Chapter 12, Problem 12.17P

(a)

To determine

Find the normal and shear stress on a plane inclined at $33°$ to the major principal plane for specimen I.

Answer to Problem 12.17P

The normal stress on a plane inclined at $33°$ to the major principal plane for specimen I is $\underset{\_}{179\text{kN}/{\text{m}}^{\text{2}}}$.

The shear stress on a plane inclined at $33°$ to the major principal plane for specimen I is $\underset{\_}{70.8\text{kN}/{\text{m}}^{\text{2}}}$.

Explanation of Solution

Given information:

Specimen I:

The chamber pressure ${\sigma }_3$ is $70\text{kN}/{\text{m}}^{\text{2}}$.

The deviator stress is ${\left({\sigma }_1-{\sigma }_3\right)}_f$ is $155{\text{kN/m}}^{\text{2}}$.

The inclined angle with major principal plane $\theta$ is $33°$.

Specimen II.

The chamber pressure ${\sigma }_3$ is $140\text{kN}/{\text{m}}^{\text{2}}$.

The deviator stress ${\left({\sigma }_1-{\sigma }_3\right)}_f$ is $265\text{kN}/{\text{m}}^{\text{2}}$.

Calculation:

For specimen I:

Find the major principal effective stress at failure $\left({{\sigma }^{\prime }}_1\right)$ as shown below:

$\left({{\sigma }^{\prime }}_1\right)={\sigma }_3+{\left(\Delta {\sigma }_d\right)}_f$ (1)

Substitute $70\text{kN}/{\text{m}}^{\text{2}}$ for ${\sigma }_3$ and $155\text{kN}/{\text{m}}^{\text{2}}$ for ${\left(\Delta {\sigma }_d\right)}_f$ in Equation (1).

$\begin{array}{c}\left({{\sigma }^{\prime }}_1\right)=70+155\\ =225\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the normal stress $\left({{\sigma }^{\prime }}_f\right)$ on a plane inclined at $33°$ to the major principal plane for

specimen I using the equation.

${{\sigma }^{\prime }}_f=\frac{{{\sigma }^{\prime }}_1+{{\sigma }^{\prime }}_3}{2}+\frac{{{\sigma }^{\prime }}_1-{{\sigma }^{\prime }}_3}{2}\cos 2\theta$ (2)

Here, ${{\sigma }^{\prime }}_3$ and ${{\sigma }^{\prime }}_1$ minor and major effective principal stress, ${\varphi }^{\prime }$ is angle of friction, and $c^{\prime }$ is cohesion.

Substitute $225\text{kN}/{\text{m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_1$, $70\text{kN}/{\text{m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_3$, and $33°$ for $\theta$ in Equation (2).

$\begin{array}{c}{{\sigma }^{\prime }}_f=\frac{225+70}{2}+\frac{225-70}{2}\cos 2\left(33\right)\\ =147.5+31.522\\ =179\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Thus, the normal stress on a plane inclined at $33°$ to the major principal plane for specimen I is $\underset{\_}{179\text{kN}/{\text{m}}^{\text{2}}}$.

Find the shear stress $\left({\tau }_f\right)$ on a plane inclined at $33°$ to the major principal plane for

specimen I using the equation.

${\tau }_f=\frac{{{\sigma }^{\prime }}_1-{{\sigma }^{\prime }}_3}{2}\sin 2\theta$ (3)

Substitute $225\text{kN}/{\text{m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_1$, $70\text{kN}/{\text{m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_3$, $33°$ for $\theta$ in Equation (3).

$\begin{array}{c}{\tau }_f=\frac{225-70}{2}\sin 2\left(33\right)\\ =77.5\times 0.9135\\ =70.8\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Thus, the shear stress on a plane inclined at $33°$ to the major principal plane for specimen I is $\underset{\_}{70.8\text{kN}/{\text{m}}^{\text{2}}}$.

(b)

To determine

Find the normal and shear stress on the failure plane at failure for specimen II.

Answer to Problem 12.17P

The normal stress on the failure plane at failure for specimen II is $\underset{\_}{214.25\text{kN}/{\text{m}}^{\text{2}}}$.

The shear stress on the failure plane at failure for specimen II is $\underset{\_}{119\text{kN}/{\text{m}}^{\text{2}}}$.

Explanation of Solution

Calculation:

Show the Mohr-Coulomb failure expression to find the major effective principal stress for specimen I as shown below:

${{\sigma }^{\prime }}_1={{\sigma }^{\prime }}_3{\tan }^2\left(45+\frac{{\varphi }^{\prime }}{2}\right)+2c^{\prime }\left(45+\frac{{\varphi }^{\prime }}{2}\right)$ (4)

Here, ${{\sigma }^{\prime }}_3$ and ${{\sigma }^{\prime }}_1$ minor and major effective principal stress, ${\varphi }^{\prime }$ is angle of friction, and $c^{\prime }$ is cohesion.

Substitute ${\sigma }_3+{\left(\Delta {\sigma }_d\right)}_f$ for ${{\sigma }^{\prime }}_1$ in Equation (4).

${\sigma }_3+{\left(\Delta {\sigma }_d\right)}_f={{\sigma }^{\prime }}_3{\tan }^2\left(45+\frac{{\varphi }^{\prime }}{2}\right)+2c^{\prime }\left(45+\frac{{\varphi }^{\prime }}{2}\right)$ (5)

Substitute $70\text{kN}/{\text{m}}^{\text{2}}$ for ${\sigma }_3$ and $155\text{kN}/{\text{m}}^{\text{2}}$ for ${\left(\Delta {\sigma }_d\right)}_f$ in Equation (5).

$\begin{array}{l}70+155=70{\tan }^2\left(45+\frac{{\varphi }^{\prime }}{2}\right)+2c^{\prime }\left(45+\frac{{\varphi }^{\prime }}{2}\right)\\ 225=70{\tan }^2\left(45+\frac{{\varphi }^{\prime }}{2}\right)+2c^{\prime }\left(45+\frac{{\varphi }^{\prime }}{2}\right)\end{array}$ (6)

Express the Mohr-Coulomb failure for specimen II using Equation (5).

Substitute $140\text{kN}/{\text{m}}^{\text{2}}$ for ${\sigma }_3$ and $265\text{kN}/{\text{m}}^{\text{2}}$ for ${\left(\Delta {\sigma }_d\right)}_f$ in Equation (5).

$\begin{array}{l}140+265=140{\tan }^2\left(45+\frac{{\varphi }^{\prime }}{2}\right)+2c^{\prime }\left(45+\frac{{\varphi }^{\prime }}{2}\right)\\ 405=140{\tan }^2\left(45+\frac{{\varphi }^{\prime }}{2}\right)+2c^{\prime }\left(45+\frac{{\varphi }^{\prime }}{2}\right)\end{array}$ (7)

Subtract the Equation (6) from (7).

$\begin{array}{c}405-225=70{\tan }^2\left(45+\frac{{\varphi }^{\prime }}{2}\right)\\ 180=70{\tan }^2\left(45+\frac{{\varphi }^{\prime }}{2}\right)\\ {\left(\frac{180}{70}\right)}^{\frac{1}{2}}=\tan \left(45+\frac{{\varphi }^{\prime }}{2}\right)\\ 1.60=\tan \left(45+\frac{{\varphi }^{\prime }}{2}\right)\end{array}$

$\begin{array}{c}{\tan }^{-1}1.60=\left(45+\frac{{\varphi }^{\prime }}{2}\right)\\ 58.0497=\left(45+\frac{{\varphi }^{\prime }}{2}\right)\\ 13.04=\frac{{\varphi }^{\prime }}{2}\\ {\varphi }^{\prime }=26.09°\end{array}$

Calculate the angle of inclination on the failure plane using the equation:

$\theta =45+\frac{{\varphi }^{\prime }}{2}$ (8)

Substitute $26.09°$ for ${\varphi }^{\prime }$ in Equation (8).

$\begin{array}{c}\theta =45+\frac{26.09}{2}\\ =45+13.045\\ =58.04°\end{array}$

Find the major principal effective stress at failure $\left({{\sigma }^{\prime }}_1\right)$ for specimen II as shown below:

Substitute $140\text{kN}/{\text{m}}^{\text{2}}$ for ${\sigma }_3$ and $265\text{kN}/{\text{m}}^{\text{2}}$ for ${\left(\Delta {\sigma }_d\right)}_f$ in Equation (1).

$\begin{array}{c}\left({{\sigma }^{\prime }}_1\right)=140+265\\ =405\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the normal stress $\left({{\sigma }^{\prime }}_f\right)$ on the failure plane at failure for specimen II.

Substitute $405\text{kN}/{\text{m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_1$, $140\text{kN}/{\text{m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_3$, $58.04°$ for $\theta$ in Equation (2).

$\begin{array}{c}{{\sigma }^{\prime }}_f=\frac{405+140}{2}+\frac{405-140}{2}\cos 2\left(58.04\right)\\ =272.5+132.5\left(-0.439\right)\\ =272.5-58.16\\ =214.25\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Thus, the normal stress on the failure plane at failure $\left({{\sigma }^{\prime }}_f\right)$ for specimen 2 is $\underset{\_}{214.25\text{kN}/{\text{m}}^{\text{2}}}$.

Find the shear stress $\left({\tau }_f\right)$ on the failure plane at failure for specimen 2.

Substitute $405\text{kN}/{\text{m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_1$, $140\text{kN}/{\text{m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_3$, $58.04°$ for $\theta$ in Equation (3).

$\begin{array}{c}{\tau }_f=\frac{405-140}{2}\sin 2\left(58.04\right)\\ =132.5\times 0.898\\ =119\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Thus, the shear stress on the failure plane at failure $\left({\tau }_f\right)$ for specimen 2 is $\underset{\_}{119\text{kN}/{\text{m}}^{\text{2}}}$.