Hi,
I have a problem based on "../sfepy/examples/multi_physics/biot.py". In the attached scripy 'block2.py', I have calculated the stress and strain according to the displacement, at the 111 and 112 lines you can see it.
Now I want to output the K of every cell, I gave a value of k at the 64 line in this script, it is a 3*3 array. I don't know which function or equation to use, so I was stucked at the 113 line. if I change the K value, will where be some difference results?
Thank you advance
K.
在 2016年12月26日星期一 UTC+8下午8:52:44，Kid Guo写道：
Hi,
I have a problem based on "../sfepy/examples/multi_physics/biot.py". In the attached scripy 'block_hc.py', I have calculated the stress and strain according to the displacement, at the 111 and 112 lines you can see it.
Now I want to output the K of every cell, I gave a value of k at the 64 line in this script, it is a 3*3 array. I don't know which function or equation to use, so I was stucked at the 113 line. if I change the K value, will where be some difference results?
Thank you advance
K.
Where Ki is the permeability of the i-th compartment and Kij is the general diffusion permeability. how to ouput the Kij of each cell.
在 2016年12月26日星期一 UTC+8下午8:52:44，Kid Guo写道：
Hi,
I have a problem based on "../sfepy/examples/multi_physics/biot.py". In the attached scripy 'block2.py', I have calculated the stress and strain according to the displacement, at the 111 and 112 lines you can see it.
Now I want to output the K of every cell, I gave a value of k at the 64 line in this script, it is a 3*3 array. I don't know which function or equation to use, so I was stucked at the 113 line. if I change the K value, will where be some difference results?
Thank you advance
K.
Hi Kid,
you can use
hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m.k, p)', m=m,
mode='el_avg')...
out['hydraulic_conductivity'] = Struct(name='output_data', mode='cell',
data=hydraulic_conductivity)Essentially, ev_integrate_mat can be used to integrate any material parameter, check [1] to see what the mode argument means.
r.
[1] http://sfepy.org/doc-devel/src/sfepy/terms/terms_basic.html#sfepy.terms.term...
On 12/28/2016 09:01 AM, Kid Guo wrote:
Where Ki is the permeability of the i-th compartment and Kij is the general diffusion permeability. how to ouput the Kij of each cell.
在 2016年12月26日星期一 UTC+8下午8:52:44，Kid Guo写道：
Hi,
I have a problem based on "../sfepy/examples/multi_physics/biot.py". In the attached scripy 'block2.py', I have calculated the stress and strain according to the displacement, at the 111 and 112 lines you can see it.
Now I want to output the K of every cell, I gave a value of k at the 64 line in this script, it is a 3*3 array. I don't know which function or equation to use, so I was stucked at the 113 line. if I change the K value, will where be some difference results?
Thank you advance
K.
Hi Robert,
Thank you very much, I have output the value of Kij in my script. In the script, I have give a initial value of K, of course , I assumed it is a homogeneous material, so it has the same value of every cell. Now I have new thinking, if it is a heterogeneous material, that Kij value are different of every cell. I want to give different K value in every cell, for example, in the cell 1, the Kij value is K1 = nm.array([[1.0, 0.0, 0.0], [0.0, 1.0, 0.0], [0.0, 0.0, 1.0]], dtype = nm.float64) in the cell 2, the Kij value is K2 = nm.array([[1.1, 0.0, 0.0], [0.0, 1.2, 0.0], [0.0, 0.0, 10.1]], dtype = nm.float64) etc. The new Kij like this: K2 = nm.array([[[[ 1., 0., 0.], [ 0, 1., 0.], [ 0., 0., 1.]]],
[[[ 1.1, 0., 0.],
[ 0, 1.2, 0.],
[ 0., 0., 10.1]]]], dtype = nm.float64)when I give the new value to the problem,it has a error. "ValueError: material parameter array must have three dimensions! ('k' has 4)"
or the K is
K3 = nm.array([[[1.0, 0.0, 0.0],
[1.0, 0.0, 0.0],
[0.0, 0.0, 1.0]],
[[1.0, 0.0, 0.0],
[0.0, 2.0, 0.0],
[0.0, 0.0, 3.0]]], dtype = nm.float64)it said:
"ValueError: incompatible shapes! ((27, 8, 3, 8), out: (27, 1, 3, 3), arr: (54, 8, 3, 3))"
what should I do?
Any idea? Thank you advance.
K.
在 2016年12月28日星期三 UTC+8下午9:14:36，Robert Cimrman写道：
Hi Kid,
you can use
hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m.k, p)', m=m, mode='el_avg')...
out['hydraulic_conductivity'] = Struct(name='output_data',mode='cell', data=hydraulic_conductivity)
Essentially, ev_integrate_mat can be used to integrate any material parameter, check [1] to see what the mode argument means.
r.
[1]
http://sfepy.org/doc-devel/src/sfepy/terms/terms_basic.html#sfepy.terms.term...
On 12/28/2016 09:01 AM, Kid Guo wrote:
Where Ki is the permeability of the i-th compartment and Kij is the general diffusion permeability. how to ouput the Kij of each cell.
在 2016年12月26日星期一 UTC+8下午8:52:44，Kid Guo写道：
Hi,
I have a problem based on "../sfepy/examples/multi_physics/biot.py". In the attached scripy 'block2.py', I have calculated the stress and strain according to the displacement, at the 111 and 112 lines you can
see
it.
Now I want to output the K of every cell, I gave a value of k at the 64 line in this script, it is a 3*3 array. I don't know which function or equation to use, so I was stucked at the 113 line. if I change the K value, will where be some difference results?
Thank you advance
K.
Hi Kid,
If you want to have the permeability varying with position, you have to define the material parameter using a function [1]- it seems that you assigned your new K2 directly to the material constructor, right?
r. [1[ https://groups.google.com/forum/?_escaped_fragment_=topic/sfepy-devel/6QWAwC...
On 12/29/2016 09:50 AM, Kid Guo wrote:
Hi Robert,
Thank you very much, I have output the value of Kij in my script. In the script, I have give a initial value of K, of course , I assumed it is a homogeneous material, so it has the same value of every cell. Now I have new thinking, if it is a heterogeneous material, that Kij value are different of every cell. I want to give different K value in every cell, for example, in the cell 1, the Kij value is K1 = nm.array([[1.0, 0.0, 0.0], [0.0, 1.0, 0.0], [0.0, 0.0, 1.0]], dtype = nm.float64) in the cell 2, the Kij value is K2 = nm.array([[1.1, 0.0, 0.0], [0.0, 1.2, 0.0], [0.0, 0.0, 10.1]], dtype = nm.float64) etc. The new Kij like this: K2 = nm.array([[[[ 1., 0., 0.], [ 0, 1., 0.], [ 0., 0., 1.]]],
[[[ 1.1, 0., 0.], [ 0, 1.2, 0.], [ 0., 0., 10.1]]]], dtype = nm.float64)when I give the new value to the problem,it has a error. "ValueError: material parameter array must have three dimensions! ('k' has 4)"
or the K is
K3 = nm.array([[[1.0, 0.0, 0.0],
[1.0, 0.0, 0.0], [0.0, 0.0, 1.0]], [[1.0, 0.0, 0.0], [0.0, 2.0, 0.0], [0.0, 0.0, 3.0]]], dtype = nm.float64)it said:
"ValueError: incompatible shapes! ((27, 8, 3, 8), out: (27, 1, 3, 3), arr: (54, 8, 3, 3))"
what should I do?
Any idea? Thank you advance.
K.
在 2016年12月28日星期三 UTC+8下午9:14:36，Robert Cimrman写道：
Hi Kid,
you can use
hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m.k, p)', m=m, mode='el_avg')...
out['hydraulic_conductivity'] = Struct(name='output_data',mode='cell', data=hydraulic_conductivity)
Essentially, ev_integrate_mat can be used to integrate any material parameter, check [1] to see what the mode argument means.
r.
[1]
http://sfepy.org/doc-devel/src/sfepy/terms/terms_basic.html#sfepy.terms.term...
On 12/28/2016 09:01 AM, Kid Guo wrote:
Where Ki is the permeability of the i-th compartment and Kij is the general diffusion permeability. how to ouput the Kij of each cell.
在 2016年12月26日星期一 UTC+8下午8:52:44，Kid Guo写道：
Hi,
I have a problem based on "../sfepy/examples/multi_physics/biot.py". In the attached scripy 'block2.py', I have calculated the stress and strain according to the displacement, at the 111 and 112 lines you can
see
it.
Now I want to output the K of every cell, I gave a value of k at the 64 line in this script, it is a 3*3 array. I don't know which function or equation to use, so I was stucked at the 113 line. if I change the K value, will where be some difference results?
Thank you advance
K.
Hi Robert,
Thank you for your answer. The initial K value is
K = nm.array([[2.0, 0.2, 0.0], [0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], dtype = nm.float64).
Assuming my mesh file has 211 tetrahedra element,
Is it for the every element in the material, or the actual K total array like this
K = nm.array([[[2.0, 0.2, 0.0],
[0.2, 1.0, 0.0],
[0.0, 0.0, 0.5]],
[[2.0, 0.2, 0.0],
[0.2, 1.0, 0.0],
[0.0, 0.0, 0.5]],
......,
[[2.0, 0.2, 0.0],
[0.2, 1.0, 0.0],
[0.0, 0.0, 0.5]]],
dtype = nm.float64).K.shape = (211,3,3)
Is it right?
If it is right, now I want to give a total K array. for example.
K = nm.array([[[2.0, 0.2, 0.0],
[0.2, 1.0, 0.0],
[0.0, 0.0, 0.5]],
[[3.0, 0.2, 0.0],
[0.2, 2.0, 0.0],
[0.0, 0.0, 1.0]],
......,
[[10.0, 0.2, 0.0],
[0.2, 2.0, 0.0],
[0.0, 0.0, 0.4]]],
dtype = nm.float64).K.shape = (211,3,3)
I change the tensor k value in every tetrahedra element. then re-solve this problem. Is it possible?How can I do it?
Thank you advance. Regards, Kid
在 2016年12月30日星期五 UTC+8上午4:26:02，Robert Cimrman写道：
Hi Kid,
If you want to have the permeability varying with position, you have to define the material parameter using a function [1]- it seems that you assigned your new K2 directly to the material constructor, right?
r. [1[
https://groups.google.com/forum/?_escaped_fragment_=topic/sfepy-devel/6QWAwC...
Hi Robert,
Thank you very much, I have output the value of Kij in my script. In the script, I have give a initial value of K, of course , I assumed it is a homogeneous material, so it has the same value of every cell. Now I have new thinking, if it is a heterogeneous material, that Kij value are different of every cell. I want to give different K value in every cell, for example, in the cell 1, the Kij value is K1 = nm.array([[1.0, 0.0, 0.0], [0.0, 1.0, 0.0], [0.0, 0.0, 1.0]], dtype = nm.float64) in the cell 2, the Kij value is K2 = nm.array([[1.1, 0.0, 0.0], [0.0, 1.2, 0.0], [0.0, 0.0, 10.1]], dtype = nm.float64) etc. The new Kij like this: K2 = nm.array([[[[ 1., 0., 0.], [ 0, 1., 0.], [ 0., 0., 1.]]],
[[[ 1.1, 0., 0.], [ 0, 1.2, 0.], [ 0., 0., 10.1]]]], dtype = nm.float64)when I give the new value to the problem,it has a error. "ValueError: material parameter array must have three dimensions! ('k' has 4)"
or the K is
K3 = nm.array([[[1.0, 0.0, 0.0],
[1.0, 0.0, 0.0], [0.0, 0.0, 1.0]], [[1.0, 0.0, 0.0], [0.0, 2.0, 0.0], [0.0, 0.0, 3.0]]], dtype = nm.float64)it said:
"ValueError: incompatible shapes! ((27, 8, 3, 8), out: (27, 1, 3, 3), arr: (54, 8, 3, 3))"
what should I do?
Any idea? Thank you advance.
K.
在 2016年12月28日星期三 UTC+8下午9:14:36，Robert Cimrman写道：
Hi Kid,
you can use
hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m.k, p)',m=m,
mode='el_avg')...
out['hydraulic_conductivity'] = Struct(name='output_data',mode='cell', data=hydraulic_conductivity)
Essentially, ev_integrate_mat can be used to integrate any material parameter, check [1] to see what the mode argument means.
r.
[1]
http://sfepy.org/doc-devel/src/sfepy/terms/terms_basic.html#sfepy.terms.term...
On 12/28/2016 09:01 AM, Kid Guo wrote:
Where Ki is the permeability of the i-th compartment and Kij is the general diffusion permeability. how to ouput the Kij of each cell.
在 2016年12月26日星期一 UTC+8下午8:52:44，Kid Guo写道：
Hi,
I have a problem based on "../sfepy/examples/multi_physics/biot.py". In the attached scripy 'block2.py', I have calculated the stress and strain according to the displacement, at the 111 and 112 lines you
can
see
it.
Now I want to output the K of every cell, I gave a value of k at the 64 line in this script, it is a 3*3 array. I don't know which function or equation to use, so I was stucked at
On 12/29/2016 09:50 AM, Kid Guo wrote: the
113 line. if I change the K value, will where be some difference results?
Thank you advance
K.
Hi Kid,
first, if you do (e.g. before pb.solve() call in your block_hc.py script):
pb.update_materials()
you could see the following:
ipdb> pb.get_materials()['m'].datas[('Omega', 2)]['k'].shape (211, 4, 3, 3) ipdb> pb.get_materials()['m'].datas[('Omega', 1)]['k'].shape (211, 1, 3, 3)
the first shape corresponds to the order 2 integral, the second shape to the order 1 integral. -> You may want to use two materials instead, one for each order of integration, so that the data are not uselessly duplicated.
This also means, that the material does not require a value per cell, but a value per quadrature point. Put the following stub after your definition of 'm'
def get_pars(ts, coors, mode=None, **kwargs):
if mode == 'qp':
out = {}
print(coors.shape)
print(domain.shape.n_el)
n_qp = coors.shape[0] // domain.shape.n_el
print n_qp
k = nm.zeros((coors.shape[0], 3, 3))
# Fill your values here...
out['k'] = k
return out
m2 = Material('m2', function=get_pars)edit it to use your values, and use m2.k instead of m.k everywhere (remove k from m).
r.
On 12/30/2016 10:00 AM, Kid Guo wrote:
Hi Robert,
Thank you for your answer. The initial K value is
K = nm.array([[2.0, 0.2, 0.0], [0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], dtype = nm.float64).
Assuming my mesh file has 211 tetrahedra element,
Is it for the every element in the material, or the actual K total array like this
K = nm.array([[[2.0, 0.2, 0.0],
[0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], [[2.0, 0.2, 0.0], [0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], ......, [[2.0, 0.2, 0.0], [0.2, 1.0, 0.0], [0.0, 0.0, 0.5]]], dtype = nm.float64).K.shape = (211,3,3)
Is it right?
If it is right, now I want to give a total K array. for example.
K = nm.array([[[2.0, 0.2, 0.0],
[0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], [[3.0, 0.2, 0.0], [0.2, 2.0, 0.0], [0.0, 0.0, 1.0]], ......, [[10.0, 0.2, 0.0], [0.2, 2.0, 0.0], [0.0, 0.0, 0.4]]], dtype = nm.float64).K.shape = (211,3,3)
I change the tensor k value in every tetrahedra element. then re-solve this problem. Is it possible?How can I do it?
Thank you advance. Regards, Kid
在 2016年12月30日星期五 UTC+8上午4:26:02，Robert Cimrman写道：
Hi Kid,
If you want to have the permeability varying with position, you have to define the material parameter using a function [1]- it seems that you assigned your new K2 directly to the material constructor, right?
r. [1[
https://groups.google.com/forum/?_escaped_fragment_=topic/sfepy-devel/6QWAwC...
Hi Robert,
Thank you very much, I have output the value of Kij in my script. In the script, I have give a initial value of K, of course , I assumed it is a homogeneous material, so it has the same value of every cell. Now I have new thinking, if it is a heterogeneous material, that Kij value are different of every cell. I want to give different K value in every cell, for example, in the cell 1, the Kij value is K1 = nm.array([[1.0, 0.0, 0.0], [0.0, 1.0, 0.0], [0.0, 0.0, 1.0]], dtype = nm.float64) in the cell 2, the Kij value is K2 = nm.array([[1.1, 0.0, 0.0], [0.0, 1.2, 0.0], [0.0, 0.0, 10.1]], dtype = nm.float64) etc. The new Kij like this: K2 = nm.array([[[[ 1., 0., 0.], [ 0, 1., 0.], [ 0., 0., 1.]]],
[[[ 1.1, 0., 0.], [ 0, 1.2, 0.], [ 0., 0., 10.1]]]], dtype = nm.float64)when I give the new value to the problem,it has a error. "ValueError: material parameter array must have three dimensions! ('k' has 4)"
or the K is
K3 = nm.array([[[1.0, 0.0, 0.0],
[1.0, 0.0, 0.0], [0.0, 0.0, 1.0]], [[1.0, 0.0, 0.0], [0.0, 2.0, 0.0], [0.0, 0.0, 3.0]]], dtype = nm.float64)it said:
"ValueError: incompatible shapes! ((27, 8, 3, 8), out: (27, 1, 3, 3), arr: (54, 8, 3, 3))"
what should I do?
Any idea? Thank you advance.
K.
在 2016年12月28日星期三 UTC+8下午9:14:36，Robert Cimrman写道：
Hi Kid,
you can use
hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m.k, p)',m=m,
mode='el_avg')...
out['hydraulic_conductivity'] = Struct(name='output_data',mode='cell', data=hydraulic_conductivity)
Essentially, ev_integrate_mat can be used to integrate any material parameter, check [1] to see what the mode argument means.
r.
[1]
http://sfepy.org/doc-devel/src/sfepy/terms/terms_basic.html#sfepy.terms.term...
On 12/28/2016 09:01 AM, Kid Guo wrote:
Where Ki is the permeability of the i-th compartment and Kij is the general diffusion permeability. how to ouput the Kij of each cell.
在 2016年12月26日星期一 UTC+8下午8:52:44，Kid Guo写道：
Hi,
I have a problem based on "../sfepy/examples/multi_physics/biot.py". In the attached scripy 'block2.py', I have calculated the stress and strain according to the displacement, at the 111 and 112 lines you
can
see
it.
Now I want to output the K of every cell, I gave a value of k at the 64 line in this script, it is a 3*3 array. I don't know which function or equation to use, so I was stucked at
On 12/29/2016 09:50 AM, Kid Guo wrote: the
113 line. if I change the K value, will where be some difference results?
Thank you advance
K.
Hi Robert,
First thanks for your reply. In my scripy, I have replaced the m.k with new m2.k, it worked. At the 98line of the script, the new term is "t22 = Term.new('dw_diffusion(m2.k, q, p)', integral1, omega, m2=m2, q=q, p=p)" Now I output the 'hydraulic_conductivity' and save it in the 'block_pointload_script_hc_strain.vtk'. I can view the result through the 142 and 143 line in the script. I have a new question, if I output 'hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m.k, p)', m=m, mode='el_avg')' ,all k value is the same and the result can view it,the hydraulic_conductivity is 0.104, you can see it in the attached file. and if I output 'hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m2.k, p)', m2=m2, mode='el_avg')' ,the output value should be different, but the hydraulic_conductivity is 0,you can see it in the attached file. What is wrong?
Thank you advance. Regards Kid
在 2016年12月30日星期五 UTC+8下午6:14:38，Robert Cimrman写道：
Hi Kid,
first, if you do (e.g. before pb.solve() call in your block_hc.py script):
pb.update_materials()
you could see the following:
ipdb> pb.get_materials()['m'].datas[('Omega', 2)]['k'].shape (211, 4, 3, 3) ipdb> pb.get_materials()['m'].datas[('Omega', 1)]['k'].shape (211, 1, 3, 3)
the first shape corresponds to the order 2 integral, the second shape to the order 1 integral. -> You may want to use two materials instead, one for each order of integration, so that the data are not uselessly duplicated.
This also means, that the material does not require a value per cell, but a value per quadrature point. Put the following stub after your definition of 'm'
def get_pars(ts, coors, mode=None, **kwargs): if mode == 'qp': out = {} print(coors.shape) print(domain.shape.n_el) n_qp = coors.shape[0] // domain.shape.n_el print n_qp k = nm.zeros((coors.shape[0], 3, 3)) # Fill your values here... out['k'] = k return out m2 = Material('m2', function=get_pars)edit it to use your values, and use m2.k instead of m.k everywhere (remove k from m).
r.
Hi Robert,
Thank you for your answer. The initial K value is
K = nm.array([[2.0, 0.2, 0.0], [0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], dtype = nm.float64).
Assuming my mesh file has 211 tetrahedra element,
Is it for the every element in the material, or the actual K total array
On 12/30/2016 10:00 AM, Kid Guo wrote: like this
K = nm.array([[[2.0, 0.2, 0.0],
[0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], [[2.0, 0.2, 0.0], [0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], ......, [[2.0, 0.2, 0.0], [0.2, 1.0, 0.0], [0.0, 0.0, 0.5]]], dtype = nm.float64).K.shape = (211,3,3)
Is it right?
If it is right, now I want to give a total K array. for example.
K = nm.array([[[2.0, 0.2, 0.0],
[0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], [[3.0, 0.2, 0.0], [0.2, 2.0, 0.0], [0.0, 0.0, 1.0]], ......, [[10.0, 0.2, 0.0], [0.2, 2.0, 0.0], [0.0, 0.0, 0.4]]], dtype = nm.float64).K.shape = (211,3,3)
I change the tensor k value in every tetrahedra element. then re-solve this problem. Is it possible?How can I do it?
Thank you advance. Regards, Kid
在 2016年12月30日星期五 UTC+8上午4:26:02，Robert Cimrman写道：
Hi Kid,
If you want to have the permeability varying with position, you have to define the material parameter using a function [1]- it seems that you assigned your new K2 directly to the material constructor, right?
r. [1[
On 12/29/2016 09:50 AM, Kid Guo wrote:
Hi Robert,
Thank you very much, I have output the value of Kij in my script. In the script, I have give a initial value of K, of course , I assumed it is a homogeneous material, so it has the same value of every cell. Now I have new thinking, if it is a heterogeneous material, that Kij value are different of every cell. I want to give different K value in every cell, for example, in the cell 1, the Kij value is K1 = nm.array([[1.0, 0.0, 0.0], [0.0, 1.0, 0.0], [0.0, 0.0, 1.0]], dtype = nm.float64) in the cell 2, the Kij value is K2 = nm.array([[1.1, 0.0, 0.0], [0.0, 1.2, 0.0], [0.0, 0.0, 10.1]], dtype = nm.float64) etc. The new Kij like this: K2 = nm.array([[[[ 1., 0., 0.], [ 0, 1., 0.], [ 0., 0., 1.]]],
[[[ 1.1, 0., 0.], [ 0, 1.2, 0.], [ 0., 0., 10.1]]]], dtype = nm.float64)when I give the new value to the problem,it has a error. "ValueError: material parameter array must have three dimensions! ('k' has 4)"
or the K is
K3 = nm.array([[[1.0, 0.0, 0.0],
[1.0, 0.0, 0.0], [0.0, 0.0, 1.0]], [[1.0, 0.0, 0.0], [0.0, 2.0, 0.0], [0.0, 0.0, 3.0]]], dtype = nm.float64)it said:
"ValueError: incompatible shapes! ((27, 8, 3, 8), out: (27, 1, 3, 3), arr: (54, 8, 3, 3))"
what should I do?
Any idea? Thank you advance.
K.
在 2016年12月28日星期三 UTC+8下午9:14:36，Robert Cimrman写道：
Hi Kid,
you can use
hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m.k, p)',m=m,
mode='el_avg')...
out['hydraulic_conductivity'] = Struct(name='output_data',mode='cell',
data=hydraulic_conductivity)
Essentially, ev_integrate_mat can be used to integrate any material parameter, check [1] to see what the mode argument means.
r.
[1]
http://sfepy.org/doc-devel/src/sfepy/terms/terms_basic.html#sfepy.terms.term...
On 12/28/2016 09:01 AM, Kid Guo wrote:
Where Ki is the permeability of the i-th compartment and Kij is the general diffusion permeability. how to ouput the Kij of each cell.
在 2016年12月26日星期一 UTC+8下午8:52:44，Kid Guo写道： > > Hi, > > I have a problem based on
"../sfepy/examples/multi_physics/biot.py".
> In the attached scripy 'block2.py', I have calculated the stress and > strain according to the displacement, at the 111 and 112 lines you can see > it. > > Now I want to output the K of every cell, I gave a value of k at
https://groups.google.com/forum/?_escaped_fragment_=topic/sfepy-devel/6QWAwC... the
> line in this script, it is a 3*3 array. > I don't know which function or equation to use, so I was stucked at
64 the
> 113 line. > if I change the K value, will where be some difference results? > > Thank you advance > > K. > > >
Hi,
This is the supplement. If I set 't22 = Term.new('dw_diffusion(m.k, q, p)', integral1, omega, m=m, q=q, p=p) and hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m.k, p)', m=m, mode= 'el_avg')', the output result is the m_k.png file. then if I set 't22 = Term.new('dw_diffusion(m2.k, q, p)', integral1, omega, m2=m2, q=q, p=p) *and*
hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m2.k, p)', m2=m2, mode='el_avg')
' the output result is the m2_k.png file.
By comparing the two file, I can see the displacement,cauchy_strain and hydraulic_conductivity is different, that is to say, when I get the m2.k, the Kij is different of each element.it works.But the view result is 0.
Why? Thank you advance. Regards, Kid
在 2016年12月30日星期五 UTC+8下午6:14:38，Robert Cimrman写道：
Hi Kid,
first, if you do (e.g. before pb.solve() call in your block_hc.py script):
pb.update_materials()
you could see the following:
ipdb> pb.get_materials()['m'].datas[('Omega', 2)]['k'].shape (211, 4, 3, 3) ipdb> pb.get_materials()['m'].datas[('Omega', 1)]['k'].shape (211, 1, 3, 3)
the first shape corresponds to the order 2 integral, the second shape to the order 1 integral. -> You may want to use two materials instead, one for each order of integration, so that the data are not uselessly duplicated.
This also means, that the material does not require a value per cell, but a value per quadrature point. Put the following stub after your definition of 'm'
def get_pars(ts, coors, mode=None, **kwargs): if mode == 'qp': out = {} print(coors.shape) print(domain.shape.n_el) n_qp = coors.shape[0] // domain.shape.n_el print n_qp k = nm.zeros((coors.shape[0], 3, 3)) # Fill your values here... out['k'] = k return out m2 = Material('m2', function=get_pars)edit it to use your values, and use m2.k instead of m.k everywhere (remove k from m).
r.
Hi Robert,
Thank you for your answer. The initial K value is
K = nm.array([[2.0, 0.2, 0.0], [0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], dtype = nm.float64).
Assuming my mesh file has 211 tetrahedra element,
Is it for the every element in the material, or the actual K total array
On 12/30/2016 10:00 AM, Kid Guo wrote: like this
K = nm.array([[[2.0, 0.2, 0.0],
[0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], [[2.0, 0.2, 0.0], [0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], ......, [[2.0, 0.2, 0.0], [0.2, 1.0, 0.0], [0.0, 0.0, 0.5]]], dtype = nm.float64).K.shape = (211,3,3)
Is it right?
If it is right, now I want to give a total K array. for example.
K = nm.array([[[2.0, 0.2, 0.0],
[0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], [[3.0, 0.2, 0.0], [0.2, 2.0, 0.0], [0.0, 0.0, 1.0]], ......, [[10.0, 0.2, 0.0], [0.2, 2.0, 0.0], [0.0, 0.0, 0.4]]], dtype = nm.float64).K.shape = (211,3,3)
I change the tensor k value in every tetrahedra element. then re-solve this problem. Is it possible?How can I do it?
Thank you advance. Regards, Kid
在 2016年12月30日星期五 UTC+8上午4:26:02，Robert Cimrman写道：
Hi Kid,
If you want to have the permeability varying with position, you have to define the material parameter using a function [1]- it seems that you assigned your new K2 directly to the material constructor, right?
r. [1[
On 12/29/2016 09:50 AM, Kid Guo wrote:
Hi Robert,
Thank you very much, I have output the value of Kij in my script. In the script, I have give a initial value of K, of course , I assumed it is a homogeneous material, so it has the same value of every cell. Now I have new thinking, if it is a heterogeneous material, that Kij value are different of every cell. I want to give different K value in every cell, for example, in the cell 1, the Kij value is K1 = nm.array([[1.0, 0.0, 0.0], [0.0, 1.0, 0.0], [0.0, 0.0, 1.0]], dtype = nm.float64) in the cell 2, the Kij value is K2 = nm.array([[1.1, 0.0, 0.0], [0.0, 1.2, 0.0], [0.0, 0.0, 10.1]], dtype = nm.float64) etc. The new Kij like this: K2 = nm.array([[[[ 1., 0., 0.], [ 0, 1., 0.], [ 0., 0., 1.]]],
[[[ 1.1, 0., 0.], [ 0, 1.2, 0.], [ 0., 0., 10.1]]]], dtype = nm.float64)when I give the new value to the problem,it has a error. "ValueError: material parameter array must have three dimensions! ('k' has 4)"
or the K is
K3 = nm.array([[[1.0, 0.0, 0.0],
[1.0, 0.0, 0.0], [0.0, 0.0, 1.0]], [[1.0, 0.0, 0.0], [0.0, 2.0, 0.0], [0.0, 0.0, 3.0]]], dtype = nm.float64)it said:
"ValueError: incompatible shapes! ((27, 8, 3, 8), out: (27, 1, 3, 3), arr: (54, 8, 3, 3))"
what should I do?
Any idea? Thank you advance.
K.
在 2016年12月28日星期三 UTC+8下午9:14:36，Robert Cimrman写道：
Hi Kid,
you can use
hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m.k, p)',m=m,
mode='el_avg')...
out['hydraulic_conductivity'] = Struct(name='output_data',mode='cell',
data=hydraulic_conductivity)
Essentially, ev_integrate_mat can be used to integrate any material parameter, check [1] to see what the mode argument means.
r.
[1]
http://sfepy.org/doc-devel/src/sfepy/terms/terms_basic.html#sfepy.terms.term...
On 12/28/2016 09:01 AM, Kid Guo wrote:
Where Ki is the permeability of the i-th compartment and Kij is the general diffusion permeability. how to ouput the Kij of each cell.
在 2016年12月26日星期一 UTC+8下午8:52:44，Kid Guo写道： > > Hi, > > I have a problem based on
"../sfepy/examples/multi_physics/biot.py".
> In the attached scripy 'block2.py', I have calculated the stress and > strain according to the displacement, at the 111 and 112 lines you can see > it. > > Now I want to output the K of every cell, I gave a value of k at
https://groups.google.com/forum/?_escaped_fragment_=topic/sfepy-devel/6QWAwC... the
> line in this script, it is a 3*3 array. > I don't know which function or equation to use, so I was stucked at
64 the
> 113 line. > if I change the K value, will where be some difference results? > > Thank you advance > > K. > > >
Hi Kid,
It is a mayavi tensor visualization issue. In paraview I can see the Kij different in each cell (with the mySample.mesh you sent previously).
The problem is, that in mayavi, tensors as a whole can be displayed only with the following three modes: component, determinant, effective_stress. SfePy uses effective_stress, which is zero for a diagonal tensor.
You can workaround that, for example, by computing the matrix norm of Kij yourself and store that in the results file, maybe alongside the full Kij:
out['hc_norm'] = Struct(name='output_data', mode='cell',
data=nm.linalg.norm(hydraulic_conductivity,
axis=(2,3), keepdims=True))BTW. you can select data to display, as:
view(is_scalar_bar=True, is_wireframe=True, is_3d=True,
only_names=['hc_norm'])r.
On 01/09/2017 07:12 AM, Kid Guo wrote:
Hi,
This is the supplement. If I set 't22 = Term.new('dw_diffusion(m.k, q, p)', integral1, omega, m=m, q=q, p=p) and hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m.k, p)', m=m, mode= 'el_avg')', the output result is the m_k.png file. then if I set 't22 = Term.new('dw_diffusion(m2.k, q, p)', integral1, omega, m2=m2, q=q, p=p) *and*
hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m2.k, p)', m2=m2, mode='el_avg')
' the output result is the m2_k.png file.
By comparing the two file, I can see the displacement,cauchy_strain and hydraulic_conductivity is different, that is to say, when I get the m2.k, the Kij is different of each element.it works.But the view result is 0.
Why? Thank you advance. Regards, Kid
在 2016年12月30日星期五 UTC+8下午6:14:38，Robert Cimrman写道：
Hi Kid,
first, if you do (e.g. before pb.solve() call in your block_hc.py script):
pb.update_materials()
you could see the following:
ipdb> pb.get_materials()['m'].datas[('Omega', 2)]['k'].shape (211, 4, 3, 3) ipdb> pb.get_materials()['m'].datas[('Omega', 1)]['k'].shape (211, 1, 3, 3)
the first shape corresponds to the order 2 integral, the second shape to the order 1 integral. -> You may want to use two materials instead, one for each order of integration, so that the data are not uselessly duplicated.
This also means, that the material does not require a value per cell, but a value per quadrature point. Put the following stub after your definition of 'm'
def get_pars(ts, coors, mode=None, **kwargs): if mode == 'qp': out = {} print(coors.shape) print(domain.shape.n_el) n_qp = coors.shape[0] // domain.shape.n_el print n_qp k = nm.zeros((coors.shape[0], 3, 3)) # Fill your values here... out['k'] = k return out m2 = Material('m2', function=get_pars)edit it to use your values, and use m2.k instead of m.k everywhere (remove k from m).
r.
Hi Robert,
Thank you for your answer. The initial K value is
K = nm.array([[2.0, 0.2, 0.0], [0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], dtype = nm.float64).
Assuming my mesh file has 211 tetrahedra element,
Is it for the every element in the material, or the actual K total array
On 12/30/2016 10:00 AM, Kid Guo wrote: like this
K = nm.array([[[2.0, 0.2, 0.0],
[0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], [[2.0, 0.2, 0.0], [0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], ......, [[2.0, 0.2, 0.0], [0.2, 1.0, 0.0], [0.0, 0.0, 0.5]]], dtype = nm.float64).K.shape = (211,3,3)
Is it right?
If it is right, now I want to give a total K array. for example.
K = nm.array([[[2.0, 0.2, 0.0],
[0.2, 1.0, 0.0], [0.0, 0.0, 0.5]], [[3.0, 0.2, 0.0], [0.2, 2.0, 0.0], [0.0, 0.0, 1.0]], ......, [[10.0, 0.2, 0.0], [0.2, 2.0, 0.0], [0.0, 0.0, 0.4]]], dtype = nm.float64).K.shape = (211,3,3)
I change the tensor k value in every tetrahedra element. then re-solve this problem. Is it possible?How can I do it?
Thank you advance. Regards, Kid
在 2016年12月30日星期五 UTC+8上午4:26:02，Robert Cimrman写道：
Hi Kid,
If you want to have the permeability varying with position, you have to define the material parameter using a function [1]- it seems that you assigned your new K2 directly to the material constructor, right?
r. [1[
On 12/29/2016 09:50 AM, Kid Guo wrote:
Hi Robert,
Thank you very much, I have output the value of Kij in my script. In the script, I have give a initial value of K, of course , I assumed it is a homogeneous material, so it has the same value of every cell. Now I have new thinking, if it is a heterogeneous material, that Kij value are different of every cell. I want to give different K value in every cell, for example, in the cell 1, the Kij value is K1 = nm.array([[1.0, 0.0, 0.0], [0.0, 1.0, 0.0], [0.0, 0.0, 1.0]], dtype = nm.float64) in the cell 2, the Kij value is K2 = nm.array([[1.1, 0.0, 0.0], [0.0, 1.2, 0.0], [0.0, 0.0, 10.1]], dtype = nm.float64) etc. The new Kij like this: K2 = nm.array([[[[ 1., 0., 0.], [ 0, 1., 0.], [ 0., 0., 1.]]],
[[[ 1.1, 0., 0.], [ 0, 1.2, 0.], [ 0., 0., 10.1]]]], dtype = nm.float64)when I give the new value to the problem,it has a error. "ValueError: material parameter array must have three dimensions! ('k' has 4)"
or the K is
K3 = nm.array([[[1.0, 0.0, 0.0],
[1.0, 0.0, 0.0], [0.0, 0.0, 1.0]], [[1.0, 0.0, 0.0], [0.0, 2.0, 0.0], [0.0, 0.0, 3.0]]], dtype = nm.float64)it said:
"ValueError: incompatible shapes! ((27, 8, 3, 8), out: (27, 1, 3, 3), arr: (54, 8, 3, 3))"
what should I do?
Any idea? Thank you advance.
K.
在 2016年12月28日星期三 UTC+8下午9:14:36，Robert Cimrman写道：
Hi Kid,
you can use
hydraulic_conductivity = ev('ev_integrate_mat.3.Omega(m.k, p)',m=m,
mode='el_avg')...
out['hydraulic_conductivity'] = Struct(name='output_data',mode='cell',
data=hydraulic_conductivity)
Essentially, ev_integrate_mat can be used to integrate any material parameter, check [1] to see what the mode argument means.
r.
[1]
http://sfepy.org/doc-devel/src/sfepy/terms/terms_basic.html#sfepy.terms.term...
On 12/28/2016 09:01 AM, Kid Guo wrote: > Where Ki is the permeability of the i-th compartment and Kij is the general > diffusion permeability. > how to ouput the Kij of each cell. > > 在 2016年12月26日星期一 UTC+8下午8:52:44，Kid Guo写道： >> >> Hi, >> >> I have a problem based on
"../sfepy/examples/multi_physics/biot.py".
>> In the attached scripy 'block2.py', I have calculated the stress and >> strain according to the displacement, at the 111 and 112 lines you can see >> it. >> >> Now I want to output the K of every cell, I gave a value of k at
https://groups.google.com/forum/?_escaped_fragment_=topic/sfepy-devel/6QWAwC... the
>> line in this script, it is a 3*3 array. >> I don't know which function or equation to use, so I was stucked at
64 the
>> 113 line. >> if I change the K value, will where be some difference results? >> >> Thank you advance >> >> K. >> >> >> >
participants (2)
-
Kid Guo -
Robert Cimrman