1. Implementation of 2D stress-strain Finite Element Modeling on MATLAB
Xingzhou Tu

2. What is stress? Stress is a tensor
𝑇 𝑥 𝑇 𝑦 𝑇 𝑧 = 𝜎 𝑥𝑥 𝜎 𝑥𝑦 𝜎 𝑥𝑧 𝜎 𝑦𝑥 𝜎 𝑦𝑦 𝜎 𝑦𝑧 𝜎 𝑧𝑥 𝜎 𝑧𝑦 𝜎 𝑧𝑧 𝑛 𝑥 𝑛 𝑦 𝑛 𝑧

3. What is strain? . . Strain is also a tensor𝑟 .
𝑟 + 𝑢 .
Deformation
𝜀 𝑖𝑗 = 1 2 ( 𝜕 𝑢 𝑖 𝜕 𝑟 𝑗 + 𝜕 𝑢 𝑗 𝜕 𝑟 𝑖 )

4. 2D case Plane stress – no stress in z-direction
Things need to consider:
𝜎 𝑥𝑥 𝜎 𝑦𝑦 𝜎 𝑥𝑦
𝜀 𝑥𝑥 𝜀 𝑦𝑦 𝜀 𝑥𝑦

5. Stress-Strain Relation in 2D case
E: Young’s Modulus
V: Poisson Ratio 𝜀
𝜎 𝑥𝑥 𝜎 𝑦𝑦 𝜎 𝑥𝑦 = 𝐸 1− 𝑣 𝑣 0 𝑣 −𝑣 𝜀 𝑥𝑥 𝜀 𝑦𝑦 2𝜀 𝑥𝑦 𝜎

6. FEM: Turner Triangle Three nodes
𝑢 = 𝑢 1,𝑥 𝑢 1,𝑦 𝑢 2,𝑥 𝑢 2,𝑦 𝑢 3,𝑥 𝑢 3,𝑦 𝑓 = 𝑓 1,𝑥 𝑓 1,𝑦 𝑓 2,𝑥 𝑓 2,𝑦 𝑓 3,𝑥 𝑓 3,𝑦

7. FEM: Linear Interpolation
Use linear interpolation to evaluate displacement at other points
𝑢 𝑥 (𝑥,𝑦) 𝑢 𝑦 (𝑥,𝑦) = 𝑁 1 (𝑥,𝑦) 0 𝑁 2 (𝑥,𝑦) 0 𝑁 1 (𝑥,𝑦) 𝑁 2 (𝑥,𝑦) 0 𝑁 2 (𝑥,𝑦) 0 𝑁 2 (𝑥,𝑦) 𝑢
𝑁 1 𝑥,𝑦 = 2 𝑥 2 𝑦 3 − 𝑥 3 𝑦 2 +𝑥 (𝑦 2 − 𝑦 3 )+𝑦( 𝑥 3 − 𝑥 2 ) 2𝐴
𝑁 2 𝑥,𝑦 = 2 𝑥 3 𝑦 1 − 𝑥 1 𝑦 3 +𝑥 (𝑦 3 − 𝑦 1 )+𝑦( 𝑥 1 − 𝑥 3 ) 2𝐴
𝑁 1 𝑥,𝑦 = 2 𝑥 1 𝑦 2 − 𝑥 2 𝑦 1 +𝑥 (𝑦 1 − 𝑦 2 )+𝑦( 𝑥 2 − 𝑥 1 ) 2𝐴
A is the area of the triangle.

8. FEM: stress and strain vectors
Differentiate the displacement to get the strain vector
Use the stress-strain relation to get the stress vector
𝜀 =𝐵 𝑢 ,
𝐵= 1 2𝐴 𝑦 2 − 𝑦 𝑦 3 − 𝑦 𝑥 3 − 𝑥 𝑥 3 − 𝑥 2 𝑦 2 − 𝑦 3 𝑥 1 − 𝑥 𝑦 1 − 𝑦 𝑥 1 − 𝑥 𝑥 2 − 𝑥 1 𝑦 3 − 𝑦 1 𝑥 2 − 𝑥 1 𝑦 1 − 𝑦 2
𝜎 =𝐸 𝜀 =𝐸𝐵 𝑢 , 𝐸= 𝐸 1− 𝑣 𝑣 0 𝑣 −𝑣 2

9. FEM: static equilibrium
Static equilibrium <-> Minimize Potential Energy
𝑊= ℎ 𝜎 𝑇 𝜀 𝑑𝐴− 𝑢 𝑇 𝑓= ℎ 𝑢 𝑇 𝐵 𝑇 𝐸𝐵𝑢 𝑑𝐴− 𝑢 𝑇 𝑓= 1 2 𝑢 𝑇 (𝐴ℎ 𝐵 𝑇 𝐸𝐵)𝑢− 𝑢 𝑇 𝑓
h is the thickness of the 2D domain in the z-direction.

10. FEM: Stiffness Matrix 𝑓 =𝐴ℎ 𝐵 𝑇 𝐸𝐵 𝑢 =𝐾 𝑢 , 𝐾=𝐴ℎ 𝐵 𝑇 𝐸𝐵,
Minimize Potential Energy -> 𝛿𝑊=0
K is the stiffness matrix of the element
𝑓 =𝐴ℎ 𝐵 𝑇 𝐸𝐵 𝑢 =𝐾 𝑢 , 𝐾=𝐴ℎ 𝐵 𝑇 𝐸𝐵,

11. Implementation: Problem
Cantilever beam
Beam Dimension: 100mm*10mm*10mm
10N load at the end of the beam
Aluminum:
Young’s modulus = 70000N/mm^2
Poisson ratio = 0.33

12. Implementation: Meshing
4000 elements
2111 nodes
1mm

13. Implementation: System Stiffness Matrix
2111 nodes:
Each node has two degrees of freedom
4222*4222 system stiffness matrix
Build a 4222-by-4222 zero matrix
Traversing each element:
Calculate the stiffness matrix of this element
Adding the matrix into the corresponding row and column of the system matrix.

14. Implementation: Boundary Condition
Two kinds of Boundary Condition:
Fixed Support:
zero displacement at the fixed node
External load:
10N load at the end of the beam
0N load inside the beam
Unknown load (reaction force) at the fixed node

15. Implementation: Solving
4222 unknowns, 4222 equations.
? : : ? : : − = 𝐾 ∗ : : 0 ? : : ?
Unknown reaction force at fixed nodes
Zero load inside the beam
10N load on the –y direction at the end of the beam
Unknown displacement at other nodes
Zero displacement at fixed nodes

16. Implementation: Result (Displacement plot)
Maximum 𝛿𝑦=− 𝜇𝑚

17. Implementation: Theory Prediction
FEM simulation result: 𝛿= 𝜇𝑚
Theory Prediction:
Matches perfectly!
𝛿= 𝐹 𝐿 3 3𝐸𝐼 = 10𝑁∗ 100𝑚𝑚 3 3∗70000𝑁 𝑚𝑚 −2 ∗ 1 12 ∗10𝑚𝑚∗ 10𝑚𝑚 3 =57.143𝜇𝑚

18. Thanks for your listening!
Questions?
Thanks for your listening!