Question about the effect of acceleration on the force of a mass

[diggedy]

Ive got a query which I am thinking should be simple but I can't for the life of me think how to solve it.

I have a mass of 500 kg resting on a beam. It is subjected to vibration of 2m/s² directly down, how would I calculate the extra force subjected on the beam due to this acceleration?

Is it as simple as working out the force of the mass while stationary then working out the force applied due to acceleration (m*a) and then adding them together?

 

[mfb]

A constant acceleration is not a vibration. For the momentary force: if all parts of the system are completely rigid and incompressible, then it is as simple as you described. Otherwise it gets more complicated.

 

[diggedy]

Cheers for the reply. I've probably been a bit vague in the question so ill give a bit more info.

Its a mass on a fabricated bracket attached a moving piece of machinery. I am treating it as a simple steel beam to avoid over complicating it. As the machine moves along the ground, imperfections in the surface cause the mass to bounce. 2m/s² is the peak acceleration so I was just taking this as a worst case scenario. All I am wanting to do is determine the maximum force this mass exerts on the beam so I can use it for FEA analysis on the beam and use it to design a new bracket as the current one is failing.

 

[mfb]

If you measured the acceleration directly at the object, the corresponding force is just F=m*a.

 

[diggedy]

Thanks, I guess I am just over thinking it :)

 

[A.T.]

It is subjected to vibration of 2m/s² directly down,

Is this acceleration relative to free fall or relative to rest?

All I am wanting to do is determine the maximum force this mass exerts on the beam

If the mass sits on the the beam, you should look for the maximal upwards acceleration.

 

[diggedy]

The accelerometer i had access to didnt give me both direction, only the maximum vibration along that axis. I am getting access to some better equipment next week which may give me better info. I was treating the peak as the maximum either way and its relative to rest