Chapter 9, Problem 9.4P

(a)

To determine

Find the change in effective stress ${\sigma }^{\prime }$ at point C when the water table drops by 2 m.

Answer to Problem 9.4P

The change in effective stress at point C when the water table drops by 2 m is $\underset{\_}{12.64{\text{kN/m}}^{\text{2}}}$.

Explanation of Solution

Given information:

The thickness $H_1$ of soil layer 1 is 5 m.

The thickness $H_2$ of soil layer 2 is 8 m.

The thickness $H_3$ of soil layer 3 is 3 m.

The void ratio (e) of soil in the first layer is 0.7.

The specific gravity $G_s$ of the soil in the first layer is 2.69.

The void ratio (e) of soil in the second layer is 0.55.

The depth (h) of water table drop is 2 m.

Calculation:

Determine the dry unit weight ${\gamma }_d$ of the soil in the first layer using the relation.

${\gamma }_{d\left(\text{layer}\text{1}\right)}=\frac{G_s{\gamma }_w}{1+e}$

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

Take the unit weight of the water as $9.81{\text{kN/m}}^{\text{3}}$.

Substitute 2.69 for $G_s$, $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, and 0.7 for e.

$\begin{array}{c}{\gamma }_{d\left(\text{layer}\text{1}\right)}=\frac{2.69\times 9.81}{1+0.7}\\ =15.52{\text{kN/m}}^{\text{3}}\end{array}$

Determine the dry unit weight ${\gamma }_{d\left(\text{layer}\text{2}\right)}$ of the soil in the second layer if the water table drops by 2 m using the relation.

${\gamma }_{d\left(\text{layer}\text{2}\right)}=\frac{G_s{\gamma }_w}{1+e}$

Substitute 2.7 for $G_s$, $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, and 0.55 for e.

$\begin{array}{c}{\gamma }_{d\left(\text{layer}\text{2}\right)}=\frac{2.70\times 9.81}{1+0.55}\\ =17.08{\text{kN/m}}^3\end{array}$

Determine the saturated unit weight ${\gamma }_{\text{sat}}$ of the soil in the second layer using the relation.

${\gamma }_{\text{sat}\left(\text{layer}2\right)}=\frac{{\gamma }_w\left(G_s+e\right)}{1+e}$

Substitute $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, 2.7 for $G_s$, and 0.55 for e.

$\begin{array}{c}{\gamma }_{\text{sat}\left(\text{layer}2\right)}=\frac{9.81\left(2.7+0.55\right)}{1+0.55}\\ =20.57{\text{kN/m}}^{\text{3}}\end{array}$

Calculate the total stress at point C (13 m) using the relation.

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

Substitute $15.52{\text{kN/m}}^{\text{3}}$ for ${\gamma }_d$, 5 m for $H_1$, $20.57{\text{kN/m}}^{\text{3}}$ for ${\gamma }_{\text{sat}}$, and 8.0 m for $H_2$.

$\begin{array}{c}\sigma =15.52\times 5.0+20.57\times 8.0\\ =242.16{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the pore water pressure at point C (13 m) using the relation.

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

Substitute $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$ and 8 m for $H_2$.

$\begin{array}{c}u=9.81\times 8\\ =78.48{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the effective stress at point C (13 m) using the relation.

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

Substitute $242.16{\text{kN/m}}^{\text{2}}$ for $\sigma$ and $78.48{\text{kN/m}}^{\text{2}}$ for u.

$\begin{array}{c}{\sigma }^{\prime }=242.16-78.48\\ =163.68{\text{kN/m}}^{\text{2}}\end{array}$

Water table drops by 2 m:

Calculate the total stress at point C when the water level drops by 2 m using the relation.

$\sigma ={\gamma }_d\times H_1+{\gamma }_{d\left(\text{layer2}\right)}\times h+{\gamma }_{\text{sat}}\times H_2$

Substitute $15.52{\text{kN/m}}^{\text{3}}$ for ${\gamma }_d$, 5 m for $H_1$, $17.08{\text{kN/m}}^3$ for ${\gamma }_{d\left(\text{layer2}\right)}$, 2 m for h, $20.57{\text{kN/m}}^{\text{3}}$ for ${\gamma }_{\text{sat}}$, and 6.0 m for $H_2$.

$\begin{array}{c}\sigma =15.52\times 5.0+17.08\times 2.0+20.57\times 6.0\\ =235.18{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the pore water pressure at point C when the water table drops by 2 m using the relation.

$u={\gamma }_w\times \left(H_2-h\right)$

Substitute $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, 8 m for $H_2$, and 2.0 m for h.

$\begin{array}{c}u=9.81\times \left(8-2\right)\\ =58.86{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the effective stress at point C when the water table drops by 2 m using the relation.

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

Substitute $235.18{\text{kN/m}}^{\text{2}}$ for $\sigma$ and $58.86{\text{kN/m}}^{\text{2}}$ for u.

$\begin{array}{c}{\sigma }^{\prime }=235.18-58.86\\ =176.32{\text{kN/m}}^{\text{2}}\end{array}$

Determine the change in effective stress when the water level drops by 2 m from the original position using the relation;

$\text{Change}\text{in}\text{effective}\text{stress}=\left(\begin{array}{l}\text{Effective}\text{stress}\text{at}\text{changed}\text{water}\text{level}\\ -\text{Effective}\text{stress}\text{at}\text{initial}\text{water}\text{level}\end{array}\right)$

Substitute $176.32{\text{kN/m}}^{\text{2}}$ for effective stress at changed water level and $163.68{\text{kN/m}}^{\text{2}}$ for effective stress at initial water level.

$\begin{array}{c}\text{Change}\text{in}\text{effective}\text{stress}=176.32-163.68\\ =12.64{\text{kN/m}}^{\text{2}}\end{array}$

Thus, the change in effective stress at point C when the water table drops by 2 m is $\underset{\_}{12.64{\text{kN/m}}^{\text{2}}}$.

(b)

To determine

The change in effective stress ${\sigma }^{\prime }$ at point C when the water rises to the surface up to point A.

Answer to Problem 9.4P

The change in effective stress at point C rises to the surface up to point A is $\underset{\_}{28.85{\text{kN/m}}^{\text{2}}}$.

Explanation of Solution

Given information:

The thickness $H_1$ of soil layer 1 is 5 m.

The thickness $H_2$ of soil layer 2 is 8 m.

The thickness $H_3$ of soil layer 3 is 3 m.

The void ratio (e) of soil in the first layer is 0.7.

The specific gravity $G_s$ of the soil in the first layer is 2.69.

The void ratio (e) of soil in the second layer is 0.55.

Calculation:

Determine the saturated unit weight ${\gamma }_{\text{sat}}$ of the soil in the first layer using the relation.

${\gamma }_{\text{sat}\left(\text{layer}\text{1}\right)}=\frac{{\gamma }_w\left(G_s+e\right)}{1+e}$

Substitute $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, 2.69 for $G_s$, and 0.7 for e.

$\begin{array}{c}{\gamma }_{\text{sat}\left(\text{layer}\text{1}\right)}=\frac{9.81\left(2.69+0.7\right)}{1+0.7}\\ =19.56{\text{kN/m}}^{\text{3}}\end{array}$

Calculate the total stress at point C (13 m) using the relation.

$\sigma ={\gamma }_{\text{sat}\left(\text{layer}\text{1}\right)}\times H_1+{\gamma }_{\text{sat}\left(\text{layer}2\right)}\times H_2$

Substitute $19.56{\text{kN/m}}^{\text{3}}$ for ${\gamma }_{\text{sat}\left(\text{layer}\text{1}\right)}$, 5 m for $H_1$, $20.57{\text{kN/m}}^{\text{3}}$ for ${\gamma }_{\text{sat}\left(\text{layer}2\right)}$, and 8.0 m for $H_2$.

$\begin{array}{c}\sigma =19.56\times 5.0+20.57\times 8.0\\ =262.36{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the pore water pressure at point C (13 m) using the relation.

$u={\gamma }_w\times \left(H_1+H_2\right)$

Substitute $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, 5 m for $H_1$, and 8 m for $H_2$.

$\begin{array}{c}u=9.81\times \left(5+8\right)\\ =127.53{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the effective stress at point C (13 m) using the relation.

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

Substitute $262.36{\text{kN/m}}^{\text{2}}$ for $\sigma$ and $127.53{\text{kN/m}}^{\text{2}}$ for u.

$\begin{array}{c}{\sigma }^{\prime }=262.36-127.53\\ =134.83{\text{kN/m}}^{\text{2}}\end{array}$

Water table rises to the surface up to point A:

Determine the change in effective stress when the water table rises to the surface up to point A using the relation;

$\text{Change}\text{in}\text{effective}\text{stress}=\left(\begin{array}{l}\text{Effective}\text{stress}\text{at}\text{initial}\text{water}\text{level}\\ -\text{Effective}\text{stress}\text{at}\text{point}C\end{array}\right)$

Substitute $163.68{\text{kN/m}}^{\text{2}}$ for effective stress at initial water level and $134.83{\text{kN/m}}^{\text{2}}$ for effective stress at point C.

$\begin{array}{c}\text{Change}\text{in}\text{effective}\text{stress}=163.68-134.83\\ =28.85{\text{kN/m}}^{\text{2}}\end{array}$

Thus, the change in effective stress at point C rises to the surface up to point A is $\underset{\_}{28.85{\text{kN/m}}^{\text{2}}}$.

(c)

To determine

Find the change in effective stress ${\sigma }^{\prime }$ at point C when the water level rises 3 m above point A due to flooding.

Answer to Problem 9.4P

The change in effective stress at point C when the water level rises 3 m above point A due to flooding is $\underset{\_}{28.85{\text{kN/m}}^{\text{2}}}$.

Explanation of Solution

Given information:

The thickness $H_1$ of soil layer 1 is 5 m.

The thickness $H_2$ of soil layer 2 is 8 m.

The thickness $H_3$ of soil layer 3 is 3 m.

The void ratio (e) of soil in the first layer is 0.7.

The specific gravity $G_s$ of the soil in the first layer is 2.69.

The depth (h) of water rises above point A is 3.0 m.

Calculation:

Calculate the total stress at point C (16 m) using the relation.

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

Substitute $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, 3.0 m for h, $19.56{\text{kN/m}}^{\text{3}}$ for ${\gamma }_{\text{sat}}$, 5 m for $H_1$, $20.57{\text{kN/m}}^{\text{3}}$ for ${\gamma }_{\text{sat}}$, and 8.0 m for $H_2$.

$\begin{array}{c}\sigma =9.81\times 3.0+19.56\times 5.0+20.57\times 8.0\\ =291.79{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the pore water pressure at point C (16 m) using the relation.

$u={\gamma }_w\times \left(h+H_1+H_2\right)$

Substitute $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, 3 m for h, 5 m for $H_1$, and 8 m for $H_2$.

$\begin{array}{c}u=9.81\times \left(3+5+8\right)\\ =156.96{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the effective stress at point C (16 m) using the relation.

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

Substitute $291.79{\text{kN/m}}^{\text{2}}$ for $\sigma$ and $156.96{\text{kN/m}}^{\text{2}}$ for u.

$\begin{array}{c}{\sigma }^{\prime }=291.79-156.96\\ =134.83{\text{kN/m}}^{\text{2}}\end{array}$

Water level rises 3 m above point A due to flooding:

Determine the change in effective stress when the water level rises 3 m above point A due to flooding using the relation;

$\text{Change}\text{in}\text{effective}\text{stress}=\left(\begin{array}{l}\text{Effective}\text{stress}\text{at}\text{initial}\text{water}\text{level}\\ -\text{Effective}\text{stress}\text{at}\text{point}C\end{array}\right)$

Substitute $163.68{\text{kN/m}}^{\text{2}}$ for effective stress at initial water level and $134.83{\text{kN/m}}^{\text{2}}$ for effective stress at point C.

$\begin{array}{c}\text{Change}\text{in}\text{effective}\text{stress}=163.68-134.83\\ =28.85{\text{kN/m}}^{\text{2}}\end{array}$

Thus, the change in effective stress at point C when the water level rises 3 m above point A due to flooding is $\underset{\_}{28.85{\text{kN/m}}^{\text{2}}}$.