Chapter 14, Problem 14.8P

(a)

To determine

Find the Rankine’s active force per unit length of the wall.

Find the location of the resultant.

Answer to Problem 14.8P

The Rankine’s active force per unit length of the wall is $\underset{\_}{34.31\text{kN/m}}$.

The location of the resultant is $\underset{\_}{0.89\text{m from bottom}}$.

Explanation of Solution

Given information:

The height (H) of the retaining wall is 3.05 m.

The depth $\left(H_1\right)$ of water table is 1.52 m.

The unit weight $\left({\gamma }_1\right)$ of the soil is $16.51{\text{kN/m}}^3$.

The saturated unit weight $\left({\gamma }_2\right)$ of the soil is $19.18{\text{kN/m}}^3$.

The angle $\left({{\varphi }^{\prime }}_1\right)$ of internal friction above water table is $30°$.

The angle $\left({{\varphi }^{\prime }}_2\right)$ of internal friction below water table is $30°$.

The surcharge $\left(q\right)$ on the soil is 0.

Calculation:

Find the submerged unit weight $\left({\gamma }^{\prime }\right)$ of the soil using the equation:

${\gamma }^{\prime }={\gamma }_2-{\gamma }_w$

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

The unit weight $\left({\gamma }_w\right)$ of the water is $\text{9}\text{.81}{\text{kN/m}}^{\text{3}}$.

Substitute $19.18{\text{kN/m}}^{\text{3}}$ for ${\gamma }_2$ and $\text{9}\text{.81}{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$.

$\begin{array}{c}{\gamma }^{\prime }=19.18-9.81\\ =9.37{\text{kN/m}}^2\end{array}$

The angle of internal friction above and below the water table is same. Therefore,

${{\varphi }^{\prime }}_1={{\varphi }^{\prime }}_2={\varphi }^{\prime }=30°$.

Find the coefficient of Rankine’s active earth pressure for frictionless wall $\left(K_a\right)$ using the equation:

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

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

$\begin{array}{c}{K}_a={\tan }^2\left(45°-\frac{30°}{2}\right)\\ =\frac{1}{3}\end{array}$

Find the active earth pressure $\left({{\sigma }^{\prime }}_a\right)$ at 0 m depth using the equation:

${{\sigma }^{\prime }}_{a0}={\gamma }_{1}z{K}_a$

Substitute $\text{16}\text{.51}{\text{kN/m}}^{\text{3}}$ for ${\gamma }_1$, 0 m for z, and $\frac{1}{3}$ for $K_a$.

$\begin{array}{c}{{\sigma }^{\prime }}_{a0}=16.51\times 0\times \frac{1}{3}\\ =0{\text{kN/m}}^{\text{2}}\end{array}$

Find the active earth pressure $\left({{\sigma }^{\prime }}_{a1}\right)$ at water table (1.52 m depth from top) using the equation:

${{\sigma }^{\prime }}_{a1}={\gamma }_1{H}_1{K}_a$

Substitute $\text{16}\text{.51}{\text{kN/m}}^{\text{3}}$ for ${\gamma }_1$, 1.52 m for $H_1$, and $\frac{1}{3}$ for $K_a$.

$\begin{array}{c}{{\sigma }^{\prime }}_{a1}=16.51\times 1.52\times \frac{1}{3}\\ =8.37{\text{kN/m}}^{\text{2}}\end{array}$

Find the active earth pressure $\left({{\sigma }^{\prime }}_{a2}\right)$ at bottom (3.05 m depth) using the equation:

${{\sigma }^{\prime }}_{a2}=\left[{\gamma }_1{H}_1+{\gamma }^{\prime }\left(H-H_1\right)\right]K_a$

Substitute $\text{16}\text{.51}{\text{kN/m}}^{\text{3}}$ for ${\gamma }_1$, $9.37{\text{kN/m}}^2$ for ${\gamma }^{\prime }$, 1.52 m for $H_1$, $3.05\text{m}$ for $H$, and $\frac{1}{3}$ for $K_a$

$\begin{array}{c}{{\sigma }^{\prime }}_{a2}=\left[\left(16.51\right)\left(1.52\right)+\left(9.37\right)\left(3.05-1.52\right)\right]\left(\frac{1}{3}\right)\\ =13.15{\text{kN/m}}^2\end{array}$

Find the pore water pressure $\left(u\right)$ using the equation:

$u={\gamma }_w\left(H-H_1\right)$

Substitute $\text{9}\text{.81}{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, $3.05\text{m}$ for $H$, and 1.52 m for $H_1$.

$\begin{array}{c}u=9.81\times \left(3.05-1.52\right)\\ =15{\text{kN/m}}^{\text{2}}\end{array}$

Sketch the active earth pressure diagram as shown in Figure 1.

Find the active earth pressure per unit length at component 1 $\left(P_{a1}\right)$ using the equation:

$P_{a1}=\frac{1}{2}{{\sigma }^{\prime }}_{a1}\times H_1$

Substitute $\text{8}\text{.37}{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{a1}$ and 1.52 m for $H_1$.

$\begin{array}{c}{P}_{a1}=\frac{1}{2}\times 8.37\times 1.52\\ =6.36\text{kN/m}\end{array}$

Find the active earth pressure per unit length at component 2 $\left(P_{a2}\right)$ using the equation:

$P_{a2}={{\sigma }^{\prime }}_{a1}\times \left(H-H_1\right)$

Substitute $\text{8}\text{.37}{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{a1}$, $3.05\text{m}$ for $H$, and 1.52 m for $H_1$.

$\begin{array}{c}{P}_{a2}=8.37\times 1.53\\ =12.81\text{kN/m}\end{array}$

Find the active earth pressure per unit length at component 3 $\left(P_{a3}\right)$ using the equation:

$P_{a3}=\frac{1}{2}\left({{\sigma }^{\prime }}_{a2}-{{\sigma }^{\prime }}_{a1}\right)\times \left(H-H_1\right)$

Substitute $\text{13}\text{.15}{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{a2}$, $\text{8}\text{.37}{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{a1}$, $3.05\text{m}$ for $H$, and 1.52 m for $H_1$.

$\begin{array}{c}{P}_{a3}=\frac{1}{2}\left(13.15-8.37\right)\times \left(3.05-1.52\right)\\ =3.66\text{kN/m}\end{array}$

Find the active earth pressure per unit length at component 4 $\left(P_{a4}\right)$ using the equation:

$P_{a4}=\frac{1}{2}\times u\times \left(H-H_1\right)$

Substitute $\text{15}\text{.0}{\text{kN/m}}^{\text{2}}$ for u, $3.05\text{m}$ for $H$, and 1.52 m for $H_1$.

$\begin{array}{c}{P}_{a4}=\frac{1}{2}\times 15\times \left(3.05-1.52\right)\\ =11.48\text{kN/m}\end{array}$

Find the total active earth pressure per unit length $\left(P_a\right)$ using the equation:

$P_a=P_{a1}+P_{a2}+P_{a3}+P_{a4}$

Substitute $6.36\text{kN/m}$ for $P_{a1}$, $12.81\text{kN/m}$ for $P_{a2}$, $3.66\text{kN/m}$ for $P_{a3}$, and $11.48\text{kN/m}$ for $P_{a4}$.

$\begin{array}{c}{P}_a=6.36+12.81+3.66+11.48\\ =34.31\text{kN/m}\end{array}$

Therefore, the Rankine’s active force per unit length of the wall is $\underset{\_}{34.31\text{kN/m}}$.

Refer Figure 1.

Find the location $\left(z_1\right)$ of active earth pressure per unit length at component 1 from bottom of the wall using the equation:

$z_1=\left(H-H_1\right)+\frac{H_1}{3}$

Substitute $3.05\text{m}$ for $H$ and 1.52 m for $H_1$.

$\begin{array}{c}{z}_1=\left(3.05-1.52\right)+\frac{1.52}{3}\\ =2.04\text{m}\end{array}$

Find the location $\left(z_2\right)$ of active earth pressure per unit length at component 2 from bottom of the wall using the equation:

$z_2=\frac{\left(H-H_1\right)}{2}$

Substitute $3.05\text{m}$ for $H$, and 1.52 m for $H_1$.

$\begin{array}{c}{z}_2=\frac{\left(3.05-1.52\right)}{2}\\ =0.765\text{m}\end{array}$

Find the location $\left(z_3\right)$ of active earth pressure per unit length at component 3 from bottom of the wall using the equation:

$z_3=\frac{\left(H-H_1\right)}{3}$

Substitute $3.05\text{m}$ for $H$ and 1.52 m for $H_1$.

$\begin{array}{c}{z}_3=\frac{\left(3.05-1.52\right)}{3}\\ =0.51\text{m}\end{array}$

Find the location $\left(z_4\right)$ of active earth pressure per unit length at component 4 from bottom of the wall using the equation:

$z_4=\frac{\left(H-H_1\right)}{3}$

Substitute $3.05\text{m}$ for $H$ and 1.52 m for $H_1$.

$\begin{array}{c}{z}_4=\frac{\left(3.05-1.52\right)}{3}\\ =0.51\text{m}\end{array}$

Find the location of resultant force $\left(\overline{z}\right)$:

Take moment about the bottom of the wall.

$P_a\times \overline{z}=\left(P_{a1}\times z_1\right)+\left(P_{a2}\times z_2\right)+\left(P_{a3}\times z_3\right)+\left(P_{a4}\times z_4\right)$

Substitute $34.31\text{kN/m}$ for $P_a$, $6.36\text{kN/m}$ for $P_{a1}$, $2.04\text{m}$ for $z_1$, $12.81\text{kN/m}$ for $P_{a2}$, $0.765\text{m}$ for $z_2$, $3.66\text{kN/m}$ for $P_{a3}$, $0.51\text{m}$ for $z_3$, $11.48\text{kN/m}$ for $P_{a4}$, and $0.51\text{m}$ for $z_4$.

$\begin{array}{c}34.31\times \overline{z}=\left(6.36\times 2.04\right)+\left(12.81\times 0.765\right)+\left(3.66\times 0.51\right)+\left(11.48\times 0.51\right)\\ 34.31\overline{z}=12.974+9.799+1.867+5.855\\ \overline{z}=\frac{30.495}{34.31}\\ =0.89\text{m}\end{array}$

Therefore, the location of the resultant is $\underset{\_}{0.89\text{m from bottom}}$.

(b)

To determine

Find the Rankine’s active force per unit length of the wall.

Find the location of the resultant.

Answer to Problem 14.8P

The Rankine’s active force per unit length of the wall is $\underset{\_}{141.13\text{kN/m}}$.

The location of the resultant is $\underset{\_}{2.04\text{m from bottom}}$.

Explanation of Solution

Given information:

The height (H) of the retaining wall is 6 m.

The depth $\left(H_1\right)$ of water table is 3 m.

The unit weight $\left({\gamma }_1\right)$ of the soil is $15.5{\text{kN/m}}^2$.

The saturated unit weight $\left({\gamma }_2\right)$ of the soil is $19.0{\text{kN/m}}^2$.

The angle $\left({{\varphi }^{\prime }}_1\right)$ of internal friction above water table is $30°$.

The angle $\left({{\varphi }^{\prime }}_2\right)$ of internal friction below water table is $36°$.

The surcharge $\left(q\right)$ on the soil is $15{\text{kN/m}}^2$.

Calculation:

Find the submerged unit weight $\left({\gamma }^{\prime }\right)$ of the soil using the equation:

${\gamma }^{\prime }={\gamma }_2-{\gamma }_w$

The unit weight $\left({\gamma }_w\right)$ of the water is $\text{9}\text{.81}{\text{kN/m}}^{\text{3}}$.

Substitute $19.0{\text{kN/m}}^2$ for ${\gamma }_2$ and $\text{9}\text{.81}{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$.

$\begin{array}{c}{\gamma }^{\prime }=19-9.81\\ =9.19{\text{kN/m}}^2\end{array}$

Find the coefficient of Rankine’s active earth pressure for frictionless wall $\left(K_{a1}\right)$ above water table using the equation:

$K_{a1}={\tan }^2\left(45-\frac{{{\varphi }^{\prime }}_1}{2}\right)$

Substitute $30°$ for ${{\varphi }^{\prime }}_1$.

$\begin{array}{c}{K}_a={\tan }^2\left(45°-\frac{30°}{2}\right)\\ =\frac{1}{3}\end{array}$

Find the coefficient of Rankine’s active earth pressure for frictionless wall $\left(K_{a1}\right)$ below water table using the equation:

$K_{a2}={\tan }^2\left(45-\frac{{{\varphi }^{\prime }}_2}{2}\right)$

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

$\begin{array}{c}{K}_{a2}={\tan }^2\left(45-\frac{36}{2}\right)\\ =0.26\end{array}$

Find the active earth pressure $\left({{\sigma }^{\prime }}_a\right)$ at 0 m depth using the equation:

${{\sigma }^{\prime }}_{a0}={\gamma }_{1}z{K}_{a1}+q{K}_{a2}$

Substitute $\text{15}\text{.50}{\text{kN/m}}^{\text{3}}$ for ${\gamma }_1$, 0 m for z, $\frac{1}{3}$ for $K_{a1}$, and $\text{15}{\text{kN/m}}^{\text{2}}$ for q.

$\begin{array}{c}{{\sigma }^{\prime }}_{a0}=\left(15.50\times 0\times \frac{1}{3}\right)+\left(15\times \frac{1}{3}\right)\\ =5{\text{kN/m}}^{\text{2}}\end{array}$

Find the active earth pressure $\left({{\sigma }^{\prime }}_{a1}\right)$ at water table (3.0 m depth from top) using the equation:

${{\sigma }^{\prime }}_{a1}={\gamma }_1{H}_1{K}_a{}_1+q{K}_{a1}$

Substitute $\text{15}\text{.50}{\text{kN/m}}^{\text{3}}$ for ${\gamma }_1$, 3.0 m for $H_1$, $\frac{1}{3}$ for $K_{a1}$, and $\text{15}{\text{kN/m}}^{\text{2}}$ for q.

$\begin{array}{c}{{\sigma }^{\prime }}_{a1}=\left(15.5\times 3\times \frac{1}{3}\right)+\left(15\times \frac{1}{3}\right)\\ =20.5{\text{kN/m}}^{\text{2}}\end{array}$

Find the active earth pressure $\left({{\sigma }^{\prime }}_{a3}\right)$ at bottom (3.0 m depth) using the equation:

${{\sigma }^{\prime }}_{a3}={\gamma }_1{H}_1{K}_a{}_2+q{K}_{a2}$

Substitute $\text{15}\text{.50}{\text{kN/m}}^{\text{3}}$ for ${\gamma }_1$, 3.0 m for $z_1$, 0.26 for $K_{a2}$, and $\text{15}{\text{kN/m}}^{\text{2}}$ for q.

$\begin{array}{c}{{\sigma }^{\prime }}_{a3}=\left(15.5\times 3\times 0.26\right)+\left(15\times 0.26\right)\\ =15.99{\text{kN/m}}^{\text{2}}\end{array}$

Find the active earth pressure $\left({{\sigma }^{\prime }}_{a2}\right)$ at bottom (6.0 m depth) using the equation:

${{\sigma }^{\prime }}_{a4}=\left[{\gamma }_1{H}_1+{\gamma }^{\prime }\left(H-H_1\right)\right]K_{a2}+q{K}_{a2}$

Substitute $\text{15}\text{.5}{\text{kN/m}}^{\text{3}}$ for ${\gamma }_1$, $19{\text{kN/m}}^{\text{3}}$ for ${\gamma }_2$, $\text{9}\text{.19}{\text{kN/m}}^{\text{3}}$ for ${\gamma }^{\prime }$, 3 m for $H_1$, $6\text{m}$ for $H$, and 0.26 for $K_{a2}$.

$\begin{array}{c}{{\sigma }^{\prime }}_{a4}=\left[\left(15.5\right)\left(3\right)+\left(9.19\right)\left(6-3\right)\right]0.26+\left[15\times 0.26\right]\\ =19.26+3.9\\ =23.16{\text{kN/m}}^2\end{array}$

Find the pore water pressure $\left(u\right)$ using the equation:

$u={\gamma }_w\left(H-H_1\right)$

Substitute $\text{9}\text{.81}{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$ and $6\text{m}$ for $H$, and 3 m for $H_1$.

$\begin{array}{c}u=9.81\times \left(6-3\right)\\ =29.43{\text{kN/m}}^{\text{2}}\end{array}$

Sketch the active earth pressure diagram as shown in Figure 2.

Find the active earth pressure of component 1 per unit length due to surcharge load $\left(P_{a1}\right)$ using the equation:

$P_{a1}={{\sigma }^{\prime }}_{a0}\times H$

Substitute $\text{5}\text{.0}{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{a0}$ and 6.0 m for H.

$\begin{array}{c}{P}_{a1}=6\times 5\\ =30\text{kN/m}\end{array}$

Find the active earth pressure per unit length at component 2 $\left(P_{a2}\right)$ using the equation:

$P_{a2}=\frac{1}{2}\left({{\sigma }^{\prime }}_{a1}-{{\sigma }^{\prime }}_{a0}\right)\times H_1$

Substitute $20.5{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{a1}$, $\text{5}{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{a0}$, and 3.0 m for $H_1$.

$\begin{array}{c}{P}_{a2}=\frac{1}{2}\times \left(20.5-5\right)\times 3\\ =23.25\text{kN/m}\end{array}$

Find the active earth pressure per unit length at component 3 $\left(P_{a3}\right)$ using the equation:

$P_{a3}=\left({{\sigma }^{\prime }}_{a3}-{{\sigma }^{\prime }}_{a0}\right)\left(H-H_1\right)$

Substitute $\text{15}\text{.99}{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{a3}$, $\text{5}{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{a0}$, 6.0 m for $H$, and 3.0 m for $H_1$.

$\begin{array}{c}{P}_{a3}=\left(15.99-5\right)\times \left(6-3\right)\\ =32.97\text{kN/m}\end{array}$

Find the active earth pressure per unit length at component 4 $\left(P_{a4}\right)$ using the equation:

$P_{a4}=\frac{1}{2}\left({{\sigma }^{\prime }}_{a4}-{{\sigma }^{\prime }}_{a3}\right)\times \left(H-H_1\right)$

Substitute $\text{23}\text{.16}{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{a2}$, $\text{15}\text{.99}{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{a1}$, 6.0 m for $H$, and 3.0 m for $H_1$.

$\begin{array}{c}{P}_{a4}=\frac{1}{2}\times \left(23.16-15.99\right)\times \left(6-3\right)\\ =10.76\text{kN/m}\end{array}$

Find the active earth pressure per unit length at component 5 $\left(P_{a5}\right)$ using the equation:

$P_{a5}=\frac{1}{2}\times u\times \left(H-H_1\right)$

Substitute $\text{29}\text{.43}{\text{kN/m}}^{\text{2}}$ for u, 6.0 m for $H$, and 3.0 m for $H_1$.

$\begin{array}{c}{P}_{a5}=\frac{1}{2}\times 29.43\times \left(6-3\right)\\ =44.15\text{kN/m}\end{array}$

Find the total active earth pressure per unit length $\left(P_a\right)$ using the equation:

$P_a=P_{a1}+P_{a2}+P_{a3}+P_{a4}+P_{a5}$

Substitute $30\text{kN/m}$ for $P_{a1}$, $23.25\text{kN/m}$ for $P_{a2}$, $32.97\text{kN/m}$ for $P_{a3}$, $\text{10}\text{.76}\text{kN/m}$ for $P_{a4}$, and $\text{44}\text{.15}\text{kN/m}$ for $P_{a5}$.

$\begin{array}{c}{P}_a=30+23.25+32.97+10.76+44.15\\ =141.13\text{kN/m}\end{array}$

Therefore, the Rankine’s active force per unit length of the wall is $\underset{\_}{141.13\text{kN/m}}$.

Refer Figure 1.

Find the location $\left(z_1\right)$ of active earth pressure per unit length at component 1 from bottom of the wall using the equation:

$z_1=\frac{H}{2}$

Substitute $6\text{m}$ for $H$.

$\begin{array}{c}{z}_1=\frac{6}{2}\\ =3\text{m}\end{array}$

Find the location $\left(z_2\right)$ of active earth pressure per unit length at component 2 from bottom of the wall using the equation:

$z_2=\left(H-H_1\right)+\frac{H_1}{3}$

Substitute $6\text{m}$ for $H$ and 3 m for $H_1$.

$\begin{array}{c}{z}_2=\left(6-3\right)+\frac{3}{3}\\ =4\text{m}\end{array}$

Find the location $\left(z_3\right)$ of active earth pressure per unit length at component 3 from bottom of the wall using the equation:

$z_3=\frac{\left(H-H_1\right)}{2}$

Substitute $6\text{m}$ for $H$ and 3 m for $H_1$.

$\begin{array}{c}{z}_3=\frac{\left(6-3\right)}{2}\\ =1.5\text{m}\end{array}$

Find the location $\left(z_4\right)$ of active earth pressure per unit length at component 4 from bottom of the wall using the equation:

$z_4=\frac{\left(H-H_1\right)}{3}$

Substitute $6\text{m}$ for $H$ and 3 m for $H_1$.

$\begin{array}{c}{z}_4=\frac{\left(6-3\right)}{3}\\ =1\text{m}\end{array}$

Find the location $\left(z_5\right)$ of active earth pressure per unit length at component 5 from bottom of the wall using the equation:

$z_5=\frac{\left(H-H_1\right)}{3}$

Substitute $6\text{m}$ for $H$ and 3 m for $H_1$.

$\begin{array}{c}{z}_5=\frac{\left(6-3\right)}{3}\\ =1\text{m}\end{array}$

Find the location of resultant force $\left(\overline{z}\right)$.

Take moment about the bottom of the wall.

$P_a\times \overline{z}=\left(P_{a1}\times H_1\right)+\left(P_{a2}\times H_2\right)+\left(P_{a3}\times H_3\right)+\left(P_{a4}\times H_4\right)+\left(P_{a5}\times H_5\right)$

Substitute $141.13\text{kN/m}$ for $P_a$, $30\text{kN/m}$ for $P_{a1}$, $3\text{m}$ for $z_1$, $23.25\text{kN/m}$ for $P_{a2}$, $4\text{m}$ for $z_2$, $32.97\text{kN/m}$ for $P_{a3}$, $1.5\text{m}$ for $z_3$, $10.76\text{kN/m}$ for $P_{a4}$, $1\text{m}$ for $z_4$, $44.15\text{kN/m}$ for $P_{a5}$, and $1\text{m}$ for $z_5$.

$\begin{array}{c}141.13\times \overline{z}=\left[\begin{array}{l}\left(30\right)\times 3+\left(23.25\right)\times 4+\left(32.97\right)\times \left(1.5\right)\\ +\left(10.76\right)\times \left(1\right)+\left(44.15\right)\left(1\right)\end{array}\right]\\ \overline{z}=\frac{90+93+49.46+10.76+44.15}{141.13}\\ =2.04\text{m}\end{array}$

Therefore, the location of the resultant is $\underset{\_}{2.04\text{m from bottom}}$.