Chapter 7.1, Problem 15P

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. P7.15

To determine

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

Answer to Problem 15P

The normal stresses are $\underset{\_}{{\sigma }_{x^{\prime }}=9.03\text{ksi}}$ and $\underset{\_}{{\sigma }_{y^{\prime }}=-13.03\text{ksi}}$.

The shear stress is $\underset{\_}{{\tau }_{x^{\prime }{y}^{\prime }}=3.80\text{ksi}}$.

Explanation of Solution

Given information:

The stress component along x direction as ${\sigma }_x=8\text{ksi}$.

The stress component along y direction as ${\sigma }_y=-12\text{ksi}$.

The shear stress component as ${\tau }_{xy}=-6\text{ksi}$.

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 $8\text{ksi}$ for ${\sigma }_x$, $-12\text{ksi}$ for ${\sigma }_y$, $-6\text{ksi}$ for ${\tau }_{xy}$, and $-25°$ for $\theta$ in Equation (1).

$\begin{array}{c}{\sigma }_{x^{\prime }}=\frac{8-12}{2}+\frac{8-\left(-12\right)}{2}\cos \left(2\times -25°\right)+\left(-6\right)\sin \left(2\times -25°\right)\\ =-2+10\cos \left(-50°\right)-6\sin \left(-50°\right)\\ =-2+6.43+4.60\\ =9.03\text{ksi}\end{array}$

Hence, the normal stress as $\underset{\_}{{\sigma }_{x^{\prime }}=9.03\text{ksi}}$.

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 $8\text{ksi}$ for ${\sigma }_x$, $-12\text{ksi}$ for ${\sigma }_y$, $-6\text{ksi}$ for ${\tau }_{xy}$, and $-25°$ for $\theta$ in Equation (2).

$\begin{array}{c}{\sigma }_{y^{\prime }}=\frac{8-12}{2}-\frac{8-\left(-12\right)}{2}\cos \left(2\times -25°\right)-\left(-6\right)\sin \left(2\times -25°\right)\\ =-2-10\cos \left(-50°\right)+6\sin \left(-50°\right)\\ =-2-6.43-4.60\\ =-13.03\text{ksi}\end{array}$

Hence, the normal stress as $\underset{\_}{{\sigma }_{y^{\prime }}=-13.03\text{ksi}}$.

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 $8\text{ksi}$ for ${\sigma }_x$, $-12\text{ksi}$ for ${\sigma }_y$, $-6\text{ksi}$ for ${\tau }_{xy}$, and $-25°$ for $\theta$ in Equation (3).

$\begin{array}{c}{\tau }_{x^{\prime }{y}^{\prime }}=-\frac{8-\left(-12\right)}{2}\sin \left(2\times -25°\right)+\left(-6\right)\cos \left(2\times -25°\right)\\ =-10\sin \left(-50°\right)-6\cos \left(-50°\right)\\ =7.66-3.86\\ =3.80\text{ksi}\end{array}$

Therefore, the shear stress as $\underset{\_}{{\tau }_{x^{\prime }{y}^{\prime }}=3.80\text{ksi}}$.

To determine

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

Answer to Problem 15P

The normal stresses are $\underset{\_}{{\sigma }_{x^{\prime }}=5.35\text{ksi}}$ and $\underset{\_}{{\sigma }_{y^{\prime }}=-9.35\text{ksi}}$.

The shear stress is $\underset{\_}{{\tau }_{x^{\prime }{y}^{\prime }}=-9.06\text{ksi}}$.

Explanation of Solution

Given information:

The stress component along x direction as ${\sigma }_x=8\text{ksi}$.

The stress component along y direction as ${\sigma }_y=-12\text{ksi}$.

The shear stress component as ${\tau }_{xy}=-6\text{ksi}$.

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 $8\text{ksi}$ for ${\sigma }_x$, $-12\text{ksi}$ for ${\sigma }_y$, $-6\text{ksi}$ for ${\tau }_{xy}$, and $10°$ for $\theta$ in Equation (1).

$\begin{array}{c}{\sigma }_{x^{\prime }}=\frac{8-12}{2}+\frac{8-\left(-12\right)}{2}\cos \left(2\times 10°\right)+\left(-6\right)\sin \left(2\times 10°\right)\\ =-2+10\cos 20°-6\sin 20°\\ =-2+9.40-2.05\\ =5.35\text{ksi}\end{array}$

Hence, the normal stress as $\underset{\_}{{\sigma }_{x^{\prime }}=5.35\text{ksi}}$.

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

Substitute $8\text{ksi}$ for ${\sigma }_x$, $-12\text{ksi}$ for ${\sigma }_y$, $-6\text{ksi}$ for ${\tau }_{xy}$, and $10°$ for $\theta$ in Equation (2).

$\begin{array}{c}{\sigma }_{y^{\prime }}=\frac{8-12}{2}-\frac{8-\left(-12\right)}{2}\cos \left(2\times 10°\right)-\left(-6\right)\sin \left(2\times 10°\right)\\ =-2-10\cos 20°+6\sin 20°\\ =-2-9.40+2.05\\ =-9.35\text{ksi}\end{array}$

Hence, the normal stress as $\underset{\_}{{\sigma }_{y^{\prime }}=-9.35\text{ksi}}$.

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

Substitute $8\text{ksi}$ for ${\sigma }_x$, $-12\text{ksi}$ for ${\sigma }_y$, $-6\text{ksi}$ for ${\tau }_{xy}$, and $10°$ for $\theta$ in Equation (3).

$\begin{array}{c}{\tau }_{x^{\prime }{y}^{\prime }}=-\frac{8-\left(-12\right)}{2}\sin \left(2\times 10°\right)+\left(-6\right)\cos \left(2\times 10°\right)\\ =-10\sin 20°-6\cos 20°\\ =-3.42-5.64\\ =-9.06\text{ksi}\end{array}$

Therefore, the shear stress as $\underset{\_}{{\tau }_{x^{\prime }{y}^{\prime }}=-9.06\text{ksi}}$.