Transcript Beam element - Politecnico di Milano

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)
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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)
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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)
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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
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
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 

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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
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Exercise 1 – Modeling geometry and
property
1
8
3
2
5
4
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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


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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
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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
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Exercise 3 - results
12
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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
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