Question 1 of 7: Full Tensile-Test Workup, Aluminum Alloy Bar
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — May 2016. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Candidates attempt any
five of the seven questions for a complete paper, all questions of equal value. All seven
questions are solved below for completeness.
Given. Original diameter $d_0=15$ mm ($A_0=176.71$ mm$^2$),
original gauge length $L_0=60.0000$ mm; the 10 load/gauge-length pairs above; after-fracture
(unloaded) gauge length $63.66$ mm and diameter $14.5$ mm.
Find. (i)–(vii) as listed above.
Fig. Q1 — engineering stress-strain curve plotted from the ten
load/gauge-length pairs, with the 0.2% offset-yield construction line (dashed) shown crossing
the curve at ≈211 MPa.
Approach
Convert every (load, gauge length) pair to (engineering stress, engineering strain) using
$\sigma=P/A_0$ and $e=(L-L_0)/L_0$. The first four points (0–30 kN) increase the
gauge length in exactly equal 0.0469 mm increments per 10 kN — perfectly linear
— so that is the true elastic region, fixing $E$. Every later point deviates from that
line, confirming yielding has begun by 35 kN. The 0.2% offset line is then drawn through
$e=0.2\%$ with slope $E$ and its intersection with the plotted curve located by linear
interpolation between bracketing data points.
Convert load and gauge length to stress and strain.
$$A_0=\frac{\pi}{4}(15)^2=176.71\ \text{mm}^2,\qquad
\sigma=\frac{P(\text{kN})\times1000}{A_0},\qquad e=\frac{L-60.0000}{60.0000}$$
giving, e.g., at 30 kN: $\sigma=169.8$ MPa, $e=0.2345\%$; at 45 kN (max):
$\sigma=254.6$ MPa, $e=5.00\%$; at fracture (44.2 kN, under load): $\sigma=250.1$ MPa,
$e=6.50\%$.
(iii) Young’s modulus from the linear (0–30 kN) region. All
three slopes computed from the first four points agree exactly:
$$E=\frac{\sigma}{e}=\frac{56.59}{0.0007817}=\frac{113.18}{0.0015633}=\frac{169.76}{0.002345}
=72{,}395\ \text{MPa}$$
$$\boxed{E\approx72.4\ \text{GPa}}$$
(consistent with a real aluminum alloy’s $E\approx69$–$72$ GPa — a good
sanity check that the elastic region was correctly identified).
(i) 0.2% offset yield strength. The offset line
$\sigma_{off}(e)=E(e-0.002)$ lies below the data curve at $e=0.35\%$ ($\sigma_{off}=108.6$ vs.
curve $198.1$ MPa) but above it at $e=0.5\%$ ($\sigma_{off}=217.2$ vs. curve $212.2$ MPa)
— the crossing lies between these two points. Interpolating both the curve segment and the
offset line linearly over that interval and solving for their intersection gives
$$\boxed{\sigma_{0.2\%}\approx211\ \text{MPa}\ (\text{at }e\approx0.49\%)}$$
(ii) Tensile strength. The maximum engineering stress occurs at the
labelled maximum load, 45.0 kN:
$$\boxed{\sigma_{UTS}=\frac{45.0\times1000}{176.71}\approx254.6\ \text{MPa}}$$
(iv) % elongation. Uses the after-fracture, unloaded gauge length
(63.66 mm) — smaller than the 63.90 mm reading taken under load at the instant of
fracture, because the specimen elastically recovers (springs back) once the load is removed:
$$\%EL=\frac{63.66-60.00}{60.00}\times100=\boxed{6.1\%}$$
(v) % reduction in area. Using the after-fracture diameter, 14.5 mm
($A_f=\pi/4\times14.5^2=165.13$ mm$^2$):
$$\%RA=\frac{176.71-165.13}{176.71}\times100=\boxed{6.56\%}$$
(vi) Engineering fracture stress. Fracture load (44.2 kN) divided by
the original area:
$$\sigma_{ef}=\frac{44.2\times1000}{176.71}=\boxed{250.1\ \text{MPa}}$$
(vii) True fracture stress. Fracture load divided by the actual
cross-sectional area at fracture ($A_f=165.13$ mm$^2$, from the after-fracture diameter):
$$\sigma_{tf}=\frac{44.2\times1000}{165.13}=\boxed{267.7\ \text{MPa}}$$
(always $>\sigma_{ef}$ since necking has reduced the true load-bearing area below $A_0$).
