12
 ( 7 .5 )

4
4
 ( 7 .5 )
]  2[
4
4
]  314 , 985
mm
4
Thus, we write
(  b ) max 
M ( 25  10

Mc
It
314 , 985 ( 10
3
)
 12
 120 (10
)
6
M  1 . 512
),
kN  m
Similarly,
(  s ) max 
3
2 M ( 7 . 5  10

nMc
It
314 , 985 ( 10
 12
)
 140 (10
)
6
M  2 . 94 kN  m
),
SOLUTION (3.24)
n 
Eb
Ea
 1.5
Ib 
Transform to brass:

64
[d
2
 ( d2 ) ] 
4
15  d
4
1024
Aluminum core: Each element of area of area has its width reduced in ratio n .
Ia 
1 
n 64
( d2 )

4
 d
4
1024 n
and
It  Ib  Ia 
 d
4
1024
( 15 
1
n
)
We have

b

Mc
It
M 
,
 bIt
d 2

 d
512
3
 b ( 15 
1
n
)
Substituting numerical values:
M 
 ( 0 .0 5 )
512
3
(3 5 0  1 0 )(1 5 
6
1
1 .5
)  4 2 0 5 .7
31
N  m = 4 .2 1 k N  m
(governs)
SOLUTION (3.25)
  1 0 5 . Apply Eq.(3.31):
o
( a )  x' 
1
2
(25  15) 
1
2
(  2 5  1 5 ) c o s 2 1 0  1 0 sin 2 1 0
o

o
x’
  5  17 . 32  5  17 . 32 MPa
 x'y'  
1
2
(  2 5  1 5 ) s in 2 1 0
 10 cos 210
o
(20)  (10)
2
 x'y'
15
o
o
x
y’
  10  8 . 66   1 . 34 MPa
( b )  1, 2   5 
x'
  5  2 2 .3 6
2
or
 1  17 . 36 MPa
 p ' '  13 . 28

  27 . 36 MPa
2
o
1

2
 p''
SOLUTION (3.26)
70
(a)
y’
 m ax
x 
30
o
x’
( b )  max 
1
2

30 70
2
cos 2(  55 )
o
  20  17 . 1   37 . 1 kPa
x
55
30 70
2

y

xy
  20  17 . 1   2 . 9 kPa
 
30  70
2
sin 2 (  55 )  47 kPa
o
[ 30  (  70 )]  50 kPa
SOLUTION (3.27)
y
50(0.866)
x’
(0.866) 
  35  90  125
50(1)
x


60

x
,

y
 0,

xy
 50 MPa
0 .8 6 6  x  0 .5 ( 5 0 )  0 .5 ( 5 0 )
F x  0:
o
o
x
 57 . 74 MPa
(C )
50(0.5)
( a ) Equations (3.31) with   1 2 5 :
o

x'

x'y'
  2 8 .8 7  2 8 .8 7 c o s 2 5 0  5 0 s in 2 5 0
o
 28 . 87 sin 250
o
 50 cos 250
o
o
  6 5 .9 8
M Pa
  44 . 23 MPa
(CONT.)
32
3.27 (CONT.)
1
2
( b )  1 , 2   2 8 .8 7  [(  2 8 .8 7 )  5 0 ]   2 8 .8 7  5 7 .7 4
2
 1  2 8 .8 7 M P a ,

2
2
 p '' 
  8 6 .6 1 M P a ,
1
2
ta n
1
x’
50
2 8 .8 7
 30
o
x
44.23
 p''
35
x’
o

x
y’
1
2
65.99
SOLUTION (3.28)
From Solution of Prob. 3.27: 

x
 

xy

6
57 . 74 ( 10
200 ( 10
2 (1  0 .3 )
200 ( 10
9
)
9
)
)

  289 μ ,
( 50  10
6

  5 7 .7 4 M P a ,
x
 
y
x
y
 0 , and 
 50 M P a.
xy
 87 μ
)  650 μ
(  x ) AC 
C
B

40
34.6
L A C  52.915
60
A
40
4 0 .8 6
289 87
2
650
2

28987
2
o
c o s 2 ( 4 0 .8 6 )
o
s in 2 ( 4 0 .8 6 )
  1 0 1  2 7 .0 7  3 2 1.6 1
 194 μ
o
D
20
o
( a ) (  x ' ) A B   1 0 1  1 8 8 c o s 2 (1 2 0 )  3 2 5 s in 2 (1 2 0 )   2 8 8 μ
o
o
 L A B  4 0 (  2 8 8  )   0 .0 1 1 5 m m
( b )  L A C  5 2 .9 1 5 (1 9 4  1 0
6
)  0 .0 1 0 3 m m
SOLUTION (3.29)

y
(a) 
y
y
  7 0 ( 0 .5 )   3 5 M P a
60.6(0.866)=52.5
 x y   7 0 ( 0 .8 6 6 )   6 0 .6 M P a
( 0 .5 ) 
30
o
140(1)
x
x

0 . 5
Fx  0 :
(b) 1 
175  35
2
x
 [( 175 2 35 )
 140  52 . 5  0  
x
 175
M Pa
 p'
1
2
 (  60 . 6 ) ] 2  70  121 . 23  191 . 23 MPa
2
 2   5 1 .2 3 M P a
p'
1
2
ta n
1

2 ( 6 0 .6 )
175 35
  15
x
x’
o
1

33
2
SOLUTION (3.30)
From Solution of Prob. 3.29:

9
2 0 0 1 0
2 (1  0 .2 5 )
G 
x 

y

x'
 175 M Pa, 
x

(
1
E
10
 80  10
 y ) 
x
6
9
2 0 0 (1 0 )
  3 5 M P a , and 

Pa,
10
6
9
2 0 0 (1 0 )
xy


xy
G

xy
 6 0 .6  1 0
8 0 1 0
6
9
  6 0 .6 M P a .
 758 
(1 7 5  3 .7 5 )  9 1 9  ,
(  3 5  4 3 .7 5 )   3 9 4 
919  394
2

9
y

919  394
2
cos 2 (141 . 5 ) 
o
o
758
2
sin 2 (141 . 5 )
 2 6 2 .5  1 4 7 .6 8  3 6 9 .2 9  7 7 9 .4 7 
x’
L B D  0.0416 m
B
C
0.05 m
A
=141.5o
x
0.075 m D
Thus
 L B D  0 .0 4 1 6 ( 7 7 9 .4 7  1 0
6
)  0 .0 3 1 3 m m
SOLUTION (3.31)
  40

x

xy
 
o
y
 100 cos 45
 100 sin 45
o
o
 70 . 71 MPa
 70 . 71 MPa
Equations (3.31):

x'


y'
 70 . 71  0  69 . 64  1 . 07 MPa

x'y'
70 . 71  70 . 71
2
 0  70 . 71 sin 80
  0  70 . 71 cos 80
y’
o

x'
40

 140 . 3 MPa
 12 . 28 MPa
x’
 x'y'
o
o
x
y'
34
SOLUTION (3.32)
  40  90  130

y
o

  40 MPa ,

 50 MPa ,
x
 25 MPa
xy
( a ) Equations (3.31):
50 40
2
 x' 

50 40
2
c o s 2 (1 3 0 )  2 5 sin 2 (1 3 0 )
o
o
 5  7 . 814  24 . 62   27 . 43 MPa
(b) 

x'y'

y'
 [(
max
 p'
1
2
  45 sin 260
o
 5  7 . 814  24 . 62  37 . 434
50  40
2
)
 44 . 316  4 . 341  39 . 975
MPa
MPa
1
 25 ] 2  51 . 48 MPa
2
2
 1 25
45
tan
 25 cos 260
o
 14 . 53
o
x’


x'
y'
x’

 x'y'
 m ax
x
 p'
x
x
y’
SOLUTION (3.33)
x”
B
AC  BD  90 . 139
x’
C
75 mm
G  E 2 .6
x 
50 mm
3 3 .6 9
1 4 6 .3
o

xy

1
E
(
 xy
G

o
1
E
( x    y )
10
E
6
[ 5 0  0 .3 (  4 0 ) ] 
6 2 1 0
E
6
x
A
y 
mm
D
y
  x ) 
6

2 5 1 0
E 2 .6
x'


10
E
6
6 5 1 0
E
[  4 0  ( 0 .3 ) ( 5 0 ) ]  
5 5 1 0
E
6
6
We have

6
10
2E
[( 62  55 )  ( 62  55 ) cos 2 ( 33 . 69
 ( 65 ) sin 2 ( 33 . 69

x"

6
10
2E
o
)] 
89 . 85
E
(10
6
o
)
)
[117  ( 7 ) cos 2 (146 . 3 )  ( 65 ) sin 2 (146 . 3 )] 
o
o
Thus
 L A C   x ' (9 0 .1 3 9 ) 
 L B D   x " (9 0 .1 3 9 ) 
8 9 .8 5  1 0
2 1 0 1 0
6
9
2 9 .8 4  1 0
2 1 0 1 0
9
6
(9 0 .1 3 9 )  0 .0 3 8 6
mm
(9 0 .1 3 9 )  0 .0 1 2 8
mm
35
29 . 84
E
6
(10 )
SOLUTION (3.34)
Vessel is thin walled.
1.2 kN/m
B
0.9 m
M
V
1.5 m
VQ
 xy 
A
 0 at A and C
Ib
V  2 .7  1 .2 (1 .5 )  0 .9 k N
M  2 .7 (1 .5 ) 
C
2.7 kN
 2 .7
1
2
(1 .2 )(1 .5 )
2
kN  m
Point A

x'

Mc
I
(a)
  
y
(b)
 m ax 
1
2

2 .7 ( 0 .4 5 )
3
 ( 0 .4 5 ) ( 3  1 0
 2
a
3
)
 6 .3 M P a   1 ,
( 1   2 ) 
1
2

  1 .4 1 5 M P a ,

2
a

pr
2t

4 2 ( 0 .4 5 )
2 ( 3 1 0
3
)
 3 .1 5 M P a
 3 .1 5  1 .4 1 5  1 .7 3 5
( 6 .3  1 .7 3 5 )  2 .2 8 2 5
M Pa
M Pa
Point C
(a)

(b)
 m ax 
x'
 2  1 .4 1 5  3 .1 5  4 .5 6 5 M P a ,
 1 .4 1 5 M P a ,
1
2
( 1   2 ) 
1
2
( 6 .3  4 .5 6 5 )  0 .8 6 8
 1  6 .3 M P a
M Pa
SOLUTION (3.35)
Refer to Solution of Prob. 3.34.

(a)

x
xy



 1, 2 
Thus
x'
VQ
Ib
0 .9
 rt



a
 0  3 .1 5  3 .1 5 M P a ,
0 .9 (  r t )(
2r

0 .9
 ( 0 .4 5 ) ( 3  1 0
3
)
 0 .2 1 2 2 M P a
1
 [( 6 .3 23 .1 5 )  ( 2 .1 2 2 ) ] 2  4 .7 2 5  1 .5 8 9
2
 1  6 .3 1 4 M P a
1
2
    6 .3 M P a
(see Table A-3)
2
or
( b )  m ax 
y
)
3
 r t(2t)
6 .3  3 .1 5
2

 2  3 .1 3 6 M P a
( 1   2 )  1 .5 8 9 M P a
36
SOLUTION (3.36)
 
x’
P
A

Tr
J

3
1 1 0 (1 0 )
2
2
 ( 0 .0 2 5  0 .0 1 2 5 )
 7 4 .7 M P a
74.7

 
x
3
2 .3 (1 0 )( 0 .0 2 5 )

4
 9 9 .9 6 M P a
4
( 0 .0 2 5  0 .0 1 2 5 )
2
99.96
  50  90  140
o
x’

Equations (3.31):
x'
140

o
 x'y'
x'


7 4 .7
2
7 4 .7
2
c o s 2 8 0  9 9 .9 6 s in 2 8 0
o
o
 1 4 2 .3 M P a
x
 x'y'  
7 4 .7
2
s in 2 8 0  9 9 .9 6 c o s 2 8 0
o
o
 1 9 .4 2 M P a
y’
SOLUTION (3.37)
410 mm
y
A
120 18
z
24 kN
M
40
50
120
24
3
4
18 kN
y

A  40 (120 )  4 . 8  10
x
A
I 
x
 xy
1
12
( 40 )( 120 )
3
3
mm
 5 . 76  10
2
6
mm
4
M  18 ( 0 . 12 )  24 ( 0 . 41 )  12 kN  m
We have at A:
x  
P
A

Mc
I
3
1 8 (1 0 )
 
4 . 8 (1 0
3
)
 (  3 . 75  104 . 17 )  10
 xy  
VQ
Ib
 
3
2 4 1 0 ( 2 2 1 0
5 .7 6  1 0
6
100 . 4
2
2
6
)
( 0 .0 4 )
3
1 2 (1 0 )( 0 . 0 5 )

5 . 7 6 (1 0
6
6
)
Q  4 0 (1 0 )(5 5 )  2 2 (1 0 ) m m
 100 . 4 MPa
  2 .2 9 M P a
Thus

or
1,2

100 . 4
2
 [(
)
 1  1 0 0 .5 M P a
2
 2   0 .0 5 M P a
 m a x  5 0 .2 8 M P a
and
tan 2  p ' 
or
 p '   1 . 31
2
xy

x
o

1
 (  2 . 29 ) ] 2  50 . 2  50 . 25
2 (  2 . 29 )
100 . 4
 1 . 31
37
3
3
SOLUTION (3.38)
y
y
B
B
16
M
P
yo
P=40
kN
C
62
z
A
Z
10 mm
y
A
8
45
From Z axis:
6 2  1 6 (3 9 )  4 5  8 ( 4 )
y 
 2 9 .7 m m
62  16  45  8
y 0  1 0  y  3 9 .7 m m
A  1 6  6 2  4 5  8  1 .3 5 (1 0 ) m m
3
2
About the z axis:
I 
1 6 (6 2 )
3
 1 6  6 2 ( 9 .3 ) 
2
4 5 (8 )
12
3
 4 5  8 ( 2 5 .7 )
2
12
 1 0 ( 0 .3 1 8  0 .0 8 6  0 .0 0 2  0 .2 3 8 )  0 .6 4 4  1 0
6
6
mm
4
M  4 0  0 .0 3 9 7  1 .5 8 8 k N  m
Thus,

a

P

A
40  10
Mc
( b ) B 
3
1 .3 5  1 0

I
(  b ) A  9 9 .4
3
 2 9 .6 M P a
 1 5 8 8 ( 0 .0 4 0 3 )
0 .6 4 4  1 0
2 9 .7
6
  9 9 .4 M P a
 7 3 .3 M P a
4 0 .3

A
 2 6 .3  7 3 .3  9 9 .6 M P a

B
 2 6 .3  9 9 .4   7 3 .1 M P a
SOLUTION (3.39)
A  b h  2 5 (1 0 0 )  2 .5 (1 0 ) m m
3
I 
2
2 5 (1 0 0 )
12
P  50 kN
M  5 0  1 0 ( 0 .0 5 )  2 .5 k N  m
3
( a ) At the top fibers:
t 
P
A

Mc
I

5  10
2 .5 (1 0
3
3
2 .5  1 0 ( 0 .0 5 )
3

)
2 .0 8 (1 0
6
 20  60  40 M Pa
At the bottom fibers:
 b  20  60  80 M Pa
38
)
40.3
3
 2 .0 8 (1 0 ) m m
6
4
SOLUTION (3.40)
A  b h  2 5 (1 5 0 )  3 .7 5 (1 0 ) m m
3
I  bh
1 2  2 5 (1 5 0 )
3
2
1 2  7 .0 3(1 0 ) m m
3
6
4
At the bottom fibers:

P

b
Mc

A
P

3 .7 5 (1 0
I
3

0 .0 7 5 P ( 0 .0 7 5 )
)
7 .0 3 (1 0
6
)
Solving,
120  10
 2 6 6 .7 P  8 0 0 .1 4 P
6
Pa ll  1 1 2 .5 k N
SOLUTION (3.41)
J 

[( 0 .0 6 )  ( 0 .0 5 0 ) ]  1 0 .5 4 (1 0
4
2
c  0 .0 6
4

m,
45
6
) m
 8 0 M P a   m ax 
o
4
Tc
J
Thus
T ( 0 .0 6 )
8 0 (1 0 ) 
6
1 0 .5 4 (1 0
6
)
T  1 4 .0 5 k N  m
,
SOLUTION (3.42)
( a )  1,2 
500  800
2
1
   [( 5 0 0 2 8 0 0 )  ( 3 52 0 ) ] 2  6 5 0   2 3 0 μ
2
2
or
 1  880 μ ,
( b )  max  G 
max
 2  420 μ ,
3
70 ( 10 )

2 (1  0 .3 )

max
 460 μ
( 460 )  12 . 38 MPa
SOLUTION (3.43)
(a) 
max
 2 [(
200  600
2
A
1
)
2
 ( 400
) ] 2  566 μ
2
2
  149
( b ) x' 
200  600
2

200 600
2

400
2
o
c o s 2 (1 4 9 )
x
C
s in 2 (1 4 9 )  1 3 0 μ
o
o
L A . C  2 7 .3 3 3 m m
Thus
 L A C   x ' L A C  1 3 0 (1 0
6
)( 2 7 .3 3 3 )  3 .5 5  1 0
3
mm
SOLUTION (3.44)
 1,2 
50  250
2
 [( 5 0 22 5 0 )  ( 
2
150
2
1
) ]2  150 μ  125 μ
2
or
1  275   2  25 
(CONT.)
39
3.44 (CONT.)
Apply Hooke’s Law (with  z  0 ):
6
2 7 5 (1 0
and
2 5 (1 0
6
) 
) 
( 1    2 )
1
9
2 1 0 (1 0 )
(1)
( 2    1 )
1
9
2 1 0 (1 0 )
(2)
Solving
 1  65 . 2 MPa ,

 24 . 8 MPa
2
SOLUTION (3.45)

a

 
pr
2t
pr
t

p (500 )

P
A
2 (1 0 )
0 .0 2 
2  ( 0 .5 )( 0 .0 1 )
 25 p  2 M Pa
x


 50 p
y
50 20
s in 4 0
A=1
40
o
cos 40
5 0  ( 2 5 p  2 ) s in
2
40
o
 50 p cos 40
2
p all  1 . 281 MPa
or
25p-2
o
F y  0:

F x  0:
2 0   ( 2 5 p  2 ) s in 4 0 c o s 4 0
o
 5 0 p c o s 4 0 s in 4 0
o
o
50 p
p all  1 . 546
or
o
MPa
SOLUTION (3.46)
( a )  x y   (  x '   y ' ) s in (  2  )   x ' y ' c o s (  2  )
o
o
0   ( 2 4 0  4 1 0 ) s in (  6 8 )  
o
Then
1
x 
2
1

( x '  
y'
)
1
2
( x '  
1
(240  410) 
2

cos(68 ) ,
o
x'y'
1
) c o s (  2 ) 
o
y'
x'y'

2
( 2 4 0  4 1 0 ) co s( 6 8 ) 
o
2
 421 μ
s in (  2  )
o
x'y'
1
( 4 2 1) s in (  6 8 )
o
2
 9 8 .2 μ

y
1

( x '  
2
1

y'
)
1
2
(240  410) 
2
( x '  
1
1
) c o s (  2 ) 
o
y'
2
( 2 4 0  4 1 0 ) co s( 6 8 ) 
o
2
( b ) Hooke’s Law, with  t   y ,  a   x , and  t  2  a

So
y
x


x
(1  2 )
E
552
9 8 .2
 5 .6 2 

y

2 
1  2

x
(2   )
E
  0 .3 5
,
40
s in (  2  )
o
x'y'
1
2
 552 μ
x 

o
( 4 2 1) s in ( 6 8 )
o
o
SOLUTION (3.47)
 a  880 
( a  0 ),
 b  320 
o
( b   6 0 ),
 a   x c o s  a   y s in  a  
2
Thus
8 8 0 (1 0
6
xy
)   x c o s 0   y s in 0  
2
o
( c   1 2 0 )
o
s in  a c o s  a
2
2
 c  60 
o
o
o
 x  880 μ
o
s in 0 c o s 0 ,
xy
Likewise,
 b   x c o s  b   y s in  b   x y s in  b c o s  b
2
3 2 0 (1 0
3 2 0 (1 0
6
6
2
)   x c o s (  6 0 )   y s in (  6 0 )  
2
2
o
)  0 .2 5  x  0 .7 5 
y
 0 .4 4 3 
 c   x c o s  c   y s in  c  
2
and,
 6 0 (1 0
6
 6 0 (1 0
6
o
2
xy
s in (  6 0 ) c o s (  6 0 )
o
xy
(1)
xy
s in  c c o s  c
)   x c o s (  1 2 0 )   y s in (  1 2 0 )  
2
2
o
)  0 .2 5  x  0 .7 5 
y
o
 0 .4 4 3 
o
s in (  1 2 0 ) c o s (  1 2 0 )
o
xy
o
(2)
xy
Subtract Eq. (2) from Eq. (1):
So
3 8 0 μ   0 .8 8 6 
xy

and then from Eq. (1):
xy
  4 2 8 .9 μ



y
  1 0 2 .1 μ

2
880
A
388
2
O
 avg 
x
p
288.1
'
 ()
C
R
y

1
1
2
( x   y )
(8 8 0  1 0 2 .1)  3 8 9 μ
2
R  [ (8 8 0  3 8 9 )  ( 
2
4 2 8 .9
2
 5 3 5 .8 μ
 1,2   a vg  R
 1  3 8 9  5 3 5 .8  9 2 4 .8 μ
 2   1 4 6 .8 μ
ta n 2 
p
'
2 8 8 .1
880  389
 p '  1 6 .8
,
x’
o
x
16.8o
924.8 
y
A
y’
146.8 
41
2
1
) ]2
SOLUTION (3.48)
y
 xy 

A
x
Tc
y
2
( x   y )  0
    a v g  R s in 2  
 xy

 0

3

xy
 xy
G
x  y  0

Mohr’s circle for strain:
 avg 
c
2
J


    25
o
x

1
J 
Tc
s i n 2 
2G
R 
1

2
1

2
xy

2

x
2
xy
A’ x’
R
s in 2 
  
O
s i n 2
y
2G J
This gives
2G J  
T 
c s in 2 
Substitute the given data:
 (8 0 .5  1 0 )( 0 .0 4 4 ) ( 6 0 0  1 0
9
T 
3
( 0 .0 4 4 ) s in 5 0
6
o
)
 2 2 5 .0 3 8 k N  m
SOLUTION (3.49)
Using Eqs. (3.39), we have
 x  300 μ,

y

 150 μ,
xy
 2(375)  (300  150)  600 μ
Equations (3.38) are thus,
300  150
 1, 2 
p

(
450
2
1  300 μ ,


1
2
ta n
1
)  (
2
2
600
)
2
 175  375
2
 2  450 μ
[
600
300  150
]  2 6 .6
o
 x '   1 7 5  2 2 5 c o s 5 3 .1  3 0 0 s in 5 3 .1   4 5 0    2
o
o
Thus,
 p "  2 6 .6
y’
300 
o
450 
x’
26.6o
x
42
SOLUTION (3.50)
Given: D  5 0 m m , t  1 5 m m , P  2 5 k N . Refer to Figs. P3.50 and C.1:
d
r
r d
D d
K
32
33
34
35
9
8.5
8
7.5
0.28
0.26
0.24
0.21
1.56
1.52
1.47
1.43
1.64
1.66
1.62
1.7
t
We have
6
 a ll  K t
P
A
;
150 (10 )
1.9
 Kt
3
25 (10 )
0 .0 1 5 h
In this equation, minimum h is reached when K t is minimum. Thus, use
h  34 mm
SOLUTION (3.51)
D
d

45
30
 1.5 ,

r
d
6
30
 0 .2
Figure C.1, K t  1.7 2 .

Hence

and
Pall  A 
nom

max
1 . 72

nom
210 1 . 5
 81 . 4 MPa
1 . 72
 ( 30  12 )( 81 . 4 )  29 . 3 kN
SOLUTION (3.52)
At the notch and hole:

nom

P
( D  d h  2 r )t

12  10
3
( 0 .0 9  0 .0 1 5  2  0 .0 0 7 5 ) ( 0 .0 1)
 20 M Pa
For the notch : (see Fig. C.1):
D

90
 1 .2
75
d
r
7 .5

 0 .1,
75
d
K t  1 .7 8
 m a x  K t  n o m  1 .7 8 ( 2 0 )  3 5 .6 M P a
For the hole (see Fig. C.5):
dh

15
 0 .1 6 7 ,
90
D
K t  2 .5
 m a x  K t  n o m  2 .5 ( 2 0 )  5 0 M P a
Hence
SOLUTION (3.53)
Kt 

m ax

nom

180
 1 .6 3 6 and D d  1 .5
110
From Fig. C.1: r d  0 .2 4 . Then
D  2r  d
(CONT.)
43
3.53 (CONT.)
4 0 .6 2 5  2 ( 0 .2 4 d )  d  1 .4 8 d
gives
or
d  2 7 .4 5 m m
r  6 .5 9 m m
SOLUTION (3.54)
For
r
 0 .2 0 :
d
D  2r  d ;
4 0 .6 2 5  2 ( 0 .2 d )  d  1 .4 d
or d  2 9 .0 2 m m We thus have
D
 1 .4
d
Figure C.1 gives K t  1 .7 . Hence,

 3 5 0 (1 0 )  1 .7
Pa ll
6
m ax
(1 2 .5  1 0
3
) ( 4 0 .6 2 5  1 0
3
)
Pa ll  1 0 4 .5 k N
or
SOLUTION (3.55)
(ksi)

avg
1

(
2

1
x
 y)
(1 6 8  8 4 )  1 2 6 M P a
2
3
O 
R 
C

(
x

y
1 (MPa)

(
168  84
2
 5 9 .4 M P a
avg =59.4
( a )  1,2  
 R
avg
 1  1 2 6  5 9 .4  1 8 5 .4 M P a
 2  6 6 .6 M P a
 3  21 M Pa
( b ) ( m a x ) a 

1
2
1
( 1   3 )
[1 8 5 .4  (  2 1) ]  1 0 3 .2 M P a
2
44
) 
2
2
2
xy
)  (42)
2
2
SOLUTION (3.56)
(MPa)

avg
avg
1

(
2

1
x
 y)
(5 0  0 )  2 5 M P a
2
R
3
O
1
C

R 
(
x

y
) 
2
2
(MPa)


50  0
(
2
xy
)  (25)
2
2
2
 3 5 .3 6 P a
( a )  1,2  
 R
avg
 1  2 5  3 5 .3 6  6 0 .3 6 M P a
 2   1 0 .3 6 M P a
 3  60 M Pa
( b ) ( m a x ) a 
1
( 1   3 )
2

1
[ 6 0 .3 6  (  6 0 ) ]  6 0 .1 8 M P a
2
SOLUTION (3.57)
(ksi)
avg =42

avg
1

(
2
35

1
x
 y)
(7 0  1 4 )  4 2 M P a
2

3
C
1
(MPa)
R 

(
x

y
) 
2
2

(
70  14
2
xy
)  (56)
2
2
2
 6 2 .6 M P a
( a )  1,3  
avg
 R
 1  4 2  6 2 .6  1 0 4 .6 M P a
 3  4 2  6 2 .6   2 0 .6 M P a
 2  35 M Pa
(CONT.)
45
3.57 (CONT.)
( b ) ( m a x ) a 

1
2
1
( 1   3 )
[1 0 4 .6  (  2 0 .6 ) ]  6 2 .6 M P a
2
SOLUTION (3.58)
( a ) In the yz plane:
 ' =35
70 MPa
(MPa)
y
21 MPa
z
21
C
3
R
 (MPa)
O 1
70
We have
R 
35  21
2
2
 4 0 .8 2 M P a
Thus
 1  R   '  4 0 .8 2  3 5  5 .8 2 M P a
 3   R   '   4 0 .8 2  3 5   7 5 .8 2 M P a
 2  28 M Pa
( b ) Using Eqs. (3.58) :
 oct 
1
3
1
[ ( 5 .8 2  2 8 )  (  2 8  7 5 .8 2 )  (  7 5 .8 2  5 .8 2 ) ] 2
2
2
2
 3 3 .4 9 M P a
 oct 
1
3
(5 .8 2  2 8  7 5 .8 2 )   3 2 .6 7 M P a
From Eq. (3.50),
 m a x  12 (5 .8 2  7 5 .8 2 )  4 0 .8 2 M P a
SOLUTION (3.59)
( a ) Using Eq. (3.47):
(70   i )
0
56
0
(1 4   i )
0
56
0
(1 4   i )
 0
(CONT.)
46
3.59 (CONT.)
Expanding,
(1 4   i )[( i  7 0 )(  i  1 4 )  3 1 3 6 ]  0
or
 1  42 M Pa,
 2  14 M Pa,
 3  98 M Pa
( b ) From Eq. (3.50),
 m a x  12 ( 4 2  9 8 )  7 0 M P a
acts on planes shown in Fig. 3.41.
( c ) Using Eqs. (3.52), we have
 o c t  13 ( 4 2  1 4  9 8 )   1 4 M P a
 oct 
1
3
1
[( 4 2  1 4 )  (1 4  9 8 )  (  9 8  4 2 ) ] 2
2
2
2
 6 0 .4 9 M P a
They act on planes depicted in Fig. 3.43.
SOLUTION (3.60)
The principal stresses are  1  8 4 M P a ,  2  6 3 M P a , and  3   1 2 6 M P a .
( a ) By Eq. (3.50),
 m ax 
1
2
(8 4  1 2 6 )  1 0 5 M P a
acts on planes shown in Fig. 3.41.
( b ) Applying Eqs. (3.52):
 o c t  13 (8 4  6 3  1 2 6 )  7 M P a
 oct 
1
3
1
[ (8 4  6 3 )  ( 6 3  1 2 6 )  (  1 2 6  8 4 ) ] 2  9 4 .4 4 M P a
2
2
2
They act on planes shown in Fig. 3.43.
SOLUTION (3.61)
The principal stresses are  1  2 9 7 .5 M P a ,  2  3 6 .8 2 M P a , and  3   5 4 .7 4 M P a .
The direction cosines:
l  cos 40
o
m  cos 60
n  cos 66 . 2
o
o
Thus, by Eqs. (3.51), we obtain
  2 9 7 .5 (c o s 4 0 )  3 6 .8 2 (c o s 6 0 )  5 4 .7 4 (c o s 6 6 .2 )
o
2
o
2
o
2
= 174.9 MPA
  [( 2 9 7 .5  3 6 .8 2 ) (c o s 4 0 ) (c o s 6 0 )
2
2
o
o
2
 (3 6 .8 2  5 4 .7 4 ) (c o s 6 0 ) (c o s 6 6 .2 )
2
o
2
o
2
1
 (  5 4 .7 4  2 9 7 .5 ) (c o s 6 6 .2 ) (c o s 4 0 ) ] 2
2
o
2
o
2
 1 4 8 .9 M P a
End of Chapter 3
47
CHAPTER 4
DEFLECTION AND IMPACT
SOLUTION (4.1)
( a ) We have A req . 
A req . 
and
 48 mm
3
3
10 ( 10 )( 6  10 )

PL
)
250 1 . 2
all
E
3
10 ( 10

P

5 ( 200  10
3
2
 60 mm
)
2
Since 60  48 mm ,
d 
(b) k 
AE
L


( 6 0 1 0
6
4 ( 60 )

4A
 8 . 74 mm .

9
)( 2 0 0  1 0 )
 2 (1 0 ) k N m
3
6
SOLUTION (4.2)
Refer to Example 4.4. Use numerical values for bar AB.
Cross-sectional area: A A B  1 2  8  9 6 m m .
2
Stress:


AB

F
A
4 0 1 0
9 6 (1 0
3
6
 4 1 6 .7 M P a  4 3 5 M P a
)
OK.
Deflection:
 AB 
FL
AE
k AB 


 2 .3 8 m m
3
3
4 0 (1 0 )

F
3
4 0 (1 0 )( 0 .6 )
( 9 6 )(1 0 5 )(1 0 )
2 .3 8 (1 0
3
 1 6 .8 1(1 0 ) k N m
3
)
SOLUTION (4.3)
The cross-sectional area:

A 
4
(D
2
d )
2
D
4
2
[1  ( Dd ) ]  0 .5 0 2 7 D
2
2
Also
A 
7 5 1 0

P

3
1 4 0 1 0
6
 0 .5 3 6  1 0
3
m
2
Equating these,
D
2
 1 .0 6 6 2  1 0
3
D  0 .0 3 2 7 m
2
m ,
It follows that
 
k 

PL
AE
P


3
7 5  1 0 ( 0 .3 7 5 )
( 0 .5 3 6  1 0
7 5 1 0
3
0 .7 2 9  1 0
3
3
9
)( 7 2  1 0 )
 0 .7 2 9  1 0
3
m
 1 0 2 .8 8 M N m
SOLUTION (4.4)
Statics:
RA
36 kN
R A  RB  36
RB
kN
(1)
Deformations and Compatibility: Assume gap closes.
(CONT.)
48
4.4 (CONT.)
0 .3 5  1 0
3

1

(1 2 .5  1 0
3
2
9
) (1 2 0  1 0 )
( 0 .2 R A  0 .2 5 R B )
4
R A  1 .2 5 R B  2 5, 7 7 0 .8 7 7
or
(2)
Solving Eqs.(1) and (2):
R B  4 .5 4 6
kN
R A  3 1 .4 5 4
kN
Since the answer for R B is positive, the gap closes, as assumed.
SOLUTION (4.5)
Increase in length due to  T (unrestrained):
t 
  (T )L
 (12  10
6
)( 120
o
)( 250 )  ( 23  10
6
)( 120
o
)( 300 )  1 . 188 mm
( a ) Compressive axial force P
 P   t  1  1 . 188  1  0 . 188 mm
(1)
But

P 
PL
AE
P ( 0 .2 5 )

5 0 0 (1 0
6
9
)( 2 1 0  1 0 )
P ( 0 .3 )

1 0 0 0 (1 0
6
(2)
9
)( 7 0  1 0 )
Equating Eqs.(1) and (2):
0 .1 8 8 (1 0
3
)  2 .3 8 1(1 0
9
) P  4 .2 8 6 ( 1 0
9
)P
P  28 . 2 kN
or
( b ) Change in length of aluminum bar
 a  (  t ) a  (  P ) a   a (  T ) L a  4 .2 8 6 ( 1 0
 ( 23  10
6
)( 120
o
)( 0 . 3 )  4 . 28 (10
9
9
)P
)( 28 . 2  10 )  0 . 707
3
mm
SOLUTION (4.6)
Refer to Solution of Prob.4.5:
( a )  P   t  1 . 188 mm
and
1 . 188 (10
( b )  a  0 .8 2 8 (1 0
3
3
)  ( 2 . 381  4 . 286 )10
)  4 .2 8 (1 0
9
9
P , P  178 . 2 kN
)(1 7 8 .2  1 0 )  0 .0 6 5 3 m m
3
SOLUTION (4.7)
Let the compressive axial force in the pipe be R.
Moment equilibrium about point A:
R 
Pa
b

1 2 (1 .3 )
 4 4 .5 7 k N
0 .3 5
a
P
b
C
A
R
C

B
B
(CONT.)
49
4.7 (CONT.)
The beam can rotate only about A
and the deflections  B and  C can be
described by the rotational angle  as
 B  a  and  C  b
a
B 
Hence,
b
C
(1)
The contraction of pipe  C is as follows:
C 
RL
RL

(  4 )( D
AE
2
(2)
 d )E
2
Inserting Eq.(2) into Eq.(1) gives
B 
 b(D  d )E
2
2
 0 .3 2 9  1 0
4 ( 4 4 .5 7  1 0 )( 0 .6 2 5 )(1 .3 )
3
4RLa
3

 ( 0 .3 5 )( 0 .1 0 5  0 .0 9 5 )( 2 0 0  1 0 )
2
2
9
m = 0 .3 2 9 m m
SOLUTION (4.8)
(a) b  
P Lb
P La
a  
Eb A
La
a  b;

Ea
Ea A
Lb
(1)
Eb
La  Lb  L
(2)
Solving,


Eb
1
)  L
Lb  L (
E
Ea  Eb

1 a

Eb



,



La  L  Lb
Introducing the given data:




1
L b  0 .6 
  0 .3 6 7 m ,
70
1

110 

(b)  

PL
P

A

AE
3

1 0 0 (1 0 )
(  4 ) ( 0 .0 4 )
2
(
Li
Ei
L a  0 .2 3 3 m
 a   b
0 .3 6 7
110

0 .2 3 3
70
)
1
10
9
 0 .2 6 5  0 .2 6 5  0 .5 3 m m
50
SOLUTION (4.9)
C
FD
Q=12 kN
A

2b
 A  2 D
b
PCD = -20 kN
D
FD =24 kN
D
B
D
0.15 m
PDE =4 kN

M
B
 0 : 2b Q  b FD
E
P=4 kN
FD  2 Q  2 4 k N
(a) D 

A
0.3 m
 2 0 (1 0 )(3 0 0 )
3
PL

AE
130  10
6
(7 0  1 0 )
9
 0 .6 6 m m 
 2  D  1 .2 2 m m 
(b) E 

PL


AE
1
130  70
10
130  10
6
3
(7 0  1 0 )
9
( 0 .6  6 )   0 .5 9  1 0
[ 4  0 .1 5  2 0  0 .3 ]
3
m  0 .5 9 m m 
SOLUTION (4.10)
PD E   4 k N
(a) D 

A
PD C   2 8 k N
 2 8 (1 0 )(3 0 0 )
3
PL

AE
130  10
6
(7 0  1 0 )
9
 0 .9 2 m m 
 2  D  1 .8 4 m m 
(b) E 

PL

AE
1
130  70

10
130  10
6
3
(7 0  1 0 )
9
[  4  0 .1 5  2 8  0 .3 ]
(  0 .6  8 .4 )   9 8 9  1 0
3
m  0 .9 9 m m 
SOLUTION (4.11)
F
C
120 mm
F
B
180 mm
(CONT.)
51
4.11 (CONT.)
T CD  F ( 0 . 12 )  500
N m
or
F  4 . 167
( a )  AB 
7 5 0 (1 .2 )

TL
GJ
T AB  4 . 167 ( 0 . 18 )  750
kN
9
7 9 (1 0 )

N m
 0 .0 2 8 ra d
( 0 .0 2 2 5 )
4
2
CD 
5 0 0 (1 .8 )
9
( 7 9 1 0 )

 0 .0 7 7 ra d
( 0 .0 1 7 5 )
4
2
 D   CD  1 . 5  AB  0 . 119
Thus
(b) 
AB

2TAB
 c
2 (750 )

3
 ( 0 .0 2 2 5 )
rad  6 . 82
o
 4 1 .9 2 M P a
3
SOLUTION (4.12)
 all 
TC D 
FC 
 125
150
1 .2
3
2
 a ll (  d )
T CD
6

3
1 2 5  1 0 (  0 .0 6 5 )

16
( r gear ) C
MPa
 6 .7 4 k N  m
16
 56 . 167
6 . 74
0 . 24 2
kN  m
Hence, we have
T AB  56 . 167 ( 0 .236 )  10 . 11 kN  m
 all 
Thus
16 T AB

3
d1
d1 
3
,
3
16 ( 10 . 11  10 )
6
 ( 125  10 )
or
d 1  74 . 4 mm
SOLUTION (4.13)
( a ) Polar moment of inertia for a hollow cylinder is
J 

(c  b ) 
4
2
4
c
4
2
[1  ( bc ) ]
4
(1)
From Eq. (4.10), we have

J
c
T
 m ax

4 .5  1 0
3
1 4 0 1 0
6
 3 2 .1 4 3  1 0
6
m
2
 3 2 .1 4 3 m m
Note that, b  0 .5 c . Using Eqs. (1) and (2) then
c
2
3
[1  ( 0 .5 ) ]  3 2 .1 4 3,
c  2 .7 9 5 m m
4
So,
D  2 c  5 .5 9 m m ,
(b)  
TL
GJ
(c) k 
T



d  0 .5 D  2 .7 9 5 m m
3
( 4 .5  1 0 )( 0 .2 5 )
9
( 7 9  1 0 )( 3 2 .1 4 3  1 0
3
4 .5 (1 0 )
0 .1 5 8 5
6
)( 2 .7 9 5  1 0
3
)
 0 .1 5 8 5 ra d  9 .0 8 1
 2 8 .3 9 k N  m ra d
52
o
2
(2)
SOLUTION (4.14)
TBC  5 6 0 N  m
TC D   8 4 0 N  m
( a ) Shaft BC
J BC 

d


4
32
(3 4 )  1 3 1, 1 9 4 m m
4
LBC  6 2 5 m m
4
32
TBC  5 6 0 N  m
So
 BC 
TBC L BC

G J BC
5 6 0 (6 2 5  1 0
3
)
( 7 9  1 0 ) (1 3 1, 1 9 4  1 0
9
 0 .0 3 4 r a d  1 .9 5
12
o
)
( b ) Shaft CD
J CD 


d2 
4
32
(2 5)  3 8, 3 5 0 m m
4
4
32
LCD  7 5 0 m m
TC D   8 4 0 N  m
CD 
TC D LC D

G J CD
 8 4 0 (7 5 0  1 0
3
)
(7 9  1 0 )(3 8, 3 5 0  1 0
9
12
  0 .2 0 8 r a d   1 1 .9 2
o
)
Hence
 B D   B C   C D  0 .0 3 4  0 .2 0 8   0 .1 7 4 ra d   9 .9 7
o
SOLUTION (4.15)
The polar moment of inertia for the shaft is
J 

32
 d ) 
4
(D
4

(5 0  3 5 )  4 6 6 , 2 6 9 m m
4
32
4
4
Equilibrium Condition. From the free-body diagram of appropriate portions of
the shaft:
T A C   1 .1 k N  m
T C D  1 .9 k N  m
T D E  2 .6 k N  m
The results are shown on the torque diagram in Fig. S4.15
( a ) Angle of twist. The shear modulus of elasticity G is a constant for the entire shaft.
Through the use of Eq. (4.9), we obtain
A 



TL
GJ
1
GJ
10
( T A C L1  T C D L 2  T D E L 3 )
3
9
( 7 9 1 0 ) ( 4 6 6 , 2 6 9 1 0
12
)
[ (  1 .1) ( 0 .4 5 )  (1 .9 ) ( 0 .3 7 5 )  ( 2 .6 ) ( 0 .6 2 5 ) ]
 0 .0 5 0 0 ra d = 2 .8 6
o
Comments: A positive result means that the gear will rotate in the direction of the
applied torque at free end.
( b ) Maximum shear stress. The largest stress takes place in region DE, where
magnitude of the highest torque occurs (Fig. S4.15) and J is a constant for the
shaft. Applying the torsion formula:
 m ax 
TD E (
J
D
2
)

3
( 2 .6  1 0 )( 2 5  1 0
4 6 6 , 2 6 9 1 0
3
12
)
 1 3 9 .4 M P a
(CONT.)
53
4.15 (CONT.)

Hence, n 
y
 m ax

210
1 3 9 .4
 1 .5
Comments: For the situation under consideration, a safety factor of 1.5 to 2 is to
be selected (see Sec. 1.6)
2.6
T
(kN  m)
1.9
A
C
D
-1.1
x
E
Figure S4.15
SOLUTION (4.16)
 

or
TL
GJ
1.5
180 
;
0 . 02618


T ( 0 .5  h )
GJs
1000
79 ( 10
9
)
[


Th
GJh
0 .5  h
( 0 . 02 )
4

h
4
( 0 . 02 )  ( 0 . 011 )
4
]
2
Solving h  1 9 7 m m
SOLUTION (4.17)
T
Assume T A as redundant.
TA
x
 TB
Tx  TA 
x
T
0
1
d x  T A  T1 x
Deformation:

A 
Tx dx
GJ

TA
GJ
 A  0;
Geometry:
T A  T B  T1 L ,
Statics:
L
 dx
T1

GJ
0
TA 
1
2

L
xdx 
0
TAL
GJ

T1 L
2
2GJ
T1 L
TB 
1
2
T1 L
SOLUTION (4.18)
Conditions of equilibrium gives: R A  P and M
For portion AC
E Iv1 ''  P x
E I v1 ' 
E I v1 
1
2
1
6
B
 P L 2 as shown in Fig. S4.18.
P
A
Px
RA=P
x
P x  C1
B
C
Px
PL/2
x
2
L/2
P x  C1x  C 2
3
L/2
Figure S4.18
(CONT.)
54
4.18 (CONT.)
For portion CB
1
E I v 2 '' 
E I v1 ' 
1
PLx  C3
2
1
E Iv 2 
PL
2
PLx  C3x  C4
2
4
Boundary conditions lead to
v1 ( 0 )  0 :
C2  0
v2 (0 )  0 :
C3  0
Then
v1 (
L
)  v2 (
2
v 1 '(
L
L
2
)  v 2 '(
2
P
):
(
6
L
2
P
):
2
Solving C 1  
L
(
2
3
PL
)  C1(
3
L
2
L
2
2
11
C4  
8
L
PL(
4
)  C1  
2
P
) 
4
P
L(
2
L
)  C4
2
)
2
PL
3
48
At B(x=0):
vB  v2 (0 )  
11
11
PL 
3
48
PL 
3
48
SOLUTION (4.19)
4
EI
d v
dx
4
 EIv ' ' ' '   w 0
x
L
w0 x
EIv ' ' '  
,
2L
2
 c1 ,
EIv ' '  
Boundary Conditions:
v ' ' (0)  0 : c2  0,
v ' ' ( L )  0 : c1 
w0L
6
Then
EIv '  
E Iv  
w0 x
4
24 L
w0x
5
120 L


w 0 Lx
2
12
w0Lx
3
36
 c3
 c3 x  c4
Boundary Conditions:
v ( 0 )  0:
c4  0,
v ( L )  0:
c3  
Thus, we obtain
and
Solving,
w0x
v  
3 6 0 E IL
v'  
360 EIL
w0
(7 L  10 L x
4
2
2
 3x )
4
( 7 L  30 L x 1  15 x 1 )  0
x1  L 1 
4
8
15
2
2
4
 0 .5 1 9 3 L
Hence
v m a x  v ( x 1 )  0 .0 0 6 5 2 2
w0L
EI
4

55
7 w0L
360
3
w0 x
6L
3
 c1 x  c 2
SOLUTION (4.20)
M
Segment AC: EIv 1 ' ' 
0
2
EIv 1 ' 
,
M
Segment CB: EIv 2 ' '  
EIv
x
L
0
L
M
' 
0
x
2
 c1
2L
y
(L  x)
A
2
Mo/L
( Lx 
0
L
M
x
2
)  c2
a
Mo
C
B
x
Mo/L
L
Boundary Conditions:
v1 ' ( a )  v 2 ' ( a ) :
M 0x
E Iv1 
Then
Also, we have
EIv
2
3
c 2  c1  M 0 a
 c1 x  c 3 ,
6L
M
' M 0x 
0
2
x
2L
v 1 ( 0 )  0:
c3  0
 c1  M 0 a ,
EIv
 
2
M
0
x
2
2

M
0
x
6L
3
 c 1 x  M 0 ax  c 4
Boundary Conditions:
v 2 ( L )  0;
c4   M 0 L(a 
v 1 ( a )  v 2 ( a );
M ox
v1 
Thus
M
c1 
0
6L
2
(3a
L
3
)  c1 L
 6 aL  2 L )
2
(6 aL  3a  2 L  x )
2
6 E IL
2
2
SOLUTION (4.21)
We have  m a x 
1 6 I
w 
or
(wL

Mc
I
2
8 )( h 2 )
I
m ax
2
L h
Therefore
4
v m ax 
5
5L
384 EI
(
1 6 I
5
) 
m ax
2
L h
m ax
L
2
24 hE
1
2 4 h E v m ax
Solving, L  (
4

5wL
384 EI
)2
m ax
Introducing the given data:
L [
9
2 4 ( 0 .3 1 5 )( 2 0 0  1 0 )( 3 .1  1 0
3
)
6
5 ( 7 0 1 0 )
1
] 2  3 .6 6 m
SOLUTION (4.22)
x
A
x1
E
P
C
B
R B  P (1 
( a ) EIv ' '  P
a
L
x,
EIv ' 
1
2
a
L
RC  P
)
P
a
L
a
L
y
x
2
 c1 ,
EIv 
1
6
P
a
L
x
3
 c1 x  c 2
1
6
PaL
Boundary Conditions:
v (0)  0 :
Thus
v 
PaLx
6 EI
[(
c 2  0,
x
L
)
v(L)  0 :
c1  
 1]
2
( b ) The v m a x occurs at E where v '  0 . Hence
0 
PaL
6 EI
[3(
x1
L
)
2
 1 ],
x1 
1
3
L  0 . 577 L
(CONT.)
56
4.22 (CONT.)
v m ax 
and
( c ) v m a x  0 .0 6 4 1
2
PaL
6 EI
[( 0 .5 7 7 )  0 .5 7 7 ]  0 .0 6 4 1
3
3
2 5 (1 0 )( 0 .5 )( 2 )
9
2 0 0 (1 0 )( 5 .1 2  1 0
2
6
)
PaL
EI
2

 3 .1 3 m m 
SOLUTION (4.23)
F
y
B
A
L
x
C
L/2
3F/2
F/2
Figure S4.23
Free-body diagram of shaft is in Fig. S4.23. Since reactions at supports A and B,
two differential equations must be written for portion AB and BC.
( a ) We have
M1  
M
(0  x  L )
P
2
 Px 
2
(L  x 
3 PL
2
3 PL
2
)
Integrating twice, the preceding leads to the following expression:
For segment AB
E Iv1 ''  
E Iv 1 '  
E Iv 1  
For segment AB
E Iv 2 ''  P x 
Px
2
Px
4
2
3
Px
12
 C1
E Iv 2 ' 
E Iv 2 
 C1 x  C 2
Px
2
Px
6
2
3


3 PL
2
3 PLx
2
3 PLx
4
2
 C3
 C3x  C4
Using the boundary and the continuity conditions, we find
v1 ( 0 )  0 :
C2  0
v1 '( L )  v 2 '( L ) :
C3 
5 PL
6
2
2
v1 ( L )  0 :
C1 
v2 ( L )  0 :
C4  
PL
12
PL
4
3
The result elastic curves of the beam are
v1 
Px
12 EI
(L  x )
v2 
P
12 EI
(3L  10 L x  9 L x  2 x )
2
(0  x  L )
2
3
2
2
3
(L  x 
3L
2
(P4.23a)
)
( b ) The deflection at the free end of the beam is readily found by introducing x=3L/2
into the second of Eqs. (P4.23a):
vc  
3
PL
8 EI

3
PL
8 EI

(P4.23b)
Comment: Observe that the deflection of the shaft is downward between B and C
and upward between A and B.
57
SOLUTION (4.24)
Refer to Table A.8:
P
B
E 1 I1
A
Beam AB:
L/2
L/2
R
c'
D
3
5 PL
48 E 1 I 1
Beam CD:

3
RL
24 E 1 I 1
 c''

3
RL
24 E 2 I 2

C
E 2 I2
Condition of compatibility,  c '   c ' ' :
3
5 PL
48 E 1 I 1

3
RL
24 E 1 I 1
3

RL
24 E 2 I 2
; R 
5P (E2I2 )
2 ( E1 I1  E 2 I 2 )
SOLUTION (4.25)
From a free body diagram of beam BC, we observe that it has vertical reactions 2P/3 and P/3 at
ends B and C, respectively. Thus, beam AB is in the condition of a cantilever beam under a
uniform load of intensity w and a concentrated load B equal to 2P/3.
The deflection of the hinge:
vB 
4
wa
8 EI

2 Pa
9 EI
3
as obtained by Cases 3 and 1 of Table A.8, respectively.
SOLUTION (4.26)
M
C C
A
RA
A
W
A
F
C
v 'C
B
(a)
W
(b)
B
C
A
(c)
F
B
v ''C
Figure S4.26: (a) Free-body diagram of beam; (b) deflection due to P; (c)
deflection due to reactive force F.
Consider F as redundant and to release the rod from the beam at C (Fig. S4.26a)
and then reapplied (Figs. S4.26 b and c).
(CONT.)
58
4.26 (CONT.)
Refer to Table A.8:
2
vc ' 
3
v c '' 
(2 L  3a ) 
WL
6 EI

FL
3 EI
Equation of compatibility at point C:
 c  v c '  v c ''

F
k
or
2
(2 L  3a ) 
WL
6 EI
3
FL
3 EI
Solving,
2
WL k (2 L3a )
F 
Q.E.D.
3
2 ( kL  3 E I )
SOLUTION (4.27)
y
w
x
M
A
B
F y  0:

M
A
RA  RB  wL
 0:
M
A
 RB L 
(1)
1
2
wL
2
RB
RA
E Iv "  R A x 
1
2

E Iv ' 
RAx
E Iv 
1
6
RAx 
2
2
wx
1
2
3
v ' ( 0 )  0:

1
6
 M
A
w x  M A x  c1
1
24
3
wx

4
c1  0 ,
1
2
 c1 x  c 2
2
M Ax
v ( 0 )  0:
c2  0
and
E Iv 
1
6
RAx 
3
1
24
The condition that v ( L )  0

4
wx
1
2
M Ax
2
(2)
RAL 
gives
1
4
wL
2
 3M
A
 0
(3)
Solving Eqs.(1) and (3):
RA 
5
8
wL 
M

A
1
8
wL
RB 
2
3
8
wL 
SOLUTION (4.28)
w0L
Due to symmetry: R A  R B 
,
4
M
A
 M
B
We have, for 0  x  L 2 :
M  RAx  M
A

w0x
3
w0Lx

3L
4
 M
A

w0x
3
3L
Therefore
E Iv "   M
E Iv '   M
A

x 
A
w0Lx
4
w0Lx

2
8
Boundary conditions:
  v ' (0)  0 :
v ' ( L2 )  0 :
w0x
3
3L

w0x
4
 c1
12 L
c1  0
M
A

5 w0L
2
96
(CONT.)
59
4.28 (CONT.)
Hence
2
5 w0L x
EIv  
2
3
w 0 Lx

192
w0 x

24
5
60 L
 c2
Boundary condition:
v ( 0 )  0:
c2  0
Thus, we obtain
v 
w0 x
2
(  25 L  40 L x  16 x )
3
960 LEI
2
( for 0  x  L 2 )
3
and
v max  v ( L2 ) 
7 w0L
4

3840 EI
SOLUTION (4.29)
y
M
P
A
By virtue of symmetry:
B
M
A
C
L/2
x
B
M
A
  M
RA  RB 
L/2
B
P
2
RB
RA
Segment AC
EIv ' ' ' '  0 ,
E Iv ' 
1
2
c1 x
EIv ' ' '  c 1 ,
2
 c2 x  c3 ,
EIv ' '  c 1 x  c 2
E Iv 
1
6
c1 x 
3
1
2
c2 x
2
 c3 x  c4
Boundary Conditions:
EIv ' ' ' ( 0 )  c 1  V  
v ' (0)  0 :
P
2
c3  0,
v ( 0 )  0:
EIv ' ' ( 0 )  M
,
v' (L 2)  0 :
A
c2  M
,
M
A
 M
B
A

PL
8
c 4  0.
Equation (1) is thus
v  
2
(3L  4 x )
Px
48 EI
SOLUTION (4.30)
Let the reaction R B is selected as redundant and the corresponding constraint is removed.
Then v ' B   w L 8 E I , from Case 3 of Table A.8.
4
The deflection caused by the redundant is
v"B  R B L
3
3EI
The total deflection must be zero;
vB  
4
wL
8 EI

RB L
3
3EI
 0
or
RB 
3wL
8
Now, applying equations of statics, we obtain
RA 
5wL
8
M
A

wL
8
2
60
(1)
SOLUTION (4.31)
w
A
a
C
a
D
M
EI
a
B
x2
x1
 wa
2
 3w a
2
2EI
A2
2EI
A3
x
Spandrel
parabola
A1
x3
Tangent at A
A
vB
D
B
B
A1 
 
bh
3EI
A2  
3
3a
4
2
a

3
 
wa
2 EI
A3  
wa
2 EI
x1  a 
1
3
wa
6 EI
a(wa )  
1
2
x2 
7a
4
(App. A.3)
2
1 wa
2 EI
5a
2
x3 
3
8a
3
( a )  B A   B   A  A1  A 2  A 3
B  
3
wa
12 EI
(2  6  6) 
7 wa
6 EI
3
( b ) t B A  v B  A1 x 1  A 2 x 2  A 3 x 3
vB  
4
wa
24 EI
(7  30  32)  
69 w a
24 EI
4

23 w a
EI
8
4

SOLUTION (4.32)
P
4EI
(a)
EI
A
L/2
B
L/2
C
P
2
P
2
M/EI
PL/4EI
Px/2E
I
PL/16EI
A
A2
A1
C
x
B
D
x
B
A
v m ax
B
t BD
Tangent
at B
1
PL
2 16 EI
(2) 
A2 
1 PL
2 4 EI
(2) 
L
L
2
1 PL
64 EI
2
1 PL
16 EI
t A B  A1 ( 3 )  A 2 (
L
D
t AB
A1 
Tangent
at D
61
B 
1
L
t AB 
2L
3
) 
3
3 PL
64 EI
2
3 PL
64 EI
(CONT.)
4.32 (CONT.)
( b )  BD   B   D   B  0 .
B 
Also
1 Px
2 2 EI
2
(x) 
Px
4 EI
Hence
2
3 PL
64 EI
2

Px
4 EI
x 
,
3
4
L
2x
3
) 
Thus,
v m ax  t B D 

2
1 Px
2 2 EI
3 3L
P
6 EI
64
3
(
3
Px
6 EI
3 P L3
128 EI


SOLUTION (4.33)
M
Tangent at A
A
2a
A
P
Let R B be redundant.
a
B
RA
A  0
C
RB
A1 
M
M
1
2
M A (2a )  M Aa
A2  
1
2
Pa(2a )   Pa
A3  
1
2
Pa
2
2
4a/3
v B  t BA  0 
A
B
A1
A
A2
or
C x
A3
Pa
M
A

Statics: R B 
2a/3
1
EI
1
2
7
4
4a
3
[ A1 (
)  A2 (
2a
3
)]
Pa
P 
RA 
3
4
P 
SOLUTION (4.34)
Assume R c as redundant.
w
A
tCB
Tangent
at B
RA
M
wL
2
B L/2
L
t AB
C
RC
RB
L/2
L/3
8
A1 
A1
A2
2L/3
x
A3
R C L/2
2 wL
3
8
2
(L) 
A2  
1
2
Rc L
A3  
1
2
Rc L
2
2
wL
12
3
(App. A.3)
(L)  
Rc L
(2)  
Rc L
L
2
4
2
8
(CONT.)
62
4.34 (CONT.)
E I t A B  A1 ( 2 )  A 2 (
L
2L
3
Rc L
E It CB  A3 ( 3 )  
L
4
) 
Rc L

wL
24
3
6
3
24
Since
2 t CB   t AB :
or
Rc 
1
6
wL 
Statics: R B 
3
4
wL 
RcL
3

12
wL
24
RA 
4
RcL

3
6
wL 
5
12
SOLUTION (4.35)
S
 m ax 

y
n
2
2
W
g
v E
AL
gS y AL
W a ll 
,
2
2
n v E
Substitute given data:
6
2
9 .8 1 ( 2 5 0  1 0 ) [
W a ll 

2
( 0 .0 2 ) ( 2 )]
 1 6 .6 N
4
2
2
9
( 3 ) ( 3 .5 ) ( 2 1 0  1 0 )
SOLUTION (4.36)
P=mg
1.25=a
0 .7 5
2
x
B
P
 st 
Table A.8 ( Case 6 ).
b=0.75
A
1. 2 5
2
L=2
(L  b
2
Pbx
6 LEI
20 ( 9 . 81 )( 0 . 75 )( 1 . 25 )
 x ) 
2
P
2
6 ( 2 )( 210  10
9
)( 0 . 06  0 . 08
3
12 )
(2
2
 0 . 75
2
 1 . 25 )  0 . 0535
2
mm
Max. moment is under load. Thus

M

st
m ax
c
We have K  1 
( 0 .6 2 5  0 .7 5  2 0  9 .8 1 ) ( 0 .0 4 )

I
0 .0 6  0 .0 8
1
3
12
 1  [1 
2h
 st
 1 .4 3 7 M P a
1
2 ( 0 .5 )
0 . 0535  10
3
] 2  137 . 7
( a )  m a x   s t K  7 .3 7 m m
( b )  m ax   st K  1 9 8 M P a
SOLUTION (4.37)

A 

4
max
( 2 5 )  1 5 6 .2 5  m m
2

W
A
[1 
1

from which
2h


W

Solving W 
2(
hE
L
[

st
2
2
m ax A
2
2
m ax
];
2
Since  s t  W L A E , thus:
2
m ax A
W
m ax
W
A
 1]
2
1 1
 1
2 hAE
WL
A

6
)
2
( 2 5 0  1 0 ) (1 5 6 .2 5   1 0
6
)
9
2[
st
(1)
(P4.37)
m ax
Substituting given data:
W 
2h

(1 .1 )( 2 0 0  1 0 )
 3 1 9 .0 7 N
6
 2 5 0 1 0 ]
4 .6
63
SOLUTION (4.38)
Refer to Solution of Prob. 4.37. From Eq.(1), we obtain
A 
2W

[
2
m ax
hE
L
6
(P4.38)
]
m ax
9
2 (90 )

 
(1 2 5  1 0 )
2
[
1 .2 ( 2 0 0  1 0 )
 1 2 5  1 0 ]  1 .8 4 5  1 0
6
1 .5
3
m
2
Thus
d
2
 1 .8 4 5  1 0
4
3
d  0 .0 4 8 5 m = 4 8 .5 m m
,
SOLUTION (4.39)
A   (10 )
 314 . 2 mm
2
From Eq. (P4.37):
L
Solving, h 
2W (
  m ax )   m ax A
2
hE
L
( A  m ax  2W )
m ax
2W E
3 ( 350  10

2
6
)
2 ( 500 )( 170  10
9
)
(P4.39)
[ 314 . 2  350  2  500 ]  673
mm
SOLUTION (4.40)
We have W  2 4  9 .8 1  2 3 5 .4 N and M
I 
1
12
6
( 0 .0 4 )( 0 .0 6 )  0 .7 2  1 0
3
m
m ax

1
4
WL
4
W
h
C
A
L/2
B
k
L/2
W=196.2 N
R
 st 
R
R
k
Using Table A.9:
 st 
or
(W  R ) L
3
3
R

48EI
650  10
6
 R(
1
L

R

k
)  10
6
R[
48EI
7 8 .2
180  10
k
3
6 5 0  R ( 5  3 .3 0 7 ) ,
 st 
RL

48EI
k
or
Hence
3
 0 .6 5  1 0
3
1
0 .2
R  7 8 .2 N
(2)

3
]
4 8 ( 0 .0 7 )( 0 .7 2 )
and
W  R  1 5 7 .2 N
 0 .4 3 4 m m
Maximum stress occurs at midspan:

st

Mc
I

1 5 7 .2 ( 2 ) ( 0 .0 3 )
4 ( 0 .7 2  1 0
6
 3 .2 8 M P a
)
(CONT.)
64
4.40 (CONT.)
K 1
2 ( 0 .0 5 )
1
0 .4 3 4  1 0
 1 6 .2
3
( a ) v m a x  1 6 .2 ( 0 .4 3 4 )  7 .0 m m
( b )  m a x  1 6 .2 (3 .2 8 )  5 3 .1 M P a
SOLUTION (4.41)
Refer to Example 4.14.
EkG
 m ax  2
or
AL
4 1 6 .6  7 9  1 0
2 1 0  1 0  2[
6
;
44100  10
12
4 .4 1 8  1 0
3
9
L m in
]
12
 4 ( 7 4 4 9L.3 9 1 0 )
m im
or
L m in  0 .6 7 6 m
Thus, we obtain
2 Ek L
 m ax 
2 ( 4 1 6 .6 )( 0 .6 7 6 )

GJ
 0 .0 4 8 ra d = 2 .7 5
3
7 9  1 0 ( 3 .1 1 )
o
SOLUTION (4.42)
The E k in wheel B must be absorbed by the shaft. We have
J 

32
(1 2 5 )  2 .4 (1 0
4
5
) m
4
Substitute Eqs.(4.42a) and (4.42b) into (4.41):
Ek 
1
4
 b t 
4
2 Ek L
(a)
 m ax 
(b)
 m ax  2
GJ
EkG
AL
2
[


( 0 . 075 ) ( 0 . 025 )( 1800 )( 150060 2  )
4
4
1
2 ( 2 7 .5 9 )( 0 .3 )
9
(1 9  1 0 )( 2 .4  1 0
 2[
5
)
9
( 2 7 .5 9 )(1 9  1 0 )

] 2  0 .0 0 6 0 2 5
 27 . 59 N  m
2
ra d  0 .3 5
o
1
] 2  1 1 9 .3 3 M P a
2
( 0 .0 2 5 ) ( 0 .3 )
4
SOLUTION (4.43)
The E k in wheel A must be absorbed by the shaft. Refer to Solution of Prob.4.42. We have
Ek 
1
4
 b t 
(a)
 m ax 
(b)
 m ax  2
4
2 Ek L
GJ
EkG
AL
2
[


( 0 . 0625 ) ( 0 . 025 )( 1800 )( 120060 2  )
4
4
1
2 ( 8 .5 2 )( 0 .3 )
9
(1 9  1 0 )( 2 .4  1 0
5
)
9
( 8 .5 2 )(1 9  1 0 )
 2[ 
] 2  0 .0 0 3 3 5 ra d  0 .1 9
1
] 2  6 6 .3 1 M P a
2
( 0 .0 2 5 ) ( 0 .3 )
4
65
o
2
 8 . 52 N  m
SOLUTION (4.44)
Let r  r y , then r x  r xy   ,
Hence, Eq. (4.46) for z 


max
Et
2 ( 1 
2
t
2
1
) r
M  M
,
y
M
 M
x
xy
 0.
:
;
126 (10
6
200  10
) 
9
( 3 . 2  10
3
) 1
r
2
2 (1  0 .3 )
or
r  2 .7 9 1 m and D  5 .5 8 2 m
Equations (4.50) lead to

6M

max
t
m ax
;
2
6
126 (10
6M
) 
m ax
( 3 . 2  10
3
)
2
or
M
 2 1 5 .0 4
m ax
N m m
SOLUTION (4.45)
(a)
y
Dw ' ' ' '  p 0 sin
Dw ' ' '   ( b ) p 0 cos
b
y
Dw ' '   (  ) p 0 sin
2
b
 c1 y  c 2
b
y
b
 c1
Dw '  ( b ) p 0 cos
3
y

b
1
2
c1 y
2
 c2 y  c3
and
y
D w  (  ) p 0 s in
4
b

b
1
6

3
c1 y
1
2
c2 y
2
 c3 y  c4
(1)
Boundary conditions:
w ' ( 0 )  0:
c 3   (  ) P0 ,
w ( b )  0:
c2  
1
3
w ( 0 )  0:
3
b
c1 b 
2

p0b ,
3
c4  0
w ' ( b )  0:
2
c1  0 ,
c2 
2

3
p0b
2
Equation (1) becomes
p0b
w 
D
4
4
( s in
y


b
b
2
y

2

b
(2)
y)
( b ) At y=0:
 m ax 
(c) D 
Et
3
1 2 (1  
2
)

6M
t
m ax
2

9
7 0  1 0 (1 0  1 0
2
6
t
3
)
d w
[D
2
3
dy
2
]y0 
6
t
2
[0  0 
2 p0b

3
2
]
12 p0

3
b
(t)
2
 6 .4 1 k N  m
2
1 2 (1  0 .3 )
Thus
w m a x  w ( b2 ) 
p0b
4
4
 D
(1 

) 
4
3
3 5  1 0  ( 0 .5 )

4
3
( 6 .4 1  1 0 )
and

m ax

3
1 2 ( 3 5 1 0 )

3
( 00.0.51 )  3 3 .8 6
2
4
M Pa
66
(1 

4
)  0 .7 5 2  1 0
3
m = 0 .7 5 2 m m
SOLUTION (4.46)
( a ) Dw ' ' ' '  p 0 ,
Dw' 
Dw ' ' '  p 0 y  c 1 ,
p0 y 
3
1
6
1
2
2
 c2 y  c3
4

c1 y
Dw ' ' 
1
2
p0 y
2
 c1 y  c 2
and
Dw 
1
24
p0 y
c1 y 
3
1
6
1
2
c2 y
2
 c3 y  c4
(1)
Boundary Conditions:
w (0)  0 : c4  0,
w ' ' (0)  0 : c2  0
w ( b )  0 a n d w ' ( b )  0:
c1  
3
8
p0 b,
c3 
1
48
p0b
3
Equation (1) is therefore
p0b
w 
2
(b)
d w
dy
2

p0
2D
(y
4
48 D
2

y
y
y
[ ( b )  3( b )  2 ( b ) ]
3 yb
4
3
4
(2)
)
At y=b:
M
m ax
2
 D
 p0
d w
dy
2
2
b
8
 m ax 
,
6M
t
m ax
2
 0 .7 5 p 0 ( t )
b
2
( c ) From Solution of Prob. 4.45, D  6 4 .1 k N  m
Thus
w m ax  w ( 2 ) 
b

p0b
4
192 D

p0b
4
[ ( 2 )  3( 2 )  2 ( 2 ) ]
1
48 D
1
3
( 3 5  1 0 )( 0 .5 )
3
4
1 9 2 ( 6 .4 1  1 0 )
3
1
4
 0 .0 0 1 8 m = 1 .8 m m
and

 0 .7 5 p 0 ( bt )  0 .7 5 (3 5  1 0 )( 51000 )
2
m ax
3
2
 6 5 .6 3 M P a
End of Chapter 4
67
CHAPTER 5
ENERGY METHODS AND STABILITY
SOLUTION (5.1)
Axial strain energy
Use Eq. (5.11), circular Part:
2
U
P L

C
2
2
P L

2 AE
d
2(
2P L

2
d E
2
)E
4
Square Part:
2
U
P L

S
2
2
P (L 2)

2
2 AE
P L

2a E
2
4a E
Requirement::
2
U
 US;
C
d
2
2P L
P L

d E
2
2
4a E
8


a
SOLUTION (5.2)
Refer to solution 5.1:
U 
2

P L
2
4a E
2
2P L
2
d E

2
P L
E
(
1
4a
2

2
d
(1)
)
2
Using Eq. (5.27),
U 
1
2
P
(2)
Equating Eqs. (1) and (2), we obtain
  PEL ( 2  4 )
a
2
d
2
SOLUTION (5.3)
T A B  2 .5 k N  m
T B C  1 .5 k N  m
Segment AB
J AB 

( 4 5 )  4 0 2 .5 8  1 0
4
AB

m
4
( 2 .5  1 0 ) ( 0 .5 4 )
2
U
9
32
T L
3

2G J
2
2 ( 4 0  1 0 )( 4 0 2 .5 8  1 0
9
9
 1 0 4 .8 J
)
Segment BC
J BC 

(3 0 )  7 9 .5 2  1 0
4
BC

m
4
32
2
U
9
T L
2G J
(1 .5  1 0 ) ( 0 .3 6 )
3

2
2 ( 4 0  1 0 )( 7 9 .5 2  1 0
9
9
 7 5 .8 3 J
)
(CONT.)
68
5.3 (CONT.)
Total strain energy
U  1 0 4 .8  1 2 7 .3  2 3 2 .1 J
SOLUTION (5.4)
T AB  3 k N  m
TBC  5 k N  m
30 mm
20 mm
A
B TB=2 kN  m
45 mm
C
TC=5 kN  m
0.36 m
0.54 m
Segment AB
J AB 

( 4 5  2 0 )  3 8 6 .9  1 0
4
4
AB
m
4
(3  1 0 ) ( 0 .5 4 )
2
U
9
32
3
T L


2
2 ( 4 0  1 0 )(3 8 6 .9  1 0
9
2G J
9
 1 5 7 .0 2 J
)
Segment BC
J BC 

(3 0 )  7 9 .5 2  1 0
4
BC
m
4
32
(5  1 0 ) ( 0 .3 6 )
3
U
9

2
2 ( 4 0  1 0 )( 7 9 .5 2  1 0
9
9
 1, 4 1 5 J
)
Total strain energy
U  1 5 7 .0 2  1 4 1 5  1 5 7 2 J
SOLUTION (5.5)
See solution of Prob. 5.3:
J A B  4 0 2 .5 8  1 0
J B C  7 9 .5 2  1 0
9
9
m
m
4
4
Therefore
2
2
U 
C 


T L

2G J
TL
TC
[
0 .5 4

2 ( 2 8 ) 4 0 2 .5 8

TC
0 .5 4
[

2 8 4 0 2 .5 8
GJ
0 .3 6
]  1 0 4 .8  1 0
6
7 9 .5 2
0 .3 6
]  2 0 9 .6  1 0
7 9 .5 2
6
2
TC
TC
or
 C (1 0 )
6
TC 
2 0 9 .6
Then
U 
1 0 4 .8  1 0
6
( C  1 0
( 2 0 9 .6 )
2
2
12
)
 2 3 8 5  C  2 3 8 5 ( 0 .0 6 )
2
69
2
 8 .5 8 6 J
SOLUTION (5.6)
See solution of Prob. 5.3:
J A B  4 0 2 .5 8  1 0
TC
U 
Thus,
2
2G
L AB
(
J
3
(1 .4  1 0 )

LBC

2
2 (80 )
m
J B C  7 9 .5 2  1 0
4
9
m
4
)
J BC
AB
9
( 4 00 .52 .54 8 
0 .3 6
7 9 .5 2
)
 7 1 .8 9 J
Equation (5.29): U 
1
2
TC  C ;
1 4 0 0 C
7 1 .8 9 
2
from which
 C  0 .1 0 2 7 r a d  5 .8 8
o
SOLUTION (5.7)
P
B
A
 
C
a
V AB  P ,
6
5
V BC 
Pa
L
L
Pa/L
V
x
P
U
s


L
2
Vx
dx 
2 AG
0
3
5 AG
[


P dx 
2
0
L
2
P a
L
0
2
2
dx ] 
2
3P a
5 AGL
(L  a)
SOLUTION (5.8)
Vx 
U
s



 
 wx
wL
2
L
0
2
Vx
2 AG
2
2
3w
5 AG
L
4
2
dx 
x 
3w
5 AG
2
x L
2


6
5
L
(
0
L
3
x
3
 x ) dx
2
L
2
2

3
3w
L
5 AG 12
0
2

1 w L
20 AG
3
SOLUTION (5.9)
I  bh
P
A
B
C
P
U
a
L
x’
x
(a) U
U
b

t

1
2 EI
1
2E
[
a
M
0
2
AB
 dx  
2
dx 
dA 

L
M
0
1
2E
2
CB
(
P
A
M
2
dx' ] 
My
I
12
2
P (La )
2 AE

2
P (La )
2 Ebh
We have
Pa/L


a
3
2
P a
6 EI
2
) dA
AB
(a  L ) 
 Px
2
2P a
Ebh
2
3
M
CB

Pa
L
x'
(a  L )
(1)
(CONT.)
70
5.9 (CONT.)
But

Thus
U
2
P (La )

t
bh
U
3
12


0 ,m ax

 U b.
a
2
a
6 Pa
bh
 U
t
[1  4 ( h ) ]
2 Ebh
Pa ( h 2 )
( b )  b ,m ax 
Hence
y d A  0 , and Eq. (1) becomes U
2
 a ,m ax 
,
2
m ax

2E
2
P
2
2 Eb h
2
(1 
P
bh
6a
h
)
 m ax 
,
P
bh
(1 
6a
h
)
2
SOLUTION (5.10)
M 
( Lx  x )
2
w
2
A
x
wL
2
U
0 ,m ax
U 
b
w
B


1
2 EI
2
m ax
2E
2

9w L
2
m ax

I  bh
1
8
3
 m ax 
wL
2
c  h 2
12 ,
M
m ax c
I

3wL
4 bh
2
2
4
2
32 Eb h
M dx 
h
wL
2
L

M
(1)
4

1
2 EI
L
w
2
(
0
) ( Lx  x ) dx 
2
2
2
2
w L
5
20 Ebh
(2)
3
It is required to obtain C : U 0 , m a x  C U V ,
or
C  U
V
0 ,m ax U
(3)
Substituting Eqs.(1) and (2) into (3): C  4 5 8 .
Thus
U

0 ,m ax
45 U
8 V
SOLUTION (5.11)
From Solutions of Probs. 5.9 and 5.7:
U
b

2
2
(a  L )
P a
6 EI
U 
Thus
or
2
p a
2
6 EI
U
2

s
(a  L ) 
3P a
5 AGL
2
(a  L )
(a  L ) 
3P a
5 AGL
v A  2 P a ( a  L )[ 6 E I 
a
3
5 AGL
1
2
Pv A
]
SOLUTION (5.12)
x
w
x
A
2
U 

M dx
2EI

1
2EI
L

L
0
(
1
2
B
2
w x ) dx 
2
2
1 w L
40
71
EI
5
M
x
 
1
2
wx
2
SOLUTION (5.13)
P
P
A
a
2a
C
B
D a
MAC=Px
MCD=Pa
P
P
Pa
M
x
Segment AC
U
AB

a

2
M dx
0
P

2EI
2

2EI
2
a
x dx 
2
P a
0
3
6EI
Segment CD
U
CD


2
P a
3a
0
2
2
dx 
2EI
By symmetry: U
3
P a
EI
U
AC
BD
.
Total strain energy :
Pa
U  2
3
2

P a
6EI
3
2
4 P a

3
EI
3
EI
SOLUTION (5.14)
x
w
b
h
A
L
6
We have  
Vx  wx
5
U
s



L
0
V
2
6
dx 
2 AG
3 w
2
x
5 AG
B
1
5 2 AG
3
L
2

3
1 w L

L
( w x) dx
2
0
3
5 AG
0
SOLUTION (5.15)
We have
V AC  V BD  P
  6 5
VCD  0
Thus
U
s


 V
2
2 AG
a
dx  2 
0
2
P dx
 2(
2 AG
2

6 P a
5 AG
72
6
5
2
)
P a
2 AG
SOLUTION (5.16)
See solution of Probs. 5.13 and 5.15:
2
U
b

3
3
4 P a
3
s

5 AG
2
6 P a

1

5 AG
EI
vC  2 P a (
or
U
6 P a
(due to symmetry)
2
U 
2
EI
vC  v D
Thus
3
4 P a
2a
2
3

3EI
1
P (vCb  vCs ) 
2
P vC
)
5G A
SOLUTION (5.17)
( a ) Axial strain energy in bolt.
2
U
b
2
P L

T L

2 AE
2 AE
2
T ( 0 .0 3 )


2[
 2 .6 5 2 6 (1 0
9
)T
2
(6 ) ( 2 0 0  1 0 )]
2
3
4
 2 .6 5 2 6 (1 0
Ub 
Thus
1
2
T ;
9
)( 6 3 0 )  1 .0 5 2 8 J
2
1 .0 5 2 8 N  m 
1
2
( 6 3 0 )
  3 .3 4 2 m m
( b ) Bending strain energy in link.
I  bh
U
b


L
3
12  216 m m
2
0
2EI
2EI
{[
0 .0 2 5
(420 x) dx 
2
0

0 .0 5
2
( 2 1 0 x ') d x '}
0
A
EI
B
Substituting the data,
U
b

50 mm
25 mm
1 .3 8

4
M dx
1

1 2  (1 2 )( 6 )
3
3
6
630
420
( 2 0 0  1 0 )( 2 1 6 )
 3 1 9 (1 0
x’
x
1 .3 8
210
) J
SOLUTION (5.18)
a
x
Q
P
B
A
C
M
AC
 Qx
M
CB
 Qx  P(x  a )
L
(CONT.)
73
5.18 (CONT.)
Thus
M
vA 
1
EI


1
EI
[ ( Q x )( x )d x 
M
i
dx
Q
i
a
L
 [Q x
0
 P ( x  a ) ]( x ) d x
a
Set Q  0 , and integrate:
vA 

1
EI
L
P ( x  a )xdx 
( 2 L  3a L  a ) 
3
P
6 EI
a
2
3
SOLUTION (5.19)
M
BC
 Px
M
B 
Thus

1
EI
[
 P L  P R s in 
CA
L
M
0
M

BC
P
BC

dx 
2
M
M
Rd ]
CA
P
CA
0
( 4 L  6  R L  2 4 R L  3 R )
3
P
12 EI
2
2
3
SOLUTION (5.20)
A a
P a
a
Pa/2
C
P/2
M
B
D
P/2
Pa/2
M
+

AD
P
2

vD 
1
EI
vD 
P
2 EI
( x  a ),
M
[
M
i
2a
i
P
(xa )
2
0
M
BD

P
2
dx
2
dx 

a
0
2
x
2
dx
x
Pa/2
Integrating,
3
vD 

Pa
4 EI
SOLUTION (5.21)
B
C
M
A
RA
L/2
AB
Consider R A as redundant.
L
RB
x
M
0
 RAx
M
CB
RC 
x’
 (
M
0
L
RA

2
M
0
L

RA
2
) x ' M 0 ,
vA  0 
1
2
Thus
L
vA 

2
0
( R A x )xdx 

L
(
0
M
0
L

RA
2
) x ' M 0 ]
x'
2
After integrating
RA 
Then
RC 
2
3
M
4
3
M
0

0

L
L
For the entire beam,

F y  0:
RB  2
74
M
L
0

dx  0

M
M
i
i R A
dx
x
SOLUTION (5.22)
U
a

D 
2
P (2L)
2 AE
U
P
U
a

2 PL
AEa


PL
3EsI
2 PL
AEa
or
Solving P 
3

3
wL
8

1
EsI
2
M
2Es
0
L
(Px 
0
4
 0
a
)
wL
8 EsI
3 AE
(
Ls


s
2
AE a L  6 E s I
M  Px 
dx
wx
2
2
wx
2
2
U  U
a
U
s
)xdx  0
SOLUTION (5.23)
w
M
AB

1
2
M
BC

3x
5
1
EI
M
Q
x
x A
B

A
4

3
C

1
EI
L AB  2 a
2
wx
( 2 wa )  2 wa
{
M
Q
i
i

4x
5
Q
2

L BC  5 a
dx
2a
2
1
( 2 w x )( 0 )d x
0

2

5a
0
( 65x wa  2 wa
4x
5
Q )(
4x
5
) dx }
Set Q  0 and integrate:

A
 60
wa
EI
4

SOLUTION (5.24)
C
M
P
x
A
AB
 Px
M
BC
 Pa
a
B
x
L
A 
1
EI

M
M
i
dx 
i
P
a
1
EI
{ ( P x ) xd x 
0
L
 ( P a )( a )d x}
0
Integrating,
A 
2
(a  3L )
Pa
3EI
SOLUTION (5.25)
Introduce a fictitious horizontal force Q at end B. Vertical reactions at supports are P/2.
We have M
Thus
BC
 R (1  c o s  )
 2
B 
2
EI


2
EI

M
0
 2
M
BC
BC
Q
P
2
 ( R sin  ) Q
P
2
 M
BC
 ( R s in  ) Q ] [ R s in  ] R d 
Setting Q=0 and integrating:
B 
CA
Rd
[ R (1  c o s  )
0
M
3
PR
2 EI
75
SOLUTION (5.26)
(a)
Let fictitious couple C=0:
x
C
B
a
2a
a
x
A
P
M
AB
 Px
M
BC
 Pa
M
DC
 Px  Pa
C
x
D
P
Pa+C
A 
Thus

M
M
i
dx
P
i
a
 P  x dx  Pa
2
2
0


4
3
2 Pa
EI
3
 P
0
2a
(x
 2ax  a )dx
2
2
0
 Pa ( 3  4  2)
3
Pa
a
 dx
3
8

( b ) We now have
M
AB
A 
 Px  C
[ M
1
EI
M
M
i
M
DC
 Px  Pa  C
dx ]
C
i
 Pa  C
BC
Hence, after setting C=0:
A 
a
P
EI
a
[ xdx  a  dx 
0
0
2a
 (x
 a )dx ] 
0
3 Pa
2 EI
2
SOLUTION (5.27)
M
(a) 
A
AB


  Pz,
M
a
1
EI
[ M
P
EI
(
0
3
a
3

M
P
AB
3
L
3
AB
)
  Px,
BC

dz 
L
0
1
GJ
(
T0 L
2
T B C  T 0 (1 
M
M
BC
P
BC
dx ] 

1
GJ
x
L
L
0
)  Pa
[ T 0 (1 
x
L
)  Pa ]( a ) dx
 Pa L)
2
( b ) Introduce a fictitious couple C about x axis at point B.
T B C  C  T 0 (1 
x
L
)  Pa
Hence
B 
1
EI
[
a
M
0
 0  0 
M
1
GJ

AB
C
AB
L
0
dz 

L
M
0
[ C  T 0 (1 
x
L
M
BC
1
GJ
(
T0 L
2
dx ] 
)  P a ] (1 ) d x
Setting C=0 and integrating:
B 
BC
C
 PaL )
76
1
GJ

L
0
TBC
TBC
C
dx
SOLUTION (5.28)
Due to Symmetry: F A B  F C D ,
Statics: R A x  6 0 k N
F AC  FBD
,
R A y  1 2 .5
Method of joints:
Joint A
kN
,
R D y  1 2 .5
Joint B
F AC  30 kN ( T )
F BC   12 . 5 kN ( C )
F AB  32 . 5 kN ( T )
Total strain energy:
U 

2
Fi L i
6

2 AE
[ 3 2 .5 (1.3 ) 2  3 0 (1.2 ) 2  1 2 .5 ( 0 .5 ) ]
2
10
2 AE
2
2
This gives, substituting the given data,
U  4 7 .4 7 N  m
Hence
U  W;
4 7 .4 7 

mm
1
2
( 6 0  1 0 ) D
3
or
D
 1 . 582

SOLUTION (5.29)
2P
B
Introduce a fictitious force Q at C.
P
Statics:
0.9 m
A
R Ax  P  Q 
C
R Ax
R A y  0 .2 4 P 
Q
R Ay
1.2 m
0.675 m
R C  1.7 6 P 
RC
Apply method of joints at C and B:
F B C  2 .2 P
(C )
F A B  0 .4 P
(C )
F A C  1.3 2 P  Q
Thus
C 
1
AE

Fi
 Fi
Q
Li
Differentiating and setting Q=0:
C 
1 . 32 P ( 1 )
AE
(1 . 875 ) 
2 . 475 P
AE

77
(T )
kN

SOLUTION (5.30)
A
P
B
Introduce Q  at C.
By method of joints:
L
F AB  0,
L
F AC  ( P  Q )
P
C
D
FBC   P
FC D   Q
2,
Q
We write
( C ) v 

1
AE

Fi
 Fi
Q
Li
[ 0  (  P )( 0 )  ( P  Q )
1
AE
2(
2)
2  (  Q )( 1 )] L
This yields, for Q  0:
( C ) v 
2
( C ) h 
1
AE

( C ) h 
1
AE
[0  (  P ) (  1 )  ( P )
2PL
AE
 2 .8 2 8
PL
AE

Similarly
or

1 2 2
AE
Fi
 Fi
P
Li
P L  3.8 2 8
PL
AE
2(
2)
2 ]L

SOLUTION (5.31)
Introduce Q  at point C.
F.b.d. - Entire truss ( from

M  0,

F  0 ):
R A y  1.2 5 P  0 .9 3 7 5 Q 
R Ax  Q 
R B y  2 .2 5 P  0 .9 3 7 5 Q 
Joint A:
F AC
13
Q
12
F A C  3 .2 5 P  2 .4 3 7 5 Q
5
F A B  3 P  1. 2 5 Q
F AB
A
(T )
(C )
R Ay
F B C  3 .7 5 P  1.5 6 2 5 Q
Joint B:
(C )
Thus, we have
( C ) h 
1
AE

Fi L i
( C ) v 
1
AE

Fi Li
 Fi
Q

P
AE
[1 2  6 1.7 9 1  2 9 .2 9 6 9 ]  1 0 3.0 8 8

P
AE
[ 3 ( 3 . 2 )( 3 )  3 . 25 ( 7 . 8 )( 3 . 25 )
P
AE
and
 Fi
P
 ( 3 . 75 )( 3 . 75 )( 5 )]  181 . 5
78
P
AE


SOLUTION (5.32)
F BD  0 .
C
32
Q
D
B
32
Introduce Q at point D.
Reactions, as found by statics, are shown in the
figure. We shall apply the method of joints, as
needed.
32 kN C
Joint C
15
FBC
42-Q/2
Joint E
17
E
A
8
42+Q/2
E
4
FDE   7 0 
5
6
F AE  5 6 
Q

Thus
D
(C )
F AD
A
3
56+2Q/3
42-Q/2
(C )
Q
F AD   3 0 
(T )


1
AE

10
AE

2 8 6 .1 3 3  1 0
AE
3
5
4
32
42+Q/2
2
3
(32 )  68 kN
F BC  60 kN ( T )
60
5
3
17
8
Joint A
FDE
F AE
FC D  
FC D
F jL j
(C )
Q
F j
Q
[0  0  0  0  2 (  7 0 ) ( 
3
5
6
)  3 .2 ( 5 6 ) ( 3 )  2 (  3 0 ) ( 
5
6
2
5
6
)]

SOLUTION (5.33)
Joint B
FBD
FBC
3
FBA
4
FBD  
5
4
FBA
(1)
B
FBC  P 
3
4
FBA
P
A  0 
1
AE


1
AE
[( P 
FjL j
3
4
F j
 FBA
F B A )(
3L
4
)( 4 )  ( 
3
 0 .5 6 2 5 P L  3 .3 7 5 F B A L  0 ,
5
4
F B A )(
5
24
P
FBC 
7
8
)( 
5
4
)  F B A ( L )(1 )]
F B A   0 .1 6 6 P   P 6
Then Eqs.(1) give
FBD 
5L
4
P
79
SOLUTION (5.34)
M  F R s in   P R (1  c o s  ) ,
v  0 
1
EI

M
M
F
dx
Therefore
v 

1
EI
[ FR sin   PR ( 1  cos  )]( R sin  ) Rd 
0
3
FR



2 EI
2 PR
EI
3
 0,
F  
4P



4P

SOLUTION (5.35)
P
C
M 
Px
2
c o s   Q x s in 
L
x

A
Q
B
P/2
P/2
Line of symmetry
( a ) With Q  0:
B 

2
EI
3

L
c o s  ) ( x s in  ) d x
Px
2
(
0
s in  c o s  
PL
3EI
3
PL
6 EI
s in 2 
( b ) With Q   R B h :
B  0 
2
EI

L
c o s   R B h x s in  ) (  x s in  ) d x
Px
2
(
0
or
R Bh 
P cot  
1
2
SOLUTION (5.36)
Statics: F B C 
2P
FC D 
,
3
P
F AB 
,
3
P
3
,
M
AB
 Px
Thus
U 


B
A
2
FAB
2 AE
1
2 AE

dx 
2L
0
 ( 6 AE 
11L
2
P
3

M
)P
2
AB
2 EI
A
dx 
3
4L
3EI
B
1
2 RI
dx 

2L

C
D
2
FBC
2 AE
P x dx 
2
2
0
2
We have
C 
U
P

PL
3E
( 11A 
8L
I
2
dx 
)
80

1
2 AE
D
2
FC D
dx
2 AE
C

2L
0
4
3
P dx 
2
1
2 AE

L
0
2
P
3
dx
SOLUTION (5.37)
Let the load at B be designated by Q. Locate origin of coordinates at A:
V AB  P
M
 Px
AB

V AB
 1,
P
M
AB
P
 x
Locate origin of coordinates at B:
V BC  P  Q
M
 P(x 
BC
L
2
)  Qx

V BC
 1,
P
M
BC
P
 x 
L
2
( a ) Equation(5.38), with Q=P, gives
vA 
M
1
EI
M
i
P
i
dx 
L



L
2
0
3
7 PL
16 EI

( x )dx 
Px
EI

 V
1
AG
P(2xL 2)
2
EI
0
Vi
dx
P
L
(x 
L
2
)dx 

L
2
0
1.2 P
AG
(1 ) d x 

2
0
1.2 ( 2 P )
AG
1 .8 P L
AG
(1 ) d x
(P5.37)
( b ) From Table B.1: E=200 GPa and G=79 GPa. Given L/h=5.
We have A=bh and I  b h 1 2 . Eq.(P5.25) is rewritten as
3
Ebv A
P
3

7 (1 2 ) L

7 (1 2 )( 5 )
16 h
3

3
1 .8 E L
Gh

16
1.8 ( 2 0 0 ) ( 5 )
79
 656 . 25  22 . 78  679 . 03
Error:
22 . 78
679 . 03
(100 )  3 . 35 %
SOLUTION (5.38)
Inasmuch as horizontal displacement at C,  h is zero, Eq.(5.41) gives
h 
U
H
 0 
1
EI


[ HR sin   FR ( 1  cos  )] R (sin  ) Rd 
0
2R

1
EI
 [( H
 P ) x  2 F R ]x d x
0
 H ( 2 
8
3
)  2F 
8P
3
or
4 . 2375 H  2 F  2 . 6667 P  0
(1)
Similarly, vertical displacement at C is zero:
v 
U
F
 0 
1
EI



[ HR sin   FR ( 1  cos  )][  R ( 1  cos  ) Rd 
0
2R
1
EI
 [( H
 P ) x  2 F R ]2 R d x
0
 2H  F(
3
2
 8)  4P
or
2 H  1 2 .7 1 2 4 F  4 P  0
(2)
Solving Eqs.(1) and (2):
H  0 . 5193 P
F  0 . 2329 P
81
SOLUTION (5.39)
The structure is statically indeterminate to the first degree. Select R, the reaction at B, as
redundant. From the equilibrium of forces at joint D,
R
with F A D  F D C  F :
F
F
4
3
F 
D
5
8
(W  R )
(1)
W
Substituting Eq.(1), into Eq.(5.45), together with Eq. (5.38) we have
B  0 

L
F
R
[2  F
1
AE
0
1
AE
 
h
(W  R ) L 
25
32 AE
R
R
R
0
dx ]
(W  R )(  85 ) L  0 . 8 RL ]
5
8
[( 2 )

dx 
0 .8 R L
AE
Solving, R  0 .4 9 4 W 
Thus
F A D  F C D  0 .3 1 6 W
F B D  0 .4 9 4 W
and
SOLUTION (5.40)
We write M   M
Thus
U
R A
vA 

U
M A
A 

6.
M
R A
M
dx 

1
EI
L
( M
0
 RAx 
A
1
6
kx )xdx  0
(1)
k x )(  1) d x  0
(2)
3
 0

1
EI
3
 0

1
EI
 R A x  kx
A
M
M A
M
dx 

1
EI
L
( M
0
A
 RAx 
1
6
3
Integrating and simplifying Eqs.(1) and (2) we obtain, respectively:

1
2
M
A

1
3
RAL 
M
A

1
2
RAL  
1
30
kL
1
24
3
kL
3
Solving
RA 
3
20
kL
2
M
A

1
30
kL
3
SOLUTION (5.41)
U 
W 
Thus

EI
2

L
(d
2
dx
0
v
2
) dx 
2
L
0
EI
2

L
0
L
w v d x  w 0  a s in
0
U  W :
E I
2L
3
4
a 
( L ) a sin
4
2 x
L
w0L
2
dx 
w0
EI
L
4
(  ) s in
w0L
2
a 
,
We have
v 
2 x
L
2
x
L
82
dx 
a
w0
EI
L
( )
4
4
EI 
4L
3
a
2
SOLUTION (5.42)
L A D  L C D  3 .4 6 m
U 
AL i E 
1
2

2
1
2
AL i E (  v cos  / L i )
2
Vertical load of the joint, by Eq.(5.46):
3
U
 v
W 


E jAj
Lj
 v cos 
2
1
or
W  E A v [
2
cos 30
L AD
o
2

o
 E A  v [ c o3s .4360 
2
cos 30
LC D
2
o
cos 30
3 .4 6

o
1
LBD
]
 13 ]  0 .7 6 6 9 E A  v
Substitute the given data
W  0 .7 6 6 9 ( 0 .0 0 6 )( 2 0 0  1 0 )( 6 2 5  1 0
9
6
)  5 7 5 .2 k N  m
SOLUTION (5.43)
v 
(a)
2
ax
2L
U 
3
EI
2
(3L  x )

L
v"  3
(v" ) dx 
2
0
9 EI
2L
6
a
2
a
L

3
(L  x)
L
( L  x ) dx 
2
0
3EI
2L
3
a
2
(1)
We have
W  P  v A
(2)
Virtual work principle,  U   W , is thus
P  a 
and
3EI
2L
3
(2a  a )
a  PL 3EI
3
v 
2
Px
6 EI
(3L  x )
(3)
( b ) At x=L, Eq.(3) gives
v m ax 
3
PL
3EI
Using Eq.(3):  
and at x=L:
 m ax 
(4)
dv
dx

Px
2 EI
(2 L  x )
2
PL
2 EI
(5)
SOLUTION (5.44)
We have
v  ax ( L  x )  axL  ax ,
v'  aL  2ax,
2
So,
Also
U 
EI
2

L
( v " ) d x  2 a E IL
2
2
0
W  P  v A  P  (acL  ac )
2
From  u   W , it follows that
E IL ( 4 a   a )  ( P c L  P c )  a ,
2
a 
Pc ( L c )
4 E IL
Hence, at x=c:
vA 
2
Pc ( Lc )
2
4 E IL
83
v"  2a
SOLUTION (5.45)
I 
 d
4
64
2
 d
A 
,
r 
,
4
I A  d 4 , Le  L
We have


cr
2
 E

P
A
4P
;
2
( Le r )
 d
2
2
2
 Ed

16 L
2
d 
,
4
64 PL
2
3
 E
Substituting given data:
d  [
Check:

L
r
3
64 ( 50  10 )( 1 . 2 )
3
( 210  10

4 ( 1. 2 )

4L
d
9
0 .0 2 9
1
2
] 4  29 mm
)
 1 6 5 .5 . Also  
P
A
3
4 ( 50  10 )

 ( 0 . 029 )
2
 75 . 7 MPa
 600
MPa.
OK.
SOLUTION (5.46)
Substituting the given data: ( L e r ) c  
E S y  45 . 7
Try Johnson’s formula
Pc r
 S
A
S

y
2
y
2
4 E
(
Le
r
)
2
or
3
2 2 0 (1 0 )
 d
Check
Le
2

d 4
 5 2 0 (1 0 ) 
6
( 5 2 0 1 0 )
6
4
4
2
2
2
( 0 .2 5 ) 1 6
9
(1 1 0  1 0 )
d  2 5 .7 m m
;
2
d
 38 . 9 OK.
250 ( 4 )
25 . 7
Try Euler formula:
d [
and
4 Le

d
2
6 4 P Le
3
 E
1
]4  [
3
6 4 ( 2 2 0  1 0 )( 0 .2 5 )
3
2
1
] 4  2 2 .5
9
 (1 1 0  1 0 )
 44 . 4  45 . 7
4 ( 250 )
22 . 5
mm
 does not apply.
SOLUTION (5.47)
( a ) By Eq.(5.61) with A   d
d  [
64
2
Pc r L e
2
 E
]
4 and r  d 4 :
2
1
4
( b ) Equation (5.61) with A  b h ,
b 
I  bh
3
12 ,
r
2
 h
2
1 2:
2
1 2 Pc r L e
2
 Eh
3
SOLUTION (5.48)
( a ) Same area:

(D
4
2
 d )  ao  ai
2
ai  ao 
2
2

4
2
(D
2
2
 d )  50 
2
2

(5 0  3 5 )
2
2
4
(CONT.)
84
5.48 (CONT.)
a i  3 8 .7 m m ,
or
1
t 
2
( a o  a i )  1 1 .3 m m
( b ) Circular bar
I 

4
(D
 d ) 
4
64

( 5 0  3 5 )  2 3 3 .1  1 0
4
4
9
m
4
64
 EI
 ( 7 2 ) ( 2 3 3 .1)
2
Pc r 
2

2
 3 4 .2 k N
2
( 2 .2 )
Le
Square bar
I 
1
4
12
1
(ao  ai ) 
4
( 5 0  3 8 .7 )  3 3 3 .9  1 0
4
9
m
4
12
 EI
 ( 7 2 ) ( 3 3 3 .9 )
2
Pc r 
4
2

2
2
( 2 .2 )
Le
 49 kN
SOLUTION (5.49)
I 
d
 (8 )
4

64
A 
d
4
 2 0 1 .0 6 m m
4
64
2

 (8 )
4
2
 5 0 .2 7 m m
2
4
 EI
2
P A  Pc r 
L
2
 ( 2 0 0  1 0 )( 2 0 1 .0 6  1 0
2

9
( 0 .4 )
2
12
)
 2 .4 8 k N
The corresponding stress is

cr
Pc r


A
2480
5 0 .2 7  1 0
6
 4 9 .3 M P a  2 5 0 M P a OK .
We have

M
 0:
C
( a  b ) Q  b Pc r
1 8 0 Q  3 0 ( 2 4 8 0 ),
or
Therefore
Q a ll 
Q

4 1 3 .3
Q  413 N
 295 N
1 .4
n
SOLUTION (5.50)
Pc r  n P  2 .6 ( 2 2 )  5 7 .2 k N ,
I 

d
A 
4
d
64
2
r 
4
L e  0 .7 L  0 .7 (1)  0 .7 m
d
4
Equation (5.61) gives
2
d
4

6 4 Pc r L e
 E
3
6 4 (5 7 .2  1 0 )( 0 .7 )
3

 (200  10 )
3
9
2
,
d  0 .0 2 3 m = 2 3 m m
(CONT.)
85
5.50 (CONT.)
Hence
Le
0 .7

 1 2 1 .7
0 .0 2 3 4
r
Equation (5.62):
Le
(
r
E
)c  
 
Sy
200  10
9
250  10
6
 8 8 .8 6  1 2 1 .7
Euler formula is valid, Therefore
d  23 m m
SOLUTION (5.51)
Refer to Solution of Prob. 5.50. Now we have L e  0 .7 ( 6 2 5 )  4 3 7 .5 m m
and Pc r  2 .6 (1 2 5 )  3 2 5 k N . Equation (5.61):
2
d
4

6 4 Pc r L e
 E
3
6 4 (3 2 5  1 0 )( 0 .4 3 7 5 )
3

2
3
d  0 .0 2 8 m
,
 (200  10 )
9
and
Le

r
4 3 7 .5
 6 2 .5  1 2 1 .7
28 4
Euler formula does not apply.
Apply Johnson formula, Eq. (5.66):
d  2(
or
2
Pc
 Sy

S y Le
 E
2
1
2
)
325  10
 2[
3
  250  10
2 5 0  1 0 ( 0 .4 3 7 5 )
6
6

 (200  10 )
2
9
2
1
]
2
d  0 .0 4 1 9 m = 4 1 .9 m m
SOLUTION (5.52)
L B C  0 .6 5 m
F AB 
5
12
FBC 
P,
13
12
P
P
B
F AB
12 13
5
FBC
Bar AB
( F A B ) cr 
2
 EI
2
Le

2

9
( 2 1 0  1 0 )[

( 5 1 0
3
( 0 .4 )
4
) ]
4
2
 6 .3 5 9 k N 
5
12
Pc r ,
Pc r  1 5 .2 6 k N
Bar BC
( F B C ) cr 

2
9
( 2 1 0  1 0 )[

( 7 .5  1 0
4
( 0 .6 5 )
2
3
4
) ]
 1 2 .1 9 k N 
Choose the small value, Pc r  1 1.2 5 with n  2 .5 . Thus
Pall 
Pcr
n

11 . 25
2 .5
 4 . 5 kN
86
13
12
Pc r ,
Pc r  1 1 .2 5 k N
SOLUTION (5.53)
( a ) Applying the method of joints at A: F AB  40 kN ( C ) and F B C  2 2 0 k N ( C )
L A B  2 .5 m .

cr
FAB

cr
6
 d
2
d  1 4 .3 m m
,
4
I A  d 4 and Euler’s formula:
We have r 

3
4 0 (1 0 )
2 5 0 (1 0 ) 
;
A
2
 E

(L r)
2
; 250 (10

) 
6
2
( 210  10
9
( 2 . 5 0 . 25 d )
)
, d  109 . 8 mm
2
Use, a commercial size of :
d  110 mm diameter
( b ) L B C  1 .8 7 5 m

FBC

cr
3
2 2 0 (1 0 )
2 5 0 (1 0 ) 
6
;
A
 d
2
d  3 3 .5 m m
,
4
Euler formula:

cr
2
 E

(L r)
2
6
; 250 (10

) 
2
( 210  10
9
)
( 1 . 875 0 . 25 d )
2
, d  82 . 4 mm
Use
8 3  m m diameter
SOLUTION (5.54)
I 
r 

64
( 0 . 04 )
I
A

d
4
9
 125 . 66  10
4
 10 mm
L
r

m
1600
10
4
 160
Euler’s formula:
2
P cr 
 EI
F a ll 
1 .0
0 .5
nL
2


2
( 200  125 . 66 )
1 .5 (1 .6 )
2
 64 . 6 kN
Thus
Pc r  2 ( 6 4 .6 )  1 2 9 .2 k N
SOLUTION (5.55)
Two angles:
I x  2 I x  2 ( 4 2 6  1 0 )  8 5 2 (1 0 ) m m
3
3
4
I y  2 ( I y  A x )  2[1 5 3 (1 0 )  7 4 4 (1 0 .5 ) ]  4 7 0 , 0 5 2 m m
2
I m in  4 7 0 , 0 5 2
Use smaller I :
2
Pc r 
Pa ll 
3
 EI y
2
Le
Pc r
n



2
1 2 2 .7
2 .5
mm
9
4
( 2 0 0  1 0 )( 4 7 0 , 0 5 2  1 0
( 2 .7 5 )
2
We have L e  2 .7 5 m Thus
12
)
2
 4 9 .1 k N
87
 1 2 2 .7 k N
4
SOLUTION (5.56)
A  d
I 
2
4   ( 60 )

 (60 )
4
 d
64
P cr 

2
L
4
64


EI
2
2
2
4  2 ,827
 6 3 6 ,1 7 3 m m
( 210  10
9
1
)( 636  10
 12
)
mm
4
r 
 1319
2
2

I
A
d
4

60
4
 15 mm
kN
Using Eq.(5.72):
Py
350 ( 10 ) 
6
3
2 . 827  10
0 . 002 ( 2 . 827  10
[1 
636 ,173 ( 10
 12
3
)
1
) 0 . 03 1 
]
Py
3
1319 ( 10 )
or
P y  2 6 6 0 1 0 4 .4 P y  1.3 0 5  1 0
2
Solving, this quadratic gives: P y  649
12
 0
kN . Thus, Pa ll  6 4 9 3  2 1 6 .3 k N
SOLUTION (5.57)
I m in 
1
(1 6 0  8 0  1 3 0  5 0 )  5 .4 7 2 5  1 0
3
3
A  1 6 0  8 0  1 3 0  5 0  6 .3  1 0
rm in 
6
mm
4
12
A  2 9 .5 m m
I m in
3
mm
3
L e  0 .5 L  2 .7 5 m
L e r  2 7 5 0 2 9 .5  9 3 .2
Hence,
 E
 (7 2  1 0 )
2

cr

( Le r )
2

2
9
( 9 3 .2 )
2
 8 1 .8 M P a
SOLUTION (5.58)
L e  0 .7 L  3 .8 5 m . From solution of Prob.5.57: rm in  2 9 .5 m m .
We now have
L e r  3 8 5 0 2 9 .5  1 3 0 .5
Hence,
 E
2

cr

 (7 2  1 0 )
2

( Le r )
9
(1 3 0 .5 )
2
 4 1 .7 M P a
SOLUTION (5.59)
A   (5 0  4 4 )  1, 7 7 2
2
2
mm
2
Equation (5.68):
r 
1
4
D
2
 d
2

1
4
100
2
 88
2
 33 . 3 mm ,
L
r

2000
33 . 3
 60 . 1
(CONT.)
88
5.59 (CONT.)
r
12(50 )

ec
(a)
2
( 3 3 .3 )
Py
Refer to Fig.5.23b:
r
2
( 3 3 .3 )
Refer to Fig.5.23b:
 160 M Pa,
A
9( 50 )

ec
(b)
 0 .5 4 1
2
P y  1 6 0 (1, 7 7 2 )  2 8 3 .5
kN
 0 .4 0 6
2
Py
 175 M Pa,
A
P y  1 7 5 (1, 7 7 2 )  3 1 0 .1 k N
SOLUTION (5.60)
We have

I 
(D
 d ) and A 
4
4

(D
2
 d );
2
r 
1
I A 
4
64
D
2
 d
2
4

A 
P  n ( 4 4 0 )  1 .5 ( 4 4 0 )  6 6 0 k N ,
(200  175 )  7363 m m
2
2
2
4
I 

( 2 0 0  1 7 5 )  3 2 .5 (1 0 ) m m
4
4
6
4
64
r 
1
200  175
2
2
 6 6 .4 4 m m
4
Hence,
P

A
ec
r
2

660  10
7363  10
3
6

r
0 .1 e
( 6 6 .4 4  1 0
L
 8 9 .6 M P a
3
)
2
3 .6 6
6 6 .4 4  1 0
 5 5 .1
3
 2 2 .6 5 e
Substitute these into Eq. (5.74a):


2 1 0  8 9 .6 1  2 2 .6 5 e s e c  2 7 .5 5



8 9 .6  1 0
190  10
6
9



 
from which
e  0 .0 4 9 0 2 m = 4 9 .0 2 m m
SOLUTION (5.61)
From solution of Prob. 5.60:
P
A  7363 m m
2
I  3 2 .5 (1 0 ) m m
6
4
Hence
 EI
2
Pc r 
v m ax
L
2
 ( 2 0 0  1 0 )(3 2 .5  1 0
2

9
( 4 .6 )
e
6
)
2
 3 .0 2 3 M N
Figure S5.61
(CONT.)
P
89
5.61 (CONT.)
( a ) Using Eq. (5.73):


1 .2 5  e  s e c 
 2


45  10
3
3 .0 3 2  1 0
6


o
  1   e  s e c (1 0 .9 6 )  1  ,



e  6 7 .2 8 m m
( b ) Referring to Fig S5.61:
M  P ( v m a x  e )  4 5 (1 .2 5  6 7 .2 8 )  3 0 8 0 N  m
Hence, 
P

m ax

Mc
A

I
45  10
3
7363  10

6
3 0 8 0 ( 0 .1)
3 2 .5  1 0
6
 1 5 .5 9 M P a
SOLUTION (5.62)
L e  2 ( 4 .6 )  9 .2 m . Refer to solution of Prob. 5.61
 EI
2
Pc r 
 ( 2 0 0  1 0 )(3 2 .5  1 0
2

2
9
(9 .2 )
Le
2
6
)
 7 5 7 .9 k N
( a ) Equation (5.73):


1 .2 5  e  s e c 
 2




o
  1   e [s e c ( 2 1 .9 3 )  1] ,
7 5 7 .9 

45
e  1 6 .0 2 m m
( b ) M  P ( v m a x  e )  4 5 (1 .2 5  1 6 .0 2 )  7 7 7 .2 N  m Therefore,

m ax

P
A


Mc
I
45  10
3
7363  10
6

7 7 7 .2 ( 0 .1)
3 2 .5  1 0
6
 6 .1 1 2  2 .3 9 1  8 .5 M P a
SOLUTION (5.63)
A
L e  2 L  3 .6 m
75 mm C
15 mm
A  150  75  120  45
 5 .8 5  1 0
P
150 mm
D
B
1
I 
3
mm
2
(1 5 0  7 5  1 2 0  4 5 )
3
3
12
P
 4 .3 6 2  1 0
r 
L
6
mm
4
I A  2 7 .3 m m
e  c  3 7 .5 m m
Thus,
ec
r
2

3 7 .5  3 7 .5
( 2 7 .3 )
2
 1 .8 8 7
Le
 6 5 .9 3
2r
(CONT.)
90
5.63 (CONT.)
Use Eq.(5.74b) with L  L e :

m ax

160  10
5 .8 5  1 0


 1  1 .8 8 7 s e c  6 5 .9 3



3
3

  9 9 .3 M P a

3
9
2 0 0  1 0 (5 .8 5  1 0 )  

160  10
3
SOLUTION (5.64)
150 mm
C
L e  2 L  3 .6 m
D
P
A  150  75  120  45
75 mm
B
A
P
 5 .8 5  1 0
3
mm
2
15 mm
1
I 
(7 5  1 5 0  4 5  1 2 0 )
3
3
12
 1 4 .6 1  1 0
6
mm
4
e  c  75 m m
r 
I A  4 9 .9 7 m m
Therefore,
ec
r
2
75  75

( 4 9 .9 7 )
 2 .2 5
2
Le
 3 6 .0 2
2r
Apply Eq.(5.74b) with L  L e :

m ax

160  10
5 .8 5  1 0
3
3


 1  2 .2 5 s e c  3 6 .0 2




   9 4 .8 M P a
3
9
2 0 0  1 0 (5 .8 5  1 0 )  

160 10
3
SOLUTION (5.65)
We have
A  ao  ai  120  100
2
1
I 
2
(ao  ai ) 
4
12
ao
3
2
(1 2 0  1 0 0 )  8 .9 5 (1 0 ) m m
4
4
6
4
12
8 .9 5 (1 0 )

 4 5 .1 m m
4 .4
A
c 
1
4
 4 .4 (1 0 ) m m
2
3
I
r 
2
 60 m m
2
Le  2 L  2 (2 )  4 m
Hence
ec
r
2

(5 5 )(6 0 )
( 4 5 .1)
 EI
2
2
Pc r 
2
Le
3

A
 ( 7 0  1 0 )8 .9 5 (1 0
2

P
 1 .6 2 2 ,
9
(4)
2
2 0 0 (1 0 )
4 .4 (1 0
6
)
6
 4 5 .5 M P a
)
 3 8 6 .5 k N
(CONT.)
91
5.65 (CONT.)
P
200

 0 .5 1 7
3 8 6 .5
Pc r
( a) Apply Eq. (5.73):
v m ax  e[ s e c (

P
2
Pc r
)  1]

 (5 5 )[se c (
0 .5 1 7 )  1]  7 3 .7 6 m m
2
( b ) Use Eq.(5.74a)


m ax
P
[1 
A
ec
r
sec(
2

P
2
Pc r
)]
 ( 4 5 .5 ) [1  (1 .6 2 2 ) s e c (

0 .5 1 7 ) ]  2 1 8 .3 M P a
2
SOLUTION (5.66)
From solution of Prob. 5.65:
A  a o  b i  4 .4 (1 0 ) m m
2
1
I 
2
3
( a o  a i )  8 .9 5 (1 0 ) m m
4
12
2
4
6
L e  2 ( L )  2 (1 .9 )  3 .8 m
 EI
2
Pc r 
P
2
Pc r
9
( 3 .8 )
Le
300

c  60 m m
 ( 2 0 0  1 0 ) (8 .9 5  1 0
2

4
6
)
2
 1, 2 2 3 k N
 0 .2 4 5
1223
( a ) Equation (5.73):
v m ax  e[ s e c (

P
2
Pc r
)  1]
or
1 5  e[ s e c (

0 .2 4 5 )  1] ,
e  3 7 .2 m m
2
(b) M

m ax
m ax
 P ( e  v m a x )  3 0 0 (3 7 .2  1 5 )  1 5 .6 6 k N  m

P
A

M
m ax
I
c

300  10
4 .4 (1 0
3
3
)
1 5 .6 6  1 0 ( 0 .0 6 )
3

4 .9 5 (1 0
 1 7 3 .2 M P a
92
6
)
SOLUTION (5.67)
Figure 5.17a: L e  2 L  2 ( 2 )  4 m .
( a ) Cylindrical tube:
A  500 m m

2

4
2
(D
 d )
2
or
d 
D
2


4A

40 
2
4 (500 )

 31 m m
Thus
I 

64
(D

d ) 
4
4
[ 4 0  3 1 ]  8 .0 3 3 (1 0 ) m m
4
64
4
4
4
and
r 
4
( 8 .0 3 3 )(1 0 )

I
A
 1 2 .6 8 m m
500
It follows that
L e r  4 0 0 0 1 2 .6 8  3 1 5 .5
Since
L e r  2 0 0 , the Euler formula applies. Hence
Pc r 
2
 EI
2
Le


2
9
(1 0 5  1 0 )( 8 .0 3 3  1 0
(4)
8
)
 5 .2 0 3 k N
2
( b ) Square tube.
The cross-sectional area: A  a o  a i . Inner diameter is
2
ai 
ao  A 
2
2
4 0  5 0 0  3 3 .1 7 m m
2
Then
I 
1
12
( a o  bi ) 
4
1
12
r 
I
A
 10
1 1 .2 5
500
 15 m m ,
4
( 4 0  3 3 .1 7 )  1 1 .2 5  1 0
4
4
4
mm
4
and
2
Le r  4 0 0 1 5  2 6 7
Since
L e r  2 0 0 , the Euler formula is valid. Therefore
Pc r 
2
 EI
2
Le


2
9
(1 0 5  1 0 )( 0 .1 1 2 5  1 0
(4)
2
6
)
 7287 N
Comment: Hollow square has a critical load that is 1.4 times more than for a
hollow circular section.
SOLUTION (5.68)
From Table B.1: E  2 0 0 G P a
S y  250 M Pa
The properties of area are
A  b h  (3 5 )(1 0 )  3 5 0 m m ,
2
I 
1
12
bh 
3
1
(3 5 )(1 0 )  2 9 1 7 m m
3
4
12
(CONT.)
93
5.68 (CONT.)
I
r 
2917

 2 .8 8 7 m m ,
(
r
350
A
Le
E
)c  
200  10
 
3
 8 8 .9
250
Sy
( a ) From Fig. 5.17c, L e  0 .7 L  0 .7 (1 8 0 )  1 2 6 m m . Hence
Le
126

 4 3 .6
2 .8 8 7
r
Since 4 3 .6  8 8 .9 , Johnson Formula should be used. Thus:
S y ( Le r )
Pc r  A S y [1 
2
]
4 E
2
2
2 5 0 ( 4 3 .6 )
 ( 3 5 0 ) ( 2 5 0 ) [1 
4
 200  10
2
3
]  8 2 .2 3 k N
( b ) Now we have
Le
0 .7 ( 5 0 0 )

 1 2 1 .2  8 8 .9
2 .8 8 7
r
Euler formula applies. So
 EI
 ( 2 0 0  1 0 )( 2 9 1 7  1 0
2
Pc r 
2

2
9
( 0 .3 5 )
Le
12
2
)
 47 kN
SOLUTION (5.69)
( a ) Cross-sectional area:
A 

4
2
(D

 d ) 
2
( 6 2 .5  6 0 )  2 4 0 .5 m m
2
4
2
2
Moment of inertia:

I 
64
4
(D

 d ) 
4
64
( 6 2 .5  6 0 )  1 1 2 , 8 4 1 .5 m m
4
4
4
and
r 
Le
I
A

r
750
2 1 .7

 2 1 .7 m m
1 1 2 ,8 4 1 .5
2 4 0 .5
 3 4 .5 6
Also
(
Le
r
2
 E
)c 
Sy


2
9
( 7 0 1 0 )
2 7 0 1 0
6
 5 0 .5 8
Since L e r  5 0 .9 6 , the Johnson formula applies. Thus
Pc r  A S y [1 
Sy
2
4 E
(
Le
r
) ]  2 4 0 .5  1 0
2
6
( 2 7 0  1 0 )[1 
6
6
2 7 0  1 0 ( 3 4 .5 6 )
4
2
9
( 7 0 1 0 )
2
]
 5 7 .3 6 k N
( b ) We have C  D 2 and:
ce
r
2

3125 (3)
( 2 1 .7 )
2
 0 .2
A  2 4 0 .5 m m
2
P  16 kN
(CONT.)
94
5.69 (CONT.)
Equation (5.74a) gives then


m ax
[1 
P
A
2
3
2 4 0 .5  1 0
P
Pc r
)]
[1  0 .2 s e c ( 2
1 6 (1 0 )

s e c ( 2
ec
r
6
16
5 7 .3 6
)]
 7 9 .8 M P a
SOLUTION (5.70)
From Prob.5.55 for both angles: I m in  4 7 0 , 0 5 2 m m
r 
1,4 8 8
2
2 E
Sy
Cc 
Le
 1 7 .7 7 m m ,
4 7 0 ,0 5 2
2
2

( 200  10
240  10
9
)
6
r

4
and A  7 4 4  2  1 4 8 8 m m
2 .7 5  1 0
1 7 .7 7
3
 1 5 4 .8
 128
Since C c  1 5 4 .8, use Eqs. (5.77b):



a ll
2
9
( 2 0 0 1 0 )
1 .9 2 (1 5 4 .8 )
 4 2 .9 M P a
2
Hence
Pa ll  
a ll
A  4 2 .9 (1 4 8 8 )  6 3 .8 4 k N
SOLUTION (5.71)
Cc 
2
2 E
Sy
 [
2
2
1
3
( 200  10 )
Le
] 2  106 ,
350

r
0 . 65 ( 3 )
d 4

7 .8
d
(Eq.c of Sec. 6.2)
Equation (5.77b):

Le
Check:
3
5 0 (1 0 )

a ll
 d

r
2
4
6
0 . 0441
2


9
( 2 0 0 1 0 )
1 .9 2 ( 7 .8 d )
2
d  4 4 .1 m m
,
 136  C c
OK.
SOLUTION (5.72)
L e  0 . 5 L  2 m . Assume 1 1  L e d  2 6 and use Eq.(5.80b). Thus

a ll

P
A

3
1 0 0 (1 0 )
d
2
 8 .2 7 [1 
1
3
(
Le d
26
2
) ]1 0
6
This gives
1 0 0 (1 0 )  8 .2 7 d
3
2
 0 .0 2 4 5,
d  109 m m
Then
 
P
A

3
1 0 0 (1 0 )
( 0 .1 0 9 )
Check: 8 .4 2  1 0
2
 8 .4 2 M P a
O K .,
L e d  2 0 .1 0 9  1 8 .3
95
OK.
2
Thus
SOLUTION (5.73)
A 

(3 5 0  3 0 0 )  2 5 , 5 2 5 m m ,
2
4
2
L e  0 .7 L  0 .7 ( 6 .1)  4 .2 7 m .
2
Equation (5.68):
r 
1
4
 d
2
D

2
350  300
2
1
4
Le
 1 1 5 .2 4 m m ,
2
r

4 .2 7  1 0
1 1 5 .2 4
3
 3 7 .0 5
Using Eq.(5.79b):
 a ll  [1 4 0  0 .8 7 (3 7 .0 5 )]  1 0 7 .7 7 M P a
Hence
Pa ll  1 0 7 .7 7  1 0 ( 2 5 , 5 2 5  1 0
6
6
)  2751 kPa
SOLUTION (5.74)
0 .1 2  0 .0 8
12
I m in 
3
6
 5 .1 2 (1 0
A  0 .0 0 9 6 m ,
rm in 
2
Cc 
4
) m ,
I m in
2
2 E
Sy
A  2 3 .0 9 m m ,
 1 2 1 .7
Le r 
3,5 0 0
2 3.0 9
 1 5 1.6
By Eq.(5.77b):

a ll
2
 E

1 .9 2 ( L e r )
2
2


9
( 2 1 0 1 0 )
1 .9 2 (1 5 1 .6 )
2
 4 6 .9 7 M P a
We have
600
9 .6
 62 . 5 MPa
 280
MPa
SOLUTION (5.75)
Table A.6: A  27 . 5  10
3
mm
r z  101 . 1 mm
2
Buckling in xy plane:
C c  [ 2
2
( 200  10
1
3)
280 ] 2  119 , L e  0 . 7 L  4 . 2 m
and
L e r z  4 .2 0 .1 0 1 1  4 1.5  C c . Apply Eq.(5.77a):
n 
5
3


all

3
8
( 119 ) 
1
8
( 1 1 9 )  1.7 9
1
2
.5
( 41
) ]  146 . 9 MPa
119
4 1.5
280  10
1 . 79
6
[1 
3
4 1 .5
2
and
Pall  146 . 9 ( 27 . 5 )  4 , 040
kN
Buckling in xz plane:
L e  0 . 5 (12 )  6 m
n 
5
3


a ll

3
8
( 3171.59 ) 
2 8 0 1 0
1 .7 7
6
[1 
r y  161
mm ;
1
8
( 31 71 .39 )  1 .7 7
1
2
( 31 71 .39 ) ]  1 5 0 .4 M P a ,
L r y  37 . 3  C c
2
2
96
Pa ll  1 5 0 .4 ( 2 7 .5 )  4 ,1 3 6 k N
SOLUTION (5.76)
A  a
r 
2
a
I A 
2
a
I  a
L
a
4
r
12
0 .5

3
a (2

1 .7 3 2
a
3)
Assume: 9 .5  L e r  6 6 .
Using Eq. (5.79b):

a ll
250  10

a
3
 [1 4 0  0 .8 7 (
2
1 .7 3 2
) ]1 0
6
a
or
3
a  1 0 .7 6  1 0
2
a  1 .7 8 6  1 0
3
 0,
a  48 m m .
So,
r  a 2
3  48 2
3  1 3 .9 m m
L r  5 0 0 1 3 .9  3 6  6 6
Our assumption was correct. Use a  4 8 m m
SOLUTION (5.77)
A  200  100  160  60  10, 400 m m
I m in 
rm in 
1
2
( 2 0 0  1 0 0  1 6 0  6 0 )  1 3, 7 8 6 , 6 6 6 .6 7 m m
3
3
4
12
A  3 6 .4 1 m m ,
I m in
L e rm in  4 .3  1 0
3
3 6 .4 1  1 1 8 .1
Use Eq. (5.79c):

a ll

350  10
(1 1 8 .1)
3
 2 5 .0 9 M P a
2
Pa ll  2 5 .0 9  (1 0 , 4 0 0 )  2 6 0 .9 4 k N
SOLUTION (5.78)
For the situation described L e  2 L (see Fig. 5.17a) and d=62 mm So,
Le
2 (1 .2 2  1 0 )
3

 3 9 .4
62
d
Since 39.4 > 26, use Eq. (5.80c).
0 .3 (1 2  1 0 )
9

a ll

(3 9 .4 )
2
 2 .3 2 M P a
and
Pa ll  
a ll
A  ( 2 .3 2 ) (8 8  6 2 )  1 2 .6 6 k N
97
SOLUTION (5.79)
We now have L e  2 L  2 ( 7 5 0 )  1 5 0 0 m m and d=62 mm
Le
1500

 2 4 .2
62
d
Since 24 .2< 26, apply Eq. (5.80b).

1 2 4 .2 2
(
) ]  5 .8 8 M P a
3 26
 8 .2 7[1 
a ll
Thus
Pa ll  
a ll
A  ( 5 .8 8 ) (8 8  6 2 )  3 2 .0 8 k N
SOLUTION (5.80)
Boundary conditions: v ( 0 )  0 , v ( L )  0 ,
With w=0, the solution of Eq.(5.82b):
M (0)   M 0 ,
M (L)   M
0
v  A sin k x  B c o s k x  C x  D
(1)
v "   A k s in k x  B k
(2)
2
2
cos kx
Substituting boundary conditions into these equations, we have
v (0 )  B  D  0,
v ( L )  A s in k L  B c o s k L  C L  D  0
and since E Iv "  M ,
M ( 0 )   B E Ik
Solving, and setting k
M
A 
  M 0,
M ( L )   A E Ik sin k L  B E Ik c o s k L   M
2
2
 P E I as needed,
2
1 c o s k L
s in k L
0
P
2
B  D 
,
M
0
P
C  0
,
Equation (1) is thus
M
v 
0
1 c o s k L
s in k L
(
P
s in k x  c o s k x  1 )
SOLUTION (5.81)
Apply Eq.(4.14)
2
EI
dx
M   Pv 
 M
d v
2
2
dx
2
(x
2
 Lx )
P
or
d v
w
2
 k x 
2
w
2
(x
2
wL
2
 Lx )
A
B
x
wL
2
Figure S5.81
We have, the deflection:
v  vh  vP
(1)
v h  A s in k x  B c o s k x
(2)
Here
It can be shown that, for the case of uniform loading w (Fig. S5.81):
vP 
w
2 E Ik
2
(x
2
 Lx 
2
k
2
(3)
)
Boundary conditions v h ( 0 )  0 and v h ( L )  0 give A and B. In doing so, Eq.(1) results in:
v 
w
E Ik
4
[(1  c o s k L )
s in k x
s in k L
 cos kx 
98
2
k
2
(x
2
 L x )  1]
0
SOLUTION (5.82)
Refer to Example 5.23.

v 

m x
L
a m sin
(1)
m 1
Hence
U 
W 

EI
2
1
2
L
(d
0

2
dx
v
2
L
4
 EI
) dx 
2
P ( v ') d x 
2
0
4L

L
3

4
w vdx 
0
2
m am

2
 P
4L
m am 
2
2
2wL


am
1
m
Applying  U   W , we obtain
am 
4 wL
5
4
 EI
1
3
2
m (m b)
where b 
PL
2
2
 EI
Substitution of this into Eq.(1) gives the solution.
End of Chapter 5
99
Section II FAILURE PREVENTATION
CHAPTER 6
STATIC FAILURE CRITERIA AND RELIBILITY
SOLUTION (6.1)
Table 6.2: K c  5 9 1 0 0 0 M P a
mm
S y  1503 M Pa
and
  1 . 01
Case A of Table 6.1: with a w  0 . 1 ,
From Eq.(6.3), with n  1 :
 
Kc

 a

59
 2 0 8 .4
1000
(1 .0 1 )
 ( 25 )
M Pa
Thus, we have
P   ( 2 w t )  2 0 8 .4 ( 0 .5  0 .0 2 5 )1 0  2 , 6 0 5
6
kN
The nominal stress at fracture
 
6
2 .6 0 5 (1 0 )
 2 3 1 .6
0 .0 2 5 ( 0 .5  0 .0 5 )
M Pa
This is well below the yield strength of 1503 MPa.
SOLUTION (6.2)
Py  
all
A net 
( 350  15 )  2 . 844
650
1 .2
Case B of Table 6.1: with a w  0 .0 7 ,
 
By Eq.(6.3),
Kc
n
 a

MN
  1.1 2
100 1000
1 .2 (1 .1 2 )
 ( 25 )
 2 6 5 .5 M P a
Since 265.5 < 650, fracture controls. Thus
P f  265 . 5 ( 350  15 )  1 , 394
kN
SOLUTION (6.3)
From Table 6.2: K c  23 1000
MPa
Case B of Table 6.1: a w  0 .1 6 ,
mm
and S
  1.1 2
By Eq.(6.3), with n  1 :
 
Kc

 a

 8 1 .9 3 M P a
23 1000
1 .1 2
 ( 20 )
It follows that
P   ( wt )  81 . 93 (125  25 )  256
Then
 
P
( w  a )t

3
256 ( 10 )
( 0 . 125  0 . 02 ) 0 . 025
 97 . 5 MPa
 S
 97 . 5 MPa
y
100
kN
y
 444
MPa
SOLUTION (6.4)
Table 6.1, Case A:   1 .0 3 .
Table 6.2, K c  5 9 M P a
S y  1503 M Pa .
m,
We have
 
Sy
n

 6 0 1 .2 M P a
1503
2 .5
Equation (6.1) gives
K  
 a  (1 .0 3 )( 6 0 1 .2 )  ( 2  1 0
3
)  4 9 .0 8 M P a
Using Eq. (6.2), we find
n 
Kc

K
59
4 9 .0 8
 1 .2
SOLUTION (6.5)
Table 6.1, Case B:   1 .3 7 .
Table 6.2, K c  3 1 M P a
m,
S y  392 M Pa .
We have
 
Sy
n

 140 M Pa
392
2 .8
Equation (6.1):
K  
 a  1 .3 7 (1 4 0 )  ( 0 .0 0 5 )  2 4 .0 4 M P a
Applying Eq. (6.2),
n 
Kc

K
31
2 4 .0 5
 1 .2 9
SOLUTION (6.6)
Table 6.1: a w  5 0 5 0 0  0 .1,
Table 6.2: K c  1 1 1 M P a
m,
  1 .0 1 .
S y  798 M Pa .
( a ) Equation (6.1),
K  
 a  1 .0 1(1 5 0 )  ( 0 .0 5 )  6 0 M P a
Equation (6.2) gives the safety factor for fracture as,
n 
Kc

K
111
60
 1 .8 5
Safety factor for yielding is
n 
Sy


798
150
 5 .3 2
( b ) Using Eq. (6.3) with n=1, fracture stress is

f

Kc

a

111
(1 .0 1 )
 ( 0 .0 5 )
 2 7 7 .3 M P a
101
m
m
m
SOLUTION (6.7)
Case A of Table 6.1 and Table 6.2:
K
c
 59
1000
  1 . 01
MPa
S y  1503 M Pa
mm
(assumed)
By Eq.(6.3):
 
Kc
n
(a)
 
pfr
(b)
 
pfr
2t
t

 a
59
 3 1 0 .7
1000
 (5)
(1 .5 )(1 .0 1 )
M Pa  S y
,
p f  2 ( 4 9 .7 1)  9 9 .4 2 M P a
,
pf 
t
r
3 1 0 .7 ( 4 )

25
 4 9 .7 1 M P a
SOLUTION (6.8)
Table 6.2: K c  31 MPa
Case D of Table 6.1:
m
y
 392
MPa
   1.3 2
 0 .4
a
w
S
Using Eq.(6.3) with n  1 and   6 M tw :
2
Kc  
a ;
6M
tw
2
3 1(1 0 )  1.3 2
6
6M
0 . 0 2 5 ( 0 .1 )
2
 ( 0 .0 4 )
Solving
M  2 . 76 kN  m
SOLUTION (6.9)
( a ) K c  59 M Pa
m
S y  1 5 0 3 M P a (Table 6.2).
and
Case B of Table 6.1: with a=12.5 mm and w=125 mm:
a
w
 0 . 1    1 . 12
Eq.(6.3) with n  1 :
5 9  (1 .1 2 ) 
 (1 2 .5  1 0
3
),
  2 6 5 .8 M P a
Therefore
Pa ll  ( w t )  (1 2 5  2 5 )( 2 6 5 .8 )  8 3 0 .6 k N
Note that the nominal stress at fracture
 
8 3 0 .6
2 5 (1 2 5  1 2 .5 )
 2 9 5 .3 M P a  S y
( b ) Table 6.2: K c  6 6 M P a
m
and S y  1149 M P a
Thus
66  10
6
 (1 . 12 ) 
 a  (1 . 12 )(
Solving
a  0 .0 1 5 m = 1 5 .6 m m
Comment: This value of a satisfies Table 6.2
102
830 . 6  10
125  25  10
3
6
) a
SOLUTION (6.10)
a
w
Refer to Example 6.3. We now have
 a  2 .1 1

 0 .4 :
15
37 . 5
 b  1.3 2
and
Equation (6.5) is therefore
  ( 2 . 11 )
Table 6.2: K c  7 7 M P a
m
6 ( 0 . 175 P )
 (1 . 32 )
P
0 . 0375 ( 0 . 0125 )
0 . 0125 ( 0 . 0375 )
2
 83 , 349 P
and S y  690 M P a
Note that both a and t satisfy Table 6.2. Using Eq.(6.3),
 
Kc
 a
n
7 7 1 0
8 3, 3 4 9 P 
:
2
3
 ( 0 .0 1 5 )
Solving
P  2 .1 2 8 k N
The nominal stress at fracture,
( 2 .1 2 8  1 0 ) [ 0 .0 1 2 5 ( 0 .0 3 7 5  0 .0 1 5 )]  7 .5 7 M P a  S y .
3
SOLUTION (6.11)
By Table 6.2, K c  5 9 M P a
m and S y  1503
M P a . Note that values of a and t satisfy
Table 6.2.
a
w
From Table 6.1:

a ll
Kc

n
 a
   1.1 2 . Therefore,
 0 .1
5 9 1 0

(1 .4 )(1 .1 2 )
6
 ( 0 .0 0 5 )
 3 0 0 .2 2 M P a
and
Pa ll  
a ll
( w t )  3 0 0 .2 2 ( 0 .0 5  0 .0 2 5 )  0 .3 7 5 M N = 3 7 5 k N
Comment: The nominal stress at the fracture
375 kN
t(2w2a)

3
3 7 5 1 0
( 0 .0 2 5 )( 0 .1  0 .0 1 )
 1 6 6 .7 M P a  S y
SOLUTION (6.12)
(a)
(b)

x


xy

S
y
n
S
y
n
32 M
3
D
16 T
D
3
 (
x
 (
x
2
2


4P
D
2
3
32 ( 5  10 )

 ( 0 .1 )
3
16 ( 8  10 )
 ( 0 .1 )
3
2
3
4 ( 50  10 )
 ( 0 .1 )
2
 57 . 3 MPa
1
260
n
 [ 5 7 .3  3 ( 4 0 .7 4 ) ] 2 ,
260
n
 [ 5 7 .3  4 ( 4 0 .7 4 ) ] 2 ,
1
 4 xy ) 2 ;

 40 . 74 MPa
1
 3 x y ) 2 ;
2
3
2
2
103
2
2
1
n  2 .8 6
n  2 .6 1
SOLUTION (6.13)
Table B.1: S u  2 4 0 M P a and S u '  6 5 0 M P a
From the solution of Prob. 6.12, we have 
 5 7 .3 M P a and 
x
xy
 4 0 .7 5 M P a .
Thus




x
2
1 ,2
(

x
2
) 
2
2
xy
 2 8 .6 5 
( 2 8 .6 5 )  ( 4 0 .7 5 )
2
2
or
 1  7 8 .4 6 M P a
 2   2 1 .1 6 M P a
Thus
1


u n
 1;
2
u n
7 8 .4 6
240

 2 1 .1 6
650

1
n
or
n  2 .7 8
SOLUTION (6.14)
y
A
F
B
0.8 m

A
P
T
We have
T
D
M
P=20F,
T=0.4F,
P
x

M=0.8F
Stresses at fixed end:

b


32 M
a 
 D
3
32 ( 0 . 8 F )
 ( 0 . 04 )
20 F
 ( 0 .0 4 )
2
4
3
 
 127 , 324 F

 1 5 , 9 1 5 .5 F
x
16 T
 D
3

16 ( 0 . 4 F )
 ( 0 . 04 )
3
 31 , 831 F
  a   b  1 4 3 , 2 3 9 .5 F
Apply Eq.(6.16):
S
y
n
1
 [ x  3 ] 2 ;
2
2
6
2 5 0 (1 0 )
1.4
2
or
F  1 . 163
kN
SOLUTION (6.15)
Refer to solution of Prob. 6.14. We have
 m ax 
(
 [(

x
2
) 
2
1 4 3 , 2 3 9 .5 F
2
2
)  (3 1, 8 3 1 F ) ]
2
2
Hence,
 m ax 
Sy
2n
;
7 8 , 3 7 4 .7 F 
1
 F [(1 4 3 , 2 3 9 .5 )  3 ( 3 1, 8 3 1 ) ] 2
6
2 5 0 (1 0 )
2 (1 .4 )
or
F  1 .1 3 9 k N
104
1
2
 7 8 , 3 7 4 .7 F
2
x
SOLUTION (6.16)
T  0 .4 F
At the fixed end A:
M
 0 .8 F
z
Vy  F
Thus,


x
32M
D
A
1 6T
  
D
 ( 0 .0 6 )
 ( 0 .0 6 )
 3 7 .7 2 6 F
3
1 6 ( 0 .4 F )
 
3
3 2 ( 0 .8 F )

3
  9 .4 3 1 F
3
Then
 1, 2 
3 7 .7 2 6 F
3 7 .7 2 6 F
)  (  9 .4 3 1 F )
2
2
2
 1  3 9 .9 5  1 0 F
140  10
(
2

3
(a)

 m a x  2 1 .0 9  1 0 F
  2 .2 3  1 0 F
3
2
3
6
 2 1 .0 9  1 0 F ,
F  4 .1 5 k N
3
1 .6
( b )  1   1 2   2  ( S y n )
2
2
2
[3 9 .9 5  3 9 .9 5 (  2 .2 3 )  (  2 .2 3 ) ]
2
2
1
2
(1 0 ) F  ( 2 6 0 1 .6 )1 0
3
6
or
F  3 .9 5 k N
SOLUTION (6.17)
Q B  102 (10 . 3 )( 74 . 85 )  78 . 6  10
3
3
mm
Q C  Q B  6 . 6 ( 69 . 7 )( 34 . 85 )  94 . 6  10
3
mm
3
We have

A

Mc
I

VQB

3
24 ( 10 )( 0 . 08 )
13 . 4 ( 10
6
)
 143 . 3 MPa ,

B

(143 . 3 )  124 . 9 MPa
69 . 7
80
and
B 
Ib
3
6 0 (1 0 )( 7 8 .6  1 0
1 3 .4 (1 0
6
6
)
)( 0 .0 0 6 6 )
 5 3 .3 M P a ,
c 
VQc
Ib
 6 4 .2 M P a
Thus
(  1, 2 ) B 
1 2 4 .9
2
 [(
1 2 4 .9
2
 1 B  1 4 4 .6 M P a
or
(
max
1
)  5 3.3 ] 2  6 2 .5  8 2 .1
2
2
 2 B   1 9 .6 M P a
) B  82 . 1 MPa
Hence
( m ax ) B 
S
y
2(2 )

320
4
8 2 .1  8 0
;
 F a ils
Alternatively, using Eq.(6.11), we have
[1 2 4 .9
2
1
 4 ( 5 3.3 ) ] 2 
2
320
2
;
1 6 4 .2  1 6 0
105
 F a ils
SOLUTION (6.18)
From Solution of Prob.6.17, at point B:
 1  1 4 4 .6 M P a
 2   1 9 .6 M P a
Thus
1
S
[ 1   1 2   2 ] 2 
2
2

y
n
320
2
 160
1
or
2
[( 144 . 6 )
 (144 . 6 )(  19 . 6 )  (19 . 6 ) ] 2  155 . 3  160
2
 No failure
Alternatively, by Eq.(6.16):
1
 3 ( 53 . 3 ) ] 2 
2
[124 . 9
2
; 155 . 3  160
320
2
 No failure
SOLUTION (6.19)
1   
(a)
1 
(b)
 1   1

3
pr

t
Sy
n
2
2
3 .5 ( 0 .2 5 )
0 .8 7 5
t
;


t

250
1 .5
S
 (
2
2
0 .8 7 5
t
2

0 .4 3 7 5
t
) ; [(
0 . 875
t
)
2
3  0
,
t  5 .2 5  1 0
,
2
y
n

,
3
m = 5 .2 5 m m
2
t
or
1
1
t
[ 0 .7 5 8 ] 2 
250
1 .5
t  4 .5 4 8  1 0
,
3
m = 4 .5 4 8 m m
SOLUTION (6.20)
From Solution of Prob.6.19:
 1  0 . 875
,
2 
t
1 
(a)
Su
;
n
0 .8 7 5
t
0 . 4375
t

350
1 .5

,
3
 0
t  3 .7 5  1 0
,
3
 3 .7 5 m m
( b ) Using Eq.(6.26):
1 
Su

Su
0 .8 7 5
t
;
n

350
1 .5
t  3 .7 5  1 0
,
3
m = 3 .7 5 m m
and
2

n
0 .4 7 5
t
;

350
1 .5
,
t  1 .8 7 5  1 0
3
SOLUTION (6.21)
From Solution of Prob.6.17, at point B:
 1  1 4 4 .6 M P a
 2   1 9 .6 5 M P a
(a)

1
( b ) By Eq.(6.25),

Su
n
,
n 
n 
Su
1

280
144 . 6
280
144 . 6  19 . 65 ( 280 620 )
 1 . 94
 1 . 82
106
1
 ( 0 . 875  20 . 4375 )  ( 0 . 4375
) ]2 
t
m = 1 .8 7 5 m m
250
1 .5
SOLUTION (6.22)


M
A
 0:
Fy  0 :
1 7 5 ( 2 .4 6 )( 2 .4 6 2 )  R B (1 .7 )  0 ,
R B  3 1 1 .5 k N
R A  119 kN
y
175 kN/m
z
A
1.7 m
119 kN
2b
0.76 m C
B
b
311.5 kN
133
V, kN
119
x
-178.5
M, kN  m
x
-50.54
At B at the upper outermost fiber:

m ax

M
m ax
c
5 0 .5 4  1 0 b
3

1
I
b (2b )

7 5 .8 1  1 0
b
3
3
3
12
Thus,

m ax

7 5 .8 1  1 0
;
a ll
b
3
 140  10 ,
6
3
b  0 .0 8 2 m = 8 2 m m
At B at the neutral axis axis:
 m ax 
3V

2A
3 1 7 8 .5  1 0
2
2b
3

2
1 3 3 .8 7 5  1 0
b
3
2
And
 m ax   1    2
Thus,
 m ax  
;
a ll
1 3 3 .8 7 5  1 0
b
2
3
 140  10 ,
6
Use a 8 2 m m by 1 6 4 m m rectangular beam.
107
b  0 .0 3 1 m = 3 1 m m
SOLUTION (6.23)
We now have
N.A.
A  0 .3  0 .1 2  0 .0 3 6 m
B
120 mm
A
I 
1
12
( 0 .1 2 )( 0 .3 )  0 .2 7  1 0
3
M  0 .4  0 .1 5  0 .5 5 P
300 mm
c A  c B  0 .1 5 m
Referring to Example 6.9:


P
A

McA
A


P
A

M cB
B
 2 7 .7 7 8 P  3 0 5 .5 5 6 P
I
 2 7 .7 7 8 P  3 0 5 .5 5 6 P
I
Thus
 1  3 3 3 .3 3 4 P
2  0
 2   2 7 7 .7 7 8 P
1  0
It follows that
3 3 3 .3 3 4 P 
6
1 7 0 (1 0 )
2 .5
P  204 kN
,
6
6 5 0 (1 0 )
 2 7 7 .7 7 8 P 
P  936 kN
,
2 .5
SOLUTION (6.24)
S u  170
S uc  650
MPa
(Table B.1).
MPa
Thus
 1, 2 
and
(a)
100 50
2
 [(
 1  127 . 6 MPa
 127 . 6 , n  1 . 33
170
n
  7 7 .6 ,
1
)  (70 ) ]2
2
2

170
n
n 
(b)
100 50
2
2
  77 . 6 MPa
n  2 .1 9
170
1 2 7 .6  7 7 .6 (1 7 0 6 5 0 )
 1.1 5
SOLUTION (6.25)
Table B.1: S u  170
 1, 2 
or
(a)
(b)
120 60
2
 1  140
S uc  650
MPa
 [(
MPa ,
120 60
2

)  40 ]
2
2
 140 ,
n  1 . 21
170
n
  80 ,
n  2 . 13
170
1 4 0  8 0 (1 7 0 6 5 0 )
MPa
1
2
 40 MPa ,
170
n
n 
2
 1.0 6
108
2

3
  80 MPa
3
m
3
SOLUTION (6.26)
The circumferential, axial, and radial stresses are given by
1 
pr

 24 p
t
2
pr

2t
3  0
 12 p
Insertion of these expression into Eqs. (6.6) and (6.14) provide the critical
pressures.
( a ) For the maximum shearing stress theory:
24 p  0 
(250  10 )
6
1
1 .4
p  7 .4 4 M P a
( b ) For the maximum energy of distortion theory:
p (24  24  12  12 )
2
2
1 2

1
1 .2
(250  10 )
6
p  1 0 .0 2 M P a
Comment: The permissible value of the internal pressure is conservatively limited
to 7.44 MPa.
SOLUTION (6.27)
Maximum shear stress criterion, substituting the given expressions for 
n 
Sy
 1 

2
S yt
( 3 . 56  1 . 70 )  a
 0 . 19
2
1
and 
2
into Eq.(6.6):
S yt
 a
2
Maximum distortion energy criterion, using Eq.(6.14) together with the given expressions for
 1 and  2 :
Sy
n 
2
[  1   1
2

2
2
S yt

1
 0 . 215
1
2
2
[ 3 . 56  ( 3 . 56 )(  1 . 70 )  (  1 . 70 ) ] 2  a
]2
2
S yt
 a
2
The factor of safety based on energy of distortion theory is therefore 11.6 % larger than
that based on the maximum shear stress theory. This indicates that the maximum energy of
distortion criterion is less conservative, as expected.
SOLUTION (6.28)
Principle stress criterion, carrying the given expressions for 
n 
Su
1
Su

3 . 56  a
2
t
 0 . 281
1
and 
2
into Eq.(6.22):
Sut
 a
2
Coulomb-Mohr criterion, applying Eq.(6.25) together with the given expressions for 
1
and  2 :
n 
Su
 1 
2
S u S uc

Sut
[ 3 . 56  (  1 . 7 )( 1 5 )]  a
2
 0 . 256
Sut
 a
2
The n according to the principle stress criterion is thus 8.9 % larger than that on the basis of
The Coulomb-Mohr criterion. This indicates that the Coulomb-Mohr theory is more conservative,
particularly when S u c   S u .
109
SOLUTION (6.29)
Table B.1; S y  2 5 0 M P a


x
4P

,
2
 D

xy
16T
3
 D
Equation (6.11) results in
 xy 
1
2
Sy
[(
1
)   x ]2 
2
n
2
6
{ ( 2 5 01 .51 0 )  [
1
2
2
3
4 ( 4 5 1 0 )
 ( 0 .0 5 )
2
1
] } 2  8 2 .5 M P a
2
Thus
T 
D
3
 ( 0 .0 5 )
 xy 
16
3
16
(8 2 .5  1 0 )  2 .0 2 5 k N  m
6
SOLUTION (6.30)
Refer to solution of Prob.6.29, Equation (6.16) gives
 xy 
1
[(
3
Sy
n
) x]
2
2
1
3
6
{ ( 2 5 01 .51 0 )  [
2
3
4 ( 4 5 1 0 )
 ( 0 .0 5 )
2
1
] } 2  9 5 .3 1 M P a
2
and
T 
D
3
 xy 
16
 (2)
3
16
(9 5 .3 1  1 0 )  2 .3 3 9 k N  m
6
SOLUTION (6.31)
We have  1   ,
 2   .
Table B.2: S u  1 5 0 M P a ,
S uc  5 7 5 M P a
( a ) Equation (6.22):
 
Su

n
 107 . 1 MPa
150
1 .4
( b ) Equation (6.24) is thus

150


575

1
1 .4
or
  84 . 98 MPa
SOLUTION (6.32)
We have n=2, Table B.1: S u  340 MPa ,

1, 2

1
2
(  102  0 )
(  10  0 ) 
2
S uc  620

Equation (6.24):
1
Su



2
S uc
1
n
Thus
5
25  
340
2

5
25  
620
2

1
2
Solving
  111 . 1 MPa
110
2
 5 
MPa
25  
2
SOLUTION (6.33)
We have n=2. Table B.1: S y  3 4 5 M P a
( a ) Equation (6.11):
S
 (
y
n
345
2
or
1
 4 xy ) 2
2
2
x
1
  8 6 .1 M P a
1
  9 9 .4 3 M P a
 [(  1 0 )  4  ] 2 ,
2
2
( b ) Equation (6.16):
S
 (
y
n
345
2
or
1
 3 x y ) 2
2
2
x
 [ (  1 0 )  3 ] 2 ,
2
2
SOLUTION (6.34)
We have 
y
 xy  4 5 M P a
 30 M Pa

x
 0.
Thus

 1, 2  
y


y
(
2
) 
2
2
 1  3 2 .4 M P a
2
xy
 
30

2
(
30
)  (45)
2
2
2
 2   6 2 .4 M P a
( a ) Maximum principal stress theory:
1
 1;
3 2 .4  5 5
 1;
6 2 .4  5 5
Su
But

2
(failure occurs)
Su
( b ) Coulomb-Mohr Theory:
1  2

Su
32
 1;

55
S uc
 6 2 .4
1
160
gives
0 .5 8  0 .3 9  0 .9 7  1
(no fracture)
SOLUTION (6.35)

x

32M
D
3
 xy 
1 6T
D
3
M  7 5 0 ( 0 .3 )  2 2 5 N  m
(CONT.)
111
6.35 (CONT.)
350
( S u ) a ll 
( S u c ) a ll 
Also
 140 M Pa
2 .5
630
 252 M Pa
2 .5

 1, 2 

x
(

2
x
) 
2
2

2
xy
16
D
3
(M 
M
T
2
2
)
Substituting the given data
16
 1,2 
or
D
(225 
3
2

3
D
2
D
4 7 9 3 .7 9
1 
1
225  680 ) 
 
2
3
(1 1 4 5 .9 2  3 6 4 7 .8 7 )
2 5 0 1 .9 5
D
3
( a ) Maximum Principal Stress Theory
4 7 9 3 .7 9
 1 4 0 (1 0 )
6
3
D
( b ) Coulomb-Mohr Theory:
1
2

( S u ) a ll
or
or
D  0 .0 3 2 5 m = 3 2 .5 m m
4 7 9 3 .7 9
 1;
( S u c ) a ll
140 D
3

 2 5 0 1 .9 5
252 D
3
 10
6
D  0 .0 3 5 3 m = 3 4 .3 m m
SOLUTION (6.36)
M  (W  F ) L  ( 9  2 )  0 .2 5  2 .7 5 k N  m
T  W a  9  0 .3  2 .7 k N  m
An element at point A:

x
32M

t 
D
1 6T
D

3
 ( 0 .0 5 )
1 6 ( 2 .7 )

3
3 2 ( 2 .7 5 )
 ( 0 .0 5 )
3
 224 M Pa
3
y
A
T
 110 M Pa
z
So
 m ax 

(

x
2
(
224
) t 
2
2
Sy
)  (1 1 0 )
2
C
B
W+F
n
2
 157 M Pa
y
2

210
x
n
A
Solving,
n  1 .3 4

B
x
t
t
An element at point B:
z
x
Figure (a)
(CONT.)
112
6.36 (CONT.)
4 (W  F )
 m ax   d   t 
1 6T

D
3A
4 (1 1  1 0 )
3
3

3  ( 0 .0 2 5 )
2
 1 1 0 (1 0 )  1 1 7 .4 7 M P a 
6
210
n
or
n  1 .7 9
SOLUTION (6.37)
Table B.1: S y  250
and   7 . 86 Mg
MPa
m , n  2 .1
3
State of stress, at a point C at bottom surface on midspan:
x 
C
 
32 M
D
3
16T
D
3
We have
w  7 . 86 ( 9 . 81 )(  D
M
Thus

 wL
m ax
2
32( w L

x
D
2
D
8)
2 .7 8 1 0
D
6
7 .7 2 8 1 0
6
2
8D
2
3
3

6
2 .7 8 (1 0 )
D
3
D
3
2
)  3( 
2
) ( 6 )1 0
2 .0 4 1 0
D
2
3
3
 (S
)  (
2
2
y
n) :
2 5 0 1 0
2 .1
6
or
D
2

 1 4 1.7  1 0
1 2 .4 8 4 8
D
6
8
Solving, by trial and error:
D  34 . 34 mm
Use a 35-mm diameter shaft.
SOLUTION (6.38)
Applying Eq.(6.34), we have
z 
From Fig.6.15:
s l

2
s

2
l

kN
2 . 0 4 (1 0 )
 x  3
Equation (6.16),
(
3 2 ( 6 0 .6 D

 
3
2
8
3
16( 400 )
  
4 )  60 . 6 D
2
400  250
2
30  35
2
 3 . 25
R  9 9 .9 4 % .
113
)
2
m
SOLUTION (6.39)
The state of stress is  
 
 d
 
 d
4P
. Thus
2
 d
l 
4 ( 200 )
2
l 
4 (30 )
2
4
4
 d
 d

2

2
2 5 4 .6
d
3 8 .2
d
kPa
2
2
kPa
Figure 6.16, for R=99.7 %: z  2 . 75
Equation (6.34), in rearranged form:
z (
2
s

2

1
)2   s  
1
2 .7 5[ ( 3 5  1 0 )  ( 3 8 2.2 ) ] 2  3 5 0  1 0 
3
2
2
3
2 5 4 .6
d
d
2
from which
(3 5  1 0 )  ( 3 8 2.2 )
3
2
2
d
3
 (1 2 7 .2 7 3  1 0 
3
9 2 .5 8 2
d
2
)
2
6
or
d  1 .5 7 4 (1 0 ) d  0 .4 7 5 (1 0 )  0
Solving,
d  0 .0 3 4 2 m
and d  3 4 .2 m m
4
2
SOLUTION (6.40)
Apply Eqs.(6.30a) and (6.30b):
12
x 

1
12
12
xi 
912
12
y 
 76,
i 1
1
12

yi 
936
12
 78
i 1
12
x  [
1
11

1
1
( x i   x ) ] 2  [ 111 ( 2 2 0 4 )] 2  1 4 .1 5 5
2
i 1
12

 [ 11
1
y

1
1
( y i   y ) ] 2  [ 1 1 (1, 3 5 1.3 4 )] 2  1 1.0 8 4
2
1
i 1
SOLUTION (6.41)
(a) x
n( xi   )
2
n
nx
520
7
3640
7( -44.78 )
2
460
2
920
2(-104.78 )
2
430
5
2150
5(-134.78 )
2
545
5
2725
5( -19.78 )
2
570
10
5700
10(
5.22 )
2
575
18
10350
18( 10.22 )
2
595
8
4760
8( 30.22 )
2
600
3
1800
3( 35.22 )
2
620
6
3720
6( 55.22 )
2
660
-----
4
----68
2640
4( 95.22 )
------------ ---------------38405
196521.69
2
(CONT.)
114
6.41 (CONT.)
Thus, Eq.(6.30a):
 
 5 6 4 .7 8 M P a
38 ,405
68
Eq.(6.30b):
n

  [ 617
1
( x i  5 6 4 .7 8 ) ] 2          5 4 .1 5 9 M P a
2
i 1
( b ) Eq.(6.34),
z 
Figure 6.16:
s  l

5 2 5  5 6 4 .7 8
5 4 .1 5 9

m
 0 .7 3 5
R  76 %
SOLUTION (6.42)
Maximum load:  l  2 5 k N ,
 l  3 kN
Strength of part:  s  3 0 k N ,
 s  2 kN
Equation (6.33a):  m   s   l  3 0  2 5  5 k N
Equation (6.33b):



m
2
s

2
l

2 3
2
2
 3 .6 k N
Thus, failure impends at
m
z 
Figure 6.16:

m

5
3 .6
 1 .3 8 9
R  92 %
SOLUTION (6.43)
( a ) We have

nom
Pn o m

A

4 (35 )
 ( 0 .0 1 2 5 )
2

35
1 .2 2 7 2  1 0
4
 2 8 5 .2 M P a
So
n 
350
2 8 5 .2
 1 .2 3
( b ) Mean stress value equals  l   n o m  2 8 5 .2 M P a .
Estimated standard deviation equals
2 .5
 2 0 .3 7 2 M P a
l 
1 .2 2 7 2  1 0
4
The margin of safety, Eq. (6.34), is thus
z 
s  l

2
s
2
 l

3 5 0  2 8 5 .2
2
2 8  2 0 .3 7 2
2
 1 .8 7
(CONT.)
115
6.43 (CONT.)
From Fig. 6.16, reliability corresponding to z=1.87 is
R  97 %
Hence, failure percentage equals 100-97=3%.
Comment: In foregoing calculations, statistical variability of dimensions is
omitted.
SOLUTION (6.44)
Equation (6.34):
Figure 6.16:
z 
25  20
2
2 .5  3
2
 1 . 28
R  90 %
Thus failure percentage is 10 %.
End of Chapter 6
116
CHAPTER 7
FATIGUE FAILURE CRITERIA
SOLUTION (7.1)
Use Figure 7.5 for steel (1020).
(a)
At infinite life, S e '  2 3 0 M P a
The maximum stress in the beam is

m ax

M
m ax c
I
32 M

D
m ax
3
from which
D 
3
32 M

(1)
m ax
m ax
Letting  m a x  S e ' ,
D 
(b)
3
3
3 2 ( 4 1 0 )
6
 ( 2 3 0 1 0 )
 5 6 .2 m m
5
At 1 0 cycles, S  3 1 0 M P a . Equation (1) gives
Then
D 
3
3
3 2 ( 4 1 0 )
6
 ( 3 1 0 1 0 )
 5 0 .8 m m
SOLUTION (7.2)
Use Figure 7.5 for aluminum alloy (2024).
(a)
Endurance strength, S n '  1 3 5 M P a
The maximum stress in the beam:

m ax

32 M
D
or D 
m ax
3
3
32 M

(1)
m ax
m ax
With  m a x  S n '  1 3 5 M P a ,
D 
(b)
3
3
3 2 (1 .5  1 0 )
6
 (1 3 5  1 0 )
 4 8 .4 m m
7
At 1 0 cycles, S  1 6 5 M P a . Equation (1) results in
D 
3
3
3 2 (1 .5  1 0 )
6
 (1 6 5  1 0 )
 4 5 .2 m m
SOLUTION (7.3)
To determine the K t , we use Fig.C.1. Structural steel: S u  400
MPa
(Table B.1).
At section C:
D
d

38
30
 1.2 6 7
(a) 
r
d

4
30
 0 .1 3 3
( b ) Figure 7.9a:
max
 1 .7
3
15 ( 10 )
0 . 03 ( 0 . 01 )
 85 MPa
(CONT.)
117
7.3 (CONT.)
 K
t
 1. 7
r  4 mm : q  0 . 78
K
 1  0 .7 8 (1.7  1 )  1.5 5
f
(Eq.7.13b)
Similarly, at D:
D
d

r
d

38
34
2
34
 K
t
 1.1 1 8
(a) 
 0 .0 5 9
( b ) Figure 7.9a:
 1.8
3
15 ( 10 )
 1 .8
max
 79 . 41 MPa
0 . 034 ( 0 . 01 )
r  2 mm : q  0 . 72
K
 1  0 .7 2 (1.8  1 )  1.5 8
f
SOLUTION (7.4)
S u  400
Table B.1:
MPa ,
S e  C f C rC sC t
where
K
f
S
Se
 1.5 8 at D (from Solution of Prob.7.3)
S e  0 .4 5 S u  1 8 0
M Pa
 A S u  272( 400 )
b
f
MPa
'
1
K f
'
C
 250
y
 0 .9 9 5
 0 .7
C r  0 .8 7 (Table 7.3)
C s  1 (axial loading)
C t  1  0 .0 0 5 8 ( 4 7 5  4 5 0 )  0 .8 5 5
Thus
S e  ( 0 .7 )( 0 .8 7 )(1)( 0 .8 5 5 )
1
1 .5 8
(1 8 0 )  5 9 .3 M P a
SOLUTION (7.5)
S e  C f C r C s C t (1 K f ) S e
'
We have
r
d

t
 2 .5 (Fig.C.3)
K
2
26
(1)
 0 .0 7 7 ,
 1.1 5 4
D
d
Table B.4: S u  655 MPa ,
S e  0 .4 5 S u  2 9 4 .8
'
H
B
 197
M Pa
Table 7.3: C r  0 .8 9
Fig.7.9a: q  0 . 8 ,
K
f
 1  0 .8 ( 2 .5  1)  2 .2
Table 7.2: C f  A S u  4 .5 1( 6 5 5 )
b
Use C s  1 (axial loading)
 0 .2 6 5
 0 .8 0 9
Ct  1
Equation (1) is therefore
S e  ( 0 .8 0 9 )( 0 .8 9 )(1)(1)( 21.2 )( 2 9 4 .8 )  9 6 .4 8
118
M Pa
SOLUTION (7.6)
S u  626
Table B.4:
MPa
H
B
 179
From Eq.(7.1):
S e  0 . 5 S u  313
'
MPa
(Note: by Eq.(2.22): S u  3500 (179 )  626 . 5 MPa. )
C s  0 .8 5
From Eq.(7.9):
By Eq.(7.7):
Ct  1
C r  0 .8 9
Using Table 7.3:
C
 A S u  4 .5 1( 6 2 6
b
f
 0 .2 6 5
)  0 .8 1 9
For Fillet:
r
d

4
25
 0 .1 6

D
d
35
25
 1.4
K t  1.4 5
Hence, from Fig.C.9:
From Fig.7.9a, q=0.82
Equation (7.13b):
K f  1  0 .8 2 (1.4 5  1 )  1.3 7
Thus
S e  C f C r C s C t ( K1 ) S e  ( 0 . 819 )( 0 . 89 )( 0 . 85 )( 1 )( 1 .137 ) 313  141 . 6 MPa
'
f
SOLUTION (7.7)
Table B.3: S u  4 7 0 M P a . We apply
S e  C f C r C sC t (
1
K f
'
)Se
where
K
f
 2 .5 ,
S e  0 .5 S u  2 3 5
'
C s  0 .7 ,
C r  0 .8 4 (Table 7.3),
C
f
 AS
b
u
M Pa
 57 . 7 ( 470 )
 0 . 718
Ct  1
 0 . 696
Thus
S e  ( 0 .6 9 6 )( 0 .8 4 )( 0 .7 )(1)( 21.5 )( 2 3 5 )  3 8 .5 M P a
SOLUTION (7.8)
S e  C f C r C s C t (1 K f ) S e '
(1)
We have
D
d

30
25
 1 .2
1
d

2
25
 0 .0 8
Hence, from Fig. C.12, K t  1 .9 5
Table B.4: S u  6 5 8 M P a ,
H
B
 192
Equation (7.1): S e '  0 .5 S u  3 2 9 M P a
Figure 7.9a: r  2 m m ;
Therefore,
C
 A S u  4 .5 1( 6 5 8
b
f
q  0 .8 2
 0 .2 6 5
)  0 .8 0 8
(CONT.)
119
7.8 (CONT.)
Table 7.3: C r  0 .8 7
K
 1  q ( K t  1)  1  0 .8 2 (1 .9 5  1)  1 .7 8
f
Equation (7.9): C s  0 .8 5
Ct  1
(T  4 5 0 C )
o
Equation (1) results in then
S e  ( 0 .8 0 8 )( 0 .8 7 )( 0 .8 5 )(1)( 1 .71 8 )(3 2 9 )  1 1 0 .4 M P a
SOLUTION (7.9)
Refer to solution of Prob. 7.8, Now, Fig. C.11 give
 1 .2 ,
D
d
 0 .0 8,
r
d
Table B.3: S u  7 7 0 M P a ;
r  2 mm;
Figure 7.9b:
K t  1 .5
H
B
 229
q  0 .9 8
K
f
 1  q ( K t  1)  1  0 .9 8 (1 .5  1)  1 .4 9
C
f
 A S u  5 7 .7 ( 7 7 0
Also
b
C r  0 .8 7 ,
 0 .7 1 8
C t  0 .5 6 5,
)  0 .4 8 8
C s  0 .8 5
S e '  0 .5 S u  3 8 5 M P a
Thus
S e  C f C r C s C t (1 K f ) S e '
 ( 0 .4 8 8 )( 0 .8 7 )( 0 .8 5 )( 0 .5 6 5 )( 1 .41 9 )(3 8 5 )
 5 2 .7 M P a
SOLUTION (7.10)
Refer to definitions given by Eqs. (7.14).
( a )  m  12 ( m a x   m in )  12 (8 4  8 4 )  0
a 
1
2
(
R 
Thus,

m ax

m in

m ax
m in

)
84
84
1
2
(8 4  8 4 )  8 4 M P a
 1
and

A 

a
m

12
0
 
(b) m 
1
2
(8 4  1 4 )  3 5 M P a
a 
1
2
(8 4  1 4 )  4 9 M P a
R 
14
84
 
(c) m a 
R 
0
84
 0
A 
42
42
1
1
6
1
2
,
A 
7
5
(8 4  0 )  4 2 M P a
120
SOLUTION (7.11)
M
max
 M
 M , 
min
m
 0, 
a

 651 , 898 M
32 M
 ( 0 . 025 )
3
Equation (7.7):
C
f
 AS
b
u
 4 . 51 ( 700 )
 0 . 265
 0 . 794
Also
Ct  1
C r  0 .8 7
Table 7.3:
Equation (7.9):
C s  0 .8 5
Equation (7.1):
S e  0 .5 ( 7 0 0 )  3 5 0
'
M Pa
From Fig. C.9, with D d  1 . 5 , r d  0 . 05 :
K t  2 .1
q  0 .7 7
By Fig.7.9a:
and K
Hence
 1  0 . 77 ( 2 . 1  1 )  1 . 85
f
S e  C f C r C sC t (
1
K f
'
)Se
 ( 0 .7 9 4 )( 0 .8 7 )( 0 .8 5 )(1)( 1 .81 5 )(3 5 0 )  1 1 1 .1 M P a
Thus, Eq.(7.24):
n 
Se

6
1 1 1 .1  1 0
6 5 1 ,8 9 8 .6 M
1 .5 
;
a
or
M  1 1 3 .6 N  m
SOLUTION (7.12)
Table B.3: S u  4 7 0 M P a
H
B
 131
 S e  0 .4 5 S u  2 1 1 .5
'
M Pa
Tensile area through the hole:
2 ( R  r ) t  2 (10  4 )( 2 . 5 )  30 mm
m  a 
and
F
2A

F
2( 30 )

2
F
60
(1)
We have
C r  0 .7 0 (Table 7.3)
C
 A S u  4 .5 1( 4 7 0 )
b
f
Ct  1
 0 .2 6 5
 0 .8 8
C s  1 (axial loading)
From Fig.C.5:
d
D
 0 .4 ,
q  0 .8
By Fig.7.9a:
Hence
K
f
K t  2 .8
 1  0 . 8 ( 2 . 8  1 )  2 . 44
Therefore,
S e  C f C r C s C t (1 K f ) S e
'
 ( 0 .8 8 )( 0 .7 )(1)(1)(1 2 .4 4 )( 2 1 1 .5 )  5 3 .4
M Pa
(CONT.)
121
7.12 (CONT.)
By Eq.(7.20):

m
4 7 0 1 .4

470
(1 )(
 3 4 .2 5
(2)
M Pa
) 1
5 3 .4
From Eqs.(1) & (2):
34 . 25 
F
60
or
F  2 .0 6
kN
SOLUTION (7.13)
Refer to solution of Prob. 7.12.
We have,  m   a  F 6 0 and
C r  0 .8 7
(Table 7.3)
C t  1  0 .0 0 5 8 ( T  4 5 0 )
(Eq. 7.11)
 1  0 .0 0 5 8 ( 5 4 5  4 5 0 )  0 .4 5
Endurance limit becomes
S e  5 3 .4 ( 00.8.77 )( 0 .4 5 )  2 9 .8 7 M P a
The SAE criterion from Eq. (7.20) with S u  S

m

S


a
m
n
f
S
f
Se
1

4 1 5 1 .2
(1)
415
2 9 .8 7
1
f
is
 2 3 .2 2 M P a
Thus
F 6 0  2 3 .2 2 ,
F  1 .3 9 k N
SOLUTION (7.14)
Table B.3: S y  3 9 0 M P a
Refer to solution of Prob. 7.12.
We have,  m   a  F 6 0 and
C r  0 .8 9
(Table 7.3)
C t  1  0 .0 0 5 8 ( T  4 5 0 )
(Eq. 7.11)
 1  0 .0 0 5 8 ( 5 4 0  4 5 0 )  0 .4 8
Endurance limit is then
S e  5 3 .4 ( 00.8.79 )( 0 .4 8 )  3 2 .5 9 M P a
The Soderberg criterion from Eq. (7.20) with S u  S f :

m

Sy n


a
Sy
m
Se
1

3 9 0 2 .2
(1)
390
3 2 .5 9
1
 1 3 .6 7 M P a
Thus
F
60
 1 3 .6 7 ,
F  8 2 0 .2 N
122
SOLUTION (7.15)
A  10 ( 25  5 )  200
Pm 
1
2
mm
( 5  25 )  15 kN ,
Pa  10 kN
( a ) Stress concentration factor is neglected for ductile materials under static loading. Thus


max
Pmax
3
25 ( 10 )

A
S
n 

6
200 ( 10

y
m ax
)
580
125
 125
MPa
 4 .6 4
( b ) We now have
 0 .2 ,
d
D
S u  690
K t  2 .4 5 (Fig.C.5)
MPa ,
S
y
 580
MPa ,
H
 197
B
(Table B.3)
q  0 .8 3 (Fig 7.9a)
K
 1  0 .8 3 ( 2 .4 5  1 )  2 .2
f
Ct  1
C r  0 .8 4 (Table 7.3)
C s  1 (axial loading)
 A S u  4 .5 1( 6 9 0 )
b
C
f
 0 .2 6 5
 0 .7 9 8
S e  0 .4 5 S u  3 1 0 .5 M P a
'
Hence
S e  ( 0 .7 9 8 )( 0 .8 4 )(1)(1)( 21.2 )(3 1 0 .5 )  9 4 .6 1 M P a
We have


m
Pm
A
3
15 ( 10 )

200 ( 10
6
)

 75 MPa ,
a
 50 MPa
Equation (7.22) gives
n 
75 
 1 . 57
690
690
( 50 )
94 . 61
SOLUTION (7.16)
Refer to Solution of Prob.7.15
(a)
n 
S


y
m ax
580
125
 4 .6 4
( b ) We now have
Pm 
1
2
[ 25  (  5 )]  10 kN ,
Pa  15 kN
Hence

a
 75 MPa ,

m
 50 MPa
Thus
n 
50 
690
690
 1 . 16
( 75 )
94 . 61
123
SOLUTION (7.17)
D
A 
4
2
 ( 0 .0 5 3 1 2 5 )

2
2 .2 1 7  1 0
4
( a ) Su  670 M Pa
H
3
m
 1 9 7 (Table B.4)
B
P  S u A  6 7 0 ( 2 .2 1 7  1 0
and
2
3
)  1 .4 8 5 M N
( b ) S e  C f C r C s C t (1 K f ) S e
'
where
K
f
1
C r  0 .8 7 (Table 7.3)
C s  1 (axial load)
C
f
 AS
 4 . 51 ( 670 )
b
u
 0 . 265
 0 . 804
C t  1  0 . 0058 ( 480  450 )  0 . 826
S e '  0 .4 5 S u  0 .4 5 ( 6 7 0 )  3 0 1 .5
M Pa
S e  ( 0 .8 0 4 )( 0 .8 7 )(1)( 0 .8 2 6 )(1 / 1)(3 0 1 .5 )  1 7 4 .2
and
M Pa
Thus
P  A S e  ( 2 .2 1 7  1 0
3
)(1 7 4 .2 )  3 8 6 .2 k N
SOLUTION (7.18)
Refer to Solution of Prob.7.17. We now have A   d
2
4   ( 0 .0 5 )
2
4  1 .9 6 3  1 0
3
m
2
( a ) For a static fracture of a ductile material, the groove has little effect. Hence,
P  S u A  ( 6 7 0  1 0 )  (1 .9 6 3  1 0
6
(b)
r
d
 0 .0 2 5,
D
d
 1 .0 6 2 5
3
)  1 .3 1 5 M N
 K t  2 .6 (Fig.C.10)
From Fig.7.9a, with S u  6 7 0 M P a and r  1 .2 5 m m
and
K
f
 q  0 .7 5
 1  q ( K t  1 )  1  0 . 75 ( 2 . 6  1 )  2 . 2
We now have
S e  1 7 4 .2 2 .2  7 9 .1 8
M Pa
Thus
P  A S e  (1 .9 6 3  1 0
3
)( 7 9 .1 8  1 0 )  1 5 5 .4 3 k N
6
SOLUTION (7.19)
From Table B.4: S u  519
MPa ,
S
y
 353
MPa ,
H
B
 149
By Eq.(6.20), S ys  0 . 577 S y  203 . 7 MPa
( a ) Thus, S y s  1 6 T  d :
3
T 
3
 ( 0 . 025 ) ( 203 . 7  10
16
6
)
 624 . 9 N  m
(CONT.)
124
7.19 (CONT.)
( b ) S es  C f C r C s C t (1 K f ) S es
'
where
Ct  1
C r  0 .8 4 (Table 7.3)
C s  0 .8 5 (Eq.7.9)
C
f
 AS
 1 . 58 ( 519 )
b
u
S e  0 .2 9 S u  1 5 0 .5
'
From Fig.C.8, with
From Fig.7.9b: q  0 .9 ,
K
f
 0 . 929
D
d
(Eq.7.7)
(Eq.7.4)
M Pa
 0 . 05 and
r
d
 0 . 085
 2
 K t  1 . 72
 1  q ( K t  1 )  1.6 5
Hence
S e s  ( 0 .9 2 9 )( 0 .8 4 )( 0 .8 5 )(1)( 1 .61 5 )(1 5 0 .5 )  6 0 .5
M Pa
Refer to Eq.(7.24): S es  16 T  d .
3
Thus
T 
3
6
 ( 0 .0 2 5 ) ( 6 0 .5  1 0 )
16
 1 8 5 .6
N m
SOLUTION (7.20)
Refer to Solution of Prob.7.19.
( a ) A  d
2
4   ( 25 )
4  490 . 874
2
mm
2
, and S
y
 353
MPa . Thus
P  S y A  490 . 874 ( 353 )  173 . 3 kN
( b ) We now have with
r
d
 0 .0 5 ,
D
d
 2
From Fig. C.7: K t  2 .5 2
Figure 7.9a: q  0 .7 and
K
f
 1  q ( K t  1 )  1  0 .7 ( 2 .5 2  1 )  2 .0 6
By Eq.(7.3): S e  0 .4 5 S u  2 3 3 .6 M P a
Hence
S e  ( 0 .9 2 9 )( 0 .8 4 )( 0 .8 5 )(1)( 2 .01 6 )( 2 3 3 .6 )  7 5 .2 2
M Pa
Thus
P  S e A  ( 7 5 .2 2 )( 4 9 0 .8 7 4 )  3 6 .9 2
kN
SOLUTION (7.21)
From Table B.3: S u  830
MPa ,
S
y
 460
MPa ,
H
B
 248
By Eq.(6.20),
S
ys
 0 . 577 S
y
 265 . 4 MPa
Refer to Solution of Prob.7.19.
(CONT.)
125
7.21 (CONT.)
3
 d S ys
(a) T 
3
6
 ( 0 .0 2 5 ) ( 2 6 5 .4  1 0 )

16
16
( b ) C f  A S u  4 .5 1( 8 3 0 )
b
 8 1 4 .2 N  m
 0 .2 6 5
 0 .7 6
S e s  0 .2 9 S u  2 4 0 .7 M P a
'
S es  C f C r C s C t (1 K
and
'
f
) S es
 ( 0 .7 6 )( 0 .8 4 )( 0 .8 5 )(1)( 1 .61 5 )( 2 4 0 .7 )  7 9 .1 6 M P a
Therefore
T 
3
 d S es
3
6
 ( 0 .0 2 5 ) ( 7 9 .1 6  1 0 )

16
16
 2 4 2 .9 k N  m
SOLUTION (7.22)
Su  658 M Pa
H
 1 9 2 (Table B.4)
B
S e  C f C r C s C t (1 K f ) S e
'
where
C r  0 .8 1 (Table 7.3)
C
 AS
f
b
u
C s  0 .8 5 (Eq.7.9)
 1 . 58 ( 658 )
 0 . 085
 0 . 91
(Table 7.2)
S e  0 .5 S u  3 2 9 M P a
'
 0 . 125 ,
d
D
and
K t  2 . 18
From Fig.7.9a: q  0 . 79 ,
K
f
Fig.C.13)
 1  0 . 79 ( 2 . 18  1 )  1 . 93
Thus
S e  ( 0 .9 1)( 0 .8 1)( 0 .8 5 )(1)( 1 .91 3 )(3 2 9 )  1 0 6 .8 M P a
We have
M

m
m


1
2
(1 5 8 .2  5 6 .5 )  1 0 7 .3 5 N  m ,
M
( D
3
m
32 ) ( dD
2
6)

M
1 0 7 .3 5
 ( 0 .0 2 5 )
3
3 2  [ ( 0 .0 0 3 1 2 5 ) ( 0 .0 2 5 )
2
6]
a
 5 0 .8 5 N  m
 8 8 .8 3 M P a
(Fig.C.13)
 a  8 8 .8 3( 15007.8.355 )  4 2 .0 8 M P a
Equation (7.22):
n 
8 8 .8 3 
658
658
 1 .8 9
( 4 2 .0 8 )
1 0 6 .8
SOLUTION (7.23)
Refer to Solution of Prob.7.22. We now have
Su  680 M Pa,
H
B
 2 0 1 (Table B.3)
q  0 . 8 (Fig.7.9a)
Also
d
D
 0 .1 2 5
K
f
K t  2 .6 5 (Fig.C.13)
 1  0 . 8 ( 2 . 65  1 )  2 . 32
S e  0 .4 5 S u  0 .4 5 ( 6 8 0 )  3 0 6 M P a
'
(CONT.)
126
7.23 (CONT.)
S e  ( 0 .9 1)( 0 .8 1)( 0 .8 5 )(1)( 2 .31 2 )(3 0 6 )  8 2 .6 4 M P a
and
We write
Pm 
1
2
(60  20)  40 kN ,
Pa  2 0 k N
A  D
2
4  Dd
(Fig.C.13)
and


m
 9 6 .9 1 M P a ,
40
 ( 0 .0 2 5 )
2
4  [( 0 .0 2 5 )( 0 .0 0 3 1 2 5 )]
Equation (7.22) is therefore
n 
 1 .3 7
680
680
9 6 .9 1 
( 4 8 .4 6 )
8 2 .6 4
SOLUTION (7.24)
From Table B.2: S u  5 2 0 M P a ,
H
B
 149
C r  0 .7
Table 7.3:
Equation (7.7) and Table 7.2:
C
f
 4 .5 1(5 2 0
 0 .2 6 5
)  0 .8 6
S e '  0 .5 S u  2 6 0 M P a
Thus
S e  C f C r C s C t (1 K f ) S e '
 ( 0 .8 6 )( 0 .7 )(1)(1)(1 1)( 2 6 0 )  1 5 6 .5 M P a
We have


32 M
a


Pm
m
d
 ( 0 .0 2 5 )
3
4 (1 5  1 0 )

A
3 2 (1 5 0 )

3
 ( 0 .0 2 5 )
2
3
 9 7 .7 8 M P a
 3 0 .5 6 M P a
Goodman criterion

a
Se


m
Su

1
n
9 7 .7 8
1 5 6 .5
;

3 0 .5 6
520

1
n
Solving,
n  1 .4 6
SOLUTION (7.25)
Refer to solution of Prob. 7.22.
We have
S y  380 M Pa
(Table B.4)
C r  0 .8 9
(Table 7.3)
C t  1  0 .0 0 5 8 ( T  4 5 0 )
(Eq. 7.11)
 1  0 .0 0 5 8 ( 4 5 5  4 5 0 )  0 .9 7
Endurance limit is therefore
S e  1 0 6 .8 ( 00 .8.8 91 )( 0 .9 7 )  1 1 3 .8 3 M P a
The Soderberg criterion (by Table 7.4)

a
Se


m
Sy

1
n
Inserting the data, we have
4 2 .0 8
1 1 3 .8 3

8 8 .8 3
380

1
n
,
n  1 .6 6
127

a
 9 6 .9 1
20
40
 4 8 .4 6 M P a
SOLUTION (7.26)
1 
pr
t

,
a
pa

m
pr

2

pm
2
3

,
2t
 0.
3
pm  1260 kPa,
pa  840 kPa
.
Replace S u with S y in Eq.(7.30):
S
 [1 
y
n
 1e 
or
1

1
2
] 2  1 e  0 .8 6 6  1 e
1
4
Sy
(1)
0 .8 6 6 n
Similarly, Eq.(7.25):

 
1e

S

1m
y
Se

(1260 
1
t

1a
280
210
r
t
S
( pm 
 840 ) 
y
pa )
Se
2380
t
(2)
By Eq.(1) and (2):
280
( 0 .8 6 6 ) 2 .5

2380
t
,

2
t  0 .0 1 8 4 m = 1 8 .4 m m
SOLUTION (7.27)
1 
pr
t
,

pr
2t

,
3
 0.
p m  1 . 7 MPa ,
p a  1 . 1 MPa
Refer to Solution of Prob.7.26:
 1e 
Su
(1)
0 .8 6 6 n
Use Eq.(7.25):
Su
 1e   1m 
Se
 1a 
r
t
( pm 
Su
Se
p a )  3 0[1 .7 
From Eqs.(1) and (2):
 128,
350
0 .8 6 6 n
n  3.1 6
SOLUTION (7.28)
b  0 .0 1 m ,
M
max
M
a

1
4
 M
M  PL 4.
( 20 )( 0 . 1)  0 . 5 N  m ,

 0 . 25 N  m ,
m
6M
m 
We have,
L  0 .1 m ,
m
2
bh

6 ( 0 .2 5 )
0 .0 1 h
2

m

150
h
2
Equation (7.20) gives

m

980 4
(1 )
98
 7 1 .0 1 M P a
1
40
Thus
h 
150
71 . 01  10
6
 1 . 45 mm
128
M
a
min
 0
35
15
(1 .1)]  1 2 8 M P a
(2)
SOLUTION (7.29)
From Solution of Prob.7.28, we have
a  m 
150
n
.
2
(1)
Equation (7.20) by replacing S u with S y :

m

S

a

m
n
y
S
y
1
Se
Substituting the given data gives

m
620 4

62
(1 )(
 6 0 .7 8 M P a
(2)
) 1
40
By Eqs.(1) and (2),
6 0 .7 8 ( 1 0 ) 
6
150
h
2
or
h  1 . 57 mm
SOLUTION (7.30)
At fixed end M
M  M

m
 P L  2 .2 5 ( 0 .0 3 7 5 )  0 .0 8 4 4 N  m Hence
m ax

 M
m

a
bh
2

Equation (7.20):
m ax
M
m in
2
6 ( 0 .0 4 2 2 )

6M
M

a
( 0 .0 0 3 1 2 5 ) h

m
2
1 0 5 0 1 .2
(1 )
1050
 0 .0 4 2 2 N  m

81
h
2
 2 8 3 .7 8 M P a
1
504
Thus
 2 8 3 .7 8  1 0
81
h
2
6
Pa,
h  0 .5 3 4  1 0
3
m = 0 .5 3 4 m m
SOLUTION (7.31)
Refer to solution of Prob. 7.30.
Equation (7.11):
C t  1  0 .0 0 5 8 ( T  4 5 0 )  1  0 .0 0 5 8 ( 4 7 0  4 5 0 )  0 .8 8
and
S e  5 0 4 ( 0 .8 8 )  4 4 3 .5 2 M P a
Equation (7.20), replacing S u by S f , becomes

m

S

a

m
n
f
S
f
Se
1
Inserting numerical values:

m

6 9 0 1 .5
(1)( 4 4639.50 2 )  1
 1 7 9 .9 9 M P a
Therefore,
81
h
2
 1 7 9 .9 9  1 0 ,
6
h  0 .6 7 1  1 0
129
3
m = 0 .6 7 1 m m
SOLUTION (7.32)
P0
Pm  Pa 

Eq.(7.20):
,
2
m
M
850 4

85
(1 )
 0 . 2 Pm  0 . 1 P0 ,
m
Pa Pm  1

a
 3 6 .2 8 M P a
1
1 7 .5
6M
m 
Also
bh
6 ( 0 .1 ) P0

m
2
0 . 0 0 5  1 0 0 (1 0
 1.2 (1 0 ) P0
6
6
)
Thus
1 . 2 P0  36 . 28 ,
P0  30 . 23 N
SOLUTION (7.33)
M
M
 0 .2 P0 ,
m ax
m
P0

M
m in
 0 .2 (  0 .5 P0 )   0 .1 P0
( 0 .2  0 .1 )  0 .0 5 P0 ,
2
Equation (7.20):  m 
850 2
1 .5 ( 3 )
85
M
 3 5 .6 2 9
a

P0
2
( 0 .2  0 .1 )  0 .1 5 P0
M Pa
1
35
6M
m 
Also
bh
m
2
6 ( 0 . 0 5 P0 )

0 . 0 0 5  1 0 0 (1 0
 0 .6 (1 0 ) P0
6
6
)
Thus
0 . 6 P0  35 . 629 ,
P0  59 . 38 N
SOLUTION (7.34)
I 
bh
12
3
24 ( 4 )

3
 128
12
mm
4
. Table B.3: S u  690
MPa
For a wide cantilever beam (see Secs 4.4 and 4.10, and Case 1 of Table A.8):
3
  (1   )
2
This gives Pmin 
3 EI
0 . 91 L
 0 .9 1
PL
3EI
3


min
3
PL
3EI
9
3 ( 200  10 )( 128  10
0 . 91 ( 0 . 3 )
3
 12
)
( 0 . 01 )
and hence Pmax  62 . 52 N
 31 . 26 N
Thus
Pm  46 . 89 N ,
Pa  15 . 63 N and
46 . 89 ( 0 . 3 )( 2  10
3

m


a
. 63
 219 . 8 ( 15
)  73 . 27 MPa
46 . 89
0 . 128 ( 10
9
)
)
 219 . 8 MPa
Eq. (7.22):
Su
n 

m

Su
Se


a
219 . 8 
690
690
 1 . 63
( 73 . 27 )
250
130

m
1
SOLUTION (7.35)
Table B.4: S
 500
y
MPa
Refer to Solution of Prob.7.34. Replacing S u by S y in Eq.(7.22):
S
n 


m

y
S
y

Se
219 . 8 
a
500
500
 1 . 36
( 73 . 27 )
250
SOLUTION (7.36)
Refer to solution of Prob. 7.34. We now have
(Table 7.3)
C r  0 .7 5
C t  1  0 .0 0 5 8 ( T  4 5 0 )
(Eq. 7.11)
 1  0 .0 0 5 8 ( 4 9 0  4 5 0 )  0 .7 7
Endurance limit becomes
S e  2 5 0 ( 0 .7 5 )( 0 .7 7 )  1 4 4 .4 M P a
The Gerber criterion (Table 7.4):

 (
a
Se

) 1
2
m
Su
Substitution of the numerical values gives
 ( 2 1699.80 n )  1
2
7 3 .2 7 n
1 4 4 .4
or
0 .5 0 7 4 n  0 .3 1 8 6 n  1  0
2
Solving this quadratic equation:
 0 .5 0 7 4 
n 
2
( 0 .5 0 7 4 )  4 ( 0 .3 1 8 6 )(  1 )
2 ( 0 .3 1 8 6 )
or
n  1 .1 5
SOLUTION (7.37)
I 

4
( 0 . 03 )
4
 0 . 636  10

m


a
 23 . 6 MPa
M
m
c
I
6
m
2000 ( 0 . 5 )( 0 . 03 )

0 . 636  10
6
4
Pm  4 kN ,
,
Pa  2 kN
 47 . 2 MPa
Eq. (7.16), with replacing S u by S y :
S
y
n
 
m

S
y
Se
a;
280
2 .5
 4 7 .2 
280
150 K
f
( 2 3 .6 )
(CONT.)
131
7.37 (CONT.)
Solving K
f
 1.4 7 .
From Fig.7.9a: q=0.8
and
K
f
 1  q ( K t  1 );
1.4 7  1  0 .8 ( K t  1 ),
K t  1.5 9
Then from Fig. C.9:
K
t
r
d

 1.5 9
9
60


 0 .1 5 
D
 3
d
and
D  3 d  3 ( 60 )  180
mm
SOLUTION (7.38)
Table B.4: S u  634
I 
bh
12
3
MPa ,
0 . 02 ( 0 . 06 )

3
 1 . 8 kN  m
D
d

120
60
r
d

4
60
 192
B
 0 . 36  10
12
m
M
H
M
6
m
4
 1 . 2 kN  m
a
Figure C.2:
 2


 0 .0 6 7 
K t  2 .1
Figure 7.9a: q  0 . 82 and K
f
 1  0 .8 2 ( 2 .1  1)  1 .9 0 2 .
Se 
400
1 .9 0 2
 2 1 0 .3 M P a
We have

m


a
 150 ( 11 .. 28 )  100
M
m
c
I

1800 ( 0 . 03 )
0 . 36  10
6
 150
MPa
MPa
Equation (7.22):
Su
n 

m

Su
Se


a
150 
634
634
 1 .4
( 100 )
210 . 3
SOLUTION (7.39)
Refer to solution of Prob. 7.38.
We now have
M
m

1
2
(2200  200)  1200 N  m
M
a

1
2
(2200  200)  1000 N  m
Thus,
 m  1 5 0 ( 11 28 00 00 )  1 0 0 M P a ,
 a  1 0 0 ( 11 02 00 00 )  8 3 .3 M P a
(CONT.)
132
7.39 (CONT.)
The Gerber criterion (by Table 7.4).

(
a
Se

m
Su
) 
2
1
n
Inserting the numerical values, results in
8 3 .3 n
2 1 0 .3
 ( 160304n )  1,
0 .3 9 6 1 n  0 .0 2 4 9 n  1  0
2
2
Solving this quadratic equation:
 0 .0 2 4 9 
n 
2
( 0 .0 2 4 9 )  4 ( 0 .3 9 6 1 )(  1 )
2 ( 0 .3 9 6 1 )
or
n  1 .5 6
SOLUTION (7.40)
Refer to solution of Prob. 7.38.
We now have
S
 434 M Pa
f
(from Table B.4)
C t  1  0 .0 0 5 8 ( 4 7 5  4 5 0 )  0 .8 6
(by Eq. 7.11))
 1  0 .0 0 5 8 ( 4 9 0  4 5 0 )  0 .7 7
C r  0 .8 7
(Table 7.3)
and
S e  2 1 0 .3( 0 .8 7 )( 0 .8 6 )  1 5 7 .3 M P a
Also
M
m
 1 .5 k N  m
M
a
 0 .5 k N  m
and
 m  1 5 0 ( 11 58 00 00 )  1 2 5 M P a ,
 a  1 0 0 ( 1520000 )  4 1 .6 7 M P a
The SAE criterion from Table 7.4:

a
Se


m
S
f

1
n
Introducing the data, we obtain
4 1 .6 7
1 5 7 .3

125
434

1
n
Solving,
n  1 .8 1
SOLUTION (7.41)
See Tables 7.5, 6.2, and 6.1. We have
A  5 .6  1 0
12
n  3 .2 5
K c  77 M Pa
m
S y  690 M Pa
  1 .1 4 for a w  0 .4 (Case B, Table 6.1)
(CONT.)
133
7.41 (CONT.)
Note that values of a and t satisfy Table 6.2. The stresses are:

m ax

Pm a x

2wt
3
9 5 0 (1 0 )
2 ( 0 .0 8 )( 0 .0 3 4 )

 1 7 4 .6
m in
 7 9 .4 M P a
The range stress is then    1 7 4 .6  7 9 .4  9 5 .2 M P a
The final crack length at fracture, by Eq. (7.39):
af 
1

Kc
( 
) 
2
m ax
( 1 .1 4 7 17 7 4 .6 )  0 .0 4 7 6 m  4 7 .6 m m
2
1

Equation (7.41) results in then
N 
0 .0 4 7 6
5 .6 (1 0
12
 0 .6 2 5
 0 .0 3 2
 0 .6 2 5
) (  0 .6 2 5 ) [ (1 .7 7 ) (1 .1 4 ) ( 9 5 .2 ) ]
3 .2 5
 2 0 , 4 5 4 cycles
For a period of 20 second, approximate fatigue life L is
L 
20 ,454 ( 20 )
60 (60 )
 1 1 3 .6 h
End of Chapter 7
134
CHAPTER 8
SURFACE FAILURE
SOLUTION (8.1)
From Table 8.2, we have
K  10
2
(steel on steel)
5
K  2  10
K  10
(lead on steel)
3
(brass on steel)
7
K  10
(polyurethylene on steel)
The Brinell hardness numbers are in MPa:
Steel
Su  58  10
H
s
3
 58  10
(Table B.1)
psi
3
500  116 Bhn
(Eq. 2.22)
 1 1 6 ( 9 .8 1)  1 1 3 8 M P a
Lead
H l  3(9 .8 1)  2 9 .4 M P a
Brass
H b  8 (9 .8 1)  7 8 .5 M P a
Polyurethylene
H
p
 7 (9 .8 1)  6 8 .7 M P a
Thus referring to Eq. (a) of Sec. 8.5:
Lead
H K  2 9 .4 (1 0 ) 2 (1 0
Brass
H K  7 8 .5 (1 0 ) 1 0
6
6
3
5
)  1470 G Pa
 7 8 .5 G P a
Polyurenthylene H K  6 8 .7 (1 0 ) 1 0
6
7
 687, 000 G Pa
2
Steel
H K  1 1 3 8 (1 0 ) 1 0  1 1 3 .8 G P a
Comment: Polyurethylene gives much longer life than any other material.
6
SOLUTION (8.2)
Through the use of Eq. (8.2), wear volume is
V  K
WL
H
K  10
4
where
W  40 N
(by Table 8.3),
H b  9 .8 1( 6 0 )  5 8 9 M P a ,
L  80  (2
mm
c y c le
) (1 5 0 0
c y c le s
s ix m o n th s
H
s
)  240, 000 m m
Therefore
Vb 
(1 1 0
Vs 
(1 1 0
4
)( 4 0 )( 2 4 0 )
6
5 8 9 (1 0 )
4
)( 4 0 )( 2 4 0 )
6
1 0 3 0 (1 0 )
 9 .8 1(1 0 5 )  1 0 3 0 M P a
 1 .6 3 m m
3
 0 .9 3 m m
3
135
SOLUTION (8.3)
From Eq. (8.2), we have
V  K
WL
H
Here
K  3  10
H
s
H
f
5
W  45 N
(by Table 8.3),
 9 .8 1(1 6 0 )  1 5 6 9 .6 M P a
 9 .8 1( 4 5 0 )  4 4 1 4 .5 M P a
L  (3 7 .5  2
mm
c yc le
)( 6 0 0 0
c yc le s
m o n th
)(1 2 m o n th s )  5 , 4 0 0 , 0 0 0 m m
Hence,
V s  (3  1 0
V f  ( 4 .6 )
5
)
( 4 5 )( 5 , 4 0 0 )
6
1 5 6 9 .6 (1 0 )
1 5 6 9 .6
4 4 1 4 .5
 4 .6 m m
 1 .6 4 m m
3
3
SOLUTION (8.4)
Follow procedure of Example 8.1. We have
Copper disk: 110 Vickers hardness, V c  0 .9 8 m m
Aluminum pin: 95 Brinell hardness, V a  4 .1 m m
3
3
Contact force: W  2 5 N at a radius R  2 4 m m
Test duration: t  1 8 0 m in at a speed n  1 2 0 r p m
Total length of sliding is then
L  2 R n t
 2  ( 2 4 )(1 2 0 )(1 8 0 )  3 .2 6  1 0
6
mm
The hardness of pin and disk are
H a  9 .8 1(9 5 )  9 3 2 M P a
H c  9 .8 1(1 1 0 )  1 0 7 9 M P a
From Eq. (8.2); K  V H W L . Thus
4 .1 ( 9 3 2 )
Ka 
2 5 ( 3 .2 6  1 0 )
Kc 
2 5 ( 3 .2 6  1 0 )
6
0 .9 8 (1 0 7 9 )
6
 4 .9 6  1 0
 1 .3  1 0
5
5
SOLUTION (8.5)
Follow procedure of Example 8.1. Given data:
Steel disk: 248 Brinell hardness, V s  0 .9 8 m m
Copper pin: 85 Vickers hardness, V c  4 .1 m m
3
3
Contact force: P  3 5 N at a radius R  2 4 m m
Test duration: t  1 8 0 m in at a speed n  1 2 0 r p m
Total length of sliding equals
L  2 R n t
 2  ( 2 4 )(1 2 0 )(1 8 0 )  3 .2 6  1 0
6
mm
(CONT.)
136
8.5 (CONT.)
Hardness of pin and disks are
H c  9 .8 1(8 5 )  8 3 4 M P a ,
H s  9 .8 1( 2 4 8 )  2 4 3 3 M P a
Using Eq. (8.2); K  V H W L . Therefore
4 .1 ( 8 3 4 )
Kc 
3 5 ( 3 .2 6  1 0 )
Ks 
3 5 ( 3 .2 6  1 0 )
 3  10
6
0 .9 8 ( 2 4 3 3 )
5
 2 .1  1 0
6
5
SOLUTION (8.6)
Case B ( 1st column ), Table 8.4 with r1  r2 ,
(a)  
1
E


1
E
a  0 . 88
3
( b ) p 0  1 .5
 0 . 88
r
E
F
 1 .5
P
a
m 
2
E
2

1
r
210 ( 10
500
2
 ( 0 . 624 ) ( 10

2
r
 0 . 624
0 . 15
500
3
1
r
6
)
E1  E 2 .
9
)
mm
 613 . 1 MPa
( c ) Equations (8.4) at z=0:

x
 

z
  p 0   613 . 1 MPa

yz

  p 0 [( 1   ) 
y

xz
1
2
(
1
2
] 
 z)  
x
p0
2
1  2
2
p 0   0 . 8 ( 613 . 1 )   490 . 5 MPa
( 0 . 8  1 )  61 . 31 MPa
SOLUTION (8.7)
Refer to Table 8.4 (Case C, Column 2).

n
( a ) a  0 .8 8 3 F
where
  2 E  1(1 0
n 
1
r1

1
r2

11
1
0 .0 0 6
F  2 .2 k N
)

1
0 .0 0 6 0 5
)
1
 1 .3 7 7 4
Thus
and
a  0 .8 8[
2 2 0 0 (1 0
p o  1 .5
F
11
1 .3 7 7 4
a
2
 1 .5
]
3
 2 .2 1 6 2 m m
3
2 .2 (1 0 )
2
 ( 2 .2 1 6 2 ) (1 0
Since 2 1 4 M P a  4 3 6 M P a
6
 214 M Pa
)
OK
( b )   0 .7 7 5 F  n
3
2
2
 0 .7 7 5[( 2 2 0 0 ) (1 0
2
11
2
1
) (1 .3 7 7 4 )]
3
 0 .0 0 6 7 7 m m
( c ) From Fig. 8.9a,  m a x is at about z  0 .4 a and  m a x p o  0 .3 2 . Therefore
z  0 .4 ( 2 .2 1 6 2 )  0 .8 8 6 m m
 m a x  0 .3 2 ( 2 1 4 )  6 8 .4 8 M P a
137
SOLUTION (8.8)
See Table 8.4 (Case B, Column 3).
We have
11
  2 E  1(1 0
( a ) a  1 .0 7 6
m 
),
1
0 .0 2 5

1
0 .0 7 5
 5 3 .3 3 3 3

Lm
F
2 2 0 (1 0
11
)
 1 .0 7 6[ ( 0 .0 2 5 )( 5 3 .3 3 3 3 ) ]
1
 4 .3 7 0 7 (1 0
2
5
) m  0 .0 4 3 7 m m
Then
po 
2

F
aL

220
2

5
( 4 .3 7 0 7  1 0
)( 0 .0 2 5 )
 1 2 8 .2 M P a  3 2 0 M P a
( b ) From Fig. 8.9b, 
OK
is at z  0 .7 5 a and 
yz ,m ax
p o  0 .3 .
yz ,m ax
Hence,
z  0 .7 5 ( 0 .0 4 3 7 )  0 .0 3 2 8 m m

yz ,m ax
 0 .3 (1 2 8 .2 )  3 8 .4 6 M P a
SOLUTION (8.9)
Refer 2nd column of C, Table 8.4.
E 1  E 2  E  210
GPa
 
1
r1  7 mm
r 2  45 mm
F 1  200
kN
m
Hence
( a ) a  1 . 076 [
( b ) p0 
2

F
aL

2
E
2
9
2 1 0 (1 0 )
105 ( 10

)( 120 . 635 )
3
0 . 135 ( 10
n 
,
] 2  0 . 135
200 ( 10 )
2

9
1 0 5 (1 0 )
1
0 .0 0 7

1
0 .0 4 5
 1 2 0 .6 3 5
1
3
200 ( 10 )
9

3
mm
 943 . 1 MPa
)
SOLUTION (8.10)
We have r1 '   . Substitution of the numerical values into Eqs.(8.6) through (8.10)gives
m 
A 
4
2
0 .2 5

2
0 .0 3 4 9
n 
 0 .0 3 4 9
1
0 .0 0 9 3 7 5
 5 7 .3 0 6 6 ,
B 
1
2
9
4 ( 2 0 0 1 0 )
3 ( 0 .9 1 )
 2 9 3 .0 4 (1 0 )
9
1
[( 0 .0 019 3 7 5  0 )  ( 0 )  ( 0 ) ] 2  5 3 .3 3 3 3
2
2
2
From Eq.(8.9)
cos   
5 3 .3 3 3 3
5 7 .3 0 6 6
  2 1 .4 6
 0 .9 3 0 7 ,
o
Then, interpolating in Table 8.5, we have c a  3 .6 2 5 1 and c b  0 .4 2 0 4 . Through the
use of Eq.(8.7):
3
a  3 .6 2 5 1[
2 .2 5  1 0  ( 0 .0 3 4 9 )
b  0 .4 2 0 4[
2 .2 5  1 0 ( 0 .0 3 4 9 )
9
2 9 3 .0 4 (1 0 )
3
9
2 9 3 .0 4 (1 0 )
1
] 3  2 .3 3 7 1 m m
1
] 3  0 .2 7 1 0 m m
(CONT.)
138
8.10 (CONT.)
The maximum contact pressure, by Eq.(8.6), is thus
2 .2 5  1 0
p 0  1 .5
3
 1 6 9 6 .2 M P a
 ( 2 .3 3 7 1 )( 0 .2 7 1 0 )
Thus stress may be satisfactory for the steel used.
SOLUTION (8.11)
In this case: r2  r2 '   as well as r1 '   . Substituting w=L=6.25 mm. and given data into the
equations on the second column of Case A of Table 8.4:
a  1 .0 7 6
p0 
Thus
2

F
aw

3
2 .2 5  1 0
0 .0 0 6 2 5
r1   1 .0 7 6
F
w
3
2 ( 2 .2 5  1 0 )
 (1 .9 7 6 7  1 0
4
)( 0 .0 0 6 2 5 )
( 0 .0 0 9 3 7 5 )(
2
2 0 0 1 0
9
)  1 .9 7 6 7  1 0
4
m
 1 1 5 9 .4 2 3 M P a
Comments: With the flat-faced follower inaccuracies in machining, misalignment, and shaft
deflection may cause edge contact and hence higher actual stress than 1159.423. This cannot
occur with a spherical face follower, however.
SOLUTION (8.12)
Refer to Table 8.4 (Case A, Column 2).
r2   and   2 E
a  0 .8 8 3 F r1 
Thus
 0 .8 8[3 6 0 ( 0 .0 5 6 2 5 )(
(a)
p o  1 .5
 1 .5
1
2
2 0 0 1 0
)]
9
3
 5 .1 6 7 6  1 0
4
m = 0 .5 1 6 7 6 m m
F
a
2
360
 ( 5 .1 6 7 6  1 0
4
)
2
 6 4 3 .7 M P a
2 2
r1
( b )   0 .7 7 5 3 F
 0 .7 7 5[(3 6 0 ) (
2
 4 .7 5 1 1  1 0
6
1
2
2
2 0 0 1 0
9
) ( 0 .0 516 2 5 )]
m = 4 .7 5 1  1 0
3
3
mm
SOLUTION (8.13)
See Table 8.4 (Case C, Column 3).
  2 E and n  1 r1  1 r2  1 0 .0 1 5  1 0 .0 1 6 2 5  5 .1 2 8 2
( a ) a  1 .0 7 6
F
Ln
 9 .0 1 5  1 0
po 
2F
 aL

 1 .0 7 6[
4
3
1 3 .5  1 0 ( 2 )
1
9
( 2 0 0  1 0 )( 0 .0 3 7 5 )( 5 .1 2 8 2 )
]
2
m = 0 .9 0 1 5 m m
3
2 (1 3 .5  1 0 )
 ( 0 .9 0 1 5 )( 0 .0 3 7 5 )
 2 5 4 .2 M P a
 2 5 4 .2 M P a  3 1 7 M P a
OK.
(CONT.)
139
8.13 (CONT.)
( b ) Equations (8.5) at z=0:
 x   2 p o   2 ( 0 .3)( 2 5 4 .2 )   1 5 2 .5 M P a

y
  p o   2 5 4 .2 M P a

z
  p o   2 5 4 .2 M P a
 xy 
1
2
(  1 5 2 .5  2 5 4 .2 )  5 0 .9 M P a
 xz 
1
2
(
x
  z )  5 0 .9 M P a

1
2
(
y
 z)  0
yz

( c ) From Fig. 8.9b, 
yz ,m ax
is at z  0 .7 5 a and 
p o  0 .3 .
yz ,m ax
So,
z  0 .7 5 (9 .0 1 5 5  1 0

4
)  6 .7 6 2  1 0
4
m = 0 .6 7 2 m m
 0 .3 ( 2 5 4 .2 ) M P a  7 6 .2 6 M P a
yz ,m ax
SOLUTION (8.14)
Refer to Table 8.4 (Case C, Column 2).
  2 E  2 (200  10 )  1 10
9
n 
( a ) a  0 .8 8 3 F

1
r1

n
F
a
2
1
0 .0 5

1
0 .0 5 5
 0 .8 8[5 6 0
 1 .2 8 0 3 6  1 0
( b ) p o  1 .5

1
r2
3
 1 .5
11
 1 .8 1 8 2
11
1 1 0
1 .8 1 8 2
1
]
3
m = 1 .2 8 0 3 6 m m
560
 (1 .2 8 0 3 6  1 0
3
)
2
 1 6 3 .1 M P a
( c )   0 .7 7 5 F  n  0 .7 7 5[(5 6 0 ) (1  1 0
3
 2 .9 8 3  1 0
2
2
2
6
m = 2 .9 8 3  1 0
3
11
1
2
mm
3
( d ) At z  0 .3 7 5 a  0 .3 7 5 (1 .2 8 0 3 6  1 0 )  0 .4 8 0 1  1 0

yz ,m ax
3
) (1 .8 1 8 2 )]
3
m = 0 .4 8 0 1 m m (Fig. 8.9a):
 0 .3 2 p o  0 .3 2 (1 6 3 .1)  5 2 .1 9 2 M P a
SOLUTION (8.15)
See Table 8.4 (Column 2). We have   2 E  2 ( 2 0 0  1 0 )  1  1 0
9
11
( a ) Case A: r2   . Thus
a  0 .8 8 3 F r1   0 .8 8[(5 6 0 )( 0 .0 5 )(1  1 0
 5 .7 5 7 1  1 0
p o  1 .5
F
a
2
4
 1 .5
11
1
)]
3
m
560
 ( 5 .7 5 7 1  1 0
4
)
2
 8 0 6 .7 1 8 M P a
(CONT.)
140
8.15 (CONT.)
( b ) Case B:
r1  r2  1 0 0 m m ,

m
a  0 .8 8 3 F
p o  1 .5
 0 .8 8[(5 6 0 )
 1 .5
F
a
2
m  2 r  2 0 .1  2 0
560
 ( 5 .7 5 7  1 0
4
)
11
1 1 0
20
2
1
]
3
 5 .7 5 7  1 0
4
m = 0 .5 7 5 7 m m
 8 0 6 .7 M P a
SOLUTION (8.16)
Use Table 8.4 (Case A, Column 3).
  2 E  2 (200 10 )  1  10
9
a  1 .0 7 6
11
3
r   1 .0 7 6[ 1 .80.11 0 ( 0 .0 1 2 5 )(1  1 0
F
L
 5 .1 0 3 9  1 0
5
11
1
)]
2
m = 0 .0 5 1 0 4 m m
4
2 a  1 .0 2 0 7 8  1 0
m = 0 .1 0 2 1 m m
Therefore
 

2 r1
( 13  ln
0 .5 7 9 F
EL
3
0 .5 7 9 (1 .8  1 0 )
9
( 2 0 0  1 0 )( 0 .1 )
a
)
2 ( 0 .0 1 2 5 )
[ 13  ln
 3 .4 0 1 4  1 0
7
5
( 5 .1 0 3 9 1 0
)
]
m = 3 .4 0 1 4  1 0
-4
mm
SOLUTION (8.17)
We have r1  r2  r . Thus, Eqs. (8.8) and (8.9) become
m 
 2 r  2 ( 0 .2 2 )  0 .4 4
4
(1 r )  (1 r )
(1 r )  (1 r )
cos  
(1 r )  (1 r )
  90
 0,
o
From Table 8.5 it can be concluded that surface of contact has a circular
boundary: c a  c b  1 . Then
n 
9
4 ( 2 0 6 1 0 )
2
3 (1  0 .2 5 )
 2 .9 2 9 7 8 (1 0 )
11
Eqs. (8.7) gives,
a  b  1[
3
2 (1 0 ) ( 0 .4 4 )
2 .9 2 9 7 8  1 0
11
1
3
]
 1 .4 4 3 m m
Hence, Eq. (8.6):
p o  1 .5
6
2 (1 0 )
 (1 .4 4 3 )(1 0
3
)
 6 6 1 .8 M P a
SOLUTION (8.18)
We have r1  0 .5 m , r2  0 .3 5 m , r1 '   , r2 '   , and   90 . Thus, using Eqs.(8.6)
through (8.10):
o
m 
A 
4
1
0 .5
2
m
,

1
0 . 35
n 
 0 . 824
B  
1
2
( r1 
1
1
r2
9
4 ( 206  10 )
3 ( 1  0 . 09 )
 3 . 0183 (10
11
)
)
(CONT.)
141
8.18 (CONT.)
cos  
 
B
A
1 0 .5  1 0 .3 5
1 0 .5  1 0 .3 5
  7 9 .8 6
 0 .1 7 6 ,
Interpolating from Table 8.5: c a  1.1 2 8 ,
3
5 ( 10 )( 0 . 824 )
a  1 . 128 [
3 . 0183 ( 10
p 0  1 .5
)
3
5 ( 10 )( 0 . 824 )
b  0 . 893 [
Thus
11
3 . 0183 ( 10
11
c b  0 .8 9 3 . Hence
1
] 3  2 . 6958
mm
1
] 3  2 . 1342
mm
3
5 ( 10 )
 1 .5
F
 ab
)
o
 ( 2 . 6958  2 . 1342  10
6
 414 . 9 MPa
)
SOLUTION (8.19)
We now have F 
 300
600
2
S
r2   5 .2 m m ,
and r2 '   3 0 m m .
r1  r1 '  5 m m ,
and
 1500
N for each row,
y
MPa
We proceed as in Example 8.3.
(a) m 
4

2
0 . 005

1
0 . 0052
1
0 . 030
 0 . 0229 ,
n  293 . 0403  10
9
and
A 
 87 . 3362
2
0 . 0229
B 
1
2
[( 0 )
 (
2
1
0 . 0052

1
0 . 03
)
2
1
 2 ( 0 ) ] 2  79 . 4872
2
Using Eq.(8.9),
cos   
Table 8.5: c a  3.3 3 3 0
Hence
b  0 . 4441 [
o
c b  0 .4 4 4 1
1
300  0 . 0229
a  3 . 3330 [
  24 . 48
 0 . 9101
79 . 4872
87 . 3362
293 . 0403  10
9
mm
1
300  0 . 0229
293 . 0403  10
] 3  0 . 9539
9
] 3  0 . 1271
mm
 1 ,181
MPa
Then
p 0  1 .5
(b) n 
300
 ( 0 . 9539  0 . 1271  10
6
)
Fy
(1)
F
It is required that S y  1.5 F y  a b
S
y
1 .5

or
Fy
3
cacb (m n )
Substituting the data given
2
3
1 .5
1500 ( 10 ) 
6
3
 ( 3 . 3330  0 . 4441 )(
Fy
Solving
F y  614
N
Equation (1):
n 
614
300
2
0 . 0229
293 . 0403  10
 2 .0 5
142
9
)
3
SOLUTION (8.20)
Refer to Table 8.4 (Case A, Column 3).
a  1 .0 7 6
r1 
F
L
where
F  5 kN ,
 

2
E
L  25 m m ,
2 0 6 1 0
r1  5 0 0 m m
 0 .9 7 1(1 0
2
9
11
)
Therefore
3
( 5 1 0 )
a  1 .0 7 6[
( 0 .5 ) ( 0 .9 7 1  1 0
0 .0 2 5
11
1
]
 1 .0 6 0 3 m m
2
Hence,
po 
2
3
2 ( 5 1 0 )

F
aL

 (1 .0 6 0 3  1 0
3
 120 M Pa
)( 0 .0 2 5 )
SOLUTION (8.21)
See Table 8.4 (Case C, Column 3).
 
2

F
aL

F
Ln
( a ) a  1 .0 7 6
( b ) po 
2
E

1
 1 .0 7 6[
1
0 .0 1
3 0 0 1 0
3
9
1 0 0 (1 0 )( 8 4 )

1
0 .0 6 2 5
1
 0 .2 0 3 3 m m
]
2
 84,
 0 .2 0 3 3
( c ) From Fig. 8.9b, z  0 .7 5 a  0 .1 5 2 5 m m

F L  300 kN
 9 3 9 .4 M P a
300
2
n 
,
9
1 0 0 (1 0 )
yz ,m ax
and
 0 .3 p o  0 .3 (9 3 9 .4 )  2 8 2 M P a
SOLUTION (8.22)
Given: r1  r1 '  0 .0 2 m ,
Equation (8.8),
m 
4
1
0 .0 2

1
0 .0 2

1
0 .0 2 5

1
0 .1
r2   0 .0 2 5 m ,
 0 .0 8 ,
n 
  0 .3,
r2 '   0 .1 m ,
9
4 ( 2 0 0 1 0 )
2
3 (1  0 .3 )
E  200 G Pa
 2 .9 3 (1 0 )
11
Also
A 

2
0 .0 8
  cos
1 15
25
2
m
 2 5,
B  
 5 3 .1 3 0 1
1
2
[( 0 )  ( 
2
1
0 .0 2 5

1
0 .1
2
o
From Table 8.5, we find
c a  1 .6 6 4 5
c b  0 .6 6 4 2
The semiaxes are then
a  1 .6 6 4 5[
(1 2 0 0 ) ( 0 .0 8 )
b  0 .6 6 4 2[
(1 2 0 0 ) ( 0 .0 8 )
2 .9 3 (1 0
11
2 .9 3 (1 0
11
)
)
1
]
3
 0 .0 0 1 1 4 7 m  1 .1 5 m m
3
 0 .0 0 0 4 6 m  0 .4 6 m m
1
]
Thus
p o  1 .5
F
 ab
 1 .5
1200
 (1 .1 5  0 .4 6 )(1 0
6
)
 1083 M Pa
143
1
)  2 ( 0 )] 2   1 5
SOLUTION (8.23)
We have
r1  r1 '  0 .0 1 8 m ,
r2   0 .0 2 m ,
r2 '   0 .1 m ,
  0 .3,
E  200 G Pa
Equations (8.8),
m 
4
1
0 .0 1 8

1
0 .0 1 8

1
0 .0 2

1
0 .1
 0 .0 7 8 3,
n 
9
4 ( 2 0 0 1 0 )
2
3 (1  0 .3 )
 2 .9 3 (1 0 )
11
Equations(8.10)
A 
2
m
B  
 2 5 .5 4 2 8
1
2
  cos
[( 
1
1
0 .0 2

20
2 5 .5 4 2 8
1
0 .1
2
 3 8 .4 6
From Table 8.5, c a  2 .2 1 6 4
Equations (8.7):
a  2 .2 1 6 4[
b  0 .5 5 5 6[
1
) ]2  20
c b  0 .5 5 5 6
( 9 0 0 ) ( 0 .0 7 8 4 )
2 .9 3 (1 0
11
)
( 9 0 0 ) ( 0 .0 7 8 3 )
2 .9 3 (1 0
11
o
)
1
]
3
 0 .0 0 1 3 7 9 m  1 .3 8 m m
3
 0 .0 0 0 3 5 m  0 .3 5 m m
1
]
It follows that
p o  1 .5
F
 ab
 1 .5
900
 (1 .3 8  0 .3 5 )(1 0
6
)
 890 M Pa
End of Chapter 8
144
Section III
APPLICATIONS
CHAPTER 9
SHAFTS AND ASSOCIATED PARTS
SOLUTION (9.1)
T 
9549 kW
n
9 5 4 9 (1 0 )

 3 8 .1 9 6 N  m
2500
and

Also
c 
3
2
 a ll
L

T
 a ll
T
GJ

3 8 .1 9 6
6
1 5 0 1 0
3
c  7 .8 6 4
,
2 ( 1 8 0 ) 
;
3 8 .1 9 6 ( 2 )
9
8 0 (1 0 )  c
4
mm
c  9 .6 6 m m
,
A 20-mm dia. shaft would probably be selected.
SOLUTION (9.2)
Refer to Example 9.1. We now have S u  3 6 5 M P a (T a b le B .2 ), n  2 .5 , and S y is
replaced by S u .
Equation (9.9) gives
D  [ 1 6Sn ( M
u
c

M
2
c
 Tc ]
2
1
3
Substituting the given data,
D [
1 6 ( 2 .5 )
6
 ( 3 6 5 1 0 )
( 2 0 2 .5 
( 2 0 2 .5 )  (1 6 2 ) ]
2
2
1
3
 0 .0 2 5 2 5 6 m = 2 5 .2 6 m m
SOLUTION (9.3)
T AC 
(a)
9 5 4 9 ( 7 .5 )

9549 kW
n
1200
 5 9 .6 8 N  m
and

2
c 
3
T
 a ll

5 9 .6 8
6
2 1 0 (1 0 ) 2
c  7 .1 2 6
,
 D A C  1 4 .2 5
mm
Similarly
TC B 

2
c 
9 5 4 9 ( 2 2 .5 )
1200
3
 179
179
c  1 0 .2 8
,
6
2 1 0 (1 0 ) 2
N m
( b ) We have   T L G J , with J   c
Thus
 AC 
5 9 .6 8 ( 2 )

4
mm
 D C B  2 0 .5 6
2.
 0 .3 5 9 6 ra d  2 0 .6
4
9
( 0 .0 0 7 1 2 5 ) ( 8 2  1 0 )
2
 BC 
179 ( 4 )

4
 0 .4 9 8 ra d  2 8 .5 3
9
( 0 .0 1 0 2 8 ) ( 8 2  1 0 )
2
Hence
 A B   B C   A C  7 .9 3
o
145
o
o
mm
mm
SOLUTION (9.4)
G a  28 GPa ,
Table B.1:
S u  520
Table B.3:
MPa ,
We have  a   s but L a  L s ,
Js
Ja
Ga

 0 . 354
Gs
G s  79 GPa
S
y
 440
MPa
Ta  Ts
where J 

D
32
4
Hence
Ds
Da
Wa
Also
Ws
 0 .7 7 1
or
Va a
(  D a 4 ) a

V s
s

D a  1. 2 9 6 D s
4
2
(Ds
4 )

s
2
Da a
2
Ds 
s
 a  2 .8 M g m ,
3
Table B.1:
 s  7 .8 6 M g m
3
Thus
Wa
Ws
2
(1. 2 9 6 D s ) ( 2 . 8 )

2
Ds
 0 .5 9 8
( 7 .8 6 )
SOLUTION (9.5)
( a ) Use Eq.(9.5):
6
2 6 0 (1 0 )
n

3
4 (1 0 )
 ( 0 .1 )
3
1
[( 8  5  5 0  0 .1 )  ( 8  8 ) ] 2
2
2
n  2 .6 1
or
( b ) Apply Eq.(9.6):
6
2 6 0 (1 0 )
n

3
4 (1 0 )
 ( 0 .1 )
3
1
[ ( 8  5  5 0  0 .1 )  4 8 ( 8 ) ] 2
2
2
or
n  2 .8 6
SOLUTION (9.6)

F y  0 : 0 and

M
A
 0 yield
960
Mz
(N  m)
480
x
A
My
(N  m)
A
R By  8 0 0 N
R Az  5 0 0 N
R Bz  1 0 0 0 N
M

960  300
2
C
 1006 N  m
M

600  480
2
D
 7 6 8 .4 k N  m
B
C
300
R Ay  1 6 0 0 N
600
2
T  1 .2 5 k N  m
x
D
2
B
Critical section is at C.
(CONT.)
146
9.6 (CONT.)
We have
1 6T
 
(a) 
n 
32M

m ax

D
m ax
Sy
( b )  m ax 
n 
D

(
1 6 (1, 2 5 0 )
 ( 0 .0 4 5 )
 ( 0 .0 4 5 )
3
3
 6 9 .9 M P a
 1 1 2 .5 M P a
 3 .0 7
345
1 1 2 .5


3 2 (1, 0 0 6 )

3
3
) 
2
2

(5 6 .2 5 )  ( 6 9 .9 )
2
 8 9 .7 M P a
2
2
 m ax

S ys
 2 .3 4
210
8 9 .7
SOLUTION (9.7)
y
0 . 6 kN  m
0 . 6 kN  m
x
A
B
C
z
6 kN
3.75 kN
Critical point is just to
the right of C.
2.25 kN
1 . 125
kN  m
M
x
0 . 6 kN  m
T
x
S
(a)
y
n

32
D
3
2
M
 T
2
6
2 5 0 (1 0 )
;
1 .5

3
3 2 (1 0 )
D
3
[1.1 2 5
2
1
 0 .6 ] 2
2
or
D  42 . 71 mm
250 ( 10
(b)
1 .5
6
)

3
32 ( 10 )
D
3
[1 . 125
2

3
4
2
1
( 0 .6 ) ] 2
Solving D  4 2 .3 m m
SOLUTION (9.8)
We have S y  3 4 0 M P a and n  1 .5 .
The reactions and the torques, as found by the equations of statics, are marked in
Fig. S9.8a. Moment and torque diagrams are shown in Fig. S9.89b. Critical sections is at C.
(CONT.)
147
9.8 (CONT.)
y
A
2m
90 N
C
7.86N  m
2m
z
360 N
D
7.86N  m.
450 N
360 N
1m
720 N
B
x
360 N
(a)
720
Mz
(N  m)
A
C
My
(N  m)
x
D
Figure S9.8
360
180
B
x
T
(N  m)
-7.86
C
x
B
(b)
Eq. (9.7) is thus
D 
32n
3
 Sy
M
 M
2
z
2
y
T
2
Inserting the numerical values:
[
or
3 2 (1 .5 )
 (3 4 0  1 0 )
6
(
7 2 0  1 8 0  (  7 .8 6 ) ) ]
2
2
2
1
3
D  0 .0 3 2 2 m = 3 2 .2 m m
SOLUTION (9.9)
Refer to solution of Prob. 9.8. Now apply Eq. (9.8):
D 
[
32n
3
 Sy
M
3 2 (1 .2 )
 (3 4 0  1 0 )
6
 M
2
z
(
2
y

3
T
2
4
720  180 
2
2
3
4
or
D  0 .0 2 9 9 m = 2 9 .9 m m
148
(  7 .8 6 ) ) ]
2
1
3
SOLUTION (9.10)
Table B.1: S
 520
y
 xy 
We have M  0 ,
MPa
16T
D
3
By Eq.(1.15),
T 

9549 kW
n
9549 (70 )
110
 6 .0 7 7 k N  m
( a ) Equation (6.11)
Sy
n
 2
xy
,
D

3
32Tn
 Sy
32 (6077 ) 4

6
 ( 5 2 0 1 0 )
D  7 8 .1 m m
,
( b ) Equation (6.16):
Sy
n
3

xy
,
D
3

16

3T n
 Sy
16
3 (6077 ) 4
6
 ( 5 2 0 1 0 )
D  7 4 .4 m m
or
SOLUTION (9.11)
We have S u  5 9 0 M P a and n  1 .4
Ay  B y  8 kN
From statics:
8 kN
y
8 kN
0.5 m
A
z
1m
0.5 m
C
0.48 kN  m
Ay
D
0.48 kN  m
B
x
(a)
By
8  0.5=4 kN  m
Mz
x
T
0.48 kN  m
x
(b)
(c)
Figure S9.11
The critical section is between C and D. Thus
M
2
z
T
2

4  0 .4 8
2
2
 4 .0 2 9 k N  m
Apply Eq. (9.9):
1 6  1 0 (1 .4 )
3
D 
3
 (5 9 0  1 0 )
6
( 4  4 .0 2 9 )  0 .0 4 6 m
Use a 4 6 m m diameter shaft.
SOLUTION (9.12)
Now Fig. S9.11a (see: Solution of Prob. 9.11) becomes as shown below.
From statics:
(CONT.)
149
9.12 (CONT.)
Ay  2 kN
B y  6 kN
Az  6 k N
Bz  2 kN
y
A
1m
0.5 m
D
0.48 kN  m
C
0.48 kN  m
Az Ay
z
8 kN
8 kN
0.5 m
B
x
(a)
x
(b)
x
(c)
x
(d)
By
Bz
Mz
2  0.5=1 kN  m
2  1.5=3 kN  in.
3 kN  m
My
1 kN  m
T
0.48 kN  m
The critical section is just to the right of C (or just to the left of D). Thus
M
2
z
 M
2
y
T
2

3  1  0 .4 8
2
2
2
 3 .1 9 9 k N  m
Applying Eq. (9.9):
1 6  1 0 (1 .5 )
3
D 
3
 (5 9 0  1 0 )
6
( 3  3 .1 9 9 )  0 .0 4 3 1 m
Use a 4 4 m m diameter shaft.
SOLUTION (9.13)
y
A
Az
0 . 6 kN  m
0.3 m
C
5.2 kN
3 kN
Ay
z
x
0 . 6 kN  m
0.5 m
Bz
30o
975
M
B y  1 . 95 kN ,
B
B z  1 . 125
A y  3 . 25 kN
kN , A z  1 . 875
kN
By
N m
z
x
M
563
Critical point is just to the right of C.
N m
M
y
x
T
600
1
 [ 9 7 5  5 6 3 ] 2  1 .1 2 6 k N  m
2
C
2
N m
x
150
(CONT.)
9.13 (CONT.)
Thus
M
 0
m
M
 1 . 126
a
kN  m
T m  600
N m
From Sec. 8.6: C t  1  0 .0 0 5 8 ( 5 0 0  4 5 0 )  0 .7 1
S e  0 . 5 S u  260
'
MPa
C
f
 AS
b
u
 4 . 51 ( 520
 0 . 265
)  0 . 86
C r  0 .8 9
Assume D  51 mm and C s  0 .7 0 .
Thus
S e  C f C r C s C t ( K1 ) S e  ( 0 . 86 )( 0 . 89 )( 0 . 70 )( 0 . 71 )( 11. 2 )( 260 )  82 . 42 MPa
'
f
Substitute the data and K s b  K s t  1.5 (Table 9.1) into Eq.(9.13):
6
520 (10 )
1 .5

[ 1.5 ( 8 2 . 4 2  1 ,1 2 6 )
32
D
520
3
2
1
 1.5 ( 6 0 0 ) ] 2
2
Solving, D  63 . 5 mm
SOLUTION (9.14)
Statics: R A  9 .4 5 k N ,
R B  6 .7 5
kN ,
M
D
 2 .5 3 k N  m  M
m ax
Refer to Secs. 7.6 and 7.7:
C s  0 .7 K
C r  0 .8 1 (Table 7.3)
C
 AS
f
b
u
 0 . 085
 1 . 58 (1260
 1, S e  0 . 5 S u  630
'
f
)  0 . 861
S e  ( 0 .8 6 1)( 0 .8 1)( 0 .7 )(1)( 6 3 0 )  3 0 7 .5 6 M P a
and
We have M
m
 0
M
a
 2 .5 3 k N  m
T m  2 .2 6 k N  m
T a  0 .2 2 6 k N  m
Using Eq.(9.12):
6
1 2 6 0 (1 0 )
n

3
3 2 (1 0 )
 ( 6 2 .5  1 0
3
[( 310276.506  2 .5 3 ) 
2
)
3
3
4
( 2 .2 6 
1260
3 0 7 .5 6
1
 0 .2 2 6 ) ] 2
2
n  2 .8 2
or
SOLUTION (9.15)
Statics: R A  5 .2 2 k N ,
M
m
 0,
R B  8 .2 8 k N ,
T m  3 .4
kN  m ,
From Secs. 7.6 and 7.7: C r  0 .8 9
M
 2 .6 1 k N  m  M
C
a
. We have
T a  0 .6 8 k N  m
C s  0 .7
K
f
1
C t  1  0 . 0058 ( 510  450 )  0 . 652
S e  0 .5 S u  7 0 0
'
M Pa
C
 A S u  5 7 .7 (1 4 0 0
b
f
 0 .7 1 8
S e  ( 0 .3 1 8 )( 0 .8 9 )( 0 .7 )( 0 .6 5 2 )(1)( 7 0 0 )  9 0 .4 M P a
and
Apply Eq.(9.13), with replacing S u by S y and K s t  2 (Table 9.1):
910 ( 10
n
Solving,
6
)

3
32 ( 10 )
 ( 87 . 5  10
3
)
3
[(
910
90 . 4
 2 . 61 )
2
n  1 .9 9
151
 2 (3 .4 
910
90 . 4
1
 0 . 68 ) ] 2
2
)  0 .3 1 8
SOLUTION (9.16)
From Fig. C.13, for d D  0 .2 :
K t  1 .5 (torsion)
K t  2 .2 (bending)
Figures (7.9) : q  0 .9 5 (torsion) q  0 .8 (bending)
The endurance limit is
From Eq. 7.13b we have
K
fb
 1  q ( K t  1)  1  0 .8 ( 2 .2  1)  1 .9 6
K
ft
 1  0 .9 5 (1 .5  1)  1 .4 8
Endurance limit:
S e  C f C r C s C t (1 K t ) S e '
where
C
 4 .5 1(5 2 0
f
 0 .2 6 5
)  0 .8 6
C r  0 .8 1
(Table 7.3)
C s  0 .8 5
(Eq. 7.9)
C t  1  0 .0 0 5 8 ( 4 9 0  4 5 0 )  0 .7 7
S e '  0 .5 S u  2 6 0 M P a
(Eq. 7.11)
(Eq. 7.1)
Thus,
S e  ( 0 .8 6 )( 0 .8 1)( 0 .8 5 )( 0 .7 7 )(1 1 .9 6 )( 2 6 0 )
 6 0 .5 M P a
Also
M
a
 120 N  m ,
M
 0,
m
Tm  6 0 0 N  m ,
Ta  0
Equation (9.16a):
D
3

32 n
 Su
S
[( S u M a ) 
2
3
4
e
2
1
2
Tm ]
Substituting the given data, we have
3
( 4 0 ) (1 0
9
) 
[ ( 6502.50  1 2 0 ) 
2
32 n
6
 ( 5 2 0 1 0 )
2
1
2
1
3
4
(6 0 0 ) ]
3
4
(6 0 0 ) ]
2
Solving,
n  2 .8 3
SOLUTION (9.17)
Refer to solution of Prob.9.16.
Apply Eq. (9.16b):
D
3

32 n
 Sy
[(
Sy
Se
M a) 
2
3
4
2
Tm ]
1
2
Inserting the given numerical values,
3
( 4 0 ) (1 0
9
) 
[ ( 6404.50  1 2 0 ) 
2
32 n
6
 ( 4 4 0 1 0 )
From which
n  2 .7 2
152
2
SOLUTION (9.18)
M
S
 200
a
N m,
 290
y
T m  500
N  m,
S u  455
MPa ,
M
 0,
m
Ta  0,
K
st
 2
MPa
Refer to Secs. 7.6 and 7.7:
C
 4 . 51 ( 455
f
 0 . 265
Assume C s  0 . 7 ,
)  0 . 89 ,
C r  0 . 87 ,
S e  ( 0 . 89 )( 0 . 87 )( 0 . 7 )( 1)(
S e  0 . 5 S u  227 . 5 MPa
'
1
2 .2
)( 227 . 5 )  56 . 05 MPa
Through the use of Eq.(9.14), we have
6
4 5 5 (1 0 )
1 .5

32
D
[1( 5 6 . 0 5  2 0 0 ) 
2
455
3
D  38 . 8 mm  51 mm
or
3
4
1
2
( 2 )(5 0 0 ) ] 2
 Assumption is incorrect.
Assume C s  0 .8 5 :
Se 
0 . 85
0 .7
( 56 . 05 )  68 . 06 MPa
Equation (9.14):
6
4 5 5 (1 0 )
1.5

32
D
[1( 6 8 . 0 6  2 0 0 ) 
2
455
3
D  36 . 7 mm
or
3
4
1
2
( 2 )(5 0 0 ) ] 2
 Assumption is correct.
SOLUTION (9.19)
Refer to Secs. 7.6 and 7.7: q  0 .9 2
K
 1  q ( K t  1 )  1  0 .9 2 (1.8  1 )  1.7 4
f
S e  0 . 5 S u  500
'
C
Thus
 AS
f
b
u
MPa ,
 57 . 7 (1000
 0 . 718
C s  0 . 85 ,
C r  0 . 84
)  0 . 405
S e  ( 0 .4 0 5 )( 0 .8 4 )( 0 .8 5 )(1)( 1 .71 4 )(5 0 0 )  8 3 .0 9 M P a
We have: M
 0,
m
M
a
 500
N  m,
T m  600
N  m,
T a  90 N  m
Apply Eq.(9.11), with replacing S u by S y :
6
600 ( 10 )
n

32
 ( 0 . 05 )
3
[(
600
83 . 09
 500 )  ( 600 
2
600
83 . 09
1
 90 ) ] 2
2
or
n  1 . 93
SOLUTION (9.20)
The endurance limit is expressed as
S e  C f C r C s C t (1 K f ) S e '
Here
C
f
 4 .5 1( 6 9 0
 0 .2 6 5
)  0 .7 9 8
(Eq. 7.7 and Table 7.2)
C r  0 .8 7
(Table 7.3),
C t  1 (Eq. 7.11)
C s  0 .7 0
(Assuming D > 51 mm, Eq. 7.9)
(CONT.)
153
9.20 (CONT.)
S e '  0 .5 S u  0 .5 ( 6 9 0 )  3 4 5 M P a
(Eq. 7.1)
Therefore
S e  ( 0 .7 9 8 )( 0 .8 7 )( 0 .7 )(1)(1 1 .2 )(3 4 5 )  1 3 9 .7 M P a
Equation (9.13) with T a  0 becomes

3
D
32 n
 Sy
[K s (M
m

Su
Se
M a )  K sT m ]
2
2
1
2
Introducing the given numerical values, we have
3 2 ( 3 .5 )

3
D
6
 ( 6 9 0 1 0 )
6
( 6 9 0 1 0 )
{1 .5[ 2 0 0 
6
(1 3 9 .7  1 0 )
( 6 0 0 )]  1 .5 (3 6 0 ) }
2
2
1
2
Solving,
D  0 .0 5 8 6 m  5 8 .6 m m
Since D  5 1 m m , our assumption is correct.
SOLUTION (9.21)
We now have K s  2 (from Table 9.1) and S u replaced by S y .
See solution of Prob. 9.20. Equation (9.14) becomes

3
D
32 n
 Sy
[K s (M
m

Sy
Se
M a )  K s ( 34 T m )]
2
2
1
2
or
3 2 ( 3 .5 )

3
D
6
 ( 5 8 0 1 0 )
{ 2[ 2 0 0 
580
1 3 9 .7
(6 0 0 )] 
2
3
2
1
2
(3 6 0 ) }
2
Solution is
D  0 .0 6 1 7 4 m  6 1 .7 m m
Since D  5 1 m m , our assumption is correct.
SOLUTION (9.22)
2 kN  m
M
0 . 8 kN  m
y
Critical section is just to the left of point D
x
C
A
M
D
B
z
600
T
2 .5 7 k N  m
M
D
 [ 2 . 57
x
M
a
 2 . 692
T m  600
N m
1
2
 0 . 8 ] 2  2 . 692
2
kN  m ,
N m,
M
 0
m
T a  30 N  m
x
Refer to Secs. 7.6 and 7.7:
S e  0 . 5 S u  330
'
S e  C f C rC sC t (
MPa ,
1
K f
C
f
 AS
b
u
 4 . 51 ( 660
 0 . 265
)  0 . 807
) S e  ( 0 . 807 )( 1 )( 0 . 7 )( 1 )( 1 )( 330 )  186 . 4 MPa
'
Equation (9.14):
6
6 6 0 (1 0 )
n

32
 ( 0 .0 7 5 )
[1.5 ( 1 8 6 . 4  2 , 6 9 2 ) 
2
660
3
Solving n  2 .3 4
154
3
4
(1.5 ) ( 6 0 0 
660
1 8 6 .4
kN  m
1
 30 ) ]2
2
SOLUTION (9.23)
From Solution of Prob. 9.22
M
 0,
m
M
 2 . 692
a
kN  m ,
T a  30 N  m ,
T m  600
N m
S e  1 8 6 .4 M P a
Now we have C r  0 .8 9 . Thus, S e  1 8 6 .4 ( 0 .8 9 )  1 6 5 .9 M P a
Through the use of Eq.(9.13), with replacing S u by S y :
6
550 ( 10 )
n

550
[1 . 5 ( 165
 2 , 692 )  1 . 5 ( 600 
.9
2
32
 ( 0 . 075 )
3
550
165 . 9
1
 30 ) ] 2
2
n  2 . 08
or
SOLUTION (9.24)
R B y  1 .0 3 8 k N ,
0.45 kN
y
R A y  6 .1 6 2 k N R B z  3 .7 3 8 k N ,
1.35 kN
z
C
M
D
B
M
( kN  m)
B
6.3 kN R B z
2.7 kN
9 kN
x
D
C
R Az
R By
z
x
A
x
A
R Ay
R A z  0 .1 3 8 k N
y
(kN  m)
0.208
x
0.021
0.748
0.924
We have T  0 .9 k N  m and
1
M
D
 [ 0 .2 0 8  0 .7 4 8 ] 2  0 .7 7 6 k N  m
M
C
 [ 0 .9 2 4  0 .0 2 1 ] 2  0 .9 2 4 k N  m
M
m
 0,
2
2
2
2
1
Hence
M
a
 0 .9 2 4 k N  m ,
Ta  0 ,
T m  0 .9 k N  m
From Table B.3: S u  5 9 0 M P a .
Refer to Secs. 7.6 and 7.7:
C r  0 .8 7 ,
C
and
f
 AS
C s  0 .7 (assumed D>50 mm)
 4 . 51 ( 590 )
b
u
 0 . 265
 0 . 83
S e  ( 0 .8 3)( 0 .8 7 )( 0 .7 )(1)( 11.8 )( 0 .5  5 9 0 )  8 2 .8 4 M P a
Applying Eq.(9.12):
5 9 0 1 0
1 .6
6

[ ( 8529.804  9 2 4 ) 
2
32
D
3
3
4
2
1
(9 0 0 ) ] 2
D  0 .0 5 6 8 m = 5 6 .8 m m
155
SOLUTION (9.25)
R B y  2 .0 8 k N ,
R A y  6 .9 2 k N
9 kN
y
C
B
D
M
x
( kN  m )
We have T  0 .9 k N  m and M
C
 M
x
0.021
D
1
2
 [ (1 .0 3 8 )  ( 0 .0 2 1) ]  1 .0 3 8 k N  m  M
2
3.74 kN
y
(kN  m )
0.416
B
D
0.748
2.08 kN
1.038
C
C
0.14 kN
z
M
x
A
x
6.92 kN
R A z  0 .1 4 k N
6.3 kN
2.7 kN
z
A
M
R B z  3 .7 4 k N ,
2
Refer to Solution of Prob. 9.24: C f  0 .8 3 ,
C s  0 .7
a
From Table B-3: S y  4 9 0 M P a
From Sec. 7.6:
S e  C f C r C s C t ( K1 ) S e  ( 0 .8 3 )( 0 .7 5 )( 0 .7 )(1)( 11.2 )( 0 .5  5 9 0 )  1 0 7 .1 2 M P a
'
f
Using Eq.(9.11), with S u replaced by S y :
D
or
3

32 n
 Sy
[(
Sy
Se
1
32 ( 2 )
M a )  Tm ] 2 
2
2
D  0 .0 5 8 6 m = 5 8 .5 m m
SOLUTION (9.26)
Refer to Case 6 of Table A.8:
 
We have I 
Pbx
6 LEI

4
(L  b  x )
2
(12 . 5 )
2
4
2
 19 . 175  10
3
mm
4
W D  W E  15  9 . 81  147 . 2 N
Deflection at D:
D '
 D '' 
2
2
2
1 4 7 .2 (1 )( 0 .4 )(1 .4  1  0 .4 )
6 (1 4 0 0 )( 2 1 0  1 9 .1 7 5 )
2
 1 .3 9 3 m m
2
2
1 4 7 .2 ( 0 .4 )( 0 .4 )(1 .4  0 .4  0 .4 )
6 (1 4 0 0 )( 2 1 0  1 9 .1 7 5 )
 1 .1 4 2 m m
and
 D   E  1 .3 9 3  1 .1 4 2  2 .5 3 5 m m
Equation (9.18) results in
n cr 
1
2
 594
2 gW 
2W 
2

1
2
g


1
2
9 . 81
2 . 535  10
rpm
156
1
[( 1 0479.10 2  1, 0 3 8 )  9 0 0 ] 2
2
6
 ( 4 9 0 1 0 )
3
 9 . 901 cps
2
SOLUTION (9.27)
Refer to Case 6 of Table A.8:
 
Pbx
6 LEI
(L  b  x )
2
2
2
We have W D  W E  15  9 . 81  147 . 2 N
At midspan C:
C 
0 . 5 ( 10
3
)
2

147 . 2 ( 0 . 4 )( 0 . 7 )
(1 . 4
6 ( 1 . 4 ) EI
2
 0 .4
2
 0 .7 ) 
2
or
EI  25 . 711 (10 ) N  m
3
2
Deflection at D:
D '
 D '' 
2
2
2
1 4 7 .2 (1 )( 0 .4 )(1 .4  1  0 .4 )
 0 .2 1 8
3
6 (1 .4 )( 2 5 .7 1 1  1 0 )
2
2
2
1 4 7 .2 ( 0 .4 )( 0 .4 )(1 .4  0 .4  0 .4 )
3
6 (1 .4 )( 2 5 .7 1 1  1 0 )
mm
 0 .1 7 9
mm
and

D
 
E
 0 . 218  0 . 179  0 . 397
mm
Equation (9.18) gives
n cr 
g
1
2


9 .8 1
1
2
0 .3 9 7  1 0
3
 2 5 .0 2
cps
SOLUTION (9.28)
We have
I 

4
(15 )
 39 . 76  10
4
3
mm
4
Refer to Case10 of Table A.8:
a
P
C
b
A
B
L
2
C 
Pb L
3 EI
C 
150 ( 0 . 4 )( 1000 )
Thus
2
3 ( 210  39 . 76 )
 0 . 958
mm
Equation (9.18) results in
n cr 
1
2


1
2
0 .9 5 8  1 0
g
9 .8 1
3
 1 6 .1 1 c p s  9 6 7
157
rp m
6 . 428
EI
SOLUTION (9.29)
See Table A.8 (Case 6) and Example 9.5.
We have
I 
4
d
 ( 0 .0 5 6 2 5 )

64
4
 0 .4 9 1 4 (1 0
64
6
) m
4
( a ) Deflection at C owing to 450 N:
C ' 
4 5 0 ( 2 .2 5 )( 0 .6 2 5 )
( 2 .8 7 5  0 .6 2 5  2 .2 5 )  1 .0 5 m m
2
9
6 ( 2 .8 7 5 )( 2 0 0  1 0 )( 0 .4 9 1 4  1 0
6
)
2
2
Deflection at C due to 270 N:
C " 
2 7 0 (1 )( 0 .6 2 5 )
( 2 .8 7 5  0 .6 2 5  1 )  0 .6 8 4 m m
2
9
6 ( 2 .8 7 5 )( 2 0 0  1 0 )( 0 .4 9 1 4 1 0
6
)
2
2
Total deflection at C:
 C  1 .0 5  0 .6 8 4  1 .7 3 4 m m
Deflection at D owing to 450 N:
4 5 0 ( 0 .6 2 5 )( 2 .8 7 5  1 .8 7 5 )
D '
9
6 ( 2 .8 7 5 )( 2 0 0  1 0 )( 0 .4 9 1 4  1 0
[ 2 ( 2 .8 7 5 )(1 .8 7 5 )  0 .6 2 5  1 .8 7 5 ]  1 .1 4 1 m m
2
6
)
2
Deflection at D due to 270 N:
D " 
2 7 0 (1 )(1 .8 7 5 )
( 2 .8 7 5  1 .8 7 5  1 )  1 .1 2 m m
2
9
6 ( 2 .8 7 5 )( 2 0 0  1 0 )( 0 .4 9 1 4 1 0
6
)
2
2
Total deflection at D:
 D  1 .1 4 1  1 .1 2  2 .2 6 1 m m
Applying Eq. (9.18), we obtain
n cr 
1
2
[
9 .8 1[( 4 5 0 )(1 .7 3 4  1 0
4 5 0 (1 .7 3 4  1 0
3
3
)  ( 2 7 0 )( 2 .2 6 1  1 0
2
)  2 7 0 ( 2 .2 6 1  1 0
3
)
3
)]
2
1
]
2
 1 1 .2 4 c p s  6 7 4 .4 r p m
( b ) Apply Eq. (9.21):
n c r ,C n c r , D
n cr 
where
Hence
2
2
n c r ,C  n c r , D
g
n cr ,C 
1
2
C '
n cr , D 
1
2
D "
n cr 

g

( 9 2 3 )( 8 9 4 )
2
1
2
(923)  (894 )
2
1
2
9 .8 1
1 .0 5  1 0
3
9 .8 1
1 .1 2  1 0
3
 1 5 .3 8 c p s  9 2 3 rp m
 1 4 .9 c p s  8 9 4 rp m
 6 4 2 rp m
SOLUTION (9.30)
Refer to Table A.8 (Case 6) and Solution of Prob. 9.29.
We now have I   D
4
6 4   ( 0 .0 8 7 5 )
4
6 4  2 .8 7 7 4  1 0
6
m
4
( a ) Deflection at C owing to 450 N:
C ' 
4 5 0 ( 2 .2 5 )( 0 .6 2 5 )
9
( 2 .8 7 5  0 .6 2 5  2 .2 5 ) = 0 .1 7 9 3 m m
2
6 ( 2 .8 7 5 )( 2 0 0  1 0 )( 2 .8 7 7 4 4  1 0
6
)
2
2
Deflection at C due to 270 N:
C " 
2 7 0 (1 )( 0 .6 2 5 )
9
6 ( 2 .8 7 5 )( 2 0 0  1 0 )( 2 .8 7 7 4  1 0
( 2 .8 7 5  0 .6 2 5  1 )  0 .1 1 6 9 m m
2
6
)
2
2
Total deflection at C:
 C  0 .1 7 9 3  0 .1 1 6 9  0 .2 9 6 2 m m
(CONT.)
158
9.30 (CONT.)
Deflection at D owing to 450 N:
D '
4 5 0 ( 0 .6 2 5 )( 2 .8 7 5  1 .8 7 5 )
[ 2 ( 2 .8 7 5 )(1 .8 7 5 )  0 .6 2 5  1 .8 7 5 ]  0 .1 9 4 8 m m
2
9
6 ( 2 .8 7 5 )( 2 0 0  1 0 )( 2 .8 7 7 4 4  1 0
6
)
2
Deflection at D due to 270 N:
D " 
2 7 0 (1 )(1 .8 7 5 )
( 2 .8 7 5  1 .8 7 5  1 )  0 .1 9 1 2 m m
2
9
6 ( 2 .8 7 5 )( 2 0 0  1 0 )( 2 .8 7 7 4 4 1 0
6
)
2
2
Total deflection at D:
 D  0 .1 9 4 8  0 .1 9 1 2  0 .3 8 6 m m
Applying Eq. (9.18), we find
n cr 
1
2
9 .8 1[( 4 5 0 )( 0 .2 9 6 2  1 0
[
4 5 0 ( 0 .2 9 6 2  1 0
3
3
)  ( 2 7 0 )( 0 .3 8 6 1 0
3
2
)  2 7 0 ( 0 .3 8 6  1 0
)
3
)]
2
1
]
2
 2 7 .2 1 c p s  1 6 3 3 r p m
( b ) Apply Eq. (9.21):
n c r ,C n c r , D
n cr 
where
Hence
2
2
n c r ,C  n c r , D
g
n c r ,C 
1
2
C '
n cr , D 
1
2
D "
g


2
( 2234 )  ( 2163)
0 .1 7 9 3  1 0
2
3
9 .8 1
1
2
( 2 2 3 4 )( 2 1 6 3 )
n cr 
 3 7 .2 3 c p s  2 2 3 4 rp m
9 .8 1
1
2
0 .1 9 1 2  1 0
3
 3 6 .0 5 c p s  2 1 6 3 rp m
 1 5 5 4 rp m
SOLUTION (9.31)
Refer to Table A.8 (Case 6) and Example 9.5.
We have
I 
4
d

64
 ( 0 .0 4 6 9 )
4
64
 0 .2 3 7 5  1 0
6
m
4
( a ) Deflection at C owing to 340 N:
C ' 
3 4 0 (1 .1 2 5 )( 2 5 0 .6 2 5 )
(1 .7 5  1 .1 2 5  0 .6 2 5 )  1 .2 8 3 9 m m
2
9
6 (1 .7 5 )(1 0 5  1 0 )( 0 .2 3 7 5  1 0
6
)
2
2
Deflection at C due to 500 N:
C " 
5 0 0 ( 0 .3 7 5 )( 0 .6 2 5 )
(1 .7 5  0 .3 7 5  0 .6 2 5 )  1 .1 3 2 9 m m
2
9
6 (1 .7 5 )(1 0 5  1 0 )( 0 .2 3 7 5  1 0
6
)
2
2
Total deflection at C:
 C  1 .2 8 3 9  1 .1 3 2 9  2 .4 1 6 8 m m
Deflection at D owing to 340 N:
D '
3 4 0 ( 0 .6 2 5 )(1 .7 5  1 .3 7 5 )
9
6 (1 .7 5 )(1 0 5  1 0 )( 0 .2 3 7 5 1 0
[ 2 (1 .7 5 )(1 .3 7 5 )  0 .6 2 5  1 .3 7 5 ]  0 .7 7 0 3 m m
2
6
)
2
Deflection at D due to 500 N:
D " 
5 0 0 ( 0 .3 7 5 )(1 .3 7 5 )
9
(1 .7 5  1 .3 7 5  0 .3 7 5 )  1 .0 1 5 4 m m
2
6 (1 .7 5 )(1 0 5  1 0 )( 0 .2 3 7 5  1 0
6
)
2
2
Total deflection at D:
 D  0 .7 7 0 3  1 .0 1 5 4  1 .7 8 5 7 m m
Applying Eq. (9.18), we obtain
n cr 
1
2
[
9 .8 1 ( 3 4 0  2 .4 1 6 8  1 0
3 4 0  ( 2 .4 1 6 8  1 0
3
3
 5 0 0  1 .0 1 5 4 1 0
2
)  1 1 0  (1 .0 1 5 4 1 0
3
3
)
)
2
1
]
2
 1 1 .4 9 c p s  6 9 0 r p m
(CONT.)
159
9.31 (CONT.)
( b ) From Eq. (9.21), we have
n c r ,C n c r , D
n cr 
where
2
2
n c r ,C  n c r , D
g
n c r ,C 
1
2
C '

n cr , D 
1
2
D "
g
9 .8 1
1
2

1 .2 8 3 9  1 0
3
9 .8 1
1
2
1 .0 1 5 4  1 0
3
 1 5 .9 1 c p s  8 3 5 rp m
 1 5 .6 5 c p s  9 3 9 rp m
It follows that
( 8 3 5 )( 9 3 9 )
n cr 
2
 6 2 4 rp m
2
(835 ) (939 )
SOLUTION (9.32)
Shaft


Sy
a ll
Key


Sy
a ll
 a ll 
n
n
S ys
n

580
2 .3
 2 5 2 .2 M P a

3 0 .8
2 .3

154
2 .3
 1 3 3 .9 1 M P a
 6 6 .9 5 M P a
Torque:
9549 kW
T 
9549 (90 )

n
 9 5 4 .9 N  m
900
Force at the shaft surface is then
F 
(a) L 
(b) L 
(c) L 

F
(w 2)

a ll

F
(w 2)

a ll
F
 a ll ( w )


T
r
9 5 4 .9
1 8 .7 5
 5 0 .9 3 k N
5 0 .9 3  1 0
3
2 5 2 .2 ( 9 .3 5  1 0
3
5 0 .9 3  1 0
2)
3
6
1 3 3 .9 1  1 0 ( 9 .3 5  1 0
5 0 .9 3  1 0
3
6
6 6 .9 5  1 0 ( 9 .3 5  1 0
 0 .0 4 3 2 m = 4 3 .2 m m
3
)
3
2)
 0 .0 8 1 4 m = 8 1 .4 m m
 0 .0 8 1 4 m = 8 1 .4 m m
SOLUTION (9.33)
From Table B.3: For shaft: S y  S y c  4 4 0 M P a For key: S y  S y c  3 9 0 M P a n  2 .
Shaft:
 a ll 
T 
and
F 
0 .5 8 S y
n
 a ll J
r

1 0 .5 7
0 .0 7 5 2

0 .5 8 ( 4 4 0 )
2
 a ll (  D
4
D 2
32 )
 1 2 7 .6 M P a

 (1 2 7 .6 ) ( 0 .0 7 5 )
3
16
 1 0 .5 7
kN  m
 2 8 1 .9 k N
Key length:
Based on bearing on shaft: L 
2 8 1 .9  1 0

F
( S y n )( h 2 )
(
4 4 0 1 0
6
)(
3
9 .3 7 5  1 0
2
Based on bearing on key: L 
F
( S y n )( h 2 )
2 8 1 .9  1 0

(
3 9 0 1 0
6
Based on shear in key: L 
)(
(
)w
n
(
0 .5 8  3 9 0  1 0
2
160
 0 .3 0 8 4 m = 3 0 8 .4 m m
3
)
2
2 8 1 .9  1 0

)
3
9 .3 7 5  1 0
2
F
0 .5 8 S y
 0 .2 7 3 4 m = 2 7 3 .4 m m
3
2
3
6
)(1 8 .7 5  1 0
 0 .1 3 3 m = 1 3 3 m m
3
)
SOLUTION (9.34)
We have F 

T
r
0 .4  1 0
3 7 .5  1 0
 a ll w L
By Eq.(9.23): n 
3
3
6
( 7 0  1 0 ) ( 9 .3 7 5  1 0

2F
 2 1 .3 3 k N
2
3
) ( 7 5 1 0
3
)
3
2 ( 2 1 .3 3  1 0 )
 1 .1 5
SOLUTION (9.35)

Shaft
Sy

a ll

460
4

460 2

340
4
n
S ys
 a ll 
n
 115 M Pa
 5 7 .5 M P a
4
Key

Sy

a ll
n
Sy 2
 a ll 
 85 M Pa
 4 2 .5 M P a
n
Torque in shaft. We have J   D
3 2   (6 0 )
4
4
3 2  1 .2 7 2  1 0
6
mm
4
Hence,
 a ll J
T 
r
5 7 .5 (1 .2 7 2 )

 2 .4 4 k N  m
0 .0 3
Force at the shaft surface:
F 
(a) L 
(b) L 
(c) L 

F
(w 2)

a ll

F
(w 2)

a ll
F

 a ll ( w 2 )

T
r
2 .4 4
0 .0 3
 2 .4 4 k N  m
 7 0 .7 m m
81300
6
1 1 5 (1 0 )( 0 .0 1 )
 9 5 .6 m m
81300
6
8 5 (1 0 )( 0 .0 1 )
81300
6
4 2 .5 (1 0 )( 0 .0 2 )
 9 5 .6 m m
SOLUTION (9.36)
Shear force in each bolt:
F 
T
RN
Average shear stress in each bolt,
 a ll 
Sy
n

F
A

T RN
 db 4

4T
2
 db RN
Solving
db 
4Tn
(P9.36)
 RNS y
Introducing the given data:
db  [
3
4 ( 5  1 0 )(1 .2 )
1
6
 ( 0 .0 8 )( 6 )( 2 6 0 )(1 0 )
]
2
 7 .8 2 m m
161
SOLUTION (9.37)
From Eq. (1.15),
9549 kW
T 

n
9549 (30 )
 2 .3 8 7 k N  m
120
Force at shaft surface
F 
T
r

 
F
L(w 2)
 7 9 .6 k N
2 .3 8 7
0 .0 3
( a ) Key
 
7 5 ( 5 )(1 0
6
3
7 9 .6 (1 0 )

F
L(w)
3
7 9 .6 (1 0 )

7 5 (1 0 )(1 0
6
)
 212 M Pa;
)
n 
 106 M Pa;
n 
S ys

Sy


 1 .9 8
420
212

210
106
 1 .9 8

210
3 1 .3
 6 .7 1

420
2 4 .6
 1 7 .1
( b ) Bolts
Force at bolt circle
F 
T
Db 2

2 .3 8 7
0 .0 7 2
 3 3 .1 5 k N
Sear Stress in bolts
 
F
2
6 ( d b
4)

 3 1 .3 M P a ;
n 
S ys
 2 4 .6 M P a ;
n 
Sy
3 3 .1 5 ( 4 )
6  ( 0 .0 1 5 )
2

( c ) Bearing on bolts in flange
 
F
6 d b Ft

3 3 .1 5
6 (1 5 )(1 5 )1 0
6

SOLUTION (9.38)
(a)
T 
9549 kW
n
F 
T
r

9549 ( 45 )

200
2148
0 .0 2 5
 2 .1 4 8 k N  m
 8 5 .9 2 k N
3
3
Area in shear in key  (1 4 .0 6 2 5  1 0 )(8 7 .5  1 0 )  1 .2 3  1 0
3
3
3
m
2
Area in bearing for key  [(1 4 .0 6 2 5 2 )  1 0 ](8 7 .5  1 0 )  6 .1 5  1 0
4
m
2
Shear stress in key:
 
8 5 .9 2  1 0
1 .2 3  1 0
3
3
 6 9 .8 5 M P a
Bearing stress in key:  
(b)
8 5 .9 2  1 0
6 .1 5  1 0
3
4
 1 3 9 .7 M P a
Area in shear for bolts  6 ( 4 )(9 .3 7 5 )  4 1 4 .1 7 m m
2
2
Force at bolt circle:
F 
2 .1 4 8
0 .0 7 5
 2 8 .6 k N
Shear stress in bolts:  
28600
4 1 4 .1 7
 6 9 .0 5 M P a
(CONT.)
162
9.38 (CONT.)
( c ) Area in bearing for bolts  6 (9 .3 7 5 )( 2 1 .8 7 5 )  1 2 3 0 .5 m m
2
Bearing stress in bolts:
 
28600
1 2 3 0 .5 (1 0
6
)
 2 3 .2 M P a
( d ) Area in shear at edge of hub   D h t f   (1 0 0 )( 2 1 .8 7 5 )  6 8 7 2 .2 m m
Force at edge of hub: F 
Shear stress in web:  
T
Dh 2

4 2 .9 6 0  1 0
6 8 7 2 .2 (1 0
2148
0 .0 5
3
6
)
2
 4 2 .9 6 k N
 6 .2 5 M P a
End of Chapter 9
163
CHAPTER 10
BEARINGS AND LUBRICATION
SOLUTION (10.1)
We have
H  9 .8 1( 6 5 )  6 3 8 M P a
Equation (8.3') will be used.
 
KWl
HAp
(1)
where
A p  D L  2 4 (1 2 )  2 8 8 m m
2
l  n  D t  1 8  ( 2 4 )( 6 0  1 2 0 0 )  9 7 .7  1 0
( a ) Good Lub: K  2  1 0
 
( 2 1 0
6
6
6
mm
(Table 8.3), Equation (1) gives:
6
)(1 5 0 )( 9 7 .7  1 0 )
6
6
( 6 3 8  1 0 )( 2 8 8  1 0
)
 0 .1 6 m m
( b ) Excellent Lub: K  1  1 0
 
(1  1 0
7
7
(Table 8.3), Equation (1) is then
6
)(1 5 0 )( 9 7 .7  1 0 )
6
( 6 3 8  1 0 )( 2 8 8  1 0
6
)
 0 .0 0 8 m m
SOLUTION (10.2)
Refer to solution of Prob. 10.1.
H  9 .8 1(1 0 5 )  1 0 3 0 M P a
(Table B.6 and Sec. 8.5)
Use Eq. (8.3'),
 
KWl
HAp
(1)
Here
A p  D L  ( 2 5 )( 2 5 )  6 2 5 m m
2
l  n  D t  1 8  ( 2 5 )( 6 0  1 5 0 0 )  1 2 7 .2  1 0
( a ) Good Lub: K  2  1 0
 
( 2 1 0
6
6
6
(Table 8.3), Then, Eq. (1) result in
6
)(1 1 5 )(1 2 7 .2  1 0 )
6
(1 0 3 0  1 0 )( 6 2 5  1 0
6
)
 0 .0 4 5 m m
( b ) Excellent Lub: K  1  1 0
 
(1  1 0
7
7
(Table 8.3), Equation (1) gives
6
)(1 1 5 )(1 2 7 .2  1 0 )
6
(1 0 3 0  1 0 )( 6 2 5  1 0
6
)
 0 .0 0 2 m m
164
mm
SOLUTION (10.3)
We have
H  9 .8 1( 6 0 )  5 8 8 .6 M P a
Equation (8.3) will be used,
  KH WA l
(1)
a
where
  0 .1 5 m m ,
7
K  1 10
W  450 N
(Table 8.3)
l  n  D t  n  ( 2 5 )( 6 0  2 4  3 6 5  1 .2 )  4 9 .5 4  1 0 n m m
6
A p  ( 2 5 )( 2 5 )  6 2 5 m m
2
Equation (1) is thus
(1  1 0
0 .1 5 
7
6
)( 4 5 0 )( 4 9 .5 4  1 0 n )
6
( 5 8 8 .6  1 0 )( 6 2 5  1 0
6
)
Solving,
n  2 4 .8 r p m
SOLUTION (10.4)
Equation (10.10):
Tf 

159 kW
n
1 5 9 (1 .5 )
1200 60
 1 1 .9 N  m
Equation (10.7):
Tfc
 
4
2
3
Lr n

11 . 9 ( 0 . 09 ) 10
4
t  80
From Fig.10.8:
o
2
3
3
( 0 . 22 )( 0 . 08 ) ( 20 )
 12 . 04 mPa  s
C
SOLUTION (10.5)
(a) Tf 
2
f 

c
Tf n
( b ) kW 
(c)
3
4  Lr n
159
Tf

Wr

4
2
( 4 .1 4  1 0
3
3
)( 0 .1 )( 0 .0 3 7 5 ) ( 2 4 0 0 0 6 0 )
0 .0 7 5 (1 0
4 .6 ( 4 0 0 )
159
4 .6
2 2 5 0 ( 0 .0 3 7 5 )
3
)
 4 .6 4 N  m
 1 1 .6
 0 .0 5 5
SOLUTION (10.6)
T
f
(a)  
 fWr  0 . 01 ( 8000 )( 0 . 06 )  4 . 8 N  m
Tfc
4
2
3
Lr n

4 . 8 ( 0 . 05 ) 10
4
2
3
3
( 0 . 12 )( 0 . 06 ) ( 10 )
 23 . 45 mPa  s
( b ) Figure 10.8: SAE 40 oil
165
SOLUTION (10.7)
P 
W
DL

12 ( 10
3
)
 0 . 768
( 0 . 125 )( 0 . 125 )
f 
MPa,
F
W

80
12  10
3
 0 . 667 (10
2
)
Equation (10.9):
 
fP
2
2
0 . 667 ( 10
( cr ) 
n
2
)( 0 . 768 )( 10
2
2
(4)
6
)
( 0 . 0004 )  25 . 95 mPa  s
SOLUTION (10.8)
From Fig. 10.8:   1 1 m P a  s
(a) P 
W
DL

 400
2250
( 0 .1 5 )( 0 .0 3 7 5 )
n 
kPa
1500
60
 25
rp s
Equation (10.9):
2 n r
P c
f  2
 2
2 11 ( 10
3
)( 25 )
400  10
3
1
0 . 001
 0 . 014
( b ) T f  fW D 2  ( 0 .0 1 4 )( 2 .2 5 )( 7 5 )  2 .3 6
Tf n
kW 
2 .3 6 ( 2 5 )

159
159
N m
 0 .3 7
SOLUTION (10.9)
( a ) From Eq. (10.10):
Tf 

159 kW
n
1 5 9 (1 .4 )
2100 60
 6 .3 6 N  m
Equation (10.7) gives then
 
Tfc
2

3
4 Lr n
( 6 .3 6 )( 0 .1 7 5 )
4
2
3
( 0 .2 )( 0 .0 7 5 ) ( 3 5 )
 9 .5 5 m P a  s
Using Fig. 10.8:
t  90 C
o
( b ) Equation (10.8) results in
f 
Tf

Wr
6 .3 6
2 (75 )
 0 .0 4 2
SOLUTION (10.10)
Figure 10.8:   8 .4 m P a  s ,
Figure 10.14:
h0
c

0 . 025
0 . 05
 0 .5
 10
6 8 .4 (1 0
c  0 .0 0 1 r  0 .0 0 1(5 0 )  0 .0 0 5 m m ,
 S  0 . 52
Equation (10.17):
S  ( cr )
2 n
P
3
P
)( 3 0 )
 0 .5 2 ,
and
W  P D L  4 8 4 .6 ( 0 .1) ( 0 .0 5 )  2 .4 2 k N
166
P  4 8 4 .6 k P a
L D 1 2
SOLUTION (10.11)
P 
W
DL
3
1 2 (1 0 )

 0 .7 6 8
( 0 .1 2 5 )( 0 .1 2 5 )
f 
M Pa,
F
W

80
3
1 2 (1 0 )
 0 .6 6 7 (1 0
Thus
f 
r
c
( 0 .6 6 7  1 0
1
0 .0 0 0 4
2
)  1 6 .7
From Fig. 10.15: S  0 .8 .
The viscosity is therefore
 

S
P
n ( r c )2
0 .8 ( 0 .7 6 8 )1 0
4 (1 0 .0 0 0 4 )
6
2
 0 .0 2 4 5 8
P a  s  2 4 .5
mPa s
SOLUTION (10.12)
( a ) From Solution of Prob.10.8:
  1 1 m P a  s , P  4 0 0 k P a , n  2 5 rp s , L D  1 4
We thus have
S  ( cr )
2 n
P
1
 ( 0 . 001
)
2 11 ( 10
3
)( 25 )
 0 . 69
3
400 ( 10 )
From Fig. 10.15:
f  20,
r
c
( b ) Thus
fW D
Tf 
2
Tfn
kW 
159
f  2 0 ( r )  2 0 ( 0 .0 0 1 )  0 .0 2
c
 0 .0 2 ( 2 .2 5 )( 0 .0 7 5 )  3 .3 7 5 N  m

3 . 375 ( 25 )
159
 0 . 531
SOLUTION (10.13)
( a ) We have
P 
W
DL

1 .4 5
( 0 .0 3 )( 0 .0 3 )
 1 .6 1 M P a
L D  30 30  1
Somerfield number:
S  ( cr ) (
2
n
P
)  ( 01.05 3 ) [
2
2 0 .7 (1 0
3
)( 4 0 )
6
1 .6 1 (1 0 )
]  0 .1 2 9
Then, Fig. 10.14 gives, h o c  0 .4 2 and   1  h o c  0 .5 8 .
It follows that
e   c  0 .5 8 ( 0 .0 3 )  0 .0 1 7 4 m m
( b ) Figure 10.15. With S  0 .1 2 9 , L D  1;
( r c ) f  3 .5 .
So,
f  3 .5 ( c r )  3 .5 ( 0 .0 3 ) (1 5 )  0 .0 0 7 N  m
Friction torque,
T  fW ( D 2 )  0 .0 0 7 (1 .4 5 )(1 5 )  0 .1 5 2 N  m
Thus,
kW 
Tn
159

0 .1 5 2 ( 4 0 )
159
 0 .0 3 8 2
( c ) Figure 10.16 ( S  0 .1 2 9 with L D  1 ) gives
P
P
 0 .4 2
m ax
or
Pm a x  P 0 .4 2  1 .6 1(1 0 ) 0 .4 2  3 .8 3 M P a
6
167
2
),
L D 1
SOLUTION (10.14)
P 

W
DL
3
1 5 (1 0 )
 2 .0 8 3 M P a
6
(1 2 0 )( 6 0 )1 0
Figure 10.8:   1 6 m P a  s
Apply Eq. (10.17):
n
S  ( cr ) (
2
P
o
(for SAE 40 oil at 8 0 C )
)
or
0 .1 5  (
120 2
c
)
2 (1 6  1 0
3
)(1 5 0 0 6 0 )
6
2 .0 8 3 (1 0 )
Solving,
c  0 .0 6 7 9 m m
It follows that
c r  0 .0 6 7 9 6 0  0 .0 0 1 1
OK (See Sec. 10.8)
Figure 10.14:
(for S  0 .1 5 and L D  1 2 )
h o c  0 .2 8
from which
h o  0 .2 8 ( 0 .0 6 7 9 )  0 .0 1 9 m m
SOLUTION (10.15)
h0
c

S  ( cr )
Figure 10.14: S  0 . 13 ,
 0 .4 ,
0 . 0025
0 . 00625
2 n
P
 ( 0 .05006 2 5 )
( a ) Fig.10.8:
t  80 C
( b ) Fig.10.15:
r
c
2  (900 60 )
3
7 0 0 (1 0 )
,
L D 1
  9 .4 5 m P a  s
o
f  3 , f  3 ( 0 . 00625
)  0 . 00375
50
( c ) T f  fW r  0 .0 0 3 7 5 ( 7 0 0  0 .1  0 .1)( 0 .5 )  1 3 .1 N  m
kW 
Tfn
1050

13 . 1 ( 15 )
159
 1 . 24
SOLUTION (10.16)
P 
W
DL
S  ( cr )
( a ) Figure 10.14:
( b ) Figure 10.15:
T
f

1500
0 .0 2 5  0 .0 2 5
2 n
P
h0
c
r
c
 2 .4 M P a ,
 ( 0 .010 0 8 )
2 5 0 (1 0
3
6
Tf n
159

f  1 1,
0 .1 6 5 (1 6 . 6 7 )
159
 0 .5 4 3,
 0 . 75 ; h 0  0 . 75 ( 0 . 01 )  0 . 008
2 )  0 . 165
 0 .0 1 7
168
L D 1
c  0 .0 1 m m
mm
f  1 1( 0 .0 0 0 8 )  0 .0 0 8 8
 fWr  ( 0 . 0088 )( 1500 )( 0 . 025
kW 
)(1 6 .6 7 )
2 .4 (1 0 )
n  1 6 .6 7 rp s ,
N m
SOLUTION (10.17)
P 
W
DL

 1 .2 5 M P a ,
4000
0 .0 8  0 .0 4
S  (c)
r
( a ) Figure 10.14:
h0
( b ) Figure 10.16:
P
p m ax
c
c  0 .0 8
rp s ,
mm,
L D 1 2
  30 mPa  s
Fig.10.8:
Thus
n  10
2 n
P
 ( 0 .0 0 2 )
1
 0 .1 6 ,
2 3 0 (1 0
3
)(1 0 )
6
1 . 2 5 (1 0 )
 0 .0 6
h 0  0 .1 6 ( 0 .0 8 )  0 .0 1 3 m m
 0 .2 6 ,
p m ax 
1 .2 5
0 .2 6
 4 .8 0 8 M P a
SOLUTION (10.18)
  8 mPa s
Figure 10.8:
T 
119 hp

n
Try S=0.03:
Fig.10.15:
1 1 9 ( 0 .1 6 )
2 7 .2
 0 .7 N  m ,
f  1 . 45 ,
r
c
n  1 6 3 0 6 0  2 7 .2
F 
T
r

0 .7
0 .0 2 5
rp s ,
L D 1
 28 N
f  1 . 45 ( 0 . 001 )  0 . 00145
From Eq.(10.17):
P  ( cr )
Thus
2 n
S
2 8 1 0
 ( 0 .01 0 1 )
3
( 2 7 .2 )
0 .0 3
 7 .2 5 M P a
W  P D L  7 .2 5 (1 0 )( 0 .0 5 )( 0 .0 5 )  1 8 .1 3 k N
6
F  fW  0 .0 0 1 4 5 (1 8 .1 3  1 0 )  2 6 .3 N
3
Since
28 > 26.3, assumption S=0.03 is incorrect.
Try S=0.025:
Fig.10.15:
P  ( 0 .01 0 1 )
r
c
f  1.3 ,
2 ( 8 1 0
3
)( 2 7 .2 )
0 .0 2 5
f  1.3 ( 0 .0 0 1 )  0 .0 0 1 3
 8 .7 M P a
W  P D L  8 .7 (1 0 )( 0 .0 5 )( 0 .0 5 )  2 1 .8
6
kN
F  fW  0 .0 0 1 3( 2 1 .8  1 0 )  2 8 .3 N ,
3
Assumption is correct.
( a ) W  2 1 .8 k N
( b ) Fig.10.14:
h0
c
 0 . 13 , h 0  0 . 13 ( 0 . 001  25 )  0 . 0033
( c ) Equation (10.20):  1 
h0
c
mm
 1  0 . 13  0 . 87
SOLUTION (10.19)
h0
c  0 .0 0 1 5 (5 0 )  0 .0 7 5 m m ,
Figure 10.14:
P 
c

0 .0 2 5
0 .0 7 5

1
3
,
L D 1 2
S=0.022
W
DL

8000
( 0 . 1 )( 0 . 05 )
 1 . 6 MPa
n  900
60  15 rps
(CONT.)
169
10.19 (CONT.)
( a ) Equation (10.17):
 
SP
n
0 . 22 ( 1 . 6  10
6
)
15
f  6 .5 ,
r
c
( b ) Figure 10.15:

2
( cr )
( 0 . 0015 )
kW 
( 80  0 . 050 ) 15

159
 52 . 8 mPa  s
f  6 .5 ( 0 .0 0 1 5 )  0 .0 1
F  fW  0 . 01 ( 8000 )  80
Tfn
2
N
 0 . 377
159
SOLUTION (10.20)
P 
W
DL

 1 .2 M P a ,
6000
0 .1 0 .0 5
r  50 m m ,
From Sec. 10.10: A  1 2 .5 D L  1 2 .5 (1 0 0  5 0 )  1 0
Table 10.3: C  7 . 4 watts
Equation (10.17):
S  ( cr )
2 n
P
1
0 .0 0 1
)
L D 1 2
 0 .0 6 2 5
2
m ,
 C
o
2 ( 2 0 1 0
3
)( 5 )
6
1.2 (1 0 )
f  3.2 ,
r
c
Figure 10.15:
 (
m
2
6
 0 .0 8 3
f  3 .2 ( 0 .0 0 1 )  0 . 0 0 3 2
Through the use of Eq.(10.24), we have
H  fW ( 2  r n )  0 .0 0 3 2 ( 6 , 0 0 0 ) ( 2   0 .0 5  5 )  3 0 .1 6 w a tts
Thus, by Eq.(10.23):
t0  ta 
 20 
H
CA
30 . 16
7 . 4 ( 0 . 0625 )
 20  65 . 2  85 . 2
o
C
SOLUTION (10.21)
Refer to Solution of Prob.10.20. We have
C  8 . 5 watts
m
Using Eq.(10.23): t 0  t a 
2
 C (Table 10.3)
o
 30 
H
CA
30 . 16
8 . 5 ( 0 . 0625 )
 30  56 . 8  86 . 8
o
SOLUTION (10.22)
C  14 kN
Table 10.5:
Fa
Table 10.7:
V Fr

3
1.2 ( 2 )
C s  6 . 95 kN
Fa
 1.2 5  e
X  0 .5 6
Cs

n  1500
3
6 .9 5
 0 .4 3 2
Y  1.0 3 7 (by interpolation)
Thus, we obtain
P  XVF
r
 YF a  0 . 56 (1 . 2 )( 2 )  1 . 037 ( 3 )  4 . 455
P  VF
 1 . 2 ( 2 )  2 . 4 kN
or
r
Hence
L1 0 
6
10
60 n
( CP )
a

6
10
6 0 (1 5 0 0 )
( 4 .41 45 5 )  3 4 4 .8 h
3
170
kN
C
n  5 rp s
SOLUTION (10.23)
Table 10.5:
C  1 4 .8 k N
Table 10.8:
V Fr
Fa
C s  7 .6 5
 1.2 5  e  0 .9 5 ;
X  0 .3 7 ,
Y  0 .6 6
We have then
P  XVF
r
 YF a  0 . 37 (1 . 2 )( 2 )  0 . 66 ( 3 )  2 . 868
P  VF
 1 . 2 ( 2 )  2 . 4 kN
kN
or
r
Thus
L1 0 
6
10
60 n
( CP )
a

6
10
6 0 (1 5 0 0 )
( 21.84 .86 8 )  1 5 2 6 .9
3
h
SOLUTION (10.24)
Refer to Solution of Prob. 10.22:
C  14 kN ,
C s  6 .9 5
Fa
kN ,
Cs
 0 .4 3 2
Table 10.7:
Fa
VF

r
3
1( 2 )
 1 .5  e ;
X  0 . 56
Y  1 . 037
(by interpolation)
It follows that
P  K s [ X V F r  Y F a ]  1 .5[( 0 .5 6 )1( 2 )  1 .0 3 7 (3)]  6 .3 4 7 k N
or
P  K s VF r  (1 . 5 )1 ( 2 )  3 kN
Thus
L1 0 
6
10
60 n
( CP )
a

6
10
6 0 (1 5 0 0 )
( 6 .31 44 7 )  1 1 9 .2
3
h
SOLUTION (10.25)
Table 10.5:
C  5 5 .9 k N
C s  3 5 .5 k N
Table 10.8:
Fa
V Fr

6 .7 5
(1 )( 2 2 .5 )
iF a
 0 .3  e ,
Cs

1 ( 6 .7 5 )
3 5 .5
 0 .1 9
and
X  1,
Y  0 .9 2
Thus
P  X V F r  Y F a  1(1)( 2 2 .5 )  ( 0 .9 2 )( 6 .7 5 )  2 9 k N
or
P  V F r  (1)( 2 2 .5 )  2 2 .5 k N
Hence
L1 0 
6
10
60 n
( CP ) 
3
6
10
60 (700 )
( 5259.9 )  1 7 0 .5 h
3
171
SOLUTION (10.26)
Table 10.5: C  1 4 k N
C s  6 .9 5 k N
Table 10.7:
Fa
Cs

Fa
 0 .2 9  e ,
2
6 .9 5
V Fr

2
(1 )( 4 )
 0 .5  e
and
X  0 .5 6 ,
Y  1 .1 4 2
(by interpolation)
Hence
P  X V F r  Y F a  0 .5 6 (1)( 4 )  (1 .1 4 2 )( 2 )
 4 .5 2 k N
or
P  V F r  (1)( 4 )  4 k N
Thus
L1 0 
6
10
60 n
( CP )
a

6
10
60 (3500 )
( 41.542 )  1 4 1 .5 h
3
From Fig. 10.26: K r  0 .6 2 .
So
L 5  K r L1 0  0 .6 2 (1 4 1 .5 )  8 7 .7 h
SOLUTION (10.27)
Variation factor: V  1 (Eq. 10.25)
C  1 9 .5 k N
Table 10.5:
C s  10 kN
Table 10.7:
Fa
Cs

1 .7
10
 0 .1 7 ,
Fa
e  0 .3 4 ;
V Fr

1 .7
(1 )( 4 .5 )
 0 .3 8  e
and
X  0 .5 6 ,
Y  1 .3 1
Thus
P  X V F r  Y F a  0 .5 6 (1)( 4 .5 )  (1 .3 1)(1 .7 )  4 .7 4 7 k N
P  V F r  (1)( 4 .5 )  4 .5 k N
or
It follows that
L1 0 
6
10
60 n
( CP )
a

6
10
60 (600 )
( 41.79 .54 7 )  1 9 2 6 h
3
SOLUTION (10.28)
K s  2 .5
Table 10.9:
V  1 .2
and
Fa
V Fr

1 .7
(1 .2 )( 4 .5 )
 0 .3 1 5  e ;
X  0 .5 6 ,
Y  1 .3 1
Therefore, Eq. (10.27):
P  K s ( X V F r  Y F a )  2 .5[( 0 .5 6 )(1 .2 )( 4 .5 )  (1 .3 1)(1 .7 )]
 1 3 .1 3 k N
(CONT.)
172
10.28 (CONT.)
or
P  K sV F r  2 .5 (1 .2 )( 4 .5 )  1 3 .5 k N
Hence
L1 0 
6
10
60 n

a
( CP )
6
10
60 (600 )
( 11 39 .5.5 )  8 3 .7 h
3
From Fig. 10.26: K r  0 .6 2 . So,
L 5  K r L1 0  0 .7 (8 3 .7 )  5 8 .6 h
SOLUTION (10.29)
P2 
3
3
L '10 P1
2 L " 10
 0 . 5 P1 ,
P 2  0 . 794 P1
3
 20 . 6 %
SOLUTION (10.30)
10
10 3
P2

L '10 P1 3
2 L " 10
10
 0 . 5 P1 3 ,
P 2  0 . 812 P1
18 . 8 %
SOLUTION (10.31)
Table 10.5:
C  5 5 .9 k N ,
Table 10.8:
VF
Fa

1 .5
1 .2 ( 5 )
r
C s  3 5 .5 k N
iF a
 0 . 25  e ,
Cs

2 ( 1 .5 )
35 . 5
 0 . 085
and
X  1,
Y  1.3 8 6 (by interpolation)
Then we obtain
P  XVF
r
 YF a  1(1 . 2 )( 5 )  1 . 386 (1 . 5 )  8 . 079
or
P  V F r  1 .2 (5 )  6 k N
Thus
L1 0 
6
10
60 n
( CP )
a

6
10
6 0 (1 0 0 0 )
( 85.05 7.99 )  5 5 2 1 h
3
and
5 L1 0  5 (5 5 2 1)  2 7 , 6 0 5 h
SOLUTION (10.32)
P  K s [V X F r  0 ]  1 .7[1 .2 (1)5 ]  1 0 .2
35 mm-02-series (Table 10.6):
L1 0 
6
10
60 n
( CP ) 
a
6
10
60 ( 2400 )
10
[ 13 01 .2.9 ] 3  3 1 0 .6 5 h
35 mm-03-series (Table 10.6):
L1 0 
6
10
60 ( 2400 )
10
[ 14 04 .2.6 ] 3  9 4 9 .3 h
173
kN
kN
SOLUTION (10.33)
P  K sV F r  1(1)( 2 5 )  2 5
(Eq.10.26)
kN
75 mm-02-series (Table 10.6):
L1 0 
6
10
60 n
( CP )
a
6

10
10
60 ( 2000 )
[ 9215.3 ] 3  6 2 5 .1 h
and 5 L1 0  3,1 2 5 h r
75 mm-03-series (Table 10.6):
L1 0 
10
6
10
60 ( 2000 )
[ 12853 ] 3  6 3 4 6 h
and 5 L1 0  3 1, 7 3 0 h
SOLUTION (10.34)
P  K sV F r  2 (1  1  1 2 .5 )  2 5
(Eq.10.26)
kN
We have
L1 0 
6
10
60 n
6
24 
a
( CP ) :
02-series (Table 10.6):
03-series (Table 10.6):
10
60 ( 2400 )
3
[ 2C4 ] ,
C  3 6 .2 8 6 k N
40-mm-bore bearing
30-mm-bore bearing
SOLUTION (10.35)
Fa
Table 10.7:
Cs
 0 , use 0.014: X  1 ,
Y  0
We have, by Eq.(10.27),
P  K s XVF
r
 2 . 5 (1)( 1 . 2 ) 8  24 kN
Thus
L 10 
6
10
60 n
( CP )
From Table 10.5:
a
40
5
:

6
10
60 ( 900 )
C
[ 24
] , C  18 . 1 kN
3
30-mm-bore bearing
SOLUTION (10.36)
Refer to Solution of Prob.10.22: C  14 kN ,
Figure 10.26: K r  0 .6 2 .
P  4 . 455
We now have n  1200
rpm .
Hence
L5  K r
6
10
60 n
( CP )  0 .6 2
3
6
10
6 0 (1 2 0 0 )
( 4 .41 45 5 )  2 6 7 h
3
174
kN
SOLUTION (10.37)
Figure 10.26:
K
r
 0 .3 2 ,
Refer to Solution of Prob.10.32:
35 mm-02-series:
L 5  K r L1 0  0 .3 2 (3 1 0 .6 5 )  9 9 .4 1 h
35 mm-03-series:
L 5  K r L1 0  0 .3 2 (9 4 9 .3)  3 0 3 .8 h
End of Chapter 10
175
CHAPTER 11
SPUR GEARS
SOLUTION (11.1)
Equation (11.4): m 
d
N
d  6 (3 2 )  1 9 2 m m ,
,
r 
d
2
 96 m m
Table 11.1:
a  m  6 mm ,
h  2 .2 5 ( 6 )  1 3 .5 m m ,
hk  2 ( 6 )  1 2 m m
Equation (11.9) of Sec.11.4:
rb  r c o s   9 6 c o s 2 0
 9 0 .2 1 1 m m ,
o
r0  r  a  9 6 .2 1 m m
SOLUTION (11.2)
N1
Equation (11.8): r s 
N
2

or N 1 
1
3
N
2
3
Equations (11.6) and (11.5b):
N1 N 2
c  r1  r2 

2P
m
2
(N1  N 2) 
3
2
(
N
2
3
 N2) 
3
2
(
4
3
N 2 )  360
or
N
 1 8 0 teeth.
2
Thus N 1 
180
3
 6 0 teeth
SOLUTION (11.3)
( a ) Pitch circle is
P   m  1 0   3 1 .4 2 m m
Pitch diameters are
d
 N
p
p
 1 8 (1 0 )  1 8 0 m m ,
d g  4 2 (1 0 )  4 2 0 m m
The center distance:
d
c 
p
dg
2

1
2
(1 8 0  4 2 0 )  3 0 0 m m
Base circle are
Pb p  r p c o s   9 0 c o s 2 0
Pb g  2 1 0 c o s 2 0
(b) c 
d
p
dg
o
o
 8 4 .5 7 m m
 1 9 7 .3 4 m m
 300  8  308 m m
2
(1)
The velocity ratio does not change. Hence
d
p
dg

18
42
(2)
Solving Eqs. (1) and (2),
d
p
 1 8 4 .8 2 m m
d g  4 3 1 .1 9 m m
Since rb  r c o s  , the new pressure angle is
 new  c o s
1
rb p
d
p
2
 cos
1
8 4 .5 7
1 8 4 .8 2 2
 2 3 .7 7
176
o
SOLUTION (11.4)
Addendum and circular pitch:
a  1 m  1 (4)  4 m m
(Table 11.1)
p   m   ( 4 )  1 2 .5 5 6 m m ; (Eq. 11.5a)
Tooth thickness,
t  1 .1 5 7 1 m  1 .5 7 1( 4 )  6 .2 8 4 m m
(Table 11.1)
SOLUTION (11.5)
Refer to Table 11.1;
m 1 P :
with
Dedendum,
b d  1 .2 5 m  1 .2 5 (1 2 )  1 5 m m
Clearance,
f  0 .2 5 m  0 .2 5 (1 2 )  3 m m
h k  2 m  2 (1 2 )  2 4 m m
Working depth,
t  1 .5 7 1 m  1 .5 7 1(1 2 )  1 8 .8 5 2 0 m m
Thickness,
Circular pitch, by Eq. (11.3):
p   m   (1 2 )  3 7 .6 9 9 1 m m
SOLUTION (11.6)
d g  4d
p
c 
p
(Eq. 11.8)
We have
1
2
(d
 d g )  2 0 0;
5d
p
 400
or
and
m  d
p
N
 4N
p
p
,
N
p
 d
p
m  80 5  16
Thus,
N
g
 64
SOLUTION (11.7)
Equation (11.8),
rs 
N
p
N
g

1200
2400

1
2
,
N
g
 2N
p
 2 (1 8 )  3 6
Equation (11.4),
d
p
 N p m  1 8 (3 )  5 4 m m
and
c 
1
2
(d
p
 dg) 
1
2
(5 4  3 6 )  4 5 m m
Equation (11.5a) gives
p   m   ( 3 )  9 .4 2 4 8 m m
177
d
p
 80 m m
SOLUTION (11.8)
rs 
Equation (11.8):
m 
d
p


p


0 . 5234
 mN
p
N
p
N
g

1
4
N

p
84
,
N
p
 2 1 teeth
 4
13 . 1942

 4 ( 2 1)  8 4 m m ,
dg  mN
p
 4 (8 4 )  3 3 6 m m
Therefore
c 
1
2
(d g  d p ) 
1
2
(3 3 6  8 4 )  2 1 0 m m
SOLUTION (11.9)
( a ) From Eq. (11.4), pitch diameters are
d g  N g m  6 0 (8 )  4 8 0 m m
d
p
 2 4 (8 )  1 9 2 m m
The center distance
c 
1
2
(d
p
 dg) 
1
2
(1 9 2  4 8 0 )  3 3 6 m m
The base circles are
rb p 
rb g 
d
p
2
480
2
cos  
cos 20
192
2
o
cos 20
o
 9 0 .2 1 m m
 2 2 5 .5 2 m m
( b ) When center distance is increased by 6 mm, we have c=342.
This corresponds to a 1.79% increased in center distance. However,
d
p
 d g  2 (3 4 2 )  6 8 4 m m
(1)
and
N
p
d
p

N
g
dg
;

24
dp
60
dg
(2)
Solving Eqs. (1) and (2):
d g  4 8 8 .6 m m
d
p
 1 9 5 .4 m m
We have
m 
d
p
N
p

1 9 5 .4
24
 8 .1 4 2 m m
Circular pithch
p   m  8 .1 4 2   2 5 .5 8 m m
Changing center distance does not affect the base circle. Thus,
 new  c o s
1
(
rb p
rp
)  cos
1
( 1 99 50 .4.2 12 )  2 2 .6
178
o
SOLUTION (11.10)
Equation (11.8): rs 
N
p
N
g

1
4
N
 4N
g
p
 4 ( 22 )  88 teeth
Equation (11.4):
d g  N g m  88(4)  352
d
p
mm
 N p m  22 ( 4 )  88 mm
Hence,
c 
(3 5 2  8 8 )  2 2 0
1
2
mm
SOLUTION (11.11)
From Eq. (11.5b), the modulus is
1
m

N
p
d
p
or d p  N p m  2 5 (3 )  7 5 m m
The speed at the contact point, V p  V g  V :
V  w p rb p  w p
d
p
2
cos 20
o
 ( 3 46 00 0 ) 2  ( 725 ) c o s 2 0
o
 1 2 , 5 4 7 m m s  1 2 .5 5 m s
SOLUTION (11.12)
From Eq. (11.5a), the circular pitch,
p   m   ( 2 )  6 .2 8 m m
By Eqs. (11.5b) and (11.6), the center distance:
c 
1
2
(N
p
 N g )m 
1
2
(3 0  1 0 0 )( 2 )  1 3 0 m m
The pitch circular radii are
rp 
1
2
N pm 
1
2
(3 0 )( 2 )  3 0 m m
rg 
1
2
N gm 
1
2
(1 0 0 )( 2 )  1 0 0 m m
Using Eq. (11.9), The base radii:
rb p  r p c o s   3 0 c o s 2 0
o
 2 8 .1 9 m m
rb g  rg c o s   1 0 0 c o s 2 0  9 3 .9 7 m m
o
The addendum is equal to a p 
1
p
 m  2 mm
(CONT.)
179
11.12 (CONT.)
The order radii are therefore
ro p  r p  a  3 0  2  3 2 m m
ro g  1 0 0  2  1 0 2 m m
and
ro p  ro g  1 3 4
SOLUTION (11.13)
Refer to Solution of Prob. 11.9. We have
p   m   (8 )  2 5 .1 3 3 m m ,
rg  4 8 0 2  2 4 0 m m ,
rb p  9 0 .2 1 m m ,
  20
o
rp  1 9 2 2  9 6 m m
rb g  2 2 5 , 5 2 m m ,
c  336 m m
From Table 11.1, the addendum for gears are
a  1 p  8 mm
The outside radii are then
ro p  r p  a  9 6  8  1 0 4 m m
ro g  2 4 0  8  2 4 8 m m
Substituting the data into Eq. (11.14), the contact ratio:
Cr 
[ 1 0 4  9 0 .2 1 
2
1
2 5 .1 3 3 c o s 2 0
o
2
 1 .6 9 4
SOLUTION (11.14)
We have
e  (
N1
N3
N
)( N 3 )  
4
24
96
  0 .2 5
Substituting into Eq. (11.18)
 0 .2 5 
0  n2
120  n2
,
n 2  2 4 rp m
Direction is the same as that of the sun gear.
180
2 4 8  2 2 5 .5 2 ] 
2
2
3 3 6 ta n 2 0
2 5 .1 3 3
o
SOLUTION (11.15)
kW 
Tn
9549
T1 
,
22 (9549 )
d 1  2 7 (8 )  2 1 6 m m ,
( a ) Ft1 
T1
r1

5 2 .5 2
0 .1 0 8
d 2  4 8 (8 )  3 8 4 m m ,
d 3  3 6 (8 )  2 8 8 m m
(Eq.11.2)
 486 N
Equation (11.19): F r  F t ta n 
( b ) RC 
 5 2 .5 2 N  m
4000
486  227
2
2
o
 4 8 6 ta n 2 5
o
 2 2 7 lb
T C  4 8 6 ( 0 .1 4 4 )  7 0 N  m
 536 N ,
486
227
A
486
B
486
2 2 .9
o
C
RC
486
TC
SOLUTION (11.16)
T1 

9549 kW
n
22 (9549 )
4000
 5 2 .5 2 N  m
d 1  2 7 (5 )  1 3 5 m m ,
( a ) Ft 1 
T1
r1

5 2 .5 2
0 .0 6 7 5
d 2  4 8 (5 )  2 4 0 m m ,
d 3  3 6 (5 )  1 8 0 m m
 778 N
Equation (11.19): F r 1  F t 1 ta n 2 0  2 8 3 N
o
( b ) RC 
778  283
2
2
T C  7 7 8 ( 0 .0 9 )  7 0 N  m
 828 N ,
778
B
A
45
283
o
778
778
20
778
o
C
RC
TC
181
(Eq.11.4)
SOLUTION (11.17)
Equation (11.4): d 1  2 0 ( 6 )  1 2 0 m m ,
d 2  40(6)  240 m m ,
d 3  20(6)  120 m m ,
( a ) T1 
9549 kW
n
Ft 1 

9549 (37 )
 2 9 4 .4 N  m
1200
F r 1  4 .9 1 ta n 2 0
 4 .9 1 k N ,
2 9 4 .4
0 .0 6
d 4  60(6)  360 m m
We have 4 .9 1( 0 .1 2 )  0 .0 6 F t 3 ,
o
 1 .7 9 k N
F t 3  9 .8 2 k N ,
(Eq.11.19)
F r 3  9 .8 2 ta n 2 0  3 .5 7 k N
o
2
( b )  F x  0 : F x  4 .9 1  3 .5 7  1 .3 4 k N
B
3
3.57

9.82
o
9 .5
RB 
4.91
RB
F y  0 : F y  9 .8 2  1 .7 9  8 .0 3 k N
8 .0 3  1 .3 4
2
 8 .1 4 k N
2
TB  0
1.79
SOLUTION (11.18)
( a ) T1 
9549 kW
n
T1
F t1 

r1
9549 ( 80 )

1600
477 . 5
0 . 054
 8 . 843
T1
2
1

F r 1  8 . 843 tan 20
kN ,
3
B
A
mN
r1 
2
8843
1
 477 . 5 N  m ,
TC
8843
20
2
o
 54 mm
 3 . 219
( b ) r3 
C
3219
6 ( 18 )
(Eq.11.4)
(Eq.11.19)
kN
mN
2
3

6 ( 50 )
 150
2
mm
T C  8 . 843 ( 0 . 15 )
o
 1 .3 2 6 k N  m
RC 
RC
8 . 843
 9 . 411
2
 3 . 219
2
kN
SOLUTION (11.19)
9549 kW
( a ) T1 
n
9549 ( 80 )

1600
Equation (11.4): r1 
F r 1  6632 tan 25
( b ) r3 
mN3
RC 
2

8 (50 )
2
o
 477 . 5 N  m
mN
1
2
8 ( 18 )
2
 3 . 093
 200
6632  3093
2

 72 mm .
F t1 
r1

477 . 5
0 . 072
 6 . 632
kN
kN
mm
2
 7 .3 1 8
T C  6 6 3 2 ( 0 .2 0 0 )  1 .3 2 6 k N  m
kN ,
6632
B
A
T1
C
T1
TC
3093
25
o
6632
RC
182
SOLUTION (11.20)
Equation (11.4): d 1  N 1 m  1 5 (5 .2 )  7 8 m m ,
d 3  2 0 (5 .2 )  1 0 4 m m ,
( a ) T1 

9549 kW
n
Ft 1 
9 5 4 9 ( 7 .5 )
1500
d 4  4 0 (5 .2 )  2 0 8 m m
 4 7 .5 7 N  m
F r 1  1 .2 2 ta n 2 5
 1 .2 2 k N  F t 2 ,
4 7 .5 7
0 .0 3 9
d 2  3 5 (5 .2 )  1 8 2 m m ,
o
 0 .5 6 k N  F r 2
T 2  1 .2 2 ( 0 .0 9 1)  1 1 1 N  m
Ft 3 
 2 .1 3 k N ,
111
0 .0 5 2
F r 3  2 .1 3 ta n 2 5
o
 0 .9 9 k N
(b)
4
2
3
2.13
RC
TC
B
65
0.99
0.56
C
0.99
1.22
o
T C  2 .1 3( 0 .1 0 4 )
 2 2 1 .5 N  m
RC 
SOLUTION (11.21)
d 3  18(4)  72 m m ,
( a ) T1 

9549 kW
n
Ft 1  Ft 2 
9 5 4 9 (1 5 )
1800
7 9 .5 8
0 .0 4 8
 7 9 .5 8
d 2  36(4)  144 m m ,
d 4  42(4)  168 m m
N m
 1 .6 6 k N
F r 1  1 .6 6 ta n 2 5
o
 0 .7 7
kN
T 2  T 3  1 .6 6 ( 0 .0 7 2 )  1 1 9 .5 N  m
Ft 3 
 3 .3 2
1 1 9 .5
0 .0 3 6
kN ,
F r 3  3 .3 2 ta n 2 5
( b ) R x  1 . 66  1 . 55  0 . 11 kN 
RB 
R x  R y  2 .5 5
2
2
0.77
RB
1.66
2
2 .5
o
B
o
 1 .5 5 k N
R y  3 . 32  0 . 77  2 . 55 kN 
kN
1.55
3
3.32
183
2
 2 .3 5 k N
2.13
Equation (11.4): d 1  2 4 ( 4 )  9 6 m m ,
0 .9 9  2 .1 3
2
SOLUTION (11.22)
Equations (11.2) and (11.8): d 1  1 5 (5 .2 )  7 8 m m ,
n 2  n1
N1
 1 5 0 0 ( 13 55 )  6 4 2 .9
N2
 0b Ym
K
1
f

rp m
Table 11.4:  0  1 7 2 M P a and 2 5 0 B h n
( a ) Table 11.3: Y=0.443,
Fb 
d 2  3 5 (5 .2 )  1 8 2 m m ,
6
1 7 2  1 0 ( 0 .0 1 2 ) 0 .4 4 3 ( 0 .0 0 5 2 )
1 .6
1
( b ) Table 11.10: K  1 . 117 ,
Q 
 2 .9 7 2
2 ( 35 )
 1 .4
15  35
(Eq.11.33)
kN
(Eq.11.40)
F w  d p b Q K  7 8 (1 2 )(1 .4 )(1 .1 1 7 )  1 .4 6 4 k N
( c ) V2 
 dn

12
 ( 0 .1 8 2 )( 6 4 2 .9 )
60
 6 .1 2 m s ,
Fd 
Thus
2 .9 7 2  3 .0 0 7 F t ,
Ft  9 8 8
and
1 .4 6 4  3 .0 0 7 F t ,
Ft  4 8 7 N
(Eq.11.38)
3 .0 5  6 .1 2
3 .0 5
F t  3 .0 0 7 F t
(Eq.11.24a)
N
SOLUTION (11.23)
Equations (11.4) and (11.8):
d 3  mN
V3 
3
 dn
 5 ( 20 )  100
  ( 0 .1)( 1 16 20 5 )  5 .8 9
60
( a ) Table 11.3: Y  0 .3 2 0 ,
Fb 
mm ,
 0 bYm
K

f
1
1 .5
( c ) Fd 
3 . 0 5  5 .8 9
3.0 5
 1500 ( 34 )  1125
N3
rpm
m s
Table 11.4: 
0
 172
and 250 Bhn
MPa
(172 )( 15 )( 0 . 320 )( 5 )  2 . 75 kN
( b ) Table 11.10: K  0 .9 0 3 M P a ,
F w  d 3 bQK
N1
n 3  n1
Q 
2N4
N3 N4
(Eq.11.33)

 100 (15 )( 43 )( 0 . 903 )  1 . 81 kN
2 ( 40 )
20  40

4
3
(Eq.11.40)
(Eq.11.38)
F t  2 .9 3 1 F t
Thus
2 . 75  10
and
1 . 81  10
3
3
 2 . 931 F t ,
F t  938
 2 . 931 F t ,
F t  617 . 5 N
N
SOLUTION (11.24)
Refer to Solution of Prob. 11.23.
Equation (11.4) and (11.8):
d 1  m N 1  5 (1 5 )  7 5 m m ,
V1 
 dn
60
n 2  n1
N1
N2
 1 5 0 0 ( 13 55 )  6 4 2 .9 rp m
  ( 0 .0 7 5 )( 1 56 00 0 )  5 .8 9 m s
(CONT.)
184
11.24 (CONT.)
( a ) Table 11.3: Y  0 .2 8 9 ,
Fb 
 obYm
K

1
1 .5
f
Table 11.4: 1 7 2 M P a at 2 5 0 B h n
(1 7 2 ) (1 5 ) ( 0 .2 8 9 ) ( 5 )  2 .4 9 k N
( b ) Table 11.10: K  0 .9 0 3 M P a ,
2N2
(Eq. 11.40)
F w  d 1 b Q K  7 5 (1 5 )(1 .4 )( 0 .9 0 3)  1 .4 2 k N
(Eq. 11.38)
N1  N 2

2 (35 )
 1 .4
3 .0 5  5 .8 9
3 .0 5
( c ) Fd 
Q 
(Eq. 11.33)
15 35
F t  2 .9 3 1 F t
Thus
2 .4 9  1 0  2 .9 3 1 F t ,
F t  8 4 9 .5 N
1 .4 2  1 0  2 .9 3 1 F t ,
F t  4 8 4 .5 N
3
and
3
SOLUTION (11.25)
We have d  N m  2 2 ( 2 .5 )  5 5 m m
( a )  o  2 2 1 M P a (Table 11.4), Y  0 .3 3 (Table 11.3)
Equation (11.33):
Fb 
(b) V 
 dn
60

 obYm
K
2 2 1 ( 3 0 ) ( 0 .3 3 ) ( 2 .5 )

1 .5
f
 ( 0 .0 5 5 )(1 5 0 0 )
 4 .3 2 m p s
60
Fd 
Equation (11.24a):
 3 .6 5 k N
3 .0 5  4 .3 2
3 .0 5
F t  2 .4 2 F t
Hence
3 .6 5  2 .4 2 F t ,
F t  1 .5 0 8 k N
Equation (11.22) is then
kW 
Ft V
1000

1 5 0 8 ( 4 .3 2 )
1000
 6 .5 1
SOLUTION (11.26)
Equations (11.4), (11.8), and (11.20'):
d 2  36(4)  144 m m ,
n 2  1 8 0 0 ( 23 )  1 2 0 0
rpm ,
d 3  18(4)  72
V 
 dn2
60

mm
 ( 0 .1 4 4 )(1 2 0 0 )
60
 9 .0 5
m ps
(CONT.)
185
11.26 (CONT.)
( a ) Table 11.3: Y  0 .4 4 6 (by interpolation), Table 11.4:  0  1 7 2 M P a (for steel of 200 Bhn)
 0b Ym
Fb 
K

1
f
1 7 2 (1 0 ) ( 0 .4 4 6 ) ( 4 )
1 .4
 2 .1 9 2 k N (Eq.11.33)
( b ) Table 11.10: K  0 .6 7 6 M P a ,
2 ( 42 )
Q 
18 42

F w  d 3 b Q K  7 2 (1 0 )( 2 1 1 5 )( 0 .6 7 6 )  6 8 1 .4
( c ) Fd 
3 . 05  9 . 05
3 . 05
F t  3 . 97 F t
(Eq.11.40)
(Eq.11.38)
N
( Eq.11.24a)
2 .1 9 2  3 .9 7 F t ,
Thus
21
15
Ft  5 5 2
N
SOLUTION (11.27)
Equation (11.4): d 1  24 (10 )  240
( a ) Table 11.3: Y  0 .3 9 3 (interpolated).
 0 bYm
Fb 
K

f
( b ) Table 11.10:
F w  d 1 bQK
1
1 .5
mm
Table 11.4: 
0
 172
[172 (15 )( 0 . 393 )( 10 )]  6 . 76 kN
K  0 . 545
Q 
MPa,
2 ( 36 )
24  36

(for steel of 200 Bhn)
MPa
(Eq.11.33)
6
5
(Eq.11.40)
 240 (15 )( 65 )( 0 . 545 )  2 . 35 kN
(Eq.11.38)
SOLUTION (11.28)
Refer to Solution of Prob. 11.27.
Equation (11.4): d 1  2 4 (1 0 )  2 4 0 m m ,
( a ) Table 11.3: Y  0 .3 3 7 , Table 11.4:
Fb 
 obYm
K

f
1
1 .5
d 3  1 8 (1 0 )  1 8 0 m m
 o  1 7 2 M P a (for steel of 200 Bhn)
[1 7 2 (1 5 ) ( 0 .3 3 7 ) (1 0 ) ]  5 .8 k N (Eq. 11.33)
( b ) Table 11.10: K  0 .5 4 5 M P a ,
Q 
2 N3
N3 N4

2 (1 8 )
18 42
 0 .6
(Eq. 11.40)
F w  d 3 b Q K  1 8 0 (1 5 )( 0 .5 4 5 )  8 8 3 N
(Eq. 11.38)
SOLUTION (11.29)
d
p
 N p m  24(2)  48
V 
 d pnp
Kv 
60

6 . 1  4 . 02
6 .1
 ( 0 .0 4 8 )(1 6 0 0 )
60
 1 . 659
mm,
d g  60(2)  120
 4 .0 2 m p s
mm
(Eq.11.4)
(Eq.11.20')
(Fig.11.15)
(CONT.)
186
11.29 (CONT.)
Ft 
1000 kW
V

1 0 0 0 ( 0 .9 )
 2 2 3 .9
4 .0 2
N
J  0 .2 6
Pinion: (Fig.11.16):
Table 11.7: S t  1 9 7 .5 M P a
(mid-point).
Equation (11.35'):
  Ft K 0 K v
KsKm
1
bm
 2 2 3 .9 (1) (1 .6 5 9 )
J
1 (1 .6 )
1
0 .0 1 5 ( 0 .0 0 2 ) 0 .2 6
 7 6 .2 M P a
Equation (11.36):

a ll
St K L

KT K R
1 9 7 .5 (1 )

1 (1 .2 5 )
 158
J  0 .3 0 ,
Gear: (Fig.11.16):
  2 2 3 .9 (1)(1 .6 5 9 )


M Pa
1
0 .0 1 5 ( 0 .0 0 2 )
0 .3
 ,
Yes.
S t  58 . 6 ksi
Table 11.7:
1 (1 .6 )
a ll
 66 M Pa
and

a ll

5 8 .6 (1 )
1 (1 .2 5 )
 4 6 .9

M Pa

a ll
 ,
No.
SOLUTION (11.30)
From Solution of Prob.11.29: d p  4 8 m m ,
F t  2 2 3 .9 N ,
C H  1 . 001
mG  d g d
p
d g  120 m m ,
 2 .5 ,
ng  np
N
p
N
g
V  4 .0 2 m p s ,
24
 1600 ( 60
)  640
(from Eq.11.46)
Pinion:
 c  C p ( Ft K 0 K v
where
C
p
I 
 166
Ks
KmC
bd
I
f
1
)
(Eq.11.42)
2
M P a (Table 11.11)
sin  cos 
mG
2
m G 1
 0 . 161
60
84
 0 . 115
Thus
 c  1 6 6  1 0 [ 2 2 3 .9 (1)(1 .6 5 9 )
3
S c  620 . 5 MPa
 c , a ll 
ScC LC H

KT K R
1 .6 (1 )
1
( 0 .0 1 5 )( 0 .0 4 8 ) 0 .1 1 5
(Table 11.12)
6 2 0 .5 (1 )(1 .0 0 1 )
1 (1 .2 5 )
1
] 2  4 4 4 .7 M P a
(mid-point)
 4 9 6 .9 M P a
(Eq.11.44)
Hence
 c , a ll   c
Gear:
Table 11.12:
Yes.
S c  4 8 5 .5
(mid-point)
M Pa
 c  1 6 6  1 0 [ 2 2 3 .9 (1)(1 .6 5 9 )
3
 c , a ll 
4 8 2 .5 (1 )(1 )
1 (1 .2 5 )
1 .6 (1 )
1
( 0 .0 1 5 )( 0 .1 2 ) 0 .1 1 5
 386 M Pa
Hence
 c , a ll   c
Yes.
187
1
] 2  2 8 1 .3 M P a
rpm
SOLUTION (11.31)
d
p
 N p m  2 0 (3 )  6 0 m m ,
V 
 d pnp
 ( 0 .0 6 )(1 2 0 0 )

60
Ft 
1000 kW
V
(Eq. 11.4)
 3 .7 7 m p s
60
K   1 .6 1 8
d g  5 0 (3 )  1 5 0 m m
(Fig. 11.15)

1 0 0 0 (1 .0 )
 2 6 5 .3 N
3 .7 7
J  0 .2 4 5 (mid-point)
Pinion, (Fig. 11.16):
Table 11.7: S t  2 8 6 M P a
Equation (11.35'):
  Ft K o K v
KsKm
1
bm
 2 6 5 .3 (1)(1 .6 1 8 )
J
1 (1 .7 )
1
( 0 .0 1 5 )( 0 .0 0 3 ) 0 .2 4 5
 6 6 .2 M P a
Equation (11.36):


a ll
St K L

KT K R
2 8 6 (1 )
1 (1 .3 )
Table 11.7: S t  8 9 .6 M P a
  2 6 5 .3 (1)(1 .6 1 8 )

8 9 .6 (1 )

a ll
1 (1 .3 )
  , Yes.
a ll
Gear, (Fig. 11.16): J  0 .3 ,
and
 
 220 M Pa
1 (1 .7 )
1
( 0 .0 1 5 )( 0 .0 0 3 )
0 .3
 
 6 8 .9 M P a
a ll
 5 4 .1 M P a
  , Yes.
SOLUTION (11.32)
Refer to Solution of Prob. 11.31
d
p
 60 m m ,
ng  n p
N
p
N
g
d g  150 m m ,
V  3 .7 7 m p s ,
 1 2 0 0 ( 52 00 )  4 8 0 rp m ,
C H  1 .0 0 8
F t  2 9 6 .7 N ,
mG  d g d
p
 2 .5
K   4 .2
(from Eq. 11.46).
Pinion:

where
C
p
I 
Then

c
Ks
 C p [ Ft K o K v
 166  10
3
s in  c o s 
mG
2
m G 1
 1 .6 1
50
70
KT K R

7 8 9 (1 )(1 .0 0 8 )
1 (1 .3 )
1
]
2
(Eq. 11.42)
 1 .1 5
S c  8 7 9 M P a (Table 11.12)
ScC LC H
f
(Table 11.11)
 1 6 6  1 0 [ 2 9 6 .7 (1) ( 4 .2 )
 c , a ll 
I
M Pa
3
c
K mC
bd
1 .7 (1 )
1
( 0 .0 1 5 )( 0 .0 6 ) 1 .1 5
1
]
2
 2 3 7 .5 M P a
(mid-point)
 6 8 1 .6 M P a
(Eq.11.44)
(CONT.)
188
11.32 (CONT.)
Hence,  c , a ll   c
Yes.
Gear:
Table 11.12: S c  5 5 1 .5 M P a

 1 6 6  1 0 [ 2 9 6 .7 (1) ( 4 .2 )
3
c
 c , a ll 

So
(mid-point)
5 5 1 .5 (1 )(1 )
1
]
2
 1 5 0 .2 M P a
 4 2 4 .2 M P a
1 (1 .3 )

c , a ll
1 .7 (1 )
1
( 0 .0 1 5 )( 0 .1 5 ) 1 .1 5
Yes.
c
SOLUTION (11.33)
N
p
 20,
2( 40 )
Q 
20 40
Thus
N

4
3
 40,
g
d
 2 0 (5 )  1 0 0
p
Table 11.10: S e  6 2 1 k s i,
,
b  50 m m
mm,
K  1 .8 2 1 M P a
F w  d p b Q K  1 0 0 (5 0 )( 43 )(1 .8 2 1)  1 2 .1 4
(Eq.11.38)
kN
Equations (11.20') and (11.24a):
 dn
V 
60
 ( 0 .1 )( 9 0 )

60
 0 .4 7 m p s ,
Then, we obtain 1 2 .1 4  1 .1 5 4 F t ,
Fd 
3 .0 5  0 .4 7
3 .0 5
F t  1 .1 5 4 F t
F t  1 0 .5 2 k N
Hence
kW 
Ft V
1000
1 0 ,5 2 0 ( 0 .4 7 )

1000
 4 .9 5
SOLUTION (11.34)
From Solution of Prob.11.33:
N
p
 20, N
g
 40, d
ScCLC H
 c , a ll 
KT K
 1 0 0 m m , b  5 0 m m , V  0 .4 7 m p s , m G  N
p
where S c  5 5 1 .5 M P a (Table 11.12)
,
R
C
L
Then
5 5 1 .5 (1 .1 )(1 )
6
F t  ( 6 0 6 C.7  1 0 )
2
1
K0Kv
p
where
C
p
 149
Kv 
M Pa
5 . 56  0 . 47
5 . 56
K 0  1,
I 
 6 0 6 .7
(1 )1
K
 1,
sin  cos 
mG
2
m G 1
I
K mC
p
 2
(mid-point)
K T  1,
K
R
 1
M Pa
(Eq.11.42)
f
(Table 11.11)
 1 . 08
s
bd
Ks
N
 1.1 (Fig.11.19)
C H  1,
 c , a ll 
g
(Curve B, Fig.11.15)
K m  1 .3,
C
f
 1
 0 . 107
(CONT.)
189
11.34 (CONT.)
6
F t  ( 6 0 6 .7 1 03 )
Thus
0 .0 5 ( 0 .1 ) 0 .1 0 7
1
1 (1 .0 8 )
1
1 .3 (1 )
2
1 4 9 1 0
 6 .3 1 8 k N
and
Ft V
kW 
1000

6 3 1 8 ( 0 .4 7 )
 2 .9 7
1000
SOLUTION (11.35)
N
 2 5,
p
2 (50 )
Q 
25 50
 50,
g
4
3
, Table 11.10: S e  6 2 1 M P a ,

d
 N m  2 5 (5 )  1 2 5 m m ,
N
p
b  40 m m
K  1 .8 2 1 M P a
Thus,
F w  d p b Q K  0 .1 2 5 ( 0 .0 4 )( 43 )(1 .8 2 1  1 0 )  1 2 .1 4 k N
6
(Eq. 11.38)
Equations (11.20') and (11.24a) with 6 0 0 fp m  6 0 0 1 9 6 .8  3 .0 5 m s :
 dn
V 

60
 ( 0 .1 2 5 )(1 2 0 )
 0 .7 8 5 m s ,
60
Fd 
3 .0 5  0 .7 8 5
3 .0 5
F t  1 .2 5 7 F t
Then, we have
1 2 .1 4  1 .2 5 7 F t ,
F t  9 .6 6 k N
Hence,
kW 
Ft V
1000

9 6 6 0 ( 0 .7 8 5 )
1000
 7 .5 8
SOLUTION (11.36)
N
 N
p
ng
g
 60
np
Table 11.4:

240
600
 24,
 8 7 .2
0g
Y p  0 .3 3 7 ,
M Pa,

0 p
Y g  0 .4 2 1
 172
M Pa
Hence
Y p
0 p
 0 .3 3 7 (1 7 2 )  5 8
Yg
0g
 0 .4 2 1(8 2 .7 )  3 4 .8 2
 Gear is weaker
Thus
Fb 
 0bYm
K

8 2 .7 ( 9 0 ) ( 0 .4 2 1 ) ( 4 )
1 .4
f
 8 .9 5 3
kN
We have
d g  60(4)  240 m m ,
V 
 dn
60

 ( 0 .2 4 )( 2 4 0 )
60
 3 .0 2 m p s
Hence
Fd 
3 .0 5  3 .0 2
3 .0 5
F t  1 .9 9 F t
 8 .9 5 3  1 .9 9 F t ,
and
kW 
Ft V
1000

4 5 0 0 ( 3 .0 2 )
1000
 1 3 .5 9
190
F t  4 .5 k N
SOLUTION (11.37)
N
 N
p
ng
 60
np
g
Table 11.4: 
 24,
240
600
Y p  0 .3 3 7 ,
 82 . 7 MPa ,
0g
Y p 0 p  Yg 0 g

Y g  0 .4 2 1
 172
0 p
MPa
 the gear is weaker
Thus
 0 bYm
Fb 
K

82 . 7 ( 80 )( 0 . 421 )( 4 )
 7 . 958
1 .4
f
kN
We have the quantities:
d
g
 N g m  60 ( 4 )  240
mm
V   d n 6 0   ( 0 .2 4 )( 2 4 0 6 0 )  3 .0 2 m p s
3 .0 5  3 .0 2
3 .0 5
Fd 
F t  1 .9 9 F t
(from Eq.11.24a)
and
7 , 9 5 8  1 .9 9 F t ,
Ft  4 k N
Equation (1.15) results in
Ft V
kW 
1000

4 0 0 0 ( 3 .0 2 )
1000
 1 2 .1
SOLUTION (11.38)
Gear is weaker and can transmit lower hp.
d g  60(4)  240 m m ,
K
 KT  K
L
s
 1,
Table 11.7: S t  3 9 .3 M P a ,
V 
 d g ng
60

 ( 0 .2 4 ) 2 4 0
60
 3 .0 2 m p s
We have

St K L

a ll

KtKR
3 9 .3 (1 )
0 .8 5 (1 )
 4 6 .2
M Pa
Also
K
v
3 . 56 

3 . 02
3 . 56
 1 . 49 (Curve C, Fig.11.15)
m  4 m m , b  9 0 m m . (given)
J  0 .3 0 (from Fig.11.16, N
K
0
 1.2 5 (Table 11.5)
K
s
 1 (standard gear),
K
m
g
 60 )
 1.7 (Table 11.6)
Thus, Eq.(11.35'):
Ft 
4 6 .2 ( 9 0 )( 4 )( 0 .3 0 )
1 .2 5 (1 .0 )(1 .7 )(1 .4 9 )
 1 .5 7 6 k N
Hence,
kW 
Ft V
1000

1 5 7 6 ( 3 .0 2 )
1000
 4 .7 6
191
Table 11.9: K
R
 0 .8 5
SOLUTION (11.39)
N
N

g
Q 
V 

p
rs
2N
N
p
 dn
60

g
N
 3 5,
28
4 5
70
28 35
g
d

70
63
 ( 0 .1 6 8 )( 6 0 0 )

 N p m  2 8(6 )  1 6 8 m m ,
p
K  1 .8 6 2
,
b  50 m m
M P a (Table 11.10)
 5 .2 8 m p s
60
We calculate that
F w  d p b Q K  1 6 8 (5 0 )( 76 03 )(1 .8 6 2 )  1 7 .3 8
kN
and
3 .0 5  5 .2 8
3 .0 5
Fd 
F t  2 .7 3 1 F t ,
1 7 .3 8  2 .7 3 1 F t
 F t  6 .3 6 4 k N
Hence
Ft V
kW 
1000
6 3 6 4 ( 5 .2 8 )

1000
 3 3 .6
SOLUTION (11.40)
N
g
 35 ,
2 ( 35 )
Q 
63
N

70
63
 28 ,
p
p
K  1 . 862
,
F w  d p bQK
d
 28 ( 6 )  168
 168 ( 60 )( 1 . 862 )(
b  60 mm
(Table 11.10)
MPa
70
63
mm ,
)  20 . 85 kN
We obtain
V   d n 6 0   ( 0 .1 6 8 )( 6 0 0 6 0 )  5 .2 8 m p s
Using Eq.(11.24a),
3 .0 5  5 .2 8
3 .0 5
Fd 
F t  2 .7 3 F t
2 0 .8 5  1 0  2 .7 3 F t
 F t  7 .6 3 7 k N
3
Equation (1.15) is thus
Ft V
kW 
1000
7 6 3 7 ( 5 .2 8 )

1000
 4 0 .3
SOLUTION (11.41)
From Solution of Prob. 11.39:
N
g
 3 5,
V  5 .2 8
N
p
 2 8,
d
p
 168 m m ,
b  50 m m ,
mG  N
g
N
p
 1 .2 5
m ps
Apply Eq. (11.44):
 c , a ll 
ScCLC H
KT K
R
Here S c  1 0 2 0 M P a (mid-point, Table 11.12)
K
v

5 . 56 
5 . 28
5 . 56
 1 . 19
K 0  K s  K T  1,
CL  C
f
 1,
C H  1 (Eq.11.46)
(Curve A, Fig.11.15)
K
R
 1. 2 5
(Table 11.9)
(CONT.)
192
11.41 (CONT.)
Thus
 c , a ll 
1 0 2 0 (1 )(1 )
 816 M Pa
(1 )(1 .2 5 )
(Eq.11.44)
We find that
C
p
 191 M Pa
C
f
 1, K m  1 . 3
I 
sin  cos 
mG
2
m G 1
(from Table 11.11)
(from Table 11.6)
 0 . 161
35
28  35
 0 . 089
Hence, from Eq.(11.45):
Ft  (

c , a ll
C
p
)
2
6
 ( 8 1 6 1 03 )
1 9 1 1 0
1
KsKv
2
bd
Ks
I
KmC
f
( 0 .0 5 )( 0 .1 6 8 ) 0 .0 8 9
1
1 (1 .1 9 )
1
1 .3 (1 )
 8 .8 2 k N
and
kW 
Ft V
1000

8 8 2 0 ( 5 .2 8 )
1000
 4 6 .6
End of Chapter 11
193
CHAPTER 12
HELICAL, BEVEL, AND WORM GEARS
SOLUTION (12.1)
mn
( a ) Use Eqs.(12.1) and (12.2'). m 
c o s

p   m   ( 4 .6 1 9 )  1 4 .5 1 1 m m ,
4
cos 30
o
p a  p c o t   1 4 .5 1 1 c o t 3 0
  ta n
( b ) P  1 m  1 4 .6 1 9  0 .2 1 6 m m  5 .4 8 6 in ,
-1
(c) dp 
Nmn

c o s
20 ( 4 )
cos 30
 9 2 .4 m m ,
o
p n   m n  1 2 .5 6 6 m m
 4 .6 1 9 m m ,
40 ( 4 )
dg 
cos 30
o
 1 ta n  n
c o s
o
 ta n
 2 5 .1 3 4 m m
 1 ta n 2 5 o
cos 30
o
 2 8 .3
o
 1 8 4 .7 5 m m
Gear
(d)
30
Thrust
o
Pinion, R.H.
SOLUTION (12.2)
( a ) Apply Eqs.(12.1) through (12.2').
m 
mn
c o s

3 .1 7 5
cos 30
o
 3 .6 6 6 m m , p n   m n  9 .9 7 5 m m ,
p   m   (3 .6 6 6 )  1 1 .5 1 7 m m , p a  p c o t   1 1 .5 1 7 c o t 2 0
( b ) P  1 m  1 3 .6 6 6  0 .2 7 3 m m  6 .9 3 4 in . ,   ta n
-1
(c) dp 
Nmn
c o s

1 8 ( 3 .1 7 5 )
cos 20
o
 6 0 .8 1 8 m m ,
dg 
5 5 ( 3 .1 7 5 )
cos 20
o
 1 ta n  n
c o s
o
 ta n
 3 1 .6 4 m m
 1 ta n 1 4 .5 o
cos 20
o
 1 5 .3 9
o
 1 8 5 .8 3 2 m m
Gear
(d)
Pinion, L.H.
SOLUTION (12.3)
( a ) d  N m  3 0 (3 .1 7 5 )  9 5 .2 5 m m , m n  m c o s 3 0  2 .7 5 m m ,
o
p   m  9 .9 7 5 m m
p n  p c o s   9 .9 7 5 c o s 4 0
o
 7 .6 4 m m
p a  p c o t   9 .9 7 5 c o t 4 0
o
 1 1 .8 9 m m
(CONT.)
194
12.3 (CONT.)
( b ) Pn  1 m n  1 2 .7 5  0 .3 6 4 m m  9 .2 5 in .
  ta n
1
ta n  n
(
)  ta n
c o s
1
o
( ta n 1 4 .5o )  1 8 .6 5
-1
o
cos 40
SOLUTION (12.4)
rs 
c 
1
4

Pn
N
2n
N
p
N
g
N
p

g
c o s
18
Ng
;
N
g
1 5 .6 2 5 1 8  7 2
2
c o s
250 
;
 72
Solving, c o s   0 .8 9 5,
  2 6 .5
Comment: The helix angle of 2 6 .5
o
o
o
o
and
P  N d . Thus
is in usual range of 1 5 to 3 0 .
SOLUTION (12.5)
cos   P Pn where m n  1 Pn
Equation (12.2):
Nmn
cos  

d
32( 6 )
260
  4 2 .4
 0 .7 3 8 5 ,
o
F t  F n c o s  n c o s   1 0 (c o s 2 0 )(c o s 4 2 .4 )  6 .9 4 N
o
kW 
Thus
(  dn ) F t
60

o
 ( 0 . 26 )( 800 )( 6 . 94 )
60
(Eq.12.10)
 75 . 58
SOLUTION (12.6)
1
( a )  n  ta n
(ta n  c o s  )  ta n
1
o
o
(ta n 2 0 c o s 3 0 );
 n  1 7 .5
o
N '  N c o s   3 5 c o s 3 0  5 3 .8 9
3
3
o
( b ) A super gear of equal strength would have 53.89 teeth and   1 7 .5
o
SOLUTION (12.7)
1
3

N
p
N
g

40
Ng
,
N
Equation (12.5)
c 
Pn
2
g
 3(4 0 )  1 2 0
P  Pn c o s  :
with

N
p
N
c o s
g

14
2
40 120
cos15
o
 3 6 9 .1 m m
SOLUTION (12.8)
( a )  n  ta n
1
(ta n  c o s  )  ta n
1
o
o
(ta n 2 0 c o s 2 2 );
 n  1 8 .6
N '  N c o s   3 5 c o s 2 2  4 3 .9 1 teeth
3
3
o
( b ) A super gear of equal strength would have 43.91 teeth and   1 8 .6
195
o
o
SOLUTION (12.9)
a p  a g  a  1 P  m  1 .5 m m
( a ) Addendum:
p   m  1 .5   4 .7 1 2 m m
Circular Pitch:
c 
1
2
rp 
(d
1
2
 dg) 
p
N Pm 
1
2
N
1
2
p
N

g
P
1
2
(N
p
 N g )m 
( 2 0 )(1 .5 )  1 5 m m ,
rg 
1
2
1
2
( 2 0  1 2 0 )(1 .5 )  1 0 5 m m
(1 2 0 )(1 .5 )  9 0 m m
The radii of base circles:
rb p  r p c o s   1 5 c o s 2 0
 1 4 .0 9 5 m m
o
rb g  rg c o s   9 0 c o s 2 0  8 4 .5 7 2 m m
o
Outside radii:
ro p  r p  a  r p  m  1 5  1 .5  1 6 .5 m m
ro g  r g  a  9 0  1 .5  9 1 .5 m m
From Eq. (11.14) the contact ratio is therefore
Cr 

1
p cos 
[
( rp  a )  ( rp c o s  )
2
1
( 4 .7 1 2 ) c o s 2 0
[
o
2

(1 6 .5 )  (1 4 .0 9 5 ) 
2
2
( rg  a )  ( rg c o s  ) ] 
2
2
(9 1 .5 )  (8 4 .5 7 2 ) ] 
2
2
c ta n 
p
1 0 5 ta n 2 0
4 .7 1 2
o
 1 .7 1 5
The total contact ratio is
C r t  C r  rr a  1 .7 1 5 
4 0 ta n 3 0
4 .7 1 2
o
 6 .6 1 6
( b ) To obtain a total contact ratio of 4.0, the helix angle has to be
C ra 
b ta n 
p
ta n  
 4 .0  1 .7 1 5  2 .2 8 5
2 .2 8 5 ( 4 .7 1 2 )
40
  1 5 .0 7
;
o
SOLUTION (12.10)
299
(a)
153.9
  tan
299
 1 tan  n
cos 
 tan
 1 tan 20 o
cos 45
 27 . 24
o
o
(Eq. 12.3)
Ft  F n c o s  n c o s   4 5 0 c o s 2 0 c o s 4 5
o
F a  F t ta n   2 9 9 ta n 4 5
o
o
 299 N
F r  F t ta n   2 9 9 ta n 2 7 .2 4
o
(Eqs.12.10)
 1 5 3 .9 N
( b ) m  m n c o s   2 .5 c o s 4 5  3 .5 3 6 m m
o
d g  N g m  6 0 (3 .5 3 6 )  2 1 2 .1 6 m m , d
Thus
 299 N
T p  2 9 9 ( 0 .1 123 1 5 )  1 6 .9 2 N  m
T g  2 9 9 ( 0 .2 122 1 6 )  3 1 .7 2 N  m
196
p
(Eq.12.2')
 3 2 (3 .5 3 6 )  1 1 3 .1 5 m m
(Eq.12.3)
SOLUTION (12.11)
T2
(a)
2
1178
T3
B
549
C
1
549
1178
429
3
2
T1
429
T1 
9549 kW
n1
9 5 4 9 (1 5 )

 9 5 .4 9 N  m
1500
m  m n c o s   6 .3 5 c o s 2 0
o
 6 .7 5 8 m m ,
d 1  N 1 m  2 4 ( 6 .7 5 8 )  1 6 2 .1 9 m m
(Eq.12.2)
We have
( rs ) 1  2 

24
36
d1
( rs ) 1  3  0 .5 
Ft1 
T1

d1 2
d1

2
3
d3
d2
d2 
,
d3 
,
1 6 2 .1 9
2 3
1 6 2 .1 9
0 .5
 2 4 3 .3 m m
 3 2 4 .4 m m
 1 .1 7 8 k N  F t 2  F t 3
9 5 .4 9
0 .1 6 2 .1 9 2
F r 1  F t 1 ta n   1 .1 7 8 ta n 2 5
 5 4 9 N  Fr 2  Fr 3
o
F a 1  F t 1 ta n   1 .1 7 8 ta n 2 0
o
 429 N
T2  0 ,
( b ) T1  9 5 .4 9 N  m ,
(Eqs.12.10)
T 3  9 5 .4 9 ( 12 )  1 9 1 N  m
SOLUTION (12.12)
Equations (12.7b) and (12.2'):
N '
d
p
N
p
3
cos 

 2 9 .6 , m  m n c o s   3 .1 2 c o s 2 5
22
3
cos 25
o
2 2 ( 3 .4 4 )
 N pm 
3
cos 25
o
o
 3 .4 4 m m
 1 0 1 .6 6 m m
Table 11.4:  o  1 7 2 M P a
Table 11.3: Y  0 .3 5 7 ,
We have
 0b Ym n
Fb 
K
1
f
 d pnp
V 
60


1 7 2 ( 5 0 ) 0 .3 5 7 ( 3 .1 2 )
1 .5
1
 ( 0 .1 0 1 6 6 )(1 8 0 0 )
60
 6 .3 8 6
kN
(Eq.11.33, modified)
 9 .5 8 m p s
Also
Fd 
or
and
5 .5 6 
9 .5 8
5 .5 6
Fb  F d :
Ft 
1000 kW
V
F t  1 .5 5 7 F t
(1)
6 .3 8 6  1 .5 5 7 F t ,

1000 ( 22 )
9 .5 8
 2 .2 9 6 k N
Hence, by Eqs.(2) and (3):
n 
4 .1 0 1
2 .2 9 6
 1 .7 9
197
F t  4 .1 0 1 k N
(2)
(3)
SOLUTION (12.13)
( a ) Speed ratio: 37 05 
 5
50
25
Center distance is
c  m(N
p
mn
 Ng) 
c o s
(N
p
 Ng)
(1)
Thus, for equal center distance:
(3 0  7 5 ) 
4
c o s 2 9 .8
(b)
4
c o s 2 9 .8
o
(3 0  7 5 ) 
5 .5
c o s
  3 1 .5
( 2 5  5 0 );
5 .5
c o s
  5 3 .8
( 2 0  3 2 );
o
o
SOLUTION (12.14)
( a )   ta n
 1 ta n  n
 ta n
c o s
 1 ta n 2 5 o
cos 30
m  m n co s  4 co s 3 0
d
p
 2 8 .3
o
o
o
 4 .6 2 m m
 N p m  2 2 ( 4 .6 2 )  1 0 1 .6 m m
( b ) V   d n   ( 0 .1 0 1 6 )( 2 4 0 0 6 0 )  1 2 .8 m p s
Ft 
1000 kW
V

1 0 0 0 (1 .5 )
 117 N
1 2 .8
( c ) F r  F t ta n   1 1 7 ta n 2 8 .3  6 3 N
o
F a  F t ta n   1 1 7 ta n 3 0
Fn 
Ft
c o s  n c o s

117
o
cos 25 cos 30
o
o
 6 7 .5 N
 1 4 9 .1 N
SOLUTION (12.15)
Refer to Solution of Prob.12.12. We now have
Equation (11.40): Q 
and
Fw 
d pbQ K
2
cos 
2( 40 )
22 40

40
31
, Table 11.10: K  0 .9 0 3 M P a
 (1 0 1 .6 6 )(5 0 )( 43 01 )( 0 .9 0 3 )(
Equation (1) for F w  F d :
)  7 .2 1 k N
1
2
cos 25
o
7 .2 1  1 .5 5 7 F t ;
(Eq.11.38, modified)
F t  4 .6 3 1 k N
(4)
Thus, by Eqs.(3) and (4):
n 
4 .6 3 1
2 .2 9 6
 2 .0 2
SOLUTION (12.16)
Equation (12.2'), (12.4) and (12.7b):
m  m n c o s   4 .2 c o s 3 5
o
 5 .1 2 7 m m ,
d 2  6 5 ( 5 .1 2 7 )  3 3 3 .3 m m ,
N1 
N1
'
3
cos 
d 1  N 1 m  3 0 (5 .1 2 7 )  1 5 3 .8 m m

30
3
cos 35
o
 5 4 .5 8
(CONT.)
198
12.16 (CONT.)
and
Table 11.4:  0  1 2 4 M P a
Table 11.3: Y  0 .4 8 3 ,
Thus
Ym n
Fb   0 b
0 .4 8 3 ( 4 .2 )
 1 2 4 (3 8 )
1
 9 .5 6 k N
1
(Eq.11.33, modified)
From Table 11.10: K  0 . 3 5 2 M P a
We have the quantities:
2N
Q 
N
Fd 
2
cos 35
19 . 33
5 . 56
F t  1 . 791 F t
4 .1 9  1 .7 9 1 F t ,
Hence
 4 .1 9 k N
o
(Eq.11.38, modified)
 1 9 .3 3 m p s
60
5 . 56 
(Eq.11.40)
1 5 3 .8 ( 3 8 ) (1 .3 6 8 ) ( 0 .3 5 2 )
 ( 0 .1 5 3 8 )( 2 4 0 0 )

60
 1.3 6 8
30 65

2
cos 
 d 1 n1
V 
g
d 1b Q K
Fw 
and
2( 65 )

g
N
p
(Eq.11.24c, modified)
F t  2 .3 4 k N
Therefore
Ft V
kW 
2 ,3 4 0 (1 9 .3 3 )

1000
1000
 4 5 .2
SOLUTION (12.17)
( a ) Equations (12.2'), (12.5), and (12.7b):
rs 
1
3
N1

N
d1

c 
,
d2
2
(N1  N 2 )
m
2
or
N 1  44
and
d 1  4 4 (1 .7 )  7 4 .8 m m ,
N1 
'
N1
cos
3


and N
3
30
1 .7
2
(N1  3N1)
 132
2
d 2  2 2 4 .4 m m
 74 . 8 mm , m n  m cos   (1 . 7 ) cos 30
44
cos
150 
o
Using 69.28 teeth, Y  0 .4 2 8 (interpolated, Table 11.3).
Fb 
 0b Ym n
K

1
f
1 7 2 ( 6 4 ) 0 .4 2 8 (1 .4 7 )
1 .4
1
 4 .9 5 k N
o
 1 . 47 mm
By Table 11.4:  0  1 7 2 M P a
(Eq.11.33, modified)
Similarly, Table 11.10: K  0 .5 4 5 M P a
and
Q 
2N
N

g
N
g
d 1b Q K
Fw 
V 
p

2
cos 
 d 1 n1
60

5 .5 6 
2 ( 132 )
44  132
 1 .5
( 7 4 .8 ) ( 6 4 ) (1 .5 ) ( 0 .5 4 5 )
2
cos 30
 ( 0 .0 7 4 8 )( 9 0 0 )
60
Also
Fd 
and
4 .9 5  1 .3 3 7 F t ,
3 .5 2
5 .5 6
o
(Eq.11.40)
 5 .2 2 k N
(Eq.11.38, modified)
 3 .5 2 m p s
F t  1 .3 3 7 F t
(Eq.11.24c, modified)
F t  3 .7 k N
Thus
kW 
Ft V
1000

3 7 0 0 ( 3 .5 2 )
1000
 1 3 .0 2
(CONT.)
199
12.17 (CONT.)
( b ) Equation (11.36):
St K
 a ll 
L
KT KR
We have
S t  2 1 0 M P a (by interpolation of mid-point values, Table 11.7)
K
L
 1
K
R
 0 . 85
KT  1
(indefinite life)
(Table 11.9)
and

a ll
2 1 0 (1 )

 247 M Pa
(1 )( 0 .8 5 )
Equation (11.35):
 bJm
K0K sK mKv
Ft 
(where    a ll )
where
K 0  1.2 5
Ks  1
3 .5 6 
Kv 
3 .5 2
3 .5 6
K m  1.4
 1 .2 4 (Fig.11.15)
(Table 11.6)
J  0 .4 8 (Fig.12.6a)
J  multiplier
 1 . 01
(Fig.12.6b)
and
J  0 .4 8 ( 1.0 1 )  0 .4 8 5
Hence
Ft 
2 4 7 ( 6 4 )( 0 .4 8 5 )(1 .7 )
(1 .2 5 )(1 )(1 .4 )(1 .2 4 )
 6 kN
Then
kW 
Ft V
1000

6 0 0 0 ( 3 .5 2 )
1000
 2 1 .1
SOLUTION (12.18)
( a ) d p  N p m  2 0 (3 )  6 0 m m ,
( b )  p  ta n
1 20
42
 2 5 .4 6 ,
o
 g  ta n
d g  4 2 (3 )  1 2 6 m m
1 42
20
 6 4 .5 4
o
(Eq.12.12a)
(Eq.12.12b)
(c)
2

60
1
L  [ 6 0  1 2 6 ] 2 2  1 3 9 .6 m m
L
2
Equation ( a ) of Sec.12.6:
p
g
L
3
 46 . 5 mm ;
Thus
b  30 m m
126
( d ) Equation ( b ) of Sec.12.6:
c  0 .1 8 8 m  0 .0 0 5  0 .1 8 8 ( 3 )  0 .0 0 5  0 .5 6 9 m m
200
10 m  30
SOLUTION (12.19)
Equation (12.13): d p  2 0 0 m m ,
ta n 
g

N
g
N
p

 2;
1
rs
d
dg 
 g  6 3 .4 3 ,

o
p
rs
p

200
0 .5
 400 m m
 9 0  6 3 .4 3  2 6 .5 7
o
Therefore
rp ,avg  rp 
b
2
s in 
r g , a v g  rg 
b
2
s in  g  2 0 0  3 2 .5 s in 6 3 .4 3  1 7 0 .9 m m
Vp 
d
p ,a v g
np
o
 8 5 .4 6 m m
o

60
 1 0 0  3 2 .5 s in 2 6 .5 7
p
 ( 0 .1 7 0 9 )( 5 0 0 )
60
 4 .4 7 m p s
Hence
Ft 
1 0 0 0 (1 1 )

1000 kW
Vp
4 .4 7
 2 .4 6 k N
Equations (12.17):
F a p  F t ta n  s in 
p
 2 .4 6 (ta n 2 0 )(s in 2 6 .5 7 )  4 0 0 N  F r g
F r p  F t ta n  c o s 
p
 2 .4 6 (ta n 2 0 )(c o s 2 6 .5 7 )  8 0 0 .8 N  F a g
o
o
2460
800.8
o
o
Gear
400
800.8
Pinion
SOLUTION (12.20)
From Solution of Prob.12.19: V p  2 .4 7 m p s ,
F t  2 .4 6 k N ,
b  65 m m
Thus
Fd 
N
p
3 .0 5  4 .4 7
3 .0 5
 d
( 2 .4 6 )  6 .0 7 k N
m  2 0 0 3 .5  5 7 ,
p
Hence, from Table 11.3:
(Eq.11.24a)
N
p
'
N

p
cos 
p
57
c o s 2 6 .5 7
(using N p '  6 3 .7 , interpolated)
Y  0 .4 2 4
 0  124 M Pa
Also, from Table 11.4:
Hence,
Fb 
Since
 0b Ym
K
f
Fb  Fd ,
1

1 2 4 ( 6 5 ) 0 .4 2 4 ( 3 .5 )
1 .4
1
o
 8 .5 4 k N
the gears are safe.
201
(Eq.11.33)
 6 3 .7
SOLUTION (12.21)
From Solution 12.20:
F d  6 .0 7
kN ,
 114,
N
N
p
 57,
N
p
'  6 3 .7
Hence
N
g
N


p
rs
2N
Q '
N
p
g
57
0 .5
'
' N
2 ( 2 5 4 .9 )

'
g
 1 .6
6 3 .7  2 5 4 .9
g
'
N

g
cos 
114
c o s 6 3 .4 3
g
o
 2 5 4 .9
(Eq.11.40, modified)
From Table 11.10: K  0 .6 9 M P a Thus
0 .7 5 d p b K Q '
Fw 
Since
cos 

0 .7 5 ( 2 0 0 )( 6 5 )( 0 .6 9 )(1 .6 )
c o s 2 6 .5 7
p
Fw  Fd ,
o
 1 2 .0 4 k N
(Eq.11.38, modified)
gears are safe.
SOLUTION (12.22)
Table 11.10: K  0 .5 4 5 M P a ,
Table 11.4:  0  1 7 2 M P a ,
N
p
' N
c o s  1  3 0 c o s 2 0  3 1 .9 3
o
p
Table 11.3: Y  0 . 364 (using N p '  31 . 93 , interpolated)
We have
r1 
N 1m

2
 1  tan
30 (8 )
 1 120
240
r1 , a v g  1 2 0 
V1 
 d 1 , a v g n1
N2
 26 . 6 , 
2
60 (8 )
2
60
60
cos 63 . 4
o
 240 m m
 90  26 . 6  63 . 4
s in  1  1 0 4 .3 m m ,
 ( 0 .2 0 8 7 )( 7 2 0 )

cos  2
r2 
o
70
2

60
N2'
 120 m m ,
2
o
(Eq.12.12b)
r2 , a v g  2 4 0 
70
2
s in  2  2 0 8 .7 m m
 7 .8 7 m p s
 134 ,
N 1' 
30
cos 26 . 6
o
 33 . 6
(Eq.12.15)
Hence
Fb 
Q'
 0b Ym
K
1 7 2 ( 7 0 ) 0 .3 6 4 ( 8 )

1
f
2N2'
N 1 ' N 2 '
1 .5
 1 .6
1
 2 3 .4 k N
(Eq.11.33)
(Eq.11.40, modified)
It follows that
Fw 
0 .7 5 d 1 b K Q '
cos  1

0 .7 5 ( 2 4 0 )( 7 0 )( 0 .5 4 5 )(1 .6 )
c o s 2 6 .6
o
 1 2 .2 9 k N
Since F w  F b , power capacity depends on F w . We find that
Fd 
and
3 . 05  7 . 87
3 . 05
F t  3 . 58 F t
1 2 .2 9  3 .5 8 F t ,
(Eq.11.24a)
F t  3 .4 3 k N
Therefore
kW 
Ft V
1000

3 4 3 0 ( 7 .8 7 )
1000
 27
202
(Eq.11.38, modified)
SOLUTION (12.23)
( a ) d p  N p m  30(4)  120 m m ,

d
g
 ta n
1
1
dg
( d )  ta n
p
p ,a v
 d
p
V av   d
Ft 
 b s in 
o
 1 2 0  4 5 s in 1 8 .4
p
7 4 5 .7 ( 4 0 )

Vav

( 13 26 00 )  7 1 .6 ,
1500
500
 360 m m
 9 0  7 1 .6
o
p
o
o
 1 8 .4
o
 1 0 5 .8 m m
n   ( 0 .1 0 5 8 )(1 5 0 0 6 0 )  8 .3 1 m p s
p ,av
7 4 5 .7 h p
d g  (1 2 0 )
 3 .5 9 k N
8 .3 1
( b ) F a  F t ta n  s in 
p
 3 .5 9 ta n 2 0 s in 1 8 .4
F r  F t ta n  c o s 
p
 3 .5 9 ta n 2 0 c o s 1 8 .4
o
o
 0 .4 1 2 k N
o
o
 1 .2 4 k N
Equation (1.16):
Tp 
9549 kW
Tg 
9549 kW
ng
np

9549 (30 )

9549 (30 )
1500
500
 191 N  m
 573 N  m
SOLUTION (12.24)
r1 
1
2
r2 
1
2
mN
mN
 1  ta n
1

2
( 8 . 5 )( 30 )  127 . 5 mm
1
2

 1 1 2 7 .5
255
o
Table 11.4: 
70
2
70
2
sin 26 . 6
sin 63 . 4
 172
0
o
(Eq.11.4)
mm
 2  6 3.4
 2 6 .6 ,
r1 , av  127 . 5 
r 2 , av  255 
( 8 . 5 )( 60 )  255
1
2
o
(Eq.12.12b)
 111 . 8 mm
o
 223 . 7 mm
Table 11.10: K  0 . 545
MPa ,
MPa
Note that number of teeth is as same as in Prob.12.14. Hence, from
Solution of Prob.12.22: Q '  1.6 , Y  0 . 364
Therefore, we have:
Fb 
 0b
K
Ym 
0 . 75 d 1 bKQ '
Fw 
6
1 7 2 (1 0 )( 0 .0 7 )
1 .5
f
cos  1

( 0 .3 6 4 )( 0 .0 0 8 5 )  2 4 .8 3 k N
0 . 75 ( 0 . 255 )( 0 . 07 )( 0 . 545  10
cos 26 . 6
o
6
)( 1 . 6 )
 13 . 06 kN
(Eq.11.33)
(Eq.11.38, modified)
Also
V 1   d 1 n1 6 0   ( 0 .2 2 3 6 )( 7 2 0 6 0 )  8 .4 3 m s
Ft 
3 .0 5  8 .4 3
3 .0 5
F t  3 .7 6 4 F t ;
1 3 .0 6  3 .7 6 4 F t ,
Hence
kW 
F V1
1000

3 , 4 7 0 ( 8 .4 3 )
1000
 2 9 .2 5
203
F t  3 .4 7
kN
(Eq.11.24a)
SOLUTION (12.25)
Worm:
Vg
ta n  
d
 tan  w
g
c 
1
2
( d g 2 )wg

Vw
(d
2 )w
p
dw
 tan 35
rs
dg
dw
o
(Eq.12.21)
rs
dw
 28 d w
0 . 025
( d w  d g )  150 ,
29 d w  300,
and

p
dw  d
 300
g
d w  1 0 .3 4 5 m m
d g  2 8 (1 0 .3 4 5 )  2 8 9 .7 m m
Then
L   d w ta n  w   (1 0 .3 4 5 )(ta n 3 5 )  2 2 .7 6 m m
o
Gear:
pg 
p ng

c o s
 1 1 .4 8 m m (Eq.12.1)
9 .4
cos 35
o
But p g  p w .
Hence
Nw 

L
pw
2 2 .7 6
1 1 .4 8
 1 .9 8 or 2 teeth
(Eq.12.20)
SOLUTION (12.26)
(T w ) i 
9549 kW
n
9549 (30 )

1800
 1 5 9 .2 N  m
( T w ) o  1 5 9 .2  0 .9 0  1 4 3 .3 N  m
ng  nw
N
w
N
g
 1800 ( 802 )  45 rpm
(T g ) o  (T w ) o
nw
(Eq.12.18)
 1 4 3 .3 ( 1 84 05 0 )  5 .7 3 2
ng
kN  m
Thus
(Tg )0
Ft 
d 2

5 .7 3 2
0 .2 5 2
 4 5 .8 6 k N
SOLUTION (12.27)
n 3   ( 22 00 )( 550 )( 2 5 0 )   2 5 rp m
Gear 3 rotates
counterclockwise (negative) at 25 rpm.
SOLUTION (12.28)
( a ) N g N w  40 2  20,
d g  4 0 (3 .2 )  1 2 8 m m ,
L  2 (1 0 .0 5 )  2 0 .1 m m ,
(b)
ta n  
c 
1
2
L
 dw

d w  3 .5 (1 0 .0 5 )  3 5 .1 8 m m
2 0 .1
 ( 3 5 .1 8 )
(d w  d g ) 
1
2
p   (3 .2 )  1 0 .0 5 m m
,
  1 0 .3
o
(3 5 .1 8  1 2 8 )  8 1 .5 9 m m
204
SOLUTION (12.29)
( a ) dw  c
2  210
0 .8 7 5
ta n  

L
 dw
0 .8 7 5
m
 ( 5 3 .8 )

2  5 3 .8 m m ,
6
5 3 .8
  6 .3 6
;
d g  2 c  d w  4 2 0  5 3 .8  3 6 6 .2 m m
o
( b ) V w   d w n w   ( 0 .0 5 3 8 )(1 2 0 0 6 0 )  3 .3 8 m s
where
d g  3 6 6 .2 m m ,
b  25 m m
K w  1 5 0 p s i  1 5 0 ( 6 8 9 5 )  1 .0 3 4 M P a
(Table 12.2)
Thus,
F w  (3 6 6 .2 )( 2 5 )(1 .0 3 4 )  9 .4 7 k N
SOLUTION (12.30)
p g   m   ( 4 )  1 2 .5 7 m m  p w
( a ) L  N w p w  4 (1 2 .5 7 )  5 0 .2 8 m m
( b )   ta n
1
L
 dw
 ta n
 1 5 0 .2 8
 (60 )
 1 4 .9 4
(Eq.12.20)
o
(Eq.12.21)
( c ) d g  N g m  90(4)  360 m m
c 
(d w  d g ) 
1
2
1
2
(6 0  3 6 0 )  2 1 0 m m
SOLUTION (12.31)
(a)
rs 
c 

1
20
1
2
pg 
N
w
N
g
,
N
g
 3( 2 0 )  6 0
(d g  d w )  175 
d g
N
 ( 275 )

(d g  75)
 d g  275 m m
 14 . 4 mm  p w
60
g
1
2
Then, Eqs.(12.20) and (12.21):
L  p w N w  1 4 .4 (3)  4 3 .2 m m
  tan
1
L
d w
 tan
1
43 . 2
 ( 75 )
 10 . 39
o
(b)
Vw 
 d w nw
60

Fw t  F ga 
 ( 0 .0 7 5 )(1 0 0 0 )
60
1000 kW
Vw

 3 .9 3 m p s
1 0 0 0 ( 7 .5 )
3 .9 3
 1 .9 1 k N
(CONT.)
205
12.31 (CONT.)
(c)
Vs 
Vw

cos 
 4 m ps
3 .9 3
c o s 1 0 .3 9
o
f  0 . 024
Hence
(Eq.12.28)
(from Table 12.3)
Thus, Eq.(12.27):
c o s  n  f ta n 
e 
o
o
o
o
c o s 2 0  0 . 0 2 4 ( ta n 1 0 . 3 9 )

cos  n  f cot 
c o s 2 0  0 . 0 2 4 ( c o t 1 0 .3 9 )
 0 .8 7 4
or
8 7 .4 %
Power delivered to machine
( k W ) m  0 .8 7 4 ( 7 .5 )  6 .5 6 k W
SOLUTION (12.32)
p w  1 4 .4 m m ,
From Solution of Prob.12.31:

pn 
We have
Table 11.4:
Fb 
p w c o s


1 4 .4 c o s 1 0 .3 9
 0  172 M Pa
 0b Ym n
K

1
f
o
 0 .2 2 2
o
mn 
d g  275 m m
 4 .5 1 m m
1
pn
Table 12.1: Y  0 .3 9 2
1 7 2 ( 3 8 ) 0 .3 9 2 ( 4 .5 1 )
1 .4
  1 0 .3 9 ,
1
 8 .2 5 k N
(Eq.11.33, modified)
Also
Vg 
T 
Ft 
Hence
Fd 
 d g ng
60
9540 kW
ng

T
dg 2
6 .1  V g
6 .1

 ( 0 .2 7 5 )( 5 0 )

9 5 4 0 ( 7 .5 )
60
50
1 .4 3
0 .2 7 5 2

 0 .7 2 m p s
 1 .4 3 k N  m
 1 0 .4 k N
6 .1  0 .7 2
6 .1
F t  1 .1 2 (1 0 .4 )  1 1 .6 5 k N
(Eq.11.24b)
Since
Fb  Fd
gears are not safe.
SOLUTION (12.33)
From Solution of Prob.12.32:
Table 12.2:
K
w
F d  1 1 .6 5 k N
 1 . 295 MPa
Therefore
F w  d g b K w  2 7 5 (3 8 )(1 .2 9 5 )  1 3 .5 3 k N
Since
Fw  Fd
gears are safe.
206
(Eq.12.22)
SOLUTION (12.34)
We have c  1 7 5 m m  7 in . and p o w e r  7 .5 k W  1 0 h p
Equation (12.24): A  0 .3( 7 )
1 .7
 8 .2 ft
2
e  8 7 .4 % , (from Solution of Prob.12.31)
By Fig.12.17:
o
C  5 6 ft  lb m in  ft  F
2
Hence, Eqs. (12.25) and (12.26):
Since
(hp )d 
C At
3 3 ,0 0 0
( hp ) i 
( hp ) d
1 e


5 6 ( 8 .2 )(1 0 0 )
3 3 ,0 0 0
1 . 392
1  0 . 874
 1 .3 9 2 h p = 1 .0 4 4 k W
 11 . 05  8 . 288 kW
( hp ) i  10 hp or ( k W ) i  7 .5 k W , overheating will not be a problem.
End of Chapter 12
207
CHAPTER 13
BELTS, CHAINS, CLUTCHES, AND BRAKES
SOLUTION (13.1)
( a ) w  (5 )(1 0 0 )(1 0 , 8 0 0 )  5 .4 N m
V 
 dn
 ( 0 .1 2 5 )(1 5 0 0 )

60
60
F1  F 2 
Fc 
w
g

2
V
5 .4
9 .8 1
[
 7 6 3 .7 N
9 .8 2
 ( 0 .1 2 5 )(1 5 0 0 )
60
    2   1 8 0  2 s in
o
  e
Thus
f
 e
0 .3 ( 2 .9 7 8 )
1500
 4 7 .7 5 N  m
 9 .8 2 m p s
1 0 0 0 ( 7 .5 )

1000 kW
V
9 5 4 9 ( 7 .5 )
T1 
]  5 3 .0 5 N
1
2

T1
r1
(Eq.13.13)
( 1 8 71.55 2 56 2 .5 )  1 7 0 .6
 2 .4 4 3
F1  F c  (   1 )
(Eq.13.3)
o
 2 .9 7 8 ra d
(Eq.13.21)
 5 3 .0 5  ( 12 .4.4 44 33 )
4 7 .7 5
0 .0 6 2 5
 1 .3 4 7 k N
(Eq.13.20)
F 2  1, 3 4 7  7 6 3 .7  5 8 3 .3 N
( b ) L  2 c   ( r1  r2 ) 
( r2  r1 )
2
(Eq.13.9)
c
 2 (1, 5 2 4 )   ( 6 2 .5  1 8 7 .5 ) 
(1 2 5 )
2
1525
 3 8 4 5 .6 m m .
SOLUTION (13.2)
( a ) T1 
9549 kW
9549 ( 10 )

n1
2800
( b ) T1  ( F1  F 2 ) r1 ,
r1
r2

n1
n2
 34 . 1 N  m
F1  F 2 
T1
r1

 F 2  2 2 7 .3  F 2
3 4 .1
0 .1 5
n
r2  r1 ( n 1 )  0 .1 5 ( 12 68 00 00 )  0 .2 6 2 5
,
(Eq.13.19)
m
2
r2  r1
s in  

c
0 . 2 6 2 5  0 .1 5
0 .7
    2   2 . 819
  9 .2 5
 0 .1 6 1 ,
o
(Eq.13.6)
rad
( c ) V   d n 6 0   ( 0 .3)( 2 68 00 0 )  4 3 .9 8 m p s
w  25 , 000 ( 0 . 06  0 . 0005 )  0 . 75 N m
Fc 
w
g
V
2

0 . 75
9 . 81

T1
F1  F c  (   1 )
r1
( 43 . 98 )
2
 147 . 9 N ,
 1 4 7 .9  ( 10 .7.7 55 77 )
3 4 .1
0 .1 5
  e
0 . 2 ( 2 . 819 )
 6 7 5 .5 N
 1 . 757
(Eq.13.20)
F 2  6 7 5 .5  2 2 7 .3  4 4 8 .2 N
We have K s  1.5 (Table 13.5). Therefore
F max  1 . 5 ( 675 . 5 )  1 . 013

max

1 , 013
60  0 . 5
kN
 33 . 77 MPa
208
(Eq.13.22)
SOLUTION (13.3)
We have r2  2 0 0 m m and f  0 .2 5 . The remaining data are the same.
Equation (13.6) gives
1
  s in [
r2  r1
c
]  s in
1
[ 2 01 09 506 2 ]  4 .0 5 8
o
Then
    2   1 8 0  2 ( 4 .0 5 8 )  1 7 1 .9
o
e
f
 e
o
( 0 .2 5 ) (1 7 1 .9 ) (  1 8 0 )
o
 2 .1 1 7
and
F1  2 1 4
F2  2 1 4
 2 .1 1 7
or
F1  2 .1 1 7 F 2  2 3 9
(1)
Also
F1  F 2  6 2 5 N
Solving,
F1  1, 3 9 8 .5 N ,
F 2  7 7 3 .5 N
Length of the belt, by Eq. (13.9):
L  2 c   ( r2  r1 ) 
1
C
( r2  r1 )
2
 2 (1 .9 5 )   ( 0 .2 0 0  0 .0 6 2 ) 
1
1 .9 5
( 0 .2 0 0  0 .0 6 2 )
2
 4 .7 3 3 m
SOLUTION (13.4)
Pulley A
Using Eq. (13.16), with F1  2 .5 k N ,
F1
F2
 e
f
,
2 .5
F2
 e
0 .1 5 (1 2 0 )(  1 8 0 )
f  0 .1 5 ,
 1 .3 6 9 ;
Fc  0 ,
  120 :
F 2  1 .8 2 6 k N
Pulley B
Then, equilibrium condition
T
 0 applied to free-body diagram of
pulley B gives
T
B
 0;
T B  2 .5 ( 0 .1 5 )  1 .8 2 6 ( 0 .1 5 )  0
or
TB  1 0 1 N  m
B
TB
F1  2 .5 k N
0.15 m
F 2  1 .8 2 6 k N
Similarly, we have
T
A
 0; T A  2 .5 ( 0 .0 2 )  1 .8 2 6 ( 0 .0 2 )  0 ,
209
TA  13 N  m
o
SOLUTION (13.5)
( a ) V1 
2  r1 n1

60
2  ( 0 .0 3 7 5 )( 2 5 0 0 )
 9 .8 2 m p s
60
From Eq. (1.15); with F  F1  F 2 :
kW 
FV
1000
1 .5 
;
( F1  F 2 ) 9 .8 2
1000
F1  F 2  1 5 2 .7 5 N
(1)
( b ) Evaluate V 2 as,
V1 
or
2  r2 n 2
60
9 .8 2 
;
r2  9 3 .8 in .,
2  r2 (1 0 0 0 )
60
d 2  1 8 7 .6 m m
and
1
    2     2 s in [
   2 s in
1
r2  r1
c
]
[ 9 3 .86 2 53 7 .5 ]  2 .9 6 ra d
( c ) Centrifugal force, using Eq. (13.13),
Fc 
w
g
V

2
1 .7 5
9 .8 1
(9 .8 2 )  1 7 .2 N
2
Then, Eq. (13.16):
F1  1 7 .2
 e
F 2  1 7 .2
0 .3 5 ( 2 .9 6 )
 2 .8 1 8
(2)
Solving Eqs. (1) and (2), we have
F1  2 5 4 N
F 2  1 0 1 .2 N
SOLUTION (13.6)
We have
w g  8 .8 9 .8 1  0 .8 9 7
and
P  V ( F1 
w
g
V )  V ( 2 5 0 0  0 .8 9 7 V )
2
2
Thus
P
V
 0  5 0 0  0 .8 9 7 (3 )V ;
2
So
V  2 r n ;
Solving
V  3 0 .4 8 m p s
3 0 .4 8  2  ( r )( 4 2 0 0 6 0 )
r  6 9 .3 m m
SOLUTION (13.7)
By Eq. (13.13):
Fc 
w
g
V
2

1 .4
9 .8 1
[ 2  ( 0 .0 8 )(3 0 0 0 6 0 )]  9 0 .1 N
2
From Eq. (13.16),
1 1 0 0  9 0 .1
F 2  9 0 .1
 e
o
( 0 .2 5 ) ( s in 1 8 ) (1 6 0  1 8 0 )
 1 .2 4 1
Solving
F 2  9 0 3 .9 N
(CONT.)
210
13.7 (CONT.)
Then
T A  ( F1  F 2 ) r1  (1 1 0 0  9 0 3 .9 )( 0 .0 8 )  1 5 .7 N  m
and
P 
3 0 0 0 (1 5 .7 )

VT
9549
9549
 4 .9 3 k W
SOLUTION (13.8)
Fc 
w
g

2
V
F1  F c
F2  Fc
8
9 .8 1
[  ( 0 .2 )( 1 66 00 0 )]  2 2 8 .9 N
2
f  s in 
 e
3 , 0 0 0  2 2 8 .9
;
F2  2 2 8 .9
 e
  1 7 0  2 .9 6 7 ra d
o
0 .1 5 ( 2 . 9 6 7 ) s in 1 9
o
 3.9 2 4
Solving F 2  935 . 1 N
Thus
T  ( F1  F 2 ) r  (3  0 .9 3 1)  1 0 ( 0 .1)  2 0 6 .9
N m
3
kW 

Tn
9549
2 0 6 .9 (1 6 0 0 )
 3 4 .7
9549
SOLUTION (13.9)
( a ) Fc 
w
g
F1  F c
2
V
3
9 . 81
f  s in 
 e
8 0 0  Fc

[  ( 0 . 3 )(
 e
3000
60
)]
2
0 . 2 5 ( 2 . 7 9 3 ) s in 1 9
o
 679 . 1 N ,
  160
o
 2 . 793
rad
 8 .5 4
or
F1  679 . 1
 8 . 54 , F 1  1, 711 . 6 N
800  679 . 1
Also
T  ( F 1  F 2 ) r1  (1711 . 6  800 )( 0 . 15 )  136 . 7 N  m
kW 

Tn
9549
136 . 7 ( 3000 )
9549
 42 . 95
(b)

F1

max

A
1 , 711 . 6
6
145 ( 10
)
 11 . 8 MPa
SOLUTION (13.10)
r2
r1

n1
n2
2700
, r 2  0 . 1 ( 1800
)  0 . 15 m
sin  
T1 
9549 kW
  e
Fc 
0 . 15  0 . 1
0 .5
2700

w
g
V

o
9 5 4 9 (1 5 )
2700
0 . 2 ( 2 . 9 4 1 ) s in 1 7
2
  5 . 74 ;
,
2 .5
9 . 81
o
  180
 5 3 .0 5 N  m
o
 2 ( 5 . 74 )  2 . 941
rad
V   d n   ( 0 .2 )( 2 67 00 0 )  2 8 .2 7 m p s
 7 .4 7 7
( 28 . 27 )
2
 203 . 7 N
(CONT.)
211
13.10 (CONT.)

Thus F 1  F c  (   1 )
T1
r1
 203 . 7  ( 76 .. 477
)
477
53 . 05
0 .1
 816 . 1 N
Table 13.5: K s  1.4
and
F max  K s F 1  1 . 4 ( 816 . 1)  1 . 143
kN
SOLUTION (13.11)
N 1 n1
(a) N2 
2 2 (1 4 0 0 )

n2
 4 4 teeth
700
( b ) H d  H r K 1K 2
where
H r  2 6 .6 h p = 1 9 .8 k N
K 1  1 .5
N 1 P n1
1000 H
60
1 0 0 0 ( 5 0 .6 )

d
V1
So,
2 2 ( 0 .0 1 9 0 5 )(1 4 0 0 )

60
F1 
K 2  1 .7 by Table 13.10)
(from Table 13.9,
H d  2 6 .6 (1 .5 )(1 .7 )  6 7 .8 h p = 5 0 .6 k W
Hence
( c ) V1 
(by Table 13.8)
9 .7 8
n 
F a ll
n 
6 2 .6
5 .1 7
 9 .7 8 m p s
 5 .1 7 k N
F a ll  3 1 .3 ( 2 )  6 2 .6 k N
F1
(by Table 13.7)
and
 1 2 .1
SOLUTION (13.12)
N 1 n1
(a) N2 
n2

1 8 (1 6 0 0 )
640
 4 5 teeth
( b ) H d  H r K 1K 2
where
H r  1 6 .1 h p = 1 2 k W
K 1  1 .2
(from Table 13.8)
(by Table 13.9),
K 2  3 .3 (from Table 13.10)
and
H d  1 6 .1(1 .2 )(3 .3)  6 3 .8 h p = 4 7 .6 k W
( c ) V1  N 1 P n1  1 8 ( 0 .0 1 9 0 5 )(1 6 0 0 6 0 )  9 .1 4 m p s
7 4 5 .7 H
F1 
(d) n 
d
V1
F a ll
F1
,

7 4 5 .7 ( 6 3 .8 )
9 .1 4
 5 .2 1 k N
where F a ll  2 (3 1 .3)  6 2 .6 k N
Thus,
r2 
6 2 .6
5 .2 1
 12
212
(by Table 13.7)
SOLUTION (13.13)
n1
n2

4
1
 3
 Use c  2 ( r2  r1 )
(Sec.13.6)
We have
r1 
N1P

2
2 2 (1 6 )
2
 5 6 .0 2 m m ,
r2 
n1
n2
r1 
4
1
(5 6 .0 2 )  2 2 4 .0 8 m m
Thus
c  2 ( 2 2 4 .0 8  5 6 .0 2 )  3 3 6 .1 2 m m
r1  r2  2 8 0 .1 m m
Since
sprocket will clear.
SOLUTION (13.14)
n1
n2

r1 
 2 . 19  3  Use c  2 ( r1  r2 )
4600
2100
N1P

2
1 4 (1 4 )
2
 3 1 .1 9 m m ,
n1
r2 
n2
(Sec. 13.6)
r1  2 .1 9 (3 1 .1 9 )  6 3 .3 1 m m
c  ( r1  r2 )  3 1 .1 9  6 3 .3 1  9 4 .5 m m
c  2 ( 9 4 .5 )  1 8 9 m m
Use c  1 9 0 m m
SOLUTION (13.15)
( a ) Speed ratio is ; 3:1. Thus
N 2  3( 2 3)  6 9 teeth
( b ) H d  H r K 1K 2
where
H r  1 9 .5 h p
(from Table 13.8)
K 1  1 .3
(from Table 13.9),
K 2  1 .7 (by Table 13.10)
H d  1 9 .5 (1 .3)(1 .7 )  4 3 .1 h p
and
( c ) V 1  N 1 P n1  2 3( 0 .0 1 9 )(1 8 0 0 6 0 )  1 .3 1 m p s
7 4 5 .7 H
F1 
(d) n 
d
V1
F a ll
F1

7 4 5 .7 ( 4 3 .1 )
1 3 .1
 2 .4 5 k N
where F a ll  4 (3 1 .3)  1 2 5 .2 k N
,
(Table 13.7)
So,
n 
1 2 5 .2
2 .4 5
 5 1 .1
SOLUTION (13.16)
( a ) N 2  3(3 5 )  1 0 5 teeth
( b ) H d  H r K 1K 2
where
H r  3 6 .6 h p = 2 7 .1 k W
(by Table 13.8)
(CONT.)
213
13.16 (CONT.)
K 1  1 .3 ,
K 2  1 .7
(by Table 13.9 & 13.10)
and
H d  3 5 (1 .3)(1 .7 )  7 7 .3 5 h p = 5 7 .6 8 k W
( c ) V 1  N 1 P n1  3 0 ( 0 .0 1 9 0 5 )(9 0 0 6 0 )  8 .5 7 m p s
F1 
(d) n 
7 4 5 .7 H
F a ll
F1
7 4 5 .7 ( 7 7 .3 )

d
V1
 6 .7 3 k N
8 .5 2
where F a ll  2 (3 1 .3)  6 2 .6 k N
,
Hence
n 
 9 .3
6 2 .6
6 .7 3
SOLUTION (13.17)
(a)
p m ax 
T 
2 Fa
2(6)

 d (D d )
 ( 0 .1 5 )( 0 .2 5  0 .1 5 )
Fa f ( D  d ) 
1
4
1
4
 2 5 4 .6 k P a
(Eq.13.28)
( 6 , 000 )( 0 . 3 )( 0 . 25  0 . 15 )  180
N m
(Eq.13.30)
(b)
p max 
T 
1
3
4 Fa
 (D
2
d
2
3
Fa f
D d
D
2

)
d
3
4(6)
2
 ( 0 . 25  0 . 15

2
2
)
 191
3
1
3
( 6 , 000 )( 0 . 3 )
(Eq.13.32)
kPa
0 . 25  0 . 15
0 . 25
2
 0 . 15
3
2
 183 . 8 N  m
SOLUTION (13.18)
(a) T 
9549 kW
n
9549 (30 )

500
 5 7 2 .9 N  m
N  2
and
T 
1
12
 fp
max
3
(D
 577 , 267 . 7 d
 d ) 
3

3

( 0 . 25 )( 140  10 )[ 64 d
3
12
3
 d ]
3
(Eq.13.33)
572 . 9
2
Solving d  7 9 .2 m m and D  4 d  3 1 6 .8 m m
( b ) Fa 
1
4
 p m ax ( D  d ) 
2
2

4
(1 4 0  1 0 )[1 6 d
3
2
 d ]  1 .6 4 9 (1 0 ) d
2
6
2
 1 0 .3 4
kN
SOLUTION (13.19)
( a ) T  5 7 2 .9 N  m
We now have
T 
1
8
N  2 (from solution of Prob. 13.18)
 fp m a x d ( D  d ) 
 206 ,167 d
2
3

2

8
( 0 .2 5 )(1 4 0  1 0 ) d (1 6 d
3
2
d )
2
(Eq.13.29)
572 . 9
2
Solving, d  1 1 1 .6 m m and D  4 d  4 4 6 .4 m m
(CONT.)
214
13.19 (CONT.)
( b ) Fa 
1
2
 p max d ( D  d )


( 0 .1 4 )(1 1 1 .6 )[ 4 4 6 .4  1 1 1 .6 ]  8 .2 2 k N
2
(Eq.13.28)
SOLUTION (13.20)
( a ) From Eq. (13.29), with a factor of safety n :
d (D
d ) 
2
2
8 nT
;
 fPm a x
2
0 .0 5 ( D
 0 .0 0 2 5 ) 
8 (1 .6 )(1 3 5 .6 )
6
 ( 0 .3 )(1 .6  1 0 )
Solving D  1 5 9 .8 m m
( b ) From Eq. (13.30), we have
Fa 
4 nT
f (Dd )
4 (1 .6 )(1 3 5 .6 )

( 0 .3 )(1 5 9 .8  5 0 )
 1 3 .8 k N
SOLUTION (13.21)
(a) T 
9549 kW
n
9 5 4 9 ( 3 7 .5 )

 8 9 5 .2 N  m
400
and
T 
 fp
1
8
 0 .8 (1 0
Fa 

2
( b ) p avg 
d (D
max
3
Fa
2

(D d
2
)
 d ) 
2
) p m ax 

p m ax d ( D  d ) 

2
2
8
( 0 . 2 ) p max 0 . 15 [ 0 . 3
2
 0 . 15
2
2
 ( 0 .3  0 .1 5 )
 1 8 6 .6 k P a
4
SOLUTION (13.22)
( a ) Use Eq. (13.29) by multiplying  3 6 0 :
T 

360
[ 81  fPm a x d ( D
2
 d )]
2
from which
p m ax 
8 (360  )T
2
 fd ( D  d
2
)
8 ( 3 6 0 8 0 ) (1 8 0 0 )

2
2
 ( 0 .3 ) ( 0 .2 ) ( 0 .2 8  0 .2 )
 8 9 5 2 .5 k P a
( b ) From Eq. (13.28) by multiplying  3 6 0 :
Fa 

360
[ 12  p m a x d ( D  d )] 
80
360
[ 2 (8 9 5 2 .5 )( 0 .2 )( 0 .4 8 )]
 300 kN
( c ) Each cylinder supplies a force F a 2 . Thus
p hyd 
Fa 2
d
2
4

3 0 0 ,0 0 0 2
 ( 0 .2 )
2
4
]
(Eq.13.29)
p m a x  2 7 9 .8 k P a
,
( 0 .2 7 9 8 )(1 5 0 )(1 5 0 )  9 .8 9 k N
9890 ( 4 )
2
8 9 5 .2
4

 4 .7 7 M P a
215
(Eq.13.28)
SOLUTION (13.23)
( a ) Use Eq. (13.33) by multiplying  3 6 0 :

T 
[ 12 fPm a x ( D  d )]
3
360
3
from which
12 (360  )T
p m ax 
3
3
 f (D d )

1 2 ( 3 6 0 8 0 ) (1 8 0 0 )
3
3
 ( 0 .3 ) ( 0 .2 ) ( 0 .2 8  0 .2 )
 7 .3 9 M P a
( b ) From Eq. (13.32), by multiplying by  3 6 0 :

Fa 
(c)
p hyd 
Fa 2
d
2
360

4
[ 4 p m a x ( D
2 ( 49500 )
 ( 0 .0 4 )
 d )] 
2
2
80
360
[ 4 ( 7 , 3 9 0 )( 0 .2 8  0 .2 )]  4 9 .5 k N
2
2
 1 9 .7 M P a
2
SOLUTION (13.24)
T 
9549 (35 )
800
and T 
 4 1 7 .8 N  m
 fp m a x
1 2 s in 
 d ) 
3
(D
3
R is e 
3
 ( 0 .3 ) 4 2 0 (1 0 )
1 2 s in 8
o
1
2
( D  d )  w s in 
[ 0 .2 5  d ]  4 1 7 .8
3
3
(Eq.13.41)
Solving d  2 4 0 .2 m m
Therefore
w 
1 D d
2 s in 

2 5 0  2 4 0 .2
2 s in 8
 3 5 .2 1 m m
o
SOLUTION (13.25)
T  4 1 7 .8 N  m (from Solution of Prob. 13.24)
Now we have
T 
 fp m a x d
8 s in 
(D
Solving 0 .0 6 2 5 d  d
2
 d ) 
2
 0 .0 0 1 2
3
3
 ( 0 .3 ) 4 2 0 (1 0 )
8 s in 8
o
[ 0 .0 6 2 5 d  d ]  4 1 7 .8
3
or d  2 4 0 m m (by trial and error)
Thus, we have
w 
1 D d
2 s in 

250  240
2 s in 8
o
 3 5 .9 m m
SOLUTION (13.26)
( a ) Rise  w sin   80 sin 10
o
 13 . 89 mm
D  5 0 0  1 3 .8 9  5 1 3 .8 9 m m
d  5 0 0  1 3 .8 9  4 8 6 .1 1 m m
Equation (13.38):
T 
 ( 0 . 2 )( 500 )( 0 . 48611 )
8 sin 10
o
[ 0 . 51389
2
 0 . 48611
2
]  3 . 05 kN  m
Equation (13.37):
F a  12  (5 0 0 )( 0 .4 8 6 1 1)[ 0 .5 1 3 8 9  0 .4 8 6 1 1]  1 0 .6 1 k N
( b ) kW 
Tn
9549

3 ,0 5 0 ( 5 0 0 )
9549
 1 5 9 .7
216
(Eq.13.38)
SOLUTION (13.27)
Fa 

4
2
p max ( D
 d ),
2
D
2
 d
4 Fa

2
 p max

 0 . 016
4(5)
 ( 400 )
We have
1
2
( D  d )  0 .2 5
D  d  0 .5
or
(1)
Thus
 d
2
D
2
 ( D  d )( D  d )  0 .5 ( D  d )  0 .0 1 6
or
D  d  0 .0 3 2
(2)
D  266
From Equation (1) & (2):
d  234
mm ,
mm
Equation (13.41):
T 
5000 ( 0 . 2 ) 0 . 266
3
 0 . 234
3
o
2
 0 . 234
2
3 sin 12
0 . 266
 602
N m
SOLUTION (13.28)
d Fn
Fn 

d/2

D/2
r

D 2
d 2
p max
F h  0:
2  rdr
sin 
F a  F n s in 


D 2
d 2
p max (
2  rdr
sin 
) sin 
Fa
This, after integration, yields Eq.(13.40).
T  fF n r 
Similarly,

D 2
d 2
fp
max
(
2  rdr
sin 
)r
Integrating gives Eq.(13.41).
SOLUTION (13.29)
F 1  wrp
max
 ( 0 . 1)( 0 . 2 )( 700 )  14 kN
f   0 .3 ( 2 4 0 
2
360
(Eq.13.45)
)  1.2 5 7
We have
F2 
F1
e
f

14
e
1 .2 5 7
 3 .9 8 3 k N
Thus
T  ( F 1  F 2 ) r  (14  3 . 983 )( 0 . 2 )  2 . 003
and
kW 
Tn
9549

2 , 0 0 3(1 5 0 )
9549
 3 1.4 6
217
kN  m
SOLUTION (13.30)
  2 1 0  3 .6 6 5 ra d
T  I   2 .3( 2 0 0 )  4 6 0 N  m
o
We have
F1  F 2 
F1
F2
 e
f

T
r
 e
460
0 .1 2 5
 3 .6 8 k N
0 .3 ( 3 . 6 6 5 )
(Eq.13.42)
 3 .0 0 3
Solving
F 1  3 . 003 F 2  3 . 003 ( F 1  3 . 68 )
or
F1  5 .5 2 k N
F 2  1 .8 4 k N
Thus
Fa  F2
 1, 8 4 0 ( 13 20 50 )  7 6 7 N
r
a
(Eq.13.43)
SOLUTION (13.31)
T 
9549 kW
n

9549 ( 40 )
 636 . 6 N  m
600
(1)
Also
F1  F 2 e
f
 F2 e
 5 .7 2 7 F 2
0 . 4 ( 4 .3 6 3 )
and
T  r ( F1  F 2 )  0 .2 5 ( 5 .7 2 7 F 2  F 2 )  1.1 8 2 F 2
Equation (1) and (2) give
F 2  538 . 6 N
F 1  3 , 085
N
SOLUTION (13.32)
  240
o
 4 . 189
rad
(a)
F 1  p max wr  600 ( 0 . 075 )( 0 . 15 )  6 . 75 kN
Also
F1
F2
 e
f
 e
0 . 4 ( 4 .1 8 9 )
 5 .3 4 2
Thus
F1  5 .3 4 2 F 2
Fa  F2
r
a
and
F 2  1 . 264
 (1 . 264  10 )
3
150
400
 474
( b ) T  r ( F 1  F 2 )  0 . 15 ( 6 . 75  1 . 264 )  10
kW 
Tn
9549

823( 200 )
9549
kN
 1 7 .2 4
218
3
N
 823
N m
(2)
SOLUTION (13.33)
(a)
F 1  p max wr  800 ( 0 . 06 )( 0 . 15 )  7 . 2 kN
T 
9549 ( 10 )

9549 kW
n
 434
220
T  ( F1  F 2 ) r ,
N m
F 2  F1 
 7 .2 
T
r
0 . 434
0 . 15
 4 . 307
kN
We have
e
f

F1
F2

f   ln 1.6 7 2  0 .5 1 4
 1.6 7 2 ,
7 .2
4 .3 0 7
Thus
 
0 .5 1 4
0 .1 4
(b)
 3 .6 7 1 ra d  2 1 0 .3
o
129.4
From triangle ABC:
3 0 .4
s  2 0 6 .6 c o s 3 0 .4
 178 . 2 mm
o
o
O 150
75.9
200
124.1
3 0 .4
o
C
A
s
B
206.6
72.8
SOLUTION (13.34)
e
f
 e
F1  e
0 . 12 ( 3 . 665 )
f
 1 . 552
  210
o
 3 . 665
rad
F 2  1.5 5 2 F 2
F2
A

M
A
 0:
F1 s  F 2 c  F a a
F1
or
1 . 552 F 2 ( 80 )  50 F 2  1 . 5 ( 300 )
from which
F 2  6 .0 7 k N , and F1  9 .4 2 k N
Thus
kW 
( F1  F 2 ) r n
9549

( 9 .4 2  6 .0 7 )( 4 )(1 0 0 )
9549
 1 0 .5 2
219
S
Fa
c
a
SOLUTION (13.35)
T 
F1  F 2 
F2
900

T
r
F1
Also
9 5 4 9 (1 5 )

9549 kW
n
f
o
ra d
 7 9 5 .7 5 N
1 5 9 .1 5
0 .2
 e
  2 1 0  3 .6 6 5
 1 5 9 .1 5 N  m
 e
0 .4 ( 3.6 6 5 )
(1)
 4 .3 3 2
or
F1  4 .3 3 2 F 2
(2)
From Eqs.(1) and (2): F 2  2 3 8 .8 N
F1  1 0 3 4 .6
N
We therefore obtain
Fa 
( cF 2  sF 1 ) 
1
a
1
250
  111 . 4 N
[100 ( 238 . 8 )  50 (1034 . 6 )]
Yes. Self-locking
SOLUTION (13.36)
From Solution of Prob.13.35:
T  159 . 15 ,
F 2  238 . 8 N ,
F 1  1034 . 5 N
Now we have
A
F2
F1
c

Fa
M
A
 0:
a
( a ) From Eq. (13.45):
F1  p m a x w r  5 0 0 (1 0 )( 0 .0 2 )( 0 .1)  1 k N
3
Equation (13.44),
 1(1 0 ) e
3
 0 .2 5 ( 2 6 5  1 8 0 )
 315 N
Equation (13.42):
T  r ( F1  F 2 )  ( 0 .1)(1 0 0 0  3 1 5 )  6 8 .5 N  m
( b ) Fa 
c F 2  s F1
a

4 0 ( 3 1 5 )  1 0 (1 0 0 0 )
200
 13 N
If F a  0 , the brake will self-lock:
s 
c F2
F1

[ c F1  s F 2 ]
No. Not self-locking
SOLUTION (13.37)
 f
1
a
 366 . 04
S
F 2  F1 e
Fa 
40 (315 )
1000
 1 2 .6 m m
220
N
SOLUTION (13.38)
( a ) From Eq. (13.45):
F1  w r p m a x  7 5 ( 2 5 0 )( 0 .4 9 )  9 .1 9 k N
Equation (13.44),
F 2  F1 e
 0 .2 5 ( 4 .5 3 8 )
 9 .1 9 ( 0 .3 2 2 )  2 .9 6 k N
Using Eq. (13.42):
T  ( 9 .1 9  2 .9 6 ) ( 2 5 0 )  1 .5 5 8 N  m
( b ) Using Eq. (13.46):
2 .9 6 (1 5 0 )  9 .1 9 ( 3 5 )
Fa 
 1 9 5 .8 N
0 .6 2 5
From Eq.(13.46): F a  0 for s  2 .9 6 (1 5 0 ) 9 .1 9  4 8 .3 m m
The brake is self-locking (for f  0 .2 5 ), if:
s  4 8 .3 m m
SOLUTION (13.39)
(a) T 
9549 kN
n
Fn 
T
fr

9549 ( 25 )

800
298 . 4
0 . 25 ( 0 . 3 )
Hence F a 
Fn
 298 . 4 N  m
 3979
N
( b  fc ) 
a
3979
1
( 0 . 4  0 . 25  0 . 05 )  1 . 542
kN
No. Not self-locking
( b ) R Ax  fF n  0 . 25 ( 3979 )  994 . 8 N 
R Ay  F n  F a  3979  1542
R A  [ 994 . 8
 2437
2
 2437
N 
1
2
] 2  2 . 632
kN
SOLUTION (13.40)
( a ) From Solution of Prob.13.39: F n  3979
N
We now have
Fa 
Fn
a
( b  fc ) 
3979
1
( 0 . 4  0 . 25  0 . 05 )  1, 641
N
No. Not self-locking
( b ) From Solution of Prob.13.39:
R A  2 . 632
kN
SOLUTION (13.41)

M
A
 0:
Fn 
4 ( 0 .4 5 )
0 .2
 9 kN
We now use
p avg 
Fn
2 ( r s in


)w
9 ,0 0 0
o
2 ( 0 .1 5 )(s in 4 5 )( 0 .0 7 5 )
 0 .5 6 6 M P a
2
(CONT.)
221
13.41 (CONT.)
T  fF n r  ( 0 .3 5 )(9  1 0 )( 0 .1 5 )  4 7 2 .5 N  m
3
Thus
and
kW 
Tn
9549

 12 . 37
472 . 5 ( 250 )
9549
Comment: Short-Shoe analysis overestimates pressure, torque, and power capacity of the brake.
SOLUTION (13.42)
( a ) From Eq.(13.47):
F n  Pm a x [ 2 ( r s in

2
)] w  (8 0 0 )(1 0 )[ 2 ( 0 .1 5 s in
3
and
fF n  0 .2 5 ( 4 4 9 5 )  1 1 2 4 N
So

M
 0:
A
44
2
o
)]( 0 .0 5 )  4 4 9 5 N
0 .5 F a  0 .0 2 5 (1 1 2 4 )  4 4 9 5 ( 0 .2 )  0 ,
F a  1 .7 4 2 k N
and
T  fF n r  1 1 2 4 ( 0 .1 5 )  1 6 8 .6 N
(b)
R A  [( 4 4 9 5  1 7 4 2 )  1 1 2 4 ]
2
2
1
 2 .9 7 4 k N
2
Fa
300
200
A
25
fF n
Fn
SOLUTION (13.43)
( a ) From Eq. (13.48):
T  fF n r ,
(b)

or
M
Fn 
T
fr

250
( 0 .4 )( 0 .3 5 )
 1 .7 9 k N
 ( 0 .3 5 )[ 0 .4  1 .7 9 (1 0 )]  ( 0 .9 F a )  0 .3 2[1 .7 9 (1 0 )]  0
3
o
3
Fa  9 1 5 N
SOLUTION (13.44)
Equation (13.47):
F n  p m a x [ 2 ( r s in

2
)] w  7 0 0[ 2 ( 0 .1  s in
30
2
o
 1 .4 5 k N
Then, Eg.(13.49):
Fa 
Fn
a
( b  fc ) 
3
1 .4 5 (1 0 )
0 .2
[ 0 .1 2  0 .2 ( 0 .0 3 ) ]
 827 N
222
)]( 0 .0 4 )
SOLUTION (13.45)
( a ) From Eq. (13.57),
T ( s in  ) m
p m ax 
(1)
2
fw r ( c o s  1  c o s  2 )
Substituting Eq. (13.52) and the data:
p m ax 
1 3 6 s in 9 0
o
2
o
 2 .7 6 M P a
o
( 0 .3 5 ) ( 0 .0 2 5 ) ( 0 .0 7 5 ) ( c o s 0  c o s 9 0 )
( b ) Introducing Eq. (13.52) and the data into Eq. (1); we have
p m ax 
1 3 6 s in 6 5
o
2
o
o
( 0 .3 5 ) ( 0 .0 2 5 ) ( 0 .0 7 5 ) ( c o s 2 0  c o s 6 5 )
 4 .8 4 M P a
SOLUTION (13.46)
c (cos 2  2  cos 2  1 )  4 r (cos  2  cos  1 )
Equation (13.54):
(1)
where 1  0 ,  2  9 0
Equation (1) thus
o
AO  c
o
c (  1  1)  4 r (  1)
c  2r
or
and
b  c cos 45
o
 2 r cos 45
o
 1 . 414 r
SOLUTION (13.47)
Refer to Figs. P13.47, 13.22, and 13.23.
1
c  [ 2 5 0  1 7 5 ] 2  3 0 5 .2 m m
2
  tan
2
 1 250
175
 1  10 . 01 ,  2  100 . 01
 55 . 01 ;
o
We have  2  9 0
o
o
o
hence ( s in  ) m  1.
( a ) Equation (13.53):
M
n

6
( 0 .0 5 )( 0 .2 )( 0 .3 0 5 2 )( 0 .9 1 0 )
[ 2 ( 2 )  s in 2 0 0 .0 2  s in 2 0 .0 2 ]  2 6 2 7 .5 N  m
o
4 (1 )
o
Equation (13.54):
M
f

6
( 0 . 3 )( 0 . 05 )( 0 . 2 )( 0 . 9  10 )
4 (1 )
[ 0 . 3052 (cos 200 . 02
o
 cos 20 . 02 )
o
 4 ( 0 .2 )(c o s 1 0 0 .0 1  c o s 1 0 .0 1 )]  2 3 8 .5 N  m
o
o
Equation (13.55):
Fa 
1
a
(M
n
 M
f
) 
1
0 .5 5
[ 2 3 8 9 ]  4 .3 4 4 k N
( b ) Equation (13.57):
T 
2
6
( 0 .3 )( 0 .0 5 )( 0 .2 )( 0 .9  1 0 )
1
[ c o s 1 0 .0 1  c o s 1 0 0 .0 1 ]  6 2 5 .6 N  m
o
Thus
kW 
Tn
9549

625 . 6 ( 600 )
9549
 39 . 31
223
o
SOLUTION (13.48)
Refer to Figs.P13.48 and 13.22.
1  1 5 ,
2  120  90
o
c 
150
s in 3 0
o
o
 300
o
and (s in  ) m  1,
  105
a  300 cos 30  250  510
o
m,
o
mm
( a ) Equations (13.53), (13.54), and (13.55):
6
M

( 0 .0 6 )( 0 .2 )( 0 .3 )( 0 .8  1 0 )
n
M

( 0 .3 )( 0 .0 6 )( 0 .2 )( 0 .8  1 0 )
f
4 (1 )
6
4 (1 )
[ 2 (1 .8 3 3 )  s in 2 4 0  s in 3 0 ]  3, 6 2 3 N  m
o
o
[ 0 .3 (  0 .5  0 .8 6 6 )  4  0 .2 (  0 .5  0 .9 6 6 )]  5 4 9 N  m
Thus
Fa 
(b) T 
1
a
(M
n
 M
2
6
( 0 .3 )( 0 .0 6 )( 0 .2 )( 0 .8  1 0 )
1
f
) 
1
0 .5 1
[3, 6 2 3  5 4 9 ]  6 .0 2 7 k N
[ 0 .9 6 6  0 .5 ]  8 4 4 N  m
(Eq.13.57)
Hence
kW 

Tn
9549
( 844 )( 500 )
9549
 44 . 19
SOLUTION (13.49)
( a ) We have  2  4 5 , Apply Eq. (13.58) to obtain
o
a 
4 r s in 
2
2  2  s in  2

4 (1 2 5 ) s in 4 5
o
2 ( 4 5 )(  1 8 0 )  s in 4 5
o
 1 5 5 .2 m m
( b ) Applying Eqs. (13.59) and (13.60):
R Ax 
2wr
a
2
Pm a x s in  2 
2 ( 0 .0 5 )( 0 .1 2 5 )
0 .1 5 5 2
2
(1 .5 5  1 0 ) s in 4 5
6
o
 1 1 .0 3 k N
R A y  fR A x  0 .3 (1 1 .0 3 )  3 .3 1 k N
( c ) From Eq. (13.61b):
T  R A y a  (3 .3 1)( 0 .1 5 5 2 )  0 .5 1 4 k N  m
End of Chapter 13
224
CHAPTER 14
MECHANICAL SPRINGS
SOLUTION (14.1)

(a) J 
32
 
TL
GJ
(8 )
4
; 80
 402 . 124
mm
 1 . 396
rad 
o
2
T ( 1 . 25 )
79 ( 10
9
)( 402 . 124  10
 12
)
or
T  35 . 48 N  m
(b)  
16T
 d
3
1 6 ( 3 5 .4 8 )

 ( 8 1 0
3
)
 353 M Pa
3
SOLUTION (14.2)
(a) d
3

3
1 6 ( 2  1 0 )( 0 .1 5 )

16 PR
  a ll
 ( 3 5 0 1 0
6
d  1 8 .7 1 m m
,
1 .5 )
( b ) Equation (14.2):
L 
4
 d G
4
 (1 8 .7 1 ) ( 7 9 )( 4 0 )

2
32 PR
3
3 2 ( 2  1 0 )(1 5 0 )
 0 .8 4 5 m  8 4 5 m m
2
SOLUTION (14.3)
Angle of twist,   T L J G ;
T   J G L , where
  2 0 (  1 8 0 )  0 .3 4 9 ra d
o
J d
3 2   (1 2 )
4
3 2  2 .0 3 6 (1 0 ) m m
4
3
4
 2 .0 3 6 (1 0
Thus,
0 .3 4 9 ( 2 .0 3 6  1 0
T 
9
9
)( 7 9  1 0 )
 4 6 .8 N  m
1 .2
From Eq.(14.1):
  1 6T  d
3
1 6 ( 4 6 .8 )

 ( 0 .0 1 2 )
3
 1 3 7 .9 M P a
SOLUTION (14.4)
D  Cd  5 ( 7 )  35 mm
 a ll  K s
8 PC
 d
2
450 
;
8 P2 ( 5 )
 (7 )
2
(1 
0 .6 1 5
5
),
Thus
k 
P2  P1

 2 1
1 5 4 2 1 0 0 0
20
 2 7 .1 N m m
and
N
a

dG
3
8C k

3
7 ( 29  10 )
3
8 ( 5 ) ( 27 . 1 )
 7 . 49 coils
225
P2  1 .5 4 2 k N
9
) m
4
SOLUTION (14.5)
Apply Eq. (14.10):
P 
4
d G
3
8D Na
where
 N a  p  d  1 2 .5  9 .5  3 m m
So,
P 
4
9
( 0 .0 0 9 5 ) ( 7 9  1 0 )( 0 .1 2 5 )
8 ( 0 .0 5 )
 1 .9 3 k N
3
By Eq. (14.6),
  8 PD K s
d
For
3
C  D d  5 0 9 .5  5 .2 6 3
Ks 1
 1 .1 1 7
0 .6 1 5
5 .2 6 3
(Eq. 14.7)
It follows that
 
8 (1 .9 3 )( 0 .0 5 )
 ( 0 .0 0 9 5 )
(1 .1 1 7 )  3 2 0 .1 M P a
3
From Table (14.3), S y s  0 .4 S u
By Eq. (14.12),
Su  Ad
 1 6 1 0 (9 .5 )
b
 0 .1 9 3
 1 0 4 2 .6 M P a
Hence
S y s  0 .4 5 (1 0 4 2 .6 )  4 6 9 .2 M P a
Since
3 2 0 .1  4 6 9 .2
 Yes.
SOLUTION (14.6)
From Eq. (14.11), k 
4

P

d G
3
8D Na
As d =constant and k proportional to d
4
3
D , The largest active coil will have the smallest
value of k .
That is, the bottom coil will deflect to zero pitch first.
Using Eq. (14.10), with  N a  p  6 m m :
P 
4
d G

3
8D Na
4
9
( 0 .0 4 ) ( 7 9  1 0 )( 0 .0 0 6 )
3
8 ( 0 .0 6 ) (1 )
 7 0 .2 N
The total deflection is
  P N a  6 (5 )  3 0 m m
SOLUTION (14.7)
( a ) From Eq. (14.11), k 
4
d G
3
8D Na
Outer spring:
ko 
4
(7 ) (79000 )
3
8 ( 40 ) ( 4 )
 9 2 .6 2 N m m
Inner spring:
ki 
4
( 4 .5 ) ( 7 9 0 0 0 )
3
8 ( 22 ) (8 )
 4 7 .5 4 N m m
(CONT.)
226
14.7 (CONT.)
By Eq. (14.11), k  W  or   W k . Hence
 
 1 4 .2 7 m m
2000
( 4 7 .5 4  9 2 .6 2 )
( b ) Force on each spring:
W o  k o   (9 2 .6 2 )(1 4 .2 7 )  1 3 2 2 N
W i  k i   ( 4 7 .5 4 )(1 4 .2 7 )  6 7 8 N
For outer spring: C  4 0 7  5 .7 1, K s  1 
0 .6 1 5
5 .7 1
For inner spring: C  2 2 4 .5  4 .8 9 , K s  1 
 1 .1 1
0 .6 1 5
4 .8 9
 1 .1 3
  (8 w D  d ) K s
3
Apply Eq. (14.6):
i 
8 (1 3 2 2 )( 4 0 )
i 
8 ( 6 7 8 )( 2 2 )
 (7 )
3
 ( 4 .5 )
(1 .1 1)  4 3 5 M P a
(1 .1 3 )  4 7 1 M P a
3
SOLUTION (14.8)
C  15 3  5
S u  Ad
b
 1510 ( 3
Table 14.3:
( a)
k 
3
Nt  N
S
 0 . 42 S u  509
MPa
3
A  1510
 23 . 7 N mm
3
 2  12 , h s  ( N t  1) d  13 ( 3 )  39 mm
8 Pmax C
( b ) S ys  K s
 d
2
Pm a x 
Thus
MPa
8 ( 5 ) ( 10 )
a
a
ys
 0 . 201
)  1 , 211
3 ( 79  10 )

dG
8C N
b   0 .2 0 1 ,
Table 14.2:
Pmax 
,
 ( 5 0 9 )( 3 )
8 ( 5 )(1 
2
0 .6 1 5
2
S ys  d
8 CK
s
 3 2 0 .4 N
)
5
SOLUTION (14.9)
(a) d
2

(1 
8 PCn
 S ys
0 . 615
C
3
8 ( 2  10 )( 5 )( 1 . 3 )
) 
6
 ( 500  10 )
(1 
0 . 615
5
)
or
d  8 . 62 mm
( b ) D  8 .6 2 ( 5 )  4 3 .1 m m ,
N

a
dG
hs  (N
(c)
c
hf

3
8C K
a
2 4 .4 4
106

3
( 8 .6 2 )( 7 9 1 0 )
3
8(5 ) (90 )
 c  1 .1 0
hf
D
90
 2 4 .4 4 m m
 7 .5 7
 2 ) d  82 . 49 mm,
 0 .2 3
3
2 (1 0 )

1 0 6 .9
4 3 .1
h f  24 . 44  82 . 49  106 . 9 mm
 2 .4 8
 Spring is safe (Curve A, Fig.14.10)
227
SOLUTION (14.10)
(a) Nt 
hs

d
 12
2 1.6
1.8
C 
N a  12  2  10
( S ys n )  d
P 
k 
2

8 K sC
3
 ( 9 0 0 2 )(1 .8 )
( 1 . 8 )( 79  10 )
 1 .1 5
P
k
 1.0 7 4
0 .6 1 5
8 .3 3
 64 N
 3 . 075
3
8 ( 8 . 33 ) ( 10 )
a
2
8 (1 .0 7 4 )( 8 .3 3 )
 s  1 .1 5
Thus
Ks  1
3

dG
8C N
 8 .3 3
15
1 .8
N mm
64
3 .0 7 5
 2 3 .9 3 m m
h f  h s   s  2 1 .6  2 3 .9 3  4 5 .5 3
(b)
s
h
h
 0 . 53
 3 . 04
f
D
f
mm
 Spring is safe (Curve B, Fig.14.10)
SOLUTION (14.11)
d

2
8CP
 S ys n
(1 
0 .6 1 5
C
) 
8 ( 8 )( 2 0 0 )
6
 ( 4 2 0  1 0 ) 2 .5
(1 
0 .6 1 5
8
),
d  5 .1 1 m m
D  5 .1 1(8 )  4 0 .8 8 m m
N
a

5 . 11 ( 79  10

dG
3
8C k
3
8 ( 8 ) ( 9  10
6
3
)
)
 10 . 95
N t  1 0 .9 5  2  1 2 .9 5,
 s  1 .2 0
h s  (1 2 .9 5 )(5 .1 1)  6 6 .2 m m
 2 6 .7 m m
200
9000
h f  6 6 .2  2 6 .7  9 2 .9 m m
Therefore
s
h
h
 0 . 287 ,
f
D
f
 2 . 23  Spring is safe (Curve A, Fig.14.10)
SOLUTION (14.12)
( a )  a ll 
8 PD
 d
3
,
d
3

8 PD
  a ll

8 ( 2 )75
6
 ( 5 2 5 1 0 )
,
d  8 .9 9 m m
( b ) C  D d  7 5 8 .9 9  8 .3 4
Na 
dG
3
8C k
 s  1 .1 0
(c)
s
hf
 0 .5 3,

P
k
3
8 .9 9 ( 7 9  1 0 )
 7 .2 9 ,
3
8 ( 8 .3 4 ) ( 2 1 )
 1 .1 0
hf
D
2
21
h s  ( N a  2  1)8 .9 9  9 2 .5 1 m m
 1 0 4 .8 m m ,
 2 .6 3
h f  h s   s  1 9 7 .3 m m
 Spring is safe (Curve B, Fig.14.10)
228
SOLUTION (14.13)
We have C  D d  2 0 2 .5  8 .0
( a ) From Table 14.3: S y s  0 .4 0 S u
where
Su  Ad
b
 2 0 6 0 ( 2 .5 )
 0 .1 6 3
 1774 M Pa
and
 m a x  0 .4 0 (1 7 7 4 )  7 0 9 .6 M P a
By Eq. (14.8):
2
 d  m ax
Pm a x 
8CK
where K w  1 
w
2
 ( 2 .5 ) ( 7 0 9 .6 )

0 .6 1 5
8
 1 .0 7 7
(Eq. 14.7)
 2 0 2 .1
8 ( 8 )(1 .0 7 7 )
Use Eq. (14.11),
k 
4

d G
3
8D Na
4
( 2 .5 ) ( 7 9 0 0 0 )
3
8 ( 2 0 ) (1 1 )
 4 .3 8 N m
( b )   Pm a x k  2 0 2 .1 4 .3 8  4 6 .1 4 m m
By Fig. 14.8c: h s  N t d
h s  ( N a  2 ) d  (1 1  2 )  3 2 .5 m m
Thus,
h f  3 2 .5  4 6 .1 4  7 8 .6 4 m m
( c ) h f D  7 8 .6 4 2 0  3 .9 3 2
 h f  4 6 .1 4 7 8 .6 4  0 .5 8 7
From Fig. 14.10, for Case B:
Yes, Buckling occurs.
SOLUTION (14.14)
( a ) Use Eq. (14.12) and Table 14.2.
Su  Ad
b
 1 6 1 0 ( 0 .9 )
 0 .1 9 3
 1643 M Pa
Table (14.3):
S y s  0 .4 5 S u  0 .4 5 (1 6 4 3 )  7 3 9 .4 M P a
( b ) Spring index, with C  D d  1 0 0 .9  1 1 .1 :
Ks 1
0 .6 1 5
C
 1
0 .6 1 5
1 1 .1
 1 .0 5 5
(Eq. 14.7)
Rearrange Eq. (14.6), let   S y s , and solve for P .
3
P 
 d S ys
8KsD

3
 ( 0 .9 ) ( 7 3 9 .9 )
8 (1 .0 5 5 )(1 0 )
 2 0 .1 N
From Fig. 14.7, N a  1 4 .5  2  1 2 .5 . Equation (14.11) is then
k 
Gd
3
4
8D Na

( 7 9 , 0 0 0 )( 0 .9 )
3
8 (1 0 ) (1 2 .5 )
4
 0 .5 1 8 N m m
(CONT.)
229
14.14 (CONT.)
(c) s 
P
k

 3 8 .8 m m
2 0 .1
0 .5 1 8
From Fig. 14.7c:
h s  ( N a  3) d  (1 2 .5  3)( 0 .9 )  1 3 .9 5 m m
h f   s  h s  3 8 .8  1 3 .9 5  5 2 .7 5 m m
Figure 14.7:
p 
(d)
s
hf

3 8 .8
1 3 .9 5
hf 2d
5 2 .7 5  2 ( 0 .9 )

Na
1 2 .5
hf
 2 .7 8 ,

D
5 2 .7 5
10
 4 .0 7 6 m m
 5 .2 7 5
Figure 14.10, Curve A: No buckling failure.
SOLUTION (14.15)
( a ) Equation (14.4), C  D d  1 4 1 .5  9 .3 3 3
Equation (14.7):
Ks 1
0 .6 1 5
9 .3 3 3
 1 .0 6 6
From Fig. 14.7:
N t  N a  2  1 6  2  1 8,
h s  d N t  1 .5 (1 8 )  2 7 m m
The pitch equals
P  ( h f  2 d ) N a  [3 5  2 (1 .5 )] 1 6  2 m m
Solid deflection
 s  h f  hs  3 5  2 7  8 m m
( b ) Equation (14.12) and Table 14.2:
S u  1 6 1 0 (1 .5 )
 0 .1 9 3
 1498 M Pa
From Table (14.3):
S y s  0 .4 5 S u  0 .4 5 (1 4 8 9 )  6 7 0 M P a
From Eq. (14.11):
P 
Gds
3
8C N a
( 2 9 , 0 0 0 ) (1 .5 ) ( 8 )

 3 .3 4 4 N
3
8 ( 9 .3 3 3 ) (1 6 )
and
 m ax 
8DKsP
8 (1 4 ) (1 .0 6 6 ) ( 3 .3 4 4 )

d
3

670
3 7 .6 5
 (1 .5 )
3
 3 7 .6 5 M P a
Then
n 
(c)
s
hf

8
3 .5
 a ll
 m ax
 0 .2 2 9 ,
hf
D
 1 7 .8

35
14
 2 .5
From Fig. 14.10  No buckling.
230
SOLUTION (14.16)
( a ) Pm 
42
2
4 C 1
4C4
Kw 
a

m
 3 kN
Kw
Pa
Ks
Pm
42
2
Pa 

0 .6 1 5
C
 1.3 1 1

1 . 311
1 . 123
1
3
 1 kN
C 
Ks  1
15
3
 5
 1.1 2 3
0 .6 1 5
C
 0 . 389
Equation (14.24), with replacing S u s by S y s :
m 
5 0 0 1 .3
( 0 .3 8 9 )( 2  5 0 0  2 8 0 )
 1 9 2 .3 M P a
1
280
and
d
2
8 Pm C
 Ks
3
8 ( 3  1 0 )( 5 )
 1 .1 2 3
 m
6
 (1 9 2 .3  1 0 )
 s  1 . 10
( b ) D  14 . 94 ( 5 )  74 . 7 mm
N

a
dG
hs  (N
(c)
s
h
f
3
8C k
a

3
1 4 .9 4 ( 7 9  1 0 )
h
f
D
mm
 48 . 89 mm
4000
90
 1 3.1 1
3
8( 5 ) ( 90 )
 2 ) d  225 . 7 mm
 0 .1 8 ,
d  1 4 .9 4
,
 3 .6 8
h f  274 . 6 mm
 Spring is safe (Curve A, Fig.14.10)
Also
fn 
356000 d
2
D N
356 , 620 ( 14 . 94 )

2
( 74 . 7 ) ( 13 . 11 )
a
 72 . 8 cps  4 , 370
cpm
SOLUTION (14.17)
From Eq. (14.20): S e s '  4 6 5 M P a ,
 P  (400  0) 2  200 N
k  P   200 10  20 N m m
By Eq. (14.8),  m a x 
8 Pm a x D
d
Kw;
3
465 
8 ( 4 0 0 )( 4 0 )
d
3
(1 .3 ) .
Solving
d  4 .8 5 m m
Equation (14.11): k  d G 8 D N a or
4
Na 
4
d G
3
8D k

3
4
( 4 .8 5 ) ( 7 9 , 0 0 0 )
3
8 ( 40 ) ( 20 )
 4 .2 7
Crash allowance =15 %
 s  1 .1 5 42000  2 3 m m
h s  ( N a  3) d  ( 4 .2 7  3)( 4 .8 5 )  3 5 .2 6 m m
h f  3 5 .2 6  2 3  5 8 .2 6 m m
231
SOLUTION (14.18)
From Eq. (14.20): S e s '  3 1 0 M P a
k  P   200 10  20 N m m
From Eq. (14.8):
8 Pm a x D
 m ax 
310 
Kw;
3
d
8 ( 4 0 0 )( 4 0 )
d
3
(1 .3 )
Solving
d  5 .5 5 m m
Equation (14.11): k  d G 8 D N a or
4
4
Na 
4
( 5 .5 5 ) ( 7 9 , 0 0 0 )

d G
3
8D k
3
3
8 ( 40 ) ( 20 )
 7 .3 2
Crash allowance =8 %
 s  1 .0 8 42000  2 1 .6 m m
h s  ( N a  3) d  ( 7 .3 2  3)(5 .5 5 )  5 7 .2 8 m m
h f  5 7 .2 8  2 1 .6  7 8 .8 8 m m
SOLUTION (14.19)
Refer to Solution of Prob. 14.15.
( a ) We now have
4 C 1
4C 4
Kw 

0 .6 1 5
C
Pm a x  Pm in
Pa 
14  4
2

2
4 ( 9 .3 3 3 )  1

4 ( 9 .3 3 3 )  4

 5 N,
0 .6 1 5
9 .3 3 3
Pm 
 1 .1 5 6
14  4
2
 9 N
From Eq. (14.8):
8CK
a 
Pm
m a(
(b) n 
Pa
S ys
Pa
2

8 ( 9 .3 3 3 ) (1 .1 5 6 ) ( 5 )
 (1 .5 )
2
 6 1 .0 5 M P a
)  6 1 .0 5 ( 95 )  1 0 9 .9 M P a

 a  m
w
d
 3 .9 2
670
6 1 .0 5  1 0 9 .9
( c ) From Table 14.3 the modified endurance limit is
S e s '  0 .2 2 S u  0 .2 2 (1 4 8 9 )  3 2 7 .6 M P a
Thus
S es '
n 
a

3 2 7 .6
6 1 .0 5
 5 .3 7
Pm 
180  30
2
SOLUTION (14.20)
( a ) C  15 3  5
Kw 
a
m

4 C 1
4C 4
Kw
Pa
Ks
Pm


0 . 615
C
 1 . 311
 105
Pa 
N
Ks  1
0 . 615
C
180  30
2
 75 N
 1 . 123
 0 . 834
1 . 311 75
1 . 123 105
Thus

m
 K
8 Pm C
s
d
2
 1 . 123
8 ( 105 )( 5 )
 ( 3  10
3
)
2
 166 . 8 MPa
 a  0 . 834 (166 . 8 )  139 . 1 MPa
(CONT.)
232
14.20 (CONT.)
Table 14.2 and Eq.(14.12):
S u  Ad
b
 2060 ( 3
 0 . 163
Table 14.3: S es  0 . 23 S u  396
'
)  1 , 722
MPa
Equation (7.5a): S us  0 . 67 S u  1,154
MPa
Equation (14.25) results in
1 ,1 5 4 ( 3 9 6 )
n 
( b ) hs  ( N a  2 ) d  7 2 m m ,
 s  1 .1 0
(c)
fn 
(d)
h
s
 1.3 8
1 3 9 .1 ( 2  1 ,1 5 4  3 9 6 )  1 6 6 . 8  3 9 6
k 
 1 8 .3 8 m m ,
180
1 0 .7 7
1 356 , 620 d
2
2
D Na
 0 .2 ,

356 , 620 ( 3 )
h
 6 . 03
f
D
f
 108
2
2 ( 15 ) ( 22 )
dG
3
8C N a

3
3 ( 7 9 1 0 )
3
8 (5 ) ( 22 )
 1 0 .7 7 N m m
h f  7 2  1 8 .3 8  9 0 .4
mm
cps  6 , 480
(for fixed-free ends)
cpm
 Buckling will occur (Curve B, Fig.14.10)
SOLUTION (14.21)
D  C d  8 (5 )  4 0 m m
Kw 
4 C 1
4C4
m  Ks

0 .6 1 5
C
8 Pm C
 d
0 .6 1 5
8
 1 .0 7 7
 1.1 8 4
8 ( 5 0 0 )( 8 )
 1 .0 7 7
2
Ks 1
 (5)
2
 4 3 8 .8 M P a
Table 14.3 and Eq.(14.12):
Su  Ad
b
 2 0 6 0 (5
 0 .1 6 3
)  1585 M Pa
Equation (7.5a): S u s  0 .6 7 (1 5 8 5 )  1 0 6 2 M P a
Table 14.3: S e s  0 .2 3 (1 5 8 5 )  3 6 4 .6 M P a
'
Equation (14.23) gives
a 
1
2
( 3 6 4 .6 )(1 0 6 2 1 .2  4 3 8 .8 )
1062  (
1
2
)( 3 6 4 .6 )
 9 2 .4 7 M P a
Thus
Pa 
Ks
a
Kw m
Pm 
1 .0 7 7 9 2 .4 7
1 .1 8 4 4 3 8 .8
(5 0 0 )  9 5 .8 N
and
Pm in  5 0 0  9 5 .8  4 0 4 .2 N
Pm a x  5 0 0  9 5 .8  5 9 5 .8 N
233
MPa
SOLUTION (14.22)
( a ) C  24 5  4 . 8
4 C 1
4C4
Kw 

Pm 
 320
640
2
 1.3 2 5
0 .6 1 5
C
Pa 
N
Ks  1
0 .6 1 5
C
160
2
 80 N
 1.1 2 8
Table 14.2 and Eq.(14.12):
S u  Ad
 0 . 163
 2060 ( 5
b
S es  0 . 23 S u  365
'
Table 14.3:
)  1 , 585
MPa
MPa
Equation (7.5a): S us  0 . 67 S u  1, 062
MPa
Therefore we obtain

m
 K

a

8 Pm C
s
K
w
Pa
K
s
Pm

8 ( 320 )( 4 . 8 )
 1 . 128
2
d

m
1 . 325
1 . 128
 ( 0 . 005 )
80
320
 176 . 5 MPa
2
(176 . 5 )  51 . 83 MPa
Equation (14.25):
n 
1 , 062 ( 365 )
51 . 83 ( 2  1 , 062  365 )  176 . 5 ( 365 )
 2 . 49
( b ) Load of 160 N causes deflection of 72-65=7 mm.
N
G d

a
8 PC

3
7 ( 7 9 , 0 0 0 )( 5 )
8 (1 6 0 )( 5 )
 1 7 .3
3
SOLUTION (14.23)
90  20
2
( a ) Pa 
4 C 1
4C4
Kw 
a
m

 35 N
K
w
Pa
K
s
Pm

0 .6 1 5
C

1 . 253
1 . 103
90  20
2
Pm 
 1.2 5 3
 55 N
Ks  1
0 .6 1 5
C
 1.1 0 3
 0 . 723
35
55
Use Eq.(14.24) with replacing S u s by S y s .
m 
Thus
5 6 0 1 .6
0 .7 2 3 ( 2  5 6 0  3 1 5 )
 1 2 2 .9 M P a
1
315
d
2
 K
8 Pm C
 m
s
 1 .1 0 3
8 ( 5 5 )( 6 )
6
 (1 2 2 .9  1 0 )
,
d  2 .7 5 m m
( b ) D  d C  2 .7 5 ( 6 )  1 6 .5 m m
k 
P

Na 

90  20
0 .0 7 5  0 .0 6 5
dG
3
8C k

 7 kN m
3
2 .7 5 ( 2 9  1 0 )
3
8(6) (7 )
 6 .6 ,
h s  8 .6 ( 2 .7 5 )  2 3 .7 m m ,
N t  6 .6  2  8 .6
h f  0 .0 7 5  1 .1 0 ( 7 02 00 0 )  7 8 .1 4 m m
 s  h f  h s  7 8 .1 4  2 3 .7  5 4 .4 m m
(c)
(d)
fn 
s
hf
3 5 6 ,6 2 0 d
2
D Na
 0 .7 ,

3 5 6 , 6 2 0 ( 2 .7 5 )
2
(1 6 .5 ) ( 6 .6 )
hf
D
 4 .7
 5 4 5 .8
cps  32, 748
cpm
 Spring is safe (Curve A, Fig.14.10)
234
SOLUTION (14.24)
4 C 1
4C4
(a) Kw 

 1.2 5 3
0 .6 1 5
C
Pa 
5 0 1 0
2
 20 N
a
K


m
w Pa
K s Pm
1 . 253
1 . 103
m 
Thus
Ks  1
5 0 1 0
2
Pm 
0 .6 1 5
C
 1.1 0 3
 30 N
 0 . 757
20
30
5 6 0 1 .6
 1 1 9 .3 M P a
0 .7 5 7 ( 2  5 6 0  3 1 5 )
1
315
Hence
2
d
8 Pm C
 K
 1 .1 0 3
 m
s
8 ( 3 0 )( 6 )
6
 (1 1 9 .3  1 0 )
d  2 .0 6 m m
or
( b ) D  2 .0 6 ( 6 )  1 2 .3 6 m m
5 0 1 0
0 .1 2 5  0 .1 0 5
k 
P


N

dG
a
3
2 . 06 ( 29  10 )

3
8C k
 2 kN m
 17 . 29
3
8(6) (2)
N t  17 . 29  2  19 . 29
h s  1 9 .2 9 ( 2 .0 6 )  3 9 .7 4 m m
h f  1 2 5  1 .1 0 ( 2 10 00 0 )  1 2 5 .0 0 6 m m
 s  h f  h s  1 2 5 .0 0 6  3 9 .7 4  8 5 .2 7 m m
(c)
(d)
fn 
s
3 5 6 , 6 2 0 ( 2 .0 6 )
hf
 0 .6 8 ,
hf
 2 7 8 .1 c p s  1 6 , 6 8 6 c p m
2
(1 2 .3 6 ) (1 7 .2 9 )
 1 0 .1
D
 Spring will buckle (Curve A, Fig.14.10)
SOLUTION (14.25)
( a ) Pm  400
Kw 
Pa  100
N
4 C 1
4C4

a 
K
1
s
0 . 615
C
 1.2 5 3
0 .6 1 5
C
m  K s
N
8 Pm C
d
K
w
Pa
K
s
Pm
2
 1.1 0 3
m 
1.2 5 3
1 .1 0 3
8 ( 4 0 0 )( 6 )
1
( 4 )(
6,741
d
2
6,741

2
 (d )
d
) 
2
1,9 1 4
d
2
Equation (14.23) leads
1,9 1 4
d
2
(720 
330
2
) 
330
2
720
1. 6
(

6,741
d
2
)
or
6 , 438
d
2
 450 
6 , 741
d
2
,
d  5 . 41 mm
( b ) D  5 . 41 ( 6 )  32 . 46 mm
N
a

dG 
8 PC
3

5 .4 1( 7 9 , 0 0 0 )( 8 )
8 ( 5 0 0  3 0 0 )( 6 )
3
 9 .8 9
235
 1 . 103
SOLUTION (14.26)
( a ) Pm  470
Pa  130
N,
C  30 6  5
N,
Table 14.2 and Eq.(14.12):
S u  Ad
b
 0 . 201
 1510 ( 6
)  1053
MPa
Table 14.3:
S
 0 . 42 (1053 )  442
ys
Ks  1
P2  P1
We have k 
N


a
 1G d
8 P1 C
 K
m
a 
3
 2 1
d
Kw
Pa
Ks
Pm

600  340
13
1 7 ( 2 9 0 0 0 )( 6 )

8 ( 3 4 0 )( 5 )
8 Pm C
s
 1.1 2 3 ,
0 .6 1 5
C
3
m 
1 .3 1 1
1 .1 2 3
'
Kw 
4 C 1
4C4

0 .6 1 5
C
1 
 20 N mm ,
340
20
MPa
 1.3 1 1
 17 mm
 8 .7
8 ( 470 )( 5 )
 1 . 123
2
S es  0 . 21 (1053 )  221
MPa ,
 (6)
2
 186 . 7 MPa
( 14 37 )(1 8 6 .7 )  6 0 .2 9
M Pa
It follows that
n 
(b)
hs  ( N
k 

s
3
a
 1 . 20
fn 
(d)
hf
6 ( 79000 )
(Eq.14.25)
 54 . 48 N mm
3
8 ( 5 ) ( 8 .7 )
 13 . 22 mm ,
600
54 . 48
h
356 , 620 ( 6 )

1 356 , 620 d
2
2
D Na
 0 .1 7 ,
 1.2
 2 ) d  (10 . 7 ) 6  64 . 2 mm

dG
8C N
(c)
s
a
442( 221)
6 0 .2 9 ( 2  4 4 2  2 2 1 ) 1 8 6 .7 ( 2 2 1 )
2
2 ( 30 ) ( 8 . 7 )
 2 .6
f
D
h f  64 . 2  13 . 22  77 . 42 mm
 136 . 6 cps  8 ,196
cpm
(fixed-free ends)
 Spring is safe (Curve B, Fig.14.10)
SOLUTION (14.27)
Refer to Example 14.6.
The critical torsional shear stress in the hook is from Eq. (14.34b):
B  (
rm
ri
)
8 PD
d
3
 S ys
where
S y s  0 .4 0 S u  0 .4 0 (1 2 5 6 )  5 0 2 .4 M P a
rm  3 .7 5 m m
D  7 .5 m m
ri  3 .7 5  ( 2 .5 2 )  2 .5 m m
d  2 .5 m m
Thus,
5 0 2 .4  ( 32.7.55 )
8 P (1 2 .5 )
 ( 2 .5 )
3
,
P  1 6 4 .4 N
The larger load is PA  2 3 2 .1 N , as found in Part (b) of Example 14.6.
This shows that failure will occur first by shear stress in the hook.
236
SOLUTION (14.28)
k 
4
d1
3
D1
Gd
4
Since k 1  k 2 :
3
8D Na
d

4
2
3
D2
d 2  d1
4
,
3
D2
4
3
D1
 3 ( 81 ) ,
4
d 2  14 . 27 mm
3
SOLUTION (14.29)
C 
4 C 1
4C 4
Kw 
a
m

 10
5
0 .5
K
w
Pa
K
s
Pm

K
1
s
0 .6 1 5
C
 1 . 0615
0 . 615
C
 1 .1 4 5,
Pm  3 N ,
m  Ks
 0 .7 1 9
8 Pm C
d
2
Pa  2 N
 1 .0 6 1 5
8 ( 3 ) (1 0 )
 ( 0 .5 )
2
 3 0 5 .6 M P a
 a  0 .7 1 9 (3 0 5 .6 )  2 1 9 .7 M P a
Table 14.2 and Eq.(14.12):
Su  Ad
 2 0 6 0 ( 0 .5
b
 0 .1 6 3
)  2306
M Pa
Table 14.3: S e s  0 .2 3 S u  5 3 0 .4 M P a
Equation (7.5a): S u s  0 .6 7 S u  1 5 4 5 M P a
'
Equation (14.25) is therefore
n 
1545 ( 530 . 4 )
 1 . 13
219 . 7 ( 2  1545  530 . 4 )  305 . 6 ( 530 . 4 )
SOLUTION (14.30)
( a ) We have C  D d  2 .4 0 .6  4 . So,
4 C 1
4C 4
Kw 

0 .6 1 5
C

4 ( 4 ) 1

4(4)4
0 .6 1 5
4
 1 .2 5  0 .1 5 3 7 5  1 .4 0 3 7 5
From Eq. (14.8):
 m ax  K w
8 PD
d
3
8 ( 6 )( 2 .4 )(1 0
 1 .4 0 3 7 5
 ( 0 .6 )(1 0
3
9
)
)
 2 3 8 .3 M P a
( b ) Use Eq.(14.34). Here ( ri ) A  ( rm ) A  d 2  1 .2  0 .3  0 .9 m m . Hence
(
A
)c 
3
1 .2 1 6 ( 6 )( 2 .4  1 0 )
9
0 .9
 ( 0 .6 )(1 0 )

4 (5)
2
 ( 0 .6 ) (1 0
6
)
 4 5 2 .7  1 7 .6 8  4 7 0 .3 8 M P a
From Eq. (14.12) and Table 14.2 for music wire:
Su  Ad
 2 0 6 0 ( 0 .6 )
b
 0 .1 6 3
 2239 M Pa
We also have
S y  S y s 0 .5 7 7  ( 0 .4 0 S u 0 .5 7 7  0 .6 9 3 S u  0 .6 9 3 ( 2 2 3 9 )  1, 5 5 2 M P a
Thus
n 
Sy
(
A
)c

1552
4 7 0 .3 8
 3 .3
SOLUTION (14.31)
( a ) I  bh
 
ML
EI
Hence
3
12  (1 . 2 )
L 
,
N
a

L
D
 EI
M

4

12  0 . 1728
( 16  2  )( 207  10
14 ,983
 (1 8 )
240
mm
3
)( 0 . 1728 )
4
,
C  D d  18 1 . 2  15
 14 , 983
mm
 265
(CONT.)
237
14.31 (CONT.)
( b ) P a  1 5 (1 6 )  2 4 0 N  m m
Ki 
2
3 C  C  0 .8
3 C ( C 1)
i  Ki
 1 .0 4 6 ,
6 Pa
bh
2
 1 .0 4 6
6 ( 240 )
(1 .2 )
3
 8 7 1 .7 M P a
SOLUTION (14.32)
The spring index is C  D d  1 5 1 .5  1 0 .
From Eq. (14.36):
2
4 C  C 1
4 C ( C 1 )
Ki 

2
4 (1 0 )  1 0  1
 1 .0 8 1
4 (1 0 )(1 0  1 )
Equation (14.12) and Table 14.2 for hard-drawn wire:
Su  Ad
 1 5 1 0 (1 .5 )
b
 0 .2 0 1
 1392 M Pa
S y  S y  S y s 0 .5 7 7  ( 0 .4 2 0 .5 7 7 ) S u  1 0 1 3 M P a  1 0 1 3 N m m
From Eq. (14.39):
3
M 
 d Sy
3
 (1 .5 ) (1 0 1 3 )

32 K i
3 2 (1 .0 8 1 )
 3 1 0 .5 N  m m
We have
I d
6 4   (1 .5 )
4
4
6 4  0 .2 4 9 m m
4
Equation (14.41), with  r a d  1 .2 , is thus
M lw
1 .2 

EI
3 1 0 .5 [  (1 5 ) N a ]
3
( 2 1 0  1 0 ) ( 0 .2 4 9 )
N a  4 .2 8 8
,
SOLUTION (14.33)
From Eq.(14.12) and Table 14.2 for oil-tempered wire:
Su  Ad
b
 1610(2)
 0 .1 9 3
 1 4 0 8 .4 M P a
( a ) Equation (7.5b) and Table 14.3:
S y  S y s 0 .5 7 7  0 .4 5 S u 0 .5 7 7  0 .7 8 S u
 0 .7 8 (1 4 0 8 .4 )  1 0 9 8 .6 M P a
Spring index: C  D d  1 2 .5 2  6 .2 5 .
Equation (14.36) :
4 ( 6 .2 5 )  6 .2 5  1
Ki 
4 ( 6 .2 5 )( 6 .2 6  1 )
Equation (14.39) with 
i 
32 Pa
d
3
a ll
 0 .1 3 5
 S y n  1 0 9 8 .6 1 .8  6 1 0 .3 M P a :
6 1 0 .3 (1 0 ) 
6
K i;
3 2 P ( 0 .0 2 8 )
 ( 0 .0 0 2 )
3
( 0 .1 3 5 )
Solving, P  1 2 6 .8 N
( b ) L w   D N a   (1 2 .5 )(3 .5 )  1 3 7 .4 m m
I d
4
64   (2)
4
6 4  0 .7 8 5 4 m m
4
Equation (14.41), with M  P a  1 2 6 .8 ( 0 .0 2 8 )  3 .5 5 N  m
Hence
 rad 
M lw
EI

( 3 .5 5 ) ( 0 .1 3 7 4 )
9
2 0 0  1 0 ( 0 .7 8 5 4  1 0
12
)
 3 .1 1 r a d
238
2
SOLUTION (14.34)
( a ) We have  a ll   m a x n  8 0 0 2 .5  3 2 0 M P a
From Eq. (14.42):

a ll

6 PL
nbh
3 2 0 (1 0 ) 
6
;
2
3
6 ( 4 0  1 0 )( 0 .7 )
8 ( 6 )( 0 .0 2 2 )
b  136 m m
,
2
( b ) Eq. (14.43):
  (1   )
2
( hl )  (1  0 .3 )
3
6P
Enb
3
6 ( 4 0 1 0 )
2
9
2 0 0 (1 0 )( 8 )( 0 .1 3 6 )
( 0 0.0.72 2 )
3
 0 .0 3 2 3 m  3 2 .3 m m
SOLUTION (14.35)
( a ) C  0 .6 0 .0 8  7 .5
Table 14.2 and Eq.(14.12):
Su  Ad
S ys
Table 14.3: S y 
 0 .1 9 3
 1610(2
b
0 .4 5 S u

0 .5 7 7
 1 0 9 8 .4 M P a
0 .5 7 7
2
4 C  C 1
4 C ( C 1 )
Equation (14.36): K i 
)  1 4 0 8 .4 M P a
 1 . 11
Equation (14.39) is therefore
3
 ( 2 ) (1 .0 9 8 4 ) 1 .4
M  Pa 
3 2 (1 .1 1 )
( b ) L   (1 5 ) 6  2 8 2 .7 m m ,
 
ML
EI
0 .5 5 5 ( 0 .2 8 2 7 )

9
( 2 0 0  1 0 )( 0 .7 8 5 4  1 0
12
)
 0 .5 5 5 N  m
I 
4
d
64
 0 .7 8 5 4 m m
 0 .9 9 9 ra d  5 7 .3
4
o
SOLUTION (14.36)
( a ) P m  700
m 
6 Pm L
bh
6 ( 7 0 0 )( 0 . 4 )

2

Also
P a  400
N

m

2
40 h ( h )
h

1
'
m
( Se K
f
a
m

Pa
Pm

4
7
3
1 4 0 0 1 .2

Su
a


42
Su n

N
(
4
7
)
1400
)(
 3 6 0 .1 M P a
) 1
5 0 0 1 .4
Thus we obtain
360 . 1 (10
) 
42
h
3
,
h  4 . 89 mm
b  195 . 6 mm
Hence
(b) k 
6
Ebh
3
3
3 L ( 1 
2
)

3
( 207  10 )( 195 . 6 )( 4 . 89 )
3
3 ( 400 ) ( 0 . 91 )
3
 27 . 1 N mm
End of Chapter 14
239
CHAPTER 15
POWER SCREWS, FASTENERS, AND
CONNECTIONS
SOLUTION (15.1)
( a ) By Fig.15.4b:
Thus
thread depth=7.5 mm,
thread width=7.5 mm
d m  75  7 . 5  67 . 5 mm ,
( b ) By Fig.15.4a:
Hence
d r  75  15  60 mm ,
thread depth=7.5 mm,
d m  67 . 5 mm ,
L  p  15 mm
thread width at pitch line=7.5 mm
d r  60 mm ,
L  p  15 mm
SOLUTION (15.2)
 4 , p  6 . 35
25 . 4
p
Table 15.3:
( a ) L  n p  2 ( 6 .3 5 )  1 2 .7 m m
p
dm  d 
ta n  
2
Fig.15.4a:   1 4 .5
 3 8 .1  3 .1 7 5  3 4 .9 2 5 m m
L
 dm

1 2 .7
 ( 3 4 .9 2 5 )
  6 .6
;
o
ta n  n  c o s  ta n   c o s 6 .6 (ta n 1 4 .5 );
 n  1 4 .4 1
o
( b ) For starting, we have f 
Tu 

Td 
(c) e 
(d) F 
Wd
f  cos  n tan 
m
cos  n  f tan 
2

Wf
4
3
( 0 .1 )  0 .1 3 ,
o
o
o
(c o s 1 4 .4 1 )  ( 0 .1 3 ) ta n 6 .6
7 .5 ( 3 4 .9 2 5 )
0 .1 3  (c o s 1 4 .4 1 ) ta n 6 .6
a

5 4 .2
0 .1 9

7 .5 ( 0 .1 1 )( 5 0 .8 )

7 .5 ( 0 .1 1 ) ( 5 0 .8 )
o
o
(c o s 1 4 .4 1 )  ( 0 .1 3 ) ta n 6 .6
cos  n  f cot 
Tu
o
o
c o s  n  f ta n 
o
o
o
o
o
c o s 1 4 .4 1  ( 0 .1 3 ) ta n 6 .6

4
3
( 0 .0 8 )  0 .1 1
2
2
2
fc 
o
cd c
0 .1 3  (c o s 1 4 .4 1 ) ta n 6 .6
7 .5 ( 3 4 .9 2 5 )
o
c o s 1 4 .4 1  ( 0 .1 3 ) c o t 6 .6
2
2
 5 4 .2 N  m
 2 3 .3 N  m
 0 .4 6  4 6 %
 2 8 5 .3 N
SOLUTION (15.3)
f  cos  n tan 
T o  0  d m [ cos 
Solving, ta n  
  1 2 .5
or
Then, Eq.(15.12):
n
 f tan 
]  d c fc
f  fc cos  n ( d c d m )
cos  n  f fc ( d c d m )
dm [
o
0 .1  0 .0 8 (c o s 1 4 .4 1 )(1 .4 5 4 5 )
o
c o s 1 4 .4 1  ( 0 .1 )( 0 .0 8 )(1 .4 5 4 5 )
o
d m ta n 
e 

(Eq.15.11)
f  c o s  n ta n 
c o s  n  f ta n 
o
3 4 .9 2 5 (ta n 1 2 .5 )

] d c fc
o
3 4 .9 2 5 [
o
o
c o s 1 4 .4 1  0 .1 (ta n 1 2 .5 )
240
 0 .4 9 4  4 9 .4 %
o
0 .1  (c o s 1 4 .4 1 )(ta n 1 2 .5 )
]  5 0 .8 ( 0 .0 8 )
SOLUTION (15.4)
d m  32  2  30 mm ,
  ta n
1
1
 ta n
L
d m
Wd
Tu 
 4 .8 5
8
 ( 30 )
f  tan 
m
1  f tan 
2
 n  0
L  2 ( 4 )  8 mm ,
Wf

cdc
2
o
6 ( 30 )

2
[
0 . 1  tan 4 . 85
1  0 . 1 tan 4 . 85
o
o
]
6 ( 0 . 15 ) 50
2
 39 . 28 N  m
We have
n 

V
L
40
8
 5 rps  300
rpm
Thus
kW 

Tn
9549
 1 . 23
39 . 28 ( 300 )
9549
SOLUTION (15.5)
   n  0,
c o s  n  1,
L  25 m m ,
V  0 .1 5  6 0  9 m m in
T 
9549 kW
n

9549 ( 4 )
360
n 

V
L
9
0 .0 2 5
ta n  
 360
L
 dm

25
 ( 45 )
rp m
 1 0 6 .1 N  m
Hence, by Eq.(15.9):
Tu 
Wd
f  tan 
m
1  f tan 
2
; 106 . 1 
10 ( 45 )
f  0 . 1768
2
1  f ( 0 . 1768 )
Solving
f  0 . 272
SOLUTION (15.6)
N  2,
( a ) Table 15.3:
p 
2 5 .4
2
 1 2 .7
m m th re a d ,
V  0 .0 1 m p s  0 .6 m m in ,
( b ) We have
fc  0,
  ta n
dm  d 
1
p
 ta n
 dm
1
p
2
n 
 70 
1 2 .7
2
[  1( 623.7.6 ) ]  3 .6 4
0 .6
0 .0 1 2 7
n  0
 4 7 .2 rp m
 6 3 .6 m m
o
Then, we obtain
Tu 
d mW
kW 
2
f  c o s  n ta n 
[ cos 
Tu n
9549

n
 f ta n 
]
1 7 1 5 ( 4 7 .2 )
9549
6 3 .6 ( 2 5 0 )
2
[
0 .1 5  (1 ) ta n 3 .6 4
o
1  ( 0 .1 5 ) ta n 3 .6 4
o
 8 .4 8
Hence
( k W ) req 
8 .4 8
0 .8 5
 9 .9 8
241
]  1 .7 1 5 k N  m
 0 .1 7 6 8
SOLUTION (15.7)
( a ) From Fig. 15.4b,
d  dm 
P
4
 24 
 2 5 .5 m m
6
4
Equation (15.6), with    n  0 ,
Wdm
T 
f  dm  L
(
2
 d m  fL
100 ( 24 )

2
)
c o s  n  1, ta n   L  d m ,
l  p , becomes:
W fc d c
2
( 0 .0 1 )  ( 2 4 )  6
[  ( 2 4 )  ( 0 .1 1 )( 6 ) ] 
1 0 0 ( 0 .1 1 )( 3 6 )
2
 2 2 9 .5  1 9 8  4 2 7 .5 N  m
( b ) The screw is self locking, if:
f 

L
 dm
6
 ( 24 )
 0 .0 8
SOLUTION (15.8)
( a ) Because of the triple-threaded screw,
L  3 p  3 (8 )  2 4 m m
From Fig. 15.4a: thread depth  0 .5 p  0 .5 (8 )  4 m m
d m  d  0 .5 p  5 0  4  4 6 m m
From Eq. (15.2),
  ta n
1
L
 dm
 ta n
1
( 4264 )  9 .4 3
o
( b ) For starting, we have
fc 
( 0 .1 2 )  0 .1 6
4
3
f 
4
3
( 0 .1 3)  0 .1 7
From Eq. (15.8):
 n  ta n
1
(ta n  c o s  )  ta n
1
(ta n 1 4 .5 c o s 9 .4 3 )  1 4 .3 1
o
o
o
By Eq. (15.7):
Td 

Wdm
f  c o s  n ta n 
2
c o s  n  f ta n 
15 ( 46 )
2
W fc d c

2
o
0 .1 7  c o s 1 4 .3 1 t a n 9 .4 3
o
o
o
c o s 1 4 .3 1  ( 0 .1 7 ) ta n 9 .4 3

1 5 ( 0 .1 6 ) ( 6 8 )
2
 3 4 5 ( 0 .0 0 9 1)  8 1 .6
 8 4 .7 N  m
SOLUTION (15.9)
( a ) We have
n 
0 .0 2 ( 6 0 )( m m in )
 5 0 rp m
0 .0 2 4 ( m r e v )
From Eq. (15.6), with f c  0 .1 2 and f  0 .1 3 :
Tu 

Wdm
f  c o s  n ta n 
2
c o s  n  f ta n 
W fc d c

2
o
o
1 5 ( 4 6 ) ( 0 .1 2 )  c o s 1 4 .3 1 ta n 9 .4 3
2
o
c o s 1 4 .3 1  ( 0 .1 3 ) ta n 9 .4 3
o

1 5 ( 0 .1 2 )( 6 8 )
2
 3 4 5 ( 0 .2 9 6 5 )  6 1 .2
 1 6 3 .5 N  m
(CONT.)
242
15.9 (CONT.)
Power:
( h p ) out 
( h p ) in 
FV
7 4 5 .7
Tn
7121

1 5 0 0 0 ( 0 .0 2 )

 0 .4 0 2 h p
7 4 5 .7
1 6 9 .4 ( 5 0 )
7121
 1 .1 8 9 h p
Efficiency is thus
e 
 0 .3 4  3 4 %
0 .4 0 2
1 .1 8 9
( b ) From Eq. (15.10):
f  c o s  n ta n   c o s 1 4 .3 1 ta n 9 .4 3
o
o
 0 .1 6 1
Since f  0 .1 3  0 .1 6 1  the screw is not self-locking and overhauling
SOLUTION (15.10)
   n  0,
e 
or
ta n
c o s  n  1.
1  f ta n 
1  0 .1 2 ta n 
1  0 .1 2 c o t 
0 .7 0 
;
1 f c o t 
 [
1  0 .1 2 ta n 
ta n   0 .1 2
] ta n 
  2 .5 ta n   0 .7 0  0
2
Solving
tan   0 . 3213 or 2 . 1783 ;   17 . 81
o
or 65 . 34
o
Thus
Wd
T0 
m
2
 f  tan 
[ 1 f
tan 
]
50 ( 30 )
o
[  0 . 12  tan 17 . 81o ]  145 . 3 N  m
1  0 . 12 tan 17 . 8
2
SOLUTION (15.11)
   n  0,
ta n  
p
 dm
cos  n  1
p   d m ta n    ( 4 6 .8 ) ta n 
,
V  0 .2 ( 6 0 )  1 2 m m in  n p
(n in rp m )
(1)
(2)
From Eqs.(1) and (2):
n 
V
p

12
 ( 0 . 0468 ) tan 
We have the torque expressed as
and
Tu 
9549 kW
n
Tu 
Wd
2
m

9549 ( 3 . 75 )  ( 0 . 0468 ) tan 
f  tan 
1  f tan 
12
 438 . 7 tan 
 438 . 7 tan 
Thus

438 . 7 tan   12 ( 23 . 4 )[ 10. 015. 15 tan
]
tan 
or
438 . 7 tan   65 . 8 tan
2
  42 . 12  280 . 8 tan 
or
157 . 9 tan   65 . 8 tan
2
  42 . 12
or
tan
2
  2 . 4 tan   0 . 64  0
(CONT.)
243
15.11 (CONT.)
Solving
ta n  1 , 2 
2
b
b  4 ac

2a
2 .4  1 .7 9
2
Use ta n   0 .3 0 5 . Then, Eq.(1) gives
p  4 6 .8  ( 0 .3 0 5 )  4 4 .8 4 m m
SOLUTION (15.12)
Table 15.1 and Fig.15.3: A t  3 5 3 m m ,
2
h  d  d r  3 .6 8 m m ,
b 
(a)  
(b) b 
P
A
 2 h ta n 3 0
3
8
3
6 0 (1 0 )

3 5 3 (1 0
P
 d m hne
6
 170
)
ne 
;
p  3 mm,
dm 
d dr
2

2 4  2 0 .3 2
2
 0 .3 7 5  2 (3 .6 8 ) ta n 3 0
o
d r  d - 1 .2 2 7 p = 2 0 .3 2 m m
o
 2 2 .1 6 m m
 4 .6 2 m m
M Pa

P
 d m h
b
3
6 0 (1 0 )
 ( 2 2 .1 6 ) ( 3 .6 8 ) 7 0
 3 .3 4 6
th r e a d engaged
Minimum nut length is thus
L n  p n e  (3 )(3 .3 4 6 )  1 0 .0 4 m m
e
 
( c ) Screw:
 
Nut:
3P
2  d r neb
3P
2  d neb

p
8
2
3
3 ( 6 0 1 0 )

2  ( 2 2 .1 6 )( 4 .6 2 )(1 0
6
)( 3 .3 4 6 )
3
3 ( 6 0 1 0 )
2  ( 2 4 )( 4 .6 2 )(1 0
6
 8 3 .6 M P a
 7 7 .2 M P a
)( 3 .3 4 6 )
SOLUTION (15.13)
d m  46 m m ,
(a)  
(b) b 
W
A
h 

2
W
2
 dr
W
 d m h ne
;
4


 4 mm,
3
4 (1 5  1 0 )
 ( 0 .0 4 2 )
ne 
2
d r  d  p  50  8  42 m m ,
 1 0 .8 M P a
W
 b d m h

3
1 5 (1 0 )
6
1 0 (1 0 )  ( 0 .0 4 6 )( 0 .0 0 4 )
 2 .6 th re a d engaged
L n  p n e  8 ( 2 .6 )  2 0 .8
Thus, minimum length of nut:
e
( c ) Screw:
Nut:
 
 
3W
2  d r neb
3W
2  d neb


b 
3
3 (1 5  1 0 )
2  ( 0 .0 4 2 ) ( 2 .6 ) ( 0 .0 0 4 )
3
3 (1 5  1 0 )
2  ( 0 .0 5 ) ( 2 .6 ) ( 0 .0 0 4 )
 1 6 .4 M P a
 1 3 .8 M P a
244
mm
p
2

8
2
 4 mm
SOLUTION (15.14)
kb 
kp 
2
d E s
0 . 58  50  0 . 5  15
0 . 58  50  2 . 5  15
kb

kb  k
 3.5 3 4 (1 0
4 ( 0 .0 5 )
0 . 58  ( E s 3 )( 0 . 015 )
2 ln[ 5
C 
2
 ( 0 .0 1 5 ) E s

4L
 4 . 512 (10
3
)E s
(Eq.15.31a)
)E s
(Eq.15.34)
]
 0 . 439
3 . 534
3 . 534  4 . 512
p
3
We have the preload:
Fi 
T
0 .2 d

72
0 . 2 ( 0 . 015 )
 24 kN
Thus
F b  C P  F i  0;
0 .4 3 9 P  2 4 ,
P  5 4 .6 7
kN
SOLUTION (15.15)
At  3 5 3 m m ,
2
Table 15.1:
Table 15.4: S u  5 2 0 M P a
We have Pm  6 0 k N ,
C r  0 .8 7
Table 7.3:
Table 15.6: K
f
Pa  2 0 k N ,
 3
n  1 .4
Equation (15.39) gives
S e  ( 0 .8 7 )( 13 )( 0 .4 5  5 2 0 )  6 7 .8 6
( a ) kb 
2
 d Es
2
 ( 0 .0 2 4 ) E s
 0 .0 0 9 E s
4 ( 0 .0 5 )
0 .5 8  ( E s 3 )( 0 .0 2 4 )
kp 
2 ln [ 5
C 

4L
0 .5 8 ( 0 .0 5 )  0 .5 ( 0 .0 2 4 )
0 .5 8 ( 0 .0 5 )  2 .5 ( 0 .0 2 4 )
kb

kb  k p
0 .0 0 9
0 .0 0 9  0 .0 0 8 7
M Pa
 0 .0 0 8 7 E s
]
 0 .5 0 8
F b a  C Pa  0 .5 0 8 ( 2 0 )  1 0 .1 6
 ba 
kN ,
1 0 ,1 7 0
353
Therefore we obtain
Su
n
or
520
1 .4
 

bm
bm


Su
Se

520
67 . 86
ba
( 28 . 8 ), 
bm
 150 . 74 MPa
F b m   b m A t  1 5 0 .7 4 (3 5 3)  5 3 .2 1 k N
Also
F bm  CP m  F i ;
53 . 21  0 . 509 ( 60 )  F i
or
F i  2 2 .7 k N
( b ) T  K d F i  0 .2 ( 2 0 )( 2 2 .7 )  9 0 .8 N  m
245
 2 8 .8 M P a
SOLUTION (15.16)
Table 15.1: A t  245 mm
2
,
K
 3 . 8 (Table 15.6),
f
C r  0 .8 9 (Table 7.3),
Ct  1
Using Eq.(15.39):
S e  ( 0 . 89 )( 1)(
1
3 .8
)( 0 . 45  750 )  79 . 05 MPa
( a ) F bm  160 ( 245 )  39 . 2 kN
kb 
k
p
2
 d Es
4L
2
 ( 0 . 02 ) E s
0 . 58 ( 50 )  0 . 5 ( 20 )
2 ln[ 5
0 . 58 ( 50 )  2 . 5 ( 20 )
kb

kb  k p
6 .2 8 3
6 .2 8 3 6 .7 2 2
3
 6 . 283 (10
4 ( 0 . 05 )
0 . 58  ( E s 3 )( 0 . 02 )

C 

 6 . 722 ( 10
)E s
3
)E s
]
 0 .4 8 3
Hence, we obtain
F b m  C Pm  F i  0 . 4 8 3 P m  2 5
or
39 . 2  0 . 483 Pm  25 , Pm  29 . 4 kN
and
S
y
n
  bm 
or

Also
F ba  CP a ;
ba
S
y
Se
 ba ;
620
2 .2
 15 . 53 MPa
and
 160 
620
7 9 .0 5
 ba
F ba  15 . 53 ( 245 )  3 . 8 kN
3 . 8  0 . 483 Pa ,
Pa  7 . 867
kN
Then
Pmax  Pm  Pa  29 . 4  7 . 867  37 . 27 kN
Pmin  Pm  Pa  29 . 4  7 . 867  21 . 53 kN
( b ) T  0 . 15 ( 20 )( 25 )  75 N  m
SOLUTION (15.17)
( a ) From Eq. (15.23):
Fb  C P  Fi  ( k
kb
b
 4 kb
)5  4 .2  5 .2 k N
By Eq. (15.24):
4k
F P  (1  C ) P  F i  ( 5 k b )5  4 .2   0 .2 k N
b
( b ) There is a compression (of- 0 .2 k N ) in parts. Thus they do not separate
under 5 k N load.
SOLUTION (15.18)
( a ) Each bolt supports P  1 1 .6 2  5 .8 k N .
From Eqs. (15.23) and (15.24)
Fb  C P  Fi  ( 3 k
kb
b
kp
) ( 5 .8 )  F i  1 4 .5  F i
2 kb
F p  (1  C ) P  F i  ( 3 k
b
kp
) ( 5 .8 )  F i  2 .9  F i
(1)
(2)
(CONT.)
246
15.18 (CONT.)
Joint separates when F p  0 . Equation (2) is then
F p  2 .9  F i  0 ,
F i  2 .9 k N
( b ) Using Eq. (1):
F b  1 .4 5  2 .9  4 .3 5 k N
Fb
From Table 15.4: S p 
3 1 0 (1 0 ) 
6
;
At
4 ,3 5 0
At
,
A t  1 4 .0 3 m m
2
So, by Table 15.1, select: I S O 5  0 .8 steel bolt
(with tensile stress area closest to A t  1 4 .0 3 m m .)
2
SOLUTION (15.19)
( a ) kb 
2
 d Es
4L
3
4 ( 0 .0 3 2 )
0 .5 8  E c d
kp 
2 ln [ 5
C 
2
 (1 8 ) ( 2 0 0  1 0 )

0 .5 8 L  0 .5 d
0 .5 8 L  2 .5 d
kb
kb  k

p
N m
0 .5 8  (1 6 5  1 0 )(1 8 )
2 ln [ 5
1 . 59
1 . 59  3 . 49
9
6

]
 1 .5 9  1 0
0 .5 8 ( 0 .0 3 2 )  0 .5 ( 0 .0 1 8 )
0 .5 8 ( 0 .0 3 2 )  2 .5 ( 0 .0 1 8 )
 3 .4 9  1 0
9
N m
]
 0 . 313
At  2 1 6 m m ,
2
Tables 15.1 and 15.4:
S p  600 M Pa
F i  0 .9 S p A t  0 .9 ( 6 0 0 )( 2 1 6 )  1 1 6 .6 4 k N
Equation (15.20):
Then
F b  C P  F i  0 .3 1 3( 860 )  1 1 6 .6 4  1 2 0 .8 k N
b 
(b)
Fb
At

120800
216
 5 5 9 .3 M P a
T  0 .2 F i d  0 .2 (1 1 6 .6 4 )(1 8 )  4 1 9 .9 N  m
SOLUTION (15.20)
Pm  Pa  1 0 k N ,
Su  830 M Pa ,
Table 15.6: K
f
 3,
Equation (7.11): C t  1  0 .0 0 3 2 ( 9 0 0  8 4 0 )  0 .8 1
Refer to Solution of Prob.15.19:
C  0 .3 1 3
At  2 1 6 m m
2
Equation (15.39): S e  ( 0 .8 4 )( 0 .8 1)( 13 )( 0 .4 5  8 3 0 )  8 4 .7 M P a
Thus, Eq.(15.40) results in
n 
( 8 3 0 )( 2 1 6 )  1 1 6 , 6 4 0
( 0 .3 1 3 )(1 0 , 0 0 0 )[(
830
 1 .8 5
) 1]
8 4 .7
247
C r  0 .8 4 (Table 7.3)
SOLUTION (15.21)
( a ) Compression of the parts is lost when F p  0 . Thus,
from Eq. (15.24):
F i  (1  C ) P  F P  ( k
 (k
4 kb
b
kp
kp
b
) P  FP
)(3 5 )  0  2 8 k N
 4 kb
( b ) Minimum force in parts occurs when fluctuating load is maximum.
Using Eq. (15.24):
Fp  ( k
kp
b
) P  Fi 
kp
4
5
(3 5 )  3 8   1 0 k N
SOLUTION (15.22)
( a ) Compression of the parts is lost when F p  0 . Therefore,
by Eq. (15.24):
Fi  ( k
 (k
kp
b
)P  Fp
kp
2 kb
b
)(3 5 )  0  1 7 .5 k N
 3 kb
( b ) Minimum force in parts takes place when fluctuating load is maximum.
From Eq. (15.24):
Fp  ( k
kp
b
kp
)P  Fp 
1
2
(3 5 )  3 8   2 0 .5 k N
SOLUTION (15.23)
( a ) From Eq. (15.24):
F p  (1  C ) P  F i  ( k
2 .5  ( 3 3 1 ) P  1 3,
kp
kp
b
) P  Fi
P  2 0 .6 7 k N
Fb  Fi  1 3 k N
( b ) Load off:
Load on: F b 
1
4
( 2 0 .6 7 )  1 3  1 8 .1 7 k N
Hence,
Pm 
1 3  1 8 .1 7
2
 1 5 .5 9 k N
Pa 
1 8 .1 7  1 3
2
 2 .5 8 5 k N
SOLUTION (15.24)
( a ) From Eq. (15.24):
F p  (1  C ) P  F i  ( k
600  ( 2k
kp
p
kp
kp
p
) P  8000,
 kb
) P  Fi
P  2 5, 8 0 0 N
(CONT.)
248
15.24 (CONT.)
Fb  Fi  8 0 0 0 N
( b ) Load off:
Load on: F b 
1
3
( 2 5, 8 0 0 )  8 0 0 0  1 6 , 6 0 0 N
So,
8 0 0 0 1 6 ,6 0 0
Pm 
 1 2 .3 k N
2
1 6 ,6 0 0  8 0 0 0
Pa 
 4 .3 k N
2
SOLUTION (15.25)
( a ) By Eq. (15.24):
F p  (1  C ) P  F i  ( k
1800 
4
5
P  6000,
kp
p
 kb
) P  Fi
P  9700 N
Fb  Fi  6 0 0 0 N
( b ) Load off:
Load on: F b  6 0 0 0 
(9 7 5 0 )  7 9 5 0 N
1
5
Thus
Fm 
6000  7950
2
Fa 
7950  6000
2
 6975 N
 975 N
SOLUTION (15.26)
Table 15.4:
d  20 mm ,
A t  245
2
mm
( a ) We have F b  0 . 9 S y A t  0 . 9 ( 630 )( 245 )  138 . 915
and
kb 
kp 
AE
s
L
2
 d Es

4L

2
 ( 20 ) E s
4 ( 60 )
0 . 58  ( E s 2 )( 20 )
2 ln[ 5
0 . 58 ( 60 )  0 . 5 ( 20 )
0 . 58 ( 60 )  2 . 5 ( 20 )
]
F i  0 . 75 F b  104 . 2 kN
kN
 5 . 236 E s
 9 . 379 E s
C 
5 . 236
5 . 236  9 . 379
 0 . 358
Thus
Pb  C P  F i  0 .3 5 8 ( 4 0 )  1 0 4 .2  1 1 8 .5
(b)
T  KdF
i
kN
 0 . 15 ( 20 )( 104 . 2 )  312 . 6 N  m
SOLUTION (15.27)
Tables 15.1, 15.4, and 15.6:
A t  8 4 .3 m m ,
K
S p  380 M Pa,
S y  420
2
f
 2 .2
M Pa ,
Su  520 M Pa
Table 7.3: C r  0 .8 9
Equation (15.39): S e  ( 0 .8 9 )(1)( 21.2 )( 0 .4 5  5 2 0 )  9 4 .7 M P a
(CONT.)
249
15.27 (CONT.)
20  4
2
( a ) We have Pa 
 8 kN ,
Pm 
20 4
2
 12 kN
Thus
a 
( MPa )
m 
 9 4 .9 M P a ,
8
8 4 .3
12
8 4 .3
 1 4 2 .3 M P a
 Not safe. (Fig. S15.27)
a
(a) No preload

140
Soderberg Line
a
Se
105
Modified Goodman Line
70
 ba
Figure S15.27
35

0
(b) With preload
m
210 280
140
70
45o

0
S

420 490 520
350
bm
Su
m
( MPa )
y
( b ) F i  0 .7 5 (3 8 0  8 4 .3)  2 4 k N
kb 
2
 d Es
2
 ( 0 .0 1 2 ) E s

4L
 0 .0 0 2 3 E s
4 ( 0 .0 5 )
0 .5 8  ( E s 2 )( 0 .0 1 2 )
kp 
2 ln [ 5
0 .5 8 ( 0 .0 5 )  0 .5 ( 0 .0 1 2 )
0 .5 8 ( 0 .0 5 )  2 .5 ( 0 .0 1 2 )
 0 .0 0 5 E s ,
C 
]
0 .0 0 2 3
0 .0 0 2 3  0 .0 0 5
We have F b m  C Pm  F i  0 .3 1 5 (1 2 )  2 4  2 7 .8 k N ,
 ba 
F b a  0 .3 1 5 (8 )  2 .5 2 k N ,
2520
8 4 .3
 0 .3 1 5
 bm 
2 7 ,8 0 0
8 4 .3
 3 2 9 .8 M P a
 2 9 .9 M P a
It is seen from Fig. S15.27 that the joint fails according to the Soderberg theory, while it is
safe on the basis of Goodman criteria.
( c ) Use Eq.(15.37):
5 2 0 ( 8 4 .3 )  2 4 , 0 0 0
n 
520
( 0 .3 1 5 )[ 8 0 0 0 (
 1 .1 3
) 1 2 ,0 0 0 ]
9 4 .7
( d ) Applying Eq.(15.28), we have
ns 
 1 . 75
24
20 ( 1  0 . 315 )
SOLUTION (15.28)
We have
S
p
 600
A t  115

i
mm
 0 . 75 S
Pa  Pm 

a
 
m
(Table 15.5)
MPa
(Table 15.2)
 0 . 75 ( 600 )  450
p
MPa
 5 kN
10
2

2
3
5 ( 10 )
115 ( 10
6
)
 43 . 48 MPa
Equation (15.39):
Se  CrCt (
1
K f
) ( 0 .4 5 S u )
(CONT.)
250
15.28 (CONT.)
where
C r  0 .8 7
(Table 7.3)
C t  1  0 .0 0 5 8 ( 4 9 0  4 5 0 )  0 .7 7
K
 3 .8
f
(Table 15.6)
S u  830
Hence
(Table 15.5)
MPa
S e  ( 0 . 87 )( 0 . 77 )(
We have 

a
S u 
n 
C
a
[(
(Eq.7.11)
1
3 .8
)( 0 . 45  830 )  65 . 84 MPa
. With preload, Eq.(15.38) gives
m
i
Su
) 1]
Sy
830  450

( 43 . 48 )( 0 . 31 )(
 2 . 07
830
1)
65 . 84
Without preload, C=1 and  i  0 :
n 
 1 . 40
830
830
1)
( 43 . 48 )(
65 . 84
Comment: The presence of preload is beneficial.
SOLUTION (15.29)
Refer to Solution of Prob.15.28. We have A t  1 1 5 m m ,
2
Hence,
C  0 .3 1,
S p  600 M Pa.
F i  0 . 75 ( A t S p )  0 . 75 (115  600 )  51 . 75 kN
Apply Eq.(15.27),
S p At  Fi
P 
Cn

600 ( 115 )  51 . 75  10
( 0 . 31 )( 2 )
3
 27 . 82 kN
Using Eq.(15.28b):
P 

Fi
n s (1 C )
 37 . 5 kN
51 . 75
2 ( 1  0 . 31 )
Comment: Failure owing to separation will not take place before bolt failure.
SOLUTION (15.30)
( a ) We have
 380
MPa
At  1 5 7
mm
Pa  Pm 
12
2
S
p
S u  520
2
MPa
Se  100 M Pa
(Table 15.1),
 6
(Table 15.4)
k N b o lt
F i  0 .9 S p A t  0 .9 (3 8 0 )(1 5 7 )  5 3 .6 9 k N

a


i
 0 .9 S
m
6 1 0

1 5 7 (1 0
p
3
6
)
 3 8 .2 2
M Pa
 0 . 9 ( 380 )  342
MPa
(CONT.)
251
15.30 (CONT.)
kb
C 
We have 
S u 
C
a
m
. Eq.(15.38), with preload:
i
Su
[(
 0 . 167
kb 5kb

a
n 
kb

kb  kb
) 1]
Sy
520  342

( 0 . 167 )( 38 . 22 )(
520
 4 .5
1)
100
Without preload, C=1 and  i  0 :
n 
 2 . 19
520
520
1)
( 38 . 22 )(
100
( b ) Equation (15.28b):
Fi
ns 
P (1 C )

 5 . 37
53 . 69
12 ( 1  0 . 167 )
Comments: The presence of preload is very beneficial. Joint will separate before bolts fail.
SOLUTION (15.31)
Repeating section
L
P
P
We have L=w=50 mm and total number of rivets in the joint n=2. Therefore
 
P
n (d
2
4)

b

P
ndt

t

P
( w  d e )t

3
4 ( 32  10 )

2 (  )( 0 . 019 )
3
32 ( 10 )
2 ( 0 . 019 )( 0 . 01 )
 56 . 43 MPa
2
 84 . 21 MPa
3
32 ( 10 )

[ 50  ( 19  3 )] 10 ( 10
6
)
 114 . 3 MPa
SOLUTION (15.32)
Sketch is the same as that given in Solution of Prob.15.31 and n=2. The allowable loads are:
F s   a ll n  d
Fb  
b , a ll
Ft  
t , a ll
2
4  1 0 5 ( 2 )(  )(1 8 )
4  5 3 .4 4 k N
n d t  3 3 0 ( 2 )(1 8 )(1 0 )  1 1 8 .8
kN
( w  d e ) t  1 5 0[ 6 0  (1 8  1 .5 )]1 0  6 0 .7 5
Thus
e 
2
6 0 .7 5
(1 5 0 )( 6 0 )(1 0 )
 6 7 .5 %
252
kN
SOLUTION (15.33)
For members, by Table B.3: S y  4 6 0 M P a ,
S y s  0 .5 7 7 ( 4 6 0 )  2 6 5 .4
M Pa
For bolts, from Table 15.4: S y  9 4 0 M P a ,
S y s  0 .5 7 7 (9 4 0 )  5 4 2 .4
M Pa
from Table 15.1: d r  1 8  1 .2 2 7 ( 2 .5 )  1 4 .9 3 m m
 (1 4 .9 3 )
As  2
Shear on bolts:
As S ys
Fs 
3 5 0 .1 4 ( 5 4 2 .4 )

n
2
Ab S y
360 (940 )

n
 3 5 0 .1 4 m m
2
 95 kN
A b  2 (1 8 )(1 0 )  3 6 0 m m
Bearing on bolts:
Fb 
2
4
2
 1 1 2 .8 k N
3
Bearing on members:
360 ( 460 )
Fb 
 6 6 .2
2 .5
A t  ( 6 0  1 8 )1 0  4 2 0 m m
Tension in members:
420 ( 460 )
Ft 
kN
2
 5 5 .2 k N  Pa ll
3 .5
SOLUTION (15.34)
For members, using Table B.3:
S
 210
y
MPa
For bolts, from Table 15.1: d r  1 4  1 .2 2 7 ( 2 )  1 1 .5 5 m m
 
2  (1 1 .5 5 )
As 
Shear of bolt:

P
As
2 0 ,0 0 0
2 0 9 .5
 2 0 9 .5 m m
 119 M Pa,
2 0 ,0 0 0
168
2
 9 5 .4 7 M P a ,
n 
A b  2 ( 6 )( 14 )  168
Bearing on bolt:
b 
2
4
mm
n 
640
119
 3 .8 8
370
9 5 .4 7
2
 5 .3 8
Bearing on members:
n 
 1 . 76
210
119
A t  ( 60  2  14 ) 6  192
Tension on members:

t

20 , 000
192
 104 . 2 MPa ,
n 
210
104 . 2
mm
2
 2 . 02
SOLUTION (15.35)
Table 15.1: A t  8 4 .3 m m ,
d r  1 2  1 .2 2 7 (1 .7 5 )  9 .8 5 3 m m
2
Hence, A s 

4
(9 .8 5 3 )  7 6 .2 5 m m
2
 
2
P 4
As

P
4 ( 7 6 .2 5 )
 P 305
Pivots about point A. Row 1 is the highest loaded. Apply Eq.(15.43) with j=2 and e=250 mm:
F1 
Thus

t ,m ax
M r1
2

2
r1  r2


2
t

250 P ( 225 )
2
2
2 ( 225 ) 2 (75 )
(

2
t
) 
2
t 
 0 .5 P ,
2

P
3 3 7 .2

0 .5 P
8 4 .3
( 3 3P7 .2 )  ( 3 P0 5 )
2
 ( 0 .0 0 3  0 .0 0 4 4 ) P  0 .0 0 7 4 P
Hence, 0 .0 0 7 4 P  1 4 0;
or
 m a x  0 .0 0 4 4 P  8 4 :
Pa ll  1 8 .9 2 k N
P  1 9 .0 9
253
 P 1 6 8 .6
kN
2
SOLUTION (15.36)
The maximum design load is Pm a x  n P  ( 2 .3)( 2 5 )  5 7 .5 k N .
From geometry: F 2  ( 4 1 5 ) F1 . Here F1 and F 2 are tensile forces in bolts 1 and 2, respectively.
We have

M
 0:
A
1 0 0 0 (5 7 .5 )  1 0 0 ( 145 F1 )  3 7 5 F1 ;
F1  1 4 3 .2 k N
The required tensile stress area is then
At 
143,200
600
 239 m m
2
From Table 15.1: The required thread size is about M 2 0  2 .5  C .
SOLUTION (15.37)
( a ) Using Table 15.4, S p  3 1 0 M P a ,
S y  340 M Pa
S y s  0 .5 7 7 S y  1 9 6 .2 M P a
The tensile stress area:
( F o r c e )( n )
At 
S

2 ,5 0 0 ( 3 )
310
p
 2 4 .2 m m
2
From Table 15.1, select M 7  1  C thread with A t  2 8 .9 m m
2
( b ) Table 15.1 and Fig. 15.3:
p  1 mm,
h 
1
2
d  7 mm,
d r  d  1 .2 2 7 p  5 .7 7 m m ,
( d  d r )  0 .6 1 5 m m
d m  d  0 .6 5 p  6 .3 5 m m ,
b 
p
8
 2 h ta n 3 0
o
 0 .8 3 5 m m
Apply Eq. (15.19a):
3K tPp
2  dbLn

S ys
3 ( 4 )( 2 5 0 0 )(1 )
n
2  ( 7 )( 0 .8 3 5 ) L n

1 9 6 .2
3
Solving,
L n  1 2 .5 m m
SOLUTION (15.38)
(a)
Refer to Solution of Prob. 15.37.
Using Table 15.4:
S p  310 M Pa,
S y  340 M Pa ,
F o rce ( n )
 6 4 .5 2 m m
At 
Sp

4000 (5 )
310
S y s  0 .5 7 7 S y  1 9 6 .2 M P a
2
From Table 15.1:
Select M 1 2  1 .7 5  C thread with A t  8 4 .3 m m
(b)
2
Table 15.1 and Fig. 15.3:
d  12 m m ,
p  1 .7 5 m m
d m  d  0 .6 5 p  1 0 .8 6 m m
d r  d  1 .2 2 7 p  9 .8 5 m m
h 
1
2
( d  d r )  1 .0 7 5 m m ,
b  1 .7 5 8  2 h ta n 3 0
o
 1 .4 6 m m
(CONT.)
254
15.38 (CONT.)
Using Eq. (15.19a):
3 ( 4 ) ( 4 0 0 0 ) (1 .7 5 )

2  (1 2 ) (1 .4 6 ) L n
1 9 6 .2
5
from which
L n  1 9 .4 m m
SOLUTION (15.39)
Table 15.1:
At  2 4 5 m m ,
Shear area:
As 
d r  2 0  1 .2 2 7 ( 2 .5 )  1 6 .9 3 m m
2
 
P 3


As
(1 6 .9 3 )  2 2 5
2
4
2
 1 , 481 P
P
3 ( 225 )  10
mm
6
Pivots about point A. Bolt 1 is the highest loaded.
M r1
F1 
r1  r2  r3
t 
2 4 5 (1 0
2
2
)
From Mohr’s circle: 
2
2
140  90  40
2
 0 .5 9 7 P
 2 ,4 3 7 P
0 .5 9 7 P
6
1 2 5 P (1 4 0 )

2


t , max
t
2

(

t
2
) 
2
2

2 , 437 P
2

(
2 , 437 P
2
)  ( 1 , 481 P )
2
2
 ( 1 , 219  1 , 918 ) P  3 ,137 P
Hence 3,1 3 7 P  1 4 5  1 0 ,
P  4 6 .2 2
6
or
1, 9 1 8 P  8 0  1 0 ,
kN
Pa ll  4 1 .7 1 k N
6
SOLUTION (15.40)
Vertical and horizontal components of the force P=10 kN are 8 kN and 6 kN at B, respectively.
Rivet B is most heavily loaded. The centroid of the group of rivets is at C. We have
FB 
M rB

8  3 0 (1 5 0 )

2
rj
2
2
2
2[ 3 0  9 0 1 5 0 ]
 0 .5 7 1 k N
6/6=1 kN
B
1
V B  [1  1 . 904
2
2
] 2  2 . 15 kN
(Fig.a)
8/6+0.571
=1.904 kN
Therefore
B 

B

VB
 d
2
VB
4

dt

2 ,1 5 0
 ( 20 )
2 ,1 5 0
2 0 1 5
2
4
 6 .8 4 4
 7 .1 6 7
M Pa
VB
M Pa
Figure S15.40
SOLUTION (15.41)
Rivet A is the most heavily loaded.
As 

Vd 
P
5
4
(15 )

50
5
2
 176 . 715
 10
kN
mm
2
V a ll  1 0 0 (1 7 6 .7 1 5 )  1 7 .6 7 2
kN
Centroid C of the line AB, with respect to A, is determined from:
5 0 x  7 0 (1 0 )  2 5 0 (1 0 )  3 2 0 (1 0 )  3 9 0 (1 0 ) ,
x
A
x  206
mm
C
d
e
B
(CONT.)
P
255
15.41 (CONT.)
Thus
Mr
FA 

A
2
rj

50 , 000 e ( 206 )
206
2
2
 136
2
 44  114
2
2
 184
 93 . 875 e
V A  1 0 , 0 0 0  9 3 .8 7 5 e  1 7 , 6 7 2 ,
e  8 1 .7 m m
and
d  54 . 3 mm
SOLUTION (15.42)
The A is the highest loaded point.
A s   (15 )
4  176 . 715
2
Thus
FA 

A
2
rj

46 P ( 206 )
206
2
 136
2
2
x  206
Refer to Solution of Prob. 15.41:
Per
mm
2
 44  114
2
2
 184
F 
,
mm,
P
5
 0 .2 P
e  206  70  90  46 mm
 0 . 086 P
Hence V A  ( 0 .2  0 .0 8 6 ) P  0 .2 8 6 P
and

A

 100  10 ;
P  6 1 .7 9 k N
6
0 .2 8 6 P
1 7 6 .7 1 5 (1 0
6
)
SOLUTION (15.43)
A  0 . 707 hL  0 . 707 ( 7 )( 60 )  297
P 
S
ys
A
n

200 ( 297 )
2 .5
mm
2
 23 . 76 kN
Cross-sectional area of one plate A p  40 (10 )  400
mm
2
. Then
P  ( S y n ) A p  ( 2 5 0 2 .5 ) 4 0 0  4 0 k N
Hence the capacity of the plate significantly exceeds that of the weld.
SOLUTION (15.44)
Table B.4:
for plates,
Table 15.8:
for weld,
S y  427 M Pa .
S y  345 M Pa .
Since 3 4 5  4 2 7 , the weld should yield first.
The maximum load that can be applied equals
P 
SyA
n

3 4 5 (1 5  8 7 .5 )
4
 1 1 3 .2 k N
SOLUTION (15.45)
Refer to Solution of Prob. 15.44.
S y  0 .5 S y  0 .5 ( 4 2 7 )  2 1 3 .5 M P a
From Eq. (15.44):
(for plate)
S y s  0 .5 S y  0 .5 ( 3 4 5 )  1 7 2 .5 M P a
(for weld)
(CONT.)
256
15.45 (CONT.)
Since
1 7 2 .5  2 1 3 .5
 The weld should yield first.
Hence,
S ys A
P 
1 7 2 .5 (1 5  8 7 .5 )

n
3
 7 5 .4 7 k N
SOLUTION (15.46)
For plates: S y  4 6 0 M P a
(Table B.3), S y s  0 .5 S y  2 3 0 M P a
For weld: S y  3 7 9 M P a
(Table 15.8), S y s  0 .5 S y  1 8 9 .5 M P a
(Eq. 15.44)
(Eq. 15.44)
Since
1 8 9 .5  2 3 0
 weld would yield first.
Thus, Fig. 15.27a:
S ys
P 

n
1 8 9 .5 ( 0 .7 0 7 )( 8 )( 2  7 0 )
 4 2 .8 7 k N
3 .5
SOLUTION (15.47)
For plates: S y  5 3 0 M P a
(Table B.3),
S y s  0 .5 S y  2 7 5 M P a
For weld: S y  4 1 4 M P a
(Table 15.8),
S y s  0 .5 S y  2 1 2 M P a
(Eq. 15.44)
(Eq. 15.44)
Since
212  275
 weld would yield first.
Thus, Fig. 15.26a:
S ys
P 

n
2 1 2 ( 0 .7 0 7 )( 5 )( 2  6 0 )
4
 2 2 .5 k N
SOLUTION (15.48)
A p  75  10  750


mm
2
P  A
,
all
 750 (140 )  105
kN
R 1  21 kN 
(along AB)
M
D
 0:
R 1 ( 75 )  105 (15 )  0 ;
M
A
 0:
R 2 ( 75 )  105 ( 75  15 )  0 ;
Check:
Hence
R 1  R 2  105
L1 
R1
Sw

21
1 .2
R 2  84 kN 
(along DE)
kN
 17 . 5 mm ,
L2 
R2
Sw

84
1 .2
 70 mm
SOLUTION (15.49)
I x  2 (1 0 0 t )( 6 2 .5 ) 
2
3
2 (1 2 5 ) t
12
 1 .1 0 7 (1 0 ) t
A  2 (1 0 0 t  1 2 5 t )  4 5 0 t m m
We have
6
1
2
  [( 1455000t0 )  ( 3 .71 .15 0672t .5 ) ] 
Thus
   a ll ;
and
h 
0 .1 3 4
0 .7 0 7
2
2 1 4 .3
t
 5 5,
 a ll  5 5 M P a
M  1 5 ( 2 5 0 )  3 .7 5 M N  m m
2
2
4
mm ,
2 1 4 .3
t
t  3 .9 m m
 5 .5 2 m m
257
SOLUTION (15.50)
I x  1 .1 0 7 (1 0 ) t m m ,
6
From Solution of Prob.15.49:
We have Pm  1 5 k N ,
Pa  5 k N ,
K
Refer to Example 15.12. We obtain C
f
A  450t m m
4
 1.5
(Table 15.9)
 A Su  272(420)
b
f
2
 0 .9 9 5
 0 .6 6 7 .
Equation (7.5a): S u s  0 .6 7 ( S u )  0 .6 7 ( 4 2 0 )  2 8 1 .4 M P a
S e  C s C f (1 K f ) S e  ( 0 .7 )( 0 .6 6 7 )( 11.5 )( 0 .5  4 2 0 )  6 5 .3 7 M P a
'
Then, from Eq.(7.21):
S us n
m 
(
a
)
m
Su
2 8 1 .4 2 .5

1
 3 5 .8 3 M P a
1
420
1
)
3 6 5 .3 7
(
Se
Also, from Solution of Prob.15.49:  m 
Thus
2 1 4 .3
t
 3 5 .8 3
and
h 
5 .9 8
0 .7 0 7
1 . 072
t
t  5 .9 8 m m
or
 8 .4 6 m m
SOLUTION (15.51)
Table 15.8: S y  4 1 4 M P a
S y s  0 .5 S y  2 0 7 M P a
We have t  0 . 707 (12 )  8 . 484
mm
A one  weld  8 . 484 L
T  100 ( 60 )  6 MN  mm
Total
3
J  2
 2
tL
12
8 .4 8 4 L
12
3
 1.4 1 4 L
3
At point A:
A 
S ys
and
P
A

n

207
2 .5
Tr
J

100 ,000

1 6 .9 6 8 L
 82 . 8  
6
6 (1 0 )( L 2 )
1. 4 1 4 L
3

5 ,893
L
2 ,1 2 2 , 0 0 0

L
(1)
2
(2)
A
From Eqs.(1) and (2):
82 . 8 L  5 ,893 L  2 ,122 , 000  0 ;
2
L  71 . 17 L  25 , 628  0
2
Solving
L 
1
2
[ 7 1 .1 7 
7 1 .1 7  4 ( 2 5 , 6 2 8 ) ]  1 9 9 .6
2
mm
SOLUTION (15.52)
Pm  100
Pa  20 kN
kN ,
Table 15.9: K
f
 1.5
Refer to Solution of Prob.15.51:
m 
Also
5 ,893

L
S u  496
C
f
S
ys
 AS
2 ,1 2 2 , 0 0 0
L
MPa ,
b
u
(1)
2
S
y
 272 ( 496 )
 0 . 5 ( 414 )  207
 414
 0 . 995
MPa
MPa
(Table 15.8),
 0 . 566
Refer to Example 15.12:
S e  (0 .7 )(0 .5 6 6 )( 11.5 )(0 .5  4 9 6 )  6 5 .5 1 M P a
(CONT.)
258
15.52 (CONT.)
We have, from Eq.(7.21) with S y s and S y replacing S u s and S u :
S ys n
m 
(
a
m
)
Sy
2 0 7 2 .5

1
(
Se
 3 6 .5 7 M P a
1
414
1
)
5 6 5 .5 1
(2)
Equations (1) and (2) are therefore
L  161 . 1 L  58 , 025 . 7  0
2
L 
or
1
2
[1 6 1 .1 
1 6 1 .1  4 (5 8 , 0 2 5 .7 ) ]  3 3 4 .5
2
mm
SOLUTION (15.53)
S u s  0 .6 7 S u  2 8 6 .1 M P a ;
Table 15.8: S u  4 2 7 M P a;
We have C
f
 1 .5
A b o th  2[ 0 .7 0 7 ( 6 .5 ) 2 5 0 ]  2 2 9 7 .8
mm
 A Su  272(427)
b
f
 0 .9 9 5
K
(Table 15.9)
 0 .6 5 7
Refer to Example 15.12:
S e  ( 0 .7 )( 0 .6 5 7 )( 11.5 )( 0 .5  4 2 7 )  6 5 .5 M P a
We have
Pm  Pa 
3
J  2

tL
12
Pm a x ,
AL
12
2
2 , 2 9 7 .8 ( 2 5 0 )

2
12
 1 2 (1 0 )
6
We have Pm  0 .5 Pm a x ,
At point A:
m 
and
1
2
Pm

A
Tm r
J

Pm a x
4
T m  7 5 Pm  3 7 .5 Pm a x
3 7 .5 Pm a x (1 2 5 )

2 ( 2 2 9 7 .8 )
mm
2
6
1 2 (1 0 )
 0 .0 0 0 6 Pm a x
Also, from Eq.(7.21):
S us n
m 
(
a
m
)
Su
2 8 6 .1 2

1
(1 )
427
 1 9 .0 2 M P a
1
6 5 .5
Se
Thus
0 .0 0 0 6 Pm a x  1 9 .0 2 ,
Pm a x  3 1 .7 k N
SOLUTION (15.54)
150 mm
y
A
B
8
200 mm
C
6
x
T
30 ( 10
A
D
E
40 ( 10
A
3
)

3
)

T ( 200 )
Figure S15.54
J
T ( 75 )
J
Inspection of Fig. S15.54 shows that point E has the highest stress. We write
T  4 0 (1 7 5 )  3 0 (1 0 0 )  4 k N  m
A b o th  2 (1 5 0 t )  3 0 0 t m m
2
(CONT.)
259
15.54 (CONT.)
t (1 5 0 )
I x  2[
3
 1 5 0 t (1 0 0 ) ]  2 ( 0 .2 8 1 t  1 .5 t )(1 0 )  3 .5 6 2 (1 0 ) t m m
2
12
6
I y  2[ 0  1 5 0 t ( 7 5 ) ]  1 .6 8 7 (1 0 ) t m m ,
2
6
6
J  5 .2 4 9 (1 0 ) t m m
4
6
4
4
We have
3
v 
and
4 0 (1 0 )
300 t
4 (75 )


6
5 .2 4 9 (1 0 ) t
1
2
 E  [ v   h ] 
2
2
1 3 3 .3
t
1 6 6 .6
t
h 
,
3
3 0 (1 0 )
300 t

4 (1 0 0 )
6
5 .2 4 9 (1 0 ) t

100
t
M Pa
Therefore, by Eq.(15.44),
n E  0 .5 S y ;
3(
1 6 6 ,6
t
)  0 .5 (3 5 0 ),
t  2 .8 6 m m
and
h 
 4 .0 5 m m
2 .8 6
0 .7 0 7
SOLUTION (15.55)
J  5 .2 4 9 (1 0 ) t m m ,
6
From Solution of Prob.15.54:
From Table 15.9:
C
K
 2 .7 ;
f
 A Su  272(427)
b
f
A b o th  3 0 0 t m m
4
2
S u s  0 .6 7 S u  0 .6 7 ( 4 2 7 )  2 8 6 M P a
 0 .9 9 5
 0 .6 5 7
Refer Example 15.12: S e  ( 0 .7 )( 0 .6 5 7 )( 21.7 )( 0 .5  4 2 7 )  3 6 .4 M P a
We have Pm  3 0 k N ,
Pa  2 0
a m  2 3
kN ,
Thus, by Eq.(7.21):
S us n
m 
(
a
m
)
Su
2 8 6 1 .5

1
(
Se
2
427
1
)
3 3 6 .4
 2 1 .6 M P a
(1)
At point E: Use the method of Solution of Prob.15.54, with P  Pm  3 0 k N Then
Pm v  2 4 k N
Pm h  1 8 k N
T m  2 4 (1 7 5 )  1 8 (1 0 0 )  2 .4
kN  m
Therefore
and
3
 vm 
2 4 (1 0 )
 hm 
1 8 (1 0 )
300 t
3
300 t
2 .4 ( 7 5 )

6
5 .2 4 9 (1 0 ) t
2 .4 (1 0 0 )

6
5 .2 4 9 (1 0 ) t
1
2
 m  ( v m   h m ) 
2
2

80
t

60
t
100
t
(2)
Equations (1) and (2) give
100
t
 2 1 .6 ,
t  4 .6 m m
and
h 
4 .6
0 .7 0 7
 6 .5 1 m m
End of Chapter 15
260
CHAPTER 16
MISSELLANEOUS MACHINE COMPONENTS
SOLUTION (16.1)
Equation (16.17) and Eq.(16.16a) at r=a:
 1    , max  p i
(a) 1 
 Sy;
2
2
2
2
2
a b
b a

5
4

pi
 
2
p i  (  p i )  260 ,
5
4
  pi
r , max
p i  115 . 6 MPa
( b )  1   1 2   2  S y
2
2
2
1
p i [( 54 )
 ( 54 )(  1 )  (  1 ) ] 2  260 ,
2
p i  133 . 2 MPa
2
SOLUTION (16.2)
T 
 b
3

2
 ( 100 ) b
3
 157 . 08 b
2
3
Also
T  2  b fp l  2  b fp ( 3b )  6  fb p
and
6  fb p  157 . 08 b ,
2
2
3
 
Hence
p 
3
h 
We have E h  E s  E ,
 
3
2 bpc
2
2
2
2 bp ( 4 b )

2
E (c b )
2
 0 . 706 (10
8 ( 55 . 56 ) b
3
3 ( 210  10 )
(Eq.16.27b)
 55 . 56 MPa
  , and a=0. Equation (16.25) becomes
s
8 pb

2
E (4b b )
157 . 08
6  ( 0 . 15 )
3E
3
)b
SOLUTION (16.3)
( a ) From Table B.1: S y  2 5 0 M P a ,
  0 .3 .
E  200 G Pa,
Applying Eq. (16.17),
p i    ,m ax
2
2
2
2
b a
b a
 250
2
2
2
2
0 .1 8  0 .1 2
0 .1 8  0 .1 2
 9 6 .1 5 M P a
Equation (16.16c) at r  a :
u m ax 
2
Pi a
2
( b2 a2   ) 
b a
E
6
9 6 .1 5 (1 0 )
9
2 0 0 (1 0 )
2
2
( 0 .1 2 )( 0 .1 8 2  0 .1 2 2  0 .3 )
0 .1 8  0 .1 2
 0 .1 6 7 m m
( b ) Equation (16.19):
2
Po     , m a x
b a
2b
2
2
 250
2
0 .1 8  0 .1 2
2
2 ( 0 .1 8 )
2
  6 9 .4 4 M P a
SOLUTION (16.4)
  ,m ax  p i
(a) Su 
and
5
3
pi ,
2
2
2
2
b a
b a

pi 
Su  pi ,
pi  1,
5
3
3
5
 r ,m ax   p i   2
( 350 )  210
p i  350
MPa
(governs)
MPa
(CONT.)
261
16.4 (CONT.)
(b)
1
Su


2
S uc
5 pi
 1;
 pi

3( 350 )
 1
650
p i  158 . 7 MPa
or
SOLUTION (16.5)
( a ) Use Eq.(16.17) with p i  p ,
  ,m ax  p
2
2
2
2
c b
c b
a  b,
b  c:
6 2 .5 p
1 0 0 1 0
9
[
6 0 1 0
p
2
2
c b
By Eq.(16.25), with a  0 ,
0 .0 5 
2
c b
 60 M Pa,
6
2
(Fig. 16.6b)
6
6 0 (1 0 )

  0 . 05 mm :
b  62 . 5 ,
 0 .3 ] 
6 2 .5 p
2 0 0 1 0
(1)
p
[1  0 .3 ]
9
or
50  10
6
 37 . 5  10
6
 0 . 188 p  0 . 219 p ,
p  30 . 71 MPa
( b ) Equation (1) becomes
2
2
2
2
c  0 .0 6 2 5

60
3 0 .7 1
c  0 .0 6 2 5
c  110 m m
,
2c  220 m m
SOLUTION (16.6)
( a ) a=0, b=12.5 mm, c=50 mm
F 
T
b

 12 kN
150
0 . 0125
Thus
1 2 ( 1 0 )  2  b fp l  2  ( 0 .0 1 2 5 )( 0 .1 5 )( 0 .0 5 ) p
or
p  20 . 37 MPa
3
(Eq.16.27a)
Equation (16.25) gives then
 
2
3
1 2 .5 ( 2 0 .3 7 )
5 0  1 2 .5
1 0 0 (1 0 )
( b )   ,m ax  p
2
[ 5 0 2  1 2 .5 2  0 .3 ] 
1 2 .5 ( 2 0 .3 7 )
2
2
2
2
c b
c b
2 0 0 1 0
2
3
(1  0 .3 )  ( 3 .6 5  0 .8 9 1)1 0
3
 0 .0 0 5
mm
2
 2 0 .3 7[ 5 0 2  1 2 .5 2 ]  2 3 .0 9
M Pa
5 0  1 2 .5
SOLUTION (16.7)
We have a=15 mm, b=25 mm, c=50 mm.
Equation (16.25):
0 . 025 
2
25 p
210  10
9
2
[ 50 2  25 2  0 . 3 ] 
50  25
2
25 p
105  10
9
2
[ 25 2  15 2  0 . 3 ],
25  15
Steel, Eq.(16.17):
  ,m ax  p
2
2
2
2
c b
c b
 3 7 .3 9
2
2
2
2
50  25
50  25
 6 2 .3 2
M Pa
Bronze, Eq.(16.19):
  ,m ax   2 p
b
2
2
b a
2
  2 (3 7 .3 9 )
25
2
2
2 5 1 5
262
2
  1 1 6 .8
M Pa
p  37 . 39 MPa
SOLUTION (16.8)
Equation (16.25) with a=0, b=50 mm, c=150 mm.
 
bp
Ec
2
2
b c
bp
[ c2 b2   c ] 
2
150  50
50 p
0 .0 3 
12010
9
Es
[1   s ]
2
50 p
[ 1 5 0 2  5 0 2  0 .2 5 ] 
21010
9
( 0 .7 )
Solving p  3 7 .8 9 M P a
Shaft:     r   p   3 7 .8 9 M P a
Cylinder:
  ,m ax  p

2
2
2
2
c b
c b
2
150  50
  p   3 7 .8 9
r ,m ax
2
 3 7 .8 9[ 1 5 0 2  5 0 2 ]  4 7 .3 6
M Pa
M Pa
SOLUTION (16.9)
Equation (16.23):

 (u d ) rb 
2
bp
E
2
b c
2
bp
[ c2 b2   ] 
E
( 3  0 .3 )  1.9 6 7
5
bp
E
or
p 
E
3 .9 3 4 b
Then, Eq.(16.24) with a=0,
us 
bp
E
(1   ) 
0 .7 
3 .9 3 4
 0 .1 7 8 
Therefore,
 d s  0 .3 5 6 
SOLUTION (16.10)
Equation (16.31a) with (  r ) p  0 and b  c :

r


3 
8
(a  b 
2
3  0 .3 4
8
2
2
a b
r
2
2
 r )
2
[( 0 .0 2 )  ( 0 .0 4 ) 
2
2
2
 p
2
( 0 .0 2 ) ( 0 .0 4 )
( 0 .0 3 )
2
2
]8 5 0 0 
2
 9 0 (1 0 )
6
Solving,
  8 , 0 7 5 .5 tp s  4 8 , 4 5 3 r p m
SOLUTION (16.11)
 m ax 
3600
60
 2   1 2 0  ra d s ,
 m in  1 1 4 
ra d s ,
p  0
( a ) Equation (16.32), with p  0 ;
  ,m ax 

4
2
[ (1   ) a
2
 ( 3   )b ]
2
(CONT.)
263
16.11 (CONT.)
Let a=b and b=c=4b:
75  10
7 , 800 ( 120  )

6
2
 3 .3( 4 b ) ]
2
[ 0 .7 b
4
2
Solving,
b  7 1 .1 2 m m ,
c  2 8 4 .4 m m
( b ) Equation (16.37) of Sec.16.5:
I 
 l
7 .8  ( 5 0 )
(c  b ) 
4
2
4
( 0 .2 8 4 4  0 .0 7 1 1 2 )  3 .9 9 2 1 N  m  s
4
2
4
2
Thus
Ek 
I (  m a x   m in ) 
2
1
2
2
(3 .9 9 2 1)(1 2 0  1 1 4 ) 
2
1
2
2
2
 2 7 .6 5 9
kN  m
SOLUTION (16.12)
  2 4 0 0 ( 2  6 0 )  2 5 1 .3 ra d s
( a ) Equation (16.35) with a=b and b=c:
 
bp
E
2
b
2
[ b2  c 2  1] 
c b

20 ( 10

20
E
3
2
4E
2
[ 0 .7 b
2
[ 100 2  20 2  1 ] 
)p
E
2
20 ( 7 . 8 ) 
 20
100
 3 .3 c ]
2
[ 2 . 083 p  64 . 896  ]( 10
2
For p  3 . 2 MPa
  1 .0 2 5  1 0
[ 0 . 7 ( 0 . 02 )
3
2
 3 .3( 0 .1) ]
2
(1)
)
  2 5 1 .3 ra d s , Eq.(1) gives.
and
3
2
4E
mm
We have at   0:
3
1 .0 2 5  1 0

20
2 1 0 1 0
p  5 .1 6 7 M P a
( 2 .0 8 3 p ) ,
9
( b ) Thus
  p
2
2
2
2
b c
c b
 5 .1 6 7 (1 .0 8 3 )  5 .5 9 6 M P a
SOLUTION (16.13)
( a ) Equation (16.26) with a=0:
p 
2
E
b
c b
2

2
2c
200 ( 10
9
2
)( 0 . 02 ) 300
50
 50
2 ( 300 )
2
2
 38 . 89 MPa
Then

 , max
 p
2
2
2
2
c b
c b
 38 . 89
300
300
2
 50
2
2
 50
2
 41 . 11 MPa
( b ) Equation (16.35) with b=0.05 m, c=0.3 m, p=0,   7 .8 k N  m , and E  2 0 0 G P a :
0 . 02 ( 10
3
)   
b 
4E
2
[ 0 .7 b
2
 3 . 3 c ]  145 . 64  10
2
Solving,
  3 7 0 .6
ra d s . Thus
n  3 7 0 .6 ( 6 0 2  )  3 5 3 9 rp m
264
2
 12
SOLUTION (16.14)
( a ) From Eq. (16.17), we have
p    ,m ax
2
2
2
2
c b
b c
 30
2
2
2
2
( 0 .1 2 )  ( 0 .0 6 )
( 0 .0 6 )  ( 0 .1 2 )
 18 M Pa
( b ) Using Eq. (16.27a):
F  2  b p fl  2  ( 0 .0 6 )(1 8  1 0 )( 0 .1 8 )( 0 .2 )
6
 2 4 4 .3 k N
( c ) By Eq. (16.27b):
T  F b  2 4 4 .3 ( 0 .0 6 )  1 4 .6 6 k N  m
SOLUTION (16.15)
   4
I 

2
 m a x  2 4 0 0 ( 2  6 0 )  2 5 1 .3 ra d s

(b  a )l  
4
4
2
 m in  1 2 5 .7 ra d s
[ 0 .2  0 .0 5 ]( 6 0 )( 7 .8 )  1 .1 7 2 N  m  s
4
4
2
Therefore
T   
I (  m a x   m in )
2
1
2
2
or
T 
2
2
1 .1 7 2 ( 2 5 1 .3  1 2 5 .7 )
2 (4 )
 2 .2 0 8 k N  m
SOLUTION (16.16)
(a) I 

2

(b  a )l  
 m ax 
4
4
3000
60
[ 0 .2 5  0 .0 2 5 ]( 7 .8 )( 6 0 )  2 .8 7 1 N  m  s
4
2
4
( 2 )  1 0 0 ,
2
 m in  0 .9 (1 0 0  )  9 0  ra d s
Equation (16.32) with p=0:
  , max 

( b ) E k 
1
2

4
2
[( 1   ) a
7 ,8 0 0 (1 0 0  )
2
2
 (3   )b ]
2
[ 0 .7 ( 0 .0 2 5 )  3 .3 ( 0 .2 5 ) ]  3 9 .7 8 M P a
2
4
I (  max   min ) 
2
2
1
2
2
( 2 . 871 )( 100
2
 90
2
)
 26 . 919
2
kN  m
SOLUTION (16.17)
Dimensions are in millimeters.
A2
25
50
B
A
25
A1
r A  ri  2 1 5 m m ,
A  12 , 500
r 
150
50
A1 r1  2 A 2 r2
A1  2 A 2
mm

rB  4 1 5 m m
2
( 5 0  1 0 0 )( 2 1 5  2 5 )  2 ( 2 5 1 5 0 )( 4 1 5  7 5 )
5 0  1 0 0  2 ( 2 5 1 5 0 )
 300 m m
r
(CONT.)
265
16.17 (CONT.)
Equation (16.50):
R 
A

12, 500

dA
r

265
dr
r
100
215


 2 8 8 .4 3 9 6 m m
415
25
265
dr
r
e  r  R  3 0 0  2 8 8 .4 3 9 6  1 1 .6 0 4 m m
Equations (16.55a):
(  ) A  
100  
P
A
[1 
P
1 2 ,5 0 0
r ( R  rA )
e rA
[1 
]
3 0 0 ( 2 8 8 .4 3 9 6  2 1 5 )
(1 1 .5 6 0 4 )( 2 1 5 )
]
or
P  1 2 6 .7 k N
(  ) B  
100  
P
A
[1 
P
1 2 ,5 0 0
r ( R  rB )
[1 
e rB
]
3 0 0 ( 2 8 8 .4 3 9 6  4 1 5 )
(1 1 .5 6 0 4 )( 4 1 5 )
]
or
P  1 8 0 .8 k N
SOLUTION (16.18)
ro  ri  h  5 0  4 0  9 0 m m
A  b h  ( 2 0 )( 4 0 )  8 0 0 m m
2
We have
h
R 
ln
ro
40

ln
1
 6 8 .0 5 1 9
50
ri
r 
90
( ri  r )  7 0 m m
2
e  r  R  1 .9 4 8 1 m m
( a ) Using Eq.(16.52):
i  

(b) 
7 0 0 ( 6 8 .5 1 9  5 0 )
8 0 0 (1 0
6
7 0 0 ( 6 8 .0 5 1 9  9 0 )
o
 
o
  i 
8 0 0 (1 0
6
Mc
I
  1 6 2 .2 M P a
) (1 .9 4 8 1) ( 0 .0 5 )
 1 0 9 .5 M P a
) (1 .9 4 8 1) ( 0 .0 9 )

7 0 0 ( 0 .0 2 )
1
( 0 .0 2 ) ( 0 .0 4 )
 1 3 1 .3 M P a
3
12
266
SOLUTION (16.19)
h
ri  r 
 2 0 0  3 2 .5  1 6 7 .5 m m
2
h
ro  r 
 2 0 0  3 2 .5  2 3 2 .5 m m
2
A  b h  ( 4 5 )( 6 5 )  2 9 2 5 m m
2
Therefore
h
R 
ro
ln
65

ln
2 3 2 .5
 1 9 8 .2 2 7 m m
1 6 7 .5
ri
e  r  R  2 0 0  1 9 8 .2 2 7  1 .7 7 3 m m
( a ) Applying Eq.(16.52):
i  
M ( R  ri )
1 .5  1 0 (1 9 8 .2 2 7  1 6 7 .5 )
3
 
  5 3 .0 6 M P a
( 2 9 2 5 ) (1 .7 7 3 ) (1 6 7 .5 )
A e ri
( b ) Use Eq. (16.52):

o
 
M ( R  ro )
1 .5  1 0 (1 9 8 .2 2 7  2 3 2 .5 )
3
 
 4 2 .6 4 M P a
( 2 9 2 5 ) (1 .7 7 3 ) ( 2 3 2 .5 )
A e ro
SOLUTION (16.20)
45 mm
rA  1 2 5 m m ,
5 mm
5 mm
A  325
A 25 mm
B
r 
A1  A 2

2
r
Equation (16.50):
A1 r1  A 2 r2
mm
A1
A2
R 
rB  1 7 0 m m
A

325

dA
r

130
25
125
dr
r
2 5 ( 5 )(1 2 7 .5 )  ( 4 0  5 )(1 5 0 )
125 200


 1 3 9 .9 7 5 3 m m
170
5
130
dr
r
 1 4 1 .3 4 6 2 m m
e  r  R  1 .3 7 9 8 m m
Equations (16.53), with M   P ( r  2 5 )   1 6 6 .3 4 6 2 P and P   P :
(  ) B 
P
A
120 
P
325
[1 
[1 
( r  2 5 )( R  rB )
e rB
]
1 6 6 .3 4 6 2 (1 3 9 .9 7 5 3  1 7 0 )
(1 .3 7 9 8 )(1 7 0 )
]
or
P  1 .9 2 2 k N
267
SOLUTION (16.21)
h
R 
ln
60

ro
ln
1
2
 2 4 8 .7 9 5 4 m m
220
ri
r 
280
1
( ri  ro ) 
(220  280)  250 m m
2
A  ( 6 0 )( 6 0 )  3 6 0 0 m m
2
e  r  R  2 5 0  2 4 8 .7 9 5 4  1 .2 0 4 6 m m
Using Eq.(16.52):
i  
M ( R  ri )
M ( 2 4 8 .7 9 5 4  2 2 0 )
 1 5 0 (1 0 )  
6
;
3 6 0 0 (1 0
A e ri
6
, M  4 .9 7 k N  m
)(1 .2 0 4 6 )( 0 .2 2 )
Similarly,

o
 
M ( R  ro )
M ( 2 4 8 .7 9 5 4  2 8 0 )
; 1 5 0 (1 0 )  
6
3 6 0 0 (1 0
A e ro
6
,
M  5 .8 4 k N  m
)(1 .2 0 4 6 )( 0 .2 8 )
Therefore
M
a ll
 4 .9 7 k N  m
SOLUTION (16.22)
A  5 0 b  2 (1 5 0  2 5 )  5 0 b  7 5 0 0
We have rA  1 5 0 m m and rB  3 5 0 m m . Applying Eq. (16.52):

A
 
B
 
(  M )( R  r A )

(  M )( R  rB )
A e rA
A e rB
from which
 rB ( R  r A )  r A ( R  rB ) ,
or
R  210 m m
Then
R 
A

or
350(R  150)  150(R  350)
A  210
dA
r
 200 b d r

 1 5 0
r


350
200
2(25)dr 
 210

r

 6 0 .4 1 3 2 b  5 8 7 5 .9 6 5 8
Hence
or
5 0 b  7 5 0 0  6 0 .4 1 3 2 b  5 8 7 5 .9 6 5 8
b  156 m m
268
4
7

b ln  4 0 ln


3
4

SOLUTION (16.23)
r  150 m m
A  5000 m m
2
M  ( R  d )P
Table 16.1, Case A:
R 

h
ln
ro
r2
100
ln
2
1
 1 4 4 .2 6 9 5 m m
e  r  R  1 5 0  1 4 4 .2 6 9 5
 5 .7 3 0 5 m m
Equation (16.53):
(  ) A 
P
A

( r  d ) P ( R  rA )
A e rA
]
P
A
( r  d )( R  r A )
[1 
e rA
]
or
3
2 5 (1 0 )
80 
5000
[1 
(1 5 0  d )(1 4 4 .2 6 9 5  1 0 0 )
( 5 .7 3 0 5 )(1 0 0 )
]
1 5 ( 5 7 3 .0 5 )  (1 5 0  d ) ( 4 4 .2 6 9 5 )  6 , 6 4 0 .4 2 5  4 4 .2 6 9 5 d
Solving,
d  4 4 .1 7 m m
SOLUTION (16.24)
Locate centroid :
120 mm
50 mm
A2
r 
C A1
O
75 mm

80 mm
A1 r1  A 2 r2
A1  A 2
4 5 0 0 (1 2 0 )  ( 3 0 0 0 ) (1 6 0 )
4500  3000
 136 m m
r
1
2
R 
h ( b1  b 2 )
2
( b1 ro  b 2 ri ) ln
ro
ri
 h ( b1  b 2 )
( 0 .5 ) (1 2 0 ) ( 7 5  5 0 )
2

[ ( 7 5 ) ( 2 0 0 )  ( 5 0 ) (8 0 ) ] ln
200
 1 2 7 .1 3 3 0 m m
P
 (1 2 0 ) ( 7 5  5 0 )
80
e  r  R  8 .8 6 7 m m
( a ) Use Eq.(16.55a),

A
 
P
[1 
r ( R  rA )
A
A
B
]
O
M=P r
e rA
P
r

(1 3 6 ) (1 2 7 .1 3 3 0  8 0 ) 
 
1 
   1 0 0 .4 M P a
6
7 5 0 0 (1 0 ) 
(8 .8 6 7 ) (8 0 )

75000
(CONT.)
269
16.24 (CONT.)
( b ) From Eq.(16.55b):

P
 
B
r ( R  rB )
[1 
A
75000
 
]
e rB
7 5 0 0 (1 0
6

(1 3 6 ) (1 2 7 .1 3 3 0  2 0 0 ) 
1 
   4 5 .9 M P a
)
(8 .8 6 7 ) ( 2 0 0 )

SOLUTION (16.25)
A d
r  100 m m
4  7854 m m
2
2
From Table 16.1, Case B:
R 

A
2 (r 
r
2
c
2
 9 3 .3 0 1 5
7854
2  (1 0 0 
)
2
100 50
2
)
e  r  R  6 .6 9 8 5 m m
(a)
Equations (16.55a):
(  ) A  
P
A
[1 
r ( R  rA )
e rA
]
or
150 
P
7854
(  ) B 
(b)
P
A
1 0 0 ( 9 3 .3 0 1 5  5 0 )
[1 
( 6 .6 9 8 5 )( 5 0 )
[1 
r ( R  rB )
e rB
]
P  8 4 .5 8 k N
],
3
8 4 .5 8 (1 0 )
7 8 5 4 (1 0
6
)
[1 
1 0 0 ( 9 3 .3 0 1 5  1 5 0 )
( 6 .6 9 8 5 )(1 5 0 )
]
 50 M Pa
SOLUTION (16.26)
rA  1 2 5 m m ,
r 
1
rB  2 2 9 m m ,
1
( ro  ri ) 
2
a  100 m m ,
b  50 m m
(229  125)  177 m m
2
From Table 16.1
A   a b   (1 0 0 )(5 0 )  5 0 0 0  m m

dA

r
2 b
[r 
r
2
2
P
 a ]
2
a

2 (5 0 )
[1 7 7 
1 7 7  1 0 0 ]  9 7 .2 4 9 5 m m
2
2
r
A
100
Hence
A
R 

dA
A

5 0 0 0
 1 6 1 .5 2 1 5 m m
9 7 .2 5
P
M=-P r
B
r
(CONT.)
270
16.26 (CONT.)
e  r  R  1 5 .4 7 8 5 m m
Equation (16.55a) with M   P r :


A
P
[1 
r ( R  rA )
A
3

1 7 7 (1 6 1 .5 2 1 5  1 2 5 ) 
1 

) 
(1 5 .4 7 8 5 ) (1 2 5 )

1 2 5 (1 0 )
]
5 0 0 0 (1 0
e rA
6
 3 4 .5 4 M P a


B
P
[1 
r ( R  rB )
A
3

1 7 7 (1 6 1 .5 2 1 5  2 2 9 ) 
1 

) 
(1 5 .4 7 8 5 ) ( 2 2 9 )

1 2 5 (1 0 )
]
5 0 0 0 (1 0
e rB
6
  1 8 .8 6 M P a
SOLUTION (16.27)
rA  1 3 0 m m
r 
1
rB  2 0 0 m m
( ro  ri )  1 6 5 m m
2
A   a b   ( 7 0 )(3 5 )  2 4 5 0  m m

dA

2 b
r
[r 
r
2
2
 a ]
2
P
a

2 (3 5 )
[1 6 5 
1 6 5  7 0 ]  4 8 .9 6 0 1 m m
2
2
70
A
R 

dA
A

r
2 4 5 0
A
 1 5 7 .2 0 7 6 m m
4 8 .9 6 0 1
P
r
e  r  R  7 .7 9 2 4 m m
M=-P r
B
Using Eq.(16.55) with M   P r :

B

P
[1 
A
r ( R  rB )
]
e rB
 9 0 (1 0 ) 
6
Pa ll
2 4 5 0  (1 0
6

(1 6 5 ) (1 5 7 .2 0 7 6  2 0 0 ) 
1 

) 
( 7 .7 9 2 4 ) ( 2 0 0 )

Solving
Pa ll  1 9 6 .2 k N
271
SOLUTION (16.28)
1
A 
bh
2
The section width w varies linearly with r. Thus
(1)
w  c 0  c1 r
Since
w  b
( a t r  ri )
w  0
dr
(2)
( a t r  ro )
r
Substituting Eq.(1) into Eq.(2);
c1  
b
c0 
h
w
O
b
b ro
h
ri
h
r
Then

dA
A


r
ro
ri
w
c 0  c1 r
dr 
r
dr
r
Inserting c1 and c 0 into this, after integrating and rearranging,
we have
dA

 b(
ro
r
ln
h
ro
 1)
ri
Therefore
1
A
R 


dA
r
2
ro
ln
h
h
ro
1
ri
SOLUTION (16.29)
r
A  c
2
r
Through use of the polar coordinates we write:
w  2 c s in 
r  r  c cos 
d r  c s in  d 
C
d A  w d r  2 c s in  d 
2
2
w

O
c

dA
r


2 c s in 
0
r  c cos 
 2
2
2
d

c (1  c o s  )
0
r  c cos 

2
ccos
2
 2

 c cos   (r  c )
2
2
2
2
r  c cos 
0
 2  (r  c cos  )d   2(r
0
2
r
dr
2
 c )

2
0
d
r  c cos 
(CONT.)
272
16.29 (CONT.)
r c
2

 2r
0
 2 c s in 

 2(r
0
2
2
c )
2
ta n
r c
2
2
ta n


2
1
r c
2
0
This gives

dA
 2 ( r 
r c )
2
2
r
SOLUTION (16.30)
A 
1
2
( b1  b 2 ) h
ri
h
w
The section width w varies
linearly with r as
b2
w  c 0  c1 r
(1)
( a t r  ri )
w  b2
( a t r  ro )
O
r
dr
We have
w  b1
b1
r
(2)
Introduce Eq.(1) into Eq.(2), then solve for c 0 and c1
c0 
ro b1  ri b 2
c1  
h
b1  b 2
(3)
h
Then, we write

dA
r


ro
w
ri
dr 
r

ro
c 0  c1 r
ri
r
d r  c 0 ln
ro
ri
 c 1 ( ro  ri )
This gives, substituting Eqs.(3):

dA
ro b1  ri b 2

r
h
ln
ro
ri
 ( b1  b 2 )
Hence
1
A
R 

dA
r

2
h ( b1  b 2 )
2
( ro b1  ri b 2 ) ln
ro
ri
 ( b1  b 2 )
SOLUTION (16.31)
As in Example 16.7, we take c 1  c 2  0 in Eq.(16.60). Then, substituting Eq.(16.60) into
(16.56a) we obtain an expression for the radial moment M r .
(CONT.)
273
16.31 (CONT.)
Boundary conditions:
w  0
M
 0
r
(r  a )
now give
p0a
 c4 
2
c3a
4
 0
64 D
p0a
c3  
2
32 D
3 
1 
from which
c 4  p 0 a ( 5   ) 6 4 (1   ) D
4
The plate deflection is thus
p0a
w 
4
r
p0a
w m ax 
At r=0:
4
3 
1 
( a4  2
64 D
4
r
2
2
a
5
1 

(1)
)
5
1 
64 D
Substituting Eq.(1) into Eq.(16.56b):
p0
M 
[( 3   ) a
16
2
 ( 1  3 ) r ]
2
At r=a:
6 M
  ,m ax 
t
3 (1  ) p 0

2
4
( at )
2
SOLUTION (16.32)
Refer to Example 16.7. At r=a:

r

6M
t
 
r
2
S
( a ) 1   2 
or
y
n
1
2
p0a
3
4
2
  2,
2
t
p0 ( t ) (
a
S

2
2
2
p 0 ( t ) [ 16 
4
y
1
S
y
)
n
1
4

3
4
6M
t
S
) 
S
n  2
,
n
( b )  1   1 2   2  (
or
a
2
a
;
p0 ( t )
2
 
y
 
4t
2
2
 
a
2
p0a
1
y
n
t
(a)
P0

2
2
2
3
16

t
2
9
16
] (
Sy
2
7
4
) ;
n
p0 ( t )

Sy
n
Solving,
n  1.5 1 2
S
y
p0
(a)
SOLUTION (16.33)
Table 16.2:
w m ax 
1 .2 5 
Also

or
m ax
3
16
p0a
(1   ) ( 5   )
Et
2
3
6
3
16
( 0 .7 )(5 .3 )
 Sy 
3
8
3 .5 (1 0 )( 4 0 )
9
2 0 0 (1 0 ) t
(3   ) p 0
a
t
2
3
2
t  0 .2 5 m m
,
5 6 0  1 .2 3 8
;
t  0 .5 6 m m
274
3 .5 ( 4 0 )
t
2
SOLUTION (16.34)
From Example 16.7
1  
p0
a
(t)
4
2  
2
3
4
a
p0 ( t )
S
We have  1   1 2   2  (
2
2
p 0 ( at ) [ 161 
2
or
4
4 ( 10
or
10
4
)a
( 10 ) ( 10
4
 12

3
16
y
)
n
2
6
]  ( 150 2 10 ) ;
9
16
 12 , 857 . 14 (10
)
2
12
p 0 ( at ) ( 167 )  5 , 625 ( 10
2
2
4
12
)
), a  238 . 1 mm
SOLUTION (16.35)
r  1 7 5  1  1 7 4
r  7 5  1  7 4
F
mm,
mm,
F   ri p z ,
p   pz,
2
t  2 mm
ri  7 3 m m
  90 ,
o
p  100 kPa
N
Equation (16.64b):
N  

and


2
 ( 0 .0 7 3 ) (  1 0 0 )
 
F
2  r s in 
2  ( 0 .0 7 4 ) (1 )
 3 .6
kN
 1 . 8 MPa
3 , 600
0 . 002
Then, Eq.(16.64a):

r
or


r
6
pz
 
1 .8 ( 1 0 )
;
t
0 .1 7 4


0 .0 7 4

3
1 0 0 (1 0 )
0 .0 0 2
   2 . 93 MPa
SOLUTION (16.36)
N  is maximum at A (    9 0 ):
o
  ,m ax 
pa ( b a 2 )
( b a )t
S

y
n
240
1. 2
;
2 . 2 ( 5 0 )( 2 5 0  2 5 )

( 25050 )t
or
t  0 . 619

Similarly
n

mm
pr
240
1 .2
;
2t

2 . 2 ( 50 )
, t  0 . 275
2t
mm
SOLUTION (16.37)
1   
pa
2  x 
t
Pa
2t
( a ) Since  1 and  2 are of the same sign (  3  0 ) :
1  0 
S
y
n
;
pa
t

S
y
n
,
t 
pan
S
y
(CONT.)
275
16.37 (CONT.)
S
( b )  1   1 2   2  (
2
(
(c)
2
pa
) [1 
2
t
1


Su

2
S uc

1
2
1
n
1
4
S
] (
y
n
pa
;
y
)
n
) , t 
2
pa

tS u
2
3
2

4 tS u
1
n
pan
S
y
3 pan
t 
,
4 Su
SOLUTION (16.38)
(a)
p  200   x  200  15 (16 )  440


all
pa
t
pr
; t 

3
440 ( 10 )( 5 )

150 ( 10
all
6
)
 14 . 67 mm
( b ) p  200  15 [ 3 (16 4 )]  380
t 
pr


3
380 ( 10 )( 5 )
all
150 ( 10
6
kPa
 12 . 67 mm
)
( c ) Top-end plate, with p  200 kPa
 m ax 
3
4
t  158
or
a
(Fig.16.18):
1 5 0 (1 0 ) 
2
6
p( t ) ;
3
4
( 2 0 0  1 0 )( t )
2
3
4
( 4 4 0  1 0 )( t )
2
3
5
mm
Bottom-end plate, with p  440
 m ax 
kPa
3
4
a
t
kPa :
1 5 0 (1 0 ) 
2
6
p( ) ;
3
5
or
t  235
mm
SOLUTION(16.39)
We have r  a . The loading is p   p z   a ( 1  c o s  ) .
The resultant force F for the part intercepted by  :
F  2 a
2


 a ( 1  c o s  ) s in  c o s  d 
0
 2a  [ 6 
3
1
1
2
c o s  (1 
2
2
3
c o s  )]
Substitution of this into Eqs.(16.65) and rearranging yield the equations quoted
in this problem.
SOLUTION (16.40)
( a ) L w  1 .7 2 a t  1 .7 2 7 5 0  1 2  1 6 3 .2 m m
(b) 
cr
 0 . 605
Et
a
 0 . 605
210  10
9
750
( 12 )
 2 , 033
MPa
No failure in buckling, since  c r  S y and material would yield.
276
SOLUTION (16.41)

cr
 0 .6 0 5
Et
a
 0 .6 0 5
9
2 0 0  1 0 (1 0 )
600
 2, 017 M P a
Thus
Pc r  2  a t  c r
 2  ( 6 0 0 ) (1 0 ) ( 2 0 1 7 )  7 6 , 0 3 9 k N
End of Chapter 16
277
CHAPTER 17
FINITE ELEMENT ANALYSIS IN DESIGN
SOLUTION (17.1)
We have (
AE
L
)1,2 
A(2 E )
6 AE

L 3
(
L
AE
L
)3 
2 A(E )

L 3
6 AE
L
There are four displacement components ( u 1 , u 2 , u 3 , u 4 ) and so the order of the
system matrix is 4x4. Using Eq. (17.1):
[ k ]1  [ k ] 2  [ k ] 3 
 1

1 
6 AE  1

L 1
( a ) System matrix, [ K ]  [ k ]1  [ k ] 2  [ k ] 3 , by superposition is thus
u1
u2
 1

6 AE 1

[K ] 
L  0

 0
u3
1
0
(1  1)
1
1
(1  1)
0
1
u4
u1
u2
u3
u4
0 

0
6 AE
 
 1
L

1 
 1

1

 0

 0
1
0
2
1
1
2
0
1
0 

0

 1

1 
( b ) Refer to Eq. (17.7b):
 F1 x

 F2 x

F
 3x

 F4 x


AE

 
L



 1

1

 0

 0
1
0
2
1
1
2
0
1
0 

0

 1

1 
 u1 
 
u2 
 
u
 3
u 
 4
(1)
Boundary conditions are u 1  u 4  0 and F x 3  P . Equation (1) is then
0
AE  2
  

L 1
P
Solving u 2 
PL
9 AE
 1  u 2 
 
2  u3 
u3 
PL
18 AE
( c ) Equations (1) result in
 F1 x

 F2 x

F
 3x

 F4 x


6 AE

 
L



 1

1

 0

 0
1
0
2
1
1
2
0
1
0 

0

 1

1 
The reactions are
R1 
2
3
P 
R4 
1
P 
3
278


 PL

PL



2 P 3



9 AE  L
P


 


18 AE  AE
0



 P 3 
0



0
SOLUTION (17.2)
We have
(
AE
L
)1 
AE
(
AE
L
L
[ k ]1  [ k ] 2 
AE  1

L 1
AE
)2 
(
AE
L
L
)3 
4 AE
 0 .8
5L
AE
L
Equation (17.1):
 1

1 
[ k ]3 
A E  0 .8

L   0 .8
 0 .8 

0 .8 
( a ) System matrix, [ K ]  [ k ]1  [ k ] 2  [ k ] 3 , by superposition is then
 1

AE 1

[K ] 
L  0

 0
1
0
(1  1)
1
1
(1  0 .8 )
0
 0 .8


0

 0 .8 

0 .8 
0
( b ) Refer to Eq. (17.7b):
 F1 x

 F2 x

F
 3x

 F4 x


AE

 
L



u1
u2
u3
 1

1

 0

 0
1
0
2
1
1
1 .8
0
 0 .8
u4


0

 0 .8 

0 .8 
0
 u1 
 
u2 
 
u
 3
u 
 4
Boundary conditions are u 1  u 4  0 , We have F x 3  P . Thus
0
AE  2
  

L 1
P
1  u2 
 
1 .8   u 3 
Solving
PL
u2  
2 .6 A E
u3 
PL
1 .3 A E
( c ) Equations (1) yield
 F1 x

 F2 x

F
 3x

 F4 x


AE

 
L



 1

1

 0

 0
1
0
2
1
1
1 .8
0
 0 .8


0

 0 .8 

0 .8 
0
The reactions are
R1 
1
2 .6
P 
R4 
0 .8
P 
1 .3
279


P

P



  P 2 .6 



2 .6  L
0


 


1 .3  A E
P





0 
  1 .6 P 2 .6 
0
(1)
* SOLUTION (17.3)
We have A E L the same for all elements. Thus, Eq. (17.1):
 1

1 
AE  1

L 1
[ k ]1  [ k ] 2  [ k ] 3  [ k ] 4 
( a ) By superposition, we have
 1

1
AE 
 0
[K ] 
L 
 0
 0

1
0
0
2
1
0
1
2
1
0
1
2
0
0
1
0

0

0

 1
1 
(b)
 F1 x

F
 2x

 F3 x
F
 4x
 F 5 x



AE

 
L



 1

1

 0

 0
 0

1
0
0
2
1
0
1
2
1
0
1
2
0
0
1
0

0

0

 1
1 
 u1 
 
u
 2
 
u3 
(1)
u 
 4
 u 5 
Boundary conditions: u 1  0 and u 5   . We have F 2 x  F 3 x  F 4 x  0 .
Hence,
 0
AE


20 
L
 0


1

0

 0
2
1
0
1
2
1
0
1
2
0

0

 1 
 0
 
u
 2
 
 u3 
 
u
 4
  
This equation may be rewritten, after transposing the product of the appropriate
stiffness coefficients by the known displacement (  ) to the left side. In so doing,

 0

 0
 AE 

 L



AE
 
L



 2

1

 0
1
2
1
0

1

2 
u2 
 
u3 
 
u4 
Solving
u2 

4
u3 

2
u4 
3
4
(CONT.)
280
17.3 (CONT.)
(c)
Equations (1) give then
 F1 x

F
 2x

 F3 x



AE

 
L



F
 4x
 F 5 x
 1

1

 0

 0

 0
1
0
0
2
1
0
1
2
1
0
1
2
0
0
1
 0   4

 

 4
0

 
 AE

 

  2    0 

 
 L
3 4
0

 

     4 
0

0

0

 1
1 
The reaction is
R1 
1
AE 
4L
SOLUTION (17.4)
We have
AE
L

1
(1 2 0 0  1 0
6
)(7 2  1 0 )  5 4  1 0
9
6
N m
1 .6
c  cos 30
o

s  s in 3 0
3 2
o
1 2
3
4

( a ) Through the use of Eq. (17.14), we have
3

4

3
 4
6
[ k ]1  5 4 (1 0 ) 
3
  4

3
 4

3
4

1
4


3
3
4
3
4
4
3
1
4
4


 14 

3
4 
1 
4 
3
4
or
0 .7 5


0 .4 3 3
6
[ k ]1  5 4 (1 0 ) 
  0 .7 5

  0 .4 3 3
0 .4 3 3
 0 .7 5
0 .2 5
 0 .4 3 3
 0 .4 3 3
0 .7 5
 0 .2 5
0 .4 3 3
 0 .4 3 3 

 0 .2 5
 N m
0 .4 3 3 

0 .2 5 
( b ) Equation (17.11b), { } e  [ T ]{ } e :
 u1 
 0 .8 6 6
 

 0 .5
 v1 

  
 0
u
 3

 
 0
 v3 
0 .5
0
0 .8 6 6
0
0
0 .8 6 6
0
 0 .5


0

0 .5 

0 .8 6 6 
0
 1 .5   1 .8 9 9 

 

 1 .2   0 .2 8 9 

  
 mm
  2 .2    1 .9 0 5 
 0   1 .1 0 0 

 

(CONT.)
281
17.4 (CONT.)
( c ) Inserting the given data into (17.15) with i=1 and j=3, we obtain the axial force
6
F1  F1 3  5 4 (1 0 ) 

3
1
2
2
  2 .2  1 .5 
3

 (1 0 )   2 0 5 .4 k N

 0  1 .2 
So, axial stress in the element equals  1  F1 A   1 7 1 .2 M P a
Comment: The negative sign denotes compression.
SOLUTION (17.5)
We have E  2 0 0 G P a (by Table B.1). Refer to Solution of Prob. 17.4.
AE
L
1

(1 2 0 0  1 0
6
)( 2 0 0  1 0 )  1 5 0  1 0
9
6
N m
1 .6
c  cos 60
o
1 2
s  s in 6 0

o
3 2
( a ) From Eq. (17.14):
 1
4

3
 4
6
[ k ]1  1 5 0 (1 0 ) 
1
 4
 3
 4
3
4
3
4
3
4
3
4
0 .2 5


0 .4 3 3
6
 1 5 0 (1 0 ) 
  0 .2 5

  0 .4 3 3


1
4


 43 

3
4 
3 
4 

3
4
1
4
3
4
3
4
0 .4 3 3
 0 .2 5
0 .7 5
 0 .4 3 3
0 .4 3 3
0 .2 5
0 .7 5
0 .4 3 3
 0 .4 3 3 

0 .7 5

0 .4 3 3 

0 .7 5 
( b ) Equation (17.11b), { } e  [ T ]{ } e :
 u1 
 0 .5
 

 0 .8 6 6
 v1 

  

u
0
 3

 
0

 v3 
( c ) F1  F1 3  1 5 (1 0 )  12
6

3
2
0 .8 6 6
0
0 .5
0
0
0 .5
0
 0 .8 6 6


0

0 .8 6 6 

0 .5 
0
 1 .5   1 .7 8 9 

 

 1 .2    0 .6 9 9 

  
 mm
  2 .2    1 .1 

 

 0   1 .9 0 5 
  2 .2  1 .5 
3

 (1 0 )   4 3 3 .4 k N

 0  1 .2 
 1  F1 A   3 6 1 .2 M P a
282
SOLUTION (17.6)
Table 17.6 Data for truss of Fig. P17.6.
E le m e n t

o
1
45
2
315
o
3
0
4
90
5
90
o
o
o
c
s
2 2
2 2

2 2
c
2 2
2
s
2
cs
1 2
1 2
1 2
1 2
1 2
1 2
1
0
1
0
0
0
1
0
1
0
0
1
0
1
0
Equation (17.14)
u1
[ k ]1 
 1

1
2 AE

2L 1

1
v1
u2
2
1 2
1 2
2
1 2
1 2
2
1 2
1 2
2
1 2
1 2
v2
1 2

1 2

1 2

1 2
Similarly, for elements 2, 3, 4, and 5, we obtain
u1
[ k ]2 
 1

2 AE 1

2L 1

 1
u1
 1

0
AE

[ k ]3 
L 1

 0
v1
v1
u3
2
1 2
1 2
2
1 2
1 2
2
1 2
1 2
2
1 2
1 2
u4
v4
0
1
0
0
0
1
0
0
0

0

0

0
u4
v4
u2
v2
0

AE 0

[ k ]4 
L 0

0
0
0
1
0
0
0
1
0
0

1

0

1
v3
1 2

1 2

1 2 

1 2
u3
0

AE 0

[ k ]5 
L 0

0
283
v3
u4
0
0
1
0
0
0
1
0
v4
0

1

0

1
SOLUTION (17.7)
Table P17.7 Data for the truss of Fig P17.7
E le m e n t
L e n g th ( m )

1
7 .5
3 6 .9
2
6
0
3
4 .5
90
4 .5
0
4 .5
2
s
0 .8
0 .6
1
0
o
4
5
o
c
o
135
2
2
cs
s
0 .6 3 9
0 .4 8
0 .3 6
1
0
0
0
1
0
0
1
1
0
1
0
0
 0 .7 0 7
0 .7 0 7
0 .5
 0 .5
0 .5
o
o
c
Apply Eq.(17.14):


AE

[ k ]1 
7 .5  




AE

[ k ]2 
6 .0  




AE

[ k ]4 
4 .5  


u1
v1
u2
v2
0 . 639
0 . 48
 0 . 639
0 . 48
0 . 361
 0 . 48
0 . 639
 0 . 48
0 . 639
0 . 48
 0 . 361
0 . 48
u1
v1
u3
1
0
1
0
0
0
1
0
1
0
0
0
u3
v3
1
0
1
0
0
0
1
0
1
0
0
0
u4
 0 . 48  u 1

 0 . 361 v 1

0 . 48  u 2

0 . 361  v 2
v3
u2
0 

0

0 

0 
0

AE 0

[ k ]3 
4 .5  0

0
u1
v1
u3
v3
v4
v2
u3
v3
0
0
1
0
0
0
1
0
0 

1

0 

1 
u2
0 

0

0 

0 



AE

[ k ]5 
4 .5 2  


u3
v3
u4
v4
v2
u2
v2
u3
v3
u4
0 .5
 0 .5
 0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
 0 .5
 0 .5
v4
0 .5  u 2

 0 .5 v 2

 0 .5  u 4

0 .5  v 4
SOLUTION (17.8)
Table P17.8 Data for the truss of Fig.P17.8
E le m e n t

1
0
2
o
60
o
3
120
4
0
5
o
o
60
o
c
s
c
1
0
0 .5
2
2
cs
s
1
0
0
0 .8 6 6
0 .2 5
0 .4 3 3
0 .7 5
 0 .5
0 .8 6 6
0 .2 5
 0 .4 3 3
0 .7 5
1
0
1
0
0
0 .5
0 .8 6 6
0 .2 5
0 .4 3 3
0 .7 5
(CONT.)
284
17.8 (CONT.)
( a ) Use Eq.(17.14):


AE

[ k ]1 
L 


u1
v1
u4
1
0
1
0
0
0
1
0
1
0
0
0
u1


AE

[ k ]2 
L 




AE 

[ k ]3 
L 




AE

[ k ]4 
L 


v4
0 

0

0 

0 
u1
v1
u4
v4
v1
u2
v2
0 .2 5
0 .4 3 3
 0 .2 5
0 .4 3 3
0 .7 5
 0 .4 3 3
0 .2 5
 0 .4 3 3
0 .2 5
0 .4 3 3
 0 .7 5
0 .4 3 3
 0 .4 3 3 

 0 .7 5

0 .4 3 3 

0 .7 5 
u4
v4
u2
v2
0 .2 5
 0 .4 3 3
 0 .2 5
0 .4 3 3
0 .7 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .2 5
 0 .7 5
0 .4 3 3
u2
v2
u3
1
0
1
0
0
0
1
0
1
0
0
0
v1
u2
v2
0 .4 3 3 

 0 .7 5

 0 .4 3 3 

0 .7 5 
 0 .4 3 3
v3
0 

0

0 

0 
u1
u2
v2
u4
v4
u3
u2
v2
u3
v3


AE

[ k ]5 
L 


v3
u4
v4
0 .2 5
0 .4 3 3
 0 .2 5
0 .4 3 3
0 .7 5
 0 .4 3 3
0 .2 5
 0 .4 3 3
0 .2 5
0 .4 3 3
 0 .7 5
0 .4 3 3
 0 .4 3 3 

 0 .7 5

0 .4 3 3 

0 .7 5 
u3
v3
u4
v4
( b ) Global Stiffness Matrix [ K ] 
u1





AE 
L 





v1
u2
v2
u3
v3
u4
1. 2 5
0 .4 3 3
 0 .2 5
 0 .4 3 3
0
0
1
0 .4 3 3
0 .7 5
 0 .4 3 3
 0 .7 5
0
0
0
0 .2 5
 0 .4 3 3
1.5
0
1
0
0 .4 3 3
 0 .7 5
0
1.5
0
0
 0 .2 5
0 .4 3 3
0
0
1
0
1. 2 5
0 .4 3 3
 0 .2 5
0
0
0
0
0 .4 3 3
0 .7 5
 0 .4 3 3
1
0
 0 .2 5
 0 .4 3 3
1.5
0
0
 0 .4 3 3
 0 .7 5
0
 0 .2 5
0 .4 3 3
0 .4 3 3
 0 .7 5
v4


0

0 .4 3 3 

 0 .7 5

 0 .4 3 3 

 0 .7 5 

0

1.5 
0
u1
v1
u2
v2
u3
v3
u4
v4
(CONT.)
285
17.8 (CONT.)
 R1 x

R
 1y
 0

 P

Q

 0

R
 4x
R
 4y
{ F }  [ K ]{  };

 0 

 
0

 

u2 

 

v2 
  K  
u

 3

v 
3

 

 0 

 0 
 

SOLUTION (17.9)
Table P17.9 Data for the truss of Fig.P17.9
E le m e n t
L e n g th ( in . )

1
7 .2 1 1
5 6 .3 1
2
3
o
o
3
90
5
 3 6 .8 7
o
7 .2 1 1
 5 6 .3 1
o
4
5
3 6 .8 7
5
o
c
s
0 .5 5 5
0 .8 3 2
0 .8
c
2
2
cs
s
0 .3 0 8
0 .4 6 2
0 .6 9 2
0 .6
0 .6 4
0 .4 8
0 .3 6
0
1
0
0
1
0 .8
 0 .6
0 .6 4
 0 .4 8
0 .3 6
0 .5 5 5
 0 .8 3 2
0 .3 0 8
 0 .4 6 2
0 .6 9 2
( a ) Use Eq. (17.14):
u1
v1
0 .3 0 8
0 .4 6 2
 0 .3 0 8
0 .4 6 2
0 .6 9 2
 0 .4 6 2
0 .3 0 8
 0 .4 6 2
0 .3 0 8
0 .4 6 2
 0 .6 9 2
0 .4 6 2
u1
v1
0 .6 4
0 .4 8
 0 .6 4
0 .4 8
0 .3 6
 0 .4 8
0 .6 4
 0 .4 8
0 .6 4
0 .4 8
 0 .3 6
0 .4 8


AE

[ k ]1 
7 .2 1 1  




AE

[ k ]2 
5 


u2
0

AE 0

[ k ]3 
3 0

0
u2
u3
v2
u3
v3
0
0
1
0
0
0
1
0
0 

1

0 

1 
v2
 0 .4 6 2 

 0 .6 9 2

0 .4 6 2 

0 .6 9 2 
u1
v1
u2
v2
v3
 0 .4 8 

 0 .3 6

0 .4 8 

0 .3 6 
u1
v1
u3
v3
u2
v2
u3
v3
(CONT.)
286
17.9 (CONT.)
u3
[k ]4


AE 


5 


v3
u4
0 . 64
 0 . 48
 0 . 64
0 . 48
0 . 36
0 . 48
0 . 64
0 . 48
0 . 64
0 . 48
 0 . 36
 0 . 48
u2



AE

[ k ]5 
7 .2 1 1  


v4
0 . 48  u 3

 0 . 36 v 3

 0 . 48  u 4

0 . 36  v 4
v2
u4
0 .3 0 8
 0 .4 6 2
 0 .3 0 8
0 .4 6 2
0 .6 9 2
0 .4 6 2
0 .3 0 8
0 .4 6 2
0 .3 0 8
0 .4 6 2
 0 .6 9 2
 0 .4 6 2
v4
0 .4 6 2 

 0 .6 9 2

 0 .4 6 2 

0 .6 9 2 
u2
v2
u4
v4
( b ) Global Stiffness Matrix [ K ] 







AE






u1
v1
u2
v2
0 .1 7 1
0 .1 6
 0 .0 4 3
 0 .0 6 4
 0 .1 2 8
 0 .0 9 6
0
0 .1 6
0 .1 6 8
 0 .0 6 4
 0 .0 9 6
 0 .0 9 6
 0 .0 7 2
0
0 .0 4 3
 0 .0 6 4
0 .0 8 5
0
0
0 .0 6 4
 0 .0 9 6
0
0 .5 2 5
0
0 .1 2 8
 0 .0 9 6
0
0
0 .2 5 6
0
0 .0 9 6
 0 .0 7 2
0
0
0 .4 7 7
0 .0 9 6
0 .1 7 1
 0 .3 3 3
u3
v3
0
 0 .3 3 3
0
0
 0 .0 4 3
0 .0 6 4
 0 .1 2 8
0 .0 9 6
0
0
0 .0 6 4
 0 .0 9 6
0 .0 9 6
 0 .0 7 2
{ F }  [ K ]{  };
 R1 x

R
 1y
 P

 0

0

 Q

R
 4x
R
 4y

 0 

 
0

 

u2 

 

v2 
  K  
u

 3

v 
3

 

 0 

 0 
 

287
u4
 0 .0 4 3
0 .0 6 4
 0 .1 2 8
 0 .1 6
v4


0

0 .0 6 4 

 0 .0 9 6

0 .0 9 6 

 0 .0 7 2 
 0 .1 6 

0 .1 6 8 
0
u1
v1
u2
v2
u3
v3
u4
v4
SOLUTION (17.10)
Table P17.10 Data for the truss of Fig.P17.10
E le m e n t
L e n g th

1
1. 2
0
2
3
c
s
c
1
0
0 .9 2 3
0
o
1.3
2 2 .6 2
0 .5
 90
o
o
o
4
1. 2
0
5
1.3
2 2 .6 2
o
2
2
cs
s
1
0
0
0 .3 8 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
1
0
0
1
1
0
1
0
0
0 .9 2 3
0 .3 8 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
( a ) Use Eq.(17.14);


AE

[ k ]1 
1.2  


uA
vA
uC
1
0
1
0
0
0
1
0
1
0
0
0
uA


AE

[ k ]2 
1.3  


0 

0

0 

0 
uA
vA
uC
vC
vA
uB
vB
 0 .3 5 5 

 0 .1 4 8

0 .3 5 5 

0 .1 4 8 
0 .8 5 2
0 .3 5 5
 0 .8 5 2
0 .3 5 5
0 .1 4 8
 0 .3 5 5
0 .8 5 2
 0 .3 5 5
0 .8 5 2
0 .3 5 5
 0 .1 4 8
0 .3 5 5
uB
0

AE 0

[ k ]3 
0 .5  0

0
uB


AE

[ k ]4 
1.2  


vC
vB
uC
vC
0
0
1
0
0
0
1
0
0 

1

0 

1 
vB
uD
1
0
1
0
0
0
1
0
1
0
0
0
uA
vA
uB
vB
uB
vB
uC
vC
vD
0 

0

0 

0 
uB
vB
uD
vD
(CONT.)
288
17.10 (CONT.)


AE

[ k ]5 
1.3  


uC
vC
uD
vD
0 .8 5 2
0 .3 5 5
 0 .8 5 2
0 .3 5 5
0 .1 4 8
 0 .3 5 5
0 .8 5 2
 0 .3 5 5
0 .8 5 2
0 .3 5 5
 0 .1 4 8
0 .3 5 5
 0 .3 5 5 

 0 .1 4 8

0 .3 5 5 

0 .1 4 8 
uC
vC
uD
vD
( b ) Global Stiffness Matrix [ K ] 







AE






uA
vA
uB
vB
uC
vC
uD
1. 4 8 9
0 .2 7 3
 0 .6 5 5
 0 .2 7 3
0
0
0 .2 7 3
0 .1 1 4
 0 .2 7 3
 0 .1 1 4
0
0
0
0 .6 5 5
 0 .2 7 3
1. 4 8 9
0 .2 7 3
0
0
0 .2 7 3
 0 .1 1 4
0 .2 7 3
2 .1 1 4
0
 2
 0 .8 3 3
 0 .8 3 3
0
0 .8 3 3
0
0
0
1. 4 8 9
0 .2 7 3
 0 .6 5 5
0
0
0
 2
0 .2 7 3
2 .1 1 4
 0 .2 7 3
0
0
0
 0 .6 5 5
 0 .2 7 3
1. 4 8 9
0
0
0
 0 .2 7 3
 0 .1 1 4
0 .2 7 3
 0 .8 3 3
0
 R Ax

R
 Ay
 0

 0

0

 0

P

R
 Dy
{ F }  [ K ]{  };






 






 0

0

uB

vB
K 
u
 C
v
C

uD
 0

vD


0


0

0

 0 .2 7 3 

 0 .1 1 4 
0 .2 7 3 

0 .1 1 4 
0













SOLUTION (17.11)
We have AE=30 MN.
Table P17.11 Data for the truss of Fig.P17.11
E le m e n t
L e n g th

1
5
5 3 .1 3
2
4
0
o
o
c
s
0 .6
0 .8
1
0
c
2
2
cs
s
0 .3 6
0 .4 8
0 .6 4
1
0
0
( a ) Use Eq.(17.14):
(CONT.)
289
uA
vA
uB
vB
uC
vC
uD
vD
17.11 (CONT.)


AE

[ k ]1 
5 




AE

[ k ]2 
4 


u1
v1
0 .3 6
0 .4 8
 0 .3 6
0 .4 8
0 .6 4
 0 .4 8
0 .3 6
 0 .4 8
0 .3 6
0 .4 8
 0 .6 4
0 .4 8
u2
v2
u3
1
0
1
0
0
0
1
0
1
0
0
0
u2
v2
 0 .4 8 

 0 .6 4

0 .4 8 

0 .6 4 
u1
v1
u2
v2
v3
0 

0

0 

0 
u2
v2
u3
v3
( b ) Global Stiffness Matrix




[K ]  AE 





u1
v1
u2
v2
u3
0 .0 7 2
0 .0 9 6
 0 .0 7 2
 0 .0 9 6
0
0 .0 9 6
0 .1 2 8
 0 .0 9 6
 0 .1 2 8
0
0 .0 7 2
 0 .0 9 6
0 .3 2 2
0 .0 9 6
0 .0 9 6
 0 .1 2 8
0 .0 9 6
0 .1 2 8
0
0
0 .2 5
0
0
0
0
0
0
 0 .2 5 0
0
v3
 0 .2 5
( c ) Boundary Conditions: u 1  v 1  u 3  v 3  0 .
 F2 x 

  AE
F
 2y 
 0 .3 2 2

 0 .0 9 6
u2 
1  4
  

AE 3
 v2 
 F1 x

 F1 y
(d) 
F
 3x
 F3 y




  AE



0 .0 9 6   u 2 
 
0 .1 2 8   v 2 
3
0

  0 .0 0 1 0 
  
 m

1 0 .0 6 3    1 0 , 0 0 0    0 .0 0 3 4 
  0 .0 7 2

 0 .0 9 6

  0 .2 5 0

0

 0 .0 9 6 
 7 .6 3 2 



 0 .1 2 8  0 .0 0 1 0   1 0 .1 7 6 



 
 kN
   0 .0 0 3 4 

0
7
.5
0
0





0
0



( e ) Use Eq.(17.16)
F1 2 
F23 
AE
5
AE
4
 0 .6
1
 0 .0 0 1 
0 .8  
   1 2 .7 2 k N
  0 .0 0 3 4 
  0 .0 0 1 
0
   7 .5 k N
 0 .0 0 3 4 
(C )
290
(C )
0 

0

0 

0

0 

0 
u1
v1
u2
v2
u3
v3
SOLUTION (17.12)
We have E  2 0 0 G P a
A  2 (1 0 ) m m
3
2
Table P17.12 Data for the truss of Fig.P17.12
E le m e n t
L e n g th ( m m )

1
3000
90
2
3000
3
2
45
3000
0
c
s
c
0
1
0 .7 0 7
1
o
o
o
2
2
cs
s
0
0
1
0 .7 0 7
0 .5
0 .5
0 .5
0
1
0
0
( a ) Use Eq.(17.14)
u1
[ k ]1 
v1
0

0

0

0
AE
3000
u2
0
0
1
0
0
0
1
0
u1



2 


AE
[k ]2 
3000
u1
[k ]3 
AE
3000






v2
0  u1

 1 v1

0 u2

1  v2
v1
0 .5
 0 .5
0 .5
0 .5
 0 .5
0 .5
 0 .5
0 .5
0 .5
 0 .5
0 .5
u4
0
1
0
0
0
1
0
1
0
0
0
(10
11
0  u1

0 v1

0 u4

0  v4
) . Adding zero’s in proper locations and adding :
u1
[K ] 
4
3
(10
11
 1 . 354

0 . 354

 0

0
)
  0 . 354

  0 . 354
1

 0
 0 .5  u 1

 0 .5 v 1

0 .5  u 3

0 .5  v 3
v4
1
4
3
v3
0 .5
v1
( b ) The common factor
u3
v1
u2
v2
u3
v3
u4
0 . 354
0
0
 0 . 354
 0 . 354
1
1 . 354
0
1
 0 . 354
 0 . 354
0
0
0
0
0
0
0
1
0
1
0
0
0
 0 . 354
0
0
0 . 354
0 . 354
0
 0 . 354
0
0
0 . 354
0 . 354
0
0
0
0
0
0
1
0
0
0
0
0
0
v4
0  u1

0 v1

0 u2

0  v2
0 u3

0  v3
0 u4

0  v 4
(CONT.)
291
17.12 (CONT.)
(c)















u1 

 
25000
v

 1

F2x
0 

 
F2 y

0 
   K  
F3x

0 

0 
F3 y

 
F4x

0 

0 
F4 y
 

0
Eliminate rows and columns in [K] corresponding to zero displacements:
0


4
(10

 
3
  25000 
11
 u1 
1
  
11
( 4 3 ) (1 0 )
 v1 
0 . 354   u 1 
 
1 . 354   v 1 
 1 . 354
)
 0 . 354
 0 .2 0 7  
0
  0 .0 3 8 9 
6
  
 (1 0 ) m m

0 .7 9 3    2 5 0 0 0    0 .1 4 8 6 
 0 .7 9 3

  0 .2 0 7
(d)
 R2 y

 R3x

R
 3y

 R4x


4

11
(1 0 )
 
3



1
 0

 0 .3 5 4

  0 .3 5 4

1

 1 9 .8 1 3 



 0 .3 5 4  u 1   5 .1 7 8 

   
 kN
 0 .3 5 4   v 1   5 .1 7 8 



0

  5 .1 8 7 
( e ) Use Eq.(17.16):
AE
F1 2 
L1
F1 3 
  u1 
2 (1 0 )  2 0 0 (1 0 )
1 
0
 
3000
  v1 
3
0
AE
L2
2 (1 0 )  2 0 0 (1 0 )
F1 4 
AE
L3
  0 .0 3 8 9 
6
0 .7 0 7  
 (1 0 )
 0 .1 4 8 6 
9
3000
 0 .7 0 7
2
  u1 
2 (1 0 )  2 0 0 (1 0 )
0
1
 
3000
  v1 
3
1
  0 .0 3 8 9 
6
1 
 (1 0 )  1 9 .8 1 3 k N ( T )
 0 .1 4 8 6 
  u1 
0 .7 0 7  

  v1 
 0 .7 0 7
3

9
 7 .3 1 2 2 k N
(T )
  0 .0 3 8 9 
6
0
 (1 0 )   5 .1 8 6 7 k N ( C )
 0 .1 4 8 6 
9
SOLUTION (17.13)
Table P17.13 Data for the truss of Fig.P17.13
E le m e n t
L e n g th

1
5
5 3 .1 3
2
4
180
o
o
c
s
0 .6
0 .8
1
0
c
2
2
cs
s
0 .3 6
0 .4 8
0 .6 4
1
0
0
(CONT.)
292
17.13 (CONT.)
( a ) Use Eq.(17.14):


AE

[ k ]1 
5 




AE

[ k ]2 
4 


u1
v1
0 .3 6
0 .4 8
 0 .3 6
0 .4 8
0 .6 4
 0 .4 8
0 .3 6
 0 .4 8
0 .3 6
0 .4 8
 0 .6 4
0 .4 8
u2
v2
1
0
1
0
0
0
1
0
1
0
0
0
u2
u3
v2
 0 .4 8 

 0 .6 4

0 .4 8 

0 .6 4 
u1
v1
u2
v2
v3
0 

0

0 

0 
u2
v2
u3
v3
( b ) Global Stiffness Matrix




[K ]  AE 





u1
v1
u2
v2
u3
0 .0 7 2
0 .0 9 6
 0 .0 7 2
 0 .0 9 6
0
0 .0 9 6
0 .1 2 8
 0 .0 9 6
 0 .1 2 8
0
0 .0 7 2
 0 .0 9 6
0 .3 2 2
0 .0 9 6
0 .0 9 6
 0 .1 2 8
0 .0 9 6
0 .1 2 8
0
0
0 .2 5
0
0
0
0
0
0
 0 .2 5
0
 0 .2 5
v3
0 

0

0 

0

0 

0 
u1
v1
u2
v2
u3
v3
(c)
0 . 096   u 2 
 
0 . 128   v 2 
0


 0 . 322

  AE 
  60000 
 0 . 096
u2 
1
  
6
 v 2  1 0 (1 0 )
 4

3
3
0

  18 
3
  
 (1 0 )

1 0 .0 6 3    6 0 0 0 0    6 0 .4 
m
(d)
 R1 x

 R1 y

 R3x


  AE


  0 .0 7 2

 0 .0 9 6

  0 .2 5
 0 .0 9 6 
 4 5 .0 2 4 

 u2  
 0 .1 2 8     6 0 .0 3 2 
 v
 2 

0

45.

kN
( e ) Use Eq.(17.16)
F1 2  A E  0 .6
u2 
6
0 .8     1 0 (1 0 )  0 .6
v
 2
 0 .0 1 8 
0 .8  
   3 7 5 .2
  0 .0 6 0 4 
kN
(C )
(CONT.)
293
17.13 (CONT.)
u2 
6)
0
  1 0 (1 0   1
v

2 

F23  A E   1
  0 .0 1 8 
0
  180
 0 .0 6 0 4 
kN
(T )
SOLUTION (17.14)
We have AE  125
MN .
Table P17.14 Data for the truss of Fig.P17.14
E le m e n t
L e n g th

1
5
1 4 3 .1 3
2
5
3
o
3 6 .8 7
5
1 4 3 .1 3
o
o
c
s
c
 0 .8
0 .6
0 .8
 0 .8
2
2
cs
s
0 .6 4
 0 .4 8
0 .3 6
0 .6
0 .6 4
0 .4 8
0 .3 6
0 .6
0 .6 4
 0 .4 8
0 .3 6
( a ) Apply Eq.(17.14):


AE 

[ k ]1 
25L 




AE

[ k ]2 
25L 


u1
v1
u2
16
 12
 16
12
9
12
16
12
16
12
 9
 12
u2
v2
u3
16
12
 16
12
9
 12
16
 12
16
12
 9
12
v2
12 

 9

 12 

9 
u1
v1
u2
v2
v3
 12 

 9

12 

9 


AE 

[ k ]3 
25 L 


u2
v2
u3
v3
u2
v2
u4
16
 12
 16
12
9
12
16
12
16
12
 9
 12
v4
12 

 9

 12 

9 
u2
v2
u4
v4
( b ) Global Stiffness Matrix






AE 
[K ] 
25L 





u1
v1
u2
v2
u3
v3
u4
v4
16
 12
 16
12
0
0
0
9
12
 9
0
0
0
16
12
48
 12
 16
 12
 16
12
 9
 12
27
 12
 9
12
0
0
 16
 12
16
12
0
0
0
 12
 9
12
9
0
0
0
 16
12
0
0
16
0
0
12
 9
0
0
 12
0 

0

12 

 9

0 

0 
 12 

9 
12
 25000
AE  48
 

25L  12
 40000
(c) 
u1
v1
u2
v2
u3
v3
u4
v4
 12  u2 
 
27  v2 
(CONT.)
294
17.14 (CONT.)
12    25000 
  1 . 0026 
  
 (10

48    40000 
  1 . 927 
u 2 
 27
25 L
  

AE (1152 )  12
v2 
3
) m
(d)
 F1 x

F
 1y
 F 3 x

F
 3y
 F4x

 F 4 y




1
 
1152




  16

12

  16

  12
  16

 12
( e ) Use Eq.(17.16):
We have u 1  v 1  0 ;
F 12 
F 32 
F 42 
AE

L
AE
u 2 
0 . 6   
v2 
0 . 8
L
AE

u3  v 3  0;
u 2 
0 . 6   
v2 
0 .8
L
1
1152
1
1152




 kN




u 4  v 4  0.

 27
0 . 6 
 12
0 .8

 27
0 . 6 
 12
0 .8
12    25000 
   8 . 854

48    40000 
12    25000 
   48 . 958

48    40000 
 27
0 . 6 
 12
0 . 8
1
1152
u 2 
0 . 6   
v2 
0 .8
  7 . 083

5 . 313

 39 . 167
12    25000 
  

48    40000 
 29 . 375
  7 . 083

 5 . 313
12 

 9

 12   27

 9   12
12 

 9 
kN
12    25000 
   8 . 854

48    40000 
kN
(C )
(C )
kN
(C )
SOLUTION (17.15)
We have E  210 GPa
4
A  5  10
m
2
Table P17.15 Data for the truss of Fig.P17.15
E le m e n t
L e n g th

1
5
5 3 .1 3
2
4
90
o
c
s
0 .6
0 .8
0
1
o
c
2
2
cs
s
0 .3 6
0 .4 8
0 .6 4
0
0
1
( a ) Apply Eq.(17.14):


AE

[ k ]1 
5 


u1
v1
u2
0 .3 6
0 .4 8
 0 .3 6
0 .4 8
0 .6 4
 0 .4 8
0 .3 6
 0 .4 8
0 .3 6
0 .4 8
 0 .6 4
0 .4 8
v2
 0 .4 8 

 0 .6 4

0 .4 8 

0 .6 4 
u1
v1
u2
v2
(CONT.)
295
17.15 (CONT.)
u1
v1
u3
v3
0

AE 0

[ k ]2 
4 0

0
0
0
1
0
0
0
1
0
0 

1

0 

1 
u1
v1
u3
v3
( b ) Global Stiffness Matrix
u1
v1
 0 .7 5 6

1.0 0 8

  0 .7 5 6
7
[K ]  10 
 1.0 0 8

 0

 0
u2
v2
u3
1.0 0 8
 0 .7 5 6
 1.0 0 8
0
3 .9 6 9
 1. 0 0 8
 1.3 4 4
0
 1.0 0 8
0 .7 5 6
1.0 0 8
0
 1.3 4 4
1. 0 0 8
1.3 4 4
0
0
0
0
0
0
0
0
 2 .6 2 5
v3


 2 .6 2 5


0

0


0

2 .6 2 5 
0
u1
v1
u2
v2
u3
v3
(c)
 F1 x

100000

  10

100  10
3
1 . 008    0 . 025 


v1
3 . 969  

 0 . 756

 1 . 008
7
 (1 . 008  10 )(  0 . 025 )  3 . 696  10 v 1
7
F 1 x  ( 0 . 756  10
7
7
)(  0 . 025 )  1 . 008  10
7
v1
or
or
v 1  0 . 00887
m
F 1 x   99 . 6 kN
( d ) Support reactions:
 F2x

 F2 y

F
 3x
 F3 y




  10



7







0 . 756
1 . 008
0
0
99 . 59
 1 . 008 



 1 . 344   0 . 025

 132 . 79

  

  0 . 00887 
0
0


  232 . 84
 2 . 625 




 kN



( e ) Use Eq.(17.16):
We have u 2  v 2  0 ,
F 12 
F 13 
AE
0 . 6
L1
AE
L2
0
u3  v 3  0.
 u1 
105  10
0 . 8 
 
5
  v1 
 u1 
105  10
1 
 
4
  v1 
6
0
6
0 . 6
 0 . 025

0 . 8 
  166
  0 . 00887 
 0 . 025

1 
   232 . 8 kN
  0 . 00887 
296
kN
(C )
(T )
SOLUTION (17.16)
We have AE  20 MN .
Table P17.16 Data for the truss of Fig.P17.16
E le m e n t
L e n g th

1
5
3 6 .8 7
2
3
8
0
5
c
s
0 .8
0 .6
1
 0 .8
o
o
1 4 3 .1 3
o
c
2
2
cs
s
0 .6 4
0 .4 8
0 .3 6
0
1
0
0
0 .6
0 .6 4
 0 .4 8
0 .3 6
( a ) Apply Eq.(17.14):


AE

[ k ]1 
5 




AE

[ k ]2 
8 


u1
v1
0 .6 4
0 .4 8
 0 .6 4
0 .4 8
0 .3 6
 0 .4 8
0 .6 4
 0 .4 8
0 .6 4
0 .4 8
 0 .3 6
0 .4 8
u1
v1
1
0
1
0
0
0
1
0
1
0
0
0
u2


AE 

[ k ]3 
5 


u3
u2
v2
 0 .4 8 

 0 .3 6

0 .4 8 

0 .3 6 
u1
v1
u2
v2
v3
v2
0 

0

0 

0 
u1
v1
u3
v3
u3
v3
0 .6 4
 0 .4 8
 0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
0 .6 4
0 .4 8
 0 .3 6
 0 .4 8
0 .4 8 

 0 .3 6

 0 .4 8 

0 .3 6 
u2
v2
u3
 0 .1 2 5
u2
v2
u3
v3
( b ) Global Stiffness Matrix




[K ]  AE 





u1
v1
0 .2 5 3
0 .0 9 6
 0 .1 2 8
 0 .0 9 6
0 .0 9 6
0 .0 7 2
 0 .0 9 6
 0 .0 7 2
0 .1 2 8
 0 .0 9 6
0 .2 5 6
0
0 .0 9 6
 0 .0 7 2
0
0 .1 1 4
0 .0 9 6
0
 0 .1 2 8
0 .1 2 5
0
 0 .1 2 8
0 .0 9 6
0 .2 5 3
0
0
0 .0 9 6
 0 .0 7 2
 0 .0 9 6
v3


0

0 .0 9 6 

 0 .0 7 2

 0 .0 9 6 

0 .0 7 2 
0
u1
v1
u2
v2
u3
v3
(CONT.)
297
17.16 (CONT.)
(c)
 F2 x

 F2 y

 F3 x


  AE


  
0
0 .1 1 4
0 .0 9 6
 0 .1 2 8   u 2 
 
0 .0 9 6   v 2 
 
0 .2 5 3   u 3 
1
K  F
1
AE
u2 
 
 v2  
 
u3 
 0 .2 5 6

 0
  0 .1 2 8

1
2 0 1 0
6
 F1 x 


( d )  F1 y   A E
F 
 3x 
 3 .8 9 2 9
 6 .2 1 7 7

 3 .8 9 2 7

 4 .6 2 2 9
  0 .1 2 8

 0 .0 9 6

 0 .0 9 6
1 5 .3 2 8 5
 7 .7 8 5 9
 0 .0 9 6
 0 .0 7 2
 0 .0 7 2
4 .6 2 2 9 

 7 .7 8 5 9

9 .2 4 5 2 
 0 .1 2 5 

0.

 0 .0 9 6 
 40000

80000

0

  0 .0 2 8 0 
 

    0 .0 6 9 1  m
 

  0 .0 4 0 4 
 0 .0 2 8 0   4 0 .0 0 8 

 

  0 .0 6 9 1    4 5 .7 4 4 

 

 0 .0 4 0 4   7 5 .6 9 6 
kN
( e ) Use Eq.(17.16):
We have u 1  v 1  v 3  0 .
F1 2 
F1 3 
F3 2 
AE
 0 .0 2 8 0 
0 .6  
   7 6 .2 4
  0 .0 6 9 1 
 0 .8
5
AE
1
8
AE
5
 0 .0 4 0 4 
0
  1 0 1 .0
0


  0 .8
kN
kN
(C )
(T )
 0 .0 2 8 0  0 .0 4 0 4 
0 .6  
   1 2 6 .1 6 0
 0 .0 6 9 1


kN
(C )
SOLUTION (17.17)
Due to symmetry, only one-half of the beam need be considered.
 12

6L
EI
[ k ]1  3 
L 12

 6L
P/2
L
2
6L
4L
2
12
6L
6 L
2L
12
2
6L
0
0
0
0
0
0
0
0
6L
2 
2L

6 L 
2 
4L 
1
1
2
k/2
0

 3
E I kL E I
[ k ]2  3 
0
L 

0

0

0

0

0
(CONT.)
298
17.17 (CONT.)
Therefore
 12

6L
EI 
[K ]  3 
L 12

 6 L
12
6L
4L
6L
2 
2L


6L 

2
4 L 
6 L
2
6L
kL
12 
3
EI
2L
6 L
2
( a ) Boundary conditions are v1  0 and  2  0 . Equation (17.19a)
with F 2 y   P 2 and M 1  0 :
 4L
 0 
EI 

  3
L 6L
 P 2

6L 
  1 
3
kL   
12 
 v2 
E I 
2
Introduce the data and solve:
 1   5 .1(1 0
3
v 2   1 3 .5 (1 0
) ra d
3
) m
(b)
 12
 F1 y 

6L


0
EI 



  3 
F
L 12
2
y



M 
 2
 6 L
12
6L
4L
6 L
2
6L
12 
kL
3
EI
2L
6 L
2
6L
2 
2L


6L 

2
4 L 
 0 


  5 .1 
3

 (1 0 )
  1 3 .5 
 0 


or
 F1 y   7 .4 2 5 k N

 
 F 2 y     9 .8 5 5 k N
 M   3 0 .1 5 0 k N  m
 2 
F s p r in g  1 8 0 (1 3 .5 )  2 .4 3 k N





(C )
From symmetry: F1 y  F 3 y  7 .4 2 5 k N 
SOLUTION (17.18)
L  6 .7 m ,
P = 9 kN ,
v1
 12

6L
EI
[ k ]1  3 
L 12

 6L
1
2
6 L
2L
5
u2
6L
4L
E I = 6 5 (1 0 ) N  m ,
2
12
6L
12
6L
4
2
k  210 kN m
v2
6L
2 
2L

6 L 
2 
4L 
0

 3
E I kL E I
[ k ]2  3 
0
L 

0

2
u3  3
0
0
0
0
0
0
0
0
0

0

0

0
(CONT.)
299
17.18 (CONT.)
Thus
 12

6L
EI 
[K ]  3 
L 12

 6 L
12
6L
4L
6L
6L
2 
2L


6L 

2
4 L 
6 L
2
kL
12 
3
EI
2L
6L
2
(1)
( a ) Boundary conditions are v1  0 and  1  0 . Equation (17.19a), with F 2 y   P
and M
2
 0:

kL
 P 
E I 1 2 
EI

  3
L 
 0 
6 L


 6 L   v2 
 
 
2
 2
4 L 
3
Substituting the given numerical values and solving, we have
v 2   0 .0 3 2 7 m
 2   0 .0 0 7 3 3 ra d
(b)
 12
 F1 y 

6L


EI 
M1

  3 
F
L 12
y
2





M 2 
 6 L
12
6L
4L
6L
2
6L
12 
kL
3
EI
2L
6L
2
6L
2 
2L


6 L 

2
4 L 
0




0




  0 .0 3 2 7 
  0 .0 0 7 3 3 


Introducing the data and multiplying:
2 .1 3 8 k N
 F1 y  

 
 M 1   1 4 .2 4 5 k N  m

  
F2 y

   9 .0 0 5 k N
 M    0 .0 8 1 k N  m
 2 







The spring force is Ps p r in g  2 1 0 ( 0 .0 3 2 7 )  6 .8 6 7 k N
(C )
SOLUTION (17.19)
We have E I  7 0  1 0 N  m ,
( a ) Use Eqs.(17.19):
v1
1
v2
4
 12

18
EI
[ k ]1  3 
L 12

 18
2
18
12
36
18
18
12
18
18
L  3 m,
P  50 kN
2
18 

18

18 

36 
v1
1
v2
2
(CONT.)
300
17.19 (CONT.)
v2
2
v3
 12

18
EI
[ k ]2  3 
L 12

 18
18
12
36
18
18
12
18
18
( b ) Global Stiffness Matrix
v1
1
 12

18

EI 12
[K ]  3 
L  18

0

0

3
18 

18

18 

36 
v2
2
v3
3
v2
2
v3
3
18
12
18
0
36
18
18
0
18
24
0
12
18
0
72
18
0
12
18
12
0
18
18
18
0 

0

18 

18 
18 

36 
v1
1
v2
2
v3
3
Use Eq.(17.20a):
 F2 y 
 24
EI 


M 2   3  0
L
M 
 18
 3
  
3
L
EI
18   v 2

18   2

36    3
0
72
18





1
K  F 
 v2 
3
L
 


 2
EI
 
 3
 0 .0 7 2 9 1 7

0 .0 1 0 4 1 7

  0 .0 4 1 6 6 7
0 .0 1 0 4 1 7
0 .0 1 7 3 6 1
 0 .0 1 3 8 8 9
 0 .0 4 1 6 6 7 

 0 .0 1 3 8 8 9

0 .0 5 5 5 5 6 
50 10

0


0

3
   0 .1 4 0 6 m 
 

    0 .0 2 0 1 r a d 
  0 .0 8 0 4 r a d 

 
(c)
 F1 y 
EI


M1  3
L
F 
3
y


12

18

  1 2
18
18
18
0 

0

 1 8 
 v 2   3 4 .3 6 2 k N
  
  2    5 6 .2 3 3 k N  m
  
  3   1 5 .6 0 2 k N





From a free-body diagram of element 2: ( M 2 ) 2  4 6 .8 0 6 k N  m and
( F 2 ) 2   1 5 .6 0 2 k N
(CONT.)
301
17.19 (CONT.)
(d)
50 kN
2
1
1
1.5 m
3
1.5 m
2
34.362
V
+
(kN )
x
15.602
M
46.806
(kN  m )
+
x
56.28
SOLUTION (17.20)
( a ) The element stiffness matrices are, from Eq. (17.19a):
v1
1
v2
2
v3
 12

6L

EI 12
[ k ]1  3 
L  6L
 0

 0
6L
12
6L
0
6 L
2L
12
6 L
6 L
4L
0
0
0
0
0
0
0
0
0

0

0
EI 
[ k ]2  3 0
L 

0

 0
4L
2
6 L
2L
2
2
0
2
0
0
0
0
0
0
0
0
0
12
6L
12
0
6L
4L
12
12 
kL
3
EI
0
6L
2L
6 L
2


0

6L 
2 
2L 

6 L 

2
4 L 
0
6 L
2
6 L
0

0

0

0
0

0
0
0
0
3
(CONT.)
302
17.20 (CONT.)
(b)
 12

6L

12

[K ]  6L


 0

 0
12
6L
4L
2
6L
2L
24
0
0
8L
6L
2L
2
12
0
6L
0
2
0
12
6L
2
6 L
12 
kL
3
EI
0
6L
2L
2


0

6L 
2 
2L 

6 L 

2
4 L 
0
6 L
( c ) Boundary conditions are: v1  0 ,  1  0 , v 2  0
System governing equations, by Eqs. (17.19a), after rearrangement:

 8 L2
 0 
EI 


2
 0   3  2L
L
 P 



6L

2L
4L
2
2
6L

6L 

6L 
3 
kL

12 
EI 
 2 
 
 3 
v 
 3
Solving,
v3  
2  
where k 1  k L
7PL
3
(
EI
3PL
3
EI
1
1 2  7 k1
2
(
1
1 2  7 k1
)
),
 3  3 2
EI
SOLUTION (17.21)
Boundary conditions are
v1   1  v 3   3
v 2    (given)
Equation (17.19):
 24

EI 12 L
[ k ]1  3 
L 24

1 2 L
12 L
8L
2
12 L
4L
2
24
12 L
24
12 L
12 L 
2 
4L

12 L 
2 
8L 
 12

6L
EI
[ k ]2  3 
L 12

 6L
6L
4L
2
6L
2L
2
12
6L
12
6L
6L
2 
2L

6L 
2 
4L 
After assembling [ K ] and considering the boundary conditions, the pertinent
equations are found as
(CONT.)
303
17.21 (CONT.)
 P 
EI

  3
L
 0 
 24  12

1 2 L  6 L
12 L  6 L    

2
2  
8L  4L   2 
Multiplying,
EI
P 
(  3 6   6 L 2 )
(1)
(6 L   12 L  2 )
(2)
L
0 
EI
L
3
3
2
( a ) Equation (1) is then
P  33
EI
L
3

( b ) Equation (2) gives
2 

2L
SOLUTION (17.22) through (17.26)
It is important to take into account any conditions of symmetry which may exist. Use a 2-D finite
element program such as ANSYS.
End of Chapter 17
304
CHAPTER 18
CASE STUDIES IN MACHINE DESIGN
SOLUTION (18.1)
a2=0.16 m
a1=2.6 m
B
P=15 kN
L=2.5 m
C
H
D
40o
a3=1.0 m
80o
FBG
A
FC F
60
o
o
50
Figure S18.1 Free Body diagram of loader arm
FAE
( a ) Member forces Dismember the arm ABD. It is assumed that the links and
hydraulic cylinder are all in tension. The conditions of moment equilibrium are
applied to Fig. S18.1:

M
 0:
B
 F C F c o s 1 0 ( 0 .1 6 )  F A E s in 5 0 (1 .0 )  1 5 ( 2 .7 6 )  0
o
o
F A E  0 .2 0 6 F C F  5 4 .0

Fx  0 :
(1)
F A E c o s 6 0  F B G c o s 4 0  F C F s in 1 0  0
o
o
o
F B G   3 5 .2 4 6  0 .3 6 1 F C F

Fy  0 :
(2)
 F A E s in 6 0  F B G c o s 4 0  F C F c o s 1 0  1 5  0
o
o
o
FC F   4 2 k N ( C )
Substitution of this into Eqs. (1) and (2) result in
F A E  4 5 .3 5 k N ( T )
F B G   2 0 .0 8 k N ( C )
Comments: Since the result obtained for F A E is positive. A negative sign means
that the sense of the force is opposite to that taken originally.
( b ) Diameters of pins at A, B, and C in double shear. We have
S ys
n

F 2
d
2
4
d  [ 2 SF n ]
,
1
2
ys
Thus,
dA  [
2 ( 4 5 .3 5 ) ( 2 .4 )
dB  [
2 ( 2 0 .0 8 ) ( 2 .4 )
dC  [
3
 (1 5 0  1 0 )
3
 (1 5 0  1 0 )
2 ( 4 2 ) ( 2 .4 )
3
 (1 5 0  1 0 )
1
]
 2 1 .5 m m
2
 1 4 .3 m m
1
]
1
]
2
2
 2 0 .7 m m
305
SOLUTION (18.2)
The location of the critical point is at K, where the maximum moment and the shear force are
M  P L  1 5 ( 2 .5 )  3 7 .5 N  m
V  P  15 kN
The cross-sectional area properties:
A   ( c 2  c 1 )   ( 7 5  5 0 )  9 .8 1 7 5  1 0
2

I 
4
2
2
( c 2  c1 ) 
4
4

2
3
( 7 5  5 0 )  1 9 .9 4 1 8  1 0
4
4
4
mm
6
2
mm
4
The maximum tensile stress due to the bending occurs at point K. Therefore,

m ax
M c2

I

3
3 7 .5 (1 0 )( 0 .0 7 5 )
1 9 .9 4 1 8 (1 0
6
)
 141 M Pa
The shearing stress is zero,   0 , at point K. The maximum shearing stress is at the neutral axis
z and parallel to y axis. From third case, Table 3.2:
 m ax  2
V
A
3
1 5 (1 0 )
 2
9 .8 1 7 5 (1 0
3
)
 3 .0 6 M P a
This is a very low stress for the specified material. The bending stress vanishes at the neutral axis,
 H  0 . The factor of safety is thus
n 
 3 .4
480
141
SOLUTION (18.3)
A sketch of Mohr's circle is shown in Fig. S18.3 constructed by obtaining the position of point C
at (  x   y ) 2  4 0 0 μ on the horizontal axis and of point at
( x ,  
xy
2 )  (1 0 0 0 μ ,  3 5 0 μ ) from the origin O.
The principal strains are represented by points A1 and B1 . Thus, referring to the figure:

( 1 0 0 02 2 0 0 )  3 5 0
 400 
2
1 ,2
2

'=400
D
y
B(-200, 350)
B1 O
”s
C
’p
A1

A(1000, -350)
x
E
Figure S18.3
or
 1  1 0 9 4 .6 μ ,
 2   2 9 4 .6 μ
(CONT.)
306
18.3 (CONT.)
As a check, note that  x   y   1   2  8 0 0  . The planes of principal strains are
2
1 350
600
'  ta n
p
 3 0 .3
o
2  p "  3 0 .3  1 8 0  2 1 0 .3
and
o
and
 p '  1 5 .1 5
o
 p "  1 0 5 .1 5
and
o
(1)
From Mohr's circle,  p ' locates the  1 direction. The maximum shearing strains are given by
points D and E:
 m ax   2
( 1 0 0 02 2 0 0 )  3 5 0
2
 1389 
2
Alternatively,    1   2  1 0 9 4 .6  2 9 4 .6  1 3 8 9 μ .
SOLUTION (18.4)
From Prob. 18.3, we have
 1  1 0 9 4 .6 μ ,  2   2 9 4 .6 μ ,  m a x  1 3 8 9 μ
The first two of Eqs. (2.7) together with (1) give
1 


2
2 1 0 1 0
3
1  ( 0 .2 8 )
2
2 1 0 1 0
3
1  ( 0 .2 8 )
(1)
[1 0 9 4 .6  0 .2 8 (  2 9 4 .6 ) ]  2 3 1 M P a
2
[  2 9 4 .6  0 .2 8 (1 0 9 4 .6 ) ]  2 .7 1 M P a
From the last of Eqs. (2.7):
 m ax 
E
2 (1   )
 m ax 
3
2 1 0 1 0
2 (1  0 .2 8 )
 p '  1 5 .1 5
From Prob. 18.3:
o
(1 3 8 9 )  1 1 4 M P a
 s  6 0 .1 5
and
o
Using Eq. (3.35),
 '  12 ( 2 3 1  2 .7 1)  1 1 6 .9 M P a
SOLUTION (18.5)
Use Eq. (3.40):
 a   x c o s  a   y s in  a  
2
1 1 0 4 (1 0
6
2
xy
s in  a c o s  a
)   x c o s 0   y s in 0  
2
2
o
o
o
xy
 x  1104 μ
o
s in 0 c o s 0 ,
Similarly,
 b   x c o s  b   y s in  b  
2
4 3 2 (1 0
4 3 2 (1 0
6
6
2
xy
s in  b c o s  b
)   x c o s (  6 0 )   y s in (  6 0 )  
2
2
o
)  0 .2 5  x  0 .7 5 
y
 0 .4 3 3 
o
xy
s in (  6 0 ) c o s (  6 0 )
o
xy
o
(1)
and
 c   x c o s  c   y s in  c  
2
 9 6 (1 0
6
2
)  0 .2 5  x  0 .7 5 
y
xy
s in  c c o s  c
 0 .4 3 3 
Subtract Eq. (2) from Eq. (1):
5 2 8    0 .8 6 6 
Thus

xy
 610 μ ,
xy

y
 144 μ
307
xy
(2)
SOLUTION (18.6)
Equation (5.78a):
2 E
Sy
Cc 
2

2
3
( 2 1 0 1 0 )
 1 2 8 .8
250
For the 1.6 m link column,
L B G r  1 6 0 0 1 0 .6 9  1 4 9 .7  C c
and Eq. (5.77b) is used. Hence,

a ll
2
 E
1 .9 2 ( L B G r )



2
9
( 2 1 0 1 0 )
1 .9 2 (1 4 9 .7 )
2
 4 8 .2 M P a
Comment: This stress is very low compared to 250 MPa; The link will not yield.
SOLUTION (18.7)
From Prob. 18.6: A  3 1 8 m m ,
E  200 G Pa ,
2
r  1 0 .6 9 m m .
The required value of  a ll :

a ll
FB G


A
3
1 1 .3 4 (1 0 )
3 1 8 (1 0
6
 3 5 .7 M P a
)
(1)
Assume L m r  C c . Equation (5.77b):

a ll
2
 E

1 .9 2 ( L m r )
2


2
9
( 2 0 0 1 0 )
1 .9 2 ( L m r )
Lm
Equating Eqs. (1) and (2), we obtain
(2)
2
r
 1 6 9 .7
Since L m r  C c , our assumption is OK. Thus
Lm
r

Lm
1 0 .6 9
 1 6 9 .7
or
L m  1 .8 1 4 m
Comment: If this link is more than 1.814 m in length, it will buckle.
SOLUTION (18.8)
( a ) Central cross brace. Stain energy due to bending. Equation (5.18) with M  W x 2 :
U  2[ 
c 2
M
2 EI
0
dx] 
1
EI

c 2
0
W
4
2
x dx 
2
2
3
W c
96 EI
Due to shear, Eq. (5.23) with V  W 2 :
U 

c
0
2
3V
5GA
dx 
2
3W c
20 G A
Side supports. Strain energy owing to bending, with M  W x 2 :
U  4[ 
b 2
0
2
M
2 EI
dx] 
2
EI

b 2
0
2
W
16
x dx 
2
2
3
W b
384 EI
(CONT.)
308
18.8 (CONT.)
Due to shear with V  W 4 :
U  2
b
2
2
dx 
3V
5GA
0
3W b
40 G A
Total strain energy is then
Ut 
2
3

W c
96 EI
2
3

W b
384 EI
2
2

3W c
20 G A
2W b
40 G A
or
3
3
U t  W [ E1I ( 9c 6 
2
b
384
)
(c 
3
20 G A
b
2
Q.E.D.
)]
(P18.8a)
( b ) Introducing Eq. (P18.8a) into Eq. (5.31), we obtain
 st 
U t
3
 W [ 4 81E I ( c 
3
W
b
4
)
(c 
3
10 G A
b
2
(P18.8b)
)]
SOLUTION (18.9)
( a ) Substituting the given data into Eq. (P18.8b) result in
 s t  W [ 4 81E I ( c 
3
W{
)
3
4 8 ( 2 0 0  1 0  0 .2 7 8 )
6
6
(c 
3
10 G A
[( 0 .8 ) 
3
1
 W (1 0
 10
3
b
4
b
2
)]
(1 .2 )
3
4
]
3
3
1 0 ( 7 9 1 0  8 1 9 )
[( 0 .8 ) 
1 .2
2
]}
){0 .3 7 4 7[ 0 .5 1 2 0  0 .4 3 2 ]  [ 0 .0 0 4 6 4[1 .4 ]}
W {0 .3 5 3 7  0 .0 0 6 5}  1 0
6
W ( 0 .3 6 0 2 )
or
 s t  0 .3 6 0 2 (1 0
 5 .4 0 3(1 0
3
6
) W  ( 0 .3 6 0 2  1 0
6
)(1 5  1 0 )
3
) m
( b ) The impact factor, by Eq. (4.32):
K  1
1
2h
 st
 1
2 ( 250 )
1
 1 0 .6 7
5 .4 0 3
Equation (4.35) is therefore
 m a x  K  s t  1 0 .6 7 (5 .4 0 3)  5 7 .7 m m
SOLUTION (18.10)
( a ) Thin-walled cylinder is in biaxial stress (  3  0 ) with     1 and  a  2 .
From Eqs. (3.6) at r  a :
1 
pa
t
20 (350 )

t


7000
t
,
Sy
2
2

1
2

3500
t
(1)
Equation (6.14):
 1   1
2
2

2
2
 (
n
)
(2)
Substituting Eqs. (1) into (2), we have
6
[( 7 )  ( 7 )(3 .5 )  (3 .5 ) ] 1 02  ( 5 55 2 ) ,
2
2
2
t
t  55 m m
Hence,
b  a  t  405 m m
( b ) We have a t  6 .3 6 : the thin-walled analysis does not apply. So, the solution is not valid.
309
SOLUTION (18.11)
Refer to Prob. 18.10.
For a closed-ended thick-walled cylinder under internal pressure, the critical
section where the maximum stresses occur is at r  a (see Fig. 16.3). The
stresses, from Eqs. (16.16a), 16.16b), and (16.15) with p  p i and p o  0 :
r  p
  p

 p
a
2
2
2
2
b a
b a
a
(1)
2
2
b a
2
We observe that     1 ,  a   2 , and  r   3 , where algebraically  1   2   3 .
With a safety factor included, Eq. (6.13) appears as
[ ( 1   2 )  (
2
  3 )  (
2
2
  1) ]  2(
2
3
Sy
n
(2)
)
Substituting Eq. (1) into (2), we have
[(b  a  a )  ( a  b  a )  (  b  a  b  a ) ](
2
or
4
3b p
2
2
2
2
2
 (b  a ) (
2
2
2
Sy
n
2
)
2
2
2
2
2
2
p
2
2
b a
)  2(
2
2
2
Sy
n
)
2
(3)
By introducing the given data ( a  0 .3 5 m , S y  5 5 2 M P a , p  2 0 M P a , n  5 ) and
simplifying Eq. (3) becomes:
9 .1 5 7 b  2 .4 8 8 b  0 .1 5 2  0
4
2
Let x  b and solve the resulting quadratic equation to find x  0 .1 7 7 8 .
The outer radius is then
2
b  0 .4 2 1 7 m = 4 2 1 .7 m m
Hence, t  b  a  4 2 1 .7  3 5 0  7 1 .7 m m
Comment: The outer diameter equals 2 b  8 4 4 m m . A standard cylinder with about
8 4 5  m m outer diameter and 7 0 0  m m inner diameter should be selected.
SOLUTION (18.12)
( a ) Element Stiffness Matrix. Referring to Fig. P18.12 we sketch Fig. S18.12.
F1 y , v 1
L
1
F1 x , u 1
 45o

F3 x , u 3
2

F2 x , u 2
F2 y , v 2
3
F3 y , v 3
Figure S18.12 Finite element model
(CONT.)
310
18.12 (CONT.)
Using Eq. (17.4), Fig. S18.12, Table P18.1, and the given data:
u1
v1
 c

cs
7
[ k ]1  4 (1 0 ) 
c2

 c s
2
u1
0

0
7
[ k ] 2  4 (1 0 ) 
0

0
u2
c
cs
s
 cs
c
s
cs
2
u3
0
0
1
0
0
0
1
0
u1
 cs 
2 
s 
cs 

2
s 
2
 cs
2
v1
v2
2
 1

v1
0
7
 4 (1 0 ) 
1
u2

v2
 0
u1
v3
v1
u2
v2
0
1
0
0
0
1
0
0
0

0

0

0
u2
0 

1

0 

1 
 0 .5

 0 .5
7
2 (1 0 ) 
  0 .5

 0 .5
u1
v1
[ k ]3  2
u3
v3
u1
v1
u2
v2
v2
u3
v3
 0 .5
 0 .5
0 .5
0 .5
0 .5
0 .5
 0 .5
 0 .5
0 .5 

 0 .5

 0 .5 

0 .5 
u2
v2
u3
v3
In the foregoing, the column and row of each stiffness matrix are labeled according to the nodal
displacements associated with them. Observe that displacements u 3 and v 3 are not involved in
element 1; u 2 and v 2 are not involved in element 2; u 1 and v1 are not involved in element 3.
Thus, before adding [ k ]1 , [ k ] 2 , and [ k ] 3 to form the system matrix, two row and columns of zero
must be added to each of the element matrices to account for the absence of these displacements.
7
In so doing, and using a common factor 1 0 , the element stiffness matrices become:
u1
v1
u2
v2
u3
 4

0

4
7
[ k ]1  (1 0 ) 
 0
 0

 0
0
4
0
0
0
0
0
0
0
4
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0

0

0

0
0

0
u3
v3
u1
v1
u2
v2
0

0

0
7
[ k ] 2  (1 0 ) 
0
0

0
0
0
0
0
4
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
4
0
0
0
v3
0 

4

0 

0 
0 

4 
u1
v1
u2
v2
u3
v3
u1
v1
u2
v2
u3
v3
(CONT.)
311
18.12 (CONT.)
u1
0

0

0
7
[ k ] 3  (1 0 ) 
0

0

0
v1
u2
v2
u3
0
0
0
0
0
0
0
0
0
0
0

2

2
v3
2
2
0

2
2
2

2
0

2
2
2

2
0

2

2
2
2










u1
v1
u2
v2
u3
v3
System Stiffness Matrix. There are a total of six components of displacement for the truss before
boundary constraints are imposed. Hence, the order of the truss stiffness matrix must be 6  6 .
Before addition of the terms from each element stiffness matrix into their corresponding locations
in [ K ] , we obtain the global matrix for the truss:
u1
 4

0

4
7
[ K ]  (1 0 ) 
 0

 0

 0
v1
u2
v2
u3
0
4
0
0
4
0
0
0
4
0

2
2
2
0

2
2
2
4


2
2
2




4

2 

 2 

 2 

4 2
0
0
2
v3
2
2
u1
v1
u2
(1)
v2
u3
v3
System Force-Displacement Relationship. From Fig. P18.12, the
boundary conditions are u 1  0 , v1  0 , and u 3  0 . In addition F 3 y  0 .
Equation (17.17 is therefore
 R1 x

R
 1y
 P

W
R
3x

0





  [K




 0 
 
0
 
 u 2 
] 
v
 2
 0 
 
 v 3 
(2)
where [ K ] is given by Eq. (1).
( b ) Displacements. To find u 2 , v 2 , and v 3 only part of Eq. (2) relating to these
displacements is considered:
(CONT.)
312
18.12 (CONT.)
(4  2 )
  24, 000 

 


7
    3 6 , 0 0 0   (1 0 )   2
 


0
2
 


P

W
 0

2
2




 2 

(4  2 )

2
2
u2 
 
 v2 
v 
 3
(3)
Inverting,
u2 
 
8
 v 2   2 .5 (1 0 )
 
 v3 
1
 1

1

 0
0

1

1 
4 .8 2 8
1
 24 
 1 .5 




3
  3 6  (1 0 )    4 .9 4  m m




 0 
  0 .9 
Reactions. Inserting of the preceding values of u 2 , v 2 , and v 3 into Eq. (2) gives
the reactional forces:
 R1 x

 R1 y
R
 3x
 4

 0

 2


7
  10


0
0
2
0   1 .5 
60




3
 4    4 .9 4  (1 0 )   3 6  k N

 36 

2    0 .9 


The results may be verified by applying the equations of equilibrium to the free-body diagram of
the entire truss, Fig. SP18.12.
Axial Forces in Bars. Using Eqs. (17.16), (3) and Table P18.12,
we obtain
F1  F1 2 
AE
L
[1
F 2  F1 3 
AE
L
[0
F3  F 2 3 
AE
2L
u2 
7
0 ]    4 (1 0 ) [1
 v2 
 1 .5 
3
0] 
 (1 0 )  6 0 k N
  4 .9 4 
0 
7
 1]    4 (1 0 ) [ 0
v
 3
 u2 
 1] 
  2
 v3  v2 
[1
 0 
3
 1] 
 (1 0 )  3 6 k N
  0 .9 
2 (1 0 ) [  1
7
 1 .5


3
 1] 
 (1 0 )
  0 .9  4 .0 4 
  7 1 .8 k N
Stresses in Bars. Driving the element forces by the cross-sectional area results in
1 
3
6 0 (1 0 )
4 8 0 (1 0
6
)
 125 M Pa

2
 1 2 5 ( 36 60 )  7 5 M P a
 3  1 2 5 (  7610.8 )   1 4 9 .6 M P a
The negative sign means a compressive stress.
Factor of safety against yielding. Dividing the yield strength of S y  2 5 0 M P a
(from Table B.1) by each stress, we obtain
n1 
250
125
 2
n2 
250
75
 3 .3 3
n3 
250
1 4 9 .6
 1 .6 7
Comments: The bar axial stresses found are relatively low for the well known material
considered (see Sec. 1.6).
End of Chapter 18
313
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