Dynamic problem in 3D thermoelastic half-space with rotation in context of G-N type II and type III

S. Santra1 , N. C. Das3 , A. Lahiri2

1Gargi Memorial Institute of Technology, Kolkata, 700144, India

3, 2Department of Mathematics, Jadavpur University, Kolkata, 700032, India

1Corresponding author

Mathematical Models in Engineering, Vol. 3, Issue 1, 2017, p. 58-70. https://doi.org/10.21595/mme.2017.18143
Received 28 December 2016; received in revised form 16 January 2017; accepted 17 January 2017; published 30 June 2017

Copyright © 2017 JVE International Ltd.

Abstract.

In this paper, comparison between G-N model of type II (without energy dissipation) and G-N model of type III (with energy dissipation) has been shown in a three dimensional thermoelastic half space with rotation subjected to time dependent heat source on the traction free boundary. Eigenvalue methodology has been adopted to solve the equations resulting from the application of the Normal mode analysis to the non-dimensional coupled equations. Variation of the numerically computed values of thermal stresses and temperature with and without rotation has been illustrated graphically.

Keywords: anaisotropic half space, G-N model II and III, normal mode analysis, eigenvalue approach.

1. Introduction

The inconsistency of heat conduction equation of classical uncoupled theory of thermoelasticity with the experimental results is due to the fact that i. no elastic term is included to account for elastic changes producing heat effects; ii. parabolic nature of heat conduction equation indicating infinite speed of propagation of heat waves means that thermal disturbances (with infinite speed) and elastic disturbances (with finite speed) from the classical theory of thermoelasticity, are coupled together. This suggests that every solution of the equations has a part which extends to infinity.

Biot [1] developed a theory of irreversible thermodynamics and gave a satisfactory derivation of the linear theory of coupled thermoelasticity. In order to obtain a wave type heat conduction equation the concept of generalized thermoelasticity was introduced modifying CCTE and later extended by Dhaliwal and Sherief [3] for anisotropic body, and the uniqueness of the solutions was proved by Ignaczak [2, 4]. Green-Naghdi proposed a generalized thermoelasticity by modifying the energy equation. There are three types of constitutive relations in G-N model [6-8]. Type-I leads to classical heat conduction equation. Type-II provides solutions for thermal waves propagating finite speed without energy dissipation (TEWOED) and type-III also confirms propagation of thermal waves of finite speed with energy dissipation (TEWED). Several investigations with these extensions have been studied by Abd-Alla and Abo-Dahab [11], Kar and Kanoria [9] and Yousef [10]. Pal et al. [5, 14] studied the effect of homogeneity of the surface waves in anisotropic media.

In this paper, the stress distributions and temperature variation has been depicted in an anisotropic triclinic half space for G-N model II and III both with rotation.

2. Basic equations

For a linear thermoelastic anisotropic body subjected to rotation the field equations are as follows:

The equation of motion in the absence of inner heat source:

(1)
τ i j , i = ρ u ¨ i + - × - × u - i + 2 - × u - ˙ i .

Heat – conduction equation in absence of body force:

(2)
k * θ , i j + k i j θ ˙ , i j = ρ c e θ ¨ + θ 0 β i j u ¨ i , j .

The Duhamel-Neumann stress-strain relations are:

(3)
τ i j = c i j k l e k l - β i j θ δ i j   ,

where βij=cijklαkl  i,j,k,l=1, 2, 3.

Strain – displacement relation:

(4)
e i j = 1 2 u i , j + u j , i   ,       u i , j = u i x j   ,

where the differentiation with respect to space variable xi is denoted by and that with respect to time is denoted by notation.

3. Formulation of the problem

Let us consider a linear anisotropic thermoelastic half space within x1,x2,x3: 0x1<,  0x2<, 0x3< for a time dependent heat source on the boundary plane to the surface x1= 0. The surface x1=0 is assumed to be traction free and the body is assumed initially at rest. The components of displacement vectors of three dimensional plane waves in anisotropic elastic medium, are given as:

(5)
u i = u i x 1 , x 2 , x 3 , t   ,         i = 1 ,   2 ,   3 ,

where t is the time variable and xi, i=1, 2, 3 denotes the respective orthogonal Cartesian co-ordinate axes. The elastic medium is now considered as rotating uniformly with an angular velocity Ω=Ωn^, where n^ is the unit vector along the direction of rotation. The equation of motion of the rotating frame contains two additional terms: (Ω×(Ω×u) representing the centripetal acceleration due to time varying motion only and (2Ω×u˙) representing the coriolis acceleration and n^ = (1, 0, 0).

Using Hook’s law, the stress- strain- temperature relations can be written as follows:

(6a)
τ 11 = c 11 e 11 + c 12 e 22 + c 13 e 33 + 2 c 14 e 23 + c 15 e 13 + c 16 e 12 - β 11 θ ,
(6b)
τ 22 = c 21 e 11 + c 22 e 22 + c 23 e 33 + 2 c 24 e 23 + c 25 e 13 + c 26 e 12 - β 22 θ ,
(6c)
τ 33 = c 31 e 11 + c 32 e 22 + c 33 e 33 + 2 c 34 e 23 + c 35 e 13 + c 36 e 12 - β 33 θ ,
(6d)
τ 23 = c 41 e 11 + c 42 e 22 + c 43 e 33 + 2 c 44 e 23 + c 45 e 13 + c 46 e 12 ,
(6e)
τ 13 = c 51 e 11 + c 52 e 22 + c 53 e 33 + 2 c 54 e 23 + c 55 e 13 + c 56 e 12 ,
(6f)
τ 12 = c 61 e 11 + c 62 e 22 + c 63 e 33 + 2 c 64 e 23 + c 65 e 13 + c 66 e 12 .

In absence of inner heat source and body force the equation of motion are given as:

(7a)
τ 11,1 + τ 12,2 + τ 13,3 = ρ u ¨ 1 ,
(7b)
τ 21,1 + τ 22,2 + τ 23,3 = ρ u ¨ 2 - 2 u 2 - 2 u ˙ 3 ,
(7c)
τ 31,1 + τ 32,2 + τ 33,3 = ρ u ¨ 3 - 2 u 3 + 2 u ˙ 2 .

With the help of Eqs. (5) and (6), the equations of motions Eq. (7) become:

(8a)
c 11 u 1,11 + c 66 u 1,22 + c 55 u 1,33 + 2 c 16 u 1,12 + c 15 u 1,13 + c 56 u 1,23 + c 16 u 2,11 + c 26 u 2,22 + c 45 u 2,33 + c 12 + c 66 u 2,12 + c 14 + c 56 u 2,13 + c 46 + c 25 u 2,23 + c 15 u 3,11 + c 46 u 3,22 + c 35 u 3,33 + c 14 + c 56 u 3,12 + c 13 + c 55 u 3,13 + c 36 + c 45 u 3,23 - β 11 θ , 1 = ρ u ¨ 1 ,
(8b)
c 16 u 1,11 + c 26 u 1,22 + c 45 u 1,33 + c 12 + c 66 u 1,12 + c 14 + c 56 u 1,13 + c 46 + c 25 u 1,23 + c 66 u 2,11 + c 22 u 2,22 + c 44 u 2,33 + 2 c 26 u 2,12 + c 46 u 2,13 + c 24 u 2,23 + c 56 u 3,11 + c 24 u 3,22 + c 34 u 3,33 + c 46 + c 25 u 3,12 + c 36 + c 45 u 3,13 + c 23 + c 44 u 3,23 - β 22 θ , 2 = ρ u ¨ 2 - 2 u 2 - 2 u ˙ 3 ,
(8c)
c 15 u 1,11 + c 46 u 1,22 + c 35 u 1,33 + c 14 + c 56 u 1,12 + c 13 + c 55 u 1,13 + c 36 + c 45 u 1,23 + c 56 u 2,11 + c 24 u 2,22 + c 34 u 2,33 + c 46 + c 25 u 2,12 + c 36 + c 45 u 2,13 + c 23 + c 44 u 2,23 + c 55 u 3,11 + c 44 u 3,22 + c 33 u 3,33 + 2 c 45 u 3,12 + c 35 u 3,13 + c 34 u 3,23 - β 33 θ , 3 = ρ u ¨ 3 - 2 u 3 + 2 u ˙ 2 .

The generalized heat-conduction Eq. (2) is written as:

(9)
k * + k 11 t θ , 11 + k * + k 22 t θ , 22 + k * + k 33 t θ , 33 = ρ c e θ ¨ + θ 0 β 11 u ¨ 1,1 + β 22 u ¨ 2,2 + β 33 u ¨ 3,3 .

The following non- dimensional variables are introduced to transform the above equations in non-dimensional form:

(10)
x ' i = 1 l x i ,       u ' i = c 11 l β 11 θ 0 u i ,       t ' = c 11 ρ   1 l t ,       θ ' = θ θ 0 ,       c 1 2 = c 11 ρ ,         = c 1 l ' ,

where l  is some standard length.

Using Eq. (10), the non dimensional forms of the Eqs. (8a)-(9) reduces to (omitting primes for convenience):

(11a)
u 1,11 + c 66 c 11 u 1,22 + c 55 c 11 u 1,33 + 2 c 16 c 11 u 1,12 + c 15 c 11 u 1,13 + c 56 c 11 u 1,23 + c 16 c 11 u 2,11 + c 26 c 11 u 2,22 + c 45 c 11 u 2,33 + c 12 + c 66 c 11 u 2,12 + c 14 + c 56 c 11 u 2,13 + c 46 + c 25 c 11 u 2,23 + c 15 c 11 u 3,11 + c 46 c 11 u 3,22 + c 35 c 11 u 3,33 + c 14 + c 56 c 11 u 3,12 + c 13 + c 55 c 11 u 3,13 + c 36 + c 45 c 11 u 3,23 - θ , 1 = u ¨ 1 ,
(11b)
c 16 c 11 u 1,11 + c 26 c 11 u 1,22 + c 45 c 11 u 1,33 + c 12 + c 66 c 11 u 1,12 + c 14 + c 56 c 11 u 1,13 + c 46 + c 25 c 11 u 1,23 + c 66 c 11 u 2,11 + c 22 c 11 u 2,22 + c 44 c 11 u 2,33 + 2 c 26 c 11 u 2,12 + c 46 c 11 u 2,13 + c 24 c 11 u 2,23 + c 56 c 11 u 3,11 + c 24 c 11 u 3,22 + c 34 c 11 u 3,33 + c 46 + c 25 c 11 u 3,12 + c 36 + c 45 c 11 u 3,13 + c 23 + c 44 c 11 u 3,23 - β 2 θ , 2 = u ¨ 2 - 2 u 2 - 2 u ˙ 3 ,
(11c)
c 15 c 11 u 1,11 + c 46 c 11 u 1,22 + c 35 c 11 u 1,33 + c 14 + c 56 c 11 u 1,12 + c 13 + c 55 c 11 u 1,13 + c 36 + c 45 c 11 u 1,23 + c 56 c 11 u 2,11 + c 24 c 11 u 2,22 + c 34 c 11 u 2,33 + c 46 + c 25 c 11 u 2,12 + c 36 + c 45 c 11 u 2,13 + c 23 + c 44 c 11 u 2,23 + c 55 c 11 u 3,11 + c 44 c 11 u 3,22 + c 33 c 11 u 3,33 + 2 c 45 c 11 u 3,12 + c 35 c 11 u 3,13 + c 34 c 11 u 3,23 - β 3 θ , 3 = u ¨ 3 - 2 u 3 + 2 u ˙ 2 ,
(12)
ϵ 3 + ϵ 4 t θ , 11 + ϵ 3 + ϵ 4 k 2 t θ , 22 + ϵ 3 + ϵ 4 k 3 t θ , 33 = θ ¨ + ϵ 0 u ¨ 1,1 + ϵ 1 u ¨ 2,2 + ϵ 2 u ¨ 3,3 ,

where:

ϵ 0 , ϵ 1 , ϵ 2 = β 11 θ 0 ρ c e c 11 β 11 , β 22 , β 33 ,         ϵ 3 , ϵ 4 = 1 ρ c e k * c 1 2 , k 11 c 1 l ,      
k 2 = k 22 k 11 ,       k 3 = k 33 k 11 ,           β 2 = β 22 β 11 ,       β 3 = β 33 β 11 .

Non – dimensional stress components can be calculated as:

(13a)
τ 11 = 1 c 11 c 11 e 11 + c 12 e 22 + c 13 e 33 + 2 c 14 e 23 + c 15 e 13 + c 16 e 12 - θ ,
(13b)
τ 22 = 1 c 11 c 21 e 11 + c 22 e 22 + c 23 e 33 + 2 c 24 e 23 + c 25 e 13 + c 26 e 12 - β 2 θ ,
(13c)
τ 33 = 1 c 11 c 31 e 11 + c 32 e 22 + c 33 e 33 + 2 c 34 e 23 + c 35 e 13 + c 36 e 12 - β 3 θ ,
(13d)
τ 23 = 1 c 11 c 41 e 11 + c 42 e 22 + c 43 e 33 + 2 c 44 e 23 + c 45 e 13 + c 46 e 12 ,
(13e)
τ 13 = 1 c 11 c 51 e 11 + c 52 e 22 + c 53 e 33 + 2 c 54 e 23 + c 55 e 13 + c 56 e 12 ,
(13f)
τ 12 = 1 c 11 c 61 e 11 + c 62 e 22 + c 63 e 33 + 2 c 64 e 23 + c 65 e 13 + c 66 e 12 .

4. Normal mode analysis

The physical variables are decomposed in terms of normal modes to obtain the solution of (11) and (12) (Sarkar and Lahiri [13]) in the following form:

(14)
u i , θ , τ i j x 1 , x 2 , x 3 , t = u i *   , θ *   , τ i j * x i e ω t + i a x 2 + b x 3 ,

where i=-1, ω being angular frequency and a, b are the wave numbers are a and b along x2 and x3 directions respectively.

Using Eq. (14) in Eqs. (11)-(13), we obtained (omitting ‘*’ for convenience):

(15)
u 1,11 + a 11 u 1,1 + a 12 u 1 + a 21 u 2,11 + a 22 u 2,1 + a 23 u 2
            + a 31 u 3,11 + a 32 u 3,1 + a 33 u 3 - θ , 1 = 0 ,
(16)
b 11 u 1,11 + b 12 u 1,1 + b 13 u 1 + u 2,11 + b 21 u 2,1 + b 22 u 2
            + b 31 u 3,11 + b 32 u 3,1 + b 33 u 3 - b 34 θ , 1 = 0 ,
(17)
m 11 u 1,11 + m 12 u 1,1 + m 13 u 1 + m 21 u 2,11 + m 22 u 2,1
            + m 23 u 2 + u 3,11 + m 31 u 3,1 + m 32 u 3 - m 33 θ , 1 = 0 ,
(18)
d 41 u 1,1 + d 44 u 2 + d 46 u 3 + d 48 θ = θ , 11 ,
(19)
τ 11 = u 1,1 + h 12 u 2,1 + h 13 u 3,1 + h 14 u 1 + h 15 u 2 + h 16 h 3 - θ ,
(20)
τ 22 = h 21 u 1,1 + h 22 u 2,1 + h 23 u 3,1 + h 24 u 1 + h 25 u 2 + h 26 u 3 - β 2 θ ,
(21)
τ 33 = h 31 u 1,1 + h 32 u 2,1 + h 33 u 3,1 + h 34 u 1 + h 35 u 2 + h 36 u 3 - β 3 θ ,
(22)
τ 23 = h 41 u 1,1 + h 42 u 2,1 + h 43 u 3,1 + h 44 u 1 + h 45 u 2 + h 46 u 3 ,
(23)
τ 13 = h 51 u 1,1 + h 52 u 2,1 + h 53 u 3,1 + h 54 u 1 + h 55 u 2 + h 56 u 3 ,
(24)
τ 12 = h 61 u 1,1 + h 62 u 2,1 + h 63 u 3,1 + h 64 u 1 + h 65 u 2 + h 66 u 3 ,

where aij, bij, mij, dij and hij(i,j=1, 2, 3) are given in Appendix I.

Equations (15)- (18) can be written in the vector- matrix differential equations (Sarkar and Lahiri [13]) as follows:

(25)
d A ~ d x = A ~   V ~ ,

where:

V ~ = u 1 , u 2 , u 3 , θ , u 1,1 , u 2,1 , u 3,1 , θ , 1 ,
A ~ = L 11 L 12 L 21 L 22 ,

where L11 and L12 are null matrix and identity matrix of order 4×4 and L21 and L22 are given in the Appendix I.

5. Solution of the vector- matrix differential equation: eigenvalue approach

Applying eigenvalue approach method as in Santra et al. [15] to solve the vector-matrix differential equation (25), we get the characteristic equation of the matrix A~ as:

(26)
A ~ - λ I = 0 .

The roots of the Eq. (26) are of the form:

λ = ± λ i ,         i = 1,2 , 3,4 .

The eigenvector X~ corresponding to the eigenvalue λ can be calculated as:

(27)
X ~ = δ 1 = f 24 f 13 - f 14 f 23 f 22 f 33 - f 32 f 23 - f 34 f 23 - f 24 f 33 f 12 f 23 - f 22 f 13 δ 2 = f 34 f 23 - f 24 f 33 f 11 f 23 - f 21 f 13 - f 24 f 13 - f 14 f 23 f 21 f 33 - f 31 f 23 δ 3 = f 12 f 21 - f 11 f 22 f 21 f 34 - f 31 f 24 - f 22 f 31 - f 21 f 32 f 11 f 24 - f 14 f 21 δ 4 = f 11 f 23 - f 21 f 13 f 22 f 33 - f 32 f 23 - f 12 f 23 - f 22 f 13 f 21 f 33 - f 31 f 23 λ δ 1 λ δ 2 λ δ 3 λ δ 4 ,

where fij, (i,j=1, 2, 3) are given in the Appendix I.

The eigenvector X~i [i=1(1)8] corresponding to the eigenvalue λ=λi [i=118] can be calculated from equation (27). For our further reference, we use the following notations:

(28)
X ~ i = X ~ λ = λ i + 1 2   ,         i = 1 2 7 , X ~ λ = λ - i 2   ,         i = 2 2 8 .

As in Lahiri et al. [12], the general solution of Eq. (25) which is regular as xi+ can be written as:

(29)
V ~ = i = 1 4 A i X ~ 2 i e - λ i   x 1 ,             x 1 0 ,

where the arbitrary constants Ai are to be determined from the boundary conditions of the problem and because of regularity condition of the solution at + the terms containing exponential of growing nature in the space variables xi have been neglected.

Thus, the field variables can be written from the Eq. (29) for x10 as:

(30)
u 1   , u 2   , u 3   , θ   x i = i = 1 4 A i δ 1   , δ 2   , δ 3   , δ 4 λ = - λ i   e - λ i   x 1 ,
(31)
τ 11 = i = 1 4 h 14 - λ i δ 1 λ = - λ i   + h 15 - λ i h 12 δ 2 λ = - λ i   + h 16 - λ i h 13 δ 3   λ = - λ i   - δ 4 λ = - λ i   A i e - λ i   x 1 ,
(32)
τ 22 = i = 1 4 h 24 - λ i h 21 δ 1 λ = - λ i   + h 25 - λ i h 22 δ 2 λ = - λ i   + h 26 - λ i h 23 δ 3 λ = - λ i   - β 2 δ 4 λ = - λ i   A i e - λ i   x 1 ,
(33)
τ 33 = i = 1 4 h 34 - λ i h 31 δ 1 λ = - λ i   + h 35 - λ i h 32 δ 2 λ = - λ i   + h 36 - λ i h 33 δ 3   λ = - λ i   - β 3 δ 4 λ = - λ i   A i e - λ i   x 1 ,
(34)
τ 23 = i = 1 4 h 44 - λ i h 41 δ 1 λ = - λ i   + h 45 - λ i h 42 δ 2 λ = - λ i   + h 46 - λ i h 43 δ 3 λ = - λ i   A i e - λ i   x 1 ,
(35)
τ 13 = i = 1 4 h 54 - λ i h 51 δ 1 λ = - λ i   + h 55 - λ i h 52 δ 2 λ = - λ i   + h 56 - λ i h 53 δ 3 λ = - λ i   A i e - λ i   x 1 ,
(36)
τ 12 = i = 1 4 h 64 - λ i h 61 δ 1 λ = - λ i   + h 65 - λ i h 62 δ 2 λ = - λ i   + h 66 - λ i h 63 δ 3 λ = - λ i   A i e - λ i   x 1 .

6. Boundary conditions

To determine the arbitrary constants Ai, boundary conditions are considered at the surface on the half space x1=0 as in Sarkar and Lahiri [13].

Mechanical boundary condition: On the traction, free boundary of the half-space:

(37)
x 1 = 0 ,         τ 11 0 ,   x 2 ,   x 3 ,   t = τ 22 0 ,   x 2 ,   x 3 ,   t = τ 33 0 ,   x 2 ,   x 3 ,   t = 0 .

Thermal boundary condition:

(38)
q n + ν θ = r 0 ,   x 2 ,   x 3 ,   t .

Since, qn=-θ/n, we get νθ*-θ/x1=r* at x1=0, where qn is the normal components of the heat flux vector, ν is the Biot’s number, ν0 corresponding thermally insulated boundary, ν+ corresponding to isothermal boundary. r0, x2, x3, t is the intensity of the applied heat source.

With the help of Eq. (14), Eqs. (37) and (38) become (omitting ‘*’ for convenience):

τ 11 0 ,   x 2 ,   x 3 ,   t = τ 22 0 ,   x 2 ,   x 3 ,   t = τ 33 0 ,   x 2 ,   x 3 ,   t = 0 ,
ν θ - θ x 1 = 0 ,         x 1 = 0 .

7. Numerical analysis

For the purpose of illustrating the problem, we now consider a numerical example for which computational results are presented. Since ω is complex, we take ω=ω0+iς, for studying the effect of wave propagation, we use the following physical parameters in SI units given in the following.

The numerical constants are given by:

c 11 = 16.248 GPa ,       c 14 = - 1.152 GPa ,       c 25 = 1.608 GPa ,       c 22 = 11.88 GPa ,
c 15 = 0 GPa ,       c 26 = 1.248 GPa ,       c 33 = 12.216 GPa ,       c 16 = 0.561 GPa ,
c 32 = 1.032 GPa ,       c 12 = 1.48 GPa ,       c 21 = 1.48 GPa ,       c 34 = 0.672 GPa ,
c 13 = 2.4 GPa ,       c 23 = 1.032 GPa ,       c 35 = 0.216 GPa ,       c 41 = - 1.152 GPa ,
c 24 = 0.912 GPa ,       c 36 = 0.216 GPa ,       c 42 = 0.912 GPa ,       c 51 = 0 GPa ,
c 61 = 0.561 GPa ,       c 43 = - 0.672 GPa ,       c 52 = 1.608 GPa ,       c 62 = 1.248 GPa ,
c 44 = 5.64 GPa ,       c 53 = 0.216 GPa ,       c 63 = - 0.216 GPa ,       c 45 = 2.16 GPa
c 54 = 2.16 GPa ,       c 64 = 0 GPa ,       c 46 = 0 GPa ,       c 55 = 5.88 GPa ,       c 65 = 0 GPa ,
c 56 = 0 GPa ,       c 66 = 6.91 GPa ,       c E = 0.787 ,       β 11 = 7.042 ,       β 11 = 7.046 ,
β 11 = 6.09 ,       k 11 = 0.0921 ,       k 11 = 0.0963 ,       k 11 = 0.0917 ,       θ 0 = 293 .

Variations of the stress components and temperature distribution has been graphically shown from Fig. 1 to Fig. 7 for G-N model of type III comparing with G-N model of type II with rotation and without rotation with respect to x1 for the constant values of t= 0.5 and ω= 0.2.

(1) The nature of the stress component τ11 is contraction in nature for G-N model of type II and III with rotation = 25 whereas without rotation stress is parallel to the x1 axis. Stresses for G-N model type II and III are same and parallel in 0<x1<3 and then gradually decreasing.

(2) Fig. 2 shows that the stress component τ22 is extensive in nature. The maximum value occurs for G-N model of type III with rotation. Stresses are gradually increasing in 0<x1<0.2 and then decreasing.

(3) Nature of the stress τ33 is same as τ22.

Fig. 1. Variation of τ11 with respect to x1 with rotation and without rotation

Fig. 2. Variation of τ22 with respect to x1 with rotation and without rotation

Fig. 3. Variation of τ33 with respect to x1 with rotation and without rotation

(4) In Fig. 4 nature of the stress component τ12 is same for G-N model II and III without ritation. Small variation can be seen for the case of with rotation in the range 0<x1<0.2. and then stresses are gradually decreasing for all the four cases.

(5) In Fig. 5 nature of τ13 is same as τ12 for the case of without rotation. And for the case of with rotation the stresses are parallel to the x1 axis except for a small variation in the range 0.3<x1<0.5 in case of G-N model type III.

Fig. 4. Variation of τ12 with respect to x1 with rotation and without rotation

Fig. 5. Variation of τ13 with respect to x1 with rotation and without rotation

Fig. 6. Variation of τ23 with respect to x1 with rotation and without rotation

(6) Fig. 6 shows that initial value of τ23 for G-N model type II and III with rotation is 0 for x1=0 whereas without rotation the initial values are 2 and 2.2 for G-N model II and III respectively. In the range 0<x1<1, the stress component τ23 monotonically decreases without rotation whereas it increases initially starting from 0, attains a maximum value and then gradually decreases x1>0.3 with rotation.

(7) From Fig. 7 we can see that the temperature is gradually increasing in 0<x1<0.2 and then maximum value occurs at x1= 0.2 for the case of G-N model type III with rotation and after that temperature for all the four cases are gradually decreasing.

Fig. 7. Variation of θ with respect to x1 with rotation and without rotation

8. Conclusions

Thermal stresses and temperature on a traction free boundary in a half space for G-N model type II as well as type III due to time dependent heat source shows significant dependence on rotation as evident from the above curves.

Various stress components deviate under rotation from their rotation less counter parts. Certain stress component possesses non zero initial value when subject to rotation whereas for temperature a converse effect has been found.

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