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**TERMS:**

Slip Plane: is the plane that has the densest atomic packing—that is, has the greatest planar density.

Slip Direction: corresponds to the direction in this plane that is most closely packed with atoms—that is, has the highest linear density.

**NOMENCLATURE:**

θ = angle of the slip plane as measured from cross section of material

λ = angle of that the applied force makes with the slip direction

φ = it has been said that is the angle between the normal vector and the applied force

θ = angle of the slip plane as measured from the cross section of the material

φ = θ =?; would this be so?

[A][/o] = Area of the materials cross section

A = Area of the slip plane

**ASSUMPTIONS:**-uniaxial tensile stress of a material with moderate ductility

-quasi-static

**QUESTION:**Can you show the relationship between shear stress for a typical uniaxial tensile stress (mechanics of materials) to the resolved shear stress geometrically?

It is derived from mechanics of materials principles in Chapter 6 that shear stress is τ = σsinθcosθ. Following that chapter 7 introduces resolved shear stress τ=σcosλcosφ.

I can do both derivations but I cannot relate the two.

**ATTEMPT:**__1) Derivation of Resolved Shear Stress__

-ensure that slip direction lies on slip plane via dot product of the normal vector of the slip plane and the slip direction == 0 (orthogonal)

-Load in direction of slip direction:

V = σcosλ

-The shear stress is that load across its surface area:

τ=σcosλ/A;

A

_{o}= Acosθ

=> A = A

_{o}/cosθ

-The Shear Stress

τ = V/A

=> τ=σcosλcosφ.

__2) Derivation of Shear Stress in General:__

I know how to derive via a force balance (equations of equilibrium) that gave the following result:

=> τ = σsinθcosθ

**SO:**

is it true that τ=σcosλcosφ == τ = σsinθcosθ

**END.**

**TEXTBOOK:**Materials Science and Engineering: An Introduction: William D. Callister. Chapter 6 & 7.

Thank you!

Thank you!

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