Soil mechanics miscellaneous
- A non-homogeneous soil deposit consists of a silt layer sandwiched between a fine-sand layer at top and a clay layer below.
Permeability of the silt layer is 10 times the permeability of the clay layer and one-tenth of the permeability of the sand layer.
Thickness of the silt layer is 2 Times the thickness of the sand layer and two-third of the thickness of the clay layer. The ratio of equivalent horizontal and equivalent vertical permeability of the deposit is ________.
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Given, k2 = 10 k3 and k2 = 1/10 k1
d2 =2d1 and d2 = 2/3 d3
∴ k1 = 10k2, k3 = 1/10 k2
d1 = 1/2 d2,d3 = 3/2 d2kH = k1d1 + k2d2 + k3d3 d1 + d2 + d3 = (10k2 × 1 d2) + k2d2 + 
1 k2 × 3 d2 
2 10 2 1 d2 + d2 + 3 d2 2 2 = k2d2 (5 + 1 + 3/20) d2 3 = k2 × 123 = 2.05 k2 60 
= 60 k2 = 0.187 k2 321 ∴ kH = 2.05k2 = 10.9675 kV 0.187k2 Correct Option: B
Given, k2 = 10 k3 and k2 = 1/10 k1
d2 =2d1 and d2 = 2/3 d3
∴ k1 = 10k2, k3 = 1/10 k2
d1 = 1/2 d2,d3 = 3/2 d2kH = k1d1 + k2d2 + k3d3 d1 + d2 + d3 = (10k2 × 1 d2) + k2d2 + 1 k2 × 3 d2 2 10 2 1 d2 + d2 + 3 d2 2 2 = k2d2 (5 + 1 + 3/20) d2 3 = k2 × 123 = 2.05 k2 60 = 60 k2 = 0.187 k2 321 ∴ kH = 2.05k2 = 10.9675 kV 0.187k2
- A 20m thick clay layer is sandwiched between a silty sand layer and a gravelly sand layer. The layer experiences 30 mm settlement in 2 years.
Where Tv is the time factor and U is the degree of consolidation in% If the coefficient of consolidation of the layer is 0.003 cm2 /s, the deposit will experience a total of 50 mm settlement in the next ________ years.
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4.43
Tv = Cv.t d2 = 0.003 × (2 × 86400 × 365) 
300 × 100 2 150 π × u2 = 0.189 4
⇒ u = 0.49 ≤ 60%Consolidation = 30 = 61.22 mm 0.49
For 50 mm settlementu = 50 = 0.817 = 81.7% 61.22
TV = 1.784 – 0.933 log10(100 – u) = 0.608Tv = Cv.t = 0.608 d2 Correct Option: D
4.43
Tv = Cv.t d2 = 0.003 × (2 × 86400 × 365) 300 × 100 2 150 π × u2 = 0.189 4
⇒ u = 0.49 ≤ 60%Consolidation = 30 = 61.22 mm 0.49
For 50 mm settlementu = 50 = 0.817 = 81.7% 61.22
TV = 1.784 – 0.933 log10(100 – u) = 0.608Tv = Cv.t = 0.608 d2
- A water tank is to be constructed on the soil deposit shown in the figure below. A circular footing of diameter 3 m and depth of embedment 1m has been designed to support the tank. The total vertical load to be taken by the footing is 1500 kN. Assume the unit weight of water as 10 kN/m3 and the load dispersion pattern as 2V:1H. The expected settlement of the tank due to primary consolidation of the clay layer is ______mm.
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53.236
Settlement,∆H = Cc H0 1 + e0 log σ0 + ∆σ σ0
σ0 = (15 × 2) + (18 – 10) × 6 + (18 – 10) × 5
= 118 kN/m2∆σ0 = 1500 = 8.48kN/m2 π (3 + 6 + 6)2 4 ∴ ∆H = 0.3 × 10 log10 118 + 8.488 1 + 0.7 118
= 0.0532 m
= 53.2 mmCorrect Option: B
53.236
Settlement,∆H = Cc H0 1 + e0 log σ0 + ∆σ σ0
σ0 = (15 × 2) + (18 – 10) × 6 + (18 – 10) × 5
= 118 kN/m2∆σ0 = 1500 = 8.48kN/m2 π (3 + 6 + 6)2 4 ∴ ∆H = 0.3 × 10 log10 118 + 8.488 1 + 0.7 118
= 0.0532 m
= 53.2 mm
- An infinitely long slope is made up of a c-φ soil having the properties: cohesion (c) = 20 kPa, and dry unit weight (γd) = 16 kN/m3. The angle of inclination and critical height of the slope are 40° and 5 m, respectively. To maintain the limiting equilibrium, the angle of internal friction of the soil (in degrees) is __________________
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21 to 23
Correct Option: A
21 to 23
- A certain soil has the following properties: Gs = 2.71, n = 40% and w = 20%. The degree of saturation of the soil (rounded off to the nearest percent) is ________
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81 to 81.5
e = WG Sr also, e = η 1 - η
η = porosity= 0.4 = 0.67 1 - 0.4 S = 20 × 2.71 = 81.3% 0.67 Correct Option: C
81 to 81.5
e = WG Sr also, e = η 1 - η
η = porosity= 0.4 = 0.67 1 - 0.4 S = 20 × 2.71 = 81.3% 0.67
