statics and mechanics of materials- chapter 10
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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 $'c urrl it witM
pcrmirlbn.
8/10/2019 Statics and Mechanics of materials- Chapter 10
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probrem
0
lfl*J"di$*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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MATERIAL'
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rrc.
All
rights
eserved.
o
part
of
this Manualmaybc displayed, eproduced,
or
distributed
n
any.fo11
o1
by
any
means,
withou
tlrc
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",ritlrn
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or used
beyond h€ imited distribution to
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y
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ou^" pi"p"lttn.
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(
J
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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]
=
$r
c;
=
TTcz
c?=&c
=
fr.f".s?,t)=
o,433orgi,n,
?3
Cr
3
*
ar.
=
O.
216s0g
in,
J
=
{(c"*-
€,t)
=
{
(o..+
Svrtg+
o-
?r
6s06+),
0.os\7?a
int
r=ry=W
PROPRIETA*Y
MATERIAL. O 20ll Tlre McGraw-Hill
Comprnkx,
trc.
All
rights cscrved.
o
part
of
this Menual
may
h disphycd
rcproduccd,
or distributod
n eny mm or by any mcaru,without thc
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wrificn
pcrmiecion
of the
publisbr,
or used
bcyond
hc
limitcd
dicfibution
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atd
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crmitted
y McGraw-Hill or their ndividual ourse
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sing his manual
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Probfem
O.T
l0'7
The
torques
shown
are
exerted
on
pulleys
A, B, andC.
Ifuowing that both sheft$
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
MATERIAL.
O 20ll The
McGraw-Hill
Companies,
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All rights cserved.
o
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.
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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.
o
part
of this Manual
may
bc displayed,
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orm
or by any means,
ithout
he
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r used
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imited
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pdrrnalnrc
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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.r^e
vq]oe
",f
sl^ers
-
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,
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MATERIAL.
O 201I
Thc McGraw-Hill Compenics,
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eservcd.
o
part
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Manual
may
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displayc4
tprcducod,
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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.?;.
r^E
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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eserved.
o
part
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14nn*,
*",
o,
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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
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sing
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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.
O 201
The McGraw-Hill
Courpanies,
nc. All r ights
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
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ermitted
y
McGraw-Hill
or
their ndividual
ours€
r€paration.
tudcnts sing
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re
using t without
permission.
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Problem 0.10
10.16 r turhlc *ctoig ffis tu 0 fi$r in &r hn 6d f ed ffi'ffir tueE
nldn
Fd
tr" l(rronkg
&m
a
frryr sfqffi f-
l1,S H
r;
ru l rppfld
n
"{,
dcitmbc
&r ruquird
diamwr of
{n}
rd*I,
{*f
matr
*f
=
Tar
(aJ
RaJ
AB:
L
=
L€-rSrlo-3
r,'r
=
2fi.
f5,
h^h^
dfr=
R.e
=
5O.S
rr",* <
r
T^+=
F
n _
2 T
\'
li-z;*
=
I5-7tS
,,
ldo
t*tt
rR€/
Tr
zsx
to.)
c
=
3[.€qxjcr
-
r.a
=
3l-€4 an
du.=
?
e
=
63.tf
*rn
<*
F'X ffiTfty lltTf,'tlAL O 2011 Thc McGrffi-Hill Coryroia,
hp. All
righfir
rffivld. No
pN'rt
f 6is
Mrnnl
my bc 61ryh1q4
rqil0erc64
or ffi h
ry ful r
V
sl
nr',
?tu
tb
Fio.
?rfr
Firh
of
fu
p*trfr,
a
* b.]Era
L
ffi d|rfrlbi
n
rj*a
f
d|cr,m*r
prrnittad
by Mc{tsw-Hill
fsr
their fudivkl|trl
ccrrc
puprr*ion.
Sndcm ruing frie
rnrmal rl
Blng
it
wffurn
paminirn.
'
8/10/2019 Statics and Mechanics of materials- Chapter 10
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problem
0.17
**
P;.'ll'$"*?*:*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
d^u
--
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'
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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$a)
c- --ff i = o'o2129'lt^r
C
=
O.3olo5 in ,
do .
=
?C
:
O.GO1-
n.
<
Sh',+l-
FGH
e
=
9)f
'T' l
=
o.o727s-?
,3
TI
(f
o5ao)
C
=
O.
t+
71 ir,.
dno
=
?C
--
O.BS5
n,
PROPR.IETARY
MATERIAL.
O 201
The McGraw-Hill Companies, Inc.
All rights
reserved.
No
part
of
this Manual
may be
displayed,
reproduced,
or distributed
in
any
form or by any
nreans,
without
thc
prior
writtcn
permission
of the
publishcr,
or
used bcyond
thc linitcd
distribution
to teachn
ard
wi
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Problem
0.21
Probfem
0.22
f0.21
A torqucof
magnitudeT=
000Nr rn s
+ptiduD
asshown.
Knowing hat
the
diamcer
of
shaft AB is
56 mm and that
the diamaer
of shaft
CD
is
42 mm,
detennine
hc
maximum
hearing
tressn
(a)
shaftl.B,
(b)
shaftCD.
T"o
=
foOO N.n^
T^.
f;.
-i;o
=
Jff
(rooo)
?foo
N.m
e
=
Ia
=
o.o?B
4
1-zF
:
??.50
x
lD6
7?.5
MPa
.-
ShacF+
C:
C- - *d
=
O.o?O
h ' l
/= F= +*=ffi= 68'7*to
68.7 t'l?a
.G
10.22A
torqueof magnituden-
1000
N
;:m is
applied
at
D as
shown.
Knowing
hat
the allowable
hearing tress
s 60 MPa
n each
shaft,
determine
he
requircd
diameter
of
(a)
shaftl.B,
(D)
shaftCD.
T.o
=
looo N'q
lie
=
t T;
=
ffClooo)
=
?wlrl rryr
6JSh4++
AB
tdt
= r.o
xlos
Pa
t-- Lg
=
2I
.r=?T-
{
- l lcs
=
Tt
C
3
?q.82
x
6r
:
21.?2
)nha
T
:
II.XXJ
t
.
rtr
D
G
Xlsoo)
rr eoxro")
d
=
2c
n.C.SA6r
D-
6
h4
=
5?.6
vnrl
2T
Trc
a
lo-3
,q,
10.610 lo- ' ln3
+3.
I
vn
?)
0$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.
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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.
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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.
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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
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it
'i&nt
p'brbn
8/10/2019 Statics and Mechanics of materials- Chapter 10
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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-
$tdGffi ucng
hb mnuel arc
rriry t
wifror*
prmirrftro
8/10/2019 Statics and Mechanics of materials- Chapter 10
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ProMm
10.28
I
000
_i
t
Q= I I
JGir
( -
I
I-s
T
\l
I
L
f rc., = (l)(tTr) = l?. 566
-fooa
ff
=
6oooo
;n.
y.nd , Q,= *d - - t f -O in . ,
f=
6o
o
oer
=
7-38t6rfos
ps,:
=
?38
ksi
{r*
fROFnIETfRy
MATEruAL.
O Z0ll
ThcMcOnr-HiU
Coqnicr,
IEE. ll righirruronnd.Hoprt of thirHrmd
ryy Ft
di+ry.4 '@
r ffi
i
ry
lr
c ly
lf n,
rtrt ln
f.L
*,rti.
frlloo
oe U
f*firUr,
or urd
bclm{d
6. licittd
dl*rfoatiol
to
brEb3E
e&rato6
Fnr.itt6,i
by Mccni-Hilt
ftr fuir idividrul
courrc
cn*i*i.rn
Str&nu
unnc
hir
rnffitl
EEulrng
t
rifuu
pcrbin'
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Problem 0.29
10,29
The
orques
hown
are exerted n
pulleys
A andB. Knowing hat
he
shafts
are
solidandmadeof steel G : 77 GPa),determineheangleof twist betweena) A and
B,
(b)
A
and
C.
T{
'r,r
30 mm
= - l O ( ) \ . n r
(o)
1le
=
3OO
N-.,
L^u
:
c).Q
''
r
Jr"
=
*(o.o,s) l
=
71.5?a
,o-1
w+
90.=#
=
Qre
=
C3.
=
{d
- 3 '
x
lo
14,
r
1gs
por
=
C^u.
al
=
O-otS
:
4?.
o?S
lo
t
rud
?.530
d
=
O . O { g
k } r
3 . {?o <
(b)
Ti"
T
I JBA
-
Qa.=
A
I
0.9
m
I
8 l
I
0.75
m
=
3OO
+
Ll}o
=
?OO
N.h,
Lsc
=
O.7Srv,r
,
*
t
o .o lg) t
=
lgq.s?gv td?
t
F$.'
=
,= '?o= 9:rt)=
.
=
ts,
il
GJ".
-
QZxOtffi)
-
9A.
=
%.
+
(ee.
=
g1
606
PROPRIETARY MATERIAL.
@
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
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publisher,
or used bcyond
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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.
€) 201
The McGraw-Hill Companies,
Inc. All
rights
reserved.
No
part
of
this
Manual
may be displayed,
rcproduced;
or dis6ibuted in any
form or by
any means,
without the
prior
writicn
permission
of the
publisher,
or used
beyond
the
limited
distribution
to
tcachcrs
and
McGraw-Hill
for
their individual course
preparation.
Students
using
this
manual
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"
PRoPRTETARY
MATERTAL'
o
20ll
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Companies,
nc..All
rights
esewed.
o
part
of this Manual
may
be displayed,
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
educators
ermitted
y
McGraw-Hill
or
their
ndividual
our"" pr"p#tion-
Students
sing
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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
PROPRIETARY MATERIAL. O 20ll The McGraw-Hill
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nc. All rights eserved. o
part
of
this
Manualmay
be
displayed,eproduced,
or distributed
n
any
orm or by any npans,witlrout
he
prior
writtenpcrmission
f
the
publishcr,
r used eyond lre imited distribution o tcachcnand
educators
ermitted
y McGraw-Hill or their
ndividual
ourse
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tudents sing
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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"
.{
PROPRIETARY
MATERHL.
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The
McGrarv-Hill
Companies,
nc.
All rights
eserved.
o
part
of this Manual
may
be displayed,
eproduced,
or distributcd
n any
form
or
by any means,
withou
the
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writln permission
of the
publisher,
or
usedbeyord
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distribution
o teschcrs
ard
educators
ermitted
y McGraw-Hill
or
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ndividual
ourse rcparation.
tudents
sing
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.,
uring t without
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0s)
(o.gao'1
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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{"
PR0PRIETARY
MATERIAL'
o
201I
The
McGraw-Hill
Companies,
nc.
All
rights
eserved.
o
part
of
this
Manual
may
be displayed,
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"=Tc^u*=*(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
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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
,^o4,
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<
PROPRIETARY
MATERIAL.
@
20ll
The
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esewed. o
part
of
this Manualmay
be displayed,eproduced,
or distributed
n any form
or by any means,without
tbe
prior
writtcn permission
of the
publisher,
or usedbeyond tre imited distribuiion
to
teactrenc16
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ermitted
y
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ndividual
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Co.l
cu,
h
on
ol twqves
Tan
qe^Iiol
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
+"
PROPRIETARY
MATERHL.
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All rights
eserved.
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part
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displayed,eproduced,
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disfiibuted
n
any form
or
by any
means,without
thc
prior
writtcn permission
of the
publisher,
or
usedbeyond
1e imited
distribution
to teachers
nd
educators
ermitted
y McGraw-Hill
or
their
ndividual
oursie
r€paration.
tudents
sing
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manual
rc uiing it without
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. t
I n
l
=
29 .7 t3w(o - t
r * J
( t t . lxoc
)GLo(3
x/o 's
)
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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
PROPRIETARY
MATERIAL.
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McGraw-Hill
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escrved.
o
part
of
this Manual
may be
,lisplayed,
eproduced,
or distributed
n
any orm or
by any
means,
ithout
he
prior
written
permission
f the
publisher,
r usedbeyond he
imit€d
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ermitted
y McGraw-Hill or
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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
MATERIAL.
O 20ll The McGraw-Hill
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All rights esewed. o
part
of this Manualmay be displayed,
eproduced,
or
dirtibuted in rny
form or by any mcans,
without
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prior
writtcn
permission
of the
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or usedbeyond he imited distribution
o teachers
nd
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ermitted
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Students sing
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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.
O 201I
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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
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teacbrs nnd
educators
crmittcd
y McGraw-Hill
or their ndividual ours€
ttpsration.
Students sing
his manual reusing
t without
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|
=
?,
h i
=
ixlDs
ps, '
r=F=
#
"=W
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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
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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
Ju^i,. 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.^
str^ess
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.(u^inu^
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.
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eserved.
o
part
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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
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for
their
ndividual
cource
prcparation.
Students sing
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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
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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
{
PROPRIETARY MATERIAL.
O 201l Tlp McGraw-Hill
Compenies,
rrc.
All rights
eserved. o
part
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Manualmay bc
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eny form
or by any mcans,
witlrout
thc
prior
wrificn
permission
of the
publislrer,
or uscdbcyond
hc limitcd
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ermittcd
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aourcc
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Students sing his
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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)
T^U=
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.
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permission.
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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-?
r^o).
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^=T^o=
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.
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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.
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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)"
=
PROPRIETARY
MATERIAL.
O
20ll The
McGraw-Hill
Companies,
nc.
All rights
eserved. o
part
of this Manualmay
bc displayed, eproduced,
or disributed
in any
form
or by any
rneans,without
the
prior
written pcrmission
of
the
publishcr,
or
usedbeyond he limited
distribution to tcachcrsard
educators
crmittcd
by McGraw-Hill
for their individual
coune
prepEntion.
Studenc
using his manual
are using
t
without
permission.
Gear
A
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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)
PROPRIETARY
MATEnHL.
O 201I
The McGraw-Hill Companies,nc. All
rights
eserved.
o
part
of
this
Manual
may be
displayed,
eproduced,
or distributcd
n any form or
by any
rrcene,
without
thc
prior
writtcn
permission
of thc
publishcr,
or uscd
bcyond
hc
lirnited distribution
o
tcachcrr
and
educators
ermitted
y McGraw-Hill
or
their ndividual ource
rcpamtion.
tudents
sing
his
manual reusing
t without
permission.
G
6
l
Gear
A
(q,)
14
,ri',.r*'
s
hec,ri
i
stl^<s
in
sh.f+
CD
:
(b)
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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
MATERIAL.
A 20n The
McGraw-Hill
Companies,
nc. All
rights eserved.
o
part
of this Manual
may be displayed,
eproduced,
or disFibutedn
any
orm
or by any
mcans,
without
he
prior
written
permission
f the
publisher,
r used eyond
he imited
distribution
o teachers nd
educatorsermitted y McGraw-Hill or their ndividual ourse reparation. tudents sing his manual reusing t withoutpermission.
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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,^o,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
PROPRIETARY MATERIAL. O 20ll The
McGraw-HillCourpanies,rrc.All rights eserved. o
part
of this Manual
mey be displayed,
cproduced,
or
distributed
n any form
or
by any means,without
thc
prior
written
permission
of
the
publisher,
or
usedbeyond he limited
distribution to
tcrchcrs and
educators
ermitted
y McGraw-Hill or their nd ividual
ourse
reparation.
tudents
sing his manual reusing
t without
permission.
4
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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 , ' , ,
T^on=
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 a^e4.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
PROPRIETARY MATERIAL.
O 201I The
McGraw-HillCompanies,nc.
All rights eserved.
o
part
of
this Manual
may be displayed,
eproduced,
or distributed n any orm or by any means, ithout hepriorwrittenpermissionf thepublisher, r usedbeyond he imiteddistribution o teachersnd
educators
ermitted
y McGraw-Hill or their
ndividual
ourse
reparation.
tudents
sing his manual re
using
t
without
permission.
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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
PROPRIETAR.Y MATERHL. O 201I TheMcGraw-Hill Conpanics,
rrc. All
righb reserved.
No
part
of this Manuel may
be dispbyed, rc,producc4
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
using
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
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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
PROPRIETARY
MATERIAL.
@ 201I
The
McGraw-Hill
Companies,nc. All rights eserved. o partof
'\is
Manualmaybe displayed,eproduced,or distributedn
any
orm
or
by any
neans,
witboulth€ prior
writien
pcrmission
fihe
publisher,
r used
eyond
he imited
distribution
o teachers
rd
educators
ermitted
y
McGraw-Hill
or their
ndividual
ourse repration.
students
sing
his
manual
reusing t
without
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Fo*.
St*f,es
:
A.0 ".*b.0e va,0uq
"l
f^
is
*he
s^olls.
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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r usedbeyoud he imited
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ermitted
y McGraw-Hill
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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
PROPRIETAR.Y
M.ATERIAL.
O 2011 The
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Companies,
Inc.
All rights reserved.
No
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of
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be displayed, reproduced,
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in
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or by any
means, witlout
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written pcrmission
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or used
beyond tlre limitcd
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ard
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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-^e-,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
{
PROPRIETARY
MATERTAL'
o
2011
he
McGraw-Hitl
companies,
nc..All
r1s-tts
e19.ryea.
o part
of
this
Manuar
may
be disprayed,
eproduced,
r
dishibutcd
n
any
form
or
by
any
mcans,
without
.fu.
n.lo.
*tirti p-t
i*ion
of
the
pubtisher,
or
uscd
beyond
be imited
dishibution
to
tcachcn
andducators
ermitted
y
McGraw-Hitt
or
tleir
inaiviaul
cou"" p-r"p"[ri"".
students
sing
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re
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0.2 m-
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Problem
0.59
sha$t
AB
f = l ie , Lm=O.6rn, C =tC = O.Olf ,vn
J^-
=
T.c*=
$(o.olS)t
=
11.52x,o-1
,,u
1-
=
Io,
L*=
O,i
m,
c
=tCe
e.ol8n
r
J*=
TC*
=
{(o.orE)t
.t
= |
6?.
816
x
o-1
tn
.1
s
Gr
J-
",
-Qn'ror)(
tc't-s1(,lo-t
"u=
ff%
='
V.
M
#oh;'
3
**.,lio.
o,l
ll,
c
F.l
"h
3
c, ge
=
c/,.
:
g
T"la
I.,r,tvc
o.'t
S,/ong..
J.
=
T^u+T"o
=
Soo
N-rne
5oo
=(to.Tosx losr
r . l
toEtot )g
."
g=?o.s6sx/orr . , " l
T ie=
0o-2of" /o3)(zo-sc5' lot )
=
2o1.87
N-n
TLo
=
(lr .
fog' tot)(2o.sss
r
lo-t
)
:
z io,.
3
N-
r. ,
Maari*, ". shearJrne
Strtrs
iv?
AB
:
10.59
Two
solid
stcel
shafu are
fitted with
flanges
hat are then connecteO
i ntt.a
boltc
so
fr*
thcre is no relatlve rotdim betrveen he flanges.
Knowing thnt G
=
77
GPq d€fi€rmine
hc maximum
hearirg
sEessn erch chcft
when a torque
of magnitude
I=
500N
I
m is
applied o flangeB.
-
tn=
Q.g
:
(2?r.g-?X9=ors)
=
si.sqttos
4
*Irs
7ot.f
2rlo-
q
Morai*,1'n
LcarinT
s{r.erg
i* CD
'n
E
Tilc
:
(?ta.rg)(o.ora)
-L
16+'896
lo'q
3?,€
HP+
G
=
3t .€?y lo t
?o
3l .7MPa-<
PROPRTETARY
MATERIAL
@
201I
The
McGraw-Hill
companies,
rrc..All
lshts
re,s9wed.
o part
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Manual
may
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eproduced,
or
distributed
n
any
form
or
by any
mearu,
without
le-prior
writLn
p".rnirrion
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publisher,
or
used
beyond
he limited
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o
tcachcrs
end
ducators ermitM
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or
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ndividual
ouo"
ptue"oltioolstudents
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Problem0.60
10.60
The
steel
acket
CD
has
been
attached
o
the
40-mm-diameter teel
shaft E by
rneansof rigid flangeswcldedto the acke md to therod. The outer diameterof the
jacka
is
E0 mm and ts wall
thickness
s
4
mm.
If 500
N
q
m torquos
are appliedas
shown,
etermine
he
maximum hearing
tress
n the
acket.
5".0rJ
slra$
:
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
<