Chapter 15, Problem 15.12P

To determine

Find the factor of safety $F_s$ with respect to sliding.

Answer to Problem 15.12P

The factor of safety $F_s$ with respect to sliding is $\underset{\_}{2.25}$.

Explanation of Solution

Given information:

The slope with an inclination $\beta$ is $55°$.

The unit weight $\gamma$ of the soil is $121{\text{lb/ft}}^{\text{3}}$.

The angle of friction ${\varphi }^{\prime }$ is $17°$.

The cohesion $c^{\prime }$ is $1,200{\text{lb/ft}}^{\text{2}}$.

The height (H) of the retaining wall is 45 ft.

Calculation:

Trial 1:

Consider the factor of safety as 2.

Determine the cohesion ${c^{\prime }}_d$ develop along the potential failure surface using the relation.

${c^{\prime }}_d=\frac{c^{\prime }}{F_s}$

Substitute $1,200{\text{lb/ft}}^{\text{2}}$ for $c^{\prime }$ and 2.0 for $F_s$.

$\begin{array}{c}{c^{\prime }}_d=\frac{1,200}{2.0}\\ =600{\text{lb/ft}}^2\end{array}$

Determine the angle ${{\varphi }^{\prime }}_d$ of friction that develops along the potential failure surface using the relation.

${{\varphi }^{\prime }}_d={\tan }^{-1}\left(\frac{\tan {\varphi }^{\prime }}{F_s}\right)$

Substitute $17°$ for ${\varphi }^{\prime }$ and 2.0 for $F_s$.

$\begin{array}{c}{{\varphi }^{\prime }}_d={\tan }^1\left(\frac{\tan 17°}{2.0}\right)\\ ={\tan }^{-1}\left(0.153\right)\\ =8.69°\end{array}$

Determine height of the slope that will have a factor of safety of 2.0 against sliding using the formula.

$H_{\text{cr}}=\frac{4{c^{\prime }}_d}{\gamma }\left[\frac{\sin \beta \cos {{\varphi }^{\prime }}_d}{1-\cos \left(\beta -{{\varphi }^{\prime }}_d\right)}\right]$

Substitute $600{\text{lb/ft}}^2$ for ${c^{\prime }}_d$, $121{\text{lb/ft}}^{\text{3}}$ for $\gamma$, $55°$ for $\beta$, and $8.69°$ for ${{\varphi }^{\prime }}_d$.

$\begin{array}{c}{H}_{\text{cr}}=\frac{4\left(600\right)}{121}\left[\frac{\sin 55°\cos 8.69°}{1-\cos \left(55°-8.69°\right)}\right]\\ =19.835\left(\frac{0.81}{0.31}\right)\\ =51.83\text{ft}\end{array}$

The height of the retaining wall is not equal to the calculated height of the wall.

$51.83\text{ft}\ne 45\text{ft}$

Hence, the assumption is incorrect.

Trial 2:

Consider the factor of safety as 2.5.

Determine the cohesion ${c^{\prime }}_d$ develop along the potential failure surface using the relation.

${c^{\prime }}_d=\frac{c^{\prime }}{F_s}$

Substitute $1,200{\text{lb/ft}}^{\text{2}}$ for $c^{\prime }$ and 2.5 for $F_s$.

$\begin{array}{c}{c^{\prime }}_d=\frac{1,200}{2.5}\\ =480{\text{lb/ft}}^{\text{2}}\end{array}$

Determine the angle ${{\varphi }^{\prime }}_d$ of friction that develops along the potential failure surface using the relation.

${{\varphi }^{\prime }}_d={\tan }^{-1}\left(\frac{\tan {\varphi }^{\prime }}{F_s}\right)$

Substitute $17°$ for ${\varphi }^{\prime }$ and 2.5 for $F_s$.

$\begin{array}{c}{{\varphi }^{\prime }}_d={\tan }^1\left(\frac{\tan 17°}{2.5}\right)\\ ={\tan }^{-1}\left(0.1222\right)\\ =6.97°\end{array}$

Determine height of the slope that will have a factor of safety of 2.0 against sliding using the formula.

$H_{\text{cr}}=\frac{4{c^{\prime }}_d}{\gamma }\left[\frac{\sin \beta \cos {{\varphi }^{\prime }}_d}{1-\cos \left(\beta -{{\varphi }^{\prime }}_d\right)}\right]$

Substitute $480{\text{lb/ft}}^{\text{2}}$ for ${c^{\prime }}_d$, $121{\text{lb/ft}}^{\text{3}}$ for $\gamma$, $55°$ for $\beta$, and $6.97°$ for ${{\varphi }^{\prime }}_d$.

$\begin{array}{c}{H}_{\text{cr}}=\frac{4\left(480\right)}{121}\left[\frac{\sin 55°\cos 6.97°}{1-\cos \left(55°-6.97°\right)}\right]\\ =15.867\left(\frac{0.81}{0.33}\right)\\ =38.94\text{ft}\end{array}$

The height of the retaining wall is not equal to the calculated height of the wall.

$38.94\text{ft}\ne 45\text{ft}$

Hence, the assumption is incorrect.

Trial 3:

Consider the factor of safety as 2.25.

Determine the cohesion ${c^{\prime }}_d$ develop along the potential failure surface using the relation.

${c^{\prime }}_d=\frac{c^{\prime }}{F_s}$

Substitute $1,200{\text{lb/ft}}^{\text{2}}$ for $c^{\prime }$ and 2.25 for $F_s$.

$\begin{array}{c}{c^{\prime }}_d=\frac{1,200}{2.25}\\ =533.33{\text{lb/ft}}^{\text{2}}\end{array}$

Determine the angle ${{\varphi }^{\prime }}_d$ of friction that develops along the potential failure surface using the relation.

${{\varphi }^{\prime }}_d={\tan }^{-1}\left(\frac{\tan {\varphi }^{\prime }}{F_s}\right)$

Substitute $17°$ for ${\varphi }^{\prime }$ and 2.25 for $F_s$.

$\begin{array}{c}{{\varphi }^{\prime }}_d={\tan }^1\left(\frac{\tan 17°}{2.25}\right)\\ ={\tan }^{-1}\left(0.136\right)\\ =7.74°\end{array}$

Determine height of the slope that will have a factor of safety of 2.25 against sliding using the formula.

$H_{\text{cr}}=\frac{4{c^{\prime }}_d}{\gamma }\left[\frac{\sin \beta \cos {{\varphi }^{\prime }}_d}{1-\cos \left(\beta -{{\varphi }^{\prime }}_d\right)}\right]$

Substitute $533.33{\text{lb/ft}}^{\text{2}}$ for ${c^{\prime }}_d$, $121{\text{lb/ft}}^{\text{3}}$ for $\gamma$, $55°$ for $\beta$, and $7.74°$ for ${{\varphi }^{\prime }}_d$.

$\begin{array}{c}{H}_{\text{cr}}=\frac{4\left(533.33\right)}{121}\left[\frac{\sin 55°\cos 7.74°}{1-\cos \left(55°-7.74°\right)}\right]\\ =17.631\left(\frac{0.812}{0.32}\right)\\ =44.74\text{ft}\\ \simeq \text{45}\text{ft}\end{array}$

The height of the retaining wall is equal to the calculated height of the wall.

$45\text{ft}=45\text{ft}$

Thus, the factor of safety with respect to sliding is $\underset{\_}{2.25}$.