﻿ 不同温度下含裂纹压气机叶片的模态特征和振动特性分析
 舰船科学技术  2023, Vol. 45 Issue (12): 93-98    DOI: 10.3404/j.issn.1672-7619.2023.12.017 PDF

Analysis of modal characteristics and vibration characteristics of cracked compressor blades at different temperatures
ZHOU Zhen-ting, HE Xing, LIU Yong-bao
School of Power Engineering, Naval University of Engineering, Wuhan 430033, China
Abstract: In this paper, a cracked cantilever beam is selected to simulate a cracked compressor blade. Based on a massless torsion spring and a breathing crack stiffness model, a method for analyzing the modal characteristics and vibration characteristics of a cracked compressor blade at different ambient temperatures is obtained. The elastic modulus is introduced into the temperature module, and the characteristic equation about the cracked beam is obtained by using the continuous condition of the massless torsion spring, and the influence of the ambient temperature and crack depth on the natural frequency of the cracked beam is analyzed; the forced bending vibration equation of the cantilever beam is used to introduce the breathing The crack stiffness model was used to change the excitation force frequency to analyze the influence of ambient temperature and excitation frequency on the vibration displacement response of the beam with cracks. The results show that the higher the ambient temperature, the smaller the natural frequency of the beam with cracks, and the greater the vibration displacement response of the beam; at the same time, the selection of the excitation frequency also has a certain influence.
Key words: compressor blade     breathing crack     temperature     modal characteristics     vibration characteristics
0 引　言

1 模型建立和模态分析 1.1 含裂纹悬臂梁模型建立

 图 1 含裂纹叶片等效模型 Fig. 1 A equivalent model of cracked compressor blade

 $\frac{1}{{{K_{\text{T}}}}} = \frac{{72{\text{π}} L(1 - {\eta ^2})(1 - {\nu ^2})}}{{{E_{\text{T}}}b{h^2}}}\phi ，$ (1)
 $\begin{split} \phi =\; & 0.629{{\text{r}}^2} - 1.047{r^3} + 4.602{r^4} - 9.975{r^5} + 20.285{r^6}- \\&32.993{r^7} + 47.041{r^8} - 40.693{r^9} + 19.6{r^{10}} 。\end{split}$ (2)

1.2 模态分析

 ${E}_{T}I\frac{{\partial }^{4}y（x,t）}{\partial {x}^{4}}+\rho A\frac{{\partial }^{2}y(x,t)}{\partial {x}^{2}}=0 。$ (3)

 $\frac{{{\rm{d}}y}}{{{\rm{d}}x}} - {k^4}y = 0 。$ (4)

 $\left\{ {\begin{array}{l}\begin{split} {Y_1} \left( x \right) =& {A_1}{\rm cos}\left( {kx} \right) + {B_1}{\rm sin}\left( {kx} \right) + \\&{C_1}{\rm cos}h\left( {kx} \right) + {D_1}{\rm sin}h\left( {kx} \right) ，\end{split}\\\begin{split} {Y_2}\left( x \right) =& {A_2}{\rm{cos}}\left( {kx} \right) + {B_2}{\rm{sin}}\left( {kx} \right) +\\& {C_2}{\rm{cos}}h\left( {kx} \right) + {D_2}{\rm{sin}}h\left( {kx} \right)。\end{split} \end{array}} \right.$ (5)

Y(x)表示位移，A1,B1,C1,D1表示裂纹左侧梁与边界条件有关的未知参数，A2,B2,C2,D2表示裂纹右侧梁与边界条件有关的未知参数。

 ${Y_1}(0) = 0 \text{，} \frac{{\partial {Y_1}(x)}}{{\partial x}} = 0 。$

 $\frac{{\partial Y_3^2(x)}}{{\partial {x^2}}} = 0 \text{，} \frac{{\partial Y_2^3(x)}}{{\partial {x^3}}} = 0 。$

 ${Y_1}(x) = {Y_2}(x) 。$

 $\frac{{\partial {Y_1}(x)}}{{\partial x}} + \frac{1}{{{K_{\text{T}}}}}\frac{{\partial {Y_1}(x)}}{{\partial {x^2}}} = \frac{{\partial {Y_2}(x)}}{{\partial x}} ，$

 $\frac{{\partial Y_1^2(x)}}{{\partial {x^2}}} = \frac{{\partial Y_2^2(x)}}{{\partial {x^2}}} ，$

 $\frac{{\partial Y_1^3(x)}}{{\partial {x^3}}} = \frac{{\partial Y_2^3(x)}}{{\partial {x^3}}}。$ (6)

 $\det ({\boldsymbol{S}}) = 0 ^{ } 。$ (7)

 ${\begin{split} S =& \left[ {\begin{array}{*{20}{c}} 1 & 0 & 1 \\ 0 & 1 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0\\ {\cos (kR)} & {\sin (kR)} & {\cosh (kR)}\\ { - \sin (kR) - k/{K_T}\cos (kR)} & {\cos (kR) - k/{K_T}\sin (kR)} & {\sinh (kR) + k/{K_T}\cosh (kR)} \\ { - \cos (kR)} & { - \sin (kR)} & {\cosh (kR)} \\ {\sin (kR)} & { - \cos (kR)} & {\sinh (kR)} \end{array}} \right. \\ &\left. \begin{array}{*{20}{c}} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & { - \cos (kl)} & { - \sin (kl)} & {\cosh (kl)} & {\sinh (kl)} \\ 0 & {\sin (kl)} & { - \cos (kl)} & {\sinh (kl)} & {\cosh (kl)} \\ {\sinh (kR)} & { - \cosh (kR)} & { - \sin (kR)} & { - \cosh (kR)} & { - \sinh (kR)} \\ {\cosh (kR) + k/{K_T}\sinh (kR)} & {\sin (kR)} & { - \cosh (kR)} & { - \sinh (kR)} & { - \cosh (kR)} \\ {\sinh (kR)} & {\cos (kR)} & {\sin (kR)} & { - \cosh (kR)} & { - \sinh (kR)} \\ {\cosh (kR)} & { - \sin (kR)} & {\cos (kR)} & { - \sinh (kR)} & { - \cosh (kR)} 。\end{array} \right] 。\end{split}}$

2 振动特性分析 2.1 建立呼吸式裂纹梁弯曲振动方程

 图 2 含裂纹悬臂梁模型 Fig. 2 The cracked cantilever beam model

 ${E_T}I\frac{{{\partial ^4}y(x,t)}}{{\partial {x^4}}} + c\frac{{\partial y(x,t)}}{{\partial t}}dx + \rho A\frac{{{\partial ^2}y(x,t)}}{{\partial {t^2}}} = p\cos ({\omega _{\text{J}}}t)\delta (x - {l_{\text{p}}})。$ (8)

 $Y(x) = \left[ {(\sin kx - \sinh kx) + {\alpha _n}(\cos kx - \cosh kx)} \right]，$ (9)

$y\left( {x,t} \right) = Y\left( x \right)T(t)$ 代入式（8）中，左右两端同时乘U(x)，并在0～L上积分可得：

 $m^* \frac{\partial^{2} T(t)}{\partial t^{2}}+c^{*} \frac{\partial T(t)}{\partial t}+k^{*} T(t)=F^{*}。$ (10)

 ${m^*} = \rho A\int_0^L {{Y^2}(x){\text{d}}x} ，$ (11)
 ${c^*} = c\int_0^L {Y{}^2(x){\text{d}}x}，$ (12)
 ${k^*} = {E_T}I\int_0^L {\frac{{{\partial ^4}Y(x)}}{{\partial {x^4}}}Y(x){\text{d}}x}。$ (13)
 ${F^*} = \cos (\omega t)\int_0^L {p\delta (x - {l_{\text{p}}})Y(x){\text{d}}x} 。$ (14)

 ${k_{{{br}}}} = {k_{{o}}} + \frac{{{k^*} - {k_{{o}}}}}{2}(1 + \cos (\omega t)) 。$ (16)

 ${k_{\text{o}}} = \frac{{{k_{{T}}}{k^*}}}{{{k_{{T}}} + {k^*}}} 。$ (17)

 ${m^*}\frac{{{\partial ^2}T(t)}}{{\partial {t^2}}} + {c^*}\frac{{\partial T(t)}}{{\partial t}} + {k_{{{br}}}}T(t) = {F^*} 。$ (18)
2.2 振动特性分析

3 结果与讨论

3.1 环境温度对裂纹梁模态特征的影响

 图 3 固有频率对环境温度变化曲线 Fig. 3 The natural frequency vs. ambient temperature

3.2 环境温度对裂纹梁振动特性的影响

 图 4 400 Hz频率下振动特性图 Fig. 4 The vibration performance with 400 Hz frequency

 图 6 4000 Hz下振动特性图 Fig. 6 The vibration performance with 4000 Hz frequency

 图 5 固有频率下振动特性图 Fig. 5 The vibration performance with natural frequency

4 结　语

1）环境温度对梁的一阶固有频率具有一定的影响，随着环境温度的升高，梁的一阶固有频率下降，而且下降的幅度较裂纹深度带来的影响更大。

2）在低频及固有频率激振力作用下，随着环境温度的升高，梁的位移响应越大，但在高频激振力作用下，出现了不同的现象。这是所选激振频率处于不同温度下各阶频率的距离远近导致的。

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