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FE analysis with bar elements E. Tarallo, G. Mastinu

FE analysis with bar elements E. Tarallo, G. Mastinu POLITECNICO DI MILANO, Dipartimento di Meccanica. Contact information and reference. Ermes Tarallo Tel : (02 2399) 8667 Email : ermes.tarallo@mail.polimi.it References: ABAQUS 6.9-1 documentation www.simulia.com

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FE analysis with bar elements E. Tarallo, G. Mastinu

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  1. FE analysis with bar elements E. Tarallo, G. Mastinu POLITECNICO DI MILANO, Dipartimento di Meccanica

  2. Contact information and reference • Ermes Tarallo • Tel: (02 2399) 8667 • Email: ermes.tarallo@mail.polimi.it • References: • ABAQUS 6.9-1 documentation • www.simulia.com • Robert D.Cook, Finite Element Modeling for Stress Analysis, 1995 • Other FEM Books: • Klaus J. Bathe, Finite Element Procedures, 1996 • O.C. Zienkiewicz Finite Element Method, Vol 1+2+3, 2000

  3. Summary • Subjects covered in this tutorial • An introduction to Abaqus / AbaqusCAE for beginners • A guided example to construct the finite element model of a simple structure • Comparison analytical vs numerical solutions • Other few exercises • Lab –course structure: • In each lesson one example will be presented and explained • At the end of each lesson, few exercises will be proposed in order to evaluate comprehension • For the final evaluation it’s necessary to produce the collection of the proposed exercises solved and discussed

  4. Bar element - topic • Bar (truss in AbaqusCAE) elements are one-dimensional rods that are assumed to deform by axial stretching only. • They are pin jointed at their nodes, and so only translational displacements are used in the discretization (2 dof); • A bar elements is well defined by: length L, elastic modulus E and section area A • Shape function: • Stiffness matrix:

  5. Bar element - limitations It can represent only a constant state of strain In terms of generalized coordinates If axial force are applied only at nodes, the element agree exactly with a mathematical model that represents the bar as straight line having constant A and E between locations where axial forces are applied If axial force are distributed along all part of the length or if the bar is tapered, then the element is only approximate Distributed load can still be applied, in the form of equivalent forces applied to nodes

  6. Exercise 1 – data problem Geometry: L=1 m; A1,2=6.0x10-4 m2; A3=6√2x10-4 m2 Material: E=210 GPa; ν=0.3 Load: P=1000 kN Build the stiffness matrix and solve the equations (find nodal displacements)

  7. Exercise 1 – Analytic Results m kN

  8. AbaqusCAE structure • AbaqusCAE is divided into the following modules: • Part – Create individual parts (geometry) • Property – Create and assign material properties; for beam and bar elements it allows to create and assign the transversal section • Assembly – Create and place all parts in instances (assemblies): it allows to translate,to rotate, to duplicate each part • Step – Define all analysis steps and the results you want (output): the analysis may be static, dynamic, frequency • Interaction – Define any contact information or special constraints • Load – Define and place all loads (force, moment, pressure, body gravity,…) and boundary conditions (encastre, pin,…) • Mesh – Define nodes and elements of the discretized structure • Job – Create the input file for abaqus solver; submitdirectly the job for analysis (abaqus solver embedded) • Visualization –View your results

  9. Exercise 1 – Modeling geometry • Module Part • Sketch the part as 2D planar deformable wire • Note: in module part it’s possible to create 1D, 2D and 3D geometry as like as another CAD software • AbaqusCAE is not so powerful: if you want to create complex geometry, it’s better to use other CAD system (CATIA, ProE, SolidWorks,…) and import the model in AbaqusCAE

  10. Exercise 1 – Defining property • Module Property • Define material property • Create section with area A and material just defined • Assign the section defined to the relative segment of the structure

  11. Exercise 1 – Assembly and CSYS • Module Assembly • Assembly the part!! Even if there’s only one part it must be create an assembly (instance) • Assembly may be created as dependent (the mesh will be created on each part) or independent (the mesh will be created on the assembly-instance) • Create a new CSYS rectangular located in 3 and directed as segment 3 (use 3 point method)

  12. Exercise 1 – Step • Module Step • Create a step in which it’s defined the analysis type (static, dynamic, frequency) • For this ex: static, general

  13. Exercise 1 – Load and boundary • Module Load • Define the force in 2 • Define boundary condition in 1, 2 and 3 (for bc_3 use local CSYS)

  14. Exercise 1 - Mesh • Module Mesh • Assign mesh type (solver standard-static, order linear, family truss) • Define mesh seed (partition of the edge): seed by size (assign dimension of element); seed by number (define number of elem. on the edge) • Create mesh of the part (automatic process)

  15. Exercise 1 - Job • Module Mesh • Write input file (launch the solver from external shell) • Data check (evaluate inp file – find error) • Submit: launch solver (write files in work directory: see .dat!) • Monitor: pop-up window with warnings, errors and process messages • Results: load the solution of the analysis (it brings you to module visualization)

  16. Exercise 1 – FEM results Deformed shape U1=0.01190m U2=0 CF1=1000kN RF1=0 RF2=0 U1=0.003968m U2=0.003968m RF1=-500kN RF2=500kN Undeformed shape U1=0 U2=0 RF1=-500kN RF2=-500kN

  17. Excercise 2 - data Geometry: L=1m, A1=400mm2, A2=225mm2, A3=100mm2 Material: E=210000 MPa, ν=0.3 (steel) Load: P1=10 kN,P2=5kN Build the stiffness matrix and solve the equations

  18. Exercise 2 - results

  19. Excercise 3 P L H Geometry: L=1m, H=0.2 m, A=400mm2 Material: E=210000 MPa, ν=0.3 (steel) Load: P=10 kN Solve the problem (find stress and max displacement) Find other solutions Compare max stress and displacement (look at total weight and total cost) of each solution

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