浙江大学学报(工学版), 2020, 54(4): 778-786 doi: 10.3785/j.issn.1008-973X.2020.04.017

土木工程、交通工程

基于Chebyshev-Ritz法分析多裂纹梁自振特性

赵佳雷,, 周叮, 张建东, 胡朝斌,

Free vibration characteristics of multi-cracked beam based on Chebyshev-Ritz method

ZHAO Jia-lei,, ZHOU Ding, ZHANG Jian-dong, HU Chao-bin,

通讯作者: 胡朝斌,男,讲师. orcid.org/0000-0002-0313-8553. E-mail: huchaobin@njtech.edu.cn

收稿日期: 2019-04-7  

Received: 2019-04-7  

作者简介 About authors

赵佳雷(1995—),男,硕士生,从事结构动力学的研究.orcid.org/0000-0001-6175-9473.E-mail:zhaojialei1995@njtech.edu.cn , E-mail:zhaojialei1995@njtech.edu.cn

摘要

基于弹性力学平面应力理论,利用Chebyshev-Ritz法分析多裂纹梁的自振特性. 根据裂纹情况将裂纹梁分成若干个梁段,用边界函数与第一类Chebyshev多项式的乘积构造各梁段的位移函数,具有很好的收敛性,能够适用于不同的几何边界条件. 用Ritz法得到各梁段的振动方程,根据各梁段之间的位移连续条件整合方程,建立整个裂纹梁的振动特征方程. 计算结果与有限元分析和相关文献数据吻合很好. 分析裂纹深度和位置对自振特性的影响. 随着裂纹深度的增大,裂纹梁的频率减小,振型的幅值变大,且影响的程度会受裂纹的位置影响.

关键词: 弹性力学 ; Chebyshev-Ritz法 ; 裂纹 ; 自由振动 ; 位移连续条件

Abstract

The free vibration characteristics of multi-cracked beam were analyzed based on the plane stress theory of elasticity by using Chebyshev-Ritz method. The cracked beams were divided into several sections according to their cracks. The products of boundary functions and Chebyshev polynomials were taken as the functions of the displacement, which had good convergence, making the method applicable for different geometric boundary conditions. The vibration equation of each sub-beam could be obtained by using Ritz method. The vibration characteristic equation of the whole cracked beam was established by the continuity conditions of displacements between adjacent sub-beams. The calculation results accorded well with those available from the literature and the finite element analysis. The effects of the structural parameters such as crack depth and location on the natural vibration characteristics of the beam were analyzed. As the crack depth increases, the natural frequency of the cracked beam decreases, the amplitude of the mode shape increases, and the degree of influence is affected by the location of the crack.

Keywords: elasticity ; Chebyshev-Ritz method ; crack ; free vibration ; continuity conditions of displacement

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本文引用格式

赵佳雷, 周叮, 张建东, 胡朝斌. 基于Chebyshev-Ritz法分析多裂纹梁自振特性. 浙江大学学报(工学版)[J], 2020, 54(4): 778-786 doi:10.3785/j.issn.1008-973X.2020.04.017

ZHAO Jia-lei, ZHOU Ding, ZHANG Jian-dong, HU Chao-bin. Free vibration characteristics of multi-cracked beam based on Chebyshev-Ritz method. Journal of Zhejiang University(Engineering Science)[J], 2020, 54(4): 778-786 doi:10.3785/j.issn.1008-973X.2020.04.017

在机械和土木等工程中,梁作为一种受力构件有着广泛的应用. 由于自身初始缺陷和长期的受力尤其是在动力荷载的作用下,梁会产生裂纹,动力特性会受到影响,如车间和料场中桥式起重机的吊车梁等. Chondros等[1]用无质量弹簧来模拟裂纹,利用局部柔度法计算带裂纹结构的频率和振型. Khaji等[2]用局部柔度法分析Timoshenko梁理论下的裂纹梁,计算得到1~4阶频率,但误差随着阶次的提高而增大. 马一江等[3]使用这种弹簧模拟裂纹模型分析多裂纹梁的动力特性,仅对1阶频率和振型进行分析. Zhao等[4-5]结合格林函数与无质量弹簧模型,对基于Euler梁理论的不同边界条件下裂纹梁的振动问题进行求解. Kim等[6]将裂纹对梁的作用等效为力的作用,将裂纹所在位置看成变截面进行求解. 杨鄂川等[7]将裂纹梁等效为弯曲刚度沿梁长度变化的梁,分析开口裂纹Euler梁的振动特性. 马爱敏等[8]基于这种连续抗弯刚度模型,将裂纹等效为截面刚度的削弱,对裂纹简支梁进行分析. 这种连续刚度模型仅对于Euler梁理论下的裂纹梁具有较好的精度. 李兆军等[9]将能量法结合有限元位移模式,应用有限单元模型分析裂纹梁. 郁杨天等[10]将有限单元法结合近场动力学,对裂纹简支梁进行分析. 这种基于有限元的分析模型具有较高的精度,但是需要将分析对象划分为许多个单元,基于该方法所编制的程序运算量较大.

胡霖远等[11]用能量变分法分析各种边界条件下的夹芯梁,所得自振频率与有限元结果基本吻合. 蒋杰等[12]应用能量变分原理,将两端有裂纹的固支梁划分为3层分析振动特性,与有限元解吻合很好. 本文基于弹性力学平面应力理论,用能量变分法分析裂纹在任意位置处的多裂纹梁的自由振动特性,无需假设模型或计算裂纹处截面刚度的削弱程度,用Chebyshev-Ritz法[13-14]建立每个梁段的振动特征方程,根据各梁段的位移连续条件得到整个裂纹梁的振动特征方程. 求解该方程得到裂纹梁的频率参数和模态系数,对各种边界条件下的细长梁和短梁都有很高的精度.

1. 裂纹梁分析模型和振动方程

1.1. 带有两条不同深度裂纹的梁

1.1.1. 分析模型

考虑如图1所示的下端带有2条不同深度裂纹的矩形截面梁,为了便于分析,假设裂纹深度沿梁的宽度方向不变,裂缝方向与梁的轴线垂直且裂缝侧面光滑. 只研究裂纹梁作微幅自由振动时的固有振动特性,假设梁的上表面承受静载,自由振动时裂纹始终处于开口状态. 该梁长度为L,高度为h,宽度为b. 2条裂纹的深度分别为c1c2,裂纹到梁左端的距离分别为d1d2. 将裂纹梁分成5个梁段,建立5个局部坐标系.

图 1

图 1   带有2条不同深度裂纹的梁的分析模型

Fig.1   Analytical model of beam with two cracks of different depths


各梁段的线弹性应变能为

${V_q} = \frac{1}{2}\int_{ - {{{a_q}}}/{2}}^{{{{a_q}}}/{2}} {\int_{ - {{{h_q}}}/{2}}^{{{{h_q}}}/{2}} {(\sigma _x^q\varepsilon _x^q + \sigma _z^q\varepsilon _z^q + \tau _{xz}^q\gamma _{xz}^q)b{\rm d}x{\rm d}z} } .$

式中: $\sigma _x^q$$\varepsilon _x^q$为第q个梁段在x方向上的正应力和正应变, $\sigma _z^q$$\varepsilon _z^q$为第q个梁段在z方向上的正应力和正应变, $\tau _{xz}^q$$\gamma _{xz}^q$为切应力和切应变,其中q=1,2,3,4,5,a1=La2=d1a3=Ld1a4=d2d1a5=Ld2h1=hc1h2=c1h3=c1c2h4=h5=c2.

在平面应力理论中,应力可以由应变表示为

$\left. \begin{aligned} &\sigma _x^q = {E}({\varepsilon _x} + v{\varepsilon _z}) /({{1 - {v^2}}}),\\ &\sigma _z^q = {E}(v{\varepsilon _x} + {\varepsilon _z}) /({{1 - {v^2}}}),\\ &\tau _{xz}^q = \frac{E}{{2(1 + v)}}\gamma _{xz}^q .\\ \end{aligned} \right\}$

式中:E为弹性模量,v为泊松比.

将式(2)代入式(1),可得

$ \begin{split} {V^q} =& \frac{{bE}}{{2(1 - {v^2})}}\int_{ - {{{a_q}}}/{2}}^{{{{a_q}}}/{2}} {\int_{ - {{{h_q}}}/{2}}^{{{{h_q}}}/{2}} {[{{(\varepsilon _{xx}^q)}^2} + {{(\varepsilon _{zz}^q)}^2} +}} \\&{{2v\varepsilon _{xx}^q\varepsilon _{zz}^q} } + \frac{{1 - v}}{2}{(\gamma _{xz}^q)^2}]{\rm d}x{\rm d}z. \end{split} $

式中: $\varepsilon _{xx}^q = {{\partial {u_q}}}/{{\partial x}},$ $\varepsilon _{zz}^q = {{\partial {{{w}}_q}}}/{{\partial z}},$ $ \gamma _{xz}^q = {{\partial {{{u}}_q}}}/{{\partial z}} +$ $ {{\partial {{{w}}_q}}}/$ ${{\partial x}} $uqwq分别为第q个梁段位移在xz方向上的分量.

各梁段的动能Tq

${T^q} = {{\rho b}}/{2}\int_{ - {{{a_q}}}/{2}}^{{{{a_q}}}/{2}} {\int_{ - {{{h_q}}}/{2}}^{{{{h_q}}}/{2}} {\left[{{\left(\frac{{\partial {{{u}}_q}}}{{\partial t}}\right)}^2} + {{\left(\frac{{\partial {{{w}}_q}}}{{\partial t}}\right)}^2}\right]{\rm d}x{\rm d}z} }. $

式中:ρ为密度,t为时间变量.

用无量纲坐标替换原坐标:

${\xi _q} = {{2x} / {{a_q}}},\;{\zeta _q} = {{2z} / {{h_q}}}.$

1.1.2. 梁的位移函数

各梁段自由振动的位移函数在ξqζq方向的分量表示为

$\left.\begin{split}&{{{u}}_q}(\xi ,{\zeta _q},t) = {{{U}}_q}(\xi ,{\zeta _q}){\exp\;({{\rm j}\omega t})},\\& {{{w}}_q}(\xi ,{\zeta _q},t) = {{{W}}_q}(\xi ,{\zeta _q}){\exp\;({{\rm j}\omega t})}.\end{split}\right\}$

式中:UqWq分别为第q个梁段在ξqζq方向的振幅函数,j为单位虚数,ω为振动的角速度.

将式(6)代入式(4)、(3),可得第q个梁段的最大应变能和最大动能:

$\begin{split} V_{\max}^q = &\frac{{bE}}{{2{\lambda _q}(1 - {v^2})}}\int_{ - 1}^1 {\int_{ - 1}^1 {\bigg[{{\bigg({\lambda _q}\frac{{\partial {{{U}}_q}}}{{\partial {\xi _q}}}\bigg)}^2} + {{\bigg(\frac{{\partial {{{W}}_q}}}{{\partial {\zeta _q}}}\bigg)}^2}} } {\rm{ + }}\\ &2\nu {\lambda _q}\frac{{\partial {{{U}}_q}}}{{\partial {\xi _q}}}\frac{{\partial {{{W}}_q}}}{{\partial {\zeta _q}}} + \frac{{1 - v}}{2}{\bigg(\frac{{\partial {{{U}}_q}}}{{\partial {\zeta _q}}} + {\lambda _q}\frac{{\partial {{{W}}_q}}}{{\partial {\xi _q}}}\bigg)^2}\bigg]{\rm d}{\xi _q}{\rm d}{\zeta _q}, \end{split}$

$T_{\rm max}^q = \frac{{\rho b{a_q}{h_q}{\omega ^2}}}{8}\int_{ - 1}^1 {\int_{ - 1}^1 {[{{{{{U}}_q}}^2} + {{{{{W}}_q}}^2}]{\rm d}{\xi _q}{\rm d}{\zeta _q}} } .$

式中: ${\lambda _q} = {h_q}/{a_q}$.

用边界函数和第一类切比雪夫多项式构造位移振幅函数:

$\left. {\begin{array}{*{20}{l}} {{{{U}}_q}({\xi _q},{\zeta _q}) = f_u^q({\xi _q})\varphi _u^q({\zeta _q})\displaystyle\sum\limits_{i = 1}^{{I_q}} {\sum\limits_{j = 1}^{{J_q}} {A_{ij}^q} {P_i}({\xi _q})} {P_j}({\zeta _q})}, \\ {{{{W}}_q}({\xi _q},{\zeta _q}) = f_w^q({\xi _q})\varphi _w^q({\zeta _q})\displaystyle\sum\limits_{l = 1}^{{L_q}} {\sum\limits_{m = 1}^{{M_q}} {B_{lm}^q} {P_l}({\xi _q})} {P_m}({\zeta _q})} . \end{array}} \right\}$

${P_s}(x) = \cos\;[(s - 1)\arccos\;x];\;\;s = 1,2,3, \cdots .$

各梁段具有不同的几何边界条件,将每个梁段视为互相独立的梁来构造边界函数. 在平面应力问题中,若梁的支承端为固支,则限制该处ξqζq方向的位移. 若支承端为简支,则仅限制该处ζq方向的位移. 自由端无需限制其位移. 相应的边界条件可以用边界函数表示如下.

$\left. \!\!\!\!\!\!\!{\begin{array}{*{20}{l}} \quad\quad {\text{两端固支:}}\\ \phi _u^q({\xi _q}) = \phi _w^q({\xi _q}) = 1 - {\xi _q}^{\rm{2}},\;\varphi _u^q({\zeta _q}) = \varphi _w^q({\zeta _q}) = 1;\\ \quad\quad{\text{两端简支:}}\\ \phi _u^q({\xi _q}) = 1,\;\phi _w^q({\xi _q}) = 1 - {\xi _q}^{\rm{2}},\;\varphi _u^q({\zeta _q}) = \varphi _w^q({\zeta _q}) = 1;\\ \quad\quad{\text{两端自由:}}\\ \phi _u^q({\xi _q}) = \phi _w^q({\xi _q}) = 1,\;\varphi _u^q({\zeta _q}) = \varphi _w^q({\zeta _q}) = 1;\\ \quad\quad{\text{左端固支右端自由:}}\\ \phi _u^q({\xi _q}) = \phi _w^q({\xi _q}) = 1 + {\xi _q},\;\varphi _u^q({\zeta _q}) = \varphi _w^q({\zeta _q}) = 1;\\ \quad\quad{\text{左端简支右端自由:}}\\ \phi _u^q({\xi _q}) = 1,\;\phi _w^q({\xi _q}) = 1 + {\xi _q},\;\varphi _u^q({\zeta _q}) = \varphi _w^q({\zeta _q}) = 1. \end{array}} \!\!\!\right\}$

1.1.3. 梁的特征方程

利用Ritz法,可得

$\frac{{\partial (V_{\rm max}^q - T_{\rm max}^q)}}{{\partial A_{ij}^q}} = 0,\;\frac{{\partial (V_{\rm max}^q - T_{\rm max}^q)}}{{\partial B_{lm}^q}} = 0.$

将式(7)、(8)代入式(12),可得第q个梁段的振动特征方程:

${{{K}}_q} - {{{\varOmega}} ^2}{{{M}}_q}{{{X}}_q} = 0.$

式中:KqMq分别为第q个梁段的刚度矩阵和质量矩阵,

其中

$ \left. {\begin{array}{*{20}{l}} {{{K}}_{uuq}} = {\lambda _q}D_{uiu\overline i }^{11q}G_{uju\overline j }^{00q} + \dfrac{{1 - v}}{{2{\lambda _q}}}D_{uiu\overline i }^{00q}G_{uju\overline j }^{11q},\\ {{{K}}_{uwq}} = vD_{uiwl}^{10q}G_{ujwm}^{01q} + \dfrac{{1 - v}}{2}D_{uiwl}^{01q}G_{ujwm}^{10q},\\ {{{K}}_{wwq}} = \dfrac{{1 - v}}{2}{\lambda _q}D_{wlw\overline l }^{11q}G_{wmw\overline m }^{00q} + \dfrac{1}{{{\lambda _q}}}D_{wlw\overline l }^{00q}G_{wmw\overline m }^{11q},\\ {{{M}}_{uuq}} = \dfrac{{1 - {v^2}}}{4}D_{uiu\overline i }^{00q}G_{uju\overline j }^{00q},{{{M}}_{wwq}} = \dfrac{{1 - {v^2}}}{4}D_{wlw\overline l }^{00q}G_{wmw\overline m }^{00q}, \end{array}} \right\} $

$ \left. {\begin{array}{*{20}{l}} D_{\alpha \delta \beta \theta }^{s\overline s q} = \displaystyle\int_{ - 1}^1 {\left\{ {\dfrac{{{{\rm d}^s}[f_\alpha ^q(\xi ){P_\delta }(\xi )]}}{{{\rm d}{\xi ^s}}}\dfrac{{{{\rm d}^{\overline s }}[f_\beta ^q(\xi ){P_\theta }(\xi )]}}{{{\rm d}{\xi ^{\overline s }}}}} \right\}} {\rm d}\xi ,\\ G_{\alpha \delta \beta \theta }^{s\overline s q} = \displaystyle\int_{ - 1}^1 {\left\{ {\dfrac{{{{\rm d}^s}[f_\alpha ^q({\zeta _q}){P_\delta }({\zeta _q})]}}{{{\rm d}{\zeta _q}^s}}\dfrac{{{{\rm d}^{\overline s }}[f_\beta ^q({\zeta _q}){P_\theta }({\zeta _q})]}}{{{\rm d}{\zeta _q}^{\overline s }}}} \right\}} {\rm d}{\zeta _q};\\ {s,\overline s = 0,1;\;\alpha ,\beta = u,w;\;\theta = \overline i ,\overline j ,\overline l ,\overline m ;\;\delta = i,j,l,m} ; \end{array}} \right\} $

${{\varOmega}} = \omega \sqrt {{{\rho hL} / E}} ,\;\,{{{X}}_q} = {\left[ {\begin{array}{*{20}{c}} {{{{A}}^q}}&{{{{B}}^q}} \end{array}} \right]^{\rm T}},$

其中AqBq为第q个梁段特征方程中未知系数组成的列向量,

$\begin{split} {{{A}}^q} =& {{\rm{[}}\begin{array}{*{20}{c}} {A_{11}^q}&{A_{12}^q}& \cdots & {A_{1{J_q}}^q}&{A_{21}^q}& \cdots &{A_{{I_q}{J_q}}^q} \end{array}{\rm{]}}^{\rm T}},\\ {{{B}}^q} =& {{\rm{[}}\begin{array}{*{20}{c}} {B_{11}^q}&{B_{12}^q}& \cdots & {B_{1{J_q}}^q}&{B_{21}^q}& \cdots &{B_{{I_q}{J_q}}^q} \end{array}{\rm{]}}^{\rm T}}. \end{split}$

合并5个梁段的特征方程:

${{K}} - {{{\varOmega}} ^2}{{MX}} = {\rm{0}}.$

式中:

$ \left. {\begin{array}{*{20}{l}} {{K}} = \left[ {\begin{array}{*{20}{c}} {{{{K}}^1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}} \\ {\bf{0}}&{{{{K}}^2}}&{\bf{0}}&{\bf{0}}&{\bf{0}} \\ {\bf{0}}&{\bf{0}}&{{{{K}}^{\rm{3}}}}&{\bf{0}}&{\bf{0}} \\ {\bf{0}}&{\bf{0}}&{\bf{0}}&{{{{K}}^{\rm{4}}}}&{\bf{0}} \\ {\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{{{{K}}^{\rm{5}}}} \end{array}} \right],\\ {{M}} = \left[ {\begin{array}{*{20}{c}} {{{{M}}^1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}} \\ {\bf{0}}&{{{{M}}^2}}&{\bf{0}}&{\bf{0}}&{\bf{0}} \\ {\bf{0}}&{\bf{0}}&{{{{M}}^{\rm{3}}}}&{\bf{0}}&{\bf{0}} \\ {\bf{0}}&{\bf{0}}&{\bf{0}}&{{{{M}}^{\rm{4}}}}&{\bf{0}} \\ {\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}&{{{{M}}^{\rm{5}}}} \end{array}} \right],\\ {{X}} = {\begin{array}{*{20}{c}} {[{{{X}}^{\rm{1}}}}&{{{{X}}^{\rm{2}}}}&{{{{X}}^{\rm{3}}}}&{{{{X}}^{\rm{4}}}}&{{{{X}}^{\rm{5}}}]} \end{array}^{\!\!\!\!{\rm T}}}. \end{array}} \right\} $

考虑梁段1和梁段2、3之间,梁段3和梁段4、5之间的位移连续条件:

$\left. \begin{array}{l} {U_1}({\xi _{\rm{1}}}, - 1) = {U_2}({\xi _2},1),\; - 1 \leqslant {\xi _1} < {\rm{2}}{a_2}{\rm{/}}{a_1} - 1; \\ {W_1}({\xi _1}, - 1) = {W_2}({\xi _{\rm{2}}},1),\; - 1 \leqslant {\xi _1} < {\rm{2}}{a_2}{\rm{/}}{a_1} - 1; \\ {U_1}({\xi _1}, - 1) = {U_{\rm{3}}}({\xi _3},1),\;{\rm{ 2}}{a_2}{\rm{/}}{a_1} - 1 \leqslant {\xi _1} \leqslant 1; \\ {W_1}({\xi _1}, - 1) = {W_{\rm{3}}}({\xi _3},1),\;{\rm{ 2}}{a_2}{\rm{/}}{a_1} - 1 \leqslant {\xi _1} \leqslant 1 .\\ \end{array} \right\}$

$\left. \begin{array}{l} {U_{\rm{3}}}({\xi _{\rm{3}}}, - 1) = {U_{\rm{4}}}({\xi _{\rm{4}}},1),\; - 1 \leqslant {\xi _{\rm{3}}} < {\rm{2}}{a_{\rm{4}}}{\rm{/}}{a_{\rm{3}}} - 1 ;\\ {W_{\rm{3}}}({\xi _{\rm{3}}}, - 1) = {W_{\rm{4}}}({\xi _{\rm{4}}},1),\; - 1 \leqslant {\xi _{\rm{3}}} < {\rm{2}}{a_{\rm{4}}}{\rm{/}}{a_{\rm{3}}} - 1 ;\\ {U_{\rm{3}}}({\xi _{\rm{3}}}, - 1) = {U_{\rm{5}}}({\xi _{\rm{5}}},1),{\rm{ 2}}{a_{\rm{4}}}{\rm{/}}{a_{\rm{3}}} - 1 \leqslant {\xi _{\rm{3}}} \leqslant 1 ;\\ {W_{\rm{3}}}({\xi _{\rm{3}}}, - 1) = {W_{\rm{5}}}({\xi _{\rm{5}}},1),{\rm{ 2}}{a_{\rm{4}}}{\rm{/}}{a_{\rm{3}}} - 1 \leqslant {\xi _{\rm{3}}} \leqslant 1 .\\ \end{array} \right\}$

${\xi _1}{\text{、}}{\xi _2}{\text{、}}{\xi _3}$满足条件:

$\left. \begin{array}{l} {\xi _2} = ({a_1}{\rm{/}}{a_2}){\xi _1} + ({a_1}{\rm{/}}{a_2} - 1),{\kern 1pt} \; - 1 \leqslant {\xi _1} < {\rm{2}}{a_{\rm{2}}}{\rm{/}}{a_{\rm{1}}} - 1; \\ {\xi _3} = ({a_1}{\rm{/}}{a_{\rm{3}}}){\xi _1} - ({a_1}{\rm{/}}{a_{\rm{3}}} - 1),{\rm{ 2}}{a_{\rm{2}}}{\rm{/}}{a_{\rm{1}}} - 1 \leqslant {\xi _1} \leqslant 1. \\ \end{array} \right\}$

${\xi _3}{\text{、}}{\xi _{\rm{4}}}{\text{、}}{\xi _{\rm{5}}}$满足条件:

$\left. \begin{array}{l} {\xi _{\rm{4}}} = ({a_{\rm{3}}}{\rm{/}}{a_{\rm{4}}}){\xi _{\rm{3}}} + ({a_{\rm{3}}}{\rm{/}}{a_{\rm{4}}} - 1),{\kern 1pt} \; - 1 \leqslant {\xi _{\rm{3}}} < {\rm{2}}{a_{\rm{4}}}{\rm{/}}{a_{\rm{3}}} - 1; \\ {\xi _{\rm{5}}} = ({a_{\rm{3}}}{\rm{/}}{a_{\rm{5}}}){\xi _3} - ({a_{\rm{3}}}{\rm{/}}{a_{\rm{5}}} - 1),\;{\rm{ 2}}{a_{\rm{4}}}{\rm{/}}{a_{\rm{3}}} - 1 \leqslant {\xi _{\rm{3}}} \leqslant 1 .\\ \end{array} \right\}$

切比雪夫多项式具有如下性质:

$\int_{ - 1}^1 {\frac{{{P_s}(\xi ){P_t}(\xi )}}{{\sqrt {1 - {\xi ^2}} }}} {\rm d}\xi = \left\{ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} 0,&{\;\;\;\;\;s \ne t;} \end{array}} \\ {\begin{array}{*{20}{c}} {{{\text{π}} }/{2}},&{s = t \ne 1;} \end{array}} \\ {\begin{array}{*{20}{c}} {\text{π}}, &{s = t = 1.} \end{array}} \end{array}} \right\}$

将式(22)、(23)分别代入式(20)、(21),等式两边分别同时乘以PIξq)(I=1,2,3, $\cdot \cdot \cdot $Iq),PLξq)(L=1,2,3, $ \cdot \cdot \cdot$Lq),并将式(24)代入,在区间(−1,1)对ξq积分(对不同的梁段之间ξq分别取ξ2ξ3ξ4ξ5):

$ \left. {\begin{array}{*{20}{l}} {\text{π}}\displaystyle\sum\limits_{j = 1}^{{J_2}} {A_{ij}^2} = \displaystyle\sum\limits_{\overline i = 1}^{{I_1}} {\displaystyle\sum\limits_{j = 1}^{{J_1}} {A_{\overline i j}^1} } {( - 1)^{j - 1}}{e_{i\overline i }},\;\;\;i = 1;\\ \dfrac{{\text{π}}}{2}\displaystyle\sum\limits_{j = 1}^{{J_2}} {A_{ij}^2} = \displaystyle\sum\limits_{\overline i = 1}^{{I_1}} {\displaystyle\sum\limits_{j = 1}^{{J_1}} {A_{\overline i j}^1} } {( - 1)^{j - 1}}{e_{i\overline i }},\;\;\;i = 2,3, \cdots, {I_2};\\ {\text{π}}\displaystyle\sum\limits_{m = 1}^{{M_2}} {B_{lm}^2} = \displaystyle\sum\limits_{\overline l = 1}^{{L_1}} {\displaystyle\sum\limits_{m = 1}^{{M_1}} {B_{\overline l m}^1} } {( - 1)^{m - 1}}{E_{l\overline l }},\;\;\;l = 1;\\ \dfrac{{\text{π}}}{2}\displaystyle\sum\limits_{m = 1}^{{M_2}} {B_{lm}^2} = \displaystyle\sum\limits_{\overline l = 1}^{{L_1}} {\displaystyle\sum\limits_{m = 1}^{{M_1}} {B_{\overline l m}^1{{( - 1)}^{m - 1}}} } {E_{l\overline l }},\;\;\;l = 2,3, \cdots, {L_2};\\ {\text{π}}\displaystyle\sum\limits_{j = 1}^{{J_3}} {A_{ij}^3} = \displaystyle\sum\limits_{\overline i = 1}^{{I_1}} {\displaystyle\sum\limits_{j = 1}^{{J_1}} {A_{\overline i j}^1} } {( - 1)^{j - 1}}{f_{i\overline i }},\;\;\;i = 1;\\ \dfrac{{\text{π}}}{2}\displaystyle\sum\limits_{j = 1}^{{J_3}} {A_{ij}^3} = \displaystyle\sum\limits_{\overline i = 1}^{{I_1}} {\displaystyle\sum\limits_{j = 1}^{{J_1}} {A_{\overline i j}^1} } {( - 1)^{j - 1}}{f_{i\overline i }},\;\;\;i = 2,3, \cdots, {I_3};\\ {\text{π}}\displaystyle\sum\limits_{m = 1}^{{M_3}} {B_{lm}^3} = \displaystyle\sum\limits_{\overline l = 1}^{{L_1}} {\displaystyle\sum\limits_{m = 1}^{{M_1}} {B_{\overline l m}^1} } {( - 1)^{m - 1}}{F_{l\overline l }},\;\;\;l = 1;\\ \dfrac{{\text{π}}}{2}\displaystyle\sum\limits_{m = 1}^{{M_3}} {B_{lm}^3} = \displaystyle\sum\limits_{\overline l = 1}^{{L_1}} {\displaystyle\sum\limits_{m = 1}^{{M_1}} {B_{\overline l m}^1{{( - 1)}^{m - 1}}} } {F_{l\overline l }},\;\;\;l = 2,3, \cdots, {L_3}; \end{array}} \right\} $

$ \left. {\begin{array}{*{20}{l}} {\text{π}}\displaystyle\sum\limits_{j = 1}^{{J_{\rm{4}}}} {A_{ij}^{\rm{4}}} = \displaystyle\sum\limits_{\overline i = 1}^{{I_{\rm{3}}}} {\displaystyle\sum\limits_{j = 1}^{{J_{\rm{3}}}} {A_{\overline i j}^{\rm{3}}} } {( - 1)^{j - 1}}{g_{i\overline i }},\;\;\;i = 1;\\[-2pt] \dfrac{{\text{π}}}{2}\displaystyle\sum\limits_{j = 1}^{{J_{\rm{4}}}} {A_{ij}^{\rm{4}}} = \displaystyle\sum\limits_{\overline i = 1}^{{I_{\rm{3}}}} {\displaystyle\sum\limits_{j = 1}^{{J_{\rm{3}}}} {A_{\overline i j}^{\rm{3}}} } {( - 1)^{j - 1}}{g_{i\overline i }},\;\;\;i = 2,3, \cdots, {I_{\rm{4}}};\\[-2pt] {\text{π}}\displaystyle\sum\limits_{m = 1}^{{M_{\rm{4}}}} {B_{lm}^4} = \displaystyle\sum\limits_{\overline l = 1}^{{L_{\rm{3}}}} {\displaystyle\sum\limits_{m = 1}^{{M_{\rm{3}}}} {B_{\overline l m}^{\rm{3}}} } {( - 1)^{m - 1}}{G_{l\overline l }},\;\;\;l = 1;\\[-2pt] \dfrac{{\text{π}}}{2}\displaystyle\sum\limits_{m = 1}^{{M_{\rm{4}}}} {B_{lm}^{\rm{4}}} = \displaystyle\sum\limits_{\overline l = 1}^{{L_{\rm{3}}}} {\displaystyle\sum\limits_{m = 1}^{{M_{\rm{3}}}} {B_{\overline l m}^{\rm{3}}{{( - 1)}^{m - 1}}} } {G_{l\overline l }},\;\;\;l = 2,3, \cdots, {L_{\rm{4}}};\\[-2pt] {\text{π}}\displaystyle\sum\limits_{j = 1}^{{J_{\rm{5}}}} {A_{ij}^{\rm{5}}} = \displaystyle\sum\limits_{\overline i = 1}^{{I_{\rm{3}}}} {\displaystyle\sum\limits_{j = 1}^{{J_{\rm{3}}}} {A_{\overline i j}^{\rm{3}}} } {( - 1)^{j - 1}}{h_{i\overline i }},\;\;\;i = 1;\\[-2pt] \dfrac{{\text{π}}}{2}\displaystyle\sum\limits_{j = 1}^{{J_{\rm{5}}}} {A_{ij}^{\rm{5}}} = \displaystyle\sum\limits_{\overline i = 1}^{{I_{\rm{3}}}} {\displaystyle\sum\limits_{j = 1}^{{J_{\rm{3}}}} {A_{\overline i j}^{\rm{3}}} } {( - 1)^{j - 1}}{h_{i\overline i }},\;\;\;i = 2,3, \cdots, {I_{\rm{5}}};\\[-2pt] {\text{π}}\displaystyle\sum\limits_{m = 1}^{{M_{\rm{5}}}} {B_{lm}^{\rm{5}}} = \displaystyle\sum\limits_{\overline l = 1}^{{L_{\rm{3}}}} {\displaystyle\sum\limits_{m = 1}^{{M_{\rm{3}}}} {B_{\overline l m}^{\rm{3}}} } {( - 1)^{m - 1}}{H_{l\overline l }},\;\;\;l = 1;\\[-2pt] \dfrac{{\text{π}}}{2}\displaystyle\sum\limits_{m = 1}^{{M_{\rm{5}}}} {B_{lm}^{\rm{5}}} = \displaystyle\sum\limits_{\overline l = 1}^{{L_{\rm{3}}}} {\displaystyle\sum\limits_{m = 1}^{{M_{\rm{3}}}} {B_{\overline l m}^{\rm{3}}{{( - 1)}^{m - 1}}} } {H_{l\overline l }},\;\;\;l = 2,3, \cdots, {L_{\rm{5}}}. \end{array}} \right\} $

其中,在固支梁中,

$ \left. {\begin{array}{*{20}{l}} {e_{st}} = {E_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{{2{k_{\rm{1}}} - {\xi _2} - 1}}{{{k_{\rm{1}}}^2\sqrt {1 - {\xi _2}^2} }}} {P_s}\left[\dfrac{{{\xi _2} - ({k_{\rm{1}}} - 1)}}{{{k_{\rm{1}}}}}\right]{P_t}({\xi _2}){\rm d}{\xi _2},\\[-2pt] {f_{st}} = {F_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{{2{k_{\rm{2}}}^2 + {\xi _3} - 1}}{{{k_{\rm{2}}}^2\sqrt {1 - {\xi _3}^2} }}} {P_s}\left[\dfrac{{{\xi _3} + ({k_{\rm{2}}} - 1)}}{{{k_{\rm{2}}}}}\right]{P_t}({\xi _3}){\rm d}{\xi _3},\\[-2pt] {g_{st}} = {G_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{{2{k_{\rm{3}}} - {\xi _{\rm{4}}} - 1}}{{{k_{\rm{3}}}\sqrt {1 - {\xi _{\rm{4}}}^2} }}} {P_s}\left[\dfrac{{{\xi _{\rm{4}}} - ({k_{\rm{3}}} - 1)}}{{{k_{\rm{3}}}}}\right]{P_t}({\xi _{\rm{4}}}){\rm d}{\xi _{\rm{4}}},\\[-2pt] {h_{st}} = {H_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{1}{{{k_{\rm{4}}}\sqrt {1 - {\xi _{\rm{5}}}^2} }}} {P_s}\left[\dfrac{{{\xi _{\rm{5}}} + ({k_{\rm{4}}} - 1)}}{{{k_{\rm{4}}}}}\right]{P_t}({\xi _{\rm{5}}}){\rm d}{\xi _{\rm{5}}}. \end{array}} \right\} $

在简支梁中,

$ \left. {\begin{array}{*{20}{l}} {e_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{1}{{\sqrt {1 - {\xi _2}^2} }}} {P_s}\left[\dfrac{{{\xi _2} - ({k_{\rm{1}}} - 1)}}{{{k_{\rm{1}}}}}\right]{P_t}({\xi _2}){\rm d}{\xi _2},\\[-2pt] {E_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{{2{k_{\rm{1}}} - {\xi _2} - 1}}{{{k_{\rm{1}}}^2\sqrt {1 - {\xi _2}^2} }}} {P_s}\left[\dfrac{{{\xi _2} - ({k_{\rm{1}}} - 1)}}{{{k_{\rm{1}}}}}\right]{P_t}({\xi _2}){\rm d}{\xi _2},\\[-2pt] {f_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{{\rm{1}}}{{\sqrt {1 - {\xi _3}^2} }}} {P_s}\left[\dfrac{{{\xi _3} + ({k_{\rm{2}}} - 1)}}{{{k_{\rm{2}}}}}\right]{P_t}({\xi _{\rm{3}}}){\rm d}{\xi _{\rm{3}}},\\[-2pt] {F_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{{2{k_{\rm{2}}} + {\xi _3} - 1}}{{{k_{\rm{2}}}^2\sqrt {1 - {\xi _3}^2} }}} {P_s}\left[\dfrac{{{\xi _3} + ({k_{\rm{2}}} - 1)}}{{{k_{\rm{2}}}}}\right]{P_t}({\xi _{\rm{3}}}){\rm d}{\xi _{\rm{3}}},\\[-2pt] {g_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{1}{{\sqrt {1 - {\xi _{\rm{4}}}^2} }}} {P_s}\left[\dfrac{{{\xi _{\rm{4}}} - ({k_{\rm{3}}} - 1)}}{{{k_{\rm{3}}}}}\right]{P_t}({\xi _{\rm{4}}}){\rm d}{\xi _{\rm{4}}},\\[-2pt] {G_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{{2{k_{\rm{3}}} - {\xi _{\rm{4}}} - 1}}{{{k_{\rm{3}}}\sqrt {1 - {\xi _{\rm{4}}}^2} }}} {P_s}\left[\dfrac{{{\xi _{\rm{4}}} - ({k_{\rm{3}}} - 1)}}{{{k_{\rm{3}}}}}\right]{P_t}({\xi _{\rm{4}}}){\rm d}{\xi _{\rm{4}}},\\[-2pt] {h_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{{\rm{1}}}{{\sqrt {1 - {\xi _5}^2} }}} {P_s}\left[\dfrac{{{\xi _5} + ({k_{\rm{4}}} - 1)}}{{{k_{\rm{4}}}}}\right]{P_t}({\xi _{\rm{5}}}){\rm d}{\xi _{\rm{5}}},\\[-2pt] {H_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{1}{{{k_{\rm{4}}}\sqrt {1 - {\xi _{\rm{5}}}^2} }}} {P_s}\left[\dfrac{{{\xi _{\rm{5}}} + ({k_{\rm{4}}} - 1)}}{{{k_{\rm{4}}}}}\right]{P_t}({\xi _{\rm{5}}}){\rm d}{\xi _{\rm{5}}}. \end{array}} \right\}$

在悬臂梁中,

$ \left. {\begin{array}{*{20}{l}} {e_{st}} = {E_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{1}{{{k_1}\sqrt {1 - {\xi _2}^2} }}} {P_s}\left[\dfrac{{{\xi _2} - ({k_1} - 1)}}{{{k_1}}}\right]{P_t}({\xi _2}){\rm d}{\xi _2},\\ {f_{st}} = {F_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{{2{k_2} + {\xi _3} - 1}}{{{k_2}\sqrt {1 - {\xi _3}^2} }}} {P_s}\left[\dfrac{{{\xi _3} + ({k_2} - 1)}}{{{k_2}}}\right]{P_t}({\xi _3}){\rm d}{\xi _3},\\ {g_{st}} = {G_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{{\rm{1}}}{{\sqrt {1 - {\xi _4}^2} }}} {P_s}\left[\dfrac{{{\xi _4} - ({k_{\rm{3}}} - 1)}}{{{k_{\rm{3}}}}}\right]{P_t}({\xi _4}){\rm d}{\xi _4},\\ {h_{st}} = {H_{st}} = \displaystyle\int_{ - 1}^1 {\dfrac{{\rm{1}}}{{\sqrt {1 - {\xi _5}^2} }}} {P_s}\left[\dfrac{{{\xi _5} + ({k_{\rm{4}}} - 1)}}{{{k_{\rm{4}}}}}\right]{P_t}({\xi _5}){\rm d}{\xi _5}. \end{array}} \right\} $

式中:

由式(25)、(26)可知,式(19)系数列向量X中的系数是线性相关的. 由式(25)可得矩阵S1,由式(26)可得矩阵S2. 矩阵S= S2S1可以消去X中的多余系数:

${{X}} = {{S}}\overline {{X}} .$

将式(30)代入式(18),可得

$\overline {{K}} - {{{\varOmega}} ^2}\overline {{M}}\, \overline {{X}} = 0.$

式中: $\overline {{K}} $为整个裂纹梁的刚度矩阵, $ \overline {{K}} = {{{S}}^{\rm{T}}}{{KS}}$$\overline {{M}} $为裂纹梁的质量矩阵, $\overline {{M}} = {{{S}}^{\rm{T}}}{{MS}} $$\overline {{X}} $为裂纹梁的模态系数.

1.2. 不同裂纹参数的梁

若梁中的裂纹等长,则可以减少划分梁段的数目. 如图2所示,将其划分为4个梁段,利用梁段1和梁段2、3、4之间的位移连续条件整合各梁段的振动特征方程进行求解.

图 2

图 2   带有2条相同深度裂纹的梁的分析模型

Fig.2   Analytical model of beam with two cracks of same depth


分析带有3条不同深度裂纹的梁,如图3所示,将其划分为7个梁段,利用梁段1和梁段2、3,梁段2和梁段4、5,梁段3和梁段6、7之间的位移连续条件整合各梁段的方程求解.

图 3

图 3   带有3条不同深度裂纹的梁的分析模型

Fig.3   Analytical model of beam with three cracks of different depths


本文方法理论上可以推广至带有更多条裂纹的梁的动力特性分析,仅需增加梁段的数目和所使用的位移连续条件. 梁段的数量越多,则位移函数中切比雪夫多项式的项数越多,解的收敛性越差. 由于采用了局部的无量纲坐标,支承条件相同的梁段的特征矩阵和梁段之间的位移连续性条件不需要重复计算,这提高了程序的计算效率,减少了运行时间. 一般多裂纹梁的计算流程图如图4所示.

图 4

图 4   多裂纹梁的计算流程图

Fig.4   Flow chart for calculation of multi-cracked beam


2. 特征频率的收敛性和比较研究

以带有2条裂纹的梁为例进行分析,考虑裂纹深度c1c2与梁高度之比c1/hc2/h分别为0.3和0.2,裂纹到梁左端的距离与梁的长度之比d1/Ld2/L分别为0.5和0.8的固支梁. 使用MATLAB进行编程计算,对每个梁段的振幅函数UW取相同的级数项,可以有效地减少程序的运行时间. 表1给出不同高跨比(h/L=0.1、0.2、0.3)下,固支裂纹梁无量纲频率参数Ω的收敛情况. 当取mn=60×15时,前8阶无量纲特征频率参数的精度可以达到3位有效数字.

表 1   固支梁前8阶的频率参数Ω的收敛性

Tab.1  Convergence of first-eighth frequency parameters Ω of fixed beam

h/L mn Ω1 Ω2 Ω3 Ω4 Ω5 Ω6 Ω7 Ω8
0.1 40×10 0.183 7 0.488 4 0.831 2 0.987 1 1.341 8 1.774 1 1.949 6 2.397 6
0.1 50×10 0.183 3 0.488 2 0.828 5 0.986 8 1.341 2 1.769 8 1.948 7 2.397 0
0.1 50×15 0.183 3 0.488 2 0.828 5 0.986 8 1.341 1 1.769 8 1.948 7 2.397 0
0.1 60×15 0.183 1 0.488 0 0.826 7 0.986 6 1.340 7 1.766 8 1.948 1 2.396 6
0.2 40×10 0.436 8 1.055 1 1.379 3 1.611 6 2.536 7 2.620 4 3.272 4 4.115 9
0.2 50×10 0.436 3 1.054 7 1.378 7 1.608 7 2.535 7 2.617 3 3.271 2 4.113 8
0.2 50×15 0.436 2 1.054 6 1.378 6 1.608 5 2.535 7 2.617 1 3.271 1 4.113 6
0.2 60×15 0.435 8 1.054 4 1.378 2 1.606 6 2.535 0 2.614 9 3.270 3 4.112 2
0.3 40×10 0.669 8 1.483 6 1.684 9 2.128 9 3.150 1 3.347 0 4.017 4 4.273 0
0.3 50×10 0.669 2 1.482 9 1.684 4 2.125 6 3.147 7 3.344 5 4.016 5 4.271 3
0.3 50×15 0.669 2 1.482 8 1.684 3 2.125 2 3.147 4 3.344 4 4.016 4 4.271 1
0.3 60×15 0.668 8 1.482 3 1.684 0 2.123 0 3.145 7 3.342 5 4.015 7 4.270 0

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用ANSYS软件进行有限元分析,单元类型选用Solid183,ρ=7 860 kg/m3E=2.1×1011 Pa,v=0.3. 以h/L=0.1、d1/L=0.5、d2/L=0.8的裂纹固支梁为例,在坐标轴上分别绘制长为50、30和20,高均为10的梁段,以0.5的网格宽度对其进行划分,共有4 000个单元. 在裂缝位置处用Merge items命令将连续点合并,限制梁两端节点的位移和转角为0. 建立的分析模型如图5所示.

图 5

图 5   裂纹固支梁的ANSYS分析模型

Fig.5   Analytical model of cracked fixed beam in ANSYS


考虑d1/L=0.5,d2/L=0.8,高跨比h/L=0.1的裂纹梁. 表2~4分别给出不同深度的裂纹下固支梁、简支梁和悬臂梁前8阶无量纲频率参数Ω与ANSYS有限元解的比较. 由表2~4可知,最大相对误差仅为1.463%,验证了不同裂纹深度下本文分析方法的有效性. 与有限元法相比,该方法的不足之处在于编制的MATLAB程序运行时间较长.

表 2   固支梁计算结果与有限元分析的对比(h/L=0.1,d1/L=0.5,d2/L=0.8)

Tab.2  Comparison of results of fixed beam with those from FEA(h/L=0.1,d1/L=0.5,d2/L=0.8)

参数 方法 Ω1 Ω2 Ω3 Ω4 Ω5 Ω6 Ω7 Ω8
c1/h=0.2,c2/h=0.1 本文方法 0.188 1 0.491 9 0.866 2 0.993 5 1.353 7 1.825 4 1.968 2 2.407 0
c1/h=0.2,c2/h=0.1 有限元法 0.187 5 0.491 8 0.862 2 0.993 2 1.353 2 1.817 9 1.965 8 2.406 6
c1/h=0.3,c2/h=0.2 本文方法 0.183 1 0.488 0 0.826 7 0.986 6 1.340 7 1.766 8 1.948 1 2.396 6
c1/h=0.3,c2/h=0.2 有限元法 0.182 1 0.487 7 0.820 9 0.986 1 1.339 6 1.755 8 1.945 8 2.395 1
c1/h=0.4,c2/h=0.2 本文方法 0.176 9 0.487 6 0.797 4 0.986 3 1.338 1 1.693 4 1.932 7 2.389 9
c1/h=0.4,c2/h=0.2 有限元法 0.175 7 0.487 3 0.791 1 0.985 9 1.336 5 1.680 7 1.930 7 2.387 2

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表 3   简支梁计算结果与有限元分析的对比(h/L=0.1,d1/L=0.5,d2/L=0.8)

Tab.3  Comparison of results of simply-supported beam with those from FEA(h/L=0.1,d1/L=0.5,d2/L=0.8)

参数 方法 Ω1 Ω2 Ω3 Ω4 Ω5 Ω6 Ω7 Ω8
c1/h=0.2,c2/h=0.1 本文方法 0.085 6 0.336 8 0.692 1 0.978 5 1.177 9 1.668 3 1.977 9 2.259 1
c1/h=0.2,c2/h=0.1 有限元法 0.085 1 0.336 5 0.688 3 0.976 1 1.177 5 1.662 8 1.977 2 2.258 3
c1/h=0.3,c2/h=0.2 本文方法 0.081 3 0.329 0 0.655 5 0.953 6 1.169 6 1.632 8 1.954 3 2.243 9
c1/h=0.3,c2/h=0.2 有限元法 0.080 6 0.328 4 0.649 7 0.949 6 1.168 7 1.626 5 1.952 4 2.242 0
c1/h=0.4,c2/h=0.2 本文方法 0.076 3 0.328 7 0.623 6 0.926 1 1.167 4 1.599 1 1.954 1 2.237 6
c1/h=0.4,c2/h=0.2 有限元法 0.075 2 0.328 1 0.616 6 0.921 1 1.166 2 1.593 5 1.952 4 2.234 7

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表 4   悬臂梁计算结果与有限元分析的对比(h/L=0.1,d1/L=0.5,d2/L=0.8)

Tab.4  Comparison of results of cantilevered beam with those from FEA(h/L=0.1,d1/L=0.5,d2/L=0.8)

参数 方法 Ω1 Ω2 Ω3 Ω4 Ω5 Ω6 Ω7 Ω8
c1/h=0.2,c2/h=0.1 本文方法 0.031 6 0.185 2 0.493 4 0.501 0 0.886 0 1.386 9 1.476 7 1.881 9
c1/h=0.2,c2/h=0.1 有限元法 0.031 6 0.184 2 0.492 7 0.500 7 0.881 6 1.385 8 1.474 4 1.876 0
c1/h=0.3,c2/h=0.2 本文方法 0.031 3 0.177 0 0.485 2 0.493 4 0.839 7 1.359 1 1.449 0 1.840 2
c1/h=0.3,c2/h=0.2 有限元法 0.031 2 0.175 4 0.483 9 0.492 6 0.833 3 1.356 5 1.444 8 1.834 0
c1/h=0.4,c2/h=0.2 本文方法 0.030 7 0.166 8 0.475 9 0.491 9 0.812 4 1.355 6 1.420 3 1.809 4
c1/h=0.4,c2/h=0.2 有限元法 0.030 6 0.164 7 0.473 8 0.491 1 0.806 2 1.352 4 1.414 8 1.804 3

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为了检验当裂纹在不同位置时该方法解的精度,将本文的解和Lourdes等[15]给出的解与有限元结果进行比较. 表5给出裂纹在不同位置处(d2/L=0.35,0.45,0.5),d1/L=0.25时, h/L=0.1的简支梁1阶频率参数Ω1的比较. 可知,该方法的计算结果更接近有限元结果,该方法对于分析裂纹在不同位置的裂纹梁都有很好的精度,且无需事先确定裂缝的等效弹簧刚度,分析简单直接.

表 5   第1阶频率参数Ω1与Lourdes[15]解的对比

Tab.5  Comparison of first frequency parameter Ω1 with Lourdes’s[15] results

参数 方法 Ω1
d2/L=0.35 d2/L=0.45 d2/L=0.5
d1/L=0.25,c1/h=0.05 有限元解 0.087 8 0.087 7 0.087 7
d1/L=0.25,c1/h=0.05 本文解法 0.088 1 0.088 0 0.088 0
d1/L=0.25,c2/h=0.10 Lourdes解 0.089 7 0.089 6 0.089 6
d1/L=0.25,c2/h=0.10 有限元解 0.083 7 0.082 9 0.082 8
c1/h=0.15,c2/h=0.25 本文解法 0.084 6 0.084 0 0.083 9
c1/h=0.15,c2/h=0.25 Lourdes解 0.087 4 0.087 0 0.086 9

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3. 参数分析

分析裂纹深度对裂纹梁自振频率的影响,考虑裂纹固支梁高跨比h/L=0.1. 图6(a)给出当d1/L=0.5,d2/L=0.75,c2/h=0.2时,c1/h的变化对前8阶频率参数的影响(其中Ωn'为c1=0时裂纹梁的第n阶频率参数). 图6(b)给出当d1/L=0.5,d2/L=0.75,c1/h=0.2时,c2/h的变化对前8阶频率参数的影响(其中Ωn'为c2=0时裂纹梁的第n阶频率参数). 由图6可知,随着裂纹深度的增大,裂纹梁的频率逐渐减小,各阶频率减小的速度不同.

图 6

图 6   裂纹固支梁前8阶频率参数随裂纹深度的变化图

Fig.6   First-eighth frequency parameters of cracked fixed beams with different crack depths


分析不同的裂纹深度对裂纹梁上表面振型的影响. 考虑裂纹固支梁高跨比h/L=0.1,d1/L=0.5,d2/L=0.75. 为了便于对比分析,以c1/h=0或c2/h=0时第1阶和第3阶振型中梁的中点处以及第2阶振型中梁距离左端1/4处的位移为−1绘制振型图.图7给出当c2/h=0.2,c1/h=0、0.2、0.4、0.6时W位移的振型w图. 可知,随着裂纹深度的增大,各阶振型的位移逐渐增大. 其中2阶振型的变化很小,1阶和3阶受裂纹的影响更大. 图8给出当c1/h=0.2,c2/h=0、0.2、0.4、0.6时W位移的振型图. 其中1阶振型的变化很小,2阶和3阶受裂纹的影响更大. 从图8可知,当裂纹位于振型位移的峰值处时,裂纹对振型的影响较大. 当裂纹位于振型的较小位移处时,裂纹对振型的影响较小. 这解释了图6中不同位置的裂纹对各阶频率影响不同的情况.

图 7

图 7   不同c1下固支梁W的前3阶振型

Fig.7   First-third modal shapes of W of fixed beams with different c1


图 8

图 8   不同c2下固支梁W的前3阶振型

Fig.8   First-third modal shapes of W of fixed beams with different c2


4. 结 论

(1)本文基于弹性力学能量法,提出分析裂纹在任意位置处的多裂纹梁自由振动的方法. 根据裂纹情况将多裂纹梁分段,用Chebyshev-Ritz法建立其振动特征方程,使用位移连续条件整合方程求解. 与用无质量弹簧模拟裂缝和引入等效刚度函数的常规方法相比,具有分析简单直接的特点,无须事先确定裂缝的等效弹簧刚度或整个梁的刚度变化函数,放弃使用平截面假定,采用弹性力学平面应力理论,无需专门考虑剪切效应的影响,适用于各种边界条件下的细长梁和短梁,分析简单精度高且误差可控.

(2)裂纹梁的各阶频率随着裂纹深度的增大而降低,各阶振型的位移随着裂纹深度的增大而增大.

(3)不同位置处的裂纹对各阶频率和振型的影响不同. 当裂纹位于振型位移的峰值处时,该阶振型和频率受裂纹的影响极大,此时裂纹深度的变化会对振型的位移产生很大扰动. 振型和频率受裂纹的影响随着裂纹到振型位移峰值处位置的距离的增大而减小. 当裂纹位于振型的位移零点处时,该阶振型及频率受裂纹的影响极小. 以此为依据可以开展裂纹位置的探测工作.

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