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

Given:
Soil: Normally Consolidated Clay

Effective stress increase:

Case 1:
100
→
200
 
kN/m
2
100→200kN/m
2


Settlement = 10 mm

Case 2:
200
→
400
 
kN/m
2
200→400kN/m
2


Settlement = ?

🔑 Concept (Very Important):
For normally consolidated clay,

Settlement
∝
log
⁡
(
σ
2
′
σ
1
′
)
Settlement∝log(
σ
1
′
​

σ
2
′
​

​
)
Case 1:
log
⁡
(
200
100
)
=
log
⁡
(
2
)
log(
100
200
​
)=log(2)
Settlement = 10 mm

Case 2:
log
⁡
(
400
200
)
=
log
⁡
(
2
)
log(
200
400
​
)=log(2)
👉 Stress ratio same hai (2)
👉 Log value same hai
👉 Clay aur thickness same hai

✅ Final Answer:
Settlement
=
10
mm
Settlement=10 mm
​

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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