Chapter 12, Problem 12.1P

(a)

To determine

Find the angle of friction, ${\varphi }^{\prime }$.

Answer to Problem 12.1P

The angle of friction, ${\varphi }^{\prime }$ is $\underset{\_}{24.9°}$.

Explanation of Solution

Given information:

The diameter (d) of specimen is 71 mm.

The height (h) of specimen is 25 mm.

Shear force $\left(V_u\right)$ at the failure is 276 N.

The normal stress $\left({\sigma }^{\prime }\right)$ on the failure plane is $150\text{kN}/{\text{m}}^{\text{2}}$.

Calculation:

Show the expression of Mohr’s coulomb failure as follows:

${\tau }_f=c^{\prime }+{\sigma }^{\prime }\tan {\varphi }^{\prime }$ (1)

Here, $c^{\prime }$ is the cohesion, ${\tau }_f$ is shear strength of the dry sand, and ${\varphi }^{\prime }$ is angle of friction.

Find the shear strength $\left({\tau }_f\right)$ of the dry sand as follows:

$\begin{array}{c}{\tau }_f=\frac{S}{A}\\ =\frac{S}{\frac{\pi d^2}{4}}\end{array}$

Here, S is the shear force and $A$ is area of specimen.

Substitute 276 N for shear force and 71 mm for d.

$\begin{array}{c}{\tau }_f=\frac{276}{\frac{\pi \times {\left(71\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2}{4}}\\ =69.71\times {10}^3\text{N}/{\text{m}}^{\text{2}}\times \frac{\text{1}\text{kN}}{{\text{10}}^{\text{3}}\text{N}}\\ =69.71\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the angle of friction $\left({\varphi }^{\prime }\right)$ as follows:

Substitute $69.71\text{kN}/{\text{m}}^{\text{2}}$ for ${\tau }_f$, $150\text{kN}/{\text{m}}^{\text{2}}$ for ${\sigma }^{\prime }$, and 0 for $c^{\prime }$ in Equation (1).

The value of $c^{\prime }$ for sand and inorganic silt is 0.

$\begin{array}{l}69.71=0+150\tan \left({\varphi }^{\prime }\right)\\ 69.71=150\tan \left({\varphi }^{\prime }\right)\\ {\varphi }^{\prime }={\tan }^{-1}\left(\frac{69.71}{150}\right)\\ {\varphi }^{\prime }=24.9°\end{array}$

Thus, the angle of friction, ${\varphi }^{\prime }$ is $\underset{\_}{24.9°}$.

(b)

To determine

Find the shear force (S) required to cause failure.

Answer to Problem 12.1P

The shear force required (S) to cause failure is $\underset{\_}{367.6\text{N}}$.

Explanation of Solution

Given information:

The normal stress $\left({\sigma }^{\prime }\right)$ is $200\text{kN}/{\text{m}}^{\text{2}}$.

Calculation:

Refer part (a).

Substitute 0 for $c^{\prime }$, $200\text{kN}/{\text{m}}^{\text{2}}$ for ${\sigma }^{\prime }$, and $24.9°$ for ${\varphi }^{\prime }$ in Equation (1).

$\begin{array}{c}{\tau }_f=0+200\tan 24.9°\\ =92.84\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the shear force is required to cause failure as follows:

$\begin{array}{c}S={\tau }_f\times A_s\\ ={\tau }_f\times \left(\frac{\pi d^2}{4}\right)\end{array}$

Substitute $92.84\text{kN}/{\text{m}}^{\text{2}}$ for ${\tau }_f$ and 71 mm for d.

$\begin{array}{c}S=92.84\times \left(\frac{\pi \times {\left(71\text{mm}\times \frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)}^2}{4}\right)\\ =92.84\times 3.9592\times {10}^{-3}\\ =0.3676\text{kN}\times \frac{\text{1000}\text{N}}{\text{1}\text{kN}}\\ =367.6\text{kN}\end{array}$

Thus, the shear force (S) required to cause failure is $\underset{\_}{367.6\text{N}}$.