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21 February, 19:23

A 9,000-lb load is suspended from the roof in a shopping mall with a 16-ft-long solid aluminum rod. The modulus of elasticity of the aluminum is 10,000,000 psi. If the maximum rod elongation must be limited to 0.50 in. and the maximum stress must be limited to 30,000 psi, determine the minimum diameter that may be used for the rod (precision to 0.00).

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  1. 21 February, 20:36
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    Step-by-step explanation:

    Given:

    elongation, x = 0.50 in

    Force, f = 9000 lb

    Young modulus, E = 10,000,000 psi

    Maximum Stress, Sm = 30000 psi

    Length, L = 16 ft

    Converting ft to in,

    12 in = 1 ft

    =16 * 12 = 192 in

    Young modulus, E = stress/strain

    Stress = force/area, A

    Strain = elongation, x/Length, L

    E = f * L/A * E

    1 * 10^7 = stress / (0.5/16)

    = 26041.7 psi

    Minimum stress = 26041.7 psi

    Maximum stress = 30,000 psi

    Stress = force/area

    Area = 9000/26041.7

    = 0.3456 in^2

    Stress = force/area

    Area = 9000/30000

    = 0.3 in^2

    Using minimum area of 0.3 in^2,

    A = (pi/4) (d^2)

    0.3 in^2 = (pi/4) (d^2)

    d = 0.618 inches

    diameter, d = 0.618 inches
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