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27 February, 14:59

Assuming the torsional yield strength of a compression spring is 0.43Sut and the maximum shear stress is equal to 434MPa. What is the factor of safety if the spring is a music wire (A=2170MPa and m=0.146) with a wire diameter of 4mm?

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  1. 27 February, 17:15
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    factor of safety is 1.756

    Explanation:

    given data

    torsional yield strength = 0.43 sut

    maximum shear stress = 434 MPa

    A = 2170 MPa

    m = 0.146

    diameter = 4 mm

    to find out

    factor of safety

    solution

    we know 1 sut is equal to 1772.39 MPa

    so 0.43 sut = 0.43 * 1772.39 = 762.128 MPa

    so here

    factor of safety formula is

    factor of safety = yield strength / shear stress ... 1

    put here value

    factor of safety = 762.128 / 434

    factor of safety is 1.756
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