Result of Test-I
$\mathrm{0.0034}\mathrm{Kg}/{\mathrm{cm}}^{2}$
Shear Strenghth
Undrained Shear Strenghth of the Soil
$\mathrm{Torque(T)}=\mathrm{K}\times \frac{\mathrm{Diffrence\; In\; Degree}}{\mathrm{180}}$
$\mathrm{T}=4\times \frac{\mathrm{30}}{\mathrm{180}}={\mathrm{0.67}}^{}$
$\mathrm{Cohesion(Cu)}=\frac{T}{\mathrm{\pi}{\mathrm{D}}^{2}\left(\frac{D}{6},\mathrm{+},\frac{H}{2}\right)}$
$\mathrm{Cu}=\frac{0.67}{\mathrm{3.14}\times 14.06\left(\frac{\mathrm{3.75}}{6},\mathrm{+},\frac{\mathrm{7.5}}{2}\right)}$
$\mathrm{Cu}=\mathrm{0.0034}\mathrm{Kg}/{\mathrm{cm}}^{2}$
Result of Test-II
$\mathrm{0.0069}\mathrm{Kg}/{\mathrm{cm}}^{2}$
Shear Strenghth
Undrained Shear Strenghth of the Soil
$\mathrm{Torque(T)}=\mathrm{K}\times \frac{\mathrm{Diffrence\; In\; Degree}}{\mathrm{180}}$
$\mathrm{T}=4\times \frac{\mathrm{60}}{\mathrm{180}}={\mathrm{1.33}}^{}$
$\mathrm{Cohesion(Cu)}=\frac{T}{\mathrm{\pi}{\mathrm{D}}^{2}\left(\frac{D}{6},\mathrm{+},\frac{H}{2}\right)}$
$\mathrm{Cu}=\frac{1.33}{\mathrm{3.14}\times 14.06\left(\frac{\mathrm{3.75}}{6},\mathrm{+},\frac{\mathrm{7.5}}{2}\right)}$
$\mathrm{Cu}=\mathrm{0.0069}\mathrm{Kg}/{\mathrm{cm}}^{2}$
Result of Test-III
$\mathrm{0.0103}\mathrm{Kg}/{\mathrm{cm}}^{2}$
Shear Strenghth
Undrained Shear Strenghth of the Soil
$\mathrm{Torque(T)}=\mathrm{K}\times \frac{\mathrm{Diffrence\; In\; Degree}}{\mathrm{180}}$
$\mathrm{T}=4\times \frac{\mathrm{90}}{\mathrm{180}}={\mathrm{2.00}}^{}$
$\mathrm{Cohesion(Cu)}=\frac{T}{\mathrm{\pi}{\mathrm{D}}^{2}\left(\frac{D}{6},\mathrm{+},\frac{H}{2}\right)}$
$\mathrm{Cu}=\frac{2.00}{\mathrm{3.14}\times 14.06\left(\frac{\mathrm{3.75}}{6},\mathrm{+},\frac{\mathrm{7.5}}{2}\right)}$
$\mathrm{Cu}=\mathrm{0.0103}\mathrm{Kg}/{\mathrm{cm}}^{2}$