Report

FE analysis with beam elements E. Tarallo, G. Mastinu POLITECNICO DI MILANO, Dipartimento di Meccanica Summary 2 Subjects covered in this tutorial An introduction to beam elements A guided example to evaluate a simple structure through the use of FEM Comparison analytical vs numerical solutions Other few exercises (to include in exercises-book) Es02 Es-02 Beam element – topic 3 The element library in Abaqus contains several types of beam elements A “beam” is an element in which assumptions are made so that the problem is reduced to one dimension mathematically: the primary solution variables are functions of position along the beam axis only (as bar element) A beam must be a continuum in which we can define an axis such that the shortest distance from the axis to any point in the continuum is small compared to typical lengths along the axis The simplest approach to beam theory is the classical Euler-Bernoulli assumption, that plane cross-sections initially normal to the beam's axis remain plane, normal to the beam axis, and undistorted (called B23, B33) The beam elements in AbaqusCAE allow “transverse shear strain” (Timoshenko beam theory); the cross-section may not necessarily remain normal to the beam axis. This extension is generally considered useful for thicker beams, whose shear flexibility may be important (called B21, B22, B31, B32 and PIPE) Es02 Es-02 Beam element – shape function 4 Classic mechanical approach uses 3rd order interpolation function (elastic line theory) To follow this theory use element B23, B33 Beam defined in Abaqus CAE has linear or quadratic interpolation function (element B21, B22, B31, B32) Es02 Es-02 Beam element – topic (stiffness matrix) 5 Let us consider an Euler-Bernoulli beam: F 1 x F 1 y 1 M F 2 x F 2 y 2 T M k x 1 y 1 1 x 2 y 2 where the stiffness matrix is: EA / L 0 0 k AE / L 0 0 0 12 EJ / L 6 EJ / L 3 0 2 0 12 EJ / L 4 EJ / L 0 6 EJ / L 0 EA / L 0 0 12 EJ / L 0 6 EJ / L 6 EJ / L 2 0 12 EJ / L 6 EJ / L EA / L 0 2 3 6 EJ / L 2 2 EJ / L Es02 Es-02 2 6 EJ / L 2 EJ / L 0 2 6 EJ / L 4 EJ / L 0 3 2 3 2 2 T Exercise 1 – data problem 6 Geometry: L=1 m; A=100x100 mm Material: E=210 GPa; ν=0.3 Load: p=1 N/mm Write the relation of internal load e solve the analytic problem of the deformed shape of the isostatic beam M p (x L) 2 2 3 4 (x L)4 L L EI v ( x ) p p (x L) p 24 6 8 Es02 Es-02 Exercise 1 – Results (analytic vs numeric) 7 FEM Results Exact solution: v2=-0.07142mm θ2=-9.52381 e-5 rad M1 = 5e5Nmm F1Y=1000N Comparison btw analytic solution and FEM results Note: sensitive variables are Number of elements Linear or quadratic order Es02 Es-02 Exercise 1 – Modeling geometry and property 1 8 3 2 5 4 Es02 Es-02 Exercise 2 – data problem 9 Geometry: L=1 m; A=100x100 mm Material: E=210 GPa; ν=0.3 Load: p=1 N/mm Write the relation of internal load e solve the analytic problem of the deformed shape of the iperstatic beam M p (x L) 2 2 3 pL ( x L ) 8 x 2 (L x) (3 L 2 x ) v ( x ) p 48 EI Es02 Es-02 Exercise 2 – Results (analytic vs numeric) 10 FEM Results Exact solution: θ2=-1.1904e-5 rad F1=625N F2=375N M1 =1.25e5 Nmm Comparison btw analytic solution and FEM results Note: sensitive variables are Number of elements Linear or quadratic order Es02 Es-02 Excercise 3 - data 11 F1 A2 F2 A1 A4 A3 P A1 A2 A3 Material Property: E=210GPa Forces: F1=-20kN (Z) F2=30kN(Y) P=80N/mm(X) Note: All the written dimensions are 500mm Problem: Solve the system and report max displacement and max stress A4 Es02 Es-02 Exercise 3 - results 12 Es02 Es-02 Excercise 4 13 P L H Geometry: L=1m, H=0.2 m, Section variable Material: E=206 000 MPa, ν=0.3 (steel) Load: P=10 kN Compare max stress and displacement of the structures used in the previous lesson using beam elements Es02 Es-02