
Re: Load on an edge of a beam: Simulation
Bill McEachern Feb 7, 2013 8:44 PM (in response to Anand Kothari)if you are using solid elements pick the edge and apply the load. If you are using beam elements they have no edges  they have the topology of a line. Then add the torque.

Re: Load on an edge of a beam: Simulation
Matthew Jackson Feb 8, 2013 12:03 PM (in response to Bill McEachern)No, do not add a torque if you want a force, you need to select a node and define a plane or face for a direction.

Re: Load on an edge of a beam: Simulation
Bill McEachern Feb 8, 2013 12:18 PM (in response to Matthew Jackson)a force on the edge of a beam implies that the force is not aligned/on the nuetral axis of the beam, hence it will manifest a torque. Again if solid elements are being used it doesn't matter just put the force where it is suppose to go. IF a beam element is used the the applied force would be composed of a torque and a force. Given that there is some ambiguity in the problem statement I will leave it at that.



Re: Load on an edge of a beam: Simulation
Anand Kothari Feb 10, 2013 1:05 PM (in response to Anand Kothari)
Re: Load on an edge of a beam: Simulation
Matthew Jackson Feb 11, 2013 3:09 AM (in response to Anand Kothari)Bill,
I semi understand what you mean but think the difference in results would be negligible and not to mention you would have to run the analysis twice.
Once to determine the slope and deflection of the beam so the offset from the neutral axis and beam edge would be known, then the moment perpendicular to the neutral axis could be worked out and a second time to add the moment effects.
If I take steel square section beam 100mm x 100mm x 1000mm long, fully restrain in all DOF's and apply a 10 kN force to the neutral axis on the free end, I end up with:
Theta (Slope of Beam) = F x L^2 / 2 EI = 10e3 x 1^2 / 2 x 210e9 x (0.1^4 / 12) = 0.00286 Rads = 0.164 degrees.
To calculate the distance (x) from the neutral axis to the beam edge:
x = sin(0.164) x 0.05 = 0.143mm = 0.000143 m
The moment therefore = 10e3 x 0.000143 = 1.43 Nm
This would make no difference to the results but would take at least twice as long to work out and is therefore not neccessary.
Anand,
To answer your question, you need to select 'selected direction', this then requires a face or plane to be chosen to describe the direction of the applied force as screen shot. Although I would be more inclined to use beam elements rather than solid elements. Hope this helps.
