g level of unbalanced rotating member
g level of unbalanced rotating member
(OP)
I have what I thought was a simple question, but am having difficulty obtaining an answer. Hopefully you all can help.
If an accelerometer were mounted so it could measure the normal acceleration of a perfectly balanced rotating shaft it would read zero. Now take that same shaft and make it unbalanced by adding mass at a point. The frequency of the vibration will be 1 X RPM but how does the mass affect the amplitude? Is there an equation that relates the amount of imbalance to the amplitude of the acceleration?
JLF
If an accelerometer were mounted so it could measure the normal acceleration of a perfectly balanced rotating shaft it would read zero. Now take that same shaft and make it unbalanced by adding mass at a point. The frequency of the vibration will be 1 X RPM but how does the mass affect the amplitude? Is there an equation that relates the amount of imbalance to the amplitude of the acceleration?
JLF





RE: g level of unbalanced rotating member
2. How does the mass affect the amplitude ? It depends on the shaft stiffness as well as the mass. What you are probably after is the basic "whirling" equation, which is:
r = m*omega^2*e/(k*gc-m*omega^2)
where r = lateral deflection
e = mass eccentricity
m = imbalance mass (which will include some portion of the shaft mass as well as the added mass)
omega = angular velocity (rad/sec)
k = lateral stiffness of shaft at the mass location
gc = unit multiplier (for unit consistency)
In reality, things are usually much more complicated, and you need to study texts on rotordynamics.
RE: g level of unbalanced rotating member
RE: g level of unbalanced rotating member
1. Some call it radial acceleration, I prefer normal. I mean perpendicular to the surface. If you ignore the steady-state acceleration due to gravity and only measure the acceleration due to rotation, in a perfectly balanced shaft it should be zero.
In reality, things are usually much more complicated, and you need to study texts on rotordynamics.
I realize this, but I trying to develop a simple model. Thanks for the assistance.
RE: g level of unbalanced rotating member
Take the distance from CG of the total (including the eccentricity), relative to the axis of rotation. This is distance. Take revolutions per second and determine centrifugal force from the relative eccentric mass at this distance.
F=ma gives you the MAX acceleration (total mass here not relative eccentric mass) this eccentric mass would output to the accelerometer (not counting damping, assuming rigid shaft.)
Alex
RE: g level of unbalanced rotating member
RE: g level of unbalanced rotating member
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RE: g level of unbalanced rotating member
RE: g level of unbalanced rotating member
In the case of a prox probe mounted near the imbalance mass, then taking k to be the combined effective shaft and bearing stiffness, and ignoring shaft and bearing damping, then, to a first order of approximation, the maximum displacement amplitude output from the prox probe would be given by r from the equation in my first post. You could convert this to acceleration in g's by multiplying it by omega^2/gc, where gc depends on your unit system. In the case of an accelerometer mounted on the outer race of one of the bearings, you would have to know the bearing-to-housing stiffness, and it would also be highly affected by damping. In other words, it would be almost impossible to predict theoretically in advance.
RE: g level of unbalanced rotating member
As for electricpete and EnglishMuffin:
There are two basically two types of accelerometers. Steady-state accelerometers will measure from DC to several hundred hertz. Dynamic accelerometers measure from a few hertz to several thousand hertz.
Now let's say I have a magic rotating shaft that is perfectly balanced. It's connected to the rest of the world with massless, frictionless bearings. On the top of the massless, frictionless bearings I mount a steady-state accelerometer and a dynamic accelerometer. Through more magic I make the shaft spin. The steady-state accelerometer will read the acceleration due to gravity, or 1g. The dynamic accelerometer has no DC response so it will read 0g. Now I imbalance my shaft and spin it again. The steady-state accelerometer will read gravity + the acceleration due to the imbalance. The dynamic accelerometer will read just + the acceleration due to the imbalance. Since I trust that gravity is going to be around for awhile, I posed the question using a dynamic accelerometer since I'm only interested in the acceleration due to the imbalance.
RE: g level of unbalanced rotating member
The unbalance mass affects the vibration sensed by your accelerometer directly based on an influence coefficient related to the stiffness of the structure supporting the rotor. The amount of vibration produced by any unbalance depends of how stiff the supporting structure is.
If you change the speed, vibration will increase by the change in the speed squared provided that the supporting structure stiffness remains constant and your not near or at any point of resonance. Keep in mind that you must consider the resonant frequency of the supporting structure, the rotor and/or the entire assembly depending on how its mounted. Fc=U*RPM2.
You can coorelate unbalance to vibration. You'll need to do some tests with known test masses and come up with influenc e coefficients. If you need to know how to do this with real world parts that don't have any magic..contact me and I'll show you how to get through it.
RE: g level of unbalanced rotating member
Zhivotov: my equation is not "erroneous" and can be found in "Vibration Theory and Applications" by Thomson, among other places.
I think I will have to pass on any future responses on this thread.
RE: g level of unbalanced rotating member
EnglishMuffin:sorry, but I have never heard of such a thing as a "steady state accelerometer" which "reads the acceleration due to gravity"
I think you need to get out more.
RE: g level of unbalanced rotating member
RE: g level of unbalanced rotating member
This was the second hit on Google for the search term - accelerometer DC.
http://www.engineeringtalk.com/news/tec/tec102.html
I have seen one but never used one. I understand you can do a quick field calibration simply by turning the thing over.
M
--
Dr Michael F Platten
RE: g level of unbalanced rotating member
http://www.msiusa.com/download/pdf/english/icsensors/ac...
http://www.pcb.com/product.asp?measureType=852&productT...
RE: g level of unbalanced rotating member
Yuriy (along with others) has given you a good explanation and equations. Yuriy's equation is in agreement with Den Hartog. Pick a region above or below resonance. Solve for his "a"=displacement. Multiply by W^2 to get acceleration. By the way his MW** means M*W^2.
Thanks for the unsolicited tutorial on gravity and accelerometers. I for one just asked for clarification on a post that remains vaguely posed. But I see you have discovered what you apparently were missing (F=ma)
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RE: g level of unbalanced rotating member
RE: g level of unbalanced rotating member
Safe to say the original comment
("If you ignore the steady-state acceleration due to gravity and only measure the acceleration due to rotation, in a perfectly balanced shaft it should be zero.") was flat wrong. There was no discussion of accelerometers there, only discussion of acceleration which was plainly incorrect (unless this rotor happens to be in a freefall). EM was right to question this ridiculousness.
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RE: g level of unbalanced rotating member
RE: g level of unbalanced rotating member
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RE: g level of unbalanced rotating member
OK, it is obvious why, but nonetheless it does behave in the way that fetterlabs is describing.
Cheers
Greg Locock
RE: g level of unbalanced rotating member
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RE: g level of unbalanced rotating member
RE: g level of unbalanced rotating member
I see your point and would agree with it semantically, e.g., since there is no net motion, there is no net acceleration, but that, in itself, does not mean that you cannot measure the counterbalanced forces individually.
As described above, a proof mass that is undeflected when turned 90° from the horizontal, will deflect when rotated back to horizontal. If you've calibrated the spring constant, then you've measured the acceleration of gravity acting on the proof mass.
TTFN
RE: g level of unbalanced rotating member
Gravity does not generate an acceleration, it generates a force (=G*m1*m2/r^2)
So we are being lazy when we talk about the acceleration due to gravity, what we should say is ' the acceleration due to gravity in the absence of other forces'.
A strain gauge accelerometer does not measure acceleration, it measures the force on a small mass, equal to m*g+m*a.
I guess my real issue here is that if the accelerometer is stationary it indicates 1g, yet when it is accelerating at 1g it is indicating 0g, and if I turn it over and it is accelerating at -1g in its own reference frame it will also indicate 0g.
I agree, this /is/ semantic silliness, but I don't think it is fair to claim that this is not a possible point of confusion (if entirely irrelevant tot he original post).
Cheers
Greg Locock
RE: g level of unbalanced rotating member
I agree 100% with everything Greg said.
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RE: g level of unbalanced rotating member
going back one more time to Yuriy's equation:
For speeds of a rotor before critical speed
(1) MW^2 (d1+e) - kd1 =0
For speeds of a rotor after critical speed
(2) MW^2 (d2-e) - kd2 =0
I have changed symbol for displacement to d (d1 below, d2 above resonace).
Solve:
d1 := M*W^2*e/(k-M*W^2)
d2 := -M*W^2*e/(k-M*W^2)
Looks like same magnitude either way.
Below resonance, assume k>>M*W^2
d1 ~ M*W^2*e/k
a1 = d1*w^2 ~ M*W^4/k
i.e. Below resonance/critical in spring controlled region Funbalance = M*e*W^2 = k*d. Solve d = M*e*W^2/k.
Above resonance, assume k<<M*W^2
d2 ~ M*W^2*e/M*W^2
d2 ~ e
a2 = d2*W^2 = W^2 * e
ie. above resonance/critical in mass controlled region it is a simple case Funbalance = M*e*W^2= M*a. Solve a = e*W^2. displacement is e.
Near resonance you need to know damping.
Funbalance =
Just a different discussion of the same thing. I think maybe my Funbalance is not 100% correct since it relies on e and not e +/- d. But results still comes out same as Yuriy's with the approximations far above and far below resonance. I guess in those cases it can be assumed d<<e (?) and that is why it works. I would welcome any comment if I am off base. Otherwise I'll shut up (again).
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RE: g level of unbalanced rotating member
v = M*e*W^2/c
a = W*v =M*e*W^3/c
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RE: g level of unbalanced rotating member
TTFN
RE: g level of unbalanced rotating member
Fgravity = - m* g
The table exerts an equal and opposite force
Ftable = m*g
Ftotal = Fgravity + Ftable = 0
a = Ftotal/m =0.
Your steady state accelermoter (hey, I learned a new word!) senses the force and converts it to an indicated acceleration g using proof mass for conversion constant. Did it measure or sense the accleration of gravity? No, there is no acceleration in this problem. (Acceleration =0.) It measured the force of gravity.
Sorry to beat a dead horse.
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RE: g level of unbalanced rotating member
I found a formula which may help regarding amplitude of shaft deflection related to angular velocity of the shaft.
y= (w^2 * e)/(wc^2-w^2)
where y = deflection of the shaft from the static
deflection position at any speed w in terms of
the critical speed.
e = eccentricity of the centre of gravity of a
of a mass attached to the shaft measured
from the centre of the shaft axis
wc = critical speed of shaft = (g/x)^0.5
x = static deflection of shaft where the
mass is attached to the shaft.
g= gravitional constant.
w= angular velocity of shaft
I have a derivation for the above but its quite long winded
so I haven't posted it here.
regards desertfox.
RE: g level of unbalanced rotating member
d2 := M*W^2*e/[k-M*W^2]
Multiply top and bottom of rhs by 1/M
d2 := W^2*e/[(k/M) - W^2]
using k/M = Wc^2
d2 := W^2*e/[Wc^2 - W^2]
The derivation I believe is given above, although perhaps a little fragmented. That never stopped me from adding more fragments:
It should be easy to recognize that the numerator is proportional to the force. (F=M*W^2*e). The denominator is the transfer function which gives system response of equivalent SDOF system as a function of frequency.
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RE: g level of unbalanced rotating member