Dear Jean-François,
The problem with the expression
"0.5*sqr(Normalized(element_K).(Grad_u*Normalized(element_K)))"
was that element_K is in fact declared as a matrix ( a 2x1 matrix here)
and the dot product is restricted to vectors. I commited a fix to
transform element_K as a vector for 1D elements. You can download this
fix or use ":" instead of "." for the moment.
In fact, in that case, I think taht your expression is equivalent to
"Grad_u:Grad_Test_u" since the gradient is along the tangent.
Just a remark. It is of course allowed to use potential (it will be
derived twice to obtain the tangent matrix), but I recommend to gives
the weak formulation expression not to have some surprise, especially
for coupled problems (for instance if a coefficient is depending on
another variable).
Best regards,
Yves
Le 21/11/2017 à 00:41, Jean-François Barthélémy a écrit :
Dear Yves,
Here are some small examples in Python on a simple beam [AB] made of
10 elements considered as a plane structure (problem in 2D). The mesh
is loaded in GetFEM as a 2D mesh.
The file beam_bendmanual.py gives the solution of a beam in flexion
and tension under the Navier-Bernoulli hypothesis. The formulation is
built here by means of explicit indices in the assembly strings, which
is what I would actually like to avoid with a complex structure made
of arbitrary orientated beams. In this file, the displacement vector
as well as the rotation of point A are blocked and a force loads the
beam at point B. The stiffness are set to 1 for the sake of simplicity.
The file beam_bendgeneric.py gives the beggining of the attempt of a
generic way to deal with the same problem (exploiting then the local
direction). This file raises the exception mentioned in my previous
email when trying to assembly. I haven't added the term of bending
energy (involving the Hessian) nor the boundary condition blocking the
rotation yet since they may not be as simple. For example the rotation
should be blocked by imposing Grad_u.n=0 where n is the vector
orthogonal to the beam direction but I am not sure that such a
calculation of vectorial product 'n=ez x Normalized(element_K)' to get
the vector n is available in the high-level assemblage syntax. Is it?
Thank you for your help.
Best regards
Jean-François
2017-11-20 15:00 GMT+01:00 Yves Renard <[email protected]
<mailto:[email protected]>>:
This should work, yes. If you can send me a small program that I
can test, I can have a look to this problem.
Best regards,
Yves.
Le 20/11/2017 à 14:54, Jean-François Barthélémy a écrit :
Dear Yves,
Thank you. This is precisely the formulation I used but it raises
the following problem (in python)
0.5*sqr(Normalized(element_K).(Grad_u*Normalized(element_K)))
------------------------------------------^
The second argument of the dot product has to be a vector.
logic_error exception caught
...
RuntimeError: (Getfem::InterfaceError) -- Error in
getfem_generic_assembly.cc, line 8949 void
getfem::ga_node_analysis(const string&, getfem::ga_tree&, const
getfem::ga_workspace&, getfem::pga_tree_node, bgeot::size_type,
bgeot::size_type, bool, bool, int):
Error in assembly string
following a call such as
md.add_linear_generic_assembly_brick(mim,"0.5*sqr(Normalized(element_K).(Grad_u*Normalized(element_K)))")
Did I miss something?
I am sorry to bother you again.
Thanks
Best regards
Jean-François
2017-11-20 14:10 GMT+01:00 Yves Renard <[email protected]
<mailto:[email protected]>>:
Dear Jean-François,
For a vector variable 'u', each line of 'Grad_u' is the
gradient of the ith component of 'u', each of them is tangent
to the curve and length being the derivative with respect to
the curvilinear abscissa. The linearized deformation is a
priori Normalized(element_K).(Grad_u * Normalized(element_K))
The formulas used to compute the gradient and the Hessian can
be found here:
http://getfem.org/project/femdesc.html#geometric-transformations
<http://getfem.org/project/femdesc.html#geometric-transformations>
http://getfem.org/project/appendixA.html#derivative-computation
<http://getfem.org/project/appendixA.html#derivative-computation>
The hessian of a vector valued variable is also the hessian
of each component.
Best regards,
Yves.
Le 20/11/2017 à 02:10, Jean-François Barthélémy a écrit :
Dear Yves,
Thank you very much for your answer.
It's OK for scalar variables but I do not really understand
how Grad_u is built when u is a displacement vector field of
3 components. I thought Grad_u would represent the vector
du/ds ([dux/ds,duy/ds,duz/ds]) with s the local curvilinear
abscissa (so that Normalized(element_K).Grad_u would give
the linearized longitudinal deformation) but it seems that
Grad_u is actually a 3x3 matrix field. Then I do not see how
to build the longitudinal deformation. What would be the
best way please? And by the way, what would be the right
syntax to get the second derivative of the transverse
displacement by means of Hermite elements and the Hessian?
Thank you again for your help.
Best regards
Jean-François
2017-11-17 20:52 GMT+01:00 Yves Renard
<[email protected] <mailto:[email protected]>>:
Dear Jean-François,
There is no specific tool yet for that.
You can have access to the tangent with 'element_K' in
the generic assembly language (the unit tangent is then
'Normalized(element_K)')
If you define a scalar quantity "u" on your 1D
structure, then "Grad_u" will be the gradient of the
quantity in the sense that it is a tangent vector whose
norm is the derivative of the qunatity along the curve.
So that "Grad_u.Grad_Test_u" is still the stiffness term
for a curvilinear second derivative. For a vector
quantity "u", "Grad_u" is the componentwise gradient.
Best regard,
Yves.
----- Original Message -----
From: "Jean-François Barthélémy"
<[email protected]
<mailto:[email protected]>>
To: [email protected] <mailto:[email protected]>
Sent: Friday, November 17, 2017 6:17:13 PM
Subject: [Getfem-users] Curvilinear structures in Getfem
Dear Getfem users,
I wonder whether it is possible to model simple linear
elastic curvilinear
structures submitted to traction, bending, torsion
etc... in 2D or 3D in
Getfem. I haven't found a way to have access to the
tangential or normal
parts of vectors in the local basis of a beam and their
derivatives with
respect to the curvilinear abscissa needed to build the
formulation. Does
someone have an answer please?
Thanks in advance
Best regards
Jean-François
--
Yves Renard ([email protected]
<mailto:[email protected]>) tel : (33) 04.72.43.87.08
Pole de Mathematiques, INSA-Lyon fax : (33)
04.72.43.85.29
20, rue Albert Einstein
69621 Villeurbanne Cedex, FRANCE
http://math.univ-lyon1.fr/~renard
<http://math.univ-lyon1.fr/%7Erenard>
---------
--
Yves Renard ([email protected] <mailto:[email protected]>)
tel : (33) 04.72.43.87.08
Pole de Mathematiques, INSA-Lyon fax : (33) 04.72.43.85.29
20, rue Albert Einstein
69621 Villeurbanne Cedex, FRANCE
http://math.univ-lyon1.fr/~renard
<http://math.univ-lyon1.fr/%7Erenard>
---------
--
Yves Renard ([email protected]) tel : (33) 04.72.43.87.08
Pole de Mathematiques, INSA-Lyon fax : (33) 04.72.43.85.29
20, rue Albert Einstein
69621 Villeurbanne Cedex, FRANCE
http://math.univ-lyon1.fr/~renard
---------
