SOLUTIONS MANUAL FOR

by

MECHANICAL DESIGN

OF MACHINE COMPONENTS

SECOND EDITION: SI VERSION

ANSEL C. UGURAL

SOLUTIONS MANUAL FOR

by

Boca Raton London New York

CRC Press is an imprint of theTaylor & Francis Group, an informa business

MECHANICAL DESIGNOF MACHINE

COMPONENTSSECOND EDITION: SI VERSION

ANSEL C. UGURAL

MATLAB® is a trademark of The MathWorks, Inc. and is used with permission. The MathWorks does not warrant the accuracy of the text or exercises in this book. This book’s use or discussion of MATLAB® software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular use of the MATLAB® software.

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vi

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

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.

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 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.

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 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.

x

SAMPLE ASSIGNMENT SCHEDULE MACHINE/MECHANICAL DESIGN (3 credits.)

TEXT: A. C. Ugural, MECHANICAL DESIGN of Machine Components, 2nd

SI Version, 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 Miscellaneous Machine

Components* 16.1 to 16.5, 16.8 16.6,16.17,16.25,16.38

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; (0 .5 ) 6 8(0 .2 5) 2 8 .8 (0 .5 ) 0 , 6 2 .8 k NB

M F F

( b ) 0 : 2 8 .8 0 , 3 4 k Nx B x B x

F R F R

0 : 6 8 2 1 .6 0 , 8 9 .6 k Ny B y B y

F R R

Thus

2 2

(3 4 ) (8 9 .6 ) 9 5 .8 k NB

R

and

1 3 4

8 9 .6ta n 2 0 .8

o

SOLUTION (1.2)

Free Body: Beam ADE (Fig. S1.2)

0 : (3 .7 5 ) (2 ) 0 ,A B D

M W a F a

1 .8 7 5B D

F W

0 : 0x A x

F R

0 : 0 ,y A y B D

F R F W

0 .8 7 5A y

R W

Free-Body: Entire structure (Fig. S1.2)

0 : (3 ) (3 .7 5 ) 0 ,CA

M R a W a

1 .2 5C

R W

Free Body: Part AO (Fig. S1.2)

0 .8 7 5V W

0F

0 .8 7 5M a W

B DF

M”

A

0.875 kN

D

A

V F

O

E

W

Figure S1.2

2a 1.75a

a

A xR

A yR

B

89.6

34

BR

Figure S1.1

68 kN

C

A F

28,8 kN

0.25

21.6 kN

B xR

0.25

0.5

B

B yR

2

SOLUTION (1.3)

0 : 1 1 .0 2 k NA B

M R

0 : 4 5 .8 2 k Ny A

F R

Segment CD

21

2( 2 9 )(0 .6 ) 5 .2 2 k N m

DM

1 7 .4 k ND

V

Segment CE

3 4 .8(1 .8) 4 5 .8 2 (1 .2 ) 7 .6 5 6 k N mE

M

1 1 .0 2 k NE

V

SOLUTION (1.4)

( a ) M R RB C C 0 0 8 6 0 6 2 2 4 4 0: . ( ) . ( ) ( )

kN667.26C

R

kN334.21,kN16 CyCx

RR

Then

kN16:0 BxxRF

kN66.12:0 ByyRF

( b ) Segment CD

mkN34)2(6)5.1(12)3(334.21 D

M

kN16D

F

kN334.318334.21 D

V

C 0.6 m

1.2 m 1.2 m E

3 4 .8 k N

82.45

A

VE

ME

FD

MD

4

8

C

16

3 m D V

D

21.334

C B A E D

RB

RA

1,2 m

0.6 m

2.4 m 2.4 m

3 2 k N m 29 kN/m

C D 0.6 m

MD

VD

29 kN/m

1 m 2m

3m D

RB y

A

RC

mkN8

kN10

RB x

4

3

B 2 m

2 m

C

3

SOLUTION (1.5)

( a )

M R RA C x C y 0

3

2:

0 : 4 0 (5) 4 0 5 0 k NB C y C y

M R R

Then kN75Cx

R

1 0 k N , 7 5 k NB y B x

R R

( b )

kN85)(50)(755

4

5

3

DF

kN30)(50)(755

3

5

4

DV

75(1) 50 (0 .75) 37 .5 kN mD

M

SOLUTION (1.6)

( a ) Free body entire connection

0 : (0 .7 ) 0C A

M R T

A

RT 7.0

Segment AB 2 2

(0 .5 ) (0 .1 5 ) 0 .5 2 2A B m

0)5.0()15.0(18:0 ABRM

5 .4 k NA

R

and 3 .7 8 k N mT

( b ) 0 .5

0 .5 2 20 : 1 8 0 , 1 8 .7 2 9 k N

x A B A BF F F

SOLUTION (1.7)

Free Body: Entire Crankshaft (Fig. S1.7a)

( a ) From symmetry: A B

R R

0 : 2 k Nz A B

F R R

0 : 4 (0 .0 5) 0 ,x

M T

0 .2 k N m 2 0 0 N mT

(CONT.)

RA

FA B

0.5 m

0.15 m 18 kN

B

A

RA

P A

B

T 0.15 m

0,5 m 0.2 m

C

FD

MD

VD

D

4 3 50

75

0.75

1.0

C 5

RB y

RB x

B R

C x 40 kN

C

4 3

RC y

A

C

RC x

3 m

2 m 2 3

4 m 1

m

CyR

2 k NB

R

B

4 kN C 50 mm

A D

120 mm 120 mm

70 mm

2 k NA

R

z

y

T

x

(a)

4

1.7 (CONT.)

( b ) Cross Section at D (Fig. S1.7b)

2 k Nz

V

2 0 0 N mT

2 (0 .0 7 ) 0 .1 4 k N my

M

1 4 0 N m

SOLUTION (1.8)

Free-Body Diagram, Beam AB

7

6 50 : 6 0 0 , 6 9 .1 1 k N

x C D C DF F F

4

6 50 : 3 0 0 , 6 4 .3 k N

y A C D AF R F R

7

6 50 : 6 0 (1 .8 ) (3 ) 0 ,

A C D AM F M

7 2 k N mA

M

SOLUTION (1.9)

Free body entire frame

0)3()2.1()9.0(6.129:02

1 DyDyA

RRM

4 8 .6 k N , 2 4 .3 k ND y D x

R R

Free body BCD

0 : 2 4 .3 k Nx B x

F R

0 : 4 8 .6 k Ny B y

F R

2 2

2 4 .3 4 6 .6 5 2 .6 k NB

R

B

2.4 m

1.2 m

C

48.6

24.3 D

RB x

RB y

A

B C

D

129.6 kN

0.9 m 0.9 m 1.2 m

1

2

1.2 m

2.4 m

RD y

RD

1

2 RD y

C DF

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 0 1 5 4 0 1 5 5 0 1 5 2 0 1 5 0: ( . ) ( . ) ( . ) ( . )

or mkN6.0 T

( b ) kN8.2,0)5.2()1)(34(:0 EyEyzRRM

kN4.8,0)3)(25()5.2(:0 EzEzyRRM

kN2.4,034:0 AyEyAyyRRRF

kN4.1,025:0 AzEzAzzRRRF

Thus kN427.44.12.422

AR

kN854.84.88.222

ER

SOLUTION (1.11)

(a) Free-body Diagrams, Arm BC and shaft AB

(b) At C:

2 k N 5 0 N mV T

At end B of arm BC:

2 k N 5 0 N m 2 0 0 N mV T M

At end B of shaft AB:

2 k N 2 0 0 N m 5 0 N mV T M

At A: 2 k N 2 0 0 N m 3 0 0 N mV T M

C

D

B

E

A

z

RA z

RA y

y

x

T R

E y

RE z

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 x

F R

0 : 2 0 0 0 , 2 0 0 Ny y y

F R R

0 : 0z z

F R

Moments about point A:

0 : 2 0 0 (0 .1 5) 0 , 3 0 N mx

M T T

0 : 0y y

M M

0 : 2 0 0 (0 .3) 3 6 0 , 9 6 N mz z z

M M 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.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:

1 1 0 .6 2 (0 .3) 3 3 .2 NA B

W

1 1 0 .6 2 (0 .2 ) 2 2 .1 NB C

W

1 1 0 .6 2 (0 .1 5) 1 6 .6 NC D

W

Reactional forces at point A:

0 :x

F 0x

R

0 : 2 0 0 0 , 2 7 1 .9 Ny y A B B C C D y

F R W W W R

0 :z

F 0z

R

Moments about point A:

0 : (0 .0 7 5) 1 0 (0 .1 5) 0 , 2 .7 4 5 N mx C D

M T W T

0 :y

M 0y

M

0 : (2 0 0 )(0 .3) (0 .1 5) 3 6 0z z C D B C A B

M M W W W

1 1 2 .6 N m .z

M

SOLUTION (1.14)

( a )

Free body pulley B

kN6.1:0xx

BF

kN6.1:0yy

BF

Free body CED

kN6.0,0)15.0(6.1)4.0(:0CxCxD

RRM

kN1,06.16.0:0xxx

DDF

F R Dy C y y 0:

Free body ADB

kN8.4,kN8.4,0)5.1()5.0(:0CyyyyA

RDBDM

kN2.3,0:0AyyyAyx

RBDRF

(CONT.)

A

RA y

RA x

Dy

Dx

D B

By

Bx

1 m 0.5 m C

E

D

Dx

D

y

0.25 m

0.15 m

RC y

RC x

1.6 kN

Bx

By

B

1.6 kN 150 mm

1.6 kN

8

1.14 (CONT.)

kN6.0,0:0AxxxAxx

RBDRF

( b ) kN6.1,mN960)6.0(6.1 GG

VM

kN6.1G

F

SOLUTION (1.15)

Free body entire rod

N120),1.0(300)25.0(:0DyDyx

RRM

M Rz

CB y 0 2 0 0 0 3 5 0 2 5 3 0 0 1 2 0 0 2 0: ( . ) ( . ) ( ) ( . )

N136By

R

Free body ABE

mN2.31,0)175.0(136)275.0(200:0 zzEzMMM

N64136200:0 yyVF

SOLUTION (1.16)

Free body entire rod:

0 : (0 .2 5) 4 0 0 (0 .1) 0 , 1 6 0 Nx D y D y

M R R

0 : (0 .2 5 ) ( 4 0 0 1 6 0 )(0 .2 ) 0 , 1 9 2 Nz B y B y

C

M R R

Segment ABE

At point E:

mN6.33)175.0(192 z

M

N192y

V

B 1.6

1.6 G

VG

0.6 m MG

FG

E

y

z

x

Mz

Vy

B A

200 N

136 N

100 175

A B y

x

z

E 192 N

Mz

Vy

0.175 m

9

SOLUTION (1.17)

Side view Top view

Fig. (b): kN3,0150)05.0(:011 FFM

A

Fig. (a): kN6,0)05.0()1.0(:0221 FFFM

B

Fig. (c): 2

0 : (0 .0 5) 0 , 0 .3 k N mD d d

M F T T

SOLUTION (1.18)

Free body-entire frame

0 : (1 0 ) 1 3 .5 (6 ) 1 8(4 ) 0 , 1 5 .3 k NA y y

M R R

Free body-member BC

0)4()3(:0 yxCRRM

and 4

3(1 5 .3) 2 0 .4 k N

xR

Thus

2 2

( 2 0 .4 ) (1 5 .3) 2 5 .5 k NB C

F R

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

R

y

R

18 kN 13.5 kN

4 m 6 m

3 m

1 m

10

SOLUTION (1.19)

Free body-member AB

MA 0:

R a P a P aE

o o( ) co s ( ) s in ( )4 4 0 4 0 6 0

R PE

1 1 5 6.

F R P Px A x

o 0 4 0 0 7 6 6: co s .

F R P P R Py A y

o

A y 0 1 1 5 6 4 0 0 0 5 1 3: . s in , .

Free body-member CD

M R a R aD E C y 0 4 6 0: ( ) ( )

R PC y

0 7 7 1.

M R a R a R PC D

o

E D 0 3 0 6 2 0 0 7 7 1: s in ( ) ( ) , .

F R R Px C x D

o

0 3 0 0 6 6 8: co s .

SOLUTION(1.20)

( a ) Power= P = (p A )(L )(n /6 0 )

1 5 0 0

6 0(1 .2 )(2 1 0 0 )(0 .0 6 )( ) 3 .7 8 k W

Power required 3 .7 8

0 .94 .2 k W

P

e

( b ) Use Eq.(1.15),

9 5 4 9 ( 4 .2 )9 5 4 9

1 5 0 02 6 .7 4 N m

k W

nT

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,

1 0 0 0 1 0 0 0 (1 4 )

2 94 8 2 .8 N

kW

d VF

See: Fig. P1.21:

0 : 4 8 2 .8 Nx d

F F

It follows that

0 : 0A d f

M W a F c R L

or

1 4 4 0 0 (1 .5 ) ( 4 8 2 .8 )(0 .6 2 5 ) ( 2 .7 ) 0f

R

(CONT.)

C D

E

2a 4a R

C x

RC y

RE

RD

3 0

o

P

B

a 4 0o

A R

A x

RA y

RE

4a 2a

E

11

1.21 (CONT.)

Solving,

7 .8 8 8 k Nf

R

and

0 : 1 4 .4 7 .8 8 8 0y r

F R

or

6 .5 1 2 k Nr

R

( b )

We have 0 , 0 , 0 .d

V F F

See Fig. P1.21:

0 : 0A f

M W a R L

Thus

1 .5

2 .7(1 4 .4 ) 8 k N

a

f LR W

So, 0y

F gives

1 4 .4 8 6 .4 k Nr f

R W R

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

1 4 .4 5 .4 1 9 .8 k Nt

W

( a ) From Eq. (1.15), the drag force equals,

1 0 0 0 1 0 0 0 (1 4 )

2 94 8 2 .8 N

kW

d VF

See: Fig. 1.21 (with t

W W ):

0 : 0A t d f

M W a F c R L

1 9 8 0 0 (1 .5 ) ( 4 8 2 .8 )(0 .6 2 5 ) ( 2 .7 ) 0f

R

from which

1 0 .8 8 8 k Nf

R

and 0 : 1 9 .8 1 0 .8 8 8 0y r

F R

8 .9 1 2 k Nr

R

( b ) 0, 0 , 0 ,d

V F F as before,

0 : 0A t f

M W a R L

Then 1 .5

2 .7(1 9 .8 ) 1 1 k N

a

f LR W

So, 0y

F gives

1 9 .8 1 1 8 .8 k Nr t f

R W R

12

SOLUTION (1.23)

( a ) Free-Body Diagram: Gears (Fig. S1.23).

Applying Eq. (1.15):

9 5 5 0 ( 3 5 )9 5 5 0

5 0 06 6 8 .5 N m

P

A C nT

Therefore,

6 6 8 .5

0 .1 2 55 .3 4 8 k N

A

A

T

rF

( b ) 5, 3 4 8(0 .0 7 5) 4 0 1 .1 N mD E D

T F r

SOLUTION (1.24)

1 2

1 8 0 0 rp m , 4 2 5 rp m , 2 0n n kW

From Eq. (1.15), we obtain T = 9549 kW/n. Thus

For input shaft

9 5 4 9 ( 2 3 )

1 8 0 01 2 2 N mT

For output shaft

9 5 4 9 ( 2 0 )

4 2 54 4 9 .4 N mT

Equation (1.14) gives

2 0

2 31 0 0 8 7 %e

SOLUTION (1.25)

N=150, F=2.25 kN, s=62.5 mm, e=88%

( a ) Referring to Eq. (1.12):

power output 1 5 0

6 0 6 0( ) 2 2 5 0 0 .0 6 2 5( )

NF s

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

)(2

min

2

max2

1 IEk

(1)

Here, mass moment inertia with 5 percent added:

lddIio

)()05.1(44

32 (Table A.5)

(CONT.)

D ET A C

T

Dr

Figure S1.23

D

F

F A

Ar

13

1.26 (CONT.)

)200,7)(1.0)(3.04.0(05.144

32

2

299.1 mkg

1

m ax 6 01 2 0 0 ( ) 2 0 1 2 5 .7 rad srp s

1

m in 6 01100( ) 18 .3 115 rad srps

Equation (1) is therefore

)1157.125)(299.1(22

2

1 k

E

1, 6 7 3 J

SOLUTION (1.27)

Final length of the wire:

2 2

'( 2 ) (1 .2 6 ) 2 .3 6 3 8 m

A CL

Initial length of the wire is

2 2

( 2 ) (1 2 .5 ) 2 .3 5 8 5 mA C

L

Hence, Eq. (1.20):

' 2 .3 6 3 8 2 .3 5 8 5

2 .3 5 8 5

A C A C

A C

L L

A C L

0 .0 0 2 2 5 2 2 5 0 μ

SOLUTION (1.28)

( a ) r

r

r

rrr

c

2

2)(2

μ2000)(150

3.0

ic

μ800)(250

2.0

oc

( b ) μ1000150250

2.03.0

io

io

rr

rr

r

SOLUTION (1.29)

, 2 1 .4 1 4 2 1O B A B B C

L d L L d d

( a ) 0 .0 0 1 21 2 0 0 μ

d

O B d

( b ) dddLLCBAB

41506.1)0012.1(2

122

''

1 .4 1 5 0 6 1 .4 1 4 2 1

1 .4 1 4 2 1601 μ

A B B C

( c ) 1 1 .0 0 1 2

tan 4 5 .0 3 4 4od

dC A B

Increase in angle C A B is 4 5 .0 3 4 4 4 5 0 .0 3 4 4o

.

Thus

1 8 0

0 .0 3 4 4 6 0 0 μ

14

SOLUTION (1.30)

( a ) μ1200250

5.08.0

x 0 .4 0

2 0 02 0 0 0 μ

y

( b ) )1('xADADxADAD

LLLL

mm3.250)0012.1(250

SOLUTION (1.31)

mm)10(120150)10(80036

AB

L

mm)10(200200)10(100036

AD

L

We have

L L LB D A B A D

2 2 2

2 2 2L L L L L LB D B D A B A B A D A D

or

L L LB D

L

L A B

L

L A D

A B

B D

A D

B D (1)

mm232.010)200()120(3

250

200

250

150

SOLUTION (1.32)

mm26.42430030022 BDAC

' ' 4 2 4 .2 6 0 .5 4 2 3 .7 6 m m , ' ' 4 2 4 .2 6 0 .2 4 2 4 .4 6 m mB D A C

Geometry: '''' DABA

AD

ADDA

yx

''

12 2 2

4 2 3 .7 6 4 2 4 .4 63 0 0

2 2

3 0 03 6 3 μ

μ1651tan2246.424

276.4231

22

xy

SOLUTION (1.33)

We have

A D A D A D

L

6

8 0 0 1 0 (0 .8)

0 .0 0 0 6 4 m

(CONT.)

'B

'D

'C

'A

B

F C

C'

x

A

0.00064 m

0.33 m

B'

B

A' 1 m 0.5 m

0.005 m

15

1.33 (CONT.)

From triangles 'A A F and 'C C F :

0 .0 0 0 6 4 0 .0 0 5

1 .5, 0 .1 7 m

x xx

From triangles 'B B F and 'C C F :

0 .3 3 1 .3 3

0 .0 0 5, 0 .0 0 1 2 4 m

BB B E

(contraction)

Therefore

60 .0 0 1 2 4

11 2 4 0 (1 0 )

B E

B EB E L

1, 2 4 0

SOLUTION (1.34)

( a ) μ12050

006.0

x μ160

25

004.0

y

μ8002001000 xy

( b ) ' (1 ) 2 5 (1 0 .0 0 0 1 6 ) 2 4 .9 9 6 m mA B A B y

L L

' (1 ) 5 0 (1 0 .0 0 0 1 2 ) 5 0 .0 0 6 m mA D A D x

L L

End of Chapter 1

16

CHAPTER 2 MATERIALS

SOLUTION (2.1)

2 2 2 2

0 4 4(1 2 .5 ) 1 2 2 .7 m m , (1 2 .5 0 .0 0 6 ) 1 2 2 .6 m m

fA A

We have 0 .3 0 .0 0 6

2 0 0 1 2 .51 5 0 0 μ , 480 μ

a t

Thus 3

0

1 8 (1 0 )

1 2 2 .71 4 6 .8 M P a

P

p AS

6

6

1 4 6 .7 (1 0 )

1 5 0 0 (1 0 )9 7 .8 G P a , 0 .3 2

p t

a a

S

E

Also

0 .3

2 0 0% (1 0 0 ) 0 .1 5elonga tion

082.0)100(%7.122

6.1227.122

areainreduction

SOLUTION (2.2)

Normal stress is

2

2 2 0 0

( 3 .1 2 5 )4

2 8 6 .8 M P aP

A

This is below the yield strength of 350 MPa (Table B.1).

We have

7 .5

5 6 0 00 .0 0 1 3 3 9 1 3 3 9 μ

L

Hence

6

6

2 8 6 .8 (1 0 )

1 3 3 9 (1 0 )2 1 4 .2 G P aE

SOLUTION (2.3)

The cross-sectional area: 2

1 2 .7 (6 .1) 7 7 .4 7 m mo o

A w t

( a ) Axial strain and axial stress are

0 .0 8 4 1

6 3 .50 .001324 1324 μ

a

6

2 1,5 0 0

7 7 .4 7 (1 0 )2 7 7 .5 M P a

P

a A

Because a y

S (See Table B.1), Hooke's Law is valid.

( b ) Modulus of elasticity,

6

6

2 7 7 .5 (1 0 )

1 3 2 4 (1 0 )2 0 9 .6 G P a

a

a

E

( c ) Decrease in the width and thickness

0 .3(1 2 .7 ) 3 .8 1 m mo

w w

0 .3(6 .1) 1 .8 3 m mo

t t

17

SOLUTION (2.4)

Assume Hooke's Law applies. We have

1 .5

53 0 0 μ

t

3 0 0

0 .3 48 2 2 μ

t

a

Thus,

9 9

(1 0 5 1 0 )(8 2 2 1 0 ) 9 2 .6 1 M P aa

E

Since y

S , our assumption is valid.

So

2

(9 2 .6 1)( 4 )(5) 1 .8 1 8 k NP A

SOLUTION (2.5)

We obtain

mm21.21151522

BDACLL

μ188621.21

21.2117.21

AC

AC

L

L

x

μ47121.21

21.2122.21

BD

BD

L

L

y

( a ) GPa53)10(1886

)10(100

6

6

x

xE

( b )

y

x

4 7 1

1 8 8 6 0 2 5.

( c ) GPa2.21)25.01(2

53

G

SOLUTION (2.6)

Use generalized Hooke’s law:

x y z E x y z

1 2( ) (1)

For a constant triaxial state of stress:

x y z x y z ,

Then, Eq. (1) becomes

1 2

E. Since and must have identical signs:

1 2 0 or 1

2

SOLUTION (2.7)

We have 3

4 5 0 (1 0 )

5 0 ( 7 5 )1 2 0 M P a

x

(CONT.)

18

2.7 (CONT.)

( a ) 0 .5 0 .0 2 5

2 5 0 5 02 0 0 0 μ , 5 0 0 μ

x y

5 0 0

2 0 0 0 0 2 5.

( b ) 6

6

1 2 0 (1 0 )

2 0 0 0 (1 0 )6 0 G P a

x

x

E

( c ) 6

9

1 2 0 (1 0 )

6 0 (1 0 )0 .2 5 5 0 0 μ

x

z E

6 3

5 0 0 (1 0 )7 5 3 7 .5(1 0 ) m m ; ' 7 5 0 .0 3 7 5 7 4 .9 6 2 5 m ma a

( d ) 9

6 0 (1 0 )

2 (1 0 .2 5 )2 4 G P aG

SOLUTION (2.8)

We have

y z 0 MPa125

)10(1020

)10(25

6

3

x

Thus

y E y x z 0

1[ ( )] (1)

z E z x y 0

1[ ( )] (2)

x E x y z

1[ ( )] (3)

Equations (1) and (2) become

y z x (1’)

z y x (2’)

Adding : ( ) ( )y z x 2 1

2. Then, Eq. (3):

x E

x

1 2

1

2

Substituting the data:

μ1327)10(70

)10(125

7.0

18.03.01

9

6

x

SOLUTION (2.9)

Hooke's Law. We have 0y

and

yx z

x E E E

6

9

1 0

7 2 1 0[(8 0 ) 0 0 .3(1 4 0 )] 0 .0 0 0 5 2 8 5 2 8 μ

yx z

y E E E

6

9

1 0

7 2 1 0[ 0 .3(8 0 ) 0 0 .3(1 4 0 )] 9 1 7 μ

yx z

z E E E

6

9

1 0

7 2 1 0[ 0 .3(8 0 ) 0 1 4 0 ] 1 6 1 1 μ

(CONT.)