Chapter 7.9, Problem 156P

The given state of plane stress is known to exist on the surface of a machine component. Knowing that E = 200 GPa and G = 77.2 GPa, determine the direction and magnitude of the three principal strains (a) by determining the corresponding state of strain [use Eq. (2.43) and Eq. (2.38)] and then using Mohr's circle for strain, (b) by using Mohr's circle for stress to determine the principal planes and principal stresses and then determining the corresponding strains.

Fig. P7.156

To determine

Find the direction and magnitude of the three principal strains.

Answer to Problem 156P

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:

Find the state of strain and construct the Mohr’s circle for strain.

The modulus of elasticity of the material is $E=200\text{GPa}$.

The modulus of rigidity of the material is $G=77.2\text{GPa}$.

Calculation:

The normal stress in x-axis is ${\sigma }_x=0$.

The normal stress in y-axis is ${\sigma }_y=-150\text{MPa}\times \frac{{\text{10}}^{\text{6}}\text{Pa}}{\text{1}\text{MPa}}=-150\times {10}^6\text{Pa}$.

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

The shearing stress in xy-plane is ${\tau }_{xy}=-75\text{MPa}\times \frac{{\text{10}}^{\text{6}}\text{Pa}}{\text{1}\text{MPa}}=-75\times {10}^6\text{Pa}$.

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

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

Substitute 200 GPa for E and 77.2 GPa for G.

$\begin{array}{c}77.2=\frac{200}{2\left(1+\nu \right)}\\ \nu =0.29534\end{array}$

Find the normal strain in x-axis $\left({\epsilon }_x\right)$ as follows;

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

Substitute 200 GPa for E, 0 for ${\sigma }_x$, 0.29534 for $\nu$, $-150\times {10}^6\text{Pa}$ for ${\sigma }_y$, and 0 for ${\sigma }_z$.

$\begin{array}{c}{\epsilon }_x=\frac{1}{200\text{GPa}\times \frac{{\text{10}}^{\text{9}}\text{Pa}}{\text{1}\text{GPa}}}\left(0-0.29534\left(-150\times {10}^6\right)-0.29534\left(0\right)\right)\\ =2.21505\times {10}^{-4}\end{array}$

Find the normal strain in y-axis $\left({\epsilon }_y\right)$ as follows;

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

Substitute 200 GPa for E, 0 for ${\sigma }_x$, 0.29534 for $\nu$, $-150\times {10}^6\text{Pa}$ for ${\sigma }_y$, and 0 for ${\sigma }_z$.

$\begin{array}{c}{\epsilon }_y=\frac{1}{200\text{GPa}\times \frac{{\text{10}}^{\text{9}}\text{Pa}}{\text{1}\text{GPa}}}\left(-150\times {10}^6-0.29534\left(0\right)-0.29534\left(0\right)\right)\\ =-7.5\times {10}^{-4}\end{array}$

Find the shearing strain $\left({\gamma }_{xy}\right)$ using the relation.

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

Substitute $-75\times {10}^6\text{Pa}$ for ${\tau }_{xy}$ and 77.2 GPa for G.

$\begin{array}{c}{\gamma }_{xy}=\frac{-75\times {10}^6}{77.2\text{GPa}\times \frac{{\text{10}}^9\text{Pa}}{\text{1}\text{GPa}}}\\ =-9.7150\times {10}^{-4}\end{array}$

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

$\begin{array}{c}\frac{{\gamma }_{xy}}{2}=\frac{-9.7150\times {10}^{-4}}{2}\\ =-4.8575\times {10}^{-4}\end{array}$

Construct the Mohr circle as follows;

• Plot the point X by $2.21505\times {10}^{-4}$ to the right of the $\frac{\gamma }{2}$ axis and $4.8575\times {10}^{-4}$ above the $\epsilon$ axis.
• Plot the point Y by $7.5\times {10}^{-4}$ to the left of the $\frac{\gamma }{2}$ axis and $4.8575\times {10}^{-4}$ below 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 1.

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

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

Substitute $2.21505\times {10}^{-4}$ for ${\epsilon }_x$ and $-7.5\times {10}^{-4}$ for ${\epsilon }_y$.

$\begin{array}{c}{\epsilon }_{\text{ave}}=\frac{\left(2.21505\times {10}^{-4}\right)+\left(-7.5\times {10}^{-4}\right)}{2}\\ =-2.642475\times {10}^{-4}\end{array}$

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 $-9.7150\times {10}^{-4}$ for ${\gamma }_{xy}$, $2.21505\times {10}^{-4}$ for ${\epsilon }_x$, and $-7.5\times {10}^{-4}$ for ${\epsilon }_y$.

$\begin{array}{l}\tan 2{\theta }_a=\frac{-9.7150\times {10}^{-4}}{2.21505\times {10}^{-4}-\left(-7.5\times {10}^{-4}\right)}\\ {\theta }_a=-22.5°\end{array}$

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

${\theta }_b={\theta }_a+90°$

Substitute $-22.5°$ for ${\theta }_a$.

$\begin{array}{c}{\theta }_b=-22.5°+90°\\ =67.5°\end{array}$

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 $2.21505\times {10}^{-4}$ for ${\epsilon }_x$, $-7.5\times {10}^{-4}$ for ${\epsilon }_y$, and $-9.7150\times {10}^{-4}$ for ${\gamma }_{xy}$.

$\begin{array}{c}R=\sqrt{{\left(\frac{2.21505\times {10}^{-4}-\left(-7.5\times {10}^{-4}\right)}{2}\right)}^2+{\left(\frac{-9.7150\times {10}^{-4}}{2}\right)}^2}\\ =6.86956\times {10}^{-4}\end{array}$

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

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

Substitute $-2.642475\times {10}^{-4}$ for ${\epsilon }_{\text{ave}}$ and $6.86956\times {10}^{-4}$ for R.

$\begin{array}{c}{\epsilon }_a=-2.642475\times {10}^{-4}+6.86956\times {10}^{-4}\\ =4.227\times {10}^{-4}\times \frac{1\mu }{{10}^{-6}}\\ =423\mu \end{array}$

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

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

Substitute $-2.642475\times {10}^{-4}$ for ${\epsilon }_{\text{ave}}$ and $6.86956\times {10}^{-4}$ for R.

$\begin{array}{c}{\epsilon }_b=-2.642475\times {10}^{-4}-6.86956\times {10}^{-4}\\ =-9.512\times {10}^{-4}\times \frac{1\mu }{{10}^{-6}}\\ =-951\mu \end{array}$

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

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

Substitute 200 GPa for E, 0 for ${\sigma }_x$, 0.29534 for $\nu$, $-150\times {10}^6\text{Pa}$ for ${\sigma }_y$, and 0 for ${\sigma }_z$.

$\begin{array}{c}{\epsilon }_c=\frac{1}{200\text{GPa}\times \frac{{\text{10}}^{\text{9}}\text{Pa}}{\text{1}\text{GPa}}}\left(0-0.29534\left(0\right)-0.29534\left(-150\times {10}^6\right)\right)\\ =-2.21505\times {10}^{-4}\times \frac{1\mu }{{10}^{-6}}\\ =-222\mu \end{array}$

Therefore,

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 $\underset{\_}{{\theta }_b=67.5°}$.

To determine

Find the direction and magnitude of the three principal strains.

Answer to Problem 156P

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

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

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 stress.

The modulus of elasticity of the material is $E=200\text{GPa}$.

The modulus of rigidity of the material is $G=77.2\text{GPa}$.

Calculation:

The normal stress in x-axis is ${\sigma }_x=0$.

The normal stress in y-axis is ${\sigma }_y=-150\text{MPa}\times \frac{{\text{10}}^{\text{6}}\text{Pa}}{\text{1}\text{MPa}}=-150\times {10}^6\text{Pa}$.

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

The shearing stress in xy-plane is ${\tau }_{xy}=-75\text{MPa}\times \frac{{\text{10}}^{\text{6}}\text{Pa}}{\text{1}\text{MPa}}=-75\times {10}^6\text{Pa}$.

Construct the Mohr circle as follows;

• Plot the point X by 0 on the $\tau$ axis and 75 MPa above the $\sigma$ axis.
• Plot the point Y by 150 MPa to the left of the $\tau$ axis and 75 MPa below 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 2.

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

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

Substitute 0 for ${\sigma }_x$ and ­150 MPa for ${\sigma }_y$.

$\begin{array}{c}{\sigma }_{\text{ave}}=\frac{0+\left(-150\right)}{2}\\ =-75\text{MPa}\end{array}$

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 0 for ${\sigma }_x$, ­150 MPa for ${\sigma }_y$, and –75 MPa for ${\tau }_{xy}$.

$\begin{array}{c}R=\sqrt{{\left(\frac{0-\left(-150\right)}{2}\right)}^2+{\left(-75\right)}^2}\\ =106.066\text{MPa}\end{array}$

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

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

Substitute –75 MPa for ${\sigma }_{ave}$ and 106.066 MPa for R.

$\begin{array}{c}{\sigma }_a=-75+106.066\\ =31.066\text{MPa}\end{array}$

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

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

Substitute –75 MPa for ${\sigma }_{ave}$ and 106.066 MPa for R.

$\begin{array}{c}{\sigma }_b=-75-106.066\\ =-181.066\text{MPa}\end{array}$

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

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

Substitute 200 GPa for E and 77.2 GPa for G.

$\begin{array}{c}77.2=\frac{200}{2\left(1+\nu \right)}\\ \nu =0.29534\end{array}$

Find the maximum principal strain $\left({\epsilon }_a\right)$ as follows;

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

Substitute 200 GPa for E, 31.066 MPa for ${\sigma }_a$, 0.29534 for $\nu$, and –181.066 MPa for ${\sigma }_b$.

$\begin{array}{c}{\epsilon }_a=\frac{1}{200\text{GPa}\times \frac{{\text{10}}^3\text{MPa}}{\text{1}\text{GPa}}}\left(31.066-0.29534\left(-181.066\right)\right)\\ =4.23\times {10}^{-4}\times \frac{1\mu }{{10}^{-6}}\\ =423\mu \end{array}$

Find the intermediate principal strain $\left({\epsilon }_b\right)$ as follows;

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

Substitute 200 GPa for E, 31.066 MPa for ${\sigma }_a$, 0.29534 for $\nu$, and –181.066 MPa for ${\sigma }_b$.

$\begin{array}{c}{\epsilon }_b=\frac{1}{200\text{GPa}\times \frac{{\text{10}}^3\text{MPa}}{\text{1}\text{GPa}}}\left(-181.066-0.29534\left(31.066\right)\right)\\ =-9.51\times {10}^{-4}\times \frac{1\mu }{{10}^{-6}}\\ =-951\mu \end{array}$

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 0 for ${\sigma }_x$, ­150 MPa for ${\sigma }_y$, and –75 MPa for ${\tau }_{xy}$.

$\begin{array}{l}\tan 2{\theta }_a=\frac{2\left(-75\right)}{0-\left(-150\right)}\\ {\theta }_a=-22.5°\end{array}$

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

${\theta }_b={\theta }_a+90°$

Substitute $-22.5°$ for ${\theta }_a$.

$\begin{array}{c}{\theta }_b=-22.5°+90°\\ =67.5°\end{array}$

Therefore,

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

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

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

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