Question

Site exploration was done and it was found a thin layer of loamy soil exists at a depth of 8 m below the surface of the ground. Water table is located at a depth of 2 m below ground level and with the help of undisturbed sampling following data were obtained.
$C_u=22kN/m^2{C}^{\prime }=15kN/m^2$
${\varphi }_u={15}^{\circ }{\varphi }^{\prime }={25}^{\circ }$
If the shear stress develops very slowly, the shear resistance of loamy soil on horizontal plane is_____ (kPa).
$(Assume{\gamma }_d=18kN/m^3,w_{sat}=15\%{\gamma }_w=9.81kN/m^3)$

• A. 62.26
• B. 15.17
• C. 39.28
• D. 76.25

Answer 1

The correct option is A 62.26


${\gamma }_{sat}={\gamma }_d(1+w)$

$=18\times (1+0.15)$

$=20.7kN/m^3$

${\gamma }^{\prime }={\gamma }_{sat}-{\gamma }_w=10.89kN/m^3$

When shear stress develops very slowly, effective stress analysis should be done

$\therefore S=C^{\prime }+{\sigma }^{\prime }\tan \varphi ,$

$\therefore {\sigma }^{\prime }={\gamma }_d\times 2+{\gamma }^{\prime }\times 6m$

$\therefore {\sigma }^{\prime }=18\times 2+10.89\times 6=101.34kN/m^2$

$S=15+101.34\times \tan {25}^{\circ }=62.26kN/m^2$