# Efficient Analytical Model for Long-Term Performance of Laminated Tube with Viscoelastic Interlayer by Quasi-Elastic Approximation

Authors: Yubing Qin; Kong Yue; Peng Wu
Journal: Mathematical Problems in Engineering
 (2023)

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## Abstract

An efficient analytical method is proposed so as to study the long-term performance of laminated tubes with neighboring layers glued by viscoelastic interlayers bearing inhomogeneous long-duration loads. In the analysis, the formulations for describing the mechanical performance of each tube layer are based on the elasticity theory, while the viscoelastic behavior of the glue interlayer is modeled by the generalized Maxwell model. Making use of the quasielastic approximation and the recursive matrix method, the analytical solution for the long-term performance of the laminated tube is solved efficiently. The present analytical solution is verified by comparing it with other solutions. The influences of the interlayer thickness, the material property, as well as the load distribution on the long-term performance of the laminated tube are discussed.


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## Body

## 1. Introduction

In the past decades, with the progress of structure research, the performance of layered structure has been greatly improved. The high bearing capacity, lightweight, and corrosion resistance characteristics make the layered structures have more possibilities in the structural engineering, for example, structural insulating panels, glued layered timbers, laminated tubes, and many more [1–3]. Besides the layered beams and plates, the applications about laminated tubes are increasingly applied in infrastructures [4–7]. Generally, a laminated tube is composed of two or more subcomponents of different materials, and the adjacent members are usually connected discontinuously by studs, nails, and dowels or continuously glued by adhesives [8]. Due to the characteristics of the constituent materials, practical problems exist for the laminated structure under different external conditions [9–12]. One of them is that the interfacial slip is inevitable to occur because the kinds of connections can hardly be rigid. In addition, the adhesives to bond layers are usually viscoelastic materials, which lead to long-term performance for layered systems. In practical engineering, concrete-filled fiber-reinforced polymer tubes are employed in structural applications including marine constructions and bridges. However, weak adhesion will reduce the rigidity of the pipes, resulting in the interfacial slip [13]. Additionally, the fluid-conveying pipes, which are made up of two reinforced composite layers with a viscoelastic interlayer, are widely applied in the aircraft and mechanical systems. When the pipes are subjected to sustained pressure, the viscoelasticity of the interlayer can affect the long-term performance of the pipe [14]. These phenomena deserve further study and discussion.A lot of works about the mechanical performance of laminated tubes have been reported in the literature. Xia et al. [15] proposed three-dimensional elasticity solutions to investigate the stress and deformation fields in laminated tubes under internal pressures. Zheng et al. [16] proposed a theoretical model to predict the short-term burst pressure of polyethylene tubes with winding steel wires under different temperatures. The 3-D vibration performance of a periodic composite tube was analyzed by Shen et al. [17] by the use of the transfer matrix method. Based on the three-dimensional anisotropic elasticity, Bakaiyan et al. [18] gave the elasticity solutions for thermal stress and displacement fields of the laminated tubes under temperature environment, and they found that the shear stress and strain in the tube strongly depend on the layer stacking sequence. Hashemian and Mohareb [19] developed a finite difference model to study the buckling of sandwich tubes under external pressure. By using the finite element (FE) modeling, Das and Baishya [20] analyzed the failure modes of layer tubes with socket joints. Exact solutions were proposed by Xia et al. [21] for filament wound fiber reinforced laminated tubes under pure bending load. By employing the FE method, Xu et al. [22] studied the collapse performance of sandwich tubes under external pressures. Zhu et al. [23] analyzed a new GFRP laminated tube with a stiffened core under burial condition in a theoretical way, and they also found a way to improve the strength-to-weight ratio of the tubes by optimizing the geometric parameters. Fernandez-Valdes et al. [24] presented the analysis of friction and initial imperfection of the sandwich tubes by virtue of the FE method. Closed-form analytical solutions of sandwich tubes containing functionally graded interlayers were presented by Sburlati and Kashtalyan [25] based on the elasticity theory, and the results indicated the graded interlayers can remarkably reduce the hoop stress in the tube.In most of the existing literature for laminated tubes, the neighboring layers are considered to be perfectly connected or have static stiffness, while the long-term performance of viscoelastic adhesives to bond adhesive layers is neglected. Moreover, in most cases, only the homogeneous loads are considered for the laminated tube. In reality, laminated tubes commonly bear inhomogeneous radial loads; for instance, the underground tubes are subjected to inhomogeneous soil pressures along the circumference of the tube [26–28].In the proposed work, an analytical model is presented for studying laminated tubes with infinite length, with neighboring layers glued by viscoelastic interlayers and subjected to inhomogeneous radial loads along the circumferential direction. The formulations for describing the mechanical performance of each tube layer are based on the elasticity theory, while the viscoelastic behavior of the glue interlayer is modeled by the generalized Maxwell model. By means of the quasielastic approximation as well as the recursive matrix method, the solutions for the long-term behavior of the laminated tube are achieved analytically. The present solution based on the elasticity theory has advantages over other solutions. Compared with the Ritz solution, the present solution has higher accuracy because the Ritz method has inherent approximation, and the accuracy of its results depends greatly on the trial functions assumed [29]. The Navier method is often used to solve the mechanical problems of plates and shells, but it has a restrictive condition for boundary [30]. In addition, compared with the Euler–Bernoulli theory [31], the present solution is more accurate since the elasticity theory introduces no assumption of transverse shear deformation.


## 2. Analytical Model for the Laminated Tube

The reference structure presented in Figure1 is a laminated tube with inside radius Rin, outside radius Rout, thickness H and infinite length, composed of p elastic tube layers glued by viscoelastic interlayers. The thickness of the i-th layer and the interlayer are, respectively, hi and ∆h, where the subscript i means the layer index, and ∆h is thin enough in comparison with hi. The tube bears inhomogeneous radial loads qinθ and qoutθ, respectively, acting on the inner and outer surfaces.Figure 1
Laminated tube glued by viscoelastic interlayer.


### 2.1. Equations for an Elastic Tube Layer

The referential tube can be seen as the two-dimensional plan strain problem, because the material properties, geometric shape, and external loads remain unchanged in the longitudinal direction. Based on the elasticity theory [32], the constitutive equations for the i-th (i = 1, 2, …, p) tube layer can be given by the following equation:(1)σri=1−μiEi1+μi1−2μiεri+μi1−μiεθi,σθi=1−μiEi1+μi1−2μiμi1−μiεri+εθi,τrθi=Ei21+μiγrθi,in which, σri, σθi, and τrθi are the radial, circumferential, and shear stresses, respectively, while εri, εθi, and γrθi are the corresponding strains. The strain-displacement relationships for the i-th tube layer are(2)εri=∂uri∂r,εθi=urir+1r∂uθi∂θ,γrθi=1r∂uri∂θ+∂uθi∂r−uθir,where uri and uθi are the radial and circumferential displacements, respectively. The stress components keep the equilibrium equations as follows:(3)∂σri∂r+1r∂τrθi∂θ+σri−σθir=0,1r∂σθi∂θ+∂τrθi∂r+2τrθir=0.Substituting equation (2) into equation (1), one has(4)σri=1−μiEi1+μi1−2μi∂uri∂r+μi1−μiurir+μi1−μi1r∂uθi∂θ,σθi=1−μiEi1+μi1−2μiμi1−μi∂uri∂r+urir+1r∂uθi∂θ,τrθi=Ei21+μi1r∂uri∂θ+∂uθi∂r−uθir.Then, by combining equations (3) and (4) and eliminating the stress components, we have(5)12∂2uθi∂r2+121r∂uθi∂r−12uθir2+121−2μi1r∂2uri∂r∂θ+3−4μi21−2μi1r2∂uri∂θ+1−μi1−2μi1r2∂2uθi∂θ2=0,1−μi1−2μi∂2uri∂r2+1−μi1−2μi1r∂uri∂r−1−μi1−2μiurir2+121−2μi1r∂2uθi∂r∂θ−3−4μi21−2μi1r2∂uθi∂θ+121r2∂2uri∂θ2=0.In order to solve the partial differential equations, the displacement components are beforehand expanded into Fourier series, i.e.,(6)uθir,θ,t=Θi0r,t+∑m=1∞Θim1r,tsinmθ+∑m=1∞Θim2r,tcosmθ,urir,θ,t=Ri0r,t+∑m=1∞Rim1r,tcosmθ+∑m=1∞Rim2r,tsinmθ.Substitution of equation (6) into equation (5) yields(7)1−2μ2d2Θi0dr2+1rdΘi0dr−1r2Θi0+∑m=1∞1−2μi2d2Θim1dr2+1−2μi21rdΘim1dr−1−μim2+1−2μi21r2Θim1+m21rdRim2dr+m3−4μi21r2Rim2sinmθ+∑m=1∞1−2μi2d2Θim2dr2+1−2μi21rdΘim2dr−1−μim2+1−2μi21r2Θim2−m21rdRim1dr−m3−4μi21r2Rim1cosmθ=0,1−2μi2d2Ri0dr2+1rdRi0dr−1r2Ri0+∑m=1∞1−μid2Rim2dr2+1−μi1rdRim2dr−1−μi+m221−2μi1r2Rim2−m21rdΘim1dr+m3−4μi21r2Θim1cosmθ+∑m=1∞1−μid2Rim1dr2+1−μi1rdRim1dr−1−μi+m221−2μi1r2Rim1+m21rdΘim2dr−m3−4μi21r2Θim2sinmθ=0.According to the orthogonality of the Fourier series, a set of ordinary differential equations ofΘi0, Θimn, Ri0, and Rimn is obtained, as follows:(8)d2Θi0dr2+1rdΘi0dr−1r2Θi0=0,d2Ri0dr2+1rdRi0dr−1r2Ri0=0,1−2μi2d2Θim1dr2+1−2μi21rdΘim1dr−1−μim2+1−2μi21r2Θim1+m21rdRim2dr+m3−4μi21r2Rim2=0,1−μid2Rim2dr2+1−μi1rdRim2dr−1−μi+m221−2μi1r2Rim2−m21rdΘim1dr+m3−4μi21r2Θim1=0,1−2μi2d2Θim2dr2+1−2μi21rdΘim2dr−1−μim2+1−2μi21r2Θim2−m21rdRim1dr−m3−4μi21r2Rim1=0,1−μid2Rim1dr2+1−μi1rdRim1dr−1−μi+m221−2μi1r2Rim1+m21rdΘim2dr−m3−4μi21r2Θim2=0.By solving the group of differential equations and substituting the results into equation (4), one obtains the general solutions of displacements and stresses.(9)urir,θ,t=rAi0t+r−1Bi0t+∑m=1∞cosmθαim1rm+1Aim1t+r−m−1Bim1t−rm−1Cim1t+αim2r−m+1Dim1t+∑m=1∞sinmθ−αim1rm+1Aim2t−r−m−1Bim2t+rm−1Cim2t−αim2r−m+1Dim2tuθir,θ,t=∑m=1∞sinmθrm+1Aim1t+r−m−1Bim1t+rm−1Cim1t+r−m+1Dim1t+∑m=1∞cosmθrm+1Aim2t+r−m−1Bim2t+rm−1Cim2t+r−m+1Dim2tσrir,θ,t=ρi1Ai0t+ρi2r−2Bi0t+∑m=1∞cosmθβim1rmAim1t+βim2r−m−2Bim1t+βim3rm−2Cim1t+βim4r−mDim1t+∑m=1∞sinmθ−βim1rmAim2t−βim2r−m−2Bim2t−βim3rm−2Cim2t−βim4r−mDim2tσθir,θ,t=ρi1Ai0t−ρi2r−2Bi0t+∑m=1∞cosmθχim1rmAim1t+χim2r−m−2Bim1t+χim3rm−2Cim1t+χim4r−mDim1t+∑m=1∞sinmθ−χim1rmAim2t−χim2r−m−2Bim2t−χim3rm−2Cim2t−χim4r−mDim2tτrθir,θ,t=∑m=1∞sinmθζim1rmAim1t+ζim2r−m−2Bim1t+ζim3rm−2Cim1t+ζim4r−mDim1t+∑m=1∞cosmθζim1rmAim2t+ζim2r−m−2Bim2t+ζim3rm−2Cim2t+ζim4r−mDim2t,where(10)αim1=−m+4μi−2m−4μi+4,αim2=m−4μi+2m+4μi−4,ρi1=Ei1+μi1−2μi,ρi2=−Ei1+μiβim1=−m+1m−2Ei1+μim−4μi+4,βim2=−m+1Ei1+μi,βim3=−m−1Ei1+μi,βim4=−m−1m+2Ei1+μim+4μi−4,χim1=m+1m+2Ei1+μim−4μi+4,χim2=m+1Ei1+μiχim3=m−1Ei1+μirm−2,χim4=m−1m−2Ei1+μim+4μi−4,ζim1=mm+1Ei1+μim−4μi+4,ζim2=−m+1Ei1+μi,ζim3=m−1Ei1+μiζim4=−mm−1Ei1+μim+4μi−4,and Ai0t, Bi0t, Aimnt, Bimnt, Cimnt, and Dimnt are unknown coefficients.


### 2.2. Equations for a Viscoelastic Interlayer

It is verified that the generalized Maxwell model (Figure2) is appropriate for the viscoelastic bonding interlayer [31, 33], with the modulus given by the following equation:(11)E∗t=E∞∗+∑j=1MEj∗e−t/θE,j∗,G∗t=G∗t21+μ∗,where the mark ∗represents the variable belonging to the interlayer; μ∗, E∞∗, Ej∗, and θE,j∗ denote Poisson’s ratio, the long-term moduli, the relaxation moduli, and the relaxation time. By the use of the quasielastic approximation, the constitutive equations of the i-th (i = 1, 2, …, p−1) interlayer are given by [34] the following equation:(12)σr∗iθ,t=E∗tεr∗iθ,t,τrθ∗iθ,t=G∗tγrθ∗iθ,t.Figure 2
Schematic diagram of the generalized Maxwell model.The strain-displacement relationships for the thin interlayer are(13)εr∗iθ,t=uri+1r0i+1,θ,t−urirhi,θ,tΔh,γrθ∗iθ,t=uθi+1r0i+1,θ,t−uθirhi,θ,tΔh+1rhi∂urirhi,θ,t∂θ−uθirhi,θ,trhi,where r0i=∑j=1i−1hj and rhi=∑j=1ihj. In addition, the equilibrium relations of the interlayer and the neighboring tube layers are(14)σri+1r0i+1,θ,t=σr∗iθ,t=σrirhi,θ,t,τrθi+1r0i+1,θ,t=τrθ∗iθ,t=τrθirhi,θ,t.By combining equations (11)–(13) and eliminating the stresses and strains of the interlayer, one obtains(15)σrirhi,θ,t=E∗t−urirhi,θ,tΔh+uri+1r0i+1,θ,tΔhτrθirhi,θ,t=G∗t1rhi∂urirhi,θ,t∂θ−1Δh+1rhiuθirhi,θ,t+uθi+1r0i+1,θ,tΔhσrirhi,θ,t−σri+1r0i+1,θ,t=0τrθirhi,θ,t−τrθi+1r0i+1,θ,t=0,i=1,2,…,p−1.


### 2.3. Coefficient Determination by Recursive Matrix Method

In this section, the coefficients in the general solutions of equation (9) are determined by the recursive matrix method. The loads acting on the tube can be expressed by the following equation:(16)σr1=−qinθ,τrθ1=0,atr=Rin,σrp=−qoutθ,τrθp=0,atr=Rout,in which loads functions are expanded as(17)qinθ=qin0+∑m=1∞qinm1cosmθ+∑m=2∞qinm2sinmθ,qoutθ=qout0+∑m=1∞qoutm1cosmθ+∑m=2∞qoutm2sinmθ,where(18)qin0=12π∫−ππqinθdθ,qout0=12π∫−ππqoutθdθ,qinm1=1π∫−ππqinθcosmθdθ,qoutm1=1π∫−ππqoutθcosmθdθ,qinm2=1π∫−ππqinθsinmθdθ,qoutm2=1π∫−ππqoutθsinmθdθ.The out-of-plane displacements and stresses in equation (12) are rearranged by matrix form.(19)uθir,θ,turir,θ,tσrir,θ,tτrθir,θ,t=0Ri0r,tYi0r,t0+∑m=2∞Θim1r,tsinmθRim1r,tcosmθYim1r,tcosmθZim1r,tsinmθ+∑m=2∞Θim2r,tcosmθRim2r,tsinmθYim2r,tsinmθZim2r,tcosmθ,in which the functions Ri0, Yi0, Θimn, Rimn, Yimn, and Zimn (n = 1,2) are also written into matrix form as(20)Ωi0r,t=Mi0rXi0t,Ωimnr,t=MimnrXimnt,where(21)Ωi0r,t=Ri0r,tYi0r,t,Xi0t=Ai0tBi0t,Mi0r=rr−1ρi1−ρi2r−2,Ωimnr,t=Θimnr,tRimnr,tYimnr,tZimnr,t,Ximnt=AimntBimntCimntDimnt,Mimnr=rm+1r−m−1rm−1r−m+1−1nαim1rm+1−1n+1r−m−1−1nrm−1−1n+1αim2r−m+1−1nβim1rm−1nβim2r−m−2−1nβim3rm−2−1nβim4r−mζim1rm−ζim2r−m−2ζim3rm−2−ζim4r−m.By assigningr in the above equation as r0i and rhi, respectively, we have(22)Ωi0r0i,t=Mi0r0iXi0t,Ωi0rhi,t=Mi0rhiXi0t,Ωimnr0i,t=Mimnr0iXimnt,Ωimnrhi,t=MimnrhiXimnt.By eliminatingXi0t and Ximnt in equation (22), one obtains(23)Ωi0rhi,t=Mi0rhiMi0r0i−1Ωi0r0i,t,Ωimnrhi,t=MimnrhiMimnr0i−1Ωimnr0i,t.Equation (15) can also be rearranged by matrix form:(24)Ωi0r0i+1,t=Ki0tΩi0rhi,t,Ωimnr0i+1,t=KimntΩimnrhi,t,where(25)Ki0t=1ΔhE∗t01,Kimnt=1+Δhrhi−1n+1mΔhrhi0ΔhG∗t01ΔhE∗t000100001.Combining equation (23) and equation (24), one obtains(26)Ωp0rhp,t=∏i=p2Mi0rhiMi0r0i−1Ki0tM10rh0M10r01−1Ω10r01,tΩpmnrhp,t=∏i=p2MimnrhiMimnr0i−1KimntM1mnrh1M1mnr01−1Ω1mnr01,t.The products of the matrixes in equation (26) are then defined by 2 × 2 partitioned matrixes:(27)S110tS120tS210tS220t=∏i=p2Mi0rhiMi0r0i−1Ki0tM10rh1M10r01−1Sm,11ntSm,12ntSm,21ntSm,22nt=∏i=p2MimnrhiMimnr0i−1KimntM1mnrh1M1mnr01−1.By inserting equation (27) into equation (26), the equations become(28)Rp0rhp,tYp0rhp,t=S110tS120tS210tS220tR10r01,tY10r01,tΘpmnrhp,tRpmnrhp,tYpmnrhp,tZpmnrhp,t=Sm,11ntSm,12ntSm,21ntSm,22ntΘ1mnr01,tR1mnr01,tY1mnr01,tZ1mnr01,t.By decomposing the above matrix equations, one has(29)Rp0rhp,t=S110tR10r01,t+S120tY10r01,tYp0rhp,t=S210tR10r01,t+S220tY10r01,tΘpmnrhp,tRpmnrhp,t=Sm,11ntΘ1mnr01,tR1mnr01,t+Sm,12ntY1mnr01,tZ1mnr01,tYpmnrhp,tZpmnrhp,t=Sm,21ntΘ1mnr01,tR1mnr01,t+Sm,22ntY1mnr01,tZ1mnr01,t.Substitution of equation (19) into equations (16) and (17) yields(30)Y10r01,t=−qin0,Z10r01,t=0,Yp0rhp,t=−qout0,Zp0rhp,t=0,Y10r01,t=−qinmn,Z1m0r01,t=0,Ypmnrhp,t=−qoutmn,Zpmnrhp,t=0.The solutions of the second and fourth equations of equation (29) are(31)R10r01,t=−Q0+S220tq0S210tΘ1mnr01,tR1mnr01,t=−Qmn0−Sm,22nt−qmn0Sm,21nt−1.Therefore,Ωi0r01,t and Ωimnr01,t are determined. By reusing equations (23) and (24), Ωi0rhi,t and Ωimnrhi,t are obtained:(32)Ωi0rhi,t=∏j=i2Mj0rhjMj0r0j−1Kj0tM10rh1M10r01−1Ω10r01,tΩimnrhi,t=∏j=i2MjmnrhjMjmnr0j−1KjmntM1mnrh1M1mnr01−1Ω1mnr01,t.At last, the coefficients are determined as(33)Xi0t=Mi0rhi−1Ωi0rhi,t,Ximnt=Mimnrhi−1Ωimnrhi,t.What needs to be pointed out is that by the use of the recursive matrix method, the coefficients are determined by calculating a series of small matrixes instead of solving a large matrix problem; therefore, the present analytical model is very efficient.


## 2.1. Equations for an Elastic Tube Layer

The referential tube can be seen as the two-dimensional plan strain problem, because the material properties, geometric shape, and external loads remain unchanged in the longitudinal direction. Based on the elasticity theory [32], the constitutive equations for the i-th (i = 1, 2, …, p) tube layer can be given by the following equation:(1)σri=1−μiEi1+μi1−2μiεri+μi1−μiεθi,σθi=1−μiEi1+μi1−2μiμi1−μiεri+εθi,τrθi=Ei21+μiγrθi,in which, σri, σθi, and τrθi are the radial, circumferential, and shear stresses, respectively, while εri, εθi, and γrθi are the corresponding strains. The strain-displacement relationships for the i-th tube layer are(2)εri=∂uri∂r,εθi=urir+1r∂uθi∂θ,γrθi=1r∂uri∂θ+∂uθi∂r−uθir,where uri and uθi are the radial and circumferential displacements, respectively. The stress components keep the equilibrium equations as follows:(3)∂σri∂r+1r∂τrθi∂θ+σri−σθir=0,1r∂σθi∂θ+∂τrθi∂r+2τrθir=0.Substituting equation (2) into equation (1), one has(4)σri=1−μiEi1+μi1−2μi∂uri∂r+μi1−μiurir+μi1−μi1r∂uθi∂θ,σθi=1−μiEi1+μi1−2μiμi1−μi∂uri∂r+urir+1r∂uθi∂θ,τrθi=Ei21+μi1r∂uri∂θ+∂uθi∂r−uθir.Then, by combining equations (3) and (4) and eliminating the stress components, we have(5)12∂2uθi∂r2+121r∂uθi∂r−12uθir2+121−2μi1r∂2uri∂r∂θ+3−4μi21−2μi1r2∂uri∂θ+1−μi1−2μi1r2∂2uθi∂θ2=0,1−μi1−2μi∂2uri∂r2+1−μi1−2μi1r∂uri∂r−1−μi1−2μiurir2+121−2μi1r∂2uθi∂r∂θ−3−4μi21−2μi1r2∂uθi∂θ+121r2∂2uri∂θ2=0.In order to solve the partial differential equations, the displacement components are beforehand expanded into Fourier series, i.e.,(6)uθir,θ,t=Θi0r,t+∑m=1∞Θim1r,tsinmθ+∑m=1∞Θim2r,tcosmθ,urir,θ,t=Ri0r,t+∑m=1∞Rim1r,tcosmθ+∑m=1∞Rim2r,tsinmθ.Substitution of equation (6) into equation (5) yields(7)1−2μ2d2Θi0dr2+1rdΘi0dr−1r2Θi0+∑m=1∞1−2μi2d2Θim1dr2+1−2μi21rdΘim1dr−1−μim2+1−2μi21r2Θim1+m21rdRim2dr+m3−4μi21r2Rim2sinmθ+∑m=1∞1−2μi2d2Θim2dr2+1−2μi21rdΘim2dr−1−μim2+1−2μi21r2Θim2−m21rdRim1dr−m3−4μi21r2Rim1cosmθ=0,1−2μi2d2Ri0dr2+1rdRi0dr−1r2Ri0+∑m=1∞1−μid2Rim2dr2+1−μi1rdRim2dr−1−μi+m221−2μi1r2Rim2−m21rdΘim1dr+m3−4μi21r2Θim1cosmθ+∑m=1∞1−μid2Rim1dr2+1−μi1rdRim1dr−1−μi+m221−2μi1r2Rim1+m21rdΘim2dr−m3−4μi21r2Θim2sinmθ=0.According to the orthogonality of the Fourier series, a set of ordinary differential equations ofΘi0, Θimn, Ri0, and Rimn is obtained, as follows:(8)d2Θi0dr2+1rdΘi0dr−1r2Θi0=0,d2Ri0dr2+1rdRi0dr−1r2Ri0=0,1−2μi2d2Θim1dr2+1−2μi21rdΘim1dr−1−μim2+1−2μi21r2Θim1+m21rdRim2dr+m3−4μi21r2Rim2=0,1−μid2Rim2dr2+1−μi1rdRim2dr−1−μi+m221−2μi1r2Rim2−m21rdΘim1dr+m3−4μi21r2Θim1=0,1−2μi2d2Θim2dr2+1−2μi21rdΘim2dr−1−μim2+1−2μi21r2Θim2−m21rdRim1dr−m3−4μi21r2Rim1=0,1−μid2Rim1dr2+1−μi1rdRim1dr−1−μi+m221−2μi1r2Rim1+m21rdΘim2dr−m3−4μi21r2Θim2=0.By solving the group of differential equations and substituting the results into equation (4), one obtains the general solutions of displacements and stresses.(9)urir,θ,t=rAi0t+r−1Bi0t+∑m=1∞cosmθαim1rm+1Aim1t+r−m−1Bim1t−rm−1Cim1t+αim2r−m+1Dim1t+∑m=1∞sinmθ−αim1rm+1Aim2t−r−m−1Bim2t+rm−1Cim2t−αim2r−m+1Dim2tuθir,θ,t=∑m=1∞sinmθrm+1Aim1t+r−m−1Bim1t+rm−1Cim1t+r−m+1Dim1t+∑m=1∞cosmθrm+1Aim2t+r−m−1Bim2t+rm−1Cim2t+r−m+1Dim2tσrir,θ,t=ρi1Ai0t+ρi2r−2Bi0t+∑m=1∞cosmθβim1rmAim1t+βim2r−m−2Bim1t+βim3rm−2Cim1t+βim4r−mDim1t+∑m=1∞sinmθ−βim1rmAim2t−βim2r−m−2Bim2t−βim3rm−2Cim2t−βim4r−mDim2tσθir,θ,t=ρi1Ai0t−ρi2r−2Bi0t+∑m=1∞cosmθχim1rmAim1t+χim2r−m−2Bim1t+χim3rm−2Cim1t+χim4r−mDim1t+∑m=1∞sinmθ−χim1rmAim2t−χim2r−m−2Bim2t−χim3rm−2Cim2t−χim4r−mDim2tτrθir,θ,t=∑m=1∞sinmθζim1rmAim1t+ζim2r−m−2Bim1t+ζim3rm−2Cim1t+ζim4r−mDim1t+∑m=1∞cosmθζim1rmAim2t+ζim2r−m−2Bim2t+ζim3rm−2Cim2t+ζim4r−mDim2t,where(10)αim1=−m+4μi−2m−4μi+4,αim2=m−4μi+2m+4μi−4,ρi1=Ei1+μi1−2μi,ρi2=−Ei1+μiβim1=−m+1m−2Ei1+μim−4μi+4,βim2=−m+1Ei1+μi,βim3=−m−1Ei1+μi,βim4=−m−1m+2Ei1+μim+4μi−4,χim1=m+1m+2Ei1+μim−4μi+4,χim2=m+1Ei1+μiχim3=m−1Ei1+μirm−2,χim4=m−1m−2Ei1+μim+4μi−4,ζim1=mm+1Ei1+μim−4μi+4,ζim2=−m+1Ei1+μi,ζim3=m−1Ei1+μiζim4=−mm−1Ei1+μim+4μi−4,and Ai0t, Bi0t, Aimnt, Bimnt, Cimnt, and Dimnt are unknown coefficients.


## 2.2. Equations for a Viscoelastic Interlayer

It is verified that the generalized Maxwell model (Figure2) is appropriate for the viscoelastic bonding interlayer [31, 33], with the modulus given by the following equation:(11)E∗t=E∞∗+∑j=1MEj∗e−t/θE,j∗,G∗t=G∗t21+μ∗,where the mark ∗represents the variable belonging to the interlayer; μ∗, E∞∗, Ej∗, and θE,j∗ denote Poisson’s ratio, the long-term moduli, the relaxation moduli, and the relaxation time. By the use of the quasielastic approximation, the constitutive equations of the i-th (i = 1, 2, …, p−1) interlayer are given by [34] the following equation:(12)σr∗iθ,t=E∗tεr∗iθ,t,τrθ∗iθ,t=G∗tγrθ∗iθ,t.Figure 2
Schematic diagram of the generalized Maxwell model.The strain-displacement relationships for the thin interlayer are(13)εr∗iθ,t=uri+1r0i+1,θ,t−urirhi,θ,tΔh,γrθ∗iθ,t=uθi+1r0i+1,θ,t−uθirhi,θ,tΔh+1rhi∂urirhi,θ,t∂θ−uθirhi,θ,trhi,where r0i=∑j=1i−1hj and rhi=∑j=1ihj. In addition, the equilibrium relations of the interlayer and the neighboring tube layers are(14)σri+1r0i+1,θ,t=σr∗iθ,t=σrirhi,θ,t,τrθi+1r0i+1,θ,t=τrθ∗iθ,t=τrθirhi,θ,t.By combining equations (11)–(13) and eliminating the stresses and strains of the interlayer, one obtains(15)σrirhi,θ,t=E∗t−urirhi,θ,tΔh+uri+1r0i+1,θ,tΔhτrθirhi,θ,t=G∗t1rhi∂urirhi,θ,t∂θ−1Δh+1rhiuθirhi,θ,t+uθi+1r0i+1,θ,tΔhσrirhi,θ,t−σri+1r0i+1,θ,t=0τrθirhi,θ,t−τrθi+1r0i+1,θ,t=0,i=1,2,…,p−1.


## 2.3. Coefficient Determination by Recursive Matrix Method

In this section, the coefficients in the general solutions of equation (9) are determined by the recursive matrix method. The loads acting on the tube can be expressed by the following equation:(16)σr1=−qinθ,τrθ1=0,atr=Rin,σrp=−qoutθ,τrθp=0,atr=Rout,in which loads functions are expanded as(17)qinθ=qin0+∑m=1∞qinm1cosmθ+∑m=2∞qinm2sinmθ,qoutθ=qout0+∑m=1∞qoutm1cosmθ+∑m=2∞qoutm2sinmθ,where(18)qin0=12π∫−ππqinθdθ,qout0=12π∫−ππqoutθdθ,qinm1=1π∫−ππqinθcosmθdθ,qoutm1=1π∫−ππqoutθcosmθdθ,qinm2=1π∫−ππqinθsinmθdθ,qoutm2=1π∫−ππqoutθsinmθdθ.The out-of-plane displacements and stresses in equation (12) are rearranged by matrix form.(19)uθir,θ,turir,θ,tσrir,θ,tτrθir,θ,t=0Ri0r,tYi0r,t0+∑m=2∞Θim1r,tsinmθRim1r,tcosmθYim1r,tcosmθZim1r,tsinmθ+∑m=2∞Θim2r,tcosmθRim2r,tsinmθYim2r,tsinmθZim2r,tcosmθ,in which the functions Ri0, Yi0, Θimn, Rimn, Yimn, and Zimn (n = 1,2) are also written into matrix form as(20)Ωi0r,t=Mi0rXi0t,Ωimnr,t=MimnrXimnt,where(21)Ωi0r,t=Ri0r,tYi0r,t,Xi0t=Ai0tBi0t,Mi0r=rr−1ρi1−ρi2r−2,Ωimnr,t=Θimnr,tRimnr,tYimnr,tZimnr,t,Ximnt=AimntBimntCimntDimnt,Mimnr=rm+1r−m−1rm−1r−m+1−1nαim1rm+1−1n+1r−m−1−1nrm−1−1n+1αim2r−m+1−1nβim1rm−1nβim2r−m−2−1nβim3rm−2−1nβim4r−mζim1rm−ζim2r−m−2ζim3rm−2−ζim4r−m.By assigningr in the above equation as r0i and rhi, respectively, we have(22)Ωi0r0i,t=Mi0r0iXi0t,Ωi0rhi,t=Mi0rhiXi0t,Ωimnr0i,t=Mimnr0iXimnt,Ωimnrhi,t=MimnrhiXimnt.By eliminatingXi0t and Ximnt in equation (22), one obtains(23)Ωi0rhi,t=Mi0rhiMi0r0i−1Ωi0r0i,t,Ωimnrhi,t=MimnrhiMimnr0i−1Ωimnr0i,t.Equation (15) can also be rearranged by matrix form:(24)Ωi0r0i+1,t=Ki0tΩi0rhi,t,Ωimnr0i+1,t=KimntΩimnrhi,t,where(25)Ki0t=1ΔhE∗t01,Kimnt=1+Δhrhi−1n+1mΔhrhi0ΔhG∗t01ΔhE∗t000100001.Combining equation (23) and equation (24), one obtains(26)Ωp0rhp,t=∏i=p2Mi0rhiMi0r0i−1Ki0tM10rh0M10r01−1Ω10r01,tΩpmnrhp,t=∏i=p2MimnrhiMimnr0i−1KimntM1mnrh1M1mnr01−1Ω1mnr01,t.The products of the matrixes in equation (26) are then defined by 2 × 2 partitioned matrixes:(27)S110tS120tS210tS220t=∏i=p2Mi0rhiMi0r0i−1Ki0tM10rh1M10r01−1Sm,11ntSm,12ntSm,21ntSm,22nt=∏i=p2MimnrhiMimnr0i−1KimntM1mnrh1M1mnr01−1.By inserting equation (27) into equation (26), the equations become(28)Rp0rhp,tYp0rhp,t=S110tS120tS210tS220tR10r01,tY10r01,tΘpmnrhp,tRpmnrhp,tYpmnrhp,tZpmnrhp,t=Sm,11ntSm,12ntSm,21ntSm,22ntΘ1mnr01,tR1mnr01,tY1mnr01,tZ1mnr01,t.By decomposing the above matrix equations, one has(29)Rp0rhp,t=S110tR10r01,t+S120tY10r01,tYp0rhp,t=S210tR10r01,t+S220tY10r01,tΘpmnrhp,tRpmnrhp,t=Sm,11ntΘ1mnr01,tR1mnr01,t+Sm,12ntY1mnr01,tZ1mnr01,tYpmnrhp,tZpmnrhp,t=Sm,21ntΘ1mnr01,tR1mnr01,t+Sm,22ntY1mnr01,tZ1mnr01,t.Substitution of equation (19) into equations (16) and (17) yields(30)Y10r01,t=−qin0,Z10r01,t=0,Yp0rhp,t=−qout0,Zp0rhp,t=0,Y10r01,t=−qinmn,Z1m0r01,t=0,Ypmnrhp,t=−qoutmn,Zpmnrhp,t=0.The solutions of the second and fourth equations of equation (29) are(31)R10r01,t=−Q0+S220tq0S210tΘ1mnr01,tR1mnr01,t=−Qmn0−Sm,22nt−qmn0Sm,21nt−1.Therefore,Ωi0r01,t and Ωimnr01,t are determined. By reusing equations (23) and (24), Ωi0rhi,t and Ωimnrhi,t are obtained:(32)Ωi0rhi,t=∏j=i2Mj0rhjMj0r0j−1Kj0tM10rh1M10r01−1Ω10r01,tΩimnrhi,t=∏j=i2MjmnrhjMjmnr0j−1KjmntM1mnrh1M1mnr01−1Ω1mnr01,t.At last, the coefficients are determined as(33)Xi0t=Mi0rhi−1Ωi0rhi,t,Ximnt=Mimnrhi−1Ωimnrhi,t.What needs to be pointed out is that by the use of the recursive matrix method, the coefficients are determined by calculating a series of small matrixes instead of solving a large matrix problem; therefore, the present analytical model is very efficient.


## 3. Examples and Discussion

In order to investigate the effect of inhomogeneous degree of the load distribution, an elliptical load is employed and applied on the external surface of the tube, which is expressed by(34)qoutθ=kbk2cos2θ+sin2θ,where k = b/a; a and b are half lengths of semiminor axis and major axis in the ellipse, respectively, as shown in Figure 3. The internal load is fixed at qinθ = 0.2 N/mm2. Some representative stresses and displacements are defined in advance:(35)σm=σθ1atθ=0.5π,r=Rin,τm=τrθ1atθ=0.25π,r=rh1urm=ur1atθ=0.5π,r=Rin.Figure 3
Schematic diagram of elliptical load functionqoutθ.Besides, the symbol denotes the absolute value, e.g., σθi, and the subscript max means the maximum value, e.g., σθimax.


### 3.1. Verification of the Present Solution

For the solution of stresses and displacements expressed in series form, the convergence characteristics should be checked firstly. For this purpose, a three-layer tube glued by polyvinyl butyral (PVB) is considered as an example, with parameters fixed atk = 2, E1 =  E3 = 100 GPa, E2 = 50 GPa, μ1 =  μ2 =  μ3 =  μ∗ = 0.3, Rin = 900 mm, h1 = h3 = 50 mm, h2 = 100 mm, and Δh = 0.3 mm, and the viscoelastic constants of PVB are listed in Table 1.Table 1
Viscoelastic parameters for the PVB.
IndexjGj∗ (MPa)θG,j (s)175.64263.256 × 10−11237.06774.949 × 10−93137.15527.243 × 10−8433.51409.864 × 10−65126.60482.806 × 10−3642.19501.644 × 10−1714.21622.265 × 10083.58223.536 × 10190.45389.368 × 103100.19126.414 × 105110.28934.135 × 107∞0.0880—Table2 gives the values of the stresses and displacements for different numbers of terms N = 5, 10, 15, 20, respectively. It can be observed that the stress and displacement components are convergent rapidly with the precision of three significant figures when N = 10.Table 2
Values of the stresses and displacements for different numbers of terms.
tNσθ1 (MPa)σr1 (MPa)τrθ2 (MPa)uθ2 (mm)ur1 (mm)1 day2−310.391.0706−6.3890−17.544−37.7754−331.050.94732−6.3890−17.544−38.2556−332.050.91449−6.3702−17.543−38.2648−333.080.90963−6.3702−17.543−38.26410−333.090.90906−6.3702−17.543−38.2641 year2−424.950.34484−3.2129−22.299−48.2744−445.030.20970−3.2129−22.299−48.7166−446.120.19083−3.2069−22.299−48.7278−446.170.18923−3.2069−22.299−48.72710−446.170.18910−3.2069−22.299−48.72710 years2−492.78−0.29633−1.1232−24.895−54.0484−505.39−0.38511−1.1232−24.895−54.3226−505.65−0.38970−1.1222−24.895−54.3258−505.66−0.38991−1.1222−24.895−54.32510−505.66−0.38992−1.1222−24.895−54.325Note. The values of σθ1, σr1, ur1 are given at θ = 0.5π, r = 900 mm, and the values of τrθ2, uθ2 are given at θ = 0.25π, r = 900 mm.Then, to validate the present analytical solution, the FE solution from the software ANSYS is used to make a comparison. Due to the symmetry of the structure, only 1/4 part of the tube is modeled, and the FE mesh size is fine enough, as illustrated in Figure4. Here, both tube layers and interlayers in the tube are simulated by the PLANE-183 element. Figure 5 displays the comparison of σθi, σri, and τrθi along the radial direction as well as ur1 along the circumferential direction. It can be obtained that the FE solution matches the present one well (see Figures 5(a)–5(d). However, it should be mentioned that the calculation of the FE solution is time-consuming due to fine mesh division.Figure 4
Schematic diagram of the FE model of 1/4 part of the tube.Figure 5
Comparisons between the present and FE solutions. (a)σθi at θ = 0.5π. (b) σri at θ = 0.5π. (c) τrθi at θ = 0.25π. (d) ur1 at r = 900 mm.
(a)(b)(c)(d)


### 3.2. Parametric Analysis

In this section, a sandwich tube with PVB interlayer under radial loads is considered withRin = 900 mm, h1 = h3 = 50 mm, h2 = 100 mm, μ1 =  μ2 =  μ3 =  μ∗ = 0.3, while the other parameters are variable.Figure6 presents the variations of σm, τm, and urm with t; meanwhile, E1 =  E3 = 100 GPa, E2 = 50 GPa, k = 2. It can be found from Figure 6 that σm and urm increase and remain unchanged eventually as t increases (see Figures 6(a) and 6(c)), while τm decreases gradually with t (see Figure 6(b)). This is mainly resulted from that G∗t is degenerated with t and approaches the fixed value G∞∗ as t⟶ ∞. Moreover, the thicker interlayer leads to the increase of σm and urm (see Figures 6(a) and 6(c)), and by contrast, τm decreases with the interlayer thickness (see Figure 6(b)). The main reason for this phenomenon is that the shear stiffness between layers, i.e., G∗t/∆h, decreases as Δh increases.Figure 6
Variations ofσm, τm, and urm with respect to the time. (a) σm. (b) τm. (c) urm.
(a)(b)(c)Figure7 illustrates the effects of k and t on σm, τm, and urm, with Δh = 0.3 mm, E1 =  E3 = 100 GPa, and E2 = 50 GPa. In the mass, σm, τm, and urm increase with t and remain unchanged eventually, while for k = 1, σm, τm, and urm only have small and positive values at any time (see Figures 7(a)–7(c)). For any of t, σm, τm, and urm increase with k significantly (see Figures 7(a)–7(c)). In other words, the inhomogeneous degree of the load distribution has a large effect on the stresses and displacements in the tube.Figure 7
Effects ofk and t on σm, τm, and urm. (a) σm. (b) τm. (c) urm.
(a)(b)(c)


## 3.1. Verification of the Present Solution

For the solution of stresses and displacements expressed in series form, the convergence characteristics should be checked firstly. For this purpose, a three-layer tube glued by polyvinyl butyral (PVB) is considered as an example, with parameters fixed atk = 2, E1 =  E3 = 100 GPa, E2 = 50 GPa, μ1 =  μ2 =  μ3 =  μ∗ = 0.3, Rin = 900 mm, h1 = h3 = 50 mm, h2 = 100 mm, and Δh = 0.3 mm, and the viscoelastic constants of PVB are listed in Table 1.Table 1
Viscoelastic parameters for the PVB.
IndexjGj∗ (MPa)θG,j (s)175.64263.256 × 10−11237.06774.949 × 10−93137.15527.243 × 10−8433.51409.864 × 10−65126.60482.806 × 10−3642.19501.644 × 10−1714.21622.265 × 10083.58223.536 × 10190.45389.368 × 103100.19126.414 × 105110.28934.135 × 107∞0.0880—Table2 gives the values of the stresses and displacements for different numbers of terms N = 5, 10, 15, 20, respectively. It can be observed that the stress and displacement components are convergent rapidly with the precision of three significant figures when N = 10.Table 2
Values of the stresses and displacements for different numbers of terms.
tNσθ1 (MPa)σr1 (MPa)τrθ2 (MPa)uθ2 (mm)ur1 (mm)1 day2−310.391.0706−6.3890−17.544−37.7754−331.050.94732−6.3890−17.544−38.2556−332.050.91449−6.3702−17.543−38.2648−333.080.90963−6.3702−17.543−38.26410−333.090.90906−6.3702−17.543−38.2641 year2−424.950.34484−3.2129−22.299−48.2744−445.030.20970−3.2129−22.299−48.7166−446.120.19083−3.2069−22.299−48.7278−446.170.18923−3.2069−22.299−48.72710−446.170.18910−3.2069−22.299−48.72710 years2−492.78−0.29633−1.1232−24.895−54.0484−505.39−0.38511−1.1232−24.895−54.3226−505.65−0.38970−1.1222−24.895−54.3258−505.66−0.38991−1.1222−24.895−54.32510−505.66−0.38992−1.1222−24.895−54.325Note. The values of σθ1, σr1, ur1 are given at θ = 0.5π, r = 900 mm, and the values of τrθ2, uθ2 are given at θ = 0.25π, r = 900 mm.Then, to validate the present analytical solution, the FE solution from the software ANSYS is used to make a comparison. Due to the symmetry of the structure, only 1/4 part of the tube is modeled, and the FE mesh size is fine enough, as illustrated in Figure4. Here, both tube layers and interlayers in the tube are simulated by the PLANE-183 element. Figure 5 displays the comparison of σθi, σri, and τrθi along the radial direction as well as ur1 along the circumferential direction. It can be obtained that the FE solution matches the present one well (see Figures 5(a)–5(d). However, it should be mentioned that the calculation of the FE solution is time-consuming due to fine mesh division.Figure 4
Schematic diagram of the FE model of 1/4 part of the tube.Figure 5
Comparisons between the present and FE solutions. (a)σθi at θ = 0.5π. (b) σri at θ = 0.5π. (c) τrθi at θ = 0.25π. (d) ur1 at r = 900 mm.
(a)(b)(c)(d)


## 3.2. Parametric Analysis

In this section, a sandwich tube with PVB interlayer under radial loads is considered withRin = 900 mm, h1 = h3 = 50 mm, h2 = 100 mm, μ1 =  μ2 =  μ3 =  μ∗ = 0.3, while the other parameters are variable.Figure6 presents the variations of σm, τm, and urm with t; meanwhile, E1 =  E3 = 100 GPa, E2 = 50 GPa, k = 2. It can be found from Figure 6 that σm and urm increase and remain unchanged eventually as t increases (see Figures 6(a) and 6(c)), while τm decreases gradually with t (see Figure 6(b)). This is mainly resulted from that G∗t is degenerated with t and approaches the fixed value G∞∗ as t⟶ ∞. Moreover, the thicker interlayer leads to the increase of σm and urm (see Figures 6(a) and 6(c)), and by contrast, τm decreases with the interlayer thickness (see Figure 6(b)). The main reason for this phenomenon is that the shear stiffness between layers, i.e., G∗t/∆h, decreases as Δh increases.Figure 6
Variations ofσm, τm, and urm with respect to the time. (a) σm. (b) τm. (c) urm.
(a)(b)(c)Figure7 illustrates the effects of k and t on σm, τm, and urm, with Δh = 0.3 mm, E1 =  E3 = 100 GPa, and E2 = 50 GPa. In the mass, σm, τm, and urm increase with t and remain unchanged eventually, while for k = 1, σm, τm, and urm only have small and positive values at any time (see Figures 7(a)–7(c)). For any of t, σm, τm, and urm increase with k significantly (see Figures 7(a)–7(c)). In other words, the inhomogeneous degree of the load distribution has a large effect on the stresses and displacements in the tube.Figure 7
Effects ofk and t on σm, τm, and urm. (a) σm. (b) τm. (c) urm.
(a)(b)(c)


## 4. Conclusions

According to the present work, some conclusions can be summarized:(1)
The present solution has high accuracy and matches well with the FE solution. However, the FE method is time-consuming due to fine mesh division.(2)
For the long-term behavior of the laminated tube, the circumferential stresses as well as the radial deformation increase with the time, while the shear stress in the interlayer decreases with the time. All the stresses and deformations tend to remain unchanged eventually because the material property in the interlayer degenerates into a constant value in the long term.(3)
The long-term behavior of the laminated tube is largely affected by the inhomogeneous degree of the load distribution. When the load acting on the tube surface is homogeneous, the values of stresses and displacements are very small, while they significantly increase with the inhomogeneous degree of load distribution.



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*Source: 8534406-2023-01-23.xml*