Question 3 of 8: Full Tensile-Test Workup

Question 3: Full Tensile-Test Workup (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $L_0=2.00$ in, $d_0=0.505$ in (standard 0.505-in tensile coupon, $A_0=0.2003$ in$^2$); $d_f=0.423$ in after failure; the four load/gauge-length pairs in the table above.

Find. Yield strength $\sigma_y$, tensile strength (UTS), modulus of elasticity $E$, %reduction of area, %elongation.

Approach

Each row converts directly to one engineering-stress/engineering-strain point via $\sigma=P/A_0$ (constant, undeformed area) and $e=(L-L_0)/L_0$. The first row is explicitly labelled all-elastic, so it alone defines $E$; the second row is explicitly labelled the first point of plastic deformation, so it is taken as the yield point; the third (maximum load) row gives the tensile strength; %RA uses the separately-given post-fracture diameter, and %EL uses the fourth row's post-fracture gauge length.

1. Cross-sectional area. $$A_0=\frac{\pi}{4}d_0^2=\frac{\pi}{4}(0.505)^2=0.2003\ \text{in}^2.$$
2. Modulus of elasticity, from the elastic row. At $P=2000$ lb, $L=2.001$ in: $$\sigma=\frac{2000}{0.2003}=9985\ \text{psi},\qquad e=\frac{2.001-2.000}{2.000}=0.0005,$$ $$E=\frac{\sigma}{e}=\frac{9985}{0.0005}=\boxed{1.997\times10^{7}\ \text{psi}\ (\approx138\ \text{GPa})}.$$
3. Yield strength, from the first plastic row. At $P=6000$ lb (the table labels this row's deformation "all plastic," i.e. the specimen has just yielded by this load): $$\sigma_y=\frac{6000}{0.2003}=\boxed{29{,}960\ \text{psi}\ (\approx206\ \text{MPa})}.$$
4. Tensile strength, from the maximum-load row. $$\text{UTS}=\frac{P_{\max}}{A_0}=\frac{8500}{0.2003}=\boxed{42{,}440\ \text{psi}\ (\approx293\ \text{MPa})}.$$
5. %Reduction of area, from the post-fracture diameter. With $A_f=\tfrac{\pi}{4}(0.423)^2=0.1405$ in$^2$: $$\%RA=\frac{A_0-A_f}{A_0}\times100=\frac{0.2003-0.1405}{0.2003}\times100=\boxed{29.8\%}.$$
6. %Elongation, from the post-fracture gauge length. The load has already dropped to 7800 lb by the moment of fracture (necking reduces the load-bearing area faster than the material's remaining work-hardening capacity can compensate); the paired gauge length 2.450 in is given as the final, "after failure" measurement, so it is used directly (the table gives no separate under-load reading to correct for elastic springback here): $$\%EL=\frac{L_f-L_0}{L_0}\times100=\frac{2.450-2.000}{2.000}\times100=\boxed{22.5\%}.$$