Soil mechanics miscellaneous
- A unit volume of a mass of saturated soil is subjected to horizontal seepage. The saturated unit weight is 22 kN/m3 and the hydraulic gradient is 0.3. The resultant body force on the soil mass is
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Seepage pressure per unit volume = i = γw
= 0.3 × 9.81 = 2.94 3 kN
Total vertical force = (2.2 kN/m2) × vol = 22 kN
∴ Resultant = √222 + 2.9432
= 22.19 kNCorrect Option: D
Seepage pressure per unit volume = i = γw
= 0.3 × 9.81 = 2.94 3 kN
Total vertical force = (2.2 kN/m2) × vol = 22 kN
∴ Resultant = √222 + 2.9432
= 22.19 kN
- The ratio of saturated unit weight to dry unit weight of dry unit weight is 1.25. If the specific gravity of solids (Gs) is 2.56, the void ratio of the soil is
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γsat = 1.25 γw γsat = γw(G + e) 1 + e γd = Gγw 1 + e ∴ γd = γw G 1 + e ∴ γsat = γd (G + e) G ⇒ γsat = G + e γd G ⇒ 1.25 = 2.65 + e 2.65
∴ e = 0.663Correct Option: B
γsat = 1.25 γw γsat = γw(G + e) 1 + e γd = Gγw 1 + e ∴ γd = γw G 1 + e ∴ γsat = γd (G + e) G ⇒ γsat = G + e γd G ⇒ 1.25 = 2.65 + e 2.65
∴ e = 0.663
- A braced cut, 5 m wide and 7.5 m deep is proposed in a cohesion less soil deposit having effective cohesion c’ = 0 and effective friction angle, φ’ = 36°. The first row of struts is to be installed a depth of 0.5 m below ground surface and spacing between the struts should be 1.5 m. If the horizontal spacing of struts is 3 m and unit weight of the deposit is 20kN/m3, the maximum strut load will be
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Number of struts
= 75. - 0.5 1.5
= 4.67 ≈ 5
Maximum force will be at centre.
Active earth pressure,
σ1 = σ3 = tan α – 2 c tan α
γz = α3 tan2 α20 × 7.5 = σ3 tan2 
45 + Φ 
2 = σ3 tan2 45 + 36 2
⇒ 150 = σ3 × 3.85
∴ σ3 = 38.96 kN/m2
Uniform pressure = 0.65 × σ3
= 0.65 × 38.96 = 25.32
∴ Maximum force on a street at middle
= 25.32 × 3 × 1.5 = 113.9 kNCorrect Option: C
Number of struts
= 75. - 0.5 1.5
= 4.67 ≈ 5
Maximum force will be at centre.
Active earth pressure,
σ1 = σ3 = tan α – 2 c tan α
γz = α3 tan2 α20 × 7.5 = σ3 tan2 45 + Φ 2 = σ3 tan2 45 + 36 2
⇒ 150 = σ3 × 3.85
∴ σ3 = 38.96 kN/m2
Uniform pressure = 0.65 × σ3
= 0.65 × 38.96 = 25.32
∴ Maximum force on a street at middle
= 25.32 × 3 × 1.5 = 113.9 kN
- A double draining clay layer, 6 m thick, settles by 30 mm in three years under the influence of a certain loads. Its final consolidation settlement has been estimated to be 120 mm. If a thin layer of sand having negligible thickness is introduced at a depth of 1.5 m below the top surface, the final consolidation settlement of clay layer will be
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Rate of settlement will not affect final settlement.
Rate of settlement of clay will increase 4 times
But total settlement will be unaffected.
∴ final settlement = 120 mmCorrect Option: B
Rate of settlement will not affect final settlement.
Rate of settlement of clay will increase 4 times
But total settlement will be unaffected.
∴ final settlement = 120 mm
- A masonry dam is founded on previous sand having porosity equal to 45% and specific gravity of sand particles is 2.65. For a desired factor of safety of 3 against sand boiling, the maximum permissible upward gradient will be
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Hydraulic gradient,
ic = G - 1 1 + e
= (G – 1) (1 – x)
= (2.65 – 1) (1 – 0.45) = 0.9075
Factor of safety,F.S = = critical hydraulic gradient permissible upward gradient 3 = 0.9075 ipu
∴ permissible upward gradient = 0.3025Correct Option: B
Hydraulic gradient,
ic = G - 1 1 + e
= (G – 1) (1 – x)
= (2.65 – 1) (1 – 0.45) = 0.9075
Factor of safety,F.S = = critical hydraulic gradient permissible upward gradient 3 = 0.9075 ipu
∴ permissible upward gradient = 0.3025
