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Question

An undisturbed soil sample was taken from the middle of a clay layer (i.e., 1.5 m below GL), as shown in figure. The water table was at the top of clay layer. Laboratory test result are as follows:

Natural water content of clay : 25%

Preconsolidation pressure of clay : 60 kPa

Compression index of clay : 0.50

Recompression index of clay : 0.05

Specific gravity of clay : 2.70

Bulk unit weight of sand : 17kN/m3

A compacted fill of 2.5 m height with unit weight of 20kN/m3 is placed at the ground level.

Assuming unit weight of water as 10kN/m3, the ultimate consolidation settlement (expressed in mm) of the clay layer is .

Natural water content of clay : 25%

Preconsolidation pressure of clay : 60 kPa

Compression index of clay : 0.50

Recompression index of clay : 0.05

Specific gravity of clay : 2.70

Bulk unit weight of sand : 17kN/m3

A compacted fill of 2.5 m height with unit weight of 20kN/m3 is placed at the ground level.

Assuming unit weight of water as 10kN/m3, the ultimate consolidation settlement (expressed in mm) of the clay layer is .

- 37.04

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Solution

The correct option is **A** 37.04

¯¯¯σ=17×1+0.5×γsubclay

For clay

w = 0.25

S = 1

g = 2.7

⇒Se=Gw

⇒e=2.7×0.251

e = 0.675

⇒γsub=G−11+eγw=2.7−11+0.675×10

⇒γsub=10.15kN/m3

¯¯¯σ=17+0.5×10.15

=22.07kN/m2

Δ¯¯¯σ=2.5×20=50kN/m2

Alternate Method

ΔH1=CrH1+e0log6022.07

=0.05×10001.675log6022.07

ΔH1=12.966mm

ΔH2=CcH1+e′log(72.0760)

Cr=Δelog(6022.07)=0.05

Δe=0.0217

⇒e′=0.653

ΔH2=0.5×10001.653log(72.0760)

ΔH2=24.075mm

⇒ΔH=ΔH1+ΔH2=37.04mm

¯¯¯σ=17×1+0.5×γsubclay

For clay

w = 0.25

S = 1

g = 2.7

⇒Se=Gw

⇒e=2.7×0.251

e = 0.675

⇒γsub=G−11+eγw=2.7−11+0.675×10

⇒γsub=10.15kN/m3

¯¯¯σ=17+0.5×10.15

=22.07kN/m2

Δ¯¯¯σ=2.5×20=50kN/m2

Alternate Method

ΔH1=CrH1+e0log6022.07

=0.05×10001.675log6022.07

ΔH1=12.966mm

ΔH2=CcH1+e′log(72.0760)

Cr=Δelog(6022.07)=0.05

Δe=0.0217

⇒e′=0.653

ΔH2=0.5×10001.653log(72.0760)

ΔH2=24.075mm

⇒ΔH=ΔH1+ΔH2=37.04mm

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