# Resolved Shear Stress VS General Shear Stress -Contradiction?

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;
Ao = Acosθ
=> A = Ao/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!

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## Answers and Replies

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