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FRONT ALUMINUM PLATE
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Originally posted by Obioban View Post
If your torqued the plate to spec, which is necessary for full torsional rigidity, you’d have an issue using them repeatedly.
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Originally posted by jayjaya29 View PostJust hypothesizing here. May be for crash protection. If the bolts are stretched more then once they have a lower load rating and may fail during a front-end collision, allowing the engine to submarine.
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Originally posted by bigjae46 View Post
I don't think you're stretching a M10? class 10.9 bolt with the OE torque spec.
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Originally posted by jayjaya29 View Post
Torque spec for the bolts are 44 ft-lbs + 90° angle (+/- 30°). Nominal torque spec for an M10 10.9 grade bolt is 49 ft-lbs. I think the added 90° angle torque is significantly stretching the bolt more then an additional 5 ft-lbs would.
In oil and gas, it has been awhile, I think we used an ASME standard which is not necessarily a torque to yield spec. But we would never re-use fasteners because of corrosion, paint, and the consequences of a fastener failure - massive leak and a large fireball.
Either way, I just caught a couple feet of air going sideways at 60mph and broke a control arm ball joint. The bolts didn't break or shear although I am replacing those bolts because of the impact. So I'm inclined to stick with my blast them on with an impact method.
In the end, we are arguing about a relatively minor detail in a procedure. How many E46 M3s were produced? Now how many times has the front thrust plate been removed? Now how many times have the bolts been replaced? Now how many people have had their front plates fall off or even found a broken bolt? I'm sure there would be plenty of stupid questions about it on facebook complaining about cheap BMW bolts if it was happening.
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Originally posted by bigjae46 View Post
I thought the torque spec for an M10 bolt is closer to 55 ft/lb to 60ft/lb? Probably close to that spec with a 44ft/lb + 90 deg. Torque spec isn't necessarily stretch to yield. Depends on the standard. Someone would need to measure a bolt to be sure. Wheel fasteners are torqued to spec but aren't replaced each time.
In oil and gas, it has been awhile, I think we used an ASME standard which is not necessarily a torque to yield spec. But we would never re-use fasteners because of corrosion, paint, and the consequences of a fastener failure - massive leak and a large fireball.
Either way, I just caught a couple feet of air going sideways at 60mph and broke a control arm ball joint. The bolts didn't break or shear although I am replacing those bolts because of the impact. So I'm inclined to stick with my blast them on with an impact method.
In the end, we are arguing about a relatively minor detail in a procedure. How many E46 M3s were produced? Now how many times has the front thrust plate been removed? Now how many times have the bolts been replaced? Now how many people have had their front plates fall off or even found a broken bolt? I'm sure there would be plenty of stupid questions about it on facebook complaining about cheap BMW bolts if it was happening.
44 + 90 deg is 100-120 ft lbs.
AFAIK, BMW only specs steel bolts be replaced for torque to yield locations.
2005 IR/IR M3 Coupe
2012 LMB/Black 128i
2008 Black/Black M5 Sedan
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Originally posted by Obioban View Post
If you're gunning them on, you're likely not weakening them. If you're torquing them to spec, you're almost certainly in the plastic deformation range of the bolt.
44 + 90 deg is 100-120 ft lbs.
AFAIK, BMW only specs steel bolts be replaced for torque to yield locations.
let’s let them do like they want too
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Originally posted by Obioban View Post
If you're gunning them on, you're likely not weakening them. If you're torquing them to spec, you're almost certainly in the plastic deformation range of the bolt.
44 + 90 deg is 100-120 ft lbs.
AFAIK, BMW only specs steel bolts be replaced for torque to yield locations.
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Fortunately this is pretty easy to calculate. Since I’m on a phone, let’s make some assumptions and approximate to see if simple math and assumptions tells us the answer.
Assume:
1) 50% stretch is in the bolt and 50% compression is in the joint
2) Effective length (L) is 30mm
3) Thread pitch 1.25mm and 90 degree turn = .3125mm = dl
4) Elastic modulus of steel = E = 200000 N/mm^2
5) Cross sectional area of an M10 bolt = A = 64.5mm^2
6) Tension in fastener is linearly proportional to torque
Case 1: Assume contact but no preload from initial torque
Case 2: Assume linear force proportional to proof torque, add case 1 for total
Case 1 (seat the fastener, turn 90 degrees):
dl = FL/EA
F = E * A * (dl/2) / L (dl is halved because of assumption 1)
F = 200000 * 64.5 * .3125 / 30
F = 67,188 N
Case 2 (torque to 44, turn 90 degrees):
Proof load (pure elastic) of an M10 10.9 fastener is ~ 53500 N @ 82 ft-lb
Tension at 44 ft-lb therefore ~ 29000 N (due to assumption 6)
Add Case 1 and tension (if we were still elastic, which we’re not) is above 95,000 N.
If our elastic region ends at a tensile load of 53,500 N, then anything above that is causing plastic deformation and permanent stretch. Ultimate tensile load of an M10 fastener is minimum specced to 67,100 N, so the rough math very much agrees that this fastener is going to be yielding, even if you barely torqued it before the 90 degree turn. If you torque it to yield over and over, IT WILL FAIL.
Why is it torqued to yield? I don’t know but I’d guess for frontal impact more than torsional rigidity. If you want to skip torquing to yield, I’d go to 70-80 ft-lb instead, that’s better than failing the bolts from repeated yielding. I replace them and torque to yield, so I recommend torquing to yield. Because I want my car to absorb energy if I smash into something. Because I like safety.
‘02 332iT / 6 | ‘70 Jaguar XJ6 electric conversion
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Maths hurt brain.
I think the safety argument would be a more compelling one. However, does anyone have legitimate information on how important the plate is to the safety of the car? Remember, the e36 m3, including the Euro version that is about as heavy as the e46, had no such plate. Of course, it's a much less safe car as well, but I do wonder what role this would really play in a crash scenario.
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Originally posted by 01SG View PostMaths hurt brain.
I think the safety argument would be a more compelling one. However, does anyone have legitimate information on how important the plate is to the safety of the car? Remember, the e36 m3, including the Euro version that is about as heavy as the e46, had no such plate. Of course, it's a much less safe car as well, but I do wonder what role this would really play in a crash scenario.‘02 332iT / 6 | ‘70 Jaguar XJ6 electric conversion
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Originally posted by Bry5on View Post
If it does, I'd bet it helps particularly in a front corner impact where the front longitudinal frame rail is trying to bend inward. This is one of the trickiest crashes to design around. Buckling the aluminum plate in shear takes extra energy (same reason if helps with chassis stiffness). The non-m also has this plate, but it is not braced as well to the control arms and frame rails behind the firewall. We do have an automotive safety engineer here on the forum that might be able to make more educated responses than I can. At one point in my life, I did work at an automotive OEM and was arms length involved in some of the energy absorption simulation testing as I worked mostly on body structure, I'm no expert though.
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