Drive screw to position Piston during assembly to remove fluid to compensate for thermal coef @ 300F

[johnchrc]

I know there are many post about Power Screws but the results seem too low. And I want to use a 1" ot 1-1/4 SHCS x 6" Long

I have hydraulic reservoir SS that has spring return Piston when custom Nut (Ampco 45)is assembles to cylinder and will remain assembled. The screw will be backed out prior ro asembling hydraulic pressure line from pump. Spring Free Length is 22.71". Assembly length is 15" @ 168# witehe Piston h 18.1 #/in spring rate. At final position l = 10.38 at 223#. The purpose of the screw is to drain fluid fro cylinder to compendsate for DP at 300F. Te reason can't have pressure increase is there are three (call them pistons) that will come out when pressure is applied from pump. So I hold the so called pistons doen with C=Clamps during operation.

I the past the Nut wasn't in the assembly so you had no hydraulic disconnect. The piston against the nut at 15" keeps fluid retained. Fluid id filled through parker plug ports (#2 ORB).

My question has to do with the clamping pressure of screw at P1 and P2 and the resulting torque. I calculate 1700 ft-bf torque to get to postion 2 because I begin compressing hydraulic fluid to 1000 psi hich equates to 2100 # load.

Should I purchase a rolled acme thread and modify it with split gauge for P1 and Welded Nut for applying torque and shouldering at position 2.

I have to make to nut. It has 1.69 OD with. Nut must be larger that 1.69 to shoulder at final position. However, P1 is most importatnt. Will put in autoclave at 300F and make sure no psitons expand as final test.

Use thermal coef of epansion for Univis HVI DH = .00039/F. Find DP then use Buls Modulus to determine DV to drain.

My concern is the torque requirement. I'll add a tommy bar as required. Off the shelf parts desired. Nut must be custom made due to connections at top and bottom.

I did attach a sketch that will give you some idea of what this is

- CJ

 

[Screwman1]

I think that you are off on your torque by at least a factor of 10 (too high). If I use the old T-= KDW formula for threaded fasteners and set the K factor at .05 (reducing it from the normal .15 because you will have no underhead friction), I get a total 131 in. lb to get 2100 lb. force.
If we change the joint a little bit and look at it as a regular threaded fastener where about about 50% of the input torque is absorbed in underhead friction and you are trying to create clamp load; at your 1700 ft. lb. torque you would be pulling around 81,000 lb. of clamp force in a joint.

 

[johnchrc]

Screwman,

There MAY be shoulder torque. You have a piston shouldered against a spring which will apply a shouldering torque unless I put a bearing in the cylinder. Also, as the piston o-ring pressure inceased, frictional forces at 1,000 psi will act to keep it from spinning.

I have a meeting at 9:00 but I'll try to explain better. I read my ost and it is confusing because I was in a big hurry. I agree it is like a clamping force but simpler in some regards. It is a straight thread made up through Ampco 45 fixed collar and hits a free floating piston with ~ 138# sprig force. That is at L1 of 15". It is then torqued down to L2 and compressing spring and hyd. fluid which generates a force of ~ 2100 lbf.

I have some spring calcs that show 332 in-lbf (~30 ft-lbf). Seems very low.

The L1 position is just to bleed oil to compensate for temp. The unit is then plugged. The pistons are released and pressured to ensure they come out and no leaks. Screw is removed and pistons should come back to original position.

I attached my spreadsheet for you to look at. Maybe there is an error in the calcs. It is very simple.

- CJ