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A two-dimensional problem for a half-space in magneto-thermoelasticity with thermal relaxation
International Journal of Engineering Science 40 (2002) 587–604 www.elsevier.com/locate/ijengsci
A two-dimensional problem for a half-space in magneto-thermoelasticity with thermal relaxation Hany H. Sherief *, Kamal A. Helmy Department of Mathematics, Faculty of Sciences, University of Alexandria, Alexandria, Egypt Received 29 November 1999; accepted 11 February 2000 (Communicated by E.S. S ß UHUBI_ )
Abstract In this work we study a two-dimensional problem in electromagneto-thermoelasticity for a half-space whose surface is subjected to a non-uniform thermal shock and is stress free in the presence of a transverse magnetic field. The problem is in the context of the theory of generalized thermoelasticity with one relaxation time. Laplace and exponential Fourier transform techniques are used to obtain the solution by a direct approach. The solution of the problem in the physical domain is obtained by using a numerical method for the inversion of the Laplace transforms based on Fourier series expansions. The distributions of the temperature, the displacement, the stress and the induced magnetic and electric fields are obtained. The numerical values of these functions are represented graphically. Ó 2002 Elsevier Science Ltd. All rights reserved.
1. Introduction The theory of generalized thermoelasticity with one relaxation time was introduced by Lord and Shulman [1] for isotropic materials. This theory was extended by Sherief [2] and by Dhaliwal and Sherief [3] to include the effects of anisotropy. In this theory a modified law of heat conduction including both the heat flux and its time derivative replaces the conventional Fourier’s law. The heat equation associated with this theory is a hyperbolic one and hence automatically eliminates the paradox of infinite speeds of propagation inherent in both the uncoupled and the coupled theories of thermoelasticity. For many problems involving steep heat gradients and when short time effects are sought this theory is indispensable.
*
Corresponding author. E-mail address: [email protected] (H.H. Sherief).
0020-7225/02/$ - see front matter Ó 2002 Elsevier Science Ltd. All rights reserved. PII: S 0 0 2 0 - 7 2 2 5 ( 0 0 ) 0 0 0 9 3 - 8
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Due to the complexity of the partial differential equations of this theory, the work done in this field is unfortunately limited in number. Among the theoretical contributions to the subject are the proofs of uniqueness theorems under different conditions by Ignaczak [4,5] and by Sherief [6]. Anwar and Sherief [7] did the state space formulation for problems not containing heat sources and the boundary element formulation was done by Anwar and Sherief [8]. Some concrete problems have also been solved. The fundamental solutions for spherically symmetric spaces were obtained by Sherief [9]. Sherief and Anwar have solved some two-dimensional problems in [10,11], while Sherief and Hamza have solved some two-dimensional problems and studied the wave propagation in this theory in [12,13], respectively. Increasing attention is being devoted to the interaction between magnetic fields and strain in a thermoelastic solid due to its many applications in the fields of geophysics, plasma physics and related topics. In the nuclear field, the extremely high temperatures and temperature gradients as well as the magnetic fields originating inside nuclear reactors influence their design and operations [14]. Usually, in these investigations the heat equation under consideration is taken as the uncoupled or the coupled equation, not the generalized one. This attitude is justified in some situations where the solutions obtained using any of these equations differ little quantitatively. However, when short time effects are considered, the full, generalized system of equations has to be used or a great deal of accuracy is lost [1]. A comprehensive review of the earlier contributions to the subject can be found in [15]. Among the authors who considered the generalized magneto-thermoelastic equations are Nayfeh and Nemat-Nasser [16] who studied the propagation of plane waves in a solid under the influence of an electromagnetic field. They have obtained the governing equations in the general case and the solution for some particular cases. Choudhuri [17] extended these results to rotating media. Sherief [18] has solved a problem for a solid cylinder, while Sherief and Ezzat [19] have solved a thermal shock half-space problem using asymptotic expansions.
2. Formulation of the problem We consider the problem of a thermoelastic half-space (x P 0). A magnetic field with constant intensity H0 acts parallel to the bounding plane (taken as the direction of the z-axis). The surface of the half-space is subjected at time t ¼ 0 to a thermal shock that is a function of y and t. Thus, all the quantities considered will be functions of the time variable t and of the coordinates x and y. Due to the application of the primary magnetic field H0 , there results an induced magnetic field h and an induced electric field E. We assume that both h and E are small in magnitude in accordance with the assumptions of the linear theory of thermoelasticity. Thus, the displacement vector will have the components ux ¼ uðx; y; tÞ;
uy ¼ vðx; y; tÞ;
uz ¼ 0:
The components of the magnetic intensity vector are Hx ¼ 0;
Hy ¼ 0;
Hz ¼ H0 þ hðx; y; tÞ:
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
589
The electric intensity vector is normal to both the magnetic intensity and the displacement vectors. Thus, E has the components Ex ¼ E1 ;
Ey ¼ E2 ;
Ez ¼ 0:
The current density vector J must be parallel to E, thus Jx ¼ J1 ;
Jy ¼ J2 ;
Jz ¼ 0:
Maxwell’s equations in vector form can be written as curl h ¼ J þ
curl E ¼
div B ¼ 0;
oD ; ot
ð1Þ
oB ; ot
ð2Þ
div D ¼ qe ;
B ¼ l0 ðH0 þ hÞ;
D ¼ e0 E;
ð3Þ ð4Þ
where 0 and l0 are the electric and magnetic permeabilities, respectively, and qe is the density of electric charges. The above equations are supplemented by Ohm’s law, namely ou J ¼ r0 E þ l0 H0 ; ð5Þ ot where r0 is the electric conductivity. Ohm’s law (5) after linearization gives ov J1 ¼ r0 E1 þ l0 H0 ; ot J2 ¼ r0
ou E2 l0 H0 : ot
Eqs. (1) and (6a), (6b) in our case give the two equations oh ov oE1 ¼ r0 E1 þ l0 H0 ; þ e0 oy ot ot oh ou oE2 ¼ r0 E2 l0 H0 e0 : ox ot ot
ð6aÞ
ð6bÞ
ð7Þ
ð8Þ
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H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
Eqs. (2) and (4) have one non-vanishing component, namely oE1 oE2 oh ¼ l0 : ot oy ox
ð9Þ
The strain components are given by, ou exx ¼ ; ox
ov eyy ¼ ; oy
1 exy ¼ 2
ou ov þ ; oy ox
exz ¼ eyz ¼ ezz ¼ 0:
The cubical dilatation e is thus given by e ¼ exx þ eyy þ ezz ¼
ou ov þ : ox oy
ð10Þ
Stress components are given by the tensor relation rij ¼ 2leij þ kedij cðT T0 Þdij ; which has the Cartesian components rxx ¼ ðk þ 2lÞ
ou ov þ k cðT T0 Þ; ox oy
ð11aÞ
ryy ¼ ðk þ 2lÞ
ov ou þ k cðT T0 Þ; oy ox
ð11bÞ
ou ov þ ; rxy ¼ k oy ox
ð11cÞ
where k and l are Lame’s modulii, T the absolute temperature of the medium, and c is a material constant given by c ¼ ð3k þ 2lÞat ; at being the coefficient of linear thermal expansion. T0 is a reference temperature assumed to be such that jðT T0 Þ=T0 j 1 . Equations of motion have the form, ui ; rji;j þ Fi ¼ q€
ð12Þ
where q is the density and F is the Lorentz force given by, F ¼ J B:
ð13Þ
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
From Eqs. (13) and (6a), (6b), we obtain ou Fx ¼ l0 r0 H0 E2 l0 H0 ; ot ov ; Fy ¼ l0 r0 H0 E1 þ l0 H0 ot
591
ð13aÞ ð13bÞ
Fz ¼ 0:
ð13cÞ
Substituting from Eqs. (13) into Eq. (10), we obtain the equations of motion in the form, o2 u oe oT ou 2 þ l0 r0 H0 E2 l0 H0 ; ð14Þ q 2 ¼ ðk þ lÞ þ lr u c ot ox ox ot o2 v oe oT ov 2 q 2 ¼ ðk þ lÞ þ lr v c l0 r0 H0 E1 þ l0 H0 : ot oy oy ot
ð15Þ
The equation of heat conduction has the form, 2
kr T ¼
o o2 þ s0 2 ðqcE T þ cT0 eÞ; ot ot
ð16Þ
where k is the thermal conductivity of the medium, cE the specific heat at constant strain and s0 is the relaxation time. For simplifications we shall use the following non-dimensional variables: x ¼ c1 gx;
y ¼ c1 gy;
u ¼ c1 gu;
t ¼ c21 gt;
s 0 ¼ c21 gs0 ;
E1 ¼
v ¼ c1 gv;
gE1 ; r0 l20 H0 c1
E2 ¼
r ij ¼
rij ; l
gE2 ; r0 l20 H0 c1
cðT T0 Þ ; k þ 2l gh h ¼ ; r0 l0 H0 h¼
where g ¼ qcE =k; c21 ¼ ðk þ 2lÞ=q: Using the above non-dimensional variables, the governing equations reduce to (dropping the asterisks for convenience) oe ou o2 u 2 2 oh 2 2 ð17Þ þ b me2 mE2 ¼ b2 2 ; ðb 1Þ þ r u b ox ox ot ot oe ov o2 v 2 oh 2 2 ðb 1Þ þ r v b b me2 mE1 þ ¼ b2 2 ; oy oy ot ot 2
2
r h¼
o o2 þ s0 2 ðh þ e1 eÞ; ot ot
ð18Þ
ð19Þ
592
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
oh oE1 ov ¼ mE1 þ V 2 þ ; oy ot ot
ð20Þ
oh oE2 ou ¼ mE2 V 2 þ ; ox ot ot
ð21Þ
oE1 oE2 oh ¼ ; ot oy ox
ð22Þ
where m¼
r0 l0 ; g
e2 ¼
l0 H02 ; k þ 2l
e1 ¼
T0 c2 ; qcE ðk þ 2lÞ
V ¼
c1 ; c
c2 ¼
1 ; e0 l0
b2 ¼
k þ 2l : l
The constitutive equations reduce to rxx ¼ 2
ou 2 b 2 e b2 h; ox
ð23aÞ
ryy ¼ 2
ov 2 b 2 e b2 h; oy
ð23bÞ
rxy ¼
ou ov þ : oy ox
ð23cÞ
3. Formulation and solution in the Laplace transform domain Taking the Laplace transform defined by the relation fðx; y; sÞ ¼
Z
1
est f ðx; y; tÞ dt 0
of both sides of Eqs. (17)–(22), (23a)–(23c), we get ðb2 1Þ
o e oh þ r2 u b2 þ b2 me2 mE2 s u ¼ b2 s2 u; ox ox
ð24Þ
ðb2 1Þ
o e oh þ r2 v b2 b2 me2 mE1 þ s v ¼ b2 s2 v; oy oy
ð25Þ
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
r2 h ¼ s þ s0 s2 h þ e1 e ;
593
ð26Þ
oh ¼ m þ V 2 s E1 þ s v; oy
ð27Þ
oh ¼ m þ V 2 s E2 þ s u; ox
ð28Þ
oE1 oE2 ¼ sh; oy ox
ð29Þ
rxx ¼ 2
o u 2 b 2 e b2 h; ox
ð30aÞ
ryy ¼ 2
o v 2 b 2 e b2 h; oy
ð30bÞ
rxy ¼
o u o v þ : oy ox
ð30cÞ
Differentiating Eq. (24) with respect to x and Eq. (25) with respect to y, adding, we obtain upon using Eq. (29) 2 r s2 e2 ms e r2 h e2 m2 sh ¼ 0:
ð31Þ
Eliminating E1 and E2 between Eqs. (27)–(29), we get 2 r sðm þ V 2 sÞ h ¼ s e:
ð32Þ
Eliminating h and h between Eqs. (26), (31) and (32), we obtain the following sixth-order partial differential equation satisfied by e ðr6 Ar4 þ Br2 CÞ e ¼ 0; where A ¼ s sðe1 s0 þ s0 þ V 2 þ 1Þ þ e1 þ e2 m þ m þ 1 ;
ð33Þ
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H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
B ¼ s2 s2 ½e1 s0 V 2 þ s0 ðV 2 þ 1Þ þ V 2 þ s½e1 ðms0 þ V 2 Þ þ e2 mðs0 þ V 2 Þ þ mðs0 þ 1Þ þ V 2 þ 1 þ e1 m þ e2 m þ m ; C ¼ s4 ð1 þ s0 sÞðsV 2 þ e2 mV 2 þ mÞ: In a similar manner we can show that h and h satisfy the equations ðr6 Ar4 þ Br2 CÞh ¼ 0;
ð34Þ
ðr6 Ar4 þ Br2 CÞh ¼ 0:
ð35Þ
Eq. (33) can be factorized as ðr2 k12 Þðr2 k22 Þðr2 k32 Þ e ¼ 0;
ð36Þ
where k12 ; k22 and k32 are the roots of the characteristic equation k 6 Ak 4 þ Bk 2 C ¼ 0:
ð37Þ
The solution of Eq. (36) has the form e ¼
3 X
ð38Þ
ei ;
i¼1
where ei is the solution of the equation ðr2 ki2 Þ ei ¼ 0;
i ¼ 1; 2; 3:
ð39Þ
In order to solve Eq. (39), we shall use the exponential Fourier transform with respect to the variable y (denoted by an asterisk) and defined by the relation Z 1 1
p ffiffiffiffiffi ffi f ðx; q; sÞ ¼ eiqy f ðx; y; sÞ dy; 2p 1 whose inversion is given by the relation Z 1 1 eiqy f ðx; q; sÞ dq: f ðx; y; sÞ ¼ pffiffiffiffiffiffi 2p 1 Applying the exponential Fourier transform to both sides of Eq. (39), we get e i ¼ 0; ðD2 q2i Þ where D ¼ o=ox and q2i ¼ q2 þ ki2 ; i ¼ 1; 2; 3.
ð40Þ
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
595
The solution of Eq. (40), which is bounded for x > 0, is given by e i ðx; q; sÞ ¼ Ai ðq; sÞeqi x ; where Ai ðq; sÞ is a parameter depending on q and s. Thus, e is given by the relation e ðx; q; sÞ ¼
3 X
Ai ðq; sÞeqi x :
ð41Þ
i¼1
In a similar manner, we can write h ðx; q; sÞ ¼
3 X
A0i ðq; sÞeqi x ;
ð42Þ
A00i ðq; sÞeqi x ;
ð43Þ
i¼1
h ðx; q; sÞ ¼
3 X i¼1
where A0i ðq; sÞ and A00i ðq; sÞ are parameters depending on q and s. Substituting from Eqs. (41)–(43) into the Fourier transform of Eqs. (26) and (32) we get the following relations: A0i ¼
A00i ¼
e1 sð1 þ s0 sÞ Ai ; sð1 þ s0 sÞ
ki2
ki2
s Ai ; sðm þ V 2 sÞ
i ¼ 1; 2; 3;
i ¼ 1; 2; 3:
ð44Þ
ð45Þ
Substituting Eqs. (44) and (45) into Eqs. (42) and (43), respectively, we obtain h ðx; q; sÞ ¼
h ðx; q; sÞ ¼
3 X e1 sð1 þ s0 sÞ Ai eqi x ; 2 k sð1 þ s sÞ 0 i i¼1
3 X
ki2 i¼1
s Ai eqi x : sðm þ V 2 sÞ
ð46Þ
ð47Þ
In order to obtain the displacement component u, we take the exponential Fourier transform of both sides of Eqs. (24) and (28) with respect to the variable y, to get u ¼ b2 ðD2 q2 b2 s2 b2 me2 sÞ
oh
o e
b2 m2 e2 E2 ðb2 1Þ ; ox ox
ð48Þ
596
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
# "
1 o h : E2 ¼ s u m þ V 2s ox
ð49Þ
Eliminating E2 between Eqs. (48) and (49) and using Eqs. (41), (46) and (47), we obtain the following equation satisfied by u
3 X e1 sð1 þ s0 sÞ e2 m2 s 2 D n u ¼ 1þb þ 1 Ai eqi x ; 2 2 sÞðk 2 sðm þ V 2 sÞÞ sð1 þ s sÞ k ðm þ V 0 i i i¼1 2
2
ð50Þ
where sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b2 s2 me2 V 2 : n ¼ q2 þ b2 s2 þ m þ V 2s In order to simplify the right-hand side of Eq. (50), we substitute from Eqs. (41), (46) and (47) into the Fourier transform of Eq. (31), to get the compatibility condition e1 sð1 þ s0 sÞ ki2 s2 e2 ms e2 m2 s2 ¼ : ki2 sð1 þ s0 sÞ ki2 ki2 ðki2 sðm þ V 2 sÞÞ
ð51Þ
Using Eq. (51), Eq. (50) reduces to
3 X b2 s2 ðm þ V 2 s þ e2 mV 2 Þ 1 qi Ai eqi x : D n u ¼ 2 2 sÞ k ðm þ V i i¼1 2
2
ð52Þ
The solution of Eq. (52), bounded as x ! 1 is given by u ðx; q; sÞ ¼ Bq2 enx
3 X qi Ai qi x e ; ki2 i¼1
where B ¼ Bðq; sÞ is some parameter depending on q and s. Taking the Laplace and exponential Fourier transforms of Eq. (10), we get # "
i o u : e v ¼ q ox
ð53Þ
ð54Þ
Substituting from Eqs. (41) and (53) into the right-hand side of Eq. (54), we obtain "
# 3 X A i qi x v ðx; q; sÞ ¼ iq nBenx : e 2 k i i¼1
ð55Þ
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
Substituting from Eqs. (47) and (53) into the right-hand side of Eq. (49), we get " # 3 X s A q i i E2 ¼ Bq2 enx þ sðm þ V 2 sÞ eqi x : 2 2 2 sÞÞ m þ V 2s k ð k sðm þ V i i i¼1
597
ð56Þ
Taking the exponential Fourier transform of both sides of Eq. (27), we obtain E1 ¼
i 1 h
iq h s v : m þ V 2s
Substituting from Eqs. (47) and (55) into Eq. (57), we obtain " # 3 X iqs A i nBenx þ sðm þ V 2 sÞ E1 ¼ eqi x : 2 2 2 sÞÞ m þ V 2s k ð k sðm þ V i i i¼1
ð57Þ
ð58Þ
Substituting from Eqs. (46), (47), (53) and (55) into Eqs. (30a)–(30c), we obtain the transformed stress components in the form ( ) 2 3 X 2q2 b ki2 sð1 þ s0 sÞðe1 þ 1Þ
2 nx Ai eqi x ; þ ð59Þ rxx ðx; q; sÞ ¼ 2nBq e þ 2 2 k k sð1 þ s sÞ 0 i i i¼1 (
) 3 X q A i i qi x ðn2 þ q2 ÞBenx : r xy ðx; q; sÞ ¼ iq 2 e 2 k i i¼1
ð60Þ
In order to apply the boundary conditions of the problem to find the unknown parameters, we need to find the magnetic and electric field intensities in free space. We denote these by h0 , E10 and E20 , respectively. These quantities satisfy the non-dimensional field equations oh0 oE10 ¼V2 ; oy ot
ð61Þ
oh0 oE20 ¼ V 2 ; ox ot
ð62Þ
oh0 oE10 oE20 ¼ : ot oy ox
ð63Þ
Taking the Laplace and the exponential Fourier transforms of both sides of Eqs. (51)–(63), and solving the resulting equations, we obtain the solutions bounded for x < 0 as h 0 ¼ Cemx ;
ð64Þ
598
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
iq E10 ¼ 2 Cemx ; V s
ð65Þ
m
¼ 2 Cemx ; ð66Þ E20 V s pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi where m ¼ q2 þ V 2 s2 . We have thus obtained the solution of the problem in the transformed domain in terms of five unknown parameters A1 ; A2 ; A3; B and C. These parameters are to be determined from the boundary conditions at x ¼ 0 which are: (1) Thermal boundary condition that the temperature at the surface of the half-space is known hð0; y; tÞ ¼ f ðy; tÞ: (2) Mechanical boundary conditions that the surface of the half-space is traction free rxx ð0; y; tÞ ¼ rxy ð0; y; tÞ ¼ 0: (3) The transverse components of the electric field intensity are continuous across the surface of the half-space E2 ð0; y; tÞ ¼ E20 ð0; y; tÞ: (4) The transverse components of the magnetic field intensity are continuous across the surface of the half-space hð0; y; tÞ ¼ h0 ð0; y; tÞ: The above conditions, in the transformed, domain can be written as 3 X i¼1
1 f ðq; sÞ ¼ A ; i ki2 sð1 þ s0 sÞ e1 sð1 þ s0 sÞ
2nq2 B þ
3 X
(
i¼1
ðn2 þ q2 ÞB þ 2
3 X i¼1
) 2 2q2 b ki2 sð1 þ s0 sÞðe1 þ 1Þ þ Ai ¼ 0; ki2 sð1 þ s0 sÞ ki2 3 X qi Ai ¼ 0; ki2 i¼1
Ai qi Bq2 mC þ 2 3 ¼ 0; þ 2 2 2 2 ki ðki sðm þ V sÞÞ sðm þ V sÞ V s
ð67Þ
ð68Þ
ð69Þ
ð70Þ
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
s
3 X
ki2 i¼1
1 Ai C ¼ 0: sðm þ V 2 sÞ
599
ð71Þ
Eqs. (67)–(71) constitute a system of five linear equations with five unknown parameters A1 ; A2 ; A3; B and C.
4. Inversion of the double Laplace–Fourier transform We shall now outline the method used to invert the double Laplace–Fourier transforms in the above equations. Let f ðx; q; sÞ be the Laplace–Fourier transform of a function f ðx; y; tÞ. We shall first use the inversion formula of the Fourier transform mentioned above to obtain a function fðx; y; sÞ that is the Laplace transform of the function f ðx; y; tÞ. Next, the inversion formula for Laplace transforms can be written as 1 f ðx; y; tÞ ¼ 2pi
Z
dþi1
est fðx; y; sÞ ds; di1
where d is an arbitrary real number greater than all the real parts of the singularities of fðx; y; sÞ. Taking s ¼ d þ iz, the above integral takes the form edt f ðx; y; tÞ ¼ 2p
Z
1
eitz fðx; y; d þ izÞ dz:
1
Expanding the function hðx; y; tÞ ¼ expðdtÞf ðx; y; tÞ in a Fourier series in the interval ½0; 2L , we obtain the approximate formula [20] f ðx; y; tÞ ¼ f1 ðx; y; tÞ þ ED ; where 1 X 1 f1 ðx; y; tÞ ¼ c0 þ ck 2 k¼1
for 0 6 t 6 2L
ð72Þ
and ck ¼
i edt h ikpt=L Re e f ðx; y; d þ ikp=LÞ ; L
ð73Þ
ED , the discretization error, can be made arbitrarily small by choosing d large enough [20]. As the infinite series in (72) can only be summed up to a finite number N of terms, the approximate value of f ðx; y; tÞ becomes
600
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604 N X 1 fN ðx; y; tÞ ¼ c0 þ ck 2 k¼1
for 0 6 t 6 2L:
ð74Þ
Using the above formula to evaluate f ðx; y; tÞ, we introduce a truncation error ET that must be added to the discretization error to produce the total approximation error. Two methods are used to reduce the total error. First, the ‘Korrecktur’ method is used to reduce the discretization error. Next, the e-algorithm is used to reduce the truncation error and therefore to accelerate convergence. The Korrecktur method uses the following formula to evaluate the function f ðx; y; tÞ 0 ; f ðx; y; tÞ ¼ f1 ðx; y; tÞ e2dL f1 ðx; y; 2L þ tÞ þ ED 0 ED . where the discretization error ED Thus, the approximate value of f ðx; y; tÞ becomes
fNK ðx; y; tÞ ¼ fN ðx; y; tÞ e2dL fN 0 ðx; y; 2L þ tÞ:
ð75Þ
N 0 is an integer such that N 0 < N . We shall now describe the e-algorithm that is used to accelerate the convergence of the series in (74). Let N be an odd natural number and let sm ¼
m X
ck ;
k¼1
be the sequence of partial sums of (74). We define the e-sequence by e0;m ¼ 0;
e1;m ¼ sm ; m ¼ 1; 2; 3; . . . :
and enþ1;m ¼ en1;mþ1 þ
1 ; en;mþ1 en;m
n; m ¼ 1; 2; 3; . . .
It can be shown that [20] the sequence e1;1 ; e3;1 ; . . . ; eN ;1 converges to f ðx; y; tÞ þ ED c0 =2 faster than the sequence of partial sums sm ;
m ¼ 1; 2; 3; . . .
The actual procedure used to invert the Laplace transforms consists of using Eq. (75) together with the e-algorithm. The values of d and L are chosen according the criteria outlined in [20].
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
601
5. Numerical results The copper material was chosen for purposes of numerical evaluations. The constants of the problem were taken as e1 ¼ 0:0168;
e2 ¼ 0:0008;
V ¼ 1:39ð10Þ5 ; 10
k ¼ 7:76ð10Þ ;
q ¼ 9854;
b2 ¼ 4;
T0 ¼ 293;
g ¼ 8886:73; 10
l ¼ 3:86ð10Þ ;
m ¼ 0:008;
cE ¼ 381; 9
e0 ¼ ð10Þ =ð36pÞ;
s0 ¼ 0:02;
at ¼ 1:78ð10Þ5 ; 7
r0 ¼ 5:7ð10Þ ;
k ¼ 386;
l0 ¼ 4pð10Þ7 :
During numerical computations, the function f ðy; tÞ was taken of the form f ðy; tÞ ¼ h0 H ða jyjÞ;
ð76Þ
where H ðÞ is the Heaviside unit step function and h0 is a constant. This means that heat is applied on the surface of the half-space on a narrow band of width 2a surrounding the y-axis to keep it at temperature h0 , while the rest of the surface is kept at zero temperature. Taking the Laplace and exponential Fourier transforms of both sides of Eq. (76), we get rffiffiffi 2 sinðqaÞ½1 ipqdðqÞ : f ðq; sÞ ¼ h0 p qs
ð77Þ
The computations were carried out for two values of time, namely for t ¼ 0:05 and t ¼ 0:15. The non-dimensional temperature h, horizontal displacement u, stress component rxx , induced magnetic field h and induced electric field component E2 distributions were evaluated on the x-axis ðy ¼ 0Þ. They are shown in Figs. 1–5, respectively. Solid lines represent the case when t ¼ 0:15 and dotted lines represent the case when t ¼ 0:05. We note that, due to symmetry, the displacement component v and the induced electric field component E1 vanish identically when y ¼ 0.
Fig. 1. Temperature distribution for y ¼ 0.
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Fig. 2. Horizontal displacement distribution for y ¼ 0.
Fig. 3. Stress distribution for y ¼ 0.
Fig. 4. Induced magnetic field distribution for y ¼ 0.
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
603
Fig. 5. Induced electric field distribution for y ¼ 0.
The fact that in generalized thermoelasticity, the waves propagate with finite speeds is evident in all these figures. We note that all the functions vanish identically outside some finite interval that changes with the passage of time. For example for t ¼ 0:05, all the functions vanish identically when x > 0:69. This is not the case in coupled thermoelasticity, where the considered function have non-vanishing values for all values of x due to the infinite speed of propagation of heat waves.
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14]
H. Lord, Y. Shulman, A generalized dynamical theory of thermo-elasticity, J. Mech. Phys. Solid 15 (1967) 299–309. H.H. Sherief, On generalized thermoelasticity, Ph.D. thesis, University of Calgary, Canada, 1980. R.S. Dhaliwal, H.H. Sherief, Generalized thermoelasticity for anisotropic media, Quart. Appl. Math. 33 (1980) 18. J. Ignaczak, Uniqueness in generalized thermoelasticity, J. Thermal Stresses 2 (1979) 171. J. Ignaczak, A note on uniqueness in thermoelasticity with one relaxation time, J. Thermal Stresses 5 (1982) 257–263. H.H. Sherief, On uniqueness and stability in generalized thermoelasticity, Quart. Appl. Math. 45 (1987) 773–778. M.N. Anwar, H.H. Sherief, State space approach to generalized thermoelasticity, J. Thermal Stresses 11 (1988) 353–365. M.N. Anwar, H.H. Sherief, Boundary integral equation formulation of generalized thermoelasticity in a Laplace transform domain, Appl. Math. Modeling 12 (1988) 161–166. H.H. Sherief, Fundamental solution of the generalized thermoelastic problem for short times, J. Thermal Stresses 9 (1986) 151–164. H.H. Sherief, M.N. Anwar, Two dimensional problem of a moving heated punch in generalized thermoelasticity, J. Thermal Stressess 9 (1986) 325–343. H.H. Sherief, M.N. Anwar, A two dimensional generalized thermoelasticity problem for an infinitely long cylinder, J. Thermal Stresses 17 (1994) 213–227. H.H. Sherief, F.A. Hamza, Generalized thermoelastic problem of a thick plate under axisymmetric temperature distribution, J. Thermal Stresses 17 (1994) 435–453. H.H. Sherief, F.A. Hamza, Generalized two dimensional thermoelastic problems in spherical regions under axisymmetric distributions, J. Thermal Stresses 19 (1996) 55–76. J.L. Nowinski, Theory of Thermoelasticity with Applications, Sijthoff & Noordhoff International Publishers, Alphen Aan Den Rijn, 1978.
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[15] G. Paria, Magneto-elasticity and magneto-thermoelasticity, Adv. Appl. Mech. 10 (1967) 73–112. [16] A. Nayfeh, S. Nemat-Nasser, Electromagneto-thermoelastic plane waves in solids with thermal relaxation, J. Appl. Mech. Series E 39 (1972) 108–113. [17] S. Choudhuri, Electro magneto-thermoelastic plane waves in rotating media with thermal relaxation, Int. J. Engrg. Sci. 22 (1984) 519–530. [18] H.H. Sherief, Problem in electromagneto-thermoelasticity for an infinitely long solid conducting circular cylinder with thermal relaxation, Int. J. Engrg. Sci. 32 (1994) 1137–1149. [19] H.H. Sherief, M.A. Ezzat, A thermal-shock problem in magneto-thermoelasticity with thermal relaxation, Int. J. Solids Structures 33 (1996) 4449–4459. [20] G. Honig, U. Hirdes, A method for the numerical inversion of the laplace transform, J. Comp. Appl. Math 10 (1984) 113–132.
A two-dimensional problem for a half-space in magneto-thermoelasticity with thermal relaxation Hany H. Sherief *, Kamal A. Helmy Department of Mathematics, Faculty of Sciences, University of Alexandria, Alexandria, Egypt Received 29 November 1999; accepted 11 February 2000 (Communicated by E.S. S ß UHUBI_ )
Abstract In this work we study a two-dimensional problem in electromagneto-thermoelasticity for a half-space whose surface is subjected to a non-uniform thermal shock and is stress free in the presence of a transverse magnetic field. The problem is in the context of the theory of generalized thermoelasticity with one relaxation time. Laplace and exponential Fourier transform techniques are used to obtain the solution by a direct approach. The solution of the problem in the physical domain is obtained by using a numerical method for the inversion of the Laplace transforms based on Fourier series expansions. The distributions of the temperature, the displacement, the stress and the induced magnetic and electric fields are obtained. The numerical values of these functions are represented graphically. Ó 2002 Elsevier Science Ltd. All rights reserved.
1. Introduction The theory of generalized thermoelasticity with one relaxation time was introduced by Lord and Shulman [1] for isotropic materials. This theory was extended by Sherief [2] and by Dhaliwal and Sherief [3] to include the effects of anisotropy. In this theory a modified law of heat conduction including both the heat flux and its time derivative replaces the conventional Fourier’s law. The heat equation associated with this theory is a hyperbolic one and hence automatically eliminates the paradox of infinite speeds of propagation inherent in both the uncoupled and the coupled theories of thermoelasticity. For many problems involving steep heat gradients and when short time effects are sought this theory is indispensable.
*
Corresponding author. E-mail address: [email protected] (H.H. Sherief).
0020-7225/02/$ - see front matter Ó 2002 Elsevier Science Ltd. All rights reserved. PII: S 0 0 2 0 - 7 2 2 5 ( 0 0 ) 0 0 0 9 3 - 8
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Due to the complexity of the partial differential equations of this theory, the work done in this field is unfortunately limited in number. Among the theoretical contributions to the subject are the proofs of uniqueness theorems under different conditions by Ignaczak [4,5] and by Sherief [6]. Anwar and Sherief [7] did the state space formulation for problems not containing heat sources and the boundary element formulation was done by Anwar and Sherief [8]. Some concrete problems have also been solved. The fundamental solutions for spherically symmetric spaces were obtained by Sherief [9]. Sherief and Anwar have solved some two-dimensional problems in [10,11], while Sherief and Hamza have solved some two-dimensional problems and studied the wave propagation in this theory in [12,13], respectively. Increasing attention is being devoted to the interaction between magnetic fields and strain in a thermoelastic solid due to its many applications in the fields of geophysics, plasma physics and related topics. In the nuclear field, the extremely high temperatures and temperature gradients as well as the magnetic fields originating inside nuclear reactors influence their design and operations [14]. Usually, in these investigations the heat equation under consideration is taken as the uncoupled or the coupled equation, not the generalized one. This attitude is justified in some situations where the solutions obtained using any of these equations differ little quantitatively. However, when short time effects are considered, the full, generalized system of equations has to be used or a great deal of accuracy is lost [1]. A comprehensive review of the earlier contributions to the subject can be found in [15]. Among the authors who considered the generalized magneto-thermoelastic equations are Nayfeh and Nemat-Nasser [16] who studied the propagation of plane waves in a solid under the influence of an electromagnetic field. They have obtained the governing equations in the general case and the solution for some particular cases. Choudhuri [17] extended these results to rotating media. Sherief [18] has solved a problem for a solid cylinder, while Sherief and Ezzat [19] have solved a thermal shock half-space problem using asymptotic expansions.
2. Formulation of the problem We consider the problem of a thermoelastic half-space (x P 0). A magnetic field with constant intensity H0 acts parallel to the bounding plane (taken as the direction of the z-axis). The surface of the half-space is subjected at time t ¼ 0 to a thermal shock that is a function of y and t. Thus, all the quantities considered will be functions of the time variable t and of the coordinates x and y. Due to the application of the primary magnetic field H0 , there results an induced magnetic field h and an induced electric field E. We assume that both h and E are small in magnitude in accordance with the assumptions of the linear theory of thermoelasticity. Thus, the displacement vector will have the components ux ¼ uðx; y; tÞ;
uy ¼ vðx; y; tÞ;
uz ¼ 0:
The components of the magnetic intensity vector are Hx ¼ 0;
Hy ¼ 0;
Hz ¼ H0 þ hðx; y; tÞ:
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
589
The electric intensity vector is normal to both the magnetic intensity and the displacement vectors. Thus, E has the components Ex ¼ E1 ;
Ey ¼ E2 ;
Ez ¼ 0:
The current density vector J must be parallel to E, thus Jx ¼ J1 ;
Jy ¼ J2 ;
Jz ¼ 0:
Maxwell’s equations in vector form can be written as curl h ¼ J þ
curl E ¼
div B ¼ 0;
oD ; ot
ð1Þ
oB ; ot
ð2Þ
div D ¼ qe ;
B ¼ l0 ðH0 þ hÞ;
D ¼ e0 E;
ð3Þ ð4Þ
where 0 and l0 are the electric and magnetic permeabilities, respectively, and qe is the density of electric charges. The above equations are supplemented by Ohm’s law, namely ou J ¼ r0 E þ l0 H0 ; ð5Þ ot where r0 is the electric conductivity. Ohm’s law (5) after linearization gives ov J1 ¼ r0 E1 þ l0 H0 ; ot J2 ¼ r0
ou E2 l0 H0 : ot
Eqs. (1) and (6a), (6b) in our case give the two equations oh ov oE1 ¼ r0 E1 þ l0 H0 ; þ e0 oy ot ot oh ou oE2 ¼ r0 E2 l0 H0 e0 : ox ot ot
ð6aÞ
ð6bÞ
ð7Þ
ð8Þ
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Eqs. (2) and (4) have one non-vanishing component, namely oE1 oE2 oh ¼ l0 : ot oy ox
ð9Þ
The strain components are given by, ou exx ¼ ; ox
ov eyy ¼ ; oy
1 exy ¼ 2
ou ov þ ; oy ox
exz ¼ eyz ¼ ezz ¼ 0:
The cubical dilatation e is thus given by e ¼ exx þ eyy þ ezz ¼
ou ov þ : ox oy
ð10Þ
Stress components are given by the tensor relation rij ¼ 2leij þ kedij cðT T0 Þdij ; which has the Cartesian components rxx ¼ ðk þ 2lÞ
ou ov þ k cðT T0 Þ; ox oy
ð11aÞ
ryy ¼ ðk þ 2lÞ
ov ou þ k cðT T0 Þ; oy ox
ð11bÞ
ou ov þ ; rxy ¼ k oy ox
ð11cÞ
where k and l are Lame’s modulii, T the absolute temperature of the medium, and c is a material constant given by c ¼ ð3k þ 2lÞat ; at being the coefficient of linear thermal expansion. T0 is a reference temperature assumed to be such that jðT T0 Þ=T0 j 1 . Equations of motion have the form, ui ; rji;j þ Fi ¼ q€
ð12Þ
where q is the density and F is the Lorentz force given by, F ¼ J B:
ð13Þ
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
From Eqs. (13) and (6a), (6b), we obtain ou Fx ¼ l0 r0 H0 E2 l0 H0 ; ot ov ; Fy ¼ l0 r0 H0 E1 þ l0 H0 ot
591
ð13aÞ ð13bÞ
Fz ¼ 0:
ð13cÞ
Substituting from Eqs. (13) into Eq. (10), we obtain the equations of motion in the form, o2 u oe oT ou 2 þ l0 r0 H0 E2 l0 H0 ; ð14Þ q 2 ¼ ðk þ lÞ þ lr u c ot ox ox ot o2 v oe oT ov 2 q 2 ¼ ðk þ lÞ þ lr v c l0 r0 H0 E1 þ l0 H0 : ot oy oy ot
ð15Þ
The equation of heat conduction has the form, 2
kr T ¼
o o2 þ s0 2 ðqcE T þ cT0 eÞ; ot ot
ð16Þ
where k is the thermal conductivity of the medium, cE the specific heat at constant strain and s0 is the relaxation time. For simplifications we shall use the following non-dimensional variables: x ¼ c1 gx;
y ¼ c1 gy;
u ¼ c1 gu;
t ¼ c21 gt;
s 0 ¼ c21 gs0 ;
E1 ¼
v ¼ c1 gv;
gE1 ; r0 l20 H0 c1
E2 ¼
r ij ¼
rij ; l
gE2 ; r0 l20 H0 c1
cðT T0 Þ ; k þ 2l gh h ¼ ; r0 l0 H0 h¼
where g ¼ qcE =k; c21 ¼ ðk þ 2lÞ=q: Using the above non-dimensional variables, the governing equations reduce to (dropping the asterisks for convenience) oe ou o2 u 2 2 oh 2 2 ð17Þ þ b me2 mE2 ¼ b2 2 ; ðb 1Þ þ r u b ox ox ot ot oe ov o2 v 2 oh 2 2 ðb 1Þ þ r v b b me2 mE1 þ ¼ b2 2 ; oy oy ot ot 2
2
r h¼
o o2 þ s0 2 ðh þ e1 eÞ; ot ot
ð18Þ
ð19Þ
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oh oE1 ov ¼ mE1 þ V 2 þ ; oy ot ot
ð20Þ
oh oE2 ou ¼ mE2 V 2 þ ; ox ot ot
ð21Þ
oE1 oE2 oh ¼ ; ot oy ox
ð22Þ
where m¼
r0 l0 ; g
e2 ¼
l0 H02 ; k þ 2l
e1 ¼
T0 c2 ; qcE ðk þ 2lÞ
V ¼
c1 ; c
c2 ¼
1 ; e0 l0
b2 ¼
k þ 2l : l
The constitutive equations reduce to rxx ¼ 2
ou 2 b 2 e b2 h; ox
ð23aÞ
ryy ¼ 2
ov 2 b 2 e b2 h; oy
ð23bÞ
rxy ¼
ou ov þ : oy ox
ð23cÞ
3. Formulation and solution in the Laplace transform domain Taking the Laplace transform defined by the relation fðx; y; sÞ ¼
Z
1
est f ðx; y; tÞ dt 0
of both sides of Eqs. (17)–(22), (23a)–(23c), we get ðb2 1Þ
o e oh þ r2 u b2 þ b2 me2 mE2 s u ¼ b2 s2 u; ox ox
ð24Þ
ðb2 1Þ
o e oh þ r2 v b2 b2 me2 mE1 þ s v ¼ b2 s2 v; oy oy
ð25Þ
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
r2 h ¼ s þ s0 s2 h þ e1 e ;
593
ð26Þ
oh ¼ m þ V 2 s E1 þ s v; oy
ð27Þ
oh ¼ m þ V 2 s E2 þ s u; ox
ð28Þ
oE1 oE2 ¼ sh; oy ox
ð29Þ
rxx ¼ 2
o u 2 b 2 e b2 h; ox
ð30aÞ
ryy ¼ 2
o v 2 b 2 e b2 h; oy
ð30bÞ
rxy ¼
o u o v þ : oy ox
ð30cÞ
Differentiating Eq. (24) with respect to x and Eq. (25) with respect to y, adding, we obtain upon using Eq. (29) 2 r s2 e2 ms e r2 h e2 m2 sh ¼ 0:
ð31Þ
Eliminating E1 and E2 between Eqs. (27)–(29), we get 2 r sðm þ V 2 sÞ h ¼ s e:
ð32Þ
Eliminating h and h between Eqs. (26), (31) and (32), we obtain the following sixth-order partial differential equation satisfied by e ðr6 Ar4 þ Br2 CÞ e ¼ 0; where A ¼ s sðe1 s0 þ s0 þ V 2 þ 1Þ þ e1 þ e2 m þ m þ 1 ;
ð33Þ
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B ¼ s2 s2 ½e1 s0 V 2 þ s0 ðV 2 þ 1Þ þ V 2 þ s½e1 ðms0 þ V 2 Þ þ e2 mðs0 þ V 2 Þ þ mðs0 þ 1Þ þ V 2 þ 1 þ e1 m þ e2 m þ m ; C ¼ s4 ð1 þ s0 sÞðsV 2 þ e2 mV 2 þ mÞ: In a similar manner we can show that h and h satisfy the equations ðr6 Ar4 þ Br2 CÞh ¼ 0;
ð34Þ
ðr6 Ar4 þ Br2 CÞh ¼ 0:
ð35Þ
Eq. (33) can be factorized as ðr2 k12 Þðr2 k22 Þðr2 k32 Þ e ¼ 0;
ð36Þ
where k12 ; k22 and k32 are the roots of the characteristic equation k 6 Ak 4 þ Bk 2 C ¼ 0:
ð37Þ
The solution of Eq. (36) has the form e ¼
3 X
ð38Þ
ei ;
i¼1
where ei is the solution of the equation ðr2 ki2 Þ ei ¼ 0;
i ¼ 1; 2; 3:
ð39Þ
In order to solve Eq. (39), we shall use the exponential Fourier transform with respect to the variable y (denoted by an asterisk) and defined by the relation Z 1 1
p ffiffiffiffiffi ffi f ðx; q; sÞ ¼ eiqy f ðx; y; sÞ dy; 2p 1 whose inversion is given by the relation Z 1 1 eiqy f ðx; q; sÞ dq: f ðx; y; sÞ ¼ pffiffiffiffiffiffi 2p 1 Applying the exponential Fourier transform to both sides of Eq. (39), we get e i ¼ 0; ðD2 q2i Þ where D ¼ o=ox and q2i ¼ q2 þ ki2 ; i ¼ 1; 2; 3.
ð40Þ
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
595
The solution of Eq. (40), which is bounded for x > 0, is given by e i ðx; q; sÞ ¼ Ai ðq; sÞeqi x ; where Ai ðq; sÞ is a parameter depending on q and s. Thus, e is given by the relation e ðx; q; sÞ ¼
3 X
Ai ðq; sÞeqi x :
ð41Þ
i¼1
In a similar manner, we can write h ðx; q; sÞ ¼
3 X
A0i ðq; sÞeqi x ;
ð42Þ
A00i ðq; sÞeqi x ;
ð43Þ
i¼1
h ðx; q; sÞ ¼
3 X i¼1
where A0i ðq; sÞ and A00i ðq; sÞ are parameters depending on q and s. Substituting from Eqs. (41)–(43) into the Fourier transform of Eqs. (26) and (32) we get the following relations: A0i ¼
A00i ¼
e1 sð1 þ s0 sÞ Ai ; sð1 þ s0 sÞ
ki2
ki2
s Ai ; sðm þ V 2 sÞ
i ¼ 1; 2; 3;
i ¼ 1; 2; 3:
ð44Þ
ð45Þ
Substituting Eqs. (44) and (45) into Eqs. (42) and (43), respectively, we obtain h ðx; q; sÞ ¼
h ðx; q; sÞ ¼
3 X e1 sð1 þ s0 sÞ Ai eqi x ; 2 k sð1 þ s sÞ 0 i i¼1
3 X
ki2 i¼1
s Ai eqi x : sðm þ V 2 sÞ
ð46Þ
ð47Þ
In order to obtain the displacement component u, we take the exponential Fourier transform of both sides of Eqs. (24) and (28) with respect to the variable y, to get u ¼ b2 ðD2 q2 b2 s2 b2 me2 sÞ
oh
o e
b2 m2 e2 E2 ðb2 1Þ ; ox ox
ð48Þ
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# "
1 o h : E2 ¼ s u m þ V 2s ox
ð49Þ
Eliminating E2 between Eqs. (48) and (49) and using Eqs. (41), (46) and (47), we obtain the following equation satisfied by u
3 X e1 sð1 þ s0 sÞ e2 m2 s 2 D n u ¼ 1þb þ 1 Ai eqi x ; 2 2 sÞðk 2 sðm þ V 2 sÞÞ sð1 þ s sÞ k ðm þ V 0 i i i¼1 2
2
ð50Þ
where sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b2 s2 me2 V 2 : n ¼ q2 þ b2 s2 þ m þ V 2s In order to simplify the right-hand side of Eq. (50), we substitute from Eqs. (41), (46) and (47) into the Fourier transform of Eq. (31), to get the compatibility condition e1 sð1 þ s0 sÞ ki2 s2 e2 ms e2 m2 s2 ¼ : ki2 sð1 þ s0 sÞ ki2 ki2 ðki2 sðm þ V 2 sÞÞ
ð51Þ
Using Eq. (51), Eq. (50) reduces to
3 X b2 s2 ðm þ V 2 s þ e2 mV 2 Þ 1 qi Ai eqi x : D n u ¼ 2 2 sÞ k ðm þ V i i¼1 2
2
ð52Þ
The solution of Eq. (52), bounded as x ! 1 is given by u ðx; q; sÞ ¼ Bq2 enx
3 X qi Ai qi x e ; ki2 i¼1
where B ¼ Bðq; sÞ is some parameter depending on q and s. Taking the Laplace and exponential Fourier transforms of Eq. (10), we get # "
i o u : e v ¼ q ox
ð53Þ
ð54Þ
Substituting from Eqs. (41) and (53) into the right-hand side of Eq. (54), we obtain "
# 3 X A i qi x v ðx; q; sÞ ¼ iq nBenx : e 2 k i i¼1
ð55Þ
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
Substituting from Eqs. (47) and (53) into the right-hand side of Eq. (49), we get " # 3 X s A q i i E2 ¼ Bq2 enx þ sðm þ V 2 sÞ eqi x : 2 2 2 sÞÞ m þ V 2s k ð k sðm þ V i i i¼1
597
ð56Þ
Taking the exponential Fourier transform of both sides of Eq. (27), we obtain E1 ¼
i 1 h
iq h s v : m þ V 2s
Substituting from Eqs. (47) and (55) into Eq. (57), we obtain " # 3 X iqs A i nBenx þ sðm þ V 2 sÞ E1 ¼ eqi x : 2 2 2 sÞÞ m þ V 2s k ð k sðm þ V i i i¼1
ð57Þ
ð58Þ
Substituting from Eqs. (46), (47), (53) and (55) into Eqs. (30a)–(30c), we obtain the transformed stress components in the form ( ) 2 3 X 2q2 b ki2 sð1 þ s0 sÞðe1 þ 1Þ
2 nx Ai eqi x ; þ ð59Þ rxx ðx; q; sÞ ¼ 2nBq e þ 2 2 k k sð1 þ s sÞ 0 i i i¼1 (
) 3 X q A i i qi x ðn2 þ q2 ÞBenx : r xy ðx; q; sÞ ¼ iq 2 e 2 k i i¼1
ð60Þ
In order to apply the boundary conditions of the problem to find the unknown parameters, we need to find the magnetic and electric field intensities in free space. We denote these by h0 , E10 and E20 , respectively. These quantities satisfy the non-dimensional field equations oh0 oE10 ¼V2 ; oy ot
ð61Þ
oh0 oE20 ¼ V 2 ; ox ot
ð62Þ
oh0 oE10 oE20 ¼ : ot oy ox
ð63Þ
Taking the Laplace and the exponential Fourier transforms of both sides of Eqs. (51)–(63), and solving the resulting equations, we obtain the solutions bounded for x < 0 as h 0 ¼ Cemx ;
ð64Þ
598
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
iq E10 ¼ 2 Cemx ; V s
ð65Þ
m
¼ 2 Cemx ; ð66Þ E20 V s pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi where m ¼ q2 þ V 2 s2 . We have thus obtained the solution of the problem in the transformed domain in terms of five unknown parameters A1 ; A2 ; A3; B and C. These parameters are to be determined from the boundary conditions at x ¼ 0 which are: (1) Thermal boundary condition that the temperature at the surface of the half-space is known hð0; y; tÞ ¼ f ðy; tÞ: (2) Mechanical boundary conditions that the surface of the half-space is traction free rxx ð0; y; tÞ ¼ rxy ð0; y; tÞ ¼ 0: (3) The transverse components of the electric field intensity are continuous across the surface of the half-space E2 ð0; y; tÞ ¼ E20 ð0; y; tÞ: (4) The transverse components of the magnetic field intensity are continuous across the surface of the half-space hð0; y; tÞ ¼ h0 ð0; y; tÞ: The above conditions, in the transformed, domain can be written as 3 X i¼1
1 f ðq; sÞ ¼ A ; i ki2 sð1 þ s0 sÞ e1 sð1 þ s0 sÞ
2nq2 B þ
3 X
(
i¼1
ðn2 þ q2 ÞB þ 2
3 X i¼1
) 2 2q2 b ki2 sð1 þ s0 sÞðe1 þ 1Þ þ Ai ¼ 0; ki2 sð1 þ s0 sÞ ki2 3 X qi Ai ¼ 0; ki2 i¼1
Ai qi Bq2 mC þ 2 3 ¼ 0; þ 2 2 2 2 ki ðki sðm þ V sÞÞ sðm þ V sÞ V s
ð67Þ
ð68Þ
ð69Þ
ð70Þ
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
s
3 X
ki2 i¼1
1 Ai C ¼ 0: sðm þ V 2 sÞ
599
ð71Þ
Eqs. (67)–(71) constitute a system of five linear equations with five unknown parameters A1 ; A2 ; A3; B and C.
4. Inversion of the double Laplace–Fourier transform We shall now outline the method used to invert the double Laplace–Fourier transforms in the above equations. Let f ðx; q; sÞ be the Laplace–Fourier transform of a function f ðx; y; tÞ. We shall first use the inversion formula of the Fourier transform mentioned above to obtain a function fðx; y; sÞ that is the Laplace transform of the function f ðx; y; tÞ. Next, the inversion formula for Laplace transforms can be written as 1 f ðx; y; tÞ ¼ 2pi
Z
dþi1
est fðx; y; sÞ ds; di1
where d is an arbitrary real number greater than all the real parts of the singularities of fðx; y; sÞ. Taking s ¼ d þ iz, the above integral takes the form edt f ðx; y; tÞ ¼ 2p
Z
1
eitz fðx; y; d þ izÞ dz:
1
Expanding the function hðx; y; tÞ ¼ expðdtÞf ðx; y; tÞ in a Fourier series in the interval ½0; 2L , we obtain the approximate formula [20] f ðx; y; tÞ ¼ f1 ðx; y; tÞ þ ED ; where 1 X 1 f1 ðx; y; tÞ ¼ c0 þ ck 2 k¼1
for 0 6 t 6 2L
ð72Þ
and ck ¼
i edt h ikpt=L Re e f ðx; y; d þ ikp=LÞ ; L
ð73Þ
ED , the discretization error, can be made arbitrarily small by choosing d large enough [20]. As the infinite series in (72) can only be summed up to a finite number N of terms, the approximate value of f ðx; y; tÞ becomes
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for 0 6 t 6 2L:
ð74Þ
Using the above formula to evaluate f ðx; y; tÞ, we introduce a truncation error ET that must be added to the discretization error to produce the total approximation error. Two methods are used to reduce the total error. First, the ‘Korrecktur’ method is used to reduce the discretization error. Next, the e-algorithm is used to reduce the truncation error and therefore to accelerate convergence. The Korrecktur method uses the following formula to evaluate the function f ðx; y; tÞ 0 ; f ðx; y; tÞ ¼ f1 ðx; y; tÞ e2dL f1 ðx; y; 2L þ tÞ þ ED 0 ED . where the discretization error ED Thus, the approximate value of f ðx; y; tÞ becomes
fNK ðx; y; tÞ ¼ fN ðx; y; tÞ e2dL fN 0 ðx; y; 2L þ tÞ:
ð75Þ
N 0 is an integer such that N 0 < N . We shall now describe the e-algorithm that is used to accelerate the convergence of the series in (74). Let N be an odd natural number and let sm ¼
m X
ck ;
k¼1
be the sequence of partial sums of (74). We define the e-sequence by e0;m ¼ 0;
e1;m ¼ sm ; m ¼ 1; 2; 3; . . . :
and enþ1;m ¼ en1;mþ1 þ
1 ; en;mþ1 en;m
n; m ¼ 1; 2; 3; . . .
It can be shown that [20] the sequence e1;1 ; e3;1 ; . . . ; eN ;1 converges to f ðx; y; tÞ þ ED c0 =2 faster than the sequence of partial sums sm ;
m ¼ 1; 2; 3; . . .
The actual procedure used to invert the Laplace transforms consists of using Eq. (75) together with the e-algorithm. The values of d and L are chosen according the criteria outlined in [20].
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
601
5. Numerical results The copper material was chosen for purposes of numerical evaluations. The constants of the problem were taken as e1 ¼ 0:0168;
e2 ¼ 0:0008;
V ¼ 1:39ð10Þ5 ; 10
k ¼ 7:76ð10Þ ;
q ¼ 9854;
b2 ¼ 4;
T0 ¼ 293;
g ¼ 8886:73; 10
l ¼ 3:86ð10Þ ;
m ¼ 0:008;
cE ¼ 381; 9
e0 ¼ ð10Þ =ð36pÞ;
s0 ¼ 0:02;
at ¼ 1:78ð10Þ5 ; 7
r0 ¼ 5:7ð10Þ ;
k ¼ 386;
l0 ¼ 4pð10Þ7 :
During numerical computations, the function f ðy; tÞ was taken of the form f ðy; tÞ ¼ h0 H ða jyjÞ;
ð76Þ
where H ðÞ is the Heaviside unit step function and h0 is a constant. This means that heat is applied on the surface of the half-space on a narrow band of width 2a surrounding the y-axis to keep it at temperature h0 , while the rest of the surface is kept at zero temperature. Taking the Laplace and exponential Fourier transforms of both sides of Eq. (76), we get rffiffiffi 2 sinðqaÞ½1 ipqdðqÞ : f ðq; sÞ ¼ h0 p qs
ð77Þ
The computations were carried out for two values of time, namely for t ¼ 0:05 and t ¼ 0:15. The non-dimensional temperature h, horizontal displacement u, stress component rxx , induced magnetic field h and induced electric field component E2 distributions were evaluated on the x-axis ðy ¼ 0Þ. They are shown in Figs. 1–5, respectively. Solid lines represent the case when t ¼ 0:15 and dotted lines represent the case when t ¼ 0:05. We note that, due to symmetry, the displacement component v and the induced electric field component E1 vanish identically when y ¼ 0.
Fig. 1. Temperature distribution for y ¼ 0.
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Fig. 2. Horizontal displacement distribution for y ¼ 0.
Fig. 3. Stress distribution for y ¼ 0.
Fig. 4. Induced magnetic field distribution for y ¼ 0.
H.H. Sherief, K.A. Helmy / International Journal of Engineering Science 40 (2002) 587–604
603
Fig. 5. Induced electric field distribution for y ¼ 0.
The fact that in generalized thermoelasticity, the waves propagate with finite speeds is evident in all these figures. We note that all the functions vanish identically outside some finite interval that changes with the passage of time. For example for t ¼ 0:05, all the functions vanish identically when x > 0:69. This is not the case in coupled thermoelasticity, where the considered function have non-vanishing values for all values of x due to the infinite speed of propagation of heat waves.
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