CHAPTER 4
PROBLEM 4.1
Knowing that the couple shown acts in a vertical plane, determine the
stress at (a) point A, (b) point B.
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.
σA = −
My A
(25)(2.75)
=−
I
28.854
(b)
yB = 0.75 in.
σB = −
MyB
(25)(0.75)
=−
I
28.854
σ A = −2.38 ksi 
σ B = −0.650 ksi 
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PROBLEM 4.2
Knowing that the couple shown acts in a vertical plane, determine
the stress at (a) point A, (b) point B.
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
σA = −
My A
(15 × 103 )(0.040)
=−
= −61.6 × 106 Pa
I
9.81333 × 10−6
σ A = −61.6 MPa 
(b)
yB = −60 mm = −0.060 m
σB = −
MyB
(15 × 103 )(−0.060)
=−
= 91.7 × 106 Pa
−6
I
9.81333 × 10
σ B = 91.7 MPa 
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PROBLEM 4.3
Using an allowable stress of 16 ksi, determine the largest couple that can be
applied to each pipe.
SOLUTION
(a)
I =
π
(r
4
4
o
)
− ri4 =
π
4
(0.64 − 0.54 ) = 52.7 × 10−3 in 4
c = 0.6 in.
σ =
Mc
:
I
M =
σI
c
=
(16)(52.7 × 10−3 )
0.6
M = 1.405 kip ⋅ in 
(b)
π
(0.7 4 − 0.54 ) = 139.49 × 10−3 in 4
4
c = 0.7 in.
I =
σ =
Mc
:
I
M =
σI
c
=
(16)(139.49 × 10−3 )
0.7
M = 3.19 kip ⋅ in 
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PROBLEM 4.4
A nylon spacing bar has the cross section shown. Knowing that the
allowable stress for the grade of nylon used is 24 MPa, determine the
largest couple Mz that can be applied to the bar.
SOLUTION
I = I rect − I circle =
=
1 3 π 4
bh − r
12
4
1
π
(100)(80)3 − (25) 4 = 3.9599 × 106 mm 4
12
4
= 3.9599 × 10−6 m
c=
80
= 40 mm = 0.040 m
2
σ =
Mc
:
I
Mz =
σI
c
=
(24 × 106 )(3.9599 × 10−6 )
= 2.38 × 103 N ⋅ m
0.040
M z = 2.38 kN ⋅ m 
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PROBLEM 4.5
A beam of the cross section shown is extruded from an aluminum
alloy for which σ Y = 250 MPa and σ U = 450 MPa. Using a factor
of safety of 3.00, determine the largest couple that can be applied to
the beam when it is bent about the z-axis.
SOLUTION
Allowable stress.
=
σU
F.S .
=
450
= 150 MPa
3
= 150 × 106 Pa
Moment of inertia about z-axis.
1
(16)(80)3 = 682.67 × 103 mm 4
12
1
I 2 = (16)(32)3 = 43.69 × 103 mm 4
12
I 3 = I1 = 682.67 × 103 mm 4
I1 =
I = I1 + I 2 + I 3 = 1.40902 × 106 mm 4 = 1.40902 × 10−6 m 4
1
Mc
with c = (80) = 40 mm = 0.040 m
2
I
I σ (1.40902 × 10−6 )(150 × 106 )
=
= 5.28 × 103 N ⋅ m
M=
0.040
c
σ=
M = 5.28 kN ⋅ m 
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PROBLEM 4.6
Solve Prob. 4.5, assuming that the beam is bent about the y-axis.
PROBLEM 4.5 A beam of the cross section shown is extruded from an
aluminum alloy for which σ Y = 250 MPa and σ U = 450 MPa. Using a
factor of safety of 3.00, determine the largest couple that can be applied to
the beam when it is bent about the z-axis.
SOLUTION
=
Allowable stress:
σU
F.S .
=
450
= 150 MPa
3.00
= 150 × 106 Pa
Moment of inertia about y-axis.
1
(80)(16)3 + (80)(16)(16) 2 = 354.987 × 103 mm 4
12
1
I 2 = (32)(16)3 = 10.923 × 103 mm 4
12
I 3 = I1 = 354.987 × 103 mm 4
I1 =
I = I1 + I 2 + I 3 = 720.9 × 103 mm 4 = 720.9 × 10−9 m 4
1
Mc
with c = (48) = 24 mm = 0.024 m
2
I
I σ (720.9 × 10−9 )(150 × 106 )
=
= 4.51 × 103 N ⋅ m
M=
0.024
c
σ=
M = 4.51 kN ⋅ m 
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PROBLEM 4.7
Two W4 × 13 rolled sections are welded together as shown. Knowing that for the steel
alloy used σ Y = 36 ksi and σ 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.
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.
M all
σU
58
= 19.333 ksi
F .S . 3.0
σ I (19.333) (55.74)
= all =
c
4.16
σ all =
=
σ=
Mc
I
M all = 259 kip ⋅ in 
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PROBLEM 4.8
Two W4 × 13 rolled sections are welded together as shown. Knowing that for the steel
alloy used σ Y = 36 ksi and σ 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.
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.
σU
58
=
= 19.333 ksi
F .S . 3.0
σ I (19.333) (39.286)
= all =
4.060
c
σ all =
M all
σ=
Mc
I
M all = 187.1 kip ⋅ in 
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PROBLEM 4.9
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.
SOLUTION
A1 =
π
2
r2 =
π
2
(25) 2 = 981.7 mm 2
A2 = bh = (50)(25) = 1250 mm 2
y =
4r
(4)(25)
=
= 10.610 mm
3π
3π
h
25
y2 = − = −
= −12.5 mm
2
2
y1 =
A1 y1 + A2 y2
(981.7)(10.610) + (1250)(−12.5)
=
= −2.334 mm
A1 + A2
981.7 + 1250
I1 = I x1 − A1 y12 =
π
r 4 − A1 y12 =
π
(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
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
I2 =
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
M − Pa = 0 :
σ top =
σ bot =
−Mytop
M = Pa = (4 × 103 )(300 × 10−3 ) = 1200 N ⋅ m
(1200)(0.027334)
= −81.76 × 106 Pa
401.16 × 10−9
σ top = −81.8 MPa 
−Mybot
(1200)(−0.022666)
=−
= 67.80 × 106 Pa
I
401.16 × 10−9
σ bot = 67.8 MPa 
I
=−
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PROBLEM 4.10
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.
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
= 25 mm
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
σ top = −
σ bot = −
M ytop
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
I
512.5 × 10−9
σ top = −102.4 MPa (compression) 
σ bot = 73.2 MPa (tension) 
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PROBLEM 4.11
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.
SOLUTION
A
y0
A y0

8
7.5
60

6
4
24

4
0.5
Σ
18
Yo =
2
86
86
= 4.778 in.
18
Neutral axis lies 4.778 in. above the base.
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.
I1 =
M − Pa = 0
M = Pa = (25)(20) = 500 kip ⋅ in
σ top = −
σ bot = −
Mytop
I
=−
(500)(3.222)
155.16
Mybot
(500)(−4.778)
=−
I
155.16
σ top = −10.38 ksi (compression) 
σ bot = 15.40 ksi (tension) 
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PROBLEM 4.12
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 top flange.
SOLUTION
The stress distribution over the entire cross section is given by the bending stress formula:
σx = −
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 = σ x dA = −
My
dA
I
The total force on the shaded area is then

F = dF = −

My
M
dA = −
I
I
 ydA = −
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
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PROBLEM 4.12 (Continued)
Moment of inertia of entire cross section:
1
1
b1h13 + A1d12 = (216)(36)3 + (216)(36)(36)2 = 10.9175 × 106 mm4
12
12
1
1
3
2
I 2 = b2 h2 + A2 d 2 = (72)(108)3 + (72)(108)(36)2 = 17.6360 × 106 mm 4
12
12
I = I1 + I 2 = 28.5535 × 106 mm4 = 28.5535 × 10−6 m 4
I1 =
For the shaded area,
A* = (216)(36) = 7776 mm 2
y * = 36 mm
A* y * = 279.936 × 103 mm3 = 279.936 × 10−6 m3
F =−
MA* y * (6 × 103 )(279.936 × 10−6 )
=
I
28.5535 × 10−6
= 58.8 × 103 N
F = 58.8 kN 
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PROBLEM 4.13
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.
SOLUTION
The stress distribution over the entire cross section is given by the bending stress formula:
σx = −
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 = σ x dA = −
My
dA
I
The total force on the shaded area is then

F = dF = −

My
M
dA = −
I
I
 ydA = −
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
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PROBLEM 4.13 (Continued)
Moment of inertia of entire cross section:
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
I1 =
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 
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PROBLEM 4.14
Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is 50 kip ⋅ in., determine the total force acting
(a) on the top flange, (b) on the shaded portion of the web.
SOLUTION
The stress distribution over the entire cross-section is given by the bending stress formula:
σx = −
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 = σ x dA = −
My
dA
I
The total force on the shaded area is then

F = dF = −

My
M
dA = −
I
I
 ydA = −
M * *
y A
I
where y * is the centroidal coordinate of the shaded portion and A* is its area.
Calculate the moment of inertia.
1
1
(6 in.)(7 in.)3 − (4 in.)(4 in.)3 = 150.17 in 4
12
12
M = 50 kip ⋅ in
I =
(a)
Top flange:
A* = (6 in.)(1.5 in.) = 9 in 2
F =
(b)
Half web:
50 kip ⋅ in
(9 in 2 )(2.75 in.) = 8.24 kips
150.17 in 4
A* = (2 in.)(2 in.) = 4 in 2
F =
y * = 2 in. + 0.75 in. = 2.75 in.
F = 8.24 kips 
y * = 1 in.
50 kip ⋅ in
(4 in 2 )(1 in.) = 1.332 kips
150.17 in 4
F = 1.332 kips 
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PROBLEM 4.15
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.
SOLUTION
A, mm 2
y0 , mm
A y0 , mm3

600
22.5
13.5 × 103

300
7.5
2.25 × 103
Σ
900
Y0 =
15.75 × 103
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
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
I1 =
|σ | =
My
I
M=
σI
y
Top: (tension side)
M=
(24 × 106 )(61.875 × 10−9 )
= 118.8 N ⋅ m
0.0125
Bottom: (compression)
M=
(30 × 106 )(61.875 × 10−9 )
= 106.1 N ⋅ m
0.0175
Choose smaller value.
M = 106.1 N ⋅ m 
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PROBLEM 4.16
Solve Prob. 4.15, assuming that d = 40 mm.
PROBLEM 4.15 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.
SOLUTION
A, mm 2
y0 , mm
A y0 , mm3

600
32.5
19.5 × 103

500
12.5
6.25 × 103
Σ
1100
Y0 =
25.75 × 103
25.75 × 103
= 23.41 mm
1100
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
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
I1 =
|σ | =
My
I
M=
σI
y
Top: (tension side)
M=
(24 × 106 )(146.383 × 10−9 )
= 212 N ⋅ m
0.01659
Bottom: (compression)
M=
(30 × 106 )(146.383 × 10−9 )
= 187.6 N ⋅ m
0.02341
Choose smaller value.
M = 187.6 N ⋅ m 
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PROBLEM 4.17
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.
SOLUTION
A
y0

2.25
1.25
2.8125

2.25
0.25
0.5625
4.50
Y=
A y0
3.375
3.375
= 0.75 in.
4.50
The neutral axis lies 0.75 in. above bottom.
ytop = 2.0 − 0.75 = 1.25 in.,
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
I 2 = b2 h22 + A2 d 22 = (4.5)(0.5)3 + (2.25)(0.5)2 = 0.609375 in 4
12
12
I = I1 + I 2 = 1.59375 in 4
I1 =
σ =
My
I
M =
σI
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 
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PROBLEM 4.18
Knowing that for the casting shown the allowable stress is 5 ksi in
tension and 18 ksi in compression, determine the largest couple M
that can be applied.
SOLUTION
Locate the neutral axis and compute the moment of inertia.
Y =
 Ai yi
 Ai
I =
1
bi hi3 for rectangle
12
I = ( Ai di2 + I )
di = yi − Y
I , in 4
Part
A, in 2
yi , in.
Ai yi , in 3
1
1.5
1.25
1.875
0.3333
0.1667
0.03125
2
1.0
0.75
0.75
0.1667
0.0277
0.02083
3
0.5
0.25
0.125
0.6667
0.2222
0.01042
Σ
3.0
0.4166
0.0625
Y =
 Ay 2.75
=
= 0.9167 in.
A
3.0
Ai di2 , in 4
di , in.
2.75
I = ( I + Ad 2 ) = 0.4166 + 0.0625 = 0.479 in 4
Allowable bending moment.
σ =
Tension at A:
Mc
I
or
M =
σI
c
σ A ≤ 5 ksi
c A = 0.583 in.
M ≤
Compression at B:
(5)(0.479)
= 4.11 kip ⋅ in
0.583
σ B ≤ 18 ksi
M ≤
The smaller value is the allowable value of M.
cB = 0.917 in.
(18)(0.479)
= 9.40 kip ⋅ in
0.917
M = 4.11 kip ⋅ in 
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PROBLEM 4.19
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.
SOLUTION
 rectangle
 circular cutout
A1 = (150)(250) = 37.5 × 103 mm 2
A2 = −π (50) 2 = − 7.85398 × 103 mm 2
A = A1 + A2 = 29.64602 × 103 mm 2
y1 = 0 mm
y2 = −50 mm
Y =
ΣA y
ΣA
(37.5 × 103 )(0) + (−7.85393 × 103 )(−50)
29.64602 × 103
= 13.2463 mm
Y =
I X ′ = Σ( I + Ad 2 ) = I1 − I 2
1

=  (150)(250)3 + (37.5 × 103 )(13.2463) 2 
12

π

−  (50) 4 + (7.85398 × 103 )(50 + 13.2463) 2 
4

= 201.892 × 106 − 36.3254 × 106 = 165.567 × 106 mm 4
= 165.567 × 10−6 m 4
Top: (tension side)
c = 125 − 13.2463 = 111.7537 mm = 0.11175 m
σ=
Bottom: (compression side)
M=
I σ (165.567 × 10−6 )(120 × 106 )
=
c
0.11175
= 177.79 × 103 N ⋅ m
c = 125 + 13.2463 = 138.2463 mm
= 0.13825 m
σ=
Choose the smaller.
Mc
I
Mc
I
M=
I σ (165.567 × 10−6 )(150 × 106 )
=
c
0.13825
= 179.64 × 103 N ⋅ m
M = 177.8 × 103 N ⋅ m
M = 177.8 kN ⋅ m 
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PROBLEM 4.20
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.
SOLUTION
A, mm 2
y0 , mm
A y0 , mm3
d, mm

2160
27
58320
3

1080
36
38880
3
Σ
3240
Y =
97200
= 30 mm
3240
97200
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 =
|σ | =
My
I
|M | =
σI
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 
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PROBLEM 4.21
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.
SOLUTION
Band blade thickness:
t = 0.018 in.
Radius of pulley:
r=
1
d = 4.000 in.
2
Radius of curvature of centerline of blade:
1
2
ρ = r + t = 4.009 in.
c=
1
t = 0.009 in.
2
c
0.009
= 0.002245
4.009
Maximum strain:
εm =
Maximum stress:
σ m = Eε m = (29 × 106 )(0.002245)
ρ
=
σ m = 65.1 × 103 psi
σ m = 65.1 ksi 
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PROBLEM 4.22
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.
ρ =
1
1
d = (0.30) = 0.15 in.
2
2
π
4
r4 =
π
4
(0.15) 4 = 397.61 × 10−6 in 4
1
D = 30 in.
2
c = r = 0.15 in.
(a)
σ max =
(b)
M =
EI
ρ
Ec
ρ
=
=
(29 × 106 )(0.15)
= 145 × 103 psi
30
(29 × 106 )(397.61 × 10−6 )
30
σ max = 145 ksi 
M = 384 lb ⋅ in 
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PROBLEM 4.23
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 .
SOLUTION
When the rod is bent into a full circle, the circumference is 900 mm. Since the circumference is equal to
2π times ρ, the radius of curvature, we get
ρ =
900 mm
= 143.24 mm = 0.14324 m
2π
σ = Eε =
Stress:
Ec
or
ρ
c=
ρσ
E
For σ = 420 MPa and E = 200 GPa,
c=
(a)
(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
ρ
(200 × 109 )(145.16 × 10−15 )
= 0.203 N ⋅ m
0.14324
M = 0.203 N ⋅ m 
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PROBLEM 4.24
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.
SOLUTION
(a)
Bending about z-axis.
1 3 1
bh = (12)(20)3 = 8 × 103 mm4 = 8 × 10−9 m 4
12
12
20
c=
= 10 mm = 0.010 m
2
I=
σ=
1
ρ
(b)
=
Mc (60)(0.010)
=
= 75.0 × 106 Pa
I
8 × 10−9
M
60
=
= 37.5 × 10−3 m −1
EI (200 × 109 )(8 × 10 −9 )
σ = 75.0 MPa 
ρ = 26.7 m 
Bending about y-axis.
1 3 1
bh = (20)(12)3 = 2.88 × 103 mm4 = 2.88 × 10−9 m4
12
12
12
c=
= 6 mm = 0.006 m
2
Mc (60)(0.006)
=
= 125.0 × 106 Pa
σ=
I
2.88 × 10−9
I=
1
ρ
=
M
60
=
= 104.17 × 10−3 m −1
−9
9
EI (200 × 10 )(2.88 × 10 )
σ = 125.0 MPa 
ρ = 9.60 m 
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PROBLEM 4.25
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.
SOLUTION
1 3 1 3 a4
bh = aa =
12
12
12
a
c=
2
I=
σ max
1
ρ
a
Mc M 2
=
= 4
I
a
12
=
M
M
=
EI E a 4
12
σ max =
1
ρ
=
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
)
a4
24
I = I1 + I 2 =
c=
(
3
 a 
a4
2a 
=

24
 2
a4
12
σ max =
1
ρ
=
Mc Ma/ 2 6 2 M
= 4
=
I
a /12
a3
M
M
=

EI E a 4
12
σ max =
1
ρ
8.49M

a3
=
12M

Ea 4
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PROBLEM 4.26
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 σ m = kσ 0 , where σ 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.
SOLUTION
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
σ=
For the original square,
3M
Mc
Mh
=
=
3
4
4
I
(4h 0 − 3h) h 2
h h −h
3 0
h = h0 , c = h0 .
σ0 =
3M
3M
= 3
2
(4h0 − 3h0 )h0
h0
h03
h03
σ
=
=
= 0.950
σ 0 (4h0 − 3h) h 2 (4h0 − (3)(0.9)h0 )(0.9 h02 )
σ = 0.950 σ 0
k = 0.950 
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PROBLEM 4.27
In Prob. 4.26, determine (a) the value of h for which the maximum stress
σ m is as small as possible, (b) the corresponding value of k.
PROBLEM 4.26 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 σ m = kσ 0 , where σ 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.
SOLUTION
I = 4 I1 + 2 I 2
 1 
1
= (4)   hh3 + (2)   (2h0 − 2h) h3
 12 
3
1 4 4
4
4
= h − h 0 h3 − h3 = h 0 h3 − h 4
3
3
3
3
I 4
c=h
= h 0 h 2 − h3
c 3
I
d 4

is maximum at
h 0 h 2 − h3  = 0.

c
dh  3

8
h 0 h − 3h 2 = 0
3
2

3
I 4 8  8 
256 3
= h0  h0  −  h0  =
h0
729
c 3 9  9 
For the original square,
h = h0
c = h0
σ0 =
h=
σ=
8
h 0 
9
Mc 729M
=

I
256h03
I0 1 3
= h0
c0 3
Mc0 3M
= 2
I0
h0
σ 729 1 729
=
⋅ =
= 0.949
σ 0 256 3 768
k = 0.949 
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PROBLEM 4.28
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 σm will be as small as possible,
(b) the radius of curvature of the beam will be maximum.
SOLUTION
Let D be the diameter of the log.
D2 = b2 + d 2
I =
(a)
1 3
bd
12
d 2 = D 2 − b2
c=
σm is the minimum when
1
d
2
I
is maximum.
c
I
1
= b( D 2 − b 2 )
c
6
d I 1 2
 = D −
db  c  6
d =
(b)
ρ =
I
1
= bd 2
c
6
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
ρ is maximum when I is maximum,
1 3
bd is maximum, or b 2d 6 is maximum.
12
( D 2 − d 2 )d 6 is maximum.
6D 2d 5 − 8d 7 = 0
b=
D2 −
3 2
1
D = D
4
2
d =
3
D
2
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PROBLEM 4.29
For the aluminum bar and loading of Sample Prob. 4.1, determine (a) the radius of curvature ρ ′ of a
transverse cross section, (b) the angle between the sides of the bar that were originally vertical. Use
E = 10.6 × 106 psi and v = 0.33.
SOLUTION
From Sample Prob. 4.1,
I = 12.97 in 4
1
ρ
(a)
=
M
103.8 × 103
=
= 755 × 10−6 in −1
EI (10.6 × 106 )(12.97)
1
1
= v = (0.33) (755 × 10−6 ) = 249 × 10−6 in −1
ρ′
ρ
ρ ′ = 4010 in.
(b)
M = 103.8 kip ⋅ in
θ=
length of arc b
3.25
=
=
= 810 × 10−6 rad
ρ ′ 4010
radius
ρ ′ = 334 ft 
θ = 0.0464° 
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PROBLEM 4.30
For the bar and loading of Example 4.01, determine (a) the radius of curvature ρ , (b) the radius of curvature
ρ ′ 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)
(c)
1
ρ
=
I = 1.042 in 4
M
(30 × 103 )
=
= 993 × 10−6 in −1
EI (29 × 106 )(1.042)
ε 1 = vε =
vc
ρ
=v
ρ = 1007 in. 
c
ρ′
1
1
= v = (0.29)(993 × 10−6 )in.−1 = 288 × 10−6 in.−1
ρ′
ρ
ρ ′ = 3470 in. 
length of arc b
0.8
=
=
= 230 × 10−6 rad
ρ ′ 3470
radius
θ = 0.01320° 
θ=
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PROBLEM 4.31
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 ρ , (b) the radius of curvature ρ ′ of a transverse cross
section.
SOLUTION
For W 200 × 31.3 rolled steel section,
I = 31.4 × 106 mm4
= 31.4 × 10−6 m 4
(a)
(b)
1
ρ
=
M
45 × 103
=
= 7.17 × 10−3 m −1
EI (200 × 109 ) (31.4 × 10−6 )
1
1
= v = (0.29) (7.17 × 10−3 ) = 2.07 × 10−3 m −1
ρ′
ρ
ρ = 139.6 m 
ρ ′ = 481 m 
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PROBLEM 4.32
It was assumed in Sec. 4.3 that the normal stresses σ 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 σ y as a function of y, (b) show that
( σ y ) max = −(c/2 ρ )(σ x )max and, thus, that σ 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 σ x is still linear.)
SOLUTION
Denote the width of the beam by b and the length by L.
θ=
Using the free body diagram above, with
Σ Fy = 0 :
σy = −
(a)
σ y bL + 2
θ
2
sin
L
2
σ x = −(σ x )max
But,
(σ )
σ y = x max
ρc

y
−c
cos
(σ )
y2
ydy = x max
ρc
2

y
−c

θ
2
y
−c
L
ρ
≈1
σ x bdy sin
σ x dy ≈ −
θ
L

θ
2
y
−c
=0
σ x dy = −
1
ρ

y
−c
σ x dy
y
c
y
σy =
−c
(σ x )max 2
( y − c 2 ) 
2ρ c
The maximum value σ y occurs at y = 0 .
(b)
(σ y ) max = −
(σ x )max c 2
(σ ) c
= − x max 
2ρ c
2ρ
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PROBLEM 4.33
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
Modulus of elasticity
Allowable stress
Brass
70 GPa
105 GPa
100 MPa
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 figure.
For the transformed section,
n1
1.0
b1 h13 =
(8) (32)3 = 21.8453 × 103 mm 4
12
12
n
1.5
I 2 = 2 b2 h23 =
(32)(32)3 = 131.072 × 103 mm 4
12
12
I 3 = I1 = 21.8453 × 103 mm 4
I1 =
I = I1 + I 2 + I 3 = 174.7626 × 103 mm 4 = 174.7626 × 10−9 m 4
|σ | =
Aluminum:
σI
ny
σ = 100 × 106 Pa
(100 × 106 )(174.7626 × 10−9 )
= 1.0923 × 103 N ⋅ m
(1.0)(0.016)
n = 1.5, | y | = 16 mm = 0.016 m,
M=
Choose the smaller value.
M=
n = 1.0, | y | = 16 mm = 0.016 m,
M=
Brass:
nM y
I
σ = 160 × 106 Pa
(160 × 106 )(174.7626 × 10−9 )
= 1.1651 × 103 N ⋅ m
(1.5)(0.016)
M = 1.092 × 103 N ⋅ m
M = 1.092 kN ⋅ m 
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PROBLEM 4.34
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
Brass
70 GPa
105 GPa
100 MPa
160 MPa
Modulus of elasticity
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,
n1
1.5
(8)(32)3 = 32.768 × 103 mm 4
b1h13 =
12
12
n
1.0
I 2 = 2 b2 H 23 − h23 =
(32)(323 − 163 ) = 76.459 × 103 mm 4
12
12
I 3 = I1 = 32.768 × 103 mm 4
I1 =
(
)
I = I1 + I 2 + I 3 = 141.995 × 103 mm 4 = 141.995 × 10−9 m 4
|σ | =
Aluminum:
M=
σI
ny
n = 1.0, | y | = 16 mm = 0.016 m, σ = 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, σ = 160 × 106 Pa
M=
Choose the smaller value.
(160 × 106 )(141.995 × 10−9 )
= 946.63 N ⋅ m
(1.5)(0.016)
M = 887 N ⋅ m 
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PROBLEM 4.35
For the composite bar indicated, determine the largest permissible bending
moment when the bar is bent about a vertical axis.
PROBLEM 4.35 Bar of Prob. 4.33.
Modulus of elasticity
Allowable stress
Aluminum
Brass
70 GPa
105 GPa
100 MPa
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 figure.
For the transformed section,
n1
1.0
(32) (8)3 + (1.0)[(32)(8)](20) 2 = 103.7653 × 103 mm 4
h1 b13 + n1 A1d12 =
12
12
n
1.5
(32)(32)3 = 131.072 × 103 mm 4
I 2 = 2 h2b23 =
12
12
I 3 = I1 = 103.7653 × 103 mm 4
I1 =
I = I1 + I 2 + I 3 = 338.58 × 103 mm 4 = 338.58 × 10−9 m 4
|σ | =
n My
I
Aluminum:
M=
σ = 100 × 106 Pa
(100 × 106 )(338.58 × 10−9 )
= 1.411 × 103 N ⋅ m
(1.0)(0.024)
n = 1.5, | y | = 16 mm = 0.016 m,
M=
Choose the smaller value.
ny
n = 1.0, | y | = 24 mm = 0.024 m,
M=
Brass:
σI
σ = 160 × 106 Pa
(160 × 106 )(338.58 × 10−9 )
= 2.257 × 103 N ⋅ m
(1.5)(0.016)
M = 1.411 × 103 N ⋅ m
M = 1.411 kN ⋅ m 
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PROBLEM 4.36
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.
Modulus of elasticity
Allowable stress
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,
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
(
I1 =
)
I = I1 + I 2 + I 3 = 354.99 × 103 mm 4 = 354.99 × 10−9 m 4
|σ | =
Aluminum:
M=
σI
ny
n = 1.0, | y | = 16 mm = 0.016 m, σ = 100 × 106 Pa
M=
Brass:
nMy
I
(100 × 106 )(354.99 × 10−9 )
= 2.2187 × 103 N ⋅ m
(1.0)(0.016)
n = 1.5 | y | = 24 mm = 0.024 m σ = 160 × 106 Pa
M=
(160 × 106 )(354.99 × 10−9 )
= 1.57773 × 103 N ⋅ m
(1.5)(0.024)
Choose the smaller value.
M = 1.57773 × 103 N ⋅ m
M = 1.578 kN ⋅ m 
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PROBLEM 4.37
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.
Modulus of elasticity:
Allowable stress:
Wood
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,
n1
1.0
b1h13 =
(3)(10)3 = 250 in 4
12
12
n
14.5  1 
3
4
I 2 = 2 b2 h23 =
  (10) = 604.17 in
12
12  2 
I1 =
I 3 = I1 = 250 in 4
I = I1 + I 2 + I 3 = 1104.2 in 4
σ =
Wood:
nMy
I
∴ M =
n = 1.0,
M =
Steel:
ny
y = 5 in., σ = 2000 psi
(2000)(1104.2)
= 441.7 × 103 lb ⋅ in
(1.0)(5)
n = 14.5,
M =
σI
y = 5 in., σ = 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 
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PROBLEM 4.38
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.
Modulus of elasticity:
Allowable stress:
Wood
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
3
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
I = I1 + I 2 + I 3 = 2498.7 in 4
σ =
Wood:
nMy
I
∴ M =
n = 1.0,
M =
Steel:
ny
y = 5 in., σ = 2000 psi
(2000)(2499)
= 999.5 × 103 lb ⋅ in
(1.0)(5)
n = 14.5,
M =
σI
y = 5.5 in., σ = 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 
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PROBLEM 4.39
A steel bar and an aluminum bar are bonded together to form the composite
beam shown. The modulus of elasticity for aluminum is 70 GPa and for steel is
200 GPa. Knowing that the beam is bent about a horizontal axis by a couple of
moment M = 1500 N ⋅ m, determine the maximum stress in (a) the aluminum,
(b) the steel.
SOLUTION
Use aluminum as the reference material.
For aluminum,
n =1
For steel,
n = Es /Ea = 200/70 = 2.8571
Transformed section:
Part
1
2
Σ
Y0 =
A, mm2
600
1200
nA, mm2
1714.3
1200
2914.3
109714
= 37.65 mm
2914.3
yo , mm
50
20
nAyo , mm3
85714
24000
109714
d, mm
12.35
17.65
d = y0 − Y0
n1
2.8571
(30)(20)3 + (1714.3)(12.35) 2 = 318.61 × 103 mm 4
b1 h13 + n1 A1 d12 =
12
12
n
1
I 2 = 2 b2 h23 + n2 A2 d 22 = (30)(40)3 + (1200)(17.65) 2 = 533.83 × 103 mm 4
12
12
3
I = I1 + I 2 = 852.44 × 10 mm 4 = 852.44 × 10−9 m 4
I1 =
M = 1500 N ⋅ m
Stress:
(a)
σ =−
Aluminum:
n My
I
n = 1,
σa = −
(b)
Steel:
y = −37.65 mm = −0.03765 m
(1)(1500)(−0.03765)
= 66.2 × 106 Pa
852.44 × 10−9
n = 2.8571,
σs = −
σ a = 66.2 MPa 
y = 60 − 37.65 = 22.35 mm = 0.02235 m
n My
(2.8571)(1500) (0.02235)
=−
= −112.4 × 106 Pa
I
852.44 × 10−9
σ s = −112.4 MPa 
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PROBLEM 4.40
A steel bar and an aluminum bar are bonded together to form the composite
beam shown. The modulus of elasticity for aluminum is 70 GPa and for steel is
200 GPa. Knowing that the beam is bent about a horizontal axis by a couple of
moment M = 1500 N ⋅ m, determine the maximum stress in (a) the aluminum,
(b) the steel.
SOLUTION
Use aluminum as the reference material.
For aluminum,
n =1
For steel,
n = Es /Ea = 200/70 = 2.8571
Transformed section:
A, mm2
Part
nA, mm2
yo , mm
nAyo , mm3
d, mm
1
600
600
50
30000
25.53
2
1200
3428.5
20
68570
4.47
4028.5
Σ
Y0 =
98570
= 24.47 mm
4028.5
98570
d = y0 − Y0
n1
1
b1 h13 + n1 A1 d12 = (30)(20)3 + (600)(25.53)2 = 411.07 × 103 mm 4
12
12
n
2.8571
I 2 = 2 b2 h23 + n2 A2 d 22 =
(30)(40)3 + (3428.5)(4.47) 2 = 525.64 × 103 mm 4
12
12
I = I1 + I 2 = 936.71 × 103 mm 4 = 936.71 × 10−9 m 4
I1 =
M = 1500 N ⋅ m
Stress:
(a)
Aluminum:
σ =−
nM y
I
n = 1,
σa = −
(b)
Steel:
y = 60 − 24.47 = 35.53 mm = 0.03553 m
(1)(1500)(0.03553)
= −56.9 × 106 Pa
−9
936.71 × 10
n = 2.8571,
σs = −
σ a = −56.9 MPa 
y = −24.47 mm = −0.02447 m
(2.8571)(1500) (−0.02447)
= 111.9 × 106 Pa
−9
936.71 × 10
σ s = 111.9 MPa 
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PROBLEM 4.41
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.
SOLUTION
Use wood as the reference material.
For wood,
n =1
For steel,
n = Es / Ew = 29 /1.8 = 16.1111
Transformed section:
421.931
112.278
= 3.758 in.
Yo =
 = wood


 = steel
A, in 2
nA, in 2
72
72
2.5
40.278
yo
6
−0.25
112.278
nA yo , in 3
432
−10.069
421.931
The neutral axis lies 3.758 in. above the wood-steel interface.
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
I1 =
σ =−
M = 450 kip ⋅ in
(a)
Wood:
n = 1,
y = 12 − 3.758 = 8.242 in
σw = −
(b)
Steel:
(1) (450) (8.242)
= −1.979 ksi
1873.77
n = 16.1111,
σs = −
nMy
I
σ w = −1.979 ksi 
y = −3.758 − 0.5 = −4.258 in
(16.1111) (450) (−4.258)
= 16.48 ksi
1873.77
σ s = 16.48 ksi 
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PROBLEM 4.42
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.
SOLUTION
Use wood as the reference material.
For wood,
n =1
For steel,
n=
Es 29 × 106
=
= 16.1111
Ew 1.8 × 106
For C8 × 11.5 channel section,
A = 3.38 in 2 , t w = 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
y , in.
0.571
nAy , in 3
31.091
d, in.
3.216
72
6.22
447.84
2.433
478.93
126.456
Σ
Y0 =
nA, in2
54.456
478.93 in 3
= 3.787 in.
126.456 in 2
d = y0 − Y0
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
n2
1
b2 h23 + n2 A2 d 22 = (6)(12)3 + (72)(2.433)2 = 1290.20 in 4
12
12
4
I = I1 + I 2 = 1874.69 in
I2 =
M = 450 kip ⋅ in
(a)
Wood:
n = 1,
σw = −
(b)
Steel:
n My
I
y = 12 + 0.220 − 3.787 = 8.433 in.
σ =−
(1)(450)(8.433)
= −2.02 ksi
1874.69
n = 16.1111,
σs = −
σ w = −2.02 ksi 
y = −3.787 in.
(16.1111) (450) (−3.787)
= 14.65 ksi
1874.67
σ s = 14.65 ksi 
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PROBLEM 4.43
For the composite beam indicated, determine the radius of curvature caused by
the couple of moment 1500 N ⋅ m.
Beam of Prob. 4.39.
SOLUTION
See solution to Prob. 4.39 for the calculation of I.
I = 852.44 × 10−9 m 4
1
ρ
=
Ea = 70 × 109 Pa
M
1500
=
= 0.02513 m −1
9
EI (70 × 10 )(852.44 × 10−9 )
ρ = 39.8 m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.44
For the composite bar indicated, determine the radius of curvature caused by the
couple of moment 1500 N ⋅ m.
Beam of Prob. 4.40.
SOLUTION
See solution to Prob. 4.40 for calculation of I.
I = 936.71 × 10−9 m 4
1
ρ
=
Ea = 70 × 109 Pa
M
1500
=
= 0.02288 m −1
9
EI (70 × 10 )(936.71 × 10−9 )
ρ = 43.7 m 
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PROBLEM 4.45
For the composite beam indicated, determine the radius of curvature caused by the
couple of moment 450 kip ⋅ in.
Beam of Prob. 4.41.
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
ρ
=
M
450 × 103
=
= 133.42 × 10−6 in −1
6
EI (1.8 × 10 )(1873.77)
ρ = 7495 in. = 625 ft 
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PROBLEM 4.46
For the composite beam indicated, determine the radius of curvature caused by the
couple of moment 450 kip ⋅ in.
Beam of Prob. 4.42.
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
ρ
=
M
450 × 103
=
= 133.36 × 10−6 in −1
6
EI (1.8 × 10 )(1874.69)
ρ = 7499 in. = 625 ft 
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PROBLEM 4.47
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 ⋅
π
π 
d 2 = (4)   (25) 2 = 1.9635 × 103 mm 2
4
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
nMy
σ =−
I
(a)
Steel:
y = −302.45 mm = −0.30245 m
σ =−
(b)
Concrete:
(8.0)(175 × 103 )(−0.30245)
= 212 × 106 Pa
−3
1.9966 × 10
σ = 212 MPa 
y = 177.87 mm = 0.17787 m
σ =−
(1.0)(175 × 103 )(0.17787)
= −15.59 × 106 Pa
1.9966 × 10−3
σ = −15.59 MPa 
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PROBLEM 4.48
Solve Prob. 4.47, assuming that the 300-mm width is increased to
350 mm.
PROBLEM 4.47 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
Ec
25 GPa
As = 4
π
π 
d 2 = (4)   (25)2
4
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
σ =−
I
(a)
Steel:
y = −312.52 mm = −0.31252 m
σ =−
(b)
Concrete:
(8.0)(175 × 103 )(−0.31252)
= 210 × 106 Pa
2.0823 × 10−3
σ = 210 MPa 
y = 167.48 mm = 0.16748 m
σ =−
(1.0)(175 × 103 )(0.16748)
= −14.08 × 106 Pa
−3
2.0823 × 10
σ = −14.08 MPa 
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PROBLEM 4.49
A concrete slab is reinforced by 16-mm-diameter steel rods placed on
180-mm centers as shown. The modulus of elasticity is 20 GPa for
concrete and 200 GPa for steel. Using an allowable stress of 9 MPa for
the concrete and of 120 MPa for the steel, determine the largest bending
moment in a portion of slab 1 m wide.
SOLUTION
n=
Es 200 GPa
=
= 10
20 GPa
Ec
Consider a section 180-mm wide with one steel rod.
As =
π
d2 =
π
(16)2 = 201.06 mm 2
4
4
nAs = 2.0106 × 103 mm 2
Locate the neutral axis:
x
− (100 − x)(2.0106 × 103 ) = 0
2
90 x 2 + 2.0106 × 103 x − 201.06 × 103 = 0
180 x
Solving for x:
−2.0106 × 103 + (2.0106 × 103 ) 2 + (4)(90)(201.06 × 103 )
(2)(90)
x = 37.397 mm 100 − x = 62.603 mm
x=
1
I = (180) x3 + (2.0106 × 103 )(100 − x) 2
3
1
= (180)(37.397)3 + (2.0106 × 103 )(62.603) 2
3
= 11.018 × 106 mm 4 = 11.018 × 10−6 m 4
Concrete:
σI
σ =−
nMy
I
n = 1,
y = 37.397 mm = 0.037397 m,
M=
∴ M=
ny
σ = 9 × 106 Pa
(9 × 106 )(11.018 × 10−6 )
= 2.6516 × 103 N ⋅ m
(1.0)(0.037397)
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PROBLEM 4.49 (Continued)
n = 10,
Steel:
M=
Choose the smaller value.
y = 62.603 mm = 0.062603 m, σ = 120 × 106 Pa
(120 × 106 )(11.018 × 10−6 )
= 2.1120 × 103 N ⋅ m
(10)(0.062603)
M = 2.1120 × 103 N ⋅ m
The above is the allowable positive moment for a 180-mm wide section. For a 1-m = 1000-mm width,
1000
= 5.556
mutiply by
180
M = (5.556)(2.1120 × 103 ) = 11.73 × 103 N ⋅ m
M = 11.73 kN ⋅ m 
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PROBLEM 4.50
A concrete slab is reinforced by 16-mm-diameter steel rods placed on
180-mm centers as shown. The modulus of elasticity is 20 GPa for
concrete and 200 GPa for steel. Using an allowable stress of 9 MPa
for the concrete and of 120 MPa for the steel, determine the largest
allowable positive bending moment in a portion of slab 1 m wide.
Solve Prob. 4.49, assuming that the spacing of the 16-mm-diameter
rods is increased to 225 mm on centers.
SOLUTION
n=
Es 200 GPa
=
= 10
20 GPa
Ea
Consider a section 225-mm wide with one steel rod.
As =
π
d2 =
π
(16)2 = 201.06 mm 2
4
4
nAs = 2.0106 × 103 mm 2
Locate the neutral axis:
x
225 x − (100 − x)(2.0106 × 103 ) = 0
2
112.5 x 2 + 2.0106 x − 201.06 × 103 = 0
Solving for x:
−2.0106 × 103 + (2.0106 × 103 ) 2 + (4)(112.5)(201.06 × 103 )
(2)(112.5)
x = 34.273 mm
100 − x = 65.727
1
I = (225) x3 + 2.0106 × 103 (100 − x) 2
3
1
= (225)(34.273)3 + (2.0106 × 103 )(65.727) 2
3
= 11.705 × 106 mm 4 = 11.705 × 10−6 m 4
x=
|σ | =
Concrete:
nMy
I
n = 1,
M=
∴ M =
σI
ny
y = 34.273 mm = 0.034273 m, σ = 9 × 106 Pa
(9 × 106 )(11.705 × 10−6 )
= 3.0738 × 103 N ⋅ m
(1)(0.034273)
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PROBLEM 4.50 (Continued)
n = 10,
Steel:
M=
Choose the smaller value.
y = 65.727 mm = 0.065727 m, σ = 120 × 106 Pa
(120 × 106 )(11.705 × 10−6 )
= 2.1370 × 103 N ⋅ m
(10)(0.065727)
M = 2.1370 × 103 N ⋅ m
The above is the allowable positive moment for a 225-mm-wide section. For a 1-m = 1000-mm section,
1000
= 4.4444
multiply by
225
M = (4.4444)(2.1370 × 103 ) = 9.50 × 103 N ⋅ m
M = 9.50 kN ⋅ m 
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PROBLEM 4.51
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.
SOLUTION
n=
Es 29 × 10 6
=
= 9.67
Ec
3 × 106
As = 3
2
π
 π  7 
d 2 = (3)    = 1.8040 in 2
4
 4  8 
nAs = 17.438 in 2
x
− (17.438)(14 − x) = 0
2
4 x 2 + 17.438 x − 244.14 = 0
Locate the neutral axis:
8x
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
8x + nAs (14 − x)2 = (8)(5.6326)3 + (17.438)(8.3674) 2 = 1697.45 in 4
3
3
σ =
nMy
I
∴ M=
σI
ny
n = 1.0,
Concrete:
M =
M =
Choose the smaller value.
σ = 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., σ = 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 
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PROBLEM 4.52
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.
SOLUTION
n=
Es
29 × 106
=
= 8.0
Ec
3.625 × 106

π 
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.
1
b1h13 +
12
1
I 2 = b2 x3 =
3
I1 =
A1d12 =
1
(24)(4)3 + (24)(4)(3.150) 2 = 1080.6 in 4
12
1
(12)(1.150)3 = 6.1 in 4
3
I 3 = nA3d32 = (25.133)(12.350) 2 = 3833.3 in 4
I = I1 + I 2 + I 3 = 4920 in 4
σ =−
(a)
Steel:
nMy
I
n = 8.0
y = −12.350 in.
σs = −
(b)
Concrete:
n = 1.0,
σc = −
where M = 100 kip ⋅ ft = 1200 kip ⋅ in.
(8.0)(1200)(−12.350)
4920
σ s = 24.1 ksi 
y = 4 + 1.150 = 5.150 in.
(1.0)(1200)(5.150)
4920
σ c = −1.256 ksi 
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PROBLEM 4.53
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 σ s and σ c .
Show that to achieve a balanced design the distance x from the top of the beam to the
neutral axis must be
d
x=
1+
σ s Ec
σ c Es
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
σc =
I
I
σ s n( d − x )
d
=
=n −n
σc
x
x
σs =
Eσ
d
1 σs
=1+
=1+ c s
x
n σc
E sσ c
x=

d
Ecσ s
1+
E sσ c
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PROBLEM 4.54
For the concrete beam shown, the modulus of elasticity is 3.5 × 106 psi for the concrete and
29 × 106 psi for the steel. Knowing that b = 8 in. and d = 22 in., and using an allowable
stress of 1800 psi for the concrete and 20 ksi 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.)
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 σ s and σ c .
SOLUTION
Es 29 × 10 6
=
= 8.2857
Ec 3.5 × 106
nM ( d − x)
Mx
σs =
σc =
I
I
σ s n( d − x )
d
=
=n −n
σc
x
x
n=
1 σs
1
20 × 103
d
=1+
=1+
⋅
= 2.3410
8.2857 1800
x
n σc
x = 0.42717 d = (0.42717)(22) = 9.398 in.
Locate neutral axis:
(a)
As =
bx
d − x = 22 − 9.398 = 12.602 in .
x
− nAs ( d − x) = 0
2
bx 2
(8)(9.398) 2
=
= 3.3835 in 2
2n(d − x) (2)(8.2857)(12.602)
As = 3.38 in 2 
1
1
I = bx3 + nAs ( d − x) 2 = (8)(9.398)3 + (8.2857)(3.3835)(12.602)2 = 6665.6 in 4
3
3
n My
σI
M=
σ=
I
ny
(b)
Steel:
Concrete:
n = 1.0
y = 9.398 in. σ c = 1800 psi
M=
(1800)(6665.6)
= 1.277 × 106 lb ⋅ in
(1.0)(9.398)
n = 8.2857 | y | = 12.602 in. σ s = 20 × 103 psi
M=
(20 × 103 )(6665.6)
= 1.277 × 106 lb ⋅ in
(8.2857)(12.602)
Note that both values are the same for balanced design.
M = 106.4 kip ⋅ ft 
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PROBLEM 4.55
Five metal strips, each 40 mm wide, are bonded together to form the
composite beam shown. The modulus of elasticity is 210 GPa for the
steel, 105 GPa for the brass, and 70 GPa for the aluminum. Knowing that
the beam is bent about a horizontal axis by a couple of moment 1800 N ⋅
m, 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 = 1 in aluminum.
n = Es / Ea = 210 / 70 = 3 in steel.
n = Eb / Ea = 105 / 70 = 1.5 in brass.
Due to symmetry of both the material arrangement and the geometry, the
neutral axis passes through the center of the steel portion.
For the transformed section,
n1
1
b1h13 + n1 A1d12 = (40)(10)3 + (40)(10)(25) 2 = 253.33 × 103 mm 4
12
12
n2
1.5
3
2
(40)(10)3 + (1.5)(40)(10)(15)2 = 140 × 103 mm 4
I 2 = b2 h2 + n2 A2 d 2 =
12
12
n
3.0
(40)(20)3 = 80 × 103 mm 4
I 3 = 3 b3 h33 =
12
12
I 4 = I 2 = 140 × 103 mm 4
I 5 = I1 = 253.33 × 103 mm 4
I1 =
I=
(a)
= 866.66 × 103 mm 4 = 866.66 × 10−9 m 4
nMy
where M = 1800 N ⋅ m
I
n = 1.0
y = −30 mm = 0.030 m
Aluminum:
σ =−
σa =
(1.0)(1800)(0.030)
= 62.3 × 106 Pa
−9
866.66 × 10
Brass:
σb =
Steel:
σs =
(b)
I
n = 1.5
y = −20 mm = −0.020 m
(1.5)(1800)(0.020)
= 62.3 × 106 Pa
−9
866.66 × 10
n = 3.0
σ b = 62.3 MPa 
y = −10 mm = −0.010 m
(3.0)(1800)(0.010)
= 62.3 × 106 Pa
−9
866.66 × 10
Radius of curvature.
σ a = 62.3 MPa 
1
ρ
=
M
1800
=
= 0.02967 m −1
9
Ea I (70 × 10 )(866.66 × 10−9 )
σ s = 62.3 MPa 
ρ = 33.7 m 
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PROBLEM 4.56
Five metal strips, each of 40 mm wide, are bonded together to form the
composite beam shown. The modulus of elasticity is 210 GPa for the
steel, 105 GPa for the brass, and 70 GPa for the aluminum. Knowing
that the beam is bent about a horizontal axis by a couple of moment
1800 N ⋅ m, 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 = 1.0 in aluminum.
n = Es / Ea = 210 / 70 = 3 in steel.
n = Eb / Ea = 105 / 70 = 1.5 in brass.
Due to symmetry of both the material arrangement and the geometry, the neutral axis
passes through the center of the aluminum portion.
For the transformed section,
n1
1.5
b1h13 + n1 A1d12 =
(40)(10)3 + (1.5)(40)(10)(25) 2 = 380 × 103 mm 4
12
12
n2
3.0
3
2
I 2 = b2 h2 + n2 A2 d 2 =
(40)(10)3 + (3.0)(40)(10)(15)2 = 280 × 103 mm 4
12
12
n3
1.0
I 3 = b3 h33 =
(40)(20)3 = 26.67 × 103 mm 4
12
12
I 4 = I 2 = 280 × 103 mm 4
I 5 = I1 = 380 × 103 mm 4
I1 =
I=
(a)
σ =−
nMy
I
Aluminum:
σa =
Steel:
σs =
(b)
where
6
mm 4 = 1.3467 × 10−6 m 4
M = 1800 N ⋅ m
n = 1,
y = −10 mm = −0.010 m
(1.0)(1800)(0.010)
= 13.37 × 106 Pa
1.3467 × 10−6
Brass:
σb =
 I = 1.3467 × 10
n = 1.5,
y = −30 mm = −0.030 m
(1.5)(1800)(0.030)
= 60.1 × 106 Pa
1.3467 × 10−6
n = 3.0,
σ b = 60.1 MPa 
y = −20 mm = −0.020 m
(3.0)(1800)(0.020)
= 80.1 × 106 Pa
1.3467 × 10−6
Radius of curvature.
σ a = 13.37 MPa 
1
ρ
=
M
1800
=
= 0.01909 m −1
9
−6
Ea I (70 × 10 )(1.3467 × 10 )
σ s = 80.1 MPa 
ρ = 52.4 m 
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PROBLEM 4.57
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,
yo =
r = 0.8 in.
π
A=
2
4r
(4)(0.8)
=
= 0.33953 in.
3π
3π
r 2 = 1.00531 in 2 ,
I base =
π
8
r 4 = 0.160850 in 4
I = I base − Ayo2 = 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
yo , in.

1.00531
1.50796
0.33953
0.51200

1.00531
1.00531
−0.33953
−0.34133
2.51327
Σ
nAyo , in 3
Yo =
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.
σ =−
(b)
Aluminium:
n = 1.0,
nMy
(1.5)(8)(0.73209)
=−
I
0.39054
σ = −22.5 ksi 
y = −0.8 − 0.06791 = −0.86791 in.
σ =−
nMy
(1.0)(8)(−0.86791)
=−
I
0.39054
σ = 17.78 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.58
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.
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
π
4
π
(r
4
o
(r
4
4
o
π 
− ri4 = (2.857)   (164 − 104 ) = 124.62 × 103 mm 4
4
)
π 
− 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
σa =
na Mc (1.0)(500)(0.019)
=
= 54.1 × 106 Pa
−9
I
175.5 × 10
σ a = 54.1 MPa 
(b)
Steel:
c = 16 mm = 0.016 m
σs =
ns Mc (2.857)(500)(0.016)
=
= 130.2 × 106 Pa
−9
I
175.5 × 10
σ s = 130.2 MPa 
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PROBLEM 4.59
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
σ =−
(b)
Compressive stress:
n = 1,
σ =−
y = −58.579 mm = −0.058579 m
nMy
(0.5)(600)(−0.058579)
=−
= 6.15 × 106 Pa
I
2.8595 × 10−6
σ t = 6.15 MPa 
y = 41.421 mm = 0.041421 m
nMy
(1.0)(600)(0.041421)
=−
= −8.69 × 106 Pa
I
2.8595 × 10−6
σ c = −8.69 MPa 
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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
ρ
=
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
h
x=
I trans
1
ρ
n +1
nh
h−x=
n +1
3
 h  1 3 
n 3 1
n   3
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 =
n
(
)
n +1
2
bh3
1 3
bh
12
Er I = Et I trans
Er =
=
Et I trans 12
= 3 × Et ×
I
bh
3
(
4 Et Ec /Et
=
) (
Ec /Et + 1
2
n
(
)
n +1
4 Et Ec
Ec + Et
)
2
bh3
2
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PROBLEM 4.61
Semicircular grooves of radius r must be milled as shown in the sides
of a steel member. Using an allowable stress of 60 MPa, determine the
largest bending moment that can be applied to the member when
(a) r = 9 mm, (b) r = 18 mm.
SOLUTION
(a)
d = D − 2r = 108 − (2)(9) = 90 mm
D 108
=
= 1.20
d
90
From Fig. 4.32,
r
9
=
= 0.1
d 90
K = 2.07
1
(18)(90)3
12
= 1.0935 × 106 mm 4
I=
= 1.0935 × 10−6 m 4
1
d = 45 mm = 0.045 m
2
KMc
σ=
I
c=
M=
(b)
σI
Kc
=
(60 × 106 )(1.0935 × 10−6 )
= 704
(2.07)(0.045)
M = 704 N ⋅ m 
d = 108 − (2)(18) = 72 mm
D 108
r 18
=
= 1.5
=
= 0.25
d
72
d 72
1
c = d = 36 mm = 0.036 m
2
From Fig. 4.32,
K = 1.61
1
I = (18)(72)3 = 559.87 × 103 mm 4 = 559.87 × 10−9 m 4
12
M=
σI
Kc
=
(60 × 106 )(559.87 × 10−9 )
= 580
(1.61)(0.036)
M = 580 N ⋅ m 
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PROBLEM 4.62
Semicircular grooves of radius r must be milled as shown in the sides
of a steel member. Knowing that M = 450 N ⋅ m, determine the
maximum stress in the member when the radius r of the semicircular
grooves
is
(a) r = 9 mm, (b) r = 18 mm.
SOLUTION
(a)
d = D − 2r = 108 − (2)(9) = 90 mm
D 108
=
= 1.20
d
90
From Fig. 4.32,
r
9
=
= 0.1
d 90
K = 2.07
1 3 1
bh = (18)(90)3
12
12
= 1.0935 × 106 mm 4
I=
= 1.0935 × 10−6 m 4
1
c = d = 45 mm = 0.045 m
2
KMc (2.07)(450)(0.045)
=
I
1.0935 × 10−6
= 38.3 × 106 Pa
σ max =
(b)
σ max = 38.3 MPa 
d = D − 2r = 108 − (2)(18) = 72 mm
D 108
=
= 1.5
d
72
From Fig. 4.32,
r 18
=
= 0.25
d 72
K = 1.61
1
d = 72 mm = 0.036 m
2
1
I = (18)(72)3 = 559.87 × 103 mm 4 = 559.87 × 10−9 m 4
12
c=
σ max =
KMc (1.61)(450)(0.036)
=
= 46.6 × 106 Pa
I
559.87 × 10−9
σ max = 46.6 MPa 
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PROBLEM 4.63
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.
SOLUTION
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
I =
D 80 mm
=
= 2.00
40 mm
d
(a)
r
8 mm
=
= 0.2
d
40 mm
σ max = K
Mc
I
From Fig. 4.31,
M =
σ max I
Kc
=
K = 1.50
(90 × 106 )(42.667 × 10−9 )
(1.50)(0.020)
M = 128 N ⋅ m 
(b)
r
12 mm
=
= 0.3
d
40 mm
From Fig. 4.31,
M =
K = 1.35
(90 × 106 )(42.667 × 10−9 )
(1.35)(0.020)
M = 142 N ⋅ m 
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PROBLEM 4.64
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.
SOLUTION
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
I =
D 80 mm
=
= 2.00
40 mm
d
(a)
r
4 mm
=
= 0.10
d
40 mm
σ max = K
(b)
K = 1.87
Mc (1.87)(250)(0.020)
=
= 219 × 106 Pa
I
42.667 × 10−9
r
8 mm
=
= 0.20
d
40 mm
σ max = K
From Fig. 4.31,
From Fig. 4.31,
σ max = 219 MPa 
K = 1.50
Mc
(1.50)(250)(0.020)
=
= 176 × 106 Pa
I
42.667 × 10−9
σ max = 176 MPa 
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PROBLEM 4.65
The allowable stress used in the design of a steel bar is
12 ksi. 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 = 34 in. as shown in
Fig. a, (b) if the bar is redesigned by removing the
material above and below the dashed lines as shown in
Fig. b.
SOLUTION
7
in. = 0.875 in.
8
D = 7.5 in.
3
in. = 0.75 in.
4
d = 5 in.
r
0.75
=
= 0.15
d
5
D 7.5
=
= 1.5
d
5
t =
Dimensions:
Stress concentration factors:
Figs. 4.32 and 4.31
Configuration (a):
K = 1.92
Configuration (b):
K = 1.58
I =
Moment of inertia:
c=
1
d = 2.5 in.
2
σm =
KMc
I
r =
1 3
1
td =
(0.875)(5)3 = 9.115 in 4
12
12
σ m = 12 ksi
M =
Iσ m
Kc
(a)
M =
(9.115)(12)
(1.92)(2.5)
M = 22.8 kip ⋅ in 
(b)
M =
(9.115)(12)
(1.58)(2.5)
M = 27.7 kip ⋅ in 
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PROBLEM 4.66
A couple of moment M = 20 kip ⋅ in. 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 = 12 in., as shown in Fig. a, (b) if the bar
is redesigned by removing the material above and below the
dashed line as shown in Fig. b.
SOLUTION
7
in. = 0.875 in.
8
D = 7.5 in.
1
in. = 0.5 in.
2
d = 5 in.
r
0.5
=
= 0.10
d
5
D 7.5
=
= 1.5
d
5
t =
Dimensions:
Stress concentration factors:
Figs. 4.32 and 4.31
Configuration (a):
K = 2.22
Configuration (b):
K = 1.80
I =
Moment of inertia:
c=
1
d = 2.5 in.
2
Maximum stress:
r =
1 3
1
td =
(0.875)(5)3 = 9.115 in 4
12
12
M = 20 kip ⋅ in
σm =
KMc
I
(a)
σm =
(2.22)(20)(2.5)
9.115
σ m = 12.2 ksi 
(b)
σm =
(1.80)(20)(2.5)
9.115
σ m = 9.9 ksi 
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PROBLEM 4.67
The prismatic bar shown is made of a steel that is assumed to be
elastoplastic with σ 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.
SOLUTION
(a)
I=
1 3 1
bh = (12)(8)3 = 512 mm 4
12
12
= 512 × 10−12 m 4
c=
1
h = 4 mm = 0.004 m
2
(300 × 106 )(512 × 10−12 )
0.004
c
= 38.4 N ⋅ m
MY =
(b)
yY =
1
(4) = 2 mm
2
σY I
=
yY 2
= = 0.5
c
4
 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 
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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 σ 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.
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
(300 × 106 )(1.152 × 10−9 )
c
0.006
= 57.6 N ⋅ m
MY =
(b)
yY =
1
(4) = 2 mm
2
σY I
=
M Y = 57.6 N ⋅ m 
yY 2 1
= =
c
6 3
M=
 1  y 2 
3
M Y 1 −  Y  
2
 3  c  
 1  1 2 
3
= (57.6) 1 −   
2
 3  3  
= 83.2 N ⋅ m
M = 83.2 N ⋅ m 
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PROBLEM 4.69
The prismatic bar shown, made of a steel that is assumed to be elastoplastic
with E = 29 × 106 psi and σ Y = 36 ksi, is subjected to a couple of
1350 lb ⋅ in. parallel to the z axis. Determine (a) the thickness of the elastic
core, (b) the radius of curvature of the bar.
SOLUTION
3
(a)
I=
1  1  5 
4
−3
   = 10.1725 × 10 in
12  2  8 
c=
15
  = 0.3125 in.
28
(36 × 103 )(10.1725 × 10−3 )
0.3125
c
= 1171.872 lb ⋅ in
MY =
σY I
=
3
M = MY
2
 1  y 3 
1 −  Y  
 3  c  
yY
(2)(1350)
M
= 3−2
= 3−
= 0.83426
1171.872
c
MY
yY = (0.83426)(0.3125) = 0.26071 in.
Thickness of elastic core:
(b)
yY = ε Y ρ =
σY
2 yY = 0.521 in. 
ρ
E
y E (0.26071)(29 × 106 )
= 210.02 in.
ρ= Y =
σY
36 × 103
ρ = 17.50 ft 
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PROBLEM 4.70
Solve Prob. 4.69, assuming that the 1350-lb ⋅ in. couple is parallel to the
y axis.
PROBLEM 4.69 The prismatic bar shown, made of a steel that is assumed
to be elastoplastic with E = 29 × 106 psi and σ Y = 36 ksi, is subjected to a
couple of 1350 lb ⋅ in parallel to the z axis. Determine (a) the thickness of
the elastic core, (b) the radius of curvature of the bar.
SOLUTION
3
(a)
I=
1  5  1 
−3 4
   = 6.5104 × 10 in
12  8  2 
 1  1 
c =    = 0.25 in.
 2  2 
(36 × 103 )(6.5104 × 10−3 )
0.25
c
= 937.5 lb ⋅ in
MY =
σY I
=
2
3  1  yY  
M = 1 − 

2  3  c  
yY
(2)(1350)
M
= 3−2
= 3−
= 0.34641
937.5
c
MY
yY = (0.34641)(0.25) = 0.086603 in.
Thickness of elastic core:
(b)
yY = ε Y ρ =
σY
2 yY = 0.1732 in. 
ρ
E
y E (0.086603)(29 × 106 )
= 69.763 in.
ρ= Y =
σY
36 × 103
ρ = 5.81 ft 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.71
A bar of rectangular cross section shown is made of a steel that is
assumed to be elastoplastic with E = 200 GPa and σ Y = 300 MPa.
Determine the bending moment M for which (a) yield first occurs,
(b) the plastic zones at the top and bottom of the bar are 12 mm thick.
SOLUTION
1
(30)(40)3 = 160 × 103 mm 4 = 160 × 10−9 m 4
12
1
c = (40) = 20 mm = 0.020 m
2
I =
(a)
First yielding:
MY =
σ =
Mc
= σY
I
Iσ Y
(160)(10−9 )(300 × 106 )
=
= 2400 N ⋅ m
c
0.020
M Y = 2.40 kN ⋅ m 
(b)
Plastic zones are 12 mm thick:
yY = 20 mm − 12 mm = 8 mm
yY
8 mm
=
= 0.4
c
20 mm
M =
=
2

3
1 y  
M Y 1 −  Y  
2
3  c  

3
1


(2400) 1 − (0.4)2  = 3408 N ⋅ m
2
3


M = 3.41 kN ⋅ m 
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PROBLEM 4.72
Bar AB is made of a steel that is assumed to be elastoplastic with
E = 200 GPa and σ Y = 240 MPa. Determine the bending moment M
for which the radius of curvature of the bar will be (a) 18 m, (b) 9 m.
SOLUTION
1
(30)(40)3 = 160 × 103 mm 4 = 160 × 10−9 m 4
12
1
c = (40) = 20 mm = 0.020 m
2
I =
Iσ Y
(160 × 10−9 )(240 × 106 )
=
= 1920 N ⋅ m
c
0.020
M
1
1920
= Y =
= 0.0600 m −1
EI
ρY
(200 × 109 )(160 × 10−9 )
MY =
(a)
1
ρ = 18 m:
ρ
= 0.05556 m −1 < 0.0600 m −1
The bar is fully elastic.
M =
EI
ρ
=
(200 × 109 )(160 × 10−9 )
= 1778 N ⋅ m
18
M = 1.778 kN ⋅ m 
(b)
1
ρ = 9 m:
ρ
= 0.11111 m −1 > 0.0600 m −1
The bar is partially plastic.
σY =
EyY
ρ
yY =
ρσ Y
E
(9)(240 × 106 )
= 0.0108 m = 10.8 mm
200 × 109
yY
10.8 mm
=
= 0.54
c
20 mm
yY =
M =
3
MY
2
2

1 y   3
1


1 −  Y   = (1920) 1 − (0.54)2  = 2600 N ⋅ m
c
3
2
3
  



M = 2.60 kN ⋅ m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.73
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E = 200 GPa and σ Y = 240 MPa. For bending about the zaxis, 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.
SOLUTION
(a)
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
σ I (240 × 106 )(3.645 × 10−6 )
MY = Y =
= 19.44 × 10 N ⋅ m
c
0.045
I=
M Y = 19.44 kN ⋅ m 
R1 = σ Y A1 = (240 × 106 )(0.060)(0.030)
= 432 × 103 N
y1 = 15 mm + 15 mm = 0.030 m
1
1
R2 = σ Y A2 =   (240 × 106 )(0.060)(0.015)
2
2
= 108 × 103 N
2
y2 = (15 mm) = 10 mm = 0.010 m
3
(b)
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 
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PROBLEM 4.74
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E = 200 GPa and σ 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.
SOLUTION
(a)
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
I rect =
= 3.5775 × 10−6 mm 4
1
h = 45 mm = 0.045 m
2
σ I (240 × 106 )(3.5775 × 10−6 )
MY = Y =
0.045
c
3
= 19.08 × 10 N ⋅ m
c=
M Y = 19.08 kN ⋅ m 
R1 = σ Y A1 = (240 × 106 )(0.060)(0.030) = 432 × 103 N
y1 = 15 mm + 15 mm = 30 mm = 0.030 m
1
1
σ 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 
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PROBLEM 4.75
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E = 29 × 106 psi and σ 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.
SOLUTION
(a)
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
I1 =
I = I1 + I 2 + I 3 = 357.75 in 4
c = 4.5 in.
MY =
σY I
c
=
(42)(357.75)
4.5
M Y = 3339 kip ⋅ in 
R1 = σ Y A1 = (42)(6)(3) = 756 kip
y1 = 1.5 + 1.5 = 3 in.
1
1
R2 = σ Y A2 = (42)(3)(1.5)
2
2
= 94.5 kip
2
y2 = (1.5) = 1.0 in.
3
(b)
M = 2( R1 y1 + R2 y2 ) = 2[(756)(3) + (94.5)(1.0)]
M = 4725 kip ⋅ in 
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PROBLEM 4.76
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E = 29 × 106 psi and σ 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.
SOLUTION
(a)
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
I1 =
I = I1 + I 2 + I 3 = 188.5 in 4
c = 4.5 in.
σ I (42)(188.5)
MY = Y =
c
4.5
M Y = 1759 kip ⋅ in 
R1 = σ Y A1 = (42)(3)(3) = 378 kip
y1 = 1.5 + 1.5 = 3.0 in.
1
1
R2 = σ 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)]
M = 2646 kip ⋅ in 
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PROBLEM 4.77
For the beam indicated (of Prob. 4.73), determine (a) the fully plastic
moment M p , (b) the shape factor of the cross section.
SOLUTION
From Problem 4.73,
E = 200 GPa and σ Y = 240 MPa.
A1 = (60)(45) = 2700 mm 2
= 2700 × 10−6 m 2
R = σ Y A1
= (240 × 106 )(2700 × 10−6 )
= 648 × 103 N
d = 45 mm = 0.045 m
(a)
(b)
M p = Rd = (648 × 103 )(0.045) = 29.16 × 103 N ⋅ m
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
I=
MY =
σY 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 
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PROBLEM 4.78
For the beam indicated (of Prob. 4.74), determine (a) the fully plastic moment
M p , (b) the shape factor of the cross section.
SOLUTION
From Problem 4.74,
(a)
E = 200 GPa and σ Y = 240 MPa.
R1 = σ Y A1
= (240 × 106 )(0.060)(0.030)
= 432 × 103 N
y1 = 15 mm + 15 mm = 30 mm
= 0.030 m
R2 = σ 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 
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
I rect =
= 3.5775 × 10−9 m 4
1
h = 45 mm = 0.045 m
2
σ 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
c=
k = 1.443 
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PROBLEM 4.79
For the beam indicated (of Prob. 4.75), determine (a) the fully plastic moment
M p , (b) the shape factor of the cross section.
SOLUTION
From Problem 4.75,
(a)
E = 29 × 106 and σ Y = 42 ksi.
R1 = σ Y A1 = (42)(6)(3) = 756 kip
y1 = 1.5 + 1.5 = 3.0 in.
R2 = σ 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 = 4819.5 kip ⋅ in 
1
1
b1h13 + A1d12 = (6)(3)3 + (6)(3)(3)2 = 175.5 in 4
12
12
1
1
I 2 = b2 h23 = (3)(3)3 = 6.75 in 4
12
12
I 3 = I1 = 175.5 in 4
I1 =
I = I1 + I 2 + I 3 = 357.75 in 4
c = 4.5 in.
σY I
(42)(357.75)
= 3339 kip ⋅ in
4.5
4819.5
=
k=
MY
3339
MY =
c
Mp
=
k = 1.443 
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PROBLEM 4.80
For the beam indicated (of Prob. 4.76), determine (a) the fully plastic moment
M p , (b) the shape factor of the cross section.
SOLUTION
From Problem 4.76,
(a)
E = 29 × 106 psi and σ Y = 42 ksi.
R1 = σ Y A1 = (42)(3)(3) = 378 kip
y1 = 1.5 + 1.5 = 3.0 in.
R2 = σ 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 = 2835 kip ⋅ in 
1
1
b1h13 + A1d12 = (3)(3)3 + (3)(3)(3)2 = 87.75 in 4
12
12
1
1
3
I 2 = b2 h2 = (6)(3)3 = 13.5 in 4
12
12
I 3 = I1 = 87.75 in 4
I1 =
I = I1 + I 2 + I 3 = 188.5 in 4
c = 4.5 in.
σY I
(42)(188.5)
= 1759.3 kip ⋅ in
c
4.5
Mp
2835
k=
=
MY 1759.3
MY =
=
k = 1.611 
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PROBLEM 4.81
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:
π
2
r 2;
r =
4r
3π
R = σY A
Resultant force on semicircular section:
Resultant moment on entire cross section:
M p = 2 Rr =
Data:
4
σY r3
3
σ 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 
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PROBLEM 4.82
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
Total area:
A = (50)(90) − (30)(30) = 3600 mm 2
1
A = 1800 mm 2
2
1 A
1800
x= 2 =
= 36 mm
b
50
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
A3 = (20)(30) = 600 mm 2 ,
y3 = 29 mm,
A3 y3 = 17.4 × 103 mm3
A4 = (50)(10) = 500 mm 2 ,
y4 = 49 mm,
A4 y4 = 24.5 × 103 mm3
A1 y1 + A2 y2 + A3 y3 + A4 y4 = 79.2 × 103 mm3 = 79.2 × 10−6 m3
M p = σ Y ΣAi yi = (240 × 106 )(79.2 × 10−6 ) = 19.008 × 103 N ⋅ m
M p = 19.01 kN ⋅ m 
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PROBLEM 4.83
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
Total area:
A=
1
(30)(36) = 540 mm 2
2
By similar triangles,
b 30
=
y 36
Since
A1 =
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
y=
A3 = A − A1 − A2 = 46.324 mm 2 = 46.324 × 10−6 m 2
Ri = σ 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 
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PROBLEM 4.84
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
 π  4c   π  4c 
=  c12  1  −  c22  2 
 2  3π   2  3π 
3
2 3
=
c1 − c2
3
(
A2 y2 = A1 y1 =
)
(
2 3
c1 − c 22
3
)
M p = σ Y ( A1 y1 + A2 y2 ) =
Data:
(
4
σ Y c13 − c32
3
)
σ Y = 240 MPa = 240 × 106 Pa
c1 = 60 mm = 0.060 m
c2 = 40 mm = 0.040 m
4
(240 × 106 )(0.0603 − 0.0403 )
3
= 48.64 × 103 N ⋅ m
Mp =
M p = 48.6 kN ⋅ m 
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PROBLEM 4.85
Determine the plastic moment M p of the cross section shown, assuming the steel
to be elastoplastic with a yield strength of 36 ksi.
SOLUTION
A = (1.8)(0.4) + (0.6)(1.2) = 1.44 in 2
Total area:
1
A = 0.72 in 2
2
1
A 0.72
x= 2 =
= 1.2 in.
b
0.6
Neutral axis lies 1.2 in. below the top.
1
1
A = 0.72 in 2 , y1 = (1.2) = 0.6 in.
2
2
1
1
2
A2 = A = 0.72 in , y2 = (0.4) = 0.2 in.
2
2
M p = σ Y ( A1 y1 + A2 y2 )
A1 =
= (36)[(0.72)(0.6) + (0.72)(0.2)] = 20.7 kip ⋅ in
M p = 20.7 kip ⋅ in 
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PROBLEM 4.86
Determine the plastic moment M p of the cross section shown, assuming the steel to
be elastoplastic with a yield strength of 36 ksi.
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,
A1 y1 = 1.50 in 3
A2 = 0.25 in 2 ,
y2 = 0.25,
A2 y2 = 0.0625 in 3
A3 = 1.25 in 2 ,
y3 = 1.25,
A3 y3 = 1.5625 in 3
A4 = 1.00 in 2 ,
y4 = 2.75,
A4 y4 = 2.75 in 3
M p = σ Y ( A1 y1 + A2 y2 + A3 y3 + A4 y4 )
= (36)(1.50 + 0.0625 + 1.5625 + 2.75)
M p = 212 kip ⋅ in 
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PROBLEM 4.87
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 = 45 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
σ′ =
Mp c
M max y
=
I
I
UNLOADING
LOADING
σ′ =
at y = c = 45 mm
RESIDUAL STRESSES
(29.16 × 103 )(0.045)
= 360 × 106 Pa
−6
3.645 × 10
σ res = σ ′ − σ Y = 360 × 106 − 240 × 106
= 120 × 106 Pa
σ res = 120.0 MPa 
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PROBLEM 4.88
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 = 45 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
c = 0.045 m
σ′ =
Mp c
M max y
=
I
I
σ′ =
(27.54 × 103 )(0.045)
= 346.4 × 106 Pa
−6
3.5775 × 10
UNLOADING
at
y =c
RESIDUAL STRESSES
σ res = σ ′ − σ Y = 346.4 × 106 − 240 × 106 = 106.4 × 106 Pa
σ res = 106.4 MPa 
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PROBLEM 4.89
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 = 4.5in. , (b) the points
where the residual stress is zero, (c) the radius of curvature corresponding
to the permanent deformation of the bar.
SOLUTION
See solution to Problem 4.75 for bending couple and stress distribution.
M = 4725 kip ⋅ in
σ Y = 42 ksi
(a)
yY = 1.5 in.
I = 357.75 in 4
c = 4.5 in.
Mc (4725)(4.5)
=
= 59.43 ksi
I
357.75
MyY
(4725)(1.5)
σ ′′ =
=
= 19.81 ksi
I
357.75
At y = c, σ res = σ ′ − σ Y = 59.43 − 42 = 17.43 ksi
σ′ =
At
σ res = 0 ∴
y0 =
(c)
At

y = yY , σ res = σ ′′ − σ Y = 19.81 − 42 = −22.19 ksi
UNLOADING
LOADING
(b)
E = 29 × 106 psi = 29 × 103 ksi
RESIDUAL STRESSES
My0
− σY = 0
I
Iσ Y
(357.75)(42)
=
= 3.18 in.
M
4725
Answer:
y0 = −3.18in., 0, 3.18 in. 
y = yY , σ res = −22.19 ksi
σ =−
Ey
ρ
∴ ρ =−
Ey
σ
=
(29 × 103 )(1.5)
= 1960 in.
22.19
ρ = 163.4 ft. 
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PROBLEM 4.90
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 = 4.5in., (b) the points
where the residual stress is zero, (c) the radius of curvature corresponding
to the permanent deformation of the bar.
SOLUTION
See solution to Problem 4.76 for bending couple and stress distribution during loading.
M = 2646 kip ⋅ in
I = 188.5 in 4
σ Y = 42 ksi
(a)
(c)
E = 29 × 106 psi = 29 × 103 ksi
c = 4.5 in.
Mc (2646)(4.5)
=
= 63.17 ksi
I
188.5
MyY
(2646)(1.5)
σ ′′ =
=
= 21.06 ksi
I
188.5
σ′ =
At
y = c, σ res = σ ′ − σ Y = 63.17 − 42 = 21.17 ksi
At
y = yY , σ res = σ ′′ − σ Y = 21.06 − 42
σ res = −20.94 ksi 
UNLOADING
LOADING
(b)
yY = 1.5in.
My0
= σY
I
Iσ Y
(188.5)(42)
y0 =
=
= 2.992 in.
M
2646
RESIDUAL STRESSES
σ res = 0 ∴
At
y = yY ,
σ =−
Ey
ρ
Answer:
y0 = −2.992 in.,
0,
2.992 in. 
σ res = −20.94 ksi
∴ ρ =−
Ey
σ
=
(29 × 103 )(1.5)
= 2077 in.
20.94
ρ = 173.1 ft 
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PROBLEM 4.91
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.
SOLUTION
See solution to Problem 4.73 for bending couple and stress distribution during loading:
M = 28.08 × 103 N ⋅ m
I = 3.645 × 10−6 m 4
σ Y = 240 MPa
(a)
E = 200 GPa
c = 0.045 m
σ′ =
Mc (28.08 × 103 )(0.045)
=
= 346.7 × 106 Pa = 346.7 MPa
I
3.645 × 10−6
σ ′′ =
M yY
(28.08 × 103 )(0.015)
=
= 115.6 × 106 Pa = 115.6 MPa
−6
I
3.645 × 10
At y = c, σ res = σ ′ − σ Y = 346.7 − 240
σ res = 106.7 MPa 
At y = yY , σ res = σ ′′ − σ Y = 115.6 − 240
σ res = −124.4 MPa
UNLOADING
LOADING
(b)
yY = 15 mm = 0.015 m
σ res = 0 ∴
y0 =
RESIDUAL STRESSES
M y0
− σY = 0
I
Iσ Y
(3.645 × 10−6 )(240 × 106 )
=
= 31.15 × 10−3 m = 31.15 mm
M
28.08 × 103
Answer: y0 = −31.15 mm, 0, 31.15 mm 
(c)
At
y = yY , σ res = −124.4 × 106 Pa
σ =−
Ey
ρ
∴ ρ =−
Ey
σ
=
(200 × 109 )(0.015)
−124.4 × 106
ρ = 24.1 m 
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PROBLEM 4.92
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E = 29 × 106 psi and σ Y = 42 ksi . A bending couple is
applied to the beam about 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.
SOLUTION
See solution to Problem 4.76 for bending couple and stress distribution during loading.
M = 406 kip ⋅ in
σ Y = 42 ksi
(a)
yY = 1.0 in.
I = 14.6667 in 4
E = 29 × 106 psi = 29 × 103 ksi
c = 2 in.
Mc (406)(2)
=
= 55.36 ksi
I
14.6667
MyY
(406)(1.0)
σ ′′ =
=
= 27.68 ksi
I
14.6667
σ′ =
σ res = σ ′ − σ Y = 55.36 − 42
At y = c,
At y = yY , σ res = σ ′′ − σ Y = 27.68 − 42
UNLOADING
LOADING
(b)
σ res = 0 ∴
y0 =
(c)
σ =−
Ey
ρ
σ res = −14.32 ksi 
RESIDUAL STRESSES
M y0
− σY = 0
I
Iσ Y
(14.6667)(42)
=
= 1.517 in.
M
406
At y = yY ,
σ res = 13.36 ksi 
Answer: y0 = −1.517 in., 0, 1.517 in. 
σ res = −14.32 ksi
∴ ρ =−
Ey
σ
=
(29 × 103 )(1.0)
= 2025 in.
14.32
ρ = 168.8 ft 
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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 ρ R .
Denoting by ρY the radius of curvature of the bar at the onset of yield, show that the radii of curvature satisfy
the following relation:
1
ρR
2 

1
3 ρ 
1  ρ  
1 − 
= 1 −
 
ρ
2 ρY 
3  ρY   



SOLUTION
1
ρY
Let m denote
=
MY
,
EI
M =

3
1 ρ2 
M Y 1 −
,
2
3 ρY2 

M
:
MY
M
3
ρ2 
ρ2
= 1 − 2  ∴
= 3−2m
MY
2
ρY 
ρY2
1
1 M
1 mM Y
1
m
=
−
=
−
=
−
ρR
ρ EI
ρ
EI
ρ ρY
m=
=
1
ρ  1  3 ρ 
1 ρ 2  
m
−
=
−
−
1
1
1





ρ
ρY  ρ  2 ρY 
3 ρY2  

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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 ρY , 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)
1
ρY
=
m=
MY
,
EI
M =

3
1 ρ2 
M Y 1 −

2
3 ρY2 

M
3
1 ρ2 
= 1 −

MY
2
3 ρY2 
Let m =
M
= 1.25
MY
ρ
= 3 − 2m = 0.70711
ρY
ρ = 0.70711ρY 
(b)
1
ρR
=
=
1
ρ
−
M
1 mM Y
1
m
1
1.25
=
−
=
−
=
−
EI
EI
0.70711ρY
ρ
ρ ρY
ρY
0.16421
ρY
ρ R = 6.09ρY 
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PROBLEM 4.95
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.
SOLUTION
M =

3
1 ρ2 
M Y 1 −

2
3 ρY2 


1 ρ 2σ Y2 
1 −

3 E 2c 2 

=
3 σY I
2 c
=
3 σ Y b(2c)3 
1 ρ 2σ Y2 
1 −

2 12c 
3 E 2c 2 

1 ρ 2σ Y2 
= σ Y bc 2 1 −

3 E 2c 2 

(a)

ρ 2σ Y2 
bc 2σ y 1 −
=M
3E 2c 2 

Cubic equation for σ Y
Data:
E = 200 × 109 Pa
M = 420 N ⋅ m
ρ = 2.4 m
b = 20 mm = 0.020 m
c=
1
h = 8 mm = 0.008 m
2
(1.28 × 10−6 ) σ Y 1 − 750 × 10−21σ Y2  = 350
σ Y 1 − 750 × 10−21σ Y2  = 273.44 × 106
Solving by trial,
(b)
yY =
σY ρ
E
=
σ Y = 292 × 106 Pa
σ Y = 292 MPa 
(292 × 106 )(2.4)
= 3.504 × 10−3 m = 3.504 mm
200 × 109
thickness of elastic core = 2 yY = 7.01 mm 
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PROBLEM 4.96
The prismatic bar AB is made of an aluminum alloy for
which the tensile stress-strain diagram is as shown.
Assuming that the σ -ε 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 σ versus y and use an approximate method
of integration.)
SOLUTION
(a)
σ m = 250 MPa = 250 × 106 Pa
ε m = 0.0064 from curve
1
h = 30 mm = 0.030 m
2
b = 40 mm = 0.040 m
c=
1
ρ
(b)
=
ε m 0.0064
=
= 0.21333 m −1
c
0.030
ρ = 4.69 m 
ε = −ε m
Strain distribution:
M = −
Bending couple:
y
= −ε mu
c
c
−c
u =
where
c
0
y
ε
1
yσ bdy = 2b  y |σ | d y = 2bc 2  0 u |σ | du = 2bc 2 J
1
where the integral J is given by  0 u |σ | du
Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is
J =
Δu
Σwu |σ |
3
where w is a weighting factor.
Using Δu = 0.25, we get the values given in the table below:
u
0
0.25
0.5
0.75
1.00
|ε |
0
0.0016
0.0032
0.0048
0.0064
|σ |, (MPa)
0
110
180
225
250
u |σ |, (MPa)
0
27.5
90
168.75
250
w
1
4
2
4
1
wu |σ |, (MPa)
0
110
180
675
250
1215
← Σwu |σ |
(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
J =
M = 7.29 kN ⋅ m 
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PROBLEM 4.97
The prismatic bar AB is made of a bronze alloy for which the tensile
stress-strain diagram is as shown. Assuming that the σ − ε 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.)
SOLUTION
(a)
ρ = 100 in., b = 0.8 in., c = 0.6 in.
εm =
(b)
c
ρ
=
0.6
= 0.006
100
Strain distribution:
Bending couple:
From the curve,
ε = −ε m
M = −
y
= −ε mu
c
c
−c
where u =
c
0
σ m = 43 ksi 
y
ε
1
y σ bdy = 2b  y |σ | dy = 2bc 2  0 u |σ | du = 2bc 2 J
1
where the integral J is given by  0 u |σ | du
Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is
J =
Δu
Σwu|σ |
3
where w is a weighting factor.
Using Δu = 0.25, we get the values given the table below:
u
|ε |
0
0.25
0.5
0.75
1.00
0
0.0015
0.003
0.0045
0.006
|σ |, ksi
0
25
36
40
43
u |σ |, ksi
0
6.25
18
30
43
J =
w
1
4
2
4
1
wu |σ |, ksi
0
25
36
120
43
224
← Σwu |σ |
(0.25) (224)
= 18.67 ksi
3
M = (2)(0.8)(0.6)2 (18.67)
M = 10.75 kip ⋅ in 
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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
ε = kσ n for σ > 0 and ε = − kσ n for σ < 0 . If a couple M is
applied to the bar, show that the maximum stress is
σm =
1 + 2n Mc
3n
I
SOLUTION
Strain distribution:
ε = −ε m
y
= −ε mu where
c
u =
y
c
Bending couple:
c
c
c
M = −  − c y σ bdy = 2b  0 y |σ | dy = 2bc 2  0
y
dy
|σ |
c
c
1
= 2 bc 2  0 u |σ | du
For
ε = Kσ n ,
ε m = Kσ m
 σ 
ε
=u =

εm
 σm 
Then
n
∴
1
| σ | = σ mu n
1
1
1 1+ 1n
du
M = 2bc 2  0 uσ mu n du = 2bc 2σ m  0 u
2+ 1
u n
= 2bc σ m
2 + 1n
2
σm =
1
2n
2
 0 = 2n + 1 bc σ m
2n + 1 M
2 bc 2
Recall: that
I
1 b(2c)3
2
1
2 c
=
= bc 2 ∴
=
2
3
3 I
c 12 c
bc
Then
σm =
2n + 1 Mc
3n
I
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PROBLEM 4.99
A short wooden post supports a 6-kip axial load as shown. Determine the
stress at point A when (a) b = 0, (b) b = 1.5in., (c) b = 3in.
SOLUTION
A = π r 2 = π (3) 2 = 28.27 in 2
π
π
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
I =
(a)
b=0
σ =−
(b)
M =0
P
6
=−
= −0.212 ksi
A
28.27
σ = −212 psi 
b = 1.5 in. M = (6)(1.5) = 9 kip ⋅ in
σ =−
P M
6
9
−
=−
−
= −0.637 ksi
28.27 21.206
A
S
σ = −637 psi 
(c)
b = 3 in.
σ =−
M = (6)(3) = 18 kip ⋅ in
P M
6
18
−
=−
−
= −1.061 ksi
28.27 21.206
A
S
σ = −1061 psi 
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PROBLEM 4.100
As many as three axial loads, each of magnitude P = 10 kips, can be
applied to the end of a W8 × 21 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.
SOLUTION
For W8 × 21 Appendix C gives
A = 6.16 in 2 , d = 8.28 in., I x = 75.3 in 4
At point A,
(a)
Centric loading:
1
d = 4.14 in.
2
F My
−
σ =
A
I
y =
F = 30 kips, M = 0
σ =
(b)
Eccentric loading:
30
6.16
σ = 4.87 ksi 
F = 2 P = 20 kips
M = −(10)(3.5) = −35 kip ⋅ in
σ =
20
(−35)(4.14)
−
6.16
75.3
σ = 5.17 ksi 
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PROBLEM 4.101
Knowing that the magnitude of the horizontal force P is 8 kN,
determine the stress at (a) point A, (b) point B.
SOLUTION
A = (30)(24) = 720 mm 2 = 720 × 10−6 m 2
e = 45 − 12 = 33mm = 0.033 m
1 3
1
(30)(24)3 = 34.56 × 103 mm 4 = 34.56 × 10−9 m 4
bh =
12
12
1
c = (24 mm) = 12 mm = 0.012 m
P = 8 × 103 N
2
I =
M = Pe = (8 × 103 )(0.033) = 264 N ⋅ m
(a)
σA = −
P Mc
8 × 103
(264)(0.012)
−
=−
=
= −102.8 × 106 Pa
A
I
720 × 10−6
34.56 × 10−9
σ A = −102.8 MPa 
(b)
σB = −
P Mc
8 × 103
(264)(0.012)
+
=−
+
= 80.6 × 106 Pa
−6
A
I
720 × 10
34.56 × 10−9
σ B = 80.6 MPa 
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PROBLEM 4.102
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.
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)
σA =
P Mc
20 × 103
(4.8 × 103 )(0.040)
+
=
+
= 112.7 × 106 Pa
A
I
2.4 × 10−3
1.84 × 10−6
σ A = 112.7 MPa 
(b)
σB =
P Mc
20 × 103
(4.8 × 103 ) (0.040)
−
=
−
= −96.0 × 106 Pa
A
I
2.4 × 10−3
1.84 × 10−6
σ B = −96.0 MPa 
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PROBLEM 4.103
Solve Prob. 4.102, assuming that t = 8 mm.
PROBLEM 4.102 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.
SOLUTION
Rectangular cutout is 64 mm × 44 mm.
A = (80) (60) − (64)(44) = 1.984 × 103 mm 2
= 1.984 × 10−3 mm 2
I =
1
1
(60)(80)3 − (44)(64)3 = 1.59881 × 106 mm 2
12
12
= 1.59881 × 10−6 m 4
c = 40 mm = 0.004
e = 200 + 40 = 240 mm = 0.240 m
3
P = 20 × 10 N
M = Pe = (20 × 103 )(0.240) = 4.8 × 103 N ⋅ m
(a)
σA =
P Mc
20 × 103
(4.8 × 103 )(0.040)
+
=
+
= 130.2 × 106 Pa
A
I
1.984 × 10−3
1.59881 × 10−6
(b)
σB =
P Mc
20 × 103
(4.8 × 103 ) (0.040)
−
=
−
= −110.0 × 106 Pa
A
I
1.984 × 10−3
1.59881 × 10−6
σ A = 130.2 MPa 
σ B = −110.0 MPa 
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PROBLEM 4.104
Determine the stress at points A and B, (a) for the loading
shown, (b) if the 60-kN loads are applied at points 1 and 2 only.
SOLUTION
(a)
Loading is centric.
P = 180 kN = 180 × 103 N
A = (90)(240) = 21.6 × 103 mm 2 = 21.6 × 10−6 m 2
σ =−
At A and B:
P
180 × 103
=
= −8.33 × 106 Pa
−3
A 21.6 × 10
σ A = σ B = −8.33 MPa 
(b) Eccentric loading.
P = 120 kN = 120 × 103 N
M = (60 × 103 )(150 × 10−3 ) = 9.0 × 103 N ⋅ m
1 3
1
bh =
(90)(240)3 = 103.68 × 106 mm 4 = 103.68 × 10−6 m 4
12
12
c = 120 mm = 0.120 m
I =
At A:
σA = −
At B:
σB −
P Mc
120 × 103
(9.0 × 103 )(0.120)
−
=−
−
= −15.97 × 106 Pa 
A
I
21.6 × 10−3
103.68 × 10−6
P Mc
120 × 103
(9.0 × 103 )(0.120)
+
=−
+
= 4.86 × 106 Pa
A
I
21.6 × 10−3
103.68 × 10−6
σ A = −15.97 MPa 
σ B = 4.86 MPa 
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PROBLEM 4.105
Knowing that the allowable stress in section ABD is 10 ksi,
determine the largest force P that can be applied to the bracket
shown.
SOLUTION
Statics: M = 2.45 P
Cross section: A = (0.9)(1.2) = 1.08 in 2
c=
1
(0.9) = 0.45 in.
2
I =
1
(1.2)(0.9)3 = 0.0729 in 4
12
At point B: σ = −10 ksi
P Mc
−
A
I
P
(2.45P)(0.45)
−10 = −
−
= −16.049P
1.08
0.0729
σ =−
P = 0.623 kips
P = 623 lb 

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PROBLEM 4.106
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.
SOLUTION
The maximum stress occurs at point B.
σ B = −15 ksi = −15 × 103 psi
σB = −
1 ec
+
A
I
K =
where
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
−3
0.25 5.2083 × 10
P=−
(b)
(b)
c=
σB
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=−
σB
K
=−
(−15 × 103 )
71.882
P = 209 lb 
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PROBLEM 4.107
The four forces shown are applied to a rigid plate supported by a solid
steel post of radius a. Knowing that P = 100 kN and a = 40 mm,
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 = π a2
σ =−
a4
M x = − Pa,
σ =−
F
4P
=− 2
A
πa
Mz = 0
F M xz
3P
(− Pa)(−a)
7P
−
=− 2 −
=− 2
2
π
A
I
a
πa
πa
4
Forces at C and D are removed:
F = 2 P,
Resultant bending couple:
σ =−
M =
M x = − Pa,
M x2 + M z2 =
F Mc
2P
−
=− 2 −
A
I
πa
M z = − Pa
2 Pa
2 Pa a
2+4 2 P
=−
= −2.437 P/a 2
2
π a2
π
a
4
P = 100 × 103 N,
Numerical data:

4
Force at D is removed:
F = 3P,
(b)
π
F = 4 P, M x = M z = 0
Centric force:
(a)
I =
a = 0.040 m
Answers:
(a)
σ =−
(7) (100 × 103 )
= −139.3 × 106 Pa
2
π (0.040)
σ = −139.3 MPa 

(b)
σ =−
(2.437)(100 × 103 )
= −152.3 × 106 Pa
(0.040) 2
σ = −152.3 MPa 
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PROBLEM 4.108
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 σ 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 ,
I =
1
ad 3 ,
12
c=
a d
−
2 2
P Mc
P
6Ped
σ = +
=
+
A
I
ad
ad 3
3 P (a − d )
P
σ =
+
= KP
ad
ad 2
1
d
2
e=
Data:
a = 30 mm = 0.030 m
K =
P=
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
σ
K
=
60 × 106
= 14.40 × 103 N
3
4.1667 × 10
P = 14.40 kN 
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PROBLEM 4.109
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
σ all = 135 MPa, determine the smallest allowable depth d of the
milled portion of the bar.
SOLUTION
A = ad ,
e=
I =
1
ad 3 ,
12
c=
1
d
2
a d
−
2 2
P 1 (a − d ) 1 d
P Mc
P
Pec
P
2 = P + 3P ( a − d )
=
+
=
+ 2
σ = +
1
A
I
ad
I
ad
ad
ad 2
ad 3
12
3P 2P
2P
d − 3P = 0
σ = 2 −
or σ d 2 +
ad
a
d
2


1   2P 
2P 
d =
 

 + 12 Pσ −
2σ   a 
a 


Solving for d,
a = 0.030 m,
Data:
d =
P = 18 × 103 N, σ = 135 × 106 Pa
2


1
(2) (18 × 103 ) 
  (2) (18 × 103 ) 
3
6
+
×
×
−
12(18
10
)
(135
10
)
 


0.030
0.030
(2) (135 × 106 )  




= 16.04 × 10−3 m
d = 16.04 mm 
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PROBLEM 4.110
A short column is made by nailing two 1 × 4-in. planks to a 2 × 4-in.
timber. Determine the largest compressive stress created in the column by a
12-kip load applied as shown in the center of the top section of the timber if
(a) the column is as described, (b) plank 1 is removed, (c) both planks are
removed.
SOLUTION
(a)
Centric loading: 4 in. × 4 in. cross section
σ =−
(b)
P
12
=−
A
16
Eccentric loading: 4 in. × 3 in. cross section
1
c =   (3) = 1.5 in.
2
I =
σ = −0.75 ksi 
A = (4) (3) = 12 in 2
e = 1.5 − 1.0 = 0.5 in.
1 3
1
bh =
(4) (3)3 = 9 in 4
12
12
σ =−
(c)
A = (4)(4) = 16 in 2
P Pec
12 (12) (0.5) (1.5)
−
=− −
A
I
12
9
Centric loading: 4 in. × 2 in. cross section
σ =−
P
12
=−
A
8
σ = −2.00 ksi 
A = (4) (2) = 8 in 2
σ = −1.50 ksi 
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PROBLEM 4.111
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.
SOLUTION
For centric loading,
σc =
P
A
For eccentric loading,
σe =
P Phc
+
A
I
Given
σe = 5 σc
P Phc
P
+
=5
A
I
A
Phc
P
=4
I
A
∴
π

(4)  d 4 
4I
64

 = 1d
h=
=
cA  d  π 2  2
  d 
 2  4 
h = 0.500 d 
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PROBLEM 4.112
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.
SOLUTION
1
d = 0.375 in.
2
c1 = c − t = 0.375 − 0.08 = 0.295 in.
c=
(
)
A = π c 2 − c12 = π (0.3752 − 0.2952 )
= 0.168389 in 2
I =
π
(c
4
4
)
− c14 =
π
4
(0.3754 − 0.2954 )
−3
= 9.5835 × 10 in 4
For centric loading,
σ cen =
P
A
For eccentric loading,
σ ecc =
P Phc
+
A
I
σ ecc = 4 σ cen
hc
3
=
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. 
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PROBLEM 4.113
A steel rod is welded to a steel plate to form the machine
element shown. Knowing that the allowable stress is 135 MPa,
determine (a) the largest force P that can be applied to the
element, (b) the corresponding location of the neutral axis.
Given: The centroid of the cross section is at C and
I z = 4195 mm 4 .
SOLUTION
(a)
A = (3)(18) +
π
4
(6)2 = 82.27 mm 2 = 82.27 × 10−6 m 2
I = 4195 mm 4 = 4195 × 10−12 m 4
e = 13.12 mm = 0.01312 m
Based on tensile stress at y = −13.12 mm = −0.01312 m
σ =
P Pec  1 ec 
+
=  +  P = KP
A
I
I 
A
K =
1 ec
1
(0.01312)(0.01312)
+
=
+
= 53.188 × 103 m −2
−6
−12
A
I
82.27 × 10
4195 × 10
P=
(b)
σ
K
=
135 × 106
= 2.538 × 103 N
3
53.188 × 10
Location of neutral axis.
P = 2.54 kN 
σ =0
σ =
P My
P Pey
−
=
−
=0
A
I
A
I
ey
1
=
I
A
y =
I
4195 × 10−12
=
= 3.89 × 10−3 m
−6
Ae (82.27 × 10 )(0.01312)
y = 3.89 mm 
The neutral axis lies 3.89 mm to the right of the centroid or 17.01 mm to the right of the line of action
of the loads.
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PROBLEM 4.114
A vertical rod is attached at point A to the cast iron hanger
shown. Knowing that the allowable stresses in the hanger
are σ all = +5 ksi and σ all = −12 ksi , determine the
largest downward force and the largest upward force that
can be exerted by the rod.
SOLUTION
Ay
(1 × 3)(0.5) + 2(3 × 0.25)(2.5)
X = 
=
A
(1 × 3) + 2(3 × 0.75)

X =
12.75 in 3
= 1.700 in.
7.5 in 2
A = 7.5 in 2
σ all = +5 ksi
σ 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: σ D = +
P Mc
+
A
I
P
(3.20) P(1.70)
+
7.5
10.825
+ 5 = P(+0.6359)
+ 5 ksi =
At E: σ E = +
P = 7.86 kips ↓ 
P Mc
−
A
I
P
(3.20) P(2.30)
−
7.5
10.825
− 12 = P(−0.5466)
− 12 ksi =
We choose the smaller value.
P = 21.95 kips ↓
P = 7.96 kips ↓ 
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PROBLEM 4.114 (Continued)
Upward Force.
M = P(1.5 in. + 1.70 in.) = (3.20 in.) P
At D: σ D = +
P Mc
−
A
I
P
(3.20) P(1.70)
−
7.5
10.825
−12 = P(−0.6359)
−12 ksi = −
At E: σ E = +
P = 18.87 kips ↑ 
P Mc
+
A
I
P
(3.20) P(2.30)
+
7.5
10.825
+5 = P(+0.5466)
+5 ksi =
We choose the smaller value.
P = 9.15 kips ↑ 
P = 9.15 kips ↑ 
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PROBLEM 4.115
Solve Prob. 4.114, assuming that the vertical rod is attached
at point B instead of point A.
PROBLEM 4.114 A vertical rod is attached at point A to the
cast iron hanger shown. Knowing that the allowable stresses
in the hanger are σ all = +5 ksi and σ all = −12 ksi ,
determine the largest downward force and the largest upward
force that can be exerted by the rod.
SOLUTION
Ay
(1 × 3)(0.5) + 2(3 × 0.25)(2.5)
=
X = 
A
(1 × 3) + 2(3 × 0.75)

X =
12.75 in 3
= 1.700 in.
7.5 in 2
A = 7.5 in 2
σ all = +5 ksi
σ 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.
σ all = +5 ksi
σ all = −12 ksi
M = (2.30 in. + 1.5 in.) = (3.80 in.) P
At D: σ D = +
P Mc
−
A
I
P
(3.80) P(1.70)
−
7.5
10.825
− 12 = P(−0.4634)
−12 ksi = +
At E: σ E = +
P = 25.9 kips ↓
P Mc
+
A
I
P
(3.80) P(2.30)
+
7.5
10.825
+5 = P(+0.9407)
+5 ksi = +
We choose the smaller value.
P = 5.32 kips ↓ 
P = 5.32 kips ↓ 
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PROBLEM 4.115 (Continued)
Upward Force.

σ all = +5 ksi

M = (2.30 in. + 1.5 in.)P = (3.80 in.)P 


σ all = −12 ksi 
At D: σ D = −
P Mc
+

A
I
P
(3.80) P(1.70)
+
7.5
10.825
5 = P(+0.4634)

5 ksi = −

At E: σ E = −
P = 10.79 kips ↑ 
P Mc
−
A
I
P
(3.80) P(2.30)
−
7.5
10.825
−12 = P(−0.9407)
−12 ksi = −
We choose the smaller value.
P = 12.76 kips ↑ 
P = 10.79 kips ↑ 
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PROBLEM 4.116
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.
SOLUTION
(a)
Centric loading:
σ =−
P
A
A = (3)(150)(25) = 11.25 × 103 mm2 = 11.25 × 10−3 m 2
P = −σ 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
A y (103 mm3 )

3.75
87.5
328.125
76.5625

3.75
0
0
10.9375

2.50
−87.5
−218.75
98.4375
Σ
10.00
d , mm
109.375
Y =
ΣAy 109.375 × 103
=
= 10.9375 mm
ΣA
10.00 × 103
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PROBLEM 4.116 (Continued)
The centroid lies 10.9375 mm from the midpoint of the web.
1
1
b1h13 + A1d12 =
(150)(25)3 + (3.75 × 103 )(76.5625) 2 = 22.177 × 106 mm 4
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
I3 =
b3h33 + A3d32 =
(100)(25)3 + (2.50 × 103 ) (98.4375)2 = 24.355 × 106 mm 4
12
12
I1 =
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
σ =−
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
+
=
+
−6
−3
A
I
10.00 × 10
54.012 × 10
P=−
σ
K
=−
(−100 × 106 )
= 817 × 103 N
122.465
P = 817 kN 
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PROBLEM 4.117
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.
SOLUTION
Locate centroid.
Part
A, in 2
y , in.

12
5

8
2
Σ
20
Eccentricity of load: e = 5 − 3.8 = 1.2 in.
I1 =
Ay , in 3
60
16
76
1
(6)(2)3 + (12)(1.2)2 = 21.28 in 4
12
y =
I2 =
ΣAy
76
=
= 3.8 in.
ΣA
20
1
(2)(4)3 + (8)(1.8) 2 = 36.587 in 4
12
I = I1 + I 2 = 57.867 in 4
(a)
Stress at A: c A = 3.8 in.
σA = −
(b)
σ A = 0.576 ksi 
Stress at B: cB = 6 − 3.8 = 2.2 in.
σB = −
(c)
P Pec A
20 20(1.2)(3.8)
+
=−
+
20
57.867
A
I
P PecB
20 20(1.2)(2.2)
+
=−
−
20
57.867
A
I
σ B = −1.912 ksi 
P Pea
ea
1
+
=0 ∴
=
A
I
I
A
I
57.867
a =
=
= 2.411 in.
Ae (20)(1.2)
Location of neutral axis: σ = 0
σ =−
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. 
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PROBLEM 4.118
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.
SOLUTION
Locate centroid.
A, in 2


Σ
Eccentricity of load:
12
8
20
e = y − 3.8 in.
I1 =
y , in.
Ay , in 3
5
2
60
16
76
y =
ΣAi yi
76
=
= 3.8 in.
ΣAi
20
y = e + 3.8 in.
1
(6)(2)3 + (12)(1.2)2 = 21.28 in 4
12
1
(2)(4)3 + (8)(1.8)2 = 36.587 in 4
12
I2 =
I = I1 + I 2 = 57.867 in 4
If stress at A equa ls zero, c A = 3.8 in.
σA = −
e=
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.
If stress at B equa ls zero. cB = 6 − 3.8 = 2.2 in.
P PecB
ecB
1
−
=0
∴
=−
A
I
I
A
I
57.867
e=−
=−
= −1.315 in.
AcB
(20)(2.2)
σB = −
y = −1.315 + 3.8 = 2.485 in.
Answer: 2.485 in. < y < 4.561 in. 
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PROBLEM 4.119
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.
SOLUTION
Cross section: Rectangle  + Circle 
A1 = (20 mm)(4 mm) = 80 mm 2
y1 =
1
(20 mm) = 10 mm
2
A2 = π (2 mm)2 = 4π mm 2
y2 = 20 − 2 = 18 mm
Ay
(80)(10) + (4π )(18)
cB = y = 
=
= 11.086 mm
80 + 4π
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
π
4
(2) 4 + (4π )(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
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PROBLEM 4.119 (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:
σA =
P Mc
400
(16.3656)(8.914 × 10−3 )
+
=
+
−6
A
I
92.566 × 10
3.374 × 10−9
= 4.321 × 106 + 43.23 × 106 = 47.55 × 106 Pa
(b)
Point B:
σB =
P Mc
400
(16.3656)(11.086)
−
=
−
A
I
92.566 × 10−6
3.374 × 10−9
= 4.321 × 106 − 53.72 × 106 = −49.45 × 106 Pa
(c)
σ A = 47.6 MPa
Neutral axis:
σ 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
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PROBLEM 4.120
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,
σA = −
Aec A 
P Pec A
P
−
= − 1 +
A
I
A
I 
At B,
σB = −
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  
σ A = − 1 +
1 2


A

a



12




 2  1  1  

 (a )  a  a  
 2  2  − 1
σ = P 
 B A
1 2

a



12



a
π

2
2
, I 2 = c4 , e = c
 A2 = π c = a ∴ c =
4
π





P  (π c 2 )(c)(c) 
σ A = − 1 +


π 4
A2 

c


4







P  (π c 2 )(c)(c) 
σ B =
− 1

π 4
A2 


c


4


σ A = −4
P

A1
σB = 2
P

A1
σ A = −5
P

A2
σB = 3
P

A2
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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


σ A = − 1 +

1 4
A3

a


12








 2  2  

a 
 ( a 2 ) 
 2 a  

2 
P

 −1
σ B =

1 4
A3 

a



12





σ A = −7
P

A3

σB = 5
P

A3
  3 2   s  s  
s  
 


P   4
 3  3  

σ A = − 1 +

A4
3 4


s


96


σ A = −9
P

A4
  3 2   s  s  
s  


 
P   4
 3  2 3  

σB = 
− 1
A4
3
4


s


96


σB = 3
P

A4
A4 =
1  3 
3 2
(s) 
s =
s
2  2  4
3
1  3 
3 4
I4 =
s
s =
s


36  2 
96
cA =
2 3
s
=e
s=
3 2
3
cB = s
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PROBLEM 4.121
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 μ,
determine the magnitude P of the forces. Use E = 200 GPa.
SOLUTION
At the strain gage location,
σ = Eε = (200 × 109 )(450 × 10−6 ) = 90 × 106 Pa
A = (40)(40) = 1600 mm 2 = 1600 × 10−6 m 2
1
(40)(40)3 = 213.33 × 103 mm 4 = 213.33 × 10−9 m 4
12
e = 80 + 20 = 100 mm = 0.100 m
I =
c = 20 mm = 0.020 m
σ =
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=
σ
K
=
90 × 106
= 9.00 × 103 N
10.00 × 103
P = 9.00 kN 
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PROBLEM 4.122
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
ε A = +350 μ
ε B = −70 μ
Knowing that E = 200 GPa, determine (a) the distance d, (b) the
magnitude of the force P.
SOLUTION
1
h = 45 mm = 0.045 m
2
A = bh − (25)(90) = 2.25 × 103 mm 2 = 2.25 × 10−3 m 2
h = 15 + 45 + 30 = 90 mm
b = 25 mm
c=
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
I=
Stresses from strain gages at A and B:
σ A = Eε A = (200 × 109 )(350 × 10−6 ) = 70 × 106 Pa
σ B = Eε B = (200 × 109 )( −70 × 10−6 ) = −14 × 106 Pa
Subtracting,
σA −σB = −
M =−
σA =
P MyA
−
A
I
(1)
σB =
P MyB
−
A
I
(2)
M ( y A − yB )
I
I (σ A − σ B )
(1.51875 × 10−6 )(84 × 106 )
=−
= −2835 N ⋅ m
0.045
y A − yB
Multiplying (2) by y A and (1) by yB and subtracting,
P=
(a)
(b)
y Aσ B − yBσ A = ( y A − yB )
P
A
A( y Aσ B − yBσ A ) (2.25 × 10−3 )[(0.015)(−14 × 106 ) − (−0.030)(70 × 106 )]
=
= 94.5 × 103 N
0.045
y A − yB
M = − Pd ∴ d = −
−2835
M
=−
= 0.030 m
P
94.5 × 103
d = 30.0 mm 
P = 94.5 kN 
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PROBLEM 4.123
Solve Prob. 4.122, assuming that the measured strains are
ε A = +600 μ
ε B = +420 μ
PROBLEM 4.122 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
ε A = +350μ
ε B = −70 μ
Knowing that E = 200 GPa, determine (a) the distance d, (b) the
magnitude of the force P.
SOLUTION
1
h = 45 mm = 0.045 m
2
A = bh = (25)(90) = 2.25 × 103 mm 2 = 2.25 × 10−3 m 2
h = 15 + 45 + 30 = 90 mm
b = 25 mm
c=
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
I=
Stresses from strain gages at A and B:
σ A = Eε A = (200 × 109 )(600 × 10−6 ) = 120 × 106 Pa
σ B = Eε B = (200 × 109 )(420 × 10−6 ) = 84 × 106 Pa
Subtracting,
σA =
P My A
−
A
I
(1)
σB =
P MyB
−
A
I
(2)
σA −σB = −
M =−
M ( y A − yB )
I
I (σ A − σ B )
(1.51875 × 10−6 )(36 × 106 )
=−
= −1215 N ⋅ m
0.045
y A − yB
Multiplying (2) by y A and (1) by yB and subtracting,
P=
y Aσ B − yBσ A = ( y A − yB )
P
A
A( y Aσ B − yBσ A ) (2.25 × 10−3 )[(0.015)(84 × 106 ) − ( −0.030)(120 × 106 )]
=
= 243 × 103 N
y A − yB
0.045
M = − Pd
(a)
(b)
∴ d =−
−1215
M
=−
= 5 × 10−3 m
3
P
243 × 10
d = 5.00 mm 
P = 243 kN 
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PROBLEM 4.124
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
ε A = −550 × 10−6 in./in.
ε B = −680 × 10−6 in./in.
Knowing that E = 29 × 106 psi, determine the magnitude of each load.
SOLUTION
Strains:
ε A = −550 × 10−6 in./in.
εC =
Stresses:
ε B = −680 × 10−6 in./in.
1
1
(ε A + ε B ) = (−550 − 680)10−6 = −615 × 10−6 in./in.
2
2
σ A = Eε A = (29 × 106 psi)(−550 × 10−6 in./in.) = −15.95 ksi
σ C = Eε C = (29 × 106 psi)(−615 × 10−6 in./in.) = −17.935 ksi
A = 9.13 in 2
W8 × 31
S = 27.5 in 3
M = (4.5 in.)( P − Q)
At point C:
σC = −
P+Q
P+Q
; − 17.835 ksi = −
A
9.13 in 2
P + Q = 162.83 kips (1)
At point A:
σA = −
P+Q M
−
A
S
−15.95 ksi = −17.835 ksi −
Solve simultaneously,
(4.5 in.)( P − Q)
;
27.5 in 3
P − Q = −11.52 kips (2)
P = 25.7 kips Q = 87.2 kips
P = 25.7 kips ↓ 
Q = 87.2 kips ↓ 
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PROBLEM 4.125
Solve Prob. 4.124, assuming that the measured strains are
ε A = +35 × 10−6 in./in. ε B = −450 × 10−6 in./in.
PROBLEM 4.124 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
ε A = −550 × 10−6 in./in.
ε B = −680 × 10−6 in./in.
Knowing that E = 29 × 106 psi, determine the magnitude of each load.
SOLUTION
See solution and figures of Prob. 4.124.
ε A = +35 × 10−6 in./in.;
εC =
ε B = −450 × 10−6 in./in.
1
1
(ε A + ε B ) = (35 − 450)10−6 in./in. = −207.5 × 10−6 in./in.
2
2
Stresses:
σ A = Eε A = (29 × 106 psi)(+35 × 10−6 in./in.) = +1.015 ksi
σ C = Eε C = (29 × 106 psi)(−207.5 × 10−6 in./in.) = −6.0175 ksi
At point C:
σC = −
P+Q
P+Q
; − 6.0175 ksi = −
A
9.13 in 2
P + Q = 54.94 kips (1)
At point A:
σA = −
P+Q M
−
A
S
+1.015 ksi = −6.0175 −
(4.5 in.)( P − Q)
27.5 in 3
P − Q = −42.98 kips (2)
Solve simultaneously,
P = 5.98 kips
Q = 49.0 kips 
P = 5.98 kips ↓ 
Q = 49.0 kips ↓ 
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PROBLEM 4.126
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.
SOLUTION
1
bd 3
12
1
1
c= d
e= d −a
2
2
P Pec
σA = +
A
I

1
1 
P  1 12 2 d − a 2 d  P  4 6a 
σA =  +
=  − 2
b d
d3
 b d d 


A = bd
I=
(
(a)
)( )
Depth d for maximum σ A : Differentiate with respect to d.
dσ A P  4 12a 
= − 2 + 3  = 0
dd
b d
d 
(b)
σA =
60 × 103
40 × 10−3
4
(6)(25 × 10−3 ) 

6
−

 = 40 × 10 Pa
−3
(75 × 10−3 ) 2 
 75 × 10
d = 3a
d = 75 mm 
σ A = 40 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.127
The couple M is applied to a beam of the cross section shown in a
plane forming an angle β with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
1
(80)(100)3 = 6.6667 × 106 mm 4 = 6.6667 × 10−6 m 4
12
1
I y = (100)(80)3 = 4.2667 × 106 mm 4 = 4.2667 × 10−6 m 4
12
y A = − yB = − yD = 50 mm
Iz =
z A = z B = − z D = 40 mm
M y = −250 sin 30° = −125 N ⋅ m
M z = 250 cos 30° = 216.51 N ⋅ m
(a)
σA = −
M z yA M y zA
(216.51)(0.050) (−125)(0.040)
+
=−
+
Iz
Iy
6.6667 × 10−6
4.2667 × 10−6
= −2.80 × 106 Pa
(b)
σB = −
M z yB M y z B
(216.51)(−0.050) ( −125)(0.040)
+
=−
+
Iz
Iy
6.6667 × 10−6
4.2667 × 10−6
= 0.452 × 103 Pa
(c)
σD = −
σ A = −2.80 MPa 
σ B = 0.452 MPa 
M z yD M y z D
(216.51)(−0.050) (−125)(−0.040)
+
=−
+
Iz
Iy
6.6667 × 10−6
4.2667 × 10−6
= 2.80 × 106 Pa
σ D = 2.80 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.128
The couple M is applied to a beam of the cross section shown in a plane
forming an angle β with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
SOLUTION
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
Iz =
z A = − z B = − z D = 40 mm
M y = 300 cos 30° = 259.81 N ⋅ m
(a)
σA = −
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
(b)
σB = −
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
= −18.60 × 106 Pa
(c)
σD = −
σ A = −3.37 MPa 
σ B = −18.60 MPa 
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
σ D = 3.37 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.129
The couple M is applied to a beam of the cross section shown in a
plane forming an angle β with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
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.
(a)
σA = −
Iz =
1
π

(10)(6)3 − 2  (1)2  = 178.429 in 4
12
4


Iy =
1
π

(6)(10)3 − 2  (1)4 + π (1)2 (2.5) 2  = 459.16 in 4
12
4


M z yA M y zA
( −38.567)(3) (45.963)(5)
+
=−
+
178.429
459.16
Iz
Iy
= 1.149 ksi
(b)
σB = −
M z yB M y z B
(−38.567)(3) (45.963)(−5)
+
=−
+
178.429
459.16
Iz
Iy
= 0.1479
(c)
σD = −
σ A = 1.149 ksi 
σ B = 0.1479 ksi 
M z yD M y z D
(−38.567)(−3) (45.963)(−5)
+
=−
+
178.429
459.16
Iz
Iy
= −1.149 ksi
σ D = −1.149 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.130
The couple M is applied to a beam of the cross section shown in a plane
forming an angle β with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
SOLUTION
Locate centroid.
A, in 2
z , in.
Az , in 3

16
−1
−16

8
2
16
Σ
24
0
The centroid lies at point C.
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.
Iz =
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)
σA = −
M z yA M y zA
(9.3969)(1) (3.4202)(−4)
+
=−
+
88
64
Iz
Iy
(b)
σB = −
M z yB M y z B
(9.3969)( −1) (3.4202)(−4)
+
=−
+
88
64
Iz
Iy
σ B = −0.107 ksi 
(c)
σD = −
M z yD M y z D
(9.3969)(−4) (3.4202)(0)
+
=−
+
88
64
Iz
Iy
σ D = 0.427 ksi 
σ A = 0.321 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.131
The couple M is applied to a beam of the cross section shown in a
plane forming an angle β with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
M y = 25sin 15° = 6.4705 kN ⋅ m
M z = 25cos15° = 24.148 kN ⋅ m
1
1
(80)(90)3 + (80)(30)3 = 5.04 × 106 mm 4
12
12
−6
4
I y = 5.04 × 10 m
Iy =
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)
σA =
σ =
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)
σB =
(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)
σD =
σ A = −29.3 MPa 
σ B = −144.8 MPa 
(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
σ D = −125.9 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.132
The couple M is applied to a beam of the cross section shown in a
plane forming an angle β with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
Flange:
1
(8)(0.5)3 + (8)(0.5)(4.75) 2
12
= 90.333 in 4
Iz =
Iy =
1
(0.5)(8)3 = 21.333 in 4
12
1
(0.3)(9)3 = 18.225 in 4
12
1
I y = (9)(0.3)3 = 0.02025 in 4
12
Iz =
Web:
I z = (2)(90.333) + 18.225 = 198.89 in 4
Total:
I y = (2)(21.333) + 0.02025 = 42.687 in 4
y A = yB = − yD = 5 in.
z A = − z B = − zC = 4 in.
M z = 250 cos 30° = 216.51 kip ⋅ in
M y = −250 sin 30° = −125 kip ⋅ in
(a)
σA = −
M z yA M y zA
(216.51)(5) (−125)(4)
+
=−
+
198.89
42.687
Iz
Iy
(b)
σB = −
M z yB M y z B
(216.51)(5) (−125)(−4)
+
=−
+
198.89
42.687
Iz
Iy
(c)
σD = −
M z yD M y z D
(216.51)( −5) ( −125)(−4)
+
=−
+
198.89
42.687
Iz
Iy
σ A = −17.16 ksi 
σ B = 6.27 ksi 
σ D = 17.16 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.133
The couple M is applied to a beam of the cross section shown in a
plane forming an angle β with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
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.
Iz =
z A = − z B = − z D = 2.4 in.
M z = 75sin15° = 19.4114 kip ⋅ in
M y = 75cos15° = 72.444 kip ⋅ in
(a)
σA = −
M z yA M y zA
(19.4114)(1.2) (72.444)(2.4)
+
=−
+
4.1643
13.5851
Iz
Iy
σ A = 7.20 ksi 
(b)
σB = −
M z yB M y z B
(19.4114)(1.2) (72.444)(−2.4)
+
=−
+
4.1643
13.5851
Iz
Iy
σ B = −18.39 ksi 
(c)
σD = −
M z yD M y z D
(19.4114)(−1.2) (72.444)(−2.4)
+
=−
+
4.1643
13.5851
Iz
Iy
σ D = −7.20 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.134
The couple M is applied to a beam of the cross section shown in a
plane forming an angle β with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
π
2
8  4
 π  4r   π
I z = r −  r 2 
 = −
r
8
 2  3π   8 9π 
4
= (0.109757)(20) 4 = 17.5611 × 10−3 mm 4 = 17.5611 × 10−9 m 4
Iy =
π
8
r4 =
π (20)4
8
= 62.832 × 10−3 mm 4 = 62.832 × 10−9 m 4
4r
(4)(20)
=−
= −8.4883 mm
3π
3π
yB = 20 − 8.4883 = 11.5117 mm
y A = yD = −
z A = − z D = 20 mm
zB = 0
M z = 100 cos 30° = 86.603 N ⋅ m
M y = 100sin 30° = 50 N ⋅ m
(a)
σA = −
(86.603)(−8.4883 × 10−3 ) (50)(20 × 10−3 )
M z yA M y z A
+
=−
+
Iz
Iy
17.5611 × 10−9
62.832 × 10−9
= 57.8 × 106 Pa
(b)
σB = −
σ A = 57.8 MPa 
−3
(86.603)(11.5117 × 10 )
(50)(0)
M z yB M y z B
+
=−
+
−9
Iz
Iy
17.5611 × 10
62.832 × 10−9
= −56.8 × 106 Pa
(c)
σD = −
σ B = −56.8 MPa 
−3
3
(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
= 25.9 × 106 Pa
σ D =25.9 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.135
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.
SOLUTION
For C 200 × 17.1 rolled steel shape,
I z = 0.538 × 106 mm 4 = 0.538 × 10−6 m 4
I y = 13.4 × 106 mm 4 = 13.4 × 10−6 m 4
zE = zD = − z A = − zB =
yD = yB = −14.4 mm
1
(203) = 101.5 mm
2
yE = y A = 57 − 14.4 = 42.6 mm
M z = (2.8 × 103 ) cos 10° = 2.7575 × 103 N ⋅ m
M y = (2.8 × 103 ) sin 10° = 486.21 N ⋅ m
(a)
Angle of neutral axis.
tan ϕ =
Iz
0.538
tan θ =
tan10° = 0.007079
13.4
Iy
ϕ = 0.4056°
α = 10° − 0.4056°
(b)
α = 9.59° 
Maximum tensile stress occurs at point D.
σD = −
M z yD M y z D
(2.7575 × 103 )(−14.4 × 10−3 ) (486.21)(0.1015)
+
=−
+
Iz
Iy
0.538 × 10−6
13.4 × 10−6
= 73.807 × 106 + 3.682 × 106 = 77.5 × 106 Pa
σ D = 77.5 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.136
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.
SOLUTION
For W 310 × 38.7 rolled steel shape,
I z = 85.1 × 106 mm 4 = 85.1 × 10−6 m 4
I y = 7.27 × 106 mm 4 = 7.27 × 10−6 m 4
1
y A = yB = − yD = − yE =   (310) = 155 mm
2
1
z A = z E = − z B = − z D =   (165) = 82.5 mm
2
M z = (16 × 103 ) cos 15° = 15.455 × 103 N ⋅ m
M y = (16 × 103 ) sin 15° = 4.1411 × 103 N ⋅ m
(a)
Angle of neutral axis.
tan ϕ =
Iz
85.1 × 10−6
tan θ =
tan15° = 3.1365
Iy
7.27 × 10−6
ϕ = 72.3°
α = 72.3° − 15°
(b)
α = 57.3° 
Maximum tensile stress occurs at point E.
σE = −
=−
M z yE M y z E
+
Iz
Iy
(15.455 × 103 )(−155 × 10−3 ) (4.1411 × 103 )(82.5 × 10−3 )
+
85.1 × 10−6
7.27 × 10−6
= 75.1 × 106 Pa
σ E = 75.1 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.137
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.
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 ϕ =
I z′
21.4
tan θ =
tan (−45°) = 3.1751
6.74
I y′
ϕ = −72.5°
α = 72.5° − 45°
α = 27.5° 



(b)
The maximum tensile stress occurs at point D.
σD = −
M z yD M y z D
(10.6066)(−0.25) (−10.6066)(−3.141)
+
=−
+
21.4
6.74
Iz
Iy
= 0.12391 + 4.9429
σ D = 5.07 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.138
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.
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 ϕ =
I z′
176.9 × 10−9
tan θ =
⋅ tan 30° = 0.36346
I y′
281 × 10−9
ϕ = 19.97°
α = 30° − 19.97°
(b)
α = 10.03° 
Maximum tensile stress occurs at point E.
σE = −
M y ′ z′E
(346.41)(−18.57 × 10−3 ) (200)(25 × 10−3 )
M z′ y′E
+
=−
+
I z′
I y′
176.9 × 10−9
281 × 10−9
= 36.36 × 106 + 17.79 × 106 = 54.2 × 106 Pa
σ E = 54.2 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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PROBLEM 4.139
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.
SOLUTION
Bending moments:
M y′ = −35sin15° = −9.059 kip ⋅ in
M z ′ = −35cos15° = 33.807 kip ⋅ in
Moments of inertia:
1
1
(2.4)(4)3 − (2 × 0.4)(2)3 = 12.267 in 4
12
12
1
1
(2)(2.4)3 +
(2)(1.6)3 = 2.987 in 4
=
12
12
I y′ =
I z′
(a)
Neutral axis:
tan φ =
I z′
2.987 in 4
tan θ =
tan(−15°) = −0.06525
I y′
12.267 in 4
φ = 3.73°
α = 15° − φ = 15° − 3.73° = 11.27°
α = 11.3°
(b)

Maximum tensile stress at D:
y′D = −1.2 in.
σD =
M y′ z D
I y′
z′D = −2 in.
−
(−9.059 kip ⋅ in)(−2 in.) (33.807 kip ⋅ in)(−1.2 in.)
M z ′ y′D
=
−
I z′
12.267 in 4
2.987 in 4
= 1.477 ksi + 13.582 ksi = 15.059 ksi
σ D = 15.06 ksi 
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PROBLEM 4.140
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.
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.
θ = 20°.
tan ϕ =
I z′
I y′
tan θ =
53.6 × 10−9
tan 20° = 1.32084
14.77 × 10−9
ϕ = 52.871°
α = 52.871° − 20°
(b)
α = 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′
σ 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
σ E = 61.4 MPa 
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PROBLEM 4.141
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine the stress at point A.
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)
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PROBLEM 4.141 (Continued)
DY 49.766
=
DE 66.355
2θ m = 36.87° θ m = 18.435°
tan 2θ m =
2
2
R = DE + DY = 82.944 in 4
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.
ν A = −4.8 sin 18.435° + 2.4 cos 18.435° = 0.7589 in.
M u
Mν
σA = − ν A + u A
Iν
=−
Iu
(118.585)(5.3126) (39.529)(0.7589)
+
215.654
49.766
= −2.32 ksi
σ A = −2.32 ksi 
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PROBLEM 4.142
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine the stress at point A.
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
FZ 8.3
=
= 1.0506
EF 7.9
I v = 16.6 + 11.46 = 28.06 in 4
tan 2θ m =
θ m = 23.2° I u = 16.6 − 11.46 = 5.14 in 4
M u = M sin θ m = (60) sin 23.2° = 23.64 kip ⋅ in
M v = M cos θ m = (60) cos 23.2° = 55.15 kip ⋅ in
u A = y A cos θ m + z A sin θ m = −3.92cos 23.2° − 1.08 sin 23.2° = −4.03 in.
v A = z A cos θ m − y A sin θ m = −1.08cos 23.2° + 3.92 sin 23.2° = 0.552 in.
σA = −
M vu A M u vA
(55.15)( −4.03) (23.64)(0.552)
+
+
=−
28.06
5.14
Iv
Iu
σ A = 10.46 ksi 
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PROBLEM 4.143
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine the stress at point A.
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
2
2
R = EF + FZ = 0.6402 + 0.8002 × 10−6 = 1.0245 × 106 mm 4
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
θ m = 25.67°
=
= 1.25
FE 0.640 × 106
M v = M cos θ m = (1.2 × 103 ) cos 25.67° = 1.0816 × 103 N ⋅ m
tan 2θ m =
M u = − M sin θ m = −(1.2 × 103 ) sin 25.67° = −0.5198 × 103 N ⋅ m
u A = y A cos θ m − z A sin θ m = 45 cos 25.67° − 45 sin 25.67° = 21.07 mm
v A = z A cos θ m + y A sin θ m = 45 cos 25.67° + 45 sin 25.67° = 60.05 mm
σA = −
M vu 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
= 113.0 × 106 Pa
σ A = 113.0 MPa 
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PROBLEM 4.144
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.
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.
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
Iz =
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)
σA =
70 × 103
(2625)(−0.0625) (−525)(0.0375)
P M z yA M y zA
−
+
=
−
+
−3
A
Iz
Iy
4.224 × 10
7.8283 × 10−6
3.2781 × 10−6
= 31.524 × 106 Pa
σB =
σ A = 31.5 MPa 
70 × 103
(2625)(0.0625) (−525)(0.0375)
P M z yB M y z B
−
+
=
−
+
−3
A
Iz
Iy
4.224 × 10
7.8283 × 10−6
3.2781 × 10−6
= −10.39 × 106 Pa
(b)
σ B = −10.39 MPa 
Let point H be the point where the neutral axis intersects AB.
z H = 0.0375 m,
0=
yH = ?, σ 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
2625
I y 
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. 
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PROBLEM 4.145
Solve Prob. 4.144, assuming that the 28-kN force at point E is
removed.
PROBLEM 4.144 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.
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.
1
1
(75)(125)3 − (51)(101)3 = 7.8283 × 106 mm 4 = 7.8283 × 10−6 m 4
12
12
1
1
3
I y = (125)(75) − (101)(51)3 = 3.2781 × 106 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
Iz =
Resultant force and bending couples:
P = 14 + 28 = 42 kN = 42 × 103 N
M z = −(62.5 mm)(14 kN) + (62.5 mm)(28 kN) = 875 N ⋅ m
M y = −(37.5 mm)(14 kN) + (37.5 mm)(28 kN) = 525 N ⋅ m
(a)
σA =
42 × 103
(875)(−0.0625) (525)(0.0375)
P M z yA M y zA
−
+
=
−
+
−3
A
Iz
Iy
4.224 × 10
7.8283 × 10−6
3.2781 × 10−6
= 22.935 × 106 Pa
σB =
σ A = 22.9 MPa 
P M z yB M y z B
42 × 103
(875)(0.0625) (525)(0.0375)
−
+
=
−
+
−3
A
Iz
Iy
4.224 × 10
7.8283 × 10−6
3.2781 × 10−6
= 8.9631 × 106 Pa
(b)
σ B = 8.96 MPa 
Let point K be the point where the neutral axis intersects BD.
z K = ?,
0=
yK = 0.0625 m, σ H = 0
P M z yH M y z H
−
+
A
Iz
Iy
I y  M z yH P  3.2781 × 10−6  (875)(0.0625)
42 × 103 
− =
−



−6
M y  Iz
A
525
4.224 × 10−3 
 7.8283 × 10
= −0.018465 m = −18.465 mm
zH =
37.5 + 18.465 = 56.0 mm
Answer: 56.0 mm to the right of point B. 
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PROBLEM 4.146
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 θ = 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 (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
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
Ix =
A = (200)(150) = 30 × 103 mm 2 = 30 × 10−3 m 2
(a)
σA = −
P M x zA M z xA
4 × 103
(−250)(75 × 10−3 ) (−433)(−100 × 10−3 )
−
+
=−
−
+
−3
A
Ix
Iz
30 × 10
56.25 × 10−6
100 × 10−6
σ A = 633 × 103 Pa = 633 kPa 
(b)
σB = −
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
σ B = −233 × 103 Pa = −233 kPa 
(c)
Let G be the point on AB where the neutral axis intersects.
σG = 0
zG = 75 mm
σG = −
P M x zG M z xG
−
+
=0
A
Ix
Iz
xG =
xG = ?
I z  P M x Z G  100 × 10−6
 +
=
Mz  A
Ix 
−433
= 46.2 × 10−3 m = 46.2 mm
(−250)(75 × 10−3 ) 
 4 × 103
+


−3
56.25 × 10−6 
 30 × 10
Point G lies 146.2 mm from point A 
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PROBLEM 4.147
4.147 In Prob. 4.146, 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.
4.146 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 θ = 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 θ = −500sin θ
M x = − PR cos θ = −500cos θ
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
Ix =
A = (200)(150) = 30 × 103 mm 2 = 30 × 10−3 m 2
σ =−
 1 Rz sin θ
P M xz M z x
Rx cos θ 
−
+
= −P  −
+

A
Ix
Iz
Ix
Iz 
A
For σ to be a maximum,
dσ
=0
dθ
with z = z D , x = xD

dσ D
Rz cos θ
Rx sin θ 
= − P 0 + D
+ D
=0
dθ
Ix
IZ


sin θ
I z
(100 × 10−6 )(−75 × 10−3 )
4
= tan θ = − z D = −
=
−6
−3
cos θ
I x xD
(56.25 × 10 )(100 × 10 ) 3
sin θ = 0.8,
(b)
σA = −
cos θ = 0.6,
θ = 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
σ A = 700 kPa 
σ B = (−0.13333 + 0.53333 − 0.300) × 106 Pa = 0.100 × 106 Pa
σ B = 100 kPa 
σ C = (−0.13333 + 0 + 0) × 106 Pa
σ D = (−0.13333 − 0.53333 − 0.300) × 106 Pa = − 0.967 × 106 Pa
σ C = −133.3kPa 
σ D = −967 kPa 
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PROBLEM 4.148
Knowing that P = 90 kips, determine the largest distance a for
which the maximum compressive stress dose not exceed 18 ksi.
SOLUTION
A = (5 in.)(6 in.) − 2(2 in.)(4 in.) = 14 in 2
1
1
(5 in.)(6 in.)3 − 2 (2 in.)(4 in.)3 = 68.67 in 4
12
12
1
1
I z = 2 (1 in.)(5 in.)3 +
(4 in.)(1 in.)3 = 21.17 in 4
12
12
Ix =
Force-couple system at C:
P= P
For P = 90 kips:
P = 90 kips M x = (90 kips)(2.5 in.) = 225 kip ⋅ in
Maximum compressive stress at B:
σB = −
−18 ksi = −
M x = P(2.5 in.)
M z = Pa
M z = (90 kips) a
σ B = −18 ksi
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. 
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PROBLEM 4.149
Knowing that a = 1.25 in., determine the largest value of P that
can be applied without exceeding either of the following
allowable stresses:
σ ten = 10 ksi
σ comp = 18 ksi
SOLUTION
A = (5 in.)(6 in.) − (2)(2 in.)(4 in.) = 14 in 2
1
1
(5 in.)(6 in.)3 − 2 (2 in.)(4 in.)3 = 68.67 in 4
12
12
1
1
I z = 2 (1 in.)(5 in.)3 + (4 in.)(1 in.)3 = 21.17 in 4
12
12
Ix =
For a = 1.25 in.,
Force-couple system at C:
P = P M x = P(2.5 in.)
M y = Pa = (1.25in.)
Maximum compressive stress at B:
σB = −
−18 ksi = −
σ B = −18 ksi
P M x (3 in.) M z (2.5 in.)
−
−
A
Ix
Iz
P
P(2.5 in.)(3 in.) P(1.25in.)(2.5in.)
−
−
2
14 in
68.67 in 4
21.17 in 4
−18 = − 0.0714P − 0.1092P − 0.1476 P
−18 = 0.3282 P
P = 54.8 kips
Maximum tensile stress at D:
σD = −
σ D = +10 ksi
P M x (3 in.) M z (2.5 in.)
+
+
A
Ix
Iz
+10 ksi = − 0.0714P + 0.1092P + 0.1476P
10 = 0.1854 P
P = 53.9 kips
The smaller value of P is the largest allowable value.
P = 53.9 kips 
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PROBLEM 4.150
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°
θ = 64.3°
I
tan ϕ = v tan θ
Iu
2.28 × 10−6
tan 64.3°
0.23 × 10−6
= 20.597
=
ϕ = 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
σB = −
80 × 106 = −
M vu B M u v B
+
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 
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PROBLEM 4.151
Solve Prob. 4.150 assuming that the couple M 0 acts in a horizontal
plane.
PROBLEM 4.150 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 ,
and 64.3°
.
principal axes 25.7°
I min = 0.23 × 10−6 mm 4 ,
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°
θ = 64.3°
tan ϕ =
Iv
tan θ
Iu
0.23 × 10−6
tan 64.3°
2.28 × 10−6
= 0.20961
=
ϕ = 11.84°
Points D and E are farthest from the neutral axis.
uD = 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
σD = −
M v u D M u vD
(M D cos 64.3°)(−24.02 × 10−3 )
+
=−
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 × 103 N ⋅ m
M 0 = 1.323 kN ⋅ m 
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PROBLEM 4.152
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 )
SOLUTION
M u = M 0 sin 45° = 0.70711 M 0
M v = M 0 cos 45° = 0.7071 M 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.
σB = −
 u
M v u B M u vB
v 
+
= −0.70711 M 0  − B + B 
Iv
Iu
I max 
 I min
 (−1.866) 3.182 
= 0.70711 M 0  −
+
= 0.4124 M 0
4.59
18.01 

M0 =
σB
0.4124
=
12
0.4124
M 0 = 29.1 kip ⋅ in 
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PROBLEM 4.153
Solve Prob. 4.152, assuming that the couple M 0 acts in a horizontal
plane.
PROBLEM 4.152 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
2
and I min + I max = I y + I z )
axes. Use the relations I min = Akmin
SOLUTION
M u = M 0 cos 45° = 0.70711 M 0
M v = − M 0 sin 45° = − 0.70711 M 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 D = 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.
σD = −
 u
M vu D M u v D
v 
+
= −0.70711 M 0  − D + D 
Iv
Iu
I max 
 I min
 (−1.866) 3.182 
= 0.70711 M 0  −
+
= 0.4124 M 0
4.59
18.01 

M0 =
σD
0.4124
=
12
0.4124
M 0 = 29.1 kip ⋅ in 
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PROBLEM 4.154
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°
θ = 29.4°
tan ϕ =
Iv
0.427
tan θ =
tan 29.4°
Iu
0.957
= 0.2514
ϕ = 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.
σA = −
M vu A
MV
(M cos 29.4°)(−1.0216) ( M 0 sin 29.4°)(−0.2852)
+ u A =− 0
+
Iv
Iu
0.427
0.957
= 1.9381 M 0
M0 =
σA
1.9381
=
12
1.9381
M 0 = 6.19 kip ⋅ in 
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PROBLEM 4.155
A couple M 0 acting in a vertical plane is applied to a W12 × 16 rolled-steel beam,
whose web forms an angle θ with the vertical. Denoting by σ 0 the maximum stress
in the beam when θ = 0, determine the angle of inclination θ of the beam for which
the maximum stress is 2σ 0.
SOLUTION
For W 12 × 16 rolled steel section,
I z = 103 in 4
I y = 2.82 in 4
d = 11.99 in.
b f = 3.990 in.
yA = −
tan ϕ =
d
2
zA =
bf
2
Iz
103
tan θ =
tan θ = 36.52 tan θ
Iy
2.82
Point A is farthest from the neutral axis.
M y = M 0 sin θ
σA = −
For
θ = 0,
M z = M 0 cos θ

M 0b f
I zb f
M z yA M y zA
M d
M d
+
= 0 cos θ +
sin θ = 0 1 +
tan θ  cos θ

Iz
Iy
I yd
2I z
2I y
2 I z 

σ0 =
M 0d
2I z



σ A = σ 0 1 +
I 2b f
I yd
tan θ =

tan θ  cos θ = 2σ 0

I yd

I zb f
(103)(3.990)  2

− 1

(2.82)(11.99)  cos θ

 2

− 1
tan θ = 0.082273 
 cos θ

Assuming cos θ ≈ 1, we get
θ = 4.70° 
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PROBLEM 4.156
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.
SOLUTION
b
h
M z = M cos θ , M z = M sin θ
tan θ =
Iz =
1 3
bh
12
Iy =
1 3
hb
12
1 3
bh
Iz
b
h
tan ϕ =
tan θ = 12
⋅ =
1
Iy
b
hb3 h
12
The neutral axis passes through corner A of the diagonal AD. 
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PROBLEM 4.157
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
σA =
zI z − yI yz
2
I y I z − I yz
My
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 σ A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are
centroidal axes:
σ A = C1 y + C2 z
where C1 and C2 are constants.
M z = −  yσ AdA = −C1 y 2dA − C2  yzdA
= − I zC1 − I yzC2 = 0
C1 = −
I yz
Iz
C2
M y =  zσ AdA = C1 yz dA + C2  z 2dA
= I yzC1 + I yC2
− I yz
(
I yz
Iz
C2 + I yC2
)
2
I z M y = I y I z − I yz
C2
C2 =
IzM y
2
I y I z − I yz
C1 = −
σA =
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

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PROBLEM 4.158
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
σA =
yI y − zI yz
2
I y I z − I yz
Mz
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 σ A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are
centroidal axes:
σ A = C1 y + C2 z
where C1 and C2 are constants.
M y =  zσ AdA = C1  yzdA + C2  z 2dA
= I yzC1 + I yC2 = 0
C2 = −
I yz
Iy
C1
M z = −  yσ Adz = −C1 y 2dA + C2  yzdA
I yz
C1
= − I Z C1 − I yz
Iy
(
)
2
I y M z = − I y I z − I yz
C1
C1 = −
C2 = +
σA = −
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

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PROBLEM 4.159
(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
 zA 
 xA 
 2  x +  2  z = −1
 rz 
 rx 
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
σE = −
=−
Ix
I
, rz2 = z
A
A
M z = − Px A
P M z xE M x z E
P Px A xE
Pz z
+
−
=− −
− A 2E
2
A
Iz
Ix
A
Arz
Arx
z  
P   xA 
1 +  2  xE +  A2  z E  = 0
A   rz 
 rx  
if E lies on neutral axis.
z 
x 
1 +  A2  x +  A2  z = 0,
 rz 
 rx 
(b)
M x = PzE
σA = −
 zA 
 xA 
 2  x +  2  z = −1
 rz 
 rx 

M z = − PxE
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).

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PROBLEM 4.160
(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
(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
1 3
1 3
hb
Ix =
bh
12
12
h
b
zA = −
xA = −
2
2
Iz =
A = bh
Let P be the load point.
M z = − PxP
σA = −
σ A = 0,
1−
M x = Pz P
P M z xA M x z A
+
−
A
Iz
Ix
( )
=−
(− PxP ) − b2
( Pz P − 2h )
P
+
−
3
1 hb
1 bh3
bh
12
12
=−
P 
x
z 
1− P − P 

bh 
b/6 h/6 
x
z
−
=0
b/6 h/6
(a)
For
(b)
At point E:
z = 0 ∴ xE = b/6
At point F:
x = 0 ∴ z F = h /6
x
z
+
=1
b/6 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 σ B = 0, σ C = 0, and σ D = 0, the other portions producing tensile stresses are identified.
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PROBLEM 4.161
For the machine component and loading shown, determine the
stress at point A when (a) h = 2 in., (b) h = 2.6 in.
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
r1 = 3 − 2 = 1 in.
R=
2
= 1.8205 in.
ln 13
At point A:
σA =
r =
1
(3 + 1) = 2 in.
2
e = 2 − 1.8205 = 0.1795 in.
r = r1 = 1 in.
M (r − R) (−4)(1 − 1.8205)
=
= 12.19 ksi
Aer
(1.5)(0.1795)(1)
σ A = 12.19 ksi 
(b)
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:
σA =
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
Aer
(1.95)(0.4096)(0.4)
σ A = 11.15 ksi 
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PROBLEM 4.162
For the machine component and loading shown, determine the
stress at points A and B when h = 2.5 in.
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.
1
1
(r1 + r2 ) = (0.5 + 3.0) = 1.75 in.
2
2
2.5
h
R = r = 3.0 = 1.3953 in.
2
ln 0.5
ln
r =
r1
e = r − R = 1.75 − 1.3953 = 0.3547 in.
At point A:
r = r1 = 0.5 in.
σA =
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.)
σ A = 10.77 ksi 
r = r2 = 3 in.
σB =
M (r − R )
(−4 kip ⋅ in)(3 in. − 1.3953 in.)
=
Aer
(0.75 in. × 2.5 in.)(0.3547 in.)(3 in.)
σ B = −3.22 ksi 
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PROBLEM 4.163
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.
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=
h
ln
r2
r1
Also e = r − R
The maximum compressive stress occurs at point A. It is given by
σA = −
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
R − r1 = 10.8288 mm
P = 3 × 103 N ⋅ m, σ A = −150 × 106 Pa
(−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 =−
σ AA
=−
a = 93.37 − 32.5
a = 60.9 mm 
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PROBLEM 4.164
The curved portion of the bar shown has an inner radius of 20 mm.
Knowing that the line of action of the 3-kN force is located at a distance
a = 60 mm from the vertical plane containing the center of curvature
of the bar, determine the largest compressive stress in the bar.
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=
h
ln
r2
r1
.
Also e = r − R
The maximum compressive stress occurs at point A. It is given by
σA = −
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:
h = 25 mm, r1 = 20 mm, r2 = 45 mm, r = 32.5 mm
R=
(a + r )( R − r1)
er1
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
a = 60 mm, a + r = 92.5 mm, R − r1 = 10.8288 mm
(92.5)(10.8288)
K = 1+
= 30.968 P = 30 × 103 N
(1.6712)(20)
σA =
KP
(30.968)(3 × 103 )
=−
= −148.6 × 106 Pa
A
625 × 10−6
σ A = −148.6 MPa 
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PROBLEM 4.165
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.
SOLUTION
h = 40 mm, r1 = 15 mm, r2 = 55mm
A = (60)(40) = 2400 mm 2 = 2400 × 10−6 m 2
R=
r =
h
40
=
= 30.786 mm
r2
55
ln
ln
r1
40
1
(r1 + r2 ) = 35 mm
2
e = r − R = 4.214 mm
At r = 15mm,
σ =−
σ =−
My
Aer
y = 30.786 − 15 = 15.756 mm
(120)(15.786 × 10−3 )
= −12.49 × 10−6 Pa
(2400 × 10−6 )(4.214 × 10−3 )(15 × 10−3 )
σ = −12.49MPa 
At r = 55mm,
σ =−
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 )
σ = 5.22 MPa 
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PROBLEM 4.166
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.
SOLUTION
h = 40 mm, A = (60)(40) = 2400 mm2 = 2400 × 10−6 m 2 , M = 120 N ⋅ m
1 3
1
bh =
(60)(40)3 = 0.32 × 106 mm 4 = 0.32 × 10 −6 mm 4 ,
12
12
I =
c=
1
h = 20 mm
2
Assuming that the bar is straight
σs = −
(a)
Mc
(120)(20 × 10−8 )
=−
= 7.5 × 106 Pa = 7.5 MPa
I
(0.32 × 10−6 )
r1 = 20 mm r2 = 60 mm
h
40
=
= 36.4096 mm
60
r2
ln
ln
r1
20
R=
r =
1
(r1 + r2 ) = 40 mm
2
σa =
r1 − R = −16.4096 mm
e = r − R = 3.5904 mm
M (r1 − R)
(120)(−16.4096 × 10−3 )
=
= −11.426 × 106 Pa = −11.426 MPa
Aer
(2400 × 10−6 )(3.5904 × 10−3 )(20 × 10−3 )
% 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
σ , MPa
−11.426
−7.982
−7.546
% error
−34.4 %
6.0 % 
0.6 % 
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PROBLEM 4.167
The curved bar shown has a cross section of 30 × 30 mm. Knowing
that a = 60 mm, determine the stress at (a) point A, (b) point B.
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
h
.
r
ln 2
r1
e=r −R
The maximum compressive stress occurs at point A. It is given by
σA = −
P My A
P P (a + r ) y A
−
=− −
A Aer1
A
Aer1
= −K
Thus,
P
A
with
K =1+
y A = R − r1
(a + r )( R − r1)
er1
h = 30 mm, r1 = 20 mm, r2 = 50 mm, r = 35 mm
Data:
R=
30
= 32.7407 mm, e = 35 − 32.7407 = 2.2593 mm
50
ln
20
b = 30 mm,
A = bh = (30)(30) = 900 mm 2 = 900 × 10−6 m 2
a = 60 mm, a + r = 95 mm, R − r1 = 12.7407 mm
K =1+
(95)(12.7407)
= 27.786
(2.2593)(20)
P = 5 × 103 N
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PROBLEM 4.167 (Continued)
(a)
σA = −
KP
(27.786)(5 × 103 )
=−
= −154.4 × 106 Pa
A
900 × 10−6
σ A = −154.4 MPa 
(b)
At point B,
yB = r2 − R = 50 − 32.7407 = 17.2953 mm
P MyB
P
(a + r ) y B 
+
=  −1 +

A Aer2
A
er2

K ′P
(a + r ) y B
where K ′ =
=
−1
A
er2
σB = −
K′ =
(95)(17.2953)
− 1 = 13.545
(2.2593)(50)
σB =
(13.545)(5 × 103 )
= 75.2 × 106 Pa
−6
900 × 10
σ B = 75.2 MPa 
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PROBLEM 4.168
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.
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=
h
.
r
ln 2
r1
Also e = r − R
The maximum compressive stress occurs at point A . It is given by
σA = −
P My A
P P (a + r ) y A
−
=− −
A Aer1
A
Aer1
P
with y A = R − r1
A
(a + r )( R − r1)
Thus, K = 1 +
er1
= −K
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 = ?
σ A = −175 MPa = −175 × 106 Pa, P = 5 kN = 5 × 103 N
K =−
Solving (1) for
a+r =
Aσ A
(900 × 10−6 )(−175 × 106 )
=−
= 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 
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PROBLEM 4.169
Steel links having the cross section shown are available with different
central angles β . Knowing that the allowable stress is 12 ksi, determine
the largest force P that can be applied to a link for which β = 90°.
SOLUTION
Reduce section force to a force-couple system at G; the centroid of the cross section AB.
β

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=
h
ln
r2
r1
.
Also e = r − R
At point A the tensile stress is
K =1+
where
P=
Data:
σA =
P My A
P Pay A
P
ay 
P
−
=
+
= 1 + A  = K
A Aer1
A
Aer1
A
er1 
A
ay A
er1
and
y A = R − r1
Aσ A
K
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)
(0.24)(12)
= 0.65544 kips
4.3940
P = 655 lb 
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PROBLEM 4.170
Solve Prob. 4.169, assuming that β = 60°.
PROBLEM 4.169 Steel links having the cross section shown are
available with different central angles β. Knowing that the allowable
stress is 12 ksi, determine the largest force P that can be applied to a
link for which β = 90°.
SOLUTION
Reduce section force to a force-couple system at G, the centroid of the cross section AB.
β

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=
h
r2
r1
ln
.
Also e = r − R
At point A, the tensile stress is
σA =
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
and
y A = R − r1
Aσ A
K
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)
(0.24)(12)
= 1.128 kips
2.5525
P = 1128 lb 
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PROBLEM 4.171
A machine component has a T-shaped cross section that is orientated
as shown. Knowing that M = 2500 N ⋅ m, determine the stress at
(a) point A, (b) point B.
SOLUTION
Properties of the cross section.
r , mm
Part
b, mm
R=
A
 bi hi
 Ai
=
=
1
ri +1
r
  dA  bi ln
 bi ln i +1
r
ri
ri
h, mm
40
90
110
1
20
2
60

2200
R=
= 77.852 mm,
28.2588
A, mm 2
bi ln
50
20
ri +1
, mm
ri
r =
 Ai ri
 Ai
ri , mm
1000
16.2186
65
1200
12.0402
100
2200
28.2588
185000
r =
= 84.091 mm, e = r − R = 6.239 mm
2200
Ai ri , mm3
65,000
120,000
185,000
Stresses.
(a)
Point A:
σA =
(b)
M (r − R )
(2.5 kN ⋅ m)(0.040 m − 0.0778517 m)
=
Aer
(2.2 × 10−3 m 2 )(6.2392 × 10−3 m)(0.040 m)
Point B:
σB =
r = rA = 40 mm
σ A = −172.4 MPa 
r = rB = 110 mm
M (r − R )
(2.5 kN ⋅ m)(0.110 m − 0.0778517 m)
=
Aer
(2.2 × 10−3 m 2 )(6.2392 × 10−3 m)(0.110 m)
σ B = 53.2 MPa 
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PROBLEM 4.172
Assuming that the couple shown is replaced by a vertical 10-kN force
attached at point D and acting downward, determine the stress at
(a) point A, (b) point B.
SOLUTION
Properties of the cross section. R =
r , mm
40
90
110
Part
b, mm
1
20
2
60

2200
R=
= 77.852 mm,
28.2588
A
 bi hi
 Ai
=
=
1
r
r
  dA  bi ln i +1
 bi ln i +1
r
ri
ri
h, mm
50
20
A, mm 2
ri +1
, mm
ri
16.2186
12.0402
28.2588
bi ln
1000
1200
2200
185000
r =
= 84.091 mm,
2200
r =
 Ai ri
 Ai
ri , mm
Ai ri , mm3
65
100
65,000
120,000
185,000
e = r − R = 6.239 mm
Force-couple system at the centroid E.
P = 10 kN
M = (10 kN)(100 mm + 84.0909 mm) = 1840.9 N ⋅ m
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PROBLEM 4.172 (Continued)
Stresses.
(a)
Point A:
σA = −
r = rA = 40 mm
P M ( r − R)
10 kN
(1840.9 N ⋅ m)(0.040 m − 0.0778517 m)
+
=−
+
−3
2
A
Aer
2.2 × 10 m
(2.2 × 10−3 m 2 )(6.2392 × 10−3 m)(0.040 m)
= −4.545 MPa − 126.913 MPa
σ A = −131.5 MPa 
(b)
Point B:
σB = −
r = rB = 110 mm
P M ( r − R)
10 kN
(1840.9 N ⋅ m)(0.110 m − 0.0778517 m)
+
=−
+
−3
2
A
Aer
2.2 × 10 m
(2.2 × 10−3 m 2 )(6.2392 × 10−3 m)(0.110 m)
= −4.545 MPa + 39.196 MPa
σ B = 34.7 MPa 
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PROBLEM 4.173
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.
SOLUTION
R=
r =
r
3
3.5
5.5
6
Part
ΣA
Σbi hi
ΣA
=
=
1
r
Σ  r dA Σb ln ri +1
Σbi ln i +1
i
ri
ri
ΣAri
ΣA
b
h
A
b ln
ri +1
ri
r
Ar

3
0.5
1.5
0.462452
3.25
4.875

0.5
2
1.0
0.225993
4.5
4.5

2
0.5
1.0
0.174023
5.75
5.75
3.5
0.862468
Σ
15.125
3.5
15.125
= 4.05812 in., r =
= 4.32143 in.
0.862468
3.5
e = r − R = 0.26331 in.
R=
e = 0.263 in. 
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PROBLEM 4.174
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.
SOLUTION
R=
r =
r
3
3.5
5.5
6
Part
b
ΣA
Σbi hi
ΣA
=
=
1
r
r
Σ  r dA Σb ln i +1
Σbi ln i +1
i
ri
ri
ΣAri
ΣA
h
A
b ln
r
Ar

3
0.5
1.5
0.462452
3.25
4.875

0.5
2
1.0
0.225993
4.5
4.5

2
0.5
1.0
0.174023
5.75
5.75
3.5
0.862468
Σ
3.5
= 4.05812 in.,
0.862468
e = r − R = 0.26331 in.
R=
(a)
ri +1
ri
15.125
15.125
= 4.32143 in.
3.5
M = −8 kip ⋅ in
r =
y A = R − r1 = 4.05812 − 3 = 1.05812 in.
σA = −
My A
(−8)(1.05812)
=−
Aer1
(3.5)(0.26331)(3)
(b)
yB = R − r2 = 4.05812 − 6 = −1.94188 in. 

σB = −
(c)
yC = R − r = −e
σC = −
MyB
(−8)(−1.94188)

=−
Aer2
(3.5)(0.26331)(6)
MyC
Me
M
−8
=−
=−
=−
Aer
Aer
Ar
(3.5)(4.32143)
σ A = 3.06 ksi 
σ B = −2.81 ksi  
σ C = 0.529 ksi 
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PROBLEM 4.175
The split ring shown has an inner radius r1 = 20 mm and a
circular cross section of diameter d = 32 mm. For the loading
shown, determine the stress at (a) point A, (b) point B.
SOLUTION
1
d = 16 mm, r1 = 20 mm, r2 = r1 + d = 52 mm
2
r = r1 + c = 36 mm
c=
1
1
r + r 2 − c 2  = 36 + 362 − 162 



 2 
2
= 34.1245 mm
R=
e = r − R = 1.8755 mm
A=
π
4
d2 =
π
4
(32)2 = 804.25 mm 2 = 804.25 × 10−6 m 2
P = 2.5 × 103 N
(a)
Point A:
M = Pr = (2.5 × 103 )(36 × 10−3 ) = 90 N ⋅ m
y A = R − r1 = 34.1245 − 20 = 14.125 mm
σA = −
P My A
2.5 × 103
(90)(14.1245 × 10−3 )
−
=−
−
A Aer1
804.25 × 10−6 (804.25 × 10−6 )(1.8755 × 10−3 )(20 × 10−3 )
= −45.2 × 106 Pa
(b)
Point B:
σ A = −45.2 MPa 
yB = R − r2 = 34.1245 − 52 = −17.8755 mm
σB = −
P MyB
2.5 × 103
(90)(−17.8755 × 10−3 )
−
=−
−
A Aer2
804.25 × 10−6 (804.25 × 10−6 )(1.8755 × 10−3 )(52 × 10−3 )
= 17.40 × 106 Pa
σ B = 17.40 MPa 
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PROBLEM 4.176
The split ring shown has an inner radius r1 = 16 mm and a
circular cross section of diameter d = 32 mm. For the loading
shown, determine the stress at (a) point A, (b) point B.
SOLUTION
1
d = 16 mm, r1 = 16 mm, r2 = r1 + d = 48 mm
2
r = r1 + c = 32 mm
c=
1
1
r + r 2 − c 2  = 32 + 322 − 162 



 2 
2
= 29.8564 mm
R=
e = r − R = 2.1436 mm
A=
π
4
d2 =
π
4
(32)2 = 804.25 mm 2 = 804.25 × 10−6 m 2
P = 2.5 × 103 N
(a)
Point A:
M = Pr = (2.5 × 103 )(32 × 10−3 ) = 80 N ⋅ m
y A = R − r1 = 29.8564 − 16 = 13.8564 mm
σA = −
P My A
2.5 × 103
(80)(13.8564 × 10−3 )
−
=−
−
A Aer1
804.25 × 10−6 (804.25 × 10−6 )(2.1436 × 10−3 )(16 × 10−3 )
= −43.3 × 106 Pa
(b)
Point B:
σ A = −43.3 MPa 
yB = R − r2 = 29.8564 − 48 = −18.1436 mm
σB = −
P MyB
2.5 × 106
(80)(−18.1436 × 10−3 )
−
=−
−
A Aer2
804.25 × 10−6 (804.25 × 10−6 )(2.1436 × 10−3 )(48 × 10−3 )
= 14.43 × 106 Pa
σ B = 14.43 MPa 
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PROBLEM 4.177
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.
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
σ max occurs at A, which lies at the inner radius.
It is given by σ max =
My A
from which
Aer1
M =
Aer1 σ max
Also,
A = π c 2 = π (16) 2 = 804.25 mm 2
Data:
y A = R − r1 = 25.4891 − 12 = 13.4891 mm
M =
yA
.
(804.25 × 10−6 )(2.5109 × 10−3 )(12 × 10−3 )(60 × 106 )
13.4891 × 10−3
M = 107.8 N ⋅ m 
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PROBLEM 4.178
The bar shown has a circular cross section of 0.6 in. diameter.
Knowing that a = 1.2 in., determine the stress at (a) point A,
(b) point B.
SOLUTION
1
d = 0.3 in.
r = 0.5 + 0.3 = 0.8 in.
2
1
1
R =  r + r 2 − c 2  = 0.8 + 0.82 − 0.32 
 2 

2 
= 0.77081 in.
c=
e = r − R = 0.02919 in.
A = π c 2 = π (0.3)2 = 0.28274 in 2
P = 50 lb
M = − P(a + r ) = −50(1.2 + 0.8) = −100 lb ⋅ in
y A = R − r1 = 0.77081 − 0.5 = 0.27081 in.
yB = R − r2 = 0.77081 − 1.1 = −0.32919 in.
(a)
σA =
P My A
50
(−100)(0.27081)
−
=
−
= 6.74 × 103 psi
A Aer1
0.28274 (0.28274)(0.02919)(0.5)
σ A = 6.74 ksi 
(b)
σB =
P MyB
50
(−100)(−0.32919)
−
=
−
= −3.45 × 103 psi
A Aer2
0.28274 (0.28274)(0.02919)(1.1)
σ B = −3.45 ksi 
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PROBLEM 4.179
The bar shown has a circular cross section of 0.6 in. diameter.
Knowing that the allowable stress is 8 ksi, determine the largest
permissible distance a from the line of action of the 50-lb forces
to the plane containing the center of curvature of the bar.
SOLUTION
1
d = 0.3 in.,
r = 0.5 + 0.3 = 0.8 in.
2
1
1
R =  r + r 2 − c 2  = 0.8 + 0.82 − 0.32 





2
2 
= 0.77081 in.
e = r − R = 0.02919 in.
c=
A = π c 2 = π (0.3)2 = 0.28274 in 2
M = − P (a + r )
y A = R − r1 = 0.77081 − 0.5 = 0.27081 in.
σA =
=
P My A
P P( a + r ) y A
P
(a + r ) y A 
−
=
+
= 1 +

A Aer1
A
Aer1
A
er1

KP
A
where
K =1+
(a + r ) y A
er1
(8 × 103 )(0.28274)
= 45.238
P
50
( K − 1)er1
(44.238)(0.02919)(0.5)
a+r =
=
= 2.384 in.
yA
0.27081
K =
σ AA
=
a = 2.384 − 0.8
a = 1.584 in. 
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PROBLEM 4.180
Knowing that P = 10 kN , determine the stress at (a) point A, (b) point B.
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=
=
=
r2 r2
190 190
ln − 1
ln
− 1 0.355025
h r1
90 100
R = 126.752 mm
e = r − R = 130 mm − 126.752 mm = 3.248 mm
(a)
Point A:
σA = −
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)
= −2.778 MPa − 29.743 MPa
(b)
Point B:
σB = −
σ A = −32.5 MPa 
rB = 190 mm = 0.190 m
P M (rB − R)
10 kN
(1300 N ⋅ m)(0.190 m − 0.126752 m)
+
=−
+
−6 2
A
AerB
3600 × 10 m
(3600 × 10−6 m 2 )(3.248 × 10−3 m)(0.190 m)
= −2.778 MPa + 37.01 MPa
σ B = +34.2 MPa 
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PROBLEM 4.181
Knowing that M = 5 kip ⋅ in., determine the stress at (a) point A,
(b) point B.
SOLUTION
1
1
bh = (2.5)(3) = 3.75 in 2
2
2
r = 2 + 1 = 3.00000 in.
A=
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
=
(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.
(a)
y A = R − r1 = 0.84548 in.
σA = −
(b)
M = 5 kip ⋅ in
My A
(5) (0.84548)
=−
Aer1
(3.75)(0.15452) (2)
σ A = −3.65 ksi 
yB = R − r2 = −2.15452 in.
σB = −
MyB
(5)(−2.15452)
=−
Aer2
(3.75)(0.15452)(5)
σ B = 3.72 ksi 
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PROBLEM 4.182
Knowing that M = 5 kip ⋅ in, determine the stress at (a) point A,
(b) point B.
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
b2 = 2.5 in.,
r2 = 5 in.
b1 = 0.
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)
y A = R − r1 = 1.85466 in.
σA = −
(b)
My A
(5) (1.85466)
=−
Aer1
(3.75)(0.14534) (2)
σ A = −8.51 ksi 
yB = R − r2 = −1.14534 in.
σB = −
MyB
(5) (−1.14534)
=−
Aer2
(3.75)(0.14534) (5)
σ B = 2.10 ksi 
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PROBLEM 4.183
For the curved beam and loading shown, determine the stress at
(a) point A, (b) point B.
SOLUTION
Locate centroid.
A, mm 2
r , mm
Ar , mm3

600
45
27 × 103

300
55
16.5 × 103
Σ
900
r =
R=
=
43.5 × 103
43.5 × 103
= 48.333 mm
900
1
2
h 2 (b1 + b2 )
(b1r2 − b2r1) ln
r2
r1
− h(b2 − b1)
(0.5)(30) 2 (40 + 20)
= 46.8608 mm
− (30)(40 − 20)
[(40)(65) − (20)(35)]ln 65
35
e = r − R = 1.4725 mm
(a)
y A = R − r1 = 11.8608 mm
σA = −
(b)
M = −250 N ⋅ m
My A
(−250)(11.8608 × 10−3 )
=−
= 63.9 × 106 Pa
Aer1
(900 × 10−6 )(1.4725 × 10−3 )(35 × 10−3 )
σ A = 63.9 MPa 
yB = R − r2 = −18.1392 mm
σB = −
MyB
(−250)(−18.1392 × 10−3 )
=−
= −52.6 × 106 Pa
Aer2
(900 × 10−6 )(1.4725 × 10−3 )(65 × 10−3 )
σ B = −52.6 MPa 
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PROBLEM 4.184
For the crane hook shown, determine the largest tensile stress in
section a-a.
SOLUTION
Locate centroid.
A, mm 2
r , mm
Ar , mm3

1050
60
63 × 103

750
80
60 × 103
Σ
1800
r=
Force-couple system at centroid:
103 × 103
103 × 103
= 63.333 mm
1800
P = 15 × 103 N
M = − Pr = −(15 × 103 )(68.333 × 10−3 ) = −1.025 × 103 N ⋅ m
1
2
R=
=
h 2 (b1 + b2 )
(b1r2 − b2 r1 ) ln
r2
r1
− h(b1 − b2 )
(0.5)(60) 2 (35 + 25)
= 63.878 mm
[(35)(100) − (25)(40)]ln 100
− (60)(35 + 25)
40
e = r − R = 4.452 mm
Maximum tensile stress occurs at point A.
y A = R − r1 = 23.878 mm
σA =
=
P My A
−
A Aer1
15 × 103
−(1.025 × 103 )(23.878 × 10−3 )
−
1800 × 10−6 (1800 × 10 −6 )(4.452 × 10−3 )(40 × 10−3 )
= 84.7 × 106 Pa
σ A = 84.7 MPa 
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PROBLEM 4.185
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.
SOLUTION
Locate centroid.
A, in 2
r , in.
Ar , in 3

10.5
6
63

7.5
8
60
Σ
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)
σA = − A = −
(18)(0.4452)(4)
Aer1
yB = R − r2 = 6.3878 − 10 = −3.6122 in.
My
(80)( −3.6122)
σB = − B = −
(18)(0.4452)(10)
Aer2
M = 80 kip ⋅ in
σ A = −5.96 ksi 
σ B = 3.61 ksi 
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PROBLEM 4.186
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.
SOLUTION
Locate centroid.
A, in 2
r , in.
Ar , in 3

7.5
6
45

10.5
8
84
Σ
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)
y A = R − r1 = 2.7168 in.
σA = −
(b)
M = 80 kip ⋅ in
My A
(80)(2.7168)
=−
(18)(0.4499)(4)
Aer1
σ A = −6.71 ksi 
yB = R − r2 = −3.2832 in.
σB = −
MyB
(80)( −3.2832)
=−
(18)(0.4499)(10)
Aer2
σ B = 3.24 ksi 
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PROBLEM 4.187
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
R=
A
 r   r b2  r b3 
ln  2   3   4  
 r1   r2   r3  


b1
where A is the total area of the cross section.
SOLUTION
R=
=
Note that for each rectangle,

ΣA
A
=
1
Σ dA Σ b ln ri +1
i
r
r

i
A
r 
Σ ln  i +1 
 ri 
bi
=
A
 r b1  r b2  r b3 
ln  2   3   4  
 r1   r2   r3  



ri + 1 dr
1
d A=
bi
ri
r
r
ri + 1 dr
ri +1
= bi
= bi ln
r2 r
ri


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PROBLEM 4.188
Using Eq. (4.66), derive the expression for R given in Fig. 4.73 for a circular cross section.
SOLUTION
Use polar coordinate β as shown. Let w be the width as a function of β
w = 2 c sin β
r = r − c cos β
dr = c sin β d β
dA = w dr = 2c 2 sin 2 β d β

dA
=
r

dA
=
r


=2
=2
2 c 2 sin β
dβ
r − c cos β
π
0
c 2 (1 − cos 2 β )
dβ
r − c cos β
π
0


r 2 − c 2 cos 2 β − (r 2 − c 2 )
dβ
r − c cos β
π
0
π
0
(r + c cos β ) d β − 2(r 2 − c 2 )
= 2r β
π
0
2
+ 2c sin β
2
− 2( r − c )

π
0
dr
r − c cos β
π
0
2
r 2 − c2
tan
−1
( )
r 2 − c 2 tan 1 β
2
r +c
π
0
π

= 2r (π − 0) + 2c(0 − 0) − 4 r 2 − c 2 ⋅  − 0 
2

= 2π r − 2π r 2 − c 2
A = π c2
R=

1
=
2
A
c2
r + r 2 − c2
1
π c2
=
=
×
dA 2π r − 2π r 2 − c 2 2 r − r 2 − c 2 r + r 2 − c 2
r
(
c 2 r + r 2 − c2
2
2
2
r − (r − c )
) = 1 c (r +
2
2
r 2 − c2
c
2
)=1 r+
2(
)
r 2 − c2 

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PROBLEM 4.189
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
r2 b1 − 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
r2b1 − r1b2 r2 b1 − b2
=
ln −
h
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
(r2 b1 − r1b2 ) ln r2 − h(b1 − b2 )
r


1
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PROBLEM 4.190
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
r
r
2 w
2 c + c r
dA
0
1
dr =
dr
=
r
r
r
r
1 r
1
r2
r2
= c0 ln r + c1 r


r1
r1
r
= c0 ln 2 + c1 (r2 − r1 )
r1
br
r b
= 2 ln 2 − 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
2
R=
=
= r 2r
r
r
2
dA
ln r2 − 1
b h2 ln r2 − 1
h
r
1

(
1
)

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PROBLEM 4.191
For a curved bar of rectangular cross section subjected to a
bending couple M, show that the radial stress at the neutral
surface is
r1
M 
R
σr =
1 − − ln 
Ae 
R
r1 
and compute the value of σ r for the curved bar of Examples
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
σr =
At radial distance r,

H = σ r dA =

R
r1
σ r bdr
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 

=


Fr = σ r cos β dA
Resultant of σ n :
=
θ /2
θ
− /2
σ r cos β (bRd β ) = σ r bR
θ /2
= σ r bR sin β
For equilibrium: Fr − 2 H sin
2σ r bR sin
θ
2
θ
2
−θ /2
θ /2
θ
− /2
cos β d β
= 2σ r bR sin θ
2
=0
−2
θ
MbR  r1
R
1 − − ln  sin = 0
Ae  R
r1 
2
σr =
M  r1
R
1 − − ln 
Ae 
R
r1 

Using results of Examples 4.10 and 4.11 as data,
M = 8 kip ⋅ in,
σr =
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 
σ r = −0.54 ksi 
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PROBLEM 4.192
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.
SOLUTION
A
y0
A y0

18
5
90

18
1
18
Σ
36
Y0 =
108
108
= 3 in.
36
Neutral axis lies 3 in. above the base.
1
1
b1h13 + A1d12 = (3)(6)3 + (18)(2) 2 = 126 in 4
12
12
1
1
I 2 = b2 h23 + A2 d 22 = (9)(2)3 + (18)(2) 2 = 78 in 4
12
12
I = I1 + I 2 = 126 + 78 = 204 in 4
I1 =
ytop = 5 in. ybot = −3 in.
M − Pa = 0
M = Pa = (15)(40) = 600 kip ⋅ in
σ top = −
σ bot = −
M ytop
I
=−
(600)(5)
204
M ybot
(600)(−3)
=−
I
204
σ top = −14.71 ksi (compression) 
σ bot = 8.82 ksi (tension) 
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PROBLEM 4.193
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
1
d,
2
ρ = radius of curvature of centerline of rods when bent.
d = diameter of rod,
ρ=
I=
(a)
σ max =
(b)
M=
Ec
EI
ρ
ρ
=
=
c=
1
1
1
1
D − d = (1.25) − (6 × 10−3 ) = 0.622 m
2
2
2
2
π
4
c4 =
π
4
(0.003) 4 = 63.617 × 10−12 m 4
(200 × 109 )(0.003)
= 965 × 106 Pa
0.622
(200 × 109 )(63.617 × 10−12 )
= 20.5 N ⋅ m
0.622
σ = 965 MPa 
M = 20.5 N ⋅ m 
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PROBLEM 4.194
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.
SOLUTION
 = rectangle
 = semi-circular cutout
A1 = (2.4)(1.2) = 2.88 in 2
A2 =
π
2
(0.75) 2 = 0.8836 in 2
A = 2.88 − 0.8836 = 1.9964 in 2
y1 = 0.6 in.
4r
(4)(0.75)
=
= 0.3183 in.
3π
3π
ΣA y
(2.88)(0.6) − (0.8836)(0.3183)
Y =
=
= 0.7247 in.
ΣA
1.9964
y2 =
Neutral axis lies 0.7247 in. above the bottom.
Moment of inertia about the base:
1
π
1
π
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.
|σ | =
My
I
M=
σI
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 
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PROBLEM 4.195
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.
Ea
70
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 ) − 67,375mm 4 = 45,125mm 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)
ys = 13mm = 0.013m
ns Mys
(2.857)(300)(0.013)
=
= 46.9 × 106 Pa
I
237.615 × 10−9
Maximum stress in aluminum:
σa =
(c)
ns = 2.857
Maximum stress in steel:
σs =
(b)
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: ρ =
EI
M
ρ =
σ s = 46.9 MPa 
(70 × 109 )(237.615 × 10−9 )
300
σ a = 18.94 MPa 
ρ = 55.4 m 
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PROBLEM 4.196
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:
∈A = −500μ
∈B = −1000μ
∈C = −200μ
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.
SOLUTION
1
(5)(3)3 = 11.25 in 4
12
M x = Pz
M z = − Px
Ix =
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.
σ A = Eε A = (29 × 106 )(−500 × 10−6 ) = −14,500 psi = −14.5 ksi
σ B = Eε B = (29 × 106 )(−1000 × 10−6 ) = −29,000 psi = −29 ksi
σ C = Eε C = (29 × 106 )(−200 × 10−6 ) = −5800 psi = −5.8 ksi
σA = −
P M x z A M z xA
−
+
= −0.06667 P − 0.13333 M x − 0.08 M z
A
Ix
Iz
(1)
σB = −
P M x z B M z xB
−
+
= −0.06667 P − 0.13333 M x + 0.08 M z
A
Ix
Iz
(2)
σC = −
P M x zC
M x
−
+ z C = −0.06667 P + 0.13333 M x + 0.08 M z
A
Ix
Iz
(3)
Substituting the values for σ A , σ B , and σ C into (1), (2), and (3) and solving the simultaneous equations
gives
M x = 87 kip ⋅ in, M z = −90.625 kip ⋅ in,
(a) P = 152.25 kips 

x=−

z =
Mz
−90.625
=−

P
152.25
Mx
87
=

P
152.25
σD = −

(c)
(b) x = 0.595 in. 
z = 0.571 in. 
P M x z D M z xD
−
+
= −0.06667 P + 0.13333 M x − 0.08 M z
A
Ix
Iz
= −(0.06667)(152.25) + (0.13333)(87) − (0.08)(−90.625) = 8.70 ksi 
Strain at hidden edge:
ε =
σD
E
=
8.70 × 103
29 × 106
 
ε = 300 u 
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PROBLEM 4.197
For the split ring shown, determine the stress at (a) point A,
(b) point B.
SOLUTION
r1 =
1
40 = 20 mm,
2
r2 =
A = (14)(25) = 350 mm 2
1
(90) = 45 mm h = r2 − r1 = 25 mm
2
25
h
R = r = 45 = 30.8288 mm
2
ln 20
ln r
1
1
r = (r1 + r2 ) = 32.5 mm
2
e = r − R = 1.6712 mm
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 )
6
= −82.4 × 10 Pa
σA = −
(b)
Point B:
rB = 45 mm
σ A = −82.4 MPa 
yB = 30.8288 − 45 = −14.1712 mm
2500
(81.25)( −14.1712 × 10−3 )
P MyB
−
=−
−
A AerB
350 × 10−6 (350 × 10−6 )(1.6712 × 10−3 )(45 × 10−3 )
= 36.6 × 106 Pa
σB = −
σ B = 36.6 MPa 
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PROBLEM 4.198
A couple M of moment 8 kN ⋅ m acting in a vertical plane is
applied to a W200 × 19.3 rolled-steel beam as shown. Determine
(a) the angle that the neutral axis forms with the horizontal plane,
(b) the maximum stress in the beam.
SOLUTION
For W200 × 19.3 rolled steel sector,
I z′ = 16.6 × 106 mm 4 = 16.6 × 10−6 m 4
I y′ = 1.15 × 106 mm 4 = 1.15 × 10−6 m 4
203
= 101.5 mm
2
102
= zE =
= 51 mm
2
y A = yB = − yD = − yE =
z A = − zB = − zD
M z = (8 × 103 ) cos 5° = 7.9696 × 103 N ⋅ m
M y = −(8 × 103 )sin 5° = −0.6972 × 103 N ⋅ m
(a)
tan ϕ =
Iz
16.6 × 10−6
tan θ =
tan(−5°) = −1.2629
Iy
1.15 × 10−6
ϕ = −51.6°
α = 51.6° − 5°
(b)
α = 46.6° 
Maximum tensile stress occurs at point D.
σD = −
(7.9696 × 103 )(−101.5 × 10−3 ) (−0.6972 × 103 )(−51 × 10−3 )
M z yD M y z D
+
=−
+
Iz
Iy
16.6 × 10−6
1.15 × 10−6
= 79.6 × 106 Pa
σ D = 79.6 MPa 
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PROBLEM 4.199
Determine the maximum stress in each of the two machine
elements shown.
SOLUTION
For each case, M = (400)(2.5) = 1000 lb ⋅ in
At the minimum section,
1
(0.5)(1.5)3 = 0.140625 in 4
12
c = 0.75 in.
I =
(a)
D/d = 3/1.5 = 2
r/d = 0.3/1.5 = 0.2
From Fig 4.32,
σ 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,
σ max =
K = 1.75
σ 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
σ max = 8.00 ksi 
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PROBLEM 4.200
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.
SOLUTION
Moment of inertia about centroid:
 a 
1
2 2t 

12
 2
1
= ta3
12
(
I=
Area:
)
3
 a 
a
A = 2 2t 
 = 2at , c =
2 2
 2
(
)
(a)
P
P Pec
P
σA = −
=
−
A
I
2at
(b)
P
P Pec
P
σB = +
=
+
2at
A
I
(c)
σC = σ A
( )( )
a
2 2
1
ta
12
a
2 2
3
( )( )
a
2
1
ta 3
12
a
2
σA = −
P

2at
σB = −
2P

at
σC = −
P

2at
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PROBLEM 4.201
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 σ 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.
SOLUTION
A1 = (120)(10) = 1200 mm 2
R1 = σ Y A1 = (300 × 106 )(1200 × 10− 6 ) = 360 × 103 N
A2 = (30)(10) = 300 mm 2
R2 = σ Y A2 = (300 × 106 )(300 × 10−6 ) = 90 × 103 N
A3 = (30)(10) = 300 mm 2
1
1
R3 = σ 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)}
= 56.7 × 103 N ⋅ m
(b)
y3 = 20 mm = 20 × 10−3 m
yY
ρ
=
σY
E
ρ=
EyY
σY
M = 56.7 kN ⋅ m 
=
(200 × 109 )(30 × 10−3 )
300 × 106
ρ = 20 m 
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PROBLEM 4.202
A short column is made by nailing four 1 × 4-in. planks to a 4 × 4-in.
timber. Determine the largest compressive stress created in the column by a
16-kip load applied as shown in the center of the top section of the timber 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.
SOLUTION
(a)
M =0
Centric loading:
σ =−
P
A
A = (4)(4) + (4)(1)(4) = 32 in 2
σ =−
(b)
Eccentric loading:
M = Pe
σ =−
16 × 103
32
σ = −500 psi 
P Pec
−
A
I
A = (4) (4) + (3)(1)(3) = 28 in 2
e= y
ΣAy (1)(4)(2.5)
y=
=
= 0.35714 in.
28
A
I = Σ( I + Ad 2 ) =
+
1
(4)(1)3 + (4)(1)(2.14286) 2 = 53.762 in 4
12
σ =−
(c)
Centric loading:
1
(6)(4)3 + (6) (4)(0.35714) 2
12
16 × 103 (16 × 103 ) (0.35714) (2.35714)
−
28
53.762
M =0
σ =−
σ = −822 psi 
P
A
A = (6) (4) = 24 in 2
σ =−
16 × 103
24
σ = −667 psi 
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PROBLEM 4.202 (Continued)
(d)
Eccentric loading:
M = Pe
σ =−
P Pec
−
A
I
A = (4) (1) = (1)(4)(1) = 20 in 2
x = 2.5 − 2 = 0.5 in.
1
I = (4)(5)3 = 41.667 in 4
12
σ =−
(e)
Centric loading:
M =0
16 × 103 (16 × 103 ) (0.5) (2.5)
−
20
41.667
σ =−
e=x
σ = −1280 psi 
P
A
A = (4)(4) = 16 in 2
σ =−
16 × 103
16
σ = −1000 psi 
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PROBLEM 4.203
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 σ1 the maximum stress and
by ρ1 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
M 1c σ M1
=
I
bt 2
1
M
12 M1
= 1 =
ρ1 EI1
Et 3
σ1 =
After M1 is applied to each of the strips, the stresses are those given in the sketch above. They are
σ A = −σ1, σ B = σ1, σ C = −σ1, σ D = σ 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
σ′ = −
M ′y
2M y
3M1 y
= − 2 13 = −
I
bt
bt 2
3
σ ′A = −
3M1
1
3M 1 1
= − σ1, σ ′B = σ C′ = 0, σ ′D =
= σ1
2
2
2
bt
bt 2
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PROBLEM 4.203 (Continued)
(a)
Final stresses:
1
2
σ A = −σ1 − (− σ 1)
1
2
σ A = − σ1 
σ B = σ1 
σ C = −σ1 
1
2
σ D = σ1 − σ1
1
ρ′
(b)

=
σD =
1
σ1 
2
M′
2M
3M 1 1 1
= 2 13 =
=
EI ′
4 ρ′
E 3 bt
Et 3
Final radius:
1
ρ
=
1
ρ1
−
1
1
1 1
3 1
=
−
=
ρ ′ ρ1 4 ρ1 4 ρ1
ρ =
4
ρ1 
3
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