Dynamic Modeling and Characteristic Analysis of Compliant Mechanisms With Inflection Point
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摘要:
为分析含拐点的大变形柔性梁的动态响应，建立含拐点的柔顺杆的动力学模型，基于拉格朗日方程推导其动力学方程组，在MATLAB软件中求解出其转角随时间变化的特性曲线，并与ADAMS软件仿真所得角位移曲线进行比较.研究表明：含拐点柔顺机构的拐点关节依靠各杆间的动力学耦合运动，拐点关节的转角变化很大，这体现了含拐点的柔顺机构动力学模型与不含拐点的柔顺机构动力学模型的区别，柔顺机构动力学模型能精确地反映柔顺机构的运动和变形.
Abstract:To analyze large deflection beams with inflection point in compliant mechanisms, a dynamic model of the compliant mechanisms with inflection point was established, and the kinetic equations were deduced based on the Lagrange equation. The curves of the angle of rotation were obtained in MATLAB software and compared with the simulation results of ADAMS software. Results show that the rotation angle of the joint at the inflection point varies significantly with the dynamic coupling between the rods. This reflects the difference between the dynamic model of the compliant mechanism with inflection point and the dynamic model of the compliant mechanism without inflection point. The dynamic model can reflect the complicated deformation and motion of the compliant mechanisms exactly.
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Keywords:
- compliant mechanisms /
- pseudo-rigid-body model /
- inflection point /
- dynamics
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由于柔性构件产生的弹性变形能传递和转换运动与力，柔顺机构可以实现传统机构的功能，并且因其具有免安装、无间隙和免润滑的特点，柔顺机构在精度、可靠性和成本方面具有更高的优势[1-2]，近年来，柔顺机构被广泛应用到MEMS产品、机器人微操作系统等许多前沿领域中，已经成为机构学研究和发展的新方向[3].
目前，在柔顺机构运动学和设计方面国内外学者所做的研究比较多.为了将刚性机构的理论应用到柔顺机构的研究中，Howell等[4]提出用扭簧链接的刚体杆件来模拟柔性杆件的变形，这就是伪刚体模型法，伪刚体模型法的提出加快了柔顺机构发展的进程. Edwards等[5]提出的1R伪刚体模型计算简单，能在一定的柔性角范围内保证模型的正确性和有效性.为提高伪刚体模型的精度，Su[6]和冯忠磊等[7]分别提出了3R伪刚体模型和2R伪刚体模型，3R伪刚体模型和2R伪刚体模型均比1R伪刚体模型的精确度高，但是3R伪刚体模型比2R伪刚体模型的参数计算复杂得多，因此2R伪刚体模型相比于另外2种伪刚体模型具有明显的优越性.在含拐点的大变形梁运动学研究方面，余跃庆等[8]提出了3种能模拟柔顺杆拐点和末端位置变化的伪刚体运动学模型，并对伪刚体模型和柔顺杆的拐点与末端位置进行误差分析.
柔顺机构动力学特性分析对机械系统的运动精度和是否能满足系统要求至关重要. Boyle等[9]应用伪刚体模型法对柔顺常力机构进行了动力学建模和分析，并对模型的可靠性进行了实验验证.同样基于伪刚体模型法，Scott等[10]分析了简单的平行导向柔顺四杆机构的动力学特性.除了伪刚体模型法，还可以用有限元法分析柔顺机构的动力学，Li等[11]基于此法分析了特征参数对柔顺机构位形精确度的影响.根据1R伪刚体运动学模型和等效原理，Yu等[12]建立了1R伪刚体动力学模型，另外还开展了平面柔顺机构动力学研究.李茜等[13]推导了2R和3R伪刚体动力学新模型的动力学方程，研究表明2R伪刚体动力学模型具有计算简单、精度高的特性.余跃庆等[14]对考虑了柔顺杆的轴向运动的PR和PRR伪刚体模型进行了动力学建模及相关研究，并将PR、PRR分别和1R、2R伪刚体动力学模型进行比较，得出各个模型的优缺点[14].
目前，对于含拐点的柔顺机构的研究并没有涉及动力学方面.为丰富相关理论研究，本文建立含拐点的柔顺机构伪刚体动力学模型，推导其动力学方程.然后应用MATLAB软件画出4条转角曲线，最后在ADAMS软件里进行仿真计算并与MATLAB求解的曲线对比.
1. RRHR动力学模型
图 1(a)为一般载荷情况下含拐点的柔顺杆，I为柔顺杆的拐点，图 1(b)为对应的四自由度RRHR伪刚体动力学模型，该伪刚体模型具有3个带扭簧的转动副和1个不带扭簧的铰链，“R”表示具有扭簧的转动副，“H”代表铰链(Hinge)，表示模型中没有扭簧的转动副.其中柔顺杆的杆长为l，扭簧常数分别为K，各刚性杆件0杆、1杆、2杆、3杆、4杆的杆长分别为li=γil(i=0, 1, 2, 3, 4)，质量分别为mi=γim(i=0, 1, 2, 3, 4)，γi为5个刚性杆件的特征半径系数，且γ0+γ1+γ2+γ3+γ4=1.特征半径系数和扭簧刚度系数的计算是伪刚体运动学建模的内容，而这部分内容已经在文献[8]中完成，本文在文献[8]建立的模型基础上进行动力学方程组的推导及动力学特性的分析，因此在本文中只给出了文献[8]所建立模型的特征参数的结果，特征参数的具体数值如表 1所示[8].
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图 1 含拐点的柔顺杆和RRHR伪刚体模型Figure 1. Compliant beam with inflection point and the RRHR PRBM表 1 RRHR伪刚体模型的特征参数Table 1. Characteristic parameters of RRHR PRBMγ0 γ1 γ2 γ3 γ4 Kθ1 Kθ2 Kθ4 0.047 0.294 0.235 0.317 0.107 6.106 3.229 3.835 1.1 动能分析
设图 1(b)中各刚性杆件质心点坐标分别为(xi, yi)(i=0, 1, 2, 3, 4)，则刚性杆件1杆、2杆、3杆、4杆的质心点位置可以表示为
$$ \left\{ \begin{array}{l} {x_1} = {\gamma _0}l + \frac{1}{2}{\gamma _1}l\cos {\theta _1}\\ {y_1} = \frac{1}{2}{\gamma _1}l\sin {\theta _1} \end{array} \right. $$ (1) $$ \left\{ \begin{array}{l} {x_2} = {\gamma _0}l + {\gamma _1}l\cos {\theta _1} + \frac{1}{2}{\gamma _2}l\cos \left( {{\theta _1} + {\theta _2}} \right)\\ {y_2} = {\gamma _1}l\sin {\theta _1} + \frac{1}{2}{\gamma _2}l\sin \left( {{\theta _1} + {\theta _2}} \right) \end{array} \right. $$ (2) $$ \left\{ \begin{array}{l} {x_3} = {\gamma _0}l + {\gamma _1}l\cos {\theta _1} + {\gamma _2}l\cos \left( {{\theta _1} + {\theta _2}} \right) + \\ \;\;\;\;\;\;\;\frac{1}{2}{\gamma _3}l\cos \left( {{\theta _1} + {\theta _2} + {\theta _3}} \right)\\ {y_3} = {\gamma _1}l\sin {\theta _1} + {\gamma _2}l\sin \left( {{\theta _1} + {\theta _2}} \right) + \\ \;\;\;\;\;\;\;\frac{1}{2}{\gamma _3}l\sin \left( {{\theta _1} + {\theta _2} + {\theta _3}} \right) \end{array} \right. $$ (3) $$ \left\{ \begin{array}{l} {x_4} = {\gamma _0}l + {\gamma _1}l\cos {\theta _1} + {\gamma _2}l\cos \left( {{\theta _1} + {\theta _2}} \right) + \\ \;\;\;\;\;\;\;{\gamma _3}l\cos \left( {{\theta _1} + {\theta _2} + {\theta _3}} \right) + \\ \;\;\;\;\;\;\;\frac{1}{2}{\gamma _4}l\cos \left( {{\theta _1} + {\theta _2} + {\theta _3} - {\theta _4}} \right)\\ {y_4} = {\gamma _1}l\sin {\theta _1} + {\gamma _2}l\sin \left( {{\theta _1} + {\theta _2}} \right) + \\ \;\;\;\;\;\;\;{\gamma _3}l\sin \left( {{\theta _1} + {\theta _2} + {\theta _3}} \right) + \\ \;\;\;\;\;\;\;\frac{1}{2}{\gamma _4}l\sin \left( {{\theta _1} + {\theta _2} + {\theta _3} - {\theta _4}} \right) \end{array} \right. $$ (4) 对方程式(1)~(4)分别求导得到各质心点的速度，进而可以求出各个杆的动能，系统的总动能等于1杆、2杆、3杆、4杆的动能之和，可以表示为
$$ \begin{array}{*{20}{c}} {T = {T_1} + {T_2} + {T_3} + {T_4} = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{J_1}\dot \theta _1^2 + \frac{1}{2}{m_2}v_2^2 + }\\ {\frac{1}{2}{J_2}{{\left( {{{\dot \theta }_1} + {{\dot \theta }_2}} \right)}^2} + \frac{1}{2}{m_3}v_3^2 + \frac{1}{2}{J_3}{{\left( {{{\dot \theta }_1} + {{\dot \theta }_2} + {{\dot \theta }_3}} \right)}^2} + }\\ {\frac{1}{2}{m_4}v_4^2 + \frac{1}{2}{J_4}{{\left( {{{\dot \theta }_1} + {{\dot \theta }_2} + {{\dot \theta }_3} - {{\dot \theta }_4}} \right)}^2}} \end{array} $$ (5) 1.2 变形能分析
当柔顺杆末端受载荷作用时，伪刚体模型各个杆之间的扭簧产生变形，扭簧常数可表示为
$$ {K_i} = {K_{{\theta _i}}}\frac{{EI}}{l} $$ (6) 所以柔顺杆的总变形能为
$$ U = \frac{1}{2}\sum\limits_i {{K_i}\theta _i^2} = \frac{{EI}}{{2l}}\left( {{K_{{\theta _1}}}\theta _1^2 + {K_{{\theta _2}}}\theta _2^2 + {K_{{\theta _4}}}\theta _4^2} \right) $$ (7) 1.3 动力学方程
由分析力学可知，拉格朗日方程式为
$$ \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {\frac{{\partial T}}{{\partial {{\dot \theta }_i}}}} \right) - \frac{{\partial T}}{{\partial {\theta _i}}} + \frac{{\partial U}}{{\partial {\theta _i}}} = 0,i = 1,2,3,4 $$ (8) 由于本文研究的是末端载荷释放后伪刚体模型的自由振动过程，此时系统没有外力作用，因此方程右端广义力为零.将式(5)(7)分别代入式(8)，可以得到RRHR伪刚体模型的动力学方程，系数写成矩阵的形式
$$ {\mathit{\boldsymbol{Q}}_1}\mathit{\boldsymbol{\ddot \theta }} + {\mathit{\boldsymbol{Q}}_2}{{\mathit{\boldsymbol{\dot \theta }}}^2} + {\mathit{\boldsymbol{Q}}_3}{{\mathit{\boldsymbol{\dot \theta }}}_i}{{\mathit{\boldsymbol{\dot \theta }}}_j} + {\mathit{\boldsymbol{Q}}_4}{\mathit{\boldsymbol{\theta }}_i} = 0 $$ (9) $$ {\mathit{\boldsymbol{Q}}_1} = \left[ {\begin{array}{*{20}{c}} {{D_{11}}}&{{D_{12}}}&{{D_{13}}}&{{D_{14}}}\\ {{D_{21}}}&{{D_{22}}}&{{D_{23}}}&{{D_{24}}}\\ {{D_{31}}}&{{D_{32}}}&{{D_{33}}}&{{D_{34}}}\\ {{D_{41}}}&{{D_{42}}}&{{D_{43}}}&{{D_{44}}} \end{array}} \right] $$ $$ {\mathit{\boldsymbol{Q}}_2} = \left[ {\begin{array}{*{20}{c}} {{D_{111}}}&{{D_{122}}}&{{D_{133}}}&{{D_{144}}}\\ {{D_{211}}}&{{D_{222}}}&{{D_{233}}}&{{D_{244}}}\\ {{D_{311}}}&{{D_{322}}}&{{D_{333}}}&{{D_{344}}}\\ {{D_{411}}}&{{D_{422}}}&{{D_{433}}}&{{D_{444}}} \end{array}} \right] $$ $$ {\mathit{\boldsymbol{Q}}_3} = \left[ {\begin{array}{*{20}{c}} {{D_{112}}}&{{D_{113}}}&{{D_{114}}}&{{D_{123}}}&{{D_{124}}}&{{D_{134}}}\\ {{D_{212}}}&{{D_{213}}}&{{D_{214}}}&{{D_{223}}}&{{D_{224}}}&{{D_{234}}}\\ {{D_{312}}}&{{D_{313}}}&{{D_{314}}}&{{D_{323}}}&{{D_{324}}}&{{D_{334}}}\\ {{D_{412}}}&{{D_{413}}}&{{D_{414}}}&{{D_{423}}}&{{D_{424}}}&{{D_{434}}} \end{array}} \right] $$ $$ {\mathit{\boldsymbol{Q}}_4} = \left[ {\begin{array}{*{20}{c}} {{D_1}}&{}&{}&{}\\ {}&{{D_2}}&{}&{}\\ {}&{}&{{D_3}}&{}\\ {}&{}&{}&{{D_4}} \end{array}} \right] $$ $$ \mathit{\boldsymbol{\ddot \theta = }}\left[ {\begin{array}{*{20}{c}} {{{\ddot \theta }_1}}\\ {{{\ddot \theta }_2}}\\ {{{\ddot \theta }_3}}\\ {{{\ddot \theta }_4}} \end{array}} \right] $$ $$ {{\mathit{\boldsymbol{\ddot \theta }}}^2}\mathit{\boldsymbol{ = }}\left[ {\begin{array}{*{20}{c}} {\ddot \theta _1^2}\\ {\ddot \theta _2^2}\\ {\ddot \theta _3^2}\\ {\ddot \theta _4^2} \end{array}} \right] $$ $$ {{\mathit{\boldsymbol{\dot \theta }}}_i}{{\mathit{\boldsymbol{\dot \theta }}}_j} = \left[ {\begin{array}{*{20}{c}} {{{\dot \theta }_1}{{\dot \theta }_2}}\\ {{{\dot \theta }_1}{{\dot \theta }_3}}\\ {{{\dot \theta }_1}{{\dot \theta }_4}}\\ {{{\dot \theta }_2}{{\dot \theta }_3}}\\ {{{\dot \theta }_2}{{\dot \theta }_4}}\\ {{{\dot \theta }_3}{{\dot \theta }_4}} \end{array}} \right] $$ $$ {\mathit{\boldsymbol{\theta }}_i} = \left[ {\begin{array}{*{20}{c}} {{\theta _1}}\\ {{\theta _2}}\\ {{\theta _3}}\\ {{\theta _4}} \end{array}} \right] $$ 式中各符号的具体表达式和物理意义分述如下：
1) 等效惯量Dii
$$ \begin{array}{*{20}{c}} {{D_{11}} = \left[ {\frac{1}{3}\gamma _1^3 + \frac{1}{3}\gamma _2^3 + \frac{1}{3}\gamma _3^3 + \frac{1}{3}\gamma _4^3 + } \right.}\\ {\gamma _1^2{\gamma _2} + \gamma _1^2{\gamma _3} + \gamma _1^2{\gamma _4} + \gamma _2^2{\gamma _3} + \gamma _2^2{\gamma _4} + \gamma _3^2{\gamma _4} + }\\ {{\gamma _1}\gamma _2^2\cos {\theta _2} + 2{\gamma _1}{\gamma _2}{\gamma _3}\cos {\theta _2} + 2{\gamma _1}{\gamma _2}{\gamma _4}\cos {\theta _2} + }\\ {{\gamma _2}\gamma _3^2\cos {\theta _3} + 2{\gamma _2}{\gamma _3}{\gamma _4}\cos {\theta _3} + {\gamma _3}\gamma _4^2\cos {\theta _4} + }\\ {{\gamma _1}\gamma _3^2\cos \left( {{\theta _2} + {\theta _3}} \right) + 2{\gamma _1}{\gamma _3}{\gamma _4}\cos \left( {{\theta _2} + {\theta _3}} \right) + }\\ {\left. {{\gamma _1}\gamma _4^2\cos \left( {{\theta _2} + {\theta _3} - {\theta _4}} \right) + {\gamma _2}\gamma _4^2\cos \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ \begin{array}{*{20}{c}} {{D_{22}} = \left[ {\frac{1}{3}\gamma _2^3 + \frac{1}{3}\gamma _3^3 + \frac{1}{3}\gamma _4^3 + \gamma _2^2{\gamma _3} + \gamma _2^2{\gamma _4} + \gamma _3^2{\gamma _4} + } \right.}\\ {{\gamma _2}\gamma _3^2\cos {\theta _3} + 2{\gamma _2}{\gamma _3}{\gamma _4}\cos {\theta _3} + {\gamma _3}\gamma _4^2\cos {\theta _4} + }\\ {\left. {{\gamma _2}\gamma _4^2\cos \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ {D_{33}} = \left( {\frac{1}{3}\gamma _3^3 + \frac{1}{3}\gamma _4^3 + \gamma _3^2{\gamma _4} + {\gamma _3}\gamma _4^2\cos {\theta _4}} \right)m{l^2} $$ $$ {D_{44}} = \frac{1}{3}\gamma _4^3m{l^2} $$ 式中：Dii表示刚性杆件i的等效惯量，刚性杆件i的加速度${{{\ddot{\theta }}}_{i}}$在刚性杆件i上作用一个惯性力矩${{D}_{ii}}{{{\ddot{\theta }}}_{i}}$.
2) 耦合惯量Dij(i≠j)
$$ \begin{array}{*{20}{c}} {{D_{12}} = {D_{21}} = \left[ {\frac{1}{3}\gamma _2^3 + \frac{1}{3}\gamma _3^3 + \frac{1}{3}\gamma _4^3 + \gamma _2^2{\gamma _3} + \gamma _2^2{\gamma _4} + \gamma _3^2{\gamma _4} + } \right.}\\ {\frac{1}{2}{\gamma _1}\gamma _2^2\cos {\theta _2} + {\gamma _1}{\gamma _2}{\gamma _3}\cos {\theta _2} + {\gamma _1}{\gamma _2}{\gamma _4}\cos {\theta _2} + }\\ {{\gamma _2}\gamma _3^2\cos {\theta _3} + 2{\gamma _2}{\gamma _3}{\gamma _4}\cos {\theta _3} + {\gamma _3}\gamma _4^2\cos {\theta _4} + }\\ {\frac{1}{2}{\gamma _1}\gamma _3^2\cos \left( {{\theta _2} + {\theta _3}} \right) + {\gamma _1}{\gamma _3}{\gamma _4}\cos \left( {{\theta _2} + {\theta _3}} \right) + }\\ {\left. {\frac{1}{2}{\gamma _1}\gamma _4^2\cos \left( {{\theta _2} + {\theta _3} - {\theta _4}} \right) + {\gamma _2}\gamma _4^2\cos \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ \begin{array}{*{20}{c}} {{D_{13}} = {D_{31}} = \left[ {\frac{1}{3}\gamma _3^3 + \frac{1}{3}\gamma _4^3 + \gamma _3^2{\gamma _4} + \frac{1}{2}{\gamma _2}\gamma _3^2\cos {\theta _3} + } \right.}\\ {{\gamma _2}{\gamma _3}{\gamma _4}\cos {\theta _3} + {\gamma _3}\gamma _4^2\cos {\theta _4} + \frac{1}{2}{\gamma _1}\gamma _3^2\cos \left( {{\theta _2} + {\theta _3}} \right) + }\\ {{\gamma _1}{\gamma _3}{\gamma _4}\cos \left( {{\theta _2} + {\theta _3}} \right) + \frac{1}{2}{\gamma _1}\gamma _4^2\cos \left( {{\theta _2} + {\theta _3} - {\theta _4}} \right) + }\\ {\left. {\frac{1}{2}{\gamma _2}\gamma _4^2\cos \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ \begin{array}{*{20}{c}} {{D_{14}} = {D_{41}} = \left[ { - \frac{1}{3}\gamma _4^3 - \frac{1}{2}{\gamma _3}\gamma _4^2\cos {\theta _4} - } \right.}\\ {\frac{1}{2}{\gamma _1}\gamma _4^2\cos \left( {{\theta _2} + {\theta _3} - {\theta _4}} \right) - }\\ {\left. {\frac{1}{2}{\gamma _2}\gamma _4^2\cos \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ \begin{array}{*{20}{c}} {{D_{23}} = {D_{32}} = \left[ {\frac{1}{3}\gamma _3^3 + \frac{1}{3}\gamma _4^3 + \gamma _3^2{\gamma _4} + \frac{1}{2}{\gamma _2}\gamma _3^2\cos {\theta _3} + } \right.}\\ {{\gamma _2}{\gamma _3}{\gamma _4}\cos {\theta _3} + {\gamma _3}\gamma _4^2\cos {\theta _4} + }\\ {\left. {\frac{1}{2}{\gamma _2}\gamma _4^2\cos \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ \begin{array}{*{20}{c}} {{D_{24}} = {D_{42}} = \left[ { - \frac{1}{3}\gamma _4^3 - \frac{1}{2}{\gamma _3}\gamma _4^2\cos {\theta _4} - } \right.}\\ {\left. {\frac{1}{2}{\gamma _2}\gamma _4^2\cos \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ {D_{34}} = {D_{43}} = \left[ { - \frac{1}{3}\gamma _4^3 - \frac{1}{2}{\gamma _3}\gamma _4^2\cos {\theta _4}} \right]m{l^2} $$ 式中：Dij(i≠j)表示刚性杆件i和j间的耦合惯量，刚性杆件i或j的加速度${{{\ddot{\theta }}}_{i}}$或${{{\ddot{\theta }}}_{j}}$在刚性杆件j或i上分别作用一个惯性力矩${{D}_{ij}}{{{\ddot{\theta }}}_{i}}$或${{D}_{ij}}{{{\ddot{\theta }}}_{j}}$.
3) 向心加速度系数Dijj
$$ \begin{array}{*{20}{c}} {{D_{111}} = 0,{D_{222}} = 0,{D_{333}} = 0,{D_{444}} = 0}\\ {{D_{122}} = - {D_{211}} = }\\ {\left[ {\left( { - \frac{1}{2}{\gamma _1}\gamma _2^2 - {\gamma _1}{\gamma _2}{\gamma _3} - {\gamma _1}{\gamma _2}{\gamma _4}} \right)\sin {\theta _2} + } \right.}\\ {\left( { - \frac{1}{2}{\gamma _1}\gamma _3^2 - {\gamma _1}{\gamma _2}{\gamma _4}} \right)\sin \left( {{\theta _2} + {\theta _3}} \right) - }\\ {\left. {\frac{1}{2}{\gamma _1}\gamma _4^2\sin \left( {{\theta _2} + {\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ \begin{array}{*{20}{c}} {{D_{133}} = - {D_{311}} = \left[ {\left( { - \frac{1}{2}{\gamma _2}\gamma _3^2 - {\gamma _2}{\gamma _3}{\gamma _4}} \right)\sin {\theta _3} + } \right.}\\ {\left( { - \frac{1}{2}{\gamma _1}\gamma _3^2 - {\gamma _1}{\gamma _3}{\gamma _4}} \right)\sin \left( {{\theta _2} + {\theta _3}} \right) - }\\ {\frac{1}{2}{\gamma _1}\gamma _4^2\sin \left( {{\theta _2} + {\theta _3} - {\theta _4}} \right) - }\\ {\left. {\frac{1}{2}{\gamma _2}\gamma _4^2\sin \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ \begin{array}{*{20}{c}} {{D_{144}} = {D_{411}} = }\\ {\left[ {\frac{1}{2}{\gamma _3}\gamma _4^2\sin {\theta _4} - \frac{1}{2}{\gamma _1}\gamma _4^2\sin \left( {{\theta _2} + {\theta _3} - {\theta _4}} \right) - } \right.}\\ {\left. {\frac{1}{2}{\gamma _2}\gamma _4^2\sin \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ \begin{array}{*{20}{c}} {{D_{233}} = - {D_{322}} = \left[ {\left( { - \frac{1}{2}{\gamma _2}\gamma _3^2 - {\gamma _2}{\gamma _3}{\gamma _4}} \right)\sin {\theta _3} - } \right.}\\ {\left. {\frac{1}{2}{\gamma _2}\gamma _4^2\sin \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ {D_{244}} = {D_{422}} = \left[ {\frac{1}{2}{\gamma _3}\gamma _4^2\sin {\theta _4} - \frac{1}{2}{\gamma _2}\gamma _4^2\sin \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2} $$ $$ {D_{344}} = {D_{433}} = \frac{1}{2}{\gamma _3}\gamma _4^2\sin {\theta _4}m{l^2} $$ 式中${{D}_{ijj}}\dot{\theta }_{j}^{2}$项表示刚性杆件j的速度${{{\dot{\theta }}}_{j}}$在刚性杆件i处的向心力.
4) 科氏加速度系数Dijk
$$ \begin{array}{*{20}{c}} {{D_{112}} = \left[ {\left( { - {\gamma _1}\gamma _2^2 - 2{\gamma _1}{\gamma _2}{\gamma _3} - 2{\gamma _1}{\gamma _2}{\gamma _4}} \right)\sin {\theta _2} + } \right.}\\ {\left( { - {\gamma _1}\gamma _3^2 - 2{\gamma _1}{\gamma _3}{\gamma _4}} \right)\sin \left( {{\theta _2} + {\theta _3}} \right) - }\\ {\left. {{\gamma _1}\gamma _4^2\sin \left( {{\theta _2} + {\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ \begin{array}{*{20}{c}} {{D_{113}} = {D_{123}} = \left[ {\left( { - {\gamma _2}\gamma _3^2 - 2{\gamma _2}{\gamma _3}{\gamma _4}} \right)\sin {\theta _3} + } \right.}\\ {\left( { - {\gamma _1}\gamma _3^2 - 2{\gamma _1}{\gamma _3}{\gamma _4}} \right)\sin \left( {{\theta _2} + {\theta _3}} \right) - }\\ {\left. {{\gamma _1}\gamma _4^2\sin \left( {{\theta _2} + {\theta _3} - {\theta _4}} \right) - {\gamma _2}\gamma _4^2\sin \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ \begin{array}{*{20}{c}} {{D_{114}} = {D_{124}} = {D_{134}} = \left[ { - {\gamma _3}\gamma _4^2\sin {\theta _4} + } \right.}\\ {{\gamma _1}\gamma _4^2\sin \left( {{\theta _2} + {\theta _3} - {\theta _4}} \right) + }\\ {\left. {{\gamma _2}\gamma _4^2\sin \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ {D_{212}} = {D_{313}} = {D_{323}} = {D_{414}} = {D_{424}} = {D_{434}} = 0 $$ $$ \begin{array}{*{20}{c}} {{D_{213}} = {D_{223}} = \left[ {\left( { - {\gamma _2}\gamma _3^2 - 2{\gamma _2}{\gamma _3}{\gamma _4}} \right)\sin {\theta _3} - } \right.}\\ {\left. {{\gamma _2}\gamma _4^2\sin \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ \begin{array}{*{20}{c}} {{D_{214}} = {D_{224}} = {D_{234}} = \left[ { - {\gamma _3}\gamma _4^2\sin {\theta _4} + } \right.}\\ {\left. {{\gamma _2}\gamma _4^2\sin \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ \begin{array}{*{20}{c}} {{D_{312}} = - {D_{213}} = \left[ {\left( {{\gamma _2}\gamma _3^2 + 2{\gamma _2}{\gamma _3}{\gamma _4}} \right)\sin {\theta _3} + } \right.}\\ {\left. {{\gamma _2}\gamma _4^2\sin \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2}} \end{array} $$ $$ {D_{314}} = {D_{324}} = {D_{334}} = - {\gamma _3}\gamma _4^2\sin {\theta _4}m{l^2} $$ $$ {D_{412}} = - {D_{214}} = \left[ {{\gamma _3}\gamma _4^2\sin {\theta _4} - {\gamma _2}\gamma _4^2\sin \left( {{\theta _3} - {\theta _4}} \right)} \right]m{l^2} $$ $$ {D_{413}} = {D_{423}} = - {D_{314}} = {\gamma _3}\gamma _4^2\sin {\theta _4}m{l^2} $$ 式中(${{D}_{ijk}}{{{\dot{\theta }}}_{j}}{{{\dot{\theta }}}_{k}}+{{D}_{ikj}}{{{\dot{\theta }}}_{k}}{{{\dot{\theta }}}_{j}}$)项表示刚性杆件j和k处的速度${{{\dot{\theta }}}_{j}}$和${{{\dot{\theta }}}_{k}}$作用于刚性杆件i的科氏力.
5) 变性能项Di
$$ {D_i} = \frac{{EI}}{l}{K_{{\theta _i}}},i = 1,2,4,{D_3} = 0 $$ 式中Di表示扭簧i处的变性能影响项.
由方程组的组成可以看出，RRHR伪刚体动力学模型是以θ1、θ2、θ3、θ4为变量的变系数二阶微分方程组. RRHR伪刚体动力学模型的刚性杆件之间存在向心力、科氏力和惯性力矩，并且除2、3杆之外其他杆件之间均存在扭簧的变形回复力，这体现了刚性杆件之间存在耦合作用，即反映了动力学模型模拟柔顺杆变形的有效性.由于拐点H副处没有扭簧，因此该模型属于欠驱动系统，其本质是二阶非完整动力学系统，依靠各杆间的动力学耦合实现运动，本文采用MATLAB软件对此非线性微分方程组进行数值求解.下面对含拐点的RRHR动力学模型动态响应进行分析.
2. 动态响应分析
设柔顺杆材料为聚丙烯，杆长l=300 mm，宽度b=5.841 mm，厚度h=1.12 mm，在MATLAB数学软件中求解RRHR伪刚体模型动力学方程组，得到模型的4条转角位移曲线，分别如图 2(a)、图 3(a)、图 4(a)、图 5(a)所示. ADAMS在多体系统动力学模型建模及求解方面功能十分强大，采用的建模方法是欧拉-拉格朗日方法，求解方法是刚性积分算法.本文在ADAMS刚体模型建模，仿真求解关节转角曲线，文中给出了ADAMS模型图，如图 6所示，由于篇幅所限，没有对ADAMS原理和计算方法展开表述.应用ADAMS软件进行仿真分析得出RRHR模型的转角位移曲线如图 2(b)、图 3(b)、图 4(b)、图 5(b)所示.
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图 2 RRHR伪刚体模型关节1转角变化曲线Figure 2. Joint 1 angular displacement curves of the RRHR PRBDM![]()
图 3 RRHR伪刚体模型关节2转角变化曲线Figure 3. Joint 2 angular displacement curves of the RRHR PRBDM![]()
图 4 RRHR伪刚体模型关节3转角变化曲线Figure 4. Joint 3 angular displacement curves of the RRHR PRBDM![]()
图 5 RRHR伪刚体模型关节4转角变化曲线Figure 5. Joint 4 angular displacement curves of the RRHR PRBDM对比图 2(a)和(b)、图 3(a)和(b)、图 4(a)和(b)、图 5(a)和(b)可知，4条位移曲线的变化趋势、周期和最大最小幅值基本一致，下面列出表格对理论计算结果和ADAMS仿真计算结果进行具体的误差分析，包括最大幅值、最小幅值、最大最小幅值差、平均值、均方差，结果如表 2~5所示.
表 2 θ1位移曲线的误差Table 2. Error of angular displacement curve θ1计算方法 最大幅值 最小幅值 最大最小幅值差 均方差 理论/rad 0.601 0 -0.615 1 1.216 1 0.141 1 仿真/rad 0.605 1 -0.616 2 1.221 3 0.139 2 误差/% -0.67 -0.19 -0.42 1.42 表 3 θ2位移曲线的误差Table 3. Error of angular displacement curve θ2计算方法 最大幅值 最小幅值 最大最小幅值差 均方差 理论/rad 0.577 5 -0.610 2 1.187 7 0.100 4 仿真/rad 0.577 5 -0.591 3 1.168 8 0.090 6 误差/% 0 3.19 1.61 10.83 表 4 θ3位移曲线的误差Table 4. Error of angular displacement curve θ3计算方法 最大幅值 最小幅值 最大最小幅值差 均方差 理论/rad 2.981 5 3.004 6 6.379 9 3.444 6 仿真/rad -3.398 4 -3.527 1 6.531 7 3.606 3 误差/% -0.77 -3.65 -2.32 -4.49 表 5 θ4位移曲线的误差Table 5. Error of angular displacement curve θ4计算方法 最大幅值 最小幅值 最大最小幅值差 均方差 理论/rad 0.533 9 -0.546 4 1.080 4 0.125 1 仿真/rad 0.547 2 -0.547 3 1.094 4 0.116 7 误差/% -2.42 -0.15 -1.29 7.24 由图 2~5可看出，θ1、θ2的振动频率较低，且θ2对θ1有明显的依附性，θ4则在做高频振动，另外可以看出θ1、θ2、θ3、θ4在大角度转动的同时还存在自身的局部振荡，这体现了柔顺杆的整体大幅度低频振荡以及柔顺杆的局部高频振动和变形. θ3角度的变化范围很大并且没有严格的规律性，这是由于2杆、3杆之间没有扭簧，3杆承受0杆、1杆, 1杆、2杆, 3杆、4杆之间扭簧的回复力耦合作用的结果. θ3角度变化的特殊性体现了具有单拐点大变形柔顺杆的变形特性和RRHR伪刚体模型的优越性，这是以前1R、2R、3R、PR、PRR伪刚体模型所不具备的.
根据RRHR伪刚体模型动力学方程得到该模型各时刻点的θ1、θ2、θ3、θ4取值，从而得到模型振动过程中的位形图，更加清楚地反映出柔顺杆的实际振荡的位置和形状变化.在ADAMS软件中对RRHR伪刚体模型进行仿真分析，可以得到不同时刻的RRHR模型所运动的位置.理论计算与ADAMS仿真计算对应t=0.22 s时刻的位形图如图 7(a)(b)所示，位形转角误差结果如表 6所示，由误差分析结果可以看出，基于动力学微分方程求出的RRHR模型的位形图与ADAMS软件仿真的结果吻合.
表 6 t=0.22 s时RRHR伪刚体模型转角误差Table 6. Angular error of dynamic calculation and simulation calculation at t=0.22 s计算方法 θ1 θ2 θ3 θ4 理论/rad 0.325 4 0.509 8 1.622 7 -0.104 9 仿真/rad 0.331 6 0.524 1 1.676 8 -0.099 1 误差/% -1.89 2.73 -0.323 5.87 3. 结论
本文根据柔顺杆四自由度RRHR运动学模型提出了含拐点的柔顺机构的RRHR动力学模型，推导并求解其动力学方程组，在MATLAB软件中求解出其转角随时间变化的特性曲线，并与ADAMS软件仿真所得到的角位移曲线进行了比较.该模型具有以下特性：
1) RRHR伪刚体动力学模型的方程是含有4个变量的变系数二阶微分方程组，由于2、3杆之间没有扭簧即刚度系数为零，第3个方程的系数矩阵欠秩.
2) RRHR伪刚体动力学模型杆3在振动过程中承受0杆、1杆, 1杆、2杆, 3杆、4杆之间扭簧回复力的耦合作用，因此杆3可以实现大角度摆动，并且2杆、3杆之间转角变化没有严格的周期性.
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图 1 含拐点的柔顺杆和RRHR伪刚体模型
Figure 1. Compliant beam with inflection point and the RRHR PRBM
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图 2 RRHR伪刚体模型关节1转角变化曲线
Figure 2. Joint 1 angular displacement curves of the RRHR PRBDM
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图 3 RRHR伪刚体模型关节2转角变化曲线
Figure 3. Joint 2 angular displacement curves of the RRHR PRBDM
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图 4 RRHR伪刚体模型关节3转角变化曲线
Figure 4. Joint 3 angular displacement curves of the RRHR PRBDM
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图 5 RRHR伪刚体模型关节4转角变化曲线
Figure 5. Joint 4 angular displacement curves of the RRHR PRBDM
表 1 RRHR伪刚体模型的特征参数
Table 1 Characteristic parameters of RRHR PRBM
γ0 γ1 γ2 γ3 γ4 Kθ1 Kθ2 Kθ4 0.047 0.294 0.235 0.317 0.107 6.106 3.229 3.835 
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表 2 θ1位移曲线的误差
Table 2 Error of angular displacement curve θ1
计算方法 最大幅值 最小幅值 最大最小幅值差 均方差 理论/rad 0.601 0 -0.615 1 1.216 1 0.141 1 仿真/rad 0.605 1 -0.616 2 1.221 3 0.139 2 误差/% -0.67 -0.19 -0.42 1.42
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表 3 θ2位移曲线的误差
Table 3 Error of angular displacement curve θ2
计算方法 最大幅值 最小幅值 最大最小幅值差 均方差 理论/rad 0.577 5 -0.610 2 1.187 7 0.100 4 仿真/rad 0.577 5 -0.591 3 1.168 8 0.090 6 误差/% 0 3.19 1.61 10.83
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表 4 θ3位移曲线的误差
Table 4 Error of angular displacement curve θ3
计算方法 最大幅值 最小幅值 最大最小幅值差 均方差 理论/rad 2.981 5 3.004 6 6.379 9 3.444 6 仿真/rad -3.398 4 -3.527 1 6.531 7 3.606 3 误差/% -0.77 -3.65 -2.32 -4.49
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表 5 θ4位移曲线的误差
Table 5 Error of angular displacement curve θ4
计算方法 最大幅值 最小幅值 最大最小幅值差 均方差 理论/rad 0.533 9 -0.546 4 1.080 4 0.125 1 仿真/rad 0.547 2 -0.547 3 1.094 4 0.116 7 误差/% -2.42 -0.15 -1.29 7.24
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表 6 t=0.22 s时RRHR伪刚体模型转角误差
Table 6 Angular error of dynamic calculation and simulation calculation at t=0.22 s
计算方法 θ1 θ2 θ3 θ4 理论/rad 0.325 4 0.509 8 1.622 7 -0.104 9 仿真/rad 0.331 6 0.524 1 1.676 8 -0.099 1 误差/% -1.89 2.73 -0.323 5.87
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