Chapter 7.2, Problem 31P

Solve Probs. 7.5 and 7.9, using Mohr's circle.

7.5 through 7.8 For the given state of stress, determine (a) the principal planes, (b) the principal stresses.

7.9 through 7.12 For the given state of stress, determine (a) the orientation of the planes of maximum in-plane shearing stress, (b) the maximum in-plane shearing stress, (c) the corresponding normal stress.

Fig. P7.5 and P7.9

To determine

The principal planes of the state of stress using Mohr’s circle.

Answer to Problem 31P

The principal planes of the state of stress using Mohr’s circle is $\underset{\_}{{\theta }_a=-37°\text{and }{\theta }_b=53°}$.

Explanation of Solution

Given information:

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

The stress component along y direction ${\sigma }_y=-40\text{MPa}$.

The shear stress component ${\tau }_{xy}=35\text{MPa}$.

Calculation:

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

• Find the center 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 $-60\text{MPa}$ for ${\sigma }_x$ and $-40\text{MPa}$ for ${\sigma }_y$.

$\begin{array}{c}{\sigma }_{\text{avg}}=\frac{-60+\left(-40\right)}{2}\\ =\frac{-100}{2}\\ =-50\text{MPa}\end{array}$

The centre of the circle is $C=-50\text{MPa}$.

Coordinates of the reference point X.

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

Substitute $-60\text{MPa}$ for ${\sigma }_x$ and $35\text{MPa}$ for ${\tau }_{xy}$.

$X=\left(-60\text{MPa},-35\text{MPa}\right)$

Coordinates of the reference point Y.

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

Substitute $-40\text{MPa}$ for ${\sigma }_x$ and $35\text{MPa}$ for ${\tau }_{xy}$.

$Y=\left(-40\text{MPa},35\text{MPa}\right)$

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 $-40\text{MPa}$ for ${\sigma }_x$, $-50\text{MPa}$ for ${\sigma }_{\text{avg}}$, and $35\text{MPa}$ for ${\tau }_{xy}$.

$\begin{array}{c}R=\sqrt{{\left(-40-\left(-50\right)\right)}^2+{\left(35\right)}^2}\\ =\sqrt{1,325}\\ =36.4\text{MPa}\end{array}$

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

Refer to Figure 1.

Calculate the angle $\beta$ as shown below.

$\begin{array}{l}\tan \beta =\frac{GX}{CG}\\ \tan \beta =\frac{35}{10}\\ \beta ={\tan }^{-1}\left(3.5\right)\\ \beta =74.05°\end{array}$

Calculate the angle $\alpha$ as shown below.

$\alpha =180°-\beta$

Here, $\beta$ is the angle of CX with respect to BC.

Substitute $74.05°$ for $\beta$.

$\begin{array}{c}\alpha =180°-74.05°\\ =105.95°\end{array}$

Calculate the principal plane $\left({\theta }_a\right)$ of the state of stress as shown below.

${\theta }_b=-\frac{1}{2}\beta$

Substitute $74.05°$ for $\beta$.

$\begin{array}{c}{\theta }_b=-\frac{74.05}{2}\\ =-37.03°\end{array}$

Calculate the principal planes $\left({\theta }_b\right)$ of the state of stress as shown below.

${\theta }_a=\frac{1}{2}\alpha$

Substitute $105.95°$ for $\alpha$.

$\begin{array}{c}{\theta }_a=\frac{1}{2}\times 105.95°\\ =52.975°\\ =53°\end{array}$

Hence, the principal planes of the state of stress using Mohr’s circle is $\underset{\_}{{\theta }_a=-37°\text{and }{\theta }_b=53°}$.

To determine

The principal stresses of the state of stress using Mohr’s circle.

Answer to Problem 31P

The maximum principal stress is $\underset{\_}{{\sigma }_{\max }=-13.6\text{MPa}}$.

The minimum principal stress is $\underset{\_}{{\sigma }_{\min }=-86.4\text{MPa}}$.

Explanation of Solution

Given information:

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

The stress component along y direction ${\sigma }_y=-40\text{MPa}$.

The shear stress component ${\tau }_{xy}=35\text{MPa}$.

Calculation:

Refer to part (a).

The average stress is ${\sigma }_{\text{avg}}=-50\text{MPa}$.

The radius of the Mohr’s circle is $R=36.4\text{MPa}$.

Calculate the principal stresses $\left({\sigma }_{\max }\text{and }{\sigma }_{\min }\right)$ as shown below.

${\sigma }_{\max ,\min }={\sigma }_{\text{avg}}\pm R$

Substitute $-50\text{MPa}$ for ${\sigma }_{\text{avg}}$ and $36.4\text{MPa}$ for $R$.

${\sigma }_{\max ,\min }=-50\pm 36.4$

Calculate the maximum principal stress as shown below.

$\begin{array}{c}{\sigma }_{\max }=-50+36.4\\ =-13.6\text{MPa}\end{array}$

Hence, the maximum principal stress is $\underset{\_}{{\sigma }_{\max }=-13.6\text{MPa}}$.

Calculate the minimum principal stress as shown below.

$\begin{array}{c}{\sigma }_{\min }=-50-36.4\\ =-86.4\text{MPa}\end{array}$

Hence, the minimum principal stress is $\underset{\_}{{\sigma }_{\min }=-86.4\text{MPa}}$.

To determine

The orientation of the planes of maximum in-plane shearing stress using Mohr’s circle.

Answer to Problem 31P

The orientation of the planes of maximum in-plane shearing stress is $\underset{\_}{{\theta }_d=8°\text{and }{\theta }_e=\text{98}°}$.

Explanation of Solution

Given information:

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

The stress component along y direction ${\sigma }_y=-40\text{MPa}$.

The shear stress component ${\tau }_{xy}=35\text{MPa}$.

Calculation:

Refer to part (a).

The principal planes ${\theta }_a=53°$ and ${\theta }_b=-37°$.

Calculate the orientation of the planes of maximum in-plane shearing stress $\left({\theta }_d\right)$ as shown below.

${\theta }_d={\theta }_b+45°$

Substitute $-37°$ for ${\theta }_b$.

$\begin{array}{c}{\theta }_d=-37°+45°\\ =8°\end{array}$

Calculate the orientation of the planes of maximum in-plane shearing stress $\left({\theta }_e\right)$ as shown below.

${\theta }_e={\theta }_a+45°$

Substitute $53°$ for ${\theta }_a$.

$\begin{array}{c}{\theta }_e=53°+45°\\ =98°\end{array}$

Hence, the orientation of the planes of maximum in-plane shearing stress is $\underset{\_}{{\theta }_d=8°\text{and}{\theta }_e=98°}$.

To determine

The maximum in-plane shearing stress using Mohr’s circle.

Answer to Problem 31P

The maximum in-plane shearing stress is $\underset{\_}{{\tau }_{\max }=36.4\text{MPa}}$.

Explanation of Solution

Given information:

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

The stress component along y direction ${\sigma }_y=-40\text{MPa}$.

The shear stress component ${\tau }_{xy}=35\text{MPa}$.

Calculation:

Refer to part (a).

The maximum in-plane shearing stress is ${\tau }_{\max }=R$$=36.4\text{MPa}$.

Hence, the maximum in-plane shearing stress is $\underset{\_}{{\tau }_{\max }=36.4\text{MPa}}$.

To determine

The normal stress using Mohr’s circle.

Answer to Problem 31P

The normal stress is $\underset{\_}{{\sigma }^{\prime }=-50\text{MPa}}$.

Explanation of Solution

Given information:

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

The stress component along y direction ${\sigma }_y=-40\text{MPa}$.

The shear stress component ${\tau }_{xy}=35\text{MPa}$.

Calculation:

Refer to part (a).

The average normal stress is ${\sigma }_{\text{avg}}=-50\text{MPa}$.

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

${\sigma }^{\prime }={\sigma }_{\text{avg}}$

Substitute $-50\text{MPa}$ for ${\sigma }_{\text{avg}}$.

${\sigma }^{\prime }=-50\text{MPa}$

Therefore, the normal stress is $\underset{\_}{{\sigma }^{\prime }=-50\text{MPa}}$.