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  • 1 十年前
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    2a) Cross sectional area of the thicker bar = 52 x π = 78.5 mm2

    Cross sectional area of the thinner bar = 32 x π = 28.2 mm2

    Hence stress in the thicker bar = 104/78.5 = 127.3 N/mm2

    Stress in the thinner bar = 104/28.2 = 353.7 N/mm2

    b) Converting the units of stresses:

    Thicker bar = 1.273 x 108 N/m2

    Thinner bar = 3.537 x 108 N/m2

    So strain in the thicker bar = 1.273 x 108/(2 x 1011) = 6.37 x 10-4

    with extension = 6.37 x 10-4 x 500 = 0.318 mm

    Hence extension of the thinner bar = 0.4 - 0.318 = 0.082 mm

    Thus,

    0.082/x = 3.537 x 108/(2 x 1011)

    Giving x = 72.2 mm

    3) Effective area for stress = 62 x π/sin 30 = 226.2 mm2

    Dividing the force components:

    In perpendicular to the joint surface = 8 x 104 x sin 30 = 4 x 104 N

    Hence tensile stress = 4 x 104/226.2 = 176.8 N/mm2

    In parallel to the joint surface = 8 x 104 x cos 30 = 6.93 x 104 N

    Hence tensile stress = 6.93 x 104/226.2 = 306.3 N/mm2

    資料來源: Myself
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