A specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s ratio of 0.30

Question

A specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s ratio of 0.30. If this specimen is pulled in tension with a force of 60,000 N, what is the change in width if deformation is totally elastic?

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RI SƠ 4 years 2021-08-20T19:01:41+00:00 1 Answers 7 views 0

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  1. MichaelMet
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    2021-08-20T19:02:53+00:00
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    Answer:

    There’s a decrease in width of 2.18 × 10^(-6) m

    Explanation:

    We are given;

    Shear Modulus;E = 207 GPa = 207 × 10^(9) N/m²

    Force;F = 60000 N.

    Poisson’s ratio; υ =0.30

    We are told width is 20 mm and thickness 40 mm.

    Thus;

    Area = 20 × 10^(-3) × 40 × 10^(-3)

    Area = 8 × 10^(-4) m²

    Now formula for shear modulus is;

    E = σ/ε_z

    Where σ is stress given by the formula Force(F)/Area(A)

    While ε_z is longitudinal strain.

    Thus;

    E = (F/A)/ε_z

    ε_z = (F/A)/E

    ε_z = (60,000/(8 × 10^(-4)))/(207 × 10^(9))

    ε_z = 3.62 × 10^(-4)

    Now, formula for lateral strain is;

    ε_x = – υ × ε_z

    ε_x = -0.3 × 3.62 × 10^(-4)

    ε_x = -1.09 × 10^(-4)

    Now, change in width is given by;

    Δw = w_o × ε_x

    Where w_o is initial width = 20 × 10^(-3) m

    So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)

    Δw = -2.18 × 10^(-6) m

    Negative means the width decreased.

    So there’s a decrease in width of 2.18 × 10^(-6) m

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