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

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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