SOLUTIONS MANUAL
, 2.................................................................................................................................. 1
4 9
5 25
6 41
8................................................................................................................................ 51
9................................................................................................................................ 61
10 ............................................................................................................................... 65
11 ............................................................................................................................... 79
12 ............................................................................................................................... 91
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, Chapter 2
Spring constant, k The change in the force per unit length change of the spring.
Coefficient of subgrade reaction, k:
k
Spring constant divided by the foundation contact area, k =
A
Undamped natural circular frequency: n = rad/s
W
where m = mass =
g
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.
Period, T: The time required for the motion to begin repeating itself.
n
Resonance: Resonance occurs when =1
Critical damping coefficient: cc = 2 km
W
where k = spring constant; m = mass =
g
c c
Damping ratio: D = =
cc 2 km
where c = viscous damping coefficient; cc = critical damping coefficient
Damped natural frequency:
d = n 1 − D2
fd = 1 − D2 fn
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, 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 = = =
2 m 2 40.77
1 1
From Eq. (2.18), T = = =
fn 7.88
Weight of machine + foundation, W = 400 kN
Spring constant, k = 100,000 kN/m
Static deflection of foundation is [Eq. (2.2)]
W
z = =
400
= 4 10−3 m =
s
k 100,000
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
Mass of the machine + foundation, m = = = 18.145
g 9.81 m s2
k
n = = = 62.11 rad/s
m
2 2
T= = =
n 62.11
The frequency of loading, f = 13.33 Hz
= 2 f = 2 (13.33) = 83.75 rad/s
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© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.