Question 1 of 8: Magnesium-Alloy Tensile Test — Full Stress–Strain Analysis
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — May 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
eight questions are solved below for completeness.
Given. Original diameter $d_0=12$ mm, original gauge length $L_0=30.00$ mm;
load–gauge-length table above; post-fracture (unloaded) measured gauge length
$L_f=32.61$ mm and diameter $d_f=11.74$ mm.
Find. (i) 0.2% offset yield strength, (ii) tensile strength, (iii) modulus of
elasticity, (iv) % elongation, (v) % reduction in area, (vi) engineering stress at fracture,
(vii) true stress at fracture.
Fig. Q1 — engineering stress–strain curve plotted from the load–gauge-length
table, with the 0.2% offset line (slope $=E$), the UTS (maximum-load) point, and the fracture point marked.
Approach
Every load/gauge-length pair converts to one engineering stress–strain point using the
original cross-section and gauge length ($\sigma=P/A_0$, $\varepsilon=(L-L_0)/L_0$); the
resulting curve is plotted once and every other quantity is read from or derived from it. The
elastic modulus comes from the slope of the initial straight-line portion; the 0.2% offset yield
point is where a line of that same slope, shifted to start at $\varepsilon=0.002$, re-intersects
the curve; % elongation and % reduction in area use the separately-measured post-fracture
specimen dimensions (which include some elastic springback recovery); true stress at fracture
divides the fracture load by the actual (necked) area at fracture rather than the original area.
Convert every row to stress and strain. Original area
$A_0=\frac{\pi}{4}d_0^2=\frac{\pi}{4}(12)^2=113.10\ \text{mm}^2$. For each row,
$\varepsilon=(L-30.00)/30.00$ and $\sigma=P(\text{kN})\times1000/A_0\ \text{MPa}$. The first four
rows (up to $P=15$ kN) give $\sigma/\varepsilon\approx44{,}800$ MPa consistently — this is
the elastic region; the $P=20$ kN row already departs from that ratio, marking the onset of
plastic flow.
(iii) Modulus of elasticity. Averaging the consistent elastic-region ratios
(rows at 5, 10, 15 kN),
$$E=\overline{\left(\frac{\sigma}{\varepsilon}\right)}=\boxed{44.8\ \text{GPa}}$$
— close to the accepted value for magnesium alloys ($\approx45$ GPa), a useful
self-consistency check on the plotted slope.
(i) 0.2% offset yield strength. Draw a line of slope $E$ starting at
$\varepsilon=0.002$ (parallel to the elastic portion, offset by the 0.2% permanent-strain
convention) and find where it re-crosses the stress–strain curve. Between the $P=20$ kN
point ($\varepsilon=0.00500$, $\sigma=176.8$ MPa) and the $P=25$ kN point ($\varepsilon=0.0170$,
$\sigma=221.0$ MPa), solving the offset line against that segment gives
$$\varepsilon_y\approx0.0060,\qquad\sigma_{0.2}=\boxed{180.6\ \text{MPa}}.$$
(ii) Tensile strength. The tensile strength (UTS) is the engineering stress at
the maximum load, $P=27$ kN ($\varepsilon=0.0500$):
$$\sigma_{ts}=\frac{27{,}000}{113.10}=\boxed{238.7\ \text{MPa}}.$$
Beyond this point the load drops (26.5 kN, then 25 kN) even as the gauge length keeps increasing
— the classic signature of necking, where the local cross-section shrinks faster than the
material work-hardens.
(iv) % elongation. Using the separately-measured post-fracture gauge length
$L_f=32.61$ mm (the two broken halves fitted back together, after elastic springback):
$$\%EL=\frac{L_f-L_0}{L_0}\times100=\frac{32.61-30.00}{30.00}\times100=\boxed{8.70\%}.$$
(Note this differs slightly from the last table row's $\varepsilon=0.093$, which was recorded
under load at the instant of fracture and therefore still includes elastic strain that
recovers on unloading.)
(v) % reduction in area. Post-fracture diameter $d_f=11.74$ mm gives
$A_f=\frac{\pi}{4}(11.74)^2=108.29\ \text{mm}^2$:
$$\%RA=\frac{A_0-A_f}{A_0}\times100=\frac{113.10-108.29}{113.10}\times100=\boxed{4.29\%}.$$
(vi) Engineering stress at fracture. The last table row records $P=25$ kN as
the load at the instant of fracture; dividing by the original area,
$$\sigma_{f,eng}=\frac{25{,}000}{113.10}=\boxed{221.0\ \text{MPa}}.$$
(vii) True stress at fracture. True stress divides the same fracture load by
the actual (necked) area at the fracture location — the post-fracture measured area
$A_f$ from step 6:
$$\sigma_{f,true}=\frac{25{,}000}{108.29}=\boxed{230.9\ \text{MPa}}.$$
True stress at fracture exceeds engineering stress at fracture because the load is now carried
over a smaller, necked cross-section.
