12 Variational Formulation of Bar Element IFEM Ch 12 Slide 1 - PDF document

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM 12 Variational Formulation of Bar Element IFEM Ch 12 – Slide 1

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Bar Member - Variational Derivation y �� �� Cross section �� �� z �� �� P Longitudinal axis x IFEM Ch 12 – Slide 2

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Bar Member (cont'd) cross section x axial rigidity EA ��� u(x) q(x) P L IFEM Ch 12 – Slide 3

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM The Bar Revisited - Notation Quantity Meaning Longitudinal bar axis ∗ x (.) ′ d (.)/ dx u ( x ) Axial displacement q ( x ) Distributed axial force, given per unit of bar length L Total bar length E Elastic modulus Cross section area; may vary with x A E A Axial rigidity e = du / dx = u ′ Infinitesimal axial strain σ = Ee = Eu ′ Axial stress p = A σ = E A e = E Au ′ 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 12 – Slide 4

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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 12 – Slide 5

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Potential Energy of the Bar Member (before discretization) Internal energy (= strain energy) � L � L � L ( E Au ′ ) u ′ dx = 1 u ′ E Au ′ dx U = 1 pe dx = 1 2 2 2 0 0 0 External work � L W = qu dx 0 Total potential energy � = U − W IFEM Ch 12 – Slide 6

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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 12 – Slide 7

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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: u = u ∗ δ� = δ U − δ W = 0 iff with respect to admissible variations u = u + δ u of the exact * * displacement solution u ( x ) IFEM Ch 12 – Slide 8

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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 12 – Slide 9

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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) 1 2 3 4 5 End node 1 assumed fixed u u 5 u(x) u 4 u 2 u 3 x u = 0 1 Axial displacement plotted normal to x for visualization convenience IFEM Ch 12 – Slide 10

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Element Shape Functions j i (e) x ℓ = L ( e ) 1 N (e) i 0 1 N (e) j 0 IFEM Ch 12 – Slide 11

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Total Potential Energy Principle and Decomposition over Elements u = u ∗ δ� = δ U − δ W = 0 (exact solution) iff � = � ( 1 ) + � ( 2 ) + . . . + � ( N e ) But δ� = δ� ( 1 ) + δ� ( 2 ) + . . . + δ� ( N e ) = 0 and From fundamental lemma of variational calculus, each component variation must vanish, giving δ� ( e ) = δ U ( e ) − δ W ( e ) = 0 IFEM Ch 12 – Slide 12

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Element Shape Functions (cont'd) Linear displacement interpolation: � u ( e ) � N ( e ) u ( e ) ( x ) = N ( e ) u ( e ) + N ( e ) j u ( e ) = [ N ( e ) = Nu ( e ) i ] j u ( e ) i i j i j in which x − x i x − x i = 1 − ℓ = 1 − ζ , = = ζ N ( e ) N ( e ) i j ℓ � x − x ζ = i dimensionless (natural) coordinate ℓ IFEM Ch 12 – Slide 13

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Displacement Variation Process Yields the Element Stiffness Equations { U ( e ) = 1 2 ( u ( e ) ) T K ( e ) u ( e ) (e) Π = U − W (e) (e) W ( e ) = ( u ( e ) ) T f ( e ) K ( e ) u ( e ) − f ( e ) � δΠ = 0 ☞ δ u ( e ) � T � (e) � = 0 (Appendix D) since δ u is arbitrary [...] = 0 ( e ) K ( e ) u ( e ) = f ( e ) the element stiffness equations IFEM Ch 12 – Slide 14

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM The Bar Element Stiffness � ℓ U ( e ) = 1 e E A e dx e = u' 2 0 � − 1 � 1 � u ( e ) � ℓ � ] 1 U ( e ) = 1 [ u ( e ) u ( e ) i ℓ [ − 1 1 ] dx u ( e ) i j 2 1 ℓ 0 j � u ( e ) � ℓ � � � E A 1 − 1 u ( e ) � T K ( e ) u ( e ) U ( e ) = 1 2 [ u ( e ) u ( e ) i = 1 � ] dx u ( e ) i j 2 ℓ 2 − 1 1 0 j � ℓ � ℓ � � E A 1 − 1 K ( e ) = E A B T B dx = dx ℓ 2 − 1 1 0 0 If EA is constant over element � � K ( e ) = E A 1 − 1 − 1 1 ℓ IFEM Ch 12 – Slide 15

Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM The Consistent Nodal Force Vector � 1 − ζ � ℓ � ℓ u ( e ) � T � ℓ � u ( e ) � T f ( e ) W ( e ) = ( e ) T N T � � ( u ) q qu dx = dx = q dx = ζ 0 0 0 � 1 − ζ � ℓ � f ( e ) = q dx ζ 0 � x − x ζ = i in which ℓ IFEM Ch 12 – Slide 16

Department of Engineering Mechanics PhD. TRUONG Tich Thien 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 12 – Slide 17