Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stress is applied along a [112] direction. If slip occurs on a (111) plane and in a [011] direction, and the crystal yields at a stress of \(5.12 \mathrm{MPa}\), compute the critical resolved shear stress.

We will first calculate the dot product and magnitudes of the direction vectors. The stress direction vector is [112] and the slip direction vector is [011]. Calculate the magnitudes and dot product using the following formulas: - Magnitude of a vector: \(\sqrt{x^2 + y^2 + z^2}\) - Dot product of two vectors: \(a \cdot b = a_x b_x + a_y b_y + a_z b_z\) For the stress direction vector [112], the magnitude is: \[magnitude_{[112]} = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}\] For the slip direction vector [011], the magnitude is: \[magnitude_{[011]} = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}\] The dot product of the two-direction vectors [112] and [011] is: \[dot_{[112] \cdot [011]} = (1)(0) + (1)(1) + (2)(1) = 3\]

Now, calculate the angle between the stress direction and the slip direction using the dot product and magnitudes. Use the following formula: Cosine of the angle: \(\cos{\phi} = \frac{a \cdot b}{|a||b|}\) So, the cosine of angle between the stress direction and the slip direction is: \[\cos{\phi} = \frac{3}{\sqrt{6} \cdot \sqrt{2}} = \frac{3}{2 \sqrt{3}}\] Now find the angle \(\phi\): \[\phi = \arccos{ \frac{3}{2 \sqrt{3}}}\]

The slip plane is (111), which gives us a normal vector of [111]. Calculate the dot product and magnitude for the stress direction vector [112] and slip plane normal vector [111]: For the slip plane normal vector [111], the magnitude is: \[magnitude_{[111]} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}\] The dot product of the two-direction vectors [112] and [111] is: \[dot_{[112] \cdot [111]} = (1)(1) + (1)(1) + (2)(1) = 4\] Now calculate the angle between the stress direction and slip plane normal by using the dot product and magnitudes: Cosine of the angle: \(\cos{\lambda} = \frac{a \cdot b}{|a||b|}\) So, the cosine of angle between the stress direction and the slip direction is: \[\cos{\lambda} = \frac{4}{\sqrt{6} \cdot \sqrt{3}} = \frac{4}{3 \sqrt{2}}\] Now find the angle \(\lambda\): \[\lambda = \arccos{ \frac{4}{3 \sqrt{2}}}\]

Schmid's Law states that the critical resolved shear stress (\(\tau_{CRSS}\)) can be calculated using the following equation: \(\tau_{CRSS} = \sigma \cdot \cos{\phi} \cdot \cos{\lambda}\) Where \(\sigma\) is the applied stress, and \(\phi\) and \(\lambda\) are the angles we calculated in step 2 and 3. Substitute the values into the equation: \[\tau_{CRSS} = (5.12 \,\mathrm{MPa})\left(\frac{3}{2\sqrt{3}}\right)\left(\frac{4}{3\sqrt{2}}\right)\] Now solve for the critical resolved shear stress: \[\tau_{CRSS} \approx 1.48 \,\mathrm{MPa}\] Therefore, the critical resolved shear stress for this hypothetical metal under the given conditions is approximately 1.48 MPa.