Question 6 of 8: Mechanical Deformation (20 marks)

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.

VI.1 — Yield vs. ultimate strength; engineering vs. true stress

(a) The yield strength marks the stress at which the material's response first departs measurably from linear-elastic behaviour and permanent (plastic) deformation begins — in practice usually taken as the 0.2% offset stress. The ultimate tensile strength is the maximum engineering stress reached anywhere on the stress–strain curve, corresponding to the onset of necking in a ductile metal; beyond it the engineering stress falls even though the true stress in the necked region keeps rising, because the nominal (original) cross-sectional area used to compute engineering stress no longer represents the load-bearing area.

(b) Engineering stress/strain are computed against the specimen's original, undeformed dimensions ($\sigma_e = F/A_0$, $\epsilon_e = \Delta L/L_0$). True stress/strain are computed against the specimen's instantaneous (current) dimensions ($\sigma_T = F/A_i$, $\epsilon_T=\ln(L_i/L_0)$). The two coincide only at small strain; once necking begins the instantaneous area shrinks faster than the engineering formula assumes, so true stress runs measurably above engineering stress for the same load.

VI.2 — Rod under load: stress, strain, yield and stored energy

Given.

Find. The final length (a); true stress and strain (b); engineering stress and strain (c); the yield strength and elastic strain energy to yield, taking the 2.2% yield-elongation datum as a separately supplied value for that sub-part (d).

Approach. Use volume conservation to get the deformed length and hence true strain directly; get engineering values from the original dimensions; get the yield strength from Hooke's law applied at the stated yield strain, and the stored elastic energy from the strain-energy-density integral up to yield.

1. (a) Final length from volume conservation. $A_0=\pi(d_0/2)^2=0.04909\ \text{cm}^2$, $A_f=\pi(d_f/2)^2=0.04155\ \text{cm}^2$; $A_0L_0=A_fL_f$: $$L_f = L_0\frac{A_0}{A_f} = 25\times\frac{0.04909}{0.04155} = \boxed{29.5\ \text{cm}}$$
2. (b) True stress and true strain. True strain from the length ratio (equivalently $\ln(A_0/A_f)$, same result by volume conservation): $$\epsilon_T = \ln\!\left(\frac{L_f}{L_0}\right) = \ln\!\left(\frac{29.54}{25}\right) = \boxed{0.167}$$ True stress uses the current (necked) area, $A_f = 4.155\times10^{-6}\ \text{m}^2$: $$\sigma_T = \frac{F}{A_f} = \frac{2000}{4.155\times10^{-6}} = \boxed{481\ \text{MPa}}$$
3. (c) Engineering stress and strain. Both referenced to the original dimensions, $A_0 = 4.909\times10^{-6}\ \text{m}^2$: $$\epsilon_e = \frac{L_f-L_0}{L_0} = \frac{29.54-25}{25} = \boxed{0.181\ (18.1\%)}$$ $$\sigma_e = \frac{F}{A_0} = \frac{2000}{4.909\times10^{-6}} = \boxed{407\ \text{MPa}}$$
4. (d) Yield strength and elastic energy to yield. Linear-elastic behaviour up to the stated 2.2% yield strain gives, by Hooke's law, $$\sigma_y = E\epsilon_y = (210\,000\ \text{MPa})(0.022) = \boxed{4620\ \text{MPa}}$$ The elastic strain energy stored up to yield is the area under the linear elastic line, $u=\tfrac12\sigma_y\epsilon_y$ per unit volume, times the original volume $V_0=A_0L_0$ ($L_0=0.25\ \text{m}$, $V_0=1.227\times10^{-6}\ \text{m}^3$): $$U = \tfrac12\sigma_y\epsilon_y V_0 = \tfrac12(4.62\times10^9)(0.022)(1.227\times10^{-6}) = \boxed{62.4\ \text{J}}$$