Question

A designer used plate load test to obtain the value of the bearing capacity factor \( N_t \). A circular plate of 1 m diameter was placed on the surface of a dry sand layer extending very deep beneath the ground. The unit weight of the sand is 16.66 kN/m\(^3\). The plate is loaded to failure at a pressure of 1500 kPa.
Considering Terzaghi's bearing capacity theory, the bearing capacity factor \( N_t \) is ......... (round off to the nearest integer).

Hint:

When using Terzaghi's bearing capacity theory, ensure you account for the material properties such as cohesion, unit weight, and the depth of the foundation when calculating the bearing capacity factors. For sand, the cohesion is often zero.

Answer 1

We know the formula for a circular plate: \[ q_u = 1.3 C_N C + \gamma D_f N_q + 0.3 B N_t \] For sand, the cohesion \( c = 0 \), so the formula simplifies to: \[ q_u = 0.3 \times 1 \times 16.66 \times N_t = 1500 \] Where:
- \( \gamma = 16.66 \, {kN/m}^3 \) (unit weight of sand)
- \( D_f = 1 \, {m} \) (depth of foundation)
- \( N_t \) is the bearing capacity factor
Now, solving for \( N_t \): \[ N_t = \frac{1500}{0.3 \times 16.66} = 300.12 \] Thus, the bearing capacity factor \( N_t \) is approximately \( 300 \).