MANE 4240 & CIVL 4240
Introduction to Finite Elements
Reading assignment:
Lecture notes
Summary:
Prof. Suvranu De
FEM Discretization of
2D Elasticity
• 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
Example
Recap: 2D Elasticity
y
2
Volume
element dV
Xb dV
Xa dV
v
y
py
px
Volume (V)
u
ST
x
Su
x
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)
2
1
3
4
2
x
For the square block shown above, determine u and v for the
following displacements
Case 2: Pure shear
y
y
Case 1: Stretch
1/2
2
4
2
Examples: concept of displacement field
1
x
1
Solution
Recap: 2D Elasticity
u=x
Case 1: Stretch
y
v=−
2
Check that the new coordinates (in the deformed configuration)
x' = x + u = 2x
y
y' = y + v =
2
Case 2: Pure shear
u = y/4
v=0
Check that the new coordinates (in the deformed configuration)
x' = x + u = x + y / 4
y' = y + v = y
⎧u (x, y)⎫
u=⎨
⎬
⎩ v (x, y)⎭
⎤
⎡∂
Stress - Strain Law σ = Dε = D∂u
0⎥
⎢
⎥
⎢ ∂x
⎧εx ⎫
⎧σ x ⎫
∂⎥
∂=⎢0
⎪ ⎪
⎪ ⎪
⎢
ε = ⎨ε y ⎬
σ = ⎨σ y ⎬
∂y ⎥
⎢∂
⎪γ ⎪
⎪τ ⎪
∂⎥
⎥
⎢
⎩ xy ⎭
⎩ xy ⎭
⎣⎢ ∂y ∂x ⎦⎥
For plane stress
For plane strain
(3 nonzero stress components) (3 nonzero strain components)
⎤
⎡
⎤
⎡
ν
0 ⎥
1 ν
0 ⎥
⎢1 − ν
E
E ⎢
⎥
⎢
D
ν
ν
1
−
0
=
⎢ν 1
D=
0 ⎥
(1 + ν )(1 − 2ν ) ⎢
1 − 2ν ⎥
1 −ν 2 ⎢
1 −ν ⎥
0
0
⎥
⎢
⎥
⎢0 0
⎣
2 ⎦
⎣
2 ⎦
Displacement field u = u ( x, y )
Strain - Displacement Relation ε = ∂u
Strong formulation
Principle of Minimum Potential Energy (2D)
Equilibrium equations
∂ σ + X = 0 in V
T
Boundary conditions
1. Displacement boundary conditions: Displacements are specified on
portion Su of the boundary
u =u
specified
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 ε and stresses σ, the potential
energy Π is defined as the strain energy minus the potential energy
of the loads (X and TS)
on S u
2. Traction (force) boundary conditions: Tractions are specified on
portion ST of the boundary
Now, how do I express this mathematically?
Π=U-W
But in finite element analysis we DO NOT work with the strong
formulation (why?), instead we use an equivalent Principle of
Minimum Potential Energy
2
Volume
element dV
Xb dV
Xa dV
v
y
Strain energy of the elastic body
py
px
Volume (V)
u
ST
x
x
1
1
σ T ε dV = ∫ ε T D ε dV
2 ∫V
2 V
In 2D plane stress/plane strain
1
σ T ε dV
2 ∫V
T
1
σ T ε dV
2 ∫V
W = ∫ u X dV + ∫ u T S dS
T
V
U=
U=
Su
U=
Using the stress-strain law σ = D ε
T
ST
Principle of minimum potential energy: Among all admissible
displacement fields the one that satisfies the equilibrium equations
also render the potential energy Π a minimum.
“admissible displacement field”:
1. first derivative of the displacement components exist
2. satisfies the boundary conditions on Su
⎧σ x ⎫ ⎧ ε x ⎫
1 ⎪ ⎪ ⎪ ⎪
= ∫ ⎨σ y ⎬ ⎨ ε y ⎬ dV
2 V⎪ ⎪ ⎪ ⎪
⎩τ xy ⎭ ⎩γ xy ⎭
1
= ∫ (σ xε x + σ yε y + τ xyγ xy ) dV
2 V
Finite element formulation for 2D:
Step 1: Divide the body into finite elements connected to each
other through special points (“nodes”)
py
v3
3
px
4
3
u3
⎧u 1 ⎫
v4
2
⎪v ⎪
v2
Element ‘e’
v
⎪ 1⎪
1
4
⎪u 2 ⎪
u
u4
ST
u2
⎪ ⎪
v1
2
⎪v 2 ⎪
y
d=⎨ ⎬
x
y
⎪u 3 ⎪
Su
u1
⎪v 3 ⎪
1
⎪ ⎪
v
x
x
⎪u 4 ⎪
u
⎪v ⎪
⎩ 4⎭
3
Step 2: Describe the behavior of each element (i.e., derive the
stiffness matrix of each element and the nodal load vector).
Total potential energy
Π=
1
σ T ε dV − ∫ u T X dV − ∫ u T T S dS
V
ST
2 ∫V
Potential energy of element ‘e’:
1
T
T
T
Π e = ∫ e σ ε dV − ∫ e u X dV − ∫ e u T S dS
ST
V
2 V
This term may or may not be present
depending on whether the element is
actually on ST
Total potential energy = sum of potential energies of the elements
Π = ∑Πe
e
Recall
Strain - Displacement Relation ε = ∂u
Stress - Strain Law σ = Dε = D∂u
⎧ε x ⎫
⎪ ⎪
ε = ⎨ε y ⎬
⎪γ ⎪
⎩ xy ⎭
⎧σ x ⎫
⎪ ⎪
σ = ⎨σ y ⎬
⎪τ ⎪
⎩ xy ⎭
⎡∂
⎢
⎢ ∂x
∂=⎢0
⎢
⎢∂
⎢
⎣⎢ ∂y
⎤
0⎥
⎥
∂⎥
∂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 ε and σ (recall the case of a 1D bar)
Inside the element ‘e’
v3
(x3,y3) 3
v4
v2
Displacement at any point x=(x,y)
⎧u (x, y)⎫
u=⎨
⎬
⎩ v (x, y)⎭
(x4,y4)
Nodal displacement vector
4
u4
⎧u 1 ⎫
u2
v
v1
2
⎪v ⎪ where
u (x2,y2)
⎪ 1 ⎪ u1=u(x1,y1)
y
⎪u 2 ⎪ v =v(x ,y )
1 1
u
⎪ ⎪ 1
1(x ,y ) 1
⎪v 2 ⎪ etc
1 1
d=⎨ ⎬
x
⎪u 3 ⎪
⎪v 3 ⎪
⎪ ⎪
⎪u 4 ⎪
⎪v ⎪
⎩ 4⎭
u3
TASK 1: APPROXIMATE THE DISPLACEMENTS WITHIN
EACH ELEMENT
Displacement approximation in terms of shape functions
u=Nd
TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN
EACH ELEMENT
Strain approximation
ε=Bd
Stress approximation
σ = DB d
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
This is accomplished in the following 3 Tasks in the next slide
4
TASK 1: APPROXIMATE THE DISPLACEMENTS
WITHIN EACH ELEMENT
Displacement approximation in terms of shape functions
v3
3
u3
v4
v2 Displacement approximation within element ‘e’
4
u4
u2
v1
2
u (x,y)≈ N1(x,y)u1 + N2(x,y)u2 + N3(x,y)u3 + N4(x,y)u4
y
u1
v (x,y)≈ N1(x,y)v1 + N2(x,y)v2 + N3(x,y) v3 + N4(x,y)v4
1
v
x
u
u (x,y)≈ N1(x,y)u1 + N2(x,y)u2 + N3(x,y)u3 + N4(x,y)u4
v (x,y)≈ N1(x,y)v1 + N2(x,y)v2 + N3(x,y) v3 + N4(x,y)v4
⎧u (x, y)⎫ ⎡ N1
u=⎨
⎬=⎢
⎩ v (x, y)⎭ ⎣ 0
u=Nd
0
N2
0
N3
0
N4
N1
0
N2
0
N3
0
⎧u 1 ⎫
⎪v ⎪
⎪ 1⎪
⎪u 2 ⎪
⎪ ⎪
0 ⎤ ⎪v 2 ⎪
⎨ ⎬
N 4 ⎥⎦ ⎪u 3 ⎪
⎪v 3 ⎪
⎪ ⎪
⎪u 4 ⎪
⎪v ⎪
⎩ 4⎭
We’ll derive specific expressions of the shape functions for
different finite elements later
TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN
EACH ELEMENT
Approximation of the strain in element ‘e’
∂N (x, y)
∂N 2 (x, y)
∂N (x, y)
∂u (x, y) ∂N1(x, y)
u1 +
u2 + 3
u3 + 4
u4
≈
∂x
∂x
∂x
∂x
∂x
∂N (x, y)
∂N 2 (x, y)
∂N (x, y)
∂v (x, y) ∂N1(x, y)
εy =
≈
v1 +
v2 + 3
v3 + 4
v4
∂y
∂y
∂y
∂y
∂y
∂N (x, y)
∂u (x, y) ∂v (x, y) ∂N1(x, y)
γ xy =
+
≈
u1 + 1
v1 + ......
∂y
∂x
∂y
∂x
εx =
⎧εx ⎫
⎪ ⎪
ε = ⎨ε y ⎬
⎪γ xy ⎪
⎩ ⎭
⎧u1 ⎫
⎪v ⎪
⎤⎪ 1 ⎪
⎡ ∂N1(x, y)
∂N3(x, y)
∂N 2 (x, y)
∂N 4 (x, y)
0
0
0
0
⎥⎪u 2 ⎪
⎢
∂x
∂x
∂x
⎥⎪ ⎪
⎢ ∂x
∂N3(x, y)
∂N1(x, y)
∂N 2 (x, y)
∂N 4 (x, y)⎥⎪v 2 ⎪
0
0
0
=⎢ 0
⎨ ⎬
⎢
∂y
∂y
∂y
∂y ⎥⎪u 3 ⎪
⎢ ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y)⎥
3
3
1
2
2
4
4
⎥⎪v3 ⎪
⎢ 1
∂x
∂y
∂x
∂y
∂x
∂y
∂x ⎥⎪ ⎪
⎢⎣ ∂y
⎦⎪u 4 ⎪
B
⎪v ⎪
⎩ 4⎭
ε=Bd
5
Compact approach to derive the B matrix:
Displacement field u = N d
Strain - Displacement Relation ε = ∂u = (∂ N )d = Bd
TASK 3: DERIVE THE STIFFNESS MATRIX OF EACH
ELEMENT USING THE PRINCIPLE OF MININUM
POTENTIAL ENERGY
Potential energy of element ‘e’:
1
σ T ε dV − ∫ e u T X dV − ∫ e u T T S dS
ST
V
2 ∫V e
Lets plug in the approximations
u=Nd
Π e (d ) ≈
Stress - Strain Law σ = Dε
⇒σ = D Bε
B=∂ N
Πe =
Stress approximation within the element ‘e’
ε=Bd
σ = DB d
1
(DB d )T (B d ) dV − ∫V e (N d )T X dV − ∫S e (N d )T T S dS
T
2 ∫V e
Rearranging
1 T
T
T
T
T
T
Π e (d) ≈ d ⎛⎜ ∫ e B D B dV ⎞⎟ d − d ⎛⎜ ∫ e N X dV ⎞⎟ − d ⎛⎜ ∫ e N T S dS ⎞⎟
⎠
⎝V
⎠
⎝ ST
⎠
2 ⎝V
1 T⎛
T⎛
T
T
T
⎞
⎞
= d ⎜ ∫ e B D B dV ⎟ d − d ⎜ ∫ e N X dV + ∫ e N T S dS ⎟
ST
⎝V
2 ⎝V
⎠
⎠
k
f
1 T
T
⇒ Π e (d ) = d k d − d f
2
From the Principle of Minimum Potential Energy
∂Π e (d)
=k d− f =0
∂d
Discrete equilibrium equation for element ‘e’
kd= f
6
Element stiffness matrix for element ‘e’
k = ∫ e B D B dV
T
V
If the element is of thickness ‘t’
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?
k = ∫ e t B D B dA
T
A
dA
t
f = ∫ e t N X dA + ∫ e t N T S dl
A
lT
f = ∫ e N X dV + ∫ e N T S dS
V
ST
f
T
b
f
S
STe
T
T
f
f
b
S
e
Due to body force
Due to body force
dV=tdA
Element nodal load vector
Element nodal load vector
T
For a 2D element, the size of the k matrix is
2 x number of nodes of the element
Due to surface traction
Due to surface traction
The properties of the element stiffness matrix
Computation of the terms in the stiffness matrix of 2D elements
v4
k = ∫ e B D B dV
T
v3
4
u4
The B-matrix (strain-displacement) corresponding to this element is
3
u3
V
u1
v
1. The element stiffness matrix is singular and is therefore noninvertible
2. The stiffness matrix is symmetric
3. Sum of any row (or column) of the stiffness matrix is zero!
(why?)
y
v1 (x,y)
v2
u
2
1 u1
u2
⎡ ∂N1 (x,y)
⎢
∂x
⎢
⎢
0
⎢
⎢
⎢ ∂N1 (x,y)
⎢
∂y
⎢⎣
v1
u2
v2
0
∂N 2 (x,y)
∂x
0
∂N1 (x,y)
∂y
∂N1 (x,y)
∂x
0
∂N 2 (x,y)
∂y
∂N2 (x,y)
∂y
∂N2 (x,y)
∂x
u3
∂N 3 (x,y)
∂x
0
∂N 3 (x,y)
∂y
v3
0
∂N 3 (x,y)
∂y
∂N 3 (x,y)
∂x
u4
v4
∂N4 (x,y)
∂x
⎤
0
⎥
⎥
∂N4 (x,y) ⎥
⎥
∂y
⎥
∂N4 (x,y) ⎥
⎥
∂x
⎥⎦
0
∂N4 (x,y)
∂y
x
We will denote the columns of the B-matrix as
B u1
⎡
⎤
⎡ ∂ N 1 (x,y) ⎤
⎢
⎥
0
⎢
⎥
⎢
⎥
∂x
⎢
⎥
∂ N 1 (x,y) ⎥
⎢
0
; and so on...
=⎢
⎥ ; B v1 = ⎢
⎥
∂y
⎢ ∂ N (x,y) ⎥
⎢
⎥
1
⎢
⎥
∂ N 1 (x,y) ⎥
⎢
∂y
⎣⎢
⎦⎥
⎢⎣
⎥⎦
∂x
7
The stiffness matrix corresponding to this element is
k = ∫ e B D B dV
T
Step 3: Assemble the element stiffness matrices into the global
stiffness matrix of the entire structure
which has the following form
V
u1
v2
u3
u4
v4
k12
k 22
k13
k 23
k14
k 24
k15
k 25
k16
k 26
k17
k 27
k 32
k 42
k 33
k 43
k 34
k 44
k 35
k 45
k 36
k 46
k 37
k 47
k 52
k 62
k 53
k 63
k 54
k 64
k 55
k 65
k 56
k 66
k 57
k 67
k 72
k 82
k 73
k 83
k 74
k 84
k 75
k 85
k 76
k 86
k 77
k 87
k18 ⎤
k 2 8 ⎥⎥
k 38 ⎥
⎥
k 48 ⎥
k 58 ⎥
⎥
k 68 ⎥
k 78 ⎥
⎥
k 8 8 ⎦⎥
v1
⎡ k11
⎢k
⎢ 21
⎢ k 31
⎢
k
k = ⎢ 41
⎢ k 51
⎢
⎢ k 61
⎢k
⎢ 71
⎢⎣ k 8 1
u2
v3
For this create a node-element connectivity chart exactly as in 1D
v3
u1
Element #1
v1
v2
v3
V
u4
1
2
3
2
2
3
4
u2
2
v
u
T
V
4
1
Element #2
v4
k11 = ∫ e Bu1 D Bu1 dV; k12 = ∫ e Bu1 D Bv1 dV; k13 = ∫ e Bu1 D Bu2 dV,...
T
u1
y
u4
u3 ELEMENT Node 1 Node 2 Node 3
v4
1
v2
u3
The individual entries of the stiffness matrix may be computed as follows
T
3
v1
u2
x
V
k 21 = ∫ e Bv1 D Bu1 dV; k21 = ∫ e Bv1 D Bv1 dV;.....
T
V
T
V
Stiffness matrix of element 1
u1 v1 u2 v2 u3 v3
k
(1 )
⎡
⎢
⎢
⎢
= ⎢
⎢
⎢
⎢
⎣⎢
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦⎥
u1
v1
u2
v2
u3
v3
Stiffness matrix of element 2
u v u v u
⎡ 2 2 3 3 4 v4 ⎤
⎥
⎢
⎥
⎢
⎥
⎢
(2)
k
= ⎢
⎥
⎥
⎢
⎥
⎢
⎥
⎢
⎦⎥
⎣⎢
There are 6 degrees of freedom (dof) per element (2 per node)
Global stiffness matrix
u2
v2
u3
v3
u4
v4
k
(1)
u1 v1 u2 v2 u3 v3 u v4
4
⎡
⎢
⎢
⎢
K = ⎢
⎢
⎢
⎢
⎣⎢
⎤ u1
⎥ v1
⎥ u
⎥ 2
⎥ v2
⎥ u3
⎥ v3
⎥ u4
⎦⎥ 8v×48
k
( 2)
How do you incorporate boundary conditions?
Exactly as in 1D
8
Physical interpretation of the stiffness matrix
Finally, solve the system equations taking care of the
displacement boundary conditions.
Consider a single triangular element. The six corresponding
equilibrium equations ( 2 equilibrium equations in the x- and ydirections at each node times the number of nodes) can be written
symbolically as
kd= f
v1
u1
1
v2
3
y
u2
2
x
Choose u1 = 1 and rest of the nodal displacements = 0
⎧k11⎫ ⎧f1x ⎫
⎪k ⎪ ⎪f ⎪
⎪ 21⎪ ⎪ 1y ⎪
⎪⎪k31⎪⎪ ⎪⎪f2x ⎪⎪
⎨ ⎬=⎨ ⎬
⎪k41⎪ ⎪f2y ⎪
⎪k51⎪ ⎪f3x ⎪
⎪ ⎪ ⎪ ⎪
⎩⎪k61⎭⎪ ⎩⎪f3y ⎭⎪
v3
u3
⎡k11
⎢k
⎢ 21
⎢k31
⎢
⎢k41
⎢k51
⎢
⎣⎢k61
k12
k22
k32
k42
k52
k62
k13
k23
k33
k43
k53
k63
k14
k24
k34
k44
k54
k64
k15
k25
k35
k45
k55
k65
k16 ⎤⎧u1 ⎫ ⎧f1x ⎫
⎪ ⎪
k26⎥⎥⎪⎪v1 ⎪⎪ ⎪f1y ⎪
k36⎥⎪⎪u2 ⎪⎪ ⎪⎪f2x ⎪⎪
⎥⎨ ⎬ = ⎨ ⎬
k46⎥⎪v2 ⎪ ⎪f2y ⎪
k56⎥⎪u3 ⎪ ⎪f3x ⎪
⎥⎪ ⎪ ⎪ ⎪
k66⎦⎥⎩⎪v3 ⎭⎪ ⎩⎪f3y ⎭⎪
“Force” at d.o.f ‘i’ due to unit displacement at d.o.f ‘j’
k ij = keeping
all the other d.o.fs fixed
u1=1
1
3
y
Now consider the ith row of the matrix equation k d = f
[ki1
2
x
ki2 ki3 ki4 ki5 ki6 ] d = fix
This is the equation of equilibrium at the ith dof
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
9
Consistent and Lumped nodal loads
Example
p per unit area
y
Recall that the nodal loads due to body forces and surface tractions
f b = ∫ e N X dV ;
T
f
V
S
=∫
ST e
1
b
T
N T S dS
2
x
b
3
These are known as “consistent nodal loads”
1. They are derived in a consistent manner using the Principle of
Minimum Potential Energy
2. The same shape functions used in the computation of the stiffness
matrix are employed to compute these vectors
We’ll see later that
N1 =
y(b + y)
b2 − y2
y(b − y)
; N2 = 2 ; N3 = −
2
2b
b
2b2
N1
The consistent nodal loads are
N2
The lumped nodal loads are
y(b + y)
pb
F1x = ∫ p N1dy = ∫ p
dy =
−b
−b
3
2b2
b
b
4pb
b2 − y2
F2x = ∫ p N2dy = ∫ p 2 dy =
−b
−b
3
b
b
b
y(b − y)
pb
F3x = ∫ p N3dy = −∫ p
dy =
−b
−b
3
2b2
b
Traction distribution on the 12-3 edge
px= p
py= 0
pb
2
F2x = pb
N3
y
F1x =
b
pb
F3x =
2
b
1
pb/2
2
pb
b
3
x
pb/2
Lumping produces poor results and will not be pursued further
y
b
1
pb/3
2
4pb/3
b
3
x
pb/3
10
Summary: For each element
Displacement approximation in terms of shape functions
u=Nd
Strain approximation in terms of strain-displacement matrix
ε=Bd
Stress approximation
σ = DB d
Element stiffness matrix
k = ∫ e B D B dV
T
V
Element nodal load vector
f = ∫ e N X dV + ∫ e N T S dS
V
ST
T
f
T
b
f
S
11