A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the \([110]\) direction. If the critical resolved shear stress for this material is \(1.75 \mathrm{MPa}\), calculate the magnitude(s) of applied stress(es) necessary to cause slip to occur on the (111) plane in each of the [1\overline{110], [10\overline{1} ] } \text { and } [ 0 1 \overline { 1 } ] \text { directions. }

To find the angle between the normal to the slip plane (111) and the applied stress direction [110], we will use the dot product. The dot product of two unit vectors is equal to the cosine of the angle between the two vectors. Therefore, \(\cos \alpha = \frac{\text{(111)} \cdot \text{(110)}}{|\text{(111)}| \cdot |\text{(110)}|}\) \(\cos \alpha = \frac{1+ 0 + 0}{\sqrt{1^2+1^2+1^2} \cdot \sqrt{1^2+1^2+0^2}} \implies \cos \alpha = \frac{\sqrt{2}}{2}\) So, the angle \(\alpha\) between the normal to the slip plane and the applied stress is: \(\alpha = \cos^{-1} \frac{\sqrt{2}}{2} = 45^\circ\).

For each slip direction, we will use the dot product to find the angle, \(\beta\). (1) Slip direction [1\(\overline{110}\)]: \(\cos \beta_1 = \frac{\text{[110]} \cdot \text{[1\)\overline{110}\(]}}{|\text{[110]}| \cdot |\text{[1\)\overline{110}\(]}|} = \frac{0}{2} \implies \cos \beta_1 = 0\) \(\beta_1 = \cos^{-1}(0) = 90^\circ\) (2) Slip direction [10\(\overline{1}\)]: \(\cos \beta_2 = \frac{\text{[110]} \cdot \text{[10\)\overline{1}\(]}}{|\text{[110]}| \cdot |\text{[10\)\overline{1}\(]}|} = \frac{1}{2} \implies \cos \beta_2 = \frac{1}{2}\) \(\beta_2 = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ\) (3) Slip direction [ 0 1 \(\overline{1}\)]: \(\cos \beta_3 = \frac{\text{[110]} \cdot \text{[ 0 1 \)\overline{1}\(]}}{|\text{[110]}| \cdot |\text{[ 0 1 \)\overline{1}\(]}|} = \frac{-1}{2} \implies \cos \beta_3 = \frac{-1}{2}\) \(\beta_3 = \cos^{-1}\left(\frac{-1}{2}\right) = 120^\circ\)

For each of the slip directions, apply the Schmid's Law equation: \(\tau_{cr} = 1.75\,\mathrm{MPa}\) (1) For slip direction [1\(\overline{110}\)]: \(\sigma_1 = \frac{\tau_{cr}}{\cos \alpha \cos \beta_1} = \frac{1.75}{\frac{\sqrt{2}}{2} \cdot 0} = \text{undefined}\) (No slip will occur in this direction). (2) For slip direction [10\(\overline{1}\)]: \(\sigma_2 = \frac{\tau_{cr}}{\cos \alpha \cos \beta_2} = \frac{1.75}{\frac{\sqrt{2}}{2} \cdot \frac{1}{2}} \implies \sigma_2 = 3.50\,\mathrm{MPa}\) (Slip will occur with this applied stress). (3) For slip direction [ 0 1 \(\overline{1}\)]: \(\sigma_3 = \frac{\tau_{cr}}{\cos \alpha \cos \beta_3} = \frac{1.75}{\frac{\sqrt{2}}{2} \cdot (-\frac{1}{2})} \implies \sigma_3 = -3.50\,\mathrm{MPa}\) (Slip will occur with this applied stress). In conclusion, the applied stresses necessary to cause slip on the (111) plane in the given slip directions are undefined for direction [1\(\overline{110}\)], \(3.50\,\mathrm{MPa}\) for direction [10\(\overline{1}\)], and \(-3.50\,\mathrm{MPa}\) for direction [ 0 1 \(\overline{1}\)].