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31 March, 14:13

Consider a metal single crystal oriented such that the normal to the slip plane and the slip directions are at angles of 43.1° and 47.9°, respectively, with the tensile axis. If the critical resolved shear stress is 20.7 MPa (3,000 psi), will an applied stress of 45 mPa (6,500 psi) cause the single crystal to yield? If not, what stress will be necessary?

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  1. 31 March, 17:06
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    resolve shear stress = 22 MPa

    Explanation:

    Given data

    slip plane α = 43.1°

    slip directions β = 47.9°

    shear stress = 20.7 MPa (3,000 psi)

    applied stress = 45 mPa (6,500 psi)

    to find out

    what stress will be necessary

    solution

    we know that

    resolve shear stress = aplied stress * cosα * cosβ

    resolve shear stress = 45 * cos (43.1) * cos (47.9)

    resolve shear stress = 22 MPa

    we can say that here single cristal will be yield

    because resolve shear stress is bigger than critical shear stress
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