Two Semi-Infinite Elastic Plates of Identical Properties

Introduction

We present here the solution of Evans and Porter 2005 for the simple case of a single crack with waves incident from normal (they also considered multiple cracks and waves incident from different angles). The solution of Evans and Porter 2005 expresses the potential [math]\displaystyle{ \phi }[/math] in terms of a linear combination of the incident wave and certain source functions located at the crack. Along with satisfying the field and boundary conditions, these source functions satisfy the jump conditions in the displacements and gradients across the crack.

Governing Equations

We consider the entire free surface to be occupied by a Floating Elastic Plate with a single discontinuity at [math]\displaystyle{ x=0 }[/math]. The equations are the following

[math]\displaystyle{ \nabla^2 \phi = 0, -H\lt z\lt 0, }[/math] [math]\displaystyle{ \frac{\partial \phi}{\partial z} =0, z=-H, }[/math] [math]\displaystyle{ {\left( \beta \frac{\partial^4}{\partial x^4} - \gamma\alpha + 1\right)\frac{\partial \phi}{\partial z} - \alpha \phi} = 0, z=0, x\neq 0. }[/math]

Solution using the Free-Surface Green Function for a Floating Elastic Plate

We then use Green's second identity If φ and ψ are both twice continuously differentiable on U, then

[math]\displaystyle{ \int_U \left( \psi \nabla^2 \varphi - \varphi \nabla^2 \psi\right)\, dV = \oint_{\partial U} \left( \psi {\partial \varphi \over \partial n} - \varphi {\partial \psi \over \partial n}\right)\, dS }[/math]

If we then substitiute the Free-Surface Green Function for a Floating Elastic Plate which satisfies the following equations (plus the Sommerfeld Radiation Condition far from the body)

[math]\displaystyle{ \nabla^2 G = 0, -H\lt z\lt 0, }[/math] [math]\displaystyle{ \frac{\partial G}{\partial z} =0, z=-H, }[/math] [math]\displaystyle{ {\left( \beta \frac{\partial^4}{\partial x^4} - \gamma\alpha + 1\right)\frac{\partial G}{\partial z} - \alpha G} = \delta(x), z=0, }[/math]

for ψ we obtain

[math]\displaystyle{ 0 = \int_{-\infty}^{\infty}\left( G_{n}\left( x,x^{\prime }\right) \phi \left( x ^{\prime }\right) -G\left( x,x^{\prime }\right) \phi _{n}\left( x^{\prime }\right) \right) dx^{\prime } }[/math]

We then substitute [math]\displaystyle{ G = \phi }[/math] to obtain

[math]\displaystyle{ \int_{-\infty}^{\infty}\left( G_{n}\left( x,x^{\prime }\right) \frac{1}{\alpha} \left( \beta \partial_x^4 - \gamma\alpha + 1\right)\phi_{n}\left( x^{\prime }\right) -G\left( x,x^{\prime }\right) \phi _{n}\left( x^{\prime }\right) \right) dx^{\prime } = 0 }[/math]

plus the vertical integrals at the ends (which will give a contribution). [math]\displaystyle{ \phi_n^{In} }[/math]

Integration by parts

Integrate by parts says that

[math]\displaystyle{ \int_a^b (\partial_x u) v dx= -\int_a^b u(\partial_x v) dx + u(b)v(b) - u(a)v(a) }[/math]

For the second derivative

[math]\displaystyle{ \int_a^b (\partial_x^2 u) v dx= -\int_a^b (\partial_x u) (\partial_x v) dx + \partial_x u(b)v(b) - \partial_x u(a)v(a) }[/math]

[math]\displaystyle{ = \int_a^b u (\partial_x^2 v) dx + \partial_x u(b)v(b) - \partial_x u(a)v(a) + u(b)\partial_x v(b) - u(a)\partial_x v(a) }[/math]

For the fourth derivative

[math]\displaystyle{ \int_a^b (\partial_x^4 u) v dx = \int_a^b u (\partial_x^4v) dx - u(b)(\partial_x^3v(b)) + u(a)(\partial_x^3v(a)) + (\partial_xu(b))(\partial_x^2v(b)) }[/math] [math]\displaystyle{ - (\partial_xu(a))(\partial_x^2v(a)) - (\partial_x^2u(b))\partial_xv(b) + (\partial_x^2u(a))\partial_xv(a) + (\partial_x^3u(b))v(b) - (\partial_x^3u(a))v(b) }[/math]

new section

[math]\displaystyle{ \int_{-\infty}^{0} G_{n}\left( x,x^{\prime }\right) \partial_{x^{\prime }}^4 \phi_n^{-}(x^{\prime })dx^{\prime } =G(x,0)\partial_x^3\phi_n(0) - \partial_x G(x,0)\partial_x^2\phi_n^{-}(0) +\partial_x^2 G(x,0)\partial_x\phi_n^{-}(0) -\partial_x^3G(x,0)\phi_n^{-}(0) }[/math]

old section

We now integrate by parts remembering that [math]\displaystyle{ \phi_n }[/math] is continuous everywhere except at [math]\displaystyle{ x^\prime = 0 }[/math] so that

[math]\displaystyle{ \int_{-\infty}^\infty(\partial_x^4\phi_n)G_n dx = \int_{-\infty}^0(\partial_x^4\phi_n)G_n dx + \int_0^\infty(\partial_x^4\phi_n)G_n dx }[/math]

and obtain

[math]\displaystyle{ \int_{-\infty}^{\infty}\left\{ \frac{1}{\alpha} \left( \beta \partial_x^4 - \gamma\alpha + 1\right)G_{n}\left( x,x^{\prime }\right) - G\left( x,x^{\prime }\right)\right\} \phi_{n}\left( x^{\prime }\right) dx^\prime }[/math]

[math]\displaystyle{ + \frac{\beta}{\alpha}\left(-\partial_x^3G_n(x,0)[\phi_n] + \partial_x^2 G_n(x,0)\partial_x[\phi_n] - \partial_x G_n(x,0)\partial_x^2[\phi_n] + G(x,0)\partial_x^3[\phi_n] \right) =0 }[/math]

where [] denotes the jump in the function at [math]\displaystyle{ x^{\prime}=0 }[/math].

The integral can be simplified using the delta function property of the Green function to give us IF I'M RIGHT, THE LAST TWO TERMS SHOULD BE OPPOSITE SIGNS. AGREE????

I doubt this - but check it out. The signs should alternate + - + i etc.

Exactly, that's why the last two terms of the equation below should be +, - rather than - +? Oh just realised your final equation is correct. I guess this one was just a typo. I've changed it now.

[math]\displaystyle{ \phi_{n}\left( x\right) = \beta \left(\partial_x^3 G_n [\phi_n] - \partial_x^2 G_n [\partial_x\phi_n] + \partial_x G_n [\partial_x^2\phi_n] - G_n [\partial_x^3\phi_n]\right) }[/math]

We can write the equation in terms of [math]\displaystyle{ \phi }[/math] as was done by Porter and Evans 2005 but there is no real point because the boundary conditions are given in terms of [math]\displaystyle{ \phi_n }[/math] since this represents the displacement.

We are missing the incident wave which comes into the equations through the boundary conditions at infinity.

The correct equation is

[math]\displaystyle{ \phi_{n}\left( x\right) = \phi_{n}^{\mathrm{In}}\left( x\right) + \beta\left(\partial_x^3 G_n [\phi_n] - \partial_x^2 G_n [\partial_x\phi_n] + \partial_x G_n [\partial_x^2\phi_n] - G_n [\partial_x^3\phi_n]\right) }[/math]

which can be solved by applying the edge conditions at x = 0 and z = 0

[math]\displaystyle{ \partial_x^2\phi_n=0,\,\,\, {\rm and}\,\,\,\, \partial_x^3\phi_n=0. }[/math]

Porter and Evans 2005 formulation

Consequently, the source functions for a single crack at [math]\displaystyle{ x=0 }[/math] can be defined as

[math]\displaystyle{ \psi_s(x,z)= \beta\chi_{xx}(x,z),\,\,\, \psi_a(x,z)= \beta\chi_{xxx}(x,z),\,\,\,(2) }[/math]

It can easily be shown that [math]\displaystyle{ \psi_s }[/math] is symmetric about [math]\displaystyle{ x = 0 }[/math] and [math]\displaystyle{ \psi_a }[/math] is antisymmetric about [math]\displaystyle{ x = 0 }[/math].

Substituting (1) into (2) gives

[math]\displaystyle{ \psi_s(x,z)= { -\frac{\beta}{\alpha} \sum_{n=-2}^\infty \frac{g_n\cos{(k_n(z+h))}}{2k_{xn}C_n}e^{k_n|x|} }, \psi_a(x,z)= { {\rm sgn}(x) i\frac{\beta}{\alpha}\sum_{n=-2}^\infty \frac{g_n'\cos{(k_n(z+h))}}{2k_{xn}C_n}e^{k_n|x|}}, }[/math]

where

[math]\displaystyle{ g_n = ik_n^3 \sin{k_n h},\,\,\,\, g'_n= -k_n^4 \sin{k_n h}. }[/math]

We then express the solution to the problem as a linear combination of the incident wave and pairs of source functions at each crack,

[math]\displaystyle{ \phi(x,z) = e^{-k_0 x}\frac{\cos(k_0(z+h))}{\cos(k_0h)} + (P\psi_s(x,z) + Q\psi_a(x,z))\,\,\,(3) }[/math]

where [math]\displaystyle{ P }[/math] and [math]\displaystyle{ Q }[/math] are coefficients to be solved which represent the jump in the gradient and elevation respectively of the plates across the crack [math]\displaystyle{ x = a_j }[/math]. The coefficients [math]\displaystyle{ P }[/math] and [math]\displaystyle{ Q }[/math] are found by applying the edge conditions and to the [math]\displaystyle{ z }[/math] derivative of [math]\displaystyle{ \phi }[/math] at [math]\displaystyle{ z=0 }[/math],

[math]\displaystyle{ \frac{\partial^2}{\partial x^2}\left. \frac{\partial \phi}{\partial z}\right|_{x=0,z=0}=0,\,\,\, {\rm and}\,\,\,\, \frac{\partial^3}{\partial x^3}\left. \frac{\partial \phi}{\partial z}\right|_{x=0,z=0}=0. }[/math]

The reflection and transmission coefficients, [math]\displaystyle{ R }[/math] and [math]\displaystyle{ T }[/math] can be found from (3) by taking the limits as [math]\displaystyle{ x\rightarrow\pm\infty }[/math] to obtain

[math]\displaystyle{ R = {- \frac{\beta}{\alpha} (g'_0Q + ig_0P)} }[/math]

and

[math]\displaystyle{ T= 1 + {\frac{\beta}{\alpha}(g'_0Q - ig_0P)} }[/math]

Other boundary conditions

More complicated boundary conditions can be treated using this formulation.

What about entering these - you know what they are and it would be useful.