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 哈尔滨工程大学学报  2017, Vol. 38 Issue (8): 1303-1308  DOI: 10.11990/jheu.201607045 0

### 引用本文

YANG Bo, XIA Yan, LI He, et al. Dynamic characteristics analysis of TBM thrust system[J]. Journal of Harbin Engineering University, 2017, 38(8), 1303-1308. DOI: 10.11990/jheu.201607045.

### 文章历史

TBM推进系统动力学特性分析

Dynamic characteristics analysis of TBM thrust system
YANG Bo, XIA Yan, LI He, WEN Bangchun
School of Mechanical Engineering and Automation, Northeastern University, Shenyang 110819, China
Abstract: To research the response of the TBM's thrust system to initial conditions, a multiple-degree-of-freedom coupling dynamic model, which contains the main frame and a pair of grippers, was established considering the external stochastic excitations and variable stiffness between the gripper and the rock. Dynamic responses were calculated by using a numerical method and were analyzed in different initial conditions. The X and Y directions strongly influence the attitude of the thrust system. The deflection of the thrust system was up to 0.18 m when the initial angle was 0.01 m and up to 1.4° when the initial angle was 1°. This condition presented a great challenge to the adjustment of the angle of the TBM attitude and even caused the machine to deviate from the planned route and be unable to complete the excavation work. This research provides theoretical guidance for TBM attitude adjustment.
Key words: tunnel boring machine(TBM)    thrust system    variable stiffness    dynamic model    dynamic response    sensitivity

1 TBM推进系统动力学建模

 图 1 TBM部件结构图 Fig.1 The component diagram of TBM

 图 2 TBM动力学模型 Fig.2 The dynamic model of TBM

 $\left\{ \begin{array}{l} \mathit{\boldsymbol{M\ddot q}} + \mathit{\boldsymbol{C\dot q}} + \mathit{\boldsymbol{Kq}} = \mathit{\boldsymbol{F}}\\ \mathit{\boldsymbol{J\ddot \theta }} + \mathit{\boldsymbol{C\dot \theta }} + \mathit{\boldsymbol{K\theta }} = \mathit{\boldsymbol{T}} \end{array} \right.$ (1)

 $\begin{array}{*{20}{c}} {T = \frac{1}{2}{m_1}\dot x_1^2 + \frac{1}{2}{m_1}\dot z_1^2 + \frac{1}{2}{m_2}\dot x_2^2 + \frac{1}{2}{m_2}\dot z_2^2 + }\\ {\frac{1}{2}{m_3}\dot x_3^2 + \frac{1}{2}{m_3}\dot z_3^2 + \frac{1}{2}{J_1}\dot \theta _1^2 + \frac{1}{2}{J_2}\dot \theta _2^2 + \frac{1}{2}{J_3}\dot \theta _3^2} \end{array}$ (2)

 $\begin{array}{*{20}{c}} {U = \frac{1}{2}{k_{j12}}{{\left( {\frac{{{z_2} - {z_1}}}{{\sin {{30}^ \circ }}}} \right)}^2} + \frac{1}{2}{k_{j13}}{{\left( {\frac{{{z_3} - {z_1}}}{{\sin {{30}^ \circ }}}} \right)}^2} + }\\ {\frac{1}{2}{k_{12x}}\left( {{x_2} - {x_1}} \right)2 + \frac{1}{2}{k_{2dx}}x_2^2 + \frac{1}{2}{k_{2dz}}z_2^2 + }\\ {\frac{1}{2}{k_{13x}}\left( {{x_3} - {x_1}} \right)2 + \frac{1}{2}{k_{3dx}}x_3^2 + \frac{1}{2}{k_{3dz}}z_3^2 + }\\ {\frac{1}{2}{k_{12q}}{{\left( {{{\dot \theta }_2} - {{\dot \theta }_1}} \right)}^2} + \frac{1}{2}{k_{2rq}}\dot \theta _2^2 + }\\ {\frac{1}{2}{k_{13q}}{{\left( {{{\dot \theta }_3} - {{\dot \theta }_1}} \right)}^2} + \frac{1}{2}{k_{3rq}}\dot \theta _3^2} \end{array}$ (3)

 $\begin{array}{*{20}{c}} {D = \frac{1}{2}{C_{j12}}{{\left( {\frac{{{{\dot z}_2} - {{\dot z}_1}}}{{\sin {{30}^ \circ }}}} \right)}^2} + \frac{1}{2}{C_{j13}}{{\left( {\frac{{{{\dot z}_3} - {{\dot z}_1}}}{{\sin {{30}^ \circ }}}} \right)}^2} + }\\ {\frac{1}{2}{C_{2dx}}\dot z_2^2 + \frac{1}{2}{C_{3dx}}\dot z_3^2 + \frac{1}{2}{C_{12x}}{{\left( {{{\dot x}_2} - {{\dot x}_1}} \right)}^2} + }\\ {\frac{1}{2}{C_{2dx}}x_2^2 + \frac{1}{2}{C_{13x}}{{\left( {{{\dot x}_3} - {{\dot x}_1}} \right)}^2} + \frac{1}{2}{C_{3dx}}x_3^2 + }\\ {\frac{1}{2}{C_{12q}}{{\left( {{{\dot \theta }_2} - {{\dot \theta }_1}} \right)}^2} + \frac{1}{2}{C_{2rq}}\dot \theta _2^2 + }\\ {\frac{1}{2}{C_{13q}}{{\left( {{{\dot \theta }_3} - {{\dot \theta }_1}} \right)}^2} + \frac{1}{2}{C_{3rq}}\dot \theta _3^2} \end{array}$ (4)

 $\frac{{\rm{d}}}{{{\rm{d}}t}}\frac{{\partial T}}{{\partial {{\dot q}_i}}} - \frac{{\partial T}}{{\partial {q_i}}} + \frac{{\partial U}}{{\partial {q_i}}} + \frac{{\partial D}}{{\partial {{\dot q}_i}}} = {F_i}\left( t \right)$ (5)

 $\begin{array}{*{20}{c}} {{m_1}{{\ddot x}_1} + \left( {{k_{12x}} + {k_{13x}}} \right){x_1} - {k_{12x}}{x_2} - {k_{13x}}{x_3} + }\\ {\left( {{c_{12x}} + {c_{13x}}} \right){{\dot x}_1} - {c_{12x}}{{\dot x}_2} - {c_{13}}{{\dot x}_3} = {F_x}} \end{array}$ (6)
 $\begin{array}{*{20}{c}} {{m_2}{{\ddot x}_2} - {k_{12x}}{x_1} + \left( {{k_{2dx}} + {k_{12x}}} \right){x_2} - {c_{12x}}{{\dot x}_1} + }\\ {\left( {{c_{2dx}} + {c_{12x}}} \right){{\dot x}_2} = 0} \end{array}$ (7)
 $\begin{array}{*{20}{c}} {{m_3}{{\ddot x}_3} - {k_{13x}}{x_1} + \left( {{k_{13x}} + {k_{3dx}}} \right){x_3} - {c_{13x}}{{\dot x}_1} + }\\ {\left( {{c_{13x}} + {c_{3dx}}} \right){{\dot x}_3} = 0} \end{array}$ (8)
 $\begin{array}{*{20}{c}} {{m_1}{{\dot z}_1} + 4\left( {{k_{j12}} + {k_{j13}}} \right){z_1} - 4{k_{j12}}{z_2} - 4{k_{j13}}{z_3} + }\\ {4\left( {{c_{j12}} + {c_{j13}}} \right){{\dot z}_1} - 4{c_{j12}}{{\dot z}_2} - 4{c_{j13}}{{\dot z}_3} = {F_z}} \end{array}$ (9)
 $\begin{array}{*{20}{c}} {{m_2}{{\ddot z}_2} - 4{k_{j12}}{z_1} + \left( {{k_{j13}} + {k_{2dx}}} \right){z_2} - 4{c_{j12}}{{\dot z}_1} + }\\ {\left( {4{c_{j13}} + {c_{2dz}}} \right){{\dot z}_2} = 0} \end{array}$ (10)
 $\begin{array}{*{20}{c}} {{m_3}{{\ddot z}_3} - 4{k_{j13}}{z_1} + \left( {4{k_{j13}} + {k_{3dz}}} \right){z_3} - 4{c_{j13}}{{\dot z}_1} + }\\ {\left( {4{c_{j13}} + {c_{3dz}}} \right){{\dot z}_3} = 0} \end{array}$ (11)
 $\begin{array}{*{20}{c}} {{J_1}{{\ddot \theta }_1} + \left( {{k_{12q}} + {k_{13q}}} \right){\theta _1} - {k_{12q}}{\theta _2} - {k_{13q}}{\theta _3} + }\\ {\left( {{c_{12q}} + {c_{13q}}} \right){{\dot \theta }_1} - {c_{12q}}{{\dot \theta }_2} - {c_{13q}}{{\dot \theta }_3} = {T_y}} \end{array}$ (12)
 $\begin{array}{*{20}{c}} {{J_2}{{\ddot \theta }_2} - {k_{12q}}{\theta _1} + \left( {{k_{12q}} + {k_{2rq}}} \right){\theta _2} - {c_{12q}}{{\dot \theta }_1} + }\\ {\left( {{c_{12q}} + {c_{2rq}}} \right){{\dot \theta }_2} = 0} \end{array}$ (13)
 $\begin{array}{*{20}{c}} {{J_3}{{\ddot \theta }_3} - {k_{13q}}{\theta _1} + \left( {{k_{13q}} + {k_{3rq}}} \right){\theta _2} - {c_{13q}}{{\dot \theta }_1} + }\\ {\left( {{c_{13q}} + {c_{3rq}}} \right){{\dot \theta }_2} = 0} \end{array}$ (14)
2 动力学模型参数计算 2.1 外部激励计算

 图 3 滚刀布置图 Fig.3 Arrangement drawing of cutter

 图 4 正滚刀和边滚的受力 Fig.4 Forces acting on a normal cutter and a gauge cutter

 ${F_x} = {F_{vx\Sigma }} + {F_{rx\Sigma }} + {F_{sx\Sigma }}$ (15)
 ${F_z} = \sum\limits_{i = 1}^n {{F_{vi}}} + \sum\limits_{j = 1}^m {{F_{vj}}} + \sum\limits_{k = 1}^p {\left( {{F_{vk}} + {F_{sk}}\sin {\beta _k}} \right)}$ (16)
 ${T_y} = \sum\limits_{i = 1}^n {{F_{vi}}{l_{yi}}} + \sum\limits_{j = 1}^m {{F_{vj}}{l_{yj}}}$ (17)

2.2 撑靴接触面刚度计算

TBM掘进过程中，撑靴和岩石表面的摩擦力提供驱动力和扭转反向力。由于刀盘阻力和推进载荷具有较大波动，所以撑靴和岩石截面间的接触刚度特性对其动力学特性非常重要。

 $\begin{array}{*{20}{c}} {{k_x} = \frac{4}{{3\sqrt {2{\rm{\pi }}} }}E\frac{{\left( {4 - D} \right)\left( {D - 1} \right)}}{{\left( {3 - D} \right)\left( {2 - D} \right)}}\left( {\alpha _L^{'1/2} - } \right.}\\ {\left. {\alpha _L^{'\left( {D - 1} \right)/2}\alpha _c^{'\left( {2 - D} \right)/2}} \right)} \end{array}$ (18)
 $\left\{ \begin{array}{l} {k_z} = \frac{{8Gr'}}{{\left( {2 - v} \right)}}{\left[ {\mu \frac{{{\rm{d}}P}}{{{\rm{d}}Q}} + \left( {1 - \frac{{{\rm{d}}P}}{{{\rm{d}}Q}}} \right){{\left( {1 - \frac{Q}{{\mu P}}} \right)}^{ - 1/3}}} \right]^{ - 1}},\\ \;\;\;\;\;\;\;\;\;\;0 \le \frac{{{\rm{d}}P}}{{{\rm{d}}Q}} \le \frac{1}{\eta }\\ {k_z} = \frac{{8Gr'}}{{\left( {2 - v} \right)}},\;\;\;\;\;\frac{{{\rm{d}}P}}{{{\rm{d}}Q}} \ge \frac{1}{\eta } \end{array} \right.$ (19)

2.3 外液压缸刚度计算

 ${k_j} = E\left( {\frac{{A_1^2}}{{{V_{L1}} + {V_1}}} + \frac{{A_2^2}}{{{V_{L2}} + {V_2}}}} \right)$ (20)

2.4 阻尼计算

 $c = 2\xi \sqrt {{m_e}{k_e}}$ (21)

3 动力学模型求解

 ${{\dot x}_{t + \Delta t}} = {{\dot x}_t} + \left[ {\left( {1 - \delta } \right){{\ddot x}_t} + \delta {{\ddot x}_{t + \Delta t}}} \right]\Delta t$ (22)
 ${x_{t + \Delta t}} = {x_t} + {{\dot x}_t}\Delta t + \left[ {\left( {\frac{1}{2} - \beta } \right){{\ddot x}_t} + \beta {{\ddot x}_{t + \Delta t}}} \right]\Delta {t^2}$ (23)

 ${{\dot x}_{t + \Delta t}} = \frac{1}{{\beta \Delta {t^2}}}\left( {{x_{t + \Delta t}} - {x_t}} \right) - \frac{1}{{\beta \Delta {t^2}}}{{\dot x}_t} - \left( {\frac{1}{{2\beta }} - 1} \right)\Delta t{{\ddot x}_t}$ (24)
 ${{\ddot x}_{t + \Delta t}} = \frac{\delta }{{\beta \Delta t}}\left( {{x_{t + \Delta t}} - {x_t}} \right) + \left( {1 - \frac{\delta }{\beta }} \right){{\dot x}_t} + \left( {1 - \frac{\delta }{{2\beta }}} \right)\Delta t{{\ddot x}_t}$ (25)

 $M{{\ddot x}_{t + \Delta t}} + C{{\dot x}_{t + \Delta t}} + K{x_{t + \Delta t}} = {R_{t + \Delta t}}$ (26)

 $\dot K{x_{t + \Delta t}} = {{\dot R}_{t + \Delta t}}$ (27)

 $\dot K = K + \frac{\delta }{{\beta \Delta t}}C + \frac{1}{{\beta \Delta {t^2}}}M$ (28)
 $\begin{array}{l} {{\dot R}_{t + \Delta t}} = {R_{t + \Delta t}} + M\left[ {\frac{1}{{\beta \Delta {t^2}}}{x_t} + \frac{\delta }{{\beta \Delta t}}{{\dot x}_t} + \left( {\frac{1}{{2\beta }} - 1} \right){{\ddot x}_t}} \right] + \\ \;\;\;\;\;\;\;\;\;\;\;C\left[ {\frac{\delta }{{\beta \Delta t}}{x_t} + \left( {\frac{\delta }{\beta } - 1} \right){{\dot x}_t} + \left( {\frac{\delta }{{2\beta }} - 1} \right)\Delta t{{\ddot x}_t}} \right] \end{array}$ (29)
4 动力学模型的实际案例适用分析 4.1 动力学响应求解

 图 5 时程载荷图 Fig.5 Loads times history

 图 6 系统的响应 Fig.6 Dynamic responses of system
4.2 变初始条件下敏捷度分析

 $\left\{ \begin{array}{l} {v_{11}}\left( t \right) = \frac{{{x_1}\left( t \right)}}{{{x_0}\left( t \right)}},{v_{12}}\left( t \right) = \frac{{{x_2}\left( t \right)}}{{{x_0}\left( t \right)}}\\ {v_{21}}\left( t \right) = \frac{{{z_1}\left( t \right)}}{{{z_0}\left( t \right)}},{v_{22}}\left( t \right) = \frac{{{z_2}\left( t \right)}}{{{z_0}\left( t \right)}}\\ {v_{31}}\left( t \right) = \frac{{{\theta _1}\left( t \right)}}{{{\theta _0}\left( t \right)}},{v_{32}}\left( t \right) = \frac{{{\theta _2}\left( t \right)}}{{{\theta _0}\left( t \right)}} \end{array} \right.$ (30)

 图 7 系统的敏感度时程图 Fig.7 Times history chart of sensitivity
5 结论

1）推进系统的稳态响应具有明显的周期性，并且Z方向响应周期短于摆动方向的周期。

2）主机在X方向和绕Y方向的偏航会导致刀盘波动幅值的较大变化，从而导致掘进过程出现较大误差。

3）撑靴在初始条件下的偏航位移和偏转角度对整机的影响较小，整机的角度调节仍要以调整两侧推进液压缸为主。

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