Introduction to FEM The Direct Stiffness Method Part I IFEM Ch 2 – Slide 1
Introduction to FEM The Direct Stiffness Method (DSM) Importance: DSM is used by all major commercial FEM codes A democratic method, works the same no matter what the element: Bar (truss member) element, 2 nodes, 4 DOFs Tricubic brick element, 64 nodes, 192 DOFs Obvious decision: use the truss to teach the DSM IFEM Ch 2 – Slide 2
Introduction to FEM Model Based Simulation (a simplification of diagrams of Chapter 1) IDEALIZATION DISCRETIZATION SOLUTION FEM Discrete Discrete Physical Mathematical solution model system model Solution error Discretization + solution error Modeling + discretization + solution error VERIFICATION & VALIDATION IFEM Ch 2 – Slide 3
Introduction to FEM The Direct Stiffness Method (DSM) Steps Starting with: Idealization Disconnection Breakdown Localization (Chapter 2) Member (Element) Formation Globalization Assembly & Merge Solution Application of BCs (Chapter 3) Solution Recovery of Derived Quantities IFEM Ch 2 – Slide 4
Introduction to FEM A Physical Plane Truss member support joint Too complicated to do by hand. We will use a simpler one to illustrate DSM steps IFEM Ch 2 – Slide 5
Introduction to FEM The Example Truss: Physical Model (Loads not shown) IFEM Ch 2 – Slide 6
Introduction to FEM The Example Truss - FEM Model: Nodes, Elements and DOFs f , u y 3 y 3 f , u x 3 x 3 3 √ (3) L = 10 2 (2) √ = 10 L (3) (3) E A = 200 2 (2) (2) E A = 50 (3) (2) y f , u f , u x (1) x 1 x 1 x 2 x 2 2 1 (1) L = 10 f , u f , u y 1 y 1 y 2 y 2 (1) (1) E A = 100 IFEM Ch 2 – Slide 7
Introduction to FEM The Example Truss - FEM Model BCs: Applied Loads and Supports Saved for Last f y 3 = 1 3 f x 3 = 2 y x 1 2 �� �� �� �� �� �� IFEM Ch 2 – Slide 8
Introduction to FEM Master (Global) Stiffness Equations f x 1 u x 1 f y 1 u y 1 f x 2 u x 2 f = u = f y 2 u y 2 f x 3 u x 3 f y 3 u y 3 Linear structure: f x 1 K x 1 x 1 K x 1 y 1 K x 1 x 2 K x 1 y 2 K x 1 x 3 K x 1 y 3 u x 1 f y 1 K y 1 x 1 K y 1 y 1 K y 1 x 2 K y 1 y 2 K y 1 x 3 K y 1 y 3 u y 1 f x 2 K x 2 x 1 K x 2 y 1 K x 2 x 2 K x 2 y 2 K x 2 x 3 K x 2 y 3 u x 2 = f y 2 K y 2 x 1 K y 2 y 1 K y 2 x 2 K y 2 y 2 K y 2 x 3 K y 2 y 3 u y 2 f x 3 K x 3 x 1 K x 3 y 1 K x 3 x 2 K x 3 y 2 K x 3 x 3 K x 3 y 3 u x 3 f y 3 K y 3 x 1 K y 3 y 1 K y 3 x 2 K y 3 y 2 K y 3 x 3 K y 3 y 3 u y 3 Nodal Master stiffness matrix Nodal forces displacements f = K u or IFEM Ch 2 – Slide 9
Introduction to FEM Member (Element) Stiffness Equations _ ¯ f = K ¯ u ¯ ¯ ¯ ¯ K xixi K xiyi K xix j K xiyj ¯ f xi u xi ¯ ¯ ¯ ¯ ¯ ¯ K yixi K yiyi K yix j K yiyj f yi u yi ¯ = ¯ ¯ ¯ ¯ ¯ f x j u x j ¯ K x jxi K x jyi K x jx j K x jyj ¯ f yj u yj ¯ ¯ ¯ ¯ ¯ K yjxi K yjyi K yjx j K yjyj IFEM Ch 2 – Slide 10
Introduction to FEM First Two Breakdown Steps: Disconnection and Localization 3 _ (3) _ (3) x y _ (2) x (3) (2) _ (2) y y _ (1) y _ (1) x x 1 2 (1) These steps are conceptual (not actually programmed) IFEM Ch 2 – Slide 11
Introduction to FEM The 2-Node Truss (Bar) Element _ _ _ _ f , u _ f , u yi yi yj yj y _ _ _ _ _ f , u x f , u (e) xj xj xi xi j i Equivalent k = EA / L s spring stiffness − F F L d IFEM Ch 2 – Slide 12
Introduction to FEM Truss (Bar) Element Formulation by Mechanics of Materials (MoM) F = k s d = E A F = ¯ f x j = − ¯ , f xi , d = ¯ u x j − ¯ u xi L d Exercise 2.3 ¯ f xi 1 0 − 1 0 u xi ¯ ¯ = E A f yi 0 0 0 0 u yi Element stiffness ¯ equations in local ¯ f x j − 1 0 1 0 u x j ¯ L coordinates ¯ f yj 0 0 0 0 u yj ¯ from which 1 0 − 1 0 Element stiffness K = E A 0 0 0 0 matrix in local − 1 0 1 0 L coordinates 0 0 0 0 IFEM Ch 2 – Slide 13
Introduction to FEM Where We Are So Far in the DSM we are done with this ... Disconnection Breakdown Localization (Chapter 2) Member (Element) Formation we finish Chapter 2 with Globalization Assembly & Merge Solution Application of BCs (Chapter 3) Solution Recovery of Derived Quantities IFEM Ch 2 – Slide 14
Introduction to FEM Globalization: Displacement Transformation u yj u yj ¯ u x j ¯ y ¯ u x j x ¯ j y u yi u xi ¯ ϕ u yi ¯ x u xi i Node displacements transform as u xi = u xi c + u yi s , u yi = − u xi s + u yi c γ ¯ ¯ u x j = u x j c + u yj s , ¯ u yj = − u x j s + u yj c γ ¯ in which s = sin ϕ c = cos ϕ IFEM Ch 2 – Slide 15
Introduction to FEM Displacement Transformation (cont'd) In matrix form u xi c s 0 0 u xi ¯ u yi ¯ − s c 0 0 u yi = 0 0 u x j ¯ c s u x j u yj ¯ 0 0 − s c u yj Note: global on RHS, local on LHS e γ = T e u e u γ or ¯ IFEM Ch 2 – Slide 16
Introduction to FEM Globalization: Force Transformation f yj ¯ f yj ¯ f x j f x j j y f yi ¯ f yi ¯ ϕ f xi x f xi i Node forces transform as ¯ f xi f xi c − s 0 0 Note: ¯ f yi f yi s c 0 0 = global on LHS, ¯ f x j 0 0 c − s f x j local on RHS ¯ f yj 0 0 s c f yj f e = ( T e ) T ¯ e f IFEM Ch 2 – Slide 17
Introduction to FEM Globalization: Congruential Transformation of Element Stiffness Matrices f e e ¯ ¯ u e K = T e u e f e = T e T f e u e = ¯ ( ) ¯ Exercise 2.8 = T e T ¯ e T K e e ( ) K c 2 − c 2 sc − sc K e = E e A e s 2 − s 2 sc − sc − c 2 c 2 L e − sc sc − s 2 s 2 − sc sc IFEM Ch 2 – Slide 18
Introduction to FEM The Example Truss - FEM Model (Recalled for Convenience) f , u y 3 y 3 f , u x 3 x 3 3 √ (3) L = 10 2 (2) √ L = 10 (3) (3) E A = 200 2 (2) (2) E A = 50 (3) (2) Insert the geometric & y physical properties of this model into f , u x f , u (1) x 1 x 1 x 2 x 2 the globalized member 2 1 (1) stiffness equations L = 10 f , u f , u y 1 y 1 y 2 y 2 (1) (1) E A = 100 IFEM Ch 2 – Slide 19
Introduction to FEM We Obtain the Globalized Element Stiffness Equations of the Example Truss (1) (1) f x 1 u x 1 1 0 − 1 0 (1) (1) f y 1 u y 1 0 0 0 0 = 10 (1) (1) − 1 0 1 0 f x 2 u x 2 (1) 0 0 0 0 (1) f y 2 u y 2 (2) (2) f x 2 u x 2 0 0 0 0 In the next class (2) (2) f y 2 u y 2 0 1 0 − 1 we will put these = 5 (2) (2) 0 0 0 0 to good use f x 3 u x 3 (2) 0 − 1 0 1 (2) f y 3 u y 3 (3) (3) f x 1 u x 1 0 . 5 0 . 5 − 0 . 5 − 0 . 5 (3) (3) f y 1 u y 1 0 . 5 0 . 5 − 0 . 5 − 0 . 5 = 20 (3) (3) − 0 . 5 − 0 . 5 0 . 5 0 . 5 f x 3 u x 3 (3) − 0 . 5 − 0 . 5 0 . 5 0 . 5 (3) f y 3 u y 3 IFEM Ch 2 – Slide 20
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