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 Authors : D. Raju
 Paper ID : IJERTV1IS5298
 Volume & Issue : Volume 01, Issue 05 (July 2012)
 Published (First Online): 02082012
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Thermoelastic Disturbances In A Halfspace Without Energy Dissipation For The Temperature
D. Raju
Vidya Jyothi Institute of Technology, AzizNagar Gate, Hyderabad500 075, Andhra Pradesh, India.
0020
Abstract
In this paper, we have solved the problem of one dimensional thermoelastic disturbance in a Half Space without energy dissipation due to suddenly applied constant temperature on the boundary which is rigidly fixed. Using the Laplace transform technique, exact expressions in closed form, for the temperature field is obtained. The results are illustrated graphically.
AMS Mathematics Subject Classification: 34 B 40.
Key Words: Entropy production, Thermoelastic disturbances, Energy dissipation, elastic Dilatational, Thermoelastic Theory.

INTRODUCTION:
In the conventional approach to thermomechanical theories the constitutive equations are formulated upon the basis of the equation of balance of energy and an entropy production inequality. The theory of thermoelasticity without energy dissipation was formulated by Green A.E and Naghdi P.M [4]. They have suggested an alternative procedure that is significantly different from the conventional one. In this procedure, the constitutive equations are formulated upon the basis of a reduced equation of balance of energy which is a blend of the equation of balance of energy and an equation of balance of entropy. A novel feature of this procedure is that an entropy production inequality is not employed in the process of obtaining the constitutive equations. The inequality is utilized to improve additional restrictions, if any, on the constitutive variables only after the constitutive equations have been derived.
In this we study the problem of onedimensional thermoelastic disturbances in a half space without energy dissipation due to suddenly applied constant temperature on the boundary which is rigidly fixed. Using the Laplace transform technique, exact expressions, in closed form, for the displacement, temperature and stress fields are obtained. The results are illustrated through graphs.

FORMULATION OF THE PROBLEM:
Consider onedimensional thermoelastic disturbances propagating along
the xdirection in the halfspace x 0. The displacement vector associated with these
disturbances are supposed to have only one nonzero component u in the xdirection, and this displacement component and temperature are supposed to depend only on x and t. It is assumed that the body force and heat sources are absent.
As given by Chandrasekharaiah, D.S [1, 2], the equation of motion and the equation of heat transport and other equations of thermoelastic theory without energy dissipation, in dimensionless form are
2 2 2 2
c c c
2
divU c
t U
s U p s p
T
c 2 2 Q
divU
1 T
E U U
2
( A)
and
2 2
cs
T 1 2 2
c p
cs
divU I 2
c p
T
U U I
Where,
2
2 2 2 2 k 0
c p 2 , cs 2 , cT 2 ,
v v cv c 2
Here cp and cs respectively represent the dimensionless speeds of purely elastic dilatational and Shear waves, and cT represents the dimensionless speed of purely thermal waves. is the thermoelastic coupling parameter.
Here for one dimensional problem, first, second and fourth equations of (A) reduces to,
2 2
2 u u
c p 2 2
x x t
(1)
2 2 3
2 u
cT 2 2 2
x t x t
(2)
and,
u
(3)
x
Here
T11 is the normal stress in the xdirection.
We suppose that initially the halfspace is at rest in its undeformed state and has
its temperature change and temperature rate equal to zero. Then the following homogeneous initial conditions hold.
u x, 0
u
x, 0
x, 0
x, 0 0 , x
0 (4)
t t
If the disturbances are caused by the boundary loads (on x=0), then the effects are pronounced only in the vicinity of the boundary, as such, we suppose that the following regularity conditions hold.
u x, t x, t x, t
0 as
x for t
0 (5)

SOLUTION OF THE PROBLEM:
Applying Laplace Transform to equations (1) (2) and (3), and using initial conditions (4), we get
2
2 d
c p 2
2 2 d
s u c p
(6)
dx dx
2
2 d
cT 2
2 2 du
s s
(7)
dx dx
du
and , (8)
dx
Here, is the Laplace transforms of respectively and s is the Laplace transform parameter.
Eliminating from equations (6) and (7), we get the following equation satisfied by
u ,
4 2
2 2 d 2 2 2 d 4
c pcT 4 s cT 1 c p 2 s u
0 (9)
dx dx
Once we determine u by solving this fourth order ordinary linear differential equation, then can be determined by interchanging equation (6). So equation (9) serves as the central equation of the problem.
Using the first of the regularity condition (5), the general solution of equation (9) is given by,
u A e
m1x
A e m2x
(10)
1 2
Where m1 and m2 are roots with positive real parts of the biquadratic equation
2 2 4 2 2 2 2 4
c pcT m s cT 1 c p m s 0 (11)
and A1 and A2 are functions of s that may be determined by the specified boundary conditions (i.e., on x=0). For u to be non trivial, A1 and A2 both cannot be zero.
Solving the biquadratic equation (11), we find that
s
mk , k
vk
1, 2 (12)
Where,
1 2 2
1
k 1 2
vk cT 1 c p
2
1 (13)
With
2 2 2
1
2 2 2
cT 1 c p
4 c p cT
(14)
2 2
v1 v2
(15)
In view of equation (12), equation (10) can be written as,
x s x s
v1
u A e
A e v2 (16)
1 2
Substituting u from (16) in equation (6) and integrating the resulting equation with respect to x, we obtain
2 2 x s 2 2 x s
s v1 c p
v1 c p v2 v2
2
c p v1
A1e A2e
v2
(17)
Once A1 and A2 are determined by using the specified boundary conditions, equations (17) can be inverted to obtain solutions for in terms of x and t.

PROBLEM OF CONSTANAT STEP IN TEMPERATURE ON THE RIGID BOUNDARY:
Here we consider the case where the boundary x=0 is held rigidly fixed for
all time t 0 and the disturbances are caused by the sudden application of a constant
step in temperature on this boundary at time t 0 . Then the boundary conditions are
u 0, t
0, t
0 (18)
0, t H t , t 0 (19)
Here is constant and H (t) is Unit Step function defined by
H (t ) 0 , t 0
1 , t 0
Taking the Laplace transform of the boundary conditions (19), we get,
0, s
S
(20)
Using above conditions, from equations (16) and (17), we get the following two linear equations in A1 and A2.
A1 A2 0
(21)
2 2 2 2 2
v1 c p
A1
v1
v2 c p v2
c p
A2 2
s
Solving the above equations, we get, A1 and A2 as,
k 1 2
A
1 c pv1v2
p
k s2 c2
v1v2
v1 v2
, k 1, 2 (22)
Substituting these values in equations, (16), (17) we get
2 2 x s 2 2 x s
v1v2
v1 c p 1 v1 c p v2 1 v2
2 e e
(23)
c p v1v2
v1 v2
v1 s v2 s
Taking inverse Laplace transform of above equations, we get as,
2 2 x 2 2 x
2
c p v1v2
v1 v2
v2 v1
c p H t v1 c p v
v2 H t
v2
(24)

DISCUSSION OF THE RESULTS:
From the solution given above by equation (23), we observe that is
identically zero for
x tV1
. This means that at a given instant of time t
0 , the points
of the half space that lie beyond the faster wave front
x t V1
do not experience any
disturbance. This phenomenon is a characteristic feature of all hyperbolic thermoelasticity theories. Therefore thermoelasticity without energy dissipation is a hyperbolic thermoelasticity theory.
We can compute the discontinuities experienced by u at the wave fronts
x
t , k
Vk
1, 2
from equation (24). This discontinuity is
2 2
k 1 v3 k vk c p
k 1 2
(25)
c p v1v2 v1 v2
Here … k denotes the discontinuity of the function across the wave front
x
t , k
vk
1, 2 .
Equation (25) shows that the temperature is discontinuous at both the wave fronts.

NUMERICAL EVALUATION OF THE RESULTS:
To evaluate the results numerically, we consider a material for which
2 2 1
c p 1 , cT
0.05
, 0.0168. Dhaliwal, R.S and Sherief, H.H [3] also considered the
same material for which the non dimensional relaxation time
1
0 0.05
. By using
expressions (13) and (14), we get the dimensionless speeds of wave and ewave as
v1 4.474113 and v2
0.999558
respectively. Therefore wave is faster than ewave. We
analyze the behavior of displacement, temperature and stress at dimensionless time t=
0.25. At this instant of time, the faster wave front ( – wave front) is positioned at
x x1
t v1
1.1185 and the slower wave front (e wave front) at
x x2
tv2
0.2499 .
We have computed the values of at time t= 0.25 for x
0 by using equations
(24). These are depicted in Figure. Figure shows that the temperature is discontinuous at both the wave fronts as predicted by theoretical results we also observe that assume
constant value in each of the intervals 0
x v2 , v2
x v1
a n d v1 x
. At all points
beyond the location of the faster wave front both vanish identically.
REFERENCES:

Chandrasekaraiah, D.S A Uniqueness Theorem In The Theory Of Thermoelasticity Without Energy Dissipation, J.Thermal Stresses, Vol. 19, pp. 267 – 272, 1996.

Chandrasekaraiah, D.S Thermoelastic Plane Waves Without Energy Dissipation, Mechanics Research Communications, Vol. 23, 5, pp. 549, 1996.

Dhaliwal, R.S and Sherief, H.H, Generalized Thermoelasticity For Anisotropic Media, Quar.Appl.Maths, Vol. 1, pp. 1, 1980.

Green, A.E and Naghdi, P.M, Thermoelasticity without Energy Dissipation, J.Elast, Vol. 31, pp. 189 208, 1993.