Resolved Shear Stress Compared w/ Shear Stress-Contradictioni

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Main Question or Discussion Point

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

-uniaxial tensile stress of a material with moderate ductility
-quasi-static
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

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.

ATTACHMENTS: 1(

BEGIN:
-Derivation of resolved shear stress
0) 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)

1) Load in direction of slip direction

τ=σcosλ/A;
[A][/o] = Acosθ
=> τ=σcosλcosφ.


-Derivation of shear stress in general:
I can derive via a force balance, sum of forces resulting that resulted to

=> τ = σsinθcosθ


Thank you!
 

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