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Research on the Principle of Lorry Rollover and Anti-Rollover Device Dynamics Analysis
DOI: 10.12677/IJM.2021.103016, PDF, HTML, XML, 下载: 259  浏览: 832  国家自然科学基金支持

Abstract: Based on two major cases of truck rollover, this paper constructs a dynamic model of Dongfeng heavy duty truck turning through reasonable simplification, then through the rigid body mechanics theory and MATLAB, the main reasons for the truck rollover are analyzed. In order to solve the problem of truck rollover, a safety valve and gyroscope are designed to prevent truck rollover. The main conclusions include: 1) Rigid body mechanics can simplify the analysis of the rollover problem of trucks due to turning and tilting of the carriage. The conclusions are consistent with the actual situation. 2) The centrifugal force and the angle of rotation are approximately linear; the faster the vehicle speed is, the greater the angle of rotation is, and the greater the deflection angle of the carriage is, the more easily the vehicle will roll over. 3) Both the safety valve and the gyroscope can provide effective torque to prevent rollover, which has very important guiding significance for the design of anti-rollover structure, optimization design and safety control of large trucks.

1. 引言

(a) 东风重型载货车 (b) 东风重型载货车结构图

Figure 1. Dongfeng heavy-duty truck [2]

2. 货车受力模型与解析

(a为俯视图，b为主视图，其中b图两侧的“弹簧”仅代表车厢可以发生倾斜，不参与运算)

Figure 2. Dynamic model of truck rollover

Figure 3. Simplified model of truck turning

$R=\frac{L}{\mathrm{tan}\alpha }\sqrt{4+{\mathrm{tan}}^{2}\alpha }$ (1)

$F=\frac{m{v}^{2}}{R}=\frac{m{v}^{2}\mathrm{tan}\alpha }{L\sqrt{4+{\mathrm{tan}}^{2}\alpha }}$ (2)

Figure 4. The relationship between centrifugal force and rotation angle

$F=5.6245×{10}^{4}×\alpha -2.81063×{10}^{5}$ (3)

(a) (b)(c)

Figure 5. Fitting analysis. (a) The conventional variance relationship between percentile and centrifugal force; (b) The relationship between the conventional residual of centrifugal force and the fitted centrifugal force; (c) The relationship between the conventional residual of centrifugal force and the wheel angle

Figure 6. Simplified model of truck load

$\sum {M}_{A}=0$ (4)

$Fh+{N}_{B}b-mg\left(\frac{b}{2}-h\mathrm{tan}\beta \right)=0$ (5)

${M}_{mg\to A}=mg\left(\frac{b}{2}-h\mathrm{tan}\beta \right)$ (6)

Figure 7. The relationship between the moment of gravity to A and the deflection angle

${N}_{B}=mg\left(\frac{1}{2}-\frac{h}{b}\mathrm{tan}\beta \right)-\frac{m{v}^{2}h\mathrm{tan}\alpha }{Lb\sqrt{4+{\mathrm{tan}}^{2}\alpha }}$ (7)

Figure 8. The relationship between the change in the moment of gravity and the deflection angle

Figure 9. The relationship between supporting force and deflection angle and wheel rotation angle

3. 货车防侧翻装置概念设计

3.1. 安全气阀

${z}_{1}+\frac{{p}_{1}}{\rho g}+\frac{{v}_{1}^{2}}{2g}={z}_{2}+\frac{{p}_{2}}{\rho g}+\frac{{v}_{2}^{2}}{2g}$ (8)

$v=\sqrt{\frac{2p}{\rho }}$ (9)

${F}_{i}={Q}_{m}v+PA$ (10)

Figure 10. Schematic diagram of the position of the jet reaction force

${F}_{T}=\sum {F}_{i}$ (11)

$M={F}_{T}H$ (12)

$Fh+{{N}^{\prime }}_{B}b-mg\left(\frac{b}{2}-h\mathrm{tan}\beta \right)-{F}_{T}H=0$ (13)

${N}_{B}=mg\left(\frac{1}{2}-\frac{h}{b}\mathrm{tan}\beta \right)-\frac{m{v}^{2}h\mathrm{tan}\alpha }{Lb\sqrt{4+{\mathrm{tan}}^{2}\alpha }}+\frac{{F}_{T}H}{b}$ (14)

3.2. 陀螺仪原理

Figure 11. Schematic diagram of gyroscope installation

$\tau ={I}_{Z}\stackrel{˙}{\theta }×\omega$ (15)

(a) 陀螺仪内外环 (b) 陀螺工作状态

Figure 12. (a) Schematic diagram of the gyroscope; (b) Schematic diagram of the gyro rotor

Figure 13. Schematic diagram of the rotation of the gyro rotor

Figure 14. Schematic diagram of the rotation of the gyro rotor disturbed by external forces

Figure 15. Torque offset diagram

${\tau }_{x}=2I\stackrel{˙}{\theta }\omega \mathrm{sin}\theta$ (16)

4. 结论

1) 在货车车速恒定的条件下：① 在车厢不发生偏转时，离心力的大小与转弯角度近似成线性关系，拟合结果为： $F=5.6245×{10}^{4}×\alpha -2.81063×{10}^{5}$，可以通过此公式大致判断车辆在行驶转弯时离心力的大小。② 在车轮不发生偏转，车厢偏转角度 $\beta \approx 32.5˚$ 时，重力对A的矩为 ${M}_{A}\approx \text{0}\text{\hspace{0.17em}}\text{N}\cdot \text{m}$，此时右侧车轮受地面支持力 ${N}_{B}\approx 0\text{\hspace{0.17em}}\text{N}$，这表明此时已经达到了货车侧翻的临界值。

2) 在货车行驶过程中，假设车厢相对于底盘不发生偏转，定义右侧车轮受到地面的支持力为 ${N}_{B}=0\text{\hspace{0.17em}}\text{N}$ (即达到货车侧翻的临界条件)时，车速越快，则车轮转角越小。这意味着货车在行驶中如果车速较快，即便转角很小也可能会发生侧翻。

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