New Cylinder AD's released by FAA
So, an engineering question from a mere academic. 
When a doctor puts one pound per square inch pressure on the thumb plate of a syringe, what is the pressure at the tip of the needle?
When the pressure inside a cylinder of an IO 520 is 800 pounds per square inch, what is the pressure on the surface area of the thread between the head and the barrel?
When a doctor puts one pound per square inch pressure on the thumb plate of a syringe, what is the pressure at the tip of the needle?
When the pressure inside a cylinder of an IO 520 is 800 pounds per square inch, what is the pressure on the surface area of the thread between the head and the barrel?
If I have it right. There is about 900psi evenly loaded on the thread. 30000psi shear for Al and things are pretty good. Given 6in bore, at least ten winds of thread...more winds equals less pressure, and 1/8th thread depth.
Does that sound about right?
If its an 8in bore that goes out to about 1300psi evenly applied.
Does that sound about right?
If its an 8in bore that goes out to about 1300psi evenly applied.
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Turn the pressure on the head into a force (F) in Newtons or Lbs, whichever is your flavour of choice then apply the force to the thread area. Less any opposing forces of course.
Another way is by comparing areas but then you miss the value of knowing the forces.
Another way is by comparing areas but then you miss the value of knowing the forces.
Train wheels don't fall off because both components, the hub and the steel tyre, are made of the same metal.
Not quite true, the axles and hubs/wheels tend to be something like EN40B, while the removable tires are an SG iron. If I remember correctly, the expansion co-efficient of the hub/wheel is actually higher than the SG iron, so heating during running actually tightens the interference fit.
Tootle pip!!
Frankly, I’m jiggered if I know!
My understanding of Mr Atkinson’s point is that it doesn’t matter. My understanding of his point is that the ‘squeeze’ caused by the interference fit is what holds the head on, rather than the strength of the threads.
If I get a bolt and clamp it in a vice and tighten the vice as far as I can make it go, it ain’t the threads on the bolt that stop me pulling that bolt out of the vice …..
My understanding of Mr Atkinson’s point is that it doesn’t matter. My understanding of his point is that the ‘squeeze’ caused by the interference fit is what holds the head on, rather than the strength of the threads.
If I get a bolt and clamp it in a vice and tighten the vice as far as I can make it go, it ain’t the threads on the bolt that stop me pulling that bolt out of the vice …..
No Hoper.
I am certainly not an engineer.
I think Walter said roughly 20000 lbs of force is applied to the threads as a result of combustion?
You said,
Why did you misquote him?
Correct me again if I am wrong, but I think combustion results in a force being applied to something, be it pistons or whatever?
This is an engineering discussion. 20000 pounds is not a force it is mass or weight relatively speaking
I think Walter said roughly 20000 lbs of force is applied to the threads as a result of combustion?
You said,
Where did the 20000 psi come from?
Correct me again if I am wrong, but I think combustion results in a force being applied to something, be it pistons or whatever?
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It's beneficial to understand how the head and barrel are mated. The head is heated to expand and the barrel is cooled to contract. The head is then screwed on to the proper alignment. As the two metals reach temperature equilibrium, the interference fit is attained. The design is to allow for adequate strength absent the threads (which will shear at a much lower pressure) by the interference fit up to the design redline temp. The shear strength of aluminum is fairly low (can't recall the exact value). BTW, the interference fit is NOT at the same place as the threads. It is below the threads. The threads do not hold the head on during operation.
As Yr Right as alluded to but not exactly explained accurately, the cyclical fatigue on the aluminum is the issue that results in failure over time--NOT the time in service. If the cylinder is operated below about 380dF, the time in service will be MUCH longer than if the metal gets hotter. In addition, if the internal cylinder pressures are controlled (and by our best estimate based on a LOT of test data) and kept under about 800 psi peak pressures, the longevity is vastly improved.
Remember, the enemies of metal are heat and pressure. Control those and the design parameters are adequate. Operating at high CHTs (above about 380dF) and with high ICPs and the metal will fatigue and fail much earlier.
ROUGH, back of the napkin calculations:
Diameter of the cylinder approx. 5"
Area of cylinder head approx. 20 sq. in.
ICPs during a various mixtures range from 600psi LOP to 1100 psi at 50dF ROP.
20 x 600 = 12,000 pounds of force down on piston and up on head.
20 x 1100 = 22,000 pounds of force down on the piston and up on the head.
NOTE: These are peak pressures (stress) not mean pressures (HP).
Both of the above stress scenarios can actually produce the same HP!!
The former will have CHTs lower by about 50dF.
In which scenario is the cylinder going to last longer?
As Yr Right as alluded to but not exactly explained accurately, the cyclical fatigue on the aluminum is the issue that results in failure over time--NOT the time in service. If the cylinder is operated below about 380dF, the time in service will be MUCH longer than if the metal gets hotter. In addition, if the internal cylinder pressures are controlled (and by our best estimate based on a LOT of test data) and kept under about 800 psi peak pressures, the longevity is vastly improved.
Remember, the enemies of metal are heat and pressure. Control those and the design parameters are adequate. Operating at high CHTs (above about 380dF) and with high ICPs and the metal will fatigue and fail much earlier.
ROUGH, back of the napkin calculations:
Diameter of the cylinder approx. 5"
Area of cylinder head approx. 20 sq. in.
ICPs during a various mixtures range from 600psi LOP to 1100 psi at 50dF ROP.
20 x 600 = 12,000 pounds of force down on piston and up on head.
20 x 1100 = 22,000 pounds of force down on the piston and up on the head.
NOTE: These are peak pressures (stress) not mean pressures (HP).
Both of the above stress scenarios can actually produce the same HP!!
The former will have CHTs lower by about 50dF.
In which scenario is the cylinder going to last longer?
AH, but have a think about what would happen if the jaws of the vice had a thread of the same diameter and thread form in them as the bolt, and the bolt was clamped into this thread.
Now assume that the vice is tensioned up to clamp onto the thread so now you have the frictional resistance caused by the pressure of the vice on the thread and also the shearing force required to strip out the thread, to be applied before you can pull the bolt from the vice.
I am neither an engineer nor a LAME but I suspect that, in the case of the cylinder heads under discussion, both the thread and the shrink fit are required to ensure that the cylinder head remains on the cylinder barrel. I always believed that the thread was the primary way that the head was secured with the shrink fit ensuring that the thread both on the barrel and in the head remained very tight so that there was no tendency for the head to fret or move during operation, especially as they are subject to such varying temperatures.
Again see my disclaimer above and I certainly will be happy if I have this wrong and someone with accurate information (data) sets me straight.
The 20000 pounds of force mentioned by Walter is just an approximation of the measure of the surface area of the cylinder in square inches multiplied by the maximum cylinder pressure in PSI and is the force being applied to the cylinder head during each power stroke. This is the force that the head to barrel connection must resist, with the appropriate safety margins to allow for changes in temperatures etc.
Edit: I posted this while Walter was posting. I have just learned some more. Thanks Walter.
Now assume that the vice is tensioned up to clamp onto the thread so now you have the frictional resistance caused by the pressure of the vice on the thread and also the shearing force required to strip out the thread, to be applied before you can pull the bolt from the vice.
I am neither an engineer nor a LAME but I suspect that, in the case of the cylinder heads under discussion, both the thread and the shrink fit are required to ensure that the cylinder head remains on the cylinder barrel. I always believed that the thread was the primary way that the head was secured with the shrink fit ensuring that the thread both on the barrel and in the head remained very tight so that there was no tendency for the head to fret or move during operation, especially as they are subject to such varying temperatures.
Again see my disclaimer above and I certainly will be happy if I have this wrong and someone with accurate information (data) sets me straight.
The 20000 pounds of force mentioned by Walter is just an approximation of the measure of the surface area of the cylinder in square inches multiplied by the maximum cylinder pressure in PSI and is the force being applied to the cylinder head during each power stroke. This is the force that the head to barrel connection must resist, with the appropriate safety margins to allow for changes in temperatures etc.
Edit: I posted this while Walter was posting. I have just learned some more. Thanks Walter.
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Forces on a thread are not distributed evenly across the whole of the thread face. Meaning say the nut at the closest point to the surface it mates against takes more load than at the end where the nyloc is. This is why cly crack at the start of the thread and not at the end
This is very important:
This, too, is very important:
Thank you Mr Atkinson!
BTW, the interference fit is NOT at the same place as the threads. It is below the threads. The threads do not hold the head on during operation.
ICPs during a various mixtures range from 600psi LOP to 1100 psi at 50dF ROP.
20 x 600 = 12,000 pounds of force down on piston and up on head.
20 x 1100 = 22,000 pounds of force down on the piston and up on the head.
NOTE: These are peak pressures (stress) not mean pressures (HP).
Both of the above stress scenarios can actually produce the same HP!!
The former will have CHTs lower by about 50dF.
20 x 600 = 12,000 pounds of force down on piston and up on head.
20 x 1100 = 22,000 pounds of force down on the piston and up on the head.
NOTE: These are peak pressures (stress) not mean pressures (HP).
Both of the above stress scenarios can actually produce the same HP!!
The former will have CHTs lower by about 50dF.
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Fatigue is a matter of time and / or cycles. You may have a high time engine with low hours or a low time engine engine with high cycles it evens out. That's why Qantas aircraft at high hours low cycles. That's why we log starts cycles hours.
Fatigue is a matter of time and / or cycles.
However, let’s also assume that one engine is always run in the cruise at a mixture setting that produces internal cylinder pressures of around 600 PSI, and the other is always run in the cruise at a mixture setting that produces internal cylinder pressures of around 1100 PSI.
Which of those two engines is more likely to have a cylinder failure first? (Fingers crossed and breath bated for a potential breakthrough…. )
As a rule of thumb the first 3 threads of any fastener bear pretty much all the load. This is pretty much fastener design 101.
If a part fails at the beginning of the thread my first assumption would be that it is caused by a stress concentrator resulting from the manufacturing process of the thread forming. I assume that cylinder head threads are cut rather than rolled (bolts are rolled threads about 99.9% of the time). Rolling a thread is a process analogous to forging and produces a stronger thread and doesn't introduce stress concentrators from the thread cutting process.
There are formulae to calculate the thread strength. But, I doubt that its an issue because its worked successfully in aircraft engines for decades. I'd be pretty confident that the AD is the result of a material or manufacturing defect.
Properly torqued threaded fasteners do not come loose. Full stop. But typically we use fasteners tightened below the proper torque. Hence the need for Nylok nuts, lock wires, loctite, etc.
My guess would be that the shrink fit of the cylinder head to barrel is more to do with sealing than it is mechanical strength.
If a part fails at the beginning of the thread my first assumption would be that it is caused by a stress concentrator resulting from the manufacturing process of the thread forming. I assume that cylinder head threads are cut rather than rolled (bolts are rolled threads about 99.9% of the time). Rolling a thread is a process analogous to forging and produces a stronger thread and doesn't introduce stress concentrators from the thread cutting process.
There are formulae to calculate the thread strength. But, I doubt that its an issue because its worked successfully in aircraft engines for decades. I'd be pretty confident that the AD is the result of a material or manufacturing defect.
Properly torqued threaded fasteners do not come loose. Full stop. But typically we use fasteners tightened below the proper torque. Hence the need for Nylok nuts, lock wires, loctite, etc.
My guess would be that the shrink fit of the cylinder head to barrel is more to do with sealing than it is mechanical strength.
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As in life there are no guarantees. If the engine is run as per the book you generally don't have problems. As I said before this is really not a lop issue. The engine is designed around the px in the chambers. The bearings will fail before the head. Rapid cooling is something else which is hard on an engine. Flight schools and meat bomb aircraft suffer the worst cracking. That's why you have reduced cruise power settings.
So which will crack first. If you have an aircraft that is px to a max diff of 8.5 psi and you fly it un pressurized will it crack more at max diff or with no px in the cabin.
So which will crack first. If you have an aircraft that is px to a max diff of 8.5 psi and you fly it un pressurized will it crack more at max diff or with no px in the cabin.
No, that won’t do at all. Learning difficulties are no excuse for deliberate obfuscation.
So, let’s assume we have two identical model piston aero engines that were manufactured at the same time, are fitted to the same aircraft type and have exactly the same number of cycles and exactly the same time in service with the same operator.
However, let’s also assume that one engine is always run in the cruise at a mixture setting that produces internal cylinder pressures of around 600 PSI, and the other is always run in the cruise at a mixture setting that produces internal cylinder pressures of around 1100 PSI.
Which of those two engines is more likely to have a cylinder failure first?
It’s not a hard question, and it won’t hurt you to answer it.
Fatigue is a matter of time and / or cycles.
However, let’s also assume that one engine is always run in the cruise at a mixture setting that produces internal cylinder pressures of around 600 PSI, and the other is always run in the cruise at a mixture setting that produces internal cylinder pressures of around 1100 PSI.
Which of those two engines is more likely to have a cylinder failure first?
It’s not a hard question, and it won’t hurt you to answer it.
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They both may or may not make o/h. If you gong to continue with personal abuse two things will happen you decide. If the engine is run to its book it will and should make o/h. Cly px have not much bearing on head failure period. Detenation which has the most extreme px will damage the bearing first