STATIC FEM ALGORITHM FOR A TRUSS

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1 STATIC FEM ALGORITHM FOR A TRUSS Piotr Pluciński p.plucinski@l5.pk.edu.pl Jerzy Pamin jpamin@l5.pk.edu.pl

2 Scope of lecture Truss as a set of nodes and bars (truss elements) Truss element description Example computations for 2D truss (MathCAD)

3 Computational model - discretization Computational model In the computational model we must guarantee: continuity of displacements (at nodes, where elements are connected) satisfaction of kinematic constraints satisfaction of equilibrium equations for the structure and any substructure (e.g. a node or element)

4 Computational model - discretization Discretization process - mesh generation Bar structure is idealized as a discrete system of elements and nodes Numbering nodes Numbering elements Specification of relations between elements and nodes (topology of discrete model) e i j

5 Truss element description (2D) Definitions of variables describing the displacements, strain and cross-section force in the bar under tension/compression u(x) = {u(x)}, e(x) = {ε 0 (x)}, s(x) = {N(x)} Kinematic and physical equation for a point P(x, y, z) = P(x, 0, 0) = P(x) on bar axis ε 0 = du dx e = Lu, L = [ ] d dx N = EA ε 0 s = De, D = [EA]

6 Truss element description (2D) Displacement approximation in element (local) coordinate system 0x Number of local dofs of a node ndof w = 1 and element ndof e = 2. 1 q e 1 = ue 1 e L e 2 x e, u ξ = x e L e q e 2 = ue 2 where ξ = x e q w = {u w } [1 1] q e = {q1, e q2} e = {u1, e u2} e [2 1] u(ξ) [1 1] = {u(ξ)} = N(ξ) q e = [1 2] [2 1] ] = [ (1 ξ) ξ ] [ q e 1 q e 2 normalized coordinate Le

7 Truss element description (2D) Strain, normal force and stiffness matrix in element (local) coordinate system 0x e(ξ) [1 1] = {ε 0 (ξ)} = LN e (ξ) q e = B(ξ) q e = [ [ ] 1 1 q e L e L 1 e [1 2] [2 1] q2 e s(ξ) = {N(ξ)} = D [1 1] k e [2 2] = Le 0 B T DBdx = [1 1] B(ξ) [1 2] ( ) e [ EA L q e = [EA] e [ [ ] 1 1 q e L e L 1 e [2 1] q2 e ] ] ]

8 Truss element description (2D) Description of truss element in global coordinate set 0XY y Number of global dofs of a node NDOF w = 2 and element NDOF e = 4 (c = cos α e, s = sin α e ) Q e 4 q e 2 Q e 1 q e 1 Q e 2 [ q e q1 = [2 1] q 2 α e ] e = Q e = T et [4 1] [4 2] qe [2 1] x Q e 3 [ c s c s ] e Q w = {U w, V w } [2 1] Q e = {Q 1, Q 2, Q 3, Q 4 } = [4 1] Q 1 Q 2 Q 3 Q 4 = {U 1, V 1, U 2, V 2 } e = T e [2 4] Qe [4 1]

9 Truss element description (2D) Global stiffness matrix: (c = cos α e, s = sin α e ) K e [4 4] = (TT kt) e = ( EA L ) e cc cs cc cs cs ss cs ss cc cs cc cs cs ss cs ss e

10 Flowchart of FEM algorithm for statics Discretization Computation of stiffness matrices and nodal load vectors for elements Assembly Consideration of boundary conditions Computation of nodal displacement vector and reaction vector Return to each element to compute nodal force vectors

11 Problem definition and discretization Q 6 Q Q 1 Q 2 1 Q 4 Q 3

12 Input data Stiffness matrix k e Incidence matrix Transformation matrix T e Longitudinal stiffness

13 Computation of length, values of trigonometric functions and transformation matrix for element 1

14 Stiffness matrix of element 1 in local and global coordinate set

15 Computation of length, values of trigonometric functions and transformation matrix for element 2

16 Stiffness matrix of element 2 in local and global coordinate set

17 Computation of length, values of trigonometric functions and transformation matrix for element 3

18 Stiffness matrix of element 3 in local and global coordinate set

19 Assembly of stiffness matrix for element 1 into global stiffness matrix

20 Assembly of stiffness matrix for element 2 into global stiffness matrix

21 Assembly of stiffness matrix for element 3 into global stiffness matrix

22 Definition of load vector with point load 10 kn Q 6 Q 5

23 Computation of substitute load vector due to imposed displacement Q 4 Q 3 = m

24 Imposition of boundary conditions K K w, F F w Q 6 Q 5 = 0 K 11 K 12 K 13 K 14 K 15 K 16 K 21 K 22 K 23 K 24 K 25 K 26 K 31 K 32 K 33 K 34 K 35 K 36 K 41 K 42 K 43 K 44 K 45 K 46 K 51 K 52 K 53 K 54 K 55 K 56 K 61 K 62 K 63 K 64 K 65 K Q 3 0 Q 6 = F s6 + R 1 R 2 0 R 4 R 5 0 Q 1 = 0 Q 4 = Q 3 Q 2 = 0

25 Computation of nodal displacements Q 6 = Q 3 = Q 4 =

26 Determination of support reactions R 5 = R 1 = R 2 = R 4 = 2.222

27 Return to element to compute nodal forces - diagram of nodal forces Element 1 Element 2 Element 3

The elements used in commercial codes can be classified in two basic categories:

The elements used in commercial codes can be classified in two basic categories: CHAPTER 3 Truss Element 3.1 Introduction The single most important concept in understanding FEA, is the basic understanding of various finite elements that we employ in an analysis. Elements are used for