Let the length and width of the four-sided fixed membrane roof be a and b respectively; The pretension along the length direction is N0x, and the pretension along the width direction is N0y; The wind blows parallel to the membrane surface, as shown in Figure 1 . Single-mode divergent instability characterized by large vibration will occur in the state of relaxation or insufficient pre-tension of roof membrane [2] . According to the test results carried out by Usenatsu, Kimoto et al. [11] , it is found that the film does not tear when the single-mode divergent instability occurs, but only the vibration displacement amplitude jumps. Therefore, this paper assumes that the film is still in the linear elastic stage when the divergent instability occurs.

When the roof membrane vibrates, the generalized external load includes wind load acting on the membrane surface, structural damping force and inertia force [2] [3] . In the process of membrane vibration, ignoring the influence of shear stress [12] , the dynamic governing equations of roof membrane are as follows:

$\{\begin{array}{l}\left(N_{0y}+h\frac{{\partial }^2\phi }{\partial x^2}\right)k_y+\left(N_{0x}+h\frac{{\partial }^2\phi }{\partial y^2}\right)k_x+p-2\rho c\frac{\partial w}{\partial t}-\rho \frac{{\partial }^{2}w}{\partial t^2}=0\\ \frac{1}{E_1}\frac{{\partial }^4\phi }{\partial y^4}+\frac{1}{E_2}\frac{{\partial }^4\phi }{\partial x^4}={\left(\frac{{\partial }^{2}w}{\partial x\partial y}\right)}^2-\frac{{\partial }^{2}w}{\partial x^2}\frac{{\partial }^{2}w}{\partial y^2}-k_{0x}\frac{{\partial }^{2}w}{\partial y^2}-k_{0y}\frac{{\partial }^{2}w}{\partial x^2}\end{array}$ (1)

where h is the thickness of the film; $E_1$ and $E_2$ represent Young’s modulus of elasticity in the x-direction and y-direction, respectively; w represents the normal vibration displacement of the membrane material; $\rho$ is the surface density of the film material; c represents the damping coefficient of membrane material; $\phi =\phi \left(x,y,t\right)$ denotes the stress function.

The membrane roof is fixed on four sides, so the boundary conditions are:

$\{\begin{array}{l}w\left(0,y,t\right)=0\\ w\left(a,y,t\right)=0\end{array}$, $\{\begin{array}{l}w\left(x,0,t\right)=0\\ w\left(x,b,t\right)=0\end{array}$ (2)

Since the initial surface function of the membrane roof is $z_0\left(x,y\right)=0$, the surface equation of the roof under wind load is

$z\left(x,y,t\right)=w\left(x,y,t\right)$ (3)

The initial curvature of the roof membrane and the curvature during vibration are

$\{\begin{array}{l}{k}_{0x}=\frac{{\partial }^2{z}_0}{\partial x^2}=0,k_x=\frac{{\partial }^{2}z}{\partial x^2}=\frac{{\partial }^{2}w}{\partial x^2}\\ k_{0y}=\frac{{\partial }^2{z}_0}{\partial y^2}=0,k_y=\frac{{\partial }^{2}z}{\partial y^2}=\frac{{\partial }^{2}w}{\partial y^2}\end{array}$ (4)

Let the solution of the governing equation be [12]

$w\left(x,y,t\right)=T\left(t\right)\sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b}$ (5)

In reference [2] , the partial differential equation of wind-induced vibration of membrane roof is obtained as follows:

$\begin{array}{l}\left(\rho W+\frac{{\rho }_0}{\pi }{\gamma }_1\right)\frac{{\text{d}}^{2}T\left(t\right)}{\text{d}{t}^2}+\left[\frac{{\rho }_{0}V}{2\pi }\left({\gamma }_2-{\gamma }_4\right)+2\rho cW\right]\frac{\text{d}T\left(t\right)}{\text{d}t}\\ -\left(N_{0x}\frac{{\partial }^{2}W}{\partial x^2}+N_{0y}\frac{{\partial }^{2}W}{\partial y^2}+\frac{{\rho }_0{V}^2}{2\pi }{\gamma }_3\right)T\left(t\right)\\ -h\left(\frac{{\partial }^2\Phi }{\partial y^2}\frac{{\partial }^{2}W}{\partial x^2}+\frac{{\partial }^2\Phi }{\partial x^2}\frac{{\partial }^{2}W}{\partial y^2}\right)T^3\left(t\right)=0\end{array}$ (6)

where,

$\begin{array}{l}{\gamma }_1={\displaystyle \underset{Ra}{\iint }\frac{1}{r}}{\left(W\right)}_{\begin{array}{l}x=\xi \\ y=\eta \end{array}}\text{d}\xi \text{d}\eta ={\displaystyle \underset{Ra}{\iint }\frac{1}{r}\sin \frac{m\pi \xi }{a}}\sin \frac{n\pi \eta }{b}\text{d}\xi \text{d}\eta \\ {\gamma }_2={\displaystyle \underset{Ra}{\iint }\frac{1}{r}}{\left(\frac{\partial W}{\partial x}\right)}_{\begin{array}{l}x=\xi \\ y=\eta \end{array}}\text{d}\xi \text{d}\eta =\frac{m\pi }{a}{\displaystyle \underset{Ra}{\iint }\frac{1}{r}\cos \frac{m\pi \xi }{a}}\sin \frac{n\pi \eta }{b}\text{d}\xi \text{d}\eta \\ {\gamma }_3={\displaystyle \underset{Ra}{\iint }\frac{1}{r^3}}{\left(\frac{\partial W}{\partial x}\right)}_{\begin{array}{l}x=\xi \\ y=\eta \end{array}}\left(x-\xi \right)\text{d}\xi \text{d}\eta =\frac{m\pi }{a}{\displaystyle \underset{Ra}{\iint }\frac{1}{r^3}}\left(x-\xi \right)\cos \frac{m\pi \xi }{a}\sin \frac{n\pi \eta }{b}\text{d}\xi \text{d}\eta \\ {\gamma }_4={\displaystyle \underset{Ra}{\iint }\frac{1}{r^3}}{\left(W\right)}_{\begin{array}{l}x=\xi \\ y=\eta \end{array}}\left(x-\xi \right)\text{d}\xi \text{d}\eta ={\displaystyle \underset{Ra}{\iint }\frac{1}{r^3}\left(x-\xi \right)\sin \frac{m\pi \xi }{a}}\sin \frac{n\pi \eta }{b}\text{d}\xi \text{d}\eta \end{array}$

Equation (6) is integrated using the Bubnov-Galerkin method, yields,

$\begin{array}{l}{\displaystyle \underset{S}{\iint }\{\left(\rho W+\frac{{\rho }_0}{\pi }{\gamma }_1\right)\frac{{\text{d}}^{2}T\left(t\right)}{\text{d}{t}^2}+\left[\frac{{\rho }_{0}V}{2\pi }\left({\gamma }_2-{\gamma }_4\right)+2\rho cW\right]\frac{\text{d}T\left(t\right)}{\text{d}t}}\\ -\left(N_{0x}\frac{{\partial }^{2}W}{\partial x^2}+N_{0y}\frac{{\partial }^{2}W}{\partial y^2}+\frac{{\rho }_0{V}^2}{2\pi }{\gamma }_3\right)T\left(t\right)\\ -h\left(\frac{{\partial }^2\Phi }{\partial y^2}\frac{{\partial }^{2}W}{\partial x^2}+\frac{{\partial }^2\Phi }{\partial x^2}\frac{{\partial }^{2}W}{\partial y^2}\right)T^3\left(t\right)\}W\left(x,y\right)\text{d}x\text{d}y=0\end{array}$ (7)

The integration region in Equation (7) is $S\in \left\{0\le x\le a,0\le y\le b\right\}$.

Simplifying Equation (7), yields,

$\begin{array}{l}\left(\frac{\rho \pi ab+4{\rho }_0{\alpha }_1}{4\pi }\right)\frac{{\text{d}}^{2}T\left(t\right)}{\text{d}{t}^2}+\left(\frac{{\rho }_0\pi mV{\alpha }_2-a{\rho }_{0}V{\alpha }_4+\rho \pi c{a}^{2}b}{2\pi a}\right)\frac{\text{d}T\left(t\right)}{\text{d}t}\\ +\left(\frac{m^2{\pi }^2{b}^2{N}_{0x}+n^2{\pi }^2{a}^2{N}_{0y}-2{\rho }_{0}m{V}^{2}b{\alpha }_3}{4ab}\right)T\left(t\right)\\ +\frac{h{\pi }^4\left(E_2{a}^4{n}^4+E_1{b}^4{m}^4\right)}{64a^3{b}^3{m}^2}{T}^3\left(t\right)=0\end{array}$ (8)

where

$\begin{array}{l}{\alpha }_1={\displaystyle \underset{S}{\iint }\left({\displaystyle \underset{Ra}{\iint }\frac{1}{r}\sin \frac{m\pi \xi }{a}}\sin \frac{n\pi \eta }{b}\text{d}\xi \text{d}\eta \right)\sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b}\text{d}x\text{d}y}\\ {\alpha }_2={\displaystyle \underset{S}{\iint }\left({\displaystyle \underset{Ra}{\iint }\frac{1}{r}\cos \frac{m\pi \xi }{a}}\sin \frac{n\pi \eta }{b}\text{d}\xi \text{d}\eta \right)\sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b}\text{d}x\text{d}y}\\ {\alpha }_3={\displaystyle \underset{S}{\iint }\left({\displaystyle \underset{Ra}{\iint }\frac{1}{r^3}}\left(x-\xi \right)\cos \frac{m\pi \xi }{a}\sin \frac{n\pi \eta }{b}\text{d}\xi \text{d}\eta \right)\sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b}\text{d}x\text{d}y}\\ {\alpha }_4={\displaystyle \underset{S}{\iint }\left[{\displaystyle \underset{Ra}{\iint }\frac{1}{r^3}\left(x-\xi \right)\sin \frac{m\pi \xi }{a}}\sin \frac{n\pi \eta }{b}\text{d}\xi \text{d}\eta \right]\sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b}\text{d}x\text{d}y}\end{array}$

In Equation (8), $\frac{\rho \pi ab+4{\rho }_0{\alpha }_1}{4\pi }\ne 0$, Let $x=x\left(t\right)=T\left(t\right)$, according to the

calculation principle of improved multi-scale method [13] , take the perturbation

parameters as $\epsilon =\frac{h{\pi }^5\left(E_2{a}^4{n}^4+E_1{b}^4{m}^4\right)}{16a^3{b}^3{m}^2\left(\rho \pi ab+4{\rho }_0{\alpha }_1\right)}$, then, Equation (8) can be simplified

as:

$\ddot{x}+{\omega }_0^{2}x+\epsilon \left(\mu \stackrel{\dot{}}{x}+x^3\right)=0$ (9)

where,

${\omega }_0^2=\frac{\pi \left(m^2{\pi }^2{b}^2{N}_{0x}+n^2{\pi }^2{a}^2{N}_{0y}-2{\rho }_{0}m{V}^{2}b{\alpha }_3\right)}{ab\left(\rho \pi ab+4{\rho }_0{\alpha }_1\right)}$

$u=\frac{2\left({\rho }_0\pi mV{\alpha }_2-a{\rho }_{0}V{\alpha }_4+\rho \pi c{a}^{2}b\right)}{a\epsilon \left(\rho \pi ab+4{\rho }_0{\alpha }_1\right)}$

Let ω be the vibration frequency of the roof membrane, and expand ${\omega }^2$ near ${\omega }_0^2$ into a power series of ε, that is,

${\omega }^2={\omega }_0^2+\epsilon {\omega }_1+{\epsilon }^2{\omega }_2+\cdots$ (10)

Letting,

$\alpha =\frac{\epsilon {\omega }_1}{{\omega }_0^2+\epsilon {\omega }_1}$ (11)

Then,

$\epsilon =\frac{{\omega }_0^2\alpha }{{\omega }_1\left(1-\alpha \right)}$ (12)

${\omega }_0^2+\epsilon {\omega }_1=\frac{{\omega }_0^2}{1-\alpha }$

Expand ω² into a power series of ε [14] ,

$\begin{array}{c}{\omega }^2={\omega }_0^2+\epsilon {\omega }_1\left[1+\frac{1}{{\omega }_0^2+\epsilon {\omega }_1}\left({\epsilon }^2{\omega }_2+{\epsilon }^3{\omega }_3+\cdots \right)\right]\\ =\frac{{\omega }_0^2}{1-\alpha }\left(1+{\delta }_2{\alpha }^2+{\delta }_3{\alpha }^3+\cdots \right)\end{array}$ (13)

$\omega ={\omega }_0\left[1+\frac{1}{2}\alpha +\left(\frac{3}{8}+\frac{{\delta }^2}{2}\right){\alpha }^2+\cdots \right]$ (14)

The solution of Equation (9) can be assumed to be

$x\left(t,\alpha \right)=x_0\left(T_0,T_1\right)+\alpha x_1\left(T_0,T_1\right)+{\alpha }^2{x}_2\left(T_0,T_1\right)+\cdots$ (15)

where $T_0=t,T_1=\alpha t$.

The differential operators can be obtained as follows:

$\begin{array}{l}\frac{\text{d}}{\text{d}t}=D_0+\alpha D_1+{\alpha }^2{D}_2+\cdots ,\\ \frac{{\text{d}}^2}{\text{d}{t}^2}=D_0^2+2\alpha D_0{D}_1+{\alpha }^2\left(D_1^2+2D_0{D}_2\right)+\cdots \end{array}$ (16)

By substituting Equations (11)-(13) and (16) into Equation (9) and collating, we can obtain:

$\begin{array}{l}\left(1-\alpha \right)\left[D_0^2+2\alpha D_0{D}_1+{\alpha }^2\left(D_1^2+2D_0{D}_2\right)\right]\left(x_0+\alpha x_1+{\alpha }^2{x}_2+\cdots \right)\\ +\left(1-\alpha \right){\omega }_0^2\left(x_0+\alpha x_1+{\alpha }^2{x}_2+\cdots \right)+\frac{\alpha {\omega }_0^2}{{\omega }_1^2}[{\left(x_0+\alpha x_1+{\alpha }^2{x}_2+\cdots \right)}^3\\ +\left(D_0+\alpha D_1+{\alpha }^2{D}_2\right)\cdot \left(x_0+\alpha x_1+{\alpha }^2{x}_2+\cdots \right)]=0\end{array}$ (17)

Expand Equation (17), let the coefficients of each power of $\alpha$ in the equation be zero, yields,

$\begin{array}{l}\begin{array}{cc}{\alpha }^0& D_0^2{x}_0+{\omega }_0^2{x}_0=0\end{array}\\ \begin{array}{cc}{\alpha }^1& D_0^2{x}_1+{\omega }_0^2{x}_1+2D_0{D}_1{x}_0+\frac{{\omega }_0^2}{{\omega }_1}\left(D_0{x}_0+x_0^3\right)=0\end{array}\\ \begin{array}{cc}{\alpha }^2& D_0^2{x}_2+x_2+2D_0\end{array}{D}_1{x}_1+\left(D_1^2+2D_0{D}_2\right)x_0+\frac{{\omega }_0^2}{{\omega }_1}3x_0^2{x}_1=0\end{array}$ (18)

The solution of the first equation in the system of Equation (18) can be

$x_0=A\left(T_1\right){\text{e}}^{i{\omega }_0{T}_0}+\bar{A}\left(T_1\right){\text{e}}^{-i{\omega }_0{T}_0}$ (19)

By substituting Equation (19) into ${\alpha }^1$, yields:

$D_0^2{x}_1+{\omega }_0^2{x}_1+\left(2i{\omega }_0{D}_{1}A+3\frac{{\omega }_0^2}{{\omega }_1}{A}^2\bar{A}+i\mu \frac{{\omega }_0^3}{{\omega }_1}A\right){\text{e}}^{i{T}_0}+\frac{1}{{\omega }_1}{A}^3{\text{e}}^{3i{T}_0}+cc=0$ (20)

Where cc is a conjugate complex number, let:

$2D_{1}A+3\frac{{\omega }_0}{{\omega }_1}{A}^2\bar{A}+i\mu \frac{{\omega }_0^3}{{\omega }_1}A=0$ (21)

Solving Equations (21), yields

$x_1=\frac{1}{8{\omega }_1}\left(A^3{\text{e}}^{3i{T}_0}+{\bar{A}}^3{\text{e}}^{-3i{T}_0}\right)$ (22)

Let $A=\frac{1}{2}f{\text{e}}^{i\varphi }$, then substitute it into Equation (22) and separate the imaginary

and real parts,

$\begin{array}{cc}\frac{\text{d}f}{\text{d}{T}_1}=-\frac{1}{2}\mu \frac{{\omega }_0^2}{{\omega }_1}f,& f\frac{\text{d}\varphi }{\text{d}{T}_1}=-\frac{3{\omega }_0}{8{\omega }_1}\end{array}{f}^3$ (23)

substitute $A=\frac{1}{2}f{\text{e}}^{i\varphi }$ into Equation (19), yields:

$x_0=f\cos \left({\omega }_0{T}_0+\varphi \right)=f\cos \left(\omega t+{\varphi }_0\right)$ (24)

It can be obtained from Equations (24) and (14) that under the first-order approximation condition

$\frac{\text{d}\varphi }{\text{d}{T}_1}=\frac{{\omega }_0}{2}$ (25)

Combining Equations (23) and (25), yields:

${\omega }_1=\frac{3}{4}{f}^2$ (26)

By substituting Equation (26) into Equation (23) and omitting the higher-order terms, yields:

$\omega =\sqrt{{\omega }_0^2+\frac{3}{4}\epsilon f^2}$ (27)

The existing research results show that when the membrane roof reaches the critical state of instability, the frequency of the characteristic equation of the system approaches zero [2] [3] . i.e.:

${\omega }_0^2+\frac{3}{4}\epsilon f^2=0$ (28)

where f is the vibration amplitude of the membrane material.

${\omega }_0^2=\frac{\pi \left(m^2{\pi }^2{b}^2{N}_{0x}+n^2{\pi }^2{a}^2{N}_{0y}-2{\rho }_{0}m{V}^{2}b{\alpha }_3\right)}{ab\left(\rho \pi ab+4{\rho }_0{\alpha }_1\right)}$.

From Equation (28), the critical wind speed when the roof membrane material is instable can be obtained as follows:

$V_{cr}=\sqrt{\frac{4\pi \left(m^2{\pi }^2{b}^2{N}_{0x}+n^2{\pi }^2{a}^2{N}_{0y}\right)+3\epsilon f^2\left(\rho \pi a^2{b}^2+4{\rho }_{0}ab{\alpha }_1\right)}{8\pi b{\rho }_{0}m{\alpha }_3}}$ (29)

where f is the vibration amplitude, that is, the vibration amplitude corresponding to the critical wind speed of instability. It shows that the critical wind speed is related to the vibration amplitude. Because the geometric nonlinear influence of membrane is considered, the stiffness of membrane will change with the change of amplitude in the vibration process, which will affect the stability of the structure to a certain extent, which is consistent with the existing research results. When f → 0, the critical wind speed calculated according to the small deflection theory can be obtained.