受限亚音速气流中倒置悬臂壁板静气弹稳定性的理论及实验研究 1)

图1

Fig. 1

(1)$ \begin{eqnarray}\label{1} {\rho _s}h\frac{{{\partial ^2}w}}{{\partial {t^2}}} + D\frac{{{\partial ^4}w}}{{\partial {x^4}}} +\Delta p=0 \end{eqnarray}$

及如下的边界条件

(2)$ \begin{eqnarray}\label{2} {\left. {\frac{{{\partial ^2}w}}{{\partial {x^2}}}} \right|_{x = 0}} = {\left. {\frac{{{\partial ^3}w}}{{\partial {x^3}}}} \right|_{x = 0}} = {\left. w \right|_{x = L}} = {\left. {\frac{{\partial w}}{{\partial x}}} \right|_{x = L}} = 0 \end{eqnarray}$

其中, $w$, $\rho_s$, $h$分别为壁板的横向变形, 密度及厚度, $D = E{h^3}/12\left( {1 - {\nu ^2}} \right)$ 为壁板的抗弯刚度, $\nu$为壁板的泊松比, $\Delta p= {p_{{H^ + }}} - {p_{{H^ - }}}$为壁板上下表面扰动压力差(向下为正).

考虑理想欧拉流体, 通过引入等熵条件、小扰动假设和小扰动速度势可得壁板上下表面压力的近似线化表达式^[28]

(3)$ \begin{eqnarray}\label{3} {p_{{H^ \pm }}} = {\left. { - {\rho _\infty }\left( {\frac{{\partial \varphi }}{{\partial t}} + U_\infty\frac{{\partial \varphi }}{{\partial x}}} \right)} \right|_{z = {H^ \pm }}} \end{eqnarray}$

式中, $\rho _\infty$ 为气流密度, $\varphi$ 为扰动速度势函数, 其满足如下的线化速度势方程

(4)$ \begin{eqnarray}\label{4} \frac{{{\partial ^2}\varphi }}{{\partial {t^2}}} + 2{M_\infty }{a_\infty }\frac{{{\partial ^2}\varphi }}{{\partial x\partial t}} = a_\infty ^2{\beta ^2}\frac{{{\partial ^2}\varphi }}{{\partial {x^2}}} + a_\infty ^2\frac{{{\partial ^2}\varphi }}{{\partial {z^2}}} \end{eqnarray}$

其中${\beta ^2} = {1 - M_\infty ^2}$, 马赫数 $M_\infty$定义为气流来流速度$U_\infty$与声速${a_\infty }$之比(${M_\infty } = {U_\infty}/{a_\infty }$). 势方程(4)需满足如下的边界条件

(5a)$ \begin{align} {\left. {\frac{{\partial \varphi }}{{\partial z}}} \right|_{z = 0}} = 0 \end{align} $

(5b)$ \begin{align} {\left. {\frac{{\partial \varphi }}{{\partial z}}} \right|_{z = H}} = v\left(z=H \right) = w_a,\ x \in \left[ {0,L} \right] \end{align} $

(5c)$ \begin{align} \Delta p=0,\ \ x \notin \left[ {0,L} \right] \end{align} $

(5d)$ \begin{align} \mathop {\lim }\limits_{x \to {L^ - }} \Delta p = 0 \end{align} $

式(5a)为刚性壁面约束条件; 式(5b)为气固相容条件, 其中

$\begin{eqnarray*} {w_a} = \frac{{\partial w}}{{\partial t}} + U_\infty\frac{{\partial w}}{{\partial x}} \end{eqnarray*}$

定义为壁板表面的法洗速度; 式(5c)和式(5d)分别为流动的连续及Kutta-Joukowski条件. 方程(4)的定解还需要给定远场扰动量的条件和流场的初始条件

(6a)$\begin{align} &\mathop {\lim }\limits_{\left| x \right| \to \infty ,z \to \infty } u,\ \ v = 0 \end{align}$

(6b)$\begin{align} {\left. {\frac{{\partial \varphi }}{{\partial t}}} \right|_{t = 0}} = \varphi \left( {x,z,0} \right) = 0 \end{align}$

2 Possio积分方程及其求解

注意到方程(4)为关于时间及空间的偏微分方程, 直接求解有很大困难. 将其左右两边关于时间作用${\cal L}$(Laplace)变换, 并

利用初始条件(6b), 可得

(7)$ \begin{eqnarray}\label{7} {\lambda ^2}\hat \varphi + 2{M_\infty }{a_\infty }\lambda \frac{{\partial \hat \varphi }}{{\partial x}} = a_\infty ^2{\beta ^2}\frac{{{\partial ^2}\hat \varphi }}{{\partial {x^2}}} + a_\infty ^2\frac{{{\partial ^2}\hat \varphi }}{{\partial {z^2}}} \end{eqnarray}$

其中, $\hat \varphi = \int_0^{ + \infty } {\varphi {{\rm e}^{ - \lambda t}}{\rm d}t} $.

考虑到$\hat \varphi$在空间的变化, 故在式(7)两边关于$x$作用${\cal F}$(Fourier)变换, 得

(8)$ {\lambda ^2}\hat{\hat{\varphi}} + 2{M_\infty }{a_\infty }{\rm i}\omega \lambda \hat{\hat{\varphi}} = - {\omega ^2}a_\infty ^2{\beta ^2}\hat{\hat{\varphi}} + a_\infty ^2\frac{\partial ^2\hat{\hat{\varphi}}}{{\partial {z^2}}} $

其中, $\hat {\hat \varphi} = \int_{ - \infty }^{ + \infty } {\hat \varphi {{\rm e}^{ - {\rm i}\omega x}}{\rm d}x}$.

整理式(8)并将其写为

(9)$ \begin{eqnarray}\label{9} \frac{{\partial ^2}\hat{\hat \varphi}}{{\partial {z^2}}} = \left[ {M_\infty ^2{{\left( {k + {\rm i}\omega } \right)}^2} + {\omega ^2}} \right]\hat{\hat \varphi} \triangleq A\left( {\omega ,k} \right)\hat{\hat \varphi} \end{eqnarray}$

其中$k = \lambda /{U_\infty }$ 定义为系统的减缩频率.

对本文研究的亚音速流体而言, $A$具有正实部. 考虑边界条件(6a), 方程(9)的解可以写作

(10)$ \begin{eqnarray}\label{e3.4} {\hat {\hat \varphi}_1} \left( {k,\omega ,z} \right) = \left\{ {\begin{array}{*{20}{l}} {{\hat {\hat \varphi}}^+{{\rm e}^{ - \sqrt A \left( {z - H} \right)}},\ \ z > H}\\ {{\hat {\hat \varphi}}^-{{\rm e}^{+\sqrt A \left( {z - H} \right)}},\ \ z < H} \end{array}} \right. \end{eqnarray}$

其中, ${\hat {\hat \varphi}}^\pm={\hat {\hat \varphi}}\left( z=H^\pm\right)$. 为满足壁面约束条件(5a), 如图1(a)所示, 在$z=-H$处布置壁板的镜像, 其产生的势函数为

(11)$ \begin{eqnarray}\label{e3.5} {{\hat {\hat \varphi}_2}}\left( {k,\omega ,z} \right) = \left\{ {\begin{array}{*{20}{l}} {{\hat {\hat \varphi}}^-{{\rm e}^{ - \sqrt A \left( {z + H} \right)}},\ \ z > - H}\\ {{\hat {\hat \varphi}}^+{{\rm e}^{+\sqrt A \left( {z + H} \right)}},\ \ z < - H} \end{array}} \right. \end{eqnarray}$

系统势函数$\hat {\hat \varphi}$可由叠加原理获得

(12)$ \begin{eqnarray}\label{e3.5a} \hat {\hat \varphi}={\hat {\hat \varphi} }_1+ {\hat {\hat \varphi} _2} \end{eqnarray}$

由式(5b)及式(12)可得壁板表面气流的横向速度

(13)$ \begin{eqnarray}\label{e3.7} \left. {\begin{array}{*{20}{l}} {{{\hat {\hat v}}^ + } = - \sqrt A \left( \hat{\hat{\varphi}}^ + + \hat{\hat{\varphi}}^ - {{\rm e}^{ - 2\sqrt A H}} \right)}\\ {{{\hat {\hat v}}^ - } = + \sqrt A \left( \hat{\hat{\varphi}}^ - - \hat{\hat{\varphi}}^ - {{\rm e}^{ - 2\sqrt A H}} \right)} \end{array}} \right\} \end{eqnarray}$

其中, ${\hat {\hat v}}^\pm={\hat {\hat v}}\left( z=H^\pm\right)$, 且由流场速度的连续条件可知$\hat {\hat v}^+=\hat {\hat v}^-=\hat {\hat v}$.

由式(5b)及式(13)可得壁板上下表面势函数之差

(14)$ \begin{eqnarray}\label{14} \begin{aligned} \Delta {\hat {\hat \varphi}} \equiv \hat {\hat \varphi}^+ - \hat {\hat \varphi}^- = - \frac{{ 2\hat {\hat v}}}{{\sqrt A \left( {1 - {{\rm e}^{ - 2\sqrt A H}}} \right)}} \end{aligned} \end{eqnarray}$

进一步考虑式(3)后可得

(15)$ \begin{eqnarray}\label{15} \Delta {\hat {\hat p}} = {\rho _\infty U_\infty}{\frac{{2\hat {\hat v}\left( {k + {\rm i}\omega } \right)}}{{\sqrt A \left( {1 - {{\rm e}^{ - 2\sqrt A H}}} \right)}}} \end{eqnarray}$

虽然式(15)给出了任意运动时壁板表面上的压力, 但其求解仍很困难. 对于本文研究的壁板静态失稳问题, 可令$k=0$并由式(15)解得

(16)$ \begin{eqnarray}\label{16} \hat {\hat v}(H) = \frac{1}{{{\rho _\infty}U_\infty}}\left[ {\frac{{\beta \left| \omega \right|\left( {1 - {{\rm e}^{ - c\left| \omega \right|}}} \right)}}{{2{\rm i}\omega }}\Delta \hat {\hat p}} \right] \end{eqnarray}$

其中, $c=2H\beta$. 注意到函数$f(t)$的${\cal H}$(Hilbert)变换的${\cal F}$变换

\begin{eqnarray*} {\cal F}\left[ {{\cal H}\left[ {f(t)} \right]} \right] = \frac{{\left| \omega \right|}}{{{\rm i}\omega }}{\cal F}\left[ {f\left( t \right)} \right] \end{eqnarray*}

的形式后, 相应定义如下变换^[29]

(17)$ \begin{eqnarray}\label{17} {{\cal Q}\left[ {f(t)} \right]} = \frac{1}{{\pi c}}\int_{ - \infty }^{ + \infty } {\frac{{f\left( t \right)}}{{1 + {{\left( {\dfrac{{t - \tau }}{c}} \right)}^2}}}{\rm d}\tau } \end{eqnarray}$

在式(16)两边首先作用${\cal L}$和${\cal F}$的逆变换, 然后考虑式(5b), 式(5c), 应用投影运算${\cal P}_R $向$[0,L]$投影可得如下的积分方程

(18)$ \begin{eqnarray}\label{18} {w_a} = \frac{{\beta {{\cal P}_R}{\cal H}\left( {\cal{I - Q}} \right)\Delta p}}{{2{\rho _\infty }{U_\infty }}} \end{eqnarray}$

由式(5d), 有${\cal P}_R \Delta p=\Delta p$, 考虑算子${\cal QH}$满足交换律(简要证明见附录), 则式(18)变为

(19)$ \begin{eqnarray}\label{19} {w_a} = \frac{{\beta \left( {{{\cal P}_R}{\cal H}{{\cal P}_R} - {{\cal P}_R}{\cal QH}{{\cal P}_R}} \right)\Delta p}}{{2{\rho _\infty }{U_\infty }}} \end{eqnarray}$

其中

\begin{eqnarray*} &&{{\cal P}_R}{\cal H}{{\cal P}_R}[f(t)]= \frac{1}{\pi }\int_{ 0 }^{ L } {\frac{{f\left( \tau \right)}}{{t - \tau }}{\rm d}\tau },\ \ t\in [0,L]\\ &&{{\cal P}_R}{\cal QH}{{\cal P}_R}[f(t)]= \frac{1}{{\pi c}}\int_0^L\frac{ {f\left( \tau \right)} {\left( {\dfrac{{\tau - t}}{c}} \right)}}{{1 + {{\left( {\dfrac{{\tau - t}}{c}} \right)}^2}}}{\rm d}\tau ,\ \ t \in [0,L] \end{eqnarray*}

上式即为Possio积分方程^[29-31].

考虑式(5d)给出的压力条件, 参考薄翼理论中的Söghen解^[28], 引入移位Tricomi算子${\cal T}$^[29-30]

(20)$ \begin{eqnarray}\label{20} {\cal T}\left[ {f\left( x \right)} \right] = \frac{1}{\pi }\sqrt {\frac{{L - x}}{x}} \int_0^L {\sqrt {\frac{\tau }{{L - \tau }}} } \frac{{f\left( \tau \right)}}{{\tau - x}}{\rm d}\tau \end{eqnarray}$

将算子${\cal T}$作用在式(19)的两边, 可得压力和法洗速度之间的关系^[29,31]

(21)$ \begin{eqnarray}\label{21} {2{\rho _\infty}U _\infty}{\cal T}\left[ {{w_a}} \right] = \beta\left( {\cal I}- {\cal P}\right)\Delta p \end{eqnarray}$

其中, ${\cal I}$为单位算子, 与未受限流体相关; 而壁面效应则表征为一复合算子$\cal{P}$, 定义为

\begin{eqnarray*}\label{e3.16} &&{\cal P}[f(x)]\triangleq{\cal T}{{\cal P}_R}{\cal QH}{{\cal P}_R}[f(x)]= \frac{1}{{{\pi ^2}c}}\sqrt {\frac{{L - x}}{x}}\cdot \\&&\qquad \int_0^L {\sqrt {\dfrac{t}{{L - t}}} }\frac{1}{{t - x}}\int_0^L\dfrac{ {f\left( \tau \right)} {\left( {\dfrac{{\tau - t}}{c}} \right)}}{{1 + {{\left( {\dfrac{{\tau - t}}{c}} \right)}^2}}}{\rm d}\tau {\rm d}t \end{eqnarray*}

由式(21)可得

(22)$ \begin{eqnarray}\label{e3.17} \frac{\Delta p}{D} = 2\alpha {\left( \cal {I - P} \right)^{ - 1}}{\cal T}\left( {{\frac{{\partial w}}{{\partial x}}}} \right) \end{eqnarray}$

其中, $\alpha = {{{\rho _\infty }U_\infty ^2}}/({{D\beta }})$.

3 失稳方程及其最小二乘解

考虑壁板的发散失稳方程

(23)$ \begin{eqnarray}\label{23} \frac{{{\partial ^4}w}}{{\partial {x^4}}} = - 2\alpha {\left( \cal {I - P} \right)^{ - 1}}{\cal T}\left[ {\frac{{\partial w}}{{\partial x}}} \right] \end{eqnarray}$

上式为关于$w$的四阶偏微分方程, 将其扩阶写作

(24)$ \begin{eqnarray}\label{24} &&\frac{{\partial {\cal W}}}{{\partial x}} = \left[ {\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0 \end{array}} \right]{\cal W} + \left[ {\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&\cal G&0&0 \end{array}} \right]{\cal W} \\ &&\qquad \triangleq {\bf B}{\cal W} + {\cal {B}}{\cal W} \end{eqnarray}$

其中

\begin{eqnarray*} && {\cal W} = {\left[ {w,\frac{{\partial w}}{{\partial x}},\frac{{{\partial ^2}w}}{{\partial {x^2}}},\frac{{{\partial ^3}w}}{{\partial {x^3}}}} \right]^{\rm T}}\\ &&{\cal G}[f(x)]= - 2\alpha {\left( \cal {I - P} \right)^{ - 1}}{\cal T}\left[ f(x)\right] \end{eqnarray*}

采用常数变易法求解(24)可得

(25)$ \begin{eqnarray}\label{25} { {{\cal D}}} \left[ {{\cal W}\left( x \right)} \right] = {{\rm e}^{{\bf B} x}}{\cal W}\left( 0 \right) \end{eqnarray}$

其中

\begin{eqnarray*} \begin{aligned} &{\cal D} = \left[ {\begin{array}{*{20}{c}} {\cal I}&{ - {{\cal R}_1}}&0&0\\ 0&{{\cal I} - {{\cal R}_2}}&0&0\\ 0&{ - {{\cal R}_3}}&{\cal I}&0\\ 0&{ - {{\cal R}_4}}&0&{\cal I} \end{array}} \right]\\ &{{\cal R}_n}\left[ {f\left( x \right)} \right] = \int_0^x {\frac{{{{\left( {x - s} \right)}^{4 - n}}}}{{\left( {4 - n} \right)!}}{\cal G}\left[ {f\left( s \right)} \right]} {\rm d}s \end{aligned} \end{eqnarray*}

在式(25)两端作用$\cal D$的逆算子可得

(26)$ \begin{eqnarray}\label{e4.5} {\cal W}\left( x \right) = {{\cal D} ^{ - 1}}{{\rm e}^{{B}x}}{\cal W}\left( 0 \right) \end{eqnarray}$

其中

$\begin{eqnarray*} {\cal D}^{-1} = \left[ {\begin{array}{c@{\ \ }c@{\ \ }c@{\ \ }c} {\cal I}&{ {{\cal R}_1}({\cal I-R}_2)^{-1}}&0&0\\ 0&({\cal I-R}_2)^{-1}&0&0\\ 0&{ {{\cal R}_3}({\cal I-R}_2)^{-1}}&{\cal I}&0\\ 0&{ {{\cal R}_4}({\cal I-R}_2)^{-1}}&0&{\cal I} \end{array}} \right] \end{eqnarray*}$

将壁板的边界条件(2)代入式(26)后整理得到

(27)$ \begin{eqnarray}\label{27} {E}{\left. {\left( {{{\cal D}^{-1}}{{\rm e}^{{B}x}}} \right)} \right|_{x = L}}{{E}^{\rm T}}{\left[ {\begin{array}{*{20}{c}} w\\ {\dfrac{{\partial w}}{{\partial x}}} \end{array}} \right]_{x = 0}} = \left[ {\begin{array}{*{20}{c}} 0\\0 \end{array}} \right] \end{eqnarray}$

其中, ${E}=\left[ {\begin{array}{*{20}{c}}{{{{I}}_{2 \times 2}}}&{{{\bf 0}_{2 \times 2}}}\end{array}} \right]$.

式(27)仅有零解对应系统的稳定状态; 当其系数行列式等于零时系统出现非零解, 意味着系统处于失稳状态, 即

(28)$ \begin{eqnarray}\label{28} {\left( { {\cal I} - {{\cal R}_2}} \right)^{ - 1}}{\left[ {y(x)} \right]_{x = L}}= 0 \end{eqnarray}$

其中,函数$y(x)$满足$y(x)\equiv 1$.

因算子${ {\cal I} - {{\cal R}_2}}$是紧的^[31], 对所求参数$\alpha$, 其逆都是存在的, 那么

(29)$ \begin{eqnarray}\label{29} ({\cal I-R}_2)^{-1}= {\cal I} + \sum\limits_{n = 1}^{\infty } {{{\left( {{{\cal R}_2}} \right)}^n}} \end{eqnarray}$

由于${\cal R}_2$中含有需要确定的参数$\alpha$, 直接采用式(29)求解并不利于计算. 但注意到

(30)$ \begin{eqnarray}\label{30} {\left( { {\cal I - {R}}_2} \right)^{ - 1}}\left[ {y(L)} \right] = f(x = L) = 0 \end{eqnarray}$

上式表明方程(30)的解等价于寻找目标函数$f(x)$使

(31)$ \begin{eqnarray}\label{31} \left( {{\cal I - {R}}_2} \right)\left[ {f\left( x \right)} \right] \equiv 1,\ \ f(x=L)= 0 \end{eqnarray}$

将${\cal R}_2$代入式(31)有

(32)$ \begin{eqnarray}\label{e4.12} f\left( x \right) + \alpha \int_0^x {{{\left( {x - s} \right)}^2}{{\left( {\cal {I - P}} \right)}^{ - 1}}{\cal T}\left[ {f\left( s \right)} \right]}{\rm d}s = 1 \end{eqnarray}$

因$(\cal {I - P})$为压缩映射^[29], $\left( \cal {I - P} \right)^{ - 1}{\cal T}\left[ {{f(x)}} \right]$可近似展开为

(33)$ \begin{eqnarray}\label{e4.13} {\left( \cal {I - P} \right)^{ - 1}}{\cal T}\left[ {{f(x)}} \right]\simeq\left( {{\cal I} + \sum\limits_{p = 1}^P {{ {\cal P}^p}} } \right){\cal T}\left[ {{f(x)}} \right] \end{eqnarray}$

假设所求函数$f(x)$在$ \left[ {0,L} \right]$区间内是连续的, 并注意到$f(L)=0$, 由Weierstrass定理可将$f(x)$近似展开为

(34)$ \begin{eqnarray}\label{34} f\left( x \right) \simeq\sum\limits_{n = 1}^{ N } {{a_n}{{\left( {L - x} \right)}^n}} \end{eqnarray}$

考虑式(32)在$[0,L]$区间内任意$M,(M\gg N)$个离散点上均成立. 那么求解方程式(32)等价于求解如下的矛盾方程组

(35)$ \begin{eqnarray}\label{e4.14} {{{\tilde{A}}}_{M \times N}}{{a}_{N \times 1}} = {{\bf{1}}_{M \times 1}} \end{eqnarray}$

其中

$ \begin{eqnarray*} &&{{\tilde A}_{ji}} = \left( {L - x_j} \right)^i+ \alpha \int_0^{x_j} \left( {x_j - s} \right)^2\left( {{\cal I} + \sum\limits_{p = 1}^P {{ {\cal P}^p}} } \right)\cdot \\&&\qquad{\cal T}\left[ \left( L - x_j \right)^i \right]{\rm d}s \end{eqnarray*} $

采用最小二乘方法可获得方程组(35)的最优解为

(36)$ \begin{eqnarray}\label{36} {{{a}}_{N \times 1}} = \left({{{\tilde{A}}}^{\rm T}{{\tilde{A}}}}\right)^{-1}{{\tilde{A}}}^{\rm T}\cdot {{\bf{1}}_{M \times 1}} \end{eqnarray}$

为寻找临界参数$\alpha_{\rm c}$, 定义关于$\alpha$ 的误差函数

(37)$ \begin{eqnarray}\label{37} {\cal E}(\alpha) = \sum\limits_{m = 1}^M {\frac{{\left| {1 - {{{{\tilde{A}}}}_{M \times N}}{{{a}}_{N \times 1}}}(m) \right|}}{M}} \end{eqnarray}$

改变$\alpha$使得误差$|{\cal E}|\leqslant 1.0{\rm e}^{-4}$可得临界参数$\alpha_{\rm c}$及其对应的临界气流速度.

4 理论结果及对比验证

在下面计算中选取无量纲参数$\xi = x/L,{r_H} =H/L$. 为避免间隙$H$过小而导致较强的粘性作用, 本文仅考虑$r_H>0.05$的情况. 基于图2给出分析及计算流程进行算例分析. 首先考察${r_H} \to \infty$, 即无壁面的情况. 此时, $\left\| {\cal P}\right\|\to 0,{\left( \cal {I - P} \right)^{ - 1}}\to {\cal I}$. 图3(a)给出了此工况下, $M=200$时, 误差函数${\cal E}$随截断阶数$N$的变化关系. 由该图可知随着$N$的增加误差会最终会在$\alpha_{\rm c}=1.84$处满足设置的条件(进入绿色阴影区域), 此参数即为系统的临界参数. 下面考察式(33)和式(34)中截断参数$P$和$N$对本文解收敛性的影响. 图3(b)给出了不同${r_H}$时截断阶数$P$对结果收敛性的影响. 由图3可知, 选择$P=N=7,M=200$可以保证本文计算结果的收敛性. 图4给出了不同$r_H$对应的临界参数$\alpha_{\rm c}$. 与预期一致, $\alpha_{\rm c}$随$r_H$的增大而增大并最终稳定在$\alpha_{\rm c}\simeq1.84$.

图2

图2 本文理论分析和计算的流程

Fig. 2 The flowchart of the present theoretical analysis

图3

图3

Fig. 3

图5给出了不同$r_H$下壁板的失稳模态$\psi(\xi)$. 从图可知, 尽管不同$r_H$下的临界动压不同, 但失稳模式都以倒置悬臂梁一阶模态$\varphi_1$为主导. 考察失稳模态下壁板表面气动压力的分布, 定义压强系数

图4

图4 本文理论解与其他理论及数值结果的对比分析

Fig. 4 Comparison of the present theoretical results with other existing theory and the numerical method

图5

图5 系统的失稳模态(上)及压强系数(下)

Fig. 5 The instability model (top) and the pressure coefficient (bottom) of the plate

(38)$ \begin{eqnarray}\label{38} {C_{\rm p}}\left( \xi \right) = - \frac{4}{\beta }\left( {{\cal I} + \sum\limits_{p = 1}^P {{{{\cal P}}^p}} } \right){\cal T}\left[ {\frac{{\partial \psi}}{{\partial \xi}}} \right] \end{eqnarray}$

图5给出了不同$r_H$对应的压力系数(由于前缘压差为无穷大, 图中已做截断处理). 由图可知, 壁板前缘附近为正压区而尾缘附近为负压区; 随$r_H$的减小正压区扩大且压力值增加, 而负压区缩小且压力值减小. 壁板上合力增大且力矩中心前移, 对应的临界失稳动压也就越低.

为验证本文的理论结果, 考虑无壁面情况, 此时满足Kutta-Joukowski条件的壁板的压强解 (Theodorsen解^[28])为

(39)$ \begin{eqnarray}\label{39} \Delta p(x)= \frac{{2{\rho _\infty }U_\infty ^2}}{{\pi L}}\frac{{f\left( \theta \right) - f\left( 0 \right)}}{{\sin \theta }} \end{eqnarray}$

其中

$\begin{eqnarray*} &&f\left( \theta \right) = \int_0^\pi {\frac{{\partial w}}{{\partial \zeta }}\frac{{{{\sin }^2}\varphi }}{{\cos \varphi - \cos \theta }}{\rm d}\varphi }\\ && x = \frac{L}{2}\left( {1 + \cos \theta } \right),\ \zeta = \frac{L}{2}\left( {1 + \cos \varphi } \right) \end{eqnarray*}$

取倒置悬臂梁模态函数$\varphi_k,(k=1,2,...,K)$, 应用伽辽金方法求解方程(1), 通过特征值分析获得系统临界参数. 如图3(b)中虚线所示, 选取前三阶模态便可得收敛结果$\alpha_{\rm c}\simeq1.8464$. 本文理论结果与经典Theodorsen解得到的结果吻合很好.

为了进一步考察本文理论方法对$r_H$的适用性, 采用如图1(b)所示的离散面元模型进行数值求解. 壁板上的面元分布由线性分布的连续涡来表示以计算满足边界条件的流体势函数, 在质量点处采用有限差分方法求解壁板结构运动方程, 详细的离散和计算过程请参见文献^[12]. 本文采用该离散模型计算获得了不同$r_H$时的临界参数, 结果如图4所示. 由图可知本文理论与数值结果的变化趋势一致且吻合较好(两者相对误差保持在5%之内, 图中绿色区域内). 上述对比结果均表明了本文气动力理论及失稳分析方法的有效性.

5 风洞实验验证

采用风洞实验研究壁板失稳的临界速度并验证本文理论的正确性. 实验中使用直径1.2 m的开口式风洞, 其具有5 $\sim$ 40 m/s的风速调节范围和低于0.3%的湍流度. 实验中壁板模型采用长宽比为2/3 (20 cm $\times$ 30 cm)的镀锌铁板($\rho_{\rm p}=7.85$ g/cm$^3$, $E_{\rm p}=210$ GPa). 如图6所示, 壁板竖直安装, 并用两块厚钢板将其夹持固定在刚性支架上以保证固支约束并进行实验, 夹持段壁板的长度约为3 cm; 壁板一侧竖直放置厚木板以模拟壁面对气流的约束限制. 靠近壁板固定端的根部位置贴有应变片, 实验开始前采用敲击法测试了壁板在无风状态下的自振频率(7.9 Hz), 并与理论结果(8.2 Hz)进行了对比, 检验了固支约束实现及测试设备的可靠性. 实验中缓慢增加风速至目标风速, 稳定至少一分钟后进行数据采集, 以保证实验数据的可靠性.

静态失稳不能像颤振这类动态失稳一样可以依据信号等幅周期变化的特征而直接判定. 本文借鉴材料力学压杆失稳原理设计了一种等效拉力测试方法, 测试原理如图7(a)所示. 理论上给定壁板初始变形量为失稳模态的任意小倍数, 当气流速度小于临界值时, 壁板需要外部作用(实验中依靠张紧的细绳提供拉力)才能维持该状态; 而当流速达到临界值时, 则可不依靠任何外部作用. 因此可由是否需要细绳提供拉力而维持给定的初始变形来等效判定系统是否达到了临界状态. 本文的理论分析表明壁板将呈现一阶悬臂梁模态失稳模式, 而该模态形式的初始壁板静变形在实验中却不易精确给出. 依据文献^[26]给出的悬臂梁一阶模态与其端部受集中载荷产生的静变形形态相类似这一结论, 本文实验采用在壁板自由端施加集中力(张紧细绳使其具有一定预拉力)而实现壁板的初始变形.

图6

图6 风洞中模型安装图和不同风速下细绳的状态图

Fig. 6 Pictures of the setup of the model in the wind tunnel, and of the string states with different flow velocities

如图6及图7(a)所示, 悬臂板紧靠前缘正中心位置连接有细棉绳, 细绳的另一端连接在拉力传感器上(量程5 N, 灵敏度0.01 N). 调整细绳的长度可使壁板前缘产生不同的初始挠度$\Delta$, 其在传感器上表现为不同的预拉力值. 当流速小于临界值时, 细绳会处于张紧状态并为壁板提供拉力而维持其初始变形, 如图6(a)所示; 而当流速接近于临界值时, 细绳则会处于松弛状态, 如图6(b)所示. 注意到细绳仅受气流作用也会导致拉力($F_{\rm w}$)的存在, 传感器的实测拉力($F$)则包含($F_{\rm w}$)及壁板对绳子的拉力($F_{\rm p}$). 图 7(b)给出了风速12 m/s时时间1 s内两种拉力的实测值. 测试拉力$F_{\rm w}$时, 将细绳保持自然状态, 两端分别固定在刚性支架及夹持端并单独置于风洞中进行测试. 从图7(b)可知, 虽然两种拉力均表现出明显的波动性, 但$F_{\rm w}$的变化幅值较小($\pm 0.005$ N之内). 若考虑两种拉力之间的弱关联性则可由下式

图7

图7

Fig. 7

(40)$ \begin{eqnarray}\label{e6.1} F^{\rm m}\triangleq \min(F)\leqslant {\rm mean}(F_{\rm w})\triangleq F_{\rm w}^{\rm m} \end{eqnarray}$

作为临界状态的近似判定条件.

理论上, 对于任意的自由端部挠度$\varDelta$, 实验中均应测试得到相同的临界风速. 然而实验中发现, 较小的值会因较差的抗干扰性而导致测量精度不易满足, 而较大的$\varDelta$则会导致壁板产生几何大变形非线性^[26]. 因此, 实验中首先以无壁面的情形进行多组重复测试, 通过对比已有理论及计算结果来确定最佳的$\varDelta$给定范围. 图8(a)给出了不同$\varDelta$值下$F$及$F_{\rm w}$在不同风速下的测试值. 依据式(40)可判定当拉力$F^{\rm m}$曲线处于图8(a)的绿色阴影区域时, 系统处于失稳状态. 依据图8(a)得到临界速度$U_{\rm {cr}}$与$\varDelta$之间的变化关系如图8(b)所示. 通过与理论值(39)及数值结果(图4)对比可知, 当$\varDelta$在0.3 $\sim $ 0.5 cm之间取值时, 实验结果与理论及数值结果吻合较好(实验值与理论值(39)相差2%之内, 图中绿色区域内), 因此本文实验选取该区间的值作为拉出挠度值完成了不同$r_H$下的实验, 结果如图9所示. 图9(a)中无风时细绳中的初始拉力并不完全相等, 这是由于为了保证数据的多样性而在0.3 $\sim $ 0.5 cm内给定不同的$\varDelta$值而导致的.

图10给出了本文理论与试验结果的对比, 由图可知两者吻合很好, 各风速下的实验结果与理论结果之间的相对误差均保持在2%之内(绿色区域内), 这充分表明了本文的理论计算及风洞试验的有效性和准确性.

图8

图8

Fig. 8

图9

图9

Fig. 9

图10

图10 本文理论结果与试验结果及已有理论结果的对比

Fig. 10 Comparison of the present theoretical results with the experiment and the other existing theory

6 结语

本文考虑壁面效应对亚音速气流中倒置悬臂壁板的静态失稳问题进行了理论及实验研究. 文中采用镜像函数法描述壁面约束条件, 基于算子理论对壁板上的气动力进行了理论分析, 获得了以Possio积分方程为表征的气动力表达式, 壁面效应则表征为一包含$\cal{T}$, $\cal{Q}$及$\cal{H}$算子的复合算子; 壁板的失稳方程转化为了定区间上的函数逼近问题, 并利用Wererstrass定理及最小二乘方法得到了最优逼近函数确定临界动压, 无需进行特征值计算; 给出了壁面效应对失稳的定量影响并对壁面影响失稳的原理进行了探讨; 设计了静态失稳测试方法, 其相比于动态实验对风洞品质要求更低且鲁棒性较高; 风洞实验分析结果与本文理论结果吻合较好, 验证了本文理论方法的正确性.

本文中以Possio积分方程表征的气动力物理含义清晰, 依据微分算子理论将失稳问题转化为了定区间上的函数逼近问题, 为该类问题的求解提供了新的思路. 值得指出的是, 本文方法在壁板颤振问题中的推广及应用正是下一步计划开展的工作.

附录:算子${\cal QH}$的交换性

由本文算子的定义可知

(A1)$ \begin{eqnarray} \left. {\begin{array}{*{20}{l}} {H\left[ {Q\left[ {f(x)} \right]} \right] = \dfrac{1}{{{\pi ^2}c}}\int_{ - \infty }^\infty {f\left( \tau \right){g_1}\left( {x,\tau } \right)} {\rm d}\tau }\\ {Q\left[ {H\left[ {f(x)} \right]} \right] = \dfrac{1}{{{\pi ^2}c}}\int_{ - \infty }^\infty {f\left( \tau \right){g_2}\left( {x,\tau } \right)} {\rm d}\tau } \end{array}} \right\} \end{eqnarray} $

其中

(A2)$ \begin{eqnarray} \left. {\begin{array}{*{20}{l}} {g_1}\left( {x,\tau } \right) = \int_{ - \infty }^{ + \infty } {\dfrac{1}{{x - t}}\dfrac{1}{{1 + {{\left( {\dfrac{{t - \tau }}{c}} \right)}^2}}}{\rm d}t}\\ {g_2}\left( {x,\tau } \right) =\int_{ - \infty }^{ + \infty } {\dfrac{1}{{t - \tau }}\dfrac{1}{{1 + {{\left( {\dfrac{{x - t}}{c}} \right)}^2}}}{\rm d}t} \end{array}} \right\} \end{eqnarray} $

分别令$t = \tau + p$及$t=x+q$, 则上式变作

(A3)$ \begin{eqnarray} \left. {\begin{array}{*{20}{l}} {g_1}\left( {x,\tau } \right) = \int_{ - \infty }^{ + \infty } {\dfrac{1}{{\left( {x - \tau } \right) - p}}\dfrac{1}{{1 + {{\left( {\dfrac{p}{c}} \right)}^2}}}{\rm d}p}\\ {g_2}\left( {x,\tau } \right) = \int_{ - \infty }^{ + \infty } {\dfrac{1}{{\left( {x - \tau } \right) + q}}\dfrac{1}{{1 + {{\left( {\dfrac{q}{c}} \right)}^2}}}{\rm d}q} \end{array}} \right\} \end{eqnarray} $

上式中的积分为瑕积分, 注意到$c>0$, 利用围道积分进行计算后可得

(A4)$ \begin{eqnarray} {g_1}\left( {x,\tau } \right) = {g_2}\left( {x,\tau } \right) = \dfrac{{\pi c\left( {x - \tau } \right)}}{{{{\left( {x - \tau } \right)}^2} + {c^2}}} \end{eqnarray} $

即得

(A5)$ \begin{eqnarray} {\cal H}\left[ {{\cal Q}\left[ {f(x)} \right]} \right] = {\cal Q}\left[ {{\cal H}\left[ {f(x)} \right]} \right] \end{eqnarray} $

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