ch03 PHISICS SOLUTION

Prévia do material em texto

3-1 
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 
 
6( 500 10 )(800) 0.4AB AB ABL mm        
 From triangles B’BF and EE’F: 
 
0.4 2.4
, 214.3
1500
x mm
x x
 

 
 From triangles DD’F and EE’F: 
 
285.7 1285.7
, 0.533
2.4
D
D
mm  
 
 Thus 
 
0.533
533
1000
D
CD
CDL
   
 
 
________________________________________________________________________ 
PROBLEM (3.2) A spherical balloon changes its diameter from 200 to 201 mm when 
pressurized. Determine the average circumferential strain. 
 
SOLUTION 
 
 
0
0
1
5000
200
fd d
d
  

  
 
 
________________________________________________________________________ 
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 
 
 (a) 
2 ( ) 2
2
c
r r r r
r r
  
   
 
 
 
max
0.25
( ) ( ) 2500
100
c c i    
 
 
0.15
( ) 750
200
c o  
 
 
1000 mm 
F 
2.4 mm 
0.4 mm 
D’ 
B’ 
D 
E 
B 
x 
E’ 
500 mm 
D
 
 3-2 
 (b) 
0.25 0.15
1000
200 100
o i
o i
r r
r r
      
 
 
 
________________________________________________________________________ 
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: 
(a) u = (
2x
 /L)
310
. 
(b) u = L(
310
) sin ( 

x/2L). 
 
SOLUTION 
 
 (a) 2
1000 500
x
x du x
u
L dx L
  
 
 At x=L: 
 
max( ) 2000x 
 
 
 (b) 
sin cos
1000 2 2000 2
x
L x du x
u
L dx L
    
 
 At x=0: 
 
max( ) 1570
2000
x
  
 
 
________________________________________________________________________ 
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: 
 
0.125
40 60 80
CE DF
 
 
 
Solving, 
 
0.0625 . 0.09375 .CE DFin in  
 
Normal Strains: 
 
0.0625
1786
35
CE
CE
CEL
   
 
 
0.09375
1875
50
DF
DF
DFL
   
 
 
 
 
 
 
B C 
D
F 
D 
E 
CE 
A 
B 
Figure (a) 
F 
20 in. 
35 in. 
50 in. 
20 in. 40 in. 
 3-3 
________________________________________________________________________ 
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]: 
 
40 60 80
CE DF B
  
 
 
Normal Strains: 
 
0.0015(35) 0.0525 .CE all CEL in   
 
 
8
2(0.0525) 0.105 .
4
B CE in   
 
 Also 
 
0.0015(50) 0.075 .DF all DEL in   
 
 
8 4
(0.075) 0.1 .
6 3
B DF allin     
 
 
________________________________________________________________________ 
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 
AB
, if the rod elongates 0.05 in. 
(c) The normal strain 
BC
, if the strut shortens 0.025 in. 
Given: P = 3.5 kips. The cross-sectional areas of the members: 
BCA
= 0.25 
2.in
 and 
 
BCA
 =0.4 
2.in
 
 
*SOLUTION 
 
 We have 
2 260 45 75 .ABL in  
 
 
2 260 25 65 .BCL in  
 
 Equilibrium: 
 
4 12
0 : 0
5 13
x AB BCF F F   
 
 
3 5
0 : 3.5 0
5 13
y AB BCF F F   
 
 
 
 
 
B 
C D 
E 
A 
F 
50 in. 
35 in. 
20 in. 20 in. 40 in. 
A 
C BCF
 
ABF
 
P 
B 
45 
25 
12 
60 
5 
 
3 
4 
13 
 
5 
 3-4 
 
 Solving, 
 
3.25 ( ) 3.75 ( )BC ABF kips C F kips T 
 
 
 (a) 
3.75 3.25
15 8.125
0.25 0.4
AB BCksi ksi      
 
 (b) 
0.05
667
75
AB  
 
 (c) 
0.025
385
65
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 
 
 
0.002
50
40
x
u
x
     

 
0.005
250
20
y
v
y
   

 
400 1200 1600xy
v u
x y
      
 
 
 
________________________________________________________________________ 
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 
 
 
0.001
25
40
x
u
x
   

 
0.005
250
20
y
v
y
   

 
1600 1200 400xy
v u
x y
        
 
 
 
 
 
 
 
 
40 mm 
20 mm 
A 
D 
y 
C B 
x 
 3-5 
________________________________________________________________________ 
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 
 
 
6400(10 )(6)ABL
 
32.4 10 .in 
 
 
6600(10 )(8)ADL
 
34.8 10 .in 
 
 
2 2 2
AC AB ADL L L 
 
 
 
2 2 2AC AC AB AB AD ADL L L L L L    
 
 or 
 
AB AD
AC AB AD
AC AC
L L
L L L
L L
    
 (1) 
 Substituting the numerical values: 
 
36 8[ (2.4) (4.8)]10
10 10
ACL
  
35.28 10 .in 
 
________________________________________________________________________ 
PROBLEM (3.11) Redo Prob.3.10, with the plate in biaxial compression for which
x
= -200

 
and 
y
= - 100

. 
 
SOLUTION 
 
 
6 3100(10 )6 0.6 10 .ABL in
      
 
 
6 3200(10 )8 1.6 10 .ADL in
      
 
 
 Using Eq. (1) of Solution of Prob. 3.8: 
 
36 8[ ( 0.6) ( 1.6)]10
10 10
ACL
    
31.64 10 .in  
 
________________________________________________________________________ 
PROBLEM (3.12) Determine the normal strain in the members AB and CB of the pin-
connected 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 
 
3
1500
AB
AB
AB
L
L
   
 
 
2000  
 
 
 
0.003cos
2.5
BC
BC
BC
L
L
   
 
8 in. 
6 in. 
A 
D 
y 
C B 
x 
10 in. 
3 mm 
2.5 m 
A 
C 
B 
1..5 m 
2 m 

 BCL
 
 3-6 
 
0.003(1.5 2.5)
2.5


720  
 
________________________________________________________________________ 
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) 
 
2
2dx h
y
dy b
  
2
2(0.0005)
(0.2)
y
 
 or 
 
0.025 y 
 (1) 
 
 (b) From Eq. (1) at 
2 :y b
 
0.025(0.1) 2500  
 
 
 (c) From Eq. (1) at 
0 :y 
 
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 
 
 (a) 
0.4
1600
250
h
b
   
 
 (b) 
0.2
1600
125
  
 
 (c) 
1600 
 
________________________________________________________________________ 
PROBLEM (3.15) A 100 mm by 100 mm square plate is deformed into a 100 mm by 100.2 
mm rectangle as shown by the dashed lines in Fig. P3.15. Determine the positive shear strain 
between its diagonals. 
 
SOLUTION 
 
 
2

  
 
 
1 1002 tan
2 100.2
  
 
 
1998 
 
 
dx 
y 
x 
dy 
100 mm 0.2 mm 

 
2
 
100 mm 
A D 
C B 
 3-7 
________________________________________________________________________ 
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 
 
 
 
2

  
 
 
1 1.00052 tan
2 0.9995
L
L
  
 
 
1000  
 
 
________________________________________________________________________ 
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 
 
 
2 2400 400 565.69AC BD mm   
 
 
 
' ' 565.69 0.4A C  
565.29 mm
 
 
' ' 565.69 0.2B D  
565.89 mm
 
 
 Geometry: 
' ' ' 'A B A D
 
 Thus, 
 
' '
x y
A B AD
AD
   
1
2 2 2
565.29 565.89
400
2 2
400
    
     
      
 
0.0001685 169    
 
 
1 565.89 2' 2 tan
2 2 565.29 2
xy
      
 
 
0.001061 1061    
 
 
________________________________________________________________________ 
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: 
-0.0005L 
0.0005L 

 
2
 
L 
L 
A' 
D' 
y 
C' B' 
x 

'
 3-8 
 
2 220 15 25 .AC BD in   
 
 
2 2' (20.12) (15.1) 25.156 .AC in  
 
 
2 2' ' (20.02) (15) 25.016 .B D in  
 
 
 Normal strains: 
 
' 25.156 25
6240
25
AC
AC AC
AC
    
 
 
' ' 25.016 25
640
25
BD
B D BD
BD
    
 
________________________________________________________________________ 
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 
 
 Geometry (for small angles): 
 
0.05
2,491
20.07
rad  
 
 
0.05
3,322
15.05
rad  
 
 Shear strains: 
 
( ) ( ) 2,491 3.222xy A xy C       
 
 
5,813 rad
 
 
( ) ( ) ( ) 5,813xy B xy D rad         
 
________________________________________________________________________ 
PROBLEM (3.20) The pin-connected structure ABCD is deformed into a shape AB'C'D, as 
shown by the dashed lines in Fig. P3.20. Calculate the average normal strains in member BC and 
AC. 
 
SOLUTION 
 
 
' 5 64 . ' 5 32CC in BB 
 
 
5 64 5 32
6 12
BC
BC
BC
L
L
   

 
 
1085  
 
 
AC
AC
AC
L
L



 
 
(5 64)cos 45
543
72 12
o
 

 
 
________________________________________________________________________ 
A 
D 
y 
C B 
x 

 
  

0.05 in. 
20 in. 
0.05 in 
15 in. 
0.05 in 
0.07 in 
0.12 in. 
6 ft 
6 ft A 
C 
B 
45o 
ACL
 
C" 
C' 
72 ft
 
B’ 
 3-9 
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. 
0
0BCBC
BC BC
L
L L
   
 
 
(3 16)cos 45
1302
72 12
o
AC
AC
AC
L
L
   

 
 
________________________________________________________________________ 
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). 
Assuming that a force P causes a parabolic deflection (x = 
2ky
) of the rubber when the brakes 
are applied (Fig. P3. 22b), determine the shearing strain in the rubber. 
 
SOLUTION 
 
 
2x ky
 
 
2ka 
 
 
2k a
 
 
We have 
 
2
2 2dx kydy ydy
a

 
 
Thus, 
2
2( )
dx
y
dy a

  
 
________________________________________________________________________ 
PROBLEM (*3.23) The thin, triangular plate ABC shown in Fig. P3.23 is uniformly 
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 
 
 (a) 
2.4
1200
2000
x  
 
 
1.5
1500
1000
y 

  
 
 
0xy 
 
 
 (b) 
90oACB 
 
x 
y 

 
x=ky2 
a
 
 3-10 
 
1 1.0012' ' ' 2 tan 90.1547
0.9985
oA C B  
 
 
390 90.1547 0.1547 2.7 10o o o rad        
 
2700  
 
________________________________________________________________________ 
PROBLEM (3.24) Redo Prob. 3.23 for the case in which the plate ABC is uniformly 
deformed into a shape ABC'. 
 
SOLUTION 
 
 (a) 
0 0x xy  
 
 
1.5
1500
1000
y    
 
 (b) 
90oACB 
 
 
1 1' 2 tan 90.086
0.9985
oAC B  
 
 
390 90.086 0.086 1.5 10o o o rad        
 
1500  
 
 
________________________________________________________________________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) 
0.0014 1400y  
 
 
0.28 280,000f  
 
 (b) 
50 50(0.28) 64fL mm  
 
 % Elongation=
64 50
(100)
50
 28 %
 
________________________________________________________________________ 
PROBLEM (3.26) A 10 mm by 10 mm square ABCD is drawn on a member prior to loading. 
After loading, the square becomes the rhombus shown in Fig. P3.26. Determine: 
(a) The modulus of elasticity. 
(b) Poisson's ratio. 
 
SOLUTION 
 
 
2 210 10 14.14AC BDL L mm   
 
 
14.11 14.14
2122
14.14
AC
x
AC
L
L
     
 
 3-11 
 
14.15 14.14
707
14.14
BD
x
BD
L
L
    
 
 
 (a) 6
6
85(10 )
40
2122(10 )
x
x
E GPa

 

  

 
 (b) 
707
0.33
2122
y
x
 
 
   
 
________________________________________________________________________ 
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. 
 315(10 )
1
2 2(0.05)(0.15)
P
MPa
bh
   
 
 
2
0.08 80,000
25
rad
a
    
 
 61(10 )
12.5
0.08
G GPa


  
 
________________________________________________________________________ 
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 
2 .A rh
 Hence 
2
P P
A rh


 
 
 
 (b) Displacement: Applying Hook’s Law 
 
2
P
G rhG



 
 
 Then, 
dy dr 
 at 
:y 
 
 3-12 
 
2
b b
a a
P
dr dr
rhG
    
 
 
ln
2 2
b
a
P dr P b
hG r hG a  
 
 
________________________________________________________________________ 
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: 
 2
2
2 ;
2
a
y
a y
   
 (1) 
 Also, 
 
2
P P
G AG bLG
   
 (2) 
 For 
y a
, Eqs. (1) & (2) gives 
 2
2 4
a Pa
a bGL

  
 
________________________________________________________________________ 
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 
 
 (a) 
3 64
651
6 12
x    

 
 30.4(10 )
200
2
y 

 
 
 
 Thus, 
 
200
0.31
651



 
 
 
 (b) 3
2
40(10 )
12.73
(2) 4
x ksi 

  
 
 3-13 
 3
6
6
12.73(10 )
19.6 10
651(10 )
x
x
E psi

    
 
 
 (c) 619.6 10
2(1 0.31) 2.62
E
G

 

67.48 10 psi 
 
________________________________________________________________________ 
PROBLEM (3.31) Figure P3.31 shows a steel block subjected to an axial compression load of 400 
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 
 
 
 
 
 
 
 
 3400(10 )
166.7
0.04 0.06
x MPa    

 
 
 (a) 
0.3
1500
200
x    
 
 
0.02
500
40
y  
 
 
500 1
1500 3



 
 
 
 (b) 6
6
166.7(10 )
1500(10 )
x
x
E

 
 
111 GPa
 
 
 (c) 6
9
1 166.7 10
( ) 500
111 10 3
x
z
E
    

 
 
6500(10 )(60)a   0.03 mm
 
 
60 0.03 60.03fa mm  
 
 
 (d) 9111 10
2(1 ) 8 3
E
G


 

41.6 GPa
 
L=200 mm 
a=60 mm 
x 
y 
z 
400 kN 
b=40 mm 
 3-14 
________________________________________________________________________ 
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 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 (a) 6240(10 )
200
0.0012
E GPa
 
 
 (b) 
255y MPa 
 
 
 (c) 
240p MPa 
 
 
 (d) 
450u MPa 
 
 
 
0 0.1 0.2 
0.002 0.001 
0.4 
450 
)(MPa
 
M 

 
N 
0.3 
100 
200 
400 
255 
450 
240 
A 
300 
 3-15 
________________________________________________________________________ 
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 
 
 
 
 
 
 
 
 
 
 
 
 
 
 (a) 6242(10 )
69
0.0035
E GPa 
 
 
 (b) 
242p MPa 
 
 
 (c) 
275y MPa 
 
Et 
275 
310 
Es 
0 0.05 0.1 
0.004 0.002 
0.2 
326 
)(MPa
 
M 
 N 
0.15 
100 
200 
0.006 
242 
300 
 3-16 
 
 (d) 
326u MPa 
 
 
 (e) 
50 0.2(50) 50
(100) 20 %
50
 

 
 
 (f) 
2 2
2
(12.7) (8.8)
4 4 (100) 52 %
(12.7)
4
 


 
 
 (g) 
2
2
(12.7)
4( ) 326 679
(8.8)
4
u t MPa



  
 
 (h) 6(330 290)10
0.8
0.05
tE GPa

 
 
 6310 10
10
0.031
sE GPa

 
 
________________________________________________________________________ 
PROBLEM (3.34) A stepped bar of diameters 
ABd
 and 
BCd
 made of an aluminum alloy, is 
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., 
ABd
= 7/8 in., 
BCd
= 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 
 
 (a) Stresses 
 
2
16
26.6
7
( )
4 8
AB
AB
P
ksi
A
   
 
 
2
16
52.2
5
( )
4 8
BC
BC
P
ksi
A
   
 
 
 
 
 
 
- diagram 

(ksi) 
F 
0 
 
 
 
y = 40 
0.04 
60 
E 20 
G 
0.02 0.03 
D 
 3-17 
 From 
 
 diagram, material in region 0D is strained elastically, since 
 
40 26.6 .y ksi  
By Hooke’s Law, 
 3
6
26.6(10 )
2,660
10 10
AB
AB
E
   

 
 Material in region DF is strained plastically, since 
40 52.2 .y ksi  
 
 From 
 
 curve, for 
52.2BC ksi 
: 
 
0.03 30,000BC  
 
 
 The approximate elongation of the bar is thus 
 
6[(2,660 20) (30,000 15)]10 0.5 .L 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: 
 3
6
52.2(10 )
5,000
10(10 )
BC
rec
E
   
 
The remaining plastic strain in part BC is 
 
30,000 5,000 25,000OG   
 
Thus, when the load is removed the bar is elongates 
 
0.025(15) 0.375 .p OG BCL 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 
 
 
 
 
 
 
 3 3
6
2 2
4(10 ) 16(10 )
150 10
4
x
d d
     
 
 
min 0.0058 5.8d m mm 
 
 Also, 
 6
9150 10 200 10
0.0025
E
L

  
 
 or 
 
min 3.333L m
 
 
L 
d 
4 kN 
2.5 mm 
4 kN 
 3-18 
________________________________________________________________________ 
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., L = 4 in., E = 
610 10
 psi, 

= 1/3 
 
 
*SOLUTION 
 34(10 )
0 14.22
(3 4)(3 8)
y z x ksi     
 
 
1
0 [ ( )]y y x z
E
       
 (1) 
 
1
0 [ ( )]z z x y
E
       
 (2) 
 
1
0 [ ( )]x x y z
E
       
 (3) 
 
 The first two expressions become 
 
y z x   
 (1’) 
 
z y x   
 (2’) 
 
 Adding: 
2( ) 2 (1 ).y z x       
 
 
 Equation (3) now reduces to 
 21 2
1
x
x
E
 

 


 
 
 Substituting the data: 
 
3
6
1 2
1
14.22(10 )3 9
1 10 10
1
3
x
 



948 
 
 
________________________________________________________________________ 
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 
 3-19 
 
 
y
x x
d AE
P A EA EA
d
   

   
 
 2
4 4
d d d
E dE
d
 
 
 
 
 
 Inserting the given numerical values: 
 3
60.5 10 (1)(29 10 )
4 3
P 

 
34.16 kips
 
 
________________________________________________________________________ 
PROBLEM (3.38) A short cylindrical pipe of cold-rolled (510) bronze having an outside diameter 
D and thickness t is placed in acompression machine and sequeezed until the axial load applied is P 
with a safety factor of ns against yielding. Determine the minimum required Dmin 
Given: t = D/6, P = 250 kips, 
sn
 = 2, 
y
 = 75 ksi (from Table B.4) 
 
SOLUTION 
 
 2
2 2 5[ ( ) ]
4 3 36
D D
A D D
 
   
 
 
,
y s
all
s all y
PnP
A
n

 
  
 
 Hence 
 2
min
5
, 6
36 5
s s
y y
Pn PnD
D

  
 
 Introducing the data given 
 
min
250 2
6 1.954 .
5 (75)
D in

 
 
 
________________________________________________________________________ 
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 = 
610 10
 psi , 
y
= 38 ksi (from Table B.4). 
 
SOLUTION 
 
 Normal stress: 
 
2
4(8)
13.304 38
7
( )
8
P
ksi ksi
A


   
 
 and hence Hooke’s law applies. 
 
 Normal Strain: 
 3-20 
 
3
13,304
1,330
10(10 )E
   
 
 So 
 
6
0 1,330(10 )(5 12) 0.08 .L L in      
 
0 60 0.08 60.08 .fL L L 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 
 
500(30) 15s sE ksi   
 
 Hence 
 
215[ (1.5 )] 26.5
4
P A kips
  
 
 
 (b) Axial strain in aluminum equals 
 3
6
15(10 )
1,500
10(10 )
a
a
aE
   
 
 Therefore 
 
a a s sL L   
 
 
6[1,500(15) 500(45)]10  0.045 .in
 
 
________________________________________________________________________ 
PROBLEM (3.41) A cold-rolled yellow brass specimen see (Table B.4) has a diameter 
od
 and a 
gage length 
oL
 (Fig. P3.41). When a force P elongates the gage length 

, determine: 
(a) The modulus of elasticity E. 
(b) The contraction of diameter 
0d
. 
Given: 
od
= 0.4 in., 
oL
 = 2 in., 

= 6 (
310
), P = 5.6 kips 
 
SOLUTION 
 
 From Table B.4; 
63y ksi 
and 
65.6 10G ksi 
 
 
(a) Modulus of elasticity. 
 3-21 
 
2
5.6
45
(0.4)
4
P
ksi
A
   
 
 3
0
6(10 )
0.003 3,000
2
a
L
 

   
 
 Since
63 ,y ksi  
 material behaves elastically. 
 So 
 3
645(10 ) 15 10
0.003a
E ksi

   
 
 
(b) Poisson’s ratio: 
 6
6 15 10; 5.6 10 , 0.34
2(1 ) 2(1 )
E
G  

   
 
 
 Then 
 
; 0.34 , 0.00102
0.003
t t
t
a
       
 
 Hence 
 
4
0 0 (0.00102)(0.4) 4.08(10 ) .td d 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: 
 
1 ( )
[ 1]
1 ( )
a E
b E



  

 (P3.42) 
where alb is the initial slope. Calculate the value of 

 when 

 = 120 MPa. 
Given: a = 25 mm, b = 50 mm, 

 = 0.3, E = 70 GPa 
 
*SOLUTION 
 
 
a t
b a
b a
E E
         
 
 
1 ( )
( )
1 ( )
t
f
a
a a E
slope
b b E
 
 
 
 
 
 
 Thus, 
 
1 ( )
( ) ( )
1 ( )
f i
a E a
slope slope
b E b
 

    

 
 
 or 
 
1 ( )
[ 1]
1 ( )
a E
b E



  

 Q.E.D. 
 
 3-22 
 Introducing the data 
 6 9
6 9
25 1 (0.3 120 10 70 10 )
[ 1]
50 1 (120 10 70 10 )
      
  
 
 
0.0011 0.06orad 
 
 
________________________________________________________________________ 
PROBLEM (3.43) Rework Prob.3.42, assuming that the plate is subjected to a uniaxial 
tensile stress 
.
 
 
SOLUTION 
 
 
a t
b a
b a
E E
         
 
 
1 ( )
( )
1 ( )
t
f
a
a a E
slope
b b E
 
 
 
 
 
 
 Thus, 
 
1 ( )
( ) ( ) [ 1]
1 ( )
f i
a E
slope slope
b E
 

    

 
 
 Inserting the given numerical values 
 3
3
25 1 (0.3 120 70 10 )
1
50 1 (120 70 10 )
       
  
 
 
0.011 0.06orad   
 
 
________________________________________________________________________ 
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 limit, 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) 3
2
8(10 )
26.1
(5 8) 4
p ksi 
 
 
 314.4(10 )
2880
5
p 

 
 
 
 (b) 3
6
6
26.08 10
9.1 10
2880(10 )
E psi


  
 
 
 3-23 
 (c) 30.6(10 )
960
5 8
t 

  
 
 
960 1
2880 3
t
a


   
 
 
 (d) 
63 (9.1 10 )
2(1 ) 8
E
G

  

63.4 10 psi 
 
 
________________________________________________________________________ 
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 
 
 
2 22( 50 4) 1250A mm    
 
 (a) 3
6
250 (10 )
200
1250 (10 )
P
MPa
A
    
 
 
 Since 
y 
, the result is valid. 
 Thus, 
 6
9
200(10 )
952
210(10 )E
   
 
 
3 66(10 )(952 10 )L L    5.71 mm 
 
 (b) 
60.3(952)(10 ) 286t        
 
6286(10 )(50)d    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 
yP
 on the specimen that causes tielding. 
(c) The ultimate load 
uP
 the specimen supports. 
(d) Percent elongation in 2.00 in. and percent reduction in area. 
 
SOLUTION 
 
 3-24 
 (a) Refer to 
 
 diagram in Fig. P3.46. 
 
40 ksi 
 when 
0.001 1000  
 
 Hence 
 
640 0 40 10
0.001 0
E ksi

  

 
 
 (b) We have 
40p y ksi  
( by 
 
 curve). 
 
240[ (0.51) ] 8.17
4
y yP A kips
  (c) From 
 
 curve, 
76u ksi 
 
276[ (0.51) ] 15.52
4
uP kips

 
 
(d) Percent elongation
0
0
(100)
fL L
L


 
 
2.4 2
(100) 20 %
2

 
 
 Percent reduction 
iA
 area
0
0
(100)
fA A
A


 
 2 2
2
(0.51) (0.36)
(100) 50.2 %
(0.51)

 
 
 
________________________________________________________________________ 
PROBLEM (*3.47) For many materials, the stress-strain curve may be described by the 
Ramberg-Osgood equation of the form (Reference 3.7): 
 
nk
E

  
 (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 
 
670(10 )p Pa 
 is at 
31 (10 )
3
p

. Thus 
 
210p pE GPa  
 
 We also have 
 
280 MPa 
 at 
0.1, 420 MPa  
 at 
0.3 
 
 
 3-25 
 Applying Eq. (P3.47): 
 
3
3
280
0.1(10 ) (280)
210(10 )
nk  
 
 or 
 
0.012333 (280)nk 
 (1) 
 Similarly, 
 
3
3
420
0.3(10 ) (420)
210(10 )
nk  
 
 or 
 
0.001700 (420)nk 
 (2) 
 Solving, Eqs.(1) and (2), we obtain 
 
60.7915 14.26(10 )n k   
 
 
________________________________________________________________________ 
PROBLEM (3.48) A short cylindrical rod of ASTM – A36 structural steel, having an 
original diameter of 
od
 and length 
oL
 is placed in a compression machine and sequeezed until 
its length becomes 
fL
. Determine the new diameter of the rod. 
Given: d = 30 mm, 
oL
= 50 mm, 
fL
= 49.7 mm, 

= 0.3 
 
SOLUTION 
 
 Axial strain: 
 
0
0
49.7 50
6,000
50
f
a
L L
L
     
 
 Transverse strain: 
 
(0.3)( 0.006) 1,800t a      
 
 Therefore, 
 
0 0.0018(30) 0.054td d mm   
 
 
0 30 0.054 30.054d d d 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): 
 
0
(1 2 ) x
V
V
 

 
 (1) 
 Upon following a procedure similar to that of Sec. 3.10, the final area: 
 
[(1 ) (1 ) ]f y zA dy dz    
 
 3-26 
 
0[1 ( )]y z dydz A A     
 
 and 
0
2y z x
A
A
      
 (2) 
 It is required that 
 
0
0
1 2 5
2 8
V V
A A


 
   
 
 
 Solving, 
 
4
13
 
 
________________________________________________________________________ 
PROBLEM (3.50) A tensile test is performed on a 2024-T6 aluminum alloy specimen of 
diameter d and gage length 
oL
, depicted in Fig. P3.50. After the loading reaches a value of P 
= 5 kips, the distance 
oL
 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., 
oL
= 2 in., 

= 0.33 
 
SOLUTION 
 (a) 3
34.8 100.33( )(0.5) 0.396 10 .
2
d d in       
 (b) 
0
0
PLP A
E
L A

  
  
 
 3
6
2 3
5(10 )(2)
10.6 10
1
( ) (4.8 10 )
4 2
psi 
  

 
 (c) 34.8 10
(1 2 ) (1 2 0.33)( )
2
e  

    
30.816 10 
 
________________________________________________________________________ 
PROBLEM (3.51) A 2-in.-diameter solid brass bar (E = 15 x 106 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: 
 3
2
40 10
12.73
(2) 4
x ksi 

  
 
 and 
 3
6
12.73 10
0.3 255
15 10
x
y
E
     

 
 The diametral increase of the bar is 
 3-27 
 
6 3255(10 )2 0.51 10 .yd d in     
 
 The required internal diameter of the tube is thus 
 
32 0.51 10iD d d
    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, P = 200 kN, E = 70 GPa, 

= 0.3 
Assumption: Buckling does not occur. 
 
SOLUTION 
 
 3
9 2 2
200(10 )
70(10 )( 4)(0.13 0.1 )
P
E AE
 

  

527  
 
 
 (a) 
6 3527(10 )(0.5 10 )L L     0.264 mm  
 
 (b) 
6( ) 0.3(527)(10 )130D D     0.021 mm 
 
 (c) 
6( ) 0.3(527)(10 )15t t     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 
 
 3
9 2 2
200(10 )
105(10 )( 4)(0.13 0.1 )
P
E AE
    
351 
 
 
 (a) 
6351(10 )500L L    0.176 mm
 
 
 (b) 
6( ) 0.3(351)(10 )130D D      0.014 mm  
 
 (c) 
6( ) 0.3(351)(10 )15t t       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. 
 
 3-28 
SOLUTION 
 
 
 
 
 
 
2 6 3(0.05 )(1.2) 2355(10 )
4
oV m
  
 
 3
2
200(10 )
101.9
(0.05 ) 4
x MPa  
 
 6
9
101.9(10 )
1456
70(10 )
x  
 
 
60.3(1456)(10 ) 437t  
 
 
6437(10 )(0.05) 0.022td d mm     
 
max 50.022d mm
 
 Also, 
 
6 6(1 2 ) 0.4(1456)(10 ) 582(10 )xe        
 
6 6582(10 )(2355)(10 )oV eV
   
6 31.371 10 m 
 
 
6
max (2355 1.371)10oV V V
   
3 32356.371(10 ) mm
 
 
________________________________________________________________________ 
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: 
 
6 30.022 2355(10 )d mm V m  
 
6 31.3710V m  
 
 Thus, 
 
min 50 0.022 49.978d mm  
 
 
6
min (2355 1.371)10V
 
3 32353.629(10 ) mm
 
________________________________________________________________________ 
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 
 
 
 
 
 100 mm 
20 mm 
50 kN 
0.2 mm 
50 kN 
1.2 m 
200 kN 
0.05 m 
 3-29 
 
 3
2
50(10 )
159.2
(0.02) 4
x MPa  
 
 (a) 
0.2
2000 ln(1 0.002) 1998
100
x t       
 
 (b) 6159.2(10 )
79.6
0.002
E GPa 
 
________________________________________________________________________ 
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. 
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
610
 psi, 

= 0.3 
 
*SOLUTION 
 
 
 
 
 
 
2 2
4
4
x
x x t x
P P
d E d E
     

    
 
 
 
4
t
P
d d
dE



  
 
 or 
4
d
P dE



 
 Introducing the given number values: 
 
60.002 ( 1 11 10 )
4 0.3
P    

57.6 kips
 
________________________________________________________________________ 
PROBLEM (3.58) A cast-iron bar of diameter d and length L is subjected to an axial 
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=1 in. 
P 
d=1.002 in. 
P 
0.5 m 
d=75 mm 
200 kN 200 kN 
 3-30 
 3
2
200(10 )
45.3
(0.075) 4
x MPa 

  
 
 6
9
45.3(10 )
566
80 10
x
x
E
     

 
 
 (a) 
6( ) 0.3(566)(10 )75xd d    0.013 mm
 
 
 (b) 
6566(10 )500xL L     0.283 mm 
 
 
 (c) 
6(1 2 ) 0.4( 566)(10 )xe      6226.4(10 ) 
 
 
6 2
4
226.4(10 )[ (0.075 )(0.5)]oV eV
   
 
 
6 3 30.5(10 ) 500m mm   
 
________________________________________________________________________ 
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 = 30x106 psi, 

= 1/3 
 
SOLUTION 
 
 
 
 
 
 
 
2 3(1) 4 12.566 .oV in 
 
 3
6
7.2(10 )
240
30 10
x  

 
 
 (a) 
6 3240(10 )4 0.96 10 .xL L in       
 
 (b) 
(1 2 ) x oV V   
 
 
6 31 (240)(10 )(12.566) .
3
in
3 31.005 10 .in 
 
________________________________________________________________________ 
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: 
7.2 ksi 2 in. 
4 in. 
 3-31 
 
1 2
( )x y z 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 
 
 or 
1
2
 
 
________________________________________________________________________ 
PROBLEM (3.61) The rectangular concrete block shown in Fig. P3.61 is subjected to loads that 
have the resultants 
xP
= 100 kN, 
yP
= 150 kN, and 
zP
= 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 
 
 
 
 
 
 
 
 
 3100(10 )
20
0.1 0.05
x MPa

  

 
 3150(10 )
15
0.2 0.05
y MPa

  

 
 350(10 )
2.5
0.2 0.1
z MPa

  

 
 
 Thus, 
 6
9
10
[20 0.2(15 2.5)] 688
24(10 )
x      
 
 6
9
10
[15 0.2(22.5)] 438
24(10 )
y     
 
 6
9
10
[2.5 0.2(35)] 188
24(10 )
z    
 
 
0.05 m 
0.1 m 
0.2 m 
A 
D 
E 
C 
B 
150 kN 
50 kN 
100 kN 
z 
y 
x 
 3-32 
 (a) 
6438(10 )100 0.044ABL mm
    
 
 
6688(10 )200 0.138BCL mm
    
 
 
6188(10 )50 0.009CEL mm
  
 
 
 (b) 
6
9
438(10 )
0.2 0.05(24 10 )
y y
y
P P
AE
      
 
 
 or 
 
105.1yP kN
 
 
________________________________________________________________________ 
PROBLEM (3.62) Redo Prob. 3.61 for the x faces of the block to be free of stress. 
 
SOLUTION 
 
 We have 
0x 
. From Solution of Prob. 3.61: 
 
15 2.5y zMPa MPa    
 
 Thus, 
 6
9
10
[ 0.2(17.5)] 146
24(10 )
x    
 
 6
9
10
[15 0.2(2.5)] 604
24(10 )
y     
 
 6
9
10
[2.5 0.2(15)] 21
24(10 )
z    
 
 
 (a) 
6604(10 )100 0.604ABL mm
    
 
 
6146(10 )200 0.0292BCL mm
    
 
 
621(10 )50 0.001CEL mm
  
 
 
 (b) 
6
9
604(10 )
0.2 0.05(24 10 )
y
y
P    
 
, 
145yP 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. 
 
 3-33 
*SOLUTION 
 
(a) According to assumption 1, the rubber is in triaxial stress: 
2
2
4
,
4
x z y
R R
p
d
d
          
 
 
 Strains are:
0.x z  
 The first of Eqs. (3.17) gives 
 
1
0 [ ( )]x x y z
E
       
 
 or 
 
2
4
0 ( )
R
p p
d


   
 
 Solving, 
 
2
4
(1 )
R
p
d

 


 
 
(b) Substituting the data, 
3
2
4(0.45)(4 10 )
1.157 ( )
(0.06) (1 0.45)
p MPa C

 

 
________________________________________________________________________ 
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 
xw
 and 
yw
 (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 1000
3000 2000
x y      
 
 970 10
27
2(1 0.3)
G GPa

 

 
 
 Equations (3.15): 
 
6
9
1
500(10 ) ( 0.3 )
70(10 )
x y   
 
 
6
9
1
1000(10 ) ( 0.3 )
70(10 )
y x   
 
 or 
 
635(10 ) 0.3x y  
 (1) 
 
670(10 ) 0.3y x  
 (2) 
 Solve Eqs. (1) and (2): 
x
 
z
 
y
 
 3-34 
 
61.5 88.5x yMPa MPa  
 
 Thus, 
 
661.5(10 )(0.005) 308xw kN m 
 
 
688.5(10 )(0.005) 443yw 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
750 500
200 4000
x y      
 
 9210(10 )
80.8
2(1 0.3)
G GPa 

 
 
 Equations (3.15): 
 
6
9
1
750(10 ) ( 0.3 )
210(10 )
x y   
 
 
6
9
1
500(10 ) ( 0.3 )
210(10 )
y x   
 
 or 
 
6157.5(10 ) 0.3x y  
 (1) 
 
6105(10 ) 0.3y x  
 (2) 
 Solve Eqs. (1) and (2): 
 
207.3 167.2x yMPa MPa  
 
 Hence, 
 
6207.3(10 )(0.006) 1244xw kN m 
 
 
6167.2(10 )(0.006) 1003yw 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 
 
 
 
 
 
x y z      
 
 
p=120 MPa= - 
d=200 mm 
 3-35 
 
3 3 6 34 4 (100 ) 4.19(10 )
3 3
oV r mm
  
 
 
 (a) 
1 [ ( )] (1 2 )x E E
            
 6
9
120(10 )
(1 0.6) 686
70 10
   

 
 
 
6686(10 )200xd d     0.137 mm 
 
 Decrease in circumference: 
 
( ) 0.137 0.43d mm      
 
 (b) 
(1 2 ) x oV V   
 
 
6 6(0.4)( 686)(10 )(4.19 10 )  31150 mm 
 
________________________________________________________________________ 
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
610
 psi, 

= 1/3 
 
SOLUTION 
 
 
2 3(1) 4 12.566 .oV in 
 
 
1
[ (2 )] (1 2 )x
E E
         
 
 3
3
7.2(10 )(1 3)
80
30(10 )
   
 
 
 (a) 
6 380(10 )4 0.32 10 .L in      
 
 
 (b) 
(1 2 ) x oV V   
 
 
6 30.4( 80)(10 )(12.566) .in  3 30.402 10 .in  
 
________________________________________________________________________ 
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) The length of edge AB. 
 (b) The length of edge AD. 
 (c) The length of diagonal BD. 
 (d) The thickness. 
 
Given: a = 160 mm, b = 200 mm, t = 5 mm, 
x
= 120 MPa, 
y
= 90 MPa 
 E = 200 GPa, 

= 1/3 
 3-36 
 
*SOLUTION 
 
 
 
2 2160 200BDL   256 mm
 
 
 
 
 
 
 6
9
1 10 90
( ) (120 ) 450
200(10 ) 3
x x y
E
       
 
 6
9
1 10 120
( ) (90 ) 250
200(10 ) 3
y y x
E
       
 
 
 (a) 
6200(250)(10 ) 0.05ABL mm
  
 
 
 (b) 
6160(450)(10 ) 0.072ADL mm
  
 
 
 (c) 
2 2 2
BD AB ADL L L 
 
 
2 2 2BD BD AB AB AD ADL L L L L L   
 
 or 
 
AB AD
BD AB AD
BD BD
L L
L L L
L L
    
 (1) 
 
200 160
(0.05) (0.072)
256 256
BDL  
0.084 mm
 
 
 (d) 
1
[0 ( )]z x y
E
     
 
 6
9
10 210
( ) 350
200(10 ) 3
   
 
 
6350(10 )5 0.002zt t mm       
 
________________________________________________________________________ 
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 
 
 6
9
10 60
(100 ) 600
200 10 3
x   

 
160 mm 
200 mm 
A 
D 
y 
C B 
90 MPa 
120 MPa 
x 
 3-37 
 6
9
10 100
( 60 ) 467
200 10 3
y     

 
 6
9
10 100 60
(0 ) 67
200 10 3
z    

 
 
 (a) 
6200( 467)(10 ) 0.093ABL mm
    
 
 
 (b) 
6160(600)(10 ) 0.096ADL mm
  
 
 
 (c) From Eq. (1) of solution of Prob. 3.68: 
 
200 160
( 0.093) (0.096)
256 256
BDL   
0.013 mm
 
 
 (d) 
667(10 )5zt t     30.34(10 ) 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: 
 6190(10 )
190
0.001
E GPa 
 
 
245y MPa 
 
 (a) 2 6 2
0 9 3
(245 10 )
158
2 2(190 10 )
y kJ
U
E m
 
  

 
 
 (b) Total area under 
 
 diagram: 
 
6
3
350 10 (0.28) 98t
MJ
U
m
  
 
________________________________________________________________________ 
PROBLEM (3.71) From the stress-strain curve of a magnesium alloy seen in Fig. P3.71, find 
the approximate values of: 
(a) The modulus of resilience. 
(b) The modulus of toughness. 
 
SOLUTION 
 
 Referring Fig. P3.71, we have 
 6100 10
45
0.0022
E GPa

 
 
200y MPa 
 
 
 3-38 
 (a) 2 6 2
0 9
(200 10 )
2 2(45 10 )
y
U
E
 
 

3
444
kJ
m

 
 
 (b) Total area under 
 
 diagram: 
 
6250(10 )(0.176)tU 
3
44
MJ
m

 
________________________________________________________________________ 
PROBLEM (3.72) Calculate the modulus of resilience for two grades of steel (see Table 
D.4): 
(a) ASTM-A242. 
(b) Cold-rolled, stainless steel (302). 
 
SOLUTION 
 
 (a) ASTM-A242 
 
200 345yE GPa MPa 
 
 2 6 2
0 9 3
(345 10 )
298
2 2(200 10 )
y kJ
U
E m
 
  

 
 
3298 43.2 . .
6.895
in lb in  
 
 
 (b) Stainless (302) 
 
190 520yE GPa MPa 
 
 2 6 2
0 9 3
(520 10 )
712
2 2(190 10 )
y kJ
U
E m
 
  

 
 
3712 103 . .
6.895
in lb in  
 
________________________________________________________________________ 
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: 
6 31 (280)(10 )(0.001) 140
2
oU m kN m  
 
 
(b) Toughness modulus is represented by the total area under 
 
 curve that can be 
Estimated by counting the numberof square from the diagram. From the diagram, 
The number of square equals about 45. 
Thus, 
 
6 345(10 )(70 0.04) 126tU m MN m   
 
 3-39 
________________________________________________________________________ 
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 
 
72 410yE GPa MPa 
 
 2 6 2
0 9
(410 10 )
2 2(72 10 )
y
U
E
 
 

 
 
3
3
1167
1167 169 . .
6.895
kJ
in lb in
m
   
 
 
 (b) Yellow, annealed, brass 
 
105 105yE GPa MPa 
 
 2 6 2
0 9
(105 10 )
2 2(105 10 )
y
U
E
 
 

 
 
3
3
52.5
52.5 7.61 . .
6.895
kJ
in lb in
m
   
 
 
________________________________________________________________________ 
PROBLEM (3.75) A
3
41
-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 
 
 
31.75 1.75 4 12 147 .V in    
 
 
 (a) 2
2
p
sn U V
E

 
 
 Solving, 
 6
2 (2 ) 2(1500)(2 30 10 )
147
s
p
n U E
V
    812.24 10 
 
 or 
35p ksi 
 
 
 (b) 2 8
3
6
12.24 10
20.4 . .
2 2(30 10 )
p
oU in lb in
E
 
   

 
 
 
 3-40 
________________________________________________________________________ 
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 cross-
sectional areas AAB , and ABC. 
 
SOLUTION 
 
tan sin
AB BC
P P
F F  
 
 for 
:uP P
 
 350 10
28.87
tan 60
AB o
F kips

 
 
 350 10
57.74
sin 60
BC o
F kips

 
 
 Thus, 
 3
2
3
28.87 10
0.722 .
40 10
AB
AB
y
F
A in

  

 
 3
2
3
57.74 10
1.444 .
40 10
BC
BC
y
F
A in

  

 
 
________________________________________________________________________ 
PROBLEM (3.77) Solve Prob. .3.75 for Pu = 20 kips and 

 = 30°. 
 
SOLUTION 
 
 We have 
 
 320 10
34.64
tan tan 30
u
AB o
P
F kips

  
 
 320 10
40
sin sin 30
u
BC o
P
F kips

  
 
 Thus, 
 3
2
3
34.64 10
0.866 .
40 10
BC
BC
y
F
A in

  

 
 3
2
3
40 10
1.0 .
40 10
BCA in

 

 
 
________________________________________________________________________ 
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° 
 
 3-41 
SOLUTION 
 
 Since 
BC ABA A
 and 
sin tan 
: 
 
338 10 (0.9) 34.2BC yp BCF A kips   
 
 
 Thus, 
 
3sin 34.2 10 (sin 40 )ou BCP F   21.98 kips
 
 
 
 End of Chapter 3