Calculating Reaction Forces in ANSYS: Step-by-Step Guide"

[Laurry]

Hey,

I have a simple ANSYS model shown in link:
https://dl.dropboxusercontent.com/u/104865119/hand_calc.PNG .

Ansys reaction force in z axis is +121 N and -121 N. Can someone clarify how do you
get to this result using hand calculations?

 

[haruspex]

Hey,

I have a simple ANSYS model shown in link:
https://dl.dropboxusercontent.com/u/104865119/hand_calc.PNG .

Ansys reaction force in z axis is +121 N and -121 N. Can someone clarify how do you
get to this result using hand calculations?

I need some description of what the diagram represents. It looks like a block somehow subjected to a z-axis torque of 100Nm, while restrained at two points 1.1m apart. But in that case the restraining forces would obviously be 100/1.1N, or about 91N.

 

[Laurry]

Using the global coordinate system found in the picture the moment is about the y axis, My = 100Nm.

https://dl.dropboxusercontent.com/u/104865119/hand_calc_2.PNG

 

[haruspex]

Using the global coordinate system found in the picture the moment is about the y axis, My = 100Nm.

https://dl.dropboxusercontent.com/u/104865119/hand_calc_2.PNG

Ok, but same result. 91N, as you say.

 

[Laurry]

Does anyone have an idea what could cause this discrepancy (121 vs 91 N) between hand calc and Ansys?

 

[JBA]

The problem appears to me to be an apparent offset of the point at which you are applying the moment from the centerline of the shaft (for which you are showing no dimension). Since there is no x restraint at moment application point then the moment due to that offset must be carried by the shaft as opposed to being a x direction vector force on the shaft restraints.

 

[haruspex]

The problem appears to me to be an apparent offset of the point at which you are applying the moment from the centerline of the shaft (for which you are showing no dimension). Since there is no x restraint at moment application point then the moment due to that offset must be carried by the shaft as opposed to being a x direction vector force on the shaft restraints.

Are you suggesting we should take the forces applied at the shaft end points as being normal to the line joining them to the torque axis?
That does not seem to help. It makes the forces less, not more.

 

[Laurry]

I made the simple model even simpler using even numbers in order to find out if dimensions can somehow explain the difference.

https://dl.dropboxusercontent.com/u/104865119/ansys2.PNG

Did also a parametric study in Ansys and found out that as the length R decreases, the relative difference between results decreases.

 

[haruspex]

I made the simple model even simpler using even numbers in order to find out if dimensions can somehow explain the difference.

https://dl.dropboxusercontent.com/u/104865119/ansys2.PNG

Did also a parametric study in Ansys and found out that as the length R decreases, the relative difference between results decreases.

I have no idea what ANSYS is. Can you vary other parameters to get a clue as to what affects it? Can you get more precision out of it - the graph is not very smooth.

 

[Nidum]

Pinned v fully fixed .

 

[Laurry]

Thanks for the answer. In hand calculations I assumed only forces in the ends of the shaft but for it to be fully fixed there should also be a moment in each end.
By allowing the computer model's surfaces to rotate around the vertical axis, representing a pinned support, I got the same result.

 

[Laurry]

A continuation to this problem. I drew the correct FBD of the system, shown in the link:

http://imgur.com/a/AkTci

How is this statically indeterminate system solved? My first guess was to split it into beams which are table cases.