CHAPTER 4
20
40
20
PROBLEM 4.1
20
A
M 5 15 kN · m
Knowing that the couple shown acts in a vertical plane, determine
the stress at (a) point A, (b) point B.
80
20
B
Dimensions in mm
SOLUTION
For rectangle:
I 
Outside rectangle:
I1 
1 3
bh
12
1
(80)(120)3
12
I1  11.52  106 mm 4  11.52  106 m 4
Cutout:
I2 
1
(40)(80)3
12
I 2  1.70667  106 mm 4  1.70667  106 m 4
Section:
(a)
(b)
I  I1  I 2  9.81333  106 m 4
y A  40 mm  0.040 m
yB  60 mm  0.060 m
A  
My A
(15  103 )(0.040)

 61.6  106 Pa
6
I
9.81333  10
B  
MyB
(15  103 )(0.060)

 91.7  106 Pa
I
9.81333  106
 A  61.6 MPa 
 B  91.7 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
447
2 in. 2 in. 2 in.
PROBLEM 4.2
M 25 kip · in.
A
B
Knowing that the couple shown acts in a vertical plane, determine
the stress at (a) point A, (b) point B.
2 in.
1.5 in.
2 in.
SOLUTION
For rectangle:
I
1 3
bh
12
For cross sectional area:
I  I1  I 2  I 3 
1
1
1
(2)(1.5)3  (2)(5.5)3  (2)(1.5)3  28.854 in 4
12
12
12
(a)
y A  2.75 in.
A  
My A
(25)(2.75)

I
28.854
(b)
yB  0.75 in.
B  
(25)(0.75)
MyB

28.854
I
 A  2.38 ksi 
 B  0.650 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
448
200 mm
PROBLEM 4.3
12 mm
y
C
x
Using an allowable stress of 155 MPa, determine the largest bending
moment M that can be applied to the wide-flange beam shown. Neglect the
effect of fillets.
220 mm
M
8 mm
12 mm
SOLUTION
Moment of inertia about x-axis:
I1 
1
(200)(12)3  (200)(12)(104)2
12
I2 
1
(8)(196)3  5.0197  106 mm 4
12
 25.9872  106 mm 4
I 3  I1  25.9872  106 mm 4
I  I1  I 2  I 3  56.944  106 mm 4  56.944  106 m 4
 
Mc
1
with c  (220)  110 mm  0.110 m
I
2
I
M 
with   155  106 Pa
c
Mx 
(56.944  106 )(155  106 )
 80.2  103 N  m
0.110
M x  80.2 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
449
200 mm
PROBLEM 4.4
12 mm
y
Solve Prob. 4.3, assuming that the wide-flange beam is bent about the y axis
by a couple of moment My.
C
x
220 mm
PROBLEM 4.3. Using an allowable stress of 155 MPa, determine the
largest bending moment M that can be applied to the wide-flange beam
shown. Neglect the effect of fillets.
M
8 mm
12 mm
SOLUTION
Moment of inertia about y axis:
I1 
1
(12)(200)3  8  106 mm 4
12
1
(196)(8)3  8.3627  103 mm 4
I2 
12
I 3  I1  8  106 mm 4
I  I1  I 2  I 3  16.0084  106 mm 4  16.0084  106 m 4
 
1
Mc
with c  (200)  100 mm  0.100 m
2
I
I
with   155  106 Pa
My 
c
My 
(16.0084  106 )(155  106 )
 24.8  103 N  m
0.100
M y  24.8 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
450
0.1 in.
PROBLEM 4.5
0.5 in.
Using an allowable stress of 16 ksi, determine the largest couple that can be
applied to each pipe.
M1
(a)
0.2 in.
0.5 in.
M2
(b)
SOLUTION
(a)
I 
4
o
c  0.6 in.
 
(b)
r
4

I 
Mc
:
I

 ri4 

M 
4
(0.64  0.54 )  52.7  103 in 4
I
c

(16)(52.7  103 )
0.6
M  1.405 kip  in. 

(0.74  0.54 )  139.49  103 in 4
4
c  0.7 in.
 
Mc
:
I
M 
I
c

(16)(139.49  103 )
0.7
M  3.19 kip  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
451
r 5 20 mm
A
PROBLEM 4.6
M = 2.8 kN · m
30 mm
B
30 mm
Knowing that the couple shown acts in a vertical plane,
determine the stress at (a) point A, (b) point B.
120 mm
SOLUTION
I 
1
1 

(0.120 m)(0.06 m)3  2   (0.02 m) 4 
12
12
4


 2.1391  106 mm 4
(a)
(b)
A  
B 
(2.8  103 N  m)(0.03 m)
M yA

I
2.1391  106 mm 4
 A  39.3 MPa 
(2.8  103 N  m)(0.02 m)
M yB

I
2.1391  106 m 4
 B  26.2 MPa 

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
452
PROBLEM 4.7
y
Two W4  13 rolled sections are welded together as shown. Knowing that for the steel
alloy used  Y  36 ksi and  U  58 ksi and using a factor of safety of 3.0, determine the
largest couple that can be applied when the assembly is bent about the z axis.
z
C
SOLUTION
Properties of W4  13 rolled section.
(See Appendix C.)
Area  3.83 in 2
Width  4.060 in.
I y  3.86 in 4
For one rolled section, moment of inertia about axis b-b is
I b  I y  Ad 2  3.86  (3.83)(2.030)2  19.643 in 4
For both sections,
I z  2 I b  39.286 in 4
c  width  4.060 in.
 all 
M all
U

58
 19.333 ksi
F .S . 3.0
 I (19.333)(39.286)
 all 
4.060
c

Mc
I
M all  187.1 kip  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
453
y
PROBLEM 4.8
C
Two W4  13 rolled sections are welded together as shown. Knowing that for the steel
alloy used  U  58 ksi and using a factor of safety of 3.0, determine the largest couple
that can be applied when the assembly is bent about the z axis.
z
SOLUTION
Properties of W4  13 rolled section.
(See Appendix C.)
Area  3.83 in 2
Depth  4.16 in.
I x  11.3 in 4
For one rolled section, moment of inertia about axis a-a is
I a  I x  Ad 2  11.3  (3.83)(2.08)2  27.87 in 4
For both sections,
I z  2 I a  55.74 in 4
c  depth  4.16 in.
 all 
M all
U

58
 19.333 ksi
F .S . 3.0
 I (19.333)(55.74)
 all 
4.16
c

Mc
I
M all  259 kip  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
454
PROBLEM 4.9
3 in. 3 in. 3 in.
Two vertical forces are applied to a beam of the cross section shown. Determine
the maximum tensile and compressive stresses in portion BC of the beam.
6 in.
2 in.
A
15 kips
15 kips
B
C
40 in.
60 in.
D
40 in.
SOLUTION



A
y0
A y0
18
5
90
18
1
18
36
Y0 
108
108
 3 in.
36
Neutral axis lies 3 in. above the base.
I1 
1
1
b1h13  A1d12  (3)(6)3  (18)(2) 2  126 in 4
12
12
1
1
3
2
I 2  b2 h2  A2 d 2  (9)(2)3  (18)(2) 2  78 in 4
12
12
I  I1  I 2  126  78  204 in 4
ytop  5 in. ybot  3 in.
M  Pa  0
M  Pa  (15)(40)  600 kip  in.
 top  
M ytop
 bot  
M ybot
(600)(3)

I
204
I

 top  14.71 ksi (compression) 
(600)(5)
204
 bot  8.82 ksi (tension) 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
455
PROBLEM 4.10
8 in.
1 in.
Two vertical forces are applied to a beam of the cross section shown.
Determine the maximum tensile and compressive stresses in portion
BC of the beam.
6 in.
1 in.
1 in.
4 in.
A
25 kips
25 kips
B
C
20 in.
60 in.
D
20 in.
SOLUTION




A
y0
A y0
8
7.5
60
6
4
24
4
0.5
18
Yo 
2
86
86
 4.778 in.
18
Neutral axis lies 4.778 in. above the base.
I1 
1
1
b1h13  A1d12  (8)(1)3  (8)(2.772) 2  59.94 in 4
12
12
1
1
I 2  b2 h23  A2 d 22  (1)(6)3  (6)(0.778)2  21.63 in 4
12
12
1
1
3
2
I 3  b3 h3  A3 d3  (4)(1)3  (4)(4.278)2  73.54 in 4
12
12
I  I1  I 2  I 3  59.94  21.63  73.57  155.16 in 4
ytop  3.222 in. ybot  4.778 in.
M  Pa  0
M  Pa  (25)(20)  500 kip  in.
 top  
 bot  
Mytop
I

(500)(3.222)
155.16
Mybot
(500)(4.778)

I
155.16
 top  10.38 ksi (compression) 
 bot  15.40 ksi (tension) 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
456
10 mm
PROBLEM 4.11
10 mm
10 kN
10 kN
B
50 mm
C
A
D
Two vertical forces are applied to a beam of
the cross section shown. Determine the
maximum tensile and compressive stresses
in portion BC of the beam.
10 mm
50 mm
150 mm
250 mm
150 mm
SOLUTION




A, mm 2
y0 , mm
A y0 , mm3
600
30
600
30
18  103
300
5
1500
Y0 
37.5  103
 25mm
1500
18  103
1.5  103
37.5  103
Neutral axis lies 25 mm above the base.
I1 
1
(10)(60)3  (600)(5)2  195  103 mm 4 I 2  I1  195 mm 4
12
1
I 3  (30)(10)3  (300)(20) 2  122.5  103 mm 4
12
I  I1  I 2  I 3  512.5  103 mm 4  512.5  109 m 4
ytop  35 mm  0.035 m
ybot  25 mm  0.025 m
a  150 mm  0.150 m P  10  103 N
M  Pa  (10  103 )(0.150)  1.5 103 N  m
(1.5  103 )(0.035)
 102.4  106 Pa
512.5  109
 top  
Mytop
 bot  
M ybot
(1.5  103 )(0.025)

 73.2  106 Pa
I
512.5  109
I

 top  102.4 MPa (compression) 
 bot  73.2 MPa (tension) 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
457
216 mm
y
36 mm
54 mm
z
PROBLEM 4.12
C
108 mm
Knowing that a beam of the cross section shown is bent about a horizontal
axis and that the bending moment is 6 kN  m, determine the total force
acting on the shaded portion of the web.
72 mm
SOLUTION
The stress distribution over the entire cross section is given by the bending stress formula:
x  
My
I
where y is a coordinate with its origin on the neutral axis and I is the moment of inertia of the entire cross
sectional area. The force on the shaded portion is calculated from this stress distribution. Over an area
element dA, the force is
dF   x dA  

The total force on the shaded area is then
F  dF  

My
M
dA  
I
I
My
dA
I
 y dA  
M * *
y A
I
where y * is the centroidal coordinate of the shaded portion and A* is its area.
d1  54  18  36 mm
d 2  54  36  54  36 mm
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
458
PROBLEM 4.12 (Continued)
Moment of inertia of entire cross section:
I1 
1
1
b1h13  A1d12  (216)(36)3  (216)(36)(36)2  10.9175  106 mm 4
12
12
1
1
I 2  b2 h23  A2 d 22  (72)(108)3  (72)(108)(36)2  17.6360  106 mm 4
12
12
I  I1  I 2  28.5535  106 mm 4  28.5535  106 m 4
For the shaded area,
A*  (72)(90)  6480 mm 2
y *  45 mm
A* y *  291.6  103 mm3  291.6  106 m
F
MA* y * (6  103 )(291.6  106 )

I
28.5535  106
 61.3  103 N
F  61.3 kN 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
459
20 mm
12 mm
PROBLEM 4.13
20 mm
y
12 mm
24 mm
Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is 4 kN  m, determine the total force acting on
the shaded portion of the beam.
20 mm
z
C
20 mm
24 mm
SOLUTION
Dimensions in mm:
Iz 
1
1
(12  12)(88)3  (40)(40)3
12
12
 1.3629  106  0.213  106
 1.5763  106 mm 4  1.5763  106 m 4
For use in Prob. 4.14,
Iy 
1
1
(88)(64)3  (24  24)(40)3
12
12
6
 1.9224  10  0.256  106
 1.6664  106 mm 4  1.6664  106 m 4
Bending about horizontal axis.
M z  4 kN  m
M z c (4 kN  m)(0.044 m)

 111.654 MPa
Iz
1.5763  10 6 m 4
M c (4 kN  m)(0.020 m)
B  z 
 50.752 MPa
Iz
1.5763  10 6 m 4
A 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
460
PROBLEM 4.13 (Continued)
A  (44)(12)  528 mm 2  528  10 6 m 2
Portion (1):
 avg   A  (111.654)  55.83 MPa
1
2
1
2
Force1   avg A  (55.83 MPa)(528  106 m 2 )  29.477 kN
A  (20)(20)  400 mm 2  400  10 6 m 2
Portion (2):
 avg 
1
1
 B  (50.752)  25.376 MPa
2
2
Force2   avg A  (25.376 MPa)(400  106 m 2 )  10.150 kN
Total force on shaded area  29.477  10.150  39.6 kN

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
461
20 mm
12 mm
PROBLEM 4.14
20 mm
y
12 mm
Solve Prob. 4.13, assuming that the beam is bent about a vertical axis by a
couple of moment 4 kN  m.
24 mm
PROBLEM 4.13. Knowing that a beam of the cross section shown is bent
about a horizontal axis and that the bending moment is 4 kN  m, determine the
total force acting on the shaded portion of the beam.
20 mm
z
C
20 mm
24 mm
SOLUTION
M y  4 kN  m
Bending about vertical axis.
See Prob. 4.13 for sketch and
D 
E 
I y  1.6664  106 m 4
Mc (4 kN  m)(0.032 m)

 76.81 MPa
Iy
1.6664  106 m 4
Mc (4 kN  m)(0.020 m)

 48.01 MPa
Iy
1.6664  106 m 4
A  (44)(12)  528 mm 2  528  10 6 m 2
Portion (1):
 avg  ( D   E )  (76.81  48.01)  62.41 MPa
1
2
1
2
Force1   avg A  (62.41 MPa)(528  106 m 2 )  32.952 kN
A  (20)(20)  400 mm 2  400  106 m 2
Portion (2):
 avg 
1
1
 E  (48.01)  24.005 MPa
2
2
Force2   avg A  (24.005 MPa)(400  106 m 2 )  9.602 kN
Total force on shaded area  32.952  9.602  42.6 kN

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
462
0.5 in.
0.5 in.
0.5 in.
PROBLEM 4.15
1.5 in.
1.5 in.
Knowing that for the extruded beam shown the allowable stress is 12 ksi
in tension and 16 ksi in compression, determine the largest couple M that
can be applied.
1.5 in.
0.5 in.
M
SOLUTION


Ay0
A
y0
2.25
1.25
2.8125
2.25
0.25
0.5625
4.50
Y
The neutral axis lies 0.75 in. above bottom.
ytop  2.0  0.75  1.25 in.,
I1 
3.375
3.375
 0.75 in.
4.50
ybot  0.75 in.
1
1
b1h13  A1d12  (1.5)(1.5)3  (2.25)(0.5)2  0.984375 in 4
12
12
1
1
2
2
I 2  b2 h2  A2 d 2  (4.5)(0.5)3  (2.25)(0.5)2  0.609375 in 4
12
12
I  I1  I 2  1.59375 in 4
 
My
I
M 
I
y
Top: (compression)
M 
(16)(1.59375)
 20.4 kip  in.
1.25
Bottom: (tension)
M 
(12)(1.59375)
 25.5 kip  in.
0.75
Choose the smaller as Mall.
M all  20.4 kip  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
463
PROBLEM 4.16
40 mm
The beam shown is made of a nylon for which the allowable stress is
24 MPa in tension and 30 MPa in compression. Determine the largest
couple M that can be applied to the beam.
15 mm
d 30 mm
20 mm
M
SOLUTION


Σ
Y0 
A, mm2
y0 , mm
A y0 , mm3
600
22.5
300
7.5
2.25  103
13.5  103
15.75  103
900
15.5  103
 17.5 mm
900
The neutral axis lies 17.5 mm above the bottom.
ytop  30  17.5  12.5 mm  0.0125 m
ybot  17.5 mm  0.0175 m
I1 
1
1
b1h13  A1d12  (40)(15)3  (600)(5)2  26.25  103 mm 4
12
12
1
1
I 2  b2 h23  A2 d 22  (20)(15)3  (300)(10)2  35.625  103 mm 4
12
12
I  I1  I 2  61.875  103 mm 4  61.875  109 m 4
| | 
My
I
Top: (tension side)
M
Bottom: (compression)
M
M
I
y
(24  106 )(61.875  109 )
 118.8 N  m
0.0125
(30  106 )(61.875  109 )
 106.1 N  m
0.0175
Choose smaller value.
M  106.1 N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
464
PROBLEM 4.17
40 mm
Solve Prob. 4.16, assuming that d  40 mm.
15 mm
d 30 mm
PROBLEM 4.16 The beam shown is made of a nylon for which the
allowable stress is 24 MPa in tension and 30 MPa in compression.
Determine the largest couple M that can be applied to the beam.
20 mm
M
SOLUTION
A, mm2
y0 , mm
A y0 , mm3
600
32.5
500
12.5
6.25  103


Σ
Y0 
25.75  103
1100
25.75  103
 23.41 mm
1100
19.5  103
The neutral axis lies 23.41 mm above the bottom.
ytop  40  23.41  16.59 mm  0.01659 m
ybot  23.41 mm  0.02341 m
I1 
1
1
b1h13  A1d12  (40)(15)3  (600)(9.09) 2  60.827  103 mm 4
12
12
1
1
I 2  b2 h22  A2 d 22  (20)(25)3  (500)(10.91) 2  85.556  103 mm 4
12
12
I  I1  I 2  146.383  103 mm 4  146.383  109 m 4
| | 
M
My
I
Top: (tension side)
M
Bottom: (compression)
M
I
y
(24  106 )(146.383  109 )
 212 N  m
0.01659
(30  106 )(146.383  109 )
 187.6 N  m
0.02341
Choose smaller value.
M  187.6 N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
465
PROBLEM 4.18
2.4 in.
0.75 in.
1.2 in.
Knowing that for the beam shown the allowable stress is 12 ksi in tension and
16 ksi in compression, determine the largest couple M that can be applied.
M
SOLUTION
  rectangle
  semi-circular cutout
A1  (2.4)(1.2)  2.88 in 2
A2 

2
(0.75)2  0.8836 in 2
A  2.88  0.8836  1.9964 in 2
y1  0.6 in.
y2 
4r
(4)(0.75)

 0.3183 in.
3
3
(2.88)(0.6)  (0.8836)(0.3183)
A y

 0.7247 in.
Y 
1.9964
A
Neutral axis lies 0.7247 in. above the bottom.
Moment of inertia about the base:
1
1


I b  bh3  r 4  (2.4)(1.2)3  (0.75) 4  1.25815 in 4
3
8
3
8
Centroidal moment of inertia:
I  I b  AY 2  1.25815  (1.9964)(0.7247) 2  0.2097 in 4
ytop  1.2  0.7247  0.4753 in.,
ybot  0.7247 in.
| | 
M
My
I
I
y
Top: (tension side)
M
(12)(0.2097)
 5.29 kip  in.
0.4753
Bottom: (compression)
M
(16)(0.2097)
 4.63 kip  in.
0.7247
Choose the smaller value.
M  4.63 kip  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
466
PROBLEM 4.19
80 mm
Knowing that for the extruded beam shown the allowable stress is
120 MPa in tension and 150 MPa in compression, determine the largest
couple M that can be applied.
54 mm
40 mm
M
SOLUTION


Σ
Y 
A, mm2
y0 , mm
2160
27
58,320
3
1080
36
38,880
3
A y0 , mm3
3240
97, 200
 30 mm
3240
d, mm
97,200
The neutral axis lies 30 mm above the bottom.
ytop  54  30  24 mm  0.024 m
I1 
ybot  30 mm  0.030 m
1
1
b1h13  A1d12  (40)(54)3  (40)(54)(3)2  544.32  103 mm 4
12
12
1
1
1
I 2  b2 h22  A2 d 22  (40)(54)3  (40)(54)(6)2  213.84  103 mm 4
36
36
2
3
4
I  I1  I 2  758.16  10 mm  758.16  109 m 4
| | 
My
I
|M | 
I
y
Top: (tension side)
M
Bottom: (compression)
M
Choose the smaller as Mall.
(120  106 )(758.16  109 )
 3.7908  103 N  m
0.024
(150  106 )(758.16  109 )
 3.7908  103 N  m
0.030
M all  3.7908  103 N  m
M all  3.79 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
467
PROBLEM 4.20
48 mm
Knowing that for the extruded beam shown the allowable stress is
120 MPa in tension and 150 MPa in compression, determine the largest
couple M that can be applied.
48 mm
36 mm
48 mm
36 mm
M
SOLUTION
 Solid rectangle
 Square cutout
Σ
Y 
A, mm2
y0 , mm
4608
48
221,184
–1296
30
–38,880
3312
182,304
 55.04 mm
3312
A y0 , mm3
182,304
Neutral axis lies 55.04 mm above bottom.
ytop  96  55.04  40.96 mm  0.04096 m
ybot  55.04 mm  0.05504 m
I1 
1
1
b1h13  A1d12  (48)(96)3  (48)(96)(7.04)2  3.7673  106 mm 4
12
12
1
1
I 2  b2 h32  A2 d 22  (36)(36)3  (36)(36)(25.04) 2  0.9526  106 mm 4
12
12
I  I1  I 2  2.8147  106 mm 4  2.8147  106 m 4
| | 
My
I
 M 
I
y
(120  106 )(2.8147  106 )
 8.25  103 N  m
0.04096
Top: (tension side)
M
Bottom: (compression)
M
Mall is the smaller value.
M  7.67  103 N  m
(150  106 )(2.8147  106 )
 7.67  103 N  m
0.05504
7.67 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
468
PROBLEM 4.21
Straight rods of 6-mm diameter and 30-m length are stored by coiling the
rods inside a drum of 1.25-m inside diameter. Assuming that the yield
strength is not exceeded, determine (a) the maximum stress in a coiled rod,
(b) the corresponding bending moment in the rod. Use E  200 GPa.
SOLUTION
D  inside diameter of the drum,
Let
d  diameter of rod,
c
1
d,
2
  radius of curvature of center line of rods when bent.

I
(a)
 max 
(b)
M 


Ec
EI


1
1
1
1
D  d  (1.25)  (6  103 )  0.622 m
2
2
2
2

4
c4 

4
(0.003) 4  63.617  1012 m 4
(200  109 )(0.003)
 965  106 Pa
0.622
  965 MPa 
(200  109 )(63.617  1012 )
 20.5 N  m
0.622
M  20.5 N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
469
M'
PROBLEM 4.22
M
8 mm
A 900-mm strip of steel is bent into a full circle by two couples applied
as shown. Determine (a) the maximum thickness t of the strip if the
allowable stress of the steel is 420 MPa, (b) the corresponding moment
M of the couples. Use E  200 GPa.
t
r
900 mm
SOLUTION
When the rod is bent into a full circle, the circumference is 900 mm. Since the circumference is equal to
2 times , the radius of curvature, we get
 
900 mm
 143.24 mm  0.14324 m
2
  E 
Stress:

Ec
or
c
For   420 MPa and E  200 GPa,
c
(a)

E
(0.14324)(420  106 )
 0.3008  103 m
200  109
Maximum thickness:
t  2c  0.6016  103 m
t  0.602 mm 
Moment of inertia for a rectangular section.
I 
(b)
bt 3
(8  103 )(0.6016  103 )3

 145.16  1015 m 4
12
12
Bending moment:
M
M 

EI
(200  109 )(145.16  1015 )
 0.203 N  m
0.14324
M  0.203 N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
470
PROBLEM 4.23
5 ft
Straight rods of 0.30-in. diameter and 200-ft length are sometimes
used to clear underground conduits of obstructions or to thread wires
through a new conduit. The rods are made of high-strength steel and,
for storage and transportation, are wrapped on spools of 5-ft
diameter. Assuming that the yield strength is not exceeded,
determine (a) the maximum stress in a rod, when the rod, which is
initially straight, is wrapped on a spool, (b) the corresponding
bending moment in the rod. Use E  29  106 psi .
SOLUTION
Radius of cross section:
r 
Moment of inertia:
I 
D  5 ft  60 in.
c  r  0.15 in.
(a)
 max 
(b)
M 

EI

Ec


 
1
1
d  (0.30)  0.15 in.
2
2

4
r4 

4
(0.15) 4  397.61  106 in 4
1
D  30 in.
2
(29  106 )(0.15)
 145.0  103 psi
30
(29  106 )(397.61  106 )
30
 max  145.0 ksi 
M  384 lb  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
471
PROBLEM 4.24
12 mm
y
60 N · m
A 60-N  m couple is applied to the steel bar shown. (a) Assuming that
the couple is applied about the z axis as shown, determine the maximum
stress and the radius of curvature of the bar. (b) Solve part a, assuming
that the couple is applied about the y axis. Use E  200 GPa.
20 mm
z
SOLUTION
(a)
Bending about z-axis.
I
1 3 1
bh  (12)(20)3  8  103 mm 4  8  109 m 4
12
12
20
c
 10 mm  0.010 m
2


1
(b)

Mc (60)(0.010)

 75.0  106 Pa
9
I
8  10
60
M

 37.5  103 m 1
EI (200  109 )(8  109 )
  75.0 MPa 
  26.7 m 
Bending about y-axis.
I
1 3 1
bh  (20)(12)3  2.88  103 mm 4  2.88  109 m 4
12
12
12
 6 mm  0.006 m
c
2
Mc (60)(0.006)


 125.0  106 Pa
9
I
2.88  10

1

60
M

 104.17  103 m 1
9
EI (200  10 )(2.88  109 )
  125.0 MPa 
  9.60 m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
472
10 mm
M
(a) Using an allowable stress of 120 MPa, determine the largest couple
M that can be applied to a beam of the cross section shown. (b) Solve
part a, assuming that the cross section of the beam is an 80-mm square.
80 mm
C
PROBLEM 4.25
10 mm
80 mm
5 mm
5 mm
SOLUTION
(a)
I  I1  4 I 2 , where I1 is the moment of inertia of an 80-mm square and I2 is the moment of inertia of
one of the four protruding ears.
I1 
1 3 1
bh  (80)(80)3  3.4133  106 mm 4
12
12
I2 
1 3
1
bh  Ad 2  (5)(10)3  (5)(10)(45)2  101.667  103 mm 4
12
12
I  I1  4 I 2  3.82  106 mm 4  3.82  106 mm 4 ,
c  50 mm  0.050 m

(b)
Without the ears:
Mc
I
M
I
c

 9.168  103 N  m
I  I1  3.4133  106 m 2 ,
M
I
c

(120  106 )(3.82  106 )
0.050
 9.17 kN  m 
c  40 mm  0.040 m
(120  106 )(3.4133  106 )
 10.24  103 N  m
0.040
 10.24 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
473
0.1 in.
PROBLEM 4.26
0.2 in.
A thick-walled pipe is bent about a horizontal axis by a couple M. The
pipe may be designed with or without four fins. (a) Using an allowable
stress of 20 ksi, determine the largest couple that may be applied if the
pipe is designed with four fins as shown. (b) Solve part a, assuming that
the pipe is designed with no fins.
1.5 in.
M
0.75 in.
SOLUTION
I x of hollow pipe:
I x of fins:
Ix 

(1.5 in.) 4  (0.75 in.) 4   3.7276 in 4
4

1

1

I x  2  (0.1)(0.2)3  (0.1  0.2)(1.6)2   2  (0.2)(0.1)3 
12
12




 0.1026 in 4
(a)
Pipe as designed, with fins:
I x  3.8302 in 4 ,
 all  20 ksi,
(b)
Pipe with no fins:
 all  20 ksi,
M   all
c  1.7 in.
M   all
Ix
3.8302 in 4
 (20 ksi)
1.7 in.
c
I x  3.7276 in 4 ,
Ix
3.7276 in 4
 (20 ksi)
1.5 in.
c
M  45.1 kip  in. 
c  1.5 in.
M  49.7 kip  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
474
PROBLEM 4.27
A couple M will be applied to a beam of rectangular cross section that
is to be sawed from a log of circular cross section. Determine the ratio
d/b for which (a) the maximum stress m will be as small as possible,
(b) the radius of curvature of the beam will be maximum.
M
M'
d
b
SOLUTION
Let D be the diameter of the log.
D 2  b2  d 2
I 
(a)
1
bd 3
12
c
m is the minimum when
d 2  D 2  b2
I
is maximum.
c
I
1
 b( D 2  b 2 )
6
c
d I 1 2
  D 
db  c  6
d 
(b)
 
I
1
 bd 2
c
6
1
d
2
D2 

1 2
1
D b  b3
6
6
3 2
b 0
6
1 2
D 
3
b
1
D
3
2
D
3
d

b
2 
d

b
3 
EI
M
 is maximum when I is maximum,
1
bd 3 is maximum, or b2d 6 is maximum.
12
( D 2  d 2 )d 6 is maximum.
6D2d 5  8d 7  0
b
D2 
d 
3 2
1
D  D
4
2
3
D
2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
475
PROBLEM 4.28
h0
M
h
C
h0
SOLUTION
h
A portion of a square bar is removed by milling, so that its cross section
is as shown. The bar is then bent about its horizontal axis by a couple M.
Considering the case where h  0.9h0, express the maximum stress in the
bar in the form  m  k 0 , where  0 is the maximum stress that would
have occurred if the original square bar had been bent by the same
couple M, and determine the value of k.
I  4 I1  2 I 2
 1 
1
 (4)   h h3  (2)   (2h0  2h)(h3 )
 12 
3
1
4
4
4
 h 4  h 0 h3  h h3  h 0 h3  h 4
3
3
3
3
ch
Mc
Mh
3M



4 h h3  h 4 (4h 0  3h) h 2
I
3 0
For the original square,
h  h0 , c  h0 .
0 
3M
3M
 3
2
h0
(4h0  3h0 )h0
h03
h03



 0.950
 0 (4h0  3h)h 2 (4h0  (3)(0.9)h0 )(0.9 h02 )
  0.950  0
k  0.950 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
476
PROBLEM 4.29
h0
M
In Prob. 4.28, determine (a) the value of h for which the maximum stress
 m is as small as possible, (b) the corresponding value of k.
h
PROBLEM 4.28 A portion of a square bar is removed by milling, so that
its cross section is as shown. The bar is then bent about its horizontal
axis by a couple M. Considering the case where h  0.9h0, express the
maximum stress in the bar in the form  m  k 0 , where  0 is the
maximum stress that would have occurred if the original square bar had
been bent by the same couple M, and determine the value of k.
C
h
h0
SOLUTION
I  4 I1  2 I 2
 1 
1
 (4)   hh3  (2)   (2h0  2h) h3
12
 
3
1
4
4
4
 h 4  h 0 h3  h3  h 0 h 3  h 4
3
3
3
3
I 4
2
3
ch
 h0 h  h
c 3
d 4
I

is maximum at
h 0 h 2  h3   0.

c
dh  3

8
h 0 h  3h 2  0
3

256 3
I 4 8  8 
 h0  h0    h0  
h0
729
c 3 9  9 
For the original square,
2
h  h0
3
c  h0
0 
h

Mc 729M


I
256h03
8
h 0 
9
I0 1 3
 h0
c0 3
Mc0 3M
 2
I0
h0
 729 1 729

 
 0.949
 0 256 3 768
k  0.949 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
477
PROBLEM 4.30
For the bar and loading of Concept Application 4.1, determine (a) the radius of curvature  , (b) the radius of
curvature   of a transverse cross section, (c) the angle between the sides of the bar that were originally
vertical. Use E  29  106 psi and v  0.29.
SOLUTION
M  30 kip  in.
From Example 4.01,
(a)
(b)

1

(30  103 )
M

 993  106 in.1
EI (29  106 )(1.042)
 1  v 

vc
v
I  1.042 in 4
  1007 in. 

c
1
1
 v  (0.29)(993  106 )in.1  288  106 in.1


(c)

length of arc b
0.8


 230  106 rad
  3470
radius
   3470 in. 
  0.01320 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
478
PROBLEM 4.31
y
A
z
M
A W200  31.3 rolled-steel beam is subjected to
45 kN  m. Knowing that E  200 GPa and v 
radius of curvature  , (b) the radius of curvature
section.
a couple M of moment
0.29, determine (a) the
  of a transverse cross
C
x
SOLUTION
For W 200  31.3 rolled steel section,
I  31.3  106 mm 4
 31.3  10 6 m 4
(a)
(b)

1

45  103
M

 7.1885  103 m 1
EI (200  109 )(31.3  106 )
1
1
 v  (0.29)(7.1885  103 )  2.0847  103 m 1


  139.1 m 
   480 m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
479
PROBLEM 4.32
y
␪
2
y ⫽ ⫹c
␴y
It was assumed in Sec. 4.1B that the normal stresses  y in a
member in pure bending are negligible. For an initially straight
elastic member of rectangular cross section, (a) derive an
approximate expression for  y as a function of y, (b) show that
( y )max  (c /2  )( x ) max and, thus, that  y can be neglected in
all practical situations. (Hint: Consider the free-body diagram of
the portion of beam located below the surface of ordinate y and
assume that the distribution of the stress  x is still linear.)
␪
2
␴y
␴x
␴x
␪
2
␪
2
y ⫽ ⫺c
SOLUTION
Denote the width of the beam by b and the length by L.

cos
Using the free body diagram above, with
Fy  0 :
y  
But,
(a)
y 
( x )max
c

 y bL  2
2

sin
2
L
 x  ( x )max
y
c
y dy 
( x )max y 2
c 2

y
c


2
y
c

L
1
 x b dy sin
 x dy  
L


2
y
c
0
 x dy  

1
y
c
 x dy
y
c
y 
y
c
The maximum value  y occurs at y  0 .
( y )max  
(b)
( x ) max 2
( y  c 2 ) 
2 c
( x )max c 2
( ) c
  x max 
2c
2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
480
PROBLEM 4.33
Brass
6 mm
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the
composite bar is bent about a horizontal axis.
Aluminum
30 mm
6 mm
Aluminum
Brass
70 GPa
105 GPa
100 MPa
160 MPa
Modulus of elasticity
30 mm
Allowable stress
SOLUTION
Use aluminum as the reference material.
n  1.0 in aluminum
n  Eb /Ea  105/70  1.5 in brass
For the transformed section,
I1 
n1
b1h13  n1 A1d 12
12
1.5

(30)(6)3  (1.5)(30)(6)(18)3  88.29  103 mm 4
12
I2 
n2
1.0
b2h23 
(30)(30)3  67.5  103 mm 4
12
12
I 3  I1  88.29  103 mm 4
I  I1  I 2  I 3  244.08  103 mm 4
 
Aluminum:
M 
I
ny
y  15 mm  0.015 m,   100  106 Pa
(100  106 )(244.08  10 9 )
 1.627  103 N  m
(1.0)(0.015)
n  1.5,
M
Choose the smaller value
nMy
I
n  1.0,
M
Brass:
 244.08  109 m4
y  21 mm  0.021 m,   160  106 Pa
(160  106 )(244.08  10 9 )
 1.240  103 N  m
(1.5)(0.021)
M  1.240  103 N  m
1.240 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
481
PROBLEM 4.34
8 mm
8 mm
32 mm
8 mm
32 mm
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the
composite bar is bent about a horizontal axis.
8 mm
Brass
Aluminum
Brass
70 GPa
105 GPa
100 MPa
160 MPa
Modulus of elasticity
Aluminum
Allowable stress
SOLUTION
Use aluminum as the reference material.
For aluminum, n  1.0
For brass, n  Eb /Ea  105/70  1.5
Values of n are shown on the sketch.
For the transformed section,
I1 
n1
1.5
b1h13 
(8)(32)3  32.768  103 mm 4
12
12
n2
1.0
I 2  b2 H 23  h23 
(32)(323  163 )  76.459  103 mm 4
12
12
I 3  I1  32.768  103 mm 4


I  I1  I 2  I 3  141.995  103 mm 4  141.995  109 m 4
| | 
Aluminum:
M
I
ny
n  1.0, | y |  16 mm  0.016 m,   100  106 Pa
M
Brass:
nMy
I
(100  106 )(141.995  109 )
 887.47 N  m
(1.0)(0.016)
n  1.5, | y |  16 mm  0.016 m,   160  106 Pa
M
(160  106 )(141.995  109 )
 946.63 N  m
(1.5)(0.016)
Choose the smaller value.
M  887 N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
482
PROBLEM 4.35
Brass
6 mm
For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
30 mm
PROBLEM 4.35. Bar of Prob. 4.33.
Aluminum
Modulus of elasticity
6 mm
Allowable stress
30 mm
Aluminum
Brass
70 GPa
105 GPa
100 MPa
160 MPa
SOLUTION
Use aluminum as reference material.
n  1.0 in aluminum
n  Eb /Ea  105/70  1.5 in brass
For transformed section,
I1 
n1
b1h13
12
1.5
(6)(30)3  20.25  103 mm 4

12
I2 
n2
b2h23
12
1.0
(30)(30)3  67.5  103 mm 4

12
I3  I1  20.25  103 mm 4
I  I1  I 2  I 3  108  103 mm 4  108  10 9 m 4
 
Aluminum:
n  1.0,
M
Brass:
nMy
I
I
ny
y  15 mm  0.015 m,   100  106 Pa
(100  106 )(108  10 9 )
 720 N  m
(1.0)(0.015)
n  1.5,
M
 M 
y  15 mm  0.015 m,   160  106 Pa
(160  106 )(108  109 )
 768 N  m
(1.5)(0.015)
Choose the smaller value.
M  720 N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
483
PROBLEM 4.36
8 mm
8 mm
32 mm
8 mm
For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
PROBLEM 4.36 Bar of Prob. 4.34.
32 mm
8 mm
Brass
Modulus of elasticity
Allowable stress
Aluminum
Aluminum
70 GPa
100 MPa
Brass
105 GPa
160 MPa
SOLUTION
Use aluminum as the reference material.
For aluminum, n  1.0
For brass, n  Eb /Ea  105/70  1.5
Values of n are shown on the sketch.

For the transformed section,
I1 

n1
1.5
(32)(483  323 )  311.296  103 mm 4
h1 B13  b13 
12
12
n2
1.0
I 2  h2b23 
(8)(32)3  21.8453  103 mm 4
12
12
I 3  I 2  21.8453  103 mm 4
I  I1  I 2  I 3  354.99  103 mm 4  354.99  109 m 4
| | 
Aluminum:
I
ny
(100  106 )(354.99  109 )
 2.2187  103 N  m
(1.0)(0.016)
n  1.5 | y |  24 mm  0.024 m   160  106 Pa
M
Choose the smaller value.
M
n  1.0, | y |  16 mm  0.016 m,   100  106 Pa
M
Brass:
nMy
I
(160  106 )(354.99  109 )
 1.57773  103 N  m
(1.5)(0.024)
M  1.57773  103 N  m
M  1.578 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
484
PROBLEM 4.37
1
5 3 2 in.
Wooden beams and steel plates are securely bolted together to form the composite
member shown. Using the data given below, determine the largest permissible bending
moment when the member is bent about a horizontal axis.
10 in.
Wood
1
5 3 2 in.
Modulus of elasticity:
6 in.
Allowable stress:
Steel
2  106 psi
29  106 psi
2000 psi
22 ksi
SOLUTION
Use wood as the reference material.
n  1.0 in wood
n  Es /Ew  29/2  14.5 in steel
For the transformed section,
I1 
n1
b1h13  n1 A1d12
12
14.5  1 
1
(5)    (14.5)(5)   (5.25)2  999.36 in 4
12
2
 
2
n
1.0
(6)(10)3  500 in 4
I 2  2 b2 h22 
12
12
I 3  I1  999.36 in 4

3
I  I1  I 2  I 3  2498.7 in 4
 
Wood:
Steel:
nMy
I
 M 
n  1.0,
M 
ny
y  5 in.,   2000 psi
(2000)(2499)
 999.5  103 lb  in.
(1.0)(5)
n  14.5,
M 
I
y  5.5 in.,   22 ksi  22  103 psi
(22  103 )(2499)
 689.3  103 lb  in.
(14.5)(5.5)
Choose the smaller value.
M  689  103 lb  in.
M  689 kip  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
485
PROBLEM 4.38
10 in.
Wooden beams and steel plates are securely bolted together to form the composite
member shown. Using the data given below, determine the largest permissible bending
moment when the member is bent about a horizontal axis.
Wood
3 in.
1
2
3 in.
Modulus of elasticity:
in.
Allowable stress:
Steel
2  106 psi
29  106 psi
2000 psi
22 ksi
SOLUTION
Use wood as the reference material.
n  1.0 in wood
n  Es /Ew  29/2  14.5 in steel
For the transformed section,
I1 
n1
1.0
(3)(10)3  250 in 4
b1h13 
12
12
n
14.5  1 
I 2  2 b2 h23 
(10)3  604.17 in 4


12
12  2 
I 3  I1  250 in 4
 
Wood:
Steel:
nMy
I
 M 
n  1.0,
M 
ny
y  5 in.,   2000 psi
(2000)(1104.2)
 441.7  103 lb  in.
(1.0)(5)
n  14.5,
M 
I
I  I1  I 2  I 3  1104.2 in 4
y  5 in.,   22 ksi  22  103 psi
(22  103 )(1104.2)
 335.1  103 lb  in.
(14.5)(5)
Choose the smaller value.
M  335  103 lb  in.
M  335 kip  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
486
PROBLEM 4.39
A copper strip (Ec = 105 GPa) and an aluminum strip (Ea = 75 GPa)
are bonded together to form the composite beam shown. Knowing
that the beam is bent about a horizontal axis by a couple of moment
M = 35 N  m, determine the maximum stress in (a) the aluminum
strip, (b) the copper strip.
SOLUTION
Use aluminum as the reference material.
n  1.0 in aluminum
n  Ec /Ea  105/75  1.4 in copper
Transformed section:


A, mm2
nA, mm 2
y0 , mm
nAy0 , mm3
144
144
9
1296
144
201.6
3
Σ
Y0 
345.6
604.8
1900.8
1900.8
 5.50 mm
345.6
The neutral axis lies 5.50 mm above the bottom.
I1 
n1
1.0
b1h13  n1 A1d12 
(24)(6)3  (1.0)(24)(6)(3.5) 2  2196 mm 4
12
12
n
1.4
(24)(6)3  (1.4)(24)(6)(2.5)2  1864.8 mm 4
I 2  2 b2 h23  n2 A2 d 22 
12
12
I  I1  I 2  4060.8 mm 4  4.0608  109 m 4
(a)
(b)
Aluminum:
Copper:
n  1.0,
 
nMy
(1.0)(35)(0.0065)

 56.0  106 Pa  56.0 MPa
9
I
4.0608  10
n  1.4,
 
y  12  5.5  6.5 mm  0.0065 m
y  5.5 mm  0.0055 m
  56.0 MPa 
nMy
(1.4)(35)(0.0055)

 66.4  106 Pa  66.4 MPa
I
4.0608  109
  66.4 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
487
PROBLEM 4.40
A copper strip (Ec = 105 GPa) and an aluminum strip (Ea = 75 GPa)
are bonded together to form the composite beam shown. Knowing
that the beam is bent about a horizontal axis by a couple of moment
M = 35 N  m, determine the maximum stress in (a) the aluminum
strip, (b) the copper strip.
SOLUTION
Use aluminum as the reference material.
n  1.0 in aluminum
n  Ec /Ea  105/75  1.4 in copper
Transformed section:


A, mm2
nA, mm 2
Ay0 , mm
nAy0 , mm3
216
216
7.5
1620
72
100.8
1.5
Σ
Y0 
316.8
151.8
1771.2
1771.2
 5.5909 mm
316.8
The neutral axis lies 5.5909 mm above the bottom.
I1 
n1
1.0
b1h13  n1 A1d12 
(24)(9)3  (1.0)(24)(9)(1.9091) 2  2245.2 mm 4
12
12
n
1.4
I 2  2 b2 h23  n2 A2 d 22 
(24)(3)3  (1.4)(24)(3)(4.0909)2  1762.5 mm 4
12
12
I  I1  I 2  4839 mm 4  4.008  109 m 4
(a)
(b)

Aluminum:
Copper:

n  1.0,
 
y  12  5.5909  6.4091 mm  0.0064091
nMy
(1.0)(35)(0.0064091)

 56.0  106 Pa
I
4.008  109
 56.0 MPa
n  1.4, y  5.5909 mm  0.0055909 m

nMy
(1.4)(35)(0.0055909)

 68.4  106 Pa
 
9
I
4.008  10
 68.4 MPa
  56.0 MPa 

  68.4 MPa 

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
488
PROBLEM 4.41
6 in.
M
12 in.
53
1
2
The 6  12-in. timber beam has been strengthened by bolting to it the steel
reinforcement shown. The modulus of elasticity for wood is 1.8  106 psi and for
steel, 29  106 psi. Knowing that the beam is bent about a horizontal axis by a
couple of moment M  450 kip  in., determine the maximum stress in (a) the
wood, (b) the steel.
in.
SOLUTION
Use wood as the reference material.
For wood,
For steel,
Transformed section:
421.931
Yo 
112.278
 3.758 in.
n 1
n  Es /Ew  29 / 1.8  16.1111
  wood

  steel
A, in 2
nA, in 2
72
72

2.5
40.278
y0
6
0.25
112.278
nA y0 , in 3
432
10.069
421.931
The neutral axis lies 3.758 in. above the wood-steel interface.
I1 
n1
1
b1h13  n1 A1d12  (6)(12)3  (72)(6  3.758) 2  1225.91 in 4
12
12
n2
16.1111
I 2  b2 h23  n2 A2 d 22 
(5)(0.5)3  (40.278)(3.578  0.25)2  647.87 in 4
12
12
I  I1  I 2  1873.77 in 4
M  450 kip  in.
(a)
Wood:
n  1,
 
y  12  3.758  8.242 in.
w  
(b)
Steel:
(1)(450)(8.242)
 1.979 ksi
1873.77
n  16.1111,
s  
nMy
I
y  3.758  0.5  4.258 in.
(16.1111)(450)(4.258)
 16.48 ksi
1873.77
 w  1.979 ksi 
 s  16.48 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
489
PROBLEM 4.42
6 in.
M
12 in.
The 6  12-in. timber beam has been strengthened by bolting to it the steel
reinforcement shown. The modulus of elasticity for wood is 1.8  106 psi and for
steel, 29  106 psi. Knowing that the beam is bent about a horizontal axis by a
couple of moment M  450 kip  in., determine the maximum stress in (a) the
wood, (b) the steel.
C8 11.5
SOLUTION
For wood,
Use wood as the reference material.
For steel,
n 1
n
Es 29  106

 16.1111
Ew 1.8  106
For C8  11.5 channel section,
A  3.38 in 2 , tw  0.220 in., x  0.571 in., I y  1.32 in 4
For the composite section, the centroid of the channel (part 1) lies 0.571 in. above the bottom of the section.
The centroid of the wood (part 2) lies 0.220  6.00  6.22 in. above the bottom.
Transformed section:
A, in2
3.38
Part
1
72
2

Y0 
nA, in2
54.456
y , in.
0.571
nAy , in 3
31.091
d, in.
3.216
72
6.22
447.84
2.433
478.93
126.456
d  y0  Y0
478.93 in 3
 3.787 in.
126.456 in 2
The neutral axis lies 3.787 in. above the bottom of the section.
I1  n1 I1  n1 A1d12  (16.1111)(1.32)  (54.456)(3.216) 2  584.49 in 4
I2 
n2
1
b2 h23  n2 A2 d 22  (6)(12)3  (72)(2.433) 2  1290.20 in 4
12
12
I  I1  I 2  1874.69 in 4
M  450 kip  in
(a)
(b)
Wood:
Steel:
n  1,
w  
 
n My
I
y  12  0.220  3.787  8.433 in.
(1)(450)(8.433)
 2.02 ksi
1874.69
n  16.1111, y  3.787 in.
s  
(16.1111)(450)(3.787)
 14.65 ksi
1874.67
 w  2.02 ksi 
 s  14.65 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
490
PROBLEM 4.43
Aluminum
Copper
6 mm
6 mm
For the composite beam indicated, determine the radius of curvature
caused by the couple of moment 35 N  m.
Beam of Prob. 4.39.
24 mm
SOLUTION
See solution to Prob. 4.39 for the calculation of I.

1

35
M

 0.1149 m 1
9
Ea I (75  10 )(4.0608  109 )
  8.70 m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
491
PROBLEM 4.44
Aluminum
9 mm
Copper
3 mm
For the composite beam indicated, determine the radius of curvature
caused by the couple of moment 35 N  m.
Beam of Prob. 4.40.
24 mm
SOLUTION
See solution to Prob. 4.40 for the calculation of I.

1

M
35

 0.1164 m 1
Ea I (75  109 )(4.008  109 )
  8.59 m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
492
6 in.
PROBLEM 4.45
M
12 in.
For the composite beam indicated, determine the radius of curvature caused by the
couple of moment 450 kip  in.
Beam of Prob. 4.41.
53
1
2
in.
SOLUTION
See solution to Prob. 4.41 for calculation of I.
I  1873.77 in 4
Ew  1.8  106 psi
M  450 kip  in  450  103 lb  in.

1

M
450  103

 133.421  106 in.1
6
EI (1.8  10 )(1873.77)
  7495 in.  625 ft 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
493
6 in.
PROBLEM 4.46
M
12 in.
For the composite beam indicated, determine the radius of curvature caused by the
couple of moment 450 kip  in.
Beam of Prob. 4.42.
C8 11.5
SOLUTION
See solution to Prob. 4.42 for calculation of I.
I  1874.69 in 4
Ew  1.8  106 psi
M  450 kip  in.  450  103 lb  in.

1

M
450  103

 133.355  106 in.1
EI (1.8  106 )(1874.69)
  7499 in.  625 ft 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
494
PROBLEM 4.47
5
8
-in. diameter
4 in.
5.5 in.
5.5 in.
A concrete slab is reinforced by 85 -in.-diameter steel rods
placed on 5.5-in. centers as shown. The modulus of elasticity is
3  106 psi for the concrete and 29  106 psi for the steel.
Using an allowable stress of 1400 psi for the concrete and
20 ksi for the steel, determine the largest bending moment in a
portion of slab 1 ft wide.
5.5 in.
6 in.
5.5 in.
SOLUTION
n
Es 29  106

 9.6667
Ec
3  106
Consider a section 5.5 in. wide.
As 

4
d s2 
 5
 0.3068 in 2
4  8 
2
nAs  2.9657 in 2
Locate the natural axis.
5.5 x
x
 (4  x )(2.9657)  0
2
2.75 x 2  2.9657 x  11.8628  0
Solve for x.
x  1.6066 in.
4  x  2.3934 in.
1
I  (5.5) x3  (2.9657)(4  x)2
3
1
 (5.5)(1.6066)3  (2.9657)(2.3934) 2  24.591 in 4
3
 
nMy
I
Concrete: n  1,
M 
M 
I
ny
y  1.6066 in.,   1400 psi
(24.591)(1400)
 21.429  103 lb  in.
(1.0)(1.6066)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
495
PROBLEM 4.47 (Continued)
Steel:
n  9.6667,
M 
y  2.3934 in.,   20 ksi=20  103 psi
(24.591)(20  103 )
 21.258  103 lb  in.
(9.6667)(2.3934)
Choose the smaller value as the allowable moment for a 5.5 in. width.
M  21.258  103 lb  in.
For a 1 ft = 12 in. width,
M 
12
(21.258  103 )  46.38  103 lb  in.
5.5
M  46.38 kip  in.
3.87 kip  ft 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
496
PROBLEM 4.48
5
8
-in. diameter
Solve Prob. 4.47, assuming that the spacing of the 85 -in.-diameter
steel rods is increased to 7.5 in.
4 in.
PROBLEM 4.47 A concrete slab is reinforced by 85 -in.-diameter
steel rods placed on 5.5-in. centers as shown. The modulus of
elasticity is 3 × 106 psi for the concrete and 29  106 psi for the
steel. Using an allowable stress of 1400 psi for the concrete and
20 ksi for the steel, determine the largest bending moment in a
portion of slab 1 ft wide.
5.5 in.
5.5 in.
5.5 in.
6 in.
5.5 in.
SOLUTION
n
Number of rails per foot:

Es 29  106 psi

 9.667
Ec
3  106 psi
12 in.
 1.6
7.5 in.
5
 5
Area of -in.-diameter bars per foot: 1.6  
4 8
8
As  0.4909 in 2
2
Transformed section, all concrete.
First moment of area:
x
12 x    4.745(4  x)  0
2
x  1.4266 in.
nAs  9.667(0.4909)  4.745 in 2
1
I NA  (12)(1.4266)3  4.745(4  1.4266)2  43.037 in 4
3
For concrete:
 all  1400 psi c  x  1.4266 in.
M 
For steel:
43.037 in 4
1.4266 in.
 all  20 ksi c  4  x  4  1.4266  2.5734 in.
M
We choose the smaller M.
I
 (1400 psi)
c
 steel I
n

c

20 ksi 43.042 in 4

9.667 2.5734 in.
M  34.60 kip  in.
M  42.24 kip.in. 
M  34.60 kip  in. 
M  2.88 kip  ft 
Steel controls.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
497
PROBLEM 4.49
540 mm
The reinforced concrete beam shown is subjected to a positive bending
moment of 175 kN  m. Knowing that the modulus of elasticity is 25 GPa
for the concrete and 200 GPa for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
25-mm
diameter
60 mm
300 mm
SOLUTION
n
Es 200 GPa

 8.0
25 GPa
Ec
As  4 

4
 
d 2  (4)   (25) 2  1.9635  103 mm 2
4
nAs  15.708  103 mm 2
Locate the neutral axis.
x
 (15.708  103 )(480  x)  0
2
150 x 2  15.708  103 x  7.5398  106  0
300 x
Solve for x.
15.708  103  (15.708  103 ) 2  (4)(150)(7.5398  106 )
(2)(150)
x  177.87 mm,
480  x  302.13 mm
x
1
I  (300) x3  (15.708  103 )(480  x) 2
3
1
 (300)(177.87)3  (15.708  103 )(302.13) 2
3
 1.9966  109 mm 4  1.9966  103 m 4
 
(a)
Steel:
y  302.45 mm  0.30245 m
 
(b)
Concrete:
nMy
I
(8.0)(175  103 )(0.30245)
 212  106 Pa
3
1.9966  10
  212 MPa 
y  177.87 mm  0.17787 m
 
(1.0)(175  103 )(0.17787)
 15.59  106 Pa
1.9966  103
  15.59 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
498
PROBLEM 4.50
540 mm
25-mm
diameter
60 mm
300 mm
Solve Prob. 4.49, assuming that the 300-mm width is increased to 350 mm.
PROBLEM 4.49 The reinforced concrete beam shown is subjected to a
positive bending moment of 175 kN  m. Knowing that the modulus of
elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine
(a) the stress in the steel, (b) the maximum stress in the concrete.
SOLUTION
n
Es 200 GPa

 8.0
25 GPa
Ec
As  4

4
 
d 2  (4)   (25) 2
4
 1.9635  103 mm 2
nAs  15.708  103 mm 2
Locate the neutral axis.
x
 (15.708  103 )(480  x)  0
2
175 x 2  15.708  103 x  7.5398  106  0
350 x
Solve for x.
15.708  103  (15.708  103 ) 2  (4)(175)(7.5398  106 )
(2)(175)
x  167.48 mm, 480  x  312.52 mm
x
1
I  (350) x3  (15.708  103 )(480  x)2
3
1
 (350)(167.48)3  (15.708  103 )(312.52) 2
3
 2.0823  109 mm 4  2.0823  103 m 4
nMy
 
I
(a)
Steel:
y  312.52 mm  0.31252 m
 
(b)
Concrete:
(8.0)(175  103 )(0.31252)
 210  106 Pa
3
2.0823  10
  210 MPa 
y  167.48 mm  0.16748 m
 
(1.0)(175  103 )(0.16748)
 14.08  106 Pa
2.0823  103
  14.08 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
499
4 in.
24 in.
PROBLEM 4.51
Knowing that the bending moment in the reinforced concrete beam is
100 kip  ft and that the modulus of elasticity is 3.625  106 psi for the
concrete and 29  106 psi for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
20 in.
1-in.
diameter
2.5 in.
12 in.
SOLUTION
n
29  106
Es

 8.0
Ec
3.625  106
 
As  (4)   (1)2  3.1416 in 2
4
nAs  25.133 in

2
Locate the neutral axis.
 x
(24)(4)( x  2)  (12 x)    (25.133)(17.5  4  x)  0
2
96 x  192  6 x 2  339.3  25.133x  0
x
Solve for x.
or
6 x 2  121.133x  147.3  0
121.133  (121.133)2  (4)(6)(147.3)
 1.150 in.
(2)(6)
d3  17.5  4  x  12.350 in.
I1 
1
1
b1h13  A1d12 
(24)(4)3  (24)(4)(3.150) 2  1080.6 in 4
12
12
1 3 1
I 2  b2 x  (12)(1.150)3  6.1 in 4
3
3
I 3  nA3d32  (25.133)(12.350) 2  3833.3 in 4
I  I1  I 2  I 3  4920 in 4
 
nMy
I
(a)
Steel:
n  8.0
(b)
Concrete:
n  1.0,
c  
where M  100 kip  ft  1200 kip  in.
s  
y  12.350 in.
(8.0)(1200)(12.350)
4920
 s  24.1 ksi 
y  4  1.150  5.150 in.
(1.0)(1200)(5.150)
4920
 c  1.256 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
500
PROBLEM 4.52
7
8
16 in.
-in. diameter
2 in.
A concrete beam is reinforced by three steel rods placed as shown. The
modulus of elasticity is 3  106 psi for the concrete and 29  106 psi for the
steel. Using an allowable stress of 1350 psi for the concrete and 20 ksi for the
steel, determine the largest allowable positive bending moment in the beam.
8 in.
SOLUTION
n
Es 29  10 6

 9.67
Ec
3  106
As  3

   7 
d 2  (3)     1.8040 in 2
4
 4  8 
2
nAs  17.438 in 2
x
 (17.438)(14  x)  0
2
4 x 2  17.438 x  244.14  0
8x
Locate the neutral axis:
Solve for x.
x
17.438  17.4382  (4)(4)(244.14)
 5.6326 in.
(2)(4)
14  x  8.3674 in.
I 
 
1 3
1
8 x  nAs (14  x)2  (8)(5.6326)3  (17.438)(8.3674)2  1697.45 in 4
3
3
nMy
I
 M
I
ny
Concrete:
n  1.0,
M 
M 
Choose the smaller value.
  1350 psi
(1350)(1697.45)
 406.835  103 lb  in.  407 kip  in.
(1.0)(5.6326)
n  9.67,
Steel:
y  5.6326 in.,
y  8.3674 in.,   20  103 psi
(20  103 )(1697.45)
 419.72 lb  in.  420 kip  in.
(9.67)(8.3674)
M  407 kip  in.
M  33.9 kip  ft 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
501
PROBLEM 4.53
d
The design of a reinforced concrete beam is said to be balanced if the maximum stresses
in the steel and concrete are equal, respectively, to the allowable stresses  s and  c .
Show that to achieve a balanced design the distance x from the top of the beam to the
neutral axis must be
x
b
1
 s Ec
 c Es
d
where Ec and Es are the moduli of elasticity of concrete and steel, respectively, and d is
the distance from the top of the beam to the reinforcing steel.
SOLUTION
nM (d  x)
Mx
c 
I
I
 s n(d  x)
d

n n
x
x
c
s 
E
d
1 s
1
1 c s
x
n c
Es  c
x

d
Ec s
1
Es  c
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
502
PROBLEM 4.54
d
For the concrete beam shown, the modulus of elasticity is 25 GPa for the concrete and
200 GPa for the steel. Knowing that b = 200 mm and d = 450 mm, and using an
allowable stress of 12.5 MPa for the concrete and 140 MPa for the steel, determine
(a) the required area As of the steel reinforcement if the beam is to be balanced, (b) the
largest allowable bending moment. (See Prob. 4.53 for definition of a balanced beam.)
b
SOLUTION
Es 200  109

 8.0
Ec
25  109
nM (d  x)
Mx
s 
c 
I
I
 s n( d  x )
d

n n
x
x
c
n
d
1 s
1 140  106
1
1

 2.40
x
n c
8.0 12.5  106
x  0.41667d  (0.41667)(450)  187.5 mm
bx
Locate neutral axis.
As 
(a)
x
 nAs (d  x)
2
bx 2
(200)(187.5) 2

 1674 mm 2
2n(d  x) (2)(8.0)(262.5)
As  1674 mm 2 
1
1
I  bx3  nAs (d  x)2  (200)(187.5)3  (8.0)(1674)(262.5) 2
3
3
9
4
 1.3623  10 mm  1.3623  103 m 4
nMy
I
M

I
ny
(b)
n  1.0
Concrete:
M
M
  12.5  106 Pa
y  262.5 mm  0.2625 m
  140  106 Pa
(1.3623  103 )(12.5  106 )
 90.8  103 N  m
(1.0)(0.1875)
n  8.0
Steel:
y  187.5 mm  0.1875 m
(1.3623  103 )(140  106 )
 90.8  103 N  m
(8.0)(0.2625)
Note that both values are the same for balanced design.
M  90.8 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
503
Aluminum
0.5 in.
Brass
0.5 in.
Steel
0.5 in.
Brass
0.5 in.
Aluminum
0.5 in.
1.5 in.
PROBLEM 4.55
Five metal strips, each a 0.5 × 1.5-in. cross section, are bonded together to
form the composite beam shown. The modulus of elasticity is 30 × 106 psi
for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by a couple of
moment 12 kip  in., determine (a) the maximum stress in each of the
three metals, (b) the radius of curvature of the composite beam.
SOLUTION
Use aluminum as the reference material.
n
Es 30  106

 3.0 in steel
Ea 10  106
Eb 15  106

 1.5 in brass
Ea 10  106
n  1.0 in aluminum
n
For the transformed section,
I1 
n1
1
b1h13  n1 A1d12  (1.5)(0.5)3  (0.75)(1.0) 2
12
12
4
 0.7656 in
I2 
n2
1.5
(1.5)(0.5)3  (1.5)(0.75)(0.5)2  0.3047 in 4
b2 h23  n2 A2 d 22 
12
12
n
3.0
(1.5)(0.5)3  0.0469 in 4
I 3  3 b3 h33 
12
12
I 4  I 2  0.3047 in 4
I 5  I1  0.7656 in 4
I
(a)
i
 2.1875 in 4
1
Aluminum:

nMy (1.0)(12)(1.25)

 6.86 ksi
I
2.1875

Brass:

nMy (1.5)(12)(0.75)

 6.17 ksi
I
2.1875

Steel:

nMy (3.0)(12)(0.25)

 4.11 ksi
I
2.1875


1
(b)

I
5


12  103
M

 548.57  106 in.1
Ea I (10  106 )(2.1875)
  1823 in. = 151.9 ft

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
504
Steel
0.5 in.
Aluminum
0.5 in.
Brass
0.5 in.
Aluminum
0.5 in.
Steel
0.5 in.
1.5 in.
PROBLEM 4.56
Five metal strips, each a 0.5 × 1.5-in. cross section, are bonded together to
form the composite beam shown. The modulus of elasticity is 30 × 106 psi
for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by a couple of
moment 12 kip · in., determine (a) the maximum stress in each of the
three metals, (b) the radius of curvature of the composite beam.
SOLUTION
Use aluminum as the reference material.
n
Es 30  106

 3.0 in steel
Ea 10  106
Eb 15  106

 1.5 in brass
Ea 10  106
n  1.0 in aluminum
n
For the transformed section,
I1 
n1
b1h13  n1 A1d12
12
3.0

(1.5)(0.5)3  (3.0)(0.75)(1.0)2
12
 2.2969 in 4
I2 
n2
1.0
(1.5)(0.5)3  (1.0)(0.75)(0.5) 2  0.2031 in 4
b2 h23  n2 A2 d 22 
12
12
n
1.5
(1.5)(0.5)3  0.0234 in 4
I 3  3 b3 h33 
12
12
I 4  I 2  0.2031 in 4
I 5  I1  2.2969 in 4
I
(a)
5
i
 5.0234 in 4
1
Steel:

nMy (3.0)(12)(1.25)

 8.96 ksi
I
5.0234

Aluminum:

nMy (1.0)(12)(0.75)

 1.792 ksi
I
5.0234

Brass:

nMy (1.5)(12)(0.25)

 0.896 ksi
I
5.0234


1
(b)

I


12  103
M

 238.89  106 in.1
Ea I (10  106 )(5.0234)
  4186 in.  349 ft

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
505
PROBLEM 4.57
Brass
0.8 in.
Aluminum
The composite beam shown is formed by bonding together a brass rod and an
aluminum rod of semicircular cross sections. The modulus of elasticity is
15  106 psi for the brass and 10  106 psi for the aluminum. Knowing that the
composite beam is bent about a horizontal axis by couples of moment
8 kip  in., determine the maximum stress (a) in the brass, (b) in the aluminum.
SOLUTION
For each semicircle,
y0 
r  0.8 in.

A
2
4r
(4)(0.8)

 0.33953 in.
3
3
r 2  1.00531 in 2 ,
I base 

8
r 4  0.160850 in 4
I  I base  Ay02  0.160850  (1.00531)(0.33953) 2  0.044953 in 4
Use aluminum as the reference material.
n  1.0 in aluminum
n
Eb 15  106

 1.5 in brass
Ea 10  106
Locate the neutral axis.



A, in2
nA, in2
1.00531
1.50796
1.00531
1.00531
Y0 
nAy0 , in 3
y0 , in.
0.33953
0.51200
0.33953
0.34133
2.51327
0.17067
 0.06791 in.
2.51327
The neutral axis lies 0.06791 in.
above the material interface.
0.17067
d1  0.33953  0.06791  0.27162 in., d 2  0.33953  0.06791  0.40744 in.
I1  n1I  n1 Ad12  (1.5)(0.044957)  (1.5)(1.00531)(0.27162) 2  0.17869 in 4
I 2  n2 I  n2 Ad 22  (1.0)(0.044957)  (1.0)(1.00531)(0.40744)2  0.21185 in 4
I  I1  I 2  0.39054 in 4
(a)
Brass:
n  1.5,
y  0.8  0.06791  0.73209 in.
 
(b)
Aluminium:
n  1.0,
nMy
(1.5)(8)(0.73209)

I
0.39054
  22.5 ksi 
nMy
(1.0)(8)(0.86791)

I
0.39054
  17.78 ksi 
y  0.8  0.06791  0.86791 in.
 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
506
y
Aluminum
PROBLEM 4.58
3 mm
A steel pipe and an aluminum pipe are securely bonded together to form
the composite beam shown. The modulus of elasticity is 200 GPa for the
steel and 70 GPa for the aluminum. Knowing that the composite beam is
bent by a couple of moment 500 N  m, determine the maximum stress
(a) in the aluminum, (b) in the steel.
Steel
6 mm
z
10 mm
38 mm
SOLUTION
Use aluminum as the reference material.
n  1.0 in aluminum
n  Es / Ea  200 / 70  2.857 in steel
For the transformed section,
Steel:
I s  ns
Aluminium:
I a  na
r
4

4
o
r
4

4
o

 
 ri4  (2.857)   (164  104 )  124.62  103 mm 4
4

 
 ri4  (1.0)   (194  164 )  50.88  103 mm 4
4
I  I s  I a  175.50  103 mm 4  175.5  109 m 4
(a)
Aluminum:
c  19 mm  0.019 m
a 
(b)
Steel:
na Mc (1.0)(500)(0.019)

 54.1  106 Pa
9
I
175.5  10
c  16 mm  0.016 m
s 
ns Mc (2.857)(500)(0.016)

 130.2  106 Pa
9
I
175.5  10
 a  54.1 MPa 
 s  130.2 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
507
M
PROBLEM 4.59
Et 100 mm
1
2
Ec
50 mm
Ec
The rectangular beam shown is made of a plastic for which the
value of the modulus of elasticity in tension is one-half of its
value in compression. For a bending moment M  600 N  m,
determine the maximum (a) tensile stress, (b) compressive stress.
SOLUTION
n
1
on the tension side of neutral axis
2
n  1 on the compression side
Locate neutral axis.
n1bx
x
hx
 n2 b(h  x)
0
2
2
1 2 1
bx  b(h  x) 2  0
2
4
x2 
1
1
(h  x) 2 x 
(h  x)
2
2
1
x
h  0.41421 h  41.421 mm
2 1
h  x  58.579 mm
1
1
I1  n1 bx3  (1)   (50)(41.421)3  1.1844  106 mm 4
3
 3
1
 1  1 
I 2  n2 b(h  x)3     (50)(58.579)3  1.6751  106 mm 4
3
 2  3 
I  I1  I 2  2.8595  106 mm 4  2.8595  106 m 4
(a)
Tensile stress:
n
1
,
2
 
(b)
Compressive stress:
n  1,
 
y  58.579 mm  0.058579 m
nMy
(0.5)(600)(0.058579)

 6.15  106 Pa
6
I
2.8595  10
 t  6.15 MPa 
nMy
(1.0)(600)(0.041421)

 8.69  106 Pa
6
I
2.8595  10
 c  8.69 MPa 
y  41.421 mm  0.041421 m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
508
PROBLEM 4.60*
A rectangular beam is made of material for which the modulus of elasticity is Et in tension and Ec in
compression. Show that the curvature of the beam in pure bending is

1

M
Er I
where
Er 
4 Et Ec
( Et  Ec ) 2
SOLUTION
Use Et as the reference modulus.
Then Ec  nEt .
Locate neutral axis.
x
hx
 b( h  x )
0
2
2
nx 2  (h  x) 2  0
nx  (h  x)
nbx
x
I trans

1
h
n 1
hx
nh
n 1
3
 h  1 3 
  3
n 3 1
n
3
 bx  b(h  x)   
  bh
 
 3  n  1   n  1  
3
3





1 n  n3/ 2
1 n 1 n
1
3

bh
bh3  
3
3
3 n 1
3 n 1
3

M
M

Et I trans Er I



where I 
Er I  Et I trans
Er 

Et I trans 12
 3  Et 
I
bh
3

4 Et Ec /Et
 
Ec /Et  1
2




n 1
2
bh3
1 3
bh
12
n

n 1
4 Et Ec

n
Ec  Et

2
bh3
2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
509
8 mm
PROBLEM 4.61
Knowing that M  250 N  m, determine the maximum stress in the beam
shown when the radius r of the fillets is (a) 4 mm, (b) 8 mm.
r
M
80 mm
40 mm
SOLUTION
I 
1 3
1
bh 
(8)(40)3  42.667  103 mm 4  42.667  109 m 4
12
12
c  20 mm  0.020 m
D 80 mm

 2.00
d
40 mm
(a)
4 mm
r

 0.10
d
40 mm
 max  K
(b)
Mc (1.87)(250)(0.020)

 219  106 Pa
I
42.667  109
8 mm
r

 0.20
d
40 mm
 max  K
K  1.87
From Fig. 4.27,
 max  219 MPa 
K  1.50
From Fig. 4.27,
Mc (1.50)(250)(0.020)

 176.0  106 Pa
I
42.667  109
 max  176.0 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
510
8 mm
PROBLEM 4.62
r
M
80 mm
Knowing that the allowable stress for the beam shown is 90 MPa, determine
the allowable bending moment M when the radius r of the fillets is (a) 8 mm,
(b) 12 mm.
40 mm
SOLUTION
I 
1 3
1
(8)(40)3  42.667  103 mm 4  42.667  109 m 4
bh 
12
12
c  20 mm  0.020 m
D 80 mm

 2.00
d
40 mm
(a)
8 mm
r

 0.2
d
40 mm
 max  K
(b)
Mc
I
12 mm
r

 0.3
d
40 mm
K  1.50
From Fig. 4.27,
M 
 max I
Kc

(90  106 )(42.667  109 )
(1.50)(0.020)
K  1.35
From Fig. 4.27,
M 
(90  106 )(42.667  109 )
(1.35)(0.020)
M  128.0 N  m 
M  142.0 N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
511
r
4.5 in.
3
4
M
in.
PROBLEM 4.63
Semicircular grooves of radius r must be milled as shown in the sides of
a steel member. Using an allowable stress of 8 ksi, determine the largest
bending moment that can be applied to the member when (a) r = 83 in.,
(b) r = 34 in.
SOLUTION
(a)
3
d  D  2r  4.5  (2)    3.75 in.
8
D
r
4.5
0.375

 1.20

 0.10
d
d
3.75
3.75
From Fig. 4.28,
K  2.07
1 3
1 3
1
3
4
bh 
c   1.875 in.
  (3.75)  3.296 in
12
12  4 
2
(8)(3.296)
I
Mc
 K
 M 

 6.79 kip  in.
I
Kc (2.07)(1.875)
I 
(b)
3
d  D  2r  4.5  (2)    3.0
4
From Fig. 4.28,
K  1.61
c
D
4.5

 1.5
d
3.0
I 
1
d  1.5 in.
2
  6.79 kip  in. 
r
0.75

 0.25
d
3.0
1 3
1 3
3
4
bh 
  (3.0)  1.6875 in
12
12  4 
M 
I
Kc

(8)(1.6875)
 5.59 kip  in.
(1.61)(1.5)

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
512
3
4
r
in.
Semicircular grooves of radius r must be milled as shown in the sides
of a steel member. Knowing that M = 4 kip · in., determine the
maximum stress in the member when the radius r of the semicircular
grooves is (a) r = 83 in., (b) r = 34 in.
M
4.5 in.
PROBLEM 4.64
SOLUTION
3
d  D  2r  4.5  (2)    3.75 in.
8
4.5
0.375
D
r

 1.20

 0.10
3.75
3.75
d
d
(a)
From Fig. 4.28,
K  2.07
1 3
1 3
3
4
bh 
  (3.75)  3.296 in
12
12  4 
Mc (2.07)(4)(1.875)

 4.71 ksi
 K
I
3.296
I 
(b)
3
d  D  2r  4.5  (2)    3.00 in.
4
From Fig. 4.28,
K  1.61
I 
D
4.5

 1.50
d
3.00
1 3
1 3
3
4
bh 
  (3.00)  1.6875 in
12
12  4 
 K
c
Mc (1.61)(4)(1.5)

 5.72 ksi
I
1.6875
c
1
 1.875 in.
2
  4.71 ksi 
r
0.75

 0.25
d
3.0
1
d  1.5 in.
2
  5.72 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
513
M
PROBLEM 4.65
M
100 mm
100 mm
150 mm
150 mm
18 mm
18 mm
(a)
A couple of moment M = 2 kN · m is to be
applied to the end of a steel bar. Determine the
maximum stress in the bar (a) if the bar is
designed with grooves having semicircular
portions of radius r = 10 mm, as shown in Fig. a,
(b) if the bar is redesigned by removing the
material to the left and right of the dashed lines
as shown in Fig. b.
(b)
SOLUTION
For both configurations,
D  150 mm
d  100 mm
r  10 mm
D 150

 1.50
100
d
10
r

 0.10
d 100
For configuration (a),
Fig. 4.28 gives
For configuration (b),
K a  2.21.
Fig. 4.27 gives K b  1.79.
I 
1 3
1
(18)(100)3  1.5  106 mm 4  1.5  106 m 4
bh 
12
12
1
c  d  50 mm  0.05 m
2
(a)
(b)
 
KMc (2.21)(2  103 )(0.05)

 147.0  106 Pa  147.0 MPa
6
I
1.5  10
 
KMc (1.79)(2  103 )(0.05)

 119.0  106 Pa  119.0 MPa
I
1.5  106
  147.0 MPa 
  119.0 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
514
M
PROBLEM 4.66
M
100 mm
The allowable stress used in the design of a
steel bar is 80 MPa. Determine the largest
couple M that can be applied to the bar (a) if
the bar is designed with grooves having
semicircular portions of radius r  15 mm, as
shown in Fig. a, (b) if the bar is redesigned by
removing the material to the left and right of
the dashed lines as shown in Fig. b.
100 mm
150 mm
150 mm
18 mm
18 mm
(a)
(b)
SOLUTION
For both configurations,
D  150 mm
r  15 mm
d  100 mm
D 150

 1.50
d
100
r
15

 0.15
d 100
For configuration (a), Fig. 4.28 gives K a  1.92.
For configuration (b), Fig. 4.27 gives Kb  1.57.
I 
1 3
1
(18)(100)3  1.5  106 mm 4  1.5  106 m 4
bh 
12
12
1
c  d  50 mm  0.050 m
2
(a)
(a)
 
M 
KMc

I
I
Kc

M 
I
Kc

(80  106 )(1.5  106 )
 1.250  103 N  m
(1.92)  (0.05)
 1.250 kN  m
(80  106 )(1.5  106 )
 1.530  103 N  m  1.530 kN  m
(1.57)(0.050)


PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
515
M
PROBLEM 4.67
The prismatic bar shown is made of a steel that is assumed to be
elastoplastic with  Y  300 MPa and is subjected to a couple M
parallel to the x axis. Determine the moment M of the couple for which
(a) yield first occurs, (b) the elastic core of the bar is 4 mm thick.
x
z
12 mm
8 mm
SOLUTION
(a)
I
1 3 1
bh  (12)(8)3  512 mm 4
12
12
 512  1012 m 4
1
c  h  4 mm  0.004 m
2
Y I
(300  106 )(512  1012 )
c
0.004
 38.4 N  m
MY 
(b)
yY 
1
(4)  2 mm
2

yY 2
  0.5
4
c
 1  y 2 
3
M  M Y 1   Y  
2
 3  c  
3
 1

 (38.4) 1  (0.5) 2 
2
 3

 52.8 N  m
M Y  38.4 N  m 
M  52.8 N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
516
M
PROBLEM 4.68
Solve Prob. 4.67, assuming that the couple M is parallel to the z axis.
PROBLEM 4.67 The prismatic bar shown is made of a steel that is
assumed to be elastoplastic with  Y  300 MPa and is subjected to a
couple M parallel to the x axis. Determine the moment M of the couple
for which (a) yield first occurs, (b) the elastic core of the bar is 4 mm
thick.
x
z
12 mm
8 mm
SOLUTION
(a)
I
1 3 1
bh  (8)(12)3  1.152  103 mm 4
12
12
 1.152  109 m 4
1
c  h  6 mm  0.006 m
2
Y I
(300  106 )(1.152  109 )
c
0.006
 57.6 N  m
MY 
(b)
yY 
1
(4)  2 mm
2

yY 2 1
 
6 3
c
M

3
MY
2
M Y  57.6 N  m 
 1  y 2 
1   Y  
 3  c  
 1  1 2 
3
(57.6) 1    
2
 3  3  
 83.2 N  m
M  83.2 N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
517
PROBLEM 4.69
A solid square rod of side 0.6 in. is made of a steel that is assumed to be elastoplastic with E = 29 × 106 psi
and  Y  48 ksi. Knowing that a couple M is applied and maintained about an axis parallel to a side of the
cross section, determine the moment M of the couple for which the radius of curvature is 6 ft.
SOLUTION
I 
MY 
 max 
1
(0.6)(0.6)3  10.8  103 in 4
12
c
1
h  0.3 in.
2
I Y
(10.8  103 )(48  103 )

 1728 lb  in.
c
0.3

c

0.3
 4.16667  103
72
Y 
yY

1.65517  103
 Y 
 0.39724
c
M
4.16667  103
Y
E

  6 ft  72 in.
48  103
 1.65517  103
29  106
2

1  YY   3
1
3


M  M Y 1      (1728) 1  (0.39724)2 
3 c   2
3
2



M  2460 lb  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
518
PROBLEM 4.70
For the solid square rod of Prob. 4.69, determine the moment M for which the radius of curvature is 3 ft.
PROBLEM 4.69. A solid square rod of side 0.6 in. is made of a steel that is assumed to be elastoplastic with
E = 29 × 106 psi and  Y  48 ksi. Knowing that a couple M is applied and maintained about an axis parallel
to a side of the cross section, determine the moment M of the couple for which the radius of curvature is 6 ft.
SOLUTION
I 
MY 
 max 
1
(0.6)(0.6)3  10.8  103 in 4
12
1
h  0.3 in.
2
(10.8  103 )(48  103 )
I Y

 1728 lb  in.
0.3
c

c

0.3
 8.3333  103
36
Y 
1.65517  10
yY

 Y 
 0.19862
c
 max
8.3333  103
M 
c
3
Y
E

  3 ft  36 in.
48  103
 1.65517  103
29  106
2

3
1Y   3
1


M Y 1   Y    (1728) 1  (0.19862)2 
2
3
2
3
c






M  2560 lb  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
519
y
PROBLEM 4.71
18 mm
The prismatic rod shown is made of a steel that is assumed to be elastoplastic with
E  200 GPa and  Y  280 MPa . Knowing that couples M and M of moment
525 N · m are applied and maintained about axes parallel to the y axis, determine
(a) the thickness of the elastic core, (b) the radius of curvature of the bar.
M
24 mm
M⬘
x
SOLUTION
I 
1 3
1
bh 
(24)(18)3  11.664  103 mm 4  11.664  103 m 4
12
12
1
c  h  9 mm  0.009 m
2
MY 
M 
yY

c
Y I
c
Y 
(280  106 )(11.664  109 )
 362.88 N  m
0.009

3
1 yY 2 
M Y 1 

2
3 c2 

3
(a)
(b)


yY

yY

c
or
(2)(525)
 0.32632
362.88
Y
E
  
Y
EyY
32
M
MY
yY  0.32632c  2.9368 mm

(200  109 )(2.9368  103 )
 2.09 m
280  106
tcore  2 yY  5.87 mm 

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
520
y
PROBLEM 4.72
Solve Prob. 4.71, assuming that the couples M and M are applied and
maintained about axes parallel to the x axis.
18 mm
M
24 mm
M⬘
x
PROBLEM 4.71 The prismatic rod shown is made of a steel that is assumed to
be elastoplastic with E  200 GPa and  Y  280 MPa . Knowing that couples
M and M of moment 525 N · m are applied and maintained about axes parallel
to the y axis, determine (a) the thickness of the elastic core, (b) the radius of
curvature of the bar.
SOLUTION
I 
1 3
1
(18)(24)3  20.736  103 mm 4  20.736  109 m 4
bh 
12
12
1
c  h  12 mm  0.012 m
2
MY 
M 
yY

c
Y I
c
Y 
(280  106 )(20.736  109 )
 483.84 N  m
0.012

3
1 yY 2 
M Y 1 

2
3 c2 

3
(a)
(b)


yY

yY

c
or
(2)(525)
 0.91097
483.84
Y
E
  
Y
EyY
32
M
MY
yY  0.91097

c  10.932 mm
(200  109 )(10.932  103 )
 7.81 m
280  106
tcore  2 yY  21.9 mm 

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
521
y
z
PROBLEM 4.73
C
90 mm
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E  200 GPa and  Y  240 MPa. For bending about the z axis,
determine the bending moment at which (a) yield first occurs, (b) the plastic
zones at the top and bottom of the bar are 30 mm thick.
60 mm
SOLUTION
(a)
I
1 3 1
bh  (60)(90)3  3.645  106 mm 4  3.645  106 m 4
12
12
1
c  h  45 mm  0.045 m
2
 I (240  106 )(3.645  106 )
 19.44  10 N  m
MY  Y 
0.045
c
M Y  19.44 kN  m 
R1   Y A1  (240  106 )(0.060)(0.030)
 432  103 N
y1  15 mm  15 mm  0.030 m
1
1
R2   Y A2    (240  106 )(0.060)(0.015)
2
2
 108  103 N
y2 
(b)
2
(15 mm)  10 mm  0.010 m
3
M  2( R1 y1  R2 y2 )  2[(432  103 )(0.030)  (108  103 )(0.010)]
 28.08  103 N  m
M  28.1 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
522
y
PROBLEM 4.74
30 mm
30 mm
C
z
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E  200 GPa and  Y  240 MPa. For bending about the
z axis, determine the bending moment at which (a) yield first occurs,
(b) the plastic zones at the top and bottom of the bar are 30 mm thick.
30 mm
15 mm
15 mm
30 mm
SOLUTION
(a)
I rect 
1 3 1
bh  (60)(90)3  3.645  106 mm 4
12
12
1
1
I cutout  bh3  (30)(30)3  67.5  103 mm 4
12
12
I  3.645  106  67.5  103  3.5775  106 mm 4
 3.5775  106 mm 4
c
1
h  45 mm  0.045 m
2
 I (240  106 )(3.5775  106 )
MY  Y 
0.045
c
3
 19.08  10 N  m
M Y  19.08 kN  m 
R1   Y A1  (240  106 )(0.060)(0.030)  432  103 N
y1  15 mm  15 mm  30 mm  0.030 m
1
1
 Y A2  (240  106 )(0.030)(0.015)  54  103 N
2
2
2
y2  (15 mm)  10 mm  0.010 m
3
R2 
(b)
M  2( R1 y1  R2 y2 )
 2[(432  103 )(0.030)  (54  103 )(0.010)]
 27.00  103 N  m
M  27.0 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
523
PROBLEM 4.75
y
3 in.
C
z
3 in.
A beam of the cross section shown is made of a steel that is assumed to
be elastoplastic with E  29  106 psi and  Y  42 ksi. For bending
about the z axis, determine the bending moment at which (a) yield first
occurs, (b) the plastic zones at the top and bottom of the bar are 3 in.
thick.
3 in.
1.5 in.
1.5 in.
3 in.
SOLUTION
(a)
I1 
1
1
b1h13  A1d12  (3)(3)3  (3)(3)(3) 2  87.75 in 4
12
12
1
1
I 2  b2 h23  (6)(3)3  13.5 in 4
12
12
I 3  I1  87.75 in 4
I  I1  I 2  I 3  188.5 in 4
c  4.5 in.
 I (42)(188.5)
MY  Y 
c
4.5
M Y  1759 kip  in. 
R1   Y A1  (42)(3)(3)  378 kip
y1  1.5  1.5  3.0 in.
1
1
R2   Y A2  (42)(6)(1.5)
2
2
 189 kip
2
y2  (1.5)  1.0 in.
3
(b)
M  2( R1 y1  R2 y2 )  2[(378)(3.0)  (189)(1.0)]  2646 kip  in.
M  2650 kip  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
524
PROBLEM 4.76
y
3 in.
3 in.
C
z
A beam of the cross section shown is made of a steel that is assumed to
be elastoplastic with E  29  106 psi and  Y  42 ksi. For bending
about the z axis, determine the bending moment at which (a) yield first
occurs, (b) the plastic zones at the top and bottom of the bar are 3 in.
thick.
3 in.
1.5 in.
3 in.
1.5 in.
SOLUTION
(a)
I1 
1
1
b1h13  A1d12  (6)(3)3  (6)(3)(3) 2  175.5 in 4
12
12
1
1
3
I 2  b2 h2  (3)(3)3  6.75 in 4
12
12
I 3  I1  175.5 in 4
I  I1  I 2  I 3  357.75 in 4
c  4.5 in.
MY 
Y I
c

(42)(357.75)
 3339 kip  in.
4.5
M Y  3340 kip  in. 
R1   Y A1  (42)(6)(3)  756 kip
y1  1.5  1.5  3 in.
1
1
R2   Y A2  (42)(3)(1.5)
2
2
 94.5 kip
y2 
(b)
M  2( R1 y1  R2 y2 )  2[(756)(3)  (94.5)(1.0)]  4725 kip  in.
2
(1.5)  1.0 in.
3
M  4730 kip  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
525
PROBLEM 4.77
y
For the beam indicated (of Prob. 4.73), determine (a) the fully plastic moment
M p , (b) the shape factor of the cross section.
z
90 mm
C
60 mm
SOLUTION
E  200 GPa and  Y  240 MPa
From Problem 4.73,
A1  (60)(45)  2700 mm 2
 2700  106 m 2
R   Y A1
 (240  106 )(2700  106 )
 648  103 N
d  45 mm  0.045 m
(a)
M p  Rd  (648  103 )(0.045)  29.16  103 N  m
(b)
I
M p  29.2 kN  m 
1 3 1
bh  (60)(90)3  3.645  106 mm 4  3.645  106 m 4
12
12
c  45 mm  0.045 m
MY 
Y I
c

(240  106 )(3.645  106 )
 19.44  103 N  m
0.045
k
Mp
MY

29.16
19.44
k  1.500 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
526
y
PROBLEM 4.78
30 mm
30 mm
C
z
For the beam indicated (of Prob. 4.74), determine (a) the fully plastic moment
M p , (b) the shape factor of the cross section.
30 mm
15 mm
15 mm
30 mm
SOLUTION
From Problem 4.74,
(a)
R1   Y A1
E  200 GPa and  Y  240 MPa
 (240  106 )(0.060)(0.030)
 432  103 N
y1  15 mm  15 mm  30 mm
 0.030 m
R2   Y A2
 (240  106 )(0.030)(0.015)
 108  103 N
1
y2  (15)  7.5 mm  0.0075 m
2
M p  2( R1 y1  R2 y2 )  2[( 432 103 )(0.030)  (108 103 )(0.0075)]
 27.54  103 N  m
(b)
M p  27.5 kN  m 
I rect 
1 3 1
bh  (60)(90)3  3.645  106 mm 4
12
12
1 3 1
I cutout  bh  (30)(30)3  67.5  103 mm 4
12
12
I  I rect  I cutout  3.645  106  67.5  103  3.5775  103 mm 4
c
 3.5775  109 m 4
1
h  45 mm  0.045 m
2
 I (240  106 )(3.5775  109 )
 19.08  103 N  m
MY  Y 
0.045
c
M p 27.54
k

MY 19.08
k  1.443 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
527
PROBLEM 4.79
y
3 in.
3 in.
C
z
For the beam indicated (of Prob. 4.75), determine (a) the fully plastic
moment M p , (b) the shape factor of the cross section.
3 in.
1.5 in.
1.5 in.
3 in.
SOLUTION
From Problem 4.75,
(a)
E  29  106 psi and  Y  42 ksi.
R1   Y A1  (42)(3)(3)  378 kip
y1  1.5  1.5  3.0 in.
R2   Y A2  (42)(6)(1.5)  378 kip
y2 
1
(1.5)  0.75 in.
2
M p  2( R1 y1  R2 y2 )  2[( 378)( 3.0)  ( 378)(0. 75)]
(b)
M p  2840 kip  in. 
I1 
1
1
b1h13  A1d12  (3)(3)3  (3)(3)(3) 2  87.75 in 4
12
12
1
1
I 2  b2 h23  (6)(3)3  13.5 in 4
12
12
I 3  I1  87.75 in 4
I  I1  I 2  I 3  188.5 in 4
c  4.5 in.
MY 
Y I
(42)(188.5)
 1759.3 kip  in
4.5
2835

k
MY 1759.3
c
Mp

k  1.611 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
528
PROBLEM 4.80
y
3 in.
C
z
For the beam indicated (of Prob. 4.76), determine (a) the fully plastic
moment M p , (b) the shape factor of the cross section.
3 in.
3 in.
1.5 in.
3 in.
1.5 in.
SOLUTION
From Problem 4.76,
(a)
E  29  106 and  Y  42 ksi
R1   Y A1  (42)(6)(3)  756 kip
y1  1.5  1.5  3.0 in.
R2   Y A2  (42)(3)(1.5)  189 kip
y2 
1
(1.5)  0.75 in.
2
M p  2( R1 y1  R2 y2 )  2[( 756)( 3.0)  (189)(0. 75)]
(b)
M p  4820 kip  in. 
I1 
1
1
b1h13  A1d12  (6)(3)3  (6)(3)(3) 2  175.5 in 4
12
12
1
1
3
I 2  b2 h2  (3)(3)3  6.75 in 4
12
12
I 3  I1  175.5 in 4
I  I1  I 2  I 3  357.75 in 4
c  4.5 in.
MY 
Y I

(42)(357.75)
 3339 kip  in.
c
4.5
M p 4819.5
k

MY
3339
k  1.443 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
529
PROBLEM 4.81
r 18 mm
Determine the plastic moment M p of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
A
For a semicircle,
Resultant force on semicircular section:

2
r 2;
r 
R  Y A
4r
3
Resultant moment on entire cross section:
M p  2 Rr 
Data:
4
Y r3
3
 Y  240 MPa  240  106 Pa,
Mp 
r  18 mm  0.018 m
4
(240  106 )(0.018)3  1866 N  m
3
M p  1.866 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
530
PROBLEM 4.82
50 mm
30 mm
Determine the plastic moment M p of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
10 mm
10 mm
10 mm
30 mm
SOLUTION
A  (50)(90)  (30)(30)  3600 mm 2
Total area:
1
A  1800 mm 2
2
1
A 1800
x 2 
 36 mm
50
b
A1  (50)(36)  1800 mm 2 ,
y1  18 mm,
A1 y1  32.4  103 mm3
A2  (50)(14)  700 mm 2 ,
y2  7 mm,
A2 y2  4.9  103 mm3
A4  (50)(10)  500 mm 2 ,
y4  49 mm,
A4 y4  24.5  103 mm3
A3  (20)(30)  600 mm 2 ,
y3  29 mm,
A3 y3  17.4  103 mm3
A1 y1  A2 y2  A3 y3  A4 y4  79.2  103 mm3  79.2  106 m3
M p   Y Ai yi  (240  106 )(79.2  106 )  19.008  103 N  m
M p  19.01 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
531
PROBLEM 4.83
36 mm
Determine the plastic moment M p of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
30 mm
SOLUTION
Total area:
A
1
(30)(36)  540 mm 2
2
By similar triangles,
Since
b 30

y 36
A1 
y
b
Half area:
1
A  270 mm 2  A1
2
5
y
6
1
5 2
by 
y ,
2
12
y2 
12
A1
5
12
(270)  25.4558 mm
5
b  21.2132 mm
1
A1  (21.2132)(25.4558)  270 mm 2  270  106 m 2
2
A2  (21.2132)(36  25.4558)  223.676 mm 2  223.676  106 m 2
A3  A  A1  A2  46.324 mm 2  46.324  106 m 2
Ri   Y Ai  240  106 Ai
R1  64.8  103 N, R2  53.6822  103 N, R3  11.1178  103 N
1
y1  y  8.4853 mm  8.4853  103 m
3
1
y2  (36  25.4558)  5.2721 mm  5.2721  103 m
2
2
y3  (36  25.4558)  7.0295 mm  7.0295  103 m
3
M p  R1 y1  R2 y2  R3 y3  911 N  m
M p  911 N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
532
0.4 in.
PROBLEM 4.84
1.0 in.
Determine the plastic moment Mp of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 42 ksi.
1.0 in.
0.4 in.
0.4 in.
SOLUTION
A  (1.0 in.)(0.4 in.)  2(1.4 in.)(0.4 in.)  1.52 in 2
x
1
2
(1.52)
2(0.4)
 0.95 in.
R1  (1.0)(0.4)  0.4 in 2
y1  1.2  0.95  0.25 in.
R2  2(0.4)(1.4  0.95)  0.36 in 2
y2 
1
(1.4  0.95)  0.225 in.
2
R3  2(0.4)(0.95)  0.760 in 2
y3 
1
(0.95)  0.475 in.
2
M p  ( R1 y1  R2 y 2  R3 y 3 )( y )
 [(0.4)(0.25)  (0.36)(0.225)  (0.760)(0.475)](42)
M p  22.8 kip  in . 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
533
5 mm
PROBLEM 4.85
5 mm
80 mm
t = 5 mm
120 mm
Determine the plastic moment Mp of the cross section shown when the beam is
bent about a horizontal axis. Assume the material to be elastoplastic with a
yield strength of 175 MPa.
SOLUTION
For M p , the neutral axis divides the area into two equal parts.
Total area  (100  100  120)t  320t
Shaded area  2at 
1
(320)t
2
a  80 mm
80
(80 mm)  64 mm
b
100
A1  2at  2(0.08 m)(0.005 m)  800  106 m 2
A2  2(100  a )t  2(0.02 m)(0.005 m)  200  106 m 2
A3  (120 mm)t  (0.120 m)(0.005 m)  600  106 m 2
R1   Y A1  (175 MPa)(800  106 m 2 )  140 kN
R2   Y A2  (175 MPa)(200  106 m 2 )  35 kN
R3   Y A3  (175 MPa)(600  106 m 2 )  105 kN
 M z : M p  R1 (32 mm)  R2 (8 mm)  R3 (16 mm)
 (140 kN)(0.032 m)  (35 kN)(0.008 m)  (105 kN)(0.016 m)
M p = 6.44 k N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
534
PROBLEM 4.86
4 in.
1
2
1
2
in.
in.
3 in.
1
2
in.
Determine the plastic moment M p of a steel beam of the cross section shown, assuming
the steel to be elastoplastic with a yield strength of 36 ksi.
2 in.
SOLUTION
Total area:
1 1
1
A  (4)      (3)  (2)    4.5 in 2
2 2
2
1
A  2.25 in 2
2
A1  2.00 in 2 ,
y1  0.75,
A2  0.25 in 2 ,
y2  0.25,
A4  1.00 in 2 ,
y4  2.75,
A3  1.25 in 2 ,
y3  1.25,
A1 y1  1.50 in 3
A2 y2  0.0625 in 3
A3 y3  1.5625 in 3
A4 y4  2.75 in 3
M p   Y ( A1 y1  A2 y2  A3 y3  A4 y4 )
 (36)(1.50  0.0625  1.5625  2.75)
M p  212 kip  in . 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
535
PROBLEM 4.87
y
z
C
90 mm
For the beam indicated (of Prob. 4.73), a couple of moment equal to the full
plastic moment M p is applied and then removed. Using a yield strength of
240 MPa, determine the residual stress at y  45mm .
60 mm
SOLUTION
M p  29.16  103 N  m
See solutions to Problems 4.73 and 4.77.
I  3.645  106 m 4
c  0.045 m
 
LOADING
Mp c
M max y

I
I
UNLOADING
 
at y  c  45 mm
RESIDUAL STRESSES
(29.16  103 )(0.045)
 360  106 Pa
6
3.645  10
 res      Y  360  106  240  106
 120  106 Pa
 res  120.0 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
536
y
PROBLEM 4.88
30 mm
z
30 mm
C
For the beam indicated (of Prob. 4.74), a couple of moment equal to the
full plastic moment M p is applied and then removed. Using a yield
strength of 240 MPa, determine the residual stress at y  45mm .
30 mm
15 mm
15 mm
30 mm
SOLUTION
M p  27.54  103 N  m (See solutions to Problems 4.74 and 4.78.)
I  3.5775  106 m 4 ,
 
 
LOADING
Mp c
M max y

I
I
c  0.045 m
at
y c
(27.54  103 )(0.045)
 346.4  106 Pa
6
3.5775  10
UNLOADING
RESIDUAL STRESSES
 res      Y  346.4  106  240  106  106.4  106 Pa
 res  106.4 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
537
PROBLEM 4.89
y
A bending couple is applied to the bar indicated, causing plastic zones
3 in. thick to develop at the top and bottom of the bar. After the couple
has been removed, determine (a) the residual stress at y  4.5 in.,
(b) the points where the residual stress is zero, (c) the radius of
curvature corresponding to the permanent deformation of the bar.
3 in.
3 in.
C
z
Beam of Prob. 4.75.
3 in.
1.5 in.
1.5 in.
3 in.
SOLUTION
See solution to Problem 4.75 for bending couple and stress distribution during loading.
M  2646 kip  in.
(a)
 Y  42 ksi
 
I  188.5 in 4
At
c  4.5 in.
y  c,  res      Y  63.167  42  21.167 ksi
y  yY ,  res      Y  21.056  42  20.944 ksi
UNLOADING
LOADING
(c)
E  29  106 psi  29  103 ksi
Mc (2646)(4.5)

 63.167 ksi
I
188.5
MyY
(2646)(1.5)
  

 21.056 ksi
I
188.5
At
(b)
yY  1.5 in.
My0
 Y
I
(188.5)(42)
I Y
y0 

 2.99 in.
2646
M
 res  20.9 ksi 
RESIDUAL STRESSES
 res  0 
At
 
y  yY ,

Ey
Answer:
y0  2.99 in.,
0,
2.99 in. 
 res  20.944 ksi
  

Ey

(29  103 )(1.5)
 2077 in.
20.944
  173.1 ft 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
538
PROBLEM 4.90
y
A bending couple is applied to the bar indicated, causing plastic zones
3 in. thick to develop at the top and bottom of the bar. After the couple
has been removed, determine (a) the residual stress at y  4.5 in.,
(b) the points where the residual stress is zero, (c) the radius of
curvature corresponding to the permanent deformation of the bar.
3 in.
3 in.
C
z
Beam of Prob. 4.76.
3 in.
1.5 in.
3 in.
1.5 in.
SOLUTION
See solution to Problem 4.76 for bending couple and stress distribution.
M  4725 kip  in.
 Y  42 ksi
(a)
yY  1.5 in.
I  357.75 in 4
 
   59.4 ksi 
y  yY ,  res      Y  19.8113  42  22.189 ksi
UNLOADING
LOADING
 res  0 
y0 
(c)
c  4.5 in.
Mc (4725)(4.5)

 59.434 ksi
357.75
I
(4725)(1.5)
MyY
  

 19.8113 ksi
357.75
I
At y  c,  res      Y  59.434  42  17.4340 ksi
At
(b)
E  29  106 psi  29  103 ksi
At
RESIDUAL STRESSES
My0
 Y  0
I
(357.75)(42)
I Y

 3.18 in.
4725
M
Answer:
y0  3.18 in., 0, 3.18 in. 
y  yY ,  res  22.189 ksi
 

Ey
  

Ey

(29  103 )(1.5)
 1960.43 in.
22.189
  163.4 ft. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
539
PROBLEM 4.91
y
z
90 mm
C
A bending couple is applied to the beam of Prob. 4.73, causing plastic zones
30 mm thick to develop at the top and bottom of the beam. After the couple has
been removed, determine (a) the residual stress at y  45 mm, (b) the points
where the residual stress is zero, (c) the radius of curvature corresponding to
the permanent deformation of the beam.
60 mm
SOLUTION
See solution to Problem 4.73 for bending couple and stress distribution during loading.
M  28.08  103 N  m
 Y  240 MPa
(a)
 
  
yY  15 mm  0.015 m
I  3.645  106 m 4
E  200 GPa
c  0.045 m
Mc (28.08  103 )(0.045)

 346.67  106 Pa  346.67 MPa
I
3.645  106
(28.08  103 )(0.015)
M yY

 115.556  106 Pa  115.556 MPa
I
3.645  106
At y  c,  res      Y  346.67  240  106.670 MPa
At y  yY ,  res      Y  115.556  240  124.444 MPa
LOADING
(b)
 res  0 
y0 
(c)
At
UNLOADING
 res  106.7 MPa 
 res  124.4 MPa
RESIDUAL STRESSES
My0
 Y  0
I
(3.645  106 )(240  106 )
I Y

 31.15  103 m  31.15 mm
M
28.08  103
Answer: y0  31.2 mm, 0, 31.2 mm 
y  yY ,  res  124.444  106 Pa
 

Ey
  

Ey

(200  109 )(0.015)
124.444  106
  24.1 m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
540
PROBLEM 4.92
y
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E  29 × 106 psi and  Y = 42 ksi. A bending couple is
applied to the beam about the z axis, causing plastic zones 2 in. thick to
develop at the top and bottom of the beam. After the couple has been removed,
determine (a) the residual stress at y  2 in., (b) the points where the residual
stress is zero, (c) the radius of curvature corresponding to the permanent
deformation of the beam.
1 in.
z
C
2 in.
1 in.
1 in.
1 in.
1 in.
SOLUTION
Flange:
I1 
Web:
I2 
1
1
b1h13  A1d12  (3)(1)3  (3)(1)(1.5)2  7.0 in 4
12
12
1
1
b2 h23  (1)(2)3  0.6667 in 4
12
12
4
I 3  I1  7.0 in
I  I1  I 2  I 3  14.6667 in 4
c  2 in.
MY 
Y I
c

(42)(14.6667)
2
MY  308 kip  in . 
R1   Y A1  (42)(3)(1)  126 kips
y1  1.0  0.5  1.5 in.
1
1
R2   Y A2  (42)(1)(1)
2
2
 21 kips
y2 
2
(1.0)  0.6667 in.
3
M  2( R1 y1  R2 y 2 )  2[(126)(1.5)  (21)(0.6667)]
M  406 kip  in. yY  1.0 in. E  29  106  29  103 ksi
 Y  42 ksi
(a)
M  406 kip  in . 
I  14.6667 in 4 c  2 in.

Mc (406)(2)
 55.364
I 14.6667
  
MyY (406)(1.0)
 27.682
I
14.662

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
541
PROBLEM 4.92 (Continued)
At y  c,  res      Y  55.364  42  13.3640 ksi
At y  yY ,  res      Y  27.682  42  14.3180 ksi
(b)
(c)
My0
 Y  0
I
I
(14.667)(42)
y0  Y 
 1.517 in
M
406
 res  13.36 k si 
 res  14.32 k si 
 res  0 
At y  yY ,
 

Ey
Answer : y0  1.517 in., 0, 1.517 in. 
 res  14.3180 ksi
  

Ey

(29  103 )(1.0)
 2025.42 in.
14.3180
  168.8 ft 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
542
PROBLEM 4.93*
A rectangular bar that is straight and unstressed is bent into an arc of circle of radius by two couples of
moment M. After the couples are removed, it is observed that the radius of curvature of the bar is  R .
Denoting by Y the radius of curvature of the bar at the onset of yield, show that the radii of curvature satisfy
the following relation:
R
1

2 

1
3  
1     
1
1




 
2 Y 
3  Y   




SOLUTION
Y
1
Let m denote
M
.
MY

MY
,
EI
M 

3
1 2 
M Y 1 
,
2
3 Y2 

3
2 
2
M
 1  2  
 32m
2
MY
Y 
Y2
1
1 M
1 mM Y
1
m






R
 EI

 Y
EI
m

1
1  2  
  1  3  




1
1
1
m





3 Y2  

Y    2 Y 

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
543
PROBLEM 4.94
A solid bar of rectangular cross section is made of a material that is assumed to be elastoplastic. Denoting by
M Y and Y , respectively, the bending moment and radius of curvature at the onset of yield, determine (a) the
radius of curvature when a couple of moment M  1.25 M Y is applied to the bar, (b) the radius of curvature
after the couple is removed. Check the results obtained by using the relation derived in Prob. 4.93.
SOLUTION
(a)
Y
1

m
(b)
R
1


M 
MY
,
EI

3
1 2 
M Y 1 

2
3 Y2 

3
1 2 
M
 1 

2
3 Y2 
MY

1

Let m 
M
 1.25
MY

 3  2m  0.70711
Y
M
m
1 mM Y
1
1
1.25






EI

EI
 Y
Y
0.70711Y
Y
0.16421
  0.707Y 
 R  6.09Y 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
544
PROBLEM 4.95
M
The prismatic bar AB is made of a steel that is assumed to be elastoplastic and
for which E  200 GPa . Knowing that the radius of curvature of the bar is
2.4 m when a couple of moment M  350 N  m is applied as shown,
determine (a) the yield strength of the steel, (b) the thickness of the elastic
core of the bar.
B
A
20 mm
16 mm
SOLUTION
M 


(a)

 2 Y2 
bc 2 Y 1 
M
3E 2c 2 

Data:

3
1 2 
M Y 1 

2
3 Y2 

3 Y I
2 c

1  2 Y2 
1 

3 E 2c 2 

3  Y b(2c)3 
1  2 Y2 
1 

2 12c 
3 E 2c 2 

1  2 Y2 
  Y bc 2 1 

3 E 2c 2 

Cubic equation for  Y
E  200  109 Pa
M  420 N  m
  2.4 m
b  20 mm  0.020 m
c
1
h  8 mm  0.008 m
2
(1.28  106 )  Y 1  750  1021 Y2   350
 Y 1  750  1021 Y2   273.44  106
Solving by trial,
(b)
yY 
Y 
E

 Y  292  106 Pa
(292  106 )(2.4)
 3.504  103 m  3.504 mm
9
200  10
 Y  292 MPa 
thickness of elastic core  2 yY  7.01 mm 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
545
PROBLEM 4.96
(MPa)
300
The prismatic bar AB is made of an aluminum
alloy for which the tensile stress-strain diagram
is as shown. Assuming that the  - diagram is
the same in compression as in tension,
determine (a) the radius of curvature of the bar
when the maximum stress is 250 MPa, (b) the
corresponding value of the bending moment.
(Hint: For part b, plot  versus y and use an
approximate method of integration.)
B
40 mm
M'
200
M
60 mm
A
100
0
0.005
0.010
⑀
SOLUTION
(a)
 m  250 MPa  250  106 Pa
 m  0.0064 from curve
c
1
h  30 mm  0.030 m
2
b  40 mm  0.040 m

1
(b)

m
c

  4.69 m 
0.0064
 0.21333 m 1
0.030
   m
Strain distribution:
y
  mu
c
u 
where

y
M    c y b dy  2b 0 y | | dy  2bc 2  0 u | | du  2bc 2 J
c
Bending couple:
c
1
where the integral J is given by  0 u | | du
1
Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is
J 
u
wu | |
3
where w is a weighting factor.
Using u  0.25, we get the values given in the table below:
u
0
0.25
0.5
0.75
1.00
| |
0
0.0016
0.0032
0.0048
0.0064
J 
| |, (MPa)
0
110
180
225
250
u | |, (MPa)
0
27.5
90
168.75
250
w
1
4
2
4
1
wu | |, (MPa)
0
110
180
675
250
1215
 wu | |
(0.25)(1215)
 101.25 MPa  101.25  106 Pa
3
M  (2)(0.040)(0.030)2 (101.25  106 )  7.29  103 N  m
M  7.29 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
546
PROBLEM 4.97
0.8 in.
The prismatic bar AB is made of a bronze alloy for which the tensile
stress-strain diagram is as shown. Assuming that the  - diagram is
the same in compression as in tension, determine (a) the maximum
stress in the bar when the radius of curvature of the bar is 100 in.,
(b) the corresponding value of the bending moment. (See hint given
in Prob. 4.96.)
B
M
␴ (ksi)
1.2 in.
A
50
40
30
20
10
0
0.004
⑀
0.008
SOLUTION
(a)
  100 in., b  0.8 in., c  0.6 in.
m 
(b)

c

0.6
 0.006
100
Strain distribution:
Bending couple:
From the curve,
   m
M  
y
  mu
c
c
c
where u 
 m  43.0 ksi 

y
y  b dy  2b  y | | dy  2bc 2  0 u | | du  2bc 2 J
c
0
where the integral J is given by  0 u | | du
1
1
Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is
J 
u
wu| |
3
where w is a weighting factor.
Using u  0.25, we get the values given the table below:
u
| |
0
0.25
0.5
0.75
1.00
0
0.0015
0.003
0.0045
0.006
| |, ksi
0
25
36
40
43
u | |, ksi
0
6.25
18
30
43
w
1
4
2
4
1
J 
(0.25)(224)
 18.67 ksi
3
M  (2)(0.8)(0.6) 2 (18.67)
wu | |, ksi
0
25
36
120
43
224
 wu | |
M  10.75 kip  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
547
PROBLEM 4.98
␴
A prismatic bar of rectangular cross section is made of an alloy for
which the stress-strain diagram can be represented by the relation
  k n for   0 and    k n for   0 . If a couple M is
applied to the bar, show that the maximum stress is
⑀
m 
M
1  2n Mc
3n
I
SOLUTION
Strain distribution:
Bending couple:
   m
y
  mu
c
where
u 
y
c
M    c y  b dy  2b  0 y | | dy  2bc 2  0
c
c
c
 2 bc 2  0 u | | du
y
dy
| |
c
c
1
For
Then
  K n ,
 m  K m
  

u 

m
 m 

n
2 1
m 
1 1 1n
du
1
1
u n
 2bc  m
2  1n
Then
1
M  2bc 2  0 u mu n du  2bc 2 m  0 u
2
Recall that
| |   mu n
2n  1 M
2 bc 2
 0  2n  1 bc  m
1
2n
2
I
1 b(2c)3
2
1
2 c

 bc 2 

2
c 12 c
3
3 I
bc
m 
2n  1 Mc
3n
I
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
548
PROBLEM 4.99
30 mm
Knowing that the magnitude of the horizontal force P is 8 kN,
determine the stress at (a) point A, (b) point B.
B
24 mm
A
D
P
45 mm
15 mm
SOLUTION
A  (30)(24)  720 mm 2  720  106 m 2
e  45  12  33 mm  0.033 m
I 
1 3
1
bh 
(30)(24)3  34.56  103 mm 4  34.56  109 m 4
12
12
1
c  (24 mm)  12 mm  0.012 m
P  8  103 N
2
(a)
(b)
M  Pe  (8  103 )(0.033)  264 N  m
A  
P Mc
8  103
(264)(0.012)



 102.8  106 Pa
6
A
I
720  10
34.56  109
B  
P Mc
8  103
(264)(0.012)



 80.6  106 Pa
6
A
I
720  10
34.56  109
 A  102.8 MPa 
 B  80.6 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
549
PROBLEM 4.100
y
b
3 in.
6 kips
A short wooden post supports a 6-kip axial load as shown. Determine the
stress at point A when (a) b  0, (b) b  1.5 in., (c) b  3 in.
C
A
z
x
SOLUTION
A   r 2   (3) 2  28.27 in 2
I 
(a)
(b)
b0
 

r 4  (3) 4  63.62 in 4
4
4
I
63.62
S  
 21.206 in 3
c
3
P  6 kips
M  Pb
P
6

 0.212 ksi
A
28.27
  212 psi 
b  1.5 in. M  (6)(1.5)  9 kip  in.
 
(c)
M 0

6
9
P M



 0.637 ksi
A
S
28.27 21.206
b  3 in.
 
M  (6)(3)  18 kip  in.
6
18
P M



 1.061 ksi
A
S
28.27 21.206
  637 psi 
  1061 psi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
550
P
P
r
PROBLEM 4.101
r
Two forces P can be applied separately or at the same time to a plate that is welded
to a solid circular bar of radius r. Determine the largest compressive stress in the
circular bar, (a) when both forces are applied, (b) when only one of the forces is
applied.
SOLUTION
For a solid section,
Compressive stress
(a)
Both forces applied.
(b)
One force applied.
A   r 2, I 

4
F Mc
  
A
I
F
4M
 2  3
r
r
r 4, c  r
F  2 P, M  0
F  P, M  Pr
F
4P
   2  r2
r
r
 
2P

 r2
 
5P

 r2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
551
30 mm
PROBLEM 4.102
y
60 mm
A short 120 × 180-mm column supports the three axial loads
shown. Knowing that section ABD is sufficiently far from the
loads to remain plane, determine the stress at (a) corner A,
(b) corner B.
30 kN
20 kN
100 kN
C
z
x
A
D
90 mm
90 mm
B
120 mm
SOLUTION
A  (0.120 m)(0.180 m)  21.6  103 m 2
1
S  (0.120 m)(0.180 m)2  6.48  104 m 2
6
M  (30 kN)(0.03 m)  (100 kN)(0.06 m)  5.10 kN  m
(a)
A  
P M
150  103 N
5.10  103 N  m



A
S
21.6  103 m 2
6.48  104 m3
(b)
B  
P M
150  10 N
5.10  10 N  m



2
3
A
S
21.6  10 m
6.48  104 m3
3
 A  0.926 MPa 
3
 B  14.81 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
552
80 mm
80 mm
P
P
P
C
1
2
3
PROBLEM 4.103
As many as three axial loads, each of magnitude P  50 kN, can be applied to
the end of a W200 × 31.1 rolled-steel shape. Determine the stress at point A
(a) for the loading shown, (b) if loads are applied at points 1 and 2 only.
A
SOLUTION
For W200  31.3 rolled-steel shape.
A  3970 mm 2  4.000  103 m 2
c
1
1
d  (210)  105 mm  0.105 m
2
2
I  31.3  106 mm 4  31.3  106 m 4
(a)
Centric load:
3P  50  50  50  150 kN  150  103 N
 
(b)
Ececentric loading:
3P
150  103

 37.783  106 Pa  37.8 MPa
3
A
3.970  10
e  80 mm  0.080 m

2 P  50  50  100 kN  100  103 N
M  Pe  (50  103 )(0.080)  4.0  103 N  m
A  
2P Mc
100  103
(4.0  103 )(0.105)



 38.607  106 Pa  38.6 MPa
A
I
3.970  103
31.3  106

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
553
PROBLEM 4.104
y
10 mm
10 mm
Two 10-kN forces are applied to a 20 × 60-mm rectangular bar as
shown. Determine the stress at point A when (a) b  0, (b) b  15 mm,
(c) b  25 mm.
A
30 mm
z
30 mm
10 kN
C
b
10 kN
x
25 mm
SOLUTION
A  (0.060 m)(0.020 m)  1.2  103 m 2
S 
(a)
1
(0.020 m)(0.060 m)2  12  106 m3
6
b  0, M  (10 kN)(0.025 m)  250 N  m
 A  (20 kN)/(1.2  103 m 2 )  (250 N  m)/(12  106 m3 )
 16.667 MPa  20.833 MPa
(b)
b  15 mm, M  (10 kN)(0.025 m  0.015 m)  100 N  m
 A  (20 kN)/(1.2  103 m 2 )  (100 N  m)/(12  106 m3 )
 16.667 MPa  8.333 MPa
(c)
 A  4.17 MPa 
b  25 mm, M  0:  A  (20 kN)/(1.2  103 m 2 )
 A  8.33 MPa 
 A  16.67 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
554
P
P'
P
PROBLEM 4.105
P'
Portions of a 12  12 -in. square bar have been bent
to form the two machine components shown.
Knowing that the allowable stress is 15 ksi,
determine the maximum load that can be applied
to each component.
1 in.
(a)
(b)
SOLUTION
The maximum stress occurs at point B.
 B  15 ksi  15  103 psi
B  
K 
where
1 ec

A
I
P Mc
P Pec

 
  KP
A
I
A
I
e  1.0 in.
A  (0.5)(0.5)  0.25 in 2
I 
1
(0.5)(0.5)3  5.2083  103 in 4 for all centroidal axes.
12
(a)
(a)
c  0.25 in.
K 
1
(1.0)(0.25)

 52 in 2
0.25 5.2083  103
P
(b)
(b)
c
B
K

(15  103 )
52
P  288 lb 
0.5
 0.35355 in.
2
K 
1
(1.0)(0.35355)

 71.882 in 2
0.25 5.2083  103
P
B
K

(15  103 )
71.882
P  209 lb 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
555
PROBLEM 4.106
P
Knowing that the allowable stress in section ABD is 80 MPa,
determine the largest force P that can be applied to the bracket shown.
A
D
B
18 mm
40 mm
12 mm
12 mm
SOLUTION
A  (24)(18)  432 mm 2  432  106 m 2
I 
1
(24)(18)3  11.664  103 mm 4  11.664  109 m 4
12
1
c  (18)  9 mm  0.009 m
2
1
e  (18)  40  49 mm  0.049 m
2
On line BD, both axial and bending stresses are compressive.
Therefore
Solving for P gives
 max 
P Mc

A
I

P Pec

A
I
P
P
 max
 1 ec 
 

I 
A
80  106 Pa

1
(0.049 m)(0.009 m) 



6 2
11.664  109 m 4 
 432  10 m
P  1.994 kN 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
556
PROBLEM 4.107
P'
a
d
P
a
A milling operation was used to remove a portion of a solid bar
of square cross section. Knowing that a  30 mm, d  20 mm,
and  all  60 MPa, determine the magnitude P of the largest
forces that can be safely applied at the centers of the ends of the
bar.
SOLUTION
A  ad ,
e
I 
1
ad 3 ,
12
c
a d

2 2
P Mc
P
6 Ped


  
A
I
ad
ad 3
3P (a  d )
P

 KP
 
ad
ad 2
Data:
a  30 mm  0.030 m
K 
P
1
d
2
where
K 
1
3(a  d )

ad
ad 2
d  20 mm  0.020 m
1
(3)(0.010)

 4.1667  103 m 2
(0.030)(0.020) (0.030)(0.020)2

K

60  106
 14.40  103 N
4.1667  103
P  14.40 kN 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
557
PROBLEM 4.108
P'
A milling operation was used to remove a portion of a solid bar of
square cross section. Forces of magnitude P  18 kN are applied at
the centers of the ends of the bar. Knowing that a  30 mm and
 all  135 MPa, determine the smallest allowable depth d of the
milled portion of the bar.
a
d
P
a
SOLUTION
A  ad ,
e
I 
a d

2 2
1
ad 3 ,
12
c
1
d
2
P 1 (a  d ) 1 d
P Mc
P
Pec
P
2  P  3P(a  d )




 2
1
A
I
ad
I
ad
ad
ad 2
ad 3
12
P
3P 2P
2
  2 
d  3P  0
or  d 2 
ad
a
d
 
2


1   2P 
2P 
P


12

 


a 
2   a 


Solving for d,
d 
Data:
a  0.030 m,
P  18  103 N,   135  106 Pa
2


1
(2)(18  103 ) 
  (2)(18  103 ) 
3
6
d 
 

  12(18  10 )(135  10 ) 
0.030
0.030
(2)(135  106 )  




 16.04  103 m
d  16.04 mm 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
558
12 kips
5 in.
P
PROBLEM 4.109
The two forces shown are applied to a rigid plate supported by a steel pipe of 8-in.
outer diameter and 7-in. inner diameter. Determine the value of P for which the
maximum compressive stress in the pipe is 15 ksi.
SOLUTION
 all  15 ksi
I NA 

4
(4 in.)4 

4
(3.5 in.)4  83.2 in 4
A   (4 in.) 2   (3.5 in.)2  11.78 in 2
Max. compressive stress is at point B.
B  
12  P
(5P)(4.0 in.)
Q Mc



2
A
I
11.78 in
83.2 in 4
15 ksi  1.019  0.085P  0.240 P
13.981  0.325P
P  43.0 kips 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
559
PROBLEM 4.110
d
P'
P
An offset h must be introduced into a solid circular rod of
diameter d. Knowing that the maximum stress after the offset
is introduced must not exceed 5 times the stress in the rod
when it is straight, determine the largest offset that can be
used.
h
P'
P
d
SOLUTION
For centric loading,
c 
For eccentric loading,
e 
Given
P
A
P Phc

A
I
 e  5 c
P Phc
P

5
A
I
A
Phc
P
4
I
A



(4)  d 4 
4I
1
64 


 d
h

d
2
cA  
2
  d 
 2  4 
h  0.500 d 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
560
PROBLEM 4.111
d
P'
P
h
P'
P
An offset h must be introduced into a metal tube of 0.75-in. outer
diameter and 0.08-in. wall thickness. Knowing that the maximum
stress after the offset is introduced must not exceed 4 times the
stress in the tube when it is straight, determine the largest offset
that can be used.
d
SOLUTION
c
1
d  0.375 in.
2
c1  c  t  0.375  0.08  0.295 in.


A   c 2  c12   (0.3752  0.2952 )
 0.168389 in 2
I 
c
4

4

 c14 

4
 9.5835  103 in 4
For centric loading,
 cen 
For eccentric loading,
 ecc 
(0.3754  0.2954 )
P
A
P Phc

A
I
 ecc  4  cen
3
hc

I
A
or
h
P Phc
P

4
A
I
A
3I
(3)(9.5835  103 )

Ac (0.168389)(0.375)
h  0.455 in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
561
PROBLEM 4.112
16 kips
2
4
1
A short column is made by nailing four 1 × 4-in. planks to a 4 × 4-in. timber. Using
an allowable stress of 600 psi, determine the largest compressive load P that can be
applied at the center of the top section of the timber column as shown if (a) the
column is as described, (b) plank 1 is removed, (c) planks 1 and 2 are removed,
(d) planks 1, 2, and 3 are removed, (e) all planks are removed.
3
SOLUTION
(a)
M 0
Centric loading:
 
P
A
A  (4  4)  4(1)(4)  32 in 2
 
(b)
P
A

P  (0.600 ksi)(32 in 2 )
 19.20 kips

Eccentric loading:
M  Pe
 
P Pec

A
I
A  (4)(4)  (3)(1)(3)  28 in 2
y 
e y
 Ay
(1)(4)(2.5)

 0.35714 in.
A
28
I  ( I  Ad 2 )

1
1
(6)(4)3  (6)(4)(0.35714)2 
(4)(1)3  (4)(1)(2.14286)2  53.762 in 4
12
12
 1 ec 
 
I 
A
  P
(c)
Centric loading:
 P
M 0
(0.600 ksi)
 11.68 kips
1
(0.35714)(2.35714)



 28

53.762
 

P
A
A  (6)(4)  24 in 2
P  (0.600 ksi)(24)  14.40 kips

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
562
PROBLEM 4.112 (Continued)
(d)
Eccentric loading:
M  Pe
 
P Pec

A
I
A  (4)(4)  (1)(4)(1)  20 in 2
x  2.5  2  0.5 in.
I 
P
(e)
Centric loading:
e x
1
(4)(5)3  41.667 in 4
12
(0.600 ksi)
 7.50 kips
(0.5)(2.5) 
1
 20  41.667 
M 0
 
P
A
A  (4)(4)  16 in 2


P  (0.600 ksi)(16.0 in 2 )  9.60 kips

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
563
1 in.
a
1.5 in.
PROBLEM 4.113
3 in.
a
0.75 in.
1.5 in.
A
3 in.
0.75 in.
B
Section a–a
SOLUTION
A vertical rod is attached at point A to the cast
iron hanger shown. Knowing that the allowable
stresses in the hanger are  all  5 ksi and
 all  12 ksi , determine the largest downward
force and the largest upward force that can be
exerted by the rod.
(1  3)(0.5)  2(3  0.25)(2.5)
Ay

X  
(1  3)  2(3  0.75)
A
X 
12.75 in 3
 1.700 in.
7.5 in 2
A  7.5 in 2
 all  5 ksi
 all  12 ksi
 1

I c    bh3  Ad 2 
 12

1
1

(3)(1)3  (3  1)(1.70  0.5)2 
(1.5)(3)3  (1.5  3)(2.5  1.70)2
12
12
I c  10.825 in 4
Downward force.
M  P(1.5 in.+1.70 in.)  (3.20 in.) P
At D:  D  
P Mc

A
I
 5 ksi 
P
(3.20) P(1.70)

7.5
10.825
 5  P(0.6359)
At E:  E  
P  7.86 kips 
P Mc

A
I
 12 ksi 
P
(3.20) P(2.30)

7.5
10.825
 12  P(0.5466)
We choose the smaller value.
P  21.95 kips 
P  7.86 kips  
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
564
PROBLEM 4.113 (Continued)
Upward force.
M  P(1.5 in.  1.70 in.)  (3.20 in.) P
At D:  D  
P Mc

A
I
12 ksi  
P
(3.20) P(1.70)

7.5
10.825
12  P(0.6359)
At E:  E  
P  18.87 kips 
P Mc

A
I
5 ksi 
P
(3.20) P(2.30)

7.5
10.825
5  P(0.5466)
We choose the smaller value.
P  9.15 kips 
P  9.15 kips  
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
565
PROBLEM 4.114
1 in.
a
1.5 in.
3 in.
a
1.5 in.
A
Solve Prob. 4.113, assuming that the vertical
rod is attached at point B instead of point A.
0.75 in.
3 in.
0.75 in.
B
Section a–a
SOLUTION
PROBLEM 4.113 A vertical rod is attached at
point A to the cast iron hanger shown. Knowing
that the allowable stresses in the hanger are
 all  5 ksi and  all  12 ksi, determine the
largest downward force and the largest upward
force that can be exerted by the rod.
(1  3)(0.5)  2(3  0.25)(2.5)
Ay

X  
(1  3)  2(3  0.75)
A

X 
12.75 in 3
 1.700 in.
7.5 in 2
A  7.5 in 2
 all  5 ksi
 all  12 ksi
 1

I c    bh3  Ad 2 
12


1
1

(3)(1)3  (3  1)(1.70  0.5)2 
(1.5)(3)3  (1.5  3)(2.5  1.70)2
12
12
I c  10.825 in 4
Downward force.
 all  5 ksi
 all  12 ksi
M  (2.30 in.  1.5 in.)  (3.80 in.) P
At D:  D  
P Mc

A
I
12 ksi  
P
(3.80) P(1.70)

7.5
10.825
 12  P(0.4634)
At E:  E  
P  25.9 kips 
P Mc

A
I
5 ksi  
P
(3.80) P(2.30)

7.5
10.825
5  P(0.9407)
We choose the smaller value.
P  5.32 kips  
P  5.32 kips  
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
566
PROBLEM 4.114 (Continued)
Upward force.

 all  5 ksi

M  (2.30 in.  1.5 in.)P  (3.80 in.)P 


 all  12 ksi 
At D:  D  
P Mc


A
I
5 ksi  
P
(3.80) P(1.70)

7.5
10.825
5  P(0.4634)


At E:  E  
P  10.79 kips 
P Mc

A
I
12 ksi  
P
(3.80) P(2.30)

7.5
10.825
12  P(0.9407)
We choose the smaller value.
P  12.76 kips 
P  10.79 kips  
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
567
2 mm radius
A
P
PROBLEM 4.115
P'
32 mm
20 mm
a
B
a
Knowing that the clamp shown has been tightened until
P  400 N, determine (a) the stress at point A,
(b) the stress at point B, (c) the location of the neutral
axis of section a  a.
4 mm
Section a–a
SOLUTION
Cross section:
Rectangle   Circle 
A1  (20 mm)(4 mm)  80 mm 2
y1 
1
(20 mm)  10 mm
2
A2   (2 mm)2  4 mm 2
y2  20  2  18 mm
Ay
(80)(10)  (4 )(18)
cB  y  

 11.086 mm
80  4
A
c A  20  y  8.914 mm
d1  11.086  10  1.086 mm
d 2  18  11.086  6.914 mm
I1  I1  A1d12 
I 2  I 2  A2d 22 
1
(4)(20)3  (80)(1.086) 2  2.761  103 mm 4
12

4
(2) 4  (4 )(6.914)2  0.613  103 mm 4
I  I1  I 2  3.374  103 mm 4  3.374  109 m 4
A  A1  A2  92.566 mm 2  92.566  106 m 2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
568
PROBLEM 4.115 (Continued)
e  32  8.914  40.914 mm  0.040914 m
M  Pe  (400 N)(0.040914 m)  16.3656 N  m
(a)
Point A:
A 
400
(16.3656)(8.914  103 )
P Mc



6
A
I
92.566  10
3.374  109
 4.321  106  43.23  106  47.55  106 Pa
(b)
Point B:
B 
400
(16.3656)(11.086)
P Mc



A
I
92.566  106
3.374  109
 4.321  106  53.72  106  49.45  106 Pa
(c)
 A  47.6 MPa 
Neutral axis:
 B  49.4 MPa 
By proportions,
a
20

47.55 47.55  49.45
a  9.80 mm
9.80 mm below top of section 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
569
PROBLEM 4.116
P
a
a
The shape shown was formed by bending a thin steel plate. Assuming that the
thickness t is small compared to the length a of a side of the shape, determine the
stress (a) at A, (b) at B, (c) at C.
90⬚
t
B
C
A
P'
SOLUTION
Moment of inertia about centroid:


 a 
1
2 2t 

12
 2
1
 ta3
12
I
Area:
(a)
(b)
(c)

3

 a 
a
A  2 2t 
  2at , c 
2 2
 2
A 
P
P Pec
P



2at
A
I
P
P Pec
P
B  


2at
A
I
  
a
2 2
a
2 2
3
A  
P

2at
a
2
B  
2P

at
C  
P

2at
  
1
ta
12
a
2
1
ta3
12
C   A
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
570
PROBLEM 4.117
P
Three steel plates, each of 25  150-mm cross section, are welded
together to form a short H-shaped column. Later, for architectural
reasons, a 25-mm strip is removed from each side of one of the
flanges. Knowing that the load remains centric with respect to the
original cross section, and that the allowable stress is 100 MPa,
determine the largest force P (a) that could be applied to the original
column, (b) that can be applied to the modified column.
50 mm
50 mm
SOLUTION
(a)
Centric loading:
 
P
A
A  (3)(150)(25)  11.25  103 mm 2  11.25  103 m 2
P   A  (100  106 )(11.25  103 )
 1.125  106 N
(b)
P  1125 kN 
Eccentric loading (reduced cross section):
A, 103 mm 2
y , mm
3.75
87.5
328.125
76.5625
3.75
0
0
10.9375




A y (103 mm3 )
87.5
2.50
10.00
218.75
d , mm
98.4375
109.375
Y 
A y 109.375  103

 10.9375 mm
A
10.00  103
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
571
PROBLEM 4.117 (Continued)
The centroid lies 10.9375 mm from the midpoint of the web.
I1 
1
1
(150)(25)3  (3.75  103 )(76.5625)2  22.177  106 mm 4
b1h13  A1d12 
12
12
1
1
I2 
b2h23  A2d 22 
(25)(150)3  (3.75  103 )(10.9375)2  7.480  106 mm 4
12
12
1
1
(100)(25)3  (2.50  103 )(98.4375) 2  24.355  106 mm 4
I3 
b3h33  A3d32 
12
12
I  I1  I 2  I 3  54.012  106 mm 4  54.012  106 m 4
c  10.9375  75  25  110.9375 mm  0.1109375 m
M  Pe
 
K 
where
e  10.4375 mm  10.4375  103 m
P Mc
P Pec

 
  KP
A
I
A
I
A  10.00  103 m 2
1 ec
1
(101.9375  103 )(0.1109375)
 122.465 m 2



A
I
10.00  103
54.012  106
P

K

(100  106 )
 817  103 N
122.465
P  817 kN 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
572
y
P
y
B
3 in.
y
x
PROBLEM 4.118
3 in.
B
A vertical force P of magnitude 20 kips is applied
at point C located on the axis of symmetry of the
cross section of a short column. Knowing that
y  5 in., determine (a) the stress at point A,
(b) the stress at point B, (c) the location of the
neutral axis.
2 in.
C
A
4 in.
A
2 in.
x
2 in.
1 in.
(a)
(b)
SOLUTION
Locate centroid.
Part
A, in 2
y , in.

12
5

8
2

20
Eccentricity of load: e  5  3.8  1.2 in.
I1 
1
(6)(2)3  (12)(1.2) 2  21.28 in 4
12
I  I1  I 2  57.867 in 4
(a)
1
(2)(4)3  (8)(1.8)2  36.587 in 4
12
P Pec A
20 20(1.2)(3.8)



A
I
20
57.867
 A  0.576 ksi 
P PecB
20 20(1.2)(2.2)



A
I
20
57.867
 B  1.912 ksi 
Stress at B : cB  6  3.8  2.2 in.
B  
(c)
I2 
Ay
76

 3.8 in.
A
20
Stress at A : c A  3.8 in.
A  
(b)
y 
Ay , in 3
60
16
76
Location of neutral axis:   0
 
P Pea
ea
1

0 

A
I
I
A
I
57.867
a

 2.411 in.
Ae (20)(1.2)
Neutral axis lies 2.411 in. below centroid or 3.8  2.411  1.389 in. above point A.
Answer: 1.389 in. from point A 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
573
y
P
y
B
3 in.
y
x
PROBLEM 4.119
3 in.
B
A vertical force P is applied at point C located
on the axis of symmetry of the cross section of
a short column. Determine the range of values
of y for which tensile stresses do not occur in
the column.
2 in.
C
A
4 in.
A
2 in.
x
2 in.
1 in.
(a)
(b)
SOLUTION
Locate centroid.



Eccentricity of load:
A, in 2
12
8
20
e  y  3.8 in.
I1 
y , in.
Ay , in 3
5
2
60
16
76
y  e  3.8 in.
1
(6)(2)3  (12)(1.2) 2  21.28 in 4
12
I  I1  I 2  57.867 in 4
If stress at A equals zero,
Ai yi
76

 3.8 in.
20
Ai
1
(2)(4)3  (8)(1.8) 2  36.587 in 4
12
c A  3.8 in.
A  
e
If stress at B equals zero,
I2 
y 
P Pec A

0
A
I

ec A
1

I
A
I
57.867

 0.761 in.
Ac A
(20)(3.8)
y  0.761  3.8  4.561 in.
cB  6  3.8  2.2 in.
B  
1
P PecB
ecB

0


A
I
I
A
57.867
I

 1.315 in.
e
(20)(2.2)
AcB
y  1.315  3.8  2.485 in.
Answer: 2.49 in.  y  4.56 in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
574
PROBLEM 4.120
P
P
P
P
The four bars shown have the same cross-sectional area. For the given
loadings, show that (a) the maximum compressive stresses are in the
ratio 4:5:7:9, (b) the maximum tensile stresses are in the ratio 2:3:5:3.
(Note: the cross section of the triangular bar is an equilateral triangle.)
SOLUTION
Stresses:
At A,
A  
At B,
B  
Aec A 
P Pec A
P

  1 
A
I
A
I 
P PecB P  AecB


 
 1
A
I
A I

1 4
1
1

2
 A1  a , I1  12 a , c A  cB  2 a, e  2 a



 1  1  
(a 2 )  a  a  


P
 2  2  
 A   1 
1 2


A

a



12




 2  1  1  

 (a )  a  a  
P
 2  2   1
  
 B A
1 2

a



12



 4
a

2
2
 A2   c  a  c   , I 2  4 c , e  c





P  ( c 2 )(c)(c) 
 A   1 


 4
A2 

c


4







P  ( c 2 )(c)(c) 
 B 
 1

 4
A2 


c


4


 A  4
P

A1
B  2
P

A1
 A  5
P

A2
B  3
P

A2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
575
PROBLEM 4.120 (Continued)

2
1
2
a I3  a4 e  c
 A3  a c 
2
12



 2  2  

a 
a 
 (a 2 ) 
2 
2  

P


 A   1 

1 4
A3

a


12








 2  2  

a 
a 
 (a 2 ) 


2 
2  
P

 1
 B 
1 4
A3 

a



12





A4 
 A  7
P

A3

B  5
P

A3
 A  9
P

A4
B  3
P

A4
1  3 
3 2
(s) 
s 
s
2  2  4
1  3 
3 4
I4 
s
s 
s


36  2 
96
3
cA 
2 3
s
e
s
3 2
3
cB  s
  3 2   s  s  
s  
 


P   4
 3  3  

 A   1 

A4
3 4


s


96


  3 2   s  s  
s  


 
P   4
 3  2 3  

 1
B  
A4
3
4


s


96


PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
576
PROBLEM 4.121
25 mm
An eccentric force P is applied as shown to a steel bar of
25  90-mm cross section. The strains at A and B have been
measured and found to be
30 mm
 A  350 
A
90 mm
B
45 mm
 B  70 
Knowing that E  200 GPa, determine (a) the distance d, (b) the
magnitude of the force P.
P
d
15 mm
SOLUTION
h  15  45  30  90 mm
b  25 mm
c
1
h  45 mm  0.045 m
2
A  bh  (25)(90)  2.25  103 mm 2  2.25  103 m 2
I
1 3 1
bh  (25)(90)3  1.51875  106 mm 4  1.51875  106 m 4
12
12
y A  60  45  15 mm  0.015 m
yB  15  45  30 mm  0.030 m
Stresses from strain gages at A and B:
Subtracting,
 A B  
M 
 A  E A  (200  109 )(350  106 )  70  106 Pa
 B  E B  (200  109 )(70  106 )  14  106 Pa
A 
P MyA

A
I
(1)
B 
P MyB

A
I
(2)
M ( y A  yB )
I
I ( A   B )
(1.51875  106 )(84  106 )

 2835 N  m
y A  yB
0.045
Multiplying (2) by y A and (1) by yB and subtracting,
P
(a)
y A B  yB A  ( y A  yB )
P
A
A( y A B  yB A ) (2.25  103 )[(0.015)(14  106 )  (0.030)(70  106 )]

 94.5  103 N
y A  yB
0.045
M   Pd  d  
M
2835

 0.030 m
P
94.5  103
(b)
d  30.0 mm 
P  94.5 kN 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
577
PROBLEM 4.122
25 mm
Solve Prob. 4.121, assuming that the measured strains are
 A  600 
30 mm
PROBLEM 4.121 An eccentric force P is applied as shown to a
steel bar of 25  90-mm cross section. The strains at A and B have
been measured and found to be
A
90 mm
B
45 mm
 B  420 
P
 A  350
d
 B  70
Knowing that E  200 GPa, determine (a) the distance d, (b) the
magnitude of the force P.
15 mm
SOLUTION
h  15  45  30  90 mm
b  25 mm
c
1
h  45 mm  0.045 m
2
A  bh  (25)(90)  2.25  103 mm 2  2.25  103 m 2
I
1 3 1
bh  (25)(90)3  1.51875  106 mm 4  1.51875  106 m 4
12
12
y A  60  45  15 mm  0.015 m yB  15  45  30 mm  0.030 m
Stresses from strain gages at A and B:
A 
P My A

A
I
B 
P MyB

A
I
Subtracting,
 A  E A  (200  109 )(600  106 )  120  106 Pa
 B  E B  (200  109 )(420  106 )  84  106 Pa
(1)
(2)
 A B  
M 
M ( y A  yB )
I
I ( A   B )
(1.51875  106 )(36  106 )

 1215 N  m
y A  yB
0.045
Multiplying (2) by y A and (1) by yB and subtracting,
P
A
A( y A B  yB A ) (2.25  103 )[(0.015)(84  106 )  (0.030)(120  106 )]

 243  103 N
0.045
y A  yB
M
1215
d  5.00 mm 

 5  103 m
M   Pd  d  
3
P
243  10
P
(a)
y A B  yB A  ( y A  yB )
P  243 kN 
(b)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
578
P'
PROBLEM 4.123
The C-shaped steel bar is used as a dynamometer to determine the magnitude P
of the forces shown. Knowing that the cross section of the bar is a square of side
40 mm and that strain on the inner edge was measured and found to be 450,
determine the magnitude P of the forces. Use E  200 GPa.
40 mm
80 mm
P
SOLUTION
At the strain gage location,
  E  (200  109 )(450  106 )  90  106 Pa
A  (40)(40)  1600 mm 2  1600  106 m 2
I 
1
(40)(40)3  213.33  103 mm 4  213.33  109 m 4
12
e  80  20  100 mm  0.100 m
c  20 mm  0.020 m
 
P Mc
P Pec



 KP
A
I
A
I
K 
1 ec
1
(0.100)(0.020)



 10.00  103 m 2
6
A
I
1600  10
213.33  109
P

K

90  106
 9.00  103 N
10.00  103
P  9.00 kN 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
579
P
y
6 in.
PROBLEM 4.124
6 in.
10 in.
Q
B
A
x
x
z
A
z
A = 10.0 in2
Iz = 273 in4
A short length of a rolled-steel column supports a rigid plate
on which two loads P and Q are applied as shown. The strains
at two points A and B on the centerline of the outer faces of the
flanges have been measured and found to be
 A  400  106 in./in.
 B  300  106 in./in.
Knowing that E  29  106 psi, determine the magnitude of
each load.
SOLUTION
Stresses at A and B from strain gages:
 A  E A  (29  106 )(400  106 )  11.6  103 psi
 B  E B  (29  106 )(300  106 )  8.7  103 psi
F  PQ
Centric force:
M  6 P  6Q
Bending couple:
c  5 in.
A  
F Mc
P  Q (6 P  6Q)(5)



10.0
273
A
I
11.6  103  0.00989 P  0.20989Q
B  
F Mc
P  Q (6 P  6Q )(5)



10.0
273
A
I
8.7  103  0.20989 P  0.00989Q
Solving (1) and (2) simultaneously,
P  44.2  103 lb  44.2 kips
Q  57.3  103 lb  57.3 kips
(1)
(2)


PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
580
PROBLEM 4.125
y
P
A single vertical force P is applied to a short steel post as shown.
Gages located at A, B, and C indicate the following strains:
z
 A  500
x
 C  200
Knowing that E  29  106 psi, determine (a) the magnitude of
P, (b) the line of action of P, (c) the corresponding strain at the
hidden edge of the post, where x  2.5 in. and z  1.5 in.
C
A
 B  1000
B
3 in.
5 in.
SOLUTION
Ix 
1
(5)(3)3  11.25 in 4
12
M x  Pz
M z   Px
Iz 
1
(3)(5)3  31.25 in 4
12
A  (5)(3)  15 in 2
x A  2.5 in., xB  2.5 in., xC  2.5 in., xD  2.5 in.
z A  1.5 in., z B  1.5 in., zC  1.5 in., z D  1.5 in.
 A  E A  (29  106 )(500  106 )  14,500 psi  14.5 ksi
 B  E B  (29  106 )(1000  106 )  29, 000 psi  29 ksi
 C  E C  (29  106 )(200  106 )  5800 psi  5.8 ksi
A  
B  
C  
P M x z A M z xA


 0.06667 P  0.13333M x  0.08M z
A
Ix
Iz
(1)
P M x z B M z xB


 0.06667 P  0.13333M x  0.08M z
A
Ix
Iz
P M x zC
M x

 z C  0.06667 P  0.13333M x  0.08M z
A
Ix
Iz
(2)
(3)
Substituting the values for  A ,  B , and  C into (1), (2), and (3) and solving the simultaneous equations
gives
M x  87 kip  in.

x

z 
(c)
(a) P  152.3 kips 
Mz
90.625


152.25
P
(b) x  0.595 in. 
Mx
87


152.25
P
D  

M z  90.625 kip  in.
z  0.571 in. 
P M x z D M z xD


 0.06667 P  0.13333M x  0.08M z
A
Ix
Iz
 (0.06667)(152.25)  (0.13333)(87)  (0.08)(90.625)  8.70 ksi
Strain at hidden edge:
 
D
E

8.70  103
29  106
 
  300 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
581
PROBLEM 4.126
b ⫽ 40 mm
A
a ⫽ 25 mm
D
d
B
P
The eccentric axial force P acts at point D, which must be located
25 mm below the top surface of the steel bar shown. For P  60 kN,
determine (a) the depth d of the bar for which the tensile stress at
point A is maximum, (b) the corresponding stress at point A.
C
20 mm
SOLUTION
A  bd
I
1
bd 3
12
1
1
c d
e d a
2
2
P Pec
A  
A
I

1
1 
P  1 12 2 d  a 2 d  P  4 6a 
A   
   2
b d
d3
 b d d 



(a)
(b)
 
Depth d for maximum  A : Differentiate with respect to d.
A 
60  103
40  103
d A P  4 12a 
  2  3   0
dd
b d
d 
d  3a

4
(6)(25  103 ) 

 40  106 Pa

3
3 2 
75
10
(75
10
)




d  75 mm 
 A  40 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
582
PROBLEM 4.127
y
M 300 N · m
60
A
z
B
16 mm
The couple M is applied to a beam of the cross section shown in a plane
forming an angle  with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
C
16 mm
D
40 mm
40 mm
SOLUTION
Iz 
1
(80)(32)3  218.45  103 mm 4  218.45  109 m 4
12
1
I y  (32)(80)3  1.36533  106 mm 4  1.36533  106 m 4
12
y A  yB   yD  16 mm
z A   z B   z D  40 mm
M y  300cos 30  259.81 N  m
(a)
(b)
(c)
A  
B  
D  
M z  300sin 30  150 N  m
M z yA M y zA
(150)(16  103 ) (259.81)(40  103 )



Iz
Iy
218.45  109
1.36533  106
 3.37  106 Pa
 A  3.37 MPa 
 18.60  106 Pa
 B  18.60 MPa 
M z yB M y z B
(150)(16  103 ) (259.81)(40  103 )



Iz
Iy
218.45  109
1.36533  106
M z yD M y z D
(150)(16  103 ) (259.81)(40  103 )



Iz
Iy
218.45  109
1.36533  106
 3.37  106 Pa
 D  3.37 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
583
PROBLEM 4.128
y
␤ ⫽ 30⬚
A
The couple M is applied to a beam of the cross section shown in a plane
forming an angle  with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
B
M ⫽ 400 lb · m
0.6 in.
z
C
0.6 in.
D
0.4 in.
SOLUTION
Iz 
1
(0.4)(1.2)3  57.6  10 3 in 4
12
1
I y  (1.2)(0.4)3  6.40  103 in 4
12
y A  yB   yD  0.6 in.
1
z A   z B   z D    (0.4)  0.2 in.
2
M y  400 cos 60  200 lb  in.,
(a)
(b)
(c)
A  
B  
D  
M z  400 sin 60  346.41 lb  in.
M z yA M y zA
(346.41)(0.6) (200)(0.2)



Iz
Iy
57.6  103
6.40  103
 9.86  103 psi  9.86 ksi

 2.64  103 psi  2.64 ksi

 9.86  103 psi  9.86 ksi

M z yB M y zB
(346.41)(0.6) (200)(0.2)



Iz
Iy
57.6  106
6.4  103
M z yD M y z D
(346.41)(0.6) (200)(0.2)



Iz
Iy
57.6  103
6.40  103
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
584
y
PROBLEM 4.129
15
A
The couple M is applied to a beam of the cross section shown in a
plane forming an angle  with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
B
M 25 kN · m
80 mm
C
z
20 mm
80 mm
D
30 mm
SOLUTION
M y  25sin 15  6.4705 kN  m
M z  25cos15  24.148 kN  m
Iy 
1
1
(80)(90)3  (80)(30)3  5.04  106 mm 4
12
12
6
4
I y  5.04  10 m
1
1
1
I z  (90)(60)3  (60)(20)3  (30)(100)3  16.64  106 mm 4  16.64  106 m 4
3
3
3
Stress:
(a)
A 
 
M yz
Iy

Mzy
Iz
(6.4705 kN  m)(0.045 m) (24.148 kN  m)(0.060 m)

5.04  106 m 4
16.64  106 m 4
 57.772 MPa  87.072 MPa
(b)
B 
(6.4705 kN  m)(0.045 m) (24.148 kN  m)(0.060 m)

5.04  106 m 4
16.64  106 m 4
 57.772 MPa  87.072 MPa
(c)
D 
(6.4705 kN  m)(0.015 m) (24.148 kN  m)(0.100 m)

5.04  106 m 4
16.64  106 m 4
 19.257 MPa  145.12 MPa
 A  29.3 MPa 
 B  144.8 MPa 
 D  125.9 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
585
y
PROBLEM 4.130
20
M 10 kip · in.
A
z
3 in.
The couple M is applied to a beam of the cross section shown in a plane
forming an angle  with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
2 in.
C
B
3 in.
D
2 in.
4 in.
SOLUTION
Locate centroid.
A, in 2


Σ
The centroid lies at point C.
z , in.
16
1
Az , in 3
8
2
16
16
24
0
Iz 
1
1
(2)(8)3  (4)(2)3  88 in 4
12
12
1
1
I y  (8)(2)3  (2)(4)3  64 in 4
3
3
y A   yB  1 in.,
yD  4 in.
z A  z B  4 in.,
zD  0
M z  10 cos 20  9.3969 kip  in.
M y  10 sin 20  3.4202 kip  in.
(a)
A  
(b)
B  
(c)
D  
M z yA M y zA
(9.3969)(1) (3.4202)(4)



88
64
Iz
Iy
M z yB M y z B
(9.3969)( 1) (3.4202)( 4)



88
64
Iz
Iy
M z yD M y z D
(9.3969)(4) (3.4202)(0)



88
64
Iz
Iy
 A  0.321 ksi 
 B  0.107 ksi 
 D  0.427 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
586
PROBLEM 4.131
y
M 5 60 kip · in.
b 5 508
A
B
3 in.
The couple M is applied to a beam of the cross section shown in a
plane forming an angle  with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
C
z
3 in.
D
1 in.
1 in.
2.5 in. 2.5 in.
5 in.
5 in.
SOLUTION
M z  60 sin 40  38.567 kip  in.
M y  60 cos 40  45.963 kip  in.
y A  yB   yD  3 in.
z A   z B   z D  5 in.
Iz 
Iy 
(a)
A  
1


(10)(6)3  2  (1) 2   178.429 in 4
12
4

1


(6)(10)3  2  (1) 4   (1) 2 (2.5) 2   459.16 in 4
12
4


M z yA M y zA
(38.567)(3) (45.963)(5)



Iz
Iy
178.429
459.16
 1.149 ksi
(b)
(c)
B  
D  
M z yB M y z B
(38.567)(3) (45.963)(5)



Iz
Iy
178.429
459.16
 A  1.149 ksi 
 0.1479
 B  0.1479 ksi 
 1.149 ksi
 D  1.149 ksi 
M z yD M y z D
(38.567)(3) (45.963)(5)



178.429
459.16
Iz
Iy
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
587
y
PROBLEM 4.132
M 5 75 kip · in.
b 5 758
A
2.4 in.
B
1.6 in. z
The couple M is applied to a beam of the cross section shown in a
plane forming an angle  with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
C
D
4 in.
4.8 in.
SOLUTION
Iz 
1
1
(4.8)(2.4)3  (4)(1.6)3  4.1643 in 4
12
12
1
1
I y  (2.4)(4.8)3  (1.6)(4)3  13.5851 in 4
12
12
y A  yB   yD  1.2 in.
z A   z B   z D  2.4 in.
M z  75sin15  19.4114 kip  in.
M y  75cos15  72.444 kip  in.
(a)
A  
(b)
B  
(c)
D  
M z yA M y zA
(19.4114)(1.2) (72.444)(2.4)



Iz
Iy
4.1643
13.5851
M z yB M y z B
(19.4114)(1.2) (72.444)(2.4)



4.1643
13.5851
Iz
Iy
M z yD M y z D
(19.4114)(1.2) (72.444)(2.4)



4.1643
13.5851
Iz
Iy
 A  7.20 ksi 
 B  18.39 ksi 
 D  7.20 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
588
y
PROBLEM 4.133
␤ ⫽ 30⬚
M ⫽ 100 N · m
The couple M is applied to a beam of the cross section shown in a
plane forming an angle  with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
B
z
C
D
A
r ⫽ 20 mm
SOLUTION
Iz 

8
   4r   
 
r 4   r 2 

8
 2  3   8 9
2
 4
r

 (0.109757)(20) 4  17.5611  103 mm 4  17.5611  109 m 4
Iy 

8
r4 
y A  yD  
 (20) 4
8
 62.832  103 mm 4  62.832  109 m 4
4r
(4)(20)

 8.4883 mm
3
3
yB  20  8.4883  11.5117 mm
z A   z D  20 mm
zB  0
M z  100 cos30  86.603 N  m
M y  100sin 30  50 N  m
(a)
(b)
(c)
A  
B  
D  
(86.603)(8.4883  103 ) (50)(20  103 )
M z yA M y zA



Iz
Iy
17.5611  109
62.832  109
 57.8  106 Pa
(86.603)(11.5117  10 )
(50)(0)
M z yB M y z B



9
Iz
Iy
17.5611  10
62.832  109
3
 56.8  106 Pa
(86.603)(8.4883  10 ) (50)(20  10 )
M z yD M y z D



Iz
Iy
17.5611  109
62.832  109
3
 25.9  106 Pa
 A  57.8 MPa 
 B  56.8 MPa 
3
 D  25.9 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
589
W310 ⫻ 38.7 15
PROBLEM 4.134
B
A
C
310 mm
The couple M is applied to a beam of the cross section shown in a plane
forming an angle  with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
M ⫽ 16 kN · m
D
E
165 mm
SOLUTION
For W310  38.7 rolled steel shape,
I z  84.9  106 mm 4  84.9  106 m 4
I y  7.20  106 mm 4  7.20  106 m 4
1
1
yD   yE    (310)  155 mm
2
2
1
z A  z E   z B   z D    (165)  82.5 mm
2
y A  yB 
M z  (16  103 ) cos 15  15.455  103 N  m
M y  (16  103 ) sin 15  4.1411  103 N  m
(a)
tan  
Iz
84.9  106
tan  
tan 15  3.1596
I
7.20  106
  72.4
  72.4  15  57.4
(b)

Maximum tensile stress occurs at point E.
E  
M z yE M y z E
(15.455  103 )(155  103 ) (4.1411  103 )(82.5  103 )



Iz
Iy
84.9  106
7.20  106
 75.7  106 Pa  75.7 MPa

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
590
PROBLEM 4.135
208
S6 3 12.5
A
M 5 15 kip · in.
B
C
E
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
3.33 in.
6 in.
D
SOLUTION
For S6  12.5 rolled steel shape,
I z  22.0 in 4
I y  1.80 in 4
zE   z A   zB  zD 
1
(3.33)  1.665 in.
2
1
y A  yB   yD   yE  (6)  3 in.
2
M z  15 sin 20  5.1303 kip  in.
M y  15 cos 20  14.095 kip  in.
(a)
tan  
Iz
22.0
tan  
tan (90  20)  33.58
Iy
1.80
  88.29
  88.29  70  18.29
(b)

Maximum tensile stress occurs at point D.
D  
M z yD M y z D
(5.1303)(3) (14.095)(1.665)



 13.74 ksi
Iz
Iy
22.0
1.80

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
591
PROBLEM 4.136
5⬚
C150 ⫻ 12.2
B
A
M ⫽ 6 kN · m
152 mm
C
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
13 mm
E
D
48.8 mm
SOLUTION
M y  6 sin 5  0.52293 kN  m
M z   6 cos 5  5.9772 kN  m
C150  12.2
I y  0.286  106 m 4
I z   5.45  106 m 4
(a)
Neutral axis:
tan  
I z
5.45  106 m 4
tan  
tan 5
I y
0.286  106 m 4
tan   1.66718
  59.044
    5
(b)
  54.0
Maximum tensile stress at E: y E  76 mm, zE  13 mm
E  
M y zE
I y

M z  yE (0.52293 kN  m)(0.013 m) (5.9772 kN  m)(0.076 m)


I z
0.286  106 m 4
5.45  106 m 4

 E  107.1 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
592
y'
PROBLEM 4.137
30⬚
B
50 mm
A
5 mm
5 mm
C
M 400 N · m
z'
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
D
E
18.57 mm
5 mm
50 mm
Iy' ⫽ 281 ⫻ 103 mm4
Iz' ⫽ 176.9 ⫻ 103 mm4
SOLUTION
I z  176.9  103 mm 4  176.9  109 m 4
I y  281  103 mm 4  281  109 m 4
yE  18.57 mm, z E  25 mm
M z  400 cos 30  346.41 N  m
M y  400sin 30  200 N  m
(a)
tan  
I z
176.9  109
tan  
 tan 30  0.36346
I y
281  109
  19.97
  30  19.97
(b)
  10.03 
Maximum tensile stress occurs at point E.
E  
(346.41)(18.57  103 ) (200)(25  103 )
M z  yE M y zE



I z
I y
176.9  109
281  109
 36.36  106  17.79  106  54.2  106 Pa
 E  54.2 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
593
458
PROBLEM 4.138
y'
B
z'
0.859 in.
M 5 15 kip · in.
C
A
1
2
in.
4 in.
D
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
4 in.
4 in.
Iy' 5 6.74 in4
Iz' 5 21.4 in4
SOLUTION
I z  21.4 in 4
I y  6.74 in 4
z A  z B  0.859 in.
y A  4 in.
z D  4  0.859 in.  3.141 in.
yB  4 in.
yD  0.25 in.
M y  15 sin 45  10.6066 kip  in.
M z  15 cos 45  10.6066 kip  in.
(a)
Angle of neutral axis:
tan  
I z
21.4
tan  
tan (45)  3.1751
6.74
I y
  72.5
  72.5  45

  27.5 


(b)
The maximum tensile stress occurs at point D.
D  
M z yD M y z D
(10.6066)(0.25) (10.6066)(3.141)



Iz
Iy
21.4
6.74
 0.12391  4.9429
 D  5.07 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
594
y'
20⬚
PROBLEM 4.139
B
10 mm
A
The couple M acts in a vertical plane and is applied to a beam
oriented as shown. Determine (a) the angle that the neutral
axis forms with the horizontal, (b) the maximum tensile stress
in the beam.
6 mm
M 120 N · m
C
z'
10 mm
D
6 mm
E
Iy' ⫽ 14.77 ⫻ 103 mm4
Iz' ⫽ 53.6 ⫻ 103 mm4
10 mm
10 mm
SOLUTION
I z  53.6  103 mm 4  53.6  109 m 4
I y  14.77  103 mm 4  14.77  109 m 4
M z  120 sin 70  112.763 N  m
M y  120 cos 70  41.042 N  m
(a)
Angle of neutral axis:
  20
tan  
I z
53.6  109
tan  
tan 20  1.32084
I y
14.77  109
  52.871
  52.871  20
(b)
  32.9 
The maximum tensile stress occurs at point E.
yE  16 mm  0.016 m
z E  10 mm  0.010 m
M y  M y z E
 E   z E 
I z
I y

(112.763)(0.016) (41.042)(0.010)

53.6  109
14.77  109
 61.448  106 Pa
 E  61.4 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
595
PROBLEM 4.140
208
A
90 mm
M 5 750 N · m
C
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine (a) the angle that the neutral axis forms with the horizontal,
(b) the maximum tensile stress in the beam.
B
30 mm
25 mm
25 mm
SOLUTION
M y  750 sin 20
M y  256.5 N  m
M z   750 cos 20
M z   704.8 N  m
1

I y  2  (90)(25)3   0.2344  106 mm 4
12


I y  0.2344  106 m 4
I z 
(a)
Neutral axis:
tan  
1
(50)(90)3  1.0125  106 mm 4  1.0125  106 m 4
36
I z
1.0125  106 m 4
tan  
tan 20
I y
0.2344  106 m 4
tan   1.5724
  57.5

    20  57.5  30  37.5
(b)
Maximum tensile stress, at D: yD  30 mm
D 
M y zD
I y

z  25 mm
M z  yD (256.5 N  m)(0.030 m) (704.8 N  m)(0.025 m)


I z
0.2344  106 m 4
1.0125  106 m 4
 32.83 MPa  17.40 MPa  50.23 MPa
  37.5

 D  50.2 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
596
A
PROBLEM 4.141
y
40 mm
z
10 mm
C
M 1.2 kN · m
10 mm
40 mm
70 mm
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine the stress at point A.
10 mm
Iy 1.894 ⫻ 106 mm4
Iz ⫽ 0.614 ⫻ 106 mm4
Iyz ⫽ ⫹0.800 ⫻ 106 mm4
SOLUTION
Using Mohr’s circle, determine the principal axes and principal moments of inertia.
Y : (1.894, 0.800)  106 mm 4
Z : (0.614, 0.800)  106 mm 4
E : (1.254, 0)  106 mm 4
R  EF  FZ  0.6402  0.8002  106  1.0245  106 mm 4
2
2
I v  (1.254  1.0245)  106 mm 4  0.2295  106 mm 4  0.2295  106 m 4
I u  (1.254  1.0245)  106 mm 4  2.2785  106 mm 4  2.2785  106 m 4
FZ 0.800  106

 1.25
 m  25.67
FE 0.640  106
M v  M cos  m  (1.2  103 ) cos 25.67  1.0816  103 N  m
tan 2 m 
M u   M sin  m  (1.2  103 ) sin 25.67  0.5198  103 N  m
u A  y A cos  m  z A sin  m  45 cos 25.67  45 sin 25.67  21.07 mm
v A  z A cos  m  y A sin  m  45 cos 25.67  45 sin 25.67  60.05 mm
A  
M v u A M u vA
(1.0816  103 )(21.07  103 ) ( 0.5198  103 )(60.05  103 )



Iu
Iv
0.2295  106
2.2785  106
 A  113.0 MPa 
 113.0  106 Pa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
597
y
PROBLEM 4.142
A
2.4 in.
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine the stress at point A.
2.4 in.
z
M 125 kip · in.
C
2.4 in.
2.4 in.
2.4 in. 2.4 in.
SOLUTION
1

I y  2  (7.2)(2.4)3   66.355 in 4
3

1

I z  2  (2.4)(7.2)3  (2.4)(7.2)(1.2) 2   199.066 in 4
12

I yz  2 (2.4)(7.2)(1.2)(1.2)  49.766 in 4
Using Mohr’s circle, determine the principal axes and principal moments of inertia.
Y : (66.355 in 4 , 49.766 in 4 )
Z : (199.066 in 4 ,  49.766 in 4 )
E : (132.710 in 4 , 0)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
598
PROBLEM 4.142 (Continued)
tan 2 m 
DY 49.766

DE 66.355
2 m  36.87  m  18.435
R  DE  DY  82.944 in 4
2
2
I u  132.710  82.944  49.766 in 4
I v  132.710  82.944  215.654 in 4
M u  125sin18.435  39.529 kip  in.
M v  125cos18.435  118.585 kip  in.
u A  4.8 cos 18.435  2.4 sin 18.435  5.3126 in.
 A  4.8 sin 18.435  2.4 cos 18.435  0.7589 in.
M u
M
A    A  u A

I
Iu
(118.585)(5.3126) (39.529)(0.7589)

215.654
49.766
 2.32 ksi
 A  2.32 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
599
y
PROBLEM 4.143
1.08 in.
0.75 in.
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine the stress at point A.
2.08 in.
z
C
M 5 60 kip · in.
6 in.
0.75 in.
A
4 in.
Iy 5 8.7 in4
Iz 5 24.5 in4
Iyz 5 18.3 in4
SOLUTION
Using Mohr’s circle, determine the principal axes and principal moments of inertia.
Y : (8.7, 8.3) in 4
Z : (24.5,  8.3) in 4
E : (16.6, 0) in 4
EF  7.9 in 4
FZ  8.3 in 4
R  7.92  8.32  11.46 in 4
tan 2 m 
 m  23.2 I u  16.6  11.46  5.14 in 4
FZ 8.3

 1.0506
EF 7.9
I v  16.6  11.46  28.06 in 4
M u  M sin  m  (60) sin 23.2  23.64 kip  in.
M v  M cos  m  (60) cos 23.2  55.15 kip  in.
u A  y A cos  m  z A sin  m  3.92cos 23.2  1.08 sin 23.2  4.03 in.
v A  z A cos  m  y A sin  m  1.08cos 23.2  3.92 sin 23.2  0.552 in.
A  
M vu A M u vA
(55.15)(4.03) (23.64)(0.552)



28.06
5.14
Iv
Iu
 A  10.46 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
600
PROBLEM 4.144
D
H 14 kN
B
The tube shown has a uniform wall thickness of 12 mm. For the
loading given, determine (a) the stress at points A and B, (b) the point
where the neutral axis intersects line ABD.
G
28 kN
125 mm
E
A
F
28 kN
75 mm
SOLUTION
Add y- and z-axes as shown. Cross section is a 75 mm  125-mm rectangle with a 51 mm  101-mm rectangular
cutout.
Iz 
1
1
(75)(125)3  (51)(101)3  7.8283  106 mm 4  7.8283  106 m 4
12
12
1
1
I y  (125)(75)3  (101)(51)3  3.2781  103 mm 4  3.2781  106 m 4
12
12
A  (75)(125)  (51)(101)  4.224  103 mm 2  4.224  103 m 2
Resultant force and bending couples:
P  14  28  28  70 kN  70  103 N
M z  (62.5 mm)(14 kN)  (62.5 mm)(28kN)  (62.5 mm)(28 kN)  2625 N  m
M y  (37.5 mm)(14 kN)  (37.5 mm)(28 kN)  (37.5 mm)(28 kN)  525 N  m
(a)
A 
P M z yA M y zA
70  103
(2625)(0.0625) ( 525)(0.0375)





3
A
Iz
Iy
4.224  10
7.8283  106
3.2781  106
 A  31.5 MPa 
 31.524  106 Pa
B 
P M z yB M y z B
70  103
(2625)(0.0625) (525)(0.0375)





3
A
Iz
Iy
4.224  10
7.8283  106
3.2781  106
 10.39  106 Pa
(b)
 B  10.39 MPa 
Let point H be the point where the neutral axis intersects AB.
z H  0.0375 m,
0
yH  ?,  H  0
P M z yH M y z H


A
Iz
Iy
 P Mz H  7.8283  106  70  103
(525)(0.0375) 

 




3
A
I y 
2625
3.2781  106 
 4.224  10

 0.03151 m  31.51 mm
yH 
Iz
Mz
31.51  62.5  94.0 mm
Answer: 94.0 mm above point A. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
601
PROBLEM 4.145
A horizontal load P of magnitude 100 kN is applied to the beam
shown. Determine the largest distance a for which the maximum
tensile stress in the beam does not exceed 75 MPa.
y
20 mm
a
20 mm
O
z
x
P
20 mm
60 mm
20 mm
SOLUTION
Locate the centroid.


∑
Move coordinate origin to the centroid.
Coordinates of load point:
Bending couples:
Ix 
X P  a,
M x  yP P
A, mm2
y , mm
2000
10
1200
3200
–10
Ay , mm3
20  103
12  103
8  103
Ay
Y 
A
8  103
3200
 2.5 mm

yP  2.5 mm
M y  aP
1
1
(100)(20)3  (2000)(7.5) 2 
(60)(20)3  (1200)(12.5) 2  0.40667  106 mm 4
12
12
 0.40667  106 m 4
Iy 
1
1
(20)(100)3 
(20)(60)3  2.0267  106 mm 4  2.0267  106 m 4
12
12
P M y M x
   x  y
 A  75  106 Pa, P  100  103 N
A
Ix
Iy
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
602
PROBLEM 4.145 (Continued)
My 
My 

Iy P Mxy


 
x A
Ix

For point A,
x  50 mm, y  2.5 mm
2.0267  106  100  103
(2.5)(100  103 )(2.5  103 )


 75  106 

3
6
6
50  10
0.40667  10
 3200  10

2.0267  106
{31.25  1.537  75}  106  1.7111  103 N  m
50  103
a
My
P

(1.7111  103 )
 17.11  103 m
3
100  10
a  17.11 mm 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
603
1 in.
PROBLEM 4.146
1 in.
4 in.
Knowing that P  90 kips, determine the largest distance a for
which the maximum compressive stress dose not exceed 18 ksi.
5 in.
P
1 in.
a
2.5 in.
SOLUTION
A  (5 in.)(6 in.)  2(2 in.)(4 in.)  14 in 2
Ix 
1
1
(5 in.)(6 in.)3  2 (2 in.)(4 in.)3  68.67 in 4
12
12
1
1
(4 in.)(1 in.)3  21.17 in 4
I z  2 (1 in.)(5 in.)3 
12
12
Force-couple system at C:
For P  90 kips:
P P
18 ksi  
M z  Pa
P  90 kips M x  (90 kips)(2.5 in.)  225 kip  in.
Maximum compressive stress at B:
B  
M x  P(2.5 in.)
 B  18 ksi
M z  (90 kips) a
P M x (3 in.) M z (2.5 in.)


A
Ix
Iz
90 kips (225 kip  in.)(3 in.) (90 kips) a (2.5 in.)


14 in 2
68.67 in 4
21.17 in 4
18  6.429  9.830  10.628a
1.741  10.628 a
a  0.1638 in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
604
1 in.
PROBLEM 4.147
1 in.
4 in.
Knowing that a  1.25 in., determine the largest value of P that
can be applied without exceeding either of the following
allowable stresses:
5 in.
P
1 in.
 ten  10 ksi
a
2.5 in.
 comp  18 ksi
SOLUTION
A  (5 in.)(6 in.)  (2)(2 in.)(4 in.)  14 in 2
Ix 
1
1
(5 in.)(6 in.)3  2 (2 in.)(4 in.)3  68.67 in 4
12
12
1
1
(4 in.)(1 in.)3  21.17 in 4
I z  2 (1 in.)(5 in.)3 
12
12
Force-couple system at C:
P  P M x  P(2.5 in.)
For a  1.25 in.,
M y  Pa  (1.25 in.)
Maximum compressive stress at B:
B  
18 ksi  
 B  18 ksi
P M x (3 in.) M z (2.5 in.)


A
Ix
Iz
P
P(2.5 in.)(3 in.) P(1.25 in.)(2.5 in.)


2
14 in
68.67 in 4
21.17 in 4
18   0.0714P  0.1092 P  0.1476 P
18  0.3282P
P  54.8 kips
Maximum tensile stress at D:  D  10 ksi
D  
P M x (3 in.) M z (2.5 in.)


A
Ix
Iz
10 ksi   0.0714 P  0.1092 P  0.1476 P
10  0.1854 P
The smaller value of P is the largest allowable value.
P  53.9 kips
P  53.9 kips 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
605
y
R 125 mm
C
z
P 4 kN
PROBLEM 4.148
E
A rigid circular plate of 125-mm radius is attached to a solid
150  200-mm rectangular post, with the center of the plate directly above
the center of the post. If a 4-kN force P is applied at E with   30,
determine (a) the stress at point A, (b) the stress at point B, (c) the point
where the neutral axis intersects line ABD.
x
A
D
B
200 mm 150 mm
SOLUTION
P  4  103 N (compression)
M x   PR sin 30  (4  103 )(125  103 )sin 30  250 N  m
M z   PR cos 30  (4  103 )(125  103 ) cos 30  433 N  m
Ix 
1
(200)(150)3  56.25  106 mm 4  56.25  106 m 4
12
1
(150)(200)3  100  106 mm 4  100  106 m 4
Iz 
12
 x A  xB  100 mm z A  z B  75 mm
A  (200)(150)  30  103 mm 2  30  103 m 2
(a)
A  
P M x z A M z xA
4  103
(250)(75  103 ) (433)(100  103 )





A
Ix
Iz
30  103
56.25  106
100  106
 A  633  103 Pa  633 kPa 
(b)
B  
P M x z B M z xB
4  103
(250)(75  103 ) (433)(100  103 )





A
Ix
Iz
30  103
56.25  106
100  106
 B  233  103 Pa  233 kPa 
(c)
Let G be the point on AB where the neutral axis intersects.
G  0
G  
xG 
zG  75 mm
P M x zG M z xG


0
A
Ix
Iz
xG  ?
I z  P M x zG  100  106  4  103
(250)(75  103 ) 

 



Mz  A
Ix 
433  30  103
56.25  106 
 46.2  103 m  46.2 mm
Point G lies 146.2 mm from point A. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
606
y
R 125 mm
C
z
P 4 kN
PROBLEM 4.149
E
In Prob. 4.148, determine (a) the value of θ for which the stress at D
reaches it largest value, (b) the corresponding values of the stress at A, B,
C, and D.
x
A
D
B
200 mm 150 mm
PROBLEM 4.148 A rigid circular plate of 125-mm radius is attached to a
solid 150  200-mm rectangular post, with the center of the plate directly
above the center of the post. If a 4-kN force P is applied at E with
  30, determine (a) the stress at point A, (b) the stress at point B,
(c) the point where the neutral axis intersects line ABD.
SOLUTION
P  4  103 N
(a)
PR  (4  103 )(125  103 )  500 N  m
M x   PR sin   500sin 
Ix 
M x   PR cos   500 cos 
1
(200)(150)3  56.25  106 mm 4  56.25  106 m 4
2
1
I z  (150)(200)3  100  106 mm 4  100  106 m 4
2
xD  100 mm
z D  75 mm
 
A  (200)(150)  30  103 mm 2  30  103 m 2
 1 Rz sin 
P M xz M zx
Rx cos  


 P  


A
Ix
Iz
Ix
Iz 
A
For  to be a maximum,
d
0
d
with z  z D , x  xD

d D
Rz cos 
RxD sin  
  P 0  D

0
d
Ix
IZ


sin 
(100  106 )(75  103 )
4
I z
 tan    z D  

6
3
cos 
I x xD
(56.25  10 )(100  10 ) 3
(b)
A  
sin   0.8,
cos   0.6
  53.1 
P M x z A M z xA
4  103
(500)(0.8)(75  103 ) (500)(0.6)(100  103 )





A
Ix
Iz
30  103
56.25  106
100  106
 (0.13333  0.53333  0.300)  106 Pa  0.700  106 Pa
 B  (0.13333  0.53333  0.300)  106 Pa  0.100  106 Pa
 C  (0.13333  0  0)  106 Pa
 D  (0.13333  0.53333  0.300)  106 Pa   0.967  106 Pa
 A  700 kPa 
 B  100 kPa 
 C  133.3 kPa 
 D  967 kPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
607
y
PROBLEM 4.150
0.5 in.
1.43 in.
z
M0
C
5 in.
0.5 in.
1.43 in.
A beam having the cross section shown is subjected to a couple M 0 that
acts in a vertical plane. Determine the largest permissible value of the
moment M 0 of the couple if the maximum stress in the beam is
not to exceed 12 ksi. Given: I y  I z  11.3 in 4 , A  4.75 in 2 ,
kmin  0.983 in. (Hint: By reason of symmetry, the principal axes form
2
an angle of 45 with the coordinate axes. Use the relations I min  Akmin
and I min  I max  I y  I z .)
5 in.
SOLUTION
M u  M 0 sin 45  0.70711M 0
M v  M 0 cos 45  0.70711M 0
2
 (4.75)(0.983)2  4.59 in 4
I min  Akmin
I max  I y  I z  I min  11.3  11.3  4.59  18.01 in 4
u B  yB cos 45  z B sin 45  3.57 cos 45  0.93 sin 45  1.866 in.
vB  z B cos 45  yB sin 45  0.93 cos 45  (3.57) sin 45  3.182 in.
B  
 u
M vu B M u vB
v 

 0.70711M 0   B  B 
Iv
Iu
I max 
 I min
 (1.866) 3.182 
 0.70711M 0  

 0.4124M 0
4.59
18.01 

M0 
B
0.4124

12
0.4124
M 0  29.1 kip  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
608
PROBLEM 4.151
y
0.5 in.
Solve Prob. 4.150, assuming that the couple M 0 acts in a horizontal
plane.
1.43 in.
z
M0
C
5 in.
0.5 in.
1.43 in.
5 in.
SOLUTION
PROBLEM 4.150 A beam having the cross section shown is subjected
to a couple M 0 that acts in a vertical plane. Determine the largest
permissible value of the moment M 0 of the couple if the maximum
stress in the beam is not to exceed 12 ksi. Given: I y  I z  11.3 in 4,
A  4.75 in 2, kmin  0.983 in. (Hint: By reason of symmetry, the
principal axes form an angle of 45 with the coordinate axes. Use the
2
relations I min  Akmin
and I min  I max  I y  I z .)
M u  M 0 cos 45  0.70711M 0
M v  M 0 sin 45   0.70711M 0
2
 (4.75)(0.983)2  4.59 in 4
I min  Akmin
I max  I y  I z  I min  11.3  11.3  4.59  18.01 in 4
uD  yD cos 45  z D sin 45  0.93 cos 45  (3.57 sin 45)  1.866 in.
vD  z D cos 45  yD sin 45  (3.57) cos 45  (0.93) sin 45  3.182 in.
D  
 u
M v u D M u vD
v 

 0.70711M 0   D  D 
Iv
Iu
I max 
 I min
 (1.866) 3.182 
 0.70711M 0  

 0.4124M 0
4.59
18.01 

M0 
D
0.4124

12
0.4124
M 0  29.1 kip  in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
609
PROBLEM 4.152
y
z
M0
40 mm
10 mm
C
10 mm
40 mm
70 mm
10 mm
The Z section shown is subjected to a couple M 0 acting in a vertical
plane. Determine the largest permissible value of the moment M 0 of the
couple if the maximum stress is not to exceed 80 MPa. Given:
I max  2.28  106 mm 4 ,
I min  0.23  106 mm 4 , principal axes
25.7
and 64.3
.
SOLUTION
I v  I max  2.28  106 mm 4  2.28  106 m 4
I u  I min  0.23  106 mm 4  0.23  106 m 4
M v  M 0 cos 64.3
M u  M 0 sin 64.3
  64.3
tan  
Iv
tan 
Iu
2.28  106
tan 64.3
0.23  106
 20.597

  87.22
Points A and B are farthest from the neutral axis.
uB  yB cos 64.3  z B sin 64.3  (45) cos 64.3  (35) sin 64.3
 51.05 mm
vB  z B cos 64.3  yB sin 64.3  (35) cos 64.3  (45) sin 64.3
 25.37 mm
B  
80  106  
M v u B M u vB

Iv
Iu
(M 0 cos 64.3)(51.05  103 ) (M 0 sin 64.3)(25.37  103 )

0.23  106
2.28  106
 109.1  103 M 0
M0 
80  106
109.1  103
M 0  733 N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
610
PROBLEM 4.153
y
z
M0
40 mm
10 mm
C
10 mm
40 mm
70 mm
10 mm
Solve Prob. 4.152 assuming that the couple M 0 acts in a horizontal
plane.
PROBLEM 4.152 The Z section shown is subjected to a couple M 0
acting in a vertical plane. Determine the largest permissible value of the
moment M 0 of the couple if the maximum stress is not to exceed 80 MPa.
Given: I max  2.28  106 mm 4 , I min  0.23  106 mm 4 , principal axes
and 64.3
.
25.7
SOLUTION
I v  I min  0.23  106 mm 4  0.23  106 m 4
I u  I max  2.28  106 mm 4  2.28  106 m 4
M v  M 0 cos 64.3
M u  M 0 sin 64.3
  64.3
tan  
Iv
tan 
Iu
0.23  106
tan 64.3
2.28  106
 0.20961

  11.84
Points D and E are farthest from the neutral axis.
u D  yD cos 25.7  z D sin 25.7  (5) cos 25.7  45 sin 25.7
 24.02 mm
vD  z D cos 25.7  yD sin 25.7  45cos 25.7  (5) sin 25.7
 38.38 mm
D  
(M D cos 64.3)(24.02  103 )
M vu D M u vD


Iv
Iu
0.23  106

(M 0 sin 64.3)(38.38  103 )
2.28  106
80  106  60.48  103 M 0
M 0  1.323 kN  m 
M 0  1.323  103 N  m
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
611
PROBLEM 4.154
y
0.3 in.
M0
C
z
0.3 in.
1.5 in.
0.6 in. 1.5 in. 0.6 in.
An extruded aluminum member having the cross section shown
is subjected to a couple acting in a vertical plane. Determine the
largest permissible value of the moment M 0 of the couple if the
maximum stress is not to exceed 12 ksi. Given:
I max  0.957 in 4 , I min  0.427 in 4 , principal axes 29.4 and
60.6 
SOLUTION
I u  I max  0.957 in 4
I v  I min  0.427 in 4
M u  M 0 sin 29.4, M v  M 0 cos 29.4
  29.4
tan  
0.427
Iv
tan  
tan 29.4
Iu
0.957
 0.2514
  14.11
Point A is farthest from the neutral axis.
y A  0.75 in., z A  0.75 in.
u A  y A cos 29.4  z A sin 29.4  1.0216 in.
v A  z A cos 29.4  y A sin 29.4  0.2852 in.
A  
M0 
(M cos 29.4)(1.0216) (M 0 sin 29.4)(0.2852)
M vu A
MV
 u A  0

0.427
0.957
Iv
Iu
A
1.9381
 1.9381 M 0

M 0  6.19 kip  in. 
12
1.9381
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
612
PROBLEM 4.155
y
20 mm
z
A beam having the cross section shown is subjected to a couple M0
acting in a vertical plane. Determine the largest permissible value of
the moment M0 of the couple if the maximum stress is not to exceed
100 MPa. Given: I y  I z  b 4 /36 and I yz  b 4 /72.
M0
C
b = 60 mm
20 mm
b = 60 mm
SOLUTION
I y  Iz 
I yz
b 4 604

 0.360  106 mm 4
36 36
b 4 604


 0.180  106 mm 4
72 72
Principal axes are symmetry axes.
Using Mohr’s circle, determine the principal moments of inertia.
R  I yz  0.180  106 mm 4
Iv 
I y  Iz
R
2
 0.540  106 mm 4  0.540  106 m 4
I y  Iz
R
Iu 
2
 0.180  106 mm 4  0.180  106 m 4
M u  M 0 sin 45  0.70711M 0 ,
  45
  71.56
Point A:
uA  0
A  
M0  
tan  
M v  M 0 cos 45  0.70711M 0
Iv
0.540  106
tan  
tan 45  3
Iu
0.180  106
v A  20 2 mm
M vu A M u vA
(0.70711M 0 )(20 2  103 )

0
 11.11  103 M 0
Iv
Iu
0.180  106
A
111.11  103

100  106
 900 N  m
111.11  103
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
613
PROBLEM 4.155 (Continued)
Point B:
uB  
60
2
mm,
vB 
20
2

mm


(0.70711M 0 )  602  103
(0.70711M 0 ) 202  103
M v u B M u vB



B  
Iv
Iu
0.540  106
0.180  106

 111.11  103 M 0
M0 
B
111.11  103

100  106
 900 N  m
111.11  103
Choose the smaller value.
M 0  900 N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
614
b
PROBLEM 4.156
B
A
Show that, if a solid rectangular beam is bent by a couple applied in a plane containing
one diagonal of a rectangular cross section, the neutral axis will lie along the other
diagonal.
h
M
C
D
E
SOLUTION
tan  
b
h
M z  M cos  , M z  M sin 
Iz 
1 3
bh
12
Iy 
1 3
hb
12
1 3
bh b
Iz
h
 
tan  
tan   12
1
Iy
b
hb3 h
12
The neutral axis passes through corner A of the diagonal AD. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
615
PROBLEM 4.157
y
D
A
A
D
P
B
x
z
C
h
6
x
z
(a) Show that the stress at corner A of the prismatic
member shown in part a of the figure will be zero if
the vertical force P is applied at a point located on the
line
x
z

1
b /6 h /6
h
B
b
(a)
(b)
C
b
6
(b) Further show that, if no tensile stress is to occur in
the member, the force P must be applied at a point
located within the area bounded by the line found in
part a and three similar lines corresponding to the
condition of zero stress at B, C, and D, respectively.
This area, shown in part b of the figure, is known as
the kern of the cross section.
SOLUTION
Iz 
1 3
1 3
hb
Ix 
bh
12
12
h
b
zA  
xA  
2
2
Let P be the load point.
M z   PxP
A  


 A  0,
(a)
For
(b)
At point E:
At point F:
1
x
z

0
b/6 h/6
z  0  xE  b /6
A  bh
M x  Pz P
P M z xA M x z A


A
Iz
Ix
 
( PxP )  b2
( Pz P  h2 )
P


3
1 hb
1 bh3
bh
12
12
P 
x
z 
1 P  P 

bh 
b /6 h/6 
x
z

1
b/6 h/6
x  0  z F  h/6
If the line of action ( xP , z P ) lies within the portion marked TA , a tensile stress will occur at corner A.
By considering  B  0,  C  0, and  D  0, the other portions producing tensile stresses are identified.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
616
PROBLEM 4.158
y
A beam of unsymmetric cross section is subjected to a couple M 0 acting in the
horizontal plane xz. Show that the stress at point A, of coordinates y and z, is
z
A 
A
y
C
z
zI z  yI yz
2
I y I z  I yz
My
x
where Iy, Iz, and Iyz denote the moments and product of inertia of the cross section
with respect to the coordinate axes, and My the moment of the couple.
SOLUTION
The stress  A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are
centroidal axes:
 A  C1 y  C2 z
where C1 and C2 are constants.
M z    y AdA  C1 y 2dA  C2  yz dA
  I zC1  I yzC2  0
C1  
I yz
Iz
C2
M y   z AdA  C1 yz dA  C2  z 2dA
 I yzC1  I yC2
 I yz

I yz
Iz
C2  I y C2

2
I z M y  I y I z  I yz
C2
C2 
IzM y
2
I y I z  I yz
C1  
A 
I yz M y
2
I y I z  I yz
I z z  I yz y
2
I y I z  I yz
My

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
617
PROBLEM 4.159
y
A beam of unsymmetric cross section is subjected to a couple M 0 acting in the
vertical plane xy. Show that the stress at point A, of coordinates y and z, is
z
A 
A
y
C
z
yI y  zI yz
2
I y I z  I yz
Mz
x
where I y , I z , and I yz denote the moments and product of inertia of the cross
section with respect to the coordinate axes, and M z the moment of the couple.
SOLUTION
The stress  A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are
centroidal axes:
 A  C1 y  C2 z
where C1 and C2 are constants.
M y   z AdA  C1  yz dA  C2  z 2dA
 I yzC1  I yC2  0
C2  
I yz
Iy
C1
M z   y Adz  C1 y 2dA  C2  yz dA
  I z C1  I yz

I yz

Iy
C1
2
I y M z   I y I z  I yz
C1
C1  
C2  
A  
I yM z
2
I y I z  I yz
I yz M z
2
I y I z  I yz
I y y  I yz 2
2
I y I z  I yz
Mz

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
618
PROBLEM 4.160
y
(a) Show that, if a vertical force P is applied at point A of the section
shown, the equation of the neutral axis BD is
D
B
P
C
A
z
zA
xA
 zA 
 xA 
 2  x   2  z  1
 rz 
 rx 
x
where rz and rx denote the radius of gyration of the cross section with
respect to the z axis and the x axis, respectively. (b) Further show that, if
a vertical force Q is applied at any point located on line BD, the stress at
point A will be zero.
SOLUTION
Definitions:
rx2 
(a)
M x  Pz A
E  

M z   Px A
P M z xE M x z E
P Px A xE Pz A z E


 

A
Iz
Ix
A
Arz2
Arx2
z  
P   xA 
1   2  xE   A2  z E   0
A   rz 
 rx  
if E lies on neutral axis.
(b)
M x  Pz E
A  
Ix
I
, rz2  z
A
A
z 
x 
1   A2  x   A2  z  0,
 rz 
 rx 
M z   PxE
 zA 
 xA 
 2  x   2  z  1
 rz 
 rx 
P M z xA M x z A
P Px x
Pz z


   E 2 A  E 2A
A
Iz
Iy
A
Arz
Arx
 0 by equation from part (a).


PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
619
24 mm
B
PROBLEM 4.161
B
For the curved bar shown, determine the stress at point A when
(a) h  50 mm, (b) h  60 mm.
h
A
600 N · m
C
A
50 mm
600 N · m
SOLUTION
(a)
h  50 mm,
r1  50 mm,
A  (24)(50)  1.200  103 mm
R 
r 
r2  100 mm
h
50

 72.13475 mm
r2
ln 100
50
ln
r1
1
(r1  r2 )  75 mm
2
e  r  R  2.8652 mm
y A  72.13475  50  22.13475 mm
A  
(b)
(600)(22.13475  103 )
My A

 77.3  106 Pa
AerA
(1.200  103 )(2.8652  103 )(50  103 )
h  60 mm,
R
rA  50 mm
r1  50 mm,
r2  110 mm,
h
60

 76.09796 mm
110
r2
ln
50
ln r
1
A  (24)(60)  1.440  103 mm 2
77.3 MPa 
r 
1
e  r  R  3.90204 mm
(r1  r2 )  80 mm
2
y A  76.09796  50  26.09796 mm
rA  50 mm
A  
M yA
(600)(26.09796  103 )

 55.7  106 Pa
3
3
3
AerA
(1.440  10 )(3.90204  10 )(50  10 )
55.7 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
620
24 mm
B
B
A
A
PROBLEM 4.162
For the curved bar shown, determine the stress at points A and B
when h  55 mm.
h
600 N · m
C
50 mm
600 N · m
SOLUTION
h  55 mm,
r1  50 mm,
A  (24)(55)  1.320  10 mm
3
r2  105 mm
2
R
50
h

 74.13025 mm
105
r2
ln
ln
50
r1
1
r  (r1  r2 )  77.5 mm
2
e  r  R  3.36975 mm
y A  74.13025  50  24.13025 mm
A  
(600)(24.13025  103 )
M yA

 65.1  106 Pa
AerA
(1.320  103 )(3.36975  103 )(50  103 )
yB  74.13025  105  30.86975 mm
B 
rA  50 mm
rB  105 mm
(600)(30.8697  103 )
MyB

 39.7  106 Pa
AerB
(1.320  103 )(3.36975  103 )(105  103 )
65.1 MPa 
39.7 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
621
4 kip · in.
4 kip · in.
PROBLEM 4.163
C
3 in.
For the machine component and loading shown, determine the
stress at point A when (a) h  2 in., (b) h  2.6 in.
h
A
0.75 in.
B
SOLUTION
M  4 kip  in.
Rectangular cross section:
r 
(a)
A  bh r2  3 in. r1  r2  h
1
h
(r1  r2 ), R 
, er R
r
2
ln 2
r1
h  2 in.
A  (0.75)(2)  1.5 in 2
r 
r1  3  2  1 in.
R 
2
 1.8205 in.
ln 13
(b)
e  2  1.8205  0.1795 in.
r  r1  1 in.
At point A:
A 
M (r  R) (4)(1  1.8205)

 12.19 ksi
(1.5)(0.1795)(1)
Aer
h  2.6 in.
A  (0.75)(2.6)  1.95 in 2
r 
r1  3  2.6  0.4 in.
R 
2.6
 1.2904 in.
3
ln 0.4
At point A:
A 
1
(3  1)  2 in.
2
 A  12.19 ksi 
1
(3  0.4)  1.7 in.
2
e  1.7  1.2904  0.4906 in.
r  r1  0.4 in.
M (r  R )
(4)(0.4  1.2904)

 11.15 ksi
(1.95)(0.4096)(0.4)
Aer
 A  11.15 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
622
4 kip · in.
4 kip · in.
PROBLEM 4.164
C
0.75 in.
3 in.
For the machine component and loading shown, determine the
stress at points A and B when h  2.5 in.
h
A
B
SOLUTION
M  4 kip  in.
Rectangular cross section:
h  2.5 in. b  0.75 in. A  1.875 in.2
r2  3 in. r1  r2  h  0.5 in.
r 
1
1
(r1  r2 )  (0.5  3.0)  1.75 in.
2
2
h
2.5
R  r  3.0  1.3953 in.
ln 0.5
ln r2
e  r  R  1.75  1.3953  0.3547 in.
1
At point A:
r  r1  0.5 in.
A 
At point B:
M (r  R)
(4 kip  in.)(0.5 in.  1.3953 in.)

Aer
(0.75 in.)(2.5 in.)(0.3547 in.)(0.5 in.)
 A  10.77 ksi 
M (r  R )
(4 kip  in.)(3 in.  1.3953 in.)

Aer
(0.75 in.  2.5 in.)(0.3547 in.)(3 in.)
 B  3.22 ksi 
r  r2  3 in.
B 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
623
PROBLEM 4.165
r1
40 mm
The curved bar shown has a cross section of 40  60 mm and an inner
radius r1  15 mm. For the loading shown, determine the largest tensile
and compressive stresses.
60 mm
120 N · m
SOLUTION
h  40 mm, r1  15 mm, r2  55 mm
A  (60)(40)  2400 mm 2  2400  106 m 2
R
r 
40
h

 30.786 mm
55
r2
ln
ln
r1
40
1
(r1  r2 )  35 mm
2
e  r  R  4.214 mm
At r  15 mm,
 
y  30.786  15  15.756 mm
My
Aer
(120)(15.786  103 )
 12.49  10 6 Pa
(2400  10 6 )(4.214  103 )(15  103 )
At r  55 mm,
 
 
y  30.786  55  24.214 mm
(120)(24.214  103 )
 5.22  106 Pa
(2400  106 )(4.214  103 )(55  103 )
  12.49 MPa 
  5.22 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
624
PROBLEM 4.166
r1
40 mm
For the curved bar and loading shown, determine the percent
error introduced in the computation of the maximum stress by
assuming that the bar is straight. Consider the case when
(a) r1  20 mm, (b) r1  200 mm, (c) r1  2 m.
60 mm
120 N · m
SOLUTION
h  40 mm, A  (60)(40)  2400 mm 2  2400  106 m 2 , M  120 N  m
I 
1 3
1
(60)(40)3  0.32  106 mm 4  0.32  106 mm 4 ,
bh 
12
12
c
1
h  20 mm
2
Assuming that the bar is straight,
s  
(a)
Mc
(120)(20  108 )

 7.5  106 Pa  7.5 MPa
I
(0.32  106 )
r1  20 mm r2  60 mm
R
40
h

 36.4096 mm
60
r2
ln
ln
r1
20
r 
1
(r1  r2 )  40 mm
2
a 
r1  R  16.4096 mm
e  r  R  3.5904 mm
(120)(16.4096  103 )
M (r1  R)

 11.426  106 Pa  11.426 MPa
6
3
3
Aer
(2400  10 )(3.5904  10 )(20  10 )
% error 
11.426  (7.5)
 100%  34.4%
11.426

For parts (b) and (c), we get the values in the table below:
(a)
(b)
(c)
r1, mm
r2 , mm
20
200
2000
60
240
2040
R, mm
36.4096
219.3926
2019.9340
r , mm
40
220
2020
e, mm
3.5904
0.6074
0.0660
 , MPa
11.426
7.982
7.546
% error
34.4 %
6.0 % 
0.6 % 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
625
0.3 in.
B
0.4 in.
P9
PROBLEM 4.167
B
P
0.4 in.
Steel links having the cross section shown are available with
different central angles  . Knowing that the allowable stress is
12 ksi, determine the largest force P that can be applied to a
link for which   90.
0.8 in.
A
A
0.8 in.
b
1.2 in.
C
C
SOLUTION
Reduce section force to a force-couple system at G, the centroid of the cross section AB.


a  r 1  cos 
2

The bending couple is M   Pa.
For the rectangular section, the neutral axis for bending couple only lies at
R
Also, e  r  R
h
ln
r2
r1
At point A, the tensile stress is
A 
K 1
where
P
Data:
.
P My A
P Pay A
P
ay 
P



 1  A   K
A Aer1
A
Aer1
A
er1 
A
ay A
er1
A A
K
y A  R  r1
and
r  1.2 in., r1  0.8 in., r2  1.6 in., h  0.8 in., b  0.3 in.
0.8
A  (0.3)(0.8)  0.24 in 2
R  1.6  1.154156 in.
ln 0.8
e  1.2  1.154156  0.045844 in.,
y A  1.154156  0.8  0.35416 in.
a  1.2(1  cos 45)  0.35147 in.
K 1
P
(0.35147)(0.35416)
 4.3940
(0.045844)(0.8)
P  655 lb 
(0.24)(12)
 0.65544 kips
4.3940
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
626
PROBLEM 4.168
0.3 in.
B
0.4 in.
P9
Solve Prob. 4.167, assuming that   60.
B
0.8 in.
P
0.4 in.
A
PROBLEM 4.167 Steel links having the cross section shown are
available with different central angles . Knowing that the
allowable stress is 12 ksi, determine the largest force P that can be
applied to a link for which   90.
A
0.8 in.
b
1.2 in.
C
C
SOLUTION
Reduce section force to a force-couple system at G, the centroid of the cross section AB.
The bending couple is M   Pa.


a  r 1  cos 
2

For the rectangular section, the neutral axis for bending couple only lies at
R
Also, e  r  R
At point A, the tensile stress is
A 
ln
P
r2
r1
.
P My A
P Pay A
P
ay 
P



 1  A   K
A Aer1
A
Aer1
A
er1 
A
K 1
where
Data:
h
ay A
er1
A A
K
and
y A  R  r1
r  1.2 in., r1  0.8 in., r2  1.6 in., h  0.8 in., b  0.3 in.
0.8
A  (0.3)(0.8)  0.24 in 2
R  1.6  1.154156 in.
ln 0.8
e  1.2  1.154156  0.045844 in.
y A  1.154156  0.8  0.35416 in.
a  (1.2)(1  cos 30)  0.160770 in.
K 1
P
(0.160770)(0.35416)
 2.5525
(0.045844)(0.8)
P  1128 lb 
(0.24)(12)
 1.128 kips
2.5525
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
627
5 kN
PROBLEM 4.169
a
The curved bar shown has a cross section of 30  30 mm. Knowing
that the allowable compressive stress is 175 MPa, determine the largest
allowable distance a.
30 mm
B
A
20 mm
20 mm
C
30 mm
5 kN
SOLUTION
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the
section. The bending couple is
M  P(a  r )
For the rectangular section, the neutral axis for bending couple only lies at
R 
Also, e  r  R
h
.
r2
ln
r1
The maximum compressive stress occurs at point A . It is given by
A  
P My A
P P (a  r ) y A

 
A Aer1
A
Aer1
 K
P
with y A  R  r1
A
(a  r )( R  r1)
Thus, K  1 
er1
h  30 mm, r1  20 mm, r2  50 mm, r  35 mm, R 
Data:
(1)
30
 32.7407 mm
ln 50
20
e  35  32.7407  2.2593 mm, b  30 mm, R  r1  12.7407 mm, a  ?
 A  175 MPa  175  106 Pa, P  5 kN  5  103 N
K 
Solving (1) for
ar 
(900  106 )(175  106 )
A A

 31.5
P
5  103
a  r,
ar 
( K  1)er1
R  r1
(30.5)(2.2593)(20)
 108.17 mm
12.7407
a  108.17 mm  35 mm
a  73.2 mm 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
628
PROBLEM 4.170
2500 N
For the split ring shown, determine the stress at (a) point A,
(b) point B.
90 mm
40 mm
B
A
14 mm
SOLUTION
r1 
1
40  20 mm,
2
r2 
A  (14)(25)  350 mm 2
1
r  (r1  r2 )  32.5 mm
2
1
(90)  45 mm h  r2  r1  25 mm
2
25
h
R  r  45  30.8288 mm
2
ln r ln 20
e  r  R  1.6712 mm
1
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the cross
section. The bending couple is
M  Pa  P r  (2500)(32.5  103 )  81.25 N  m
(a)
Point A:
rA  20 mm
y A  30.8288  20  10.8288 mm
P My A
2500
(81.25)(10.8288  103 )



A AeR
350  106 (350  106 )(1.6712  103 )(20  103 )
 82.4  106 Pa
A  
(b)
Point B:
rB  45 mm
yB  30.8288  45  14.1712 mm
P MyB
2500
(81.25)(14.1712  103 )



A AerB
350  106 (350  106 )(1.6712  103 )(45  103 )
 36.6  106 Pa
B  
 A  82.4 MPa 
 B  36.6 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
629
2 in.
PROBLEM 4.171
B
0.5 in.
0.5 in.
Three plates are welded together to form the curved
beam shown. For M  8 kip  in., determine the stress
at (a) point A, (b) point B, (c) the centroid of the cross
section.
2 in.
0.5 in.
A
3 in.
M
M'
3 in.
C
SOLUTION
R
r 
r
3
3.5
5.5
6
Part




b
A
bi hi
A


r
  1r dA b ln ri 1
bi ln i 1
i
ri
ri
Ari
A
Ar
0.462452
3.25
4.875
1.0
0.225993
4.5
4.5
1.0
0.174023
5.75
5.75
3.5
0.862468
A
3
0.5
1.5
0.5
2
2
0.5
b ln
R
3.5
 4.05812 in.,
0.862468
e  r  R  0.26331 in.
(a)
y A  R  r1  4.05812  3  1.05812 in.
A  
My A
(8)(1.05812)

Aer1
(3.5)(0.26331)(3)
15.125
r 
15.125
 4.32143 in.
3.5
M  8 kip  in.
 A  3.06 ksi 
(b)
yB  R  r2  4.05812  6  1.94188 in. 

B  
(c)
yC  R  r  e
C  
ri 1
ri
r
h
MyB
(8)(1.94188)


Aer2
(3.5)(0.26331)(6)
 B  2.81 ksi  
8
MyC
Me
M



(3.5)(4.32143)
Aer
Aer
Ar
 C  0.529 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
630
2 in.
PROBLEM 4.172
B
0.5 in.
0.5 in.
2 in.
0.5 in.
A
3 in.
M
M'
3 in.
Three plates are welded together to form the curved beam
shown. For the given loading, determine the distance e
between the neutral axis and the centroid of the cross
section.
C
SOLUTION
R
r 
r
3
3.5
5.5
6
Part
A
bi hi
A


r
  1r dA b ln ri 1
bi ln i 1
i
ri
ri
Ari
A
b




ri 1
ri
r
Ar
0.462452
3.25
4.875
1.0
0.225993
4.5
4.5
1.0
0.174023
5.75
5.75
3.5
0.862468
h
A
3
0.5
1.5
0.5
2
2
0.5
b ln
R
15.125
3.5
15.125
 4.05812 in., r 
 4.32143 in.
0.862468
3.5
e  r  R  0.26331 in.
e  0.263 in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
631
C
M
PROBLEM 4.173
Mʹ
A
150 mm
A
Knowing that the maximum allowable stress is 45 MPa, determine
the magnitude of the largest moment M that can be applied to the
components shown.
45 mm
135 mm
B
B
36 mm
SOLUTION
R
r 
A
bi hi
Ai


1
r
r
  r dA b 2 ln i 1
bi ln i 1
i
ri
ri
Ai ri
Ai
r, mm
150
Part bi , mm h, mm


195
330
bi ln
A, mm 2
ri 1
, mm
ri
r , mm
Ar , mm3
838.35  103
108
45
4860
28.3353
172.5
36
135
4860
18.9394
262.5
9720
47.2747
∑
R
9720
 205.606 mm
47.2747
e  r  R  11.894 mm
r 
1275.75  103
2114.1  103
2114.1  103
 217.5 mm
9720
y A  R  r1  205.606  150  55.606 mm
A  
M 

My A
Aer1
 A Aer1
yA
(45  106 )(9720  106 )(11.894  103 )(150  103 )
 14.03 kN  m
(55.606  103 )
yB  R  r2  205.606  330  124.394 mm
B  
M 

MyB
Aer2
 B Aer2
yB
(45  106 )(9720  106 )(11.894  103 )(330  103 )
(124.394  103 )
 13.80 kN  m
M  13.80 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
632
C
M
PROBLEM 4.174
Mʹ
A
150 mm
A
Knowing that the maximum allowable stress is 45 MPa, determine
the magnitude of the largest moment M that can be applied to the
components shown.
135 mm
45 mm
B
B
36 mm
SOLUTION
R
r 
A
bi hi
Ai


r
  1r dA b ln ri 1
bi ln i 1
i
ri
ri
Ai ri
Ai
r, mm
150

285

330
∑
R
bi , mm h, mm
A, mm 2
bi ln
ri 1
, mm
ri2
r , mm
36
135
4860
23.1067
217.5
108
45
4860
15.8332
307.5
9720
38.9399
9720
 249.615 mm,
38.9399
e  r  R  12.885 mm,
r 
2.5515  106
 262.5 mm
9720
Ar , mm3
1.05705  106
1.49445  106
2.5515  106
M  20  103 N  m
y A  R  r1  249.615  150  99.615 mm
A  
M 

My A
Aer1
 A Aer1
yA
(45  106 )(9720  106 )(12.885  103 )(150  103 )
 8.49 kN  m
(99.615  103 )
yB  R  r2  249.615  330  80.385 mm
B  
M 

MyB
Aer2
 B Aer2
yB
(45  106 )(9720  106 )(12.885  103 )(330  103 )
(80.385  103 )
 23.1 kN  m
M  8.49 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
633
120 lb
120 lb
The split ring shown has an inner radius r1  0.8 in. and a circular cross
section of diameter d  0.6 in. Knowing that each of the 120-lb forces is
applied at the centroid of the cross section, determine the stress (a) at point A,
(b) at point B.
r1
A
PROBLEM 4.175
d
B
SOLUTION
rA  r1  0.8 in.
rB  rA  d  0.8  0.6  1.4 in.
r 
1
(rA  rB )  1.1 in.
2
c
A   c 2   (0.3)2  0.28274 in 2
1
d  0.3 in.
2
for solid circular section
1
1
r  r 2  c 2   1.1  (1.1)2  (0.3)2   1.079150 in.
 2 

2 
e  r  R  1.1  1.0791503  0.020850 in.
R
Q  120 lb
  Fx  0: P  Q  0
 M 0  0: 2rQ  M  0
(a)
(b)
r  rA  0.8 in.
A 
P  120 lb
M  2rQ  (2)(1.1)(120)  264 lb  in.
P M (rA  R)
120
(264)(0.8  1.079150)



 16.05  103 psi
A
AerA
0.28274 (0.28274)(0.020850)(0.8)
r  rB  1.4 in.
B 
 A  16.05 ksi 
120
(264)(1.4  1.079150)
P M (rB  R)



0.28274 (0.28274)(0.020850)(1.4)
A
AerB
 9.84  103 psi
 B  9.84 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
634
120 lb
120 lb
PROBLEM 4.176
Solve Prob. 4.175, assuming that the ring has an inner radius r1 = 0.6 in. and
a cross-sectional diameter d  0.8 in.
r1
A
PROBLEM 4.175 The split ring shown has an inner radius r1  0.8 in. and a
circular cross section of diameter d  0.6 in. Knowing that each of the 120-lb
forces is applied at the centroid of the cross section, determine the stress
(a) at point A, (b) at point B.
d
B
SOLUTION
rA  r1  0.6 in.
rB  rA  d  0.6  0.8  1.4 in.
r 
1
(rA  rB )  1.0 in.
2
c
A   c 2   (0.4)2  0.50265 in 2
1
d  0.4 in.
2
for solid circular section.
1
1
r  r 2  c 2   1.0  (1.0) 2  (0.4)2   0.958258 in.



 2 
2
e  r  R  1.0  0.958258  0.041742 in.
R
Q  120 lb
 M 0  0:
(a)
  Fx  0
2rQ  M  0
r  rA  0.6 in.
A 
PQ  0
P  120 lb
M  2rQ  (2)(1.0)(120)  240 lb  in.
120
(240)(0.6  0.958258)
P M (rA  R)



0.50265 (0.50265)(0.041742)(0.6)
A
AerA
 7.069  103 psi
(b)
r  rB  1.4 in.
B 
120
(240)(1.4  0.958258)
P M (rB  R)



0.50265 (0.50265)(0.041742)(1.4)
A
AerB
 3.37  103 psi
 A  7.07 ksi 
 B  3.37 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
635
220 N
B
PROBLEM 4.177
The bar shown has a circular cross section of 14-mm diameter. Knowing
that a  32 mm, determine the stress at (a) point A, (b) point B.
A C
220 N
a
16 mm
12 mm
SOLUTION
c
1
d  8 mm
r  12  8  20 mm
2
1
1
R   r  r 2  c 2    20  202  82   19.1652 mm



 2 
2
e  r  R  20  19.1652  0.83485 mm
A   r 2   (8) 2  201.06 mm 2
P  220 N
M   P(a  r )  220(0.032  0.020)  11.44 N  m
y A  R  r1  19.1652  12  7.1652 mm
yb  R  r2  19.1652  28  8.8348 mm
(a)
A 

(b)
B 

P My A

A Aer1
220
(11.44)(7.1652  103 )

201.06  106 (201.06  106 )(0.83485  103 )(12  103 )
 A  41.8 MPa 
P MyB

A Aer2
220
(11.44)(8.8348  103 )

201.06  106 (201.06  106 )(0.83485  103 )(28  103 )
 B  20.4 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
636
220 N
B
PROBLEM 4.178
The bar shown has a circular cross section of 14-mm diameter. Knowing
that the allowable stress is 38 MPa, determine the largest permissible
distance a from the line of action of the 220-N forces to the plane
containing the center of curvature of the bar.
A C
220 N
a
16 mm
12 mm
SOLUTION
c
1
d  8 mm
r  12  8  20 mm
2
1
1
R   r  r 2  c 2    20  202  82   19.1652 mm



 2 
2
e  r  R  20  19.1652  0.83485 mm
A   r 2   (8) 2  201.06 mm 2
P  220 N
M   P (a  r )
y A  R  r1  19.1652  12  7.1652 mm
(a  r ) y A 
P My A
P P(a  r ) y A
P



 1 

A Aer1
A
Aer1
A
er1

(a  r ) y A
KP
where

K 1
A
er1
A 
 AA
(38  106 )(201.06  106 )
 34.729
P
220
( K  1)er1
ar 
yA
K 


(34.729  1)(0.83485  103 )(12  103 )
(7.1652  103 )
 0.047158 m
a  0.047158  0.020  0.027158
a  27.2 mm 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
637
C
M
PROBLEM 4.179
16 mm
The curved bar shown has a circular cross section of 32-mm
diameter. Determine the largest couple M that can be applied to
the bar about a horizontal axis if the maximum stress is not to
exceed 60 MPa.
12 mm
SOLUTION
c  16 mm
r  12  16  28 mm
1
r  r 2  c2 



2
1
  28  282  162   25.4891 mm

2 
R
e  r  R  28  25.4891  2.5109 mm
 max occurs at A, which lies at the inner radius.
It is given by  max 
My A
from which
Aer1
M 
Aer1  max
yA
A   c 2   (16)2  804.25 mm 2
Also,
.
y A  R  r1  25.4891  12  13.4891 mm
Data:
M 
(804.25  106 )(2.5109  103 )(12  103 )(60  106 )
13.4891  103
M  107.8 N  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
638
P
PROBLEM 4.180
90 mm
Knowing that P  10 kN, determine the stress at (a) point A, (b) point B.
80 mm
B
A
100 mm
SOLUTION
Locate the centroid D of the cross section.
r  100 mm 
90 mm
 130 mm
3
Force-couple system at D.
P  10 kN
M  Pr  (10 kN)(130 mm)  1300 N  m
Triangular cross section.
A
1
1
bh  (90 mm)(80 mm)
2
2
 3600 mm 2  3600  106 m 2
1
1
h
(90)
45 mm
2
2
R


190 190
r2 r2
ln  1
ln
 1 0.355025
90 100
h r1
R  126.752 mm
e  r  R  130 mm  126.752 mm  3.248 mm
(a)
Point A:
A  
rA  100 mm  0.100 m
P M (rA  R)
10 kN
(1300 N  m)(0.100 m  0.126752 m)



6 2
A
AerA
3600  10 m
(3600  106 m 2 )(3248  103 m)(0.100 m)
 A  32.5 MPa 
 2.778 MPa  29.743 MPa
(b)
Point B:
B  
rB  190 mm  0.190 m
P M (rB  R)
10 kN
(1300 N  m)(0.190 m  0.126752 m)



A
AerB
3600  106 m 2 (3600  106 m 2 )(3.248  103 m)(0.190 m)
 B  34.2 MPa 
 2.778 MPa  37.01 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
639
M
2.5 in.
A
PROBLEM 4.181
B
M
Knowing that M  5 kip  in., determine the stress at (a) point A,
(b) point B.
C
3 in.
2 in.
2 in.
3 in.
SOLUTION
A
1
1
bh  (2.5)(3)  3.75 in 2
2
2
r  2  1  3.00000 in.
b1  2.5 in., r1  2 in., b2  0, r2  5 in.
Use formula for trapezoid with
b2  0.
1 h 2 (b  b )
1
2
2
R
r2
(b1r2  b2r1) ln  h(b1  b2 )
r1

(a)
(b)
y A  R  r1  0.84548 in.
A  
(0.5)(3)2 (2.5  0)
 2.84548 in.
[(2.5)(5)  (0)(2)] ln 52  (3)(2.5  0)
e  r  R  0.15452 in.
My A
(5)(0.84548)

Aer1
(3.75)(0.15452)(2)
 A  3.65 ksi 
MyB
(5)(2.15452)

Aer2
(3.75)(0.15452)(5)
 B  3.72 ksi 
yB  R  r2  2.15452 in.
B  
M  5 kip  in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
640
M
PROBLEM 4.182
B
Knowing that M  5 kip  in., determine the stress at (a) point A,
(b) point B.
2.5 in.
A
3 in.
M
C
2 in.
2 in.
3 in.
SOLUTION
A  1 (2.5)(3)  3.75 in 2
2
r  2  2  4.00000 in.
b1  0, r1  2 in.,
Use formula for trapezoid with
b1  0.
b2  2.5 in.,
r2  5 in.
1 h 2 (b  b )
1
2
2
R
r2
(b1r2  b2r1) ln  h (b1  b2 )
r1
(0.5)(3) 2 (0  2.5)
 3.85466 in.
[(0)(5)  (2.5)(2)] ln 5  (3)(0  2.5)
2
e  r  R  0.14534 in.
M  5 kip  in.

(a)
(b)
y A  R  r1  1.85466 in.
A  
My A
(5)(1.85466)

Aer1
(3.75)(0.14534)(2)
 A  8.51 ksi 
MyB
(5)(1.14534)

Aer2
(3.75)(0.14534)(5)
 B  2.10 ksi 
yB  R  r2  1.14534 in.
B  
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
641
80 kip · in.
b
B
A
B
A
PROBLEM 4.183
Knowing that the machine component shown has a trapezoidal cross
section with a  3.5 in. and b  2.5 in., determine the stress at
(a) point A, (b) point B.
C
a
6 in. 4 in.
SOLUTION
Locate centroid.
A, in 2
r , in.
Ar , in 3
10.5
6
63
7.5
8
60



18
r 
R

123
123
 6.8333 in.
18
1 2
h (b1  b2 )
2
(b1r2  b2 r1 ) ln
r2
r1
 h(b1  b2 )
(0.5)(6)2 (3.5  2.5)
 6.3878 in.
[(3.5)(10)  (2.5)(4)]ln 104  (6)(3.5  2.5)
e  r  R  0.4452 in.
(a)
(b)
y A  R  r1  6.3878  4  2.3878 in.
My
(80)(2.3878)
A   A  
(18)(0.4452)(4)
Aer1
M  80 kip  in.
 A  5.96 ksi 
yB  R  r2  6.3878  10  3.6122 in.
My
(80)(3.6122)
B   B  
(18)(0.4452)(10)
Aer2
 B  3.61 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
642
80 kip · in.
b
B
A
B
A
PROBLEM 4.184
Knowing that the machine component shown has a trapezoidal cross
section with a  2.5 in. and b  3.5 in., determine the stress at
(a) point A, (b) point B.
C
a
6 in. 4 in.
SOLUTION
Locate centroid.



A, in 2
r , in.
Ar , in 3
7.5
6
45
10.5
8
84
18
r 
R

129
129
 7.1667 in.
18
1 2
h (b1  b2 )
2
(b1r2  b2 r1 ) ln
r2
r1
 h(b1  b2 )
(0.5)(6) 2 (2.5  3.5)
 6.7168 in.
[(2.5)(10)  (3.5)(4)]ln 104  (6)(2.5  3.5)
e  r  R  0.4499 in.
(a)
(b)
M  80 kip  in.
y A  R  r1  2.7168 in.
A  
 A  6.71 ksi 
My A
(80)(2.7168)

(18)(0.4499)(4)
Aer1
yB  R  r2  3.2832 in.
B  
MyB
(80)(3.2832)

(18)(0.4499)(10)
Aer2
 B  3.24 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
643
B
a
20 mm
PROBLEM 4.185
B
A
For the curved beam and loading shown, determine the stress at
(a) point A, (b) point B.
30 mm
a
A
250 N · m
250 N · m
35 mm
40 mm
Section a–a
SOLUTION
Locate centroid.



A, mm 2
r , mm
Ar , mm3
600
45
300
55
16.5  103
900
r 
R

43.5  103
 48.333 mm
900
1
2
(b1r2  b2r1) ln
r2
r1
 h(b2  b1)
(0.5)(30) 2 (40  20)
 46.8608 mm
65  (30)(40  20)
[(40)(65)  (20)(35)]ln 35
(b)
y A  R  r1  11.8608 mm
A  
M  250 N  m
(250)(11.8608  103 )
My A

 63.9  106 Pa
Aer1
(900  106 )(1.4725  103 )(35  103 )
yB  R  r2  18.1392 mm
B  
43.5  103
h 2 (b1  b2 )
e  r  R  1.4725 mm
(a)
27  103
MyB
(250)(18.1392  103 )

 52.6  106 Pa
Aer2
(900  106 )(1.4725  103 )(65  103 )
 A  63.9 MPa 
 B  52.6 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
644
PROBLEM 4.186
For the crane hook shown, determine the largest tensile stress in
section a-a.
35 mm
25 mm
60 mm
40 mm
a
a
60 mm
Section a–a
15 kN
SOLUTION
Locate centroid.



A, mm 2
r , mm
Ar , mm3
1050
60
750
80
60  103
63  103
103  103
1800
r
Force-couple system at centroid:
103  103
 63.333 mm
1800
P  15  103 N
M   Pr  (15  103 )(68.333  103 )  1.025  103 N  m
R

1
2
h 2 (b1  b2 )
(b1r2  b2 r1 ) ln
r2
r1
 h(b1  b2 )
(0.5)(60)2 (35  25)
 63.878 mm
 (60)(35  25)
[(35)(100)  (25)(40)]ln 100
40
e  r  R  4.452 mm
Maximum tensile stress occurs at point A.
y A  R  r1  23.878 mm
A 

P My A

A Aer1
(1.025  103 )(23.878  103 )
15  103

1800  106 (1800  106 )(4.452  103 )(40  103 )
 84.7  106 Pa
 A  84.7 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
645
PROBLEM 4.187
Using Eq. (4.66), derive the expression for R given in Fig. 4.61 for a circular cross section.
SOLUTION
Use polar coordinate  as shown. Let w be the width as a function of 
w  2c sin 
r  r  c cos 
dr  c sin  d 
dA  w dr  2c 2 sin 2  d 


dA

r


0

dA

r
2
2


0

2c 2 sin 
d
r  c cos 
c 2 (1  cos 2  )
d
r  c cos 
r 2  c 2 cos 2   (r 2  c 2 )
d
r  c cos 

0

0
(r  c cos  ) d   2(r 2  c 2 )
 2r 

0
 2c sin 
 2(r  c )
2
2

0
2
r c
2
2
tan
1


0
dr
r  c cos 
 
r 2  c 2 tan 1 
2
r c


 2r (  0)  2c(0  0)  4 r 2  c 2    0 
2



1

2
0
 2 r  2 r 2  c 2
A   c2
R

A
c2
r  r 2  c2
1
 c2



dA 2 r  2 r 2  c 2 2 r  r 2  c 2 r  r 2  c 2
r

c2 r  r 2  c2
r  (r  c )
2
2
2
  1 c r 
r 2  c2
2
2
c
2
1 r
2

r 2  c2 

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
646
PROBLEM 4.188
Using Eq. (4.66), derive the expression for R given in Fig. 4.73 for a trapezoidal cross section.
SOLUTION
The section width w varies linearly with r.
w  c0  c1r
w  b1 at r  r1 and w  b2 at r  r2
b1  c0  c1r1
b2  c0  c1r2
b1  b2  c1 (r1  r2 )  c1h
b b
c1   1 2
h
r2b1  r1b2  (r2  r1 )c0  hc0
r b  rb
c0  2 1 1 2
h
r2 w
r2 c  c r
dA
0
1
dr 
dr

r1 r
r1
r
r
r2
r2
 c0 ln r  c1 r



r1
r1
r
 c0 ln 2  c1 (r2  r1 )
r1
r2 b1  r1b2 r2 b1  b2
h
ln 

h
r1
h
r b  rb
r
 2 1 1 2 ln 2  (b1  b2 )
h
r1
1
A  (b1  b2 ) h
2
1 2
h (b1  b2 )
A
2

R
r
dA
( r2b1  r1b2 ) ln r2  h(b1  b2 )
r


1
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
647
PROBLEM 4.189
Using Equation (4.66), derive the expression for R given in Fig. 4.73 for a triangular cross section.
SOLUTION
The section width w varies linearly with r.

w  c0  c1r
w  b at r  r1 and w  0 at r  r2
b  c0  c1r1
0  c0  c1r2
b  c1 (r1  r2 )  c1h
br
b
c1  
and c0  c1r2  2
h
h
r2 w
r2 c  c r
dA
0
1

dr 
dr
r1 r
r1
r
r
r2
r2
 c0 ln r  c1 r


r1
r1
r
 c0 ln 2  c1 (r2  r1 )
r1
br2 r2 b
ln  h

h
r1 h

r
br
r
r
 2 ln 2  b  b  2 ln 2  1
h
r1
 h r1

1
A  bh
2
1
1
bh
h
A
R
 r 2 r
 r 2r
2
dA
ln r2  1
b h2 ln r2  1
h
r
1


1


PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
648
PROBLEM 4.190
b2
b3
Show that if the cross section of a curved beam consists of two or more
rectangles, the radius R of the neutral surface can be expressed as
b1
R
r1
r2
r3
r4
A
 r b1  r b2  r b3 
ln  2   3   4  
 r1   r2   r3  


where A is the total area of the cross section.
SOLUTION
R

Note that for each rectangle,

A
A

1
 dA  b ln ri 1
i
r
r

A
r 
 ln  i 1 
 ri 

bi

i
A
 r b1  r b2  r b3 
ln  2   3   4  
 r1   r2   r3  



ri 1 dr
1
dA 
bi
ri
r
r
ri  1 dr
ri 1
 bi
 bi ln
r2 r
ri

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
649
PROBLEM 4.191
C
␪
2
␪
2
r1
␴x
␴x
b
␴r
␴r
R
For a curved bar of rectangular cross section subjected to a
bending couple M, show that the radial stress at the neutral
surface is
M  r1
R
r 
1   ln 
Ae 
R
r1 
and compute the value of  r for the curved bar of Concept
Applications 4.10 and 4.11. (Hint: consider the free-body
diagram of the portion of the beam located above the neutral
surface.)
SOLUTION
M (r  R) M MR


Aer
Ae Aer
For portion above the neutral axis, the resultant force is
r 
At radial distance r,

H   r dA 



R
r1
 r b dr

Mb R
MRb R dr
dr 
Ae r1
Ae r1 r
r1
Mb
MRb R MbR 
R

( R  r1 ) 
ln 
1   ln 
Ae
Ae
r1
Ae 
R
r1 
Resultant of  n :


 /2
 /2

Fr   r cos  dA
 r cos  (bR d  )   r bR
  r bR sin 
For equilibrium: Fr  2 H sin
2 r bR sin

2

2
 /2
 /2
0
2

 /2
 /2
 2 r bR sin 
2
cos  d 
r1
MbR 
R

1   ln  sin  0
Ae 
R
r1 
2
r 
r1
M 
R
1   ln 
Ae 
R
r1 

Using results of Examples 4.10 and 4.11 as data,
M  8 kip  in.,
r 
A  3.75 in 2 , R  5.9686 in., e  0.0314 in., r1  5.25 in.
8
5.25
5.9686 

 ln
1

(3.75)(0.0314) 
5.9686
5.25 
 r  0.536 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
650
PROBLEM 4.192
25 mm
25 mm
4 kN
A
Two vertical forces are applied to a beam of the cross section shown. Determine
the maximum tensile and compressive stresses in portion BC of the beam.
4 kN
B
300 mm
C
300 mm
SOLUTION
A1 

2
r2 

2
(25)2  981.7 mm 2
A2  bh  (50)(25)  1250 mm 2
y 
y1 
4r
(4)(25)

 10.610 mm
3
3
25
h
 12.5 mm
y2    
2
2
(981.7)(10.610)  (1250)(12.5)
A1 y1  A2 y2

 2.334 mm
981.7  1250
A1  A2
I1  I x1  A1 y12 

r 4  A1 y12 

(25)4  (981.7)(10.610) 2  42.886  106 mm 4
8
8
d1  y1  y  10.610  (2.334)  12.944 mm
I1  I1  A1d12  42.866  103  (981.7)(12.944) 2  207.35  103 mm 4
I2 
1 3 1
bh  (50)(25)3  65.104  103 mm 4
12
12
d 2  y2  y  12.5  (2.334)  10.166 mm
I 2  I 2  A2 d 22  65.104  103  (1250)(10.166) 2  194.288  103 mm 4
I  I1  I 2  401.16  103 mm 4  401.16  109 m 4
ytop  25  2.334  27.334 mm  0.027334 m
ybot  25  2.334  22.666 mm  0.022666 m
 top 
 bot 
 Mytop
I
M  Pa  0:

M  Pa  (4  103 )(300  103 )  1200 N  m
(1200)(0.027334)
 81.76  106 Pa
401.16  109
Mybot
(1200)(0.022666)

 67.80  106 Pa
I
401.16  109
 top  81.8 MPa 
 bot  67.8 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
651
PROBLEM 4.193
A steel band saw blade that was originally straight passes over 8-in.-diameter pulleys
when mounted on a band saw. Determine the maximum stress in the blade, knowing that
it is 0.018 in. thick and 0.625 in. wide. Use E  29  106 psi.
0.018 in.
SOLUTION
t  0.018 in.
Band blade thickness:
r
Radius of pulley:
1
d  4.000 in.
2
Radius of curvature of centerline of blade:
  r  t  4.009 in.
c
1
2
1
t  0.009 in.
2
Maximum strain:
m 
Maximum stress:
 m  E m  (29  106 )(0.002245)

c

0.009
 0.002245
4.009
 m  65.1  103 psi
 m  65.1 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
652
PROBLEM 4.194
M
A couple of magnitude M is applied to a square bar of side a. For
each of the orientations shown, determine the maximum stress
and the curvature of the bar.
M
a
(a)
(b)
SOLUTION
I
1 3 1 3 a4
bh  aa 
12
12
12
a
c
2
 max

1

a
Mc M 2

 4
I
a
12

M
M

EI E a 4
12

 max 

1

6M

a3
12M

Ea 4
For one triangle, the moment of inertia about its base is
I1 
1 3 1
bh 
12
12
I 2  I1 

a
2
3
a4
24
I  I1  I 2 
c
 a 
a4

2a 

24
 2
a4
12
 max 

1

Mc Ma/ 2 6 2 M
 4

I
a /12
a3
M
M


EI E a 4
12
 max 

1
8.49M

a3

12M

Ea 4
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
653
40 mm
60 mm
PROBLEM 4.195
Determine the plastic moment M p of a steel beam of the cross section shown, assuming
the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
Let c1 be the outer radius and c2 the inner radius.
A1 y1  Aa ya  Ab yb
   4c     4c 
  c12  1    c22  2 
 2  3   2  3 
3
2 3

c1  c2
3

A2 y2  A1 y1 


2 3
c1  c 22
3

M p   Y ( A1 y1  A2 y2 ) 
Data:

4
 Y c13  c32
3
 Y  240 MPa  240  106 Pa

c1  60 mm  0.060 m
c2  40 mm  0.040 m
Mp 
4
(240  106 )(0.0603  0.0403 )
3
 48.64  103 N  m
M p  48.6 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
654
PROBLEM 4.196
M 300 N · m
26 mm
30 mm
46 mm
50 mm
In order to increase corrosion resistance, a 2-mm-thick cladding of
aluminum has been added to a steel bar as shown. The modulus
of elasticity is 200 GPa for steel and 70 GPa for aluminum. For a
bending moment of 300 N  m, determine (a) the maximum stress
in the steel, (b) the maximum stress in the aluminum, (c) the radius
of curvature of the bar.
SOLUTION
Use aluminum as the reference material.
n  1 in aluminum
n
Es
200

 2.857 in steel
70
Ea
Cross section geometry:
Steel: As  (46 mm)(26 mm)  1196 mm 2
Is 
1
(46 mm)(26 mm)3  67,375 mm 4
12
Aluminum: Aa  (50 mm)(30 mm)  1196 mm 2  304 mm 2
Ia 
1
(50 mm)(30 mm3 )  67,375 mm 4  45,125 mm 4
12
Transformed section.
I  na I a  ns I s  (1)(45,125)  (2.857)(67,375)  237, 615 mm 4  237.615  109 m 4
Bending moment.
(a)
Maximum stress in steel:
s 
(b)
ns  2.857
ys  13 mm  0.013 m
ns Mys
(2.857)(300)(0.013)

 46.9  106 Pa
I
237.615  109
Maximum stress in aluminum:
a 
(c)
M  300 N  m
na  1,
ya  15 mm  0.015 m
na Mya
(1)(300)(0.015)

 18.94  106 Pa
I
237.615  109
Radius of curvative:  
EI
M
 
 s  46.9 MPa 
(70  109 )(237.615  109 )
300
 a  18.94 MPa 
  55.4 m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
655
PROBLEM 4.197
t
P
P'
a
a
t
60 mm
A
80 mm
200 mm
80 mm
The vertical portion of the press shown consists of a rectangular
tube of wall thickness t  10 mm. Knowing that the press has
been tightened on wooden planks being glued together until P 
20 kN, determine the stress at (a) point A, (b) point B.
B
Section a–a
SOLUTION
Rectangular cutout is 60 mm  40 mm.
A  (80)(60)  (60)(40)  2.4  103 mm 2  2.4  103 m 2
I 
1
1
(60)(80)3  (40)(60)3  1.84  106 mm 4
12
12
 1.84  106 m 4
c  40 mm  0.040 m e  200  40  240 mm  0.240 m
P  20  103 N
M  Pe  (20  103 )(0.240)  4.8  103 N  m
(a)
A 
(b)
B 
P Mc
20  103
(4.8  103 )(0.040)



 112.7  106 Pa
A
I
2.4  103
1.84  106
P Mc
20  103
(4.8  103 )(0.040)



 96.0  106 Pa
3
6
A
I
2.4  10
1.84  10
 A  112.7 MPa 
 B  96.0 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
656
P y
PROBLEM 4.198
P
P
P
B
C
a
A
D
x
z
The four forces shown are applied to a rigid plate supported by a solid steel
post of radius a. Knowing that P  24 kips and a  1.6 in., determine the
maximum stress in the post when (a) the force at D is removed, (b) the forces
at C and D are removed.
SOLUTION
For a solid circular section of radius a,
A   a2
Centric force:
(a)
Force at D is removed.
 
M x   Pa,
Mz  0
F
4P
 2
A
a
M x   Pa,
M z   Pa
F M xz
3P
( Pa)(a)
7P

 2 
 2
2

A
I
a
a
a
4
M 
F Mc
2P

 2 
A
I
a
M x2  M z2 
2 Pa
2 Paa
24 2 P

 2.437 P/a 2
2
2

a
a
4

P  24.0 kips,
Numerical data:
Answers:
a4
F  3P,
F  2P,
 
4
Mx  Mz  0
Forces at C and D are removed.
Resultant bending couple:

F  4P,
 
(b)
I 
(a)
 
(b)
 
a  1.6 in.
(7)(24.0)
 (1.6)2
(2.437)(24.0)
(1.6)2
  20.9 ksi 
  22.8 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
657
a
r 20 mm
PROBLEM 4.199
P 3 kN
The curved portion of the bar shown has an inner radius of 20 mm.
Knowing that the allowable stress in the bar is 150 MPa, determine the
largest permissible distance a from the line of action of the 3-kN force to
the vertical plane containing the center of curvature of the bar.
25 mm
25 mm
SOLUTION
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the
section. The bending couple is
M  P(a  r )
For the rectangular section, the neutral axis for bending couple only lies at
R
Also, e  r  R
h
r2
r1
ln
The maximum compressive stress occurs at point A. It is given by
A  
P My A
P P (a  r ) y A
P

 
 K
A Aer1
A
Aer1
A
with
y A  R  r1
Thus,
K 1
Data:
(a  r )( R  r1)
er1
h  25 mm, r1  20 mm, r2  45 mm, r  32.5 mm
R
25
 30.8288 mm, e  32.5  30.8288  1.6712 mm
45
ln 20
b  25 mm,
A  bh  (25)(25)  625 mm 2  625  106 m 2
P  3  10 N  m,  A  150  10 Pa
3
 AA
(150  106 )(625  106 )
 31.25
P
3  103
( K  1)er1 (30.25)(1.6712)(20)
ar 

 93.37 mm
R  r1
10.8288
K 
R  r1  10.8288 mm
6

a  93.37  32.5
a  60.9 mm 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
658
Dimensions in inches
400 lb
PROBLEM 4.200
400 lb
400 lb
2.5
Determine the maximum stress in each of the two machine
elements shown.
400 lb
2.5
3
0.5
3
r 5 0.3
r 5 0.3
1.5
0.5
0.5
1.5
0.5
SOLUTION
For each case, M  (400)(2.5)  1000 lb  in.
At the minimum section,
I 
1
(0.5)(1.5)3  0.140625 in 4
12
c  0.75 in.
(a)
D/d  3/1.5  2
r/d  0.3/1.5  0.2
From Fig 4.32,
 max 
(b)
KMc (1.75)(1000)(0.75)

 9.33  103 psi
I
0.140625
D/d  3/1.5  2
From Fig. 4.31,
 max 
K  1.75
 max  9.33 ksi 
r/d  0.3/1.5  0.2
K  1.50
KMc (1.50)(1000)(0.75)

 8.00  103 psi
I
0.140625
 max  8.00 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
659
120 mm
PROBLEM 4.201
10 mm
M
120 mm
10 mm
Three 120  10-mm steel plates have been welded together to form the
beam shown. Assuming that the steel is elastoplastic with
E  200 GPa and  Y  300 MPa, determine (a) the bending moment
for which the plastic zones at the top and bottom of the beam are
40 mm thick, (b) the corresponding radius of curvature of the beam.
10 mm
SOLUTION
A1  (120)(10)  1200 mm 2
R1   Y A1  (300  106 )(1200  10 6 )  360  103 N
A2  (30)(10)  300 mm 2
R2   Y A2  (300  106 )(300  106 )  90  103 N
A3  (30)(10)  300 mm 2
1
1
R3   Y A2  (300  106 )(300  106 )  45  103 N
2
2
y1  65 mm  65  103 m y2  45 mm  45  103 m
(a)
M  2( R1 y1  R2 y2  R3 y3 )  2{(360)(65)  (90)(45)  (45)(20)}
(b)
yY
 56.7  103 N  m


Y
E

Y
EyY

(200  109 )(30  103 )
300  106
y3  20 mm  20  103 m
M  56.7 kN  m 
  20.0 mm 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
660
P
4.5 in.
4.5 in.
A
PROBLEM 4.202
Q
A short length of a W8  31 rolled-steel shape supports a rigid plate on which two
loads P and Q are applied as shown. The strains at two points A and B on the
centerline of the outer faces of the flanges have been measured and found to be
 A  550  106 in./in.
B
 B  680  106 in./in.
Knowing that E  29  106 psi, determine the magnitude of each load.
SOLUTION
Strains:
 A  550  106 in./in.
C 
Stresses:
 B  680  106 in./in.
1
1
( A   B )  (550  680)106  615  106 in./in.
2
2
 A  E A  (29  106 psi)(550  106 in./in.)  15.95 ksi
 C  E C  (29  106 psi)(615  106 in./in.)  17.835 ksi
W8  31:
A  9.12 in 2
M  (4.5 in.)( P  Q)
At point C:
At point A:
C  
A  
PQ
PQ
;  17.835 ksi  
A
9.12 in 2
PQ M

A
S
15.95 ksi  17.835 ksi 
Solve simultaneously.
S  27.5 in 3
(4.5 in.)( P  Q)
;
27.5 in 3
P  75.6 kips Q  87.1 kips
P  Q  162.655 kips (1)
P  Q  11.5194 kips (2)
P  75.6 kips  
Q  87.1 kips  
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
661
PROBLEM 4.203
␴1
M1
A
M'1
M1
M'1
␴1
B
C
D
␴1
␴1
Two thin strips of the same material and same cross
section are bent by couples of the same magnitude
and glued together. After the two surfaces of
contact have been securely bonded, the couples are
removed. Denoting by 1 the maximum stress and
by 1 the radius of curvature of each strip while
the couples were applied, determine (a) the final
stresses at points A, B, C, and D, (b) the final radius
of curvature.
SOLUTION
Let b  width and t  thickness of one strip.
Loading one strip, M  M1
I1 
1 3
1
bt , c  t
12
2
M1c  M 1

I
bt 2
M
1
12 M 1
 1 
1 EI1
Et 3
1 
After M1 is applied to each of the strips, the stresses are those given in the sketch above. They are
 A  1,  B  1,  C  1,  D  1
The total bending couple is 2M1.
After the strips are glued together, this couple is removed.
M   2 M 1, I  
1
2
b(2t )3  bt 3
12
3
ct
The stresses removed are
  
2M y
3M1 y
M y
  2 13  
I
bt
bt 2
3
 A  
3M1
1
3M1 1
  1,  B   C  0,  D 
 1
2
2
2
bt
bt 2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
662
PROBLEM 4.203 (Continued)
(a)
Final stresses:
 A   1  ( 1)
 A   1 
1
2
1
2
 B  1 
 C  1 
 D  1  1
1
2M
3M 1
1 1
M

 2 13 

3
4 
  EI  E 3 bt
Et
(b)
Final radius:

1

1
1

D 
1
2
1
1
1 1
3 1



  1 4 1 4 1

 
1
1 
2
4
1 
3
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
663
PROBLEM 4.C1
Two aluminum strips and a steel strip are to be bonded together to form
a composite member of width b  60 mm and depth h  40 mm. The
modulus of elasticity is 200 GPa for the steel and 75 GPa for the
aluminum. Knowing that M  1500 N  m, write a computer program
to calculate the maximum stress in the aluminum and in the steel for
values of a from 0 to 20 mm using 2-mm increments. Using appropriate
smaller increments, determine (a) the largest stress that can occur in the
steel, (b) the corresponding value of a.
Aluminum
a
Steel
h ⫽ 40 mm
a
b ⫽ 60 mm
SOLUTION
Transformed section:
n
(all steel)
I 
Esteel
Ealum

1 3 1  1
bh   2   (nb  b)  (h  2a )3
12
12   2 

M  h2 
At Point 1:
 alum 
At Point 2:
 steel  n
I
M  h2  a 
I
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
664
PROBLEM 4.C1 (Continued)
For a  0 to 20 mm using 2-mm intervals: compute: n, I ,  alum ,  steel .
b  60 mm h  40 mm M  1500 N  m
Steel  200 GPa Aluminum  75 GPa
Moduli of elasticity:
Program Output
a
mm
I
m 4 /106
Sigma
Aluminum
MPa
Sigma
Steel
MPa
0.000
0.8533
35.156
93.750
2.000
0.7088
42.325
101.580
4.000
0.5931
50.585
107.914
6.000
0.5029
59.650
111.347
8.000
0.4352
68.934
110.294
10.000
0.3867
77.586
103.448
12.000
0.3541
84.714
90.361
14.000
0.3344
89.713
71.770
16.000
0.3243
92.516
49.342
18.000
0.3205
93.594
24.958
20.000
0.3200
93.750
0.000
Find ‘a’ for max. steel stress and the corresponding aluminum stress.
6.600
0.4804
62.447
111.572083
6.610
0.4800
62.494
111.572159
6.620
0.4797
62.540
111.572113
Max. steel stress  111.6 MPa occurs when a  6.61 mm.
Corresponding aluminum stress  62.5 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
665
PROBLEM 4.C2
tf
y
x
tw
d
bf
A beam of the cross section shown, made of a steel that is assumed to be
elastoplastic with a yield strength  Y and a modulus of elasticity E, is bent
about the x axis. (a) Denoting by yY the half thickness of the elastic core, write
a computer program to calculate the bending moment M and the radius of
curvature  for values of yY from 12 d to 16 d using decrements equal
to 12 t f . Neglect the effect of fillets. (b) Use this program to solve Prob. 4.201.
SOLUTION
Compute moment of inertia I x .
Ix 
Maximum elastic moment:
1
1
b f d 3  (b f  tw )(d  2t f )3
12
12
MY  Y
Ix
(d/2)
For yielding in the flanges,
(Consider upper half of cross section.)
Stress at junction of web and flange:
A 
c
d
2
(d/2)  t f
yY
Y
Resultant forces:
Detail of stress diagram:
a1 
1
(c  yY )
2
1
a2  yY   yY  (c  t f ) 
3
2
a3  yY  [ yY  (c  t f ) ]
3
2
a4  (c  t f )
3
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
666
PROBLEM 4.C2 (Continued)
Bending moment.
M 2
R a
4
n 1
n n
Radius of curvature.
yY   Y  
Y
E
;  
Y
yY E
(Consider upper half of cross section.)
For yielding in the web,
1
a5  c  t f
2
1
a6  [ yY  (c  t f )]
2
2
a7  yY
3
R a
Bending moment.
M 2
Radius of curvature.
yY   Y  
7
n 5
n n
Y
E

Program: Key in expressions for an and Rn for n  1 to 7.

Y
yY E
For yY  c to (c  tf ) at  tf /2 decrements, compute M  2  Rn an for n  1 to 4 and  
Y
yY E
For yY  (c  tw ) to c /3 at  tf /2 decrements, compute M  2Rn an for n  5 to 7 and  
, then print.
Y
yY E
, then print.
Input numerical values and run program.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
667
PROBLEM 4.C2 (Continued)
Program Output
For a beam of Problem 4.201,
Depth d  140.00 mm
Thickness of flange t f  10.00 mm
I  0.000011600 m to the 4th
Width of flange b f  120.00 mm
Thickness of web tw  10.00 mm
Yield strength of steel sigmaY  300 MPa
Yield moment M Y  49.71 kip  in.
yY (mm)
M (kN  m)
 (m)
For yielding still in the flange,
70.000
49.71
46.67
65.000
52.59
43.33
60.000
54.00
40.00
For yielding in the web,

60.000
54.00
40.00
55.000
54.58
36.67
50.000
55.10
33.33
45.000
55.58
30.00
40.000
56.00
26.67
35.000
56.38
23.33
30.000
56.70
20.00
25.000
56.97
16.67
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
668
y
␤
PROBLEM 4.C3
0.4
0.4
A
B
1.2
z
C
0.4
␤
1.2
M
D
E
0.8 0.4
1.6
An 8 kip  in. couple M is applied to a beam of the cross section shown
in a plane forming an angle  with the vertical. Noting that the
centroid of the cross section is located at C and that the y and z axes
are principal axes, write a computer program to calculate the stress at
A, B, C, and D for values of  from 0 to 180° using 10° increments.
(Given: I y  6.23 in 4 and I z  1.481 in 4 . )
0.4 0.8
Dimensions in inches
SOLUTION
Input coordinates of A, B, C, D.
z A  z (1)  2
y A  y (1)  1.4
z B  z (2)  2
yB  y (2)  1.4
z D  z (4)  1
yD  y (4)  1.4
zC  z (3)  1
Components of M.
yC  y (3)  1.4
M y   M sin 
M z  M cos 
Equation 4.55, Page 305:
 ( n)  
M z y ( n) M y z ( n)

Iz
Iy
Program: For   0 to 180 using 10° increments.
For n  1 to 4 using unit increments.
Evaluate Equation 4.55 and print stresses.
Return
Return
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
669
PROBLEM 4.C3 (Continued)
Program Output
M  8.00 kip  in.
Moment of couple:
I y  6.23 in 4
Moments of inertia:
I z  1.481 in 4
Coordinates of Points A, B, D, and E:
Point A: z (1)  2:
y (1)  1.4
Point B: z (2)  2: y (2)  1.4
Point D: z (3)  1:
Po int E: z (4)  1:
Beta


y (3)  1.4
y (4)  1.4
---Stress at Points--A
B
ksi
ksi
D
ksi
E
ksi
0
–7.565
–7.565
7.565
7.565
10
–7.896
–7.004
7.673
7.227
20
–7.987
–6.230
7.548
6.669
30
–7.836
–5.267
7.193
5.909
40
–7.446
–4.144
6.621
4.970
50
–6.830
–2.895
5.846
3.879
60
–6.007
–1.558
4.895
2.670
70
–5.001
–0.174
3.794
1.381
80
–3.843
1.216
2.578
0.049
90
–2.569
2.569
1.284
–1.284
100
–1.216
3.843
–0.049
–2.578
110
0.174
5.001
–1.381
–3.794
120
1.558
6.007
–2.670
–4.895
130
2.895
6.830
–3.879
–5.846
140
4.144
7.446
–4.970
–6.621
150
5.267
7.836
–5.909
–7.193
160
6.230
7.987
–6.669
–7.548
170
7.004
7.896
–7.227
–7.673
180
7.565
7.565
–7.565
–7.565
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
670
PROBLEM 4.C4
b
B
B
A
A
h
M'
M
r1
C
Couples of moment M  2 kN  m are applied as shown to a curved bar
having a rectangular cross section with h  100 mm and b  25 mm. Write
a computer program and use it to calculate the stresses at points A and B for
values of the ratio r1/h from 10 to 1 using decrements of 1, and from 1
to 0.1 using decrements of 0.1. Using appropriate smaller increments,
determine the ratio r1/h for which the maximum stress in the curved bar is
50% larger than the maximum stress in a straight bar of the same cross
section.
SOLUTION
h  100 mm,
b  25 mm,
M  2 kN  m
Input:
For straight bar,
 straight 
M
S
6M
 2
hb
 48 MPa
Following notation of Section 4.15, key in the following:
r2  h  r1 ; R  h /ln (r2  r1 ); r  r1  r2 : e  r  R; A  bh  2500
 A  1  M (r1  R)( Aer1 )
Stresses:
 B   2  M (r2  R)/(Aer2 )
Since h  100 mm, for r1/h  10, r1  1000 mm. Also, r1/h  10, r1  100
Program:
(I)
(II)
For r1  1000 to 100 at 100 decrements,
using equations of Lines I and II, evaluate r2 , R, r , e,  1 , and  2
Also evaluate ratio   1/ straight
Return and repeat for r1  100 to 10 at 10 decrements.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
671
PROBLEM 4.C4 (Continued)
Program Output
M  Bending moment  2 kN  m h  100.000 in. A  2500.00 mm 2
Stress in straight beam  48.00 MPa
1
2
r1
mm
rbar
mm
R
mm
e
mm
MPa
MPa
r1 /h
–
ratio
–
1000
1050
1049
0.794
–49.57
46.51
10.000
–1.033
900
950
949
0.878
–49.74
46.36
9.000
–1.036
800
850
849
0.981
–49.95
46.18
8.000
–1.041
700
750
749
1.112
–50.22
45.95
7.000
–1.046
600
650
649
1.284
–50.59
45.64
6.000
–1.054
500
550
548
1.518
–51.08
45.24
5.000
–1.064
400
450
448
1.858
–51.82
44.66
4.000
–1.080
300
350
348
2.394
–53.03
43.77
3.000
–1.105
200
250
247
3.370
–55.35
42.24
2.000
–1.153
100
150
144
5.730
–61.80
38.90
1.000
–1.288
=====================================================
100
150
144
5.730
–61.80
38.90
1.000
–1.288
90
140
134
6.170
–63.15
38.33
0.900
–1.316
80
130
123
6.685
–64.80
37.69
0.800
–1.350
70
120
113
7.299
–66.86
36.94
0.700
–1.393
60
110
102
8.045
–69.53
36.07
0.600
–1.449
50
100
91
8.976
–73.13
35.04
0.500
–1.523
40
90
80
10.176
–78.27
33.79
0.400
–1.631
30
80
68
11.803
–86.30
32.22
0.300
–1.798
20
70
56
14.189
–100.95
30.16
0.200
–2.103
10
60
42
18.297
–138.62
27.15
0.100
–2.888
===================================================================
Find r1/h for ( max ) /( straight )  1.5
52.70
52.80
103
103
94
94
8.703
8.693
–72.036
–71.998
35.34
35.35
0.527
0.528
–1.501
–1.500
52.90
103
94
8.683
–71.959
35.36
0.529
–1.499
Ratio of stresses is 1.5 for r1  52.8 mm or r1/h  0.529.
[Note: The desired ratio r1/h is valid for any beam having a rectangular cross section.]
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
672
bn
PROBLEM 4.C5
hn
M
h2
The couple M is applied to a beam of the cross
section shown. (a) Write a computer program that,
for loads expressed in either SI or U.S. customary
units, can be used to calculate the maximum tensile
and compressive stresses in the beam. (b) Use this
program to solve Probs. 4.9, 4.10, and 4.11.
b2
h1
b1
SOLUTION
Input:
Bending moment M.
For n  1 to n,
Enter bn and hn
Area  bn hn
an  an 1  (hn 1 )/2  hn /2
(Print)
[Moment of rectangle about base]
m  (Area)an
[For whole cross section]
m  m  m; Area  Area  Area
Location of centroid above base.
y  m/Area
(Print)
Moment of inertia about horizontal centroidal axis.
For n  1 to n,
an  an 1  (hn 1 )/2  hn /2
 I  bn hn3 /12  (bn hn )( y  an )2
I  I  I
Computation of stresses.
Total height:
(Print)
For n  1 to n,
H  H  hn
Stress at top:
M top   M
Hy
I
(Print)
Stress at bottom:
M bottom  M
y
I
(Print)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
673
PROBLEM 4.C5 (Continued)
Problem 4.9
Summary of cross section dimensions:
Width (in.)
Height (in.)
9.00
2.00
3.00
6.00
Bending moment  600.000 kip  in.
Centroid is 3.000 mm above lower edge.
Centroidal moment of inertia is 204.000 in4.
Stress at top of beam  14.706 ksi
Stress at bottom of beam  8.824 ksi
Problem 4.10
Summary of cross section dimensions:
Width (in.)
Height (in.)
4.00
1.00
1.00
6.00
8.00
1.00
Bending moment  500.000 kip  in.
Centroid is 4.778 in. above lower edge.
Centroidal moment of inertia is 155.111 in4.
Stress at top of beam  10.387 ksi
Stress at bottom of beam  15.401 ksi
Problem 4.11
Summary of cross section dimensions:
Width (mm)
Height (mm)
50
10
20
50
Bending moment  1500.0000 N  m
Centroid is 25.000 mm above lower edge.
Centroidal moment of inertia is 512,500 mm4.
Stress at top of beam  102.439 MPa
Stress at bottom of beam  72.171 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
674
PROBLEM 4.C6
y
Dy
y
c
M
z
A solid rod of radius c  1.2 in. is made of a steel that is assumed to be
elastoplastic with E  29,000 ksi and  Y  42 ksi. The rod is subjected to a
couple of moment M that increases from zero to the maximum elastic
moment M Y and then to the plastic moment M p . Denoting by yY the half
thickness of the elastic core, write a computer program and use it to calculate
the bending moment M and the radius of curvature  for values of yY from
1.2 in. to 0 using 0.2-in. decrements. (Hint: Divide the cross section into 80
horizontal elements of 0.03-in. height.)
SOLUTION
MY  Y

c3  (42 ksi)

(1.2 in.)3  57 kip  in.
4
4
4
4
M p   Y c3  (42 ksi) (1.2 in.)3  96.8 kip  in.
3
3
Consider top half of rod.
Let
i  Number of elements in top half.
c
h  Height of each element: h 
L
For n  0 to i  1, Step 1:


y  n(h)

z  c 2  {(n  0.5)h}2   z at midheight of element 


If y  yY go to 100

(n  0.5)h

 Stress in elastic core
 E  Y

y

 Repeat for yY  1.2 in.
go to 200

to yY  0
100
 Stress in plastic zone 
 E  Y
 At 0.2-in. decrements
200 Area  2 z (h)


Force   E (Area)

Moment  Force (n  0.5)h


M  M  Moment


P  yY E / Y

Print yY , M , and  .


Next

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
675
PROBLEM 4.C6 (Continued)
Program Output
Radius of rod  1.2 in.
Yield point of steel  42 ksi
Yield moment  57.0 kip  in.
Plastic moment  96.8 kip  in.
Number of elements in half of the rod  40
For yY  1.20 in.,
M  57.1 kip  in.
For yY  0.80 in.,
M  76.9 kip  in.
For yY  0.40 in.,
M  91.6 kip  in.
Radius of curvature  276.19 in.
For yY  0.00 in.,
M  infinite
Radius of curvature  zero
For yY  1.00 in.,
For yY  0.60 in.,
For yY  0.20 in.,

Radius of curvature  828.57 in.
M  67.2 kip  in.
Radius of curvature  690.48 in.
M  85.2 kip  in.
Radius of curvature  414.29 in.
M  95.5 kip  in.
Radius of curvature  552.38 in.
Radius of curvature  138.10 in.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
676
PROBLEM 4.C7
2 in.
C
3 in.
The machine element of Prob 4.178 is to be redesigned by removing part
of the triangular cross section. It is believed that the removal of a small
triangular area of width a will lower the maximum stress in the element.
In order to verify this design concept, write a computer program to
calculate the maximum stress in the element for values of a from 0 to
1 in. using 0.1-in. increments. Using appropriate smaller increments,
determine the distance a for which the maximum stress is as small as
possible and the corresponding value of the maximum stress.
B
A
2.5 in.
a
SOLUTION
See Figure 4.79, Page 289.
M  5 kip  in. r2  5 in. b2  2.5 in.
For a  0 to 1.0 at 0.1 intervals,
h  3 a
r1  2  a
b1  b2 (a /(h  a ))
Area  (b1  b2 )(h/2)
1
1

x  a +  b1h(h/3)  b2 h  2h/3  Area
2
2

r  r2  (h  x )
R
1
2
h 2 (b1  b2 )
(b1r2  b2 r1 ) ln
er R
r2
r1
 h(b1  b2 )
 D  M (r1  R)/[Area (e)(r1 )]
 B  M (r2  R) /[Area (e)(r2 )]
Print and Return
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
677
PROBLEM 4.C7 (Continued)
Program Output
a
in.
R
in.
0.00
3.855
0.10
3.858
0.20
3.869
0.30
3.884
0.40
3.904
0.50
3.928
0.60
3.956
0.70
3.985
0.80
4.018
0.90
4.052
1.00
4.089
D
B
ksi
ksi
b1
r
e
7.7736
2.1014
0.00
4.00
0.145
2.1197
0.08
4.00
0.144
6.9260
2.1689
0.17
4.01
0.140
6.7004
2.2438
0.25
4.02
0.134
2.3423
0.33
4.03
0.127
6.5143
2.4641
0.42
4.05
0.119
2.6102
0.50
4.07
0.111
2.7828
0.58
4.09
0.103
2.9852
0.67
4.11
0.094
3.2220
0.75
4.14
0.086
3.4992
0.83
4.17
0.078
8.5071
7.2700
6.5683
6.5296
6.6098
6.7541
6.9647
Determination of the maximum compressive stress that is as small as possible.
a
in.
R
in.
0.620
3.961
0.625
3.963
0.630
3.964
D
B
ksi
ksi
b1
r
e
6.51185
2.6425
0.52
4.07
0.109
2.6507
0.52
4.07
0.109
2.6591
0.52
4.07
0.109
6.51198
6.51188
Answer: When a  625 in., the compressive stress is 6.51 ksi.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
678