I need clarification on a hydraulic cylinder applied force issue. 2

[T53mechanic]

Hello everyone. I am currently studying to become an Aerospace engineer. Due to school, I ran into a dilemma at my job. I am afraid it may be a safety of flight issue. I did propose this question to my class, but before I go to our CEO about this issue, I need to confirm my assumptions are correct. I want to reiterate that this is NOT for school work. This is an actual, real life issue that is unrelated to school.

I build jet engines for a living. These engines have turbine blades that require typically 66 pins be pressed into slots to retain their position. This cylinder is a single actuating cyl much like that of a bottle jack or hydraulic press. The specified pressures are between 2,300 and 2,450 psi to deform the pin to keep it in place. There is a dispute between another tech and I about the actual pressure being applied to the retaining pin. That leads me into the equation that I would like some insight on.
In the attached image, it mimics what the cylinder is doing. Diameter 1 is 0.997in (this is the actual hydraulic piston). There is a head attached to this piston it retains a driving pin with a diameter (D2) of 0.086in. We are currently applying 2300psi. Does this directly translate to 2300psi of force at the 0.086in pin? or is it somewhere around 1808 psi of force?

 

[hydtools]

Do the math. Pressure times area equals force. And force is expressed in pounds, not pounds per square inch, psi.

Ted

 

[T53mechanic]

Ok, so if you wouldn't mind confirming for me, 2300psi * (π*0.043)²=13.363 lb/F?
So the diameter of the piston does not matter in the equation?

 

[IRstuff]

No, your equation is incorrect pi*r^2, not (pi*r)^2, but the result is correct

Diameter probably matters, but you've not really described what you mean by "deform the pin." Are you referring to a rivet deformation, or a press-fit deformation? In the former, diameter dictates volume and stiffness, which determine buckling, while the latter requires displacement of material along the diameter.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[T53mechanic]

I hope this clarifies what I am trying to figure out.

https://postimg.cc/gallery/1eoq6dqm0/

 

[TimSchrader2]

Hello

The pressure of 2300PSI is on the .997 Dia piston and develops a force of 1795lbs that goes thru the rod. If all that force is on the head of the .086IN dia pin then that is a pressure of about 309,482psi on the pin. Which is enough to deform it.

 

[PNachtwey]

There is the side of the piston where pressure is applied but what about the pressure on the rod side of the piston. The opposing pressure/force will reduce the net force.
You should be thinking in terms of net force. Pressure is just a means of creating that force. It is the force that will insert the pin.

Are you sure that obtaining a pressure or force is really what you want to do? Isn't driving the pin to the right position most critical? The force or pressure necessary to drive the pin to the correct position should be relatively consistent and if it isn't then something is wrong.

Peter Nachtwey
Delta Computer Systems

http://www.deltamotion.com

http://forum.deltamotion.com/

 

[T53mechanic]

Our manual states the following:
"provide 2375 to
2425 pounds of force for driving the blade retaining pin."

 

[IRstuff]

OK, so a serious failure to communicate. The force that's applied goes into displacing material in the rotor wheel so that the pin will not come back out on its own, and will be essentially unremovable. Presumably, the pin cannot be pushed into the hole by hand. The pressure is irrelevant, you need a certain level of force to overcome friction and push the rotor material aside.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[T53mechanic]

What brought this issue about is the manual gives instructions for a different type of machine. It uses air pressure to drive hydraulic pressure to deliver the 2375 to
2425 pounds of force. I will post the whole section of the manual, however please keep in mind that I have not seen this machine in person, nor do I know exactly how it works. I am merely trying to verify that we are using the same amount of force that is required in the manual.

https://postimg.cc/gallery/2tgfak5iw/

This is all the data I have for the machine. Once again, this is not the same machine that I took pictures of earlier in the thread.

 

[IRstuff]

How is it possible to remove the blade after the procedure? If you can manually remove the blade, what's the point of the pin?

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[T53mechanic]

The blade is driven out with a hammer and brass punch at the root to sheer the pin. Then the machine that presses in the pin, also presses the remainder of the pin out after the blade is removed. Everything gets NDI after the parts are disassembled.

 

[hydtools]

Back to the math. Pressure required to develop your stated required insertion force = force/piston area.
Piston area = (pi*0.997^2)/4 = 0.783 sq in
2375/.783 = 3033 psi


Ted

 

[IRstuff]

But your procedure says that the blade is re-installed. Is that actually allowable?

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[T53mechanic]

Thank you Ted, that was one of the answers I came up with as well. I was afraid we were applying too little pressure to properly seat the pins.

IRstuff, Yes it is allowable, the blades are much harder than the pin. As I stated before, the blades go through a florescent penetrant inspection (FPI) as a form of non-destructive testing (NDT)each time they are removed. The manual does not call for it to be done each time, but we do so since we have in house NDT.

 

[LittleInch]

T53 mechanic,

I'm still getting the feeling here that you fundamentally are confusing pressure and force. They use similar units, but you really need to be clear what you're describing.

Force is, well a force. It is measured in your case in Lbs force. In SI units this would be Newtons.

Pressure on the other hand is force per unit area, Lbs per square inch.

So if you want to develop a force of say 1000 lbs and you have a piston with a square area of say 4" ( piston diameter 2.25") then you need a pressure of 1000 / 4 = 250 psi.

Now how your air driven machine actually works is a bit of a mystery as it seems to use relatively low pressure air ( 40 psi) to somehow create a pressure on the "high pressure guage" of 990 to 10010 psi which presumably acts on a piston to create your requires 2400 lbs force (+/- 25 lbs).

So looking at your photos you MUST know very accurately the diameter of your piston in order to convert the applied pressure from your hand pump into lbs force.

2400 psi on the hand pump is NOT 2400 lbs force ( unless the piston area is EXACTLY 1 square inch).

Apologies if you have this understood, but your post continually uses psi when it should be lbs(F)

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[T53mechanic]

Thank you for clarifying LittleInch. I know that Ted (hydtools) gave the math, but in your explanation, does the 0.086 dia pin that is transferring the power from the piston matter in the equation? That has been the conflict in this whole thing it seems.

 

[LittleInch]

Does it matter - No.

The only item in terms of the small pin is whether it can transmit 2400 lbs of force ( just over a ton) without buckling or deforming the pin that is being inserted.

It is a lot of force on a small thing, but if this is how its been done and it works then who are we to criticise....

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[hydtools]

Littleinch, air over hydraulic pumps can be had to produce 1000psi or more hydraulic pressure using low pressure air. See Enerpac products.

T53, the size of the final push pin does not enter into the output force calculation for your single-acting piston. It just has to tolerate the insertion push of 2400lbs. The instructions say 1000psi will produce 2400 lbs push force. That means that hydraulic piston diameter is about 1.75 inches. If you change to a smaller hydraulic piston, you must provide more hydraulic pressure to produce the required 2400 lbs. insertion force.

Ted

 

[T53mechanic]

Ok, I went and talked to my boss and we did agree that the amount of force was not enough. I did more digging around and found that the actual piston size is 0.888. I only had the information I could get with calipers. So upon using the formula for force (P*A=F) I found that 2700psi is needed. Therefore, .888*2700psi= 2397.6lb/f.

I want to thank all of you for helping me with this. Even though I did the math all different ways, I am glad that we found the correct answer. I sincerely appreciate your responses!