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5 November, 04:28

A 6x50 mm flat alloy bar clongates 1.22 mm in a length of 1.5 m under a total axial load of 35.0 NThe proportional limit of this alloy is 300 MPa Determine: '.50" 35 kN 35 kN 6 x 50 A. The axial stress in the bar. B. The modulus of elasticity of the material. C. The change in the two lateral dimensions if Poisson's ratio for the material is 0.32

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  1. 5 November, 07:11
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    A. Axial stress is 116.66 MPa

    B. Modulus of elasticity is 143.5 GPa

    C. The changes in two lateral dimensions are 1.3 μm and 0.156 μm

    Explanation:

    The following information is given:

    Axial Load = P = 35 KN = 35000 N

    Original Length = L = 1.5 m

    Change in length = ΔL = 1.22 mm = 1.22 x 10^-3 m

    width = w = 50 mm = 0.05 m

    breadth = b = 6 mm = 0.006 m

    A.

    The axial stress is given as:

    σ = P/A

    σ = P/wb

    σ = (35000 N) / (0.05 m) (0.006 m)

    σ = 116.66 MPa

    B.

    The modulus of elasticity is given in the proportional limit as:

    Modulus of Elasticity = E = Stress/Strain = σ/∈

    Strain = ∈ = ΔL/L

    ∈ = 0.00122 m/1.5 m

    ∈ = 8.13 x 10^-4 mm/mm

    Therefore,

    E = (116.66 MPa) / 8.13 x 10^-4)

    E = 143.5 GPa

    C.

    Poisson's Ratio = n = - (Lateral Strain) / (Axial Strain)

    n = 0.32 = - (Lateral Strain) / 8.13 x 10^-4

    Lateral Strain = - 2.6 x 10^-4 mm/mm

    Here, the negative sign indicates the decrease in dimensions.

    For change in width (ΔW):

    Lateral Strain = - 2.6 x 10^-4 mm/mm = ΔW/W

    ΔW = ( - 2.6 x 10^-4 mm/mm) (50 mm)

    ΔW = - 130 x 10^-4 mm

    ΔW = - 1.3 μm

    Here, the negative sign indicates the decrease in width.

    For change in breadth (Δb):

    Lateral Strain = - 2.6 x 10^-4 mm/mm = Δb/b

    ΔW = ( - 2.6 x 10^-4 mm/mm) (6 mm)

    ΔW = - 15.6 x 10^-4 mm

    ΔW = - 0.156 μm

    Here, the negative sign indicates the decrease in breadth.
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