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 30 106 0.27 490
6
Cast iron 26 10 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)
6
E (1 ) 30 10 (1 0.27)
M 37.5 106 psi
(1 2 )(1 ) [1 2(0.27)](1 0.27)
M 37.5 106 (144)
vc 18,838 ft/s 5.74 km/s
.2
and similarly for the cast iron and reinforced concrete rods,
6
E (1 ) 26 10 (1 0.25)
M 31.2 106 psi
(1 2 )(1 ) [1 2(0.25)](1 0.25)
M 31.2 106 (144)
vc 17,271 ft/s 5.26 km/s
.2
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) u0 0 x 1
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
1 x vc t
u(x, t)
2v v0 ( )d
c x v ct
which may be considered as the superposition of the two displacement waves
x vct x vct
u(x, t) 1 1
2v v0 ( )d v0 ( )d
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,
x 0.5t
u(x, t) Ad A[ ]x 0.5t 0 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
xl 0.5t 2.0 or 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,
x 0.5t
u(x, t) Ad A[ ]x 0.5t 0 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
xu 0.5t 2.0 or xu 2.0 0.5t xl
xl 0.5t 2.0 or 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.
u
2A 2A
t =0s
x
-2A -2A
2A 2A
t =1s
x
-2A -2A
2A 2A
t =2s
x
-2A -2A
2A 2A
t =3s
x
-2A -2A
2A
t =4s
-5 -4 -3 -2 -1 0 1 2 3 4 5 x
-2A -2A
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
3
E 7.85 10
vc 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 v ct
u(x, t)
2v v0 ( )d
c x v ct
where ζ is a dummy variable. Hence, for the 1case under consideration
1
xt
u(x, t) cos d [sin ]x t
xt
2 xt
2
provided
x t
2 2
and
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, x t
2 2
Otherwise,
u(x, t) 0
Since for x = - /3 and t = 2 /3 one has that
2
x t
3 3 3
and
2
x t (less than - )
3 3 2
then
2 1 1 1
u( , ) [sin ]x t [sin ] / 3 [sin( ) sin( )] 0.933 m
3 3 2 xt
2 /2
2 3 2
Problem 4.5
During the process of driving a 40-m long reinforced concrete pile into a foundation soil, a pile- driving
hammer imparts a force impulse that can be assumed to vary as a half sine wave with an amplitude of
2500 kN and a duration of 0.012 seconds. To monitor the stresses in the pile during the driving process,
strain gages are attached at the middle of the pile and at 5 m from its tip.
Determine the maximum tensile and compressive stresses recorded by the gauges (a) when the pile is
being driven through a soft soil that offers no resistance to the penetration of the pile; and
(b) when the pile tip encounters rigid bedrock. Consider a modulus of elasticity of 20,000 MPa, a diameter
of 500 mm, and a density of 2300 kg/m3.
Solution:
The compression force imparted to the pile top is
P(t) 2500 sin( t)
0.012
which, since the wave velocity in the pile is equal to
6
E 20,000 10
v 2,949 m/s
c 2,300
may also be written as
P( vct) 2500 sin[ ( vct)]
0.012(2,949)
or as
P(x vct) 2500 sin[ (x 2949t)]
35.39
Hence, the equation for the stress wave is of the form
P P
(x v t) 12,732 sin[ (x 2949t)]
c 2
A (0.5) / 4 35.39
Note, also, that the length of the pulse is
Pulse length vc (0.012) 2949(0.012) 35.39 m
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,(a) Soft soil
i) Pile mid-height
The maximum compressive stress is attained when the incident wave is positioned as shown in the left-
hand side of Figure P4.5. Similarly, since the reflected wave is a tension wave when the pile tip encounters
no resistance, and since the incident wave has completely disappeared when the center of the reflected
wave reaches the mid-height of the pile, the maximum tension stress is attained when the reflected wave is
positioned as shown in the right-hand side of Figure P4.5. In consequence,
Maximum compressive stress 12,732 kN/m2 Maximum
tension stress 12,732 kN/m2
2.3 m
20 m 20 m
17.7 m
max max
(-) (+)
17.7 m
20 m 20 m
2.3 m
Compressionstresses Tensionstresses
Figure P4.5a. Position of incident and reflected waves at the time of maximum tension and compression stresses
at pile’s midheight
35 m
(+)
(-) max
(+)
5m
Compression stresses Tension stresses
Figure P4.5b. Position of incident and reflected waves at the time of maximum tension and compression stresses
at strain gage 5 m above pile tip
ii) 5m from pile tip
At any time before the tail of the incident wave reaches the strain gauge 5 m above the pile tip, the incident
and reflected waves are positioned as shown in Figure P4.5b. Therefore, the maxi-
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,mum compression stress at 5 m from the pile tip is given by the superposition of the incident and reflected
waves. To compute this maximum stress, consider that, in accordance with the equa- tion developed
above, the stress at x = 35 m induced by the incident wave is given by
(x v t) 12,732 sin[ (35 2949t)] 12,732 sin(261.8t 3.11)
i c
35.39
Similarly, if one defines x and t as
x x 40
t t 0.014
the equation for the reflected stress wave (tension) is
(x vct ) 12,732 sin[ (x 2949t )]
35.39
and thus at x = 5 m, the stress induced by the reflected wave is
r (5, t ) 12,732 sin[ (5 2949t )] 12,732 sin(261.8t 0.444)
35.39
which in terms of t may be written as
(5, t) 12,732 sin(261.8t 4.11)
r
It should be noted that the time at which the reflected wave appears for the first time is
40
t 0.014 s
2949
and the equation for the reflected wave is thus only valid for t 0.014 s.
As a result, the stress induced by the incident and reflected waves is given by
(35, t) 12,732[sin(261.8t 3.11) sin(261.8t 4.11)] which may also
be expressed as
(35, t) 12,732[sin(261.8t) cos(3.11) cos(261.8t) sin(3.11)
sin(261.8t) cos 4.11) cos(261.8t) sin 4.11)]
12,732[ 0.433 sin(261.8t) 0.856 cos(261.8t)
12,732 0.4332 0.8562 cos(261.8t )
12,214 cos(261.8t ), if t 0.014 s
where
0.856
tan 1 1.102 rad
0.433
It may be seen, thus, that the total stress is maximum when
261.8t 1.102 0 or t 0.004 s 0.014 s (not possible)
or when
261.8t 1.102 or t 0.016 s 0.014 s
In consequence,
max (35, t) -12,214 kN/m2 (compression)
The maximum tensile stress occurs when the incident wave completely disappears; that is, when the
tail of the incident wave reaches the pile tip, which, in turn, occurs when t = 40/2949
+0.012 = 0.026 s (see right-hand side of Figure P4.5b). Therefore,
2
max (5,
r t) 12,732 sin[261.8(0.026) 4.11] 5,478 kN/m (tension)
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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 30 106 0.27 490
6
Cast iron 26 10 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)
6
E (1 ) 30 10 (1 0.27)
M 37.5 106 psi
(1 2 )(1 ) [1 2(0.27)](1 0.27)
M 37.5 106 (144)
vc 18,838 ft/s 5.74 km/s
.2
and similarly for the cast iron and reinforced concrete rods,
6
E (1 ) 26 10 (1 0.25)
M 31.2 106 psi
(1 2 )(1 ) [1 2(0.25)](1 0.25)
M 31.2 106 (144)
vc 17,271 ft/s 5.26 km/s
.2
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) u0 0 x 1
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.
@
@SSeeisismmi3
,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
1 x vc t
u(x, t)
2v v0 ( )d
c x v ct
which may be considered as the superposition of the two displacement waves
x vct x vct
u(x, t) 1 1
2v v0 ( )d v0 ( )d
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,
x 0.5t
u(x, t) Ad A[ ]x 0.5t 0 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
xl 0.5t 2.0 or 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,
x 0.5t
u(x, t) Ad A[ ]x 0.5t 0 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
xu 0.5t 2.0 or xu 2.0 0.5t xl
xl 0.5t 2.0 or 2.0 0.5t
@
@SSeeisismmi5
,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.
u
2A 2A
t =0s
x
-2A -2A
2A 2A
t =1s
x
-2A -2A
2A 2A
t =2s
x
-2A -2A
2A 2A
t =3s
x
-2A -2A
2A
t =4s
-5 -4 -3 -2 -1 0 1 2 3 4 5 x
-2A -2A
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
3
E 7.85 10
vc 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 v ct
u(x, t)
2v v0 ( )d
c x v ct
where ζ is a dummy variable. Hence, for the 1case under consideration
1
xt
u(x, t) cos d [sin ]x t
xt
2 xt
2
provided
x t
2 2
and
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@SSeeisismmi6
, x t
2 2
Otherwise,
u(x, t) 0
Since for x = - /3 and t = 2 /3 one has that
2
x t
3 3 3
and
2
x t (less than - )
3 3 2
then
2 1 1 1
u( , ) [sin ]x t [sin ] / 3 [sin( ) sin( )] 0.933 m
3 3 2 xt
2 /2
2 3 2
Problem 4.5
During the process of driving a 40-m long reinforced concrete pile into a foundation soil, a pile- driving
hammer imparts a force impulse that can be assumed to vary as a half sine wave with an amplitude of
2500 kN and a duration of 0.012 seconds. To monitor the stresses in the pile during the driving process,
strain gages are attached at the middle of the pile and at 5 m from its tip.
Determine the maximum tensile and compressive stresses recorded by the gauges (a) when the pile is
being driven through a soft soil that offers no resistance to the penetration of the pile; and
(b) when the pile tip encounters rigid bedrock. Consider a modulus of elasticity of 20,000 MPa, a diameter
of 500 mm, and a density of 2300 kg/m3.
Solution:
The compression force imparted to the pile top is
P(t) 2500 sin( t)
0.012
which, since the wave velocity in the pile is equal to
6
E 20,000 10
v 2,949 m/s
c 2,300
may also be written as
P( vct) 2500 sin[ ( vct)]
0.012(2,949)
or as
P(x vct) 2500 sin[ (x 2949t)]
35.39
Hence, the equation for the stress wave is of the form
P P
(x v t) 12,732 sin[ (x 2949t)]
c 2
A (0.5) / 4 35.39
Note, also, that the length of the pulse is
Pulse length vc (0.012) 2949(0.012) 35.39 m
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@SSeeisismmi7
,(a) Soft soil
i) Pile mid-height
The maximum compressive stress is attained when the incident wave is positioned as shown in the left-
hand side of Figure P4.5. Similarly, since the reflected wave is a tension wave when the pile tip encounters
no resistance, and since the incident wave has completely disappeared when the center of the reflected
wave reaches the mid-height of the pile, the maximum tension stress is attained when the reflected wave is
positioned as shown in the right-hand side of Figure P4.5. In consequence,
Maximum compressive stress 12,732 kN/m2 Maximum
tension stress 12,732 kN/m2
2.3 m
20 m 20 m
17.7 m
max max
(-) (+)
17.7 m
20 m 20 m
2.3 m
Compressionstresses Tensionstresses
Figure P4.5a. Position of incident and reflected waves at the time of maximum tension and compression stresses
at pile’s midheight
35 m
(+)
(-) max
(+)
5m
Compression stresses Tension stresses
Figure P4.5b. Position of incident and reflected waves at the time of maximum tension and compression stresses
at strain gage 5 m above pile tip
ii) 5m from pile tip
At any time before the tail of the incident wave reaches the strain gauge 5 m above the pile tip, the incident
and reflected waves are positioned as shown in Figure P4.5b. Therefore, the maxi-
@
@SSeeisismmi8
,mum compression stress at 5 m from the pile tip is given by the superposition of the incident and reflected
waves. To compute this maximum stress, consider that, in accordance with the equa- tion developed
above, the stress at x = 35 m induced by the incident wave is given by
(x v t) 12,732 sin[ (35 2949t)] 12,732 sin(261.8t 3.11)
i c
35.39
Similarly, if one defines x and t as
x x 40
t t 0.014
the equation for the reflected stress wave (tension) is
(x vct ) 12,732 sin[ (x 2949t )]
35.39
and thus at x = 5 m, the stress induced by the reflected wave is
r (5, t ) 12,732 sin[ (5 2949t )] 12,732 sin(261.8t 0.444)
35.39
which in terms of t may be written as
(5, t) 12,732 sin(261.8t 4.11)
r
It should be noted that the time at which the reflected wave appears for the first time is
40
t 0.014 s
2949
and the equation for the reflected wave is thus only valid for t 0.014 s.
As a result, the stress induced by the incident and reflected waves is given by
(35, t) 12,732[sin(261.8t 3.11) sin(261.8t 4.11)] which may also
be expressed as
(35, t) 12,732[sin(261.8t) cos(3.11) cos(261.8t) sin(3.11)
sin(261.8t) cos 4.11) cos(261.8t) sin 4.11)]
12,732[ 0.433 sin(261.8t) 0.856 cos(261.8t)
12,732 0.4332 0.8562 cos(261.8t )
12,214 cos(261.8t ), if t 0.014 s
where
0.856
tan 1 1.102 rad
0.433
It may be seen, thus, that the total stress is maximum when
261.8t 1.102 0 or t 0.004 s 0.014 s (not possible)
or when
261.8t 1.102 or t 0.016 s 0.014 s
In consequence,
max (35, t) -12,214 kN/m2 (compression)
The maximum tensile stress occurs when the incident wave completely disappears; that is, when the
tail of the incident wave reaches the pile tip, which, in turn, occurs when t = 40/2949
+0.012 = 0.026 s (see right-hand side of Figure P4.5b). Therefore,
2
max (5,
r t) 12,732 sin[261.8(0.026) 4.11] 5,478 kN/m (tension)
@
@SSeeisismmi9
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