- #1

ltkach2015

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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.

**TEXTBOOK:**Materials Science and Engineering: An Introduction: William D. Callister.

**Chapter 6 & 7.**

ASSUMPTIONS:

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!*