# Inflation of an Artery with reference to Zero-Stress State

### Description

• Many older biomechanical studies utilized the no-load state as a reference. However, vascular tissue experiences residual strain in the no-load state. This is evident when a vessel or ring of tissue opens up when radially cut; this new state is called the zero-stress state. This tutorial explores the difference between the no-load state and zero-stress state by modeling and comparing different arteries.

### Create Mesh

• Select cylindrical polar in the Global Coordinates: pop-up menu

• Click OK to submit Coordinate Form

• Use the equation below to write dY_dMatl, the material coordinate transformation from cylindrical polar to rectangular cartesian.

• dY_dMatl = Matrix([[cos(X[1]), -sin(X[1]),0],[sin(X[1]),cos(X[1]),0],[0,0,1]])
• Click OK to submit the Material Coordinate model

• Choose Lagrange Basis Function→3D→Linear-Linear-Linear

• Click OK to submit Basis Form

• Click Insert Node in the left panel to create a total of 8 nodes

• In the Value fields under Coordinate 1Coordinate 2, and Coordinate 3 enter the following (R,Theta,Z) nodal coordinates:

• Node 1: (0.99, 0, 0)
• Node 2: (0.99, 0, 1)
• Node 3: (1.16, 0, 0)
• Node 4: (1.16, 0, 1)
• Node 5: (0.99, 59, 0)
• Node 6: (0.99, 59, 1)
• Node 7: (1.16, 59, 0)
• Node 8: (1.16, 59, 1)
• Replace all default node labels with the appropriate as shown below.

• Node 1: inner_proximal1…
• Node 2: inner_distal1…
• Node 5: inner_proximal2…
• Node 6: inner_distal2…
• Enable a Field Variable by selecting the Field Vector 1 tab and choosing Linear-Linear-Linear Lagrange 3*3*3 from the Select Basis Number menu below Field variable 1.

• Enter 6, 2, 5, 1, 8, 4, 7, 3 in the Global Node Numbers boxes.

• Click OK to submit Element Form

• Next, Render the nodes and elements.

• The current model should look similar to the one shown below.

### Refine the Mesh

To get a sufficiently converged result using linear elements, it is necessary to use multiple elements.

• Decrease the New Element per old element in xi1 and xi2 to 1. Increase the New Element per old element in xi3 to 10

• Click OK to refine the mesh

• Render the new set of nodes and elements.

A biomechanics problem requires material properties and boundary conditions before it can be solved. The boundary conditions will define how the biological testing is preformed.

##### Incorporate a Published Constitutive Model

Several arteries including the thoracic, abdominal, femoral, carotid, and pulmonary arteries will be modeled (Am J Physiol Heart Circ Physiol 282:H622-H629, 2002). An exponential constitutive model template from the continuity database will serve as the basis for the model. However, specific values, constants, and the strain energy function will be used from the literature.

• Select model 1097: BM_InflateTube_ConstitutiveModel_Basic

• In the toolbar above the models select Biomechanics→Constitutive Model

• Select Export and save the tabled material coordinates model (File→Save…)

• Import the constitutive model saved from the Continuity database

• Import… the constitutive model and update

• Edit the exponential term of the strain energy function.
• Q = <a1>*E[1,1]*E[1,1] + <a2>*E[2,2]*E[2,2] + <a4>*2.0*E[1,1]*E[2,2]

• Next output variables need to be added. Right-click on the variable stress_out in the left panel of the Edit Equations tab and select Insert variable here…

• Add any output variables that may be informative using the type temporary variable. Some examples are provided below:

• F_out = F#* – Deformation Gradient Tensor wrt material coordinates

• C_out = C#* – Right Cauchy-Green Deformation Tensor wrt material coordinates

• J_out = J#* – Determinant of Deformation Gradient Tensor

• E_out = E#* – Lagrangian Green’s Strain Tensor wrt material coordinates

• T = (F*stress*F.T)/J#* – Cauchy Stress Tensor

• N = stress*F.T#* – Nominal Stress Tensor

• Compile the model

• Click the Set parameters tab to change the Value for each coefficient. See literature for specific values.

• a1 – x

• a2 – y

• a4 – z

• C – 8.92 (4.46 – stress_scaling_coefficient)

• bulk_modulus – 350

• Compile the model

#### Define the Boundary Conditions

The boundary conditions for this model will first deform and close the open blood vessel and then apply a pressure on the inner wall causing inflation of the tube. We will need to make sure to account for rigid body motion.

• Click on the Deformed Coordinate 2 tab

• Click the Insert Nodes button

• Enter .*proximal1 under Nodes(s): and 0.0 under Value:

• Click the Insert Nodes button

• Enter .*distal1 under Nodes(s): and 0.0 under Value:

• Click the Insert Nodes button

• Enter .*proximal2 under Nodes(s): and θ under Value:

• Click the Insert Nodes button

• Enter .*distal2 under Nodes(s): and θ under Value:

• Click on the Deformed Coordinate 3 tab

• Click the Insert Nodes button

• Enter .*proximal1 under Nodes(s): and 0.0 under Value:

• Click the Insert Nodes button

• Enter .*distal1 under Nodes(s): and 0.0 under Value:

• Click the Insert Nodes button

• Enter .*proximal2 under Nodes(s): and 0.0 under Value:

• Click the Insert Nodes button

• Enter .*distal2 under Nodes(s): and 0.0 under Value:

• Click on the External Pressure tab

• Change Select Pressure Type to Incremental

• Specify the pressure increment to be x on the Inner Surface. See literature for specific values.

• The units of the pressure increment depend on your mesh scale and constitutive equation parameters.
• It is a good idea to now go back through the Boundary Conditions Form to double check the parameters you just set up
• Click OK

### Solve Biomechanics Problem

• View→Edit Dimensions…

• Click on Apply Marked Recommendations

• For Time Step, enter 0.1

• For Number of Steps, enter 10

• Click the Solve button, and wait for the solver to finish its job

• Note that the time counter (Initial time) updates to 1.0 after the solve. For ever 1.0 increase in the time counter, the boundary conditions are implemented once. For example, the pressure is incrementally increased to 0.4 during this simulation. If a displacement boundary condition of 0.1 had been given and the simulation run to a total time of 3.0 (Time Step*Number of Steps), a total displacement of 0.3 would be applied.

### Calculate and Render Strains

• Click the lines radio button

• This time select the deformed radio button

• Click Render to display deformed mesh lines

• The mesh should look like the screenshot below

• In the Element List, enter 1 to render the inner surface of the element

• Click the surfaces radio button

• Click the deformed radio button

• Click Render to display inner mesh surfaces (at Xi(3)=0.)

• Change the pop-up menu choice after At Xi to 2 and change Location to 0.5

• Select the deformed radio button

• under Variables select T – Cauchy Stress Tensor

• Click OK to render the fiber (circumferential) strain

• The mesh should now look similar to the screenshot below

• Unselect All Variables and reselect X and T

• In the Locations tab, select Number of Points and change xi1xi2 and xi3 to 1. This will export the solutions at the center of each element (xi1xi2 and `xi3′ = 0.5) and provide the radial stress distribution. Since this model uses only a small number of linear finite elements, the center of each element has the most accurate solution

• Click OK to display a listing of the selected Output Variables in the Table Manager

• This table can be saved (File→Save…) and graphed using external software
• Biomechanics→List→Nodal Solutions…
• The nodal solutions table contains the final node locations and the residual values
• Since Continuity minimizes these residuals when solving the equilibrium equations, they will be very small. However, if a displacement boundary condition was enforced, the residual will be non-zero. The residual will be the force experienced by the body to achieve that displacement
• In this model, the residuals in the Z direction are non-zero. This is a result of the zero displacement boundary conditions in the Z direction
• The opposite of the residuals could be applied as a force boundary condition to achieve the same displacement