Chapter 2, Problem 2.95P

(a)

To determine

Engineering stress and engineering strain.

Answer to Problem 2.95P

The value of engineering stress is $233.86MPa.$

Hence the value of engineering strain is 0.96.

Explanation of Solution

Given data:

  $\begin{array}{l}\text{Tensile force applied at the end of solid bar, }F=9\text{kN,}\\ \text{Initial diamter of solid bar, }{d}_i=7\text{mm,}\\ \text{Final diameter of solid bar, }{d}_o=5\text{mm}\text{.}\end{array}$

The engineering stress and engineering strain.

A tensile load is applied to a solid bar. The bar deforms uniformly and the volume remains constant.

Calculating the initial area of the solid bar,

  $\begin{array}{l}{A}_i=\frac{\pi }{4}{D}^2\\ A_i=\frac{\pi }{4}{\left(7\right)}^2\\ A_i=38.484{\text{mm}}^2\end{array}$

Calculating the final area of the solid bar,

  $\begin{array}{l}{A}_o=\frac{\pi }{4}{d}_o^2\\ A_o=\frac{\pi }{4}{\left(5\right)}^2\\ A_o=19.635{\text{mm}}^2\end{array}$

  $\begin{array}{l}\text{Volume remains constant,}\\ V=A_i{L}_i=A_o{L}_o\\ \Rightarrow A_o{L}_o=A_i{L}_i\\ \Rightarrow \frac{L_o}{L_i}=\frac{A_i}{A_o}\\ \Rightarrow \frac{L_o}{L_i}-1=\frac{A_i}{A_o}-1\\ \Rightarrow \frac{L_o-L_i}{L_i}=\frac{A_i-A_o}{A_o}\\ \Rightarrow \frac{\Delta L}{L_i}=\frac{d_i^2-d_o^2}{d_o^2}\mathrm{..........}(1)\end{array}$

Writing the formula for engineering stress,

  $\begin{array}{l}\text{Engineering stress, }\sigma =\frac{\text{Applied Force}}{\text{Original Area}}\\ \sigma =\frac{\text{9}\times 1000}{38.484\times {10}^{-6}}\\ \sigma =233.86\text{MPa}\end{array}$

  $\text{The value of engineering stress is 233}\text{.86 MPa}\text{.}$

Writing the formula for engineering strain,

  $\begin{array}{l}\text{Engineering strain, }\epsilon =\frac{\text{Change in length}}{\text{Original length}}\\ \epsilon =\frac{\Delta L}{L_i}\\ \text{From equation (1),}\\ \frac{\Delta L}{L_i}=\frac{d_i^2-d_o^2}{d_o^2}\\ \epsilon =\frac{\Delta L}{L_i}=\frac{d_i^2-d_o^2}{d_o^2}\\ \epsilon =\frac{7^2-5^2}{5^2}\\ \epsilon =0.96\end{array}$

Hence the value of engineering strain is 0.96.

(b)

To determine

True stress and true strain

Answer to Problem 2.95P

  $\text{The value of true stress is 458 MPa}\text{.}$

The value of true strain is 0.673.

Explanation of Solution

Writing the formula for true stress,

  $\begin{array}{l}\text{True stress, }{\sigma }_t=\sigma \left(1+engineeringstrain\right)\\ {\sigma }_t=\sigma \left(1+\epsilon \right)\\ {\sigma }_t=233.86\left(1+0.6\right)\\ {\sigma }_t=374.176\text{MPa}\text{.}\end{array}$

  $\text{The value of true stress is 458 MPa}\text{.}$

Writing the formula for true strain,

  $\begin{array}{l}\text{True strain, }{\epsilon }_t=\ln (1+engineeringstrain)\\ {\epsilon }_t=\ln \left(1+\epsilon \right)\\ {\epsilon }_t=\ln \left(1+0.96\right)\\ {\epsilon }_t=0.673\end{array}$

Hence, the value of true strain is 0.673.

(c)

To determine

The engineering stress and engineering strain when the true stress is $\text{345 MPa}\text{.}$

Answer to Problem 2.95P

The value of engineering stress is $220.8\text{MPa}$

The value of engineering strain is 0.5625.

Explanation of Solution

Now if the given condition is −

  $\begin{array}{l}\text{True stress, }{\sigma }_t=345\text{MPa,}\\ \text{Final diamter, }{d}_o=5.6\text{mm}\end{array}$

The final area can be given as,

  $\begin{array}{l}{A}_o=\frac{\pi }{4}{d}_o^2\\ A_o=\frac{\pi }{4}{\left(5.6\right)}^2\\ A_o=24.63{\text{mm}}^2\end{array}$

From the true stress formula,

  $\begin{array}{l}{\sigma }_t=\sigma \left(1+\epsilon \right)\\ {\sigma }_t=\sigma \left(\frac{A_{{}_i}}{A_o}\right)\\ {\sigma }_t=\sigma \left(\frac{d_i^2}{A_o^2}\right)\\ 345=\sigma \left(\frac{7^2}{{5.6}^2}\right)\\ \sigma =220.8\text{MPa}\text{.}\end{array}$

Hence the value of engineering stress is $220.8\text{MPa}$

From equation (1),

  $\begin{array}{l}\frac{\Delta L}{L_i}=\frac{d_i^2-d_o^2}{d_o^2}\\ \text{And engineering strain is also given as,}\\ \epsilon =\frac{\Delta L}{L_i}=\frac{d_i^2-d_o^2}{d_o^2}\\ \epsilon =\frac{d_i^2-d_o^2}{d_o^2}\\ \epsilon =\frac{7^2-{5.6}^2}{{5.6}^2}\\ \epsilon =0.5625\end{array}$

Hence the value of engineering strain is 0.5625.