How flexible do you think a stock, trailing arm is Options

[jluetjen]

QUOTE (Racer Chris @ Jan 4 2006, 08:05 PM)

One thing the FEA didn't pick up is the amount of improvement from adding the bulkhead. Is that due to the wire frame model being different from sheet material or some other shortcoming of the software?

When I added the bulkhead, I used the same 2mm thick steel as I used for mostof the rest of the trailing arm model. While a bulkhead helps some, since most of the forces travel across the exterior of the arm, it's not going to be the best solution. Keep in mind that when modelling torsion bars, generally it doesn't make a big difference from a torsional displacement perspective if they are hollow or not since most of the forces travel across the exterior. This trailing arm really isn't much more then a stubby T-bar*. As you picked up with your experiments, most of the surface stress is in the area where the trailing arm's cross-section transitions from being an upright rectangle to being a laying down rectangle. Increasing the cross-sectional area of the trailing arm -- specifically in this area should yield the biggest improvement.

* Basic rule: The angular rate of a tube (in in/lbs/degree) is = (19700 * (OD^4-ID^4))/tube length. So if you were to generalize and say that the trailing arm is a 16 inch long tube with an OD of 4 inches (average) and an ID of (4 inches - (4mm or 3.84 inches)), then the rate of the trailing arm is 19700 * (256-217)/16 = 47.5k in/lbs/degree. If you separate it into 2 - 4 inch OD bars, one hollow bar of (16 inches - 2 mm (.08 inches) = 15.92 inches, and the other a solid bar of 2 mm in length, you get the hollow bar has a rate of 48k in/lbs/degree, while the solid bar has a rate of 63,040k in/lbs/degree. Combined using the formula 1(1/rate1 + 1/rate2) = total rate, we come up with a total spring rate for tube with a bulkhead of essentially... (wait for it!)... 48k in/lbs/degree! But, if you were to increase the OD of the the tube to 4.5 inches (still with an ID 4 mm less), you would get a rate of 19700 * (410 - 355)/16 = 68k in/lbs/degree. So you'll get much better results by increasing the cross-sectional area of the arm rather then adding internal bulkheads. This is what you did by adding the gusseting where the fabricated assembly attaches to the smaller tube.

[ChrisFoley]

QUOTE (jluetjen @ Jan 5 2006, 04:54 PM)

OK. I'll stop lurking on this thread and post some pictures of the model -- without the bulkhead.

I'm glad to see you over here John. Thanks for doing this for me!
In a previous life I was going to be a mathematician and I could have had this in hand with little more than a snap of the fingers. Instead I embarked on developing the weak side of my personality and now too much math gives me a headache. (IMG:http://www.914world.com/bbs2/html/emoticons/screwy.gif)

QUOTE

If you separate it into 2 - 4 inch OD bars, one hollow bar of (16 inches - 2 mm (.08 inches) = 15.92 inches, and the other a solid bar of 2 mm in length, you get the hollow bar has a rate of 48k in/lbs/degree, while the solid bar has a rate of 63,040k in/lbs/degree. Combined using the formula 1(1/rate1 + 1/rate2) = total rate, we come up with a total spring rate for tube with a bulkhead of essentially... (wait for it!)... 48k in/lbs/degree!

Don't you want that to be two bars of 8 inch length held together at the middle by a solid bar of 2mm length? That would drastically alter the calculation if I'm not mistaken.

[Porcharu]

Something that might be interesting would be to load up the trailing arm in your rig and use a mic or a good set of calipers and measure across the arm and see if it is bulging or puckering (cross section) under load (compared to the unloaded condition.) If it was it would explain why the simple bulkhead works so well - "pucker facter reduction technology!"

[ChrisFoley]

QUOTE (srbliss @ Jan 5 2006, 06:46 PM)

"pucker facter reduction technology!"

lol, I like that!
It would probably take a while to isolate the location of maximum pucker but I expect that would be possible to measure a change there.

[jluetjen]

QUOTE (Racer Chris @ Jan 5 2006, 03:38 PM)

QUOTE (jluetjen @ Jan 5 2006, 04:54 PM) OK.  I'll stop lurking on this thread and post some pictures of the model -- without the bulkhead. I'm glad to see you over here John. Thanks for doing this for me! In a previous life I was going to be a mathematician and I could have had this in hand with little more than a snap of the fingers. Instead I embarked on developing the weak side of my personality and now too much math gives me a headache. (IMG:http://www.914world.com/bbs2/html/emoticons/screwy.gif) QUOTE If you separate it into 2 - 4 inch OD bars, one hollow bar of (16 inches - 2 mm (.08 inches) = 15.92 inches, and the other a solid bar of 2 mm in length, you get the hollow bar has a rate of 48k in/lbs/degree, while the solid bar has a rate of 63,040k in/lbs/degree. Combined using the formula 1(1/rate1 + 1/rate2) = total rate, we come up with a total spring rate for tube with a bulkhead of essentially... (wait for it!)... 48k in/lbs/degree! Don't you want that to be two bars of 8 inch length held together at the middle by a solid bar of 2mm length? That would drastically alter the calculation if I'm not mistaken.

OK -- I was never very good at arithmetic...

3 springs in series, 2 of them have a rate of 96,037 in/lb/degree, and one of them has a rate of 63,037,000 in/lb/degree. So the formula would be...

1/(1/96,037 + 1/96,037 + 1/63,037,000) = 47,982 in/lb/degree. So the essentially solid bulkhead contributes very little to the overall stiffness because it basically replaces .5% of the springs total length with a solid piece. At best I would expect it to reduce the overall spring rate by about .5%.

Or did I get my math wrong? (IMG:http://www.914world.com/bbs2/html/emoticons/unsure.gif)

This weekend I'll try running a bulkhead diagonally through the arm (end to end) to see what it does. That may make the whole arm act like it is (more) solid.

[ChrisFoley]

Actually, I was thinking about this while I was making dinner, and I decided the reason was something else.
The formula for a torsion bar doesn't properly apply in this application. The wall thickness is so small compared to the od that non-linear deformation occurs from the amount of load applied. The sides of the box deform laterally. That''s where the "pucker factor" comes in. The bulkhead stops the puckering and actually makes the torsion bar formula more appropriate to predict the spring rate.

[jluetjen]

OK. For all of the engineers out there-- what sort of forces cause "pucker"?

* Axial Load?
* Shear Y Load?
* Shear Z Load?
* Torsion Load?
* Moment Y Load?
* Moment Z Load?

Or any of the above as a stress?

I can model all of those with my software.

[ChrisFoley]

Could it be because the arm is square as opposed to round?
That causes the corners to stretch more than the sides since they are farther from the twist axis.
On body panels puckers occur when a part of the panel is stretched, like when it gets dented.
The difference is the trailing arm isn't stretched beyond the elastic limit of the steel, unless it gets hit by another car.

[TimT]

John continues to impress...

I havent built the model put it through a FEA yet.. I have many things on my plate right now.. which are distracting me., I am working day to day, the hammer may drop tomorrow morning when I get to the office. And then its look for a new job

Im modernizing my 1918 home.. replacing windows etc..the furnace got flooded a few weeks ago....

The cross section of the trailing arm has everything to do with its ability to resist torsion and bending forces. Square sections and circular sections behave differently in regards to torsion and bending.

[ChrisFoley]

Square sections and circular sections behave differently in regards to torsion and bending.

(IMG:style_emoticons/default/agree.gif)
Square is better in bending and round is better in torsion.
For torsion, having all the material equidistant from the axis is better.
For bending, a square section puts more material in tension/compression at a greater distance from the centerline.

[jluetjen]

True. But if we have a hollow "bar" X (square or round section) which is too flexible in torsion and in flex -- what to do? ( 'Just questioning the fundimentals...)

It would seem that in decreasing order of effectiveness our choices are...

1) Make it solid and thicker
2) Filling it in (or replacing it with a unit with a solid cross section. BTW -- This is what Porsche did with the 911 trailing arms when they converted to cast aluminum pieces in 1974. And the 911 is actually a better situation since the trailing arm is triangulated by the spring plate which is a feature the 914 doesn't have room for.)
3) Increase the cross section
3a) Increase the cross section across the largest percentage of it's length that you can
4) Increase the wall thickness
5) Add latetudinal bulkheads.

The question is where do longitudinal bulkheads appear on the list? That's what I hope to test.

[michel richard]

I am not an engineer nor a mathematician etc . . .
Still, my own empirical experience has been that No5 in the list above should be much higher in that list, at least as regards tortional flex.

[crash914]

Again going back, what is the effect of adding the hollow tube from the trailing arm to the pivot arm? This should resist the lateral forces..

[jluetjen]

QUOTE (michel richard @ Jan 6 2006, 04:27 AM)

I am not an engineer nor a mathematician etc . . . Still, my own empirical experience has been that No5  in the list above should be much higher in that list, at least as regards tortional flex.

When I orginally modeled the arm for just tortional flex (-1000 lbs Z), it didn't flex by much at all. But based on input from this thread I tried modeling it with all of the forces that would be influencing it while cornering -- so -1000 lbs X (braking), -1000 lbs Y (lateral cornering forces) and 1000 lbs Z (roll), all at the same time. This resulted in the deflection shown below. This is about 13% more the tortional flex of earlier case (Z deflection), and about 1000X the deflection in the Y direction -- which impacts Toe.

Specifically...
Global Coordinate Displacements/Rotations (in inches)
Node X-Displ. Y-Displ. Z-Displ. X-Rot. Y-Rot. Z-Rot.

First, Just 1000 lbs -Z, no bulkhead.
232 1.027E-05 -4.044E-05 -1.355E-03 -3.071E-04 6.046E-05 2.470E-06

Now, same model, but with all 3 forces inflicted upon it.
232 -1.035E-03 -1.537E-02 1.325E-03 3.710E-04 -6.441E-05 1.560E-03

Now the model with the bulkhead and all 3 forces.
232 -1.034E-03 -1.534E-02 1.310E-03 3.533E-04 -6.467E-05 1.560E-03

Going back to this picture, note that the arm is bending across it's entire length (and exagerated by 100X in the illustration). Adding a latitudinal bulkhead across just 1% or 2% of it's length just isn't really going to make much difference in this situation.

(IMG:http://www.914world.com/bbs2/uploads/post-2-1136518881.gif)

I suspect it will when the forces become high enough for significant distortion to occur in the outer skin or deformation in the entire structure. Basically just short of the structure's failure. But just 1000 lbs here or there will only result in a gentle bend across it's entire length as each individiual element (molecule) deforms by 1% or 2%.

I don't think that my model is an exact replica of a 914's trailing arm, but I think that it's close enough that it can give us some ideas of how and why the arm reacts the way that it does.

crash914; I'll also try modeling what you described this weekend.

[ChrisFoley]

I tried modeling it with all of the forces that would be influencing it while cornering -- so -1000 lbs X (braking), -1000 lbs Y (lateral cornering forces) and 1000 lbs Z (roll), all at the same time.

That's a little extreme. Working with a friction circle you would find the car looses rear grip well before this situation occurred.
Real world - a torsional force of greater than 1000 ft-lbs results in less than 1 degree of twist. I didn't measure the lateral deflection with the same load but I did observe it and estimate it to be also on the order of 1 degree.
John, I can't see in the drawings where a moment arm is used to apply the torsional load of 1000 ft-lbs.

[jluetjen]

Chris, I applied all of the loads to point 232. Note that I didn't model the complete hub tube out to the hub, but rather modeled an empty cylindrical area which is closed on one side and open on the other on the plane with the outside of the trailing arm. The point in the center of the closed end is node 232. So my forces are not cantelivered as they would be in the real arm, so they should actually represent a lower level of forces at the hub. Essentially I'm assuming that there is 0 defection in the large tube and hub. You can see the arrows representing these forces radiating out of node 232 in the picture below.

(IMG:http://www.914world.com/bbs2/uploads/post-2-1136518768.gif)

I'm not sure if I'll be able to model that extra bracing tube as was requested earlier since that intersects near the hub which is essentially outside of my model. What I will try though is just adding some surface gussets down the length of the arm -- One or two gussets on either plane to essentially convert the arm into an I-Beam or an X-Beam (when viewed in cross section).

As far as the actual level of forces, I'm not yet confident that the model is precise enough to reflect that. We could reduce the forces to 500 lbs, and then you'd get half the defection. I think that the importent value of the model right now is to use it to see how a shape like this deflects when forces are imposed upon it, and what is the magnitude of changes in this deflection as the shape is changed.

If we were really good (like an F1 team, with an F1 team's resources), we'd have a more precise model which is calibrated to test results so that we could review the specific forces on each area of the piece and design our solution to have just enough material (but not to much) in every square inch of the assembly. We're not there yet! (IMG:http://www.914world.com/bbs2/html/emoticons/headbang.gif)

[Eric_Shea]

I poured molten lead into mine.

The stuff smells great when you cook it and... it only added 123lbs. per side. I can definitely notice the difference at the track (IMG:http://www.914world.com/bbs2/html/emoticons/w00t.gif)

OK... would it be that difficult to make a cast aluminum version of these (I know it wouldn't apply to the racing class)?

[URY914]

Didn't someone here fabricate a completely new control arm?

I thought I remembered pictures of it.

[Eric_Shea]

There's been a couple of those that I've seen but, taking John's cue on the 'solid' and the '911 cast' comments, I wondered how difficult it would be to make a mold and drop some aluminum in it? (IMG:http://www.914world.com/bbs2/html/emoticons/confused24.gif)

Also, aren't the cast 911 arms kinda hollow? The wall thickness is much greater than a steel one but I seem to remember mine being hollowed out.

[andys]

QUOTE (URY914 @ Jan 6 2006, 08:18 AM)

Didn't someone here fabricate a completely new control arm? I thought I remembered pictures of it.

URY914,

Yup. Here's a pic.

Andys