How to pressure and force in a telescopic cylinder are related 2

[amrbekhit]

Hello all,

I am familiar with the method of calculating the required pressure in a hydraulic cylinder in order to generate a desired force: P = F/A, where A is the area of the piston. However, looking at cutaway drawings of telescopic cylinders shows that it is more complicated internally, with multiple pistons and each one letting fluid into the next. How would I go about relating the force and pressure in that situation?

Thanks

--Amr

 

[hydtools]

The equation is the same. The equation is used multiple times as each section becomes inactive. The area changes as each stage reaches its limit and stops on extension. Start with the main cylinder and reduce the area to those sections that have not reached their stops.

see for ref:

http://www.hydraulicspneumatics.com/200/Issue/Article/False/21653/Issue

Ted

 

[amrbekhit]

Hi Ted,

Thanks for your post. Let's see if I understood this. According to the link you provided, the first stage, along with all the sub stages extend as one unit first. When the first stage hits its stop, the second stage and all its sub stages extend as one unit. This continues until the cylinder has been fully extended.

If we consider the cylinder shown in the attached image, with the cylinder fully retracted, would the force equation be as follows:

F = P*A2 + P*(A3-A5) - P*A4 + P*A6?

Likewise, with the first stage fully extended, would the equation be as follows:

F = P*(A3-A4) + P*A6 ?

Also, I do not understand why the cylinder extends in the way described by the article and not simultaneously. In the example I have attached, with the cylinders fully retracted, won't the fluid pressure exert forces on A3 and A6 at the same time? Will this not cause the second stage to extend independently of the first stage? Perhaps A6 is always less than A2?

--Amr

 

[hydtools]

Amr,
Consider only the areas of the seal diameters.
In the case of the example call the area of the first stage seal A1 and it is found to be the area of the first stage cylinder outside diameter D1. Call the area of the second stage A2 and it is found to be the area of the second stage cylinder outside diameter D2.

When fully retracted and with a load F find the pressure by
P1 = F/A1. When the first stage reaches its stop find the pressure by P2 = F/A2. A1 is always greater than A2. If there is a load F, then all the stages move at some low pressure P. Pressure P is applied to all external seal areas which add together to efffectively equal the first stage cylinder area. If there is no load F, then which ever stage has the least seal friction will move at some low pressure P and that may be the last stage cylinder which has the smallest seal.

If you want to think in terms of annulus areas, then consider the area of the first stage to be (A1-A2). Then when the cylinders are fully retracted the pressure is found by P1 = F/((A1-A2)+A2)) = F/A1. When the first stage stops subtract (A1-A2) from the total area (A1-A2)+A2 and find the pressure by P2=F/A2.

I did not use your definitions of the areas. You are including internal areas which cancel each other when the internal pressure distribution is considered. The net force areas are the external seal areas.

Ted

 

[amrbekhit]

Hi Ted,

I'm (hopefully!) beginning to see how it all works. Here's another diagram to clarify:

A1 would be calculated as Pi*D1^2/4. A2 would be calculated as Pi*D2^2/4. When fully retracted, F = P*A1. When fully extended, F = P*A2. So as the cylinder extends its stages, the force it can produce decreases, correct?

--Amr

 

[desertfox]

Hi amrbekhit

Perhaps I am missing something but from your diagram area's A2 and A6 would be the area equivalent to a solid diameter of the larger piston, so wouldn't the maximum force just be that equivalent area multiplied by the pressure.

regards

desertfox

 

[Compositepro]

Amr's second drawing is correct. Each successive stage has a lower maximum force due to the smaller piston areas. That is also why stage two will not extend until stage one has fully extended.

 

[rmw]

AMR,

To go back to the question you asked at the end of your second post, it would extend all the way simultaneously if you had a pump capable of pumping an infinite amout of fluid all at once. The rate of extension is a function of your pump's ability to fill the cylinder(s), or to say it differently it's pumping rate. The force produced at any given time is a function of the reaction to the pump's output pressure. If you encounter a force that is more than your pump's ability to put up the pressure to overcome it, the ram will stop moving (and open the overpressure protection valves, I hope).

rmw

 

[hydtools]

Amr,
No. In your sketch, A1 = pi*D2^2/4. The seal for the first stage is at diamter D2. If you notice in the picture, pressure is communicated to the space between your D1 and D2. That annular space just swaps oil out and in as the cylinders move out and back. D1 is simply a bearing diameter to stabilize the cylinder. the effective push diameter for stage 1 is the seal diameter D2.
The same is true for stage 2. Let's call the seal diameter for stage 2 D3. A2 = pi*D3^2/4. Again oil pressure is communicated to the space between D2 and D3. Your D2 is a bearing diameter to stabilize the cylinder and not a seal diameter. In your simplified shetch you have neglected a wall thickness for the cylinders so your D2 cannot be both stage 1's seal diameter and stage 2's bearing diameter, but the thought is there.

Ted

 

[Compositepro]

I have to agree that hydtools is correct. The first drawing you posted appears to be for a single acting cylinder (oil pressure cannot retract the cylinder), so the piston area is the cylinder rod area. There are double acting telescopic cylinders.

 

[amrbekhit]

Hello all,

One last diagram! Again, want to make sure that I have fully understood this.

Am I there yet?

--Amr

 

[hydtools]

Looks like you are there.

Ted