Question

A vane shear test was conducted to determine undrained cohesion of soft clay. The vane was 11.25 cm high and 7.5 cm across the blades. The equivalent torque recorded at the torque head at failure was 312.5 kg-cm. What will be the undrained cohesion of the soft clay ?

• A. $0.128kg/c{m}^2$
• B. $0.196kg/c{m}^2$
• C. $0.224kg/c{m}^2$
• D. $0.257kg/c{m}^2$

Answer 1

The correct option is D $0.257kg/c{m}^2$

Given,
H = 11.25 cm
D = 7.5 cm
T = 312.5 kg-cm

Undrained cohesion

$C_u=\frac{T}{\pi D^2(\frac{H}{2}+\frac{D}{6})}$

$=\frac{312.5}{\pi \times {7.5}^2(\frac{11.25}{2}+\frac{7.5}{6})}$

$=0.257kg/c{m}^2$