Question 6 of 8: Mechanical Deformation

Question VI: 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.

Part 1(a). The yield strength is the stress at which a material first exhibits significant plastic (permanent) deformation — conventionally identified by the 0.2% offset method for metals with no sharp yield point. The ultimate (tensile) strength is the maximum engineering stress the specimen sustains anywhere on the stress-strain curve, reached after strain hardening has raised the load-carrying capacity beyond yield and before necking causes the engineering stress to fall. Yield marks the onset of permanent deformation; ultimate strength marks the peak load capacity, always at or beyond yield.

Part 1(b). Engineering stress divides the instantaneous load by the original, undeformed cross-sectional area, $\sigma_{\text{eng}}=F/A_0$; true stress divides it by the actual, current (instantaneous) cross-sectional area, $\sigma_{\text{true}}=F/A_i$. Because the cross-section shrinks as a ductile specimen elongates, $\sigma_{\text{true}}>\sigma_{\text{eng}}$ once the load is applied; the two coincide only in the limit of zero strain. The same distinction applies to strain: engineering strain $\varepsilon_{\text{eng}}=(L-L_0)/L_0$ uses the original gauge length, true strain $\varepsilon_{\text{true}}=\ln(L/L_0)$ integrates the incremental strain over the changing length.

Part 2 — Given.

Find. (a) final length (volume conserved), (b) true stress/strain, (c) engineering stress/strain, (d) yield strength and elastic energy stored to yield.

Approach. The diameter change under a modest 2 kN load is far too large to be purely elastic, so the specimen has been loaded onto the perfectly-plastic plateau (past yield); plastic deformation conserves volume, which fixes the current length and area directly from the measured diameter.

1. (a) Final length from volume conservation. $$A_0=\frac{\pi D_0^2}{4}=\frac{\pi(0.25)^2}{4}=0.04909\ \text{cm}^2,\qquad A_f=\frac{\pi D_f^2}{4}=\frac{\pi(0.23)^2}{4}=0.04155\ \text{cm}^2,$$ $$V_0=A_0L_0=0.04909\times25=1.2272\ \text{cm}^3,\qquad L_f=\frac{V_0}{A_f}=\frac{1.2272}{0.04155}=\boxed{29.54\ \text{cm}}.$$
2. (b) True stress and true strain. True stress uses the current area directly: $$\sigma_{\text{true}}=\frac{F}{A_f}=\frac{2000}{0.04155\times10^{-4}}=\boxed{481\ \text{MPa}}, \qquad \varepsilon_{\text{true}}=\ln\!\frac{L_f}{L_0}=\ln\!\frac{29.54}{25}=\boxed{0.1668}.$$ (Cross-check: volume conservation gives the equivalent $\varepsilon_{\text{true}}=\ln(A_0/A_f)=\ln(0.04909/ 0.04155)=0.1668$, matching — confirming internal consistency.)
3. (c) Engineering stress and strain. $$\sigma_{\text{eng}}=\frac{F}{A_0}=\frac{2000}{0.04909\times10^{-4}}=\boxed{407\ \text{MPa}}, \qquad \varepsilon_{\text{eng}}=\frac{L_f-L_0}{L_0}=\frac{29.54-25}{25}=\boxed{0.1815\ (18.15\%)}.$$ As expected, $\sigma_{\text{true}}>\sigma_{\text{eng}}$ and $\varepsilon_{\text{true}} <\varepsilon_{\text{eng}}$, consistent with Part 1(b).
4. (d) Yield strength and stored elastic energy.

5. Yield strength from the elastic strain at yield. With $\varepsilon_y=0.002$ purely elastic up to yield (linear elastic-perfectly-plastic model), $$\sigma_y=E\varepsilon_y=210\times10^3\times0.002=\boxed{420\ \text{MPa}}.$$
6. Elastic energy stored to yield. The modulus of resilience is the area under the linear-elastic portion of the curve, $$U_r=\tfrac12\sigma_y\varepsilon_y=\tfrac12(420\times10^6)(0.002)=4.20\times10^{5}\ \text{J/m}^3.$$ Multiplying by the (original, still-elastic) volume $V_0=1.2272\times10^{-6}\ \text{m}^3$, $$E_{\text{stored}}=U_rV_0=4.20\times10^{5}\times1.2272\times10^{-6}=\boxed{0.515\ \text{J}}.$$