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ALL 12 CHAPTERS COVERED
SOLUTIONS MANUAL
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CONTENTS
Chapter Page
2 ............................................................................................................................. 1
4 9
5 25
6 41
8 ........................................................................................................................... 51
9 ........................................................................................................................... 61
10 ........................................................................................................................... 65
11 ........................................................................................................................... 79
12 ........................................................................................................................... 91
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Chapter 2
2.1 a. Spring constant, k : The change in the force per unit length change of the spring.
b. Coefficient of subgrade reaction, k :
k
Spring constant divided by the foundation contact area, k
A
c. Undamped natural circular frequency: n rad/s
W
where m = mass =
g
d. Undamped natural frequency: f n (in Hz)
2 m
Note: Circular frequency defines the rate of oscillation in term of radians per unit
time; 2π radians being equal to one complete cycle of rotation.
e. Period, T: The time required for the motion to begin repeating itself.
n
f. Resonance: Resonance occurs when 1
g. Critical damping coefficient: cc 2 km
W
where k = spring constant; m = mass =
g
h. c c
Damping ratio: D =
cc 2 km
where c = viscous damping coefficient; cc = critical damping coefficient
i. Damped natural frequency:
d n 1 D2
fd 1 D2 fn
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2.2 Weight of machine + foundation, W = 400 kN
Spring constant, k = 100,000 kN/m
W 400 kN
Mass of the machine + foundation, m = = = 40.77
g 9.81 m s2
Natural frequency of undamped free vibration is [Eq. (2.19)]
1 k
fn = = 7.88 Hz
2 m 2 40.77
1 1
From Eq. (2.18), T = = = 0.127 s
fn 7.88
2.3 Weight of machine + foundation, W = 400 kN
Spring constant, k = 100,000 kN/m
Static deflection of foundation is [Eq. (2.2)]
z 400
4 10 3 m = 4 mm
W 100,000
s
k
2.4 External force to which the foundation is subjected, Q 35.6sin t kN
f = 13.33 Hz
Weight of the machine + foundation, W = 178 kN
Spring constant, k = 70,000 kN/m
For this foundation, let time t = 0, z = z0 = 0, zɺ = v0 = 0
W 178 kN
a. Mass of the machine + foundation, m = = = 18.145
g 9.81 m s2
k
n = 62.11 rad/s
m
2 2
T = 0.101 s
n 62.11
b. The frequency of loading, f = 13.33 Hz
2 f 2 (13.33) 83.75 rad/s
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Force due to forced part, F1 k Q0 2 k 2 sin t
1 n
35.6 70, 000
(70, 000) 2 2
sin(83.75t)
1 83.75 62.11
43.51sin(83.75t ) kN
See the plot below for F1 vs. t
c. Force due to free part, F k Q0 k sin t
n
n 2 35.6 70,
2 2
1 n
000 83.75
70, 000 sin(62.11t)
1 83.75 62.112
2
62.11
58.67sin(62.11t ) kN
See the plot above in Part b for F2 vs. t.
d. Total dynamic force on the subgrade:
F F1 F2 43.51sin(83.75t ) 58.67(62.11t ) kN
The plot of variation of the dynamic force on the subgrade of the foundation due
to (a) forced part, (b) free part, and (c) total of the response for time t = 0 to t = 2T
is shown in the figure above (Part b).
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2.5 The natural frequency of the undamped free vibration of the spring mass system is given by
1 keq
fn where keq = equivalent stiffness of the spring system
2 m
For springs attached in series, the equivalent stiffness is given by
1 1 1 k1k2
, or
keq 1 keq k1
k1 k2 k2
The natural frequency of the given undamped free vibration spring mass system is
1 k1k2
fn
2π k1 k2 m
2.6 The natural frequency of the undamped free vibration of the spring mass system is given by
1 keq
fn where keq = equivalent stiffness of the spring system
2 m
For springs attached in parallel, the equivalent stiffness is given by
keq = k1 + k2
The natural frequency of the given undamped free vibration spring mass system is
1 (k1 k2 )
fn
2π m
2.7 The natural frequency of the undamped free vibration of the spring mass system is given by
1 keq
fn where keq = equivalent stiffness of the spring system
2 m
In the given spring-mass system, springs with stiffness k1 and k2 are in series.
Hence, their equivalent stiffness is
k1k2 100 20,000
keq(1,2) 66.67 N/mm
k 200 300
k 100
200
1 2
Similarly, springs with stiffness k4 and k5 are in series. Hence, their equivalent stiffness is
k4 k5 100 150
keq (4,5) 60 N/mm
k4 100 150
k5
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Now, the given spring system can be reduced to three springs in series.
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The resulting system will be three springs in parallel. Their equivalent stiffness is given by
keq keq (1,2) k3 keq ( 4,5) 66.67 150 60 276.67 N/mm 276.67 kN/m
The natural frequency of the undamped free vibration of the spring mass system is given by
1 keq 1 267.67 1000
fn = 8.37 Hz
2 m 2 100
Time period T (1/fn ) (1/ 8.37) 0.119 s
2.8 Sinusoidal–varying force, Q 50 sin t N; Q0 50 N; ω = 47 rad/s
keq 276.67 1000
n = 52.6 rad/s
m 100
Amplitude of vibration = static deflection zs × magnification M
Q0 50
z s = 0.1807 mm
k eq 276.67
From Eq. (2.34),
1 1
M = 4.96
1 ( n ) 2
Amplitude of vibration = 0.1807 4.96 = 0.896 mm
2.9 Weight of the body, W = 135 N
Mass of the body, m W g 135 9.81 13.76 kg
Spring constant, k = 2600 N/m
Dashpot resistance, c 0.7 (60 1000) 11.67 N-s/m
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a. Damped natural frequency [Eq. (2.67)]
fd 1 D2 fn
c c 11.67
D = 0.031
2 km
cc
1 k
fn = 2.19 Hz
2 m 2 13.76
fd 1 0.0312 (2.19) = 2.18 Hz
b. Damping ratio [Eq. (2.47b)],
c c 11.67
D 0.031
2 km
cc
c. Ratio of successive amplitudes of the body is given by [Eq. (2.70)],
Zn
e
Zn 1
Zn 2 2 0.031
D
where = 0.195
ln 1 0.0312
Z
n 1 1 D2
Zn
e e0.195 = 1.215
Zn 1
d. At time t = 0 s, amplitude Z0 = 25 mm.
After n cycles of disturbance
1 Z0 2 D Z0 2 nD
ln ; ln
1 D2 1 D2
n Zn Zn
With n = 5,
Z0 2 5 D 2 5 0.031
ln = 0.974
Z5 1 D2 1 0.0312
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Z0 25
e0.974 = Z5 = 9.44 mm
2.649
2.649;
Z5
After 5 cycles of disturbance, the amplitude of vibration = 9.44 mm
ALL 12 CHAPTERS COVERED
SOLUTIONS MANUAL
, @LECTSOLUTIONSSTUVIA
CONTENTS
Chapter Page
2 ............................................................................................................................. 1
4 9
5 25
6 41
8 ........................................................................................................................... 51
9 ........................................................................................................................... 61
10 ........................................................................................................................... 65
11 ........................................................................................................................... 79
12 ........................................................................................................................... 91
, @LECTSOLUTIONSSTUVIA
Chapter 2
2.1 a. Spring constant, k : The change in the force per unit length change of the spring.
b. Coefficient of subgrade reaction, k :
k
Spring constant divided by the foundation contact area, k
A
c. Undamped natural circular frequency: n rad/s
W
where m = mass =
g
d. Undamped natural frequency: f n (in Hz)
2 m
Note: Circular frequency defines the rate of oscillation in term of radians per unit
time; 2π radians being equal to one complete cycle of rotation.
e. Period, T: The time required for the motion to begin repeating itself.
n
f. Resonance: Resonance occurs when 1
g. Critical damping coefficient: cc 2 km
W
where k = spring constant; m = mass =
g
h. c c
Damping ratio: D =
cc 2 km
where c = viscous damping coefficient; cc = critical damping coefficient
i. Damped natural frequency:
d n 1 D2
fd 1 D2 fn
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2.2 Weight of machine + foundation, W = 400 kN
Spring constant, k = 100,000 kN/m
W 400 kN
Mass of the machine + foundation, m = = = 40.77
g 9.81 m s2
Natural frequency of undamped free vibration is [Eq. (2.19)]
1 k
fn = = 7.88 Hz
2 m 2 40.77
1 1
From Eq. (2.18), T = = = 0.127 s
fn 7.88
2.3 Weight of machine + foundation, W = 400 kN
Spring constant, k = 100,000 kN/m
Static deflection of foundation is [Eq. (2.2)]
z 400
4 10 3 m = 4 mm
W 100,000
s
k
2.4 External force to which the foundation is subjected, Q 35.6sin t kN
f = 13.33 Hz
Weight of the machine + foundation, W = 178 kN
Spring constant, k = 70,000 kN/m
For this foundation, let time t = 0, z = z0 = 0, zɺ = v0 = 0
W 178 kN
a. Mass of the machine + foundation, m = = = 18.145
g 9.81 m s2
k
n = 62.11 rad/s
m
2 2
T = 0.101 s
n 62.11
b. The frequency of loading, f = 13.33 Hz
2 f 2 (13.33) 83.75 rad/s
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Force due to forced part, F1 k Q0 2 k 2 sin t
1 n
35.6 70, 000
(70, 000) 2 2
sin(83.75t)
1 83.75 62.11
43.51sin(83.75t ) kN
See the plot below for F1 vs. t
c. Force due to free part, F k Q0 k sin t
n
n 2 35.6 70,
2 2
1 n
000 83.75
70, 000 sin(62.11t)
1 83.75 62.112
2
62.11
58.67sin(62.11t ) kN
See the plot above in Part b for F2 vs. t.
d. Total dynamic force on the subgrade:
F F1 F2 43.51sin(83.75t ) 58.67(62.11t ) kN
The plot of variation of the dynamic force on the subgrade of the foundation due
to (a) forced part, (b) free part, and (c) total of the response for time t = 0 to t = 2T
is shown in the figure above (Part b).
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2.5 The natural frequency of the undamped free vibration of the spring mass system is given by
1 keq
fn where keq = equivalent stiffness of the spring system
2 m
For springs attached in series, the equivalent stiffness is given by
1 1 1 k1k2
, or
keq 1 keq k1
k1 k2 k2
The natural frequency of the given undamped free vibration spring mass system is
1 k1k2
fn
2π k1 k2 m
2.6 The natural frequency of the undamped free vibration of the spring mass system is given by
1 keq
fn where keq = equivalent stiffness of the spring system
2 m
For springs attached in parallel, the equivalent stiffness is given by
keq = k1 + k2
The natural frequency of the given undamped free vibration spring mass system is
1 (k1 k2 )
fn
2π m
2.7 The natural frequency of the undamped free vibration of the spring mass system is given by
1 keq
fn where keq = equivalent stiffness of the spring system
2 m
In the given spring-mass system, springs with stiffness k1 and k2 are in series.
Hence, their equivalent stiffness is
k1k2 100 20,000
keq(1,2) 66.67 N/mm
k 200 300
k 100
200
1 2
Similarly, springs with stiffness k4 and k5 are in series. Hence, their equivalent stiffness is
k4 k5 100 150
keq (4,5) 60 N/mm
k4 100 150
k5
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Now, the given spring system can be reduced to three springs in series.
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The resulting system will be three springs in parallel. Their equivalent stiffness is given by
keq keq (1,2) k3 keq ( 4,5) 66.67 150 60 276.67 N/mm 276.67 kN/m
The natural frequency of the undamped free vibration of the spring mass system is given by
1 keq 1 267.67 1000
fn = 8.37 Hz
2 m 2 100
Time period T (1/fn ) (1/ 8.37) 0.119 s
2.8 Sinusoidal–varying force, Q 50 sin t N; Q0 50 N; ω = 47 rad/s
keq 276.67 1000
n = 52.6 rad/s
m 100
Amplitude of vibration = static deflection zs × magnification M
Q0 50
z s = 0.1807 mm
k eq 276.67
From Eq. (2.34),
1 1
M = 4.96
1 ( n ) 2
Amplitude of vibration = 0.1807 4.96 = 0.896 mm
2.9 Weight of the body, W = 135 N
Mass of the body, m W g 135 9.81 13.76 kg
Spring constant, k = 2600 N/m
Dashpot resistance, c 0.7 (60 1000) 11.67 N-s/m
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a. Damped natural frequency [Eq. (2.67)]
fd 1 D2 fn
c c 11.67
D = 0.031
2 km
cc
1 k
fn = 2.19 Hz
2 m 2 13.76
fd 1 0.0312 (2.19) = 2.18 Hz
b. Damping ratio [Eq. (2.47b)],
c c 11.67
D 0.031
2 km
cc
c. Ratio of successive amplitudes of the body is given by [Eq. (2.70)],
Zn
e
Zn 1
Zn 2 2 0.031
D
where = 0.195
ln 1 0.0312
Z
n 1 1 D2
Zn
e e0.195 = 1.215
Zn 1
d. At time t = 0 s, amplitude Z0 = 25 mm.
After n cycles of disturbance
1 Z0 2 D Z0 2 nD
ln ; ln
1 D2 1 D2
n Zn Zn
With n = 5,
Z0 2 5 D 2 5 0.031
ln = 0.974
Z5 1 D2 1 0.0312
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Z0 25
e0.974 = Z5 = 9.44 mm
2.649
2.649;
Z5
After 5 cycles of disturbance, the amplitude of vibration = 9.44 mm
