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Torsion angle of square bar

Torsion angle of square bar

Torsion angle of square bar

(OP)
I want to verify the torsion angle on a square bar with FEA software (NX 6). The bar has dimensions of 50mmx50mmx100mm. The bar is fixed on one end and on the other a moment of 200Nm is exerted. Is it possible to calculate the torsion angle with the nodal displacements using an approximating formula? And if so, which formula?

RE: Torsion angle of square bar

yes ...

twist due to torque is in just about any structures text book ... theta (radians) = TL/(G*(a^4/6)) ...

i have to admit the "/6" is my interpretation of the polar moment of inertia, my reference has 0.141 (not quite 0.1667)

RE: Torsion angle of square bar

(OP)
Well I know that's the theoretical formula but now I want to physically verify the angle that I get from that formula with the model in my FEA software, to see if my model is correct. I want to use the nodal displacements of the top plane which is under torsion and compare them with the fixed plane. I can't use the cosine formula because the edges of the planes don't stay straight.   

RE: Torsion angle of square bar

Your torsion bar is extremely stocky, end effects will dominate. The fixed end is forced to stay planar, whereas the free end will do something more interesting.

Try using a much more slender bar, if you are just trying to validate your method.

The torsional elastic constant for non circular shapes is a rather exciting topic, in general.

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Torsion angle of square bar

the formula defines the twist on the section, so how much has your section twisted ?

you could take the four corner nodes and see how they've deflected relative to the center node (or relative to their average deflection).

 

RE: Torsion angle of square bar

i'd be careful with the drawing, "s" is a generalised vector; yes, theoretically it may be tangential.

could you calculate the angle between a line on the undeformed beam and the deformed section ?

RE: Torsion angle of square bar

(OP)
Mesh(mm) Torsion angle (rad)      τmax(Mpa)          ∆s(m)
25          0,0023904              1,882E-08        2,99E-06
               
12,5          0,0026496              1,298                3,31E-06
               
5          0,0027616              7,636                3,45E-06

Theoretical values:
thèta = 0,0028
tau max = 7,7 MPa

By the way, the torsion angle values are thèta = s/(0.5a/0.5L)
muliplied by 2. Could you explain why this formula could be correct?

RE: Torsion angle of square bar

you could express theta as (s/(a/2))/L ... this is twist per inch (length) ... note, this is not quite the equation you wrote, check brackets and divisions

RE: Torsion angle of square bar

(OP)
With your formula I get a thèta of 0.00138 rad which is half of the theoretical solution. I should still get a factor 2 from somewhere :)

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