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13 June, 15:03

A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy and maximum-shear-stress theories determine the factors of safety for the following plane stress states: (a) σx = 100 MPa, σy = 50 MPa (b) σx = 100 MPa, τxy = - 75 MPa (c) σx = - 50 MPa, σy = - 75 MPa, τxy = - 50 MPa (d) σx = 100 MPa, σy = 20 MPa, τxy = - 20 MPa

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  1. 13 June, 15:08
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    See explanation

    Explanation:

    a)

    MSS: σ1 = 100 MPa; σ2 = 50 MPa; σ3 = 0

    n = 350 / (100 - 0) = 3.5 Ans.

    DE: σ' = 100² - 100 * (50) + 50²) ^0.5 = 86.6 MPa

    n = 350/86.6 = 4.04 Ans.

    b) σA, σB = 100/2 ± √ ((100/2) ² + (-75) ²) = 140,

    σ1 = 140, σ2 = 0, σ3 = - 40 MPa

    MSS: n = 350 / (140 - (-40)) = 1.94 Ans.

    DE: σ' = [100² + 3 * (-75) ²] ^0.5 = 164 MPa

    n = 350/164 = 2.13 Ans.

    c) σA,σB = ((-50 - 75) / 2) ± √ (((-50 + 75) / 2) ² + (-50) ²)

    σ1 = 0, σ2 = - 11.0, σ3 = - 114.0 MPa

    MSS: n = 350 / (0 - (-1142.0)) = 3.07 Ans.

    DE: σ' [ (-50) ² - (-50) (-75) + (-75) ² + 3 (-50) ²]^0.5 = 105.0 MPa

    n = 350/109.0 = 3.21 Ans.

    d) σA,σB = (100 + 20) / 1 ± √ (((100-20) / 2) ² + (-20) ²)

    σ1 = 104.7, σ2 = 15.3, σ3 = 0 MPa

    MSS: n = 350 / (140.7 - 0) = 3.34 Ans.

    DE: σ' = [100² - 100 (20) + 20² + 3 (-20) ²]^0.5 = 98.0 MPa

    n = 350 / 98.0 = 3.57 Ans.
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