11 Variational Formulation of Bar Element IFEM Ch 11 Slide 1 - PDF document

Introduction to FEM 11 Variational Formulation of Bar Element IFEM Ch 11 – Slide 1

Introduction to FEM Bar Member - Variational Derivation y �� �� Cross section �� �� z �� �� P Longitudinal axis x IFEM Ch 11 – Slide 2

Introduction to FEM Bar Member (cont'd) cross section x axial rigidity EA ��� u(x) q(x) P L IFEM Ch 11 – Slide 3

Introduction to FEM The Bar Revisited - Notation Quantity Meaning x Longitudinal bar axis * (.)' d(.)/dx u(x) Axial displacement q(x) Distributed axial force, given per unit of bar length L Total length of bar member E Elastic modulus A Cross section area, may vary with x EA Axial rigidity e = du/dx = u' Infinitesimal axial strain σ = E e = E u' Axial stress p = A σ = EA e = EA u' Internal axial force P Prescribed end load _ * x is used in this Chapter instead of x (as in Chapters 2-3) to simplify the notation IFEM Ch 11 – Slide 4

Introduction to FEM Tonti Diagram of Governing Equations Displacement Prescribed Distributed Axial BCs end axial load displacements displacements q(x) u(x) Kinematic e=u' p'+q=0 Equilibrium Axial Axial p = EA e Force BCs Prescribed force strains end loads Constitutive p(x) e(x) given (problem data) unknown IFEM Ch 11 – Slide 5

Introduction to FEM Potential Energy of the Bar Member (before discretization) Internal energy (= strain energy) � L � L � L U = pe dx = ( E Au' ) u' dx = 1 1 1 u'E Au' dx 2 2 2 0 0 0 External work � L W = qu dx 0 Total potential energy Π = U − W IFEM Ch 11 – Slide 6

Introduction to FEM Concept of Kinematically Admissible Variation u(x)+ δ u(x) δ u(x) u u(L) u(x) x u (0) = 0 L δ u(x) is kinematically admissible if u(x) and u(x) + δ u(x) u ( x ) ∈ C 0 in x ∈ [0 , L ]. (i) are continuous over bar length, i.e. (ii) satisfy exactly displacement BC ; in the figure, u (0) = 0 IFEM Ch 11 – Slide 7

Introduction to FEM The Minimum Potential Energy (MPE) Principle The MPE principle states that the actual displacement solution u ( x ) that satisfies the governing equations is that which renders * the TPE functional Π[ u ] stationary: δΠ = δ U − δ W = 0 iff u = u* with respect to admissible variations u = u + δ u of the exact * * displacement solution u ( x ) IFEM Ch 11 – Slide 8

Introduction to FEM FEM Discretization of Bar Member u 1 , f 1 u 2 , f 2 u 3 , f 3 u 4 , f 4 u 5 , f 5 (1) (2) (3) (4) 1 2 3 4 5 IFEM Ch 11 – Slide 9

Introduction to FEM FEM Displacement Trial Function u 1 , f 1 u 2 , f 2 u 4 , f 4 u 5 , f 5 u 3 , f 3 (1) (2) (3) (4) 4 1 2 3 5 End node 1 assumed fixed u u 5 u(x) u 4 u 3 u 2 x u = 0 1 Axial displacement plotted normal to x for visualization convenience IFEM Ch 11 – Slide 10

Introduction to FEM Element Shape Functions (e) 2 1 L e ℓ = - γ x = x - x 1 - 1− x/ ℓ 1 N e i 0 - x/ ℓ 1 N e j 0 IFEM Ch 11 – Slide 11

Introduction to FEM Total Potential Energy Principle and Decomposition over Elements δΠ = δ U − δ W = 0 iff u = u* (exact solution) (2) ( N ) Π = Π (1) e + Π + ... + Π But (1) ( N ) (2) e δΠ = δΠ + δΠ + ... + δΠ = 0 and From fundamental lemma of variational calculus, each element variation must vanish , giving δΠ = δ U − δ W = 0 e e e IFEM Ch 11 – Slide 12

Introduction to FEM Element Shape Functions (cont'd) Linear displacement interpolation: e u 1 e e e e e e e e u (x) = N u + N u = [ N N ] = N u e 2 1 1 2 2 1 u 2 in which x − x 1 x − x 1 e N e = 1 − ℓ = 1 − ζ , = = ζ N 1 2 ℓ � x − x ζ = 1 dimensionless (natural) coordinate ℓ IFEM Ch 11 – Slide 13

Introduction to FEM Displacement Variation Process Yields the Element Stiffness Equations { 2 ( u ) T K e e e 1 e U = u e Π = U − W e e = ( u ) T f e e e W K u − f � = 0 δΠ = 0 ☞ � T � e δ e e e e � u (Appendix D) since δ u is arbitrary [...] = 0 e e e e K u = f the element stiffness equations IFEM Ch 11 – Slide 14

Introduction to FEM The Bar Element Stiffness � ℓ e = 1 U e E A e dx e = u' 2 0 � − 1 � 1 � u 1 � ℓ e � 1 e = 1 [ u 1 u 2 ] U ℓ [ − 1 1 ] dx e 2 1 ℓ u 2 0 � u 1 � ℓ e � � � E A 1 − 1 u � T K e e U = 1 e = 1 � e u 2 ] e e 2 [ u 1 dx u e 2 ℓ 2 − 1 1 u 2 0 � ℓ � ℓ � � E A 1 − 1 e E A B T B dx = K = dx ℓ 2 − 1 1 0 0 If EA is constant over element � � K = E A 1 − 1 e − 1 1 ℓ IFEM Ch 11 – Slide 15

Introduction to FEM The Consistent Nodal Force Vector � 1 − ζ � ℓ � ℓ � T � ℓ � � T f W e e e T N T e � e � ( u ) u u = q u dx = q dx = q dx = ζ 0 0 0 Since u is arbitrary e � 1 − ζ � ℓ � e f = q dx ζ 0 � x − x ζ = 1 in which ℓ IFEM Ch 11 – Slide 16

Introduction to FEM Bar Consistent Force Vector (cont'd) If q is constant along element � 1 − ζ 1/2 � ℓ � � f e = � = q ℓ q dx 1/2 ζ 0 the same result as with EbE load lumping (i.e., assigning one half of the total load to each node) IFEM Ch 11 – Slide 17