Moment of interia of an armature 2

[bernardg]

Hello All,

We make 3.0" DC motors... and for some blower applications, we mount the wheels to the shaft and assemble the motor-wheel into blower housings.
Because of the tight clearance that we maintain between the wheel and the housing... even when the motor shaft flexes a little due to the armature imbalance, the wheels start to hit the housing and damages it.

Our customer wanted us to find the moment of interia of the armature. (Not the moment of interia of the armature while it is running in the motor... but just the armature as is). How would I do that?

Your input would very much be appreciated.

Thanks,
Bernie

 

[Skogsgurra]

I see no clear relation between the flexing problem and the inertia issue.

However, the moment of inertia is based on the fact that a piece of steel (or any other material) needs some force to be accelerated from one speed to another. The force needed is proportional to mass (kg) multiplied by acceleration (radians/seconds squared) and distance from center (torque arm). Integrating the basic relationships results in the equation:

I = 0.5*m*r^2

Where m is mass in kg and r is radius in metres. I is given in kgm^4.

You can calculate I for different parts like shaft and rotor body and then add them to get a total I.

Search

http://www.martindalecenter.com/Calculators4.html

for calculators to help evaluating your specific configuration.

 

[electricpete]

When we talk about acceleration we use the moment of inertia described by skogsgurra.

When we talk about bending, there is a different quantity called moment of inertia which describes resistance to bending. That is probably the one of interest.

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[electricpete]

Mark's Standard Handbook for Mechanical Engineer's 8th edition Chapter 5, Table 6 gives

I = PI * D^4 / 64 for solid cylinder of radius D

I = PI * (D^4 - d^4) / 64 for hollow cylinder with i.d.=d and o.d. = D


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[itsmoked]

bernardg:
The customer probably wants to know the inertia because this blower may be moved while operating!! This will cause large gyroscopic effects on the blower fan and spinning armature. These forces will follow the Right Hand Rule causing large deflections of the motor shaft that will probably allow the blower to contact the housing in *some* normally expected situations. With the inertial info it would be possible to calculate those deflections and make some recommendations towards a better minumum clearance.

Follow electricpetes equations. Take the wired armature and weigh it, measure it, use the formula. You can ignore the fact that the shaft is different material from the iron/copper windings because the outer part is the big contributor to inertia anyway.

 

[bernardg]

Thanks for all your prompt replies. I very much appreciate it.

As I've mentioned before, all these discussions are on the armature as is... and not inside a motor.

The mtr-wheel config would look kinda similar to this...

___________________________________________blower housing
___ |-----------| ___
|___|----------------Motor----------------|___|
blower wheel-1 |-----------| shaft blower wheel-2
___________________________________________blower housing


My customer also shared an important piece of information this morning...
When the wheels are mounted on to the motor and ran for 200 hours just as mtr-wheel (and not inside the blower housing).... the blades broke only on the wheel that is mounted on left hand side of the motor (opposite-commutator end). This is consistant with what we observed in the failures we had inside the blower housing. This proves that there is no rubbing of the wheels against the housing.

What I suspect is that there could be some moulding(fill) issues. But, my customer still wants to explore all the potential causes... and the loading on the wheel because of the inertia of the armature being one of them.

I know the mass of the armature (with coil and everything).
(i) How can I calculate the acceleration?
My customer suggested that we can roll down the armature in a small ramp and calculate the acceleration.... and in turn calculate the moment of inertia of the armature... and use that to see how much of the load that the armature would impart on the wheels when it is spinning at 4000 rpm.

The things that I don't understand are...
say, if I build a small ramp with rails... on which I can roll the armature with ofcourse the shafts sliding on the rails... I can calculate the linear acceleration.
(ii) But how can I calculate the angular acceleration?
(iii) How is acceleration related to the moment of inertia? What equation do I have to use to calculate it.
(iv) Even if I calculate the moment of inertia, how am I going to tie it up as a load @ x-rpm?

Kindly help me with your expertise.

Thanks in advance for all your help.

Bernard G.

 

[bernardg]

Thanks for all your prompt replies. I very much appreciate it.

As I've mentioned before, all these discussions are on the armature as is... and not inside a motor.

The mtr-wheel config would look kinda similar to this...

___________________________________________blower housing
___ |-----------| ___
|___|----------------Motor----------------|___|
blower wheel-1 |-----------| shaft blower wheel-2
___________________________________________blower housing


My customer also shared an important piece of information this morning...
When the wheels are mounted on to the motor and ran for 200 hours just as mtr-wheel (and not inside the blower housing).... the blades broke only on the wheel that is mounted on left hand side of the motor (opposite-commutator end). This is consistant with what we observed in the failures we had inside the blower housing. This proves that there is no rubbing of the wheels against the housing.

What I suspect is that there could be some moulding(fill) issues. But, my customer still wants to explore all the potential causes... and the loading on the wheel because of the inertia of the armature being one of them.

I know the mass of the armature (with coil and everything).
(i) How can I calculate the acceleration?
My customer suggested that we can roll down the armature in a small ramp and calculate the acceleration.... and in turn calculate the moment of inertia of the armature... and use that to see how much of the load that the armature would impart on the wheels when it is spinning at 4000 rpm.

The things that I don't understand are...
say, if I build a small ramp with rails... on which I can roll the armature with ofcourse the shafts sliding on the rails... I can calculate the linear acceleration.
(ii) But how can I calculate the angular acceleration?
(iii) How is acceleration related to the moment of inertia? What equation do I have to use to calculate it.
(iv) Even if I calculate the moment of inertia, how am I going to tie it up as a load @ x-rpm?

Kindly help me with your expertise.

Thanks in advance for all your help.

Bernard.

 

[electricpete]

It looks like the moment of inertia you are after is the one described by skogsgurra after all.

One quick comment - load will inertia will only affect load during transients.

I'm pretty sure you can calculate your inertia using physics.

F = m * g.

T = F * d *cos(theta) where d*cos(theta) will describe a lever arm formed as if the force is acting directly down at the center of mass of the rotor and pivoting about the contact point which is not directly below the center. To determine this part is probably the toughest but a diagram and some trigonometry should do it.

J = T / alpha where J is moment of inerta, T is torque and alpha is angular acceleration.

alpha = 2*theta / t^2 where theta is the total radian rotation in time t. Can calculate as theta = d / (2*Pi) where d is linear distance traveled.

Hope that'll get you started. I may have made an error in there somewhere.


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[bernardg]

Electricpete,

I am going to take this one step at a time. First i am gonna start to calculate the angular acceleration of the armature:

Looking at the equation that you wrote for finding out the angular acceleration of the armature when slided down a small ramp...

We can control the linear distance that the armature will travel when slided down the ramp and calculate the time 't' for the travel.

one thing that i do not see in your equation for calculation of angular acceleration, is the radius 'r' of the armature.

did you miss that? or am I interpretting it wrong?

kindly clarify.

bernard

 

[itsmoked]

A couple of points.
From what you've said, the blower is shedding blades.
This is THE problem.
Also the blower is shedding blades at speed, after a while.

This begs the question: Why do you care about the inertia of ANY PART of the motor? Further more, why do care about start up acceleration?

The forces on the blower cage and only the forces on the blower cage are causing the problem. You need to find the forces on the blades that are present at full speed.

1) Find, accurately, the full speed RPM.
2) Saw a blade off at the root, weigh it, and do the calculus that distributes the Force on the blade over the blade's length.

This will tell you what the forces are where the blade is failing. Then you go back to the design of the blade or the exact plastic the blade is made from and see if you have a design failure or a materials failure.

One note. If the blower has an outer ring around the blades, (a common item), you need to saw the ring off, weigh it also. Divide its weight by the number of blades.
Mathematically break the ring into little chunks, one for each blade.
Do the calculus one one chunk. Add it to the blade. THEN you will have all the forces in the blade.

 

[bernardg]

itssmoked,

As I mentioned before,
the last testing was done just with the blower-wheel assembly, without putting them into the double scroll blower.
The wheels have both the inner ring and the outer ring (inner ring has the shaft hole into which the armature shaft goes into)... and all the 20 blades are attached to both the rings.

At any given speed, the wheel is going to push air.. thus, would see the forces on the blades. This might be, as you mentioned, the big contributor for failure if either the design is faulty or if there are any material fill issues.

The thing to note is that,
All failure happens at the blade to outer ring interface starting at the same blade breaking first every time... and moving on to the next blade and so on... (we see a consistant pattern here).

with that said...
Though insignificant it might seem, won't the inertia of the armature put some loading to the blower wheel?

bernard.

 

[IRstuff]

nope. the inertia of the armature is a load to the motor. The blower wheel sees nothing but the acceleration and drive from the motor.

TTFN

 

[itsmoked]

Hi bernardg. Irstuff is correct.

The largest force in this picture is the radial forces trying to throw the blades from the axle. Or in your case throw the outer ring from the blade tips. NOT the forces of the air the blade is accelerating. This sounds like your material or the blade/ring cross section, (too small), is the main problem. What I said above will come down to X number of pounds *force* trying to separate the outer ring from the blade tips. The question becomes, "is there enough cross section at the joint, or is the material strong enough?"

Other points. A blade end ring joint is a major stress area. The material is changing from a down the blade to "around" the ring.

The same blade failing each time *first* is telling. If this is a plastic injected part, does maybe the injection port location cause the lowest plastic density near that blade area? Is the mold slightly wrong there? Is there a sharper transition line in the mold?

Having a second blade fail next is not very telling as once one blade is sheared the system will grossly overload other areas and generate complex vibrational failure modes.

 

[electricpete]

I said before- inertia only important during transients. I should add it would also be important for calculating torsional resonance.

Where did r go? r is used to calculate d. d is used to calculate theta. Theta is used to calculate alpha.

My equation alpha = 2*theta / t^2 is the rotating analogy of the constant linear acceleration equation a = 2*d/t^2 or rearranged as d = a * t^2 / 2. This is derived by integrating d twice, starting with 0 initial position and zero initial velocity.



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[electricpete]

last sentence should be integrating a twice (not integrating d twice)

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[electricpete]

a minor correction: theta = d / (1*Pi) (not (d/2*pi))

a major correction: your equation needs to consider both rotational acceleration and linear acceleration.

There is one accelerating force. It is the component of the weight force in the direction parallel to the inclined plane. If the angle from horizontal is theta, then that component of force is m*g*sin(theta).

Part of that force goes into rotational acceleration. Part goes into linear acceleraiton.

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[electricpete]

I think the equation to start with would be

m*g*sin(theta) = J* alpha/r + m*a

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[electricpete]

OK, I used theta for two different things... one for angle from horizontal and one for rotation angle. Let's leave rotation angle called theta but rename the angle from horizointal as delta.

Also,d = r*theta. d referring here to arc length not diameter.

F = m*g*sin(delta) = J* alpha/r + m*a

Convert a to alpha using a = r*alpha

m*g*sin(delta) = J* alpha/r + r*m*alpha

Solve for J
J = [m*g*sin(delta) - r*m*alpha]*r/alpha

alpha computed as above
alpha = 2*theta / t^2 = 2*(d/r) / t^2

You would have to start with the armature a small way down the incline where you have to hold it to prevent it from rolling, not at the top where it wants to sit still.

You really better check my math.

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[sreid]

Two items may be significant here.

1) The time to failure is 200 hours

2) The failure is always at one end of the shaft.

This suggets to me that a fatigue failure may be involved perhaps due to torsional resonance that is different at the shaft ends. DC motor shafts are not as stiff as one might expect from inspection in that the rotor laminations are not rigidly attached to the shaft.

 

[itsmoked]

Hi sreid,
I'd say the only thing significant is that your blower design has some short comings. Why? Because unless you are operating it in a Class 100 clean room, you are eventually going to get the classic dust build up on the blades/blower/shaft/bearings and that is going to cause lots of dang near immpossible to predict forces hither and yon. If your blower cage can't take the minor resonances caused by the motor running *normally* you will have continuing failures that will consume a great deal of time and energy and your problem will not be solved in a timely manner.

Further, the torsional resonances that may be present change greatly with small changes in speed. Your motors will never all run the exact same speed so a tiny fix that works uncaged in your lab will have no bearing on the next unit coming off the production line.

Further suggestions. If you have consistant failures with your test setup reduce the motor speed by 5%, and run the test again. Increase the speed by 5% over normal. If your failures suddenly take farrrrr longer or far shorter you have your proof the blower can't cut its normal operating speed and needs modification.