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

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

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

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

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

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

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) 0 0 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.

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 3 0 𝑦 3 − 𝑦 1 0 𝑥 3 − 𝑥 2 0 𝑥 3 − 𝑥 2 𝑦 2 − 𝑦 3 𝑥 1 − 𝑥 3 0 𝑦 1 − 𝑦 2 0 𝑥 1 − 𝑥 3 0 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 2 1 𝑣 0 𝑣 1 0 0 0 1−𝑣 2

FEM: static equilibrium Static equilibrium <-> Minimize Potential Energy 𝑊= 1 2 ℎ 𝜎 𝑇 𝜀 𝑑𝐴− 𝑢 𝑇 𝑓= 1 2 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.

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,

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

Implementation: Meshing 4000 elements 2111 nodes 1mm

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.

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

Implementation: Solving 4222 unknowns, 4222 equations. ? : : ? 0 : : 0 0 −10 = 𝐾 ∗ 0 : : 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

Implementation: Result (Displacement plot) Maximum 𝛿𝑦=−57.124 𝜇𝑚

Implementation: Theory Prediction FEM simulation result: 𝛿=57.124 𝜇𝑚 Theory Prediction: Matches perfectly! 𝛿= 𝐹 𝐿 3 3𝐸𝐼 = 10𝑁∗ 100𝑚𝑚 3 3∗70000𝑁 𝑚𝑚 −2 ∗ 1 12 ∗10𝑚𝑚∗ 10𝑚𝑚 3 =57.143𝜇𝑚

Thanks for your listening! Questions? Thanks for your listening!