A tension test was performed on a steel specimen having an original diameter of 0.503 in. and gage length of 2.00 in. The data is listed in the table. Plot the stress–strain diagram and determine approximately the modulus of elasticity, the yield stress, the ultimate stress, and the fracture stress. Use a scale of 1 in. = 20 ksi and 1 in. = 0.05 in./in. Redraw the elastic region, using the same stress scale but a strain scale of 1 in. = 0.001 in./in.
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Question
The stress–strain diagram for an aluminum alloy specimen having an original diameter of 0.5 in. and a gage length of 2 in. is given in the figure. Determine approximately 3 the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support.
Solution
Step 1 of 3
Calculate the modulus of elasticity from the stress-strain diagram.
\(\mathrm{E}_{\text {approx }}=\frac{\sigma_{@ \varepsilon=0.005}-\sigma_{@ \varepsilon=0}}{\varepsilon_{@ \varepsilon=0.05-} \varepsilon_{@ \varepsilon=0}}\)
Here, \(E_{@ \varepsilon=0.005}\) is the stress of the alloy when strain is 0.005 in./in., \(E_{@ \varepsilon=0}\) is stress of the alloy when strain is 0 in./in., \(\varepsilon_{@ \varepsilon=0.005}\) is strain of the alloy when strain is 0.005 in./in., and \(\varepsilon_{@ \varepsilon=0}\) is strain of the alloy when strain is 0 in./in..
Substitute 50 ksi for \(E_{@ \varepsilon=0.005}\), \(0\ \mathrm{ksi}\) ksi for \(E_{@ \varepsilon=0}\), 0.005 in./in. for \(\varepsilon_{@ \varepsilon=0.005}\), and 0 in./in. for \(\varepsilon_{@ \varepsilon=0}\).
\(\begin{aligned}E_{\text{approx }}&=\frac{50-0}{0.005-0}\\ &=10\times10^3\mathrm{\ ksi}\end{aligned}\)
Therefore, the modulus of elasticity is \(10 \times 10^{3}\ \mathrm{ksi}\)
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