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21 March, 14:03

A pipe is subjected to a tension force of P = 50.00 kN. The pipe outside diameter is 32.00 mm, the wall thickness is 6.00 mm, and the elastic modulus is E = 150.00 GPa. Determine the normal strain in the pipe. Express the strain in mm/mm.

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  1. 21 March, 14:18
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    Strain = 6.8 x 10^-4 mm/mm

    Explanation:

    First we need to calculate normal stress on the pipe. For that we have formula:

    Normal Stress = σ = F/A

    where,

    F = Normal Force = 50 KN = 50000 N

    A = Area normal to the force = Area of outer circle - Area of inner circle = Ao - Ai

    Now, we find outer area and inner area:

    Ao = π (outside radius) ²

    Ao = π (0.032m/2) ²

    Ao = 8.04 x 10^-4 m²

    Ai = π (inside radius) ²

    Ai = π (0.032m/2 - 0.006 m) ²

    Ai = 3.14 x 10^-4 m²

    Now, the Area normal to force is calculated as:

    A = Ao - Ai = 8.04 x 10^-4 m² - 3.14 x 10^-4 m²

    A = 4.9 x 10^-4 m²

    Now the stress can be calculated by given formula:

    σ = 50000 N/4.9 x 10^-4 m²

    σ = 0.102 GPa

    Now, for the strain we have formula:

    Elastic Modulus = E = σ/strain

    Strain = σ/E

    Strain = (0.102 GPa) / (150 GPa)

    Strain = 6.8 x 10^-4 mm/mm
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