Introduction to Finite Element Method

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

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* Application Examples of FEM z Structural Analysis (steady, time-dependent 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

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: 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

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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

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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 self-obvious 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

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⎡ 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 x-axis 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

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Example 2: Energy Conservation (1-D 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

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⎡ 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

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⎤ ⎧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 well-posed 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 well-posed problem? You will definitely find the similarities among the

three examples.

© 2005 by T. H. Kwon

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Example 4: Direct Stiffness Methods for Truss Elements

Next example is concerned about a two-dimensional analysis of truss structures with the so-called 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

Force-Deformation 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

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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 end-nodes 1 and 2 of the element to be added. Second, add 4x4 element matrix components to the corresponding rows and columns of 2i-1, 2i, 2j-1, 2j in the global stiffness matrix. 2i-1

2i-1 2i

2j-1 2j

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣⎢

2j-1

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

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⎤⎧ ⎫ ⎧ ⎥⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎥⎪ ⎪ ⎪ ⎥ ⎪ 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 well-posed 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

“Displacement-based FEM” ii)

[ℑ]{F}={x}

ℑij

with [ℑ] being flexibility matrix

: influence coefficient which represents displacement xi due to unit force of Fj

“Force-based FEM”

© 2005 by T. H. Kwon

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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 self-obvious 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 2-D

⎧ u1′ ⎫ ⎨ ⎬ ⎩u ′2 ⎭

Local system

y y′ (x,y) or (x′,y′) j

i′

j′

α

x

i

© 2005 by T. H. Kwon

x′

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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)

pre-multiply Eq. (2.6) by [Φ ]

−1

© 2005 by T. H. Kwon

(a generalized inverse matrix of [Φ ] ).

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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

⇔

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− 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

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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 (stress-strain relationship).

2-D 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)t-τxy(y3- y1) t

σy

σy (x2- x3)t+τxy(y3- y2) τxy

σx(y3- y1)t-τxy(x3- x1)t

t

σx(y3- y2)t+τxy(x2- x3)t

σx f2 -σx (y1- y2)t-τxy(x2-x1) 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 end-nodes 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 mid-side 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 Betti-Maxwell 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 Q-th 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 partially-filled (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 : half-bandwidth 2b-1 : 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 i-th 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.143-146] (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.110-116]

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.86-87, 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

: Euler-Lagrange 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)

Euler-Lagrange 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)

Euler-Lagrange 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 Euler-Lagrange 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 Euler-Lagrange 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 Euler-Lagrange 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 (T-V) 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 Euler-Lagrange 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 Euler-Lagrange 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

Euler-Lagrange equation: B.C.:

T

dy =0 dx

or

© 2005 by T. H. Kwon

d ⎛ dy ⎞ ⎜ T ⎟ + w = 0, dx ⎝ dx ⎠

0< x
δy = 0 (i.e. y is prescribed) at x = 0 and x = l

47

Equivalence

i)

⎧ d ⎛ dy ⎞ 0< x
ii)

Minimize

2 ⎤ l ⎡ 1 ⎛ dy ⎞ Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy ⎥dx 0 2 ⎢⎣ ⎝ dx ⎠ ⎥⎦

or l ⎡ dy ⎛ dy ⎞ ⎤ δΠ = ∫ ⎢T δ⎜ ⎟ − wδy ⎥dx = 0 for any δy 0 ⎣ dx ⎝ dx ⎠ ⎦

Now, consider a case where one has to introduce a natural boundary condition at one end of the string as depicted in the following figure.

x

w(x) y

y(x)

T

T

If

T

dy dy =T (prescribed downward load) dx dx

dy is prescribed at x = l , the functional should be modified as follows: dx 2 ⎤ l⎡ 1 dy ⎛ dy ⎞ Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy ⎥dx − T y 0 2 dx l ⎥⎦ ⎢⎣ ⎝ dx ⎠

(2.26)

It should be noted that the additional term represents the potential energy due to the dy . In this case, the variation of functional becomes slightly dx different from the previous case to yield the same Euler-Lagrange equation with

external load T

© 2005 by T. H. Kwon

48

different set of boundary condition as derived below. l

⎤ ⎡ d ⎛ dy ⎞ dy dy δy − T δy δΠ = ∫ ⎢− ⎜ T ⎟ − w⎥δydx + T 0 dx 0 dx l ⎦ ⎣ dx ⎝ dx ⎠ l

l⎡ ⎛ dy ⎤ d ⎛ dy ⎞ dy ⎞ dy − T ⎟⎟δy − T δy = ∫ ⎢− ⎜ T ⎟ − w⎥δydx + ⎜⎜ T 0 dx dx dx ⎦ ⎣ dx ⎝ dx ⎠ 0 ⎝ ⎠ l = 0 for any δy

which implies that ⎧ d ⎛ dy ⎞ 0< x
l ⎡ dy ⎛ dy ⎞ ⎤ dy δΠ = ∫ ⎢T δ⎜ ⎟ − wδy ⎥dx − T δy = 0 for any δy 0 dx ⎣ dx ⎝ dx ⎠ ⎦

(2.27)

In other words, as long as the total potential energy is correctly found including the effect of natural boundary conditions, one does not have to worry about the detailed procedure of introducing the natural boundary condition in the FEM formulation any more. This point is the merit of FEM over other numerical analysis techniques in dealing with the natural boundary conditions which involve higher order derivatives than essential boundary conditions. This aspect of merits will be discussed later again. The tight string problem described above is analogous to other physical problems represented by one-dimensional second order differential equation, for instance, onedimensional steady heat conduction in particular. See the next summary for this analogy.

© 2005 by T. H. Kwon

49

Analogy between Tight String problem and Heat Conduction

Tight String

d ⎛ dy ⎞ ⎜ T ⎟ + w = 0, dx ⎝ dx ⎠

Heat Conduction

d ⎛ dθ ⎞ ⎜ k ⎟ + q = 0, dx ⎝ dx ⎠

0< x
2 ⎤ l ⎡ 1 ⎛ dy ⎞ Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy ⎥dx 0 2 ⎥⎦ ⎢⎣ ⎝ dx ⎠

2 ⎤ l ⎡ 1 ⎛ dθ ⎞ Π (θ( x) ) = ∫ ⎢ k ⎜ ⎟ − qθ⎥dx 0 2 ⎥⎦ ⎢⎣ ⎝ dx ⎠

y = y at x = 0 and x = l

when

0< x
θ = θ at x = 0 and x = l

when

2 2 ⎤ ⎤ l ⎡ 1 ⎛ dy ⎞ l ⎡ 1 ⎛ dθ ⎞ dθ dy Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy ⎥dx − T y Π (θ( x) ) = ∫ ⎢ k ⎜ ⎟ − qθ⎥dx − k θ 0 2 0 2 dx l dx l ⎥⎦ ⎥⎦ ⎢⎣ ⎝ dx ⎠ ⎢⎣ ⎝ dx ⎠

y = y at x = 0

when

T

θ = θ at x = 0 when

dy dy =T at x = l dx dx

k

dθ dθ =k at x = l dx dx

2 2 ⎤ ⎤ l ⎡ 1 ⎛ dy ⎞ l ⎡ 1 ⎛ dθ ⎞ dy dθ Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy ⎥dx + T y Π (θ( x) ) = ∫ ⎢ k ⎜ ⎟ − qθ⎥dx + k θ 0 2 0 2 dx 0 dx 0 ⎥⎦ ⎥⎦ ⎢⎣ ⎝ dx ⎠ ⎢⎣ ⎝ dx ⎠

y = y at x = l

when

T

θ = θ at x = l when

dy dy =T at x = 0 dx dx

k

dθ dθ =k at x = 0 dx dx

Read: [Bathe Ex. 3.18, Ex. 3.19, Ex. 3.20] for natural boundary conditions. Question: Can one find a functional for a variational principle for any differential equations? The answer is No. It is not always found. (One may try to reverse the derivation of Euler-Lagrange equation to obtain the corresponding functional, even though not always successful.)

© 2005 by T. H. Kwon

50

2.8.4 Finite Element Method versus Rayleigh-Ritz Method Find φ(x) to minimize Π (φ(x) ) . One wants to obtain an approximate solution to minimize a functional Π (φ(x) ) . One of the historically famous approximate methods for this kind of problem is Rayleigh-Ritz Method, and the other modern method is the Finite Element Method. Here we will discuss both methods with the comparison in mind. i) Rayleigh-Ritz Method: n ~ φ ( x) = ∑ c i ψ i ( x)

: approximate solution satisfying the essential B.C.

i =1

where

ψ i (x) : trial functions (defined over the whole domain)

Then, ~ Π ( φ ) = Π (c1 , c 2 , L , c n )

Therefore

~ ∂Π ( φ ) = 0, ∂ci

to be minimized w.r.t. ci.

i = 1,L , n

: n equations for n unknown ci’s.

This method is very simple and easy to understand. However, it is not easy to find a family of trial functions for the entire domain satisfying the essential boundary conditions when geometry is complicated. The solution to this troublesome point can be found in the Finite Element Method.

ii) Finite Element Method ~ φ ( x) = N i ( x)φ i ⎧ φ i : nodal values ⎪ ⎨ ⎪ N (x) : shape functions ⎩ i

© 2005 by T. H. Kwon

51

Then, ~ Π ( φ ) = Π (φ1 , φ 2 , L , φ n ) to be minimized w.r.t. φi.

∴

∂Π = 0, ∂φ i

i = 1,L, n

: n equations for n unknown φi’s.

In this case, the shape functions can be found more easily than the trial functions without having to worry about satisfying the essential boundary conditions, which makes FEM much more useful than Rayleigh-Ritz Method. In this regard, the Finite Element Method is a modernized approximation method suitable for computer environment.

Example: Tight string problem via two methods

w(x) x

y(x) y 2 ⎤ l ⎡ 1 ⎛ dy ⎞ Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy ⎥dx 0 2 ⎥⎦ ⎢⎣ ⎝ dx ⎠

Special case: w(x) = w : constant

to be minimized.

i) Rayleigh-Ritz Method πx 2πx ~ + A2 sin y = A1 sin l l (Note that trial functions satisfy essential B.C.) 2 ⎤ ⎡ 1 ⎛ d~ y⎞ y ⎥dx Π = Π ( A1 , A2 ) = ∫ ⎢ T ⎜ ⎟ − w~ 0 2 ⎦⎥ ⎣⎢ ⎝ dx ⎠ l

= −2wA1 ∂Π =0 ∂A1 ∂Π =0 ∂A2

© 2005 by T. H. Kwon

→ →

-2w

2 2 l 1 2 π 2 π + TA1 + TA2 π 4 l l

l 1 π2 + TA1 =0 π 2 l

π2 2TA2 =0 l

52

⇒

A1 =

4l 2 w , A2 = 0 π3T

∴

πx 4l 2 w ~ y = 3 sin l πT wl ⎛1⎞ y ⎜ ⎟ = 0.129 Note: ~ T ⎝ 2⎠

wl ⎛1⎞ y⎜ ⎟ = 0.125 T ⎝ 2⎠

2

vs.

2

Question: When the natural boundary condition is applied at the right end, what should be done in this method? (Read [Bathe Example 3.22]) i.e.,

⎤ ⎡ 1 ⎛ dy ⎞ 2 dy Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy ⎥dx − T y 0 2 dx l ⎦⎥ ⎣⎢ ⎝ dx ⎠ l

is to be minimized.

Hint: One may introduce, for instance, 2πx πx ~ + A3 x y = A1 sin + A2 sin l l which satisfies the essential B.C. at the left end. You should recognize that the approximate solution can satisfy the natural B.C. only approximately. ii) Finite Element Method ⎤ ⎡ 1 ⎛ dy ⎞ 2 Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy ⎥dx 0 2 ⎦⎥ ⎣⎢ ⎝ dx ⎠ l

to be minimized

is equivalent to l⎡ ⎤ dy ⎛ dy ⎞ δΠ = ∫ ⎢T δ⎜ ⎟ − wδy ⎥dx = 0 for any δy. 0 ⎦ ⎣ dx ⎝ dx ⎠

Let us introduce elements to the system as depicted below.

1

①

y

y1 e

2

②

xl

③

3

4

x

ⓘ : element number i : node number

xl : local coordinate

y2 e

le Introduce the approximate solution via interpolation functions (or shape functions) for

© 2005 by T. H. Kwon

53

each element. y e ( x) = N i ( x) y ie Then dy e dN i e = yi dx dx δy ( x) = N i δy ie

1

⎛ dy ⎞ d (δy ) = dN i δyie δ⎜ ⎟ = dx ⎝ dx ⎠ dx and the variation of functional over each element is summed to result in the variation of the whole system, i.e., δΠ = ∑ δΠ e = δΠ (1) + δΠ ( 2 ) + δΠ ( 3) e

Let us consider δΠ e for any element. le ⎡ dy ⎛ dy ⎞ ⎤ δΠ e = ∫ ⎢T δ⎜ ⎟ − wδy ⎥dxl 0 ⎣ dx ⎝ dx ⎠ ⎦

(henceforth xl → x for convenience)

le ⎡ dN j ⎤ dN i e δΠ e = ∫ ⎢T δy i − wN i δy ie ⎥dx y ej 0 dx ⎣ dx ⎦ ⎛ le ⎡ dN j dN i ⎤ ⎞ le = δy ie ⎜⎜ ∫ ⎢T dx y ej − ∫ wN i dx ⎟⎟ ⎥ 0 0 ⎝ ⎣ dx dx ⎦ ⎠

= δy ie (K ije y ej − f i e ) ⎧δy e ⎫ = ⎨ 1e ⎬ ⎩δy 2 ⎭

T

(in an indicial form)

⎧⎪ e ⎧ y1e ⎫ ⎧ f1e ⎫⎫⎪ ⎨[K ] ⎨ e ⎬ − ⎨ e ⎬⎬ ⎪⎩ ⎩ y 2 ⎭ ⎩ f 2 ⎭⎪⎭

(in a matrix form)

where le ⎡ dN j dN ⎤ i K ije = ∫ ⎢T ⎥dx 0 ⎣ dx dx ⎦ le

f i e = ∫ wN i dx 0

1

: element stiffness matrix : work - equivalent nodal force

Note that δy(x) is also interpolated by the same shape function with the variation at the nodes, δy i . e

In fact, this specific δy(x) limits the arbitrariness of δy in the variational principle so that the solution becomes just an approximation, not an exact one.

© 2005 by T. H. Kwon

54

Assembly procedure is a matter of summation of scalar quantities yielding δΠ = ∑ δΠ e

2

e

= δy i (K ij y j − Fi ) = 0

for any δy i

which provide us with the following global matrix equation:

[K ]{y} = {F } {F }

being the work equivalent nodal force3. Consider specifically a linear element for simplicity. y e ( x) = N 1 ( x) y1e + N 2 ( x) y 2e

N 1 (ξ) = 1 − ξ

1

N 2 (ξ) = ξ with

ξ≡

∴

1 ξ

xl called a normalized coordinate. le dN i dN i dξ 1 dN i = = dx dξ dx l e dξ

N2(ξ)

N1(ξ)

where the shape functions can be found as

ξ=0 dN 1 1 =− , dx le

⇒

ξ=1

dN 2 1 = dx le

1 − 1⎤ l e ⎣− 1 1 ⎥⎦

[K ]e = T ⎡⎢ { f }e = ∫0

le

⎧N ⎫ wl ⎧1⎫ w⎨ 1 ⎬dx = e ⎨ ⎬ 2 ⎩1⎭ ⎩N 2 ⎭

3

In this particular case, three elements yields the global matrix equation as follows:

2

The assembly procedure is in fact the same as before, identifying the global nodal number and add the element stiffness matrix coefficient to the corresponding row and column in the global stiffness matrix. 3 The work equivalent nodal force obtained in this way is to be regarded as a concentrated external load replacing the distributed force effect. For a constant distributed force, in the case of a linear element, the total force applied to the element is equally divided to two adjacent nodes as the work equivalent nodal force. For the case of other type elements, variational formulation tell us how to distribute the total force to nodes associated with the element in contrast to direct approach. As an exercise, find yourself the work equivalent nodal forces at three nodes for quadratic element.

© 2005 by T. H. Kwon

55

0 ⎤ ⎧ y1 ⎫ ⎡ 1 −1 0 ⎧1 / 2⎫ ⎢− 1 2 − 1 0 ⎥ ⎪ y ⎪ ⎪ ⎪ 3T ⎢ ⎥ ⎪⎨ 2 ⎪⎬ = wl ⎪⎨ 1 ⎪⎬ l ⎢ 0 − 1 2 − 1⎥ ⎪ y 3 ⎪ 3 ⎪ 1 ⎪ ⎢ ⎥ ⎪⎩1 / 2⎪⎭ 0 − 1 1 ⎦ ⎪⎩ y 4 ⎪⎭ ⎣0

(singular as it is)

Boundary conditions (y1 = y4 =0) are introduced to yield the solution: y 2 = y3 =

1 wl 2 9 T

: happen to be the same as exact solution

Notes: 1. 1st and 4th equations are not to be used (or obtained more precisely) since δy1 and δy4 are not arbitrary, but zero. As a matter of fact, however, introduction of boundary conditions replaces those equations. The reaction force F1 and F4 can be obtained from the 1st and 4th equations, respectively. 2. Advantages of variational approach over the direct one: Use of scalar quantity (energy) versus vectors Ease in treatment of distributed load 3. Treatment of concentrated loads: F c

1

y

yc

The functional should incorporate the effect of the concentrated force: ⎤ ⎡ 1 ⎛ dy ⎞ 2 Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy ⎥dx − Fc y c 0 2 ⎦⎥ ⎣⎢ ⎝ dx ⎠ l

l ⎡ dy ⎛ dy ⎞ ⎤ δΠ = ∫ ⎢T δ⎜ ⎟ − wδy ⎥dx − Fc δy c = 0 0 ⎣ dx ⎝ dx ⎠ ⎦

or ⎤ ⎡ 1 ⎛ dy ⎞ 2 Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy − Fc δ( x − xc ) y ⎥dx 0 2 ⎦⎥ ⎣⎢ ⎝ dx ⎠ l

or

⎡ dy ⎛ dy ⎞ ⎤ δΠ = ∫ ⎢T δ⎜ ⎟ − wδy − Fc δ( x − x c )δy ⎥dx = 0 0 ⎦ ⎣ dx ⎝ dx ⎠

(the same result)

l

(One can introduce a node at the concentrated load point. Otherwise, what happens?)

© 2005 by T. H. Kwon

56

2.8.5 FEM for Second -Order Elliptic Partial Differential Equation There are many physical problems described by a 2nd order elliptic partial differential equation. Most notably, steady state heat conduction, flow though porous media, torsion, etc. This problem may be stated as follows:

nˆ

S2 z

S1 y

x ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞ ⎟ + ⎜kz ⎟ = f ( x, y , z ) ⎜kx ⎟ + ⎜ky ∂x ⎝ ∂x ⎠ ∂y ⎜⎝ ∂y ⎟⎠ ∂z ⎝ ∂z ⎠ B.C.

φ = φ ( x, y , z ) ∂φ ∂φ kx

∂x

nx + k y

∂y

in Ω

(2.28)

on S1 ny + kz

∂φ n z + g ( x, y, z ) + h( x, y, z )φ = 0 on S 2 ∂z

The above partial differential equation with boundary conditions is equivalent to the following variational principle: Minimize 2 2 ⎡ 1 ⎧⎪ ⎛ ∂φ ⎞ 2 ⎤ ⎛ ∂φ ⎞ ⎛ ∂φ ⎞ ⎫⎪ J (φ , φ x , φ y , φ z ) = ∫ ⎢ ⎨k x ⎜ ⎟ + k y ⎜⎜ ⎟⎟ + k z ⎜ ⎟ ⎬ + fφ ⎥ dV 2 ∂x ⎥⎦ ⎝ ∂z ⎠ ⎪⎭ ⎝ ∂y ⎠ Ω⎢ ⎪ ⎣ ⎩ ⎝ ⎠ 1 ⎛ ⎞ + ∫ ⎜ gφ + hφ 2 ⎟dS 2 ⎠ S2 ⎝

i.e.

(2.29)

δJ = 0 for any δφ .

(You are requested to verify this equivalence yourself.) With many elements introduced, one can sum the contributions of each element to the

© 2005 by T. H. Kwon

57

functionals as described below: Ωe

J = ∑ Je e

δJ = ∑ δJ e e

Introduce the approximate solution in terms of shape functions and nodal values of φ over each element: Ωe

φ = Niφ e

e i

S2

∂Ω

δφ e = N i δφ ie

2 2 ⎡ 1 ⎧⎪ ⎛ ∂φ ⎞ 2 ⎤ ⎛ ∂φ ⎞ ∂φ ⎞ ⎫⎪ ⎛ J e = ∫ ⎢ ⎨k x ⎜ ⎟ + k y ⎜⎜ ⎟⎟ + k z ⎜ ⎟ ⎬ + fφ⎥ dV ∂x 2 ⎥⎦ ⎝ ∂z ⎠ ⎪⎭ ⎝ ∂y ⎠ Ωe ⎢ ⎣ ⎪⎩ ⎝ ⎠ 1 ⎛ ⎞ + ∫ ⎜ gφ + hφ 2 ⎟dS 2 ⎠ ∂Ω e ∩ S 2 ⎝

δJ e =

⎡⎧ ∂φ ⎛ ∂φ ⎞ ⎤ ∂φ ⎛ ∂φ ⎞ ∂φ ⎛ ∂φ ⎞⎫ ⎜ ⎟ δ δ + δφ δ + + k k k f ⎜ ⎟ ⎜ ⎟ ⎢ ⎥ dV ⎨ ⎬ x y z ∫ ⎢⎩ ∂x ⎝ ∂x ⎠ ∂y ⎜⎝ ∂y ⎟⎠ ∂z ⎝ ∂z ⎠⎭ Ωe ⎣ ⎦⎥

∫ (gδφ + hφδφ)dS

+

∂Ω e ∩ S 2

∂N j e ∂N i e ∂N j e ∂N i e ⎫ ⎧ ∂N j e ∂N i e = ∫ ⎨k x φj δφ i + k y φj δφ i + k z φj δφ i ⎬dV ∂x ∂x ∂y ∂y ∂z ∂z ⎭ Ωe ⎩ +

∫ fN δφ dV e i

i

Ωe

+

∫ gN δφ dS + ∫ hN e i

i

∂Ω e ∩ S 2

j

φ ej N i δφ ie dS

∂Ω e ∩ S 2

∂N j ∂N i ∂N j ∂N i ⎫ ⎧ ∂N j ∂N i e δJ e = δφ ie ∫ ⎨k x + ky + kz ⎬dV φ j ∂x ∂x ∂y ∂y ∂z ∂z ⎭ Ωe ⎩ + δφ ie

∫ fN dV i

Ωe

+ δφ ie

∫ gN dS + δφ ∫ hN i

∂Ω e ∩ S 2

© 2005 by T. H. Kwon

e i

j

N i dSφ ej

∂Ω e ∩ S 2

58

e e δJ e = δφ ie (K Cij φ ej + K Sij φ ej + R efi + RSie )

where ⎧ ∂N i ∂N j ∂N i ∂N j ∂N i ∂N j ⎫ e = ∫ ⎨k x + ky + kz K Cij ⎬dV : stiffness matrix ∂x ∂x ∂y ∂y ∂y ∂y ⎭ Ωe ⎩ (due to conduction)

∫ hN

e K Sij =

j

N i dS

:

stiffness matrix

∂Ω e ∩ S 2

(due to convection)

R efi =

∫ fN dV

: forcing matrix

i

Ωe

(due to distributed heat sink) RSie =

∫ gN dS

: forcing matrix

i

∂Ω e S 2

(due to distributed heat outflux) After the assembly procedure, one can obtain

δ J = ∑ δ Je e

(

= δφi KCijφ j + K Sijφ j + R fi + RSi =0

∴

∴ or with

)

for any arbitray δφi , with δφ = 0 for nodes on S1.

[K C ]{φ } + [K S ]{φ } + {R f }+ {RS } = 0

(2.30)

[K ]{φ } = {F }

[K ] = [K C ] + [K S ] {F } = −{R f }− {RS }

:

forcing matrix due to heat source and heat influx

© 2005 by T. H. Kwon

59

2.8.6 Variational Principle for Deformation A. Principle of Virtual Displacement (Work)

t i = σ ij n j

equilibrium state

σ ij , j + ρf i = 0

δu i : virtual displacement For a given statically admissible stress field σ ij , consider any kinematically admissible virtual displacement δu i and external work due to the virtual displacement.

δWext = ∫ t i δu i dS + ∫ ρf i δu i dV ∂Ω

Ω

= ∫ σ ij n j δu i dS + ∫ ρf i δu i dV ∂Ω

= ∫ σ ij Ω

Ω

⎡ ∂σ ij ⎤ ∂δu i dV + ∫ ⎢ + ρf i ⎥δu i dV ∂x j Ω⎢ ⎣ ∂x j ⎦⎥

= ∫ σ ij δε ij dV Ω

where 1 ⎛ ∂δu i ∂δu j + δε ij ≡ ⎜⎜ ∂xi 2 ⎝ ∂x j ∂δu i = δε ij + δω ij , ∂x j ∴

⎞ ⎛ ∂δu j ⎟, δω ij ≡ 1 ⎜ ∂δu i − ⎟ ∂xi 2 ⎜⎝ ∂x j ⎠

σ ij δω ij = 0

δWext = ∫ t i δu i dS + ∫ ρf i δu i dV = ∫ σ ij δε ij dV ∂Ω

© 2005 by T. H. Kwon

⎞ ⎟ ⎟ ⎠

Ω

Ω

60

(2.31)

The converse can be proved, i.e., if the external work is equal to

∫σ

ij

δε ij dV for any

Ω

kinematically admissible virtual displacement δu i for a certain assumed field σ ij , then the stress field σ ij satisfies the equilibrium equation, i.e. σ ij , j + ρf i = 0 is satisfied. The physical meaning of equation (2.31) might be restated as follows:

∫σ

Ω

ij

δε ij dV − ∫ t i δu i dS − ∫ ρf i δu i dV = 0 ∂Ω

(2.32)

Ω

for any kinematically admissible virtual displacement δu i if σ ij , t i , f i are in equilibrium. That is, equation (2.32) can be regarded as a stationary condition of the following functional for an equilibrium: δΠ = ∫ σ ij δε ij dV − ∫ t i δu i dS − ∫ ρf i δu i dV = 0 Ω

with

∂Ω

Ω

Π = ∫ σ ij ε ij dV − ∫ t i u i dS − ∫ ρf i u i dV Ω

∂Ω

(with σ ij , t i , ρf i fixed)

(2.33)

Ω

Notes:

1. Kinematically admissible virtual displacement δu i →

δu i = 0 where u i = u i

prescribed. 2. Statically admissible stress field satisfies not only the equilibrium equation but also the prescribed traction boundary condition (i.e., natural boundary condition), i.e. t i = σ ij n j = t i 3. Principle of virtual velocity (power)

δε ij δWext δu δu = ∫ t i i dS + ∫ ρf i i dV = ∫ σ ij dV δt δt δt δt ∂Ω Ω Ω δvi

© 2005 by T. H. Kwon

δvi

δDij

61

4. If inertial term − ρu&&i is included in the body force term, the principle of virtual displacement can be extended to fictitious equilibrium state.

∫ t δu dS + ∫ ρ ( f i

i

∂Ω

Ω

i

− v&i )δu i dV = ∫ σ ij δε ij dV Ω

u i , ε ij can be replaced with vi , Dij , respectively.

B. Principle of Minimum Potential Energy

Principle of virtual displacement : ∴

δWext = ∫ ti δui dS + ∫ ρf i δui dV = ∫ σ ij δε ij dV ∂Ω

Ω

Ω

For an elastic body, there exists strain energy density, Uo, such that

σ ij =

then

∫σ

Ω

ij

∂U o ∂ε ij

(a constitutive equation) ⎛ ⎞ ∂U o δε ij dV = ∫ δU o dV = δ ⎜⎜ ∫ U o dV ⎟⎟ = δU ∂ε ij Ω Ω ⎝Ω ⎠

δε ij dV = ∫

where

⎛ ⎞ U ≡ ⎜⎜ ∫ U o dV ⎟⎟ ⎝Ω ⎠

∴

δWext = δU

:

strain energy

Now, let us define the potential energy V as ⎛ ⎞ V = −⎜⎜ ∫ t i u i dS + ∫ ρf i u i dV ⎟⎟ Ω ⎝ ∂Ω ⎠

with fixed ti and ρfi.

δWext = −δV Therefore, δ (U + V ) = 0 Then

Defining the total potential energy Π p as Π p = U +V

© 2005 by T. H. Kwon

62

yields

δΠ p = 0 .

In summary, the deformation of any elastic body (linear or nonlinear) is governed by minimizing the functional, the total potential energy, Π p (u i ) = ∫ U o dV − ∫ t i u i dS − ∫ ρf i u i dV Ω

with

σ ij =

∂Ω

(2.34)

Ω

∂U o 1 and ε ij = (u i , j + u j ,i ) given. 2 ∂ε ij

Note: Principle of Virtual Work, equations (2.31)-(2.33) is valid for any material, whereas Principle of Minimum Potential Energy is valid only for elastic materials. C. Principle of Complementary Virtual Work

Consider a variation of statically admissible stress field σ ij and external forces f i , t i while keeping kinematically admissible displacement u i .

δσ ij , j + δ (ρf i ) = 0

in Ω

δσ ij n j = δt i

on S

(∂Ω)

Define a complementary virtual work δW * as

δW*

dε

Ω

= ∫ u i (δσ ij )n j dS + ∫ u i δ(ρf i )dV ∂Ω

dσ

δWext

δW * = ∫ u i δt i dS + ∫ u i δ(ρf i )dV ∂Ω

σ

Ω

[

ε

]

= ∫ u i , j δσ ij dV + ∫ δ σ ij , j + ρf i u i dV Ω

Ω

= ∫ ε ij δσ ij dV Ω

∴

δW * = ∫ u i δt i dS + ∫ u i δ(ρf i )dV = ∫ ε ij δσ ij dV ∂Ω

© 2005 by T. H. Kwon

Ω

Ω

63

(2.35)

Now, as a counterpart to the Principle of Minimum Potential Energy for an elastic material, consider the case for an elastic material for which a complementary strain energy density exists as ε ij ≡

∂U o* ∂σ ij

(a constitutive equation)

Then we can rewrite the right hand side of the expression for δW * as ⎛ ⎞ ∂U o* δW = ∫ ε ij δσ ij dV = ∫ δσ ij dV = ∫ δU o* dV =δ⎜⎜ ∫ U o* dV ⎟⎟ = δU * ∂σ ij Ω Ω Ω ⎝Ω ⎠ *

and the left hand side can be rewritten in terms of the complementary potential energy as follows δV * = − ∫ u i δt i dS − ∫ u i δ(ρf i )dV ∂Ω

Ω

V * = − ∫ u i t i dS − ∫ u i (ρf i )dV

with

∂Ω

(keeping u i fixed)

Ω

Therefore, one can have Π * (σ ij , t i , f i ) = U * + V *

δΠ * = 0

and

:

Total complementary potential energy

for any statically admissible stress, force system

Note:

Π p (u i )

: leads to displacement-based FEM yielding a stiffness matrix

Π * (σ ij , t i , f i )

:

© 2005 by T. H. Kwon

leads to equilibrium-based FEM yielding a flexibility matrix

64

2.8.7 Displacement-based FEM for Elasticity

[Bathe P.120-125]

Let us apply the principle of virtual displacement (or principle of minimum potential energy) to the two-dimensional elastic deformation problem. Principle of Virtual Displacement (Work):

δΠ p = ∫ σ ij δε ij dV − ∫ t i δu i dS − ∫ ρf i δu i dV = 0

(2.36)

Π p = ∫ σ ij ε ij dV − ∫ t i u i dS − ∫ ρf i u i dV

(2.37)

Ω

∂Ω

Ω

Ω

∂Ω

(with σ ij , t i , ρf i fixed)

Ω

Principle of Minimum Potential Energy:

Π p (u i ) = ∫ U o dV − ∫ t i u i dS − ∫ ρf i u i dV Ω

σ ij =

with

∂Ω

(2.38)

Ω

∂U o 1 and ε ij = (u i , j + u j ,i ) given. 2 ∂ε ij

In this section, let us include the inertia force and concentrated forces as the most general FEM formulation for elasticity. So, consider the following variational expression as the starting form.

δΠ p = ∫ σ ij δε ij dV − ∫ t i δu i dS − ∫ ( ρf i − ρu&&i )δu i dV − ∑ Fi ( k )δu i( k ) = 0 Ω

∂Ω

(2.36)’

k

Ω

Displacement approximation via shape functions

{ur} = ⎧⎨

u ( x, y ) ⎫ ⎬ = [N ]{V } ⎩ v ( x, y ) ⎭

r ⎧u ⎫ ⎡ N e.g. {u } = ⎨ ⎬ = ⎢ 1 ⎩v ⎭ ⎣ 0

{}

0

N2

0

N3

N1

0

N2

0

r ⎧u&&⎫ Acceleration u&& = ⎨ ⎬ = [N ] V&& ⎩v&&⎭ Strain matrix:

© 2005 by T. H. Kwon

{}

65

⎧ u1 ⎫ ⎪v ⎪ ⎪ 1⎪ 0 ⎤ ⎪⎪u 2 ⎪⎪ ⎨ ⎬ N 3 ⎥⎦ ⎪v 2 ⎪ 1 ⎪u 3 ⎪ ⎪ ⎪ ⎪⎩ v3 ⎪⎭

3

2

⎧ε x ⎫ {ε } = ⎪⎨ ε y ⎪⎬ = [N ′]{V } ⎪γ ⎪ ⎩ xy ⎭

e.g.

⎡ ∂N 1 ⎢ ⎢ ∂x [N ′] = ⎢⎢ 0 ⎢ ∂N ⎢ 1 ⎣⎢ ∂y

∂N 2 ∂x

0 ∂N 1 ∂y ∂N 1 ∂x

0 ∂N 2 ∂y

⎤ 0 ⎥ ⎥ ∂N 3 ⎥ ∂y ⎥ ∂N 3 ⎥ ⎥ ∂x ⎥⎦

∂N 3 ∂x

0 ∂N 2 ∂y ∂N 2 ∂x

0 ∂N 3 ∂y

Stress-Strain relation (constitutive law): ⎧σ x ⎫ {σ } = ⎪⎨σ y ⎪⎬, ⎪τ ⎪ ⎩ xy ⎭

⎧ε x ⎫ {ε } = ⎪⎨ ε y ⎪⎬ ⎪γ ⎪ ⎩ xy ⎭

(Superscript (e) is omitted for simplicity.)

{σ } = [C ]{ε } Remind that we already had equations (2.12), (2.13): ⎡ ⎤ 1 ν 0 ⎥ ⎢ [C ] ≡ E 2 ⎢ν 1 0 ⎥ 1 −ν ⎢ 1 −ν ⎥ ⎢0 0 ⎥ 2 ⎦ ⎣

for plane stress

⎡ ν ⎢1 − ν E ⎢ ν 1 −ν [C ] ≡ (1 + ν )(1 − 2ν ) ⎢ 0 ⎢ 0 ⎣

⎤ 0 ⎥ 0 ⎥ 1 − 2ν ⎥ ⎥ 2 ⎦

for plane strain

[Note: With this notation, the strain energy can be represented as 1 T U = ∫ {ε } [C ]{ε }dV 2 Ω

and the total potential energy becomes 1 T Π p = ∫ {ε } [C ]{ε }dV − ∫ t i u i dS − ∫ ( ρf i − ρu&&i )u i dV − ∑ Fi ( k ) u i( k ) . 2 k Ω ∂Ω Ω corresponding to Bathe Example 4.4, P.160

δΠ p = ∫ {δε }T [C ]{ε }dV − ∫ t i δu i dS − ∫ ( ρf i − ρu&&i )δu i dV − ∑ Fi ( k )δu i( k ) = 0 ] Ω

© 2005 by T. H. Kwon

∂Ω

k

Ω

66

[Note: In case of initial strain {ε } , {ε } is to be replaced with {ε } − {ε } . In case init

init

of initial stress, {σ } = [C ]{ε } + {σ init }.] Define the force matrices as follows: ⎧t ⎫ ⎩t y ⎭

{t} = ⎨ x ⎬

: traction force

⎧ fx ⎫ ⎬ ⎩ fy⎭

: body force

⎧⎪ Fx( k ) ⎫⎪ = ⎨ (k ) ⎬ ⎪⎩ Fy ⎪⎭

: concentrated force applied at k-th position

{f } = ⎨

{F } (k )

Virtual displacement and corresponding strain:

δu ( x, y ) ⎫ ⎬ = [N ]{δV } ⎩δv( x, y ) ⎭

{δur} = ⎧⎨

⎧δε xx ⎫ {δε } = ⎪⎨δε yy ⎪⎬ = [N ′]{δV } ⎪δγ ⎪ ⎩ xy ⎭ The variation of total potential energy is summation of contributions of each element and concentrated forces. That is, T

⎧δu ( k ) ⎫ δΠ p = ∑ δΠ − ∑ ⎨ ( k ) ⎬ {F ( k ) } k ⎩δv e ⎭ e p

= ∑ δΠ ep − ∑ {δV } [N ] T

e

T

k

{F } (k )

x( k )

Now, for an element:

δΠ ep =

∫ {δε } [C ]{ε }dV − ∫ t δu dS − ∫ (ρf T

i ∂Ω e ∩ S 2

Ωe

=

i

i

− ρu&&i )δu i dV

Ωe

r r r r ∫ {δε } [C ]{ε }dV − ∫ {δu} {t}dS − ∫ {δu} ρ{ f }dV + ∫ {δu} ρ {u&&}dV T

Ωe

© 2005 by T. H. Kwon

T

T

∂Ω e ∩ S 2

Ωe

67

T

Ωe

Introducing the matrix notations for approximated displacement field and constitutive law and so on into the above equation gives:

∫ [N ′] [C ][N ′]dV {V } − {δV } ∫ [N ] {t}dS − {δV } ∫ [N ] ρ{ f }dV

δΠ ep = {δV }

T

T

T

Ωe

+ {δV }

T

T

∂Ω e ∩ S 2

T

T

Ωe

∫ [N ] ρ[N ]dV {V&&} T

Ωe

= {δV }

eT

{[K ] {V } − {F } − {F } + [m] {V&&} } e

e

e

d

e

e

e

b

After assembly, one finally obtains:

δΠ p = {δV }T {[K ]{V } − {Fd } − {Fb } − {Fc } + [M ]{V&&}} = 0 for any {δV } .

[M ]{V&&}+ [K ]{V } = {Fd } + {Fb } + {Fc } = {F } 1

∴

(2.39)

where

[m]e = ∫ [N ] ρ [N ]dV

: element mass matrix

[K ]e = ∫ [N ′] [C ][N ′]dV

: element stiffness matrix

T

Ωe

T

Ωe

{Fd }

e

∫ [N ] {t}dS

: work equivalent nodal force

∫ [N ] ρ { f }dV

: work equivalent nodal force

T

=

∂Ω e ∩ S 2

{Fb }e

=

T

Ωe

{FC }e = ∑ [N ]T k ∩e

{F } (k )

xc

: concentrated force

Question: How can the damping force effect be incorporated in the formulation?

1

1 {V }T [K ]{V } − {V }T {F } is the functional to be minimized. δΠ p = 0 results in 2 [K ]{V } = {F } .

Note:

Πp =

© 2005 by T. H. Kwon

68

2.9 Method of Weighted Residual for Finite Element Formulation [Bathe P.116-119], [Cook et al. Chap. 15], [Huebner Chap. 4]

Method of Weighted Residual (MWR) is a mathematical approach to obtain an approximate solution to differential equations and provides another means of finite element formulation. Consider the problem stated below:

L(φ)=f

S2

Ω

z

∂Ω=S1+S2

S1 y

x PDE:

L(φ ) = f B.C.:

in Ω

r

φ = φ (x) r B(φ ) = g ( x )

(2.40)

on S1 on S 2

For the exact solution:

R D = L(φ ) − f = 0 r R B = B (φ ) − g ( x ) = 0

in Ω on S 2

If one assumes an approximate solution

~

n

r

φ ≈ φ = ∑ Ciψ i ( x )

(or

i =1

where

r

~

n

i =1

ψ i (x )

are linearly independent trial functions

Ci

are constants to be determined.

© 2005 by T. H. Kwon

69

r

φ = ∑ N i ( x )φ i

)

Then

~ R D = L(φ ) − f ≠ 0

in Ω

r ~ R B = B(φ ) − g( x ) ≠ 0

: nonzero residual indicating error in PDE

on S 2

: nonzero residual indicating error in natural boundary condition

MWR seeks to determine n unknown parameters Ci’s in such a way that the weighted average of the residual vanishes over the domain (or including on S2) For the time being, let us limit ourselves to RD only without RB on S2 . r Taking n linearly independent weighting functions, Wi (x ) , one wants to have:

∫ [L(φ ) − f ]W dΩ = ∫ R ~

i

Ω

D

Wi dΩ = 0,

Ω

i = 1, L , n

(2.41)

2.9.1 Several Kinds of Methods of Weighted Residual First let us discuss general MWR as a mathematical approximation method. There are many of them depending on the choice of the weighting functions: [Bathe P.118-119]

a) Subdomain method b) Collocation method c) Least square method (least square collocation method) d) Method of moments e) Galerkin’s method a) Subdomain Method Ω i : small subdomains ⎧1 Wi = ⎨ ⎩0

© 2005 by T. H. Kwon

i = 1, L, n

r x ∈ Ωi r x ∉ Ωi

70

Ωi

b) Collocation method r r Wi = δ (x − xi )

∫R

D

i = 1,L, n

r Wi dΩ = R D ( xi ) = 0

residuals at n collocation points are zero.

Ω

*orthogonal collocation method (using Chebyshev polynomial) [Finlayson ‘The method of weighted residuals and variational principles’, P.97]

c) Least square method I (C i ) = ∫ R D dΩ 2

to be minimized w.r.t. C i

Ω

i.e.

∂I (C i ) ∂R = ∫ 2 R D D dΩ = 0 ∂C i ∂C i Ω Wi = 2

i = 1, L , n

∂R D ∂C i

c’) Least square collocation method

[Cook, Malkus & Plesha, P.460]

When there are collocation points more than the number of Ci’s: r R D ( xi ) ≠ 0,

i = 1, L, N

N r 2 I = ∑ R D ( xi )

with N > n

to be minimized.

i =1

∂I (C i ) =0 ∂C i

yields n equations as follows :

{RD }N x 1 = [Q]N x n {C}n x 1 − {d }N x 1 ≠ {0} then I = {RD } {RD } = {C} [Q ] [Q ]{C} − 2{C} [Q ] {d } + {d } {d } T

© 2005 by T. H. Kwon

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T

71

T

T

T

⎧ ∂I ⎫ = {0}n x 1 ⎨ ⎬ ⎩ ∂C ⎭ n x 1

yields

[Q ]Tn x N [Q ]N x n {C}n x 1 = [Q ]Tn x N {d }N x 1 Note that the above equation is also obtainable from

[Q]T {[Q]{C} − {d }} = {0} .

d) Method of moments Wi = 1, x, x 2 , x 3 ,L, x n −1 i.e.

∫R

Ω

D

dΩ = ∫ xRD dΩ = ∫ x 2 R D dΩ = L = 0 Ω

Ω

(integral method: Von Karman, Polhausen for a boundary layer problem) e) Galerkin’s method Wi = ψ i the weighting functions are taken to be the trial functions (or shape functions). [Example of various MWR] : Solve the following differential equation with essential boundary conditions. d 2u +u+ x = 0 dx 2 B.C. u=0

at x = 0

u=0

at x = 1

The exact solution for this problem:

u exact =

sin x −x, sin 1

(Set u = a1 sin x + a 2 cos x − x with B.C.;

© 2005 by T. H. Kwon

72

a1 =

⎛1⎞ u exact ⎜ ⎟ = 0.06975 ⎝2⎠

1 , a2 = 0 ) sin 1

i) Collocation method One can consider the approximate solution of the form

u~ = x (1 − x )(α 1 + α 2 x + L ) Let us take only two terms among them as the first approximate solution. That is, u~ = x(1 − x)(α 1 + α 2 x)

d 2 u~ ~ + u + x = x + (−2 + x − x 2 )α 1 + (−6 x + 2 + x 2 − x 3 )α 2 . dx 2 To determine two parameters, one has to introduce two collocation points: Then

R=

1 1 , and x = 4 2

x=

Two equations are obtained: 116 35 1 α1 + α 2 + = 0 64 64 4 14 7 1 − α1 − α 2 + = 0 8 8 2

6 ≈ 0.1935 31 40 α2 = ≈ 0.1843 217

α1 =

−

∴

⇒

x(1 − x) u~ = (42 + 40 x) 217

⇒

⎛1⎞ u~⎜ ⎟ = 0.0714 ⎝2⎠

ii) Least square method ∂ ∂α 1 ∂ ∂α 2

1

∫R

2

0

1

∫R 0

dx = 0

1

⇒ 2

dx = 0

∫ R ( −2 + x + x 1

∫ R (2 − 6 x + x 0

⎛1⎞ u~⎜ ⎟ = 0.0686 ⎝2⎠ iii) Method of moment

© 2005 by T. H. Kwon

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0

73

2

)dx = 0 − x 3 )dx = 0

⇒

α 1 = 0.192 α 2 = 0.165

1

1

∫ R ⋅ 1dx = 0.,

∫ R ⋅ xdx = 0.

0

⎡ 11 ⎢6 ⎢ 11 ⎢ ⎣ 22

0

11 ⎤ ⎧1 ⎫ α ⎥ 6 ⎧ 1 ⎫ = ⎪2⎪ 19 ⎥ ⎨⎩α 2 ⎬⎭ ⎨ 1 ⎬ ⎪ ⎪ ⎥ 20 ⎦ ⎩3⎭

⇒

122 649 110 α2 = 649

α1 =

⎛1⎞ u~⎜ ⎟ = 0.0681 ⎝2⎠ iv) Galerkin’s method u~ = α 1 x(1 − x) + α 2 x 2 (1 − x) = α 1 N 1 ( x) + α 2 N 2 ( x)

1

∫

0 1

Rx(1 − x)dx = 0

∫ Rx 0

2

(1 − x)dx = 0

⇒

3 ⎤ ⎡3 ⎧1⎫ ⎢ 10 20 ⎥ ⎧α 1 ⎫ ⎪ 12 ⎪ = ⎢3 13 ⎥ ⎨⎩α 2 ⎬⎭ ⎨ 1 ⎬ ⎪ ⎪ ⎢ ⎥ ⎩ 20 ⎭ ⎣ 20 105 ⎦

⇒

71 369 7 α2 = 41

α1 =

⎛1⎞ u~⎜ ⎟ = 0.0694 ⎝2⎠ v) Subdomain method 1 2 0

∫

1

∫

1 2

Rdx = 0 Rdx = 0

⇒

⎡ 11 ⎢12 ⎢ 11 ⎢ ⎣12

− 53 ⎤ ⎧1 ⎫ α ⎥ ⎪ ⎪ ⎧ ⎫ 1 192 = ⎨8 ⎬ ⎨ ⎬ ⎥ 3 229 ⎩α 2 ⎭ ⎪ ⎪ ⎥ 192 ⎦ ⎩8 ⎭

⇒

α 1 = 0.188 α 2 = 0.170

⎛1⎞ u~⎜ ⎟ = 0.06825 ⎝2⎠

Gelerkin’s method for finite element formulation Now, let us discuss Gelerkin’s method for finite element formulation. The trial functions

© 2005 by T. H. Kwon

74

can be restated as shape functions if the function is expressed in terms of nodal value of the function and corresponding shape function for finite element treatment. That is, the weighting functions are taken to be the shape functions: Wi = N i

(Bubnov Galerkin)

(For upwinding scheme: Wi ≠ N i

Petrov-Galerkin) [Huebner & Thornton P. 445-448]

In summary, the Galerkin’s method of residual for finite element formulation begins with: r

n

~

φ ≈ φ = ∑ N i ( x )φ i i =1

∫ [L(φ ) − f ]N dΩ = 0, ~

i

Ω

(Cf .

~ ∂φ Wi = ∂φ i

(2.41)’ i = 1, L , n

Wi = 2

for Galerkin method vs

∂RD ∂φ i

for least square method)

This technique can be used for the entire domain as well as each element. For each element:

φ = φ e = ⎣N i ⎦1e x r {φ i }er x 1 ~

∫ [L(φ ) − f ]N e

Ω

e

e

i

dΩ = 0,

i = 1, L , r

If we stop here, i) trial function must satisfy all boundary conditions ii) one may not obtain a symmetric matrix. Therefore, one can introduce integral by parts to obtain: i) boundary conditions of higher order (i.e., natural boundary condition) ii) a symmetric matrix [Bathe P.123-129]

© 2005 by T. H. Kwon

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Symbolically, one can obtain

∫ N L(φ)dΩ = ∫ B′(φ, N )dS − ∫ A(φ, N )dΩ i

i

Ω

i

∂Ω

Ω

Resulting equation becomes

∫ A(φ , N )dΩ = − ∫ N i

Ω

Ω

i

fdΩ + ∫ B ′(φ , N i )dS

(2.42)

∂Ω

[K ]{φ } [K ]{φ } = {F }

Example: L includes 2m-th order of derivatives (I-D)

∫ L(φ )N dx e

i

⇓

∫

R

L

Ni

d 2m N j dx 2 m

φ j dx

2 m −1 Nj d 2 m −1 N j dN i d =−∫ φ j dx + N i φj L dx dx 2 m −1 dx 2 m −1

R

R

L

M

{

}

m d m Ni d N j =∫ φ j dx + N i ( 0 ) N j ( 2 m −1) − N i (1) N j ( 2 m −2 ) + L + N i ( m −1) N j ( m ) φ j m m L dx dx R

There are pairs of boundary conditions as noted below:

© 2005 by T. H. Kwon

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R L

⎧ N i ( 0) , ⎪ ⎨ ⎪ N ( 2 m −1) , i ⎩

(1)

N i ,L, Ni

( 2m−2)

Ni

( m −1)

, L, N i

: essential boundary conditions

(m)

: natural boundary conditions [Bathe P.110-111]

Now, let us consider the more general case including the residuals of natural boundary condition.

∫ [L(φ ) − f ]W dΩ = 0 ~

i

Ω

∫ [B(φ ) − g]⋅ W dS = 0 (

~

i

S2

One can generally introduce the combination of the weighted residuals as follows:

∫ [L(φ ) − f ]W dΩ + α ∫ [B(φ ) − g]⋅ W dS = 0 ~

(

~

i

Ω

i

S2

where α can be any real number in general. (α is at your disposal. The larger α is, the larger weighting in the natural boundary condition. Different α results in different approximate solution.) Specific α may result in the same expression as the variational approach when it is available. For instance, when L includes 2m-th order of derivatives, the natural boundary ~ condition, B(φ ) − g = 0 , may look like the following:

φ (m) = g m

on S 2

φ ( m +1) = g m +1

on S 2

φ ( m+ 2) = g m + 2

on S 2

M

φ

( 2 m −1)

= g 2 m −1

on S 2

B(φ) and g might be rewritten as

© 2005 by T. H. Kwon

77

⎧φ ( 2 m −1) ⎫ ⎪ ( 2 m−2) ⎪ ⎪ ⎪φ B(φ ) = ⎨ ⎬, M ⎪ ⎪ ⎪⎩ φ ( m ) ⎪⎭

⎧φ ( 2 m −1) ⎫ ⎪ ( 2 m−2) ⎪ ⎪ ⎪φ g=⎨ ⎬, M ⎪ ⎪ ⎪⎩ φ ( m ) ⎪⎭

⎧ Wi ⎫ ⎪ dWi ⎪ ⎪⎪ ⎪⎪ ( with choosing Wi = ⎨ dx ⎬ ⎪ mM −1 ⎪ ⎪ ⎪d ⎪⎩ dx m −1 ⎪⎭

Then, with a special choice of α=-1, one may introduce

[∫ L(φ~ ) − f ]W dΩ − ∫ [B(φ~ ) − g]⋅ W( dS = 0 i

i

Ω

S2

Integral by parts for the first term results in

∫ A(φ ,W )dΩ = − ∫ W fdΩ + ∫ g ⋅ W dS + ∫ B′(φ ,W )dS − ∫ B(φ )⋅ W dS (

~

i

Ω

i

Ω

~

i

∂Ω

S2

~

i

(

1

i

S2

( where Wi are chosen in such a way that the last two terms in the right hand side could

be cancelled out. Then the following equation results:

∫ A(φ ,W )dΩ = − ∫ W fdΩ + ∫ g ⋅ W dS (

~

i

Ω

i

Ω

i

(2.43)

S2

Comparing equations (2.42) and (2.43), one can recognize that the last term in the right hand side of equation (2.42) is replaced with the last term in the right hand side of equation (2.43). This results is equivalent to disregarding the residual in the natural boundary condition and just introducing the natural boundary condition, wherever appropriate terms appear, as if the natural boundary condition is fully satisfied (i.e. with no residuals). It may be noted that, in the derivation of equation (2.43), the essential boundary condition is assumed to be satisfied completely, which is accomplished by setting the corresponding weighting functions zero on the surface S1. The example of this procedure will be demonstrated later with the tight string problem with a natural boundary condition involved.

1

The contribution of surface S1 on the third term in the right hand side can be dropped out since the corresponding value of Wi appearing in the term are set to zero in the weighted residual method.

© 2005 by T. H. Kwon

78

2.9.2. String Problem via Method of Weighted Residual

d ⎛ dy ⎞ ⎜T ⎟ + w = 0 dx ⎝ dx ⎠ y=0 at x = 0 T

dy dy =T dx dx

at x = l

x

w(x) y y(x)

T T

dy dx

i) Variational Approach (review) ⎡ 1 ⎛ dy ⎞2 ⎤ dy Π ( y ( x) ) = ∫ ⎢ T ⎜ ⎟ − wy ⎥dx − T y 0 2 dx l ⎣⎢ ⎝ dx ⎠ ⎦⎥ l

l ⎡ dy ⎛ dy ⎞ ⎤ dy δΠ = ∫ ⎢T δ⎜ ⎟ − wδy ⎥dx − T δy = 0 0 dx l ⎣ dx ⎝ dx ⎠ ⎦

(2.44A)

for any δy

δΠ = ∑ δΠ e + δΠ B = 0 e

⎡ dy ⎛ dy ⎞ ⎤ dy δ ⎜ ⎟ − wδy ⎥dx − T δ y = 0 dx ⎣ dx ⎝ dx ⎠ ⎦ l

δΠ = ∑ ∫ ⎢T le

0

e

Let us consider δΠ e for any element. le

⎡ dy ⎛ dy ⎞ ⎤ δ ⎜ ⎟ − wδy ⎥dx ⎣ dx ⎝ dx ⎠ ⎦

δΠ e = ∫ ⎢T 0

le ⎡ dN j ⎤ dN i e = ∫ ⎢T δy i − wN i δy ie ⎥dx y ej 0 dx ⎣ dx ⎦ ⎛ le ⎡ dN j dN i ⎤ ⎞ le = δy ie ⎜⎜ ∫ ⎢T dx y ej − ∫ wN i dx ⎟⎟ ⎥ 0 0 ⎝ ⎣ dx dx ⎦ ⎠

= δy ie (K ije y ej − f Die )

where

© 2005 by T. H. Kwon

79

(2.44B)

le ⎡ dN j dN ⎤ i K ije = ∫ ⎢T ⎥dx 0 ⎣ dx dx ⎦

: element stiffness matrix

le

f Die = ∫ wN i dx

: work - equivalent nodal force

0

⎛

δΠ = δy i (K ij y j − FDi ) + δy l ⎜⎜ − T = {δy}

T

⎝

{[K ]{y} − {FD } − {FB }}

dy ⎞ ⎟ dx ⎟⎠

=0

[K ]{y} = {FD } + {FB } = {F } where

{FD } {FB }

(2.45)

: work equivalent force due to distributed load : external load applied at the boundary node

In this case, one can proceed to obtain the Euler-Lagrange equation and associated boundary condition: l

⎛ dy ⎤ d ⎛ dy ⎞ dy ⎞ dy − T ⎟⎟δy − T δy = 0 ⎜ T ⎟ − w⎥δydx + ⎜⎜ T dx ⎠ l dx 0 ⎦ ⎣ dx ⎝ dx ⎠ ⎝ dx ⎡

δΠ = ∫ ⎢ − 0

ii) Galerkin’s Method Method of Weighted Residual:

One can start with the residual in the domain as follows. ⎡ d ⎛ d~ ⎤ y⎞ ∫0 ⎢⎣ dx ⎜⎝ T dx ⎟⎠ + w⎥⎦Wi dx = 0 l

l l d~ l d~ y y dWi =T Wi − ∫ T dx + ∫ wWi dx 0 0 dx dx dx 0 ~ l dy dW l d~ y d~ y i =T Wi − T Wi − ∫ T dx + ∫ wWi dx = 0 0 0 dx dx dx dx l 0

© 2005 by T. H. Kwon

80

for any δy (2.44C)

(Remind:

∫ A(φ , N )dΩ = − ∫ N i

Ω

i

Ω

fdΩ + ∫ B ′(φ , N i )dS ) ∂Ω

Substituting the natural boundary condition at x=l for T ⎛ dy d ~y ⎞ ⎟W equivalent to adding ⎜⎜ T − T dx dx ⎟⎠ ⎝

d~ y dx

in the above equation is l

.

i l

The most appropriate form for the MWR is thus ⎛ dy ⎡ d ⎛ d~ ⎤ d~ y⎞ y⎞ ⎜ ⎟Wi = 0 + + − T w W dx T T ⎜ ⎟ i ∫0 ⎢⎣ dx ⎝ dx ⎠ ⎥⎦ ⎜ dx ⎟ dx ⎝ ⎠ l l

l

=−∫ T 0

(2.44D)

l d~ y d~ y dWi dy dx + ∫ wWi dx + T Wi − T Wi 0 dx dx dx l dx

=0 0

i.e.

∫

l

0

T

l d~ y d~ y dWi dy dx − ∫ wWi dx − T Wi + T Wi 0 dx dx dx l dx

=0

(2.44E)

0

The above residual expression is valid for all MWR’s. Now, consider specifically the Galerkin’s Method which is appropriate for FEM. Galerkin’s Method

The weighting function for a node is chosen such that it takes the shape function for each element which includes the node as one of its nodes as defined below: N1(I+1)

N3(I)

Wi for end node (I)

(I+1) N2(I) Wi for mid-node

(I)

© 2005 by T. H. Kwon

81

Note that the displacement in an element is approximated by means of shape functions: e ~ y e = N i yi

For the end node, equation (E) becomes

∫

le

0

T l

dN j

( I +1)

(I )

dx

yj

(I )

(I ) dN j le dN i dx ( I ) + ∫ T 0 dx dx l

− ∫ wN i dx ( I )

− ∫ wN i

(I )

0

0

( I +1)

yj

( I +1)

( I +1)

dN i dx

dx ( I +1)

dx ( I +1)

=0

For the mid-node, equation (E) becomes

∫

le

0

T

dN j

(I )

dx

(I )

yj

(I )

l dN i (I ) dx ( I ) − ∫ wN i dx ( I ) = 0 0 dx

For the right end node for the rightmost element, (E) becomes

∫

le

0

T

dN j

(I )

dx

(I )

yj

(I )

l dN i dy (I ) ⋅1 = 0 dx ( I ) − ∫ wN i dx ( I ) − T 0 dx dx

For the left end node for the leftmost element, (E) becomes

∫

le

0

T

dN j

(I )

dx

yj

(I )

(I ) l dN i d~ y (I ) dx ( I ) − ∫ wN i dx ( I ) + T 0 dx dx

⋅1 = 0 x =0

However, the last equation is not going to be used since the essential boundary condition is to be imposed for this node. That is, y=0 at x=0 is to be introduced in place of the corresponding equation. Therefore, you do not have to worry about d~ y term. Note also that one can set Wi=0 at the node where the essential T dx x =0 boundary condition is applied. For all the meaningful equations, W=0 on S1. Noting that one has a global nodal numbering system associated with each element nodal numbering system, i.e.

© 2005 by T. H. Kwon

82

yj

→

(I )

element

yj global

The above equations are equivalent to the final global matrix equations by assembling the element matrix equations K ije y ej = f Die

where le ⎡ dN j dN ⎤ i K ije = ∫ ⎢T ⎥dx 0 ⎣ dx dx ⎦ le

f Die = ∫ wN i dx 0

: element stiffness matrix : work - equivalent nodal force

which are exactly the same as the results by the variational approach. Moreover, the external force matrix {FB } due to the natural boundary condition is also included in the equation for the right end node of the rightmost element. Therefore, one ends up with the same global matrix equation as the variational approach.

[K ]{y} = {FD } + {FB } = {F }

Note: Relationship between Variational Approach and Method of Weighted Residual

∏ minimized

δΠ = 0 in a balance form w.r.t. order of derivatives

Euler-Lagrange equation

⇔

Balanced form

⇒ Weak form

( R W d Ω − R ⋅ W D i B i dS = 0 ∫ ∫

Ω

∂Ω

Mathematical approximation to DE

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2.9.3 Weak Form of Problems

[Hughes Sect. 1.3] [Bathe P.126-127]

Apparently there is equivalence between variational approach and MWR:

⎛ dy ⎤ d ⎛ dy ⎞ dy ⎞ dy T ⎟ − w⎥δydx + ⎜⎜ T − T ⎟⎟δy − T δy = 0 ⎜ 0 dx ⎠ dx 0 ⎣ dx ⎝ dx ⎠ ⎦ ⎝ dx l l

⎡

δΠ = ∫ ⎢ −

for any δy

c

⎛ dy ⎡ d ⎛ d~ ⎤ y⎞ d~ y⎞ ⎜ T w W dx T + + − T ⎜ ⎟ ∫0 ⎢⎣ dx ⎝ dx ⎠ ⎥⎦ i ⎜⎝ dx dx ⎟⎟⎠Wi = 0 for arbitrary Weighting function Wi l l

δy ↔ Wi δy = 0 at x = 0

↔

1

Wi = 0 at x = 0

⎡ dy ⎛ dy ⎞ ⎤ dy δ ⎜ ⎟ − wδy ⎥dx − T δ y = 0 0 dx ⎣ dx ⎝ dx ⎠ ⎦ l l

δΠ = ∫ ⎢T

for any δy

c

y dWi dy d~ y ⎞ ⎛ d~ T wW dx − T W + T Wi − ⎟ ⎜ i i ∫0⎝ dx dx dx l dx ⎠ l

=0 0

Both approaches are nothing more than the solution procedure for the “Weak Form” of the problem. The strong form and weak form of the tight string problem can be stated, as an illustration, as follows:

1

It may be noted that, in MWR, one can further relax the essential boundary condition by introducing the residual of the essential boundary condition with an appropriate weighting function. There will be no further discussion for this subject. Those who are interested in this topic may refer to [Brebbia] for details. © 2005 by T. H. Kwon

84

Strong Form (classical) Find y such that d ⎛ dy ⎞ 0< x
dy dy =T dx dx

at x = l

Weak Form Find y ∈℘ such that for all W ∈ ℑ l

⎛

dy dWi dy ⎞ + wWi ⎟dx + T Wi = 0 dx dx l ⎠

∫ ⎜⎝ − T dx 0

where

{ {

}

℘ = y | y ∈ H 1 , y (0) = y ℑ = W | W ∈ H 1 , W (0) = 0

}

: :

Trial function y Weighting function W

One can easily find that both the variational principle and Method of Weighted Residual belong to the weak form of the problem. Definition of Sobolev Space:

{

L 2 = L2 ( Ω ) = W |

l

∫W 0

2

dx < ∞

}

: square integrable

H k = H k (Ω) = {W| W ∈ L2 ; W , x ∈ L2 ; W , xx ∈ L2 ; L; W , xxLx ∈ L2 } :Sobolev space of degree k k times

© 2005 by T. H. Kwon

85

2.9.4 Galerkin’s Method for 2nd Order Elliptic PDE ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞ ⎟ + ⎜kz ⎟ = f ( x, y , z ) ⎜kx ⎟ + ⎜ky ∂x ⎝ ∂x ⎠ ∂y ⎜⎝ ∂y ⎟⎠ ∂z ⎝ ∂z ⎠

in Ω

(2.45)

B.C. ⎧ φ = φ ( x, y , z ) on S1 ⎪ φ φ φ ∂ ∂ ∂ ⎨ ⎪⎩k x ∂x n x + k x ∂x n x + k x ∂x n x + g ( x, y, z ) + h( x, y, z )φ = 0 on S 2 Ω

S1 (I)

S2 3 (I) 2

i

i = NODE(I,J) = global node No. of J-th node for I-th element

1

~

φ e = N keφ ke

for each element

( Wi = N Je for element e)

Appropriate form of Weighted Residual is as follows: ⎡ ∂ ⎛ ∂φ~ ⎞ ∂ ⎛ ∂φ~ ⎞ ∂ ⎛ ∂φ~ ⎞ ⎤ ∫Ω ⎢⎢ ∂x ⎜⎜⎝ k x ∂x ⎟⎟⎠ + ∂y ⎜⎜⎝ k y ∂y ⎟⎟⎠ + ∂z ⎜⎜⎝ k z ∂z ⎟⎟⎠ − f ⎥⎥Wi dΩ ⎣ ⎦ ~ ~ ~ ⎡ ∂φ ∂φ ∂φ ~⎤ - ∫ ⎢k x nx + k y ny + kz n z + g + hφ ⎥Wi dS = 0 ∂x ∂y ∂z S2 ⎣ ⎦

(2.46)

Note that Wi = 0 on S1 for nodes which do not belong to S2. (One does not have to introduce the weighted residual for nodes belonging to S1.) Integral by parts results in

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~ ~ ~ ⎛ ∂φ ∂Wi ∂φ ∂Wi ∂φ ∂Wi ⎞ ⎟dΩ − ∫ fWi dΩ − ∫ ⎜⎜ k x + ky + kz ∂x ∂x ∂y ∂y ∂z ∂z ⎟⎠ Ω⎝ Ω ~ ~ ~ ⎛ ∂φ ∂φ ∂φ ⎞ + ∫ ⎜⎜ k x nx + k y ny + kz n z ⎟Wi dS ∂x ∂y ∂z ⎟⎠ S1 + S 2 ⎝ ~ ~ ~ ⎛ ∂φ ∂φ ∂φ ⎞ ~ − ∫ ⎜⎜ k x nx + k y ny + kz n z ⎟⎟Wi dS − ∫ g + hφ Wi dS ∂x ∂y ∂z ⎠ S2 ⎝ S2 =0

(

)

(Wi =0 on S1.) ~ ~ ~ ⎛ ∂φ ∂Wi ∂φ ∂Wi ∂φ ∂Wi ⎞ ⎜ ∫Ω ⎜⎝ k x ∂x ∂x + k y ∂y ∂y + k z ∂z ∂z ⎟⎟⎠dΩ + Ω∫ fWi dΩ ~ + ∫ g + hφ Wi dS

(

)

S2

=0

(Note that one can arrive at the same result with

∫R

D

Wi dΩ = 0 and replacing

Ω

~ ~ ~ ⎛ ∂φ ~ ∂φ ∂φ ⎞ ⎜kx ⎟ ⎜ ∂x n x + k y ∂y n y + k z ∂z n z ⎟ with − g + hφ after integral by parts.) ⎝ ⎠

(

)

(Note also that the above MWR expression is equal to δJ = 0 for the variational approach when Wi is identified with δφ . In this regard, Wi = 0 on S1 is equivalent to

δφ = 0 on S1.) When the domain is discretized, the volume integral is replaced with the summation of integral over each element, and the surface integral on S2 is replaced with the summation of surface integral over each element surface on S2, while Wi is replaced with N ie with an appropriate connectivity information for the assembly procedure. Then one obtains the following:

© 2005 by T. H. Kwon

87

~ ~ ~ ⎛ ∂φ ∂N i ∂φ ∂N i ∂φ ∂N i ⎞ ∑e ∫ ⎜⎜ k x ∂x ∂x + k y ∂y ∂y + k z ∂z ∂z ⎟⎟dΩ + ∑e ∫ fN i dΩ Ωe ⎝ Ωe ⎠ ~ + ∑ ∫ g + hφ N i dS = 0.

(

)

e ∂Ω e ∩ S 2

~ where superscript e is omitted over φ , N i for brevity.

After this, the procedure is exactly the same as the variational approach. The final result is ⎧ ∂N i ∂N j ∂N i ∂N j ∂N i ∂N j ⎫ e K Cij = ∫ ⎨k x + ky + kz ⎬dV : ∂x ∂x ∂x ∂x ∂x ∂x ⎭ Ωe ⎩

stiffness matrix (due to conduction)

∫ hN

e K Sij =

j

N i dS

:

stiffness matrix

∂Ω e S 2

(due to convection)

R efi =

∫ fN dV

: forcing matrix

i

Ωe

(due to distributed heat sink) RSie =

∫ gN dS

: forcing matrix

i

∂Ω e S 2

(due to distributed heat outflux)

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Notes on MWR:

1) Integral by parts reduces the order of derivative in the integrand. → lower the order of shape function required. → Easier representation of the appropriate solution 2) Integral by parts makes the stiffness matrix symmetric. (This statement is valid only for self-adjoint operator:

∫ vL(u)dΩ = ∫ uL(v)dΩ)

3) Integral by parts generates the natural boundary condition terms → It becomes easy to introduce boundary condition by a simple substitution. → More rigorous mathematical interpretation of this substitution is that the residual on the natural B.C. is incorporated into the weighted residual, i.e.

∫ [L(φ ) − f ]W dΩ − ∫ [B(φ ) − g]⋅ W dS = 0 ~

~

(

i

Ω

i

S2

Natural boundary condition terms: Energy equation → energy flux on control volume surface Static deformation → traction vector on control volume surface Mass conservation → mass flux on control volume surface (Such physical quantity appears due to integral by parts.) 4)

) R W B ∫ i dS is not included in the MWR expression. Why?

S1

→ φ i = φ i is imposed in the final global matrix equation by one of the methods described earlier for the essential B.C. → Wi = 0 or δy = 0 on S1 .

5) Equivalent nodal force (heat flux) is derived naturally as opposed to the direct method. (Variational approach has the same merit.) 6) When the variational principle exists, MWR can become exactly the same as the variational approach. → Both approaches boil down to “Weak form” solution.

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3. SHAPE FUNCTIONS AND DISCRETIZATION We will discuss the element types, shape functions and discretization in this chapter. It is important to be able to select an element type which is most suitable for the problem of interest, and to determine the shape functions for the chosen element type. Finally, automatic mesh generation techniques are, in practical sense, also important to finite element analysis applications.

3.1 Element Types i)

One-dimensional Elements

linear

ii)

quadratic

cubic

Two-dimensional Elements

linear triangular

Linear Rectangle

quadratic triangular

cubic triangular

linear quadrilateral

© 2005 by T. H. Kwon

quadratic quadrilateral

90

cubic quadrilateral

iii)

Three-dimensional Elements

Tetrahedron

Right Prism

Hexahedron

Element types depend on the followings: z Shape (1-D, 2-D, 3-D, triangular, quadrilateral, tetrahedron, etc.) z Number and type of nodes (3-node, 4-node, etc.) ∂φ ∂φ z Type of nodal variables (φ, , , etc.) ∂x ∂y z Type of interpolation functions

It may also be noted that degree of freedom for each node can vary depending on the problem, for instance,

φ

: 1 d.o.f. per each node

∂φ ∂φ , ∂x ∂y u, v

φ,

: 3 d.o.f. per each node : 2 d.o.f. per each node

u , v, p u , v, σ xx , σ xy , σ yy

: 3 d.o.f. per each node : 5 d.o.f. per each node

Sometimes, one may want to introduce somewhat special element type for which each node may have different d.o.f. (e.g. 2 d.o.f. for vertex node, 1 d.o.f. for mid-side node).

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3.2 Shape Functions, Interpolation 5 6 1

4 2

3

φ e = N iφ i i : element nodal numbering φ e : approximation of φ over an element

φ i : nodal value of φ N i : shape function (xi , yi ) : nodal coordinates

3.2.1 Polynomial Approximation 1-Dimensional case:

φ = α 0 + α1 x + α 2 x 2 + L + α n x n

: complete n-th order polynomial in 1-D

n

or

Pn ( x) = ∑ α i x i i =0

Tn(1) = n + 1

: number of terms for complete n-th order polynomial in 1-D

2-Dimensional case:

φ = α 1 + α 2 x + α 3 y + α 4 x 2 + α 5 xy + α 6 y 2 + L + α k x i y j + L

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Tn( 2 )

or

Pn ( x, y ) = ∑ α k x i y j ,

i+ j ≤n:

complete n-th order

k =1

polynomial in 2-D Tn( 2) =

(n + 1)(n + 2) 2

: number of terms for complete n-th order polynomial in 2-D

Pascal Triangle

order 1 x 2

x x3 x4

y y2

xy x2 y

xy2

x3y x2y2 xy3 x3y2 x2y3 x3y3

y3 y4

0 1 2 3 4 5 6 7

name

Tn( 2)

1 linear 3 quadratic 6 cubic 10 quartic 15 quintic 21 hexatic 28 septic 36

It may be noted that Pascal triangle is applied to triangular elements whereas the diamond is applied to quadrilateral elements. The number of parameters, α i , involved in the complete n-th order polynomial, represented by Pascal triangle, ( Tn( 2) ), is the same as the number of nodes of the corresponding triangular type of element. The number of parameters in the diamond is also the same as the number of nodes of the corresponding quadrilateral element. Compare the terms in the figure above with the nodes in the figures below.

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3-Dimensional case: Tn( 3 )

or

Pn ( x, y, z ) = ∑ α l x i y j z k ,

i + j + k ≤ n:

complete n-th order

l =1

polynomial in 2-D Tn( 3) =

(n + 1)(n + 2)(n + 3) 2

: number of terms for complete n-th order polynomial in 3-D

Generalized Coordinate

α i is called “generalized coordinate”, which has no physical interpretation. In the subsequent section, it will be explained that the shape function is better to be used than this generalized coordinate for the interpolation over an element. At this point, one should think about how one selects the element type for a certain problem, in particular, with respect to the order of polynomial in interpolation, in other words, with respect to the number of nodes within each element. The higher the order is, more accurate finite element analysis result is expected, but at the cost of computation time. Certainly, there is a trade-off between the accuracy and the computational cost. We will discuss this matter in more detail later. Geometric Isotropy

Field variable representation should remain unchanged under a coordinate transformation. This fact is called “Geometric Isotropy” This concept is applied to, for instance, temperature, displacement, strain field, etc. The guideline to ensure the geometry isotropy is as follows: 1. complete polynomial 2. incomplete polynomial with appropriate terms to preserve “symmetry” e.g. P ( x, y ) = α 1 + α 2 x + α 3 y + α 4 x 2 + α 5 xy + α 6 y 2 + α 7 x 3 + α 10 y 3 P ( x, y ) = α 1 + α 2 x + α 3 y + α 4 x 2 + α 5 xy + α 6 y 2 + α 8 x 2 y + α 9 xy 2

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3.2.2 Relationship between Generalized Coordinate and Shape Function Field variable was represented in terms of polynomial series with “Generalized Coordinate” having no physical meaning. From this representation, one can derive a representation of a field variable in terms of interpolation functions and physical degree of freedom. y

y

x

x

Tn( 2 )

φ = ∑α k x y , i + j ≤ n i

φ=

j

k =1

( n +1) 2

∑α k =1

Tn( 2 )

= ∑ N i ( x, y )φ i

=

x i y j , i ≤ n, j ≤ n

( n +1) 2

∑ N ( x, y)φ k =1

k =1

k

i

i

⎛ e.g. φ = α 1 + α 2 x + α 3 y + α 4 xy ⎞ ⎜⎜ ⎟⎟ ⎣P ⎦ = ⎣1 x y xy ⎦ ⎝ ⎠

φ = ⎣P ⎦{α } φ i at (xi , y i ) is the nodal value at node i :

⎛ ⎡1 x1 ⎜ ⎢ ⎜ e.g. [G ] = ⎢M M ⎜ ⎢⎣1 x 4 ⎝

{φ } = [G ]{α } {α }

can be determined by the inverse:

{α } = [G ]−1 {φ } ∴

φ = ⎣P ⎦[G ]−1 {φ }

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y1 M y4

x1 y1 ⎤ ⎞ ⎟ M ⎥⎥ ⎟ x 4 y 4 ⎥⎦ ⎟⎠

φ = ⎣N ⎦{φ }

i.e.

(= N iφ i )

⎣N ⎦ = ⎣P ⎦[G ]

−1

with

where ⎣N ⎦ is called “Shape Function” or “Interpolation function” There is a more systematic procedure to obtain ⎣N ⎦ without having to find [G ] . −1

We will cover this method extensively later. In any case, there are important characteristics of shape functions based on which one can generate formulae for various shape functions.

3.2.3 Characteristics of Shape Functions

⎧1 N i ( x, y ) = ⎨ ⎩0

1)

(Q

at node i, i.e. at ( xi , y i ) at any other node

φ i = N j ( xi , yi )φ j = φ i )

φ = N jφ j ,

5

5 6

6 1

1

4

1

1

3

2

N 1 ( x, y )

2)

∑N

i

=1

4 2

3

N 2 ( x, y )

(for C0 continuity, piecewise continuous element)

i

For a constant field variable φ, i.e. φ = constant, φ k = constant for any k . (∴

φ ( x, y ) = ∑ N iφ i = φ k ∑ N i = φ k = constant ) i

© 2005 by T. H. Kwon

i

96

3) Note that φ along one side of an element should depend only on the nodal values associated with the side due to the continuity and compatibility.

k j i

j

k

i

In general, nodes along a side of an element should be shared with a side of an adjacent element. However, for some reasons, one may introduce incompatible [Hughes P.159] element which, for instance, looks like the following figure.

This kind of mesh is somehow useful with a special care for transient region from a coarse one to a finer one. We will discuss a special coordinate system, namely natural coordinates, which is more suitable for representing shape functions than a global coordinate system, before explaining the method of finding shape functions in detail.

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3.2.4 Natural Coordinates Note that N i ( x, y ) in our current form is represented in terms of the nodal coordinates (xi , y i ) and a global coordinate ( x, y ) . One can have a better form in terms of so-called “Natural Coordinate”, in particular for triangular type of elements (or “Normalized Coordinate” for a quadrilateral type of elements ) 1) one-dimensional case l2

l1 xp

x1

x

x2

l

l1 l x1 + 2 x 2 l l = L1 x1 + L2 x 2

xp =

→

Natural coordinate L1 , L2 “Length coordinate”

L1 + L2 = 1

L1

L2 (looks like a linear interpolation)

1

1 x2

x1

2) two-dimensional case 3 A2 L2

A1

L1

Li =

L3 1

2 A3

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Ai : “Area coordinate” A

(L1 , L2 , L3 )

↔

( x,y )

⎧ x = Li xi ⎨ ⎩ y = Li y i

The relationship between the natural coordinate and global coordinate system for 2dimensional case is as follows: x = L1 x1 + L2 x 2 + L3 x3 y = L1 y1 + L2 y 2 + L3 y 3

⎡ x1 ⎢y ⎢ 1 ⎢⎣ 1

→

1 = L1 + L2 + L3

x3 ⎤ ⎧ L1 ⎫ ⎧ x ⎫ ⎪ ⎪ ⎪ ⎪ y 3 ⎥⎥ ⎨ L2 ⎬ = ⎨ y ⎬ 1 ⎥⎦ ⎪⎩ L3 ⎪⎭ ⎪⎩1 ⎪⎭

x2 y2 1

solve for Li : ⎡ b1 ⎧ L1 ⎫ ⎪ ⎪ 1 ⎢ ⎨ L2 ⎬ = ⎢b2 ⎪ L ⎪ 2∆ ⎢b ⎩ 3⎭ ⎣ 3

c1 c2 c3

a1 ⎤ ⎧ x ⎫ ⎪ ⎪ a 2 ⎥⎥ ⎨ y ⎬ a 3 ⎥⎦ ⎪⎩ 1 ⎪⎭

where ai, bi, ci’s have already been defined when φ = N iφ i for a triangular element in plane elasticity. 3) three-dimensional case 4 V1

V2

3

Li =

Vi : “Volume coordinate” V

x = Li xi y = Li y i

1

z = Li z i V3

© 2005 by T. H. Kwon

V4

2

4

1 = ∑ Li i =1

99

Normalized coordinate for a rectangular element:

η ξ

(1,1)

(0,1)

ξ=0

ξ=1 (0,0)

or

ξ

(1,0)

η ξ ξ=-1

(-1,1)

(1,1)

ξ

ξ=1 (-1,-1)

(1,-1)

Similar normalized coordinate system can be defined for 3-Dimensional case.

3.2.5 Classical Interpolation Functions A) Lagrange Polynomials (Lagrange Interpolation Functions) B) Hermitian Polynomials A) Lagrange Polynomials

x

0

1

2

k

n

x0

x1

x2

xk

xn

xi being nodal coordinates. n-th order polynomial needs (n+1) points.

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n

φ ( x) = ∑ Lk ( x)φ k k =0

One should find Lk (x) such that when i ≠ k , satisfying the characteristics of shape function. when i = k

⎧0 Lk ( x i ) = ⎨ ⎩1

It is easily proved that the formula below satisfies the requirement described by the equation above.

n

Lk ( x ) = ∏ m =0 m≠ k

x − xm ( x − x 0 )( x − x1 ) L ( x − x k −1 )( x − x k +1 ) L ( x − x n ) = x k − x m ( x k − x0 )( x k − x1 ) L ( x k − x k −1 )( x k − x k +1 ) L ( x k − x n )

(3.1)

It may be noted that Lk (x) is an n-th order polynomial. n

Question: Is

∑L k =0

k

( x) = 1 ?

∑N

i.e.

i

=1

i

Answer: yes. n

∑L k =0

k

( x) is an n-th order polynomial and is equal to unity at (n+1) points.

The only n-th order polynomial being equal to unity at (n+1) points is just one over the whole domain of x. The interpolation via Lagrange polynomials is demonstrated below:

φ0 x0

© 2005 by T. H. Kwon

φ1 x1

101

φ2 x2

φ3 x3

3

φ ( x) ≈ φ ( x) = ∑ Li ( x)φ i = ⎣L ⎦{φ } = ⎣N ⎦{φ } ~

i =0

∴ N i ( x) = Li ( x) Li(x) look like the following figures.

L0(x) 1

L1(x) 1

L2(x) 1

L3(x) 1

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It may be noted that, instead of global coordinate x, one can use normalized (local) coordinate ξ in the Lagrange interpolation function.

0

ξ

ξ0

1

2

k

n

ξ1

ξ2

ξk

ξn

The Lagrange interpolation for 1-dimensional case can be easily extended to 2dimensional case:

i=14 (I,J)=(3,2)

J=n N i = N IJ = LI (ξ ) L J (η ) 14 6 J=0 1

2 3

I=0

4 5 I=n

3-dimensional extension: N i = N IJK = LI (ξ ) LJ (η ) LK (ς ) B) Hermitian Interpolation Basic type of interpolation used to represent a function in terms of values of both the function itself and derivatives of the function. One method uses end values at only (2 nodes) (e.g. beam bending analysis). General form for two end nodes and derivatives up to order m-1:

φ = N 1φ1 + N 2φ1′ + N 3φ1′′ + L + N mφ1( m −1) + N m +1φ 2 + N m + 2φ 2′ + N m + 3φ 2′′ + L + N 2 mφ 2( m −1)

(3.2)

which is good for Cm-1-interelement continuity. Note that this interpolation function is

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(2m-1)-th order polynomials. Alternatively, one can introduce m points and make use of function and its first order derivatives at each node:

φ = N 1φ1 + N 2φ1′ + N 3φ 2 + N 4φ 2′ + L + N 2 m −1φ m + N 2 mφ m′

(3.3)

which is good for C1-interelement continuity. More generally, one can further introduce n points and make use of function and its derivatives up to (m-1)-th order at each node as depicted below:

φ n , φ n′ , L, φ n( m −1)

φ1 , φ1′,L , φ1( m −1) 1

2

n

In this case, the interpolation can be expressed by

φ = N 1φ1 + N 2φ1′ + L + N mφ1( m −1) + N m +1φ 2 + N m + 2φ 2′ + L + N 2 mφ 2( m −1) + LL + N ( n −1) m +1φ n + N ( n −1) m + 2φ n′ + L + N nmφ n( m −1)

(3.4)

In this case, there are total degree of freedom of nm and there are nm number of shape functions. Each shape function should satisfy nm requirements1 for each degree of freedom. In this regard, each shape function is described in terms of (nm-1)-th order polynomial. Because of the variety of strategies as described above, no fixed formulae for Hermitian interpolation are available. Instead, just use generalized parameters approach as described below. N i = a1 + a 2 x + a3 x 2 + L + a nm x nm −1 which can be determined by characteristics of shape functions. [Huebner & Thornton Sect. 5.6.2]

1

Note that the shape function corresponding to a degree of freedom should provide itself at the node when the corresponding derivative is taken for the expression of φ. Other order of derivatives at the node should be zero and any order of derivatives are zero at other nodes. As a result, total nm requirements

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Example : Beam bending analysis (m=2, C1 –continuity)

θ1

θ2 ξ

x

φ1 = y1 , dy φ1′ = , dx 1 = θ1

y1

y2

ξ=0

ξ=1

d 1 d = dx l e dξ

φ 2 = y2 dy φ 2′ = dx 2 = θ2

y (ξ ) = N 1φ1 + N 2φ1′ + N 3φ 2 + N 4φ 2′ N i (ξ ) = a1(i ) + a 2( i )ξ + a3(i )ξ 2 + a 4(i )ξ 3 Requirements:

for i=1 :

N 1 (0) = 1,

1 dN 1 (1) 1 dN 1 (0) = 0, N 1 (1) = 0, =0 l e dξ l e dξ

for i=2 :

N 2 (0) = 0,

1 dN 2 (1) 1 dN 2 (0) = 1, N 2 (1) = 0, =0 l e dξ l e dξ

for i=3 :

N 3 (0) = 0,

1 dN 3 (1) 1 dN 3 (0) = 0, N 3 (1) = 1, =0 l e dξ l e dξ

for i=4 :

N 4 (0) = 0,

1 dN 4 (1) 1 dN 4 (0) = 0, N 4 (1) = 0, =1 l e dξ l e dξ

∴ N1

1⎞ ⎛ N 1 = 2(ξ − 1) ⎜ ξ + ⎟ 2⎠ ⎝

N3

2

N 2 = l e (ξ − 1) ξ

N2

2

N 3 = ξ 2 (−2ξ + 3) N 4 = l eξ 2 (ξ − 1)

N4

* Check if this interpolation function can describe the rigid body rotation.

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3.2.6 Shape Functions for Typical Elements In this section, we will describe methods of finding shape functions for several important element types: a) Triangular elements b) Quadrilateral elements c) Serendipity elements a) Triangular elements

L1 =

α =0

α n

3

γ =4

α =1

γ =3 γ =2

α =2

γ =1

α =3

γ n

γ =0 2

α =4 1

β =0

L3 =

β =2

β =1

β =4

β =3

(α , β , γ ) L2 =

β n

α + β +γ = n N αβγ (L 1 , L 2 , L 3 ) = N α ( L 1 ) N β ( L 2 ) N γ ( L 3 ) where ⎧ α ⎛ nL 1 − i + 1 ⎞ ⎟, ⎪∏ ⎜⎜ ⎟ ⎪i = 1 ⎝ i ⎠ N α (L1 ) = ⎨ ⎪ ⎪⎩ 1 ,

© 2005 by T. H. Kwon

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α ≥1 α =0

(3.5)

It should be noted that N αβγ (L1 , L2 , L3 ) is n-th order polynomial while N α (L1 ) is

α-th order polynomial. Note also that the followings are the basis of the formula above: α ⎧ ⎪ 1 at L1 = n N α ( L1 ) = ⎨ α −1 1 2 ⎪0 at L1 = 0, , , L , n n n ⎩ One can prove this easily since the following holds:

at (α , β , γ ) →

(L1 , L2 , L3 ) = ⎛⎜ α , β , γ ⎞⎟

⎝ n n n⎠ ⎛ α ⎞ − i +1⎟ α ⎜n⋅ ⎛α ⎞ ⎟ = α ⋅ α −1 ⋅ α − 2L 1 = 1 N α (L1 ) = N α ⎜ ⎟ = ∏ ⎜ n 3 i α ⎟ 1 2 ⎝ n ⎠ i =1 ⎜ ⎜ ⎟ ⎝ ⎠ m α ⎛ m ⎞ m m −1 m −α +1 =0 L for L1 = < , N α ⎜ ⎟ = ⋅ 2 n n α ⎝n⎠ 1 since 0 ≤ m ≤ α − 1 Example of linear element:

3 (0,0,1)

n =1

⎧ 1 L1 − 1 + 1 = L1 , α = 1 ⎪∏ 1 ⎪ i =1 N α ( L1 ) = ⎨ ⎪1 α =0 ⎪ ⎩

1 (1,0,0)

2 (0,1,0)

N 1 = N 100 (L1 , L2 , L3 ) = N 1 ( L1 ) N 0 ( L2 ) N 0 ( L3 ) = L1 ⋅ 1 ⋅ 1 = L1 i.e.

N 1 = L1 , N 2 = L2 , N 3 = L3

N1

3

1 1

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2

Example of quadratic element:

α =0

n=2

(0,0,2) 5

α =1 (1,0,1) 6

4

α =2 1 (2,0,0)

2 (1,1,0)

(0,1,1)

3 (0,2,0)

N 1 = N 200 = N 2 (L1 )N 0 (L2 )N 0 (L3 ) 2 L1 2 L1 − 1 ⋅ = L1 (2 L1 − 1) 1 2 N 0 (L2 ) = N 0 (L3 ) = 1 N 2 (L1 ) =

∴ N 1 = L1 (2 L1 − 1)

N 2 = N 110 = N 1 (L1 )N 1 (L2 )N 0 (L3 ) N 1 (L1 ) = 2 L1 ,

N 1 (L2 ) = 2 L2 , N 0 (L3 ) = 1

∴ N 2 = 4 L1 L2 In summary: N 1 = L1 (2 L1 − 1) N 2 = 4 L1 L2 N 3 = L2 (2 L2 − 1) N 4 = 4 L2 L3 N 5 = L3 (2 L3 − 1) N 6 = 4 L3 L1

5

5

6

6 1 1

4

4 1 2

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108

1 2

3

b) Quadrilateral elements One makes use of Lagrange interpolation for quadrilateral elements:

ξ −ξm m =0 ξ k − ξ m

η4

n

Lk (ξ ) = ∏

η3

m≠ k

η2

N k (ξ ,η ) = N IJ = LI (ξ ) LJ (η )

η1 η0

6 1

9 2 3

k=9 (I,J)=(3,1)

4 5

ξ0 ξ1 ξ2 ξ3 ξ4 e.g.

N 9 = L3 (ξ) L1 (η) =

(ξ - ξ 0 )(ξ - ξ1 )(ξ - ξ 2 )(ξ - ξ 4 ) (η - η0 )(η - η 2 )(η - η3 )(η - η 4 ) ⋅ (ξ 3 - ξ 0 )(ξ 3 - ξ1 )(ξ 3 - ξ 2 )(ξ 3 - ξ 4 ) (η1 - η0 )(η1 - η 2 )(η1 - η3 )(η1 - η 4 )

Example of linear quadrilateral element

y

η

η1=1

3

ξ

4 2 1

3

⇒ η0=0

x

1

ξ0=0

ξ −1 η −1

= (1 − ξ )(1 − η ) ⋅ 0 −1 0 −1 ξ − 0 η −1 ⋅ = ξ (1 − η ) N 2 = L1 (ξ ) L0 (η ) = 1− 0 0 −1 N 3 = ξη N 1 = L0 (ξ ) L0 (η ) =

N 4 = (1 − ξ )η Example of quadratic quadrilateral element:

© 2005 by T. H. Kwon

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109

Find Ni’s yourself.

2

ξ1=1

c) Serendipity elements

[Zienkiewicz P.155]

It is usually most convenient to make the functions dependent on nodal values placed on the element boundary (i.e. without internal nodes) The systematic approach to determining Ni for serendipity elements is explained below. First, one of most popular serendipity elements, eight-node quadrilateral element, is taken as an illustration to help understand the general approach for general cases.

7

4

3

η 8

6

ξ

5

1

2

First, determine shape functions for mid-nodes. Consider the node number 5. N5 should behave quadratically in ξ, and linearly in η. Therefore, one can easily suggest the following:

Similarly,

(

)

N5 =

1 1 − ξ 2 (1 − η ) 2

: (quadratic in ξ ) × (linear in η )

N8 =

1 (1 − ξ ) 1 − η 2 2

(

: (quadratic in η ) × (linear in ξ )

)

Likewise, one can get N6 and N7. Those shape functions for mid-nodes look like the following figures.

© 2005 by T. H. Kwon

110

4

4

3

3

8

1

1

2

5

2 N8

N5

Secondly, consider the corner nodes, say node number 1. determine N1. Step 1)

4

There are three steps to

3

0.5 0.5

1.0

Nˆ 1 = (1 − ξ )(1 − η ) / 4

1

2

Step 2) 4

3

1 Nˆ 1 − N 5 2

1.0 1 Step 3)

2 4

3

1 1 N 1 = Nˆ 1 − N 5 − N 8 2 2

1.0 1

2

In a similar manner, one can determine shape functions for other corner nodes.

© 2005 by T. H. Kwon

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Question: Is the requirement,

∑N

i

, satisfied by the shape functions of serendipity

i

so obtained? For a general case of serendipity element shown below, there are (m+1) nodes along ξ-direction, while there are (n+1) nodes along η-direction.

The procedure to determine shape functions is as follow: First, determine shape functions for mid-nodes by taking (m-th order Lagrange interpolation)× (1-st order Lagrange interpolation), etc. Second, determine shape functions for corner nodes by having a bilinear function, Nˆ i , as a starting step and subsequently subtracting appropriate fractions of

shape functions for mid-nodes such that the shape function so obtained becomes zero at mid-nodes which belong to a side associated with the corner node of interest. A more general case is when there is a so-called bubble node at the center of the element as indicated below: m

c

η

b

ξ

The rule of the procedure is as follows:

© 2005 by T. H. Kwon

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Determine the shape function for the bubble node first, and then mid-nodes, and subsequently corner nodes as described by the following equations with b, m and c denoting bubble node, mid-node and corner node, respectively.

(

)(

)

Nb = 1 − ξ 2 1 −η 2 Nˆ m : (m - th order in ξ ) × (linear in η ) N = Nˆ − Nˆ (ξ ,η )N m

m

m

b

b

b

Nˆ c : (linear in ξ ) × (linear in η ) N c = Nˆ c − ∑ Nˆ c (ξ m ,η m )N m − Nˆ c (ξ b ,η b )N b m

Shape functions for several quadrilateral elements are summarized below:

η =1

1) linear quadrilateral element

ξ = −1

Define new variables: ξ o = ξξ i , η o = ηη i

η

ξ =1

ξ

1 N i = (1 + ξ o )(1 + η o ) 4

η = −1 2) quadratic quadrilateral element

corner nodes:

mid-nodes:

Ni =

η

1 (1 + ξ o )(1 + η o )(ξ o + η o − 1) 4

1 ⎧ 2 ⎪⎪ξ i = 0 : N i = 2 (1 − ξ )(1 + η o ) ⎨ ⎪η = 0 : N = 1 (1 − η 2 )(1 + ξ ) i o ⎪⎩ i 2

ξ

3) cubic quadrilateral element

corner nodes:

mid-nodes:

Ni =

[

(

1 (1 + ξ o )(1 + η o ) − 10 + 9 ξ 2 + η 2 32

)]

1 9 ⎧ 2 ⎪⎪ξ i = ±1, η i = ± 3 : N i = 32 (1 + ξ o )(1 − η )(1 + 9η o ) ⎨ ⎪η = ±1, ξ = ± 1 : N = 9 (1 + η )(1 − ξ 2 )(1 + 9ξ ) i i o o ⎪⎩ i 3 32

© 2005 by T. H. Kwon

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η ξ

As an exercise, you are suggested to derive shape functions, given below, for a quadratic quadrilateral element indicated below: N 1 = 1 ⋅ (1 − ξ )(1 − η )(1 − 2ξ − 2η )

η

N 2 = 4 ⋅ ξ (1 − ξ )(1 − η )

η =1 6

5

7

N 3 = −1 ⋅ ξ (1 − η )(1 − 2ξ + 2η ) N 4 = 4 ⋅ ξη (1 − η )

ξ =0

N 5 = −1 ⋅ ξη (3 − 2ξ − 2η )

8

ξ =1

4

N 6 = 4 ⋅ ξ (1 − ξ )η N 7 = −1 ⋅ (1 − ξ )η (1 + 2ξ − 2η ) N 8 = 4 ⋅ (1 − ξ )η (1 − η )

1

3

2

ξ

η=0

Question: Is there a systematic way of determining shape functions for a serendipity element of triangular type as demonstrated below? The answer is Yes. You are suggested to find shape functions for the element below.

3 6

1

4

2

5

(One might refer to [Bathe, P. 374] which presents a different method from what has been described here. It is recommended to use the method presented in this lecture note in terms of natural coordinate system for triangular type serendipity elements also.)

© 2005 by T. H. Kwon

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3.2.7 Condensation

[Bathe P.718]

internal nodes

serendipity element

Interelement connectivity depends only on boundary nodes, i.e. external nodes. Internal nodes may be useful for complete of better representation of field within an element. However, the internal nodal values have nothing to do with other elements and thus there will be no contribution to the internal nodes from other element. Therefore elimination of the internal nodes is possible after stiffness matrix coefficients of each element are calculated. This elimination is called “Condensation”, which will reduce the total degree of freedom in the global matrix. The condensation procedure might be presented as below:

[K ]{x} = {R}

for each element

Among {x}, {x 1 } is for external nodes of an element and {x 2 } is for internal nodes of the element One can break down the element matrix equation as follows: ⎡[K 11 ] ⎢[K ] ⎣ 21

then

[K12 ]⎤ ⎧{x1 }⎫ ⎧{R1 }⎫ = [K 22 ]⎥⎦ ⎨⎩{x2 }⎬⎭ ⎨⎩{R2 }⎬⎭

[K11 ]{x1 } + [K12 ]{x2 } = {R1 } [K 21 ]{x1 } + [K 22 ]{x2 } = {R2 }

From the latter, one can solve {x 2 } , which is substituted into the former resulting in

[[K

11

] − [K12 ][K 22 ]−1 [K 21 ]]{x1 } = {R1 } − [K12 ][K 22 ]−1 {R2 }

© 2005 by T. H. Kwon

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One can rewrite this equation as

[K~ ]{x } = {R~} 1

which is to be used for element matrix equation for the assembly procedure. In this way, the condensed form for degree of freedoms associated with external nodes for interelement connectivity is obtained. The global assembly is still based on the compatibility and balance. It may be noted that the condensed form is different from what one will obtain when a serendipity element is used. In this condensed form, more flexible interpolation is being used with equal external degree of freedoms, but with more computation like

[K 22 ]−1

involved than the serendipity.

After

{x1 }

is obtained,

{x 2 } can be determined from the second equation above and then the interpolation can be used to evaluate the field variable at any other points.

3.2.8 Substructuring Four triangular elements make one quadrilateral element and one internal node as illustrated below:

Internal node

This internal node can be eliminated by the condensation procedure explained above after the four elements have been added and just before the assembly. Substructuring is also useful when dealing with a very large system, breaking down the system into several substructures. (Refer to Huebner and Thornton Fig. 5.23.)

© 2005 by T. H. Kwon

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3.3 Criteria of Selection of Element Type As explained in the previous section, there are many different types of elements, for instance, with respect to the shape and the interpolation. It is very important to select one of the many possible choice of an element. In general, higher order interpolation will provide more accurate analysis result, but at the cost of computational time. In this respect, there is a kind of trade between the accuracy and cost. One has to choose an element which can provide accurate enough result within a reasonable computing time. In selecting an element, there are two important criteria: 1.

Continuity (Compatibility, Conformability )

2. Completeness In this section, those criteria are explained.

3.3.1 Continuity Criteria

[ Bathe P.234-235, Sect. 4.3.3]

φi

φ y

x

Piecewise continuous in φ derivatives of φ are not continuous

φ ,

C 0 continuity (most common)

∂φ ∂φ , are continuous ∂x ∂y

C 1 continuity (required in some problems)

φ,

∂φ ,L higher up to m-th derivative ∂x

C m continuity

(generally

not considered)

© 2005 by T. H. Kwon

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Question: What order of continuity should be chosen? We would like to answer this question as systematically as possible throughout this subsection.

In the following discussion, let’s assume that the highest order of derivative in the integrand for the stiffness matrix is m-th order. First of all, there are two types of continuity one should think about: a) Internal continuity The models must be single valued & differentiable to the highest order derivatives appearing in the variational functional or the integrand in the weak form statement.

⇒ This will satisfy internal continuity. ( C m - intraelement continuity ) b) Interelement continuity One needs some minimum order of interelement continuity to evaluate the integral functional uniquely over the entire domain. A minimum order of continuity is a degree less than the highest order derivatives in

the functional. ( C m −1 - interelement continuity)

Types of problem

Highest order in J

Interelement continuity

Truss, string, elasticity for plane strain or plane stress viscous flow, Energy eq.

1

C0

Beam, plates, shells

2

C1

Example: Heat conduction

K

d 2T +θ = 0 → dx

dT dN i dx : dx

∫ K dx

Square integrable ( H 1 : Sobolev Space of degree 1)

dT in the integrand → m = 1 → dx

C 0 : interelement continuity, C 1 : intraelement continuity.

© 2005 by T. H. Kwon

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T(x)

Essential

x

B.C

dT dx

x Natural B.C. not exactly

Piecewise integrable

satisfied

Example: Beam Analysis Cubic polynomial

Essential B.C. with

y

y , y'

x y′

x y ′′

x

Piecewise integrable

© 2005 by T. H. Kwon

Natural B.C with

119

y ′′ and y ′′′ not exactly satisfied

d2 d2y ( EI ) − w( x) = 0 → dx 2 dx 2

∫ EI

d 2 y d 2 Ni dx dx 2 dx 2

Square integrable ( H 2 : Sobolev space of degree 2)

d2y in the integrand → m = 2 → dx 2

C 1 interelement continuity C 2 intraelement continuity

Now, consider the more general cases. It might be reminded that if the highest derivative in the functional is of order m, then the order of derivatives in the essential B.C. is, at most, ( m – 1 ) the order of derivatives in the natural B.C. is m to ( 2m – 1 ) and the E-L equation has (2m) – th order of derivatives. L (φ ) = 0 : highest order derivative = 2m integral by parts m-times

∫L

' m

(φ)L''m ( w)dv = (B.C. terms)

with the boundary terms paired as follows: 0 , 1 , …. , m-1 : essential B.C.

2m-1, 2m-2, … , m

: natural B.C.

In this case, one needs C m−1 interelement continuity, C m intraelement continuity. In other words, C m−1 interelement continuity can satisfy the essential boundary conditions with no problem. And the natuaral boundary conditions is satisfied approximately with proper additional terms in the functional or in the weighted residual expression for natural B.C. It might be interesting to compare Ritz method with finite element method with regard to the continuity requirement:

© 2005 by T. H. Kwon

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Ritz method : Trial functions over the entire domain satisfying the essential B.C up to (m-1)-th derivatives ⇓

C

m

continuity in the trial function

FEM

C m−1 interelement requirement (satisfying essential B.C.) Cm

intraelement requirement → C m continuity for Ritz

∴ FEM is a modern application of Ritz analysis Treatment of discontinuity in FEM It is interesting to pay attention to how discontinuity is treated in the context of

finite element method. conditions:

Let us consider a beam analysis with the following loading

w(x) x

Mo G

y

In this case, M = − EI

k

d2y is discontinuous at x = L / 2. dx 2

The total potential energy

becomes: L− 2 0

πp = ∫

2 ⎤ ⎡1 ⎛ d 2 y ⎞2 ⎤ L ⎡1 ⎛d2y⎞ ⎢ EI ⎜⎜ 2 ⎟⎟ − wy ⎥ dx + ∫L + ⎢ EI ⎜⎜ 2 ⎟⎟ − wy ⎥ dx ⎥⎦ ⎢⎣ 2 ⎝ dx ⎠ ⎥⎦ 2 ⎢2 ⎝ dx ⎠ ⎣

1 ⎛ dy ⎞ + G⎜ ⎟ 2 ⎝ dx ⎠

2

+ L 2

1 2 ky 2

+ Mo L

dy dx

L

which will lead to Euler-Lagrange equation as well as natural boundary condition at x = L together with discontinuity equation at x = L / 2 i.e.

© 2005 by T. H. Kwon

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M + − M − = − EI

d2y dx 2

L+ 2

+ EI

d2y dx 2

L− 2

= −G

dy dx

L 2

that is, there is a discontinuity in the 2nd order derivative at the center. As far as Cmintraelement continuity is concerned, one puts a node at the center and allows d2y at the interface between elements. Then Cm-1-interelement 2 dx continuity holds with Cm- intraelement continuity. This versatility is the advantage of finite element method compared with Rayleigh-Ritz method. [Rayleigh-Ritz method cannot handle this problem with one domain, but should introduce two regions discontinuity in

−

+

L L and < x < L ) over which two different solutions should be obtained 2 2 with compatibility conditions at the center taken care of at the final solution.]

(0 < x <

Question: Try to find yourself how the discontinuity of other types, as depicted in the following figure, can be treated in the finite element method. You are supposed to be

able to describe how the condition at the center will affect the global matrix equation. x

Linear spring at the center

k y Concentrated downward force at the center

Fc x

Concentrated moment at the center x

Mc

From these examples of discontinuity, you will find that finite element method has no trouble in dealing with these various kinds of discontinuity, as long as Cm-1 – interelement continuity is ensured by introducing φ ( m −1) as a degree of freedom at end nodes via shape functions determined by Hermitian interpolation method.

© 2005 by T. H. Kwon

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3.3.2 completeness criteria

z z z z

[Bathe Sect. 4.3.2, P.229-233]

As for the completeness, there are several basic criteria of the following kinds: Rigid body mode Uniform behavior state Non-uniform behavior state Completeness with respect to geometric isotropy We will discuss those completeness criteria below.

a) completeness of rigid body modes

Zero strain energy should be evidenced when an element is subject to a rigid body motion. Hence, one possilble specialization of the models chosen must be rigid body motion. “eigen value test” can be used to ascertain whether this conditon is satisfied. Eigen value Test

[also, Gallagher Sect.2.9]

It is well known that n-th order system has n vectors with principal directions. However, if two of them are collinear, linear dependence exists, and one of the diagonal terms will become zero. (In other words, there is one less principal direction than was indicated by the size of the original n sets.) Let’s transform the stiffness matrix into diagonal form (in principal directions); the number of resulting zero diagonal terms will equal the number of such rigid body motions. And it may be noted that the number of rigid body modes is equal to the number of total degree of freedom minus the number of straining modes, i.e.: No. of rigid body modes

= No. of zero strain energy modes = No. of zero diagonal terms = No. of D.O.F – No. of straining modes

Number of zero diagonal terms can be found from the characteristic equation of the stiffness matrix [K ] as follows:

[K ]{x} = {F } [K ]{d i } = wi {d i } © 2005 by T. H. Kwon

, i = 1, …. , n

123

det( [K ] − w[I ] ) = 0:

: Characteristic equation of

[K ]

from which one can determine the eigenvalues wi and corresponding eigenvectors {d i } . One can define the orthonormal eigenvectors such that

{d i }T {d j } = δ ij then Defining

: orthonormality

{d i }T [K ]{d i } = wi

[Γd ] ≡ [{d1 }{d 2 }⋅ ⋅ ⋅ {d i }⋅ ⋅ ⋅ {d n }] ,

one has ⎡ w1 ⎢ ⎢ [K m ] = [Γd ]T [K ][Γd ] = ⎢ ⎢ ⎢ ⎢⎣ 0

w2 w3 ⋅⋅⋅

0⎤ ⎥ ⎥ ⎥ : ⎥ ⎥ wn ⎥⎦

Modal stiffness matrix

where the number of zero diagonal terms is equal to the number of rigid body modes, and the number of independent equation in [K ] is evidenced by (n – number of zero terms) in the modal stiffness matrix [K m ] which indicates the straining modes. x, u

Example 1. 1-D bar

F1 , u1

F2 , u 2

AE ⎡ 1 − 1⎤ ⎧ u1 ⎫ ⎧ F1 ⎫ ⎨ ⎬=⎨ ⎬ L ⎢⎣− 1 1 ⎥⎦ ⎩u 2 ⎭ ⎩ F2 ⎭

AE −w L AE − L

AE L =0 AE −w L −

∴ wi = 0 ,

[Γd ] =

→

2 AE L

1 ⎡1 1 ⎤ ⎢ ⎥ 2 ⎣1 − 1⎦

© 2005 by T. H. Kwon

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w2 −

2 AE w=0 L

The first and second column indicates the rigid body mode and compression mode, respectively. The modal stiffness matrix becomes:

[K m ] =

AE ⎡0 0⎤ L ⎢⎣0 2⎥⎦

where (1,1) component represents the rigid body motion. Note that {d i } [K ]{d i } = wi is related to strain energy T

For a rigid body mode, energy is zero. i.e. no deformation. Example 2: Beam bending problem

→ x-, y-translations, rotation

No. of rigid body modes = 3 (rank deficiency 3 :

∑F = ∑F x

y

= ∑M = 0)

Example 3 : Plane strain, stress :

[Bathe P.232, Cook P.190]

Isoparametric quadrilateral element

v u

( 4 nodes, bilinear model)

[K ]

w8 =1.43 Uniform expansion mode

: 8 × 8 matrix → 8 eigenvalues

w7 =0.769

w6 =0.769

Tensile mode

Shear mode

Constant strain © 2005 by T. H. Kwon

125

⎧ t =1 ⎪ with ⎨ E = 1.0 ⎪ν = 0.3 ⎩

w4 =0.495

w5 =0.495

Flexural modes (linear strain) 3 rigid body modes w1 =0

w3 =0

w2 =0

Rigid body modes There are five strain modes and three rigid body modes. 1 – D : 1 rigid body modes 2 – D : 3 rigid body modes 3 – D : 5 rigid body modes b) Uniform(constant) bebavior states (constant strain states)

The assumed model must be capable of representing constant strain, stress, (heat flux) states, etc. In the limit of mesh refinements, as the element approaches infinitesimal, each of the elementary behavior types must be still represented. If one or more are excluded, FEM analysis might converge to an incorrect solution. Uniform behavior states can be checked by the so-called “patch test” Patch Test

[Bathe P.263-268]

For a constant strain of elements, take an assemble of elements of arbitrary configurations with at least one internal modes as depicted in the figure below. Apply loads or forces (or displacements) to the boundary nodes which are consistent with a constant strain state. Internal nodes are neither constrained nor loaded.

© 2005 by T. H. Kwon

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That is, one can introduec either one of the two different ways: i) uniform traction is applied on the right side, as depicted in the figure above (or one can calculate the work-equivalent nodal forces corresponding to the uniform traction and apply them on the nodes), then check if the displacement at the center node is consistent with the uniform state. ii) uniform nodal displacement is applied at the right side nodes, and then check if force applied at the center node is zero. The computed strains, displacements, and stresses must be consistent with the constant strain. If not, the element is invalid. (In this case, the element is incompatible. Even if each element can represent a constant strain, incompatibility can cause the failure of patch test ) This patch test is sometimes called as “completeness condition on an element assemblage”. [Bathe P.263 - 268] [Cook et.al. P.126 –131] [Hughes P.237 - 238]

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One can sometimes introduce the so-called incompatible model as demonstrated below. z Incompatible model

[Bathe Sect.4.4.1] [Hughes P.243 –246]

Consider the 4-node rectangular element with shape functions as shown below.

y

1 (1 + x)(1 + y ) 1 4 1 N 2 = (1 − x)(1 + y ) 4 x 1 N 3 = (1 − x)(1 − y ) 4 4 3 1 N 4 = (1 + x)(1 − y ) 4 Now, consider a incompatible element for which displacement field is represented N1 =

2

by 4

u = ∑ N i u i + α 1φ1 + α 2φ 2 i =1 4

v = ∑ N i vi + α 3φ 3 + α 4φ 4 i =1

with φ1 = φ 3 = 1 − x 2 , φ 2 = φ 4 = 1 − y 2

: incompatible modes

Note that α’s are allowed to be different for neighboring elements so that φ’s are responsible for incompatible modes. For this incompatible element, let us check the patch test with one element : (using strain energy concept) The following displacement field represents a uniform strain state:

u c = ax + b ∴

Strain :

ε xxc = a , ε yyc = 0 , γ xyc = 0

Stress :

σ xxc =

E E c (ε xxc + νε yy )= a 2 1 −ν 1 −ν 2

σ yyc =

E E c (νε xxc + ε yy )= νa 2 1 −ν 1 −ν 2

© 2005 by T. H. Kwon

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σ xy =

E (1 − ν )γ xyc = 0 2 1 −ν

One can calculate the strain due to the incompatible model with given displacement field :

ε xx =

1 1 ∂u 1 1 = (1 + y )u1 + (− )(1 + y )u 2 + (− )(1 − y )u 3 + (1 − y )u 4 + α 1 (−2 x) 4 4 ∂x 4 4

1 1 ∂v 1 1 = (1 + x)v1 + (1 − x)v 2 + (− )(1 − x)v3 + (− )(1 + x)v 4 + α 4 (−2 y ) 4 4 ∂x 4 4 ∂u ∂v = + = Similarly obtained ∂y ∂x

ε yy = γ xy

One imposes u1 , u 2 , u 3 & u 4 as follows u1 = a + b , u 2 = − a + b , u 3 = u 2 ,

u 4 = u1

v1 = v2 = v3 = v 4 = 0 then 1 1 4 4 = a + α 1 (−2 x)

1 4

ε xx = [ (1 + y ) + (1 − y )](a + b) + [− −

1 1 1 y − + y ](−a + b) + α 1 (−2 x) 4 4 4

ε yy = −α 4 2 y γ xy = α 2 (−2 y ) + α 3 (−2 x)

Strain energy due to the strain above with imposed stress field : U = ∫ σ ijc ε ij dV = ∫∫

E Ev a[a + α 1 (−2 x)] + a(−α 4 2 y )]dxdy 2 1− v 1− v2

= ∫∫

E E Ev a 2 dxdy + α 1 ∫∫ a(−2 x)dxdy + α 4 a(−2 y )dxdy 2 1− v 1− v 1 − v 2 ∫∫

= ∫∫

E 2 a dxdy + 0 + 0 1− v

= real strain energy

© 2005 by T. H. Kwon

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∴ The incompatible modes do not contribute to the strain energy ∴ patch test passed ! However, the pass of the patch test is for the rectangular element or a parallelogram. For an arbitrary quadilateral element, the patch test fails. The representation of velocity field in each element :

u ( x, y ) ⎫ ⎡ N 1 ⎬=⎢ ⎩ v ( x, y ) ⎭ ⎣

{v} = ⎧⎨

N2

N3

N4

φ1 φ 2

0

0

0

0

0

N1

N2

N3

N4

φ3

⎧ u1 ⎫ ⎪M ⎪ ⎪ ⎪ ⎪u 4 ⎪ ⎪ ⎪ ⎪α 1 ⎪ 0 ⎤ ⎪⎪α 2 ⎪⎪ ⎨ ⎬ φ 4 ⎥⎦ ⎪ v1 ⎪ ⎪M ⎪ ⎪ ⎪ ⎪ v4 ⎪ ⎪α ⎪ ⎪ 3⎪ ⎪⎩α 4 ⎪⎭

12 D.O.F One can reduce the 12 d.o.f to 8 d.o.f by eliminating α1 , α 2 , α 3 & α 4 by the static condensation, as was done for the internal node elimination, during the assembly procedure. These incompatible elements may result in decrease of analysis cost considerably with accurate result as long as one can use rectangular for the finite element mesh. It is not always easy to generate a compatible element. In some practive, incompatible element results in a satisfactory analysis. As opposed to Ritz method, interelement discontinuity does not contribute to the strain energy (which is singular in the discontinuous boundary) in finite element analysis. These contributions are not calculated at all and these are ignored in the assembly procedure. However, the monotonic convergence is not assured. Rather, one wants to assure nonmonotonic convergence.

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c) non-uniform behavior states of lower order than the highest represented.

x, u

: u = a1 + a 2 x

Linear model

: constant behavior

Non-linear model : u = a1 + a 2 x + a3 x 2 + a 4 x 3

This omission of intermediate terms disrupts the rate of convergence →

% error

to be discussed later.

Linear model Cubic model

No. of D.O.F

d) completeness w.r.t. geometric isotropy

Element should develop same strain energy for some modal displacements regardless of orientation. (One might have a complete polynomial terms in x,y,z, for isotropy. It has been discussed earlier.) Test : eigenvalue test should give the same eigenvalues for all orientation

4

3

3

3

2 4

1

2

4

1

2 1

It may be noted that cetain flexibility of element formulation sometimes may violate this

© 2005 by T. H. Kwon

131

criteria. However, this violation is seldom encountered if local coordinate system is used for element formulation. (e.g. beam, shell etc. have special treatment in directions.)

3.4 convergence rate (with displacement based FEM) [Bathe P.236-244, 376-385] [Zienkiewicz P.203, P.281]

It is important to consider a convergence rate of solution of displacement based finite element method for a problem of which the total potential energy fuctional ( π p ) has the highest order derivative of m. In this case, as already discussed, shape functions satisfying C m −1 interelement continuity are to be used: ∆ = [N ]{∆}

with C m −1 interelement continuity,

where usually ∆ : displacement vector fields (e.g. (u, v, w( x, y, z )) ) [N ] : shape fuctions {∆} : nodal values Now, to consider the convergence of finite element solution, let us define the followings: ∆ : “exact” displacement field ~ ∆ : finite element interpolant with exact nodal displacement {∆} ~ i.e. ∆ = [N ]{∆} ∆ : finite element solution with finite element nodal displacement {∆} i.e. ∆ = [N ]{∆}

~ ∆

{∆}

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∆ : Finite element solution

132

Exact ∆

For linear elastic structure, π p is a quadratic functional of ∆ : ∴ π p (∆ + δ ∆ )= π p (∆) + δπ p + δ 2π p For a kinematically admissible variation

δ ∆ , δπ p = 0

and δ 2π p = δ 2 (U + V ) = δ 2U = U (δ ∆) > 0 ∴ π p (∆ + δ ∆) − π p = U (δ ∆)

V is linear in ∆ ∴ δ 2V = 0

π (∆) =

1 K∆2 − F∆ , 2 U

V

π ( ∆ + δ∆ ) − π ( ∆ ) =

1 1 K (∆ + δ∆) 2 − F (∆ + δ∆) − ( K∆2 − F∆) 2 2 1 K (δ∆) 2 = δπ + δ 2 π 2

= ( K∆ − F )δ∆ +

δπ = 0

→

δ 2π = U (δ∆)

K∆ = F

For a FEM solution ∆ ;

π p ( ∆ ) = π p (∆) + U ( ∆ − ∆) ~ For ∆

( i.e. δ ∆ = ∆ − ∆ )

;

π p (∆% ) = π p (∆) + U (∆% − ∆ ) Since ∆ is chosen to minimize π p for a given mesh system, ~

π p (∆) ≥ π p ( ∆ ) ~ ∴ U (∆ − ∆) ≥ U ( ∆ − ∆) ≥ 0 [Bathe P.228]

This inequality indicates that the error in the strain energy by the finite element solution

© 2005 by T. H. Kwon

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is smaller than that by the finite element interpolant ! ~ Therefore, as mesh is refined, U (∆ − ∆) → 0 , thus U ( ∆ − ∆) → 0 as mesh is refined. This proves the convergence of FEM solution. In fact, for any other ∆ˆ defined in the same mesh system, one can write: π p (∆ˆ ) = π p (∆) + U (∆ˆ − ∆) . ~ And for the same reason for ∆ , one can also have: U (∆ˆ − ∆) ≥ U ( ∆ − ∆) which indicates that finite element analysis result has the best approximate property w.r.t. strain energy. Find u ∈ V such that

: provides an exact solution u

a (u , v) = ( f , v) ∀ v ∈ V Find u h ∈ V h such that a (u , v ) = ( f , v ) ∀ v ∈ V h

h

h

h

h

Then there hold the following theorems:

: provides a finite element solution uh [Hughes P.186] or Propety 1-3 in [Bathe P.241]

Theorem a. a ( w h , e) = 0 ∀ w h ∈V h b. a(U h − u,U h − u ) ≥ a(e, e) ⇒ U (∆ˆ − ∆) ≥ U ( ∆ − ∆) with

e ≡ uh − u

where u h ,U h and u represent ∆, ∆ˆ and ∆ , repectively. That is, the strain energy due to the error of finite element solution is minimum of all the errors of other approximate solutions. corollary. (Pythagorean Theorem) a(u, u ) = a(u h , u h ) + a(e, e) ⇒ a(u h , u h ) ≤ a(u, u ) i.e. U ( ∆ ) ≤ U (∆) The finite element approximation underestimates the strain energy.

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Assume, for simplicity of discussion, that ∆ is a scalar with a spatial coordinate x ∆( x) = a0 + a1 x + a 2 x 2 + L + a n x n + Rn +1 in which Rn +1 < Mh n +1

where

h M

: mesh size : a bound depending on the maximum value of the (n+1)-th derivative of ∆ within elements.

~ If ∆ = ⎣N ⎦{∆} consists of a complete polynomial of degree n,

then

~ ∆ − ∆ ≤ M 0 h n +1 → 0 as h → 0

if

n +1 > 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 m-th 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 )

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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 one-dimensional integration is described below:

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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 (2n-1)-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 one-dimensional integration can be easily extended to two-dimensional and three-dimensional cases as described below:

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For two-dimensional 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]

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4.2 Minimum Order of Integration

[Bathe: Sect. 5.5.5] [Cook: P.190-193] [Zienkiewicz: P.202-205] [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 trade-off 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, (2n-1), 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

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)

Therefore, with a polynomial of order n for the approximation of field variable, the integral higher than the order of 2(n-m) 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 zero-energy 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 two-dimensional elastic deformation problem as indicated below.

Note that there are three essential boundary conditions.)

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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

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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 four-node 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 so-called “Hourglass” mode (or termed Mechanism, Kinematic mode, zero-energy 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.

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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 zero-strain 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 zero-strain 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 eight-node quadrilateral element, a serendipity element, does not have the first two zero-strain energy modes since it does not include the ξ 2η 2 term. If one employs 3×3 Gauss quadrature points, there is no spurious zero-strain 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

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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

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(independent d.o.f. : needed) (independent conditions : provided)

4.2.3 Reduced Numerical Integration

[Bathe P.476-478] [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 over-tightened [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.476-477] in terms of mixed formulation. Related topic is discussed via penalty function formulation in [Zienkiewicz, P.284-296].

4.3 Interpretation of Stress in Finite Element Analysis

[Bathe P.254-259]

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 stress-strain 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:

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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

2-D array

1-D array

Both 2-D array and 1-D 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.

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5.1.2 Skyline

[Bathe P.985-986]

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) (half-bandwidth – 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. 439-448]

[S ]{x} = {F ′}

→

upper triangular matrix

L−n1−1 L L−21 L1−1 K = S © 2005 by T. H. Kwon

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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 off-diagonal 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 ⎥⎦

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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.985-986] for a computer implementation for the skyline reduction method.

© 2005 by T. H. Kwon

163

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 Gauss-Seidel Iteration Method K ij x j = Fi One may suggest the simplest iterative scheme, namely Jacobi iterative method:

© 2005 by T. H. Kwon

164

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 “Gauss-Seidel 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 Gauss-Seidel 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

165

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

166

{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.749-752]

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, McGraw-Hill Book Co., 1984. Also [Bathe P.749-752]

© 2005 by T. H. Kwon

167

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

168

(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.272-283] [Bathe P.110-113] [Hughes P.194-197]

© 2005 by T. H. Kwon

170

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

171

4) Injection Molding

5) Extrusion Process

6) Hot embossing

Stamp Polymer substrate Solid

Pre-heating

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

172

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. 374-380]

( 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 (Non-Newtonian 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.

© 2005 by T. H. Kwon

174

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 Euler-Lagrange 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 Euler-Lagrange 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

175

⎛⎜ 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 2-D flow ⎪ ′⎪ ⎪⎩σ xy ⎪⎭

{}

⎧D ′ ⎫ ⎪⎪ xx ′ ⎪⎪ {D} = ⎨ D yy ⎬ for 2-D 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 2-D 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 so-called Babuška-Brezzi 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.207-209, P.210-217, 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 3-dimensional 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 k-th node) and taking, in equation (6.14), N pk as W pk with p = N pk p k ( p k being pressure at k-th 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.410-418]

Crank-Nicholson 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. Petrov-Galerkin method might be recommended for this purpose. [Bathe P. 687-688, 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, Streamline-Upwind PetrovGalerkin Method To attain a stable numerical solution for advection-diffusion 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: Streamline-Upwind Petrov-Galerkin (SUPG) Method Galerkin Least Square (GLS) Method, Streamline-Upwind 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 least-squares 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/Least-Squares method for advection-diffusive equations,”

Computer method in applied mechanics and engineering, 73, 173-189(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, 235-255 (2000). In this section, we may just briefly describe the essence of the method via a scalardiffusion problem and a steady incompressible Navier-Stokes equation. 1) Scalar-diffusion 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 Navier-stokes 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 Navier-Stokes 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

193

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 Displacement-based 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 Time-Dependent 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 time-dependent 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

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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. 61-62]

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 i-th 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.243-248]

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

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[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 <θ <1

t = tθ

[C ]{φ&}θ + [K ]{φ }θ = {R(tθ )} of which the terms can be approximated as:

{φ&} = {φ } {φ }θ

− {φ }n ∆t = (1 − θ ){φ }n + θ {φ }n +1

{R}θ

= (1 − θ ){R}n + θ {R}n +1

θ

and

n +1

© 2005 by T. H. Kwon

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[Huebner P.293]

Then it becomes 1 1 ⎡ ⎤ ⎡ ⎤ ⎢⎣θ [K ] + ∆t [C ]⎥⎦{φ }n +1 = ⎢⎣− (1 − θ )[K ] + ∆t [C ]⎥⎦{φ }n + (1 − θ ){R}n + θ {R}n +1

(7.19)

known values

Unknown for marching some special cases:

θ = 0 : forward difference (Euler Integral ) 1 : Crank – Nicolson method θ= 2

θ=

2 3

θ =1

: Galerkin method : backward difference [Huebner p 293 – 299] [Bathe Chapter 9, in particular Sect. 9.2]

© 2005 by T. H. Kwon

204

8. BOUNDARY ELEMENT METHOD 8.1 Boundary Integral Equation Formulation for Poisson Equation Consider the following problem. PDE

r ∇ 2φ = f ( x )

in Ω

(8.1)

B.C. i)

φ =φ

on S1

∂φ ∂φ = ∂n ∂n

ii)

iii) − k

:

on S 2

Dirichlet condition :

∂φ = h(φ − φ ∞ ) on S 3 ∂n

Neumann condition

: Mixed condition

There are basically two approaches toward BIE formulation : : using Green’s 2nd identity : related to strong form, weak form, inverse form via integral by parts.

Direct formulation Weighted Residual Method

We’ll be following the direct formulation in this lecture. i) Green’s 2nd Identity, Green’s Solution

For any regular function (twice differentiable) ψ 1 and ψ 2

∫ (ψ ∇ ψ 2

1

Ω

2

∂ψ 1 ⎞ ⎛ ∂ψ 2 − ψ 2 ∇ 2ψ 1 dΩ = ∫ ⎜ψ 1 −ψ 2 ⎟dS ∂n ∂n ⎠ ∂Ω⎝

)

(8.2)

holds. The solution satisfying the following equation r r ∇ 2 G = δ ( x − x ′)

(8.3)

is called Green’s solution (or fundamental solution) in an infinite space singular at

© 2005 by T. H. Kwon

205

r point p (x ) , and can be found as below:

1 ⎧ ⎪ − 4πr G=⎨ 1 ⎪ ln r ⎩ 2π

where

For 3-D

r r

r x′

For 2-D r x

r r r = x′ − x .

(8.4)

One can prove that the function for G expressed above, depending on the dimension of the problem, satisfies the governing equation as follows: : in cylindrical coordinate

⎧ 1 ∂ ⎛ ∂G ⎞ 1 ∂ 2 G ∂ 2 G + ⎟+ ⎜r ⎪⎪ r ∂r ⎝ ∂r ⎠ r 2 ∂θ 2 ∂z 2 2 ∇ G=⎨ ∂ 2G 1 1 ∂ ⎛ ∂G ⎞ ⎪ 1 ∂ ⎛⎜ r 2 ∂G ⎞⎟ + ⎟+ ⎜ sin θ ⎪⎩ r 2 ∂r ⎝ ∂r ⎠ r 2 sin θ ∂θ ⎝ ∂θ ⎠ r 2 sin 2 θ ∂φ 2

: in spherical coordinate

Note that

⎛1⎞ ∇ 2 ⎜ ⎟ = 0 except r = 0 ⎝r⎠ ∇ 2 (ln r ) = 0 except r = 0

i.e. ∇ 2 G = 0

except

r=0

nˆ

Consider the behavior of ∇ 2 G near r = 0 :

ε

r

∂Ω ε Ωε

∫ ∇ GdΩ = ∫ ∇ GdΩ = ∫ ∇G ⋅ nˆdS = ∫ 2

Ω

2

Ωε

∂Ωε

∂Ωε

∂G ∂G dS = ∫ dS ∂r ∂n ∂Ωε

⎧ 1 1 1 4πε 2 ⋅ = = =1 dS dS ⎪∫ 2 4πε 2 ∂Ω∫ε 4πε 2 ⎪∂Ωε 4π r =⎨ 1 1 1 2πε ⎪ =1 dS = dS = ∫ ∫ 2πε ∂Ωε 2πε ⎪⎩ ∂Ωε 2π r

QED.

© 2005 by T. H. Kwon

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: 3-D : 2-D

ii) BIE formulation r Let ψ 1 = φ and ψ 2 be the fundamental solution G singular at point p ( x ) , i.e.

r r ∇ 2 G = δ ( x − x ′)

Then the Green’s 2nd identity can be rewritten as ∂φ ⎞

⎛ ∂G

∫ (φ∇ G − G∇ φ )dΩ = ∫ ⎜⎝φ ∂n − G ∂n ⎟⎠dS 2

Ω

∂Ω

r

r f (x ′)

r

δ ( x − x ′)

∂φ

∂G

∫ φ∇ GdΩ = ∫ fGdΩ + ∫ (φ ∂n − G ∂n )dS 2

i.e.

Ω

Ω

r

φ (x) ⋅

: an integral e.q.

∂Ω

r subtended angle ≡ aφ ( x ) total angle 1442443

a ∴

r ∂φ ⎞ ⎛ ∂G aφ ( x ) = ∫ fGdΩ + ∫ ⎜ φ − G ⎟dS ∂n ∂n ⎠ Ω ∂Ω⎝

(8.5).

Which is an integral equation with the value of a depending on the position as described below: 1

r x′

a=

r p( x )

1 at smooth surface 2

α α or at sharp corner 2π 4π

r x

2D

The details of formulae above will be shown below.

© 2005 by T. H. Kwon

at internal point

207

3D

Consider

∫ φ∇ GdΩ 2

:

Ω

a) for an internal point r

r

r

∫ φ (x ′)∇ G(x − x ′)dx′dy ′dz ′ 2

r x

Ω

r r r r r = ∫ φ (x )∇ 2 G ( x − x ′)dΩ = φ ( x ) ∫ ∇ 2 GdΩ = φ ( x ) Ωε

∂Ω ε

Ωε

Ωε

∴a = 1 b) for a point on a surface

Sε1

Excluding a small volume Ω ε :

r x

Ωε

ε Ω

∂ (Ω − Ω ε ) = S r + S ε 2

Sε 2

∂φ ⎞ ⎛ ∂G −G ⎜φ ⎟dS ∂n ∂n ⎠ S r + Sε 2 ⎝

∫ φ∇ GdΩ = ∫ fGdΩ + ∫ 2

Ω − Ωε

Ω − Ωε

Sr

consider

⎛ ∂G

∂φ ⎞

∫ ⎜⎝φ ∂n − G ∂n ⎟⎠dS

ε

Sε 2

r ∂G ∂φ = φ (x ) ∫ dS − ∫ G dS ∂n ∂n Sε 2 Sε 2 r ⎛ ∂G ⎞ ∂φ = φ (x ) ∫ ⎜ − dS ⎟dS − ∫ G ∂r ⎠ ∂n Sε 2 ⎝ Sε 2 where

⎧ Sε 2 αε 2 α − = − =− = −a 2 ⎪⎪ 4πε 2 ⎛ ∂G ⎞ 4 π 4 πε dS − = ⎜ ⎟ ∫ ∂r ⎠ ⎨ Sε 2 αε α Sε 2 ⎝ ⎪ − =− =− = −a 2πε 2π ⎩⎪ 2πε

© 2005 by T. H. Kwon

208

α nˆ

Sε 2

α : Subtended angle

and

⎧ ∂φ 1 ⋅ ⋅ αε 2 → 0 as ε → 0 (3D) ⎪ ∂φ ∂φ ⎪ ∂n 4πε ∫S G ∂n dS ≤ ∂n max ⋅ S∫ GdS = ⎨ ∂φ max ln ε ⎪ ε2 ε2 ⋅ ⋅ αε → 0 as ε → 0 (2D) ⎪⎩ ∂n max 2π One finally obtains

⎛ ∂G

r ∴ aφ ( x ) =

Ω − Ωε

where a ≡

∂φ ⎞

∫ fGdΩ + ∫ ⎜⎝φ ∂n − G ∂n ⎟⎠dS

(8.5)

Sr

subtended angle (α ) Total angle (4π or 2π ) (a =

1 2

on smooth surface)

Equation (8.5) is an integral equation to be used in Boundary Integral Equation Method.

∫ ( )dΩ , ∫ ( )dS

z Note that we exclude the singular region on

Ω − Ωε

!!

Sr

ii) Boundary Element Method (with discretization)

i

φi ,

∂φ ∂n i

r r

10

Ωj

9

r r NEL 1

i=1

2 i=2 3

8

4

5

i=NEL

© 2005 by T. H. Kwon

209

6

7

Let us take an example with constant element in 2-D case: Discretization along a boundary : i

i

Discretization over a domain : On each element, φ i and

= 1, NEL = 1, NDOMAIN

∂φ are approximated as constant. ∂n i

Equation (5) provides NDOMAIN NEL ⎡ 1 1 1 ∂ ln r ∂φ φ i = ∑ ∫ f ⋅ ln rdΩ + ∑ ⎢φ j ∫ dS − 2 2π 2π ∂n ∂n j =1 j =1 ⎢ Ωj ⎣ Sj

j

⎤ 1 ⎥ ln rdS ∫ 2π ⎥⎦ Sj

i.e.

πφ i =

Let

NEL ⎡⎛ ⎞ ⎛ ⎞ ⎜ ∂ ln r dS ⎟ − ⎜ ln rdS ⎟ ∂φ ⎢ ln f rd ⋅ Ω + ∑ ∑ ∫ ⎟ ∂n ⎜∫ ⎟ ⎜ S∫ j =1 Ω j j =1 ⎢ S j ∂n ⎠ ⎝ j ⎠ ⎣⎝

NDOMAIL

⎤ ⎥ j⎥ ⎦

∂ ln r Hˆ ij = − ∫ dS ∂n Sj Gij = − ∫ ln rdS Sj

F1i =

∴ H ij φ j = Gij

NDOMAIN

∑ ∫ f ⋅ ln rdΩ j =1 Ω j

∂φ + F1i ∂n j

or

[H ]{φ } = [G ]⎧⎨ ∂φ ⎫⎬ + {F1 } ⎩ ∂n ⎭

with H ij = Hˆ ij + πδ ij

© 2005 by T. H. Kwon

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(8.6)

8.2 Boundary Integral Equation Formulation for Elastic Deformation [T.A. Cruse, Boundary Element Analysis in Computational Fracture Mechanics, Kluwer Academic Publication]

8.2.1 Review of elasticity

Equilibrium equation:

σ ij , j + X i = 0

(8.7)

Hooke’s law:

σ ij =

2 µν δ ij ε mm + 2 µε ij 1 − 2ν

µ : Shear Modulus ν : Poisson’s ratio

(8.8)

1 2

ε ij = (u i , j + u j ,i )

(8.9)

Navier equation 1 1 u i ,ij + ∇ 2 u j + X j = 0 1 − 2ν µ Galerkin vector

(8.10)

Gi :

u i = ∇ 2 Gi −

1

2(1 − ν ) Navier equation in terms of Gi :

∇ 4 Gi +

xi

µ

(8.11)

G j , ij

=0

(8.12)* y

8.2.2 Fundamental Solution

q(x,y) r(p,q)

A point body force distribution can be stated as X i = δ ( p − q)ei

(8.13)

p(x,y) x

Fundamental solution for the point body force satisfying ∇ 4 Gi∗ + δ ( p − q)ei / µ = 0

(8.14)

is as follows:

© 2005 by T. H. Kwon

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for 3-D case :

Gi∗ =

for 2-D case :

Gi∗ =

1

r ( p, q)ei

(8.15)

1 r 2 (1 + log )ei 8πµ r

(8.16)

8πµ 1

( r 2 term represents rigid body displacement) Note that r,i =

∂r = ( xi , q − xi , p ) / r ∂xi ,q

= −

(8.17)

∂r ∂xi , p

u i∗ can be obtained from (8.11) and either (8.15) or (8.16).

[

]

u i∗ ( p, q ) = (3 − 4ν )δ ij + r,i r, j e j /[16πµ (1 − ν )r ]

: 3-D

⎡ ⎤ ⎛1⎞ u i∗ ( p, q) = ⎢(3 − 4ν )δ ij log⎜ ⎟ + r,i r, j ⎥ e j /[8πµ (1 − ν )] ⎝r⎠ ⎣ ⎦

: 2-D

(8.18)*

(8.19)

or u i∗ ( p, q ) = U ji ( p, q )e j

: definition of U ij

(8.20)

Then traction can be found as follows: t i = σ ij n j

(8.21)

t i∗ = T ji ( p, q )e j

[

]

[

[

]

]

⎧ dr (1 − 2ν )δ ij + 3r,i r, j − (1 − 2ν ) n j r,i − ni r, j ⎫⎬e j / 8π (1 − ν )r 2 : 3-D (8.22) = −⎨ ⎭ ⎩ dn ⎧ dr (1 − 2ν )δ ij + 2r,i r, j − (1 − 2ν ) n j r,i − ni r, j ⎫⎬e j /[4π (1 − vr )] :2-D (8.223) = −⎨ ⎭ ⎩ dn

[

*

]

[

Equations (8.12) and (8.18) are derived below.

© 2005 by T. H. Kwon

212

]

Proof for Eq. (18) Gi∗ =

1 8πµ

r ( p, q )ei

u i∗ = ∇ 2 Gi∗ −

1 G ∗j ,ij 2(1 − ν )

∂ ( x1 q − x1 p ) 2 + ( x 2 q − x 2 p ) 2 + ( x3 q − x3 p ) 2 ∂r r,i = = ∂xi q ∂xi q

=

xi q − xi p r

⎛ ∂r ∂r Note also that ⎜ =− ⎜ ∂xi p ∂xi q ⎝

r,ij =

=

δ ij r

−

( xi q − xi p )

r2

⎞ ⎟ ⎟ ⎠

r, j

δ ij

1 − r,i r, j r r

∇ 2 r = r,ll =

(

)(

3 1 3 1 xi q − xi p xi q − xi p − r,i r,i = − ⋅ ⋅ r r r r r r

)

3 1 r2 2 = − ⋅ 2 = r r r r ⎡ 1 (re j )ij ⎤⎥ / 8πµ ∴ u i∗ = ⎢∇ 2 (rei ) − 2(1 − ν ) ⎣ ⎦

=

⎫ ⎤ 1 ⎡2 1 ⎧δ ij 1 ⎢ ei − ⎨ − r,i r, j ⎬e j ⎥ 8πµ ⎣⎢ r 2(1 − ν ) ⎩ r r ⎭ ⎦⎥

=

1 4(1 − ν )ei − (δ ij − r,i r, j )e j 8πµr 2(1 − ν )

=

1 (3 − 4ν )δ ij + r,i r, j e j 16πµ (1 − ν )

[

© 2005 by T. H. Kwon

[

]

213

]

( ei = δ ij e j ) / QED

Proof for Eq. (8.12) 1 1 u i ,ij + ∇ 2 u j + X j = 0 µ 1 − 2ν u i = ∇ 2 Gi −

(8.10)

1 G j ,ij 2(1 − ν )

(8.11)

Substituting (8.11) to (8.10) yields 1 1 − 2ν

⎛ 2 ⎞ ⎛ ⎞ 1 1 1 ⎜⎜ ∇ Gi − G j ,ij ⎟⎟ + ∇ 2 ⎜⎜ ∇ 2 Gl − G j ,lj ⎟⎟ + X l = 0 2(1 − ν ) 2(1 − ν ) ⎝ ⎠ ,il ⎝ ⎠ µ

1 1 1 1 (Gi ,kkil − G j ,ij ) ,il + Gl ,iijj − G j ,ljkk + X l = 0 µ 1 − 2ν 2(1 − ν ) 2(1 − ν ) 0 ⎡ 1 1 1 ⎤ 1 ⎢1 − 2ν − (1 − 2ν )2(1 − ν ) − 2(1 − ν ) ⎥Gi ,ilkk + Gl ,iijj + µ X l = 0 ⎣ ⎦ ∴ ∇ 2 (∇ 2 Gl ) +

8.2.3

1

µ

Xl = 0

→ Eq. (8.10)

/ QED

Somigliana Identities

Betti’s reciprocal work theorem: ( basis of BIE formulation )

∫σ

V

1 ij

ε ij2 dV = ∫ σ ij2 ε ij1 dV

(8.24)

V

Using the constitutive equation

σ ij = Cijkl ε kl the reciprocal work theorem can be easily proved.

© 2005 by T. H. Kwon

214

Eq. (8.24) can be rewritten as

∫σ

1 ij

V

u i2, j dV = ∫ σ ij2 u i1, j dV

∫σ

⇒

V

ij

u i∗, j dV = ∫ σ ij∗ u i , j dV

(8.25)

Let state–1 be a well-posed boundary value problem and let state–2 be the fundamental solutions.

∫σ

ij

V

(

u i∗, j dV = ∫ σ ij u i∗

)

,j

dV − ∫ σ ij , j u i∗ dV = ∫ t i u i∗ dS + ∫ X i u i∗ dV S

(8.26)

V

It may be mentioned that X i = 0 usually. Similarly, one can obtain the following.

∫σ

V

∗ ij

u i , j dV = ∫ t i∗ u i dS + ∫ δ ( p − q )ei u i dV S

V

= ∫ t i∗ u i dS + u i ( p)ei

(for interior point p )

(8.27)

S

Equating (8.26) to (8.27) and using (8.20), (8.22) or (8.23) yields

u i ( p ) = ∫ U ij ( p, q )t j (q)dS − ∫ Tij ( p, q )u j (q)dS S

S

⎛ ⎞ + ⎜⎜ ∫ U ij ( p, q) X i (q)dV ⎟⎟ (8.28) ⎝V ⎠ usually 0

which is the Somigliana identity serving as an integral equation.

u i ,k ( p) = ∫

∂U ij ( p, q) ∂x k P

S

t j (q)dS − ∫

∂Tij ( p, q)

S

∂x k p

u j (q)dS

= − ∫ U ij ,k ( p, q )t j (q )dS + ∫ Tij ,k ( p, q )u j (q)dS

(8.29)

S

where the following Kernels are defined as U ij ,k ≡

∂U ij ( p, q) ∂x k q

© 2005 by T. H. Kwon

=−

∂U ij ( p, q) ∂x k p

,

215

Tij ,k ≡

∂Tij ( p, q ) ∂x k q

=−

∂Tij ( p, q) ∂x k p

Hooke’s law with (8.29) result in σ σ σ ik ( p) = ∫ U ikj ( p, q )t j (q )dS − ∫ Tikj ( p, q )u j (q )dS S

+ ∫ U ikj ( p, q ) X i (q )dV (8.30)

S

V

: Somigliana identity for stress from which one can determine stress distribution once the traction and displacement on the surface determined. 8.2.4

Boundary Integral Equation

δ

Sε

p approaches the boundary surface, δ → 0

p

S − Sε Regular integral : Principal value integration → <S>

ui ( p) =

∫U

ij

( p, q)t j (q)ds −

S − Sε

∫T

ij

( p, q )u j (q)dS

S − Sε

+ ∫ U ij ( p, q )t j (q)dS − ∫ Tij ( p, q)u j (q)ds Sε

(8.31)

Sε

The first two terms on the right hand side correspond to a regular integral and is called by Principal value of the integration that is denoted by <S>. 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 well-posed boundary value problem.

© 2005 by T. H. Kwon

217

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

218

∂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) 3-node 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&

Navier-Poisson 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) power-law fluid model

µ (γ& , T ) = K (T )γ& n −1 K (T ) = A exp(Ta / T ) ↓ as T ↑

2) Modified-Cross 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

Kelvin-Voigt 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 rubber-like model z Phan Thien- Tanner model z Doi-Edward model There are many other models proposed in the literature.

Some special phenomena of viscoelastic materials: Kelvin-Voigt 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) Levy-Mises 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) Prandtl-Reuss 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 Levy-Mises

⎛⎜ 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 F1<0 →

D xy = 0

if F1>0 →

⎛ K ⎞⎟ σ 2ηD xy = ⎜1 − ⎟ xy ⎜ σ xy ⎠ ⎝

Generalized in 3-dimensional 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. Elasto-viscoplasticity

The elasto-viscoplastic 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