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EMAE 415 LecturesFinite Element Analysis -
Basic Concepts
Based in part on slides posted by
Dr. Ara Arabyan, University of Arizona;
Dr. Kenneth Youssefi, San Jose State University
Basics of Finite Element Analysis
FEA in Biomechanics• FEA models have proven to be powerful tools
in analyzing biomechanical structures and to evaluate designs for implants, prostheses and musculoskeletal constructs.
• The advantages are the ability to account for complex geometries and material behaviors.
• Always remember, FEA is a numerical technique. It’s answers are only as good as the formulation of the problem.
Basics of Finite Element Analysis
FEA is a mathematical representation of a physical system and the solution of that mathematical representation.
Always remember, FEA is a numerical technique. It’s answers are only as good as the formulation of the problem.
GIGO (Garbage in-garbage out)
FEA Stress Analysis
After approximating the object by finite elements, each node is associated with the unknowns to be solved.
For a two dimensional solid (e.g. a beam) the displacements in x and y would be the unknowns.
This implies that every node has two degrees of freedom and the solution process has to solve 2n degrees of freedom.
Once the displacements have been computed, the strains are derived by partial derivatives of the displacement function and then the stresses are computed from the strains.
Finite Element Analysis
Consider the cantilever beam
shown.
Finite element analysis starts with an approximation of the region of interest into a number of meshes (triangular elements). Each mesh is connected to associated nodes (black dots) and thus becomes a finite element.
Displacement Fields
• The displacements of different points in the structure set up a displacement field
• The displacement field expresses the displacement of any point in the structure as a function of its position measured in a reference frame
Strains versus Displacement Fields(2-dimensional case)
Knowing the displacement fields allows you to determine the strain fields.
Stiffness of a Uniform Rod
• Recall from elementary mechanics of solids that a uniform rod of length L, cross sectional area A, and elasticity modulus E can modeled as a linear spring of stiffness keq
with
PP P PA, E
L eqk
eq
AEk
L
Nodal Displacements, Forces
• Consider a linear spring of stiffness k. Let the displacements of its two ends, called nodes, be denoted by ui and uj, known as nodal displacements. Let the forces acting at its two ends, called nodal forces, be denoted by fi and fj.
i jk
fjfi
Nodes
y
x
ujui
Nodal force
Nodal displacement
Reference frame
Force-Displacement Relations
• The relationships between the nodal forces and displacements (as shown below) are given by:
i i j
j j i
f k u u
f k u u
k
fjfi
ujui
Element Stiffness Matrix
• These relations can be written in matrix form as
or more briefly as
i i
j j
u fk ku fk k
ku f
Element Stiffness Matrix (cont’d)
• In this relation
is known as the element stiffness matrix (always symmetric);
is known as the element nodal displacement vector; and
k k
k k
k
i
j
u
u
u
Element Stiffness Matrix (cont’d)
is known as the element nodal force vector.
• The element nodal displacements are also known as element nodal degrees of freedom (DOF)
i
j
f
f
f
Singularity of Element Stiffness Matrix
The equation ku = f cannot be solved for the nodal displacements for arbitrary f because the matrix k is singular.
Physically this means that, in static equlibrium, the displacements of the endpoints of a spring cannot be determined uniquely for an arbitrary pair of forces acting at its two ends.
One of the ends must be fixed or given a specified displacement; the displacement of the other end can then be determined uniquely.
Solution for Single Rod Element
• If node i is fixed (i.e. its displacement is set to 0) then ku = f reduces to
and the displacement of node j is easily determined as
which is the expected solution
j jku f
jj
fu
k
k
fj
ujui=0
Matrix Reduction
• Note that when a displacement or DOF is set to zero, rows and columns of k associated with that displacement are eliminated and only the remaining set is solved
i i
j j
u fk ku fk k
Row(s) associated with ui
Column(s) associated with ui
Multiple Elements
• Now consider two springs of different stiffness linked to each other
1k
F1
2k1 2 3
1 2
(1)1 iu u (1) (2)
2 j iu u u (2)3 ju u
F2 F3
Globally numbered nodes
Globally numbered elements
Continuity Relations
• When two elements are joined together the joined nodes become one and must have the same displacement
where the subscript denotes the global node number, the superscript denotes the global element number, and i and j denote local element numbers
(1) (2)2 j iu u u
Force Balance Relations
• The external nodal forces acting at each node must equal the sum of the element nodal forces at all nodes
where F1, F2, F3 are external nodal forces numbered globally
(1)1
(1) (2)2
(2)3
i
j i
j
F f
F ff
F f
Assembly of Equations
• When these continuity and force balance relations are imposed the resulting global equilibrium equations are
• More briefly this can be written as
1 1 1 1
1 1 2 2 2 2
2 2 3 3
0
0
k k u F
k k k k u F
k k u F
KU F
Global Stiffness Matrix
• In this relation
is known as the global stiffness matrix (always symmetric);
is known as the global nodal displacement vector; and
1 1
1 1 2 2
2 2
0
0
k k
k k k k
k k
K
1
2
3
u
u
u
U
Global Stiffness Matrix (cont’d)
is known as the global nodal force vector or the global load vector
• The global nodal displacements are also known as global degrees of freedom (DOF)
1
2
3
F
F
F
F
Assembly of Global Stiffness Matrix
• Note that the global stiffness matrix is assembled from element matrices as follows
1 1
1 1 2 2
2 2
0
0
k k
k k k k
k k
K
Stiffness matrix from element 1
Stiffness matrix from element 2
Stiffness terms from two matrices add at coinciding DOF
Singularity of Global Stiffness Matrix
As in the case of individual element matrices the global stiffness matrix K is singular. (You can check this out for this small example by calculating the determinant of the matrix; the result will be zero.)
Some nodes of the structure need to be constrained (i.e. fixed or given known displacements) to make it statically determinate or overconstrained. Then the remaining DOF can be determined.
Constraining some nodes in the structure corresponds to applying boundary conditions.
Solution for Global Structure
• If node 1 is fixed (i.e. its displacement is set to 0) then the equilibrium equations reduce to
1 2 2 2 2
2 2 3 3
k k k u F
k k u F
1k 2k1 2 3
1 21 0u 2u
3u
F2 F3
Solution for Global Structure (cont’d)
• The displacements of nodes 2 and 3 can now be found from
• It can be shown that constrained global stiffness matrix is not singular
12 1 2 2 2
3 2 2 3
u k k k F
u k k F
Matrix Reduction
• Note that when a DOF is set to zero rows and columns of K associated with that DOF are eliminated and only the remaining set is solved
1 1 1 1
1 1 2 2 2 2
2 2 3 3
0
0
k k u F
k k k k u F
k k u F
Row(s) associated with u1
Column(s) associated with u1
Rod Elements
• The spring models introduced thus far constitute one class of finite elements and are known as rod, spar (ANSYS), or truss (ALGOR) elements
• In the form shown these elements can be used to model only unidimensional (one dimensional) problems
• The more general form of these elements can be used to model two or three dimensional problems
Example
• An aluminum (E = 10.4 106 psi) rod of variable cross section is subjected to a point load of 1000 lb at its narrower end. Determine the deflection of the rod at its loaded end.
12 in
1000 lbAl = 0.250 in2
Ar = 0.125 in2
Example, contd.
• The variable cross section rod can be approximated as a number of rods of constant cross section. For this example let us choose three rods of equal length that span the length of the original rod.
1000 lb
4 in
4 in
4 in
A1 A2 A3
Example, contd.
• The cross-sectional area of each rod can be assumed to be the average of the segment each rod spans:
21
14 0.229in
2 12l r
l l
A AA A A
22
14 8 0.188in
2 12 12l r l r
l l
A A A AA A A
23
18 0.146in
2 12l r
l r
A AA A A
Example, contd.
• The equivalent stiffness of each rod can now be computed from
which results in
i ii
i
A Ek
L
51
52
53
5.95 10 lb/ in
4.89 10 lb/ in
3.80 10 lb/ in
k
k
k
Example, contd.
• The spring (finite element) model of the problem is now
1000 lb1k 2k
1 2 3
1 21 0u 2u
3u4u
43k
3
Example, contd.
• The equilibrium equations can now be written as:
1 1 1 1
1 1 2 2 2
2 2 3 3 3
3 3 4
0 0
0 0
0 0
0 0 1000
k k u R
k k k k u
k k k k u
k k u
Row and column associated with u1
Example, contd.
• Solving the remaining equations we obtain
• Since the deflection of the right end is required the answer to the problem is
1
25 3
3
4
10.84 4.89 0 0 1.68
10 4.89 8.69 3.80 0 3.73 10 in
0 3.80 3.80 1000 6.36
u
u
u
34 6.36 10 inru u
Comparison with Exact Results
• Using “exact” analytical methods the displacement of any point (u(x)) on the rod is given by
ln l
l rl rl
APLu x
A AE A A A xL
u x
x
1000 lb
• Using this expression the displacement of the three nodes in the finite element model can be computed as
• Evidently the results from the finite element method are very close to those produced by “exact” methods
Comparison with Exact Results (cont’d)
3
3
3
4 1.68 10 in
8 3.74 10 in
12 6.40 10 in
u
u
u
Comparison with Exact Results (cont’d)
• The plot below charts the variation of displacements across the bar for the two solutions
0 2 4 6 8 10 120
1
2
3
4
5
6
7x 10
-3
x (in)
Dis
pla
ce
me
nt
(in
)
Exact solutionFE solution
u1
u2
u3
Formulation of the Finite Element Method
f B – Body forces (forces distributed over the volume of the body; gravitational, inertia, or magnetic forces)
f S – surface forces (pressure of one body on another, or hydrostatic pressure)
f i – Concentrated external forces
Formulation of the Finite Element MethodDenote the displacements of any point (X, Y, Z) of the object from the unloaded configuration as UT
The displacement U causes the strains
and the corresponding stresses
The goal is to calculate displacement, strains, and stresses from the given external forces.
Example – a plate under load Derive and solve the system equations of a plate loaded as
shown.Plate thickness is 1 cm and the applied load Py is
constant.
using two triangular elements,
Example, cont’dDisplacement within the triangular element with three
nodes can be assumed to be linear.
Example, cont’dDisplacement for each node,
Example, cont’dSolve the equations simultaneously for α and β,
Example, cont’dSubstitute x1= 0, y1= 0, x2=10, y2= 0, x3= 0, y3=4 to
obtain displacements u and v for element 1,
Element 1
Example, cont’dRewriting the equations in the matrix form,
Example, cont’d
Similarly the displacements within element 2 can be
expresses as
Example, cont’d
The next step is to determine the strains using 2D
strain-displacement relations,
Example, cont’d
Differentiate the displacement equation to obtain the strain,
Example, cont’dElement 2
Example, cont’d
Using the stress-strain relations for homogeneous, isotropic
plane-stress, we have
Formulation of the Finite Element Method
Equilibrium condition and principle of virtual displacements
The left side represents the internal virtual work done and the right side represents the external work done by the actual forces as they go through the virtual displacement. The above equation is used to generate finite element equations. And by approximating the object as an assemblage of discrete finite elements, these elements are interconnected at nodal points.
Formulation of the Finite Element MethodThe equilibrium equation can be expressed using matrix notations for m elements.
where
B(m) represents the rows of the strain displacement matrix C(m) is the elasticity matrix of element m H(m) is the displacement interpolation matrix U is a vector of the three global displacement components at all nodes F is a vector of the external concentrated forces applied to the nodes
Formulation of the Finite Element MethodThe previous equation can be rewritten as follows,
The above equation describes the static equilibrium problem. K is the stiffness matrix.
Stiffness matrix for element 1
Example, cont’dCalculating the stiffness matrix for element 2.
Example, cont’dThe stiffness of the structure as a whole is obtained by combing the two matrices.
Example, cont’dThe load vector R, equals Rc because only concentrated
loads act on the nodes
where Py is the known external force and F1x, F1y, F3x, and F3y are
the unknown reaction forces at the supports.
Example, cont’d
The following matrix equation can be solved for nodal point displacements
Example, cont’dThe solution can be obtained by applying the boundary conditions
Example, cont’dThe equation can be divided into two parts,
The first equation can be solved for the unknown nodal displacements, U3, U4, U7, and U8. And substituting these values into the second equation to obtain unknown reaction forces, F1x, F1y, F3x, and F3y .
Once the nodal displacements have been obtained, the strains and stresses can be calculated.
Finite Element Analysis
• FEA involves three major steps– Pre-Processing
– Solving Matrix (solver)
– Post-Processing
Summary of Pre-Processing
• Build the geometry
• Make the finite-element mesh
• Add boundary conditions; loads and constraints
• Provide properties of material
• Specify analysis type (static or dynamic, linear or non-linear, plane stress, etc.)
FEA Pre-Processing
Mesh Development
The FEA mesh is your way of communicating geometry to the solver, the accuracy of the solution is primarily dependent on the quality of the mesh.
The better the mesh looks, the better it is.
A good-looking mesh should have well-shaped elements, and the transition between densities should be smooth and gradual without skinny, distorted elements.
FEA Pre-Processing - Example
Coarse mesh Refined mesh - better?…probably…depends on the loading and boundary conditions
FEA Pre-Processing
Finite elements supported by most finite-element codes:
FEA Pre-Processing
Material Properties
Material properties can be specified for element regions, elements or even within elements in most large-scale FEA codes.
The material properties required for an isotropic, linear static FEA are: Young’s modulus (E), Poisson’s ratio (v), and shear modulus (G).
G = E / 2(1+v)
Only two of the three properties can be specified independently.
FEA Pre-Processing
Nonlinear Material Properties
A multi-linear model requires the input of stress-strain data pairs to essentially communicate the stress-strain curve from testing to the FE model
Highly deformable, low stiffness, incompressible materials, such as rubber and other synthetic elastomers require distortional and volumetric constants or a more complete set of tensile, compressive, and shear force versus stretch curve.
A creep analysis requires time and temperature dependent creep properties. Plastic parts are extremely sensitive to this phenomenon
FEA Pre-ProcessingBoundary Conditions
In FEA, the name of the game is “boundary condition”, that is calculating the load and constraints that each component experiences in its working environment.
The results of FEA should include a complete discussion of the boundary conditions
Boundary Conditions
Loads
Loads are used to represent inputs to the system. They can be in the forms of forces, moments, pressures, temperature, or accelerations.
Constraints
Constraints are used as reactions to the applied loads. Constraints can resist translational or rotational deformation induced by applied loads.
Boundary Conditions
Degrees of Freedom
Spatial DOFs refer to the three translational and three rotational modes of displacement that are possible for any part in 3D space. A constraint scheme must remove all six DOFs for the analysis to run.
Elemental DOFs refer to the ability of each element to transmit or react to a load. The boundary condition cannot load or constrain a DOF that is not supported by the element to which it is applied.
Post-ProcessingView AnimatedDisplacements
View DisplacementFringe Plot
View Stress Fringe Plot
View Results SpecificTo the Analysis
Review BoundaryConditions
Review Load Magnitudesand Units
Review Mesh Density and Quality of Elements
Does the shape of deformations make sense?
Are magnitudes in line with your expectations?
Is the quality and mag. Of stresses acceptable?
Yes
Yes
No
No
No
Yes
