Chapter 7, Problem 164RP

The state of plane stress shown occurs in a machine component made of a steel with σ_y = 30 ksi. Using the maximum- distortion-energy criterion, determine whether yield will occur when (a) τ_XV = 6 ksi, (b) τ_XV = 12 ksi, (c) τ_XV = 14 ksi. If yield does not occur, determine the corresponding factor of safety.

Fig. P7.164

To determine

Check the yield will occur for the given condition or not?.

Find the corresponding factor of safety for not occurring the yield.

Answer to Problem 164RP

The yielding will $\underset{\_}{\text{not}}$ occur and the factor of safety is $\underset{\_}{1.286}$.

Explanation of Solution

Given information:

The normal stress in x-axis is ${\sigma }_x=24\text{ksi}$.

The normal stress in y-axis is ${\sigma }_y=14\text{ksi}$.

The shearing stress in xy-plane is ${\tau }_{xy}=6\text{ksi}$.

The allowable yield strength of the steel is ${\sigma }_Y=30\text{ksi}$.

Use maximum distortion-energy theory.

Calculation:

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

The minimum principal stress is ${\sigma }_{\min }={\sigma }_z=0$.

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

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

Substitute 24 ksi for ${\sigma }_x$ and 14 ksi for ${\sigma }_y$.

$\begin{array}{c}{\sigma }_{\text{ave}}=\frac{24+14}{2}\\ =19\text{ksi}\end{array}$

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

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

Substitute 24 ksi for ${\sigma }_x$, 14 ksi for ${\sigma }_y$, and 6 ksi for ${\tau }_{xy}$.

$\begin{array}{c}R=\sqrt{{\left(\frac{24-14}{2}\right)}^2+{\left(6\right)}^2}\\ R=7.81\text{ksi}\end{array}$

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

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

Substitute 19 ksi for ${\sigma }_{\text{ave}}$ and 7.81 ksi for R.

$\begin{array}{c}{\sigma }_{\max }=19+7.81\\ =26.81\text{ksi}\end{array}$

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

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

Substitute 19 ksi for ${\sigma }_{\text{ave}}$ and 7.81 ksi for R.

$\begin{array}{c}{\sigma }_b=19-7.81\\ =11.19\text{ksi}\end{array}$

Check the yielding condition using the Maximum-distortion-energy criteria as follows;

$\sqrt{{\sigma }_a^2-{\sigma }_a{\sigma }_b+{\sigma }_b^2}<{\sigma }_Y$

Substitute 26.81 ksi for ${\sigma }_a$, 11.19 ksi for ${\sigma }_b$, and 30 ksi for ${\sigma }_Y$.

$\begin{array}{l}\sqrt{{26.81}^2-\left(26.81\times 11.19\right)+{11.19}^{{}^2}}<30\\ 23.324\text{ksi}<3\text{0}\text{ksi}\end{array}$

The yielding will not occur.

Find the factor of safety (FOS) using the relation.

$FOS=\frac{{\sigma }_Y}{\sqrt{{\sigma }_a^2-{\sigma }_a{\sigma }_b+{\sigma }_b^2}}$

Substitute 30 ksi for ${\sigma }_Y$ and 23.324 ksi for $\sqrt{{\sigma }_a^2-{\sigma }_a{\sigma }_b+{\sigma }_b^2}$.

$\begin{array}{c}FOS=\frac{30}{23.324}\\ =1.286\end{array}$

Therefore, the yielding will $\underset{\_}{\text{not}}$ occur and the factor of safety is $\underset{\_}{1.286}$.

To determine

Check the yield will occur for the given condition or not?.

Find the corresponding factor of safety for not occurring the yield.

Answer to Problem 164RP

The yielding will $\underset{\_}{\text{not}}$ occur and the factor of safety is $\underset{\_}{1.018}$.

Explanation of Solution

Given information:

The normal stress in x-axis is ${\sigma }_x=24\text{ksi}$.

The normal stress in y-axis is ${\sigma }_y=14\text{ksi}$.

The shearing stress in xy-plane is ${\tau }_{xy}=12\text{ksi}$.

The allowable yield strength of the steel is ${\sigma }_Y=30\text{ksi}$.

Use maximum distortion-energy theory.

Calculation:

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

The minimum principal stress is ${\sigma }_{\min }={\sigma }_z=0$.

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

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

Substitute 24 ksi for ${\sigma }_x$ and 14 ksi for ${\sigma }_y$.

$\begin{array}{c}{\sigma }_{\text{ave}}=\frac{24+14}{2}\\ =19\text{ksi}\end{array}$

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

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

Substitute 24 ksi for ${\sigma }_x$, 14 ksi for ${\sigma }_y$, and 12 ksi for ${\tau }_{xy}$.

$\begin{array}{c}R=\sqrt{{\left(\frac{24-14}{2}\right)}^2+{\left(12\right)}^2}\\ R=13\text{ksi}\end{array}$

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

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

Substitute 19 ksi for ${\sigma }_{\text{ave}}$ and 13 ksi for R.

$\begin{array}{c}{\sigma }_{\max }=19+13\\ =32\text{ksi}\end{array}$

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

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

Substitute 19 ksi for ${\sigma }_{\text{ave}}$ and 13 ksi for R.

$\begin{array}{c}{\sigma }_b=19-13\\ =6\text{ksi}\end{array}$

Check the yielding condition using the Maximum-distortion-energy criteria as follows;

$\sqrt{{\sigma }_a^2-{\sigma }_a{\sigma }_b+{\sigma }_b^2}<{\sigma }_Y$

Substitute 32 ksi for ${\sigma }_a$, 6 ksi for ${\sigma }_b$, and 30 ksi for ${\sigma }_Y$.

$\begin{array}{l}\sqrt{{32}^2-\left(32\times 6\right)+6^{{}^2}}<30\\ 29.462\text{ksi}<3\text{0}\text{ksi}\end{array}$

The yielding will not occur.

Find the factor of safety (FOS) using the relation.

$FOS=\frac{{\sigma }_Y}{\sqrt{{\sigma }_a^2-{\sigma }_a{\sigma }_b+{\sigma }_b^2}}$

Substitute 30 ksi for ${\sigma }_Y$ and 29.462 ksi for $\sqrt{{\sigma }_a^2-{\sigma }_a{\sigma }_b+{\sigma }_b^2}$.

$\begin{array}{c}FOS=\frac{30}{29.462}\\ =1.018\end{array}$

Therefore, the yielding will $\underset{\_}{\text{not}}$ occur and the factor of safety is $\underset{\_}{1.018}$.

To determine

Check the yield will occur for the given condition or not?.

Find the corresponding factor of safety for not occurring the yield.

Answer to Problem 164RP

The yielding will occur.

Explanation of Solution

Given information:

The normal stress in x-axis is ${\sigma }_x=24\text{ksi}$.

The normal stress in y-axis is ${\sigma }_y=14\text{ksi}$.

The shearing stress in xy-plane is ${\tau }_{xy}=14\text{ksi}$.

The allowable yield strength of the steel is ${\sigma }_Y=30\text{ksi}$.

Use maximum distortion-energy theory.

Calculation:

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

The minimum principal stress is ${\sigma }_{\min }={\sigma }_z=0$.

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

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

Substitute 24 ksi for ${\sigma }_x$ and 14 ksi for ${\sigma }_y$.

$\begin{array}{c}{\sigma }_{\text{ave}}=\frac{24+14}{2}\\ =19\text{ksi}\end{array}$

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

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

Substitute 24 ksi for ${\sigma }_x$, 14 ksi for ${\sigma }_y$, and 14 ksi for ${\tau }_{xy}$.

$\begin{array}{c}R=\sqrt{{\left(\frac{24-14}{2}\right)}^2+{\left(14\right)}^2}\\ R=14.866\text{ksi}\end{array}$

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

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

Substitute 19 ksi for ${\sigma }_{\text{ave}}$ and 14.866 ksi for R.

$\begin{array}{c}{\sigma }_{\max }=19+14.866\\ =33.866\text{ksi}\end{array}$

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

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

Substitute 19 ksi for ${\sigma }_{\text{ave}}$ and 14.866 ksi for R.

$\begin{array}{c}{\sigma }_b=19-14.866\\ =4.134\text{ksi}\end{array}$

Check the yielding condition using the Maximum-distortion-energy criteria as follows;

$\sqrt{{\sigma }_a^2-{\sigma }_a{\sigma }_b+{\sigma }_b^2}<{\sigma }_Y$

Substitute 33.866 ksi for ${\sigma }_a$, 4.134 ksi for ${\sigma }_b$, and 30 ksi for ${\sigma }_Y$.

$\begin{array}{l}\sqrt{{33.866}^2-\left(33.866\times 4.134\right)+6^{{}^2}}<30\\ 32.294\text{ksi}>3\text{0}\text{ksi}\end{array}$

The yielding will occur.

Therefore, the yielding will occur.