Calculating the Volumetric Strain for a Rectangular Steel Bar

[cicatriz]

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^{2} and Poissons ratio = 0.26

The Attempt at a Solution

I have calculated \sigmax = -0.036 kN/mm^{2}, \sigmay = 0.072 kN/mm^{2} and \sigmaz = 0.036 kN/mm^{2}.

With these values, using Hooke's Law: \epsilon = 1/E[\sigma1 - \upsilon(\sigma2 + \sigma3)] I have calculated:

\epsilonx = 3.6x10^{-4}, \epsilony = -3.204x10^{-4} and \epsilonz = 1.332x10^{-4}.

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

I would appreciate any guidance with this.

 

[Mapes]

Hi cicatriz, welcome to PF. Try using this equation for change in volume:

\Delta V=V_0\left[(1+\epsilon_1)(1+\epsilon_2)(1+\epsilon_3)-1\right]

I'm not sure how the other one was derived (hydrostatic pressure maybe?) but it doesn't look right. I checked your strain values and they look fine. The change in volume should be about twice what you got previously.