Question 2 of 7: Tensile Test — Yield Strength, Tensile Strength, Modulus, %RA, %EL

Question 2: Tensile Test — Yield Strength, Tensile Strength, Modulus, %RA, %EL (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. Cylindrical tensile specimen: $L_0=2.00$ in, $d_0=0.505$ in, $d_f=0.423$ in (measured after failure). Four load/gauge-length pairs as tabulated, the first row labelled fully elastic, the next two labelled fully plastic, and the last row (maximum load reached, $8500$ lb) the point at which the specimen necked and separated at $7800$ lb.

Find. Yield strength, tensile strength (UTS), modulus of elasticity $E$, $\%RA$, $\%EL$.

Approach

Convert every load to engineering stress using the constant original area $A_0$, and every gauge length to engineering strain using $L_0$. $E$ follows from the one row explicitly marked "all elastic deformation" (a straight line through the origin). Because only four sparse data points are given (not a continuous curve), a graphical 0.2%-offset construction is not possible; the first row marked "all plastic deformation" is taken as the practical onset of yielding and used as the (approximate) yield strength — flagged below. UTS uses the maximum load; $\%EL$ and $\%RA$ use the standard post-fracture gauge length/diameter.

1. Original and final cross-sectional area. $$A_0 = \frac{\pi}{4}d_0^2 = \frac{\pi}{4}(0.505)^2 = 0.2003\ \text{in}^2,\qquad A_f = \frac{\pi}{4}d_f^2 = \frac{\pi}{4}(0.423)^2 = 0.1405\ \text{in}^2$$
2. Modulus of elasticity, from the elastic row (2000 lb, $L=2.001$ in). $$\sigma = \frac{2000}{0.2003} = 9985\ \text{psi}, \qquad \varepsilon = \frac{2.001-2.000}{2.000} = 5.0\times10^{-4}$$ $$E = \frac{\sigma}{\varepsilon} = \frac{9985}{5.0\times10^{-4}}$$ $$\boxed{E \approx 1.997\times10^{7}\ \text{psi} \approx 20.0\times10^{6}\ \text{psi}}$$
3. Approximate yield strength, from the first plastic row (6000 lb). $$\sigma_y \approx \frac{6000}{0.2003}$$ $$\boxed{\sigma_y \approx 29{,}960\ \text{psi} \approx 30.0\ \text{ksi}}$$
4. Tensile strength (UTS), from the maximum load (8500 lb). $$\sigma_{UTS} = \frac{8500}{0.2003}$$ $$\boxed{\sigma_{UTS} \approx 42{,}440\ \text{psi} \approx 42.4\ \text{ksi}}$$
5. Percent elongation, from the post-fracture gauge length (2.450 in). $$\%EL = \frac{L_f-L_0}{L_0}\times100 = \frac{2.450-2.000}{2.000}\times100$$ $$\boxed{\%EL = 22.5\%}$$
6. Percent reduction of area, from $A_0$ and $A_f$. $$\%RA = \frac{A_0-A_f}{A_0}\times100 = \frac{0.2003-0.1405}{0.2003}\times100$$ $$\boxed{\%RA \approx 29.8\%}$$