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]

Thank you for letting us know what you decided.

Ted

 

[tmschrader]

Hello
I am gonna assume that you measured the piston diamater then calculated a area of .88 sqin. IF you measured a .888in with the caliper the calc needs redoing.

P x A = Force
3947psi x .608sqin= 2399lbs of force (close enough)

PHI x R Squared = AREA
R=.888/2=.444
PHI x .444 squared= .608sqin

 

[chicopee]

So now that we have the matter of pressure and force clarifies, I do have a gripe about the free body diagram as the direction of the force should be opposing the action of the pressure on the head of the piston.