Created: 20 Januari 2021 Published: 20 Januari 2021 Last updated: 26 Julyi 2022

The E-modulus of an isotropic material is well understood. But in a tensile test, are we pulling on matter or on the volume where the matter resides in? 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* 10^{23} atoms. One atom exists out of two sizes of particles, larger 26 protons and 30 neutrons and a 1000 times smaller, 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^{-15} m)*(56*6*10^{23})^{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} m^{3} for the resulting sphere of only (protons AND neutrons AND electrons).

The density of iron is 7.874 g/cm^{3}. Thus 55,85 (g/mol)/7,784 (g/cm^{3})= 7,2 (cm^{3}/mol) = 7,2 * 10^{-6} m^{3}/mol.

Finally we get for the upper limit of the volume of matter inside the volume of iron 1,9* 10 ^{-18} m^{3} / 7,2 * 10^{-6} m^{3} = 2,7 * 10^{-13} or 0,000000000027 % .

When considering only the iron atoms with a diameter of 252 * 10^{-12} m then the resulting radius is still 2,12 * 10^{-5} m, resulting in a sphere volume of 2,0*10^{-11} m^{3} and an upper limit of the volume of matter inside the volume of iron to be 2,8*10^{-6} or 0,0000028 %.

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

This must be some reality or do you think that in a tensile test only the atoms or particles are being stretched?