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 $18kN/m^3$ and $20kN/m^3$ respectively. Consolidation test on clay gave the following relation between void ratio and pressure.
$e=1.5-0.5lo{g}_{10}P$

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 $\gamma ={\gamma }_{sat}and{\gamma }_w=10kN/m^3$ )

• A. 20.78
• B. 15.19
• C. 10.25
• D. 32.17

Answer 1

The correct option is A 20.78

Rock

Given
${\gamma }_{clay}=18kN/m^3$

$e=15-0.5lo{g}_{10}P$

${\gamma }_{sand}=20kN/m^3$

In sand, above w.t, Assume $\gamma ={\gamma }_{sat}$

At centre of clay layer

${\sigma }_0^{\prime }={\gamma }_{sand}\times 3+{\gamma }_{sand}^{\prime }\times 3+{\gamma }_{clay}^{\prime }\times 3$

$\therefore {\sigma }_0^{\prime }=20\times 3+10\times 3+8\times 3$

${\sigma }_0^{\prime }=114kPa$

Due to foundation of building

Average pressure increment $=\frac{75+25}{2}=50kPa$

By excavation, average pressure release = 20 kPa

$\therefore$ So, net average pressure increment, $\Delta {\sigma }^{\prime }=50-20=30kPa$

Average effective stress $={\sigma }^{\prime }={\sigma }_0^{\prime }+\Delta {\sigma }^{\prime }=114+30=144kPa$

$\therefore e_0=1.5-0.5\times lo{g}_{10}114=0.472$
$e=1.5-0.5\times lo{g}_{10}144=0.421$

$\Delta e=e_0-e=0.051$

$\frac{\Delta H}{H}=\frac{\Delta e}{1+e_0}$

$\Delta H=H\times \frac{\Delta e}{1+e0}$

$\Delta H=6\times \frac{0.051}{1+0.472}$

$\Delta H=0.2078m$

$\Delta H=20.78cm$