A normally consolidated clay settled 10 mm when effective stress was increased from 100 kN/m2 to 200 kN/m2. If the effective stress is further increased from 200 kN/m2 to 400 kN/m2 , then the settlement of the same clay is
A. 10 mm
B. 20 mm
C. 40 mm
D. none of the above
Answer: Option A
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A. directly proportional to time and inversely proportional to drainage path
B. directly proportional to time and inversely proportional to square of drainage path
C. directly proportional to drainage path and inversely proportional to time
D. directly proportional to square of drainage path and inversely proportional to time
Amol and nikhil are wrong mv is constant only for a particular stress range you cant cancel it out. Shabirs approach is right
Nikhil dongre ur right but make mistake in calculation at end
∆H1/∆H2=Δσ1/Δσ2
10/ΔH2= 100/200
10/ΔH2= 0.5
10/0.5=∆H2
20mm = ∆H2
ΔH1=10mm
Δσ1= σ2-σ1= 200-100=100 kn/m²
Δσ2= 400-200=200 kn/m²
We know that: ΔH=mv.σ.h
ΔH1/ΔH2=mv.Δσ1.H/mv.Δσ2.h
ΔH1/ΔH2= Δσ1/Δσ2
10/ΔH2= 100/200
10/ΔH2= 0.5
10×0.5= H2
H2= 5cm=50mm
We know that, Change in height ΔH is directly proportional to Log base 10 of (Increase in pressure/ initial pressure)
Given that, height of settlement = 10mm, when pressure increased from 100 to 200 kn/m^2
Let height of new settlement be 'x', when pressure increased from 200 to 400 kn/m^2
So, lets equate the ratio of heights and pressures.
10 / x = Log base 10 ( 200/100) / Log base 10 (400/200)
x = 10 / 1
x = 10 mm = 1 cm
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