Lec9 finite element_beam_structures 1
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This document provides an introduction to using the finite element method to analyze beam structures. It discusses the basic theory behind discretizing beams into finite elements, including defining the element geometry, determining the shape functions, and assembling the element stiffness matrix. It then provides examples of using the method to calculate deflections and rotations of beams under different loading conditions. Tutorial problems are included to have students apply the concepts by modeling beam problems in Abaqus finite element software.
Lec9 finite element_beam_structures 1
- 1. B Y D R . M AH D I D AM G H AN I 2 0 1 7 - 2 0 1 8 Structural Design and Inspection- Finite Element Method (Beams) 1
- 2. Suggested Readings Reference 1 Reference 2 2
- 3. Objective(s) Familiarisation with Finite Element equations for beam structures; Ability to assemble global stiffness matrix for a beam shape structure; Familiarisation with Finite Element Modelling (FEM) of beam structures using ABAQUS CAE (Tutorial). 3
- 5. Introduction Refer to chapter 6 of Reference 1; Refer to chapter 5 of Reference 2; This method is used in most of commercially available FE based software to solve structural problems; Some typical software in aerospace industry are; Altair HyperWorks (mostly for optimisation purposes) MSC Nastran (mostly for linear analysis) Abaqus (mostly for non-linear analysis) Ansys (mostly for non-linear analysis) 5
- 6. Introduction 6 Unlike truss elements that are joined together by hinges, beams are connected by welding so they transfer both bending moment and forces across each other; In this lecture, we would be using Euler-Bernoulli beam theory that is applicable to thin beams.
- 7. Beam element 7 The length of element Node 1 has only 2 DOF (vertical displacement and rotation) Therefore, this beam element has 4 DOFs in total Node 2 has only 2 DOF (vertical displacement and rotation) Local coordinate system with origin at the middle of beam This slide shows positive direction of shear forces, bending moments, displacements and rotations.
- 8. The Finite Element Analysis (FEA) process 8
- 9. Displacement in FEM 9 In finite element methods, the displacement for an element is written in the form; e h xxu dN )()( Approximated displacement within the element Shape function Vectors of displacements at the two nodes of the element This function approximates displacements within the element by just having displacements at the two nodes, i.e. de Question: What should be N(x)???
- 10. Note 10 It is convenient if the shape functions are derived from a special set of local coordinates (natural coordinate system); The natural coordinate system is dimensionless, has its origin at the centre of the element and is defined from −1 to +1.
- 11. Note 11 The relationship between the natural coordinate system and the local coordinate system can simply be given as a scaling function, i.e. a x
- 12. Shape function 12 Since we have four DOFs for the element we need to find 4 shape functions; We assume displacement within element has the form where ; Four unknowns in the form of 4x1 matrix Vector of polynomial basis functions This matrix is 1x4 because we have a 4-DOF element ax/
- 13. Shape function 13 We also know; x v axa x x v 1 v a 1 2 321 32 1 a a/2
- 14. Note 14 v a 1 2 321 32 1 a 3211 32101 32 1 )1( )1( a v 3212 32102 32 1 )1( )1( a v a/2
- 15. Shape function 15 a/2 e T v dPP 1 )()(
- 16. Shape function 16 Shape functions for displacement are functions of natural coordinates Shape functions for displacement are functions of natural coordinates Shape functions for rotations are functions of natural coordinates and length of element Shape functions for displacement are functions of natural coordinates These cubic shape (or interpolation) functions are known as Hermite cubic interpolation (or cubic spline) functions.
- 17. Shape functions 17 -0.2 0 0.2 0.4 0.6 0.8 1 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 ξ Shape functions along the length of beam element N1 N2 N3 N4
- 18. Strain calculation 18 B is called strain matrix L is differential operator
- 19. Note 19 First derivative of the shape function is important as it will be used to obtain rotations within each element (we will see this in some examples); How first derivative of shape function (dN/dx) is obtained? d d adx d d d dx d NNN 1 21 1 4 3 adx dN 22 321 4 1 dx dN 23 3 1 4 dN dx a 24 321 4 1 dx dN
- 20. Reminder 20 x 2x 1x x Fibre ST has shortened in length whilst NQ increased in length so they have gone through strains Remember that the length of neutral axis does not change and remains as x LengthOriginal LengthinChange xx x RyR xx R y R RyR xx Positive y gives negative strain, i.e. compression x xyR xx
- 21. Reminder 21 How did the following come about? Euler-Bernoulli beam curvature (1/R) is used for this formula; e N B e e Bdd NNd xx dx d y xx dx d y dx d y dx yd y R y 2 2 2 2 2 2 2 2
- 23. Element local stiffness matrix 23 Element stiffness matrix (see chapter 3 of Ref. 2) Ec dAdxdV Aiscrosssectional area Eismodulusof elasticity(material constant)
- 24. Further elaboration 24 Some might ask where did the following come from? See below; ''111 1 22 2 2 2 2 NNN NNN aaxa xaxxxxx N
- 25. Further elaboration 25 V T e dxdA a y E a y '''' 22 NNk ''2 NB a y adx d ax a a T A e dxdAy a E 1 ,/ 2 4 '''' NNk 1 1 2 2 2 2 4 ad d Nd d Nd I a E T zek 1 1 3 '''' dNNI a E T zek
- 26. Element local stiffness matrix 26
- 27. Elements mass matrix 27 Following similar procedure as stiffness matrix; Students are advised to familiarise themselves with chapter 3 of ref. 2
- 28. Fs1 Fs2 Ms1 Ms2fy Nodal forces 28 Surface force (applied at the node) Surface force (applied at the node) Body forces (applied between nodes) Note that in FEA, body forces are always transferred to the nodes
- 29. Nodal forces 29
- 30. Example 1 30 Obtain tip deflection and rotation of the beam shown below using FEA. (E=69.0 GPa and Poisson’s ratio is 0.33)
- 31. Solution 31 For practical purposes, a beam must be divided into more than 1 element to obtain precise enough results; However, for the sake of showing the process and simplicity, let’s divide the beam into one element only.
- 33. Solution 33 eedKF Reaction forces at clamped boundary that are unknown Applying BC, i.e. v1 =0 and θ1=0
- 34. Solution 34 The reduced stiffness matrix becomes;
- 35. Example 2 35 In previous example, what are the deflection and rotation at mid-point of beam structure? 3 1.0064 10 3 1.0064 10
- 36. Tutorials 3a 36 See T3a.pdf
- 37. Tutorial 3b 37 Model beam problem of example 1 in Abaqus and extract deflection and rotation at the tip and mid point of the structure.
- 38. Tutorial 3c 38 See T3c.pdf
Editor's Notes
- #13: The complete cubic displacement function is appropriate because there are four total degrees of freedom (a transverse displacement and a small rotation at each node). The cubic function also satisfies the basic beam differential equation (y’’=M/EI and Eiy’’’’=0)— further justifying its selection. In addition, the cubic function also satisfies the conditions of displacement and slope continuity at nodes shared by two elements.