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Posted:
1 decade ago
31 lug 2014, 04:05 GMT-4
Damping and piezo exert a force (vector) on your structure. Take the inner product of this force and the displacement (vector) and you find the work done by this force. Force and displacement should be in your resulting data. Integrate the result of this inner product over the surface or vollume on which the force is exerted and over the complete displacement path, and you find the total work done by this force.
Damping and piezo exert a force (vector) on your structure. Take the inner product of this force and the displacement (vector) and you find the work done by this force. Force and displacement should be in your resulting data. Integrate the result of this inner product over the surface or vollume on which the force is exerted and over the complete displacement path, and you find the total work done by this force.
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Posted:
1 decade ago
31 lug 2014, 06:01 GMT-4
Hi Pieter,
Thanks for your reply. Do you know how to get the force exerted by the damping and piezo? It seems that I could not find the variables defining these two forces.
Thanks a lot.
Regards,
Deng
Hi Pieter,
Thanks for your reply. Do you know how to get the force exerted by the damping and piezo? It seems that I could not find the variables defining these two forces.
Thanks a lot.
Regards,
Deng
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Posted:
1 decade ago
31 lug 2014, 06:13 GMT-4
Can you maybe upload your model?
Can you maybe upload your model?
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Posted:
1 decade ago
31 lug 2014, 07:17 GMT-4
Please have a look. Thanks.
Please have a look. Thanks.
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Posted:
1 decade ago
31 lug 2014, 09:17 GMT-4
Assuming that all work done by the piezo on the beam is done by a force perpendicular to the contact surface, I think you can calculate the total work done by the piezo on the beam by making a surface integral (subnode in derived values) and selecting the surface where the two materials touch, and putting as the expression 'pzd.sz*pzd.u_tZ', and integrating the resulting data series.
The energy that has been dissipated by damping is the energy present in the beam at the start of the simulation minus the energy present at the end of the simulation, plus the energy supplied to the beam during the simulation. The energy present at any time can be found by doing a volume integral of the kinetic+elastic energy. The energy supplied is just calculated above.
Assuming that all work done by the piezo on the beam is done by a force perpendicular to the contact surface, I think you can calculate the total work done by the piezo on the beam by making a surface integral (subnode in derived values) and selecting the surface where the two materials touch, and putting as the expression 'pzd.sz*pzd.u_tZ', and integrating the resulting data series.
The energy that has been dissipated by damping is the energy present in the beam at the start of the simulation minus the energy present at the end of the simulation, plus the energy supplied to the beam during the simulation. The energy present at any time can be found by doing a volume integral of the kinetic+elastic energy. The energy supplied is just calculated above.
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Posted:
1 decade ago
31 lug 2014, 10:15 GMT-4
The energy that has been dissipated by damping is the energy present in the beam at the start of the simulation minus the energy present at the end of the simulation, plus the energy supplied to the beam during the simulation. The energy present at any time can be found by doing a volume integral of the kinetic+elastic energy. The energy supplied is just calculated above.
Thanks for your reply. I did not quite understand the energy present in the beam at the start of the simulation minus the energy present at the end of the simulation. As it is a time dependent study, I can calculate the energy at any time in terms of kinetic and elastic strain energy as you suggested, but I'm still not quite clear about the energy dissipated by damping. Does 'start' mean the energy at t0?
Thanks again.
[QUOTE]
The energy that has been dissipated by damping is the energy present in the beam at the start of the simulation minus the energy present at the end of the simulation, plus the energy supplied to the beam during the simulation. The energy present at any time can be found by doing a volume integral of the kinetic+elastic energy. The energy supplied is just calculated above.
[/QUOTE]
Thanks for your reply. I did not quite understand the energy present in the beam at the start of the simulation minus the energy present at the end of the simulation. As it is a time dependent study, I can calculate the energy at any time in terms of kinetic and elastic strain energy as you suggested, but I'm still not quite clear about the energy dissipated by damping. Does 'start' mean the energy at t0?
Thanks again.
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Posted:
1 decade ago
1 ago 2014, 03:24 GMT-4
You can do it for any time frame, so suppose you want to know the energy dissipated through damping between time t1 and t2, and the total kinetic and elastic energy in the beam at time t are Ek(t) and Ee(t) respectively, the dissipated energy will be Wd=Ek(t2)+Ee(t2)-Ek(t1)-Ee(t1)+Wp, where Wp is the work done by the piezo between t1 and t2 (calculated by integrating stress*velocity (as described above) integrated over the contact surface and integrated from t1 to t2). (Here I have taken the convention that Wp is positive when the piezo has supplied more energy to the beam than that it has removed from it, so when the force exerted on the beam was generally in the same direction as the velocity of that part of the beam.)
You can do it for any time frame, so suppose you want to know the energy dissipated through damping between time t1 and t2, and the total kinetic and elastic energy in the beam at time t are Ek(t) and Ee(t) respectively, the dissipated energy will be Wd=Ek(t2)+Ee(t2)-Ek(t1)-Ee(t1)+Wp, where Wp is the work done by the piezo between t1 and t2 (calculated by integrating stress*velocity (as described above) integrated over the contact surface and integrated from t1 to t2). (Here I have taken the convention that Wp is positive when the piezo has supplied more energy to the beam than that it has removed from it, so when the force exerted on the beam was generally in the same direction as the velocity of that part of the beam.)
