Two-dimensional generalized thermo-elastic problem for anisotropic half-space

Debkumar Ghosh1 , Ibrahim A. Abbas3 , Abhijit Lahiri2

1, 2Department of Mathematics, Jadavpur University, Kolkata, 700032, India

3Department of Mathematics, Sohag University, Sohag, Egypt

1Corresponding author

Mathematical Models in Engineering, Vol. 3, Issue 1, 2017, p. 27-40. https://doi.org/10.21595/mme.2017.18236
Received 7 February 2017; accepted 11 March 2017; published 30 June 2017

Copyright © 2017 JVE International Ltd.

Abstract.

This paper concerns with the study of wave propagation in fibre reinforced anisotropic half space under the influence of temperature and hydrostatic initial stress. Lord-Shulman theory is applied to the heat conduction equation. The resulting equations are written in the form of vector matrix differential equation by using Normal Mode technique, finally which is solved by Eigen value approach.

Keywords: eigenvalue, generalized thermoelasticity, normal mode, vector-matrix differential equation.

1. Introduction

Fibre-reinforced composite(FRC) materials are usually low weight and high strength used in construction engineering. The physical property of FRC material is governed by the theory of elasticity for different materials with the direction along the direction of fibre. Green [1] studied wave propagation in anisotropic elastic plates. Abbas and Othman [2] discussed the distribution of wave propagation under hydrostatic initial stress of fibre-reinforced materials in anisotropic half-space. Baylies and Green [3] analyse the flexural waves in fibre-reinforced laminated plates. Rogerson [4] discussed effect of penetration in a six-ply composite laminates.

Most of the thermoelasticity and generalized thermoelasticity (coupled or uncoupled) problems have been solved by potential function approach. This method is not always suitable as discussed by Dhaliwal and Sherief [5] and Sherief and Anwar [6]. These may be summarized by the initial conditions and the boundary conditions for physical problems which are directly concern with the material quantities under consideration and not with the potential function. Also, the potential function representations are not convergent always while the physical problems in natural variables constitute convergent solution. So, the alternative method of potential function approach is eigenvalue approach. In this method, we obtain a vector-matrix differential equation from the basic equations which reduces finally to an algebraic eigenvalue problem and the solutions for the field variables are obtained by determining the eigenvalues and eigenvectors from the corresponding coefficient matrix. In this theory, body forces and/or heat sources are also accommodated as in Das and Lahiri [7], Bachher et al. [8]. Now, two different models of generalized thermoelastisities are extensively used. One is Lord and Shulman (L-S) [9] theory and the other is Green and Lindsay (G-L) [10] theory. Introducing one relaxtion time parameter in L-S theory the heat conduction equation becomes hyperbolic type without violating conventional Fourier’s law. Whereas the G-L theory modified the heat conduction equation as well as the equation of motion in coupled thermoelasticitywo relaxation time parameters. There are other three models (Model I, II and III by Green and Nagdhi [11-13]) for generalized thermoelasticity concerned to the theory of with or without energy dissipation.

2. Development of governing equations

The stress-strain relation and the governing equations of motion without body forces and heat sources are written as follow:

(1)
σ i j , j - P ω i j , j = ρ u ¨ i ,
(2)
σ i j = λ e k k δ i j + 2 μ T e i j + α a k a m e k m δ i j + a i a j e k k
      + 2 μ L - μ T a i a k e k j + a j a k e k i + β a k a m e k m a i a j - β i j T - T 0 δ i j ,       i , j , k , m = 1,2 , 3 ,
(3)
e i j = 1 2 u i , j + u j , i ,
(4)
ω i j = 1 2 u j , i - u i , j ,
(5)
K i j T i j = ρ c e T ˙ + t 0 T ¨ + T 0 u ˙ i , j u ¨ i , j ,         i , j = 1,2 , 3 .

We consider the problem of a elastic half-space (x0) in fibre-reinforced anisotropic material with a(a1,a2,a3) where a12+a22+a32=1 as in I. A. Abbas [14], where the displacements are given:

(6)
u = u x = u x , y , t ,       v = u y = v x , y , t ,       w = u z = 0 .

We consider the direction of fibre as a(1,0,0) with x-axis as prefered direction, and Eqs. (1-5), reduces as given:

(7)
σ 11 = λ + 2 α + 4 μ L - 2 μ T + β u x + λ + α v y - β 11 T - T 0 ,
(8)
σ 22 = λ + 2 μ T v y + α + λ u x - β 22 T - T 0 ,
(9)
σ 12 = μ L v x + u y ,
(10)
A 11 2 u x 2 + A 12 + μ L - P 2 2 v x y + μ L + P 2 2 u y 2 - β 11 T x = ρ 2 u t 2 ,
(11)
A 22 2 v y 2 + A 12 + μ L - P 2 2 u x y + μ L + P 2 2 v x 2 - β 22 T y = ρ 2 v t 2 ,
(12)
K 11 2 T x 2 + K 22 2 T y 2 = t + t 0 2 t 2 ρ c e T + T 0 β 11 u x + T 0 β 22 v y ,

with:

A 11 = λ + 2 α + μ L + 4 μ L - μ T + β ,           A 12 = α + λ ,           A 22 = λ + 2 μ T ,
β 11 = 2 λ + 3 α + 4 μ L - 2 μ T + β α 11 + λ + α α 22 ,         β 22 = 2 λ + α α 11 + λ + 2 μ T α 22 ,

where α11, α22 are linear thermal expansion coeeficients.

To transform the above governing equations in non-dimensional forms, we introduce the non-dimensional variables as follows:

(13)
x ' , y ' , u ' , v ' = c 1 χ x , y , u , v ,         t ' = c 1 2 χ t ,         T ' = β 11 T - T 0 ρ c 1 2 ,         χ = ρ c e K 11 ,
σ ' 11 , σ ' 12 , σ ' 22 = 1 ρ c 1 2 σ 11 , σ 12 , σ 22 ,           c 1 2 = A 11 ρ .

Using non-dimensional Eq. (13), the governing equations reduces to (eleminating primes for convenience):

(14)
σ 11 = u x + B 1 v y - T ,
(15)
σ 22 = B 1 u x + B 2 v y - B 3 T ,
(16)
σ 12 = B 4 v x + u y ,
(17)
2 u x 2 + B 1 + B 4 - R p 2 2 v x y + B 4 + R p 2 2 u y 2 - T x = 2 u t 2 ,
(18)
B 2 2 v y 2 + B 1 + B 4 - R p 2 2 u x y + B 4 + R p 2 2 v x 2 - B 3 T y = 2 v t 2 ,
(19)
2 T x 2 + ε 1 2 T y 2 = t + t 0 2 t 2 T + ε 2 u x + ε 3 v y ,

where:

B 1 , B 2 , B 3 = 1 A 11 A 12 , A 22 , μ 1 ,           B 3 = β 22 β 11 ,           R p = P A 11 ,
ε 2 , ε 3 = T 0 β 11 A 11 ρ c e β 11 , β 22 ,         ε 1 = K 11 K 22 .

3. Solution procedure

3.1. Normal mode analysis: formulation of vector-matrix differential equation

For the solution of the Eqs. (14-19), physical variables can be decomposed using normal modes Eq. (20) in the following form:

(20)
u , v , T , σ 11 , σ 12 , σ 22 x , y , t = u * , v * , T * , σ 11 * , σ 12 * , σ 22 * x e ω t + i a y ,

where i=-1, ω is the angular frequency and a is the wave number along x-axis.

Using Eq. (20), Eqs. (14-19) reduces to omitting ‘*’ for convenience:

(21)
σ 11 = d u d x + B 1 i a v - T ,
(22)
σ 22 = B 1 d u d x + B 2 i a v - B 3 T ,
(23)
σ 12 = B 4 d v d x + B 4 i a u ,
(24)
d 2 u d x 2 = M 41 . u + 0 . v + 0 . T + 0 . u ' + M 45 . v ' + T ' ,
(25)
d 2 v d x 2 = 0 . u + M 52 . v + M 53 . T + M 54 . u ' + 0 . v ' + 0 . T ' ,
(26)
d 2 T d x 2 = 0 . u + M 62 . v + M 63 . T + M 64 . u ' + 0 . v ' + 0 . T ' ,

where:

M 41 = a 2 B 4 + R p 2 + ω 2 ,           M 45 = - i a B 1 + B 4 - R p 2 ,       M 52 = a 2 B 2 + ω 2 B 4 + R p 2 ,
M 53 = R p 2 - B 1 - B 4 B 4 + R p 2 ,           M 54 = i a B 3 B 4 + R p 2 ,           M 62 = i a ε 3 ( ω + t 0 ω 2 ) ,
M 63 = ω + t 0 ω 2 + ε 2 a 2 ,           M 64 = ε 2 2 ω + t 0 ω 2 .

Eqs. (24-26) can be written in the form of vector-matrix differential equation as [2, 8]:

(27)
d W d x = A W ,

where W =u      v      T      u'        v'      T'    T and A=L11L12L21L22. Where L11 is null matrix and L12 identity matrix of order 3×3 respectively and L21 and L22 are given by:

L 11 = 0 0 0 0 0 0 0 0 0 ,           L 12 = 1 0 0 0 1 0 0 0 1 ,          
L 21 = M 41 0 0 0 M 52 M 53 0 M 62 M 63 ,           L 22 = 0 M 45 1 M 54 0 0 M 64 0 0 .

3.2. Solution of the vector-matrix differential equation

To solve the vector-matrix differential Eq. (27), we apply the method of eigenvalue approach,

The characteristic equation of the matrix A is given by:

(28)
A - λ I = 0 .

The roots of the characteristic Eq. (28) are λ=λi, i= 1, 2, 3 which are of the form λ=±λ1, λ=±λ2 and λ=±λ3 and they are also eigenvalues of the matrix.

The eigenvector, W corresponding to the eigenvalue λ can obtained as:

(29)
X λ = δ 1             δ 2             δ 3         λ δ 1                 λ δ 2         λ δ 3 T ,

where δ1=a2b3-a3b2, δ2=a3b1-a1b3, δ3=a1b2-a2b1.

As in Lahiri et al. [7], the general solution of Eq. (27) which is regular as can be written as:

(30)
W = i = 1 3 A i X i e - λ i x ,           x 0 .

Hence the field variables can be written as the following:

u = A 1 x 11 e - λ 1 x + A 2 x 21 e - λ 2 x + A 3 x 31 e - λ 3 x ,
v = A 1 x 12 e - λ 1 x + A 2 x 22 e - λ 2 x + A 3 x 32 e - λ 3 x ,
T = A 1 x 13 e - λ 1 x + A 2 x 23 e - λ 2 x + A 3 x 33 e - λ 3 x .

The simplified form of Eqs. (21-23) can be written as:

σ 11 = A 1 R 11 x + A 2 R 12 x + A 3 R 13 x , σ 22 = A 1 R 21 x + A 2 R 22 x + A 3 R 23 x , σ 33 = A 1 R 31 x + A 2 R 32 x + A 3 R 33 x ,

where:

R 11 x = - λ 1 x 11 + B 1 i a x 12 - x 13 e - λ 1 x ,
R 12 x = - λ 2 x 21 + B 1 i a x 22 - x 23 e - λ 2 x ,
R 13 x = - λ 3 x 31 + B 1 i a x 32 - x 33 e - λ 3 x ,
R 21 x = - λ 1 B 1 x 11 + B 2 i a x 12 - B 3 x 13 e - λ 1 x ,
R 21 x = - λ 2 B 1 x 21 + B 2 i a x 22 - B 3 x 23 e - λ 2 x ,
R 21 x = - λ 3 B 1 x 31 + B 2 i a x 32 - B 3 x 33 e - λ 3 x ,
R 31 x = B 4 i a x 11 - λ 1 x 12 e - λ 1 x ,
R 32 x = B 4 i a x 21 - λ 2 x 22 e - λ 2 x ,
R 33 x = B 4 i a x 31 - λ 3 x 32 e - λ 3 x .

4. Boundary conditions

Considering the problem of a half-space ϕ, defined as follows:

ϕ = x , y , z : 0 x ,         - y ,           - z .

In order to determine the arbitrary constants Ai's, i= 1, 2, 3, we consider the boundary conditions as follows.

4.1. Case 1

a) Mechanical Boundary condition:

For stress-free surface x=0, σ11=0, σ12=0.

b) Thermal Boundary condition:

(31)
ν T - d T d x = r ,

where ν is Biot’s number.

4.2. Case 2

a) Mechanical Boundary condition:

For stress-free surface x= 0, σ11=-P1+P2eωt+iay, σ12=0.

b) Thermal Boundary condition:

(32)
T = P 3 e ω t + i a y .

5. Numerical analysis

5.1. Case 1

5.1.1. Distribution of different stress components

Fig. 1 represents distribution of normal stress σ11 for y= 0.3.

For fixed time t, σ11 gradually increases as x increases. For fixed x numerical values of σ11 gradually decreases as t increases.

Fig. 2 represents distribution of normal stress σ12 for y= 0.2.

For fixed time t, σ12 gradually decreases as x increases. For fixed x numerical values of σ12 gradually increases as t increases.

Fig. 3 represents distribution of normal stress σ22 for y= 0.5.

For fixed time t, σ22 gradually decreases as x increases. For fixed x numerical values of σ12 gradually increases as t increases.

Fig. 1. Stress component σ11 at y= 0.3 for different values of t verses x

Fig. 2. Stress component σ12 at y= 0.2 for different values of t verses x

Fig. 4 represents distribution of normal stress σ11 for different values of x and y for fixed t= 0.1 and ω= 0.5.

x numerical value of σ11 gradually decreases as y increases. For fixed y the numerical value of σ11 gradually increases as x increases. σ11 is maximum when x= 1 and y= 0.

Fig. 5 represent distribution of normal stress σ12 for different values of x and y for fixed t= 0.4 and ω= 5.

For fixed x numerical value of σ12 gradually decreases as y increases. For fixed y the numerical value of σ12 gradually decreases as x increases. Significant changes occur in the region 0.2 x 0.6 and 0.6 y 1.0.

Fig. 6 represent distribution of stress component σ22 at for different values of x and y for fixed t= 0.1 and 0.1.

Fig. 4. Stress component σ11 at t= 0.1 and ω= 0.5 verses x and y

Fig. 5. The variation of stress component σ12 at t= 0.4 and ω= 3 verses x and y

Fig. 6. Stress component σ22 at t= 0.1 and ω= 0.1 verses x and y

For fixed x numerical value of σ22 gradually increases as y increases. For fixed y numerical value of σ22 gradually increases as x increases.

Fig. 7 represent distribution of normal stress σ12 for different values of x and t for fixed y= 0.2 and 1.

For fixed x, nominal decreasing of numerical values of σ12 has been seen as t increases, while For fixed t, numerical values of σ12 decreases gradually as x increases. numerical values of σ12 minimum at x= 1 and 0.02 t 0.1

Fig. 8 represent distribution of normal stress σ22 for different values of x and t for fixed y= 0.5 and 2.

For fixed x, nominal decreasing of σ22 has been seen as t increases. For fixed t, numerical values of σ12 decreases as x increases. Also, significant changes occur in the region 0.6 x 1.0 and 0 t 1.0.

Fig. 7. Stress component σ12 at y= 0.2 and ω= 1 verses x and t

Fig. 8. The distribution of stress component σ22 at y= 0.5 and ω= 2 verses x and t

5.1.2. Distribution of temperature

Fig. 9 represent distribution of temperature, T for different values of x and y for fixed t= 0.3 and 2.

For fixed x numerical value of T gradually decreases as y increases. For fixed y numerical value of T gradually increases as x increases. T in minimum at x= 0 and significant changes occurs in the region 0.6 x 1.0 and 0 y 1.0

Fig. 10 represent distribution of temperature, T for different values of x and t for fixed y= 0.1 and 1.5.

Fig. 9. The variation of T at t= 0.3 and ω= 2 verses x and y

For fixed t numerical value of T nominally increases as x increases. For fixed x numerical value of T gradually increases as t increases.

Fig. 11 represent distribution of temperature, T for different values of y and t for fixed x= 0.5 and 3.

For fixed t numerical value of T nominally increases as y increases. For fixed x numerical values of T decreases as t increases.

Fig. 10. Variation of T at y= 0.1 and ω= 1.5 verses x and t

Fig. 11. Variation of T at x= 0.5 and ω= 3 verses y and t

5.2. Case 2

5.2.1. Distribution of different stress components

Fig. 12 represents distribution of normal stress σ11 for y= 0.3, t= 0.01 and 5 for different numerical values of Rp.

For fixed Rp, σ11 gradually decreases as x increases. For fixed x numerical values of σ11 gradually decreases as Rp increases.

Fig. 13 represents distribution of normal stress σ11 for y= 0.3, t= 0.01 and 5 for different fractional values of Rp.

For fixed time Rp, σ11 gradually increases as x increases. For fixed x numerical values of σ11 gradually decreases as Rp increases. Significant changes occurred for 0 x 0.4.

Fig. 14 represents the distribution of stress component σ12 at y= 0.2, t= 0.03 and 0.3 for different fractional values Rp of verses x for P1= 1.

For fixed time Rp, σ11 gradually decreases as x increases. For fixed x numerical values of σ11 increases as Rp increases. Significant changes occurred for 0 x 0.4.

Fig. 15 represents the distribution of stress component σ12 at y= 0.3, t= 0.1 and 4 for different integral values Rp for P1= 0.

For fixed Rp, σ12 gradually decreases as x increases. For fixed x numerical values of σ12 increases as Rp increases.

Fig. 16 represents the distribution of stress component σ22 at y= 0.2, t= 0.6 and 0.4 for different values Rp for P1= 0.

For fixed Rp= 0.9, σ22 gradually increases as x increases but for other fixed values of Rp, σ22 gradually decreases as x increases. For fixed x, σ22 gradually increases as x increases for Rp= 0.9 but for other fixed values of Rp, σ22 gradually decreases as x increases.

Fig. 17 represents the distribution of stress component σ11 for different values of t for fixed y= 0.2 and 1.5.

Fig. 12. Stress component σ11 at y= 0.3, t= 0.01 and ω= 5 for different values Rp of verses x

Fig. 13. Stress component σ11 at y= 0.2 and ω= 0.2 for different values Rp of verses x

Fig. 14. Stress component σ12 at y= 0.2, t= 0.03 and ω= 0.3 for different values Rp of verses x for P1= 1

Fig. 15. Stress component σ12 at y= 0.3, t= 0.1 and ω= 4 for different values Rp of verses x for P1= 0

For fixed t, the numerical value of σ11 gradually increases as x increases. For fixed x, the numerical value of σ11 gradually increases as t increases.

Fig. 18 The distribution of stress component σ12 for fixed y= 0.3 and 0.5 for different values of t.

For fixed t, the numerical value of σ12 gradually decreases as x increases. For fixed x, the numerical value of σ12 gradually increases as t increases.

Fig. 16. Stress component σ22 at y= 0.2, t= 0.6 and ω= 0.4 for different values Rp of verses x for P1= 0

Fig. 17. Stress component σ11 at y= 0.2 and ω= 1.5 for different values of t verses x

Fig. 18. Stress component σ12 at y= 0.3 and ω= 0.5 for different values of t verses x

Fig. 19. Stress component σ22 at y= 0.5 and ω= 0.2 for different values of t verses x

Fig. 19 represents the distribution of stress component σ22 at fixed y= 0.5 and 0.2 for different values of t.

For fixed t, the numerical value of σ12 gradually increases as x increases, but for fixed x, the numerical value of σ12 gradually decreases as t increases.

Fig. 20 represents distribution of normal stress σ11 for different values of x and for fixed t= 0.3 and y= 0.3.

For fixed x numerical value of σ11 remain constant as increases, but for fixed ω the numerical value of σ11 gradually decreases as x increases.

Fig. 21 represents distribution of normal stress σ12 for different values of x and for fixed y= 0.3 and t= 0.5.

For fixed x, nominal increasing of numerical values of σ12 has been seen as increases, while for fixed ω, numerical values of σ12 increases gradually as x increases after x= 0.6 (approx.). Numerical values of σ12 minimum at x= 0.5 (approx.) and 1 t 3.

Fig. 22 represents distribution of stress component σ22 at for different values of x and for fixed t= 0.1 and y= 0.4.

For fixed ω numerical value of σ22 gradually decreases as x increases. Numerical values of σ22 minimum at x= 1 and 0 ω 1.

Fig. 20. Stress component σ11 at y= 0.3 and t= 0.3 verses x and ω

Fig. 21. Stress component σ12 at y= 0.3 and t= 0.5 verses x and ω

Fig. 22. Stress component σ22 at y= 0.4 and t= 0.1 verses x and ω

5.2.2. Distribution of temperature

Fig. 23 represents distribution of temperature, T for different values of x and ω for fixed t= 0.1 and y= 0.4.

For fixed x numerical value of T gradually increases in the region 0 ω 0.3 (approx.) For fixed ω numerical value of T nominally increases as x increases.

Fig. 23. The variation of T at y=0.4 and t= 0.1 verses x and ω

6. Conclusion

We consider the physical parameters in SI units given in Dhaliwal and Singh following below to obtain the numerical result to observe the effect of wave propagation:

ρ = 2660 k g / m 3 ,       λ = 5.65 × 1 0 10 N / m 2 ,
μ T = 2.46 × 1 0 10 N / m 2 ,       μ L = 5.66 × 1 0 10 N / m 2 ,
α = - 1.28 × 1 0 10 N / m 2 ,       β = 220.90 × 1 0 10 N / m 2 ,
α 11 = 0.017 × 1 0 - 4 d e g - 1 ,       α 22 = 0.015 × 1 0 - 4 d e g - 1 ,
l = 0.5 ,       T 0 = 293 K ,       c e = 0.787 × 1 0 3 J K g - 1 d e g - 1
K 11 = 0.0921 × 1 0 10 J m - 1 s - 1 d e g - 1
K 22 = 0.0963 × 1 0 10 J m - 1 s - 1 d e g - 1
P 1 = 0   o r   1 ,       P 2 = 0.1 ,         P 3 = 0.2 .

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