A steel specimen is tested in tension. The specimen is 1 in. wide by 0.5 in. thick in the test region. By monitoring the load dial of the testing machine, it was found that the specimen yielded at a load of 36 kips and fractured at 48 kips.

a. Determine the tensile stress at yield and at fracture.

b. If the original gauge length was 4 in., estimate the gauge length when the specimen is stressed to 1/2 the yield stress.

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Step 1Given:The width of the specimen is 1 in.The thickness of the specimen is 0.5 in.The yield load of the specimen is 36 kips.The fracture load of the specimen is 48 kips.The original gauge length of the specimen is 4 in.Step 2The formula to calculate the stress is,\sigma=\frac{\text { Load }}{\text { Area }}The formula to calculate the change in length of the specimen is,\Delta L=\varepsilon LThe formula to calculate the strain is,\varepsilon=\frac{\sigma}{E}Here,\sigma is the stressE is the modulus of elasticity \varepsilon is the strainStep 3(a)Calculate the tensile stress.\begin{aligned} \sigma_{y} & =\frac{\mathrm{Load}}{\text { Area }} \\ & =\frac{36}{1 \times 0.5} \\ & =72 \mathrm{ksi}\end{aligned}Calculate the fracture stress.\begin{aligned} \sigma_{f} & =\frac{\mathrm{Load}}{\text { Area }} \\ & =\frac{48}{1 \times 0.5} \\ & =96 \mathrm{ksi}\end{aligned}Step 4Assume the modulus of elasticity is 30 x 106 psi.Calculate the strain.\begin{aligned} \varepsilon & =\frac{\sigma}{E} \\ & =\frac{1}{2}\left(\frac{\sigma_{y}}{E}\right) \\ & =\frac{1}{2}\left(\frac{72 \times 10^{3}}{30 \times 10^{6}}\right) \\ & =0.0012\end{aligned}Calculate the change in length of the specimen.\begin{aligned} \Delta L & =\varepsilon L \\ & =0.0012(4) \\ & =0.0048 \mathrm{in}\end{aligned}Calculate the final gauge length of the specimen.\begin{aligned} L_{f} & =4+0.0048 \\ & =4.0048 \mathrm{in}\end{aligned}Step 5Answer:(a) The tensile stress is 72 ksi and the fracture stress is 96 ksi.(b) The final gauge length of the specimen is 4.0048 in. ...