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
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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
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website, in whole or in part.
, Soil B: % of gravel = 24, % of sand = 69, % of fines = 7
D10 = 0.17 mm, D30 = 0.82 mm, D60 = 2.6 mm Cu = 15.3; Cc = 1.5
LL = 42, PL = 22, PI = 20 plots above the A-line; hence, clay
The soil can be described as well graded clayey gravelly sand with a
group symbol of SW-SM.
Soil C: % of gravel = 1, % of sand = 99, % of fines = 0
D10 = 0.7 mm, D30 = 1.2 mm, D60 = 1.6 mm Cu = 2.3; Cc = 1.3
The soil can be described as poorly (uniformly) graded sand with a
group symbol of SP.
Soil D: % of gravel = 0, % of sand = 12, % of fines = 88
LL = 75, PL = 31, PI = 44 plots above the A-line; hence, clay.
The soil can be described as sandy clay of high plasticity with a group
symbol of CH.
2.8 The head loss from the reservoir to the ditch, ∆h = 38 − 28 = 10.0 m
The length of the sand seam in the direction of the flow, L = 200 cos10 = 203.1 m
The hydraulic gradient, i = 10.0 203.1 = 0.0492
By Darcy’s law [Eq. (2.35)], v = ( 2.6 × 10 −5 m/s)( 0.0492 ) = 0.128 × 10 −5 m/s
The cross section of the sand seam through which the flow takes place is
1.0 × 500.0 = 500.0 m2
The flow rate = (0.128 × 10 −5 m/s)(500.0 m 2 ) = 64.0 × 10 −5 m 3/s
Volume of water flowing into the ditch per day:
= (64.0 × 10 −5 m 3/s)(24 )( 3600) = 55.3 m 3
2.9 For the flow net shown in Figure P2.9, Nf = 3 and Nd = 10
Total head loss from right to left, hmax = 5.0 m
The flow rate is given by [Eq. (2.46)]
Nf 3
q = khmax = (1.5 × 10−5 )(5.0) = 2.25 × 10−5 m 3/s/m length
Nd 10
= ( 2.25 × 10− 5 )(50.0)( 24)( 3600 m 3/day) = 97.2 m 3 /day
3
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website, in whole or in part.
, 2.10 On top of the soft clay layer (i.e., at 10 m depth), initially:
σ ′ = 1 × 17.0 + 9( 20 − 9.81) = 108.7 kN/m 2
After the water table is lowered,
σ ′ = 3 × 17.0 + 7( 20 − 9.81) = 122.3 kN/m 2
By lowering the water table, the effective stress has increased by
(122.3 – 108.7) = 13.6 kN/m2
2.11 The soil below the water table can be assumed to be fully saturated (i.e., S = 1).
e = wGs = 0.25 × 2.70 = 0.675
The saturated unit weight can be computed as
(Gs + e)γw (2.70 + 0.675) × 9.81
γsat = = = 19.8 kN/m3
1+ e 1 + 0.675
At a depth of 5 m into the sandy clay,
σ = 4 × 9.81 + 5 × 19.8 = 138.3 kN/m2
u = 9 × 9.81 = 88.3 kN/m2
σ ′ = σ − u = 138.3 − 88.3 = 50 kN/m2
2.12 Refer to the figure.
a. The compression index Cc is given by [Eq. (2.53)],
e1 − e2 1.10 − 0.85
Cc = = = 0.441
log σ 2′ − log σ1′ log 240 − log 65
4
© 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.
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.
, Soil B: % of gravel = 24, % of sand = 69, % of fines = 7
D10 = 0.17 mm, D30 = 0.82 mm, D60 = 2.6 mm Cu = 15.3; Cc = 1.5
LL = 42, PL = 22, PI = 20 plots above the A-line; hence, clay
The soil can be described as well graded clayey gravelly sand with a
group symbol of SW-SM.
Soil C: % of gravel = 1, % of sand = 99, % of fines = 0
D10 = 0.7 mm, D30 = 1.2 mm, D60 = 1.6 mm Cu = 2.3; Cc = 1.3
The soil can be described as poorly (uniformly) graded sand with a
group symbol of SP.
Soil D: % of gravel = 0, % of sand = 12, % of fines = 88
LL = 75, PL = 31, PI = 44 plots above the A-line; hence, clay.
The soil can be described as sandy clay of high plasticity with a group
symbol of CH.
2.8 The head loss from the reservoir to the ditch, ∆h = 38 − 28 = 10.0 m
The length of the sand seam in the direction of the flow, L = 200 cos10 = 203.1 m
The hydraulic gradient, i = 10.0 203.1 = 0.0492
By Darcy’s law [Eq. (2.35)], v = ( 2.6 × 10 −5 m/s)( 0.0492 ) = 0.128 × 10 −5 m/s
The cross section of the sand seam through which the flow takes place is
1.0 × 500.0 = 500.0 m2
The flow rate = (0.128 × 10 −5 m/s)(500.0 m 2 ) = 64.0 × 10 −5 m 3/s
Volume of water flowing into the ditch per day:
= (64.0 × 10 −5 m 3/s)(24 )( 3600) = 55.3 m 3
2.9 For the flow net shown in Figure P2.9, Nf = 3 and Nd = 10
Total head loss from right to left, hmax = 5.0 m
The flow rate is given by [Eq. (2.46)]
Nf 3
q = khmax = (1.5 × 10−5 )(5.0) = 2.25 × 10−5 m 3/s/m length
Nd 10
= ( 2.25 × 10− 5 )(50.0)( 24)( 3600 m 3/day) = 97.2 m 3 /day
3
© 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.
, 2.10 On top of the soft clay layer (i.e., at 10 m depth), initially:
σ ′ = 1 × 17.0 + 9( 20 − 9.81) = 108.7 kN/m 2
After the water table is lowered,
σ ′ = 3 × 17.0 + 7( 20 − 9.81) = 122.3 kN/m 2
By lowering the water table, the effective stress has increased by
(122.3 – 108.7) = 13.6 kN/m2
2.11 The soil below the water table can be assumed to be fully saturated (i.e., S = 1).
e = wGs = 0.25 × 2.70 = 0.675
The saturated unit weight can be computed as
(Gs + e)γw (2.70 + 0.675) × 9.81
γsat = = = 19.8 kN/m3
1+ e 1 + 0.675
At a depth of 5 m into the sandy clay,
σ = 4 × 9.81 + 5 × 19.8 = 138.3 kN/m2
u = 9 × 9.81 = 88.3 kN/m2
σ ′ = σ − u = 138.3 − 88.3 = 50 kN/m2
2.12 Refer to the figure.
a. The compression index Cc is given by [Eq. (2.53)],
e1 − e2 1.10 − 0.85
Cc = = = 0.441
log σ 2′ − log σ1′ log 240 − log 65
4
© 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.
