Lotus Europa Community
Lotus Europa Forums => Garage => Topic started by: Sparkrite on Sunday,November 17, 2024, 10:35:05 AM
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I decided to remove the gearbox output shaft seals on my tcs in order to try and eliminate the small leak from the seal nut threads. In order to do this I drifted out the split pins that hold the half shaft/axle. Upon inspection they were cracked (as can be seen in the photo). This crack would only worsen until the wheel/axle would come adrift from the output shafts with dire consequences. These pins were new about 1000 miles ago,and I had the smaller one inside the larger pin. The shaft was properly shimmed so the only thing I can think of is the poor quality of the pins or maybe I should have used the spiral pins instead.
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The photo doesn't show it but was the smaller pin also cracked?
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They crack because they are not shimmed tight enough which allows flex, causing the cracking and breakages. It is not a big deal as they will run just fine without any pins installed as long as you don't jack the car up our Keanu Reeves your car off hill crests.
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One of the inner pins has one crack where's the outer pins had two cracks each.
The way I see it is the correct shims only work in compression,when the road wheel is being pulled outwards there are stresses on the pins,hence the cracks.
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It's good to be reminded that proper shimming of the inner u-joint is important.
The TC and TCS use spirol pins which are rolled spring steel. The hardness (and therefore brittleness) of spring steel and the thin section of that steel in the spirol pins encourages cracking. I would expect that the u-joint being loose also encourages cracking because shock loading would be more likely to cause a crack than simple shear caused by tension or compression forces on the u-joint.
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I recently removed the the inner half-shafts to replace leaking seals and found a failed spiral pin. I know the shimming was correct but still the failure. I ordered new spiral pins from Grainger. Comes in a package of 10 so should last until I'm no longer able drive my TCS! O:-)
https://www.grainger.com/product/Spring-Pin-41KN49
BTW, no more leaks from the seals as of yet! :trophy:
Gerry
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Wonder if stainless steel pins might be worth a try? They area bit more ductile.
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I'm pretty sure the right move here is jumping directly to a hardened shoulder bolt. May need to grind a small taper on the end to hammer it in nicely, and likely will require match drilling to get everything just perfect without slop. I do this with pretty much every cross pin in anything, except the ones designed to fail and protect other components.
I've thought through a few ways of doing this, will post with part numbers if/when I implement it.
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If we sit down behind the rear left wheel and look at the suspension, it should be observed that the load path travels from the tyre patch, up to the wheel hub and then along the stub axle, half-shaft and then to the gearbox side bearings.
This load is accepted by the spring but constrained by the lower link because it connects to the upright about 6 inches or so below the axle.
In a right hand cornering manoeuvre, the additional load is transferred to the spring which acts, in turn, on that lower pivot point of the upright.
Therefor, the axle assembly is always* in compression and the lower link is always in tension.
*unless performing the antics JB outlined.
At least I think that's right and someone will unceremoniously beat me up if it isn't. ;D
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"Therefore, the axle assembly is always in compression and the lower link is always in tension."
100% correct. It is not an issue.
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I must be wrong as I see it different to the experts.
But if you're viewing the car as Gavin described, then on hard right cornering the left tire contact patch moves inwards as the centrifugal force is pushing the car/wheel outwards to the left.This means that there is compression on the lower link and tension on the half shaft surely?.
On a simplified view, if I just grab the lower tire and push it inwards towards the gearbox,then the lower link is compressed and axle tensioned.
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G'day Sparky,
Think of it this way:
Presumably it's accepted that the load path dictated by gravity travels as first laid out - upwards from the tyre etc.
The spring, in reaction, resists that load but the important part to note is that the spring is connected below the stub axle and, as such, follows a different path to the gravity induced load.
The spring is actually trying to rotate the stub axle clockwise with the U-joint acting as the pivot in a kinda bellcrank effect.
This is the case even when the car is static in the garage.
This could be tested.
Jack up the car to relieve suspension loads. Place a dolly or garage skate under the wheel. Disconnect the lower link from the upright and slowly lower the car till the tyre starts to contact the dolly.
Which way does the wheel move?
So, yes, in a right hand corner, the car will distort the tyre contact patch inwards but that's because the spring is effectively 'pushing' the lower pivot point of the upright outwards. The centrifugal force induces weight transfer and is translated as increased spring force, not a separate one.
If the lower pivot point of the spring was at axle height, the bellcrank effect would be negated.
In a wilder moment it occurred to me one could jury-rig some lower links from wire rope and still be OK, but don't try this at home. ;)
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How much does the gyroscopic effect of the wheel have on all this?
When a wheel and tyre are revolving at say 40 mph, and the wheel is affected by turning a long right turn, what gyro effect will that wheel have on the suspension.
Asking for a friend .....
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It's pretty simple really, people have driven their Europas hard with no pins in place and no issues. However, if you jack up the car with the rear wheels hanging, the half-shaft may slide off and would be a right pain to get back on. By all means use the pins but just don't worry about them cracking.
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I always thought that the pin was important so that the drive shaft does not slam repeatedly into the output shaft/bearing of the trans.
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I appreciate this discussion, and thanks Gavin for the explanation. The subject of axel loads in tension (and the fact that in normal driving, this is never the case) has long puzzled me. My understanding has been the same as Mr Sparkrite's, though I have always known this thinking is flawed as a 5 mm (or whatever size) steel pin could never reliably retain those loads. I am somewhat embarrassed to admit that I am still having trouble getting my head wrapped around it, because this isn't rocket science. In fact, it should be very simple to grasp.
This weekend I shall do just as Gavin suggested; have a sit behind the car, with his explanation in hand, and stay there until I can 'see it'.
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Rather than sitting in a cold garage floor, how about this....
I think of it as in the image below. This is the S2 layout whereas the TC has the damper/lower link on the same bolt, but it's close enough.
The weight of the car is the blue line through the spring/damper. The reaction from the ground through the tyre is yellow arrow A, which is outboard from the lower mount of the damper which forms a pivot point. If there were no top link/driveshaft then the wheel would pivot and the top go inwards as shown on yellow arrow B. This means the driveshaft is always going to be in compression as long as there's a ground reaction at A. The more the force on the wheel, the more compression on the driveshaft and in theory the lower link should go in tension.
Even when you're cornering fast and the tyres sliding, unless you lift the opposite wheel clear of the ground, there's always going to be some compression in the driveshaft. If you do manage to corner fast enough to lift the opposite wheel then yes, the driveshaft could go into tension but I think the load would only be at most the weight of the wheel trying to go into positive camber
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Yep.
The fundamental forces in play in my super simple pic are gravity (red arrow) and the reaction from the spring (green arrow) – Isaac Newton's third law.
The core principal of the upright's function is that the spring acts on a lower pivot point and, thus, aways maintains a leverage advantage.
My slide-rule is in for service but someone smarter than me could calculate the leverage effect.
As a side note, the spring is laid over somewhat, so the actual spring force at the lower pivot will be something less.
So, if Colin walked into his engineers office with this upright, how would they stop it from rotating clockwise?
It's getting on for summer here, but to avoid a cold backside in the Northern hemisphere, I'd sacrifice a cornflakes packet to some cardboard aided design.
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I have now seen for myself. With the car jacked up and the suspension at full extension and the slot pins removed there is in fact tension on the axle as it separates from the gearbox by a quarter inch or so. But the minute you raise the wheel about half inch, the axle butts up nicely against the gearbox.
So unless you go over a hump at speed and your car gets sufficient "air" then the axle will be in compression.
I do still wonder though if any of the driving forces might be sufficient to overcome the set up and momentarily apply tension force to the axle and thus stress the pins.
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I do still wonder though if any of the driving forces might be sufficient to overcome the set up and momentarily apply tension force to the axle and thus stress the pins.
Unless you lift the wheel I don't think you will move the driveshaft into tension and even then, the load won't be a great deal. But you've had an obvious failure on a pin and apart from not being fully shimmed in place I am struggling to work out a mechanism.
I'm not convinced it's a poor quality part, these things are in shear and I would guess that plain mild steel bar would do the job because the drive is supposed to be transmitted by splines, not the pin.
So I go back to shimming. Mine is so tight that I have to really drive the pins in and getting them out is equally tough. If yours drove out relatively easily then I'd add a bit more shimming.
The other, very blue sky thinking idea, was if it was possible to go from a very high compressive load to much lower stress and the cycling effect was enough to initiate a fatigue failure. That's very close to your proposal but I'm struggling with that one because you've always got some compressive stress.
But if you are driving fast enough to lift wheels on our potholed roads, you deserve a medal alongside a free set of super high tensile steel pins !
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But the minute you raise the wheel about half inch, the axle butts up nicely against the gearbox.
So unless you go over a hump at speed and your car gets sufficient "air" then the axle will be in compression.
Or, to look at it another way, if the dampers were ~ half an inch shorter, there would be no situation where the yokes would escape the gearbox output shafts.
Dunno the answer, but did the stock dampers permit this at full droop?
I do still wonder though if any of the driving forces might be sufficient to overcome the set up and momentarily apply tension force to the axle and thus stress the pins.
These are 50 year old cars.
I wouldn't be surprised to find a few thou of rotational play in the yokes - that'd do it, methinks.
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Thanks much Brian and Gavin. Things are very much clearer. I see that my earlier thinking was not so much flawed, as it was the imagining of incorrect values to some of the force loads, and not giving enough credit to the lever action of the shape & orientation of the upright. This is an issue that has been gnawing at me for several years now. I'm going to sleep much better tonight. Cheers.