20
40
PROBLEM 4.1
20
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 
1 3
bh
12
Outside rectangle:
I1 
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)
I  I1  I 2  9.81333  106 m 4
y A  40 mm  0.040 m
A  
My A
(15  103 )(0.040)

 61.6  106 Pa
I
9.81333  106
 A  61.6 MPa 
(b)
yB  60 mm  0.060 m
B  
MyB
(15  103 )(0.060)

 91.7  106 Pa
6
I
9.81333  10
 B  91.7 MPa 
PROPRIETARY
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447
443
BeerMOM_ISM_C04.indd 443
12/23/2014 3:58:19 PM
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
18  103

600
30
18  103

300
5
1.5  103

1500
Y0 
37.5  103
37.5  103
 25mm
1500
Neutral axis lies 25 mm above the base.
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
I1 
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
 top  
 bot  
Mytop
I

(1.5  103 )(0.035)
 102.4  106 Pa
512.5  109
M ybot
(1.5  103 )(0.025)

 73.2  106 Pa
9
I
512.5  10
 top  102.4 MPa (compression) 
 bot  73.2 MPa (tension) 
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457
453
BeerMOM_ISM_C04.indd 453
12/23/2014 3:58:23 PM
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
A, mm2
y0 , mm

2160
27
58,320
3

1080
36
38,880
3
Σ
3240
Y 
97, 200
 30 mm
3240
A y0 , mm3
d, mm
97,200
The neutral axis lies 30 mm above the bottom.
ytop  54  30  24 mm  0.024 m
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
I1 
| | 
My
I
|M | 
I
y
Top: (tension side)
M
(120  106 )(758.16  109 )
 3.7908  103 N  m
0.024
Bottom: (compression)
M
(150  106 )(758.16  109 )
 3.7908  103 N  m
0.030
Choose the smaller as Mall.
M all  3.7908  103 N  m
M all  3.79 kN  m 
PROPRIETARY
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467
463
BeerMOM_ISM_C04.indd 463
12/23/2014 3:58:26 PM
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.
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
I

1

(b)

Mc (60)(0.010)

 75.0  106 Pa
I
8  109
M
60

 37.5  103 m 1
9
9
EI (200  10 )(8  10 )
  75.0 MPa 
  26.7 m 
Bending about y-axis.
1 3 1
bh  (20)(12)3  2.88  103 mm 4  2.88  109 m 4
12
12
12
c
 6 mm  0.006 m
2
Mc (60)(0.006)

 125.0  106 Pa

9
I
2.88  10
I
1


M
60

 104.17  103 m 1
EI (200  109 )(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.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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472
468
BeerMOM_ISM_C04.indd 468
12/23/2014 3:58:29 PM
PROBLEM 4.31
y
A
z
M
A W200  31.3 rolled-steel beam is subjected to a couple M of moment
45 kN  m. Knowing that E  200 GPa and v  0.29, determine (a) the
radius of curvature  , (b) the radius of curvature   of a transverse cross
section.
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


M
45  103

 7.1885  103 m 1
6
9
EI (200  10 )(31.3  10 )
1
1
 v  (0.29)(7.1885  103 )  2.0847  103 m 1


  139.1 m 
   480 m 
PROPRIETARY
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479
475
BeerMOM_ISM_C04.indd 475
12/23/2014 3:58:32 PM
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,
n1
b1h13  n1 A1d 12
12
1.5

(30)(6)3  (1.5)(30)(6)(18)3  88.29  103 mm 4
12
I1 
I2 
1.0
n2
(30)(30)3  67.5  103 mm 4
b2 h23 
12
12
I 3  I1  88.29  103 mm 4
I  I1  I 2  I 3  244.08  103 mm 4
 244.08  109 m4
 
Aluminum:
n  1.0,
M
Brass:
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
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 
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481
477
BeerMOM_ISM_C04.indd 477
12/23/2014 3:58:33 PM
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.
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
I1 
(a)
Aluminum:
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
  56.0 MPa 
(b)

Copper:

n  1.4, y  5.5909 mm  0.0055909 m

nMy
(1.4)(35)(0.0055909)
 

 68.4  106 Pa
I
4.008  109

 68.4 MPa
  68.4 MPa 

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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488
484
BeerMOM_ISM_C04.indd 484
12/23/2014 3:58:36 PM
PROBLEM 4.52
A concrete beam is reinforced by three steel rods placed as shown. The
modulus of elasticity is 20 GPa for the concrete and 200 GPa for the
steel. Using an allowable stress of 9 MPa for the concrete and 140 MPa
for the steel, determine the largest allowable positive bending moment in
the beam.
SOLUTION
n=
Es 200 GPa
=
= 10
Ec 20 GPa
π
π
d 2 = 3 (22) 2 = 1140 mm 2
4
4
2
nAs = 11400 mm
As = 3
x
− (11400)(350 − x) = 0
2
100 x 2 + 11400 x − 3,990, 00 = 0
Locate neutral axis
200 x
x=
Solve for x:
−11400 + 114002 + (4)(100)(3,990,000)
= 150.72 mm
(2)(100)
350 − x = 199.28 mm
I =
1
1
200 x3 + nAs (350 − x) 2 = (200)(150.72)3 + (11400)(199.28) 2
3
3
= 681 × 106 mm 4
σ =
nMy
I
∴ M=
n = 1.0,
Concrete:
M =
σI
ny
y = 150.72 in,
(9 × 106 )(681 × 10−6 )
= 40664 N ⋅ m = 407 kN ⋅ m
(1.0)(0.15072)
n = 10,
Steel:
M=
Choose the smaller value.
σ = 9 MPa
y = 199.28, σ = 140 MPa
(140 × 106 )(681 × 10−6 )
= 478.42 N ⋅ m = 47.8 kN ⋅ m
(10)(0.19928)
M = 40.7 kN ⋅ m

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BeerMOM_ISM_C04.indd 500
500
12/23/2014 3:58:43 PM
PROBLEM 4.59
M
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
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
x2 
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
(0.5)(600)( 0.058579)
nMy

 6.15  106 Pa
6
I
2.8595  10
 t  6.15 MPa 
y  41.421 mm  0.041421 m
(1.0)(600)(0.041421)
nMy

 8.69  106 Pa
6
I
2.8595  10
 c  8.69 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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508
509
BeerMOM_ISM_C04.indd 509
12/23/2014 3:58:46 PM
M
100 mm
150 mm
PROBLEM 4.65
M
100 mm
150 mm
18 mm
(a)
18 mm
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
d
100
r
10

 0.10
d 100
For configuration (a),
Fig. 4.28 gives
K a  2.21.
For configuration (b),
Fig. 4.27 gives K b  1.79.
1 3
1
bh 
(18)(100)3  1.5  106 mm 4  1.5  10 6 m 4
12
12
1
c  d  50 mm  0.05 m
2
I 
(a)
 
KMc (2.21)(2  103 )(0.05)

 147.0  106 Pa  147.0 MPa
I
1.5  106
  147.0 MPa 
(b)
 
KMc (1.79)(2  103 )(0.05)

 119.0  106 Pa  119.0 MPa
I
1.5  106
  119.0 MPa 
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