Soil mechanics miscellaneous
- In a falling head permeability test the initial head of 1.0 m dropped to 0.35 m in 3 hours, the diameter of the stand pipe being 5 mm. The soil specimen was 200 mm long and of 100 mm diameter. The coefficient of permeability of the soil is:
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a = π × (0.5)2 4
= 0.196 cm2Ax = π × (10)2 4
= 78.5 cm2
L = 20 cm
h1 = 100 cm, h2 = 35 cm
t = 3 hoursk = aL . log h1 At h2 = 0.196 × 20 In 
100 
78.5 × 3 35
= 0.0175 cm/hr
= 4.86 × 10–6 cm/sCorrect Option: B
a = π × (0.5)2 4
= 0.196 cm2Ax = π × (10)2 4
= 78.5 cm2
L = 20 cm
h1 = 100 cm, h2 = 35 cm
t = 3 hoursk = aL . log h1 At h2 = 0.196 × 20 In 100 78.5 × 3 35
= 0.0175 cm/hr
= 4.86 × 10–6 cm/s
- In a triaxial test carried, out on a cohesion less soil sample with a cell pressure of 20 kPa, the observed value of applied stress at the point of failure was 40 kPa. The angle of internal friction of the soil is
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C = 0
σ1 = σ3 tan2 α + 2 c tan α
σ3 = 20 kN/m2
σd = 40 kN/m2
σ1 = σ3 + σd = 20 + 40
= 60 kN/m2
⇒ 60 = 20 tan2 α + 0
∴ tan2 α = 3
∴ α = 60°α = 45 + Φ 2 ⇒ 60 = 45 + Φ 2
∴ Φ = 30°Correct Option: D
C = 0
σ1 = σ3 tan2 α + 2 c tan α
σ3 = 20 kN/m2
σd = 40 kN/m2
σ1 = σ3 + σd = 20 + 40
= 60 kN/m2
⇒ 60 = 20 tan2 α + 0
∴ tan2 α = 3
∴ α = 60°α = 45 + Φ 2 ⇒ 60 = 45 + Φ 2
∴ Φ = 30°
- The time for a clay layer to achieve 85% consolidation is 10 years. If the layer was half as thick, 10 times more permeable and 4 times more compressible then the time that would be required to achieve the same degree of sonsolidation is
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Co-eff. of consolidation, Cv
Cv = k ⇒ Cv ∝ k mv.γw mv
Also, Time factor,Tv = Cv.t d2 ∴ Cv ∝ d2 t ∴ = k ∝ d2 mv t ∴ t ∝ d2mv k ∴ t2 = d2 2 k1 mv2 t1 d1 k2 mv1
given, d2 = 0.5 d1
kz = 10 k1
mv2 = v1 4m
t1 = 10 years∴ = t2 = (0.5)2 × 1 × 4 10 10 ∴ t2 = 0.25 × 4 × 10 = 1 Year 10 Correct Option: A
Co-eff. of consolidation, Cv
Cv = k ⇒ Cv ∝ k mv.γw mv
Also, Time factor,Tv = Cv.t d2 ∴ Cv ∝ d2 t ∴ = k ∝ d2 mv t ∴ t ∝ d2mv k ∴ t2 = d2 2 k1 mv2 t1 d1 k2 mv1
given, d2 = 0.5 d1
kz = 10 k1
mv2 = v1 4m
t1 = 10 years∴ = t2 = (0.5)2 × 1 × 4 10 10 ∴ t2 = 0.25 × 4 × 10 = 1 Year 10
- If the effective stress strength parameters of a soil are c’ = 10kPa and Φ’ = 30°, the shear strength on a plane within the saturated soil mass at a point where the total normal stress is 300kPa and pore water pressure is 150 kPa will be
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S = C' + σ' tan Φ
= C' + (σ – μ) tan Φ
= 10 + (300 – 150) tan 30°
= 96.6 kN/m2Correct Option: B
S = C' + σ' tan Φ
= C' + (σ – μ) tan Φ
= 10 + (300 – 150) tan 30°
= 96.6 kN/m2
- An infinite slope is to be constructed in a soil. The effective stress strength parameters of the soil are c’ = 0 and Φ’ = 30°. The saturated unit weight of the slope is 20kN/m3 and the unit weight of water is 10kN m3. Assuming that seepage is occurring parallel to the slope, the maximum slope angle for a factor of safety of 1.5 would be
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Factor of safety,
F = γ' . tanΦ' γsat tan i
γ' = γsat - γw
= (20 – 10) = 10⇒ 1.5 = 10 . tan30° 20 tan i ∴ tan i = tan 30° = 0.192 1.5 × 2
∴ i = 10.89°.Correct Option: A
Factor of safety,
F = γ' . tanΦ' γsat tan i
γ' = γsat - γw
= (20 – 10) = 10⇒ 1.5 = 10 . tan30° 20 tan i ∴ tan i = tan 30° = 0.192 1.5 × 2
∴ i = 10.89°.
