Chapter 7, Problem 7.7.1P

An element of material in plain strain has the following strains:

${\epsilon }_x$ = 0.001 and ${\epsilon }_y$ = 0.0015.

(a) Determine the strains for an element oriented at an angle θ = 250.

(b) Find the principal strains of the element. Confirm the solution using Mohr’s circle for

plane strain.

To determine

The strain for an element inclined at an angle $25°$.

Answer to Problem 7.7.1P

The rotated strain along $x_1$-direction is $-5.53\times {10}^{-4}$.

The rotated strain along $y_1$-direction is $1.053\times {10}^{-3}$.

The rotated shear strain along $x_1{y}_1$plane is $1.915\times {10}^{-3}$.

Explanation of Solution

Given information:

The strain along x-direction is $-0.001$, the strain along y-direction is $0.0015$,

The angle of orientation is $25°$.

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)

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

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 $-0.001$for ${\epsilon }_x$, $0.0015$for ${\epsilon }_y$, $0$for ${\gamma }_{xy}$, $25°$for $\theta$in Equation (I).

$\begin{array}{c}{\epsilon }_{x_1}=\frac{-0.001+0.0015}{2}+\frac{-0.001-0.0015}{2}\cos \left(2\times 25°\right)+\frac{0}{2}\sin \left(2\times 25°\right)\\ =2.5\times {10}^{-4}-8.03\times {10}^{-4}\\ =-5.53\times {10}^{-4}\end{array}$

Substitute, $-0.001$for ${\epsilon }_x$, $0.0015$for ${\epsilon }_y$, $0$for ${\gamma }_{xy}$, $25°$for $\theta$in Equation (II).

$\begin{array}{c}{\epsilon }_{y_1}=\frac{-0.001+0.0015}{2}+\frac{-0.001-0.0015}{2}\cos \left(2\times \left(25°+90°\right)\right)+\frac{0}{2}\sin \left(2\times \left(25°+90°\right)\right)\\ =2.5\times {10}^{-4}+8.03\times {10}^{-4}\\ =1.053\times {10}^{-3}\end{array}$

Substitute, $-0.001$for ${\epsilon }_x$, $0.0015$for ${\epsilon }_y$, $0$for ${\gamma }_{xy}$, $25°$for $\theta$in Equation (III).

$\begin{array}{c}{\gamma }_{x_1{y}_1}=2\left\{-\frac{-0.001-0.0015}{2}\sin \left(2\times 25°\right)+\frac{0}{2}\cos \left(2\times 25°\right)\right\}\\ =1.915\times {10}^{-3}.\end{array}$

Conclusion:

The rotated strain along $x_1$-direction is $-5.53\times {10}^{-4}$.

The rotated strain along $y_1$-direction is $1.053\times {10}^{-3}$.

The rotated shear strain along $x_1{y}_1$plane is $1.915\times {10}^{-3}$.

To determine

The principal strains of the element.

The Mohr’s circle for plane strain.

Answer to Problem 7.7.1P

The maximum principal strain is $1.5\times {10}^{-3}$.

The minimum principal strain is $-1\times {10}^{-3}$.

The maximum shear strain is $2.5\times {10}^{-3}$.

The principal angle is $0°$.

The radius of Mohr’s circle is $1.25\times {10}^{-3}$.

Explanation of Solution

Write the expression for maximum principal strain.

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

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.

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

Write the expression for maximum shear strain.

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

Write the expression for principal angle.

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

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:

Draw a horizontal axis and consider it as $\epsilon$

1. axis.

Draw a vertical axis and consider it to be ${\gamma }_{xy}/2$

Mark the point $C\left({\epsilon }_{\text{avg}},0\right)$.

With the help of radius $R=1.25\times {10}^{-3}$and center $C$draw a circle.

Thus, the Mohr’s circle is obtained.

Calculation:

Substitute $-0.001$for ${\epsilon }_x$, $0.0015$for ${\epsilon }_y$, $0$for ${\gamma }_{xy}$in Equation (IV).

$\begin{array}{c}{\epsilon }_1=\frac{-0.001+0.0015}{2}+\sqrt{{\left(\frac{-0.001-0.0015}{2}\right)}^2+{\left(\frac{0}{2}\right)}^2}\\ =2.5\times {10}^{-4}+1.25\times {10}^{-3}\\ =1.5\times {10}^{-3}.\end{array}$

Substitute $-0.001$for ${\epsilon }_x$, $0.0015$for ${\epsilon }_y$, $0$for ${\gamma }_{xy}$in Equation (V).

$$$\begin{array}{c}{\epsilon }_2=\frac{-0.001+0.0015}{2}-\sqrt{{\left(\frac{-0.001-0.0015}{2}\right)}^2+{\left(\frac{0}{2}\right)}^2}\\ =2.5\times {10}^{-4}-1.25\times {10}^{-3}\\ =-1\times {10}^{-3}\end{array}$

Substitute $-0.001$for ${\epsilon }_x$, $0.0015$for ${\epsilon }_y$, $0$for ${\gamma }_{xy}$in Equation (VI).

$\begin{array}{c}{\gamma }_{\max }=2\sqrt{{\left(\frac{-0.001-0.0015}{2}\right)}^2+{\left(\frac{0}{2}\right)}^2}\\ =2\left(1.25\times {10}^{-3}\right)\\ =2.5\times {10}^{-3}.\end{array}$

Substitute $-0.001$for ${\epsilon }_x$, $0.0015$for ${\epsilon }_y$, $0$for ${\gamma }_{xy}$in Equation (VII).

$\begin{array}{c}{\theta }_p=\frac{1}{2}{\tan }^{-1}\left(\frac{0}{-0.001-0.0015}\right)\\ =0°\end{array}$

Substitute $-0.001$for ${\epsilon }_x$, $0.0015$for ${\epsilon }_y$, $0$for ${\gamma }_{xy}$in Equation (VIII).

$\begin{array}{c}R=\sqrt{{\left(\frac{-0.001-0.0015}{2}\right)}^2+{\left(\frac{0}{2}\right)}^2}\\ =1.25\times {10}^{-3}.\end{array}$

Substitute $-0.001$for ${\epsilon }_x$, $0.0015$for ${\epsilon }_y$in Equation (IX).

$\begin{array}{c}{\epsilon }_{\text{avg}}=\frac{-0.001+0.0015}{2}\\ =2.5\times {10}^{-4}.\end{array}$

The below figure shows the Mohr’s circle.

    

    Figure-(1)

Substitute $2.5\times {10}^{-4}$for ${\epsilon }_{\text{avg}}$, $1.25\times {10}^{-3}$for $R$, $25°$for $\theta$and $0°$for ${\theta }_p$in Equation (X).

$\begin{array}{c}{\epsilon }_{x_1}=2.5\times {10}^{-4}-1.25\times {10}^{-3}\cos \left(2\times 25°-2\times 0°\right)\\ =2.5\times {10}^{-4}-8.03\times {10}^{-4}\\ =-5.53\times {10}^{-4}\end{array}$

Substitute $2.5\times {10}^{-4}$for ${\epsilon }_{\text{avg}}$, $1.25\times {10}^{-3}$for $R$, $25°$for $\theta$and $0°$for ${\theta }_p$in Equation (XI).

$\begin{array}{c}{\epsilon }_{y_1}=2.5\times {10}^{-4}+1.25\times {10}^{-3}\cos \left(2\times 25°-2\times 0°\right)\\ =2.5\times {10}^{-4}+8.03\times {10}^{-4}\\ =1.053\times {10}^{-3}\end{array}$

Substitute $1.25\times {10}^{-3}$for $R$, $25°$for $\theta$and $0°$for ${\theta }_p$in Equation (XII).

$\begin{array}{c}{\gamma }_{x_1{y}_1}=2\times 1.25\times {10}^{-3}\times \sin \left(2\times 25°-2\times 0°\right)\\ =1.915\times {10}^{-3}\end{array}$

Substitute $2.5\times {10}^{-4}$for ${\epsilon }_{\text{avg}}$, $1.25\times {10}^{-3}$for $R$in Equation (XIII).

$\begin{array}{c}{\epsilon }_1=2.5\times {10}^{-4}+1.25\times {10}^{-3}\\ =1.5\times {10}^{-3}\end{array}$

Substitute $2.5\times {10}^{-4}$for ${\epsilon }_{\text{avg}}$, $1.25\times {10}^{-3}$for $R$in Equation (XIV).

$\begin{array}{c}{\epsilon }_2=2.5\times {10}^{-4}-1.25\times {10}^{-3}\\ =-1\times {10}^{-3}\end{array}$

Substitute $1.25\times {10}^{-3}$for $R$in Equation (XV).

$\begin{array}{c}{\gamma }_{\max }=2\left(1.25\times {10}^{-3}\right)\\ =2.5\times {10}^{-3}.\end{array}$

Conclusion:

The maximum principal strain is $1.5\times {10}^{-3}$.

The minimum principal strain is $-1\times {10}^{-3}$.

The maximum shear strain is $2.5\times {10}^{-3}$.

The principal angle is $0°$.

The radius of Mohr’s circle is $1.25\times {10}^{-3}$.