I want to know what terms to add to the differential equation for structural vibrations to account for structural damping. Not external damping, but internal damping. I consider the transfer of energy into sound waves, or air drag as external damping. I want to focus on the energy dissipation mechanism of a typical metal (crystalline) structure.

Two scenarios.

  1. Firstly, a long rod with longitudinal waves has the following dynamics in terms of deflection $u(x,t)$

    $$ \left( E A \frac{\partial^2 u}{\partial x^2} - \rho A \frac{\partial^2 u}{\partial t^2} \right) {\rm d}x = 0 $$

    which is derived from the balance of forces on a thin slice ${\rm d}x$ with mass ${\rm d}m = \rho A {\rm d}x$, with $\rho$ the density, $A$ the cross sectional area and $E$ the elastic modulus.


    My question is how do you account for structural damping with the above equation. Do you add a term with both curvature and time, like $ \frac{\partial^2}{\partial x^2}\frac{\partial}{\partial t} u$ ?

    The tension-deflection relationship for a rod is $T = E A \frac{\partial}{\partial x} u$.

  2. Secondly, what about the transverse vibrations of a beam. The undamped equation is

    $$ \left( E I \frac{\partial^4 u}{\partial x^4} + \rho A \frac{\partial^2 u}{\partial t^2} \right) {\rm d}x = 0 $$

    where $I$ is the 2nd moment of area. What does the structural damping term look like? is it like $ \frac{\partial^3}{\partial x^3}\frac{\partial}{\partial t} u$ which can be interpreted as the time rate of the shear force?

    The shear force-deflection relensionship is $S = E I \frac{\partial^3}{\partial x^3} u$ and the moment-deflection relationship $M = E I \frac{\partial^2}{\partial x^2} u$.

Ultimately what I want to examine is the effect of damping on the natural frequency of structural vibrations.

PS. External damping would add a term proportional to speed only $\frac{\partial u}{\partial t}$ and the shape of the structure does not play any role in it. I am not interested in this.

  • $\begingroup$ For beams I am inclined to have bending moment being proportional to time change of slope. Something like $$M = -\alpha \frac{\partial}{\partial t}\frac{\partial u}{\partial x}$$ and also the moment-deflection relatioship $$M = E I \frac{\partial^2 u}{\partial x^2}$$ $\endgroup$ Jan 3, 2021 at 21:04
  • $\begingroup$ It depends on the type of absorption you are thinking about. Accounting for viscoelasticity will look different from accounting for thermoelasticity, which in turn looks different from accounting for relaxation. There are many ways to represent wave attentuation. I don't know for sure which mechanism is dominant, but my guess is that you should look into viscoelastic beams and bars. $\endgroup$
    – Michael M
    Jan 6, 2021 at 12:44
  • $\begingroup$ @MichaelM how is hysteresis (viscoelastic) loss different from thermoelastic loss. They both result in heat being generated. Also, I am not talking about acoustic/air drag losses, but simply losses due to repeated bending/stretching as it happens in real life. $\endgroup$ Jan 6, 2021 at 13:26
  • $\begingroup$ There are entire engineering textbook devoted to this topic... you’d be better searching there. $\endgroup$ Jan 6, 2021 at 13:30
  • $\begingroup$ I have a master's in engineering, and I have been searching for this for years. @ZeroTheHero care to nudge me in the right direction? $\endgroup$ Jan 7, 2021 at 5:39

1 Answer 1


The simplest answer is to consider the stress-strain relation as:

$σ = Eε + ηε_t$

This is the stress relation for a simple damped spring (Kelvin Voigt).

Then you can define your equation of motion as:

$ρAy_{tt} = M_{xx}$

Where Moment can be derived from the stress equation by substituting $I*y_{xx}$ (bend) for the strain.

Thus you have:

$M = (-EIy_{xx} - ηIy_{xxt})_{xx}$

$M_{xx} = -EIy_{xxxx} - ηIy_{xxxxt}$

And your final equation is:

$ρAy_{tt} = -EIy_{xxxx} - Iηy_{xxxxt}$

I conceptualize this as the damping that would occur from bending a stiff spring with a rotational viscous damper in parallel.

If you want to get nuts, this article takes it a lot further applying nonlocal effects and this type of damping to a Timoshenko beam.

  • $\begingroup$ Is $ε_t = \tfrac{\partial}{\partial t} ε$? So $M = -E I \frac{\partial^2}{\partial x^2} y$ becomes $$M = -E I \frac{\partial^2}{\partial x^2} y + \eta I \frac{\partial}{\partial t} \frac{\partial^2}{\partial x^2}y$$$ $\endgroup$ Feb 2, 2021 at 2:06
  • $\begingroup$ Yes exactly John. I just like using subscripts for my derivatives. Personal preference. You've got it. Try it out. You should find it works reasonably enough. Except they are both negative terms. Ie. $M = -EIy_{xx} - ηIy_{xxt}$. This is the common approach for wave equation modeling. I'm wondering if maybe you can help me back out on a related question I just made here: physics.stackexchange.com/questions/611830/… I would like to calculate the total force at any one point in the finite difference beam. $\endgroup$
    – mike
    Feb 2, 2021 at 4:34
  • $\begingroup$ Also FYI, you can read about various viscoelastic models here: en.wikipedia.org/wiki/Viscoelasticity Any of these can theoretically be used to damp a beam in a similar fashion but the Kelvin-Voigt which I shared is the simplest decent approach to it. More complex models become harder mathematically and may not be necessary for your needs. $\endgroup$
    – mike
    Feb 2, 2021 at 4:36

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