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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? (A) Increase in width of 3.62  10−6 m (B) Decrease in width of 7.24  10−6 m (C) Increase in width of 7.24  10−6 m (D) Decrease in width of 2.18  10−6 m

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Rau7star

Answer:

(D) Decrease in width of 2.18 x [tex]10^{-6[/tex]m

Explanation:

Given:

force 'F'= 60,000 N

elastic modulus 'E' = 207 GPa => 2.07 x  [tex]10^{11[/tex]N/m²

cross section area ' [tex]A_{0}[/tex]'= 20 mm x 40 mm => 800mm² =>8 x [tex]10^{-4}[/tex] m²

∈z = б/E => (F/ [tex]A_{0}[/tex])/E => F/ [tex]A_{0}[/tex]E

∈z = 60,000/(8 x [tex]10^{-4}[/tex] x 2.07 x  [tex]10^{11[/tex])

∈z =3.62 x  [tex]10^{-4}[/tex]

Lateral strain is given by,

∈x= -v∈z => -(0.30)(3.62 x  [tex]10^{-4}[/tex])

∈x=1.09  x  [tex]10^{-4}[/tex]

Next is to calculate the change in width

ΔW= Wo x ∈x =>20 x 1.09  x  [tex]10^{-4}[/tex]

ΔW= -2.18 x  [tex]10^{-6[/tex] m

Therefore, the correct option is 'D'

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