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    Journals >Infrared and Laser Engineering >Volume 51 >Issue 6 >Page 20210670 > Article
    • Infrared and Laser Engineering
    • Vol. 51, Issue 6, 20210670 (2022)
    Optimization design of large-aperture lens mixed flexible support structure
    Xiaodi Wang1、2, Yuyan Cao1、*, Fuguo Wang1, Hongliang Chu1, and Yanwei Li3
    Author Affiliations
  • 1Changchun Institute of Optics, Fine Mechanics and Physics, Chinese Academy of Sciences, Changchun 130033, China
  • 2University of Chinese Academy of Sciences, Beijing 100049, China
  • 3Ji Hua Laboratory, Foshan 528200, China
    • show less
    DOI: 10.3788/IRLA20210670 Cite this Article
    Xiaodi Wang, Yuyan Cao, Fuguo Wang, Hongliang Chu, Yanwei Li. Optimization design of large-aperture lens mixed flexible support structure[J]. Infrared and Laser Engineering, 2022, 51(6): 20210670 Copy Citation Text show less

    Abstract

    A novel mixed flexible support structure was proposed to satisfy the requirements of surface and position accuracy of a large-aperture lens. First, Castigliano’s second theorem was used to analyze the various flexible hinges and the whole flexibility model of the support component was established. Then, the objective function of the total deformation energy of the flexible support assembly was used to establish the constraint equation based on the position accuracy and the actual space requirements, and a structural optimization design model was established. It was determined that the radially flexible support structure was the most sensitive to the flexibility of the flexible support assembly, and its stiffness was verified. Finally, the finite element analysis of the optimized whole structure of the lens assembly was carried out, and the surface accuracy was obtained by using the geometric fitting method. The simulation results show that the surface shape accuracy of the new hybrid flexible support structure is better than λ/20 (λ=632.8 nm) under various conditions. The novel mixed flexible support structure and its theoretical analysis process can provide a reference for the supporting technology of high precision large-aperture lens.

    Keywords

    $ {t_1}(x) = {t_1} + 2{r_1} - 2\sqrt {x(2{r_1} - x)} {\text{ }},{\text{ }}0 < x < 2{r_1} $(1)

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    $ {t_2}(x) = \frac{{{t_2}}}{2} + {r_2} - \sqrt {x(2{r_2} - x)} {\text{ }},{\text{ }}0 < x < 2{r_2} $(2)

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    $ \begin{split} \left[ {\begin{array}{*{20}{c}} {{u_{{x_1}}}} \\ {{u_{{y_1}}}} \\ {{\theta _{{z_1}}}} \end{array}} \right] = &[{{\boldsymbol{C}}_1}]\left[ {\begin{array}{*{20}{c}} {{F_{{x_1}}}} \\ {{F_{{y_1}}}} \\ {{M_{{z_1}}}} \end{array}} \right] =\\ & \left[ {\begin{array}{*{20}{c}} {{C_{1,x - {F_x}}}}&0&0 \\ 0&{{C_{1,y - {F_y}}}}&{{C_{1,y - {M_z}}}} \\ 0&{{C_{1,{\theta _z} - {F_y}}}}&{{C_{1,{\theta _z} - {M_z}}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{F_{{x_1}}}} \\ {{F_{{y_1}}}} \\ {{M_{{z_1}}}} \end{array}} \right] \end{split}$(3)

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    $ {u}_{x}=\frac{\partial U}{\partial {F}_{x}}\text{，}{u}_{y}=\frac{\partial U}{\partial {F}_{y}}\text{，}{\theta }_{z}=\frac{\partial U}{\partial {M}_{z}} $(4)

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    $ U = \frac{1}{2}\int\nolimits_0^l {\left(\frac{{{F_x}^2}}{{EA(x)}} + \frac{{{M_z}^2}}{{EI(x)}}\right)}{\rm{d}}x $(5)

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    $ \left\{ {\begin{array}{*{20}{l}} {{u_{{x_1}}} = \dfrac{{\partial U}}{{\partial {F_x}}} = \dfrac{{{F_{{x_1}}}}}{{{E_1}{b_1}}}\displaystyle \int\nolimits_0^{2{r_1}} {\dfrac{1}{{{t_1}(x)}}} {\rm{d}}x}\\ {{{u_{{y_1}}} = \dfrac{{\partial U}}{{\partial {F_y}}} = \dfrac{{12}}{{{E_1}{b_1}}}\left[{F_{{y_1}}}\displaystyle \int\nolimits_0^{2{r_1}} {\dfrac{{{{(2{r_1} - {x_1})}^2}}}{{{t_1}^3(x)}}} {\rm{d}}x - {M_e}\displaystyle \int\nolimits_0^{2{r_1}} {\dfrac{{2{r_1} - {x_1}}}{{{t_1}^3(x)}}} {\rm{d}}x\right]}} \\ {{\theta _{{z_1}}} = \dfrac{{\partial U}}{{\partial {M_z}}} = \dfrac{{12}}{{{E_1}{b_1}}}\left[{M_e}\displaystyle \int\nolimits_0^{2{r_1}} {\dfrac{1}{{{t_1}^3(x)}}} {\rm{d}}x - {F_{{y_1}}}\displaystyle \int\nolimits_0^{2{r_1}} {\dfrac{{2{r_1} - {x_1}}}{{{t_1}^3(x)}}} {\rm{d}}x\right]{\text{ }}} \end{array}} \right. $(6)

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    $ \left\{ \begin{gathered} {C_{1,x - {F_x}}} = \frac{1}{{{E_1}{b_1}}}\int\nolimits_0^{2{r_1}_{}} {\frac{1}{{{t_1} + 2{r_1} - 2\sqrt {x(2{r_1} - x)} }}} {\rm{d}}x \\ {C_{1,y - {F_y}}} = \frac{{12}}{{{E_1}{b_1}}}\int\nolimits_0^{2{r_1}} {\frac{{{{(2{r_1} - x)}^2}}}{{{{[{t_1} + 2{r_1} - 2\sqrt {x(2{r_1} - x)} ]}^3}}}} {\rm{d}}x \\ {C_{1,y - {M_z}}} = {C_{1,{\theta _z} - {F_y}}} = \frac{{12}}{{{E_1}{b_1}}}\int\nolimits_0^{2{r_1}} {\frac{{2{r_1} - x}}{{{{[{t_1} + 2{r_1} - 2\sqrt {x(2{r_1} - x)} ]}^3}}}} {\rm{d}}x \\ {C_{1,{\theta _z} - {M_z}}} = \frac{{12}}{{{E_1}{b_1}}}\int\nolimits_0^{2{r_1}} {\frac{1}{{{{[{t_1} + 2{r_1} - 2\sqrt {x(2{r_1} - x)} ]}^3}}}} {\rm{d}}x \end{gathered} \right. $(7)

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    $ \begin{split} \left[ {\begin{array}{*{20}{c}} {{u_{{x_2}}}} \\ {{u_{{y_2}}}} \\ {{\theta _{{z_2}}}} \end{array}} \right] =& [{{\boldsymbol{C}}_2}]\left[ {\begin{array}{*{20}{c}} {{F_{{x_2}}}} \\ {{F_{{y_2}}}} \\ {{M_{{z_2}}}} \end{array}} \right] =\\ &\left[ {\begin{array}{*{20}{c}} {{C_{2,x - {F_x}}}}&0&0 \\ 0&{{C_{2,y - {F_y}}}}&{{C_{2,y - {M_z}}}} \\ 0&{{C_{2,{\theta _z} - {F_y}}}}&{{C_{2,{\theta _z} - {M_z}}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{F_{{x_2}}}} \\ {{F_{{y_2}}}} \\ {{M_{{z_2}}}} \end{array}} \right] \end{split}$(8)

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    $ \left\{ \begin{gathered} {C_{2,x - {F_x}}} = \frac{1}{{{E_2}{b_2}}}\int\nolimits_0^{2{r_2}} {\frac{1}{{{t_2} + {r_2} - \sqrt {x(2{r_2} - x)} }}} {\rm{d}}x \\ {C_{2,y - {F_y}}} = \frac{{12}}{{{E_2}{b_2}}}\int\nolimits_0^{2{r_2}} {\frac{{{{(2{r_2} - x)}^2}}}{{{{[{t_2} + {r_2} - \sqrt {x(2{r_2} - x)} ]}^3}}}} {\rm{d}}x \\ {C_{2,y - {M_z}}} = {C_{2,{\theta _z} - {F_y}}} = \frac{{12}}{{{E_2}{b_2}}}\int\nolimits_0^{2{r_2}} {\frac{{2{r_2} - x}}{{{{[{t_2} + {r_2} - \sqrt {x(2{r_2} - x)} ]}^3}}}} {\rm{d}}x\\ {C_{2,{\theta _z} - {M_z}}} = \frac{{12}}{{{E_2}{b_2}}}\int\nolimits_0^{2{r_2}} {\frac{1}{{{{[{t_2} + {r_2} - \sqrt {x(2{r_2} - x)} ]}^3}}}} {\rm{d}}x \end{gathered} \right.{\text{ }} $(9)

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    $ {\overline M _{{X_1}}}(x) = l - x $(10)

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    $ {\overline M _{{X_2}}}(x) = 0 $(11)

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    $ {\overline M _{{X_3}}}(x) = 1 $(12)

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    $ {\overline M _{{X_4}}}(x) = \left\{ {\begin{array}{*{20}{c}} 1&{0 \leqslant x \leqslant \dfrac{l}{2}} \\ 0&{\dfrac{l}{2} \leqslant x \leqslant l} \end{array}} \right. $(13)

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    $ {\overline M _D}(x) = \left\{ {\begin{array}{*{20}{c}} {\dfrac{l}{2} - x}&{0 \leqslant x \leqslant \dfrac{l}{2}} \\ {0{\text{ }}}&{\dfrac{l}{2} \leqslant x \leqslant l} \end{array}} \right. $(14)

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    $ {\Delta _i} = {\delta _{i1}}{X_1} + {\delta _{i2}}{X_2} + {\delta _{i3}}{X_3} + {\delta _{i4}}{X_4} + {\Delta _{iF}} = 0 $(15)

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    $ {\delta }_{ij}={\displaystyle {\int _{l}}\frac{{\overline{M}}_{i}(x){\overline{M}}_{j}(x)}{EI}}{\rm{d}}x\text{   }(i=1,2,3,4\text{；}j=1,2,3,4) $(16)

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    $ {\Delta }_{iF}={\displaystyle {\int _l}\frac{{F}_{D}{\overline{{M}_{D}}}\;{\overline{{M}_{i}}}}{EI}}{\rm{d}}x (i=1,2,3,4) $(17)

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    $ \left( {\begin{array}{*{20}{c}} {\dfrac{{{l^3}}}{{3EI}}}&0&{\dfrac{{{l^2}}}{{2EI}}}&{\dfrac{{3{l^2}}}{{8EI}}} \\ 0&0&0&0 \\ {\dfrac{{{l^2}}}{{2EI}}}&0&{\dfrac{l}{{EI}}}&{\dfrac{l}{{2EI}}} \\ {\dfrac{{3{l^2}}}{{8EI}}}&0&{\dfrac{l}{{2EI}}}&{\dfrac{l}{{2EI}}} \end{array}} \right)\left\{ {\begin{array}{*{20}{c}} {{X_1}} \\ {{X_2}} \\ {{X_3}} \\ {{X_4}} \end{array}} \right\} + \left\{ {\begin{array}{*{20}{c}} { - \dfrac{{5{l^3}{F_D}}}{{48EI}}} \\ 0 \\ { - \dfrac{{{l^2}{F_D}}}{{8EI}}} \\ { - \dfrac{{{l^2}{F_D}}}{{8EI}}} \end{array}} \right\} = 0 $(18)

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    $ {X_1} = \frac{{{F_D}}}{2}\text{；}{X_3} = - \frac{{l{F_D}}}{8}\text{；}{X_4} = 0{\text{ }} $(19)

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    $ \begin{split} M(x) =& \left\{ {\begin{array}{*{20}{c}} {{X_1}(l - x) + {X_3} + {X_4} - {F_D}\left(\dfrac{l}{2} - x\right)}&{0 \leqslant x \leqslant \dfrac{l}{2}} \\ {{X_1}(l - x) + {X_3}{\text{ }}}&{\dfrac{l}{2} \leqslant x \leqslant l} \end{array}} \right. \hfill \\ {\text{ = }}& \left\{ {\begin{array}{*{20}{c}} {\dfrac{{{F_{Dx}}}}{2} - \dfrac{{l{F_D}}}{8}{\text{ }}}&{0 \leqslant x \leqslant \dfrac{l}{2}} \\ { - \dfrac{{{F_D}}}{2}x + \dfrac{{3l{F_D}}}{8}}&{\dfrac{l}{2} \leqslant x \leqslant l} \end{array}} \right. \end{split} $(20)

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    $ \begin{split}{\delta }_{D}=&{\displaystyle {\int_{l} }\dfrac{M(x)}{{E}_{2}I}\cdot \dfrac{\partial M(x)}{\partial {F}_{D}}}{\rm{d}}x \text{=}\\ &\dfrac{{F}_{D}}{{E}_{2}I}{\displaystyle \int\nolimits_0^{\frac{l}{2}} {\left(\dfrac{x}{2}-\dfrac{l}{8}\right)}^{2}}{\rm{d}}x+\dfrac{{F}_{D}}{{E}_{2}I}{\displaystyle \int\nolimits_0^{\frac{l}{2}}{\left(\dfrac{3l}{8}-\dfrac{x}{2}\right)}^{2}}{\rm{d}}x \text{=}\\ &\dfrac{{F}_{D}{l}^{3}}{192{E}_{2}I}\end{split} $(21)

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    $ {C_D} = \frac{{{\delta _D}}}{{{F_D}}} = \frac{{{l^3}}}{{16{E_2}h{t^3}}} $(22)

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    $ {C_{3x}} = \dfrac{1}{{\dfrac{1}{{2{C_{2,x - {F_x}}}}} + \dfrac{1}{{2{C_{3,x - {F_x}}}}} + \dfrac{1}{{2{C_{2,x - {F_x}}}}}}} = \dfrac{{2{C_{2,x - {F_x}}}{C_{3,x - {F_x}}}}}{{{C_{3,x - {F_x}}} + 2{C_{2,x - {F_x}}}}} $(23)

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    $ {u}_{x}={C}_{1,x-{F}_{x}}\cdot{F}_{{x}_{1}}+{C}_{D}\cdot{F}_{D}+{C}_{3x}\cdot{F}_{{x}_{2}} $(24)

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    $ {C_x} = {C_{1,x - {F_x}}} + {C_D} + {C_{3x}} $(25)

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    $ \begin{split}& \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\min \frac{1}{2}n{K_x}\delta _{^T}^2 \hfill \\ &{\rm{s.t.}}\left\{ {\begin{array}{*{20}{l}} {{u_x} < 0.1{\text{ }}} \\ {{\sigma _f} \leqslant {\sigma _{f\max }}} \\ {{\sigma _g} \leqslant {\sigma _{g\max }}} \\ {50 \leqslant {b_1} \leqslant 70,{\text{ }}20 \leqslant {b_2} \leqslant 40,{\text{ }}5 \leqslant {b_3} \leqslant 12} \\ {0.1 \leqslant {r_1} \leqslant 2,{\text{ }}0.1 \leqslant {r_3} \leqslant 3} \\ \begin{gathered} 30 < l < 50{\text{ }} \hfill \\ 10 < h < 30{\text{ }} \hfill \\ \end{gathered} \\ {0.5 < t < 2{\text{ }}} \\ {0.1 < {t_1} < 6,{\text{ }}0.1 < {t_2} < 10} \end{array}} \right. \end{split} $(26)

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    $ {R^2} = \left\| {{{\boldsymbol{X}}_i} - {{\boldsymbol{X}}_c}} \right\|, ({X}_{i}\ne {X}_{c}) $(27)

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    $ {\boldsymbol{X}}_i^{'} = {{\boldsymbol{X}}_c} + R\frac{{{{\boldsymbol{X}}_i} - {{\boldsymbol{X}}_c}}}{{\left\| {{{\boldsymbol{X}}_i} - {{\boldsymbol{X}}_c}} \right\|}} ,{\text{ }}i = 1,2\cdots,m $(28)

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    $ {{\boldsymbol{a}}^{\rm{T}}} = (R,{\boldsymbol{X}}_{_c}^{\rm{T}}) = (R,{{\boldsymbol{X}}_c},{{\boldsymbol{Y}}_c},{{\boldsymbol{Z}}_c}) $(29)

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    $ \begin{split} {{\boldsymbol{J}}_{{\boldsymbol{X}}_i^{'},{\boldsymbol{a}}}} =& \frac{{\partial X_i^{'}}}{{\partial {\boldsymbol{a}}}} = \frac{\partial }{{\partial a}}[{{\boldsymbol{X}}_c} + R\frac{{{{\boldsymbol{X}}_i} - {{\boldsymbol{X}}_c}}}{{\left\| {{{\boldsymbol{X}}_i} - {{\boldsymbol{X}}_c}} \right\|}}] \hfill {\text{ = }}\\ &\frac{{\partial {{\boldsymbol{X}}_c}}}{{\partial {\boldsymbol{a}}}} + \frac{{{{\boldsymbol{X}}_i} - {{\boldsymbol{X}}_c}}}{{\left\| {{{\boldsymbol{X}}_i} - {{\boldsymbol{X}}_c}} \right\|}}\frac{{\partial R}}{{\partial a}} - \frac{R}{{\left\| {{{\boldsymbol{X}}_i} - {{\boldsymbol{X}}_c}} \right\|}} \hfill \times\\ &{\text{ }} \left[ {{\boldsymbol{I}} - \frac{{({{\boldsymbol{X}}_i} - {{\boldsymbol{X}}_c}){{({{\boldsymbol{X}}_i} - {{\boldsymbol{X}}_c})}^{\rm{T}}}}}{{{{\left\| {{{\boldsymbol{X}}_i} - {{\boldsymbol{X}}_c}} \right\|}^2}}}} \right]\frac{{\partial {{\boldsymbol{X}}_c}}}{{\partial {\boldsymbol{a}}}} \hfill \\ \end{split} $(30)

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    $ {\boldsymbol{X}}_i^{''} = {{\boldsymbol{X}}_i} - {\boldsymbol{X}}_i^{'} = {\boldsymbol{J}}\Delta {\boldsymbol{a}} $(31)

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    Xiaodi Wang, Yuyan Cao, Fuguo Wang, Hongliang Chu, Yanwei Li. Optimization design of large-aperture lens mixed flexible support structure[J]. Infrared and Laser Engineering, 2022, 51(6): 20210670
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