﻿ 基于双层梁模型的电磁轨道发射身管动力学研究
 舰船科学技术  2019, Vol. 41 Issue (3): 101-107 PDF

Transient resonance in electromagnetic launchers on double-deck rail
CAI Xi-yuan, LU Jun-yong, TAN Sai, ZHANG Yong-sheng, LI Bai, JIANG Yuan-zhi
National Key Laboratory for Science and Technology on Vessel Integrated Power System, Naval University of Engineering, Wuhan 430033, China
Abstract: The dynamic response of electromagnetic launchers influences the contact between the rail and the armature, fatigue life of the rail and the firing accuracy. A double-deck Euler-Bernoulli rail on Winkler elastic foundation is used to simplify the bolt pre-tightening electromagnetic launcher. The analytical solution of the dynamic response model is deduced. A finite element model of the housing is developed to simulate and analysis the vibration process of electromagnetic launcher in the view of time and space. The influence mechanism of different structural parameters and material properties on the transient resonance is analyzed from two aspects: the vibration deflection of the rail and the relative displacement between the rail and the containment. These results provide a foundation for further research of trajectory analysis, as well as transient resonance in electromagnetic launchers under repeat firing.
Key words: electromagnetic launch     dynamic response     double-deck rail     elastic wave
0 引　言

 图 1 发射装置模型图 Fig. 1 Electromagnetic launcher model

1 身管动力学双层梁模型的建立 1.1 身管动力学模型简化

 图 2 身管截面 Fig. 2 Cross section of electromagnetic launcher tube

 图 3 身管动力学分析模型 Fig. 3 Structural analysis model of electromagnetic launcher tube
1.2 身管动力学微分方程的建立

 $Mx'' + Cx' + Kx = F(t)\text{。}$ (1-1)

 $\begin{split} & {E_1}{I_1}\frac{{{\partial ^4}{w_1}}}{{\partial {x^4}}} + {m_1}\frac{{{\partial ^2}{w_1}}}{{\partial {t^2}}} + {k_1}\left( {{w_1} - {w_2}} \right) + \\ & {c_1}\left( {\frac{{\partial {w_1}}}{{\partial t}} - \frac{{\partial {w_2}}}{{\partial t}}} \right) = f\left( {x,t} \right)\text{，} \end{split}$ (1-2)
 $\begin{split} \!\!{E_2}{I_2}\frac{{{\partial ^4}{w_2}}}{{\partial {x^4}}} + {m_2}\frac{{{\partial ^2}{w_2}}}{{\partial {t^2}}} \!+\! {k_2}{w_2}\! +\! {c_2}\frac{{\partial {w_2}}}{{\partial t}} = \!-\! {k_1}\left( {{w_2} \!-\! {w_1}} \right) \text{。} \end{split}$ (1-3)

 $f\left( {x,t} \right) = - qH\left( {{L_a} - x} \right) - F\delta \left( {x - {L_a}} \right)\text{，}\;\;\left( {0 < t < t'} \right)\text{。}$ (1-4)

 ${k_1} = \frac{{{E_3}L{b_3}}}{{{h_3}}}\text{，}$ (1-5)

 ${k_2} = \frac{{{n_1}{E_4}\text{π} {r^2}}}{l}\text{。}$ (1-6)

2 身管动力学双层梁模型的理论求解 2.1 双层梁动力学振动微分方程的求解

 $\begin{split} & {E_1}{I_1}\frac{{{\partial ^4}{w_1}}}{{\partial {x^4}}} + {m_1}\frac{{{\partial ^2}{w_1}}}{{\partial {t^2}}} + {k_1}\left( {{w_1} - {w_2}} \right) +\\ & \quad {c_1}\left( {\frac{{\partial {w_1}}}{{\partial t}} - \frac{{\partial {w_2}}}{{\partial t}}} \right) = 0 \text{，} \end{split}$ (2-1)
 $\begin{split} {E_2}{I_2}\frac{{{\partial ^4}{w_2}}}{{\partial {x^4}}} \! + {m_2}\frac{{{\partial ^2}{w_2}}}{{\partial {t^2}}} \!+ \! {k_2}{w_2} \!+ {c_2}\frac{{\partial {w_2}}}{{\partial t}} \!= \!- \! {k_1}\left( {{w_2} \!- \!{w_1}} \right) \text{，} \end{split}$ (2-2)

 ${w_1} = \frac{{{E_2}{I_2}}}{{{k_1}}}\frac{{{\partial ^4}{w_2}}}{{\partial {x^4}}} + \frac{{{m_2}}}{{{k_1}}}\frac{{{\partial ^2}{w_2}}}{{\partial {t^2}}} + \frac{{{c_2}}}{{{k_1}}}\frac{{\partial {w_2}}}{{\partial t}} + \frac{{{k_1} + {k_2}}}{{{k_1}}}{w_2}\text{，}$ (2-3)
 $\begin{split} & \frac{{{E_1}{I_1}{E_2}{I_2}}}{{{k_1}}}\frac{{{\partial ^8}{w_2}}}{{\partial {x^8}}} + \frac{{{m_1}{E_2}{I_2} + {E_1}{I_1}{m_2}}}{{{k_1}}}\frac{{{\partial ^6}{w_2}}}{{\partial {x^4}\partial {t^2}}} + \\ & \frac{{{E_1}{I_1}{c_2} + {c_1}{E_2}{I_2}}}{{{k_1}}}\frac{{{\partial ^5}{w_2}}}{{\partial t\partial {x^4}}} + \\ & \left( {\frac{{{k_1} + {k_2}}}{{{k_1}}}{E_1}{I_1} + {E_2}{I_2}} \right)\frac{{{\partial ^4}{w_2}}}{{\partial {x^4}}} + \\ & \frac{{{m_1}{m_2}}}{{{k_1}}}\frac{{{\partial ^4}{w_2}}}{{\partial {t^4}}} + \frac{{{m_1}{c_2} + {c_1}{m_2}}}{{{k_1}}}\frac{{{\partial ^3}{w_2}}}{{\partial {t^3}}} + \\ & \left( {\frac{{{k_1} + {k_2}}}{{{k_1}}}{m_1} + {m_2} + \frac{{{c_1}{c_2}}}{{{k_1}}}} \right)\frac{{{\partial ^2}{w_2}}}{{\partial {t^2}}} + \\ & \frac{{{c_1}{k_2} + {c_2}{k_1}}}{{{k_1}}}\frac{{\partial {w_2}}}{{\partial t}} + {k_2}{w_2} = 0 \text{。} \end{split}$ (2-4)

 ${w_2} = \sum\limits_{i = 1}^\infty {{X_i}\left( {{A_i}\sin {\omega _i}t + {B_i}\cos {\omega _i}t} \right)} = \sum\limits_{i = 1}^\infty {{X_i}{T_i}} \text{，}$ (2-5)

 $\left\{ {\begin{split} & {{w_{1,2}}\left( 0 \right) = 0} \text{，}\\ & {{w_{1,2}}\left( L \right) = 0} \text{，}\\ & {{{\left. {\frac{{{\partial ^2}{w_{1,2}}}}{{\partial {x^2}}}} \right|}_{x = 0}} = 0} \text{，}\\ & {{{\left. {\frac{{{\partial ^2}{w_{1,2}}}}{{\partial {x^2}}}} \right|}_{x = L}} = 0} \text{。} \end{split}}\right.$ (2-6)

 $\begin{split} {w_1} = & \sum\limits_{i = 1}^\infty [ \frac{{{E_2}{I_2}}}{{{k_1}}}{X_i}^{\left( 4 \right)}{T_i} + \\ & \left( {\frac{{{k_1} + {k_2}}}{{{k_1}}} - \frac{{{m_2}}}{{{k_1}}}\omega _i^2} \right){X_i}{T_i} + \frac{{{c_2}}}{{{k_1}}}{X_i}{{T'}_i}] = \\ & \sum\limits_{i = 1}^\infty {\left[ {{P_i} + {Q_i} + {R_i}} \right]} \text{，}\end{split}$ (2-7)
 ${w_2} = \sum\limits_{i = 1}^\infty {\sqrt {\frac{2}{{{m_t}}}} } \sin {\beta _i}x{T_i}\text{。}$ (2-8)

 ${\beta _i} = \frac{i}{l}\text{π} \;\;\;(i = 1,2,3 \cdots )\text{，}$ (2-9)
 ${P_i} = \frac{{{E_2}{I_2}}}{{{k_1}}}\sqrt {\frac{2}{{{m_t}}}} \beta _i^4\sin {\beta _i}x \times \left( {{A_i}\sin {\omega _i}t + {B_i}\cos {\omega _i}t} \right) \text{，}$ (2-10)
 $\begin{split} {Q_i} =& \left( {\frac{{{k_1} + {k_2}}}{{{k_1}}} - \frac{{{m_2}}}{{{k_1}}}\omega _i^2} \right)\sqrt {\frac{2}{{{m_t}}}} \times \\ & \sin {\beta _i}x\left( {{A_i}\sin {\omega _i}t + {B_i}\cos {\omega _i}t} \right) \text{，} \end{split}$ (2-11)
 $\begin{split} {R_i} = \frac{{{c_2}}}{{{k_1}}}\sqrt {\frac{2}{{{m_t}}}} \sin {\beta _i}x \times \left( {{A_i}{\omega _i}\cos {\omega _i}t - {B_i}{\omega _i}\sin {\omega _i}t} \right)\text{。} \end{split}$ (2-12)
2.2 身管动态响应

 ${\int_0^L {\left( {\sin {\beta _i}x} \right)} ^2}{\rm d}x = \frac{L}{2}\text{，}$ (2-13)
 $\int_0^L {\sin {\beta _i}x\cos {\beta _i}x} {\rm d}x = 0\text{，}$ (2-14)

 $\mu \frac{{\partial {T_i}}}{{\partial t}} + \eta {T_i} = {F_i}\left( {x,t} \right)\text{。}$ (2-15)

 $\begin{split} \mu = & \frac{{{c_1}{E_2}{I_2} + {c_2}{E_1}{I_1}}}{{{k_1}}}\sqrt {\frac{2}{{{m_t}}}} \beta _i^4\frac{L}{2} - \\ & \frac{{{m_1}{c_2} + {c_1}{m_2}}}{{{k_1}}}\sqrt {\frac{2}{{{m_t}}}} \omega _i^2\frac{L}{2} + \frac{{{c_1}{k_2} + {k_1}{c_2}}}{{{k_1}}}\sqrt {\frac{2}{{{m_t}}}} \frac{L}{2}\text{，} \end{split}$ (2-16)
 $\begin{split} \eta = & \frac{{{E_1}{I_1}{E_2}{I_2}}}{{{k_1}}}\sqrt {\frac{2}{{{m_t}}}} \beta _i^8\frac{L}{2} + {k_2}\sqrt {\frac{2}{{{m_t}}}} \frac{L}{2}+ \\ & \frac{{\left( {{k_1} + {k_2}} \right){E_1}{I_1} + {k_1}{E_2}{I_2}}}{{{k_1}}} \times \sqrt {\frac{2}{{{m_t}}}} \beta _i^4\frac{L}{2} - \\ & \frac{{{m_2}{E_1}{I_1} + {m_1}{E_2}{I_2}}}{{{k_1}}}\omega _i^2\sqrt {\frac{2}{{{m_t}}}} \beta _i^4\frac{L}{2} - \\ & {m_1}\left( {\frac{{{k_1} + {k_2}}}{{{k_1}}} - \frac{{{m_2}}}{{{k_1}}}\omega _i^2 - \frac{{{c_1}{c_2}}}{{{k_1}}} - {m_2}} \right) \times \sqrt {\frac{2}{{{m_t}}}} \omega _i^2\frac{L}{2} \text{，} \end{split}$ (2-17)
 ${F_i}\left( {x,t} \right) = \int_0^L {f\left( {x,t} \right)\sin {\beta _i}x{\rm d}x}\text{，}$ (2-18)

 ${T_i} = {e^{ - \frac{\eta }{\mu }t}}\int {\frac{{{F_i}}}{\mu }{e^{\frac{\eta }{\mu }t}}{\rm d}t}\text{。}$ (2-19)

 $\begin{split} {w_1} = & \sum\limits_{i = 1}^\infty {\bigg[\bigg(\frac{{{E_2}{I_2}}}{{{k_1}}}\beta _i^4 + \frac{{{k_1} + {k_2}}}{{{k_1}}} - \frac{{{m_2}}}{{{k_1}}}\omega _i^2} - \\ & \frac{{{c_2}\eta }}{{{k_1}\mu }}\bigg) \times {e^{ - \frac{\eta }{\mu }t}}\int {\frac{{{F_i}}}{\mu }{e^{\frac{\eta }{\mu }t}}{\rm d}t} + \frac{{{c_2}{F_i}}}{{{k_1}\mu }}\bigg]{X_i} \text{，} \end{split}$ (2-20)
 ${w_2} = \sum\limits_{i = 1}^\infty {\sqrt {\frac{2}{{{m_t}}}} } {e^{ - \frac{\eta }{\mu }t}}\int {\frac{{{F_i}}}{\mu }{e^{\frac{\eta }{\mu }t}}{\rm d}t} \sin {\beta _i}x\text{。}$ (2-21)
3 身管动力学仿真分析

 图 4 导轨和外封装板挠度随位置和时间的变化 Fig. 4 Deflection of the rail and containment with time and location

 图 5 发射过程中身管挠度最大值 Fig. 5 The maximum deflection of railgun during the launching process

 图 6 发射时间历程中身管振动状态 Fig. 6 Vibration of railgun during the launching process

 图 7 导轨距起始端1.30 m和1.80 m处振动状态 Fig. 7 Vibration of the rail at 1.30 m and 1.80 m location

 图 8 导轨变形过程 Fig. 8 The deformation process of rail

 图 9 预紧螺栓对身管振动的影响 Fig. 9 The influence of the bolt on the vibration of railgun

 图 10 绝缘压板和外封装板对身管振动的影响 Fig. 10 The influence of the insulation clamping and containment on the vibration of railgun

4 结　语

1）发射过程中导轨和外封装板的挠度变形在整个时空分布情况基本一致，导轨和外封装板主要向外侧变形；

2）发射初始阶段，电枢运动距离较短，身管振动随着电枢前进迅速增大。随着电枢前进，外加载荷作用区域增大，挠度最大点也在向前推进，最大变形值基本不变，在出口阶段出现波动。任何一个时刻，导轨与外封装板间隙的最大值位置发生在导轨处于最大挠度处；

3）集中载荷和分布载荷共同作用和应力波传播的影响，对于导轨上任何位置处，在电枢行进到该位置处之前，导轨该位置处已经开始振动，振幅逐渐增大，并在电枢驶过该位置一段时间后，该位置振动才达到最大值，之后振动不断衰减；

4）综合增加第1层和第2层弹性层等效刚度，提高外封装板的抗弯刚度，如减小绝缘压板厚度并选用高模量材料做绝缘压板，增加预紧螺栓半径和数量，增加外封装板厚度并选用高模量材料做外封装板都可以有效减小电磁轨道发射身管振动。

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