技術(shù)特征：

1.一種基于譜有限元的矩形板振動(dòng)模態(tài)計(jì)算方法，包括以下步驟：

步驟一：基于能量函數(shù)變分原理得到板振動(dòng)的邊界值問題模型，再基于變量分離方法得到兩個(gè)方向的自由振動(dòng)方程；

步驟二：確定矩形板兩個(gè)方向的邊界條件，在預(yù)設(shè)定的模態(tài)階數(shù)下采用不同方向相互迭代來計(jì)算振動(dòng)頻率和振型，直至兩個(gè)方向計(jì)算得到的振動(dòng)頻率誤差在某個(gè)指定范圍時(shí)停止；

步驟三：利用最終得到的兩個(gè)方向的振動(dòng)模態(tài)進(jìn)行疊加得到矩形板振動(dòng)模態(tài)。

2.根據(jù)權(quán)利要求1所述的基于譜有限元的矩形板振動(dòng)模態(tài)計(jì)算方法，其特征在于：所述步驟一具體步驟為

建立矩形板直角坐標(biāo)系(x,y,z)，分別定義x,y,z三個(gè)方向的位移多元函數(shù)u(x,y,t)，v(x,y,t)，w(x,y,t)如下:

u(x,y,t)＝zθ_y(x,y,t)；

v(x,y,t)＝-zθ_x(x,y,t)；

w(x,y,t)＝w₀(x,y,t)；

其中w₀(x,y,t)表示板(x,y)位置在z方向t時(shí)刻的振動(dòng)幅值，θ_x表示板(x,y)位置在t時(shí)刻繞x軸的旋轉(zhuǎn)角，θ_y表示板(x,y)位置在t時(shí)刻繞y軸的旋轉(zhuǎn)角；

定義能量函數(shù)：

Π＝U+T，

$\begin{array}{c}U=\frac{1}{2}{\∫}_0^{a}dx{\∫}_0^{b}dy\[D\left({\left(\frac{\∂{\mathrm{\θ}}_y}{\∂x}\right)}^2-v\frac{\∂{\mathrm{\θ}}_y}{\∂x}\frac{\∂{\mathrm{\θ}}_x}{\∂y}\right)+D\left({\left(\frac{\∂{\mathrm{\θ}}_x}{\∂y}\right)}^2-v\frac{\∂{\mathrm{\θ}}_x}{\∂y}\frac{\∂{\mathrm{\θ}}_y}{\∂x}\right)\\ +\frac{{\mathrm{Gh}}^3}{12}{\left(\frac{\∂{\mathrm{\θ}}_y}{\∂y}-\frac{\∂{\mathrm{\θ}}_x}{\∂x}\right)}^2+Gh{\left({\mathrm{\θ}}_y+\frac{\∂w}{\∂x}\right)}^2+Gh{\left(\frac{\∂w}{\∂y}-{\mathrm{\θ}}_x\right)}^2\]\end{array}$

$T=\frac{1}{2}{\∫}_A(\mathrm{\ρ}h{\stackrel{\·}{w}}^2+\mathrm{\ρ}I{{\stackrel{\·}{\mathrm{\θ}}}_x}^2+\mathrm{\ρ}I{{\stackrel{\·}{\mathrm{\θ}}}_y}^2)dA$

其中表示w(x,y,t)的對(duì)時(shí)間的導(dǎo)數(shù)表示θ_x(x,y,t)對(duì)時(shí)間的導(dǎo)數(shù)表示θ_y(x,y,t)對(duì)時(shí)間的導(dǎo)數(shù)常數(shù)常數(shù)

參數(shù)定義為：E：板抗彎剛度，a：板長(zhǎng)度，b：板寬度，h：板厚度，v：泊松比，ρ:板材料密度；

對(duì)w(x,y,t)，θ_x(x,y,t)和θ_y(x,y,t)進(jìn)行時(shí)空變量分離：

w(x,y,t)＝W(x,y)e^jωt,

${\mathrm{\θ}}_x(x,y,t)={\stackrel{\‾}{\mathrm{\θ}}}_x(x,y)e^{j\mathrm{\ω}t},$

${\mathrm{\θ}}_y(x,y,t)={\stackrel{\‾}{\mathrm{\θ}}}_y(x,y)e^{j\mathrm{\ω}t},$

其中二元變量W(x,y)表示板(x,y)位置的振動(dòng)幅值，二元變量表示板(x,y)位置繞x軸的旋轉(zhuǎn)角，二元變量表示板(x,y)位置繞y軸的旋轉(zhuǎn)角；ω:圓頻率(rad/s)；

求Π的變分且令變分等于0，即

$\begin{array}{c}\mathrm{\δ}\mathrm{\Π}=0\\ ={\∫}_0^{a}dx{\∫}_0^{b}dy\[-{\mathrm{\ω}}^2\mathrm{\ρ}hW-Gh\left(\frac{{\∂}^{2}W}{\∂x^2}+\frac{{\∂}^{2}W}{\∂y^2}\right)+Gh\frac{\∂{\stackrel{\‾}{\mathrm{\θ}}}_x}{\∂y}-Gh\frac{\∂{\stackrel{\‾}{\mathrm{\θ}}}_y}{\∂x}\]\mathrm{\δ}W\\ +{\∫}_0^{b}dy\[Gh\left({\stackrel{\‾}{\mathrm{\θ}}}_y+\frac{\∂W}{\∂x}\right)\mathrm{\δ}W{|}_0^a\]\\ -{\∫}_0^{b}dy\[\frac{D\left(1-v\right)}{2}\left(\frac{\∂{\stackrel{\‾}{\mathrm{\θ}}}_y}{\∂y}-\frac{\∂{\stackrel{\‾}{\mathrm{\θ}}}_x}{\∂x}\right)\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_x{|}_0^a\]\\ +{\∫}_0^{b}dy\[D\left(\frac{\∂{\stackrel{\‾}{\mathrm{\θ}}}_y}{\∂x}-v\frac{\∂{\stackrel{\‾}{\mathrm{\θ}}}_x}{\∂y}\right)\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_y{|}_0^a\]\end{array}$

$\begin{array}{c}+{\∫}_0^{a}dx{\∫}_0^{b}dy\[-{\mathrm{\ω}}^2\mathrm{\ρ}I{\stackrel{\‾}{\mathrm{\θ}}}_x-\frac{D\left(1-v\right)}{2}\frac{{\∂}^2{\stackrel{\‾}{\mathrm{\θ}}}_x}{\∂x^2}-D\frac{{\∂}^2{\stackrel{\‾}{\mathrm{\θ}}}_x}{\∂y^2}+\frac{D\left(1+v\right)}{2}\frac{{\∂}^2{\stackrel{\‾}{\mathrm{\θ}}}_y}{\∂x\∂y}-Gh\left(\frac{\∂W}{\∂y}-{\stackrel{\‾}{\mathrm{\θ}}}_x\right)\]\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_x\\ +{\∫}_0^{a}dx{\∫}_0^{b}dy\[-{\mathrm{\ω}}^2\mathrm{\ρ}I{\stackrel{\‾}{\mathrm{\θ}}}_y-D\frac{{\∂}^2{\stackrel{\‾}{\mathrm{\θ}}}_y}{\∂x^2}-\frac{D\left(1-v\right)}{2}\frac{{\∂}^2{\stackrel{\‾}{\mathrm{\θ}}}_y}{\∂y^2}+\frac{D\left(1+v\right)}{2}\frac{{\∂}^2{\stackrel{\‾}{\mathrm{\θ}}}_x}{\∂x\∂y}+Gh\left(\frac{\∂W}{\∂x}+{\stackrel{\‾}{\mathrm{\θ}}}_y\right)\]\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_y\\ +{\∫}_0^{a}dx\[Gh\left(\frac{\∂W}{\∂x}-{\stackrel{\‾}{\mathrm{\θ}}}_x\right)\mathrm{\δ}W{|}_0^b\]\\ +{\∫}_0^{a}dx\[D\left(\frac{\∂{\stackrel{\‾}{\mathrm{\θ}}}_x}{\∂y}-v\frac{\∂{\stackrel{\‾}{\mathrm{\θ}}}_y}{\∂x}\right)\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_x{|}_0^b\]\\ +{\∫}_0^{a}dx\[\left(\frac{D\left(1-v\right)}{2}\left(\frac{\∂{\stackrel{\‾}{\mathrm{\θ}}}_y}{\∂y}-\frac{\∂{\stackrel{\‾}{\mathrm{\θ}}}_x}{\∂x}\right)\right)\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_y{|}_0^b\]\end{array}$

對(duì)變量W(x,y),進(jìn)行二元變量分離有：

W(x,y)＝W_x(x)W_y(y)

${\stackrel{\‾}{\mathrm{\θ}}}_x(x,y)={\stackrel{\‾}{\mathrm{\θ}}}_{xx}\left(x\right){\stackrel{\‾}{\mathrm{\θ}}}_{xy}\left(y\right)$

${\stackrel{\‾}{\mathrm{\θ}}}_y(x,y)={\stackrel{\‾}{\mathrm{\θ}}}_{yx}\left(x\right){\stackrel{\‾}{\mathrm{\θ}}}_{yy}\left(y\right)$

其中W_x(x)，W_y(y)分別表示W(wǎng)(x,y)在x,y方向的分離模態(tài)函數(shù)，分別表示在x,y方向的分離模態(tài)函數(shù)，分別表示在x,y方向的分離模態(tài)函數(shù)

假定已知y向振動(dòng)模態(tài)函數(shù)W_y,利用如下變分公式：

δW(x,y)＝W_yδW_x

$\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_x(x,y)={\stackrel{\‾}{\mathrm{\θ}}}_{xy}\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_{xx}$

$\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_y(x,y)={\stackrel{\‾}{\mathrm{\θ}}}_{yy}\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_{yx}$

得到關(guān)于x向三個(gè)變量的常微分方程組Ⅰ：

$\left\{\begin{array}{c}\frac{d^2{w}_x}{{\mathrm{dx}}^2}+a_1\frac{d{\stackrel{\‾}{\mathrm{\θ}}}_{yx}}{dx}+a_2{\stackrel{\‾}{\mathrm{\θ}}}_{xx}+a_3{w}_x=0\\ \frac{d^2{\stackrel{\‾}{\mathrm{\θ}}}_{yx}}{{\mathrm{dx}}^2}+c_1\frac{d{\stackrel{\‾}{\mathrm{\θ}}}_{xx}}{dx}+c_2\frac{{\mathrm{dw}}_x}{dx}+c_3{\stackrel{\‾}{\mathrm{\θ}}}_{yx}=0\\ \frac{d^2{\stackrel{\‾}{\mathrm{\θ}}}_{xx}}{{\mathrm{dx}}^2}+b_1\frac{d{\stackrel{\‾}{\mathrm{\θ}}}_{yx}}{dx}+b_2{w}_x+b_3{\stackrel{\‾}{\mathrm{\θ}}}_{xx}=0\end{array}\right.---\left(1\right)$

其中

$a_0={\∫}_0^b{W}_y^{2}dy;a_1=\frac{{\∫}_0^b{\stackrel{\‾}{\mathrm{\θ}}}_{yy}{W}_{y}dy}{a_0};a_2=\frac{-{\∫}_0^b\frac{d{\stackrel{\‾}{\mathrm{\θ}}}_{xy}}{dy}{W}_{y}dy}{a_0};a_3=\frac{{\∫}_0^b\frac{d^2{W}_y}{{\mathrm{dy}}^2}{W}_{y}dy}{{\∫}_0^b{W}_y^{2}dy}+\frac{{\mathrm{\ω}}^2\mathrm{\ρ}h}{Gh}$

$b_0=\frac{D(1-v)}{2}{\∫}_0^b{\stackrel{\‾}{\mathrm{\θ}}}_{xy}^{2}dy;b_1=\frac{-\frac{D\left(1+v\right)}{2}{\∫}_0^b\frac{d{\stackrel{\‾}{\mathrm{\θ}}}_{yy}}{dy}{\stackrel{\‾}{\mathrm{\θ}}}_{xy}dy}{b_0};b_2=\frac{Gh{\∫}_0^b\frac{{\mathrm{dW}}_y}{dy}{\stackrel{\‾}{\mathrm{\θ}}}_{xy}dy}{b_0}$

$b_3=\frac{{\mathrm{\ω}}^2\mathrm{\ρ}I{\∫}_0^b{\stackrel{\‾}{\mathrm{\θ}}}_{xy}^{2}dy+D{\∫}_0^b\frac{d^2{\stackrel{\‾}{\mathrm{\θ}}}_{xy}}{{\mathrm{dy}}^2}{\stackrel{\‾}{\mathrm{\θ}}}_{xy}dy-Gh{\∫}_0^b{\stackrel{\‾}{\mathrm{\θ}}}_{xy}^{2}dy}{b_0}$

$c_0=D{\∫}_0^b{\stackrel{\‾}{\mathrm{\θ}}}_{yy}^{2}dy;c_1=\frac{-\frac{D(1+v)}{2}{\∫}_0^b\frac{d{\stackrel{\‾}{\mathrm{\θ}}}_{xy}}{dy}dy}{c_0};c_2=\frac{-Gh{\∫}_0^b{W}_y{\stackrel{\‾}{\mathrm{\θ}}}_{yy}dy}{c_0};$

$c_3=\frac{{\mathrm{\ω}}^2\mathrm{\ρ}I{\∫}_0^b{\stackrel{\‾}{\mathrm{\θ}}}_{yy}^{2}dy+\frac{D(1-v)}{2}{\∫}_0^b\frac{d^2{\stackrel{\‾}{\mathrm{\θ}}}_{yy}}{{\mathrm{dy}}^2}{\stackrel{\‾}{\mathrm{\θ}}}_{yy}dy-Gh{\∫}_0^b{\stackrel{\‾}{\mathrm{\θ}}}_{yy}^{2}dy}{c_0}$

假定已知x向振動(dòng)模態(tài)函數(shù)W_x,則利用如下變分公式：

δW(x,y)＝W_xδW_y

$\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_x(x,y)={\stackrel{\‾}{\mathrm{\θ}}}_{xx}\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_{xy}$

$\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_y(x,y)={\stackrel{\‾}{\mathrm{\θ}}}_{yx}\mathrm{\δ}{\stackrel{\‾}{\mathrm{\θ}}}_{yy}$

得到關(guān)于y向三個(gè)變量的常微分方程組Ⅱ：

$\left\{\begin{array}{c}\frac{d^2{w}_y}{{\mathrm{dy}}^2}+e_1\frac{d{\stackrel{\‾}{\mathrm{\θ}}}_{xy}}{dy}+e_2{\stackrel{\‾}{\mathrm{\θ}}}_{yy}+e_3{w}_y=0\\ \frac{d^2{\stackrel{\‾}{\mathrm{\θ}}}_{xy}}{{\mathrm{dy}}^2}+f_1\frac{d{\stackrel{\‾}{\mathrm{\θ}}}_{yy}}{dy}+f_2\frac{{\mathrm{dw}}_y}{dy}+f_3{\stackrel{\‾}{\mathrm{\θ}}}_{xy}=0\\ \frac{d^2{\stackrel{\‾}{\mathrm{\θ}}}_{yy}}{{\mathrm{dy}}^2}+g_1\frac{d{\stackrel{\‾}{\mathrm{\θ}}}_{xy}}{dy}+g_2{w}_y+g_3{\stackrel{\‾}{\mathrm{\θ}}}_{yy}=0\end{array}\right.---\left(2\right)$

其中

$e_0={\∫}_0^a{W}_x^{2}dx;e_1=\frac{-{\∫}_0^a{\stackrel{\‾}{\mathrm{\θ}}}_{xx}{W}_{x}d}{e_0};e_2=\frac{{\∫}_0^a\frac{d{\stackrel{\‾}{\mathrm{\θ}}}_{yx}}{dx}{W}_{x}dx}{e_0};e_3=\frac{{\∫}_0^a\frac{d^2{W}_x}{{\mathrm{dx}}^2}{W}_{x}dx}{e_0}+\frac{{\mathrm{\ω}}^2\mathrm{\ρ}h}{Gh}$

$f_0=D{\∫}_0^a{\stackrel{\‾}{\mathrm{\θ}}}_{xx}^{2}dx;f_1=\frac{-\frac{D\left(1+v\right)}{2}{\∫}_0^a\frac{d{\stackrel{\‾}{\mathrm{\θ}}}_{yx}}{dx}{\stackrel{\‾}{\mathrm{\θ}}}_{xx}dx}{f_0};f_2=\frac{Gh{\∫}_0^a{W}_x{\stackrel{\‾}{\mathrm{\θ}}}_{xx}dx}{f_0}$

$f_3=\frac{{\mathrm{\ω}}^2\mathrm{\ρ}I{\∫}_0^a{\stackrel{\‾}{\mathrm{\θ}}}_{xx}^{2}dx+\frac{D(1-v)}{2}{\∫}_0^a\frac{d^2{\stackrel{\‾}{\mathrm{\θ}}}_{xx}}{{\mathrm{dx}}^2}{\stackrel{\‾}{\mathrm{\θ}}}_{xx}dx-Gh{\∫}_0^a{\stackrel{\‾}{\mathrm{\θ}}}_{xx}^{2}dx}{f_0}$

$g_0=\frac{D(1-v)}{2}{\∫}_0^a{\stackrel{\‾}{\mathrm{\θ}}}_{yx}^{2}dx;g_1=\frac{-\frac{D(1+v)}{2}{\∫}_0^a\frac{d{\stackrel{\‾}{\mathrm{\θ}}}_{xx}}{dx}{\stackrel{\‾}{\mathrm{\θ}}}_{yx}dx}{g_0};g_2=\frac{-Gh{\∫}_0^a\frac{{\mathrm{dW}}_x}{dx}{\stackrel{\‾}{\mathrm{\θ}}}_{yx}dx}{g_0};$

$g_3=\frac{{\mathrm{\ω}}^2\mathrm{\ρ}I{\∫}_0^a{\stackrel{\‾}{\mathrm{\θ}}}_{yx}^{2}dx+D{\∫}_0^b\frac{d^2{\stackrel{\‾}{\mathrm{\θ}}}_{yx}}{{\mathrm{dx}}^2}{\stackrel{\‾}{\mathrm{\θ}}}_{yx}dx-Gh{\∫}_0^a{\stackrel{\‾}{\mathrm{\θ}}}_{yx}^{2}dx}{g_0}.$

3.根據(jù)權(quán)利要求1所述的基于譜有限元的矩形板振動(dòng)模態(tài)計(jì)算方法，其特征在于：所述步驟二中，

針對(duì)x方向的邊界條件定義為三種：

簡(jiǎn)支邊界：W_x＝0,M_xx＝0；

夾緊邊界：W_x＝0,

自由邊界：V_xz＝0,M_xx＝0,M_xy＝0；

其中力矩

經(jīng)過計(jì)算后得到如下的計(jì)算用邊界條件：

簡(jiǎn)支邊界：W_x＝0,

夾緊邊界：W_x＝0,

自由邊界：

針對(duì)y方向的邊界條件定義為三種：

簡(jiǎn)支邊界：W_y＝0,M_yy＝0；

夾緊邊界：W_y＝0,

自由邊界：V_yz＝0,M_yy＝0,M_xy＝0；

其中力矩

經(jīng)過計(jì)算后得到如下的計(jì)算用邊界條件：

簡(jiǎn)支邊界：W_y＝0,

夾緊邊界：W_y＝0,

自由邊界：

4.根據(jù)權(quán)利要求3所述的基于譜有限元的矩形板振動(dòng)模態(tài)計(jì)算方法，其特征在于：所述步驟二中，采用不同方向相互迭代計(jì)算振動(dòng)頻率和振型函數(shù)的具體步驟如下

2-1)設(shè)定y向?yàn)榈趍階模態(tài)，根據(jù)y向邊界條件選擇相應(yīng)自由振動(dòng)梁的模態(tài)函數(shù)作為y向的模態(tài)函數(shù)；

2-2)利用給定的y向模態(tài)函數(shù)計(jì)算得到x向的常微分方程組Ⅰ，根據(jù)常微分方程特征值的情況確定x向模態(tài)函數(shù)的形式，根據(jù)邊界條件建立以模態(tài)函數(shù)未知系數(shù)為變量的線性方程組，對(duì)線性方程組6×6系數(shù)矩陣行列式值為0的非線性方程進(jìn)行優(yōu)化求解，得到頻率值w_x,再利用w_x代入線性方程組進(jìn)行求解得到模態(tài)函數(shù)系數(shù)，從而得到x向的模態(tài)函數(shù)；

2-3)利用步驟2-2中計(jì)算得到的x向的模態(tài)函數(shù)，計(jì)算得到y(tǒng)向的常微分方程組Ⅱ，根據(jù)常微分方程特征值的情況確定y向模態(tài)函數(shù)的形式，根據(jù)邊界條件建立以模態(tài)函數(shù)未知系數(shù)為變量的線性方程組，利用系數(shù)矩陣的行列式為0進(jìn)行優(yōu)化求解得到頻率值w_y，再利用w_y代入線性方程組進(jìn)行求解得到模態(tài)函數(shù)系數(shù)，從而得到y(tǒng)向的模態(tài)函數(shù)；

2-4)比較步驟2-2和2-3得到的頻率值的大小，如果滿足|w_x-w_y|≤ε，則退出迭代，ε為誤差值，取值為0.0001；如果不滿足條件，則將步驟2-3中得到的y向的模態(tài)函數(shù)作為給定的步驟2-2中模態(tài)函數(shù)進(jìn)行迭代計(jì)算。

5.根據(jù)權(quán)利要求4所述的基于譜有限元的矩形板振動(dòng)模態(tài)計(jì)算方法，其特征在于：所述步驟2-2中，基于已知y向模態(tài)函數(shù)計(jì)算得到頻率值w_x和x向的模態(tài)函數(shù)方法如下：

根據(jù)常微分方程組Ⅰ，令表示微分算子d/dx，則有如下的式子：

展開式(3)中的行列式得到如下的方程:

其中ψ＝w_x,

d₁＝a₃+b₃+c₃-b₁c₁-a₁c₂；

d₂＝a₃b₃+a₃c₃+b₃c₃+a₂b₁c₂+a₁b₂c₁-a₃b₁c₁-a₂b₂-a₁b₃c₂；

d₃＝a₃b₃c₃-a₂b₂c₃；

將試驗(yàn)解ψ＝e^λ代入方程(4)，由于表示微分算子，可產(chǎn)生如下的方程：

λ⁶+d₁λ⁴+d₂λ²+d₃＝0 (5)

令μ＝λ²，則

μ³+d₁μ²+d₂μ+d₃＝0 (6)

令判別式Δ＝18d₁d₂d₃-4d₁³d₃+d₁²d₂²-4d₂³-27d₃²，

定義w_x(x)與函數(shù)表達(dá)式中的系數(shù)關(guān)系數(shù)值δ₁,δ₂,δ₃，與函數(shù)表達(dá)式中的系數(shù)關(guān)系數(shù)值γ₁,γ₂,γ₃如下

${\mathrm{\δ}}_i=\frac{Gk(12Ghk-h^3{\mathrm{\ρ\ω}}^2-6D(-1+v\left)\right({\mathrm{\α}}^2-r_i^2\left)\right)}{6r_i(2G^2{\mathrm{hk}}^2-{\mathrm{DGk\α}}^2(1+v)+D(1+v){\mathrm{\ρ\ω}}^2+DGk(1+v\left)r_i^2\right)};$

${\mathrm{\γ}}_i=\frac{\left(\right(12{\mathrm{Dr}}_i^2-12Ghk+6{\mathrm{D\α}}^2\left(v-1\right)+h^3{\mathrm{\ρ\ω}}^2\left)\right(Gk\left({\mathrm{\α}}^2-r_i^2\right)-{\mathrm{\ρ\ω}}^2)-12G^2{\mathrm{hk}}^2{r}_i^2)}{6{\mathrm{\αr}}_i(-2G^2{\mathrm{hk}}^2+D(1+v\left)\right(-{\mathrm{\ρ\ω}}^2+Gk\left({\mathrm{\α}}^2-r_i^2\right)\left)\right)}$

其中i＝1,2,3，k為調(diào)節(jié)系數(shù)，為展開式用到的正弦函數(shù)頻率值，根據(jù)展開式不同的項(xiàng)來進(jìn)行取整數(shù)值，m＝1,2,…,∞；r_i為根據(jù)式(6)的解計(jì)算得到的值；根據(jù)判別式的符號(hào)以及式(6)中解μ₁,μ₂,μ₃的情形得到如下10種x向振動(dòng)模態(tài)函數(shù)形式：

(1)判別式小于0，μ₁>0：實(shí)根，μ₂,μ₃：共軛復(fù)根

r₂,r₃為共軛復(fù)數(shù)，Re表示實(shí)部，Im表示虛部；

w_x(x)＝-δ₁B₂cosh(r₁x)-δ₁B₁sinh(r₁x)-δ₂B₄cosh(Re(r₂)x)-δ₂B₃sinh(Re(r₂)x)

-δ₃B₆cos(Im(r₂)x)-δ₃B₅sin(Im(r₂)x)

$\begin{array}{c}{\stackrel{\‾}{\mathrm{\θ}}}_{yx}\left(x\right)=B_1\cosh \left(r_{1}x\right)+B_2\sinh \left(r_{1}x\right)+B_3\cosh \left(\mathrm{Re}\right(r_2\left)x\right)+B_4\sinh \left(\mathrm{Re}\right(r_2\left)x\right)\\ +B_5\cos \left(\mathrm{Im}\left(r_2\right)x\right)+B_6\sin \left(\mathrm{Im}\left(r_2\right)x\right)\end{array}$

$\begin{array}{c}{\stackrel{\‾}{\mathrm{\θ}}}_{xx}\left(x\right)=-{\mathrm{\γ}}_1{B}_2\cosh \left(r_{1}x\right)-{\mathrm{\γ}}_1{B}_1\sinh \left(r_{1}x\right)-{\mathrm{\γ}}_2{B}_4\cosh \left(\mathrm{Re}\left(r_2\right)x\right)-{\mathrm{\γ}}_2{B}_3\sinh \left(\mathrm{Re}\left(r_2\right)x\right)\\ -{\mathrm{\γ}}_3{B}_6\cos \left(\mathrm{Im}\left(r_2\right)x\right)-{\mathrm{\γ}}_3{B}_5\sin \left(\mathrm{Im}\left(r_2\right)x\right)\end{array}$

(2)判別式小于0，μ₁<0：實(shí)根，μ₂,μ₃：共軛復(fù)根

令r₂,r₃為共軛復(fù)數(shù)，Re表示實(shí)部，Im表示虛部；

w_x(x)＝-δ₁B₂cos(r₁x)-δ₁B₁sin(r₁x)-δ₂B₄cosh(Re(r₂)x)-δ₂B₃sinh(Re(r₂)x)

-δ₃B₆cos(Im(r₂)x)-δ₃B₅sin(Im(r₂)x)

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yx}\left(x\right)=B_{1}cos\left(r_{1}x\right)+B_{2}sin\left(r_{1}x\right)+B_3\cosh \left(\mathrm{Re}\right(r_2\left)x\right)+B_4\sinh \left(\mathrm{Re}\right(r_2\left)x\right)\\ +B_5\cos \left(\mathrm{Im}\left(r_2\right)x\right)+B_6\sin \left(\mathrm{Im}\left(r_2\right)x\right)\end{array}$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xx}\left(x\right)=-{\mathrm{\&gamma;}}_1{B}_{2}cos\left(r_{1}x\right)-{\mathrm{\&gamma;}}_1{B}_{1}sin\left(r_{1}x\right)-{\mathrm{\&gamma;}}_2{B}_4\cosh \left(\mathrm{Re}\right(r_2\left)x\right)-{\mathrm{\&gamma;}}_2{B}_3\sinh \left(\mathrm{Re}\right(r_2\left)x\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cos \left(\mathrm{Im}\left(r_2\right)x\right)-{\mathrm{\&gamma;}}_3{B}_5\sin \left(\mathrm{Im}\left(r_2\right)x\right)\end{array}$

(3)判別式大于0，μ₁>0：實(shí)根，μ₂>0：實(shí)根，μ₃>0：實(shí)根

令

$\begin{array}{c}{w}_x\left(x\right)=-{\mathrm{\&delta;}}_1{B}_2\cosh \left(r_{1}x\right)-{\mathrm{\&delta;}}_1{B}_1\sinh \left(r_{1}x\right)-{\mathrm{\&delta;}}_2{B}_4\cosh \left(r_{2}x\right)-{\mathrm{\&delta;}}_2{B}_3\sinh \left(r_{2}x\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cosh \left(r_{3}x\right)-{\mathrm{\&delta;}}_3{B}_5\sinh \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xx}\left(x\right)=-{\mathrm{\&gamma;}}_1{B}_2\cosh \left(r_{1}x\right)-{\mathrm{\&gamma;}}_1{B}_1\sinh \left(r_{1}x\right)-{\mathrm{\&gamma;}}_2{B}_4\cosh \left(r_{2}x\right)-{\mathrm{\&gamma;}}_2{B}_3\sinh \left(r_{2}x\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cosh \left(r_{3}x\right)-{\mathrm{\&gamma;}}_3{B}_5\sinh \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yx}\left(x\right)=B_1\cosh \left(r_{1}x\right)+B_2\sinh \left(r_{1}x\right)+B_3\cosh \left(r_{2}x\right)+B_4\sinh \left(r_{2}x\right)\\ +B_5\cosh \left(r_{3}x\right)+B_6\sinh \left(r_{3}x\right)\end{array};$

(4)判別式大于0，μ₁>0：實(shí)根，μ₂>0：實(shí)根，μ₃<0：實(shí)根

$r_1=\sqrt{{\mathrm{\&mu;}}_1},r_2=\sqrt{{\mathrm{\&mu;}}_2},r_3=\sqrt{-{\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_x\left(x\right)=-{\mathrm{\&delta;}}_1{B}_2\cosh \left(r_{1}x\right)-{\mathrm{\&delta;}}_1{B}_1\sinh \left(r_{1}x\right)-{\mathrm{\&delta;}}_2{B}_4\cosh \left(r_{2}x\right)-{\mathrm{\&delta;}}_2{B}_3\sinh \left(r_{2}x\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cos \left(r_{3}x\right)-{\mathrm{\&delta;}}_3{B}_5\sin \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xx}\left(x\right)=-{\mathrm{\&gamma;}}_1{B}_2\cosh \left(r_{1}x\right)-{\mathrm{\&gamma;}}_1{B}_1\sinh \left(r_{1}x\right)-{\mathrm{\&gamma;}}_2{B}_4\cosh \left(r_{2}x\right)-{\mathrm{\&gamma;}}_2{B}_3\sinh \left(r_{2}x\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cos \left(r_{3}x\right)-{\mathrm{\&gamma;}}_3{B}_5\sin \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yx}\left(x\right)=B_1\cosh \left(r_{1}x\right)+B_2\sinh \left(r_{1}x\right)+B_3\cosh \left(r_{2}x\right)+B_4\sinh \left(r_{2}x\right)\\ +B_5\cos \left(r_{3}x\right)+B_6\sin \left(r_{3}x\right)\end{array};$

(5)判別式大于0，μ₁>0：實(shí)根，μ₂<0：實(shí)根，μ₃>0：實(shí)根

$r_1=\sqrt{{\mathrm{\&mu;}}_1},r_2=\sqrt{-{\mathrm{\&mu;}}_2},r_3=\sqrt{{\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_x\left(x\right)=-{\mathrm{\&delta;}}_1{B}_2\cosh \left(r_{1}x\right)-{\mathrm{\&delta;}}_1{B}_1\sinh \left(r_{1}x\right)-{\mathrm{\&delta;}}_2{B}_4\cos \left(r_{2}x\right)-{\mathrm{\&delta;}}_2{B}_3\sin \left(r_{2}x\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cosh \left(r_{3}h\right)-{\mathrm{\&delta;}}_3{B}_5\sinh \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xx}\left(x\right)=-{\mathrm{\&gamma;}}_1{B}_2\cosh \left(r_{1}x\right)-{\mathrm{\&gamma;}}_1{B}_1\sinh \left(r_{1}x\right)-{\mathrm{\&gamma;}}_2{B}_4\cos \left(r_{2}x\right)-{\mathrm{\&gamma;}}_2{B}_3\sin \left(r_{2}x\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cosh \left(r_{3}x\right)-{\mathrm{\&gamma;}}_3{B}_5\sinh \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yx}\left(x\right)=B_1\cosh \left(r_{1}x\right)+B_2\sinh \left(r_{1}x\right)+B_{3}cos\left(r_{2}x\right)+B_{4}sin\left(r_{2}x\right)\\ +B_5\cosh \left(r_{3}x\right)+B_6\sinh \left(r_{3}x\right)\end{array}$

(6)判別式大于0，μ₁<0：實(shí)根，μ₂>0：實(shí)根，μ₃>0：實(shí)根

$r_1=\sqrt{-{\mathrm{\&mu;}}_1},r_2=\sqrt{{\mathrm{\&mu;}}_2},r_3=\sqrt{{\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_x\left(x\right)=-{\mathrm{\&delta;}}_1{B}_{2}cos\left(r_{1}x\right)-{\mathrm{\&delta;}}_1{B}_{1}sin\left(r_{1}x\right)-{\mathrm{\&delta;}}_2{B}_4\cosh \left(r_{2}x\right)-{\mathrm{\&delta;}}_2{B}_3\sinh \left(r_{2}x\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cosh \left(r_{3}x\right)-{\mathrm{\&delta;}}_3{B}_5\sinh \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xx}\left(x\right)=-{\mathrm{\&gamma;}}_1{B}_2\cos \left(r_{1}x\right)-{\mathrm{\&gamma;}}_1{B}_1\sin \left(r_{1}x\right)-{\mathrm{\&gamma;}}_2{B}_4\cosh \left(r_{2}x\right)-{\mathrm{\&gamma;}}_2{B}_3\sinh \left(r_{2}x\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cosh \left(r_{3}x\right)-{\mathrm{\&gamma;}}_3{B}_5\sinh \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yx}\left(x\right)=B_{1}cos\left(r_{1}x\right)+B_{2}sin\left(r_{1}x\right)+B_3\cosh \left(r_{2}x\right)+B_4\sinh \left(r_{2}x\right)\\ +B_5\cosh \left(r_{3}x\right)+B_6\sinh \left(r_{3}x\right)\end{array};$

(7)判別式大于0，μ₁>0：實(shí)根，μ₂<0：實(shí)根，μ₃<0：實(shí)根

$r_1=\sqrt{{\mathrm{\&mu;}}_1},r_2=\sqrt{-{\mathrm{\&mu;}}_2},r_3=\sqrt{-{\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_x\left(x\right)=-{\mathrm{\&delta;}}_1{B}_2\cosh \left(r_{1}x\right)-{\mathrm{\&delta;}}_1{B}_1\sinh \left(r_{1}x\right)-{\mathrm{\&delta;}}_2{B}_4\cos \left(r_{2}x\right)-{\mathrm{\&delta;}}_2{B}_3\sin \left(r_{2}x\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cos \left(r_{3}x\right)-{\mathrm{\&delta;}}_3{B}_5\sin \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yx}\left(x\right)=B_1\cosh \left(r_{1}x\right)+B_2\sinh \left(r_{1}x\right)+B_{3}cos\left(r_{2}x\right)+B_{4}sin\left(r_{2}x\right)\\ +B_5\cos \left(r_{3}x\right)+B_6\sin \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xx}\left(x\right)=-{\mathrm{\&gamma;}}_1{B}_2\cosh \left(r_{1}x\right)-{\mathrm{\&gamma;}}_1{B}_1\sinh \left(r_{1}x\right)-{\mathrm{\&gamma;}}_2{B}_{4}cos\left(r_{2}x\right)-{\mathrm{\&gamma;}}_2{B}_{3}sin\left(r_{2}x\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cos \left(r_{3}x\right)-{\mathrm{\&gamma;}}_3{B}_5\sin \left(r_{3}x\right)\end{array}$

(8)判別式大于0，μ₁<0：實(shí)根，μ₂>0：實(shí)根，μ₃<0：實(shí)根

$r_1=\sqrt{-{\mathrm{\&mu;}}_1},r_2=\sqrt{{\mathrm{\&mu;}}_2},r_3=\sqrt{-{\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_x\left(x\right)=-{\mathrm{\&delta;}}_1{B}_{2}cos\left(r_{1}x\right)-{\mathrm{\&delta;}}_1{B}_{1}sin\left(r_{1}x\right)-{\mathrm{\&delta;}}_2{B}_4\cosh \left(r_{2}x\right)-{\mathrm{\&delta;}}_2{B}_3\sinh \left(r_{2}x\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cos \left(r_{3}x\right)-{\mathrm{\&delta;}}_3{B}_5\sin \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yx}\left(x\right)=B_1\cos \left(r_{1}x\right)+B_2\sin \left(r_{1}x\right)+B_3\cosh \left(r_{2}x\right)+B_4\sinh \left(r_{2}x\right)\\ +B_5\cos \left(r_{3}x\right)+B_6\sin \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xx}\left(x\right)=-{\mathrm{\&gamma;}}_1{B}_{2}cos\left(r_{1}x\right)-{\mathrm{\&gamma;}}_1{B}_{1}sin\left(r_{1}x\right)-{\mathrm{\&gamma;}}_2{B}_4\cosh \left(r_{2}x\right)-{\mathrm{\&gamma;}}_2{B}_3\sinh \left(r_{2}x\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cos \left(r_{3}x\right)-{\mathrm{\&gamma;}}_3{B}_5\sin \left(r_{3}x\right)\end{array}$

(9)判別式大于0，μ₁<0：實(shí)根，μ₂<0：實(shí)根，μ₃>0：實(shí)根

$r_1=\sqrt{-{\mathrm{\&mu;}}_1},r_2=\sqrt{-{\mathrm{\&mu;}}_2},r_3=\sqrt{{\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_x\left(x\right)=-{\mathrm{\&delta;}}_1{B}_{2}cos\left(r_{1}x\right)-{\mathrm{\&delta;}}_1{B}_{1}sin\left(r_{1}x\right)-{\mathrm{\&delta;}}_2{B}_{4}cos\left(r_{2}x\right)-{\mathrm{\&delta;}}_2{B}_{3}sin\left(r_{2}x\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cosh \left(r_{3}x\right)-{\mathrm{\&delta;}}_3{B}_5\sinh \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yx}\left(x\right)=B_1\cos \left(r_{1}x\right)+B_2\sin \left(r_{1}x\right)+B_3\cos \left(r_{2}x\right)+B_4\sin \left(r_{2}x\right)\\ +B_5\cosh \left(r_{3}x\right)+B_6\sinh \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xx}\left(x\right)=-{\mathrm{\&gamma;}}_1{B}_{2}cos\left(r_{1}x\right)-{\mathrm{\&gamma;}}_1{B}_{1}sin\left(r_{1}x\right)-{\mathrm{\&gamma;}}_2{B}_{4}cos\left(r_{2}x\right)-{\mathrm{\&gamma;}}_2{B}_{3}sin\left(r_{2}x\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cosh \left(r_{3}x\right)-{\mathrm{\&gamma;}}_3{B}_5\sinh \left(r_{3}x\right)\end{array}$

(10)判別式大于0，μ₁<0：實(shí)根，μ₂<0：實(shí)根，μ₃<0：實(shí)根

$r_1=\sqrt{-{\mathrm{\&mu;}}_1},r_2=\sqrt{-{\mathrm{\&mu;}}_2},r_3=\sqrt{{-\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_x\left(x\right)=-{\mathrm{\&delta;}}_1{B}_2\cos \left(r_{1}x\right)-{\mathrm{\&delta;}}_1{B}_1\sin \left(r_{1}x\right)-{\mathrm{\&delta;}}_2{B}_4\cos \left(r_{2}x\right)-{\mathrm{\&delta;}}_2{B}_3\sin \left(r_{2}x\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cos \left(r_{3}x\right)-{\mathrm{\&delta;}}_3{B}_5\sin \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yx}\left(x\right)=B_1\cos \left(r_{1}x\right)+B_2\sin \left(r_{1}x\right)+B_3\cos \left(r_{2}x\right)+B_4\sin \left(r_{2}x\right)\\ +B_5\cos \left(r_{3}x\right)+B_6\sin \left(r_{3}x\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xx}\left(x\right)=-{\mathrm{\&gamma;}}_1{B}_2\cos \left(r_{1}x\right)-{\mathrm{\&gamma;}}_1{B}_1\sin \left(r_{1}x\right)-{\mathrm{\&gamma;}}_2{B}_4\cos \left(r_{2}x\right)-{\mathrm{\&gamma;}}_2{B}_3\sin \left(r_{2}x\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cos \left(r_{3}x\right)-{\mathrm{\&gamma;}}_3{B}_5\sin \left(r_{3}x\right)\end{array}$

將得到的模態(tài)函數(shù)表達(dá)式代入邊界條件，整理以后得到如下形式的方程組：

其中表示線性方程組系數(shù)矩陣，A₁₁～A₆₆表示對(duì)應(yīng)位置的系數(shù)值；令采用matlab中的非線性優(yōu)化函數(shù)fsolve進(jìn)行求解得到ω_x，再將ω_x代入式(7)中得到線性方程組，對(duì)線性方程組求解得到B₁,B₂,…,B₆的值，從而得到x向的振動(dòng)模態(tài)函數(shù)的表達(dá)式。

6.根據(jù)權(quán)利要求5所述的基于譜有限元的矩形板振動(dòng)模態(tài)計(jì)算方法，其特征在于：所述步驟2-3中基于已知x向模態(tài)函數(shù)計(jì)算得到頻率值w_y和y向模態(tài)函數(shù)方法如下：

根據(jù)常微分方程組Ⅱ，令表示微分算子d/dy，則有如下的式子：

展開式(8)中的行列式得到如下的方程:

其中

j₁＝e₃+f₃+g₃-e₁f₂-g₁f₁；

j₂＝g₃f₃+e₃g₃+e₃f₃+e₁f₁g₂+e₂f₂g₁-e₃g₁f₁-e₂g₂

j₃＝e₃g₃f₃-e₂g₂f₃-e₁g₃f₂；

將試驗(yàn)解ψ＝e^λ代入方程(9)，由于表示微分算子，則產(chǎn)生如下的方程：

λ⁶+j₁λ⁴+j₂λ²+j₃＝0 (10)

令μ＝λ²，則

μ³+j₁μ²+j₂μ+j₃＝0 (11)

令判別式Δ＝18j₁j₂j₃-4j₁³j₃+j₁²j₂²-4j₂³-27j₃²，

定義w_y(y)與函數(shù)表達(dá)式中的系數(shù)關(guān)系數(shù)值仍表示為δ₁,δ₂,δ₃，與函數(shù)表達(dá)式中的系數(shù)關(guān)系數(shù)值仍表示為γ₁,γ₂,γ₃如下

${\mathrm{\&delta;}}_i=\frac{Gk(12Ghk-h^3{\mathrm{\&rho;\&omega;}}^2-6D(-1+v\left)\right({\mathrm{\&alpha;}}^2-r_i^2\left)\right)}{6r_i(2G^2{\mathrm{hk}}^2-{\mathrm{DGk\&alpha;}}^2(1+v)+D(1+v){\mathrm{\&rho;\&omega;}}^2+DGk(1+v\left)r_i^2\right)};$

${\mathrm{\&gamma;}}_i=\frac{\left(\left(12{\mathrm{Dr}}_i^2-12Ghk+6{\mathrm{D\&alpha;}}^2\left(v-1\right)+h^3{\mathrm{\&rho;\&omega;}}^2\right)\left(Gk\left({\mathrm{\&alpha;}}^2-r_i^2\right)-{\mathrm{\&rho;\&omega;}}^2\right)-12G^2{\mathrm{hk}}^2{r}_i^2\right)}{6{\mathrm{\&alpha;r}}_i\left(-2G^2{\mathrm{hk}}^2+D\left(1+v\right)\left(-{\mathrm{\&rho;\&omega;}}^2+Gk\left({\mathrm{\&alpha;}}^2-r_i^2\right)\right)\right)}$

其中i＝1,2,3，k為調(diào)節(jié)系數(shù)，為展開式用到的正弦函數(shù)頻率值，根據(jù)展開式不同的項(xiàng)來進(jìn)行取整數(shù)值，m＝1,2,…,∞；r_i為根據(jù)式(11)的解計(jì)算得到的值；根據(jù)判別式的符號(hào)以及式(11)中解μ₁,μ₂,μ₃的情形得到如下10種y向振動(dòng)模態(tài)函數(shù)形式：

(11)判別式小于0，μ₁>0：實(shí)根，μ₂,μ₃：共軛復(fù)根

r₂,r₃為共軛復(fù)數(shù)，Re表示實(shí)部，Im表示虛部；

w_y(y)＝-δ₁B₂cosh(r₁y)-δ₁B₁sinh(r₁y)-δ₂B₄cosh(Re(r₂)y)-δ₂B₃sinh(Re(r₂)y)

-δ₃B₆cos(Im(r₂)y)-δ₃B₅sin(Im(r₂)y)

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xy}\left(y\right)=B_1\cosh \left(r_{1}y\right)+B_2\sinh \left(r_{1}y\right)+B_3\cosh \left(\mathrm{Re}\left(r_2\right)y\right)+B_4\sinh \left(\mathrm{Re}\left(r_2\right)y\right)\\ +B_5\cos \left(\mathrm{Im}\left(r_2\right)y\right)+B_6\sin \left(\mathrm{Im}\left(r_2\right)y\right)\end{array}$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yy}\left(y\right)=-{\mathrm{\&gamma;}}_1{B}_2\cosh \left(r_{1}y\right)-{\mathrm{\&gamma;}}_1{B}_1\sinh \left(r_{1}y\right)-{\mathrm{\&gamma;}}_2{B}_4\cosh \left(\mathrm{Re}\left(r_2\right)y\right)-{\mathrm{\&gamma;}}_2{B}_3\sinh \left(\mathrm{Re}\left(r_2\right)y\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cos \left(\mathrm{Im}\left(r_2\right)y\right)-{\mathrm{\&gamma;}}_3{B}_5\sin \left(\mathrm{Im}\left(r_2\right)y\right)\end{array}$

(12)判別式小于0，μ₁<0：實(shí)根，μ₂,μ₃：共軛復(fù)根

令r₂,r₃為共軛復(fù)數(shù)，Re表示實(shí)部，Im表示虛部；

w_y(y)＝-δ₁B₂cos(r₁y)-δ₁B₁sin(r₁y)-δ₂B₄cosh(Re(r₂)y)-δ₂B₃sinh(Re(r₂)y)

-δ₃B₆cos(Im(r₂)y)-δ₃B₅sin(Im(r₂)y)

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xy}\left(y\right)=B_1\cos \left(r_{1}y\right)+B_2\sin \left(r_{1}y\right)+B_3\cosh \left(\mathrm{Re}\left(r_2\right)y\right)+B_4\sinh \left(\mathrm{Re}\left(r_2\right)y\right)\\ +B_5\cos \left(\mathrm{Im}\left(r_2\right)y\right)+B_6\sin \left(\mathrm{Im}\left(r_2\right)y\right)\end{array}$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yy}\left(y\right)=-{\mathrm{\&gamma;}}_1{B}_2\cos \left(r_{1}y\right)-{\mathrm{\&gamma;}}_1{B}_1\sin \left(r_{1}y\right)-{\mathrm{\&gamma;}}_2{B}_4\cosh \left(\mathrm{Re}\left(r_2\right)y\right)-{\mathrm{\&gamma;}}_2{B}_3\sinh \left(\mathrm{Re}\left(r_2\right)y\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cos \left(\mathrm{Im}\left(r_2\right)y\right)-{\mathrm{\&gamma;}}_3{B}_5\sin \left(\mathrm{Im}\left(r_2\right)y\right)\end{array}$

(13)判別式大于0，μ₁>0：實(shí)根，μ₂>0：實(shí)根，μ₃>0：實(shí)根

令

$\begin{array}{c}{w}_y\left(y\right)=-{\mathrm{\&delta;}}_1{B}_2\cosh \left(r_{1}y\right)-{\mathrm{\&delta;}}_1{B}_1\sinh \left(r_{1}y\right)-{\mathrm{\&delta;}}_2{B}_4\cosh \left(r_{2}y\right)-{\mathrm{\&delta;}}_2{B}_3\sinh \left(r_{2}y\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cosh \left(r_{3}y\right)-{\mathrm{\&delta;}}_3{B}_5\sinh \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yy}\left(y\right)=-{\mathrm{\&gamma;}}_1{B}_2\cosh \left(r_{1}y\right)-{\mathrm{\&gamma;}}_1{B}_1\sinh \left(r_{1}y\right)-{\mathrm{\&gamma;}}_2{B}_4\cosh \left(r_{2}y\right)-{\mathrm{\&gamma;}}_2{B}_3\sinh \left(r_{2}y\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cosh \left(r_{3}y\right)-{\mathrm{\&gamma;}}_3{B}_5\sinh \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xy}\left(y\right)=B_1\cosh \left(r_{1}y\right)+B_2\sinh \left(r_{1}y\right)+B_3\cosh \left(r_{2}y\right)+B_4\sinh \left(r_{2}y\right)\\ +B_5\cosh \left(r_{3}y\right)+B_6\sinh \left(r_{3}y\right)\end{array}$

(14)判別式大于0，μ₁>0：實(shí)根，μ₂>0：實(shí)根，μ₃<0：實(shí)根

$r_1=\sqrt{{\mathrm{\&mu;}}_1},r_2=\sqrt{{\mathrm{\&mu;}}_2},r_3=\sqrt{-{\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_y\left(y\right)=-{\mathrm{\&delta;}}_1{B}_2\cosh \left(r_{1}y\right)-{\mathrm{\&delta;}}_1{B}_1\sinh \left(r_{1}y\right)-{\mathrm{\&delta;}}_2{B}_4\cosh \left(r_{2}y\right)-{\mathrm{\&delta;}}_2{B}_3\sinh \left(r_{2}y\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cos \left(r_{3}y\right)-{\mathrm{\&delta;}}_3{B}_5\sin \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yy}\left(y\right)=-{\mathrm{\&gamma;}}_1{B}_2\cosh \left(r_{1}y\right)-{\mathrm{\&gamma;}}_1{B}_1\sinh \left(r_{1}y\right)-{\mathrm{\&gamma;}}_2{B}_4\cosh \left(r_{2}y\right)-{\mathrm{\&gamma;}}_2{B}_3\sinh \left(r_{2}y\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cos \left(r_{3}y\right)-{\mathrm{\&gamma;}}_3{B}_5\sin \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xy}\left(y\right)=B_1\cosh \left(r_{1}y\right)+B_2\sinh \left(r_{1}y\right)+B_3\cosh \left(r_{2}y\right)+B_4\sinh \left(r_{2}y\right)\\ +B_5\cos \left(r_{3}y\right)+B_6\sin \left(r_{3}y\right)\end{array};$

(15)判別式大于0，μ₁>0：實(shí)根，μ₂<0：實(shí)根，μ₃>0：實(shí)根

$r_1=\sqrt{{\mathrm{\&mu;}}_1},r_2=\sqrt{-{\mathrm{\&mu;}}_2},r_3=\sqrt{{\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_y\left(y\right)=-{\mathrm{\&delta;}}_1{B}_2\cosh \left(r_{1}y\right)-{\mathrm{\&delta;}}_1{B}_1\sinh \left(r_{1}y\right)-{\mathrm{\&delta;}}_2{B}_{4}cos\left(r_{2}y\right)-{\mathrm{\&delta;}}_2{B}_{3}sin\left(r_{2}y\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cosh \left(r_{3}y\right)-{\mathrm{\&delta;}}_3{B}_5\sinh \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yy}\left(y\right)=-{\mathrm{\&gamma;}}_1{B}_2\cosh \left(r_{1}y\right)-{\mathrm{\&gamma;}}_1{B}_1\sinh \left(r_{1}y\right)-{\mathrm{\&gamma;}}_2{B}_{4}cos\left(r_{2}y\right)-{\mathrm{\&gamma;}}_2{B}_{3}sin\left(r_{2}y\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cosh \left(r_{3}y\right)-{\mathrm{\&gamma;}}_3{B}_5\sinh \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xy}\left(y\right)=B_1\cosh \left(r_{1}y\right)+B_2\sinh \left(r_{1}y\right)+B_{3}cos\left(r_{2}y\right)+B_{4}sin\left(r_{2}y\right)\\ +B_5\cosh \left(r_{3}y\right)+B_6\sinh \left(r_{3}y\right)\end{array}$

(16)判別式大于0，μ₁<0：實(shí)根，μ₂>0：實(shí)根，μ₃>0：實(shí)根

$r_1=\sqrt{-{\mathrm{\&mu;}}_1},r_2=\sqrt{{\mathrm{\&mu;}}_2},r_3=\sqrt{{\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_y\left(y\right)=-{\mathrm{\&delta;}}_1{B}_2\cos \left(r_{1}y\right)-{\mathrm{\&delta;}}_1{B}_1\sin \left(r_{1}y\right)-{\mathrm{\&delta;}}_2{B}_4\cosh \left(r_{2}y\right)-{\mathrm{\&delta;}}_2{B}_3\sinh \left(r_{2}y\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cosh \left(r_{3}y\right)-{\mathrm{\&delta;}}_3{B}_5\sinh \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yy}\left(y\right)=-{\mathrm{\&gamma;}}_1{B}_{2}cos\left(r_{1}y\right)-{\mathrm{\&gamma;}}_1{B}_{1}sin\left(r_{1}y\right)-{\mathrm{\&gamma;}}_2{B}_4\cosh \left(r_{2}y\right)-{\mathrm{\&gamma;}}_2{B}_3\sinh \left(r_{2}y\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cosh \left(r_{3}y\right)-{\mathrm{\&gamma;}}_3{B}_5\sinh \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xy}\left(y\right)=B_1\cos \left(r_{1}y\right)+B_2\sin \left(r_{1}y\right)+B_3\cosh \left(r_{2}y\right)+B_4\sinh \left(r_{2}y\right)\\ +B_5\cosh \left(r_{3}y\right)+B_6\sinh \left(r_{3}y\right)\end{array};$

(17)判別式大于0，μ₁>0：實(shí)根，μ₂<0：實(shí)根，μ₃<0：實(shí)根

$r_1=\sqrt{{\mathrm{\&mu;}}_1},r_2=\sqrt{-{\mathrm{\&mu;}}_2},r_3=\sqrt{-{\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_y\left(y\right)=-{\mathrm{\&delta;}}_1{B}_2\cosh \left(r_{1}y\right)-{\mathrm{\&delta;}}_1{B}_1\sinh \left(r_{1}y\right)-{\mathrm{\&delta;}}_2{B}_{4}cos\left(r_{2}y\right)-{\mathrm{\&delta;}}_2{B}_{3}sin\left(r_{2}y\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cos \left(r_{3}y\right)-{\mathrm{\&delta;}}_3{B}_5\sin \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xy}\left(x\right)=B_1\cosh \left(r_{1}y\right)+B_2\sinh \left(r_{1}y\right)+B_{3}cos\left(r_{2}y\right)+B_{4}sin\left(r_{2}y\right)\\ +B_5\cos \left(r_{3}y\right)+B_6\sin \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yy}\left(y\right)=-{\mathrm{\&gamma;}}_1{B}_2\cosh \left(r_{1}y\right)-{\mathrm{\&gamma;}}_1{B}_1\sinh \left(r_{1}y\right)-{\mathrm{\&gamma;}}_2{B}_{4}cos\left(r_{2}y\right)-{\mathrm{\&gamma;}}_2{B}_{3}sin\left(r_{2}y\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cos \left(r_{3}y\right)-{\mathrm{\&gamma;}}_3{B}_5\sin \left(r_{3}y\right)\end{array}$

(18)判別式大于0，μ₁<0：實(shí)根，μ₂>0：實(shí)根，μ₃<0：實(shí)根

$r_1=\sqrt{-{\mathrm{\&mu;}}_1},r_2=\sqrt{{\mathrm{\&mu;}}_2},r_3=\sqrt{-{\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_y\left(y\right)=-{\mathrm{\&delta;}}_1{B}_2\cos \left(r_{1}y\right)-{\mathrm{\&delta;}}_1{B}_1\sin \left(r_{1}y\right)-{\mathrm{\&delta;}}_2{B}_4\cosh \left(r_{2}y\right)-{\mathrm{\&delta;}}_2{B}_3\sinh \left(r_{2}y\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cos \left(r_{3}y\right)-{\mathrm{\&delta;}}_3{B}_5\sin \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xy}\left(y\right)=B_1\cos \left(r_{1}y\right)+B_2\sin \left(r_{1}y\right)+B_3\cosh \left(r_{3}y\right)+B_4\sinh \left(r_{2}y\right)\\ +B_5\cos \left(r_{3}y\right)+B_6\sin \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yy}\left(y\right)=-{\mathrm{\&gamma;}}_1{B}_{2}cos\left(r_{1}y\right)-{\mathrm{\&gamma;}}_1{B}_{1}sin\left(r_{1}y\right)-{\mathrm{\&gamma;}}_2{B}_4\cosh \left(r_{2}y\right)-{\mathrm{\&gamma;}}_2{B}_3\sinh \left(r_{2}y\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cos \left(r_{3}y\right)-{\mathrm{\&gamma;}}_3{B}_5\sin \left(r_{3}y\right)\end{array}$

(19)判別式大于0，μ₁<0：實(shí)根，μ₂<0：實(shí)根，μ₃>0：實(shí)根

$r_1=\sqrt{-{\mathrm{\&mu;}}_1},r_2=\sqrt{-{\mathrm{\&mu;}}_2},r_3=\sqrt{{\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_y\left(y\right)=-{\mathrm{\&delta;}}_1{B}_{2}cos\left(r_{1}y\right)-{\mathrm{\&delta;}}_1{B}_{1}sin\left(r_{1}y\right)-{\mathrm{\&delta;}}_2{B}_{4}cos\left(r_{2}y\right)-{\mathrm{\&delta;}}_2{B}_{3}sin\left(r_{2}y\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cosh \left(r_{3}y\right)-{\mathrm{\&delta;}}_3{B}_5\sinh \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xy}\left(y\right)=B_1\cos \left(r_{1}y\right)+B_2\sin \left(r_{1}y\right)+B_3\cos \left(r_{2}y\right)+B_4\sin \left(r_{2}y\right)\\ +B_5\cosh \left(r_{3}y\right)+B_6\sinh \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yy}\left(y\right)=-{\mathrm{\&gamma;}}_1{B}_2\cos \left(r_{1}y\right)-{\mathrm{\&gamma;}}_1{B}_1\sin \left(r_{1}y\right)-{\mathrm{\&gamma;}}_2{B}_4\cos \left(r_{2}y\right)-{\mathrm{\&gamma;}}_2{B}_3\sin \left(r_{2}y\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cosh \left(r_{3}y\right)-{\mathrm{\&gamma;}}_3{B}_5\sinh \left(r_{3}y\right)\end{array}$

(20)判別式大于0，μ₁<0：實(shí)根，μ₂<0：實(shí)根，μ₃<0：實(shí)根

$r_1=\sqrt{-{\mathrm{\&mu;}}_1},r_2=\sqrt{-{\mathrm{\&mu;}}_2},r_3=\sqrt{-{\mathrm{\&mu;}}_3}$

$\begin{array}{c}{w}_y\left(y\right)=-{\mathrm{\&delta;}}_1{B}_{2}cos\left(r_{1}y\right)-{\mathrm{\&delta;}}_1{B}_{1}sin\left(r_{1}y\right)-{\mathrm{\&delta;}}_2{B}_{4}cos\left(r_{2}y\right)-{\mathrm{\&delta;}}_2{B}_{3}sin\left(r_{2}y\right)\\ -{\mathrm{\&delta;}}_3{B}_6\cos \left(r_{3}y\right)-{\mathrm{\&delta;}}_3{B}_5\sin \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{xy}\left(y\right)=B_1\cos \left(r_{1}y\right)+B_2\sin \left(r_{1}y\right)+B_3\cos \left(r_{2}y\right)+B_4\sin \left(r_{2}y\right)\\ +B_5\cos \left(r_{3}y\right)+B_6\sin \left(r_{3}y\right)\end{array};$

$\begin{array}{c}{\stackrel{\&OverBar;}{\mathrm{\&theta;}}}_{yy}\left(y\right)=-{\mathrm{\&gamma;}}_1{B}_{2}cos\left(r_{1}y\right)-{\mathrm{\&gamma;}}_1{B}_{1}sin\left(r_{1}y\right)-{\mathrm{\&gamma;}}_2{B}_{4}cos\left(r_{2}y\right)-{\mathrm{\&gamma;}}_2{B}_{3}sin\left(r_{2}y\right)\\ -{\mathrm{\&gamma;}}_3{B}_6\cos \left(r_{3}y\right)-{\mathrm{\&gamma;}}_3{B}_5\sin \left(r_{3}y\right)\end{array}$

將得到的表達(dá)式代入邊界條件，整理以后得到如下形式的方程組：

令系數(shù)矩陣的行列式采用matlab中的非線性優(yōu)化函數(shù)fsolve進(jìn)行求解得到ω_y，再將ω_y代入方程(12)中得到線性方程組，對(duì)線性方程組求解得到B₁,B₂,…,B₆的值，從而得到y(tǒng)向的振動(dòng)模態(tài)函數(shù)的表達(dá)式。