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13 September, 02:43

Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 43.1 degrees and 47.9 degrees, respectively, with the tensile axis. If the critical resolved shear stress is 20.7 MPa (3000 psi), will an applied stress of 45 MPa (6500 psi) cause the single crystal to yield? If not, what stress will be necessary?

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  1. 13 September, 02:46
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    resolved shear stress = 22.0 MPa

    so we can say that here single crystal will yield because critical resolved shear stress i. e 20.7 MPa is less than resolved shear stress i. e 22.0 MPa

    Explanation:

    given data

    angles φ = 43.1 degrees

    angles λ = 47.9 degrees

    shear stress = 20.7 MPa (3000 psi)

    stress σ = 45 MPa (6500 psi)

    solution

    we have given shear stress so first we calculate here resolved shear stress that is express as

    resolved shear stress = σ cosφ cosλ ... 1

    here σ is stress and φ and λ are angles given

    so put here value we get

    resolved shear stress = σ cosφ cosλ

    resolved shear stress = 45 cos (43.1) cos (47.9)

    resolved shear stress = 22.0 MPa

    so we can say that here single crystal will yield because critical resolved shear stress i. e 20.7 MPa is less than resolved shear stress i. e 22.0 MPa
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