Chapter 7, Problem 7.7.18P

(a)

To determine

The strains for an element oriented at $\theta =50°$.

Answer to Problem 7.7.18P

The normal strain in x direction is $-14.69\times {10}^{-4}$.

The normal strain along y direction is $-9.07\times {10}^{-4}$.

The shear strain is $-7.18\times {10}^{-4}$in x-y plane.

Explanation of Solution

Given:

Stress along x direction is $-150\text{MPa}$and stress along y direction is $-210\text{MPa}$and shear stress along x-y plane is $-16\text{MPa}$.

Write the Equation for strain along x axis

${\epsilon }_x=\frac{1}{E}\left({\sigma }_x-\nu {\sigma }_y\right)$(I)

Here, the normal strain along x direction is ${\epsilon }_x$, direct stress along x direction is ${\sigma }_x$and direct stress along y direction is ${\sigma }_y$Poissons ratio is $\nu$and Young modulus is $E$.

Write the Equation for strain along y axis

${\epsilon }_y=\frac{1}{E}\left({\sigma }_y-\nu {\sigma }_x\right)$(II)

Here, the normal strain along the y direction is ${\epsilon }_y$.

Write the expression for the shear strain ix-y plane

${\gamma }_{xy}=\frac{2{\tau }_{xy}\left(1+\nu \right)}{E}$(III)

Here, modulus of rigidity is $G$and the shear strain along x-y plane is ${\gamma }_{xy}$.

Write the expression for strain along x direction

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

Write the expression for strain along y direction

${\epsilon }_{y_1}={\epsilon }_x+{\epsilon }_y-{\epsilon }_{x_1}$(V)

Write the expression for shear strain

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

Here, the shear strain along x-y plane is ${\gamma }_{xy}$.

Calculation:

Substitute $-150\text{MPa}$for ${\sigma }_x$, $-210\text{MPa}$for ${\sigma }_y$and $0.34$for $\nu$and $100\text{GPa}$for $E$in Equation (I).

$\begin{array}{c}{\epsilon }_x=\frac{\left[\left(-150\text{MPa}\right)-\left(\left(0.34\right)\times \left(-210\text{MPa}\right)\right)\right]}{100\text{GPa}}\\ =\frac{\left(-150\text{MPa}\right)-\left(\left(0.34\right)\times \left(-210\text{MPa}\right)\right)}{100\text{GPa}\left(\frac{1000\text{MPa}}{1\text{GPa}}\right)}\\ =\frac{1}{{10}^5\text{MPa}}\left[\left(-150\text{MPa}\right)-\left(\left(0.34\right)\times \left(-210\text{MPa}\right)\right)\right]\\ =-7.86\times {10}^{-4}\end{array}$

Substitute $-150\text{MPa}$for ${\sigma }_x$, $-210\text{MPa}$for ${\sigma }_y$and $0.34$for $\nu$an $100\text{GPa}$for $E$in Equation (II).

$\begin{array}{c}{\epsilon }_y=\frac{\left(-210\text{MPa}\right)-\left(\left(0.34\right)\times \left(-150\text{MPa}\right)\right)}{100\text{GPa}}\\ =\frac{1}{100\text{GPa}\left(\frac{1000\text{MPa}}{1\text{GPa}}\right)}\left[\left(-210\text{MPa}\right)-\left(\left(0.34\right)\times \left(-150\text{MPa}\right)\right)\right]\\ =\frac{-159\text{MPa}}{{10}^5\text{MPa}}\\ =-15.9\times {10}^{-4}\end{array}$

Substitute $-16\text{MPa}$for ${\tau }_{xy}$, $0.34$for $\nu$and $100\text{GPa}$for $E$in Equation (III).

$\begin{array}{c}{\gamma }_{xy}=\frac{2\left(-16\text{MPa}\right)\left(1+0.34\right)}{100\text{GPa}}\\ =\frac{2\left(-16\text{MPa}\right)\left(1+0.34\right)}{100\text{GPa}\left(\frac{1000\text{MPa}}{1\text{GPa}}\right)}\\ =\frac{42.88\text{MPa}}{{10}^5\text{MPa}}\\ =-4.29\times {10}^{-4}\end{array}$

Substitute $-7.86\times {10}^{-4}$for ${\epsilon }_x$, $-15.9\times {10}^{-4}$for ${\epsilon }_y$and $-4.29\times {10}^{-4}$for ${\gamma }_{xy}$in Equation (IV).

$\begin{array}{c}{\epsilon }_{x_1}=\left(\begin{array}{l}\frac{\left(-7.86\times {10}^{-4}-15.9\times {10}^{-4}\right)}{2}\\ +\frac{\left(-7.86\times {10}^{-4}+15.9\times {10}^{-4}\right)}{2}\cos \left(2\times \left(50°\right)\right)\\ +\frac{\left(-4.29\times {10}^{-4}\right)}{2}\sin \left(2\times \left(50°\right)\right)\end{array}\right)\\ =\left(-11.88\times {10}^{-4}-0.7\times {10}^{-4}-2.11\times {10}^{-4}\right)\\ =-14.69\times {10}^{-4}\end{array}$

Substitute $-7.86\times {10}^{-4}$for ${\epsilon }_x$, $-15.9\times {10}^{-4}$for ${\epsilon }_y$and $-14.69\times {10}^{-4}$for ${\epsilon }_{x_1}$in Equation (V).

$\begin{array}{c}{\epsilon }_{y_1}=-7.86\times {10}^{-4}-15.9\times {10}^{-4}+14.69\times {10}^{-4}\\ =-23.76\times {10}^{-4}+14.69\times {10}^{-4}\\ =-9.07\times {10}^{-4}\end{array}$

Substitute $-7.86\times {10}^{-4}$for ${\epsilon }_x$, $-15.9\times {10}^{-4}$for ${\epsilon }_y$and $-4.29\times {10}^{-4}$for ${\gamma }_{xy}$in Equation (VI).

$\begin{array}{c}\frac{{\gamma }_{x_1{y}_1}}{2}=-\left[\begin{array}{r}\left(\frac{\left(-7.86\times {10}^{-4}+15.9\times {10}^{-4}\right)}{2}\sin \left(2\times 50°\right)\right)\\ +\left(\left(\frac{-4.29\times {10}^{-4}}{2}\right)\cos \left(2\times 50°\right)\right)\end{array}\right]\\ \frac{{\gamma }_{x_1{y}_1}}{2}=-3.59\times {10}^{-4}\\ {\gamma }_{x_1{y}_1}=-7.18\times {10}^{-4}\end{array}$

In below figure the normal strains are shown

:

    

    Figure (1)

Conclusion:

The normal strain along x direction is $-14.69\times {10}^{-4}$.

The normal strain along y direction is $-9.07\times {10}^{-4}$.

The shear strain along x-y plane is $-7.18\times {10}^{-4}$.

(b)

To determine

The principal strains.

Answer to Problem 7.7.18P

The maximum principal strain is $-7.32\times {10}^{-4}$, and the minimum principal strain is $-16.44\times {10}^{-4}$.

Explanation of Solution

Write expression for the principal strains.

${\epsilon }_{1,2}=\left(\frac{{\epsilon }_x+{\epsilon }_y}{2}\right)+\sqrt{{\left(\frac{{\epsilon }_x-{\epsilon }_y}{2}\right)}^2+{\left(\frac{{\gamma }_{xy}}{2}\right)}^2}$(VII)

Here, the maximum principal strain is ${\epsilon }_1$and the minimum principal strain is ${\epsilon }_2$.

Write the expression for the principal angle.

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

Here, ${\theta }_p$is the principal angle.

Calculation:

Substitute $-7.86\times {10}^{-4}$for ${\epsilon }_x$, $-15.9\times {10}^{-4}$for ${\epsilon }_y$and $-4.29\times {10}^{-4}$for ${\gamma }_{xy}$in Equation (VII).

$\begin{array}{c}{\epsilon }_{1,2}=\left(\frac{\left(-7.86\times {10}^{-4}-15.9\times {10}^{-4}\right)}{2}\right)\pm \\ \sqrt{{\left(\frac{\left(-7.86\times {10}^{-4}+15.9\times {10}^{-4}\right)}{2}\right)}^2+{\left(\frac{-4.29\times {10}^{-4}}{2}\right)}^2}\\ =-11.88\times {10}^{-4}\pm \sqrt{\left(16.16\times {10}^{-8}\right)+\left(4.60\times {10}^{-8}\right)}\\ =-11.88\times {10}^{-4}\pm 4.556\times {10}^{-4}\end{array}$

While taking positive sign you get maximum principal strain:

$\begin{array}{c}{\epsilon }_1=\left(-11.88+4.556\right)\times {10}^{-4}\\ =-7.32\times {10}^{-4}\end{array}$

While taking negative sign you get minimum principal strain.

$\begin{array}{c}{\epsilon }_1=\left(-11.88-4.556\right)\times {10}^{-4}\\ =-16.44\times {10}^{-4}\end{array}$

Substitute $-7.86\times {10}^{-4}$for ${\epsilon }_x$, $-15.9\times {10}^{-4}$for ${\epsilon }_y$and $-4.29\times {10}^{-4}$for ${\gamma }_{xy}$in Equation (VIII).

$\begin{array}{c}\tan 2{\theta }_p=\frac{-4.29\times {10}^{-4}}{\left(-7.86\times {10}^{-4}+15.9\times {10}^{-4}\right)}\\ \tan 2{\theta }_p=-0.53358\end{array}$

$\begin{array}{l}180-2{\theta }_p=28.1°\\ 2{\theta }_p=151.9°\\ {\theta }_p=75.95°\end{array}$

In below figure principal strains and principal angle are shown:

    

    Figure (2)

Conclusion:

The maximum principal strain is $-7.32\times {10}^{-4}$and minimum principal strain is $-16.44\times {10}^{-4}$.

(c)

To determine

The maximum shear strain.

Answer to Problem 7.7.18P

The maximum shear strain is $9.12\times {10}^{-4}$.

Explanation of Solution

Write the expression for maximum shear strain,

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

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

Write expression for first shear angle

${\theta }_{s_1}={\theta }_{p_1}-45°$ (X)

Here, ${\theta }_{s_1}$is the first shear angle.

Write expression for second shear angle

${\theta }_{s_2}={\theta }_{s_1}-45°$ (XI)

Here, ${\theta }_{s_2}$is the second shear angle.

Write the Equation for average strain

${\epsilon }_{avg}=\frac{{\epsilon }_x+{\epsilon }_y}{2}$(XII)

Write expression for maximum shear strain.

${\gamma }_{\min }=-{\gamma }_{\max }$ (XIII)

Calculation:

Substitute $-7.86\times {10}^{-4}$for ${\epsilon }_x$, $-15.9\times {10}^{-4}$for ${\epsilon }_y$and $-4.29\times {10}^{-4}$for ${\gamma }_{xy}$in Equation (IX).

$\begin{array}{c}\frac{{\gamma }_{\max }}{2}=\sqrt{{\left(\frac{\left(-7.86\times {10}^{-4}+15.9\times {10}^{-4}\right)}{2}\right)}^2+{\left(\frac{-4.29\times {10}^{-4}}{2}\right)}^2}\\ \frac{{\gamma }_{\max }}{2}=\sqrt{{\left(\frac{\left(-7.86+15.9\right)\times {10}^{-4}}{2}\right)}^2+{\left(2.145\times {10}^{-4}\right)}^2}\\ \frac{{\gamma }_{\max }}{2}=\left(4.56\times {10}^{-4}\right)\\ {\gamma }_{\max }=9.12\times {10}^{-4}\end{array}$

Substitute $165.95°$for ${\theta }_{p_1}$in Equation (X).

$\begin{array}{c}{\theta }_{s_1}=165.95°-45°\\ =120.95°\end{array}$

Substitute $120.95°$for ${\theta }_{s_1}$in Equation (XI)

$\begin{array}{c}{\theta }_{s_2}=120.95°-90°\\ =30.95°\end{array}$

Substitute $-7.86\times {10}^{-4}$for ${\epsilon }_x$, $-15.9\times {10}^{-4}$for ${\epsilon }_y$in Equation (XII).

$\begin{array}{c}{\epsilon }_{avg}=\frac{-7.86\times {10}^{-4}-15.9\times {10}^{-4}}{2}\\ =\frac{-23.76\times {10}^{-4}}{2}\\ =-11.88\times {10}^{-4}\end{array}$

Substitute $9.12\times {10}^{-4}$for ${\gamma }_{\max }$.in Equation (XIII).

${\gamma }_{\min }=-{\gamma }_{\max }=9.12\times {10}^{-4}$

In below figure average minimum strain and shear angle are shown:

    

    Figure (3)

Conclusion:

The maximum shear strain is $9.12\times {10}^{-4}.$