一种薄膜杨氏弹性模量的测定方法

摘要

本发明公开了一种薄膜杨氏弹性模量的测定方法：采用内半径为a的夹紧装置，将厚度为h、泊松比为ν的薄膜固定夹紧，形成一个半径为a的周边固定夹紧的圆薄膜，并对圆薄膜横向施加均布载荷q，基于这个轴对称圆薄膜静力平衡问题的精确分析，利用圆薄膜最大挠度测量值w_m，则可以得到薄膜的杨氏弹性模量E。该方法彻底放弃了薄膜小转角假设，从而彻底消除薄膜转角θ对计算误差的影响，同时也消除了横向载荷q不能过大的限制条件，为测量操作工作提供了方便。

主权项

一种薄膜杨氏弹性模量的测定方法，其特征在于：采用内半径为a的夹紧装置，将厚度为h、泊松比为ν的薄膜固定夹紧，形成一个半径为a的周边固定夹紧的圆薄膜，对圆薄膜横向施加均布载荷q，并测量圆薄膜的最大挠度值w_m，求解以下方程组$\left\{\begin{array}{c}2{\mathrm{Ehw}}_m^3=a^4{\mathrm{cqg}}^3\left(c\right)\\ \nu f\left(c\right)=2{\mathrm{cf}}^{\prime }\left(c\right)+f\left(c\right)\end{array}\right.$得到薄膜的杨氏弹性模量E，其中，c为中间变量，函数g(c)、f(c)和f′(c)分别为：$\begin{array}{c}f\left(c\right)1-\frac{1}{2}c-\frac{1}{6}(1+2\times \frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}})c^2-\frac{1}{144}(13+56\times \frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +48\times \frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}})c^3-\frac{1}{1440}(85+584\times \frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+1104\times \frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +1152\frac{a^2{q}^2}{h^2{E}^{2}c})c^4-\frac{1}{21600}(925+8904\times \frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+27104\times \frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}\\ +62592\frac{a^2{q}^2}{h^2{E}^{2}c}+23040\times \frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}})c^5-\frac{1}{907200}(30125+377688\times \frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +1619968\times \frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+6038784\frac{a^2{q}^2}{h^2{E}^{2}c}+4907520\times \frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +1382400\times \frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}})c^6-\frac{1}{203212800}(5481025+85400640\times \frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +478992416\times \frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/3}{c}^{2/3}}+2520947712\frac{a^2{q}^2}{h^2{E}^{2}c}+3305599488\times \frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +2047057920\times \frac{2^{2/3}{a}^{10/3}{q}^{10/3}}{h^{10/3}{E}^{10/3}{c}^{5/3}}+928972800\frac{a^4{q}^4}{h^4{E}^4{c}^2})c^7\end{array}$和$\begin{array}{c}{f}^{\prime }\left(c\right)=-\frac{1}{2}-\frac{1}{3}(1+2\times \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\times \frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +48\times \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\times \frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+1104\times \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\times \frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+27104\times \frac{2^{2/3}{a}^{4/3}{q}^{4/3}}{h^{4/3}{E}^{4/2}{c}^{4/3}}\\ +62592\frac{a^2{q}^2}{h^2{E}^{2}v}+23040\times \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\times \frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +1619968\times \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\times \frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +1382400\times \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\times \frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\\ +478992416\times \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\times \frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}\\ +2047057920\times \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\times \frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}+21730770208\times \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\times \frac{2^{1/3}{a}^{8/3}{q}^{8/3}}{h^{8/3}{E}^{8/3}{c}^{4/3}}+276724703232\times \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\times \frac{2^{1/3}{a}^{14/3}{q}^{14/3}}{h^{14/3}{E}^{14/3}{c}^{7/3}})c^7\end{array}.$