 Open Access
 Total Downloads : 212
 Authors : Chandrashekhar S. Sutar, B. R. Ahirrao
 Paper ID : IJERTV2IS80514
 Volume & Issue : Volume 02, Issue 08 (August 2013)
 Published (First Online): 23082013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Solution of Inverse Thermoelastic Problem of Hollow Cylinder with Internal Heat Generation by Using Integral Transforms
1 Chandrashekhar S. Sutar, 2 B. R. Ahirrao
1Assistant Professor, Department of applied science, R.C.Patel Institute of technology, Shirpur (M.S) India.
1Head and Associate Professor, Department of Mathematics, Z.B.Patil College Dhule (M.S) India.
Abstract
Where 2= 2 + 1 + 2
2 2
This paper deals with effect of the sine transform on thermoelastic problem of hollow cylinder with internal heat generation. This paper determines the temperature distribution, Thermal displacements and stress functions of hollow cylinder by sine transform and
And are the Poissons ratio and the linear coefficient of thermal expansion of the material of the cylinder and T is the temperature of the cylinder satisfying the differential equation
finite MarchiZgrablich transform.
2 + 1 + 2 + = 1
(2)
2 2
Keywords: Inverse thermoelastic problem, Temperature distribution, Stress functions.

Introduction
Wankhede P.C. and Deshmukh K.C. [5] investigated an axisymmetric inverse steadystate
Subject to initial condition
, , 0 = 0 (3)
The boundary conditions are
problem of thermoelastic deformation of finite length
+ 1
= 1 , (4)
hollow cylinder. Sierakowski and Sun [3] have studied
=
an exact solution to the elastic deformation of a finite
+
= , (5)
length hollow cylinder. This paper consist
determination of temperature distribution, Thermal
2
= 2
displacements and stress function of hollow cylinder
, , = = (, ) (Known) (6)
with internal heat generation occupying the space: , 0 . The finite Marchi
, , =0
= 3(, ) (7)
Zgrablich transforms and Fourier sine transform technique is used.

Statement of the problem
Consider a hollow cylinder of length occupying the space: , 0 . The thermoelastic displacement function as in [4] is governed by the Poissons equation
2 = (1+) (1)
The interior condition is
, , = = (, ) (Unknown) (8)
Where is thermal diffusivity of the material of the cylinder.
The radial and axial displacements and satisfying the uncoupled thermoelastic equations as in [3] are
(1)
2 + 1 2 1 = 2 1+
(9)
With = 0 at = and =
2
1
2 + 1 + 2 1 = 2 1+
(10)
12
Where = + +
is the volume dilation and
And inverse MarchiZgrablich integral transform as
=1
=1
=
(k1,k2 ,n )
(20)
=
(11)
Where
k , k , = k , +
1 2 n
n
1 n
=
The stress functions are given by
(12)
k2, n n k1, n + k2, n
=
2 2 k , k ,
2 1 2 n
, , = 0, , , = 0, , , 0 = 0
(13)
And
, , = , , , = , , 0, = 0
1 k1, k2 , n . +1(k1, k2, n )
2 k1, k2, n
1 k1 , k2, n . +1(k1, k2, n )
An operational property is given by
2 + 1 + 2 2 k , k , =
1
0
2
2
1 2 n
2 k1 , k2, n + k2
(14) k2
=
2 k1, k2, n + k1 n 2 (21)
Where 1 and 0 are the surface pressures assumed to be uniform over the boundaries of the cylinder. The
k1
Also
=
boundary conditions for the stress functions (13) and
(14) are expressed in terms of the displacements components by the following relations.
If () satisfies Dirichelets conditions in interval (0, ) and if for that range its Fourier sine transform is defined to be
= + 2 + + (15)
= ()
(22)
0
Then at each point (0, ) at which () is continuous,
= + 2 + + (16)
= 2 =1
(23)
= + 2 + + (17)
The property of sine transform is
2 = 1 +1 + (0)
0 2
2 2 (24)
= + (18)
2
By applying the finite MarchiZgrablich integral transform stated in (19) to (21) to the equations (2), (3),
(6) and (7), using (4), (5) and then applying the Fourier
Where = 2
12
is the Lames constant, G is the shear
sine transform stated in (22) to (24), by using the boundary conditions and again taking their inverses, it
modulus and U and W are the displacement components.
gets
2
2 + 2 2
2 + 2 2
The equation (1) to (18) constitutes the mathematical
= ,=1
2
2
+
formulation of the problem.
sin ( ) 0 (k1,k2 ,m ) (25)

Solution of the problem
1
Where
The finite MarchiZgrablich integral transform
= 2 k1, k2, m 2 k1, k2 , m +
of order is defined as
2 0
2 1 0 1
1 +1 + +
=
k1 , k2, n
(19)
3
2 + 2 2
2
+ 0(1 ,2, ) sin +
And 1 =
2
1
=0
1 1+
2 + 0 (1,2 , ) +
Substitute the value from (25) in (8), it gets
1 2
,=1
1
2 2 + 2 2
2 + 2 2
(1 ,2, ) +
(1 ,2, ) +
+
+
0
1
=
,=1
2
2
+
2 + 1 0 (1,2 , ) sin +
sin ( )
(k ,k ,
)
1+
2 2 + cos
1 0 1 2 m (26)
1 2
,=1
1
+ sin 22 + + sin 0(1 ,2, )

Determination of
1
2
1
(30)
Thermoelastic displacement
= + 2 1+
2 2 +
1 2
,=1
From (1) and (25), Thermoelastic
cos
+ sin 22 + +
displacement is given by
1
1
2
sin 0(1 ,2, ) + 1+
2 +
= 1
1 2
,=1
2 2 2
2 2 2
0(1 ,2, )
+ + (1 ,2, ) +
1+
2 + 2 + 2 +
1
1
1 2
,=1
(1 ,2, )
0
0
(1,2 , )
+ 1 0 + + 0 +
sin ( ) (k ,k , )
1
0 1 2 m (27)
0 (1,2 , )
(1,2, )
1
2 +
1
1
+ + 1 0
p> +
0
0
2 + (1,2, )
1
1
+ +
0(1,2, ) +
From (27), equation (11) and (12) gets the radial and
1
axial displacements as
2 + 1 0(1 ,2, ) sin +
1+
2
0 k1,k2,m
1 1+
2 + 0 (1,2 , ) +
= 1 2 ,=1
+ 1 +
1 2
,=1
1
k ,k ,
+ (1 ,2, ) +
+ 0 1 2 m +
0
1
1
2
+ 0(k1 ,k2,m ) sin ( )
2 + 1 0 (1,2 , ) sin (31)
1
(28)
= + 2 1 1+
2 +
1
1
0(1 ,2, )
1 2
,=1
(1 ,2, )
=
,=1 1
,=1 1
= 1+ 2 +
cos +
+ + 1 0
+
1 2
2
+ 0 (1,2 , ) sin +
+ sin 0 (1 ,2, )
(29)
1
1
1+
2 + 0(1,2, ) +
2 2
2 2
1 2
,=1
1
1
(1 ,2, )
1
(1,2 , )
2 2
2 2
2 +
2 +
0
0
+ 0
+ + 1 0
+
Where =
2
2
+ (1 ,2, ) + 2 + 0(1,2 , ) +
1
(1 ,2, )
1
(1 ,2, )
5. Determination of stress functions
From (27), Equation (15), (16), (17) and (18)
+ 1 0
( ,
+ 2 + 1 0
)
+
+ 0 1 ,2 +
gives
1
2 + 1 0(1 ,2, ) sin +
= + 2 1+
2 +
1+
cos
1 2
,=1
2 2 +
0(1 ,2, )
+ + (1 ,2, ) +
0
0
1 2
,=1
sin 22
1
sin
( , ,
)
1
1
+ 1
+ + 0
1 2
(1 ,2, )
(1,2 , )
2
1
+ 1 0 + + 0 +
0 (1,2 , )
1
(1,2, )
1+
2
(32)
1
1
2 +
+ + 1 0
+
, = 1 2 ,=1 + 1 +
cos
(1,2, )
0(1,2, )
2 +
+ 2 + +
2 + 1 0
+ + +
1
1
1
2 + sin
0(1,2, ) + 2 +
1
cos
sin
(1 ,2, )
(1 ,2, )
(1,2 , )
1
+ +
0 +
+ 1 0 + + 0 +
1
1
0(1,2, )
(1 ,2, )
0 (1,2 , )
(1,2, )
2 +
+ + 0 +
2 +
+ + 0 +
1
1
1
1
2 + 1
0 (1,2 , )
cos
+ 2
+
2 + (1,2, ) + +
1
1
0(1,2, ) +
0(1 ,2, )
(1 ,2, )
0
0
0(1 ,2, )
sin
1
+ + 0 +
2
+ +
1
1
1
2 + 0 (1,2 , ) sin (33)
1 1+
2 + 0 (1,2 , ) +
1
1 2
,=1
1
(1 ,2, ) +
(1 ,2, ) +
+
+
0
1

Special case
2 + 1 0 (1,2 , ) sin +
Set = ( ) (34)
1+
2 2 + cos
And = ( 0)( 0) (35)
1 2
,=1
1
+ sin 22 + + sin 0(1 ,2, )
Applying Finite MarchiZegrablich transform then
1
2
1
Fourier sine transform to (35) it gets
= + 2 1+
2 2 +
1 2
,=1
.
= (36)
cos
+ sin 22 + +
1
1
2
2 2
2 2
Substitute the value from (34) and (36) in (25) to (33) it
sin 0(1 ,2, ) + 1+
2 +
gets
1
1 2
,=1
2 2
2 2
2 2 +
2 +
0(1 ,2, )
+ + (1 ,2, ) +
0
0
=
,=1
2
2
+
1
(1 ,2, )
1
(1,2 , )
+ 1 0 + + 0 +
sin ( ) (k ,k , )
1
0 1 2 m
0 (1,2 , )
(1,2, )
1
2 +
+ + 0 +
= =
1
(1,2, )
1
0(1,2, )
2 2
2 2
2 + 1 0
+ + +
2 ,=1
2 +
2
2 +
2
+
0(1 ,2, )
1
sin
2 + 1
+
sin ( ) 0(k1 ,k2,m )
1 1+
2 + 0 (1,2 , ) +
1
1 2
,=1
1
(1 ,2, ) +
(1 ,2, ) +
+
+
0
= 1
2 2 2 2
2 + 1 0 (1,2 , ) sin
1+
2 + 2 +
,=1 2
2
2
+
1 1+ 2
1 2
= + 2
0
0
1 2
+
sin ( ) 0 (k1,k2 ,m )
,=1
,=1
0(1 ,2, ) +
+ (1 ,2, ) +
1
1
1
2 + 1 0 (1,2 , ) sin +
1+
2
0 k1,k2,m
= 1 2 ,=1
+ 1 +
1+
2 + 0(1,2, ) +
k ,k ,
0 1 2 m
0 1 2 m
1 2
,=1
(1 ,2, )
1
(1,2 , )
+ 1 +
+ 0
+ + 1 0 +
2
+ 0(k1
,k2,m
)
sin ( )
1
(1 ,2, )
0(1,2 , )
1
+ 0
+ 2 + +
1
(1 ,2, )
1
(1 ,2, )
1+ 2
cos
+ 1 0
+ 2 + 1 0 +
,=1
,=1
= =
1 2
+ 1 +
( ,
)
sin
0 (1 ,2, )
+ 0 1 ,2 +
+ 1
1
2
+ 0(1 ,2, ) sin +
1
1+
1+
= + 2
2 +
1+
2 2 + cos
1 2
,=1
1 2
,=1
1
0
0
0(1 ,2, ) + + (1 ,2, ) +
+ sin 22 + + sin 0(1 ,2, )
1
1
1
2
1
= 1+
2 + +

Numerical results
Take a=1.5m, b=2m, h=4m, = 1, k=0.041
,
1 2
,=1
1
1
1
2 + 1 cos + 2 + +
= =
2 + sin 0(1,2 , ) + 2 +
2
2 +9.85 2 2 +9.85 2 +
0
0
1
,=1
1
cos
+ +
sin
(1 ,2, ) +
1 sin(12.56)
0 (k1,k2 ,m )
1
2 +
0(1,2, )
+
+ (1 ,2, ) +
1
1
Where
0
0
2 + 1 0 (1,2 , ) cos + 2 +
= 2 2 k , k ,
(0.5 + ) sin 3.14 +
2 0
1 2 m
2 9.85 2
0 (1,2 , )
(1,2 , )
cos 3.14 0.63
+ + 0
+
1
1
3.14
0 (1,2 , )
sin
1.5 2 k , k ,
1.5
sin 3.14 + cos 3.14
2 + 1
1 0
1 2 m
1 9.85 2
3.14
0.63 + 3.14 1 +2 1.5 (
Where
1.5) +
= 2 k , k ,
(+2 ) sin ( )
0.041
2 0 1 2 m
22
cos ( )
Conclusions
2 k , k ,
1 sin ( ) cos ( )
In this paper temperature distribution,
1 0
1 2 m
22
thermoelastic displacements, thermal stresses have
+ 1 +1 (
been determined for the hollow cylinder with internal
heat generation. The temperature distribution and
) +
1 =
2 + 2 2
(+ )
sin ( )
thermal stresses have been investigated by using finite MarchiZgrablich and Fourier sine transform. The results are obtained in the form of Bessel functions and infinite series.
2
2 k1, k2, m 2
cos ( )
2 0
22
Reference:
1] Marchi E and Zgrablich G: Heat conduction in
0 1 2 m
0 1 2 m
2 k , k ,
1
2 2
2 2
1 sin ( ) cos ( )
Hollow cylinder with radiation Proc. Edingburgh Math. Soc. V1.14 Part 2(1964), pp159164.
+ 1 +1 (
2] N.M.Ozisik: Boundary value problem of heat conduction, International text book co. Scranton,
) +
And
=0
Pennsylvania, (1968) pp. 481492.
3] Sierakowski and Sun: An exact solution to the elastic deformation of a finite length hollow cylinder Journal of the Franklin institute 286(1968), pp. 99113.
2 + 2 2
4] W. Nowacki: Thermoelasticity, mass, addition
=
2 Ã—
wesley publ.co. ChapI, 1962.
2 + 2 2
+
sin ( )
5] Wankhede P.C. and Deshmukh K.C.: An
2
2 k1 , k2, m 2
axisymmetric Inverse steadystate problem of
2 0
cos ( )
22
thermoelatic deformation of a finite length Hollow cylinder, Far East J. Appl. Math. 1(3), (1997) pp. 247
2 k , k ,
1 sin ( ) cos ( )
253.
1 0 1 2 m
22
+ 1 +1 (
) +
.