Respuesta :
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
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 [tex]\phi[/tex] be the angle between A and B.
Then,
[tex]cos \phi = \dfrac{A.B}{|A||B|} \\\\cos \phi = \dfrac{[1 \; 0 \; 0].[1 \; 1 \; 1]}{(\sqrt{1^{2}+0^{2}+0^{2}}) \times(\sqrt{1^{2}+1^{2}+1^{2}})} \\\\cos \phi = \dfrac{1}{\sqrt{1} \times\sqrt{3}} \\\\cos \phi = \dfrac{1}{\sqrt{3}}[/tex]
And for the slip along the [1 1 0],
[tex]cos \lambda = \dfrac{[1 \; 1 \; 0].[1 \; 0 \; 0]}{(\sqrt{1^{2}+1^{2}+0^{2}}) \times(\sqrt{1^{2}+0^{2}+0^{2}})} \\\\cos \lambda = \dfrac{1}{\sqrt{2} \times\sqrt{1}} \\\\cos \lambda = \dfrac{1}{\sqrt{2}}[/tex]
Then the stress required to cause the slip is,
[tex]s' = \dfrac{s}{cos \lambda \times cos \phi} \\\\s' = \dfrac{2.92}{\dfrac{1}{\sqrt{2} } \times \dfrac{1}{\sqrt{3} }}\\\\s' = 7.15 \;\rm MPa[/tex]
Thus, we can conclude that the magnitude of applied stress necessary to cause slip is of 7.15 MPa.
Learn more about the crystal structures here:
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