Question

A 6 m thick clay layer beneath a building is overlain by 6 m of sand stratum and is underlain by an impervious rock. Water table is located at a depth of 3 m below ground level. The unit weight of clay and sand are 18 kN/m3 and 20 kN/m3 respectively. Consolidation test on clay gave the following relation between void ratio and pressure.

e=1.5−0.5 log10P

The Pressure increment due to foundation of building placed at 3 m below ground level at top and bottom of clay is 75 kPa and 25 kPa respectively. Due to excavation for foundation the average pressure release in clay layer is 20 kPa. The amount of settlement expected due to building is ?

(in cm, Assume γ=γsat and γw=10 kN/m3 )

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Solution

The correct option is **A** 20.78

**Rock**

Given

γclay=18 kN/m3

e=15−0.5 log10P

γsand=20 kN/m3

In sand, above w.t, Assume γ=γsat

At centre of clay layer

σ′0=γsand×3+γ′sand×3+γ′clay×3

∴σ′0=20×3+10×3+8×3

σ′0=114 kPa

Due to foundation of building

Average pressure increment =75+252=50 kPa

By excavation, average pressure release = 20 kPa

∴ So, net average pressure increment, Δσ′=50−20=30 kPa

Average effective stress =σ′=σ′0+Δσ′=114+30=144 kPa

∴e0=1.5−0.5×log10 114=0.472

e=1.5−0.5×log10144=0.421

Δe=e0−e=0.051

ΔHH=Δe1+e0

ΔH=H×Δe1+e0

ΔH=6×0.0511+0.472

ΔH=0.2078 m

ΔH=20.78 cm

Given

γclay=18 kN/m3

e=15−0.5 log10P

γsand=20 kN/m3

In sand, above w.t, Assume γ=γsat

At centre of clay layer

σ′0=γsand×3+γ′sand×3+γ′clay×3

∴σ′0=20×3+10×3+8×3

σ′0=114 kPa

Due to foundation of building

Average pressure increment =75+252=50 kPa

By excavation, average pressure release = 20 kPa

∴ So, net average pressure increment, Δσ′=50−20=30 kPa

Average effective stress =σ′=σ′0+Δσ′=114+30=144 kPa

∴e0=1.5−0.5×log10 114=0.472

e=1.5−0.5×log10144=0.421

Δe=e0−e=0.051

ΔHH=Δe1+e0

ΔH=H×Δe1+e0

ΔH=6×0.0511+0.472

ΔH=0.2078 m

ΔH=20.78 cm

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