mechanics of rigid bodies

1. Problem 2.10 2.10 A block of250-mm length and 50x40 mm cross section is to support a centric
compressive load P. The material to be used is a bronze for which E =95 GPa.
Determine the largest load which can be applied, knowing that the normal stress must
not exceed 80 MPa and that the decrease in length of the block should be at most
0.12% of its original length.
A -:: (SD) ( ~o) : zooo ""-",,, l..::- Z ><10-3 m ~
~= %0 MP~ r 80 '1-10' ?~) E ~ crSx/O"1 'P~
CO~5iJ€"";Vj JPo...raJole sfv-ess
6" = * P -= AD:: (~/o.3)( ~Ox IO~) -= 160 )(103 N
CO"'SI-~~"'iV~ ol2i>wQ...lJf2 Je.f'oJ'"~Jio"
~ ~ ~~ 'p = .£iAct) o::-(95x/o"f)(~)iIO-~)(O.OOI:t) ~
?.2S~ {O'3. Iv
5 M ~lJ eI/' 1/0..])IJe of 'P jO ver", S
po: IbOY-lo3N ::: 160.0 ktJ .."",

2. I~A
f'
18in.
il
~ =
r< eerV i ~C J
Problem 2.14
28ldps
1.5in.
2.25 in.
"C
PL
EA
(;': 1:
A
28 kips
2.14 The aluminum rod ABC (E = 10.1 x 106psi), which consists of two cylindrical
portions AB and BC, is to be replaced with a cylindrical steel rod DE (E = 29 x 106
psi) of the same overall length. Determine the minimum required diameter d of the
steel rod ifits vertical deformation is not to exceed the deformation of the aluminum
rod under the same load and if the allowable stress in the steel rod is not to exceed
24 ksi.
D
DetoV', G.J.'cW Dt Q.)v ;I1';- f'<.)J
d
'P L"a + 'P Lac. = -E (LAe + LBe )
AA8 E "'~c..E E AM? Ace.
~8)( IO~
( 12. +- J'&
)z -= 0 03/3U. j...
/0./ x 10'- "10.5)2. ~(:z.15")Z. ~
~A ':
~
E
.
Ste~j) roC:» 5 =: 0.03/37' 1'1
A- PL - (;'?8~/O"!.)(30) .1-
- E S - (J.9>tIO6)(O.O3/37t:.)
= O.'i~317.h
A - P 2.8 )( 10:3 - . 1-
- e> ': A'i )C Ido - I./ &67 . M
.
Gtv-e
~ ;~ +-k 2G.lI'j eo(' 'Ie.. i J e
cl = -IlfA ~ J (4)(t..6b67)
IT If
A = J. I (.(; 7 ;..,~
t.~/~ ;..,. ....
-

3. I.
Problem 2.19
3in.
Co) ~ ~ £,.e + ~8' ':
(61 ~8 ": - S&
2.19 Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is
made of steel (E = 29 x 106psi), and rod BC of brass (E = 15 x 106psi). Determine
(a) the total deformation of the composite rod ABC, (b) the deflection of point B.
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./0 I'}.
Pc>...h.""'Be.: PBG ~ - :(Ovr/O3 lhJ LaG -= 30 iY1.) d-= 3Ir'1..J
A 1LJ t. Tf
( )
...
7
. 2.
f - IS 10' .
I!.<.:: <t ::' 7i ~ ~ . Ob 'if6 / r. > 4c.- 'Ii f ~I
S = PBc.Lec. = (-~O)lID»(30) -= -s:c;S88>1/c/' in.
& . EI!,C
ABc. (15 )(/0' )(l.o'lI(' ) .
-c -(
17..5"1 ><LO - S. bS138)t 10 ~ ~ JI, '10" L01 ;V',~...
~ -1
018= 5.(,""'10 i",.1'-'

4. Problem 2.21
B
50 kips
f
8ft
U
""""I I
I-- 13ft-I- 13ft 1~~
F"1I
a
-:?
A ~FAO
T ~S k..p~
I -
2.21 For the steel truss (E = 29 x 106 psi) and loading shown, determine the
deformations ofmembersAB and AD, knowing that their cross-sectional areas are 4.0
in2and 2.8 in2,respectively.
5+1L~..c. s .. Reo...//t,oV
s c..l A~.J c. ~ k:p,,- t
Me""be '8D is c.. 7eoro fO-h< -e beof'.
LAB:= ';I'~/ + g'Z :;;:- 15. ';«;'1-'3 ft ". 1'g3_/72 in.
LAo -=- 13 ft :- ISb in.
~o L,.o
SAD":0 f A"D
Spa ~
~MLl!a - (-47.701)(/O~)(lg3..'7~)-
E AAt - (1.Cf~ lo~)( It.b) -
-.0.0753 ;1). ~I I
=- (40.b2S-><ld' )(15(:.)
(A.") ~ 10 c..) (-<'~g )
= 0.0780 j~. --
Use joi"t A o..s f'r-€ hoJ'I
+t L Fj = 0: 2.5 of IS" (,'f 3 F;. =: 0 13=-47. 70/ k;p"
.:t 2" Fy ::- 0 :
13
o =- 'fo. b2 S- k,',s
D - IS. ?s FA!!.::: 0

5. Problem 2.26
D~l
cJm
2.26 The length of the 2-mm-diameter steel wire CD has been adjusted so that with
no load applied, a gap of 1.5 mm exists between the end B of the rigid beam ACB and
a contact point E. Knowing that E = 200 GPa, detennine where a 20-kg block should
be placed on the beam in order to cause contact between B and E.
J
Ri~iJ Y'od ACE t"O+<vh~.5 H."'°4 h c..~~R<Z. e ~ C.P~'SIi! '4"'-1"
e -= 1.5></0.3 = 3.75 >l/o-$ '/'tA.J
o.'-to
POi",t C. ",",0v<2'5, JDv./'ro..'NQ."J.
~c. =- 0.08 e = (0.0'8)(3.75)( /0-3') ~ goo)< 10(. """
$<:'0= ~c = 300 x 10-<: WI
A 7f t~ 7/
(:
~ '2.. -, ~
GO::- 4 Of "" tj: :(') ~ 3./ '-IIb"""" -=-3.' '-II" ><
{o WI.
E " ,
q
P.
Fe.o Lc.o
= A 00)1 0 0.. ~C.D:' E: Ac.o
Fc.D -=- E t ~GD ::- (;?OO)( fO" '(g. 1"116 ></0-<'»( 300 v{O-c) -= 753. Cf'11 N
0. 'lS
O<;ebec.."""ACB "-s c.. f~e. bOdy. w= M~ ~ (~O)('i.8f)=- 19G_~ N
~ot ~1v ~-
r ~ +J2MA = 0 ~ O.Og 1=;.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. Problem 2.29 2.29 Determine the deflection of the apex A of a homogeneousparaboloid of
revolutionof heighth, densityp, andmodulusofelasticityE, dueto itsownweight.
11 b2p~+t,
~
1TbPd h ~
2E h .iTb~ t 'j rAt ::
A
Fo r +~e
,
1/<'
ej~...,e~t r =- b(i)
VV'1.. =- 7Tb2 f
= 1Tb2,P~f
lfb£ ~ S
'f
t-.P <ojJ.j =-
-
S
h~ =
- 0 E ACJ)
A-=
dP ~ P3A d)'
!? = s;oJ
p ::
C - "'" P At
c:> - L:.. E A
eJeme..-t
~
r-I:,i
k-b-"
Le-t b ':" v-J.'v'S at -+ ~e be..se CL"'~
V'::: V'e..A,'vs dsevt-""'" -.vd-I.,
?;.ooV'J,'n~+e ::!.
~h:L
LfE
~

7. Problem 2.35 2.35 An axial centric force of magnitude P = 450 kN is applied to the composite
block shown by means of a rigid end plate. Knowing that h = 10 mm, determine the
normal stress in (a) the brass core, (b) the alwninwn plates.
Brass core
(E = 105 CPa)
300 mm
Let Pb.: fov-tio", of ~"o.i +Olf"c.4t c.o.",-'N.eJ k'j
bV'A.55 ~o~
Pa.. -= patH"", c.CtV"V';e.J /0..,+""0 Ju,..,.,'>,o,)"""
'f.1 ,t e.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) = :?<too WI""'" =- 2lfoOxlO-r.. iN~
AQ.. = (~)(bD)(IO):: 1:Z00 11-11001..
-;:: /~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~=' 140.6 HPQ."
(~)
(h) 6<'.. ~ ECt.,£ -:: (70)llOIt)(/.3'6'13'1c,d1) :: q3- 7.5>'/0' PGt :;::~3.7rMP~ 4

8. Problem 2.39
Dimensions in mm
A
R:-I
e, c.
3
bO ktJ
0 E
.-~(~
'to ktJ
2.39 Two cylindrical rods, one of steel and the other of brass, are joined at C and
restrained by rigid supports at A and E. For the loading shown and knowing tha~Es
= 200 GPa and Eh = 105 GPa, determine (a) the reactions at A and E, (b) the
deflection of point C.
A to c.: 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 fo E: E':: Ios >I'0 q 'P~
JI /. 't t. . -~ ~
A =- ~ .'30) =- 706.8f, ,"'1 =- 70b.86 x /0 ~
EA; 7Lf.1~o)l/o' I'J
A +0>
B: P -= 1<" L,: 180 ""' , -= D.180 "'""-
<:: ~ -PL.. - r?" (0.180) - . -It n
0>1:,("3 E:A - Z5"I.~27)lLO6 - 716~~o}<IO K,.
8 +0 ~ ~ P = Rio - G,Ox LO~L. =- l;CO ""'WI= O./:l D Pt
S - Pl - (R,..-60xIO!)(6./2o)- _It -,
Bc- EiA - ~51.'!.~"")<lOc. - 4'47.47'1tIO RA-:?6.'g"'8~{O
C to> D: p::: RA - r;;{»l.IO~ L: 'DD,."", :0 0./00 "'"
- PL - (R,..-6Dx/U'?)(o.Ioo) - -" -4:
2;ec.
- t=: A - 7 <.f.2Z0" IO~ == I.'347~s)t Ic . R,. - 8D.'8'II )I/O
D to E: p::: R" - 100,,10"3-
~ =.Eb. -= (f!,.-loo)(/dXo_,oo) = 1 '="/7rl-r
' -'10 - '~Ll
7'Z- I
-(
~Oif £A 7't. 'l~O)( to' .;;) ;J~)( 0 1"." ;;I...JS)( 0
A fo E: 5AE
= ~A6 ~ See. -+~c.o + ~E- =- 3.~58!57 >l/0-1 R" - 24a..lf_~""'><JO-'
L~ 100 ~ 0./00 "'"
Since poi1l+ E c.c.1I1IC>+
Move re2J.ive +6 A) ~AG =: 0
(tt,)
-"I D ..(;
3.~S837><lo " 2'-l1.!i~'f~{o = 0 b2.9 k ~ <C-~
R,. ~ ,62. cg3/)tlo3 N.
(")
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/x 103) - /..b.8lf"8)(
10-'
U -I;;
:- .6.3)(10 W 46.3ph->
- ~

9. i
8in.
t
lOin.
L-A-iB
oil
2.46 The rigid bar AD is supported by two steel wires of 1~-in. diameter (E = 29 x
106psi) and a pin and bracket at D. Knowing that the wires were initially taught,
detennme (a) the additional tension in each wire when a 220-lb load P is applied at
D, (b) the corresponding deflection of point D.
Problem 2.46
c p
le..t e be +h-e V'O+~t;G)v of ba", A~C Q
~ '~i--'S~, ~
- D
l12 in.-l12 in.-l12 in.j The.,., as == I~ e
~c. = 2.'+e
~ = Pe.E"
Lsp
g AE
'P~~ E A ~ge- -= (?"C;>t'o') ¥C~ y..(I~ 8 )
L6E 10
c. D
A~
e B' 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 ~3 x Io~ e
lJs.in~~~~ boJy AB c.o
D 2"MA::0 12 ~E t- 1.If p~ - 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 ) -=
~D 4-. 8 lb,
2~7. 6 11,.
eIIi8
...
(b GD ~ u e ~ (3')( 1..9125"")( 10-3) ~
-::
G C;. I x Io-~ jV
0_06<=1' ;V1.. ...ell!

10. I
Problem 2.48 2.48 The assembly shown consists of an alwninum shell (Ea= 70 GPa, tra= 23.6 x
lO-Oj°C)fully bonded to a steel core (Es = 200 GPa, as = 11.7 x lO-Oj°C)and is
unstressed at a temperature of 20°C. Considering only axial defonnations,
detennine the stress in the aluminum shell when the temperature reaches 180°C.
Steel
core
A 'if
( )
~ ~ -4 t.
.s -= "if ~o = 31!+.159 Ion"" ": !'/~.ISCf~IO ~
AQ..:::- ~ (SOt. - ~o~) :::- /. t:;4cr34 )( /oj WlYI-
1.
oJ -:1"'-
':" /.b~.q~,)(/O """'
Let 1'5 be +h~ ~l'J -h,II'CQ c&t./'r"l'eJ by +~
I:.Glol'rie) ~r' +he alulM"ndW shell..
Totc..i QJ)Llc..i to""C'..~ p:::- P",- + 1-'5 -= 0
.:st-et'J c.o""" 0.",,) 'P~ -rho.t
~ = - 'Pc.. (1 )
Pefol"" "",Jic>o.t ..
~ =-
'PeL L
£..Ae;. + Ld.~(f1T)
- PsL
- ~ A.s -- Ld..s(AT)
: Po. p~ -
~~ - £:.5
A~ .
(CX.s
- dcJ(~ TI
Us;V
~ (n ( L. + -L- ) 'D
G.,A", Es A 1<:<"== (ol,s - cJ.o,.
) ( b. r)
!.S "
(' I
('0)( Iott)(1.6 QC,3..,)(IO-~ ) + (loo )(loq )~31~; 15""4)<
10<:) ') 'Po.
= (11.7)(./0-1:.- ::l3.(.,></o-t::)(J80
-~())
( ~'1-57:] )( 10-') Pa.. ~ - J.q0 if 'A.
/ 0 - b
'P11.':" - 77.47 I )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 = 15 x 106~Si
a = 12 X 10- jOF
Aluminum
A = 2.8 in.2
E = 10.6 X 106~Si
a = 12.9 X 10- ;OF
p -
-.t
p
I--
Bvt ~f ~ 'P Lb + 'P L..' -=-
£" A,. E:Q.A"
. 1'4
- ( +
- (lS)( /0(" )(~. '-t)
(~)E, I)Jl'~O
2.57 Determine (a) the compressive force in the bars shown after a temperature rise
of 180°F, (b) the corresponding change in length of the bronze bar.
The" ,.4R ~P""O"""':D'" ;-f +r-e.e 0-1 c.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.
CoWsfrA-ined &'X.f'lIA'SI'o..... ~ -= o. O~ ilfl
S.~o.,-4e in'j J,J~ ~ indvc.eJ GD_pre.S~;v~ ~e l'
~ -3 _3
eJp -:: 7'"2.0'36'11l0 - O.o;t ::. 52. O'&; >lio 10'"..
(~+~)?
r:"A.. £c..A....
12
)
(IO.6')(IO~)(~.8) ? =-
'9'15.3' X to-'" 'P
Cf"tS_g~)<.lo-1 'P '":' 52. Og'"I< (O-3 P -=..52. ?. 7<=t
")a
103 Jb
5:2.3 k:f-s ~
(b) ~b:- Lbdb (A
T) - "PL..
f1,A!>
~ ('-l)(I~JC'o-C)(180) - (S"~_~~)(lO3)(14)
OSx IO~)(~-tf)
1
':" Gf.Cf,)<'(o-J..",,- ~

12. Problem 2.68
y
2.68 A fabric used in air-inflated structures is subjected to a biaxial loading that
results in normal stresses Ux= 120 MPa and q, = 160 MPa. Knowing that the
properties of the fabric can be approximated as E =87 GPaand v= 0.34, determine
the change in length of (a) side AB, (b) side BC, (c) diagonal AC.
6>< = 110 x/a'" Pa. ) ~ = 0 ') 6;t -:: II:0 )/ l D~ "P(t
Ex::- tC6)( - £J S - 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
~ "370 I )/ 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'.s df V'~3~t h-t'G(.V
j Ie ABC. c...S cs b.; ~",J c.
~ t "L
c:. -= 0.. -t b
0 btc..;", J,'fte.r~",-t,'e.P s b"J CA J<:,JJV'5.
2c de = ?a.dQ.'" 2h Jb
+ .!2.db
c.
0. -= 100 vY'otv>
..; b= IDOVr""..> c: ~ /IO()'t..+.75'.-) = '~'J1M
clCl. ~ ~A~: O.O7S'l ~IVI db ~ <a& :: o. .370 v...~
~ :: ole. =- ~ (0. o7S'1f) + 2€,. - (0. IOl8) = 0 f'Z~o.
At:. t25 IlS . .
I'YII')') ~

13. Problem
2.78
f1>1
2.78 A vibration isolation unit consists of two blocks of hard rubber with a modulus
of rigidity G = 2.75 ksi bonded to a plate AB and to rigid supports as shown.
Denoting by P the magnitude of the force applied to the plate and by Ii the
corresponding deflection. determine the effective spring constant, k = Pili, of the
system.
E.J:+evfilfe Sr,l',"V1j cc>"sfc..Vt
k=f~ :<~A
D w.h~,: G::- ~.IS K ,o~ fS"
P--~ ((. )( tf) :- ~ L( It'!~
h ':" I.~S in.
k =- (2'{::l.71))t to"!.Y1~) ~
I. 'ZS IoS., )l.lO~ib /;" ...
sheCt-;'-- + 'f'I. r- S
-11
s he"' 'I .s-h-e.ss :- GY =- &
I r1
Fof'c J.p - AT-
GAS
-
- h
'P =
J..G-A
.,

14. Aij"..; "'>'>.1IeI/c...xve 01 t;:. I ..-
ANt -= Jt:: ( )( )
2. .3 "-
go IS :' '~oo..."",= 1.2o~/o M
Problem 2.94 2.94 Two holes have been drilled through a long steel bar that is subjected to a
centric axial load as shown. For P = 32 kN, detennine the maximum value of the
stress (a) at A, (b) at B.
15mm
(et.) A+ hole A V' = i (:to) -= 10 ~""
d =- D - :<r- ::- 100 -:(O -::=80 INII")
Y' ,D
d ': go
:.- O.12~
Fifo , F'-1' 2. b '10..) K =- :<.G.5
(b)
-6. ~ k P =- (2.G.S)(3l.)(IO~) =- 7D 7 )( lOt:. £)
AM~ I. ~o )0( (0. ~ - r4..
At h;)9~ 'B r:- * (50 ) -= 25" ",,) c1 =- /00- SO : So ,
A.net ~ (SD X'S) =- 750 M'" 2- =- 7~D)( /0-(; M a.
70.7 MPa.. ~
V' -
& - '2.5'"- D.S"D
- -
!i~ k' =- :<_/~
6 ~ KP -=- ('l,'~)(3l)l.lo3)
- A...t 15"0>(
'0-£
'::'" Ql. ~ )( lOr. 'P4 q~. Z MP4.. 4
PROPRIETARY MATERIAL. «:>
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15. Problem 2.105
A
,-- 9 mm diameter
1.25 m
1
FN
B
Q ~.
/
~
~AG e/
~ Q
~
2.105 Rod AB is made of a mild steel that is assumed to be elastoplastic with E =
200 GPa and ay= 345 MPa. After the rod has been attached to the rigid lever CD,
it is found that end C is 6 mm too high. A vertical force Q is then applied at C until
this point has moved to position C ~ Determine the required magnitude of Q and the
deflection 0) if the lever is to snap back to a horizontal position after Q is removed.
AA6 #. ~(qt,.:- ~$.,n "",:a.
': G3.,,"'hdu.C. "",,'"
S III')C.Q '('0 c1 A8 ,'s -to be sfve-t c:.J.,J 'f'eN'WQ.V&1+
J1.:f
)
(~A6 )M"1(. '::" AA6 E:y :. ( 103.' I? 'If:
lU-c )(~t.fS )(t OC)
= ~ I. cr4S >1103N
-+, L Hp ": 0: I.' Q. - 0.7 1=".6 :: 0
Q,..""", =- ~./7 (~/. t:N9 x/03)::" 13.9' 7)( lD~ N
~. ~7 kN ....
, -/(t=~1-4& - (~1.1'l:~xlo:!>
)(,.1.5) - -3
~~a -;- EA.e - (1.00')/lD'X(;3..bl7~/D-lt.)
- 2. IS~?.S)(IO M
I S I
e ~ ~ '::" s. 080'-1->liO-?:' r J
~I -= Lie' ~ 3. gq)t /0.3 tilt :3. ~Cf ""WI ....

16. Problem 2.109
P'
2.109 Two tempered-steel bars, each l~-in. thick, are bonded to a t -in. mild-steel
bar. This composite bar is subjected as shown to a centric axial load of magnitude
P. Both steels~ elastoplastic with E = 29 X 106psi and with yield strengths equal
to 100 ksi and 50 ksi, respectively, for the tempered and mild steel. The load P is
gradually increased from zero until the deformation of the bar reaches a maximum
value 15m
=0.04 in. and then decreasedbackto zero. Determine(0)the maximum
value of P, (b) the maximum stress in the tempered-steel bars, (c) the permanent set
after the load is removed.
Fo~ t~.e .,.,.,,'iJ s+eel A, ~ (t)(~) ~ 1.00 ,0 '.
C = L ()YI ::: (Ill- )(SD)(/o'~) :: 0 o:tU Ig g
.
C>'fl E :21 )('0 to . -, .....
='o~ t~e +-eWpe~ed.steel Az ~ ~(;l )(~) '=
$'1'1. ': LCI'L :::- ('tt )('OO)t'IO'3) =: O.Olt87.76 in.
E 1.'1 )1103
O.7S "-t~
T0+c:..l c:t1f'~C(. ~ A=AI+A?.':. 1.76'illl2.
T"'~ .1vl'.P..J~
:s+ee) yl~/J5. T-eWfe~ s+e~) 's eJ,..sfl'c.
A, f5YI = (/. ()o)(SO~IO?) '=. S'o )CID3 lh----.
£A2S"" = ("J.q)/ID~)(O.7S)(O-()lf) -:: G;l-14 Y/O3 lb.
L JLf
P = PI + ~ = 1/:l.l'1)l}o3.ib ': 1/2. J k.ps
('b) S1-v-esse.s 0; = R ': 6'", ": -S(»I }O3 pSI' ~ ~D ks;
At
6'"2. = R ':: {;2~ I~ )//0' -= g~.86J1/0b f',' == ~~.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.re",t s,,-,t ~p ': ~tIo- ~I -= O.OL/ - 0.. 030Cf,/
=-0..00'106 i~. ~

17. Problem 2.120 2.120 Bar AB has a cross-sectional area of 1200 mm2 and is made of a steel that is
assumed to be elastoplastic with E = 200 GPa and Uy= 250 MPa. Knowing that the
force F increases from 0 to 520 kN and then decreases to zero, determine (a) the
permanent deflection of point C, (b) the residual stress in the bar.
~ I
~440mm~
A 2. -'- ~
':. IZoO,.,.., = I'ZC>D~/O ..,
FcJII'~~ 'r,:> yt.JJ P°.r--+-lOv Ac: 'PAc. = A0." = ("00 1C,0')05"0 X/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.' 3333 K'to 3 WJ
- (: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
= - PcB
LCB
EA
R I
(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 3 11
= 37g./~>«O -S20.,clO ":' ""'11../1..'8'8></0 N
37g. '8~..tC
log
f ~oo xia-'
= 3/S. IS=<. )( 10 C 'Pa..
(tt )
(6'
6' = Pet = _ILfI.g/~)({O""S. -= -118 /g;?><JC/. ~
ICe- A I<t)O X'lO~ . ~
~' = (37g. f8~)(O.12o) - O../'8qO~f/ )(10-3 ~
t. (';lClb )((D")( t '20C) x 10' ) ""
-3 -.3 -3
~ = ~ - ~'-= 0 ~~3333 xtO - 0.' '!t:}oq, )cIO ~ O. /DLfZ)([0 ~
c:p Qc. de. .
-= O. I oLf'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'
R I - F L..:a -
Ac. - -
L,.c. Le.
PCB':: p- F
, p
6c. = =-
A

18. Problem2.130
LfI7.0'1 I<
LcJ ~
~
2.130 Knowing that E =29 X 106psi, detennine (a) the value of 0 for which the
deflectionofpointB isdownandto theleftalonga lineforminganangleof36° with
the horizontal,(b) the correspondingmagnitudeof the deflectionof B.
~8C. =
?sc.. -::
~AB ':
-=
a ~os 3GD
EA6c-Ssc. - (::i.~)(lOG)(O.8)~ G.oS
36-
L&.. - '-Is
LfJ7. Dq~)([D3 ~
? -=:fF AM.. ~"t.. ~ (Z9' )<10' )[1- z) $ 51""36°
AI:. l,,<... 'l.S
'?
=: 31 ~. ~D xtD £
.L ~':! S,g-~ox{c:J ~ = ';-bll e:: r3 0° ~
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
a SI'" 3'-
.....