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Question

A 10 m thick clay layer is resting over a 3 m thick sand layer and is submerged. A fill of 2 m thick sand with unit weight of 20kN/m3 is placed above the clay layer to accelerate the rate of consolidation of the clay layer. Coefficient of consolidation of clay is 9×102m2/year and coefficient of volume compressibility of clay is 2.2×104m2/kN. Assume Taylor's relations between time factor and average degree of consolidation.


The settlement (in mm, round off to two decimal places) of the clay layer, 10 years after the construction of the fill, is
  1. 18.83

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Solution

The correct option is A 18.83
Given

Coefficient consolidation (Cv)

=9×102 m2/year

Volume compressibility (mv)

=2.2× 104 m2/kN

Thickness of clay layer (H) = 10 m

Change in effective stress due to surcharge

(i.e.,sand layer) Δ¯¯¯σ = 2 × 20

=40kN/m2

(Total settlement ΔH)=mvHΔ¯¯¯σ

=2.2×104×10×40

= 0.088 m

Now, According to Taylor relation

Tv=Cvtd2

d=102=5m

(double drainage both side sand layer)

t = 10 year

Tv=9×102×1052=0.036

We know that

Tv=0.197 for 50 % Consolidation

Here Tv<0.197

So, we use

π4(U2)=Tv

π4(U2)=0.036

U = 0.214

Now U=ΔhΔH

Δh = settlement after 10 years

Δh=ΔH×0.214

=0.088×0.214

= 0.01883 m

= 18.83 mm

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