Chapter 7, Problem 7.7.31P

Solve Problem 7.7-13 by using Mohr’s circle for plane strain.

To determine

The principle strains and maximum shear strain.

Answer to Problem 7.7.31P

The maximum principle strain is $568.47\times {10}^{-6}$ .

The minimum principle strain is $-18.47\times {10}^{-6}$ .

The maximum shear strain is $586.94\times {10}^{-6}$ .

The minimum shear strain is $-586.94\times {10}^{-6}$ .

Explanation of Solution

Given information:

The normal strain along x-axis is $480\times {10}^{-6}$ , normal strain along y-axis is $70\times {10}^{-6}$ ,and shear strain along xy-plane is $420\times {10}^{-6}$ and orientation of element is $75°$ .

Explanation:

The following are the steps to draw the Mohr’s circle

1. Draw the x-axis as normal strain and y-axis as shear strain and indicate the origin as point O $\left(0,0\right)$ .
2. Write the expression to locate the center point C of the circle from the origin O(ie., OC).
$\text{OC}={\epsilon }_{avg}=\frac{{\epsilon }_x+{\epsilon }_y}{2}$ ……(I)

Here, the normal strain along x-axis is ${\epsilon }_x$ , normal strain along y-axis is ${\epsilon }_y$ , and average strain is ${\epsilon }_{avg}$ .

3. Mark the center point O: $\left({\epsilon }_{avg},0\right)$ .
4. Write the expression for the radius of Mohr’s circle
$R=\sqrt{{\left({\epsilon }_x-{\epsilon }_{avg}\right)}^2+{\left(\frac{{\gamma }_{xy}}{2}\right)}^2}$ .....(II)

Here, the radius of the Mohr’s circle is $R$ and shear strain along xy-plane is ${\gamma }_{xy}$ .

Write the expression for principle angle.

  $\alpha ={\tan }^{-1}\left(\frac{\frac{{\gamma }_{xy}}{2}}{{\epsilon }_x-{\epsilon }_{avg}}\right)$ .....(III)

Here, principle plane angle is $\alpha$ .

Write the expression for angle of oriented strain plane.

  $\beta =\alpha +180°-2\theta$ .....(IV)

Here, angle of oriented strain plane is $\beta$ and $\theta$ is the orientation of element.

Write the expression for the strain at point D.

  ${\left({\epsilon }_x\right)}_1={\epsilon }_{avg}-R\cos \beta$ .....(V)

Here, the strain at point D is ${\left({\epsilon }_x\right)}_1$

Write the expression for the strain at point $\text{<msup><mi> D</mi><mo> ′</mo></msup>}$ .

  ${\left({\epsilon }_x\right)}_2={\epsilon }_{avg}+R\cos \beta$ ...... (VI)

Here, the strain at point $\text{<msup><mi> D</mi><mo> ′</mo></msup>}$ is ${\left({\epsilon }_x\right)}_2$ .

Write the expression for the shear strain at point D.

  $\frac{{\left({\gamma }_{xy}\right)}_1}{2}=-R\sin \beta$ ...... (VII)

Here, the shear strain at point D ${\left({\gamma }_{xy}\right)}_1$

Write the expression for the shear strain at point $\text{<msup><mi> D</mi><mo> ′</mo></msup>}$

  $\frac{{\left({\gamma }_{xy}\right)}_2}{2}=-R\sin \beta$ ...... (VIII)

Here, the shear strain at point $\text{<msup><mi> D</mi><mo> ′</mo></msup>}$

  ${\left({\gamma }_{xy}\right)}_2$ .

Write the expression for the maximum principle strain.

  ${\epsilon }_1={\epsilon }_{avg}+R$ ...... (IX)

Here, the maximum principle strain is ${\epsilon }_1$ .

Write the expression for the minimum principle strain.

  ${\epsilon }_2={\epsilon }_{avg}-R$ ...... (X)

Here, the minimum principle strain is ${\epsilon }_2$ .

Write the expression for maximum shear strain.

  ${\gamma }_{\max }=2R$ ...... (XI)

Here, maximum shear strain is. ${\gamma }_{\max }$ .

Write the expression for minimum shear strain.

  ${\gamma }_{\min }=-2R$ ...... (XII)

Here, minimum shear strain is. ${\gamma }_{\min }$ .

Write the expression for the maximum principle strain plane angle.

  $2{\theta }_{p_1}=180°-\alpha$ ...... (XIII)

Here, maximum shear strain plane angle is ${\theta }_{p_1}$ .

Write the expression for the minimum principle strain plane angle.

  $2{\theta }_{p_2}=180°+2{\theta }_{p_1}$ ...... (XIV)

Here, minimum shear strain plane angle is ${\theta }_{p_2}$ .

Write the expression for maximum shear strain plane angle.

  $2{\theta }_{s_1}=90°-\alpha$ ...... (XV)

Here, maximum shear strain plane angle is ${\theta }_{s_1}$ .

Write the expression for minimum shear strain plane angle.

  $2{\theta }_{s_2}=90°+2{\theta }_{s_1}$ ...... (XVI)

Here, minimum shear strain plane angle is ${\theta }_{s_2}$ .

Calculation:

Substitute $480\times {10}^{-6}$ for ${\epsilon }_x$ , and $70\times {10}^{-6}$ for ${\epsilon }_y$ in Equation (I).

  $\begin{array}{c}{\epsilon }_{avg}=\frac{\left(480\times {10}^{-6}\right)+\left(70\times {10}^{-6}\right)}{2}\\ =\frac{550\times {10}^{-6}}{2}\\ =275\times {10}^{-6}\end{array}$

Substitute $480\times {10}^{-6}$ for ${\epsilon }_x$ , $275\times {10}^{-6}$ for ${\epsilon }_{avg}$ and $420\times {10}^{-6}$ for ${\gamma }_{xy}$ in Equation (II).

  $\begin{array}{c}R=\sqrt{{\left(480\times {10}^{-6}-275\times {10}^{-6}\right)}^2+{\left(\frac{420\times {10}^{-6}}{2}\right)}^2}\\ =\sqrt{{\left(205\times {10}^{-6}\right)}^2+{\left(210\times {10}^{-6}\right)}^2}\\ =293.47\times {10}^{-6}\end{array}$

Substitute $480\times {10}^{-6}$ for ${\epsilon }_x$ , $275\times {10}^{-6}$ for ${\epsilon }_{avg}$ and $420\times {10}^{-6}$ for ${\gamma }_{xy}$ in Equation (III).

  $\begin{array}{c}\alpha ={\tan }^{-1}\frac{\left(\frac{420\times {10}^{-6}}{2}\right)}{480\times {10}^{-6}-275\times {10}^{-6}}\\ ={\tan }^{-1}1.04\\ =45.69°\end{array}$

Substitute $45.69°$ for $\alpha$ and $75°$ for $\theta$ in Equation (IV).

  $\begin{array}{c}\beta =45.69°+180°-2\left(75°\right)\\ =225.69°-150°\\ =75.69°\end{array}$

Substitute $275\times {10}^{-6}$ for ${\epsilon }_{avg}$ , $293.47\times {10}^{-6}$ for $R$ and $75.69°$ for $\beta$ in Equation (V).

  $\begin{array}{c}{\left({\epsilon }_x\right)}_1=275\times {10}^{-6}-\left(293.47\times {10}^{-6}\cos 75.69°\right)\\ =275\times {10}^{-6}-72.54\times {10}^{-6}\\ =202.46\times {10}^{-6}\end{array}$

Substitute $275\times {10}^{-6}$ for ${\epsilon }_{avg}$ , $293.47\times {10}^{-6}$ for $R$ and $75.69°$ for $\beta$ in Equation (VI).

  $\begin{array}{c}{\left({\epsilon }_x\right)}_2=275\times {10}^{-6}+\left(293.47\times {10}^{-6}\cos 75.69°\right)\\ =275\times {10}^{-6}+72.54\times {10}^{-6}\\ =347.54\times {10}^{-6}\end{array}$

Substitute $293.47\times {10}^{-6}$ for $R$ and $75.69°$ for $\beta$ in Equation (VII).

  $\begin{array}{c}\frac{{\left({\gamma }_{xy}\right)}_1}{2}=-\left(293.47\times {10}^{-6}\sin 75.69°\right)\\ {\left({\gamma }_{xy}\right)}_1=-\left(2\times 28436\times {10}^{-6}\right)\\ =-568.72\times {10}^{-6}\end{array}$

Substitute $293.47\times {10}^{-6}$ for $R$ and $75.69°$ for $\beta$ in Equation (VIII).

  $\begin{array}{c}\frac{{\left({\gamma }_{xy}\right)}_2}{2}=-\left(293.47\times {10}^{-6}\sin 75.69°\right)\\ {\left({\gamma }_{xy}\right)}_2=-\left(2\times 28436\times {10}^{-6}\right)\\ =-568.72\times {10}^{-6}\end{array}$

Substitute $275\times {10}^{-6}$ for ${\epsilon }_{avg}$ and $293.47\times {10}^{-6}$ for $R$ in Equation (IX).

  $\begin{array}{c}{\epsilon }_1=275\times {10}^{-6}+293.47\times {10}^{-6}\\ =568.47\times {10}^{-6}\end{array}$

Substitute $275\times {10}^{-6}$ for ${\epsilon }_{avg}$ and $293.47\times {10}^{-6}$ for $R$ in Equation (X).

  $\begin{array}{c}{\epsilon }_2=275\times {10}^{-6}-293.47\times {10}^{-6}\\ =-18.47\times {10}^{-6}\end{array}$

Substitute $293.47\times {10}^{-6}$ for $R$ in Equation (VI).

  $\begin{array}{c}{\gamma }_{\max }=2\left(293.47\times {10}^{-6}\right)\\ =586.94\times {10}^{-6}\end{array}$

Substitute $337.08\times {10}^{-6}$ for $R$ in Equation (VII).

  $\begin{array}{c}{\gamma }_{\min }=-2\left(293.47\times {10}^{-6}\right)\\ =-586.94\times {10}^{-6}\end{array}$

Substitute $45.69°$ for $\alpha$ in Equation (VIII).

  $\begin{array}{c}2{\theta }_{p_1}=180°-\left(45.69°\right)\\ {\theta }_{p_1}=\frac{134.31°}{2}\\ =67.15°\end{array}$

Substitute $67.15°$ for ${\theta }_{p_1}$ in Equation (IX)

  $\begin{array}{c}2{\theta }_{p_2}=180°+67.15°\\ {\theta }_{p_2}=\frac{247.15°}{2}\\ =123.57°\end{array}$

Substitute $45.69°$ for $\alpha$ in Equation (X).

  $\begin{array}{c}2{\theta }_{s_1}=90°-45.69°\\ {\theta }_{s_1}=\frac{44.31°}{2}\\ =22.15°\end{array}$

Substitute $22.15°$ for ${\theta }_{s_1}$ in Equation (XI).

  $\begin{array}{c}2{\theta }_{s_1}=90°+22.15°\\ {\theta }_{s_1}=\frac{112.15°}{2}\\ =56.07°\end{array}$

The following figure shows the Mohr ‘circle.

  

     Figure-(1)

The following figure shows the strain on oriented plane.

  

     Figure-(2)

Conclusion:

The maximum principle strain is $568.47\times {10}^{-6}$ .

The minimum principle strain is $-18.47\times {10}^{-6}$ .

The maximum shear strain is $586.94\times {10}^{-6}$ .

The minimum shear strain is $-586.94\times {10}^{-6}$ .