Introduction to Finite Element Method
Tai Hun Kwon
DEPARTMENT OF MECHANICAL ENGINEERING POHANG UNIVERSITY OF SCIENCE &...
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Introduction to Finite Element Method
Tai Hun Kwon
DEPARTMENT OF MECHANICAL ENGINEERING POHANG UNIVERSITY OF SCIENCE & TECHNOLOGY
© 2005 by T. H. Kwon
1. INTRODUCTION
* Numerical Methods 1. Finite Difference Method (FDM) z Pointwise approximation to differential equation (DE) z Array of grid points 2. Finite Element Method (FEM) z Global approximation or integral approximation to DE z Assembly of finite elements (subdomains, subregions) 3. Boundary Element Method (BEM) z Deal with integral equation rather than differential equation z Discretization over boundary only 4. Finite Volume Method (control volume method) 5. Spectral Method (spectral element method)
* FEM is good for
i) various problems (engineering, physical problems) ii) arbitrary geometry (complicated, irregular)
difficult in implementation
© 2005 by T. H. Kwon
1
* Application Examples of FEM z Structural Analysis (steady, timedependent dynamics, eigenvalue)
Beam
Plate and Shell
z Thermal System Analysis
T1
T2
Conduction z Flow Analysis
Flow + Convection Heat transfer
z Thermomechanical Process Analysis
Forging
© 2005 by T. H. Kwon
Rolling
2
Injection Molding
* Notation Convention
[ A] m x n
: m x n matrix
⎣ A⎦1 x n
: 1 x n row matrix
{A}n x 1
: n x 1 column matrix
[C ]m x n = [A]m x l [B ]l x n
: matrix multiplication
l
C ij = ∑ Aik Bkj = Aik Bkj
(Einstein summation convention)
k =1
[A]{x} = { f }
: matrix equation
{x} = [ A]−1 { f }
: inverse matrix
[A][A]−1 = [A]−1 [A] = [I ]
: identity matrix
A ji = Aij
: transposed matrix
[A]{x} = λ{x}
: eigenvalue, eigenvector
T
[[A] − λ[I ]]{x} = 0 det[[ A] − λ[I ]] = 0
: characteristic equation
− λ3 + Iλ2 − IIλ + III = 0 where
I = Aii II =
1 (Aii A jj − Aij A ji ) 2
III = det[ A]
© 2005 by T. H. Kwon
3
: invariants of [ A]
⎧0 : i ≠ j δ ij = ⎨ ⎩1 : i = j ε ijk
: Kronecker delta
⎧ 1 : even permutation ⎪ = ⎨  1 : odd permutation ⎪0 : any other indices are repeated ⎩
det[ A] = A = ε ijk Ai1 A j 2 Ak 3 = ε pqr A1 p A2 q A3r
⎣ A⎦1 x n {B}n x 1 = scalar
{A}m x 1 ⎣B ⎦1 x n = [C ]m x n
© 2005 by T. H. Kwon
= matrix
4
: permutation symbol
* General Procedure of FEM 1. Identify the system (governing) equation.
(usually DE)
L(φ) = 0.
(1.1)
2. Introduce an integral form equation. (Weak form equation) z
Direct Approach
z
Variational Approach
z
Method of Weighted Residual Approach : weak form
∫ ψL(φ)dΩ = 0
Ω
⇒
FEM Formulation
: weak form
∫ L′(ψ) L′′(φ)dΩ = 0
(1.2)
Ω
3. Discretize the domain of interest into elements.
Element Types
y
x
4. Introduce an approximation of the field variable over an element.
Interpolation
φ3
y
φ(x) x
φ1
φ2
φ(x) = N 1 (x)φ1 + N 2 (x)φ 2 + N 3 (x)φ 3
φi Ni
© 2005 by T. H. Kwon
(1.3)
: Nodal values of the field variable : Interpolation functions, Shape functions
5
5. Evaluate the integral form over each element.
Numerical Integral
[K ]e {φ}e = { f }e
(1.4)
6. Assemble the global matrix equation.
Assembly Procedure
[K ]{φ} = {F }
(1.5)
7. Solve the matrix equation to get the unknowns
{φ} = [K ]−1 {F }
(1.6)
8. Calculate the values of interest from the approximate solution. e.g.
∂φ ∂φ , , etc. ∂x ∂x
© 2005 by T. H. Kwon
Solution Techniques
6
2. FINITE ELEMENT FORMULATIONS
2.1 Direct approach for discrete systems Direct approach has the following features: z It applies physical concept (e.g. force equilibrium, energy conservation, mass conservation, etc.) directly to discretized elements. It is easy in its physical interpretation. z It does not need elaborate sophisticated mathematical manipulation or concept. z Its applicability is limited to certain problems for which equilibrium or conservation law can be easily stated in terms of physical quantities one wants to obtain. In most cases, discretized elements are selfobvious in the physical sense. There are several examples of direct approaches as illustrated in the following. Using the first example, important features of FEM will be discussed in detail.
Example 1: Force Balance (Linear Spring System)
[Bathe P.79, Ex.3.1]
The problem of a linear spring system is depicted in the following figure. F1
F2
F3
K1
F4
K2
F5 K4
K3
F6 K5
x δ1
δ2
δ3
δ4
δ5
δ6
In this particular problem, let us assume that δ1 , F2 , L , F6 are specified. the displacements at the nodes, and the reaction force at the node number 1. One can typically follow several steps as an FEM procedure described below: One element:
f1
Node 1
Node 2
δ1
© 2005 by T. H. Kwon
δ2
7
f2
Solve
Equilibrium: f 1 = k (δ1 − δ 2 ) f 2 = k (δ 2 − δ1 ), (Note that f1 = − f 2
for force equilibrium.)
Element Matrix Equation for one element: ⎡ k ⎢− k ⎣
− k ⎤ ⎧ δ1 ⎫ ⎧ f 1 ⎫ ⎨ ⎬=⎨ ⎬ k ⎥⎦ ⎩δ 2 ⎭ ⎩ f 2 ⎭
[K ]e {φ}e = { f }e
⇔
Global matrix equation by Assembly:
[K ]{δ} = {F } where
[K] is called ‘Stiffness Matrix’ and {F} is called ‘Resultant Nodal Force Matrix’. The physical meaning of assembly procedure can be found in the force balance as described below: f2(1)
f1(2)
K1
f1(1)
K2
(1 − k1δ1 + k1δ 2 ) + (k 2 δ 2 − k 2 δ 3 ) = 44244 3 14 4244 3 f 2 (1)
f2(2)
f2
(1)
+
f1
( 2)
= F2
f1( 2 )
from element 1
from element 2
External nodal force applied at node 2
The typical assembly can be done as shown below:
© 2005 by T. H. Kwon
8
⎡ k1 ⎢ −k 1 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣⎢
− k1 k1 + k 2
− k2
− k2
k 2 + k3
− k3
− k3
k3 + k 4
− k4
− k4
k 4 + k5 − k5
0
0
− k5 k5
⎤ ⎧ δ1 ⎫ ⎧ F1 ⎫ ⎥ ⎪δ ⎪ ⎪ F ⎪ ⎥⎪ 2 ⎪ ⎪ 2 ⎪ ⎥ ⎪⎪δ 3 ⎪⎪ ⎪⎪ F3 ⎪⎪ ⎥⎨ ⎬ = ⎨ ⎬ ⎥ ⎪δ 4 ⎪ ⎪ F4 ⎪ ⎥ ⎪δ 5 ⎪ ⎪ F5 ⎪ ⎥⎪ ⎪ ⎪ ⎪ ⎦⎥ ⎪⎩δ 6 ⎪⎭ ⎪⎩ F6 ⎪⎭
(2.1)
banded and symmetric matrix Now, let us pay attention to the boundary condition in this particular case. Displacements:
δ1 = 0 : specified
δ 2 , δ 3 , δ 4 , δ 5 , δ 6 : unknown
Geometric condition, Essential boundary condition Forces:
F2 , F3 , F4 , F5 , F6 : specified
F1 : unkown reaction force
Force condition, Natural boundary condition One should recognize that for each node only one of the displacement and force can be specified as a boundary condition. Nature does not allow to specify both the displacement and force simultaneously at any node. If none of the two is known, then the problem is not well posed, in other words, one does not have a problem to solve. It should also be noted that if there is no geometry constraint at all, there is no unique solution. One can get a solution only up to a constant. (In other words, the linear spring system can be moved in xaxis without further deformation.) In this case, the stiffness matrix becomes a singular matrix. You will easily understand that equation (2.1) is singular since the summation of six rows becomes null. Think about the physical meaning of the fact that the summation of six rows becomes null. It just indicates that the force applied on the system is in balance! In this regard, it is obvious that at least one geometry constraint should be assigned in order to get a unique solution. Later, we will discuss the methods of introducing boundary conditions into equation (2.1).
© 2005 by T. H. Kwon
9
Example 2: Energy Conservation (1D Heat Conduction)
[Bathe P.80, Ex.3.2]
Another example with the energy conservation principle can be found in one dimensional heat conduction problem.
∆1
Q1
∆2
Q2
Q3
k1′ T1
∆3
∆4
k 3′
T2
T3
Q6
Q5
Q4
k 2′
∆5
k 4′
k 5′
T4
T5
T6
where Qi represents heat flux input into the system through node i. the procedure described in the previous example in the same manner.
We will follow
∆1
One element:
q1
k1′ T1
q2
T2
Energy conservation: Fourier Law:
∂T ∂x k′ Q = − 1 (T2 − T1 ) ∆1 Q = −k
= −k1 (T2 − T1 ) with k ≡
k1′ ∆1
q1 = −k (T2 − T1 )
: heat flux entering the element through node 1
q 2 = k (T2 − T1 )
: heat flux entering the element through node 2
(Note that
q1 = − q 2
for energy conservation.)
Element Matrix Equation for one element:
© 2005 by T. H. Kwon
10
⎡ k ⎢− k ⎣
− k ⎤ ⎧T1 ⎫ ⎧ q1 ⎫ ⎨ ⎬=⎨ ⎬ k ⎥⎦ ⎩T2 ⎭ ⎩q 2 ⎭
[K ]e {φ}e = { f }e
⇔
Global matrix equation by Assembly:
[K ]{T } = {Q} where [K] is called ‘Stiffness Matrix’ and {Q} is called ‘Resultant Nodal Energy Influx Matrix’. The physical meaning of assembly procedure can be found in the force balance as described below: q2(1)
q1(2)
K1
q1(1)
T1
K2
T2
q2(2)
T3
(− k1T1 + k1T2 ) + (k 2T2 − k 2T3 ) =
q2
+
(1)
into element 1
q1
( 2)
=
Q2
into element 2
External energy influx entering the system through node 2 The assembly can be done in the same manner as for the previous example:
⎡ k1 ⎢ −k 1 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣⎢
− k1 k1 + k 2
− k2
− k2
k 2 + k3
− k3
− k3
k3 + k 4
− k4
− k4
k 4 + k5
− k5
− k5
k5
0
0
Again, the global stiffness matrix is banded and symmetric.
© 2005 by T. H. Kwon
11
⎤ ⎧T1 ⎫ ⎧ Q1 ⎫ ⎥ ⎪T ⎪ ⎪Q ⎪ ⎥⎪ 2 ⎪ ⎪ 2 ⎪ ⎥ ⎪⎪T3 ⎪⎪ ⎪⎪Q3 ⎪⎪ ⎥⎨ ⎬ = ⎨ ⎬ ⎥ ⎪T4 ⎪ ⎪Q4 ⎪ ⎥ ⎪T5 ⎪ ⎪Q5 ⎪ ⎥⎪ ⎪ ⎪ ⎪ ⎦⎥ ⎩⎪T6 ⎭⎪ ⎪⎩Q6 ⎪⎭
(2.2)
Now, let us pay attention to the boundary condition for this example. There are two different types of boundary conditions for each node: temperature (Ti) and energy influx (Qi). As in the previous example of force balance case, only one of the two should be provided for a wellposed problem. Otherwise, one cannot solve the problem, or nature does not allow it. Again, the two types of boundary conditions are categorized as follows: Temperature : Essential boundary condition Energy flux : Natural boundary condition Now, one may get an idea of the similar nature of problems associated with the boundary conditions. There are pairs of essential and natural boundary conditions for each node. Only one of them should be provided. With regard to the nature of singularity of the global stiffness matrix, the same is also true as in the previous example. At least one essential boundary condition should be assigned in order to eliminate the singularity. Think about what causes the singular nature of the stiffness matrix in physical sense. Answer yourself.
Example 3: Mass Conservation (Flow network, Electric network) [Bathe P.82, Ex.3.3]
Qsource
Qfaucet
There is a water flow network as depicted above. The problem is to find water flow rate in the pipes and faucets given the water flow rate from the reservoir. The nature of the problem is almost identical to the previous two examples. Therefore, only the summary will be described below.
© 2005 by T. H. Kwon
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One element: q1
P1
P2
L, D
q2
µ : water viscosity
Mass conservation: Fluid mechanics:
( P1 − P2 ) =
128q1 Lµ πD 4
q1 = k ( P1 − P2 ) πD 4 with k ≡ 128Lµ q1 = k ( P1 − P2 )
: mass flow rate entering the element through node 1
q 2 = k ( P2 − P1 ) (Note that
: mass flow rate entering the element through node 2 q1 = − q 2
for mass conservation.)
Element Matrix Equation for one element: ⎡ k ⎢− k ⎣
− k ⎤ ⎧ P1 ⎫ ⎧ q1 ⎫ ⎨ ⎬=⎨ ⎬ k ⎥⎦ ⎩ P2 ⎭ ⎩q 2 ⎭
⇔
[K ]e {φ}e = { f }e
Global matrix equation by Assembly:
[K ]{P} = {Q} Now consider the assembly procedure. The assembly procedure is identical to the previous two examples. You are requested to find the physical meaning of the assembly procedure yourself. Also think about the boundary condition types. Which physical quantities are corresponding to essential and natural boundary condition? Can you make sure the pair characteristics of the boundary condition types for each node? Do you expect to obtain a singular matrix after the assembly? Why is it so? What condition constitutes the wellposed problem? You will definitely find the similarities among the
three examples.
© 2005 by T. H. Kwon
13
Example 4: Direct Stiffness Methods for Truss Elements
Next example is concerned about a twodimensional analysis of truss structures with the socalled truss elements. Note that truss elements are joined by pin joint so that a truss element cannot bear the bending moment and shear force in contrast to a beam element. It can bear the tensile/compressive force (i.e. longitudinal force) only. Consider the following schematic diagram for this example.
F4
F2
4
2
④ ① y
⑤
③ ②
3 F3
x
One element:
y2, fy2 2 v1, fy1 y 1
u2, fx2 E, A, L
u1, fx1 x
ForceDeformation Law:
Elastic Elongation:
© 2005 by T. H. Kwon
F=
AE ∆L L
14
: elements
i : nodes
⑥
1
ⓘ
⑦ 5
Element Matrix Equation for one element:
One should relate the deformation at nodes ( u1 , u 2 , v1 , v 2 ) with the force ( f x1 , f y1 , f x 2 , f y 2 ) applied on the nodes in the following element matrix equation form: e
⎧ f x1 ⎫ ⎧u1 ⎫ ⎪f ⎪ ⎪v ⎪ e⎪ 1⎪ [K ] ⎨ ⎬ = ⎪⎨ y1 ⎪⎬ ⎪ f x2 ⎪ ⎪u 2 ⎪ ⎪⎩ f y 2 ⎪⎭ ⎪⎩v 2 ⎪⎭
e
In order to find the coefficients of the element stiffness matrix, Kij, consider the effect of displacement v1 of unity with u1 = u2 = v2 = 0 for Ki2. F
v1 = 1
α
∆L = v1 sin α = sin α
∆L F
F=
AE sin α L f x1 = F cos α f y1 = F sin α
⇒
f x 2 = − f x1
e e K12e , K 22 , K 32e , K 42
f y 2 = − f y1
i.e.
AE sin α cos α, L = − K 12e ,
K 12e = K 32e
AE sin 2 α L e = − K 22
e K 22 = e K 42
Similarly, one can obtain the other coefficients, resulting in the following element
© 2005 by T. H. Kwon
15
stiffness matrix:
[K ]e
⎡ cos 2 α sin α cos α − cos 2 α − sin α cos α ⎤ ⎢ ⎥ 2 sin α − sin α cos α − sin 2 α ⎥ AE ⎢ sin α cos α = L ⎢ − cos 2 α cos 2 α sin α cos α ⎥ − sin α cos α ⎢ ⎥ 2 sin α cos α sin 2 α ⎥⎦ − sin α ⎢⎣− sin α cos α
(Note that
∑F = ∑F x
y
= ∑ M = 0 are satisfied automatically.
: symmetric
You have to be
able to tell its implication with regard to the singular nature of the final global stiffness matrix that will be discussed below.)
Global matrix equation by Assembly:
The assembly procedure is done in the following manner. First, identify the global nodal number i and j corresponding to the two endnodes 1 and 2 of the element to be added. Second, add 4x4 element matrix components to the corresponding rows and columns of 2i1, 2i, 2j1, 2j in the global stiffness matrix. 2i1
2i1 2i
2j1 2j
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣⎢
2j1
2i
2j
+ k11e e + k 21
+ k12e e + k 22
+ k13e e + k 23
+ k14e e + k 24
+ k 31e e + k 41
+ k 32e e + k 42
+ k 33e e + k 43
+ k 34e e + k 44
© 2005 by T. H. Kwon
16
⎤⎧ ⎫ ⎧ ⎥⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎥ ⎪ u i ⎪ ⎪+ ⎥⎪ ⎪ ⎪ ⎥ ⎪ v i ⎪ ⎪+ ⎥ ⎪⎪ ⎪⎪ ⎪⎪ ⎥⎨ ⎬ = ⎨ ⎥⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎥ ⎪u j ⎪ ⎪+ ⎥ ⎪ v j ⎪ ⎪+ ⎥⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎥⎦ ⎪⎩ ⎪⎭ ⎪⎩
⎫ ⎪ ⎪ ⎪ ⎪ ⎪ f xi ⎪ ⎪ f yi ⎪ ⎪⎪ ⎬ ⎪ ⎪ ⎪ ⎪ f xj ⎪ ⎪ f yj ⎪ ⎪ ⎪ ⎪⎭
which will result in the final global matrix equation of the following form:
[K ]{δ} = {F } You are requested to find the physical meaning of the assembly procedure yourself. Again think about the boundary condition types. Which physical quantities are corresponding to essential and natural boundary condition, respectively? Can you make sure the pair characteristics of the boundary condition types for each node? Do you expect to obtain a singular matrix after the assembly? Why is it so? What condition constitutes the wellposed problem? You will definitely find the similarity to the previous examples. Note: The global stiffness matrix has three rank deficiency. For a well posed problem, one has to remove three equations (rows) or replace them with appropriate equations associated with boundary conditions. Think about the origin of the rank deficiency yourself.
Stiffness Matrix, Flexibility Matrix For the deformation problems, there are two kinds of approaches depending on which variable is considered unknown to be solved for. i)
[K]{x}={F}
Kij
with [K] being stiffness matrix
: influence coefficient which represents force Fi due to unit displacement of xj
“Displacementbased FEM” ii)
[ℑ]{F}={x}
ℑij
with [ℑ] being flexibility matrix
: influence coefficient which represents displacement xi due to unit force of Fj
“Forcebased FEM”
© 2005 by T. H. Kwon
17
2.2 Coordinate Transformation In many cases, one can introduce a local coordinate system associated with each element in addition to a global coordinate system. A local coordinate system can be defined in many cases in a selfobvious way inherent to the element itself. It is much easier to determine the stiffness matrix with respect to the local coordinate system of an element than with respect to the global coordinate system. The stiffness matrix with respect to the local coordinate system is to be transformed to that with respect to the global coordinate system before the assembly procedure. Example:
u′2
v2 y′
x′
u2
v1 u′1 y
u1 x ⎧ u1 ⎫ ⎪v ⎪ ⎪ 1⎪ ⎨ ⎬ ⎪u 2 ⎪ ⎪⎩v 2 ⎪⎭
⇒
Global system i) Vector Transformation in 2D
⎧ u1′ ⎫ ⎨ ⎬ ⎩u ′2 ⎭
Local system
y y′ (x,y) or (x′,y′) j
i′
j′
α
x
i
© 2005 by T. H. Kwon
x′
18
V = xi + yj = x ′i ′ + y ′j′ x ′ = cos α ⋅ x + sin α ⋅ y y ′ = − sin α ⋅ x + cos α ⋅ y
⎧ x ′ ⎫ ⎡ cos α sin α ⎤ ⎧ x ⎫ ⎨ ⎬=⎢ ⎥⎨ ⎬ ⎩ y ′⎭ ⎣− sin α cos α ⎦ ⎩ y ⎭
⇒
⎧ x ⎫ ⎡cos α − sin α ⎤ ⎧ x ′ ⎫ ⎨ ⎬=⎢ ⎥⎨ ⎬ ⎩ y ⎭ ⎣ sin α cos α ⎦ ⎩ y ′⎭
or
ii) Transformation of stiffness matrix The element matrix equation can be generally represented in terms of the local coordinate system. In the discussion below, we are interested in coordinate transformation of stiffness matrix associated with vectors such as displacement and force. Suppose the element stiffness matrix is represented by the following equation:
[K ′]e {x′}e = {b′}e
(2.3)
The vector transformation of {x ′} and e
{b′}e
between the local and global coordinate
system might be
and
{x ′}e = [Φ ]{x}e
(2.4)
{b′}e = [Φ ]{b}e
(2.5)
where {x} and {b} are referenced to the global coordinate system. Then, equations e
e
(2.3)(2.5) yield
[K ′]e [Φ]{x}e = [Φ]{b}e
(2.6)
To get a matrix equation in the global coordinate system in terms of
[K ]e {x}e = {b}e
(2.7)
premultiply Eq. (2.6) by [Φ ]
−1
© 2005 by T. H. Kwon
(a generalized inverse matrix of [Φ ] ).
19
Then
[Φ]−1 [K ′][Φ]{x}e = [Φ ]−1 [Φ ]{b}e = [I ]{b}e = {b}e
(2.8)
By comparison between equation (2.8) and (2.7), one can find
[K ]e = [Φ ]−1 [K ′]e [Φ]
(2.9)
Note: 1. For an orthogonal coordinate system, the transformation matrix has the
[Φ ]−1 = [Φ ]T
characteristics of
.
[K ′]e
If
is symmetric, then
[K ]e
remains
symmetric! 2. If degree of freedom in the local coordinate system is different from that in the global coordinate system, [Φ ] is not a square matrix.
Example: transformation in Truss Element
v2, f y′
u′2, f′2 u2, fx2
v1, fy1 y
E, A, L ⇒
u1, fx1
u′1, f′1
u′1, f′1 σ = Eε, ε =
x
u′2, f′2 ∆L u 2′ − u1′ = L L
EA (u 2′ − u1′ ) L f 2′ = σA = − f1′ f1′ = −σA = −
i.e. ⎧ f1′⎫ EA ⎡ 1 − 1⎤ ⎧ u1′ ⎫ ⎨ ⎬= ⎢ ⎥⎨ ⎬ ⎩ f 2′ ⎭ L ⎣− 1 1 ⎦ ⎩u 2′ ⎭
© 2005 by T. H. Kwon
⇔
20
− 1⎤ L ⎣− 1 1 ⎥⎦
[K ′]e = EA ⎡⎢
1
0 ⎤ ⎧ u1 ⎫ ⎡cos α ⎪ v ⎪ ⎢ sin α 0 ⎥⎥ ⎧ u1′ ⎫ ⎪ 1⎪ ⎢ ⎨ ⎬= ⎨ ⎬ cos α ⎥ ⎩u 2′ ⎭ ⎪u 2 ⎪ ⎢ 0 ⎥ ⎪⎩v 2 ⎪⎭ ⎢⎣ 0 sin α ⎦
⇒ [Φ ]
⎧ u1 ⎫ 0 0 ⎤ ⎪⎪ v1 ⎪⎪ ⎧ u1′ ⎫ ⎡cos α sin α ⎨ ⎬=⎢ ⎨ ⎬ 0 cos α sin α ⎥⎦ ⎪u 2 ⎪ ⎩u 2′ ⎭ ⎣ 0 ⎪⎩v 2 ⎪⎭
⇒ [Φ ]
(note:
−1
[Φ]−1 = [Φ ]T )
To get ⎧ u1 ⎫ ⎧ f x1 ⎫ ⎪ ⎪ ⎪f ⎪ e ⎪ v1 ⎪ [K ] ⎨ ⎬ = ⎪⎨ y1 ⎪⎬ ⎪u 2 ⎪ ⎪ f x 2 ⎪ ⎪⎩v 2 ⎪⎭ ⎪⎩ f y 2 ⎪⎭
make use of the transformation rule: 0 ⎤ ⎡cos α ⎢ sin α 0 0 ⎤ 0 ⎥⎥ EA ⎡ 1 − 1⎤ ⎡cos α sin α e −1 e ⎢ ′ [K ] = [Φ ] [K ] [Φ ] = ⎢ ⎥ ⎢ ⎢ 0 0 cos α sin α ⎥⎦ cos α ⎥ L ⎣− 1 1 ⎦ ⎣ 0 ⎢ ⎥ sin α ⎦ ⎣ 0 Finally one obtains the stiffness matrix that is exactly identical to the previously derived one:
[K ]e
⎡ cos 2 α sin α cos α − cos 2 α − sin α cos α ⎤ ⎥ ⎢ 2 sin α − sin α cos α − sin 2 α ⎥ AE ⎢ sin α cos α = L ⎢ − cos 2 α cos 2 α sin α cos α ⎥ − sin α cos α ⎥ ⎢ sin α cos α sin 2 α ⎦⎥ − sin 2 α ⎣⎢− sin α cos α
© 2005 by T. H. Kwon
21
The reason for using the coordinate transformation is well demonstrated via this example.
It is much easier to evaluate [K ′] in the local coordinate system than [K ] e
e
in the global coordinate system. Note, however, that one has to calculate [K ] via the e
transformation rule before the assembly (element by element). Note: In this particular example, the coordinate transformation matrix [Φ] is is not a square matrix because of the difference in the degree of freedom between the local and global coordinate systems. Think about which part of the derivation is affected by this difference. You have to recognize the following facts:
[Φ ][Φ ]−1 = ⎡⎢
1 0⎤ ⎥ = [I ]2 x 2 ⎣0 1 ⎦
but
⎤ ⎡ cos 2 α cos α sin α 0 0 ⎥ ⎢ 2 0 0 ⎥ ≠ [I ] [Φ ]−1 [Φ ] = ⎢⎢cos α sin α sin α 4x4 2 0 0 cos α cos α sin α ⎥ ⎥ ⎢ 0 0 cos α sin α sin 2 α ⎦⎥ ⎣⎢
which is not expected in the derivation of equation (2.8), i.e., [Φ ] [Φ ]{b} = [I ]{b} = {b} . −1
But one can show that ⎧ f x1 ⎫ ⎧ f x1 ⎫ ⎪f ⎪ ⎪f ⎪ −1 [Φ ] [Φ ]⎪⎨ y1 ⎪⎬ = ⎪⎨ y1 ⎪⎬ ⎪ f x2 ⎪ ⎪ f x2 ⎪ ⎪⎩ f y 2 ⎪⎭ ⎪⎩ f y 2 ⎪⎭
if one recognizes that
f x1 cos α = , etc. f y1 sin α (i.e., cos 2 α f x1 + cos α sin α f y1 = cos 2 α f x1 + sin 2 α f x1 = f x1 , etc. )
© 2005 by T. H. Kwon
22
2.3 Direct approach for Elasticity Problem (plane stress, plane strain)
In this section, we are concerned about an elastic deformation problem in twodimensional continuous media (therefore, not a discrete system). Consider the following schematic diagram of this problem.
F1
F3
F2 First, establish element nodes for a displacement matrix {δ} and a force matrix e
v3, fy3 (x3 , y3)
u3, fx3
v1, fy1 (x1 , y1)
y
v2, fy2
u1, fx1
u2, fx2
x
{δ}e
⎧ δ1 ⎫ ⎧u1 ⎫ ⎪δ ⎪ ⎪ v ⎪ ⎪ 2⎪ ⎪ 1⎪ ⎪⎪δ ⎪⎪ ⎪⎪u ⎪⎪ = ⎨ 3 ⎬ = ⎨ 2 ⎬, ⎪δ 4 ⎪ ⎪v 2 ⎪ ⎪δ 5 ⎪ ⎪u 3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩⎪δ 6 ⎭⎪ ⎩⎪ v3 ⎭⎪
© 2005 by T. H. Kwon
(x2 , y2)
{ f }e
⎧ f1 ⎫ ⎧ f x1 ⎫ ⎪f ⎪ ⎪f ⎪ ⎪ 2 ⎪ ⎪ y1 ⎪ ⎪⎪ f ⎪⎪ ⎪⎪ f x 2 ⎪⎪ = ⎨ 3⎬ = ⎨ ⎬. ⎪ f 4 ⎪ ⎪ f y2 ⎪ ⎪ f 5 ⎪ ⎪ f x3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩⎪ f 6 ⎭⎪ ⎪⎩ f y 3 ⎪⎭
23
{ f }e .
One wants to establish the element stiffness matrix equation
[K ]e {δ}e = { f }e which is our target. In order to achieve this mission, consider the following relationships to connect each physical quantity:
force
{ f }e
⇐ stress ⇔ strain ⇒ displacement field ⇒ displacement {δ}
e
equilibrium equation
<STEP 1>
constitutive equation
straindisplacement relation
Assume the displacement field within an element. u ( x, y ) = α 1 + α 2 x + α 3 y
⇒
v ( x, y ) = α 4 + α 5 x + α 6 y <STEP 2>
Find
α’s in terms of {δ} . e
Find the displacement field in terms of nodal displacement.
We have ⎧ u1 ⎫ ⎡1 x1 ⎪ ⎪ ⎢ ⎨u 2 ⎬ = ⎢1 x 2 ⎪u ⎪ ⎢1 x 3 ⎩ 3⎭ ⎣
y1 ⎤ ⎧ α 1 ⎫ ⎪ ⎪ y 2 ⎥⎥ ⎨α 2 ⎬ . y 3 ⎥⎦ ⎪⎩α 3 ⎪⎭
solving α’s by inversion to have
⎧ α1 ⎫ ⎡ x 2 y 3 − x3 y 2 1 ⎪ ⎪ ⎢ y −y ⎨α 2 ⎬ = 3 ⎢ 2 ⎪α ⎪ 1 x1 y1 ⎢ x − x 3 2 ⎩ 3⎭ 1 x y2 ⎣ 2 1 x3 y 3 ⎡a1 1 ⎢ = b1 2∆ ⎢ ⎢⎣ c1
© 2005 by T. H. Kwon
a2 b2 c2
a 3 ⎤ ⎧u1 ⎫ ⎪ ⎪ b3 ⎥⎥ ⎨u 2 ⎬. c3 ⎥⎦ ⎪⎩u 3 ⎪⎭
24
x3 y1 − x1 y 3 y 3 − y1 x1 − x3
x1 y 2 − x 2 y1 ⎤ ⎧ u1 ⎫ ⎪ ⎪ y1 − y 2 ⎥⎥ ⎨u 2 ⎬ x 2 − x1 ⎥⎦ ⎪⎩u 3 ⎪⎭
Similarly, one has ⎧α 4 ⎫ ⎡a1 ⎪ ⎪ 1 ⎢ ⎨α 5 ⎬ = ⎢ b1 ⎪α ⎪ 2∆ ⎢ c ⎩ 6⎭ ⎣ 1
a2 b2 c2
a3 ⎤ ⎧ v1 ⎫ ⎪ ⎪ b3 ⎥⎥ ⎨v 2 ⎬. c3 ⎥⎦ ⎪⎩v3 ⎪⎭
where the followings are defined:
1 x1 1 ∆ = 1 x2 2 1 x3
y1 y 2 = area of the triangular element y3
a1 ≡ x 2 y 3 − x3 y 2 , a 2 ≡ x3 y1 − x1 y 3 , a3 ≡ x1 y 2 − x 2 y1 b1 ≡ y 2 − y 3 ,
b2 ≡ y 3 − y1 ,
b3 ≡ y1 − y 2
c1 ≡ x3 − x 2 ,
c 2 ≡ x1 − x3 ,
c3 ≡ x 2 − x1
Then the displacement can be obtained by ⎧α1 ⎫ ⎪ ⎪ u ( x, y ) = ⎣1 x y ⎦⎨α 2 ⎬ ⎪α ⎪ ⎩ 3⎭ ⎡a1 1 ⎢ = ⎣1 x y ⎦⎢b1 2∆ ⎢⎣ c1
a2 b2 c2
a3 ⎤ ⎧ u1 ⎫ ⎪ ⎪ b3 ⎥⎥ ⎨u 2 ⎬ c3 ⎥⎦ ⎪⎩u 3 ⎪⎭
⎧u1 ⎫ 1 ⎪ ⎪ = ⎣a1 + b1 x + c1 y a 2 + b2 x + c 2 y a3 + b3 x + c3 y ⎦⎨u 2 ⎬ 2∆ ⎪u ⎪ ⎩ 3⎭ ⎧ u1 ⎫ ⎪ ⎪ = ⎣N 1 ( x, y ) N 2 ( x, y ) N 3 ( x, y )⎦⎨u 2 ⎬ ⎪u ⎪ ⎩ 3⎭
u = N i ui i.e. Similarly, v = N i vi .
© 2005 by T. H. Kwon
with Ni being “Shape Function”.
25
Find strain {ε} in terms of {δ} . e
<STEP 3>
{ε}e
⎧ ∂u ⎫ ⎪ ⎪ ⎧ ε x ⎫ ⎪ ∂x ⎪ ⎧ α 2 ⎫ ⎪ ⎪ ⎪ ∂v ⎪ ⎪ ⎪ ≡ ⎨εy ⎬ = ⎨ ⎬ = ⎨ α6 ⎬ ⎪γ ⎪ ⎪ ∂y ⎪ ⎪α + α ⎪ 3⎭ ⎩ xy ⎭ ⎪ ∂v ∂u ⎪ ⎩ 5 + ⎪ ∂x ∂y ⎪ ⎩ ⎭
⎡b1 1 ⎢ = 0 2∆ ⎢ ⎢⎣c1
0
b2
0
b3
c1
0
c2
0
b1
c2
b2
c3
⎧ u1 ⎫ ⎪v ⎪ 0 ⎤⎪ 1 ⎪ ⎪⎪u ⎪⎪ c3 ⎥⎥ ⎨ 2 ⎬ v b3 ⎥⎦ ⎪ 2 ⎪ ⎪u 3 ⎪ ⎪ ⎪ ⎪⎩ v3 ⎪⎭
{ε}e = [B]e {δ}e
i.e.
(2.10)
where ⎡b1 1 ⎢ [B] ≡ ⎢ 0 2∆ ⎢⎣c1 e
<STEP 4>
0
b2
0
b3
c1
0
c2
0
b1
c2
b2
c3
0⎤ c3 ⎥⎥ b3 ⎥⎦
Introduce the constitutive equation (stressstrain relationship).
2D case:
⎧εx ⎫ ⎧σ x ⎫ ⎪ ⎪ ⎪ e⎪ ⎨σ y ⎬ = [C ] ⎨ ε y ⎬ ⎪γ ⎪ ⎪τ ⎪ ⎩ xy ⎭ ⎩ xy ⎭
{σ}e = [C ]e {ε}e
⇒
(2.11)
where [C ] depends on the problem, for instance, plane stress or plane strain, as e
represented below:
© 2005 by T. H. Kwon
26
[C ]e
⎡ ⎤ ⎢1 ν 0 ⎥ ⎥ E ⎢ ν 1 0 ≡ ⎢ ⎥ 1 −ν 2 ⎢ 1 −ν ⎥ ⎢0 0 ⎥ 2 ⎦ ⎣
[C ]e
⎡ ⎢1 − ν ν ⎢ E ν 1 −ν ≡ (1 + ν )(1 − 2ν ) ⎢⎢ 0 ⎢ 0 ⎣
for plane stress
<STEP 5> Determine the nodal force matrix
⎤ 0 ⎥ ⎥ 0 ⎥ 1 − 2ν ⎥ ⎥ 2 ⎦
(2.12)
for plane strain
(2.13)
{ f }e .
One has to determine the nodal force matrix
{ f }e
in such a way that it is statically
equivalent to a constant stress field {σ} for equilibrium. The following figure shows e
the forces applied at nodes together with those applied on the element sides. f6 f5
σy (x3 x1)tτxy(y3 y1) t
σy
σy (x2 x3)t+τxy(y3 y2) τxy
σx(y3 y1)tτxy(x3 x1)t
t
σx(y3 y2)t+τxy(x2 x3)t
σx f2 σx (y1 y2)tτxy(x2x1) t
f1
f4 f3
σy (x2 x1)t+τxy(y1 y2) t
© 2005 by T. H. Kwon
27
One will recognize a problem in carrying out this mission. coefficients of
∑F
x
{ f }e while there are three equations as equilibrium conditions, i.e.,
= 0, ∑ Fy = 0, ∑ M = 0 .
coefficients of
There are six
{ f }e .
Therefore one cannot determine uniquely six
One tricky solution to this trouble is to equally distribute forces
acting on an element side to two nodal forces applied to two endnodes associated with the side. Consider one of the sides as a representative example for this procedure.
σy (x2 x3)t τxy(x2 x3)t
σx(y3 y2)t+τxy(x2 x3)t
σx(y3 y2)t
σy (x2 x3)t+τxy(y3 y2) t τxy(y3 y2) t
With equal distribution of those midside forces to two adjacent nodes, one can have
{ f }e
⎧ f1 ⎫ ⎡ y3 − y 2 ⎪f ⎪ ⎢ 0 ⎪ 2⎪ ⎢ t ⎢ y1 − y 3 ⎪⎪ f 3 ⎪⎪ =⎨ ⎬=− ⎢ 2⎢ 0 ⎪ f4 ⎪ ⎢ y 2 − y1 ⎪ f5 ⎪ ⎢ ⎪ ⎪ ⎪⎩ f 6 ⎪⎭ ⎣⎢ 0
0 x 2 − x3 0 x3 − x1 0 x1 − x 2
= [ A] {σ} e
© 2005 by T. H. Kwon
e
28
x 2 − x3 ⎤ y 3 − y 2 ⎥⎥ ⎧σ x ⎫ x3 − x1 ⎥ ⎪ ⎪ ⎥ ⎨σ y ⎬ y1 − y 3 ⎥ ⎪ ⎪ τ xy x1 − x 2 ⎥ ⎩ ⎭ ⎥ y 2 − y1 ⎦⎥
i.e.,
{ f }e
= [ A] {σ} e
e
(2.14)
<STEP 6> Determine [K ]
e
From equations (2.10), (2.11) and (2.14) one can connect the whole bridge between the force and displacement matrices.
{ f }e = [A]e [C ]e [B]e {δ}e i.e.
[K ]e {δ}e = { f }e
⇔
[K ]e = [A]e [C ]e [B]e
(2.15)
Now, we want to describe the stiffness matrix in a local coordinate system defined in such a way that x1 = y1 = y2 = 0:
y
3
1
[K ]e
⎡ y 3 λ 1 x 23 2 + ⎢ x2 y3 ⎢ x2 ⎢ λ 2 x32 ⎢ − x2 ⎢ ⎢ y 3 λ 1 x3 x 23 + ⎢− Et x2 x2 y3 = ⎢ 2(1 − ν 2 ) ⎢ νx3 λ 1 x32 ⎢ x + x 2 ⎢ 2 λ x ⎢ − 1 23 ⎢ y3 ⎢ ⎢ −ν ⎢⎣
© 2005 by T. H. Kwon
x
2
λ y x 23 + 1 3 x2 y3 x2 νx32 λ 1 x3 + x2 x2 x3 x 23 λ 1 x3 − x2 y3 x2 2
− λ1 −
x 23 y3
29
symm. y 3 λ 1 x3 + x2 x2 y3 λ x − 2 3 x2 λ x − 1 3 y3
2
ν
λ y x3 + 1 3 x2 y3 x2 2
λ1 −
x3 y3
λ1 x2 y3
0
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ x2 ⎥ ⎥ y 3 ⎥⎦
xij = xi − x j , y ij = y i − y j
with
λ1 =
1− ν , 2
λ2 =
1+ ν 2
For the general case, it is not worthwhile to derive the formula for [K ] , which is e
too tedious. Instead one can use the coordinate transformation via
[K ]e = [Φ ]−1 [K ′]e [Φ ] . y y′
3
x′
2 1
α x
Note:
Is [K ] = [ A] [C ] [B ] symmetric? Yes, as it should be due to BettiMaxwell e
e
e
e
Reciprocal Theorem. [C ] is symmetric. And interestingly, we have the following e
relation between two other matrices:
[A]e = t∆[B ]e
T
.
2.4 Assembly Procedure We have, in every other case, an element (stiffness) matrix equation of the form:
[K ]e {x}e = { f }e . We have to establish a global matrix equation for the whole system. The assembly
© 2005 by T. H. Kwon
30
procedure is based on compatibility and conservation law (e.g. force balance, mass conservation and energy conservation). In the case of the elasticity problem, the assembly procedure is described below.
j, j′ xi′ B
A
xi
m, m′
i, i′
k, k′
D
C
y l, l′ x I , J , K , L, M : global nodal number i, i ′, j , j ′, k , k ′, l , l ′, m, m′ : degree of freedom for instance i = 2 I − 1, i ′ = 2 I , etc. in 2  D case Compatibility: xiA = xiB = xiC = xiD = xi Conservation Law (Force Balance):
①
fi
f iA
②
A
f iB
fi fi
④
③
Fi = f i A + f i B + f i C + f i D = ∑ f i Q Q
© 2005 by T. H. Kwon
Fi f iD
C
31
f iC f iD
where f i e = internal force on i  th d.o.f. for element e Fi = external force on i  th d.o.f. xi = displacement of i  th d.o.f. Consider Qth element
[K ]Q {x}Q = { f }Q
⇒
f iQ =
NDOF
∑K
Q in
x nQ with n denoting n  th d.o.f.
n
For equilibrium Fi = f i A + f i B + f i C + f i D ⎛ NDOF ⎞ ⎛ NDOF ⎞ ⎛ NDOF ⎞ = ⎜ ∑ K inA x nA ⎟ + ⎜ ∑ K inB x nB ⎟ + L + ⎜ ∑ K inD x nD ⎟ ⎝ n ⎠ ⎝ n ⎠ ⎝ n ⎠
Using the nodal compatibility xiA = xiB = xiC = xiD = xi one can rearrange the equilibrium equation as ⎛ NDOF ⎞ ⎛ NDOF ⎞ ⎛ NDOF ⎞ Fi = ⎜ ∑ K inA x nA ⎟ + ⎜ ∑ K inB x nB ⎟ + L + ⎜ ∑ K inD x nD ⎟ ⎝ n ⎠ ⎝ n ⎠ ⎝ n ⎠ A B C D A B C = K ii + K ii + K ii + K ii xi + K ii′ + K ii′ + K ii′ + K iDi′ xi′
( + (K + K + (K + 0 +( 0 + 0 +( 0 +K A ij
B ij
A ik
B im
+ 0 +K
C ik
+K
C il
+ 0
) ( ) + 0 )x + (K + K + 0 + 0 )x + 0 )x + (K + 0 + K + 0 )x + K )x + ( 0 + 0 + K + K )x + K )x + ( 0 + K + 0 + K )x D il
D im
j
A ij ′
k
A ik ′
B ij ′
C il ′
l
m
j′
C ik ′
B im ′
k′
D il ′
D im ′
l′
m′
which becomes one of the equations in the global matrix equation
[K ]{x} = {F } .
© 2005 by T. H. Kwon
32
2.5 Bandwidth The global stiffness matrix is usually a banded and partiallyfilled (sparse) matrix. The nodal numbering system will determine the bandwidth. Bandwidth is also closely related with the storage of computer memory. “How to store the matrix” is one of issues in conjunction with the solution scheme.
b
b
b : halfbandwidth 2b1 : bandwidth 3
Example: 2
6
② ①
④ 5 ⑧
③
4
⑥
⑨
9
⑦
8
⑤
7
1
b = (R+1)(NDOF) R : largest difference between the node number in a single element (all elements must be examined one by one.) NDOF : number of d.o.f. for each node.
© 2005 by T. H. Kwon
33
Minimization of bandwidth
Bandwidth depends on the nodal numbering system. bandwidth with an appropriate nodal numbering.
1
5
2
6
3
7
4
8
One can minimize
Frontal Wave Technique
During the assembly procedure, one can recognize that some nodes would be completed in the assembly, thereby no more contribution will be added to that node d.o.f.. Then, one can perform the Gauss elimination procedure for such d.o.f. before proceeding the assembly procedure. Such completed node d.o.f. is called “completed d.o.f.” whereas the other d.o.f. remain active so that such d.o.f. is called “active d.o.f.”. In this manner, one can add an element contribution to the global stiffness matrix, subsequently perform Gauss elimination procedure for the completed d.o.f. and transfer the information related to the completed d.o.f. to an auxiliary memory device such as a hard disk (or a magnetic tape) from the core memory for the global stiffness matrix. One keeps on adding element contributions one by one until the assembly is completed. After the end of assembly, one has whole information for the upper triangular matrix in the auxiliary memory device that will be recovered from it to the core memory to be used one by one for the backward substitution to obtain the solution for the unknown nodal values. This tricky method is called Frontal Wave Technique since the active nodes propagates like a wave front during the simultaneous assembly and elimination.
①
④
② ③
© 2005 by T. H. Kwon
34
2.6 Property of Stiffness Matrix
[Cook, Malkus and Plesha, P.35]
[K ]{x} = {F } . [K ]
is singular since the rigid body motion is possible. One has to introduce geometric constraints to get a unique {x} . Suppose xc , Fc are known d.o.f. and loads (force, mass, energy flux, etc.) xu , Fu are unknown d.o.f. and loads One can rearrange the matrix equation and partition the matrix as ⎡[K 11 ] ⎢[K ] ⎣ 21
[K12 ]⎤ ⎧{xu }⎫ ⎧{Fc }⎫ = [K 22 ]⎥⎦ ⎨⎩{xc }⎬⎭ ⎨⎩{Fu }⎬⎭
Note that either xi or F i is given, but not both as discussed before.
[K11 ]
is nonsingular if the prescribe d.o.f. {x c } are sufficient to prevent rigid body motion. Then the unknown {xu } can be obtained by
{xu } = [K11 ]−1 {{Fc } − [K12 ]{xc }} Then the unknown load can be determined as
{Fu } = [K 21 ]{xu } + [K 22 ]{xc } Instead of following the procedure of rearranging and partitioning matrix described above, one can find several other methods of introducing boundary conditions, which will be discussed next.
2.7 Methods of introducing boundary conditions We have a global matrix equation
[K ]{x} = {F }
© 2005 by T. H. Kwon
35
with boundary conditions as follows: for each degree of freedom, say ith either or CASE 1:
Fi xi
given given
⇒ ⇒
Natural boundary condition Essential boundary condition
when Fi is given
Just introduce Fi = Fi
in the corresponding element of {F } .
CASE 2: when xi is given
a) Elimination of the known xi’s in the matrix equation. (This method is essentially the same as the rearranging and partitioning as described in the previous section.)
The number of unknown is reduced. computational viewpoint.
It looks good, but is bad from the
b) Direct substitution xi = x i
(e.g. x1 = x1 , x3 = x3 and F2 = F2 , F4 = F4 ) ⎡1 0 ⎢0 K 22 ⎢ ⎢0 0 ⎢ ⎣0 K 42
0 ⎤ ⎧ x1 ⎫ ⎧ x1 ⎫ ⎪ ⎥ ⎪ ⎪ 0 K 24 ⎥ ⎪ x 2 ⎪ ⎪ F2 − K 21 x1 − K 23 x3 ⎪⎪ ⎨ ⎬=⎨ ⎬ 1 0 ⎥ ⎪ x3 ⎪ ⎪ x3 ⎪ ⎥⎪ ⎪ ⎪ 0 K 44 ⎦ ⎩ x 4 ⎭ ⎩ F4 − K 41 x1 − K 43 x3 ⎪⎭ 0
© 2005 by T. H. Kwon
36
c) Blasting Technique ⎡ K 11 γ ⎢K ⎢ 21 ⎢ K 31 ⎢ ⎣ K 41
K 12 K 22
K 13 K 23
K 32
K 33 γ
K 42
K 43
K 14 ⎤ ⎧ x1 ⎫ ⎧ x1 K 11 γ ⎫ K 24 ⎥⎥ ⎪⎪ x 2 ⎪⎪ ⎪⎪ F2 ⎪⎪ ⎨ ⎬=⎨ ⎬ K 34 ⎥ ⎪ x3 ⎪ ⎪ x3 K 33 γ ⎪ ⎥ K 44 ⎦ ⎪⎩ x 4 ⎪⎭ ⎪⎩ F4 ⎪⎭
with γ being very large number (e.g. 1015, 1020 etc.) d) Penalty Method ⎡ K 11 + γ ⎢ K ⎢ 21 ⎢ K 31 ⎢ ⎣ K 41
K 12 K 22
K 13 K 23
K 32
K 33 + γ
K 42
K 43
K 14 ⎤ ⎧ x1 ⎫ ⎧ x1 γ ⎫ K 24 ⎥⎥ ⎪⎪ x 2 ⎪⎪ ⎪⎪ F2 ⎪⎪ ⎨ ⎬=⎨ ⎬ K 34 ⎥ ⎪ x3 ⎪ ⎪ x3 γ ⎪ ⎥ K 44 ⎦ ⎪⎩ x 4 ⎪⎭ ⎪⎩ F4 ⎪⎭
with γ being a penalty number (very large number: e.g. 1015, 1020 etc.) The addition of the large penalty number is in fact equivalent to adding a very stiff structural element (such as a stiff spring) to the corresponding d.o.f.
γ
⇔
γ
See also [Bathe Section 3.4] (imposition of constraints) [Bathe P.143146] (Lagrange Multipliers, Penalty Method) We will discuss this subject later again.
© 2005 by T. H. Kwon
37
γ
2.8 Variational Approach in Finite Element Formulation
[Bathe P.110116]
In understanding the phenomena occurring in nature, we are quite used to differential equations to describe the phenomena mathematically based on basic physical principles, namely conservation laws in many mechanical engineering problems. Instead of this differential formulation, physical phenomena can be described in terms of minimization of total energy (or functional) associated with the problem, which is called “variational formulation”. Finite element formulation can be derived by this variational formulation as long as there exists a variational principle corresponding to the problem of interest.
2.8.1 Principle of Minimum Total Potential Energy There is a very important physical principle to describe a deformation process of an elastic body, namely Principle of Minimum Total Potential Energy, which can be summarized as below: Π = U +V
U V
:
Total Potential Energy : Strain Energy : Potential Energy due to external loads (kept constant)
Π is minimum with respect to the state variables or function variables at the equilibrium state This principle can be easily applied to deformation of elastic bodies by identifying the strain energy and potential energy due to external forces which are assumed to be fixed during the deformation. The total potential energy can be a function of state variables or a function of functions, which is called “functional”. For instance, Π = Π (u1 , u 2 , L , u n )
with ui being the state variables
⇒ Π = Π ( f1 (x), f 2 (x),L , f n (x) )
⇒
∂Π = 0, i = 1,L ,n ∂u i How to minimize with respect to functions fi(x)? ⇒ subject of calculus of variation
(It may be noted that there are other variational principles than the Principle of Minimum Total Potential Energy. In this introductory course of FEM, our discussion will be limited to this principle only.)
© 2005 by T. H. Kwon
38
Example 1: Spring Problem
[Bathe P.8687, Ex.3.6]
The deformation of a spring is taken as the simplest example to demonstrate that the variational formulation yields the same equation as the conservation law, i.e., force equilibrium equation. The problem of a linear spring system is depicted in the following figure.
x
k
P
kx = P
i) Equilibrium concept : ii) Energy concept:
1 2 kx 2
Strain energy
:
U=
Potential energy
:
V = − Px
Total potential energy
: Π=
1 2 kx − Px 2
Minimize Π with respect to the variable x :
∂Π = 0 ⇒ kx = P ∂x
Example 2: Tight String Problem This is the first example of a functional. A tight string with a tension T applied at the end walls is under a distributed load w(x) as depicted below: w(x) x
y(x) y
Find y(x) for a given w(x).
© 2005 by T. H. Kwon
39
d ⎛ dy ⎞ ⎜ T ⎟ + w( x) = 0 dx ⎝ dx ⎠
i) Equilibrium concept :
ii) Energy concept:
2
Strain energy
:
1 ⎛ dy ⎞ U = ∫ T ⎜ ⎟ dx 0 2 ⎝ dx ⎠
Potential energy
:
V = − ∫ wydx
Total potential energy
l
l
0
2 ⎤ l ⎡ 1 ⎛ dy ⎞ : Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy ⎥dx 0 2 ⎢⎣ ⎝ dx ⎠ ⎥⎦
Find y(x) to minimize Π ( y (x) ) , which is a function of a function, i.e., “Functional”
There are many interesting examples of functional. To enhance motivation, one more example will be introduced, namely Brachistochrone Problem by Bernoulli 1696: A
x
y
B
One wants to find y(x) for the minimum falling time for fixed A and B points.
2
t ( y ( x) ) = ∫
B
0
© 2005 by T. H. Kwon
B ds =∫ 0 v
⎛ dy ⎞ 1+ ⎜ ⎟ ⎝ dx ⎠ dx 2 gy
40
: a functional to be minimized.
2.8.2 Calculus of Variation Calculus of variation deals with such problems to minimize a functional. At this
point, it will be explained in a concise manner so that a beginner can start with a variational formulation for FEM with no difficulty. As the first example of calculus of variation, consider a functional of the following form:
I (φ( x) ) = ∫ F ( x, φ, φ x ,φ xx )dx x2
: functional
x1
∂φ ∂ 2φ , φ xx = 2 . ∂x ∂x Find φ(x) which minimizes I (φ( x)) . with φ x =
y
φ(x) : exact solution δφ
~ φ = φ + δφ δφ : small variation x
~ Consider an approximate solution φ which has a small variation δφ over an exact solution φ( x) . That is, ~ φ ( x) = φ( x) + δφ( x) The approximate solution is substituted to the functional expression yielding
(
) ∫
~ I φ ( x) =
x2
x1
~~ ~ F ( x, φ, φx ,φxx )dx
= I (φ) + δI In order for φ(x) to be the solution, δI = 0 for any δφ .
© 2005 by T. H. Kwon
41
x2 ⎛ ∂F ⎞ ∂F ∂F δI = ∫ ⎜⎜ δφ + δφ x + δφ xx ⎟⎟dx, x1 ∂φ x ∂φ xx ⎠ ⎝ ∂φ
(
∂F δx is not to be included.) ∂x
x2 ⎛ ∂F ⎞ ∂F d ∂F d (δφ) + (δφ x ) ⎟⎟dx = ∫ ⎜⎜ δφ + x1 ∂φ x dx ∂φ xx dx ⎠ ⎝ ∂φ
Integrate by parts yields x2 ⎡ ∂F d ⎛ ∂F δI = ∫ ⎢ δφ − ⎜⎜ x1 dx ⎝ ∂φ x ⎣ ∂φ
⎞ d ⎛ ∂F ⎟⎟δφ − ⎜⎜ dx ⎝ ∂φ xx ⎠
x2
x2
1
x1
∂F ∂F + δφ + δφ x ∂φ x ∂ φ xx x
⎤ ⎞ ⎟⎟(δφ x )⎥dx ⎠ ⎦
One more integral by parts gives x2 ⎡ ∂F d ⎛ ∂F δI = ∫ ⎢ − ⎜⎜ x1 ⎣ ∂φ dx ⎝ ∂φ x
⎞ d 2 ⎛ ∂F ⎞⎤ ⎟⎟ + 2 ⎜⎜ ⎟⎟⎥δφdx ⎠ dx ⎝ ∂φ xx ⎠⎦ x2
⎡ ∂F d ⎛ ∂F ⎞⎤ ∂F ⎟⎟⎥ δφ + +⎢ − ⎜⎜ δφ x ⎣ ∂φ x dx ⎝ ∂φ xx ⎠⎦ x1 ∂φ xx
(2.16)
x2
=0 x1
δI = 0 for arbitrary δφ implies that
∂F d ⎛ ∂F − ⎜ ∂φ dx ⎜⎝ ∂φ x
⎞ d 2 ⎛ ∂F ⎟⎟ + 2 ⎜⎜ ⎠ dx ⎝ ∂φ xx
⎞ ⎟⎟ = 0. ⎠
x1 < x < x 2
: EulerLagrange equation
and the following set of boundary conditions ∂F d ⎛ ∂F − ⎜⎜ ∂φ x dx ⎝ ∂φ xx
⎞ ⎟⎟ = 0 ⎠
or
δφ = 0
or
δφ x = 0
at x = x1 and x = x 2
and ∂F =0 ∂φ xx
Natural B.C.
© 2005 by T. H. Kwon
at x = x1 and x = x 2
Essential B.C.
42
Properties of δ operator
i) The laws of variations of sums, products, and so on are completely analogous to the corresponding laws of differentiation. δ( F + G ) = δF + δG,
δ( FG ) = (δF )G + F (δG ),
ii) interchangeability between
∫
δ ( F n ) = nF n −1δF
and δ :
x2 x2 ~ x2 x2 ~ x2 x2 x2 δ ∫ Fdx = ∫ Fdx − ∫ Fdx = ∫ ( F − F )dx = ∫ δFdx ⇒ δ ∫ Fdx = ∫ δFdx x1
x1
x1
x1
iii) interchangeability between
x1
x1
x1
d and δ : dx
~ ⎫ d φ dφ d = + (δφ)⎪ dx dx dx ⎪ ⎬ ~ d φ dφ ⎛ dφ ⎞ ⎪ = + δ⎜ ⎟ ⎪ dx dx ⎝ dx ⎠ ⎭
⇒
⎛ dφ ⎞ d (δφ) δ⎜ ⎟ = ⎝ dx ⎠ dx
Example 1 : Tight string problem 2 ⎤ l ⎡ 1 ⎛ dy ⎞ Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy ⎥dx 0 2 ⎢⎣ ⎝ dx ⎠ ⎥⎦
(2.17)
EulerLagrange equation: ∂F ∂φ ∂F ∂φ x
∴ B.C.:
∂F = −w ∂y ∂F dy =T ∂y x dx
→ →
∂F ∂φ xx
→
−w−
d ⎛ dy ⎞ ⎜T ⎟ = 0 dx ⎝ dx ⎠
T
dy =0 dx
© 2005 by T. H. Kwon
∂F =0 ∂y xx
or
: confirmed δy = 0
at x = 0 and x = l
43
Example 2 : Bending beam problem
w(x) x E: Young’s modulus I : Moment of Inertia
y(x)
y
The total potential energy can be expressed as ⎡1 ⎛ d 2 y ⎞2 ⎤ Π ( y ( x) ) = ∫ ⎢ EI ⎜⎜ 2 ⎟⎟ − wy ⎥dx 0 2 ⎢⎣ ⎥⎦ ⎝ dx ⎠ l
(2.18)
EulerLagrange equation:
−w+
d2 dx 2
⎛ d2y⎞ ⎜⎜ EI 2 ⎟⎟ = 0 dx ⎠ ⎝
Boundary conditions: d ⎛ d2y⎞ ⎜ EI 2 ⎟⎟ = 0 dx ⎜⎝ dx ⎠
δy = 0
or
at x = 0 and x = l
and d2y EI 2 = 0 dx
⎛ dy ⎞ δ⎜ ⎟ = 0 ⎝ dx ⎠
or
Natural boundary condition Remind:
at x = 0 and x = l
Essential boundary condition
w(x) M+dM
M V+dV
V
dM dV =V, = −w dx dx d2y d3y d4y EI 2 = − M , EI 3 = −V , EI 4 = w dx dx dx
dx © 2005 by T. H. Kwon
44
Summary of EulerLagrange Equations for Various Forms of Functionals
A) Functions involving higher order derivatives: x2
I = ∫ F ( x, φ, φ′, φ′′, L, φ ( n ) )dx x1
n integration by parts yields the EulerLagrange equation:
n
∑ (−1) i i =0
di dx i
⎛ ∂F ⎞ ⎜⎜ ( i ) ⎟⎟ = 0 ⎝ ∂φ ⎠
(2.19)
B) Functions involving one independent variable but several dependent variables and their first derivatives: x2
I = ∫ F ( x, φ1 , φ 2 ,L , φ n , φ1′ , φ′2 ,L, φ′n )dx x1
This involves n different variables δφ n and yields n EulerLagrange equations: ∂F d ⎛ ∂F ⎞ ⎟=0 − ⎜ ∂φ i dx ⎜⎝ ∂φ′i ⎟⎠
i = 1, L , n
(2.20)
C) Functions involving more than one independent variables and one dependent variable and its first derivatives: I=
∫
Ω
F ( x, y, φ , φ x , φ y )dxdy
Using Gauss Theorem (Green’s Theorem in plane)
∂F ∂ˆ ⎛ ∂F − ⎜ ∂φ ∂x ⎜⎝ ∂φ x ∂ˆ ⎛ ∂F ⎜ ∂x ⎜⎝ ∂φ x
where
⎞ ∂ˆ ⎛ ∂F ⎟⎟ − ⎜ ⎜ ⎠ ∂y ⎝ ∂φ y
⎞ ⎟=0 ⎟ ⎠
(2.21)
⎞ ∂2F ∂ 2 F ∂φ ∂ 2 F ∂φ x ∂ 2 F ∂φ y ⎟⎟ ≡ , etc. + + + 2 ⎠ ∂φ x ∂x ∂φ x ∂φ ∂x ∂φ x ∂x ∂φ x ∂φ y ∂x
* Gauss Theorem (Green’s Theorem in plane) r r ∇ ⋅ V d Ω = V ∫ ∫ ⋅ nˆdS Ω
∂Ω
∂φ
∂φ
∫ ∂x dxdy = ∫ φn ds, ∫ ∂y dxdy = ∫ φn ds x
A
© 2005 by T. H. Kwon
∂A
y
∂A
A
45
Example 3 : Hamilton’s Principle and Lagrange’s equation in dynamics
Dynamic motion of a rigid body system can be described in several ways. Newton’s 2 law of motion is one of them. Other methods are based on energy concept . nd
Newton’s equation of motion is represented by r r F = m&x&
Lagrange’s equation can be written as d ⎛ ∂L ⎜ dt ⎜⎝ ∂q& i
⎞ ∂L ⎟⎟ − =0 ⎠ ∂qi
with L = T − V being Lagrange function
(2.22)
(or kinetic potential)
where qi : generalized coordinate T : kinetic energy V : potential energy One can easily derive Lagrange’s equation via calculus of variation from Hamilton’s principle that is represented by the following (case B in the summary). t2 δ ⎡ ∫ (T − V )dt ⎤ = 0 ⎢⎣ t1 ⎥⎦
or
(2.23)
t2
minimize I = ∫ (T − V )dt . t1
Which implies that “actual path followed by a dynamic process is such as to make the integral of (TV) a minimum”. Hamilton’s principle for deformable body can be stated as t2 t2 δ ⎡ ∫ (T − Π )dt ⎤ = δ ⎡ ∫ Ldt ⎤ = 0 ⎢⎣ t1 ⎥⎦ ⎢⎣ t1 ⎥⎦
with
L = T − Π,
© 2005 by T. H. Kwon
Π : total potential energy (Π = U + V )
46
(2.24)
2.8.3 Boundary Conditions in Variational Principle It may be noted that a variational principle gives rise to not only EulerLagrange equations but also boundary conditions associated with the problem. In fact, the functional itself include the effect of boundary conditions accordingly. We will pay attention to the method of introducing boundary condition effect to the functional. Consider the tight string problem as an example for this purpose.
w(x) x
y(x) y The functional for this problem was 2 ⎤ l ⎡ 1 ⎛ dy ⎞ Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy ⎥dx . 0 2 ⎢⎣ ⎝ dx ⎠ ⎥⎦
(2.25)
At this time, we will derive the EulerLagrange equation and the associated boundary conditions by using the δ operator to the variation of the functional without referring to the formulae derived before.
~ y = y + δy
⇒
Π( ~ y ) = Π ( y ) + δΠ
⎡ 1 ⎛ dy ⎞ 2 ⎤ δΠ = ∫ ⎢ Tδ⎜ ⎟ − wδy ⎥dx 0 2 ⎝ dx ⎠ ⎣⎢ ⎦⎥ l
l⎡ = ∫ ⎢T 0 ⎣ l⎡ = ∫ ⎢T 0 ⎣
⎤ dy ⎛ dy ⎞ δ⎜ ⎟ − wδy ⎥dx dx ⎝ dx ⎠ ⎦ dy d (δy ) − wδy ⎤⎥dx dx dx ⎦ l
⎡ d ⎛ dy ⎞ ⎤ dy = ∫ ⎢− ⎜ T ⎟ − w⎥δydx + T δy = 0 0 dx 0 ⎣ dx ⎝ dx ⎠ ⎦ l
EulerLagrange equation: B.C.:
T
dy =0 dx
or
© 2005 by T. H. Kwon
d ⎛ dy ⎞ ⎜ T ⎟ + w = 0, dx ⎝ dx ⎠
0< x 0
~ d p∆ d p∆ − ≤ M p h n − p +1 → 0 as h → 0 p p dx dx
n +1− p > 0
if
The error in the strain energy with the highest order derivative (m) in U ~ U (∆ − ∆) ≤ M p h 2( n −m +1) → 0
as
h→0
if
n − m +1 > 0
because U is quadratic in the mth derivative of ∆ Therefore, for necessary condition for convergence, the element should satisfy: n − m +1 > 0
n ≥ m is the minimum requirement
[Bathe P. 244 254]
↓ It also defines a constant strain. m
n
Displacement( n+1 )
Energy(2 ( n – m + 1))
1
1 (min.)
O(h 2 )
O(h 2 )
1
2
O(h 3 )
O(h 4 )
2
2 (min.)
O(h 3 )
O(h 2 )
© 2005 by T. H. Kwon
135
4. NUMERICAL INTEGRATION A typical stiffness matrix coefficient looks like the following: ⎛ ∂N i ∂N j ∂N i ∂N j ⎞ ⎟⎟ J dξdη K ije = ∫ ⎜⎜ k x + ky ∂ x ∂ x ∂ y ∂ y ⎠ Ωe⎝ where the derivatives of shape functions with respect to global coordinate system are included in the integrand together with Jacobian of coordinate transformation between the normalized (or natural) coordinate and global coordinate systems. One usually needs the following conversion of the derivative of shape function since the shape functions are defined in terms of the normalized (or natural) coordinate.
⎧ ∂N i ⎫ ⎧ ∂N i ⎫ ⎪ ⎪ ⎪⎪ ∂x ⎪⎪ −1 ⎪ ∂ξ ⎪ J [ ] = ⎨ ∂N ⎬ ⎨ ∂N ⎬ ⎪ i⎪ ⎪ i⎪ ⎪⎩ ∂y ⎪⎭ ⎪⎩ ∂η ⎪⎭ ⎡ ∂x ⎢ [J] = ⎢ ∂∂ξx ⎢ ⎢⎣ ∂η
∂y ⎤ ⎡ ∂N i xi ∂ξ ⎥ ⎢ ∂ξ ⎥=⎢ ∂y ⎥ ⎢ ∂N i xi ∂η ⎥⎦ ⎢⎣ ∂η
∂N i ⎤ yi ∂ξ ⎥ ⎥ ∂N i ⎥ yi ∂η ⎥⎦
We end up with the following integral form: 1
1
∫∫
−1 −1
f (ξ ,η )dξ dη
which is too complicated to evaluate analytically thus needs numerical integration. The most convenient numerical integration is “Gauss Quadrature Integral” which is described below. In this chapter, we will introduce the Gauss quadrature integral and rule of thumb with regard to the order of accuracy requirement for numerical integral.
4.1 Gauss Quadrature Integral Gauss quadrature integral for onedimensional integration is described below:
© 2005 by T. H. Kwon
145
n
I = ∫ f (ξ )dξ = ∑ H i f (ξ i ) 1
−1
(4.1)
i =1
where Hi are weighting functions for corresponding sampling points ξi.
f ×
1
ξ1
×
×
ξ3
ξ2
×
ξ4
ξ 1
The essential part of Gauss quadrature integral is the way of determining the sampling points and corresponding weighting functions. They are determined in such a way that Gauss quadrature integral can accurately evaluate the integral of polynomial up to (2n1)th order when n sampling points are used. It can be explained easily as below:
One wants to determine Hi and ξi such that Gauss quadrature integral can exactly evaluate for any polynomial of
f (ξ ) = a1 + a 2ξ + a3ξ 2 + L + a 2 nξ 2 n −1 2 n 2a 2 i f (ξ )dξ = 2a1 + a3 + L = ∑ −1 3 i : odd i 1
Then,
I =∫
Also
I = ∑ H i f (ξ i ) = function of (a1 , a 2 ,L, a 2 n )
n
i =1
The above two integration should be identical to any ai’s so that one obtains 2n equations from which one can determine n sampling points ξi and n weighting functions Hi. In this way a set of those information is established and tabulated in many numerical analysis books. Refer, for instance, to Table 5.6 in [Bathe] or Table 8.1 [Zienkiewicz] The Gauss quadrature integral developed for onedimensional integration can be easily extended to twodimensional and threedimensional cases as described below:
© 2005 by T. H. Kwon
146
For twodimensional case: 1
I =∫
1
∫
−1 −1 1
∫
−1
f (ξ ,η )dξ dη
f (ξ ,η )dξ = ∑ H j f (ξ j ,η ) = ψ (η ) n
j =1
n n 1 ⎡n ⎤ I = ∫ ψ (η )dη = ∑ H iψ (η i ) = ∑ H i ⎢∑ H j f (ξ j ,η i )⎥ −1 i =1 i =1 ⎣ j =1 ⎦
= ∑∑ H i H j f (ξ j ,η i ) n
n
(4.2)
i =1 j =1
where H i H j plays a role of a weighting function for the sampling point denoted by (i,j). There is also similar Gauss quadrature integral formula for triangular element. I = ∫ f ( x, y )dxdy = A
=
∫ f (ξ ,η ) J dξdη
A′
(4.3)
#Q . P .
1 ∑ f (ξ n ,η n ) J n Wt (n)
2
n =1
η
y ( x, y ) ↔
(L1 , L2 , L3 )
↔ (ξ ,η ,1 − ξ − η )
ξ
x
ξ n , η n and Wt (n) are tabulated, for instance, in Table 8.2 [Zienkiewicz], or Table 5.8 [Bathe]
© 2005 by T. H. Kwon
147
© 2005 by T. H. Kwon
148
© 2005 by T. H. Kwon
149
4.2 Minimum Order of Integration
[Bathe: Sect. 5.5.5] [Cook: P.190193] [Zienkiewicz: P.202205] [Hughes: P.239 & Chap. 4]
It is of our great interest to evaluate the integration numerically as accurately as possible. More accurate numerical integration needs more Gauss integration points and more number of function evaluations, which in turn requires more computational time. There is a tradeoff between the accuracy and computing cost one has to bear in mind in choosing the order of integration. In the meanwhile, for a given polynomial function of order, for instance, (2n1), we already know that Gauss quadrature points exceeding n does not help at all. In this regard, there should be some guidance on how to choose the number of Gauss quadrature points for a satisfactory integration.
4.2.1 Rule of Thumb There is a certain “rule of thumb” for the fundamental requirement as described below: i) The integral should be accurate enough to evaluate correctly the volume integral (i.e. for constant strain case in elasticity problem) Ω = ∫ J dξ dη dς Ω
ii) A “full numerical integration” is to integrate accurately the stiffness matrix coefficient with undistorted element (i.e. J = constant ),
[K ] = ∫ [N ′] [C ][N ′] J dV T
An integration lower than the full numerical integration may result in unreliable numerical analysis. [Bathe P.469] It might be reminded that, for a general case, the error in evaluating the total potential energy is represented by
( )
(
(
)
Π p ∆% − Π p ( ∆ ) = U ∆% − ∆ = O h 2( n −m+1)
© 2005 by T. H. Kwon
150
)
Therefore, with a polynomial of order n for the approximation of field variable, the integral higher than the order of 2(nm) polynomial will not improve the order of discretization error at all.
4.2.2 Spurious Rigid Body Motion Eigenvalue test with an exact integral of [K] can show the number of rigid body modes. However, spurious rigid body mode may occur when a numerical integration is used to evaluate [K]. The spurious rigid body mode (or zeroenergy mode) may result in a singularity in [K]. The spurious rigid body mode is possible due to the presence of a deformation state for which all strain components at the Gauss quadrature points happen to be zero. There is a method to determine whether or not there is a chance of such a singularity taking place for a specific order of Gauss quadrature integral given a certain element type. It is briefly summarized as below: ⎛ Number of independent ⎞ ⎟ ⎛ Number of d.o.f. ⎞ ⎛ Number of ⎞ ⎜ ⎜⎜ ⎟⎟  ⎜⎜ ⎟⎟ = ⎜ equations needed to ⎟ ⎝ for (an) element ⎠ ⎝ rigid body modes ⎠ ⎜ avoid rank deficiency ⎟ ⎝ ⎠
⎛ This can be d.o.f. either ⎞ ⎜ ⎟ ⎜ for Assemblage of elements ⎟ ⎜ or for an element ⎟ ⎝ ⎠
⎛ Rank of ⎞ ⎟⎟ = ⎜⎜ ⎝ real [K ] ⎠
⎛ equal to number of ⎞ ⎜ ⎟ ⎜ constraints to get ⎟ ⎜ a unique solution ⎟ ⎝ ⎠
(For example, there are three rigid body modes in a twodimensional elastic deformation problem as indicated below.
Note that there are three essential boundary conditions.)
© 2005 by T. H. Kwon
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For displacement based FEM, there is one independent equation for each strain component at each Gauss points. Therefore, the number of independent conditions can be calculated by the following:
⎛ Number of independent ⎞ ⎛ Number of ⎞ ⎛ Number of strain ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ × ⎜⎜ ⎟⎟ ⎝ conditions ⎠ ⎝ Gauss points ⎠ ⎝ components ⎠
Examples of element analysis:
Ex. 1. (d.o.f.)

(R.B.M.)
2×3

3
= No. of Eqs. needed =
3
: nodes : Gauss points ⎛ No. of independent ⎞ ⎛ No. of Gauss ⎞ ⎛ No. of strain ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ × ⎜⎜ ⎟⎟ conditions points components ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 3 3=3
=
∴
×
1
3
nonsingular
Ex.2. (d.o.f.)

(R.B.M.)
2×4

3
= No. of Eqs. needed =
⎛ No. of independent ⎞ ⎛ No. of Gauss ⎜⎜ ⎟⎟ = ⎜⎜ conditions ⎝ ⎠ ⎝ points 3 = 1 Rank deficiency = 5  3 = 2 > 0 ∴ singular
© 2005 by T. H. Kwon
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5 ⎞ ⎛ No. of strain ⎞ ⎟⎟ × ⎜⎜ ⎟⎟ components ⎠ ⎝ ⎠ × 3
In this case, it is very interesting to understand where those two rank deficiency originate from. A fournode quadrilateral element has three rigid body modes, three constant strain modes and two bending modes. Two bending modes have zero strain energy at the center Gauss quadrature point, thus giving rise to two spurious zero energy modes. The following displacement shapes explain the zero strain energy modes:
u=0 v = aξη
u = aηξ v=0
With such displacement fields, the strain at the center becomes identically zero, i.e. u,ξ = u,η = v,ξ = v,η = 0 at ξ = η = 0 point
It might be mentioned here that the socalled “Hourglass” mode (or termed Mechanism, Kinematic mode, zeroenergy mode, etc.) can take place as indicated in the following figures: [Cook et al. P.191]
If one takes two Gauss quadrature points, the number of independent conditions becomes six (2×3=6), thus exceeds the number of independent equations needed, i.e. five. Therefore the stiffness matrix becomes nonsingular. Of course, four quadrature points makes nonsingular stiffness matrix, too.
© 2005 by T. H. Kwon
153
Ex.3.
[Bathe: P.472, P.479]
η
(d.o.f.)

(R.B.M.)
2×8

3
= No. of Eqs. needed =
13
ξ
⎛ No. of independent ⎞ ⎛ No. of Gauss ⎞ ⎛ No. of strain ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ × ⎜⎜ ⎟⎟ ⎝ conditions ⎠ ⎝ points ⎠ ⎝ components ⎠ 12 = 4 × 3 Rank deficiency = 13 – 12 = 1 > 0 ∴ singular In this particular case, it can be shown that the following is responsible for the spurious zerostrain energy mode: Gauss Q.P. at ξ = ±
(
1 3
,η=±
1 3
)
1 u = ξ 3η 2 − 1 2 v = η 1 − 3ξ 2
(
)
u,ξ = u,η = v,ξ = v,η = 0 at ξ = ±
1 3
,η=±
1 3
If assemblage is finished, interelement compatibility may prevent the zerostrain energy mode from happening. Ex.4.
η
(d.o.f.)

(R.B.M.)
2×9

3
= No. of Eqs. needed =
15
ξ
⎛ No. of independent ⎞ ⎛ No. of Gauss ⎞ ⎛ No. of strain ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ × ⎜⎜ ⎟⎟ conditions points components ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 12 = 4 × 3 Rank deficiency = 15 – 12 = 3 > 0 ∴ singular In this particular case, three rank deficiency can be demonstrated by the following
© 2005 by T. H. Kwon
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figures:
u = 3ξ 2η 2 − ξ 2 − η 2 v=0
(
)
1 u = ξ 3η 2 − 1 2 v = η 1 − 3ξ 2
v = 3ξ 2η 2 − ξ 2 − η 2 u=0
(
)
The eightnode quadrilateral element, a serendipity element, does not have the first two zerostrain energy modes since it does not include the ξ 2η 2 term. If one employs 3×3 Gauss quadrature points, there is no spurious zerostrain energy mode, as expected. Examples of analysis for assembled elements:
4× 2  3 = = 1× 3
5 3
(independent d.o.f. : needed) (independent conditions : provided)
singular (as before)
8 × 2  3 = 13 (independent d.o.f. : needed) = 12 (independent conditions : 4×3 provided) singular (as before)
6× 2  3 = 9 = 6 2×3 singular
© 2005 by T. H. Kwon
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(independent d.o.f. : needed) (independent conditions : provided)
13 × 2  3 = 23 = 24 8×3
(independent d.o.f. : needed) (independent conditions : provided)
nonsingular (assemblage of elements prevents the singularity from happening.)
16 × 2 16 × 3
= 32 (independent d.o.f. : needed) = 48 (independent conditions : provided)
nonsingular
48 × 2 (4 × 16) × 3
= 96 (independent d.o.f. : needed) = 192 (independent conditions : provided)
nonsingular
25 × 2  3 = 47 = 48 16 × 3 nonsingular
© 2005 by T. H. Kwon
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(independent d.o.f. : needed) (independent conditions : provided)
4.2.3 Reduced Numerical Integration
[Bathe P.476478] [Zienkiewicz P.284]
Usually, FEM leads to too much constrained system and thus [K ] is too large (i.e. overestimated) and {x} becomes too small (i.e. underestimated, in other words, resulting in lower boundness). By reducing the order of integral for [K ] , error may compensate for the overtightened [K ] otherwise obtained. This may result in a better solution (but, sometimes, not always)! This reduction in the order of integral is called “Reduced Numerical Integral”. Or one could introduce “Selective Integration” which allows different order of integration for different strain components. Direction dependent reduced numerical integration could be one of such selective integrations, while one could also introduce, for instance, 2×2 Gauss quadrature points, with just one quadrature point at the centroid for the shear components.
These kinds of numerical integrations require experience without a systematic and scientific method to rely on, thus are still kind of art yet. However, lots of research efforts are being made in this direction to improve finite element analysis results. One may further refer to [Bathe, P.476477] in terms of mixed formulation. Related topic is discussed via penalty function formulation in [Zienkiewicz, P.284296].
4.3 Interpretation of Stress in Finite Element Analysis
[Bathe P.254259]
After the approximate displacement field is obtained as a solution of displacement based finite element analysis, one may want to find the stress distribution, which requires numerical differentiation of displacement vector field with respect to the global coordinate for strain components, and then calculating the stress making use of the stressstrain relationship (i.e. constitutive equation). In this procedure, the numerical differentiation usually does not provide stress results as accurately as the displacement field itself. Therefore, one has to be careful in interpreting the stress distribution with this numerical inaccuracy in mind. For instance, when a linear triangle element is chosen in the elasticity analysis, the strain and stress becomes constant over each element as illustrated below:
© 2005 by T. H. Kwon
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stress
It may be noted that [K ] = ∫ [N ′] [C ][N ′]dV is evaluated by function evaluations at T
Ω
Gauss quadrature points. Therefore, it is usually the case that the stress components are generally accurate at those Gauss quadrature points. Is this regard, evaluation of stress at the Gauss quadrature points is usually the best choice. One may want to find stress value at nodes. In this case, the weighted average of stress at elements which include the node with the subtended angle, area, etc. used as weighting functions. This kind of weighting scheme usually provides satisfactory stress result even it the weighting looks ad hoc in its nature.
θi
(There are some other methods. used.)
© 2005 by T. H. Kwon
But the methods described above are commonly
158
5. SOLUTION TECHNIQUES Once the stiffness matrix and force matrix are evaluated, boundary conditions should be introduced to the system of equation. Then a set of matrix equation should be solved to obtain the unknown variables at nodes. From the computational viewpoint, there are two issues to consider for efficient computation for a given computer system available: i) storage technique ii) solution techniques.
5.1 Storage Technique It is of importance to minimize the storage requirement for a given computer capability. Two techniques are introduced below.
5.1.1 Bandwidth
0 ⇒
⇒
0
Banded matrix
2D array
1D array
Both 2D array and 1D array can save the storage significantly when nodal numbering is arranged in such a way that the bandwidth is minimized. However, even inside the banded array, there are many zeroes for sparse banded matrix, as is usually the case for finite element analysis. In this regard, the following storage scheme is recommended.
© 2005 by T. H. Kwon
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5.1.2 Skyline
[Bathe P.985986]
Μ k = 3 : half  bandwidth  1 ⎡k11 ⎢ ⎢ ⎢ ⎢ [K ] = ⎢⎢ ⎢ ⎢ ⎢ ⎢ ⎣⎢
k12 k 22
0 k 23
k14 0
0 0
0 0
0 0
k 33
k 34
0
k 36
0
k 44
k 45
k 46
0
k 55
k 56
0
k 66
k 67
sym.
k 77
=
i − mi
( m6 = 3, m7 = 6 , etc.) = column height of column i =
skyline
the row number of first nonzero element in column i → skyline
mi
Μk
0⎤ 0 ⎥⎥ 0⎥ ⎥ 0⎥ k 58 ⎥ ⎥ k 68 ⎥ k 78 ⎥ ⎥ k 88 ⎦⎥
(e.g. i = 6 : 6 – 3 = 3) (halfbandwidth – 1) = max ( i − mi ),
i = 1,L , n
The storage scheme is as follows: ⎡ A(1) ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣
A(3) A(2)
A(5) A(4)
A(9) A(8) A(7) A(6)
A(11) A(10)
A(15) A(14) A(13) A(12)
Diagonal: 1, 2, 4, 6, 10, 12, 16, 18 → Solution requires the operation of
→
A(I )
MAXA( I ) ⇒ k ii stored in A(MAXA( I ) )
1 2 nΜ k with storage of O(nΜ k ) . Bandwidth 2
minimization is needed.
© 2005 by T. H. Kwon
A(17) A(16)
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ A(21) ⎥ ⎥ A(20)⎥ A(19) ⎥ ⎥ A(18) ⎥⎦
160
5.1.3 Frontal Wave Technique The essential feature of Frontal Wave Technique has already been explained along with the bandwidth minimization in Section 2.5. During the assembly procedure, some nodes would be completed so that no more contribution will be added to that d.o.f. afterwards. Such d.o.f. is called “completed d.o.f. and one can immediately perform the Gauss elimination before finishing the assembly procedure. In contrast to the completed d.o.f., the other d.o.f. remains active so that such d.o.f. is called “active d.o.f.”. In this manner, one can add an element contribution to the global stiffness matrix, subsequently perform Gauss elimination procedure for the completed d.o.f. and transfer the information related to the completed d.o.f. to an auxiliary memory device from the core memory. The core memory keeps just the coefficients of the active d.o.f. In this kind of sequential assembly procedure, the active nodes will be propagated like a wave front. This assembly and elimination procedure can be done element by element with the significant saving of core memory.
①
④
② ③
5.2 Solution Technique
[Bathe Chap.8]
One wants to solve the unknown {x} in the following matrix equation.
[K ]{x} = {F }
{x} = [K ]−1 {F }
→
5.2.1 Gauss Elimination
[K ]{x} = {F }
[Bathe P. 439448]
[S ]{x} = {F ′}
→
upper triangular matrix
L−n1−1 L L−21 L1−1 K = S © 2005 by T. H. Kwon
161
where ⎡1 ⎢ 1 ⎢ ⎢ 1 ⎢ 1 ⎢ ⎢ 1 L−i 1 = i ⎢ 1 ⎢ ⎢ − l i +1,i ⎢ 0 − l i + 2 ,i ⎢ ⎢ M ⎢ ⎢⎣ − l n ,i
⎤ ⎥ ⎥ ⎥ 0 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 1 ⎥ 1 ⎥ 1 ⎥ ⎥ 1⎥⎦
with
l i + j ,i =
K i(+i )j ,i K i(,ii)
where Kij is the element of L−i −11 L−i −12 L L1−1 K . Then
K = L1 L2 L Ln −1 S = LS
(LU Decomposition)
where Li is obtained by reversing the signs of the offdiagonal elements in L−i 1 , and L becomes the following: ⎡1 ⎢l ⎢ 21 ⎢l 31 L=⎢ ⎢l 41 ⎢M ⎢ ⎢⎣l n1
⎡ S11 ⎢ ⎢ ⎢ S=⎢ ⎢ ⎢ ⎢ ⎢⎣
S12 S 22
0
1 l32 l 42 M ln2
0 1 1 1 l n ,n −1
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 1⎥⎦
⎡ ⎢1 ⎢ ⎢ ⎢ ~ ⎢ S =⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣
S13 L L S1n ⎤ S 23 L L S 2 n ⎥⎥ S 33 L L S 3n ⎥ ⎥, M M M ⎥ M M ⎥ ⎥ S nn ⎥⎦
© 2005 by T. H. Kwon
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S12 S11 1
S13 S11 S 23 S 22 1
0
S1n ⎤ S11 ⎥ ⎥ S 2n ⎥ L L S 22 ⎥ S 3n ⎥ ⎥ L L S 33 ⎥ M M M ⎥ ⎥ M M ⎥ 1 ⎥⎦ L L
⎡ S11 ⎢ ⎢ ⎢ D=⎢ ⎢ ⎢ ⎢ ⎣⎢
⎤ ⎥ 0 ⎥ ⎥ ⎥ M ⎥ ⎥ M ⎥ S nn ⎦⎥
S 22 S 33 0
i.e.
~ S = DS
then
~ K = LDS
: diagonal matrix ( Dii = S ii )
For symmetric K: ~ ~ K = K T = LDS = S T D T LT
∴
K = LDLT
~ S = LT
→
which is a unique decomposition
: LDLT decomposition
(This decomposition is good for many {F} for nonlinear problem.) Once this decomposition is done, one can get the solution in the following procedure: 1st step: Forward substitution
[L]{y} = {F }
→
calculate
{y} = [L]−1 {F } = L−n1−1 L L−21 L1−1 {F }
(In fact, this can be done at the same time as decomposition.) 2nd step: Backward substitution
[D][L]T {x} = {y} (or
→
get {x}.
[L]T {x} = [D]−1 {y})
One may refer to [Bathe P.985986] for a computer implementation for the skyline reduction method.
© 2005 by T. H. Kwon
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5.2.2 Cholesky Factorization
[Bathe P.449]
~~ K = L LT where 1
1
~ L = LD 2
1
( K = LD 2 D 2 LT = LDLT )
* It applies to positive definite [K ] only. That is, all the diagonal coefficients must be positive. * It can be used in transformation of a generalized eigenvalue problem.
Kφ = λMφ into a standard form as demonstrated as follows: Cholesky factorization of M can be stated as
M = ss T Defining a new variable:
[Bathe P.573]
(i.e. corresponding to LLT )
~
φ ≡ sTφ
the original equation can then be rewritten as
( ) ( )
T ~ ~ K s −1 φ = λss T φ = λsφ T ~ ~ s −1 K s −1 φ = λφ
i.e.
~ ~~ Kφ = λφ
with
T ~ K ≡ s −1 K (s −1 ) .
5.2.3 GaussSeidel Iteration Method K ij x j = Fi One may suggest the simplest iterative scheme, namely Jacobi iterative method:
© 2005 by T. H. Kwon
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n
Fi − ∑ K ij x (jk −1) j =1 j ≠i
xˆ i( k ) =
i = 1,L , n
,
K ii
(old values in R.H.S.)
xi( k ) = αxˆ i( k ) + (1 − α ) xi( k −1)
α being a relaxation parameter,
0 < α ≤ 1.
This iterative method requires computer storage for the present iteration as well as the previous one. Storage saving and fast converging iterative scheme is to replace the old ones by new ones as soon as they are computed during the iteration. This method is called “GaussSeidel method” which can be written as i −1
xˆ i( k ) =
Fi − ∑ K ij x (jk ) − j =1
n
∑K
j =i +1
ij
x (jk −1) ,
K ii
i = 1, L , n
This GaussSeidel iterative method might have a faster convergent solution by introducing an overrelaxation method as modified by
xi( k ) = xi( k −1) + β
i −1
n
j =1
j =i
Fi − ∑ K ij x (jk ) − ∑ K ij x (jk −1) K ii
,
i = 1, L , n
β being the overrelaxation factor, normally larger than unity (e.g.1.3 ≤ β ≤ 1.9) It might be mentioned that an underrelaxation is sometimes recommended, in which case, 0 < β ≤ 1 .
5.2.4 Conjugate Gradient Method Solving [K ]{x} = {F } in elasticity problem is essentially to minimize the total potential energy which might be described as below: Πp =
1 T {x} [K ]{x} − {x}T {F } 2
© 2005 by T. H. Kwon
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Therefore, this problem might be regarded as one of an optimization problem. So, an iterative optimization technique, as schematically explained below, can be introduced to solve this matrix equation by taking Π p as an objective function to be minimized: i) Start from an initial guess solution of {x} . (1)
ii) Find a searching direction,
{p}(k )
towards a smaller Π p
iii) Determine a constant α k to minimize Π p along the direction of the solution as {x}
( k +1)
= {x}
(k )
{p}(k ) to update
+ α k {p}
(k )
iv) Go to step ii) and iterate the process till a satisfactory convergent solution is obtained. There are many optimization techniques depending on how to choose the searching vector
{p}( k ) and constant
α k . Among them, one of the most powerful optimization
techniques is conjugate gradient method. The conjugate gradient method (Fletcher and Reeves method) can be summarized as: In starting the iteration, take {r} = {F } − [K ]{x} (1)
Set the initial searching vector,
(1)
(in fact, {r} = −∇Π p )
{p}(1) = {r}(1)
And iterate the following procedure until convergence solution is obtained: (k ) (k ) { r} {r} αk = {p}( k ) [K ]{p}( k ) T
T
{x}( k +1) = {x}( k ) + α k {p}( k ) {r}( k +1) = {F } − [K ]{x}( k +1) = {r}( k ) − α k [K ]{p}( k )
{r}( k +1) {r}( k +1) βk = {r}( k ) {r}( k ) T
T
© 2005 by T. H. Kwon
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{p}( k +1) = {r}( k +1) + β k {p}( k ) This method might be improved further. One of the possible improvements is by employing the scaling of the variables in such a way that Jacobian of the objective function has the diagonal terms of unity. The other method is a preconditioning of matrix [K] in the following fashion: Instead of solving [K ]{x} = {F } , one solves
[K~ ]{~x } = {F~} where
[K~ ] = [C
]−1 [K ][C R ]−1 {~x } = [C R ]{x} {F~} = [C L ]−1 {F } L
[K ] = [C ][C ]
The nonsingular matrix
p
L
R
[ ]
~ is called the preconditioner so that K
becomes a better conditioned matrix than [K]. Among many preconditioners proposed so far, incomplete Cholesky factors of [K] is particularly valuable. T [K ] = L~ L~ : Cholesky factorization
[ ][ ]
[K ] = [L ][L ]
T
p
: Incomplete Cholesky factorization of [K ]
i.e. Cholesky factorization of [K ]small terms neglected
[ ]
There are many possible choice of K p
for preconditioning.
[ ]
Whichever K p
[ ]
is
{}
~ ~ chosen, the conjugate gradient method algorithm for solving K {~ x } = F could be used. Of course, {x} should be evaluated later from the transformation equation. [Bathe P.749752]
It is out of scope of this coursework to explain the details of optimization techniques. Those who are interested in this method may refer to the following reference. “Numerical Optimization Techniques for Engineering Design: With Applications”, by Garret N. Vanderplaats, McGrawHill Book Co., 1984. Also [Bathe P.749752]
© 2005 by T. H. Kwon
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5.3 Treatment of Constraints There are, in most cases, constraints for the unknown variables, for instance, prescribed values as essential boundary condition or relation between variables as kinematic conditions. Consider a functional to be minimized:
Πp =
1 {V }T [K ]{V } − {V }T {F } 2
(5.1)
subject to constraints stated as
[G ]r×n {V }n×1 = {Q}r×1
( r constraints)
(5.2)
It has been shown that, without such constraints, the variational principle provides a set of equations
[K ]n×n {V }n×1 = {F }n×1
(5.3)
It is of importance to find a matrix equation taking into account the constraints. In treating various kinds of constraints, there are two important methods worth introducing: i) ii)
Lagrange Multiplier Method Penalty Function Method
5.3.1 Lagrange Multiplier Method The constraints might be stated as follows:
{R} = [G ]{V } − {Q} = 0 The augmented Functional, with the help of Lagrange Multiplier Method, can be written as:
Π ap =
1 {V }T [K ]{V } − {V }T {F } + {λ}T {[G ]{V } − {Q}} 2
Applying the variational approach to the above equation yields
© 2005 by T. H. Kwon
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(5.4)
δΠ ap = {δV }T [K ]{V } − {δV }T {F } + {λ }T [G ]{δV } + {δλ }T {[G ]{V } − {Q}}
{
}
= {δV } [K ]{V } − {F } + [G ] {λ } + {δλ } {[G ]{V } − {Q}} = 0 T
T
T
for any {δV} and any {δλ }
which implies the following matrix equation:
⎡[K ] ⎢ ⎣[G ]
[G ]T ⎤ ⎧{V }⎫ = ⎧{F }⎫ ⎨ ⎬ ⎨ ⎬ [0] ⎥⎦ ⎩{λ}⎭ ⎩{Q}⎭
(5.5)
5.3.2 Penalty Function Method Π pp =
1 {V }T [K ]{V } − {V }T {F } + 1 {R}T [α ]r×r {R} 2 2
(5.6)
with ⎡α 1 ⎢ ⎢ [α] = ⎢ ⎢ ⎢ ⎢⎣
α2 α3 0
⎤ ⎥ 0 ⎥ ⎥ ⎥ L ⎥ α r ⎥⎦
: diagonal matrix for penalty parameters
α i : very large number (like 1020) δΠ pp = {δV }T [K ]{V } − {δV }T {F } + {δR}T [α ]r×r {R} since the last term is equal to {δV } [G ] [α ]{[G ]{V } − {Q}}, the above equation can be T
T
rewritten as
{
}
δΠ pp = {δV }T [K ]{V } − {F } + [G ]T [α ][G ]{V } − [G ]T [α ]{Q} = 0 for any {δV } which implies
[[K ] + [G] [α ][G]]{V } = {F } + [G] [α ]{Q} T
© 2005 by T. H. Kwon
T
169
(5.7)
Note:
Equivalence between two methods
There is a relationship between {λ } and [α ] :
{λ} = [α ]{[G ]{V } − {Q}} or
(5.8)
{λ} = [α ]{R}
With this identity, equations (5.5) and (5.7) become identical. It might be mentioned that, in the right hand side, [α ] is very large whereas {R} is supposed to be very small to satisfy the constraints. The multiplication of [α ] and {R} results in finite values which are identical to the Lagrange multipliers, which usually have physical meaning corresponding to the constraints. It may also be reminded that the penalty function method was introduced before when treatments of boundary conditions were discussed. Check yourself whether or not the previous discussion is consistent with the present one with regard to the penalty function method. Read [Cook et al. P.272283] [Bathe P.110113] [Hughes P.194197]
© 2005 by T. H. Kwon
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6. FEM FOR THERMOMECHANICAL PROCESS There are many engineering and physical processes which undergo mechanical deformation along with heat transfer. Such process is called “Thermomechanical Process”. It is very important to analyze both the deformation as well as heat transfer simultaneously, in many cases, in a coupled manner since the deformation and heat transfer affect each other. Some examples of such thermomechanical processes are illustrated below. Examples: 1) Rolling Process
2) Extrusion (Metal or Plastics)
3) Forging (or Compression Molding)
© 2005 by T. H. Kwon
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4) Injection Molding
5) Extrusion Process
6) Hot embossing
Stamp Polymer substrate Solid
Preheating
Embossing
Demolding
Thermomechanical process simulation involves i) deformation (or flow) simulation ii) temperature field simulation which are coupled with each other. One may refer to Appendix 3 for constitutive equations for nonlinear materials.
© 2005 by T. H. Kwon
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6.1 Finite Element Formulation for Incompressible Creeping Flow [Bathe Sect. 4.4.3, Sect. 7.4] [Hughes Sect. 4.3] [Huebner et al. P. 374380]
( ρv&i ≈ 0 )
Momentum equation:
σ ij , j + ρf i = 0
in Ω
Mass conservation:
Dii = 0
in Ω ( ρ& + ρDii = 0, ρ& = 0 ) (6.2)
Dij =
1 ⎛⎜ ∂vi ∂v j + 2 ⎜⎝ ∂x j ∂xi
⎞ ⎟ ⎟ ⎠
(6.1)
: rate of deformation tensor
Boundary conditions:
σ ij n j = t i
on
v i = vi
on
S2
(6.3)
S1
(6.4)
σ ij ′ = 2µ (γ& , T )Dij
Constitutive equation:
(6.5)
Reminder
a) Generalized Newtonian Fluid (NonNewtonian viscous fluid) (for polymer processing)
σ ij ′ = 2µDij ′ µ = µ (γ& , T )
γ& ≡ (2 Dij Dij )2
:
Generalized Newtonian fluid
:
generalized shear rate
1
b) Plasticity
σ ij ′ =
K II D
(for metal forming)
Dij
© 2005 by T. H. Kwon
or
σ ij ′ =
2 σ eff Dij 3 ε&eff
173
→
σ ij ′ = 2µDij
with
K
µ=
II D
=
2 σ eff 3 ε&eff
where
σ eff ≡
3 ′ ′ σ ij σ ij 2
: effective stress
ε eff ≡
2 ′ ′ Dij Dij 3
: effective strain
J2
dσ dε
µ
1 ′ ′ Dij Dij 2
II D ≡
′
σ eff σ eff = Aε&effm µ=
σ eff ε&eff
1
1 ′ ′ = K = σ ij σ ij 2 2
ε&eff
σ eff = Aε& m −1 ε&eff
Definition of deviatoric stress tensor σ ′ ⎛ trσ ⎞ σ′ =σ −⎜ ⎟δ ⎝ 3 ⎠
σ ′ = σ + pδ
→
in our case
trσ = −p 3
: pressure
Principle of virtual power:
π p = ∫ σ ij Dij dV − ∫ t i vi dS − ∫ ρf i vi dV Ω
∂Ω
Ω
to be minimized subject to the incompressibility constraint Dii = 0 First, we will make use of the Lagrange multiplier method for the incompressibility constraint and secondly introduce the penalty function method for the finite element formulation.
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6.1.1 Finite element formulation with Lagrange multiplier method The augmented functional using the Lagrange multiplier λ (which is a function of space):
π pa = ∫ σ ij ′ Dij dV − ∫ t i vi dS − ∫ ρf i vi dV − ∫ λDii dV Ω
S2
Ω
(6.6)
Ω
δπ (vi , λ ) = 0 for any δvi and δλ a p
Now, let us find the physical meaning of the Lagrange multiplier λ by obtaining the EulerLagrange equation corresponding to equation (6.6).
δπ ap = ∫ ⎛⎜ σ ij ′ − λδ ij ⎞⎟δDij dV − ∫ t i δv i dS − ∫ ρf i δv i dV − ∫ Dii δλdV Ω
⎝
⎠
S2
Ω
Ω
using integral by parts and Gauss theorem: ⎡ ∂ ⎛ ′ ⎤ ′ ⎜ σ ij − λδ ij ⎞⎟ + ρf i ⎥δvi dV + ∫ ⎛⎜ σ ij − λδ ij ⎞⎟δvi n j dS ⎠ ⎝ ⎠ ⎢ ∂x j ⎝ S1 + S 2 Ω⎣ ⎦⎥
δπ ap = − ∫ ⎢
− ∫ t i δvi dS − ∫ Dii δλdV S2
Ω
⎡ ∂ ⎛ ′ ⎤ ′ ⎜ σ ij − λδ ij ⎞⎟ + ρf i ⎥δvi dV + ∫ ⎡⎢⎛⎜ σ ij − λδ ij ⎞⎟n j − t i ⎤⎥δvi dS ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎥⎦ Ω⎢ S2 ⎣ ∂x j ′ + ∫ ⎛⎜ σ ij − λδ ij ⎞⎟n j δvi dS − ∫ Dii δλdV ⎝ ⎠ Ω S
δπ ap = − ∫ ⎢
1
=0
for any kinematically admissible δvi and δλ .
Therefore one can obtain the EulerLagrange equation and corresponding boundary conditions as follows: ⎡ ∂ ⎛ ′ ⎤ ⎜ σ ij − λδ ij ⎞⎟ + ρf i ⎥ = 0 ⎢ ⎠ ⎣⎢ ∂x j ⎝ ⎦⎥
in Ω
(6. 1′ )
Dii = 0
in Ω
(6. 2′ )
© 2005 by T. H. Kwon
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⎛⎜ σ ′ − λδ ⎞⎟n = t ij i ⎝ ij ⎠ j
on S2
(6. 3′ )
δvi = 0 i.e. vi = vi
on S1
(6. 4′ )
Comparison between (1) – (4) and ( 1′ )( 4′ ) indicates that
λ=p
: hydrostatic pressure
(6.7)
Now, Eq. (6) can be rewritten as follows:
δπ ap = ∫ σ ij ′δDij dV − ∫ t i δvi dS − ∫ ρf i δvi dV − ∫ pδDii dV − ∫ Dii δpdV = 0 Ω
S2
Ω
Ω
(6.8)
Ω
With this variational form of the principle of virtual power, let us derive the finite element formulation. Introduce matrices for stress, strain, and constitutive relations: ⎧σ ′ ⎫ ⎪ rr ′ ⎪ ⎪σ ⎪ σ ′ = ⎨ zz ′ ⎬ ⎪σ θθ ⎪ ⎪ ′⎪ ⎩σ rz ⎭
{}
for axisymmetric flow
∂v r ⎧ ⎪ ∂r ⎧ Drr ⎫ ⎪ ∂v z ⎪D ⎪ ⎪ ∂z {D} = ⎪⎨ zz ⎪⎬ = ⎪⎨ vr ⎪ Dθθ ⎪ ⎪ r ⎪⎩2 Drz ⎪⎭ ⎪ ⎪⎛ ∂v r ∂v z ⎪⎜⎝ ∂z + ∂r ⎩
{σ ′ }= [µ ]{D}
© 2005 by T. H. Kwon
⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎞⎪ ⎟⎪ ⎠⎭
: constitutive law
176
⎧σ ′ ⎫ xx ′ ⎪⎪ ′ ⎪⎪ σ = ⎨σ yy ⎬ for 2D flow ⎪ ′⎪ ⎪⎩σ xy ⎪⎭
{}
⎧D ′ ⎫ ⎪⎪ xx ′ ⎪⎪ {D} = ⎨ D yy ⎬ for 2D flow ⎪ ′⎪ ⎪⎩2 D xy ⎪⎭
⎡2µ ⎢0 [µ ] = ⎢ ⎢0 ⎢ ⎣0
0 2µ 0 0
0 0 2µ 0
0⎤ 0 ⎥⎥ 0⎥ ⎥ µ⎦
⎡2 µ [µ ] = ⎢⎢ 0 ⎢⎣ 0
0 2µ 0
0⎤ 0 ⎥⎥ for 2D flow µ ⎥⎦
σ ij ′δDij = σ rr ′δDrr + σ zz ′δD zz + σ θθ ′δDθθ + σ rz ′δDrz + σ zr ′δD zr ′ ′ ′ ′ = σ rr δDrr + σ zz δD zz + σ θθ δDθθ + σ rz δ (2 Drz ) ′ T = {δ D} σ
{}
= {δ D} [µ ]{D} T
Introduce shape function for velocity: ⎧v r ( r , z ) ⎫ ⎬ = [N ]{V } ⎩v z (r , z )⎭
{vr} = ⎨
5 6 e.g.
4 1 2
⎧v r ⎫ ⎡ N 1 ⎬=⎢ ⎩v z ⎭ ⎣ 0
{vr} = ⎨
3
0
N2
0
N3
0 N4
0
N5
0
N6
N1
0
N2
0
N3 0
N4
0
N5
0
⎧ Drr ⎫ ⎪D ⎪ {D} = ⎪⎨ zz ⎪⎬ = [N ′]{V } ⎪ Dθθ ⎪ ⎪⎩2 Drz ⎪⎭
© 2005 by T. H. Kwon
177
⎧ v r1 ⎫ ⎪v ⎪ ⎪ z1 ⎪ ⎪v r 2 ⎪ ⎪ ⎪ ⎪v z 2 ⎪ ⎪v r 3 ⎪ ⎪ ⎪ 0 ⎤ ⎪v z 3 ⎪ ⎨ ⎬ N 6 ⎥⎦ ⎪v r 4 ⎪ ⎪v z 4 ⎪ ⎪ ⎪ ⎪v r 5 ⎪ ⎪v ⎪ ⎪ z5 ⎪ ⎪v r 6 ⎪ ⎪v ⎪ ⎩ z6 ⎭
⎡ N 1,r ⎢ 0 [N ′] = ⎢⎢ N1 ⎢ r ⎢N ⎣ 1, z
e.g.
0 N 1, z 0 N 1,r
N 2,r 0 N2 r N 2, z
0 N 2, z 0 N 2,r
L 0 L N 5, z L
0
L N 5, r
0 ⎤ N 6, z ⎥⎥ 0 ⎥⎥ N 6,r ⎥⎦
N 6,r 0 N6 r N 6, z
Define the matrices for traction and body force:
{t } = ⎧⎨ r ⎫⎬
for traction on S2
{f } = ⎧⎨ ff ⎫⎬
for body force
r
t ⎩t z ⎭
r
r
⎩
z
⎭
Virtual velocity and corresponding rate of deformation: ⎧δv r ⎫ ⎬ = [N ]{δV } ⎩δv z ⎭
{δvr} = ⎨
{δ D} = [N ′]{δV } For convenience, we introduce
⎧1⎫ ⎪1⎪ {h} ≡ ⎪⎨ ⎪⎬ such that ⎪1⎪ ⎪⎩0⎪⎭
Dii = {h} {D} T
Introduce shape function for pressure: p = ⎣N p ⎦{p},
{p}
being nodal pressure matrix 5
e.g. p = ⎣N p1
© 2005 by T. H. Kwon
N p2
⎧ p1 ⎫ ⎪ ⎪ N p 3 ⎦⎨ p 2 ⎬ ⎪p ⎪ ⎩ 3⎭
178
6 4 1 2 3 Velocity node Pressure node
Then we can rewrite the variational form of the principle of virtual power as follows:
δπ ap = ∑ δπ ap
(e)
e
a (e)
δπ p
=
r ∫ {δV } [N ′] [µ ][N ′]{V }dV − ∫ {δV } [N ] {t }dS − ∫ {δV } [N ] {f }ρdV T
T
T
Ωe
−
T
T
∂Ω e ∩ S 2
T
r
Ωe
∫ {δV } [N ′] {h}⎣N ⎦{p}dV − ∫ {δp} ⎣N ⎦ {h} [N ′]{V }dV T
T
T
T
p
Ωe
T
p
Ωe
{
= {δV } [K ] {V } − [g ]
eT
e
T
}
{p} − {Fd }e − {Fb }e − {δp}T [g ]e {V }
where
[K ]e ≡ ∫ [N ′]T [µ ][N ′]dV
:
[g ]e ≡ ∫ ⎣N p ⎦T {h}T [N ′]dV
: constraint matrix due to incompressibility
element stiffness matrix
Ωe
Ωe
{Fd }e
T
∂Ω e ∩ S 2
{Fb }e
r
: work equivalent nodal force due to load
{}
: work equivalent nodal force due to body force
∫ [N ] {t }dS
≡ ≡
r
T ∫ ρ [N ] f dV
Ωe
{F }e ≡ {Fd }e + {Fb }e After standard assembly, one obtains
{
}
δπ ap = {δV }T [K ]{V } − [G ]T {P} − {F } − {δP}T [G ]{V } =0
for any kinematically admissible {δV } and {δP}
which implies that ⎡[K ] ⎢ ⎣[G ]
[G ]T ⎤ ⎧ {V } ⎫ = ⎧{F }⎫ ⎬ ⎨ ⎬ ⎨ [0] ⎥⎦ ⎩− {P}⎭ ⎩ {0}⎭
: global matrix equation
(6.9)
Notes: i) Interpolation of pressure is one order lower than that of velocity so as to have the same order of approximation on the derivatives of pressure and velocity. (It may be noted that [N ′] and
© 2005 by T. H. Kwon
⎣N p ⎦
are involved in the matrix equation.)
179
(Note also that the momentum equation has different order of derivatives for velocity field and pressure field:
∂ ∂x j
⎡ ⎛ ∂vi ∂v j + ⎢ µ ⎜⎜ ⎢⎣ ⎝ ∂x j ∂xi
⎞⎤ ∂p ⎟⎥ − + ρf i = 0 ). ⎟⎥ ∂x i ⎠⎦
Recent research results indicate that there are some successful algorithms with equal interpolation or discontinuous pressure field. ii) The global matrix equation has zero diagonal terms. However, one can rearrange the order in the unknowns to have nonzero diagonal terms, after introducing the essential boundary condition, during the Gauss elimination procedure. iii) If
⎣N p ⎦
and [N ] does not satisfy the socalled BabuškaBrezzi condition, spurious
pressure mode (“Checkerboard” pressure) may appear to ruin the velocity field. r Incompressibility condition is ∇ ⋅ V = 0 . Constrained elements can be locked with r linear triangular elements, i.e. V = 0 whatsoever with constant pressure. [read Hughes P.207209, P.210217, Appendix 4.2]
iv) In recent years, some methods using equal order interpolation for both velocity and pressure have been successfully implemented without spurious pressure mode utilizing for instance, Galerkin Least Square method.
6.1.2 Finite element formulation with Penalty function method Finite element formulation starts with the following functional when penalty function method is used for the incompressibility constraint:
π pp = ∫ σ ij ′ Dij dV − ∫ t i vi dS − ∫ ρf i vi dV + Ω
Ω
S2
α
2 Ω∫
Dii D jj dV
δπ pp = ∫ σ ij ′δDij dV − ∫ t i δvi dS − ∫ ρf i δvi dV + α ∫ Dii δD jj dV = 0 Ω
© 2005 by T. H. Kwon
S2
Ω
180
Ω
(6.10)
First three terms were already treated already. Now, let us consider the penalty term only.
α ∫ Dii δD jj dV = α {δV }T Ωe
∫ [N ′] {h}{h} [N ′]dV {V } T
T
Ωe
Therefore one obtains
δπ pp
(e)
{
[ ] {V } − {F } − {F } }
= {δV } [K ] {V } + K p T
e
e
e
d
e
b
where
[K ]
e
p
≡ α ∫ [N ′] {h}{h} [N ′]dV T
T
Ωe
:
stiffness matrix due to incompressibility constraint
After assembly, one obtains
δπ pp = {δV }T {[[K ] + [K p ]]{V } − {F }} = 0
for any kinematically admissible {δV }
implying that
[[K ] + [K ]]{V } = {F } p
[ ]
in K p
e
r ⎣ ′ ⎦ = ⎣N trace
e
p
(6.11)
:
{h}T [N ′] = ⎢⎢ N1,r + N1
[K ]
: global matrix equation
N 1, z
N 2,r +
N2 r
′ }⎣N trace ′ ⎦dV = α ∫ {N trace Ωe
© 2005 by T. H. Kwon
181
N 2, z
L L N 6,r +
N6 r
⎥ N 6, z ⎥ ⎦
6.1.3 Equivalence between Lagrange multiplier method and penalty function method
[[K ] + [K ]]{V } = {F }
: from penalty function method
[K ]{V } − [G ]T {P} = {F }
: from Lagrange multiplier method
p
For equivalence, one should have the following equality
[G ]T {P} = −[K p ]{V } i.e.
∫ [N ′] {h}⎣N ⎦dV {P} = −α ∫ [N ′] {h}{h} [N ′]dV {V } T
T
Ωe
i.e.
Ωe
N ′]{V }dV ∫ [N ′] {h}pdV = − ∫ [N ′] {h}α {1h}4[2 43 T
T
Ωe
∴
T
p
T
Ωe
Dii
p = −αDii
(6.12)
▣
▣
Looking into the original variational forms of both methods:
δπ ap = ∫ σ ij ′δDij dV − ∫ t i δvi dS − ∫ ρf i δvi dV − ∫ pδDii dV − ∫ Dii δpdV = 0 Ω
Ω
S2
Ω
Ω
δπ pp = ∫ σ ij ′δDij dV − ∫ t i δvi dS − ∫ ρf i δvi dV + α ∫ Dii δD jj dV = 0 Ω
From {δV } part : From {δP} part :
Ω
S2
Ω
− p = αDii
∫ ⎣N ⎦D dV = 0 , p
ii
Ωe
which essentially indicates that the weighted average of incompressibility constraint is satisfied with Np as weighting functions. ▣
© 2005 by T. H. Kwon
182
▣
6.1.4 Finite element formulation for incompressible flow with inertia term fi should be replaced with f i −
⎛ ∂v Dvi ∂v = fi − ⎜ i + v j i ⎜ ∂t Dt ∂x j ⎝
⎞ ⎟ ⎟ ⎠
then δπ ap should be as follows:
δπ ap = ∫ σ ij′δ Dij dV − Ω
+∫ρ Ω
∫ tiδ vi dS − ∫ ρ fiδ vi dV − ∫ pδ Dii dV − ∫ Diiδ pdV Ω
S2
Ω
Ω
∂vi ∂v δ vi dV + ∫ ρ v j i δ vi dV ∂t ∂x j Ω
=0 The last two terms will be rewritten in the matrix form: ∂vi ⎧ ∂V ⎫ T T ρ ∫Ω ∂t ρvi dV = {δV } Ω∫ ρ [N ] [N ]dV ⎨⎩ ∂t ⎬⎭
∫ ρv
j
Ω
r ⎧ ∂V ⎫ ⎧ ∂v ⎫ ⎨ ⎬ = [N ]⎨ ⎬ ⎩ ∂t ⎭ ⎩ ∂t ⎭
∂vi δvi dV = {δV }T ∫ ρ [N ]T [L ][N ]dV {V } ∂x j Ω
where ⎡ ∂v1 ⎢ ⎢ ∂x1 [L] = ⎢⎢ ∂v2 ∂x ⎢ ∂v1 ⎢ 3 ⎢⎣ ∂x1
∂v1 ∂x 2 ∂v 2 ∂x 2 ∂v3 ∂x 2
∂v1 ⎤ ⎥ ∂x3 ⎥ ∂v 2 ⎥ ∂x3 ⎥ ∂v3 ⎥⎥ ∂x3 ⎥⎦
: velocity gradient in 3dimensional case
⎡ ∂v r ⎢ [L] = ⎢ ∂∂vr ⎢ z ⎣ ∂r
∂v r ⎤ ∂z ⎥ ∂v z ⎥ ⎥ ∂z ⎦
for axisymmetric case
Define the following matrices:
[m]e = ∫ ρ [N ]T [N ]dV
: mass matrix
[K c ]e = ∫ ρ [N ]T [L][N ]dV
: convection matrix
Ωe
Ωe
The final global matrix equations are then
[M ]⎧⎨ ∂V ⎫⎬ + [[K ] + [Kc ]]{V } − [G ]T {P} = {F } ⎩ ∂t ⎭ [G ]{V } = 0
© 2005 by T. H. Kwon
183
6.1.5 Weighted Residual Method for incompressible flow analysis ⎡
∫ ⎢⎣⎛⎜⎝σ
′
⎤ − pδ ij ⎞⎟ + ρf i ⎥Wi k dV = 0 ⎠, j ⎦
(6.13)
k p
dV = 0
(6.14)
′
− pδ ij ⎞⎟n j − t i ⎤Wi k dS = 0 ⎠ ⎦⎥
(6.15)
ij
Ω
∫D W ii
Ω
⎡
∫ ⎢⎣⎛⎜⎝σ
ij
S2
From equation (6.13) ⎡
∫ ⎢⎣⎛⎜⎝σ
′ ij
Ω
⎤ − pδ ij ⎞⎟ Wi k + ρf iWi k ⎥dV ⎠, j ⎦
′ ′ = ∫ ⎛⎜ σ ij − pδ ij ⎞⎟Wi k n j dS − ∫ ⎛⎜ σ ij − pδ ij ⎞⎟Wi ,kj dV + ∫ ρf iWi k dV ⎝ ⎠ ⎝ ⎠ ∂Ω Ω Ω ′ = ∫ t iWi k dS + ∫ t iWi k dS − ∫ ⎛⎜ σ ij − pδ ij ⎞⎟Wi ,kj dV + ∫ ρf iWi k dV ⎝ ⎠ S S Ω Ω 2
1
Appropriate combination of equations (6.13) and (6.15) is then the following: ⎡
∫ ⎢⎣⎛⎜⎝σ
Ω
′ ij
⎤ − pδ ij ⎞⎟ + ρf i ⎥Wi k dV + ∫ (t i − t i )Wi k dS = 0 ⎠, j ⎦ S
(6.16)
2
Integral by parts and Gauss theorem provides: ′ − ∫ σ ij Wi ,kj dV + ∫ t iWi k dS + ∫ ρf iWi k dV + ∫ pWi ,ki dV = 0 Ω
Ω
S2
(6.17)
Ω
Taking Wi k = 0 on S1 and Wi k = N ik with vi = N ik vik , i = 1,2,3 ( N ik being shape function for vi at kth node) and taking, in equation (6.14), N pk as W pk with p = N pk p k ( p k being pressure at kth node), one can find that equations (6.17) along
with equation (6.14) is equivalent to
δπ ap = ∫ σ ij ′δDij dV − ∫ t i δvi dS − ∫ ρf i δvi dV − ∫ pδDii dV − ∫ Dii δpdV = 0 . Ω
© 2005 by T. H. Kwon
S2
Ω
Ω
184
Ω
6.2 Finite Element Formulation for Energy Equation [Bathe Chap.7]
In solving the temperature field from the energy equation, one assumes that the velocity field is known for the finite element solution of flow. The energy equation is stated as ∂T ∂T ⎞ 1 ∂ ⎛ a ∂T ⎞ ∂ ⎛ ∂T ⎞ ⎛ ∂T + vr + vz ⎟= ⎜ r kr ⎟ + ⎜kz ⎟+Q ∂r ∂z ⎠ r a ∂r ⎝ ∂r ⎠ ∂z ⎝ ∂z ⎠ ⎝ ∂t
ρC p ⎜
(6.18)
⎧0 : 2  dimensional case (r → x, z → y ) a=⎨ ⎩ 1 : axisymmetric case
where Q can be either distributed heat source or/and viscous dissipation.
Boundary conditions : T =T ⎞ ⎛ ∂T ⎞ ⎛ ∂T − ρC p v r T ⎟nr + ⎜ k z − ρC p v z T ⎟ n z = q ⎜ kr ⎝ ∂z ⎠ ⎝ ∂r ⎠
on ST
(6.19)
on Sq
(6.20)
nˆ
Sq z
Ω ST
r q = h(T∞ − T ) q
: for convective surface : net heat flux
Initial condition : T (r , z; t = 0 ) = To (r , z ) viscous dissipation :
© 2005 by T. H. Kwon
(6.21)
Qdis = 2µDij Dij
185
(6.22)
The appropriate weighted residual is as follows: ~ ~ ~ ⎡ 1 ∂ ⎛ a ∂T~ ⎞ ∂ ⎛ ∂T~ ⎞ ⎛ ∂T ∂T ∂T ⎞⎤ ∫ ⎢⎢ r a ∂r ⎜⎜⎝ r k r ∂r ⎟⎟⎠ + ∂z ⎜⎜⎝ k z ∂z ⎟⎟⎠ + Q − ρC p ⎜⎜⎝ ∂t + vr ∂r + v z ∂z ⎟⎟⎠⎥⎥Wi dV Ω⎣ ⎦ ~ ⎡⎛ ∂T~ ⎤ ⎛ ∂T ~⎞ ~⎞ − ∫ ⎢⎜⎜ k r − ρC p v r T ⎟⎟nr + ⎜⎜ k z − ρC p v z T ⎟⎟n z − q ⎥Wi dS = 0 ⎥⎦ ⎠ ⎝ ∂z ⎠ Sq ⎢ ⎣⎝ ∂r
(6.23)
with Wi = 0 on ST. Integral by parts yields: ~ ~ ⎡ 1 ∂ ⎛ a ∂T~ ⎞ ∂ ⎛ ∂T~ ⎞⎤ ⎛ ∂T ∂Wi ∂T ∂Wi ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ∫Ω ⎢⎢ r a ∂r ⎜⎝ r k r ∂r Wi ⎟⎠ + ∂z ⎜⎝ k z ∂z Wi ⎟⎠⎥⎥dV − Ω∫ ⎜⎝ k r ∂r ∂r + k z ∂z ∂z ⎟⎟⎠dV ⎦ ⎣ ~ ~ ~ ⎛ ∂T ∂T ∂T ⎞ ⎟Wi dV (6.24) + ∫ QWi dV − ∫ ρC p ⎜⎜ + vr + vz ∂r ∂z ⎟⎠ ⎝ ∂t Ω Ω ~ ~ ⎛ ∂T ∂T ⎞ ~ ~ ⎜ nr + k z n z ⎟⎟Wi dS + ∫ ρC p v r T nr + ρC p v z T n z + q Wi dS = 0 − ∫ ⎜ kr ∂r ∂z ⎠ Sq ⎝ Sq
(
)
Since the first term can be rewritten, due to Divergence theorem, as ⎡ 1 ∂ ⎛ a ∂T~ ⎞ ∂ ⎛ ∂T~ ⎞⎤ ∫ ⎢ r a ∂r ⎜⎜⎝ r k r ∂r Wi ⎟⎟⎠ + ∂z ⎜⎜⎝ k z ∂z Wi ⎟⎟⎠⎥⎥dV Ω⎢ ⎦ ⎣ ~ ~ ⎛ ∂T ∂T ⎞ n z ⎟⎟Wi dS nr + k z = ∫ ⎜⎜ k r r z ∂ ∂ ⎠ ∂Ω⎝ ~ ~ ⎛ ∂T ∂T ⎞ nr + k z n z ⎟Wi dS = ∫ ⎜⎜ k r ∂r ∂z ⎟⎠ Sq ⎝ The last equation is obtained due to Wi = 0 on ST. Equation (6.24) becomes ~ ~ ~ ~ ~ ⎛ ∂T ∂Wi ⎛ ∂T ∂T ⎞ ∂T ∂Wi ⎞ ∂T ∫ ⎜⎜ k r ∂r ∂r + k z ∂z ∂z ⎟⎟⎠dV + Ω∫ ρC p ⎜⎜⎝ ∂t + vr ∂r + v z ∂z ⎟⎟⎠Wi dV Ω⎝ ~ = ∫ QWi dV + ∫ ρC p (v r nr + v z n z )T + q Wi dS
[
Ω
]
Sq
which is the weak form statement of the problem.
© 2005 by T. H. Kwon
186
(6.25)
(As already noted through the previous examples, equation (6.25) can be also derived by introducing integral by parts on the weighted residual of the differential equation only ~ ~ ∂T ∂T ~ nr + k z n z = ρC p (v r nr + v z n z )T + q .) and by substituting k r ∂r ∂z With finite element mesh, volume and surface integrals can be replaced with summations of integral over each element, i.e.
∫ L dV = ∑ ∫ L dV , ∫ L dS = ∑ ∫ L dS e Ωe
Ω
e Sq
S2
and Wi is reduced with N ie and assembly procedure with appropriate connectivity information is followed. nonzero Wi for elements which includes node i.
Wi = N ie for each element.
e i
Equation (6.25) can be rewritten as ~ ~ ~ ~ ~ ⎛ ∂T e ⎛ ∂T e ∂N ie ∂T e ⎞ e ∂T e ∂T e ∂N ie ⎞ ⎜ ⎜ ⎟ ρ + v + + v + dV C k k ∑e ∫ ⎜ r ∂r ∂r z ∂z ∂z ⎟ ∑e ∫ p ⎜ ∂t r ∂r z ∂z ⎟⎟ N i dV ⎝ ⎠ ⎠ Ωe ⎝ Ωe (6.26) ~e e e = ∑ ∫ QN i dV + ∑ ∫ ρC p (v r n r + v z n z )T + q N i dS
[
e Ωe
]
e S q IΩe
~ ~ T e = T e (r , z; t ) = ⎣N e ⎦{T (t ) e } = N ejT je (t )
where
(6.27)
Introducing Eq. (6.27) into Eq. (6.26), each element contributions are as follows: e ⎛ ∂N ie ∂N ej ∂T je ∂N ie ∂N j ⎞⎟ e e e ⎜ kr ∫ ⎜ ∂r ∂r + k z ∂z ∂z ⎟dV T j + Ω∫ ρC p Ni N j dV ∂t Ωe ⎝ ⎠ e
∫
Ωe
⎛
∂N ej
⎝
∂r
ρC p N ie ⎜⎜ vr
= ∫ QN dV + e i
Ωe
+ vz
∂N ej ⎞ ⎟dV T je − ρC p (vr nr + vz nz ) N ie N ej dS T je ∫ ⎟ ∂z ⎠ Sq IΩe
[
∫ qN dS e i
S q IΩe
© 2005 by T. H. Kwon
187
]
(6.28)
Define the following element matrices:
K C ij
e ⎛ ∂N ie ∂N ej ∂N ie ∂N j ⎞⎟ ⎜ = ∫ kr + kz dV ⎜ ∂r ∂r ∂z ∂z ⎟⎠ Ωe ⎝
e
⎛ ⎜ ⎝
e ∫ ρC p N i ⎜ vr
e
K V ij =
Ωe
∫ [ρC (v n
e
K S ij =
p
r
r
∂N ej ∂r
+ vz
∂N ej ⎞ ⎟dV ∂z ⎟⎠
]
+ v z n z ) N ie N ej dS
stiffness matrix due to conduction
stiffness matrix due to convection stiffness matrix due to convected loss
S q IΩe
Bije =
∫ ρC
p
N ie N ej dV
matrix due to transient term
Ωe
FQ i = ∫ QN ie dV e
forcing matrix due to heat source
Ωe
e
FS i =
∫ qN
e i
dS
forcing matrix due to heat outflux
S q IΩ e
and define
[K ]e = [K C ]e + [K V ]e − [K S ]e
Total stiffness matrix
{F }e = {FQ }e + {FS }e
Total forcing matrix
Then, after assembly procedure:
[B]{T&}+ [K ]{T } = {F }
: Global matrix equation
where
{T&} = ⎧⎨ ∂∂Tt ⎫⎬ ⎩
© 2005 by T. H. Kwon
⎭
188
(6.29)
Those matrices can also be rewritten in the matrix form as follows:
[K C ]e = ∫ [N ′]T [k ][N ′]dV Ωe
[K V ]
e
T
⎧v ⎫ = ∫ ⎣N ⎦ ρC p ⎨ r ⎬ [N ′]dV ⎩v z ⎭ Ωe T
not symmetric!
⎣N ⎦ = ⎣N 1 N 2 L N 6 ⎦ ⎡ N 1,r ⎣ N 1, z
N 2,r N 2, z
[N ′] = ⎢ [k ] = ⎡⎢
0⎤ k z ⎥⎦
kr ⎣0
[K S ]
e
L N 6,r ⎤ L N 6, z ⎥⎦
T
⎧v ⎫ ⎧n ⎫ T = ∫ ρC p ⎨ r ⎬ ⎨ r ⎬⎣N ⎦ ⎣N ⎦dS ⎩v z ⎭ ⎩n z ⎭ S q IΩe
stiffness matrix due to convected loss
[B]e = ∫ ρC p ⎣N ⎦T ⎣N ⎦dV
matrix due to transient term
Ωe
[F ] = ∫ Q ⎣N ⎦ e
Q
T
dV
forcing matrix due to heat source
Ωe
[FS ]e = ∫ q ⎣N ⎦dS
forcing matrix due to heat outflux
S q IΩe
z
Special cases on Sq: i) When v r n r + v z n z = 0 , i.e. on solid boundary Sq:
[K S ]e = 0 ii) When q = h(T∞ − T ) on boundary Sq: e
FS i =
∫ hT
∞
N ie dS
S q IΩ e
and
e
FS i =
∫ hN
e i
N ej dS
is to be added to [K ] . e
S q IΩe
© 2005 by T. H. Kwon
189
z
Steady state problem
[K ]{T } = {F } z
(6.30)
Transient problem
[Bathe P.410418]
CrankNicholson implicit finite difference form for {T& } term can be used.
{T }t + ∆t 2
{T&}
t+
∆t 2
1 {{T }t + {T }t + ∆t } 2 1 = {{T }t + ∆t − {T }t } ∆t =
At time t + ∆t / 2 , equation (12) can be written as
[B] 1 {{T }t +∆t − {T }t } + [K ] 1 {{T }t + {T }t + ∆t } = {F }t + ∆t
2 ∆t Rearranging the equation for {T }t + ∆t :
2
[2[B ] + [K ]∆t ]{T }t + ∆t = [2[B] − [K ]∆t ]{T }t + 2∆t{F }t + ∆t
(6.31)
2
Notes: i) In doing this transient calculation, care should be taken for the ratio
∆t . ∆x
ii) Numerical instability might take place due to the convection term, especially for large Péclet number. Upwinding (or forward differencing) scheme might be recommended to avoid this kind of trouble. PetrovGalerkin method might be recommended for this purpose. [Bathe P. 687688, Sec. 7.4.3]
z
[Huebner et al. P.110, 376, 445]
Coupled analysis of thermomechanical processes needs iteration between the simulation of velocity/pressure field and temperature field. Velocity/ Pressure filed
Flow analysis
© 2005 by T. H. Kwon
Q
K, µ
190
Energy Eq. Solver
Temperature field
6.3 Galerkin Least Square Method, StreamlineUpwind PetrovGalerkin Method To attain a stable numerical solution for advectiondiffusion problems, many researchers have made a great effort to develop reliable numerical algorithms in conjunction with finite element formulation. The most notable methods along this line include the following methods: StreamlineUpwind PetrovGalerkin (SUPG) Method Galerkin Least Square (GLS) Method, StreamlineUpwind Petrov Galerkin method is such that the weighing function is increased in the upwind portion while it is decreased in the downstream direction. GLS is basically to add leastsquares forms of residuals to the Galerkin Method and consequently enhance the stability of the ordinary Galerkin method. It was shown that GLS method coincides with SUPG for hyperbolic cases. For the details, one may refer to the following papers: Hughes et al., “Galerkin/LeastSquares method for advectiondiffusive equations,”
Computer method in applied mechanics and engineering, 73, 173189(1989), Ilinca et al. “On stabilized finite element formulations for incompressible advectiondiffusive transport and fluid flow problems,” Computer method in applied mechanics
and engineering, 188, 235255 (2000). In this section, we may just briefly describe the essence of the method via a scalardiffusion problem and a steady incompressible NavierStokes equation. 1) Scalardiffusion problem The problem can stated by the residual form Rφ = u ⋅∇φ − ∇ ⋅ k ∇φ − q = 0
Various methods are then represented as follows:
© 2005 by T. H. Kwon
191
Galerkin Method:
( Rφ ,ψ )Ω = 0 Streamline upwind (SU) Method
( Rφ ,ψ )Ω + ∑ ∫ ( u ⋅∇φ )τ ( u ⋅∇ψ ) d Ωk = 0 k Ωk
Streamline upwind/Petrov Galerkin (SUPG) Method
( Rφ ,ψ )Ω + ∑ ∫ Rφτ ( u ⋅∇ψ ) d Ωk = 0 k Ωk
Galerkin/Least Squares (GLS) Method
( Rφ ,ψ )Ω + ∑ ∫ Rφτ ( u ⋅∇ψ − ∇ ⋅ ( k ∇ψ ) ) d Ωk = 0 k Ωk
2) Steady incompressible Navierstokes equation This problem can also be stated by the residual form as: Ru = u ⋅ ∇u + ∇ p − ∇ ⋅ (2ν D(u )) − f = 0 R p = ∇ ⋅u = 0
GLS variational formulation of the NavierStokes equation
∫ (u ⋅ ∇u − f ) ⋅ w)dΩ + ∫ 2ν D(u ) : D(w)dΩ − ∫ p(∇ ⋅ w)dΩ + ∫ (∇ ⋅ u )qdΩ + ∑ ∫ (u ⋅ ∇u + ∇ p − ∇ ⋅ (2ν D(u )) − f ) ⋅ τ (u ⋅ ∇ w + ∇q − ∇ ⋅ (2ν D(w)))dΩ Ω
Ω
Ω
k Ωk
+∑
∫ (∇ ⋅ u ) ⋅ δ (∇ ⋅ w)dΩ
k Ωk
© 2005 by T. H. Kwon
k
= ∫ t ⋅ wdΓ Γh
192
Ω
k
where
δ = a hK ξ (Re K ), τ=
hK , 2a ⎧Re K , 0 ≤ Re K < 1, Re K ≥ 1, ⎩ 1,
ξ (Re K ) = ⎨ Re K =
m K a hK 2ν
,
with hK : the size of elements a : known velocity from the previous iteration step m K : 1/3 for linear elements, 1/12 for quadratic elements Re K : element Reynolds number τ , δ : stabilizing parameters
© 2005 by T. H. Kwon
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7. FIELD PROBLEMS 7.1 Eigenvalue problems Helmholtz Equation ∂ ∂φ ∂ ∂φ ∂ ∂φ (k x ) + (k y ) + (k z ) + λφ = 0 ∂x ∂x ∂y ∂y ∂z ∂z
(7.1)
There are many examples of Helmholtz equation in many physical problems as illustrated below: a) Seiche motion :
( Standing waves on shallow water )
∂ ∂w ∂ ∂ω 4π 2 (h ) + (h )+ w=0 ∂x ∂x ∂y ∂y gT 2
(7.2)
h : water depth at quiescent state w : elevation above the quiescent state g : gravity T : period of oscillation
B.C.
∂ω ∂ω nx + ny = 0 ∂x ∂y
i.e.
∂ω =0 ∂n
at solid boundaries.
b) Electromagnetic waves :
∂ 1 ∂φ ∂ 1 ∂φ ∂ 1 ∂φ ( )+ ( )+ ( ) + ω 2 µ 0 ε 0φ = 0 ∂x ε d ∂x ∂z ε d ∂z ∂y ε d ∂y
Permittivity of dielectric
© 2005 by T. H. Kwon
Permeability
194
Permittivity of free space
(7.3)
c) Acoustic vibration: ∂2 p ∂2 p ∂2 p ω 2 p=0 + + + ∂x 2 ∂y 2 ∂z 2 c 2
(7.4)
p : pressure excess above ambient pressure ω : wave frequency c : wave velocity in the medium.
The functional for the Helmholtz equation can be written as : 2 2 ⎡ ⎛ ∂φ ⎞ 2 ⎤ ⎛ ∂φ ⎞ ⎛ ∂φ ⎞ I (φ ) = ∫ ⎢k x ⎜ ⎟ + k y ⎜⎜ ⎟⎟ + k Z ⎜ ⎟ − λφ 2 ⎥ dΩ ∂x ⎝ ∂z ⎠ ⎥⎦ ⎝ ∂y ⎠ Ω⎢ ⎣ ⎝ ⎠
⎡
δI = 2 ∫ ⎢ k x Ω
⎣
⎤ ∂φ ⎛ ∂φ ⎞ ∂φ ⎛ ∂φ ⎞ ∂φ ⎛ ∂φ ⎞ δ ⎜ ⎟ + ky δ ⎜⎜ ⎟⎟ + k z δ ⎜ ⎟ − λφδφ ⎥ dΩ ∂x ⎝ ∂x ⎠ ∂z ⎝ ∂z ⎠ ∂y ⎝ ∂y ⎠ ⎦
(7.5)
(7.6)
⎡ ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞⎤ δφ ⎟ + ⎜ k y δφ ⎟ + ⎜ k z δφ ⎟⎥ dΩ − = 2∫ ⎢ ⎜ k x ∂x ⎝ ∂x ⎠ ∂y ⎜⎝ ∂y ⎟⎠ ∂z ⎝ ∂z ⎠⎦ Ω⎣ ⎤ ⎡ ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞ ⎟⎟ + ⎜ k z − 2∫ ⎢ ⎜ k x ⎟ + ⎜⎜ k y ⎟ + λφ ⎥δφdΩ ∂x ⎝ ∂x ⎠ ∂y ⎝ ∂y ⎠ ∂z ⎝ ∂z ⎠ Ω⎣ ⎦ ⎡ ∂φ ∂φ ∂φ ⎤ = 2 ∫ ⎢k x nx + k y ny + kz n z δφdS ∂x ∂y ∂z ⎥⎦ S ⎣ ⎡ ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞ ⎤ ⎟⎟ + ⎜ k z − 2∫ ⎢ ⎜ k x ⎟ + ⎜⎜ k y ⎟ + λφ ⎥δφdΩ ∂x ⎝ ∂x ⎠ ∂y ⎝ ∂y ⎠ ∂z ⎝ ∂z ⎠ Ω⎣ ⎦
(7.7)
δI = 0 for any δφ implies that ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞ ⎟ + ⎜kz ⎜kx ⎟ + ⎜ky ⎟ + λφ = 0 ∂x ⎝ ∂x ⎠ ∂y ⎜⎝ ∂y ⎟⎠ ∂z ⎝ ∂z ⎠
© 2005 by T. H. Kwon
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in Ω
(7.8)
kx
B.C.
∂φ ∂φ ∂φ nx + k y ny + kz nz = 0 ∂x ∂y ∂z
on ∂Ω
or
φ =φ If the flux is not equal to zero on ∂Ω , an additional term is to be included in the functional (like in the Poisson equation) Substituting
φ = N iφ i ,
∂φ ∂N i = φi , ∂x ∂x
δ(
∂N i ∂φ )= δφ i , etc ∂x ∂x
into Equation (2), one obtains the following element matrix equation
[K ](e) {φ }(e) − λ [H ](e) {φ }(e) = 0
(7.9)
with ⎛
∂N j
⎝
∂x ∂x
[K ](e ) = ∫ ⎜⎜ k x ∂N i Ω
+ ky
∂N i ∂N j ⎞ ∂N i ∂N j ⎟dΩ , + kz ∂z ∂z ⎟⎠ ∂y ∂y
like in Poisson equation case
[H ](e) = ∫ N i N j dΩ ,
like one term for convection in Poisson equation case
Ω
Finally one obtains the global matrix equation
[[K ] − λ [H ]]{φ } = 0 For a nontrivial solution to exist, the following equation should hold:
[K ] − λ [H ] = 0 λi : eigenvalue,
© 2005 by T. H. Kwon
:
{φ }i
Characteristics equation.
eigenvector, modal vector
196
It is well known that if [K ] and [H ] are positive definite,
λi are distinct, real, positive {φ }i are all independent.
7.2 Structural Dynamics As explained earlier, the principle of virtual work can be stated as
δu = δWext According to D’Alembert principle, the body force can include the fictitious force due to inertia & damping , i.e.
ρf i − ρu&&i − cu& i Then, the principle of virtual work leads to the following FEM formulation: (Refer to 2.8.7 Displacementbased FEM for elasticity for detailed derivation)
[M ]{u&&} + [D]{u&} + [K ]{u} = {Fc } + {Fd } + {Fb }
(7.10)
with matrices contributed from each element given by
[M ]e = ∫ ρ [N ]T [N ]dV
: element mass matrix
[D]e = ∫ C [N ]T [N ]dV
: element damping matrix
[K ]e = ∫ [N ′]T [C ][N ′]dV
: element stiffness matrix
Ωe
Ωe
Ωe
{F }
{FC }e = ∑ [N ]T
(k )
: concentrated load
xc
{Fd }e
∫ [N ] {t}ds
=
T
: work equivalent nodal force due to traction
∂Ω e ∩ S 2
{Fb }e
=
∫ [N ] ρ { f }dV T
: work equivalent nodal force due to body force
Ωe
© 2005 by T. H. Kwon
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7.3 TimeDependent Field Problems The time dependent field problem can be stated as ∂ ∂φ ∂ ∂φ ∂ ∂φ (k x ) + (k y ) + (k z ) = f ( x, y, z , t ) + k t φ& + k tt φ&& ∂x ∂x ∂y ∂z ∂y ∂z
(7.11)
with boundary conditions
φ = Φ ( x, y , z , t )
on
S1 ,
t>0
and kx
∂φ ∂φ ∂φ nx + k y ny + kz n z + q( z , y, z , t ), h( x, y, z , t )φ = 0 ∂x ∂y ∂z
on S 2 , t > 0
with a initial condition
φ = φ 0 ( x, y , z , )
in
Ω,
t=0
φ& = ξ ( x, y, z )
in
Ω,
t=0
and
Some examples can be found in the followings: i) Diffusion Equation
D(
∂ 2φ ∂ 2φ ∂ 2φ ∂φ + 2 + 2)= 2 ∂t ∂x ∂y ∂z
(7.12)
Thermal diffusion : Fourier law Species diffusion : Fick’s law Diffusion in porous medium : Darcy’s law ii) Wave equation ∂ 2U ∂ 2U ∂ 2U 2m(hν − ν ) ∂ 2U + 2 + 2 = ∂x 2 ∂y ∂z h 2ν 2 ∂t 2
© 2005 by T. H. Kwon
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(7.13)
The FEM formulation via Galerkin Method can be obtained as below: With an approximate field variable interpolated over an element as
φ ( e ) ( x, y, z, t ) = N i ( x, y, z )φ i (t ) = ⎣N ( x, y, z )⎦{φ (t )}( e )
(7.14)
The final assembled matrix equation becomes
[K tt ]{φ&&}+ [K t ]{φ&}+ [K ]{φ } + [K S ]{φ } + {R(t )} = {0} with element matrices defined by
[K tt ]e = ∫ k tt N i N j dV , Ωe
[K t ]e = ∫ k t N i N j dV Ωe
⎛
∂N j
⎝
∂x
[K ]e = ∫ ⎜⎜ k x ∂N i ∂x Ωe
+ ky
∂N i ∂N j ∂N i ∂N j ⎞ ⎟dV + kz ∂y ∂y ∂z ∂z ⎟⎠
[K S ]e = ∫ hN i N j dS ∂Ω ∩ S 2
{R}e
=
∫ fN dV + ∫ qN dS i
Ωe
i
∂Ω ∩ S 2
7.4 Solution Technique for timedependent equation. The finite element matrix equation for the time dependent problems can be stated as the following standard form:
[M ]{φ&&}+ [C ]{φ&}+ [K ]{φ } = {R(t )}
(7.15)
Where the first, second and third terms in the left had side are representing the inertia effect, damping capacitance effect and the stiffness of the system, respectively while the right hand side represents the time dependent external loading effect. One can treat the system of equation by the following three types of analysis:
© 2005 by T. H. Kwon
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i)
Modal analysis (or Eigenfunction analysis) in case of
[M ]{φ&&}+ [K ]{φ } = 0
With no damping and external load, the equation becomes simplified as
[M ]{φ&&}+ [K ]{φ } = 0 Let
{φ } = {Φ}e iwt ,
then
(7.16)
{φ&&} = −w {Φ}e 2
iwt
and
[K ]{Φ} − ω 2 [M ]{Φ} = 0 With such expressions, the matrix equation can be restated as
[[K ] − ω [M ]]{Φ} = {0} 2
(7.17)
For a nontrivial solution, the following equation should hold
[K ] − ω 2 [M ] = 0
: characteristics equation
(7.18)
from which one can determine the eigenvalues, ω i . For a given ω i , the original matrix equation provides the corresponding eigenvectors {Φ}i . In this way one can have a set of eigenvalues and eigenvectors as follows:
ωi
{Φ}i
: eigenvalue : eigen vector or modal vectors
For real and symmetric matrices, the eigenvectors have a special characteristics, namely, the orthogonality of eigenvectors as stated below:
⎣Φ ⎦i [K ]{Φ} j = δ ij C Ki ⎣Φ ⎦i [M ]{Φ} j = δ ij C Mi where CKi and CMi are generalized stiffness and generalized mass, respectively. Furthermore, one can normalize the eigenvectors such that CMi, in which case {Φ}i are called orthogonal eigenvectors base on [M].
© 2005 by T. H. Kwon
200
One can introduce a Cholesky factorization for a positive definite [M] such that
[M ] = [Q]T [Q] and one can transform the variable
{Φ}
to
[X ]
by the following linear
transformation:
{Φ} = [Q]−1 {X } Then one can have a standard form of eigenvalue problem as
[A]{X } = ω 2 {X } where
[A] = [Q]−T [K ][Q]−1 [Bathe P.573] [Meirovitch “Computational Method of Structural Dynamics” P. 6162]
ii) General case with
[C ] ≠ 0 , {R} ≠ 0
(with the help of Modal analysis)
Take {φ (t )} as a linear combination of eigenvectors i.e.
{φ (t )} = [{Φ}1 , {Φ}2 , {Φ}3 ,L, {Φ}n ]{Λ(t )} = [ A]{Λ (t )} Vector of unknown, modal amplitude
Substituting {Φ (t )} into the original equation and premultiplying it by [A]
T
gives
[A]T [M ][A]{Λ&& }+ [AT ][C ][Λ& ] + [A]T [K ][A]{Λ} = [A]T {R}
[M ]
[C ]
*
i.e.
[K ]
*
*
[M ]{Λ&& }+ [C ]{Λ& }+ [K ]{Λ} = {F } *
[ ]
*
*
*
[ ]
{F } *
: equations for modal analysis
where M * and K * become diagonal matrices as explained below:
© 2005 by T. H. Kwon
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Since
[K ]{Φ} j
= ω 2j [M ]{Φ} j ,
K ij* = ⎣Φ ⎦i [K ]{Φ} j = C Ki δ ij
M ij* = ⎣Φ ⎦i [M ]{Φ} j
,
= C Mi δ ij
= δ ij (if normalized)
and K ij* = ω j M ij* = C Miω j δ ij 2
2
= ω j δ ij 2
[M ] = [I ] *
[K ] *
and
(no sum on i or j)
⎡ω 12 0 0 0⎤ ⎢ ⎥ 2 0 ω2 0 0 ⎥ =⎢ ⎢0 0 ω 32 0 ⎥ ⎢ ⎥ 0 0 O⎥⎦ ⎣⎢ 0
If [C ] is proportional to either [M ] or [K ] , the equations become fully uncoupled and thus easy to solve. e.g. if
[C ] ∝ [M ]
, the typical uncoupled equation for ith mode becomes
2 * && + 2ζ ω M Λ & M ii* Λ i i i ii i + ω M ii Λ i = Fi * .
[Bathe Chapter 10, 11]
iii) Finite difference Method
[Huebner P.243248]
Finite difference method can be used for solving the original coupled set of 2nd order differential equations.
{δ&} = {δ }
− {δ }t − ∆t 2∆t
t + ∆t
t
central differencing
{δ&&} = {δ }
t + ∆t
t
− 2{δ }t + {δ }t − ∆t ∆t 2
© 2005 by T. H. Kwon
202
[M ]{δ&&}t + [C ]{δ&}t + [K ]{δ }t = {R(t )}
then
becomes
1 2 ⎡ 1 ⎤ ⎡ ⎤ ⎢⎣ ∆t 2 [M ] + 2∆t [C ]⎥⎦{δ }t + ∆t = {F (t )} − ⎢⎣[K ] − ∆t 2 [M ]⎥⎦{δ }t 1 ⎡ 1 [C ]⎤⎥{δ }t −∆t − ⎢ 2 [M ] − 2∆t ⎦ ⎣ ∆t
To be solved
known
there are variety of techniques to solve these set of equations, implicit, fully explicit, etc.
{}
iv) Zero mass case of [C ] φ& + [K ]{φ } = {R}
[C ]{φ&}+ [K ]{φ } = {R} K , t n , t n + ∆t ,K
∆t tθ = t n + θ∆t at
with
0 . On S ε : 1 U ij ~ O( ) ,
ε
∫U
Tij ~ O(
1
ε2
)
( p, q )t i (q)dS → 0 as ε → 0
(8.32)
( p, q )u j (q)dS = u j ( p) ∫ Tij ( p, q)dS
(8.33)
ij
Sε
∫T
ij
Sε
Sε
Define C ij ( p) in such a way that
© 2005 by T. H. Kwon
216
⎧ ⎫ − δ ij + C ij ( p) ≡ lim⎨lim ∫ Tij ( p, q)dS ⎬ ε →0 δ →0 ⎩ Sε ⎭
(8.34)
For a constant u i field with zero traction t j , (8.31) yields
δ ij = − ∫ Tij dS − ∫ Tij dS <S >
∴
∫ T dS = −δ ij
Sε
ij
−
∫ T dS = −δ ij
ij
+ C ij ( p ) →
<S >
Sε
C ij ( p) ≡ − ∫ Tij ds
(8.34)’
<S >
Using (8.34) and (8.31), the boundary integral equation becomes C ij u j ( p) +
∫T
ij
<S >
( p, q)u j (q )dS =
∫U
ij
( p, q )t j (q )dS
(8.35)
<S >
Note that either u i or t i is given as a boundary condition for a wellposed boundary value problem.
© 2005 by T. H. Kwon
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Appendix 1. Poisson Equations in Many Engineering Problems Field Problems analogous to Heat Conduction 1. Torsion of shaft
⎛1 ⎞ ∇ 2 ⎜ φ ⎟ = −2θ ⎝G ⎠
G : Shear Modulus θ : Twist angle per unit length ∂φ ∂φ φ : Stress function τ xz = − , τ yz = − ∂x ∂y
2. Flow through porous media ∇ 2 (KH ) = −Q
K : Permeability H : Fluid Head Q : Internal flow injection rate
3. Pressurized membranes ∇ 2 (Th ) = − P
T : Membrane tension h : Membrane displacement P : Pressure imbalance
4. Electrostatic fields ∇ 2 (εV ) = − ρ
ε : Permittivity V : Electric potential (Voltage) ρ : Density charge
5. Diffusion ∇ 2 (kC ) = −Q
k : Diffusion coefficient C : Concentration Q : Production rate
6. Slow viscous flow ∇ 2ψ = −ω
ψ : Stream function u = −
∇ 2ω = 0
r ω : Vorticity, ω = ∇ × V
© 2005 by T. H. Kwon
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∂ψ ∂ψ ,v = − ∂x ∂y
Torsion of prismatic bar y
σzy
y
σzx
x z
T
x
α
σ xx = σ yy = σ zz = σ xy = 0 ε xy = 0 Angle of twist :
α = θz
(θ : angle of twist/length)
Displacement field : warping function u = −α y = −θ zy v = α x = θ zx w = f ( x, y )
w( x, y )
(A.1)
Strain : ∂w ∂u ∂w + = − θy ∂x ∂z ∂x ∂w ∂v ∂w = + = − θx ∂y ∂z ∂y
γ xz = γ yz
(A.2)
ε xx = ε yy = ε zz = ε xy = 0
Stress :
σ xx = σ yy = σ zz = σ xy = 0 ⎛ ∂w ⎞ − θy ⎟ ⎝ ∂x ⎠ ⎛ ∂w ⎞ = G⎜⎜ − θx ⎟⎟ ⎝ ∂y ⎠
σ xz = 2Gε xz = Gγ xz = G⎜ σ yz = 2Gε yz = Gγ yz
© 2005 by T. H. Kwon
219
(A.3)
∇ ⋅σ = 0
Equilibrium :
(no body force)
∂σ xx ∂σ xy ∂σ xz + + =0 ∂x ∂y ∂z
→
∂ ⎡ ⎛ ∂w ⎞⎤ − θy ⎟⎥ = 0 G⎜ ⎢ ∂z ⎣ ⎝ ∂x ⎠⎦
=0
→
⎞⎤ ∂ ⎡ ⎛ ∂w + θx ⎟⎟⎥ = 0 ⎢G⎜⎜ ∂z ⎣ ⎝ ∂y ⎠⎦
∂σ zx ∂σ zy ∂σ zz + + =0 ∂x ∂y ∂z
→
∂2w ∂2w + =0 ∂x 2 ∂y 2
∂σ yx ∂x
+
∂σ yy ∂y
+
∂σ yz ∂z
(A.5)
Stress function: φ ( x, y ) In order to satisfy the equilibrium equation, the stress function is defined as: ∂φ ∂y ∂φ =− ∂x
σ xz = σ yz
∂γ xz ∂2w = −θ ∂y ∂x∂y ∂γ xz ∂ w = +θ ∂y ∂x∂y 2
(A.6)
∂γ yz
⇒
∂x
−
∂γ xz = 2θ ∂y
Introducing Eqs. (3) and (6) into above yields ⎛ ∂ 2φ ∂ 2φ ⎞ − ⎜⎜ 2 + 2 ⎟⎟ = 2Gθ ∂y ⎠ ⎝ ∂x
i.e.
∂ 2φ ∂ 2φ + = −2Gθ ∂x 2 ∂y 2
(A.7)
One should solve Equation (A.7) and use Equation (A.6) for obtaining σ xz , σ yz . In a more general term, Equation (A.7) can be rewritten as ∂ ⎛ 1 ∂φ ⎞ ∂ ⎛ 1 ∂φ ⎞ ⎟ = −2θ ⎜ ⎟+ ⎜ ∂x ⎝ G ∂x ⎠ ∂y ⎜⎝ G ∂y ⎟⎠
© 2005 by T. H. Kwon
220
(A.7)’
Now, let us consider the boundary conditions.
sˆ
σyz y
nˆ
nˆ = (n x , n y )
r
sˆ = (− n y , n x ) r τ = (σ xz , σ yz )
τ
σxz x
⎛ ∂φ ∂φ ⎞ ⎟ = ⎜⎜ . − ∂x ⎟⎠ ⎝ ∂y
r
on boundary :
τ ⋅ nˆ = 0 ∂φ
i.e.
∂y
nx −
∂φ ny = 0 ∂x
⇒
∇φ ⋅ sˆ = 0
or
∂φ =0 ∂s
i.e. φ = constant along the boundary surface. Let us choose φ = 0 along the boundary. T = ∫ 2φdA
Torque T :
A
⎛ ∂φ ∂φ T = ∫ (σ yz x − σ xz y )dA = ∫ ⎜⎜ − x− ∂x ∂y A A⎝ ⎡ ∂ ⎤ ∂ = ∫ ⎢− ( xφ ) − ( yφ ) + 2φ ⎥dA ∂x ∂y ⎦ A⎣ = ∫ 2φdA + A
∫ (xφn
x
+ yφn y )ds
⎞ y ⎟⎟dA ⎠
(φ = 0 along s )
∂A
= ∫ 2φdA A
For a hollow bar, the formula should be slightly modified accordingly as described below.
© 2005 by T. H. Kwon
221
nˆ
sˆ
S1 nˆ
A2
sˆ
S2
A1
T=
∫ 2φdA + ∫ (xφn
A = A1 − A2
=
x
∂A = S1 + S 2
∫ 2φdA + C ∫ (xφn 1
A = A1 − A2
+ yφn y )ds + yφn y )ds + C 2 ∫ (xφn x + yφn y )ds
x
S1
S2
⎤ ⎤ ⎡ ⎡ = ⎢ ∫ 2φdA + C1 ∫ (xφn x + yφn y )ds ⎥ − ⎢ ∫ 2φdA + C 2 ∫ (xφn x + yφn y )ds ⎥ S1 S2 ⎦⎥ ⎦⎥ ⎣⎢ A2 ⎣⎢ A1 = T1 − T2 Procedure for a hollow bar: i) Given T and geometry, ii) Assume θ and C2 (C1 =0) ⎛1 ⎞ iii) Solve ∇ ⋅ ⎜ ∇φ ⎟ = −2θ with φ = 0 on S1 and φ = C 2 on S 2 ⎝G ⎠ iv) Scale θ and C2 by making use of the following equations: T=
∫ 2φdA + C ∫ (xφn 2
A = A1 − A2
x
+ yφn y )ds
S2
1 ∂φ
1 ∂φ
∫ G ∂n ds + ∫ G ∂n ds = −2θ ( A
1
S1
− A2 )
1
S2
v) Check the convergence: Stop if converged solution obtained, Otherwise, go to the step iii) and iterate the procedure. ⎛1
⎞
∫ ∇ ⋅ ⎜⎝ G ∇φ ⎟⎠dA = ∫ (− 2θ )dA = −2θ ( A
1
1
A
− A2 )
A
=
1 ∂φ
1 ∂φ
1 ∂φ
∫ G ∂n ds = ∫ G ∂n ds + ∫ G ∂n ds
∂A
© 2005 by T. H. Kwon
S1
S2
222
Appendix 2. Example of FEM Program for Poisson Equations Heat Conduction ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞ ~ ⎟+Q = 0 ⎜kx ⎟ + ⎜ky ∂x ⎝ ∂x ⎠ ∂y ⎜⎝ ∂y ⎟⎠ r T = T (x )
on S1
kx
∂T ∂T nx + k y n y − q~ + h(T − T∞ ) = 0 ∂y ∂x
⇓
~ ( Q : heat source;
on S 2
q~ : heat influx)
[K ]{T } = {F } ⎛ ∂N i ∂N j ∂N i ∂N j ⎞ e ⎟dΩ = ∫ ⎜⎜ k x + ky K Cij ∂x ∂x ∂y ∂y ⎟⎠ Ωe ⎝
K
e Sij
=
∫ hN N i
j
⇒
[K ] = [K C ] + [K S ]
⇒
{F } = {Q} + {q} + { f ∞ }
dS
∂Ω e ∩ S 2
~ Qie = ∫ QN i dΩ Ωe
∫ q N dS ~
qie =
i
∂Ω e ∩ S 2
f ∞i =
∫ hT
∞
N i dS
∂Ω e ∩ S 2
Torsion Problem ∂ ⎛ 1 ∂φ ⎞ ∂ ⎛ 1 ∂φ ⎞ ⎟ = −2θ ⎜ ⎟+ ⎜ ∂x ⎝ G ∂x ⎠ ∂y ⎜⎝ G ∂y ⎟⎠ ∂φ σ xz = ∂y ∂φ σ yz = − ∂x θ : twist angle per unit length Torque =
∫ 2φdA A
© 2005 by T. H. Kwon
223
Evaluation of [K C ] : ⎛ ∂N i ∂N j ∂N i ∂N j ⎞ e ⎟⎟dxdy K Cij = ∫ ⎜⎜ k x + ky ∂ ∂ ∂ ∂ x x y y ⎠ Ωe ⎝ One needs to evaluate
(A.8)
∂N i , etc. ∂x
i) General case
N i = N i ( x, y ) = N i (ξ ,η ) Global coordinates (x,y) ∂N i ∂N i = ∂ξ ∂x ∂N i ∂N i = ∂η ∂x
∂x ∂N i + ∂ξ ∂y ∂x ∂N i + ∂η ∂y
↔
Local coordinates
(ξ ,η )
∂y ∂ξ ∂y ∂η
i.e. ⎧ ∂N i ⎫ ⎡ ∂x ⎪⎪ ∂ξ ⎪⎪ ⎢ ∂ξ ⎨ ∂N ⎬ = ⎢ ∂x ⎪ i⎪ ⎢ ⎪⎩ ∂η ⎪⎭ ⎢⎣ ∂η
∂y ⎤ ⎧ ∂N i ⎫ ⎧ ∂N i ⎫ ⎥ ⎪ ⎪ ⎪⎪ ⎪⎪ ∂ξ ⎪ ∂x ⎪ ⎥ ⎨ ∂N ⎬ = [J ]⎨ ∂∂Nx ⎬ ∂y ⎥ ⎪ i⎪ ⎪ i⎪ ⎪⎩ ∂y ⎪⎭ ∂η ⎥⎦ ⎪⎩ ∂y ⎪⎭
∴ ⎧ ∂N i ⎫ ⎧ ∂N i ⎫ ⎪ ⎪ ⎪⎪ ∂x ⎪⎪ −1 ⎪ ∂ξ ⎪ ⎨ ∂N ⎬ = [J ] ⎨ ∂N ⎬ ⎪ i⎪ ⎪ i⎪ ⎪⎩ ∂y ⎪⎭ ⎪⎩ ∂η ⎪⎭
Note also that
(A.9)
dxdy = J dξdη
⎛ ∂N i ∂N j ∂N i ∂N j ⎞ e ⎟⎟ J dξdη = ∫ ⎜⎜ k x + ky K Cij ∂ ∂ ∂ ∂ x x y y ⎠ Ωe ⎝
(A.10)
One needs numerical integration for the stiffness coefficients, which will be discussed later in detail.
© 2005 by T. H. Kwon
224
ii) 3node triangular element (special simple case) From the elasticity problem, we have the following shape functions:
T e = N i Ti e = ⎣N 1
Ni =
∴
N2
⎧T1e ⎫ ⎪ ⎪ N 3 ⎦⎨T2e ⎬ ⎪T e ⎪ ⎩ 3⎭
1 (ai + bi x + ci y ), 2∆
i = 1,2,3
b ∂N i = i 2∆ ∂x c ∂N i = i ∂y 2∆
where
1 x1 1 ∆ = 1 x2 2 1 x3
y1 y2 = y3
1 [(x2 − x1 )( y3 − y1 ) − (x3 − x1 )( y 2 − y1 )] 2
b1 = y 2 − y 3 ,
b2 = y 3 − y1 , b3 = y1 − y 2
c1 = x3 − x 2 ,
c 2 = x1 − x3 ,
© 2005 by T. H. Kwon
c3 = x 2 − x1
225
Appendix 3. Constitutive Equations In this Appendix 3, fundamentals of constitutive equations for various materials will be briefly summarized.
z , x3
1. Linear Elasticity u i = u i (x1 , x 2 , x3 ) ,
Displacement field:
Linear strain tensor:
1 ⎛ ∂u ∂u j ε ij = ⎜⎜ i + 2 ⎝ ∂x j ∂xi
Generalized Hook’s law:
i = 1, 2, 3
⎞ ⎟ ⎟ ⎠
y, x 2 x, x1
σ ij = Cijkl ε kl
Isotropic material:
σ ij = λε kk δ ij + 2 µε ij
(A.11)
where the Lamé constants λ and µ
µ =G=
E , 2(1 + ν )
λ =
νE (1 + ν )(1 − 2ν )
E : Young’s Modulus
ν : Poisson ratio G : Shear Modulus Deviatoric tensors:
σ ij ′ ≡ σ ij − σ kk δ ij , 1 3
ε ij ′ ≡ ε ij − ε kk δ ij 1 3
σ ij ′ = 2Gε ij ′ 1 p = − σ kk = − Kε kk 3 E K= 3(1 − 2ν )
© 2005 by T. H. Kwon
(A.12) : mean normal stress : bulk modulus (ν=0.5 → incompressibility)
226
In a matrix form:
σ 11 = σ 1 , σ 22 = σ 2 , σ 33 = σ 3 , σ 23 = σ 32 = σ 4 , σ 13 = σ 31 = σ 5 , σ 12 = σ 21 = σ 6 ε 11 = ε 1 , ε 22 = ε 2 , ε 33 = ε 3 , 2ε 23 = 2ε 32 = ε 4 , 2 ε 13 = 2ε 31 = ε 5 , 2 ε 12 = 2ε 21 = ε 6 σ K = C KM ε M
( K , M = 1, 2, L , 6)
isotropic elasticity in a matrix form: ⎧σ 1 ⎫ ⎡λ ⎪σ ⎪ ⎢ ⎪ 2⎪ ⎢ ⎪⎪σ 3 ⎪⎪ ⎢ ⎨ ⎬=⎢ ⎪σ 4 ⎪ ⎢ ⎪σ 5 ⎪ ⎢ ⎪ ⎪ ⎢ ⎪⎩σ 6 ⎪⎭ ⎢⎣
λ
λ λ
λ + 2µ
λ λ
λ
λ + 2µ
0
0
0
0
0
µ
0
0
0
0
0
µ
0
0
0
0
0
+ 2µ
0 0
0 0
0 ⎤ ⎧ε 1 ⎫ 0 ⎥⎥ ⎪⎪ε 2 ⎪⎪ 0 ⎥ ⎪⎪ε 3 ⎪⎪ ⎥⎨ ⎬ 0 ⎥ ⎪ε 4 ⎪ 0 ⎥ ⎪ε 5 ⎪ ⎥⎪ ⎪ µ ⎥⎦ ⎪⎩ε 6 ⎪⎭
plane stress, plane strain case: ⎧ ε xx ⎫ ⎧σ xx ⎫ ⎪ ⎪ ⎪ ⎪ ⎨σ yy ⎬ = [C ]⎨ ε yy ⎬ ⎪2ε ⎪ ⎪σ ⎪ ⎩ xy ⎭ ⎩ xy ⎭ ⎡ ⎤ 1 ν 0 ⎥ ⎢ [C ] = E 2 ⎢ν 1 0 ⎥ , 1 −ν ⎢ 1 −ν ⎥ ⎢0 0 ⎥ 2 ⎦ ⎣
© 2005 by T. H. Kwon
⎡ ⎤ 1 −ν 0 ⎥ ν ⎢ E ⎢ ν [C ] = 1 −ν 0 ⎥ (1 + ν )(1 − 2ν ) ⎢ 1 − 2ν ⎥ 0 ⎢ 0 ⎥ 2 ⎦ ⎣
227
2. Fluids (viscous) (Newtonian fluids, Generalized Newtonian fluids)
σ = − pδ + F(D)
: general Stoke’s assumption (1845)
∂v j 1 ⎛ ∂v Dij − ⎜ i + 2 ⎜⎝ ∂x j ∂xi
⎞ ⎟ ⎟ ⎠
linear F →
: rate of deformation tensor
Newtonian
σ ij = − pδ ij + λDkk δ ij + 2 µDij
Deviatoric tensors:
or
σ = − pδ + λ (trD)δ + 2 µD
(A.13)
σ ij ′ = σ ij + pδ ij , Dij ′ = Dij − Dkk δ ij 1 3
then
⎛ ⎝
2 ⎞ 3 ⎠
σ ij ′ = ( p − p)δ ij + ⎜ λ + µ ⎟ Dkk δ ij + 2 µDij ′ 2 ⎞ ⎛ → ( p − p) + ⎜ λ + µ ⎟ Dkk = 0 3 ⎠ ⎝
′ since σ ii = 0, ∴
σ ij ′ = 2µDij ′
p = p − κDkk = p + κ
where
σ = 2 µD
or 1
ρ
ρ&
NavierPoisson Law of a Newtonian fluid
2 : bulk viscosity 3 p : mean normal pressure
κ =λ+ µ
p : thermodynamic pressure p= p
when i) Dkk = 0 ii) κ = λ +
© 2005 by T. H. Kwon
i.e.
ρ& = 0
2 µ =0 3
(incompressible) (Stokes condition)
228
(A.14)
z Newtonian fluid
⇒
µ = µ (T )
z Generalized Newtonian fluid
⇒
µ = µ (γ& , T )
γ& ≡ (2 Dij Dij )2
(T: temperature)
1
:
generalized shear rate
Example:
ln η
1) powerlaw fluid model
µ (γ& , T ) = K (T )γ& n −1 K (T ) = A exp(Ta / T ) ↓ as T ↑
2) ModifiedCross model
ln γ&
η (T ) µ (γ& , T ) = 1− n 1 + C (η oγ& )
η o (T ) = B exp(Tb / T ) 3) Carreau model
[
µ − µ∞ 2 = 1 + (λγ& ) µo − µ∞
© 2005 by T. H. Kwon
]
( n −1) / 2
229
3. Viscoelasticity
Viscoelastic materials have a peculiar feature of deformation behavior, namely showing both the viscous flow and elastic deformation characteristics. Generally speaking, the stress of such materials is determined by the history of deformation, not just by the present deformation state. It is beyond the scope of this lecture to introduce the details of those features and constitutive theories. In this Appendix, however, we will briefly mention some of the basic features and introductory constitutive equations. One of the most interesting flow behaviors is demonstrated by a periodic shear flow case. When a sinusoidal shear stress is applied to a viscoelastic material, it undergoes a sinusoidal shear strain but with a phase shift, as indicated below: Periodic response in oscillatory simple shear: 2
σ 12 = σ o sin ωt
1
γ 12 = γ o sin(ωt − δ )
→
σo
σo
G ′′γ o
J ′σ o
δ
γo
δ
ω
G ′γ o
J ′′σ o J ′ : Storage compliance, J ′′ : Loss compliance, G ′ : Storage modulus, G ′′ : Loss modulus
© 2005 by T. H. Kwon
230
γo
ω
Linear viscoelastic model 2G 2η
2G
σ′
σ′
σ′
σ′
2η
KelvinVoigt model
Maxwell model & ′ = 1 σ& ′ + 1 σ ′ E η 2G & ′ = σ& ′ + 1 σ ′ 2GE
&′ σ ′ = 2GE′ + 2ηE &′ = 2G E′ + τE
)
(
τ
Nonlinear viscoelastic models: (out of scope) z Lodge’s rubberlike model z Phan Thien Tanner model z DoiEdward model There are many other models proposed in the literature.
Some special phenomena of viscoelastic materials: KelvinVoigt retarded elasticity (under a constant stress) E′ =
(
σo 1 − e −t / τ 2G
)
(no glassy response)
Maxwell relaxation (under a constant strain)
σ ′ = 2GE′o e − t / τ
(after instant glassy response: σ ′ = 2GE′o )
Maxwell creep (under a constant stress loading) 1 1 1 (after instant glassy response E′ = σ ′o ) E′ = σ ′o + σ ′o t 2G 2η 2G
© 2005 by T. H. Kwon
231
4. Plasticity
σ
σ
σY
σY
ε
ε Rigid perfectly Plastic
σ
σ
ε
ε
elastic perfectly plastic
rigid work hardening
elastic work hardening
Yield Stress, Yield Criteria: i) Von Mises Criteria
(σ 1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ 1 )2 = 2σ Y 2 = 6 K 2 ii) Tresca Criteria
σ1 −σ 3 = σ Y
(σ1 > σ 2 > σ 3 )
Flow Rule
′ dε ij = σ ij dλ
⎛ ⎞ ⎜ i.e. dε xx = dε yy = dε zz = dε xy = L = dλ ⎟ ⎜⎜ ⎟⎟ σ xx ′ σ yy ′ σ zz ′ σ xy ′ ⎝ ⎠
i) LevyMises perfect plasticity:
σ ij ′ =
or
K II D
σ ij ′ =
© 2005 by T. H. Kwon
Dij
(A.15)
2 σ eff Dij 3 ε&eff
(A.15)’
232
where
σ eff
σ eff ≡
3 ′ ′ σ ij σ ij 2
: effective stress
ε eff ≡
2 ′ ′ Dij Dij 3
: effective strain
II D ≡
dσ dε
µ
1 ′ ′ Dij Dij 2
σ eff = Aε&effm µ=
σ eff ε&eff
1
ε&eff
Eq.(A)
→
1 ′ ′ K2 1 J 2 ≡ σ ij σ ij = Dij Dij = K 2 II D 2 2 i.e.
J2 = K 2
Note that both Eq. (A.15) and (A.15)’ have the following form:
σ ij ′ = 2µDij
2µ =
K II D
=
2 σ eff 3 ε&eff
(A.16)
which is similar to Newtonian fluid, but with varying viscosity µ. ii) PrandtlReuss elastic plastic ′ ′ ′ dε ij = ⎛⎜ dε ij ⎞⎟ + ⎛⎜ dε ij ⎞⎟ ⎠e ⎝ ⎠p ⎝
The first and second terms in the right hand side are due to elastic and plastic deformation, respectively. ⎛⎜ dε ′ ⎞⎟ = σ ′ dλ ij ⎝ ij ⎠ p
and
as in LevyMises
⎛⎜ dε ′ ⎞⎟ = 1 dσ ′ ij ⎝ ij ⎠ e 2G
© 2005 by T. H. Kwon
233
5. Viscoplasticity
The viscoplastic flow behavior might be depicted schematically as follows:
K
σ xy
σ xy 2η Let F1 be defined as
F1 = 1 −
K
where K is an yield stress (like a Mises criteria)
σ xy
if F10 →
⎛ K ⎞⎟ σ 2ηD xy = ⎜1 − ⎟ xy ⎜ σ xy ⎠ ⎝
Generalized in 3dimensional case:
F = 1−
K J2
′
for F < 0 ⎧0 2ηDij = ⎨ ′ ⎩ Fσ ij for F > 0
which is analogous to Newtonian viscosity equation,
′ ′ 2µDij = σ ij .
Taking square of both sides gives: ′ ′ 4η 2 Dij Dij = F 2σ ij σ ij
© 2005 by T. H. Kwon
→
4η 2 II D = F 2 J 2
234
′
→
′ 2η II D J2 = F
substituting the last equation to F = 1 −
K J2
F=
′
results in:
2η II D K + 2η II D
Finally one obtains the expression for stress in terms of rate of deformation tensor: ⎛ ⎜ ⎝
σ ij ′ = ⎜ 2η +
K ⎞⎟ Dij II D ⎟⎠
for
F >0
(A.17)
The first term represents the viscous force term, which is rate dependent, whereas the second term represents the rate independent yield stress. What a similar form we end up with! Almost every constitutive equation has the form of σ ′ = (viscosity like term )D . In a computer program, one just changes the viscosity like term according to the material behavior as explained so far. 6. Elastoviscoplasticity
The elastoviscoplastic flow behavior may add a spring system to the viscoplastic flow as depicted schematically below: K 2G
σ xy
σ xy 2η The constitutive theory might be represented as E = Ee + E p 2GE xye = σ xy = 2ηE& xyp + (sign of E& xyp )K
© 2005 by T. H. Kwon
235