Discussion Closed This discussion was created more than 6 months ago and has been closed. To start a new discussion with a link back to this one, click here.

shock accelerometer

Please login with a confirmed email address before reporting spam

hi i am doing project on shock accelerometer analysis.i want know how to give shock as a input to the system.that means semi sinusoidal acceleration as a input to the system.and don't know how to find stress analysis.could you please these features.
thanking you.

6 Replies Last Post 2010年11月4日 GMT-4 03:35
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 2010年1月27日 GMT-5 02:10
Hi

well for me shock analysis is similar to vibration load analysis (i.e. spacecraft launch loads) you define an acceleration field.
Actually there is a difference: for the launch load you work in "harmonic domain" and sweep frequency, while shock loads you work in "time domain" and sweep the time.

But for the load case you apply an acceleration as a field over a volume via the local density (to match the a=[m/s^2] to the force F=[N=kg*m/s^2] to the applied pressure P=[Pa=N/m^2]).

so you define your "sine function" as a Ccomsol function of time of maximum height "1" and the desired duration and shape, and you apply a load A0 to your sub-domain (global load field, similar to static gravity) Fz=-A0*sinfun(t)*rho_smsld (for the smsld application mode, adapt if you are in another mode). You might need to normalise over the total volume, check your units.

This is just as if you want to apply gravity along Fz, you define the gravity acceleraion G=1[lbf/lb] = 9.81...[m/s^2] and apply a load to all your sub-domains of Fz = -G0*rho_smsld, and off you go, gravity is turned on ;)

With COMSOL you apply the basic physics law to your calculations (but I agree in most other FEM tools gravity is turned on by a predefined click, but then you rather limited and cannot that easily play with all the physics)

Good luck
Ivar
Hi well for me shock analysis is similar to vibration load analysis (i.e. spacecraft launch loads) you define an acceleration field. Actually there is a difference: for the launch load you work in "harmonic domain" and sweep frequency, while shock loads you work in "time domain" and sweep the time. But for the load case you apply an acceleration as a field over a volume via the local density (to match the a=[m/s^2] to the force F=[N=kg*m/s^2] to the applied pressure P=[Pa=N/m^2]). so you define your "sine function" as a Ccomsol function of time of maximum height "1" and the desired duration and shape, and you apply a load A0 to your sub-domain (global load field, similar to static gravity) Fz=-A0*sinfun(t)*rho_smsld (for the smsld application mode, adapt if you are in another mode). You might need to normalise over the total volume, check your units. This is just as if you want to apply gravity along Fz, you define the gravity acceleraion G=1[lbf/lb] = 9.81...[m/s^2] and apply a load to all your sub-domains of Fz = -G0*rho_smsld, and off you go, gravity is turned on ;) With COMSOL you apply the basic physics law to your calculations (but I agree in most other FEM tools gravity is turned on by a predefined click, but then you rather limited and cannot that easily play with all the physics) Good luck Ivar

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 2010年11月3日 GMT-4 06:07
Then how to get an output as acceleration of a point?
Then how to get an output as acceleration of a point?

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 2010年11月4日 GMT-4 02:54
Hi

have you checked what you get out with the variable "utt" or vtt or wtt defined on a point ?

"u" being the displacement direction and the "tt" subscript being the COMSOL way to say second derivative

--
Good luck
Ivar
Hi have you checked what you get out with the variable "utt" or vtt or wtt defined on a point ? "u" being the displacement direction and the "tt" subscript being the COMSOL way to say second derivative -- Good luck Ivar

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 2010年11月4日 GMT-4 03:19
Hi,
It's the first thing that I did...Then u_tt_smsld....

The result is 0 field. Obviously, it can't be true since I get velocity as a function of time. Next, I derive the acceleration by myself...

I just wonder if there is a reason that acceleration does not present as a standard output?

thank you for your respond,
Stephan
Hi, It's the first thing that I did...Then u_tt_smsld.... The result is 0 field. Obviously, it can't be true since I get velocity as a function of time. Next, I derive the acceleration by myself... I just wonder if there is a reason that acceleration does not present as a standard output? thank you for your respond, Stephan

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 2010年11月4日 GMT-4 03:28
Hi

if you are in V3.5 you need to check that the time derivatives are "on" (to save space and time these are not on by default in all cases), if I remember right it's somewhere in the solver settings

--
Good luck
Ivar
Hi if you are in V3.5 you need to check that the time derivatives are "on" (to save space and time these are not on by default in all cases), if I remember right it's somewhere in the solver settings -- Good luck Ivar

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 2010年11月4日 GMT-4 03:35
Yep!
In solver manager - output menu - should check "time derivatives".

Thank you very much,
Stephan
Yep! In solver manager - output menu - should check "time derivatives". Thank you very much, Stephan

Note that while COMSOL employees may participate in the discussion forum, COMSOL® software users who are on-subscription should submit their questions via the Support Center for a more comprehensive response from the Technical Support team.