|05-07-2007 11:47 AM|
FEA actually approximates a structure by filling it with triangular elements and calculating the effect at each corner and in the middle of each straight line, TET-10 is a preferred element, so-called because it uses tetrahedrons and computes at 10 points (6 sides 4 corners).
Engineers like triangles because you can consider the sides to be in tension or compression andsuch effects tend to be an order of magnitude more significant than the bending moments seen at the corners.
Take a look at any tube frame, such as for an aircraft fuselage or a race car, and you will tend to find triangulation. Tubes can be of smaller section when they can transmit loads by tension or compression instead of bending.
None of which really speaks much to the question at hand. The thing that can't reliably or easily be simulated by FEA is the effect of welding in something to close off a C-channel section. Weld tends to be a stress riser, especially when it can't be properly normalized. Welding tends to create an area that has been heated resulting in a higher local hardness, and higher local strength, and at the edge of the heat effected zone is where you are most likely to find fracture first (on good welds).
Which means what? As far as the question at hand, probably nothing again...
In practical terms, though, there is a problem with boxing a c-channel traingularily (if that's a word)... It's an acute corner, and that isn't usually very good for getting the end of a wire to where one piece of metal touches the other. You're more likely to create a weld that is like bridging an air gap, and that means a non-optimal weld... One thing you don't want is transmitting load through weld, you want welding to be the joining of metal from one piece to the other by melting them together.
You could do this by cutting off the unused bottom flange, but now it's more work. And a sharp cormer to hit your head on.
|05-03-2007 02:07 PM|
|05-03-2007 01:45 PM|
|chieftain||What did you use to box the frame? 1/8" plate steel?|
|05-03-2007 07:34 AM|
|deuce_454||you wont save more than 5-10 pounds by swiss-cheesing the frame.. and im sure you will accumulate that weight in road crud in the frame prety quickly.. also water wont be able to run back out.. ive attached some pics of what i did on my malibu, it added about 20 pounds and made a world of difference|
|05-03-2007 05:05 AM|
|05-02-2007 11:03 AM|
|chieftain||This is a good thread for me right now. I removed the rusted old x-frame and left with open c-channels. The front section is all boxed in and middle to rear is open 2 x 5 c-channel and towards the back 2 x 3". I'm going to box mine in and like you i'm concerned about the extra weight. My solution along the 2 x 5 section anyway is to make a nice row of 3" or so holes like a swiss cheese frame. After it's boxed I was going to add in a new trans, rear shock crossmember and another crossmember like one for a ladder bar in front of my front leaf spring mounts and add in a 2x2" ? square tubing at a 45 degree from the ladder bar crossmember to the boxed in siderails.|
|05-02-2007 10:11 AM|
|bet on black||
maybe what he is trying to say is overall flex in the frame, not just one rail of it. while i think we can all agree that a boxed rail would be stronger and less prone to flex than an open c-channel, the cross members may still flex and let the two channels no longer be parallel in side view.
it seems like there are two different situations being discussed here. i don't have any way to prove it, but a triangulated cross member would have to be less prone to flex as a simple square one would, right?
|05-01-2007 01:43 PM|
While I was looking around trying to find an answer before opening this thread, I ran across something that looks promising. The trouble is that some of it looks like math and some of it looks like some kind of programming language and almost all of it flitters by right over my head:
|05-01-2007 08:25 AM|
How long do you need it to last? I would venture that either design will outlast all of us.
Metal fatigue doesn't usually occur in structures that are never subjected to stresses that are less than 1/2 their yield point. I would think that the open leg of a channel would be subject to developing stress risers at the edge way before the same channel that was boxed. Even figuring in a crappy weld. Also, the opposite corner would be sharper of an angle, and would have less ability to resist development of stress risers than a 90 degree corner that you'd have with a fully boxed beam.
And making joints at an angled side of a beam would make a fabrication nightmare, Although there would be more of an interface area to spread the stresses out.
Do you have any examples of structures that use a triangular cross section type beam instead of a rectangle or square?
I'd think if it was stronger it would be in much greater use today.
I think many times metal fatigue occurs because it was crappy metal to begin with, and joint design and misapplication of the end use causes failure.
It's easy to say that metal fatigue was the cause of failure, when you've exceeded the design limits of the frame many times over by putting more weight, power and subjecting the frame to higher stresses than it was ever intended to resist..
BTW, I misunderstood your first post. I thought you were talking about the frame structure as a whole, not just the beam.
|05-01-2007 05:15 AM|
|BillyShope||Your list of "what if's" are the sort which require "finite element analysis" and, now that I'm retired, I don't have access to the software with that capability. I modeled with sharp corners, which certainly would be avoided. But, I believe I can safely say that, after the FEA and after every effort is made to remove stress concentrations, a structural engineer would still go with the fully boxed configuration. But, again, if you're looking for space to neatly tuck away fuel lines, for instance, the triangulation would be quite adequate.|
|05-01-2007 12:09 AM|
Does that calculation take into consideration the flexing at corners or is it based upon an ideal, theoretical, unflexing shape? I would be surprised if that CAD software was doing anything more than plugging numbers into the formulas and spitting out the results. Those results don't cover everything to be considered.
I just don't think that polar moment of inertia is the whole picture. That only tells you how much the object resists. It doesn't tell what happens to the object if it is subjected to continuously varying torsion over a long period of time. It doesn't tell what happens to the steel along the edges where the planes meet after that boxed frame has been bounced down the highway a few years.
It looks to me like the greatest stress will occur at the corners. A box can flex at the corners at levels of stress far below the failure point. Isn't that what causes metal fatigue? Especially with sharp corners, isn't boxing setting the frame up to fail from little cracks, sometime down the road?
|04-30-2007 04:00 PM|
|BillyShope||grouch, my CAD software calculates the polar inertia for me, so all I had to do was describe a "for instance." With a 4 X 2 "C" fully boxed (0.090 walls), the polar inertia is only about 10% greater than with the triangulation. So, if you have some need to triangulate a section, there's not a whole lotta difference. Note, however, that the fully boxed is the better choice. And, it's lighter.|
|04-30-2007 12:25 PM|
Yes, loading is shared among A, B, C. The bottom flange, A' is no stronger than before C was added.
Imagine clamping the frame in a vise at about mid-way along one side and then bouncing the opposite side up and down. Wouldn't a rectangular box shape distort more readily than a triangular one? Is this the type of flexing that boxing is aimed at?
Clamp it at one suspension mount point and bounce the frame at the diagonally opposite corner. Would the triangular cross-section resist that flexing more than a rectangular cross-section?
|04-30-2007 11:50 AM|
I'm no engineer but I think that your new web piece (call it C) would be transferring loads from A to B just as it does when it is not boxed by B'. In the boxed situation the loads on A are shared by B and B'.
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|04-30-2007 11:15 AM|
Ok, given a C channel of A, B, A' dimensions, typical boxing closes the open side, to yield A, B, A', B'.
If, instead of that 2nd B side, you put in a web from one corner, say the B-A corner to the upper outside, it would be sqrt(A^2 + B^2) wide. That triangular tube would be weaker than the ABA'B' tube? The shape doesn't figure into its strength?
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