Presentation is loading. Please wait.

Presentation is loading. Please wait.

Symmetry & boundary conditions Joël Cugnoni, LMAF/EPFL, 2009.

Similar presentations


Presentation on theme: "Symmetry & boundary conditions Joël Cugnoni, LMAF/EPFL, 2009."— Presentation transcript:

1 Symmetry & boundary conditions Joël Cugnoni, LMAF/EPFL, 2009

2 Using symmetries in FE models A FE model is symmetric if and only if geometry, materials and loading are symmetric !! Symmetries help to:  Reduce the model size => finer meshes => better accuracy!  Simplify the definition of isostatic boundary conditions  Reduce the post-processing effort (simpler to visualize)  Show to everybody that you master FE modelling ;-)

3 Using symmetries in FE models To use symmetries:  Extract the smallest possible geometric region with « CAD » cut operations (can have multiple symmetries!!)  Model the symmetry planes as imposed displacement / rotations: No displacement perpendicular to symm. plane No rotations (shell / beams only) along 2 axis in the symm. Plane Example: X-symmetry = symmetry wrt a plane of normal along X => U1 = UR2 = UR3 =0 ALWAYS USE SYMMETRIES WHENEVER POSSIBLE !!! (This will be check at the exams)

4 Symmetry: example U normal = 0 UR inplane = 0 Symmetry plane

5 Rigid body motions In statics, rigid body motions are responsible for singular stiffness matrices => no solution In statics, YOU MUST CONSTRAIN all 6 rigid body motions with suitable boundary conditions. If you don’t want to introduce additionnal stresses: use isostatic BC 90 % of the « the solver does not want to converge » problems come from rigid body motions !! => Always double check your boundary conditions

6 The trick Is a simple trick to set isostatic boundary conditions:  Select 3 points (forming a plane)  On a 1st point: block 3 displacements => all translation are constrained  On a 2 nd point, block 2 displacements to prevent 2 rotations  On a 3rd point, block 1 displacement to block the last rotation.

7 Isostatic BC: Example of rule U1=U2=U3=0 Loads F1 + F2 = 0 But system cannot be solved because of rigid body motions F1 F2 U2=U3=0 U2=0 Using the trick, we introduce isostatic supports which do not overconstrain the system

8 Loading: standard type of loads Pressure:  Units: force / area  Is always NORMAL to the surface  Positive towards the Inside  Non uniform distribution with analytical fields function of coordinates Surface tractions:  Units: force / area  Can be freely oriented: define Gravity:  Units: L/T^2  Defines the accelaration vector of gravity loads.  You must define a Density in material properties Acceleration, Centrifugal loads …

9 Demo & tutorials Demo of Rod FEA  Use partitions to create loading surfaces  Use surface tractions  Show rigid body motion = solver problem  Use rule to set isostatic BC Video tutorial BC-Tutorial:  Use symmetries  Use cylindrical coordinate systems to apply BC  Apply non-uniform load distributions


Download ppt "Symmetry & boundary conditions Joël Cugnoni, LMAF/EPFL, 2009."

Similar presentations


Ads by Google