Chapter 9, Problem 9.1P

To determine

Find the total stress $\sigma$ , pore water pressure $u$ , and effective stress ${\sigma }^{\prime }$ at points A, B, C, and D and plot the variations of total stress, pore water pressure, and effective stress.

Answer to Problem 9.1P

The total stress at point A is $\underset{\_}{\text{zero}}$.

The pore water pressure at point A is $\underset{\_}{\text{zero}}$.

The effective stress at point A is $\underset{\_}{\text{zero}}$.

The total stress at point B is $\underset{\_}{770{\text{lb/ft}}^{\text{2}}}$.

The pore water pressure at point B is $\underset{\_}{\text{zero}}$.

The effective stress at point B is $\underset{\_}{770{\text{lb/ft}}^{\text{2}}}$.

The total stress at point C is $\underset{\_}{2,222{\text{lb/ft}}^{\text{2}}}$.

The pore water pressure at point C is $\underset{\_}{748.8{\text{lb/ft}}^{\text{2}}}$.

The effective stress at point C is $\underset{\_}{1,473.2{\text{lb/ft}}^{\text{2}}}$.

The total stress at point D is $\underset{\_}{2,930{\text{lb/ft}}^{\text{2}}}$.

The pore water pressure at point D is $\underset{\_}{1,123.2{\text{lb/ft}}^{\text{2}}}$.

The effective stress at point D is $\underset{\_}{1,806.8{\text{lb/ft}}^{\text{2}}}$.

Explanation of Solution

Given information:

The thickness $H_1$ of soil layer 1 is 7 ft.

The thickness $H_2$ of soil layer 2 is 12 ft.

The thickness $H_3$ of soil layer 3 is 6 ft.

The dry unit weight ${\gamma }_d$ of soil in the first layer is $110{\text{lb/ft}}^{\text{3}}$.

The saturated unit weight ${\gamma }_{\text{sat}}$ of soil in the second layer is $121{\text{lb/ft}}^{\text{3}}$.

The saturated unit weight ${\gamma }_{\text{sat}}$ of soil in the third layer is $118{\text{lb/ft}}^{\text{3}}$.

Calculation:

Calculate the total stress at point A (0 ft).

$\sigma =0$

Thus, the total stress at point A is $\underset{\_}{\text{zero}}$.

Calculate the pore water pressure at point A (0 ft).

$u=0$

Thus, pore water pressure at point A is $\underset{\_}{\text{zero}}$.

Calculate the effective stress at point A (0 ft) using the relation.

${\sigma }^{\prime }=\sigma -u$

Substitute 0 for $\sigma$ and 0 for u.

$\begin{array}{c}{\sigma }^{\prime }=0-0\\ =0\end{array}$

Thus, effective stress at point A is $\underset{\_}{\text{zero}}$.

Calculate the total stress at point B (7 ft) using the relation.

$\sigma ={\gamma }_d\times H_1$

Substitute $110{\text{lb/ft}}^{\text{3}}$ for ${\gamma }_d$ and 7 ft for $H_1$.

$\begin{array}{c}\sigma =110\times 7.0\\ =770{\text{lb/ft}}^{\text{2}}\end{array}$

Thus, total stress at point B is $\underset{\_}{770{\text{lb/ft}}^{\text{2}}}$.

Calculate the pore water at point B (7 ft) using the relation.

$u=0$

Thus, the pore water pressure at point B is $\underset{\_}{\text{zero}}$.

Calculate the effective stress at point B (7 ft) using the relation.

${\sigma }^{\prime }=\sigma -u$

Substitute $770{\text{lb/ft}}^{\text{2}}$ for $\sigma$ and 0 for u.

$\begin{array}{c}{\sigma }^{\prime }=770-0\\ =770{\text{lb/ft}}^{\text{2}}\end{array}$

Thus, the effective stress at point B is $\underset{\_}{770{\text{lb/ft}}^{\text{2}}}$.

Calculate the total stress at point C (19 ft) using the relation.

$\sigma ={\gamma }_d\times H_1+{\gamma }_{\text{sat}}\times H_2$

Substitute $110{\text{lb/ft}}^{\text{3}}$ for ${\gamma }_d$, 7 ft for $H_1$, $121{\text{lb/ft}}^{\text{3}}$ for ${\gamma }_{\text{sat}}$, and 12 ft for $H_2$.

$\begin{array}{c}\sigma =110\times 7.0+121\times 12\\ =2,222{\text{lb/ft}}^{\text{2}}\end{array}$

Thus, the total stress at point C is $\underset{\_}{2,222{\text{lb/ft}}^{\text{2}}}$.

Calculate the pore water pressure at point C (19 ft) using the relation.

$u={\gamma }_w\times H_2$

Here, ${\gamma }_w$ is the unit weight of the water.

Take the unit weight of the water as $62.4{\text{lb/ft}}^{\text{3}}$.

Substitute $62.4{\text{lb/ft}}^{\text{3}}$ for ${\gamma }_w$ and 12 ft for $H_2$.

$\begin{array}{c}u=62.4\times 12\\ =748.8{\text{lb/ft}}^{\text{2}}\end{array}$

Thus, the pore water pressure at point C is $\underset{\_}{748.8{\text{lb/ft}}^{\text{2}}}$.

Calculate the effective stress at point C (19 ft) using the relation.

${\sigma }^{\prime }=\sigma -u$

Substitute $2,222{\text{lb/ft}}^{\text{2}}$ for $\sigma$ and $748.8{\text{lb/ft}}^{\text{2}}$ for u.

$\begin{array}{c}{\sigma }^{\prime }=2,222-748.8\\ =1,473.2{\text{lb/ft}}^{\text{2}}\end{array}$

Thus, the effective stress at point C is $\underset{\_}{1,473.2{\text{lb/ft}}^{\text{2}}}$.

Calculate the total stress at point D (25 ft) using the relation.

$\sigma ={\gamma }_d\times H_1+{\gamma }_{\text{sat}}\times H_2+{\gamma }_{\text{sat}}\times H_3$

Substitute $110{\text{lb/ft}}^{\text{3}}$ for ${\gamma }_d$, 7 ft for $H_1$, $121{\text{lb/ft}}^{\text{3}}$ for ${\gamma }_{\text{sat}}$, 12 ft for $H_2$, $118{\text{lb/ft}}^{\text{3}}$ for ${\gamma }_{\text{sat}}$, and 6 ft for $H_3$.

$\begin{array}{c}\sigma =110\times 7.0+121\times 12+118\times 6\\ =2,930{\text{lb/ft}}^{\text{2}}\end{array}$

Thus, the total stress at point D is $\underset{\_}{2,930{\text{lb/ft}}^{\text{2}}}$.

Calculate the pore water pressure at point D (25 ft) using the relation.

$u={\gamma }_w\times \left(H_2+H_3\right)$

Substitute $62.4{\text{lb/ft}}^{\text{3}}$ for ${\gamma }_w$, 12 ft for $H_2$, and 6 ft for $H_3$.

$\begin{array}{c}u=62.4\times \left(12+6\right)\\ =1,123.2{\text{lb/ft}}^{\text{2}}\end{array}$

Thus, the pore water pressure at point D is $\underset{\_}{1,123.2{\text{lb/ft}}^{\text{2}}}$.

Calculate the effective stress at point D (25 ft) using the relation.

${\sigma }^{\prime }=\sigma -u$

Substitute $2,930{\text{lb/ft}}^{\text{2}}$ for $\sigma$ and $1,123.2{\text{lb/ft}}^{\text{2}}$ for u.

$\begin{array}{c}{\sigma }^{\prime }=2,930-1,123.2\\ =1,806.8{\text{lb/ft}}^{\text{2}}\end{array}$

Thus, the effective stress at point D is $\underset{\_}{1,806.8{\text{lb/ft}}^{\text{2}}}$.

Show the plot between depth and total stress as in Figure 1.

Show the plot between depth and pore water pressure as in Figure 2.

Show the plot between depth and effective stress as in Figure 3.