Chapter 7.1, Problem 16P

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. P7.16

To determine

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

Answer to Problem 16P

The normal stresses are $\underset{\_}{{\sigma }_{x^{\prime }}=24\text{MPa}}$ and $\underset{\_}{{\sigma }_{y^{\prime }}=-104\text{MPa}}$.

The shear stress is $\underset{\_}{{\tau }_{x^{\prime }{y}^{\prime }}=-1.5\text{MPa}}$.

Explanation of Solution

Given information:

The stress component along x direction as ${\sigma }_x=0$.

The stress component along y direction as ${\sigma }_y=-80\text{MPa}$.

The shear stress component as ${\tau }_{xy}=-50\text{MPa}$.

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 $0$ for ${\sigma }_x$, $-80\text{MPa}$ for ${\sigma }_y$, $-50\text{MPa}$ for ${\tau }_{xy}$, and $-25°$ for $\theta$ in Equation (1).

$\begin{array}{c}{\sigma }_{x^{\prime }}=\frac{0-80}{2}+\frac{0-\left(-80\right)}{2}\cos \left(2\times -25°\right)+\left(-50\right)\sin \left(2\times -25°\right)\\ =-40+40\cos \left(-50°\right)-50\sin \left(-50°\right)\\ =-40+25.71+38.30\\ =24\text{MPa}\end{array}$

Hence, the normal stress as $\underset{\_}{{\sigma }_{x^{\prime }}=24\text{MPa}}$.

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 $0$ for ${\sigma }_x$, $-80\text{MPa}$ for ${\sigma }_y$, $-50\text{MPa}$ for ${\tau }_{xy}$, and $-25°$ for $\theta$ in Equation (2).

$\begin{array}{c}{\sigma }_{y^{\prime }}=\frac{0-80}{2}-\frac{0-\left(-80\right)}{2}\cos \left(2\times -25°\right)-\left(-50\right)\sin \left(2\times -25°\right)\\ =-40-40\cos \left(-50°\right)+50\sin \left(-50°\right)\\ =-40-25.71-38.30\\ =-104\text{MPa}\end{array}$

Hence, the normal stress as $\underset{\_}{{\sigma }_{y^{\prime }}=-104\text{MPa}}$.

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 $0$ for ${\sigma }_x$, $-80\text{MPa}$ for ${\sigma }_y$, $-50\text{MPa}$ for ${\tau }_{xy}$, and $-25°$ for $\theta$ in Equation (3).

$\begin{array}{c}{\tau }_{x^{\prime }{y}^{\prime }}=-\frac{0-\left(-80\right)}{2}\sin \left(2\times -25°\right)+\left(-50\right)\cos \left(2\times -25°\right)\\ =-40\sin \left(-50°\right)-50\cos \left(-50°\right)\\ =30.64-32.14\\ =-1.5\text{MPa}\end{array}$

Therefore, the shear stress as $\underset{\_}{{\tau }_{x^{\prime }{y}^{\prime }}=-1.5\text{MPa}}$.

To determine

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

Answer to Problem 16P

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 ${\sigma }_x=0$.

The stress component along y direction as ${\sigma }_y=-80\text{MPa}$.

The shear stress component as ${\tau }_{xy}=-50\text{MPa}$.

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 $0$ for ${\sigma }_x$, $-80\text{MPa}$ for ${\sigma }_y$, $-50\text{MPa}$ 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 as $\underset{\_}{{\sigma }_{x^{\prime }}=-45.22\text{MPa}}$.

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

Substitute $0$ for ${\sigma }_x$, $-80\text{MPa}$ for ${\sigma }_y$, $-50\text{MPa}$ 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 $0$ for ${\sigma }_x$, $-80\text{MPa}$ for ${\sigma }_y$, $-50\text{MPa}$ 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 $\underset{\_}{{\tau }_{x^{\prime }{y}^{\prime }}=53.84\text{MPa}}$.