Torque to drive an eccentric mass 4

[jrstripe]

This seems a little like elementary dynamics, but I can’t seem to find an answer anywhere in my recourses or on the web.

I work for a company that, among other things, makes vibratory plate compactors that mount on the end of backhoe booms. The compactors have a single eccentric mass that is driven by a hydraulic motor. I am designing another addition to the family and am having difficulty determining what torque is required to maintain a constant mass rpm. When I compare the existing models’ test data with the calculations in the design file the numbers are WAY off. The guy that designed them has long left the company and no one else has provided any insight. I can estimate what it will take from the other models, but I want to do better than that.

Here are the details for this particular compactor…
Mass – 8kg
RPM – 2200
I (thru CG) - .018 kg-m^2
I (thru axis of rotation) - .030 kg-m^2
Eccentricity between cg and axis of rotation – 39mm

In reviewing past test data, I know that the worst cast scenario that causes maximum inlet pressure to the motor, (and therefore maximum torque) is when the compactor is on the ground and the backhoe is pressing on it with as much force as possible. This, in effect, causes the axis of rotation to be stationary and the mass’s cg to orbit around it. Torque equals the product of the mass moment of inertia and the angular acceleration (t=I*alpha), but for the life of me I can’t find the angular acceleration at a constant RPM. Am I taking the wrong approach, or am I missing something completely? Any advice would be greatly appreciated.

JRS

 

[drawoh]

jrstripe,

If your axis of rotation is horizontal, you absolutely need enough torque to lift the mass times radius to top dead centre. In theory, you should be able to recover this energy on the way back down. If your axis is vertical, there is no power consumption here.

Your real problem is that the eccentric side load is going to load your bearings, affecting the friction torque. It sounds like you have a dirty working environment, requiring sealed bearings, again affecting the friction torque.

If your numbers are way out, they either ignored something, or they made a bad assumption somewhere. Can you examine your existing equipment and work out what is consuming power? This could be weird. A worm drive can be way below 50% efficiency for example.

JHG

 

[jrstripe]

JHG,

I know what you thinking, because I struggled with this too. At first glance, I thought constant speed meant no acceleration which meant no torque. And that is true, but a mass on frictionless bearings will consume no power once it is in steady state only if it is perfectly balanced. Any time it is out of balance it requires power to maintain speed. This is also why the calculations and test results for the other models don’t agree (.25hp calculated vs. 7.5hp tested).

Consider a point mass orbiting around a fixed axis. If you were to plot it’s velocity with respect to the axis of rotation in the x direction (Vx) you end up with a sin wave. Every time it changes direction or speed it is accelerating and requires energy. The same is true in the y direction. Mark’s Handbook sort of has a description of this if you look up “Rotation of Solid Bodies”. In the 10th edition it is on page 3-16.

Back in the real world, the mass is supported inside a sealed housing by two unsealed spherical bearings and is directly driven by the motor. The bearings are greased, so drag from an oil splash lube system is not a factor. The mass spins in the vertical plane, butI think you are right about gravity not being a factor.

JRS

 

[drawoh]

jrstripe,

I am prepared to be corrected on the dynamics by one of the graduate engineers on this site.

Meanwhile, 8kg, 2200rpm, .039m radius, I get over 16kN of centrifugal force on your bearings. Even with good lubrication, this represents a lot of friction in your plain spherical (not ball?) bearings.

If I wildly guess a friction coeffient of .05 and a bearing radius of .03m, I get 5kW power. I will check this carefully when I get home tonight, but I think I am in the ballpark of reality.

JHG

 

[Compositepro]

An oscillating sping-mass system doesn't require energy to stay in motion. The energy that is dissappated to the ground or surrounding structure depends on those external factors - not the spring-mass oscillator.

The answer to your question lies in understanding what is damping the oscillation, and that is probably pretty complex and variable.

Your compactor must be applying mainly linear oscillation perpedicular to the ground, otherwise it would be very inefficient. Counter-rotating weights are usually used to do this but I suppose that an appropriate suspension system would allow a single excentric weight to work.

 

[GregLocock]

"but a mass on frictionless bearings will consume no power once it is in steady state only if it is perfectly balanced."

Wrong. If it were mounted in truly rigid bearings the eccentricity is irrlevat.

The main sink for the energy is probably in the soil being compacted/and or the suspension/tires of the vehicle.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[rmw]

My senior lab project was similar to this and it had to be solved on an analog computer back in the day (shortly after we got electricity).

What I remember was that energy was a function of the second derivitave; first derivitave was velocity I think.

You are disapating energy with the vibrating mass and you are going to have to determine that before you can then calculate the torque at a rotational speed to do it.

I do remember being amazed at the energy that it took to vibrate that mass.

rmw

 

[drawoh]

jrstripe,

I have just gone over this more carefully. I am assuming that your motor and shaft rotate at a constant speed. This makes the moments of inertia you listed above, irrelevant. When you rotate an off-centre mass, you get centrifugal force.

Mass m = 8kg.

Rotational speed [ω] = 2200rev/min [×] 2[π]rad/rev [×] 1/60 min/s = 230rad/s.

Radius r = 39mm [×] 1/1000m/mm = .039m.

Centrifugal Force F_C = m[ω]²r = 8 kg [×] (230 rad/s)² [×] .039m = 16500N.

The unit balance works because radians are dimensionless.

16.5kN is the weight (not mass!) of a substantial car. You are applying this to your bearings. You cannot ignore friction!

I am assuming you are using spherical plain bearings. I have limited experience with these, so the best I can do is assume Newtonian friction. Above, I wildly guessed that the friction coefficient [μ]=.05. Afterwards, I checked my Machinery's Handbook, and it looks like my value if anything is optimistic. Let's stick with it.

Friction Force F_F = F_C[μ] = 16500N [×] 0.05 = 825N

This force runs tangentially at the surface of your bearing. Coupled with the radius of your bearing, you have your driving torque. I wildly guessed above that your bearing's radius was 30mm.

Bearing radius R_B = 30mm [×] 1/1000m/mm = .03m.

Torque T = F_F R_B = 825N [×] .03m = 24.7N.m.

Power P = T[ω] = 24.7N.m * 230rad/s = 5680W.

This is your 7.5HP.

If you analyze your system WRT an arbitrary XY plane, you have a sinusoidal force going back and forth. If you analyze your system WRT the motor shaft, there is a constant side force, and constant bearing torque.

You should talk to a bearing engineer about this. My gut feeling is that a set of heavy duty rolling bearings should cut your power consumption.

JHG

 

[desertfox]

Hi jrstripe

At constant RPM there is no angular acceleration so your torque requirement is how fast you want to get the eccentric mass upto running speed. The torque requirement after reaching final speed will come from the power you want to transfer but this will constant.
The consequence of the mass being eccentric will generate
a centrifugal force which will impinge on your bearings as
several postrs have already indicated and drawoh as kindly
done the number crunching to give you some idea of the forces involved.

regards
desertfox

 

[jrstripe]

We actually do have decent 65mm (mean dia) spherical roller bearings. A calculator on SKF's website estimated a frictional loss of ~171W (.23hp) for both bearings. In testing of the compactor's bigger brother, which happens to use the same bearing, they never got hot. That makes me think it is a fairly good estimate.

After reading and re-reading the previous posts, I started to think about the conservation of energy. 7.5hp is going into the system and 0.23 is being dissipated by the bearings, so the rest has to be going somewhere. If it were a truly ridged system, for one complete revolution the change in kinetic and potential energy is zero. I assumed in testing the compactor's shoe was approaching a fixed condition. I'm beginning to think that this might be the cause of my problem. A small movement at 2200 rpm has the potential to dissipate a lot of energy.

Thank you all for getting me back on track.

 

[drawoh]

jrstripe,

Can you lift your compactor up out of contact with stuff and then test for power? Your disengaged system would be a combination of a mass/spring system and a pendulum, neither of which should consume significant power. If your power consumption goes way down to around the bearing friction level, we have eliminated your mechanism as the power consumer.

JHG

 

[zekeman]

This is not a torque problem per se.To finfd the horsepower you must find the power absorbed.
What I think is happening is that this eccentric mass is rocking the arm of the backhoe according to its centripetal forces. Now as the compactor approaches the ground, and simplistically assume the rocking arm is not a factor and the center of rotation doesn't move during the impact thst follows; it compresses the earth by some force deflection law which you can determine experimentally and further let us say it acts like a spring law (although I doubt it) and you know the "spring constant", K.
As an example suppose the K=100 lb/inch, then the eccentricity being .039 meters or ( I like english units)=1.53". The energy absorbed would be 1/2*100*1.53^2= 117in lb=98 ft lb. Since the happens every revolution, and the cycle time is 60/2200=.027 sec. Then in this case the average power absorbed y the ground due to the eccentricity is
98/.027=363 ft lb/sec or about 3/4 HP.
Since, as the system approaches the ground it has stored energy of
1/2Iw^2= 1/2*.03kgm^2*(2200*2pi/60)^2=1721 ft lb ;so in my case the energy dissipated per stroke, 98 ft lb should hardly affect the velocity although it will slow down a bit just after impact.

 

[zekeman]

Correction,
Make the "spring constant" in the example, 1000lb/in and then....
energy absorbed =1170 in lb= 98 ft lbs .....

 

[jrstripe]

We did run a compactor in the air suspended by just the backhoe. I recorded only a few data points, but it seems that the operating pressure is approximately 50% of what it was on firm ground. The compactor is suspended on four rubber mounts acting in sheer which do get hot. There is another source of energy loss (source of loss – does that make sense). The problem then is estimating how much energy is absorbed by the ground and backhoe which appears to be highly variable. Luckily, the rest of our line performs well so I can oversize by the same percentage.