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14 July, 13:12

Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod with E 5 73 GPa and an ultimate strength of 140 MPa. Knowing that the distance between the gage marks is 250.28 mm after a load is applied, determine (a) the stress in the rod, (b) the factor of safety

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  1. 14 July, 13:27
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    81.76 N/mm² (MPa), 1.71233

    Explanation:

    Modulus of elasticity = stress / strain

    stress = modulus of elastic * strain

    strain = ΔL / L = 250.28 mm - 250 mm / 250 mm = 0.00112

    Modulus of elasticity E = 73 GPa = 73 * 10³ MPa where 1 MPa = 1 N/mm²

    E = 73 * 10³N/mm²

    stress = 73 * 10³N/mm² * 0.00112 = 81.76 N/mm² (MPa)

    b) Factor of safety = maximum allowable stress / induced stress = 140 MPa / 81.76 MPa = 1.71233
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