Experimentally calculating dissipated energy
2

Question

Experimentally calculating dissipated energy

RigTest (Automotive)

(OP)

29 Mar 12 16:04

Hello everyone,

I have been struggling to properly calculate energy ratios (energy dissipated/energy input) in servo-hydraulic test systems.  Specifically, the test specimens are quarter-car (2DoF) suspensions, which include tire compliances.  The rig is capable of many excitation waveforms, but I typically use a constant peak velocity swept-sine.  The rig has a load-cell under the tire pad and there are accelerometers at the pad (input), the hub, and the body, which are integrated to give relative displacements.  I am hoping to get some feedback on what I'm trying, as currently my system is breaking energy conservation(!).

Initially I performed the calculations in the time domain:

Input Energy(t) = int(Force(t)*dx(t),x)      where t is time

A swept-sine allows the time history to be broken up into cycles, and for each cycle I would determine the contained area within the Force vs. Displacement curve: one area for 1 cycle (frequency), so I manually constructing a spectral energy signal (I'm not sure this is valid).  The calculation is performed using Pad Force and 3 displacements:

-Actuator displacement (to get input energy)
-Pad to Hub displacement (to get energy dissipated by tire)
-Hub to Body displacement (to get energy dissipated by dampers)

At the body resonance, the energy dissipated by the Hub - Body is greater than the input energy (not good).  I repeated the calc using Pad - Body and it was also much greater than input near resonance.

I switched to the frequency domain and calculate a cross-spectral density with the displacement signal as input and force as output.

Energy(jw) = cross-spectral-density(Displacement(t),Force(t))

using the same 3 displacements.  I actually receive something similar to the time domain calculation though it matches my predictions better.  More importantly, the energy dissipated by the dampers is larger than the input energy(!).

So I realize that I have an issue with my calculations.  Can someone offer their opinion on the validity of calculating input energy and dissipated energy in this manner?

I appreciate any feedback.

Chris

RigTest (Automotive) · Answer 1 · 2012-03-29 16:07:20

I'll add that I expect the summed energy of the tire + damper to equal the input energy - environment losses.

E_total = E_tire + E_damper = E_input

RigTest (Automotive) · Answer 2 · 2012-03-29 16:23:15

Sorry for all the editing (I guess there's no way to edit posts).

To clarify the spectral calculation:

cpsd(x(t),F(t)) = E_real(w) + E_imag(w)*j

I take the real part, which is the dissipated energy, while the imaginary part is stored. Hopefully I'm not off base here.

GregLocock (Automotive) · Answer 3 · 2012-03-29 18:38:30

If I understand it then you are assuming that the pad force is transmitted throughout the system.

It is not.

input work =Fpad.xpad

tire work= Fpad.xpad-F(tire to hub).xhub

etc

Cheers

Greg Locock

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RigTest (Automotive) · Answer 4 · 2012-03-30 07:59:29

Thanks for the reply, Greg.

I don't know if this changes your opinion, but I am attempting to determine the energy dissipated by the damper(s).

If we simplify to a single DoF system as a tire on a pad, then the input force is transmitted directly into the tire damper/spring.  My thoughts are that the work done by the damper should be affected by both the magnitude and phase of the relative damper displacement compared to the input (actuator) displacement, and the force should be the same.  Here, I would contend that your "Fpad" and "F(tire to hub)" are the same.

I would extend this to a 2DoF system containing a body mass, by considering my enclosed system to be both tire and suspension together.  It is probably not valid to use the Hub-Body displacement, as the measured force is not input to the hub.  However, we should be able to use the entire suspension (Pad-Body) and remove the tire energy component after calculation.

E_system = E_tire + E_suspension

Or am I getting too far ahead of myself?

hacksaw (Mechanical) · Answer 5 · 2012-03-31 11:47:20

believe gL's post is more than opinion, look at what he has told you in more detail

nullius in verba

RigTest (Automotive) · Answer 6 · 2012-03-31 12:12:22

I don't want to come across as dismissive of Greg's post.  I feel that I did not properly describe the setup and the problem that I am attempting to solve.

Let's just simplify to a single spring-mass-damper system, with the spring/damper attached to the actuator pad.  It seems as Greg is stating that the energy (work) input will be Force_pad*X_pad, and the energy dissipated by the damper is:

(Force_pad - Force_hub)*(X_pad_to_hub)

am I correct?

I am attempting to use:

Force_pad*(X_pad_to_hub)

as I thought you only needed the input force to the system and the resulting response (relative displacement).

The pad force would be transmitted through the system, but with a magnitude and phase change, and maybe this is what I am failing to account for.  So to take Greg's advice I would need load cells on the hub and the body?

Thanks for your replies.

hacksaw (Mechanical) · Answer 7 · 2012-03-31 14:21:16

you might draw a free-body diagram of your setup with all the forces & masses shown, ultimately you need the force acting on the damper and the displacement(compression/extension) it experiences

GregLocock (Automotive) · Answer 8 · 2012-04-01 03:17:13

Simple though experiment - double the mass of the tire.

Does the plot of vibration amplitude vs frequency change? Yes

therefore the force in the spring must have changed as it is k.x

Yes, you need to measure forces in between the components, or more cunning experiments.

Cheers

Greg Locock

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RigTest (Automotive) · Answer 9 · 2012-04-02 14:41:02

Greg, when you say "component" do you mean "masses", and therefore you are referring to a multi-DOF system?  For SDOF, assuming ideal elements (mass-less springs/dampers and rigid masses), force is transmitted instantaneously through an ideal spring and an ideal damper, without change in magnitude or phase.  If we think of an ideal spring or damper as having two ports, the forces at each port are equal and opposite.  Note that I am talking about a Force_out/Force_in transfer function, not a Displacement/Force transfer function (displacement will change magnitude for a spring and magnitude & phase for a damper).  This can be seen in practice by placing spring perch load cells on both sides of a damper or strain gauge eyelets on both sides of a damper: any phasing between the two sensors will be small and due to its mass (non-ideal element).  So if we know the force at one port then we know the force at both.  Analogously, the spring/damper can be thought of as 'in-series' with the mass and with the actuator and forces are the same for in-series elements.

I've confirmed this in my simulation, where I can measure forces wherever I want...I only see a relative force change across a mass, never across a spring/damper.  So I agree with Greg, that the vibration amplitude vs. frequency will change when mass is doubled, but I still believe that the force measured at the pad is the only force we have available for a Single DOF system. However...

Extending this to 2 DOF with an additional spring-damper-mass in series with the first mass (quarter-car), is where I believe I had issues and perhaps this is what Greg has been saying the entire time.  I agree that I cannot use the pad load cell for the 'suspension' damper (input force has changed mag&phase).  This makes things more difficult in practice (need a more cunning experiment) but is not an issue in simulation.  I do get expected results this way, and I will have to find a way to reconfigure the setup.

Greg, thank you for your help.

GregLocock (Automotive) · Answer 10 · 2012-04-02 17:51:59

Stop writing. Draw your fbd. Write down some very simple equations. It is 5 minutes work to prove or disprove what I am saying, it must have taken you longer than that to write your frankly unread posts.

Cheers

Greg Locock

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RigTest (Automotive) · Answer 11 · 2012-04-02 19:06:16

You wrote this:

"tire work= Fpad.xpad-F(tire to hub).xhub"

Please, explain to me where 'F(tire to hub)' would be measured at?

GregLocock (Automotive) · Answer 12 · 2012-04-02 19:25:54

At the interface between them -as a thought experiment, at the wheel bearing. You could instrument the wheel disc, or the rim, or the base of coilover, to understand that part of it.

You could buy a wheel force trandsucer, but that will cause as many problems as it solves.

Cheers

Greg Locock

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hacksaw (Mechanical) · Answer 13 · 2012-04-02 22:03:37

if you intend to model the dynamics of a suspension of any kind, then you must include the masses of all moving components,

sketch up your free body diagram identifying all the forces and masses in your system, without that step no meaningful discussion is possible...

RigTest (Automotive) · Answer 14 · 2012-04-02 22:09:19

The point I have been trying to make is that F_pad = F(tire to hub).

The reason is that ideal spring and damper elements transmit force directly, so the force is unchanged from the pad up to the hub mass. I have done plenty of FBDs, I can represent this in simulation, and I have measured equivalent forces with WFTs using a rig to check calibration.

Tire_work = F_pad*X(hub to pad)

However, you were right about the suspension damper (and I thank you for your help).

Damper_work = F_susp*X(body to hub) ... F_susp is force between hub and body.

So I have the results I want/expect. I typo'd in my first post, it must be the imaginary portion above (not the real).

For the record, I didn't like the "unread" comment either.

Chris

GregLocock (Automotive) · Answer 15 · 2012-04-03 01:22:25

Ok, you win.

Cheers

Greg Locock

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hacksaw (Mechanical) · Answer 16 · 2012-04-03 09:39:48

RE: Experimentally calculating dissipated energy

hacksaw (Mechanical) 3 Apr 12 09:39

thus the consultants dilema

RigTest (Automotive) · Answer 17 · 2012-04-03 10:04:17

Fair enough, I will stop posting. GregLocock obviously doesn't agree with me, any future readers can see what we've posted and draw their own conclusions.

I appreciate the help.

GregLocock (Automotive) · Answer 18 · 2012-04-03 19:16:26

Having taken my own advice, ok I see some confusion caused by my first post

Here's the system model I'll discuss

Pad-tirespringdamper-wheelmass-suspensionspringdamper-ground

Fwheel=Fpad=F (sorry, in my first post I said they might be different, which they are in the more detailed model but not in this model)

xwheel!=xpad

work input=F.xpad

work in tire=F.(xpad-xhweel)

work in rest of system=F.xwheel

and because we've broken the system up sensibly it is easy to write the dynamic equations, K=k+jwc F and x are of the form F*sin(w.t) etc

Fpad=(xpad-xwheel).Ktire

Fwheel=Fpad

Fwheel=xhweel.(-w^2*wheelmass+Ksuspension)

Fground=xhweel*Ksuspension

A few observations

- this is an SDOF system but there are two important measured displacements

-it is a lousy model of a suspension on a car because the end of the roadspringdamper is grounded, suppressing the major ride mode at 1-2 Hz. It is OK, but not brilliant, for examining wheelhop at 10-12 Hz.

-It will not easily correlate well to on-road results because it ignores the mass of the contact patch (probably a small effect, 30 Hz or more), and the body cannot move.

Cheers

Greg Locock

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RigTest (Automotive) · Answer 19 · 2012-04-04 08:07:46

I think we're on the same page now. Good points, especially regarding contact patch mass. My tire models are more complicated than what we've discussed (ungrounded, more elements than just parallel-spring-damper) but certainly not an FTire model or anything. They are probably inadequate for tire-specific modeling, but match data well enough to focus on other items.

I have been focusing below 30Hz, as internal tires modes (first-order wheel imbalance) start to creep in, and I can't excite them properly with the rig.

hacksaw (Mechanical) · Answer 20 · 2012-04-04 19:25:17

just curious, did gL's posting resolve the initial struggle with getting the energy balance correct?

RigTest (Automotive) · Answer 21 · 2012-04-04 19:38:22

Yes, he pointed out that I was using the wrong force. Our argument was about one of the equations.

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Experimentally calculating dissipated energy
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