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2d Axial Symmetry Solution
Posted 2010/04/14 9:40 GMT-4 1 Reply
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I am running a 2d axial symmetric and I am curious to know if the solution the program finds is half of the actual force or is indeed the actual force. Any guidance would be great. Thanks
1 Reply Last Post 2010/04/14 15:51 GMT-4
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Posted:
1 decade ago
Hi
I would say, that depends on how you have set up your model.
In any case when I have doubts (and I often have when I'm mixing multiple physics, or treat new cases) I build very simple models, to check out exactly such fundamental questions.
Futhermore, for a 2D axisymmetric model, generally there is a 2*pi*r factor to consider, for half, quater or less models in 2D with one or two symmetry planes ther are often a factor 2, 4 or more, as you state.
Even worse, if you do eigenfrequency analysis you should not forget vary and resolve for all combination of symmetric and antisymmetric boundary cases to get all modes.
Set up some simple cases, such as a slender cylinder, first in 2D axi, try L=1[m] long and R=0.1[m] radius, fix the lower boundary (z=0) and add a force F=1[MN] hence a pressure P on top of the area A:
A[m^2]=pi*(R[m])^2=(pi/100)[m^2]
Pz=-Fz[MN]/(Area[m^2])=(-100/pi)[MPa],
by default the material is steel with a Young modulus of E=200[GPa]. compression stiffness along the axial direction is:
k=E*A/L=200[GPa]*(pi*(0.1[m])^2)/(1[m])=2*pi[GN/m]=2*pi*[kN/um].
Hence the axial length change "w" along z expected is:
w[um]=F/k=P[kPa]*(A[m])/(k[kN/um])=1000*(100/pi)*(pi*0.1^2)/(2*pi)=159.154943...[um]
Next step is to revolve the mesh of this 2D-axi plane into a 3D model and repeat (by the way you need a square mesh to "Mesh-revolve") then fix the bottom boundary load the upper boundary with the same pressure and solve, I notices less than 1[um] difference between 2D and 3D, and 0 difference with my formula in 2D-axi, even with "normal" hence a rather rough mesh.
If you want a better coherence you should improve your lower "fixed" boundary, if you have selected the bottom surface and fixed it in x,y,&z and then you look at the von Mises stress you will see the affect of the Poisson coefficient and the shear stress, if you block the lower surface only along y hence v=0, and then block the centre POINT to u=w=0, the two points at x=0,y=0 along w=0 and the two points z=0,y=0 along x=0 one will even gain back this last micron difference (Point constraints are not good practice but usefull in certain cases).
In conclusion: it's only a question of respecting the right units, but note the verbouse use of () when adding the units, it's quickly done to get one wrong, hope I havnt missed anything ;)
Have fun Comsoling
Ivar
I would say, that depends on how you have set up your model.
In any case when I have doubts (and I often have when I'm mixing multiple physics, or treat new cases) I build very simple models, to check out exactly such fundamental questions.
Futhermore, for a 2D axisymmetric model, generally there is a 2*pi*r factor to consider, for half, quater or less models in 2D with one or two symmetry planes ther are often a factor 2, 4 or more, as you state.
Even worse, if you do eigenfrequency analysis you should not forget vary and resolve for all combination of symmetric and antisymmetric boundary cases to get all modes.
Set up some simple cases, such as a slender cylinder, first in 2D axi, try L=1[m] long and R=0.1[m] radius, fix the lower boundary (z=0) and add a force F=1[MN] hence a pressure P on top of the area A:
A[m^2]=pi*(R[m])^2=(pi/100)[m^2]
Pz=-Fz[MN]/(Area[m^2])=(-100/pi)[MPa],
by default the material is steel with a Young modulus of E=200[GPa]. compression stiffness along the axial direction is:
k=E*A/L=200[GPa]*(pi*(0.1[m])^2)/(1[m])=2*pi[GN/m]=2*pi*[kN/um].
Hence the axial length change "w" along z expected is:
w[um]=F/k=P[kPa]*(A[m])/(k[kN/um])=1000*(100/pi)*(pi*0.1^2)/(2*pi)=159.154943...[um]
Next step is to revolve the mesh of this 2D-axi plane into a 3D model and repeat (by the way you need a square mesh to "Mesh-revolve") then fix the bottom boundary load the upper boundary with the same pressure and solve, I notices less than 1[um] difference between 2D and 3D, and 0 difference with my formula in 2D-axi, even with "normal" hence a rather rough mesh.
If you want a better coherence you should improve your lower "fixed" boundary, if you have selected the bottom surface and fixed it in x,y,&z and then you look at the von Mises stress you will see the affect of the Poisson coefficient and the shear stress, if you block the lower surface only along y hence v=0, and then block the centre POINT to u=w=0, the two points at x=0,y=0 along w=0 and the two points z=0,y=0 along x=0 one will even gain back this last micron difference (Point constraints are not good practice but usefull in certain cases).
In conclusion: it's only a question of respecting the right units, but note the verbouse use of () when adding the units, it's quickly done to get one wrong, hope I havnt missed anything ;)
Have fun Comsoling
Ivar