Chapter 2, Problem 2.98P

To determine

The stress-strain curve.

Explanation of Solution

Given:

The initial area is $A_0=3.5\times {10}^{-5}{\text{m}}^{\text{2}}$ .

The final area is $A_f=1.0\times {10}^{-5}{\text{in}}^{\text{2}}$ .

The original length is $l_0=50\text{mm}$ .

Formula used:

The true strain is given as,

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

Here, $l$ is the extension in the specimen.

The Actual area is given as,

  $A=\frac{A_0}{\epsilon }$

The actual are for the initial load is given as,

  $A=A_0$

The final length is given as,

  $l\left(\text{mm}\right)=\Delta l+l_0$

Here, $\Delta l$ is the extension in the specimen.

The stress is given as,

  $\sigma =\frac{P}{A}$

Here, P is the applied load.

Calculation:

For $P=7.10\text{kN}$ , the values can be calculated as,

The final length can be calculated as,

  $\begin{array}{l}l=\Delta l+l_0\\ l=0\text{mm}+50\text{mm}\\ l=50\text{mm}\end{array}$

The strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{l}{l_0}\right)\\ \epsilon =\ln \left(\frac{50\text{mm}}{50\text{mm}}\right)\\ \epsilon =0\end{array}$

The actual area can be calculated as,

  $\begin{array}{l}A=A_0\\ A=3.5\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The stress can be calculated as,

  $\begin{array}{l}\sigma =\frac{P}{A}\\ \sigma =\frac{7.10\text{kN}\left(\frac{1000\text{N}}{1\text{kN}}\right)}{3.5\times {10}^{-5}{\text{m}}^{\text{2}}}\left(\frac{{10}^{-6}\text{MPa}}{1\frac{\text{N}}{{\text{m}}^{\text{2}}}}\right)\\ \sigma =203\text{MPa}\end{array}$

For $P=11.1\text{kN}$ , the values can be calculated as,

The final length can be calculated as,

  $\begin{array}{l}l=\Delta l+l_0\\ l=0.5\text{mm}+50\text{mm}\\ l=50.5\text{mm}\end{array}$

The strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{l}{l_0}\right)\\ \epsilon =\ln \left(\frac{50.5\text{mm}}{50\text{mm}}\right)\\ \epsilon =0.00995\end{array}$

The actual area can be calculated as,

  $\begin{array}{l}A=\frac{A_0}{\epsilon }\\ A=\frac{3.5\times {10}^{-5}{\text{m}}^{\text{2}}}{0.00995}\\ A=3.46\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The stress can be calculated as,

  $\begin{array}{l}\sigma =\frac{P}{A}\\ \sigma =\frac{7.10\text{kN}\left(\frac{1000\text{N}}{1\text{kN}}\right)}{3.46\times {10}^{-5}{\text{m}}^{\text{2}}}\left(\frac{{10}^{-6}\text{MPa}}{1\frac{\text{N}}{{\text{m}}^{\text{2}}}}\right)\\ \sigma =321\text{MPa}\end{array}$

For $P=13.3\text{kN}$ , the values can be calculated as,

The final length can be calculated as,

  $\begin{array}{l}l=\Delta l+l_0\\ l=2\text{mm}+50\text{mm}\\ l=52\text{mm}\end{array}$

The strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{l}{l_0}\right)\\ \epsilon =\ln \left(\frac{52\text{mm}}{50\text{mm}}\right)\\ \epsilon =0.0392\end{array}$

The actual area can be calculated as,

  $\begin{array}{l}A=\frac{A_0}{\epsilon }\\ A=\frac{3.5\times {10}^{-5}{\text{m}}^{\text{2}}}{0.0392}\\ A=3.36\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The stress can be calculated as,

  $\begin{array}{l}\sigma =\frac{P}{A}\\ \sigma =\frac{7.10\text{kN}\left(\frac{1000\text{N}}{1\text{kN}}\right)}{3.36\times {10}^{-5}{\text{m}}^{\text{2}}}\left(\frac{{10}^{-6}\text{MPa}}{1\frac{\text{N}}{{\text{m}}^{\text{2}}}}\right)\\ \sigma =396\text{MPa}\end{array}$

For $P=16\text{kN}$ , the values can be calculated as,

The final length can be calculated as,

  $\begin{array}{l}l=\Delta l+l_0\\ l=5\text{mm}+50\text{mm}\\ l=55\text{mm}\end{array}$

The strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{l}{l_0}\right)\\ \epsilon =\ln \left(\frac{55\text{mm}}{50\text{mm}}\right)\\ \epsilon =0.0953\end{array}$

The actual area can be calculated as,

  $\begin{array}{l}A=\frac{A_0}{\epsilon }\\ A=\frac{3.5\times {10}^{-5}{\text{m}}^{\text{2}}}{0.0953}\\ A=3.18\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The stress can be calculated as,\

  $\begin{array}{l}\sigma =\frac{P}{A}\\ \sigma =\frac{7.10\text{kN}\left(\frac{1000\text{N}}{1\text{kN}}\right)}{3.18\times {10}^{-5}{\text{m}}^{\text{2}}}\left(\frac{{10}^{-6}\text{MPa}}{1\frac{\text{N}}{{\text{m}}^{\text{2}}}}\right)\\ \sigma =503\text{MPa}\end{array}$

For $P=18.7\text{kN}$ , the values can be calculated as,

The final length can be calculated as,

  $\begin{array}{l}l=\Delta l+l_0\\ l=10\text{mm}+50\text{mm}\\ l=60\text{mm}\end{array}$

The strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{l}{l_0}\right)\\ \epsilon =\ln \left(\frac{60\text{mm}}{50\text{mm}}\right)\\ \epsilon =0.182\end{array}$

The actual area can be calculated as,

  $\begin{array}{l}A=\frac{A_0}{\epsilon }\\ A=\frac{3.5\times {10}^{-5}{\text{m}}^{\text{2}}}{0.182}\\ A=2.92\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The stress can be calculated as,

  $\begin{array}{l}\sigma =\frac{P}{A}\\ \sigma =\frac{7.10\text{kN}\left(\frac{1000\text{N}}{1\text{kN}}\right)}{2.92\times {10}^{-5}{\text{m}}^{\text{2}}}\left(\frac{{10}^{-6}\text{MPa}}{1\frac{\text{N}}{{\text{m}}^{\text{2}}}}\right)\\ \sigma =640\text{MPa}\end{array}$

For $P=20\text{kN}$ , the values can be calculated as,

The final length can be calculated as,

  $\begin{array}{l}l=\Delta l+l_0\\ l=15.2\text{mm}+50\text{mm}\\ l=65.2\text{mm}\end{array}$

The strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{l}{l_0}\right)\\ \epsilon =\ln \left(\frac{65.2\text{mm}}{50\text{mm}}\right)\\ \epsilon =0.262\end{array}$

The actual area can be calculated as,

  $\begin{array}{l}A=\frac{A_0}{\epsilon }\\ A=\frac{3.5\times {10}^{-5}{\text{m}}^{\text{2}}}{0.262}\\ A=2.69\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The stress can be calculated as,

  $\begin{array}{l}\sigma =\frac{P}{A}\\ \sigma =\frac{7.10\text{kN}\left(\frac{1000\text{N}}{1\text{kN}}\right)}{2.69\times {10}^{-5}{\text{m}}^{\text{2}}}\left(\frac{{10}^{-6}\text{MPa}}{1\frac{\text{N}}{{\text{m}}^{\text{2}}}}\right)\\ \sigma =743\text{MPa}\end{array}$

For $P=20.5\text{kN}$ , the values can be calculated as,

The final length can be calculated as,

  $\begin{array}{l}l=\Delta l+l_0\\ l=21.5\text{mm}+50\text{mm}\\ l=71.5\text{mm}\end{array}$

The strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{l}{l_0}\right)\\ \epsilon =\ln \left(\frac{71.5\text{mm}}{50\text{mm}}\right)\\ \epsilon =0.357\end{array}$

The actual area can be calculated as,

  $\begin{array}{l}A=\frac{A_0}{\epsilon }\\ A=\frac{3.5\times {10}^{-5}{\text{m}}^{\text{2}}}{0.357}\\ A=2.45\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The stress can be calculated as,

  $\begin{array}{l}\sigma =\frac{P}{A}\\ \sigma =\frac{7.10\text{kN}\left(\frac{1000\text{N}}{1\text{kN}}\right)}{2.45\times {10}^{-5}{\text{m}}^{\text{2}}}\left(\frac{{10}^{-6}\text{MPa}}{1\frac{\text{N}}{{\text{m}}^{\text{2}}}}\right)\\ \sigma =837\text{MPa}\end{array}$

For $P=14.7\text{kN}$ , the values can be calculated as,

The final length can be calculated as,

  $\begin{array}{l}l=\Delta l+l_0\\ l=25\text{mm}+50\text{mm}\\ l=75\text{mm}\end{array}$

The strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{l}{l_0}\right)\\ \epsilon =\ln \left(\frac{75\text{mm}}{50\text{mm}}\right)\\ \epsilon =0.405\end{array}$

The actual area can be calculated as,

  $\begin{array}{l}A=\frac{A_0}{\epsilon }\\ A=\frac{3.5\times {10}^{-5}{\text{m}}^{\text{2}}}{0.405}\\ A=2.33\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

The stress can be calculated as,

  $\begin{array}{l}\sigma =\frac{P}{A}\\ \sigma =\frac{7.10\text{kN}\left(\frac{1000\text{N}}{1\text{kN}}\right)}{2.33\times {10}^{-5}{\text{m}}^{\text{2}}}\left(\frac{{10}^{-6}\text{MPa}}{1\frac{\text{N}}{{\text{m}}^{\text{2}}}}\right)\\ \sigma =631\text{MPa}\end{array}$

The stress-strain curve from the above calculation is given as,

  

Figure 1.1