Transcript MANE 4240 & CIVL 4240 Introduction to Finite Elements

MANE 4240 & CIVL 4240
Introduction to Finite Elements
Prof. Suvranu De
Convergence of analysis
results
Reading assignment:
Lecture notes
Summary:
• Concept of convergence
• Criteria for monotonic convergence :
completeness (rigid body modes + constant strain)
+
compatibility
• Incompatible elements and the patch test
• Rate of convergence
Errors that affect finite element solution results
Type of error
Source
1. Discretization error
Use of FE interpolations for
geometry and solution variables
2. Numerical integration
Evaluation of FE element
matrices and vectors using
numerical integration
This error is due to the finite
precision arithmetic used in
digital computers
3. Round off
What is “convergence”?
Physical system
Mathematical model
FE model
“Convergence” of FE solution results to the
exact solution of the mathematical model
FE scheme exhibits convergence if the
Discretization error →0 as the mesh is made
infinitely fine (i.e., element size →0)
Mesh refinement
h-refinement
p-refinement
h=element size
p=polynomial order
Convergence in energy and displacement
u : exact displacement solution to a problem that makes the
potential energy of the system a minimum
corresponding stress  (u )
 (u )
and strain
Exact strain energy of the body
1
T
U     dV
2 V
uh : FE solution (‘h’ refers to the element size)
corresponding stress h (u h )
and strain
 h (u h )
Approximate strain energy of the body
1
T
U h    h  h dV
2 V
Calculation of strain energies
Example:
Consider a linear elastic bar with varying cross section
2
2
1
x
80cm
P=3E/80
E: Young’s modulus
x 

A( x)  1   sqcm
 40 
The governing differential (equilibrium) equation
E
d 
du 
A
(
x
)
 0 for x  (0,80)
dx 
dx 
Boundary conditions
u ( x  0)  0
du
EA
dx
x 80 cm
3E
P
80
Analytical solution


3
1 
exact
u ( x)   1 

2  1 x 


40 

Eq(1)
The exact strain energy of the system is
U
2
 du ( x) 
1
1
3E
39E

Adx

EA
dx




2 x 0
2 x 0 
dx
160 2080

80
80
exact
If we discretize the problem using a single linear finite element, the stiffness
matrix is
E
80
A( x)dx  1 1
K
 1 1 
802


13E  1 1



240  1 1 
x 0
The strain energy of the FE system is
1 80
1 T
27 E
T
U h    h h Adx  d Kd 
sin ce d  0 9 /13
2 x 0
2
2080
Note
U  Uh
Convergence in strain energy
U  U h as h  0
Monotonic convergence
Nonmonotonic convergence
Convergence in displacement
u  uh
0

 u - u   v - v   dV  0 as
2
h
V
Monotonic convergence
Nonmonotonic convergence
2
h
h0
Criteria for monotonic convergence
1. COMPLETENESS
2. COMPATIBILITY
© 2002 Brooks/Cole Publishing / Thomson Learning™
CONDITION 1. COMPLETENESS
This requires that the displacement interpolation functions
must be chosen so that the elements can represent
1. Rigid body modes
2. Constant strain states
Rigid body modes
The # of rigid body modes of an element = # of zero
eigenvalues of the element stiffness matrix
Constant strain states

Strain computed using linear finite elements
Actual variation of strain
x
Mathematical implication of the two conditions (rigid body
modes + constant strain state)
Inside a finite element (of any order) in 1D
u( x)   Ni ( x)ui
i
but this is just a polynomial…
u( x)  a0  a1x  a2 x2 
Hence
u ( x)   Ni ( x)ui   Ni ( x)  a0  a1 xi  a2 xi 2 
i

i
 a0  Ni ( x)  a1  Ni ( x) xi  a2  N i ( x) xi 2 
i
i
1
i
x
 a0  a1 x  a2  Ni ( x) xi 2 
i
The requirement for completeness in 1D is that the displacement
approximation be at least a linear polynomial of degree (k=1),
ie any 2 node element and higher is complete
Mathematical implication of the two conditions (rigid body
modes + constant strain state)
Inside a finite element (of any order) in 2D
u( x)   Ni ( x, y)ui
i
but this is just a polynomial…
u( x, y)  a0  a1 x  a2 y 
Hence
u ( x, y )   N i ( x, y )ui   N i ( x, y )  a0  a1 xi  a2 yi 
i

i
 a0  Ni ( x, y )  a1  N i ( x, y ) xi  a2  N i ( x, y ) yi 
i
i
1
i
x
 a0  a1 x  a2 y 
The requirement for completeness in 1D is that the displacement
approximation be at least a linear polynomial of degree (k=1).
Mathematical implication of the two conditions (rigid body
modes + constant strain state)
The element displacement approximation must be at least a
COMPLETE polynomial of degree one
1
1
x
x

x
2
1D
y
x 2 xy

2D
k=1
y2
In 2D, the minimum displacement assumption needs to be
u  1   2 x   3 y
v  1   2 x   3 y
1  0 all other coeffs  0 Translation along x
1  0 all other coeffs  0 Translation along y
1  2  0 and 1  3  0 but 3   2  0
Rigid body rotation about z-axis
CONDITION 2. COMPATIBILITY
The assumed displacement variations are continuous within
elements and across inter-element boundaries
Ensures that strains are bounded within elements and across
element boundaries.
If ‘u’ is discontinuous across element boundaries then
the strains blow up in-between elements and this leads
to erroneous contributions to the potential energy of the
structure
Physical meaning: no gaps/cracks open up when the finite
element assemblage is loaded
Nonconforming elements and the patch test
Conforming = compatible
Nonconforming = incompatible
Ideal: Conforming elements
Observation: Certain nonconforming elements also give good
results, at the expense of nonmonotonic convergence
Nonconforming elements:
• satisfy completeness
• do not satisfy compatibility
• result in at least nonmonotonic convergence if the element
assemblage as a whole is complete, i.e., they satisfy the
PATCH TEST
PATCH TEST:
1. A patch of elements is subjected to the minimum
displacement boundary conditions to eliminate all rigid body
motions
2. Apply to boundary nodal points forces or displacements
which should result in a state of constant stress within the
assemblage
3. Nodes not on the boundary are neither loaded nor restrained.
4. Compute the displacements of nodes which do not have a
prescribed value
5. Compute the stresses and strains
The patch test is passed if the computed stresses and strains
match the expected values to the limit of computer precision.
NOTES:
1. This is a great way to debug a computer code
2. Conforming elements ALWAYS pass the patch test
3. Nodes not on the boundary are neither loaded nor restrained.
4. Since a patch may also consist of a single element, this test
may be used to check the completeness of a single element
5. The number of constant stress states in a patch test depends
on the actual number of constant stress states in the
mathematical model (3 for plane stress analysis. 6 for a full 3D
analysis)
CONVERGENCE RATE
This is a measure of how fast the discretization error goes to
zero a the mesh is refined
Convergence rate depends on the order of the complete
polynomial (k) used in the displacement approximation
1
x
x2
x
3
y
y2
xy
2
x y

xy
k=1
2
y
3
k=2
k=3
It can be shown that for (1) a sufficiently refined mesh and
(2) for problems whose analytical solution does not contain
singularities
Convergence in strain energy : order 2k
U  Uh  C h2k
Convergence in displacements : order p=k+1
u  uh
k 1

C
h
1
0
C and C1 are constants independent of ‘h’ but dependent on
1. the analytical solution
2. material properties
3. type of element used
Ex: for a domain discretized using 4 node plane stress/strain
elements (k=1)
2
U Uh  C h
u  uh
0
 C1 h 2
log U  U h
slope = 2
log h
Large C
shifts
curve up
Important property of finite element solution:
When the conditions of monotonic convergence are satisfied
(compatibility and completeness) the finite element strain energy
always underestimates the strain energy of the actual structure
Strain energy of mathematical model
Strain energy of FE model