Question

A bridge of $40\times 10m$ is to be founded on the river bed at a depth of 6 meters. The soil is clay having shear strength of $15tonnes/m^2$. The wall cohesive on the sides of pier could be taken as half of shear strength and the saturated density of 2 gm/cc. The rectangular pier is expected to carry a total load of 25,000 tonnes including its own weight. If the depth of water in river is 5 meters, the factor of safety of pier against shear failure based on skempton’s approach is

• A. 2
• B. 3
• C. 2.5
• D. 2.7

Answer 1

The correct option is B 3

$S=C_u=15tonnes/m^2$

$Q_y=25,000$

Foundation area = $10\times 40=400m^2$

$\therefore q=\frac{25000}{400}=62.5t/m^2$

Due to wall cohesion
$\therefore$ Adhesion force

$Q_s=\frac{C_u}{2}\times P\times L=\frac{15}{2}\times 2\times (10+40)\times 6$

$Q_s=4500tons$

Equivalent pressure due to cohesive =$\frac{Q_s}{A}=\frac{4500}{10\times 40}=11.25kN/m^2$

Pressure due to self weight of soil = $\gamma D=2\times 6=12t/m^2$

Pressure due to boundary = ${\gamma }_w\times 5$

${\gamma }_w=1g/cc=1t/m^3=1\times 5=5t/m^2$

Net pressure, $q_n=q-rD-WallCohesion-Buoyancypressure$

$=62.5-12-11.25-5$

$q_n=34.25t/m^2$

Design :

$q_n\le q_{ns}$

$q_n=\frac{q_{nu}}{F}$

From skempton's approach

$\therefore q_{nu}=C{N}_c=5[1+0.2\frac{D}{B}][1+0.2\frac{B}{L}]\times C$

$=5[1+0.2\frac{6}{10}][1+0.2\frac{10}{40}]\times 15$

$\therefore q_{nu}=105.84kPa$

$\therefore q_{nu}=\frac{q_{nu}}{F}$

$F=\frac{105.84}{34.25}=3.09$