Learning Forum

[Theresa Schma]

Hello Ansys Community, i have a question about connections (especially body to ground) in Ansys Mechanical.

I created a simple model with two beams, which are so placed that they have concentric drillings (see in the picture below). I added a rotational connection (body-ground) on the surface of the left drilling and a rotational connection (body-body) on the concentric drillings in the middle of the two beams where they overlap. I also added a small force of 1 N on the right surface of the right beam and limited the movement to the x-y-plane.

My aim was to check the equilibrium of forces. Therefore i looked at the results of the connection samples, but the results were really weird and the equilibrium didn´t equalize.

After i received those results i created another model. I didn´t change most of the settings except that i replaced the body-ground connection of the left beam to fixed support. After i checked the results the equilibrium of forces equalized.

 Can anyone explain to me, how Ansys is calculating this?

Thank you very much for your help!

 

[peteroznewman]

The body on the left with the fixed support is fully supported so poses no problem for solving.

The body on the right with a revolute joint between the two bodies and a force at the right end has one unsupported degree of freedom, the rotation about the joint axis. Click on that joint and notice which direction the Z axis (the revolute axis) of the Reference Coordinate System is pointing.  Is it parallel to the Global Z axis or is it parallel with the X or Y Global Axes?

If it is parallel with the Global Z axis, this is probably what you intended but would pose a problem for solving because there would be no stiffness to resist the 1 N tip force unless you added a Torsional Stiffness to the revolute joint. The solver would fail without that torsional stiffness. If you turn on Large Deflection under the Analysis Settings it may be possible to get the link to rotate up 90 degrees since the tip force does not rotate with the link, and will be pulling the link up and putting that link into tension. Is that what you intended?  

How did you limit the motion to the XY plane, was that by setting a Displacement BC of Z = 0?

If the Joint Z axis is aligned with either the Global X or Y axes, then the Displacement BC of Z = 0 means that the revolute joint can’t rotate and acts more like a fixed joint allowing the 1 N tip force to travel through the two links to ground.

[Theresa Schma]

Thanks for your answer!

The Z axis is parallel to the Global Z axis and i also added a Torsonal Stiffness of 10 Nmm/°. Large Deflection was also active. I limited the motion as you guessed by setting a Displacement BC of Z=0. 

As i mentioned before, my aim was to check the equilibrium of forces and therefore i added a small force of 1 N. I thought that this small force would probably move the model just a bit and i don´t have the problem of the force not rotating with the body. But that isn´t true isn´t it? Even if it´s a very small force, the body would always move along a circle and therefore i need to add a torque.

I also tried to solve the equilibrium of forces with the same model before but instead of adding a force i added a torque of 100 Nmm. But there i had the same problem. The results of the connection samples were far away from 100 Nmm. 

Is the problem still the unsupported degree of freedom?

Thank you for your help!

 

 

[peteroznewman]

The distance of the 1N force from the revolute axis is 450 mm so the Torque from that force is 450 N.mm.

The torsion spring on the revolute is 10 N.mm/degree so you would expect 45 degrees of rotation at equilibruim if the 1 N force rotated with the link. But the force doesn't rotate so the force produces decreasing torque as the rotation angle increases. Therefore the equilibrium angle will be less than 45 degrees.

Since you have a torsion spring on the revolute, the link does not have zero stiffness in the rotational degree of freedom so the solver can successfully solve to find the balance of forces and torques at equilibrium.

 

[Theresa Schma]

Okay, thank you for your help!