1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Resolved Shear Stress VS General Shear Stress -Contradiction?

  1. May 10, 2015 #1
    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.

    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

    -uniaxial tensile stress of a material with moderate ductility

    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.


    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:
    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

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


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

    Thank you!


    Attached Files:

    Last edited: May 10, 2015
  2. jcsd
  3. May 16, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted