Chapter 6.13, Problem 51AAP

A 20-cm-long rod with a diameter of 0.250 cm is loaded with a 5000 N weight. If the diameter decreases to 0.210 cm, determine (a) the engineering stress and strain at this load and (b) the true stress and strain at this load.

To determine

Engineering stress and strain has to be determined.

Answer to Problem 51AAP

The magnitude of engineering stress is 1019 MPa.

The magnitude of engineering strain is 0.417.

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 $0.25\text{ cm}$ for $d_0$ in equation (I).

$\begin{array}{c}{A}_0=\frac{\pi }{4}{\left(0.25\text{ cm}\right)}^2\\ =0.04909{\text{ cm}}^2\times {\left(\frac{1\text{ m}}{100\text{ cm}}\right)}^2\\ =4.909\times {10}^{-6}{\text{ m}}^2\end{array}$

Substitute $0.21\text{ cm}$ for $d_f$ in equation (II).

$\begin{array}{c}{A}_f=\frac{\pi }{4}{\left(0.21\text{cm}\right)}^2\\ =0.03464{\text{ cm}}^2\times {\left(\frac{1\text{m}}{100\text{cm}}\right)}^2\\ =3.464\times {10}^{-6}{\text{ m}}^2\end{array}$

Substitute $5000\text{N}$ for $F$ and $4.909\times {10}^{-6}{\text{ m}}^2$ for $A_0$ in equation (IV).

$\begin{array}{c}\sigma =\frac{5000\text{N}}{4.909\times {10}^{-6}{\text{ m}}^2}\\ =1019\times {10}^6\text{ Pa}\times \frac{1\text{MPa}}{{10}^6\text{ Pa}}\\ \text{=1019}\text{MPa}\end{array}$

Thus, the magnitude of engineering stress is 1019 MPa.

Substitute $4.909\times {10}^{-6}{\text{ m}}^2$ for $A_0$ and $3.464\times {10}^{-6}{\text{ m}}^2$ for $A_f$ in equation (VI).

$\begin{array}{c}\epsilon =\frac{4.909\times {10}^{-6}{\text{ m}}^2}{3.464\times {10}^{-6}{\text{ m}}^2}-1\\ =1.417-1\\ =0.417\end{array}$

Thus, the magnitude of engineering strain is 0.417.

To determine

true stress and strain has to be determined.

Answer to Problem 51AAP

The magnitude of true stress is 1443 MPa.

The magnitude of true strain is 0.349.

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 $5000\text{ N}$ for F and $3.464\times {10}^{-6}{\text{ m}}^2$ for $A_f$ in equation (VI).

$\begin{array}{c}{\sigma }_T=\frac{5000\text{ N}}{3.464\times {10}^{-6}{\text{ m}}^2}\\ =1443\times {10}^6\text{ Pa}\times \frac{1\text{ MPa}}{{10}^6\text{ Pa}}\\ =1443\text{ MPa}\end{array}$

Thus, the magnitude of true stress is 1443 MPa.

Substitute $4.909\times {10}^{-6}{\text{ m}}^2$ for $A_0$ and $3.464\times {10}^{-6}{\text{ m}}^2$ for $A_f$ in equation (VII).

$\begin{array}{c}{\epsilon }_T=\ln \left(\frac{4.909\times {10}^{-6}{\text{ m}}^2}{3.464\times {10}^{-6}{\text{ m}}^2}\right)\\ =\ln \left(1.418\right)\\ =0.349\end{array}$

Thus, the magnitude of true strain is 0.349.