The cohesion (c), angle of internal friction ($\varphi$) and unit weight ($\gamma$) of a soil are 15kPa, $20^{\circ }$ and $17.5{\text{ kN/m}}^3$, respectively. The maximum depth of unsupported excavation in the soil (in m, round off to two decimal places) is ________

In the range: 4.8 - 5

Given

- Cohesion: $$c=15\text{kPa}=15{\text{kN/m}}^2$$
- Angle of internal friction: $$\varphi =20^{\circ }$$
- Unit weight of soil: $$\gamma =17.5{\text{kN/m}}^3$$

Required: Maximum depth of unsupported vertical excavation.

Theoretical Background

For a vertical unsupported cut in a cohesive–frictional soil, the maximum safe depth is given by:

$$H_{\max }=\frac{4c}{\gamma }\tan \left(45^{\circ }+\frac{\varphi }{2}\right)$$

This expression comes from Rankine’s earth pressure theory, considering the condition when the net lateral pressure at the base becomes zero.

Step 1: Evaluate Trigonometric Term

$$45^{\circ }+\frac{\varphi }{2}=45^{\circ }+\frac{20^{\circ }}{2}=55^{\circ }$$

$$\tan 55^{\circ }\approx 1.428$$

Step 2: Substitute Numerical Values

$$H_{\max }=\frac{4\times 15}{17.5}\times 1.428$$

$$H_{\max }=\frac{60}{17.5}\times 1.428$$

$$H_{\max }=3.4286\times 1.428$$

$$H_{\max }\approx 4.90\text{m}$$

Final Answer

The maximum depth of unsupported excavation is:

$$\boxed{4.90\text{m}}$$