Chapter 15, Problem 15.27P

To determine

Find the minimum factor of safety $F_s$ using the Bishop and Morgenstern’s method.

Answer to Problem 15.27P

The the minimum factor of safety $F_s$ using the Bishop and Morgenstern’s method is $\underset{\_}{1.19}$.

Explanation of Solution

Given information:

The height (H) of the slope is 30 m.

The inclination $\beta$ of a slope $\left(3:1\text{slope}\right)$ is $18.43°$.

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

The angle of friction ${\varphi }^{\prime }$ is $23°$.

The cohesion $c^{\prime }$ is $27{\text{kN/m}}^{\text{2}}$.

The value non-dimensional quantity $r_u$ is 0.5.

Calculation:

Determine the minimum factor of safety $F_s$ from Table (15.5), using the procedure as follows:

• Step 1. Obtain the values of angle of friction, inclination of a slope, and the value of $\frac{c^{\prime }}{\gamma H}$
• Step 2. Obtain the value of nondimensional quantity.
• Step 3. Refer Table (15.5) “Values of $m^{\prime }$ and $n^{\prime }$ for $\frac{c^{\prime }}{\gamma H}=0$” in the text book. Obtain the values of $m^{\prime }$ and $n^{\prime }$ for $D=1,1.25,\text{and}1.5$.
• Step 4. calculate the factor of safety value, using the values of $m^{\prime }$ and $n^{\prime }$ for each value of D.
• Finally, the required value of factor of safety is the smallest one obtained in step 4.

Trial 1:

Determine the value of $\frac{c^{\prime }}{\gamma H}$.

Substitute $27{\text{kN/m}}^{\text{2}}$ for $c^{\prime }$, $18{\text{kN/m}}^{\text{3}}$ for $\gamma$, and 30 m for H.

$\begin{array}{c}\frac{c^{\prime }}{\gamma H}=\frac{27}{18\left(30\right)}\\ =0.05\end{array}$

Refer Table (15.5d) “Stability coefficients $m^{\prime }$ and $n^{\prime }$ for $\frac{c^{\prime }}{\gamma H}=0.05$ and $D=1.00$” in the text book.

Take the stability coefficient $m^{\prime }$ as 2.014 when the angle of friction ${\varphi }^{\prime }$ is $22.5°$.

Take the stability coefficient $m^{\prime }$ as 2.193 when the angle of friction ${\varphi }^{\prime }$ is $25°$.

Determine the stability coefficient $m^{\prime }$ when the angle of friction ${\varphi }^{\prime }$ is $23°$ using the interpolation.

$\begin{array}{c}{m}^{\prime }=\frac{\left(23-22.5\right)\left(2.193-2.014\right)}{\left(25-22.5\right)}+2.014\\ =2.049\end{array}$

Take the stability coefficient $n^{\prime }$ as 1.568 when the angle of friction ${\varphi }^{\prime }$ is $22.5°$.

Take the stability coefficient $n^{\prime }$ as 1.757 when the angle of friction ${\varphi }^{\prime }$ is $25°$.

Calculate the stability coefficient $n^{\prime }$ when the angle of friction ${\varphi }^{\prime }$ is $23°$ using the interpolation.

$\begin{array}{c}{n}^{\prime }=\frac{\left(23-22.5\right)\left(1.757-1.568\right)}{\left(25-22.5\right)}+1.568\\ =1.606\end{array}$

Calculate the factor of safety using the formula.

$F_s=m^{\prime }-n^{\prime }{r}_u$

Substitute 2.049 for $m^{\prime }$, 1.606 for $n^{\prime }$, and 0.5 for $r_u$.

$\begin{array}{c}{F}_s=2.049-1.606\left(0.5\right)\\ =1.25\end{array}$

Trial 2:

Refer Table (15.5e) “Stability coefficients $m^{\prime }$ and $n^{\prime }$ for $\frac{c^{\prime }}{\gamma H}=0.05$ and $D=1.25$” in the text book.

Take the stability coefficient $m^{\prime }$ as 2.024 when the angle of friction ${\varphi }^{\prime }$ is $22.5°$.

Take the stability coefficient $m^{\prime }$ as 2.222 when the angle of friction ${\varphi }^{\prime }$ is $25°$.

Determine the stability coefficient $m^{\prime }$ when the angle of friction ${\varphi }^{\prime }$ is $23°$ using the interpolation.

$\begin{array}{c}{m}^{\prime }=\frac{\left(23-22.5\right)\left(2.222-2.024\right)}{\left(25-22.5\right)}+2.024\\ =2.064\end{array}$

Take the stability coefficient $n^{\prime }$ as 1.690 when the angle of friction ${\varphi }^{\prime }$ is $22.5°$.

Take the stability coefficient $n^{\prime }$ as 1.897 when the angle of friction ${\varphi }^{\prime }$ is $25°$.

Determine the stability coefficient $n^{\prime }$ when the angle of friction ${\varphi }^{\prime }$ is $23°$ using the interpolation.

$\begin{array}{c}{n}^{\prime }=\frac{\left(23-22.5\right)\left(1.897-1.690\right)}{\left(25-22.5\right)}+1.690\\ =1.731\end{array}$

Determine the factor of safety using the formula.

$F_s=m^{\prime }-n^{\prime }{r}_u$

Substitute 2.064 for $m^{\prime }$, 1.731 for $n^{\prime }$, and 0.5 for $r_u$.

$\begin{array}{c}{F}_s=2.064-1.731\left(0.5\right)\\ =1.19\end{array}$

Trial 3:

Refer Table (15.5f) “Stability coefficients $m^{\prime }$ and $n^{\prime }$ for $\frac{c^{\prime }}{\gamma H}=0.05$ and $D=1.50$” in the text book.

Take the stability coefficient $m^{\prime }$ as 2.234 when the angle of friction ${\varphi }^{\prime }$ is $22.5°$.

Take the stability coefficient $m^{\prime }$ as 2.467 when the angle of friction ${\varphi }^{\prime }$ is $25°$.

Calculate the stability coefficient $m^{\prime }$ when the angle of friction ${\varphi }^{\prime }$ is $23°$ using the interpolation.

$\begin{array}{c}{m}^{\prime }=\frac{\left(23-22.5\right)\left(2.467-2.234\right)}{\left(25-22.5\right)}+2.234\\ =2.281\end{array}$

Take the stability coefficient $n^{\prime }$ as 1.937 when the angle of friction ${\varphi }^{\prime }$ is $22.5°$.

Take the stability coefficient $n^{\prime }$ as 2.179 when the angle of friction ${\varphi }^{\prime }$ is $25°$.

Calculate the stability coefficient $n^{\prime }$ when the angle of friction ${\varphi }^{\prime }$ is $23°$ using the interpolation.

$\begin{array}{c}{n}^{\prime }=\frac{\left(23-22.5\right)\left(2.179-1.937\right)}{\left(25-22.5\right)}+1.937\\ =1.985\end{array}$

Calculate the factor of safety using the formula.

$F_s=m^{\prime }-n^{\prime }{r}_u$

Substitute 2.281 for $m^{\prime }$, 1.985 for $n^{\prime }$, and 0.5 for $r_u$.

$\begin{array}{c}{F}_s=2.281-1.985\left(0.5\right)\\ =1.29\end{array}$

The required value of factor of safety is the smallest one obtained from trial 2.

Thus, the minimum factor of safety $F_s$ using the Bishop and Morgenstern’s method is $\underset{\_}{1.19}$.