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
Messages
11
Reaction score
3
TL;DR Summary
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?
 
Engineering news on Phys.org
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.
 
Posted June 2024 - 15 years after starting this class. I have learned a whole lot. To get to the short course on making your stock car, late model, hobby stock E-mod handle, look at the index below. Read all posts on Roll Center, Jacking effect and Why does car drive straight to the wall when I gas it? Also read You really have two race cars. This will cover 90% of problems you have. Simply put, the car pushes going in and is loose coming out. You do not have enuff downforce on the right...
I'm trying to decide what size and type of galvanized steel I need for 2 cantilever extensions. The cantilever is 5 ft. The space between the two cantilever arms is a 17 ft Gap the center 7 ft of the 17 ft Gap we'll need to Bear approximately 17,000 lb spread evenly from the front of the cantilever to the back of the cantilever over 5 ft. I will put support beams across these cantilever arms to support the load evenly
Thread 'What's the most likely cause for this carbon seal crack?'
We have a molded carbon graphite seal that is used in an inline axial piston, variable displacement hydraulic pump. One of our customers reported that, when using the “A” parts in the past, they only needed to replace them due to normal wear. However, after switching to our parts, the replacement cycle seems to be much shorter due to “broken” or “cracked” failures. This issue was identified after hydraulic fluid leakage was observed. According to their records, the same problem has occurred...
Back
Top