Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a \([\overline{1} 11]\) direction, compute the stress at which the crystal yields if its critical resolved shear stress is \(2.4 \mathrm{MPa}\)

First, we need to rewrite the given direction vectors, [121] and \([\overline{1} 11]\), and (101) plane into unit vectors. This is necessary for computing the angles needed in Schmid's Law equation. For the stress direction, [121]: \(\vec{d}_{\text{stress}} = [1, 2, 1]\) Normalize this vector: \(\vec{d}_{\text{stress}} = \frac{[1, 2, 1]}{\sqrt{1^2 + 2^2 + 1^2}}=\frac{1}{\sqrt{6}}[1, 2, 1]\) For the slip direction, \([\overline{1} 11]\): \(\vec{d}_{\text{slip}} = [-1, 1, 1]\) Normalize this vector: \(\vec{d}_{\text{slip}} = \frac{[-1, 1, 1]}{\sqrt{(-1)^2 + 1^2 + 1^2}}=\frac{1}{\sqrt{3}}[-1, 1, 1]\) For the slip plane, (101): The normal to the slip plane is given by vector: \([1, 0, 1]\) Normalize this vector: \(\vec{n}_{\text{plane}} = \frac{[1, 0, 1]}{\sqrt{1^2 + 0^2 + 1^2}}=\frac{1}{\sqrt{2}}[1, 0, 1]\)

Next, we need to calculate the angles between the stress direction (φ) and slip direction (λ) with the normal vector to the slip plane using the dot product formula, which is given by \(\cos(\theta) = \frac{\vec{A}\cdot\vec{B}}{\left|\vec{A}\right|\cdot\left|\vec{B}\right|}\). Since the vectors are normalized, the denominator is 1. The angle between the applied stress direction and the normal to the slip plane (φ): \(\cos(\phi) = \vec{d}_{\text{stress}}\cdot\vec{n}_{\text{plane}}=\frac{1}{\sqrt{6} \sqrt{2}} [1, 2, 1]\cdot[1, 0, 1]=\frac{2}{\sqrt{12}}\) The angle between the slip direction and the normal to the slip plane (λ): \(\cos(\lambda) = \vec{d}_{\text{slip}}\cdot\vec{n}_{\text{plane}}=\frac{1}{\sqrt{3} \sqrt{2}} [-1, 1, 1]\cdot[1, 0, 1]=\frac{2}{\sqrt{6}}\)

Finally, use Schmid's Law formula to compute the yielding stress: Schmid's Law: \(τ_{R} = σ\cos(\phi)\cos(\lambda)\) Rearrange the formula to solve for σ: \(σ = \frac{τ_{R}}{\cos(\phi)\cos(\lambda)}\) Given the critical resolved shear stress, \(τ_{R} = 2.4\,\mathrm{MPa}\), compute σ: \(σ = \frac{2.4\,\mathrm{MPa}}{\left(\frac{2}{\sqrt{12}}\right)\left(\frac{2}{\sqrt{6}}\right)} = \frac{2.4\,\mathrm{MPa} \cdot\sqrt{12}\cdot\sqrt{6}}{4} = 14.4\,\mathrm{MPa}\)

The stress at which the BCC crystal yields when the tensile stress is applied along the [121] direction, with slip occurring on a (101) plane and in a \([\overline{1} 11]\) direction, is \(14.4\,\mathrm{MPa}\).