Solution Manual to Fundamental Concepts of Earthquake Engineering 1st Edition by Roberto Villaverde PDF | Complete Step-by-Step Solutions and Detailed Explanations | Covers Seismology, Structural Dynamics, Response Spectra, Seismic Design Principles, and

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Chapters 4,6,7,8,9,10,12,17 Covered




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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 Poisson ratio Unit
elasticity weight
(psi) (pcf)
Steel 30 10 6 0.27 490
Cast iron 26 106 0.25 485
Concrete 57,000 f 0.15 150
c



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 ) 30 106 (1 0.27)
M 37.5 psi
(1 2 )(1 ) [1 2(0.27)](1 10 6
0.27)
37.5 106
vc M (144) 18,838 ft/s 5.74 km/s
 .2
and similarly for the cast iron and reinforced concrete rods,
E(1 ) 26 106 (1 0.25)
M 31.2 psi
(1 2 )(1 ) [1 2(0.25)](1 106
0.25)
M 31.2 106 (144)
vc  .2 17,271 ft/s 5.26 km/s

E(1 ) 57,000 4,000(1 0.15)
M 3.8 106 psi
(1 2 )(1 ) [1 2(0.15)](1 0.15)
M 3.8 106 (144)
vc 10,838 ft/s 3.30 km/s
 .2

Problem 4.2
A rod of infinite length is subjected to an initial longitudinal displacement given by
u0 2(1 x
x) u0 2

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0 x 1 -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
2 x


t=1s
2 x


t=2s
2 x


2 t =3 s
x

-5 -4 -3 -2 -1 0 1 2 3 4 5 t =4 s
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 -2 x 2
A v0 elsewhere
where A is a 0
constant.

Solution:
According to Equation 4.19 and a zero initial displacement, the displacement in the rod
is given by
1 x vct
u(x, t) v0 ( )d
2v
c x vct

which may be considered as the superposition of the two displacement waves
x vct x vct
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1 1

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u( x, t) v0 ( )d v0 ( )d
2v
c 0
2v c 0




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where ζ is a dummy variable. Considering first the first integral and the fact that vc =
0.5 m/s, one has that
x 0.5t

u(x, t) v0 ( )d
0

Then, for –2.0 x +0.5t
2.0, 0
x 0.5t
u(x, t) Ad A[ ]x 0.5t A(x 0.5t)
0

In this case, therefore, the displacement u varies linearly with x for a constant t, and
the upper and lower limits for which this relationship is valid are
xu 0.5t 2.0 or xu 2.0
0.5t xl 0.5t 2.0 or xl
2.0 0.5t
Hence, for the particular times of 0, 1, 2, 3, and 4 seconds, the displacement u and the
lower and upper limits are as indicated in Table P4.3.

Table P4.3a. Displacement function and integration limits for first integral at different times
t u xl xu
0 Ax -2.0 2.0
1 A(x + 0.5) -2.5 1.5
2 A(x + 1.0) -3.0 1.0
3 A(x + 1.5) -3.5 0.5
4 A(x + 2.0) -4.0 0

The shape and position of this wave at 0, 1, 2, 3 and 4 seconds (drawn with dashed
lines) are as shown in Figure P4.3.

Table P4.3b. Displacement function and integration limits for second integral at different times
t u xl xu
0 -Ax -2.0 2.0
1 -A(x - 0.5) -1.5 2.5
2 -A(x - 1.0) -1.0 3.0
3 -A(x - 1.5) -0.5 3.5
4 -A(x - 2.0) 0 4.0

Considering now the second integral, one has that
x 0.5t

u(x, t) v0 ( )d
0

Then, for –2.0 x - 0.5t
2.0, 0
x 0.5t
u(x, t) Ad A[ ]x 0.5t A(x 0.5t)
0

In this case, the displacement u also varies linearly with x for a constant t, but the upper
and low- er limits for which the relationship is valid are
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xu 0.5t 2.0 or xu 2.0
0.5t xl 0.5t 2.0 or xl
2.0 0.5t




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Hence, for the particular times of 0, 1, 2, 3, and 4 seconds, the displacement u and the
lower and upper limits are as shown in Table P4.3b. The shape and position of this
additional wave at 0, 1, 2, 3, and 4 seconds (drawn with solid lines) are also shown in
Figure P4.3.


t=0s
x
-2A -2A

t=1s
x
-2A -2A

t=2s
x
-2A -2A


t =3s
x
-2A -2A


-5 -4 -3 -2 -1 0 1 2 3 4 5 t =4 s
-2A -2A x

Figure P4.3. Shape and position of waves generated by intial velocity at different times

Problem 4.4
A long bar with a density of 7850 kg/m3 and a modulus of elasticity of 7.85 kPa is
subjected to an initial axial disturbance defined by
u0 (x) 0
cos if x /2
x
v0 (x)
0 elsewhere
where u0 and v0 respectively denote initial displacement and initial velocity. Determine the
dis- placement induced by the disturbance at a distance x = - /3 and time t = 2 /3.

Solution:
According to Equation 4.5, the velocity of propagation of longitudinal waves in the
given bar is equal to
7.85
vc
E 103 1.0 m/s
 7850
Similarly, according to Equation 4.19, the displacement induced by an initial velocity v0
and a zero initial displacement is given by
1 x vct
u(x, t) v0 ( )d
2v
c x vct

where ζ is a dummy variable. Hence, for the case under consideration
1x t 1
u(x, t) cos d 2 [sin ]x t
x t
2

d
provide
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x t
and

x t
2 2




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