Reaction force to a flat and curved surface study

[jmex]

Hello Engineers,

I have two different geometries, where i know the value of pressure. I would like to know the reaction force on both the scenarios. Applying the same pressure on both the geometries, there is not at all difference in reaction force in vertical direction (even though the surface area is different).

If this is not the way we could find the reaction force, kindly suggest one.
I thought the applying force in vertical direction on the component surface would make two different vectors (Horizontal and Vertical). But it didn't happen.
Thank you!

 

[jambaugh]

Note that the force applied to the surface by pressure is a vector normal to that surface.
Try it with a simple polygon, figure the vectors and notice how the horizontal components all cancel_(edit) and the vertical components are exactly the same as would occur if the surface had been flat. Calculus will let you calculate the limiting case where the polygon becomes a smooth curve but a simple two or three facet surface should convince you.

 

[jmex]

Note that the force applied to the surface by pressure is a vector normal to that surface.
Try it with a simple polygon, figure the vectors and notice how the horizontal components all cancel_(edit) and the vertical components are exactly the same as would occur if the surface had been flat. Calculus will let you calculate the limiting case where the polygon becomes a smooth curve but a simple two or three facet surface should convince you.

Thank you for replying, I agree with you that the pressure will be normal to the surface. It will have horizontal components (which will cancel out each other) and vertical components which will be in upward direction. I am using FEA simulation application to evaluate the reaction force due to little complex geometry to evaluate exact R_force. But it gives me the same result for flat surface and curved surface. That is the reason I got confused and had to come for help. What am I missing here.

 

[jambaugh]

Ah, My apologies for not understanding your question.

Where are you implementing the Finite Element simulation? Is it for the fluid or for the piston? If the piston then I think my answer, still works in that my "polygon approximation" is equivalent to a simple finite element decomposition and "doing it by hand" should be clarifying. If you are modeling the fluid then, if its a static problem there's nothing to consider, the fluid will have uniform pressure and density.

Now, if the piston is not symmetric, i.e you have a one-sided wedge shaped piston, then you do indeed get a net lateral force by the fluid on the piston. You then have to account for the guiding forces on the piston either by the cylinder walls or by some off-diagram constraint on the driving shaft. There is likewise a force of pressure on the piston and again these should cancel (in static case) so that the piston will not have a net lateral force on it.

If this isn't clarifying then I'd like to hear more details about what you're doing with the FEA, what you expect and what you are getting from the model.

 

[jmex]

@jambaugh thank you for your reply. Yes I'm using structural simulation software like Ansys/Prepomax. The pressure is exerted by a fluid on the piston and it is equal as per Pascal's law. The piston is symmetry, and hence there will be lateral components being cancelled out. The only problem is, in both the scenarios, the reaction force are same. How can it be possible?

 

[tech99]

Force per se does not depend on contact area. Howqever, the curved piston would seem to make very small contact area. So if you are told pressure, then as Force = Pressure x Area, the force in the two cases will be different.

 

[Tom.G]

I think "Swept Area" ( or possibly "Projected Area") is the missing concept here.

Conceptually it is the the 2-dimensional projection onto a flat surface of the object (surface) being considered.

Similiarly, a "Swept Volume" is the change in volume of, for instance, a piston moving in a cylinder. Also known as "Displacement."

Hope this helps!

Cheers,
Tom

 

[A.T.]

The only problem is, in both the scenarios, the reaction force are same.

It would be a problem if the force wasn't the same, despite same pressure. It would allow perpetual propulsion just by putting differently shaped ends on a vessel.

 

[jambaugh]

@jambaugh thank you for your reply. Yes I'm using structural simulation software like Ansys/Prepomax. The pressure is exerted by a fluid on the piston and it is equal as per Pascal's law. The piston is symmetry, and hence there will be lateral components being cancelled out. The only problem is, in both the scenarios, the reaction force are same. How can it be possible?

I'm still unclear as to what you mean by "reaction force". The piston exerts a force on the fluid and the fluid exerts a force on the piston. Assuming no acceleration these are equal (equal and opposite reaction forces). That force will be independent of the surface geometry of the fluid facing side of the piston and only depend on the pressure and cross section area orthogonal to the direction of motion.

Note that the work done will be the pressure times the displacement volume when the piston moves and with that, the differential volume as the piston moves likewise only depends on the cross sectional area not the shape of the fluid facing part of the piston. (differential work = force times differential distance = pressure times differential volume).