Chapters 4,6,7,8,9,10,12,17 Covered
SOLUTIONS
, TABLE OF CONTENTS
CHAPTER 4 ................................................................................................................................... 3
CHAPTER 6 ................................................................................................................................. 27
CHAPTER 7 ................................................................................................................................. 33
CHAPTER 8 ................................................................................................................................. 51
CHAPTER 9 ................................................................................................................................. 69
CHAPTER 10 ............................................................................................................................... 81
CHAPTER 12 ............................................................................................................................. 108
CHAPTER 17 ............................................................................................................................. 118
Problem 17.3 ............................................................................................................................... 122
Problem 17.4 ............................................................................................................................... 124
Problem 17.5 ............................................................................................................................... 126
Problem 17.6................................................................................................................................ 127
Problem 17.8................................................................................................................................ 131
Problem 17.12.............................................................................................................................. 146
Problem 17.15.............................................................................................................................. 158
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,CHAPTER 4
Problem 4.1
Determine the velocity of propagation of longitudinal waves traveling along a laterally con-
strained rod when the rod is made of (a) steel; (b) cast iron; and (c) concrete with f 'c = 4,000 psi.
Solution:
Young’s moduli, Poisson ratios, and unit weights for steel, cast iron, and concrete with
f 'c =4,000 psi are as shown in Table P4.1
Table P4.1. Properties of steel, cast iron, and concrete
Material Modulus of elasticity Poisson ratio Unit weight
(psi) (pcf)
Steel 30×10 6
0.27 490
Cast iron 26×10 6
0.25 485
Concrete 57,000 f ′ c
0.15 150
Therefore, for the steel rod, the constrained modulus of elasticity and the propagation velocity of
longitudinal waves are respectively equal to (see Equations 4.6 and 4.7)
E (1 − μ) 30 × 106 (1 − 0.27)
M= = = 37.5 × 106 psi
(1 − 2μ)(1 + μ) [1 − 2(0.27)](1 + 0.27)
M 37.5 × 106 (144)
vc = = = 18,838 ft/s = 5.74 km/s
ρ .2
and similarly for the cast iron and reinforced concrete rods,
E (1 − μ) 26 × 106 (1 − 0.25)
M= = = 31.2 × 106 psi
(1 − 2μ)(1 + μ) [1 − 2(0.25)](1 + 0.25)
M 31.2 × 106 (144)
vc = = = 17,271 ft/s = 5.26 km/s
ρ .2
E (1 − μ) 57,000 4,000 (1 − 0.15)
M= = = 3.8 × 106 psi
(1 − 2μ)(1 + μ) [1 − 2(0.15)](1 + 0.15)
M 3.8 × 106 (144)
vc = = = 10,838 ft/s = 3.30 km/s
ρ .2
Problem 4.2
A rod of infinite length is subjected to an initial longitudinal displacement given by
u0 = 2(1 − x) 0 ≤ x ≤1
u0 = 2 + x -2 ≤ x ≤ 0
Draw plots of the rod’s longitudinal displacement u against the position variable x at times t = 1,
2, 3, and 4 seconds. Consider that the velocity of propagation of longitudinal waves in the rod is
equal to 0.5 m/s.
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, Solution:
Noticing that
u0 = 0 at x = −2 and x = +1
u0 = 2 at x = 0
the form of the initial pulse is as shown below. Note also that the initial displacement generates
two identical waves traveling in opposite directions. Furthermore, since the velocity of propaga-
tion is 0.5 m/s, the distance traveled by these waves are as indicated in the Table P4.2.
Table P4.2. Distance traveled by waves at different times
Time (s) Distance (m)
1.0 0.5
2.0 1.0
3.0 1.5
4.0 2.0
Therefore, the position of the initial displacement pulse at times of 1.0, 2.0, 3.0, and 4.0 seconds
is as indicated in Figure P4.2.
u
2
t =0s
x
2
t =1s
x
2
t =2s
x
2
t =3s
x
2
t =4s
-5 -4 -3 -2 -1 0 1 2 3 4 5 x
Figure P4.2. Position of displacement pulse at various times
Problem 4.3
Repeat Problem 4.2 considering an initial longitudinal velocity instead of an initial displacement
and that this initial velocity is given by
v0 = A - 2 ≤ x ≤ 2
v0 = 0 elsewhere
where A is a constant.
Solution:
According to Equation 4.19 and a zero initial displacement, the displacement in the rod is given
by
x + vc t
1
u ( x, t ) =
2vc ∫ v (ζ)dζ
x − vc t
0
which may be considered as the superposition of the two displacement waves
x + vc t x − vc t
1 1
u ( x, t ) =
2vc ∫0 v0 (ζ)dζ − 2vc ∫ v (ζ)dζ
0
0
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SOLUTIONS
, TABLE OF CONTENTS
CHAPTER 4 ................................................................................................................................... 3
CHAPTER 6 ................................................................................................................................. 27
CHAPTER 7 ................................................................................................................................. 33
CHAPTER 8 ................................................................................................................................. 51
CHAPTER 9 ................................................................................................................................. 69
CHAPTER 10 ............................................................................................................................... 81
CHAPTER 12 ............................................................................................................................. 108
CHAPTER 17 ............................................................................................................................. 118
Problem 17.3 ............................................................................................................................... 122
Problem 17.4 ............................................................................................................................... 124
Problem 17.5 ............................................................................................................................... 126
Problem 17.6................................................................................................................................ 127
Problem 17.8................................................................................................................................ 131
Problem 17.12.............................................................................................................................. 146
Problem 17.15.............................................................................................................................. 158
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,CHAPTER 4
Problem 4.1
Determine the velocity of propagation of longitudinal waves traveling along a laterally con-
strained rod when the rod is made of (a) steel; (b) cast iron; and (c) concrete with f 'c = 4,000 psi.
Solution:
Young’s moduli, Poisson ratios, and unit weights for steel, cast iron, and concrete with
f 'c =4,000 psi are as shown in Table P4.1
Table P4.1. Properties of steel, cast iron, and concrete
Material Modulus of elasticity Poisson ratio Unit weight
(psi) (pcf)
Steel 30×10 6
0.27 490
Cast iron 26×10 6
0.25 485
Concrete 57,000 f ′ c
0.15 150
Therefore, for the steel rod, the constrained modulus of elasticity and the propagation velocity of
longitudinal waves are respectively equal to (see Equations 4.6 and 4.7)
E (1 − μ) 30 × 106 (1 − 0.27)
M= = = 37.5 × 106 psi
(1 − 2μ)(1 + μ) [1 − 2(0.27)](1 + 0.27)
M 37.5 × 106 (144)
vc = = = 18,838 ft/s = 5.74 km/s
ρ .2
and similarly for the cast iron and reinforced concrete rods,
E (1 − μ) 26 × 106 (1 − 0.25)
M= = = 31.2 × 106 psi
(1 − 2μ)(1 + μ) [1 − 2(0.25)](1 + 0.25)
M 31.2 × 106 (144)
vc = = = 17,271 ft/s = 5.26 km/s
ρ .2
E (1 − μ) 57,000 4,000 (1 − 0.15)
M= = = 3.8 × 106 psi
(1 − 2μ)(1 + μ) [1 − 2(0.15)](1 + 0.15)
M 3.8 × 106 (144)
vc = = = 10,838 ft/s = 3.30 km/s
ρ .2
Problem 4.2
A rod of infinite length is subjected to an initial longitudinal displacement given by
u0 = 2(1 − x) 0 ≤ x ≤1
u0 = 2 + x -2 ≤ x ≤ 0
Draw plots of the rod’s longitudinal displacement u against the position variable x at times t = 1,
2, 3, and 4 seconds. Consider that the velocity of propagation of longitudinal waves in the rod is
equal to 0.5 m/s.
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, Solution:
Noticing that
u0 = 0 at x = −2 and x = +1
u0 = 2 at x = 0
the form of the initial pulse is as shown below. Note also that the initial displacement generates
two identical waves traveling in opposite directions. Furthermore, since the velocity of propaga-
tion is 0.5 m/s, the distance traveled by these waves are as indicated in the Table P4.2.
Table P4.2. Distance traveled by waves at different times
Time (s) Distance (m)
1.0 0.5
2.0 1.0
3.0 1.5
4.0 2.0
Therefore, the position of the initial displacement pulse at times of 1.0, 2.0, 3.0, and 4.0 seconds
is as indicated in Figure P4.2.
u
2
t =0s
x
2
t =1s
x
2
t =2s
x
2
t =3s
x
2
t =4s
-5 -4 -3 -2 -1 0 1 2 3 4 5 x
Figure P4.2. Position of displacement pulse at various times
Problem 4.3
Repeat Problem 4.2 considering an initial longitudinal velocity instead of an initial displacement
and that this initial velocity is given by
v0 = A - 2 ≤ x ≤ 2
v0 = 0 elsewhere
where A is a constant.
Solution:
According to Equation 4.19 and a zero initial displacement, the displacement in the rod is given
by
x + vc t
1
u ( x, t ) =
2vc ∫ v (ζ)dζ
x − vc t
0
which may be considered as the superposition of the two displacement waves
x + vc t x − vc t
1 1
u ( x, t ) =
2vc ∫0 v0 (ζ)dζ − 2vc ∫ v (ζ)dζ
0
0
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