Question

A raft of $12m\times 12m$ with an intensity of $50kN/m^2$ is to be constructed on surface of soil strata as shown in figure. The number of sectors covered by plan of raft on newmark's chart is 100. If the influence factor of chart is 0.005, the ultimate settlement at centre of clay layer is in mm.
$(Assume{\gamma }_w=10kN/m^3)$

• A. 48 mm
• B. 36 mm
• C. 12 mm
• D. 45mm

Answer 1

The correct option is B 36 mm

Ultimate settlement should be calculated at centre of clay layer
n$\to$ number of sectors covered by plan
Due to raft,

$\Delta {\sigma }^{\prime }=I.n.q$

$=0.005\times 100\times 50$

$=25kN/m^2$

At centre of clay layer
${\sigma }_o^{\prime }={\gamma }_{silt}\times 2+{\gamma }_{sand}\times 5+{\gamma }_{clay}^{\prime }\times 2.5$

$=16\times 2+19\times 5+10\times 5+8\times 2.5$

${\sigma }_o^{\prime }=197kN/m^2$

$\therefore C_C=0.009(W_L-10)$

$=0.009(40-10)=0.27$

$S_f=\frac{C_C}{1+e_o}\times H\times {\log }_{10}\left(\frac{{\sigma }_o^{\prime }+\Delta {\sigma }^{\prime }}{{\sigma }_o^{\prime }}\right)$

$S_f=\frac{0.27}{1+0.9}\times 5\times {\log }_{10}\left(\frac{197+25}{197}\right)$

$\therefore S_f=0.36m=36mm$