Chapter 7.9, Problem 149P

(a)

To determine

Find the orientation and magnitude of the principal strains in the plane.

Answer to Problem 149P

The orientation of the principal strains is $\underset{\_}{{\theta }_p=30°\text{and}120°}$.

The in-plane maximum principal strain is $\underset{\_}{{\epsilon }_{\max }=560\times {10}^{-6}\text{in}\text{./in}\text{.}}$.

The in-plane minimum principal strain is $\underset{\_}{{\epsilon }_{\min }=-140\times {10}^{-6}\text{in}\text{./in}\text{.}}$.

Explanation of Solution

Given information:

The normal strain in Rosette 1 is ${\epsilon }_1=-93.1\times {10}^{-6}\text{in}\text{./in}\text{.}$.

The Rosette 1 makes an angle with the x-axis is ${\theta }_1=-75°$.

The normal strain in Rosette 2 is ${\epsilon }_2=+385\times {10}^{-6}\text{in}\text{./in}\text{.}$.

The Rosette 2 makes an angle with the x-axis is ${\theta }_2=0°$.

The normal strain in Rosette 3 is ${\epsilon }_3=+210\times {10}^{-6}\text{in}\text{./in}\text{.}$.

The Rosette 3 makes an angle with the x-axis is ${\theta }_3=75°$.

Calculation:

Write the equation for the normal strain 1 as follows;

${\epsilon }_1={\epsilon }_x{\cos }^2{\theta }_1+{\epsilon }_y{\sin }^2{\theta }_1+{\gamma }_{xy}\sin {\theta }_1\cos {\theta }_1$

Substitute $-93.1\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_1$ and $-75°$ for ${\theta }_1$.

$-93.1\times {10}^{-6}={\epsilon }_x{\cos }^2\left(-75°\right)+{\epsilon }_y{\sin }^2\left(-75°\right)+{\gamma }_{xy}\sin \left(-75°\right)\cos \left(-75°\right)$ (1)

Write the equation for the normal strain 2 as follows;

${\epsilon }_2={\epsilon }_x{\cos }^2{\theta }_2+{\epsilon }_y{\sin }^2{\theta }_2+{\gamma }_{xy}\sin {\theta }_2\cos {\theta }_2$

Substitute $+385\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_2$ and $0°$ for ${\theta }_2$.

$\begin{array}{c}385\times {10}^{-6}={\epsilon }_x{\cos }^2\left(0°\right)+{\epsilon }_y{\sin }^2\left(0°\right)+{\gamma }_{xy}\sin \left(0°\right)\cos \left(0°\right)\\ ={\epsilon }_x+0+0\\ {\epsilon }_x=385\times {10}^{-6}\text{in}\text{./in}\text{.}\end{array}$

Write the equation for the normal strain 3 as follows;

${\epsilon }_3={\epsilon }_x{\cos }^2{\theta }_3+{\epsilon }_y{\sin }^2{\theta }_3+{\gamma }_{xy}\sin {\theta }_3\cos {\theta }_3$

Substitute $+210\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_3$ and $75°$ for ${\theta }_3$.

$+210\times {10}^{-6}={\epsilon }_x{\cos }^2\left(75°\right)+{\epsilon }_y{\sin }^2\left(75°\right)+{\gamma }_{xy}\sin \left(75°\right)\cos \left(75°\right)$ (2)

Add equation (1) and (2).

$\begin{array}{c}-93.1\times {10}^{-6}+210\times {10}^{-6}=\left(\begin{array}{l}{\epsilon }_x{\cos }^2\left(-75°\right)+{\epsilon }_y{\sin }^2\left(-75°\right)+{\gamma }_{xy}\sin \left(-75°\right)\cos \left(-75°\right)\\ +{\epsilon }_x{\cos }^2\left(75°\right)+{\epsilon }_y{\sin }^2\left(75°\right)+{\gamma }_{xy}\sin \left(75°\right)\cos \left(75°\right)\end{array}\right)\\ 116.9\times {10}^{-6}=\left(\begin{array}{l}{\epsilon }_x{\cos }^2\left(-75°\right)+{\epsilon }_y{\sin }^2\left(-75°\right)-0.25{\gamma }_{xy}\\ +{\epsilon }_x{\cos }^2\left(75°\right)+{\epsilon }_y{\sin }^2\left(75°\right)+0.25{\gamma }_{xy}\end{array}\right)\\ ={\epsilon }_x{\cos }^2\left(-75°\right)+{\epsilon }_y{\sin }^2\left(-75°\right)+{\epsilon }_x{\cos }^2\left(75°\right)+{\epsilon }_y{\sin }^2\left(75°\right)\end{array}$

Substitute $385\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_x$.

$\begin{array}{c}116.9\times {10}^{-6}=\left(\begin{array}{l}\left(385\times {10}^{-6}\right){\cos }^2\left(-75°\right)+{\epsilon }_y{\sin }^2\left(-75°\right)+\left(385\times {10}^{-6}\right){\cos }^2\left(75°\right)\\ +{\epsilon }_y{\sin }^2\left(75°\right)\end{array}\right)\\ =5.158\times {10}^{-5}+1.866{\epsilon }_y\\ {\epsilon }_y=35\times {10}^{-6}\text{in}\text{./in}\text{.}\end{array}$

Substitute $385\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_x$ and $35\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_y$ in Equation (2).

$\begin{array}{c}+210\times {10}^{-6}=\left(385\times {10}^{-6}\right){\cos }^2\left(75°\right)+\left(35\times {10}^{-6}\right){\sin }^2\left(75°\right)+{\gamma }_{xy}\sin \left(75°\right)\cos \left(75°\right)\\ =2.579\times {10}^{-5}+3.266\times {10}^{-5}+0.25{\gamma }_{xy}\\ {\gamma }_{xy}=606.2\times {10}^{-6}\text{in}\text{./in}\text{.}\end{array}$

Find the orientation $\left({\theta }_p\right)$  of the principal strains using the relation.

$\tan 2{\theta }_p=\frac{{\gamma }_{xy}}{{\epsilon }_x-{\epsilon }_y}$

Substitute $606.2\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\gamma }_{xy}$, $385\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_x$, and $35\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_y$

$\begin{array}{l}\tan 2{\theta }_p=\frac{606.2\times {10}^{-6}}{385\times {10}^{-6}-35\times {10}^{-6}}\\ {\theta }_p=30°\text{and}120°\end{array}$

Therefore, the orientation of the principal strains is $\underset{\_}{{\theta }_p=30°\text{and}120°}$.

Find the average normal strain $\left({\epsilon }_{\text{ave}}\right)$ using the relation.

${\epsilon }_{\text{ave}}=\frac{{\epsilon }_x+{\epsilon }_y}{2}$

Substitute $385\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_x$ and $35\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_y$.

$\begin{array}{c}{\epsilon }_{\text{ave}}=\frac{385\times {10}^{-6}+35\times {10}^{-6}}{2}\\ =210\times {10}^{-6}\text{in}\text{./in}\text{.}\end{array}$

Find the radius of the Mohr’s circle (R) using the equation.

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

Substitute $385\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_x$, $35\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_y$, and $606.2\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\gamma }_{xy}$.

$\begin{array}{c}R=\sqrt{{\left(\frac{385\times {10}^{-6}-35\times {10}^{-6}}{2}\right)}^2+{\left(\frac{606.2\times {10}^{-6}}{2}\right)}^2}\\ =350\times {10}^{-6}\text{in}\text{./in}\text{.}\end{array}$

Find the in-plane maximum principal strain $\left({\epsilon }_{\max }\right)$ using the relation.

${\epsilon }_{\max }={\epsilon }_{\text{ave}}+R$

Substitute $210\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_{\text{ave}}$ and $350\times {10}^{-6}\text{in}\text{./in}\text{.}$ for R.

$\begin{array}{c}{\epsilon }_{\max }=210\times {10}^{-6}+350\times {10}^{-6}\\ =560\times {10}^{-6}\text{in}\text{./in}\text{.}\end{array}$

Find the in-plane minimum principal strain $\left({\epsilon }_{\min }\right)$ using the relation.

${\epsilon }_{\min }={\epsilon }_{\text{ave}}-R$

Substitute $210\times {10}^{-6}\text{in}\text{./in}\text{.}$ for ${\epsilon }_{\text{ave}}$ and $350\times {10}^{-6}\text{in}\text{./in}\text{.}$ for R.

$\begin{array}{c}{\epsilon }_{\min }=210\times {10}^{-6}-350\times {10}^{-6}\\ =-140\times {10}^{-6}\text{in}\text{./in}\text{.}\end{array}$

Therefore,

The in-plane maximum principal strain is $\underset{\_}{{\epsilon }_{\max }=560\times {10}^{-6}\text{in}\text{./in}\text{.}}$.

The in-plane minimum principal strain is $\underset{\_}{{\epsilon }_{\min }=-140\times {10}^{-6}\text{in}\text{./in}\text{.}}$.

(b)

To determine

Find the in-plane maximum shearing strain.

Answer to Problem 149P

The in-plane maximum shearing strain is $\underset{\_}{{\left({\gamma }_{\max }\right)}_{\text{in-plane}}=700\times {10}^{-6}\text{in}\text{./in}\text{.}}$.

Explanation of Solution

Refer to part (a);

The radius of the Mohr’s circle is $R=350\times {10}^{-6}\text{in}\text{./in}\text{.}$

Find the in-plane maximum shearing strain ${\left({\gamma }_{\max }\right)}_{\text{in-plane}}$ using the relation.

${\left({\gamma }_{\max }\right)}_{\text{in-plane}}=2R$

Substitute $350\times {10}^{-6}\text{in}\text{./in}\text{.}$ for R.

$\begin{array}{c}{\left({\gamma }_{\max }\right)}_{\text{in-plane}}=2\times 350\times {10}^{-6}\\ =700\times {10}^{-6}\text{in}\text{./in}\text{.}\end{array}$

Therefore, the in-plane maximum shearing strain is $\underset{\_}{{\left({\gamma }_{\max }\right)}_{\text{in-plane}}=700\times {10}^{-6}\text{in}\text{./in}\text{.}}$.