Chapter 15, Problem 15.1P

To determine

Find the height (H) of the soil-rock interface.

Answer to Problem 15.1P

The height of the soil-rock interface is $\underset{\_}{6.28\text{m}}$.

Explanation of Solution

Given information:

The slope rise at an angle $\beta$ with the horizontal line is $19°$.

The unit weight $\gamma$ of the soil is $20{\text{kN/m}}^{\text{3}}$.

The angle of friction ${\varphi }^{\prime }$ is $33°$.

The cohesion $c^{\prime }$ is $47{\text{kN/m}}^{\text{2}}$.

The factor of safety $F_s$ is 3.1.

Calculation:

Determine the height of the soil-rock interface using the formula.

$F_s=\frac{c^{\prime }}{\gamma H{\cos }^2\beta \tan \beta }+\frac{\tan {\varphi }^{\prime }}{\tan \beta }$

Substitute 3.1 for $F_s$, $47{\text{kN/m}}^{\text{2}}$ for $c^{\prime }$, $20{\text{kN/m}}^{\text{3}}$ for $\gamma$, $19°$ for $\beta$, and $33°$ for ${\varphi }^{\prime }$.

$\begin{array}{l}3.1=\frac{47}{20H{\cos }^2\left(19°\right)\tan \left(19°\right)}+\frac{\tan \left(33°\right)}{\tan \left(19°\right)}\\ 3.1=\frac{7.634}{H}+1.886\\ 3.1=\frac{7.634+1.886H}{H}\\ 3.1H=7.634+1.886H\end{array}$

$\begin{array}{l}3.1H-1.886H=7.634\\ 1.214H=7.634\\ H=\frac{7.634}{1.214}\\ H=6.28\text{m}\end{array}$

Thus, the height of the soil-rock interface is $\underset{\_}{6.28\text{m}}$.