How to add force on Nodal boundary condition

[shawn]

Hi,

I want add force on a single node. When I set as below it shows "Solution trivially zero!".

Boundary Condition 2
Target Nodes(1) = 31
Force 1 = 2000
End

I search the mailing list, and changed as below. It shows error when reading "displacement 1 load". So how can I add the force to the single node?

Boundary Condition 2
Target Nodes(1) = 31
Displacement 1 Load = 2000
end

[raback]

Hi Shawn

Your syntax has at least the problem that you should decleare the type of the condition i.e.

Code: Select all

Boundary Condition 2
  Target Nodes(1) = 31
  Displacement 1 Load = Real 2000.0
End

The reason for this is that even though 'Displacement 1' is a known keyword all the derived keywords are not defined in the SOLVER.KEYWORDS file. It would be impossible to list all combinations. You can still use unlisted keywords but you must define their type (Real, Integer, Logical,...).

BR, Peter