Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a (111) plane and in a \([\overline{1} 01]\) direction, and is initiated at an applied tensile stress of \(13.9 \mathrm{MPa}(2020 \mathrm{psi})\) compute the critical resolved shear stress.

We are given the following information: - Tensile stress applied along [001] direction - Slip on (111) plane - Slip in \([\overline{1} 01]\) direction - Applied tensile stress = \(13.9\, \mathrm{MPa}\) Our goal is to compute the critical resolved shear stress.

Schmid's Law is a formula that relates the applied tensile stress to the resolved shear stress along with two angles: the angle between the normal to the slip plane and the tensile stress direction, and the angle between the slip direction and the tensile stress direction. The formula for Schmid's Law is: $$\tau = \sigma \cos{\phi} \cos{\lambda}$$ Where \(\tau\) is the resolved shear stress (CRSS), \(\sigma\) is the applied tensile stress, \(\phi\) is the angle between the normal to the slip plane and the tensile stress direction, and \(\lambda\) is the angle between the slip direction and the tensile stress direction Given that the applied tensile stress is along [001] direction, we can express it as a vector: $$\vec{t} = [0, 0, 1]$$ And similarly, we express the normal to the slip plane (111) as a vector: $$\vec{n} = [1, 1, 1]$$ As well as the slip direction vector, \([\overline{1} 01]\): $$\vec{s} = [-1, 0, 1]$$

We can find the angles \(\phi\) and \(\lambda\) by taking the dot product of the corresponding vectors and dividing by the product of their magnitudes. For angle \(\phi\), this gives: $$\cos{\phi} = \dfrac{\vec{n} \cdot \vec{t}}{|\vec{n}||\vec{t}|} = \dfrac{[1, 1, 1] \cdot [0, 0, 1]}{\sqrt{3}\sqrt{1}} = \dfrac{1}{\sqrt{3}}$$ And for angle \(\lambda\), similarly: $$\cos{\lambda} = \dfrac{\vec{s} \cdot \vec{t}}{|\vec{s}||\vec{t}|} = \dfrac{[-1, 0, 1] \cdot [0, 0, 1]}{\sqrt{2}\sqrt{1}} = \dfrac{1}{\sqrt{2}}$$

Now that we have found the values for \(\cos{\phi}\) and \(\cos{\lambda}\), we can apply Schmid's Law to calculate the critical resolved shear stress. Remember that the applied tensile stress \(\sigma\) is \(13.9\,\mathrm{MPa}\). Then, $$\tau = \sigma \cos{\phi} \cos{\lambda} = 13.9\,\mathrm{MPa} \times \dfrac{1}{\sqrt{3}} \times \dfrac{1}{\sqrt{2}}$$ Calculating this expression, we get: $$\tau \approx 5.65\,\mathrm{MPa}$$

The critical resolved shear stress for a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction, with slip occurring on a (111) plane and a \([\overline{1} 01]\) direction, is approximately \(5.65\,\mathrm{MPa}\).