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rotation forces around a rectangle bar?

Uni Phil

New member
I have modeled a part which primarily looks like a long rectangular bar.

The bar has a thread cut into it at 90degrees which has a round bar coming out of that.



What I want to do is find out how much twisting stress is exerted on the rectangular bar when I put pressure on the round bar which is perpendicular to the rectagular bar.



I hopt that makes sense ;)
 

dr_gallup

Moderator
Quote does creating two forces perpendicular to each other constitute a rotational force???



No, a torque is the product of a force through a distance. Two forces at right angles produce a net force equal to the square root of the sum of the squares of the individual forces. The angle of the resultant force is the arctangent of the ratio of the forces.



Sounds like you need a basic course in statics.
 

Luis Aguirre

New member
Uni Phil,



Your two perpendicular forces can still create a moment depending with respect what point it is used to calculate the moment created by the forces. The only point that will not create a moment is the point at which both forces intersect, other wise you can take any other phisical point in space and take the moment with respect that location. So in reality two perpendicular forces can create a moment at every point but the point of intersection of the two forces.
 

Uni Phil

New member
Dr_gallup: I understand that by creating two forces perpendicular to each other will create a vector resultant which is not a rotational force in any means, but what I was getting at (probably not explained well enough) was, is this the way that Pro/E works out rotational force, at pro/mechanica doesn
 

Luis Aguirre

New member
Uni Phil,



If you are using solid elements, they do not have rotational degrees of freedom, therefore, you cannot apply a moment directly. However, there is a little trick that you can use. You can apply a surface load then on the load window change the Distribution from total load to Total Load At Point at specify a point that lies on the surface you are applying the moment. Then simply applied the moment component magnitude. I hope this helps!



Luis
 

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