Chapter 7.9, Problem 157P

The following state of strain has been determined on the surface of a cast-iron machine part:

${\in }_x=\text{-720}\mu$

${\in }_y=\text{-400}\mu$

${\gamma }_{xy}=+660\mu$

Knowing that E = 69 GPa and G = 28 GPa, determine the principal planes and principal stresses (a) by determining the corresponding state of plane stress [use Eq. (2.36), Eq. (2.43), and the first two equations of Prob. 2.73] and then using Mohr's circle for stress, (b) by using Mohr's circle for strain to determine the orientation and magnitude of the principal strains and then determining the corresponding stresses.

To determine

Find the direction and magnitude of the three principal strains.

Answer to Problem 157P

The maximum principal stress is is $\underset{\_}{{\sigma }_a=-29.8\text{MPa}}$.

The intermediate principal stress is $\underset{\_}{{\sigma }_b=-70.9\text{MPa}}$.

The orientation of the major principal stress is $\underset{\_}{{\theta }_a=-32.1°}$.

The orientation of the intermediate principal stress is $\underset{\_}{{\theta }_b=57.9°}$.

Explanation of Solution

Given information:

Construct the Mohr’s circle for stress.

The modulus of elasticity of the material is $E=69\text{GPa}$.

The modulus of rigidity of the material is $G=28\text{GPa}$.

The normal strain in x-axis is ${\epsilon }_x=-720\mu \times \frac{{10}^{-6}}{1\mu }=-7.20\times {10}^{-4}$.

The normal strain in y-axis is ${\epsilon }_y=-400\mu \times \frac{{10}^{-6}}{1\mu }=-4\times {10}^{-4}$.

The shearing strain in xy-plane is ${\gamma }_{xy}=+660\mu \times \frac{{10}^{-6}}{1\mu }=+6.6\times {10}^{-4}$.

Calculation:

Find the Poisson’s ratio $\left(\nu \right)$ using the relation.

$G=\frac{E}{2\left(1+\nu \right)}$

Here, the modulus of elasticity is E and the modulus of rigidity is G.

Substitute 69 GPa for E and 28 GPa for G.

$\begin{array}{c}28=\frac{69}{2\left(1+\nu \right)}\\ \nu =0.2321\end{array}$

Find the normal stress in x-axis $\left({\sigma }_x\right)$ using the relation.

${\sigma }_x=\frac{E}{1-{\nu }^2}\left({\epsilon }_x+\nu {\epsilon }_y\right)$

Substitute 69 GPa for E, 0.2321 for $\nu$, $-7.20\times {10}^{-4}$ for ${\epsilon }_x$, and $-4\times {10}^{-4}$ for ${\epsilon }_y$.

$\begin{array}{c}{\sigma }_x=\frac{69\text{GPa}\times \frac{{\text{10}}^{\text{3}}\text{MPa}}{\text{1}\text{GPa}}}{1-{0.2321}^2}\left(-7.20\times {10}^{-4}+0.2321\left(-4\times {10}^{-4}\right)\right)\\ =-59.28\text{MPa}\end{array}$

Find the normal stress in y-axis $\left({\sigma }_y\right)$ using the relation.

${\sigma }_y=\frac{E}{1-{\nu }^2}\left({\epsilon }_y+\nu {\epsilon }_x\right)$

Substitute 69 GPa for E, 0.2321 for $\nu$, $-7.20\times {10}^{-4}$ for ${\epsilon }_x$, and $-4\times {10}^{-4}$ for ${\epsilon }_y$.

$\begin{array}{c}{\sigma }_y=\frac{69\text{GPa}\times \frac{{\text{10}}^{\text{3}}\text{MPa}}{\text{1}\text{GPa}}}{1-{0.2321}^2}\left(-4\times {10}^{-4}+0.2321\left(-7.20\times {10}^{-4}\right)\right)\\ =-41.36\text{MPa}\end{array}$

The principal stress in z-axis is ${\sigma }_z=0$.

Find the shearing stress in xy-plane $\left({\tau }_{xy}\right)$ using the relation.

${\tau }_{xy}=G{\gamma }_{xy}$

Substitute 28 GPa for G and $+6.6\times {10}^{-4}$ for ${\gamma }_{xy}$.

$\begin{array}{c}{\tau }_{xy}=28\text{GPa}\times \frac{{\text{10}}^{\text{3}}\text{MPa}}{\text{1}\text{GPa}}\times 6.6\times {10}^{-4}\\ =18.48\text{MPa}\end{array}$

Construct the Mohr circle as follows;

• Plot the point X by –59.28 MPa to the left of the $\tau$ axis and 18.48 MPa below the $\sigma$ axis.
• Plot the point Y by –41.36 MPa to the left of the $\tau$ axis and 18.48 MPa above the $\sigma$ axis.
• Connect the points X and Y and the point C is known as abscissa.

Show the plotted Mohr’s circle diagram as in Figure 1.

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

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

Substitute –59.28 MPa for ${\sigma }_x$ and ­41.36 MPa for ${\sigma }_y$.

$\begin{array}{c}{\sigma }_{\text{ave}}=\frac{-59.28+\left(-41.36\right)}{2}\\ =-50.32\text{MPa}\end{array}$

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

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

Substitute –59.28 MPa for ${\sigma }_x$, ­41.36 MPa for ${\sigma }_y$, and 18.48 MPa for ${\tau }_{xy}$.

$\begin{array}{c}R=\sqrt{{\left(\frac{-59.28-\left(-41.36\right)}{2}\right)}^2+{\left(18.48\right)}^2}\\ =20.54\text{MPa}\end{array}$

Find the maximum principal stress $\left({\sigma }_a\right)$ using the relation.

${\sigma }_a={\sigma }_{ave}+R$

Substitute –50.32 MPa for ${\sigma }_{ave}$ and 20.54 MPa for R.

$\begin{array}{c}{\sigma }_a=-50.32+20.54\\ =-29.8\text{MPa}\end{array}$

Find the minimum principal stress $\left({\sigma }_b\right)$ using the relation.

${\sigma }_b={\sigma }_{ave}-R$

Substitute –50.32 MPa for ${\sigma }_{ave}$ and 20.54 MPa for R.

$\begin{array}{c}{\sigma }_b=-50.32-20.54\\ =-70.9\text{MPa}\end{array}$

Find the orientation $\left({\theta }_a\right)$ of the maximum principal axis using the relation.

$\tan 2{\theta }_a=\frac{2{\tau }_{xy}}{{\sigma }_x-{\sigma }_y}$

Substitute –59.28 MPa for ${\sigma }_x$, ­41.36 MPa for ${\sigma }_y$, and 18.48 MPa for ${\tau }_{xy}$.

$\begin{array}{l}\tan 2{\theta }_a=\frac{2\left(18.48\right)}{-59.28-\left(-41.36\right)}\\ {\theta }_a=-32.1°\end{array}$

Find the orientation of the intermediate principal axis $\left({\theta }_b\right)$ using the relation.

${\theta }_b={\theta }_a+90°$

Substitute $-32.1°$ for ${\theta }_a$.

$\begin{array}{c}{\theta }_b=-32.1°+90°\\ =57.9°\end{array}$

Therefore,

The maximum principal stress is is $\underset{\_}{{\sigma }_a=-29.8\text{MPa}}$.

The intermediate principal stress is $\underset{\_}{{\sigma }_b=-70.9\text{MPa}}$.

The orientation of the major principal stress is $\underset{\_}{{\theta }_a=-32.1°}$.

The orientation of the intermediate principal stress is $\underset{\_}{{\theta }_b=57.9°}$.

To determine

Find the direction and magnitude of the three principal strains.

Answer to Problem 157P

The maximum principal strain is $\underset{\_}{{\epsilon }_a=423\mu }$.

The intermediate principal strain is $\underset{\_}{{\epsilon }_b=-951\mu }$.

The minimum principal strain is $\underset{\_}{{\epsilon }_c=-222\mu }$.

The orientation of the major principal strain is $\underset{\_}{{\theta }_a=-22.5°}$.

The orientation of the intermediate principal strain is $\underset{\_}{{\theta }_b=67.5°}$.

Explanation of Solution

Given information:

Construct the Mohr’s circle for strain.

The modulus of elasticity of the material is $E=69\text{GPa}$.

The modulus of rigidity of the material is $G=28\text{GPa}$.

The normal strain in x-axis is ${\epsilon }_x=-720\mu$.

The normal strain in y-axis is ${\epsilon }_y=-400\mu$.

The shearing strain in xy-plane is ${\gamma }_{xy}=+660\mu$.

Calculation:

Find the Poisson’s ratio $\left(\nu \right)$ using the relation.

$G=\frac{E}{2\left(1+\nu \right)}$

Substitute 69 GPa for E and 28 GPa for G.

$\begin{array}{c}28=\frac{69}{2\left(1+\nu \right)}\\ \nu =0.2321\end{array}$

Find the value of $\frac{{\gamma }_{xy}}{2}$;

$\begin{array}{c}\frac{{\gamma }_{xy}}{2}=\frac{660}{2}\\ =330\mu \end{array}$

Construct the Mohr circle as follows;

• Plot the point X by $-720\mu$ to the left of the $\frac{\gamma }{2}$ axis and $330\mu$ below the $\epsilon$ axis.
• Plot the point Y by $-400\mu$ to the left of the $\frac{\gamma }{2}$ axis and $330\mu$ above the $\epsilon$ axis.
• Connect the points X and Y and the point C is known as abscissa.

Show the plotted Mohr’s circle diagram as in Figure 2.

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

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

Substitute $-720\mu$ for ${\epsilon }_x$ and $-400\mu$ for ${\epsilon }_y$.

$\begin{array}{c}{\epsilon }_{\text{ave}}=\frac{-720-400}{2}\\ =-560\mu \end{array}$

Find the orientation $\left({\theta }_a\right)$ of the maximum principal axis using the relation.

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

Substitute $660\mu$ for ${\gamma }_{xy}$, $-720\mu$ for ${\epsilon }_x$, and $-400\mu$ for ${\epsilon }_y$.

$\begin{array}{l}\tan 2{\theta }_a=\frac{660}{-720-\left(-400\right)}\\ {\theta }_a=-32.1°\end{array}$

Find the orientation of the intermediate principal axis $\left({\theta }_b\right)$ using the relation.

${\theta }_b={\theta }_a+90°$

Substitute $-32.1°$ for ${\theta }_a$.

$\begin{array}{c}{\theta }_b=-32.1°+90°\\ =57.9°\end{array}$

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

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

Substitute $660\mu$ for ${\gamma }_{xy}$, $-720\mu$ for ${\epsilon }_x$, and $-400\mu$ for ${\epsilon }_y$.

$\begin{array}{c}R=\sqrt{{\left(\frac{-720-\left(-400\right)}{2}\right)}^2+{\left(\frac{660}{2}\right)}^2}\\ =366.74\mu \end{array}$

Find the maximum principal strain $\left({\epsilon }_a\right)$ using the relation.

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

Substitute $-560\mu$ for ${\epsilon }_{\text{ave}}$ and $366.74\mu$ for R.

$\begin{array}{c}{\epsilon }_a=-560+366.74\\ =-193.26\mu \end{array}$

Find the intermediate principal strain $\left({\epsilon }_b\right)$ using the relation.

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

Substitute $-560\mu$ for ${\epsilon }_{\text{ave}}$ and $366.74\mu$ for R.

$\begin{array}{c}{\epsilon }_b=-560-366.74\\ =-926.74\mu \end{array}$

Find the maximum principal stress $\left({\sigma }_a\right)$ using the equation.

${\sigma }_a=\frac{E}{1-{\nu }^2}\left({\epsilon }_a+\nu {\epsilon }_b\right)$

Substitute 69 GPa for E, 0.2321 for $\nu$, $-193.26\mu$ for ${\epsilon }_a$, and $-926.74\mu$ for ${\epsilon }_b$.

$\begin{array}{c}{\sigma }_a=\frac{69\text{GPa}\times \frac{{\text{10}}^3\text{MPa}}{\text{1}\text{GPa}}}{1-{0.2321}^2}\left(-193.26\times {10}^{-6}+0.2321\left(-926.74\times {10}^{-6}\right)\right)\\ =-59.28\text{MPa}\end{array}$

Find the intermediate principal stress $\left({\sigma }_b\right)$ using the equation.

${\sigma }_b=\frac{E}{1-{\nu }^2}\left({\epsilon }_b+\nu {\epsilon }_a\right)$

Substitute 69 GPa for E, 0.2321 for $\nu$, $-193.26\mu$ for ${\epsilon }_a$, and $-926.74\mu$ for ${\epsilon }_b$.

$\begin{array}{c}{\sigma }_b=\frac{69\text{GPa}\times \frac{{\text{10}}^3\text{MPa}}{\text{1}\text{GPa}}}{1-{0.2321}^2}\left(--926.74\times {10}^{-6}+0.2321\left(-193.26\times {10}^{-6}\right)\right)\\ =-70.9\text{MPa}\end{array}$

Therefore,

The maximum principal stress is is $\underset{\_}{{\sigma }_a=-29.8\text{MPa}}$.

The intermediate principal stress is $\underset{\_}{{\sigma }_b=-70.9\text{MPa}}$.

The orientation of the major principal stress is $\underset{\_}{{\theta }_a=-32.1°}$.

The orientation of the intermediate principal stress is $\underset{\_}{{\theta }_b=57.9°}$.