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3 March, 22:43

Calculate the stress developed in the following members that are each subjected to an axial tensile load of 70 kN:

(a) steel bar, 25 mm by 50 mm

(b) 150-mm-diameter wood post (use the given diameter)

(c) 25-mm-diameter steel tie rod

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  1. 3 March, 23:03
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    (a) 56 MPa

    (b) 39.61 MPa

    (c) 142.85 MPa

    Explanation:

    The stress developed in a member by application of axial load is given by the formula:

    Stress = σ = Force/Area

    (a)

    In this case,

    Force = 70 KN = 70,000 N

    Area = 25 mm x 50 mm = 0.025 m x 0.05 m = 0.00125 m²

    Therefore,

    σ = 70,000 N/0.00125 m²

    σ = 56 x 10^6 Pa

    σ = 56 MPa

    (b)

    In this case,

    Force = 70 KN = 70,000 N

    Area = πD²/4 = π (0.15 m) ²/4 = 0.001767 m²

    Therefore,

    σ = 70,000 N/0.001767 m²

    σ = 39.61 x 10^6 Pa

    σ = 39.61 MPa

    (c)

    In this case,

    Force = 70 KN = 70,000 N

    Area = πD²/4 = π (0.025 m) ²/4 = 0.00049 m²

    Therefore,

    σ = 70,000 N/0.00049 m²

    σ = 142.85 x 10^6 Pa

    σ = 142.85 MPa
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