g level of unbalanced rotating member 2

[fetterlabs]

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

 

[electricpete]

me too.

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[GregLocock]

Well now I'm puzzled. A strain gauge accelerometer will produce an output of 1g when it is stationary and vertical, and will produce -1g when I turn it over to calibrate it.

OK, it is obvious why, but nonetheless it does behave in the way that fetterlabs is describing.

Cheers

Greg Locock

 

[electricpete]

so what's the puzzle?

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[EnglishMuffin]

Well, when there is no acceleration present I don't see how an accelerometer can be said to be measuring it. To interpret an artifact of the instrumentation as an acceleration seems a little perverse to me, especially so since no particular direction was specified by fetterlabs.

 

[IRstuff]

It's an interesting argument, but an extension of your argument is to say that a bathroom scale does not measure acceleration. But, it does so, but countering it with strain or spring compression to balance the force applied. And the force applied is from gravitational acceleration.

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

 

[GregLocock]

Well, how about this as an explanation?

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

 

[electricpete]

I didn't agree with IRStuff "If you've calibrated the spring constant, then you've measured the acceleration of gravity acting on the proof mass." You have measured gravitational force, not gravitational acceleration.

I agree 100% with everything Greg said.

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[electricpete]

I apologize if there has been any negative tone in any of my comments. Something somewhere along the line something got my goat but that's just me.

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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[electricpete]

I forgot, at resonance, |Funbalance| ~ M*e*W^2 = c*v
v = M*e*W^2/c
a = W*v =M*e*W^3/c

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[IRstuff]

Since force=mass*acceleration and the proof mass is always known, you've &quot;measured&quot; acceleration. This is no different that &quot;measuring&quot; resistance, since you apply a voltage and get a current. Through the math, you get the resistance.

TTFN

 

[electricpete]

Let's say you have an accelerometer of mass m sitting motionless on a table:

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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[desertfox]

Hi fetterlabs

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.

 

[electricpete]

Compare to what I have posted:
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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[JONTMED]

Just out of curiosity, I looked at my Machinery's Handbook, and there is a great explanation of &quot;static&quot; balance vs &quot;dynamic&quot; balance of a rotating member, and if the body is a disc or a cylinder that is being rotated. I think what you are trying to calculate is the effect of any imbalance based on speed of the rotating member, and the radius of the imbalance. Reminds me of the automotive &quot;crankshaft balance&quot; machine featured on a popular TV auto show. The machine actually generates a read-out of imbalance, and defines where to remove material. The host explained that a (.0X) pound imbalance at 3 inch radius, rotating at 6000 rpm would produce an eccentric load of (XXX) &quot;many&quot; pounds. Is that what you're looking for?? If so, look at the Machinery's Handbook explanation.