Lecture 28 Introduction to finite elements methods

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1 Fall, 2017 ME 323 Mechanics of Materials Lecture 28 Introduction to finite elements methods Reading assignment: News: Instructor: Prof. Marcial Gonzalez Last modified: 10/27/17 10:56:52 AM

2 Some announcements Undergraduate Research Scholarship 2

3 Some announcements 3

4 Finite element methods - Finite element analysis (FEA) Every structure studied in ME323 and much more 4

- 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

5 - 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 5

6 - Example 54 (review): Number of nodes: 4 Number of elements: 3 Boundary conditions: Stiffness of each element: 6

7 - 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 7

8 - 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 8

9 - Example 54, solved in 5 steps + Step #5: Recover the reaction at the supports 9

10 - 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)');

11 - Example 56 (5 nodes) clear % set number of elements N=4; %define elemental properties EA=[(1+1)/2;(1+1)/2;(1+1.5)/2;(1.5+2)/2]; L =[0.5 ;0.5 ;0.5 ;0.5 ]; %set up forcing vector F =[1;0;0;0;0]; %define boundary conditions BC=[0;0;0;0;1]; Nodal displacement Nodal force Stiffness matrix Number of nodes: 5 Number of elements: 4 11

12 - Example 56: (6 nodes) clear % set number of elements N=5; %define elemental properties EA=[1 ;1 ;(1+4/3)/2;(4/3+5/3)/2;(5/3+2)/2]; L =[0.5 ;0.5 ;1/3 ;1/3 ;1/3 ]; %set up forcing vector F =[1;0;0;0;0;0]; %define boundary conditions BC=[0;0;0;0;0;1]; Nodal displacement Nodal force Stiffness matrix Number of nodes: 6 Number of elements: 5 12

13 - Example 56: (3 to N nodes) Num Elements Displ. Node Relative error in the solution Slope of 2 (quadratic convergence!) Number of elements 13

14 - Example 57: (5 nodes) clear % set number of elements N=4; %define elemental properties (A=pi*D^2/4) EA=pi/4*[4.00^2;5.20^2;5.00^2;(5.00^2+3.00^2)/2]; L =[5.00;1.00;1.00;10.00]; %set up forcing vector F =[1; 0; 0; 0; -3]; %define boundary conditions BC=[0; 1; 0; 0; 0]; Nodal displacement Nodal force Stiffness matrix

15 - Example 57: (6 nodes) clear % set number of elements N=5; %define elemental properties (A=pi*D^2/4) EA=pi/4*[4^2;5.2^2;5^2;(5^2+4^2)/2;(4^2+3^2)/2]; L =[5.00;1.00;1.00;5.00;5.00]; %set up forcing vector F =[1; 0; 0; 0; 0; -3]; %define boundary conditions BC=[0; 1; 0; 0; 0; 0]; Nodal displacement Nodal force Stiffness matrix

16 - Example 58: Answer: 16

17 Any questions? 17

Lecture 27 Introduction to finite elements methods

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 Finite element