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Guest ratchethack

Whoa now - the 136,000 psi is pounds per square inch of bolt not psi of hydraulics. But brake lines can go over 10,000 psi during panic stops. That times the area of four or six pistons is a lot of force trying to blow the caliper apart! Joe

OK, this is where I've obviously gotten in over me head, as noted previously. :blush: I don't know how to make an equivalent conversion from PSI of bolt rating and PSI of fluid in the caliper. I suspect it may involve some horrifically complicated analysis of internal surface area (or is it volume?) of the caliper? :huh2:

 

I'm takin' meself well out of further discussion here but would sincerely appreciate an education from those with some level of mastery and experience hereabouts. :notworthy:

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OK, this is where I've obviously gotten in over me head, as noted previously. :blush: I don't know how to make an equivalent conversion from PSI of bolt rating and PSI of fluid in the caliper. I suspect it may involve some horrifically complicated analysis of internal surface area (or is it volume?) of the caliper? :huh2:

 

They are the same. Pounds per square inch.

 

So, if you have a bolt of 1sq" section, its tensile or shear rating is exactly the same as the steel's rating.

 

If you have a caliper piston with a 1sq" surface area exposed to the fluid, its pressure is exactly that of the pressure in the system.

 

Both making allowances for friction, manufacturing tolerances, etc, etc. Multiply by the fraction for sections/areas different to 1sq".

 

I suspect you are after the relationship between brake pressure, resultant rotational force trying to drag the caliper around and shear strength of the bolts holding it place? I haven't done that calculation since school. From memory, it is also going to involve the turning moment of the distance of the caliper from the wheel axle. If I get time before I go on holiday, I'll see if I can dig it up from a school textbook.

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From memory, it is also going to involve the turning moment of the distance of the caliper from the wheel axle.

A-HA! Ya see? <_>

 

Now if Brembo's engineers would write me back on the bolt specs. :angry:

 

Is it true all of Europe is on Holiday the month of August? I suppose all the engineers are off doing what they do: :nerd::drink::nerd:

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Guest ratchethack

I suspect you are after the relationship between brake pressure, resultant rotational force trying to drag the caliper around and shear strength of the bolts holding it place? I haven't done that calculation since school. From memory, it is also going to involve the turning moment of the distance of the caliper from the wheel axle. If I get time before I go on holiday, I'll see if I can dig it up from a school textbook.

Nothing so complicated as that, Mike.

 

I was attempting to use the known tensile ratings of the caliper-half joining bolts (Metric Class 10.9 = 136,000 PSI yield (shear) strength, and 151,000 PSI tensile strength) to calculate the approx. likely max fluid pressure in the calipers, which would of course be the entire brake circuit max PSI number. I b'lieve I've already correctly reasoned that since there are 4 bolts of this rating (not 2, thanks Dave) that one doesn't combine the PSI tensile ratings of each of 4 bolts -- you only consider the rating of 1 bolt of the 4.

 

I'd used the yield (shear) strength previously where I meant to use tensile, sorry. :blush:

 

So to cut to the chase, does this mean that a fair estimation of the max caliper fluid pressure would be the same as one bolt's tensile strength rating in PSI by fastener class? This would mean a max possible pressure of 151,000 PSI?

 

Enquiring minds (well, you know). ;)

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So to cut to the chase, does this mean that a fair estimation of the max caliper fluid pressure would be the same as one bolt's tensile strength rating in PSI by fastener class? This would mean a max possible pressure of 151,000 PSI?

 

Enquiring minds (well, you know). ;)

Here is my futile attempt at trying to understand this:

I think it works out to something like four bolts of 8mm diameter.

Find the total square area of the bolt cross sections and divide into one square inch to find total tensile strength.

The tensile force on the bolts from fluid is probably something like the force at the hand multiplied by leverage of the lever, times the ratio of caliper pistons total area divided by the master pistons area.

But I don't remember doing that in school, so it is just a lay man's guess.

It is surprisingly difficult to find ratings of things in easy to comprehend terms.

I found the following at wikipedia:

"Tensile yield strength is M10, property class 8.8 bolt can very safely hold a static tensile load of about 15 kN. the screw in the image should be going the other way."

15 kN is about 3500#

M8 bolts are probably about half as strong, so we have four M8 bolts with enough strength to withstand about 7000 pounds of static load.

At the hand lever I figure some gorilla fist guzzista might be able to muscle 100lbs of force when trying to get the brake lines to pop using one hand.

There might be a five to one lever ratio, so 500 pounds of pressure might hit the master piston. (that is too much pressure but just for worst case scenario, indulge me...)

The pistons are probably 16mm, 30mm, 30mm, 34mm, 34mm.

with areas 201, 707, 707, 908, 908.

so 3230 divided by 201 is 16.

So 16 times 500 is 8000 pounds, so the gorilla fisted Guzzisti might be able to brake the bolts that can only withstand 7000 pounds.

EDIT Oooops, I forgot about the second caliper, that will take away half the pressure, so 8000 pounds becomes 4000 pounds.

But since we use stronger than 8.8 bolts maybe it is OK???????

I am probably waaaaayyyyyy off, so if you see where I am wrong, feel free to point it out.

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Guest ratchethack

I am probably waaaaayyyyyy off, so if you see where I am wrong, feel free to point it out.

Um, a, well, spectacular effort, Dave -- such as it was. . . :rolleyes:

 

But to answer your question frankly and honestly, here's where it looks like you made your very first wrong turn:

Here is my futile attempt at trying to understand this:

;)

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Nothing so complicated as that, Mike.

 

I was attempting to use the known tensile ratings of the caliper-half joining bolts (Metric Class 10.9 = 136,000 PSI yield (shear) strength, and 151,000 PSI tensile strength) to calculate the approx. likely max fluid pressure in the calipers, which would of course be the entire brake circuit max PSI number. I b'lieve I've already correctly reasoned that since there are 4 bolts of this rating (not 2, thanks Dave) that one doesn't combine the PSI tensile ratings of each of 4 bolts -- you only consider the rating of 1 bolt of the 4.

 

I'd used the yield (shear) strength previously where I meant to use tensile, sorry. :blush:

 

So to cut to the chase, does this mean that a fair estimation of the max caliper fluid pressure would be the same as one bolt's tensile strength rating in PSI by fastener class? This would mean a max possible pressure of 151,000 PSI?

 

Enquiring minds (well, you know). ;)

 

I would be horrified if I believed that my brake system could produce _anywhere near_ the pressure that matched the tensile strength of the bolts holding the calipers together. In theory, you would be able to apply enough force to the brake lever to do this - in practice, a number of other things would happen first; the brake lever would break; a line or pipe would burst; a seal would give way. All of those would appear (guesstimatedly) to be a weaker spot in the system than caliper clamping bolts.

 

My guess is that there would be _at least_ a 50% safety margin in the capabilities of the bolts. I'm sure there will be an online source with a formula to calculate exactly what the force on them would be. Unfortunately I can't look for it, as I am hiding from holiday packing and expect to be found at any

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Guest ratchethack

I would be horrified if I believed that my brake system could produce _anywhere near_ the pressure that matched the tensile strength of the bolts holding the calipers together.

Moi aussi, since merely a few tens of thousands of PSI would have the capability of cutting a hole through just about anything in it's path, if a component were to spring a leak! :o

 

But lacking any real-world figures, I continue to wonder. . . :huh2:

 

Please resume your study of an estimate when time permits?

 

Many tx in advance. -_-

 

Prof. Wrenchbracket

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Moi aussi, since merely a few tens of thousands of PSI would have the capability of cutting a hole through just about anything in it's path, if a component were to spring a leak! :o

 

But lacking any real-world figures, I continue to wonder. . . :huh2:

 

Please resume your study of an estimate when time permits?

 

Many tx in advance. -_-

 

Prof. Wrenchbracket

Moi aussi, except your comment that a few thousand pounds having the ability to cut holes, has about as much to as the few pounds of force in a crossbow having the ability to cut hole.

If you look at my futile attempt, you will see that it lines up nicely with

My guess is that there would be _at least_ a 50% safety margin in the capabilities of the bolts.

Assuming my numbers that Ratchet slammed are correct, the maximum pressure might be 4000 PSI if I used two arms or had Barry Bonds' enhanced grip, 50% of that is 2000, givin a 50% margin of 6000 and I calculated bolt strength to be about 7000 static tensile pounds.

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  • 7 years later...

Please explain which bolts you are talking about using ? Stainless steel does not like anything . It needs anti-seize lubricant or it galls the threads when you tighten .

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Yes Stainless does not like anything and yes it galls it also forms an electrochemical cell with Aluminum.  A little water gets in there and the aluminum turns to aluminum oxide. This is why it is a good idea to use either anti-seize or locktite.  Either of them keeps the water out.  I am in the aerospace business and sometimes we use Titanium bolts in aluminum.  It is still wise to use anti-seize or thread locker to keep the water out.

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