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Sinusoidal Stress Time Dependent Study

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I have been attempting to model a simple 5mm cube fixed on one side and with a sinusoidal time dependent stress on the opposite side. This should produce a propagating shear wave. I want to do this in order to animate the propagating shear wave.

The problem is that the applied sinusoidal stress does not change sign (change direction). The results should be similar to frequency domain solver results. Please reference my model. Note: study 1 is the time dependent study and study 2 is the frequency domain solver.

Any help to produce the above condition would be greatly appreciated

P.S.

My frequency is 4000Hz with a period of 2.5e-4s and my step size is 5e-5 so aliasing should not be the problem.

Use link to download model and animation

www.dropbox.com/s/91ikbc5vu9kwcys/Agar_Cube_w_time_dependence.avi

www.dropbox.com/s/jbiz9ma8m50g7an/Agar_Cube_SolidMech.mph

8 Replies Last Post Oct 17, 2012, 4:40 p.m. EDT

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Posted: 1 decade ago Oct 13, 2012, 3:42 p.m. EDT
I cannot load your model. Which version of COMSOL do you use?
I cannot load your model. Which version of COMSOL do you use?

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Posted: 1 decade ago Oct 14, 2012, 5:31 p.m. EDT
I used version 4.3.

I meant to re post to this question, so thanks for replying it reminded me to do so. I did find a solution. My solution was to create a variable for sinusoidal stress and the equation looked like this:

S = S_0*exp(i*2*pi*f*t)

S_0 = Amplitude of stress, f = frequency, i = imaginary number

This gave me the sinusoidal stress I was looking for with the stress being applied positive and negative.

When I used S = S_0*sin(2*pi*f*t), for some reason I got a stress function more similar to S=S_0*(sin(2*pi*f*t))^2.

So, in summary. Solution is to use complex harmonic functions instead of real valued functions when you want a harmonic function.

Also in the advanced section of the solver you need to check the box to allow complex numbers.
I used version 4.3. I meant to re post to this question, so thanks for replying it reminded me to do so. I did find a solution. My solution was to create a variable for sinusoidal stress and the equation looked like this: S = S_0*exp(i*2*pi*f*t) S_0 = Amplitude of stress, f = frequency, i = imaginary number This gave me the sinusoidal stress I was looking for with the stress being applied positive and negative. When I used S = S_0*sin(2*pi*f*t), for some reason I got a stress function more similar to S=S_0*(sin(2*pi*f*t))^2. So, in summary. Solution is to use complex harmonic functions instead of real valued functions when you want a harmonic function. Also in the advanced section of the solver you need to check the box to allow complex numbers.

Nagi Elabbasi Facebook Reality Labs

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Posted: 1 decade ago Oct 14, 2012, 7:29 p.m. EDT
Hi Steven,

It seems that you are looking at the Mises stress. That quantity is always positive and has higher harmonics just like the sin^2 function. If you look at an individual stress component you should see a sinusoidal variation.

Nagi Elabbasi
Veryst Engineering
Hi Steven, It seems that you are looking at the Mises stress. That quantity is always positive and has higher harmonics just like the sin^2 function. If you look at an individual stress component you should see a sinusoidal variation. Nagi Elabbasi Veryst Engineering

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Posted: 1 decade ago Oct 15, 2012, 11:54 a.m. EDT
Hi Nagi,

Yes you are correct in assuming that I was looking at Mises, and you are also right that when I look at individual stress components I do see a sin variation (see fig. 1,2). However this same sin^2 profile shows up for displacement as well and that is what I'm most concerned about. Let me show you 2 sets of images the first set shows the results when I use the S_0*sin(2*pi*f*t) function, and the second set shows the results for the S_0*exp(i*2*pi*f*t) function.

Also, to be clear my goal is to produce a shear wave propagating through this material. In the pictures the left face (y-z plane) is under harmonic stress (in z-dir.) and the right face (y-z plane) is fixed.

Note: 1D plots are taken along a line just below the top surface

I am mostly concerned with the difference between figures 4 and 8. Fig. 8 more accurately resembles experimental results. Therefore my conclusion is to still use S_0*exp(i*2*pi*f*t) instead of S_0*sin(2*pi*f*t).

Steve
Hi Nagi, Yes you are correct in assuming that I was looking at Mises, and you are also right that when I look at individual stress components I do see a sin variation (see fig. 1,2). However this same sin^2 profile shows up for displacement as well and that is what I'm most concerned about. Let me show you 2 sets of images the first set shows the results when I use the S_0*sin(2*pi*f*t) function, and the second set shows the results for the S_0*exp(i*2*pi*f*t) function. Also, to be clear my goal is to produce a shear wave propagating through this material. In the pictures the left face (y-z plane) is under harmonic stress (in z-dir.) and the right face (y-z plane) is fixed. Note: 1D plots are taken along a line just below the top surface I am mostly concerned with the difference between figures 4 and 8. Fig. 8 more accurately resembles experimental results. Therefore my conclusion is to still use S_0*exp(i*2*pi*f*t) instead of S_0*sin(2*pi*f*t). Steve


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Posted: 1 decade ago Oct 16, 2012, 2:28 p.m. EDT
Actually, I have been working on a similar similation. I am trying to generate a shear wave at 50Hz in a homogeneous phantom and observe its propogation. Later, this will not be a homogeneous phantom and at that time I will try to observe reflected waves from these stiffer parts of the phantom. Therefore, I also want to observe propagation of shear waves with a movie but I have problems in time dependent study such that the wave does not propagate inside the phantom, it seems it stays at the surface. I can observe the desired result in Frequency Domain but at that case I cannot make a movie and observe the propagation of the wave. Width of the phantom is 400mm. I could not figure out why the shear wave is staying just at the surface. Stress is in positive z direction.
Actually, I have been working on a similar similation. I am trying to generate a shear wave at 50Hz in a homogeneous phantom and observe its propogation. Later, this will not be a homogeneous phantom and at that time I will try to observe reflected waves from these stiffer parts of the phantom. Therefore, I also want to observe propagation of shear waves with a movie but I have problems in time dependent study such that the wave does not propagate inside the phantom, it seems it stays at the surface. I can observe the desired result in Frequency Domain but at that case I cannot make a movie and observe the propagation of the wave. Width of the phantom is 400mm. I could not figure out why the shear wave is staying just at the surface. Stress is in positive z direction.

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Posted: 1 decade ago Oct 17, 2012, 10:58 a.m. EDT
Hard to say what your problem is. In my model waves propagate below the surface as well as at the surface. Could you post your model?
Hard to say what your problem is. In my model waves propagate below the surface as well as at the surface. Could you post your model?

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Posted: 1 decade ago Oct 17, 2012, 1:09 p.m. EDT
I solved my problem. I made a typo in applied stress equation. Now I can observe the propagating shear wave through the whole phantom. I have a question about applying stress as exp(i*2*pi*f*t), just because I have been working on a similar issue in my model, thus I am trying to decide how to apply the stress. I want to apply stress only in one direction but when I apply stress as exp(i*2*pi*t), it seems stress is applied in two directions, whereas I would also like to observe reflected waves.
I solved my problem. I made a typo in applied stress equation. Now I can observe the propagating shear wave through the whole phantom. I have a question about applying stress as exp(i*2*pi*f*t), just because I have been working on a similar issue in my model, thus I am trying to decide how to apply the stress. I want to apply stress only in one direction but when I apply stress as exp(i*2*pi*t), it seems stress is applied in two directions, whereas I would also like to observe reflected waves.

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Posted: 1 decade ago Oct 17, 2012, 4:40 p.m. EDT
exp(i*2*pi*f*t) = cos(2*pi*f*t) + i*sin(2*pi*f*t), so this is a sinusoidal type function with positve and negative values. If you want your stress to be applied only in one direction then you would need to add a positive offset so that the solution to the function is never negative.

Ex:

Tau = Tau_0*exp(i*2*pi*f*t)

change to

Tau = Tau_0*exp(i*2*pi*f*t) + Tau_0
exp(i*2*pi*f*t) = cos(2*pi*f*t) + i*sin(2*pi*f*t), so this is a sinusoidal type function with positve and negative values. If you want your stress to be applied only in one direction then you would need to add a positive offset so that the solution to the function is never negative. Ex: Tau = Tau_0*exp(i*2*pi*f*t) change to Tau = Tau_0*exp(i*2*pi*f*t) + Tau_0

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