Schmid’s Law; Loading of a Nickel Wire: 04-BS-11, May 2017, Q1

Question 1: Schmid’s Law; Loading of a Nickel Wire (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 (a) — Schmid’s law and its maximum

Fig. Q1(a) — A tensile stress $\sigma$ on a single crystal. The slip-plane normal $n$ makes angle $\phi$ with the load axis; the slip direction makes angle $\lambda$. The load resolves onto the inclined slip plane as a shear.

Given. A single crystal of cross-sectional area $A_0$ carries an axial force $F=\sigma A_0$. The slip plane’s normal is inclined at $\phi$ to the load axis; the slip direction lies at $\lambda$ to the load axis.

Find. Show $\tau=\sigma\cos\phi\cos\lambda$, then the orientation giving the largest $\tau$ and its value.

Approach. Resolve the axial force onto the slip direction, divide by the (inclined) slip-plane area, and maximise the resulting Schmid factor.

Component of force along the slip direction. The full axial force is $F=\sigma A_0$; its component along the slip direction is $$F_s=F\cos\lambda .$$
Area of the slip plane. The slip plane is tilted so its normal makes angle $\phi$ with the axis; its area is larger than the bar cross-section: $$A_s=\frac{A_0}{\cos\phi}.$$
Resolved shear stress. Dividing the resolved force by the slip-plane area, $$\tau=\frac{F_s}{A_s}=\frac{\sigma A_0\cos\lambda}{A_0/\cos\phi}=\boxed{\sigma\cos\phi\cos\lambda}$$ which is Schmid’s law; the factor $m=\cos\phi\cos\lambda$ is the Schmid factor.
Maximum resolved shear stress. For the load axis, slip normal and slip direction to be coplanar, $\phi+\lambda=90^\circ$, so $\lambda=90^\circ-\phi$ and $m=\cos\phi\sin\phi=\tfrac12\sin2\phi$. This is greatest at $2\phi=90^\circ$, i.e. $$\phi=\lambda=45^\circ\ \Rightarrow\ m_{\max}=\tfrac12,\qquad \boxed{\tau_{\max}=\tfrac12\,\sigma}.$$

Part (b) — will the wire yield or neck?

Given. Axial force $F=3\text{ tons}=3\times2000=6000$ lb (info sheet: 1 ton $=2000$ lb); wire diameter $d=\tfrac38=0.375$ in; yield strength $\sigma_y=49{,}000$ psi; tensile strength $\sigma_{UTS}=58{,}000$ psi.

Find. Whether the applied engineering stress reaches (i) the yield strength (plastic deformation) and (ii) the tensile strength (necking).

Approach. Compute the engineering stress and compare it against $\sigma_y$ (onset of plastic flow) and $\sigma_{UTS}$ (onset of necking).

Cross-sectional area. $$A=\frac{\pi}{4}d^2=\frac{\pi}{4}(0.375)^2=0.1104\ \text{in}^2.$$
Applied engineering stress. $$\sigma=\frac{F}{A}=\frac{6000}{0.1104}=\boxed{54{,}325\ \text{psi}}.$$
(i) Plastic deformation? Since $\sigma=54{,}325\ \text{psi} > \sigma_y=49{,}000\ \text{psi}$, the wire is loaded past its yield point — yes, it deforms plastically.
(ii) Necking? Necking begins only when the engineering stress reaches the tensile strength. Here $\sigma=54{,}325\ \text{psi} < \sigma_{UTS}=58{,}000\ \text{psi}$, so no, the wire does not neck (it undergoes uniform plastic strain but the load is still below the necking threshold).

Quantity	Result
Maximum Schmid factor / orientation	$m_{\max}=\tfrac12$ at $\phi=\lambda=45^\circ$
Maximum resolved shear stress	$\tau_{\max}=\sigma/2$
Applied stress on wire	54,325 psi
(i) Plastic deformation	Yes ($\sigma>\sigma_y$)
(ii) Necking	No ($\sigma<\sigma_{UTS}$)

Concept & Theory

Topic. Critical resolved shear stress for slip (Schmid’s law) and the stress landmarks of a tensile test (yield, UTS, necking).

Key relations. $\tau=\sigma\cos\phi\cos\lambda$; Schmid factor $m=\cos\phi\cos\lambda\le\tfrac12$; engineering stress $\sigma=F/A_0$.

Why this works. Slip is driven by the shear resolved onto the slip system, not by the axial stress directly. Projecting the force onto the slip direction and dividing by the enlarged inclined area introduces the two cosines; their product is largest when the plane is inclined at 45°. In the tensile test, a bar begins to deform permanently once the stress reaches $\sigma_y$, and it begins to neck (localise) only once the stress reaches the UTS, after which the true load-bearing area shrinks faster than work-hardening can compensate.

Common pitfall. Converting “3 tons” with the metric tonne (2205 lb) instead of the info-sheet short ton (2000 lb), or reading the diameter as ½ in rather than the printed ⅜ in — either error changes the yes/no conclusion.

Source. Callister & Rethwisch, 9th ed., Ch. 7 (slip systems, Schmid’s law) and Ch. 6 (tensile properties).

Question 1 of 7: Schmid’s Law; Loading of a Nickel Wire

Question 1: Schmid’s Law; Loading of a Nickel Wire (20 marks)

Part (a) — Schmid’s law and its maximum

Part (b) — will the wire yield or neck?