chapter2mechanicsof deformablebodies.pdf

mechanics of rigid bodies

1. Problem2.10
2.10Ablockof250-mmlengthand50x40mmcrosssectionistosupportacentric
compressiveloadP.ThematerialtobeusedisabronzeforwhichE
=95GPa.
Determinethelargestloadwhichcanbeapplied,knowingthatthenormalstressmust
notexceed80MPaandthatthedecreaseinlengthoftheblockshouldbeatmost
0.12%ofitsoriginallength.
A-::(SD)(~o) :zooo ""-",,,l..::- Z><10-3m~
~=%0MP~ r80'1-10'?~) E~crSx/O"1 'P~
CO~5iJ€"";V\j JPo...raJolesfv-ess
6"=*P-=AD::(~/o.3)(~OxIO~)-=160)(103N
CO"'SI-~~"'iV\~ol2i>wQ...lJf2 Je.f'oJ'"~Jio"\
~~~~ 'p=.£iAct)o::-(95x/o"f)(~)iIO-~)(O.OOI:t) ~
?.2S~{O'3.Iv
5M~lJeI/'1/0..])IJeof'PjOver",S
po:IbOY-lo3N
:::
160.0ktJ.."",

2. I~A
f'
18in.
il
~=
r<eerVi~CJ
Problem2.14
28ldps
1.5in.
2.25in.
"C
PL
EA
(;':1:
A
28kips
2.14ThealuminumrodABC(E=10.1x106psi),whichconsistsoftwocylindrical
portionsABandBC,istobereplacedwithacylindricalsteelrodDE(E=29x106
psi)ofthesameoveralllength.Determinetheminimumrequireddiameterdofthe
steelrodifitsverticaldeformationisnottoexceedthedeformationofthealuminum
rodunderthesameloadandiftheallowablestressinthesteelrodisnottoexceed
24ksi.
D
DetoV',G\.J.'cW\DtQ.)v;I1\';-'<.)J
d
'PL"a+'PLac.=-E(LAe+LBe )
AA8E "'~c..EE AM? Ace.
~8)(IO~
(
12. +-J'&
)
z-=003/3U. j...
/0./x10'-"10.5)2.~(:z.15")Z. ~
~A':
~
E
.
Ste~j)roC:» 5=:0.03/37' 1'1
A-PL -(;'?8~/O"!.)(30) .1-
-ES-(J.9>tIO6)(O.O3/37t:.)=O.'i~317.h
A
-P 2.8)(10:3- .1-
-e> ':A'i)CIdo-I./&67 .M
.
Gtv-e~;~+-k 2G.lI'jeo(' 'Ie.. iJe
cl=-IlfA ~J(4)(t..6b67)IT If
A=J.I(.(;7;..,~
t.~/~;..,. ....
-

3. I.
Problem2.19
3in.
Co)~~£,.e+~8'':
(61
~8":-S&
2.19TwosolidcylindricalrodsarejoinedatBandloadedasshown.RodABis
madeofsteel(E=29x106psi),androdBCofbrass(E=15x106psi).Determine
(a)thetotaldeformationofthecompositerodABC,(b)thedeflectionofpointB.
PoV'hol'\A'B: ~e=I.fo)</0'3.llr.,~L..s= 'fa"~')01-=2.in'J
A",s-=~J" ::'~(2.)1.~3.1'"1/6,.,..2.)EAa::'";<q"'JO'f'S"
~-'PAC3L"13-(LIO~lo?J(4())- -~.
AB-EMA.-.t-(2"h<IO")('3,J'f/G:)-\1.S6/Q)1./0I\'}.
Pc>...h.""'Be.:PBG ~-:(Ovr/O3lhJ LaG-=30iY1.)d-=3Ir'1..J
A
1L
J
t.Tf
()
...
7
.2.
f-
IS10
'.
I!.<.::<t::'7i~~.Ob'if6 /r.>4c.- 'Iif~I
S=PBc.Lec.=(-~O)lID»(30) -=-s:c;S88>1/c/'in.
&.EI!,CABc.(15)(/0')(l.o'lI(') .
-c -(
17..5"1><LO-S.bS138)t10
~~JI,'10"L01;V',~...
~ -1
018=5.(,""'10i",.1'-'

4. Problem2.21
B
50kips
f
8ft
U
""""I I
I--13ft-I-13ft1~~
F"1Ia
-:?
A~FAO
T~Sk..p~
I-
2.21Forthesteeltruss(E=29 x106psi)andloadingshown,determinethe
deformationsofmembersABandAD,knowingthattheircross-sectionalareasare4.0
in2and2.8in2,respectively.
5+1L~..c.s.. Reo...//t,oV\sc..lA~.Jc. ~k:p,,-t
Me""be'8Dis c..7eorofO-h<-ebeof'.
LAB:=';I'~/+g'Z:;;:-15.';«;'1-'3ft".1'g3_/72in.
LAo-=-13 ft:-ISbin.
~oL,.o
SAD":0fA"D
Spa~
~MLl!a-(-47.701)(/O~)(lg3..'7~)-
EAAt-(1.Cf~lo~)(It.b) -
-.0.0753 ;1).~I I
=- (40.b2S-><ld')(15(:.)
(A.")~10c.. )(-<'~g)
= 0.0780j~.--
Usejoi"t Ao..sf'r-€hoJ'I
+tLFj= 0: 2.5ofIS"(,'f3F;. =:0 13=-47.70/k;p"
.:t2"Fy::-0:
13
o=-'fo.b2S-k,',sD-\IS.?s FA!!.:::0

5. Problem2.26
D~l
cJm
2.26Thelengthofthe2-mm-diametersteelwireCDhasbeenadjustedsothatwith
noloadapplied,agapof1.5mmexistsbetweentheendBoftherigidbeamACBand
acontactpointE.KnowingthatE=200GPa,detenninewherea20-kgblockshould
beplacedonthebeaminordertocausecontactbetweenBandE.
J
Ri~iJY'odACEt"O+<vh~.5H."'°4 hc..~~R<Z.e~ C.P~'SIi!'4"'-1"
e-=1.5></0.3=3.75>l/o-$'/'tA.J
o.'-to
POi",tC.",",0v<2'5,JDv./'ro..'NQ."J.
~c.=-0.08
e=(0.0'8)(3.75)(/0-3')~goo)<10(."""
$<:'0=~c=300x10-<:WI
A
7ft~7/
(:
~ '2.. -,~
GO::-4Of""tj::(')
~3./'-IIb""""-=-3.''-II"><{oWI.
E
"
,
q
P.
Fe.oLc.o
=A00)1 0 0.. ~C.D:' E:Ac.o
Fc.D-=-E
t~GD::-(;?OO)(fO"'(g.1"116></0-<'»(300v{O-c)-=753.Cf'11 N0.'lS
O<;ebec.."""ACB"-sc..f~e.bOdy. w=M~~(~O)('i.8f)=-19G_~
N
~ot~1v ~-
r ~ +J2MA =0~O.Og1=;.0 -(O.4D-X.)W -:.0
A I (00<6)(753.Q2)
1_D."Io-(-\ 0.40-k.::-.I"tb.?...=-0-307Ll-3...,
X>=O_oq~,.",=92.b-....

6. Problem2.29 2.29DeterminethedeflectionoftheapexAofahomogeneousparaboloidof
revolutionofheighth,densityp,andmodulusofelasticityE,duetoitsownweight.
11b2p~+t,~
1TbPd h ~
2Eh.iTb~
t'jrAt::
A
For+~e
,
1/<'
ej~...,e~tr=-b(i)
VV'1..=-7Tb2 f
=1Tb2,P~f
lfb£~
S
'f
t-.P<ojJ.j =-
-
S
h~ =
-0E ACJ)
A-=
dP~P3Ad)'
!?=s;oJp ::
C-"'"PAt
c:>-L:..EA
eJeme..-t
~
r-I\:,i
k-b-"
Le-tb':"v-J.'v'S at-+~ebe..se CL"'~
V':::V'e..A,'vsdsevt-""'"-.vd-I.,
?;.ooV'J,'n~+e::!.
~h:L
LfE
~

7. Problem2.35 2.35AnaxialcentricforceofmagnitudeP=450kNisappliedtothecomposite
blockshownbymeansofarigidendplate.Knowingthath=10mm,determinethe
normalstressin(a)thebrasscore,(b)thealwninwnplates.
Brasscore
(E=105CPa)
300mm
Let
Pb.:fov-tio",of~\"o.i+Olf"c.4tc.o.",-'N.eJk'j
bV'A.55~o~
Pa..-=patH"", c.CtV"V';e.J /0..,+""0Ju,..,.,'>,\o,)"""
'f.1,te.s.
~='P.L P-£hA.~
E~A.. c. -L
~=p~L 'R
-::E...A.~
ED-A- c::... L
P=H~+'PQ.-::(EbA",~~A~ )~
~-~- p
c.-L-
EIoAe-+FdA4
Ab
::(GO)(LlO) =:?<tooWI""'"=-2lfoOxlO-r.. iN\~
AQ..=(~)(bD)(IO):: 1:Z0011-1100\1..-;::/~oo)c",o-"W\'t.
E.=
4-50)(103
()OSx{oq)(?LIooxlO.G.)+(70)(10")(,~OO)(to-')
-3
>=I.33Cf3~to
6"\0~FeE.~(toS>tIO"l)(1.33"}3x/O..3)::: 140.b)(/O"P~=' 140.6HPQ."(~)
(h)
6<'\..~ECt.,£-::(70)llOIt)(/.3'6'13'1c,d1)::
q3-7.5>'/0'PGt:;::~3.7rMP~ 4

8. Problem2.39
Dimensionsinmm
A
R:-I
e, c.
3
bOktJ
0 E
.-~(~
'toktJ
2.39Twocylindricalrods,oneofsteelandtheotherofbrass,arejoinedatCand
restrainedbyrigidsupportsatAandE.Fortheloadingshownandknowingtha~Es
=200GPaandEh=105GPa,determine(a)thereactionsatAandE,(b)the
deflectionofpoint
C.
Atoc.: E=200)(/0" 'PQ"
A=~(1fo)1=1.~Sc;,"'t><'lo::'MI'>11..=1.'l5"~)tlci'lN\.1.
EA=.2..5'-31.7;.<lot; N
C
foE:E'::Ios>I'0 q'P~
JI/.'t
t. . -~ ~
A=-~\.'30)=-706.8f,,"'1=- 70b.86x/0 ~
EA;7Lf.1~o)l/o'I'J
A+0>B:
P-=1<" L,: 180""',-= D.180 "'""-
<::~-PL..-r?"(0.180) - . -Itn
0>1:,("3 E:A-Z5"I.~27)lLO6-716~~o}<IO K,.
8+0~ ~P= Rio-G,OxLO~L. =- l;CO""'WI=O./:lDPt
S-Pl-(R,..-60xIO!)(6./2o) - _It -,
Bc-EiA-~51.'!.~"")<lOc.-4'47.47'1tIO RA-:?6.'g"'8~{O
Cto>D: p:::RA-r;;{»l.IO~ L:'DD,."",:00./00"'"
-PL-(R,..-6Dx/U'?)(o.Ioo)--" -4:
2;ec.-t=:A- 7<.f.2Z0"IO~ ==I.'347~s)tIc.R,.-8D.'8'II)I/O
DtoE: p:::R" -100,,10"3-
~=.Eb.-=(f!,.-loo)(/dXo_,oo) =1'="/7
rl-r
' -'10-'
~Ll
7
'Z-
I
-(
~Oif£A 7't.'l~O)( to' .;;) ;J~)( 01"." ;;I...JS)(0
AfoE:5AE =~A6~See.-+~c.o +~E-=- 3.~58!57>l/0-1 R"-24a..lf_~""'><JO-'
L~100 ~0./00 "'"
Sincepoi1l+Ec.c.1I\1I\C>+Movere2J.ive+6A)
~AG=:0
(tt,)
-"ID ..(;
3.~S837><lo"2'-l1.!i~'f~{o=0 b2.9k~<C-~R,.~,62.cg3/)tlo3N.
(")
R
> 3"3 '3
"E::: ,.-/00>'/0 ~f.'l.8>LlD-IOO)l.{O =.-37.Z><[O ~
'37_;:lku~ 4
(b\~c=~AB +SBC.=- Llb3G7vlo-<t 1?~-~G.g43)(/O-C.
=(/./b'3~~)(10-'1)(62.83/x103) -/..b.8lf"8)(10-'
U -I;;
:-.6.3)(10 W\ 46.3ph->-~

9. i
8in.
t
lOin.
L-A-iB
oil
2.46TherigidbarADissupportedbytwosteelwiresof1~-in.diameter(E=29x
106psi)andapinandbracketatD.Knowingthatthewireswereinitiallytaught,
detennme(a)theadditionaltensionineachwirewhena220-lbloadPisappliedat
D,(b)thecorrespondingdeflectionofpointD.Problem2.46
c
p
le..tebe+h-eV'O+~t;G)v\ofba", A~C Q~'~i--'S~, ~
- D
l12in.-l12in.-l12in.j The.,.,
as==I~e
~c.=2.'+e
~=Pe.E"Lsp
g AE
'P~~EA~ge--=(?"C;>t'o')¥C~y..(I~8)
L6E 10
c. D
A~
eB' c' C'
Ati
-Pse PCF
lJ
?
!
-
tOb.77)(IO$e
b
~= ~\~ LCF
Co EA
Pc~ ::0 ~ ~cJa=(.<q)(IO(.)~(.l-)"'(.Zl/a)
LC:f: 18
-="~L~3x Io~e
lJs.in~~~~boJyABc.o
D2"MA::0 12~Et-1.Ifp~-3'?=0
Oiy"/O6.77><[O?'e)+(:N)(118.63)clO" e)-(3{.)(;no) ::0
4,1:Z~'5)(IO'e=(3')(~1o)
e= I~'118-G".,c /0-3 I/'"J
(0.)P6E=(ID6.77)(/O'3)(r.'l,sS'X(O-3) -::'
?CF= (\1'%.63)(lo~)(I.q'8S-)«(O-3 )-=
~D4-.8 lb,
2~7.611,.
eIIi8
...
(b\
GD~ue~(3')(1..9125"")(10-3)~
-::
GC;.I xIo-~jV\
0_06<=1';V1.. ...ell!

10. I
Problem2.48 2.48Theassemblyshownconsistsofanalwninumshell(Ea=70GPa,tra=23.6x
lO-Oj°C)fullybondedtoasteelcore(Es=200GPa,as=11.7xlO-Oj°C)andis
unstressedatatemperatureof20°C.Consideringonlyaxialdefonnations,
detenninethestressinthealuminumshellwhenthetemperaturereaches180°C.
Steel
core
A
'if
()
~ ~ -4t.
.s-="if ~o=31!+.159 Ion""":!'/~.ISCf~IO ~
AQ..:::-~(SOt.-~o~) :::-/.t:;4cr34)(/ojWlYI-\1.
oJ -:1"'-
':"/.b~.q~,)(/O """'
Let1'5be+h~~l'J -h,II'CQc&t./'r"l'eJ by+~
I:.Glol'rie)~r'+healulM"ndW\shell..
Totc..iQJ)Llc..ito""C'..~ p:::-P",- +1-'5-=0
.:st-et'Jc.o"""0.",,)'P~-rho.t
~=-'Pc.. (1)
Pefo\l"""",Jic>o.t..
~=-
'PeL
L
£..Ae;.+Ld.~(f1T)
-PsL
-~A.s-\-Ld..s(AT)
:Po.p~-
~~-£:.5A~.
(CX.s-dcJ(~ TI
Us;V\~(n
(
L.+-L-) 'D
G.,A", EsA 1<:<"==(ol,s-cJ.o,.)(b.r)!.S "
(' I
('0)(Iott)(1.6QC,3..,)(IO-~)+(loo)(loq)~31~;15""4)<10<:)')'Po.
=(11.7)(./0-1:.-::l3.(.,></o-t::)(J80-~())
(~'1-57:])(10-')Pa.. ~-J.q 0if'A./0-b
'P11.':"-77.47I)t/03
N
Sfre'5s;V\.rJIJ""':Y\(,)M.shell ()~=JL":-17.Y71.)(103
A.. '.6..,.q3"4~/o-~-:--Lf7.0~/o'Ptt
-47.0/vIp",-.

11. Problem2.57
0.02in.
1r-14in.~18 in.--j
Bronze
A=2.4in.2
E=15x106~Si
a
=12X10-jOF
Aluminum
A
=2.8in.2
E
=10.6X106~Si
a
=12.9X10-;OF
p-
-.t
p
I--
Bvt
~f
~'PLb+'PL..' -=-
£"A,. E:Q.A"
. 1'4-
(
+
-(lS)(/0(")(~.'-t)
(~)E,I)Jl'~O
2.57Determine(a)thecompressiveforceinthebarsshownafteratemperaturerise
of180°F,(b)thecorrespondingchangeinlengthofthebronzebar.
The",.4R~P""O""\"':D'";-f+r-e.e0-1c.o~t",-~",,-t
S-r~LJ,d.b(AT)+L~(~r)
=-(J"+)(/~)(IO-(.)(1~O)4-(1'8)(,~.<=1)(lO-(.)(I~(»)
=72.036')(to-'!.In.
CoW\sfrA-ined&'X.f'lIA'SI'o.....
~-=o.O~ilfl
S.~o.,-4ein'jJ,J~~indvc.eJ GD_pre.S~;v~ ~el'
~ -3 _3
eJp-::7'"2.0'36'11l0-O.o;t::.52.O'&;>lio10'"\..
(~+~)?
r:"A..£c..A....
12
)
(IO.6')(IO~)(~.8)?=-
'9'15.3'Xto-'"'P
Cf"tS_g~)<.lo-1'P'":'52.Og'"I<(O-3 P-=..52.?.7<=t")a103 Jb
5:2.3k:f-s~
(b)~b:-Lb db(AT)-"PL..
f1,A!>
~(\'-l)(I~JC'o-C)(180)-(S"~_~~)(lO3)(14)
OSxIO~)(~-tf)
1
':"Gf.Cf,)<'(o-J..",,-
~

12. Problem2.68
y
2.68Afabricusedinair-inflatedstructuresissubjectedtoabiaxialloadingthat
resultsinnormalstressesUx=120MPaandq,=160MPa.Knowingthatthe
propertiesofthefabriccanbeapproximatedasE =87GPaandv=0.34,determine
thechangeinlengthof(a)sideAB,(b)sideBC,(c)diagonalAC.
6><
=110x/a'"Pa.) ~=0 ')6;t-:: II:0)/lD~"P(t
Ex::-tC6)(-£JS-V~z)
=87~IO"r[I'~D)('O' -(O.3~)("o)iIO")J
=75'1.0';)(/O-c
£z::t(-")JC-,t--V~+6;)::87~/Oq-[-(o.3't)(I~O)(IO") +IbO)lJO"J
-3
~"370I )/10
~~-::(AB)E)( -=(1DD#II"')(75'i.O~)(/6() ~
~e,c,~(13G)£2=-(.75M )(1,'37",>t.IO-~ )-::
(a.-)
(6)
(c:~:
c
de:::a.Jl1.
c
13utj
O.O7SLJ M"" ..
O.loZS -4"""'"
let-he..?_S\~-l'.sdfV'~3~th-t'G(.V\jIeABC.c...Scsb.;~",Jc.
~ t \"L
c:.-=0..-tb
0btc..;",J,'fte.r~",-t,'e.Psb"JCAJ<:,JJV'5.
2cde =?a.dQ.'"2hJb
+.!2.db
c.
0.-=100vY'otv>
..;
b=IDOVr\""..>
c:~/IO()'t..+.75'.-)='~\'J1M
clCl.~~A~: O.O7S'l ~IVI db~<a&::o..370v...~
~ ::ole.=- ~ (0. o7S'1f)+ 2€,.-(0.IOl8) =0f'Z~o.
At:. t25 IlS . .
I'YII')')~

13. Problem2.78
f1>1
2.78Avibrationisolationunitconsistsoftwoblocksofhardrubberwithamodulus
ofrigidityG=2.75ksibondedtoaplateABandtorigidsupportsasshown.
DenotingbyPthemagnitudeoftheforceappliedtotheplateandbyIithe
correspondingdeflection.determinetheeffectivespringconstant,k=Pili,ofthe
system.
E.J:+evfilfe Sr,l',"V1jcc>"\sfc..V
k=f~ :<~A
Dw.h~,:
G::-~.ISK,o~fS"
P--~((.)(tf):-~L(It'!~
h':"I.~Sin.
k=-(2'{::l.71))tto"!.Y1~)~
I.'ZS
IoS.,)l.lO~ib/;" ...
sheCt-;'--+'f'I. r-S
-11
she"'\\'I.s-h-e.s s :- GY =-&
I 1
Fof'c J.p-AT-
GAS-
-h
'P=
J..G-A
\.,

14. Aij"..;"'>'>.1Ie I/c...xve01 t;:.I..-
ANt-= Jt::()()
2. .3"-
goIS :''~oo..."",= 1.2o~/oM
Problem2.94 2.94Twoholeshavebeendrilledthroughalongsteelbarthatissubjectedtoa
centricaxialloadasshown.ForP=32kN,detenninethemaximumvalueofthe
stress(a)atA,(b)atB.
15mm
(et.)A+hole A V'=i(:to)-=10~""
d=-D-:<r-::-100-:(O -::=80INII")
Y' ,D
d
':go
:.-
O.12~
Fifo,
F'-1'2.b'10..)K=-:<.G.5
(b)
-6.~kP=-(2.G.S)(3l.)(IO~)=-7D 7)(lOt:.£)
AM~ I.~o )0((0.~ - r4..
Ath;)9~'B r:-*(50) -=25"",,)c1=-/00-SO:So ,
A.net~(SDX'S)=-750M'"2-=-7~D)(/0-(;Ma.
70.7MPa..~
V'-
&-
'2.5'"-D.S"D--
!i~
k'=-:<_/~
6 ~KP -=-('l,'~)(3l)l.lo3)
-A...t15"0>('0-£
'::'"
Ql.~)(lOr.'P4 q~.ZMP4.. 4
PROPRIETARY MATERIAL.«:>2006TheMcGraw-HiliCompanies,Inc.Allrightsreserved.NopartofthisManualmaybedisplayed,reproduced
ordistributedinanyformorbyanymeans,withoutthepriorwrittenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersand
educatorspermittedbyMcGraw-Hillfortheirindividualcoursepreparation.IfyouareastudentusingthisManual,youareusingitwithoutpermission.

15. Problem2.105
A
,--9mmdiameter
1.25m
1
FN
B
Q ~.
/
~
~AG e/
~ Q
~
2.105RodABismadeofamildsteelthatisassumedtobeelastoplasticwithE =
200GPaanday=345MPa.AftertherodhasbeenattachedtotherigidleverCD,
itisfoundthatendCis6mmtoohigh.AverticalforceQisthenappliedatCuntil
thispointhasmovedtopositionC~DeterminetherequiredmagnitudeofQandthe
deflection0)iftheleveristosnapbacktoahorizontalpositionafterQisremoved.
AA6#.
~(qt,.:-~$.,n "",:a.':G3.,,"'hdu.C."",,'"
SIII')C.Q'('0c1 A8 ,'s-tobesfve-tc:.J.,J'f'eN'W\Q.V\&1+J1.:f )
(~A6)M"1(.'::"AA6 E:y:.(103.'I?'If:lU-c)(~t.fS)(tOC)
=~I.cr4S>1103N
-+,
LHp":0:I.'Q.-0.71=".6::0
Q,..""",=-~./7(~/.t:N9x/03)::"13.9'7)(lD~N
\~.~7kN ....
,-/(t=~1-4&-(~1.1'l:~xlo:!>)(,.1.5)- -3
~~a-;-EA.e-(1.00')/lD'X(;3..bl7~/D-lt.)-2.IS~?.S)(IOM
IS I
e~~ '::"s.080'-1->liO-?:'r J
~I-=Lie'~3.gq)t /0.3tilt :3.~Cf""WI....

16. Problem2.109
P'
2.109Twotempered-steelbars,eachl~-in.thick,arebondedtoa t-in.mild-steel
bar.Thiscompositebarissubjectedasshowntoacentricaxialloadofmagnitude
P.Bothsteels~elastoplasticwithE=29
X106psiandwithyieldstrengthsequal
to100ksiand50ksi,respectively,forthetemperedandmildsteel.TheloadPis
graduallyincreasedfromzerountilthedeformationofthebarreachesamaximum
value15m
=0.04in.andthendecreasedbacktozero.Determine(0)themaximum
valueofP,(b)themaximumstressinthetempered-steelbars,(c)thepermanentset
aftertheloadisremoved.
Fo~t~.e.,.,.,,'iJs+eel A,~(t)(~) ~ 1.00,\0\'\.
C =L()YI:::(Ill-)(SD)(/o'~) :: 0o:tUIgg
.
C>'fl E :21)('0to .-, .....
\='o~t~e+-eW\pe~ed.steelAz~~(;l)(~) '=
$'1'1.':LCI'L:::-('tt)('OO)t'IO'3)=:O.Olt87.76 in.
E 1.'1)1103
O.7S"-t~
T0+c:..lc:t1f'~C(.~
A=AI+A?.':. 1.76'illl2.
T"'~.1v\l'.P..J~:s+ee)yl~/J5.T-eWe~ s+e~) 'seJ,..sfl'c.
A,f5YI=(/.()o)(SO~IO?)'=.S'o)CID3lh----.
£A2S""=("J.q)/ID~)(O.7S)(O-()lf)-::G;l-14Y/O3lb.
L JLf
P=PI+ ~=1/:l.l'1)l}o3.ib':1/2.Jk.ps
('b)S1-v-esse.s0; =R':6'",":-S(»I}O3pSI'~ ~D ks;
At
6'"2.=R'::{;2~ I~)//0' -=g~.86J1/0bf','==~~.8~kosI'
A~ O.1~
~'(\<~w..<S'f2.
(a)
.FO~c.e5.
'PI';:
'P2:.
~
...
UnPoa.l:nj
£'':PL'=(\I~.\4)tIO~)(14)-
EA (:Z'1><IO'")(1.7S-)-
O.o30cr'i ,~.
(C)
Pei'Mo.r\e",ts,,-,t~p':~tIo\- ~I-=
O.OL/-0..030Cf,/
=-0..00'106i~. ~

17. Problem2.120 2.120BarABhasacross-sectionalareaof1200mm2andismadeofasteelthatis
assumedtobeelastoplasticwithE=200GPaandUy=250MPa.Knowingthatthe
forceFincreasesfrom0to520kNandthendecreasestozero,determine(a)the
permanentdeflectionofpointC,(b)theresidualstressinthebar.
~ I
~440mm~
A
2. -'-~
':.IZoO,.,..,=I'ZC>D~/O..,
FcJII'~~'r,:>yt.JJP°.r--+-lOv\ Ac: 'PAc.= A0."=("001C,0')05"0X/Oc)
PA<. (") =300 ><IO3N
rCQ
4- ~ F~ f-:oV'e1'"~J;h '() F+'Pee.-PAc:..=-0
Pea=PA~
-F=300><IO~ -S~o~/O-S
=-:(z.o;)(10$N
Dca::
~e.LdS
EA
Pea
~=
-(Z1.0)('IO!.)(0. L/</;o-D..'Zo)~0...,z.'3333K'to3WJ
-(:zOO)lID..)(I~OO )('"10-6)
~"c))(/03 -=-183.3.33 X'O~'P~
I~oo)C"(O-4;o
~c.=-
UnPoa.Ji,,~
,
~I=~L~c.
c.
EA
I
=-PcBL CB
EA
RI(
LAc.+Lk.
)
::FLea
M:.EA EA F:A
JS20')(lO'3)(0'.«+40-O.I~o)
o.4~b
=(F-PA:')L<:.e.
EA
3
=378./S~>llo N
3 3 311
=37g./~>«O -S20.,clO ":'""'11../1..'8'8></0 N
37g.'8~..tClog
f~ooxia-'
=
3/S.IS=<.)(10C'Pa..
(tt)
(6'
6'=Pet=_ILfI.g/~)({O""S.-=-118/g;?><JC/. ~
ICe- A I <t)OX'lO~ . ~
~'=(37g.f8~)(O.12o) -O../'8qO~f/)(10-3~
t.(';lClb)((D")(t'20C)x10') ""
-3 -.3 -3
~=~-~'-= 0~~3333xtO -0.''!t:}oq,)cIO~O./DLfZ)([0~
c:p Qc. de. .
-=O.IoLf'1.~~~
6~.)~-::6'y-6;.~::2St:>x/O"-3/5../5;2,~/O' ~-6S.~)([0'-f4
~-bS-.~MP~..
::-6S:~>dc)(.'P4\
:=-bS".'~HP~ ~
6fG.>f"U~OC(3-aed~-"33.~33)c'10c;+"8.IS2xlO'
RI-FL..:a-
Ac.- -
L,.c.Le.
PCB'::p-F
,p
6c.=
=-
A

18. Problem2.130
LfI7.0'1I<LcJ~
~
2.130KnowingthatE
=29X106psi,detennine(a)thevalueof0forwhichthe
deflectionofpointBisdownandtotheleftalongalineforminganangleof36°with
thehorizontal,(b)thecorrespondingmagnitudeofthedeflectionofB.
~8C.
=
?sc..-::
~AB':
-=
a~os3GD
EA6c-Ssc.-(::i.~)(lOG)(O.8)~G.oS36-
L&..- '-Is
LfJ7.Dq~)([D3 ~
?-=:fFAM..~"t.. ~(Z9')<10')[1-z)$51""36°
AI:. l,,<... 'l.S
'?
=:31~.~DxtD £
.L~':!S,g-~ox{c:J~=';-bll e::r30° ~
r'l"'"'-//7,oqxlo3 ~. 10.
'Po;:lS)(LO~':'~(417-0C;)(IO3~)2.+(gl~.~O)rLD-3~)'" =CfI'8.?::Sxld'£
'25)(Ic)'3 .
S:: ,. '=0",027:1..""-
. q{'l>.'6&)('10.3
aSI'"3'-
.....