Transcript Bar element - Politecnico di Milano

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: [email protected]
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
Es02
Es-01
2
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
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:
 1 L   1
k  
E 

1L  L
0
L
• Stiffness matrix:
Es02
Es-01
1
AE  1  1
Adx 

L
L  1 1 
Bar element - limitations
5
1. It can represent only a constant state of strain
2. In terms of generalized coordinates
u  1   2 x
du
x 
 2
dx
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
Es02
Es-01
Exercise 1 – data problem
6
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)
Es02
Es-01
Exercise 1 – Analytic Results
7
kN
Es02
Es-01
m
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
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
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
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 assemblyinstance)
• Create a new CSYS rectangular
located in 3 and directed as
segment 3 (use 3 point method)
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
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)
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)
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)
• Monitor: pop-up window with warnings,
errors and process messages
• Results: load the solution of the analysis
(it brings you to module visualization)
Es02
Es-01
•Submit: launch solver
(write files in work
directory: see .dat!)
Exercise 1 – FEM results
U1=0.01190m
U2=0
CF1=1000kN
RF1=0
RF2=0
16
Deformed shape
Undeformed shape
U1=0
U2=0
RF1=-500kN
RF2=-500kN
Es02
Es-01
U1=0.003968m
U2=0.003968m
RF1=-500kN
RF2=500kN
Excercise 2 - data
17
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
Es02
Es-01
Exercise 2 - results
18
Es02
Es-01
Excercise 3
19
P
L
H
Geometry: L=1m, H=0.2 m, A=400mm2
Material: E=210000 MPa, ν=0.3 (steel)
Load: P=10 kN
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
Es02
Es-01