# Solution Manual for Foundation Design: Principles and Practices, 3rd Edition

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Chap. 3
3.1
Soil Mechanics
Explain the difference between moisture content and degree of saturation.
Solution
Moisture content of a soil is the ratio of the weight of its water to weight of its solids, whereas
the degree of saturation is the ratio of the volume of water to volume of voids. The degree of
saturation can range from 0 to 100%, whereas the moisture content can be larger than 100%.
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Chap. 3
3.2
Soil Mechanics
A certain saturated sand (S = 100%) has a moisture content of 25.1% and a specific gravity of
solids of 2.68. It also has a maximum index void ratio of 0.84 and a minimum index void ratio
of 0.33. Compute its relative density and classify its consistency.
Solution
First we must compute the void ratio e using the given specific gravity and moisture content. The
void ratio calculated using specific gravity and moisture content is
=
e
wGs 25.1% ร 2.68
=
= 0.67
100%
S
Using Equation 3.1, the relative density is
=
D
r
emax โ e
0.84 โ 0.67
ร 100%
=
ร100%
= 33%
0.84 โ 0.33
emax โ emin
Look up the consistency in Table 3.3 based on the calculated relative density of 33%: the soil is
classified as loose.
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Chap. 3
3.3
Soil Mechanics
Consider a soil that is being placed as a fill and compacted using a sheepsfoot roller (a piece of
construction equipment). Will the action of the roller change the void ratio of the soil? Explain.
Solution
Yes, whenever a soil is compacted, the volume of the soil decreases. Since the volume of the
solids does not change with compaction, the volume of the voids decreases; and since the void
ratio is the ratio of the volume of voids to the volume of solids, the void ratio decreases as a
result.
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Chap. 3
3.4
Soil Mechanics
A sample of soil has a volume of 0.45 ft3 and a weight of 53.3 lb. After being dried in an oven, it
has a weight of 45.1 lb. It has a specific gravity of solids of 2.70. Compute its moisture content
and degree of saturation before it was placed in the oven.
Solution
First we compute the moisture content of the sample
w=
W โ Ws
53.3 โ 45.1
ร100% =
ร100% = 18%
45.1
Ws
Then we compute the degree of saturation
S=
w
ฮณw 1
โ
ฮณ d Gs
ร 100%=
0.18
ร 100%= 71%
62.4
1
โ
( 45.1 0.45) 2.70
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Chap. 3
3.5
Soil Mechanics
A site is underlain by a soil that has a unit weight of 18.7 kN/m3 above the groundwater table and
19.9 kN/m3 below. The groundwater table is located at a depth of 3.5 m below the ground
surface. Compute the total vertical stress, pore water pressure, and effective vertical stress at the
following depths below the ground surface:
a. 2.2 m
b. 4.0 m
c. 6.0 m
Solution
Using Equation 3.3, the total vertical stresses computed at the given depths are
kN/m3 )(2.2 m) 41.1 kN/m 2
=
=
ฯ z (z = 2.2 m) (18.7
(18.7 kN/m3 )(3.5 m) + (19.9 kN/m3 )(4.0=
m โ 3.5 m) 75.4 kN/m 2
ฯ=
z (z = 4.0 m)
(18.7 kN/m3 )(3.5 m) + (19.9 kN/m3 )(6.0=
m โ 3.5 m) 115 kN/m 2
ฯ=
z (z = 6.0 m)
The pore water pressures computed at the given depths are
=
=
u (z = 2.2 m) (9.81
kN/m3 )(0 m) 0.0 kN/m 2
u=
(z = 4.0 m) (9.81 kN/m3 )(4.0=
m โ 3.5 m) 4.91 kN/m 2
u=
(z = 6.0 m) (9.81 kN/m3 )(6.0=
m โ 3.5 m) 24.5 kN/m 2
Using Equation 3.5, the effective vertical stresses computed at the given depths are
ฯ z ‘ (z = 2.2 m) = 41.1 kN/m 2 โ 0.0 kN/m 2 = 41.1 kN/m 2
ฯ z ‘ (z = 4.0 m) = 75.4 kN/m 2 โ 4.9 kN/m 2 = 70.5 kN/m 2
ฯ z ‘ (z = 6.0 m) =115 kN/m 2 โ 24.5 kN/m 2 =90.5 kN/m 2
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Chap. 3
3.6
Soil Mechanics
The subsurface profile at a certain site is shown in Figure 3.20. Compute u, ฯx, ฯz, ฯสนx, and ฯสนz at
Point A.
Solution
u = ฮณ w zw
= (62.4 lb/ft 3 )(11 ft)
= 686 lb/ft 2
ฯ z = โฮณ H
= (120 lb/ft 3 )(12 ft) + (117 lb/ft 3 )(10 ft) + (121 lb/ft 3 )(11 ft)
= 3940 lb/ft 2
ฯ=
ฯz โu
z ‘
= 3940 lb/ft 2 โ 686 lb/ft 2
= 3254 lb/ft 2
ฯ x ‘ = Kฯ z ‘
= (0.70)(3254 lb/ft 2 )
= 2278 lb/ft 2
ฯ=
ฯ x ‘+ u
x
= 2278 lb/ft 2 + 686 lb/ft 2
= 2964 lb/ft 2
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Chap. 3
3.7
Soil Mechanics
The vertical load of 300 kN is applied to a 1.5 m ร 1.5 m area at the ground surface that is level.
a. Compute the induced vertical stress, ฮฯz, at a point 2.0 m below the corner of this
square loaded area.
b. Compute the induced vertical stress, ฮฯz, at a point 2.0 m below the center of this
square loaded area.
Solution
a. This solution uses the Newman solution to the Boussinesqโs Method to compute the
induced stress; there are many other methods to solve this problem.
q=
P
=
A
300 kN
=
133 kPa
2
(1.5 m )
B 2 + L2 + z 2f = 1.52 + 1.52 + 2.02 = 8.5
B 2 L2 (1.52 )(1.52 )
=
=
1.26
z 2f
2.02
B 2 + L2 + z 2f >
B 2 L2
therefore, use Equation 3.12:
z 2f
2
2
2
2
2
2
๏ฃฎ
1 ๏ฃฏ๏ฃฌ๏ฃซ 2 BLz f B + L + z f ๏ฃถ๏ฃท ๏ฃซ B + L + 2 z f ๏ฃถ
Is =
๏ฃฌ
๏ฃท
4ฯ ๏ฃฏ๏ฃฌ z 2f ( B 2 + L2 + z 2f ) + B 2 L2 ๏ฃท ๏ฃฌ๏ฃญ B 2 + L2 + z 2f ๏ฃท๏ฃธ
๏ฃธ
๏ฃฐ๏ฃญ
+ sin
Is =
โ1
๏ฃน
๏ฃบ
z 2f ( B 2 + L2 + z 2f ) + B 2 L2 ๏ฃบ
๏ฃป
2 BLz f B 2 + L2 + z 2f
1 ๏ฃฎ๏ฃซ 2(1.5)(1.5)(2.0) 8.5 ๏ฃถ ๏ฃซ (1.5) 2 + (1.5) 2 + 2(2.0) 2 ๏ฃถ
๏ฃฏ๏ฃฌ
๏ฃท๏ฃฌ
๏ฃท
8.5
4ฯ ๏ฃฏ๏ฃฐ๏ฃฌ๏ฃญ (2.0) 2 ( 8.5 ) + (1.5) 2 (1.5) 2 ๏ฃท๏ฃธ ๏ฃญ
๏ฃธ
+ sin โ1
2(1.5)(1.5)(2.0) 8.5 ๏ฃน
๏ฃบ
(2.0) 2 ( 8.5 ) + (1.5) 2 (1.5) 2 ๏ฃป
= 0.137
โs z =
Is q
= (0.137)(133 kPa)
= 18.2 kPa
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Chap. 3
Soil Mechanics
b. This solution uses the chart method to compute the induced stress; this problem can
also be solved by other methods.
q = 133 kPa, from part a.
xf
0m
= = 0
B 1.5 m
z f 2.0 m
= = 1.33
B 1.5 m
Iฯ = 0.23, from Figure 3.6
Iฯ q
โฯ z =
= (0.23)(133 kPa)
= 30.6 kPa
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Chap. 3
3.8
Soil Mechanics
A vertical load of 20 k is applied to a 6.0 ft ร 4.0 ft area at the ground surface that is level.
a. Compute the induced vertical stress, ฮฯz, at a point 6.0 ft below the corner of this
rectangular loaded area.
b. Compute the induced vertical stress, ฮฯz, at a point 6.0 ft below the center of this
rectangular loaded area.
Solution
a. This solution uses Figure 3.8 for a chart solution; there are many other methods to solve
this problem.
P
=
A
q=
20 k
= 833 lb/ft 2
( 6.0 ft )( 4.0 ft )
B 6.0 ft
=
= 1.0
z 6.0 ft
L 4.0 ft
n= =
= 0.67
z 6.0 ft
Iฯ = 0.15, from Figure 3.8
m=
Iฯ q
โฯ z =
= (0.15)(833 lb/ft 2 )
= 125 lb/ft 2
b. This solution uses Figure 3.8 for a chart solution; there are many other methods to solve
this problem. Separate the loaded area into four equal areas, and then use superposition
to compute the induced stress at below the center of the area.
q = 833 lb/ft 2
B 3.0 ft
m= =
= 0.50
z 6.0 ft
L 2.0 ft
n= =
= 0.33
z 6.0 ft
Iฯ = 0.062, from Figure 3.8
Iฯ q
โฯ z =
= (4)(0.062)(833 lb/ft 2 )
= 207 lb/ft 2
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Chap. 3
3.9
Soil Mechanics
A 3 m ร 3 m footing is to be built on the surface of a 15 m thick layer of unsaturated sand. The
sand is underlain by a very dense gravel layer. The water table is at a great depth. The sand is
relative uniform and in situ testing indicates it has a constrained modulus of 10 MPa. The
footing load is 200 kN. Compute the settlement under the center of the footing.
Solution
The sand was separated into 8 layers each 2 m thick with the exception of layer no. 8 which was
1 m thick. Using Equation 3.14, change in the stress at the center of each layer was computed.
Using Equation 3.20, the strain of each layer was computed. The results are shown in the
following table.
Layer No.
Depth to layer
midpoint (m)
Layer
Thickness
(cm)
โฯ สน
(kPa)
M
(kPa)
ฮตv
Layer compression
(cm)
1
1
200
19.43
10,000
0.19%
0.38
2
3
200
7.22
10,000
0.07%
0.14
3
5
200
3.13
10,000
0.03%
0.06
4
7
200
1.69
10,000
0.02%
0.04
5
9
200
1.05
10,000
0.01%
0.02
6
11
200
0.71
10,000
0.01%
0.02
7
13
200
0.51
10,000
0.01%
0.02
8
14.5
100
0.41
10,000
0.00%
0.00
ฮด = 0.68
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Chap. 3
Soil Mechanics
3.10 A 3-foot square footing carries a sustained load of 10 k. It is placed on the surface of a 30 foot
thick saturated overconsolidated clay underlain by dense sand. Based on laboratory tests, the
clay can be adequately modeled using the e-log-p method. The laboratory tests provide the
following compressibility information for the clay:
ฮณ = 123 lb/ft3
Cc
= 0.06
1 + e0
Cr
= 0.002
1 + e0
ฯสนm = 900 lb/ft2
The groundwater table is located at the ground surface. Compute the settlement of the
footing.
Solution
From Example 3.5 we know it should be adequate to compute the compressibility to only a depth
of 6 feet, even though the compressible layer is much thicker. The clay later will be separated
into 3 layers, each 2 feet thick.
The compression of the first layer is computed in the following fashion.
Compute the initial vertical stress considering the groundwater table at the ground surface
=
ฯ zโฒ0 1 ft (123 lb/ft 2 ) โ 1 ft ( 62.4 lb/ft 2 )
= 60.6 lb/ft 2
Using Equation 3.14 to compute the induced stress:
2.60
1
๏ฃซ
๏ฃถ
2 ๏ฃท
๏ฃฌ
Iฯ = 1 โ ๏ฃฌ ๏ฃซ 3 ๏ฃถ ๏ฃท
๏ฃฌ 1 + ๏ฃฌ 2(1) ๏ฃท ๏ฃท
๏ฃธ ๏ฃธ
๏ฃญ ๏ฃญ
= 0.953
10 k
=
q = 1,111 lb/ft 2
(3 ft)(3 ft)
โฯ z =
Iฯ q
= (0.953)(1,111 lb/ft 2 )
= 1, 059 lb/ft 2
2
โฒ ฯ zโฒ0 + โฯ=
โฒ 60.6 lb/ft 2 + 1,059 lb/ft=
ฯ=
1,120 lb/ft 2
zf
z
ฯ cโฒ = ฯ zโฒ0 + ฯ mโฒ = 60.6 lb/ft 2 + 900 lb/ft 2 = 960 lb/ft 2
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Chap. 3
Soil Mechanics
Since ฯสนzf > ฯสนc , this is case OC-II and the layer compression is computed using Equation 3.31:
๏ฃฎ
๏ฃซ 960 ๏ฃถ
๏ฃซ 1,120 ๏ฃถ ๏ฃน
=
=
ฮด 24 in ๏ฃฏ( 0.002 ) log ๏ฃฌ
๏ฃท + ( 0.06 ) log ๏ฃฌ
๏ฃท ๏ฃบ 0.15 in
๏ฃญ 60.6 ๏ฃธ
๏ฃญ 960 ๏ฃธ ๏ฃป
๏ฃฐ
This process is repeated for the remaining layers. The following table shows the results of the
calculations.
Computed at layer midpoint
Layer
No.
1
Depth to
layer
midpoint
(ft)
1
Layer
Thickness
(ft)
2
ฯz0สน
(lb/ft2)
60.6
โฯz
(lb/ft2)
1,059
ฯzfสน
(lb/ft2)
1,120
ฯcสน
(lb/ft2)
961
Case
OC-II
2
3
2
182
489
671
1,082
3
5
2
303
223
526
1,203
Cc
1 + e0
Cr
1 + e0
0.06
0.002
ฮด
(in)
0.15
OC-I
0.06
0.002
0.03
OC-I
0.06
0.002
0.01
ฮด =0.19
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Chap. 3
Soil Mechanics
3.11 A 2m thick fill is to be placed on the soil shown in Figure 3.21. Once it is compacted this fill
will have a unit weight of 19.5 kN/m3. Compute the ultimate consolidation settlement.
Solution
Note that the sand layer is dense, its consolidation is negligible compared to that of the stiff clay
layer, and that overconsolidation was assumed for the stiff clay. Also note that since the fill is
areal, the influence factor is 1.0. Using methods similar to those used to solve Problem 3.10 and
Example 3.5, the following table shows the results of the calculations.
Computed at layer midpoint
ฯz0สน
โฯz
ฯzfสน
Layer Thickness (cm)
(kPa)
(kPa)
(kPa)
100
35.9
39
75
0.046
ฮด
(cm)
1.5
83
0.046
1.3
39
92
0.046
1.1
61.0
39
100
0.046
1.0
69.4
39
108
0.046
0.9
100
77.8
39
117
0.046
0.8
8.4
100
86.2
39
125
0.046
0.7
9.4
100
94.6
39
134
0.046
0.7
Layer No.
1
Depth to layer midpoint (m)
2.4
2
3.4
100
44.2
39
3
4.4
100
52.6
4
5.4
100
5
6.4
100
6
7.4
7
8
Cr
1 + e0
ฮด =8.0
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Chap. 3
Soil Mechanics
3.12 Estimate the effective friction angle of the following soils:
a. Silty sand with dry unit weight of 100 lb/ft 3 .
b. Poorly-graded gravel with relative density of 70%.
c. Very dense well-graded sand.
Solution
Using Figure 3.14, estimate the effective friction angle of the given soils
Soil
Silty sand with dry unit weight of 100
lb/ft 3
Poorly-graded gravel with relative
density of 70%
Very dense well-graded sand
Estimated effective friction
angle (degrees)
32-37
35-37
35-40
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Chap. 3
Soil Mechanics
3.13 Explain the difference between the drained condition and the undrained condition.
Solution
โข Drained condition is a condition where excess pore water does not exist under change in
loading. In other words, the pore water pressure is equal to the hydrostatic pore water
pressure. Drained conditions can develop in one of two ways. First, if the hydraulic
conductivity of a saturated soil is sufficiently high and that loading rate is sufficiently low,
then the excess pore water pressure can dissipate quickly. Secondly, drained conditions
can also develop given sufficient time has elapsed since the end of loading, any excess
pore water pressure will have been dissipated.
โข Undrained condition is where excess pore water pressure is generated in the saturated soil,
or when the pore water pressure is not equal to the hydrostatic pore water pressure. It
develops when water cannot flow quickly from the loaded soil.
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Chap. 3
Soil Mechanics
3.14 A soil has cสน = 5 kPa and ฯสน = 32ยฐ. The effective stress at a point in the soil is 125 kPa. Compute
the shear strength normal to this stress at this point.
Solution
Using Equation 3.32, compute the shear strength
=
s 5 kPa + (125 kPa)( tan 32
=
ยฐ) 83 kPa
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Chap. 3
Soil Mechanics
3.15 A footing with an embedment of 2 m is embedded in a sand with a unit weight of 125 lb/ft3 and a
ฯสน of 36ยฐ. If the footing is subjected to a horizontal load that causes it to move horizontally,
compute the total active and passive resultant forces acting on the footing.
Solution
Using Equation 3.38, compute the active resultant force
(125 lb/ft 3 ) (2m ร 3.28 ft/m) 2 ( tan 2 (45ยฐ โ 36ยฐ / 2))
2
= 698 lb/ft
Pa =
Using Equation 3.40, compute the passive resultant force
(125 lb/ft 3 ) (2m ร 3.28 ft/m) 2 ( tan 2 (45ยฐ + 36ยฐ / 2))
Pp =
2
= 10,360 lb/ft
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Foundation Engineering: Principles and Practices, 3rd Ed
3-17
ยฉ 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from
the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Chap. 3
Soil Mechanics
3.16 The soil profile at a certain site is as follows:
cสน (lb/ft2)
0-12
ฮณ (lb/ft3)
12-20
126
200
20
20-32
129
0
32
Depth (ft)
119
ฯสน (degree)
Su(lb/ft2)
1000
The groundwater table is at a depth of 15 ft.
Develop plots of pore water pressure, total vertical stress, effective total stress, and shear
strength on a horizontal plane vs. depth. All four of these plots should be superimposed on the
same diagram with the parameters on the horizontal axis (increasing to the right) and depth on
the vertical axis (increasing downward).
Hint: Because the cohesion and friction angle suddenly change at the strata boundaries, the shear
strength also may change suddenly at these depths.
Solution
4500
4000
3500
3000
2500
2000
1500
1000
500
0
(lb/ft2)
0.0
2.0
4.0
Pore Water Pressure
6.0
8.0
10.0
Total Vertical Stress
Depth (ft bgs)
12.0
14.0
Effective Vertical Stress
16.0
18.0
Shear Strength
20.0
22.0
24.0
26.0
28.0
30.0
32.0
34.0
Solutions Manual
Foundation Engineering: Principles and Practices, 3rd Ed
3-18
ยฉ 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from
the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Chap. 3
Soil Mechanics
3.17 Repeat problem 3.16 using the following data:
cสน (kPa)
0-5
ฮณ (kN/m3)
5-12
20.0
8.4
21
12-20
20.5
0
35
Depth (m)
18.5
ฯสน (degree)
Su(kPa)
50
The groundwater table is at a depth of 7 m.
Solution
450
400
350
300
250
200
150
100
50
0
kPa
0.0
2.0
Pore Water
Pressure
4.0
Depth ( m bgs)
6.0
Total Vertical
Stress
8.0
Effective Vertical
Stress
10.0
Shear Strength
12.0
14.0
16.0
18.0
20.0
22.0
Solutions Manual
Foundation Engineering: Principles and Practices, 3rd Ed
3-19
ยฉ 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from
the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
Chap. 3
Soil Mechanics
3.18 A 9 ft thick fill is to be placed on the soil shown in Figure 3.22. Once it is compacted this fill
will have a unit weight of 122 lb/ft3. Compute the ultimate consolidation settlement caused by
consolidation of the underlying clay.
Solution
Using methods similar to those used to solve Problem 3.10 and Example 3.5, the following table
shows the results of the calculations.
Computed at center of layer
Layer
No.
1
Depth to
layer
midpoint
(ft)
18
Layer
Thickness
(ft)
6
ฯz0สน
(lb/ft2)
1,658.6
โฯz
(lb/ft2)
1,098
ฯzfสน
(lb/ft2)
2,757
ฯcสน
(lb/ft2)
2,757
Case
NC
2
24
6
1,830.2
1,098
2,928
2,928
3
30
6
2,001.8
1,098
3,100
3,100
4
36
6
2,173.4
1,098
3,271
5
50
6
2,792.8
1,098
6
56
6
3,048.4
7
62
6
3,304
Cc
1 + e0
Cr
1 + e0
0.14
0.06
ฮด
(in)
2.22
NC
0.14
0.06
2.06
NC
0.14
0.06
1.91
3,271
NC
0.14
0.06
1.79
3,891
5,774
OC-I
0.12
0.05
0.52
1,098
4,146
6,030
OC-I
0.12
0.05
0.48
1,098
4,402
6,285
OC-I
0.12
0.05
0.45
ฮด =9.43
Solutions Manual
Foundation Engineering: Principles and Practices, 3rd Ed
3-20
ยฉ 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from
the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

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