 Open Access
 Total Downloads : 382
 Authors : Vijay B Patil, B. R. Ahirrao
 Paper ID : IJERTV2IS90232
 Volume & Issue : Volume 02, Issue 09 (September 2013)
 Published (First Online): 05092013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Inverse Thermoelastic Problem of Heat Conduction with Internal Heat Generation in Circular Plate
Vijay B Patil B. R. Ahirrao
Assistance Professor Head and Associate Professor in
in R.A.I.T.,Nerul,Navi Mumbai(M.S.)India. Department of Mathematics, Z.B. Patil college, Dhule
Abstract
This paper consist of the inverse thermoelastic problem of heat conduction with internal heat generation for the determination of unknown
temperature, displacement and stress function by
Where and thermal conductivity and thermal diffusivity of the material of the disc subject to the initial conditions
, , 0 = 0 (3)
The boundary conditions and interior condition are
using finite Hankel transform and Finite fourier cosine integral transform. The results are obtained in the form of infinite series.
Keywords: Inverse thermoelastic problem,
(,,)
(,,)
(,,)
(,,)
=0
=
=0
=
= 1 , (4)
= 2 , (5)
= 3 , (6)
= , , (Known) (7)
Temperature distribution, thin circular plate.
(, , ) = = (, , ) (Unknown) (8)
The stress functions and are given by
1. Introduction
= 2 1
(9)
Nowacki, W. investigated the state of stress in a thick
circular plate due to a temperature field. Noda etc
= 2 2
2
2
(10)
determined the thermal stesses in circular plate.Recently N.W.Khobragade and Hamna parveen investigated the thermal stresses of thick circular plate due to heat generation. N.L.Khobragade and K.C.Deshmukh determined thermal deformation in a thin circualr plate due to partially distributed heat supply.Gaikwad and Ghadle determined an inverse quasistatic thermoelastic problem in a circular plate using the Hankel transform technique. The result are obtained in series form in terms of Bessels function.
Where is the Lames constant, while each of stress function ,, , are zero within the plate in the plane state of stress.
The equations (1) to (10) constitute the mathematical formulation of the problem under consideration.
3. Solution of the problem
If f(x) satisfies Dirchlets condition in the interval (0,a) then its finite Hankel transform in that range is
defined to be
Âµ n = Âµ n
(11)
2. Statement of the problem
Where
0
e root of the transcendental equation ,
Consider a thin circular plate of thickness h occupying the space
D: 0 , 0 . The differential equation governing the displacement function (, ) as
n is th
Âµ n = 0 (12)
Then at any point of (0,a) at which the function f(x) is continuous,
2 1
= 2
Âµ n
(13)
2 + = (1 + ) (1)
2 =1 Âµ
n [ Âµ n ]2
With = 0 at = 0 and =
and are the Poissios ratio and the linear coefficient of the thermal expansion of the material
of the disc respectively and , is the temperature
Where the sum is taken over all the positive roots of the equation (12),
An operational property is given by,
2 + 1 = [ + { }]
of the disc satisfying the differential equation
2 + 1 + 2 + (,,) = 1
Âµ 2
2
Âµ1
Âµ+1
(14)
2
2
(2)
If f(z) satisfies Dirichelets condition in the interval (0,h) and if for that range its finite Fourier Cosine transform is defined to be
5. Determination of stress functions
Using (19) in (9) and (10) the stress functions are obtained as
= () (15)
4Âµ 1+
2 2
0
= [ =1 =1[
2
2 2
2 2
Then at each point (0,h) at which f(z) is continuous, Inverse finite Fourier Cosine transform is given by
2
+ 0 ]
Ã— Âµ
[ Âµ ( )]2= 0 + 2
(16)
2 2
2 2
=1
+
+
=1
=1
[2
2
+
Applying finite Hankel transform again finite cosine transform and then their inverses stated in (11) to
0
]
Ã— Âµ ( ) [ Âµ ( )]2
] (20)
(16), to equations (2) to (8) ones obtain
=
4Âµ 1+ [
[2 2
2
2
2 2
2
2
2 2 2 2
2 2
=1
=1
=
4
2 2
=1
=1
[2
2
2
2
+ 0
]
Ã—
( )2 Âµ +
[ Âµ ( )]2+
] Ã— Âµ ( )
(17)
2 2
2 2
0
[ Âµ ( )]2 [2
2
+
2 2
2 2
=1 =1
= 4
[2
2
0 ] Ã— Âµ +
2 2 =1 =1
[ Âµ ( )]2
+
] Ã— Âµ ( )
(18)
2 2
2 2
0
[ Âµ ( )]2 [2
2
+
=1 =1
Âµ ( )
Where , = [ 2 [Âµ1 1
+ Âµ+1{2}]
0 ]
Ã— [ ( )]2 +
Âµ
2 2 2 2
+(1) 3 +
]
=1
=1
[2
2
2
2
+
] Ã— Âµ ( )
] (21)
and, denotes the finite cosine transform of and 0
denotes the finite Hankel transform of . denotes
[ Âµ ( )]2
the finite cosine transform of and denotes the finite Hankel transform of .
Where, = [ [Âµ1 1 + Âµ+1{2}]
2
2
( )
+ 1 ( ) 3 + ]
4. Determination of thermoelastic
and, = [ [Âµ1 1 + Âµ+1{2}]
displacement
Substituting the value of from (17) in (1) it gets,
, , =
2
+ 1 ( ) 3 +
( )
]
(1 + )
4
2 2
=1 =1[
2 2
2
2
2 2
( )
2
+ 0 ]
Ã— Âµ
[ Âµ ( )]2(19)

Special case
, , = 0 Âµ ( ) (22)
, , = ( 0)( 0)( 0) (23)
Stress functions are given by
= 4 Âµ 1+ [ =1 =1[
2 2
2
2
Where is Dirac delta function.
Applying finite Hankel transform and then finite
2 2
2 2
cosine transform to (22) and (23) it gets,
2
+ 0 ]
Ã— Âµ
[ Âµ ( )]2Âµ +1 2
[(1) 1]
2 2
2 2
= Âµ +1
(2 + 2 2 )
(24)
+
+
=1
=1
[2
0 (29)
0 (29)
(
)
2
+
= (25)
Substituting values from (24) and (25) in (17) to (21) it obtains
] Ã— Âµ ]
[ Âµ ( )]2
=
4
2 2
=1
=1
[2 2
2
2
2 2
2
2
=
4Âµ 1+ [
[2 2
2
2 2
2
[ 2 [Âµ1 1 + Âµ+1{2}]2 2
=1
=1
Âµ +1 2
1 1
( )2 Âµ
+ 1 {
Âµ +1
} 3 +
+ 0 ]
Ã—
[ ( )]2 +2 + 2 2
2 2
Âµ
2 2
] + 0 [ 2 [Âµ1 1 + Âµ+1{2}]
=1
=1
[2
2 +
+ 1 {Âµ +1 2 Âµ +1 1 1 } +
]
Âµ
Ã— +
2 + 2 2
] Ã— Âµ ( )
3
(26)
0
2 2
[ Âµ ( )]22 2
[ Âµ ( )]2=1
=1
[2
2
+
0
]
Ã— Âµ ( ) +
[ Âµ ( )]2=
4
[2 2
2
2
2 2
2
2
2 2
2
2
2 2
2
2
+
2 2 =1
=1
=1 =1
( )
[ [2
[Âµ1 1 + Âµ+1{2}]0 ]
Ã— Âµ ] (30)
[ Âµ ( )]2Âµ +1 2 Âµ +1 1 1
2
2
+ 1 {
2 + 2 2 } 3 +
Where, = [ [Âµ1 1 + Âµ+1{2}]
] + 0 [ [Âµ1 1 + Âµ+1{2}]
( )
+ 1 {
2
Âµ +1 2 Âµ +1
1 1 } +
+ 1
( ) 3 +
]
3
3
2 + 2 2
and, = [ [Âµ1 1 + Âµ+1{2}]
] Ã— Âµ ( )
(27)
2
( )
[ Âµ ( )]2+ 1 ( ) 3 +
]
Thermoelastic displacement is given by
, , =
2 2
= Âµ +1 2 Âµ +1 [(1) 1]
(2 + 2 2 )
=
2 2
2 2
1 + 4
=1
=1
[2
2 2
2
2
[ [Âµ1 1 + Âµ+1{2}]
2
2
+ 1 {Âµ +1 2 Âµ +1 1 1 } +
2 + 2 2 3
] + 0 [ [Âµ1 1 + Âµ+1{2}]
2
+ 1 {Âµ +1 2 Âµ +1 1 1 } +
2 + 2 2
] Ã— Âµ ( )
3
(28)
[ Âµ ( )]2

Numerical Results:
Take a=3m, h=5m, =1.5m, k=0.014 then,
4
2 2
2 2
= 321.52 =1 =1[
1.52
1.52
[ +
{ }] [ 2 Âµ1 1 Âµ+1 2
+ 1 {3Âµ +152 Âµ +1 3 1 1 } +
52 + 2 2 3
0.014
] + 0 [ [Âµ1 1 + Âµ+1{2}]
2
2
+ 1 {3Âµ +152 Âµ +1 3 1 1 } +
0.014
]
0.014
52 + 2 2 3
Ã— Âµ ( ) [ Âµ (3 )]2

Conclusion:
In this paper temperature, distribution thermoelasic displacement, thermal stresses have been determined for thin circular plate with internal heat generation. Temperature distribution and thermal stresses have been investigated by using Hankel transform and finite fourier cosine transform. The results are obtained in the form of Bessels function and infinite series.

Refarences:

Nowacki W: The state of stress in thick circular plate due to temperature field. Ball. Sci. Acad. Polon Sci. Tech.5(1957)

Noda N, R.B. Hetnarski, Y Tanigawa: Thermal Stresses, second edition Taylor & Francis, New York (2003), 260.

N.W.Khobragade and Hamna Parveen: Thermal stresses of thick circular plate due to heat generation. Canadian journal of science and eng. Mathematics vol. 3 no. 2, feb 2012

N.L.Khobragade and K.C.Deshmukh: Thermal deformation in a thin circular plate due to partially distributed heat supply. Sadhana vol.3, part 4, aug 2005, pp 555563.

K.R. Gaikwad and K.P. Ghadle: An inverse quasi static thermoelastic problem in a thick circular plate. SAJPAM volume 5(2011) 1325.