Chapter 6.13, Problem 50AAP

A 0.505-in.-diameter aluminum alloy test bar is subjected to a load of 25,000 lb. If the diameter of the bar is 0.490 in. at this load, determine (a) the engineering stress and strain and (b) the true stress and strain.

To determine

Engineering stress and strain has to be determined

Answer to Problem 50AAP

The magnitude of engineering stress is 125,000 psi.

The magnitude of engineering strain is 0.060.

Explanation of Solution

Write the expression for initial area of the bar:

$A_0=\frac{\pi }{4}{d}_0^2$ (I)

Here, initial diameter of the bar is $d_0$.

Write the expression for initial area of the bar:

$A_f=\frac{\pi }{4}{d}_f^2$ (II)

Here, final diameter of the bar is $d_f$.

The volume of the bar before and after deformation is same.

$V_i=V_f$

$A_0{l}_0=A_f{l}_f$

$\frac{A_0}{A_f}=\frac{l_f}{l_0}$ (III)

Write the expression for the engineering stress as given as follows.

$\sigma =\frac{F}{A_0}$ (IV)

Here, applied force is $F$ and area of the bar is $A_0$.

Write the expression for the engineering strain as given as follows.

$\epsilon =\frac{l_f-l_0}{l_0}$

$\epsilon =\frac{l_f}{l_0}-\frac{l_0}{l_0}$ (V)

Substitute equation (III) in equation (V).

$\epsilon =\frac{A_0}{A_f}-1$ (VI)

Conclusion:

Substitute $0.505\text{ in}\text{.}$ for $d_0$ in equation (I).

$\begin{array}{c}{A}_0=\frac{\pi }{4}{\left(0.505\text{ in}\text{.}\right)}^2\\ =0.200{\text{ in}}^2\end{array}$

Substitute $0.490\text{ in}\text{.}$ for $d_f$ in equation (II).

$\begin{array}{c}{A}_f=\frac{\pi }{4}{\left(0.490\text{ in}\text{.}\right)}^2\\ =0.1886{\text{ in}}^2\end{array}$

Substitute $25,000{\text{ lb}}_f$ for $F$ and $0.200{\text{ in}}^2$ for $A_0$ in equation (IV).

$\begin{array}{c}\sigma =\frac{25,000{\text{ lb}}_f}{0.200{\text{ in}}^2}\\ =125,000\text{ psi}\end{array}$

Thus, the magnitude of engineering stress is 125,000 psi.

Substitute $0.200{\text{ in}}^2$ for $A_0$ and $0.1886{\text{ in}}^2$ for $A_f$ in equation (VI).

$\begin{array}{c}\epsilon =\frac{0.200{\text{ in}}^2}{0.1886{\text{ in}}^2}-1\\ =1.0604-1\\ =0.060\end{array}$

Thus, the magnitude of engineering strain is 0.060.

To determine

true stress and strain has to be determined

Answer to Problem 50AAP

The magnitude of true stress is 132,555.67 psi.

The magnitude of true strain is 0.0587.

Explanation of Solution

Write the expression for the true stress as given as follows:

${\sigma }_T=\frac{F}{A_f}$ (VII)

Write the expression for the true strain as given as follows:

$\epsilon \text{=ln}\left(\frac{l_f}{l_0}\right)$

$\epsilon \text{=ln}\left(\frac{A_0}{A_f}\right)$ (VIII)

Conclusion:

Substitute $25,000{\text{ lb}}_f$ for F and $0.1886{\text{in}}^2$ for $A_f$ in equation (VI)

$\begin{array}{c}{\sigma }_T=\frac{25,000{\text{ lb}}_f}{0.1886{\text{in}}^2}\\ =132,555.67\text{ psi}\end{array}$

Thus, the magnitude of true stress is 132,555.67 psi.

Substitute $0.200{\text{ in}}^2$ for $A_0$ and $0.1886{\text{ in}}^2$ for $A_f$ in equation (VII)

$\begin{array}{c}{\epsilon }_T=\ln \left(\frac{0.200{\text{ in}}^2}{0.1886{\text{ in}}^2}\right)\\ =\ln \left(1.0604\right)\\ =0.0587\end{array}$

Thus, the magnitude of true strain is 0.0587.