Enyo_Face
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Question is as follows:
We have a single crystal of Al in a tensile tester. WE are applying a tensile stress, [δ][/Ψ] perpendicular to the 212 plane and we have found that slip has occurred.
What is the shear stress that produced the slip?
not sure how to proceed, first thought was use the principal sheer stress formula, and solve for the direction cosines, where [σ][/1] = [δ][/Ψ] and [σ][/2], [σ][/3] are = 0. then solve for L m and n suing the relation that L^2 + N^2 and M^2 = 1. However I am not sure if this is the right way to proceed with this, and am having trouble getting the direction cosines to simplify down.
Any help appreciated
We have a single crystal of Al in a tensile tester. WE are applying a tensile stress, [δ][/Ψ] perpendicular to the 212 plane and we have found that slip has occurred.
What is the shear stress that produced the slip?
not sure how to proceed, first thought was use the principal sheer stress formula, and solve for the direction cosines, where [σ][/1] = [δ][/Ψ] and [σ][/2], [σ][/3] are = 0. then solve for L m and n suing the relation that L^2 + N^2 and M^2 = 1. However I am not sure if this is the right way to proceed with this, and am having trouble getting the direction cosines to simplify down.
Any help appreciated