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11 December, 01:50

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.

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  1. 11 December, 02:01
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    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

    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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