In order to estimate the fatigue damage of airfoil flutter, the flutter responses of the airfoil motion model should be analyzed first. The nondimensional variables are introduced to simplify the calculation [6,8,9]: $\xi =h/b$ is the nondimensional plunging displacement, $\chi \alpha =S\alpha /(mb)$ and $\chi \beta =S\beta /(mb)$ are the nondimensional static unbalance about the elastic axis and nondimensional static unbalance about the hinge, respectively, $\tau =Vt/b$ is the nondimensional time, *V*_{1} = *V*/(*bw*_{α}) is the nondimensional airspeed parameter, $\mu =m/(4\rho b2)$ is the nondimensional mass parameter, $r\alpha 2=I\alpha /(mb2)$ and $r\beta 2=I\beta /(mb2)$ are the nondimensional radius of gyration with respect to elastic axis and the nondimensional radius of gyration with respect to the hinge, respectively, $wh2=Kh/m$, $w\alpha 2=K\alpha /I\alpha $, and $w\beta 2=K\beta /I\alpha $ are the decouple plunging, pitching, and flapping frequencies of linearization system, respectively, and $\xi h=ch/(2mwh)$, $\xi \alpha =c\alpha /(2I\alpha w\alpha )$, and $\xi \beta =c\beta /(2I\beta w\beta )$ are the nondimensional plunging, pitching, and flapping damping ratios, respectively. Then, the nondimensional equations describing the airfoil motion can be expressed by the following form
Display Formula

(17)$MZ\xa8+HZ\u02d9+KZ=F+F1u$

where
Display Formula$Z=[\xi \alpha \beta ]T,M=[1\chi \alpha \chi \beta \chi \alpha r\alpha 2(d\u2212a)\chi \beta +r\beta 2\chi \beta (d\u2212a)\chi \beta +r\beta 2r\beta 2],H=[hij]3\xd73,K=[kij]3\xd73,F=\alpha 3\xd7[fij]3\xd71,F1=[0014\rho b4\mu V12w\alpha 2]T$

in which
Display Formula$h11=2wh\xi hV1w\alpha +\lambda MA\mu ,h12=\lambda (1\u2212a)MA\mu ,h13=\lambda (2\u2212d)24MA\mu h21=\lambda (1\u2212a)MA\mu ,h22=2\xi \alpha r\alpha 2V1+\lambda 2MA\mu (83\u22124a+2a2),h23=\lambda (d\u22123a+4)(2\u2212d)212MA\mu h31=\lambda (2\u2212d)24MA\mu ,h32=\lambda (d\u22123a+4)(2\u2212d)212MA\mu ,h33=2\xi \beta r\beta 2w\beta V1w\alpha +\lambda (2\u2212d)36MA\mu k11=wh2V12w\alpha 2,h12=\lambda MA\mu ,k13=\lambda (2\u2212d)2MA\mu k21=0,k22=r\alpha 2w\alpha 2V12w\alpha 2+\lambda (1\u2212a)MA\mu ,k23=\lambda (d\u22122a+2)(2\u2212d)4MA\mu k31=0,k32=\lambda (2\u2212d)24MA\mu ,k33=r\beta 2w\beta 2V12w\alpha 2+\lambda (2\u2212d)24MA\mu $

Display Formula$f11=\lambda 3(k+1)MA12\mu ,f21=\lambda 3(k+1)MA(1\u2212a)12\mu +Br\alpha 2V12,f31=\lambda 3(k+1)MA(2\u2212d)248\mu $