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.

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.

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..

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.



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.

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.

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.



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!

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

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

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

I thought I remembered pictures of it.

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?

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.

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

Oops, wrong pic...try this one.

Andys

QUOTE

Also, aren't the cast 911 arms kinda hollow

This is a 930 trailing arm. we got punted, and the trailing arm broke.

now back to the 914 trailing arm

OK, Here are a couple of cases that I've made up as well as the earlier data for reference.

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

First New Case:
A single gusset of the same 2mm material down the short side per the attached drawing. (Note that the key node is now node 242 after the nodes were renumbered) It appears that twist was reduced in Y and increased in X as a result of this change.
242 -1.148E-03 -1.394E-02 1.323E-03 3.504E-04 -6.904E-05 1.408E-03

I doubt that we can put a gusset on the other side without interfereing with the wheel or tire.

Click here for full size image.

OK. Now I've modified the model to include a full width bulkhead down the inside between the 2 red lines. It's better then having a bulkhead across the piece, but not as good as an external gussset. Here's the numbers...

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

First New Case:
A single gusset of the same 2mm material down the short side per the attached drawing. (Note that the key node is now node 242 after the nodes were renumbered) It appears that twist was reduced in Y and increased in X as a result of this change.
242 -1.148E-03 -1.394E-02 1.323E-03 3.504E-04 -6.904E-05 1.408E-03

Now with the longitudinal internal bulkhead (actually a "floor" since it's horizontal) of 2 mm materia. (The key node is 232 again).
232 -1.070E-03 -1.421E-02 1.317E-03 3.449E-04 -6.727E-05 1.541E-03

BTW - Keep an eye on the colors. According to this simulation there are some areas up close to the smaller tubes where axial stress on the material is in the neighborhood of 10,000 lb/sq inch.

Click here for full size image.

Another idea for reducing "pucker factor" If we know where the "pucker" or dimple is occuring would a small tube or 2 (or a bolt) welded across the area work as well as the bulkhead (or almost as well.) This would be a very easy way to go if it works - I'm a little nervous about cutting my trailing arms in half and welding them back together. Drilling a few holes and welding in some tubes would be a snap.
Steve

Steve, those were my thoughts awhile back. Drill a few holes through the entire arm at strategic points and weld in rod. Also thought the same thing for longitudinals.

Steve;
I suspect that what would would happen is that if you could know where a "pucker" were to occur -- and buttressed up that area -- that the stress would then move to another unreinforced area and pucker that area even more. It's kind of the "fat lady in the girdle" sort of excercise. All of that force needs to be transmitted through the trailing arm. If you restrain it in one area, it will pop out in another.

A point approximating the center of the wheel bearing would be the best the position to apply the forces.
Since the location of point 232 is so close to the longitudinal axis of the trailing arm, when you apply a Y axis load the torguing force (ft-lbs) is much lower than the actual force (lbs) applied. More of the load ends up being applied to the rear support point compared to a real trailing arm.
The current model works well for Z axis loads much but it does decrease both X and Y moments significantly.
Speaking of X axis forces, I think they would be much smaller than half a ton, even braking in a straight line. You could argue that since the moment is smaller the effect is similar but you also pointed out some pretty high force concentrations in one example, on the order of magnitude (1E+05) of the yield strength (4E+05) of the metal. It seems that a little more accurate model is still needed to see if there is any more likelihood of component failure.
In order to more accurately model the location of force application you could simply attach a cone to the surface and use the point of the cone instead of the current point 232.

Good morning Chris. Good thoughts. Give me a minute to upload the stuff that I just finished and then we can run a couple more simulations with the modified force vectors.

OK. Here's a spreadsheet with the different cases that we ran and then some new ones. The big changes is that I've extended the hub out by 2 inches as Chris just described. The forces are still being applied to the center of that "hub". I then ran a model adding a strut as Mugs914 showed in his pictures, and then another test with the external gusset on the short side of the arm. It looks like the gusset provides the biggest improvement in the Y-displacement. I'll upload the overhead view of each simulation in a minute.

Base model with the cantelivered hub.

Click for full size image.

With Mugs914's strut

Click here for a full size image.

And now the external gusset

Click here for full size image.

Chris; If I were to assume that your 914 corners at about 1.5 g's (round figure), what sort of braking forces should I use? We can then do some quick vector addition to come up with some more precise force levels. I think that the models that we've done so far are certainly relevent to try out the different strategies for reducing the Y displacement (aka: Toe). Then we can optimise a strategy to ensure that the stresses are kept to manageable levels.

I'll check back in a few minutes.

BTW, how fabricatable would it be to use one piece of metal as both the longitudinal "floor" and the external gusset. Say -- cut a line on both sides of the arm, insert the flat piece of metal and weld from both sides?

John,

Is the rear caliper inner adjuster tube factored into your model?

QUOTE (Eric_Shea @ Jan 7 2006, 10:30 AM)

John, Is the rear caliper inner adjuster tube factored into your model?

I checked it by adding a 1.125"x12 tube there and it didn't make any difference. The numbers were exactly the same for at least 5 digits.

John,
I was away all day working on Tangerine 356 Conversion header issues.
I just noticed where you identify your x,y,z axes. I intuitively had chosen the z to be the (real world) horizontal - lateral axis and the y axis to be the vertical, as opposed to choosing the coordinates while looking "down" on the trailing arm. I have to be more careful I guess.

I just identified a big difference betweeen a real trailing arm and the GRAPE model. The box tube is actually perpendicular to the pivot axis. This affects the position of the node where you are applying forces relative to the restrained points at the front, therefore alters the moment arm length for x axis rotations. If I'm not mistaken x axis rotation (in the model coordinate system) is what I've been measuring with my tests.
Have you tested my bulkhead using the new point 284? I would still like to place some focus on x axis rotations as something to reduce. Also, what are the units for this column?

QUOTE

I just identified a big difference betweeen a real trailing arm and the GRAPE model.  The box tube is actually perpendicular to the pivot axis.

Actually, on the piece that I have, the small tube is about 7+/- degree from perpindicular with the weld seam up the trailing arm, and the large tube is an additional 7+/- degrees from perpindicular in the opposite direction. This places the hub (I'm guessing) at about 15 degree from the axis of the smaller tube. Why do I say 15 degrees instead of 14? Because on the 911, the pivot axis for the rear semi-trailing arm assembly (from the spring plate's pivot through the trailing arm pivot) is also around 15 degrees. I wouldn't be surprised if Porsche designed the 914 rear end to use similar suspension geometry as the 911, although the way that they got there mechanically would be different. Anyhow, I digress...

QUOTE

This affects the position of the node where you are applying forces relative to the restrained points at the front, therefore alters the moment arm length for x axis rotations.

Yes -- and not really. Yes it does. That is one of the compromises I made in my model. I wasn't about to spend the time to calculate all of the coordinates 7 degrees off of square with the origin coordinates. And then do the hub an additional 7 degrees off of that. That's the kind of stuff that gave me brain cramp at the start and why it took me almost a month from when I borrowed the arm until I actually started. I just decided to simplify things and assume that everything was square to start. Once I got the basic model done, I then went back and moved the small tube about 7 degrees from the main arm. Right now the large tube with the hub is perpindicular with the main arm. ( Hmmmm, as I'm writing this it just dawned on me how I could shift the hub assembly by about 7 degrees without a huge amount of recalculation. All it will take is the time to reconstruct the web.)

But I don't think it makes a huge difference from what I've seen running the models for a couple of reasons.

1) We've got 3 vectors of force on the model which sum up to a force which is 45 degrees back (away from the small tube) and 45 degrees up. Now shift that about 15 degrees. Basically we could get the same result as moving the total force 15 degrees by increasing the force in the -Y direction, and reducing the force by a corresponding amount in the -X direction. Either way, it's not going to be a magnitude change (Moving the decimal point essentially) as we're seing between the Y and the Z distortion. That is a function of the shape of the structure.

2) One of the features of the software is that I can make an animation of the structure under load. What I can print is only the total displacement (the dotted lines). If you watch it move, you'll see that it's basically bending in a fairly even arc. Shifting the forces by a few degrees just isn't going to change that by much.

QUOTE

If I'm not mistaken x axis rotation (in the model coordinate system) is what I've been measuring with my tests.

Yes, or more specifically the Z displacement. This is what I have highlighted in the first two rows of the spreadsheet. But after running the model, one of the things that jumped out at me pretty early is that given that the pivot axis at the front, and the coil-over attachment at the tail of the arm (constraining it in the Z direction), the arm is not really twisting that much at all. But after I put the lateral and (cantelivered) -X force vectors on it, that the arm displaced a whole lot in the Y direction -- about 10X the Z displacement. You can see this on the spreadsheet. This is why I standardized on posting the pictures from the top view. If you make yourself a test fixture where you put the arm's pivot axis vertical, and you restrain it perpindicular to that so that it doesn't pivot, then apply the same force on it that you have previously in a twisting fashion. I suspect that you will find that the arm deflects about 10X what you saw on your first experiments. The model suggests (by an order of 10X) that this is where your oversteer is coming from.

QUOTE

Have you tested my bulkhead using the new point 284?    I would still like to place some focus on x axis rotations as something to reduce.

I'm going to be out all afternoon, I'll try to do this (as well as combine the external gusset with an internal "floor" ) some time this week. Unfortunately, with this software, adding features inside of a model is a real test of visual accuity.

QUOTE

Also, what are the units for this column?

The units of the spreadsheet data should be inches which is the standard units I used for the model. I had to do some off-line calculations to estimate the geometric properties of a 1-inch square by 2 mm's thick. The properties of steel are stored on a second table which I just used as is. If you (or anyone else) is interested, this is what I calculated the geometric properties to be. If someone knows this better then me -- Please speak up now!!!

[B]Geometry for a 1 inch square by 2 mm strip [B]
X-Area: .078792
J-Inertia: .000112
YY-Inertia: 4.1E-5
ZZ-Inertia: .006565
Y-Recovery: .5
Z-Recovery: .5
T-Recovery: .5

Can someone resize John's pics to 800pxls W?

I would if I could -- but I can't.

No problemo... I 'think' someone here can.

It would just be easier to read and look at all the points if they were 800. That seems to be the perfect size for a 1024 view.

I would also use 1.5g for the max limit of braking force. Most of the braking is done by the front tires, so I would estimate one rear wheel at (2000*1.5/2*0.25) 375lb max x axis force. I doubt this would ever occur simultaneously with max y or z forces.
I agree. I'm just prodding you to continue optimizing the model for your practice as well as to get the most accurate results.
That wouldn't be too hard.
It is actually 12.5 degrees. I think the inner surface of the box is more nearly perpenducular than the weld seam and most of the angular difference is in the hub support tube/box section interface.
I've thought about doing this but it would take a while to create a fixture sturdy enough. If someone gives me a believable (mathematically supported) analysis of the (real world) forces involved that suggests the lateral forces are anywhere near as high as the twisting force I'll do it.
I was curious about the units of the rotational columns. They can be in degrees or radians but not inches.

QUOTE

I was curious about the units of the rotational columns. They can be in degrees or radians but not inches.

Rotations are in radians.

QUOTE

I would also use 1.5g for the max limit of braking force. Most of the braking is done by the front tires, so I would estimate one rear wheel at (2000*1.5/2*0.25) 375lb max x axis force. I doubt this would ever occur simultaneously with max y or z forces.

Can you give me an estimated CG height for a 914? If in doubt, it's usually pretty close to the crankshaft height. The other thing is the approximate roll-center heights -- front and rear. With that information I've got another software program where I can run it through and come up with a pretty good estimate. On the other hand if you have some data that supports that estimate, that's fine too. You're estimate sounds a little low given that your car is a rear-mid-engined car -- and a fairly low one at that.

For the sake of simplicity, let's just assume that each wheel is braking a quarter of the car's mass, so we're talking about 2000 * 1.5/4 = 750. By the same token, we're talking about a quarter of the car's lateral force at the same G level, so 750 lbs lateraly. If I remember (or rather can read) my trig correctly, 750 lbs in one direction plus 750 lbs in a perpendicular direction will equal (excel calculation.... COS(RADIANS(45 (degrees)))*750 = 530.33 lbs. Right? So this is what I should use for our force vectors. Now we just need to estimate the forces in roll.

BTW, Apparently Grape's calculations tend to overestimate the displacement by about 20% (most likely as a margining factor). This is based on a model done of an actual FF frame compared to experimental measurements taken off of the real frame.

Figure the center of gravity is about 16in off the deck.

Cool, and very geeky.
Is the "Z" axis the real world vertical plane in the FEA plots, with "X" being the longitudinal (fore and aft)?

I think vectorially, 750# with a 750# perpendicular would yield a (750 x 1.414)pound resultant, not a 530#...

Since I just cut up some suspension consoles yesterday, I recall remarking to myself of how flexy they are. So, Chris' test stand is great for testing the torsional resistance of the trailing arm, but when we look at suspension dynamics and ensuing commentary on the steering and handling characteristics of the car, one should take into account the inherent flex of the chassis mount points. That is a relatively long moment arm to the two chassis mount points from the main box section.

Pat

QUOTE (pek771 @ Jan 8 2006, 07:50 PM)

Cool, and very geeky. Is the "Z" axis the real world vertical plane in the FEA plots, with "X" being the longitudinal (fore and aft)? I think vectorially, 750# with a 750# perpendicular would yield a (750 x 1.414)pound resultant, not a 530#... Since I just cut up some suspension consoles yesterday, I recall remarking to myself of how flexy they are. So, Chris' test stand is great for testing the torsional resistance of the trailing arm, but when we look at suspension dynamics and ensuing commentary on the steering and handling characteristics of the car, one should take into account the inherent flex of the chassis mount points. That is a relatively long moment arm to the two chassis mount points from the main box section. Pat

Pat,

most of the race cars have bracing on the suspension consoles and no longer have this flex issue.

brant

Thanks Brant:
I'm trying to absorb this, keeping in mind I have to put my rustbucket back together, and am looking to stiffen this area up while I have it apart. From some of the photos of the race cars, there seems to be a wide variation of rear suspension designs. I suspected most racers have this area stiffened. Intuitively, I would think the biggest load is on the inner console ear area.

Regardless of the torsional characteristics of the trailing arm, any load is ultimately transferred into the longitudinal members and appendage braces. So, would this bracing be as strong a platform as Chris test rig?

Very interesting topic.

Pat

If that were the case then brake pad wear would be equal front and rear. Pad wear is around 3-1 favoring the front.
I expect the real world value to be a little higher, like 900 to 1000 lbs, due to weight transfer. The y axis force is (in reality) applied 11.75" below the hub, drastically changing the value of the y force vector at point 284.

I've seen evidence of chassis cracks on track cars that weren't well reinforced before they upgraded the suspension/tires. I'm pretty confident that the things I do to modify/reinforce the suspension pickups eliminate most of the chass flex.

QUOTE (pek771 @ Jan 8 2006, 06:50 PM)

I think vectorially, 750# with a 750# perpendicular would yield a (750 x 1.414)pound resultant, not a 530#...

Pat;
Actually I wasn't quite accurate when I said that it's straight vector addition. We're dealing with a traction circle. If the car generates a maximum of 1.5 G's under pure braking, and 1.5 G's under pure lateral, then when you combine the two you will wind up with a situation where the braking force in X + lateral force in Y = 1.5G's in total (in a 45 degree direction 1/2 of the way between X and Y). So it's really a case of what number is X = 1.5G's at a 45 degree angle.

QUOTE (Racer Chris @ Jan 9 2006, 04:53 AM)

QUOTE (jluetjen @ Jan 8 2006, 08:52 PM) For the sake of simplicity, let's just assume that each wheel is braking a quarter of the car's mass, If that were the case then brake pad wear would be equal front and rear. Pad wear is around 3-1 favoring the front.

I'm not arguing with the brake wear that you're seeing, but that sounds like an awfully strong front bias for a mid-engine'd car with a fairly low CG like a 914. Of course there are other factors that affect this like ventelation, driver preference, shocks, springing and the whole gammut. But at first blush it doesn't sound like you're getting good use out of your rear brakes. But I digress...

I can easily adjust my bias to lock up the rears first.

Moving along....

Here's pictural output of running the model with the hub and adding a bulkhead about half way down the length of the arm. I have not changed the force vectors yet in order to keep a consistent comparison between the different strategies that we're discussing. I'll post the data after I also run a separate model with a gusset and a floor combined. BTW, I've also updated the spreadsheet to translate the radians into degrees.

Here's the spreadsheet with the result of running the model with Chris's bulkhead included, and then with a floor (horizontal bulkhead) and external gusset both added. This last version seems to be the stiffest, but as a result increases the forces at the outer end of where the trailing arm attaches to the smaller tube -- where Chris has already attached some additional gussets. I'll explain the "short" and "long" gussets in the next reply.

Here's the distortion top view and stresses for the model with an additional gusset as well as a "floor" which is essentially the gusset continued through to the other side of the arm. When I first made this update, I didn't include the portion next to the red line and was surprised to find that the results were marginally worse then the earlier example I had run with a gusset. Comparing the two models I noticed that I had stopped the gusset short of the full length. This is the "Short" gussetted model. When I added the gussetting next to the red line, the performance improved significantly -- and back to what I was expecting. I guess it just goes so show how sensitive some "small" changes in can be in relation to the model.

Here's a view of the same model from outside of the arm, so picture the hub is pointing towards your left shoulder. Note that the forces are now being channelled down the lower outside edge of the trailing arm, in addition to along the length of the gusset on the backside (in this picture).

Oops! Here's the attachment.

So after all of this clever stuff, is there a conclusion as to the best and lightest way to reinforce trailing arms? I have them sitting on the floor of my shed ready to start cutting and welding.

I agree. I love watching you guys talk teck but when do you say "yes it works" and "it is lighter".

And if you wouldn't mind...a picture of what you put inside the arm would be handy. I get the bar on the leading edge. and that the outer triangle may have added strength to the corner but nothing measurable to the flex. But I seem to be missing what you put inside.