?{}tri hN
PROBLEM 2.65
A 2.5-m length of a steel pipe of 300-mm outer diameter and 15-mm
wall thickness is used as a colunn to carry a 700-kN centric axial load.
Knowing that E = 200 GPa and v = 0.30, determine (a) the change in
length of the pipe, (b) the change in its outer diameter, (c) the change in
its wall thickness.
SOLUTION
d,=0.3m l=0.015m
d,
L=2.5m
=do-2t=0.3-2(0.015) =0.27
A=
m
P=700xl0rN
d?) =
o.z'l'z)= 13.4303 xlo 3 m2
+U: [{o.r' -
(d 6=-L-- (7ooxlo3x2'5)
E.4 (200x10'q)(13.4303x10r)
6 =-0.652 mrn
=-651.51x10*6 m
6
L
-651.51x10{
2.5
{
= -260.60 x 10-6
€r*. = -ys = -(0.30)(-260.60 x l0{ )
= 78.180x
(b)
(c)
l0{
Ld, = doero, = (300 mm)(78.180 x 10-6 )
Ld.=0.0235mm
<(
A/ = o.ool l73 mm
{
r\t =t€r.or=(15 mm)(78.180x10{)
PROPRIETARY II'IATERIAL, O 2012 The McGraw-Hill C'ompanies, Inc.
reproduced, or distributed
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reserved. No part of this Manual rray be displayed,
in any fonn or by any means, without the prior rvritten pennission of the publisher, or used beyond the limited
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it
PROBLEM 2.66
An aluminum plate (E=14GPa and v=0.33) is
subjected to a
centric axial load that causes a normal stress d Knowing that, before
loading, a line of slope 2: I is scribed on the plate, determine the slope
of the line when o = 125 MPa.
SOLUTION
The slope after deformation
is
1on6
=2(1t'r)
l+ e,
x"(
106
"' = t25
''u x"'u^
t.^- o-
E
€y
=
tu, a =
-vtx=
74x10'
+tt1
= t.6892x l0-3
-(0.33X1.6892 xl0-3; = -0.5574x10-3
2!1 -^0'9=0r0llf-a)
I + 0.00 16892
=
1.99551
tanl = t.995s l
<
PR0PRIETARY MATERIAL. O 2012 The N{cGraw-Hill 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 wrilten pemrission of the publisher,
or used beyond the limited
distribution to teachers and educators permitted by McGrarv-Hill for their individual course preparation. A student using
this manual is using it
rvithout permission.
PROBLEM 2.74
The homogeneous plate ABCD is subjected to a biaxial loading as
shown. It is known that oz - 6u and that the change in length of
the plate in the x direction must be zero, thal is, t, = 0. Denoting
by E the modulus of elasticity and by v Poisson's ratio. determine
(a) the required magnitude of o,, (6) the ratio oul e ,.
SOLUTION
0,
=
Og, 6y =0, t, =0
,.
=
vo, vo,) =
voo)
|@. - *ro. -
(.a)
ox =voo
(.h) ,,=|{-ro,-vot-+o,1=!u{-r'oo-0 *o;=\t-oo
PROPRIETARY MATERIAL, O 2012 The McGrarv-Hill Companies, Inc.
10 teachers
rvithtrut permissiorr.
a
reserved. No part o[ this Manual may be displayed"
in any form t'rr by any means, without the prior written pennission of the publishel or used beyond the limited
and educators permitted hy McGraw-Hill for tlreir individual course preparation. A student using this manual is using it
reproduced, or distributed
distribution
All rights
?=*
1
PROBLEM 2.79
The plastic block shown is bondecl to a rigid suppofi and to a vertical plate to
which a 55-kip load P is applied. Knowing that fbr the plastic used G = 150 ksi,
determine the deflection of the plate.
SOLUTION
I
1
{
I
!r
h
A = (3.2)(4.8) =
15.36 in2
P=55xl03lb
P 55xl0l-
'-7-
r-5-16
G = I50 x
t
y=L
--j580.7 psi
103 psi
3580.7
= 23.871 x l0-3
G - -::-::-150 x l0'
h=2in.
6 = hy = (2)(23.8jlx l0-3)
=
4'7
.7
x 10-3 in.
5 = 0.0417 in.
J{
PROPRIETARY M.'ITERIAL, O 2012 The illcGraw-Hitl Companies, Inc. All rights reserved. No part of this Manual
may be displayed,
reproduced. or distributed in any ltrrm or by any means, witlrout the prior written pernrission of the publisher,
or used beyond the 1imited
distribution to teachers arrd educators permitted by McGrarv-Hill tbr their individual course preparation. A student.sing this manual
is using it
rvithout permission.
PROBLEM 2.82
Two blocks of rubber with a modulus of rigidity G = 10 Iv{Pa are bonded
to rigid suppofis and to a plate l.B. Knowing that h=200 mm ancl
c =125 mm, cletermine the largest allowable load P arrd the smallest
allorvable thickness a of tlre blocks if the shearing stress in the rubber
is not to exceed 1.5 MPa and the deflection of the plate is to be at least
6 rnrn.
SOLUTION
*_-7lp
2t
D*
-
+P*-
Shearingstress:
+,P
_de
l*
t=+=*
P =2bcr =2(A.2 m)(0.125
rn)(l.5xl0r
ri.Pa)
P = 75.0
kN <
Shearingstrain: y={ =+
aG
Gd
r
tl0xl0n
I)at(0.00(r m)
4=J_r_ =j--:---:-----:=:_-:-jj-.=0.0_l
1.5
x 10" pa
rn
a = 40.0 rnm
{
PROPRIETARY \IATERIAL. (9 2{ll2 The lVlc(irarv-Hill (-'ompanies, lnc. All rights reserved. No part of this Manual may be displayed.
iu any lbrnr or by any means. without tlte prior rvritten perrnission olthe publisher. or used beyond the lirnitsd
reproduced, or distribr,rted
rvithout pernr,ission.