Lecture 27 Introduction to finite elements methods
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1 Fall, 2017 ME 323 Mechanics of Materials Lecture 27 Introduction to finite elements methods Reading assignment: News: Instructor: Prof. Marcial Gonzalez Last modified: 10/24/17 7:02:00 PM
2 Finite element methods - Finite element analysis (FEA) Every structure studied in ME323 and much more 2
3 Finite element methods - Finite element analysis (FEA) - It is an energy method that can handle any geometry and any three-dimensional states of stress. - It is not limited to elastic materials and small deformations. - FEA is a skill of high-demand in the job market and it is always considered a plus! 3
4 - Review of axial deformations and rods + Elongation of the rod + Equivalent stiffness Good approximation when 4
5 - Review of axial deformations and rods + Elastic strain energy + Stiffness matrix 5
6 - Review of axial deformations and rods + Three nodes + Three degrees of freedom + Two elements + Each element has two nodes and two degrees of freedom 6
7 - Review of axial deformations and rods + Three nodes + Three degrees of freedom + Two elements + Each element has two nodes and two degrees of freedom 7
8 - Review of axial deformations and rods + (N+1) nodes + (N) elements Strain energy 8
9 - Review of axial deformations and rods + (N+1) nodes + (N) elements Global stiffness matrix (symmetric matrix) (combination of elemental stiffness matrices) 9
10 - We obtain the equilibrium solution using an energy principle Principle of minimum potential energy For a given set of admissible displacement fields for a conservative system, an equilibrium state of the system will correspond to a state for which the total potential energy is stationary. + An admissible displacement field for a rod is one that satisfies all of the displacement boundary conditions of the problem. + The total potential energy of the system is equal to the sum of the potential of the applied external forces and the strain energy in the rod. + Stationarity of the potential energy correspond to its minimization with respect to the displacement field. for each node in the mesh 10
11 - We obtain the equilibrium solution using an energy principle Principle of minimum potential energy For a given set of admissible displacement fields for a conservative system, an equilibrium state of the system will correspond to a state for which the total potential energy is stationary. + The total potential energy of the system is equal to the sum of the potential of the applied external forces and the strain energy in the rod. 11
12 - We obtain the equilibrium solution using an energy principle Principle of minimum potential energy For a given set of admissible displacement fields for a conservative system, an equilibrium state of the system will correspond to a state for which the total potential energy is stationary. + Stationarity of the potential energy correspond to its minimization with respect to the displacement field. for each node in the mesh A linear system of N+1 equations,, 12
13 - We obtain the equilibrium solution using an energy principle Principle of minimum potential energy For a given set of admissible displacement fields for a conservative system, an equilibrium state of the system will correspond to a state for which the total potential energy is stationary. + An admissible displacement field for a rod is one that satisfies all of the displacement boundary conditions of the problem. Some displacements in are going to be zero. We will enforce these conditions after the minimization of the potential energy. (similar to Castigliano s second therorem) 13
14 - We obtain the equilibrium solution using an energy principle Principle of minimum potential energy For a given set of admissible displacement fields for a conservative system, an equilibrium state of the system will correspond to a state for which the total potential energy is stationary. + Stationarity of the potential energy correspond to its minimization with respect to the displacement field. for each node in the mesh In general: recall for each node in the mesh, is equivalent to: (a linear system of N+1 equations) 14
15 - Example 54: Number of nodes: 4 Number of elements: 3 Boundary conditions: Stiffness of each element: 15
16 - Example 54, solved in 5 steps + Step #1: Identify the degrees of freedom Number of nodes: 4 Number of elements: 3 + Step #2: Build the global stiffness matrix 16
17 - Example 54, solved in 5 steps + Step #3: Enforce boundary conditions Number of nodes: 4 Number of elements: 3 + Step #4: Solve the reduced system of linear equations 17
18 - Example 54, solved in 5 steps + Step #5: Recover the reaction at the supports 18
19 - Example 55, using MATLAB: clear % set number of elements N=3; %define elemental properties EA=[1/4;1;9/4]; L=[1;1;1]; %set up forcing vector F=[0;2;1;0]; %define boundary conditions BC=[1;0;0;1]; %set up global stiffness matrix k=ea./l; K=zeros(N+1,N+1); for ii=1:n K(ii,ii)=K(ii,ii)+k(ii); K(ii+1,ii)=K(ii+1,ii)-k(ii); K(ii,ii+1)=K(ii,ii+1)-k(ii); K(ii+1,ii+1)=K(ii+1,ii+1)+k(ii); end %enforce BC's on [K] and {F} K_reduced = K; F_reduced = F; for jj=n+1:-1:1 if BC(jj)==1 K_reduced(jj,:)=[]; K_reduced(:,jj)=[]; F_reduced(jj)=[]; end end %solve reduced system of equations u_reduced=inv(k_reduced)*f_reduced; %determine reaction at supports r=1; for jj=1:n+1 if BC(jj)==1 nodal_u(jj,1) = 0; else nodal_u(jj,1) = u_reduced(r); r=r+1; end end disp('nodal displacement'); disp(nodal_u'); disp('nodal force'); disp((k*nodal_u)');
20 - Example 55, using MATLAB: clear % set number of elements N=3; %define elemental properties EA=[1/4;1;9/4]; L=[1;1;1]; %set up forcing vector F=[0;2;1;0]; %define boundary conditions BC=[1;0;0;1]; Output: Nodal displacement Nodal force
21 - Can we use the same strategy for any other geometry? Yes! This is called a discretization of the object into elements. 21
22 Finite element methods Three-dimensional elements - Can we use the same strategy for any other geometry? Yes! This is called a discretization of the object into elements. + 8 nodes per element + 3 degrees of freedom per node + 24 degrees of freedom per element 22
23 23
24 Any questions? 24
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