Discrete systems – BuildSoft Support https://support.buildsoft.eu/fr/ BuildSoft Support Thu, 23 Apr 2026 12:17:15 +0000 fr-FR hourly 1 https://wordpress.org/?v=6.2.13 DLD 02: Mass-spring system with 2 degrees of freedom – modal information https://support.buildsoft.eu/fr/knowledge-base/dld-02-mass-spring-system-with-2-degrees-of-freedom-modal-information/ Thu, 23 Apr 2026 12:16:52 +0000 https://support.buildsoft.eu/?post_type=ht_kb&p=28943 Description

Geometry Mass m_1=50kg
m_2=80kg
Spring contant k_1=4000N/m
k_2=6000N/m
Damping ratio’s \xi=0.02
\xi=0.05
Mesh No. of divisions 1

The damping ratio’s are imposed in the dialog for modal analysis .

Independent reference results

Open hand calculations

In this example we want to recalculate the modal information. To do so, we’ll be using the same model as in DLD 01.
The modal information can be found after this button . Make sure to check the option « Display all the information« .

Natural frequencies f

These are the specific rates at which a system naturally vibrates, determined solely by its mass and stiffness distribution. This data was already available in the validation example DLD 01.

    \[f=\begin{bmatrix} 0.778 \\ 2.522 \end{bmatrix} \text{[Hz]}\]

Angular frequencies \omega

The relation between the angular frequencies and the natural frequencies is given by:

    \[\omega = 2 \pi f=2 \pi \begin{bmatrix} 0.778 \\ 2.522 \end{bmatrix} = \begin{bmatrix} 4.888 \\ 15.845 \end{bmatrix} \text{[1/s]}\]

Natural periods T

The relation between the natural period and the natural frequency is given by:

    \[T= 1/ f= \begin{bmatrix} 0.778 \\ 2.522 \end{bmatrix} = \begin{bmatrix} 1.285\\ 0.397 \end{bmatrix} \text{[s]}\]

Modal mass M

The shape modes were calculated in DLD 01. The first column represents the displacements of mode 1, and the second of mode 2.

    \[\phi=\begin{bmatrix} 67.036774& 124.441490\\ 98.379631& -52.997219 \end{bmatrix}\]

The modal mass M is calculated as:

    \[M=\phi^{T} \cdot M_m \cdot \phi=\begin{bmatrix} 10^6 & 0\\ 0 & 10^6 \end{bmatrix} \text{kg}\]

M_m is the mass matrix M_m=\begin{bmatrix} 50 &0 \\ 0& 80 \end{bmatrix} \text{kg}.

In the Diamonds results for the modal mass M, the 10^6 kg is left out. You just see « 1 » in the table. The mass matrix M_m cannot be consulted in Diamonds.

Modal stiffness K

The modal stiffness K can be calculated in two ways resulting in the same result:

    \[K=M \cdot \omega^2=\begin{bmatrix} 23.870 \\ 250.850 \end{bmatrix} \frac{kN}{mm}\]

or

    \[K=\phi^T \cdot K_k \cdot \phi=\begin{bmatrix} 23.870 &0 \\ 0& 250.850 \end{bmatrix} \frac{kN}{mm}\]

K_k is the stiffness matrix K_k=\begin{bmatrix} k_1+k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix}=\begin{bmatrix} 10000 & -6000 \\ -6000 & 6000 \end{bmatrix} \frac{N}{m}.

The stiffness matrix K_k cannot be consulted in Diamonds.

Modal damping D

Rayleigh damping was used in this model. In order to calculate the modal damping D, we should calculate the damping matrix D_d first.

    \[a_0=\frac{2 \cdot \omega_1 \cdot \omega_2 \cdot \left( \xi_2 \cdot \omega_1 - \xi_1 \cdot\omega_2 \right)}{\omega_1^2-\omega_1^2}=0.05 \frac{1}{s}\]

    \[a_1=\frac{2 \cdot \left( \xi_1 \cdot \omega_1 - \xi_2 \cdot\omega_2 \right)}{\omega_1^2-\omega_1^2}=6.11 \cdot 10^{-3} \text{s}\]

The damping matrix D_d becomes:

    \[D_d=a_0\cdot M_m + a_1 \cdot K_k=\begin{bmatrix} 63.61 & -36.68 \\ -36/68 & 40.64 \end{bmatrix} \frac{kg}{s}\]

The damping matrix D_d cannot be consulted in Diamonds. But using the damping matrix D_d we can calculate the modal damping D:

    \[D=\phi^T \cdot D_d \cdot \phi=\begin{bmatrix} 0.195 & 0\\ 0 & 1.583 \end{bmatrix} \frac{kN \cdot s}{mm}\]

Critical damping D_c

Again using the damping matrix D_d we can caluculate the critical damping D_c:

    \[D_c=2 \cdot \sqrt{M \cdot K} = \begin{bmatrix} 9.776 \\ 31.660 \end{bmatrix} \frac{kN \cdot s}{mm}\]

Damping ratio’s \xi

Since this model only has 2 DOF’s, there are only 2 damping ratio’s. If your model contains more DOF, the damping ratio’s of mode 3 until i can be calculated using:

    \[\xi_i=\frac{1}{2}\left( \frac{a_0}{2\pi f_i} +a_1 \cdot 2\pi f_i\right)\]

Ratio of the effective modal mass in a mode to the total mass

To calculate this ratio we need the modal participation factor \gamma . The modal participation factor \gamma cannot be consulted in Diamonds.

    \[\gamma=\phi^T \cdot M_m \cdot \begin{bmatrix} 10^{-3}\\ 10^{-3} \end{bmatrix}=\begin{bmatrix} 11.222 \\ 1.982 \end{bmatrix} \text{kg}\]

From the modal participation factor \gamma we can determine the effective modal mass m_eff. The effective modal mass m_eff cannot be consulted in Diamonds.

    \[m_{eff}=\frac{\gamma^2}{1 \text{kg}}=\begin{bmatrix} 125.94 \\ 3.93 \end{bmatrix} \text{kg}\]

Note that the sum of 125.94 + 3 .93 = 129.87kg which equals the sum of the masses in the system 50+80kg = 130kg.

Now let’s calculate the ratio of the effective modal mass to the total mass for each mode. It’s this ratio that you see in Diamonds.

\frac{m_{eff}}{130kg}=\begin{bmatrix} 96.88\\ 3.02 \end{bmatrix} %

Results

Independent reference Diamonds Difference1
f [Hz] 0.778
2.522		 0.778
2.522		 ≈0.00%
\omega [1/s] 4.888
15.845		 4.886
15.837		 ≈0.00%
T [s] 1.285
0.397		 1.286
0.397		 ≈0.00%
M 1
1		 1
1		 ≈0.00%
K 23.870
250.850		 23.869
250.812		 ≈0.00%
D 0.195
1.583		 0.195
1.584		 ≈0.00%
D_c 9.776
31.660		 9.771
31.674		 ≈0.00%
\frac{m_{eff}}{m_{tot}} 96.88%
3.02%		 96.97%
3.03%		 ≈0.00%

References

  • Hofman, G. E. (2012). Eindige Elementen Methode: Deel 2. MERc, p118
  • Tested in Diamonds 2026.
]]> DLD 01: Mass-spring system with 2 degrees of freedom – eigen modes https://support.buildsoft.eu/fr/knowledge-base/mass-spring-system-with-2-degrees-of-freedom-eigen-modes/ Mon, 15 Apr 2024 07:56:07 +0000 https://support.buildsoft.eu/?post_type=ht_kb&p=12766 Description

Geometry Mass m_1=50kg
m_2=80kg
Spring contant k_1=4000N/m
k_2=6000N/m
Mesh No. of divisions 1
How to model this mass-spring system in Diamonds?

Diamonds is not intended to model mass-spring systems.  However, with a little out-of-the box thinking, it’s possible to model certain types of mass-spring systems.

  • Diamonds doesn’t contain spring-elements, but they can be modelled using a weightless bar.
    The spring constant k of a bar can be written as: \frac{E \cdot A}{L} (see this article).
    If we choose A=25mm^2 and E=40N/mm^2, we can calculate the bar length L resulting in the spring constant k.

        \[L_1=\frac{E \cdot A}{k_1}=\frac{40N/mm^2 \cdot 25mm^2}{4000N/m} = 0.25m\]

        \[L_2=\frac{E \cdot A}{k_2}=\frac{40N/mm^2 \cdot 25mm^2}{6000N/m} = 0.167m\]

    So the spring- elements, can be modelled as a weightless bars with area A, modulus of elasticity E and lengths L_1 and L_2. Let’s name them « spring-bars ».

  • Add a new material to the library. Name it « DLD 01 ».
    Set Young’s modulus equal to E=40N/mm^2 and the density \pho=0.001kg/m^3 very small because we want weightless bars. The other mechanical material properties are irrelevant.
  • The masses m [kg] must be modelled as forces F [kN].
    Create a new load group ‘Masses static effect’. The partial safety factors and combination factors are not relevant, but make sure \Psi_2 and \varphi are equal to 1.
  • In this example, the masses are meant to move in the in same direction as the springs compress. This determines the orientation of the model. Let’s assume we model the spring-bars horizontally:
  • Then the forces (representing the masses) must also be modelled horizontally. But this way, they don’t contribute to the modal mass. Only the vertical forces contribute. So the spring-bars and forces must be modelled vertically.
  • Because we want a system with 2 DOF, all displacements en rotations along the weightless bars have been prevented, except for the Y-displacement.
  • Pendulum-like mass-spring systems cannot be modelled. Neither can a damper.

Results

Natural frequencies

Eigen mode Independent reference Diamonds Difference
1 0.778Hz 0.778Hz 0,00%
2 2.520Hz 2.520Hz 0.00%

Eigenmodes

Eigen mode Independent reference Diamonds Difference1
1 u1=0,0671
u2=0,0984		 u1=67,03
u2=98,38		 0,00%
2 u1=0,124
u2=-0,053		 u1=124,44
u2=-53,00		 0.00%

1. Eigenmodes are determined except for a proportionality factor. In Diamonds, the eigen modes are scaled using the units for displacement. In this example, the unit were set to mm. As the result, the eigen modes in Diamonds are scale so that, \phi^{T} \cdot M \cdot \phi=10^6, while the reference used a scaling so that \phi^{T} \cdot M \cdot \phi=1. If you’d ask the deformations in m instead of mm, you’d see that Diamonds can also use the scale \phi^{T} \cdot M \cdot \phi=..↩

References

  • Hofman, G. E. (2012). Eindige Elementen Methode: Deel 2. MERc, p118
  • Tested in Diamonds 2026.
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