Edgar J. Kaiser
Certified Consultant
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Posted:
4 years ago
2020年9月11日 GMT-4 10:00
Updated:
4 years ago
2020年9月11日 GMT-4 11:14
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
-------------------
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
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Posted:
4 years ago
2020年9月12日 GMT-4 01:32
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
Please login with a confirmed email address before reporting spam
Posted:
4 years ago
2020年9月12日 GMT-4 02:57
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.
-------------------
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.
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Posted:
4 years ago
2020年9月12日 GMT-4 05:14
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
Please login with a confirmed email address before reporting spam
Posted:
4 years ago
2020年9月12日 GMT-4 06:37
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.
Please login with a confirmed email address before reporting spam
Posted:
4 years ago
2020年11月19日 GMT-5 06:10
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.