Constraint or Internal Forces?
Constraint or Internal Forces?
(OP)
This might sound trivial but I really want to ask:
A beam is modelled with 1d elements, appropriate constraints are placed on two points and a load is applied on one end. Simulation is successfull and now the question:
If one wants to use the fea calculated forces on the constraints in order to perform some analysis on bolts that will be placed where the constraints are, which calculated forces will be used? grid point internal forces, or grid point constraint forces? Is it the constraint forces, if I assume correctly?
Thank you.
PS: In case someone asks, sum of constraint forces equals the load applied in all directions.
A beam is modelled with 1d elements, appropriate constraints are placed on two points and a load is applied on one end. Simulation is successfull and now the question:
If one wants to use the fea calculated forces on the constraints in order to perform some analysis on bolts that will be placed where the constraints are, which calculated forces will be used? grid point internal forces, or grid point constraint forces? Is it the constraint forces, if I assume correctly?
Thank you.
PS: In case someone asks, sum of constraint forces equals the load applied in all directions.





RE: Constraint or Internal Forces?
Using the three force components would be most accurate ... you ccan work out an applied shear, applied tension, work the problem to death.
Using the single (vector resultant) force against the minimum allowable (shear or tension) is simple and conservative.
another day in paradise, or is paradise one day closer ?
RE: Constraint or Internal Forces?
In both cases the resulting forces (constraint and internal) are analysed and displayed in their respective components (x,y,z). Accordingly, the resultant force can be seen. Do you agree with my thought that the constraint forces results will be used or is it the internal forces? (they are not equal in magnitude).
Kind Regards.
Screenshots:
If values are not very clear (reading nodes from left to right as you see the pictures):
first image constraint forces: Fy=4e05 Fx=6.64e04, Fy=4.42e5, Fx=1.73e05
second image internal: Applied load (left end): Fy=1.31e05, Fx=6.64e04, Fy=2e05 Fx=3.32e04, Fy=2.21e05, Fy=8.66e04, Right End of beam: Fy=2.33e-10 (negligible)
RE: Constraint or Internal Forces?
Freebody loads = -Sum(Element forces).