Chapter 7.5, Problem 97P

(a)

To determine

The normal stress ${\sigma }_0$ at which rupture of the component should be expected using Mohr’s criterion.

Answer to Problem 97P

The normal stress $\underset{\_}{{\sigma }_0=8\text{ksi}}$.

Explanation of Solution

Given information:

The state of plane stress components are ${\sigma }_x={\sigma }_0\text{, }{\sigma }_y=\frac{1}{2}{\sigma }_0\text{,}$ and ${\tau }_{xy}=0$.

The uniaxial tension for the aluminum alloy is ${\sigma }_{UT}=8\text{ksi}$.

The uniaxial compression for the aluminum alloy is ${\sigma }_{UC}=20\text{ksi}$.

Calculation:

Calculate the average normal stress $\left({\sigma }_{\text{avg}}\right)$ as shown below.

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

Substitute ${\sigma }_0$ for ${\sigma }_x$ and $\frac{1}{2}{\sigma }_0$ for ${\sigma }_y$.

$\begin{array}{c}{\sigma }_{\text{avg}}=\frac{{\sigma }_0+\frac{1}{2}{\sigma }_0}{2}\\ =\frac{3}{2}\frac{{\sigma }_0}{2}\\ =\frac{3}{4}{\sigma }_0\end{array}$

Calculate the value of ${\tau }_{xy}$ as shown below.

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

Substitute ${\sigma }_0$ for ${\sigma }_x$, $\frac{1}{2}{\sigma }_0$ for ${\sigma }_y$, and $0$ for ${\tau }_{xy}$.

$\begin{array}{c}R=\sqrt{{\left(\frac{{\sigma }_0-\frac{1}{2}{\sigma }_0}{2}\right)}^2+0}\\ =\sqrt{{\left(\frac{1}{4}{\sigma }_0\right)}^2}\\ =\frac{1}{4}{\sigma }_0\end{array}$

Calculate principal stresses $\left({\sigma }_{\max }\text{and}{\sigma }_{\min }\right)$ as shown below.

${\sigma }_{\max ,\min }={\sigma }_{\text{avg}}\pm R$

Substitute $\frac{3}{4}{\sigma }_0$ for ${\sigma }_{\text{avg}}$ and $\frac{1}{4}{\sigma }_0$ for R.

$\begin{array}{l}{\sigma }_{a,b}=\frac{3}{4}{\sigma }_0\pm \frac{1}{4}{\sigma }_0\\ {\sigma }_a={\sigma }_0\\ {\sigma }_b=\frac{1}{2}{\sigma }_0\end{array}$

Hence, stress point lies in the first quadrant.

Sketch the Mohr’s criterion for the machine component as shown in Figure 1.

Refer to Figure 1.

The principal stresses for (a) in the first quadrant of boundary.

Calculate the value of ${\sigma }_0$ s shown below.

${\sigma }_a={\sigma }_0={\sigma }_{UT}$

Substitute $8\text{ksi}$ for ${\sigma }_{UT}$.

${\sigma }_0=8\text{ksi}$

Therefore, the normal stress $\underset{\_}{{\sigma }_0=8\text{ksi}}$.

(b)

To determine

The normal stress ${\sigma }_0$ at which rupture of the component should be expected using Mohr’s criterion.

Answer to Problem 97P

The normal stress $\underset{\_}{{\sigma }_0=6.67\text{ksi}}$.

Explanation of Solution

Given information:

The state of plane stress components are ${\sigma }_x={\sigma }_0\text{, }{\sigma }_y=-\frac{1}{2}{\sigma }_0\text{,}$ and ${\tau }_{xy}=0$.

The uniaxial tension for the aluminum alloy is ${\sigma }_{UT}=8\text{ksi}$.

The uniaxial compression for the aluminum alloy is ${\sigma }_{UC}=20\text{ksi}$.

Calculation:

Calculate the average normal stress $\left({\sigma }_{\text{avg}}\right)$ as shown below.

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

Substitute ${\sigma }_0$ for ${\sigma }_x$ and $-\frac{1}{2}{\sigma }_0$ for ${\sigma }_y$.

$\begin{array}{c}{\sigma }_{\text{avg}}=\frac{{\sigma }_0+\left(-\frac{1}{2}{\sigma }_0\right)}{2}\\ =\frac{1}{2}\frac{{\sigma }_0}{2}\\ =\frac{1}{4}{\sigma }_0\end{array}$

Calculate the value of ${\tau }_{xy}$ as shown below.

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

Substitute ${\sigma }_0$ for ${\sigma }_x$, $-\frac{1}{2}{\sigma }_0$ for ${\sigma }_y$, and $0$ for ${\tau }_{xy}$.

$\begin{array}{c}R=\sqrt{{\left(\frac{{\sigma }_0-\left(-\frac{1}{2}{\sigma }_0\right)}{2}\right)}^2+0}\\ =\sqrt{{\left(\frac{3}{4}{\sigma }_0\right)}^2}\\ =\frac{3}{4}{\sigma }_0\end{array}$

Calculate principal stresses $\left({\sigma }_{\max }\text{and}{\sigma }_{\min }\right)$ as shown below.

${\sigma }_{\max ,\min }={\sigma }_{\text{avg}}\pm R$

Substitute $\frac{1}{4}{\sigma }_0$ for ${\sigma }_{\text{avg}}$ and $\frac{3}{4}{\sigma }_0$ for R.

$\begin{array}{l}{\sigma }_{a,b}=\frac{1}{4}{\sigma }_0\pm \frac{3}{4}{\sigma }_0\\ {\sigma }_a={\sigma }_0\\ {\sigma }_b=-\frac{1}{2}{\sigma }_0\end{array}$

Hence, stress point lies in the forth quadrant.

Refer to Figure 1.

The principal stresses for (b) in the forth quadrant of boundary.

Calculate the value of ${\sigma }_0$ s shown below.

$\frac{{\sigma }_a}{{\sigma }_{UT}}-\frac{{\sigma }_b}{{\sigma }_{UC}}=1$

Substitute ${\sigma }_0$ for ${\sigma }_a$, $-\frac{1}{2}{\sigma }_0$ for ${\sigma }_b$, $8\text{ksi}$ for ${\sigma }_{UT}$, and $20\text{ksi}$ for ${\sigma }_{UC}$.

$\begin{array}{l}\frac{{\sigma }_0}{8}-\frac{\left(-\frac{1}{2}{\sigma }_0\right)}{20}=1\\ {\sigma }_0\left(\frac{1}{8}+\frac{1}{40}\right)=1\\ \frac{6}{40}{\sigma }_0=1\\ {\sigma }_0=6.67\text{ksi}\end{array}$

Therefore, the normal stress $\underset{\_}{{\sigma }_0=6.67\text{ksi}}$.

(c)

To determine

The normal stress ${\sigma }_0$ at which rupture of the component should be expected using Mohr’s criterion.

Answer to Problem 97P

The normal stress $\underset{\_}{{\sigma }_0=8.89\text{ksi}}$.

Explanation of Solution

Given information:

The state of plane stress components are ${\sigma }_x=-{\sigma }_0\text{, }{\sigma }_y=\frac{1}{2}{\sigma }_0\text{,}$ and ${\tau }_{xy}=0$.

The uniaxial tension for the aluminum alloy is ${\sigma }_{UT}=8\text{ksi}$.

The uniaxial compression for the aluminum alloy is ${\sigma }_{UC}=20\text{ksi}$.

Calculation:

Calculate the average normal stress $\left({\sigma }_{\text{avg}}\right)$ as shown below.

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

Substitute $-{\sigma }_0$ for ${\sigma }_x$ and $\frac{1}{2}{\sigma }_0$ for ${\sigma }_y$.

$\begin{array}{c}{\sigma }_{\text{avg}}=\frac{-{\sigma }_0+\frac{1}{2}{\sigma }_0}{2}\\ =-\frac{1}{2}\frac{{\sigma }_0}{2}\\ =-\frac{1}{4}{\sigma }_0\end{array}$

Calculate the value of ${\tau }_{xy}$ as shown below.

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

Substitute $-{\sigma }_0$ for ${\sigma }_x$, $\frac{1}{2}{\sigma }_0$ for ${\sigma }_y$, and $0$ for ${\tau }_{xy}$.

$\begin{array}{c}R=\sqrt{{\left(\frac{-{\sigma }_0-\left(\frac{1}{2}{\sigma }_0\right)}{2}\right)}^2+0}\\ =\sqrt{{\left(-\frac{3}{4}{\sigma }_0\right)}^2}\\ =\frac{3}{4}{\sigma }_0\end{array}$

Calculate principal stresses $\left({\sigma }_{\max }\text{and}{\sigma }_{\min }\right)$ as shown below.

${\sigma }_{\max ,\min }={\sigma }_{\text{avg}}\pm R$

Substitute $-\frac{1}{4}{\sigma }_0$ for ${\sigma }_{\text{avg}}$ and $\frac{3}{4}{\sigma }_0$ for R.

$\begin{array}{l}{\sigma }_{a,b}=-\frac{1}{4}{\sigma }_0\pm \frac{3}{4}{\sigma }_0\\ {\sigma }_a=\frac{1}{2}{\sigma }_0\\ {\sigma }_b=-{\sigma }_0\end{array}$

Hence, stress point lies in the forth quadrant.

Refer to Figure 1.

The principal stresses for (b) in the forth quadrant of boundary.

Calculate the value of ${\sigma }_0$ s shown below.

$\frac{{\sigma }_a}{{\sigma }_{UT}}-\frac{{\sigma }_b}{{\sigma }_{UC}}=1$

Substitute $\frac{1}{2}{\sigma }_0$ for ${\sigma }_a$, $-{\sigma }_0$ for ${\sigma }_b$, $8\text{ksi}$ for ${\sigma }_{UT}$, and $20\text{ksi}$ for ${\sigma }_{UC}$.

$\begin{array}{l}\frac{\frac{1}{2}{\sigma }_0}{8}-\frac{\left(-{\sigma }_0\right)}{20}=1\\ {\sigma }_0\left(\frac{1}{16}+\frac{1}{20}\right)=1\\ \frac{9}{80}{\sigma }_0=1\\ {\sigma }_0=8.89\text{ksi}\end{array}$

Therefore, the normal stress $\underset{\_}{{\sigma }_0=8.89\text{ksi}}$.