ch03

CHAPTER 3
PROBLEM (3.1) A rigid bar BDE is supported by two links AB and CD as shown in Fig. P3.1.
After load P is applied, point E moves 2.4mm downward and the axial strain in bar AB equals -500  .
What is the axial strain in bar CD?
SOLUTION
The change in length of bar AB is
 AB   AB LAB  (500 106 )(800)  0.4 mm
From triangles B’BF and EE’F:
500 mm
1000 mm
0.4
2.4

,
x  214.3 mm
x 1500  x
B’
F
D
0.4 mm
From triangles DD’F and EE’F:
E
B
285.7 1285.7
D

,
 D  0.533 mm
D’

2.4
D
2.4 mm
Thus

0.533
x
 CD  D 
 533 
E’
LCD 1000
________________________________________________________________________
PROBLEM (3.2) A spherical balloon changes its diameter from 200 to 201 mm when
pressurized. Determine the average circumferential strain.
SOLUTION

 d f   d0
1

 5000 
 d0
200
________________________________________________________________________
PROBLEM (3.3) A hollow cylinder is subjected to an internal pressure that increases its 200
mm inner diameter by 0.5 mm and its 400 mm outer diameter by 0.3 mm. Calculate:
(a) The maximum normal strain in the circumferential direction.
(b) The average normal strain in the radial direction.
SOLUTION
2 (r  r )  2 r r

2 r
r
0.25
( c ) max  ( c )i 
 2500 
100
0.15
( c )o 
 750 
200
(a)  c 
3-1
(b)  
ro  ri 0.25  0.15

 1000 
ro  ri
200  100
________________________________________________________________________
PROBLEM (3.4) A prismatic bar of length L is subjected to an axial load P, as shown in Fig.
P3.4. Calculate the maximum strain  x , if the displacement along the member varies as follows:
3
2
(a) u = ( x /L) 10 .
3
(b) u = L( 10 ) sin (
 x/2L).
SOLUTION
x2
du
x
x 

1000 L
dx 500 L
At x=L:
( x ) max  2000 
(a) u 
L
x
sin
1000
2L
At x=0:
x 
(b) u 
( x ) max 

2000
du

x

cos
dx 2000
2L
 1570 
________________________________________________________________________
PROBLEM (3.5) A horizontal rod AB is supported and loaded by a force P as seen in Fig.
P3.5. Determine the permissible normal strains in the wires CE and DF, if the allowable vertical
displacement of end B is 1/8 in.
Assumption: Rod AB is rigid.
Geometry:
 CE  DF 0.125


40
60
80
Solving,
 CE  0.0625 in.
 DF  0.09375 in.
F
Normal Strains:
E
 CE 0.0625
 CE 

 1786 
LCE
35
50 in.
35 in.

0.09375
 DF  DF 
 1875 
LDF
50
A
40 in. C 20 in.D 20 in.B
B
CE
Figure (a)
3-2
D
F
________________________________________________________________________
PROBLEM (3.6) The structure shown in Fig. P3.5 consists of a horizontal rod AB supported by
two vertical wires (CE and DE) and by a pin at A. What is the the allowable vertical displacement
of the end B, if the permissible normal strain in each wire is  all = 1500  ?
Assumption:
Rod AB is rigid.
SOLUTION
Geometry [see: Figure (a) of Solution 3.5]:
 CE  DF  B


40
60 80
Normal Strains:
 CE   all LCE  0.0015(35)  0.0525 in.
8
 B   CE  2(0.0525)  0.105 in.
4
Also
 DF   all LDE  0.0015(50)  0.075 in.
8
4
 B   DF  (0.075)  0.1 in.   all
6
3
F
E
50 in.
35 in.
A
C
D
40 in.
20 in.
20 in.
B
________________________________________________________________________
PROBLEM (*3.7) The assembly of the strut BC and rod AB is used to support a vertical load
P as depicted in Fig. P3.7. Determine:
(a) The normal stresses  AB and  BC in the rod and strut.
(b) The normal strain
(c) The normal strain
Given:
 AB , if the rod elongates 0.05 in.
 BC , if the strut shortens 0.025 in.
2
P = 3.5 kips. The cross-sectional areas of the members: ABC = 0.25 in. and
ABC =0.4 in.2
*SOLUTION
We have
LAB  602  452  75 in.
LBC  60  25  65 in.
2
A
FAB
2
Equilibrium:
4
12
 Fx  0 :  5 FAB  13 FBC  0
3
5
 Fy  0 : 5 FAB  13 FBC  3.5  0
3-3
45
3
60
4
B
13
25
C
5
5
FBC
12
P
Solving,
FBC  3.25 kips (C )
FAB  3.75 kips (T )
3.75
3.25
 15 ksi
 BC 
 8.125 ksi
0.25
0.4
0.05

 667 
75
0.025

 385 
65
(a)  AB 
(b)  AB
(c)  BC
________________________________________________________________________
PROBLEM (3.8) As a result of loading, the thin 40 mm by 20 mm rectangular plate of Fig. P3.8
deforms into a parallelogram in which sides AB and CD elongate 0.005 mm and rotate 1200  rad
clockwise, while sides AD and BC shorten 0.002 mm and rotate 400  rad counterclockwise.
Calculate the strain components in the xy plane.
SOLUTION
u 0.002

 50 
x
40
v 0.005
y 

 250 
y
20
v u
 xy 

 400  1200  1600 
x y
x 
y
B
40 mm
C
20 mm
A
D
x
________________________________________________________________________
PROBLEM (3.9) Solve Prob. 3.8, assuming that sides AD and BC elongate 0.001 mm and
rotate 1600  rad clockwise and the other sides have the same extension and rotation.
SOLUTION
u 0.001

 25 
x
40
v 0.005
y 

 250 
y
20
v u
 xy 

 1600  1200  400 
x y
x 
3-4
________________________________________________________________________
PROBLEM (3.10) A thin 8 in. by 6 in. rectangular plate (see Fig. P3.10) is acted upon by a
biaxial tensile loading resulting in the uniform strains  x = 600  and  y = 400  . Calculate the
change in length of diagonal AC.
SOLUTION
y
8 in.
B
LAB  400(10 6 )(6)  2.4 103 in.
C
10 in.
LAD  600(10 6 )(8)  4.8 103 in.
6 in.
A
D
L2AC  L2AB  L2AD
x
2 LAC LAC  2 LAB LAB  2 LAD LAD
or
L
L
LAC  AB LAB  AD LAD
(1)
LAC
LAC
Substituting the numerical values:
6
8
LAC  [ (2.4)  (4.8)]103  5.28 103 in.
10
10
________________________________________________________________________
PROBLEM (3.11) Redo Prob.3.10, with the plate in biaxial compression for which  x = -200 
and  y = - 100  .
SOLUTION
LAB  100(106 )6  0.6 103 in.
LAD  200(106 )8  1.6 103 in.
Using Eq. (1) of Solution of Prob. 3.8:
6
8
LAC  [ (0.6)  (1.6)]103  1.64 103 in.
10
10
________________________________________________________________________
PROBLEM (3.12) Determine the normal strain in the members AB and CB of the pinconnected plane structure shown in Fig. 3.12 if point B moves leftward 3 mm, after load P is applied.
Assumption: Axial deformation is uniform throughout the length of each member.
SOLUTION
LAB
3

LAB
1500
 2000 
C
 AB 
2.5 m
2m
LBC

A
3 mm
B
 BC 
1..5 m
3-5
LBC 0.003cos 

LBC
2.5
0.003(1.5 2.5)
 720 
2.5
________________________________________________________________________
PROBLEM (3.13) The shear force V deforms plate ABCD into AB’C’D (Fig. P3.13).

Determine the shear strain in the plate:
(a) At any point.
(b) At the center.
(c) At the origin.
Given: b = 200 mm, h = 0.5 mm
SOLUTION
(a)
y

dy
or
dx
x
dx 2h
2(0.0005)
y
 2 y
(0.2) 2
dy b
  0.025 y
(1)
(b) From Eq. (1) at y  b 2 :   0.025(0.1)  2500 
(c) From Eq. (1) at y  0 :   0
________________________________________________________________________
PROBLEM (3.14) Redo Prob. 3. 13 for the case in which curves AB’ = DC’ are straight
lines.
Given:
b = 250 mm, h = 0.4 mm
SOLUTION
h 0.4

 1600 
b 250
0.2
 1600 
(b)  
125
(c)   1600 
________________________________________________________________________
PROBLEM (3.15) A 100 mm by 100 mm square plate is deformed into a 100 mm by 100.2
(a)  
mm rectangle as shown by the dashed lines in Fig. P3.15. Determine the positive shear strain
between its diagonals.
SOLUTION
100 mm
B
0.2 mm
C



100 mm
2


 2 tan 1
2
 1998 
 2
A

D
3-6
100
100.2
________________________________________________________________________
PROBLEM (3.16) A square plate is subjected to the uniform strains  x = -500  ,  y =
500  , and
 xy = 0. Calculate the negative shearing strain between its diagonal s.
SOLUTION
-0.0005L
0.0005L


L

 2

2


 2 tan 1
2
 1000 
L
1.0005 L
0.9995 L
________________________________________________________________________
PROBLEM (*3.17) When loaded, the 400 mm by 400 mm square plate of Fig. P3.17
deforms into a shape in which diagonal BD elongates 0.2 mm and diagonal AC contracts 0.4
mm while they remain perpendicular and side AD remains horizontal. Calculate the average
strain components the xy plane.
*SOLUTION
y
B'
AC  BD  4002  4002  565.69 mm
C'
A'

'
D'
x
A ' C '  565.69  0.4  565.29 mm
B ' D '  565.69  0.2  565.89 mm
Geometry: A ' B '  A ' D '
Thus,
 565.29  2  565.89  2 

 
 
A ' B ' AD  2   2  

x  y 
400
AD
 0.0001685  169 


565.89 2
 xy    '   2 tan 1
2
2
565.29 2
 0.001061  1061 
1
2
 400
________________________________________________________________________
PROBLEM (3.18) A 15 in. by 20 in. rectangular sheet of plastic is loaded in its own plane.
After loading the sheet distorts into a shape A’B’C’D’ as shown by the dashed lines in Fig.
P3.18. Determine the normal strains occurring along the diagonals AC and BD.
SOLUTION
Refer to Figure P3.18.
Geometry:
3-7
AC  BD  202  152  25 in.
AC '  (20.12)2  (15.1)2  25.156 in.
B ' D '  (20.02)2  (15)2  25.016 in.
Normal strains:
AC ' AC 25.156  25
 AC 

 6240 
AC
25
B ' D ' BD 25.016  25
 BD 

 640 
BD
25
________________________________________________________________________
PROBLEM (3.19) Reconsider the plastic sheet that is initially rectangular (Fig. P3.18).
Subsequent to loading the sheet deforms into a shape as indicated by the dashe d lines in the
figure. Calculate the shear strains at the corners A, B, C, and D.
SOLUTION
y
0.12 in.

Geometry (for small angles):
0.05
0.05 in.

 2, 491  rad

B 
C
20.07
0.05

 3,322  rad
15 in.

15.05
0.05 in
A
Shear strains:
x
20 in.
D
( xy ) A  ( xy )C      2, 491  3.222
0.07 in
 5,813  rad
( xy )B  ( xy )D  (   )  5,813 rad
________________________________________________________________________
PROBLEM (3.20) The pin-connected structure ABCD is deformed into a shape AB'C'D, as
0.05 in
shown by the dashed lines in Fig. P3.20. Calculate the average normal strains in member BC and
AC.
SOLUTION
A
6 ft
45
o
B
B’
6 ft
 72 ft
C"
L AC
C
C'
CC '  5 64 in.
BB '  5 32
L
5 64  5 32
 BC  BC 
LBC
6  12
 1085 
L
 AC  AC
LAC

(5 64) cos 45o
 543 
72 12
________________________________________________________________________
3-8
PROBLEM (3.21) Solve Prob. 3.20, assuming that member BC moves 3/16 in. down as a
rigid body and remains vertical - that is, BB' = CC' = 3/16 in.
SOLUTION
Refer to Solution of Prob. 3.20.
L
0
 BC  BC 
0
LBC
LBC
 AC 
LAC (3 16) cos 45o

 1302 
LAC
72 12
________________________________________________________________________
PROBLEM (3.22) The handbrakes on a bicycle consist of two blocks of hard rubber
attached to the frame of the bike, which press against the wheel during stopping (Fig. P3.22a).
2
Assuming that a force P causes a parabolic deflection (x = ky ) of the rubber when the brakes
are applied (Fig. P3. 22b), determine the shearing strain in the rubber.
SOLUTION
y

x  ky 2
x=ky2
a
x
We have
dx  2kydy  2

a2
  ka 2
k   a2
ydy
dx

 2( 2 ) y
dy
a
________________________________________________________________________
PROBLEM (*3.23) The thin, triangular plate ABC shown in Fig. P3.23 is uniformly
Thus,  
deformed into a shape A’B’C’. Calculate:
(a) The plane stress components  x ,  y , and
 xy .
(b) The shear strain between edges AC and BC.
*SOLUTION
2.4
 1200 
2000
1.5
y 
 1500 
1000
 xy  0
(a)  x 
(b)
ACB  90o
3-9
1.0012
 90.1547 o
0.9985
  90o  90.1547o  0.1547o  2.7 103 rad
 2700 
________________________________________________________________________
PROBLEM (3.24) Redo Prob. 3.23 for the case in which the plate ABC is uniformly
A ' C ' B '  2 tan 1
deformed into a shape ABC'.
SOLUTION
(a)  x  0
 xy  0
1.5
 1500 
1000
(b) ACB  90o
1
AC ' B  2 tan 1
 90.086o
0.9985
o
o
  90  90.086  0.086o  1.5 103 rad
 1500 
y  
________________________________________________________________________
PROBLEM (3.25) The stress-strain curves for a structural steel bar are shown in Fig. P3.25. Note
that, the entire diagram and its initial portion are plotted using a strain scale N and an enlarged strain scale
M in the figure, respectively. Determine:
(a) The strains at yield point and fracture of the material.
(b) The % elongation of the bar for a 50-mm gage length.
SOLUTION
(a)  y  0.0014  1400 
 f  0.28  280,000 
(b) L f  50  50(0.28)  64 mm
64  50
(100)  28 %
50
________________________________________________________________________
PROBLEM (3.26) A 10 mm by 10 mm square ABCD is drawn on a member prior to loading.
% Elongation=
After loading, the square becomes the rhombus shown in Fig. P3.26. Determine:
(a) The modulus of elasticity.
(b) Poisson's ratio.
SOLUTION
LAC  LBD  102  102  14.14 mm
L
14.11  14.14
 x  AC 
 2122 
LAC
14.14
3-10
x 
LBD 14.15  14.14

 707 
LBD
14.14
x
85(106 )

 40 GPa
(a) E 
 x 2122(106 )
 y 707 
 0.33
(b)    
 x 2122 
________________________________________________________________________
PROBLEM (3.27) Two rectangular blocks of rubber, each of width a = 25 mm, depth b = 50 mm,
and height h = 150 mm are bonded together to rigid supports and to the movable center plate (see Fig.
3.26). Calculate the shear stress  , strain  , and shear modulus of elasticity G of the rubber, if a force
P = 15 kN causes a downward deflection
=
2 mm.
SOLUTION
Refer to Example 3.8.
P
15(103 )


 1 MPa
2bh 2(0.05)(0.15)
 2
 
 0.08  80, 000  rad
a 25
 1(106 )
G 
 12.5 GPa

0.08
________________________________________________________________________
PROBLEM (*3.28) Figure P3.28 depicts a vibration isolation support that includes a steel rod of
radius a bonded to a hollow rubber cylinder of height h. Determine, in terms of the quantities a, b, P,
r, h, and G, as needed:
(a) The shear stress  in the rubber at a distance r from the center of the support.
(b) The downward displacement  of the rod.
Assumptions:
1. The displacement is so small that dy/dr = tan  =  . Clearly, the maximum value of y
occurs for r = a that the deflection is zero at r = b.
2. Hooke’s law for shear applies to the rubber.
3. The steel rod and cylinder are rigid.
*SOLUTION
(a) Shear stress: The shear area at radius r equals A  2 rh. Hence
P
P
 
A 2 rh
(b) Displacement: Applying Hook’s Law

P
 
G 2 rhG
Then, dy dr   at y   :
3-11
P
dr
2 rhG
b dr
P
P
b


ln

a
2 hG
r 2 hG a
b
b
a
a
    dr  
________________________________________________________________________
PROBLEM (3.29) For the handbrakes on the bicycle described in Prob. 3.22, express the
deflection  of the hard rubber in terms of P, L, G, a, and b (see Fig. P3.22). Here b and G,
respectively, denote the width and the shear modulus of elasticity of a n a x b x L rectangular
rubber block.
SOLUTION
From solution of Prob. 3.22:

 a2
  2 2 y;

a
2y
Also,

P
P
 

G AG 2bLG
For y  a , Eqs. (1) & (2) gives
 a2
(1)
(2)
Pa
2a 4bGL
________________________________________________________________________
PROBLEM (3.30) A 2-in.-diameter bar 6 ft long, shortens 3/64 in. under an axial load of 40 kips. If


the diameter is increased 0.4(10-3) in. during loading, calculate:
(a) Poisson's ratio.
(b) The modulus of elasticity.
(c) The shear modulus of elasticity.
SOLUTION
3 64
 651 
6 12
0.4(103 )
y 
 200 
2
(a)  x  
Thus,

200 
 0.31
651 
40(103 )
 12.73 ksi
(b)  x 
 (2) 2 4
3-12
E
 x 12.73(103 )

 19.6 106 psi
6
 x 651(10 )
E
19.6 106
 7.48 106 psi

2(1  0.31)
2.62
________________________________________________________________________
PROBLEM (3.31) Figure P3.31 shows a steel block subjected to an axial compression load of 400
(c) G 
kN. After loading, if dimensions b and L are changed to 40.02 and 199.7 mm, respectively, calculate:
(a) Poisson's ratio.
(b) The modulus of elasticity.
(c) The final value of the dimension a.
(d) The shear modulus of elasticity.
Given: a = 60 mm, b = 40 mm, L = 200 mm
SOLUTION
y
b=40 mm
x
a=60 mm
400 kN
z
x  
L=200 mm
400(103 )
 166.7 MPa
0.04  0.06
0.3
 1500 
200
0.02
y 
 500 
40
500 
1


1500  3
(a)  x  
(b) E 
 x 166.7(106 )

 111 GPa
 x 1500(106 )
 x
1
166.7 106

(
)  500 
(c)  z  
E 111109
3
a  500(106 )(60)  0.03 mm
a f  60  0.03  60.03 mm
E
111109

 41.6 GPa
(d) G 
2(1  )
83
3-13
________________________________________________________________________
PROBLEM (3.32) The data shown in the accompanying table are determined from a tensile test of a
mild steel specimen. Plot the data and determine:
(a) The modulus of elasticity.
(b) The yield point.
(c) The proportional limit.
(d) The ultimate stress.
------------------------------------------------------------------------------------------Stress, MPa
Strain
Stress, MPa
Strain
-------------------------------------------------------------------------------------------35
0.0001
245
0.009
70
0.0003
300
0.025
100
0.0005
340
0.05
135
0.0007
380
0.09
170
0.0008
435
0.15
205
0.0010
450
0.25
240
0.0012
440
0.30
255
0.0025
420
0.36
250
0.0050
325
0.40
---------------------------------------------------------------------------------------------
SOLUTION
 (MPa )
450
450
400
300
255
240
A
200
100
0
(a) E 
0.001
0.002
0.1
0.2
0.3
240(106 )
 200 GPa
0.0012
(b)  y  255 MPa
(c)  p  240 MPa
(d)  u  450 MPa
3-14
N
M
0.4

________________________________________________________________________
PROBLEM (3.33) The following data are obtained from a tensile test of a 12.7-mm-diameter
aluminum specimen having a gage length of 50 mm. After the specimen ruptures, the minimum (neck)
diameter is found to be 8.8 mm.
--------------------------------------------------------------------------------------------------Stress, MPa
Strain
Stress, MPa
Strain
--------------------------------------------------------------------- ------------------------------35
0.0005
284
0.0062
70
0.0010
305
0.02
104
0.0014
319
0. 05
139
0.0017
326
0.08
172
0.0024
312
0. 12
207
0.0030
291
0.15
242
0.0035
256
0. 20
259
0.0039
(Fracture)
---------------------------------------------------------------------------------------------------
Plot the engineering stress-strain diagram and determine:
(a) The modulus of elasticity.
(b) The proportional limit.
(c) The yield strength at 0.2 %.
(d) The ultimate strength.
(e) The percent elongation in 50 mm.
(f) The percent reduction in area.
(g) The true ultimate stress.
(h) The tangent and secant moduli at a stress level of 310 MPa.
SOLUTION
 (MPa )
Et
326
310
300
275
242
200
Es
100
0
(a) E 
0.002
0.004
0.006
0.05
0.1
0.15
242(106 )
 69 GPa
0.0035
(b)  p  242 MPa
(c)  y  275 MPa
3-15
0.2
N 
M
(d)  u  326 MPa
(e)
50  0.2(50)  50
(100)  20 %
50

(f)
4
(12.7) 2 

4

(8.8) 2
4
(12.7)
(100)  52 %
2

(g) ( u )t  326 4

4
(12.7) 2
(8.8)
 679 MPa
2
(330  290)106
 0.8 GPa
0.05
310 106
Es 
 10 GPa
0.031
________________________________________________________________________
PROBLEM (3.34) A stepped bar of diameters d AB and d BC made of an aluminum alloy, is
(h) Et 
subjected to an axial force P (Fig. P3.34a). The initial portion of the stress-strain curve is as seen in
Fig. P3.34b. Calculate:
(a) The elongation of the bar when the load is applied.
(b) The permanent elongation of the bar.
Given: a = 20 in., b = 15 in., d AB = 7/8 in., d BC = 5/8 in., P = 16 kips, E = 10x106 psi
Assumption: The weight of the bar is small compared with the loading and can be neglected.
SOLUTION

(ksi)
(a) Stresses
 AB
 BC
60
P
16


 26.6 ksi
AAB  ( 7 ) 2
D
y = 40
4 8
P
16


 52.2 ksi
E
20
ABC  ( 5 ) 2
4 8
0
F
G
0.02
- diagram
3-16
0.03
0.04

From    diagram, material in region 0D is strained elastically, since
 y  40  26.6 ksi. By Hooke’s Law,
 AB
26.6(103 )
 2, 660 
E
10 106
Material in region DF is strained plastically, since  y  40  52.2 ksi.
 AB 

From    curve, for  BC  52.2 ksi :
 BC  0.03  30, 000 
The approximate elongation of the bar is thus
    L  [(2, 660  20)  (30, 000 15)]106  0.5 in.
(b) When the load is removed, segment AB will be restored to its initial length.
But segment BC will recover elastically along line FG:

52.2(103 )
 rec  BC 
 5, 000 
E
10(106 )
The remaining plastic strain in part BC is
 OG  30, 000  5, 000  25, 000 
Thus, when the load is removed the bar is elongates
 p   OG LBC  0.025(15)  0.375 in.
________________________________________________________________________
PROBLEM (3.35) Calculate the smallest diameter and shortest length that may be selected for a
steel control rod of a machine under an axial load of 4 kN if the rod must stretch 2.5 mm.
Given:
E = 200 GPa,  all = 150 MPa
SOLUTION
d
4 kN
4 kN
L
x 
2.5 mm
4(103 ) 16(103 )

 150  106
d2 4
d2
d min  0.0058 m  5.8 mm
Also,
E
150 106
 200  109
0.0025 L
or
Lmin  3.333 m
3-17
________________________________________________________________________
PROBLEM (*3.36) Determine the axial strain in the block of Fig. P3.36 when subjected to an
axial load of 4 kips. The block is constrained against y- and z-directed contractions.
Given:
a = ¾ in.,
b = 3/8 in.,
E = 10  10 psi,
6
L = 4 in.,
 = 1/3
*SOLUTION
y  z  0
4(103 )
x 
 14.22 ksi
(3 4)(3 8)
1
[ y  ( x   z )]
E
1
 z  0  [ z  ( x   y )]
E
1
 x  0  [ x  ( y   z )]
E
y  0 
(1)
(2)
(3)
The first two expressions become
 y   z   x
(1’)
 z   y   x
(2’)
Adding:  ( y   z )  2 2 x (1  ).
Equation (3) now reduces to
1   2 2  x
x 
1 
E
Substituting the data:
1 2
1 
14.22(103 )
 948 
x  3 9
6
1
10

10
1
3
________________________________________________________________________
PROBLEM (3.37) A round steel rod of diameter d is subjected to axial tensile force P as shown
in Fig. P3.37. The decrease in diameter is  d. Compute the largest value of P.
Given: d = 1 in.,  d = 0.5(10-3 ),
E = 29x106 psi ,  = 1/3
SOLUTION
d=1 in.
P
P
3-18
P   x A   x EA 
y
d AE
EA 

d 
d  d 2
d
E
 dE
d 4
4
Inserting the given numerical values:
0.5 103
P
 (1)(29 106 )  34.16 kips
43

________________________________________________________________________
PROBLEM (3.38) A short cylindrical pipe of cold-rolled (510) bronze having an outside diameter
D and thickness t is placed in a compression machine and sequeezed until the axial load applied is P
with a safety factor of ns against yielding. Determine the minimum required Dmin
ns = 2,  y = 75 ksi (from Table B.4)
Given:
t = D/6,
P = 250 kips,
SOLUTION
A

4
 all 
D 2 5 D 2
) ]
3
36
Pn
P
A
 s
[D2  (D 
y
ns
,
 all
y
Hence
5 D 2 Pns

,
36
y
Dmin  6
Pns
5 y
Introducing the data given
250 2
Dmin  6
 1.954 in.
5 (75)
________________________________________________________________________
PROBLEM (3.39) A 5-ft long and 7/8-in. diameter bar is made of a 6061-T6 aluminum alloy.
What is the final length of the bar if it is subjected to an axial tension of 8 kips?
Given:
E = 10  10 psi ,
6
 y = 38 ksi (from Table B.4).
SOLUTION
Normal stress:
P
4(8)
 
 13.304 ksi  38 ksi
A  ( 7 )2
8
and hence Hooke’s law applies.
Normal Strain:
3-19


E

13,304
 1,330 
10(103 )
So
L   L0  1,330(106 )(5 12)  0.08 in.
Lf  L0  L  60  0.08  60.08 in.
________________________________________________________________________
PROBLEM (3.40) A 1.5–in. diameter and 60 in.– total length bar ABC is composed of an
aluminum part AB and a steel part BC as shown in Fig. P3.40. When axial force P is applied , a strain
gage attached to the steel measures normal strain at the longitudinal direction as  s = 500  .
Calculate:
(a) The magnitude of the applied force P.
(b) The total elongation of the bar if each material behaves elastically.
Given: E a = 10x106 psi,
E s = 10x106 psi
SOLUTION
(a) Axial stress in the bar  s   a   is
   s Es  500(30)  15 ksi
Hence

P   A  15[ (1.52 )]  26.5 kips
4
(b) Axial strain in aluminum equals
 a 15(103 )
a 

 1,500 
Ea 10(106 )
Therefore
   a La   s Ls
 [1,500(15)  500(45)]106  0.045 in.
________________________________________________________________________
PROBLEM (3.41) A cold-rolled yellow brass specimen see (Table B.4) has a diameter d o and a
gage length Lo (Fig. P3.41). When a force P elongates the gage length  , determine:
(a) The modulus of elasticity E.
(b) The contraction of diameter d 0 .
Given: d o = 0.4 in.,
Lo = 2 in.,
=
3
6 ( 10 ),
P = 5.6 kips
SOLUTION
From Table B.4;  y  63 ksi and G  5.6 106 ksi
(a) Modulus of elasticity.
3-20
P
5.6

 45 ksi
A  (0.4) 2
4
 6(103 )
a  
 0.003  3, 000 
L0
2
Since    y  63 ksi, material behaves elastically.

So
E
 45(103 )

 15  106 ksi
a
0.003
(b) Poisson’s ratio:
E
15 106
G
; 5.6 106 
,
  0.34
2(1  )
2(1  )
Then


   t ; 0.34   t ,
 t  0.00102
a
0.003
Hence
d0   t d0  (0.00102)(0.4)  4.08(104 ) in.
________________________________________________________________________
PROBLEM (*3.42) Verify that the change in the slope of the diagonal line AB,  , of a
rectangular plate (see Fig. P3.42) subjected to a uniaxial compression st ress  is given by the
equation:
a 1  ( E )
  [
 1]
b 1  ( E )
(P3.42)
where alb is the initial slope. Calculate the value of  when
Given: a = 25 mm, b = 50 mm,  = 0.3, E = 70 GPa
*SOLUTION
a   b  
b
 t   a  
a
E
E
a   t a 1  ( E )
( slope) f 

b   a b 1  ( E )
Thus,
  ( slope) f  ( slope)i 
or
a 1  ( E )
  [
 1]
b 1  ( E )
a 1  ( E ) a

b 1  ( E ) b
Q.E.D.
3-21
 = 120 MPa.
Introducing the data
25 1  (0.3 120 106 70 109 )
  [
 1]
50 1  (120 106 70 109 )
 0.0011 rad  0.06o
________________________________________________________________________
PROBLEM (3.43) Rework Prob.3.42, assuming that the plate is subjected to a uniaxial
tensile stress  .
SOLUTION
a   b 
b
 t   a  
a
E
E
a   t a 1  ( E )
( slope) f 

b   a b 1  ( E )
Thus,
a 1  ( E )
  ( slope) f  ( slope)i  [
 1]
b 1  ( E )
Inserting the given numerical values
25 1  (0.3 120 70 103 ) 
 
 1
50  1  (120 70 103 )

 0.011 rad  0.06o
________________________________________________________________________
PROBLEM (3.44) A 5/8 -in.-diameter bar with a 5-in. gage length is subjected to a gradually
increasing tensile load. At the proportional li mit, the value of the load is 8 kips, the gage length
increases 14.4(10 -3 ) in., and the diameter decreases 0.6(10 -3 ) in. Calculate:
(a) The proportional limit.
(b) The modulus of elasticity.
(c) Poisson's ratio.
(d) The shear modulus of elasticity.
SOLUTION
(a)  p 
p 
(b) E 
8(103 )
 26.1 ksi
 (5 8) 2 4
14.4(103 )
 2880 
5
26.08 103
 9.1106 psi
2880(106 )
3-22
0.6(103 )
 960 
58

960 1
  t 

 a 2880 3
(c)  t 
(d) G 
E
3
 (9.1106 )  3.4 106 psi
2(1  ) 8
________________________________________________________________________
PROBLEM (3.45) A 6-m-long truss member is made of two 50-mm-diameter steel bars. For
a tensile load of 250 kN, calculate:
(a) The change in the length of the member.
(b) The change in the diameter of the member.
Given: E = 210 GPa,  y = 230 MPa,  = 0.3
SOLUTION
A  2(  502 4)  1250 mm2
P
250 (103 )
(a)   
 200 MPa
A 1250 (106 )
Since    y , the result is valid.
Thus,
 200(106 )
 
 952 
E 210(109 )
L  L  6(103 )(952 106 )  5.71 mm
(b) t    0.3(952)(106 )  286 
d  286(106 )(50)  0.014 mm
________________________________________________________________________
PROBLEM (3.46) The stress-strain diagram seen in Fig. P3.46 is plotted from tensile test
data of high-strength steel. The test specimen had a diameter of 0. 51 in. and gage length of 2 in.
was used. The total elongation between the gage marks at the fracture was 0.4 in. and the
minimum diameter was 0.36 in. Determine:
(a) The modulus of elasticity
(b) The load Py on the specimen that causes tielding.
(c) The ultimate load Pu the specimen supports.
(d) Percent elongation in 2.00 in. and percent reduction in area.
SOLUTION
3-23
(a) Refer to    diagram in Fig. P3.46.
  40 ksi when   0.001  1000 
Hence
40  0
E
 40  106 ksi
0.001  0
(b) We have  p   y  40 ksi ( by    curve).

Py   y A  40[ (0.51) 2 ]  8.17 kips
4
(c) From    curve,  u  76 ksi

Pu  76[ (0.51) 2 ]  15.52 kips
4
L f  L0
(100)
(d) Percent elongation 
L0
2.4  2

(100)  20 %
2
A0  A f
(100)
Percent reduction Ai area 
A0

(0.51)2  (0.36)2
(100)  50.2 %
(0.51)2
________________________________________________________________________
PROBLEM (*3.47) For many materials, the stress-strain curve may be described by the
Ramberg-Osgood equation of the form (Reference 3.7):


E
 k n
(P3.47)
in which the parameters E, k, and n are obtained from the diagram of a given material.
Considering the    diagram of a metal shown in Fig. P3.47, determine E, k, n, and hence,
obtain a formula for the curve.
*SOLUTION
Refer to    diagram in Fig. P3.47.
The estimated proportional limit
1
 p  70(106 ) Pa is at  p  (103 ) . Thus
3
E   p  p  210 GPa
We also have
  280 MPa at   0.1,
  420 MPa at   0.3
3-24
Applying Eq. (P3.47):
280
0.1(103 ) 
 k (280)n
3
210(10 )
or
0.012333  k (280)n
Similarly,
420
0.3(103 ) 
 k (420)n
3
210(10 )
or
0.001700  k (420)n
Solving, Eqs.(1) and (2), we obtain
n  0.7915
k  14.26(106 )
(1)
(2)
________________________________________________________________________
PROBLEM (3.48) A short cylindrical rod of ASTM – A36 structural steel, having an
original diameter of d o and length Lo is placed in a compression machine and sequeezed until
its length becomes L f . Determine the new diameter of the rod.
Given: d = 30 mm,
Lo = 50 mm,
L f = 49.7 mm,
 = 0.3
SOLUTION
Axial strain:
a 
L f  L0

49.7  50
 6, 000 
50
L0
Transverse strain:
 t   a  (0.3)(0.006)  1,800 
Therefore,
d   t d0  0.0018(30)  0.054 mm
d  d0  d  30  0.054  30.054 mm
________________________________________________________________________
PROBLEM (*3.49) A prismatic bar is under uniform axial tension. Determine Poisson’s ratio 
for the material, for the case in which the ratio of the unit volume change to the unit cross-sectional
area change is -5/8.
*SOLUTION
We have  x   y   z .
From Eq.(3.14):
V
 (1  2 ) x
V0
Upon following a procedure similar to that of Sec. 3.10, the final area:
Af  [(1   y )dy  (1   z )dz]
3-25
(1)
 [1  ( y   z )]dydz  A0  A
A
  y   z  2 x
and
(2)
A0
It is required that
V V0
1  2
5


A A0
2
8
Solving,
4

13
________________________________________________________________________
PROBLEM (3.50) A tensile test is performed on a 2024-T6 aluminum alloy specimen of
diameter d and gage length Lo , depicted in Fig. P3.50. After the lo ading reaches a value of P
= 5 kips, the distance Lo has increased by  = 4.8x10 -3 in. Determine:
(a) The decrease in diameter d .
(b) The modulus of elasticity E.
(c) The dilatation e of the bar.
Given: d = ½ in., Lo = 2 in.,  = 0.33
SOLUTION
4.8 103
)(0.5)  0.396 103 in.
2
 P A PL0

(b) E  
  L0 A
(a) d   d  0.33(
5(103 )(2)
 10.6  106 psi
 1 2
( ) (4.8 103 )
4 2
4.8 103
)  0.816 103
(c) e  (1  2 )  (1  2  0.33)(
2
________________________________________________________________________
PROBLEM (3.51) A 2-in.-diameter solid brass bar (E = 15 x 10 6 psi,  = 0.3) is fitted in a

hollow bronze tube. Determine the internal diameter of the tube so that its surface and that of the
bar are just in contact, with no pressure, when the bar is subjected to an axial compressive load P
= 40 kips.
SOLUTION
Axial stress in brass bar:
40 103
x 
 12.73 ksi
 (2) 2 4
and

12.73 103
 y   x  0.3
 255 
E
15 106
The diametral increase of the bar is
3-26
d   y d  255(106 )2  0.51103 in.
The required internal diameter of the tube is thus
Di  d  d  2  0.51103  2.00051 in.
________________________________________________________________________
PROBLEM (3.52) The cast-iron pipe shown in Fig. P3.52 is under an axial compressive load
P. Determine:
(a) The change in length L.
(b) The change in d iameter D .
(c) The change in thickness t .
Given: D = 130 mm, t = 15 mm, L = 0.5 m,
Assumption: Buckling does not occur.
P = 200 kN,
E = 70 GPa,
 = 0.3
SOLUTION

P
200(103 )
 

 527 
E AE 70(109 )( 4)(0.132  0.12 )
(a) L   L  527(106 )(0.5 103 )  0.264 mm
(b) D   ( D)  0.3(527)(106 )130  0.021 mm
(c) t   ( t )  0.3(527)(106 )15  0.0024 mm
________________________________________________________________________
PROBLEM (3.53) Redo Prob. 3.52 for the case in which the axial load P is in tension and the
pipe shown in Fig. P3.52 is made of brass.
Given: D = 130 mm,
t = 15 mm,
L = 0.5 m,
P = 200 kN,
E = 105 GPa,
 = 0.3
SOLUTION


E

P
200(103 )

 351 
AE 105(109 )( 4)(0.132  0.12 )
(a) L   L  351(106 )500  0.176 mm
(b) D   ( D)  0.3(351)(106 )130  0.014 mm
(c) t   ( t )  0.3(351)(106 )15  0.0016 mm
________________________________________________________________________
PROBLEM (3.54) The aluminum rod, 50 mm in diameter and 1.2 m in length, of a hydraulic ram is
subjected to the maximum axial loads of ±200 kN. What are the largest diameter and the largest volume of
the rod during service?
Given: E = 70 GPa,  = 0.3.
3-27
SOLUTION
0.05 m
200 kN
1.2 m
Vo 
x 

4
(0.052 )(1.2)  2355(106 ) m3
200(103 )
 101.9 MPa
 (0.052 ) 4
101.9(106 )
 1456 
70(109 )
t  0.3(1456)(106 )  437 
x 
d  t d  437(106 )(0.05)  0.022 mm
d max  50.022 mm
Also,
e  (1  2 ) x  0.4(1456)(106 )  582(106 )
V  eVo  582(106 )(2355)(106 )  1.371106 m3
Vmax  Vo  V  (2355  1.371)106  2356.371(103 ) mm3
________________________________________________________________________
PROBLEM (3.55) Calculate the smallest diameter and volume of the hydraulic ramrod described
in Prob. 3.54.
SOLUTION
From solution of Prob. 3.54:
d  0.022 mm
V  2355(106 ) m3
Thus,
d min  50  0.022  49.978 mm
V  1.37 106 m3
Vmin  (2355 1.371)106  2353.629(103 ) mm3
________________________________________________________________________
PROBLEM (3.56) A 20-mm-diameter bar is subjected to tensile loading. The increase in length
resulting from the load of 50 kN is 0.2 mm for an initial length of 100 mm. Determine:
(a) The conventional and true strains.
(b) The modulus of elasticity.
SOLUTION
20 mm
50 kN
50 kN
100 mm
0.2 mm
3-28
50(103 )
 159.2 MPa
 (0.02) 2 4
0.2
 2000 
 t  ln(1  0.002)  1998 
(a)  x 
100
159.2(106 )
 79.6 GPa
(b) E 
0.002
________________________________________________________________________
PROBLEM (*3.57) A solid aluminum-alloy rod of diameter d, modulus of elasticity E and and
Poisson’s ratio  is fitted in a hollow plastic tube of 1.002-in. internal diameter, as depicted in Fig. P3.57.
x 
Determine the maximum axial compressive load P that can be applied to the rod for which its surface and
that of the tube are just in contact and under no pressure.
Given: d = 1 in.,
E = 11 10 psi,
6
 = 0.3
*SOLUTION
d=1 in.
d=1.002 in.
P
x 
P
d2 4
P
x 
x
E
 t   x 
4 P
 d 2E
4 P
 dE
d
P
 dE
or
4
Introducing the given number values:
0.002
P
( 111106 )  57.6 kips
4  0.3
________________________________________________________________________
PROBLEM (3.58) A cast-iron bar of diameter d and length L is subjected to an axial
d   t d 
compressive load P. Determine:
(a) The increase d in diameter.
(b) The decrease L in length.
(c) The change in volume V .
Given: d = 75 mm, L = 0.5 m, E = 80 GPa,
 = 0.3
SOLUTION
d=75 mm
200 kN
200 kN
0.5 m
3-29
x 
x 
200(103 )
 45.3 MPa
 (0.075) 2 4
x
E

45.3(106 )
 566 
80 109
(a) d  ( x )d  0.3(566)(106 )75  0.013 mm
(b) L   x L  566(106 )500  0.283 mm
(c) e  (1  2 ) x  0.4(566)(106 )  226.4(106 )
V  eVo  226.4(106 )[ 4 (0.0752 )(0.5)]
 0.5(106 ) m3  500 mm3
________________________________________________________________________
PROBLEM (3.59) A 2-in.-diameter and 4-in.-long solid cylinder is subjected to uniform
axial tensile stress of
x
= 7.2 ksi. Calculate:
(a) The change in length of the cylinder.
(b) The change in volume of the cylinder.
Given: E = 30x10 6 psi,  = 1/3
SOLUTION
4 in.
2 in.
7.2 ksi
Vo   (1)2 4  12.566 in.3
x 
7.2(103 )
 240 
30 106
(a) L   x L  240(106 )4  0.96 103 in.
(b) V  (1  2 ) xVo
1
 (240)(106 )(12.566) in.3  1.005 103 in.3
3
________________________________________________________________________
PROBLEM (3.60) A bar of any given material is subjected to uniform triaxial stresses. Determine
the maximum value of Poisson's ratio.
SOLUTION
Using generalized Hook’s Law:
3-30
x y z 
1  2
( x   y   z )
E
(1)
For a constant triaxial state of stress, we have
 x   y   z   and  x   y   z  
Then, Eq. (1) becomes
1  2


E
Since  and  must have identical sign: 1  2  0
1

or
2
________________________________________________________________________
PROBLEM (3.61) The rectangular concrete block shown in Fig. P3.61 is subjected to loads that
have the resultants Px = 100 kN, Py = 150 kN, and Pz = 50 KN. Calculate:
(a) Changes in lengths of the block.
(b) The value of a single force system of compressive forces applied only on the y faces that would
produce the same deflection as do the initial forces.
Given: E = 24 GPa,  = 0.2
SOLUTION
150 kN
y
0.2 m
B
C
A
z
50 kN
D
E
0.1 m
100 kN
x
0.05 m
100(103 )
 20 MPa
0.1 0.05
150(103 )
y 
 15 MPa
0.2  0.05
50(103 )
z 
 2.5 MPa
0.2  0.1
x 
Thus,
106
[20  0.2(15  2.5)]  688 
24(109 )
106
y  
[15  0.2(22.5)]  438 
24(109 )
106
z  
[2.5  0.2(35)]  188 
24(109 )
x  
3-31
(a) LAB  438(106 )100  0.044 mm
LBC  688(106 )200  0.138 mm
LCE  188(106 )50  0.009 mm
(b)  y  438(106 ) 
 Py
AE

 Py
0.2  0.05(24 109 )
or
Py  105.1 kN
________________________________________________________________________
PROBLEM (3.62) Redo Prob. 3.61 for the x faces of the block to be free of stress.
SOLUTION
We have  x  0 . From Solution of Prob. 3.61:
 y  15 MPa
 z  2.5 MPa
Thus,
106
[0.2(17.5)]  146 
24(109 )
106
y  
[15  0.2(2.5)]  604 
24(109 )
106
z  
[2.5  0.2(15)]  21 
24(109 )
x  
(a) LAB  604(106 )100  0.604 mm
LBC  146(106 )200  0.0292 mm
LCE  21(106 )50  0.001 mm
(b)  y  604(106 ) 
 Py
0.2  0.05(24 109 )
,
Py  145 kN
________________________________________________________________________
PROBLEM (*3.63) A vibration isolation unit consists of rubber cylinder of diameter d
compressed inside of a steel cylinder by a force R applied to a steel rod, as shown in Fig. P3.63.
Determine, in terms of d, R, and Poisson’s ratio  for the rubber, as required:
(a) An expression for the lateral pressure p between the rubber and the steel cylinder.
(b) The lateral pressure p between the rubber and the steel cylinder for d = 60 mm,  = 0.45,
and R = 4 kN.
Assumptions:
1. Friction between the rubber and steel can be omitted.
2. Steel cylinder and rod are rigid.
3-32
*SOLUTION
(a) According to assumption 1, the rubber is in triaxial stress:
R
4R
 x   z   p,
y  
 2
 2
d
d
4
Strains are:  x   z  0. The first of Eqs. (3.17) gives
1
 x  0  [ x  ( y   z )]
E
y
or
4R
0  p  (  2  p )
d
Solving,
z
4 R
p
2
 d (1  )
x
(b) Substituting the data,
4(0.45)(4 103 )
p
 1.157 MPa (C )
 (0.06)2 (1  0.45)
________________________________________________________________________
PROBLEM (*3.64) A rectangular aluminum plate (E = 70 GPa,  = 0.3) is subjected to uniformly
distributed loading, as shown in Fig. P3.64. Determine the values of wx and wy (in kilonewtons per
meter) that produce a change in length in the x direction of 1.5 mm and in the y direction of 2 mm.
Use a = 2 m, b = 3 m, and t = 5 mm.
*SOLUTION
1.5
2
 500 
y 
 1000 
3000
2000
70 109
G
 27 GPa
2(1  0.3)
x 
Equations (3.15):
1
( x  0.3 y )
70(109 )
1
1000(106 ) 
( y  0.3 x )
70(109 )
500(106 ) 
or
35(106 )   x  0.3 y
(1)
70(106 )   y  0.3 x
Solve Eqs. (1) and (2):
(2)
3-33
 x  61.5 MPa
 y  88.5 MPa
Thus,
wx  61.5(106 )(0.005)  308 kN m
wy  88.5(106 )(0.005)  443 kN m
________________________________________________________________________
PROBLEM (3.65) Rework Prob. 3.64, assuming that a = 4 m, b = 2 m, t = 6 mm and that the
plate is made of steel (E = 210 GPa,  = 0.3).
SOLUTION
1.5
2
 750 
y 
 500 
200
4000
210(109 )
G
 80.8 GPa
2(1  0.3)
x 
Equations (3.15):
1
( x  0.3 y )
210(109 )
1
500(106 ) 
( y  0.3 x )
210(109 )
750(106 ) 
or
157.5(106 )   x  0.3 y
(1)
105(10 )   y  0.3 x
Solve Eqs. (1) and (2):
 x  207.3 MPa
 y  167.2 MPa
Hence,
wx  207.3(106 )(0.006)  1244 kN m
6
(2)
wy  167.2(106 )(0.006)  1003 kN m
________________________________________________________________________
PROBLEM (3.66) A solid sphere of diameter d experiences a uniform pressure p, as
depicted in Fig. P3.66. Calculate:
(a) The decrease in circumference of the sphere.
(b) The decrease in volume of the sphere V .
Given: d = 200 mm, p = 120 MPa, E = 70 GPa,
 = 0.3
SOLUTION
p=120 MPa= -
 x   y   z  
d=200 mm
3-34
4
4
Vo   r 3 
(1003 )  4.19(106 ) mm3
3
3
(a)  x   E1 [  (   )]   E (1  2 )

120(106 )
(1  0.6)  686 
70 109
d   x d  686(106 )200  0.137 mm
Decrease in circumference:
 (d )  0.137  0.43 mm
(b) V  (1  2 ) xVo
 (0.4)(686)(106 )(4.19 106 )  1150 mm3
________________________________________________________________________
PROBLEM (3.67) A solid cylinder of diameter d and length L is under hydrostatic loading
with  x   y   z =-7.2 ksi. Determine:
(a) The change in length of the cylinder L .
(b) The change in volume of the cylinder V .
Given:
d = 2 in.,
L = 4 in.,
E = 30 10
6
psi,
 = 1/3
SOLUTION
Vo   (1)2 4  12.566 in.3
1

 x   [  (2 )]   (1  2 )
E
E
3
7.2(10 )(1 3)

 80 
30(103 )
(a) L  80(106 )4  0.32 103 in.
(b) V  (1  2 ) xVo
 0.4(80)(106 )(12.566) in.3  0.402 103 in.3
________________________________________________________________________
PROBLEM (*3.68) A steel plate ABCD of thickness t is subjected to uniform stresses
 x and  y (see Fig. P3.68). Calculate the change in:
(a)
(b)
(c)
(d)
The length of edge AB.
The length of edge AD.
The length of diagonal BD.
The thickness.
Given: a = 160 mm,
E = 200 GPa,
b = 200 mm,
t = 5 mm,
 = 1/3
3-35
x=
120 MPa,
 y = 90 MPa
*SOLUTION
y
90 MPa
B
C
LBD  1602  2002  256 mm
200 mm
120 MPa
A
160 mm
D
x
1
106
90
 x  ( x  y ) 
(120  )  450 
9
E
200(10 )
3
6
1
10
120
 y  ( y  x ) 
(90 
)  250 
9
E
200(10 )
3
(a) LAB  200(250)(106 )  0.05 mm
(b) LAD  160(450)(106 )  0.072 mm
(c) L2BD  L2AB  L2AD
2 LBD LBD  2 LAB LAB  2 LAD LAD
or
L
L
LBD  AB LAB  AD LAD
LBD
LBD
200
160
LBD 
(0.05) 
(0.072)  0.084 mm
256
256
(1)
1
[0  ( x   y )]
E
106
210

(
)  350 
9
200(10 )
3
t   zt  350(106 )5  0.002 mm
(d)  z 
________________________________________________________________________
PROBLEM (3.69) Redo Prob. 3.68, with the plate acted upon by biaxial loading that results
in uniform stresses  x = 100 MPa and  y = -60 MPa.
SOLUTION
106
60
x 
(100  )  600 
9
200 10
3
3-36
106
100
(60 
)  467 
9
200 10
3
106
100  60
z 
(0 
)  67 
9
200 10
3
y 
(a) LAB  200(467)(106 )  0.093 mm
(b) LAD  160(600)(106 )  0.096 mm
(c) From Eq. (1) of solution of Prob. 3.68:
200
160
LBD 
(0.093) 
(0.096)  0.013 mm
256
256
(d) t   z t  67(106 )5  0.34(103 ) mm
________________________________________________________________________
PROBLEM (3.70) Using the stress-strain diagram of a structural steel shown in Fig. P3.70,
determine:
(a) The modulus of resilience.
(b) The approximate modulus of toughness.
SOLUTION
From Fig. P3.70:
190(106 )
E
 190 GPa
0.001
 y  245 MPa
(a) U 0 
 y2
2E

(245 106 ) 2
kJ
 158 3
9
2(190 10 )
m
(b) Total area under    diagram:
MJ
m3
________________________________________________________________________
PROBLEM (3.71) From the stress-strain curve of a magnesium alloy seen in Fig. P3.71, find
U t  350 106 (0.28)  98
the approximate values of:
(a) The modulus of resilience.
(b) The modulus of toughness.
SOLUTION
Referring Fig. P3.71, we have
100 106
E
 45 GPa
0.0022
 y  200 MPa
3-37
(a) U 0 
 y2
2E

kJ
(200 106 ) 2
 444 3
9
m
2(45 10 )
(b) Total area under    diagram:
MJ
m3
________________________________________________________________________
PROBLEM (3.72) Calculate the modulus of resilience for two grades of steel (see Table
Ut  250(106 )(0.176)  44
D.4):
(a) ASTM-A242.
(b) Cold-rolled, stainless steel (302).
SOLUTION
(a) ASTM-A242
E  200 GPa
U0 
 y2
 y  345 MPa
(345 106 ) 2
kJ
 298 3
9
2 E 2(200 10 )
m
298

 43.2 in.  lb in.3
6.895

(b) Stainless (302)
E  190 GPa
 y  520 MPa
 y2
(520 106 ) 2
kJ
U0 

 712 3
9
2 E 2(190 10 )
m
712

 103 in.  lb in.3
6.895
________________________________________________________________________
PROBLEM (3.73) The stress-strain diagram for a high-strength steel bar in tension is shown
in Fig. P3.73. Determine for the material:
(a) The modulus of resilience.
(b) The approximate modulus of toughness.
SOLUTION
(a) The area under the linear portion of the    curve:
1
U o  (280)(106 )(0.001)  140 m  kN m3
2
(b) Toughness modulus is represented by the total area under    curve that can be
Estimated by counting the number of square from the diagram. From the diagram,
The number of square equals about 45.
Thus,
Ut  45(106 )(70  0.04)  126 m  MN m3
3-38
________________________________________________________________________
PROBLEM (3.74) Calculate the modulus of resilience for the following two materials (see
Table B.4):
(a) Aluminum alloy 2014-T6.
(b) Annealed yellow brass.
SOLUTION
(a) 2014-T6
E  72 GPa
 y2
 y  410 MPa
(410 106 ) 2
2 E 2(72 109 )
kJ 1167
 1167 3 
 169 in.  lb in.3
m
6.895
U0 

(b) Yellow, annealed, brass
E  105 GPa
 y  105 MPa
U0 
 y2
(105 106 ) 2
2 E 2(105  109 )
kJ
52.5
 52.5 3 
 7.61 in.lb in.3
m
6.895

________________________________________________________________________
PROBLEM (3.75) A 1 3 4 -square machine part having E=30x106 psi and length L=4 ft is to resist
an axial energy load of 1.5 in.  kip. Based on a factor of safety ns=2, calculate:
(a) The required proportional limit of steel.
(b) The corresponding modulus of resilience for the steel.
SOLUTION
V  1.75 1.75  4 12  147 in.3
(a) ns  U 
 p2
2E
Solving,
V
nsU (2 E ) 2(1500)(2  30 106 )
 12.24 108

V
147
 p  35 ksi
 p2 
or
(b) U o 
 p2
2E

12.24 108
 20.4 in.  lb in.3
2(30 106 )
3-39
________________________________________________________________________
PROBLEM (3.76) Members AB and AC of the truss shown in Fig. P3.76 are fabricated of an
elastoplastic material with  y = 40 ksi. If Pu = 50 kips and  = 60°, determine the minimum crosssectional areas AAB , and ABC.
SOLUTION
P
tan 
for P  Pu :
FAB 
FBC 
P
sin 
50 103
 28.87 kips
tan 60o
50 103

 57.74 kips
sin 60o
FAB 
FBC
Thus,
28.87 103
 0.722 in.2
3
40 10
AAB 
FAB

ABC 
FBC
57.74 103

 1.444 in.2
3
40 10
y
y
________________________________________________________________________
PROBLEM (3.77) Solve Prob. .3.75 for Pu = 20 kips and  = 30°.
SOLUTION
We have
FAB
FBC
Pu
20 103


 34.64 kips
tan  tan 30o
P
20 103
 u 
 40 kips
sin  sin 30o
Thus,
ABC 
ABC
FBC
y
34.64 103

 0.866 in.2
3
40 10
40 103

 1.0 in.2
3
40 10
________________________________________________________________________
PROBLEM (3.78) Determine the ultimate load Pu that can be carried by truss ABC (see Fig.
P3.75) if each member is made of an elastoplastic material with  y = 38 ksi.
Given:
AAB = 2ABC = 1.8 in.2 ,  = 40°
3-40
SOLUTION
Since ABC  AAB and sin   tan  :
FBC   yp ABC  38 103 (0.9)  34.2 kips
Thus,
Pu  FBC sin   34.2 103 (sin 40o )  21.98 kips
End of Chapter 3
3-41