Question 2 of 7: 7075-T5 Aluminum — Full Tensile-Test Workup
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — December 2014. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Any five questions constitute
a complete paper; only the first five questions as they appear in the answer book are marked. All
seven questions are solved below for completeness.
Given. Original diameter $d_0=0.505$ in, average final diameter
$d_f=0.390$ in, gauge length $L_0=2.0000$ in; load–gauge-length table above (initial row
implies $L_0$).
Find. (i) Modulus of elasticity, (ii) 0.2% offset yield strength,
(iii) % reduction in area, (iv) % elongation, (v) tensile strength (UTS).
Fig. Q2 — engineering stress–strain curve with the 0.2% offset
line, the UTS (maximum-load) point, and the fracture point marked.
Approach
Every load/gauge-length pair converts to one stress–strain point using the fixed original
area and gauge length; the elastic modulus comes from the slope of the initial straight-line
rows, and the 0.2% offset yield point is where a line of that slope, shifted to start at
$\varepsilon=0.002$, re-intersects the plotted curve. Because this paper gives only one final
diameter and one fracture-row gauge length (no separate post-fracture unloaded measurement), both
%EL and %RA use those single given values directly.
Convert every row to stress and strain. Original area
$A_0=\frac{\pi}{4}d_0^2=\frac{\pi}{4}(0.505)^2=0.2003\ \text{in}^2$. For each row,
$\varepsilon=(L-2.0000)/2.0000$ and $\sigma=P(\text{kips})\times1000/A_0$ psi. The rows at 4, 8,
10, and 12 kips give $\sigma/\varepsilon\approx10.0\times10^6$ psi consistently — the
elastic region; the 13-kip row is the first to depart from that ratio, marking the true elastic
limit.
(i) Modulus of elasticity. Averaging the four consistent elastic-region
ratios,
$$E=\overline{\left(\frac{\sigma}{\varepsilon}\right)}=\boxed{10.0\times10^6\ \text{psi}}$$
— in the right range for aluminum alloys (7075-T5 is typically quoted near
$10.3$–$10.4\times10^6$ psi), a useful self-consistency check on the plotted slope.
(ii) 0.2% offset yield strength. Draw a line of slope $E$ starting at
$\varepsilon=0.002$ and find where it re-crosses the curve. It intersects the segment between the
13-kip point ($\varepsilon=0.00710$, $\sigma=64{,}904$ psi) and the 14-kip point
($\varepsilon=0.01010$, $\sigma=69{,}897$ psi):
$$\varepsilon_y\approx0.00876,\qquad\sigma_{0.2}=\boxed{67{,}660\ \text{psi}}.$$
(v) Tensile strength. The tensile strength (UTS) is the engineering stress at
the maximum load, $P=16.1$ kips:
$$\sigma_{ts}=\frac{16{,}100}{0.2003}=\boxed{80{,}381\ \text{psi}}.$$
Beyond this point the load drops to 15.6 kips at fracture even as the gauge length keeps
increasing — the signature of necking.
(iv) % elongation. The last table row records the gauge length at fracture,
$L=2.1340$ in:
$$\%EL=\frac{L-L_0}{L_0}\times100=\frac{2.1340-2.0000}{2.0000}\times100=\boxed{6.70\%}.$$
(iii) % reduction in area. With the given final diameter $d_f=0.390$ in,
$A_f=\frac{\pi}{4}(0.390)^2=0.1195\ \text{in}^2$:
$$\%RA=\frac{A_0-A_f}{A_0}\times100=\frac{0.2003-0.1195}{0.2003}\times100=\boxed{40.4\%}.$$
