We consider a dislocation in an FCC crystal with the following characteristics: criticalshear stress:0.5MPa, line dislocation vector:1/sqrt(6)*[-1 -1 2] an Burgers vector:ao/2*[-1 1 0](1) Determine the slip plane for this dislocation(2) Compute the magnitude of the applied normal stress in the [0 1 0] direction

(1) The slip plane for a dislocation in a crystal is determined by the direction of the Burgers vector and the line dislocation vector. In a FCC crystal, the slip plane is usually one of the {111} planes because these are the densest and have the lowest interplanar spacing.

The Burgers vector given is ao/2*[-1 1 0] and the line dislocation vector is 1/sqrt(6)*[-1 -1 2]. The slip plane normal is the cross product of these two vectors.

So, the slip plane normal = Burgers vector x line dislocation vector

= ao/2*[-1 1 0] x 1/sqrt(6)*[-1 -1 2]

= ao/21/sqrt(6)[-2, -2, 0]

This gives the direction of the normal to the slip plane. The slip plane is perpendicular to this direction.

(2) The magnitude of the applied normal stress in the [0 1 0] direction can be calculated using the formula:

σ = τ/b cos(φ) cos(λ)

where: σ is the applied normal stress, τ is the critical shear stress, b is the magnitude of the Burgers vector, φ is the angle between the slip plane normal and the stress direction, and λ is the angle between the Burgers vector and the slip plane.

Given that τ = 0.5 MPa and b = ao/2, we need to calculate cos(φ) and cos(λ).

cos(φ) is the cosine of the angle between the slip plane normal [-2, -2, 0] and the stress direction [0 1 0]. This can be calculated using the dot product of the two vectors divided by the product of their magnitudes.

cos(λ) is the cosine of the angle between the Burgers vector [-1 1 0] and the slip plane. This can also be calculated using the dot product of the Burgers vector and the slip plane normal divided by the product of their magnitudes.

Once cos(φ) and cos(λ) are calculated, they can be substituted into the formula to find σ.