Soil mechanics miscellaneous
- The flow net constructed for the dam is shown in the figure below. Taking the coefficient of permeability as 3.8 × 10–6 m/s, the quantity of flow (in cm3 /s) under the dam per meter of dam is __________.
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7.1 to 7.85
No. of flow channels, Nf = 3
No. of equipotential lines, Nd = 10
K = 3.8 × 10–6 m/s
H = 6.3 m= 3.8 × 10–6 × 6.3 × 3 10
= 7.182 × 10–6 m3 /s/m
= 7.182 cm3 /s/m7.1 to 7.85
No. of flow channels, Nf = 3
No. of equipotential lines, Nd = 10
K = 3.8 × 10–6 m/s
H = 6.3 m= 3.8 × 10–6 × 6.3 × 3 10
= 7.182 × 10–6 m3 /s/m
= 7.182 cm3 /s/mCorrect Option: A
7.1 to 7.85
No. of flow channels, Nf = 3
No. of equipotential lines, Nd = 10
K = 3.8 × 10–6 m/s
H = 6.3 m= 3.8 × 10–6 × 6.3 × 3 10
= 7.182 × 10–6 m3 /s/m
= 7.182 cm3 /s/m
- The following data are given for the laboratory sample. σo' = 175 kPa; eo = 1.1; σo' + ∆σo' = 300 kPa; e = 0.9 If thickness of the clay specimen is 25 mm, the value of coefficient of volume comprehensibility is __________ × 10–4 m2 /kN.
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7.6 to 8
Mv = av = ∆e / ∆σ 1 + e0 1 + e0 = (1.1 - 0.9) 125 × 2.1
= 7.61 × 10–6 m2 /kN
H = 6.3 mCorrect Option: D
7.6 to 8
Mv = av = ∆e / ∆σ 1 + e0 1 + e0 = (1.1 - 0.9) 125 × 2.1
= 7.61 × 10–6 m2 /kN
H = 6.3 m
- A 588 cm3 volume of moist sand weighs 1010 gm. Its dry weight is 918 gm and specific gravity of solids, G is 2.67. Assuming density of water as 1 gm/cm3, the void ratio is ______
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0.71
γd = Gγω 1 + e γd = Ws V = 918 = 1.561g/cm³ 588
G = 2.67, γω = 1, e = ?1.561 = 2.67 × 1 1 + e 1 + e = 2.67 = 1.71 1.561
∴ e = 0.71Correct Option: C
0.71
γd = Gγω 1 + e γd = Ws V = 918 = 1.561g/cm³ 588
G = 2.67, γω = 1, e = ?1.561 = 2.67 × 1 1 + e 1 + e = 2.67 = 1.71 1.561
∴ e = 0.71
Direction: The average effective overburden pressure on 10 m thick homogeneous saturated clay layer is 150 K Pa. Consolidation test on undisturbed soil sample taken from the clay layer showed that the void ratio decreased from 0.6 to 0.5 by increasing the stress intensity from 100 kPa to 200 kPa. (G = 2.65)
- The initial void ratio of the clay layer is
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Compression Index,
Cc = e0 - e1 log10 
σ1 
σ0 = 0.6 - 0.5 = 0.209 log10 300 100
Take the case of e = 0.5 and σ = 300 Cc is constant.Cc = e0 - 0.5 log10 300 150 .209= e0 - 0.5 log10(2)
∴ e0 = 0.563Correct Option: B
Compression Index,
Cc = e0 - e1 log10 σ1 σ0 = 0.6 - 0.5 = 0.209 log10 300 100
Take the case of e = 0.5 and σ = 300 Cc is constant.Cc = e0 - 0.5 log10 300 150 .209= e0 - 0.5 log10(2)
∴ e0 = 0.563
Direction: The ground conditions at a site are as shown in the figure. The water table at the site which was initially at a depth of 5 m below the ground level got permanently lowered to a depth of 15 m below the ground level due to pumping of water over a few years. Assume the following data (i) unit weight of water = 10 kN/m3 .
(ii) unit weight of sand above water table = 18 kN/m3 .
(iii) unit weight of sand and clay below the water tabl = 20 kN/m3 .
(iv) coefficient of volume comprehensibility = 0.25 m2 /MN.
- What is the change in the effective stress in kN/m2 at mid-depth of the clay due to the lowering of the water table?
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Change in effective stress = change in water level × rw .
= (15 – 5) × 10 = 100 kN/m2.Correct Option: D
Change in effective stress = change in water level × rw .
= (15 – 5) × 10 = 100 kN/m2.
