Calculating chassi stiffness with FEA

[Big Pablo]

Hello,
just registered after spent a lot of time going through forum threads. There are a lot of discussions about chassi stiffness where you guys expressed the torsional stiffness in terms of ft*lbs/deg.

My question is: how to calculate this value with a FEA software ?

I'm actually working on a new chassi design using Autocad Inventor. Pretty complete and easy to use 3D software. It also provides with an integrated FEA module.

In my FEA analysis I "bolted" the rear side of the chassi and applied a load of 1000 N on the front whishbones brackets, in opposite directions for the left and the right side.

Everything goes well, except the fact that I have all result expressed in Mpa (von mises stress) or mm (displacement).

I haven't understood yet how to come to the absolute ft*lbs/deg value.

thanks for your help.

[Bemo]

Your 1000N*distance from wishbone brackets to center of the vehicel is the torque.
With the displacement of the wishbone brackets and that distance you can calculate the angle the chassis is twisted.
torque/angle=torsional stiffness
Like this you'll have it in Nm/deg, but as a European I think that's the better unit anyway ;-)

[Big Pablo]

ok, it starts to have more sense... I have 1000N upward on one bracket of the left front arn and 1000N downward on the right front arm bracket.

I guess the right torque would be 2000N * distance to longitudinal centerline, or I need to go straight with 1000N * ?

Regarding the twisting angle, can you please tell me more how get it from the displacement ?

many thanks.

[Chris Craig]

vertical displacement divided by distance from the centre line will give you tan of the angle, then do the maths.

If you use a smaller load then you can get neglect any movement towards the vehicle centre line, if not just look at this displacement and subtract it from the original distance

Simple bit of maths really

[Gruntguru]

Originally posted by Big Pablo:
ok, it starts to have more sense... I have 1000N upward on one bracket of the left front arn and 1000N downward on the right front arm bracket.

I guess the right torque would be 2000N * distance to longitudinal centerline, or I need to go straight with 1000N * ? Sum your moments about any point ie "About Centreline" -> 1000 x (dist to centreline) + 1000 x (dist to centreline).
or
"About the point of application of one of your forces" -> 1000 x (distance between forces) = same result.

A moment will have the same effect if applied anywhere in the same plane.(the plane perpendicular to the chassis centreline). Normally you apply a moment at the plane of the front wheels and an opposite moment at the plane of the rear wheels. Note that if these moments are applied further apart than this, the section between axles still sees the same effective moment. This is often useful when doing actual torsional testing since it is not always convenient to apply the moments precisely at the axle planes. The angular displacement however must be measured at the axle planes ie the torsional stiffness measured will be of the section of chassis between the displacement measuring planes.

Measuring angular displacement at numerous planes along the chassis is useful for diagnosing where the weaknesses lie. When doing this, it is easier to understand (and compare magnitudes) if expressed as "torsional compliance" (= 1/(torsional stiffness) because torsional compliances are additive along the CL to obtain the total compliance of the chassis.

[Big Pablo]

get closing to the result. Here an example of my calculations:

http://img522.imageshack.us/i/57698391.png/

http://img522.imageshack.us/i/57698391.png/

As far as I understood, to calculate the exact stiffness in Nm/deg I need to do:

(1000*0,5 + 1000*0,5 ) / 20 = 50Nm/deg

Is it correct ?

Now, from the stress analysis in my FEA software I've noticed that the longitudinal center axis of the frame is little bit moved after the loads (let me say, about 0.1m). This means that the vertex did not a perfet "arc" in its distorsion.

How this lateral bending of the frame is taken into consideration in the rigidity calculation ?

[RollingCamel]

One thing about the torsional rigidity, the load is transfered through the push/pull mechanism to the mounts of the damper. So assuming loads at wishbone mounts or even taking consideration wishbones stiffness isn't quite a realistic scenario? Am I making any sense here?

[Gruntguru]

Originally posted by RollingCamel:
One thing about the torsional rigidity, the load is transfered through the push/pull mechanism to the mounts of the damper. So assuming loads at wishbone mounts or even taking consideration wishbones stiffness isn't quite a realistic scenario? Am I making any sense here? Even if we are talking about a "pure" roll moment input, the forces will be reacted through all suspension attachment points (including damper/rocker/pushrod mounts as you say.)

[RobbyObby]

You also want to make sure you define the correct degrees of freedom of each joint. The default joint constraint is usually fixed (ie welded), that is if you are using beam elements and not shell elements. Since your suspension members are usually pin joints, you want to model them that way in the analysis to get proper results.
I have found that it is not necessary to even model every suspension member. For example, we are running pull rod front and push rod rear suspensions. So I modeled the upper wishbone/pullrod in the front and lower wishbone/pushrod in the rear and applied loads at the outboard push/pull rod pickups and found the results were the same as modeling the entire system.
To simplify the calculations, set up a spreadsheet. Since typically your force and moment arm lengths are constant for the analysis, you can set it up to enter the vertical displacement only and spit out the TR value. This is useful when analyzing iterations through the design process.

[Big Pablo]

Originally posted by Big Pablo:
get closing to the result. Here an example of my calculations:

http://img522.imageshack.us/i/57698391.png/

As far as I understood, to calculate the exact stiffness in Nm/deg I need to do:

(1000*0,5 + 1000*0,5 ) / 20 = 50Nm/deg

Is it correct ?

Now, from the stress analysis in my FEA software I've noticed that the longitudinal center axis of the frame is little bit moved after the loads (let me say, about 0.1m). This means that the vertex did not a perfet "arc" in its distorsion.

How this lateral bending of the frame is taken into consideration in the rigidity calculation ?

can anyone validate my thought ?

[RobbyObby]

Originally posted by Big Pablo:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Big Pablo:
get closing to the result. Here an example of my calculations:

http://img522.imageshack.us/i/57698391.png/

As far as I understood, to calculate the exact stiffness in Nm/deg I need to do:

(1000*0,5 + 1000*0,5 ) / 20 = 50Nm/deg

Is it correct ?

Now, from the stress analysis in my FEA software I've noticed that the longitudinal center axis of the frame is little bit moved after the loads (let me say, about 0.1m). This means that the vertex did not a perfet "arc" in its distorsion.

How this lateral bending of the frame is taken into consideration in the rigidity calculation ?

can anyone validate my thought ? </div></BLOCKQUOTE>

20 degrees of chassis deflection is magnitudes higher than what should be expected. Heck, the weakest tube would probably buckle before you even got to 5degrees or so. Make sure you are measuring the displacement at the points where you are applying the load. And make sure you are reading the displacement correctly. Dont know about Inventor but Solidworks reads out the displacement in scientific notation, so maybe you just arent accounting for the degrees of magnitude? Or maybe your units arent set correctly?

[billywight]

Rob,

I would suspect that your model is incorrectly constrained if you are getting the same results without one of the wishbones. If this were possible, then why have the wishbone on the physical car at all? You might want to double check. Any effort saved by not modeling one wishbones is of no benefit in comparison to getting the incorrect results.

Big Pablo,

Experiment with different methods of loading and constraints in your model. It isn't as easy as grabbing the rear and twisting the front, the correct constraints are rather complex. I made a post a while back about this if you search for it. In the end the software is just doing math, and you are giving it the equations. You need to make sure you're giving it the correct information and know the limitations of your model (assumptions, etc.).

[Big Pablo]

the values you find in my example are just "as examples", and obviously they not are referring to neither real scenario. I still have to calculate them (I'm fighting with my FEA module...).

I would just like to know if the costruction of the formula is right or not.

[billywight]

I would just like to know if the costruction of the formula is right or not.

It is not.

[JasperC]

Originally posted by Bemo:
Like this you'll have it in Nm/deg, but as an engineer I think that's the better unit anyway ;-)

I couldn't resist to fix that for you. http://fsae.com/groupee_common/emoticons/icon_smile.gif

Cheers,
Jasper

[Big Pablo]

Originally posted by billywight:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">I would just like to know if the costruction of the formula is right or not.

It is not. </div></BLOCKQUOTE>

this doesn't help too much...

[billywight]

this doesn't help too much...

The best way to learn something is through lots of trial and error. You always learn best by doing it wrong the first time. I'm not going to flat out tell you the correct way, but it has been posted here before, and I even told you how to find it. Now go do it.

[Gruntguru]

Originally posted by RollingCamel:
One thing about the torsional rigidity, the load is transfered through the push/pull mechanism to the mounts of the damper. So assuming loads at wishbone mounts or even taking consideration wishbones stiffness isn't quite a realistic scenario? Am I making any sense here? Another thought on this. I would be careful about including suspension components in the analysis of chassis torsional stiffness. The effect of compliances in these components is very different to torsional compliance in the chassis. So while it is probably most technically correct to apply the loads through the suspension attachment points, I recommend you measure angular deflection of the chassis alone.

[Crispy]

Originally posted by Gruntguru:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by RollingCamel:
One thing about the torsional rigidity, the load is transfered through the push/pull mechanism to the mounts of the damper. So assuming loads at wishbone mounts or even taking consideration wishbones stiffness isn't quite a realistic scenario? Am I making any sense here?
Another thought on this. I would be careful about including suspension components in the analysis of chassis torsional stiffness. The effect of compliances in these components is very different to torsional compliance in the chassis. So while it is probably most technically correct to apply the loads through the suspension attachment points, I recommend you measure angular deflection of the chassis alone. </div></BLOCKQUOTE>

I agree that the compliance caused by the links is different than the torsional compliance of the chassis alone, but if you leave out the effect of the links you may be missing out on a very significant portion of the total hub to hub stiffness. Even if the suspension links are rigid (with the proper DOFs released), they create local deformations of the chassis that can be very significant.

When we perform torsional FEA analysis or physical testing, we place a reference point at the outboard suspension that moves with the chassis (rigidly attached to the front bulkhead). By comparing the movement of the reference point relative to the outboard we isolate the "installed stiffness". When we are done, we have an overall hub to hub stiffness, front and rear installed stiffnesses and a back calculated chassis torsional stiffness. I'm sure there are other ways of implementing the same idea, but however you do it, I think it is important to identify and measure the different causes of hub to hub twist.

Once you have identified the various stiffnesses, it is a question of how you can you most efficiently increase overall stiffness and/or reduce weight, depending on your stiffness and/or weight targets.

@Pablo: As for Billy's comments, I think he has more of an issue with your loads and restraints than the concept of your basic stiffness calculation.

@Billy: I think I agree with Robby (mostly at least). If you apply the loads/constraints at (or near) the outer balljoints, as you suggest, and they are connected to the push/pull rods, you shouldn't have any forces other than the static loads in the other balljoints. There might be a small difference in the results, but it should be trivial. If you apply the loads/constraints at the contact patches, which I prefer, I might expect some issues depending on geometry. Does this seem reasonable?

[Hector]

Crispy, I agree with your method. In the end it's a good idea to have both a chassis stiffness and overall hub-to-hub stiffness. It allows you to decide if your chassis needs tweaking, your suspension needs tweaking, both, or neither. If you see a very stiff chassis but noodly suspension you may want to look into suspension compliance.

[billywight]

I think I agree with Robby (mostly at least). If you apply the loads/constraints at (or near) the outer balljoints, as you suggest, and they are connected to the push/pull rods, you shouldn't have any forces other than the static loads in the other balljoints. There might be a small difference in the results, but it should be trivial. If you apply the loads/constraints at the contact patches, which I prefer, I might expect some issues depending on geometry. Does this seem reasonable?

I would apply the loads and constraints at the hub centre, i.e. the middle of the upright. This will load both of the a-arms. You can load the chassis if you want, however, what is this telling you? In the end the tyres are all that matter right? My opinion is that if you use your simulations to maximize the stiffness with regards to the tyres, then you are focusing your efforts in the right areas. Whenever I do a torsional stiffness analysis, I will model the suspension as rigid members and effectively check the installation stiffness of the chassis only. You can then change the properties from rigid to the actual component properties and compare the differences to gain an understanding of the suspension member compliances.

[Crispy]

I definitely agree that applying loads/constraints directly to the chassis is not a good idea. It may be possible to do correctly, but I can't think of how it could be done simply.

As far as where on the suspension to apply the loads, it seems like applying the load/constraints at the hub is inherently no better or worse than applying the load/constraint at whichever balljoint one feels like. Depending on their relative positions, moving the load/constraint point may add or remove a load from each of the balljoints.

Like you said, in the end the tires are all that matter, so why not apply your load/constraints/measurements at the tire contact patches? This should be about as close to reality as we can possibly (or practically) get to measuring stiffness with respect to the tires.

I also very much agree with modeling the suspension both with rigid and non-rigid links to identify their contributions to the installed stiffness.

[billywight]

As far as where on the suspension to apply the loads, it seems like applying the load/constraints at the hub is inherently no better or worse than applying the load/constraint at whichever balljoint one feels like. Depending on their relative positions, moving the load/constraint point may add or remove a load from each of the balljoints.

The results would be close with either method as long as all of the suspension links are modeled and are all rigid, however, there may be a small difference due to the mechanism motions under the deformations (it will be very negligably small). I mostly prefer applying the load at the hub centre because it is most intuitive as to what's really happening and not any more difficult to model anyway.

[buggaero]

I'm sorry to say i haven't yet reached the FEA part, it's my debut as the designer,
done with the chassis, used catia though.

I read the complete thread and could almost visualize the process.
One thing that still bothers my visual understanding is, keeping the wishbones stiff, and doing hub to hub, does it take into account the motion mechanism of the arms because stiffness would come into play only after that. i hope i am clear.

[billywight]

One thing that still bothers my visual understanding is, keeping the wishbones stiff, and doing hub to hub, does it take into account the motion mechanism of the arms because stiffness would come into play only after that.

Yes, the mechanism is modeled by using beam end releases or similar. If you were to put a spring element where the shock would go you would see the mechanism's motion. (in fact this is a good technique to check if you've set things up properly)

[Crispy]

I mostly prefer applying the load at the hub centre because it is most intuitive as to what's really happening and not any more difficult to model anyway.

To each his own, I think it's easier and more intuitive to apply everything at the contact patches http://fsae.com/groupee_common/emoticons/icon_smile.gif

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