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Force/Pressure

Force/Pressure

Force/Pressure

(OP)
hi,
can we create a load on a surface other than a pressure. For example, on this surface there is a 10 kN force, and Abaqus find the equivalent uniform pressure for the selected surface ?

thanks

RE: Force/Pressure

if you knew the number of nodes and they were evenly spaced you could apply the force equally between all the nodes. This would be an approximation however as the edge nodes should have only a proportion of the load. I find it easier just to divide the force by the area.

corus

RE: Force/Pressure

"if you knew the number of nodes and they were evenly spaced you could apply the force equally between all the nodes."

- only for linear elements (i.e. no mid-side nodes) could you attempt this, and then only on a regular mesh where all element faces are identical.

RE: Force/Pressure

True john, I forget the nodal force relationship with mid-side nodes. It's maybe a quarter for each mid-side node. For linear elements where the nodes are evenly spaced the mesh is regularly spaced, obviously.

corus

RE: Force/Pressure

Corus - it's one third at each mid side node and minus one twelth at each corner node for a regular four sided element face.

RE: Force/Pressure

(OP)
thanks a lot :)

RE: Force/Pressure

Use coupling on the reference point to the surface that you want to apply the force. Use distributed one.

RE: Force/Pressure

Added a vote for johnsmith2's method - it has the benefit of being mesh independent (you can remesh the surface underneath and not have to do any recalculations manually)

RE: Force/Pressure

I don't believe this method will produce a uniform pressure as requested by the OP.

RE: Force/Pressure

A distributed coupling does not eliminate degrees of freedom at the coupling nodes. It distributes loads such that the resultant forces at the coupling nodes is equal to the forces at the reference node.

RE: Force/Pressure

You can use kinematic coupling for constant pressure.

RE: Force/Pressure

Yoman I have usually used a kinematic coupling for this. But I am not sure how the behavior would differ between kinematic and distributed coupling for this application. Could you please explain?

Thanks,
Gurmeet

RE: Force/Pressure

Answer 2216 from abaqus web site will give you some details.

"Kinematic coupling is enforced in a strict master-slave approach. Degrees of freedom (DOFs) at the coupling nodes are eliminated, and the coupling nodes will be constrained to move with the rigid body motion of the reference node. "

"Distributing coupling is enforced in an average sense. Degrees of freedom at the coupling nodes are not eliminated. Rather, the constraint is enforced by distributing loads such that:

o The resultants of the forces at the coupling nodes are equivalent to the forces and moments at the reference node, o Force and moment equilibrium of the distributed loads about the reference point is maintained.

RE: Force/Pressure

From that definition Kinematic coupling is a rigid body element (like a Nastran RBE2) - this is in no way equivalent to applying a uniform pressure !

and distributed coupling (sounds similar to a Nastran RBE3) at least it doesn't fix all the nodes together but will not apply a uniform pressure either !

RE: Force/Pressure

Opp.. JohnHors, I think you are right, sorry for the confusion.

RE: Force/Pressure

To know it is correct or wrong?

Apply the loading in three cases, applying pressure, distributed or kinetic coupling loading.

RE: Force/Pressure

And you'll get three different results !

RE: Force/Pressure

You need to check the reaction forces in all directions and you will see the difference.

RE: Force/Pressure

No, if the centre of pressure for the pressure loading is in the same position and direction as the point force used in the other two loadings then all three will produce identical reaction forces. But that doesn't mean all three types of loading are equivalent, far from it !

RE: Force/Pressure

Only one direction of reaction force is the same when you compare three cases. However, you will find the difference in other two directions. So, they are not the same reaction forces, but the difference is not much. If I were you, I prefer to apply the pressure in loading.

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