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How to do a distributed "body force"
Posted 13 nov 2010, 01:15 GMT-5 4 Replies
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Hi,
I played with the MEMS module example "ale_cantilever_beam_3d.mph" and have to do something similar but slightly different. In a nutshell, that .mph file creates expressions for the Maxwell stress tensor and then applies the force due to that stress tensor to a boundary of a cantilever, making the cantilever bend due to an electric field. In contrast, I have an expression for the force at each (interior) point of a dielectric subdomain and want to apply this force to each interior point, not just to a boundary. I can't figure out how to do this!
In more detail:
ale_cantilever_beam_3d.mph is a 300 um long cantilever, fixed at one end, charged to some voltage, 2 microns above a ground plane. This forms a capacitor, and the cantilever bends downwards because by bending it lowers the energy in the capacitor. To link the electrostatic module to the stress/strain module, first we tell it that it should calculate the maxwell stress tensor by going: "Subdomain settings--electrostatics" > "Force" tab > Enter "Fes" into the first row. That tells it it should generate things like Fes_nTx_emes etc (Maxwell stress tensor). Next, under "Solid, Stress-Strain" mode > "Boundary Settings" the bottom surface of the cantilever is given a Load on the Load tab. It is given the load of
Fx = Fes_nTx_emes
Fy = Fes_nTy_emes
Fz = Fes_nTz_emes
which are the forces calculated due to the maxwell stress tensor over that boundary. So far, simple.
However, what I want to do instead is to apply a force at **each interior point** of a given subdomain, not just at the boundary.
I have the equation for what the force at each interior point of the chosen subdomain should be. The equation is basically:
Fx = Ex_emes*d(Ex_emes,x)+Ey_emes*d(Ex_emes,y)+Ez_emes*d(Ex_emes,z)
Fy = Ex_emes*d(Ey_emes,x)+Ey_emes*d(Ey_emes,y)+Ez_emes*d(Ey_emes,z)
Fz = Ex_emes*d(Ez_emes,x)+Ey_emes*d(Ez_emes,y)+Ez_emes*d(Ez_emes,z)
...omitting some constants. This is a force on each point of a dielectric due to its immersion in a nonuniform electric field. It comes from F = ( p dot Del ) E, where F is force vector, dot is dot product, Del is del, E is electric field vector. (See for example, Griffith's Intro to Electrodynamics, 3rd ed, end of section 4.1.
Any ideas how to apply this force that acts on each interior point of a body?
Much appreciated! :-)
I played with the MEMS module example "ale_cantilever_beam_3d.mph" and have to do something similar but slightly different. In a nutshell, that .mph file creates expressions for the Maxwell stress tensor and then applies the force due to that stress tensor to a boundary of a cantilever, making the cantilever bend due to an electric field. In contrast, I have an expression for the force at each (interior) point of a dielectric subdomain and want to apply this force to each interior point, not just to a boundary. I can't figure out how to do this!
In more detail:
ale_cantilever_beam_3d.mph is a 300 um long cantilever, fixed at one end, charged to some voltage, 2 microns above a ground plane. This forms a capacitor, and the cantilever bends downwards because by bending it lowers the energy in the capacitor. To link the electrostatic module to the stress/strain module, first we tell it that it should calculate the maxwell stress tensor by going: "Subdomain settings--electrostatics" > "Force" tab > Enter "Fes" into the first row. That tells it it should generate things like Fes_nTx_emes etc (Maxwell stress tensor). Next, under "Solid, Stress-Strain" mode > "Boundary Settings" the bottom surface of the cantilever is given a Load on the Load tab. It is given the load of
Fx = Fes_nTx_emes
Fy = Fes_nTy_emes
Fz = Fes_nTz_emes
which are the forces calculated due to the maxwell stress tensor over that boundary. So far, simple.
However, what I want to do instead is to apply a force at **each interior point** of a given subdomain, not just at the boundary.
I have the equation for what the force at each interior point of the chosen subdomain should be. The equation is basically:
Fx = Ex_emes*d(Ex_emes,x)+Ey_emes*d(Ex_emes,y)+Ez_emes*d(Ex_emes,z)
Fy = Ex_emes*d(Ey_emes,x)+Ey_emes*d(Ey_emes,y)+Ez_emes*d(Ey_emes,z)
Fz = Ex_emes*d(Ez_emes,x)+Ey_emes*d(Ez_emes,y)+Ez_emes*d(Ez_emes,z)
...omitting some constants. This is a force on each point of a dielectric due to its immersion in a nonuniform electric field. It comes from F = ( p dot Del ) E, where F is force vector, dot is dot product, Del is del, E is electric field vector. (See for example, Griffith's Intro to Electrodynamics, 3rd ed, end of section 4.1.
Any ideas how to apply this force that acts on each interior point of a body?
Much appreciated! :-)
4 Replies Last Post 14 nov 2010, 02:24 GMT-5
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Posted:
1 decade ago
Hi
if yo apply a load on the "volume" domain, you need to use a "body load BC" and define your load as [N/m^3], because you are acting on a infinitesimal volume dxdydz.
Just as for gravity in structural you apply the formula
F[N/m^3] = g_const[^m/s^2] * solid.rho[kg/m^3]
--
Good luck
Ivar
if yo apply a load on the "volume" domain, you need to use a "body load BC" and define your load as [N/m^3], because you are acting on a infinitesimal volume dxdydz.
Just as for gravity in structural you apply the formula
F[N/m^3] = g_const[^m/s^2] * solid.rho[kg/m^3]
--
Good luck
Ivar
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Posted:
1 decade ago
Thanks Ivar,
Where do I enter a "body load boundary condition"?
Where do I enter a "body load boundary condition"?
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Posted:
1 decade ago
And one more thing, probably a harder problem:
When I've solved that I will have the *static* case (electrostatic and mechanically-static).
Now I want to apply an ac voltage (rather than the DC voltage I was applying before). The resulting body force profile should be just the same as in the static case but now multiplied by sin(w*t), at least to first approximation. I want to use this sine wavy force to actuate the structure and see my structure's frequency response.
In other words, so far I have done electrostatic actuation and now I know how the structure deforms under a static force I want to modulate that force and sweep the modulation frequency, thereby getting out the structure's frequency response.
Does that make sense?
Thanks so much! :-)
When I've solved that I will have the *static* case (electrostatic and mechanically-static).
Now I want to apply an ac voltage (rather than the DC voltage I was applying before). The resulting body force profile should be just the same as in the static case but now multiplied by sin(w*t), at least to first approximation. I want to use this sine wavy force to actuate the structure and see my structure's frequency response.
In other words, so far I have done electrostatic actuation and now I know how the structure deforms under a static force I want to modulate that force and sweep the modulation frequency, thereby getting out the structure's frequency response.
Does that make sense?
Thanks so much! :-)
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Posted:
1 decade ago
Hi
first of all when you have a sinus excitation, you must decide if you run a time dependent case or a frequency (harmonic) (sweep) case. The latter uses only the amplitude and phase of you excitation signal so it solves much quicker, the time series needs many steps to follow the true sinus ;)
In any case you must decide what is your driveing source and the "slave". I would expect its the electric field, and then the corresponding force will follow via COMSOL
for the body force, it is under structural and in the domain (highest orer level) of the MEMS physics node
--
Good luck
Ivar
first of all when you have a sinus excitation, you must decide if you run a time dependent case or a frequency (harmonic) (sweep) case. The latter uses only the amplitude and phase of you excitation signal so it solves much quicker, the time series needs many steps to follow the true sinus ;)
In any case you must decide what is your driveing source and the "slave". I would expect its the electric field, and then the corresponding force will follow via COMSOL
for the body force, it is under structural and in the domain (highest orer level) of the MEMS physics node
--
Good luck
Ivar