技術(shù)特征：

1.橫向均布載荷下預(yù)應(yīng)力圓薄膜最大應(yīng)力的確定方法，其特征在于：對周邊固定夾緊的預(yù)應(yīng)力圓形薄膜橫向施加一個均布載荷q，其中圓形薄膜的厚度為h、半徑為a、楊氏彈性模量為E、泊松比為ν、預(yù)應(yīng)力為σ₀，測得所施加的均布載荷q的值，由以下公式確定該圓形薄膜變形后的最大應(yīng)力σ_m：

${\mathrm{\σ}}_m=\frac{a^{2/3}{E}^{1/3}{q}^{2/3}}{2^{5/3}{h}^{2/3}{c}^{1/3}},$

而中間參量c的值由方程

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

確定，其中，

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

$\begin{array}{c}{f}^{\′}\left(c\right)=-\frac{1}{2}-\frac{1}{3}\left(1+2\×\frac{2^{1/3}{a}^{2/3}{q}^{2/3}}{h^{2/3}{E}^{2/3}{c}^{1/3}}\right)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},$

中間參量c沒有單位，其它所有參量皆采用國際單位制。