principles of computer-aided design and manufacturing second edition 2004 isbn 0-13-064631-8

Principles of Computer-Aided Design and Manufacturing

Second Edition 2004 ISBN 0-13-064631-8

Author: Prof. Farid. Amirouche

University of Illinois-Chicago

University of Illinois-Chicago

Chapter 9

Heat Conduction Analysis and the Finite Element

Method

Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN 0-13-064631-8

Author: Prof. Farid. Amirouche, University of Illinois-Chicago

CHAPTER 9 9.1 Introduction

•In most instances, the important problems of engineering involving an exchange of energy by the flow of heat are those in which there is a transfer of internal energy between two systems. In general the internal energy transfer is called Heat Transfer.

•When such exchanges of internal energy or heat take place, the first law of thermodynamics requires that the heat given up by one body must equal that taken up by the other. The second law of thermodynamics demands that the transfer of heat take place from the hotter system to the colder system.

•The three modes are conduction, convection, and radiation. Heat conduction will be the focus of this chapter. Heat conduction is the term applied to the mechanism of internal energy exchange from one body to another, or from one part of a body to another part, by the exchange of kinetic energy.

•When the relationship between force and displacement can be approximated by a linear function, the problem reduces to a one-dimensional analysis. In this chapter, we will extend the one-dimensional solution to heat conduction problems, and define the concept of shape functions for one- and two- dimensions in the finite element method.

9.1 Introduction



CHAPTER 9 9.2 One-dimensional Elements

9.2 One Dimensional elements

Now we apply the finite-element method to the solution of heat flow in some simple one dimensional steady-state heat conduction systems. Several physical shapes fall into the one-dimensional analysis, such as spherical and cylindrical systems, in which the temperature of the body is a function only of radial distance.

Consider the straight bar of Figure 9.1 where the heat flows across the end surfaces. Heat is also assumed to be generated internally by a heat source at a rate per unit volume. The temperature varies only along the axial direction x, and we suppose to formulate a finite-element technique that would yield the temperature T=T(x) along the position x in the steady-state condition.

In steady-state conditions, the net rate of heat flow into any differential element is zero. We know that for heat conduction analysis, the Fourier heat conduction equation is

This equation states that the heat flux q in direction x is proportional to the gradient of temperature in direction x.

dxdTq (9.1)

The conductivity constant is defined by .



CHAPTER 9

dx

T0

qA qA+d(Aq)

Tf

A

Figure 9.1 A typical bar with temperature T0 &Tf at each end

9.2 One-dimensional Elements



CHAPTER 9

From the differential element in Fig. 9.1, we can write the heat flux balance:

0 AqdqAfAdxqA Taking the differentiation of q , the heat flux equation becomes

0

dxA

dxdqqAfAdxqA

This reduces to a first order differential equation of the form

fdxdq

(9.3)

(9.2b)

(9.2a)

A : cross sectional areaf : heat source/unit volumeq : heat fluxT : temperature




CHAPTER 9

Substituting Equation (9.1) into equation (9.3), we get the governing differential equation for the temperature:

fdx

Td2

2

The boundary conditions for the physical problem described in Figure 9.1 are

LxatTTandxatTT f 00

Integrating (9.4) we get an explicit solution for the temperature at any point along the bar.

oof Tx

LTT

LxxfLxT

2

2

(9.4)

(9.5)

For one-dimensional problem the temperature at any point x can be found using equation 9.5




CHAPTER 9

9.3 Finite-Element Formulation

We must use either the principle of virtual work or energy to derive the necessary governing equations in finite element method. The method as shown in the previous two chapters leads to the formulation of the element stiffness and stiffness matrix. We first develop the following energy equation as

dxfA

dxTdkAI 2

2

which yields Equation (9.4) for d I = 0 using the standard manipulation of calculus of variations. Equation (9.6) could be expressed further in two parts, I1 and I2 as

L

0

L

01

1

L

02

L

01

L

0

L

0

dxdxdTAk

dxdT

dxdTTAkI

parts,by Ifunction thegIntegratin

fATdxI ,TdxdxdTA

dxdI fATdxTdx

dxdTA

dxdI

(9.6)

(9.8)




CHAPTER 9

The first term defines the boundary conditions’ contributions, which if we assume that the boundary conditions are such that

Lx TT )0(

mperatureambient te Where TTThq LLx

then the functional I becomes

L

L TThdxfATdxdxdTAk

dxdTI

0

2)(21

(9.8 a)

and

Next, consider the functional I (e) for an element rather than for the total system:

2

1

2

1

2

1

x

x

x

x

x

x

e dxfATdxdTTAdx

dxdTA

dxdTI (9.9)

eeee IIII 321 (9.10)




CHAPTER 9

To develop all the I1(e) terms we need to find an expression for the temperature T. Assume a linear

interpolation for the temperature between x1 and x2 as the distance between these two points is assumed small. A representation of the temperature is shown in Figure 9.2. where the temperature varies linearly as: baxT (9.11) At each node, the temperature is assumed to be T1 and T2 respectively we can write the temperature equation for each node becomes as

baxT 11 baxT 22

from which we can solve for a and b:

1e

121

e

12 xL

TTTbandL

TTa

(9.12)

where Le denotes the length of the element (x2-x1). Substituting the values of a and b into Equation (9.11), we get an expression for T which is written by introducing shape functions as

2211 NTNTT (9.13)




CHAPTER 9

te

X2

T2

T1

X1

Figure 9.2 Linear interpolation of the temperature

e

12

e

21 L

xxNandL

xxN

The latter are known as shape functions. These functions are linear in x and represent the characteristicof the function assumed in representing the temperature between x1 and x2.

where

(9.14)




CHAPTER 9

In matrix form, the temperature from Equation (9.13) can be expressed as

2

121 N

NTTT

We note in equation (9.10) that the time derivatives of T is also required, hence derivative of T as given by equation (9.15) takes on the following form:

dxdNdx

dN

TTdxdT

2

1

21

ee LdxdN

andLdx

dN 11 21

(9.17)

(9.16)

With

(9.15)




CHAPTER 9

The functional I(e) then becomes

TThdxNN

TTfAdxTT

dxdN

dxdNA

dxdNdx

dN

TTI L

x

x

x

x

e

212

1

2

1 2

121

2

121

2

1

21

Here the boundary conditions at both ends are defined by the last term in the above equation.Let the first term be I1

e and defined by

dxTT

dxdN

dxdN

dxdN

dxdN

dxdN

dxdN

TTAIx

x

e

2

12

212

212

1

211

2

1

Substituting (9.17) derivatives into (9.19) and integrating yield

2

1211 11

11

TT

LL

LLTTAI

ee

eee

(9.19)

(9.20)

(9.18)




CHAPTER 9

Similarly let I(e)2 denote the term defined in equation(9.18) Evaluating this term we obtain the term

which involve the contribution of the heat source .f.

1

12 21

2

1212

2

1

TTfALdxNN

TTfAI cx

x

e

Next, writing the steady-state condition for an element we get

0

e

e

TI

which yields

1111

ec L

kAk

and the element loading vector from the second term I(e)

2

11

2e

QfAL

f

(9.21)

(9.22) (9.23)

(9.24)

Combining the last equations we obtain the first step in the finite element formulation where

2fTk Xoutc (9.25)




CHAPTER 9

FTK

The global problem can be stated as

(9.26)

where [K] is the global conductivity matrix (equivalent to the global stiffness) assembled from the element conductivity matrix ke, {T} the nodal temperatures, and {F} the heat source contribution.

9.3.1 Boundary Condition Contribution

The term in the functional I in equation (9.9) deals with the convection can be written further as :

2

21)(

21

hTThThTT LLL

Where we see the last term drops out from the variational .TI

.

We see that hTL term will be added to the K matrix at the (L, L) location and hT will be added to the F vector at the L th location.

9.3.1 Boundary Condition Contribution



CHAPTER 9

The way the K & F will be formulated is shown below

hTF

FF

T

TT

hTKK

KKKK

LLLLLL

L

L

2

1

2

1

1

221

111 (9.27)

9.3.2 Handling of Additional ConstraintsThe handling of specified temperature boundary condition such as TL=T0 can be accompanied by either the elimination or penalty approach. The procedure for elimination is demonstrated below.

a)Elimination Approach

This technique works through the elimination of rows and columns of the corresponding temperature and then modifying the force vector to include the boundary. Force displacement relation as described in the finite element solution of trusses. In general, we write we the global problem as:

FKU (9.28)

9.3.2 Handling of Additional Constraints



CHAPTER 9

Consider the constraint where the displacement is defined by 11 CU The global displacement vector is array of order n x 1.

TnUUUUU 321and similarly the global force vector is

TnFFFFF 321We first start by defining the potential energy as function of elastic energy and the work associated with F.

FUKUU21 TT (9.29)

(9.30)

The energy explicit matrix form is further shown to be expressed as

NN

NNNNNNNN

NN

NN

FUFUFU

UKUUKUUKU

UKUUKUUKUUKUUKUUKU

2211

2211

2222221212

1221211111

21




CHAPTER 9

Let us substitute the boundary condition U1=C1. Then we get

NN

NNNNNNNN

NN

NN

FUFUFC

UKUUKUCKU

UKUUKUCKUUKCUKCCKC

2211

2211

2222221212

1121211111

21 (9.31)

To yield the problem at hand we need to minimize , hence

0

iU

For i = 1,2,3,……N

11

1313

1212

3

2

32

33332

22322

CKF

CKFCKF

U

UU

KKK

KKKKKK

NNNNNNN

N

N

But for i = 1, we have u1 = c1 (fixed), which yields

(9.32)




CHAPTER 9

An alternative to the elimination approach is the penalty approach. In handling constraints this might be easier to implement and works well for multiple constraints. The methods are designed to handle the boundary conditions once the global problem has been formulated. Once more let the boundaryconditions be given by the displacement at node 1 such that

b) Penalty Approach

11 CU

The total potential energy is then defined by adding an extra term to account for the additional boundary condition or simple to account for the additional energy contribution from the boundary conditions.

2112

121 cuQFukuu TT

211 CUQ21

N

2

11

N

2

1

NN2N1N

N22221

N11211

F

FQCF

U

UU

KKK

KKKKKQK

MMMMMMKK

(9.33)

So, the energy term is only significant if the value of Q is large enough to emphasize the contribution of (U1-C1)Minimization of results into

(9.34)




CHAPTER 9

We can view Q as a stiffness value whose numerical values can be defined or selected by noting the first equation so that

11NN1313212111 QCFUK...........UKUKu QK (9.35)

If we divide by Q we obtain

111

212

111 1 C

QFU

QKU

QKU

QK

NN

(9.36)

Observe how if Q is chosen to be a large volume then the equation reduces to 11 CU

which is the desired boundary condition. We also see further that Q is large in comparison to K11, K12,….,K1N, hence we need to select Q large enough to satisfy the condition of the equation above. A suggested value by previous work has been found to be

4j1 10KmaxQ (9.38)




CHAPTER 9

Determine the temperature distribution in the composite wall used to isolate the outside. Convection heat transfer on the inner surface of the wall with T=500 C is given by and h=25W/m2 o C. The following conductivity constants for each wall are κ 1=20 W/m o c ,κ2=30 W/m o c and κ3=40 W/m o c respectively. Let the cross section area A=1 m2 and L1=0.4m, L2=0.3m , L3=0.1m.

This example is used to demonstrate not only how to build the conductivity stiffness matrices and the loading vector F but how to implement the technique that describes how the boundary conditions are employed.

Example 9.1

Let the temperature at each wall be denoted by T and let the width of the wall represent the length of each element. We need to compute the local conductivity stiffness for each element. Since the conductivity constant is given per unit length, then we write

Solution :

1111

1

11

LK

1111

2

22

LK

1111

3

33

LK

(9.39)




CHAPTER 9

TC

L1 L2 L3

h=25W/m°C

Figure 9.3 Composite Wall




CHAPTER 9

880081020

02310011

50K

Global K :

Since convection occurs at node 1 , we add h=25 to (1,1) location in K which results in

880081020

02310015.1

50K

We have no heat generation or source occurring in the problem, then the F vector consists only of hT :

]0,0,0,50025[ F

Applying the boundary conditions T4=10C, can be handled by the penalty approach. Let us choose a value for Q from the previously proposed procedure where

4

4ij

101050

10KmaxQ

(9.40)

(9.41)

(9.42)

(9.43)




CHAPTER 9

As stated in the penalty function we add the Q value to the K matrix in the (4x4) location,

and in the (1,1), location Qc1 to the (1,4) location of the F vector, and QT4 to the (1 x 4) location of the

F vector resulting in

74

3

2

1

1050050025

10000880081020

02310015.1

50

TTTT

K

The solution of which is found to be

CT o0014.108979.264839.946559.229

(9.44)




CHAPTER 9 9.3.3 Finite Difference Approach

Finite difference is discussed briefly through the following example for the purpose of validating the one-dimensional solution we have derived.

9.3.3 Finite Difference Approach

Example 9.2

A special design for a construction-building wall is made of three studs containing the materials siding, sheathing, and insulation batting. The inside room temperature is maintained at 85o F and the outside air temperature is measured at 15o F. The area of the wall exposed to air is 180 ft2. Determine the temperature distribution through the wall.

Items Resistance (hr.ft2.F/Btu) U-factor (Btu/hr.ft2.F)

Outside film resistance 0.17 5.88

Siding 0.81 1.23

Sheathing 1.32 0.76

Insulation 11.0 0.091

Inside film resistance 0.68 1.47

Table 9.1 Characteristics of the wall



CHAPTER 9

The steady state condition of this system can be explained through Fourier’s law.

XTkAqx

We can express the gradient of temperature by (Ti+1 –Ti)/l and the heat transfer rate becomes

lTTkAq ii )( 1 )( 1 ii TTUAq

The heat transfer between the surface and fluid is due to convection. Newton’s Law of cooling governs the heat transfer rate between the fluid and the surface

)( fs TThAq

where h is the convection coefficient Ts is the surface temp and Tf is the fluid temp.

or where U is defined by k/l.

The heat loss through the wall due to conduction must be equal the heat loss to the surrounding cold air by convection. That is )( fs TThA

XTkA

(9.46)

(9.49)

(9.50)




CHAPTER 9

Expanding the above equation on the temperature distribution at the edge of each wall leads to the following equations.

)()( 121232 TTAUTTAU )()( 232343 TTAUTTAU )()( 343454 TTAUTTAU )()( 454565 TTAUTTAU

Expressing the above in a matrix form we get

65

11

5

4

3

2

544

4433

3322

221

00

000

000

ATu

ATu

TTTT

uuuuuuu

uuuuuuu

A or

2249100

15876

TTTT

561.1091.00091.851.76.0076.99.123.10023.111.7

180

5

4

3

2

The solution is found to be

CTTTT o5909.815205.269266.198523.155432

(9.51)




CHAPTER 9 Heat Conduction Analysis

9.4 Heat Conduction Analysis of a two-Element Rod

Let us divide our system into elements with three nodes, as shown in Figure 9.4. In the development of the connectivity Table 9.2, we list the node numbers under each element.First, we note that the global connectivity matrix K is a 3X3 matrix. The contribution of the conductivity matrices for elements 1 and 2 are

000011011

1

eij L

KaK

21

Element # 1

(1)

3

Element # 2

(2) Kij Elements1

2

1 1 2

2 2 3

Figure 9.4 Elements with three nodes. Table 9.2

(9.55)



CHAPTER 9

The global conductivity matrix is then obtained by summation:

21ijij KKK

110121

011

eLkAK

Similarly, the global heat source force vector is obtained by adding the two local force vectors:

121

2110

2011

2eee fALfALfAL

F

Thus, combining and writing in the form of Equation (9.26), we obtain

121

2fAL

TTT

110121

011

LkA e

3

2

1

(9.56)

(9.57)

(9.58)

(9.59)

Heat Conduction Analysis



CHAPTER 9

Applying the boundary conditions

0)(0)0( 31 LxTandxT

we solve for T2, which results into

2

2

0

0

110121

011

2

e

e

e

e fALfAL

fAL

TLkA

which reduces to

L/2 L wherekL f

81

kL f

21T e

22e

2

For simplicity, let mLmmF 111 3

then the temperature at node 2 becomes

F125.081T o

2

(9.60)

(9.61)

(9.62)

Heat Conduction Analysis



CHAPTER 9 9.5 Formulation of Global Stiffness Matrix

For the boundary conditions are such that Tf is zero, then we get an explicit solution of the temperature distribution for the assumed boundary conditions from simple integrating as stated in equation (9.5)

Lxx

kfLxT

2

2

where we can see that T ( x=1/2)=0.125oK checks exactly with our finite-element solution given by the above equation.

9.5 Formulation Of Global Stiffness Matrix For N ElementsThe concept of global conductivity matrix [K] in the above example is exactly the same as the global stiffness matrix that was discussed in Chapter 8. {T} and {F} now represent the nodal temperature vector and the heat source contribution vector, respectively, instead of the nodal displacement and the nodal force vectors as described in chapter 7, and 8. Table 9.1 is simply used as a guide to help in the formulation of the global conductivity matrix.

(9.63)



CHAPTER 9

321

Element # 1 Element # 2

(1) (2)

n+1n

element # n

Figure 9.5 Discretization of a heat conduction rod into N-elements

011 NTT (9.64)

Let us consider a body discredited into N one-dimensional elements, as shown in Figure 9.5. Let the boundary conditions be such that

9.5 Formulation of Global Stiffness Matrix



CHAPTER 9

ij Elements e, i, j

From kije

1 2 3…. N

1 1 2 3… N

2 2 3 4… N+1

Table 9.3 Connectivity matrix for the N-elements

The connectivity table (Table 9.3) shows that the global conductivity matrix is of the order (N+1) x (N+1).

The ascending order of elements helps the global K to have a predictable bandwidth.




CHAPTER 9

N21

N21

N21

N21

N21

N21

T

TTT

11001111.....011110

11110011

N

1N

3

2

1

MM

OK

By following the steps discussed in previous section and using the table information for inserting the local stiffness terms to the global matrix from Table 9.3, the global problem takes the following form:

By applying the boundary conditions, the problem reduces to ,

11

111

N1

TT

TT

210121

..........01210

1210012

2

N

1N

3

2

MM

OL

(9.65)

(9.66)




CHAPTER 9

EXAMPLE 9.3

For the one-dimensional heat transfer problem given by

00101102

2

TwithxAwheredx

Td

Find the temperature at x=0.2,0.4,0.6,0.8 and 1.0 m (Figure 9.6)

T1 T2 T3 T4 T5 T6

X

1 2 3 4 5

0.0 0.2 0.4 0.6 0.8 1.0

Le=L/N

Figure 9.6 One-dimensional heat transfer.




CHAPTER 9

Solution :Kij 1 2 3 4 5

1 1 2 3 4 5

2 2 3 4 5 6

Each element has an element conductivity matrix Ke of the form:

1111

e

e

LkAK

Substituting mNLLmA e 2.0

511 2

and assuming the conductivity constant to be k=1, then we evaluate the element conductivity matrix.

1111

5eK

Table 9.4 Connectivity Table

(9.67)

(9.68)




CHAPTER 9

Using the connectivity table, the global matrix [K] is obtained by summation:

110000121000

012100001210000121000011

5K

By applying the boundary conditions, the global temperature vector becomes

0

0

5

4

3

2

TTTT

T

The forcing vector for an element is shown to be

11

2ee fAL

F

Where is the heat generation per unit volume and is obtained from the relation

fdx

Td2

2

(9.69)




CHAPTER 9

Substituting =1 and d2T/dx2=-10 yields =10.Substituting into Fe those values, we get

11eF

Assembling the global forcing vector using the connectivity table yields

122221

111111111

1

F

Using the relation FTK

122221

110000121000

012100001210000121000011

5

6

5

4

3

2

1

TTTTTT

(9.73)

(9.74)

(9.75)




CHAPTER 9

By deleting the first and last rows together with their corresponding columns, and modifying the force vector we obtain Equation becomes

4.04.04.04.0

21001210

01210012

5

4

3

2

TTTT

Note that T2=T5 and T3=T4. From symmetry, we can solve equation very easily. The solutions are as follows:

C

TTTTTT

0

6

5

4

3

2

1

08.02.12.18.0

0

(9.76)

(9.77)




CHAPTER 9 9.6 2D Heat Conduction Analysis

The heat conduction problem is formulated by a variational boundary value problem as

0I

dfTTkI 221 2

and where k = thermal conductivity, which we assume is constantf = Heat sourceT = temperature gradient(T)2=T.T,”.” denotes the dot product = Domain of interest

Where

9.6 2D HEAT CONDUCTION ANALYSISIn a fashion similar to the one-dimensional analysis, the finite-element method can be used to analyze the 2D and 3D heat conduction problems. Let us examine the 2D case .

(9.79)



CHAPTER 9If domain is divided into N elements, as shown in Figure

9.5, then

N

e

eII1

e

eeee dTfTkI 221 2

Let us consider the triangular element shown in Figure 9.7. The local representation of the temperature can be expressed as

332211, NTNTNTyxT

where Ni (x, y) (i = 1, 2, 3) are the shape functions given by

ycxbaN ei

ei

ei

ei

and

The shape functions must satisfy the following conditions:

(x, y) are linear in both x and y. (x, y) have the value 1 at node i and zero at other nodes. (x, y) are zero at all points in , except those of Nei (x, y) can be written as

1 3

1

i

ieN

ieN

ieN

ieN

(9.82)

(9.83)

9.6 2D Heat Conduction Analysis



CHAPTER 9

yxcbayxN iiii

1,

Three nodes of the triangular element

yx

cbacbacba

NNN 1

333

222

111

3

2

1

4

1 23

X

Y Y

X

T2

T3

T1

Figure 9.7 Triangular element




CHAPTER 9 For node 1, following condition 1, Equation (9.85) yields

111111 1 ycxbaN 212112 0 ycxbaN 313113 0 ycxbaN

Which can be written in matrix form as

1

1

1

001

cba

A

where

33

22

11

111

yxyxyx

A

Solving for coefficients a, b, and c, we get

001

001

111

1

1

33

22

11

1

1

1

Ayxyxyx

cba

(9.51)

(9.87)

(9.88)

(9.89)




CHAPTER 9

Similarly, for the interpolation functions N2 and N3, we get

100

010

1

3

3

31

2

2

2

Acba

andAcba

The inverse of matrix A is

12211221

31133113

233223321

21

xxyyyxyxxxyyyxyxxxyyyxyx

aA

where a is the area of the triangle.

Combining (9.89) and (9.90) The inverse of A is

333

222

1111

cbacbacba

A

(9.90)

(9.91)

(9.92)




CHAPTER 9

Then the triangle element functions can be written in a more general form:

yxA

NNN

N e

11

3

2

1

233223321 21 xxyyyxyxyxa

N


N


N

Now that we have defined the shape function, we can proceed in the evaluation of the conductivity matrix of individual elements.

(9.93)

(9.94)




CHAPTER 9 9.7 Element Conductivity Matrix

From Equation (9.81), we write the variational equation in terms of elements. This defines the element equation as

dTfTkI eeee 2

21

The temperature at the nodes of the triangle element is expressed following the triangular element assumption developed in previous section where

332211, NTNTNTyxT

From Equation (9.94), we define the partial derivatives w.r.t x and y as

ii

ii c

yN

andbx

N

Hence, we can write the gradient of the temperature as follows

3

2

1

321

321

TTT

yN

yN

yN

xN

xN

xN

yTxT

T

9.7 Element Conductivity Matrix

(9.95)

(9.96)

(9.97)

(9.98)



CHAPTER 9

which, expressed in compact form, yields BTT

3

2

1

321

321

TTT

Tandcccbbb

B

yTxT

yT

xTT 2

eeTeTee TBBTT 2

where

This yields

e

dTBBTkI eeTeTee

21

1

(9.99)

(9.100)

(9.101)

(9.102)




CHAPTER 9

e

eeTeTee dTBBTkI21

1

or simply

eeTee TkTI21

1

Where [ke] denotes the element conductivity matrix:

eTee BBakK

Which takes the final form

23

2323231313

323222

221212

3131212121

21

cbccbbccbbccbbcbccbbccbbccbbcb

kaK e

and a is the area of the triangular element.

(9.103)

(9.104)

(9.105)

(9.106)




CHAPTER 9 9.8 Element Forcing Function

dTfI eee2

3

2

1

321332211

NNN

fffNfNfNff

For an arbitrary element, this equation can be written in compact matrix form:

eee Nff

3

2

1

321332211

TTT

NNNNTNTNTT

eTee TNT

(9.73)

9.8 Element-Forcing FunctionTo complete the integration of Equation (9.95), we need to evaluate the second term, Ie

2

As we have done with temperature, the heat source f can be expressed in a similar fashion: (9.107)

(9.108)

(9.109)

(9.110)



CHAPTER 9

Therefore, Ie2 after substitution becomes

eeeTeeee TgdTNNfI2

dNNfgTeeTeTe

The integrand {Ne}{Ne}T yields

1

2

2111

11

1

TTTee A

yxyyxyxxyx

AAyxyxANN

where

An alternative is to use a method developed by Eisenberg and Malvern.From this method, we have the following statement of the integral:

e

pnm apnmpnmdNNN 2

)!2(!!!

321

e

e

Teee fdNNg

(9.112)

(9.113)

(9.114)

(9.115)

(9.116)

9.8 Element Forcing Function



CHAPTER 9

Hence,

ee fdNNNNN

NNNNNNNNNN

g

232313

322212

31212

1

Which yields

ee fag

211121112

12

The element integral of the variational formulation is broken into two parts:

eee III 21

eTeeeTee TgTkTI 21

Simplifies to

(9.117)

(9.118)

(9.119)

(9.120)




CHAPTER 9

The “global integral” over the domain of the entire body becomes

eTeeeTeN

e

e TgTkTII21

1

or TFTkTIN

e

TeN

eT

11321

where TeTe gF

and TnTTTT 21Hence,

TFTkTI TT 21

Where the global conductivity matrix is defined by

N

e

ekk1

(9.121)

(9.122)

(9.123)

(9.124)




CHAPTER 9

and the global function (equivalent to the global force in the analysis of a truss) is

N

e

eFF1

The variation I = 0 is equivalent to niTI

i

,,10

Applying Equation (9.76) to Equation (9.73) gives the global equation governing the temperature distribution and the heat source:

FTk This equation is similar to our FEM application to the truss and the one-dimensional heat flow problems.The analysis of 2D heat conduction problems can be done by using the FEM procedures developed herein. One proceeds by identifying the element shape functions and then evaluating the local conductivity (stiffness) matrices. The global [K] is then assembled using Equation (9.87). The element forcing functions is computed using Equation (9.75) and then the global array {F} is assembled according to Equation (9.86).

(9.125)

(9.126)

(9.127)




CHAPTER 9Example 9.4 Temperature Distribution on a Square PlateFor the square plate shown in Figure 9.8, find element matrices [Be] and [ke] and solve for all the element conductivity matrices. Find the temperature distribution at all of the nodes shown for the boundary conditions given.

T=200

T=50 8 55

1

49

32

2

4

Y

7

7 8

6

T=0

T=0

TYPE 3TYPE 1

3

6

XTYPE 2 TYPE 4

9

1

2 3 3 3 32

12121

a)Element discretization of the plate

b)All possible element types

Figure 9.8




CHAPTER 9

There are four types of elements, as shown in Figure 9.8. The area of each triangular element is a=1/8.Figure 9.9 shows the temperature distribution along the x-axis and y-axis for the plate. Matrices [Be] for each type of element are obtained from

123123

211332

)()()()()(

21

xxxxxxyyyyyy

aB e

from which we can compute the contribution of each element. This is simply done by evaluating the Be matrix by identifying the (x, y) coordinate of each node. The element corresponding Be matrices are found to be:

0222201B

2200222B

2020223B

2022204B




CHAPTER 9

T=200

T=0

T=0

T=50

1

1

X

Y

Figure 9.9 Temperature Distribution

The element conductivity matrices are then obtained from

eTee BBkak




CHAPTER 9

which results into

440484

044

821 kkk

404044448

83 kk

844440404

84 kk

Table 9.5 Element conductivity stiffness matrix

Nodes

1 2 3 4 Elements5

6 7 8

1 1 1 2 3 4 5 5 5

2 4 2 3 5 5 7 8 6

3 5 5 5 6 7 8 9 9




CHAPTER 9

The relationship between elements and nodes is described by Table 9.5 from the boundary conditions, we get

00250

50100200125

5

9

8

7

6

5

4

3

2

1

T

TTTTTTTTT

T

Where T5 is the only unknown. Hence, from the global equation kT=f, problem becomes

9

155

iii FTk

Because there is no heat source, F5 is simply given by adding to zero the contribution from the penalty function or F5=0+ . . . . . From the relationship between [ke] and the triangles, we can easily deduce the following contribution from each element for the element conductivity stiffness matrix




CHAPTER 9

hypotenuseandjiforusenonhypotenandjifor

iandjifor

iandjifor

kk eij

04

904

9044

8

08

08

328

08

0

85kk i

Solving for T5 we obtain CT o5.625




CHAPTER 9

Example 9.5 Steady State Heat ConductionFind the temperature distribution for steady-state heat transfer conduction in a square domain, as shown in Figure 9.10, with

10;1001,0,,110,0

yxxTxTyTandyT

The boundary value for this problem is given by 02

2

2

2

yT

xT

This solution differs from the previous example in two respects: (1) there are only two types of elements used and (2) we doubled the number of elements to learn more about the temperature inside the plate. As shown in Figure 9.10, we divide this domain into 18 elements. There are two different types of triangles in the model (see Figure 9.11).

The method of numbering the elements and nodes is arbitrary. However, one has to do it systematically so as to obtain matrices that require less storage space. Once the global conductivity matrix [K] is formulated, its bandwidth will be checked to see whether its final form is mathematically sound. Let us proceed in the solution of this problem by identifying the element types and computing their corresponding [B] and [K] matrices.

Solution:




CHAPTER 9

X

Y

1 2 3 4

8

12

16151413

9

5 6 7

1110

(1)(10)

(4)(13)

(7)(16) (17)

(8)

(14)(5)

(11)(2)

(12)

(3)

(15)(6)

(18)(9)

Figure 9.10: Square domain with triangular elements. Figure 9.11 Element types for the finite element model




CHAPTER 9

The area of the two triangles is the same and is given by

181

31

31

21

a

For an arbitrary triangular element, we have

123123

211332

21

xxxxxxyyyyyy

aB e

For a type 1 element [B1] becomes

330033

31

310

031

31

1 gB

303330

310

31

31

310

gB e

The conductivity matrix is given by eTee BBkak




CHAPTER 9

For a type 1 element,

9909189

099

18kk e

For a type 2 element,

1899

990909

18kk e

The relationship between elements and nodes is given in Table 9.5Assembling the element conductivity matrices yields the global conductivity matrix:

18321 kkkkK

hypotenuseandjiforusenonhypotenandjifor

iandjifor

iandjiforkk

o

o

eij

09

909

90)99(

18




CHAPTER 9

12100

21

221001

221001

100021

210021

4100141001

200021

210021

4100141001

200021

12100

2210

221

1

K




CHAPTER 9

Table 9.6 Connectivity relations of elements and nodes

Nodes 1 2 3 Elements4

. . . 18

1 1 2 3 5 . . . 11

2 2 3 4 6 . . . 15

3 6 7 8 10 . . . 16

T1=T13=1/2 (10+0)=5 and T5=T9=10 C

0161514128432 TTTTTTTT

Therefore, the unknown nodal temperatures are T6, T7, T10, and T11.

Note that the heat source is zero thus the system of equation becomes

0Tk

00050100100005 111076 TTTTT where

9.8.3 Boundary Conditions

(9.128)




CHAPTER 9

1040404

104

11106

11107

1176

1076

TTTTTTTTT

TTT

75.34

15106 TT

25.145

117 TT

Using the boundary conditions on the global system, we obtain the equations for the unknown nodal temperatures

From the property of symmetry of the system, we know T8 = T10 and T7 = T11.

The solution is as follows:

(9.129)

(9.130)




CHAPTER 9 9.9 FEM and Optimization

9.9 FEM AND OPTIMIZATIONIn order to survive in today’s competitive industrial/scientific world, the products will have to have the following characteristic features:

1. Low cost

2.High built-in reliability of performance

3. Limited time frame for design/manufacture

The first factor is usually achieved by minimizing the volume/mass/weight of the structure component, whereas the second factor would need the various constraints defined in the problem statement to be satisfied in the process of design. The third factor emphasizes the reduction of the overall time for bringing the product into the market by using proper computational tools/manufacturing techniques, which will complete the process at higher speeds.

In recent times, state-of -the-art structural optimization algorithms and design sensitivity analysis methods have come into existence, which cover the first two points mentioned above to a considerable extent. The third point could be brought into control by utilizing a combination of hardwares and softwares.

The concepts of inherent vector and concurrent processing made possible by the recent advances in the computer architecture would assist in the design and analysis stage as well as in the numerical control machines, Group Technology and CIM architectures discussed in the latter chapters. This technology will definitely be a key to the speed of the manufacturing process.



CHAPTER 9

The structural optimization process deals with a systematic procedure of manipulating the design variables that describe the structural system while simultaneously satisfying prescribed limits on the structural response. Hence it is seen that there are three major operations integrated into the procedure of structural optimization.

These are:1.Finite-Element Analysis2.Design Sensitivity Analysis3.Optimization Algorithm.

9.9 FEM and Optimization