Chapter 7.2, Problem 38P

Solve Prob. 7.16, using Mohr's circle.

7.13 through 7.76 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 using Mohr’s circle.

Answer to Problem 38P

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:

Apply the procedure to construct the Mohr’s circle as shown below.

• Find the centre of the circle C located ${\sigma }_{\text{avg}}=\frac{{\sigma }_x+{\sigma }_y}{2}$ from the origin.
• Plot the reference points A having coordinates $A\left({\sigma }_x,{\tau }_A\right)$.
• Connect the point A with C and from the shaded triangle and find the radius R of the circle.
• Sketch the circle once R has been determined.

Construct the Mohr’s circle as shown below.

Calculate the centre of the circle $\left({\sigma }_{\text{avg}}\right)$ using average normal strain as shown below.

${\sigma }_{\text{avg}}=\frac{{\sigma }_x+{\sigma }_y}{2}$

Substitute $0$ for ${\sigma }_x$ and $-80\text{MPa}$ for ${\sigma }_y$.

$\begin{array}{c}{\sigma }_{\text{avg}}=\frac{0-80}{2}\\ =-40\text{MPa}\end{array}$

The centre of the circle is $C=-40\text{MPa}$.

Coordinates of the reference point X.

$X=\left({\sigma }_x,-{\tau }_{xy}\right)$

Substitute $0$ for ${\sigma }_x$ and $-50\text{MPa}$ for ${\tau }_{xy}$.

$\begin{array}{c}X=\left(0,-\left(-50\text{MPa}\right)\right)\\ =\left(\text{0, 50 MPa}\right)\end{array}$

Coordinates of the reference point Y.

$Y=\left({\sigma }_y,{\tau }_{xy}\right)$

Substitute $-80\text{MPa}$ for ${\sigma }_y$ and $-50\text{MPa}$ for ${\tau }_{xy}$.

$Y=\left(-80\text{MPa},-50\text{MPa}\right)$

Calculate the radius (R) of the circle as shown below.

$R=\sqrt{{\left({\sigma }_x-{\sigma }_{\text{avg}}\right)}^2+{\left({\tau }_{xy}\right)}^2}$

Substitute $0$ for ${\sigma }_x$, $-40\text{MPa}$ for ${\sigma }_{\text{avg}}$ and $-50\text{MPa}$ for ${\tau }_{xy}$.

$\begin{array}{c}R=\sqrt{{\left(0-\left(-40\right)\right)}^2+{\left(-50\right)}^2}\\ =\sqrt{4,100}\\ =64.031\text{MPa}\end{array}$

Sketch the Mohr’s circle as shown in Figure 1.

Refer to Figure 1.

Calculate the principle plane $\left({\theta }_p\right)$ as shown below.

$\begin{array}{l}\tan 2{\theta }_p=\frac{FX}{FC}\\ \tan 2{\theta }_p=\frac{50}{40}\\ 2{\theta }_p={\tan }^{-1}\left(1.25\right)\\ 2{\theta }_p=51.34°\end{array}$

Calculate the angle $\phi$ as shown below.

$\phi =2{\theta }_p-2\theta$

Substitute $51.34°$ for $2{\theta }_p$ and $25°$ for $\theta$

$\begin{array}{c}\phi =51.34°-2\times 25°\\ =1.34°\end{array}$

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

${\sigma }_{x^{\prime }}={\sigma }_{\text{avg}}+R\cos \phi$ (1)

Substitute $-40\text{MPa}$ for ${\sigma }_{\text{avg}}$, $64.031\text{MPa}$ for $R$, and $1.34°$ for $\theta$ in Equation (1).

$\begin{array}{c}{\sigma }_{x^{\prime }}=-40+64.031\cos 1.34°\\ =-40+64.01\\ =24\text{MPa}\end{array}$

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

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

${\sigma }_{y^{\prime }}={\sigma }_{\text{avg}}-R\cos \phi$ (2)

Substitute $-40\text{MPa}$ for ${\sigma }_{\text{avg}}$, $64.031\text{MPa}$ for $R$, and $1.34°$ for $\theta$ in Equation (2).

$\begin{array}{c}{\sigma }_{y^{\prime }}=-40-64.031\cos 1.34°\\ =-40-64.01\\ =-104\text{MPa}\end{array}$

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

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

${\tau }_{x^{\prime }{y}^{\prime }}=-R\sin \phi$ (3)

Substitute $64.031\text{MPa}$ for $R$ and $1.34°$ for $\theta$ in Equation (3).

$\begin{array}{c}{\tau }_{x^{\prime }{y}^{\prime }}=-64.031\sin 1.34°\\ =-1.5\text{MPa}\end{array}$

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

To determine

The normal and shearing stresses after the element has been rotated through $10°$ counter clockwise using Mohr’s circle.

Answer to Problem 38P

The normal stresses are $\underset{\_}{{\sigma }_{x^{\prime }}=-19.51\text{MPa}}$ and $\underset{\_}{{\sigma }_{y^{\prime }}=-60.49\text{MPa}}$.

The shear stress is $\underset{\_}{{\tau }_{x^{\prime }{y}^{\prime }}=-60.67\text{MPa}}$.

Explanation of Solution

Given information:

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

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

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

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

Calculation:

Refer to part (a).

Coordinates of the reference point $X=\left(0,50\text{MPa}\right)$

Coordinates of the reference point $Y=\left(-80\text{MPa},-50\text{MPa}\right)$

The principal plane $2{\theta }_p=51.34°$.

The average normal stress ${\sigma }_{\text{avg}}=-40\text{MPa}$.

The radius of the Mohr’s circle $R=64.031\text{MPa}$.

Sketch the Mohr’s circle as shown in Figure 2.

Refer to Figure 2.

Calculate the angle $\phi$ as shown below.

$\phi =2\theta +2{\theta }_p$

Substitute $51.34°$ for $2{\theta }_p$ and $10°$ for $\theta$

$\begin{array}{c}\phi =2\times 10°+51.34°\\ =71.34°\end{array}$

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

Substitute $-40\text{MPa}$ for ${\sigma }_{\text{avg}}$, $64.031\text{MPa}$ for $R$, and $71.34°$ for $\theta$ in Equation (1).

$\begin{array}{c}{\sigma }_{x^{\prime }}=-40+64.031\cos 71.34°\\ =-40+20.49\\ =-19.51\text{MPa}\end{array}$

Hence, the normal stress $\underset{\_}{{\sigma }_{x^{\prime }}=-19.51\text{MPa}}$.

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

Substitute $-40\text{MPa}$ for ${\sigma }_{\text{avg}}$, $64.031\text{MPa}$ for $R$, and $71.34°$ for $\theta$ in Equation (2).

$\begin{array}{c}{\sigma }_{y^{\prime }}=-40-64.031\cos 71.34°\\ =-40-20.49\\ =-60.49\text{MPa}\end{array}$

Hence, the normal stress $\underset{\_}{{\sigma }_{y^{\prime }}=-60.49\text{MPa}}$.

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

${\tau }_{x^{\prime }{y}^{\prime }}=-R\sin \phi$ (3)

Substitute $64.031\text{MPa}$ for $R$ and $71.34°$ for $\theta$ in Equation (3).

$\begin{array}{c}{\tau }_{x^{\prime }{y}^{\prime }}=-64.031\sin 71.34°\\ =-60.67\text{MPa}\end{array}$

Therefore, the shear stress $\underset{\_}{{\tau }_{x^{\prime }{y}^{\prime }}=-60.67\text{MPa}}$.