Chapter 16, Problem 16.1P

To determine

Find the magnitude and location of the thrust on the wall.

Answer to Problem 16.1P

The magnitude of the thrust on the wall is $\underset{\_}{7,348.5\text{lb/ft}}$.

The location of the thrust on the wall is $\underset{\_}{5\text{ft}\text{above the bottom of the wall}}$.

Explanation of Solution

Given information:

The height of the wall (H) is 15 ft.

The overconsolidated ratio (OCR) is 1.5.

The cohesion $\left(c^{\prime }\right)$ is 0.

The angle of internal friction $\left({\varphi }^{\prime }\right)$ is $33°$.

The unit weight of the soil $\left(\gamma \right)$ is $\text{115}{\text{lb/ft}}^{\text{3}}$.

Calculation:

Show the pressure distribution of the vertical wall as in Figure (1).

Find the coefficient of earth pressure at rest $\left(K_o\right)$ using the relation:

$\begin{array}{c}{K}_o=\left(1-\sin {\varphi }^{\prime }\right){\text{OCR}}^{\sin {\varphi }^{\prime }}\\ =\left(1-\sin 33°\right)\text{1}{\text{.5}}^{\sin 33°}\\ =0.568\end{array}$

Find the magnitude of the thrust on the wall $\left(P_o\right)$ using the relation:

$\begin{array}{c}{P}_o=0.5K_o\gamma H^2\\ =0.5\times 0.568\times 115\times {15}^2\\ =7,348.5\text{lb/ft}\end{array}$

Hence, the magnitude of the thrust on the wall is $\underset{\_}{7,348.5\text{lb/ft}}$.

The resultant is acting at $\frac{H}{3}$ of the wall.

Determine the location of the thrust on the wall using the relation:

$\begin{array}{c}\text{Location of the thrust on the wall}=\frac{H}{3}\\ =\frac{15}{3}\\ =5\text{ft}\end{array}$

Thus, the location of the thrust on the wall is $\underset{\_}{5\text{ft}\text{above the bottom of the wall}}$.