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Rhino
05-07-2009, 02:56 PM
When designing a rear suspension, what is the commonly accepted "safe" safety factor. I have a tendency to over build everything I touch, and I'm trying to get out of the habit. I just want to insure I don't take it too far.

Looking at a few threads on CC's I tend to see people talking about a design giving them a "good safety factor" although this never seems to be qualified with a number.

In a design I'm currently working on I estimate a 2 to 2.5x safety factor when comparing my loads to the yield strength of the native steel. I have yet to decide on rod ends, although I would look toward something with a similar safety factor.
Is 2 to 2.5 enough for a street driven vehicle? It seems a little low to me given the repeated compression/tension loads the links will see. By stepping up the tubing size slightly I can get around a 3x safety factor on all links. I feel more comfortable with that, although I then begin wondering if I'm trying to overbuild again.

David Pozzi
05-07-2009, 03:17 PM
Check Race Car Engineering & Mechanics by Van Valkenburgh, it's the only book I've seen that even mentions safety factors, the assumed load is important too.
David

Rhino
05-07-2009, 06:17 PM
Thanks David.
Interestingly enough, I was thinking about ordering that book a few days ago. I may just do so. Using Amazon's search inside feature I only found one reference to safety factor on page 19. Roughly paraphrased it mentions to typically shoot for 3 times the maximum of the other known forces.

By assumed load do you mean the calculated load an individual link may see? If so, that's what I am trying to do at this point. Essentially taking vehicle mass, link placement, and acceleration into account to determine where loads will be placed on the chassis.

Norm Peterson
05-08-2009, 03:36 AM
Here's how parts of a similar conversation held elsewhere went:



(1) I'm wanting to design some control arms for a track day car, but I'm unsure of what to use for the loads for the FEA.

The advice I got was to use 4-3-2 as maximum load multipliers, and to assume the car corners at 1.5G's (higher than the actual).

My problem is that I don't know how to calculate the maximum loads. I know my spring rates, motion ratios, unsprung mass, and static corner weights. But, I don't have the CG height, so I can't calculate steady-state load transfer using any method I know of, unless I make a guestimate as to what it is.

So, what should I do?


(2) What about impact loads? The car is street-driven to and from the track, so I need to make sure it will handle the lovely Michigan roads without breaking. Should the 4-3-2 take care of it?

And, I'm assuming that the 4-3-2 is an actual multiplier...so if, for example, my lateral load is 1000lbs, I'll use 3000lbs in my FEA. This is correct, right?



Well, not quite.

If the corner weight is say 1000 lb, and I was using 5-4-3 then we'd apply 5000 upwards at the CP, 4000 longitudinally, and 3000 laterally.

Then work out the free body diagrams for each arm, for each of the three cases. That will give you the loads for the FEA

In my opinion the wisest thing to do is to add the sum of the squares of the stresses for all 3 directions, and compare the square root of that with the yield stress.

I know some people say you just look at the most stressful case for each arm.



Something else to consider is fatigue, and page 10 of this paper (http://www.nenastran.com/newnoran/conferencePaper2/10_CPNonlinearStaticMulti-AxialFatigueAnalysisAutomotiveLowerControlArmUsing NEiNastran.pdf) gives a little idea what this is all about. The component and loading in that example is more complex than that for stick-axle rear LCAs, although even they also see torsional and bending loads in addition to the axial tension/compression.



Norm

silver69camaro
05-11-2009, 05:58 AM
What Norm posted is typically what I do.

One note I make: reduced confidence in assumed or unknown loads require higher SF figures.

Rhino
05-11-2009, 07:16 AM
Is there a method behind deciding on each component force? Why arbitrarily pick values of 5-4-3? My thought is that it would be more accurate to actually calculate the force with known weight, acceleration, and link geometry.
Does the 5-4-3 work out "close enough" for most scenarios?

Please don't take this as sounding argumentative, I'm simply trying to insure I have a full understanding of it.

Norm Peterson
05-11-2009, 07:49 AM
The problem is that you really don't know those things anywhere near that precisely. Even if you did know things like corner weights, load transfers, and tire grip down to the single-digit force numbers - as soon as you hit a bump of any sort (or hit a curb) that all goes out the window with the unknown vertical/lateral/longitudinal "g" spikes. Same thing with respect to how well the actual fabrication matches the analysis. Stress concentration effects come to mind here, and these are highly dependent on local geometry (holes for bolts or other components, weld undercuts, notches, kinked bends, etc.) and this is also something you need to at least be aware of. You also have to have a good feeling about fatigue resistance, which is even more difficult to put a solid number on. This is just good engineering practice.

Factors of safety (aka margins) exist to cover for the uncertainties that exist in the design basis "numbers" and to provide adequate risk protection for the intended use.

If you had a Formula I size budget, complete with datalogging of the specific tracks involved and where there is essentially zero risk of encountering a completely uninvolved motorist/pedestrian/animal/stray utility pole/etc., you could certainly design much lighter pieces than you would for street driving or rock-crawling.




Norm

Rhino
05-11-2009, 10:45 AM
Thanks Norm. You bring up a lot of valid points. The real world is much less cut and dry than it appears on paper.

For some reason the forum didn't take an edit I made of that post. Strangely enough, you answered most of my additional questions.
The only thing you didn't touch was in regards to the theoretical 1000# corner weight example. Is that corner weight including weight transfer through a turn?

silver69camaro
05-12-2009, 05:26 AM
The only thing you didn't touch was in regards to the theoretical 1000# corner weight example. Is that corner weight including weight transfer through a turn?


I know the question was directed to Norm, but 1000lb corner weight is typical for say, a 4000lb vehicle with 50/50 distribution and no added weight transfer. Or a 3600lb car with 55% on the nose.

For a quickie example, in our Camaro a 1G negative acceleration transfers about 750lbs on the front end. And a 1G corner puts roughly 1500lbs on the outside tires, and as little as 200lbs on the inside. At rest, there is about 800lbs on each corner.

Rhino
05-12-2009, 09:03 AM
Thanks Matt. I wasn't sure if we were talking about a light weight car with weight transfer or simply static weight.
My assumption is that you would base the corner load on the maximum weight that particular corner would ever see. (Due to weight transfer in corner or straight light acceleration)

In your example, we'd be using 1500 lbs for that corner, rather than the 800 lbs static.
Using a 5-4-3 ratio we'd determine that to be 7500 vertical, 6000 longitudinal, and 4500 lateral forces. The total force would be roughly 14,300 lbs. (square root of squares)

Now taking this into consideration...



Well, not quite.

If the corner weight is say 1000 lb, and I was using 5-4-3 then we'd apply 5000 upwards at the CP, 4000 longitudinally, and 3000 laterally.

Then work out the free body diagrams for each arm, for each of the three cases. That will give you the loads for the FEA

In my opinion the wisest thing to do is to add the sum of the squares of the stresses for all 3 directions, and compare the square root of that with the yield stress.

I know some people say you just look at the most stressful case for each arm.


This is where I get confused. If I do what is requested later in the post (using the square root of the forces^2) I only have one force to compare to. I'm not sure how to do this in a free body diagram for each arm. The OP mentions comparing this force with the yield strength of the material, although I'm not certain why this doesn't take multiple links into consideration.
In addition to this all, shouldn't I also be worried about buckling a link through a side load? not simply the compression/tension the link is under? (Triangulated 4 link)

If I take this all as a given, using Matt's example, we'd be using 1500 lbs for this particular corner of the car.
Using a 5-4-3 ratio we'd determine that to be 7500 vertical, 6000 longitudinal, and 4500 lateral forces. The component force would be roughly 14,300 lbs. (square root of squares)

Yield strength seems to be a unit of pressure. Knowing this we need to know the area we're working with in addition to the force.
To do that I'd need to find the cross sectional area of my link and calculate the psi load.

Lets say I'm using 1.25 DOM .250 wall tubing steel with a yield strength of 50,000 psi. I'd end up with a .785 in^2 cross sectional area giving us a yield force of 39,250 lbs.

Comparing the two (39,250/14,300) I come up with a safety factor of 2.7. That seems to be much smaller than I thought it would be.
Does this sound remotely correct? Should I be taking into account any side loading or number of links?

I do appreciate everyones help. I am learning a great deal about how this truly should be done.

Norm Peterson
05-14-2009, 07:50 AM
Using a 5-4-3 ratio we'd determine that to be 7500 vertical, 6000 longitudinal, and 4500 lateral forces. The total force would be roughly 14,300 lbs. (square root of squares)
In general, I'd analyze separately for the 7500 vertical, 6000 longitudinal, and 4500 lateral loads and SRSS the results rather than SRSS'ing the input forces.



The OP mentions comparing this force with the yield strength of the material, although I'm not certain why this doesn't take multiple links into consideration.
I suspect that the linked discussion was specifically concerned with A-arms. You'd do separate free-body diagrams, analyses, and comparisons against yield for each, given how the loads at the contact patch divide at the pivots on the control arms where they are first reacted.

With a stick axle you probably need to look at the whole business including all four links/control arms and the loads at both wheels, as the loads in any given link depends on the loads at both contact patches (which would not necessarily be the same).



In addition to this all, shouldn't I also be worried about buckling a link through a side load? not simply the compression/tension the link is under? (Triangulated 4 link)Strictly speaking, yes. Pure compression can and will buckle a slender cross section. I'm not sure what sort of side loads you're thinking of unless it's something that occurs when the control arm end pivots run out of angular travel (or similarly if your spring mounts to the middle of the control arm a la Fox/SN95 Mustangs). Depending on the cross section itself the buckling limit may be a non-issue due to the control arm lengths being sufficiently short and the walls either sufficiently thick or in the case of any freestanding edges that they are adequately braced (such as by a sturdy "lip").



Yield strength seems to be a unit of pressure. Knowing this we need to know the area we're working with in addition to the force.
To do that I'd need to find the cross sectional area of my link and calculate the psi load.The units look like "pressure", and I guess it could be looked at in that manner (unseen pressure inside the metal). But generally, the term "stress" is used in structural or mechanical engineering when the discussion concerns the magnitude and distribution of forces that are internal to a structure or mechanical component to distinguish it from things like contained fluid pressure.

If you can develop much torsion in the links, such as if the OE bushings have been replaced with poly or harder material and the car is capable of developing lots of roll (think dragstrip here, with a torqued-over wheelstanding launch and very low rear roll stiffness), the combined stress calculations can get rather messy. The open-section OE control arms that work well under most conditions lose their appeal when this is the case.



I hope I haven't either made things more confusing or made it sound too much like a specific procedure (it's not intended as such). FWIW, I'd probably throw everything at a space frame analysis software and let it crunch the numbers for a variety of specific conditions.


Norm

Rhino
05-15-2009, 06:58 AM
In general, I'd analyze separately for the 7500 vertical, 6000 longitudinal, and 4500 lateral loads and SRSS the results rather than SRSS'ing the input forces.

I suspect that the linked discussion was specifically concerned with A-arms. You'd do separate free-body diagrams, analyses, and comparisons against yield for each, given how the loads at the contact patch divide at the pivots on the control arms where they are first reacted.

To me it makes much more sense to work out each separately first. I believe that's where most of my confusion came from. You're also correct that the original post was about A-arms. I did do a little more searching and ran across that thread after you referenced it. I wasn't thinking in terms of the A-arm, as the original post was intended.


Strictly speaking, yes. Pure compression can and will buckle a slender cross section. I'm not sure what sort of side loads you're thinking of unless it's something that occurs when the control arm end pivots run out of angular travel (or similarly if your spring mounts to the middle of the control arm a la Fox/SN95 Mustangs). Depending on the cross section itself the buckling limit may be a non-issue due to the control arm lengths being sufficiently short and the walls either sufficiently thick or in the case of any freestanding edges that they are adequately braced (such as by a sturdy "lip").

I guess my original thought process was that the force being applied to a triangulated link during acceleration wasn't parallel to the link. After working out a few of the numbers, that really shouldn't matter as much as I originally thought.
My second reason for the question was buckling due to a bind force. However, with the small travel and moderate loads we're working with, I highly doubt this would be the case on any properly built vehicle.


The units look like "pressure", and I guess it could be looked at in that manner (unseen pressure inside the metal). But generally, the term "stress" is used in structural or mechanical engineering when the discussion concerns the magnitude and distribution of forces that are internal to a structure or mechanical component to distinguish it from things like contained fluid pressure.

Absolutely, stress is a much better term, and not quite as confusing.


I hope I haven't either made things more confusing or made it sound too much like a specific procedure (it's not intended as such). FWIW, I'd probably throw everything at a space frame analysis software and let it crunch the numbers for a variety of specific conditions.

Norm

Not at all Norm, I do completely understand where you're coming from and appreciate the input.

I've been working with a link calculator I found online and am trying to verify it's accuracy. When working with the forces on each arm, I realize that (when accelerating) the lower links should be in compression, where the uppers will be in tension. What didn't make sense, however, is the distribution of these forces.
I would have thought that the loads would have been somewhat equal, given similar vertical separations from the axle center line.
In working with various geometries I've found that my top links seem to always end up with a much smaller load. It all begins to make sense when you realize that the torque is centered at the contact patch, and not the axle center line.

I think I have a good grasp of the reasonings behind the calculations, now I just need to work it all out. Would anyone have a good source on how to correctly create the free body diagram? I understand it's need, usage, and where the forces are transmitted through, but don't have a good enough grasp to do it without further research.

Norm Peterson
05-15-2009, 12:55 PM
I guess my original thought process was that the force being applied to a triangulated link during acceleration wasn't parallel to the link. After working out a few of the numbers, that really shouldn't matter as much as I originally thought.
My second reason for the question was buckling due to a bind force. However, with the small travel and moderate loads we're working with, I highly doubt this would be the case on any properly built vehicle.
As long as both end connections of a given link are at least a decent approximations of frictionless spherical pivots, the force will remain essentially axial along that link. It's only when you start developing end moments (such as with polyurethane bushings) or connect a the spring, shock, or sta-bar to the link that you start adding bending. Technically, when you do that it isn't really a link any more, but that's neither here nor there in this discussion.

Incidentally, a pure fore/aft load such as you'd get from acceleration will develop link loads in accordance with the angle(s) that the links are installed at relative to the purely longitudinal. IOW, a link that's horizontal in side view and has 45 plan view skew will develop 1414 lbs axially in resisting a 1000 lb fore/aft load at its axle bracket. Geometry matters a lot, and not just for roll center, anti-squat, and roll steer discussions. Similarly for lateral loadings.



I've been working with a link calculator I found online and am trying to verify it's accuracy. When working with the forces on each arm, I realize that (when accelerating) the lower links should be in compression, where the uppers will be in tension. What didn't make sense, however, is the distribution of these forces.
I would have thought that the loads would have been somewhat equal, given similar vertical separations from the axle center line.
In working with various geometries I've found that my top links seem to always end up with a much smaller load. It all begins to make sense when you realize that the torque is centered at the contact patch, and not the axle center line.
Don't feel too bad about it. Lots of people, including some who have pretty decent skills as fabricators, have made the same sort of assumption.


I think I have a good grasp of the reasonings behind the calculations, now I just need to work it all out. Would anyone have a good source on how to correctly create the free body diagram? I understand it's need, usage, and where the forces are transmitted through, but don't have a good enough grasp to do it without further research.
I guess any decent college civil/structural statics class book would have what you're looking for, or perhaps there is a for-real "Cliff's Notes" for it. But basically, you draw the component that you're investigating, draw the loads that are applied to it in the appropriate directions, and replace all of the supports for that component with the forces and moments that those supports provide (in their appropriate directions). You may have to draw more than one view to see how it all goes together.

You'll likely struggle with that a bit until you've actually gone through it a few times. I have no illusion that 75 words or so from my keyboard can substitute for I-forget-how-many hours of classroom time.


Norm

modern-muscle
05-16-2009, 05:07 AM
Wow it's really refreshing to see an educated discussion about Safety on a forum that is car related.

I typically use a 4X rule for an average safety factor. In my experience most people don't care to know what goes into the design as long as it works as designed. I prefer to build to handle the X factor as well (What if factor) and design things with typical road variations and accident spec's. I'd rather have to replace sheet metal then expensive suspension components in the event of another driver not paying attention to where my car is.

mikedc
05-23-2009, 10:18 PM
I can understand the reasoning to wanna lighten up the unsprung suspension components in theory. But in practice I don't think it really earns its incurred risk a lot of the time.

IMHO just looking at other race cars with similar layouts & stresses is more valuable than any amount of paperwork. You can probably eyeball the minimum thicknesses/materials for the larger components like the A-arms and stuff just based on other racing vehicles of the type.



Tire grip is important, but just the overall vehicle weight + the potential stresses of the speeds it could attain are a bigger part of the story IMHO.

Like the other answer said, just whacking a wheel into a curb will blow all your finite element work to hell. And the governing factors for the potential forces in these cases are more about curb weight + MPH than anything else. If it can withstand the rigors of the speed and occasional rough pavements, then the traction stresses incurred by the specifics of the car should be more than covered already.




The demands on the smaller parts can be more variable. But I think the gains of trying to squeeze it down are generally not as worthwhile. (Spindles? Bearings? Half the threads on these kinds of subjects are more about struggling to strengthen these parts than lighten them.)

Norm Peterson
05-24-2009, 03:59 AM
I think this topic is more about what should be considered when you're replacing the OE suspension arrangement with something entirely different and which may not even exist in the aftermarket. Sure, dimensions from a similar component in a similar weight car represent good starting points - they may even end up being the final solution for that particular item.

But then you need to consider what it will take in terms of chassis and axle attachment details, which for a "one-off" you almost certainly don't have an adequately close example to copy. See the Satchell Link thread elsewhere in this forum. Or you may have something that's good enough up to some limits - just off the top of my head I know of one specific aftermarket suspension kit that's fine in and of itself but for which the local chassis structure in the car that it was specifically designed to be installed in becomes suspect at very high power levels and dragstrip use. One needs to find an appropriate balance between Colin Chapman's practice ("add lightness/build a race car that's just barely strong enough to make it to the finish line") and building Brooklyn Bridges here. 16 gauge steel strip reinforcements probably wouldn't be enough, but I really doubt that you'd need 1/2" plate either. That's where some reasonable safety factors come in, not as something to fine-tune so you can cut the material thickness down by one more gauge.


Norm

BillyShope
10-09-2009, 05:23 AM
When I go back to the safety factors in the machine design text I used as a student and then look at some of these suspension kits offered to the dragracers, I wonder why there aren't a lot more racing failures. The tubing used in some of the 4link kits, in particular, remind me of something Colin Chapman might have designed into one of his Lotuses (Loti?).

Which brings me to the point I would make: You're NOT designing a Lotus!! Keep that foremost in your mind. The difference in overall car weight is not all that great between the "what everyone else is running" size and something with double or triple the safety factor.

Always err on the side of safety and make up the difference by joining Weight Watchers.
http://www.racetec.cc/shope

pdq67
11-27-2009, 06:26 PM
I need to look over at the University of Missouri, Columbia here in town and see if the old ME Automotive Design class I took way back in the early '70's has a text book b/c back then the Prof. was writing his own!

He showed us how to use computer cards to figure all this out, talk about slide rule/computer stone age back then.

pdq67