Consider a single crystal of some hypothetical metal that has the BCC 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 direction, compute the stress at which the crystal yields if its critical resolved shear stress is 3.0 MPa.

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.

2.45 Mpa

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

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it has to be b, about 97 percent.