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.
Problem 2.10
A -::
(SD) ( ~o)
~=
%0 MP~
CO~5iJ€"";V\j
6"
=
*
CO"'SI-~~"'iV\~
~
:
~
~~
5 M ~lJ e I/'
zooo
r
Z
""-",,, l..::-
80 '1-10'
?~)
JPo...raJole
P -= AD::
m ~
E ~ crSx/O"1
'P~
sfv-ess
(~/o.3)(
ol2i>wQ...lJf2
~Ox IO~)
-= 160 )(103
N
Je.f'oJ'"~Jio"\
'p = .£iAct)
1/0..])IJ e of 'P
><10-3
o::-(95x/o"f)(~)iIO-~)(O.OOI:t)
jO ver", S
po:
IbOY-lo3N
?.2S ~
{O'3. Iv
~
:::
160.0
ktJ
.."",
Problem 2.14
28 kips
28 ldps
I~A
f'
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
1.5 in.
DetoV',
d
2.25 in.
~A
G\.J.'cW\ Dt
'P L"a
AA8
E
':
18in.
il
~
"C
Ste~j)
~
=
(;':
r<
eerV i ~C
+
~8)( IO~
PL
EA
.
A-
1:
A
.
P
A -- e>
;I1\';-
\f'<.)J
'P
Lac. = -E
LBe
"'~c..E
E (LAe
AM? + Ace.
("10.5)2.
12.
/0./ x 10'-
E
Q.)v
roC:»
5
+-
)z
J'&
-= 0 03/3U.
~
~(:z.15")Z.
=: 0.03/37'
)
j...
1'1
PL - (;'?8~/O"!.)(30)
.1- E S - (J.9>tIO6)(O.O3/37t:.)
= O.'i~317.h
2.8
J Gtv-e~ ;~ +-k 2 G.lI'j
cl = -IlfA
IT
~
A'i
':
eo('
)( 10:3
)C
Ido
'Ie..
iJ
J (4)(t..6b67)
If
-
- I. / &67
. 1.M
A = J. I (.(; 7 ;..,~
e
-
t.~/~
;..,.
....
I.
Problem 2.19
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.
3in.
PoV'hol'\
A'B:
A",s -= ~J"
~
(61
~8
": - S&
':
PBG
Tf
17..5"1
~ 3.1'"1/6
L..s=
,.,..2.)
'fa
"~')
01 -= 2. in'J
EAa::'" ;<q"'JO'
-
EMA.-.t- (2"h<IO")('3,J'f/G:) - \1.S6/Q
1LJ t.
<t
::' 7i
~
- :(Ovr/O3
lhJ
LaG -= 30
. 2.
...
( ~ ) ~ 7. Ob 'if6
/ r.
>
= PBc.Lec. = (-~O)lID»(30)
&
£,.e + ~8'
::' ~(2.)1.
Pc>...h.""'Be.:
S
~ ~
I.fo )</0'3.llr.,~
- 'PAC3L"13
- (LIO~lo?J(4())
AB-
AI!.<.::
Co)
~e=
. EI!,CABc.
(15 )(/0' )(l.o'lI('
-c
-(
><LO - S. bS138)t 10
f'S"
-~ .
)1./0
I\'}.
iY1.) d-= 3Ir'1..J
IS 'Ii 10 ' f ~I.
f 4c. -= -s:c;S88>1/c/'
)
in.
.
~
~
~
018=
JI , '10" L01 ;V',~...
5.(,""'10
-1
i",.1'-'
I
Problem 2.21
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
in2 and 2.8 in2, respectively.
50 kips
B
Me""be
8ft
U
LAB:=
I
I-- 13ft-I-
Reo...//t,oV\s
5+1L~..c. s ..
f
""""I
13ft 1~~
LAo
'8D
';I'~/
-=- 13
ft
-:?
A ~FAO
+t L Fj =
Spa
~
13
D -\ IS. ?s
c.
-e
ISb
".
-
~o L,.o
SAD":0 f A"D
=-
1'g3_/72
13=-47.
0
in.
-.0.0753
It. b)
(40.b2S-><ld' )(15(:.)
(A.") ~ 10 c.. ) (-<'~g )
70/
k;p"
o =- 'fo. b2 S- k,',s
FA!!.::: 0
(1.Cf~ lo~)(
t
hoJ'I
~MLl!a - (-47.701)(/O~)(lg3..'7~)E AAt
k:p,,-
in.
o..s f' r-€
=:
~
beof'.
:;;:- 15. ';«;'1-'3 ft
2.5 of IS" (,'f 3 F;.
0:
.:t 2" Fy ::- 0 :
T ~S k..p~
:-
A
c..l A~.J
c.. 7eoro fO-h<
is
+ g'Z
Use joi"t
F"1I
a
-
;1). ~I
-
=
0.0780 j~. --
I
Problem 2.26
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 A CB 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
D~l
Ri~iJ
Y'od ACE
e -= 1.5></0.3
=
o.'-to
cJm
POi",t C.
3.75
e
AGO::=
,0
A 00)1
Fc.D -=- E
t
q
~o
A
r
t
1 _D."Io-\(-\
I
/0-3') ~ goo)< 10(. """
'2..
-,
3./ '-IIb"""" -=-3.' '-II" ><{o
Fe.o Lc.o
P.
0..
~C.D:'
"-s c.. f~e.
~1v
C.P~'SIi! '4"'-1"
x 10-<: WI
~
7f t~
7/
4 Of "" tj: (::(') ~
E: Ac.o
~GD ::- (;?OO)( fO" '(g. 0. 1"116
'lS ></0-<'»( 300
O<;ebec..""" ACB
e ~
>l/o-$ '/'tA.J
= (0.0'8)(3.75)(
$<:'0= ~c = 300
"
c..~~R<Z.
",",0v<2'5, JDv./'ro..'NQ."J.
~c. =- 0.08
E
h
t"O+<vh~.5 H."'°4
bOdy.
w=
v{O-c)
-=
753.
Cf'11
N
M~ ~ (~O)('i.8f)=- 19G_~ N
~~
+J2MA
=
0.40 - k.::-
0 ~ O.Og 1=;.0 -(O.4D-X.)W
(00<6)(753.Q2)
. I"tb. ?...
-:.
0
=- 0- 307Ll-3 ...,
X>= O_oq~,.",
= 92.b
-....
~
WI.
Problem 2.29
2.29 Determine the deflection of the apex A of a homogeneous paraboloid of
revolutionof heighth, densityp, andmodulus of elasticityE, dueto its own weight.
A
eJeme..-t
~
r-I \:,i
k-b-"
Le-t b
1/<'
Fo
r
+~e
ej~...,e~t
,
A-=
VV'1..
dP ~
!?
C
c:>
P3A d)'
s; oJp
=
-
-
"'"
L:..
r =- b(i)
=- 7Tb2
::
':" v-J.'v'S
at
-+ ~e
V'::: V'e..A,'vs dsevt-""'"
be..se
CL"'~
-.vd-I.,
?;.ooV'J,' n ~+e ::!.
f
= 1Tb2,P~f
lfb£
t-.P ~ S <oj J.j
'f
=-
~
11 b2p~+t,
= 1TbPd
P At - h~
2E h
E A - S0 E ACJ)
h
.iTb~
t
~
'j
rAt
::
~h:L
LfE
~
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)
Let
Pb.:
fov-tio", of ~\"o.i
bV'A.55 ~o~
Pa.. -= patH"",
+Olf"c.4t c.o.",-'N.eJ k'j
c.CtV"V';e.J
/0..,+""0 Ju,..,.,'>,\o,)"""
'f.1 ,t e.s.
300 mm
~
=
Pc. -- £hA.~
L
'P.L
E~A..
~ = p~ L
'R -:: E...A. ~
c::...
L
ED-A-
P=
H~ + 'PQ. -:: (EbA", ~ ~A~
~- ~c.- LAb
::
AQ.. =
E. =
(GO)(LlO)
(~)(bD)(IO)::
=
:?<too
4-50
6"\0 ~ FeE.
(h)
6<'\.. ~
iN\~
11-1100\1..
-;:: /~oo)c",o-"W\'t.
)( 103
()OSx{oq)(?LIooxlO.G.)
(~)
p
EIoAe-+FdA4
WI""'" =- 2lfoOxlO-r..
1:Z00
)~
>= I. 33Cf3 ~to
+ (70 )(10" )(,~OO)( to-'
~ (toS>tIO"l)(1.33"}3x/O..3):::
ECt.,£ -:: (70)llOIt)(/.3'6'13'1c,d1)
)
140.b)(/O"P~='
::
-3
q3- 7.5>'/0'
140.6
HPQ."
PGt :;::~3.7rMP~
4
Problem 2.39
Dimensions
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.
in mm
A to c.:
A=
A
c.
e,
3
R:-I
P
<::
~
8 +0 ~
P =
JI /.
~ \.'30)
't
=-
q
'P~
706.8f,
,"'1
7Lf.1~o)l/o'
t.
70b.86
=-
.
x /0
-~
~
~
I'J
-
r?" (0.180)
-
-
Rio
-
.
Z5"I.~27)lLO6 -
716~~o}<IO
G,Ox LO~L.
=-
"'""-
-It n
K,.
l;CO ""'WI = O./:l
D Pt
SBc-- Pl
- (R,..-60xIO!)(6./2o)
_It
-,
EiA ~51.'!.~"")<lOc.
- 4'47.47'1tIO RA-:?6.'g"'8~{O
D:
p:::
RA - r;;{»l.IO~
L:
'DD,."",
:0 0./00
"'"
- PL - (R,..-6Dx/U'?)(o.Ioo)
- -"
-4:
t=:
A
p:::
D to E:
R"
5AE=
Since
poi1l+
IO~
~A6 ~ See. -+~c.o
E
L
(")
"
-"I D
=
~AB
..(;
+ SBC. =-
f.'l.8>LlD
:-
U
.6.3)(10
W\
R" -
"'"
'
-'10 0
1?~
)I/O
~Ll
1"."
7 'Z-
;;I...JS)(
I
-(
0
24a..lf_~""'><JO-'
~AG =: 0
'3
=.-37.Z><[O
~
b2.9 k ~ <C-~
'37_;:lku~
4
- ~G.g43)(/O-C.
10-'1)(62. 83/x 103)
-I;;
'
- 8D. '8'II
R,. ~ ,62. cg3/)tlo3 N.
3"3
-IOO)l.{O
Llb3G7vlo-<t
= (/./b'3~~)(
>l/0-1
R,.
.
0./00
;J~)(
+6 A)
= 0
Ic
~
= 1 '="/7 rl-r
3. ~58!57
>
~
100
~
.;;)
re2J.ive
2'-l1.!i~'f~{o
R,.-/00>'/0
"E:::
to'
+ ~E- =-
c.c.1I\1I\C>+Move
I.'347~s)t
==
- 100,,10"37't. 'l~O)(
3.~S837><lo
~c
7 <.f.2Z0"
£A
~Oif
A fo E:
-
=.Eb. -= (f!,.-loo)(/dXo_,oo)
~
(b\
N
I os >I'0
180 ""' , -= D.180
L,:
-PL..
2;ec.-
(tt,)
1<"
E:A
~
=-
E'::
EA;
.-~(~
-=
0>1:,("3
A
E
1.~Sc;,"'t><'lo::'MI'>11..=
1.'l5"~)tlci'lN\.1.
'to ktJ
bO ktJ
A +0>
B:
C to>
0
'PQ"
2..5' - 31.7;.<lot;
E:
fo
200)(/0"
~(1fo)1=
EA=.
C
=
E
- /..b.8lf"8)(10-'
46.3ph->- ~
Problem 2.46
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.
i
8in.
t
lOin.
p
c
L-A-iB
oil
-
~ '~i--'S~,
~
D
l12 in.-l12 in.-l12 in.j
c.
A~
e
as == I~ e
D
B'
c'
C'
-Pse
PCF
?
ba", A ~C Q
be +h-e V'O+~t;G)v\ of
The.,.,
~c. = 2.'+e
~
= Pe.E"
Lsp
AE
g
E A ~ge- -= (?"C;>t'o') ¥C~ y..(I~ 8 )
'P~~
lJ
Ati
e
le..t
L6E
-
e
tOb.77)(IO$
!
b
10
~ Co =
Pc~
~\~
EA
::0
-=
LCF
~ LC:f:~cJa
= (.<q)(IO(.)~(.l-)"'(.Zl/ a)
18
"~L ~3 x Io~
e
lJs.in~~~~ boJy A B c.o
D 2"MA :: 0
12 ~E t- 1.If p~
- 3'?
= 0
Oiy"/O6.77><[O?'e) + (:N)(118.63)clO" e)
4, 1:Z~'5)(IO' e
e
(0.)
(b\
=
I~
'11 8-G" .,c /0-3
= (ID6.77)(/O'3 )(r.'l,sS'X(O-3)
?CF
=
GD ~
u
e
lo~)(I.q'8S-)«(O-3
~
(3')(
::
0
= (3')(~1o)
P6E
(\1'%.63)(
-(3{.)(;no)
1..9125"")( 10-3)
)
I/'"J
lb,
-::'
~D 4-. 8
-=
2~7. 6 11,.
~
G C;. I x Io-~
-::
0_06<=1'
eIIi8
...
jV\
;V1..
...ell!
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.
'if
A.s
Steel
core
-= "if
~
AQ..:::-
Let
1'5
be
+h~
~l'J
I:.Glol'rie) ~r' +he
Totc..i
~
(~o ) =
-
(SOt.
QJ)Llc..i to""C'..~
~o~)
-4
Ion"" ":
!'/~.ISCf~IO
:::- /. t:;4cr34
)( /oj WlYI-\
1.
oJ
':" /.b~.q~,)(/O
.:st-et'J c.o""" 0.",,) 'P~
by +~
-h,II'CQ c&t./'r"l'eJ
alulM"ndW\
~
31!+.159
t.
~
-:1"'"""'
-rho.t
shell..
p:::-
P",-
+
1-'5
-=
0
~ = - 'Pc..
Pefo\l""
"",Jic>o.t
~
'PeL
=-
(1 )
..
L
£..Ae;. + Ld.~(f1T)
: Po.
p~
-
~~ - £:.5A~ .
Us;V\~
(n
(
L.
G.,A",
+ -L-
Es A!.S
-
PsL
-
~ A.s -\- L d..s(AT)
(CX.s
- dcJ(~ TI
)
'D
" - cJ.o,.) ( b. r)
1<:<"== (ol,s
('
('0)( Iott)(1.6 QC,3..,)(IO-~ ) + (loo
I
)(loq )~31~; 15""4)<
10<:) ') 'Po.
-
= (11.7)(./0-1:.- ::l3.(.,></o-t::)(J80
( ~'1- 57:] )( 10-') Pa.. ~
'P11.':" - 77.47
Sfre'5s
;V\ .rJIJ""':Y\(,)M. shell
I
- J.q0 if
)t /03
~())
'A./ 0 - b
N
()~= JL":
A..
-17.Y71.)( 103
'.6..,.q3"4~/o-~ -:--Lf7.0 ~/o'
Ptt
- 47.0/vIp",.
Problem2.57
2.57 Determine (a) the compressive force in the bars shown after a temperature rise
of 180°F, (b) the corresponding change in length of the bronze bar.
0.02in.
1r-14in.~18
in.--j
The"
,.4R
~P""O""\"':D'" ;-f +r-e.e 0-1 c.o ~t
S-r ~ LJ,d.b(AT)
Aluminum
A = 2.8 in.2
Bronze
A=2.4in.2
E = 15
a
p
= 12 xX
E
a
106~Si
10- jOF
1
+ L ~(~r)
=- (J"+)(/~)(IO-(. )(1~O) 4- (1'8)(,~.<=1)(lO-(.)(I~(»)
=
= 10.6
12.9 X
X 106~Si
10- ;OF
=
p
-
-.t
72.036
CoW\sfrA-ined
I--
')(to-'!.
In.
indvc.eJ
~
-3
eJp -:: 7'"2.0'36'11l0
- O.o;t
~f
~
'P Lb + 'P L..'
- (.
(~)E,
I)Jl'~O
E:Q.A"
£" A,.
1'4
(~+~)?
-=-
r:"A..
(IO.6')(IO~)(~.8)
(lS)( /0(" )(~. '-t)
Cf"tS_g~)<.lo-1
'P
'":' 52.
~e
GD_pre.S~;v~
l'
_3
::. 52. O'&; >lio
10'"\..
£c..A....
12
+
~ -= o. O~ ilfl
&'X.f'lIA'SI'o.....
S.~o.,-4e in'j J,J~ ~
Bvt
",-~",,-t
) ?
=-
'9'15.3'
X
to-'"
'P
P -=..52. ?. 7<=t")a103
Og'"I< (O-3
Jb
5:2.3 k:f-s ~
(b) ~b:~
Lbdb (AT) (\'-l)(I~JC'o-C)(180)
"PL..
f1,A!>
- (S"~_~~)(lO3)(14)
OSx IO~)(~-tf)
':"
Gf.Cf,)<'(o-J..",,-
~
Problem 2.68
2.68 A fabric used in air-inflated structures is subjected to a biaxial loading that
results in normal stresses Ux= 120 MPa and q, = 160 MPa. Knowing that the
properties of the fabric can be approximated as E = 87 GPa and v= 0.34, determine
the change in length of (a) side AB, (b) side BC, (c) diagonal A C.
y
=
6><
tC
Ex::-
£z:: t(-")JC-,t
"370
~
I
(a.-)
~~-::(AB)E)(
(6)
~e,c, ~ (13G)
- -V~
)/
10
-=
£2
-
6)(
S - V~z
£J
[I'~D)('O'
= 75'1.0';
)(/O-c
-
) (1,'37",
)
>t.IO-~
let-he..? _S\~-l'.s df V'~3~t h-t'G(.V\j Ie
(c:~:
c:.
-=
0 btc..;",
c
0..
t
-t
=
6;t
-::
II:0
)/
l D~
"P(t
)
(O.3~)("o)iIO")J
+ IbO)lJO"J
~
O.O7SLJ
M""
..
-::
O.loZS
"""'"
-4
ABC.
c...S cs b.; ~",J c.
\ "L
b
J,'fte.r~",-t,'e.P
2c de
')
s
b"J CA J<:,JJV'5.
?a.dQ.'" 2h Jb
a.
c Jl1. + .!2.db
c.
de:::
13ut j
~
0
87~/Oq-[ -(o.3't)(I~O)(IO")
(1DD#II"'\)(75'i.O~)(/6()
=- (. 75M
~=
Pa. )
= 87~IO"r
+ 6;)::
-3
110 x/a'"
0. -= 100
clCl. ~
~At:. ::
~A~:
ole.
vY'otv>
..;
O.O7S'l
=-
~t 25
b=
IDOVr\""..> c: ~ /IO()'t..+.75'.-)
~IVI
db
(0. o7S'1f)
+
2€,.
~
IlS
<a&
::
o.
=
.370
- (0. IOl8)
=
'~\'J1M
v...~
0 f'Z~o.
.
.
I'YII')') ~
Problem2.78
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.
sheCt-;'--
+
s he"'\
.s-h-e.ss
\'I
r- -11S
'f'I.
GY
:-
=-
I
J.p -
Fof'c
'P
E.J:+evfilfe
=
D w.h~,:
k =-
h
J..G-A
\.,
Sr,l',"V1j
G::-
\r1
GAS
AT- -
cc>"\sfc..V\t
k=f~
f1>1
&
:<~A
~.IS
K ,o~
fS"
P-- ~ ((. )( tf)
:- ~ L( It'!~
h ':" I.~S
in.
(2'{::l.71))t
I. 'ZS to"!.Y1~)
~
I oS., )l.lO~ib /;"
...
Aij"..; "'>'>.1IeI/c...xve
t;:.
01
I ..-
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.)
d
=-
D-
Y'
d
Fifo
~ k P
=- (2.G.S)(3l.)(IO~)
I. ~o
AM~
(b)
At
h;)9~ 'B
r:-
A.net ~ (SD X'S)
)0(
-= 25"
M'" 2- =- 7~D)(
V' - '2.5'"-- D.S"D
& - -!i~
6
-
~
KP
k'
-=- ('l,'~)(3l)l.lo3)
A...t
-
",,)
15"0>(
'0-£
=-
:.-
, F'-1'
7D 7
(0. ~
* (50 )
=- 750
=-
,D
go
':
)( lOt:.
c1 =- /00/0-(;
i (:to)
V' =
:<r- ::- 100 -:(O
-= Jt::
ANt
-6.
A
A+ hole
-=
10 ~""
-::= 80
INII")
( go )( IS ) :' '~oo..."",=
2.
1.2o~/o
.3
"-
M
O.12~
2. b '10..) K =- :<.G.5
70.7
£)
r4..
SO
: So
MPa..
~
,
M a.
:<_/~
'::'" Ql. ~ )( lOr. 'P4
q~. Z MP4.. 4
PROPRIETARY MATERIAL. «:>2006 The McGraw-Hili Companies, Inc. All 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 of the publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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.
Problem 2.105
A
,-- 9 mm diameter
AA6 #.
~(qt,.:-
1.25 m
S
A8
III')C.Q '('0 c1
(~A6 )M"1(. '::"
1
=
-+, L Hp
~.
~
E:y :.
"",,'"
'f'eN'W\Q.V\&1+
J1.:f)
( 103.' I? 'If:lU-c )(~t.fS
)( t OC)
~ I. cr4S >1103N
I.' Q. -
0.7
1=".6
(~/. t:N9 x/03)::"
:: 0
13.9'
7)( lD~ N
\~. ~7 kN
....
S
e ~
I
~
e/
(~1.1'l:~xlo:!>
)(,.1.5) -3
- (1.00')/lD'X(;3..bl7~/D-lt.)
- 2. IS~?.S)(IO M
~~a -;- EA.e
I
/
Q
G3.,,"'hdu.C.
, - /(t=~ 1- 4& -
Q
~~
AA6
0:
":
"",:a.':
,'s -to be sfve-t c:.J.,J
Q,..""", =- ~./7
FN
B
~AG
~$.,n
~I
-=
Lie'
'::" s. 080'-1->liO-?:' r
~
3. gq)t /0.3
tilt
J
:3. ~Cf ""WI ....
t
Problem 2.109
2.109 Two tempered-steel bars, each l~ -in. thick, are bonded to a -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 decreasedback to 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.
P'
A,
Fo~ t~.e .,.,.,,'iJ s+eel
C
=
L ()YI
E
C>'fl
t~e
\='o~
$'1'1.
~'(\ <
(a)
. FO~c.e5.
(C)
:::- ('tt
E
Az ~ ~(;l
)('OO)t'IO'3)
1.'1 )1103
:.
£A2S""
~
R
At
0;
=
'::
{;2~
O. 1~I~
£'
Pei'Mo.r\e",t s,,-,t
':
'=.
1/:l.l'1)l}o3.ib
':
)//0'
6'",
":
-=
-S(»I
}O3
pSI'
g~.86J1/0b
~I
-=
-::
':
O.OL/
-
-
1/2.
~
f','
(:Z'1><IO'"
)(1.7S-)
~p ': ~tIo\-
O.7S
"-t~
s+e~)
's eJ,..sfl'c.
S'o )CID3 lh----.
PL '= (\I~.\4)tIO~)(14) -
EA
'=
in.
T-eW\fe~
= ("J.q)/ID~)(O.7S)(O-()lf)
JLf
=
.....
1.76'illl2.
'P2
PI +
.
)(~)
O.Olt87.76
A=AI+A?.':.
A, f5YI = (/. ()o)(SO~IO?)
RA~
=:
,\0\ '\.
o:tU I g
. -, g
0
T"'~ .1v\l'.P..J~
:s+ee) yl~/J5.
L
1.00
~
to
';:
('b) S1-v-esse.s
UnPoa.l:nj
LCI'L
(t)(~)
::
'PI
P=
6'"2. =
:21 )('0
c:t1f'~C(. ~
~w.. < S 'f2.
(Ill- )(SD)(/o'~)
+-eW\pe~ed .steel
':
T0+c:..l
:::
~
G;l-14 Y/O3 lb.
~
J k.ps
~D
ks;
== ~~.8~
O.o30cr'i
...
kosI'
,~.
0.. 030Cf,/
=-0..00'106
i~.
~
2.120 Bar AB has a cross-sectional area of 1200 mm2 and is made of a steel that is
assumed to be elastoplastic with E = 200 GPa and Uy= 250 MPa. Knowing that the
force F increases from 0 to 520 kN and then decreases to zero, determine (a) the
permanent deflection of point C, (b) the residual stress in the bar.
Problem 2.120
~
I
A ':.
~440mm~
Ac:
P°.r--+-lOv\
FcJII'~~ 'r,:> yt.JJ
PA<.
2.
IZoO,.,..,
~
F
~
f-:oV'e1'"
Pea
~e.LdS
EA
~c. = -
,
= ("00
=
300
><IO3N
~J;h
'()
RI
M:.
( EA
,
6'
ICe-
~'t.
(tt )
(6'
A
=
=
F
p
6c. =
Pet
A
.. )(I~OO
1C,0')05"0
><IO~-
300
- D..'Zo)
PAc:.. =- 0
S~o ~/O-S
N
10$
;)(
X/Oc)
~
0... ,z.' 3333
K'to 3 WJ
)('"10-6)
X 'O~ 'P~
= (F - PA:') L<:.e.
EA
CB
EA
JS20
)
::
FLea
F:A
')(lO'3 )( 0'. «+40 -O.I~o)
=
o. 4~b
=
=-
PcBL
+ Lk.
EA
LAc.
I
p-
L/</;o
)(/03
-= -183.3.33
)C"(O-4;o
= -
RAc. -- F L..:a L,.c. Le.
PCB'::
)lID
= - :(z.o
~
F + 'Pee. -
-F=
PA~
..,
I
~EA L~c.
=
)('IO!.) (0.
(:zOO
~"c)
I~oo
~=
UnPoa.Ji,,~
c.
-
Pea
Dca ::
~I
- (Z1.0
=
-'-
A0."
'PAc. =
(")
rCQ
4-
= I'ZC>D~/O
3
37g./~>«O
37g. '8~..tClog
f ~oo xia-'
= _ILfI.g/~)({O""S.
I <t)O
(37g. f8~)(O.12o)
X'lO~
(';lClb )(( D")( t '20C) x 10' )
~c:p = ~Qc. - ~'-=
de.
6~.)~ -:: 6'y - 6;.~::
-
3
-S20.,clO
=
-=
":'
378./S~
>llo
""'11../1..'8'8></0
3/S.
IS=<. )( 10 C 'Pa..
-118
. /g;?><JC/. ~~
O../'8qO~f/
)(10-3
3
N
3 11
N
~
""
-3
0 . ~~3333 xtO
-.3
- 0.' '!t:}oq, )cIO
2St:>x/O" - 3/5../5;2, ~/O'
-= O. I oLf'1. ~~
-6S.~
~
~
6fG.>f"U~ OC(3- aed ~ - "33.~33)c'10c;+ "8. IS2xlO'
-3
~ O. /DLfZ)([0
~
~
)([0'- f4
-bS-.~ MP~ ..
::-6S:~>dc)(.
'P4\
:= - bS".'~ HP~
~
= 29 X 106psi, detennine (a) the value of 0 for which the
deflectionofpoint B is downandto the left alonga lineformingan angleof36° with
the horizontal,(b) the correspondingmagnitudeof the deflectionof B.
Problem2.130
2.130 Knowing that E
= a ~os
~8C.
?sc.. -::
-=
LfI7. 0'1 I<LcJ
3GD
EA6c-Ssc. - (::i.~)(lOG)(O.8)~ G.oS36L&..
'-Is
LfJ7. Dq~)([D3
~
~
~
a
SI'" 3'-
~AB
':
?
-=:fF AM.. ~"t..
AI:.
(Z9' )<10' )[1- z) $ 51""36°
~
l,,<...
'l.S
=: 31 ~. ~D xtD
.L
r'l"'"
'Po;:lS)(LO~
S ::
.
~':!
S,g-~ox{c:J
'-//7, oqxlo3
':'~(417-0C;)(IO3
'25)( Ic)'3
,.
'=
q {'l>.
'6&)('10.3
~
~.
~)2.
=
'?
';-bll
£
e::
r3
10
+(gl~.~O)rLD-3~)'"
0",027:1..
.
""-
.0°
~
= CfI'8.?::Sxld'£
.....