A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of \(65^{\circ}\) with the tensile axis. Three possible slip directions make angles of \(30^{\circ}, 48^{\circ}\), and \(78^{\circ}\) with the same tensile axis. (a) Which of these three slip directions is most favored? (b) If plastic deformation begins at a tensile stress of \(2.5 \mathrm{MPa}\) (355 psi), determine the critical resolved shear stress for zinc.

To determine the most favored slip direction, we must calculate the Schmid factor for each of the three slip directions and choose the one with the highest value. The Schmid factor is given by: $$SF = \cos(\alpha) \cos(\phi)$$ where \(\alpha\) is the angle between the slip plane normal and the tensile axis, and \(\phi\) is the angle between the slip direction and the tensile axis. We are given that the slip plane normal makes an angle of \(65^\circ\) with the tensile axis. Calculate the Schmid factor (SF) for each slip direction: 1) Angle between slip direction 1 and tensile axis is \(30^\circ\): $$SF_1 = \cos(65^\circ) \cos(30^\circ)$$ 2) Angle between slip direction 2 and tensile axis is \(48^\circ\): $$SF_2 = \cos(65^\circ) \cos(48^\circ)$$ 3) Angle between slip direction 3 and tensile axis is \(78^\circ\): $$SF_3 = \cos(65^\circ) \cos(78^\circ)$$ Now, compare the values of SF1, SF2, and SF3, and choose the one with the highest value.

To calculate the critical resolved shear stress (CRSS) for zinc, we will use the relationship between tensile stress and shear stress: $$\tau_{CRSS} = \frac{\sigma}{SF}$$ where \(\sigma\) is the tensile stress. We are given that plastic deformation begins at a tensile stress of \(2.5 \mathrm{MPa}\), and we have already determined the most favored slip direction (the one with the highest Schmid factor). Now, substitute the values of tensile stress and the highest Schmid factor into the equation: $$\tau_{CRSS} = \frac{2.5 \mathrm{MPa}}{SF_{max}}$$ Calculate the critical resolved shear stress for zinc.