Chapter 8, Problem 8.6P

To determine

Calculate the consolidation settlement under the center of mat foundation.

 </TO-DETERMINE>

The consolidation settlement under the center of mat foundation is $\underset{¯}{125\text{mm}}$.

Given information:

The length of the foundation is $L=12\text{m}$.

The width of the foundation is $B=10\text{m}$.

The depth of the foundation $\left(D_f\right)$ is $2.2\text{m}$

The load value Q is $30\text{MN}$.

The preconsolidation pressure $\left({{\sigma }^{\prime }}_c\right)$ is $105{\text{kN/m}}^{\text{2}}$.

The depth values are $x_1=2\text{m}$, $x_2=2\text{m}$ and $x_3=5.2\text{m}$

Calculation:

Find the consolidation settlement at the center of the foundation using the relation.

$S_{c(p)}=\frac{C_c{H}_c}{1+e_o}\log \left(\frac{{{\sigma }^{\prime }}_o+\Delta {{\sigma }^{\prime }}_{av}}{{{\sigma }^{\prime }}_o}\right)$

Find the stress ${{\sigma }^{\prime }}_o$

$\begin{array}{c}{{\sigma }^{\prime }}_o=\left(4.2\text{m}\times \text{16}{\text{kN/m}}^3\right)+2\text{m}\times \left(18-9.81\right)+\frac{5.2\text{m}}{2}\times \left(17.5-9.81\right)\\ =67.2+16.38+19.99\\ =103.57{\text{kN/m}}^{\text{2}}\end{array}$

Height of clay $H_c=5.2\text{m}$

$\begin{array}{l}{C}_c=0.38\\ e_o=0.88\end{array}$

Find the net load per unit area using the relation.

$\begin{array}{c}q=\frac{Q}{A}-\gamma D_f\\ =\frac{\left(30\text{MN}\times \frac{\text{1000}\text{kN}}{1\text{MN}}\right)}{10\times 12}-\left(16\times 2.2\right)\\ =214.8{\text{kN/m}}^{\text{2}}\end{array}$

Find the average stress increase in the clay layer below the corner of each rectangular area.

$\begin{array}{c}\Delta {{\sigma }^{\prime }}_{\text{av}\left(H_2/H_1\right)}=q_o\left[\frac{H_2{I}_a{}_{(H_2)}-H_1{I}_a{}_{(H_1)}}{H_2-H_1}\right]\\ =214.8\left[\frac{\left(2+2+5.2\right)I_a{}_{(H_2)}-\left(2+2\right)I_a{}_{(H_1)}}{5.2}\right]\end{array}$

For $I_a{}_{(H_2)}$,

$\begin{array}{c}{m}_2=\frac{B}{H_2}\\ =\frac{5}{2+2+5.2}\\ =0.543\end{array}$

$\begin{array}{c}{n}_2=\frac{L}{H_2}\\ =\frac{6}{2+2+5.2}\\ =0.652\end{array}$

Refer to Figure 6.11, “Influence factor $I_a$” in the textbook.

Take the influence factor $I_a{}_{(H_2)}$ as 0.186 for the value of $m_2=0.543$ and $n_2=0.652$.

For $I_a{}_{(H_1)}$,

$\begin{array}{c}{m}_2=\frac{B}{H_1}\\ =\frac{5}{2+2}\\ =1.25\end{array}$

$\begin{array}{c}{n}_2=\frac{L}{H_2}\\ =\frac{6}{2+2}\\ =1.5\end{array}$

Refer to Figure 6.11, “Influence factor $I_a$” in the textbook.

Take the influence factor $I_a{}_{(H_1)}$ as 0.23 for the value of $m_2=1.25$ and $n_2=1.5$.

$\begin{array}{c}\Delta {{\sigma }^{\prime }}_{\text{av}\left(H_2/H_1\right)}=q_o\left[\frac{H_2{I}_a{}_{(H_2)}-H_1{I}_a{}_{(H_1)}}{H_2-H_1}\right]\\ =214.8\left[\frac{\left(2+2+5.2\right)\left(0.186\right)-\left(2+2\right)\left(0.23\right)}{5.2}\right]\\ =32.68{\text{kN/m}}^{\text{2}}\end{array}$

Find the consolidation settlement at the center of the foundation using the relation.

$\begin{array}{c}{S}_{c(p)}=\frac{C_c{H}_c}{1+e_o}\log \left(\frac{{{\sigma }^{\prime }}_o+\Delta {{\sigma }^{\prime }}_{av}}{{\sigma }^{\prime }}\right)\\ =\frac{0.38\times 5.2}{1+0.88}\log \left(\frac{103.57+32.68}{103.57}\right)\\ =0.125\text{m}\\ \text{=125}\text{mm}\end{array}$

Therefore, the consolidation settlement under the center of mat foundation is $\underset{¯}{125\text{mm}}$.