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

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Presentation on theme: "FE analysis with bar elements E. Tarallo, G. Mastinu POLITECNICO DI MILANO, Dipartimento di Meccanica."— Presentation transcript:

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

2 Es02 Es-01 Contact information and reference 2 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 Es02 Es-01 Summary 3 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 Es02 Es-01 Bar element - topic 4 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 Es02 Es-01 Bar element - limitations 5 1.It can represent only a constant state of strain 2.In terms of generalized coordinates 3.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 4.If axial force are distributed along all part of the length or if the bar is tapered, then the element is only approximate 5.Distributed load can still be applied, in the form of equivalent forces applied to nodes

6 Es02 Es-01 Exercise 1 – data problem 6 Geometry: L=1 m; A 1,2 =6.0x10 -4 m 2 ; A 3 =6√2x10 -4 m 2 Material: E=210 GPa; ν=0.3 Load: P=1000 kN Build the stiffness matrix and solve the equations (find nodal displacements)

7 Es02 Es-01 Exercise 1 – Analytic Results 7 kN m

8 Es02 Es-01 AbaqusCAE structure 8 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; submit directly the job for analysis (abaqus solver embedded) Visualization – View your results

9 Es02 Es-01 Exercise 1 – Modeling geometry 9 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 Es02 Es-01 Exercise 1 – Defining property 10 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 Es02 Es-01 Exercise 1 – Assembly and CSYS 11 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 Es02 Es-01 Exercise 1 – Step 12 Module Step Create a step in which it’s defined the analysis type (static, dynamic, frequency) For this ex: static, general

13 Es02 Es-01 Exercise 1 – Load and boundary 13 Module Load Define the force in 2 Define boundary condition in 1, 2 and 3 (for bc_3 use local CSYS)

14 Es02 Es-01 Exercise 1 - Mesh 14 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 Es02 Es-01 Exercise 1 - Job 15 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 Es02 Es-01 Exercise 1 – FEM results 16 U1=0.003968m U2=0.003968m RF1=-500kN RF2=500kN U1=0.01190m U2=0 CF1=1000kN RF1=0 RF2=0 U1=0 U2=0 RF1=-500kN RF2=-500kN Deformed shape Undeformed shape

17 Es02 Es-01 17 Excercise 2 - data Build the stiffness matrix and solve the equations Geometry: L=1m, A 1 =400mm 2, A 2 =225mm 2, A 3 =100mm 2 Material: E=210000 MPa, ν=0.3 (steel) Load: P 1 =10 kN,P 2 =5kN

18 Es02 Es-01 Exercise 2 - results 18

19 Es02 Es-01 19 Excercise 3 1.Solve the problem (find stress and max displacement) 2.Find other solutions 3.Compare max stress and displacement (look at total weight and total cost) of each solution Geometry: L=1m, H=0.2 m, A=400mm 2 Material: E=210000 MPa, ν=0.3 (steel) Load: P=10 kN L H P


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