Finite Element Method Brian Hammond Ivan Lopez Ingrid Sarvis

Fundamental Concept of FEM A continuous field of a certain domain having infinite degrees of freedom is approximated by a set of piecewise continuous models with a number of finite regions called elements. The number of unknowns defined as nodes are determined using a given relationship i.e.{F}=[K]*{d}.

Fundamental Concept of FEM Red line-Continuous field over the entire domain. Blue line-Finite number of linear approximations with the finite number of elements

General Steps 1)Discretize the domain a) Divide domain into finite elements using appropriate element types (1-D, 2-D, 3-D, or Axisymmetric) 2) Select a Displacement Function a) Define a function within each element using the nodal values 3)Define the Strain/Displacement and Stress/strain Relationships 4) Derive the Element Stiffness Matrix and Equations a)Derive the equations within each element

General Steps 5)Assemble the Element Equations to Obtain the Global or Total Equations and Introduce Boundary Conditions a)Add element equations by method of superposition to obtain global equation 6)Solve for the Unknown Degrees of Freedom (i.e primary unknowns) 7)Solve for the Element Strains and Stresses 8)Interpret the Results

Applications Stress Analysis –Truss and frame analysis –Stress concentration Buckling Vibration analysis Heat transfer Fluid flow

Advantages of FEM Model irregularly shaped bodies Compute General load conditions Model bodies composed of different materials Solve unlimited numbers and kinds of boundary conditions Able to use different element sizes in places where loads or stresses are concentrated Handle non-linear behavior using linear approximations Reduce System Cost

FEM Packages Large Commercial Programs –Designed to solve many types of problems –Can be upgraded fairly easily –Initial Cost is high –Less efficient Special-purpose programs –Relatively short, low development costs –Additions can be made quickly –Efficient in solving their specific types of problems –Can’t solve different types of problems

FEM Packages Algor ANSYS COSMOS/M STARDYNE IMAGES-3D MSC/NASTRAN SAP90 GT-STRUDL

Note on Stiffness matrix For a 1-D bar, the stiffness matrix is derived from the stress/strain relationship in Hooke’s law and the definitions of stress and strain. σ x = Eε x. ; σ x = P/A ; ε x = du/dx = (d 2x – d 1x )/L By substitution: -f 1x = EA (d 2x – d 1x ) L f 1x = EA (d 1x – d 2x ) L Similarly for f 2x : f 2x = EA (d 2x – d 1x ) L Combining into matrix form, the stiffness matrix is defined as [k] = EA L f2f2 f1f1 d 1x d 2x L A E

Note on the displacement function For a given set of nodes there exists a function that approximates the displacement at any position along the bar. This function, called the displacement function, is derived from Pascal’s Triangle. A new constant is introduced into the function for every node in the discretized domain. u(x) = a 1 + a 2 x + a 3 x 2 + … For 1-D

Note on the displacement function If a 1-D bar is broken into 2 elements, the displacement function would be u(x) = a 1 + a 2 x + a 3 x 2. Putting it into matrix notation: u(x) = [1 x x 2 ] a1a2a3a1a2a3 By knowing the distances to the nodes and the displacements at those nodes, the equation becomes: u1u2u3u1u2u3 = x 2 x x 3 x 3 2 a1a2a3a1a2a3, where x 1 = 0, x 2 and x 3 are the distances to the nodes and u 1, u 2, and u 3 are the displacements. The coefficients are found by solving the equation.

Example A A L L E 2E P Determine displacements of materials a and b if the load P is applied to the end of the bar given the above information. ab

Example con’t. 1) Discretize the domain with appropriate elements. Element a 1 2 Element b 23 u1u1 u2u2 u2u2 u3u u1u1 u2u2 u3u3 f 3 = Pf1f1 f1f1 f 21 f 22

Example con’t. 2) Select a displacement function There will be new term for each element, and the terms are 1 23 u1u1 u2u2 u3u3 u(x) = a 1 + a 2 x + a 3 x 2 derived from Pascal’s triangle.

Example con’t. 3) Define stress/displacement and stress/strain relationships σ x = Eε x ε x = du/dx 4) Derive the element stiffness matrix and element equations {F} = [k]{d} [k] = stiffness matrix [k] = EA L {F} = Force {d} = displacement f 1 f 21 = L EA u1u2u1u2 f 22 f 3 = L EA u2u3u2u3, a b

Example con’t. 5)Construct Global equation and introduce boundary conditions and known variables. f 1 f 21 +f 22 f 3 = L EAu1u2u3u1u2u3 Global Equation B.C.: (x =0) u 1 = 0 Known variables: f 3 = P and f 2 = f 21 +f 22 = 0

Example con’t. 6) Solve for unknowns. f10Pf10P = L EA0u2u30u2u3 f 1 = -EAu 2 L 0 = EA(3u 2 -2u 3 ) L P = EA(2u 3 -2u 2 ) L u 2 = PL EA u 3 = 3PL 2EA f 1 = -P

Example con’t. 7) Solve for the element strains and stresses. ε a = P = u 2 EA L σ a = Eε a = P A ε b = 3P = u 3 2EA L σ a = 2E ε b = P 3 A 8) Interpret the Results After solving for the displacements, the coefficients of the displacement function can be determined.

Bibliography Logan, Daryl L. A First Course in the Finite Element Method Using Algor. Brooks/Cole, Pacific Grove, CA ml4536/ ml4536/ 525/FEM_Lecture_Notes_Liu_UC.pdf