Problem
8'1
8.1Two solid cylindricalrods l8 and BCareweldedtogethertt B andtoaded
as
shown. Ifurowingthat d; = 50 mm md d2= 30 mm, find the avcfiage
norrnslstressat
thcmidretim of (a) rodlf, (6) rodBC.
= 70
3@ mm
ILU = 7D ,t lD= N
(So)t = t.q#s*1o3-,,'l= t-ct635' /o"'^'
'>f*fl Di ' g=
3 5 - 7 x l o oP o
6e = 35't
250 mm
MPa I
= 3a* lo3 N
, (A)" = 706 -86 n,.' = lo?-96x lti' -"
loxlo3
- = +Z-4ylo"pa
> 6,gGr l o' e
Probrem
82
Oe. " +2.4 MPa. 4
3*ff'tri'ff1'',tffi:;*#3#"fi#:,Tlti::?:'ls.'1flffi$if#
:"'
rod, determinc the smallest allowable values of d1 and d2.
= 10 klJ = '?O >tto3 hJ
300 mm
'-
= 'lP
),"
Ifd,^
= 2^q
A o2,xlo-3r'
-4
-@
1m
d, = 2S'2
250 mm
h,.|4 'J\
= 3 o * 1 o sN
P-=-tr-
trd]
r/TF..@..f
d.=/#*"
Id
TTA;.-
vh
= rc.sTxto"
^n
d " =1 6 - 5 Z
pR.opRIETAR.y *|ATERIAL O 20il The McGraw-Hill comprnicl lnc. All nghts rcscrved.Ho prt of thir Mrnual.ryv-F dbpla]rcd' rcpmduccd,
the limited dlcfrution to tcechcn rd
or dictributcd in rny form or by any ncrni, without the prior r".itlo pcrmimion ofihe publisher,or usedbeyond
it without p€rmislbn'
e&lcatorr pcnnitrd uy u"ct *-ttitt for thcir individual courseprcpontion. Strdcntsusingthis manualareusing
<
--F
PrcbremB.3f###trf#fl#ffi,T#'l*.s*:,ti#iffi#
BC.
8.3 Two solid cylindrical rodsAB andBC are weldedtogetherat B and fo"A"O*
shown. Determinethe averagenormalstressat themidsectionof (a) rodl8, (b) rod
BC.
(@) R"J AB.
P = {o hps
(*e.n:io". )
A."={4=
rl
rlal-=3.1.11
-^n
T
OAn
o.e=fi-effiZ
= t?.'7
(\) R.l Bclc;ps..
O lcips
F . &lo- C?Xgo) = - 2O
i.e\.e.
iora ..
ion
20 k:
2o
lJps
oa,rpress
?s cc olrzrfv^eSs
n({':7-oG
86ina
d e t -r -r
6 - -- v-T7- Oe 86 irn"
A.<
o."-=E
Aa.
Pmblgm
^
AAB =
.3
.f
= -2o
4
E.4
Prob. 8.3,
tra In Ptob.
8.3, d€tnninc
determine thc
the magnihde
magninrde of the
the force
fonccP
which the
P for
for which
the tensile
tcnsilc sfrcss
!tscss
in rod
rod,l.8
lB has
has the
the sune
samemrgaitudc
magnitude asthc
the compressive
stressin rod
rod rC,
BC.
compressivcstr6s
8.4
l. ,-\r
) :
t'(Z
loedcdars
at B and
and loadod
togcdrGr-rt,
weldcdtogether
and BC
8C are
are welded
l8 and
rods,t,
Two lolid
solid cylindrical
cylinddcalrods
r !33 Two
(b) tod
ro
(a) rod,{A,
rod AB, (D)
of (a)
midsection of
at the
the mirlscction
stress at
normal st€ss
average nonul
the averigp
rown. Daennine
Daermine the
fiovn.
3'l\16in-
BC.
c'
P
-
fi-:l-
ksi
= -?-81
?-83 ks,'
=
7-oea6
3 in.
3
in'
= O-3tB3t P
A."= S (a)' ' 7-oese in"
- (2\(3")- P
r-v1
A^"
=
f? , _=!
7 - 6 eA g
E1',i\^?
-- O?-18;g
-- s-lr.iaz
l.{ l +?
+" Pp
o-r.r
3
6a. l. 6i".y
,)
O.3r83r P .
8 -'9t 888s33-- o . lrtyltq. {?? P
p
= tf S
8 -. 4 6( k ,: o?,s <
P=
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or dirtributcd in rny fom or by any means,without the prior written permissionof the publisher,or usedbeyondthe limitcd dbhibution b tcrch€rt ild
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Studentsusingthis manualareusingit withoutpcnnirbn.
E J Link 8O con*i*c of a ainglebar 30 mrn wide ad 12 mm thick. Ifuowing thrt
each pin hs! a lO-ilet diraldor, deterrnincthe rmximurn value of thc sy€ragG
norurelstncsg
in link BD if (a) d= 0o,(b) g= 90"
Problem8.5
Use bor A Be
as fr'ee boill ,
2o Rn
DEMI = oi
(a)
es O
( o . l S o sin 30' )(|ortor ) - ( 0.300 cos 3o') FcD =
F"o =
(b)
e a ?oo
1 1. g Q x p 3
N,
i-e. l"*eio,t
( o.gso c.,s 3o'Xaorros) - (o.3oo c'rs 3oo ) Ftr E
F6o :
- 3o r ros N J
o
\"e. ca\P1e'ssium
Areos
(al
*e"sin, I"");ntr:
fr=
( o , 0 3 o - o - o t o ) ( 0 . o t a . \i
(b)
cor'rprcrsio.,ri
ff:
(o.ogo Xo.ot?,) =
24or/d'
86o r/dG h4r
-g'lne$ses
(o\
6- E
G)
6=
1 7 . 32 x l o g
2tlo x fo'3
- 3 o Fr o s
3go r I gl-6
'
12.2 t lo3
= - I3.grfo6
7?.2Mfu
- 8A.?MP+<
problem 9.6
t.6 Iftrowing t the cenfial portion of the link 8D has a uniform cross-smtiond 1
areaof t00 mm' determinethe magnitudeof the load P for which the norrnolsfress,
BD is 50 MPa.
b MPa- = Soxlo'?a
t
Eoo ,.rnl = 8oo ,( lo- vt'
iroAuo = (t5ox,os)(too*lo') = 4o r,o* U
boil-1 ),nr4wt
,F bAf
Ats D .
+)M.= o:
Lo.+so)(#
tr") - o.rsgP = o
P = l-scg| F"o
''')
P=_(t.::::li:r
P = 6?.7 kU -{
Problem 8.7
8.2Link,{chasr uniform rectangular cross section Vein. thick and I in. widc.
Detennine the nor nal stressin the central portion of the link.
'o*e
*.oqe-*lne,.
,u.!lejs'or
4
N"ie fl'at
le",s iorn
li
lZoolL - in
covpl e +. act
ora *he bol
t-
f;"
b>r'n^
= <>:
z
._ r tB
- (12++ )( r^" ..,s 3o' ) + ( lo )( Fo. sir, 3o')
F
- - _Ql ? 6 ?
I
Ac
Aweo.
"f
f6 c a s 3 o ._ l o s i n 3 o o
l;. k AC:
Strcss i" /;nk AC i
l TOO = O
- t 35. 50 ^pb.
A = li,,"'' } ; . =
a - les
L
.r'r
-
6o-.{' = -t..
A
: -
Cj.t1*S
r
IOS{ P', = l,Ot# ksi {
8,t Two horizontal 5-kip forces are applied to pin I of the assemblyshown..
Knowing that a pin of 0,8-in. diameteris usedat eachconnection,determinethe ;
morimumvalueof the averagenormalstress(a) in lnkAB, (b) in link BC.
Problem8.8
Use joi,^f
B d,s {ne"
b"J1.
F*u
lc k'P.
c tnian Xl.
Fornc
lo
5in 75'
sin 6d
tr^" :
Fu. = 8.9658 kips.
1.j2os kips
L i nk A B is
a- *gnsiora
t'r €nn te^,
t
An"t = ( t - g - o . 8 X o - S ) = o . s i n
Mit irarJrtr
s ec+iov4qt p ia i
(*) Stress
i, AB, 6^*= ffi
F-
= ZIM
-a{
= l+.G+ ksi
is a. ooh4p Fess io"r tn €r* L en.
(1.8)(Ct-s)
C y ^ c r sS
s golict^e,) 4Fe4 is
A=
L;^ k
BC
(b) St'ressin tsC:
-Eg
(s^r= -1-\)Bc
A
-8'76s!
= O.q
=
in"
1.gG ksi
8.9 For the Pratt bridgc truss and loading shown, determinethe averagenmnral i
areaof that mcmbcris 5.87
stnees
in mcmbcr8E, knowing that the cross-sectional
in2.
Problem8.9
Use enfire, *n..
a.sf,r.. b*ly'
ZM* = o:
(qXto)+(texao)+(ztl(ao)- 36 At = o
Ar = l2o kiTs
stt kip"-
Sttkips
g
fit) kips
Use p"*i"^
Fco
of *r^.rrs *o *he /€)+
c,rl-h'na m€mbe.^s BD, BE, c+nd
af a. s ee*io*t
cF.
+f Zry = o:
l2o - 8o #n, = o
F"c
,SttHps
<{
8.10 Knowing that the average normal stress in mernber CE of the Pratt bridge truss i
shown must not srceed 2l ksi for the given loading, determine thc croes-tGctional ,
area of thet member that will yield the most economical and safe design. Assume .
that,both qrds of the me,mber will be adequately rEinforced.
rsft
Sttkips
= 8'r;2
ksi
6es=ff= #
Problem
8.10
; F." -- 5o k,i'=
Use er,*ire *r".rss 4s $n** bo*,
DZMH = o'(q\(ao)+ (r8X8o) +(azXao) sC 4
Ay -- l?o kips
s o
S{.}kips
Feo
Dsa po*ion .tF *russ h *l.te -A"fr of o seu{:ian
cultinl
rr€h bens BD, BE, enJ C F
CZMc=o:
le Eo :" (q[rao) = o
:.
F.e = 1o k'pt
10 hiy*
i
+.261 int
{
8'11 A goupleM of magnitude1500N . m is appliedto the crank
of an engine. For
the positionshown,determine(a) the force P requireato hold
the engineJystemin
equilibrium,.(D)
the avercgenormalstressin thi connectingrodBC,which has a
z+)u-rnmunrtormcrosssection.
Problem
8.11
Use pisforar rodr an) crank
(o.a8orn)t1 - l5oo f,l-pr = ()
H s .5.35?lxlos N
60 mm
'AY
Urg f;itq, 1!on,"o.,\"
2:
=
?oE .81
Frh^
P:
l?.86xfo3
P-
l-7,gc lN
lv
Fg.= lg.€+s xlos N
Rod BC
b
cor.aprae.gst'o'ri.'gr.^ben.
q.
Fn
=
Tfs
qlaa. is
+-fo p,r,rz= ?f,o xlOt
- / S- G + 3 xl o 3
= - l l . L +x r o c ? q .
+so .,;:-"
6u.= -+l-+ MPa-
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or distributedin any form or by any means,withoutthe prior writien peimissionof the publisher,
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ml
E.12 Two hydraulic cylindws are used to control the position of the robotic rrn
ABC. Knowing that the control rods attached at A and D each have a 20-mm ;
disrneter and happen to be parallel in the position shown, dctermine the avcrrge t
normal stressin (a) memberlE, (b) member DG.
Prublem
8,12
150mm
EOIIN
c
use men/lp^Age
a.s S rea bo)y.
E
200mm
(0.rso) + Fo.
DzMB?o;
Areo, df
,,,e*. Le" A E iq
(O. 6ooxgoo) = O
A = {d"=
= St+ t4 x tdL rn"-
{(no*6")'
r= = Ft. = +x1c3 =
"FE A
3lT. 16xlo--
Ee= eFrlogN
12.1?xlo' ?<-
6rp = 12.73 MPa
Use co-r b;ncJ yutpn6e,^sABC
o-) BFD 4s fnee hoJ,y,
(o,tso[#F_)- (o.eoox$
F*)
( t.osa - O.3.So)(6ao) = 11
Av e-^ in fs,i D G
Slncss
i'. noJ DC I
rs
Fo+ !i r l5oo
N
fl= fi )' = {(a"rto*)' = gtl,t6*ti' rnt
q"
a +t=
A
-r5oo - = -+.-tixlo^ pa
3. lt{l6xlce
(b)
5o* =
1.17 MPa '-
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or dirtributed in rny form or by eny means,without the prir wriucn pcrmissionofihe publisher,or usedbcfrord the limitod distributioi to tcachen md
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E /3 The woodenmembersA andB arc to bejoined by plywoodspliceplateswhich I
will be fully gluedon the surfacesin contact. As part of the designof thejoint, and i
knowing that the clearancebetweenthe ends of the membersis to be 8 mm, r
determinethe smallestallowablelengthZ if the avera1e
shearingstressin the glue is I
not to exceed800kPa.
8.13
Problem
4
L
Thene. clne Su^ se,pa.rals 4rtao-;s
oi} Xlve. ry
onea- hnus+frnansr,"i| ha-l| ,rf +l'.c--2+hN -!oJ,
)
T-h"*cfare
lfi) mm
tr = le "&U :
t/.x tos N
Sheo.""X gfress ia
. {v -t
t = a-
xlo3
Soor(I O
F
Le*
tI"
A
'/o
t
=
area
vJl = , u i l h h
a'nJ
=
8 a ( , ^ /t o s F q
| 5 >t lo'"
loo F1h : o.l
/sY to-3 =
A
I So v rrj3
o,jW
( a ) ( t s o ) + I = 3og ^nr^
y.n =
h1
)Yl
l5d2 p,q
{
8.14 Determine the diameter of the largest circular hole that can be punched into a r
of polystyrene 6 mm thick, knowing that the force exerted by the punch is 45 ;
-sfeet
kN and that a 55-MPa average shearing sffess is required to cause the material to
fail.
Problem
8.14
SOLUTION
A = TTdt
f"^ .y/i nJri .^l
;
S h" orl nfl s+Y"€ss
ETud'n? A's,
S'fvintrf.n C,
S^,'/r,ro
S t,rl q.q- -
. tvl FP
rTdt =
e=
-g
+3,1x/en
d = +3-1 wn,
Problem
8.15
8./5 Two wooden planks, each'A in. thick and 6 in. wide, are joined by the glued i
mortise joint shown. Knowing that the joint will fail when the average shearing ,
stress in the glue reaches 120 psi, determine the smallest allowable length d of the r
cuts if the joint is to withstand an axial load of magnitude P : 1200 lb.
Seven sor faces
6arY^/ *he t"*.P
1""A f = t?oa lb,,
A = cz)G)d = fr3
:- A=
A
P
L
Pa=
2,
d = 1.G43in.
r uin.-l
Fo'^
stee^0 A,= r Jt = Tf(o.oXo.q)
O.1g+O i"
Tt=
P.
4"
L i'.,'/ i"
X
v n.lu,
13.57 k ips
A" = rrJt
Fon o -er* r h U t n
j.
L
p = A,t, s (o,Zs+eY
fg)
:
'tz=fi
.{
8.16 A load P is applied to a steel rod supported as shown by un aluminum plate into '
which a 0.6-in.-diameter hole has been drilled. Knowing that the shearing stress :
must not exceed l8 ksi in the steel rod and l0 ksi in the aluminum plate, determine :
the largest load P that may be applied to the rod.
Problem
8.16
f-
-P
.p =
Art" = (t.2566xto-; F
"f
P is th.
sua+.llen'rtJve.
l?. f,? I(ips
P = t2.5'7klas
8.17 An axial load P is supported by a shon W250 x .67 column of cross-sectionalarea
I : 8580 mm2 and is distributed to a concrete foundatio:r by a square plate as shown.
Knowing that the average normal stress in the column must not excced 150 MPa and
that the bearing stresson th" concrete foundation must not excred 12.5 MPa, dctcnnine
the side a of the plate that will provide the most economical and safe design.
Problem
8.17
A,reo "F colo^n i A = &,:son,i = &fgo * 164v4
'sfness
in c o l o r . , . r , r i S = t S ? C Y ! O 6P a "
Nor rnc,l
g= F .'. p= A g' :
(gsaoxro"XlSo 'tCIa)
t.787 x l}e N
?
g s a o i n Qg I J e :
Ab=
b=
t'= JS-g
_E_n " C
Au
l.?87 ,( log
l ? . . sx I 0 6
0-T
Problem.8.18
=
Ab--i
fo"
s1oa-'.e- p./d.*e
lo&D x 1d3 a
lL=get
o, gQ-l ln
Yu,ry 4
t.!t The axial force in the columnsupportingthe timberbeamshownis P = 75 tfrI.
Determinethe smallestallowablelengthI of thebearingplateif thebearingstressin t
thetimberis not to excecd3.0 MPa.
6t=fr = _E_
Lw
Solvintrfo* L.,
I
t:
A
trlm
r
m
118.6 xlO
t
=
17 8 .6
tn'u1
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or disfributedin any form or by anymeans,withoutthe prior written permissionof the publisher,or usedbeyondthe limited distributionto teachersand
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Problem8.19
E,I9 Tltree woode,n planls arc fastened together by a series of bolts to form a
column. The diamcter of each bolt is Vz in. andthe inner diameter of each washcr ic
Vt in., which is slightly largcr than the diameter of the holes in the plankr.
Determine thd smallcst allowable outer diametq d of the washers, knowing that the
av€,ragenormal stress in the bolts is 5 ksi and that the bearing sfess betweeir the
wa$ers and the planks must not exceed 1.2 ksi.
BoWt AuJt= Sdr" = +(*)' -- o. ti63f iu'
t;te fr^". in bolf, p =,6oA -,(5 )( o.'t?63s)= o.1sl?5.kip
Q = * :- Te,n
i.srJe J;or^e-len= d; = # ir., oufsiileJ;,"",*.<^= 4o
W*sher:
P
,
^
-
r
,a
Beori^3q.*eqiA* = +( 4: - d;. )
?
E7u#;''3,+U: - di') = s't
do'=,4'++& = €)'
+ t'
TT
*^J A* =
&
o . 1 8 _ : - /J5 _ l . + g 2 3 i n z
t? .5)
do=
Problem8.20
t
l-lq7
in-
E.20Link lB, of width b = 2 in. andthicknesst = Ynin., is usedto supportthe endof
a horizontalbcarn. Knowing that the averagenormal stressin the link is -20 ksi and
that the ever.Ee*rwing stressin eachof the two pins is 12 ksi, determine(a) thc
diameterd of ilre pins, (b) the averagebearingstressin the link.
Root A B is in Corr"Pr€5s r'Dlf,
A= b t u , r l r e r c b = 2 i " , . o n o l l =
P=
Pin :
(o.)
d=ffi =
oXaxt)=
+in.
to kips
o,"d A? = +d'
=
l.O3O in. I
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or distributcd in rny forn or by rny means,without the prior writtcn permissionof thc pblirhar, or usedbeyondthe limited distribution to bechsn rnd
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Studentc
ruing thir manualut uring it withoutpermission.
E t Two horizontal 5-kip forces are applied to pin B of the assembly shown. Knowing
that a pin of 0.8-in. diarneter is used at each connection, determine the maximum value
of th_eaveragenormal stress(a) in linklB, (b) in linkBC.
Problem8.21
0.5 in.
t.21 For the assembly and loading of Prob. 8.8, determine (a) the average shearing
strss in the pin at A, (b) the averagebearing stressat A in mernber,4B.
B c.s^a$ree b oJy.
U,.. j oi'*
e l .
;) KtPs
;) Klps
J l -
0.5in.
IO IIJF
Larv
"l
lo l;ps
tnra nTfe
Fornce
S i', es
Ee
F
1.3?af J,,P,
'AB-
si"r {5'
(a-) Shear,'n st nessin pi*r a+ A .
f=
?
rrrhe+aA.' tA^ = *to-8)"= 0.5026t.4z
2Ar
-7^8 ksi
(b) 9eo^J"r # n s: a-l A
Au:
6b=
tA
F
tfirg-
Au
l" -,r,,e*,ben AB,
(o-sXo,8) = O-+
7'-32.o{ =
0-+
| g.3o
.1
l,'l
l?.3o ks; <
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educators
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usingthis manualureoiing it withoutpermission.
l
Problem8.22
t.22 The hydrauliccylinder CF, which partially controlsthe positionof rod DE, has
gb is% in. thickand is conncctcdto the
@ lgckedin the positionshown. Member
-vclticalrod by a %-in'-diaurctsbolt. Determine(c) the averageshearingsbcrs in the
bolt, (r) thebearingstressat C inmember.gD.
bo&t., a-J
Use rrnevaber BCD a-s o Se.
vrole. th of AB is a. *uo {or.e ,;L*,bef',
4oo
t{oo
cs75"
.
$tn
<!'C
/S
cr
8r-
-t_
E.2
(Qro.lu")(&fi")-
DzMe=oi
(+siq?"o"\(#n")
- (1c.rs{oo)(+uo
\2fr=o'-
-'*
C* =
-t-r-
+t z\
[*
+ Cx +
+ Cl
r\
uy
, , ^qq
'a)
- - Sheq,.i*
tloo
+ +oo sia 750
=
lcr" + C,
c',-'75') =
4r.ocosTso
'Ioo c,e.s
?So =
W
= o:
C:
sia 7.1*) - ( 7 si-r lo"Xtoo
7?97.,35 = o
3-3e€78 F^,
:i
. Ae
7f.-3'I lL.
r
+aQ
srr 75-
= Q
= ll?+.cs^gL.
ltq7.7 ,lb
ssff rcss
r c s s ii"q ff,'h te b
6.,^Pf
"!tt 5
P = ll17-?
IL
A ={a" = +(t)' ' o. troqsin'
t = * = #tr#
) B-- e a , ^ i n q s l L r . e s s# C
Au= dL=
q:
s-
A,
: lo.8+"ro3p,i
i^ -e,-'A€^BCD.:
(*Xf):
o.Ls+s?.fin'
= llq?'?- = ',tlxlosps,
o.23rt37s
lo.8{ ksf
<
p= IIq?.L !lr,
= 5- lf ks,
<
Problem8.23
t'.tJ Knowing that & = 40" and P = 9 kN, determine(a) the smallestallowcblc
$lg""f of thc pin at d if thc averageshearingstressin the pin is to not exceedl?0
MPa, (e) the correspondingaverqgebearing stressin mernberAB at B, (c)
the
correspondingaveragebearingstressin eachof the supportbracketsat.B.
16 mm{
Geor'n"fv"y: Trr'ar,Xle ABC rs
c..'r isose,les fuianSle wi]ln
e
ant3les Shorun h€re .
Use jo;nt A
c-g
o -fr.* bily Lqw of Sines npfl ;"4
lei
+" foncc *nr'r.,a1
?
]P-sinl2"
Fanc e
tu.iarrXle
l=oo = Flq
sinIlO"-- si"tfOo
F;
I
s
\
(a) Affo,.ro,Llepi', J;o*. rie.,
= &
:
2Ar -+
2Td'
nd"
(3Xa'|.zs[tgE).
=
A--- 1?et=
f= +
1lt,
n(taortD" )
wLr,.*F*e=a'J,7Jr,
lo3lJ
|gt.fgxto-c,'
d = l f . 9 5 , 1 o - 3u
ll.95 r',lrn
bl EeqrinT slne"s i" AB o+ A,
rr/o'3
Al= t ol = (0.otgX,t,l.qs
) = lA3.A6 ,lo-" h^a
Sr:
+t:
Ar
-1'l'!3xt'o3-=
/33-?(, lo-6
tgrl.ivloc
{
131.7P1Pq.
c)
A=
fir=
1 3 ? . 9v l o - " n n
qo.Oxlo'
?O,oPlP+{
E.24 Determine the largest load P that may be applied at I when 0 = 6C , knowing
that the averageshearing stressin the l0-mm-diameter pin at B must not exceed 120
MPa and that the averagebearing stressin member AB and in the bracket at I must
not exceed90 MPa.
Problem8.24
P
16mmAs
'
Gcorn.fry Tr,'qTJ" ABC is
al.r i sose--Cestnia n1)e wi*ll
ongles shou,rv, h etne.
l2O"
goi
Usej"int A as $.. " b"Jy .
l.au,r,rf s Ines ap f, h oJ {*
lonle tnia n1)e i
P
P
l 20 "
?
sin 3oo
Forre
fuianl la
P=
?=
F;
I$ shec.inX sf rsss i', pi. .t B
F^.
=Fte'=
is cri |r'co,l.l
sin 30"
Sin f?g'
Fo" =tl 3=o = g. s7735 F^.
srn
l2Oo
t*ftff" = E.
xrDt r'
|, = $d" = T(o.oro)a= ?8.Sq
(aX78,s',rto'c)(taovlot) = t8.85a*los N
Fre=
"A't,:
+ !eaninl slness in -e- ber AB "* brr.aJre*a* A is €rr't;.^i,
( 0 , o 1 6 ) ( o . o t o ) - r 1 6 o x l o ' 6r '
Ar= tJ:
F*u:
If
Ab6t = (teoxto^cXqoxlo6)=l.l.ToxlLf
A/
ber^in3 shncrs in *[,e bnaolcef a-] ts ,'s cni]i .4,
Au= 2t,d
= Rl;(o.ot2{o.oto)r
?qoylro-6;'
F r e = A r 6 o ; ( 2 1 o * 1 6 c , ) ( q o" , o " ) = 2 l - € - * / o ' N
A.0!""t.$Je
Then,
F4e is tl,e snq/J*t
'..F,,ra,^,
S4#f
"s
Prr*"
t.e. lrl .tlo x ,,esM
(0. -sz?gg )( l't.*ox lo3)
= 8 - 3 l , ( l o?' A /
83t AN
<
Problem8.25
E.25Two woodenmernbersof 3 x 6-in. uniformrectangularcrosssectionarejoined
by the simplcglued scarfsplice shown. Knowing that P = 2400 lb, determinethe
in thegluedsplice.
normalandshearingstrcsses
= 5o'
O = ?oo-4oo
Ao = (sXG) =
P:
Z t l o oj L
l8 i',r
(aroo) er3d
o
5 = #.^t"t
= 5'f. t
Pri
r=#sial.e-
= 6.t.7
fl
Problem8.26
4
joined
E.26Two woodenmernbersof 3 x 6-in. unifonn rectangularcrosssectionare
allowable
by the simple glued scarf splice shown. Knowing that the maximum
rtt"*ing stressiti ttt" gluedspliceis 90 psi, determine(a) the largestload P that can
tensilestressin the splice.
bc safelyapplied,(b) thecorresponding
o=.9 0 0 - { o o
=500
A o = (3XG)
=
b
PI =
18 f'"t
fi'Areinz€
sin Ze
= sLTo
= (axrsJi:e\
sin lr)c)'
P - 321tt lb.
(,a)
ff=
<
3?9o c.'sesd
t8
=
75-S
fl = 7€.F psi
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4'
E.27 The 6-kN load P is supportedby two woodenmembersof 75 x 125-mm
,uniform cross section that are joined by the simple glued scarf splice shown.
in the gluedsplice.
Determinethe normalandshearingstresses
Problem8.27
P 3 6 > 1b 3 l J
Ao = (o.o7SXa, l2S)
P
_6 = fre"sre
e = ?oo- 1oo = 20"
=
1,375r(lo-3hn'
{ qxlgl) c,re"eo' _ 565. x to3
$ = 56-f
(g*163) sia too
= 20 6Y/os
s i n 2 O = (a\(q.3?fx,o-s
)
t--
t
I28
Problem
hPq
= 2od kPe-
E.2ETwo wooden mernbersof 75 x 125-mm uniform cross section are joined by the
simple scarf splice shown. Knowing that the marcimum allowable tensile stress in
the glued splice is 500 kPa, determine (a) the largest load P that can be safely
supported,(D) the corresponding shearing stressin the splice.
A. = /o, o75-)(o.lzs) = 9.-g?g .td'
e = 7oo-7oo = ?O'
6
(a)
L
P s i ^ 2e =
XAt
+
(b)
rua
I
P= 5-s, kru
(E3oa5rtlo3 ) si" 9o- :
lgl.?q xroe
t
= f 8?-o kPo-
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PfOblem
8.29
8.29 A 240-kip load P is applied to the granite block shown. Determine the
resulting maximum value of (c) the normal stress, (E) the shearing stress. Sprciff
the orientation of the plane on which each of thesemaximum values occurs.
(6)( E) =
sE in'
?'
+A o cos e-
e'e
tvtc*, t nsile. ={r*.s
*+ e = ?oo
t' O
Yao.,,/r.co*-f |.45s i u€, s*rc sS
*t e = c"
t,o = #" = Affi1 = 3-33ksi
**
probrem
B.B0
<
O = t+S"
:#,**'#;f*l;#'lr':1i:j};,ffiJiiii:'f;ffr,ffH#i:l:;
ttre magnitude of P, (D) the orientation of the surface on which the marimum
shearing stres occurs, (c) the normal stress exerted on the zurface, (d) ttre maximum
alue of the normal stressin the block.
(6)(6 ) = 3a i"r"
t,^o-n= 2-f
ks;
/
sinTe=t
(c\ 5+s=
2A--10"
*, ".'""
(d) 6n-tt=#=#
+5"= ,t
\.q(-= ({ Xga )(r-s )
(ips
1
e={So
<
=
= -2.{ksi<
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usingthis manualareusingit without permission.
Students
Problem8.31
ao
ni
f" 3 *)"
o.3oo vr,,
= D' t$o
t.l
(>- t = 0. tro- O.ooG = O.l.14.Yor
A. T(yL'- h") = fl (o.,so* - o. l+{t,)
= s.St+ x, o-3 l'rt
e= n50
c.,sr&5o
6:= + coste = ?go:F
f4o
=-37. I x loo
t =
&
$- ='- 37-l HP6- a
srr ?e =
=-11.28x lo'
! = l7-?z |4fu 4
8.32 Astecl pipe of 300-mm outer diamcter is fabricated from 6-mm-thick plate by ,
we.lding along a helix that forms an angle of 25o with a plane perpardicular to the i
axis ofthe pipe. Knowing that the ma:cimum allowable normal and shearing stressesL
in the directions respectively normal and tangential to the weld are d= SOtvtpu and 1
= 30 MP4 determine the magnitude P of the largest
axial force that can be ap'plied
to the pipe.
Problem8.42
tr\Sor^a
f!=*J"'
n= t - t = O . l S o - O . c n $ = O . , + t /h ^
A.'= T ( rot- nr') = r ( o- ,soa- o. ,9+1)
do= 0.Wn
Weld
?5"
= T- S! r to-s
w1z
0=25'
MPa: 6= f, c,s'O
= (Sgt xro-=)(soxl.a6)= gA7x los
e',T
MFa.:
Z, = #-si'r 2e
= (ZXs-s{ *f o-3X3o * toe) = +g4x lo3
srh 5o'
SJfen
Vafue is *le
^.0I"-J0e. vo.l,*
"f
P. E
P= 337 KN
ttSG
i gff Link,4B is to be rd' of r *d for whichffit ultiffi mormalstrwr it
r$ll
of
cefcy
&sor
fu
sffi
AB
fu
ih" ,ror.sc!fiid rlca for
] llF.;;;ir"
thatthelink w,illlbt @telv reiahrd ffiDd &c pinslt,4 nd
I il-lSo. il;;
1t
Probhn 8.33
P = (t.?.)(e) = ?.d &tl
tr^"
t)znt"=Dl
- (o,gXF^"s,',.3s1 +
+(o-'t) (eo1
Fre = 3"1.€n kM
Es:
a
"rrg
Fre _
= ( E,s..)
6i*
=
t68-l vfo'
?
w
Ao*' t68-[ *r*t
I
Problem8.34
t
83{ MembcrABC, which is supportedby a pin and bracketat C urda cable.BD,
wasdecignd to zuppofithe 4-kip load P as shown. Knowingthat theultimatclod
for cableBD is 25 kips, determinethe factorof safetywith respectto cablefailure.
Ose ]n€rnber ABC as 4.
Sree bady. ort{ n ole
fhaf ne#ber €D is a,
{wo - Son"e rr€wr b erl8 in.
,n".J
EM" = Q:
(F e"s 4o') (gi,) + (P
40'Xrs i"-) - lfu cosgo"Xrs;".)
"i'
-(F?o sin 3o'Xt2 i.) = o
F.o= +i#
? =#
= G.87ts
h;p,
=
F.g = trnt. = =S
6'87b 3'6+ ''
R"
Fa.at. of' s*Fe+' f. c*[le tsD:
Problem8.35
E's-sKngwing Sratthc ultimate load for cableBD is 25 kips and that a factor
of
safetyof 3.2 rrith respectto cablefailure is required,determinethemagninrdeof
the
largestforceP thu canbe safelyappliedasshownto membcrABC.
A
TI
Dse Fe€n Le. ABC as a-
+*" il;-
tS in
a'^) no**
A
that v"oehber B D rs a.
two - frrd"se lrrCrqber*,
18 in.
(P"*
,,".-J
ZM. = oi
9o")(3o ;.) + ( P si",{o"X ls ;") - (Fro cosJooXlsi.)
=
,tt:'u.?"xta':
3r.62e FBD
A!,!"*A,0" !"*)
A)!"onil* IrJ
=o
O.SgZt6 F"o
'I"
F* =
Frr rrtl''be, tsD
3.Q
7.8125k.ps
P = (o.fsetc\0.8 ta5) = +.ss kr'ps
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Problem8.36
t'tf Mcmbss AB andlC of thc trussshownconsist
of barcof squarecrosssgtion
is
ofthesame
alloywas
*:i*:,y'^:PI._.It
T:,*, r|u,12!-1m+uuoi*
tcsM to failureandthatanultimate
road
l
of 120kN ** ;;rd;."?
saf€ty of 3'2 is t'o be achieved for both bars,
"-#il of
"r
determinc ttre required dimensions
th_ecross *Tion of (a) bar AB, (b)bar AC.
I en3*!'
L.4
ItI
lL
Dz
z'M
\
.e
0.85 y'n
Dse enlrne |russ ag a Srve bdyt.aAx -(AlSXeg) = e
Ar=15,(lr/
6t
'l - ot
+tz
IzF,
r!
=
o.?Seio.,t'
.l^o
I
m g , , . , L g rA g :
"f
I
Al =?8 ltN'
Ay-
l'
'
us3le jro i n,+A
a,s4^a.
Au
$ EF'= 17 kN
Fp"
_1
-t[ry=6t
A,-F*-$IFre=o
F^.= 2g - I+tJ!il.
o.Es
t
r-i
For tI" tet*
Fa. tle
bq,
r,n-tet*,0
A = (o.ozo)' =
g"
- &
vu =
A
4oo r/d6
P, = t2o x los
= ,l?o"{ol, = soo>ltae Pa,
*ooxlo-6
o
F. S. = -!-L =
Fra,
(q.\
h4'
= 20 kN
G,A = 614,"
Fre,
Fro
(3,1Xtzx/os) = / 8 1 - & 3 x l o - ' , q .
3o x rog
1 3 .1 ? x / o - 3 m
Ec
|
.Y.
P"
=J9:
|--.
rk
3oo r loL
t4-G, xro-s rn
13.47rr,rr,a<
q,A
vuf't
l-*
6, b'
=-}1!l.3
Fn"
t
2 l3 .33 )./c>- r'
19.6 I ra*,
8.37
Problem
steelbolts areto be usedto attachthe steelplate shown
8,37 Threot/, -in.-diameter
the
to a woodenbeam. Knowing that the plate will supporta}4't'tp load and that
for
safety
of
factor
the
determine
ksi,
ultimateshearingstressfor G steelusedis 52
this design.
= o'4\tsin'
Fo"ec,,,lb"Jt fi=fat=*(ff
P" = Ato = (o.4+ts)(s:)= 2?-q7l*'Ps
Pe"bo|t;
p:
t=
8k:Fs
8.38 Two plates,each 3 mm thick, are used to splice a plastic strip-as shown'
Knowing that the ultimate shearingstressofthe bondingbetweenthe surfacesis 900
kpa"d*ennine the factorof safetywith respectto shearwhenP: 1500N'
8.38
Problem
Bar/
a.n,a.: ( See f;g r^. )
[= ](a")(a")+(ls)(a.\
= l SOo -',"
= l!;Do rloe m-
LD
*-_
,5
r
Pu= 2 A,yb u : Q\(tfioox td)(qoo 'fos ) = 2zoo 1.,
: 1,'c,o,
=
Es. = tu
P
8.39 Two wooden mernbers of 3.5 x 5.5-in. uniform rectangular cross section are
joined by the simple
-rtr"rrglued scarf splice shown. Knowing that the maximum
in the glued splice is ?5 psi, determine the largest axial
shearing
allowabti
load P that can be safely applied.
A o = ( 3 . 5 x S -s ) - - | q , 2 5 i n
O = ?o"- 20'
f
=
*
P=
si"6coso
=
,fr*''"'2o
@
s in l{o"
= 4417 Ib
4.47 V,r,,
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Students
Problem8.40
t.40 A load P is supported as shown by a steel pin that has been inserted in a short
woodcn mernber hanging from the ceiling. The ultimate strength of the wood usod
is 60 MPa in tension and 7.5 MPa in shear, while the ultimate strength of the stccl is
150 MPa in shear. Knowing that the diameter of the pin is d = 16 mm and that the
magnitude of the load is P:20 kN, determine (a) the factor of safety for the pin, (D)
the required values of 6 and c if the factor of safety for the wooden member is to be
the same as that found in part a for the pin.
?o t(N =
F:
-fi = + d;-= + ('' 0'6)l 2ol'06"l6q^
(a) Pi", :
rl
b lg
Do,.r
s I e4ra
"t,=f; tu=ft-
x/03)
?Ato = (2X?ot,t6xt66Xrso
R
W= 4Or+r., ? P'O.I*O 6
?Ov IDNN
5
60,gt?x los N
Fu - 6o.3t?xros= 3 - O 2
AO trlog
F
6 0 . 3 1 1 v l l s N $ o " Sara € F, S.
w|ene
b-- C +
=
= O.O{D r',r
U = 4O w'l''r-
= +l-lxlog
('-016-1
'l/. l' tuln
b =
Pu = €O.gq
Do.rLle sAea.j
4P,
Lu
g ao[r
a-f?a- is
" lDs N
A =
$o,* s4.h^e ES.
WC
2A
?
c --
Pu
=
-
6o.9!1 x lD3 -
=
(iXo"ovo)(7.sY/o')
lOO.5 x/ds rn
C = loo.{ mn
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5
Problem8.41
t.4l A steelplate TI in. thick is embeddedin a horizontalconcreteslabandir usGd
to anchora high-shengthvertical cableas shown. Thc diameterof the hole in thc
plate is '/e in., the ultimate $trengthof the stccl used is 36 ksi, and the ultimatc
bondingstressbetwcenplate and concreteis 300 psi. Knowing that a factor of
ssfetyof 3.60is desiredwhenP = 2.5 kips, determine(a) therequiredwidth a of thc
plalc,(6) theminimumd€pth6 to which a plateof thatwidth shouldbe embcddedin
the concreteslab. (Neglcct the normal stre$cesbetwecnthe concreteand the lowcr
endof theplate.)
2.5 lrJpt
B ,s e"l ov1*ens ia^ i n p!**e',
A E (A-J) L
P, = 6'4
E,S- = Pu =
F
So.0u'ni {or
fl-= d +
Bosed or, * h e.rr
Q-)
( ns.)P
6r
1s.5o)(as)
LseXf )
(a)
o-
bet-ee.^
p,0r"ie anJ eonc,t,v..*e s.loL ".
A = pe^ no*er * Jep*h=
B= T"A = 2f"(a+t)b
Solv drr3, f"," b,
b=
GI
P
2(a+a) b
{
t, = 0.3oo }zsi
P"
FiS. = P
(3.G) ( z,5 )
(2 X/.sro*f,Xo. 3oo)
b = 8- oS ,tn.
r<t
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-nt ptl* *titlo pcrmirsion or-trrcprutirtrcr, or usedbeyondthe lirnitod dirtrfrution to tsrchon rd
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Shrdcntruring thi* manualareusing it withut pcrmirrbn
5
is uscdto
t.41 A steelplatelc in. ttrick is embeddedin a horizontalconcreteslaband
plrtc it
in
the
hole
the
of
diamefcr
Thc
anchora high-strengthvertical cablc as shown.
strcss
bonding
ultimate
the
and
ksi,
is
36
used
stcel
the
of
% in., the ultimateJtrength
OT3n plateandconcreteis:00 nsi.
Problem8.42
t.fi Detcrminethe factor of safetyfor the cableanchorin Prob. 8.41 whenP = 3
kips,knowingthata= 2in. andb = 7.5 in.
3 kfps
B osed on *ens,'ovt i'a PIa*e i
A
= (2' + X#) = 0.3106in'
Pu= 6rA
ltips
= B6 Xo.3?06)= \q.OG
F.S.=
F= *#
B o.seJ oh sheor be*weem pla[e and
-- +.Gg
concreJe sla.E3
A= penimelerr Jept[, = ?(a-+ t) b = 2(a+#)(z,s)
Tu = O . 3 o o l * i
A= 3+ . Gq int
= lo.+l kips
P"= tuA = (o,?oo)(g+"r-?')
F. S. = 3r- =
P
Ac*lrol ln in
lo' til-
?=
3'+7
o| so|+ly is *Ae snal.ler ur,P.re,
ES.= 3.17{
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Problem8.43
E dJ In the structure shown, an 8-mm-diametcr pin is used at14 and l2-mm-dirmcfcr
pins are used at B andD. Knowing that the ultimate shearing stress is 100 MPe af
all connections and the ultimate normal stress is 250 MPa in each of the two links
joining B and D, determine the allowable load I if an overall factor of safety of 3.0
Top view
8mm{
l_
--\- |
Stc'ir'cs : Use ABC
r
F^
8mm
Z M " = o> :
D
12mml
F"o
8mm
p=
l--
Side view
=
\?
ft = 4e"
etul _ (edtoox,lo.)(str.tge
tr
- -F.s.
I$
rA =
s.o
P= +F^ = 3.72,(lotN
'.).
ll oo
q
)i
o:
ZMl=
Based on do.rtle s[,eo* in pin A:
o.2D
Fn- o.rt P = e
Fe
O.20 F " o - O . 3 8 F = Q
lo
F"o
lct
= 3-sstylog N
BaseJ on Jo,rU. she.in ir,r Pi.s tuJ 8 c,!al D;
A= tCt=
Fro -== - ? M -
q(o.olt)r = ll3. lo x/o'6 ^'
- ( a X t o ox t o ' ) ( t r g - t o t l o : 6)
F.S"
P = jr" Fro r
3-o
s.q1 x r o 3 N
ks
B oseJ oh GDrra
P Fessiovr ir, ,f ;"
Fon
Eo
on? l;^lc
? 6, A
F-s.
= T.$rf x/Oj N
8D :
A=(o.oeoXo,ooE) =
(a)( ?$b '/0. X tGoxlo-c)
2A
J.
L'
P = i+-|1o= lt.o+vfosN
AlJo.ro,[,0uva.lue oF P is t-oll,qs+.
p= g,72"roEN
P E ?-jZ ki.,
Problem8.44
Top view
f.ff fn the stnrcturcshown,an 8-mm-diameter
pin is usedat A andl2-mm-diamcter
pins areusedat.B andD. Knowing that the ultimateshearingstressis 100MPa st all
conncctionsandfrreultimatenormal stressis 250 MPa in eachof the two links joining
I andD, determinethe allowableloadP if an overallfretor of safetyof 3.0 is dcsfud.
8.44In an altemativedeign for the structureof Prob.8.43,a pin of 10-mm-diamaer
is to be used at l. Aszuming that all other spocificationsremain unchangd,
dctcnninethe allowableloadP if an overallfactorof safetyof 3.0 is desircd.
Stah'cs
i
lJse ABC a,s 9"c. boJy.
A
B
l=A
Feo
ZM" = o 3
p= f p^
u mm-'l
Z M6 = O :
Bc,sed oh do.rUe slreqr inapi. A l
=
\
fi = 4 et : S (o.ofe)r J ff,-,rt. xtd6 ,,,n'\
i-A -T.s.
o.ZDFn - O.rg P = o
E
O.2oF"o- O.38P:g o
FL"
3.o
P = .f FA = 5.[12 iltDt N
BaseJ on Jorbl* sLec"nit'r Pirs LJ 8 c'nJ D:
A = + C ' = T ( o . o t ? ) r s l l 3 . l o N I D ' 6t n L
(aXr"o*to')(rrs.foxlo-g)
F
r s-o _= _ ? f r A =
= zs+ yto"N
3-o
F
P = jfr' Fro r
s.q7 x/os N
BoseJ oh corvrpFess
iou ir, ,e;r,ks 8D ;
For onc l;"t
Eo= #
l6o x/o-t *r'
A=(aoloXo,ooa)
\(tgo'ro.'
)
= (4)?.*..r/o'
?G-7x/ot f.l
P = i+Eo = l,f.o.lxfosN
AlJ"-",[.0u vol,,reoF P is
"-ol]es+,
p= g.q7 r/o$ N
P = g-qT RN
Problem8.45
E.15Link AC is madeof a steel with a 65-ksi ultimatenormal sfressand has a % x
%-in. uniform rectangulmctoss section. It is connectedto a supportat I and to
mcmberBCD at C by %-in-diameterpins, while mernberBCD is connectedto its
E
supportat B by a ft-in.-diameter pin; all of the pins aremadeof a steelwith a 25-kei
ultimateshearingstressandarein singleshear.Knowingthat a factorof safetyof 3.25
is desired,determinethe largcstload P that canbe applicdat D. Note that link lC is
not rcinforcedaroundthepin holes.
l--u
U s e S " " b " l y B C D+ J M e = o : ( 6 ) [ € F ^ " )- l o p - o
".-l--n,n
p = O.tA F*.
B"-frtr^.= o
't'zE'= o:
6?
lo'he
Bx=
-z p
3r
t.as'* (€)' P r
I
f
=
l-2,5? .-
68, -ttP
+) Mc =' O ' .
BJ=
cr)
r.€.
= o
B= ePI
P = o-705t8
l.tf ILL-t P
B
Sheo. in pi^s ,-+ A o,.) C.
h.= t Ap,,= *ffi11=e#+xsf
+ffiJ
HtxE)= O-9+a51
T.^a
' ot4
ot-1 nr*
s"4io.n
"-f
A
o-rJ C .
Fo.= 6-A; = g A".r=€*IdXi-?) ;
S**1!/e. vJu"
F.o*.(r)
p=
df Fn. i5
lc;p"
O.6Z5 k,p.
O.6eS k,p,,
(o-'ls)f o-62f)
=
o-3ooLJps
SL"o. in pi^ ,J B.
=(#ifxTxf)' ,
B = t Ae;^:
* Te'.
F^"^ (a)
P=
(o-zosaa{cl.s8?{q) = o.+le
Ailu* J,0" vJuu J P ,'s fL.-
"-&'n
v*!oe.
0.5 81qq l;p*
l<:ps
p = O - g o ok J p s
or P=
%o.th
PR'OPRIETAR'YMATERIAL. O 20ll The McGraw-Hill Companies,Inc. All rightsreserved.No part of
this Mrnuel nreyhc dr6ph)€4 rqo6rc4
or dirtributod in any forra ot by my rnstns, without the prior writien peimissionof the publisher,or usedbeyord
th limitcd dfutrftutbn to tcrcbn rnd
adrcebn pcrmitbd by McGnw-Hill for their individualioursepreparation.
Students
usingthis manualur" uiiog it witho't pormi*bn.
(z\
5
Problem8.46
to useF-i"'|.ff Solveprob. 8.45,assumingthat the structurehasbeenrodceigned
madc'
hes
been
changc
no
other
that
and
as
at.B
dimreterpins at A arriC aswcll
8.d5 Link AC is made of a stecl with a 65-ksi ultimate normal stress and has a tA x
%-in. uniform rectangular cross section. It is connected to a support at I and to
monbcr BCD at C by %-in-diamcter pins, while mernber BCD is connected to its
5
support at B by a E-in.diameter pin; all of the pins are made of a steel with a 25-ksi
ultimate shearing sffess and arc in single shear. Knowing that a factor of safety of 3.25
is desired, determine the largest load P that can be applied at D. Note that link ,dC is
not rEinforced around the pin holes.
-t
-o
in.+l+4
in-
BC o
Use f.-o. hJt
+) ZMB = o: (6)GtF*) - lo ? = o
(r)
f = O.t+g F*.
B,,- 3q.= c
+tZFr=o:
B"= Etr*. =
-GBt
+)Z trle = o
S, = -l8.rt *
Sh.on
l 3 y= - Z "
=O
'1.2s'*(€)'P''.
=
A
in pi ns
-+f
t.7€ ? -
ine Br=3e
F' D.7o583B
l.qtefJP
onJ C.
k:p"
= *, {/" =(;ftX+)€f : o-58?7q
Fk = l-Ap,^
s ea l i a .
Tcnti on on n.l
FA"= 64,'r'
S,. nflr.
F^.
p=
? o'qs?slc,'ps,
is
o. SEqqq L;p. '
O.?83 lc,'ps
B.
g,= dApr*= #
At!.-"Jol"
C.
P = (O.t3XO.St?dt?) =
S[.co. in Pi^ -I
Fr.r*. (a),
A o^l
*A^*=(f=X+)(*-F)
vo)u,e of
F,o.^ (r1.1
A
4e" =(#XFX#f . o-syqqllc;ps
( O . T o E t a X o - S E q i q ) - O . t 1 f dk i p s
vn!,te J P i. lL" s^J.!e,
uo.,fr"-
p=
c.-293 lc;ps
P=
2#lh
pRoFnIETARy MATEHAL. c) 2011rh McGnw-Hill coEFDbr, Irc. All rightsreserved.No partof thir Yry"l.ryy.bc dirphpd, mrcaucaq
bcyondthc limitcd dirtrbution to tcrchcrt ffid
or dbributcd in rny forrn or by any means,without the prinr wrien pctmiarion of-the publisher,or used
win3 it without pcrmi$ion'
are
manual
using
this
ptrpli.tion
strdents
counc
e&rcetorrFffnitcd by Mccnw-Hill for their individr,rel
Q\
Problem8.47
8.4? Eachof the two vertical tinks CF connectingthe two horizontal membss lD
crosssectionand is madeof a steel
andEG hasa l0 x 40-mmuniform rectangutar.
with an ultimatestrengthin tensiono14ffi MPA while eachof the pins at C andF
has a 20-mm diameterand is made of a steel with an ultimate strengthin shesrof
150 MPa. Determinethe overall factor of safety for the links CF and the pins
connectingthemto thehorizontalmembets.
Use ncv,tberEFC 4s fuee bJl.
\
Ozmr = o:
e o
o.qoF." -(o,g'SXe+vros)
24kN
F"" :
3? rt lor N
BnseJ on lensiEn in /;^ ks CF:
= (o.o+o -o.ol\(o-oto)=
A= (6-J)t
F, = 26u A = (Z{+o ottat) (Zoo r,o-5 ) =
Basex oh doublu'shea.*in
A = q)"
20orlo-6rt1\
(ane*^f)
l6cl.cl x los N
pinsJ
= Xb.ozol' = 3tq-,6xlD-' Y'4'
F u = 2 t , A = ( 2 \ ( t € o x 1 s e [ S t +l .€ y l o - c ) = ? + . | L t 8x / o s N
q+.2't1 *lo3 N
Aa{- oe.I Fntig sna,!,ferVahe, i.q F; =
F*{-
-"} s^{o\
71L1g llos = 2.+?
3q rlor
pR.OpnIETAny MATEruAL. @ 20ll The McGraw-Hill Companks,Inc. All rightsreserved.No part of this Manurl mey bc dnphycq npme*d'
or diatrbupd in my forrn or by any means,witlout the prior written pcrmision of the publisher,or usedbeyondthc limibd dirtributbn b brchot' rnd
c&lqtbt| pcrnitfi by Mccnw-Hill for their individual coursepreprntbn. Stndentlruing this manualareusing it without pcenilrion
<
memberr lD and
!.47 Each of the trvo vertical linlcs cF connecting the two horizontal
of a stecl with an
made
EG has a l0 H 40-mm uniform rcctangular cross scction and is
a 20-mm
andFhas
pins
at
C
the
of
each
while
ultimate sficngth in tension of 400 MPa,
MPa"
150
of.
shear
in
str$gth
ultimate
an
with
steel
a
of
diamctcr end is made
pins connecting them to
Detcrminc the overall factor of safety for the links CF and the
Problem8.48
the horizontal mernbers.
replaced by pina
8.48 Solve Prob. 8.47, assuming that the pins at C and Fhave been
with a 30-mm diameter.
Use fttember Efo
t-
a6 fre.boJ,1.
. ,t. o.?g
o^,1o
DZY, = o3
oAo F.. - (o.esXe+
x/o3) = 6
ltf kN
F., = 31 *to3 N
irn ,l;n k.
B o'tuJ orf Iensian
A
= (t - d)t
l-u ?
F
C tr :
- O.oso)(o.olo) = lOo*lO'"^" (o^, ^!,'^k\
= (O.O,lo
26rA = izlt4oo*tot)(l0axlo-c) = 1o.ox,os N
Bose) oh {ooLle. shea,r in pins:
A --
':t
S of" = T (o.o3a)a = 1oG. Br, lD'6rn'
F; = 2 t, A = (a,X\.5orf 03 XZoA.86*ra't ) = ?t? -oGx lo3
Acl'tol
Fu is sj+tallec v aloe, i-e. Fu =
Fo"lo, ol sale].v
ES.=E
-
E=
go.o >t/o3N
= -8W-:lo" = Z.as
xtDs
3q
pnOpRIETARy MATERIAL. O 20ll The McGraw-HiltConprnicc, Inc. All rightsreserved.No part of this Manualmry bc di!pl|)'od' t€ProdtH4
or dirtributod ia any form or by my reanr, without the prior nitLn pcrminion of the publisher,or usedbeyondthe limited disFfoutionto Erchcrs rd
edrrcrton pr3rgdttcdby McGnw-Hill for thcir individual coursepmprntion. Snrdcnsusingthis manualareusingit without pcrmi$hn
-4
Problem8-49
E.49 A 40-kN axial load is applied to a short wooden post that is supportcd by a
concrete footing resting on undisturbed soil. Determine (a) the morimum bearing
stress on the concrete footing, (b) the size of the footing for which the average
bearing stressin the soil is 145 kPa.
(a) Beatt",g
s*^ess o"'' corc.ste
f.,o* i"'e.
1o l4N = \o*lot N
3
A = (toaXla@)= 12* ,D3**a = l?, 16 ,n'
p=
g=
-lgi-tg:
*
s.s3xtD6 ?o,
a33
MPa- d
g= lr{5. [.P<= t]Sxlo= ?q
Llo x /o3 N
= *"t,tl.#sx,os
t
=
o-?-1s8e,,,','
A= b'
sl uafl )
't
i : t
o. 27 586
=
O.5?S
v-,
b + 525 hu I
PROPRIETARY MATERIAL. O 201I The McGraw-HitlConrpanies,
Irtc. All rightsreserved.No part of this Manurl mey bc displayed,reproduced,
or dirtributcd in eny form or by any msans,without the prior writtcn permissionof the publisher,or usedbeyondthe limitc.ddirtribution to tcrclrcrr ud
educatorspermittcdby McGraw-Hillfor their individualcourseprcparation.Studentsusingthis manualareusingit withoutpcrmission.
Problem8.50
E 50 The frameshor*nconsistsof four woodenmembers,ABC, DEF, BE,and
CF.
Knowingrthateachmcmberhas a 2 x 4-in.rectangularcrosssectionand ttru
eacl
pin hasat/z'in. diameter,detenninethe maximum
of the averagenormel strcss
(c) in memberBE,(b) in mernberCF.
"ilrr"
ls in'-1-
w h'-lc
A)e suppo.t vea,.Atov,s
+r' fig,r,n
a-S sh aurn.
Dsi"'1 entiF{. Fra*,e as $r'e< btl-
Zl+ = o:
rsin.*|I-
4o D, - LrlS+ So[+eo) = o
Dx = loo )L'
soin.-Jr
Usb ms,\be* DE F a.6f"** bdf .
IF
t-' =e :
FD/-+Dx=o
Dy = *O" = ttoo )bD
EH= = O i
{sc14Efr')
7H-= =
(eo)(S F"") -(ts)Pt =o
O:
Fee=-225o lL.
( S o + r g ) D f= o
F..g = 1So lL.
ynaep,
in Cor,rpne.ssica,r
LAn BF
Ay"ea, A = ?in,* 1;n. z 8ina
(4) S*ress
, 6es= .F,e"=
- -32so-=
g
A
(b) Siress
"zlion
64r€a- OCCuv\S
A.i,, r- (a)(eo -o-5)
\ rynit irnu*t
see*io.
4
i.a tensron rner,nbe,^CF
M i n in, ,.rp', se
7golb,
-4alpsi
6
-=F"--7'So
Q"":
;;
=
tr
= 7.o i','
"t
= tol-l ps;
P in.
Problem8.51
A+ €aaLr boll
t.Sl Two steelplatcc are to be held togetherby meansof %-in.-diamdcrhighrtrcngth gteelbolts fitting snugly inside cylindricd brassspoccrs.Knowing tr* thc
wcregc normalstressmustnot orceed30 ksi in the bolts and 18ksi in the opl€ffi,
dctcnninethe outerdiarnctcrof the spacersthat yieldsthc mosteconomicalandsafc
design.
J*olio",
*\ a uppqn p,lo.*e is polle/ donrn by *lt *usih
A+ *[,e scrh^e*;rne *le spa.c.er p,,rslres
Force R of *l,e b"!t.
+h^+ glale upwarJ wi*Ji a. c or,rpr€ssir/e fo*"e
rn a.i", * oin
P" . In
ud e n *o
e1.li /,'btiuly\,
Ps=e
F* +[. E"I*,
F-t
f
5
rl
Ite
sPa,ce"r
E 1r,*ing
Pu
Au
E =
q
R
=
A*
r+R
Td'
+R
or
9q
f
b
-
T €, drr"
or R =+G(do'-J.')
frffiIfT
il (dot-otr')
Pr arhJ P-s,
+sbdr' = Ft"(d"'
d.t =
drt .r
*
;;
d==(f + #X+)'
- rlf )
( r+$) dr'
0. ,c€e7
ds 3 O. tlO8 i',-
tlprods4
rnoFnlETARy MATERIAL. O 20ll The McGraw-Hill coryaicr, Inc. All rightsreserved.No part of this Manual.mgr-pediwhyo4
pcmieion ofihe publisher,or usedbeyondthe limited distribution to terchon ud
or dirtributed h rny form or by any means,without the prior *tlipermission.
drc.1ort eG-.tt61i by McGraw-Hillfor their individualcounc pnqdrtioo- Sudcntsusingthis manualareusingit without
E 5? Whenthe force P reached8 kN, the woodenspecimenshownfailed in shcar
along the surfaceindicatedby the dashedline. Determinethe averageshearing
stressalongthat surfaceat the time of failure.
Are". being sh een eJ :
f i = Q O n n q s rl S n * = l35o ^.t
N
Fo*"
=g
Problem8.53
= t3f,o xl(JG wl
P
S-qs*[o3 Pa=5.?3 ]aPa-<
A
8.53Knowingthat link DE is I in. wide and % in. thick, determinethe normalshess
in thecentralportionof thatlink when(a) 0= 0", (b) 0:90".
Usa hne'^te* CEF "'s ct f'""it bJY'
DEM.=
ol
= o
- l" Fo, - GYeo sino) (tel(eo .^,s6 )'
Foe = - \O sin I
D aose
A o " = (r)C*\ = o- l " S
b"Jo /;'qta:1
Ree
-:
c*
'lV
ir'"
5ot =
(al O=
6Lt
(b) e=
Ge
- go .flL.
- 6Io
P=i
E- = -9o.,0L'
= - Af,o psi
PROPRIETARY MATERIAL. @2011The McGraw-Hill Companies,
No part of this Manualmay be displayed,reproduced,
Inc. All rightsreserved.
or diskibutedin anyform or by any means,without the prior written permissionof the publisher,or usedbeyondthe limited distributionto teachersand
permittedby McGraw-Hillfor their individualcoursepreparation.
educators
usingthis manualareusingit without permission.
Students
/ i
.I
t
Probhn A.**
ctneeus, eot\
Six
bainl
A$f Trp iloo&n planks,each l2 mm 6idr nd225 mm wide, rt joind hf fu &1
mfitiF joint town. Knowin3 Ss thGuruodusedshearsoff along ib 1l*r Sr
t MPr, determinethe magnitudeI of lh nil
, Stc rrurfe drming stressre#
hd 6J will causethejoint to ftil.
16 rr"n * 12 nn
aFe
sf,eq.rcC.
A,'r*: A . (e\( lgXr*) = I l.f,?**'
?.25mm -a>
= lrge *lo'o r'r\
t'= .P,
A
p= tA i (dxlalXtf-fexlo'.)=
q.2Ayto"tt = 7.nkh}
ry tr *fhl
O Sll Th ]le{hr-|fl
b. All rights reserved.No part of this Md
l.lOtl|3TAfT
XtT[ltlL
Ceryir+
fttdt
of the publisheaor usedbeyord tb: hu
c trrH
i ry lrr * ly ry n,
r{ld
tte tr|t r'di.pnilon
tr r|lri h#rifurl m
*ffi
fy Xdlr-l*
ltrdents usingthis manualareusing it riH
Pn*f
F*a
rulbr
rr|rfu.1
d
f *
F
Problem8.55
Us" one S"" t< as a. f"".
boo,\./,
+)Z Me = c);
71 E - (zolllSrro)
E = ,?5o .!L. -->
|^5oo!6
LZF*
= o:
E+Br = O
+t
(al
& = t25o .lL *
zt=o:
B
*:
8f * 8;
_She".!h1 st*.r
fSoo = o
(t)
B
Ap;.,
BJ = lioo -lL
l?Soa + fsoo
in
pin A
Ar,.=+Ar:={hf
d( - , \=
Brr=-E
B-
= o.t q C Z S
tqs?.sc
in'
7.qq*loz
O.lq635
f)t
q-?.+k s ,
B ea.,'^3 s4 r ess a,-[ Bg-
B-=
dt
1152:-56
(+x+)
=
6,-?5*1o3y=,
6-e5
PROPRIETARY MATERIAL. O 20ll The McGraw-Hill Companies,Inc. All rightsreserved.No part of this Manualmay be disphycd reproduccd,
or distributcd in rny forn or by any rcaru, without the prior written permissionofihe publisher,or usedbeyondthc limitcd distribution to tcaihers a116
pcrmittedby McGraw-Hillfor tbeir individualcoursepreparation.Students
educatoru
usingthis manual uiing it withoutpcrmirrion
"r"
4
t/z-in.-diarnetrr#
L l/"-in-rliompf:
d-l
ffif
/R
ie
Gtr.oA tn
t'tf A
is fitted
sd lB
to rt trurnd lrolc near end C 0f ffrG
ttoodn memberCD, Fs Srplolding shown,detanrninc(e) ec maximumfifi*r1n
naill rhess in the umo{ $) SG distance6 for whi& trc mmge shearingf;
if
90 pd m the surfacesiilfi*Gd hf dre dashedlinel {c} thc lrrunge bearingstu m
ProbHn8.S
: L#
fursd.
(A)
Pl**iF u*r not"y^a.I s*ress
An"t
=
3
P
(t)
t
b
(ct
6,
P
A
P
2t,t
$= A..t
P
in J4re woo/,-
l. 8?S itn'
533 P $ ,
<
?bf
Iooo
(?X+Xqo)
7-41 in.
' 1 - - 4 1i n ,
b =
'/ooo
(*)(+)
26C7 F"i
O 201I The McGraw-Hill C,ort:*.,
hc. AI rib ruerved. No part of this Manud ry h fihrf'
?]ftlffTAfY
HATIIIAL
ef*r
of ir p-*her, or usedbeyondthe lH
or 1||'ha
L ry frr r ty ry means,without thc prior writtc f*
rrfil {ti manualareusingit without pil*r.
ly Ha0n-ilf;n for their individual courseprepalilin lth
*q5lr
tda
<
prU'
b *
d
is placedT-tlo*"
8.57A steelloop ABCD of length 1.2 mand of lO-mmdiameter
eachof l2-mm
DF,
and
BE
Cables
aluminum tod AC.
arounda 24-mm-diameter
s(ength of the
ultimate
the
that
Knowing
load
the
Q.
diameter,arc usedto uppfy
for the steel
strength
is 260 l,tpa anOthat the ultimate
aluminumused tor ih;il
thst cenb€
load
largest
the
Q
determine
MPa"
usedfor the loop andthe cablesis 480
desired'
is
of
3
safety
of
wiratl factor
applied
Problem8.57
,l
1an
Usinq ioi*t
B
osq$n.c- b"J1
anolocJr,t,clcria1 sy *-"in1T
?-t'fio - a = o
e= 8n.
[)si'.n ioi^] A as q-S". bnlu
o.J4.1. siol*h3 sy"nr'.a*""t;
F^. = o
Fo.=o
Z- $ri.
E'{A
"'
.8":;el ovrs.l**3+lr # c",Lle tsF;
rf\
*l
Q, = q,A = qr\ol ")= (+ao/166)
+ (o-oti-)'= Stl.?qxtd
Bose J on =t^e',
X++J Jee/
-!"., y:
tr^q = 36"4 = 3q,ffid)
3 r.*"
e - ^f\ 6
-6
t1
tq{u - =
.JA
-c-
\-,rr
A
n
F
= f; t +Boxto') + (o.oto)" = 45.L+*lo' N
Bose) oA sfnr" {an "F noJ AC I
c:$tr".,E+qA = +gr(+d-)
= tr (u" vlDn)
+ (o.oa't)"=
Ar*o"! uf+;^te
,l"J
O, is *he s^o.lfert, -'fl
A!,lruJ.!" ) rJ
A
r^' = JIa
trs.
45'?txf ?33
g g . 1 . 2x l o 3 M
Q,, = +-5. vtd
",t
lS. of x loo lJ
= l,S.Og k iJ
FROFRIETARY MATERIAL. O 201I The McGraw-Hill Coryldr+ lrp. Att righu reserved.No part of this Manual may bc dirpleytd, tpprudud,
or dirufr@ ia rny fonn or by rny rmans,without the prior wrinm prmirion of th publisher,or usedbeyondthe limited dirffiutioo b brch€rr rad
cdnrlofi paruittd by McGn*-Hill for their individual coursepilfrsion. Shdcffi uring thb manualareusingit withoutperni$bn,
joid bf 6c rinryle glued joirt *rorur. Knowinj S* P - 3.6 kt{ and trr ilr
*b
ril!6
of the gluo ir l.l MPr in tensiond l.{ }tPt ia shear,dcadr
hbofrfuy.
f
A = Io'-
6-5' ! f;S'
fr= 3-6 Lu =
-A.=(zs)Ctzs)
Tentioq
o4
qlue-
--
q-v-.a -
s:
P "u{a
A'
3.G * lo3 N
7.g?5 ' ki r# = 7.975 * fdo *n
(g-e"tol)(ses*s"\o
3
9- 3?5- * fo-
P""
"
E S - = C6-
3.+7
(g-€ >"ld X*t" 5o')
(A\t1.gzs' to-s)
l + ? . o 8 t x f o 3 Fc,
trs.
1 . 1x t o '
l{7- o8f r [o3
q.57
i{
Tle
s"'.o.I.l=n A**o.
J =*{"*7
3ouet'Y\s.
E s. = 3.'+1
l$fEfAlt
il*TIINAL
O t6ll Tte McGraw-Hill Companies,Inc. Atl rights reserved.No part of this hdrd
cfrlhtltryhrtyqn;witlroutthepriorwrittenpermissionofthepublisher,orusedbeyondtbHrhlebrfud
.-fir
17 lChr.lfl
fu rr.ir individual courscpr€paration.Shrdentsusingthiu manualare usingit tlh
tr*a
ry
h t;tcA
pr*n
ryrfu*
-1,
Problem8.59
r-600
8.J9 Link 8C is 6 mm thick, hasa width w = 25 mm, and is madeof a steelwith a
480-MPaultimate strengthin tension. What was the safety factor usd if the
structureshownwasdesignedto supporta l6-kN loadP?
_l
mm
Use bo* AC D as a $ne.
Fe.
V : " d y , a n J n o f e tl- ^.1
mer*"'herBD r's c L l w o fol'i,e
rt e*-Lel
.
ZMo = o i
(+eo1Fea - (6oo)
F-
6oo
Fec=E;r
^
=
=
2O t, /Os |vt
rn€raryLey.tsC :
)( o.o06)(o.025)
F.S.=
+ =
rTe
pROpnIETAR.y MATERIAL. A zfill The McGraw-Hill Companies,Inc. All rightsreserved.No part of this Manusl mey be dirplayd rrprofixr4
or distributcd in any form or by any means,without the prior writtcn permissionof the publisher,or usedbeyondthe limited distnlbutionto tcachcruand
usingthis manualareusingit withoutpermission
cdpcatorspcrmittcdby McGraw-Hillfor their individualcouneprcparation.Students
{
8.60
Problem
t.60 The two portions of membcz_AB are glued together along a plane forming an
angle d with the horizontal. Knowing that the ultimate stress for the glued joint is
2.5 l$i in te,nsionand 1.3 ksi in shear, determine the range of values of 0 for which
the factor of safetv of the members is at least 3.0.
rl
I
Ao=
(l,oJ(t.ts) ? 2.So,i^L
P= 8.+ krps
B nsel
evtte'.r i le
v^*
5a=
Jl
fSosel
g
ev, sh eanin3 slncss ;
" A'frsi', 20 = -',
Hg"..-
r)
?l-3;
g6g'@
= q& = (n-s.l(a-s?-)
= 0-868
J'u
z
7-
zl-3 0
e
?
e : 3 ? .B o
olo
Or
=
= n
sinOc-rs@
tr= &.
A,L
: (zXe-so){l.3f
_?_
2 g = 6 + -S ? '
s*res: :
jl0
c.osee
6t>s9 = O.13t€q
R = (FJ)P = 7-2.k:y,
e i - ^20
sin
o. qo?7 g
g -< 32.3o
o
PROPRIETARY MATERIAI. O 201| Tb McGraw-Hill Companies,Inc. All rightsreserved.No part of this Manualmay be displayc4reprcduced,
or dktributedin any frrm or by any tmens,withouttheprior writtenpermissionof thepublisher,or usedbeyondthe limited distributionto teachersand
cducatorspcrmittcdby McGnr'-Hitl fot thcir individual coursepreparation.Studentsusingthis manualareusingit without pcrmi$$ion.
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