CHAPTERl
PROBLEM 2.1
A nylon thread is subjected to a 8.5-N tens ion force. Know ing that E = 3.3 GPa and that the length of the
thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread.
SOLUTION
(a)
Strain:
c = f = __!_J_ = 0.0 11
Stress:
a = E& = (3.3 x 10 9 )(0 .0 11) = 36.3 x 10 6 Pa
L
100
p
a = -
A
Area:
A = !_ =
CY
Diameter:
(b)
d
85
·
= 234.1 6 x I0- 9 m 2
36.3 X 106
=~ =
(4)(234. 16
X
10- 9) = 546 X
J0- 6 m
d = 0.546 mm ◄
TC
a = 36. 3 MPa
Stress:
◄
PROPRI ETARY MATERIAL. Copy right © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
93
PROBLEM 2.2
A 4.8-ft-long steel wire of ¼-in.-diameter is subjected to a 750-lb tensile load. Knowing that E = 29
x
6
10 psi,
determin e (a) the e longation of the wire, (h) th e correspo nding norma l stress.
SOLUTION
(a) Deformation:
Area:
b = PL .
AE '
A = n(0.25 in.)2
4
6
=
(750 lb)(4.8 ft x 12 in./ft)
(4.9087 x 10- 2 in 2 )(29 x 106 psi)
b = 3.0347 X 10- 2 in.
(b) Stress:
Area:
,5 = 0.0303 in. ◄
p
a =-
A
(750 lb)
a =- - - - -2- 2
(4.9087 X 10- in
)
a= 1.52790 x 104 psi
a= l 5.28ksi ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
94
PROBLEM 2.3
An 18-m-long steel wire of 5-mm diameter is to be used in the manufacture of a prestressed concrete beam.
It is observed that the wire stretches 45 mm when a tensile force P is applied. Knowing that E =200 GPa,
determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire.
SOLUTION
(a)
o = PL
or
AE '
w ith A= ~7Z"d 2 = ~1r(0.005) 2 = 19.6350 x 10- 6 m 2
4
4
2
9
p = (0.045 m)(l 9.6350 x 10-6 m )(200x 10 N/m
18 m
2
)
=9817 _5 N
P =9.82 kN ◄
(b)
0-
P
9817.5 N
A
19.6350x l 0- m
=- =-------,6
2
6
o- = 500 MPa ◄
500 x 10 Pa
PROPRIETARY MA TERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
95
PROBLEM 2.4
Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod with E = 73 GPa and
an ultimate strength of 140 MPa. Knowing that the distance between the gage marks is 250.28 mm after a load
is applied, determine (a) the stress in the rod, (b) the factor of safety.
SOLUTION
(a)
o= L- 4i
= 250.28 mm - 250 mm
= 0.28 mm
0
Lo
ti = -
0.28 mm
250mm
= 1.11643 X J0- 4
a = Ee
= 8.1760x l07 Pa
a =8 1.8 MPa ◄
(b)
F.S. = '!.!!...
a
140 MPa
81.760 MPa
= 1.71233
F.S. = 1.7 12
◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
96
PROBLEM 2.5
An aluminum f ipe must not stretch more than 0.05 in. when it is subjected to a tens ile load. Know ing that
E == 10.1 x 10 psi and that the max imum allowable nonnal stress is 14 ksi, determine (a) the max imum
allow able length of the pipe, (b) the required area of the pipe if the tensile load is 127.5 kips .
SOLUTION
(a)
6
Thus,
L == £A o == Eo == (10.1 x 10 )(0.05)
p
0"
14 X 103
L = 36.1 in. ◄
(b)
p
a- = -
A
3
Thus,
A == !_ == 127.5 x 10
0"
14 X 10 3
A == 9.1 1 in
2
◄
PROPRIETARY MATERIAL. C opyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
97
PROBLEM 2.6
A control rod made of yellow brass must not stretch more than 3 mm when the tension in the wire is 4 kN.
Knowing that E = 105 GPa and that the maximum allowable normal stress is 180 MPa, determine (a) the
smallest diameter rod that should be used, (b) the corresponding maximum length of the rod.
SOLUTION
p
(a)
(]' = -
·
A'
Substituting, we have
⇒
d =
d =
d
{4P
v~
4(4 x 103 N)
(1 80 x I0 6 Pa) 1r
= 5.3 192 x 10- 3 m
d = 5.32 mm ◄
8
L
(b)
& = -
Substituting, we have
CJ' =
E §_
L
⇒ L = Eo
CJ'
L
= (105 x 109 Pa) (3 x 10- 3 m)
6
(180 x 10 Pa)
L = 1.750m ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
98
PROBLEM 2.7
A steel control rod is 5.5 ft long and must not stretch more than 0.04 in. when a 2-kip tensile load is applied to it.
6
Knowing that E = 29 x 10 psi, determine (a) the smallest diameter rod that should be used, (b) the
corresponding nonnal stress caused by the load.
SOLUTION
(a)
Pl
o=- :
AE
.
0.04 m.
=
(2000 lb)(5 .5x l 2 in.)
6
A (29 x 10 psi)
A = .!_ nd 2
4
d
(b)
P
a= A
2000 lb
= - - - - -2
= 0.11 3793 in2
= 0.38063 in.
d
= 0.38 1 in.
◄
a = 17.58 ksi ◄
17575.8 psi
0.11 3793 in
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
99
PROBLEM 2.8
A cast-iron tube is used to support a compressive load. Knowing that E = 10 x 10 6 psi and that the maximum
allowable change in length is 0.025%, determine (a) the maximum normal stress in the tube, (b) the minimum
wall thickness for a load of 1600 lb if the outside diameter of the tube is 2.0 in.
SOLUTION
(a)
§_ = __§_ = 0.00025
L
100
0
L
6= -
CY =£§_
L
6
0" =
(10
0" =
2.50 X 10 psi
X
10 psi)(0.00025)
3
CY= 2.50 ksi
p
(b)
O" = -
A = !:_ = 1600 lb = 0 _64 in 2
CY 2.50 x 103 psi
·
A'
2A = ~(d
4
0
df
◄
d2)
I
2
= (2.0 in.)2 - 4(0.64 in ) = 3.185 1 in 2
7[
d;
t
t
1
= -(d
2
0
-
d)
I
= 1.78469 in.
1
.
= -(2.0
m. 2
.
1.78469 m.)
= 0.1 07655 in.
t=
0.1 077 ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
IOO
PROBLEM 2.9
A 4-m-long steel rod must not stretch more than 3 mm and the normal stress must not exceed 150 MPa when
the rod is subjected to a 10-kN axial load. Knowing that E = 200 GPa, determine the required diameter of
the rod.
SOLUTION
L = 4m
0
= 3 X I0- 3 m,
a = l50 x l06 Pa
= 10 X I03 N
E = 200 x 109 Pa,
p
Stress:
a =-
p
A
3
A = !__ = lO x l0 N = 66.667 x I0- 6 m 2 = 66.667 111111 2
a 150 x 106 Pa
Deformation:
0
=
PL
AE
A = PL =
Eo
The larger value of A governs:
3
(IO x l0 )(4 )
= 66.667 x 10- 6 1112 = 66.667 mm 2
c200 x to9 )(3 x 10- 3 )
A
= 66.667 mm 2
4(66.667)
d = 9.2 1 mm ◄
7r
PROPRIETARY MA TERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
IO I
PROBLEM 2.10
A nylon thread is to be subjected to a 10-N tension. Knowing that E = 3.2 GPa, that the maximum allowable
normal stress is 40 MPa, and that the length of the thread must not increase by more than 1%, determine the
required diameter of the thread.
SOLUTION
Stress criterion:
a = 40 MPa = 40 x 10 6 Pa
p
A= p =
a =-:
a
A
A = ~d
2
:
P = 10 N
lON
= 250 X 10- 9111 2
40 x 106 Pa
d = 2H =
2✓ 250 : 10-
9
= 564.19 X 10-
6
111
d = 0.564 mm
Elongation criterion:
~= 1%= 0.0 1
L
0
= PL:
AE
9
A = PI E = 10 Nl3.2 x 10 Pa = 312 _ x 10_9 111 2
5
oil
d = 2H = 2
0.0 1
312 5 10 9
· ;
- = 630.78 X 10- 6 111 2
d = 0.63 1 mm
d = 0.63 1 mm ◄
The required diam eter is the larger value:
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
I02
PROBLEM 2.11
A block of 10-in. length and 1.8 x 1.6-in. cross section is to support a centric compressive load P. The
material to be used is a bronze for which E = 14 x 106 ps i. Determine the largest load that can be applied,
know ing that the normal stress must not exceed 18 ksi and that the decrease in length of the block should be at
most 0.12% of its original length.
SOLUTION
Considering allowable stress,
a=18 ksi or 18x JO"psi
A=(1 .8 in.)(1.6 in .) = 2.880 in
P=aA
2
Cross-sectional area:
p
O' = -
A
⇒
= (1 8 x I 03 psi)(2.880 in 2 )
=5. 1840 X I 04
lb
or 5 1.840 kips
Considering allowable deformat ion,
<5
L
- = 0.1 2%
⇒
or
0.0012 in.
P=AE(f)
P=(2. 880 in )(14x !0
2
=4.8384 X 10
4
6
psi)(0.00 12in.)
lb
or 4 8.384 kips
The smaller value for P resulting from the required deformation criteria governs.
48.4 kips ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
103
PROBLEM 2.12
A square yellow-brass bar must not stretch more than 2.5 mm when it is subjected to a tensile load. Knowi ng
that E = 105 GPa and that the allowable tensile strength is 180 MPa, determine (a) the maxi mum allowable
length of the bar, (b) the required dimensions of the cross section if the tensile load is 40 kN.
SOLUTION
o- = 180 x I 06 Pa P = 40 x 103 N
E = I 05 x 109 Pa
(a)
i5 = 2.5 x 10- 3 m
i5 = PL = o-L
AE
E
9
L = Ei5 = (105 x 10 )(2.5 x 10ol80x 106
3
)
= 1.45833
111
L = 1.458 m ◄
p
(b)
(j
=-
A
3
p
40 X I 0
')
-6
2
A= - =
6 = L22.22 X IO 111 = 222.22
(j
180 X 10
A = a2
a =✓
A = ../222.22
111111
2
a = l 4.9 lmm ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
I04
PROBLEM 2.13
Rod BD is made of steel (£ = 29 x I 0 6 psi) and is used to brace the axially
compressed member ABC. The max imum force that can be developed in member
BD is 0.02P. If the stress must not exceed 18 ksi and the maximum change in
length of BD must not exceed 0.00 I times the length of ABC, determine the
smallest-diameter rod that can be used for member BD.
~
72 in.
tu
I 54 in.
SOLUTION
F8D = 0.02P = (0.02)(130) = 2.6 kips = 2.6 x 103 lb
Considering stress,
CJ =
18 ksi = 18 x 103 psi
F
CJ = _j]Q_
A
A = FaD =~= 0. 14444 in2
CJ
18
Considering deformation, 5 = (0.001)(144) = 0 .144 in.
A = FaDLBD = (2.6x 103)(54)
£5
(29x l06 )(0.144)
0.03362 in2
Larger area governs. A = 0. 14444 in 2
d = 0.429 in. ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
105
PROBLEM 2.14
t
2.5
Ill
The 4-mm-diameter cable BC is made of a steel with E = 200 GPa. Knowing
that the maximum stress in the cable must not exceed 190 MPa and that the
elongation of the cable must not exceed 6 mm, find the maximum load P that
can be applied as shown.
t-
3.5m
l
SOLUTION
~
Lac= vo- + 4- = 7.2 11lm
Use bar AB as a free body.
4
3.5? - (6)(~ ~ Fae)= 0
7.2 111
P = 0 .'J509Fac
Considering allowable stress, a- = 190 x 106 Pa
Fae
a- = A
Considering allowable elongation, ,5 = 6 x I0- 3 m
,5 = Fac Lac
AE
Fae= AE,5
Lac
Smaller value governs. Fae
6
9
3
(12.566 x 10- )(200 x 10 )(6 x 10- )
7.2 111
2 _091 x l 03N
= 2.09 1 x I 03N
P = 0 .9509FBC = (0.9509)(2 .091x 103 ) = 1.988 x I 0 3 N
P = 1.988 kN ◄
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
I06
PROBLEM 2.15
A single axial load of magnitude P = 15 kips is applied at e nd C of the steel
6
rod ABC. Knowing that E = 30 x 10 psi, determine the diameter d of portion
BC for which the deflection of point C will be 0.05 in.
SOLUTION
<5 _
L AB
PL; _ ( PL )
L
A;E; -
C -
AE
= 4 ft = 4 8 in.;
( PL )
AB +
AE
BC
L 8 c = 3 ft = 36 in.
Jr(l.25 in.)2 = 1.227 18 in 2
4
Substituting, we have
3
005 . =( 15xl06 lb J( 48 in. 2 + 36 in. J
· m.
30 x I 0 psi 1.227 18 in
A8 c
A8 c = 0.59127 in
2
:rrd 2
Ase= - 4
or
d
=✓
d=
4
:c
4(0.59 127 in 2 )
Jr
d = 0.86766 in.
d = 0.868 in. ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
107
PROBLEM 2.16
28
36111111
A 250-mm-long aluminum tube (£ = 70 GPa) of 36-mm outer
diameter and 28-mm inner diameter can be closed at both ends by
means of single-threaded screw-on covers of 1.5-nun pitch. With
one cover screwed on tight, a solid brass rod (£ = 105 GPa) of
25-mm diameter is placed inside the tube and the second cover is
screwed on. Since the rod is slightly longer than the tube, it is
observed that the cover must be forced against the rod by rotating
it one-quarter of a turn before it can be tightly closed. Determine
(a) the average normal stress in the tube and in the rod, (b) the
deformations of the tube and of the rod .
111111
E:b, :
25
;,111~
I
1-+--- - --
250 mm -
----+-<
SOLUTION
:rr2:rr
A ,0<1 = - d
4
,:\ube
=
=
- (25) 2
4
__
rO<l -
490.87 mm 2
=
490.87 x IO- 6 m 2
:(0.250)
6 = 8.88 15 x 10-9 p
(70 X 10 )(402. 12 X 10- )
PL
£ tubeA tube
,5
=
PL
E,cd Arod
P(0.250)
(105 X 106 )(490.87 X 1o-6 )
-4_8505 x l0-9p
6
,5• = (¼ turn ) x l.5 mm = 0.375 mm = 375x 10- m
,5tube
= ,5• + o,od
or
otube -
o =o·
rod
8.88 15 X 10- 9p + 4.8505 X 10- 9p = 375 X 10- 6
0.375 X 10- 3
P = ------~= 27.308x 103 N
(8.8&15 + 4.85 05)(10- 9)
(a)
P
er
27.308x l03
P
a-,od = - A rod =
(b)
67.9x 106 Pa
= -- = ----~
tube
A tube
402.1 2x l0- 6
Otube
c;tube
= 67.9 MPa ◄
3
27. 308x l0
490.87 X 10- 6
- 55.6 x 106 Pa
= (8.88 15 X 10-9)(27.308 X 103 ) = 242.5 X 10- 6 111
o,od = - (4.8505 x 10-9)(27.308 x 103 ) = - 132.5 x 1o-6 m
a-,od = - 55.6 MPa ◄
o tube
= 0.243 111111 ◄
orod =-0. 1325 mm
◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
IOS
B 0.4 in.
A
P - 350 11, ~ -. - ~
PROBLEM 2.17
C
I ~ -.-~
l in.
The specimen shown has been cut from a ¼-in.-thick sheet
of vinyl (£ = 0.45 x 106 psi) and is subjected to a 350-lb
tensile load. Determine (a) the total deformation of the
specimen, (b) the deformation of its central portion BC.
l in.
I
1-1.6 in.-l-2 in. -1-1.6 in.~
SOLUTION
6
_ PLAB _
(350 lb)( l.6in.)
AB - EAAB - (0.45 x I 0 6 psi)(! in.)(0.25 in.)
;;BC = PLBc
EABc
6CD
(a)
=
= 4 _9?7S x l0-3in.
(350 lb)(2 in.)
= 15.5556 X 10-3 in.
(0.45 x I 0 6 psi)(0.4 in.)(0.25 in.)
= (JAB = 4.9778 X 10- 3 in.
Total defonnation:
6 = 25.51 1 X 10- 3 in.
6 = 25.5 X 10- 3 in. ◄
(b)
Deformation of portion BC:
<'iBc
= 15.56 x 10- 3 in.
◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
109
!p
✓\\
I
D
PROBLEM 2.18
.
r,
I
A
375
1 111 111
The brass tube AB (E = 105 GPa) has a cross-sectional area of
140 mm2 and is fitted with a plug at A. The tube is attached at B to a
rigid plate that is itself attached at C to the bottom of an aluminum
2
cylinder (E = 72 GPa) with a cross-sectional area of 250 mm . The
cylinder is then hung from a support at D . In order to close the
cylinder, the plug must move down through 1 mm. Detennine the
force P that must be applied to the cylinder.
111111
B
C
SOLUTION
Shortening of brass tube AB:
LAB = 375 + 1 = 376 mm = 0 .376 m
AA8 = 140 mm 2 = 140 x 10- 6 111 2
EAB = 105 x 109 Pa
P(0.376)
= 25 .578 x 10- 9 p
(I 05 x l09 )(140 x I 0-6)
Lengthening of aluminum cylinder CD:
Leo = 0.375 m
2
Aco = 250 111111 = 250x I0-6 111
en
P(0.375 )
x 109 )(250 x I o-6 )
2
9
Eco = 72 x I 0 Pa
20.833x l0- 9 P
5 A = 5A8 + Oco where 5 A = 0.001 m
Total deflection :
0.001 = (25.578 X 10- 9 + 20.833
X
P = 2 1.547 x 10 3 N
lQ- 9 )?
P = 21.5 kN ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
110
PROBLEM 2.19
p
r
Both portions of the rod ABC are made of an aluminum for which E = 70 GPa.
Knowing that the magnitude of P is 4 kN, determine (a) the value of Q so that
the deflection at A is zero, (b) the corresponding deflection of B.
A
20-mm diameter
0.4 Ill
l
B
Q
0.5111
60- mm diameter
C
SOLUTION
(a)
1!
2
1!
2
-3
A8 c = -dnc = - (0.060) = 2.8274 x 10 m
4
4
2
Force in member AB is P tension.
Elongation:
Force in member BC is Q - P compression.
Shortening:
<5
- (Q - P)Lsc (Q - P)(0.5)
EA 8 c
- (70 x l 09 )(2.8274x l0- 3 )
2.5263 x I o-9(Q - P)
BC -
For zero deflection at A , <5sc
=<5AB
2.5263 x 10- 9(Q - P) = 72.756 x 10-6
Q = 28.3 x l 03 + 4 x 103 = 32.8x l 03 N
Q = 32.8 kN ◄
<5A8 = 0.0728 mm ,J., ◄
(b)
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
111
PROBLEM 2.20
p
r
0.4
The rod ABC is made of an aluminum for which E = 70 GPa. Knowing that
P = 6 kN and Q = 42 kN, determine the deflection of (a) point A , ( b) point B .
A
20- mm diameter
Ill
l
B
Q
0.5
Ill
L
60-111111
diameter
I_.._....._..
SOLUTION
"2
Jr
4
4
2
- 62
2
-3
AA8 = - dA8 = - (0 .020) =3 14.1 6x l 0 m
Jr
2
Jr
A8 c = - d 8 c = - (0.060) = 2.8274xl0 m
4
PAB
2
4
= p =6 X 103 N
P8 c = P - Q = 6 x I 0 3 - 42 x I0 3 = - 36 >< 103N
LA8 = 0.4 m L8 c= 0.5 m
OAa = PABLAB
AABEA
<5
BC
(6 x l 03)(0.4)
= 109.135 x l0- 6m
(3 14.16 x l 0- 6)(70xl09 )
_ Pac Lac_
(- 36 x l 03)(0.5)
A8 cE - (2.8274 x 10- 3)(70 x 109 )
(a)
OA = <5A8 + 88c = 109.1 35 x 10- 6 - 90.94 7 x 10- 6111 = 18.19 x I0-6 111
(b)
88 =08 c =-90.9x l0-6 m =-0.0909 mm
or
- 90.947 x l0- 6m
oA = 0.0 18 19 mm t
◄
88 = 0 .0909 mm ,j,. ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material sole ly for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
112
PROBLEM 2.21
228 k:--1
For the steel truss (£ = 200 GPa) and load ing shown, determine
the deformation s of the membe rs AB and A D , know ing that their
cross-sectional areas are 2400 mm 2 and 1800 mm 2 , respectively.
2.5 m
~==::::W:::===--==~ u
1-4.0 m-1-4.0 m~
SOLUTION
Statics: Reactions are 114 kN upward at A and C.
Member BD is a zero force member.
L AB
= ✓4.0 2 +2.52 = 4.7 17 111
Use joint A as a free body.
t
2.5
+ "i.F = 0: 114 + - -F = 0
y
4. 717 AB
FAB =-2 15.1 0 kN
+
~r.F
= 0:
4
F AD + - - F AB
4.7 17
X
=0
(4)(- 215.1 0) = 182 .4 kN
4.717
Member AB:
s - F AB L AB
AB -
EA
AB
(- 2 15 .1 0 X 10 3 ) ( 4.71 7)
(200 X 10 9 )(2400 x 10- 6 )
S AB
=-2. 11 mm ◄
3
Member AD:
(1 82.4 X 10 )(4.0)
= 2.03 x l 0- 3 m
SAD
= 2.03
111111
◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
11 3
r
30 kips
PROBLEM 2.22
A
t-.. . .;. ,.~ =~
st--~~=""""""'~
For the steel truss (£ = 29 x 10 6 psi) and loading shown, determine the
deformations of the members BD and DE, knowing that their cross2
sectional areas are 2 in and 3 in2, respectively.
8 ft
8 ft
I
SOLUTION
Free body: Portion ABC of truss
+J"'i.M E = 0: F80 (15 ft) - (30 kips)(S ft) - (30 kips)(l 6 ft) = 0
F80 = + 48.0 kips
Free body: Portion ABEC of truss
~ "'i.F" = 0: 30 kips+ 30 kips - FM = 0
30kip;_ ~-,,,....,-'-r-:~<-r-;:,-,C
FoE = +60.0 kips
.B
ff;;D
3
680
= PL = (+48.0 x 10 lb)(8x 12 in.)
AE
(2 in 2 )(29x l 06 psi)
68 0
= +79.4 x 10-3 in. ◄
(+60.0 x 103 lb)(IS x 12 in.)
(3 in 2 )(29 x 106 psi)
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
114
PROBLEM 2.23
6ftci
75 ft
Members AB and BC are made of steel (E = 29 x 106 psi) with crosssectional areas of 0.80 in2 and 0.64 in2, respectively. For the loading
shown, determine the elongation of (a) member AB, (b) member BC.
~
A
E
/j.j
kips
SOLUTION
(a)
L AB
=
✓6 2 + 52
= 7.810 ft = 93.72 in.
U se joint A as a free body.
+h:Fy= O: F AB
5
- FAa- 28 =0
7.810
= 43 .74 kip = 43.74 x l 0 3 lb
bAB
=
3
F ABLAB
EAAB
(b)
= (43.74 x 10 )(93.72)
(29 x I 0 6 )(0.80)
b AB
= 0.1 767 in. ◄
Use joint B as a free body.
6
F8c - - - F A8 = 0
7.810
6 43 74
F8 c = ( )( - ) = 33.60 kip = 33.60 x I 03 lb
7.8 10
~"i.Fx
6
BC
= 0:
_ F8 cL 8 c _ (33 .60 x 103)(72)
EAac - (29 x 106 )(0.64)
S 8 c = 0.1 304 i n. ◄
PROPRI ETARY MA TERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
11 5
.,
r~~
PROBLEM 2.24
The steel frame (£ = 200 GPa) shown has a diagonal brace BD with
an area of 1920 mni. Determine the largest allowable load P if the
change in length of member BD is not to exceed 1.6 mm.
6 rn
L~
SOLUTION
<58 D
= l.6x l0- 3 m,
Lnn =
A8 D = 1920mm 2 = 1920x10-6 m 2
~ = 7.8 10 m,
9
Enn = 200 x 10 Pa
FnDLBD
bBD =~~~
EBDABD
FBD
=
9
= (200 x 10 )(1920 x 10-6)(1.6 x 107.81
E 8 DA8 D<5BD
L BD
3
)
= 78.67 x 103 N
Use joint Bas a free body. ~ "'i:.F, = 0:
5
7.8 10
- - FBD -
p = _ 5_ F
7.8 10 BO
= 50.4xl03 N
P
=0
3
(5)(78.67 x 10
7.810
)
P = 50.4kN ◄
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
116
PROBLEM 2.25
Link BD is made of brass (E = I 05 GPa) and has a cross-sectional area of
240 mni2. Link CE is made of aluminum (E = 72 GPa) and has a crosssectional area of 300 mni. Knowing that they support rigid member ABC,
determine the maximum force P that can be applied vertically at point A if
the deflection of A is not to exceed 0.35 mm.
2-25 mm
t - - - - - - = -C
l
- t
't-,---,-,
Bc - - - - - t g
150 llllll
E
Q
l
I125 mm~ 225mrn ~
SOLUTION
Free body member AC:
+):,Mc = 0: 0.350P - 0 .225F80 = 0
C.
F8 0 = 1.55556?
+)i:,M8
p
= 0:
0.125P - 0.225FcE = 0
FCE
(1.55556?)(0.225)
(105 X 109 )(240 X 10- 6 )
_
13 8889
X
= 0.55556?
_9 p
10
(0.55556?)(0.]50) = _
X lQ-9p
3 858 1
(72 X 109 )(300 X 10-6 )
Defonnation Diagram:
From the deformat ion diagram,
9
Slope:
_ <58 + be _ 17.7470 x 10- P _
_ x - 9p
78 876 10
L8 c
0.225
6A = 60 + LAB0
0
= I 3.8889 x 10-9 P + co. 125)(78.876 x 10- 9 P)
= 23.748 x !0- 9 P
Apply displacement limit.
<SA
= 0.35 x 10- 3 m = 23.748 x 10- 9p
P = 14 .738 1 x 103 N
P = l4. 74kN ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
11 7
-
-
C o
I
l
o
F
PROBLEM 2.26
o
E
Members ABC and DEF are joined with steel links (E = 200 GPa) .
Each of the links is made of a pair of 25 x 35-mm plates. Detennine
the change in length of(a) member BE, (b) member CF.
180 111111
Bo
I
260mm
l
--
A
D
18 k"I
1-240111111~
18 k"I
SOLUTION
Free body diagram of Member ABC:
+j''i.M 8 = 0:
(0.26m)(l8kN) - (0.18 m)FCF = 0
F CF
= 26.0 kN
+~"'i.F, =0:
18 kN + F8 E + 26.0 kN = 0
F8 E = -44.0 kN
.............. _ A
\6\<.~
Area for link made of two plates:
A = 2(0.025 m)(0.035 m) = 1.750 x 10- 3 1112
(a)
g
_ F8 EL _
EA
BE
(-44.0 x l03 N)(0.240m)
(200 x 109 Pa)(l.75 x 10- 3 111 2 )
=-30.1 7 l x l 0-6 m
88 £ = - 0.0302 mm ◄
(b)
8
(26.0 x I 03 N)(0.240 m)
EA - (200 x 109 Pa)(l. 75 x 10- 3 1112 )
_ F8 FL _
CF -
= 17.8286x l 0- 6 m
b CF
= 0.01783 111111 ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
118
': D
---~ A
I
["'
18 in.
_I
PROBLEM 2.27
Each of the links AB and CD is made of aluminum (E = I 0 .9 x 106 psi)
2
and has a cross-sectional area of 0.2 in . Knowing that they support the
rigid member BC, determine the deflection of point E.
I~
~~
i!-------1~
22
10
in. ------IC
in.
SOLUTION
Free body BC:
+):.Mc = 0: - (32)FA8 +(22)(1 x l 03 ) = 0
FAB = 687.5 lb
+hFY= 0:
e,
687.5 - l xl0 3 + FcD = 0
FcD = 312.5 lb
8 = FAB LAB
AB
EA
<5
_ FcD LCD
EA
CD -
(687 ·5)(1 3) = 5.6766 X 10- 3 in. = b
(10 .9 X )0 6 )(0.2)
B
(3 12.5)(18)
--'-----'-,-'---'-6
3
2.5803 x 10- in. = be
(I 0.9 x IO )(0.2)
Deformation diagram:
B
E.
C.
S lope 0 = bs -De = 3.0963 x 10L8 c
32
3
= 96.759 x 10- 6 rad
8£ = be + LEC0
= 2.5803 X 10- 3 + (22)(96.759 X 10--o )
= 4.7090 X 10- 3 in.
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
11 9
PROBLEM 2.28
Dl
The length of the {z -in.-diameter steel wire CD has been adjusted so that
with no load applied, a gap of~ in. exists between the end B of the rigid
beam A CB and a contact pomt E. Knowing that E = 29 x 106 psi,
determine where a 50-lb block should be placed on the beam in order to
cause contact between Band £ .
12.5 in.
C
_I
1'7
50 1b
IB
4 in.
SOLUTION
Rigid beam ACB rotates through angle 0 to close gap.
6
0 = Ill = 3 .1 25 x I 0- 3 rad
20
Point C moves downward.
Oc = 40 = 4(3.1 25x 10- 3 ) = 12.5x 10- 3 in.
oCD = Oc = 12.5 x 10- 3 in.
2
ACD = !!_di =
d
!!__(2-)
= 6.9029 x 104 32
3
in 2
o. _ FcDLcD
CD -
EA
CD
(29 X 106 )(6.9029 X 10- 3 )(12.5 X 10- 3 )
12.5
= 200.18 lb
Free body A CB:
t,..,
A-,
F,pt
t
!50 ,. .,
20 - -I-
+°'}'L M A = 0: 4FCD - (50)(20- x) = 0
20 -x = (4 )(200. l 3) = 16.0144
50
X = 3.9856 in.
x<3.99 in. ◄
For contact,
PROPRI ETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
120
PROBLEM 2.29
A ho mogeneous cable of length Land uniform cross section is suspended from one end. (a) Denoting by p the
density (mass per unit volume) of the cable and by E its modulus of elasticity, detennine the elongation of the
cable due to its own weight. (b) Show that the same e longation would be obtained if the cable were horizontal
and if a force equal to half of its weight were applied at each end.
SOLUTION
(a)
For element at point identified by coordinate y,
P
= weight of portion below the point
=pgA(L - y)
dS- Pdy _ pgA(L -y)dy
EA
EA
<5 =
I/
7N
(b)
Total weight:
t
=
pg(L - y) dy
E
pg(~-y) dy = p:( Ly-½/
)I:
P:[L-f J
2
W = pgAL
2
F _ EM _ EA . I pgL
L
L 2 E
_
1 pgAL
2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
12 1
PROBLEM 2.30
A vertical load P is applied at the center A of the upper section of a homogeneous
frustum of a circular cone of height h, minimum radius a, and maximum radius b.
Denoting by E the modulus of elasticity of the material and neglecting the effect
of its weight, detennine the deflection of point A .
SOLUTION
Extend the slant sides of the cone to meet at a point O and place the origin of the coordinate system there.
:J
b- a
tan a = - h
From geometry,
a
a, = - - ,
tan a
bi = -b-
tan a
,
r =y tan a
At coordinate point y, A = :rrr2
Defonnation of element of height dy:
dS = Pdy
AE
dt5= _!!._dy=
P
dy
E:rr r 2 :rr£tan2 a y2
Total deformation:
P
[ I
:rrEtan 2 a a,
P
b1 - a1
P(b1 - a1 )
:rrE tan 2 a
a1b1
:rrEab
I
b1
J
t5
A
= __!_!!:_j, ◄
:rrEab
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
122
PROBLEM 2.31
Denoting by t: the "engineering strain" in a tensile specimen, show that the true strain is t:1 = In (1 + e).
SOLUTION
5J= ln (l +t:)
L0 +5
t:,= ln -L = ln - = ln ( l + -
Lo
Lo
Lo
e, =ln (l +t:) ◄
Thus,
PROPRIETARY MA TERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
123
PROBLEM 2.32
The volume of a tensile specimen is essentially constant while plastic deformation occurs. If the initial
diameter of the specimen is d 1, show that when the diameter is d, the true strain is 6 1 = 2 ln(d / d).
SOLUTION
If the volume is constant,
!!....d 2 L = !!.. . df L0
4
4
2
61
= In -L= In (d
_!_ )
L0
d
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
124
Brass core
(E = 105 CPa)
Aluminum plates
(E = 70 CPa)
PROBLEM 2.33
Rigid
end plate
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 aluminum plates.
SOLUTION
o = PaL
EaAa
Therefore,
Substituting,
Pa = (EaAa{z)
PA= (EAAA + E8 A8
)(%)
0
p
E= - =~- - - -~
L
(EAAA + E 8 A8 )
(450 x 103 N)
E = - - -9- , , - - - - - - - - ~ - - - ~ - -9 , - - - - - - - - - (70 x 10 Pa)(2)(0.06 m)(0.01 m) + (105 x 10 Pa)(0.06 m)(0.04 m)
E= 1.33929
10- 3
a = Ee
Now,
(a)
X
Brass-core:
a 8 = (105x l 09 Pa)( l.33929x l 0-3 )
= 1.40625 x 108 Pa
a 8 = 140.6 M Pa
(b)
Aluminum:
9
aA = (70 x 10 Pa)(l.33929 x 10-
3
◄
)
7
= 9.3750x l0 Pa
a A = 93.8 MPa ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
125
Brass c.-ore
(E = 105 CPa)
PROBLEM 2.34
Rigid
Al un1inurn plates
(E = 70 C Pa)
end plate
For the composite block shown in Prob. 2.33, determine
(a) the value of h if the portion of the load carried by the
aluminum plates is half the portion of the load carried by
the brass core, (b) the total load if the stress in the brass is
80 MPa.
PROBLEM 2.33. 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
aluminum plates.
SOLUTION
b = b a = bb;
<5 = P,,L
EaAa
P = Pa + Pb
b = Pbl
EbAb
and
Therefore,
I
pa = -2 P.b
(a)
(EaAa)(f) = ½(EbAb)(f)
I ( Eb1
Aa = 2lEJAb
Aa = _!_( I 05 GPa J(40 mm)(60 mm)
2l 70 GPa
A" = 1800
111111
2
1800 mm 2 = 2(60 mm)(h)
h = l 5.00 mm ◄
(b)
O"h
Pb
= -
Ab
⇒
I
2
Pi, = O"hAb and P,, = - Pi,
P = Pa + Pb
PROPRIETA RY MA TERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scann ed, duplicated, forwarded, di stributed, or posted
on a website, in whole or part.
126
PROBLEM 2.34 (Continued)
P
I
= -2 (abAh) + a bAb
= (abAb)l.5
P = (80 x I0 6 Pa)(0.04 m)(0.06 m)( l.5)
P = 2.880 x 105 N
p
P = 288 kN ◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
127
PROBLEM 2.35
The 4.5-ft concrete post is reinforced with six steel bars, each with a 1½-in. diameter.
Knowing that E., = 29 x 106 psi and Ee= 4.2 x 106 psi, determine the normal stresses
in the steel and in the concrete when a 350-kip axial centric force P is applied to the
post.
SOLUTION
Let P,, = portion of axial force carried by concrete.
P., = portion carried by the six steel rods.
p
= ECACt5
L
C
p
.,
P = P,, + P,
t5
L
= E,A.,t5
L
t5
= (ECAC+ E., A.,)L
-P
Ec.( +E., A.,
e= - = - - - - 6
A = 6!!._d 2 = 1! (1.125 in.) 2 = 5.9641 in 2
s
4 s
4
Ac =!!_dc2 - A = !!._(18in.)2 - 5.9641 in 2
4
" 4
= 248.5 1in 2
L = 4.5 ft = 54 in.
3
ti =
- 350x l 0 lb
=-2 _8767 x 10- 4
(4.2 x 106 psi)(248.5 1 in 2 ) + (29x l06 psi)(5.964 1 in 2
er., = E.,e = (29 x 106 psi)(- 2.8767 x 10- 4 ) = - 8 .3424 x 103 psi
er_,= -8.34 ksi ◄
ere = £Ce = (4.2 X106 psi)(- 2.8767 X 1o-4 ) = 1.2082 1X103 psi
ere= - 1.208 ksi ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
128
PROBLEM 2.36
For the post of Prob. 2.35, determine the maximum centric force that can be applied if the
allowable nonnal stress is 20 ksi in the steel and 2.4 ksi in the concrete.
PROBLEM 2.35 The 4 .5-ft concrete post is reinforced with six steel bars, each with
a I ½ -in. diameter. Knowing that E_, = 29 x I 0 6 psi and Ee = 4.2 x I 0 6 psi, determine
the normal stresses in the steel and in the concrete when a 350-kip axial centric force P
is applied to the post.
SOLUTION
A llowable strain in each material:
Steel:
6
= as= 20xl0:psi = 6 _8966 x 10_4
"'
E,
{; = O"c
c Ee
Concrete:
29x 10 psi
2 _4 X 10: psi = 5.7 143 X 10-4
4 .2xl0 psi
c = f = 5.7 143x l0- 4
Smaller value governs.
L
Let Pc = Portion of load carried by concrete.
P., = Portion of load carried by 6 steel rods.
~ = EcAc(f )= EcAc E
P.,
A, = 6 ( !!_)d; =
.
4
= ESAS(
f) =
ESAS E
6
2
41[ (1.125 in.)2 = 5.964 1 in
.
A=(!!_)d
- A= !!_(18in.)2
- 5.964lin
4
4
2
C
C
2
= 2.485 l x l 0 2 in 2
S
p = ~ + P, = ECACE + E, A,E
P = [(4.2x l 0 6 psi)(2.485 1x 10 2 in 2 ) + (29x 10 6 psi)(5.9641 in 2 )](5.7 143 x 10-4 )
5
P = 6.9526 x 10 lb
P = 695 kips ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
129
PROBLEM 2.37
25111111
r~
1=1
,,-
'"'
Brass C..'Ore
E
= 105 C Pa
An axial force of 200 kN is applied to the assembly shown by means of
rigid end plates. Determine (a) the normal stress in th e aluminum shell ,
(b) the correspond ing deformat ion of the assembly.
300mm
- -
Aluminium shell
E = 70 CPa
--,
CJ
60111111
SOLUTION
Let Pa= Portion of axial force carried by shell.
Pb= Portion of axial force carried by core.
p
or
= EaAa o
L
a
P,,
or
= EbAb o
L
Thus,
Aa = ~ [(0.060) 2
w ith
-
4
(0.025)2] = 2.3366
X
10-3 111 2
X
109)(0.49087
Ab = ~ (0.025)2 = 0.49087 x 10- 3 1112
4
p = [(70 X 109)(2.3366
p = 2 15.1 0
0
X
lQ
6
X
X
l0- 3)]f
L
f
L
p
Ii=-=-------,-L
215 .10 x 106
Strain:
lQ- 3) + (105
200
103
X
215.10
(a)
<Ya= Ea&= (70 X l0 9 )(0.92980 x 10- 3) = 65. 1x 106 Pa
(b)
o = &L = (0 .92980 x 10- 3 )(300 mm)
X
106
= 0.92980
X
10-3
c,a
= 65. 1 MPa ◄
0 = 0.279 111111 ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
130
r-~
PROBLEM 2.38
25 mm
D
Bn1ss core
E = 105CPa
The length of the assembly shown decreases by 0.40 mm when an axial
force is applied by means ofrigid end plates. Determine (a) the magnitude
of the applied force, (b) the corresponding stress in the brass core.
300111111
- -
Alumin ium she ll
E = 70CPa
-d
60 111111
SOLUTION
Let Pa= Portion of axial force carried by shell and Pb= Portion of axial force carried by core.
or PO
= E LA",,u
0
Thus,
with
A" = .::.[(0.060) 2
4
-
(0.025) 2 ] = 2.3366 x 10- 3 m 2
Ab = .::.(0.025)2 = 0.49087
4
p = [(70
with
(a)
(b)
X
109 )(2.3366
X
X
10- 3 1112
(0- 3 ) + (105
X
109 )(0.49087
X
10- 3 ) ] ~ = 2 (5.(0 X 106 ~
L
L
<5 = 0.40 mm, L = 300 mm
P = (2 15.10 x l06 ) 0.40 = 286.8 x 103 N
300
ab =
Fi
Ab
P = 287 kN ◄
= Eb<5 = (105 x 109)(0.40 x 10- 3) = 140 x 106 Pa
L
300 X 10- 3
a b = 140.0 MPa ◄
PROPRIETARY MA TERIAL. C opyright © 2015 McG raw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in who le or part.
131
PROBLEM 2.39
A
i
1.25 in.
6 kips
B
A polystyrene rod consisting of two cylindrical portions AB and BC is restrained at
6
both ends and supports two 6-kip loads as shown. Knowing that E = 0.45 x 10 psi,
detennine (a) the reactions at A and C, (b) the normal stress in each portion of
the rod.
- 2 in.
C
SOLUTION
(a)
We express that the elongation of the rod is zero.
<5 = PAaL AB + Pac Lac
fd~ 8 E
A
=0
fdJcE
But
Substituting and simplifying,
RALAB _ Re Lac = 0
d~B
dJc
B
C
( I)
From the free body diagram,
Substituting (I) into (2),
(2)
RA + Re = 12 kips
5.2667RA = 12
RA = 2.2785 kips
RA = 2.28 kips t ◄
Re - 4 .2667 (2.2785) - 9 .7217 kips
r rom ( 1),
Re = 9.72 kips t ◄
(b)
P
+RA - __dfL _ _ _
AB - AAB - AAB -
2.2785
----
t (1.25)2
a AB = + 1.857 ks i
◄
P c - Re - 9.72 17
O-ac = - 8 -= - - = - - Ase Ase
-¼ (2)2
a 8c =-3.09 ksi
◄
(j
PROPRIETA RY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
132
I
I
A
"
r
D
e
"l
E
e
6
Three steel rods (E = 29 x I 0 psi) support an 8.5-kip load P. Each of the
rods AB and CD has a 0.32-in2 cross-sectional area and rod EF has a l -in2
cross-sectional area. Neglecting the deformation of bar BED, determine
(a) the change in length of rod EF, (b) the stress in each rod.
20in.
B
'"
PROBLEM 2.40
" C- 1
i
16 in.
I
F
'
I
SOLUTION
Use member BED as a free body.
i;.e.
B
t- -----J';p-E-- - -t~It~~
D
rtl
fEi:
By symmetry, or by I.ME = 0:
PcD = PAB
+) I. FY = 0:
PAB + PcD + PEF - p
=0
P = 2PAB + PEF
Ii
AB
= PABLAB
EA
AB
8
CD
= PcDLcD
Ii
EA
CD
EF
= PEFLEF
EA
EF
Since
Since points A , C, and Fare fixed, 8B = liAB' 8D = lico, 8 £ = liEF
Since member BED is rigid, 8 £ = 8B = lie
p
PABLAB = PEFLEF
EAAB
EAEF
_ AAB . LEF P. _ 0.32. ~P.
AB -
AEF LAB
EF -
20
EF
= 0.256PEF
P = 2PAB + PEF = 2(0.256PEF) + PEF = l .5 l 2PEF
PEF = ___!_ =
1.512
~ = 5 .62 17 kips
1.512
PAB = PcD = 0 .256(5.62 17) = 1.43916 kips
5 62 17
6
- PEFLEF - ( )(1 )
EF - EAEF - (29 x I 0 3 )(1)
0.0031016 in.
(a)
8
(b)
PAB 1.439 16
.
a AB= CYcD = - - = - - - = 4.4974 ks1
AAB
0.32
liEF = 0.003 10 in.
◄
. ◄
a A8 = CYcD = 4.50 ks1
_ PEF _ 5.62 17 _ 62 7 k .
CYEF- - - - - - - - 5. I SI
AEF
I
aEF = 5.62 ksi ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
133
Dirnensions in
PROBLEM 2.41
ff11T1
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 that E, = 200 GPa and Eh= 105 GPa, detennine
(a) the reactions at A and E, (b) the deflection of point C.
60kN
40-mm diam.
30-mm cham.
SOLUTION
9
E = 200 x 10 Pa
A to C:
A = ~(40) 2 = 1.25664 X 103 mm 2 = 1.25664 X 10- 3 m 2
4
EA = 251.327 x 106 N
E = 105 x 109 Pa
CtoE:
A = ~(30)2 = 706.86 mm 2 = 706.86 x I0- 6 m 2
4
EA = 74.220 x 10 6 N
A to B:
P = RA
L = 180 mm = 0 .180 m
8
_ PL _
AB -
RA(0 .1 80)
t.A - 251.327 x l 0 6
12
= 7 16.20 X 10- RA
B to C:
P = RA - 60x l 0 3
L = l 20mm = 0 .1 20m
3
PL
(RA - 60x l 0 )(0 .1 20)
EA
251.327 X 10 6
bac= -
12
= 447 .47 x 10- RA - 26.848 x 10 -6
PROPRI ETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, di stributed, or posted
on a website, in whole or part.
134
PROBLEM 2.41 (Continued)
C to D:
P == RA - 60 x I0
3
L == 100 mm == 0.1 00 m
<5
BC
3
_ PL _ (RA- 60xl0 )(0.I00)
- EA 74.220x !06
== 1.34735 x 10- 9 RA - 80.84 1x 1o-6
D to£:
P == RA - IOOx 103
L == 100 mm = 0.1 00 m
PL
bDE == £A
3
(RA - 100x l 0 )(0.1 00)
74.220xl06
== 1.34735 x 10- 9 RA - 134.735 x 10-
A to£:
6
<SAE == SAB + bac + bcD + <5DE
9
== 3.85837 X 10- RA - 242.424 X 10-
6
Since point E cannot move relative to A,
(a)
3.85837 x 10- 9 RA - 242.424 x 10- 6 = 0
3
3
RA = 62.83 1x 103 N
3
3
Re== RA - 100 x 10 = 62.8 x 10 - 100 x 10 = - 37.2 x 10 N
(b)
RA = 62.8 kN ~ ◄
Re = 37.2 kN ~ ◄
9
6
be ==bAB +<58c == l.1 6367x l 0- RA - 26.848xl0== (1.16369 X 10-9 )(62.83 1X 103 ) - 26.848 X 10-6
= 46.3x 10- 6 m
be = 46.3 µm ➔ ◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
135
Dirnensions in
PROBLEM 2.42
ff11T1
Solve Prob. 2.41 , assuming that rod AC is made of brass and
rod CE is made of steel.
60kN
40-mm diam.
PROBLEM 2.41 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 that E., = 200 GPa
and Eb= 105 GPa, detennine (a) the reactions at A and E, (b) the
deflection of point C.
30-mm cham.
SOLUTION
E = l05xl09 Pa
A to C:
A = ~(40) 2 = 1.25664 X I 0 3 mm 2 = 1.25664 X I 0- 3 m 2
4
EA = 13 1.947 x 10 6 N
9
C toE:
E = 200xl0 Pa
A = ~ (30)2 = 706.86 nun 2 = 706.86 x 10- 6 m 2
4
6
EA = 141.372 X 10 N
A to B:
A
P = RA
B
L = 180 mm = 0.180 m
r5
_ PL _ RA(0.1 80)
EA - 13 1.947x l0 6
Gokt-J
AB -
c..
D
--
E
'lo k IJ
= 1.364 18 x 10-9 RA
P = RA - 60x l 0 3
Bto C:
L = l 20 mm = 0. 120m
PL
EA
bac= -
1
(RA - 60x l 0 )(0.120)
131.947 x 106
= 909.456 X 10- 12 RA - 54.567 X 10-6
C toD:
P = RA - 60x l 0
3
L = I00 nun = 0.100 m
0
CD
_ PL _ (RA - 60 x 10 3 )(0.100)
- EA 14 1.3 72 X I 0 6
= 707.354 X 10- 12 RA - 42.441 X 10-6
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
136
PROBLEM 2.42 (Continued)
P = RA- 100x l 0
DtoE:
3
L = IO0mm = 0.l00m
J
_ PL _ (RA - 100x l 03 )(0.1 00)
DE- EA 14 1.372 x !0 6
= 707.354 X 10- 12 RA - 70.735 X 10- 6
A to£:
J AE= JAB + bee+ JCD + JD£
=3.68834 x 10- 9 RA -
167.743 x 10- 6
Since point E cannot move relative to A,
(a)
3.68834xl0- 9 R A - 167.743x l 0- 6 = 0
RA = 45.479 x l03 N
RE= RA - 100 x 103 = 45.479 x 103 - 100 x 103 = - 54.521 x 103
(b)
RA= 45 .5 kN f-- ◄
RE =54.5 kN f-- ◄
J c = JAB +<58 c = 2.27364 x 10-9 RA - 54.567 x 10-6
=(2.27364 X ] 0- 9 )(45.479 X 103 ) =48.8 X 10- 6 tn
54.567 X 10- 6
Jc = 4 8 .8 µm ➔ ◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
137
-,
225 1n1n
_J
t---- - -
5.50 mm
PROBLEM 2.43
Each of the rods BD and CE is made of brass (E = 105 GPa)
and has a cross-sectional area of 200 mni2. Determine the
deflection of end A of the rigid member ABC caused by the
2-kN load.
-----ii ,~I-I
I
\
75mm
100mm
SOLUTION
Let 0 be the rotation of member ABC as shown.
A'
Then
oA = 0.625(1t
But
0B
_ Pan Lan
-
AE
£Ao8
Pan= - -
(105
X
109 )(200 X 10--6)(0.0750)
0.225
L 8D
= 7 X 106 0
Free body ABC:
r~b
2 kt-1
f'ce.
J~=========ittF=:=!
S
_ PcE LcE
8c- AE
_ EAoc
PCE -
(105 x 109 )(200 x 10- 6 )(0.1 0 )
LcE
0.225
= 9.3333 x I 0 6 0
C
From free body of member ABC:
+JL Mp = 0 : (0.625)(2000)- 0.075P8 n - 0. lPcE = 0
or
6
6
(0.625)(2000) - 0.075(7 x 10 0) - 0.1(9.3333 x 10 0) = 0
0 = 0.857 14 x !0- 3 rad
and
oA = 0.6250 = 0.625(0.85714 x 10- 3 ) = 0.53571 x 10- 3 111
OA = 0.536
111111 -J.,
◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
138
PROBLEM 2.44
F
I
8 in.
t
t
The rigid bar AD is supported by two steel w ires of k -in. diameter
E
6
10 ill.
_I_A~-
B
C
(E = 29 x 10 psi) and a pin and bracket at A . Knowing that the
wires were initially taut, determine (a) the additional tension in each
w ire when a 220-lb load P is applied at D , (b) the corresponding
deflection of point D .
pi
•
SOLUTION
Let 0 be the notation of bar ABCD.
Then b 8 = 12 0
be = 24 0
1
6
(29 x l 0 ) ~ (-)\120)
4
6
P.BE -- -EAbBE
- - -- - - - -- -- - L 8£
10
= 106.765 X J03 0
l)
2
(29 X 106
- EAbCE
P,CF _ _ __
)~(..!___)
(24 0)
4 16
18
LcF
3
= l 18.628x 10 0
Using free body ABCD,
+J LMA = 0 :
12P8 £ + 24PcF -36P = 0
3
6
(12)(106. 765 x I 0 0) + (24)(11 8.628 x I 0 0) - (36)(220) = 0
4.1 283 x I 0 6 0 = (36)(220)
0 = 1.9 1847x l 0-3 rad)
(a)
(b)
P8 £ = (106.765 x 103 )(1.9 1847 x 10- 3 ) = 204 .83 lb
P8 E = 205 lb ◄
PCF = (11 8.628 x 103 )(1.91847 x 10- 3 ) = 227.58 lb
PCF = 228 lb ◄
3
3
b0 = 36 0 = (36)(1.9 1847x l0- ) = 69.l x l0- in.
0.069 1 in.
i◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
139
I
1
~L~1~L~1.
PROBLEM 2.45
,,.
The rigid bar ABC is suspended from three w ires of the same material. The crosssectional area of the w ire at B is equal to half of the cross-sectional area of the
wires at A and C. Determine the tension in each w ire caused by the load P shown.
Ac
1
u
,.
Jj
iC
~L-i
p
4
SOLUTION
3
2LPc + LP8 - - LP = 0
4
3
I
P. = -P - -P
C
8
2
B
5
- 2LPA - LP8 + - LP = 0
lf
4
5
I
P = -P - -P
A
8
2 B
Let l be the length of the wires .
<5 = PAI =
A
A
B
EA
-'-(~P-~P,)
EA 8
2
8
C
f.~
.C
B'
From the deformation diagram,
A'
SA - 6 s = 6 s - 6c
I
<58 = -(SA+ SJ
or
2
/
I / (5
I
3
I
E(A / 2)Pa = 2EA 8 P - 2Pa + 8P - 2Pa
I
PB =- P
5
)
PB = 0.200P ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
140
PROBLEM 2.45 (Continued)
P =~ P- _l_( P ) =3.!_ P
A
8
Pc
=i
8
2 5
40
p- _I_(
p)= _I_!_
p
2 5
40
PA= 0.525P
◄
Pc= 0.275P ◄
Check:
PROPRI ETARY MA TERIAL. C opyright © 2015 McG raw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be cop ied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
14 1
PROBLEM 2.46
The rigid bar AD is supported by two steel wires of k-in. diameter
6
(E = 29 x 10 psi) and a pin and bracket at D . Knowing that the wires
were initially taut, determine (a) the additional tension in each wire when
a 120-lb load Pis applied at B, (b) the corresponding deflection of point B.
TF
8 in.
l_
B
8 in .
C
D
J-s j
in.
p
SOLUTION
Let 0 be the rotation of bar ABCD.
Then
oA
oe = 80
= 240
1>
O = pAELAE
AE
A
~A
pAEt
A
6
_ EAoA
PAE -
A'
(29x 10 )f(n;-)2(240)
15
LAE
e
PcFf Dy!
lp
= 142.353xl03 0
,5 - PCF LCF
1)
C.
e - ~
EAoe
(29 x 10
6
)f( k)2 (80)
~p = - - = - - - - - ~ - -
8
LeF
= 88.97 \ X 103 0
Using free body ABCD,
- 24PAE + 16P - 8Pep
=0
3
- 24(142.353 X 10 0) + \6(\20) - 8(88.97\
X
103 0) = 0
0 = 0.465 l Ox I 0-3 radL,,
(a)
(b)
PAE = (142.353x l03 )(0.465 10 x l 0-3 )
PAE = 66.2 lb ◄
Pep = (88.97 1x I03 )(0.465 10 x I o-3 )
Pep= 4 1.4 lb ◄
Os= 160 = 16(0.46510 ><10-
3
Os = 7.44 x 10- 3 in. ,I,. ◄
)
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. Th is document may not be copied, scann ed, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
142
25
PROBLEM 2.47
111ffl
The aluminum shell is fu lly bonded to the brass core and the
assembly is unstressed at a temperature of 15°C. Considering only
axial deformations, determine the stress in the aluminum when the
temperature reaches 195°C.
Brass core
E=l05CPa
a = 20.9 X I0-6/°C
Alnminum shell
E = 70 CPa
a = 23.6 x 10-"/°C
SOLUTION
Brass core:
E = 105 GPa
a = 20.9 X 10- 6/°C
A luminum shell:
E = 70 GPa
a = 23.6 x 10- 61°c
Let L be the length of the assembly.
Free thermal expansion:
L'iT = l 95 - 15 = 180°C
Brass core:
A luminum shell:
(c:\)6 = L a 6 (L'.T)
(or
t
= La" (L'.T)
Net expansion of shell with respect to the core:
Let P be the tensile force in the core and the compressive force in the shell.
Brass core:
£6
= I 05 x 109 Pa
Ab =.'.:. (25)2 = 490.87
4
= 490.87 xl0- 6 111 2
111111
2
PL
(Op)b = - £ bAb
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
143
PROBLEM 2.47 (Continued)
Aluminum shell :
(0 ) - __!_I,__
EA
a a
0
"
-
£ 0 = 70x l 09 Pa
A0
= ~(60 2 -
25
2
)
= 2.3366 x I 03 trun 2
= 2.3366 X I 0- 3 m 2
O = (Op)b +(Op) 0
PL
PL
L(ab - a 0 )(t.T) = - - + - - = KPL
EbAb
EaAa
where
I
I
K = - -+ - -
EbAb
EaAa
I
I
=----,,------~
+ - - - ,9, - - - - - ~
6
3
9
(I 05 x 10
)( 490.87 x
1o-
)
(70 x 10 )(2.3366 x 1o-
)
= 25.516 x 10- 9 N- 1
Then
P = (ab- a 0 )(t.T)
K
(23.6 x Io-6 - 20.9 x Io-6 )(180)
25.516 x l 0- 9
= 19.047 x 103 N
Stress in aluminum:
P
19.047 x 103
a- =-- = - - - - - . , . 3.
a
A0
2.3366 X 10-
- 8.15x l06 Pa
0-0
=-8.15 MPa ◄
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
144
PROBLEM 2.48
25 m m
Solve Prob. 2.47,
assuming that the core is made of steel (£.,
6
a,= 11.7 x 10- / 0 C) instead of brass.
Brass core
E = 105 CPa
a = 20.9 X I0-6/°C
= 200
GPa,
PROBLEM 2.47 The aluminum shell is fu lly bonded to the brass core
and the assembly is unstressed at a temperatu re of 15°C. Considering
only axial deformations, determine the stress in the aluminum when the
temperature reaches 195°C.
Aluminum shell
E = 70 C Pa
a = 23.6 x l 0-"/°C
1260mm ~ I
SOLUTION
Aluminum shell:
£=70GPa a=23 .6 x l0-6/ °C
Let L be the length of the assembly.
Free thermal expansion:
!'J.T = 195 - I 5 = I 80 °C
Steel core:
(6r ), = La, (!'J.T)
Aluminum shell:
(br )a = Laa (!'J.T)
Net expansion of shell with respect to the core:
Let P be the tensile force in the core and the compressive force in the shell.
p
E, = 200 x I 0 9 Pa, A, = ¾(25) 2 = 490 .87 mnl = 490.87 x I o-6 nl
Steel core:
(S ) _ PL
P ' - EA
s
Aluminum shell:
s
Ea = 70 x !09 Pa
PL
(bp)a = - EaAa
Aa = .::_( 60
4
2
-
25)2 = 2 .3366X J03 111111 2 = 2.3366 X10- 3 1112
5 = (5p), + (5p)"
PL
PL
L(aa - a,)(!'J.T) = - - + - - = KPL
E, A, EaAa
where
K =- l- +_ I_
E, A_, EaAa
I
I
= - - -~ - - - - -~6 + - - -~
- - - -~
9
c200 x I o9 )(490.87 x I o-
)
(70 x 10 )(2.3366 x 10- 3 )
=16.2999 X10- 9 N- 1
PROPRIETARY MA TERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be cop ied, scanned, duplicated, forwarded, distribut ed, or posted
on a website, in whole or part.
145
PROBLEM 2.48 (Continued)
Then
p
= (a. -
a s)(i',T)
K
= (23 .6 x 10- 6 -
11.7 x 10-6)(180)
16.2999 x 10- 9
1.
x l03N
13 41 2
p
Stress in aluminum: a- = - - = 13 1.412 x I 033 = - 56.241x I 06Pa
a
Aa
2.3366 X 10-
0"0
= - 56.2 MPa
◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
146
PROBLEM 2.49
The brass shell (a 6 = 11.6 x I0- 6 /°F) is fully bonded to the steel core
(as = 6.5 x 1o-6 /°F). Determine the largest allowable increase in
temperature if the stress in the steel core is not to exceed 8 ksi.
Steel core /
E = 29 X 106 psi
A
llrassshe ll
E = 15 X 106 psi
SOLUTION
Let P, = axial force developed in the steel core.
For equilibrium with zero total force, the compressive force in the brass shell is
Strains:
ti
s
J>..
p
=-s-+ a (!'!.T)
EA
s
s
s
Matching:
(I)
Ab = (1.5)(1.5) - (1.0)(1.0) = 1.25 in 2
As= (1.0)(1.0) = 1.0 in
2
6
ab - as = 5. l x l 0- /°F
3
3
/>,; = a-SAS = (8 x 10 )(1.0) = 8 x 10 lb
_ l_ + _ l_ =
l
+
l
= 87.8 16x l 0- 9 lb- 1
EsAs EbAb (29x l06 )(1.0) (15x l 0 6 )(1.25)
From (I),
(87.816 x 10- 9 )(8 x 103 ) = (5 .1 x 10- 6 )(1'!.T)
l'!.T = 137.8°F ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
147
PROBLEM 2.50
The concrete post (Ee ==3.6x l06 ps i and a = 5.5 x 10- 6 /°F) is reinforced w ith six steel
bars, each of f- in. diameter(£, == 29x l 0tf psi and a, == 6.5x l 0--o/°F) . Determine the
normal stresses induced in the steel and in the concrete by a temperature rise of 65°F.
6 ft
10 in .
10 in.
SOLUTION
2
2
A == 6!!...d == 6!!...(2) == 3.6079 in
'
4
2
4 8
Ac== 102 - A,== 10 2 - 3.6079 == 96.392 in 2
Let P,, == tensile force developed in the concrete.
For equilibrium with zero total force, the compressive force in the six steel rods equals ?,, .
Strains:
t: ==- ~
'
EA
S'
s
+a (t-.T)
'
Matching: t:c == c:,
-JI',,
_ l_ +_ l
( ECAC E, A.,
== (a, - ac )(t-.T)
l
+
l
J P == (1 0 x l0--o )(65)
[ (3.6x l06 )(96.392) (29x !0 6 )(3.6079) c
I',, == 5 .2254 x 103 lb
5.2254 x l 03 ==54 .210 si
96.392
p
P,,
a =- - ='
A,
5.2254 x 10 3
3.6079
ac == 54.2 psi
a s == - 1.448 ksi
- 1448.32 psi
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
148
r
PROBLEM 2.51
A
30-mm diameter
A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel ( E., = 200 GPa, a., = 11. 7 x 1o-6/ 0 C) and
portion BC is made of brass (/:,i, = I 05 GPa, a b = 20.9 x IO 6/ C). Knowing
that the rod is initially unstressed, detennine the compressive force induced in
ABC when there is a temperature rise of 50°C.
250 mm
t-
0
B
50-mm dia1T1eter
300mm
L
C
SOLUTION
1!
2
7!
2
7!
2
7!
2
2
-62
AA8 = - d AB = - (30) = 706.86 mm = 706.86 x 10 m
4
4
3
2
-3
A8 c=-d 8 c =-(50) = 1.9635x 10 mm = 1.963 5 x 10 m
4
2
4
Free the rmal expansion:
br
p
= LA8 a .,(t1T) + L8 cah (t'iT)
= (0.250)(11.7 x I 0-6 )(50) + (0.300)(20.9 x 10- 6 )(50)
A
= 459 .75x l 0-6 m
Shortening due to induced compressive force P:
!3
5P = ~ + ~
E., AAB EbAac
0.250P
0.300P
=---~-----~
+---~
----~6
9
(200 x 1o )(706.86 x 1o9
)
(105 x 10 )(1.9635 x 10- 3 )
C
= 3.2235 x 10- 9 P
For zero net deflection, bp = br
3.2235 X J0- 9 p = 459.75 X 10- 6
3
P = 142.624 x 10 N
? = 142.6 kN ◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
149
r
-
24 in
A
7~32 ~1in.
B
C
J
- (
2¼-in. diameter
\ -
l½-in. diairieter
PROBLEM 2.52
A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel (£, = 29 x 106 psi, a , = 6.5 x 10- 6 /°F) and
portion BC is made of aluminum (Ea= 10.4 x 106 psi; a 0 = 13.3 x l0- 6 /°F).
Knowing that the rod is initially unstressed, determine (a) the normal stresses
induced in portions AB and BC by a temperature rise of 70°F, (b) the
corresponding deflection of point B .
SOLUTION
2
AAB = ~(2.25) = 3.976 1 in
2
4
ABc = ~(1.5)2 = 1.76715 in
2
4
? ~
Free thermal expansion.
(t5r) AB = LABa , (t,,.T) = (24)(6.5 x 10- 6 )(70) = 10.92 x 10- 3 in.
(<Sr ) 8 c = L 8 c a a(i',.T) = (32)(13.3 x 10- 6 )(70) = 29.792 x 10- 3 in.
3
Total:
br = (br )AB + (br )Bc = 40.7 12 x I 0- in.
Shortening due to induced compressive force P.
(6)
P
(8)
AB
= PLAB =
24P
= 208.1 4 x l0-9P
E, AAB (29x l 0 6 )(3.976 1)
32P
- PLBc Ea A8 c - (I 0.4 x I 0 6 )(1.76715)
174 1.18x l 0-9 P
P BC -
9
Total:
6p = (bp)AB + (bp)ac = 1949.32x 10- P
For zero net deflection, bp = br
(a)
(b)
1949.32 X10- 9 p
P
20.885 x I 03
~AB = - - - =- - - - - AAB
3.976 1
- 5.25 X103 psi
p
20.885 XI 0 3
~Be = - - - = - - - - - ABc
1.767 15
- I L82x l 0 3 psi
(bp )AB
=40.7 12 X10- 3
3
P =20.885 x 10 lb
~AB = - 5.25 ksi ◄
~Be =-I 1.82 ksi ◄
=(208.14 x 10- 9)(20.885 x I 03 ) =4.3470 x I 0- 3 in.
t5B = (br) AB ➔ +(bp)AB ~ = 10.92x l 0- ➔ +4.3470 x 10- ~
3
3
t5B = 6.57 x 10- in. ➔ ◄
3
or
(bp he =(174 1.1 8 x 10-9 )(20.885 x 10 3 )
=36.365 x 10- 3 in.
3
3
bB =(brhc ~+ (bp)ac ➔ = 29.792x10- ~+ 36.365 x l0- ➔ = 6.57x 10- in. ➔
3
(checks)
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
150
PROBLEM 2.53
1-24 in
-A
7-32
in.
Solve Prob. 2.52, assuming that portion AB of the composite rod is made of
aluminum and portion BC is made of steel.
-1
B
C
-
I
- (
2¼-in. diameter
\
I½-in. diameter
PROBLEM 2.52 A rod consisting of two cylindrical portions AB and BC is
restrained at both ends. Portion AB is made of steel (£, = 29 x 106 psi,
a, = 6.5 x 10-6/°F) and portion BC is made of aluminum (£ ~ 10.4 x 106 psi,
a 0 = 13.3 x 10- 6 /°F). Knowing that the rod is initially unstressed, determine
(a) the nonnal stresses induced in portions AB and BC by a temperature rise of
70°F, (b) the corresponding defl ection of point B.
0
SOLUTION
AAB = ~(2 .25)2 = 3.9761 in
4
2
Free thermal expansion.
ABc = ~(1.5)2 = 1.76715 in
4
2
t1T = 70°F
6
3
(Or )AB = LABa 0 (t1T) = (24)(13.3 x 10- )(70) = 22.344 x 10- in.
(Or h e = LBca, (t:..T) = (32)(6.5 x 10-6 )(70) = 14.56x 10- 3 in.
3
Total:
Or = (Or)AB +(Orhc = 36.904xl0- in.
Shortening due to induced compressive force P.
= PLAB =
(o )
P
AB
= PLBc =
(o)
P
BC
E, ABc
24P
(10.4 x 106 )(3.9761)
= 580.39 x 10-9P
32
p
= 624.42x l0-9 P
(29 x l06 )(1.767 15)
9
Op = (Op)AB + (op)Bc = 1204.8 1x 10- P
Total:
For zero net deflection,
(a)
E0 AAB
p
a AB =---=
AAB
1204.8 t x 10- 9P = 36.904 x 10- 3
oP = Or
P = 30.63 l x l03 lb
3
30.63 l xl0 =-7 _7 0 x l03 si
3.9761
p
a AB= - 7.70 ksi ◄
3
P
30.63 1x I 0
aBc = - - - = - - - - -
A8 c
1.767 15
3
- 17.33 x l0 psi
a 8 c =-17.33 ksi
◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
15 1
PROBLEM 2.53 (Continued)
(b)
(bp)AB = (580.39 x 10- 9 )(30.631 x 10
3
)
= l 7 .7779x 10- 3 in.
bs = (br )As ➔ + (8p)A 8 ~ = 22.344 x 10- ➔ + 17 .7779 x 10- ~
3
or
9
(bp he = (624.42 X10- )(30.63 1X10
3
)
3
= 19.1 266 X10- 3 in.
3
3
bs =(brhc ~ + (bp)sc ➔ = 14.56x10- ~+ 19.1 266x l 0- ➔ =4.57x10- in. ➔
3
(checks)
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
152
PROBLEM 2.54
The steel rails of a railroad track (£ , = 200 GPa, o.., = 11. 7 x I OT 0 /°C) were laid at a temperature of 6°C.
Determine the normal stress in the rails w hen the temperature reaches 48°C, assuming that the rails (a) are
welded to form a continuous track, (b) are 10 m long with 3-mm gaps between them.
SOLUTION
(a)
Or = a(L'.T) L = (11.7 x I o-6 )(48 - 6)(10) = 4.9 14 x I 0- 3 m
0 = Or+ Op = 4.9 14 x 10- 3 + 50x 10- 12 a = 0
a = - 98.3 x 106 Pa
(b)
a =-98.3MPa
0 = Or+ Op = 4.9 14 X , 0- 3 + 50x 10- 12 a = 3 X , 0- 3
3x l0-3 - 4.914 x l0- 3
a = - - - - -12- - -
50x 10-
= - 38.3x 106 Pa
a =-38.3 MPa
PROPRI ETARY MA TERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
153
PROBLEM 2.55
Two steel bars ( Es == 200 GPa and a s == 11.7 x I~
0-6/°C) are used to
reinforce a brass bar (Eb == 105 GPa, ab== 20.9 x 10 / °C) that is subjected
to a load P == 25 kN. When the steel bars were fabricated, the distance
between the centers of the holes that were to fit on the pins was made
0.5 mm smaller than the 2 m needed. The steel bars were then placed in
an oven to increase their length so that they would just fit on the pins.
Following fabrication, the temperature in the steel bars dropped back to
room temperature. Determine (a) the increase in temperature that was
required to fit the steel bars on the pins, (b) the stress in the brass bar
after the load is applied to it.
SOLUTION
(a)
Required temperature change for fabrication:
5r == 0 .5 1nm == 0.5 XI 0- 3 m
Temperature change required to expand steel bar by this amount:
57 == La,6 T, 0.5 x 10- 3 == (2.00)(11.7 x 10-6 )(1:-,.T),
1:-,.T == 0.5 x 10- 3 == (2)(11 .7 x 10-6 )(!:-,.T)
21.4 °C ◄
6T == 2l.368 °C
(b)
Once assembled, a tensile force p * develops in the steel, and a compressive force p * develops in the
brass, in order to elongate the steel and contract the brass.
6
A, == (2)(5)(40) == 400 mm 2 == 400 x I0- m
Elongation of steel:
P\2.00)
(5 ) _ p * L _
6
9
P ' - A, E, - (400 x l0- )(200x!0 )
2
25 x 10-9p *
Contraction of brass: A; == (40)(15) == 600 mm 2 == 600x 10- 6 m 2
P\ 2.00)
(5) - p *L 9
P b - AbEb - (600 x 10-6 )(105 x 10 )
31.746 x !0- 9P*
But (5p ), + (5p)h is equal to the initial amount of misfit:
(5p), + (5p)h == o.5 x 10-3 , 56.746 x 10- 9 p * == o.s x 10- 3
p * == 8.8 112x l0 3 N
Stresses due to fabrication:
Steel:
3
8 8 112
x I0
400 x 10-6
ci == p * == ·
'
A_,
22.028 x 10 6 Pa == 22.028 MPa
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
154
PROBLEM 2.55 (Continued)
Brass:
a~ = _!:._ = - 8 -8 112 x
Ab
103
= - 14 .6853 x 10 6 Pa = - 14.685 MPa
600x l0-6
To these stresses must be added the stresses due to the 25-kN load.
For the added load, the additional defommtion is the sam e for both the steel and the brass. Let <5' be the
additional displacement. Also, let Ps and Ph be the additional forces developed in the steel and brass,
respectively .
<5' = P,L
As£ ,
=
Pi,L
Ah£h
9
p = As£ , <5' = (400 XI 0-6 )(200 XI 0 ) <5' = 40 XI 06<5'
s
L
2.00
6
9
Pb = AbEh <5' = (600x 10- )(105 x 10 ) <5' = 3 1. 5 x 10 6<5'
L
2.00
Total:
40 X10 6t5' + 31.5 XI 0 6<5' = 25 X103
6
b' = 349.65 X 10- 6111
6
3
P, = ( 40 x 10 )(349.65 x 10- ) = 13.9860 x 10 N
Ph= (3 1.5 X106)(349.65 X10- 6) = I 1.0140 X103N
p
As
a = .....L
s
13.9860 X 103
----400
X
10-6
ll.0140x l 0 3
ab = Ab = 600 x 10- 6
Ph
34.965 x 106 Pa
18.3566 x 106Pa
Add stress due to fabrication.
Total stresses:
a s - 34.965 x 10 6 + 22.028 x 10 6 - 56. 99 1 x 10 6 Pa
6
6
6
a h = I 8.3566 x I 0 - 14.6853 x I 0 = 3.67 13 x I 0 Pa
a s - 57.0 MPa
a h= 3.67MPa ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
155
PROBLEM 2.56
15mm
1
Determine the maximum load P that may be applied to the brass bar of
Prob. 2.55 if the allowable stress in the steel bars is 30 MPa and the
allowable stress in the brass bar is 25 MPa.
I
mm
PROBLEM 2.55 Two steel bars ( £ , = 200 GPa and a .,= 11.7 x 10~/0 C)
are used to reinforce a brass bar (Eb = 105 GPa, a b= 20.9 x 10 / 0 C)
that is subjected to a load P = 25 kN. When the steel bars were fabricated,
the distance between the centers of the holes that were to fit on the pins
was made 0.5 mm smaller than the 2 m needed. The steel bars were then
placed in an oven to increase their length so that they would just fit on the
pins. Following fabrication, the temperature in the steel bars dropped back
to room temperature. Detennine (a) the increase in temperature that was
required to fit the steel bars on the pins, (b) the stress in the brass bar after
the load is applied to it.
SOLUTION
See solution to Problem 2.55 to obtain the fabrication stresses.
a-; = 22.028 MPa
a-;=14.6853 MPa
Allowable stresses:
O"s,all
= 30 MPa,
O"b,all
= 25 MPa
Available stress increase from load.
a-., = 30 - 22.028 = 7.9720 MPa
o-b = 25+ 14.6853 = 39.685MPa
Corresponding available strains.
7.9720 X 106
200x 10
(Tb
6b = £b
Smaller value governs :.
6
9
39.860x ! 0-6
6
39.685 X JO = 377 _95
9
105 X 10
X
\ 0- 6
=39.860 x 10- 6
Areas: A., = (2)(5)(40) = 400 mm 2 = 400 x 10- 6 m 2
Ab= (15)(40) = 600 mm 2 = 600x 10- 6m 2
6
3
Forces P, = E.,A_,c = (200 x 109 )(400 x 10- )(39.860 x 10-6 ) = 3. 1888 x 10 N
Pb= EbAbc= (105 x 109 )(600 x 10-6 )(39.860 x 10- 6) = 2.5 112 x 10-3 N
Total a llowable additional force:
P
= P., + f1 = 3.1 888x 103 + 2.5112 x 10 3 = 5.70 x 10 3 N
P = 5.70 kN ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
156
PROBLEM 2.57
Dimensio ns in mm
0.15
I
An aluminum rod (£0 = 70 GPa, a0 = 23 .6 x I 0-6/0 C) and a steel link
6
(Es x 200 GPa, a0 = 11.7 x 10- / 0 C) have the dimensions shown at a
temperature of 20°C. The steel link is heated until the a luminum rod
can be fitted freely into the link. The temperature of the whole
assembly is then raised to 150°C. Determine the final normal stress
(a) in the rod, (b) in the link.
200
20
30
20
7
A
A
I
I
I
I
'() Section A-A
SOLUTION
tiT = r1 -
r;
= 1so0 c - 20°c = 130°c
Unrestrained thermal expansion of each part:
Aluminum rod:
(.5rt, = Laa(l'lT)
(br )a = (0.200 111)(23.6 X 10- 6 /°C)( 130°C)
= 6.1360x10-4 m
Steel link:
( br )s = Las (l'lT)
(br), = (0.200 m)(l 1.7 x I o- 6 / C)( l 30°C)
0
= 3.0420 x I0-4 111
Let P be the compressive force developed in the aluminum rod. It is a lso the tensile force in the steel link.
A luminum rod:
P(0.200 m )
(70 x 109 Pa)(n/ 4)(0.03 111)2
= 4 .0420xl0- 9 P
Steel link:
(
,5 ) _
PL
P s -
EA
s
s
P(0.200)
9
(200 x 10 Pa)(2)(0 .02 111)2
= 1.250 X I0- 9 p
Setting the total deformed lengths in the link and rod equal gives
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
157
PROBLEM 2.57 (Continued)
(bp), + (bp)a
J.25
X
= 0.15 x 10- 3 + (br),, -
10- 9 p + 4.0420 X 10- 9 p = 0.15
X
10- 3 + 6.1 360 X 10- 4
p = 8 .68 10
(a)
(br)s
X
-
3.0420
X
10--4
4
10 N
p
Stress in rod:
CJ = -
A
- 8.6810x l 04 N
8
a R= - - - - - 2 - l.228 1 l x l 0 Pa
(n-/4)(0.030 111)
aR = - 122.8 MPa
(b)
◄
Stress in link:
a = 8 .6810 x 104N = 1.085 13 x 108Pa
L
(2)(0.020 111)2
a l=
108.5 MPa ◄
PROPRI ETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
158
0.02 in.
-1r-
14 in. - , - 1 8 in.
li
I
l
Bronze
A= 2.4 in2
6
E = 15 X l0 psi
a = 12 X l0- 6/ °F
PROBLEM 2.58
-1
Aluminum
A= 2.8in 2
E = 10.6 X 106 psi
a= 12.9 X 10-6/°F
Knowing that a 0.02-in. gap exists when the temperature is 75°F,
determine (a) the temperature at which the nonnal stress in the
aluminum bar will be equal to - 11 ksi, (b) the corresponding exact
length of the aluminum bar.
SOLUTION
a O = - 1 I ks i = - 11 x 103 psi
3
3
P =-a 0 A0 = (ll x l 0 )(2.8) = 30.8x l 0 lb
Shortening due to P:
Op= Plb + Pl0
EbAb
EaAa
(30.8 x 103 )(14) (30.8 x I 03 )(18)
= - - - -6 - ~ + - - - -6 ~ (15x 10 )(2.4) (10.6 x 10 )(2.8)
= 30.657 X 10- 3 in.
Available elongation for thermal expansion:
Or = 0.02 + 30.657 x I 0- 3 = 50.657 x I 0- 3 in.
But Or
= Lbab (t...T) + Laaa (t...T)
= (14)(12 x 10- 6 )(/'iT) + (1 8)(12.9 x 10- 6 )(/'iT) = (400.2 x 10- 6 )/'iT
Equating, ( 400.2 x 10- 6 )/'iT = 50.657 x 10- 3
(a)
(b)
t...T = 126.6°F
Thot = 20 1.6°F ◄
Toot = Tcold + 1"1T = 75 + 126.6 = 20 1.6°F
s:
PLO
u 0 = L 0 a 0 (!'1T) - - -
EaAa
3
30
= ([ 8)(1 2.9 X 10- 6 )(26.6) - ( ·8 X lQ )(1 3)
6
(I 0.6 x I 0 )(2.8)
[0.7 [2 X 10- 3 in.
Le,act = 18 + 10.712 x I 0- 3 = 18.0107 in.
L = 18.0107 in.
◄
PROPRI ETARY MA TERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
159
-11-14 - 1-18
0 .02 in.
in.
li
in.
.1
Bronze
A = 2.4 in 2
E = 15 X l06 psi
a = 12 X l0- 6/°F
PROBLEM 2.59
- 1
r
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.
Aluminum
A = 2.8 in 2
E
a
= 10.6 X 106 psi
= 12.9 X 10- 6/ °F
SOLUTION
Thermal expansion if free of constraint:
Or = Lba b(t1T) + Laaa(t1T)
= (14)(12 x 10- 6 )(180) + (18)(12.9 x 10-6 )(180)
= 72.036 X 10- 3 in.
Constrained expansion: o = 0.02 in.
Shortening due to induced compressive force P:
p
~ . _ _ __ _-1-.,..__ __ _ _ _ J ~
Op = 72.036 x 10- 3 - 0.02 = 52.036 x 10- 3 in.
o -
But
p -
PLb
PLa -( Lb
La JP
EbAb + £ 0 A0 - EbAb + £" A0
=(
Equating,
14
18
9
+
JP = 995.36 x l0- P
6
(15xl0 )(2.4) (10.6xl06 )(2.8)
995.36 X 10-9 p = 52.036 X 10- 3
3
P = 52.279 x 10 lb
P = 52.3 kips ◄
(a)
(b)
3
= (14)(12x l0-6 )(180) - (52 ·279 x l0 )(l 4 ) = 9.91 x l0- 3 in.
(15 X 106 )(2.4)
Ob = 9.9 1 X 10- 3 in. ◄
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
160
PROBLEM 2.60
At room temperature (20°C) a 0.5-mm gap exists between the ends of
the rods shown. At a later time when the temperature has reached
140°C, detennine (a) the normal stress in the aluminum rod, (b) the
change in length of the aluminum rod.
Aluminum
= 2000 Ill 1112
E = 75CPa
a = 23 X 10- 6/"C
A
Stainless steel
A= 800 mni
E
a
= 190CPa
= 17.3 X J0-6/°C
SOLUTION
t'lT = 140 - 20 = 120°C
Free thermal expansion:
6r = L0 a 0 (t'lT) + Lp,(t'lT)
= (0.300)(23 X 10- 6 )(120) + (0.250)(17.3 X 10- 6 )(120)
=J.347 XJ0-3 111
Shortening due to P to meet constraint:
6p =J.347 X10- 3 - 0.5 X10-3 =0.847X10- 3 111
6P
= PL"
£ A
0
(
0
+PL, j __!:__,,_+~JP
£ ,A, l £ A
£ , A,
0
0
0.300
0.250
=l (75 x 109 )(2000 x 10- 6 ) + (190 x 109 )(800 x 10-6 )
=3.6447 X10- 9 p
Jp
3.6447 X 10- 9 P = 0 .847 X 10- 3
Equating,
P = 232.39 x !03 N
(a)
(J
a
P
=--=
A"
232.39 x 103
2000 X10- 6
- 116.2 x I 0 6 Pa
a"
=- 116.2 MPa
◄
=0.363 111111
◄
(b)
3
= (0.300)(23 XJ0-6 )(J20) - <232 -39 XJ0 )(0.300)
9
(75 X10 )(2000 XJ0-6 )
363X J0- 6 111
6
0
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
16 1
PROBLEM 2.61
r
1
e-
/¾ in.diameter
A standard tension test is used to determine the properties of an experimental plastic.
The test specimen is a ¾-in.-diameter rod and it is subjected to an 800-lb tensile
force. Knowing that an elongation of 0.45 in. and a decrease in diameter of 0.025 in.
are observed in a 5-in. gage length, determine the modulus of elasticity, the modulus
of rigidity, and Poisson's ratio for the material.
,....
P'
SOLUTION
2
!!_(~) = 0.306796 in
A = !!_d 2 =
4
4 8
2
P = 800 lb
P
800
.
aY = - = - - - = 2.6076 x l03 psi
A
=
B
y
B
x
E=
6
Y
By
V
=-
6
Y
L
0.306796
5
= 0.4 = 0.090
5.0
= <5, = - 0 -025 = - 0.040
d
0.625
3
2 6076
= ·
xl0 = 28.973xl03 psi
0.090
Bx
By
£ = 29.0xl0 psi ◄
3
= - 0.040 = 0.44444
0.090
V
= 0.444 ◄
3
28 973
a =-E- =
·
xl0 = 10. 029 l x l03 si
2(1 + v) (2)(1+ 0.44444)
p
a = l 0.03 x 103 psi ◄
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
162
PROBLEM 2.62
A 2-m length of an aluminum pipe of 240-mm outer diameter and 10-mm
wall thickness is used as a short column to carry a 640-kN centric axial load.
Knowing that E = 73 GPa and v = 0.33, determine (a) the change in length of
the pipe, (b) the change in its outer diameter, (c) the change in its wall
thickness.
SOLUTION
d0 = 0.240
r = 0.010
L = 2 .0
3
d; = d 0 - 2( = 0.240 - 2(0.0 10) = 0.220 111 P = 640 X 10 N
A=~( d; - d;2 ) = ~(0.240 4
(a)
4
0.220) = 7.2257 x 10- 3 m2
(640 X 103 )(2.0)
<5 =- PL =
EA
= - 2.4267 X 10- 3 m
<5=-2.43mm ◄
c = §___ = - 2.4267 =- l.2 1335x 10- 3
L
2.0
Cu r = - v& = - (0.33)(- 1.21335x l 0-
3
)
= 4.0041 X 10- 4
(b)
M o = docLAT = (240 mm)(4.004 l x 10-4) = 9.6098 x 10-2 mm
0
◄
/';.f
= 0.00400 mm ◄
M = 0.0961 mm
/';.( = (&LAT
= (10 mm)(4.004 l x 10-4 ) = 4.0041 x 10- 3 mm
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
163
200 k:-1
1
150 mm
j
~ 200
n11Tl
~
PROBLEM 2.63
A line of slope 4: IO has been scribed on a cold-rolled yellow-brass
plate, 150 mm wide and 6 mm thick. Knowing that E = 105 GPa and
v = 0.34, detennine the slope of the line when the plate is subjected to
a 200-kN centric axial load as shown.
SOLUTION
A = (0.150)(0.006) = 0.9x 10- 3 m 2
3
a = !!.._ = 200 x l O = 222.22x l06 Pa
A
x
0.9x 10- 3
6
6
x
liy
O'x
222.22 XI 0
3
= - = - - - -9 - 2.1164x l0E
105x l 0
=-
Ut:x
= - (0 .34)(2.1164 X10- 3 )
=-0.7 1958x l 0- 3
tan 0 = 4( 1+ t:y )
lO(l + t:, )
~ 4 ( 1 + E1)
4(1- 0.71958x 10- 3 )
I 0(1 + 2.1164 x 10- 3 )
/O(J+E.-,.)
= 0.39887
tan 0 = 0.399 ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
164
PROBLEM 2.64
A 2.75-kN tensile load is
applied to a test coupon
made from 1.6-mm flat
steel plate (E = 200 GPa,
v = 0.30). Determine the
resulting change (a) in the
50-mm gage length, (b) in
the w idth of portion AB
of the test coupon, (c) in
the thickness of portion
AB, (d) in the crosssectional area of portion
12mm
AB.
SOLUTION
A = (1.6)(12) = 19.20 mm 2
= 19.20x 10- 6 1112
3
P = 2.75xl0 N
3
P
2.75xl0
a = - = ------,-'
A 19.20x 10--6
= 143.229x 106 Pa
a,
Ii = x
E
(a)
143.229x l06 = 716.15xl0--6
200 X 109
L = 0.050111
0.0358 mm ◄
(b)
w= 0.012m
- 0.00258 nun ◄
(c)
t = 0.0016 m
6, = f&, = (0.0016)(- 214.84 X 10-
6
)
9
= - 343.74 X J0- m
- 0.000344 mm ◄
(d)
~A = A - Ao = w0 t 0 (cy +&,+ negligible term) = 2Wofoliy
= (2)(0.0 12)(0.00 16)(- 214.84 X 10--6 ) = -8.25 X 10-9 m 2
- 0.00825 mm ◄
2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
165
PROBLEM 2.65
/ 22-mm diameter
i5 k:-1
(
i5 kN
....,.__,!-;;;;;:;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;,,_-j..►
~
200111111-I
In a standard tensile test, a steel rod of22-mm diameter is subjected to a
tension force of 75 kN. Knowing that v = 0.3 and E = 200 GPa,
determine (a) the elongation of the rod in a 200-mm gage length, (b) the
change in diameter of the rod.
SOLUTION
P
= 75 kN = 75 x 103 N
A= - d
Jr2
= -Jr (0.022) 2 = 380.13 x 10-6 m 2
a = !_ =
75 103
x
380.13x l 0-6
= 197.301 x 106 Pa
\ 97.30 1 X 106
200 x 109
= 986.5 1 X 10-6
4
A
lix
0"
E
<5,
4
= Lcx = (200 mm)(986.5 I x 10-6)
liy
= - Vlix = - (0.3)(986.5 1 X
<5Y
= dcy = (22 mm)(- 295.95 x 10-6)
10- 6 ) = - 295.95
X
(b)
= 0.1973 mm
◄
= - 0.0065 1 mm
◄
<5,
(a)
10-6
<5Y
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
166
PROBLEM 2.66
The change in diameter of a large steel bolt is carefully measured as the
nut is tightened. Knowing that E = 29 x 106 psi and v = 0.30, determine
the internal force in the bolt if the diameter is observed to decrease by
0.5 x 10- 3 in.
SOLUTION
oy = - o.5 x 10By
= By =
3
0.5
d
in.
d
X 10-3
2.5
= - 0.2 X 10- 3
- By
V
O"x
= 2.5 in.
0 ·2
X
10 -
3
= 0.66667 X 10- 3
0.3
= frx = (29 X 106 )(0.66667 X 10- 3 ) = 19.3334 X 103 psi
A
= !!_d 2 = !!__(2.5) 2 = 4.9087 in 2
F
= a-,A = (19.3334 x 103)(4 .9087) = 94.902 x 103 lb
4
4
F = 94 .9 kips ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
167
A (:.1!1- - - - - - - - - .
B
-
-,
600mm
PROBLEM 2.67
The brass rod AD is fitted with a jacket that is used to apply a
hydrostatic pressure of 48 MPa to the 240-mm portion BC of the rod.
Knowing that E = 105 GPa and v = 0.33, determine (a) the change in
the total length AD, (b) the change in diameter at the middle of the rod.
240 mm
C
_
_
l
D ......., - - - - - - ~
-I
1somm
SOLUTION
a x= a ,= - p
= -48 x 106 Pa,
CYy
=0
1
lix
= £ (O"x -
VO"y - UO"J
1
[-48 x 106 - (0.33)(0) - (0.33)(-48 x l 06 ) ]
105 x 109
= 306.29 x I o-6
=
1
€,,
=£
( - VO"x - O"y - UO"J
l
[- (0.33)(-48 x 106) + 0 - (0.33)(-48 x 106 ))
105 x l 09
= 301.7[ X J0-6
=
(a)
Change in length: only portion BC is strained. L = 240 mm
S,, = fr,,= (240)(- 301.7 1x 10- 6 ) = - 0.0724 mm
(b)
◄
Change in diameter: d = 50 mm
bx = b, = dlix = (50)(-306.29x10-6 ) = - 0.0 153 1mm
◄
PROPRI ETA RY MATERIAL. C opyright © 20 15 McGraw- Hill Education . This is proprietary material sole ly for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
168
PROBLEM 2.68
A fabric used in air-inflated structures is subjected to a biaxial
loading that results in normal stresses ax = 18 ksi and a, = 24 ksi.
Knowing that the properties of the fabric can be approximated as
6
E = 12.6 x 10 psi and v = 0.34, determine the change in length of
(a) side AB, (b) side BC, (c) diagonal AC.
SOLUTION
a x = 18 ksi
a , = 24 ks i
lix= EI (a, - vay - vaJ =
c,=
(a)
[ 18, 000 - (0.34)(24, 000) ] = 780.95 x 10 -6
_ Ix
12 6 106
~ (- vax - vay +a,) =
_~
[- (0.34)(18,000) + 24, 000] = 1.41905 x 1012 6 106
3
<5AB = (AB)iix = (4 in.)(780.95 x 10- 6 ) = 0.003 1238 in.
0.003 12 in . ◄
(b)
<58 c = (BC)ii, = (3 in.)( 1.4 1905 x Io-3 ) = 0.0042572 in .
0.00426 in. ◄
Label s ides of right triangle ABC as a, b, c.
c2 = a 2 + b 2
Then
Obtain differentials by calculus.
2cdc = 2ada + 2bdb
a
b
dc = - da + - db
C
But a = 4 in.
b = 3in.
da = <5A 8 = 0.003 1238 in.
(c)
C
c= ✓4 2 + 3 2 = Sin.
db= <58c = 0.0042572 in.
4
3
<5Ac = de = -(0.0031238) + -(0.0042572)
5
5
0.00505 in. ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
169
PROBLEM 2.69
I
I
I in.
<Tx
A I-in. square was scribed on the side of a large steel pressure vessel.
After pressurization the biaxial stress condition at the square is as
shown. Knowing that E = 29 x 106 psi and v = 0.30, detennine the
change in length of (a) side AB, (b) side BC, (e) diagonal AC.
= 12 ksi
-lin.-
SOLUTION
c , = _.!_(o-v- uoy ) =
.
E
'
I
[ 12x !03 - (0.30)(6x l03 )]
29x 106
=35 J.72 X J0-6
liy = _.!_(o-y - uox)=
E
I
[ 6 x l03 - (0.30)(12 x l03 ) ]
29x 106
= 82.759x l0- 6
(a)
OAB = (AB) 0 cx = (l.00)(351.72 x l0-6 ) = 352x l 0-6 in.
(b)
0 8 c = (BC)0 cy = (l.00)(82.759 x !0- 6 )=82.8x l0-6 in.
(e)
2
◄
◄
(AC) = ✓(AB) + (BC) = ✓(AB0 +ox) + (BC0 +oy )
2
2
2
=.Jc1 + 35 1.12 x 10- 6 ) 2 + c1 + 82.159 x 10-6 )2
=1.41452
( AC)0 =✓
2
AC - (AC) 0 =307 x 10- 6
◄
or use calculus as follows:
Label sides using a, b, and e as shown.
e2
= a2 + b2
Obtain differentials.
from which
2ede = Zada+ 2bde
a
I
de = -da
+ -b te
e
e
But a = I 00 in., b = 1.00 in., e = ✓
2 in.
da = oA8 = 35 1.72 x 10- 6 in., db = o 8 c = 82.8 x I o-6 in .
1
1
OAc = de= - ; (351.7 x 10- 6 )+ - ; (82.8 x 10- 6 )
=307 X J0- 6 in.
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
170
PROBLEM 2.70
25110~-
The block shown is made of a magnesium alloy, for which
£ = 45 GPaand v=0.35. Knowing that O"x=-180 MPa,
detennine (a) the magnitude of o-Y for which the change in
the height of the block will be zero, (b) the corresponding
change in the area of the face ABCD, (c) the corresponding
change in the volume of the block.
40 mnl____
z
100 mm
SOLUTION
a-,= 0
(a)
8y
= -EI (0"
X
- VO"y - VO"z )
O"y = VO"x = (0.35)(- ISOx 10 6 )
o-Y =-63.0 MPa
◄
I
v
(0.35)(- 243x l 0 6 )
_3
8,= - (0",- VO"x- VO"y) =- - (O"v+ O"y ) =9
= + l.890x l 0
. E E ,.
45 x 10
157.95 X 10 6
45 X 109
(b)
A0
- 3.5 10x 10- 3
= LxL,
A = Lx(I + t;x)L, (l + &, ) = LxL, (1+ t;x + 8, + &xii,)
AA = A - Ao= LxL, (&., + &, + lix&,)"" LxL, (&x + &,)
A A = (100 111111)(25 mm)(- 3.510 x I 0- 3 + 1.890 x 10-3 )
(c)
AA = -4.05 111111 ◄
2
V0 = LxLyL,
V
= Lx(l + lix)Ly(l + liy )L, (I + &,)
= ~~4(1+ ~+~+~+~~+~~+~~+~~~)
AV = V - V0 = LxLyL, (&x + &Y +&, + small tem1s)
AV = (100)(40)(25)(- 3.5 1Ox 10- 3 + 0 + 1.890 x 10- 3 )
AV =-162.0 1111113 ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
17 1
PROBLEM 2.71
I/
The homogeneous plate ABCD is subjected to a biaxial loading as
shown. It is known that a, = a 0 and that the change in length of
the plate in the x direction must be zero, that is, ex = 0. Denoting
by Ethe modulus of e lasticity and by v Poisson 's ratio, determine
(a) the required magnitude of a x, (b) the ratio o-0 / c, .
SOLUTION
(a)
(b)
1
I
2
c -_ = -E (-. va,.. - vay + a_)
- = -E (.-v a 0 - 0+ a 0 .)
2
1- v
=E-a 0
E
l -v2 ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
172
PROBLEM 2.72
P'
-x
(a )
<T' =
For a member under axial loading, express the normal strain c' in a
direction fom1ing an angle of 45° with the ax is of the load in terms of the
axial strain c, by (a) comparing the hy potenuses of the triang les shown in
Fig. 2.43, which represent, respectively, an element before and after
deformation, (b) using the values of the corresponding stresses of o-'and
a_, shown in Fig. 1.38, and the generalized Hooke's law.
.!:...
2.A
(b)
SOLUTION
•-~0
(/,)
(a)
(c)
Figure 2.49
(a)
[✓
2 (1 +&')]2 = (l +cx)2+ (1 - vcx)2
= 1+ 2cx + c_; + I - 2 vex + v2c_;
4c' + 2c' 2 = 2cx + c_; - 2vcx + v2 c_;
2(1 + 2c' + c'
2
)
Neglect squares as sma ll.
li
I
1- V
= ~ ~ [i
2
◄
X
1-sE,ts:
\ +E~
After defo~rn,a,+io"'
(A)
(B)
PROPRIETARY MA TERIAL. C opyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
173
PROBLEM 2.72 (Continued)
(b)
1o'
a'
va'
E
E
1-v P
- - ·£ 2A
1-v
2£
=--a
X
1-v
◄
= - -£
2
X
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
174
PROBLEM 2.73
<I,
In many situations, it is known that the normal stress in a given direction is zero.
For example, a 2 = 0 in the case of the thin plate shown. For this case, which is
known as plane stress, show that if the strains e., and .s;, have been determined
experimentally, we can express ax, a y, and e, as follows:
SOLUTION
a, = 0
I
ex= E(a x-va y)
(I )
(2)
Multiplying (2) by v and adding to( !),
ex+ vey
l-v2
=~
ax
E
a ,= - (e, + vey )
· 1-v2 ·
or
◄
Multiplying ( I) by v and adding to (2),
l-v2
ey+ vex = ~ ay
◄
or
◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
175
PROBLEM 2.74
In many situations, physical constraints
prevent strain from occurring in a given
direction. For example, c, = 0 in the case
shown, where longitudinal movement of
the long prism is prevented at every point.
Plane sections perpendicular to the
longitudinal axis remain plane and the
same distance apart. Show that for this
situation, which is known as plane strain,
we can express a , , ex, and cy as follows:
<J",
= v(a_,. + a y )
ii
= -[(l
- v2 )aX - v(l + v)ay ]
E
X
1
SOLUTION
◄
1
ex= E(ax -va y - va , )
1
= E[ax- vay- v
1
= £[(1-
2
(ax +a y)]
2
v )ax - v(l +v)O"y ]
1
liy = E (- vax + a 1
1
-
◄
va,)
2
= E[- vax+ a y- v (ax +ay)]
I
2
= E[(l - v )ay- v(l +v)a x]
◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
176
PROBLEM 2.75
The plastic block shown is bonded to a rigid support and to a vertical plate to
which a 55-kip load P is applied. Knowing that for the plastic used G = 150 ksi,
determine the deflection of the plate.
SOLUTION
A = (3 .2)(4.8)
= 15.36in 2
P = 55 x 103 lb
p
A
55 X 103
15.36
r
=- =
G
= 150 x 103 psi
y
= !._ =
G
3580 7
·
150 x l03
= 3580.7 psi
= 23.87 1 X 10- 3
h
= 2 in.
5
= hy = (2)(23.87 1 x 10- 3 )
= 47.7 X 10- 3 in.
5
= 0.0477 in . .J,
◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
177
PROBLEM 2.76
3.2i
What load P should be applied to the plate of Prob. 2.75 to produce a k-in.
deflection?
(1
PROBLEM 2.75 The plastic block shown is bonded to a rigid support and to a
vertical plate to which a 55-kip load P is applied. Knowing that for the plastic
used G = 150 ksi, determine the deflection of the plate.
4.8 i
2 in.
SOLUTION
s:
u
.
= - 1 m.
= 0.0625 .m.
16
h = 2 in.
r = §_ = o.0625 = o.o3 125
h
2
G
= 150 x I03 psi
r
= Gr = (150 x I03 )(0 .03 125)
= 4687.5 psi
A = (3 .2)(4.8)
= 15.36 in 2
P = TA = (4687.5)(15.36)
= 72.0 X 103 lb
72.0 kips ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
178
PROBLEM 2.77
Two blocks of rubber w ith a modulus of rigidity G = 12 MPa are
bonded to rigid supports and to a plate AB. Know ing that c = 100 mm
and P = 45 kN, delerrnim: the s mallest allowable dimensions a and b of
the blocks if the shearing stress in the rubber is not to exceed 1.4 MPa
and the deflection of the plate is to be at least 5 mm.
SOLUTION
\~\l~
--½?
--\:P
.___I_Ill°'
{P...--
Shearing strain:
5
T
a
G
r= -=6
a= G5 = (12 x I 0 Pa)(0 .005 m) = 0 _0429 111
l.4 x l 0 6 Pa
T
Shearing stress:
1- p
p
A
2bc
a= 42.9 mm
◄
T =-2- =- b =__!__ =
2cr
3
45 x l0 N
2(0.l m)(l.4 x l0 6 Pa)
0.1 607
111
b
= 160.7 mm ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
179
PROBLEM 2.78
Two blocks of rubber with a modulus of rigidity G = 10 MPa are bonded
to rigid supports and to a plate AB. Knowing that b = 200 mm and
c = 125 mm, tktermim: the largest alluwabk luau P and the smallest
allowable thickness a of the 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
6mm.
SOLUTION
J_ p
Shearing stress:
p
r =-2-=--
A
2bc
P = 2bcr= 2(0.2 m)(0.1 25 m)( l.5x 103 kPa)
Shearing strain:
6
P = 75.0 kN ◄
T
r= - = a G
GS (10 x l 0 6 Pa)(0.006m)
a=---;-=
1.5 x 10 6 Pa
0.04111
a= 40.0 mm ◄
PROPRI ETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
180
PROBLEM 2.79
An elastomeric bearing (G = 130 psi) is used to support a bridge girder
a s shown to prov ide flexibility during earthquakes. The beam must not
displace more than ¾in. when a 5-kip lateral load is applied as shown.
Knowing that the maximum a llowable shearing stress is 60 psi,
determine (a) the smallest allowable d im en sion b, (b) the smallest
required thickness a.
SOLUTION
Shearing force:
P
= 5 kips =5000 lb
Shearing stress :
r
= 60 psi
p
r =-
A'
and
or
A= p
r
= 5 000 = 83 .333 in 2
60
A = (8 in. )(b)
p
(a)
b = _i =
8
83 333
·
= 10.41 66 in.
b = 10.42 in. ◄
8
Y = !..._ = ~ = 461.54 x 10- 3 rad
a- 130
(b)
0
But y =- ,
a
or
o
0 .375 in.
a= - = ----r
a = 0. 8 13 in. ◄
46 1.54 X 10- 3
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
18 1
PROBLEM 2.80
For the elastomeric bearing in Prob. 2.79 with b = 10 in. and a = I in.,
determine the shearing modulus G and the shear stress r for a
maximum lateral load P = 5 kips and a maximum displacement
8 = 0.4 in.
SOLUTION
Shearing force:
P = 5 kips = 5000 lb
Area:
A = (8 in.)(10 in.) = 80 in
2
Shearing stress:
P
A
p
5000
80
r= - = - -
8
0.4 in.
I in.
Shearing strain:
r = - = - - = 0.400 rad
Shearing modulus:
G =!:._= 62.5
r 0.400
a
r = 62.5 psi ◄
G = 156.3 psi ◄
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
182
PROBLEM 2.81
A vibration isolation unit consists of two blocks of hard rubber bonded
to a plate AB and to rigid supports as shown. Knowing that a force of
magnilutlt: P = 25 kN causes a tldlectiun ,5 = 1.5 mm of plate AB,
determine the modulus of rigidity of the rubber used.
30mm
30mm
SOLUTION
F
= _!_ P = _!_(25 x 103 N) = 12.5 x 103 N
r
=
2
F
A
b
r
G
2
=
(12.5 x 103 N)
(0.15 mX0 .l m)
= 1.5 X 10- 3 m
h
=
833.33 x 103 Pa
= 0.03 m
= §_ = 1.5 X 10-3 = 0.05
h
0.03
= !_ = 833.33 x 103 = 16.67 x 106 Pa
r
o.o5
G
= 16 .67 MPa
◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
183
PROBLEM 2.82
A vibration isolation unit consists of two blocks of hard rubber with a
modulus of rigidity G = 19 MPa bonded to a plate AB and to rigid
supports as shown. Denoting by P lhe magnilutle of the furc.;e applied
to the plate and by o the corresponding deflection, deten11ine the
effective spring constant, k = Pi o, of the system.
30 mm
30
111111
SOLUTION
0
h
Shearing strain:
r =-
Shearing stress:
r = Gy = -
Go
.!__p = Ar =
Force:
2
h
GAo
h
or
= !:_ = 2GA
o h
Effective spring constant:
k
with
A = (0.15)(0.1) = 0.015 m2
k
= 2(19 x 10
6
2
Pa)(0.015 m )
0.03m
h = 0.03 m
= 19 _00 x 106 Ni m
k
= 19.00 x 103 kNlm
◄
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
184
PROBLEM 2.83"
A 6-in.-diameter solid steel sphere is lowered into the ocean to a point where the pressure is 7.1 ksi (about
3 miles below the surface). Knowing that £ = 29x l 06 psi and v = 0.30, detem1ine (a) the decrease in
diameter of the sphere, (b) the decrease in volume of the sphere, (c) the percent increase in the density of the
sphere.
SOLUTION
For a solid sphere,
:rr
3
Vo = -do
6
= ~(6.00) 3
6
=113.097 in 3
=- 7. J X 103 psi
I
lix
=E
(CYX - V(Yy - vaz )
(0.4)(7.1 X 103 )
(1- 2v) p
29xl06
E
= - 97.93 X 10Likewise,
liy
6
= 6 = - 97.93 X 10- 6
2
(a)
- M = - d0 lix = - (6.00)(- 97.93 x 10- 6 ) = 588x 10- 6 in.
(b)
- L'i V = - V0 e = - (113.097)(- 293.79 x 10- 6 ) = 33.2 x 10- 3 in 3
(c)
Let m =mass of sphere.
- M = 588 x l0- in. ◄
6
- fiV=33.2x l 0- 3 in 3 ◄
m =constant.
p - po = _e_- 1=
m
xVo_ l = 1- - 1
Po
Po
Vo (1 + e) m
1+ e
= (l -e +e 2 - e 3 + ···) - 1=-e+e2 -e3 +···
:,e -e =293.79 X 10- 6
p - Po x 100% = (293 .79 x 10- 6 )(100%)
0.0294% ◄
Po
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
185
r8:5_1~11l~j
PROBLEM 2.84"
(a) For the axial loading shown, determine the change in height and the
change in volume of the brass cylinder shown. (b) Solve part a,
<Ty= -5\ MPa
assuming lhal lht: loading is hy<lruslalit; with a x= a y = a ,= - 70 MPa.
E= 105CFa
V = 0.33
I
135 mm
SOLUTION
h0 = 135 mm =0.135 m
A0=!!..dJ = .:!..(85)2 = 5.6745xl03 1111112 = 5.6745xl0- 3 m 2
4
4
3
6
3
V0 = A 0 h0 = 766.06 x 10 111111 = 766.06 x 10- 111
(a)
3
6
a x= 0, a y=-58x l 0 Pa, a ,= 0
&
Y
I
58x 106
UY
= -(- vu + a - vu) = - = - - E
x
Y
'
E
105 x l09
- 552.38xl0- 6
6
t,,.h = - 0.0746 mm ◄
Mi = h 0 &Y = (135 mm)( - 552.38 x 10- )
1- 2v
E
e = - - (a +a +a) =
X
y
'
(1- 2v)a
Y
E
6
(0.34)(- 58 x 10 ) = - 187.8 Ix 10-6
9
I 05 x 10
t,,.V =- 143.9 1111113 ◄
t,,. V =V0 e =(766.06x I03 1111113 )(- 187.81x I 0-6 )
(b)
6
6
a X = a y = a Z =-70x10 Pa
=
&Y
_!__(E
vax +ay
_
vu,
a X +ay + a Z =-210 x 10 Pa
)= I-
6
= (0.34)(- 70 x 10 ) = _
2v
E
UY
105 x l 09
226 .67 x 10_6
Mi =- 0.0306 111111 ◄
6
e = l - 2v (a + a +a) = (0.34)(- 210x!0) = -680 x l0-6
E
X
y
z
I 05 XI 09
t,,.V = - 5211111113 ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
186
L
l in. c~arneter
.
PROBLEM 2.85*
\ 11 kips
,
Determine the dilatation e and the change in volume of the 8-in. length
6
of the rod shown if (a) the rod is made of steel with E = 29 x 10 psi
6
and v = 0.30, (b) the rod is made of aluminum with E = 10.6 x 10 psi
and v = 0.35.
~8in. - - l
SOLUTION
A = 1!_d 2 = 1!_(1)2 = 0.78540in2
4
4
p = II X 103 lb
Stresses :
11 X 103
.
0.78540
aY = a , = 0
p
A
3
.
a , = - = - - - = 14.0056x !0 psi
E = 29x 106 psi
V
= 0.30
e = __l_ (a - va - va) = a x= 14.0056 x 103
X
E X
y
z
E
29 XI 06
482.95x 10- 6
I
va
6
e =-(- va +a - va.)=-- x =-ve =-(0.30)(482.95x 10- )
Y
E
X
y
.
E
X
=-144.885x !0-6
I
vax
-6
e = - (-va - va + a ) =- = e =-144.885x l0
'
E
X
z
y
E
y
e =eX +ey +eZ = 193.2x10-6
◄
Av= ve = M e= (0.78540)(8)(193.2 x I 0-6 ) = 1.2 14 x 10- 3 in 3
(b)
Aluminum.
E = 10.6x 106 psi
V
◄
= 0.35
6y
3
14.0056 x l0 = 1.3 2128 x !0-3
E
10.6x106
3
6
=-vex =-(0.35)(1.32128x l 0- ) = -462.45 x l 0-
6,
= 6 y = -462.45 X 10- 6
= ax
6
X
◄
Av = ve = M e= (0.78540)(8)(396 x 10-6 ) = 2.49 x 10- 3 in 3
◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
187
PROBLEM 2.86
Determine the change in volume of the 50-mm gage length segment AB in Prob. 2.64 (a) by computing the
dilatation of the material, (b) by subtracting the original volume of portion AB from its final volume.
PROBLEM 2.64 A 2.75-kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate
(E = 200 GPa, v = 0.30). Determine the resulting change (a) in the 50-mm gage length, (b) in the width of
portion AB of the test coupon, (c) in the thickness of portion AB, (d) in the cross-sectional area of portion AB.
12 mm
SOLUTION
(a)
A0 = (12)(1.6) = 19.2mm2 = 19.2xl0- 6 m 2
V0 = L0 A 0 = (50)(19.2) = 960 mm 3
Volume
3
2 75
- x lO = l43.229 x l06 Pa
19.2x!0-6
a =.!.._=
X
Ao
ay = a, = 0
6
6 =_!_(a - va - va )=ax=
x
E
x
y
'
E
143 ·229 x !0 = 7 16.15x l 0-6
200 x 109
3
By=&, =-v ex =-(0.30)(716.15 x !0- )=-2 14.84 x !0-6
e = Bx +By +&, = 286.46 x 10-
6
6
~ v = v0 e = (960)(286 .46 x 10- ) = 0.275 mm
(b)
3
◄
From the solution to problem 2.64,
b, = 0.035808 mm
by = - 0.0025781
b, = -0.00034374 111111
The dimensions when under the 2.75-kN load are
Length:
L = l 0 + b, = 50 + 0.035808 = 50.035808 mm
Width:
w = Wo + by = 12 - 0.0025781 = 11.997422 mm
Thickness:
Volume:
t = t 0 + b, = 1.6 - 0.00034374 = 1.599656 mm
V = l wt = (50.0358 1)(11 .997422)(1 .599656) = 960.275 mm 3
~ V = V - V0 = 960.275 - 960 = 0.275 mm 3
◄
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
188
PROBLEM 2.87*
A vibration isolation support consists of a rod A of radius R 1 = IO mm and
a tube B of inner radius R 2 =25 mm bonded to an 80-mm-long hollow
rubber cylinder with a modulus of rigidity G = 12 MPa. Determine the
largest allowable force P that can be applied to rod A if its deflection is not
to exceed 2.50 mm.
: .._.__-::::;..!.
B
I
I
I
~,.-1:----
1'
--
/~:::_::_= ::.·
f/
'~
SOLUTION
Let r be a radial coordinate. Over the hollow rubber cylinder, R1 :,; r:,; R2 •
Shearing stress r acting on a cylindrical surface of radius r is
p
p
A
21rrh
r
p
r = - = --
The shearing strain is
r =-G =21rGhr
Shearing deformation over radial length dr:
do
-=
r
dr
P dr
21rGh r
do = ydr = - - -
L
~
l:- ----:::;~==4i
~
1
Total deformation.
'
l
~r
Data:
R1 = I O tmn = 0 .0 10 m,
_J
Jr
R2 = 25 mm = 0.025 m, h = 80 mm = 0.080 m
o = 2.50 x 10- 3 m
6
3
p = (21r)(l 2 x 10 ) (0.080)(2.50 x 10- ) =16.46 x 103 N
In (0.025/0.010)
16.46 kN ◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
189
PROBLEM 2.88
A vibration isolation support consists of a rod A of radius RI and a tube B of
inner radius R2 bonded to a 80-1111n-long hollow rubber cylinder with a
modulus of rigidity G = 10.93 MPa. Detennine the required value of the ratio
R2/R1 ifa 10-kN force Pis to cause a 2-mm deflection of rod A.
SOLUTION
Let r be a radial coordinate. Over the hollow rubber cylinder, R1 ::; r::; R2 .
Shearing stress -r acting on a cylindrical surface of radius r is
p
p
A
2;rrh
T =-=--
The shearing strain is
T
p
G
2;rGhr
r= - = - -
Shearing deformation over radial length dr:
d<5
=r
dr
d<5=ydr
-
L 1
,-------_:;#~t
~
r
dr<5 = - p- dr
2;rGh r
'.
I
Total deformation.
<5 =
I
r'
d<5 = -p- f R, dr
R,
2;rGh R, r
p
,- = -p-(In R - In R )
= - -In r IR
2
1
2;rGh
R,
2;rGh
= __!!___ In R 2
2;rGh
In R2
R1
R1
= 2;rGM = (2;r)(I0.93 x 106 )(0.080)(0.002)
P
10.103
1.0988
R
--2.. = exp (1 .0988) = 3.00
R1
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
190
PROBLEM 2.89"
The material constants£, G, k, and v are related by Eqs. (2.33) and (2.43). Show that any one of these constants
may be expressed in terms of any other two constants. For example, show that (a) k = GE/(9G - 3£) and
(b) v = (3k - 2G)/(6k + 2G).
SOLUTION
E
k =--3(1- 2v)
(a)
E
G =-£-
2(1 + v)
E
l +v = or v= - - 1
2G
2G
k=
E
3 [ 1_ 2 ( ~
2
(b)
and
k
G
=
-tJ]
2EG
3[2G - 2£+ 4G]
2EG
18G - 6£
k=
EG
9G - 6E
◄
2(1+ v)
3(1- 2v)
3k - 6kv= 2G + 2Gv
3k - 2G = 2G + 6k
v= 3k - 2G ◄
6k+2G
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
19 1
PROBLEM 2.90"
Show that for any given material, the ratio GIE of the modulus of rigidity over the modulus of elasticity is
always less than ½ but more than ½- [Hint: Refer to Eq . (2.43) and to Sec. 2.13.]
SOLUTION
G = - E2(1 + v)
Assume v > 0 for almost all material s, and v <
E
- = 2(1+ v)
G
or
½for a positive bulk modulus.
Applying the bounds,
Taking the reciprocals,
1
2
1
3
G
- > - >E
I
G
I
3
£
2
- < -<- ◄
or
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
192
Ex= 50 GPa
Ey = 15.2GPa
E, = 15.2 GPa
v,, = 0 .254
= 0.254
PROBLEM 2.91 *
v,y = 0.428
A composite cube with 40-mm sides and the properties shown is
made with glass polymer fibers aligned in the x direction. The cube
is cunslraim:<l against <lt:furmaliuns in tht: y an<l z <lirt:cliuns an<l is
subjected to a tensile load of 65 kN in the x direction. Determine
(a) the change in the length of the cube in the x direction, (b) the
stresses O'_n 0'1 , and O'z .
Vxy
Ex = 50 GPa
vxz = 0.254
= 15.2 GPa
Ez = 15.2 GPa
vxy
E1
v,1
= 0.254
= 0.428
SOLUTION
Stress-to-strain equations are
(I)
(2)
(3)
The constraint conditions are
,:
1
vxy
vyx
Ex
EY
vyz
vzy
EY
E,
v zx
vxz
E,
Ex
(4)
(5)
(6)
= 0 and ,: = 0.
2
Using (2) and (3) with the constraint conditions gives
(7)
vyz
- -
(J'
E1
l
0.428
15.2 (J'y - 15.20',
0.428
- 15.20'Y
+
I
_
15 2
1
(J'x
= ~ O 'x
V,,
l
E,
0 .254
=~
0 .254
O',
+-
(J'
'
or
=-
(8)
(J'
Ex
(J'y -
'
0.4280',
or - 0.4280'1 +
= 0.0772160'x
O',
= 0.07721 60'.r
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
193
PROBLEM 2.91 • (Continued)
Solving simultaneously,
Using (4) and (5) in (I),
Ex = -f-[l - (0.254)(0.1 34993) - (0.254)(0.J 34 993)]0-x
X
0.93 1420-x
Ex
A = (40)(40) = 1600
3
o- =!__ = 65 x l0
x
A
1600 x 1o-6
111111
2
= 1600x 10- 6 ni2
= 40.625x 106 Pa
3
6
X
= (0.93 142)(40.625x 10
SOX 109
(a)
c5, = L,.&x = (40 mm)(756.78x !0-6 )
(b)
O"x
)
_ Sx l 0-6
756 7
Sx = 0.0303
= 40.625 x l 0 6 Pa
nun ◄
o-x = 40.6 MPa ◄
o-Y = o-, = (0.134993)(40.625x l 06 ) = 5.48xl06 Pa
o-Y = o-, = 5.48 MPa ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
194
v,, = 0.254
= 0.254
Ex= 50GPa
Ey = 15.2 GPa
Vxy
E, = 15.2 GPa
v,y = 0.428
y
PROBLEM 2.92*
The composite cube of Prob. 2.91 is constrained against deformation in
the z direction and elongated in the x d irection by 0.035 mm due to a
Lt:nsik load in lhe:: x tlim ;liun. De::le::rmine:: (a) lhe:: s tre::sse::s a_,, a Y ' anti CY,
and (b) the change in the dimension in they direction.
Ex = 50 GPa
vx, = 0.254
Ey =15.2 GPa v~v
E,
=0.254
= 15.2 GPa v,y = 0.428
X
SOLUTION
(I )
(2)
&
v a
= - ....E....L
-
'
v ,a
...2'.:..___-2' +
Ex
EY
a
_I._
(3)
E,
(4)
(5)
(6)
Constraint condition:
Load condition :
v,,
I
0 =-- a + -a
From Equation (3),
E,
a
v,, E,
z
= --=--=a
Ex
x
'
E,
(0.254)(15.2)
50
'
0 0772 16ax
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
195
PROBLEM 2.92• (Continued)
From Equation (1) w ith cry= 0,
1
1
= -[c,x - 0.254c,z ] = - [l - (0.254)(0.0772 16)Jc,x
Ex
Ex
0.98039
c,x
Ex
CY
X
But
c =
x
Ee
=
X
X
0.98039
o, = 0.035 mm = S?Sx l 0-6
Lx
40 mm
c,x = 44.6 MPa ◄
(a)
6
6
er, = 3.45 MPa ◄
c,z = (0.077216)( 44.625x 10 ) = 3.446 x 10 Pa
From (2),
v.'J'
1
X
y
v,y
cy =-E erX + -E ery - -E erz
Z
= (0.254)(44.625x l 06 ) + 0 - (0.428)(3.446xl06 )
50 x 109
15.2x l 09
6
=-323 .73x 10(b)
oy= L,cy= (40mm)( - 323.73x l0-
6
oy= - 0.0129 mm
)
◄
PROPRI ETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
196
PROBLEM 2.93
Knowing that, fo r the plate shown, the allowable stress is 125 MPa, determine
the maximum allowable value of Pwhen (a) r = 12 mm, (b) r = 18 mm .
SOLUTION
2
A = (60)(15) = 900 mm = 900 x 10-
6
111
2
D = l 20 mm = 2 _00
d
60mm
(a)
!...=
r = l 2mm
d
From Fig. 2.60b,
12 111111 = 0.2
60 mm
K
p
= 1.92
CYmax =
KA
-6
6
P = ACYmax = (900x l 0 )(125x l 0) = 58 _6 x l03N
K
1.92
= 58.3 kN ◄
(b)
r = l 8mm ,
P = Aama,
K
!... = 18 mm
d
60 mm
= 0 _30
From Fig 2.60b,
K
= 1.75
6
= (900x l 0-6 )(125x 10 ) = 64 _Jx l03 N
1.75
= 64.3 kN ◄
PROPRIETARY MA TER IAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
197
PROBLEM 2.94
Knowing that P = 38 kN, determine the maximum stress when (a) r = 10 mm,
(b) r = 16 mm, (c) r = 18 mm.
SOLUTION
A = (60)(15) = 900 mm 2 = 900 x 10-6 m 2
D = 10 mm = _
2 00
d
60mm
(a)
r = I0mm
From Fig. 2.60b,
O"max
(b)
From Fig. 2.60b,
(c)
K = 2.06
=A
= 87.0MPa ◄
16 mm = 0.2667
!_ =
d
60mm
K = 1.78
= (1.78)(38 XI 03) = 75.2 x 106Pa
900x 10- 6
r = l8mm,
!_= 18 mm = 0_ 30
d 60mm
From Fig 2 .60b,
K = 1.75
O"max
KP
ama.x
= (2.06)(38x 103) = 87.0x 106Pa
900 X 10-6
r = l 6mm
O"max
!_= IOmm = 0. 1667
d 60mm
= 75.2 MPa ◄
3
= (l.75)(38x 10 ) = 7 3.9 x 106 Pa
6
900 X 10-
= 73.9MPa ◄
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
198
PROBLEM 2.95
A hole is to be drilled in the plate at A . The diameters of the
1
bits available to drill the hole range from ½ to 1 / 2 in. in
¼-in. increments. If the allowable stress in the plate is 21 ksi,
determine (a) the diameter d of the largest bit that can be
used if the allowable load P at the hole is to exceed that at
the fillets, (b) the corresponding allowable load P.
SOLUTION
D = 4.6875 = 1_5
3.1 25
At the fillets:
!_
d
d
K
From Fig. 2.60b,
= 0.375 = 0.1 2
3 .1 25
= 2.1 0
~,in = (3 .1 25)(0.5) = 1.5625 in
p
-
AminO"all -
all -
At the hole:
Anet
K
-
2
(1.5625)(2 1) - 15 625 k"
2. 10
.
tps
= (D - 2r)t, K from Fig. 2.60a
Ane10"a11
K
with
Hole diam.
0.5 in.
0.75 in.
1 in.
r
0.25 in.
1.5 in.
= 4.6875 in.
d
=D-
4.1 875 in.
0.375 in.
0.5 in.
1.25 in.
D
3.9375 in.
3.6875 in.
0.625 in.
0.75 in.
(a)
Largest hole with
(b)
Allowable load ~ 11
3.4375 in.
3.1 875 in.
Pall
2r
t
= 0.5 in .
2r!D
0.1 07
0.16
0.2 13
0.267
0.32
O"all
K
2.68
2.58
2.49
2.4 1
2.34
> 15.625 kips is the ¾-in.-diameter hole.
= 15.63 kips
= 21 ksi
P.11
~el
2.0938 in
2
16.4 1 kips
1.96875 in
2
16.02 kips
1.84375 in
2
15.55 kips
I. 71875 in
2
14.98 kips
1.59375 in
2
14.30 kips
◄
◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
199
PROBLEM 2.96
= 13 kips and d = ½in., determine the maximum
stress in the plate shown. (b) Solve part a, assuming that the
hole at A is not drilled.
(a) For P
SOLUTION
Maximum stress at hole:
Use Fig. 2.60a for values of K .
2
r
D
A net
= ~ = 0.01 7
4.6875
= (0.5)(4.6875 -
O"max =
K _!__
K
'
=
(
0.5)
2 -68)(! 3)
= 2.68
= 2.0938 in 2
16.64 ksi
=
2.0938
~ et
Maximum stress at fillets:
Use Fig. 2.60b for values of K.
!_
= 0.375 = 0.1 2
d
3.1 25
A min=
(0.5)(3.1 25)
O"max =
K _!_
A min
=
=
1.5625 in 2
(2.1 0)(13)
1.5625
=
17.47 ksi
(a)
With hole and fillets:
17.47 ksi ◄
(b)
Without hole:
17 .47 ks i ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
PROBLEM 2.97
9mm
Knowing that the hole has a diameter of9 mm, determine (a) the radius
r1 of the fil lets for which the same maximum stress occurs at the hole A
aml al Lht:: lillt::ls, (b) lht:: rnrrt::spun<ling maximum alluwablt:: load P if
the allowable stress is I00 MPa.
SOLUTION
For the circular hole,
r
=
(½)c9) = 4.5 mm
d = 96 - 9 = 87 mm
Anet
From Fig. 2.60a,
(a)
For fillet,
Khole
2
24
r = C -5) = 0.09375
D
96
= dt = (0.087 m)(0.009 m) = 783 X
w-6 1112
= 2.72
D = 96 mm, d = 60 mm
D
d
Am;n
= 96 = 1.60
60
= dt = (0.060 m)(0.009 m) = 540 X
6
I0- 111
2
Kfilletp
(5.40 X J0- 6 )()00 X J0 6 )
~in
28.787 X J03
CJ'max
= 1.876
From Fig. 2.60b,
rf "' 0.1 9
d
r1 ""0.1 9d = 0.1 9(60)
r1
(b)
= 11 .4 mm ◄
P = 28.8 kN ◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
20 1
PROBLEM 2.98
r88mm
l
rA=20mm
r8 = l 5mm ~
l
B il
64
For P = 100 kN, determine the minimum plate thickness t required if the
allowable stress is 125 MPa .
~
mm----------..:1
r:·
SOLUTION
At the hole:
rA = 20 mm
d A = 88 - 40 = 48 mm
2rA = 2(20) = 0.455
DA
88
From Fig. 2.60a,
K = 2.20
KP
KP
CJ'max
A.iet
t=
At the fillet:
From Fig. 2.60b,
t = --d AO"max
(2.20)(100 x 103 N) = 36.7 x
(0.048 m)( l 25 x 106 Pa)
10-3
m = 36.7 mm
!!__ = 88 = 1.375
D = 88 mm,
d 8 = 64mm
r8 = 15 mm
!JL =
_!_2_ = 0.2344
dB
64
d8
64
K = 1.70
Clmax
KP
KP
Amin
dnt
=-- = -
t=
___!!__ =
dnO"max
103
(l.70)(IOO x
N)
= 21.25 x 10- 3 m = 2 1.25 mm
(0.064 m)(l25 x 106 Pa)
The larger value is the required mi nimum plate thickness.
t = 36.7 mm ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
202
2 in.
PROBLEM 2.99
(a) Knowing that the allowable stress is 20 ksi, determine the maximum
3 in.
allowable magnitude of the centric load P. (b) Detennine the percent
change in the maximum allowable magnitude of P if the raised portions
are removed at the ends of the specimen.
SOLUTION
!....= 0.250 = 0.1 25
d
2
K = 2.08
From Fig. 2.60b,
Amin = td = (0.625)(2) = 1.25 in 2
(a)
KP
O"max = ~
mi n
(b)
p = AminO"max (1.25)(20) = 12.0 192 kips
K
2.08
P = 12.02 kips ◄
Without raised section, K = 1.00
P = ~ inO"max = (1.25)(20) = 25 kips
25- 12.02
% change = ( - -) x 100%
12.02
= 108.0% ◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
203
PROBLEM 2.100
A centric axial force is applied to the steel bar shown . Knowing that
a-a1 1 = 20 ksi, determine the maximum allowable load P.
1 in.
SOLUTION
r = 0.5 in.
At the hole:
2r
d
d
= 5 - 1= 4 in.
= 2(0.5) = 0 _2
From Fig. 2.60a,
5
K = 2.5 1
(0.75)(4) = 3 in 2
A net= td =
KP
ermax = ~
-
Anet
p = Anet O"max
K
At the fi llet :
(3 )(2 0)
2.5 1
=23.9 kips
.
D = 6.5 = 1_3
= 5 m.,
D = 6.5 in.,
d
r = 0 .5 .m.
~ = _22 = 0. 1
From Fig. 2.60b,
d
d
5
5
K
= 2.04
A.nin
= td = (0.75)(5) = 3.75 in
(J'max
=--
2
KP
Amin
p=
Amin (Jmax
K
(3 -75)(2 0)
2.04
= 36.8 kips
Smaller value for P controls.
P
= 23 .9 kips
◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
204
I
PROBLEM 2.101
I
I
The cylindrical rod AB has a length L = 5 ft and a 0.75-in. diameter; it is made of a mild steel
6
that is assumed to be elastoplastic with E = 29 x 10 psi and O"y = 36 ksi. A force P is applied
to the bar and then removed to give it a pennanent set <5p, Detennine the maximum value of
the force P and the maximum amount <5,11 by which the bar should be stretched if the desired
value of <5p is (a) 0.1 in., (b) 0.2 in.
L
A
p
SOLUTION
A =!!._d 2 =!!._ (0.75)2 = 0.44179 in 2
4
4
L
= 5 ft = 60 in.
3
<5 = Le = Lay= (60)(36 x l0
y
E
y
When
)
29 x 103
0.074483 in.
0, exceeds by, thus causing permanent stretch b P' the maximum force is
11
P, = Aay = (0.44179)(36 x 103 ) = 15.9043 x 103 lb
11
P = 15.90 kips
bP = 0,
11
-
o' = 0,
11
-
◄
by so that
(a)
<5P
=0.1 in.
b,,, = 0.1 + 0.074483 = 0.1 745 in.
◄
(b)
b" = 0.2 in.
b,,, = 0.2 + 0.074483 = 0.274 in.
◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
205
I
PROBLEM 2.102
I
• B
The cylindrical rod AB has a length L = 6 ft and a 1.25-in. diameter; it is made of a mild steel
6
that is assumed to be elastoplastic with E = 29 x 10 psi and a y = 36 ksi . A force P is
applied to the bar until end A has moved down by an amount <5,11 • Determine the maximum
value of the force P and the permanent set of the bar after the force has been removed,
know ing (a) J,11 = 0.1 25 in., (b) <5,11 = 0.250 in.
L
A
I'
SOLUTION
A = '!_d 2 = '!__(1.25)2 = 1.227 18 in 2
4
4
L = 6 ft = 72 in.
3
by = U:y = Lay= (72)(36 x l0 ) = 0.089379 in.
29 x 10 3
E
If b, 2 by,
11
P, = Aa y = (1.22718)(36 x 103)
11
= 44. l 79x 103 lb = 44.2 kips
(a)
t5,,, = 0.1 25 in. > Sy
so that P,11 = 44.2 kips
◄
t5' = P,,,L = a yL = by = 0.089379
AE
E
◄
t51, = t5,,, - t5' = 0.125 - 0.089379 = 0.356 in.
(b)
<5,11 = 0.250 in. > Sy
so that P,11 = 44.2 kips
◄
t5' = by
SP
= t5,,, - t5' =0.250 -
0.089379 = 0.1606 in.
◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
206
l
PROBLEM 2.103
Rod AB is made of a mild steel that is assumed to be elastoplastic
with E = 200 GPa and a y = 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 51 if the lever is to snap back to a horizontal position
after Q is removed.
~mm doamdc,
1.25
Ill
SOLUTION
AAB = ~(9)2 = 63 .617 mm 2 = 63.61 7x l0- 6
111
2
4
Since rod AB is to be stretched permanently,
6
6
(FAB)max = AAsO"y = (63.617 x l 0- )(345x 10
)
= 21.948 x 103N
+) LMD = 0: I.IQ - 0 .7FAB = 0
Qmax
b'
0.7
3
~
= - (2 1.948 X 10 ) = 13.9669 X 10 N
I.I
_ (FAB)ma, LAB
ABEAAB
13.97 kN ◄
(21.948 xl03)(1.25)
= 2. 15625 x l0-3m
(200 x l09 )(63.617 x l0- 6 )
6
0' = AB' = 3.0804 x 10- 3 rad
0.7
b1 = 1.10' = 3.39 X 10- 3 111
3.39 mm ◄
PROPRIETARY MATERIAL. C opyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
207
PROBLEM 2.104
Solve Prob. 2.1 03, assuming that the yie ld point of the mild steel is
250 MPa.
PROBLEM 2.103 Rod AB is made of a mild steel that is assumed to
be elastoplastic with E =200 GPa and O"y = 34 5 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 61 if the lever is to snap back to a horizontal position
after Q is removed.
SOLUTION
AAB
= ~(9)2 = 63 .6 17 mm 2 = 63.617 x 10-6
4
111
2
S ince rod AB is to be stretc hed permanently,
( FAB )max
= AAB(T y =(63.6 17 X 10- 6 )(250 X l0 6 )
=15.9043 x 103N
+) L MD = 0: I.IQ - 0 .7FAB = 0
Qmax
b'
= .2.2_(15.9043 X 103) = 10.12 X 103 N
I.I
- ( FAB )max LAB AB -
EAAE
(15.9043x l03)(1.25)
- (200x l09 )(63 .617x l 0- 6 )
10.12 kN
◄
l.5625x !0-3 111
0' = <>'An = 2.232l x l 0- 3 rad
0 .7
61
= 1. 10' = 2.46 X 10-) 111
2.46 mm ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
208
I
-,
I
,.. C
40-mm
Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild
steel that is assumed to be elastoplastic with E = 200 GPa and O"y = 250 MPa.
A force P is applied to the rod and then removed to give it a permanent set
SP= 2 mm. Determine the maximum value of the force P and the maximum
amount S111 by which the rod should be stretched to give it the desired permanent
set.
V diameter
1.2 Ill
-I
0.8111
PROBLEM 2.105
n
B
30-mm
, -diameter
L_
A
p
SOLUTION
AAB = ~ (30)2 = 706.86 mm 2 = 706.86 X I o-6 m 2
4
A8 c =~(40)2 = l.25664x l 03 mm 2 = l.25664x l 0- 3 m 2
4
Pmax = AminO"Y
= (706.86x l0-o )(250x l06 )= 176.7 !5x !03 N
Pmax
S' = P'LAB + P'L 8 c =
EAAB
EAac
3
= 176.7 kN ◄
3
(176.7 15 x l 0 )(0.8)
+
(176.7 15x l 0 )(1.2)
9
6
(200 x 10 )(706.86 x 10- ) (200 x 109 )(1.25664 x I 0- 3 )
= 1.84375 x 10- 3 m = 1.84375 mm
SP = S111 - S' or <5,,, = SP+ S' = 2 + 1.84375
<5,,, = 3.84 mm ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
209
PROBLEM 2.106
-1
J.2
40- rnrn
diameter
Ill
t-
Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild
steel that is assumed to be elastoplastic w ith E == 200 GPa and a-y == 250 MPa.
A force P is applied to the rod until its end A has moved down by an amount
o,,, == 5 mm. Determine the maximum value of the force P and the pennanent
set of the rod after the force has been removed.
B
30-rnrn
diameter
0.8111
l_
A
p
SOLUTION
AA8 == ~(30) 2 == 706.86 mm 2 == 706.86 x I 0-6 m 2
4
A8 c == ~( 40)2 == 1.25664 x I 0 3 mm 2 == 1.25644 x I 0- 3 m 2
4
Pmax
==
AminO"Y
6
== (706.86 X 10- )(250 X 10
6
)
3
== 176.715 X 10 N
Pmax
o' == P'LAB + P'Lnc
EAAB
EAnc
==
3
== 176.7 kN ◄
3
(176.7 15x 10 )(0.8)
+
(176.7 15x 10 )(1.2)
(200 x 109 )(706.68 x 1o-6 ) (200 x 109 )(1.25664 x 10- 3 )
== 1.84375 x 10- 3 m == 1.843 75 mm
op == O,,, -
O' == 5 - 1.84375 == 3. 16 mm
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
210
r
PROBLEM 2.107
,......,...
A
Rod AB consists of two cylindrical portions AC and BC, each with a cross-sectional
area of 1750 mni2. Portion AC is made of a mi ld steel with E =200 GPa and
c,r =250 MPa, and portion CB is made of a high-strength steel with E =200 GPa
and c, y =345 MPa . A load P is applied at C as shown. Assuming both steels to be
elastoplastic, determine (a) the maximum defl ection of C if Pis gradually increased
from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each
portion of the rod, (c) the penn anent deflection of C.
190111111
t-
,Q.
190 mm
p
l -
B
SOLUTION
Displaceme nt at C to cause yielding of AC.
(0.1 90)(250 X106)
LAcCTr AC
E '
0.2375 X10- 3 111
9
200x 10
FAc = Ac, y,Ac = (1750 x 1o- 6)(250 x 106) = 43 7.5 x 10 3 N
lier
, = LAccr,AC =
Corresponding force.
9
Fcs = _ EAoc = _ (200 x 10 )(1750 x 10-6)(0.2375 x 10Lc8
0. 190
3
)
-4
_ x 1N
37 5 10
For equilibrium of element at C,
FAC - (FCB + Py) = 0
Since applied load P = 975 x 10 3 N > 875 x 10 3 N, portion AC y ields.
3
3
C
3
Fc8 = FAc - P = 437.5 x 10 - 975 x 10 N =-537.5 x 10 N
(a)
0 =- FcaLCD =
c
EA
(537.5xl03)(0.1 90) = 0.29179 x lo-3m
(200x l 0 9 )(1750x l0-6 )
0.292 mm ◄
(b)
250 MPa ◄
Maximum stresses: c,AC =CTr,Ac = 250 MPa
c, = Fae =- 537.5x l 03 =-307.14 x l06 Pa =-307MPa
BC
A
1750 X10- 6
(c)
- 307 MPa ◄
Deflection and forces for unloading.
o' = P;cLAC = - P~aLcs
EA
EA
p~B = - P;c LAC = - P~c
LAB
3
3
P' = 975 x 10 = P;c - P~8 = 2P~c P~c = 487.5 x 10- N
o' =
;;p
(487.5 x 1o3)(0.1 9o)
(200 X109 )(1750 X10-6 )
=o.26464 x 1o3 m
=o,,, - o' =0_29 179 x 10- 3 =0.027 15X l0- 3 111
o.26464 x 10- 3
0.0272 mm ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
2 11
PROBLEM 2.108
-I
For the composite rod of Prob. 2.1 07, if P is gradually increased from zero until the
deflection of point C reaches a maximum value of 0,11 = 0.3 mm and then decreased
bal:k lo zc::ru, tklt:rmint: (a) lht: maximum value:: uf P, (b) lht: maximum slr<::ss in t:al:h
portion of the rod, (c) the permanent deflection of C after the load is removed.
c--
A
190mm
t-
C
......
PROBLEM 2.107 Rod AB consists of two cylindrical portions AC and BC, each with a
cross-sectional area of 1750 mni2. Portion AC is made of a mild steel w ith E = 200 GPa
and a y = 250 MPa, and portion CB is made of a high-strength steel with E = 200 GPa
and Cly= 345 MPa. A load P is applied at C as shown. Assuming both steels to be
elastoplastic, determine (a) the maximum deflection of C if P is gradually increased
from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C.
190mm
p
l - ...
B
SOLUTION
Displacement at C is
0,
11
= 0.30 mm. The corresponding strains are
3
&Ac = °'" = 0. 0
= l.5789 x l0- 3
LAC
190 111111
111111
-- -o,,,- -0.30 111111 --1
.5789 x 10-3
lies -
L c8
190 mm
Strains at initial yielding:
Cly AC
liy AC=-'-
,
E
Cly BC
liy cs = - ' -
,
(a)
6
250xl0 = l. 25 x 10_3
200 XI 0 9
345 x 106 = - 1.725 x I o- 3
200 X 10 9
E
(yielding)
(elastic)
Forces: FAc = Aay = (1750 x 10- 6)(250 x 106) = 437.5 x 10-3 N
Fc 8 = EA&c8 = (200 x 10 )(1750 x 10- )(- 15789 x 10- 3) = - 552.6 x 10- 3 N
9
For equilibrium of element at C, FAc - Fc 8
6
-
P=0
3
3
3
P = FAc - Fco = 437.5 x 10 + 552.6x 10 = 990.1x 10 N
990 kN ◄
250 MPa ◄
CB: a
_ Fcs _ _ 552.6xl03 = - 316 x l06 Pa
CB - A 1750 x l0-6
- 3 16 MPa ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
212
PROBLEM 2.108 (Continued)
(c)
Deflection and forces for unloading.
<5' = P~c LAC
=-
EA
P' = P~c - Pea
PeaLca
EA
= 2P~c =990 .1 x 10 3N
103
<5' = (495 ·05 x
)(O . I 90)
(200 X 10 9 )(1750 X 10- 6 )
<5P
= <5,,, -
Pea= - P;c LAC
LAB
:. P~ c
= - PAC
=495.05 x 103N
0.26874 x 10- 3 m = 0.26874 mm
<5' = 0.30 mm - 0.26874 mm
0.03 13 mm ◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
2 13
PROBLEM 2.109
Each cable has a cross-sectional area of 100 mm2 and is made of an
elastoplastic material for which O"y =345 MPa and E =200 GPa . A force
Q is applied at C to the rigid bar ABC and is gradually increased from O to
50 kN and then reduced to zero. Know ing that the cables were initially taut,
determine (a) the maximum stress that occurs in cable BD, (b) the
maximum deflection of point C, (c) the final displacement of point C.
(Hint: In part c, cable CE is not taut.)
SOLUTION
Elongation constraints for taut cables.
Let 0
= rotation angle of rigid bar
ABC.
(I)
Equilibrium of bar ABC.
+) M A =0: LAaFaD + LAc FcE - LAc Q =0
LAB
I
Q = FCE + - - FBD = FCE + - FBD
LAC
2
Assume cable CE is yielded.
From (2),
(2)
Fe£ = Ao-y = (I 00 x I 0-6 )(345 x I 0 6 ) = 34 .5 x 103 N
F8 D = 2(Q - FCE ) = (2)(50 x 103 - 34.5 x 103 ) = 31.0 x 103 N
Since F8 D < Ao-y =34.5 x I 0 3 N, cable BD is elastic when Q =50 kN.
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
214
PROBLEM 2.109 (Continued)
(a)
Maximum stresses.
CJcE = Cly= 345 MPa
a
(b)
00
_ Foo_ 31.0xl03
A - 100 X 10- 6
310 x l 06Pa
-
a 00 = 310 MPa
◄
Maximum of deflection of point C.
6.20 mmJ.. ◄
From(!),
Permanent elongation of cable CE: (bcE\,
= (bed
- a yLCE
E
CJyl cE
(bcE )P = (bcE )max - -E6
= 6.20x l 0-3 - ( 34S x !0 ~( 2) = 2.75x l 0- 3 m
200 X 10
(c)
Unloading. Cable CE is slack (FCE
From (2),
= 0)
at Q = 0.
Foo= 2(Q - FCE ) = 2(0 - 0) = 0
S ince cable BD remained elastic, boo
= Foo Loo = 0.
EA
O◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
2 15
PROBLEM 2.110
Solve Prob. 2. 109, assuming that the cables are replaced by rods of the
same cross-sectional area and material. Further assume that the rods are
braced so that they can carry compressive forces.
PROBLEM 2.109 Each cable has a cross-sectional area of 100 mm 2
-1 rn--1
and is made of an elastoplastic material for which O"y = 345 MPa and
E = 200 GPa. A force Q is applied at C to the rigid bar ABC and is
gradually increased from O to 50 kN and then reduced to zero. Knowing
that the cables were initially taut, determine (a) the maximum stress
that occurs in cable BD, (b) the maximum deflection of point C, (c) the
final displacement of point C. (Hint: In part c, cable CE is not taut.)
rn-1
SOLUTION
Elongation constraints.
Let 0
= rotation angle of rig id bar
ABC.
( 1)
Equilibrium of bar ABC.
+")M A= 0: LAa Fao + LAc FCE - LAc Q = 0
Q
Assume cable CE is yielded . FCE
F8 0
From (2),
Since F80 < Ao-r
LAB
I
= FCE + - FBD = FCE + - FBD
(2)
2
LAC
= Aa-y =(IOO x 10- 6 )(345 x 106 ) = 34 .5 x l 0 3 N
= 2(Q -
FCE) = (2)(50 x 103 - 34 .5 x 103 ) = 3 I.O x 10 3 N
= 34.5 x 103 N, cable BD is elastic w hen
Q = 50 kN.
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
216
PROBLEM 2.110 (Continued)
(a)
Maximum stresses.
O'c E = O'y
O'
-
BD -
(b)
= 345 MPa
Fav - 3 1.0 x l03
A - I00 x l0- 6
3 10 x l06Pa
O'80
= 310 MPa ◄
Maximum of deflection of point C.
6.20 mm ..J, ◄
From (1),
3
Unloading. Q ' = 50 x I 0 N, 8~£ = 8~
From (!),
. ,,
EA8~0 (200 x 109 )(1 00x !0- 6 )(½8~)
Elastic F80 = - - = - - - - - - - - -~ -
Lav
' _ EA8~E
FCE -
(200 x l09 )(100 x 10- 6 )(8~)
---------~
10 6 /)~
Q' = F~E + ½F; 0 = 12.5 x 10 68~
Equating expressions for Q',
(c)
(QX
2
L CE
From (2),
5 x 1068~
2
12.5 x I 0 6 8~ = 50 x I 0 3
2.20 mm ..J, ◄
Final displacement.
PROPRIETARY MA TERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
2 17
PROBLEM 2.111
½-in.
Two tempered-steel bars, each kin. thick, are bonded to a
mild-steel
bar. This composite bar is subjected as shown to a centric axial load of
6
magnitude P . Both steels are elastoplastic with E = 29 x 10 psi and with
yield strengths equal to 100 ksi and 50 ksi, respectively, for the tempered
and m ild steel. The load P is gradually increased from zero until the
deformation of the bar reaches a maximum value J,,, = 0.04 in. and then
decreased back to zero. Determine (a) the maximum value of P, (b) the
maximum stress in the tempered-steel bars, (c) the permanent set after the
load is removed .
14 in.
2.Oin.
SOLUTION
For the mild steel,
Ai = ( ½)c2) = 1.00 in 2
6
For the tempered steel, A2 = 2(i~ )c2) = 0.75 in 2
YI
,5
Yl
_ Lan _ (14)(50 x i 0 3 )
E 29xl06
0.024138 in .
= L<Jr2 = (14)(100 x l03) = 0.048276i n.
E
29xl03
Total area: A = A1 + A2 = 1.75 in 2
bn < 0,11 < bn . The mild steel y ields. Tempered steel is elastic.
(a)
Forces:
fl = A1ay 1 = (l.00)(50xl03) = 50x!03lb
3
p _ EA2'5,,, _ (29xl0 )(0.75)(0.04)
2 -
L
-
14
_ x l 03lb
62 14
P = fl + P2 = I 12. 14 x I 03 lb = I 12. I kips
62 4 103
·1 x
= 82.86 x I 0 3 psi = 82.86 ksi
0.75
P = ll2.l kips ◄
82.86 ksi ◄
3
Unloading:
(c)
6
, = PL = (l 12. 14 x 10 )(14)
EA
6
(29 x 10 )(1.75)
0.03094 in.
Permanent set: ,SP = ,5111 - ,5' = 0.04 - 0.03094 = 0.00906 in.
0.00906 in. ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scann ed, duplicated, forwarded, di stributed, or posted
on a website, in whole or part.
218
PROBLEM 2.112
~
fBin.
~
For the composite bar of Prob. 2.111, if P is gradually increased from zero
to 98 kips and then decreased back to zero, detennine (a) the maximum
deformation of the bar, (b) the maximum stress in the tempered-steel bars,
(c) the permanent set after the load is removed.
3 .
./[6111.
t~~
14 in.
f6
PROBLEM 2.111 Two tempered-steel bars, each in. thick, are bonded to
a ½-in. mild-steel bar. This compos ite bar is subjected as shown to a centric
axial load of magnitude P . Both steels are elastoplastic with E = 29 x I 06 psi
and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the
tempered and mi ld steel. The load P is gradually increased from zero until
the deformation of the bar reaches a maximum value <5,,, = 0.04 in . and
then decreased back to zero. Determine (a) the maximum value of P, (b) the
maximum stress in the tempered-steel bars, (c) the permanent set after the
load is removed .
2.0 in.
SOLUTION
Areas:
2
Mild steel:
A 1=(½)c2) = 1.00 in
Tempered steel:
A2 = 2(i3 )c2) = 0.75 in
6
2
A = A 1+ A2 = 1.75 in 2
Total:
Total force to yield the mild steel:
Py
an
=A
P > Py, therefore, mild steel yields.
Let
fl
= force carried by mild steel.
P2 = force carried by tempered steel.
Fl = A10"1 = (1.00)(50 X103 ) = 50 X103 lb
Fl + P2 = P, Pi = p - Fl = 98 XI 03 ( 48 x I 0 3 )(14)
(a)
(b)
3
0.0309 in . ◄
(29 x I06 )(0. 75)
0-2
3
50 XI 0 = 4 8 XI 0 lb
P
48 x 103
= 2- = - - - 64xl03 psi
A2
0.75
64.0 ksi ◄
3
Unloading: <5' = PL = (98 x I0 )(l 4 ) = 0.02703 in.
6
EA (29 x I 0 )(1. 75)
(c)
<Sp = <5,,, - 5' = 0.03090 - 0.02703 = 0.003870 in .
0.00387 in. ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
2 19
PROBLEM 2.113
OD
1.7
111
r
o E
lm
I
C
0
0
The rigid bar ABC is supported by two links, AD and BE, of uniform
37.5 x 6-mm rectangular cross section and made ofa mild steel that is
assumed to be elastoplastic w ith E = 200 GPa and c,r = 250 MPa.
The magnitude of the force Q applied at B is gradually increased
from zero to 260 kN. Knowing that a = 0.640 m, determine (a) the
value of the nonnal stress in each link, (b) the maximum defl ection
of point B.
SOLUTION
'iMc = 0: 0.640(Q - P8 d- 2.64PAD = 0
Statics:
bA = 2.640, 58 = a0 = 0.6400
Defonnation:
Elastic analysis:
A = (37.5)(6) = 225 mm 2 = 225 x 10- 6 ni
PAD= EA bA = (200 X 109)(225 X 10-6 ) bA = 26.47 X 1065A
LAD
1.7
6
6
= (26.47 x I 0 )(2.640) = 69.88 x 10 0
CTAD = PAD = 3 10.6x l 0 90
A
n
9
6
EA s: _ (200 X 10 )(225 X 10- ) s: _
I 06 50
ua - - - - - - - - -ua - 45 x
ua
L8 E
1.0
6
6
= (45 X 10 )(0.6400) = 28.80 X 10 0
_
raE -
P,
c,8 E = __JJ_f__ = 128x l 0 90
A
2.64
From statics, Q = PaE + - - PAD = PaE +4.125PAD
0.640
= [28.80 X 10 6 + (4 .1 25)(69.88 X 10 6 )]0 = 3 J7.06 X 1060
0yatyielding of link AD:
6
9
a AD = ay = 250x 10 = 310.6 x !0 0
0y = 804 .89x l 0--o
Qy = (3 17.06 x I 0 6)(804.89 x 10-6 ) = 255.2 x 10 3 N
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. Th is document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
220
PROBLEM 2.113 (Continued)
(a)
Since Q = 260 x I 03 > Qy, lin k AD y ields .
a AD = 250 MPa ◄
PAD= Aay = (225 x 10- 6)(250 x 10- 6) = 56.25 x 10 3 N
From statics, P8 E = Q - 4 .125PAD = 260 x I 0 3
- ( 4 .1 25)(56.25
x I 03 )
3
PBE = 27.97 X10 N
a
= PBE = 27.97x l 03 = 124.3 x l 06 Pa
A
225 x 10-6
a 8 E = 124.3 MPa ◄
(27.97 X 103)(1.0)
= 62 1.53X Io- 6111
(200 x l09 )(225 x l0-6 )
68 = 0.622 111111 ..j,. ◄
BE
(b)
<5 = PnE LBE =
8
EA
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
22 1
PROBLEM 2.114
OD
1.7
111
Solve Prob. 2.11 3, know ing that a = 1.76 m and that the magnitude of
the force Q applied at B is gradually increased from zero to 135 kN.
r
o E
l m
I
C
0
0
PROBLEM 2.113 The rigid bar ABC is supported by two links, AD
and BE, of uniform 37.5 x 6-mm rectangular cross section and made
of a mild steel that is assumed to be elastoplastic with E = 200 GPa
and O'y = 250 MPa. The magnitude of the force Q applied at B is
gradually increased from zero to 260 kN. Knowing that a= 0.640 m,
determine (a) the value of the normal stress in each link, (b) the
maximum deflection of point B.
SOLUTION
'i.Mc = 0 : l .76(Q - P8£ ) - 2.64PAD = 0
Statics:
bA = 2.640, 58 = 1.760
Defonnation :
Elastic Analysis:
A = (37.5)(6) = 225 mm 2 = 225 x 10- 6 ni
PAD= EA bA = (200 X 109)(225 X 10-6 ) bA = 26.47 X 1065A
LAD
1.7
6
6
= (26.47 x I 0 )(2.64 0) = 69.88 x 10 0
O'AD = PAD = 3 10.6x l090
A
n
9
6
EA s: _ (200 X 10 )(225 X 10- ) s: _
I 06 50
va - - - - - - - - - va - 45 x
va
L8 £
1.0
6
6
= ( 45 X J0 )(J.760) = 79.2 X 10 0
_
r s£ -
P,
a 8 £ = ___ff,_ = 352x l 090
A
2.64
From statics, Q = PB£ + - - PAD = PB£ + l .500PAD
1.76
= [73.8 X 10 6 + (1.500)(69.88 X 10 6)]0 = J 78.62 X 10 60
0r at yielding of link BE:
0y = 7 10.23 X 10- 6
Qy = (178.62 x 106)(710.23 x 10- 6) = 126.86 x 103 N
(a)
Since Q = 135 x 103 N > Qy, link BE yields.
a 8£ = O'y = 250 MPa ◄
PBE = AO'y = (225 x 10-6 )(250 x 10 6) = 56.25 x 10 3 N
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. Th is document may not be copied, scanned, duplicated, forwarded, di stributed, or posted
on a website, in whole o r part.
222
PROBLEM 2.114 (Continued)
1
From statics, P AD= - -(Q 1.500
P8 £ ) =
a
= P AD=
AD
From elastic analysis of AD,
52.5 x 103 N
0=
A
52.5 x l03 = 233.3 x l06
225x 10-6
PAD
69.88 X 10
(b)
6
a AD= 233 MPa ◄
75 1.29x l0- 3 rad
68 =J.760 =J.322 X 10-3 111
5 8 = 1.322 mm-!- ◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
223
PROBLEM 2.115*
I
1.7 m
Solve Prob. 2.113, assuming that the magnitude of the force Q
applied at B is gradually increased from zero to 260 kN and then
decreased back to zero. Knowing that a = 0.640 m, determine (a) the
residual stress in each link, (b) the final deflection of point B .
Assume that the links are braced so that they can carry compressive
forces without buckling.
" D
1
" E
Li
C
0
,,_
~
2.64 1 1 1 ~
PROBLEM 2.113 The rigid bar ABC is supported by two links, AD
and BE, of uniform 37.5 x 6-mm rectangu lar cross section and made
of a mild steel that is assumed to be elastoplastic with E = 200 GPa
and O-y = 250 MPa. The magnitude of the force Q applied at B is
gradually increased from zero to 260 kN. Knowing that
a = 0.640 m, determine (a) the value of the normal stress in each
link, (b) the maximum deflection of point B.
SOLUTION
See solution to Problem 2. 11 3 for the normal stresses in each link and the deflection of Point B after loading.
6
o-AD
= 250x 10 Pa
o-8 £
= 124.3 x l 0 Pa
6
<58 = 62 1.53xl0-6 m
The elastic analysis given in the solution to Problem 2.1 13 appl ies to the unloading.
Q' = 3 17.06 X1060'
3
Q' =
Q
3 17.06xl06
260x l O
= 820.03x l 0-6
3 17.06 X106
o-~ 0
= 310.6 x 109 0 = (3 10.6 x 109 )(820.03 x 10- 6 ) = 254.70 x 10 6 Pa
o-~£
= 128 x I 09 0 = (128 x 109 )(820.03 x 10- 6 ) = 104.96 x 106 Pa
o~ = o.6400' = 524.82 x 1o-6 111
(a)
Residual stresses.
= -4.70MPa ◄
= 19.34 MPa ◄
(b)
bB.P = bB - b~ = 62 1.53 X10- 6 - 524.82 X10- 6 = 96.7 1X10-6 111
= 0.0967 mm J, ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
224
-
PROBLEM 2.116
B
A
A uniform steel rod of cross-sectional area A is attached to rigid supports and is
unstressed at a temperature of 45°F. The steel is assumed to be elastoplastic
w ith a y = 36 ksi and E = 29 x I 0 6 psi . Knowing that a = 6.5 x I 0..,,/°F,
detennine the stress in the bar (a) when the temperature is raised to 320°F,
(b) after the temperature has returned to 45°F.
'
SOLUTION
Let P be the compressive force in the rod.
Determine temperature change to cause yielding.
a L
PL
8 =- -+La(t,,.T) =- - Y
- + L a (t,,.T) y = 0
AE
E
3
36
(t,,.T) = 1 =
x !0
= 190.98°F
Y
Ea (29x l 0 6 )(6.5x 10- 6 )
But L'1T = 320 - 4 5 = 275°F > (L'1Ty)
(a)
a =-a y =-36.0 ksi ◄
Yielding occurs.
(t,,.T)' = 275°F
,,,
,,,
,,,
P' L
(
I
u =up=ur = - - + La t,,.T) =0
AE
P'
a' = - = - Ea(t,,.T)'
A
= - (29 X 106 )(6.5 X J0- 6 )(275) = - 5 J.8375 X 10 3 psi
(b) Residual stress:
O"res
=-
O"y
- a ' = - 36x 103 + 5 1.8375 x 103 = 15.84 x 10 psi
15.84 ksi ◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
225
PROBLEM 2.117
A
= 500 mm 2
A= 3CX) !Tlrn 2
A\
\
C
I
1-150
mm
The steel rod ABC is attached to rigid supports and is unstressed
at a temperature of 25°C. The steel is assumed elastoplastic, w ith
E = 200 GP a and a Y = 250 MPa. The temperature of both
portions of the rod is then raised to 150°C. Knowing that
a = 11.7 x 10-6 /°C, determine (a) the stress in both portions of the
rod, (b) the deflection of point C.
B
I
-L 250 __j
mm
SOLUTION
AAc = 500 x 10-6 m 2
LAc= 0 .1 50m
2
Lea = 0.250 m
Ac 8 = 300 x 10-6 m
Constraint:
Sp +Sr = 0
Detennine tiTto cause y ielding in portion CB.
A
- PLAc - Plea = LAa a(tiT)
EAAc EAca
i11-. ~1---1-.......-~
tiT = -P-(LAc + Lca J
LAa Ea AAc Ace
6
6
At yielding, P = Py = Ac8 ay = (300 x 10- )(2.50 x 10
(tiT) y = ____!t____( LAc + Lea
LAB Ea , AAc Ace
=
J
3
150°C - 25°C = l 25°C > (tiT) y
Yielding occurs. For t'lT> (tiT)y ,
P = Py = 75x 103 N
Py
75 x l01
a Ac = - AAc = - 500 x l0-6
a Ac = - 150.0 MPa ◄
Py
ace =---=-a y
a c 8 = - 250 MPa ◄
Ace
(b)
3
= 75 x 10 N
75 x !0
(
0.150 + 0.250 ) = 90 _812 oc
(0.400)(200 x 109 )(I u x Io-6 ) 500 x I o-6 300 x Io- 6
Actual t'lT:
(a)
)
B
C
For tiT > (tiT)y, portion AC remains elastic.
SCJA =- PyLAc + LAca(tiT)
EAAC
75 1
150
( x t )(O.
)
+ (0.150)(11.7 x 1o-6 )(125) = I 06.9 x I o-6 m
(200 X JO )(500 X JO- 6 )
Since Point A is stationary, be= bc!A = 106.9 x 10- 6 m
bc= 0. 1069 mm ➔ ◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
226
PROBLEM 2.118*
mni
A= 500
A= 300
A\
I
111111
Solve Prob. 2.11 7, assuming that the temperature of the rod is raised to
150°C and then returned to 25°C.
2
\
C
I.
B
I
,_ 150 J -250
111111
111111 _ ,
PROBLEM 2.117 The steel rod ABC is attached to rigid supports and is
unstressed at a temperature of25°C. The steel is assumed elastoplastic,
with E = 200 GPa and a- y = 250 MPa. The temperature of both portions
of the rod is then raised to 150°C. Knowing that a = 11 .7 x I o- 6/ °C,
determine (a) the stress in both portions of the rod, (b) the deflection
of point C.
SOLUTION
AAc = 500x!0-6 m 2
Constraint:
<Sp +<5r
LAc = 0.1 50 m
Les = 0.250 m
=0
A
Determine AT to cause yielding in portion CB.
- PLAc - PLcs
EAAC EACB
13
____,J1-----,--.----.~
i711-"
= LAaa(t!.T)
t!.T =
C
_P_(
LAc + Les
LAaEa AAc ACB
J
At yielding, P = Py = Acsa y = (300 x I 0-6 )(250 x I 0 6 ) = 75 x I0 3 N
t!.T
(
3
_ ___!3,_(LA c+ Lcs J 75 x l 0
(
0.150 + 0.250 )
}y - LAsEa AAc ACB - (0.400)(200x 10 9 )(11.7xl0-6 ) 500x !0-6 300x !0-6
= 90 .812 °C
A
Actual t!.T: 150°C - 25°C = 125°C > (t!.T) y
Yielding occurs. For t!.T > (t!.T}y ,
c.
B
3
P = Py = 75 x 10 N
Cooling:
(t!.T)' =
125
ac
P' = ELAsa(AT)'
L AC
(
9
(200 x I 0 )(0.400)(11.7 x I o-6 )(125)
~ + ___fil2Q__
Len )
A,~ + Acs
Residual force: P,., = P' - Py = I 03.235 x I 0 3
500xl0-6
-
3
103.235x10 N
300xl 0-6
75 x I 0 3 = 28.235 x I0 3 N
(tension)
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
227
PROBLEM 2.11 s• (Continued)
(a)
Residual stresses.
Pres
O"Ac = A Ac =
28.235 X I 03
500x 10- 6
O"Ac
28.235 X I 03
O"cs = Acs = 300x 10-6
P,e,
(b)
Permanent deflection ofpoint C.
= 56.5 MPa ◄
O"cs = 9.41 MPa ◄
be
=0.0424 mm ➔
◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
228
PROBLEM 2.119•
For the composite bar of Prob. 2.111, determine the residual stresses in
the tempered-steel bars if P is gradually increased from zero to
98 kips and then decreased back to zero.
3 .
)6111,
3 .
J61ll.
I.
1t
2 111.
PROBLEM 2.111 Two tempered-steel bars, each
in. thick, are
bonded to a ½-in. mild-steel bar. This composite bar is subjected as
shown to a centric axial load of magnitude P. Both steels are
elastoplastic with E = 29 x 106 psi 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 defonnation of the bar
reaches a maximum value 6111 = 0.04 in. and then decreased back to
zero. Determine (a) the maximum value of P, (b) the maximum stress
in the tempered-steel bars, (c) the permanent set after the load is
removed.
14 in.
2.0in.
SOLUTION
Mild steel:
½
A 1 = ( } 2) = 1.00 in2
3
2
Tempered steel: A 2 = (2)(i } 2) = 0.75 in
6
Total: A= A 1+ A 2 = 1.75 in2
Total force to yield the mild steel: an =
Py
A
P > P r; therefore, mild steel yields.
fj = force carried by mild steel
Let
Pi = force carried by tempered steel
fj = A 1a YI = (1.00)(50 x 103 ) = 50 x 103 lb
fj + }'2 = P,
Pi
f>i
= P - fj = 98 x I 03 - 50 x I 03 = 48 x 103 lb
48 X 10
0.75
3
3
•
a 2 = - = - - ~ = 64 x 10 psi
A2
Unloading.
a' =!_ = 98x l 03 = 56x 103psi
A
1.75
Residual stresses.
Mild steel:
a,,,e, = a, - a' = 50 XI03 - 56 XI0 3 = - 6 XI 0- 3psi = - 6 ksi
8.00 ksi ◄
Tempered steel:
PROPRI ETARY MA TERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
229
PROBLEM 2.120•
For the compos ite bar in Prob. 2.111 , determine the residual stresses in
the tempered-steel bars if P is gradually increased from zero until the
deformation of the bar reaches a maximum value 0 111 = 0.04 in. and is
then decreased back to zero.
14 in.
f6
PROBLEM 2.111 Two tempered-steel bars, each in. thick, are bonded
to a ½-in. mild-steel bar. This composite bar is subjected as shown to a
centric axial load of magnitude P . Both steels are elastoplastic
with E = 29 x I 0 6 psi and with yield strengths equal to I 00 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 <5,,, = 0.04 in. and then decreased back to zero. Determine (a) the
maximum value of P, (b) the maximum stress in the tempered-steel bars,
(c) the pennanent set after the load is removed.
SOLUTION
A,=(-1 )(2)= 1.00in 2
2
For the mild steel,
o
3
_LoYl _(l 4 )(5 0x l 0 )
E
Yl -
29 X I 06
-
0.024 138 in.
3
3
2
For the tempered steel, A2 =2( )(2) = 0.75 in
16
r5 2 - LoY2 _ (l 4 )(lOO x l 0 )
y
-
E
-
29 X 106
0.048276 in.
Total area:
Sn <<5,,, <Sn
The mild steel yields. Tempered steel is elastic.
Pi = A,Sr, = (1 .00)(50 x I 03) = 50 x I 031b
Forces:
6
P- _ EAA,, _ (29 x l0 )(0.75)(0.04)
2 L 14
Stresses:
(T2
_ Pi -_ 62. 14 x I0
-
A2
3
_ x l03lb
62 14
_ 82 . 86 X I 03p Sl.
-
0.75
112 14
c;' = !!___ =
· = 64.08x l 0 3psi
A
1.75
Unloading:
Residual stresses. c;1_,es =c;1 - c;' = 50x 103 - 64 .08 x 103 =-14 .08 x 103psi =- 14 .08 ksi
c;2,res = c;2 - c;' = 82.86 x 10 3 - 64 .08 x 103 = 18.78 x 10 3psi = 18.78 ksi
◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, di stributed, or posted
on a website, in whole or part.
230
PROBLEM 2.121 *
Narrow bars of aluminum are bonded to the two sides of a thick steel plate as
shown. Initially, at 'Fi_ =70°F, all stresses are zero. Knowing that the
temperature will be slowly raised to T2 and then reduced to T 1, determine
(a) the highest temperature T2 that does not resu lt in residual stresses, (b) the
temperature T2 that w ill result in a residual stress in the aluminum equal to
58 ksi. Assume a" =12.8 x I0- 6/°F for the aluminum and a., = 6.5 x 10-6/°F
for the steel. Further assume that the aluminum is elastoplastic, with
£ = 10.9x 106 ps i and O"y = 58 ksi. (Hint: Neglect the small stresses in the
plate.)
SOLUTION
Determine temperature change to cause yielding.
PL
<5= EA + L a 0 (AT)y = L a ., (AT)y
p
A
- = a- = - E(a0 - a , )(LI.T) y = - o-y
(AT)y =
(a)
.
O"y
E(aa - a,)
53 x l03
=844.62°F
(I 0.9 x l0 )(12.8 - 6.5)(1o-6 )
6
9 1 5°F ◄
T2 y = 'Fi_ + (!'iT) y = 70+844.62 = 9 15°F
After yielding,
<5 =
a- L
; + L a 0 (AT) = La .,(AT)
Cooling:
<5' = P'L + La (AT)' = La (AT)'
AE
"
-'
The residual stress is
Set
O"res
=- a-y
-o-y = o-y - E(aa - a , )(AT)
(b)
1759°F ◄
T2 = 'Fi_+AT = 70+ 1689 = 1759°F
IfT2 > l 759°F, the aluminum bar will most likely y ield in compression.
PROPRI ETARY MA TERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
23 1
PROBLEM 2.122•
c_____
•____. F
a = 120rnrn
Bar AB has a cross-sectional area of 1200 mnl and is made of a steel
that is asswned to be elastoplastic with £ =200 GPa and ery =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.
_J
e + - - - - - - 4 40111111
SOLUTION
A = 1200 111111 2 = 1200 X 10- 6 1112
Force to yield portion AC:
PAc =Aery= (1200 x 10-6 )(250 x 10 6 )
= 300xl03 N
Pei;
4.___ _-.._:::_➔-_ F_ __.~
For equilibrium, F + Pc8
-
PAC= 0.
Pen= PAC - F = 300x 10 3 - 520 x 10 3
3
=-220xl0 N
3
<5 = _ Pc8 Lc8 = (220 x 10 )(0.440 - 0.1 20)
c
EA
c200 x I o9 )(1200 x I o-6 )
=0 .29333 X 10- 3 111
3
er
en- -220
10
-P
- -x 6
cs -
A - 1200 x 1o-
=- 183.333 x 106 Pa
PROPRI ETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
232
PROBLEM 2.122• (Continued)
Unloading:
3
(520x 10 )(0.440 - 0.120) 378 _18 x l 03N
0.440
3
3
3
P~B = P~c - F = 378.18x l 0 - 520 x l 0 =-141.820 x 10 N
P~c =
Flcs
LAC+ Les
a ' = P~c = 378.1 8x 103 = 315.150x 106Pa
AC
A
1200 x 10- 6
,
P;c
14 1.820x 103 =-118.183x 106Pa
1200 x 1O""°
3
10 )(0.1 20) = 0_189090 x !O-J m
0 , = (378.1 8x
9
c (200xl0 )(1200x 10- 6)
a sc = A=
= 0.1042 mm ◄
(a)
(b)
O'AC,res = O"y - O'~c = 250 x 106 - 315.150 x 106 = - 65.150 x 106Pa
=-65.2 MPa ◄
O'cs, res = 0-cs - O'c;s = - 183.333 x 106 + 118.1 83 x 10 6 = - 65.150 x 106 Pa
=-65.2 MPa ◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
233
PROBLEM 2.123"
A
C
Solve Prob. 2.1 22, assuming that a = 180 mm.
2
a= 120 mm
>------- 440 mm
PROBLEM 2.122 Bar AB has a cross-sectional area of 1200 mm and
is made of a steel that is assumed to be elastoplastic with E = 200 GPa
and O"y = 250 MPa. Knowing that the force F increases from 0 to
520 kN and then decreases to zero, detennine (a) the permanent
deflection of point C, (b) the residual stress in the bar.
I
------.-1
SOLUTION
A = 1200 mm 2 = 1200x l 0- 6 m2
Force to yield portion AC:
PAC= Ao-y = (1200 x 10-6 )(250 x 10
6
)
3
= 300x l 0 N
For equilibrium, F + Pen - PAc = 0.
3
3
Pen= PAC - F = 300 x 10 - 520x 10
=-220 x l03 N
Ii _ Pc 8 Lc8
c -
_
- EA -
(220x l 03 )(0.440 - 0.1 80)
(200x J09 )(1200 x l 0- 6 )
= 0.23833 X I0- 3 m
O"
-
en -
Pen
-
220x l 0
3
A - - 1200 x 1o-
6
= - 183.333 x 106 Pa
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
234
PROBLEM 2.123• (Continued)
Unloading:
<5' _ P~c LAc __ P~aLca _ (F - P~c )Lea
c -
EA
-
EA
-
EA
= P' (LAC+ Lac)= Flea
AC
5
,
c
a'
AC
EA
EA
3
(520x 10 )(0.440 - 0.180) =
_ x l 0J N
307 27
0.440
= P~c - F = 307.27xl03 - 520 x l03 =-2 12.73xl03 N
_ (307 .27 x 103)(0.180)
_
x _3
0 23045 10 111
9
6
- c200 x 10 )(1200 x I o- )
P~c =
P~a
EA
Flea
LAc+ Lca
= P~c = 307.27 x I 03 = 256.058 x 106 Pa
A
1200 x 10-6
O~a= P~a = - 2 12.73x l03 =-177.275x l06 Pa
A
1200x l0-6
(a)
Oc, p = Oc - Oc = 0.23833 x 10- 3 - 0.23045 x 10- 3 = 0 .00788 x 10- 3111
= 0.00788 111111 ◄
(b)
a AC,res = a AC - O~c = 250 x 106 - 256.058 x 106 = - 6.0580 x 106 Pa
=-6.06 MPa ◄
O ca,res
6
= Oca - Oca = - 183.333 x 10 + 177.275 x 10
6
= - 6.0580 x 106 Pa
=-6.06 MPa ◄
PROPRIETARY MATERIAL, Copyright © 2015 McGraw-Hill Education, This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
235
PROBLEM 2.124
r-11
1-----i
The uniform wire ABC, of unstretched length 2/, is attached to the
supports shown and a vertical load P is applied at the midpoint B .
Denoting by A the cross-sectional area of the w ire and by E the
modulus of elasticity, show that, for i5 « I, the deflection at the
midpoint B is
C
B
p
i5 = /3/P
~AE
SOLUTION
Use approximation.
p _ P
Pl
AB - 2sin0 "'2i5
Elongation:
i5
- PABI
AB - AE
p/2
2AEi5
Deflection:
From the right triangle,
(I - /)AB )2 = , 2 + i52
2
Y + 2/iSAB + /)~B - I
= 2/i5AB ( 1+ ½l);s }="2/i5AB
6 =
p/3
:::,--
AEiS
t53 "' P/3
AE
◄
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
236
~r:
12 in. -
-
PROBLEM 2.125
The aluminum rod ABC ( E = I 0.1 x l0 6 psi), which consists of two
cylindrical portions AB and B C, is to be replaced w ith a cylindrical steel
rod DE ( E = 29 x 106 psi) of the same overall length. Dete rmine the
minimum required diameter d of the steel rod if its 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.
1.5 in.
t-= ~'~'"
18 in.
l
C
E
SOLUTION
Deformation of aluminum rod.
oA
= PLAB + PLsc
AAsE
AscE
J
= !_( LAB + Lac
El AAB Ase
3
28xl0 [ 12
18
= 10.lxl06 f (l.5) 2 + f (2.25)2
J
= 0.031376 in.
Steel rod.
o = 0.031376 in.
A= PL =
(28 x l 03)(30)
= 0.923 17 in2
Eo (29x l06 )(0.03 1376)
p
a =A
p 28 X 103
A =-=--1.1 6667 in 7
024 X 10 3
Required area is the larger value.
A= 1.16667 in 2
D iameter:
d=f?=
( 4)(1.16667)
d = 1.2 19 in. ◄
7!
PROPRIETARY MA TERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
237
I
PROBLEM 2.126
C
Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is
made of steel (£ = 29 x 106 psi), and rod BC of brass (E = 15 x 106 psi).
Determine (a) the total deformation of the composite rod AHC, (b) the
deflection of point B.
3 in.
30 in.
j
B
l t t
3<) kips
30 kips
40 in.
j
2in.
A
SOLUTION
Portion AB:
PAB
3
= 40xl0 lb
= 40 in .
d = 2 in.
LAB
=f(2)
= fd
E A8
= 29 x 106 psi
J
= P AB L 4B
AB
Portion BC:
2
A A8
2
= 3.1416 in
2
(40xl03)(40) = 17.5619x l 0- 3in.
(29x 106 )(3.1416)
-
E A8 A A8
P8 e =-20 x l 03 lb
L8 e
= 30 in .
d = 3 in.
1[
2
2
1[
· 2
A8 e= - d =- (3) = 7.0686m
4
4
6
•
E 8 e = 15x l0 psi
(- 20x l 03)(30)
- 5.6588x 10- 3in.
6
(15 X 10 )(7.0686)
(a)
b = bAB +b8e = 17.56 19 x l0-6 - 5.6588 x 10- 6
(b)
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
238
Brass strip:
E = 105 GPa
a = 20 X 10-6/'C
PROBLEM 2.127
The brass strip AB has been attached to a fixed
support at A and rests on a rough support at B.
Knowing that the coefficient of friction is 0.60
between the strip and the support at B , deten11ine
the decrease in temperature for which slipping
will impend.
SOLUTION
Brass strip:
E = l05 GPa
a = 20 X 10-6/oc
N =W
+
-r.Fr
= 0 : P - µ N =0
PL
6 = - - + La (t-.T)= 0
P = µ W =µmg
t-.T = _!__= µmg
EA
Data:
µ
EAa
EAa
= 0.60
A = (20)(3) = 60
m
= 100 kg
g
= 9.8 1 m/s 2
111111
2
= 60 X 10- 6 1112
E = l 05xl09 Pa
t-.T =
(0.60)(100)(9.81)
(J05 X 109 ) ( 60 X 10- 6 )(20 X 106 )
t-.T =4.67°C ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
239
PROBLEM 2.128
I½-in. diameter
P'
1-in. diarneter
I-½-in. diameter
3 in.
The specimen shown is made from a 1-in.-diameter cylindrical steel rod
with two 1.5-in.-outer-diameter sleeves bonded to the rod as shown.
Knowing lhal £ =29 x 106 psi, d t::Lt:rmim: (a) lht:: luau P SU lhal lht::
total deformation is 0.002 in., (b) the corresponding deformation of the
central portion BC.
2m'i
SOLUTION
(a)
L
- 1
P = E<5I--'(
A; )
tr
A =- d2
I
4 I
L, in.
d, in.
A, in2
LIA, in- 1
AB
2
1.5
1.7671
1.1 318
BC
3
1.0
0.7854
3.8197
CD
2
1.5
1.767 1
I.1 318
6.083
P = c29 x 106 )co.002)(6.083r 1 = 9.353 x 103 1b
(b)
<Sac= Plac = p Lac
AacE
EAac
=9.5 35 x
It
~ sum
P = 9.53 kips ◄
(3.8197)
29xl0
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
240
PROBLEM 2.129
11111
Each of the four vertical links connecting the
two rigid horizontal members is made of
aluminum ( E = 70 GPa) and has a uniform
rectangular cross section of 10 x 40 mm. For
the loading shown, determine the deflection
of (a) point E, (b) pointF, (c) point G.
400 111111
250111111
-------------
SOLUTION
Statics. Free body EFG:
'~ 'J..~G
i---------~.;........_ _ _ _ _ _ G
4-oo ,.,,,.,,
---;➔..,\~<'-- 15D
MM~
'24
kl\)
+)"i:.MF = 0 : - (400)(2F8 d - (250)(24) = 0
F8 E =-7.5 kN =-7.5 x 103 N
+)"i:.M E = 0:
(400)( 2FcF ) - (650)(24) = 0
FCF = 19.5 kN = 19.5 x l 03 N
Area of one link:
A = (10)(40) = 400 1111112
= 400 X 10- 6111 2
Length: L = 300 mm = 0 .300 m
Defonnations.
r5
3
_ F8 EL _ (- 7 .5 x l 0 )(0.300)
BE -
<5
CF
EA - (70 x !09 )(400 x !0-6 )
_
_
x !0_6111
80 357
3
= FCFL = (19.5 x 10 )(0.300) = Z0S_93 x !0- 6m
EA
(70 x I0 )(400 x 10-6 )
9
PROPRIETARY MATERIAL. C opyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
24 1
PROBLEM 2.129 (Continued)
6£ = 80.4 µm t ◄
= 15BF I
(a)
Deflection of Point E.
6E
(b)
Deflection of PointF.
6p = 6cp
6p = 209 µm
,.J,.
◄
6c =390 µ m
,.J,.
◄
Geometry change.
Let 0 be the small change in slope angle.
0 = 6£ + 6 p = 80.357 X I o-6 + 208.93 X I0- 6
L EF
0.400
(c)
Deflection of Point G.
6c
723.22 X 10- 6 radians
= 6 p + Lpc 0
6c = 6p + Lpc 0 = 208.93 x 10- 6 + (0.250)(723.22 x 10-6 )
= 389.73X 10-6 111
PROPRI ETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
242
PROBLEM 2.130
1·
4
A 4-ft concrete post is reinforced with four steel bars, each with a ¾-in. diameter.
Knowing that Es =29x l06 psi and Ec =3 .6x l0 6 psi, detennine the nonnal
stresses in the steel and in the concrete whe n a 150-kip axial centric force P is
applied to the post.
ft
SOLUTION
Ac = 82 - A., = 62.233 in 2
<5 = P, L =
P, ( 4 S)
= 0.93663xl0- 6 P
' A,Es (l.76715)(29 x 10 6 )
s
<5 = JJ,,L =
P,,(4 S)
c AcEc (62.233)(3.6x l 0 6 )
But <5, = J c : 0.93663 x 10-
6
0.2 1425 x !0--o~
P., =0.21425 x 10- 6 fJ,,
P., =0.22875~
Also,
Substituting ( I) into (2),
From(! ),
P_, + ~
l.22875~
( I)
=P =I 50 kips
= 150 kips
~
=122.075 kips
P,
= 0.22875(122.075) = 27.925 kips
C,
P.,
27.925
=--=---A.,
1.76715
s
C,
c
(2)
~
=- -
Ac
er,= - 15.80 ks i
◄
122.075
=-- - -
62.233
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
243
PROBLEM 2.131
The steel rods BE and CD each have a 16-mm diameter
(E =200 GPa); the ends of the rods are single-threaded with a
pitch of 2.5 mm. Knowing that after being snugly fitted, the nut at
C is tightened one fu ll turn, determine (a) the tension in rod CD,
(b) the deflection of point C of the rigid member ABC.
150mm
9
100 m m
J_
~
, m
---+-,
m
J
SOLUTION
Let 0 be the rotation of bar ABC as shown.
Then
58 = 0 .150
But
0 = 0.
C
pCD
tum
-
Oc = 0.250
PCDL CD
£
A
CD CD
E rn A cD
=--=-=---=-(
O.um -
Oc)
L CD
c.'
(200xl09 Pa) f (0.0 16 111)2
- - - - ~ - - - (0.0025 m - 0.250)
2m
R.,_
A
3
6
= 50.265 x l0 - 5.0265xl0 0
or
p~
B
PBE -_
EaEABE s:
U3
Lo£
9
P8 E
C.
(200 x 10 Pa) f (0.016 m)2
=- - - - - - - ' - - - -(0.150)
3m
6
=2.0 106x 10 0
Pc..1>
From free body of member ABC:
+) LMA = O: 0.l 5P8 E- 0.25PcD = 0
0.15(2.0106 X 1060) - 0.25(50.265 X 103 - 5.0265 X 1060) = 0
0 =8 .0645 x 10- 3 rad
(a)
Pco
=50.265 x I 03 =9.7288 X 103
(b)
Oc
5.0265 x 106 (8.0645 x 10- 3 )
Pm = 9.73 kN ◄
N
=0.250 =0.25(8.0645 x 10- 3 )
oc =2.02 mm ~
= 2.016 l x l0- 3 m
◄
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
244
PROBLEM 2.132
The assembly shown consists of an alwninwn shell (£" = 10.6 x 106 psi,
a,, = 12.9 x 10-6/°F) fully bonded to a steel core (£ , = 29 x 106 psi,
a:, = 6.5 x I 0-6!°F) and is unstressed. Detem1ine (a) the largest allowable
change in temperature if the stress in the aluminum shell is not to exceed
6 ksi, (b) the corresponding change in length of the assembly.
Steel
l.25inv
core
SOLUTION
Since aa > a,, the s hell is in compression for a positive temperature rise.
3
Let
c,a =-6ksi=-6x l 0 psi
where Pis the tensile force in the steel core.
CY
'
=- c,aAa = (6x l 03)(0.78540) = 10.667x l 03 si
A,
0.44 179
p
(a - a )(1'1.T) =
a
'
CY., - <:Ya
E
'
E
a
10 667 103
6 103
(6.4 x 10-6 )(1'1.T) = ·
x
+
x
6
29 X 10
I 0.6 X 106
(a)
t'1.T = l45.91°F
(b)
C
=
0.93385 x 10- 3
1'1.T = 145.9°F ◄
3
10 667
3
·
X ~0 + (6.5 X 10-6 )(145.9 1) = 1.3163 X 1029 x l0
or
c=
- 6 X 103
10.6x l06
+(12.9x l0-6 )(145.9 1) = 1.3 163x !0-3
b = Le = (8.0)(1.3 163 x 10- 3 ) = 0.0 1053 in.
5= 0.01053 in. ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
245
PROBLEM 2.133
The plastic block shown is bonded to a fixed base and to a horizontal
rigid plate to which a force P is applied. Knowing that for the
pla~lit: u~t:d G = 55 k~i, dt:Lt:rmint: tht: ddkclion of lht: plalt:
when P = 9 kips.
SOLUTION
Consider the plastic block. The s hearing force carried is P
= 9 x 103 lb
The area is A = (3 .5)(5.5) = 19.25 in 2
Shearing stress:
T
Shearing strain:
r
But
p
= -
A
=
T
0
0
r =--;;
9 X 103
.
= - - = 467.52 pSl
19.25
=
467 .52
x
= o.0085006
55 103
o
= hy = (2.2)(0.0085006)
o = 0.01 870 in.
◄
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
246
PROBLEM 2.134
~
The aluminum test specimen shown is subjected to two equal and opposite
centric axial forces of magnitude P. (a) Know ing that E = 70 GPa
aml <Tall =200 MPa, dt::lt::rmim: tht:: maximum alluwabk valut:: of P and tht::
corresponding total e longation of the specimen. (b) Solve part a, assuming that
the specimen has been replaced by an a luminum bar of the same length and a
uniform 60 x 15-mm rectangular cross section.
150
t--
300
C
60
r
=6
150
~ 75
P'
Dimensions in mm
SOLUTION
6
<Tan = 200xl0 Pa
9
£ = 70x!0 Pa
Amin = (60 mm)( l 5 nun) = 900 mm 2 = 900x 10- 6 m 2
(a)
D = 75 111111, d = 60 nun, r = 6 nm1
Test specimen.
..'.:..=~= 0.1 0
d
p
K = 1.95
From Fig. 2.60b,
60
(Tmax
=KA
P = Ao-max= (900xl0-6) (200xl06)
K
1.95
92.308x l 03 N
P = 92.3 kN ◄
Wide area .4' = (75 mm)(15 nun) = I 125 nun 2 = l.125x 10- 3 m 2
3
o = I P;L; = P I!:i_ = 92.308 x l0 [
~E;
E
A;
9
70x!0
0.150
+ 0.300 + 0.150 ]
1. 125x l 0- 3 900xl0- 6 1.125x l0- 3
0 = 0.79 1 111111 ◄
= 7.9 l x l0-6 m
(b)
Uniform bar.
6
3
P = Ao-all = (900 XI 0-6 )(200 XI 0 ) = 180 X10 N
3
o = PL = (180x l 0 )(0.600)
AE
(900 x 10- 6)(70 x I 0 9 )
l. 7 ! 4 x!0-3m
P = l 80.0 kN ◄
o=1.7 14 mm
◄
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
247
PROBLEM 2.135
, . . . . _ - - - - L - - - ----<
C
~p
I
I
I
B~
~-=---,
V
V
V
V
p
The unifom1 rod BC has a cross-sectional area A and is made of a
mild steel that can be assumed to be elastoplastic with a modulus of
dastk ity E anti a yidtl sln::nglh U y - Using lht:: blut:k-antl-spring
system shown, it is desired to simulate the deflection of end C of the
rod as the axial force P is gradually applied and removed, that is, the
deflection of points C and C' should be the same for all values of P.
Denoting by µ the coefficient of friction between the block and the
horizontal surface, derive an expression for (a) the required mass m
of the block, (b) the required constant k of the spring.
SOLUTION
Force-deflection diagram for Point C or rod BC.
For
P< Py= Auy
PL
Sc= -
EA
P =-
EA
pmax
L
Sc
= Py= Auy
Force-deflection diagram for Point C' of block-and-spring system.
+ h:Fv = 0:
..±..+- I.Fx = 0 :
If block does not move, i.e., F1 < µ N
= µmg
N - mg = 0
P - F1
or
<5' =.!:...
K
then
=0
P < µmg,
or
C
If P = µmg, then slip at P = F,,,
N = mg
= µmg occurs.
If the force P is the removed, the spring returns to its initial length.
(a)
Equating Py and Fmax,
(b)
Equating slopes,
Auy
m =--
Auy = µmg
µg
k = EA
◄
◄
L
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
248
PROBLEM 2.C1
A rod cons isting of n elements, each of which is homogeneous and
of uniform cross section, is subjected to the loading shown. The length
uf d e 1m:nl i b tlenule tl by L;, il8 c.:ru88-8t:c.:liunal a rea by A; , mutlulu8 u f
elasticity by E;, and the load applied to its right end by P;, the magnitude
P; of this load being assumed to be positive if P; is directed to the right
and negative otherwise. (a) Write a computer program that can be used
to determine the average normal stress in each element, the deformation
of each element, and the total defonnation of the rod. (b) Use this
program to solve Probs. 2 .20 and 2.1 26.
SOLUTION
For each element, enter
Compute deformation
U pdate axial load
P = P +P;
Compute for each element
ai
= PI A
'
Total deformation:
Update through n elements
Program Outputs
Problem 2.20
Element
2
Stress (MPa)
Deformation (mm)
19.0986
0.1091
- 12.7324
- 0.0909
0.01 82 111111
Total Deformation =
Problem 2.1 26
Element
Stress (ks i)
2
Deformation (in.)
12.7324
0.0 176
- 2.8294
- 0.0057
0.01 190 in.
Total Deformation =
PROPRIETARY MA TERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
249
PROBLEM 2.C2
Rod AB is horizontal with both ends fixed; it consists of n elements, each
of which is homogeneous and ofunifonn cross section, and is subjected to
the loading shown. The length of element i is denoted by L;, its crosssectional area by A;, its modulus of elasticity by E; , and the load applied
to its right end by P;, the magnitude P; of this load being assumed to be
positive if P; is directed to the right and negative otherwise. (Note that
P 1 = 0.) (a) Write a computer program which can be u sed to determine the
reactions at A and B, the average normal stress in each element, and the
deformation of each element. (b) Use this program to solve Probs. 2.41
and 2.42.
A
SOLUTION
We Consider the reaction at B redundant and release the rod at B
Compute b 8 with R8
=0
For each element, enter
Update axial load
P = P+P;
Compute for each element
er; =PIA;
b; = PL/ A;E;
Update total deformation
Compute b8 due to unit load at B
U nit
Unit
CF;
= l!A;
b; = L/ A;E;
Update total unit deformation
Unit b 8
= Unit b 8 + Unit b;
Superposition
For total displacement at
B=O
Solving:
R8
Then:
= -b8 /Unit b8
RA= I:P; + Re
PROPRIETA RY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
250
PROBLEM 2.C2 (Continued)
For each element
a = a; + Ra
Unit a;
o=o; + Ra
Unit o;
Program Outputs
Problem 2.41
RA = - 62.809 kN
RB = - 37.1 91 kN
Element Stress (MPa) Deformation (mm)
- 52.6 15
- 0.05011
2
3.974
0.00378
3
2.235
0.001 34
4
49.982
0.04498
Problem 2.42
RA = -45.479 kN
RB =-54.52 1 kN
E lement Stress (l\,fPa) Deformation (mm)
-77. 131
- 0.03857
2
- 20.542
- 0.01027
3
- I 1.555
- 0.01 32 1
4
36. 19 1
0.06204
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
251
PROBLEM 2.C3
Element n
Ele ment l
~
-t°"i-
r
' -r------._
,
[
Rod AB consists of n elements, each of which is homogeneous and of
unifonn cross section. End A is fixed , while initially there is a gap o0
between end B and the fixed vertical surface on the right. The length of
element i is denoted by L;, its cross-sectional area by A; , its modulus of
elasticity by E;, and its coefficient of thermal expansion by a;, After the
temperature of the rod has been increased by t,,.T, the gap at B is closed
and the vertical surfaces exert equal and opposite forces on the rod.
(a) Write a computer program which can be used to determine the
magnitude of the reactions at A and B, the normal stress in each element,
and the deformation of each element. (b) Use this program to solve
Probs. 2.59 and 2.60.
SOLUTION
We compute the displacements at B.
Assuming there is no support at B,
enter
Enter temperature change T. Compute for each element.
Update total deformation.
Compute On due to unit load at B.
Update total unit deformation.
Unit On
= Unit On+ Unit o;
Compute reactions.
From superposition,
Then
For each element,
=-RnlA;
O; = a;L;T + Rnl;IA;E;
CY;
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
252
PROBLEM 2.C3 (Continued)
Program Ou tputs
Problem 2.59.
R = 52.279 kips
Element
2
Stress (ksi)
Deformation (10* - 3 in.)
- 21.783
9.909
- 18.67 1
10.091
Problem 2.60.
R = 232.390 kN
Element
2
Stress (MPa)
Deformation (microm)
- 11 6.195
363.220
- 290.487
136.780
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
253
PROBLEM 2.C4
Bar AB has a length L and is made of two different materials of given
cross-sectional area, modulus of elasticity, and yield strength. The bar is
subjected as shown to a load P that is gradually increased from zero until
the deformation of the bar has reached a maximum value o,,, and then
decreased back to zero. (a) Write a computer program that, for each of 25
values of 0111 equally spaced over a range extending from O to a value
equal to 120% of the deformation causing both materials to yield, can be
used to determine the maximum value P,,, of the load, the maximum
normal stress in each material, the permanent deformation op of the bar,
and the residual stress in each material. (b) Use this program to solve
Probs. 2.111 and 2.1 12.
SOLUTION
Note: The following assumes
Displacement increment
Displacements at yielding
For each displacement
If
0,,, < oA:
a 1 == o,,,E/ L
a 2 == o,,,£2 /L
P,,, == (0 /L) (A 1£ 1 + A2E2 )
111
If
oA< 0,,, < Oa:
0"1 == ( a r )1
a 2 == o,,,£ 2 /L
P,,, == A1a 1 + (o,,,/L)A2 E 2
If
o,,, > 08 :
0"1 == (ar )1
a 2 == (ay )2
P,,, == A1 a 1 + A2a 2
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
254
PROBLEM 2.C4 (Continued)
P,,,
Permanent deformations, residual stresses
Slope of first (elastic) segment
S lope = (A1£ 1 + A2 E 2 )/L
Sp =<5111 - (P,,,/Slope)
I
(0-1\e, = a-1-(E 1P,,,l(L Slope))
5LOPE
(a-2 \es =a-2 - (E2 P,,,l(L Slope))
~I
Program Outputs
Problems 2. 111 and 2. 112
DM
10** - 3 in.
0.000
2.4 14
4.828
PM
kips
0.000
8.750
17.500
SIGM (I)
ksi
SIGM (2)
ksi
DP
10** - 3 in.
SIGR ( I)
ksi
SIG (2)
ksi
0.000
5.000
10.000
0.000
5.000
10.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
7.241
26.250
15.000
15.000
0.000
0 .000
0.000
9.655
12.069
14.483
20.000
25.000
30.000
0.000
0.000
0.000
0.000
0.000
0.000
35.000
20.000
25.000
30.000
35.000
0.000
0.000
0.000
16.897
35.000
43 .750
52.500
6 1.250
0.000
0.000
0.000
19.3 10
21.724
70.000
78.750
40.000
45.000
40.000
45.000
0.000
0.000
0.000
0 .000
0.000
0.000
24.138
26.552
87.500
9 1.250
50.000
50.000
50.000
55.000
0.000
1.379
0.000
- 2.143
0.000
2.857
28.966
95.000
50.000
60.000
2.759
-4.286
5.714
31.379
98.750
50.000
65.000
4 .1 38
- 6.429
8.57 1
33.793
36.207
102.500
106.250
50.000
50.000
70.000
75.000
5.517
6.897
- 8.571
- 10.7 14
11.429
14.286
38.621
110.000
50.000
80.000
8.276
- 12.857
17. 143
4 1.034
113.750
50.000
85.000
9.655
- 15.000
20.000
43.448
117.500
50.000
90.000
11.034
- 17 .143
22. 857
45 .862
12 1.250
50.000
95.000
12.414
- 19.286
25.714
4 8.276
125.000
50.000
100.000
13.793
- 2 1.429
28.571
50.690
125.000
50.000
100.000
16.207
- 2 1.429
28.571
53.1 03
125.000
50.000
100.000
18.62 1
- 2 1.429
28.57 1
55.517
125.000
50.000
100.000
2 1.034
- 2 1.429
28.571
57.93 1
125.000
50.000
100.000
23.448
- 2 1.429
28.57 1
fm
2. 11 2
◄
2. 111
◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
255
PROBLEM 2.CS
The plate has a hole centered across the width. The stress concentration factor
for a flat bar under axial loading with a centric hole is
ta½,.
..... Ft
P'
•
I
l
D•
....
p
K = 3.00 - 3. l3(~) +3 .66(~J - 1.53 (~J
where r is the radius of the hole and Dis the width of the bar. Write a computer
program to determine the allowable load P for the given values of r, D, the
thickness t of the bar, and the allowable stress c,011 of the material. Knowing
that t = ¼in., D = 3 .0 in., and c,011 = 16 ksi, detennine the allowable load P for
values of r from 0.125 in. to 0.75 in., using 0.125 in. increments.
SOLUTION
Enter
r, D, t,
c,311
Compute K
RD = 2.0r/D
K = 3.00 - 3. l 3RD + 3.66RD 2 - l .53RD3
Compute average stress
Allowable load
P,.11 = CYave(D - 2.0r)t
Program Output
Radius
(in.)
Allowable Load
(kips)
0. 1250
3.9802
0.2500
3.8866
0.3750
3.7 154
0.5000
3.4682
0.6250
3.1523
0. 7500
2.7794
PROPRIETARY MATERIAL. Copyright © 20 15 McGraw- Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
256
PROBLEM 2.C6
A solid truncated cone is subjected to an axial force P as shown. The exact
elongation is (PL) /(21rc 2E). By replacing the cone by n circu lar cylinders
I
uf i::qual thkkni::ss, wr iti:: a cumpuli::r program thal can b i:: usi::d Lu calculalt:
I
2c:
the elongation of the truncated cone. What is the percentage error in the
answerobtainedfromtheprogramusing(a) n= 6, (b) n = l 2, (c)n = 60?
I
I
SOLUTION
For
i = l ton:
L; = (i + 0 .5)(Lln)
,; = 2c-c(LJ L)
Area:
Displacement:
o= o+ P(Lln)l (AE)
Exact displacement:
()exact
= PL/(2.01rc2 E)
Percentage error:
Percent = I 00(0 -
Oexact ) / ()exact
Program Output
Approximate
Exact
6
0.15852
0. 159 15
- 0.40083
12
U.1 5899
U.1 59 15
- 0. IUIUU
60
0.1 59 15
0.159 15
- 0.00405
n
Percent
PROPRI ETARY MATERIAL. Copyright © 2015 McGraw-Hill Education . This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
257