FEA engine

I'm using CCX finite elements analysis engine along with its pre- and post-processor CGX:




I'm looking at one example here that is using steel material:


Elastic specifications

In the example above, the declaration is done for Young’s modulus, Poisson’s ratio, and Temperature:



Mass density is initialized by:



Gravity, i.e. acceleration vector, is defined like this:


Input values

Therefore, the input values are:

  • Young’s modulus = 210000
  • Mass density = 7.85e-9
  • Gravity acceleration = 9810.


What are the units of measurement for the above example?

My guess

I know the gravity acceleration is 9.8 m/s2. So, 9810. means mm/s2.

But I'm confused about Young’s modulus and Mass density. What are their units? They should be consistent with the mm. But I cannot figure them out.


This table from the latest CalculiX solver documentation helps with unit conversion. The units of the above example are mm,N,s,K.

Table of units conversion

  • $\begingroup$ Can you look up the Young's modulus and density of steel in clearly-stated units in an independent source and compare? $\endgroup$ Mar 7, 2023 at 10:19
  • $\begingroup$ @DanielHatton I did. But for some reason, it doesn't make sense to me. Let me doublecheck... $\endgroup$
    – Megidd
    Mar 7, 2023 at 10:20
  • 2
    $\begingroup$ You might need to remember your units analysis as in metres per second is equivalent to furlongs per fortnight... $\endgroup$
    – Solar Mike
    Mar 7, 2023 at 10:21
  • $\begingroup$ user3405291 I'm not sure you can expect it to make sense, in terms of being a sane choice of units. On a quick (as in, I did it in a hurry so don't rely on it) analysis, these seem to be the numbers that would come out with a coherent system of units based on picograms for mass, femtometres for length, and microseconds for time. $\endgroup$ Mar 7, 2023 at 10:47

1 Answer 1


DISCAIMER: I haven't used calculix.

From this page, I found out that there are 3 predominant unit systems that are being used

  • m,kg,s,K : meter, kilogram, second, Kelvin
  • mm,N,s,K: millimeter, N, second, Kelvin
  • cm,g,s,K: centimeter, g, second, Kelvin

You are right in that the length units seems to be consistent with mm, so most likely the second is the one being used. If you notice though the second is the only one that used Force (N) instead of mass (kg, or g).

This is consistent with the Modulus is 210000 MPa (Mega Pascal) i.e. $\frac{N}{mm^2}$ (modulus of steel is 201000 - 210000MPa).

However it gets a bit confusing with the density. More specifically:

  • Mass density of steel should be equal to 7800 $\frac{kg}{m^3}$.

In the page mentioned above you can see that density is given by: $\frac{N\cdot s^2}{mm^4}$ in the 2nd system of measurement. The reason is that because $mass = \frac{Force}{\left(\frac{Length}{time^2}\right)}\rightarrow mass = \frac{Force\cdot time^2}{Length}$, and volume is $Length^3$, therefore the units for density are:

$$[density] = \left[\frac{Force \cdot time^2}{Length^4}\right]$$

The actual value of steel density is 7800 $\frac{kg}{m^3}\equiv \frac{N\cdot s^2}{m^4}$. So when you convert m to mm you get

$$7800\cdot \frac{N\cdot s^2}{m^4}=7800\cdot \frac{N\cdot s^2}{(1000mm)^4}=7800\cdot \frac{N\cdot s^2}{10^{12}mm^4}=7800\cdot 10^{-12}\frac{N\cdot s^2}{mm^4}$$

and finally: $$7800\cdot 10^{-12}\frac{N\cdot s^2}{mm^4}\Rightarrow 7.8\cdot 10^-9\frac{N\cdot s^2}{mm^4}$$

  • $\begingroup$ Right. Mass density is confusing. We know the steel mass density is 7.85 g/cm3 therefore 7.85e-9 means the unit is 1000 kg/mm3 or ton/mm3. I didn't expect that. But then you mentioned that the mass density unit is actually Ns2/mm4. I need to think about it... $\endgroup$
    – Megidd
    Mar 7, 2023 at 11:21
  • $\begingroup$ I have a hard time figuring this out: 7800 kg/m3 = 7.8 × 10−9 Ns2/mm4 $\endgroup$
    – Megidd
    Mar 7, 2023 at 11:33
  • $\begingroup$ The latest CalculiX solver documentation has a Units chapter. I'm studying it... $\endgroup$
    – Megidd
    Mar 7, 2023 at 11:37
  • $\begingroup$ So for your density units issue, express Newtons in terms of mass, length and time. $\endgroup$
    – Solar Mike
    Mar 7, 2023 at 14:39

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