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 $$\sigma$$x = -0.036 kN/mm$$^{2}$$, $$\sigma$$y = 0.072 kN/mm$$^{2}$$ and $$\sigma$$z = 0.036 kN/mm$$^{2}$$.

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

$$\epsilon$$x = 3.6x10$$^{-4}$$, $$\epsilon$$y = -3.204x10$$^{-4}$$ and $$\epsilon$$z = 1.332x10$$^{-4}$$.

Furthermore, using the equation for volumetric strain: $$\nabla$$V/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.