Mechanical Deformation (20 marks): 21-Mat-A4 Deformation Behaviour and Properties of Materials, May 2014, Q6

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 — Critical resolved shear stress

The critical resolved shear stress $\tau_{crss}$ is the shear stress, resolved onto a specific slip plane and slip direction, that must be reached to initiate dislocation glide (yielding) on that system. Resolving an applied tensile stress $\sigma$ onto a slip system inclined at angle $\phi$ to the stress axis (slip-plane normal) and $\lambda$ between the slip direction and the stress axis gives Schmid's law: $$\tau_R = \sigma\cos\phi\cos\lambda$$ The Schmid factor $\cos\phi\cos\lambda$ is maximised (equal to $0.5$) when $\phi=\lambda=45^\circ$ — slip begins earliest on systems oriented near $45^\circ$ to the loading axis, and $\tau_{crss}$ is reached soonest there. The macroscopic yield stress is the applied tensile stress at which the resolved shear stress on the most favourably oriented system first reaches $\tau_{crss}$: $$\sigma_y = \frac{\tau_{crss}}{(\cos\phi\cos\lambda)_{max}}$$ for a single crystal at the optimum orientation this reduces to $\sigma_y = 2\tau_{crss}$; for a polycrystal the Taylor factor (averaging over randomly oriented grains) replaces the single-crystal Schmid factor.

VI.2 — Three strengthening mechanisms

Grain-size refinement (Hall-Petch): $\sigma_y=\sigma_i+k_yd^{-1/2}$ — grain boundaries block dislocation motion (a dislocation pile-up in one grain cannot easily transmit slip across the misoriented boundary), so finer grains (more boundary area per volume) raise the yield strength. Solid-solution strengthening: substitutional or interstitial solute atoms create local lattice strain fields that interact with (impede) moving dislocations; strengthening scales roughly with solute concentration and the size/modulus mismatch between solute and solvent. Strain hardening (cold work): plastic deformation multiplies the dislocation density; the resulting increased dislocation-dislocation interactions (tangling, forest-cutting) progressively raise the stress needed for further slip, so a previously cold-worked metal is stronger (and less ductile) than its annealed state.

VI.3 — Reading the tensile stress-strain curve

Given. The printed tensile stress-strain diagram (main plot to $\varepsilon=0.15$, with an inset zoom of the elastic/early-yield region to $\varepsilon=0.005$); rod diameter $d_0=10$ mm; Poisson's ratio $\nu=0.34$.

Find. (a) $E$; (b) 0.2%-offset yield strength; (c) elastic load for a $2.5\ \mu\text{m}$ diameter reduction; (d) permanent strain after loading to 500 MPa past the UTS and unloading.

Approach. Digitize the elastic-region slope for $E$ (a); construct a line of slope $E$ offset by $0.002$ strain and find where it re-intersects the curve for the yield point (b); use $\nu$ to convert the diameter strain to axial strain and Hooke's law for the elastic load (c); read the descending (post-necking) branch of the main curve at 500 MPa for the total strain there, then subtract the elastic recovery $500/E$ (d).

Elastic-region inset: the curve's initial slope gives $E$; the 0.2%-offset construction (red, parallel to the elastic line, shifted $+0.002$ strain) re-intersects the curve just as it flattens, giving the yield point.

(a) Young's modulus. Reading the straight initial portion of the elastic-region inset gives a slope $$E \approx \boxed{250\ \text{GPa}}$$ Check: graphically read from the printed curve; expect a few percent reading uncertainty. Usefully, part (b)'s answer is not sensitive to this: repeating the offset construction for any $E$ between 200 and 300 GPa relocates the crossing strain but changes the crossing stress by less than 2%, because the curve is already nearly flat there.
(b) 0.2% offset yield strength. Drawing a line of slope $E$ through $(\varepsilon,\sigma)=(0.002,0)$ and finding where it re-crosses the digitized curve: $$\sigma_y \approx \boxed{536\ \text{MPa}}$$
(c) Elastic load for a 2.5 μm diameter reduction. Transverse (diameter) strain magnitude: $$\varepsilon_t = \frac{\Delta d}{d_0} = \frac{2.5\times10^{-6}}{0.010} = 2.5\times10^{-4}$$ Axial strain from Poisson's ratio ($\varepsilon_t = \nu\varepsilon_{ax}$): $$\varepsilon_{ax} = \frac{\varepsilon_t}{\nu} = \frac{2.5\times10^{-4}}{0.34} = 7.353\times10^{-4}$$ Axial stress and load (Hooke's law; $7.35\times10^{-4}\times250{,}000 = 184$ MPa is comfortably below the 536 MPa yield point, so "entirely elastic" is self-consistent): $$\sigma = E\varepsilon_{ax} = 250{,}000(7.353\times10^{-4}) = 183.8\ \text{MPa}, \qquad A_0=\frac{\pi d_0^2}{4}=7.854\times10^{-5}\ \text{m}^2$$ $$F = \sigma A_0 = (183.8\times10^6)(7.854\times10^{-5}) = \boxed{14.4\ \text{kN}}$$
(d) Permanent strain after loading to 500 MPa past the UTS. On the descending (post-necking) branch of the main curve, digitizing where stress falls back through 500 MPa gives a total strain there of $\varepsilon \approx 0.0901$. Unloading is elastic (parallel to the original $E$ slope), recovering $$\varepsilon_{elastic} = \frac{500}{250{,}000} = 0.0020$$ $$\varepsilon_{permanent} = 0.0901 - 0.0020 = \boxed{0.088\ (8.8\%)}$$

Full curve (digitized): loading past the UTS (~580 MPa) into the necking region, unloading elastically from 500 MPa back to zero stress leaves a permanent strain of about 0.088.

Question VI — final results
Quantity	Value
Young's modulus, $E$	$\approx 250$ GPa
0.2%-offset yield strength, $\sigma_y$	$\approx 536$ MPa
Elastic load, $F$ (2.5 μm reduction)	$14.4$ kN
Permanent strain after 500 MPa unload	$0.088\ (8.8\%)$

Topic. Schmid's law and critical resolved shear stress; strengthening mechanisms; tensile test interpretation (modulus, offset yield, elastic Poisson response, plastic-then-elastic unloading).

Key relations. $\tau_R=\sigma\cos\phi\cos\lambda$; Hall-Petch $\sigma_y=\sigma_i+k_yd^{-1/2}$; $\sigma=E\varepsilon$; $\nu=-\varepsilon_t/\varepsilon_{ax}$; 0.2% offset construction.

Why this works. Schmid's law is simply the vector resolution of an applied uniaxial force onto an inclined slip plane and direction. The 0.2% offset method exists because most engineering alloys yield gradually (no sharp elastic limit); the parallel-line construction operationally defines "yield" as a small, repeatable, permanent-set criterion. Unloading from any point on the plastic curve is elastic and parallel to the original modulus line because dislocation motion (plastic strain) is essentially irreversible on unloading, while lattice bond stretching (elastic strain) is fully reversible.

Common pitfall. Reading yield strength directly off the curve at 0.2% strain (rather than constructing the offset line and finding its intersection); using the transverse strain directly as $\varepsilon E$ instead of first converting to axial strain via $\nu$; assuming unloading returns along the original loading path rather than parallel to the elastic slope.

Source. Callister & Rethwisch, Materials Science and Engineering, 10th ed., Ch. 6 (Mechanical Properties) & Ch. 7 (Dislocations and Strengthening Mechanisms); Dieter, Mechanical Metallurgy, 3rd ed., Ch. 3 (slip and Schmid's law).