A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is 2.92 MPa, calculate the magnitude of applied stress necessary to cause slip to occur on the (111) plane in the direction.

The magnitude of applied stress necessary to cause slip is of 7.15 MPa.

Given data:

The metal structure is, FCC crystal structure.

The critical resolved shear stress of material is, s = 2.92 MPa.

The direction of stress is, A = [1 0 0]

The slipping plane is, [1 1 1].

The normal to slip plane is, B = [1 1 1]

The atomic arrangement such that each atoms is at corner of a cube has the fraction of atom with six atoms at center of each face, is known as face-centered cubic structure or FCC structure.

Let be the angle between A and B.

Then,

And for the slip along the [1 1 0],

Then the stress required to cause the slip is,

Thus, we can conclude that the magnitude of applied stress necessary to cause slip is of 7.15 MPa.

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Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa

Explanation:

Given -

Stress Direction, A = [1 0 0 ]

Slip plane = [ 1 1 1]

Normal to slip plane, B = [ 1 1 1 ]

Critical stress, Sc = 2.92 MPa

Let the direction of slip on = [ 1 1 0 ]

Let Ф be the angle between A and B

cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3

cos Ф = 1/√3

σ = Sc / cosФ cosλ

For slip along [ 1 1 0 ]

cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1

cos λ = 1/√2

Therefore,

σ = 2.92 / 1/√3 1/√2

σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa

Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa

this could be newton#'s law of cooling. the cooling curve is exponential