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How to avoid errors with the units in Abaqus

Written by Miguel

22 Feb, 2021

One of the first questions that come up when we start working with Abaqus is, what units shall I use for the density, elastic modulus or even the dimensions of my model? I mean, what units should I use for the length, time, mass, force, stress or any other magnitude.

1. Types of magnitudes

Before going on, let’s remember that there are: fundamental magnitudes and derivate magnitudes

There are 7 fundamental magnitudes from which we can define any other magnitude. They are summarized in the following table:

Fundamenal magnitud Abbreviation Units
(International System)
Length \(L\) \(m\)
Mass \(M\) \(kg\)
Time \(t\) \(s\)
Temperature \(T\) \(K\)
Amount of substance \(n\) \(mol\)
Luminous intensity \(I_v\) \(cd\)
Electric current \(I\) \(A\)

Derivate magnitudes are defined combining fundamental magnitudes, Although there is a huge number of derivate magnitudes, some of the most common in mechanical, electrical and thermal disciplines are the following:

Derivate magnitud Combination Units
(International System)
Surface \(L^2\) \(m^2\)
Volume \(L^3\) \(m^3\)
Density \(M / L^3\) \(kg /m^3\)
Velocity \(L/t\) \(m/s\)
Acceleration \(L/t^2\) \(m/s^2\)
Force \(M \cdot L/t^2\) (Newton) \(N = kg · m/s^2\)
Pressure or stress \(M /(L · t^2)\) (Pascal) \(Pa = kg /(m · s^2)\)
Energy \(M \cdot L^2/t^2\) (Joule) \(J = kg · m^2/s^2\)
Electrical charge \(I · t\) (Coulomb) \(C = A · s\)
Voltage or electric potential \(M · L^2 / (I · t^3)\) (Volt) \(V = J/C = kg · m^2 / (A · s^3)\)
Electrical resistance \(M · L^2 / (I^2 · t^3)\) (Ohm) \(\Omega = V/A = kg · m^2 / (A^2 · s^3)\)
Power \(M · L^2 / t^3\) (Watt) \(W = J / s = V · A = kg · m^2 / s^3\)
Capacitance \(I^2 · t^4 / (M · L^3)\) (Farad) \(F = C/V = A^2 · s^4 / (kg · m^3)\)
Latent heat \(L^2 / t^2\) \(J/kg = m^2 / s^2\)
Specific heat \(L^2 / (t^2 · T)\) \(J/(kg·K) = m^2 / (s^2 · K)\)
Thermal conductivity \(M · L / (t^2 · T)\) \(W/(m·K) = kg · m / (s^3 · K)\)

2. The units system of Abaqus

The first thing we need to know is that Abaqus does not oblige to use any particular units system, such as the International System or the Imperial System. Instead, it follows a coherent units system. What does it mean?

Let’s say that we have sketched our part using meters for the length (m). It involves that every derivate magnitude based on length (L), will take it in meters. In this way, we could introduce the rest of magnitudes in units of the International System (SIU), so for a material as steel we would end up like:

Elastic modulus, \(E=210·10^9~(Pa)\)

Poisson ratio, \(\nu = 0.3\) (non-dimensional)

Density, \(\rho = 7800~(kg/m^3)\)

3. Alternatives

However, if we do that, we will be working with huge quantities. Especially when dealing with stresses, which are in the order of MPa (millions of pascals).

To avoid that, we can take advantage of the coherent units system of Abaqus and make a smart choice of the fundamental magnitudes for our model. Some of the most convenient units systems in mechanics, depending on the model size,  are shown in the table below:

International System Micromechanics Mesomechanics 1 Mesomechanics 2 Mesomechanics 3 Macromechanics
Fundamental magnitudes
Length \(m\) \(\mu m\) \(mm\) \(mm\) \(mm\) \(m\)
Mass \(kg\) \(g\) \(g\) \(ton\) \(kg\) \(kton\)
Time \(s\) \(ms\) \(ms\) \(s\) \(ms\) \(s\)
Derivate magnitudes
Force \(N\) \(mN\) \(N\) \(N\) \(kN\) \(MN\)
Stress \(Pa\) \(GPa\) \(MPa\) \(MPa\) \(GPa\) \(MPa\)
Energy \(J\) \(nJ\) \(mJ\) \(mJ\) \(J\) \(MJ\)
Density \(kg/m^3\) \(10^{15}~kg/m^3\) \(10^{6}~kg/m^3\) \(10^{12}~kg/m^3\) \(10^{9}~kg/m^3\) \(10^{6}~kg/m^3\)
Water density \(10^{3}\) \(10^{-12}\) \(10^{-3}\) \(10^{-9}\) \(10^{-6}\) \(10^{-3}\)

My favourite units system is the one I called ‘Mesomechanics 2‘, because it combines length in millimeters (mm), time in seconds (s) and stress in megapascals (MPa). Hence, for the material shown before:

Elastic modulus, \(E=210·10^3~(MPa)\)

Poisson ratio, \(\nu = 0.3\) (non-dimensional)

Density, \(\rho = 7.8 · 10^{-9}~(10^{12}~kg/m^3)\)

There are very small models devoted to analyze microscopic phenomena, also known as Computational Micromechanics (CMM). In these cases, it is much more convenient to use microns for the length (μm). The best suited units system for CMM models is the one named ‘Micromechanics‘, with stresses in GPa.

On the other hand, if the model expands along dozens or hundreds of meters, as it occurs in big infrastructures (e.g. bridges, skyscrappers, wind turbines…). It is prefered using meters for the length (m). Adopting the ‘Macromechanics‘ system seems to be the most appropriate taking the mass in kilotons (1 kton = 1000 ton) and time in seconds. You can check that, like this, stress will be in MPa.

In the next video I will show you how to use different units systems in Abaqus. In addition, you will find some tips to improve results visualization.

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