Chapter 13, Problem 13.27CTP

(a)

To determine

Find the maximum shear stress developed within the soil.

Answer to Problem 13.27CTP

The maximum shear stress developed within the soil is $\underset{\_}{33.3{\text{kN/m}}^2}$.

Explanation of Solution

Given information:

The depth (H) of slope is 5 m.

The angle $\left(\beta \right)$ of bed slope is $23°$.

The unit weight of the soil $\left(\gamma \right)$ is $18.5{\text{kN/m}}^3$.

The cohesion $\left(c^{\prime }\right)$ is $15{\text{kN/m}}^2$.

The angle $\left({\varphi }^{\prime }\right)$ of friction is $20°$.

Calculation:

The maximum shear stress developed at soil rock interface (5 m depth).

Find the maximum shear stress $20°$ developed within the soil using the equation:

${\tau }_d=\gamma H\sin \beta \cos \beta$

Substitute $18.5\text{kN}/{\text{m}}^{\text{3}}$ for $\gamma$, 5 m for H, and $23°$ for $\beta$

$\begin{array}{c}{\tau }_d=18.5\times 5\times \sin 23°\times \cos 23°\\ =33.3{\text{kN/m}}^2\end{array}$

Thus, the maximum shear stress developed within the soil is $\underset{\_}{33.3{\text{kN/m}}^2}$.

(b)

To determine

Find the maximum shear strength available within the soil.

Answer to Problem 13.27CTP

The maximum shear strength available within the soil is $\underset{\_}{43.5{\text{kN/m}}^2}$.

Explanation of Solution

Given information:

The depth (H) of slope is 5 m.

The angle $\left(\beta \right)$ of bed slope is $23°$.

The unit weight of the soil $\left(\gamma \right)$ is $18.5{\text{kN/m}}^3$.

The cohesion $\left(c^{\prime }\right)$ is $15{\text{kN/m}}^2$.

The angle $\left({\varphi }^{\prime }\right)$ of friction is $20°$.

Calculation:

Find the maximum shear strength $\left({\tau }_f\right)$ available within the soil using the equation:

${\tau }_f=c^{\prime }+\gamma H{\cos }^2\beta \tan {\varphi }^{\prime }$.

Substitute $15\text{kN}/{\text{m}}^{\text{2}}$ for $c^{\prime }$, $18.5\text{kN}/{\text{m}}^{\text{3}}$ for $\gamma$, 5 m for H, $23°$ for $\beta$, and $20°$ for ${\varphi }^{\prime }$.

$\begin{array}{c}{\tau }_f=15+\left(18.5\right)\left(5\right){\cos }^2\left(23°\right)\tan \left(20°\right)\\ =15+28.5\\ =43.5{\text{kN/m}}^2\end{array}$

Thus, the maximum shear strength available within the soil is $\underset{\_}{43.5{\text{kN/m}}^2}$.

(c)

To determine

Find the factor of safety of the slope.

Answer to Problem 13.27CTP

The factor of safety of the slope is $\underset{\_}{1.31}$.

Explanation of Solution

Given information:

The depth (H) of slope is 5 m.

The angle $\left(\beta \right)$ of bed slope is $23°$.

The unit weight of the soil $\left(\gamma \right)$ is $18.5{\text{kN/m}}^3$.

The cohesion $\left(c^{\prime }\right)$ is $15{\text{kN/m}}^2$.

The angle $\left({\varphi }^{\prime }\right)$ of friction is $20°$.

Calculation:

Find the factor of safety $\left(F{S}_s\right)$ of the slope using the equation:

$F{S}_s=\frac{c^{\prime }}{\gamma H{\cos }^2\beta \tan \beta }+\frac{\tan {\varphi }^{\prime }}{\tan \beta }$

Substitute $15\text{kN}/{\text{m}}^{\text{2}}$ for $c^{\prime }$, $18.5\text{kN}/{\text{m}}^{\text{3}}$ for $\gamma$, 5 m for H, $23°$ for $\beta$, and $20°$ for ${\varphi }^{\prime }$.

$\begin{array}{c}F{S}_s=\frac{15}{\left(18.5\right)\left(5\right){\cos }^2\left(23°\right)\tan \left(23°\right)}+\frac{\tan \left(20°\right)}{\tan \left(23°\right)}\\ =0.45+0.86\\ =1.31\end{array}$

Therefore, the factor of safety of the slope is $\underset{\_}{1.31}$.

(d)

To determine

Find the maximum possible depth for the soil before it becomes unstable.

Answer to Problem 13.27CTP

The maximum possible depth for the soil before it becomes unstable is $\underset{\_}{15.8\text{m}}$.

Explanation of Solution

Given information:

The depth (H) of slope is 5 m.

The angle $\left(\beta \right)$ of bed slope is $23°$.

The unit weight of the soil $\left(\gamma \right)$ is $18.5{\text{kN/m}}^3$.

The cohesion $\left(c^{\prime }\right)$ is $15{\text{kN/m}}^2$.

The angle $\left({\varphi }^{\prime }\right)$ of friction is $20°$.

Calculation:

The slope becomes unstable then the factor of safety $\left(F{S}_s\right)$ against sliding is 1.0.

Find the maximum possible depth $\left(H_{\text{cr}}\right)$ for the soil before it becomes unstable using the equation:

$H_{cr}=\frac{c^{\prime }}{\gamma }\frac{1}{{\cos }^2\beta \left(\tan \beta -\tan {\varphi }^{\prime }\right)}$

Substitute $15\text{kN}/{\text{m}}^{\text{2}}$ for $c^{\prime }$, $18.5\text{kN}/{\text{m}}^{\text{3}}$ for $\gamma$, $23°$ for $\beta$, and $20°$ for ${\varphi }^{\prime }$.

$\begin{array}{c}{H}_{cr}=\left(\frac{15}{18.5}\right)\frac{1}{{\cos }^{2}23°\left(\tan 23°-\tan 20°\right)}\\ =0.811\left(19.51\right)\\ =15.8\text{m}\end{array}$

Thus, the maximum possible depth for the soil before it becomes unstable is $\underset{\_}{15.8\text{m}}$.

(e)

To determine

Find the factor of safety with respect to cohesion when the friction is fully mobilized.

Answer to Problem 13.27CTP

The factor of safety with respect to cohesion when the friction is fully mobilized is $\underset{\_}{3.1}$.

Explanation of Solution

Given information:

The depth (H) of slope is 5 m.

The angle $\left(\beta \right)$ of bed slope is $23°$.

The unit weight of the soil $\left(\gamma \right)$ is $18.5{\text{kN/m}}^3$.

The cohesion $\left(c^{\prime }\right)$ is $15{\text{kN/m}}^2$.

The angle $\left({\varphi }^{\prime }\right)$ of friction is $20°$.

Calculation:

The developed angle of friction is equal to the angle of friction when the friction is fully mobilized. Therefore, ${{\varphi }^{\prime }}_d={\varphi }^{\prime }$.

Find the developed cohesion in the soil using the equation:

$\begin{array}{c}{\tau }_d={c^{\prime }}_d+\gamma H{\cos }^2\beta \tan {{\varphi }^{\prime }}_d\\ {c^{\prime }}_d={\tau }_d-\gamma H{\cos }^2\beta \tan {{\varphi }^{\prime }}_d\end{array}$

Substitute ${\varphi }^{\prime }$ for ${{\varphi }^{\prime }}_d$.

${c^{\prime }}_d={\tau }_d-\gamma H{\cos }^2\beta \tan {\varphi }^{\prime }$

Substitute $33.3\text{kN}/{\text{m}}^{\text{2}}$ for ${\tau }_d$, $18.5\text{kN}/{\text{m}}^{\text{3}}$ for $\gamma$, 5 m for H, $23°$ for $\beta$, and $20°$ for ${\varphi }^{\prime }$.

$\begin{array}{c}{c^{\prime }}_d=33.3-\left(18.5\right)\left(5\right){\cos }^2\left(23°\right)\tan \left(20°\right)\\ =33.3-28.5\\ =4.8{\text{kN/m}}^2\end{array}$

Find the factor of safety $\left(F{S}_{c^{\prime }}\right)$ with respect to cohesion when the friction is fully mobilized using the equation:

$F{S}_{c^{\prime }}=\frac{c^{\prime }}{{c^{\prime }}_d}$

Substitute $15\text{kN}/{\text{m}}^{\text{2}}$ for $c^{\prime }$ and $4.8\text{kN}/{\text{m}}^{\text{2}}$ for ${c^{\prime }}_d$.

$\begin{array}{c}F{S}_{c^{\prime }}=\frac{15}{4.8}\\ =3.1\end{array}$

Thus, the factor of safety with respect to cohesion when the friction is fully mobilized is $\underset{\_}{3.1}$.