Chapter 7.1, Problem 14P

7.13 through 7.16 For the given state of stress, determine the normal and shearing stresses after the element shown has been rotated through (a) 25° clockwise, (b) 10° counterclockwise.

Fig. P7.14

To determine

The normal and shearing stresses after the element has been rotated through $25°$ clockwise.

Answer to Problem 14P

The normal stresses are $\underset{\_}{{\sigma }_{x^{\prime }}=-56.19\text{MPa}}$ and $\underset{\_}{{\sigma }_{y^{\prime }}=86.19\text{MPa}}$.

The shear stress is $\underset{\_}{{\tau }_{x^{\prime }{y}^{\prime }}=-38.17\text{MPa}}$.

Explanation of Solution

Given information:

The stress component along x direction as ${\sigma }_x=-60\text{MPa}$.

The stress component along y direction as ${\sigma }_y=90\text{MPa}$.

The shear stress component as ${\tau }_{xy}=30\text{MPa}$.

The orientation of the principal plane as $\theta =-25°$.

Calculation:

Calculate the normal stress along x direction $\left({\sigma }_{x^{\prime }}\right)$ as shown below.

${\sigma }_{x^{\prime }}=\frac{{\sigma }_x+{\sigma }_y}{2}+\frac{{\sigma }_x-{\sigma }_y}{2}\cos 2\theta +{\tau }_{xy}\sin 2\theta$ (1)

Substitute $-60\text{MPa}$ for ${\sigma }_x$, $90\text{MPa}$ for ${\sigma }_y$, $30\text{MPa}$ for ${\tau }_{xy}$, and $-25°$ for $\theta$ in Equation (1).

$\begin{array}{c}{\sigma }_{x^{\prime }}=\frac{-60+90}{2}+\frac{-60-90}{2}\cos \left(2\times -25°\right)+30\sin \left(2\times -25°\right)\\ =15-75\cos \left(-50°\right)+30\sin \left(-50°\right)\\ =15-48.21-22.98\\ =-56.19\text{MPa}\end{array}$

Hence, the normal stress as $\underset{\_}{{\sigma }_{x^{\prime }}=-56.19\text{MPa}}$.

Calculate the normal stress along y direction $\left({\sigma }_{y^{\prime }}\right)$ as shown below.

${\sigma }_{y^{\prime }}=\frac{{\sigma }_x+{\sigma }_y}{2}-\frac{{\sigma }_x-{\sigma }_y}{2}\cos 2\theta -{\tau }_{xy}\sin 2\theta$ (2)

Substitute $-60\text{MPa}$ for ${\sigma }_x$, $90\text{MPa}$ for ${\sigma }_y$, $30\text{MPa}$ for ${\tau }_{xy}$, and $-25°$ for $\theta$ in Equation (2).

$\begin{array}{c}{\sigma }_{y^{\prime }}=\frac{-60+90}{2}-\frac{-60-90}{2}\cos \left(2\times -25°\right)-30\sin \left(2\times -25°\right)\\ =15+75\cos \left(-50°\right)-30\sin \left(-50°\right)\\ =15+48.21+22.98\\ =86.19\text{MPa}\end{array}$

Hence, the normal stress as $\underset{\_}{{\sigma }_{y^{\prime }}=86.19\text{MPa}}$.

Calculate the shear stress $\left({\tau }_{x^{\prime }{y}^{\prime }}\right)$ as shown below.

${\tau }_{x^{\prime }{y}^{\prime }}=-\frac{{\sigma }_x-{\sigma }_y}{2}\sin 2\theta +{\tau }_{xy}\cos 2\theta$ (3)

Substitute $-60\text{MPa}$ for ${\sigma }_x$, $90\text{MPa}$ for ${\sigma }_y$, $30\text{MPa}$ for ${\tau }_{xy}$, and $-25°$ for $\theta$ in Equation (3).

$\begin{array}{c}{\tau }_{x^{\prime }{y}^{\prime }}=-\frac{-60-90}{2}\sin \left(2\times -25°\right)+30\cos \left(2\times -25°\right)\\ =75\sin \left(-50°\right)+30\cos \left(-50°\right)\\ =-57.45+19.28\\ =-38.17\text{MPa}\end{array}$

Therefore, the shear stress as $\underset{\_}{{\tau }_{x^{\prime }{y}^{\prime }}=-38.17\text{MPa}}$.

To determine

The normal and shearing stresses after the element has been rotated through $10°$ counter clockwise.

Answer to Problem 14P

The normal stresses are $\underset{\_}{{\sigma }_{x^{\prime }}=-45.22\text{MPa}}$ and $\underset{\_}{{\sigma }_{y^{\prime }}=75.22\text{MPa}}$.

The shear stress is $\underset{\_}{{\tau }_{x^{\prime }{y}^{\prime }}=53.84\text{MPa}}$.

Explanation of Solution

Given information:

The stress component along x direction as ${\sigma }_x=-60\text{MPa}$.

The stress component along y direction as ${\sigma }_y=90\text{MPa}$.

The shear stress component as ${\tau }_{xy}=30\text{MPa}$.

The orientation of the principal plane as $\theta =10°$.

Calculation:

Calculate the normal stress along x direction $\left({\sigma }_{x^{\prime }}\right)$ as shown below.

Substitute $-60\text{MPa}$ for ${\sigma }_x$, $90\text{MPa}$ for ${\sigma }_y$, $30\text{MPa}$ for ${\tau }_{xy}$, and $10°$ for $\theta$ in Equation (1).

$\begin{array}{c}{\sigma }_{x^{\prime }}=\frac{-60+90}{2}+\frac{-60-90}{2}\cos \left(2\times 10°\right)+30\sin \left(2\times 10°\right)\\ =15-75\cos 20°+30\sin 20°\\ =15-70.48+10.26\\ =-45.22\text{MPa}\end{array}$

Hence, the normal stress $\underset{\_}{{\sigma }_{x^{\prime }}=-45.22\text{MPa}}$.

Calculate the normal stress along y direction $\left({\sigma }_{y^{\prime }}\right)$ as shown below.

Substitute $-60\text{MPa}$ for ${\sigma }_x$, $90\text{MPa}$ for ${\sigma }_y$, $30\text{MPa}$ for ${\tau }_{xy}$, and $10°$ for $\theta$ in Equation (2).

$\begin{array}{c}{\sigma }_{y^{\prime }}=\frac{-60+90}{2}-\frac{-60-90}{2}\cos \left(2\times 10°\right)-30\sin \left(2\times 10°\right)\\ =15+75\cos 20°-30\sin 20°\\ =15+70.48-10.26\\ =75.22\text{MPa}\end{array}$

Hence, the normal stress as $\underset{\_}{{\sigma }_{y^{\prime }}=75.22\text{MPa}}$.

Calculate the shear stress $\left({\tau }_{x^{\prime }{y}^{\prime }}\right)$ as shown below.

Substitute $-60\text{MPa}$ for ${\sigma }_x$, $90\text{MPa}$ for ${\sigma }_y$, $30\text{MPa}$ for ${\tau }_{xy}$, and $10°$ for $\theta$ in Equation (3).

$\begin{array}{c}{\tau }_{x^{\prime }{y}^{\prime }}=-\frac{-60-90}{2}\sin \left(2\times 10°\right)+30\cos \left(2\times 10°\right)\\ =75\sin 20°+30\cos 20°\\ =25.65+28.19\\ =53.84\text{MPa}\end{array}$

Therefore, the shear stress as $\underset{\_}{{\tau }_{x^{\prime }{y}^{\prime }}=53.84\text{MPa}}$.