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solutions MAnuAl FoR
by
Mechanical 
Design of Machine 
coMponents
ansel c. UgURal
solutions MAnuAl FoR
by
Boca Raton London New York
CRC Press is an imprint of the
Taylor & Francis Group, an informa business
Mechanical 
Design of Machine 
coMponents
ansel c. UgURal
CRC Press
Taylor & Francis Group
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Boca Raton, FL 33487-2742
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 vi
 
CONTENTS 
 
 
 
 
 
 
 
Section 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 
 
 
Section II FAILURE PREVENTION 
 
Chapter 6 STATIC FAILURE CRITERIA AND RELIABILITY 100 
 
Chapter 7 FATIGUE FAILURE CRITERIA 117 
 
Chapter 8 SURFACE FAILURE 135 
 
 
Section 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 MECHANICAL COMPONENTS 261 
 
Chapter 17 FINITE ELEMENT ANALYSIS IN DESIGN 278 
 
Chapter 18 CASE STUDIES IN MACHINE DESIGN 308 
 
 
 vii
 
 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. 
 
The instructor has the choice of assigning problems using SI units and problems using U.S. 
customary units. 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. 
 
 
 
 
 
 
 
 viii
 
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 materialsand 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 deal with 
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. 
 
 ix
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 introduce 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 a comprehensive introduction to the finite 
element analysis in mechanical design. A variety of case studies illustrate solutions of problems involving 
structural assemblies, stress concentration factors in plates and disks, and thick-walled pressure vessels. 
Finally, case studies in preliminary design of the entire crane with winch and a high-speed cutting machine 
are introduced in Chapter 18. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 x
SAMPLE ASSIGNMENT SCHEDULE MACHINE/MECHANICAL DESIGN (3 credits.) 
 
TEXT: A. C. Ugural, Mechanical Design of Machine Components, CRC Press (T & F Group). 
Prerequisites: Courses on Mechanics of Materials and Engineering Materials. 
 
 
WEEK TOPICS SECTIONS PROBLEMS 
 
 1 Introduction 1.1 to 1.12 1.6,1.17,1.26,1.34 
 Materials* 2.1 to 2.5,2.8 to 2.11 2.7,2.11,2.15,2.20 
 
 2 A Review of Stress Analysis* 3.1 to 3.14, 4.1 to 4.9 3.12,3.34,3.45,4.24,4.37 
 5.1 to 5.12 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 Miscellaeneous Mechanical 
 Components* 16.1 to 16.5, 16.816.6,16.17,16.25,16.38 
 to 16.11 
 
15 FEA in Design and 
 Case Studies in Machine Design* 17.1 to 17.3,17.5,17.7 17.2,17.4,17.8,17.12 
 18.1 and 18.2 18.2,18.4,18.6,18.10 
 FINAL EXAM - - - - - - - - - - 
 
 * Secondary topics. The remaining, major, topics constitute the “main stream” of the machine design 
 course. 
 1
Section I BASICS 
 
CHAPTER 1 INTRODUCTION 
 
 
SOLUTION (1.1) 
 
 Free Body: Angle Bracket (Fig. S1.1) 
 ( a ) 0; (20) 15(10) 4.8(20) 0, 12.3BM F F kN= − − = = ←∑ 
 
 ( b ) 0 : 6.4 0, 5.9x Bx BxF R F R kN= + − = = →∑ 
 0 : 15 4.8 0, 19.8y By ByF R R kN= + − = = ↑∑ 
 
 Thus 
 2 2(5.9) (19.8) 20.7BR kN= + = 
 and 
 1 5.920.7tan 15.9
oα −= = 
 
 
 
 
 
 
 
 
 
 
 
 
 
SOLUTION (1.2) 
 Free Body: Beam ADE (Fig. S1.2) 
 0 : (3.75 ) (2 ) 0,A BDM W F aα= − + =∑ 
 1.875BDF W= ↑ 
 0 : 0x AxF R= =∑ 
 0 : 0,y Ay BDF R F W= − + − =∑ 
 0.875AyR W= ↓ 
 Free-Body: Entire structure (Fig. S1.2) 
 0 : (3 ) (3.75 ) 0,CAM R a W a= − =∑ 
 1.25CR W= ↑ 
 Free Body: Part AO (Fig. S1.2) 
 0.875V W= ↑ 
 0F = 
 0.875M aW= 
BDF 
M” 
A 
0.875 kN 
D 
A 
V 
F 
O 
E 
W 
Figure S1.2 
2a 1.75a 
a 
AxR 
AyR 
α 
B 
19.8 
5.9 
BR 
Figure S1.1 
15 kN 
C 
A F 
6.4 kN 
10 in. 
4.8 kN 
BxR 
10 in. 
20 in. 
B 
 2
 
SOLUTION (1.3) 
 
 
 M R kipsA B= = ↓∑ 0 2 5: . 
 F R kipsy A= = ↑∑ 0 10 5: . 
 
 
 
 Segment CD 
 
 M kip ftD = = ⋅
1
2
22 2 4( )( ) 
 V kipsD = 4 
 
 
 Segment CE 
 
 M kip ftE = − = ⋅8 6 105 4 6( ) . ( ) 
 V kipsE = 25. 
 
 
 
 
SOLUTION (1.4) 
 
 ( a ) M R RB C C∑ = − − =0 08 6 0 6 2 24 4 0: . ( ) . ( ) ( ) 
 
 ∴ =R kNC 26 667. 
 R kN R kNCx Cy= =16 21334, . 
 
 Then 
 F R kNx Bx∑ = =0 16: 
 F R kNy By∑ = =0 12 66: . 
 
 
 ( b ) Segment CD 
 
 M kN mD = − − = ⋅21334 3 12 15 6 2 34. ( ) ( . ) ( ) 
 F kND = 16 
 V kND = − =21334 18 3334. . 
 
 
 
 
 
 
 
C 
2’ 
4’ 4’ E 
kips8 
105. 
A 
VE 
M E 
FD 
M D 4 
8 
C 
16 
3 m D VD 
21.334 
C B A E D 
RB RA 
4’ 
2’ 
8’ 8’ 
24 kip ft⋅ 2 kip/ft 
C 
D 2’ 
M D 
VD 
2 kip/ft 
1 m 
2m 
3m D 
RBy 
A 
RC 
8 kN m 
10 kN 
RBx 4 
3 
B 2 m 
2 m 
C 
 3
SOLUTION (1.5) 
 
 ( a ) 
 M R RA Cx Cy∑ = =0 32: 
 
 
 
 
 
 ↑==−=∑ kNRRM CyCyB 5004)5(40:0 
 Then R kNCx = ←75 
 ←=↑= kNRkNR BxBy 75,10 
 
 ( b ) 
 
 F kND = + =75 50 85
3
5
4
5( ) ( ) 
 V kND = − =75 50 3045 35( ) ( ) 
 mkNM D ⋅=−= 5.37)75.0(50)1(75 
 
 
SOLUTION (1.6) 
 
 ( a ) Free body entire connection 
 M R TC A= − =∑ 0 28 0: ( ) 
 T RA= 28 
 
 
 Segment AB AB in= + =20 6 20882 2 . . 
 M RB A= − =∑ 0 4 6 20 0: ( ) ( ) 
 R kipsA = 12. 
 and T kip in= ⋅336. . 
 
 ( b ) F F F kipsx AB AB= − = =∑ 0 4 0 41762020 88: , .. 
 
SOLUTION (1.7) 
 Free Body: Entire Crankshaft (Fig. S1.7a) 
 ( a ) From symmetry: A BR R= 
 0 : 2z A BF R R kN= = =∑ ր 
 0 : 4(0.05) 0,xM T= − + =∑ 
 0.2 200T kN m N m= ⋅ = ⋅ 
 
 
 
 (CONT.) 
RA 
FAB 
20” 
6” 
4kip 
B 
A 
RA 
P A 
B 
T 6” 
20” 8” 
C 
FD M D 
VD D 
4 
3 50 
75 
0.75 
1.0 
C 
5 
RBy 
RBx B 
RCx 40 kN 
C 
4 
3 
RCy 
A 
C 
RCx 
3 m 
2 m 2 
3 
4 m 1 
m 
CyR 
kNRB 2= 
B 
4 kN 
C 50 mm 
A 
D 
120 mm 
120 mm 
70 mm 
2AR kN= 
z 
y 
T 
x 
(a) 
 4
 
1.7 (CONT.) 
 
 
 ( b ) Cross Section at D (Fig. S1.7b) 
 2zV kN= ր 
 200T N m= ⋅ 
 2(0.07) 0.14yM kN m= = ⋅ 
 140 N m= ⋅ 
 
SOLUTION (1.8) 
 
 
 
 Free-Body Diagram, Beam AB 
 7
65
0 : 60 0, 69.11x CD CDF F F kN= − + = =∑ 
 4
65
0 : 30 0, 64.3y A CD AF R F R kN= − − = = ↑∑ 
 7
65
0 : 60(1.8) (3) 0,A CD AM F M= − + − =∑ 
 72AM kN m= ⋅ 
 
 
 
 
 
 
 
SOLUTION (1.9) 
 
 Free body entire frame 
 M R RA Dy Dy= − − + =∑ 0 30 3 4 10 012: ( ) ( ) ( ) 
 R kips R kipsDy Dx= =1125 5625. , . 
 
 
 
 
 
 
Free body BCD 
 F R kipsx Bx= =∑ 0 5625: . 
 F R kipsy By= =∑ 0 1125: . 
 R kipsB = + =5625 1125 12 58
2 2. . . 
 
 
 
 
 
B 
8’ 
4’ 
C 
11.25 
5.625 D 
RBx 
RBy 
A 
B C 
D 
30 kips 
3’ 3’ 4’ 
1 
2 
4’ 
8’ 
RDy 
RD 
1
2 RDy 
CDF
A 
30 kN 
C 
60 kN 
B 
4 
7 
AR
AM
1.8 m 
1.8 m 
1.2 m 
Figure S 1.7 
zV 
yM 
T 
D 
(b) 
 5
SOLUTION (1.10) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ( a ) 
M Tx∑ = − + − + =0 3 015 4 015 5 015 2 015 0: ( . ) ( . ) ( . ) ( . ) 
 or T kN m= ⋅0 6. 
 ( b ) M R R kNz Ey Ey∑ = + + = = −0 4 3 1 25 0 28: ( )( ) ( . ) , . 
 M R R kNy Ez Ez= − + + = =∑ 0 25 5 2 3 0 8 4: ( . ) ( )( ) , . 
 F R R R kNy Ay Ey Ay∑ = + + + = = −0 4 3 0 4 2: , . 
 F R R R kNz Az Ez Az∑ = + − − = = −0 5 2 0 14: , . 
 Thus R kNA = + =4 2 14 4 427
2 2. . . 
 R kNE = + =28 8 4 8 854
2 2. . . 
 
SOLUTION (1.11) 
 (a) Free-body Diagrams, Arm BC and shaft AB 
 
 
 
 
 
 
 
 
 
 
 
 
 (b) At C: 
 2 50V kN T N m= − = − ⋅ 
 At end B of arm BC: 
 2 50 200V kN T N m M N m= = ⋅ = ⋅ 
 At end B of shaft AB: 
 2 200 50V kN T N m M N m= − = − ⋅ = − ⋅ 
 At A: 2 200 300V kN T N m M N m= = ⋅ = ⋅ 
 
C 
D 
B 
E 
A 
z 
RAz 
RAy 
y 
x 
T REy 
REz 
5 kN 
2 kN 
4 kN 3 kN 
0.3 m 
1 m 
1 m 
0.5 m 
0.5 m 
0.3 m 
B 
C 
A 
V 
T 
z 
y 
x 
V 
V 
M 
M 
T 
T 
M 
T 
B Figure S1.11 
V 
 6
SOLUTION (1.12) 
 
Free Body: Entire Pipe 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Reactional forces at point A: 
0 : 0x xF R= =∑ 
 0 : 200 0, 200y y yF R R N= − = =∑ 
 0 : 0z zF R= =∑ 
 
Moments about point A: 
0 : 200(0.15) 0, 30xM T T N m= − = = ⋅∑ 
 0 : 0y yM M= =∑ 
 0 : 200(0.3) 36 0, 96z z zM M M N m= − − = = ⋅∑ 
 
The reactions act in the directions shown on the free-body diagram. 
 
 
 
SOLUTION (1.13) 
 
 Free Body : Entire Pipe 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 (CONT.) 
T 
z 
Rz 
Ry 
x 
Mz 
Rx 
D 
B 
My 
A 
C 
y 
0.15 m 
0.3 m 
0.2 m 
200 N 
36 N⋅ m 
T 
z 
Rz 
Ry 
x 
Mz 
Rx 
D 
B 
My 
A 
200 N 
C 36 N⋅ m 
y 
0.15 m 
0.3 m 
0.2 m 
WCD 
WAB 
WBC 
 
0.075 m 
0.15 m 
 7
1.13 (CONT.) 
 
 We have 1 lb/ft=14.5939 N/m (Table A.1). 
 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: 
 110.62(0.3) 33.2ABW N= = 
 110.62(0.2) 22.1BCW lb= = 
 110.62(0.15) 16.6CDW N= = 
 
Reactional forces at point A: 
0 :xF =∑ 0xR = 
 0 : 200 0, 271.9y y AB BC CD yF R W W W R N= − − − − = =∑ 
 0 :zF =∑ 0zR = 
 
Moments about point A: 
0 : (0.075) 10(0.15) 0, 2.745x CDM T W T N m= − − = = ⋅∑ 
0 :yM =∑ 0yM = 
 0 : (200 )(0.3) (0.15) 36 0z z CD BC ABM M W W W= − + + − − =∑ 
 112.6 .zM N m= ⋅ 
 
SOLUTION (1.14) 
 
 ( a ) 
 
 
 
 
 
 
 
 Free body pulley B 
 F B kNx x= = →∑ 0 16: . 
 F B kNy y= = ↑∑ 0 16: . 
 Free body CED 
 M R R kND Cx Cx= − = = ←∑ 0 0 4 16 015 0 0 6: ( . ) . ( . ) , . 
 F D D kNx x x= − − + = = ←∑ 0 0 6 16 0 1: . . , 
 F R Dy Cy y= =∑ 0: 
 Free body ADB 
 M D B D kN R kNA y y y Cy= − = = = ↑∑ 0 05 15 0 4 8 48: ( . ) ( . ) , . , . 
 F R D B R kNx Ay y y Ay= − + − = = ↓∑ 0 0 32: , . 
 
 (CONT.) 
 
A 
RAy 
RAx 
Dy 
Dx D B 
By 
Bx 
1 m 0.5m 
C 
E 
D 
Dx Dy 
0.25 m 
0.15 m 
RCy 
RCx 
1.6 kN 
Bx 
By 
B 
1.6 kN 150 mm 
1.6 kN 
 8
1.14 (CONT.) 
 
 F R D B R kNx Ax x x Ax= + − = = →∑ 0 0 0 6: , . 
 
 
 ( b ) M N m V kNG G= = ⋅ =16 0 6 960 16. ( . ) , . 
 F kNG = 16. 
 
 
SOLUTION (1.15) 
 
 Free body entire rod 
 M R R Nx Dy Dy= − = ↑∑ 0 0 25 300 01 120: ( . ) ( . ), 
 ( )M Rz C By∑ = − + + − =0 200 035 0 25 300 120 0 2 0: ( . ) ( . ) ( )( . ) 
 R NBy = ↑136 
 Free body ABE 
 
 
 
 
 
 
 
 ( )M M M N mz E z z∑ = − + − = = ⋅0 200 0 275 136 0175 0 312: ( . ) ( . ) , . 
 F V Ny y= = − =∑ 0 200 136 64: 
 
 
SOLUTION (1.16) 
 
 Free body entire rod: 
 ↑==−=∑ NRRM DyDyx 160,0)1.0(400)25.0(:0 
 ( ) ↓==−+=∑ 192,0)2.0)(160400()25.0(:0 ByByCz RRM 
 
 Segment ABE 
 
 
 
 
 
 
 
 At point E: 
 M N mz = − = − ⋅192 0175 336( . ) . 
 V Ny = 192 
 
 
 
B 
1.6 
1.6 
G 
VG 0.6 m MG 
FG 
E 
y 
z 
x 
M z Vy 
B A 
200 N 
136 N 
100 
175 
A B 
y 
x 
z 
E 192 N 
M z 
Vy 
0.175 m 
 9
 
SOLUTION (1.17) 
 
 Side view Top view 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Fig. (b): M F F kNA∑ = − = =0 0 05 150 0 31 1: ( . ) , 
 Fig. (a): M F F F kNB∑ = − = =0 01 0 05 0 61 2 2: ( . ) ( . ) , 
 Fig. (c): mkNTTFM ddD ⋅==−=∑ 3.0,0)05.0(:0 2 
 
 
 
 
SOLUTION (1.18) 
 
 
 
 
 
 
 
 
 
 
 
 
 Free body-entire frame 
 M R R kipsA y y∑ = − − = =0 30 3 18 4 12 0 34: ( ) ( ) ( ) , . 
 
 Free body-member BC 
 M R RC x y∑ = − =0 9 12 0: ( ) ( ) 
 and 
R kipsx = =
4
3 34 4 533( . ) . 
Thus 
 F R kipsBC = = + =( . ) ( . ) .4533 34 5666
2 2 
 
 
B 
C 
50 mm 
100 mm 
D 
50 mm 
A 
50 mm 
F2 
F2 
F1 
F1 
Td 
N m⋅ 150 
Figure (a) Figure (c) 
Figure (b) 
C 
B 
A 
4 
3 
Rx 
Ry 
R 
4 kips 3 kips 
12 ft 18 ft 
9 ft 
3ft 
 10
SOLUTION (1.19) 
 
 Free body-member AB 
 M A∑ = 0: 
 
R a P a P aE
o o( ) cos ( ) sin ( )4 40 40 6 0− − = 
 ∴ =R PE 1156. 
 
 F R P Px Ax
o= = = ←∑ 0 40 0 766: cos . 
 F R P P R Py Ay
o
Ay= + − = = ↓∑ 0 1156 40 0 0513: . sin , . 
 
 Free body-member CD 
 
 
 M R a R aD E Cy= − =∑ 0 4 6 0: ( ) ( ) 
 ∴ = ↑R PCy 0 771. 
 
 
 M R a R a R PC D
o
E D= − = =∑ 0 30 6 2 0 0 771: sin ( ) ( ) , . 
 F R R Px Cx D
o∑ = = = →0 30 0 668: cos . 
 
 
SOLUTION(1.20) 
 
 ( a ) Power=P=(pA)(L)(n/60) 
 150060(1.2)(2100)(0.06)( ) 3.78kW= = 
 Power required 3.780.9 4.2
P
e kW= = = 
 
 ( b ) Use Eq.(1.15), 
 
9549(4.2)9549
1500 26.74
kW
nT N m= = = ⋅ 
 
 
SOLUTION (1.21) 
 
 Refer to App. A.1: 65 mph=65(5280)/60=5720 fpm. 
 ( a ) From Eq. (1.17), the drag force equals, 
 
33,000 33,000(18)
5720 103.4
hp
d VF lb= = = 
 See: Fig. P1.21: 
 0 : 103.4xF F lb= =∑ 
 It follows that 
 0 : 0A d fM Wa F c R L= − + + =∑ 
 or 
 3.2(60) (0.1034)(25) (110) 0fR− + + = 
 (CONT.) 
C D 
E 
2a 4a 
RCx 
RCy 
RE 
RD 
30o 
P 
B 
a 40
o A 
RAx 
RAy RE 
4a 2a 
E 
 11
1.21 (CONT.) 
 
 Solving, 
 1.722fR kips= 
 and 
 0 : 3.2 1.722 0y rF R= − + =∑ 
 or 
 1.478rR kips= 
 
 ( b ) 
 We have 0, 0, 0.dV F F= = = 
 See Fig. P1.21: 
 0 : 0A fM Wa R L= + =∑ 
 Thus 
 60110 (3.2) 1.745
a
f LR W kips= = = 
 So, 0yF =∑ gives 
 3.2 1.745 1.455r fR W R kips= − = − = 
 
SOLUTION (1.22) 
 
 Refer to Solution of Prob. 1.21. Now we have 3.2 1.2 4.4tW kips= + = and friction force 
 103.4F lb= acts at point A. 
 
 ( a ) See: Fig. 1.21 (with tW W= ): 
 4.4(60) (0.1034)(25) (110) 0A fM R= − + + =∑ 
 from which 
 2.377fR kips= 
 and 
 4.4 2.377 2.023rR kips= − = 
 
 ( b ) 0, 0, 0,dV F F= = = as before, 
 601100 : (4.4) 2.4
a
A f tLM R W kips= = = =∑ 
 and 
 4.2 2.4 1.8r t fR W R kips= − = − = 
 
SOLUTION (1.23) 
 
 ( a ) Free-Body Diagram: Gears (Fig. S1.23). 
 Applying Eq. (1.15): 
 
9550(35)9550
500 668.5
P
AC nT N m= = = ⋅ 
 Therefore, 
 668.50.125 5.348
A
A
T
rF kN= = = 
 (CONT.) 
 12
1.23 (CONT.) 
 
 ( b ) 5,348(0.075) 401.1DE DT Fr N m= = = ⋅ 
 
 
 
 
 
 
 
 
 
SOLUTION (1.24) 
 
 From Eq. (1.17), we obtain T=63,000 hp/n. Thus 
 
 For input shaft 
 .05.11800
)30(63000 inkipT ⋅== 
 For output shaft 
 .0.4425
)27(63000 inkipT ⋅== 
 Equation (1.14) gives 
 %909.030
27 ===e 
 
SOLUTION (1.25) 
 
 ( a ) Referring to Eq. (1.12): 
 power output )(5.2500)( 60
150
60 ×== NFs 
 sinlb .125,3 ⋅= 
 
 ( b ) Using Eq. (1.14), power transmitted by the shaft: 
 power input sinlb .551,3)88.0(125,3 ⋅== 
 
SOLUTION (1.26) 
 
 Equation (1.10) becomes 
 )( 2min
2
max2
1 ωω −=∆ IEk (1) 
 Here, mass moment inertia with 5 percent added: 
 ρπ lddI io ⋅−= )()05.1(
44
32 (Table A-5) 
 )200,7)(1.0)(3.04.0(05.1 4432 −= π 
 2299.1 mkg ⋅= 
 sradrps 7.12520)(1200 60
1
max ===ω 
 sradrps 1153.18)(1100 60
1
min ===ω 
 
 Equation (1) is therefore 
 )1157.125)(299.1( 222
1 −=∆ kE 
 J673,1= 
DET AC
T 
Dr 
Figure S1.23 
D 
F 
F A 
Ar 
 13
 
SOLUTION (1.27) 
 
 Final length of the wire: 
 2 2' (80) (50.4) 94.55242 .ACL in= + = 
 Initial length of the wire is 
 2 2(80) (50) 94.33981 .ACL in= + = 
 Hence, Eq. (1.20): 
 ' 94.55242 94.3398194.33981
AC AC
AC
L L
AC Lε
− −= = 
 0.00225 2250µ= = 
 
 
SOLUTION (1.28) 
 
 ( a ) r
r
r
rrr
c
∆−∆+ == π
ππε 2
2)(2
 
 ( ) .ε µc i = =0 3150 2000 
 ( ) .ε µc o = =0 2250 800 
 
 ( b ) ε µr
r r
r r
o i
o i
= = =−− −−
∆ ∆ 0 3 0 2
250 150 1000
. . 
 
 
SOLUTION (1.29) 
 
, 2 1.41421OB AB BCL d L L d d= = = = 
 ( a ) µε 12000012.0 == d dOB 
 
( b ) [ ] dddLL CBAB 41506.1)0012.1( 2122'' =+== 
 µεε 60141421.1 41421.141506.1 === −BCAB 
 
( c ) ( )1 1.0012tan 45.0344oddCAB −= = 
Increase in angle CAB is 45.0344 45 0.0344o− = . 
 Thus 
 ( )1800.0344 600πγ µ= = 
 
 
SOLUTION (1.30) 
 
( a ) ε µx = =−0 8 0 5250 1200. . µε 200020004.0 −== −−y 
 
( b ) )1(' xADADxADAD LLLL εε +=+= 
 = =250 10012 250 3( . ) . mm 
 
 
 14
SOLUTION (1.31) 
 
 ∆L mmAB = =
− −800 10 150 120 106 3( ) ( ) 
 ∆L mmAD = =
− −1000 10 200 200 106 3( ) ( ) 
 
 We have 
 L L LBD AB AD
2 2 2= + 
 2 2 2L L L L L LBD BD AB AB AD AD∆ ∆ ∆= + 
 or 
∆ ∆ ∆L L LBD
L
L AB
L
L AD
AB
BD
AD
BD
= + (1) 
 [ ]= + =−150250 200250 3120 200 10 0 232( ) ( ) . mm 
 
SOLUTION (1.32) 
 
 AC BD mm= = + =300 300 424 262 2 . 
 mmCAmmDB 46.4242.026.424'',76.4235.026.424'' =+==−= 
 Geometry: '''' DABA = 
 
 AD
ADDA
yx
−== ''εε 
 µ363300
300
2
46.424
2
76.423 2
1
22
−==
−













+





 
 γ β µπ πxy = − = − =
−
2 2
1 423 76 2
424 46 22 1651tan
.
. 
 
 
SOLUTION (1.33) 
 
 We have 
 AD AD ADLδ ε= 
 6800 10 (32)−= × 
 0.026 .in= 
 
 
 
 
 
 From triangles 'A AF and 'C CF : 
 0.026 0.260 , 6.9 .x x x in−= = 
 From triangles 'B BF and 'C CF : 
 13.1 53.10.2 , 0.049 .B B BEinδ δ δ= = = − (contraction) 
 Therefore 
 60.04940 1225(10 )
BE
BEBE L
δε − −= = = − 
 1,225µ= − 
 
'B 
'D 
'C 
'A 
Bδ 
F C 
C' 
x 
A 
0.026 
in. 
13.1 in. 
B' 
B 
A' 
40 in. 20 in. 
0.2 in. 
 15
 
SOLUTION (1.34) 
 
 ( a ) ε µx = =0 00650 120. ε µy = = −−0 00425 160. 
 γ µxy = − + = −1000 200 800 
 
 ( b ) mmLL yABAB 996.24)00016.01(25)1(' =−=+= ε 
 mmLL xADAD 006.50)00012.01(50)1(' =+=+= ε 
 
 
 End of Chapter 1 
 
 
 
 16 
CHAPTER 2 MATERIALS 
 
 
SOLUTION (2.1) 
 
 A in A inf0 4
2 3 2
4
2 3 205 196 35 10 05 0 00024 19616 10= = = − =− −π π( . ) . ( ) . , ( . . ) . ( ) . 
 We have ε µ ε µa t= = = =
− −12 10
8
0 24 10
0 5
3 3
1500 480( ) . ( )., 
 Thus 
S ksip
P
A= = =−0
3
3
4 10
196 35(10
20 37( )
. )
. 
 E psi
S p
a
t
a
= = = = =−εε
εν
20 37 10
1500 10
6
3
6 1358 10 0 32
. ( )
( )
. ( ) , . 
 Also 
 
 
 % ( ) .. ..reduction in area = =−196 35 196 16196 35 100 0 097 
 
 
SOLUTION (2.2) 
 
 Normal stress is 
 
2
500
(1 8)
4
4744PA ksiπσ = = = 
 This is below the yield strength of 50 ksi (Table B.1). 
 We have 
 0.318.5 12 0.001351 1351L
δε µ×= = = = 
 Hence 
 6
640,744
135(10 )
30 10E psiσε −= = = × 
 
 
SOLUTION (2.3) 
 
 The cross-sectional area: 20.5(0.25) 0.125 .o oA w t in= = = 
 ( a ) Axial strain and axial stress are 
 0.003312.5 0.01324 1324aε µ= = = 
 4.80.125 38.4
P
a A ksiσ = = = 
 Because a ySσ < (See Table B.1), Hooke's Law is valid. 
 ( b ) Modulus of elasticity, 
 6
638,400
1324(10 )
29 10a
a
E psiσε −= = = × 
 
 ( c ) Decrease in the width and thickness 
 0.3(0.5) 0.15 .ow w inν∆ = = = 
 0.3(0.24) 0.072 .ot t inν∆ = = = 
 
 
15.0)100(% 8
)10(12 3 ==
−
elongation
 17 
 
SOLUTION (2.4) 
 
 Assume Hooke's Law applies. We have 
 1.55 300tε µ= − = − 
 3000.34 822
t
a
ε
νε µ−= − = − = 
 Thus, 
 9 9(105 10 )(822 10 ) 92.61aE MPaσ ε
−= = × × = 
 Since ySσ < , our assumption is valid. 
 So 
 2(92.61)( 4)(5) 1.818P A kNσ π= = = 
 
 
 
SOLUTION (2.5) 
 
 We obtain 
 L L mmAC BD= = + =15 15 2121
2 2 . 
 
 ε µx
L
L
AC
AC
= = = −−∆ 2117 21 2121 21 1886. .. 
 ε µy
L
L
BD
BD
= = =−∆ 21 22 21 2121 21 471. .. 
 
 ( a ) E GPax
x
= = =− −
σ
ε
100
1886
6
6 53
(10 )
(10 )
 
 ( b ) ν εε= = =
y
x
471
1886 0 25. 
 ( c ) G GPa= =+532 0 25 212(1 . ) . 
 
 
SOLUTION (2.6) 
 
 Use generalized Hooke’s law: 
 ε ε ε σ σ σνx y z E x y z+ + = + +−1 2 ( ) (1) 
 For a constant triaxial state of stress: 
 ε ε ε ε σ σ σ σx y z x y z= = = = = =, 
 
 Then, Eq. (1) becomes ε σν= −1 2E . Since σ and ε must have identical signs: 
 1 2 0− ≥ν or ν = 12 
 
 
 
SOLUTION (2.7) 
 
 We have σ x ksi= =
100 10
3 2
3
16 67( )( ) . 
 
 (CONT.)