Force Boundary Condition

Hello,

I want to the use the force boundary condition to distribute 42000 N to a face set with 15 faces. Every face should have a load of 2800 N.

So I have to assign 42000 N. Is this correct?

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Hi @lap1, the force will be applied to the geometry in proportion to the surface area. So if all 15 faces are the same size you will have 2800 N applied to each face. If the faces are made up of different sizes the small faces will get less load and the large faces will get more load. The sum of all the loads on all the faces will equal the load defined in the force boundary condition (42000 N).

The applied force is not node density dependent, you can have a course or fine mesh and the load will still be applied according to surface area.

You can test this by applying a result control measure at the reaction(s). The reaction force should equal the applied load. You can also add result control measures to each load face to see how load is apportioned to each face.

Here is a simple example showing how to apply a load to multiple faces and check the reaction forces.

https://www.simscale.com/workbench?publiclink=44b37f3a-a107-4ea5-8fde-98b9e98a311b

Use the post processor to check reaction forces.

Regards, Ben

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Hi BenLewis!, I have a problem interpreting the introduction of functions to describe a non-constant distributed force. I have a rectangular plate where a section on top of it (0.05m long -i.e. x direction- by 0.1m wide -i.e. z direction) is to be loaded with a distributed force (acting down on the y direction) that in the plane xy would look like a triangle, whereas its value remains constant in the z direction. I wrote a function where fy = 2000x, fx=0, fz=0. I have not had the results expected when I try to validate the results with some simple reaction force calculations…so I guess I’m not interpreting the aforementioned functions corectly. What does fy=2000x mean from the point of view of a distributed force over a surface, as it is represented in textbook’s static analyses? The free documentation about this (Docs>Simulation>Model>Boundary Conditions>Force) shows very little detail in this sense. Thanks in advance.

Hi @embarrera,

The equation you have given (fy=2000*x) will apply a triangular shaped UDL. At an x-location of zero the load in the y-direction will be zero and at an x-location of 0.05 m the load in the y-direction will be 100 N.

The equivalent point load will be q*L/2 = 2000 * 0.05 / 2 = 50 N. Please see this video for an explanation.

This means your reaction force should be 50 N. What reaction force are you getting?

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Hi @BenLewis, thank you for your reply. Indeed, I am getting 50 N as reaction force from Simscale. I got confused because, if at x=0.05m the force is 100 N, then I thought that qL/2 meant 1000.05/2. Hence, I guess that “x” in fy=2000*x could rather be interpreted as a local coordinate for the base length of the triangle shape that represents the UDL.
Kind Regards, embarrera

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