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2D Deformation and Creep Response of Articular Cartilage

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1 2D Deformation and Creep Response of Articular Cartilage
By: Mikhail Yakhnis & Robert Zhang

2 Motivation Articular cartilage
transfers load between bones enables smooth motion along joints Cartilage has limited capacity for self repair Applications: biomaterials, prosthetics, biomedical devices

3 Problem Description Consider cartilage in an unconfined compression under constant load F Analyze the 2D elastic deformation over time Articular Cartilage F Compression plate Frictionless Supports

4 Material Background Cartilage often modeled as a viscoelastic material
Viscous and elastic by superposition Elasticity and viscosity can be linear or nonlinear Established models: Kelvin-Voigt, Maxwell, Standard-Linear Solid

5 Mathematical Model for Cartilage
We chose the Kelvin-Voigt model to focus on the creep response The constitutive equation is 𝜎=𝐷𝜀+𝜂 𝑑𝜀 𝑑𝑡 Mechanical Analogue of Kelvin-Voigt Model

6 Assumptions for Model Conditions Properties F L c B3 B4 B2 B1
Constant force F normal to boundary B3 No gravity (body force) 2D, plane stress* Confined in y-direction along B1 and B3 Confined in x-direction along B4 Properties c = 0.1m; L = 0.125m Constant cross-sectional area A Isotropic elasticity* * 𝐷= 𝐸 1−𝜈 𝜈 0 𝜈 −𝜈 2 B3 B2 B1 B4 x y

7 Experimental Data 𝐻 𝐴 =7𝑒5 𝑃𝑎 (Aggregate Modulus) 𝐸=3.37𝑒5 𝑃𝑎 𝜈=0.396
Data Book on Mechanical Properties of Living Cells, Tissues, and Organs /. Tokyo ; New York : Springer, Print.

8 Derivation of Weak Form
By definition, stress 𝜎= 𝐹 𝐴 Strain can be rewritten as gradient of displacement u 𝜀=𝛻𝑢= 𝜀 𝑥𝑥 𝜀 𝑦𝑦 𝜀 𝑥𝑦 Our constitutive equation (in strong form) becomes 𝐹=𝐴 𝐷 𝛻𝑢+𝜂 𝑑 𝑑𝑡 𝛻𝑢

9 Derivation of Weak Form
(1) Take the gradient of the force equation (which equals zero) (2) Multiply by an arbitrary displacement 𝑤 𝐴 𝐷( 𝛻 2 𝑢)+𝜂 𝑑 𝛻 2 𝑢 𝑑𝑡 𝑤 𝑑Ω =0 (3) Integrate by parts to induce symmetry of 𝑢 and 𝑤 𝐹 𝑜 𝑤+𝜂𝐴 𝑑 𝛻𝑢 𝑑𝑡 𝑤 Γ −𝐴 𝐷𝛻𝑢 𝛻𝑤+𝜂 𝑑𝛻𝑢 𝑑𝑡 𝛻𝑤 𝑑Ω =0

10 Decoupling a Transient Problem
We can decouple the formulation and assume the time and spatial variations are separate 𝑢 𝑥,𝑡 ≈ 𝑢 𝑛 𝑒 𝑥,𝑡 = 𝑗=1 𝑛 𝑢 𝑗 𝑒 (𝑡) 𝑁 𝑗 𝑒 (𝑥) where 𝑢 is a function of time only and basis function N is function of space The weak differential equation rewritten in matrix form is 𝐹 𝑜 [𝑁] 𝑇 +𝜂𝐴 𝑢 𝑛−1 − 𝑢 𝑛−2 ∆𝑡 𝐵 𝑇 (𝑥𝑥,𝑦𝑦) 𝑁 𝑇 Γ =𝐴∫ 𝐷 𝐵 𝑇 𝐸 𝐵 𝑢 𝑛 +𝜂 𝐵 𝑇 𝐵 𝑢 𝑛−1 − 𝑢 𝑛−2 ∆𝑡 𝑑Ω Reddy, J. N.. "Time-Dependent Problems." An introduction to nonlinear finite element analysis. Oxford: Oxford University Press, Print.

11 Displacement Equation for Creep Response
𝐾 = 𝐵 𝑇 𝐷 𝐵 𝐶 =𝜂 𝐵 𝑇 𝐵 At each time step n 𝐹 𝑜 [𝑁] 𝑇 +𝜂𝐴 𝑢 𝑛−1 − 𝑢 𝑛−2 ∆𝑡 𝐵 𝑇 𝑁 𝑇 =𝐴∫ 𝐵 𝑇 𝐷 𝐵 𝑢 𝑛 +𝜂 𝐵 𝑇 𝐵 𝑢 𝑛−1 − 𝑢 𝑛−2 ∆𝑡 𝑑Ω The equation for 𝑢 𝑛 becomes 𝑢 𝑛 = [𝐶] −1 𝐾 𝑢 𝑛−1 − 𝐹 𝑜 𝐴 𝑁 𝑇 +𝜂 𝑢 𝑛−1 − 𝑢 𝑛−2 ∆𝑡 𝐵 𝑇 𝑁 𝑇 Γ

12 Modeling Creep in MATLAB
Changes in Preprocessor.m Provide initial displacement Define time step Adjust boundary conditions Changes in Assemble.m Assemble the damping matrix [C] Changes in NodalSoln.m Add initial condition, damping, time inputs Modify reaction force and displacement equations

13 Modeling Creep in MATLAB
Discussion: MATLAB result converges toward experimental data farther away from initial time 10% error at 6 seconds MATLAB model reaches equilibrium faster than experimental data

14 Modeling Creep in MATLAB

15 Modeling Creep in ANSYS
A variety of models are available Differences include suitability for primary and secondary creep Usually of the form 𝜀 𝑐𝑟 = 𝑓 1 𝜎 𝑓 2 𝜀 𝑓 3 𝑡 𝑓 4 (𝑇) Examples Strain Hardening: 𝜀 𝑐𝑟 = 𝐶 1 𝜎 𝐶 2 𝜀 𝐶 3 𝑒 − 𝐶 4 /𝑇 Time Hardening: 𝜀 𝑐𝑟 = 𝐶 1 𝜎 𝐶 2 𝑡 𝐶 3 𝑒 − 𝐶 4 /𝑇 ANSYS Advanced Nonlinear Materials: Lecture 3 – Rate Dependent Creep

16 Considerations for ANSYS Model
What experimental data is available to us? Can we fit the experimental data to the model? Can we use the built-in Mechanical APDL curve fitting procedure? Is there more emphasis on primary creep or secondary creep? Does the model satisfy our constitutive equation?

17 Parameters in the ANSYS Model
Experimental data provides aggregate modulus and Poisson’s ratio Young’s Modulus can be derived from 𝐻 𝐴 = 𝐸 1−𝜈 1+𝜈 1−2𝜈 The solution for time-dependent strain in the K-V model is 𝜀 𝑡 = 𝜎 𝑜 𝐸 (1− 𝑒 −𝜆𝑡 ) We can use the Modified Exponential Function in ANSYS 𝜀 𝑐𝑟 = 𝐶 1 𝜎 𝐶 2 𝑟 𝑒 −𝑟𝑡 𝑟= 𝐶 5 𝜎 𝐶 3 𝑒 − 𝐶 4 /𝑇 where 𝐶 2 =1, 𝐶 3 = 𝐶 4 =0; we can solve for 𝐶 1 and 𝐶 5 ANSYS Advanced Nonlinear Materials: Lecture 3 – Rate Dependent Creep

18 ANSYS Results – Creep Response
Short Term Response – 30 Seconds Long Term Response – 3000 Seconds

19 Animation of Deformation in ANSYS

20 Comparison of ANSYS and Experiment
Result: Theoretical Model-Based ANSYS data tends to overshoot experimental data Error is between 30% to 40% per data point Experimental-based model performs better Discussion: Results demonstrate the limitations of ANSYS models A combined primary-secondary model is ideal Long term response in ANSYS is not accurate Function models primary response Primary + Secondary Time Hardening 𝜖 𝑐𝑟 = 𝐶 1 𝜎 𝐶 2 𝑡 𝐶 𝑒 − 𝐶 4 𝑇 𝐶 𝐶 5 𝜎 𝐶 6 𝑡 𝑒 − 𝐶 7 𝑇

21 ANSYS Model: Mesh and Time Refinement
Mesh [Nodes] Time [s] Base Case 805 Between 0.1 and 900 Refinement 15747 Between 1e-4 and 1e-2 Time % Difference w.r.t. Base Case - Mesh Mesh and Time 1 -0.459 0.000 -0.470 2 -0.367 0.025 -0.202 4 -0.294 1.447 1.145 6 -0.267 2.008 1.733 8 -0.255 2.261 2.001 10 2.732 2.384 2.136

22 Sensitivity Analysis Recall the creep model: 𝜀 𝑐𝑟 = 𝐶 1 𝜎 𝐶 2 𝑟 𝑒 −𝑟𝑡
𝜀 𝑐𝑟 = 𝐶 1 𝜎 𝐶 2 𝑟 𝑒 −𝑟𝑡 𝑟= 𝐶 5 𝜎 𝐶 3 𝑒 − 𝐶 4 /𝑇 We varied each non-zero model constant by 50%* to perform a rudimentary sensitivity analysis: Time Base Case Case C1 Difference % Case C2 Case C5 1 3.60E-03 6.64E-03 84.59 7.70E-03 114.09 6.53E-03 81.49 2 4.36E-03 7.69E-03 76.20 9.53E-03 118.54 7.39E-03 69.36 4 5.23E-03 8.92E-03 70.51 1.18E-02 124.92 8.23E-03 57.35 6 5.68E-03 9.56E-03 68.20 1.29E-02 127.25 8.55E-03 50.43 8 5.92E-03 9.89E-03 67.13 1.35E-02 128.25 8.67E-03 46.46 10 6.04E-03 1.01E-02 66.61 1.38E-02 128.74 8.71E-03 44.23 *The simulation did not converge at C2 +50% so C2 +10% was used instead

23 2D Deformation and Creep Response of Articular Cartilage
By: DJ Mikey Mike & Big Rob Zhang Thank you for listening. Questions?

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