Tension test on an aluminium bar: 22-Mec-B8 Engineering Materials, May 2018, Q8

Question 8: Tension test on an aluminium bar (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. A standard round tensile specimen taken to fracture:

Given data
Quantity	Symbol	Value
Original diameter	d₀	12.8 mm
Original gauge length	L₀	50.8 mm
Load at the onset of yielding	P_y	31 000 N
Maximum load reached	P_max	40 000 N
Final length after fracture	L_f	63.5 mm
Young’s modulus	E	69 GPa

Find. (a) the yield strength, (b) the tensile strength, (c) the elastic strain at the yield point and (d) the total elongation.

The engineering stress–strain curve implied by the test. The shape between the marked points is schematic — the exam gives four points, not a full record — but every labelled value is computed from the question’s own data. Note how small the elastic strain at yield is beside the total elongation: that ratio is what separates a metal’s stiffness from its ductility.

Approach. All four answers come from the original cross-sectional area and the original gauge length. Compute the area once, divide the two quoted loads by it to get the two strengths, apply Hooke’s law at yield for the elastic strain, and take the permanent length change over the original gauge length for the elongation.

Compute the original cross-sectional area. The specimen is cylindrical, so $$A_0 = \frac{\pi d_0^{2}}{4} = \frac{\pi (12.8\ \text{mm})^{2}}{4} = \frac{\pi (163.84)}{4}$$ $$A_0 = 128.68\ \text{mm}^{2} = 1.2868\times 10^{-4}\ \text{m}^{2}$$ Every engineering stress in this question is referred to this original area, whatever the specimen has actually necked down to at the moment the load is read. That convention is what makes the answers comparable with handbook data.
Part (a) — find the yield strength. Dividing the load at first yielding by the original area, $$\sigma_y = \frac{P_y}{A_0} = \frac{31\,000\ \text{N}}{128.68\ \text{mm}^{2}}$$ $$\boxed{\ \sigma_y = 240.9\ \text{MPa}\ }$$ recalling that one newton per square millimetre is exactly one megapascal, which is why the mm–N pairing is the convenient one for this calculation.
Part (b) — find the tensile strength. The tensile strength, or ultimate tensile strength, is the maximum load divided by the same original area: $$\sigma_{UTS} = \frac{P_{max}}{A_0} = \frac{40\,000}{128.68}$$ $$\boxed{\ \sigma_{UTS} = 310.8\ \text{MPa}\ }$$ Because both strengths ride on the same area, their ratio is simply the load ratio, 40 000/31 000 = 1.290 — a free check on the arithmetic, and a yield-to-tensile ratio of 0.78 that is typical of a heat-treatable aluminium alloy.
Part (c) — find the elastic strain at the yield point. Up to yielding the material obeys Hooke’s law, so the strain there follows from the stress just computed: $$e_y = \frac{\sigma_y}{E} = \frac{240.9\ \text{MPa}}{69\,000\ \text{MPa}}$$ $$\boxed{\ e_y = 0.003491 = 0.349\ \%\ }$$ On a 50.8 mm gauge length this is an extension of only 0.177 mm, which is why yield strain is measured with an extensometer and never with a rule.
Part (d) — find the total elongation. The elongation reported from a tension test is the permanent length change measured after the broken halves are fitted back together, expressed as a percentage of the original gauge length: $$\text{elongation} = \frac{L_f - L_0}{L_0}\times 100 = \frac{63.5 - 50.8}{50.8}\times 100 = \frac{12.7}{50.8}\times 100$$ $$\boxed{\ \text{elongation} = 25.0\ \%\ }$$
Read the four answers together. The elastic strain at yield is 0.349 % and the total elongation 25.0 %, so the specimen deformed permanently some seventy times as much as it ever deformed elastically. That contrast is the practical definition of a ductile metal, and it is what makes aluminium formable by rolling, drawing and stretch-forming. The modulus of resilience — the elastic energy stored per unit volume at the point of yielding — is ½σ_ye_y = 0.42 MJ/m³, a small fraction of the energy absorbed in the plastic region, which is the toughness.

Results of the tension test
Quantity	Symbol	Value
Original cross-sectional area	A₀	128.68 mm²
(a) Yield strength	σ_y	240.9 MPa
(b) Tensile strength	σ_UTS	310.8 MPa
(c) Elastic strain at yield	e_y	0.003491 (0.349 %)
(d) Total elongation	—	25.0 %
Elastic extension at yield	ΔL_el	0.177 mm
Yield-to-tensile ratio	σ_y/σ_UTS	0.78
Modulus of resilience	U_r	0.42 MJ/m³

Check: the elongation is quoted here on the 50.8 mm (2 inch) gauge length the question gives. Elongation is not a material constant but depends on gauge length, because the necked region contributes a fixed amount of extension regardless of how long the gauge is; a value quoted without its gauge length is incomplete, and 25 % on 50 mm would read higher on a shorter gauge. The calculation also assumes the load at first yielding is the load at the conventional 0.2 % offset proof stress, which is how a yield load is determined for an aluminium alloy since it shows no sharp yield point.

Topic. The uniaxial tension test, and the engineering properties read off it.

Key relations. Engineering stress σ = P/A₀; engineering strain e = (L − L₀)/L₀; Hooke’s law σ = Ee in the elastic region; percentage elongation = 100(L_f − L₀)/L₀; modulus of resilience U_r = σ_y²/2E; true stress σ_t = σ(1 + e) and true strain ε_t = ln(1 + e), valid up to the onset of necking.

Why this works. The tension test is deliberately arranged so that a single specimen yields the whole set of design properties, and the trick that makes it work is referring every stress to the original area. That convention is an approximation — the real area shrinks throughout the test — but it is a useful one, because a designer sizes a member from its as-drawn dimensions, not from whatever it will have deformed to under load. It also explains the shape of the curve: the engineering stress passes through a maximum and then falls, not because the material is weakening but because the load is being divided by an area that no longer exists. The maximum marks the onset of necking, where uniform deformation gives way to a local instability, and everything beyond it describes one small region of the bar rather than the bar as a whole.

Common pitfall. Dividing the maximum load by the area at fracture rather than the original area, which gives a true stress and not the tensile strength the question asks for. A second frequent error is adding the elastic strain to the elongation, or subtracting it; the elongation measured after fracture is already purely plastic, because the elastic part springs back the moment the bar breaks. A third is dropping the factor of four in the area of a circle, which puts every stress out by a factor of four at once.

Source. Askeland & Wright, Ch. 6 (mechanical properties: the tensile test); Callister & Rethwisch, Ch. 6 (mechanical properties of metals); Dieter, Mechanical Metallurgy, Ch. 8 for the true-stress treatment and the necking instability.

Question 8 of 8: Tension test on an aluminium bar

Question 8: Tension test on an aluminium bar (20 marks)