Introduction to Finite Element Method. Dr. Aamer Haque

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1 Introduction to Finite Element Method 4 th Order Beam Equation Dr. Aamer Haque ahaque7@iit.edu Illinois Institute of Technology July 1, 009

2 Outline Euler-Bernoulli Beams Assumptions Statics Kinematics 4 th Order Differential Equation Variational/Weak Form Galerkin Finite Element Method Formulation Basis/shape functions Element assembly Solution Post-processing Application

3 Euler-Bernoulli Beam Theory Plane sections remain plane after bending Small displacements Linear elastic material (i.e. Hooke's Law) Aial deformation neglected Shear deformation neglected

4 Statics of Beams M f M dm y V V dv Neglect nd order term F y =0 M z =0 V dv V f d=0 M dm M V d f d d =0 dv d = f dm d =V

5 Kinematics of Beams u y Small displacements = du d Constitutive model Geometric model M =EI = d d M =EI d d

6 4 th Order Differential Equation Statics dv d = f dm d =V d M d f =0 Kinematics = du d M =EI d d EI d u M =0 d Differential Equation (Valid for continuous regions) d d [ EI d d u ] = f

7 Common Boundary Conditions Fied End u=0, =0 Pinned End u=0, M =0 Roller End =0, V =0 Free End V =0, M =0 Static BC V, M Kinematic BC u,

8 Discontinuous Material f M - M + EI - EI + y V - F y =0 V + V - f d=0 V =0 V + M z =0 M dm M V d f d d =0 dm d =0

9 Discontinuous Load f - f + M - M + y V - d V + F y =0 M z =0 V + V - f + d f - d =0 M + M - V - d f + d d 4 f - d 3d 4 =0 V =0 dm d =0

10 Point Load f M - M + p d p y V - V + F y =0 M z =0 V + V - f d p=0 M + M - V - d f d d p d p=0 V = p dm d = p

11 Concentrated Applied Moment f M - M + M c y V - V + F y =0 M z =0 V + V - f d=0 M + M - V - d f d d M c=0 V =0 M = M c

12 Integrate with a test function Differential Equation [EI u' ' ]' ' = f Multiply by a test function and integrate over each subdomain Integrate by parts twice i [ EI u' ' ]' ' v d= i f v d i i i 1 EI u' ' v' ' d[ EI u ' ' v ' i ]i 1 [ EI u ' ' ' v ] 1 = i 1 f v d Observations [ EI u' ' v ' ] i 1 [ EI u ' ' v' ] [EI u' ' ' v ] i 1 [ EI u' ' ' v ] i 1 =M 1 v ' 1 M v ' M 1 v ' 1 1 = V 1 v 1 V v V 1 v 1

13 Sum over whole domain Sum up the subdomains to obtain the entire domain = i i =[0, L] i i [EI u ' ]' ' v d= i i f v d i i 1 EI u ' ' v ' ' d i [ EI u' ' v ' ] i 1 i [ EI u' ' ' v ] i i 1 = i 1 f v d Observe that we have the following epressions for boundary terms i [ EI u' ' v ' ] i i 1 =M 0v ' 0 i M v' M L v' L i [ EI u ' ' ' v ] i i 1 = V 0v0 i V v V Lv L Jumps are zero for all points that are not point loads/moments i [ EI u ' ' v ' ] i i 1 =M 0v ' 0 M c v ' c M L v' L i [ EI u' ' ' v ] i i 1 = V 0v0 p v p V Lv L

14 Variational/Weak Form 0 L EI u' ' v ' ' d= 0 L f v dm c v' c p v p M 0v ' 0 M Lv ' L V 0v0V L vl Trial Functions: Test Functions: u v The goal is to find a solution that satisfies this epression for all possible test functions We still need to specify the space of trial and test functions and also incorporate the boundary conditions

15 Function Spaces A classical solution to the differential equation requires u C 4 [0, L] A weak solution only requires that the solution and its first two derivatives be square integrable Continuity requirement u C 1 [0, L] v :[0, L] R H ={v : 0 L [v' ' v' v ]d } H 0 ={v H : v0=v ' 0=v L=v' L=0}

16 All Essential Boundary Conditions u0=u 0 0= 0 u L=u L L= L Find u H v H 0 0 L EI u' ' v ' ' d= 0 L f v dm c v' c p v p M 0v ' 0 M Lv ' L V 0v0V L vl

17 All Natural Boundary Conditions M 0=M 0 V 0=V 0 M L=M L V L=V L Find u H v H 0 L EI u' ' v ' ' d= 0 L f v dm c v' c p v p M 0v ' 0 M Lv ' L V 0v0V L vl

18 The Galerkin FEM Method The Galerkin method finds a solution from a finite dimensional subspace of the space of trial functions H h H Since the space is finite dimensional, every function in it can be written as a linear combination of basis functions N u h = j=1 j j We will formulate the Galerkin FEM for the natural BCs Other BCs will be enforced in the solution process Also, it is easier to formulate using shape functions instead of basis functions

19 Galerkin FEM Variational Problem M 0=M 0 V 0=V 0 M L=M L V L=V L Find u h H h v h H h 0 L EI uh ' ' v h ' ' d= 0 L f v h dm c v h ' c p v h p M 0v h ' 0 M L v h ' L V 0v h ' 0V Lv h ' L

20 Shape Functions The FEM solution has the following continuity requirement Cubic Hermite polynomials are the lowest order polynomials that satisfy the continuity requirement Consider the FEM approimation on a single element It can be epanded in terms of the shape functions 4 u h = j=1 u C 1 [0, L] j j, i =[ 1, ] Matri notation (used etensively in engineering!) i 1 u h =N N =[ ] =[u i 1 ] u i u B=[ i h ' ' =B 1 ' ' ' ' 3 ' ' 4 ' ' ] i

21 Cubic Hermite Shape Functions 1 3 = 1 h h= = = = =

22 Cubic Hermite Basis Functions 1 1

23 Local Element Variational Form Consider the variational problem over an element and ignore the boundary terms for now (they will be incorporated in the assembly process) i i 1 EI uh ' ' v h ' ' d= i 1 f v h d u h =N, u h ' ' =B, v h =N w, v h ' ' =B w i i 1 EI B B wd= i 1 f N w d i i 1 EI w T B T B d= i 1 f w T N T d w T i 1 EI B T B d=w T i 1 f N T d i i 1 EI B T B d= i 1 f N T d [ i 1 EI B T B ] d= i 1 f N T d

24 Local Element Stiffness Matri and Load Vector Components of Local Element Stiffness Matri and Load Vector K e = ij i 1 F e = i i 1 EI i ' ' j ' ' d f i d i, j=1,,3,4 K e = i 1 F e = i 1 EI B T B d f N T d Matri Notation N =[ ] B=[ 1 ' ' ' ' 3 ' ' 4 ' ' ]

25 Constant Stiffness and Load For constant material parameters and load, the integration can be performed eactly [ K e = EI 3 L e EI =const. f =w=const. 1 6 Le 1 6 Le 6 L e 4 L e 6 L e L e ] 1 6 L e 1 6 L e 6 L e L e 6 L e 4 L e =[ F e ] w Le w L e 1 w L e w L e 1

26 Cantilever Beam Elements u0=0=0 w y M L=V L=0

27 Global System [ K 1 33 K 11 K 1 34 K 1 K 13 K K 1 43 K 1 K 1 44 K K 3 K K 31 K 3 K 33 K K 41 K 4 K 43 K 44 0 u ][u0 ]=[ 1 1 u 0 0 F 1 3 F 1 F 4 1 F ] F 3 F 4

28 Results

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