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!