What is the compression force of a ring seated inside a cylinder

[SirMarx01]

Hello.
I am finding out how many normal force between ring and cylinder. I try to find around for many hours but get noting. there are 8.2 mm inner diameters than put in 8.6 mm cylinder. The hardness of sealing element is 55 ShA. Contact area is 48.63 mm^2. I have data and drawing in attachment file. There are equation for this situation or not?
Regards

P.S.- I can convert to get Young’s Modulus. I got 1.919759658 MPa

 

[Baluncore]

Welcome to PF.
The normal force applied by the ring to the cylinder wall is NOT due to the elasticity of the ring material.

On the high pressure side of the ring, gas passes between the ring and the wall of the groove in the piston. That gas enters the groove space behind the ring, which then forces the ring outwards against the cylinder wall. The normal force is a function of gas pressure, not the spring nature of the ring. Rings are made from cast iron or steel, not from spring steel. There is very little change in ring OD as it remains in contact with the cylinder wall throughout each stroke. Some elasticity is needed in the ring to make sure it initially seals against the cylinder, and that it can gradually open as it wears over the life of the engine.

My explanation is in terms of piston rings in engines. The same understanding applies to plastic and rubber hydraulic seals in the cylinder wall, such as hydraulic rod seals, and to rubber O-ring seals. They all rely on pressure on one side of the ring to make a secure seal on the other side.

 

[SirMarx01]

Baluncore
So, I'm sorry for misunderstand. I find problem about Sealing element drop from production when garb from Packaging. So me study is to find max force to pull out sealing element from production. the resistant force apply with weight and fiction.
This is my study only compound not when assembly.
Regards.

 

[Baluncore]

I do not understand the statement.
Please identify your country and what national language?

There is no groove shown on your diagram to hold the seal.
Without a groove the seal may roll on piston and lock or slide.
There is no fluid pressure shown to force the seal against the internal part.

Define parts.
Is central material that contacts ID of seal? a cylinder? or a cylindrical piston?
What is a seal? What is a grommet?
Is the seal different to the grommet? Or are both the same thing?
Is the seal one surface of a grommet? What do you call a grommet?

Does "Put in cylinder" mean insert a cylindrical piston rod, or maybe put seal inside an external cylinder?
Friction depends on materials that contact. The subject is “Tribology”.
What is seal material? Rubber, plastic or PTFE?
What is internal inserted material? Metal or plastic?

Friction is dependent on material and “surface lubrication”.
Are surfaces assembled dry? Or lubricated with oil, grease or water before assembly?

 

[JBA]

In addition to the above, the amount of pressure (bar) is being sealed with this ring and what is being sealed (gas or liquid).

 

[SirMarx01]

this sealing element, we called Grommet. But forget it. it made of fluorocarbon rubber or just soft rubber and Shore A-hardness is 55. (can convert to Young’s Modulus is 1.919759658 MPa) normal diameter is 8.2 mm and cross-section area of 1 side is 8.234 mm^2.
Cylinder is a protect sensitive part inside product made of stainless steel with outer diameter 8.6 mm.
this product will assembly with slot and there are lock clip. But the problem is before assembly sealing element is drop out in packaging from transportation or pick up force.
last one, this product used with fuel oil.

 

[JBA]

What does the receptacle look like where the grommet is inserted; and, what is the tolerance of the diameter of projection where the grommet sits?

 

[SirMarx01]

we didn't measure grommet after slide in product, So I have to measure that ?.

I need to know how to find normal force of sealing element when increases diameter.

sealing element just slide in.it didn't rotate or deform by eye.

the part is dry not use any lubricate.
I need to know how to find normal force of sealing element when increases diameter.

 

[JBA]

After some consideration I decided to analyze your problem as a thin wall cylinder with internal pressure, starting with the grommet's O.D. strain when installed on the plug, the resulting cylinder hoop stress, the resulting internal pressure, multiplied by the contact area of the grommet I.D., that resulting force x rubber on steel friction factor = resisting force of grommet to sliding off of the plug. Below is a screen print of my resulting calculation.

 

[Tom.G]

Would it be easier to open the package from the bottom and pull the part out from there?
Or turn the part over so the the grommet is at the top of the package?

 

[SirMarx01]

JBA
I don't know about Your Calculation of 1.OD Circumferential stretch 2.Pi Contact Pressure 3.Convert N to Kg and 4.where do you got Friction Factor.
Regards.

Tom.G
the product are different height. the packaging is for universal for this product by place on this way. and normally there are Plastic mold to block sealing element. But this time they don't want it, So it have a boom(Problem) happen.
Also if upside-down it can cause some foam's scrap when put in packaging. it will damage inside product.

 

[JBA]

I would have preferred to simply attach the Excel file so you could see all of the relevant calculations; but, some of the forum members are uncomfortable about the safety of posted Excel files.

1. The O.D. circumferential stretch comes from the expanding of your grommet I.D. to the size of your plug O.D. when the grommet is installed on that part. In the calculation it is the result of multiplying the uninstalled O.D. x the strain value above it. The amount of strain is the amount of stretch from the installation on the plug.

2. The contact pressure is the result of using the thin cylinder formula for the stress resulting from an internal pressure. Using the amount stretch (strain) and your E value I calculated the cylinder stress. Next using the thin cylinder formula, which is: σ = P x Do / 2 t. Where: P = the internal pressure, Do = cylinder O.D.; and t = radial wall thickness of the cylinder. By rearranging the formula to: P = 2σ / Do then I can determine the internal pressure on the grommet required to stretch it.

3. I am not sure to what you are referring, I didn't need to do any Newton to Kgf conversion for the calculation; and I didn't waste my time converting the Kgf value to Newtons when I finished; because, I see that as an unnecessary waste of my time. (Values in Newtons are totally unusable for calculations until converted to Kgf)

4. As to the source of my f factor, the reference website for that factor is on my #9 post attached calculation. (I don't know where you found the abridged copy attached to your post.)

 

[SirMarx01]

please help me to fix it.

 

[JBA]

I think the easiest way for me to do that is to post my original Excel calculation.

 

[SirMarx01]

@ all of you. Thanks a lot for your tips.
I will keep going with theory of Elastomer's Stress and stain to find it out.
and the last thing, JBA I thank you very much for calculation. If I run out of time I will use your calculation.
Best Regards.