complete question:
Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [011] direction, and the crystal yields at a stress of 3.0 MPa, compute the critical resolved shear stress.
Answer:
2.45 Mpa
Step-by-step explanation:
we will first calculate λ using:
λ = cos^-1[(u1*u2+v1*v2+w1*w2)/√(u1^2+v1^2+w1^2)(u2^2+v2^2+w2^2)
From the Problem:
u1 = 1
v1 = 2
w1 = 1
u2 = 0
v2 = 1
w2 = 1
now solve
λ = cos^-1[(1*1+2*0+1*1)/√(1^2+2^2+1^2)(0^2+1^2+1^2)
λ = cos^-1[2/√12)]
λ =30°
similarly we can find Ф
Ф= cos^-1[(u1*u2+v1*v2+w1*w2)/√(u1^2+v1^2+w1^2)(u2^2+v2^2+w2^2)
u1 = 1
v1 = 2
w1 = 1
u2 = 1
v2 = 0
w2 = 1
Ф= cos^-1[(1*1+2*0+1*1)/√(1^2+2^2+1^2)(1^2+0^2+1^2)
Ф= cos^-1[(4/√18)]
Ф= 19.5°
The yield stress is given by:
σ = T/cosФ*cos λ
however we are interested in shear stress and thus rearrange the equation to:
T=σcosФ*cos λ
= 3.0cos19.5*cos 30
= 2.45 Mpa