Soil mechanics miscellaneous
- Assuming that a river bed level does not change and the depth of water in river was 10 m, 15 m and 8 m during months of February, July and December respectively of a particular year. The average bulk density of the soil is 20 kN/m3. The density of water is 110 kN/m3. The effective stress at a depth of 10m below the river bed during these months would be
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Effective stress below riverbed does not depend upon the depth of water above bed level
Correct Option: B
Effective stress below riverbed does not depend upon the depth of water above bed level
- For a triaxial shear test conducted on a sand specimen at a confining pressure of 100 kN/m2 under drained condition,s resulted in a deviator stress (s1 – s3) at failutre of 100 kN/m2. The angle of shearing resistance of the solid would be
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σ1' = σ3' + σd'
= 100 + 100 = 200 kN/m2
C = 0
σ1' = σ3' tan2 α + 2c tan α'
α = 45 + Φ 
2
⇒ 200 = 100 tan2 α
∴ α = 54.73°
⇒ Φ = 19.47°Correct Option: B
σ1' = σ3' + σd'
= 100 + 100 = 200 kN/m2
C = 0
σ1' = σ3' tan2 α + 2c tan α'α = 45 + Φ 2
⇒ 200 = 100 tan2 α
∴ α = 54.73°
⇒ Φ = 19.47°
- The water content of a saturated soil and the specific gravity of soil solids were found to be 30% and 2.70, respectively, Assuming the unit weight of water to be 10 kN/m3, the saturated unit weight (kN/m3) and the void ratio of the soil are
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w = 0.3
G = 2.7
Sr = 1 (∵ it is saturated soil)
We know,e = wG = 0.3 × 2.7 = 0.81 Sr 1
Saturated unit weight,γsat = (g + e)γw 1 + e = (2.7 + .81) × 10 1 + .81
= 19.39 kN/m3.Correct Option: A
w = 0.3
G = 2.7
Sr = 1 (∵ it is saturated soil)
We know,e = wG = 0.3 × 2.7 = 0.81 Sr 1
Saturated unit weight,γsat = (g + e)γw 1 + e = (2.7 + .81) × 10 1 + .81
= 19.39 kN/m3.
- In its natural condition a soil sample has a mass of 1.980 kg and a volume of 0.001 m3. After being completely dried in an oven, the mass of sample is 1.800 kg. Specific gravity is 2.7. Unit wt of water is 10 kN/m3. Degree of saturation of soil is.
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Mass in natural state (m1) = 1.980 kg.
Mass in dry state (m2) = 1.800 kg.
Volume = 0.001 m3
G = 2.7 γw = 10 kN/m3γ = m1 = 1.980 = 1980 kg/m3. v .001 γd = m2 = 1.800 = 1800 kg/m3. v .001 γd = γ 1 + w 1800 = 1980 1 + w w = 1980 - 1 = 0.1 1800 γd = Gγw 1 + e 1800 = 2.7 × 1000 1 + e
(rw = 1000 kg/m3)e = 2.7 × 1000 - 1 = 0.5 1800 e = WG = 0.1 × 2.7 Sr Sr Sr = 0.1 × 2.7 = 0.54 0.5 Correct Option: C
Mass in natural state (m1) = 1.980 kg.
Mass in dry state (m2) = 1.800 kg.
Volume = 0.001 m3
G = 2.7 γw = 10 kN/m3γ = m1 = 1.980 = 1980 kg/m3. v .001 γd = m2 = 1.800 = 1800 kg/m3. v .001 γd = γ 1 + w 1800 = 1980 1 + w w = 1980 - 1 = 0.1 1800 γd = Gγw 1 + e 1800 = 2.7 × 1000 1 + e
(rw = 1000 kg/m3)e = 2.7 × 1000 - 1 = 0.5 1800 e = WG = 0.1 × 2.7 Sr Sr Sr = 0.1 × 2.7 = 0.54 0.5
- A soil is composed of solid spherical grains of identical specific gravity and diameter between 0.075 mm and 0.0075 mm. If the terminal velocity of the largest particle falling through water without flocculation is 0.5 mm/s, that for the smallest particle would be
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Terminal velocity,
v = g (s - 1)d2 18v
∴ v ∝ d2
v1 = .5 mm/s
d1 = 0.075 mm
d2 = 0.0075 mm
v2 =?v1 = d12 v2 d22 0.5 = 0.752 v2 0.00752 v2 = 0.5 = 0.005 mm/s 102 Correct Option: A
Terminal velocity,
v = g (s - 1)d2 18v
∴ v ∝ d2
v1 = .5 mm/s
d1 = 0.075 mm
d2 = 0.0075 mm
v2 =?v1 = d12 v2 d22 0.5 = 0.752 v2 0.00752 v2 = 0.5 = 0.005 mm/s 102
