Chapter 10, Problem 10.1P

(a)

To determine

Calculate the maximum and minimum principal stresses.

Answer to Problem 10.1P

The maximum principal stress $\left({\sigma }_1\right)$ is $\underset{\_}{200.72\text{kN}/{\text{m}}^2}$.

The minimum principal stress $\left({\sigma }_3\right)$ is $\underset{\_}{116.28\text{kN}/{\text{m}}^2}$.

Explanation of Solution

Given information:

The normal stress along x axis $\left({\sigma }_x\right)$ is $172\text{kN}/{\text{m}}^2$.

The normal stress along y axis $\left({\sigma }_y\right)$ is $145\text{kN}/{\text{m}}^2$.

The shear stress along xy axis $\left({\tau }_{xy}\right)$ is $+40\text{kN}/{\text{m}}^2$.

Calculation:

Find the horizontal angle as follows:

$\begin{array}{c}\theta =90°-68°\\ =22°\end{array}$

Calculate the maximum principal stress $\left({\sigma }_1\right)$ using the relation.

${\sigma }_1=\frac{{\sigma }_y+{\sigma }_x}{2}+\sqrt{{\left[\frac{{\sigma }_y+{\sigma }_x}{2}\right]}^2+{\tau }_{xy}^2}$

Substitute $172\text{kN}/{\text{m}}^2$ for ${\sigma }_x$, $145\text{kN}/{\text{m}}^2$ for ${\sigma }_y$, and $40\text{kN}/{\text{m}}^2$ for ${\tau }_{xy}$.

$\begin{array}{c}{\sigma }_1=\frac{145+172}{2}+\sqrt{{\left[\frac{145-172}{2}\right]}^2+{40}^2}\\ =158.5+\sqrt{1,782.25}\\ =158.5+42.22\\ =200.72\text{kN}/{\text{m}}^2\end{array}$

Hence, the maximum principal stress $\left({\sigma }_1\right)$ is $\underset{\_}{200.72\text{kN}/{\text{m}}^2}$.

Calculate the minimum principal stress $\left({\sigma }_3\right)$ using the relation.

${\sigma }_3=\frac{{\sigma }_y+{\sigma }_x}{2}-\sqrt{{\left[\frac{{\sigma }_y+{\sigma }_x}{2}\right]}^2+{\tau }_{xy}^2}$

Substitute $172\text{kN}/{\text{m}}^2$ for ${\sigma }_x$, $145\text{kN}/{\text{m}}^2$ for ${\sigma }_y$, and $40\text{kN}/{\text{m}}^2$ for ${\tau }_{xy}$.

$\begin{array}{c}{\sigma }_3=\frac{145+172}{2}-\sqrt{{\left[\frac{145-172}{2}\right]}^2+{40}^2}\\ =158.5-\sqrt{1,782.25}\\ =158.5-42.22\\ =116.28\text{kN}/{\text{m}}^2\end{array}$

Hence, the minimum principal stress $\left({\sigma }_3\right)$ is $\underset{\_}{116.28\text{kN}/{\text{m}}^2}$.

(b)

To determine

Calculate the normal and shear stresses on plane AB.

Answer to Problem 10.1P

The normal stress on plane AB $\left({\sigma }_n\right)$ is $\underset{\_}{176.6\text{kN}/{\text{m}}^2}$.

The shear stress on plane AB $\left({\tau }_n\right)$ is $\underset{\_}{-38.15\text{kN}/{\text{m}}^2}$.

Explanation of Solution

Given information:

The normal stress along x axis $\left({\sigma }_x\right)$ is $172\text{kN}/{\text{m}}^2$.

The normal stress along y axis $\left({\sigma }_y\right)$ is $145\text{kN}/{\text{m}}^2$.

The shear stress along xy axis $\left({\tau }_{xy}\right)$ is $+40\text{kN}/{\text{m}}^2$.

The angle $\left(\theta \right)$ is $22°$.

Calculation:

Calculate the normal stress $\left({\sigma }_n\right)$ using the relation.

${\sigma }_n=\frac{{\sigma }_y+{\sigma }_x}{2}+\frac{{\sigma }_y-{\sigma }_x}{2}\cos 2\theta +{\tau }_{xy}\sin 2\theta$

Substitute $172\text{kN}/{\text{m}}^2$ for ${\sigma }_x$, $145\text{kN}/{\text{m}}^2$ for ${\sigma }_y$, $40\text{kN}/{\text{m}}^2$ for ${\tau }_{xy}$, and $22°$ for $\theta$.

$\begin{array}{c}{\sigma }_n=\frac{145+172}{2}+\frac{145-172}{2}\cos \left(2\times 22°\right)+40\sin \left(2\times 22°\right)\\ =158.5-13.5\cos 44°+40\sin 44°\\ =176.6\text{kN}/{\text{m}}^2\end{array}$

Hence, the normal stress on plane AB $\left({\sigma }_n\right)$ is $\underset{\_}{176.6\text{kN}/{\text{m}}^2}$.

Calculate the shear stress $\left({\tau }_n\right)$ using the relation.

${\tau }_n=\frac{{\sigma }_y-{\sigma }_x}{2}\sin 2\theta -{\tau }_{xy}\cos 2\theta$

Substitute $172\text{kN}/{\text{m}}^2$ for ${\sigma }_x$, $145\text{kN}/{\text{m}}^2$ for ${\sigma }_y$, $40\text{kN}/{\text{m}}^2$ for ${\tau }_{xy}$, and $22°$ for $\theta$.

$\begin{array}{c}{\tau }_n=\frac{145-172}{2}\sin \left(2\times 22°\right)-40\cos \left(2\times 22°\right)\\ =-13.5\sin 44°-40\cos 44°\\ =-38.15\text{kN}/{\text{m}}^2\end{array}$

Therefore, the shear stress on plane AB $\left({\tau }_n\right)$ is $\underset{\_}{-38.15\text{kN}/{\text{m}}^2}$.