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## Homework Statement

A rectangular steel bar has length 250 mm, width 50 mm, and thickness 25 mm. The bar is subjected to a compressive force of 450 kN on the 250 mm x 50 mm face, a tensile force of 450 kN on the 250 mm x 25 mm face, and a tensile force of 45 kN on the 50 mm x 25 mm face.

(a) Find the change in volume of the bar under the force system.

## Homework Equations

The forces can be assumed to be uniformly distributed over the respective faces.

Take E = 200 kN/mm[tex]^{2}[/tex] and Poissons ratio = 0.26

## The Attempt at a Solution

I have calculated [tex]\sigma[/tex]x = -0.036 kN/mm[tex]^{2}[/tex], [tex]\sigma[/tex]y = 0.072 kN/mm[tex]^{2}[/tex] and [tex]\sigma[/tex]z = 0.036 kN/mm[tex]^{2}[/tex].

With these values, using Hooke's Law: [tex]\epsilon[/tex] = 1/E[[tex]\sigma[/tex]1 - [tex]\upsilon[/tex]([tex]\sigma[/tex]2 + [tex]\sigma[/tex]3)] I have calculated:

[tex]\epsilon[/tex]x = 3.6x10[tex]^{-4}[/tex], [tex]\epsilon[/tex]y = -3.204x10[tex]^{-4}[/tex] and [tex]\epsilon[/tex]z = 1.332x10[tex]^{-4}[/tex].

Furthermore, using the equation for volumetric strain: [tex]\nabla[/tex]V/Vo = [tex]\epsilon[/tex](1-2[tex]\upsilon[/tex]) I have calculated the change in volume to be 25.92 mm[tex]^{2}[/tex]. This, according to the answer I have been provided with, appears to be incorrect.

I would appreciate any guidance with this.

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