0
$\begingroup$

enter image description here

In the table above, the period of the structure $(T)$, the acceleration coefficient on the structure ($g$ in terms of gravitational acceleration) $Sa(T)$, The damping ratio is expressed as $ζ$ .

As the damping ratio of a structure increases, the earthquake design acceleration on the structure decreases. That's why the earthquake force is decreasing. But how can damping change the amount of an external force? In other words, why does the damping ratio cause the earthquake design acceleration of a structure to decrease? We know that as the period of the structure increases, the earthquake force on the structure decreases. Can you explain the theoretical reason for this? Am I misunderstanding the nature of the earthquake force?

$\endgroup$

1 Answer 1

1
$\begingroup$

Am I misunderstanding the nature of the earthquake force? - Yes.

The earthquake force is a force external to the structure and so does not change.
You have forced oscillation which means that damping changes the response of the structure to the external force.
Put another way the $Q$ value (sharpness of the movement of the structure against frequency graph) is decreases as the amount of damping increases.
Energy which, with little or no damping, went into the structure as mechanical energy is dissipated within the damping system.

Damping is not the only way to protect structures.

$\endgroup$
2
  • $\begingroup$ Thank you for the answer, but I did not understand. If the earthquake force does not change, the above graph should give the same acceleration values ​​at all damping rates, however, as the damping increases and the period values ​​increase, the earthquake force on the structure decreases, according to this graph. $\endgroup$ Commented Jun 2 at 9:57
  • $\begingroup$ The external force does work on the structure but with damping some of that work is converted to heat and the rest to mechanical energy of the structure, ie the structure's mechanical response is less, it sways and accelerates less. $\endgroup$
    – Farcher
    Commented Jun 2 at 10:00

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