introduction to fem-mongi ben ouezdou
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M. Ben Ouezdou, University of Nizwa, 2011 1
INTRODUCTION TO FINITE
ELEMENT METHOD
Professor Dr. Mongi Ben Ouezdou
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Basic Concept
Building a complicated object with simpleblocks,
or
dividing a complicated object into small
and manageable pieces.
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The structure is considered as anassemblage of a finite number of individualstructural components called elements.
These elements can be put together in a
number of ways,represent complex geometry.
Basic Concept
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FEM based on Principle of discretization
=
procedure in which a complex problem of
large extent is divided (discretized) intosmaller equivalent units.
Basic Concept
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Application
Application of this idea can be found everywhere in
everyday life and in engineering.
Examples:
Lego (kids play) aircraft
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Examples
beam bridge
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Steel frames
Concrete building
Application to buildings
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Advantages of the FEM
1- Incorporate diff properties of each element.2- No restriction for the shape of the medium;
hence arbitrary and irregular geometries
cause no difficulty.
3- Accommodation of any type of BC.
4- Handle non-linearities, time-dependant Pb.5- Valid for any engineering Pb.
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Some History of the FEM 1943: Courant (Variational Methods);
1956: Turner, Clough, Martin and Trop (Stiffness);
1960: Clough (Finite Element, plane problems);
1970s:Applications on mainframe computers;
1980s: Micocomputers, pre- and postprocessors;
1990s:Analysis of large structural systems.
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Some Terminology FE: small elements (1D, 2D, 3D) obtained by
subdividing the given domain to be analyzed.
Nodes or nodal points: intersections of the sidesof the elements.
Nodal lines and nodal planes: interfaces
between elements. Linear elements: FE with straight sides.
Higher order elements: FE with curved sides.
Primary unknowns: nodal displacements Secondary unknowns: strains, stresses,
moments, shear forces, etc.
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Basic procedure
Step 1: Discretise the continuum: subdivide into elements: generate a mesh;
Step 2: Select element displacement functions;
Step 3: Calculate element properties: stiffness matrix [k].
Step 4: obtain element load vector [F];
Step 5: Assemble element properties (element stiffnesses global stiffness,
load vector).
Step 6: Incorporate B.C. (set the element to the ground so disp = 0 or finite):the stiffness matrix developed in step 5 will be modified to realize the
condition that disp of some coordinates = 0 or finite.
[F] = [K] {u} and {u}=[K]-1{F}. [K]: global stiffness matrix, [F]: vector of known forces
and {u}: displacements. Step 7: Determine displacements, strains and stresses
Step 8: Check and iterate to eliminate precision errors if present.
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Discretization
=
Process of separating the length, area or volumeinto discrete (separate) parts or elements.
structure
1-D elements 2-D elements 3-D elements Axisymmetric
elements
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1-D elements
Used for beams or frames
Basic element
Node 1 Node 2
H-element
1 23
Curved element
3
1
2
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2-D elements
Used ,for plane-wall, diaphragm, slab, shell, etc.
1- Triangular elements
3 nodes 6 nodes
1 2
3
14 2
3
6 5
1
4
2
3
5
6
6- nodes curved triangle
- Triangular elements are the most used ones
- Curved elements for 2-D domain with curved boundaries
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2-D elements
2- Quadrilateral elements
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Used for the analysis of solid bodies (stresses underfoundation, contact stress under point loads, etc).
3-D elements
Tetrahedron Hexahedron Curved 3D element
Problem: Complex visualization and stiffness matrices size can be enormous.
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Axisymmetric elements
Used in problems that are axisymmetric in nature.
Can achieve huge simplification in axisymmetric problems.
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Guidelines for discretization Discretization is a major decision making step in FEM.
Simple structures: no problems. Most real structures: difficulties in
processing of subdividing the structure;
Numbering the nodes;
Assigning coordinates to each node;
Relating the structure coordinate numbers to elements numbers and
their coordinate number.
In most FEM Software: discretization is handled automaticallyby the preprocessing module of the software.
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Examples of discretization
1 2 3 4
1 2 3 4
Point loads:
5 nodes and 4 elements
Stepped beam:
5 nodes and 4 elements
1 2 3 4
1 23 45 5
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Examples of discretization
Nodal line
Change in loading
Material 1
Material 2
Change in material
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Size of elements when discretize
In recent past, the number of elements was limited due to
capabilities of both hardware and software. But nowadays a
sufficient fine discretization of a whole structure can easily beproduced very Quickly by graphical preprocessors.
Most new FE softwares provide automatic mesh generation;
But this tool should not be used in an uncaring manner:engineering knowledge is still required.
An inadequate modeling of apparently irrelevant details (e.g. small
cantilever slab or opening in a slab) can lead to faulty result and
unsafe design. A sufficiently FE mesh should be used in regions of high deformation
pr stress gradients.
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FE Equation
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Element stiffness matrix of a prismatic bar:
Direct method
[F] = [k] [u]
Load vector stiffness displacement
Node 1 Node 2
1, u12, u2A,E
x
=
2
1
1
1
1111
u
uEA
l
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Element stiffness matrix of a prismatic bar:
Formal approach
[F] = [k] [u]
Node 1 Node 2
1, u12, u2A,E
x
( )=V
TdVBEBk : element stiffness matrix, with
=
ll
11B
[ ]
= 11
11
l
EA
k
Same result !
Use conservation of energy: strain energy = work done by nodal forces
where
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Global stiffness matrix for a prismatic bar
1 3
E, A
x
E, 2A
P21 2
Example:
Solution: Use two 1-D bar elements
Element 1:
Find stresses in the 2 bars
[ ]
=11112
1l
EAk
u1 u2
Element 2: [ ]
=1111
2l
EAk
u2 u3
Global stiffness matrix [ ]
=
110
132
022
l
EAK
u1 u2 u3
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=
3
2
1
3
2
1
110132
022
FF
F
u
u
uEA
l
Global FE equation
Use B.C. and Loads condition: u1 = u3 = 0. and F2 = P.
=
0
0
110
132
022
F
P
F
2
3
1
uEA
l
{ } [ ] { }23 uEA
Pl
=EA
Pu
32
l=
A
P
EA
PEuu
EE 303
12
11=
=
==
l
ll
A
P
EA
PEuuEE
3302322 =
=
==
l
ll
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2211 vuvu
[ ]
2
2
1
1
22
22
22
22
v
u
v
u
scsscs
csccsc
scsscs
csccsc
EAk
= l
c = cos
s= sin
Bar element in 2-D
in global coordinates
l
x, u
Y, v
1
2
x, u02
u01
E, A
y, v01
v02
Element stiffness matrix in
global coordinates
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El t tiff t i f b
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M1, 1
2211 vv
l
=
2
2
1
1
22
3
2
2
1
1
4626612612
2646
612612
v
v
EI
M
F
M
F
llll
ll
llll
ll
l
1 2
F1, v1 F2, v2
M2, 2
x
E, I
Element stiffness matrix of a beam:
Direct method
Element stiffness equation:
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El t tiff t i f b
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Element stiffness matrix of a beam:
Formal approachStrain energy stored in, the beam element
Stiffness matrix for the simple beam element: k
=l
0
T BBk dxEI
with the strain-displacement matrix B:
+++=
232232
6212664126B
llllllll
xxxx
=
llll
ll
llll
ll
l
4626
612612
2646
61261222
3
EIkobtain the same result
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Local Stiffness matri of a general 2 D beam element
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Local Stiffness matrix of a general 2-D beam element:
=
llll
llll
ll
llll
llll
ll
EIEIEIEI
EIEIEIEI
EAEA
EIEIEIEI
EIEIEIEI
EAEA
k
460
260
612
0
612
0
0000
260
460
6120
6120
0000
22
2323
22
2323
222111 vuvu
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M2, 2
Element Stiffness matrix of a general 2-D beam
element in a global coordinate system:
l
1
2
x, u
y, v x0, u02
u01
E, A
y0, v01
v02
M1,
1
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+
+
+
+
++++
+
++++
+
+
++
=
llllll
llllllllll
llllllllll
lllllll
llllllllll
llllllllll
EIc
EIs
EIEIc
EIs
EI
cEI
cEI
sEA
csEI
csEA
cEI
cEI
sEA
csEI
csEA
sEIcsEIcsEAsEIcEAsEIcsEIcsEAsEIcEA
EIs
EIc
EAs
EIEIc
EIs
EI
cEI
csEI
sEA
csEI
csEA
cEI
cEI
sEA
csEI
csEA
sEI
csEI
csEA
sEI
cEA
sEI
csEI
csEA
sEI
cEA
k
466266
6121261212
612661212
266466
6121261212
6121261212
2222
2
2
3
2
32
2
3
2
3
23
2
2
2
23
2
3
2
2
2
222
232
322
32
3
23
2
3
2
33
2
3
2
222111
vuvu
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Computation of nodal displacements
{F} = [k] {u} {u} = [k]-1 {F}
This step needs the use of the computer (mainly if stiffness matrix exceeds 5 x 5)
because it needs to invert the matrix.
Solution: enforcing 0 displacement BC and solve by:
- Unit diagonal method;
- Large diagonal method;
- Row column delete method.
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Features of the assembled stiffness matrix
Calculation of primary unknowns
1) The stiffness matrix has its non-zeros terms along its
main diagonal (terms distant from the diagonal are 0).
2) Stiffness matrices are symmetric: advantage in storing
the matrices.
=
534000
376500
469320
053474
002763
000435
k
Half band width
=
005
037469
534
276435
k
Values to be stored
-Reduction of the
required storage memory
-Reduction of the
solution time
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C l l ti f i k
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Calculation of primary unknowns
Solution of equations {u} = [k]-1 {F}
direct scheme:
Linear problems
(Gaussian elimination)
iterative scheme:
Non Linear problems
(Jacobi, Gauss Seidal)
The most known methods of solutions are:
- Choleskys square root methods;
- Halfband Gauss elimination solution technique;
- Skyline technqiue
- Frontal solution technique.
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Calculation of secondary unknowns
nodal displacement {u}
stresses {} : (Hookes law)
strain {}
Standard software give the output in a tabularform and on graphical form.
Results include:
- Refined colored graphics;- Direct stresses x, y;
-Shear stresses xy; maximum shear, etc.
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Summary:
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Computer implementations
Preprocessing (build FE model, loads andconstraints);
FE Analysis solver (assemble and solve the
system of equations); Postprocessing (sort and display the results).
Summary:
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Conclusions:
Procedures of FEM in Structural Analysis
1) Divide structure into elements with nodes;
2) Describe the behavior of the physical quantities on eachelement;
3) Assemble (connect) the elements at the nodes to form anapproximate system of equations for the whole structure;
4) Solve the system of equations involving unknown quantitiesat the nodes (e.g., displacements);
5) Calculate desired quantities (e.g., strains and stresses) atthe selected elements.
N.B.: be aware of the limitations of the FEM: such as loadsapplication is imposed (no moving loads), and do notmisuse the FEM (it is a numerical tool).
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References1- Yijun Liu, Introduction to Finite Element Method,
Lecture notes, University of Cincinnati, Ohio, USA,1998.
2- R. Vaidyanathan, P. Perumal, ComprehensiveStructural Analysis, 2nd ed., Laxmi Publication ed.,
New Delhi, 2008.
M. Ben Ouezdou, University of Nizwa, 2011