Ask Question
2 January, 00:55

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.

+3
Answers (1)
  1. 2 January, 02:37
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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] ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers