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29 August, 06:02

The change in diameter of a large steel bolt is carefully measured as the nut is tightened. knowing that e 5 29 3 10 6 psi and n 5 0.30, determine the internal force in the bolt if the diameter is observed to decrease by 0.5 3 10 23 in. beer. mechanics of materials (mechanical engineering) (page 110). mcgraw-hill education. kindle edition.

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  1. 29 August, 06:18
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    solution:

    write the expression for the lateral strain in the bolt.

    ɛy = δ y/d

    Here, δy and d are the decrease in diameter and original diameter respectively.

    Substitute - 0.5 * 10-3 in for δy and 2.5 in for d.

    ɛy =

    = - 2.0 x 10-4

    Write the expression for the Poisson's ratio.

    V = - ɛy / ɛx

    Substitute - 2 x 10-4 for ɛy and 0.3 for v ...

    0.3 = - (2.0 x 10-4) / 0.3

    ɛx = 2.0 x 10-4/0.3

    = 6.67 x 10-4

    Write the expression for strain in - direction.

    ɛx = σx / E - vσx/E

    Substitute 0 for σy, 29 x 106 psi for and 6.67 x 10-4 for ɛx.

    6.67 x 10-4 = σx/29 x 106 psi

    σx = 6.67 x 10-4 (29 x 106)

    = 19.34 x 103 psi

    the expression for the cross-section area of the bolt.

    A =

    Here, d is the diameter of a steel bolt.

    Substitute 2.5 in for d.

    A =

    = 4.906

    Consider the axial stress.

    σx =

    Substitute 19.34 x 103 psi for σx and 4.906 in2 for A.

    19.34 x 10 psi =

    F = 19.34 x 103 (4.906)

    = 95 x 103

    The initial force in the bolt is 95 x 103 ib.
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