fem module 9

8/9/2019 FEM Module 9

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Module 9 :Lecture 26 : Axisymmetric Problems

Introduction

The problems in which the body possesses an axis of symmetry and the boundary conditions and otherparameters of the problem are symmetric with respect to this axis are called as axisymmetric problems.

The cylindrical coordinate system is found to be convenient for analyzing such problems. Inthis coordinate system, the primary and secondary variables (or their components if they happen to be

vectors and tensors) become independent of , in axisymmetric problems. Therefore, the domain ofanalysis, in axisymmetric problems, becomes just a r- zplane. Thus, axisymmetric problems are two-

dimensional problems, in some sense but not completely. Further, if primary variable is a vector, its -component becomes zero in axisymmetric problems. Because of this, and the fact that r and z

components are independent of , some of the components of the secondary variable (which happensto be a tensor in this case) become zero. Thus, the size of the problem reduces.

This lecture describes the integral formulation of a typical axisymmetric problem. Its finite element

formulation would be similar to 2-D problems, and hence, will not be discussed.

Module 9 :

Lecture 26 : Axisymmetric Problems

Model Boundary Value Problem

Figure 26.1 Domain and Boundary of a Model Boundary Value Problem

To illustrate the development of integral formulations of axisymmetric problems, we consider the

axisymmetric steady-state heat conduction problem. The domain and its boundary are shown inFig. 26.1. The thermal conductivity of the domain material is k and Q is the heat generated at point

per unit volume per unit time. In axisymmetric problems, kand Q are independent of . A part of


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the boundary, denoted by , is held at a temperature which can vary with the boundary

coordinate s. The remaining part of the boundary, denoted by , receives the heat flux which alsocan vary with the boundary coordinate s . The heat flux q , as defined in Lecture 20, is the heat flow(normal to the area) per unit area per unit time. In cylindrical coordinates, since the temperature T is

independent of , the expression forq becomes :

(26.1)

Here, is the gradient of temperature T and is the unit outward normal to the boundary.

) are the components of T while are the components of . As stated inLecture 20, q is considered positive if the heat is flowing out of the domain and vice-versa. For

axisymmetric problems, and are independent of .

Temperature T ( r , z) at point ( r , z) of the domain is governed by the following boundary valueproblem consisting of the differential equation (D.E) and boundary conditions (B.C.) :

D.E.: (26.2a)

B.C.: (i) (26.2b)

(ii) (26.2c)

The differential equation 26.2(a) represents the heat balance of small element of the domain. Theboundary condition 26.2(b) is called as the Temperature of Dirichlet boundary condition where as the

boundary condition 26.2(c) is called as the Heat Flux or Neumann boundary condition. Using thedefinition of the divergence operator and eq. (26.1), the above problem can be expressed in vectornotation :

(26.3a)

(26.3b)

(26.3c)

Here is the divergence operator.


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Module 9 :


Weak or

Weighted Residual Formulation

Consider a function , defined over the domain, which satisfies both the boundary conditions(26.3b) and (26.3c). In general, this function will not satisfy the differential equation (26.3a). Instead,

when is substituted in the left hand side of eq. (26.3a), it will lead to the following error called as

the residue and denoted by :

(26.4)

To make the function an approximate solution of the problem (26.3a, 26.3b, 26.3c), we

minimize the above residue by setting the integral of the product ofRand a weight function tozero :

(26.5)

Here, we chose the weight function wto be independent to . Note that, the integration in eq. (26.5) isover the entire volume Vof the body and not over the domain of the analysis shown in Fig 26.1. Thus, itis a 3-D integral and not 2-D integral. This is a major difference between the finite element formulationsof the axisymmetric and 2-D problems. The integral (26.5), however can be expressed as an integral

over using the property of the axisymmetric problems that all the variables are independent of .But, that will be done later.

The weight function w( r, z) must belong to a class of admissible functions. For the present problem,

this class consists of the functions which satisfy the following conditions :

1. On the boundary where Tis specified, wmust be zero. Thus, in the present problem, w= 0 on

.

2. On the boundary where is specified, w must be unconstrained. Thus, in the present

problem, wis unconstrained on .3. The function wshould be smooth enough for the integral of the weighted residue to be finite.

We rewrite eq. (26.5) as :

(26.6)

To relax the smoothness requirements on the choice of the approximation function , we use thevector identity (20.7) and the divergence theorem. In the identity (20.7),fand gare functions of (x, y).


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We now take them to be the functions of . Setting and g= wand using the identity

(20.7), the left side of eq. (26.6) becomes :

(26.7)

In Lecture 20, the divergence theorem (eq. 20.8) has been stated for 2-D problems. For 3-D problems,its statement is as follows. For a vector-valued function h, the volume integral of can be converted

to the (boundary) surface integral using the following relation :

(26.8)

where S is the boundary surface of the volume V and is the unit outward normal to S . Settingand using the divergence theorem (26.8) to convert the first right side integral to a surface

integral, eq. (26.7) becomes :

(26.9)

Using eq. (26.1), the boundary integral on the right side of eq. (26.9) can be expressed in terms of theheat flux q . Thus, we get

(26.10)

Combing equations (26.6) and (26.10) and transposing the surface integral to the other side, we get

(26.11)

Next, we express the volume integrals of eq. (26.11) as the integrals over and the surface integral of

eq. (26.11) as the integral over . Since , the first integral of eq (26.11) can beexpressed as :

(26.12)

The second step in eq. (26.12) follows from the property that k , T and w are independent of .


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Similarly, using the properly that Q is independent of , the second volume integral of eq. (26.11) can

be written as :

(26.13)

Since ( lis the coordinate along ) and q is independent of , the surface integral of eq.

(26.11) becomes :

(26.14)

Substituting equations (26.12)-(26.14) and canceling the factor from all the terms, the expression(26.11) becomes :

(26.15)

Note that the integral form of the axisymmetric problem differs from that of the 2-D problem in the sensethat each integral contains an extra multiplier, namely the coordinate r.

Finally, we simplify the boundary integral of eq. (26.15). Since , we split the boundary

integral of eq. (26.15) into two parts : (i) integral over and (ii) integral over . Further, since w= 0

is on , the integral over becomes zero. Using the Neumann boundary condition (eq. 26.2c), the

integral over can be expressed in terms of . Thus, we get

(26.16)

Combining equations (26.15) and (26.16), we get

(26.17)

This is called as the Weighted Residual Integral . This is the integral form to be used in the WeightedResidual Formulation . Now, the condition 3 of the class of admissible functions can be made explicit.

For all the integrals of eq. (26.17) to be finite, must be piecewise continuous over the domain


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with only finite discontinuities.

Module 9 :


Variational Formulation

We obtain the other integral form in a similar fashion. Let be a function which represents a

small change in the function with the constraint that wherever T is specified. Thus,

on the boundary . Such a function is called as the variation of Tand the operator is

called as the variational operator. Setting in eq. (26.17), we get

(26.18)

Using the properties of -operator from section 5 of Lecture 2 and assuming k to be independent oftemperature, eq. (26.18) can be converted to the form :

(26.19)

where

(26.20)

The functional Iis called as the Variational Functional of the boundary value problem ((26.3a), (26.3b),(26.3c)). This is the integral form to be used for the Variational Formulation . This integral form needs

to be extremized to obtain the solution of the boundary value problem.

The extremizing function T( r, z) needs to satisfy the following three conditions, which are similar to

the conditions on the weight functions w( r, z) :

1. The function T( r, z) must satisfy the temperature boundary condition (26.3b). Thus,

on . Further, the variation must satisfy the condition on .

2. The function T( r , z) and its variation must be unconstrained where the heat flux


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boundary condition (eq. (26.3c)) is specified. Thus, Tand must be unconstrained on .

3. The function T( r, z) must be smooth enough to make the functional Ifinite. Thus, Tmust be

such that is finite at every point of the domain.

The equation (26.3a) is called as the Euler equation of the functional I (eq. 26.20). Further, theboundary condition (26.3b) is called as the Essential boundary condition and the boundary condition(26.3c) is called as the Natural boundary condition .

As stated in Lecture 20, the variational functional of the steady-state heat conduction problem existsonly if the thermal conductivity k is independent of the temperature T . Further, unlike the solidmechanics problems like the bar and beam problems, the variational functional of the heat conductionproblem does not represent any physically meaningful quantity.

Since the domain of axisymmetric problems is two-dimensional, the triangular and rectangular elementsdescribed in Lectures 21-25 can be used to discretize the domain. Further, the finite elementformulation of axisymmetric problems is similar to that of the 2-D problems described in Lectures 21-22.Thus, the procedures of Lectures 21-22 can be used to obtain the element quantities, to assemble themand to apply the Dirichlet boundary condition for axisymmetric problem as well.

fem module 9

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