Question 1 of 8: Schmid's Law and a Nickel Wire Under Load
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — December 2013. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Any five questions constitute
a complete paper; only the first five questions as they appear in the answer book are marked. All
eight questions are solved below for completeness.
Given. (a) Applied tensile stress $\sigma$; slip-plane normal at angle $\phi$
to the load axis; slip direction at angle $\lambda$ to the load axis. (b) Force $P=2500$ lb; wire
diameter $d=0.25$ in; nickel yield strength $\sigma_y=45{,}000$ psi; tensile strength
$\sigma_{ts}=55{,}000$ psi.
Find. (a) Derivation of $\tau=\sigma\cos\phi\cos\lambda$ and the orientation
giving maximum resolved shear stress. (b) Whether the wire deforms plastically and/or necks.
Fig. Q1 — slip-plane geometry: applied load $P$ at angle $\phi$ to the slip-plane normal $n$ and angle $\lambda$ to the slip direction (in-plane).
Approach
Part (a) resolves the applied axial force into a shear component acting in the slip direction,
within the slip plane, by projecting the force first onto the slip-plane area and then onto the
slip direction. Part (b) compares the wire's applied stress against its yield strength (onset of
plastic flow) and tensile strength (onset of necking, i.e. the UTS).
(a) Resolve force onto the slip-plane area. Let the bar have cross-sectional
area $A$ (perpendicular to the load axis) and applied force $F=\sigma A$. The slip plane's normal
is at angle $\phi$ to the load axis, so the slip-plane's actual area is larger than $A$ by
$1/\cos\phi$: $A_{slip}=A/\cos\phi$. The component of $F$ acting perpendicular to the load axis
within that plane, resolved along the slip direction (itself at angle $\lambda$ to the load axis),
has magnitude $F\cos\lambda$.
Resolved shear stress. Dividing the resolved force component by the (larger)
slip-plane area,
$$\tau=\frac{F\cos\lambda}{A_{slip}}=\frac{F\cos\lambda}{A/\cos\phi}=\frac{F}{A}\cos\phi\cos\lambda=\boxed{\sigma\cos\phi\cos\lambda}.$$
Maximum resolved shear stress. Since $\tau=\sigma\cos\phi\cos\lambda$ is a
product of two independent cosines (the normal direction $n$ and the slip direction lie in the
same plane and are always $90^\circ$ apart, so $\phi$ and $\lambda$ cannot both shrink to zero
together), the product $\cos\phi\cos\lambda$ is maximized when $\phi=\lambda=45^\circ$:
$$\tau_{max}=\sigma\cos45^\circ\cos45^\circ=\sigma\left(\tfrac{\sqrt2}{2}\right)^2=\boxed{0.5\,\sigma}.$$
This is the well-known Schmid's-law result: slip is easiest on planes/directions oriented at
$45^\circ$ to the applied load, where the Schmid factor $m=\cos\phi\cos\lambda$ reaches its
maximum value of $0.5$.
(b) Wire cross-sectional area and applied stress.
$$A_{wire}=\frac{\pi}{4}d^2=\frac{\pi}{4}(0.25)^2=0.04909\ \text{in}^2,\qquad
\sigma_{applied}=\frac{P}{A_{wire}}=\frac{2500}{0.04909}=\boxed{50{,}930\ \text{psi}}.$$
Compare against yield and tensile strength. Since
$\sigma_{applied}=50{,}930\ \text{psi} > \sigma_y=45{,}000\ \text{psi}$, the wire is stressed
past its elastic limit: (i) it deforms plastically. Since
$\sigma_{applied}=50{,}930\ \text{psi} < \sigma_{ts}=55{,}000\ \text{psi}$, the stress has not
reached the ultimate tensile strength (the point of maximum load, where a local neck begins to
form and the true stress there runs away from the nominal stress): (ii) the wire does not
neck — it is in the uniform strain-hardening region between yield and UTS.
