International Scholarly Research Notices

International Scholarly Research Notices / 2011 / Article

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Volume 2011 |Article ID 305402 |

M. Jabbari, S. Karampour, M. R. Eslami, "Radially Symmetric Steady State Thermal and Mechanical Stresses of a Poro FGM Hollow Sphere", International Scholarly Research Notices, vol. 2011, Article ID 305402, 7 pages, 2011.

Radially Symmetric Steady State Thermal and Mechanical Stresses of a Poro FGM Hollow Sphere

Academic Editor: M. Al-Nimr
Received07 Apr 2011
Accepted05 May 2011
Published28 Jul 2011


A general solution for the one-dimensional steady-state thermal and mechanical stresses in a hollow thick sphere made of porous functionally graded material (FGPM) is presented. The temperature distribution is assumed to be a function of radius, with general thermal and mechanical boundary conditions on the inside and outside surfaces of the sphere. The material properties, except Poisson's ratio, are assumed to vary along the radius r according to a power law function. The analytical solution of the heat conduction equation and the Navier equation lead to the temperature profile, radial displacement, radial stress, and hoop stress as a function of radial direction.

1. Introduction

Functionally graded materials (FGMs) are made of a mixture with arbitrary composition of two different materials, and the volume fraction of each material changes continuously and gradually. The FGMs concept is applicable to many industrial fields such as chemical plants, electronics, and biomaterials [1]. Thick hollow sphere analysis made of FGM under mechanical and thermal loads and in asymmetric and two-dimensional (𝑟,𝜃) state was conducted investigating navier equations and using legendre polynomials [2]. The analytical solution for the stresses in spheres and cylinders made of functionally graded materials are given by Lutz and Zimmerman [3, 4]. They considered thick spheres and cylinders under radial thermal loads, where radially graded materials with linear composition of the constituent materials were considered. Obata and Noda [5] studied one-dimensional steady thermal stresses in a functionally graded circular hollow cylinder and a hollow sphere using the perturbation method. The transient thermal stresses in a plate made of FGM are presented by the same authors [6]. By introducing the theory of laminated composites, Ootao and Tanigawa derived the three-dimensional transient thermal stresses of a nonhomogeneous hollow sphere with a rotating heat source [7]. Jabbari et al. presented the analytical solution of one and two-dimensional steady state thermoelastic problems of the FGM cylinder [8]. Two-dimensional nonaxisymmetric transient mechanical and thermal stresses in a thick hollow cylinder is presented by Jabbari et al. [9]. Shao and Wang presented three-dimensional solution to obtain stress fields in an FGM cylindrical plate with a finite length which is mechanical and thermal load tests and showed the results an a graph [10]. The transient thermal stress problem in a hollow sphere with homogeneous and isotropic properties is analytically solved by Cheung et al. [11]. Their assumed thermal boundary conditions are varied along the 𝛾 direction and the problem is solved with the potential function method. Takeuti and Tanigawa have employed the potential function method to obtain the analytical solution of a homogeneous spherical vessel with rotating heat source [12]. They considered the general form of thermal boundary conditions and analyzed the problem in transient condition with the potential functions method.

Porous spheres of nanometer to micrometer dimensions are being pursued with great interest because of several possible technical applications in catalysis, drug delivery systems, separation techniques, photonics, as well as piezoelectric and other dielectric devices [13, 14]. The study of the thermomechanical response of fluid saturated porous materials is important for several branches of engineering [15–19]. Some of the important cases are the disposal of high-level radioactive waste, the extraction of oil or geothermal energy, the storage of hot fluids, and road subgrade or furnace foundations, which are usually subjected to cyclic changes of temperature. Wang and Papamichos [20, 21] presented analytical solutions for the temperature, pore pressure and stresses around a cylindrical wellbore and a spherical cavity subjected to a constant temperature change and a constant fluid flow rate by coupling the conductive heat transfer with the pore-fluid flow. Using the so-called heat source function method. Kurashige [22] analyzed the heat and fluid flow problem of a spherical cavity subjected to a constant heat flux in an impermeable medium.

Inspite of conducted studies on spherical and cylindrical vessel made of FGM to obtain mechanical displacements and mechanical and thermal stresses, there is not any study on composition of poro and FGM materials. This study investigates the effect of mechanical and thermal stresses on one-dimensional steady state in the poro FGM hollow sphere that fluid trapped in the poro medium which is located in undrained conditions.

2. Analysis

Consider a thick hollow sphere of inside radius ğ‘Ž and outside radius 𝑏 made of poro FGM. The sphere material is graded through the radial 𝑟-direction. Thus, the material properties are functions of 𝑟. Let 𝑢 be the displacement component along the radial direction. The strain-displacement relations are 𝜀𝑟𝑟=𝑢,𝑟,𝜀𝜃𝜃=𝜀𝜙𝜙=𝑢𝑟.(1) The stress-strain relations one-dimensional in the poro FGM hollow sphere with fluid trapped in the poro pure are ğœŽğ‘Ÿğ‘Ÿ=𝐶11𝜀𝑟𝑟+2𝐶12ğœ€ğœƒğœƒâˆ’ğ›¾ğ‘ğ›¿ğ‘Ÿğ‘Ÿâˆ’ğ‘§ğ‘ŸğœŽğ‘‡(𝑟),𝜃𝜃=ğœŽğœ™ğœ™=𝐶12𝜀𝑟𝑟+𝐶22+𝐶23𝜀𝜃𝜃−𝛾𝑝𝛿𝜃𝜃−𝑧𝜃𝑇(𝑟),(2) where 𝑧𝑟=𝐶11𝛼𝑟+2𝐶12𝛼𝜃,𝑧𝜃=𝐶12𝛼𝑟+𝐶22+𝐶23𝛼𝜃,𝛼𝑟=𝛼𝜃,(3) where ğœŽğ‘–ğ‘— and 𝜀𝑖𝑗(𝑖,𝑗=𝑟,𝜃) are the stress and strain tensors, 𝑇(𝑟) is the temperature distribution determined from the heat conduction equation, 𝛼𝑖 is the coefficient of thermal expansion of effective stress, and 𝐶𝑖𝑗, 𝛾, 𝑝, and 𝛿𝑖𝑖(𝑖=𝑟,𝜃) are elastic constants, Biots coefficient of effective stress, poro pressure, and kronecker delta.

Equations in a porous material are 𝑘𝑝=𝑀(𝜁−𝛾∈),𝑀=𝑢−𝑘𝛾2,𝑘𝐵=𝑢−𝑘𝛾𝑘𝑢,𝑘𝑢𝛾=𝑘1+2𝑘𝑓(1−𝛾)𝛾−𝜙𝑝𝑘𝑓+𝜙𝑝𝑘𝑓.(4) For fluid in undrained condition, 𝜀𝜁=0,𝑝=−𝛾𝑀∈=−𝛾𝑀𝑟𝑟+𝜀𝜃𝜃+𝜀𝜙𝜙,𝜀𝜃𝜃=𝜀𝜙𝜙,(5) where 𝜁, ∈, 𝑀, 𝑘𝑢, 𝑘, 𝑘𝑓, 𝐵, and 𝜙𝑝are variation of fluid content, volumetric strain, Biots modulus, undrained bulk modulus, drained bulk modulus, bulk modulus of fluid, compressibility coefficient, and porosity.

The equilibrium equation in the radial direction, disregarding the body force and the inertia term, is ğœŽğ‘Ÿğ‘Ÿ,𝑟+2ğ‘Ÿî€·ğœŽğ‘Ÿğ‘Ÿâˆ’ğœŽğœƒğœƒî€¸=0.(6) Substituting (5) into (2) leads to ğœŽğ‘Ÿğ‘Ÿ=𝑐∗1𝜀𝑟𝑟+𝑐∗2ğœ€ğœƒğœƒâˆ’ğ‘§ğ‘ŸğœŽğ‘‡(𝑟),𝜃𝜃=ğœŽğœ™ğœ™=𝑐∗3𝜀𝑟𝑟+𝑐∗4𝜀𝜃𝜃−𝑧𝜃𝑇(𝑟),(7) where 𝑐∗1=𝐶11+𝛾2𝑐𝑚,∗2𝐶=212+𝛾2𝑚,𝑐∗3=𝐶12+𝛾2𝑐𝑚,∗4=𝐶22+𝐶23+2𝛾2𝑚.(8) The sphere’s material is assumed to be described with a power law function of the radial direction as 𝑐∗𝑖=𝑐∗𝑖𝑟𝑚,𝑧𝑖=𝑧𝑖𝑟2𝑚,(9) where 𝑚 is the power law index of the material.

Using relations (1)–(9), the Navier equation in term of the displacement is 𝑢,𝑟𝑟+1𝑟(𝑚+2)𝑐∗1+𝑐∗2−2𝑐∗3𝐶∗1𝑢,𝑟+1𝑟2(𝑚+1)𝑐∗2−2𝑐∗4𝑐∗1=𝑧𝑟𝑐∗1(𝑚+1+2𝐿)𝑟𝑚−2+2𝑐∗2(𝑚+1−𝐿)𝑟2𝑚−1,(10) where 𝐿=𝑧𝜃𝑧𝑟.(11)

3. Heat Conduction Problem

The heat conduction equation in the steady-state condition for the one-dimensional problem in spherical coordinates and the thermal boundary conditions for a poro FGM hollow sphere are given, respectively, as 1𝑟2𝑟2𝑘(𝑟)ğ‘‡î…žî€¸(𝑟)î…žğ‘=0,11ğ‘‡î…ž(ğ‘Ž)+𝑐12𝑇(ğ‘Ž)=𝑓1,𝑐21ğ‘‡î…ž(𝑏)+𝑐22𝑇(𝑏)=𝑓2,(12) where 𝑘=𝑘(𝑟) is the thermal conduction coefficient and 𝑐𝑖𝑗 are either thermal conduction coefficient 𝑘, or convection coefficient ℎ, depending on the type of thermal boundary conditions. The terms 𝑓1 and 𝑓2 are known constants on the inside and outside radii. It is assumed that the thermal conduction coefficient 𝑘(𝑟) is a power function of 𝑟 as 𝑘=𝑘(𝑟)𝑟𝑚.(13) Using (13), the heat conduction equation becomes 1𝑟2𝑟𝑚+2ğ‘‡î…žî€»(𝑟)=0.(14) Integrating (12) twice yields 𝑇(𝑟)=𝑐1𝑟−(𝑚+1)+𝑐2.(15) Using the boundary conditions (16) and (17) to determine the constants 𝑐1 and 𝑐2 yields𝑐1=𝑐22𝑓1−𝑐12𝑓2𝑐12(𝑚+1)𝑐21𝑏−(𝑚+2)−𝑐22𝑏−(𝑚+1)−𝑐22(𝑚+1)𝑐11ğ‘Žâˆ’(𝑚+1)−𝑐11ğ‘Žâˆ’(𝑚+1),(16)𝑐2=𝑓1(𝑚+1)𝑐21𝑏−(𝑚+2)𝑐22𝑏−(𝑚+1)−𝑓2(𝑚+1)𝑐11ğ‘Žâˆ’(𝑚+2)−𝑐12ğ‘Žâˆ’(𝑚+1)𝑐12(𝑚+1)𝑐21𝑏−(𝑚+2)𝑐22𝑏−(𝑚+1)−𝑐22(𝑚+1)𝑐11ğ‘Žâˆ’(𝑚+2)−𝑐12ğ‘Žâˆ’(𝑚+1).(17)

4. Solution of the Navier Equation

The Navier equation for the radial displacement 𝑢 was given in (10). Equation (10) is the nonhomogeneous Euler differential equation with the general and particular solutions. The general solution, 𝑢𝑔, is obtained by assuming𝑢𝑔(𝑟)=𝐵𝑟𝜂.(18) Substituting (18) into the homogeneous form of (10) yields 𝜂2+(𝑚+1)𝑐∗1+𝑐∗2−2𝑐∗3𝑐∗1𝜂+(𝑚+1)𝑐∗2−2𝑐∗4𝑐∗1=0,(19) where (19) has two real roots 𝜂1 and 𝜂2 as𝜂1,2=−(𝑚+1)𝑐∗1+𝑐∗2+2𝑐∗3±(𝑚+1)𝑐∗1+𝑐∗2+2𝑐∗32−4𝑐∗1(𝑚+1)𝑐∗2−2𝑐∗42𝑐∗1.(20)Thus, the general solution is 𝑢𝑔(𝑟)=𝐵1𝑟𝜂1+𝐵2𝑟𝜂2.(21) The particular solution 𝑢𝑝(𝑟)is assumed to have the form 𝑢𝑝(𝑟)=𝐷1𝑟𝑚+𝐷2𝑟2𝑚+1.(22)

Substituting (22) in (10) yields 𝑚(𝑚−1)+(𝑚+2)𝑐∗1+𝑐∗2−2𝑐∗3𝑐∗1𝑚+(𝑚+1)𝑐∗2+2𝑐∗4𝑐∗1𝐷1𝑟𝑚−2+((2𝑚+1)(2𝑚)+𝑚+2)𝑐∗1+𝑐∗2+2𝑐∗3𝑐∗1×(2𝑚+1)+(𝑚+1)𝑐∗2+2𝑐∗4𝑐∗1𝐷2𝑟2𝑚−1=𝑧𝑟𝑐∗1(𝑚+1−2𝐿)𝑟𝑚−2+2𝑐∗2(𝑚+1−𝐿)𝑟2𝑚−1.(23) Equating the coefficients of the identical powers yields𝐷1=𝑧𝑟𝑐∗1(𝑚+1)−2𝐿𝑚(𝑚−1)𝑐∗1+(𝑚+2)𝑐∗1+𝑐∗2+2𝑐∗3𝑚+(𝑚+1)𝑐∗2+2𝑐∗4,(24)𝐷2=𝑧𝑟2𝑐∗2(𝑚+1−𝐿)(2𝑚+1)(2𝑚)𝑐∗1+(𝑚+2)𝑐∗1+𝑐∗2+2𝑐∗3((2𝑚+1)+𝑚+1)𝑐∗2+2𝑐∗4.(25) The complete solution for 𝑢(𝑟) is the sum of the general and particular solutions as 𝑢(𝑟)=𝑢𝑔(𝑟)+𝑢𝑝(𝑟).(26) Thus, 𝑢(𝑟)=𝐵1𝑟𝜂1+𝐵2𝑟𝜂2+𝐷1𝑟𝑚+𝐷2𝑟2𝑚+1.(27) Substituting (27) into (1) and (2), the strains and stresses are obtained as 𝜀𝑟𝑟=𝐵1𝜂1𝑟𝜂1−1+𝐵2𝜂2𝑟𝜂2−1+𝐷1𝑚𝑟𝑚−1+𝐷2(2𝑚+1)𝑟2𝑚,𝜀𝜃𝜃=𝐵1𝑟𝜂1−1+𝐵2𝑟𝜂2−1+𝐷1𝑟𝑚−1+𝐷2𝑟2𝑚,ğœŽğ‘Ÿğ‘Ÿ=𝑐∗1𝐵1𝜂1𝑟+â„Žğœ‚1−1+𝐵2𝜂2𝑟+â„Žğœ‚2−1+𝐷1(𝑚+ℎ)𝑟𝑚−1+𝐷2(2𝑚+1+ℎ)𝑟2𝑚−𝑆𝑐∗1𝑟−(𝑚+1)+𝑐∗2,ğœŽî€¸î€»ğœƒğœƒ=𝑐∗1𝐵1𝜂1𝑟+𝑓𝜂1−1+𝐵2𝜂2𝑟+𝑓𝜂2−1+𝐷1(𝑚+𝑓)𝑟𝑚−1+𝐷2(2𝑚+1+𝑓)𝑟2𝑚−𝑔𝑐∗1𝑟−(𝑚+1)+𝑐∗2,(28) where ℎ=𝑐∗2𝑐∗1,𝑆=𝑧𝑟𝑐∗1,𝑓=𝑐∗4𝑐∗3,𝑔=𝑧𝜃𝑐∗3.(29) To determine the constants 𝐵1 and 𝐵2, the boundary conditions for stresses may be used. Consider the mechanical boundary conditions on the inside and outside radii as ğœŽğ‘Ÿğ‘Ÿ(ğ‘Ž)=âˆ’ğ‘ƒğ‘Ž,ğœŽğ‘Ÿğ‘Ÿ(𝑏)=−𝑃𝑏.(30)

5. Numerical Results and Discussion

The analytical solution obtained in the previous section may be checked for a one of example. Then, consider a thick hollow sphere of inner radius ğ‘Ž = 1 m, and the outer radius 𝑏 = 1.2 m of poro FGM material with properties are given in Table 1. The boundary conditions for temperature are taken as 𝑇(ğ‘Ž)=10∘C and 𝑇(𝑏)=0∘C. The hollow sphere may be assumed to be under internal pressure of 50 MPa and zero external pressure (ğœŽğ‘Ÿğ‘Ÿ(ğ‘Ž) = −50 MPa andğœŽğ‘Ÿğ‘Ÿ(𝑏) = 0 MPa). For different values of 𝑚, temperature profile, radial displacement, radial stresses, and hoop stresses along the radial direction are plotted in Figures 1–4. Figure 1 shows that as the power law increases, the temperature decreases. Figure 2 shows that for higher values of 𝑚, radial displacement decreases. Figure 3 represents the radial stress along the radial direction and decreases as the power law index increases. The circumferential stress versus the radial direction is shown in Figure 4. It is seen that for 𝑚<1, the circumferential stress decreases along the radial direction. When 𝑚>1, the situation is reversed, and the circumferential stress increases along the radial direction. The curve associated with 𝑚=1 shows that the variation of circumferential stress along the radial direction is minor and is almost uniform across the radius. Effective porosity and compressibility on mechanical radial displacement in Figures 5 and 6, radial stress in Figures 7 and 8 and hoop stress in Figures 9 and 10 for 𝑚=1 are shown. As noted from Figures 5–10, distribution that with increasing porosity increases radial displacement, and all mechanical stresses decrease. Also, with increasing compressibility, radial displacement decreases, and all mechanical stresses increases.

Elastic constantsPoro constants

𝐶 1 1 = 1 . 8 6 7 × 1 0 1 1 𝛾 = 0 . 2 7
𝐶 1 2 = 0 . 8 1 9 × 1 0 1 1 𝑘 𝑢 = 4 . 1 × 1 0 1 0
𝐶 2 2 = 1 . 6 2 8 × 1 0 1 1 𝑘 = 3 . 5 × 1 0 1 0
𝐶 2 3 = 0 . 2 7 0 × 1 0 1 1 𝑘 𝑓 = 3 . 3 × 1 0 3
𝛼 𝑟 = 1 . 2 × 1 0 − 6 𝐵 = 0 . 4 5
𝜙 𝑝 = 0 . 0 1


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