﻿ 大口径舰炮弹药储供系统可靠性研究
 舰船科学技术  2022, Vol. 44 Issue (10): 185-189    DOI: 10.3404/j.issn.1672-7649.2022.10.040 PDF

Reliability of ammunition storage and supply system for large caliber naval gun
ZHANG Xiao-jun, YAO Zhong, FENG Lian-qi, ZHANG Jian-bin, HUI Kai
Northwest Institute of Mechanical and Electrical Engineering, Xianyang 712099, China
Abstract: After the large-caliber naval gun ammunition storage and supply system uses new technologies such as central servo lift, push centering and box replenishment, the reliability has been significantly improved compared with the technology used in active naval guns. Then constructs a fault tree for abnormal bombing with full participation according to its failure mode, the reliability distribution research is carried out. The reliability improvement strategy of the storage and supply system is given, has a certain significance for the reliability evaluation. The results show that events such as chain or gearbox failure, valve component failure, and rotation module failure are the weak links of the system. During the design, the reasons for the failure should be analyzed and the allocation of such bottom events due to high task reliability and other reasons should be small. The reliability index.
Key words: ammunition storage and supply system     large-caliber naval gun     FMRBF     reliability
0 引　言

 图 1 弹药储供系统设计方案 Fig. 1 Scheme of ammunition storage and supply system
1 弹药储供系统FTA模型构建

 图 2 弹药储供系统“非正常扬弹”故障树 Fig. 2 "Abnormal raising ammunition" fault tree of ammunition storage and supply system

2 基于模糊数学的可靠性建模

2.1 综合模糊数学与评判法分析过程

1）将模糊语言转化为模糊数

 $\begin{split}&\underset{˜}{A}(x)=\left\{\begin{array}{*{20}{c}}0,&x < a，\\ \dfrac{x-a}{b-a},&a\leqslant x < b，\\ 1,&b\leqslant x < c，\\ \dfrac{d-x}{d-c},&c\leqslant x < d，\\ 0,&x\geqslant d，\end{array}\right.\\ &\underset{˜}{A}(x)=\left\{\begin{array}{*{20}{c}}0,&x < a，\\ \dfrac{x-a}{b-a},&a\leqslant x < b，\\ \dfrac{c-x}{c-b},&b\leqslant x < c，\\ 0,&x\geqslant c。\end{array}\right.\end{split}$ (1)

2）求出模糊数区间及专家权重

 图 3 置信水平为 $\lambda$ 时的模糊数区间 Fig. 3 Fuzzy number interval when the confidence level is $\lambda$

 $\underset{˜}{A} (x) = \sum\limits_{n = 1}^k {{w_n}\underset{˜}{A}_i}(x) 。$ (2)

3）求取模糊可能性值

 $\begin{split} &{f}_{\mathrm{min}}(x)=\left\{\begin{array}{*{20}{c}}1-x,&0 < x < 1，\\ 0,& {\rm{others}}。\end{array}\right.\\ &{f}_{\mathrm{max}}(x)=\left\{\begin{array}{*{20}{c}}x,&0 < x < 1\\ 0,&{\rm{others}}。\end{array}\right.\end{split}$ (3)
 $\begin{split}&FP{S}_{左}(W)=\underset{x}{\mathrm{sup}}\left[\underset{˜}{A}(x)^{f}_{\mathrm{min}}(x)\right]，\\ &FP{S}_{右}(W)=\underset{x}{\mathrm{sup}}\left[\underset{˜}{A}(x)^{f}_{\mathrm{max}}(x)\right]，\\ &FPS(W)=\frac{FP{S}_{左}(W)-FP{S}_{右}(W)+1}{2}。\end{split}$ (4)

4）求出模糊平均故障间隔发数（FMRBF）

 $\begin{split} &FMRBF=\bigg\{\begin{array}{l} {10}^{C},(FPS\ne 0)，\\ 0,(FPS=0)。\end{array}\\ &C=2.301\times \sqrt[3]{\frac{1-FPS}{FPS}} 。\bigg.\end{split}$ (5)
2.2 专家权重及隶属函数确定

 图 4 隶属函数 $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{A} (x)$ Fig. 4 Membership function $\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{A} (x)$

 $\begin{gathered} {w_k} = ({w_1},{w_2},{w_3},{w_4},{w_5}) = (0.2232, \hfill \\ {\text{ }}0.2232,0.1964,0.1964,0.1607) 。\hfill \\ \end{gathered}$
3 基于FTA的可靠性分配建模

 $\overline{{q}_{顶}}=1-{\displaystyle \prod _{i={E}_{1}}^{{E}_{3}}(1-{q}_{i})}\text=1-{\displaystyle \prod _{i={X}_{1}}^{{X}_{15}}(1-{q}_{i})}。$ (6)
3.1 对顶事件进行分配

 ${I_g}(i) = \frac{{\partial g}}{{\partial {q_i}}} = \prod\limits_{j = 1,j \ne i}^k {(1 - {q_j})} 。$ (7)

 ${Q_x} = 1 - \prod\limits_{i = 1}^n {\left[ {1 - ({q_i} + \Delta {q_i})} \right]} 。$ (8)
3.2 对中间事件进行分配

 ${q_i} + \Delta {q_i} \leqslant 1 - {(1 - {q_0})^n} \times \prod\limits_{i = n + 1}^k {(1 - {q_i})} 。$ (9)

M1重新分配后的 ${q_i} + \Delta {q_i} = 2.8024 \times {10^{ - 4}}$ 为例进行说明计算。

$n = 1$ 时，由式（9）可得：

${q_i} + \Delta {q_i} = 1 - (1 - {q_0})(1 - {q_{{E_8}}}) = 2.8024 \times {10^{ - 4}}$ ，可得：

 ${q_0} = 1.3578 \times {10^{ - 4}} < {q_{{M_5}}} \text{；}$

$n = 2$ 时，由式（9）可得：

${q_i} + \Delta {q_i} = 1 - {(1 - {q_0})^2} = 2.8024 \times {10^{ - 4}}$ ，可得：

 ${q_0} = 1.4013 \times {10^{ - 4}} < {q_{{M_4}}} 。$

 ${\boldsymbol{A}} = \left[ {\begin{array}{*{20}{c}} 1&3&4 \\ {1/3}&1&3 \\ {1/4}&{1/3}&1 \end{array}} \right] 。$

 ${h_{{\text{ }}11}}:{h_{{\text{ }}12}}:{h_{{\text{ }}13}} = 0.4472:0.2922:0.2606 。$

 ${q_{{X_i}}} + \Delta {q_{{X_i}}} = \left( {{q_8} + \Delta {q_8}} \right) \cdot \frac{{{h_i}^{ - 1}}}{{\displaystyle\sum\limits_{i = 11}^{13} {{h_i}^{ - 1}} }} 。$ (10)

 图 5 底事件分配前后FMRBF值变化过程 Fig. 5 FMRBF value change process before and after the bottom event allocation
4 结　语

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