### Release DOF

Verfasst:

**Sa 15. Dez 2012, 15:34**Is it possible to release a nodal degree of freedom for an individual element?

thanks.

thanks.

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Verfasst: **Sa 15. Dez 2012, 15:34**

Is it possible to release a nodal degree of freedom for an individual element?

thanks.

thanks.

Verfasst: **Sa 15. Dez 2012, 15:52**

What do you mean exactly with 'releasing a DOF'? Please explain what you want to do.

Verfasst: **Sa 15. Dez 2012, 18:23**

For a beam element, is it possible to make the stiffness for any of the 6 degrees of freedom at one (or both) nodes zero (0)? This is sometimes refered to as a nodal/joint release or pin flags, or slide joints, etc...

Verfasst: **So 16. Dez 2012, 11:55**

I understand. You want to define a join to a node.

Unfortunatelly, this is in general not possible for ordinary beam elements - one needs two special beam elements, one with all dof on the left side but no bending dof on the right side, and one element just vice versa. Combining this two special elements will define a join to the middle node.

A half year ago I began trying to implement such elements but the Z88 solver operates always with symmetrical elements (my blame because I programmed the linear solver Z88R), i.e. the same number of dof for each node.

But - please give us some time, we're just programming an enhanced beam element, and perhaps we'll include this feature.

Regards

Prof. Rieg

Unfortunatelly, this is in general not possible for ordinary beam elements - one needs two special beam elements, one with all dof on the left side but no bending dof on the right side, and one element just vice versa. Combining this two special elements will define a join to the middle node.

A half year ago I began trying to implement such elements but the Z88 solver operates always with symmetrical elements (my blame because I programmed the linear solver Z88R), i.e. the same number of dof for each node.

But - please give us some time, we're just programming an enhanced beam element, and perhaps we'll include this feature.

Regards

Prof. Rieg

Verfasst: **So 16. Dez 2012, 15:38**

Is it not possible to have an interface that accesses the beam element stiffness matrices and and operate directly on those?

Many thanks.

Many thanks.

Verfasst: **Mo 17. Dez 2012, 16:21**

This is not possible at the moment. A possible workaround could be to define a beam element with custom geometry values, for example defining Iy = 0 while the element still has a crossection A != 0. You can define a custom geometry via Pre-Processor -> Element Parameters -> Section -> Various.stopwalve hat geschrieben:Is it not possible to have an interface that accesses the beam element stiffness matrices and and operate directly on those?

I hope this helps.

Regards

Felix Viebahn

Verfasst: **Mo 17. Dez 2012, 17:06**

Unfortunately, operating on the beam section properties won't work. For example, assume you have two cantilever beams with a common node:

Many thanks however. Perhaps in a future version.

Regards.

The beams are still capable of carrying bending. However the bending Degree of Freedom is zero for the common node.Many thanks however. Perhaps in a future version.

Regards.

Verfasst: **Di 18. Dez 2012, 13:43**

Now I understand the problem you are trying to solve. My "workaround" does not do you any good in this case.

Hopefully we will be able to remedy this in a future version of Z88 Aurora.

Regards

Felix Viebahn

Hopefully we will be able to remedy this in a future version of Z88 Aurora.

Regards

Felix Viebahn

Verfasst: **Di 18. Dez 2012, 15:19**

Once again, many thanks and good luck.

Verfasst: **Mi 19. Dez 2012, 14:43**

Thank you for your input, you are very welcome!

Regards

Felix Viebahn

Regards

Felix Viebahn

Verfasst: **So 27. Jan 2013, 15:32**

I’d like to offer two suggestions that might help…stopwalve hat geschrieben:Unfortunately, operating on the beam section properties won't work. For example, assume you have two cantilever beams with a common node:The beams are still capable of carrying bending. However the bending Degree of Freedom is zero for the common node.

Many thanks however. Perhaps in a future version.

Regards.

The first suggestion is if you are a theoretician and modeling exactly the example you gave and wanting the “purist” answer. If the example is the X/Y plane, then using symmetry about the center point, you can constrain the X, Z, RX and RY to zero. Leaving Y and RZ free. This would “exactly” model your situation. Exactly - defined in the theoretical FEA world.

Now, if you are more like me where you are trying to solve a real world problem and approximations are of value, especially, if you are really interested in the far-field displacements and stresses, then here is an alternative.

Instead of expecting it to hinge about the node, let an element itself hinge based on its section properties. By this I mean add a short (several orders of magnitude shorter than other elements) element to be in the location of your “hinge” point. Create a different set of section properties for this one element. Use the “user defined” properties option and set the bending properties to be several orders of magnitude smaller than other section properties. Here is your example using this method. Using this model, I got less than 1% error in the y displacement between this and the symmetric “purist” case. Note – Using zero for these “special” properties causes the stiffness matrix to fail and using too small properties and you start getting shear artifacts. You’d have to play with it for your specific example.

Notes

• The model is 200 units long

• The elements are 1 unit long

• The Iz of the elements are 0.08333 units

• The hinge element is 2.0E-5 units long

• The Iz of the hinge element is 1.0E-10

Hope this helps.

BTW – I use this method to model automobile control arms hinging off a chassis.