# Exact Solution of Thermoelastic Problem for a One-Dimensional Bar without Energy Dissipation

Authors: A. M. Abd El-Latief; S. E. Khader
Journal: ISRN Mechanical Engineering
 (2014)

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## Abstract

We consider a homogeneous isotropic thermoelastic half-space in the context of the theory of thermoelasticity without energy dissipation. There are no body forces or heat source acting on the half-space. The surface of the half-space is affected by a time dependent thermal shock and is traction free. The Laplace transform with respect to time is used. The inverse transforms are obtained in an exact manner for the temperature, thermal stress, and displacement distributions. These solutions are represented graphically and discussed for several cases of the applied heating. Comparison is made between the predictions here and those of the theory of thermoelasticity with one relaxation time.


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## Body

## 1. Introduction

In 1967, Lord and Shulman [1] introduced the theory of generalized thermoelasticity with one relaxation time for an isotropic body. This theory corrects the unrealistic conclusions of the older theories (the uncoupled and the coupled theories of thermoelasticity) that heat waves travel with infinite speeds. The Heat conduction law of this theory is the Cattaneo law which is different from Fourier’s law utilized in both the coupled and the uncoupled theories. Among the contribution to the subject are the works in [2–6]. In 1972 Green and Lindsay [7] developed the theory of generalized thermoelasticity with two relaxation times, based on a generalized inequality of thermodynamics. In this theory both the equations of motion and of heat conduction are hyperbolic. The heat conduction law is the same as Fourier’s law when the system has a centre of symmetry. Among the contributions to this theory are the works in [8, 9].Green and Naghdi [10–12] have formulated three new models of thermoelasticity. In one of these models Green and Naghdi [12] predict that the internal rate of production of entropy is identically zero; that is, there is no dissipation of thermal energy. This theory (GN theory) is known as thermoelasticity without energy dissipation theory. In the development of this theory the thermal displacement gradient is considered as a constitutive variable, whereas in the conventional development of a thermoelasticity theory, the temperature gradient is taken as a constitutive variable [12]. A couple of uniqueness theorems have been proved in [13, 14], and one-dimensional waves in a half-space and in an unbounded body have been studied in [15, 16]. In view of some experimental evidence available in favour of finiteness of heat propagation speed, generalized thermoelasticity theories are considered to be more realistic than the conventional theory in dealing with practical problems involving very large heat fluxes at short intervals, like those occurring in laser units and energy channels [17].


## 2. Formulation of the Problem

In this work, we consider a homogeneous isotropic and thermoelastic half-space in the context of the theory of thermoelasticity without energy dissipation. There are no body forces or heat source acting on the half-space taken asx≥0. The surface of the half-space is affected by a time dependent thermal shock and is traction free.From the physics of the problem, all functions will depend on the distancex and the time t only. The problem is thus one-dimensional. The displacement components have the form
(1)u_=(u(x,t),0,0).The basic equations due to Green and Nagdhi [12] in the absence of body forces and heat sources for isotropic elastic medium are given by the following:(i) equation of motion:(2)(λ+2μ)∂2u∂x2-γ∂T∂x=ρ∂2u∂t2,(ii) the equation of heat conduction:(3)k*∂2T∂x2=ρCE∂2T∂t2+γT0∂3u∂t2∂x,(iii) the constitutive equation:(4)σxx=σ=(λ+2μ)∂u∂x-γ(T-T0),
where λ and μ are Lamé’s modulii, T is the absolute temperature of the medium, and γ is a material constant given by γ=(3λ+2μ)αt, where αt is the coefficient of linear thermal expansion and T0 is a reference temperature assumed to be such that |(T-T0)/T0|≪1. k* is a material constant, characteristic of the theory and CE is the specific heat at constant strain.Let us introduce the following nondimension variables:(5)x*=c1ηx,u*=c1ηu,t*=c12ηt,σ*=σλ+2μ,θ=γ(T-T0)(λ+2μ).The governing equations (2)–(4) in nondimensional form become (dropping the asterisks for convenience)
(6)∂2u∂x2-∂θ∂x=∂2u∂t2,Ct2∂2θ∂x2=∂2θ∂t2+ε∂3u∂t2∂x,σ=∂u∂x-θ,
where Ct2=k*/ρcEc12, andε=γ2T0/ρcE(λ+2μ).


## 3. Solution of the Problem in the Laplace Transform Domain

Applying the Laplace transform with parameters defined by the relation [14]
(7)f-(s)=∫0∞f(t)e-stdt,
to both sides of (6), we obtain(8)∂2u¯∂x2-∂θ¯∂x=s2u¯,(9)Ct2∂2θ¯∂x2=s2θ¯+εs2∂u¯∂x,(10)σ¯=∂u¯∂x-θ¯.Eliminatingu¯ from (12) and (14), we get
(11){D4-(Ct2+1+ε)Ct2s2D2+s4Ct2}θ¯=0,
where D=∂/∂x.Equation (11) can be factorized as
(12)(D2-k12)(D2-k22)θ¯=0,
where k12 and k22 are the roots of the characteristic equation
(13)k4-(Ct2+1+ε)Ct2s2k2+s4Ct2=0.k12 and k22 are given by
(14)k12=s2(B+A),k22=s2(B-A),
where B=(Ct2+1+ε)/2Ct2 and A2=((Ct2+1+ε)/2Ct2)2-1/Ct2.The solutions of (12), bounded as x tends to ∞ have the form
(15)θ¯=D1e-k1x+D2e-k2x,
where D1 and D2 are some parameters depending on s only.To calculateu- using (9), we obtain
(16)∂u-∂x=Ct2εs2[D1(k12-s2Ct2)e-k1x+D2(k22-s2Ct2)e-k2x].Integrating both sides of (16) with respect to x, we get
(17)u-=-Ct2εs2[D1k1(k12-s2Ct2)e-k1x+D2k2(k22-s2Ct2)e-k2x].Substituting (15) and (17) into (10), we obtain
(18)σ¯=Ct2εs2[D1(k12-s2(1+ε)Ct2)e-k1xdddd+D2(k22-s2(1+ε)Ct2)e-k2x].The boundary conditions of the problem can be written as follows:(19)σ=0,θ=F(t),atx=0.Applying the Laplace transform with parameters to both sides of (19), we get
(20)θ¯=F¯(s),σ¯=0,atx=0.Substituting (15) and (18) into (20), we obtain the following system of linear equations in the unknown parameters D1 and D2:
(21)D1+D2=F¯(s),D1(k12-s2(1+ε)Ct2)+D2(k22-s2(1+ε)Ct2)=0.The solution of the above equations is(22)D1=EF-(s),D2=(1-E)F-(s),
where E=(1+ε-Ct2(B-A))/2Ct2A.Substituting (22) into (15), (17), and (18), we obtain
(23)θ¯=F-(s)(Ee-B+Axs+(1-E)e-B-Axs),u¯=-F-(s)Ct2εs[EB+A(B+A-1Ct2)e-B+Axsdddddddd+1-EB-A(B-A-1Ct2)e-B-Axs],σ¯=F-(s)Ct2ε[E(B+A-1+εCt2)e-B+Axsddddddd+(1-E)(B-A-1+εCt2)e-B-Axs].
The above equations can be written as
(24)θ¯=F-(s)(Ee-μ1s+(1-E)e-μ2s),(25)u¯=-F-(s)s[C1e-μ1s+C2e-μ2s],(26)σ¯=F-(s)[C3e-μ1s+C4e-μ2s],
where
(27)C1=Ct2(1-E)εB-A(B-A-1Ct2),C2=ECt2εB+A(B+A-1Ct2),C3=Ct2E(B+A-(1+ε)/Ct2)ε,C4=Ct2(1-E)(B-A-(1+ε)/Ct2)ε,μ1=B+A,μ2=B-A.
This completes the solution of the problem in the Laplace transform domain.


## 4. Exact Solution in the Physical Domain

We shall derive the temperature and stress distributions in an exact manner; by applying the inverse Laplace transform to both sides of (24) and (26), we get the following expressions:
(28)θ(x,t)=Ef(t-μ1x)H(t-μ1x)+(1-E)f(t-μ2x)H(t-μ2x),(29)σ(x,t)=C3f(t-μ1x)H(t-μ1x)+C4f(t-μ2x)H(t-μ2x).
In obtaining the above equations, we have used the following property of the Laplace Transform:
(30)L-1[F¯(s)e-αs]=f(t-α)H(t-α),
where H(t-μ) is the Heaviside unit step function; we have
(31)L-1[F¯(s)e-αss]=∫0t[L-1F¯(s)e-αs]t=τdτ=∫0tf(τ-α)H(τ-α)dτ.
According to the definition of Heaviside unit step function, we have two cases:(i)
ifμ<t, then the integral right hand side of (31) becomes ∫μtf(τ-μ)dτ;(ii)
while ifμ>t in this case the integral vanishes; thus (31) becomes
(32)L-1[F¯(s)e-αss]=H(τ-α)∫μtf(τ-α)dτ.Now, using (32) and (25), we obtain the exact form of the displacement distribution in the form
(33)u(x,t)=-[C1H(t-μ1x)∫μ1xtf(τ-μ1x)dτddd+C2H(t-μ2x)∫μ2xtf(τ-μ2x)dτ].
This completes the solution of the problem in the physical domain.


## 5. Numerical Results and Discussion

We will consider the solution of the problem for different choices for the functionf(t) as shown in Table 1, where θ0 and a are constants.Table 1
Case
f
(
t
)
1
θ
0
H
(
t
)
2
e
-
t
3
H
(
t
-
a
)
sin
(
π
t
a
)Substitutingf(t) into (29), (30), and (33), we obtain the following.Case 1.
Consider the following:(34)θ(x,t)=EH(t-μ1x)+(1-E)H(t-μ2x),σ(x,t)=C3H(t-μ1x)+C4H(t-μ2x),u(x,t)=-[C1(t-μ1x)H(τ-μ1x)d.d+C2(t-μ2x)H(τ-μ2x)].Case 2.
Consider the following:(35)θ(x,t)=Ee-(t-μ1x)H(t-μ1x)+(1-E)e-(t-μ2x)H(t-μ2x)σ(x,t)=C3e-(t-μ1x)H(t-μ1x)+C4e-(t-μ2x)H(t-μ2x)u(x,t)=-C1H(t-μ1x)⌊1-e-(t-μ1x)⌋-C2H(t-μ2x)⌊1-e-(t-μ2x)⌋.Case 3.
Consider the following:(36)θ=EH(t-μ1x)H(a-t)sin(π(t-μ1x)a)+(1-E)H(t-μ2x)H(a-t)sin(π(t-μ2x)a),σ=C3H(t-μ1x)H(a-t)sin(π(t-μ1x)a)+C4H(t-μ2x)H(a-t)sin(π(t-μ2x)a),u=aπ(π(t-μ1x)aC1H(t-μ1x)H(a-t)ddd×[cos(π(t-μ1x)a)-1]ddd+C2H(t-μ2x)H(a-t)ddd×[cos(π(t-μ2x)a)-1]).Obviously the above exact solution verifies the boundary conditions introduced in (19) for all different cases.We shall now compare, graphically, our results, where the theory of thermoelasticity without energy dissipation (GN theory) was used with the results of the solution of a similar problem in [18], where the generalized theory of thermoelasticity due to Lord-Shulman (LS theory) was used. Due to the nature of the characteristic roots of the LS theory, the authors of [18] could not obtain a closed form exact solution. They used asymptotic expansions valid for small values of time.For purposes of numerical evaluations, the calculations were performed for a copper-like material for which [19]
(37)ε=0.0168,Ct2=40,θ0=1,a=0.04.The temperature, displacement, and stress distributions are plotted for one value of time, namely, fort=0.1. In all figures, LS theory is represented by dotted lines, while GN theory is represented by solid lines.For the three cases, the temperature distributions are shown in Figures1, 4, and 7, the displacement distributions are shown in Figures 2, 5, and 8, while the stress distributions are shown in Figures 3, 6, and 9.Figure 1
Temperature distribution for Case1.Figure 2
Displacement distribution for Case1.Figure 3
Stress distribution for Case1.Figure 4
Temperature distribution for Case2.Figure 5
Displacement distribution for Case2.Figure 6
Stress distribution for Case2.Figure 7
Temperature distribution for Case3.Figure 8
Displacement distribution for Case3.Figure 9
Stress distribution for Case3.At the locationsx1=t/μ1 and x2=t/μ2 the temperature and stress distributions have two finite jumps, while the displacement is continuous everywhere having a discontinuous first derivative at the above locations. The first discontinuity is not shown in Figures 1, 4, and 7 because it is very small (E≈4(10)-4).x
1 and x2 are the locations of the wave fronts. We thus have two waves propagating into the medium from the surface of the half space. The first is mainly mechanical and has velocity equal to v1=1/μ1. The second is mainly thermal and has velocity equal to v2=1/μ2.Forx>x2 the solution for all functions is identically zero. Thus, the solution is confined to a finite region of space and does not extend to infinity. This is not the case for the coupled and the uncoupled theories of thermoelasticity, where the solution extends to infinity instantaneously signifying an infinite speed of propagation of waves. The values of the velocities and locations of the wave fronts are as follows:
(38)v1=0.999784687321833,v2=6.32591737054768,x1=0.09997846873218333,x2=0.632591737054768.The difference between the predictions of the theories of LS and GN is most apparent in the graphs of the temperature distribution. In the LS theory the temperature decreases monotonically signifying continuous dissipation of heat energy. This is not the case for GN theory.



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