21-Mat-A4 Deformation Behaviour and Properties of Materials · December 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Exams, December 2016 — 10-Met-A4, Structure of Materials. Three hours, closed book, one approved calculator (Casio or Sharp). Seven questions of 20 marks each (Roman numerals I–VII); the rubric asks for any five, with only the first five in the answer book marked. All seven are solved here, because this set is a study resource rather than an exam script. All necessary equations, constants and an error-function table are provided in the exam's own appendix (reproduced where used below).
Note on the exam title. The printed exam header reads 10-Met-A4, Structure of Materials. Only Question VI (dislocation theory) is genuinely deformation/mechanical-properties content; the paper as a whole is a broad introductory materials-science survey — electron structure, bonding, crystal structure, crystallographic directions/planes, solid solubility, XRD, microscopy and phase diagrams — and is answered as such below.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
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.
Schmid's law. Slip begins on a given slip system once the shear stress resolved onto that system's slip plane, in its slip direction, reaches a critical value (the critical resolved shear stress, CRSS, $\tau_{crss}$). For an applied uniaxial tensile stress $\sigma$, the resolved shear stress is $$\tau_R=\sigma\cos\phi\cos\lambda$$ where $\phi$ is the angle between the tensile axis and the slip plane's normal, and $\lambda$ is the angle between the tensile axis and the slip direction (which lies within the slip plane, so $\phi$ and $\lambda$ are not independent — but the formula uses both as measured from the load axis). Slip occurs on the system with the largest $\tau_R$, once $\tau_R\ge\tau_{crss}$.
Slip plane normal to the applied stress. If the slip plane is normal to the tensile axis, the tensile axis is parallel to the slip-plane normal, so $\phi=0^{\circ}$ and $\cos\phi=1$. But the slip direction lies within the slip plane, i.e. perpendicular to that same normal — so the tensile axis is also perpendicular to the slip direction, $\lambda=90^{\circ}$ and $\cos\lambda=0$. Then $$\tau_R=\sigma\cos(0^{\circ})\cos(90^{\circ})=\sigma(1)(0)=\boxed{0}$$ Since the resolved shear stress is exactly zero regardless of the applied stress magnitude $\sigma$, no slip can occur on this system for this orientation, no matter how large $\sigma$ is made (short of fracture). This is one of the two special "Schmid-factor-zero" orientations (the other being the slip direction itself parallel to the tensile axis, $\lambda=0^{\circ}$, which gives $\phi=90^{\circ}$ and the same null result).
The dominant HCP slip system, active in essentially all HCP metals, is the basal system: the $(0001)$ basal plane (the closest-packed plane in HCP) in the $\langle11\bar20\rangle$ close-packed directions (three equivalent directions in that plane) — written $\{0001\}\langle11\bar20\rangle$, giving 3 independent slip systems (1 plane × 3 directions). This is favoured because the basal plane has both the highest planar atomic density (Question III.2's APF geometry) and the shortest slip-direction repeat distance ($\langle11\bar20\rangle$, of length exactly $a$), minimising the Peierls (lattice friction) stress. Metals with a low $c/a$ ratio (e.g. Ti, Zr, below the ideal 1.633) additionally activate secondary prismatic $\{10\bar10\}\langle11\bar20\rangle$ and pyramidal $\{10\bar11\}\langle11\bar20\rangle$ systems, needed (together with basal slip and twinning) to supply the 5 independent slip systems von Mises' criterion requires for arbitrary polycrystalline ductility — a requirement basal slip alone, with only 3 systems (two of which are dependent for a given basal plane), cannot meet.
Given. Copper, FCC, lattice constant $a=3.615$ Å.
Find. Burgers vector direction and magnitude $|\vec b|$.
Approach. In FCC metals the slip (and Burgers vector) direction is $\langle110\rangle$, the face diagonal, because the shortest full lattice-translation vector runs corner-to-face-centre-to-corner along that diagonal: $\vec b=\tfrac{a}{2}\langle110\rangle$.
Given. $\tau_{crss,1}=2.10$ MPa at $\rho_1=10^5$/mm²; $G=48$ GPa; $\alpha=0.2$; $b=2.556\times10^{-10}$ m (from VI.3a); target $\rho_2=10^7$/mm².
Find. $\tau_{crss,2}$ at $\rho_2$.
Approach. Use the first data point to back out the intrinsic strength $\tau_0=\tau_{crss,1}-\alpha Gb\sqrt{\rho_1}$, then evaluate $\tau_{crss,2}=\tau_0+\alpha Gb\sqrt{\rho_2}$ (converting $\rho$ from /mm² to /m² throughout so units are consistent with $G$ in Pa and $b$ in m).
A 100-fold rise in dislocation density (from $10^5$ to $10^7$/mm², a $\sqrt{100}=10$-fold rise in $\sqrt{\rho}$) raises the CRSS by a factor of about 4.3, from 2.10 to 9.08 MPa — exactly the strain-hardening mechanism this question sets up: each new dislocation's stress field impedes the motion of others, so a heavily deformed (dislocation-dense) metal is measurably stronger than the same metal in its annealed (low-$\rho$) state.
