A problem for an infinite thermoelastic body with a spherical cavity

doi:10.1016/S0020-7225(97)00084-0

International Journal of Engineering Science

Volume 36, Issue 4, March 1998, Pages 473-487

https://doi.org/10.1016/S0020-7225(97)00084-0 Get rights and content

Abstract

In this work we study a one-dimensional problem of distribution of thermal stresses and temperature in a generalized thermoelastic infinite medium with a spherical cavity subjected to a sudden change in the temperature of its internal boundary which is assumed to be traction free. Laplace transform techniques are used. The solution of the problem in the transformed domain is obtained by using a direct approach without the customary use of potential functions. The inverse transforms are obtained in an exact manner using the complex inversion formula of the transform together with contour integration techniques. Numerical computations are carried out and represented graphically.

Introduction

Biot[1] formulated the theory of coupled thermoelasticity to eliminate the paradox inherent in the classical uncoupled theory that elastic changes have no effect on the temperature. The heat equations for both theories are of the diffusion type predicting infinite speeds of propagation for heat waves contrary to physical observations.

Lord and Shulman[2] introduced the theory of generalized thermoelasticity with one relaxation time for the special case of an isotropic body. This theory was extended[3] by Dhaliwal and Sherief to include the anisotropic case. 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 hyperbolic and hence eliminates the paradox of infinite speeds of propagation inherent in both the uncoupled and the coupled theories of thermoelasticity. Uniqueness of solution for this theory was proved under different conditions by Ignaczak4, 5, Sherief and Dhaliwal3, 6 and Sherief[7]. The state space approach to this theory was developed by Anwar and Sherief[8] and Sherief[9]. The fundamental solution for this theory was obtained by Sherief[10]. Sherief and Hamza have solved some two-dimensional problems11, 12. Ref.[11] also contains a study of the wave propagation in this theory.

Green and Lindsay[13] developed the theory of generalized thermoelasticity with two relaxation times which is based on a generalized inequality of thermodynamics. This theory does not violate Fourier's law of heat conduction when the body under consideration has a center of symmetry. In this theory both the equations of motion and of heat conduction are hyperbolic but the equation of motion is modified and differs from that of the coupled thermoelasticity theory. This theory was initiated by Müller[14]. It was further extended by Green and Laws[15]. The final form used in the present work is that of Green and Lindsay[13]. This theory was also obtained independently by Şuhubi[16]. Longitudinal wave propagation for this theory was studied by Erbay and Şuhubi[17]. Ignaczak proved a decomposition theorem for this theory[18]. Sherief has obtained the fundamental solution for this theory[19], has formulated the state space approach[20] and solved a thermomechanical shock problem[21]. The boundary integral equation formulation was done by Anwar and Sherief[22].

Section snippets

Formulation of the problem

Let (r,ϑ,ϕ) denote the radial coordinate, the co-latitude, and longitude of a spherical coordinate system, respectively. We shall consider a homogeneous, isotropic, thermoelastic medium occupying the region a≤r<∞, where a is the radius of the spherical cavity. We note that due to spherical symmetry the only non-vanishing displacement component is the radial one u_r=u(r,t).

The strain tensor has the following components[13]: $e_{rr} = ∂ u ∂ r, e_{ϑϑ} =e_{ϕϕ} = u r, e_{rϕ} =e_{rϑ} =e_{ϑϕ} =0.$

The cubical dilatation e is thus given

Solution in the Laplace transform domain

Introducing the Laplace transform defined by the formula $f ̄ (p)= ∫ 0 ∞ e^{−pt} f(t) d t$ into , , , and using the initial conditions (11), we obtain $p^{2} u ̄ = ∂ e ̄ ∂ r −c(1+νp) ∂ θ ̄ ∂ r,$ $∇^{2} θ ̄ =p(1+τp) θ ̄ +gp e ̄,$ $σ ̄_{rr} =(β^{2} −2) e ̄ +2 ∂ u ̄ ∂ r −b(1+pν) θ ̄,$ $σ ̄_{ϑϑ} = σ ̄_{ϕϕ} =(β^{2} −2) e ̄ +2 u ̄ r −b(1+νp) θ ̄ .$

Taking the divergence of Eq. (12), we obtain $(∇^{2} −p^{2}) e ̄ =c(1+νp)∇^{2} θ ̄ .$

Eliminating $θ ̄$ between , , we obtain ${∇^{4} −p[1+ε+p(1+τ+εν)]∇^{2} +p^{3} (1+τp)} e ̄ =0,$ where ε is a positive constant defined by ε=gb/β².

The above equation can be factorized as $(∇^{2} −k^{2}_{1})(∇^{2} −k^{2}_{2}) e ̄ =0,$ where k

Inversion of the Laplace transforms

In order to obtain the inverse Laplace transforms of the above functions, we shall use the complex inversion formula of the Laplace transform[24], namely $f(t)= 1 2π i ∫ d− i ∞ d+ i ∞ e^{pt} f ̄ (p) d p,$ where d is a constant greater than all the real parts of the singularities of $f ̄ (p)$ 24, 25.

In the forthcoming analysis we shall assume that $ε(ν−τ)+1>0, τ>0$ which is true for all realistic materials.

It is clear that all functions considered $θ ̄$ , $σ ̄_{rr}$ and ū have a simple pole at the point p=−α due to the term $F ̄$

Temperature distribution

We consider first the inversion of the function $θ ̄_{1} (r,p)$ . According to the inversion formula (39), the inverse Laplace transform of this function is given by $θ_{1} (r,t)= 1 2π i ∫ d− i ∞ d+ i ∞ e^{pt} θ ̄_{1} (r,p) d p.$

As mentioned above, we have $∫ Γ e^{pt} θ ̄_{1} (r,p) d p=0.$

Eq. (45)can be written as $∫ AB + ∫ BC + ∫ ZA + ∫ CD + ∫ YZ + ∫ IJ + ∫ ST + ∫ KLM + ∫ NPQ + ∫ JK + ∫ MN + ∫ QS + ∫ DE + ∫ XY + ∫ HI + ∫ TU + ∫ FG + ∫ VW + ∫ EF + ∫ GH + ∫ UV + ∫ WX =0$

From Eq. (44)it follows that $lim_{R→∞}} ∫ AB e^{pt} θ ̄_{1} (r,p) d p=2π i θ_{1} (r,t).$

On the circular arcs HI and TU, we have p=δe^iθ, and it can easily be shown that $∫$

Stress distribution

Integrating the function $σ ̄_{1} (r,p)$ along the contour Γ and discarding the integrals with a value of zero (the same as above), we obtain $∫ AB e^{pt} σ ̄_{1} (r,p) d p+ ∫ EF∪GH e^{pt} σ ̄_{1} (r,p) d p+ ∫ UV∪WX e^{pt} σ ̄_{1} (r,p) d p+ ∫ FG∪VW e^{pt} σ ̄_{1} (r,p) d p=0.$

From the inversion formula (39) it follows that $lim_{R→∞} ∫ AB e^{pt} σ ̄_{1} (r,p) d p=2π i σ_{1} (r,t).$

Integrating along the lines EF, GH, UV, and WX, we obtain $lim_{δ→0} ∫ EF e^{pt} σ ̄_{1} (r,p) d p+ ∫ GH e^{pt} σ ̄_{1} (r,p) d p+ ∫ UV e^{pt} σ ̄_{1} (r,p) d p+ ∫ WX e^{pt} σ ̄_{1} (r,p) d p = −2 i ac r^{3} ∫ 0 1/τ e^{−xt} (1−νx)[x^{2} a^{2} β^{2} +4aλ_{2} (x)+4]I_{3} (x,r) (α−x)[λ^{2}_{1} (x)+λ^{2}_{2}$

Displacement distribution

Repeating the same procedure as above, we obtain the displacement functions u₁(r, t) and u₂(r, t) in the following form: $u_{1} (r,t)= ac e^{−αt} (1−να)[α^{2} a^{2} β^{2} +4aλ_{2} (α)+4]R_{5} (α,r) r^{2} [λ_{1}^{2} (α)+λ^{2}_{2} (α)] Δ (α) − ac πr^{2} ∫ 0 1/τ e^{−xt} (1−νx)[x^{2} a^{2} β^{2} +4aλ_{2} (x)+4]I_{5} (x,r) (α−x)[λ^{2}_{1} (x)+λ^{2}_{2} (x)] Δ (x) d x,$ where $R_{5} (x,r)= [λ^{2}_{1} (x)+λ^{2}_{2} (x)](a^{2} β^{2} x^{2} +4)+4a λ_{2} (x)[λ^{2}_{1} (x)+x^{2}]+λ^{2}_{1} (x)r[λ^{2}_{2} (x)−x^{2}] cos [λ_{1} (x)(r−a)] +λ_{1} (x) [λ^{2}_{1} (x)+λ^{2}_{2} (x)]r(a^{2} β^{2} x^{2} +4)+4a λ_{2} (x)r[λ^{2}_{1} (x)+x^{2}] −[λ^{2}_{2} (x)−x^{2}] sin [λ_{1} (x)(r−a)],$ $I_{5} (x,r)= λ_{1} (x) r[λ^{2}_{1} (x)+λ^{2}_{2} (x)](a^{2} β^{2} x^{2} +4)+4a λ_{2} (x)r[λ^{2}_{1} (x)+x^{2}] −[λ^{2}_{2}$

Numerical results

The copper material was chosen for purposes of numerical evaluations. The constants of the problem were taken as ε=0.0168 and β²=3.5. The computations were carried out for three values of time, namely for t=0.1, 0.2 and 0.5. We note here that the approach used in this work to invert the Laplace transforms is valid for all values of time, not for short times only. We have also taken τ=0.02 and ν=0.02. The radius a of the cylinder and the constant α in formula (37) were both taken equal to 1.

The

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International Journal of Engineering Science

A problem for an infinite thermoelastic body with a spherical cavity

Abstract

Introduction

Section snippets

Formulation of the problem

Solution in the Laplace transform domain

Inversion of the Laplace transforms

Temperature distribution

Stress distribution

Displacement distribution

Numerical results

J. Mech. Phys. Solid

Int. J. Engng. Sci.

Int. J. Engng. Sci.

Int. J. Engng. Sci.

J. Appl. Phys.

Quart. Appl. Math.

J. Thermal Stresses

J. Thermal Stresses

J. Thermal Stresses

Quart. Appl. Math.

J. Thermal Stresses

J. Thermal Stresses