Chapter 7, Problem 7.1P

To determine

Find the average elastic settlement of the foundation.

Answer to Problem 7.1P

The average elastic settlement of the foundation is $\underset{¯}{\text{246}\text{mm}}$.

Explanation of Solution

Given information:

The width and length of the foundation is $\left(B\times L\right)=1.5\text{m}\times \text{3}\text{m}$

The depth of foundation $\left(D_f\right)$ is $\text{1}\text{.2}\text{m}$.

The depth of bottom of foundation to rigid layer H is $\text{3}\text{m}$.

The modulus of elasticity of clay $E_s$ is $\text{600}{\text{kN/m}}^{\text{2}}$

The net load per unit area of the foundation is $\left(q_o\right)$ is $\text{150}{\text{kN/m}}^{\text{2}}$.

Calculation:

Find the elastic settlement $S_e$ using the formula.

$S_e=A_1{A}_2\frac{q_{o}B}{E_s}$ (1)

Consider$A_1$ is relationship between depth of bottom of the foundation to a rigid layer H, length L and width of the foundation B.

$A_1=f\left(\frac{H}{B},\frac{L}{B}\right)$

$\begin{array}{c}\frac{H}{B}=\frac{3\text{m}}{1.5\text{m}}\\ =2\end{array}$

$\begin{array}{c}\frac{L}{B}=\frac{3\text{m}}{1.5\text{m}}\\ =2\end{array}$

Consider$A_2$ is relationship between depth of foundation and width of foundation.

$A_2=f\left(\frac{D_f}{B}\right)$

$\begin{array}{c}\left(\frac{D_f}{B}\right)=\frac{1.2\text{m}}{1.5\text{m}}\\ =0.8\end{array}$

Refer to figure 7.1 in the text book for finding $A_1$ and $A_2$ values.

Take $A_1$ value as 0.69 for corresponding value as $\frac{H}{B}=2$.

Take $A_2$ value as 0.95 for corresponding value as $\frac{D_f}{B}=0.8$.

Find the elastic settlement using Equation (1).

$\begin{array}{c}{S}_e=A_1{A}_2\frac{q_{o}B}{E_s}\\ =\frac{0.69\times 0.95\times 150{\text{kN/m}}^{\text{2}}\times 1.5}{600{\text{kN/m}}^{\text{2}}}\\ =0.246\text{m}\\ =0.246\text{m}\times \frac{1,000\text{mm}}{1\text{m}}\\ =246\text{mm}\end{array}$

Therefore, The average elastic settlement of the foundation is $\underset{¯}{\text{246}\text{mm}}$.