Solution: A tension test was performed on a steel specimen

Step 1 of 5

Calculate the original cross-sectional area of the bar.

\(A = \frac{{\pi {d^2}}}{4}\)

Here, \(d\) is the original diameter of the bar.

Substitute \(0.503\;{\rm{in}}\) for \(d\).

\(A = \frac{{\pi {{\left( {0.503} \right)}^2}}}{4}\)

\( = 0.199\;{\rm{i}}{{\rm{n}}^2}\)

Find the stress using the following equation:

\(\sigma  = \frac{P}{A}\)

Here, \(P\) is the load applied, \(A\) is the original cross-sectional area of the bar.

Find the strain using the following relation:

\(\varepsilon  = \frac{{\delta L}}{L}\)

Here, elongation is \(\delta L\) and gauge length is \(L\).

The values of stress and strain are tabulated below:

\(\begin{array}{|l|l|l|l|l|l|}
\hline \begin{array}{l}
\text { Load } \\
\text { (kips) }
\end{array} & \begin{array}{l}
\text { Elongation } \\
\text { (in.) }
\end{array} & \begin{array}{l}
\text { Area } \\
\text { (in.2) }
\end{array} & \begin{array}{l}
\text { Length } \\
\text { (in.) }
\end{array} & \begin{array}{l}
\text { Strain, } \varepsilon \\
\text { (in./in.) }
\end{array} & \begin{array}{l}
\text { Stress, } \sigma \\
\text { (ksi) }
\end{array} \\
\hline 0.00 & 0.00000 & 0.199 & 2 & 0.00000 & 0.00 \\
\hline 1.50 & 0.00050 & 0.199 & 2 & 0.00025 & 7.54 \\
\hline 4.60 & 0.00150 & 0.199 & 2 & 0.00075 & 23.12 \\
\hline 8.00 & 0.00250 & 0.199 & 2 & 0.00125 & 40.20 \\
\hline 11.00 & 0.00350 & 0.199 & 2 & 0.00175 & 55.28 \\
\hline 11.80 & 0.00500 & 0.199 & 2 & 0.00250 & 59.30 \\
\hline 11.80 & 0.00800 & 0.199 & 2 & 0.00400 & 59.30 \\
\hline 12.00 & 0.02000 & 0.199 & 2 & 0.01000 & 60.30 \\
\hline 16.60 & 0.04000 & 0.199 & 2 & 0.02000 & 83.42 \\
\hline 20.00 & 0.10000 & 0.199 & 2 & 0.05000 & 100.50 \\
\hline 21.50 & 0.28000 & 0.199 & 2 & 0.14000 & 108.04 \\
\hline 19.50 & 0.40000 & 0.199 & 2 & 0.20000 & 97.99 \\
\hline 18.50 & 0.46000 & 0.199 & 2 & 0.23000 & 92.96 \\
\hline
\end{array}\)