How much force is required to excite a vibrating machine with a load?

AI Thread Summary
The discussion centers on a formula for calculating the force required to excite a vibrating machine with a load, expressed as F = 2*S / (K*M). The variables include stroke, angular velocity, and a mass ratio of the machine and material. The participants analyze how the M variable functions as a ratio, noting that it can yield values lower than the machine weight, which may lead to incorrect force calculations. There is also curiosity about the origin of the constant 140 in the formula, with suggestions that it may relate to pre-calculated damping. The conversation concludes with a consensus on the importance of considering machine mass when the load is minimal.
slobberingant
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Need help understanding a vibrating force formula that was given to me at work.
I have been given a formula at work to use for calculating how much force is required to excite a vibrating machine with load. Only a proportion of material load on a vibrating machine is considered.

The formula is
F = 2*S / (K*M)
Where:
S = stroke (mm)
K = (140/w)^2
w = angular velocity (rad/s)
M = (M1 + (M1 + M2)) / (M1 * (M1 + M2)) (1/kg)
M1 = machine weight (kg)
M2 = material weight (kg)

I've manage to derive that the formula is based on Newton's second law and the acceleration of simple harmonic motion.
F = ma
a = Aw^2 - simple harmonic acceleration formula
Where:
A = amplitude (mm) => Stroke = 2*A

Therefore
F = mAw^2

My formula above is different in the following ways.
- w^2 ((rad/s)^2) becomes K - (140/w)^2 (1/((rad/s)^2)) but is divided rather than multiplied.
- m (kg) becomes M (1/kg) and is also divided rather than multiplied.

What does the M variable in my formula above do? - It seems like a ratio between machine and material mass. However, in isolation, the values it produces can be lower than the machine weight which would result in incorrect force values.

Where does the 140 constant in K come from?
 
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slobberingant said:
- w^2 ((rad/s)^2) becomes K - (140/w)^2 (1/((rad/s)^2)) but is divided rather than multiplied.
No. w^2 becomes 1/K

slobberingant said:
- m (kg) becomes M (1/kg) and is also divided rather than multiplied.
No. m becomes 1/M

slobberingant said:
What does the M variable in my formula above do? - It seems like a ratio between machine and material mass. However, in isolation, the values it produces can be lower than the machine weight which would result in incorrect force values.
Actually, if M1 >> M2, then 1/M tends toward 0.5*M1, which multiplied by the 2 in the equation really gives F = M1 * [acceleration].

If M2 >> M1, then 1/M tends toward M1, which gives F = 2 * M1 * [acceleration]. I'm not sure what it means though.

slobberingant said:
Where does the 140 constant in K come from?
Not sure, but I think it might have to do with some pre-calculated damping for the machine.

More info: https://www.brown.edu/Departments/E...Notes/vibrations_forced/vibrations_forced.htm
 
Thanks Jack.
Fantastic insight.
Regarding M, if there is little load relative to mass then use the mass of machine only. This makes sense as larger machines are less effected by material load. This logic works with M2 >> M1. This approach helps greatly.
I will do some more reading on damping.
Thank you.
 
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