The E-modulus of volume?

Created: 20 Januari 2021
Published: 20 Januari 2021
Last updated: 20 Januari 2021

The E-modulus of an isotropic material is well understood. But in a tensile test, are we pulling on matter or on volume? For example, only 0,000000000027% of iron volume is filled with matter.


CALCULATION

Iron has a mass of 55,85 g/mol. 1 mol contains 6* 1023 atoms. One atom contains 26 protons, 30 neutrons and 26 electrons.

When all protons and neutrons are combined as a close-packed structure of spheres then the resulting sphere has a radius of (1,2* 10 -15m)*(56*6*1023)1/3= 3,9 * 10 -7 m.

Because the resulting sphere has 26% of volume left between the protons and neutrons AND electrons are 1000 times smaller than protons and neutrons we can assume that all electrons can be contained into the volume left.

Thus we remain at a radius 3,9 * 10 -7 m, which gives a volume of 1,9 * 10 -18 m3 for the resulting sphere of only (protons AND neutrons AND electrons).

The density of iron is 7.874 g/cm3. Thus 55,85 g/mol/7,784g/cm3 = 7,2 cm3/mol = 7,2 * 10-6 m3/mol.

Finally we get for the upper limit of the volume of matter inside the volume of iron 1,9* 10 -18 m3 / 7,2 * 10-6 m3 = 2,7 * 10-13 or 0,000000000027% .


When we simply accept this number also for the cross sectional area of a tensile test specimen we must conclude we are pulling at volume, not matter. The E-modulus says something about the strength of the ‘inter res’ volume (volume between the matter).