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MANE 4240 & CIVL 4240Introduction to Finite Elements
FEM Discretization of 2D Elasticity
Prof. Suvranu De
Reading assignment:
Lecture notes
Summary:
• FEM Formulation of 2D elasticity (plane stress/strain)•Displacement approximation•Strain and stress approximation•Derivation of element stiffness matrix and nodal load vector•Assembling the global stiffness matrix
• Application of boundary conditions• Physical interpretation of the stiffness matrix
Recap: 2D Elasticity
x
y
Su
ST
Volume (V)u
v
x
px
py
Xa dV
Xb dVVolume element dV Su: Portion of the
boundary on which displacements are prescribed (zero or nonzero)
ST: Portion of the boundary on which tractions are prescribed (zero or nonzero)
Examples: concept of displacement field
x
y
3
2 1
4
2
2
Example
For the square block shown above, determine u and v for the following displacements
x
y
1
4
Case 1: Stretch Case 2: Pure sheary
2
21/2
Solution
Case 1: Stretch
2yv
xu
Check that the new coordinates (in the deformed configuration)
2
2
'
'
yvyy
xuxx
Case 2: Pure shear/ 4
0u yv
Check that the new coordinates (in the deformed configuration)
'
'
/ 4x x u x y
y y v y
y)(x,vy)(x,u
u
xy
y
x
xy
y
x0
0
xy
y
x
uDDu
yxuu
LawStrain -StressRelationnt Displaceme-Strain
),(fieldntDisplaceme
Recap: 2D Elasticity
2100
0101
1 2
ED
For plane stress(3 nonzero stress components)
22100
0101
211
ED
For plane strain(3 nonzero strain components)
VinXT 0 Equilibrium equations
Boundary conditions
1. Displacement boundary conditions: Displacements are specified on portion Su of the boundary
uspecified Sonuu
2. Traction (force) boundary conditions: Tractions are specified on portion ST of the boundaryNow, how do I express this mathematically?
Strong formulation
But in finite element analysis we DO NOT work with the strong formulation (why?), instead we use an equivalent Principle of Minimum Potential Energy
Principle of Minimum Potential Energy (2D)
Definition: For a linear elastic body subjected to body forces X=[Xa,Xb]T and surface tractions TS=[px,py]T, causing displacements u=[u,v]T and strains and stresses , the potential energy is defined as the strain energy minus the potential energy of the loads (X and TS)
U-W
TSS
T
V
T
V
T
dSTudVXu
dV
W
21U
x
y
Su
ST
Volume (V)u
v
x
px
py
Xa dV
Xb dVVolume element dV
Strain energy of the elastic body
V
T
V
T dVDdV 21
21U
DUsing the stress-strain law
In 2D plane stress/plane strain
V xyxyyyxx
V
xy
y
x
T
xy
y
x
V
T
dV
dV
dV
21
21
21U
Principle of minimum potential energy: Among all admissible displacement fields the one that satisfies the equilibrium equations also render the potential energy a minimum.
“admissible displacement field”: 1. first derivative of the displacement components exist2. satisfies the boundary conditions on Su
Finite element formulation for 2D:
Step 1: Divide the body into finite elements connected to each other through special points (“nodes”)
x
y
Su
STu
v
x
px
py
Element ‘e’
3
21
4
y
xvu
1
2
3
4
u1
u2
u3
u4
v4
v3
v2
v1
4
4
3
3
2
2
1
1
vuvuvuvu
d
Total potential energy
Potential energy of element ‘e’:
TS
ST
V
T
V
T dSTudVXudV21
eT
ee S ST
V
T
V
Te dSTudVXudV
21
Total potential energy = sum of potential energies of the elements
e
e
This term may or may not be present depending on whether the element is actually on ST
Step 2: Describe the behavior of each element (i.e., derive the stiffness matrix of each element and the nodal load vector).
Inside the element ‘e’
y
x
vu
1
2
3
4
u1
u2
u3
u4
v4
v3
v2
v1
y)(x,vy)(x,u
u
Displacement at any point x=(x,y)
4
4
3
3
2
2
1
1
vuvuvuvu
d
Nodal displacement vector
(x1,y1)
(x2,y2)
(x4,y4)
(x3,y3)
whereu1=u(x1,y1)v1=v(x1,y1)etc
uDDu
LawStrain -StressRelationnt Displaceme-Strain
xy
y
x
xy
y
x0
0
xy
y
x
If we knew u then we could compute the strains and stresses within the element. But I DO NOT KNOW u!!
Hence we need to approximate u first (using shape functions) and then obtain the approximations for and (recall the case of a 1D bar)
This is accomplished in the following 3 Tasks in the next slide
Recall
TASK 1: APPROXIMATE THE DISPLACEMENTS WITHIN EACH ELEMENT
TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN EACH ELEMENT
TASK 3: DERIVE THE STIFFNESS MATRIX OF EACH ELEMENT USING THE PRINCIPLE OF MIN. POT ENERGY
We’ll see these for a generic element in 2D today and then derive expressions for specific finite elements in the next few classes
Displacement approximation in terms of shape functionsdNu
dBD
dBε Strain approximation
Stress approximation
Displacement approximation in terms of shape functions
u
v3
y
xv1
2
3
4
u1
u2
u3
u4
v4v2
v1
TASK 1: APPROXIMATE THE DISPLACEMENTS WITHIN EACH ELEMENT
44332211
44332211
vy)(x,N vy)(x,N vy)(x,N vy)(x,Ny)(x,vu y)(x,Nu y)(x,Nu y)(x,Nu y)(x,Ny)(x,u
Displacement approximation within element ‘e’
44332211
44332211
vy)(x,N vy)(x,N vy)(x,N vy)(x,Ny)(x,vu y)(x,Nu y)(x,Nu y)(x,Nu y)(x,Ny)(x,u
4
4
3
3
2
2
1
1
4321
4321
vuvuvuvu
N0N0N0N00N0N0N0N
y)(x,vy)(x,u
u
dNu
We’ll derive specific expressions of the shape functions for different finite elements later
TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN EACH ELEMENT
...... vy)(x,N
uy)(x,Ny)(x,vy)(x,u
vy)(x,N
vy)(x,N
v y)(x,N
vy)(x,Ny)(x,v
u y)(x,N
u y)(x,N
u y)(x,N
u y)(x,Ny)(x,u
11
11
xy
44
33
22
11
y
44
33
22
11
x
xyxy
yyyyy
xxxxx
Approximation of the strain in element ‘e’
4
4
3
3
2
2
1
1
B
44332211
4321
4321
xy
vuvuvuvu
y)(x,Ny)(x,Ny)(x,Ny)(x,N y)(x,N y)(x,Ny)(x,Ny)(x,N
y)(x,N0
y)(x,N0
y)(x,N0
y)(x,N0
0y)(x,N
0y)(x,N
0 y)(x,N
0y)(x,N
xyxyxyxy
yyyy
xxxx
y
x
dBε
Compact approach to derive the B matrix:
NB
dBdNRelationnt Displaceme-StraindNufieldntDisplaceme
u
Stress approximation within the element ‘e’
DLawStrain -Stress
BD
Potential energy of element ‘e’:
TASK 3: DERIVE THE STIFFNESS MATRIX OF EACH ELEMENT USING THE PRINCIPLE OF MININUM POTENTIAL ENERGY
Lets plug in the approximations
eT
ee S ST
V
T
V
Te dSTudVXudV
21
dNu dBDdBε
eT
ee S ST
V
T
V
Te dSTdVXdV dNdNdBdBD
21)d(
fk
dSTdVXdV
dSTdVXdV
TTe
f
S ST
V
TT
k
V
TT
S STT
V
TT
V
TTe
eT
ee
eT
ee
ddd21)d(
NNddBDBd21
NdNddBDBd21)d(
Rearranging
From the Principle of Minimum Potential Energy
0dd
)d(
fke
Discrete equilibrium equation for element ‘e’ fk d
eV
k dVBDBT
Element stiffness matrix for element ‘e’
Element nodal load vector
S
eT
b
e
f
S ST
f
V
T dSTdVXf NN
Due to body force Due to surface traction
STe
e
For a 2D element, the size of the k matrix is 2 x number of nodes of the element
Question: If there are ‘n’ nodes per element, then what is the size of the stiffness matrix of that element?
If the element is of thickness ‘t’
eA
k dABDBt T
Element nodal load vector
S
eT
b
e
f
l ST
f
A
T dlTdAXf NtNt
Due to body force Due to surface traction
For a 2D element, the size of the k matrix is 2 x number of nodes of the element
t
dA dV=tdA
The properties of the element stiffness matrix
1. The element stiffness matrix is singular and is therefore non-invertible2. The stiffness matrix is symmetric3. Sum of any row (or column) of the stiffness matrix is zero! (why?)
eV
k dVBDBT
The B-matrix (strain-displacement) corresponding to this element is
We will denote the columns of the B-matrix as
Computation of the terms in the stiffness matrix of 2D elements
x
y
(x,y)
v
u
1 2
34v4
v3
v2v1
u1u2
u3u4
1 2 3 4
1 2 3 4
1 1 2 2 3 3 4 4
N (x,y) N (x,y) N (x,y) N (x,y)0 0 0 0
N (x,y) N (x,y) N (x,y) N (x,y)0 0 0 0
N (x,y) N (x,y) N (x,y) N (x,y) N (x,y) N (x,y) N (x,y) N (x,y)
x x x x
y y y y
y x y x y x y x
u1 v1 u2 v2u3 u4
v3 v4
1 1
1
1
11
N (x,y) 0N (x,y)0 ; ; and so on...
N (x,y)N (x,y)
u v
xB B
y
yx
eV
k dVBDBT
11 12 13 14 15 16 17 18
21 22 23 24 25 26 27 28
31 32 33 34 35 36 37 38
41 42 43 44 45 46 47 48
51 52 53 54 55 56 57 58
61 62 63 64 65 66 67 68
71 72 73 74 75 76 77 78
81 82 83 84 85 86 87 88
k k k k k k k kk k k k k k k kk k k k k k k kk k k k k k k k
kk k k k k k k kk k k k k k k kk k k k k k k kk k k k k k k k
u1
v1
u2
v2
u3
u4
v3
v4
u1 v1u2 v2 u3
u4v3v4
The stiffness matrix corresponding to this element is
which has the following form
1 1 1 1 1 2
1 1 1 1
T T T11 12 13
T T21 21
B D B dV; B D B dV; B D B dV,...
B D B dV; B D B dV;.....
e e e
e e
u u u v u uV V V
v u v vV V
k k k
k k
The individual entries of the stiffness matrix may be computed as follows
Step 3: Assemble the element stiffness matrices into the global stiffness matrix of the entire structure
For this create a node-element connectivity chart exactly as in 1D
ELEMENT Node 1 Node 2 Node 3
1 1 2 3
2 2 3 4
u
v3
y
xv
1
2
4
3
u2
u4
u3
u1
v1v4
v2
Element #1
Element #2
Stiffness matrix of element 1 Stiffness matrix of element 2
There are 6 degrees of freedom (dof) per element (2 per node)
)2(k
)1(k
u1
u2
v1
v2
u3
v3
u1 v1 u2 v2 u3v3 u2
u3
v2
v3
u4
v4
v2 u3 v3 u4 v4u2
88
K
Global stiffness matrix
How do you incorporate boundary conditions?Exactly as in 1D
)2(k
)1(k
u1
v1
u2
v2
u3
v3
u4
v4
u1 v1 u2 v2 u3 u4v4v3
Finally, solve the system equations taking care of the Finally, solve the system equations taking care of the displacement boundary conditionsdisplacement boundary conditions..
Physical interpretation of the stiffness matrix
3y
3x
2y
2x
1y
1x
3
3
2
2
1
1
666564636261
565554535251
464544434241
363534333231
262524232221
161514131211
ffffff
vuvuvu
d
kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk
fk
Consider a single triangular element. The six corresponding equilibrium equations ( 2 equilibrium equations in the x- and y-directions at each node times the number of nodes) can be written symbolically as
x
yu3
v3
v1
u1
u2
v2
2
3
1
Choose u1 = 1 and rest of the nodal displacements = 0
3y
3x
2y
2x
1y
1x
61
51
41
31
21
11
ffffff
kkkkkk
Hence, the first column of the stiffness matrix represents the nodal loads when u1=1 and all other dofs are fixed. This is the physical interpretation of the first column of the stiffness matrix. Similar interpretations exist for the other columns
x
y
u1=1
2
3
1
ijk = “Force” at d.o.f ‘i’ due to unit displacement at d.o.f ‘j’ keeping all the other d.o.fs fixed
Now consider the ith row of the matrix equation fk d
ix654321 fd iiiiii kkkkkk
This is the equation of equilibrium at the ith dof
Consistent and Lumped nodal loads
Recall that the nodal loads due to body forces and surface tractions
eT
e S ST
SV
T
bdSTfdVXf N;N
These are known as “consistent nodal loads”1. They are derived in a consistent manner using the Principle of Minimum Potential Energy2. The same shape functions used in the computation of the stiffness matrix are employed to compute these vectors
Example
1
2
3
x
yp per unit area
Traction distribution on the 1-2-3 edgepx= ppy= 0
We’ll see later that
232
22
221 2)(;;
2)(
bybyN
bybN
bybyN
b
b
N1 N2 N3
The consistent nodal loads are
32)(
3432
)(
233
2
22
22
211
pbdybybypdyNpF
pbdybybpdyNpF
pbdybybypdyNpF
b
b
b
bx
b
b
b
bx
b
b
b
bx
1
2
3
x
y
b
b
pb/3
pb/3
4pb/3
The lumped nodal loads are
2
2
3
2
1
pbF
pbF
pbF
x
x
x
1
2
3
x
y
b
b
pb/2
pb/2
pb
Lumping produces poor results and will not be pursued further
Displacement approximation in terms of shape functions
Strain approximation in terms of strain-displacement matrix
Stress approximation
Summary: For each element
Element stiffness matrix
Element nodal load vector
dNu
dBD
dBε
eV
k dVBDBT
S
eT
b
e
f
S ST
f
V
T dSTdVXf NN
