﻿ 双端固支谐振梁动态响应分析

# 双端固支谐振梁动态响应分析Dynamic Response Analysis of Double-Ended Resonant Beams

Abstract: The resonant sensor has attracted more attention because of its high sensitivity, small footprint, stability, compatibility with multiple-phase samples. The typical resonant sensor measures reso-nant frequency modulated by axial load to realize mass sensing. Dynamic characteristic is an im-portant factor to evaluate stability of resonant beam, and the key point is to analyze the dynamic response of the resonant beam under dynamic axial load. The article is based on infinitesimal method; mathematical model of vibration of resonant beam impacted by dynamic axial force was established through the resonant beam micro mechanical balance and analyzed through Mathieu equation. Equivalent circuit method was chosen to solve vibration model. The simulation of the circuit shows the dynamic response of resonant beam under the typical axial load. Besides, the article studied the effect of damping on the resonant beam and concluded the damping only influenced on vibration amplitude scarcely on frequency.

1. 引言

2. 谐振梁振动数学模型建立

F——梁承受的载荷；

Figure 1. Schematic diagram of bending vibration of double end fixed beam under dynamic axial force

Figure 2. Force analysis of the micro sec- tion dx of the resonant beam

$\rho$ ——梁的密度；

E——梁的弹性模量；

I——梁横截面对中性轴的惯性矩。

A、L、h、b分别是梁的横截面积、长度、高度、宽度。设 $y\left(x,t\right)$ 表示谐振梁上距原点为x处的截面在t时刻的横向位移，由力矩平衡公式与弯矩平衡公式共同联立得梁横向弯曲振动的方程为：

$EI\frac{{\partial }^{4}y}{\partial {x}^{4}}-F\frac{{\partial }^{2}y}{\partial {x}^{2}}+\rho A\frac{{\partial }^{2}y}{\partial {t}^{2}}=0$ (1)

$y\left(x,t\right)=\varphi \left(x\right)\mathrm{sin}\left(\omega t+\phi \right)$ (2)

$EI\frac{{\text{d}}^{4}Y}{\text{d}{x}^{4}}-F\frac{{\text{d}}^{2}Y}{\text{d}{x}^{2}}-\rho AY\frac{{\text{d}}^{2}P}{\text{d}{t}^{2}}=0$ (3)

${\phi }_{i}\left(x\right)=\text{ch}{\lambda }_{i}x-\mathrm{cos}{\lambda }_{i}x+{\eta }_{i}\left(\text{sh}{\lambda }_{i}x-\mathrm{sin}{\lambda }_{i}x\right)$ (4)

$\left\{\begin{array}{l}\frac{{\text{d}}^{2}Y\left(0\right)}{\text{d}{x}^{2}}=0\\ \frac{{\text{d}}^{2}Y\left(l\right)}{\text{d}{x}^{2}}=0\\ EI\left[\frac{{\text{d}}^{3}Y\left(0\right)}{\text{d}{x}^{3}}\right]=kY\left(0\right)\\ EI\left[\frac{{\text{d}}^{3}Y\left(l\right)}{\text{d}{x}^{3}}\right]=kY\left(l\right)\end{array}$ (5)

$\mathrm{cos}\lambda l\text{ }\text{ }\text{ch}\lambda l+1=-\frac{k}{EI{\lambda }^{3}}\left(\mathrm{cos}\lambda l\text{ }\text{ }\text{sh}\lambda l-\mathrm{sin}\lambda l\text{ }\text{ }\text{ch}\lambda l\right)$ (6)

Table 1. Parameters of the resonant beam

${\lambda }_{1}l=1.880$ (7)

${\eta }_{1}=\frac{\mathrm{cos}\lambda L-\text{ch}\lambda L}{\text{sh}\lambda L-\mathrm{sin}\lambda L}$ (8)

$\varphi \left(x\right)=\mathrm{ch}\lambda x+\mathrm{cos}\lambda x+0.7341\left(\mathrm{sh}\lambda x+\mathrm{sin}\lambda x\right)$ (9)

$\left(\underset{0}{\overset{l}{\int }}\rho A{\varphi }^{2}\text{d}x\right)\frac{{\text{d}}^{2}P}{\text{d}{t}^{2}}+\left(\underset{0}{\overset{l}{\int }}EI{\left({\varphi }^{″}\right)}^{2}\text{d}x-F\left(t\right)\underset{0}{\overset{l}{\int }}{\left({\varphi }^{\prime }\right)}^{2}\text{d}x\right)\cdot P=0$ (10)

$\frac{{\text{d}}^{2}P}{\text{d}{t}^{2}}+\left(A-B×F\left(t\right)\right)\cdot P=0$ (11)

$\left\{\begin{array}{l}\stackrel{˜}{K}=\frac{\underset{0}{\overset{l}{\int }}EJ{\left({\varphi }^{″}\right)}^{2}\text{d}x}{\underset{0}{\overset{l}{\int }}\rho A{\varphi }^{2}\text{d}x}\\ \stackrel{˜}{\alpha }=\frac{\epsilon \cdot \underset{0}{\overset{l}{\int }}{\left({\varphi }^{\prime }\right)}^{2}\text{d}x}{2\underset{0}{\overset{l}{\int }}\rho A{\varphi }^{2}\text{d}x}\end{array}$ (12)

$\frac{{\text{d}}^{2}P}{\text{d}{t}^{2}}+\left(\stackrel{˜}{K}-2\stackrel{˜}{\alpha }\mathrm{cos}\left(\omega \cdot t\right)\right)P=0$ (13)

$\frac{{\text{d}}^{2}P}{\text{d}{t}^{2}}+\left(3.0434×{10}^{11}-N\left(t\right)×5.2798×{10}^{12}\right)\cdot P=0$ (14)

3. 谐振梁振动模型电路模拟

4. 阻尼对谐振梁振动特性影响

(a) 载荷为0 mV (b) 载荷为−20 mV (c) 载荷频率为10 kHz幅值为10 mv (d) 载荷频率为10 kHz幅值为20 mV (e) 载荷频率为50 kHz幅度为10 mv

Figure 4. Circuit simulation results

① 等效电路方法分析

$\frac{{\mathrm{d}}^{2}P}{\mathrm{d}{t}^{2}}+c\cdot \frac{\mathrm{d}P}{\mathrm{d}t}+\left(A-B×F\left(t\right)\right)\cdot P=0$ (15)

$\left\{\begin{array}{l}\frac{-1}{{C}_{1}{R}_{1}}\underset{0}{\overset{t}{\int }}-\left(\left(A-B×F\left(t\right)\right)\cdot P+{k}^{\prime }{u}_{1}\left(t\right)\right)\text{d}t={u}_{1}\left(t\right)\\ \frac{-1}{{C}_{2}{R}_{2}}\underset{0}{\overset{t}{\int }}{u}_{1}\left(t\right)\text{d}t=P\end{array}$ (16)

$\frac{{\text{d}}^{2}P}{\text{d}{t}^{2}}+\frac{{k}^{\prime }}{{R}_{1}{C}_{1}}\cdot \frac{\text{d}P}{\text{d}t}+\frac{1}{{R}_{1}{C}_{1}{R}_{2}{C}_{2}}\left(A-B×F\left(t\right)\right)\cdot P=0$ (17)

② COMSOL方法分析

(a) 载荷为0 mV (b) 载荷为−20 mV (c) 载荷频率为10 kHz幅值为10 mv (d) 载荷频率为10 kHz幅值为20 mV (e) 载荷频率为50 kHz幅度为10 mv

Figure 6. Circuit with damping simulation results

Figure 7. Simulation of resonant beam’s amplitude by comsol

Figure 8. Simulation of resonant beam’s frequency by comsol

5. 结论

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