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How to calculate natural freq. with a spring/mass w/constant preload?
- Thread starter Scheeml
- Start date
GregLocock
Automotive
I can't find it either. The problem in the original thread is that the OP never explained what he was really up to, that I could see.
If you can sketch the problem you are trying to solve then I will have a go.
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376
If you can sketch the problem you are trying to solve then I will have a go.
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376
It moved:
I would not expect a constant force to do anything in a linear system to the natural frequency. It might change a phase angle, presuming one has an origin to measure that angle from. It moves an equation from something like F = kx to F+c = kx +c.
I would not expect a constant force to do anything in a linear system to the natural frequency. It might change a phase angle, presuming one has an origin to measure that angle from. It moves an equation from something like F = kx to F+c = kx +c.
- Thread starter
- #4
Well, really what I am trying to determine is the natural frequency of a hydraulic cylinder by treating it as a simple spring mass system (the cylinder is the spring, the mass is the load the cylinder is holding in place). The calculation we typically use treats the fluid in the cylinder as two springs (one on either side of the piston) and combines their spring constants, and then it's just ω = sqrt(k/m). However, when the cylinder is all the way extended, there is only hydraulic pressure force on one side of the piston, and so it seems like it would be better represented as a spring with preload - which is why I was looking for that link.
electricpete
Electrical
I've never done hydraulic cylinder resonance calculations. I may be off base but I'd like to offer my two cents fwiw
If you have a linear model then preload shouldn't matter for natural frequency.
Is there a non-linear model that describes how the pressure changes in response to piston movement? I guess it could involve pipe expansion. But even in that case (non-linear) what do you mean by natural frequency? Typically in a non-linear situation your approach would be to linearize about the operating point and use that in a linear natural frequency calculation.
> the mass is the load the cylinder is holding in place
Is it a mass or not? If whatever load you're talking about doesn't generate a force proportional to acceleration then I don't think you can model it as a mass.
Also is the fluid mass negligible compared to the other masses?
=====================================
(2B)+(2B)' ?
If you have a linear model then preload shouldn't matter for natural frequency.
Is there a non-linear model that describes how the pressure changes in response to piston movement? I guess it could involve pipe expansion. But even in that case (non-linear) what do you mean by natural frequency? Typically in a non-linear situation your approach would be to linearize about the operating point and use that in a linear natural frequency calculation.
> the mass is the load the cylinder is holding in place
Is it a mass or not? If whatever load you're talking about doesn't generate a force proportional to acceleration then I don't think you can model it as a mass.
Also is the fluid mass negligible compared to the other masses?
=====================================
(2B)+(2B)' ?
Hi Scheeml
The link in the old post worked for me, so I have copied the link from the site and pasted it here
, hopefully it will work.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
The link in the old post worked for me, so I have copied the link from the site and pasted it here
, hopefully it will work.“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
GregLocock
Automotive
Isn't there a bump stop inside the cylinder? or is it metal on metal?
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376
If the cylinder is against the endcap and there is enough pressure to keep it there then you would have 2 springs in parallel - the stiff metal and the soft oil spring. If there is not enough preload then you have a non linear function where the net stiffness changes suddenly.
- Thread starter
- #10
Ah - the second link not the second reply, way, way down the replies.
It looks like something from which is now at
Which has
Accession Number: ADA800851
Title: Experimental Investigation of a Preloaded Spring-Tab Flutter Model
It looks like something from which is now at
Which has
Accession Number: ADA800851
Title: Experimental Investigation of a Preloaded Spring-Tab Flutter Model
Hi Scheeml
Just done so searching and I think you only need to calculate the natural frequency of the cylinder at its mid position, this is due to the fact that at this position this will be it’s lowest natural frequency. When the cylinder is retracted or extended it’s natural frequency is higher, see link posted.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
Just done so searching and I think you only need to calculate the natural frequency of the cylinder at its mid position, this is due to the fact that at this position this will be it’s lowest natural frequency. When the cylinder is retracted or extended it’s natural frequency is higher, see link posted.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
- Thread starter
- #13
That is more what I was looking for, thanks!
About the cylinder mid position point - We use that mid value when the cylinder will be in that range during operation, this one will not. It will always be extended during operation. Doing that calc will give a natural frequency at end of stroke, but that is assuming there is always still a very small amount of fluid in the rod end.
I think that (like was said by others earlier), as long as you don't exceed the force of the cylinder, then it is essentially a rigid body, and it is best to keep the cylinder big enough to make sure it doesn't get overpowered. I think if you let the mass force exceed the cylinder force it would just result in undesirable banging. Also, I realize that allowing the force of the mass to overcome the cylinder force would be undesirable because the position of the mass would change - I should have thought of that in the first place. Hopefully I didn't waste too much of anyone's time.
Thanks all for your input and help!
About the cylinder mid position point - We use that mid value when the cylinder will be in that range during operation, this one will not. It will always be extended during operation. Doing that calc will give a natural frequency at end of stroke, but that is assuming there is always still a very small amount of fluid in the rod end.
I think that (like was said by others earlier), as long as you don't exceed the force of the cylinder, then it is essentially a rigid body, and it is best to keep the cylinder big enough to make sure it doesn't get overpowered. I think if you let the mass force exceed the cylinder force it would just result in undesirable banging. Also, I realize that allowing the force of the mass to overcome the cylinder force would be undesirable because the position of the mass would change - I should have thought of that in the first place. Hopefully I didn't waste too much of anyone's time.
Thanks all for your input and help!
Hi Scheeml
You’re very welcome.
I agree that if the piston cylinder is always extended then it natural frequency will be higher than it would be if it were centralised. I was thinking that if there are no influencing vibrations that were higher than the lowest natural frequency (ie piston in the central position) then your cylinder application would be fine.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
You’re very welcome.
I agree that if the piston cylinder is always extended then it natural frequency will be higher than it would be if it were centralised. I was thinking that if there are no influencing vibrations that were higher than the lowest natural frequency (ie piston in the central position) then your cylinder application would be fine.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
electricpete
Electrical
Interesting paper.
Sorry that in my earlier response, I got distracted/preoccupied with the word "preload", knowing that spring preload is typically irrelevant for natural frequency calcs. I'm not sure what preload even means in this context of the hydraulic cylinder.
But I see now it says constant preload so it really shouldn't have mattered either way.
tldr: regarding my earlier comments... never mind!
=====================================
(2B)+(2B)' ?
Sorry that in my earlier response, I got distracted/preoccupied with the word "preload", knowing that spring preload is typically irrelevant for natural frequency calcs. I'm not sure what preload even means in this context of the hydraulic cylinder.
But I see now it says constant preload so it really shouldn't have mattered either way.
tldr: regarding my earlier comments... never mind!
=====================================
(2B)+(2B)' ?
electricpete
Electrical
Here is desertfox' link: Interactive Analysis of Closed Loop Electro-hydraulic Control Systems
I'm going to make my own summary of that so it’s easier for me to remember later. I share it here in case it’s a useful summary / memory device to anyone else. (if it’s just a layer of useless chatter, then I apologize).
I’m going to make it simpler than the authors and NEGLECT the external volume. Assume cylinder has constant diameter D and therefore constant area A
Cylinder stroke H is dividided into H1 and H2 where H1+H2 = H.
Let’s increase H1 by a small increment dh
Assume fractional change in pressure proportional to fractional change in trapped volume
dP / P = B dv / V where B is proportionality constant
define differential forces dF in direction opposite of dh.
For each volume (V1 and V2) we have a Pressure (P1 and P2) which generates a force (F1 and F2) where the two forces F1 and F2 acting in parallel (total force = F1+F2)
We will only work with the differential forces dF1 and dF2 not total forces F1 and F2. That presumes the control system when at rest keeps the pressure on both sides of the piston the same (except for the pressure oscillations on each side associated with the effect of piston position oscillation upon individual trapped volumes on each side). So the sum of the forces from the same-static pressure acting on both sides of the piston is zero regardless of the static pressure.
dFi = A dPi = A B dv / Vi = A B dVi / Vi = A B (dh * A) / (Hi*A) = A B dh / Hi
where Hi = H1 or H2 (H2 = H-H1)
dF = dF1 + dF2 = A B dh / H1 + A B dh / (H – H1)
Stiffness = dF / dh = A B { 1/ H1 + 1/(H-H1)}
Shape of the curve now makes sense. As you get near to H1 = 0 or H1 = H (the ends of the cylinder) the stiffness blows up towards infinity (or lower than that if you added back the effects of the trapped volumes in the tubing outside of the cylinder). It is a result of fractional change in volume (for a given small movement dh) getting huge when trapped volume on either end gets small. The algebraic manipulations were probably more than was required to get there, but that bolded last part seems easy to remember / visualize to me.
=====================================
(2B)+(2B)' ?
I'm going to make my own summary of that so it’s easier for me to remember later. I share it here in case it’s a useful summary / memory device to anyone else. (if it’s just a layer of useless chatter, then I apologize).
I’m going to make it simpler than the authors and NEGLECT the external volume. Assume cylinder has constant diameter D and therefore constant area A
Cylinder stroke H is dividided into H1 and H2 where H1+H2 = H.
Let’s increase H1 by a small increment dh
Assume fractional change in pressure proportional to fractional change in trapped volume
dP / P = B dv / V where B is proportionality constant
define differential forces dF in direction opposite of dh.
For each volume (V1 and V2) we have a Pressure (P1 and P2) which generates a force (F1 and F2) where the two forces F1 and F2 acting in parallel (total force = F1+F2)
We will only work with the differential forces dF1 and dF2 not total forces F1 and F2. That presumes the control system when at rest keeps the pressure on both sides of the piston the same (except for the pressure oscillations on each side associated with the effect of piston position oscillation upon individual trapped volumes on each side). So the sum of the forces from the same-static pressure acting on both sides of the piston is zero regardless of the static pressure.
dFi = A dPi = A B dv / Vi = A B dVi / Vi = A B (dh * A) / (Hi*A) = A B dh / Hi
where Hi = H1 or H2 (H2 = H-H1)
dF = dF1 + dF2 = A B dh / H1 + A B dh / (H – H1)
Stiffness = dF / dh = A B { 1/ H1 + 1/(H-H1)}
Shape of the curve now makes sense. As you get near to H1 = 0 or H1 = H (the ends of the cylinder) the stiffness blows up towards infinity (or lower than that if you added back the effects of the trapped volumes in the tubing outside of the cylinder). It is a result of fractional change in volume (for a given small movement dh) getting huge when trapped volume on either end gets small. The algebraic manipulations were probably more than was required to get there, but that bolded last part seems easy to remember / visualize to me.
=====================================
(2B)+(2B)' ?
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