Remember young whipper snappers, Measure twice cut once, or
Its done when I say its done.
Or Shut the hell up I am thinking.

All us back yard guys are just amazed at what we are seeingn here. My brain is hurting right now.

Botomline is it what it is until we test it and proclaim it is better. But we really want to know why it is better.

Erik Madsen

You may have to "see" some things, you wont get blueprints to use. There is enough quality info and R&D here.

Chris knows what he is doing, John the nOOb is a whiz at FEA analysis....

I think people should send core arms to Chris to get the secret, "whats inside"
and I have to be honest here. More important than the stiffening, is the camber.
I'm thinking that the camber (-negative I mean) is the genius break through here!
good job chris!

brant

Ok. I think I have a better handle on the mathematics of the forces at play here.
For simplicity I reduced the problem to two dimensions.
Imagine a vertical plane, perpendicular to the direction of travel, containing the center point of the hub and the center point of the tire contact patch. Another important point is where the trailing arm axis intersects the plane. The contact patch is 7" outboard and 12" below the axis point. Two simultaneous forces occur during cornering, weight and cornering friction. The forces act on the axis point through the contact patch. At the contact patch the weight may reach about 1000lbs due to weight transfer. The cornering friction may be up to 1.5 times this amount - 1500lbs. Both forces can be split into two components - one pointing directly at the axis and the other perpendicular to the first - creating torque. The force components aiming directly at the axis can be further split into two components - horizontal and vertical - all pointing directly at the axis.
The torque components of both forces have opposite direction so one partially cancels the other, leaving a net torque of 900 ft-lbs. The sum of the two vertical components works out to about 1400lbs. The sum of the two horizontal components works out to about 820 lbs. That's more than I expected for the lateral force.

These numbers are approximately correct for a 9.5" slick on 7" rims. A narrower track width and taller tire will tend to increase the torque and decrease the lateral load, but lower grip will result in all the values being smaller anyway.

I came up with a simple way to fixture a trailing arm to my bench so I can test them using combined rotational and lateral loads.

OK Folks. And now the moment that you've been waiting for -- Updated models with the forces moved out to the area of the contact patch. Some conclusions...

1) For the most part all of the displacements (aka: bending) got worse. The only exception was...
2) The model with the external gusset and internal floor. This model's output stayed the same (I double checked it too) and as a result was the best performing. Looking at the data, it even suggests that this model may channel the forces in such a way as to twist around the X-axis positively (increasing camber) by a small amount as opposed to twisting negatively.
3) This stuff is getting pretty computationally demanding. The base model runs almost instantaniously. The model with the bulkhead took about 1 minute, the model with the gusset kept my computer busy for 2 minutes, while the model with the bulkhead and floor took 3:48. Imagine doing that by hand!

Here's the data.

Here's how I set the models up. I used the geometry for a 3x3 inch solid steel beam, straight out from point 284 for 3 inches (a total of 7 inches from the X-axis), and 12 inches down. At that point (node 308 in this model) I applied the same 1000 lb forces as before (in order to remain consistent with the earlier data). So -1000 lbs X, -1000 lbs Y, and 1000 lbs Z.

Here's the top view of the base model.

Here's the base model with the vertical bulkhead.

Here's the version with just the external gusset

And finally the model with the external gusset and the floor added.

sigh....

Now it's Chris's turn...

I'm going to graphically show how I arrived at the force numbers I posted above.
Point A represents a point along the twisting (longitudinal) axis of the trailing arm. Point B represents the center of the tire contact patch. I estimated the center of the tire to be offset 7" outboard of the axis, and 12" below. The distance between the points is 13.8". The ground is pushing up on the tire with a force of about 1000lb. At 1.5G the sideways force exerted on the tire contact patch is about 1500lb.

Working with just the sideways force for a moment, I can separate that into two vector components. One points directly at point A and the other is perpendicular to the line segment A-B.

Doing the same thing for the vertical force yields this graph. Notice the perpendicular component points the opposite direction from the previous graph.

Now both sets of force vectors are shown simultaneously.

Performing simple vector addition yields this graph.
A force of 1600 lb is aimed directly at Point A, and a force of 790 lb is perpendicular - twisting the trailing arm.

Working only with the force vector directed at point A I can further split that into a vertical and a horizontal component. Since that force is aimed directly at A I moved the origin to point A.

Since the perpendicular force vector is 13.8" from Point A it exerts a torque of 910 ft-lbs.

Now graphing everything simultaneously it can be seen that the two forces acting on the tire contact patch can be studied as three forces acting directly on the trailing arm at a point along the longitudinal twisting axis.
1) a vertical force pushing up on the trailing arm of 1390 lb.
2) a horizontal force of 810 lb trying to bend the trailing arm laterally.
3) a torque of 910 ft-lb trying to twist the trailing arm.

Any questions?

Wow -- I started having flashbacks to high-school math!

One question now that you've done all the math....

By just moving the application of the forces out to the contact patch like I did in the last round of models (based on your input -- thank you!) did I do essentially the same thing? In a couple of days I'll try modeling your forces directly on the arm and see if it results in similar outputs to what I go by moving the forces.

(BTW -- I've got no internet until tomorrow some time thanks to a tree that fell over yesterday morning in the high winds. On the way to the ground the tree also snapped off a utility pole at ground level and yanked all of the wires to our house. )

QUOTE (Racer Chris @ Jan 16 2006, 07:39 AM)

Any questions?

when do you get started for external customers?

QUOTE (jluetjen @ Jan 16 2006, 04:38 PM)

By just moving the application of the forces out to the contact patch like I did in the last round of models (based on your input -- thank you!) did I do essentially the same thing?

You moved the force application point outboard too far. You added 7" from point 284 but it should be 7" from the approximate centerline of the box. Otherwise you did the same thing.

My best estimate is sometime in the spring. I still need a proper fixture to hold the trailing arms.

Once I actually sat down and started working on the problem it turned out to be fairly simple. I think, even adding the 3rd dimension wouldn't be too complicated, but a bit tough on a sheet of graph paper.
That explains why my email to you was kicked back this morning.
I hope you at least have electricity tonight.

Can someone please resize the images?

QUOTE (Howard R @ Jan 10 2006, 09:05 PM)

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.

We're not done quite yet. I'm pretty happy with the results of my 3 piece reinforcement but the tests aren't complete. Hang in there.

Chris is pretty close on this. I thought about it while I applied Dry-Lok to the basement walls today.

My crude treatment has a twisting force of 1142 ft-lb, not counting any sway bar, and without constraint from the shock absorber laterally. This is using the 13.8" and 7", with the 1500# side load and 1000" vertical load. I did not use any of the acute angle deviations from the normal axes, as would be encountered in a real trailing arm, but I believe were incorporated in the Chris and John stuff.

If I could upload my scratch paper from the scanner, I'd post it.

Chris and John did a very masterful job on this real world engineering problem, in case anyone had any doubts.

Pat

Comcast came to the house and we're back in business today.

OK. I've gone back and updated my earlier postings with revised outputs reflecting the contact patch being only about 3 inches outside of node 284. The results are similar to the earlier outputs. The big difference is that by reducing the distance out from the center from 11 inches to 7, we actually increased the twisting (torque force) some. Playing with Chris's vector math will show why.

Thanks John!
Looking at the data -
It appears that the Y displacements are all around 20 thousandths of an inch. The corresponding rotation is 25 degrees? If you measured the angular change of the surface at node 284 (X direction in the X-Z plane) I think it would be quite small.
The Z displacements are all around 5 thousandths with a corresponding rotation of 0.5 degrees. At least that is in the same order of magnitude as the real world experiments. Do these rotations correspond with the angular change (Z direction in the X-Z plane) of the surface at node 284?
The angular change of this surface is what should closely follow my data if the model is believable.
I am surprised (disappointed) that the model doesn't reflect the improvement I see from adding the bulkhead. There must be something going on in the real world that the model doesn't reflect.
It looks like there are some nodes near the outer pickup point that have very high stress levels. This corresponds with the earlier comment about cracks in the pivot tube near this location.

Chris; Good catch about the magnitude of the y-rotation. It's not real clear to me how that manifests itself in the real world short of the hub trying to twist itself off of the arm under braking. As I mentioned, I really haven't modelled the hub that accurately since it should be modeled as a solid piece of metal as opposed to being fabricated sheet metal. This definitely gives me an appreciation for the stress that a hub has to deal with.

Here's a couple of different views of the 3 primary models -- the base model, the model with the vertical bulkhead and the model with the gusset and floor. In this case you are looking down the arm. The X and Z rotation is much clearer in the animation rather then just the distortion plot. Does anyone have any ideas on how to capture the "movie" off of my screen and post it?

First the base model

This is the model with the bulkhead.

And the model with the gusset and floor.

Here's a different view that may be helpful to understand the Y-Rotation issue. Note the very high stresses in a star pattern at the hub. I believe that this is the result of node 284 trying to spin (torque around) and thus add a lot of stress to those elements that make up the web in that area.

John,
It is apparent that the "arm" you added to move the forces out to the contact patch also is bending at node 284.
Using the values I computed can you apply the loads directly to node 284, ignoring the braking force. You will have to calculate the appropriate force to equal 900 ft-lbs torque for the vertical (down) load.

Chris;
The software only allows me to add forces in the X, Y and Z direction. There's no place to add torques. Would you agree that as a result I should be adding 1390 lb vertically in addition to (910 * 12 / 7 inches = ) 1560 lbs vertically to equal 2950 lbs vertically? And then the 810 lbs horizontally.

Does this sound right?

Not exactly. The 1390 is in the opposite direction and is not aimed at node 284. I think it could be ignored, since it shouldn't cause any bending or twisting. If included it needs to aim at a point on the twisting axis (imaginary line between the shock absorber mount and somewhere along the pivot shaft) parallel to node 284.
The 910 is foot pounds. If node 284 is only 3 inches from the twisting axis you need to multiply the force by 4 to get the same torque effect.

QUOTE (Racer Chris @ Jan 18 2006, 03:00 PM)

QUOTE (jluetjen @ Jan 18 2006, 02:00 PM) Does this sound right? Not exactly. The 1390 is in the opposite direction and is not aimed at node 284. I think it could be ignored, since it shouldn't cause any bending or twisting. If included it needs to aim at a point on the twisting axis (imaginary line between the shock absorber mount and somewhere along the pivot shaft) parallel to node 284. The 910 is foot pounds. If node 284 is only 3 inches from the twisting axis you need to multiply the force by 4 to get the same torque effect.

Duh! I wasn't thinking right. Let me try again. As far as ignoring the twisting forces -- I'm not there. Maybe I'll stick on a cone so that the point is about 12 inches from the axis and then apply the 910 lb-feet there.

I'm not suggesting ignoring any twisting forces. The 1390lb vertical up is the component that pushes straight into the trailing arm. It is the part of the "weight" that does not exert ANY twisting force. If you want to include it, break it up and apply it at several nodes on the bottom the trailing arm That would eliminate any point loading.
All the twisting force is included in the 910 ft lb. What's wrong with multiplying the torque by the fraction of a foot node 284 is from the axis? This eliminates any extraneous structure that might deform under the load.
My last graph shows all the forces being applied at what I labeled point A (an imaginary point on the longitudinal axis of the trailing arm). You just need to figure out how far node 284 is from the axis.
The horizontal force can be applied at node 284 since that is in a direct line with the axis, as viewed from the side.

Maybe this will help.
Imagine that all three arrows I drew are in the same plane as node 284.
The vertical up arrow needs to be a force of 1390lb.
The horizontal arrow needs to be a force of 810lb.
The vertical down arrow needs to be a force of 3600lb!!! That's right. It is only 3 inches from the centerline of the trailing arm box.

OK. Here you go -- although I fear that we might be going down a "rat hole" with this approach. The reason is that I fear that we're trying to kluge the computations which are already inherent in the model. At the end of the day I don't believe that it really tells us any more, but could very well confuse us more. Kind of like if you wear one watch -- you're always pretty sure what time it is, but if you wear two watches you'll never know what time it is. In this case we'll get another set of numbers, but since they ignore braking and other lateral forces -- of course they'll be different.

The picture below is looking at the arm from behind the shock mount and is a little busier then normal to show what was done.

Forces: Per Chris's diagram above.

Constraints: Both front pivots are constrained in all three directions -- X, Y and Z. So those points won't move. I had to constrain the shock mount in the vertical (Z) direction to keep the model from flying all over the place. I'm kind of concerned about his since this constraint is only a couple of inches away from where we're applying the vertical forces. Essentially I'm concerned that I'm over constraining the model.

Results: The model is still bending along it's length. While I follow Chris's vector math, I don't believe the way that the forces are being applied to the arm accurately reflect the environment the arm is in. For example, the arm never sees a positive Z force at the bottom of the arm -- but our model has such a force. So we're essentially solving a problem that doesn't exist from a mathematical perspective.

I think a better way to cross check the model and help develop a test methodology would be to do the vector math as of point 284. Basically do the math so that the answer results in a single torque being applied at point 284. This answer can then be used to mount a beam at a given angle from X/Y/Z which Chris can then fabricate and test with. I'll try giving it a shot later today.

And one more view that reflects Chris's mark-up above.

Here's the updated data table resized to fit.

I like what I see now. I do still have a couple of minor issues:
1) Can you reduce the displaced skeleton view to 10x instead of 100x. I think that it over-dramatizes the situation currently.
2) I want the outboard pivot location moved closer to node 284. The pivot axis properly should be perpendicular to the inner box surface. The distance of node 284 from the pivot axis will be approximately correct with this change.
3) I'm picking nits now, but - since the arm is mounted to the car with the pivot axis at a 12.5 degreee angle the tube connecting node 284 should be at such an angle, the lateral force should also be at the same angle, and the vertical up force should be moved forward a corresponding amount.
These changes would satisfy me that we have modeled a trailing arm as well as GRAPE can do for us within reason.
The braking force will never be applied simultaneously with the other forces, and is much smaller as well.
It does, but by the time it gets there it is spread out instead of point loaded. How much does the arm bend in this direction? Maybe this should be measured on a real trailing arm.
I'll look at that to see if it is possible, but I don't believe so. I think the three components I derived constitute the easiest way to input the actual forces into the model. (Adding the braking force would definitely complicate things unnecessarily.)

One thing to keep in mind is the shocks will provide no lateral location of the trailing arm. So restraining the shock mount will provide no usable data.

One last point for now, the rotations listed in the table don't accurately reflect the real world. I don't think this means the model isn't accurate, only that the point about which the rotations occur isn't modeled accurately. If possible, it would be better to tabulate the angular change (in 2 axes) of the surface that node 284 is at the center of.
That's why the constraint for this point is only in the Z direction.

QUOTE (Racer Chris @ Jan 19 2006, 06:31 AM)

QUOTE (jluetjen @ Jan 19 2006, 08:43 AM) OK.  Here you go -- I like what I see now. I do still have a couple of minor issues: 1) Can you reduce the displaced skeleton view to 10x instead of 100x. I think that it over-dramatizes the situation currently. 2) I want the outboard pivot location moved closer to node 284. The pivot axis properly should be perpendicular to the inner box surface. The distance of node 284 from the pivot axis will be approximately correct with this change. 3) I'm picking nits now, but - since the arm is mounted to the car with the pivot axis at a 12.5 degreee angle the tube connecting node 284 should be at such an angle, the lateral force should also be at the same angle, and the vertical up force should be moved forward a corresponding amount. These changes would satisfy me that we have modeled a trailing arm as well as GRAPE can do for us within reason.

Chris!!! Now you're "gilding lillies". How is any of this going to change the strategy that you use for modifying the trailing arm?

Duh! My bad, I should have read closer.