Statics and Mechanics of materials- Chapter 10

8/10/2019 Statics and Mechanics of materials- Chapter 10

http://slidepdf.com/reader/full/statics-and-mechanics-of-materials-chapter-10 1/48



F

-.f=

e*

$t*t**=

*(o,atnf

(1o*loo

)

=

Gqt

N,nn

{

Problam

0,2

-=

97.3

xros

Pe

97.?

MPa

fRO?[IfT.tnY

IilTIRIAL.

O ztll ThcMcCtt-lH

Ccpir*

lF.

Afl dtb rxrd. No

Ft

dtb

Menrnlmry bc diilph 64 nnoo*m,

r

fu|r:l

b

q

Lr

q

V

:l ru, *h

lb

Fh

rft

p;*

of b

l*5rr,

r rrd bqila b H ffiitn b

Fhors

ard

€fucr0or

r*a

It

Mcf.fl*'-Hill

for r€fu

adividrdcou,tn

ryantion.

$tudarils

nrng

hir melrul $'c urrl it witM

pcrmirlbn.



probrem

0

lfl*J"di$*ljl1JilffiiTfl::;}XffiHffiffi|i;ilH'#1il

iffi;;

hown

is the same

as

n

part

a.

=(*Y f '6 )

=

o '8 in '

t = ( t [a ' t ) = i .e in ,

tJ-c , t )=

E(-2,

-

o-8")=

2.c ,sg

, , r

oo

lL'ft

-€

zt

6

o

o

1.

;n

(o)

T^*=

ts" W

=qetl

r=;

L*

=

q.1?

ksr'

{

(b)

/=

Tc

.r= I .+

d=

zT

J

"

\r

2'*

-L

=

IT3E:

c3

= 3I

=

(R\(zr ,oo\

f f i

=

-T r

(gg

t?

)

=

1.39666

i , , ,3

PfOblem

10.4

10.a (a)

Determine

he

torque

hat

can be

applied

o

a solid

shaft

of

3.6 in. outer

diameter

without

exceeding

n

allowable

shearing

tress

f l0

ksi.

(D)

Solve

pafi

a,

ffl#*%ffi"H.tottd

shaft

s eplaced

yahollow

haft

f he

am"ass

ana

r

.O-

:

to) f r f l

es" / ,Js l ' ,a{ l }

c=+A=

({X3.6)=

t .g in .

S

=

I.t

=

{

(r-e) t

9-re

q

,.ng

T*,*=

F

or

:f

=

H=

(fO)( t . tGoq)=

gl.Go?

*ip. i" ,

|

=

7.63 Hf

.ft

b)

Il"//"w

s['of*:

cr

=

*d,

=

(tX

3-e)

=

t-B

n-

f i ' r

equo,l

mcgses

*4.

6pqss

sec,*r ' , ,no.

a.re.Ls -u"l

b.

.|uol.

A

=

Tr

z

= Tr

c..-

c,.

)

or

cr

=

{c,rc

ez=ff i=2..sq$ti . ,

J

:

T

(ar t

-

c ,u)

=

+q.+6q in+

T=T'

W=

t?'i.33kip.i, '

T

:

t6.17 ,F.ff

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eserved.

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this Manualmaybc displayed, eproduced,

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distributed

n

any.fo11

o1

by

any

means,

withou

tlrc

prior

",ritlrn

p.trission

of

the

publisher,

or used

beyond h€ imited distribution to

teachprs nd

educators ermitted

y

McGraw-Hill

or

heir

ndividual

ou^" pi"p"lttn.

students

sing

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re

using t witlnut

permission.

(

J



PfOblgm

10.5

I0.5

(e)

For

thc 3-in.-diameter

olid

cylinder

rnd

loding

*rown,

detennine

hc

maximumshoring stress.

(D)

Determinc

hc

inner

dimreter

of

the

hollow

rylinder,

of

T,

4-in.

outerdiurutor, for which

the manimum

$tr€ss

s the sarne

s

n

part

a.

sh4+t i c=*Jt=

|(ao1

= l ,5 in .

=

-4,

=

(L)(ryI

--

7,5+s

ksi

I

Tcg

T

( f .s ) t

(b)

(b)

Hollow

sLr*t;

cz=+d

=

{(to)

=

z.o

i,,.

{

=

$(c.{l-

c,t)

=

T

ce

c"

t**

+

)or

-

JIS.

=

?,,01

(2)(t lox3.o')

=

q.2s

inl

r

=

L- TrT^*>

-

ffi

=

1

Ct

=

l -

7+

315

i r ,

4 ,=

2C,

?

g,1?

in

PfOblem

10.6

10.6

a)

Determine

he torque hat can

be appliod o

a solid

shaftof 0.75-in.

diameter

trJiHffff

f;fi H?$:"*T:i:ffi-"il:1H.:?*:'H*il#

SOLUTION

and with an inncr

diameter equal to

half of its owll outer

diamden.

(d

Sr.0;J

sho{t:

s

=

+"1

(*Xo.zs)

--

o.3?.f

n.

J'= * c* i { (a-e7s)" = O-ogloig ir, |.*,.,= o ts,'

T

=

+

=(o€?l?FX,*-.

o.sas

k,.p.i,r

r

sitE

L.in

<

(b)

llo0 o,

sh$tt

E.

tlne sa-n-e.

ur<a

a-s

*he

so,lr/

t[o]t,

A

=

n(ci-c, ')

=

nlr|

-t*."I]

=

$r

c;

=

TTcz

c?=&c

=

fr.f".s?,t)=

o,433orgi,n,

?3

Cr

3

*

ar.

=

O.

216s0g

in,

J

=

{(c"*-

€,t)

=

{

(o..+

Svrtg+

o-

?r

6s06+),

0.os\7?a

int

r=ry=W

PROPRIETA*Y

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this Menual

may

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rcproduccd,

or distributod

n eny mm or by any mcaru,without thc

prior

wrificn

pcrmiecion

of the

publisbr,

or used

bcyond

hc

limitcd

dicfibution

to tcachcrr

atd

educators

crmitted

y McGraw-Hill or their ndividual ourse

rcparation.

tudents

sing his manual

reusing

t

without

permission.

i



Probfem

O.T

l0'7

The

torques

shown

are

exerted

on

pulleys

A, B, andC.

Ifuowing that both sheft$

are

solid,

detcrrnine

he

maximum

hearing

tress-in

a)

slraft

AB,

(b)

shaft.BC.

8 0 O

. m

(a)

Shnlt

Ats: T

=

4oo

N-

I'r

"€= ' *d=

*_r={. t

*(o.

o3o)

=

.Y

Lry,=

4[$

N.

rn

rr

(o-ors)3

t.n*=

7.5-5

MPa-{

J=

8oo

N. rn

=

O.

OtO

t'n

T.

=

2T-

kXaocr

Lriec

.

TrcE

-i-to-o#

=

63-7

x

lon

?t.

t"*= 6'3J MPo.

-

Problem

0.8

l0.E The

shaftsof the

pulley

assembly

hown

are o

be redesigned.

ftrowing

het

Sre

allowableshearing tress

n

each

shaft

s 60

MPa" determine

he

smallest

allowoble

diameter f

(a)

shaft

B,

(6)

shaft

BC.

(a)

Sh4ft

AB: T=

4oo N-w,

t..r=

6O

l4Por-

AQrlOt

Pa-

f,= T"t T**=Tc = g

\r

lt

e3

4 0 0N . m

Q =

=

l6 -n * lo - "

h ^

=

t6 -19

w , * ' ,

l*c=

2C

=

32-'f

^^ <

(h,)

haf

+

BC

f

=

8oo

A/-

t*

=

Lo

MPa"

=

6ox

lot

?a.

f=fl#=7#-,

=

?.O,

O

xlo'3

h

z

2O-{O

r"rr"r

dr"=

2c

=

4o.8

.

PROPRIETARY

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McGraw-Hill

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All rights cserved.

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part

of this Manual

may be displayed,eproduce4

or

distributed

n any form

or by

any rneans,without

thc

prior

writtcn permirsion

of the

publisher,

or usedbeyond

he limihd distribution

to tcochers nd

educators

ermittcd

y McGraw-Hill

or

their ndividual

ourse

rcpention

Students sing hismanual

reusing t

without

permission.

SL' ft

BC

:

8'f i)N. m

?T6a",otl

t.



Problem

0.9

9400Ib

in .

Problem

0.10

t f f)o

b.in.

2400Ib.in.

Shaft Ag

:

T

C-LA=

10.9

Knowing

hat

each

of

the

shafts

AB, BC,

and

CD consist

of

solid

circular

ods,

determine

a)

the

shaft

in which

the

maximum

shering

stress

occurs, (D)

the'

magnitude

f

that

stress.

800 lb . in

,300lb. in.

Sh"Pl

CD:

T

=

-Boo

+

tnoe=

-w

T

(o-e

)s

?r.loo

+

lDoo

=

AhOO

b.in

=

7e63

p" i

Ansu

e"tl

(a\

Shaft B

C

e=+e

=

0-6

n.

(b) g.

t5

ksi

I

Shaf{

BC:

f=-goo+?qoo=

160q1

L.v ,

'e

i+.A B

o.5

in.

f'.-=

S#+

=

I tqg

ps,'

I*J."+)

1q'lq

Knowing

hat

a 0.40-in.-diameter

ole

has

been

drilled

hrough

each

of the

shafts

4 , C,

and

CD,

detErmine

a)

the

shaft

n

which

he

maximum

hearing

tress

ccurs,

(6)

the

magnitude

f

that

stress.

H.'fe cr

=

+J,

=

*(aqo)

=

o.

eo

,u,.

Sh4f+ABi T= goo k. in

Ce,

*l*

=

O.9

i,,.

dcD:1.2

in.

u

.1

s

*G.t-

c,')

=

{(o.r

{-

o-eo*

=

co3' t6??

in t

t^*

=

F

=

t3:#i;uq

-

Eq

lp";

.o.-rs+)

Ca=

C^

=

O5.

in.

gf

=

*( . r t *c , t )=

L(o.Sf -o.Zo*)

=

o_oqs66t i , , r

: I :*.*-F--'# =

8g6s

"i

Fh"ft

CD

:

f

=

-goo

+

Z?oo

+

looo

=

?CoO pl. i " ,

C"=*dr=

oAin.

r f

:

T

(c . t -

c , t )

=

{ (0.61-

o.?o*)

=

o.ao '106

' r

ts**=?r#=

77sipsi

Ansverc (al

Sha.P'l-

E

(b )

8 . t19

ks i

<

c

c

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eserved.

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part

of this Manual

may

bc displayed,

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or disbibut'edn

any

orm

or by any means,

ithout

he

prior

written pcrmission

f the

publisher,

r used

beyond

he

imited

distribution

o teaclrers

nd

pdrrnalnrc



Problem

0.1

l0.II

Undcr

normal operatingconditione,

he dwuic

rnotm exerts

a torque of 2.4 kl.[

r

m on

sheft B.

Knowing hat eachshaft

s solid,

dctermine

he maximum

shearing

stress

a)

in shaft,{^8,

6)

n shaft.BC

c)

in shaft

CD.

T s = l . 2 k N . n t

46

mm /Tr-.

=

0.S

tN

.

m

46

mm

iTr.r

=

() . - l

N

. r r r

54

mm

sha$tAg:

4J

c13 =

Tas=

2.4* td

N.r " i

c=*C=

O.o27m

Iq

2T-

g\( j .r f

x roe

f

=

=

ffi=

f7-EX5*p6Pa

17.614Pa.

Tpo

=

?.1

kN-'-

-

1.7

klJ-n,

=

1.2

kh)-m,

c-=*C'

=

O.o?3

n

Ts

2r

=

Q\(.t.2

yto:)

nn

Tggx

oor.q

6n.6

Hp.-

<

. r

Ea=@=6.7.

Tr"

=

O.4

>tlo?'N.-

c

=

*J

=

e.o23

v,-,

F=T*:W

co

=

PfOblgm

10.12

10.12 n

ordcr

o

reduce

he

otal *, of ,h" essembly

f Prob.

10.1l,a new designs

being

considered

n which

the

diameter

of

shaft

BC will be smaller.

Determine he

smallest

diamaer

of shaft BC

for

which the maximum

value of the

shearingstress

n

the assembly

ill not

be ncreased.

See

sol,;*,'o.

*.,

p,.ob

le.'

/O,

ll

$on

$iqur^"

a.nJ

%"

tyl4xi

nnur,-l

"[,"o^r,q

s'Fresses

in

g.,**,t^s

AB

,

BC,

q,.C'eD

o* t{rc

sh4+t

The

1",^1es#

value fE t,rn = 17.€?3 xrd6 Pe. occ,urn',,3 n IB.

AJju=* Jto*,e-ter

uS

BC

t,,, al*ai-r

*he

sq.rê

vq]oe

",f

slêrs

-

4) Tc

2T

' -F=rTF

c3=

"+t

=

ff i

=

1.i l t t fx,oahng

C.

=

21

43

xtO

"

rr^

d

=

LC

=

+L.Ex1O-3

v',

*?.8 r.n,r,

PROPRIETARY

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Manual

may

be

displayc4

tprcducod,



Problgm

1

0.1 3

10'13

The

dlowable

shearing

tress

s

15

ksi

n

the

L5-in.-diameter

teel

cd

Arrnd

g

H:,*.lt:-iqteter

iod

BC.

Neglecting

he

etrJ

of

sfress

concenffrrions,

'-\

_

detcrnrinc

lrc

argest

orquc

hat

can

u"

-lfirrrJ.?;.

rÊ

F,

J=

* "*,

l-

=

#ct

tw

Sh4ff

Ag

:

Tun,:

tf

ksi

e

=

*d

=

o-Z5in.

|

=

E(o-zs)3(rs)

=

1-1,t

kip

- ir ,

Slr "H

BC;

tr_*=

gks;

c=

ha

=

O.j ,O)n.

T

=

{

(o. lo) ' (g)

=

q-t6

Vip. in.

Tle

*'N*"J

L.-

orl

,.re

s

l-te sno. e,

{alue.

T

=

1.l€

llip,ta

<

probrem

0.14

:,1

",flH1 'HlH:#T,'#il*:

:",T1:fj?"il

,fru

",'tr;

eglecting

the

effect

of

s[ess

concentrations,

deternrin"

tir" required

diameter

of

(a)

.

tod

AB,(b)

od..BC.

i t ee l

t n r=

P _T= I

.3=

4

J

)

\'.,

?

)

(-

E+

Brass

Sh

aPt

AB

,

T

--

lo

k,p.

n

(,_, =

lS

ks

^3-

(?Xt" )

c"=ffi = o-et2+Y''3

La)

c

=

O.

?Sl..f

in.

d

=

ec

F

t-

S-o

3

iy.-

Shcfi

BC:

7-

-

lo

kip.ia

l|,-o

=

g

ksi

a3

=

(e( f\

=

;tr

=

0.11s7?,n

(b)

c=o.q267

in.

d=Ze: t .g f3 in,

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rights

eserved.

o

part

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this

14nn*,

*",

o,

displayed,

eproduced,

or distributcd

n

any form

or

by any

trcans,

withou .th".ptiot

writln

p"ro,ir"ion

oi"dfilt#;:

".

*"a

beyond

hc

limited

distribution

to

tcachen

ard

ducators

ermitted

y McGraw-Hill

or their

ndividual

ourse

reparation.

tudents

;"t

thir;";;

aie

using

t

withoutpermission.

10'13

The

dlowable

shearing

stress

s

15

ksi

in

the

L5-in.-diameter

steel

rcd

Alrna

C

ksi

in

the

l'E-in.-diameter

rod

BC.

Neglecting

the

effEct

of

sfress

concenffltions,

detcrnrinc

lrc

largest

orquc

that

can

be

applied

atl.

,

J=*"*,

f

=

#ctr,w

I

:

Tunr:

t f

ks i

e

=

*d

=

o-z' in .

o-7s) " ( rs)

=

1-1, t

k ip- i r ,

= ;

t^ , ,=gks;

c=ha

=O. jO)n.

(o.lo)t(8

)

=

q-tG

vip-

n

.

-\s

sn

a. le,

{alue.

T

=

7.l€

ltip.ta

<

0'14

The

allowable

hearing

tress

s 15

ksi

in

the

steel

od

AB

arrdg

ksi

in

the

brass

td.BC.

Knowing

thl

a

torque

of

magnitude

T

=

l0

kip

.r in.

is

applied

st r

and

eglecting

he

effect

of

suess

concentralions,

eternrine

ire required

diameter

of

(a)

td

AB,

(6)

rod.

BC.

-

_T=4

' -3 -

2T

)

\-r

?

)

L

ff*r

j

T

--

lo

k,p. in

(,-r=

l . f

ksi

\ / , - - \

i=:+ = o-et?+Y,,3

t t . : I

?515

in.

d

=

2C

F

l-

S-o

iy.-

<

i

f - lok ip. ia

/ , -o=8ks i

t )( tcr)

b

2\

ffi

=

0.1qsl?

i,n

q267

in.

d

=

2e

:

t .853

in,

<

r'-Hill

Companies,

nc.

All

rights

e-s9.ryed.

o

part

of

this

Manua,

",

O,

displayed,

eproduced,

e-prior

wrifren

pcrmission

of

the

publisher,

or

used

beyond

mc

fimited

aistrifution

to

tcachen

ard

tl

course

reparation.

tudents

sing

his

manual

re

using

t

withoutpermission.



Problem

0.15

?ipe

CD,

T*t

=

3/lt loe

Pa

C.=

*

),

=

O.OtlS

rn,

Ct

--

C.

-

t

=

O.O't$-

O.

006

=

O.O31

tn

d

=

f

(-a.-t-

,-

)

=

{(o.o{rr-

o.ogi*

)

=

z.gols

,

o'

,','

10.15

Thc

solid

rod

AB

has

a diam

eter

dp:

60 mm

and

s

mrde

oi

u ,toi

lor

ortrich

the

allowable

hearing

tress

s

85

MPa.

The

pipe

CD,

which

has

an outer

diffneter

of

90

mm

and

a

wall

thickness

f

6

mm,

s

made

oian

alurninum

or

which

he

allowable

shearing

tress

s

54

MPa

Detennine

he argest

orque

T

that

may

be

applied

atl

g"d

Ag

t

tan

=

Sirtta'

?a.

,

s=ld

=

o.o3o

rnt

T-

st-

,arr-

#

=

*"rt*

= T2(o'o3o)t18s*to") = g'6osx los N' t''n

(?.zot3

rrdcXs+

v,

116)

=

g-g ?

4Os

N.

rtr

o-o+5

is siu4 'lu Valu".

T.U

=

slLatrrlor U.

l-,

j .37

(N ' ,n

<

PROPR.IETARY

MATERIAL.

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Courpanies,

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eserved. o

part

of

this

Manualmay be displayed,

eproduced,

or disributed

in any

form

or by

any mcans,

without

thc

prior

writtcn permission

of thc

publisher,

or uscdbcyond trc

linit€d distntution to teachcrs nd

educators

ermitted

y

McGraw-Hill

or

their ndividual

ours€

r€paration.

tudcnts sing

his

manual

re

using t without

permission.



Problem 0.10

10.16 r turhlc *ctoig ffis tu 0 fi$r in &r hn 6d f ed ffi'ffir tueE

nldn

Fd

tr" l(rronkg

&m

a

frryr sfqffi f-

l1,S H

r;

ru l rppfld

n

"{,

dcitmbc

&r ruquird

diamwr of

{n}

rd*I,

{*f

matr

*f

=

Tar

(aJ

RaJ

AB:

L

=

L€-rSrlo-3

r,'r

=

2fi.

f5,

h^h^

dfr=

R.e

=

5O.S

rr",* <

r

T^+=

F

n _

2 T

\'

li-z;*

=

I5-7tS

,,

ldo

t*tt

rR€/

Tr

zsx

to.)

c

=

3[.€qxjcr

-

r.a

=

3l-€4 an

du.=

?

e

=

63.tf

*rn

<*

F'X ffiTfty lltTf,'tlAL O 2011 Thc McGrffi-Hill Coryroia,

hp. All

righfir

rffivld. No

pN'rt

f 6is

Mrnnl

my bc 61ryh1q4

rqil0erc64

or ffi h

ry ful r

V

sl

nr',

?tu

tb

Fio.

?rfr

Firh

of

fu

p*trfr,

a

* b.]Era

L

ffi d|rfrlbi

n

rj*a

f

d|cr,m*r

prrnittad

by Mc{tsw-Hill

fsr

their fudivkl|trl

ccrrc

puprr*ion.

Sndcm ruing frie

rnrmal rl

Blng

it

wffurn

paminirn.

'



problem

0.17

**

P;.'ll'$"*?*:*H.t:.:T*,"^t*".1n'"i:::*l"T::,X*

-*ffil1

rluulElll

lu. | /

stress s

55 MPa"

Neglecting

the effect

of sfiess

conccnfiations,

detennine

smallect

diameters

d6 nd

d6s for which the allowablc

shearing

stress

s not exceeded.

5s

MP',

+

55,

tot

P4

Ts = 2T c.= l LT

J \C3

V

tE"*

\E'.

Jir

e

|1r:r:-too

=

goo

N*a

Q=

tffi

=

at.oortdsh?zt.oy',

w'rinihrur.{

dou

=

?c-

=

+2-O

hah4

4

ShaPt

Bl::

Te.=

+oo

N.nr

rE\@T

C =J

+-

-

rc-667

rrcr-3rn

?

lG. G?

hra

l

Tt

(Sfxld;

)

f t'

ttt'

I

Ttrr

"r

'b

^

minir t r rn^

d* .=

2e

=

33'3F\H4

' '

pfOblem

1 0.1

g

10.f8 Solve Prob. 10.17 assuming

hat

the

direction

of

Ts

is reversed.

I shown is formed of a brass for

which

the

allowable shearing

eglecting the effect of stress

concentrations, determine

smallest

6

for which the allowable shearing stress

s not exceeded.

il*+

lle

C;*e.]ie,

ofE'hos ee",

oJ in fJ'c S,S^ *. *[,e ,f€Ft

'

-515

4FL

t

{fxlOs

Pa

=Te=

eT

.=3[2' f-

J

'IIe

E

\J

-y

fi

C*.n

S\c.tt

AE:

'l-6s

=

l?oo

+

lao

:

16oo

N.rn',

cE

ff i

=

?G.46vtdtn

--

z6-{6+a*,

,wtirr

rourn

dû

--

U

=

52.q

mr4

-

ShaPt

BC:

Te.=

\oo

N'v,.n

3/mI-

s--

V

reil;i

=

t6-g€?vlo-=,'

le-e7

,*,

Ynivrirnur,.',

*t

=

2c-

=

33.3

urT

4

pRopRIETAny

MATERIAL.

O 20ll

The McGraw-Hill

comprnies,

nc.

All rights

eserved.

o

part

of

this

Manual

may

be

displayed,

eproduced,

or

distributed

n

any orm

or by any

rreans,

without thc

prior

writicn

pcrmissionof the

publislrer,or

used

beyond

lte

limit€d

distribution

o teachers

nd

"a"*t"^

n

n"itt"i

UvUcCra",-gitt

for

their ndividualiourse

preparation.

tudents

sing

his

manual

re

using

t

without

permission'



probrem

.1

##;j,:yt,'J,11

rTi

ffis,i_

llfll?T*fr?:"iffi

il"JTJ:ff

ksi

in

each

shaft,

determine

for the

given

data the

rcquird

diameter of

(a)

shaft CDE,

{

(D)

shafr

GH'

TF

=

12c)0L.

in

I'l

l;

=

f

r;

=

$f'loo): 3zmJb.i,,j

t"q

=

lo,5

lcsi

=

logoo

ysl

rr

l=F=#,

c==#

(a)

5 l ,o f {

CDE:

?3

\ ,

f f i=

0.1?,tOlAina

A:

O,S78q

in.

do"

=

2C=

l .

lSE in.

r -

(b)

Sh,.f

fG

H

;

.s

-

(?)Cl loo)

L

=ffi

=

o.oTT'lst

n=

C

=

O.3+t7. ,

n,

dre

=

RC

i

O.g3f

; r , ,

<

PfOblgm

10.20

10.8

and

10.20

Under normal

operating

onditions motor

exertsa torque

of

T,1ffi

;ff

1,1[lll,ffi

'#*f,:x';f

,ff

.XH#T"#,T:?(H

CDE,

D)

shaft GH.

Tp

=

-

I

?oo

lL'in

wt-'D'

''' u'a.r'

'o

T

G - -

2 ' '

\

ts

=

tr

lF

=

i ( t?oo)=

4So

6. in

t

tou=

fo,S

k:i

=

lOSOo

sl

q

I

r

=

Tc

-_?f

^3-

?T

t -

-

J

-F

r-=EZ

Sh"$+

C

Dr

,

^

3

-

A

)Cq$a)

c- --ff i = o'o2129'lt^r

C

=

O.3olo5 in ,

do .

=

?C

:

O.GO1-

n.

<

Sh',+l-

FGH

e

=

9)f

'T' l

=

o.o727s-?

,3

TI

(f

o5ao)

C

=

O.

t+

71 ir,.

dno

=

?C

--

O.BS5

n,

PROPR.IETARY

MATERIAL.

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The McGraw-Hill Companies, Inc.

All rights

reserved.

No

part

of

this Manual

may be

displayed,

reproduced,

or distributed

in

any

form or by any

nreans,

without

thc

prior

writtcn

permission

of the

publishcr,

or

used bcyond

thc linitcd

distribution

to teachn

ard

wi



Problem

0.21

Probfem

0.22

f0.21

A torqucof

magnitudeT=

000Nr rn s

+ptiduD

asshown.

Knowing hat

the

diamcer

of

shaft AB is

56 mm and that

the diamaer

of shaft

CD

is

42 mm,

detennine

hc

maximum

hearing

tressn

(a)

shaftl.B,

(b)

shaftCD.

T"o

=

foOO N.n^

T^.

f;.

-i;o

=

Jff

(rooo)

?foo

N.m

e

=

Ia

=

o.o?B

4

1-zF

:

??.50

x

lD6

7?.5

MPa

.-

ShacF+

C:

C- - *d

=

O.o?O

h ' l

/= F= +*=ffi= 68'7*to

68.7 t'l?a

.G

10.22A

torqueof magnituden-

1000

N

;:m is

applied

at

D as

shown.

Knowing

hat

the allowable

hearing tress

s 60 MPa

n each

shaft,

determine

he

requircd

diameter

of

(a)

shaftl.B,

(D)

shaftCD.

T.o

=

looo N'q

lie

=

t T;

=

ffClooo)

=

?wlrl rryr

6JSh4++

AB

tdt

= r.o

xlos

Pa

t-- Lg

=

2I

.r=?T-

{

- l lcs

=

Tt

C

3

?q.82

x

6r

:

21.?2

)nha

T

:

II.XXJ

t

.

rtr

D

G

Xlsoo)

rr eoxro")

d

=

2c

n.C.SA6r

D-

6

h4

=

5?.6

vnrl

2T

Trc

a

lo-3

,q,

10.610 lo- ' ln3

+3.

I

vn

?)

0$ba.fl

CD

Tat:

6ov

PL

Pa

^,

'T-e

-i

\t

C

=

2l.q-7

't

PRoPR'IETAn'Y

MATERIAL.

o 2011

The

McGraw-Hill

Companics,

nc.

All rights

escwerl.

No

part

of this

Manual

may

bc displaycd, eprod.ce4

or

distributed

n

any

form or

by any

means,

withou

lr"

priot

writln

permiseion

f the

publislrcr,

o, ,,*d

bcyord

tt

"

ti*it"a

aiutribution o tqacbn

ad

educators

crmitted

y

McGraw-Hill

or

their

ndividualiourrr

pt"puration.

tudents

sing his

manual

reusing

t

withoutpermission.



Probrem

23

ll'1;ffiffi;,1T#";"ll#::

lrtH

",*ffi,?XT-,##:

:Iilff'_?:

Tc

=

5

kip'"-

in.

is

applid

at

C and

hat

he

assernbly

s

in

equilibrium,

determinc

he

required

iamet€r

f

(a)

shaftBC,

b)

shaftEF.

i , , *=

85oo

ps;

=

8-5

fcs;

(4).$fu1ff--g*q-:= 5 kip-i-

L,o=

Is-

2T

^ -:[zr

J

=

F

e=n-,+

r=

: l@

=

o- l?oi ia.

'(

Tf

(a.s

)

dB.

=

2e

=

l-44'2

i,,.

4

t t

sl ' .++

s)=

g.tzsk:y. ia

Q :

l . G l 6 3

i n .

&,

=

2e

=

l -233

n.

4

Probrem

0'24

il;11ilH:ll":',i:ffir,Hf

f,t6o'iH

r,:HH,ffif,ffi'#il:Hfi3,

shafts

re, espectivcly

g

=

1.6

n. and

dsv:

1.25 n.

deterrnine

he

argest

orque

T6

=

{Cz.o) (o.62s)s

2.e81

k ip ' in

i

By

{Ji .=:

T"=tl=*(a.aeq)

=

{.3ok, 'p.,, ,

M ,.

J,0" uolu"

J

T.

is

*h-

"^J)"r.

Tl

=

+.

u

kg'it'

{

PROPRIETARY

MATERIAL.

O 20ll The McGraw-Hill

Companies,nc. All

rights

eserved.,No

art

of this Manual

may

be displayed,

eproduced,

or distributed

n any

form or by any means,without

the

prior

writtsn

permission

of

the

publisher,

or usedbeyond he

imitcd distribution o teachers nd

educators

ermitted

y McGraw-Hill

or

their ndividual

ourse

r€paration.

tudents sing his manual reusing

t without

permission.



Problem

0.25

10.25

For

the

aluminumshaft

shown

G

=

3.9

x

106

psi),

deterrnine

a)

the torque

T

that

causes

n

angleof

twist

of

5o,

6)

the angleof twist caused

y

the

saine orque

T

in

a solid

cylindrical

shaftof the

same

engthandcross-sectionalrea.

(a,)

9=*,

T=

9

:

5o

=

87-2661103

d

,

[-

=

4

ft

=

+8

jn

J

=

{(cr'-

c,*

=

f

(o:.f1-

o.S*

=

o.g?

t+

in

a

q2

t -

(3.q

"ro6

Xo.3?

Be4

;'1.

ag

*lrr

)

+8

*lo3

L-,^

=

?-83

(b)

Ho "o

sir^|f

:

M

"t-trirtn

area5

e

=7.2271

A

:

n

(cr '-

c,.

a 2

-

r t - r 7

-

( -

v 2

- \ l

S"

hJ

sl.|+

A

Q

=

0-551o r . ,

{

=

9,

ct

=

*(o.sf lo\r

=

ls3

g1g

/o-t

i , ra

ft)

=

g

_

kl"1q/p3

)(+8)

V

F

=

=

?,26.8?r lo- t

vd

=

tg-

oo'

o.?5"

O.St

=

o.3r?S, ' .4

PROPRIETARY

MATERIAL.

O 201I The McGraw-Hill

Companies,

nc. All rights

eserved.

o

part

of this Manualmay

be displayed,

eproduced,

or

distributed

n

any orm or by

any mearu,

without

he

prior

written

permission

f the

publisher,

r

used eyond he imiteddistribuiio;

to teachers nd

educators

ermitted

y McGraw-Hill

or their

ndividual

oourse

reparation.

tudents sing

his manual

reusing t

without

permission.



Prsblsm

10'26

il.

*

Fm

rc

,*

u*31'*

l:ng: la1;-rffil,ffi

ffi"S

r;*dt(c)Fmfiqselid

st€t

$nilnsurawu1v

-

I '

--fi41

ir

Mffi

affi

r

l.tr-h.

twirt

d

/.

$)

il;

e^T

"} ryY*s

ft.t

*}G

wcl

s

itnrodh*imdc$.8-in.

----===-=:.---'-t

;;:

(o'6)n;=6'

Aoss?r

ih

e

L=6Ftz72 ' ,n-

T;

?k;g. i^

L

2ooo

&' i ' *

:?a)

(3.

I

SZ

Y

lo-=

J

g = 3 .6? '

- 4

[dp

in.

ffioss?s

)

I

lrl-

c,=+a,=

.

in.

0 ,1Lg363

yr*

-

- t€.?o

3

rJ

=

E(0.6 ' { -

0.4t )

3

t?ooo).(?e' )

9=

+.5 t

o

4

?no?nlrrAny

MATE*'.L.

o

20lr

rh

Mcc,.u-Hin

oryrnie,

ry.

A'ftry

ry.rv.ad :p*t

of

hir

rrnnl.ryy-F

dtuehtt4

rooro4

or

dbtihrsc

r

ry

ftru

or

bry

mv

."$*ffi

ffi;4;J*"i*

o-'-t'

ennri*''

- Jrn"-

i

nr

r-

r

s

-

G&crbil

rrr*no

w

x.i*.i,-nru

frr uir

ina*uu.t-wrc

er+.nai"^

$hdrffi-;iq

oio

nt*i

*

ui"'

it

'i&nt

p'brbn



Problem

0.27

j___[*[

.,,e.';r

-

r--.-lf:.

i

ltst famrfuu ec kqilE*dh*&h diameterf a 3-m-lorysld md

{S

- ?? fr} if

I

ruc

*i

m hr tw*M futr{h

3Sp

without

exceeding

sherfu shnl

ilS[tra

f

:

Z*,

g:

{ f t j

=

$-23.G,xlcsra|,

f

= goxlospa

T I

9=#,

r=ff9,

t

=F

=SrJ*

=#t,

c=

#

c

=

&t*HH#"q

=

5.r?53

rd3

=

5.

?53,--,

d=

?e '

l l .q t

Hh , r

<

fff}trufTAfY HATf,I|AL

O

2011

ThGM€f,rrw-HillCoqrrirn, l&. All rigk rfi€rnod.Hopcrtof tlb lr{rnnl ury bedLpblnd,nfao*mda

bl*l h

ry

hr r ff ry

n,

rlH

L

Fir

rrh

ffrtn

*

b

fSlib,

r rlrC brlma

b

H-l.

trfii f :ifrr *

a*rcrmrr

pntrittd

by hlcftrs-Hill for

tcitr diykhd

co$nc

FlF.rtioo-

$tdGffi ucng

hb mnuel arc

rriry t

wifror*

prmirrftro



ProMm

10.28

I

000

_i

t

Q= I I

JGir

( -

I

I-s

T

\l

I

L

f rc., = (l)(tTr) = l?. 566

-fooa

ff

=

6oooo

;n.

y.nd , Q,= *d - - t f -O in . ,

f=

6o

o

oer

=

7-38t6rfos

ps,:

=

?38

ksi

{r*

fROFnIETfRy

MATEruAL.

O Z0ll

ThcMcOnr-HiU

Coqnicr,

IEE. ll righirruronnd.Hoprt of thirHrmd

ryy Ft

di+ry.4 '@

r ffi

i

ry

lr

c ly

lf n,

rtrt ln

f.L

*,rti.

frlloo

oe U

f*firUr,

or urd

bclm{d

6. licittd

dl*rfoatiol

to

brEb3E

e&rato6

Fnr.itt6,i

by Mccni-Hilt

ftr fuir idividrul

courrc

cn*i*i.rn

Str&nu

unnc

hir

rnffitl

EEulrng

t

rifuu

pcrbin'



Problem 0.29

10,29

The

orques

hown

are exerted n

pulleys

A andB. Knowing hat

he

shafts

are

solidandmadeof steel G : 77 GPa),determineheangleof twist betweena) A and

B,

(b)

A

and

C.

T{

'r,r

30 mm

= - l O ( ) \ . n r

(o)

1le

=

3OO

N-.,

Lû

:

c).Q

''

r

Jr"

=

*(o.o,s) l

=

71.5?a

,o-1

w+

90.=#

=

Qre

=

C3.

=

{d

- 3 '

x

lo

14,

r

1gs

por

=

Cû.

al

=

O-otS

:

4?.

o?S

lo

t

rud

?.530

d

=

O . O { g

k } r

3 . {?o <

(b)

Ti"

T

I JBA

-

Qa.=

A

I

0.9

m

I

8 l

I

0.75

m

=

3OO

+

Ll}o

=

?OO

N.h,

Lsc

=

O.7Srv,r

,

*

t

o .o lg) t

=

lgq.s?gv td?

t

F$.'

=

,= '?o= 9:rt)=

.

=

ts,

il

GJ".

-

QZxOtffi)

-

9A.

=

%.

+

(ee.

=

g1

606


@

201I The

McGraw-Hill

Companies, Inc. All rights reserved.

No

part

of this Manual

may be displaye4

reproduced,

or distributed in

any

form

or by any means,

without

the

prior

wrinen

permission

of the

publisher,

or used bcyond

the limiied

distribution to teaclren

and

McGraw-Hill for their

individual

course

prcparation.

Students

using this

manual

are

using

it without

permission.



Problem

0.30

30

mm

f030

The

orques

hownare

exerted

n

pulleys

B,

C, and

D. Knowing

hat he entire

shaft s madeof aluminum G - 27 GPa),determinehe angleof twist betweena\ C

andB,

(b)

D

andB.

(dt) $

half

gc:

c=*d= o,otdwt

Je.

=

4C,

=

27.SAAxro-.

',{

Lg.:

O.8rn,

G=2?rt lD"

u

=

O.

14

g

o.f

Fr.l

s g.f+o

4

C

=

+,4

=

O,O|B

J.o

=

T

a*

=

l6t.

n6rl6'q hn{

E"

€

4oo

?oo

=

-soo

U.wr

GSoo

)( f

o)

=

t

O.

le3O

na. l

Q.o

=

O,

r {?

*

-

o. l r ,3o

=

O.

oga?,4

c.d

Z. t lo

. {

(u\

5h4l+

cD:

L"o= l ,o

th

P"o

=

rt

6s

9a,

*

""

=

-,"+.a-*

PROPR.IETARY

MATERIAL.

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reserved.

No

part

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Manual

may be displayed,

rcproduced;

or dis6ibuted in any

form or by

any means,

without the

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of the

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Students

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are

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it

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Problem

0.31

10.31The

lurninum

cd BC

(G

=

3.9

"

106

si)

s bondcd

o

thebrass

od

AB

(G

=

5.6

x

106

si).

Knowing

hat

eacir od

is solid

and

has

a diameter f 0.5

n.,

determine

he

angle

of twist

(a)

at B,

(b)

at C.

I

B"th

p. , r l ions3 C=id:

o.1{ )n.

J

=

*

c '

=

6.

tgsq

x

ro-s

nt

f

=

3oo, lb .

n

Shaf t

AB:

Grc=S.6x lo6psr '

- ra= +t=4gt 'n ,

'rn ln

T

Lr"

Ba'c)

Cra)

yB

-lrAs

_

G[fl=1

a\

=

O.+11

vd

=

?rJ.O'

<

ShaI*

BC

@:

3.1

x

lo'p.

;

le*=

6

S-t

=

7X

in-

Goo)(zz\

_

(3-?

vlo6

X6-

tailt

).lo-3

)

=

S l .7

o

O.

gDSrnool

(b)

Qol^*vn

a*Ct

E

=Qu+

c\e"

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part

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eproducd

or

dishibuted

n any

orm

or

by

any

means,

ithout

he

prior

writien

peimission

f the

publishcr,

r used

bcyond

he imited

distribution

o teachers

nd

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ermitted

y

McGraw-Hill

or

their

ndividual

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Students

sing

his manual

reusing t without

permission.



Problem

0.32

1032

The

solid

brass

oad

AB

(G:

39 GPa)

s bonded

o the

solid

aluminum

oad,BC

(G= 27 GPa).Determineheangleof twist (a) at B, (b)atA.

=

*d

=

0,61€

n

Pa

f

=

l8o

7q.572vro-t

r,t ' t

0

eo)

o.2so)

|

=

O.Z5O

,-,

N'r.,r

(31xl

o

t

X

7

ct.

S

22r

o'

?)

:

l . l .Sto

xtdt*J

mm

shqlt BC: c=*

G

=

?7r lor

Fa

r

d:

* . t :

t6{ .

d

flr

3

:

T

c

t-o

o-

v l

)

,o l

IT

- l

o

''t1

)

L

= O-37o nt

w1

) fE

l8c

- l

oN.

7r

7

I

yn {

( lgo

)

(o.

=

12 q37

*

to-3

rad

4 \

\ , o t ' l

l l u . J ^ v

I

Yga

-

An"

.^,€r5

I

(al

9.

(b) qA

=@tt

=

l?-1g7

x

lo-3

wd

= A .1+t

"

Qa"

<

r-

Q".*

30 mm

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orm or by any npans,witlrout

he

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r used eyond lre imited distribution o tcachcnand

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ermitted

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ndividual

ourse

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Problem

0.33

10.33

Two

solid

steel

shafts

(G

=

77

GPa) are

connected y

the

gears

shown.

Knowing

hat

lre

radius

of

gear

B is 16

=

20 mm,

&tcrmine the

angle hroughwhich

endA

otatcs

hen

T1:75

N

em.

Co0.,r-e

.1"o".

|tryuc

s..:

Ci r

cu-&.e',h'ol

co,r*

a*t $"^'*

befveen

9

€ctrs

B

o"J

C

3

-

F:

l "

=

7""

' ( ."T

T.=A=

754/'

, -

r e D

\

h

Tnu

To

=

*ffi

(zs

--

zas

N-

nn

T,v;sf " s l,".*t CDz

Go

=

*q,"

=

I to,or l ) t

=

3a.s?2xro- '

l

L.o

--

o.

tl

oo wt

G

=

77*to1'

Pa

^ T t

( 2 1 , 5 ) ( o . g @ )

A l A A , 2 . -

- ,

9*=

r=ffiff"ffi=

3f.88Srro-

wd.

9o'

9",

=

35.

gt|') 'td3

w-J

Cireorr,&"^*fiJ

/;./n.u*."at

J co^*osf

prinf,

of

geans

B o,J

e:

S=

f .Q.

=

feQe

l?.o*o,*i

",'

o"?

/e

-*

B-:

Twis{

in

si '*f i-

AB'

Qe

i

ft

+

=

*ffi

(gs.t85xto')=

o-7.Gs'f

totnoJ

JRe

=

*cr"'

r

I

to-oo)t

=

ts.|oy

xlo'i

^r,

L le

.i

D.

So.o

rn

3 l .Oo+ t lo3u rno l

=

77

x loq

?e

' r

f l re=*

6S

(nti@

P,fJi*" ^l A:

P^ fr + {or 138.T lot*ol = ?.g1"

.{

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MATERHL.

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may

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n any

form

or

by any means,

withou

the

prior

writln permission

of the

publisher,

or

usedbeyord

the limitcd

distribution

o teschcrs

ard

educators

ermitted

y McGraw-Hill

or

their

ndividual

ourse rcparation.

tudents

sing

his manual

.,

uring t without

permission.

0s)

(o.gao'1



Problem

0.34

103'f

Solrre

Prob.

10.33,

assuming

hat a

changen

dcuign

of

the assonbly enrlted

n

thc radius

of

gcarB

being

ncreased

o 30 mm.

G=77GPa, ,

fa "3o- - r

T=

715 . i f . rv r

Dete,r', 'rrs

fhe

anlIe

i{ryo,.r+

.

Wh'r'e,h

enJ A r.o*ntcs.

Ca

su

I

ot;

o"

"t

*ar1

,.r

s

3

Cir.c.r-,T".

"-.I;ol

cor*url f*."

bcf

n/eyr

j e r . r z

8

an )Q i

t r=-bE=- I "o. - t t

f6

r.

l 'o

=

f.A"

l lne= I = 7SN.v" r

Tlo

T

tr,.

I

t l C O

- -

?

\ - C D

$

=

7'l

lDa

=

*

(o.ora

){

?o,

9*

=

O.loo

Y?r

=

23.7

23

*los

v,ae

l'a,d .

=

g2.5?2

ro

?

,r,

L.o

=

TL

Uso)(o.*oo)

=

-

GJ

(l t

ytDr

Xs?.

s72

rdl

)

Q.

=

g*

'

Z3-q

gx

lo

s

Cincdrr^F"n

enl;o,l

J.sp-0o..e-,e^{

o*

eanfa.*

foi*t

of

$e"ins

B

o-,J

C,.3

S=

Ytcl .

=

fe{o

Roi'f io..

onl, le

*

8:

48.

fr+

,

*H1g.szSrtot)=

$7.89("1o3r

Truis{ ',

she}t

AB

:

RJJi",.

"t

A'.

Qa

=

Oe

Qff i

=

18.85x/o-3

rnJ

=

+.S{"

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eprcduced,

or distnibutod

n

any form

or

by

any

mcem,

withou

thc

prkrr

writicn

pcnnission

or-tm puutlstrci,

or uscd

beyond

hc limit€d

distnbution

o

tcrcbrs

and

educators ermitted

y McGraw-Hill

or

their

ndividualiourse

reparation.

tudents

sing

his

manual

re

using

t

without

permission.

Jo"=Tcû*=*(o.oto){=

f5.=

7?

vtol ?n,

ene

#

:

(t l

"toq)C,s.' lot

xlo-r

)

t5 .7o t

x

tdt

, * " ,

(75 )(o-5oo)

Lna

=

g,.,foo

n



Problem 0.35

10'35

Two

shafts,

gch

of

3/o-in.

diameter,

re

connected

y

the gears

hown.

Knowing

that

G:

I

l'2

x

106

psi

and

that

the

shaft

at

F

is

fixed,

etermine

the

angle

hrough

which

endl

rotates

hen

a750lb

.

in.

orque

s

applied

tl.

cJrulo*io^

"{

to.r lqr€s

9,irc"*[e,

e^

;ol

co*I oi

$,f

t

e

br*

"rea"

te.*s

B

"fld.

F

i

t r rTi ,

=&

rsVt

" '

T,

=

*l lo

=g(zsa)=

T*;sl

ln

sr,o*{ FE',

=

8

i . - ,

G .= {a '=

E^(o -g?5)"=

g t -og3xto - t i , , ' )

G= l l .2 r lo '

=

E4rs

=.Sgg9-l(E).--

:

??,.?itx

o's

na.l

G

$e

-(tt. l

xto'

Xgr.

e

"

.o-3)

p"+

o*;

o^ a*

F

:

ge

=

22-

?qS'

lo-

3

n.J

T^no..

,"1

J;"

il",

*^"^t

",t

S"*

?

5

J -

=

feQ,

=

f*C2o

)

=

30.6(o"/0-3

1". - t

o*o*;."tBr

4b=

frO,

=

*(zr.qss'/o-3

Ti^ris{

in

sl'"}t

BA

:

f

in ,

4^ 'g1.0&3rk;s

int

(7so)Cr t1

Lw

Qo'

23.713

xlo- t

,ô4,

Q"I

Il

"^

nt

A

i

0^

:

AB

+

Q*u

--

to.

6€ovfo-3+

27.1t3

rlo

3

=

5q.g73

x/D'3

rna,l

:3 - l?o<

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tbe

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writtcn permission

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or usedbeyond tre imited distribuiion

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teactrenc16

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ermitted

y

McGraw-Hill

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ndividual

ourse

rcparation.

tudents

sing his manual

reusing t without

permission.



Co.l

cu,

h

on

ol twqves

Tan

qeÎiol

c,r,nla.

fig+

"

cea

qe\

Problem

0.36

10,35

Trvo

shafts,

each

of

*

-

in.

diarneter,

are connected

by

tlre

gears

shown

Ifuorrh3

&*

C

=

ll.2

*

l06pei

ard tlus

tlrc $sft et F

is

fixe4

determine he

angle

botEl

y*ioh

€rd A rotatcs

when

$ 730 lbin

torqrc

b

qi&d

at"A."

'^

1036

SolveProb.

10.35,

assuminghat

aftcr a design trangchc radius

of

gear

is 4

in.

and

he

adiusof

gear

E is

3

in.

c

t

=T*

l's

T

3

/"So)

tAG

, t

V

=

562.5

b.

i"

Twisi

in

sAo++

F'g

'.

Lps

=

4et

E

8i , . .u

Jh=

Tat

=

Ek.3?f)T '

g t ,06gx /o ' '

n*

G

=l l .2x loo

psr

J|,fu

=

,

,_-(*?:.-f,J(t

, -

=

12.

3s

to-s

,-a

G

*Tre

(t

-2

xfo3

X3t.063

*to-s

)

A"+o* io^n*E"

A"

=

I2 -?3 i116" ro / .

Qe=kg, = A

6 e, = i t,?.18s/o-3

Twrtt,;"

..*h*ft,

AB:

Jrs

=

3t.

oL3r

o-3

(7so)(

t

)

b

G

=

frQ,

=

Yr"Qe

o t . 7 o l

r / o t r ' , . *

e,

=

Qe

+

Q&

= 1.?o l x1o-s+ 2 ,3 .713/o5 = g3 .+ f Qx o3 u^d

=

1.1

+"

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n

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or

by any

means,without

thc

prior

writtcn permission

of the

publisher,

or

usedbeyond

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to teachers

nd

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ermitted

y McGraw-Hill

or

their

ndividual

oursie

r€paration.

tudents

sing

his

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rc uiing it without

permission.

. t

I n

l

=

29 .7 t3w(o - t

r * J

( t t . lxoc

)GLo(3

x/o 's

)



Problem

0.37

10.37 Thc

design spocifications

of a

l.Z-mlong

solid

transmission

shaft

rcquire

that

the angle of twist of the shaft

not exceed

4o when

a

torque

of 750

N

'

m

is applied.

Determine the required diameter

of

the shaft,

knowing

that

the

shaft

is made

of

a steel

with an

allowable shearing

stress

of 90

MPa

and

a

modulus

of

rigidity

of 77.2

GPa-

-i-

=

-?so

' t -

9o

N' t ,

M?a.

=

c =

4"

10

u

lo '

=

61 .g t3xrO-3r .JJ

l=

?a-

Q,

=

17-2.

G:?a-

=

18.oG

xlo-3

'n

B"oeJ

on

o^1.k"f

*-,#:

9

=

*

=

#,

F= ?r

=

t

( - -

1TteE

-{

Bnse)

h shear inS

tn

s'-

f t

=

F

=

#

=ry

'@

=

t?.+gxfo-3

=1fu

=V@

$e

l)s"nln-1"n

vJue,

-

(=

,8 -O

rt

lO3

w. ,

=

lg , 'OG

n

=

?r , .

36.1

nn

a

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ithout

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prior

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r usedbeyond he

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ermitted

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ours€

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reusing t

without

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Problem0.38

10.3E

he design pecifications f a

2-mJongsolid circular

ransmission

hsft equire

that hc angleof

twist of the

shaftnot excesd o

whcn a torque

of 9

kN'm

is

rppliod.

Dctcrmine he

rcquircd diemeter

of the shaft, knowing

that the shaft

s mde of

(a)

a

.

steelwith an

allowable hearing tress

f 90 MPa anda

modulusof rigidity

of 77 GPa,

(b)

a

bronzewith

an allowableshearing tress

f 35 MPa

and a modulus

of rigidity of

42

GPa.

9

=

3o

=

.r2.3gorro-3

nJ

,

A

=

I+

=

?rL

:.

,,

(JU

If

Ct

G

T:.

19

=

2-T

{

l rc \

T= Q*los hJ.M l=2,ow

a{ -

?TL

C'

=

trda

bcs,eJ n #,.r;st

*tr

/.

3 - ?T

boseJ

s lecr . .hq

s* r .ess

.

-

Fe

D6.seaf

v

t

l a

(4,)

Sfee/

shcPt

:

t

=

cro

xtoc

Pa,,

G

=

77

/

ro'

po.

Bc,s"J

.,,

,uist

n1J.'.

c{=

ffi19#ilfr,

=

?.ae

vt,-a

e, = qr.O6 , lc.t rn = +1.06 lnrr d= ?e- = gp.l rnn

Bose{

n

s[ea'J 'X

s]rcss?

C=gl-q1o:,1

=

163.7o2

lo-, ,r3

T(35

x

.lD.

)

C

=

5*.7oxlO-3 rrn

=

Sl.

?o

nm

d

=

?C

--

107..1

^,*

R.quire]

veloe

"f

J

is

fAe

flon3en,

d

=

lp?,r+

n^h4

furv

Bo;l.A

oh

s[sqr,h1 strtss

?

C3

=ffi

=

6g,6,61rlo-6

a3

C= 39 .?3xto -s ln

=

3? ,g3

v,2 rhn

d=f tC=

J? .e r , ,nn

Rr?ui*rf

v

oI* ol

d

is

*A

e

1".1

er

;

dl

=

gn.

naa

(t)

8ro"ze

shalt

:

t

=

GfxlD6

Pr.

e

=

+A

xtoq

Pa-

(2X7vtos) a.o)

Bosrl

o,',

*uis*

a"5/e

3

c{

=

4 .J t7v lq { . ( .2 .o )

_-

= S.? l0gy/o . h , , t

x (+axD"X52.34o lo-3

)

C=

47.78tr lo -3n

=

47.7go, ,n

d

=

2c

=

qS.6

t'.lv',

PROPRIETARY

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ermitted

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Students sing

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without

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Problem 0.39

qtB

Slr".+J

C

D'.

Qol.

=

9s

=

g"n=

**

$=

tns0n

=

faP"

"

10.39

The

dcsign

of

the

gear-and-shaft

ystem

hown equires

hat

steel

shaftsof

the

same

iameter

e

used

or

both

AB

andCD. ltis

further

equired

hat z,o

(

9 ksi

and

$at

the

angle

p

through

which

end

D

of shaft

CD rotatesnot

exceed

o. Knowing

that

G

=

ll.2

x

106

si,

determine

he equired

iameter

f the

shafts.

-F"o

=

5 klp-

i',

=

5x

IOE

eb-in

li" = ftr*' i (Sxlog)= tl.s"tf . L.in

F",

Aesi:rn

E+se)on

s+4g1

use-

l'e

*.

T=

l?.Sx

or

J[. in

=

O.8842

in-

<P"=

f

*.

34.Qo7Y

ldr

r*1

Lre

=

1.5

Ft

=

l8 in ,

31.1o1

,

lo-3

l[*

.

Lnu

--

&\(t7.5*

losxrs)

=

W3

r.

Gs

l-6

GJ

GJ

er^=

}3

L"o=

2-+t

?

zl in.

( 5x lo3

X2{ )

?

l?ov

toE

R"-lJto*

J

D:

GJ

=

I,ect-

f f i .

='

tq.sslovto'

)b. in

GJ

=

To.,

GJ

GJ

9o

=

9o+

Qotc

=

c=

Use

tl.

.lc*qer

vorlrte

c

=

l.

o2e7

in.

d

=

Zc,

p.oS

n.

{

PROPRIETARY

MATERJAL.

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Companics,nc. All righ ts eserved. o

part

of this Manual

may be displayed,

eproduced,

or disfibuted

in

any

form or by any

mcans,

without

the

prbr

writtcn

permission

of thc

publisher,

or usedbcyond

h€ limited

distntution to

teacbrs nnd

educators

crmittcd

y McGraw-Hill

or their ndividual ours€

ttpsration.

Students sing

his manual reusing

t without

permission.

|

=

?,

h i

=

ixlDs

ps, '

r=F=

#

"=W



Problem

0.40

10.40

In the

gear-and.shaft

system shown, the

diameter of

shafts are

d6: 2

in. and

dco=

1.5 n.

Knowing

that

G

:

lI.2

x

106

psi,

determine

he angle hrough

which end

D of

shaft

CD rotates.

C=

* ) n =

. O o ' t n .

i n r

l 2 - 18Q

v

16" r o l

"

I

o-s

no.l

$

=

VrcP"

v""

"

, / a3 )

=

31 .q73 ' l o -3

C=

* r l r =

O . ? 5

in .

CoJ.rJ.rt

io,^,

o|

*oq,r..

Cire,*ler.-^f i. .^/ .n+oA f"..o

F=

Ie

=

Io

- r

f . - -

t

-

Yb

E

f t . '

Y : r co

To=

S

lz ;p . i " ,

=

5x lo3 , l . i n

A,

=

#Ct" lo3)

=

'?'

5"

/oa

h;P' in

:

L=, .S+t

=

tg

in .

J= T.*= E( t ,oo\ I = l .S7og

gure=

+

=

f ie,s ' rq. t , ) ( re\

?

GJ

(n

z

x

t o ' ) ( t .s?o8

)

R,

olotl ion

o4

B:

Qt :

Q t / ^

f? -18q

R"l"*t

"n

"l

c

:

9u=

+?

fttn-7,q

i

L=

ef t :29 i " , .

] I / r ^ r r \ Y

e

1 ( J .

f J

J '

F

0 .+q lo t

i v r?

( 5 " t o 3 ) ( t { )

=

7 t -557Y o "

rn /

( t t .2 '

toR(o_r

?o

-

RJ

J;o^

"d

D

I

q)D

=

cp. t

CPr/.

=

-53.-f3o

y

lat,r

uI

(L

=

S.o7

n".l

T TL- r -

\ l

-

2 \ -

T - l

,"a

I

l-

(Pulc

-

-

=

\:rU

pRopRIETARy

MATERIAL.

O 20ll

The McGraw-Hill

companies,

rrc.All rights eserved.

o

part

of

this

Manual

may

be displayed,

eproduced,

or distributed

n any

orm or by

any means,

ithout he

prior

written

permission

f the

publisher,

r

used eyond

he imited

distribution

o teachers

nd

educators

ermitted

y McGraw-Hill

or their

ndividual ourse

reparation.

tudents

sing

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reusing

t

without

permission.



Problem

0.41

10.41

A torque

of

magnitu

e T

:35

kip

.

in. is

appliedat er-rd of

the composite

haft

shgwn.

Knowing

hat

he

modulus

of rigidity

is 11.2

x

l06..psi

or

the steel

and

3.9

x

10"

psi

for

the

aluminum,

etermine

a)

the

morimum

shearing tressn

the steel

corg

(6)

the

maximum

hearing

tressn

the

aluminum

acket,

c)

the

angleof

twist

atl.

|

=

8+t

=

?G

n.

f

=

3€v

bt

. l l . ; "

Stc.i

J i=Ea t=

Z . .S lc l

i 'nv

q-I

=

( t t

hvloc)(2.stg)

=

2g t8o

x

o'

Ib. in

Tlt?oe

e".,rynJ

t/

=1"1

a>re'l

I

=

Cdq

L

2f,in.

\ )u^ invn

ro.ohe- t '

,

=Hr :

l . l?5

in

,

cr .=

*J ,

=

15,* .

f*

=

5

G.*-

c,*

=

*( t .s*-

t . t2s*)

=

5,|1. lo

i* ' ,

A q = (9.1v ID 'Xr . T.3 o) = 2l ,?.o l / lD" l lo ^

Tor^1ue

c-e'.+iJ

by

Juî,. l,)w

j*.kcf

z

Tz=

q*"

fuz

T=T*-r"

=

(G,-I

+G*Jr)*

Steel

core

Aluminum

acket

28

3f

x

l6's

Tog.lB

x

to-"

voJ

;"

=8.cls

hi

G-)

Mo,orr'*rl.^

strêss

io.,

s*ea

ca,n,.e

t '=

G,Y

=

G,c

9,

=(r l

z* oL)( t . tzs)Oog.?grtdc)

= 8. q3x o

F"i

G)

l ' l "* iano*. t l rnt t

i" ,

o.(uînu^

i*=k*: l

Z

f t

=

GrT

E

Grcr f

=

(g.qy loc

(

r .s ) (zos

' )Ex ldt )

F

+.

1 .5

/O= ps i

g = Ly

T^t

=

L-lt

Lsi

<

= (qe\(TrS.TBxlo- ' ) =

6g.oq3y

to3 r*, f

@

s

3-

gt l '

PROPRIETARY

MATERIAL.

O 20ll

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McGraw-Hill

Companies,

nc.

All rights

eserved.

o

part

of tlis

Manual

may be displayed,

eproduce4

or distributed

n any

forrn or

by any nreuu,

witbout

thc

prior

writtcn

pcruririon

of thc

publisbr,

or

urcd beyord tl16 imigpddistribution

to tc*hcrs sd

educators

ermittcd

by McGraw-Hill

for

their

ndividual

cource

prcparation.

Students sing

his manual

are using

t without

permission.



Problem

0.42

L=

E+i=

?6 in .

9"p

=

L

S

r-,

o. .le*

v

J

u

*

rf

A+

9

ov e ^'ns

L q

A "*

&"

o^q^ o

ol

t-*f

'..

Q*=

=

c*-16vro

tn)

Q"tt

=

3-7

I'

PROPRIETARY

MATERIAL.

O 201I

The

McGraw-Hill

Companies,

nc.

All rights

eserved. o

part

of this Manualmay be displayed,eproduced,

or

distributed

n

any

orm

or

by any

means,

ithout

he

prior

writien permission

fihe

publisher,

r used

beyond he

imited distribution o teachers

nd

educat'orsermitted

y

McGraw-Hill

or

their ndividualiourse

preparation.

tudents sing

his manual r. uring t

without

permission.

10.42The composite

haft

shown

s

to be

twisted

by

applying

a

torque

T at

end .

Knowing

hat

he modulus

f rigidity

s

I1.2

x

106

.i

t*

iheiteel

and

3.9

x

106

si

for the aluminum,

determine

he

largest

angle

hrough

which

end

can

be

rotated

f

the

ollowing

allowable

stresses

re

not

be

exceeded:

steel=

500

psi

and

o6-;o,*

=

6500psi.

d =

LWlaH.

1te&

cove

:

t,A

=

ESoo

p.i

I I r f

A a I

i

l F

r h .

:

85

oo

GJt.*= ec*.*f

t-u

n

e^,J,

,n

Jer;o-|

G

c..*

tov

2jin.

Steel

core

Aluminum

acket

cw'

=

f ,4.

=

l '5

i ' -

Y

lo

c

,oJ / ; .

9.

L

l .

12.{ n.

tu_

L

0t

zv

o"

)

(1 .

?s

)

=

CSoO

7s

)

6 5 o o

(s -qx f

o ' ) ( r -s )

=

l - l l l l

=

61tt

.

Qo

x

lo-"

v-) /

;^

ry:

67't .6ox(o'

JI1' , ,

Lry

=

H4(ett.6oxlo-.)

L

Fto,-,;nurn

,ior*el

".

td t



Br^ss

.jo.r&at'

C2=

C,

+

t

r

0.6

+

o,L

--

o.8

in

Jr:

T(cr ' -c, ' )

=

*(o.g ' t-o.Gt)=

o-qgr

g2s

ir '

QJ.

=

(5. .6* lo tXO. lgctS2g)

=

2.469a

,< lyc

lb ' i r ' -

To

v+7

e

carri

e.4

by

6rwss

j

o,-kot

i

T,

=

G.f,"

+

ToIa,( tor?uer

T

=7,

+

T"

=

(G,q*A&)*

5xto3

L

T

i.fi

The composite

haft

shown

consists

f

a 0.2-in.-thick

rass

acket

Gt

*,

=

5'5

x

lO;

ptil

U*aJ

to

e

1.2-in.-diametcr

tecl

core

Gano1 l .2

x

106

si).

Knowing

hat

thc snan

s zubjccted

o 5

kip

.

in.

torques,

detc,rmine

a)

the

maximum

shearing

tness

in the

steelcore,

b)

the

angle

of

twist

of

I relative

o endl.

Sfeel

core'

c,=

ia

=

0.6

i, l

d

=

*c , ,

- - * (0 .6 )1

=

o .?oSS7S

n '

G,J ,

:

( t l .?x l06Xo.zogt7 f

)

=

2.28oo

r lon

, h-

n '

Torq,.r

c

z.rr;el

t/

=t"J

u)r1

{

=

C&

t

=

l.oSTl

v

ro-3

varl

;n

?.28ar

r.to3+

?.

1630

to3

(a)

t4

oec

r',.r

nr

s

I,

€c,

i,n+

-sf

es

s

i

"

b

nc"ss

j

aort#

;

T,o = G^T,* = errrf : (Ser106 (o.gX/.oftta to-" )

:

+-?Q

xlos

p" t

4.72

l{s,

(b)

M

av

i

r, r.r

^

S

,

€qn'nq

Si

ness

,

",

s

Ie*,| cort

?

f^*

=

G,

*

=

G,a,f

=

1.1r

to')(o.d(t.os.te.

"lo-')

=

T-O|x lo3

fa i

7. l l ks i

(c) finqle o$ *-is*t ( u = 6+l

q

=

Lt

=(za)(r-os+2"/o-3)

= 72 ,;)

75.9

r

lo-3

ro.4

+.gs

o

{


O 201l Tlp McGraw-Hill

Compenies,

rrc.

All rights

eserved. o

part

of this

Manualmay bc

displayed,eproduced

or distributcd

n

eny form

or by any mcans,

witlrout

thc

prior

wrificn

permission

of the

publislrer,

or uscdbcyond

hc limitcd

distribution b teeclprs and

edrrcators

ermittcd

by McGraw-Hill

for their individual

aourcc

prrparation.

Students sing his

manualare using

t without

permission.



Problem

10.44

The composite

haft

shown

consists

f

a 0.2-in.-thick

rass

ack*

(Gu'*

:

5'6

x

106

si)

bondd

to a

1.2-in.-diameter

teel

core

Grt*r

=

ll.2

x

106

si).

Knowing

hat

the shaft

s being

subjected

o

the

torques

hown,

determine

he

argest

angle

hrough

which t can

be t'wisted

f the

ollowing

allowable

tresses

re

not to

be

exceeded:

sog

=

15ksi and 6.n:

8

ksi.

T

Brass

acket

=

Gc*-.f

y-

LVr-/

-

9g

--

L

GT.-

1.2

n.

Steel

core

t"u

$a"

e4el\

n.-*ed

J

O.2in.

&C,'.**

h

Slee,t

cDr€,

:

T,u=

15

lcsj

=

lSUnpti ,

cr.r

=

le

=

0.6

ir,

r/o-3

v-.1

/;n

.l

-

lSorc

=

2-

2321

0l .?x lo tXo.a)

TÛ=

8 ksJ

=

Sooo

f";

)

Cr.r,

0,6+o.2

=

Os

=

t.1gs

7

x

k;s

ral

/i"

fur-

=

l.Tasl

rld\

v*t /;q

L

, r , t -

8

ooo

L5.6

osxo.e

Srn,

ll"n

Vo,.troe

3"u

grn

5

A//orJla

o-.l.f.

ol

**;s*

:

f=

6S[

=

1?

.f,\ t

0*

( L ) t t ?

I

=

=

Y o . ' r - L

ir,

.

(zeX

.?e

?

xto-3

--

tr t

.5?

lo-3

7 .37

0

r4J

-l

PROPRIETARY MATERIAL. O 201

The McGraw-Hill

Companies,

nc. All rights eserved. o

part

of this Manual

maybe displayed,

eproduced,

or distributed

n

any

orm

or by any

means,

ithout

lre

prior

written

permission

f the

publisher,

r used eyond

he imited

distribution

o teachers

nd

educators

ermitted

y

McGraw-Hill or their ndividual ourse

rieparation.

tudents sing his

manual reusing

t without

permission.



Problem

0.45

10.45

Two solid

steelshafts

G

=

77.2 GPa)areconnectodo a coupling

aist

f

rna

to

fixed

supports

at A and

C.

For

the

loading shown,&cnnine

(a)

the reactionet each

support

D)

he maximum

shearing ffeesn shaft

AB,

(c)

the maximumshearing

tress

in shaft

BC.

5

haft AB

r

G , T

T^"=ffiP"=

sha{t

tsC

O.

L:oo

qb=

l .{

x

los

-

?gc.

g+?

r'to"

g^

+

69,?

q

x

lo3

g*

CflA

+-GG57

v

lo-?

rô).

T*= (zsc.t ' l '? lot )( ' l . c657" to-t

Te"=

(eL.arg

v

lD'

)( .1.

6s?

x

to' )

@-\

Rea,r)

""s

at

s"f

f

oets

(b)

Mo*ivn;rn

sh

eo*ira

s*"ess

;,

A

B

:

d =

Lne' ,il

-

=

(c)

Ylorai-.'*

slr

ean'r,,

slhcss

in

BC

:

tao

=

.TLc.

-

(?q4-

?tt'

)

(

o--otq)

-

zT.

+, loq

po

J"n.

2ort-11

x

1e-t

PROPRIETARY MATERIAL. O 201 The McGrsw-Hill

Cornpenies,

nc. All rights escrved. o

part

of thi*,Manuol

may bc displayed

eproduced,

or distributed n any form

or

by

any

means,witbut

the

prior

wrifipn

pcrmission

of the

publisher,

or usedbeyond he

limited distribution o

teachers nd

educators

ermitted

y McGraw-Hill or their ndividual oun€

prcparation.

tudents sing his

manual

re

using t

without

permission.

l . t oSo6x los N-v ' ,

?qq.

q+

il.

t

T^=Tô=

Ta=

l;"

=

l lOS

N-

nn

zqs

N"

h^

to"=

15-o Yl?a-

<

t".=

27.+

MPa-

J

=

Ton,

L*

=

O.2oO*^,

C,

=

}

r l

,n, {Swr.a

O.O15v-,

Jre'

*.*

=

I

(o.o?s)t

=

G,3

si

v

t,-7

p,'

I

^

l r c L r a

Y a -

; -

15

\)rl

0

t-z

"

to')(e

3

51

x,o-q)

q"

=

296.

gq?

y

to"

ge

T

=

TLn

)

Le .=

O.XSOI ,

C

=

i t l

=

l?

nu t

?

O.OlTm

Je.=

* . *=

Ib-or?\ r

=

2.o+.7tx t6?

*

T

Gr

I17 .2x loq \ ( 1 .o ' t . 7 l " t dq \

G.-

tr

q"

=

E

63-?1{xlo

Eqr;

ibr;u^

l

coui l ine

sk

.

f

=

T^,

+

Tr.



10'46

Solve

Prob-

10.45,

assuming

hat

shaft

AB isreplaced

by

ahollow

shaft

of the

same

outer

diameter

and

of

25-mm

inner

diameter.

10'45

Two

solid

steel

shafts G

=

77.2

GPa)

are

connected

o

a coupling

disk

B

and

to

fixed

supports

at

A

and,

.

For

the

loading

shown,

determine

(a)

the

reaction

at

each

.support,

(b)

the

maximum

shearing

rl"r,

in

shaft

AB, (c)

the

maximum

shearing

stress

n

shaft

gC.

]L

7

T

=

7;s

,

Lrs

=

Q.T\Ow-t,

Cr=

25,",*r

=

O.

O?;fvq

Ct.=

f?.

Sr"rrn

E

O-OI?S

r.,r

(o.

oas*- o.o

tas+)

=

s7s,?43,

tt iq

rt, l

(tt-

2-v

o.

)

(s

?.r.

ztg

x

o-1

9"

=

22?.

9'/1

"

I

oz

Qa

. 2 ( )0

$r"

=

T(cJ-

,*)=

G

*l^t

Tne=

H9u?

sh"t_eg

T=I . ,

Lu .=

O.25ont

6=dJ=

11

' . , " .=

O.olq" ' '

4 .

=

E.u

=

X

(o

otq) t

-

lo+-?t

y

dt

wj

Eo

uil;

L,

un

ul

coopj

ne

l;sk

'.

T

=

Trc+Ts.

l . o8? - l

x

lo3

N- rn

3 l6

-La t

M

m

To"

Tr.

'

t o lo

N 'n t

=

3 lo

A/ .n

<

tM=

tt?.'t

t4Pc

tur=

28.s

M?a-

(b)

1.1

x

lot

=

2z.?-o.t, l

lD3

cpB

+

G3-2lyxl

o.

e^

Cp l6= .7O18* lo -= ro ]

Tn" : kzZ .o ' lq

x

t :3

(+ .q

78 ' ro

t )

=

G.

=

(e

s.2ry

"

uf)(r ,1o7x' / . ,

)=

(a\

Reootio^s J

srppo,ls

r

Cc)

-1 ,

=

I=

PROPRIETARY

MATERIAL.

O 201I

The

McGraw-Hill

Companies,

Inc.

Alt

rights

reserved.

No

part

of

this Manual

may

be displayed,

reproduced,

or distributed

in

any form

or

by any

means, without

the

prior

written

permission

ofihe

publisher,

or

used beyond

the limited

distribution

to teachers

and

educators

permitted

by

McGraw-Hill

for

their individual

course

preparation.

Students

using

this manual

ur.

uring it

without

permission.



Problem

0.47

10.41

At

a time

when

rotation

is

prevented

at the lower

end of

eachshaft,

a 50 N

.

m

torqr€

is

appliod

to

end { of

shaft B.

Ifuwiq

ilr*t

G

=

77

GPa or

both drofts,

detcrmine

a)

the ma:rimum

shearing

fess n

strsft

CD,

(b)

the

angleof rotrtion

at

.

Let

n

:

loryse

Trc = hryu"

Tcu

=

iorqre

S*o#ic

s

T1

TrB

-

Fh

=Q

T*-Fq=

O

=

€

ff^- Tl")

"g,.l:rJ

d

A

=

.foN,Mt

i" sh

"*l

Ats

i , ,

s [J+

CD.

=

t

g.g,tg

n

tci3

rccl

o.-171

So"

=3

Tie = 0.520 tl TA = (O.SzaTYSo = ?6-oa N.rrr

T;o

=

3(So

-?6.0a)

=

t | .qq

N. tn

(d

Ma,r,i-u:

slrcerint

slrcss

in

shaFl CD

:

t*=+S=*k=G,Xts.l3)

JcD TT

C

s

I

(o.

006

)s

(Ul A""t " "l r"tr*i"n *l A I

g^:

W

=

2r*,L=

6

Joe

ui.G

Cr.*

=

+*({3)*)r"

*l

(2

(rc.ol)

(

o.e

o

\T

??Yloq

{o.oo?.s)"

=

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form

or by any

rneans,without

the

prior

written pcrmission

of

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or

usedbeyond he limited

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crmittcd

by McGraw-Hill

for their individual

coune

prepEntion.

Studenc

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are using

t

without

permission.

Gear

A



Problem

0.48

t0.47

At a

time when otation s

prevented

t the ower end of eachshaft,a 50

N

'

m

torque.is

pplied

o endA of shaftAB. Knowing

hat G;77 GPa orboth shafu,

determine

a)

themaximum

hearing tress

n

shaftCD,(b)

theangleof rotationatl.

10.4E

Solve

Prob.

10.47, ssuminghat he 50 N'

m torque s applied o endC of shaft

CD.

Lat

Tc

3

'1ory,.

opp ;rJ

"t

C

=

5O

N'^,

'Eo

=

lereye

i,a

sl-a)+

CD,

TiG,

=

fong"e

ir,

slraf*

48,

T*s

-

f^F

=

o

I t-4" r:F=6

*(Ta

-T*)

0.

qq"

ac=Bg.

=Ago

=H

et=H=*W

'E

-T;

GJae

=((lf)t.+)J-.o ?r;

T.

=

(o-

+7?6

(rc

:

--

23'78

N

nn

-

23.

?8

)

=

31.

tt3

N'h.'

Tre

ft

(r.

r(17

*tot

Xo.oo75){

Geee

C

=

? O.

g71x

gs

=

l -

169 '

v^+ol

l*I

-

Gq.

/..lB ,

\ G " T

-l-

t

l c t r

-

l -

I A B

E

9.3

22

fi)2"

o.+7?6

?(so

2-l-

TT

CS

gl(aa'q4)

=

7r..?vroo

q

=

10.-)

hPa.

<

lr

(o.aa6

F

(zYsq.o3Xo.49)

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MATEnHL.

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n any form or

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permission

of thc

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or uscd

bcyond

hc

lirnited distribution

o

tcachcrr

and

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ermitted

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or

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rcpamtion.

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sing

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t without

permission.

G

6

l

Gear

A

(q,)

14

,ri',.r*'

s

hec,ri

i

stl^<s

in

sh.f+

CD

:

(b)



Problem 0.49

10.19

Knowing that the intemal diarneter of the hollow shaft shown is d

:

0.9 in.,

determine

the

maximum shearing strc*s caurcd

by a torque of magnitude

T

=

9 kip

..

in .

cz=

t .dr=( l l t .c l

3

o,8

n,

c

=

o.E

ho

cr

=

*C,

=

(*{(o.q)

=

O.{S

irr.

g

r

*(C,.t-

c,\)=

t(o.g*-o,tf

)=

o.slqo

iu+

dTc

-

r-hqx

f

\t

=

f2.{{}csi

{

o.s?al

o

Problem

0.50

10.50

Knowing that d

=

1.2

in., dete,l:nine

he

torsr€

T that cauts

a maximum

shearing

tress

f 7.5 ksi in

the hollow

shaftshown.

C

;

O.8

in ,

=

0.

1318

ina

(o.+gqg

7,sl

o.8

,

t ' t f

t.G)

o.8

rn.

+XLz)

=

0-6

n,

c,r)

-

+

(a.rr-o.g*)

ca

=

*J.

=(

Q=+J,=(

PROPRIETARY

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orm

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mcans,

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he

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f the

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r used eyond

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o teachers nd

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Problem

0.51

10,il

The

solid

spindle

^B has

a diameter

,

=

1.5 n.

and s

madeof

a steelwith

an

allowablc

shering

stres

of 12

ksi, while

sleeve

CD is made

of a brass

with

an

allowable

hcaring

tress

f 7 ksi.

Determine

he argest

orqueT

that can

be applied

aitA.

F

=

l/*

=

l(r.s)

=

o.?5

ir,r.

=

O . t ' l i f O l '

i n l

T.'

=

[(o.zs1'

d

u7tr4

-

o.?5

@.m7orxte")

=

Z

g.5?

Ap.i",

t

=

0.25 n.

ca=

L4"

=

t(so)

=

1,5

t

=

l-5-

C).25

=

l-25

in.

r -

c ,n

)

=

t

( , .s

{ -

l .?Sq)

*- t t7?

- ( t .

I tz j ( ) (z )

= tg .?13 fc ,p

l - 5

h[ r^tJh" va,lae

of

*oqre

T

is

*1.*

.r,ô,l/*.

T=

7-79

r,rp

rn

<{

in I

Problem

0.52

l0.SZ

The solid

spindle

AB

ismade

of

a

steel

with

an uiio*uUit

.it*ing

stress

oi

li

ksi, while

sleeve CD

is made

of a

b'rass

with

an allowa,ble

shearing

stress

of

7 ksi.

Determine

(a)

the

largest

torque

T

that

can

be

applied

at

A if the

allowable

straring

stress

s not

to be excedsd

in slceve

CD,

(b)

the

corresponding

required

value

of

the

diameter

d, of

spindleAB.

(a)

S.feeve.eD

l ,

' Cl,= E dt =

Cn

=

C" - t

E

f .5 -4 . ' L5 .

J

=+GJ-c,')=E(,-r'

T a

t^*o=

'S '

d

I

=

0.25 n.

r i

(

t?)

(?)( r .

o

G+)

t (g.o) = l-S in*

t

1-/(f

in-

- f . lS* )=

+.

1 t12

in r

=

t

g-

2,13 lt;p

*

f

=

lg . l i

k ip . . ,n

l<;y.n

Js

=

?.o l

i " .4

l=

C E

L

OO

GLI

in.

d"=

T-= Ja

(4-t,77Xz)

f r r

-

-

'

U'rr

C.

,. 5

Fo,"

eI

,;

I

,

L"J

,-

(10) s'.0.J spinJlc AB t T = lq.1 l3

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ermitted

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4



Problem0.53

(b)

Hollou

"l,aft

:

cl

For

eqoo. I

lna .sses

f0.53 (a) Determinehe torque

hat rnay

be applied o

a solid shaft of 90-mm outer

diameter

without

exceeding

n allowable

hearing

hess f 75

MPa.

(D)

Solve

psxt

d,

assuming

hat

he solid

shaft

s replaced

y a hollow

straftof the sirme

mass

and of 90-

mm nner

diameter.

(q-)

F".* les",0,Js l ,ot t r

c=

ie

=

+)(o,o?o)

=O.o+S'

*= {at

?

*(o.ol5)s :

t+3, t4

yto-s , ' , ,

Tôn=

F

, ' .

f

=

"2J

..

(zs"fo()(rra. l ,r lx lo-.)=

ro.z. /

loE

J.r

lo.-l

Ll

/cN.

rtn

{

=

*d,

=

*h

e

cross

(+

(o.

o?o

=

o.o41

n

Seo*ioutql aê4.s

haus+

be

e.lvnl

.

-

oF

Cz

=

lC,-

+

c'

17.7?

/o3

N.

t

??.9

kN.-

A

=

Tr

C'

=

rr(C;_

C,.)

r

=Jo.o4s '+o.o$sa

=

o.c ,6 '36916

4

*

(c^ t

-

c , t )

=

)?-s2g7

xtoc

u

ry==

-1

O-oC3€"a-6

4


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ermitted

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Problem

0.54

ll

Let

sh"f+

SI.ert"

*?P

- I

?.

o '

f

=

x

or

{ :4/

, . "

Y

y, i * ,os f ,ans

"* " rn* , 'ng

* r i lL

Xo

:

l .o .

.*- [m

:

l-ztr

("o., .rJ

/)

Fo.n

e1u^)

slnesses

2T

T\

,*

x{

2er

x

:

* :

t .?? l

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righb reserved.

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part

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or distributcd n

any form or by any rncaDs,

itlrcut

thc

prior

wriscn pcrmission

of

the

publisher,

or uccdbcyond hc limitod disfibution

to teachcrs nd

educators

ernitted

by McGraw-Hill for

their ndividual

cours€

prcparation.

Students sing

his manualare

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t without

permission.

10.51Tryo

solid

brass odsAB andCD a*E rsa€d o a brass legve

EF. Deterrnine

hc

rctia

djdl for

which the samcmaximum shccring fiess occurs

n thc rods and

n

the

sleeva

J,

r,

rz



Problem

0.55

10.55

hc

aluminum

od

AB

(G:27

GPa)is

ondd

to

the

bress

od

BD

(G=

39 Gpa).

Knowing

hat

portion

CD

of

the

brass

od

is

hollow

and

has

an nner

diameter

f 40

mm,

detemnr4e

he

angle

f

twist

atl.

Tn

=

16ffi

N

.'i

R"d

AB:

G

=

27r

o r %,

|

=

o . Ioo

T

=

8OO

N - y v r

( = J * d

=

O . O t g

P""t

BC:

G

Y4

m

f=

8oo

QEt.=

#

TL

(?'too1(o.e5o\

- ;

GJ

(37,r@

=

15.o68*td3r^4

**,ist

*l

A

:

9n

=

PorcrQet . *

Q. ,o

=

l05.o8o xtD-3

vual

C=*"1

=Q.C3on

{ 'c*

=

E

(o.

os6)e

=

t-27L3.(

"1dt

vn.s

l8-

lg7

Y

o3

,ne

Qt

=

6 .02 " 4

=

31

r

f

oo

?o

L=

0.375 v-,

,

f

l |oo

=

?Lloo

U.^,

_J-

=

, d2,'to"\(?-gts)

.-

=

(gqx

roe

Q.zt

?g,l

lc.1

Pc. ' t

D

1

c,

=

t) ,

=

o-

o2-o

,^

ct=*.dr= o.o3o,,- ,

L-- o.Lso,vn

* f

=

Ekr r -c , t )=

4

(o.

ogor -o.o lor )

=

l .oa,ou,11

.

,^^?

Q.rc=

'An),

"f

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orm

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rd

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ermitted

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reusing t

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Fo*.

St*f,es

:

A.0 ".*b.0e va,0uq

"l

f^

is

*he

sôlls.

n

:

O.t|Zg

R:p.in

10.56 n

ttre bevel-gear

ystern

shown,et

=

18.4311

Ifuowing

that

the

allowable

shearing tress

s 8

fsi

in

eachshaft

and

hat

the

system

s

in

equilibrium,

determine

the argest

orque

T,a hat

can

be applied

at .

5h4ft

Ai

T=

8

ksi

c=*d

=

O.tf irr

f^

=

$

=

Lti-

=

T

Co-zs)3(a)

O.nc3f

{n,;n

? f =-B ksr ' c=14 = O-gr2^fn

T

=

*

=

t

.t

(

=

*

to.gnf)t(e1

=

o.

g83f

kip.in

\=ftn

=

o-

tlT' lq

k;7.in

Problem

0.56

0.5

n.

A

3+

A.

Vro

Sh4P+

Ts'o

177.8

k|-ln

f,t

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ermitted

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ource

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Problem

0.57

At fhe, oUlu"Ll" furrsf a.n7lc Si^ I =

9

=

3.8226

*

=

O.

616;67

6

,no)-

l-

=

Pa

cos

p

=

0oo

)( /5)

"*

3-821,6o

lL+16-7

]L.;"

10.57

he solid cylindrical

teel

od BC

of length

L=24 in.

is attached

o the

rigid

lever

AB of lengtha

=

15 n. and o

the support

t C.

Design pecifications

equire hat

ttre displacement f I not

exceed

in. when

a 100-lb

force

P is applied

at l.

Dete,rrnine

herequired

iameter f

therod. G

=

Il.2

x

105

si

and?fr=

15ksi.

=o-o&s1

BoseJ

on

*-,sl

:

d--L =

Y6J

(n ' t )

n \ l r r r a ; r )

(o-out

t

6

)

Bcrsed

or1

5fress

| t' =

7-c.

- t - $

?TL

n{ =

ZTL

ffi,t

..

L'

lrdg

go.&3* t ie : :

c=

o-+ l83 in.

:#-.'-

ct=

#

Ar

L - t 5

=.2c

c3=f{ffi = c,s-sz^xto-'i,.' c

Dse

Jo"ge.

vJu.

f.r.

,lesi9

r,

C

=

O-tf

83

in.

d

( t= tsooop t i )

o .3 f f i in ,

0-937

in.

-

I

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the

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Students using

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Problem

0.58

10.5E

Two

solid

steel

shafu,

each

of

30-mm diameter,

arp connected

by

the

gffis

shown.

Ifuowing

that

G

=

77

GPa,determine

lte

angl€ hrough

which

end

rotates

when

a torque

of

magnitu

e T

:200

N.,

m is

applied

a,t

Colc,r

*fio^

o$

loqoes

C tcrr^|ere^tio conl,.ot lor.e belwe<n

9o

6

(eoo)

=

3oo

Maa

T";sf

,"

sl'ef*

De

;

Jo ,

=

E^ct

=

* (o .0 ls ) r=

z1 .S22r lo - thnu

( 3oo

) (o . s )

n,I +cl-y

/o-t

r^J.

(tz,t lo

e

Xz?.sre

,o-t

)

R"+

Iio^

"t

O

;

A"

Cbcunl",-u,fi

ol

Jislr,l,'ee-ê-,t

*l

Se..t

c;

tules

:

S-

q4"

=

luQu

Ro+J;,.

ts:

:

ft o"

:

x Ql .+q7vb'

=

36-71-f

lci'

rnot

Tvis i

in

sh"ft

A8:

L .u

=

O. f

+

0 .2+

O- t++

O-2 :

O.q n ,

(2ou

(

o-z)

J^s

=

71.

522

,

1o-'

-

A?.

9?6

x/, is

raJ

^r

:

T^qL$

=

GS'

(77

x

ou

X

7?.

St2"

ro-

)

Q^

=

Qs

+

cl*s

o*

f

io,^

o-t A:

= 36.?{ fv to-3+ 21,g1( r lo ' r -J : 66.1, (v tosnJ

;

g_71o

{

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o part

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Manuar

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eproduced,

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dishibutcd

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or

by

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.fu.

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*tirti p-t

i*ion

of

the

pubtisher,

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uscd

beyond

be imited

dishibution

to

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andducators

ermitted

y

McGraw-Hitt

or

tleir

inaiviaul

cou"" p-r"p"[ri"".

students

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manual

re

using

t withoutpermission.

0.2 m-



Problem

0.59

sha$t

AB

f = l ie , Lm=O.6rn, C =tC = O.Olf ,vn

J^-

=

T.c*=

$(o.olS)t

=

11.52x,o-1

,,u

1-

=

Io,

L*=

O,i

m,

c

=tCe

e.ol8n

r

J*=

TC*

=

{(o.orE)t

.t

= |

6?.

816

x

o-1

tn

.1

s

Gr

J-

",

-Qn'ror)(

tc't-s1(,lo-t

"u=

ff%

='

V.

M

#oh;'

3

**.,lio.

o,l

ll,

c

F.l

"h

3

c, ge

=

c/,.

:

g

T"la

I.,r,tvc

o.'t

S,/ong..

J.

=

Tû+T"o

=

Soo

N-rne

5oo

=(to.Tosx losr

r . l

toEtot )g

."

g=?o.s6sx/orr . , " l

T ie=

0o-2of" /o3)(zo-sc5' lot )

=

2o1.87

N-n

TLo

=

(lr .

fog' tot)(2o.sss

r

lo-t

)

:

z io,.

3

N-

r. ,

Maari*, ". shearJrne

Strtrs

iv?

AB

:

10.59

Two

solid

stcel

shafu are

fitted with

flanges

hat are then connecteO

i ntt.a

boltc

so

fr*

thcre is no relatlve rotdim betrveen he flanges.

Knowing thnt G

=

77

GPq d€fi€rmine

hc maximum

hearirg

sEessn erch chcft

when a torque

of magnitude

I=

500N

I

m is

applied o flangeB.

-

tn=

Q.g

:

(2?r.g-?X9=ors)

=

si.sqttos

4

*Irs

7ot.f

2rlo-

q

Morai*,1'n

LcarinT

s{r.erg

i* CD

'n

E

Tilc

:

(?ta.rg)(o.ora)

-L

16+'896

lo'q

3?,€

HP+

G

=

3t .€?y lo t

?o

3l .7MPa-<

PROPRTETARY

MATERIAL

@

201I

The

McGraw-Hill

companies,

rrc..All

lshts

re,s9wed.

o part

of this

Manual

may

be displaye4

eproduced,

or

distributed

n

any

form

or

by any

mearu,

without

le-prior

writLn

p".rnirrion

of

the

publisher,

or

used

beyond

he limited

distnbution

o

tcachcrs

end

ducators ermitM

by McGraw-Hill

or

their

ndividual

ouo"

ptue"oltioolstudents

sing

his

manual

reusing

t withoutpermission



Problem0.60

10.60

The

steel

acket

CD

has

been

attached

o

the

40-mm-diameter teel

shaft E by

rneansof rigid flangeswcldedto the acke md to therod. The outer diameterof the

jacka

is

E0 mm and ts wall

thickness

s

4

mm.

If 500

N

q

m torquos

are appliedas

shown,

etermine

he

maximum hearing

tress

n the

acket.

5".0rJ

slra$

:

J-

=

{.*

C=+A,

=

O.O?O

a^

=

T(O.ozo1t

25

33

to'I

Yn

Jarlat

,

Cz=

te

=

O.Or+O

Cr=

e2. L

=

Ji

=

5

(cr*-

,*

=

I

@.o+o'

o.ag6"

=

C.Ot lo

-O.OO.I

=

O.036

14

,<

t54

m+

To*J

|"nqu.

:

7

=

E

+

Tr

=

(J"

+

J") G

A/L

T,= Jb= T=

(LsBeg*to*)(5oo\-

'J

Js+Jr

r -

M

arai

"n

u'.rr

She,,..,

yl-

S*rreSS

in

.JAclcf

t

=

lE"_

=

(tl.ag.

Xo.:'tg)

r

J.

'

l.3saq

x

,o

-T**

1,3871

G+

g/L

G+

QIL

.'.

Eg

=

-J-

L

Js+J"'

= +2,3. t lJ . lv t

tL.?+v/oo

a t2.24

t4Pq

<

Documents

Statics and Mechanics of materials- Chapter 10