A cylinder of soil fails under an axial vertical stress of 140 kN/m², when it is laterally unconfined, the failure plane makes an angle of 45° with the horizontal. What will be the value of cohesion of soil sample? 

The cohesion of a soil sample can be determined using the Mohr-Coulomb failure criterion:

\(\tau = c + \sigma \tan \phi\)

For unconfined compression tests, the soil is laterally unconfined, and the failure plane makes an angle of \(45^\circ\) with the horizontal. The shear strength at failure can be represented as:

\(\tau = \sigma \tan (45^\circ + \frac{\phi}{2})\)

For the given problem, we know the soil fails at an axial vertical stress (\(\sigma\)) of \(140 \, \text{kN/m}^2\) and the failure plane makes an angle of \(45^\circ\) with the horizontal.

Calculation:

Given:

• Axial vertical stress, \(\sigma = 140 \, \text{kN/m}^2\)
• Angle of failure plane, \(\theta = 45^\circ\)

Since the failure plane makes an angle of \(45^\circ\) with the horizontal, we can assume the internal friction angle (\(\phi\)) is such that:

\(\phi/2 = 0^\circ\)

Thus, the Mohr-Coulomb criterion simplifies to:

\(\tau = c\)

In the unconfined compression test, the axial stress (\(\sigma\)) is twice the cohesion (\(c\)):

\(\sigma = 2c\)

Rearranging for cohesion (\(c\)):

\(c = \frac{\sigma}{2}\)

Substitute the given value:

\(c = \frac{140}{2} = 70 \, \text{kN/m}^2\)

The cohesion of the soil sample is 70 kN/m²