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Torsion problem simulation

Posted 11 sept. 2020, 06:32 UTC−4 Structural & Acoustics, Geometry, Structural Mechanics 6 Replies

Akshay Manthekar
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Hello guys, I am new to COMSOL and I want to simulate torsion of circular shaft in comsol, As there is not any option available for direct torque application, I applied opposite (point) forces in opposite points on the end face equivalent to the torque. But there was high stress concentration around area near points. So can anyone tell me how to simulate this problem properly so that stress concentration will not be significant as in the case for point force? Thank you in advance

6 Replies Last Post 19 nov. 2020, 06:10 UTC−5
Edgar J. Kaiser Certified Consultant
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Posted: 4 years ago 11 sept. 2020, 10:00 UTC−4
Updated: 4 years ago 11 sept. 2020, 11:14 UTC−4

Hi,

one option is to apply a tangential distributed boundary load L to the cylinder face. If the cylinder axis is in x=y=0 and along the z-axis it can look like Lx = const * y, Ly = -const * x, Lz = 0. Adapt the const to achieve the desired torque. There may be other options too.

Cheers Edgar

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Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
Hi, one option is to apply a tangential distributed boundary load L to the cylinder face. If the cylinder axis is in x=y=0 and along the z-axis it can look like Lx = const * y, Ly = -const * x, Lz = 0. Adapt the const to achieve the desired torque. There may be other options too. Cheers Edgar
Akshay Manthekar
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Posted: 4 years ago 12 sept. 2020, 01:32 UTC−4

Hi Edgar, I have cylinder of radius = 25mm, as you said earlier about constant, I have calculated its value and defined as parameter. While applying load, I have specified Fx = constrad and Fy = -constrad. But I am getting error in solution saying operator (range) is not valid in analysis. I am attaching screenshots of it. PLease have a look and help me.

Hi Edgar, I have cylinder of radius = 25mm, as you said earlier about constant, I have calculated its value and defined as parameter. While applying load, I have specified Fx = const*rad and Fy = -const*rad. But I am getting error in solution saying operator (range) is not valid in analysis. I am attaching screenshots of it. PLease have a look and help me.
Edgar J. Kaiser Certified Consultant
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Posted: 4 years ago 12 sept. 2020, 02:57 UTC−4

You can try the example I gave you, it works. You can't use range() to define variables. The boundary load must be defined as a 'Force per unit area' to make it a distributed load.

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Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
You can try the example I gave you, it works. You can't use range() to define variables. The boundary load must be defined as a 'Force per unit area' to make it a distributed load.
Akshay Manthekar
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Posted: 4 years ago 12 sept. 2020, 05:14 UTC−4

Hi Edgar, I think you forgot to upload the example file which you were to send me. Can you please send it ?

Hi Edgar, I think you forgot to upload the example file which you were to send me. Can you please send it ?
Edgar J. Kaiser Certified Consultant
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Posted: 4 years ago 12 sept. 2020, 06:37 UTC−4

There isn't an example file. My posts tell what to do.

-------------------
Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
There isn't an example file. My posts tell what to do.
Hu Kejun
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Posted: 4 years ago 19 nov. 2020, 06:10 UTC−5

Hi, i recently work on this problem also.

If your shaft is in elastic material, i think i will be work. for boudary condition: - fixed constraint at one side and rigid connector at another side - in rigid connector, active the applied moment around the rotation of your geometry. The total torque will be distributed uniformely on your boundary surface/ edge or points. - If you want to apply a constraint rotation motion, you can define it in section of rigid connector. But since the resolution is more easier that base on applied load than applied displacement, you can profit the option of 'global equation' to determine iteratively the good load according to every steps of displacement. - if you need to achieve a large rotation angle, which means a big shear strain, the non linearity geometry will be recommanded to be activated. Then you may need to use a auxiliary study with some configuration of your solver (based on blog of comsol, the constant Newton raphson method is recommanded with a dumping coef of 0.7-0.9) . - I did keep the model for the moment, but a I can show you a gif generated by comsol 5.5. the Bar is partially shown with a 90° sector in order to affiche the rotation. the angle large more 360° is achieved with elastic mat and NL geo activated.

Hi, i recently work on this problem also. If your shaft is in elastic material, i think i will be work. for boudary condition: - fixed constraint at one side and rigid connector at another side - in rigid connector, active the applied moment around the rotation of your geometry. The total torque will be distributed uniformely on your boundary surface/ edge or points. - If you want to apply a constraint rotation motion, you can define it in section of rigid connector. But since the resolution is more easier that base on applied load than applied displacement, you can profit the option of 'global equation' to determine iteratively the good load according to every steps of displacement. - if you need to achieve a large rotation angle, which means a big shear strain, the non linearity geometry will be recommanded to be activated. Then you may need to use a auxiliary study with some configuration of your solver (based on blog of comsol, the constant Newton raphson method is recommanded with a dumping coef of 0.7-0.9) . - I did keep the model for the moment, but a I can show you a gif generated by comsol 5.5. the Bar is partially shown with a 90° sector in order to affiche the rotation. the angle large more 360° is achieved with elastic mat and NL geo activated.

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