CN106248474A - 横向均布载荷下预应力圆薄膜最大挠度的确定方法 - Google Patents

Abstract

本发明公开了横向均布载荷下预应力圆薄膜最大挠度的确定方法：对周边固定夹紧的预应力圆形薄膜横向施加一个均布载荷q，其中圆形薄膜的厚度为h、半径为a、杨氏弹性模量为E、泊松比为ν、预应力为σ₀，基于这个轴对称变形问题的静力平衡分析，利用均布载荷q的测量值，则可以确定出薄膜变形后的最大挠度w_m。

Description

横向均布载荷下预应力圆薄膜最大挠度的确定方法

技术领域

本发明涉及横向均布载荷作用下周边固定夹紧的预应力圆形薄膜最大挠度的确定方法。

背景技术

横向均布载荷作用下周边固定夹紧的圆形薄膜轴对称变形问题的解析解，对传感器、以及仪器仪表的研制具有重要意义。从文献查新的结果看，目前有适用于薄膜转角θ较大、但薄膜中没有预应力的情形的横向均布载荷下周边固定夹紧的圆形薄膜轴对称变形问题的解析解，例如，申请人之前申报的发明专利(“一种均布载荷下大转角圆薄膜最大挠度的确定方法”，专利申请号：201510193793.8)中所采用的解析解，就是在假定圆形薄膜中没有预应力的条件下获得的，因而它只适用于薄膜中没有预应力的情形；也有适用于薄膜中带有预应力、但薄膜转角θ较小的情形的横向均布载荷下周边固定夹紧的圆形薄膜轴对称变形问题的解析解，例如，申请人之前申报的另一发明专利(“一种确定均布载荷下预应力圆薄膜最大挠度值的方法”，专利申请号：201410238568.7)中所采用的解析解，就是在假定圆形薄膜中带有预应力、但薄膜转角θ较小的条件下获得的，即这一解析解的求解过程中考虑了薄膜中带有预应力的情形、但采用了薄膜转角θ近似满足sinθ＝tanθ的假设，众所周知，因此，显而易见，只有在薄膜转角θ较小的情形下，才能有的近似成立，因而这一解析解只适用于薄膜中带有预应力、但薄膜转角θ较小的情形；然而迄今为止，还没有见到适用于薄膜转角θ较大、同时薄膜中又带有预应力的情形的横向均布载荷下周边固定夹紧的圆形薄膜轴对称变形问题的解析解。

发明内容

本发明致力于薄膜问题的解析研究，解析求解了横向均布载荷下周边固定夹紧的预应力圆形薄膜的轴对称变形问题，并获得了该问题的解析解。在求解过程中本发明考虑了薄膜中带有预应力的情形、并采用了(即放弃了sinθ＝tanθ的假设)，因而所获得的解析解适用于薄膜转角θ较大、同时薄膜中又带有预应力的情形。基于这一解析解，本发明给出了横向均布载荷下周边固定夹紧的预应力圆形薄膜最大挠度w_m的确定方法。

横向均布载荷下预应力圆薄膜最大挠度的确定方法：对周边固定夹紧的预应力圆形薄膜横向施加一个均布载荷q，其中圆形薄膜的厚度为h、半径为a、杨氏弹性模量为E、泊松比为ν、预应力为σ₀，基于这个轴对称变形问题的静力平衡分析，就可以得到该预应力圆形薄膜变形后的最大挠度w_m与所施加的均布载荷q的解析关系

$\begin{array}{c}{w}_m={\left(\frac{a^{4}qc}{2hE}\right)}^{1/3}\[1+(\frac{1}{4}+\frac{1}{2}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}})c+(\frac{5}{36}+\frac{11}{18}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+\frac{1}{2}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}})c^2\\ +(\frac{55}{576}+\frac{49}{72}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+\frac{61}{48}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{5}{8}\frac{2a^2{q}^2}{h^2{E}^{2}c})c^3+(\frac{7}{96}+\frac{2647}{3600}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +\frac{4051}{1800}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{253}{100}\frac{2a^2{q}^2}{h^2{E}^{2}c}+\frac{7}{8}\frac{2^{4/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}})c^4+(\frac{205}{3456}+\frac{101639}{129600}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +\frac{443119}{129600}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{8521}{1350}\frac{2a^2{q}^2}{h^2{E}^{2}c}+\frac{7141}{1440}\frac{2^{4/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}+\frac{21}{16}\frac{2^{5/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}})c^5\\ +(\frac{17051}{338688}+\frac{2639243}{3175200}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+\frac{866077}{181440}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{13308283}{1058400}\frac{2a^2{q}^2}{h^2{E}^{2}c}\\ +\frac{1904549}{117600}\frac{2^{4/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}+\frac{113623}{11760}\frac{2^{5/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}+\frac{33}{16}\frac{4a^4{q}^4}{h^4{E}^4{c}^2})c^6+(\frac{2864485}{65028096}\\ +\frac{5947093}{6773760}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+\frac{91705651}{14515200}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{1115512751}{50803200}\frac{2a^2{q}^2}{h^2{E}^{2}c}\\ +\frac{1366411591}{33868800}\frac{2^{4/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}+\frac{740195}{18816}\frac{2^{5/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}+\frac{33661}{1792}\frac{4a^4{q}^4}{h^4{E}^4{c}^2}\\ +\frac{429}{128}\frac{2^{7/3}{a}^{14/3}{q}^{14/3}}{h^{14/3}{E}^{14/3}{c}^{7/3}})c^7\]\end{array},$

而中间参量c的值由方程

$\[2{\mathrm{cf}}^{\′}\left(c\right)+(1-v)f\left(c\right)\]-(1-v)\frac{2^{5/3}{h}^{2/3}{c}^{1/3}}{a^{2/3}{q}^{2/3}{E}^{1/3}}{\mathrm{\σ}}_0=0$

确定，其中，

$\begin{array}{c}f\left(c\right)=1-\frac{1}{2}c-\frac{1}{6}(1+2\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}})c^2-\frac{1}{144}(13+56\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +48\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}})c^3-\frac{1}{1440}(85+584\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+1104\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +1152\frac{a^2{q}^2}{h^2{E}^{2}c})c^4-\frac{1}{21600}(925+8904\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+27104\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +62592\frac{a^2{q}^2}{h^2{E}^{2}c}+23040\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}})c^5-\frac{1}{907200}(30125+377688\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +1619968\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+6038784\frac{a^2{q}^2}{h^2{E}^{2}c}+4907520\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +1382400\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}})c^6-\frac{1}{203212800}(5481025+85400640\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +478992416\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+2520947712\frac{a^2{q}^2}{h^2{E}^{2}c}+3305599488\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +2047057920\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}+928972800\frac{a^4{q}^4}{h^4{E}^4{c}^2})c^7\end{array},$

$\begin{array}{c}{f}^{\′}\left(c\right)=-\frac{1}{2}-\frac{1}{3}(1+2\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}})c-\frac{1}{48}(13+56\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +48\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}})c^2-\frac{1}{360}(85+584\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+1104\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +1152\frac{a^2{q}^2}{h^2{E}^{2}c})c^3-\frac{1}{4320}(925+8904\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+27104\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +62592\frac{a^2{q}^2}{h^2{E}^{2}c}+23040\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}})c^4-\frac{1}{151200}(30125+377688\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +1619968\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+6038784\frac{a^2{q}^2}{h^2{E}^{2}c}+4907520\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +1382400\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}})c^5-\frac{1}{29030400}(5481025+85400640\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +478992416\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+2520947712\frac{a^2{q}^2}{h^2{E}^{2}c}+3305599488\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +2047057920\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}+928972800\frac{a^4{q}^4}{h^4{E}^4{c}^2})c^6-\frac{1}{914457600}(165851725\\ +3108156432\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+21730770208\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +149806363136\frac{a^2{q}^2}{h^2{E}^{2}c}+277407668736\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}+276724703232\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}\\ +274790154240\frac{a^4{q}^4}{h^4{E}^4{c}^2}+52022476800\×\frac{2^{1/3}{a}^{14/3}{q}^{14/3}}{h^{14/3}{E}^{14/3}{c}^{7/3}})c^7\end{array}.$

这样，只要准确测量出所施加的均布载荷q的值，就可以确定出该预应力圆形薄膜变形后的最大挠度w_m。其中，中间参量c没有单位，其它所有参量皆采用国际单位制。

附图说明

图1为均布载荷下周边固定夹紧的预应力圆薄膜的加载构造示意图，其中，1－圆形薄膜，2－夹紧装置，其中的符号为，a表示夹紧装置的内半径和圆形薄膜的半径，r表示径向坐标，w(r)表示点r处的横向坐标，q表示横向均布载荷，w_m表示圆薄膜的最大挠度，θ表示薄膜变形后的转角。

具体实施方式

下面结合图1对本发明的技术方案作进一步的详细说明：

对周边固定夹紧的预应力圆形橡胶薄膜横向施加一个均布载荷q，其中，圆形橡胶薄膜的厚度h＝0.5mm、半径a＝20mm、杨氏弹性模量E＝7.84MPa、泊松比ν＝0.47、预应力σ₀＝0.001MPa，并测得q＝0.1MPa。采用本发明所给出的方法，通过方程

$\[2{\mathrm{cf}}^{\′}\left(c\right)+(1-v)f\left(c\right)\]-(1-v)\frac{2^{5/3}{h}^{2/3}{c}^{1/3}}{a^{2/3}{q}^{2/3}{E}^{1/3}}{\mathrm{\σ}}_0=0$

则可以得到c＝0.1961775414，其中，

$\begin{array}{c}f\left(c\right)=1-\frac{1}{2}c-\frac{1}{6}(1+2\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}})c^2-\frac{1}{144}(13+56\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +48\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}})c^3-\frac{1}{1440}(85+584\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+1104\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +1152\frac{a^2{q}^2}{h^2{E}^{2}c})c^4-\frac{1}{21600}(925+8904\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+27104\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +62592\frac{a^2{q}^2}{h^2{E}^{2}c}+23040\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}})c^5-\frac{1}{907200}(30125+377688\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +1619968\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+6038784\frac{a^2{q}^2}{h^2{E}^{2}c}+4907520\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +1382400\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}})c^6-\frac{1}{203212800}(5481025+85400640\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +478992416\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+2520947712\frac{a^2{q}^2}{h^2{E}^{2}c}+3305599488\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +2047057920\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}+928972800\frac{a^4{q}^4}{h^4{E}^4{c}^2})c^7\end{array},$

$\begin{array}{c}{f}^{\′}\left(c\right)=-\frac{1}{2}-\frac{1}{3}(1+2\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}})c-\frac{1}{48}(13+56\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +48\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}})c^2-\frac{1}{360}(85+584\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+1104\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +1152\frac{a^2{q}^2}{h^2{E}^{2}c})c^3-\frac{1}{4320}(925+8904\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+27104\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +62592\frac{a^2{q}^2}{h^2{E}^{2}c}+23040\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}})c^4-\frac{1}{151200}(30125+377688\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +1619968\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+6038784\frac{a^2{q}^2}{h^2{E}^{2}c}+4907520\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +1382400\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}})c^5-\frac{1}{29030400}(5481025+85400640\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +478992416\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+2520947712\frac{a^2{q}^2}{h^2{E}^{2}c}+3305599488\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +2047057920\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}+928972800\frac{a^4{q}^4}{h^4{E}^4{c}^2})c^6-\frac{1}{914457600}(165851725\\ +3108156432\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+21730770208\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +149806363136\frac{a^2{q}^2}{h^2{E}^{2}c}+277407668736\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}+276724703232\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}\\ +274790154240\frac{a^4{q}^4}{h^4{E}^4{c}^2}+52022476800\×\frac{2^{1/3}{a}^{14/3}{q}^{14/3}}{h^{14/3}{E}^{14/3}{c}^{7/3}})c^7\end{array}.$

最后，由公式

$\begin{array}{c}{w}_m={\left(\frac{a^{4}qc}{2hE}\right)}^{1/3}\[1+(\frac{1}{4}+\frac{1}{2}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}})c+(\frac{5}{36}+\frac{11}{18}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+\frac{1}{2}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}})c^2\\ +(\frac{55}{576}+\frac{49}{72}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+\frac{61}{48}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{5}{8}\frac{2a^2{q}^2}{h^2{E}^{2}c})c^3+(\frac{7}{96}+\frac{2647}{3600}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +\frac{4051}{1800}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{253}{100}\frac{2a^2{q}^2}{h^2{E}^{2}c}+\frac{7}{8}\frac{2^{4/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}})c^4+(\frac{205}{3456}+\frac{101639}{129600}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +\frac{443119}{129600}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{8521}{1350}\frac{2a^2{q}^2}{h^2{E}^{2}c}+\frac{7141}{1440}\frac{2^{4/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}+\frac{21}{16}\frac{2^{5/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}})c^5\\ +(\frac{17051}{338688}+\frac{2639243}{3175200}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+\frac{866077}{181440}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{13308283}{1058400}\frac{2a^2{q}^2}{h^2{E}^{2}c}\\ +\frac{1904549}{117600}\frac{2^{4/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}+\frac{113623}{11760}\frac{2^{5/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}+\frac{33}{16}\frac{4a^4{q}^4}{h^4{E}^4{c}^2})c^6+(\frac{2864485}{65028096}\\ +\frac{5947093}{6773760}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+\frac{91705651}{14515200}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{1115512751}{50803200}\frac{2a^2{q}^2}{h^2{E}^{2}c}\\ +\frac{1366411591}{33868800}\frac{2^{4/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}+\frac{740195}{18816}\frac{2^{5/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}+\frac{33661}{1792}\frac{4a^4{q}^4}{h^4{E}^4{c}^2}\\ +\frac{429}{128}\frac{2^{7/3}{a}^{14/3}{q}^{14/3}}{h^{14/3}{E}^{14/3}{c}^{7/3}})c^7\]\end{array}$

则可以得到该圆形薄膜变形后的最大挠度w_m＝9.977886578mm。

此外，如果采用申请人之前申报的发明专利(“一种确定均布载荷下预应力圆薄膜最大挠度值的方法”，专利申请号：201410238568.7)中的方法，则可以得到该圆形薄膜变形后的最大挠度w_m＝9.66414626mm，由此可见，两种方法之间是有差别的。这一差别是由申请人之前申报的发明专利(“一种确定均布载荷下预应力圆薄膜最大挠度值的方法”，专利申请号：201410238568.7)中的假设条件(即假设sinθ＝tanθ)引起的。

Claims (1)

1.横向均布载荷下预应力圆薄膜最大挠度的确定方法，其特征在于：对周边固定夹紧的预应力圆形薄膜横向施加一个均布载荷q，其中圆形薄膜的厚度为h、半径为a、杨氏弹性模量为E、泊松比为ν、预应力为σ₀，测得所施加的均布载荷q的值，由以下公式确定该圆形薄膜变形后的最大挠度w_m：

$\begin{array}{c}{w}_m={\left(\frac{a^{2}qc}{2hE}\right)}^{1/3}\[1+(\frac{1}{4}+\frac{1}{2}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}})c+(\frac{5}{36}+\frac{11}{18}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+\frac{1}{2}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}})c^2\\ +(\frac{55}{576}+\frac{49}{72}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+\frac{61}{48}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{5}{8}\frac{2a^2{q}^2}{h^2{E}^{c}c})c^3+(\frac{7}{96}+\frac{2647}{3600}\frac{2^{1/2}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +\frac{4051}{1800}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{253}{100}\frac{2a^2{q}^2}{h^2{E}^{2}c}+\frac{7}{8}\frac{2^{4/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}})c^4+(\frac{205}{3456}+\frac{101639}{129600}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +\frac{443119}{129600}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{8521}{1350}\frac{2a^2{q}^2}{h^2{E}^{2}c}+\frac{7141}{1440}\frac{2^{4/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}+\frac{21}{16}\frac{2^{5/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}})c^5\\ +(\frac{17051}{338688}+\frac{2639243}{3175200}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+\frac{866077}{181440}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{13308283}{1058400}\frac{2a^2{q}^2}{h^2{E}^{2}c}\\ +\frac{1904549}{117600}\frac{2^{4/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}+\frac{113623}{11760}\frac{2^{5/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}+\frac{33}{16}\frac{4a^4{q}^4}{h^4{E}^4{c}^2})c^6+(\frac{2864485}{65028096}\\ +\frac{5947093}{6773760}\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+\frac{91705651}{14515200}\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+\frac{1115512751}{50803200}\frac{2a^2{q}^2}{h^2{E}^{2}c}\\ +\frac{1366411591}{33868800}\frac{2^{4/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}+\frac{740195}{18816}\frac{2^{5/3}{a}^{4/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}+\frac{33661}{1792}\frac{4a^4{q}^4}{h^4{E}^4{c}^2}\\ +\frac{429}{128}\frac{2^{7/3}{a}^{14/3}{q}^{14/3}}{h^{14/3}{E}^{14/3}{c}^{7/3}})c^7\]\end{array},$

而中间参量c的值由方程

$\[2{\mathrm{cf}}^{\′}\left(c\right)+(1-v)f\left(c\right)\]-(1-v)\frac{2^{5/3}{h}^{2/3}{c}^{1/3}}{a^{2/3}{q}^{2/3}{E}^{1/3}}{\mathrm{\σ}}_0=0$

确定，其中，

$\begin{array}{c}f\left(c\right)=1-\frac{1}{2}c-\frac{1}{6}(1+2\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}})c^2-\frac{1}{144}(13+56\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +48\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}})c^3-\frac{1}{1440}(85+584\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+1104\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +1152\frac{a^2{q}^2}{h^2{E}^{2}c})c^4-\frac{1}{21600}(925+8904\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+27104\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +62592\frac{a^2{q}^2}{h^2{E}^{2}c}+23040\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}})c^5-\frac{1}{907200}(30125+377688\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +1619968\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+6038784\frac{a^2{q}^2}{h^2{E}^{2}c}+4907520\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +1382400\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}})c^6-\frac{1}{203212800}(5481025+85400640\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +478992416\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+2520947712\frac{a^2{q}^2}{h^2{E}^{2}c}+3305599488\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +2047057920\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}+928972800\frac{a^4{q}^4}{h^4{E}^4{c}^2})c^7\end{array},$

$\begin{array}{c}{f}^{\′}\left(c\right)=-\frac{1}{2}-\frac{1}{3}(1+2\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}})c-\frac{1}{48}(13+56\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +48\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}})c^2-\frac{1}{360}(85+584\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+1104\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +1152\frac{a^2{q}^2}{h^2{E}^{2}c})c^3-\frac{1}{4320}(925+8904\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+27104\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +62592\frac{a^2{q}^2}{h^2{E}^{2}c}+23040\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}})c^4-\frac{1}{151200}(30125+377688\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +1619968\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+6038784\frac{a^4{q}^2}{h^2{E}^{2}c}+4907520\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +1382400\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}})c^5-\frac{1}{29030400}(5481025+85400640\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +478992416\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+2520947712\frac{a^2{q}^2}{h^2{E}^{2}c}+3305599488\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +2047057920\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}+928972800\frac{a^4{q}^4}{h^4{E}^4{c}^2})c^6-\frac{1}{914457600}(165851725\\ +3108156432\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+21730770208\×\frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +149806363136\frac{a^2{q}^2}{h^2{E}^{2}c}+277407668736\×\frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}+27624703232\×\frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}\\ +274790154240\frac{a^4{q}^4}{h^4{E}^4{c}^2}+52022476800\×\frac{2^{1/3}{a}^{14/3}{q}^{14/3}}{h^{14/3}{E}^{14/3}{c}^{7/3}})c^7\end{array},$

中间参量c没有单位，其他所有参量皆采用国际单位制。