Help! Broken front sway stud.....

Please help! What part do I need to buy and get this fixed? Or should I find a shop cut the nut and replace with a nut and bolt?

 

http://img.photobucket.com/albums/v66/ewg751/DSC00611.jpg

 

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drill it out and replace it with a nut and bolt, or have someone re-weld a stud on

so I cut the stud, and then replaced it with nut and bolt. it should hold i hope.

 

http://img.photobucket.com/albums/v66/ewg751/DSC00613.jpg

 

http://img.photobucket.com/albums/v66/ewg751/DSC00614.jpg

 

http://img.photobucket.com/albums/v66/ewg751/DSC00617.jpg

 

http://img.photobucket.com/albums/v66/ewg751/DSC00619.jpg

What size bolt did you use? (not length but diameter)

Length might also be important with respect to clearance to the rack boot under certain combinations of steering rack travel and suspension jounce/rebound.

 

 

Norm

Same crap happened to me Looks like I'll be drilling it out and replacing it

GTTuner: The diameter is about 1mm smaller than the stock stud.

 

Norm: I agree...a little too long, so I went picked up a nice bolt from Home depot.

 

Peter: have fun Getting under the car is tiring...unless you have a lift.

 

Thanks for the advices guys!

Any idea what's causing these things to fail? I can't imagine there being enough bending going on, but overtorquing the nut remains a possibility.

 

 

 

. . . unless you have a lift

you would have to go ahead and say that four-letter "L" word. The Mustang gets front struts, pads, rotors, and adjustable sta-bar today (rear stuff got done yesterday). Couldn't find a "bent over crippled" smiley, so try to imagine one . . .

 

 

Norm

what should we be torquing them to? I was cranking those on as hard as I could....with only a 3/8" socket though...

18.1 ft lbs

I did the exact same thing... i was using a Craftsman cordless Impact wrench and that stud just twisted right off... called up my subaru dealer, and they had nothing except a whole crossmember to the tune of about $400. So i got out the dremel and cut the stud flush, drilled through and replaced it with grade 8 hardware i got from lowes... Now it won't happen again, and I can really torque that sucker down!

There's no reason to torque it down. The sway bar twists in the mount. It doesn't have pull or anything on it.

^

I agree....mine came off because I torqued it. BIG MISTAKE! Just tighten it when you feel it is tight enough.

There's no reason to torque it down. The sway bar twists in the mount. It doesn't have pull or anything on it.

It's mostly free to turn in the D-block - there is going to be a little friction but that's not the problem.

 

You still need to balance the endlink forces applied to the bar arm ends, which are trying to pull the D-block and its bracket off of the car on the side of the car that's on the inside of the turn. The tensile force on each D-block is greater than the endlink force, because the D-blocks are spaced more closely than the endlinks are. Each stud takes ~half of that.

 

There is probably a little shear load as well, particularly if the endlinks are not exactly perpendicular to the bar arms.

 

You need enough torque (clamping force) to deal with these forces without being so large that poo happens to the studs.

 

 

Norm

All the forces put on the swaybar try to twist it. There are no significant forces trying to pull it out of it's shackles

Let's say the bar has a rate at each end of 300 lb/in and one end is being pushed up 1" and the other end down 1", such as what happens when the car has already rolled a couple of degrees or so in a corner and is holding steady at that roll angle.

 

One endlink is pushing up on its bar end with 300 lbs, and the other endlink is pulling down with 300 lbs.

 

Let's make it easy and say that the bar ends are 42" apart and the D-blocks are 28" on centers. There is a couple generated by the endlink forces of 300 * 42 = 12600 in-lb. It is being resisted by an equal and opposite couple whose forces are at the D-blocks. The D-block forces are plus and minus 12600 / 28 = 450 lbs. That's 225 lbs of straight pullout force per stud on the tension side and 450 lbs of compression against the frame on the other side. There is no other answer.

 

Stop and think. How could a stabilizer bar possibly limit body roll if it wasn't pushing up on one side of the car chassis and pulling down on it over on the other?

 

 

I've been paid to do structural analysis of one sort or another since 1970, so I think I can figure out a structural configuration as simple as this one is. I will listen if you can describe in detail how the 300 lb endlink forces in this example do not need to be balanced at the chassis side D-blocks, but just saying that they don't isn't going to cut it. If you don't like my simple numbers, I can toss this little problem at some structural analysis software. Or maybe I can extract similar results from a sta-bar analysis that I have already done that way.

 

What I will give you is that during pure two wheel bump mode, both endlinks are pushing or pulling in the same direction, i.e. both up or both down. In that case, there is no significant pullout occurring (there will be a tiny bit that comes from bushing friction/stiction against torsional rotation and endlink bushing/rod end resistance to movement).

 

 

Norm

re analyze it then.

 

the endlinks do see lots of torque up and down... then through the shape of the bar and the way it's held in place, that force is transfered into a twist through the straight section of the bar. there is very little force other than twist where the two mounts are that hold the bar in place.

 

you can sit here and try and argue this all day, but if the mounts had any significant force on them, the rubber bushings would be getting ripped out all the time. They don't. Also, the mounts would be a bit beefier.

Sorry for the delay, but my work computer denies me access to LegacyGT.com and the FEA softwares that I have access to won't run in my laptop's Vista. So please bear with me.

 

 

Endlink forces do not simply "go away" because the bar is bent into a "U" shape. They do cause the center section to undergo torsion, but the torsion does not replace the vertical forces. It acts in conjunction with them instead, and you end up with a combination of shear, torsion, and bending. Endlinks do not normally see torque or much bending either - if they do, it means that their bushings or end pivots are bound up. Mostly all they do see is axial force.

 

Here's a few sketches that show what I was trying to get across previously with words only. My ASCII sketch convention: the ▲ characters are the D-blocks/chassis-side brackets. The | characters are the endlinks. ↑ and ↓ are forces in the directions shown. Sorry for all the little dots; HTML doesn't like consecutive spaces. ASCII because I'm at work (see above) and my home scanner only works with the computer that's on dial-up.

 

 

Anywayyy . . . here's what a sta-bar looks like in front view, unloaded. As far as static equilibrium in front view is concerned, the amount of "U" shape going into or out of the paper does not matter (you'd consider that in side view or in some auxiliary view that shows the bar arms in true length).

 

 

E . . . . . . . . . . . . . . . . . . . E

|_______________________________________|

. . . ▲ . . . . . . . . . . . . ▲

. . . B . . . . . . . . . . . . B

 

 

 

I'll construct what's known as a "free-body diagram", with equal and opposite endlink forces applied at the endlinks. Free body means not restrained so you can determine the restraint forces that would put it into equilibrium. This problem is not in rotational equilibrium yet (it will rotate in a clockwise direction) so it is not a solution.

 

Diagrammatically, this is what you're telling me, that there are no forces at the chassis brackets due to roll, and that implies that a couple of pieces of bent coat hanger, some lamp cord wire, or even an old shoelace to hold the center of the bar up is all you really need from a strength point of view. Sorry, t'ain't so.

 

 

E . . . . . . . . . . . . . . . . . . . E

↑_______________________________________↓

 

. . . B . . . . . . . . . . . . B

 

 

 

 

Now let's add the chassis brackets back in as equal and opposite forces like I said in the earlier post. The front view free body diagram is now complete and is even statically determinate.

 

As soon as you tell me that the chassis side brackets keep the bar from rotating clockwise or that the sta-bar is responsible for limiting chassis/body roll this is EXACTLY what you've got. You'll have to imagine the bar bending up or down in the overhangs between points B and E, and that there will be a little bending curvature shape between points B. It's there, trust me.

 

E . . . . . . . . . . . . . . . . . . . E

↑_______________________________________↓

. . . ↓ . . . . . . . . . . . . ↑

. . . B . . . . . . . . . . . . B

 

 

I fully expect that there will be things crop up in the future that you are far more familiar with than I am or ever will be. It just so happens that this one is playing out on my home turf, and it's happening really early in my time as a member of this forum. Don't let my miniscule post count (<20) imply anything, as it might be misleading to do so. Time might come when it's me that has your back covered.

 

 

Norm

And because you asked, here is the loading summary for one chassis side bracket on a "car" with front and rear bars. As long as the car only moves vertically in 'heave', there are no forces on the sta-bar brackets (other than the bar's weight, which I had intentionally made zero for this study that I did for a different purpose a couple of years ago for a discussion on another forum) and zero torsion in its center section.

In roll (CASE 4), you do get forces at the chassis brackets (FY GLOBAL).

EQUIPMENT LOADING SUMMARY

 

 

POINT NO. 170 EQUIPMENT ID: RR BAR PICKUP

 

 

CALCULATED FORCES (LBF) CALCULATED MOMENTS (FT-LBF)

 

CASE 1 - WEIGHT

. . .FX FY FZ MX MY MZ

GLOBAL 0. 0. 0. 0. 0. 0.

LOCAL 0. 0. 0. 0. 0. 0.

 

CASE 2 - RIDE DISPLACEMENTS ONLY, -1" HEAVE, CHASSIS AT AXLE

. . .FX FY FZ MX MY MZ

GLOBAL 0. 0. 0. 0. 0. 0.

LOCAL 0. 0. 0. 0. 0. 0.

 

CASE 3 - RIDE DISPLACEMENTS ONLY, -1" HEAVE, CHASSIS AT CG

. . .FX FY FZ MX MY MZ

GLOBAL 0. 0. 0. 0. 0. 0.

LOCAL 0. 0. 0. 0. 0. 0.

 

CASE 4 - ROLL DISPLACEMENTS ONLY, 1 DEG ROLL

. . . FX FY FZ MX MY MZ

GLOBAL 0. -97. 0. 0. 0. 0.

LOCAL -97. 0. 0. 0. 0. 0.

 

 

Norm

It is an M8 size thread.

 

I snapped the bolt which goes into the chassis weldnut on the other side of that swaybar bracket. We just used an air chisel to punch off the weldnut then I used washers and 8.8 rating hi tensile M8 bolt + Nyloc nut. It is fine and probably stronger than the original setup.

 

We are very lucky to have easy access to the other side of that subframe section; on VW Golfs for example it is totally contained inside a chamber and you have to cut a big hole in the subframe, then weld it back up!

 

Just make sure you have used minimum 8.0 rated hardware including the washesr and Nyloc nuts or thread locker.

Norm, while your numbers are quite impressive - I'm not sure that your principles are accurate. The numbers given are only as good as the assumptions made during the data input. I disagree with your interpretation and will try to explain why below. As you know, it is not easy to do in writing

 

Basically, an anti-roll bar (sway bar is usually the term used - but inaccurate) is really nothing more than a torsional spring. Take a look at many modern 4WD trucks with independent suspension and torsion bars and notice the similarities. The only real difference is the end links - while the anti-roll bar is usually a "U", the torsion spring is usually a "Z" - but they work exactly the same. In the end - we are still talking about springs

Now on a torsion spring - the ends links are tightly controlled so no additional support in the center (to counter gravity) is needed. Both ends are usually splined and mounted in such a way that the complete weight of the spring is fully supported. When the truck tries to push down on the A-arm end of the spring (which is the only moving end), the rigid end of the spring/bar resists movement (by being directly connected to the frame) and the bar/spring exerts counteracting forces on the A-arm by TWISTING. Since the frame won't move and the A-arm has to move - the spring must twist. If you want a stiffer ride, increase the diameter of the spring (sound familiar?) There is no additional force on the center of the spring other than the twist.

 

Now lets look at the anti-roll bar (another torsional spring). This differs a bit from the truck's torsion bar in that it is not self supported on the ends and in fact - BOTH ends move. However, it could be if both A-arms had splined couplings on each side (but there is an engine and other items in the way, but I waver from may path...). Now, if both ends move in the same direction - there is no twisting of the spring/bar (as you have accurately demonstrated). But if one A-arm moves up and the other moves down - the BAR/SPRING must twist. Just like with the torsion bar, the middle is pretty much along for the ride and it is the ends that are twisting the bar. Since this is a "U" instead of a "Z" by design - there is some push and pull on the D links - but not 100% at all. In fact, I'd bet it would be closer to 15% since really it is the ends that are twisting the bar and it is the BAR that is resisting the movement.

 

There are really only two factors that are considered when selecting anti-roll bars (in our simple illustration).

 

1. diameter of the bar (resistance to twisting)
   
2. length of the levers (bar ends).
   

First let's tackle diameter;

In layman's terms; twist gets smaller as diameter gets larger - for every sense of measurement that you increase the diameter of the bar, you increase its resistance to twisting by Dia4 (or diameter to the 4th power)! Yes, a little goes a long ways. So I guess size does matter , oh, I digress again...back to the discussion.

Now let's look at length;

In the same layman's terms, the longer the "lever" used to twist the bar/spring, the easier it is to twist it. Or the longer the arm (lever) is from the center of the bar, the more torque it can apply to "twist" the bar. Thus the shorter the "lever" of the arm is from the bar, the less likely it can "twist" the bar (resulting in less body roll). So if we have an anti-roll bar with everything being equal, the one with the shortest arms from the center of the bar will be the stiffest.

 

Notice that we have not had to mention the D link yet when selecting a bar. If your presuppositions were correct, none of us would be getting larger bars, we would just need to get much stiffer D links! However, the only thing that stiffer D links provide is less wasted energy by forcing twist into the bar quicker (removing the slop).

 

But don't take my word for it, read it from a pro

http://www.spswebpage.com/Tech-Article-Archives/Improved-Handling-with-Anti-Sway-Bars.html

 

But I did like they way that you approached this with a non-emotional argument

 

Maybe if you explained each of the keys in your chart it would be easier. But it looks like you assume zero torsion in the center section which I would believe to be inaccurate since it is a spring.

 

 

And because you asked, here is the loading summary for one chassis side bracket on a "car" with front and rear bars. As long as the car only moves vertically in 'heave', there are no forces on the sta-bar brackets (other than the bar's weight, which I had intentionally made zero for this study that I did for a different purpose a couple of years ago for a discussion on another forum) and zero torsion in its center section.

 

In roll (CASE 4), you do get forces at the chassis brackets (FY GLOBAL).

 

 

 

 

Norm

I thought I was technical, but even that is too much of a mouthful for me right now he he he. Thanks for sharing the info

Gonna be a long one, even for me.

 

Norm, while your numbers are quite impressive - I'm not sure that your principles are accurate. The numbers given are only as good as the assumptions made during the data input. I disagree with your interpretation and will try to explain why below. As you know, it is not easy to do in writing

Let's get a few stiffness things out of the way.

 

First . . . to answer

 

"1. diameter of the bar (resistance to twisting)"

"2. length of the levers (bar ends)"

 

I'll add the following:

 

"3. length of the central torsional element", parameter "L" in the Comeskey link"

 

Explanation - not all different design bars for the same car are bent in quite the same shape. Just close, and close enough to fit without interfering with other things is good enough for fabrication and installation, but length L can easily vary by 5% or possibly more and has almost an inverse relation with overall bar stiffness.

 

I suspect that you may be occasionally confusing endlinks with bar arms. Correct me if I'm wrong here.

 

 

FWIW, I seriously doubt that you'll find a sta-bar stiffness calculator online that considers as many separate shear, bending, and torsional effects as the one I have developed for my own use. Not only does it consider seven separate deflection effects, but mine can handle the case where a bar has been built up from a torsional section of one OD and ID and the arms from a cross section of different diameters. It can even work with bar arms that aren't round or tubular, as well as with bushing and endlink stiffnesses in what you might consider to be a post-processing step.

 

The oldest published bar formula I know of – the one in Fred Puhn's book – only considers two effects (center torsion and arm bending) and the bar is assumed to be of uniform cross section from end to end. That's one more term than what Mr. Comeskey's sheet provides (which looks at the center torsional element and its effect only), Puhn's formula still results in stiffness numbers that are over-estimated by typically 5% - 8%, but Comeskey's formula can easily produce numbers that are 25% - 40% or more too high, and you'd never know it unless you know what his formula is missing. I'm not saying that either is wrong, just that they're at various levels of (in)completeness. Note that if the arms were significantly more rigid than the center section, as can happen with built-up bars such as for circle track cars, Comeskey's numbers would approach Puhn's.

 

FWIW #2, Comeskey does not handle the arm length effect on stiffness correctly (bar stiffness is closer to being a function of arm length SQUARED). Lots of people miss this one, even people who should know what the correlation between motion ratios and stiffness rates is.

 

FWIW #3, here's my other five little pieces of sta-bar flexibility (it's a lot easier to add up flexibilities and convert to stiffness last than it is to work with all of the little stiffnesses). Easy enough to copy/paste from my spreadsheet, but more time consuming to further explain than I care to spend. For this thread, it's enough that they have been identified as having minor effects on bar stiffness, with the above suggestion that they total up to being a 5%-8% effect for typical OE or similar bar designs.

 

DeflFromBarDeflFromArmShear

DeflFromCantileverSlopeFromShear

DeflFromCantileverSlopeFromMoment

DeflFromSimpleBeamSlopeFromMoment

DeflFromArmShear

 

 

Anyway, I don't think the issue here is bar stiffness per se. It's how that bar stiffness is reacted when bar end displacements are imposed. Since this particular discussion is all about the strength requirements at the chassis bracket fasteners, the ultimate result of importance is the magnitude of the forces that they are subject to.

 

Bar stiffness absolutely MUST react against the chassis if it is to provide a roll resistance effect to the chassis or a supplemental wheel rate in roll effect. Somehow, you have to be developing additional forces between the sprung chassis and the unsprung end of the suspension. The only way this is going to happen is via forces applied at the D-blocks, as those are the only chassis side attachments. In roll, these loads will be compressive on one side, tensile on the other. No other answer is possible. If only one wheel hits a bump, or drops off into a hole, there will also be compressive or tensile forces (respectively) applied to the D-blocks/studs on the same side as the wheel that encountered the obstacle.

 

Once again, you can't apply opposite direction forces at the bar ends without causing forces at the D-blocks. Your evaluation of the torsion bar analogy appears to be good as far as it goes but it is not complete – the vertical load doesn't just disappear (it goes straight into the A-arm's pivots). The full analogy as applied to a sta-bar would have the D-blocks right at the bends in the bar where the arms and torsion section intersect, with the wheel loads from roll going directly into them (the bar arms being analogous to the A-arms). This sort of bar would almost certainly be a rear bar, due to clearance requirements for steering up front. And while this analogy is accurate for this one specific case (which is physically impossible, precisely), it is more difficult to visualize than the way I chose to present it. Let alone draw up in ASCII.

 

So yes, you are entirely correct in saying that the previous post's sketch does not consider torsion. Shear and bending only, and done that way because using the front view of a typical front bar is the easiest way to demonstrate that the D-blocks absolutely must carry the "vertical-ish" forces that we're interested in here and give a hint as to the magnitude of these forces. Satisfying static equilibrium requires the forces at the D-blocks for the picture as shown and dimensioned. I will note that a rear bar could well have D-block forces more similar to the endlink forces since the distance between points "B" and that between points "E" would be more nearly equal. But the D-block forces will never be a small fraction of the endlink forces in any realistic bar configuration.

 

Torsion in the center section is a separate part of the analysis, but since the D-blocks are assumed to freely allow bar torsional rotation within them they pick up no load from that effect. IOW, there is nothing to show and nothing further to learn even if I could figure out how to show a "nothing" in ASCII.

 

 

 

If it's "pro" credentials that you're after, I'm BSCE (structural), 1970 (tenth in a class of 134 CE's at a largish university in Boston that you've probably heard of, Chi-Ep, TBP), and been paid as an engineer to do structural analysis of one sort or another for most of the 40 years since. There is nothing in the simplified hand-calculated analysis of a sta-bar that can't be approximated by standard structural beam formulas (Puhn's and Comesky's formula derivations started at the same place), so the fact that my engineering paychecks have never been issued from within the automotive industry is immaterial. I hope that's good enough.

 

 

 

BTW, I thought it was obvious, but

 

"B" = chassis side Bracket or D-block

"E" = Endlink

 

From exposure to a few people who do or have worked in the auto biz, I normally call this thing a "sta-bar", that being short for stabilizer bar (which is what most OE's call it).

 

 

Thanks for the compliment regarding my approach to discussion. I mean it.

 

 

Norm