THERMOELASTIC SOLUTIONS FOR SIMPLY SUPPORTED LAMINATED ARCH IN NON-UNIFORM VARIABLE TEMPERATURE ENVIRONMENT
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摘要:
具有轻质高强、可设计强、耐高温等优良特性的复合材料层合结构在我国被广泛应用于航空航天、建筑结构、公路桥梁等工程领域。该文基于热传导方程和热弹性理论，建立变温环境中层合拱结构理论计算模型，研究了变温环境中二维层合拱的热力学行为，得到结构温度、热应力和位移解析解。利用线性叠加原理，将层合拱非齐次温度边界转化为齐次温度边界，基于结构内层间温度和径向热流密度的连续性，建立内、外侧温度和热流密度关系，联合内、外侧温度边界条件，得到层合圆柱拱内温度场解析解。以位移和应力作为状态变量，建立状态空间方程。基于层间位移和径向应力的连续性，借助传递矩阵法，推导出两端简支层合圆柱拱内外侧位移和应力关系，并同时对层合拱内外侧应力边界条件进行傅里叶级数展开，最终得到位移和热应力解析解。收敛性分析和数值结果比对表明了该方法的有效性和准确性。利用该文方法，探讨了外部温度环境、结构尺寸、结构层数及材料组分，对结构内温度、热应力及位移分布的影响，为变温环境下层合拱的设计提供了理论依据。
Abstract:Due to their light weight, high strength, strong designability and high temperature resistance, composite laminated structures are widely used in aerospace, building structures, highway bridges and other engineering fields in China. Based on the heat conduction equation and thermoelasticity theory, built is a theoretical calculation model of laminated arch in variable temperature environment, investigated are the thermal behaviors for two-dimensional laminated arch, and obtained are the analytical solutions of structural temperature, thermal stress and, displacement. By using the linear superposition principle, the inhomogeneous temperature boundaries of the laminated arch is transformed into homogeneous temperature boundaries. Based on the continuities of the temperature and on heat flux at the interfaces between the adjacent layers of the structure, derived is the relationships between the inner and outer layers. The analytical solution of temperature is obtained by using the surface temperature of the laminated cylindrical arch. The state space equation is established by taking displacement and stress as state variables. Based on the continuities of displacements and stresses at the interface, deduced are the relationships between the inner and outer layers of the laminated arch by means of transfer matrix method. The analytical solutions of displacements and thermal stresses are obtained by Fourier series expansion of the surface loads on the laminated arch. The convergence analysis and comparisons of numerical results demonstrate the effectiveness and accuracy of this method. Discussed are the influences of external temperature environment, structure sizes, structure layers and materials on the distributions of temperature, thermal stress and, displacement in the structure in detail, which provides a theoretical reference for designing laminated arches under variable temperature environment.
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近年来，具有轻质高强、可设计强、耐高温、免维护等优良性质且绿色环保的纤维增强复合材料层合拱结构在我国被广泛应用于航空航天、建筑结构、轨道交通、公路桥梁等工程领域[1-4]，如建筑工程中复合材料拱形屋顶、公路桥梁工程中复合材料拱涵拱圈、机械加工领域中碳纤维拱形零件、压电无人机的拱形机翼、玻璃纤维船舶拱壳结构等[5]。但由于上述优良特性，复合材料层合结构往往被应用于恶劣的工作环境，其中以变温环境对结构的影响较为突出。因此，深入开展变化温度环境中复合材料层合结构力学特性的研究，积极探索变温环境中层合拱壳结构内温度、位移和热应力的分布规律，具有广泛的理论意义和工程实际价值。
目前，应用于层合拱壳结构力学分析的主要研究理论有：等效单层理论、锯齿理论和分层理论。等效单层理论是将非均质多层结构看作具有复杂构造的单层结构，并将结构的位移沿整个厚度方向假设成一个连续光滑的位移函数。根据位移函数形式不同，该理论可分为经典层合理论[6-8]、一阶剪切变形理论[9-11]和高阶剪切变形理论[12-15]。经典层合理论假设结构变形前中法线在变形后仍保持为法线。ASGARI等[16]基于经典理论研究了在热载荷作用下功能梯度材料圆形拱的热弹性屈曲；BATENI和ESLAMI[17]基于经典理论研究了均匀径向压力作用下功能梯度圆形浅拱的非线性稳定性行为。由于经典层合理论忽略了横向剪切变形的影响从而导致随着拱壳结构层厚的增加，误差显著增大。一阶剪切变形理论改进了经典层合理论的假设，考虑了由于剪切变形而引起的截面变形，认为结构变形前的中法线在变形后仍为直线但不再垂直于中面。KHDEIR等[18]基于一阶剪切变形理论和状态空间法，研究了任意边界条件下受温度场影响的层合拱热变形；TRINH和KIM[19]基于一阶剪切变形理论研究了功能梯度夹层壳在热机械荷载作用下的非线性稳定性。一阶剪切变形理论的局限性在于，横向剪切应变在厚度方向上是恒定的，导致其不满足拱壳结构内外表面剪切应力为0的条件，故而该理论更适用于不需要高精度计算横向剪应力和剪应变的情况。为了更精准地描述层合拱壳结构的力学行为，高阶位移分布形函数被用来描述层合结构的面内位移，从而发展出了高阶剪切变形理论。REDDY和LIU[20]利用壳体内外表面的横向剪应力为0的条件，得到了5个未知变量的三阶剪切壳理论，即Reddy型高阶剪切壳理论，它假设了横向剪切应变在壳厚度上的抛物线分布且满足壳边界表面上的无切向应力边界条件。因此，REDDY[21-22]提出的高阶剪切变形理论受到许多学者的采纳和推广。柏冬军和石广玉[23]基于3阶剪切变形理论描述了压电功能梯度层合梁的位移场，进行压电功能梯度层合梁的静力和动力分析。PUNERA等[24]基于高阶剪切变形理论，用于热-机械荷载作用下层合功能梯度夹层开口壳的热分析。SHINDE和SAYYAD[25]利用5阶剪切变形理论对层合壳进行了热弹性分析。然而，高阶剪切变形理论将位移场假设为与横向坐标无关的函数，从而忽略了结构厚度方向的横向拉伸效应。综上，等效单层理论是整体性理论，没有满足层合拱壳结构层间位移锯齿形连续性变化和应力的连续性条件，不能精准地预测层合拱内的面内位移和层间应力。
为了解决上述理论的局限性，研究者们提出了锯齿理论[26-27]和分层理论[28]。锯齿理论通过在等效单层理论中引入锯齿函数，以满足层间位移沿厚度方向呈锯齿形变化的分布状况。任晓辉和陈万吉[29]基于C0型锯齿理论，利用虚位移原理建立了受热荷载作用的复合材料梁的力学模型。DUMIR等[30]基于锯齿理论，提出了一种改进的高效锯齿理论，对热载荷下的层合圆柱壳进行分析。OH和CHO[31]基于高阶锯齿理论，研究了复合材料壳在机械、热和电联合荷载下的变形和应力。锯齿理论由于没有考虑横向正应变的影响，因而难以准确分析温度环境中层合拱壳结构内的位移和应力。分层理论是对层合结构每一层引入独立的位移假设，以保证层合拱壳结构层间位移和应力的连续性。BEG和YASIN[32]基于分层理论，对深曲率的功能梯度材料圆形弯曲梁进行静态、自由和受迫振动分析。由于分层理论中其未知变量个数依赖于层合拱壳结构的铺层数目，随着层数的增加，未知变量随之增多，计算成本大幅增加。
综上，除上述理论之外，其他一些方法也被应用于层合结构位移和应力的求解。LI等[33-37]研究了温度环境中纤维增强复合材料结构的热力学行为。吕朝锋等[38]运用状态空间法和微分求积技术的混合方法，给出了功能梯度厚梁的二维热弹性力学解。邹贵平和唐立民[39]推导出了变温作用下层合圆柱壳的状态方程，并得到半解析结果。李家宇等[40]基于状态空间法求得了四边简支压电热弹性层合正交双曲壳的精确解。丁克伟和唐立民[41]研究了层合开口柱壳的温度应力。QIAN等[42]研究了齐次温度边界下层合拱结构的热效应问题，并得到了温度、位移和应力的解析解。高铭等[43]基于弹性地基梁模型推导出I型临界能量释放率计算公式。STAMPOULOGLOU等[44]研究了径向温度场作用下等厚度径向非均匀球壳的热弹性问题。HE等[45]推导了具有双模效应弯曲梁的二维热弹性力学解。
本文摒弃了以往理论中针对层合结构力学响应中关于温度、位移或应力的人为假设，基于热传导方程和热弹性力学理论，在分析过程中能够严格保持层间位移和应力的连续性，且不随着层合结构层数的增多而增加待定未知变量。利用线性叠加原理，推导出非齐次温度边界条件下层合拱温度场的解析解，并基于得到的温度场分析了层合拱结构的热力学响应，将层合拱结构内热位移和应力的求解归结于求解2个二元一次方程组，最终得到层合拱结构热位移和应力的解析解。通过收敛性分析和有限元数值解比对验证了该方法的有效性和精确性。最后，基于本文温度、位移和应力解析解，通过3个算例，分别研究了结构表面温度、厚径比、层数以及材料性能对简支层合拱内温度、位移及应力分布的影响。
1 基本模型
图1(a)为本文两端简支p层层合拱的计算模型简图，其圆心角为θ0、内半径为r0、外半径为rp、厚度为h(h=rp−r0)。图1(b)为第i (i=1, 2,⋯, p)层单层拱的计算模型，单层厚度为hi(hi=ri−ri−1)，其材料属性分别表示为导热系数ki、弹性模量Ei、泊松比µi及热膨胀系数αi。以层合拱右端θ = 0处的温度作为整个温度场的参考点，层合拱内侧和外侧的表面温度分别记为t1(θ)和tp(θ)，左端即θ = θ0处温度记为tl。
2 温度场求解
由于层合拱温度场边界条件的复杂性，利用线性叠加原理，将层合拱内的温度场Tni(ri, θ)分为2部分求解：如图2(b)所示，第一部分温度场可被构造用于满足层合拱两端非齐次温度边界条件，记为T1i(ri, θ)；第二部分温度场如图2(c)所示，两端边界温度为0 ℃，来满足结构内外表面温度边界条件，记为T2i(ri, θ)。
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图 2 具有非齐次温度边界条件的简支层合拱温度场处理方法Figure 2. Treatment method of temperature field for the simply supported laminated arch with non-homogeneous temperature boundary conditions2.1 第一部分温度解T1i(ri, θ)
假设各层拱的材料性能不受温度的影响，各向均质同性材料的稳态热传导方程为：
∂2Tni(ri,θ)∂r2i+1ri∂Tni(ri,θ)∂ri+1r2i∂2Tni(ri,θ)∂θ2=0 (1) 式中，Tni (ri, θ)为第i层的温度。
层合拱两端的温度分别为：
Tni(ri,0)=0, Tni(ri,θ0)=tl (2) 基于层合拱两端温度，可以构造出满足温度边界条件式(2)的第一部分温度解函数：
T1i(ri,θ)=tlθθ0 (3) 2.2 第二部分温度解T2i(ri, θ)
扣除层合拱内第一部分温度场，第二部分温度场的两端温度边界可表示为：
T2i(ri,0)=0, T2i(ri,θ0)=0 (4) 利用温度边界条件式(4)，温度一般解T2i(ri, θ)可表示为：
T2i(ri,θ)=∞∑m=1tmi(ri)sin(αmθ) (5) 式中：tmi(ri)为待定未知函数；αm=mπ θ0(m=1, 2, 3, …)。
将温度一般解式(5)代入热传导微分方程式(1)，第二部分温度场解函数表达式可解出：
T2i(ri,θ)=∞∑m=1(rαmiEmi+r−αmiFmi)sin(αmθ) (6) 式中，Emi、Fmi为待定未知系数，可由第i层拱内外表面的温度和热流密度确定。
利用温度一般解式(6)，径向热流密度可表示为：
ki∂T2i(ri,θ)∂ri=∞∑m=1ki(αmrαm−1iEmi−αmr−αm−1iFmi)sin(αmθ) (7) 温度和径向热流密度可简写为矩阵形式：
\left[ \begin{matrix} {{T_{2i}}(r_i ,\theta )} \\ {{k_i}\displaystyle \frac{{\partial {T_{2i}}(r_i ,\theta )}}{{\partial r_i }}} \end{matrix} \right] = \sum\limits_{m = 1}^\infty {\sin ({\alpha _m}\theta ){{\boldsymbol\phi _{mi}}(r_i )} } (8) 联合温度一般解式(6)和径向热流密度式(7)， {{\boldsymbol\phi _{mi}}(r_i )} 可表示为：
{{\boldsymbol\phi _{mi}}(r_i )} = {{\boldsymbol\varphi _{mi}}(r_i )} {{\boldsymbol\varLambda _{mi}}} (9) 其中：
{{\boldsymbol\varphi _{mi}}(r_i )} = \left[ \begin{matrix} {r_i ^{{\alpha _m}}}&{r_i ^{ - {\alpha _m}}} \\ {{k_i}{\alpha _m}r_i ^{{\alpha _m} - 1}}&{ - {k_i}{\alpha _m}r_i ^{ - {\alpha _m} - 1}} \end{matrix} \right] (10) {{\boldsymbol\varLambda _{mi}}} = \left[ \begin{matrix} {{E_{mi}}} \\ {{F_{mi}}} \end{matrix} \right] (11) 将r=ri−1和r=ri代入系数矩阵式(9)中，经简单联立推导，即可得到第i层拱内外表面温度与热流密度的关系：
{{\boldsymbol\phi _{mi}}(r_i )} = {{\boldsymbol\varphi _{mi}}(r_i )} { {{\boldsymbol\varphi _{mi}}({r_{i{{ - 1}}}})} ^{ - 1}}{{\boldsymbol\phi _{mi}}({r_{i{{ - 1}}}})} (12) 相邻层层间温度及径向热流密度的连续性可表示为：
{T_{2(i + 1)}}(r_i ,\theta ) = {T_{2i}}(r_i ,\theta ), {k_{i + 1}}\frac{{\partial {T_{2(i + 1)}}(r_i ,\theta )}}{{\partial r_i }} = {k_i}\frac{{\partial {T_{2i}}(r_i ,\theta )}}{{\partial r_i }},\;\; i = 1, 2, \cdots, p{{ - 1}} (13) 利用式(8)，相邻层温度和热流密度系数矩阵关系可表示为：
{{\boldsymbol\phi _{mi}}(r_i )}= {{\boldsymbol\phi _{m(i + 1)}}(r_i )} (14) 利用式(12)和式(14)，第q(q=2,3,···, p)层与第1层矩阵关系可循环推导出：
{{\boldsymbol\phi _{mq}}({r_q})}= \left[ {\prod\limits_{j = q}^1 { {{\boldsymbol\varphi _{mj}}({r_j})} {{ {{\boldsymbol\varphi _{mj}}({r_{j - 1}})} }^{ - 1}}} } \right] {{\boldsymbol\phi _{m{\text{1}}}}({r_{\text{0}}})} (15) 利用式(9)，第q层与第1层的待定系数关系被推导为：
\left[ \begin{matrix} {{E_{mq}}} \\ {{F_{mq}}} \end{matrix} \right] = { {{\boldsymbol\varphi _{mq}}({r_q})} ^{ - 1}}\left[ {\prod\limits_{j = q}^1 { {{\boldsymbol\varphi _{mj}}({r_j})} {{ {{\boldsymbol\varphi _{mj}}({r_{j - 1}})} }^{ - 1}}} } \right] {{\boldsymbol\varphi _{m1}}({r_0})} \left[ \begin{matrix} {{E_{m1}}} \\ {{F_{m1}}} \end{matrix} \right] (16) 2.3 温度场待定系数求解
简支层合拱内外侧表面温度分别为t1(θ)和tp(θ)，即：
{T_{n1}}({r_0},\theta ) = {t_1}(\theta ),{\text{ }}{T_{np}}({r_p},\theta ) = {t_p}(\theta ){\text{ }} (17) 去除第一部分温度场的影响后，第二部分温度场的内外侧表面温度边界条件分别为：
{T_{21}}({r_0},\theta ) = {t_1}(\theta ) - {T_{10}}\left( {{r_0},\theta } \right), {T_{2p}}({r_p},\theta ) = {t_p}(\theta ) - {T_{1p}}\left( {{r_p},\theta } \right) (18) 将温度表达式(6)代入式(18)，两式同时乘以\sin \left(\displaystyle \frac{{m{\text{π }}}}{{{\theta _0}}}\theta \right)并积分，利用三角函数的正交性可得到：
\begin{split} &\frac{{{\theta _0}}}{2}\left( {r_0^{{\alpha _m}}{E_{m1}} + r_0^{ - {\alpha _m}}{F_{m1}}} \right) =\\[-2pt]&\qquad \int_0^{{\theta _0}} {[ {{t_1}(\theta ) - {T_{10}}\left( {{r_0},\theta } \right)} ]} \sin \left(\frac{{m{\text{π }}}}{{{\theta _0}}}\theta \right){\rm d}\theta , \\[-2pt]& \frac{{{\theta _0}}}{2}( {r_p^{{\alpha _m}}{E_{mp}} + r_p^{ - {\alpha _m}}{F_{mp}}}) = \\[-2pt]&\qquad\int_0^{{\theta _0}} {\left[ {{t_p}(\theta ) - {T_{1p}}\left( {{r_p},\theta } \right)} \right]} \sin \left(\frac{{m{\text{π }}}}{{{\theta _0}}}\theta \right){\rm d}\theta \end{split} (19) 通过将q=p代入内外层表面待定系数关系式(16)，可以得到层合拱内外层待定系数之间的关系。联立层合拱内外表面温度边界条件式(19)，可以确定拱内外层待定系数。通过将Em1和Fm1的值回代到式(16)，可以获得任意层的温度待定系数Emq和Fmq。最后，联立第一部分温度解式(3)和第二部分温度解式(6)，层合拱内任意位置处温度解均可被求出：
{T_{ni}}\left( {r_i ,\theta } \right) = {t_l}\frac{\theta }{{{\theta _0}}} + \sum\limits_{m = 1}^\infty { {\left( {r_i ^{{\alpha _m}}{E_{mi}} + r_i ^{ - {\alpha _m}}{F_{mi}}} \right)} } \sin \left( {{\alpha _m}\theta } \right) (20) 3 位移场和应力场求解
3.1 位移应力场的单层解
在二维极坐标系统下，均匀各向同性材料的热本构关系为：
\begin{split} \sigma _r^i(r_i ,\theta ){\text{ = }} &\frac{{(1 - {\mu _i}){E_i}}}{{(1 + {\mu _i})(1 - 2{\mu _i})}}\left[ {\frac{{\partial {u_i}(r_i ,\theta )}}{{\partial r_i }} + \frac{{{\mu _i}}}{{1 - {\mu _i}}}\frac{{{u_i}(r_i ,\theta )}}{{r_i }} + } \right. \\ & \left. {\frac{{{\mu _i}}}{{1 - {\mu _i}}}\frac{1}{{r_i }}\frac{{\partial {v_i}(r_i ,\theta )}}{{\partial \theta }}} \right] - \frac{{{E_i}{\alpha _i}{T_{ni}}(r_i ,\theta )}}{{1 - 2{\mu _i}}}, \end{split} \begin{split} \sigma _\theta ^i(r_i ,\theta ) = &\frac{{(1 - {\mu _i}){E_i}}}{{(1 + {\mu _i})(1 - 2{\mu _i})}}\left[ {\frac{{{\mu _i}}}{{1 - {\mu _i}}}\frac{{\partial {u_i}(r_i ,\theta )}}{{\partial r_i }} + \frac{{{u_i}(r_i ,\theta )}}{{r_i }} + } \right. \\ & \left. {\frac{1}{{r_i }}\frac{{\partial {v_i}(r_i ,\theta )}}{{\partial \theta }}} \right] - \frac{{{E_i}{\alpha _i}{T_{ni}}(r_i ,\theta )}}{{1 - 2{\mu _i}}}, \end{split} \tau _{r\theta }^i(r_i ,\theta ) = \frac{{{E_i}}}{{2(1 + {\mu _i})}}\left[ {\frac{1}{{r_i }}\frac{{\partial {u_i}(r_i ,\theta )}}{{\partial \theta }} } \right. + \frac{{\partial {v_i}(r_i ,\theta )}}{{\partial r_i }} - \left. {\frac{{{v_i}(r_i ,\theta )}}{{r_i }}} \right] (21) 式中： \sigma _r^i(r_i ,\theta ) 、 \sigma _\theta ^i(r_i ,\theta ) 和 \tau _{r\theta }^i(r_i ,\theta ) 分别为各层拱内任意点(ri, θ)的径向应力、环向应力和剪切应力； {u_i}(r_i ,\theta ) 和 {v_i}(r_i ,\theta ) 分别为任意点在径向和环向上的位移。
在没有外力的情况下，平衡微分方程为：
\frac{{\partial \sigma _r^i(r_i ,\theta )}}{{\partial r_i }} + \frac{1}{{r_i }}\frac{{\partial \tau _{r\theta }^i(r_i ,\theta )}}{{\partial \theta }} + \frac{{\sigma _r^i(r_i ,\theta ) - \sigma _\theta ^i(r_i ,\theta )}}{{r_i }} = 0, \frac{1}{{r_i }}\frac{{\partial \sigma _\theta ^i(r_i ,\theta )}}{{\partial \theta }} + \frac{{\partial \tau _{r\theta }^i(r_i ,\theta )}}{{\partial r_i }} + \frac{{2\tau _{r\theta }^i(r_i ,\theta )}}{{r_i }} = 0 (22) 将热本构关系式(21)代入平衡微分方程式(22)，能够得到用位移表示的平衡微分方程组：
\begin{split} &\frac{{(1 - {\mu _i})}}{{(1 + {\mu _i})(1 - 2{\mu _i})}}\left[ \frac{{{\partial ^2}{u_i}(r_i ,\theta )}}{{\partial r_i ^2}} + \frac{1}{{2r_i }}\frac{{{\partial ^2}{v_i}(r_i ,\theta )}}{{\partial \theta \partial r_i }} +\right.\\[-1pt]&\qquad \left. \frac{1}{{r_i }}\frac{{\partial {u_i}(r_i ,\theta )}}{{\partial r_i }} - \frac{{\left( {3 - 4{\mu _i}} \right)}}{{2r_i ^2}}\frac{{\partial {v_i}(r_i ,\theta )}}{{\partial \theta }} - \frac{1}{{r_i ^2}}{u_i}(r_i ,\theta ) \right] +\\[-1pt]&\qquad \frac{1}{{2(1 + {\mu _i})}}\frac{1}{{r_i ^2}}\frac{{{\partial ^2}{u_i}(r_i ,\theta )}}{{\partial {\theta ^2}}} - \frac{{{\alpha _i}}}{{(1 - 2{\mu _i})}}\frac{{\partial {T_{ni}}(r_i ,\theta )}}{{\partial \r_i }} = 0, \\[-1pt]&\frac{1}{{2(1 + {\mu _i})(1 - 2{\mu _i})}}\left[ \frac{1}{{r_i }}\frac{{{\partial ^2}{u_i}(r_i ,\theta )}}{{\partial r_i \partial \theta }} + \frac{{\left( {3 - 4{\mu _i}} \right)}}{{r_i ^2}}\frac{{\partial {u_i}(r_i ,\theta )}}{{\partial \theta }} +\right.\\[-1pt]&\qquad \left. \frac{{2(1 - {\mu _i})}}{{r_i ^2}}\frac{{{\partial ^2}{v_i}(r_i ,\theta )}}{{\partial {\theta ^2}}} \right] - \frac{{{\alpha _i}}}{{(1 - 2{\mu _i})}}\frac{1}{{r_i }}\frac{{\partial {T_{ni}}(r_i ,\theta )}}{{\partial \theta }} +\\[-1pt]& \qquad \frac{1}{{2(1 + {\mu _i})}}\left[ {\frac{{{\partial ^2}{v_i}(r_i ,\theta )}}{{\partial r_i ^2}} + \frac{1}{{r_i }}\frac{{\partial {v_i}(r_i ,\theta )}}{{\partial r_i }} - \frac{{{v_i}(r_i ,\theta )}}{{r_i ^2}}} \right] = 0\end{split} 简支拱两端支座处位移和应力边界条件分别为：
{u_i}(r_i ,0) = {u_i}(r_i ,{\theta _0}) = 0, \sigma _\theta ^i(r_i ,0) = \sigma _\theta ^i(r_i ,{\theta _0}) = 0 (24) 利用边界条件(24)，可以得到位移的一般解：
\begin{split} & {u_i}(r_i ,\theta ) = \sum\limits_{m = 1}^\infty {{U_{mi}}(r_i )} \sin ({\alpha _m}\theta ), \\& {v_i}(r_i ,\theta ) = \sum\limits_{m = 1}^\infty {{V_{mi}}(r_i )} \cos ({\alpha _m}\theta ) \end{split} (25) 式中，Umi(ri)和Vmi(ri)为关于坐标ri的位移函数。将温度解式(20)和位移一般解式(25)代入位移微分方程组式(23)，并简化可得：
\begin{split} &\frac{{1 - {\mu _i}}}{{1 - {\text{2}}{\mu _i}}}{U''_{mi}}\left( {r_i } \right) + \frac{{1 - {\mu _i}}}{{1 - {\text{2}}{\mu _i}}}\frac{1}{{r_i }}{U'_{mi}}\left( {r_i } \right) - \frac{1}{{r_i ^2}}\left( {\frac{{1 - {\mu _i}}}{{1 - {\text{2}}{\mu _i}}} + \frac{{\alpha _m^2}}{2}} \right)\\& \qquad{U_{mi}}\left( {r_i } \right) - \frac{1}{{2\left( {1 - {\text{2}}{\mu _i}} \right)}}\frac{{{\alpha _m}}}{{r_i }}{V'_{mi}}\left( {r_i } \right) + \frac{{\left( {3 - 4{\mu _i}} \right)}}{{2\left( {1 - {\text{2}}{\mu _i}} \right)}}\frac{{{\alpha _m}}}{{r_i ^2}} \cdot \\& \qquad{V_{mi}}\left( {r_i } \right) = \frac{{\left( {1 + {\mu _i}} \right)}}{{\left( {1 - {\text{2}}{\mu _i}} \right)}}{\alpha _i}{\alpha _m}( {r_i ^{{\alpha _m} - 1}{E_{mi}} - r_i ^{ - {\alpha _m} - 1}{F_{mi}}} ),\end{split} \begin{split}&\frac{1}{2}{V''_{mi}}\left( {r_i } \right){\text{ + }}\frac{1}{{2r_i }}{V'_{mi}}\left( {r_i } \right) - \frac{1}{{r_i ^2}}\left( {\alpha _m^2\frac{{1 - {\mu _i}}}{{1 - {\text{2}}{\mu _i}}} + \frac{1}{2}} \right){V_{mi}}\left( {r_i } \right) +\\&\qquad \frac{1}{{2\left( {1 - {\text{2}}{\mu _i}} \right)}}\frac{{{\alpha _m}}}{{r_i }}{U'_{mi}}\left( {r_i } \right) + \frac{{\left( {3 - 4{\mu _i}} \right)}}{{2\left( {1 - {\text{2}}{\mu _i}} \right)}}\frac{{{\alpha _m}}}{{r_i ^2}}{U_{mi}}\left( {r_i } \right) = \\&\qquad\frac{{\left( {1 + {\mu _i}} \right)}}{{\left( {1 - {\text{2}}{\mu _i}} \right)}}\frac{1}{{r_i }}{\alpha _i}{\alpha _m}\left[ { - \frac{{2{t_l}}}{{m{\text{π}} }}\cos \left( {m{\text{π}}} \right) + } \right.\Biggr. {{\text{ }}r_i ^{{\alpha _m}}{E_{mi}} + r_i ^{ - {\alpha _m}}{F_{mi}}} \Biggr] \end{split} (26) 通过对式(26)求解，得到了位移ui(ri, θ)和位移vi(ri, θ)的解析解：
\begin{split} {u_i}(r_i ,\theta ) =& \sum\limits_{m = 1}^\infty \sin \left( {\alpha _m}\theta \right)\left[ \frac{{{\alpha _m} + 4{\mu _i} - 2}}{{{\alpha _m} - 4{\mu _i} + 4}}r_i ^{{\alpha _m} + 1}{A_{mi}} - r_i ^{ - {\alpha _m} - 1}\cdot \right.\\&{B_{mi}} + r_i ^{{\alpha _m} - 1}{C_{mi}} + \frac{{ - {\alpha _m} + 4{\mu _i} - 2}}{{{\alpha _m} + 4{\mu _i} - 4}}r_i ^{ - {\alpha _m} + 1}{D_{mi}} +\\& \frac{{\left( {1 + {\mu _i}} \right){\alpha _i}}}{{{\alpha _m} - 4{\mu _i} + 4}}r_i ^{{\alpha _m} + 1}{E_{mi}} - \frac{{\left( {1 + {\mu _i}} \right){\alpha _i}}}{{{\alpha _m} + 4{\mu _i} - 4}}r_i ^{ - {\alpha _m} + 1}{F_{mi}} + \\&\left. {\frac{{4{t_l}\left( {1 + {\mu _i}} \right){\alpha _i}\cos \left( {m{\text{π}} } \right)}}{{\left( {\alpha _m^2 - 4} \right)\left( {1 - {\mu _i}} \right)m{\text{π}} }}r_i } \right], \end{split} \begin{split} {v_i}(r_i ,\theta ) =& \sum\limits_{m = 1}^\infty {\cos \left( {{\alpha _m}\theta } \right)} \Biggr[ {r_i ^{{\alpha _m} + 1}{A_{mi}} + r_i ^{ - {\alpha _m} - 1}{B_{mi}} + } \Biggr.\\&r_i ^{{\alpha _m} - 1}{C_{mi}} + r_i ^{ - {\alpha _m} + 1}{D_{mi}} - \frac{{\left( {1 + {\mu _i}} \right){\alpha _i}}}{{{\alpha _m} + 4{\mu _i} - 2}}r_i ^{{\alpha _m} + 1}{E_{mi}} -\\&\frac{{\left( {1 + {\mu _i}} \right){\alpha _i}}}{{{\alpha _m} - 4{\mu _i} + 2}}r_i ^{ - {\beta _m} + 1}{F_{mi}} +\\& \left. {\frac{{2{\alpha _m}{t_l}\left( {1 + {\mu _i}} \right){\alpha _i}\cos \left( {m{\text{π}} } \right)}}{{\left( {\alpha _m^2 - 4} \right)\left( {1 - {\mu _i}} \right)m{\text{π}} }}r_i } \right] \end{split} (27) 式中，Ami、Bmi、Cmi、Dmi为未知系数，可由第i层拱内外表面位移和应力的边界条件确定。将位移解析式(27)代入热本构关系式(21)，得到各应力表达式：
\begin{split} \sigma _r^i(r_i ,\theta ) =& \sum\limits_{m = 1}^\infty {\sin \left( {{\alpha _m}\theta } \right)} \left[ {\frac{{\left( {{\alpha _m} + 1} \right)\left( {{\alpha _m} - 2} \right){E_i}}}{{\left( {1 + {\mu _i}} \right)\left( {{\alpha _m} - 4{\mu _i} + 4} \right)}}} \cdot\right.\\& r_i ^{{\alpha _m}}{A_{mi}} + \frac{{\left( {{\alpha _m} + 1} \right){E_i}}}{{\left( {1 + {\mu _i}} \right)}}r_i ^{ - {\alpha _m} - 2}{B_{mi}} + \frac{{\left( {{\alpha _m} - 1} \right){E_i}}}{{\left( {1 + {\mu _i}} \right)}}\cdot\\& r_i ^{{\alpha _m} - 2}{C_{mi}} + \frac{{\left( {{\alpha _m} - 1} \right)\left( {{\alpha _m} + 2} \right){E_i}}}{{\left( {1 + {\mu _i}} \right)\left( {{\alpha _m} + 4{\mu _i} - 4} \right)}}r_i ^{ - {\alpha _m}}{D_{mi}} + \\&\frac{{\left( {3 - 4{\mu _i}} \right)\left( {2 - {\alpha _m}} \right){E_i}{\alpha _i}}}{{\left( {{\alpha _m} + 4{\mu _i} - 2} \right)\left( {{\alpha _m} - 4{\mu _i} + 4} \right)}}r_i ^{{\alpha _m}}{E_{mi}} +\\&\frac{{\left( {3 - 4{\mu _i}} \right)\left( {{\alpha _m} + 2} \right){E_i}{\alpha _i}}}{{\left( {{\alpha _m} + 4{\mu _i} - 4} \right)\left( {{\alpha _m} - 4{\mu _i} + 2} \right)}}r_i ^{ - {\alpha _m}}{F_{mi}} +\\& \left. {\frac{{2\left( {\alpha _m^2 - 2} \right){t_l}\cos \left( {m{\text{π}} } \right){E_i}{\alpha _i}}}{{\left( {\alpha _m^2 - 4} \right)\left( {1 - {\mu _i}} \right)m{\text{π}} }}} \right], \end{split} \begin{split} \sigma _\theta ^i(r_i ,\theta ) =& \sum\limits_{m = 1}^\infty {\sin \left( {{\alpha _m}\theta } \right)} \left[ { - \frac{{\left( {{\alpha _m} + 1} \right)\left( {{\alpha _m} + 2} \right){E_i}}}{{\left( {1 + {\mu _i}} \right)\left( {{\alpha _m} - 4{\mu _i} + 4} \right)}}} \right.r_i ^{{\alpha _m}}{A_{mi}} -\\& \frac{{\left( {{\alpha _m} + 1} \right){E_i}}}{{\left( {1 + {\mu _i}} \right)}}r_i ^{ - {\alpha _m} - 2}{B_{mi}} - \frac{{\left( {{\alpha _m} - 1} \right){E_i}}}{{\left( {1 + {\mu _i}} \right)}}r_i ^{{\alpha _m} - 2}{C_{mi}} -\\&\frac{{\left( {{\alpha _m} - 1} \right)\left( {{\alpha _m} - 2} \right){E_i}}}{{\left( {1 + {\mu _i}} \right)\left( {{\alpha _m} + 4{\mu _i} - 4} \right)}}r_i ^{ - {\alpha _m}}{D_{mi}} +\\& \frac{{\left( {3 - 4{\mu _i}} \right)\left( {{\alpha _m} + 2} \right){E_i}{\alpha _i}}}{{\left( {{\alpha _m} + 4{\mu _i} - 2} \right)\left( {{\alpha _m} - 4{\mu _i} + 4} \right)}}r_i ^{{\alpha _m}}{E_{mi}} + \\& \frac{{\left( {3 - 4{\mu _i}} \right)\left( {2 - {\alpha _m}} \right){E_i}{\alpha _i}}}{{\left( {{\alpha _m} + 4{\mu _i} - 4} \right)\left( {{\alpha _m} - 4{\mu _i} + 2} \right)}}r_i ^{ - {\alpha _m}}{F_{mi}} -\\& \left. {\frac{{4{t_l}\cos \left( {m{\text{π}} } \right){E_i}{\alpha _i}}}{{\left( {\alpha _m^2 - 4} \right)\left( {1 - {\mu _i}} \right)m{\text{π}} }}} \right],\end{split} \begin{split} \tau _{r\theta }^i(r_i ,\theta ) =& \sum\limits_{m = 1}^\infty {\cos \left( {{\alpha _m}\theta } \right)} \left[ {\frac{{\left( {{\alpha _m} + 1} \right){\alpha _m}{E_i}}}{{\left( {1 + {\mu _i}} \right)\left( {{\alpha _m} - 4{\mu _i} + 4} \right)}}} \right.r_i ^{{\alpha _m}}{A_{mi}} -\\& \frac{{\left( {{\alpha _m} + 1} \right){E_i}}}{{\left( {1 + {\mu _i}} \right)}}r_i ^{ - {\alpha _m} - 2}{B_{mi}} + \frac{{\left( {{\alpha _m} - 1} \right){E_i}}}{{\left( {1 + {\mu _i}} \right)}}r_i ^{{\alpha _m} - 2}{C_{mi}} - \\&\frac{{\left( {{\alpha _m} - 1} \right){\alpha _m}{E_i}}}{{\left( {1 + {\mu _i}} \right)\left( {{\alpha _m} + 4{\mu _i} - 4} \right)}}r_i ^{ - {\alpha _m}}{D_{mi}} + \end{split} \begin{split} & \frac{{\left( {4{\mu _i} - 3} \right){\alpha _m}{E_i}{\alpha _i}}}{{\left( {{\alpha _m} + 4{\mu _i} - 2} \right)\left( {{\alpha _m} - 4{\mu _i} + 4} \right)}}r_i ^{{\alpha _m}}{E_{mi}} + \\& \frac{{\left( {4{\mu _i} - 3} \right){\alpha _m}{E_i}{\alpha _i}}}{{\left( {{\alpha _m} + 4{\mu _i} - 4} \right)\left( {{\alpha _m} - 4{\mu _i} + 2} \right)}}r_i ^{ - {\alpha _m}}{F_{mi}} +\\& \left. {\frac{{2{\alpha _m}{t_l}\cos \left( {m{\text{π}} } \right){E_i}{\alpha _i}}}{{\left( {\alpha _m^2 - 4} \right)\left( {1 - {\mu _i}} \right)m{\text{π}} }}} \right] \end{split} 3.2 位移应力场的递推和待定系数求解
将位移和应力表达式(27)和式(28)表示成矩阵形式：
\left[ \begin{matrix} {{u_i}(r_i ,\theta )} \\ {{v_i}(r_i ,\theta )} \\ {\sigma _r^i(r_i ,\theta )} \\ {\tau _{r\theta }^i(r_i ,\theta )} \end{matrix} \right] = \left[ \begin{gathered} \sum\limits_{m = 1}^\infty {{U_{mi}}\left( {r_i } \right)\sin \left( {{\alpha _m}\theta } \right)} \\ \sum\limits_{m = 1}^\infty {{V_{mi}}\left( {r_i } \right)\cos \left( {{\alpha _m}\theta } \right)} \\ \sum\limits_{m = 1}^\infty {{Y_{mi}}\left( {r_i } \right)\sin \left( {{\alpha _m}\theta } \right)} \\ \sum\limits_{m = 1}^\infty {{Z_{mi}}\left( {r_i } \right)\cos \left( {{\alpha _m}\theta } \right)} \\ \end{gathered} \right] (29) 式中：Umi(ri)、Vmi(ri)为待定位移函数；Ymi(ri)、Zmi(ri)为待定应力函数。
层合拱第i层位移和应力解析式可表示为矩阵形式：
{{{W} _{mi}}\left( {r_i } \right)} = {{\boldsymbol{\varPhi} _{mi}}\left( {r_i } \right)} \left[ \begin{gathered} {A_{mi}} \\ {B_{mi}} \\ {C_{mi}} \\ {D_{mi}} \\ \end{gathered} \right] + {{{H}_{mi}}\left( {r_i } \right)} (30) 其中：
{{{W}_{mi}}\left( {r_i } \right)} = \left[ \begin{gathered} {U_{mi}}\left( {r_i } \right) \\ {V_{mi}}\left( {r_i } \right) \\ {Y_{mi}}\left( {r_i } \right) \\ {Z_{mi}}\left( {r_i } \right) \\ \end{gathered} \right] (31) {{{H}_{mi}}\left( {r_i } \right)} ={{{G}_{mi}}\left( {r_i } \right)} +{{{T}_{mi}}\left( {r_i } \right)} (32) \begin{array}{*{20}{l}} { {{{{\boldsymbol{T}}}_{mi}}\left( {r_i } \right)}{\text{ = }}} \\ {\left[ \begin{matrix} {\displaystyle \frac{{4{t_l}\left( {1 + {\mu _i}} \right){\alpha _i}\cos \left( {m{\text{π }}} \right)}}{{\left( {\alpha _m^2 - 4} \right)\left( {1 - {\mu _i}} \right)m{\text{π }}}}r_i } \\ {\displaystyle\frac{{2{\alpha _m}{t_l}\left( {1 + {\mu _i}} \right){\alpha _i}\cos \left( {m{\text{π }}} \right)}}{{\left( {\alpha _m^2 - 4} \right)\left( {1 - {\mu _i}} \right)m{\text{π }}}}r_i } \\ {\displaystyle\frac{{2\left( {\alpha _m^2 - 2} \right){t_l}\cos \left( {m{\text{π }}} \right){E_i}{\alpha _i}}}{{\left( {\alpha _m^2 - 4} \right)\left( {1 - {\mu _i}} \right)m{\text{π }}}}} \\ {\displaystyle\frac{{2{\alpha _m}{t_l}\cos \left( {m{\text{π }}} \right){E_i}{\alpha _i}}}{{\left( {\alpha _m^2 - 4} \right)\left( {1 - {\mu _i}} \right)m{\text{π }}}}} \end{matrix} \right],} \end{array} \begin{array}{*{20}{l}} { {{{{\boldsymbol{G}}}_{mi}}\left( {r_i } \right)} = {\text{ }}} \\ {{\text{ }}\left[ \begin{matrix} {\displaystyle\frac{{\left( {1 + {\mu _i}} \right){\alpha _i}}}{{{\alpha _m} - 4{\mu _i} + 4}}r_i ^{{\alpha _m} + 1}}&{ - \displaystyle\frac{{\left( {1 + {\mu _i}} \right){\alpha _i}}}{{{\alpha _m} + 4{\mu _i} - 4}}r_i ^{ - {\alpha _m} + 1}} \\ { - \displaystyle\frac{{\left( {1 + {\mu _i}} \right){\alpha _i}}}{{{\alpha _m} + 4{\mu _i} - 2}}r_i ^{{\alpha _m} + 1}}&{ -\displaystyle \frac{{\left( {1 + {\mu _i}} \right){\alpha _i}}}{{{\alpha _m} - 4{\mu _i} + 2}}r_i ^{ - {\alpha _m} + 1}} \\ {\displaystyle\frac{{\left( {3 - 4{\mu _i}} \right)\left( {2 - {\alpha _m}} \right){E_i}{\alpha _i}}}{{\left( {{\alpha _m} + 4{\mu _i} - 2} \right)\left( {{\alpha _m} - 4{\mu _i} + 4} \right)}}r_i ^{{\alpha _m}}}&{\displaystyle\frac{{\left( {3 - 4{\mu _i}} \right)\left( {{\alpha _m} + 2} \right){E_i}{\alpha _i}}}{{\left( {{\alpha _m} + 4{\mu _i} - 4} \right)\left( {{\alpha _m} - 4{\mu _i} + 2} \right)}}r_i ^{ - {\alpha _m}}} \\ {\displaystyle\frac{{\left( {4{\mu _i} - 3} \right){\alpha _m}{E_i}{\alpha _i}}}{{\left( {{\alpha _m} + 4{\mu _i} - 2} \right)\left( {{\alpha _m} - 4{\mu _i} + 4} \right)}}r_i ^{{\alpha _m}}}&{\displaystyle\frac{{\left( {4{\mu _i} - 3} \right){\alpha _m}{E_i}{\alpha _i}}}{{\left( {{\alpha _m} + 4{\mu _i} - 4} \right)\left( {{\alpha _m} - 4{\mu _i} + 2} \right)}}r_i ^{ - {\alpha _m}}} \end{matrix} \right]\left[ \begin{gathered} {E_{mi}} \\ {F_{mi}} \\ \end{gathered} \right]} \end{array} (33) \begin{gathered} {{\boldsymbol\varPhi _{mi}}\left( {r_i } \right)} {\text{ = }} \\ \left[ \begin{matrix} {\displaystyle \frac{{{\alpha _m} + 4{\mu _i} - 2}}{{{\alpha _m} - 4{\mu _i} + 4}}r_i^{{\alpha _m} + 1}}&{ - r_i ^{ - {\alpha _m} - 1}}&{r_i ^{{\alpha _m} - 1}}&{\displaystyle \frac{{ - {\alpha _m} + 4{\mu _i} - 2}}{{{\alpha _m} + 4{\mu _i} - 4}}r_i ^{ - {\alpha _m} + 1}} \\ {r_i^{{\alpha _m} + 1}}&{r_i ^{ - {\alpha _m} - 1}}&{r_i ^{{\alpha _m} - 1}}&{r_i ^{ - {\alpha _m} + 1}} \\ {\displaystyle \frac{{\left( {{\alpha _m} + 1} \right)\left( {{\alpha _m} - 2} \right){E_i}}}{{\left( {1 + {\mu _i}} \right)\left( {{\alpha _m} - 4{\mu _i} + 4} \right)}}r_i ^{{\alpha _m}}}&{\displaystyle \frac{{\left( {{\alpha _m} + 1} \right){E_i}}}{{ {1 + {\mu _i}} }}r_i ^{ - {\alpha _m} - 2}}&{\displaystyle \frac{{\left( {{\alpha _m} - 1} \right){E_i}}}{{ {1 + {\mu _i}} }}r_i ^{{\alpha _m} - 2}}&{\displaystyle \frac{{\left( {{\alpha _m} - 1} \right)\left( {{\alpha _m} + 2} \right){E_i}}}{{\left( {1 + {\mu _i}} \right)\left( {{\alpha _m} + 4{\mu _i} - 4} \right)}}r_i ^{ - {\alpha _m}}} \\ {\displaystyle \frac{{\left( {{\alpha _m} + 1} \right){\alpha _m}{E_i}}}{{\left( {1 + {\mu _i}} \right)\left( {{\alpha _m} - 4{\mu _i} + 4} \right)}}r_i ^{{\alpha _m}}}&{ - \displaystyle \frac{{\left( {{\alpha _m} + 1} \right){E_i}}}{{ {1 + {\mu _i}} }}r_i ^{ - {\alpha _m} - 2}}&{\displaystyle \frac{{\left( {{\alpha _m} - 1} \right){E_i}}}{{ {1 + {\mu _i}} }}r_i ^{{\alpha _m} - 2}}&{ - \displaystyle \frac{{\left( {{\alpha _m} - 1} \right){\alpha _m}{E_i}}}{{\left( {1 + {\mu _i}} \right)\left( {{\alpha _m} + 4{\mu _i} - 4} \right)}}r_i ^{ - {\alpha _m}}} \end{matrix} \right] \\ \end{gathered} (34) 通过分析第i层拱的位移和应力矩阵表达式(30)，可以得到第i (i=1, 2,···, p)层拱的内、外表面位移和应力的关系：
\begin{split} & {{{W}_{mi}}\left( {r_i } \right)} = {{\boldsymbol{\varPhi} _{mi}}\left( {r_i } \right)} { {{\boldsymbol\varPhi _{mi}}\left( {{r_{i - 1}}} \right)} ^{ - 1}} {{{W}_{mi}}\left( {{r_{i - 1}}} \right)} - \\ & \qquad {{\boldsymbol\varPhi _{mi}}\left( {r_i } \right)} {{{\boldsymbol\varPhi _{mi}}\left( {{r_{i - 1}}} \right)} ^{ - 1}} {{{H}_{mi}}\left( {{r_{i - 1}}} \right)} + {{{H}_{mi}}\left( {r_i } \right)} \end{split} (35) 层合拱内相邻层层间位移和应力的连续性关系可表示为：
{{W_{mi}}\left( {r_i } \right)} = {{W_{m(i + 1)}}\left( {r_i } \right)} (36) 利用式(35)和式(36)，可循环递推出层合拱内第q(q=2, 3, \cdots, p)层外侧与层合拱内表面位移和应力的关系：
\begin{split} {{{W}_{mq}}\left( {{r_q}} \right)} = & \left[ {\prod\limits_{i = q}^1 { {{\boldsymbol\varPhi _{mi}}\left( {r_i } \right)} } {{ {{\boldsymbol\varPhi _{mi}}\left( {{r_{i - 1}}} \right)} }^{ - 1}}} \right] {{{W}_{m1}}\left( {{r_0}} \right)} -\\ & \left[ {\prod\limits_{i = q}^1 { {{\boldsymbol\varPhi _{mi}}\left( {r_i } \right)} } {{{{\boldsymbol\varPhi _{mi}}\left( {{r_{i - 1}}} \right)} }^{ - 1}}} \right]\ {{{H}_{m1}}\left( {{r_0}} \right)} +\\& \sum\limits_{j = 2}^q {\left[ {\prod\limits_{i = q}^j { {{\boldsymbol\varPhi _{mi}}\left( {r_i } \right)} {{{{\boldsymbol\varPhi _{mi}}\left( {{r_{i - 1}}} \right)} }^{ - 1}}} } \right]}\cdot \\&\left[ { {{{H}_{m\left( {j - 1} \right)}}\left( {{r_{j - 1}}} \right)} } \right. - \left. { {{{H}_{mj}}\left( {{r_{j - 1}}} \right)} } \right] +{{{H}_{mq}}\left( {{r_q}} \right)} \end{split} (37) 在不考虑外部荷载的情况下，简支层合拱内、外侧的应力边界条件可表示为：
\sigma _r^i({r_0},\theta ) = \tau _{r\theta }^i({r_0},\theta ) = 0, \;\sigma _r^i({r_p},\theta ) = \tau _{r\theta }^i({r_p},\theta ) = 0 (38) 联立层合拱最内层与各层之间的关系式(37)以及应力边界条件式(38)，可得到层合拱内外侧表面位移-应力关系：
\left[ \begin{matrix} {{U_{mp}}\left( {{r_p}} \right)} \\ {{V_{mp}}\left( {{r_p}} \right)} \\ 0 \\ 0 \end{matrix} \right] = \left[ \begin{matrix} { \boldsymbol\nabla _m^{11}}&{\boldsymbol\nabla _m^{12}} \\ {\boldsymbol\nabla _m^{21}}&{\boldsymbol\nabla _m^{22}} \end{matrix} \right]\left[ \begin{matrix} {{U_{m1}}\left( {{r_0}} \right)} \\ {{V_{m1}}\left( {{r_0}} \right)} \\ 0 \\ 0 \end{matrix} \right] + \left[ \begin{matrix} {\varOmega _m^1} \\ {\varOmega _m^2} \\ {\varOmega _m^3} \\ {\varOmega _m^4} \end{matrix} \right] (39) 其中：
\left[ \begin{matrix} {\boldsymbol\nabla _m^{11}}&{\boldsymbol\nabla _m^{12}} \\ {\boldsymbol\nabla _m^{21}}&{\boldsymbol\nabla _m^{22}} \end{matrix} \right] = {\prod\limits_{i = p}^1 {{{\boldsymbol\varPhi _{mi}}\left( {r_i } \right)} } {{ {{\boldsymbol\varPhi _{mi}}\left( {{r_{i - 1}}} \right)} }^{ - 1}}} (40) \left[ \begin{matrix} {\varOmega _m^1} \\ {\varOmega _m^2} \\ {\varOmega _m^3} \\ {\varOmega _m^4} \end{matrix} \right] = - \left[ {\prod\limits_{i = p}^1 { {{\boldsymbol\varPhi _{mi}}\left( {r_i } \right)} } {{ {{\boldsymbol\varPhi _{mi}}\left( {{r_{i - 1}}} \right)} }^{ - 1}}} \right] {{{H}_{m1}}\left( {{r_0}} \right)} + \begin{split} & \sum\limits_{j = 2}^p {\left[ {\prod\limits_{i = p}^j { {{\boldsymbol\varPhi _{mi}}\left( {r_i } \right)} {{ {{\boldsymbol\varPhi _{mi}}\left( {{r_{i - 1}}} \right)} }^{ - 1}}} } \right]}\cdot\\& \left[ { {{H_{m\left( {j - 1} \right)}}\left( {{r_{j - 1}}} \right)} } - {{H_{mj}}\left( {{r_{j - 1}}} \right)} \right] + {{H_{mp}}\left( {{r_p}} \right)} \end{split} (41) 式中， \boldsymbol{\nabla} _m^{ij}(i,j = 1,2) 为2×2阶矩阵。
位移应力关系式(39)可简化为2个简单的子矩阵方程：
\left[ \begin{matrix} 0 \\ 0 \end{matrix} \right] = {\boldsymbol{\nabla}} _m^{21}\left[ \begin{matrix} {{U_{m1}}\left( {{r_0}} \right)} \\ {{V_{m1}}\left( {{r_0}} \right)} \end{matrix} \right] + \left[ \begin{matrix} {\varOmega _m^3} \\ {\varOmega _m^4} \end{matrix} \right], \left[ \begin{matrix} {{U_{mp}}\left( {{r_p}} \right)} \\ {{V_{mp}}\left( {{r_p}} \right)} \end{matrix} \right]{\text{ = }}{\boldsymbol{\nabla}} _m^{11}\left[ \begin{matrix} {{U_{m1}}\left( {{r_0}} \right)} \\ {{V_{m1}}\left( {{r_0}} \right)} \end{matrix} \right]{\text{ + }}\left[ \begin{matrix} {\varOmega _m^1} \\ {\varOmega _m^2} \end{matrix} \right] (42) 通过求解式(42)，可以非常方便确定位移函数Um1(r0)、Vm1(r0)、Ump(rp)和Vmp(rp)。借助递推关系式(37)，可以得到各层拱的外侧位移和应力待定函数Umq(rq)、Vmq(rq)、Ymq(rq)、Zmq(rq) (q=2, 3, \cdots , p−1)。利用位移和应力的矩阵表达式(30)，待定系数Ami、Bmi、Cmi和Dmi (i=1, 2, \cdots, p)可被唯一确定。最后将Ami、Bmi、Cmi和Dmi (i=1, 2, \cdots , p)代回位移解析式(27)和应力解析式(28)，可以得到层合拱内任意点的位移和热应力。另外，通过将机械荷载边界条件代入位移和应力边界条件式(38)，本文的解析解法可推广至热-机械荷载共同作用下的层合拱的热力学行为分析。
4 收敛性与对比性研究
利用本文提出的解析解法分析一个3层拱的热应力和位移，并研究本文方法的收敛性和精确性，本文所有计算结果保留3位有效数字。此三层拱内外层为钢材层，中间层为混凝土层(其材料属性如表1所示)。层合拱每层尺寸分别为r0=1.00 m、r1=1.30 m、r2=1.90 m、r3=2.20 m。拱圆心角为θ0=2π/3。以层合拱右端的温度作为温度参考点，层合拱内外侧温度分别为t1(θ)=20 ℃和tp(θ)=100 ℃，拱左端的温度为tl =10 ℃。
表 1 材料属性Table 1. Material properties材料 E/GPa μ α/(×10−5 ℃−1) k/(W/(m·℃)) 钢 200 0.3 1.2 50.0 混凝土 30 0.2 0.7 2.0 木材 10 0.1 0.8 0.1 注：E为弹性模量；μ为泊松比；α为热膨胀系数；k为导热系数。 将温度、位移和应力解析式(20)、式(27)和式(28)中的级数项m截断到N，可以获得三层拱任意两点的温度、位移和热应力结果。如表2所示，温度、位移和应力计算结果均随着级数项的增加而快速收敛，并且N=20与N=23的三位有效数字计算结果已经相等。因此在分析本文算例时，可取N=20的计算结果作为收敛结果。另外，利用有限元软件ANSYS (PLANE-13单元)分析此三层拱，通过不同的网格划分方式(如图3所示)，得到结构内温度、位移和应力分布，并与本文方法结果进行比对。表3给出了θ=π/6，r=1.075 m、1.225 m、1.450 m、1.750 m和2.125 m处5个点的温度、热应力和位移的结果对比。如表3所示，随着ANSYS模型网格密度的不断增大，有限元结果与本文解析解计算结果的误差不断减小，且最大误差为1.71%。误差存在的原因是，利用本文方法得到的温度、位移和应力计算结果是精确的，而有限元解是利用近似离散的方法得到的温度、位移和应力近似解。图4给出了本文解析解法与有限元方法在θ=π/6处沿厚度方向的结果对比。从图4中发现，有限元方法计算结果与本文方法计算结果非常接近。通过上述分析可以发现，本文方法具有较好的有效性和精确性。
表 2 温度、位移和应力收敛性分析Table 2. Convergence of temperature, displacements and stress位置 N T/(℃) u/mm v/mm σθ/MPa r=1.75 m,
θ=π/611 80.1 1.55 −0.0104 12.6 14 80.0 1.55 −0.0105 12.6 17 79.9 1.55 −0.0108 12.6 20 79.9 1.55 −0.0108 12.6 23 79.9 1.55 −0.0108 12.6 r=1.10 m,
θ=0.8011 20.6 1.73 0.3880 −25.6 14 20.6 1.73 0.3890 −25.6 17 20.6 1.73 0.3890 −25.5 20 20.6 1.73 0.3890 −25.4 23 20.6 1.73 0.3890 −25.4 注：N为迭代次数；T为温度；u为径向位移；v为环向位移；σθ为沿θ方向的正应力。 表 3 本文方法与有限元解的比较Table 3. Comparisons of the present method with the ANSYS solutions位置 方法 θ=π/6 T/(℃) u/mm v/mm σθ/MPa r=1.075 m 40×16 20.02 1.29 0.7730 −29.20 120×32 20.01 1.30 0.7770 −29.80 240×128 20.01 1.30 0.7770 −30.00 present 20.01 1.30 0.7770 −30.00 r=1.225 m 40×16 20.30 1.34 0.6080 19.00 120×32 20.30 1.35 0.6130 19.00 240×128 20.30 1.35 0.6120 18.90 present 20.30 1.35 0.6130 18.90 r=1.450 m 40×16 41.30 1.39 0.3570 10.40 120×32 41.30 1.40 0.3610 10.40 240×128 41.30 1.40 0.3600 10.40 present 41.30 1.40 0.3600 10.40 r=1.750 m 40×16 79.90 1.54 −0.0109 12.50 120×32 79.90 1.55 −0.0107 12.60 240×128 79.90 1.55 −0.0108 12.60 present 79.90 1.55 −0.0108 12.60 r=2.125 m 40×16 99.50 2.01 −0.4890 3.66 120×32 99.50 2.02 −0.4850 3.62 240×128 99.50 2.02 −0.4860 3.57 present 99.60 2.02 −0.4850 3.51 注：T为温度；u为径向位移；v为环向位移；σθ为沿θ方向的正应力。present为本文方法计算结果；40×16，120×32，240×128分别代表不同网格划分下的ANSYS解。 ![]()
图 4 本文精确解与有限元解(网格划分方式为240×128)在θ=π/6处沿厚度方向的结果比较比对Figure 4. Comparison of the results from the present method and the finite element method (the meshing method is 240×128) along the thickness direction at θ=π/65 算例分析
利用本文方法分别分析3组不同的算例，探究层合拱表面温度、结构尺寸、组成材料和层数等参数对层合结构内温度、位移和应力场分布的影响。
5.1 算例1：表面温度的影响
以圆心角θ0=2π/3和厚度h=1.20 m(r0=1.00 m、r1=1.30 m、r2=1.90 m、r3=2.20 m)的三层拱为例。三层拱内外层均为钢材层，核心层为混凝土。以θ=0支座处温度作为整个温度场的参考点，θ=2π/3拱支座处温度为tl=10 ℃。拱的内侧温度始终维持在t1(θ)=20 ℃，外侧温度分别为tp(θ)=100 ℃、130 ℃和160 ℃。假设材料属性不随温度的改变发生变化，图5给出了拱内θ=π/6处温度T、径向位移u、环向位移v和应力σθ沿拱厚方向的分布。从图中可以发现，三层拱内温度、热应力和位移均随着拱外表面温度的升高而增加。从图5(a)可以看出，在内外钢材表层，温度沿r方向变化较小；但在核心混凝土层，温度变化较大。原因是拱内外层为钢材层，导热系数远大于核心混凝土层。如图5(b)和图5(c)所示，位移u沿厚度方向不断增大，而位移v沿厚度方向不断减小，且随着外表面温度的不断提高，位移的绝对值均按比例增大。从图5(d)可以发现，随着外表面温度的增加，结构内应力σθ绝对值按比例增大，且在结构内相邻层界面处应力发生突变。
为了进一步探究层合拱相邻层界面处的应力σθ分布规律，图6给出了tp(θ)=100 ℃时在三层拱两个层间界面r=1.30 m和r=1.90 m处沿拱环向应力分布情况。从图6可以发现，三层拱两个界面处钢材层中应力σθ始终大于混凝土层中应力σθ，原因是，钢材的弹性模量远大于混凝土。
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图 5 不同温度环境中层合拱在θ=π/6处沿厚度方向的温度、位移和应力分布Figure 5. Temperature, displacement and stress distribution along thickness direction of laminated arch at θ=π/6 in different temperature environments![]()
图 6 外表面温度tp(θ)=100 ℃时层间应力σθ沿环向分布情况Figure 6. Circumferential distribution of interlaminar stress σθ when external surface temperature tp(θ)=100 ℃5.2 算例2：厚径比h/r0的影响
以3个不同尺寸的三层拱为例来探究变温环境中层合拱的厚径比对结构内温度、热应力和位移分布的影响。三层拱的内外层为钢材层，中间层为混凝土层。拱总厚度为h=r3−r0=1.20 m，各层拱厚分别为h1=h3=0.30 m，h2=0.60 m。拱的圆心角为θ0=8π/9。拱内径分别为r0=1.00 m、2.00 m和6.00 m，即三层拱的厚度与半径之比分别为h/r0=1.20、0.60和0.20。以θ=0支座处拱的温度作为整个温度场的参考点，θ=8π/9拱左端处温度为tl=20 ℃，拱的内表面温度为t1(θ)=30 ℃，外表面温度为tp(θ)=120 ℃。图7给出了不同厚径比的三层拱中，温度、位移和应力沿θ=π/5拱厚方向的分布情况。从图7(a)和图7(d)可以看出，随着厚径比的增加，拱内各层的温度和应力σθ变化并不显著，即厚径比对结构内温度T和应力σθ分布影响较小，原因是三种三层拱各层材料和高度均一致。而从图7(b)和图7(c)可以发现，位移u沿着拱厚方向几乎保持不变，且随着厚径比的增加，位移u和v显著减小，原因可由极坐标下几何方程可知，拱结构内的位移是随着半径的增大而增大。
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图 7 在不同厚径比三层拱中，温度、位移和应力沿θ=π/5拱厚方向的分布情况Figure 7. Distribution of temperature, displacement and stress along θ=π/5 arch thickness direction in three-layer arch with different thickness-radius ratio5.3 算例3：结构层数和材料组分的影响
以3个不同层合拱为例，探究同一温度环境中，层合拱的组成材料和层数对拱内温度、位移和应力分布的影响。三种不同的层合拱外部尺寸均相同，内径为r0=3.00 m，厚度为h=1.80 m，圆心角为θ0=3π/4。以θ=0支座处拱的温度作为整个温度场的参考点，θ=3π/4拱左端处温度为tl=10 ℃，层合拱内外表面的温度均分别为t1(θ)=40 ℃和tp(θ)=130 ℃。第一个拱为2层拱，厚度为h=1.80 m，各层拱半径分别为r1=3.90 m、r2=4.80 m。拱的外层和内层分别由钢材和木材制成。第二个拱为3层拱，厚度为h=1.80 m，各层拱半径分别为r1=3.60 m、r2=4.20 m、r3=4.80 m，外层拱由钢材制成，内层拱由木材制成，核心层由混凝土制成。第三个拱为4层拱，厚度为h=1.80 m，各层拱半径分别为r1=3.45 m、3.90 m、4.35 m、4.80 m，拱的内外层为钢材层、次内层为木材层、次外层为混凝土层。
图8给出了各层合拱θ=2π/9处沿拱厚方向的温度、位移和应力分布。从图8可以看出，同一外部尺寸的层合拱，层数和组成材料对拱内温度、位移和应力分布有显著的影响。从图8(a)可以看出，钢材层的温度变化相比其他材料层较小。原因是钢材的导热系数大于木材和混凝土。且四层拱的温度曲线发生了3次突变，那是因为四层拱由不同材料组成，在每一层材料中，温度的变化速率不同。从图8(b)和图8(c)可以看出，在相同温度环境中，双层拱的位移u和v均小于三层拱和四层拱，原因是，木材层的导热系数远小于钢材层和混凝土层且双层拱的木材层厚度均大于其他两种拱。从图8(d)可以看出，钢材层的应力变化梯度均远大于其他材料层的应力变化梯度，这是因为，钢材的弹性模量远大于其他层材料。
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图 8 不同层数和材料下θ=2π/9处，温度、位移和应力沿拱厚度方向的分布情况Figure 8. Distribution of temperature, displacement and stress along the direction of arch thickness at θ=2π/9 with different layers and materials6 结论
本文基于极坐标系下二维热传导基本方程和热弹性理论，研究了温度环境中简支层合拱结构的热力学行为，得到以下结论：
(1) 基于热传导方程和热弹性力学理论，本文得到了非齐次温度边界层合拱结构内温度、位移和热应力解析解。并利用收敛性分析以及与有限元对比，验证了本文方法的可行性和精确性。
(2) 随着层合拱结构服役环境温度的提高，层合拱内温度、位移和应力按比例增大。
(3) 随着层合拱厚径比h/r0的增大，径向位移u和环向位移v大幅减小。
(4) 层合拱层数和组成材料，显著影响结构内温度、位移和应力的分布。通过改变层合拱结构的层数和组成材料，可以用来设计结构内温度、位移和应力分布。
(5) 本分析方法可以扩展至热-机械荷载作用下的简支层合拱问题，亦可从二维问题扩展到三维问题。可为服役于温度环境下层合拱壳结构的数值计算结果提供理论校核，也可应用于非均匀温度环境中层合拱壳结构工程设计优化。
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图 2 具有非齐次温度边界条件的简支层合拱温度场处理方法
Figure 2. Treatment method of temperature field for the simply supported laminated arch with non-homogeneous temperature boundary conditions
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图 4 本文精确解与有限元解(网格划分方式为240×128)在θ=π/6处沿厚度方向的结果比较比对
Figure 4. Comparison of the results from the present method and the finite element method (the meshing method is 240×128) along the thickness direction at θ=π/6
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图 5 不同温度环境中层合拱在θ=π/6处沿厚度方向的温度、位移和应力分布
Figure 5. Temperature, displacement and stress distribution along thickness direction of laminated arch at θ=π/6 in different temperature environments
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图 6 外表面温度tp(θ)=100 ℃时层间应力σθ沿环向分布情况
Figure 6. Circumferential distribution of interlaminar stress σθ when external surface temperature tp(θ)=100 ℃
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图 7 在不同厚径比三层拱中，温度、位移和应力沿θ=π/5拱厚方向的分布情况
Figure 7. Distribution of temperature, displacement and stress along θ=π/5 arch thickness direction in three-layer arch with different thickness-radius ratio
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图 8 不同层数和材料下θ=2π/9处，温度、位移和应力沿拱厚度方向的分布情况
Figure 8. Distribution of temperature, displacement and stress along the direction of arch thickness at θ=2π/9 with different layers and materials
表 1 材料属性
Table 1 Material properties
材料 E/GPa μ α/(×10−5 ℃−1) k/(W/(m·℃)) 钢 200 0.3 1.2 50.0 混凝土 30 0.2 0.7 2.0 木材 10 0.1 0.8 0.1 注：E为弹性模量；μ为泊松比；α为热膨胀系数；k为导热系数。 
下载: 导出CSV
表 2 温度、位移和应力收敛性分析
Table 2 Convergence of temperature, displacements and stress
位置 N T/(℃) u/mm v/mm σθ/MPa r=1.75 m,
θ=π/611 80.1 1.55 −0.0104 12.6 14 80.0 1.55 −0.0105 12.6 17 79.9 1.55 −0.0108 12.6 20 79.9 1.55 −0.0108 12.6 23 79.9 1.55 −0.0108 12.6 r=1.10 m,
θ=0.8011 20.6 1.73 0.3880 −25.6 14 20.6 1.73 0.3890 −25.6 17 20.6 1.73 0.3890 −25.5 20 20.6 1.73 0.3890 −25.4 23 20.6 1.73 0.3890 −25.4 注：N为迭代次数；T为温度；u为径向位移；v为环向位移；σθ为沿θ方向的正应力。
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表 3 本文方法与有限元解的比较
Table 3 Comparisons of the present method with the ANSYS solutions
位置 方法 θ=π/6 T/(℃) u/mm v/mm σθ/MPa r=1.075 m 40×16 20.02 1.29 0.7730 −29.20 120×32 20.01 1.30 0.7770 −29.80 240×128 20.01 1.30 0.7770 −30.00 present 20.01 1.30 0.7770 −30.00 r=1.225 m 40×16 20.30 1.34 0.6080 19.00 120×32 20.30 1.35 0.6130 19.00 240×128 20.30 1.35 0.6120 18.90 present 20.30 1.35 0.6130 18.90 r=1.450 m 40×16 41.30 1.39 0.3570 10.40 120×32 41.30 1.40 0.3610 10.40 240×128 41.30 1.40 0.3600 10.40 present 41.30 1.40 0.3600 10.40 r=1.750 m 40×16 79.90 1.54 −0.0109 12.50 120×32 79.90 1.55 −0.0107 12.60 240×128 79.90 1.55 −0.0108 12.60 present 79.90 1.55 −0.0108 12.60 r=2.125 m 40×16 99.50 2.01 −0.4890 3.66 120×32 99.50 2.02 −0.4850 3.62 240×128 99.50 2.02 −0.4860 3.57 present 99.60 2.02 −0.4850 3.51 注：T为温度；u为径向位移；v为环向位移；σθ为沿θ方向的正应力。present为本文方法计算结果；40×16，120×32，240×128分别代表不同网格划分下的ANSYS解。
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