# Help me for calculate a Hydraulic pomp pressure

1. Dec 30, 2008

### ahrouhi

Dear friends
i am not a physics expert but a technical man. i encountered a problem in designing a machin for pipe pinching(squeesing) . the pipe has 20 mm diameter outside and 18 mm inside made by galvanized sheet (1.0350) with Re=140-300 N/mm2 yield strength. my machine use 3 pins for squeesing each side of pipe head( 6 pin boath sides) ( together 12 pin for boath heads). the pins are rounded and have 4.5mm diameter. i decided to use 4 hydraulic jack for four direction to push the pins to the pipe(every 3 pins are attached to shaft of a jack).
my question is what power of hydraulic jach should i use?
(please mention the formulas used for solving the problem)

2. Dec 31, 2008

### ahrouhi

I know the formula for yield strength :

p=2.Y. ((D/t)-1) / (D/t) pow2

The resault is 19 N/MM2 . it seems this formula is not correct for my purpose.

3. Jan 1, 2009

### ahrouhi

i am still waiting for your guidance.,,,,

4. Jan 1, 2009

### nucleus

I don’t have a formula but I would use a hydraulic cylinder and pump (even a hand pump would work) plus a pressure gauge connect in parallel to the cylinder. Connect one of your pins to the cylinder and apply pressure and pinch your pipe and read the pressure gauge. You will know the size of cylinder hence you can determine area so you can use then pressure = Force times area to determine pressure required.

5. Jan 1, 2009

### dingpud

Is there anyway that you could post a picture of your apparatus?

6. Jan 2, 2009

### ahrouhi

Dear nucleus
many thanks for your solution. your method is applicable. but i wanted to use formula and then investigate the subject in action. i wait for your new suggestions too.

dear dingpud
i am on mission, at first time i got office , i send picture, but as i am new in this forum , i dont know how to send pic....but i try to find it.

7. Jan 3, 2009

### stewartcs

When you post a message, look at your toolbar above where you are typing. There is a picture of a paper-clip. Click on it and upload your picture.

CS

8. Jan 5, 2009

### ahrouhi

the tube and the sqeezing area and the pivot housing which should be attached to both end side of tube are shown. the yellow part of pivot housing would be inserted in tube and the red part on pivot housing is the pressing area which should be pressed in to their hole.

#### Attached Files:

• ###### Tube&Pivot.JPG
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Last edited: Jan 5, 2009
9. Jan 7, 2009

### ahrouhi

i attached a proper picture for better understanding my problem. please answer.

10. Jan 10, 2009

### ahrouhi

if the picture is not clear please let me know.

11. Jan 11, 2009

### ahrouhi

i am eagerly waiting

12. Jan 11, 2009

### nvn

ahrouhi: User nucleus mentioned a good empirical method, in which you use a hydraulic cylinder with a pressure gauge attached, apply a force, then compute the applied force F, where F = p*A, p = hydraulic pressure, and A = hydraulic cylinder piston area.

But if you do not want to use an empirical method, we can only guess using an analytic method. I will talk about only the slot, for now. If your diagram is to-scale, your slot appears to be approximately 18 mm long and 6.5 mm wide. If you are pressing (into the slot) the rounded tip of only one 4.5-mm-diameter pin at a time, then I would probably apply a pin axial force of P1 = 80*Re*d1*(t^2)/a, where Re = pipe material tensile yield strength, d1 = pin diameter, t = pipe wall thickness, and a = slot width. Therefore, if Re = 140 MPa, then P1 = 80*Re*d1*(t^2)/a = 80(140 MPa)(4.5 mm)[(1.035 mm)^2]/(6.5 mm) = 8306 N. And this is if you are pressing near the center of the slot. If you are pressing near the ends of the slot, I would multiply P1 by 3.

But if Re = 300 MPa, and you are pressing near the center of the slot, then P1 = 17 800 N. If you are pressing near the ends of the slot, I would multiply P1 by 3. Keep in mind, P1 is the axial force on one pin.

Therefore, if the pipe material tensile yield strength is Re = 300 MPa, and you press near the end of a slot, a very high-strength pin material might be required (using a pin diameter of d1 = 4.5 mm), whereas if you are pressing near the center of the slot, a normal or high strength pin material could probably withstand the stress. If you are pressing with three pins per side at one time, then multiply the above force by 3.

In the above, I assumed you want to firmly press the pipe wall into the slot. If you instead want to only slightly deform the pipe wall into the slot, then the required compressive force P1 might be slightly less than estimated above.

13. Jan 12, 2009

### ahrouhi

Dear nvn
please note to attached picture. actually the pinch bar shape is not rounded. they are in oval shape. please inform me if i want to use a pinch bar for each hole(totaly 3 pinch bar), would you guide me how to calculate?

#### Attached Files:

• ###### PivotPressing.JPG
File size:
37.2 KB
Views:
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14. Jan 12, 2009

### nvn

My current guess is, for each yellow pin, I would probably apply a pin axial force of P1 = 36*Re*L*(t^2)/a, where L = slot length = 18 mm, and a = slot width = 5 mm. Therefore, P1 = 36(300 MPa)(18 mm)[(1.035 mm)^2]/(5 mm) = 41 650 N. And for the red pin, I would probably apply a pin axial force of P2 = 70*Re*L*(t^2)/a = 70(300 MPa)(7.8 mm)[(1.035 mm)^2]/(8 mm) = 21 930 N. Therefore, if Re = 300 MPa, the total force for two yellow pins and one red pin would be 2*P1 + P2 = 105 230 N.

15. Jan 13, 2009

### ahrouhi

Dear NVN. i dont understand why u use 36 in p1(better to say f1) calculation and 70 in p2(f2) calculation. menwhile the L= length of yellow pin is not 18, its 10+1.5+1.5=13 and yallow pin width a=3.3 mm. am i right? or we have to calculate the slot size in formula instead of pin size.

and the same for red pin 7.8 is ok for length but for width its not 8 , it is 5.8 mm.

16. Jan 13, 2009

### nvn

ahrouhi: The 36 and 70 are merely my estimates or guesses. I do not have a rigorous explanation of them, as they are my current guesses. The formulas use slot size, not pin size. Yes, change the names P1 and P2 to F1 and F2.