Chapter 13, Problem 13.19P

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 13.19P

The Rankine active force $P_a$ per unit length of the wall is $\underset{\_}{587.34\text{kN/m}}$.

The location $\overline{z}$ of the resultant force is $\underset{\_}{3.72\text{m}}$.

Explanation of Solution

Given information:

The height (H) of the retaining wall is 12 m.

The depth $H_1$ of sand is 4.0 m.

The unit weight ${\gamma }_1$ of the sand is $17{\text{kN/m}}^{\text{3}}$.

The sand friction angle ${{\varphi }^{\prime }}_1$ is $36°$.

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

The surcharge pressure (q) is $25{\text{kN/m}}^{\text{2}}$.

The depth $H_2$ of saturated sand is 12 m.

The saturated unit weight ${\gamma }_2$ of the sand is $23.2{\text{kN/m}}^{\text{3}}$.

The saturated sand friction angle ${{\varphi }^{\prime }}_2$ is $42°$.

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

Calculation:

For sand:

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

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

Substitute $36°$ for ${\varphi }^{\prime }$.

$\begin{array}{c}{K}_a={\tan }^2\left(45-\frac{36°}{2}\right)\\ ={\tan }^2\left(45-18\right)\\ ={\tan }^2\left(27\right)\\ =0.259\end{array}$

For saturated sand:

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

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

Substitute $42°$ for ${\varphi }^{\prime }$.

$\begin{array}{c}{K}_a={\tan }^2\left(45-\frac{42°}{2}\right)\\ ={\tan }^2\left(45-21\right)\\ ={\tan }^2\left(24\right)\\ =0.198\end{array}$

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

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

Substitute $25{\text{kN/m}}^{\text{2}}$ for q.

${{\sigma }^{\prime }}_o=25{\text{kN/m}}^{\text{2}}$

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 $25{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_o$ and 0.259 for $K_a$.

$\begin{array}{c}{{\sigma }^{\prime }}_a=25\left(0.259\right)\\ =6.47{\text{kN/m}}^{\text{2}}\end{array}$

Determine the total stress (sand) ${{\sigma }^{\prime }}_o$ at 4 m depth using the relation.

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

Substitute $25{\text{kN/m}}^{\text{2}}$ for q, $17{\text{kN/m}}^{\text{3}}$ ${\gamma }_1$, and 4.0 m for $H_1$.

$\begin{array}{c}{{\sigma }^{\prime }}_o=25+17\times 4.0\\ =93{\text{kN/m}}^2\end{array}$

Determine the total stress (saturated sand) ${{\sigma }^{\prime }}_o$ at 4 m depth using the relation.

${{\sigma }^{\prime }}_o=q+{\gamma }_2\times H_2$

Substitute $25{\text{kN/m}}^{\text{2}}$ for q, $17{\text{kN/m}}^{\text{3}}$ ${\gamma }_2$, and 4.0 m for $H_2$.

$\begin{array}{c}{{\sigma }^{\prime }}_o=25+17\times 4.0\\ =93{\text{kN/m}}^2\end{array}$

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 (sand) ${{\sigma }^{\prime }}_a$ at 3.0 m depth using the relation.

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

Substitute $93{\text{kN/m}}^2$ for ${{\sigma }^{\prime }}_o$ and 0.259 for $K_a$.

$\begin{array}{c}{{\sigma }^{\prime }}_a=93\left(0.259\right)\\ =24.08{\text{kN/m}}^2\end{array}$

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

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

Substitute $93{\text{kN/m}}^2$ for ${{\sigma }^{\prime }}_o$ and 0.198 for $K_a$.

$\begin{array}{c}{{\sigma }^{\prime }}_a=93\left(0.198\right)\\ =18.41{\text{kN/m}}^2\end{array}$

Determine the total stress ${{\sigma }^{\prime }}_o$ at 12 m depth using the relation.

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

Substitute $17{\text{kN/m}}^{\text{3}}$ ${\gamma }_1$, 4.0 m for $H_1$, $23.2{\text{kN/m}}^{\text{3}}$ ${\gamma }_2$, $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, and 8.0 m for $H_2$.

$\begin{array}{c}{{\sigma }^{\prime }}_o=17\times 4+\left(23.2-9.81\right)\times 8\\ =175.12{\text{kN/m}}^{\text{2}}\end{array}$

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 8.0 m for h.

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

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 $175.12{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_o$ and 0.198 for $K_a$.

$\begin{array}{c}{{\sigma }^{\prime }}_a=175.12\left(0.198\right)\\ =34.67{\text{kN/m}}^{\text{2}}\end{array}$

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 $6.47{\text{kN/m}}^{\text{2}}$ for b and 12 m for h.

$\begin{array}{c}{A}_1=\left(6.47\times 12\right)\\ =77.64\text{kN/m}\end{array}$

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

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

Substitute $\left(24.08-6.47\right){\text{kN/m}}^{\text{2}}$ for b and 4.0 m for h.

$\begin{array}{c}{A}_2=\frac{1}{2}\times \left(24.08-6.47\right)\times 4.0\\ =0.5\times 70.44\\ =35.22\text{kN/m}\end{array}$

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

$A_3=bh$

Substitute 8.0 m for b and $\left(18.41-6.47\right){\text{kN/m}}^{\text{2}}$ for h.

$\begin{array}{c}{A}_3=8\left(18.41-6.47\right)\\ =95.52\text{kN/m}\end{array}$

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

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

Substitute 8.0 m for b and $\left(34.67-18.41\right){\text{kN/m}}^{\text{2}}$ for h.

$\begin{array}{c}{A}_4=\frac{1}{2}\times 8.0\times \left(34.67-18.41\right)\\ =0.5\times 130.08\\ =65.04\text{kN/m}\end{array}$

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

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

Substitute 8.0 m for b and $78.48{\text{kN/m}}^{\text{2}}$ for h.

$\begin{array}{c}{A}_4=\frac{1}{2}\times 8.0\times 78.48\\ =313.92\text{kN/m}\end{array}$

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 $77.64\text{kN/m}$ for $A_1$, $35.22\text{kN/m}$ for $A_2$, $95.52\text{kN/m}$ for $A_3$, $65.04\text{kN/m}$ for $A_4$, and $313.92\text{kN/m}$ for $A_5$.

$\begin{array}{c}{P}_a=77.64+35.22+95.52+65.04+313.92\\ =587.34\text{kN/m}\end{array}$

Thus, the Rankine active force $P_a$ per unit length of the wall is $\underset{\_}{587.34\text{kN/m}}$.

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

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

Substitute $77.64\text{kN/m}$ for $A_1$, 12 m for H, 8.0 m for $H_2$, 4.0 m for $H_1$, $35.22\text{kN/m}$ for $A_2$, $95.52\text{kN/m}$ for $A_3$, $65.04\text{kN/m}$ for $A_4$, $313.92\text{kN/m}$ for $A_5$, and $587.34\text{kN/m}$ for $P_a$.

$\begin{array}{c}\overline{z}=\frac{77.64\left(\frac{12}{2}\right)+35.22\left(8+\frac{4}{3}\right)+95.52\left(\frac{8}{2}\right)+65.04\left(\frac{8}{3}\right)+313.92\left(\frac{8}{3}\right)}{587.34}\\ =\frac{456.84+328.72+382.08+173.44+837.12}{587.34}\\ =3.72\text{m}\end{array}$

Thus, the location of the resultant force is $\underset{\_}{3.72\text{m}}$.