SOLUTIONS MANUAL
For
Principles of Foundation
Engineering
Ninth Edition
BRAJA M. DAS
NAGARATNAM SIVAKUGAN
1
© 2019 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
, Contents
Chapter 2............................................................................................................................. 1
Chapter 3........................................................................................................................... 11
Chapter 5........................................................................................................................... 23
Chapter 6........................................................................................................................... 31
Chapter 7........................................................................................................................... 49
Chapter 8........................................................................................................................... 59
Chapter 9........................................................................................................................... 69
Chapter 10......................................................................................................................... 83
Chapter 11......................................................................................................................... 89
Chapter 12......................................................................................................................... 91
Chapter 13....................................................................................................................... 113
Chapter 14....................................................................................................................... 123
Chapter 15....................................................................................................................... 125
Chapter 16....................................................................................................................... 131
Chapter 17....................................................................................................................... 141
Chapter 18....................................................................................................................... 153
Chapter 19....................................................................................................................... 171
, Chapter 2
2.1 From Eq. (2.18),
Gs ρw 2450 2.80 × 1000
ρd = = = ; e = 0.0571
1+ e 0.925 1+ e
From Eq. (2.6),
e 0.0571
Porosity, n = = = 0.054
1 + e 1 + 0.0571
2.2 From Eq. (2.13), the dry density
ρ 2060
ρd = = = 1786.6 kg/m3
1 + w 1 + 0.153
Gs ρw
From Eq. (2.18), ρd =
1+ e
Gs ρw 2.70 × 1000
e= −1 = − 1 = 0.511
ρd 1786.6
Once saturated, from Eq. (2.19),
(Gs + e ) ( 2.70 + 0.511)
ρsat = ρw = × 1000 = 2125.1 kg/m 3
(1 + e) (1 + 0.511)
2.3 Let’s consider a 1-m2 area in plan. The initial volume of this soil is
V = 1 × 0.5 = 0.5 m3. Volume of the solids is Vs.
0.5 − Vs
e = 0 .9 = ; Vs = 0.2632 m3
Vs
Ws = 0.2632 × 2.68 × 9.81 = 6.919 kN
The new volume after compaction = 1 × 0.455 = 0.455 m3
6.919
The dry unit weight, γd = = 15.21 kN/m3
0.455
Gs γw
From Eq. (2.12), γd =
1+ e
1
© 2019 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
, Gs γ w 2.68 × 9.81
e= −1 = − 1 = 0.729
γd 15.21
From Eq. (2.13), γ = (15.21)(1 + 0.20) = 18.25 kN/m 3
2.4 At the compacted road base, the weight of solids,
Ws = 120,000 × 19.5 = 2,340,000 kN
γ 17.5
At the borrow pit, the dry unit weight, γd = = = 16.13 kN/m 3
1 + w 1 + 0.085
2,340,000
Volume of the pit, V = = 145,080 m3
16.13
The moisture content has to be increased from 8.5% (at the borrow pit) to 14.0%
(at the road base). The quantity of water to add,
2,340,000 × (0.14 ‒ 0.085) = 128,700 kN
128,700
Volume of water to be added = = 13,119.3 m 3
9.81
γ 110.4
2.5 γd = = = 99.9 lb/ft 3
1 + w 1 + 0.105
Gs γ w 2.65 × 62.4
γd = ; e= − 1 = 0.655
1+ e 99.9
0.870 − 0.655
From Eq. (2.23), Dr = × 100 = 60.6%
0.870 − 0.515
2.6 a. A-1-a c. A-3
b. A-1-b d. A-7-6
2.7 Soil A: % of gravel = 50, % of sand = 13, % of fines = 37
D10 = 0.035 mm, D30 = 0.061 mm, D60 = 9.8 mm Cu = 280; Cc = 0.02
LL = 58, PL = 34, PI = 24 plots below the A-line; hence, silt
The soil can be described as poorly (gap) graded sandy silty gravel
with a group symbol of GM.
2
© 2019 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
For
Principles of Foundation
Engineering
Ninth Edition
BRAJA M. DAS
NAGARATNAM SIVAKUGAN
1
© 2019 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
, Contents
Chapter 2............................................................................................................................. 1
Chapter 3........................................................................................................................... 11
Chapter 5........................................................................................................................... 23
Chapter 6........................................................................................................................... 31
Chapter 7........................................................................................................................... 49
Chapter 8........................................................................................................................... 59
Chapter 9........................................................................................................................... 69
Chapter 10......................................................................................................................... 83
Chapter 11......................................................................................................................... 89
Chapter 12......................................................................................................................... 91
Chapter 13....................................................................................................................... 113
Chapter 14....................................................................................................................... 123
Chapter 15....................................................................................................................... 125
Chapter 16....................................................................................................................... 131
Chapter 17....................................................................................................................... 141
Chapter 18....................................................................................................................... 153
Chapter 19....................................................................................................................... 171
, Chapter 2
2.1 From Eq. (2.18),
Gs ρw 2450 2.80 × 1000
ρd = = = ; e = 0.0571
1+ e 0.925 1+ e
From Eq. (2.6),
e 0.0571
Porosity, n = = = 0.054
1 + e 1 + 0.0571
2.2 From Eq. (2.13), the dry density
ρ 2060
ρd = = = 1786.6 kg/m3
1 + w 1 + 0.153
Gs ρw
From Eq. (2.18), ρd =
1+ e
Gs ρw 2.70 × 1000
e= −1 = − 1 = 0.511
ρd 1786.6
Once saturated, from Eq. (2.19),
(Gs + e ) ( 2.70 + 0.511)
ρsat = ρw = × 1000 = 2125.1 kg/m 3
(1 + e) (1 + 0.511)
2.3 Let’s consider a 1-m2 area in plan. The initial volume of this soil is
V = 1 × 0.5 = 0.5 m3. Volume of the solids is Vs.
0.5 − Vs
e = 0 .9 = ; Vs = 0.2632 m3
Vs
Ws = 0.2632 × 2.68 × 9.81 = 6.919 kN
The new volume after compaction = 1 × 0.455 = 0.455 m3
6.919
The dry unit weight, γd = = 15.21 kN/m3
0.455
Gs γw
From Eq. (2.12), γd =
1+ e
1
© 2019 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
, Gs γ w 2.68 × 9.81
e= −1 = − 1 = 0.729
γd 15.21
From Eq. (2.13), γ = (15.21)(1 + 0.20) = 18.25 kN/m 3
2.4 At the compacted road base, the weight of solids,
Ws = 120,000 × 19.5 = 2,340,000 kN
γ 17.5
At the borrow pit, the dry unit weight, γd = = = 16.13 kN/m 3
1 + w 1 + 0.085
2,340,000
Volume of the pit, V = = 145,080 m3
16.13
The moisture content has to be increased from 8.5% (at the borrow pit) to 14.0%
(at the road base). The quantity of water to add,
2,340,000 × (0.14 ‒ 0.085) = 128,700 kN
128,700
Volume of water to be added = = 13,119.3 m 3
9.81
γ 110.4
2.5 γd = = = 99.9 lb/ft 3
1 + w 1 + 0.105
Gs γ w 2.65 × 62.4
γd = ; e= − 1 = 0.655
1+ e 99.9
0.870 − 0.655
From Eq. (2.23), Dr = × 100 = 60.6%
0.870 − 0.515
2.6 a. A-1-a c. A-3
b. A-1-b d. A-7-6
2.7 Soil A: % of gravel = 50, % of sand = 13, % of fines = 37
D10 = 0.035 mm, D30 = 0.061 mm, D60 = 9.8 mm Cu = 280; Cc = 0.02
LL = 58, PL = 34, PI = 24 plots below the A-line; hence, silt
The soil can be described as poorly (gap) graded sandy silty gravel
with a group symbol of GM.
2
© 2019 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
