3
$\begingroup$

I asked this question previously on the physics board (LINK: https://physics.stackexchange.com/questions/404539/linear-elasticity-governing-equation), but I thought I'd post it here and see if anyone else can offer me another perspective since there seem to be multiple users on here familiar with modeling structural mechanics.

Transient linear elasticity is governed by the following hyperbolic PDE: \begin{align} \frac{\partial^2(\rho \boldsymbol{u})}{\partial t^2} - \nabla \cdot \boldsymbol{\sigma} = \boldsymbol{F} \end{align}

This equation is hyperbolic and and has no dampening term, so it will never dampen and won't reach equilibrium, i.e., it doesn't have a steady state form. If this equation doesn't have a steady state form, then why is the following equation frequently referred to as the steady state (or equilibrium) equations? \begin{align} - \nabla \cdot \boldsymbol{\sigma} = \boldsymbol{F} \end{align}

These equations are commonly referred to, in journal/conference papers, by people and/or on the web, as the "steady state" or "equilibrium" equations, and I don't understand why. Note that when I refer to papers, I mean computational science papers. I'm not sure what experimentalists refer to the equations as as I have not read any papers dealing with structural experiments.

$\endgroup$
  • 1
    $\begingroup$ Equilibrium is the preferred term for static problems. I would restate that these are conservation of momentum. On the other hand, equilibrium does not mean that something does not move. $\endgroup$ – nicoguaro May 30 '18 at 23:15
  • $\begingroup$ This equation makes more sense when written in conservative form (think of a finite volume discretization). The sum of forces applied to the boundaries of a control volume balances with the sum of all internal body forces (e.g. gravity) when the body reaches a static equilibrium (in the steady state limit). $\endgroup$ – Paul May 31 '18 at 3:07
  • $\begingroup$ Funny question. In reality, you always have a damping term. I'm now speaking as an engineer. $\endgroup$ – knl May 31 '18 at 8:01
6
$\begingroup$

The general idea comes from this example.

Imagine a rigid pendulum of length $l_0$ hanging a mass $m_0$. The equation of motion can be derived from a minimum principle (Lagrangian).

Denoting with $\theta$ to the angle which the pendulum forms with respect to the vertical direction we have: $$\mathcal{L}(\theta) = \int_{t_1}^{t_2}\left[\frac{1}{2}m_0l_0^2\dot{\theta}^2-m_0gl_0(1-\cos{(\theta))}\right]dt\tag{1}$$ The first term is the kinetic energy while the second is the potential energy.

It is obvious that this dynamic system (derived from $(1)$) expressed by the dynamic equation: $$\frac{d^2\theta}{dt^2}+\frac{g}{l_0}\sin(\theta)=0\tag{2}$$ will never reach the "equilibrium" (like your hyperbolic equations). But what is remarkable, is that certain special positions provide the system with sufficient "stability" to remain steady. These special positions are defined by the extrema of the potential $V(\theta)=m_0gl_0(1-\cos({\theta}))$: $$V'(\theta) = m_0gl_0\sin(\theta) = 0$$ being $\theta_1=0$ or $\theta_2=\pi$.

Since $\theta_1$ is a minimum the equilibrium will be stable (small perturbation in position leads to small responses), while $\theta_2$ offers an unstable equilibrium because it is a maximum.

These points could have been obtained just from $(2)$ setting $d_t^2\theta=0$ and solving for equilibrium. As a final remark, if we set the initial conditions for $(2)$ either $\theta(0) = \theta_1$ or $\theta(0)=\theta_2$ the system no longer moves (check it!).

This frame can be exactly applied to our case, in which we have the equation (now a PDE): $$\rho\partial_t^2\vec{u} =\textrm{div}(\sigma)+\vec{f}\tag{3}$$ in which the positions $\vec{u}_0$ given by solving "the equilibrium equation": $$-\textrm{div}(\sigma(\vec{u}_0))=\vec{f}$$ leads to a system that remains steady for any time if we use them as initial conditions $\vec{u}(0) = \vec{u}_0$ for the transient problem $(3)$.

That is the explanation (in my opinion) to the question of why these equations (without time derivatives) are named "equilibrium equations".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.