Chapter 14, Problem 14.11P

To determine

Find the magnitude and location of the passive force $P_{pe}$ due to earthquake conditions.

Answer to Problem 14.11P

The magnitude of the passive force $P_{pe}$ due to earthquake conditions is $\underset{\_}{9,823\text{kN/m}}$.

The location of the resultant measured from the bottom of the wall is $\underset{\_}{3.12\text{m}}$.

Explanation of Solution

Given information:

The magnitude of surcharge loading (q) is $95{\text{kN/m}}^{\text{2}}$.

The unit weight $\gamma$ of the backfill is $18{\text{kN/m}}^{\text{3}}$.

The height (H) of the retaining wall is 7.0 m.

The soil friction angle ${\varphi }^{\prime }$ is $40°$.

The angle of wall friction ${\delta }^{\prime }$ is $20°$.

The cohesion $c^{\prime }$ of the soil is $30{\text{kN/m}}^{\text{2}}$.

The soil-wall interfall adhesion ${c^{\prime }}_a$ is 0.

The horizontal inertial $k_h$ force is 0.2.

The vertical inertial $k_v$ force is 0.2.

Calculation:

Determine the ratio of angle of wall friction to the soil friction angle.

$\begin{array}{c}\frac{{\delta }^{\prime }}{{\varphi }^{\prime }}=\frac{20°}{40°}\\ =0.5\end{array}$

Determine the ratio of soil-wall interfall adhesion to the cohesion.

$\begin{array}{c}\frac{{c^{\prime }}_a}{c^{\prime }}=\frac{0}{30}\\ =0\end{array}$

Determine the magnitude of the passive force due to earthquake conditions using the formula.

$P_{pe}=\left[\frac{1}{2}\gamma H^2{K}_{p\gamma \left(e\right)}+qH{K}_{pq\left(e\right)}+2c^{\prime }H{K}_{p{c}^{\prime }\left(e\right)}\right]\frac{1}{\cos {\delta }^{\prime }}$ (1)

Here, $K_{p\gamma \left(e\right)}$, $K_{pq\left(e\right)}$, and $K_{p{c}^{\prime }\left(e\right)}$ are the passive earth-pressure coefficients in the normal direction.

Refer Figure (14.9b) “Variation of $K_{p\gamma \left(e\right)}$ (a) $\frac{{\delta }^{\prime }}{{\varphi }^{\prime }}=1$; (b)” in the text book.

For $\frac{{\delta }^{\prime }}{{\varphi }^{\prime }}=0.5$;

Take the value of $K_{p\gamma \left(e\right)}$ as 6.75.

Refer Figure (14.10b) “Variation of $K_{p\gamma \left(e\right)}$ (a) $\frac{{\delta }^{\prime }}{{\varphi }^{\prime }}=1$; (b)” in the text book.

For $\frac{{\delta }^{\prime }}{{\varphi }^{\prime }}=0.5$;

Take the value of $K_{pq\left(e\right)}$ as 6.67.

Refer Table (14.7) “Variation of $K_{p{c}^{\prime }\left(e\right)}$” in the text book.

For the ratio of angle of wall friction to the soil friction angle is 0.5.

The value of $K_{p{c}^{\prime }\left(e\right)}$ is 4.33

Substitute $18{\text{kN/m}}^{\text{3}}$ for $\gamma$, 7.0 m for H, 6.75 for $K_{p\gamma \left(e\right)}$, $95{\text{kN/m}}^{\text{2}}$ for q, 6.67 for $K_{pq\left(e\right)}$, $30{\text{kN/m}}^{\text{2}}$ for $c^{\prime }$, 4.33 for $K_{p{c}^{\prime }\left(e\right)}$, and $20°$ for ${\delta }^{\prime }$ in Equation (1).

$\begin{array}{c}{P}_{pe}=\left[\frac{1}{2}\left(18\right)\left(7^2\right)\left(6.75\right)+95\left(7\right)\left(6.67\right)+2\left(30\right)\left(7\right)\left(4.33\right)\right]\frac{1}{\cos 20°}\\ =9,823\text{kN/m}\end{array}$

Thus, the magnitude of the passive force $P_{pe}$ due to earthquake conditions is $\underset{\_}{9,823\text{kN/m}}$.

Determine the unit weight of the passive force $P_{pe}$ using the formula.

$\text{Unit}\text{weight}\text{of}{P}_{pe}=\frac{1}{\cos {\delta }^{\prime }}\left[\frac{1}{2}\gamma H^2{K}_{p\gamma \left(e\right)}\right]$

Substitute $20°$ for ${\delta }^{\prime }$, $18{\text{kN/m}}^{\text{3}}$ for $\gamma$, 7.0 m for H, and 6.75 for $K_{p\gamma \left(e\right)}$.

$\begin{array}{c}\text{Unit}\text{weight}\text{of}{P}_{pe}=\frac{1}{\cos 20°}\left[\frac{1}{2}\left(18\right){\left(7\right)}^2\left(6.75\right)\right]\\ =3,168\text{kN/m}\end{array}$

Determine the distance of passive earth force acting above the bottom of the wall using the relation.

$\text{Distance}=\frac{H}{3}$

Substitute 7.0 m for H.

$\begin{array}{c}\text{Distance}=\frac{7.0}{3}\\ =2.33\text{m}\end{array}$

Determine the weight of surcharge component using the formula.

$\text{Unit}\text{weight}\text{of}\text{surcharge}=\frac{1}{\cos {\delta }^{\prime }}\left[qH{K}_{pq\left(e\right)}\right]$

Substitute $20°$ for ${\delta }^{\prime }$, $95{\text{kN/m}}^{\text{2}}$ for q, 7.0 m for H, and 6.67 for $K_{pq\left(e\right)}$.

$\begin{array}{c}\text{Unit}\text{weight}\text{of}\text{surcharge}=\frac{1}{\cos 20°}\left[95\left(7.0\right)\left(6.67\right)\right]\\ =4,720\text{kN/m}\end{array}$

Determine the distance of surcharge component acting above the bottom of the wall using the relation.

$\text{Distance}=\frac{H}{2}$

Substitute 7.0 m for H.

$\begin{array}{c}\text{Distance}=\frac{7.0}{2}\\ =3.5\text{m}\end{array}$

Determine the unit weight of the cohesion component using the formula.

$\text{Unit}\text{weight}\text{of}\text{cohesion}\text{component}=\frac{1}{\cos {\delta }^{\prime }}\left[2c^{\prime }H{K}_{p{c}^{\prime }\left(e\right)}\right]$

Substitute $20°$ for ${\delta }^{\prime }$, $30{\text{kN/m}}^{\text{2}}$ for $c^{\prime }$, 7.0 m for H, and 4.33 for $K_{p{c}^{\prime }\left(e\right)}$.

$\begin{array}{c}\text{Unit}\text{weight}\text{of}\text{cohesion}\text{component}=\frac{1}{\cos 20°}\left[2\left(30\right)\left(7\right)\left(4.33\right)\right]\\ =1,935\text{kN/m}\end{array}$

Determine the distance of cohesion component acting above the bottom of the wall using the relation.

$\text{Distance}=\frac{H}{2}$

Substitute 7.0 m for H.

$\begin{array}{c}\text{Distance}=\frac{7.0}{2}\\ =3.5\text{m}\end{array}$

Determine the location of the resultant measured from the bottom of the wall using the formula.

$\overline{z}=\frac{\text{Unit}\text{weight}\text{of}{P}_{pe}\times \text{Distance}+\left(\text{Unit}\text{weight}\text{ofsurcharge}+\text{cohesion}\right)\left(\text{distance}\right)}{P_{pe}}$

Substitute $3,168\text{kN/m}$ for unit weight of $P_{pe}$, 2.33 m for distance, $4,720\text{kN/m}$ for unit weight of surcharge, for unit weight of cohesion, 3.5 m for distance, and $9,823\text{kN/m}$ for $P_{pe}$.

$\begin{array}{c}\overline{z}=\frac{3,168\times \text{2}\text{.33}+\left(\text{4,720}+1,935\right)\left(3.5\right)}{9,823}\\ =3.12\text{m}\end{array}$

Thus, the location of the resultant measured from the bottom of the wall is $\underset{\_}{3.12\text{m}}$.