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A beginner's tutorial highlighting the basics of SfePy.


This primer presents a step-by-step walk-through of the process to solve a simple mechanics problem. The typical process to solve a problem using SfePy is followed: a model is meshed, a problem definition file is drafted, SfePy is run to solve the problem and finally the results of the analysis are visualised.

Problem statement

A popular test to measure the tensile strength of concrete or asphalt materials is the indirect tensile strength (ITS) test pictured below. In this test a cylindrical specimen is loaded across its diameter to failure. The test is usually run by loading the specimen at a constant deformation rate of 50 mm/minute (say) and measuring the load response. When the tensile stress that develops in the specimen under loading exceeds its tensile strength then the specimen will fail. To model this problem using finite elements the indirect tensile test can be simplified to represent a diametrically point loaded disk as shown in the schematic.

images/primer/ITS.png images/primer/ITS2D.png

The tensile and compressive stresses that develop in the specimen as a result of the point loads P are a function of the diameter (D) and thickness (t) of the cylindrical specimen. At the centre of the specimen, the compressive stress is 3 times the tensile stress and the analytical formulation for these are, respectively:

These solutions may be approximated using finite element methods. To solve this problem using SfePy the first step is meshing a suitable model.


Assuming plane strain conditions, the indirect tensile test may be modelled using a 2-D finite element mesh. Furthermore, the geometry of the model is symmetrical about the x- and y-axes passing through the centre of the circle. To take advantage of this symmetry only one quarter of the 2-D model will be meshed and boundary conditions will be established to indicate this symmetry. The meshing program `Gmsh`_ is used here to very quickly mesh the model. Follow these steps to model the ITS:

  1. The ITS specimen has a diameter of 150 mm. Using Gmsh add three new points (geometry elementary entities) at the following coordinates: (0.0 0.0), (75.0,0.0) and (0.0,75.0).
  2. Next add two straight lines connecting the points.
  3. Next add a Circle arc connecting two of the points to form the quarter circle segment.
  4. Still under Geometry add a ruled surface.
  5. With the geometry of the model defined, add a mesh by clicking on the 2D button under the Mesh functions.

The figures that follow show the various stages in the model process.

images/primer/mesh1.png images/primer/mesh2.png images/primer/mesh3.png images/primer/mesh4.png

That's the meshing done. Save the mesh in a format that SfePy recognizes. For now use the medit .mesh format e.g. its2D.mesh.

Hint: Check the drop down in the Save As dialog for the different formats that Gmsh can save to.

If you open the its2D.mesh file using a text editor you'll notice that Gmsh saves the mesh in a 3-D format and includes some extra geometry items that should be deleted. Reformatted the mesh file to a 2-D format and delete the Edges block. Note that when you do this the file cannot be reopened by Gmsh so it is always a good idea to also save your meshes in Gmsh's native format as well (Shift-Ctrl-S). Click :download:`here <../meshes/2d/its2D.mesh>` to download the reformatted mesh file that will be used in the tutorial.


You'll notice that the mesh contains 55 vertices (nodes) and 83 triangle elements. The mesh file provides the coordinates of the nodes and the element connectivity. It is important to note that node and element numbering in SfePy start at 0 and not 1 as is the case in Gmsh and some other meshing programs.

To view .mesh files you can use a demo of `medit`_. After loading your mesh file with medit you can see the node and element numbering by pressing P and F respectively. The numbering in medit starts at 1 as shown. Thus the node at the center of the model in SfePy numbering is 0, and elements 76 and 77 are connected to this node. Node and element numbers can also be viewed in Gmsh - under the mesh option under the Visibility tab enable the node and surface labels. Note that the surface labels as numbered in Gmsh follow on from the line numbering. So to get the corresponding element number in SfePy you'll need to subtract the number of lines in the Gmsh file + 1. Confused yet? Luckily, SfePy provides some useful mesh functions to indicate which elements are connected to which nodes. Nodes and elements can also be identified by defining regions, which is addressed later.

Another open source python option to view .mesh files is the appropriately named `Python Mesh Viewer`_.

The next step in the process is coding the SfePy problem definition file.

Problem description

The programming of the problem description file is well documented in the SfePy :doc:`users_guide`. The problem description file used in the tutorial follows:

:download:`Download </../examples/linear_elasticity/>` and open the file in your favourite python editor. Note that you may wish to change the location of the output directory to somewhere on your drive. You may also need to edit the mesh file name. For the analysis we will assume that the material of the test specimen is linear elastic and isotropic. We define two material constants i.e. Young's modulus and Poisson's ratio. The material is assumed to be asphalt concrete having a Young's modulus of 2,000 MPa and a Poisson's ration of 0.4.

Note: Be consistent in your choice and use of units. In the tutorial we are using Newton (N), millimeters (mm) and megaPascal (MPa). The :doc:`sfepy.mechanics.units <src/sfepy/mechanics/units>` module might help you in determining which derived units correspond to given basic units.

The following block of code defines regions on your mesh:

regions = {
    'Omega' : ('all', {}),
    'Left' : ('nodes in (x < 0.001)', {}),
    'Bottom' : ('nodes in (y < 0.001)', {}),
    'Top' : ('node 2', {}),

Four regions are defined:

  1. Omega: all the elements in the mesh
  2. Left: the y-axis
  3. Bottom: the x-axis
  4. Top: the topmost node. This is where the load is applied.

Having defined the regions these can be used in other parts of your code. For example, in the definition of the boundary conditions:

ebcs = {
    'XSym' : ('Bottom', {'u.1' : 0.0}),
    'YSym' : ('Left', {'u.0' : 0.0}),

Now the power of the regions entity becomes apparent. To ensure symmetry about the x-axis, the vertical or y-displacement of the nodes in the Bottom region are prevented or set to zero. Similarly, for symmetry about the y-axis, any horizontal or displacement in the x-direction of the nodes in the Left region or y-axis is prevented.

The load is specified in terms of the 'Load' material as follows:

materials = {
    'Asphalt' : ({
        'lam' : lame_from_youngpoisson(young, poisson)[0],
        'mu' : lame_from_youngpoisson(young, poisson)[1],
    'Load' : ({'.val' : [0.0, -1000.0]},),

Note the dot in '.val' - this denotes a special material value, i.e., a value that is not to be evaluated in quadrature points. The load is then applied in equations using the 'dw_point_load.0.Top(Load.val, v)' term in the topmost node (region Top).

We provided the material constants in terms of Young's modulus and Poisson's ratio, but the linear elastic isotropic equation used requires as input Lamé's parameters. The lame_from_youngpoisson() function is thus used for conversion. Note that to use this function it was necessary to import the function into the code, which was done up front:

from sfepy.mechanics.matcoefs import lame_from_youngpoisson

Hint: Check out the :doc:`sfepy.mechanics.matcoefs <src/sfepy/mechanics/matcoefs>` module for other useful material related functions.

That's it - we are now ready to solve the problem.

Running SfePy

One option to solve the problem is to run the SfePy script from the command shell:

$ ./

Note: For the purpose of this tutorial it is assumed that the problem definition file ( is in the same directory as the script. If you have the file in another directory then make sure you include the path to this file as well.

SfePy solves the problem and outputs the solution to the output path (output_dir) provided in the script. The output file will be in the vtk format by default if this is not explicitly specified and the name of the output file will be the same as that used for the mesh file except with the vtk extension i.e. its2D.vtk.

The vtk format is an ascii format. Open the file using a text editor. You'll notice that the output file includes separate sections:

  • POINTS (these are the model nodes)
  • CELLS (the model element connectivity)
  • VECTORS (the node displacements in the x-, y- and z- directions.

SfePy includes a script ( to quickly view the solution. To run this script you need to have `Mayavi`_ installed. From the command line issue the following (with the correct paths):

$ ./ its2D.vtk

The script generates the image shown below, which shows by default the displacements in the model as arrows and their magnitude as color scale. Cool, but we are more interested in the stresses. To get these we need to modify the problem description file and do some post-processing.



SfePy provides functions to calculate stresses and strains. We'll include a function to calculate these and update the problem material definition and options to call this function as a post_process_hook. Save this file as :download:` </../examples/linear_elasticity/>`.

The updated file imports all of the previous definitions in The stress function (de_cauchy_stress) requires as input the stiffness tensor - thus it was necessary to update the materials accordingly. The problem options were also updated to call the stress_strain function as a post_process_hook.

Run SfePy to solve the updated problem and view the solution (assuring the correct paths):

$ ./
$ ./ its2D.vtk -b

In addition to the node displacements, the vtk output shown below now also includes the stresses and strains averaged in the elements:


Remember the objective was to determine the stresses at the centre of the specimen under a load P. The solution as currently derived is expressed in terms of a global displacement vector (u). The global (residual) force vector (f) is a function of the global displacement vector and the global stiffness matrix (K) as: f = Ku. Let's determine the force vector interactively.

Running SfePy in interactive mode

In addition to solving problems using the script you can also run SfePy interactively. This requires that `IPython`_ be installed. To run SfePy interactively, use isfepy:

$ ./isfepy

Once isfepy loads up, issue the following command:

In [1]: pb, state = solve_pde('')

The problem is solved and the problem definition and solution are provided in the pb and state variables, respectively. The solution, or in this case, the global displacement vector (u), contains the x- and y-displacements at the nodes in the 2D model:

In [2]: u = state()

In [3]: u
array([ 0.        ,  0.        ,  0.37376671, ..., -0.19923848,
        0.08820237, -0.11201528])

In [4]: u.shape
Out[4]: (110,)

In [5]: u.shape = (55, 2)

In [6]: u
array([[ 0.        ,  0.        ],
       [ 0.37376671,  0.        ],
       [ 0.        , -1.65318152],
       [ 0.08716448, -0.23069047],
       [ 0.27741356, -0.19923848],
       [ 0.08820237, -0.11201528]])

Note: We have used the fact, that the state vector contains only one variable (u). In general, the following can be used:

In [7]: u = state.get_parts()['u']

In [8]: u
array([[ 0.        ,  0.        ],
       [ 0.37376671,  0.        ],
       [ 0.        , -1.65318152],
       [ 0.08716448, -0.23069047],
       [ 0.27741356, -0.19923848],
       [ 0.08820237, -0.11201528]])

Both state() and state.get_parts() return a view of the DOF vector, that is why in Out[8] the vector is reshaped according to Out[6].

From the above it can be seen that u holds the displacements at the 55 nodes in the model and that the displacement at node 2 (on which the load is applied) is (0, -1.65318152). The global stiffness matrix is saved in pb as a `sparse matrix`_:

In [9]: K = pb.mtx_a

In [10]: K
<94x94 sparse matrix of type '<type 'numpy.float64'>'
        with 1070 stored elements in Compressed Sparse Row format>

In [11]: print K
  (0, 0)        2443.95959851
  (0, 7)        -2110.99917491
  (0, 14)       -332.960423597
  (0, 15)       1428.57142857
  (1, 1)        2443.95959852
  (1, 13)       -2110.99917492
  (1, 32)       1428.57142857
  (1, 33)       -332.960423596
  (2, 2)        4048.78343529
  (2, 3)        -1354.87004384
  (2, 52)       -609.367453538
  (2, 53)       -1869.0018791
  (2, 92)       -357.41672785
  (2, 93)       1510.24654193
  (3, 2)        -1354.87004384
  (3, 3)        4121.03202907
  (3, 4)        -1696.54911732
  (3, 48)       76.2400806561
  (3, 49)       -1669.59247304
  (3, 52)       -1145.85294856
  (3, 53)       2062.13955556
  (4, 3)        -1696.54911732
  (4, 4)        4410.17902905
  (4, 5)        -1872.87344838
  (4, 42)       -130.515009576
  :     :
  (91, 81)      -1610.0550578
  (91, 86)      -199.343680224
  (91, 87)      -2330.41406097
  (91, 90)      -575.80373408
  (91, 91)      7853.23899229
  (92, 2)       -357.41672785
  (92, 8)       1735.59411191
  (92, 50)      -464.976034459
  (92, 51)      -1761.31189004
  (92, 52)      -3300.45367361
  (92, 53)      1574.59387937
  (92, 88)      -250.325600254
  (92, 89)      1334.11823335
  (92, 92)      9219.18643706
  (92, 93)      -2607.52659081
  (93, 2)       1510.24654193
  (93, 8)       -657.361661955
  (93, 50)      -1761.31189004
  (93, 51)      54.1134516246
  (93, 52)      1574.59387937
  (93, 53)      -315.793227627
  (93, 88)      1334.11823335
  (93, 89)      -4348.13351285
  (93, 92)      -2607.52659081
  (93, 93)      9821.16012014

In [12]: K.shape
Out[12]: (94, 94)

One would expect the shape of the global stiffness matrix (K) to be (110,110) i.e. to have the same number of rows and columns as u. This matrix has been reduced by the fixed degrees of freedom imposed by the boundary conditions set at the nodes on symmetry axes. To restore the matrix, temporarily remove the imposed boundary conditions:

In [13]: pb.remove_bcs()

Now we can calculate the force vector (f):

In [14]: f = pb.evaluator.eval_residual(u)

In [15]: f.shape
Out[15]: (110,)

In [16]: f
array([ -4.73618436e+01,   1.42752386e+02,   1.56921124e-13, ...,
        -2.06057393e-13,   2.13162821e-14,  -2.84217094e-14])

Remember to restore the original boundary conditions previously removed in step [13]:

In [17]: pb.time_update()

To view the residual force vector, we can save it to a vtk file. This requires creating a state and set its DOF vector to f as follows:

In [18]: state = pb.create_state()

In [19]: state.set_full(f)

In [20]: out = state.create_output_dict()

In [21]: pb.save_state('file.vtk', out=out)

Running the script on file.vtk displays the average nodal forces as shown below.


The forces in the x- and y-directions at node 2 are:

In [22]: f.shape = (55, 2)

In [23]: f[2]
Out[23]: array([  6.20373272e+02,  -1.13686838e-13])

Great, we have an almost zero residual vertical load or force apparent at node 2 i.e. -1.13686838e-13 Newton. Let us now check the stress at node 0, the centre of the specimen.

Generating output at element nodes

Previously we had calculated the stresses in the model but these were averaged from those calculated at Gauss quadrature points within the elements. It is possible to provide custom integrals to allow the calculation of stresses with the Gauss quadrature points at the element nodes. This will provide us a more accurate estimate of the stress at the centre of the specimen located at node 0. The :download:`code </../examples/linear_elasticity/>` below outlines one way to achieve this.

The output:

Given load = 2000.00 N

Analytical solution
Horizontal tensile stress = 8.48826e+00 MPa/mm
Vertical compressive stress = 2.54648e+01 MPa/mm

FEM solution
Horizontal tensile stress = 7.57220e+00 MPa/mm
Vertical compressive stress = 2.58660e+01 MPa/mm

Not bad for such a coarse mesh! Re-running the problem using a :download:`finer mesh <../meshes/2d/big.mesh>` provides a more accurate solution:

Given load = 2000.00 N

Analytical solution
Horizontal tensile stress = 8.48826e+00 MPa/mm
Vertical compressive stress = 2.54648e+01 MPa/mm

FEM solution
Horizontal tensile stress = 8.50042e+00 MPa/mm
Vertical compressive stress = 2.54300e+01 MPa/mm

To see how the FEM solution approaches the analytical one, try to play with the uniform mesh refinement level in the :ref:`primer_example_file` file, namely lines 25, 26:

refinement_level = 0
filename_mesh = refine_mesh(filename_mesh, refinement_level)

The above computation could also be done in isfepy:

In [23]: from sfepy.fem.geometry_element import geometry_data

In [24]: gdata = geometry_data['2_3']
In [25]: nc = len(gdata.coors)
In [26]: ivn = Integral('ivn', order=-1,
   ....:                coors=gdata.coors, weights=[gdata.volume / nc] * nc)

In [27]: pb, state = solve_pde('examples/linear_elasticity/')

In [28]: stress = pb.evaluate('ev_cauchy_stress.ivn.Omega(Asphalt.D,u)',
   ....:                      mode='qp', integrals=Integrals([ivn]))
In [29]: sfield = Field('stress', nm.float64, (3,), pb.domain.regions['Omega'])
In [30]: svar = FieldVariable('sigma', 'parameter', sfield, 3,
   ....:                      primary_var_name='(set-to-None)')
In [31]: svar.data_from_qp(stress, ivn)

In [32]: print 'Horizontal tensile stress = %.5e MPa/mm' % (svar()[0][0])
Horizontal tensile stress = 7.57220e+00 MPa/mm
In [33]: print 'Vertical compressive stress = %.5e MPa/mm' % (-svar()[0][1])
Vertical compressive stress = 2.58660e+01 MPa/mm

In [34]: mat = pb.get_materials()['Load']
In [35]: P = 2.0 * mat.get_data('special', None, 'val')[1]
In [36]: P
Out[36]: -2000.0

In [37]: print 'Horizontal tensile stress = %.5e MPa/mm' % (-2.*P/(nm.pi*150.))
Horizontal tensile stress = 8.48826e+00 MPa/mm
In [38]: print 'Vertical compressive stress = %.5e MPa/mm' % (-6.*P/(nm.pi*150.))
Vertical compressive stress = 2.54648e+01 MPa/mm

To wrap this tutorial up let's explore SfePy's probing functions.


As a bonus for sticking to the end of this tutorial see the following :download:`problem definition file </../examples/linear_elasticity/>` that provides SfePy functions to quickly and neatly probe the solution.

Probing applies interpolation to output the solution along specified paths. For the tutorial, line probing is done along the x- and y-axes of the model.

Notice that the output_format has been defined as h5. To apply probing first solve the problem as usual:

$ ./

This will write the solution to the output directory indicated. Then run the SfePy script on the solution:

$ ./ <sfepy output path>/its2D.h5

The results of the probing will be written to text files and the following figures will be generated. These figures show the displacements, normal stresses and strains as well as shear stresses and strains along the probe paths. Note that you need `matplotlib`_ installed to run this script.

images/primer/its2D_0.png images/primer/its2D_01.png

The end.

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