Question 4 of 7: Tensile Test of 7075-T5 Aluminum Alloy — Full Property Workup

Question 4: Tensile Test of 7075-T5 Aluminum Alloy — Full Property 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. Initial diameter $d_0=0.505$ in, average final diameter $d_f=0.390$ in, initial gauge length $L_0=2.0000$ in; the load/gauge-length table above (1 kip $=1000$ lb).

Find. (i) Modulus of elasticity $E$. (ii) 0.2%-offset yield strength. (iii) Reduction in area (%RA). (iv) Elongation (%EL). (v) Tensile strength (UTS).

Approach

Convert every (load, gauge length) pair to (engineering stress, engineering strain) using the original cross-sectional area $A_0$ and original length $L_0$; this defines the digitized version of the plotted curve. $E$ is the slope of the initial, tightly-clustered linear (elastic) points; the first point whose slope departs noticeably from that cluster marks the onset of yielding. The 0.2%-offset yield strength is the stress at which a line of slope $E$, starting at $\varepsilon= 0.002$, intersects the actual stress–strain curve. UTS is simply the maximum engineering stress recorded. %RA and %EL use the diameter/length change at fracture; here the table gives only one fracture-condition row (no separate "after unloading, springback" measurement), so it is used directly.

1. Cross-sectional area and stress/strain table. $$A_0=\frac{\pi}{4}d_0^2=\frac{\pi}{4}(0.505)^2=0.2003\ \text{in}^2$$ $$\sigma=\frac{P}{A_0},\qquad \varepsilon=\frac{L-L_0}{L_0}$$

   P (kips)	σ (psi)	ε (%)
   0	0	0
   4	19,970	0.205
   8	39,941	0.395
   10	49,926	0.515
   12	59,911	0.570
   13	64,904	0.710
   14	69,897	1.010
   16	79,882	2.515
   16.1 (max load)	80,381	4.950
   15.6 (fracture)	77,885	6.700

2. Modulus of elasticity, $E$. The 4, 8, 10, and 12 kip points are the tightly-clustered, visibly linear (elastic) portion of the curve; the 13 kip point already shows a lower incremental slope, signalling the onset of yielding, so it is excluded from the fit. Averaging $\sigma/\varepsilon$ over the four elastic points: $$E=\frac{1}{4}\left(\frac{19{,}970}{0.00205}+\frac{39{,}941}{0.00395}+\frac{49{,}926}{0.00515} +\frac{59{,}911}{0.00570}\right)$$ $$\boxed{E\approx1.00\times10^{7}\ \text{psi}\ (10.0\ \text{Msi})}$$ (this compares well with the handbook value for 7075 aluminum, $E\approx10.3$–$10.4$ Msi — a good check that the elastic points were identified correctly).
3. 0.2%-offset yield strength. The offset line $\sigma=E(\varepsilon-0.002)$ is intersected against the piecewise-linear curve between successive table points. It crosses between the 13-kip point ($\varepsilon=0.71\%$, $\sigma=64{,}904$) and the 14-kip point ($\varepsilon= 1.01\%$, $\sigma=69{,}897$): $$\boxed{\sigma_{y,0.2\%}\approx67{,}700\ \text{psi}}\quad\text{at }\varepsilon\approx0.88\%$$
4. Tensile strength (UTS). The maximum engineering stress in the table occurs at the 16.1-kip (maximum load) point — note this is not the fracture point; the load (and hence engineering stress) drops after this point as the specimen necks, even though the gauge length keeps increasing: $$\text{UTS}=\frac{16{,}100}{0.2003}$$ $$\boxed{\text{UTS}\approx80{,}400\ \text{psi}}$$
5. Reduction in area (%RA). Using $d_0=0.505$ in and the given average final diameter $d_f=0.390$ in: $$A_0=0.2003\ \text{in}^2,\qquad A_f=\frac{\pi}{4}(0.390)^2=0.1195\ \text{in}^2$$ $$\%RA=\frac{A_0-A_f}{A_0}\times100=\frac{0.2003-0.1195}{0.2003}\times100$$ $$\boxed{\%RA\approx40.4\%}$$
6. Elongation (%EL). The table supplies only a single fracture-condition gauge length (2.1340 in, listed directly on the 15.6-kip fracture row, with no separate after-unloading measurement) — per the standard convention for a table shaped this way, this value is used directly as the final gauge length: $$\%EL=\frac{L_f-L_0}{L_0}\times100=\frac{2.1340-2.0000}{2.0000}\times100$$ $$\boxed{\%EL\approx6.7\%}$$