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由于柔性铰链的转动柔度远大于平移柔度,其优化设计以转动柔度为研究对象,以下主要对力矩作用下三个轴的转动角位移进行推导。如图1所示,假设椭圆弧柔性铰链一端固定,一端受x,y,z三轴方向力矩,其中:l为铰链长度;w为铰链宽度。
柔性铰链的正视图如图2所示,其中:a为长半轴长;b为短半轴长;t0为铰链最薄处厚度,设椭圆的离心角为θ, θ值范围为
$\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ ,则有:$h\left( \theta \right) = 2b - $ $ 2b\cos (\theta ) + {t_0}$ 。令$k = \dfrac{b}{{{t_0}}}$ ,则有:$$g(\theta ) = \frac{{h(\theta )}}{{{t_0}}} = 2k - 2k\cos (\theta ) + 1$$ (1) 以下推导中:E为弹性模量;G为剪切模量;u为泊松比;Iz, Iy分别为铰链在z,y轴的惯性矩。
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利用积分思想,当椭圆弧柔性铰链一端受z轴方向力矩
${M_Z}$ ,产生${\delta _{\textit{z}}}$ 角位移时,该方向的转动柔度为:$$\begin{split} {C_{{\textit{z}} - {M_{\textit{z}}}}} = &\dfrac{{{\delta _{\textit{z}}}}}{{{M_{\textit{z}}}}} = \int_0^{2a} {\dfrac{1}{{E{I_{\textit{z}}}(x)}}} {\rm{d}}x = \\ & \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\dfrac{{a\cos (\theta )}}{{E{I_{\textit{z}}}(\theta )}}} {\rm{d}}\theta = \dfrac{{12a}}{{Ew{t_0}^3}}\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\dfrac{{\cos (\theta )}}{{g{{(\theta )}^3}}}} {\rm{d}}\theta \end{split} $$ (2) 其中
$$ \begin{split} \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\cos (\theta )}}{{g{{(\theta )}^3}}}} {\rm{d}}\theta = \frac{{12{k^2} + 8k + 2}}{{{{(4k + 1)}^2}(2k + 1)}} + \frac{{12k(2k + 1){\arctan} {{(4k + 1)}^{0.5}}}}{{{{(4k + 1)}^{2.5}}}} \\ \end{split} $$ (3) 同理,将惯性矩中的y轴与z轴互换,得到y轴转动柔度如下:
$$\begin{split} {C_{y - {M_y}}} = &\frac{{{\delta _y}}}{{{M_y}}} = \int_0^{2a} {\frac{1}{{E{I_y}(x)}}} {\rm{d}}x = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{a\cos (\theta )}}{{E{I_y}(\theta )}}} {\rm{d}}\theta = \\ & \frac{{12a}}{{E{w^3}{t_0}}}\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\cos (\theta )}}{{g(\theta )}}} {\rm{d}}\theta \\ \end{split} $$ (4) 其中
$$ \begin{split} \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\cos (\theta )}}{{g(\theta )}}} {\rm{d}}\theta = \frac{{(8k + 4){\arctan} {{(4k + 1)}^{0.5}} - \pi {{(4k + 1)}^{0.5}}}}{{2k{{(4k + 1)}^{0.5}}}} \\ \end{split} $$ (5) -
由于矩形截面铰链发生扭转时横截面边线不再保持平直,使其关于扭转角的积分过于复杂。由卡式第二定理,当柔性铰链自由度端受x轴力矩
${M_X}$ , x轴转动柔度为:$${C_{x - {M_x}}} = \frac{{{\delta _x}}}{{{M_x}}} = \int_0^{2a} {\frac{1}{{G{I_j}(x)}}} {\rm{d}}x = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{a\cos (\theta )}}{{G{I_j}(\theta )}}} {\rm{d}}\theta $$ (6) 其中,
${I_j}(\theta )$ 与θ处厚度$h(\theta )$ 和宽度w有关,当$w \geqslant h(\theta )$ 时,由参考文献[9]给出的${I_j}(\theta )$ 近似公式如下:$${I_j}(\theta ) = wh{(\theta )^3}{I_0}\left(\frac{{h(\theta )}}{w}\right)$$ (7) $${I_0}\left(\frac{{h(\theta )}}{w}\right) = \frac{1}{3} - 0.21\frac{{h(\theta )}}{w}\left[ {1 - {{\left( {\frac{{h(\theta )}}{w}} \right)}^4}} \right]$$ (8) 由于公式(8)为五次多项式,后期积分过于复杂,为方便工程计算,使用最小二乘法对其局部进行近似:令
$\dfrac{1}{{{I_0}}}$ 为纵坐标,令$\dfrac{{h(\theta )}}{w}$ 为s,得到图3。由图3可知,s在[0,1]段变化近似于一元三次方程,因此以一元三次方程为基底,通过最小二乘法对其进行近似,近似后的决定系数为0.9999,近似结果如下:
$$\frac{1}{{{I_0}'}} = 1.387{s^3} + 0.9426{s^2} + 1.923s + 2.999$$ (9) 将近似结果代入公式(6)和(7),则当
$w \geqslant h(\theta )$ 时,x轴转动柔度表达式为: $$ \begin{split} {C_{X - {M_X}}} =& \frac{1}{G}\left(\frac{{1.387}}{{{w^4}}}{f_0} + \frac{{0.9426}}{{{w^3}}}{f_1} + \frac{{1.923}}{{{w^2}}}{f_2} + \frac{{2.999}}{w}{f_3}\right) \\ \end{split} $$ (10) 当
$w < h(\theta )$ 时,$w$ 与$h(\theta )$ 互换,所得x轴转动柔度表达式为:$$ \begin{split} {C_{X - {M_X}}} = \frac{1}{G}\left(1.387{f_4} + \frac{{0.9426}}{w}{f_3} + \frac{{1.923}}{{{w^2}}}{f_2} + \frac{{2.999}}{{{w^3}}}{f_1}\right) \\ \end{split} $$ (11) 其中
$$\left\{ {\begin{array}{*{20}{c}} {{f_0} = 2 \displaystyle\int_0^{\frac{\pi }{2}} {a\cos (\theta ){\rm{d}}\theta } } \\ {{f_1} = 2 \displaystyle\int_0^{\frac{\pi }{2}} {\dfrac{{a\cos (\theta )}}{{h(\theta )}}{\rm{d}}\theta } } \\ {{f_2} = 2 \displaystyle\int_0^{\frac{\pi }{2}} {\dfrac{{a\cos (\theta )}}{{h{{(\theta )}^2}}}{\rm{d}}\theta } } \\ \begin{gathered} {f_3} = 2 \displaystyle\int_0^{\frac{\pi }{2}} {\dfrac{{a\cos (\theta )}}{{h{{(\theta )}^3}}}{\rm{d}}\theta } \\ {f_4} = 2 \displaystyle\int_0^{\frac{\pi }{2}} {\dfrac{{a\cos (\theta )}}{{h{{(\theta )}^4}}}{\rm{d}}\theta } \\ \end{gathered} \end{array}} \right.$$ (12) 运算时还应注意:当
${t_0} < w \leqslant 2b + {t_0}$ 时,需将求解区间$\left[0,\dfrac{\pi }{2}\right]$ 分为$\left[0,{\theta _w}\right]$ 和$\left[{\theta _w},\dfrac{\pi }{2}\right]$ 两段,$\Bigg(h({\theta _w}) = w, $ $ {\theta _w} = {\rm{arccos}} \left(1 + \dfrac{{{t_0}}}{{2b}} - \dfrac{w}{{2b}}\right)\Bigg)$ ,再分别将两段区间代入公式(10)和(11)求解再求和;当$w \leqslant {t_0}$ 或$w > 2b + {t_0}$ 时,则直接将区间$\left[0,\dfrac{\pi }{2}\right]$ 代入公式(10)或(11)中求解。为方便求不定积分,令
$n = \dfrac{{{t_0}}}{{2b}}$ ,则公式(12)中的不定积分为:$$\int {\frac{{a\cos (\theta )}}{{h(\theta )}}{\rm{d}}\theta = \frac{a}{b}} \left( {\frac{{2(1{{ + n}}){\arctan} \left[ {{{\left( {\dfrac{{n + 2}}{n}} \right)}^{0.5}}\tan \left( {\dfrac{\theta }{2}} \right)} \right]}}{{{{({n^2} + 2n)}^{0.5}}}} - \theta } \right)$$ (13) $$\int {\dfrac{{a\cos (\theta )}}{{h{{(\theta )}^2}}}{\rm{d}}\theta = \dfrac{a}{{2{b^2}}}} \left( {\dfrac{{2{\arctan} \left[ {{{\left( {\dfrac{{n + 2}}{n}} \right)}^{0.5}}\tan \left( {\dfrac{\theta }{2}} \right)} \right]}}{{{{({n^2} + 2n)}^{1.5}}}} + \dfrac{{(1 + n)\sin (\theta )}}{{n(n + 2)(1 + n - \cos (\theta ))}}} \right)$$ (14) $$\int {\dfrac{{a\cos (\theta )}}{{h{{(\theta )}^3}}}{\rm{d}}\theta =\! \dfrac{a}{{4{b^3}}}} \left( {\dfrac{{2(1{{ \!+\! n}}){\arctan} \left[ {{{\left( {\dfrac{{n + 2}}{n}} \right)}^{0.5}}\tan \left( {\dfrac{\theta }{2}} \right)} \right]}}{{{{({n^2} + 2n)}^{2.5}}}} + \dfrac{{{{(\cos (\theta ) +\! 1)}^2}\tan \left(\dfrac{\theta }{2}\right)}}{{4{n^2}{{(n + 2)}^2}{{(1 +\! n -\! \cos (\theta ))}^2}}}(2{n^3} \!+\! 7{n^2} \!+\! 9n + 6){{\tan }^2}\left(\dfrac{\theta }{2}\right) + 2{n^3} + 5{n^2}\! +\! 5n} \right)$$ (15) $$ \begin{split} \int \dfrac{{a\cos (\theta )}}{{h{{(\theta )}^4}}}{\rm{d}}\theta =& \dfrac{a}{{8{b^4}}}\left[ {\dfrac{{(4{n^2} + 8n + 5){\arctan} \left[ {{{\left( {\dfrac{{n + 2}}{n}} \right)}^{0.5}}\tan \left( {\dfrac{\theta }{2}} \right)} \right]}}{{{{({n^2} + 2n)}^{3.5}}}} + } \right.\dfrac{{{{(\cos (\theta ) + 1)}^3}}}{{8{{(1 + n - \cos (\theta ))}^3}}}\left[ {\dfrac{{(2{n^3} + 8{n^2} + 16n + 11)\tan \left( {\dfrac{\theta }{2}} \right)}}{{n{{(n + 2)}^3}}} + } \right.\\ &\left. {\left. {\dfrac{{(12{n^3} + 36{n^2} + 64n + 40){{\tan }^3}\left( {\dfrac{\theta }{2}} \right)}}{{3{n^2}{{(n + 2)}^2}}} + \dfrac{{\left( {2{n^3} + 4{n^2} + 8n + 5} \right){{\tan }^5}\left( {\dfrac{\theta }{2}} \right)}}{{{n^3}(n + 2)}}} \right]} \right] \end{split} $$ (16) -
轴飘是衡量柔性铰链精度的一项重要指标。目前主要以椭圆弧柔性铰链z轴受单位力矩
${M_z}$ 时,中心点沿y轴的位移作为衡量轴飘的标准。中心点线位移可以用角位移的积分形式表示,中心点y轴位移${y_c}$ 为:$${y_c} = \int_{ - \frac{\pi }{2}}^0 {\left[ {\int_{ - \frac{\pi }{2}}^\varphi {\frac{{12a\cos (\theta )}}{{Ew{h^3}(\theta )}}{\rm{d}}\theta } } \right]} {\rm{d}}\varphi = \frac{{6{a^2}}}{{Ew{t_0}^3(1 + 2k)}}$$ (17) -
为验证上述柔度公式的正确性,选用表1中结构参数分别计算有限元解和方程解析解进行对比。其中:弹性模量
$E$ 为109 GPa;泊松比$u$ 为0.34;剪切模量G为40.67 GPa ;a为0.01 m;b为0.005 m;EQU为解析解;FEM为有限元解。表 1 解析解与有限元解对比
Table 1. Comparison of analytical solution and finite element solution
t0/mm w/mm Cz-Mz/rad·Nm−1 Cy-My/rad·Nm−1 Cx-Mx/rad·Nm−1 yc/m EQU 1 5 0.1130 8.699×10−3 7.572×10−2 1.000×10−7 FEM 1 5 0.1106 8.989×10−3 7.380×10−2 9.993×10−8 EQU 0.5 5 0.6479 1.360×10−2 0.4095 4.194×10−7 FEM 0.5 5 0.6419 1.403×10−2 0.3996 4.179×10−7 EQU 1 3 0.1884 4.028×10−2 0.1426 1.668×10−7 FEM 1 3 0.1874 4.141×10−2 0.1329 1.677×10−7 EQU 0.2 3 10.80 0.1089 4.508 4.501×10−6 FEM 0.2 3 10.51 0.1098 4.790 4.487×10−6 由表1可知,Cz-Mz,Cy-My和yc解析解与有限元解相对误差小于3.3%;Cx-Mx解析解与有限元解相对误差小于6.8%,验证了所推导柔度公式的正确性。
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某快速反射镜柔性铰链为避免共振,要求柔性铰链的基频高于控制系统伺服带宽fc(200 Hz)。对于工作轴向谐振可用伺服控制抑制,而非工作轴向谐振只能通过提高刚度的方式。快速反射镜分配到该柔性铰链的转动惯量如表2所示。
表 2 转动惯量
Table 2. Moment of inertia
Jθx/kg·m2 Jθy/kg·m2 Jθz/ kg·m2 3.46×10−6 3.46×10−6 6.33×10−6 为避免共振现象,一般要求结构谐振频率高于系统伺服带宽的
$\sqrt 2 $ 倍,三个转动轴谐振频率要求为:$${f_{x,y}} \geqslant \sqrt 2 {f_c} = 282\;{\rm{Hz}}$$ (18) $${f_z} \leqslant {f_c} = 200\;{\rm{Hz}}$$ (19) 由参考文献[10]中的快速反射镜刚度与系统谐振近似公式
${f_i} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{{K_\theta }_i}}{{{J_\theta }_i}}} $ ,可得各个方向柔度要求如公式(20)所示:$$\left\{ {\begin{array}{*{20}{c}} {{C_{Y - {M_Y}}},{C_{X - {M_X}}} \leqslant 9.21 \times {{10}^{ - 2}}\;{\rm{rad}}/{\rm{Nm}}} \\ {{C_{Z - {M_Z}}} \geqslant 0.100\;{\rm{rad}}/{\rm{Nm}}} \end{array}} \right.$$ (20) 该柔性铰链的优化函数如下:
$$\left\{ \begin{gathered} {\rm{Object}}:\max ({C_{Z - {M_Z}}}),\min ({C_{Y - {M_Y}}}),\min ({C_{X - {M_X}}}) \\ {\rm{s.t}}\begin{array}{*{20}{c}} :&\begin{gathered} 0.005\;{\rm{m}} \leqslant a \leqslant 0.01\;{\rm{m}} \\ 0.001\;{\rm{m}} \leqslant b \leqslant 0.01\;{\rm{m}} \\ 0.000\;1\;{\rm{m}} \leqslant {t_0} \leqslant 0.005\;{\rm{m}} \\ 0.005\;{\rm{m}} \leqslant w \leqslant 0.01\;{\rm{m}} \\ {y_c} \leqslant 0.000\;001\;{\rm{m}} \\ {\delta _{\max }} \leqslant \frac{{4.3 \times {{10}^8}\;{\rm{Pa}}}}{2} \\ \end{gathered} \end{array} \\ \end{gathered} \right.$$ (21) 在进行柔性铰链设计时,除了尺寸约束和中心点位移约束,还应对铰链中心处进行强度校核,以保障工作时铰链不被破坏。强度校核公式如下:
$${\delta _{\max }} = \frac{{{M_{\max }}{t_0}}}{{2{I_z}(a)}} = \leqslant \frac{{{\delta _s}}}{{{n_s}}}$$ (22) 式中:
${\delta _{\max }}$ 为铰链中心处弯曲正应力;${\delta _s}$ 为屈服强度;${n_s}$ 为安全系数;${I_z}(a)$ 为铰链中心处的z轴惯性距;${M_{\max }}$ 为铰链所受的最大弯矩。该柔性铰链材料为钛合金TC4,屈服强度${\delta _s}$ 为$4.3 \times {10^8}\;{\rm{Pa}}$ ,安全系数${n_s}$ 为2,最大弯矩${M_{\max }}$ 为0.2 Nm。 -
由于目前柔性铰链优化设计所使用的算法,如遗传算法、退火算法、粒子群算法等对多目标优化存在过早收敛、求解效率低和Pareto解分布不均等问题。为保障三轴柔度同时优化的效率和准确性,基于NSGA-II算法对椭圆弧柔性铰链进行三轴柔度优化设计,流程如图4所示。
优化问题的相关性分析是明确设计变量和目标函数之间关系,确定初始变量采样密度和提高优化效率的重要过程。基于拉丁超立方采样法[13]在设计变量区域内选取100个点,对三个轴向柔度进行相关性分析[14],所得结果如图5所示。由图5分析可知,结构参数与轴向柔度成负相关,其中t0和w的相关因子较大,a和b的相关因子较小。由于在NSGA-II算法中每个个体由30位二进制编码组成,根据相关性分析结果设置结构参数在个体中所占二进制位数比例,这里a,b的所占比例为0.1;t0和w的所占比例为0.4。
NSGA-II算法与普通优化算法的主要区别在于采用了非支配分层方法,即通过比较个体间的支配关系(若个体Xi在所有优化目标上都优于个体Xj,则称Xi支配Xj),将整个种群进行分层。每次迭代后,以支配层由高到低的顺序调用个体生成新种群。如若新种群个体数与所设个体数N之差小于当前支配层个体数,则采用拥挤度比较的方式从当前个体层中选取较优个体进行填补。每次重新迭代开始前,将父代种群和当前种群组合形成2N大小的新种群。这样有利于保证某些优良的个体在进化过程中不被丢弃。
拥挤度比较是解决同一支配层拥有相同适应值所致的解集单一性的重要方法。其过程为:首先,针对每一个优化目标,将个体进行排序,令边界的两个个体拥挤度为最大值,之后,其他个体i的拥挤度
${c_i}$ 的计算公式为:$${c_i} = \sum\limits_{j = 1}^m {\left(\left| {f_j^{i + 1} - f_j^{i - 1}} \right|\right)} $$ (23) 式中:
$f_j^{i + 1}$ 为i+1点的第j个优化目标值;$f_j^{i - 1}$ 为i-1点的第j个优化目标值;m为优化目标数。多目标优化算法所得结果一般为Pareto解集形式,而工程应用中往往只需要一个最优解,对此给每一个优化目标添加权值,使用相对排序法在Pareto解集内寻找最优解。设z,x,y轴转动柔度的权值为αz,αx,αy,解p在Pareto解集内每个方向转动柔度从小到大的排列顺序为
${R_{{\textit{z}} - {M_{\textit{z}}}}}(p)$ ,${R_{x - {M_x}}}(p)$ ,${R_{y - {M_y}}}(p)$ ,则解p在Pareto解集内的相对排序R(P)为:$$R(p) = - {\alpha _{\textit{z}}}{R_{{\textit{z}} - {M_{\textit{z}}}}}(p) + {\alpha _x}{R_{x - {M_x}}}(p) + {\alpha _y}{R_{y - {M_y}}}(p)$$ (24) 该Pareto解集内的最优解为相对排序R最小的解。
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设置NSGA-II算法种群数都为20,循环20次,迭代400次之后优化结束,以目标权值(
${\alpha _z} = 0.6, $ $ {\alpha _x} = 0.2,{\alpha _y} = 0.2$ )获取最优解。将依据经验设计的初始值与经NSGA-II算法优化后的结果对比,如表3所示。表 3 优化先后对比
Table 3. Comparison of before optimization and after optimization
a/mm b/mm t0/mm w/mm Cz-Mz/rad·Nm−1 Cy-My/rad·Nm−1 Cx-Mx/rad·Nm−1 Initial value 9 5 0.5 6 0.4859 7.079×10−3 0.2966 Optimized value 9.67 6.63 0.661 9.74 0.1390 1.343×10−3 8.870×10−2 分析可知,初始结构Cz-Mz虽然较高,但Cx-Mx不满足公式(20)的要求,经优化之后满足要求。
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为验证NSGA-II算法的优化效果,分别使用多岛遗传算法(MIGA)、粒子群算法(PSO)优化与上述NSGA-II算法的优化结果对比。其中,多岛遗传算法的种群数为10,共有四个岛,进化10次,共迭代400次;粒子群算法种群数为20,循环20次,共迭代400次,三种算法所得Pareto前沿如图6所示。以目标权值(
${\alpha _z} = 0.6,{\alpha _x} = 0.2,{\alpha _y} = 0.2$ )获取每种算法Pareto解集的最优解,再以公式(24)算出三个最优解的相对排序RC如表4所示。表 4 不同算法最优解对比
Table 4. Comparison of optimal solutions of different algorithms
Algorithm Cz-Mz/rad·Nm−1 Cy-My/rad·Nm−1 Cx-Mx/rad·Nm−1 RC Pareto solution number Pareto solution efficiency MIGA 0.1320 1.329×10−2 6.999×10−2 −0.6 153 38.2% PSO 0.1301 4.630×10−2 7.075×10−2 0.4 110 27.5% NSGA-II 0.1390 1.343×10−2 8.870×10−2 −0.8 210 52.5% 由图6和表4对比可知,在Pareto前沿分布上, NSGA-II算法Pareto解的分布相比多岛遗传算法和粒子群算法更为集中,同时求解效率较多岛遗传算法提高了14.3%,较粒子群算法提高了25%;在优化的结果上NSGA-II算法给出的最优解相对排序RC在三种算法中最低,质量最好。综上,NSGA-II算法可以有效地提高柔性铰链多目标优化效率和优化质量。
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为验证最优解的正确性,使用有限元法对NSGA-II算法给出的最优解进行验证,结果如表5所示。
表 5 最优解与有限元解对比
Table 5. Comparison of optimal solution and finite element solution
Cz-Mz/rad·Nm−1 Cy-My/rad·Nm−1 Cx-Mx/rad·Nm−1 EQU 0.1390 1.343×10−3 8.870×10−2 FEM 0.1320 1.340×10−3 8.290×10−2 由表5可知,Cz-Mz,Cy-My与有限元解误差小于5%,Cx-Mx与有限元解误差小于6.5%,证明了最优解的正确性。图7为柔性铰链最优解的有限元分析结果。
Multi-objective optimal design of elliptic flexible hinge in fast steering mirror
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摘要: 在快速反射镜柔性铰链优化设计中,为保障反射镜精度和稳定性,应尽量提高柔性铰链工作轴转动柔度和非工作轴转动刚度。以某快速反射镜的椭圆弧柔性铰链为研究对象,首先利用最小二乘法和积分思想推导了三个转动轴的转动柔度公式,与有限元法对比,二者最大相对误差小于6.8%,解决了矩形截面扭转柔度积分过于复杂的问题;其次基于改进的非支配排序遗传算法(NSGA-II)对三个转动轴进行了多目标优化设计,达到设计指标,所得最优解和求解效率较传统算法有了明显提高,其中Pareto解求解效率较多岛遗传算法(MIGA)提高了14.3%,较粒子群算法(PSO)提高了25%;最后对NSGA-II算法所得最优解进行了有限元验证,结果表明二者最大相对误差小于6.5%,吻合较好。Abstract: In the optimization design of the flexible hinge in the fast steering mirror, in order to ensure the accuracy and stability, the rotational flexibility of the working axis and the rotational rigidity of the non-working axis should be increased as much as possible. Taking the elliptic arc flexible hinge in a fast steering mirror as the research object, firstly the least squares method and the integral method were used to derive the flexibility formula of three rotation axes. Compared with the Finite Element Analysis (FEA), the relative error was less than 6.8%, the problem of complex integration calculation in torsional flexibility of rectangular section was solved. Secondly, based on the improved non-dominated sorting genetic algorithm (NSGA-II) algorithm, the elliptic flexible hinge was multi-objectively optimized, met the design target and the optimal solution and solution were efficiency improved significantly. The efficiency of Pareto solution was improved by 14.3% compared with the Multi-Island Genetic Algorithm (MIGA) method, and improved by 25% compared with Particle Swarm Optimization (PSO). Finally, the optimization results were verified by FEA. The relative error was less than 6.5%, which was a good agreement.
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Key words:
- fast steering mirror /
- elliptic flexible hinge /
- flexibility formula /
- NSGA-II algorithm
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表 1 解析解与有限元解对比
Table 1. Comparison of analytical solution and finite element solution
t0/mm w/mm Cz-Mz/rad·Nm−1 Cy-My/rad·Nm−1 Cx-Mx/rad·Nm−1 yc/m EQU 1 5 0.1130 8.699×10−3 7.572×10−2 1.000×10−7 FEM 1 5 0.1106 8.989×10−3 7.380×10−2 9.993×10−8 EQU 0.5 5 0.6479 1.360×10−2 0.4095 4.194×10−7 FEM 0.5 5 0.6419 1.403×10−2 0.3996 4.179×10−7 EQU 1 3 0.1884 4.028×10−2 0.1426 1.668×10−7 FEM 1 3 0.1874 4.141×10−2 0.1329 1.677×10−7 EQU 0.2 3 10.80 0.1089 4.508 4.501×10−6 FEM 0.2 3 10.51 0.1098 4.790 4.487×10−6 表 2 转动惯量
Table 2. Moment of inertia
Jθx/kg·m2 Jθy/kg·m2 Jθz/ kg·m2 3.46×10−6 3.46×10−6 6.33×10−6 表 3 优化先后对比
Table 3. Comparison of before optimization and after optimization
a/mm b/mm t0/mm w/mm Cz-Mz/rad·Nm−1 Cy-My/rad·Nm−1 Cx-Mx/rad·Nm−1 Initial value 9 5 0.5 6 0.4859 7.079×10−3 0.2966 Optimized value 9.67 6.63 0.661 9.74 0.1390 1.343×10−3 8.870×10−2 表 4 不同算法最优解对比
Table 4. Comparison of optimal solutions of different algorithms
Algorithm Cz-Mz/rad·Nm−1 Cy-My/rad·Nm−1 Cx-Mx/rad·Nm−1 RC Pareto solution number Pareto solution efficiency MIGA 0.1320 1.329×10−2 6.999×10−2 −0.6 153 38.2% PSO 0.1301 4.630×10−2 7.075×10−2 0.4 110 27.5% NSGA-II 0.1390 1.343×10−2 8.870×10−2 −0.8 210 52.5% 表 5 最优解与有限元解对比
Table 5. Comparison of optimal solution and finite element solution
Cz-Mz/rad·Nm−1 Cy-My/rad·Nm−1 Cx-Mx/rad·Nm−1 EQU 0.1390 1.343×10−3 8.870×10−2 FEM 0.1320 1.340×10−3 8.290×10−2 -
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