Contents
Description
- This example guides you through setting up and running a physical and material nonlinear problem of the passive (non-contracting) inflation of a truncated ellipsoid using pressure (natural) boundary conditions. This mesh is a rough approximation of a left ventricle.
- The mesh consists of three elements that are wrapped onto themselves in a prolate spheroidal coordinate system. The basis functions are linear in the circumferential direction and cubic in the longitudinal and radial directions (linear-cubic-cubic). The material chosen is a transversely isotropic exponential strain energy function. The material is stiffer in the fiber direction than perpendicular to them. The fibers whose orientation is defined by fiber angles in the node form with respect to the circumferential direction vary linearly in radial direction across the wall from -37 degrees on the epicardium to +84 degrees on the endocardium. The boundary conditions assure that all rigid body motions are suppressed. Pressure is prescribed on the endocardium to (partly) simulate the passive filling of a ventricle (diastole).
- The final (refined) model nodes and elements are defined according to the figure below:
- Stress and strain solution tables for various pressure loads and complete model files are provided at the end of this tutorial.
Building and solving the model
Start Continuity
- Launch the Continuity Client
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On the About Continuity startup screen
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Leave the mesh checkbox checked under Use Modules:
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In addition, check the biomechanics checkbox
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Click OK to bring up the main window
Create mesh
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Select the cont6 file for this tutorial ( prolatemesh.cont6 )
- To make sure you selected the right cont6 file, verify the following:
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- Check that the list contains only:
- Linear-Linear-Linear Lagrange 3*3*3
- Linear-Linear Lagrange 3*3
- Linear-Cubic-Cubic Hermite 3*3*3
- Linear-Cubic Hermite 3*3
- Cubic-Cubic Hermite 3*3
- Cubic-Linear Hermite 3*3
- Check that the list contains only:
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- Check that that there are only 8 nodes defined
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Make sure that in the Field Vector 1 tab, under Field Variable 1 the option Linear-Linear-Linear Langrange 3*3*3 is selected
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- Check that there are only 3 elements in the list
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If the Dimensions Form appears, simply click Apply Marked Recommendations and then OK
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Click the lines radio button
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Click Render to display mesh lines
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- The mesh should now look similar to the first screenshot
Refine the Mesh
To get a sufficiently converged result using linear elements, it is necessary to use multiple elements. Therefore, we will refine our single element into many elements.
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Decrease the New Element per old element in xi1 to 1. Increase the New Element per old element in xi2and xi3 to 3
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Click OK to refine the mesh
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Click the lines radio button
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Click Render to display the mesh as a wireframe
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To delete or hide previously rendered objects View→Show→OpenMesh…
- Note that the element labels are preserved
Add biomechanics data
- Load the required biomechanics model from the database
- File→Library→Search…
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In the window near the top, enter ‘lagrangian’ and hit return.
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From the listed models select BM_TI_of_Lagrangian_strains_comp_sympy by right-clicking on it and selecting ‘Load’
- When the warning window display, select the third choice: ‘Retain current problem but overwrite the following objects: [dims, renderer, matEquations]’
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Biomechanics→Update→Initial Conditions with undeformed nodes
- This command updates the biomechanics Boundary Conditions form with the values already inputted in the mesh nodes form.
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After having opened the two forms, ensure that the Boundary Conditions forms has the same basis functions as the Nodes forms
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Biomechanics→Edit→Boundary Conditions…
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Click on the Deformed Coordinates 1 tab
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Click the Insert Nodes button three times
- Edit the `Node(s)’ and ‘Derivative’ parameters of the three nodes based on the table below
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Node in Nodes List
Node(s)
Derivative
Value
1
APEX
wrt s(2)
0
2
APEX
wrt s(2)s(3)
0
3
APEX
Value
0
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Click on the Deformed Coordinates 2 tab
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Click the Insert Nodes button six times
- Edit the `Node(s)’ and ‘Derivative’ parameters of the three nodes based on the table below
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Node in Nodes List
Node(s)
Derivative
Value
1
APEX
Value
0
2
APEX
wrt s(3)
0
3
APEX
wrt s(2)s(3)
0
4
BASE
Value
0
5
BASE
wrt s(3)
0
6
BASE
wrt s(2)s(3)
0
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Click on the Deformed Coordinates 3 tab
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Click the Insert Nodes button once
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Enter BASE_EPI$ in the `Node(s)’ text field
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For Derivative, choose Value. Under Value enter 0.
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Click on the External Pressure tab
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For Element Number, enter ENDO, and hit the Enter key on your keyboard
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For the Select Pressure Type drop-down list, choose Incremental
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In the Inner surface box on the bottom, enter 1.0 for the Specify Pressure Increment field
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- It is a good idea to now go back through the Boundary Conditions Form to double check this parameters you just set up
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Click the OK button
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Solve the biomechanics
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If the Dimensions Form appears, simply click Apply Marked Recommendations and then OK
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For Time Step, enter 0.05
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Set Number of steps to 40
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Click the Solve button, and wait for the solver to finish its job
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Calculate and render stress and strain
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Click the lines radio button
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This time select the deformed radio button
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Click Render to display deformed mesh lines
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Click on 3. element lines3 in the list on the left, and enter 1,0,0 in the R,G,B entry field to change the mesh lines from blue to red.
- Press [return] and close the window
- The mesh should look like the screenshot below
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Change the pop-up menu choice after At Xi to 1 and change Location to 0.5
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Select the deformed radio button
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under Variables select T Cauchy Stress Tensor
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Click OK to render the fiber (circumferential) stress, OR select [2,2] to render the radial stress.
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Solution files for various pressure loads
- The variables pertinent to this problem are listed below:
- The model makes use of the following constitutive equations:
- The stress and strain solutions for various endocardial pressure loads are provided in the table:
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Pressure Load
Solution File
1.0 kPa
1.4 kPa
1.6 kPa
2.0 kPa
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Pre-built model
This cont6 file contains all data and parameters for this problem: bm2.cont6 (original), bm2_refined.cont6 (refined)