gemseo.problems.ode.springs package#

A problem illustrating how to couple instances of ODEDiscipline.

Consider a system of \(n\) point masses with masses \(m_1\), \(m_2\),... \(m_n\) connected in series by springs. The displacement of the point masses relative to the position at rest are denoted by \(x_1\), \(x_2\),... \(x_n\). Each spring has stiffness \(k_1\), \(k_2\),... \(k_{n+1}\).

Motion is assumed to only take place in one dimension, along the axes of the springs.

The extremities of the first and last spring are fixed. This means that by convention, \(x_0 = x_{n+1} = 0\).

For \(n=2\), the system is as follows:

|                                                                                 |
|        k1           ________          k2           ________          k3         |
|  /\      /\        |        |   /\      /\        |        |   /\      /\       |
|_/  \    /  \     __|   m1   |__/  \    /  \     __|   m2   |__/  \    /  \     _|
|     \  /    \  /   |        |      \  /    \  /   |        |      \  /    \  /  |
|      \/      \/    |________|       \/      \/    |________|       \/      \/   |
|                         |                              |                        |
                       ---|--->                       ---|--->
                          |   x1                         |   x2

The force of a spring with stiffness \(k\) is

\[\vec{F} = -kx\]

where \(x\) is the displacement of the extremity of the spring.

Newton's second law applied to any point mass \(m_i\) can be written as

\[m_i \ddot{x_i} = \sum \vec{F} = k_i (x_{i-1} - x_i) + k_{i+1} (x_{i+1} - x_i) = k_i x_{i-1} + k_{i+1} x_{i+1} - (k_i + k_{i+1}) x_i\]

This can be re-written as a system of first-order ordinary differential equations:

\[\begin{split}\left\{ \begin{cases} \dot{x_i} &= y_i \\ \dot{y_i} &= - \frac{k_i + k_{i+1}}{m_i}x_i + \frac{k_i}{m_i}x_{i-1} + \frac{k_{i+1}}{m_i}x_{i+1} \end{cases} \right.\end{split}\]

gemseo.problems.ode.springs package#

Submodules#

This Page