Chapter 13, Problem 13.18P

To determine

Find the Rankine active force $P_a$ per unit length of the wall and the location $\overline{z}$ of the resultant force.

Answer to Problem 13.18P

The Rankine active force $P_a$ per unit length of the wall is $\underset{\_}{287\text{kN/m}}$.

The location $\overline{z}$ of the resultant force is $\underset{\_}{2.49\text{m}}$.

Explanation of Solution

Given information:

The height (H) of the retaining wall is 8.0 m.

The depth $H_1$ of sand is 3.0 m.

The unit weight ${\gamma }_1$ of the sand is $13{\text{kN/m}}^{\text{3}}$.

The sand friction angle ${{\varphi }^{\prime }}_1$ is $30°$.

The saturated sand friction angle ${{\varphi }^{\prime }}_2$ is $30°$.

The cohesion ${c^{\prime }}_1$  of sand is 0.

The surcharge pressure (q) is $16{\text{kN/m}}^{\text{2}}$.

The depth $H_2$ of saturated sand is 4.0 ft.

The saturated unit weight ${\gamma }_2$ of the sand is $18.8{\text{kN/m}}^{\text{3}}$.

The cohesion ${c^{\prime }}_2$ of saturated sand is 0.

Calculation:

Determine the active earth pressure coefficient $K_a$ using the formula.

$K_a={\tan }^2\left(45-\frac{{\varphi }^{\prime }}{2}\right)$

Substitute $30°$ for ${\varphi }^{\prime }$.

$\begin{array}{c}{K}_a={\tan }^2\left(45-\frac{30°}{2}\right)\\ ={\tan }^2\left(45-15\right)\\ ={\tan }^2\left(30\right)\\ =0.333\end{array}$

Determine the total stress ${{\sigma }^{\prime }}_o$ at 0 m depth using the relation.

${{\sigma }^{\prime }}_o=q$

Substitute $16{\text{kN/m}}^{\text{2}}$ for q.

${{\sigma }^{\prime }}_o=16{\text{kN/m}}^{\text{2}}$

Determine the pore water pressure at 0 m depth using the relation.

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

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 $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$ and 0 m for h.

$\begin{array}{c}u=9.81\times 0\\ =0\end{array}$

Determine the effective active earth pressure ${{\sigma }^{\prime }}_a$ at 0 m depth using the relation.

${{\sigma }^{\prime }}_a={{\sigma }^{\prime }}_o{K}_a$

Substitute $16{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_o$ and 0.39 for $K_a$.

$\begin{array}{c}{{\sigma }^{\prime }}_a=16\left(0.333\right)\\ =5.32{\text{kN/m}}^{\text{2}}\end{array}$

Determine the total stress ${{\sigma }^{\prime }}_o$ at 3 m depth using the relation.

${{\sigma }^{\prime }}_o=q+{\gamma }_1\times H_1$

Substitute $16{\text{kN/m}}^{\text{2}}$ for q, $13{\text{kN/m}}^{\text{3}}$ ${\gamma }_1$, and 3.0 m for $H_1$.

$\begin{array}{c}{{\sigma }^{\prime }}_o=16+13\times 3.0\\ =55{\text{kN/m}}^2\end{array}$

Determine the pore water pressure at 3.0 m depth using the relation.

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

Substitute $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$ and 0 m for h.

$\begin{array}{c}u=9.81\times 0\\ =0\end{array}$

Determine the effective active earth pressure ${{\sigma }^{\prime }}_a$ at 3.0 m depth using the relation.

${{\sigma }^{\prime }}_a={{\sigma }^{\prime }}_o{K}_a$

Substitute $55{\text{kN/m}}^2$ for ${{\sigma }^{\prime }}_o$ and 0.333 for $K_a$.

$\begin{array}{c}{{\sigma }^{\prime }}_a=55\left(0.333\right)\\ =18.31{\text{kN/m}}^2\end{array}$

Determine the total stress ${{\sigma }^{\prime }}_o$ at 8 m depth using the relation.

${{\sigma }^{\prime }}_o=q+{\gamma }_1\times H_1+\left({\gamma }_2-{\gamma }_w\right)\times H_2$

Substitute $16{\text{kN/m}}^{\text{2}}$ for q, $13{\text{kN/m}}^{\text{3}}$ ${\gamma }_1$, 3.0 m for $H_1$, $18.8{\text{kN/m}}^{\text{3}}$ ${\gamma }_2$, $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, and 5.0 m for $H_2$.

$\begin{array}{c}{{\sigma }^{\prime }}_o=16+13\times 3+\left(18.8-9.81\right)\times 5\\ =99.95{\text{kN/m}}^{\text{2}}\end{array}$

Determine the pore water pressure at 8 m depth using the relation.

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

Substitute $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$ and 5.0 m for h.

$\begin{array}{c}u=9.81\times 5\\ =49.05{\text{kN/m}}^{\text{2}}\end{array}$

Determine the effective active earth pressure ${{\sigma }^{\prime }}_a$ at 8 m depth using the relation.

${{\sigma }^{\prime }}_a={{\sigma }^{\prime }}_o{K}_a$

Substitute $99.95{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_o$ and 0.333 for $K_a$.

$\begin{array}{c}{{\sigma }^{\prime }}_a=99.95\left(0.333\right)\\ =33.28{\text{kN/m}}^{\text{2}}\end{array}$

Show the variation of effective active earth pressure and pore water pressure for the respective depth as in Figure 1.

Refer Figure 1.

Determine the active earth pressure per unit length for area 1 using the relation.

$A_1=bh$

Here, b is the width and h is the depth.

Substitute $5.32{\text{kN/m}}^{\text{2}}$ for b and 8.0 m for h.

$\begin{array}{c}{A}_1=\left(5.32\times 8.0\right)\\ =42.56\text{kN/m}\end{array}$

Determine the active earth pressure per unit length for area 2 using the relation.

$A_2=\frac{1}{2}bh$

Substitute $\left(18.31-5.32\right){\text{kN/m}}^{\text{2}}$ for b and 3.0 m for h.

$\begin{array}{c}{A}_2=\frac{1}{2}\times \left(18.31-5.32\right)\times 3.0\\ =0.5\times 38.97\\ =19.48\text{kN/m}\end{array}$

Determine the active earth pressure per unit length for area 3 using the relation.

$A_3=bh$

Substitute 5.0 m for b and $\left(18.31-5.32\right){\text{kN/m}}^{\text{2}}$ for h.

$\begin{array}{c}{A}_3=5\times \left(18.31-5.32\right)\\ =64.95\text{kN/m}\end{array}$

Determine the active earth pressure per unit length for area 4 using the relation.

$A_4=\frac{1}{2}bh$

Substitute 5.0 m for b and $\left(33.28-18.31\right){\text{kN/m}}^{\text{2}}$ for h.

$\begin{array}{c}{A}_4=\frac{1}{2}\times 5.0\times \left(33.28-18.31\right)\\ =0.5\times 74.85\\ =37.42\text{kN/m}\end{array}$

Determine the active earth pressure per unit length for area 5 using the relation.

$A_5=\frac{1}{2}bh$

Substitute 5.0 m for b and $49.05{\text{kN/m}}^{\text{2}}$ for h.

$\begin{array}{c}{A}_4=\frac{1}{2}\times 5.0\times 49.05\\ =122.62\text{kN/m}\end{array}$

Determine the Rankine active force $P_a$ per unit length of the wall using the relation.

$P_a=A_1+A_2+A_3+A_4$

Substitute $42.56\text{kN/m}$ for $A_1$, $19.48\text{kN/m}$ for $A_2$, $64.95\text{kN/m}$ for $A_3$, $37.42\text{kN/m}$ for $A_4$, and $122.62\text{kN/m}$ for $A_5$.

$\begin{array}{c}{P}_a=42.56+19.48+64.95+37.42+122.62\\ =287\text{kN/m}\end{array}$

Thus, the Rankine active force $P_a$ per unit length of the wall is $\underset{\_}{287\text{kN/m}}$.

Determine the location $\overline{z}$ of the resultant force by taking the moment about the bottom of the wall.

$\overline{z}=\frac{A_1\left(\frac{H_2}{2}\right)+A_2\left(H_2+\frac{H_1}{3}\right)+A_3\left(\frac{H_2}{2}\right)+A_4\left(\frac{H_2}{3}\right)+A_5\left(\frac{H_2}{3}\right)}{P_a}$

Substitute $42.56\text{kN/m}$ for $A_1$, 8.0 m ft for H, 5.0 m for $H_2$, 3.0 m for $H_1$, $19.48\text{kN/m}$ for $A_2$, $64.95\text{kN/m}$ for $A_3$, $37.42\text{kN/m}$ for $A_4$, $122.62\text{kN/m}$ for $A_5$, and $287\text{kN/m}$ for $P_a$.

$\begin{array}{c}\overline{z}=\frac{42.56\left(\frac{8}{2}\right)+19.48\left(5+\frac{3}{3}\right)+64.98\left(\frac{5}{2}\right)+37.42\left(\frac{5}{3}\right)+122.62\left(\frac{5}{3}\right)}{287}\\ =\frac{170.24+116.88+162.45+62.37+204.37}{287}\\ =2.49\text{m}\end{array}$

Thus, the location of the resultant force is $\underset{\_}{2.49\text{m}}$.