Question 8 of 8: Question VIII — Mechanical Deformation (20 marks)

Question VIII — Mechanical Deformation (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.

VIII.1 — Representative stress–strain curves

The distinguishing features requested — yield point, plastic-strain region, fracture point, UTS — are marked directly on each curve above: (a) has only a fracture point (no yield, no plastic region); (b) has a yield point followed by a flat plastic plateau ending at fracture (no distinct UTS above $\sigma_y$); (c) has both a yield point and a rising UTS/fracture point, with a genuinely sloped (hardening) plastic region between them; (d) has neither a yield point nor a plastic region in the usual sense — it remains elastic (fully recoverable) over very large strains, with strongly increasing tangent stiffness at high strain (entropic straightening of polymer chains).

VIII.2 — 3003-H14 aluminum rod, 10 mm diameter

Given. $d_0=10$ mm, $E=70$ GPa, $\sigma_y=145$ MPa, $\nu=0.33$.

(a) Diameter under 6 kN tension

Approach. Check the stress is below yield (elastic); if so, get the axial strain from Hooke's law and the lateral (diameter) strain from Poisson's ratio.

1. Axial stress. $$\sigma=\frac{F}{A_0}=\frac{6000\ \text{N}}{78.54\times10^{-6}\ \text{m}^2}=76.4\ \text{MPa}\ (<\sigma_y=145\ \text{MPa}\Rightarrow\text{elastic})$$
2. Axial and lateral (diametral) strain. $$\varepsilon_z=\frac{\sigma}{E}=\frac{76.4\times10^6}{70\times10^9}=1.091\times10^{-3},\qquad \varepsilon_d=-\nu\varepsilon_z=-0.33(1.091\times10^{-3})=-3.60\times10^{-4}$$
3. New diameter. Tension necks the rod inward: $$d=d_0(1+\varepsilon_d)=10(1-3.60\times10^{-4})=\boxed{9.9964\ \text{mm}}$$

(b) Diameter under 6 kN compression

Approach. Same magnitude of stress, opposite sign; Poisson's effect now bulges the rod outward.

1. Diameter. By symmetry with (a) (same $|\varepsilon_d|$, opposite sign): $$d=d_0(1+3.60\times10^{-4})=\boxed{10.0036\ \text{mm}}$$

(c) Plastic strain under 15 kN tension

Approach. First check whether 15 kN keeps the rod elastic. Only $E$, $\sigma_y$, and $\nu$ are given — no strain-hardening modulus or full stress–strain curve for 3003-H14 is provided anywhere on this paper (confirmed against the exam's own appendix, which carries only generic constants, not alloy-specific data). That absence is the deliberate content of this sub-question: it is testing the elastic–perfectly-plastic idealisation from part VIII.1(b) directly, where the maximum FORCE the rod can carry in stable equilibrium is capped at $F_{yield}=\sigma_yA_0$ — beyond that, a perfectly-plastic material (zero strain-hardening) cannot generate additional stress to balance additional load, so there is no stable finite-strain equilibrium.

1. Stress at 15 kN (original area). $$\sigma=\frac{15{,}000}{78.54\times10^{-6}}=191.0\ \text{MPa}\ (>\sigma_y=145\ \text{MPa})$$
2. Maximum load-carrying capacity (elastic–perfectly-plastic model). $$F_{yield}=\sigma_yA_0=(145\times10^6)(78.54\times10^{-6})=\boxed{11.39\ \text{kN}}$$
3. Compare to applied load. $F_{applied}=15\ \text{kN} > F_{yield}=11.39\ \text{kN}$.

Because an elastic–perfectly-plastic material's stress cannot rise above $\sigma_y$ once yielding begins, the rod cannot statically equilibrate any axial load above $F_{yield}=11.39$ kN: any attempt to apply 15 kN drives the rod into unstable, unbounded plastic flow (necking runs away with no stabilising strain-hardening to arrest it) until fracture — there is no finite, well-defined "plastic strain" value for this load under the stated material model. $\boxed{\text{15 kN exceeds the rod's 11.39 kN plastic collapse load; the rod necks and fractures rather than reaching a stable strained state.}}$ A finite plastic strain at 15 kN could only be computed if a strain-hardening modulus or a full engineering stress–strain curve for 3003-H14 were supplied, which this paper does not provide.