Module 9 : Lecture 26 : Axisymmetric Problems
Introduction
The problems in which the body possesses an axis of symmetry and the boundary conditions and other parameters 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. In this coordinate system, the primary and secondary variables (or their components if they happen to be vectors and tensors) become independent of ,
Module 9 :Lecture 26 : Axisymmetric Problems
Introduction
The problems in which the body possesses an axis of symmetry and the boundary conditions and other parameters 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 bevectors and tensors) become independent of , in axisymmetric problems. Therefore, the domain of analysis, in axisymmetric problems, becomes just a
r

z
plane. Thus, axisymmetric problems are twodimensional 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 elementformulation would be similar to 2D 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 theaxisymmetric steadystate 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 pointper unit volume per unit time. In axisymmetric problems,
k
and
Q
are independent of . A part of
the boundary, denoted by , is held at a temperature which can vary with the boundarycoordinate
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
isindependent of , the expression for
q
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 viceversa. 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 theboundary 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 vector notation :(26.3a)(26.3b)(26.3c)Here is the divergence operator.
Module 9 :Lecture 26 : Axisymmetric Problems
W
eak or
W
eighted 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 asthe residue and denoted by :(26.4)To make the function an approximate solution of the problem (26.3a, 26.3b, 26.3c), weminimize the above residue by setting the integral of the product of
R
and a weight function tozero :(26.5)Here, we chose the weight function
w
to be independent to . Note that, the integration in eq. (26.5) isover the entire volume
V
of the body and not over the domain of the analysis shown in Fig 26.1. Thus, itis a 3D integral and not 2D integral. This is a major difference between the finite element formulationsof the axisymmetric and 2D problems. The integral (26.5), however can be expressed as an integralover 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
T
is specified,
w
must be zero. Thus, in the present problem,
w
= 0 on.2. On the boundary where is specified,
w
must be unconstrained. Thus, in the presentproblem,
w
is unconstrained on .3. The function
w
should 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),
f
and
g
are functions of (
x
,
y
).
We now take them to be the functions of . Setting and
g
=
w
and 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 2D problems. For 3D problems,its statement is as follows. For a vectorvalued function
h
, the volume integral of can be convertedto 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 surfaceintegral, 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 .