Chapter 7, Problem 7.7.3P

An element of material in plain strain is subjected to shear strain

   ${\gamma }_{xy}$ = 0.0003.

(a) Determine the strains for an element oriented at an angle θ= 30°.

(b) Determine the principal strains of the clement. Confirm the solution using Mohr’s circle for plane strain.

To determine

The strain for an element for an element inclined at an angle of $20°$ .

Answer to Problem 7.7.3P

The rotated strain along $x_1$ -direction is $1.398\times {10}^{-3}$ .

The rotated strain along $y_1$ -direction is $-9.76\times {10}^{-5}$ .

The rotated shear strain along $x_1{y}_1$ plane is $-8.63\times {10}^{-4}$ .

Explanation of Solution

Given information:

The strain along x-direction is $0.0015$ , the strain along y-direction is $-0.0002$ ,

The shear strain along xy-plane is $0.0003$ , the angle of orientation is $20°$ .

Explanation:

Write the expression for strain along $x_1$ -direction.

  ${\epsilon }_{x_1}=\frac{{\epsilon }_x+{\epsilon }_y}{2}+\frac{{\epsilon }_x-{\epsilon }_y}{2}\cos \left(2\theta \right)+\frac{{\gamma }_{xy}}{2}\sin \left(2\theta \right)$ ...... (I)

Write the expression for strain along $y_1$ -direction.

  ${\epsilon }_{y_1}=\frac{{\epsilon }_x+{\epsilon }_y}{2}+\frac{{\epsilon }_x-{\epsilon }_y}{2}\cos \left(2\left(\theta +90°\right)\right)+\frac{{\gamma }_{xy}}{2}\sin \left(2\left(\theta +90°\right)\right)$ ...... (II)

Write the expression for shear strain along $x_1{y}_1$ direction.

  ${\gamma }_{x_1{y}_1}=2\left\{-\frac{{\epsilon }_x-{\epsilon }_y}{2}\sin \left(2\theta \right)+\frac{{\gamma }_{xy}}{2}\cos \left(2\theta \right)\right\}$ ...... (III)

Calculation:

Substitute $0.0015$ for ${\epsilon }_x$ , $-0.0002$ for ${\epsilon }_y$ , $0.0003$ for ${\gamma }_{xy}$ , $20°$ for $\theta$ in Equation (I).

  $\begin{array}{c}{\epsilon }_{x_1}=\left[\begin{array}{l}\frac{0.0015-0.0002}{2}+\frac{0.0015+0.0002}{2}\cos \left(2\times 20°\right)\\ +\frac{0.0003}{2}\sin \left(2\times 20°\right)\end{array}\right]\\ =6.5\times {10}^{-4}+6.5114\times {10}^{-4}+9.642\times {10}^{-5}\\ =1.398\times {10}^{-3}\end{array}$

Substitute $0.0015$ for ${\epsilon }_x$ , $-0.0002$ for ${\epsilon }_y$ , $0.0003$ for ${\gamma }_{xy}$ , $20°$ for $\theta$ in Equation (II).

  $\begin{array}{c}{\epsilon }_{y_1}=\left[\begin{array}{l}\frac{0.0015-0.0002}{2}+\frac{0.0015+0.0002}{2}\cos \left(2\times \left(20°+90°\right)\right)\\ +\frac{0.0003}{2}\sin \left(2\times \left(20°+90°\right)\right)\end{array}\right]\\ =6.5\times {10}^{-4}-6.5114\times {10}^{-4}-9.642\times {10}^{-5}\\ =-9.76\times {10}^{-5}\end{array}$

Substitute $0.0015$ for ${\epsilon }_x$ , $-0.0002$ for ${\epsilon }_y$ , $0.0003$ for ${\gamma }_{xy}$ , $20°$ for $\theta$ in Equation (III).

  $\begin{array}{c}{\gamma }_{x_1{y}_1}=2\left\{-\frac{0.0015+0.0002}{2}\sin \left(2\times 20°\right)+\frac{0.0003}{2}\cos \left(2\times 20°\right)\right\}\\ =2\left\{-5.464\times {10}^{-4}+1.149\times {10}^{-4}\right\}\\ =-8.63\times {10}^{-4}\end{array}$

Conclusion:

The rotated strain along $x_1$ -direction is $1.398\times {10}^{-3}$ .

The rotated strain along $y_1$ -direction is $-9.76\times {10}^{-5}$ .

The rotated shear strain along $x_1{y}_1$ plane is $-8.63\times {10}^{-4}$ .

To determine

The principal strain of the element.

The Mohr’s circle for plan strain.

Answer to Problem 7.7.3P

The maximum principal strain is $1.513\times {10}^{-3}$ .

The minimum principal strain is $-2.13\times {10}^{-3}$ .

The maximum shear strain is $1.726\times {10}^{-3}$ .

The principal angle is $5.004°$ .

The radius of Mohr’s circle is $8.63\times {10}^{-4}$ .

Explanation of Solution

Write the expression for maximum principal strain.

  ${\epsilon }_1=\frac{{\epsilon }_x+{\epsilon }_y}{2}+\sqrt{{\left(\frac{{\epsilon }_x-{\epsilon }_y}{2}\right)}^2+{\left(\frac{{\gamma }_{xy}}{2}\right)}^2}$ ...... (IV)

Here, the strain along x and y-direction is ${\epsilon }_x$ and ${\epsilon }_y$ , the angle of orientation is $\theta$ ,

Write the expression for strain along $y_1$ -direction.

Write the expression for minimum principal strain.

  ${\epsilon }_2=\frac{{\epsilon }_x+{\epsilon }_y}{2}-\sqrt{{\left(\frac{{\epsilon }_x-{\epsilon }_y}{2}\right)}^2+{\left(\frac{{\gamma }_{xy}}{2}\right)}^2}$ ...... (V)

Write the expression for maximum shear strain.

  ${\gamma }_{\max }=2\sqrt{{\left(\frac{{\epsilon }_x-{\epsilon }_y}{2}\right)}^2+{\left(\frac{{\gamma }_{xy}}{2}\right)}^2}$ ...... (VI)

Write the expression for principal angle.

  ${\theta }_p=\frac{1}{2}{\tan }^{-1}\left(\frac{{\gamma }_{xy}}{{\epsilon }_x-{\epsilon }_y}\right)$ ...... (VII)

Write the expression for Mohr’s circle.

  $R=\sqrt{{\left(\frac{{\epsilon }_x-{\epsilon }_y}{2}\right)}^2+{\left(\frac{{\gamma }_{xy}}{2}\right)}^2}$ ...... (VIII)

Write the expression for average strain.

  ${\epsilon }_{\text{avg}}=\frac{{\epsilon }_x+{\epsilon }_y}{2}$ ...... (IX)

Write the expression for rotated strain along $x_1$ -direction.

  ${\epsilon }_{x_1}={\epsilon }_{\text{avg}}-R\cos \left(2\theta -2{\theta }_p\right)$ ...... (X)

Write the expression for rotated strain along $y_1$ -direction.

  ${\epsilon }_{y_1}={\epsilon }_{\text{avg}}+R\cos \left(2\theta -2{\theta }_p\right)$ ...... (XI)

Write the expression for rotated shear strain along $xy$ -plane.

  ${\gamma }_{x_1{y}_1}=2R\sin \left(2\theta -2{\theta }_p\right)$ ...... (XII)

Write the expression for maximum principal strain.

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

Write the expression for minimum principal strain.

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

Write the expression for shear strain.

  ${\gamma }_{\max }=2R$ ...... (XV)

Write the expression for steps to construct the Mohr’s circle:

1. Draw a horizontal axis and consider it as $\epsilon$ axis.
2. Draw a vertical axis and consider it to be ${\gamma }_{xy}/2$
3. Mark the point $C\left({\epsilon }_{\text{avg}},0\right)$ .
4. With the help of radius and centre $C$ draw a circle.
5. Thus the Mohr’s circle is obtained.

Calculation:

Substitute, $0.0015$ for ${\epsilon }_x$ , $-0.0002$ for ${\epsilon }_y$ , $0.0003$ for ${\gamma }_{xy}$ in Equation (IV).

  $\begin{array}{c}{\epsilon }_1=\frac{0.0015-0.0002}{2}+\sqrt{{\left(\frac{0.0015+0.0002}{2}\right)}^2+{\left(\frac{0.0003}{2}\right)}^2}\\ =6.5\times {10}^{-4}+8.63\times {10}^{-4}\\ =1.513\times {10}^{-3}\end{array}$

Substitute, $0.0015$ for ${\epsilon }_x$ , $-0.0002$ for ${\epsilon }_y$ , $0.0003$ for ${\gamma }_{xy}$ in Equation (V).

  $\begin{array}{c}{\epsilon }_2=\frac{0.0015-0.0002}{2}-\sqrt{{\left(\frac{0.0015+0.0002}{2}\right)}^2+{\left(\frac{0.0003}{2}\right)}^2}\\ =6.5\times {10}^{-4}-8.63\times {10}^{-4}\\ =-2.13\times {10}^{-3}\end{array}$

Substitute, $0.0015$ for ${\epsilon }_x$ , $-0.0002$ for ${\epsilon }_y$ , $0.0003$ for ${\gamma }_{xy}$ in Equation (VI).

  $\begin{array}{c}{\gamma }_{\max }=2\sqrt{{\left(\frac{0.0015+0.0002}{2}\right)}^2+{\left(\frac{0.0003}{2}\right)}^2}\\ =2\left(8.63\times {10}^{-4}\right)\\ =1.726\times {10}^{-3}\end{array}$

Substitute, $0.0015$ for ${\epsilon }_x$ , $-0.0002$ for ${\epsilon }_y$ , $0.0003$ for ${\gamma }_{xy}$ in Equation (VII).

  $\begin{array}{c}{\theta }_p=\frac{1}{2}{\tan }^{-1}\left(\frac{0.0003}{0.0015+0.0002}\right)\\ =\frac{1}{2}{\tan }^{-1}\left(\frac{0.0003}{0.0017}\right)\\ =\frac{1}{2}{\tan }^{-1}\left(0.1765\right)\\ =5.04°\end{array}$

Substitute, $0.0015$ for ${\epsilon }_x$ , $-0.0002$ for ${\epsilon }_y$ , $0.0003$ for ${\gamma }_{xy}$ in Equation (VIII).

  $\begin{array}{c}R=\sqrt{{\left(\frac{0.0015+0.0002}{2}\right)}^2+{\left(\frac{0.0003}{2}\right)}^2}\\ =\sqrt{{\left(0.00085\right)}^2+{\left(0.00015\right)}^2}\\ =8.63\times {10}^{-4}\end{array}$

Substitute, $0.0015$ for ${\epsilon }_x$ , $-0.0002$ for ${\epsilon }_y$ in Equation (IX).

  $\begin{array}{c}{\epsilon }_{\text{avg}}=\frac{0.0015-0.0002}{2}\\ =\frac{0.0013}{2}\\ =6.5\times {10}^{-4}\end{array}$

Substitute, $8.63\times {10}^{-4}$ for $R$ in expression to construct the Mohr’s circle.

The below figure shows the Mohr’s circle:

  

     Figure-(1)

Substitute, $6.5\times {10}^{-4}$ for ${\epsilon }_{\text{avg}}$ , $8.63\times {10}^{-4}$ for $R$ , $20°$ for $\theta$ and $5.004°$ for ${\theta }_p$ in Equation (X).

  $\begin{array}{c}{\epsilon }_{x_1}=6.5\times {10}^{-4}+8.63\times {10}^{-4}\cos \left(2\times 20°-2\times 5.004°\right)\\ =6.5\times {10}^{-4}+7.4744\times {10}^{-4}\\ =1.398\times {10}^{-3}\end{array}$

Substitute, $6.5\times {10}^{-4}$ for ${\epsilon }_{\text{avg}}$ , $8.63\times {10}^{-4}$ for $R$ , $20°$ for $\theta$ and $5.004°$ for ${\theta }_p$ in Equation (XI).

  $\begin{array}{c}{\epsilon }_{y_1}=6.5\times {10}^{-4}-8.63\times {10}^{-4}\cos \left(2\times 20°-2\times 5.004°\right)\\ =6.5\times {10}^{-4}-7.474\times {10}^{-4}\\ =-9.76\times {10}^{-5}\end{array}$

Substitute, $8.63\times {10}^{-4}$ for $R$ , $20°$ for $\theta$ and $5.004°$ for ${\theta }_p$ in Equation (XII).

  $\begin{array}{c}{\gamma }_{x_1{y}_1}=2\times 8.63\times {10}^{-4}\times \sin \left(2\times 20°-2\times 5.004°\right)\\ =17.26\times {10}^{-4}\times \sin \left(40°-10.008°\right)\\ =17.26\times {10}^{-4}\times \sin \left(29.992\right)\\ =8.63\times {10}^{-4}\end{array}$

Substitute, $6.5\times {10}^{-4}$ for ${\epsilon }_{\text{avg}}$ , $8.63\times {10}^{-4}$ for $R$ in Equation (XIII).

  $\begin{array}{c}{\epsilon }_1=6.5\times {10}^{-4}+8.63\times {10}^{-4}\\ =1.513\times {10}^{-3}\end{array}$

Substitute, $6.5\times {10}^{-4}$ for ${\epsilon }_{\text{avg}}$ , $8.63\times {10}^{-4}$ for $R$ in Equation (XIV).

  $\begin{array}{c}{\epsilon }_2=6.5\times {10}^{-4}-8.63\times {10}^{-4}\\ =-2.13\times {10}^{-3}\end{array}$

Substitute, $8.63\times {10}^{-4}$ for $R$ in Equation (XV).

  $\begin{array}{c}{\gamma }_{\max }=2\left(8.63\times {10}^{-4}\right)\\ =1.726\times {10}^{-3}\end{array}$

Conclusion:

The maximum principal strain is $1.513\times {10}^{-3}$ .

The minimum principal strain is $-2.13\times {10}^{-3}$ .

The maximum shear strain is $1.726\times {10}^{-3}$ .

The principal angle is $5.004°$ .

The radius of Mohr’s circle is $8.63\times {10}^{-4}$ .