The first boundary integral equations for the domain point can be derived from the Rayleigh–Green identity as follows (5-6):Display Formula

$8\pi u(x)=U(\zeta ,x)\u2212\u222bBU(s,x)v(s)dB(s)+\u222bB\Theta (s,x)m(s)dB(s)\u2212\u222bBM(s,x)\theta (s)dB(s)+\u222bBV(s,x)u(s)dB(s),x\u220a\Omega \u222aB$

(6)

where

$B$ is the boundary of the domain

$\Omega $;

$u(x)$,

$\theta (x)$,

$m(x)$, and

$v(x)$ are the displacement, slope, normal moment, and effective shear force; and

$s$ and

$x$ are the source point and field point, respectively. The kernel function

$U(s,x)$ in Eq.

6 is the fundamental solution that satisfies

Display Formula$\u22074U(s,x)=8\pi \delta (s\u2212x)$

(7)

Therefore, the fundamental solution can be obtained as follows:

Display Formulawhere

$r$ is the distance between the source point

$s$ and field point

$x$. The relationship among

$u(x)$,

$\theta (x)$,

$m(x)$, and

$v(x)$ is shown as follows:

Display Formula$\theta (x)=K\theta ,x(u(x))=\u2202u(x)\u2202nx$

(9)

Display Formula$m(x)=Km,x(u(x))=\nu \u2207x2u(x)+(1\u2212\nu )\u22022u(x)\u22022nx$

(10)

Display Formula$v(x)=Kv,x(u(x))=\u2202\u2207x2u(x)\u2202nx+(1\u2212\nu )\u2202\u2202tx[\u2202\u2202nx(\u2202u(x)\u2202tx)]$

(11)

where

$K\theta ,x(\u22c5)$,

$Km,x(\u22c5)$, and

$Kv,x(\u22c5)$ are the slope, moment, and shear force operators with respect to the point

$x$;

$\u2202\u2215\u2202nx$ is the normal derivative with respect to the field point

$x$;

$\u2202\u2215\u2202tx$ is the tangential derivative with respect to the field point

$x$; and

$\u2207x2$ is the Laplacian operator. The first null-field integral equations can be derived by moving the field point

$x$ outside the domain as follows:

Display Formula$0=U(\zeta ,x)\u2212\u222bBU(s,x)v(s)dB(s)+\u222bB\Theta (s,x)m(s)dB(s)\u2212\u222bBM(s,x)\theta (s)dB(s)+\u222bBV(s,x)u(s)dB(s),x\u220a\Omega C\u222aB$

(12)

where

$\Omega C$ is the complementary domain of

$\Omega $. For the kernel function

$U(s,x)$, it can be expanded in terms of degenerate kernel (

2,

5-

7) in a series form as shown below:

Display Formula$U(s,x)={UI(R,\theta ;\rho ,\varphi )=\rho 2(1+lnR)+R2lnR\u2212[R\rho (1+2lnR)+12\rho 3R]cos(\theta \u2212\varphi )\u2212\u2211m=2\u221e[1m(m+1)\rho m+2Rm\u22121m(m\u22121)\rho mRm\u22122]cos[m(\theta \u2212\varphi )],R\u2265\rho UE(R,\theta ;\rho ,\varphi )=R2(1+ln\rho )+\rho 2ln\rho \u2212[\rho R(1+2ln\rho )+12R3\rho ]cos(\theta \u2212\varphi )\u2212\u2211m=2\u221e[1m(m+1)Rm+2\rho m\u22121m(m\u22121)Rm\rho m\u22122]cos[m(\theta \u2212\varphi )],\rho >R}$

(13)

where the superscripts

$I$ and

$E$ denote the interior and exterior cases of

$U(s,x)$ kernel depending on the location of

$s$ and

$x$. For the annular plate clamped at the outer edge and free at the inner edge, the unknown Fourier coefficients of

$m$,

$v$ on the outer boundary and

$u$,

$\theta $ on the inner boundary can be expanded to

Display Formula$v(s)=a0+\u2211n=1M(ancosn\theta +bnsinn\theta ),s\u220aouterboundary$

(14)

Display Formula$m(s)=a\xaf0+\u2211n=1M(a\xafncosn\theta +b\xafnsinn\theta ),s\u220aouterboundary$

(15)

Display Formula$\theta (s)=p0+\u2211n=1M(pncosn\theta +qnsinn\theta ),s\u220ainnerboundary$

(16)

Display Formula$u(s)=p\xaf0+\u2211n=1M(p\xafncosn\theta +q\xafnsinn\theta ),s\u220ainnerboundary$

(17)

where

$a0$,

$an$,

$bn$,

$a\xaf0$,

$a\xafn$,

$b\xafn$,

$p0$,

$pn$,

$qn$,

$p\xaf0$,

$p\xafn$, and

$q\xafn$ are the Fourier coefficients, and

$M$ is the number of Fourier series terms in real computation. By substituting all the Fourier coefficients of boundary densities and boundary conditions, the displacement field can be obtained as shown below:

Display Formula$8\pi u(x)=U(\zeta ,x)\u2212\u222bBU(s,x)[a0+\u2211n=1M(ancosn\theta +bnsinn\theta )]dB(s)+\u222bB\Theta (s,x)[a\xaf0+\u2211n=1M(a\xafncosn\theta +b\xafnsinn\theta )]dB(s)\u2212\u222bBM(s,x)[p0+\u2211n=1M(pncosn\theta +qnsinn\theta )]dB(s)+\u222bBV(s,x)[p\xaf0+\u2211n=1M(p\xafncosn\theta +q\xafnsinn\theta )]dB(s),x\u220a\Omega \u222aB$

(18)

where

$an$,

$bn$,

$a\xafn$,

$b\xafn$,

$pn$,

$qn$,

$p\xafn$, and

$q\xafn$$(n=0,1,2,\u2026)$ are solved in Ref.

7.