Find Sheer stress that produced slip

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SUMMARY

The discussion centers on calculating the shear stress that causes slip in a single crystal of aluminum subjected to tensile stress. The user initially considers using the principal shear stress formula and direction cosines to solve the problem. They reference an MIT document on shearing stress calculations for metallic slip planes to aid their understanding. The key takeaway is that proper application of the principal shear stress formula and direction cosines is essential for determining the shear stress responsible for slip.

PREREQUISITES
  • Understanding of principal shear stress formula
  • Knowledge of direction cosines in stress analysis
  • Familiarity with slip systems in crystalline materials
  • Basic concepts of tensile testing in materials science
NEXT STEPS
  • Study the principal shear stress formula in detail
  • Learn how to calculate direction cosines for stress analysis
  • Explore slip systems and mechanisms in metallic materials
  • Review the MIT document on shearing stress calculations for practical examples
USEFUL FOR

Materials scientists, mechanical engineers, and students studying crystallography and material deformation will benefit from this discussion.

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
 
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