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24 July, 06:30

A 1.5-m-long aluminum rod must not stretch more than 1 mm andthe normal stress must not exceed 40 MPa when the rod is subjectedto a 3-kN axial load. Knowing that E = 70 GPa, determine therequired diameter of the rod.

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  1. 24 July, 06:59
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    d = 11.2 mm

    Explanation:

    L = 1.5 m

    σ = 40 * 10^6 Pa

    E = 40 * 10^9 Pa

    P = 3 * 10^3 N

    δ = 1 * 10^-3 m

    stress:

    σ = P/A

    A = P/σ

    = 3 * 10^3 N/40 * 10^6 Pa

    = 75 mm^2

    deformation:

    δ = P*L/A*E

    A = P*L/E*δ

    = 75 mm^2

    A = π/4*d^2 (1)

    putting all values in eq (1)

    d = 11.2 mm

    note:

    there maybe calculation mistake but method is correct
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