Floating Elastic Plates of Identical Properties

Introduction

We begin by presenting the solution for the case of a single crack with waves incident from normal. The solution method is derived from Squire and Dixon 2001 and Evans and Porter 2005.

We consider the entire free surface to be occupied by a Floating Elastic Plate with a single discontinuity at [math]\displaystyle{ x=x^\prime }[/math] (Fig. 1).

Image:GreenFunct.jpg

The governing equations are

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

The Free-Surface Green Function for a Floating Elastic Plate 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-x^{\prime}), z=0, }[/math]

where

[math]\displaystyle{ G(x,x^{\prime},z) = -i\sum_{n=-2}^\infty\frac{\sin{(k_n H)}\cos{(k_n(z+H))}}{2\alpha C(k_n)}e^{-k_n|x-x^{\prime}|}, }[/math] [math]\displaystyle{ C(k_n)=\frac{1}{2}\left(h - \frac{(5\beta k_n^4 + 1 - \alpha\gamma)\sin^2{(k_n H)}}{\alpha}\right), }[/math]

and [math]\displaystyle{ k_n }[/math] are the solutions of the Dispersion Relation for a Floating Elastic Plate,

[math]\displaystyle{ \beta k_n^5 \sin(k_nH) - k_n \left(1 - \alpha \gamma \right) \sin(k_nH) = -\alpha \cos(k_nH) \, }[/math]

with [math]\displaystyle{ n=-1,-2 }[/math] corresponding to the complex solutions with positive real part, [math]\displaystyle{ n=0 }[/math] corresponding to the imaginary solution with positive imaginary part and [math]\displaystyle{ n\gt 0 }[/math] corresponding to the real solutions with positive real part.

Green's Second Identity

Since φ and G are both twice continuously differentiable on U, where U represents the area bounded by the contour, S (Fig 1), the Green's second identity can be applied and gives

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

where n repressents the plane normal to the boundary, S.

Our governing equations for G and [math]\displaystyle{ \phi }[/math] imply that the L.H.S of Green's second identity is zero so that

[math]\displaystyle{ 0 = \oint_{\partial U} \left( G {\partial \phi \over \partial n} - \phi {\partial G \over \partial n}\right)\, dS }[/math]

expanding gives [math]\displaystyle{ 0 = -\int_{-\infty}^\infty \left( G {\partial \phi \over \partial z}|_{z=0} - \phi {\partial G \over \partial z}|_{z=0}\right)\, dx +\int_{-h}^0 \left( G {\partial \phi \over \partial x}|_{x=N} - \phi {\partial G \over \partial x}|_{x=N}\right)\, dz }[/math]

[math]\displaystyle{ +\int_{-\infty}^\infty \left( G {\partial \phi \over \partial z}|_{z=-h} - \phi {\partial G \over \partial z}|_{z=-h}\right)\, dx -\int_{-h}^0 \left( G {\partial \phi \over \partial x}|_{x=-N} - \phi {\partial G \over \partial x}|_{x=-N}\right)\, dz }[/math]

where we take the limit as N goes to infinity.

We now introduce the incident potential so that for x small

[math]\displaystyle{ \phi = A(e^{-k_0x} + R e^{k_0x})\cos{(k_n(z+H))} }[/math] [math]\displaystyle{ \phi_x = Ak_0(-e^{-k_0x} + R e^{k_0x})\cos{(k_n(z+H))} }[/math]

---

At the moment your incident and reflected wave are travelling in the same direction. I think you need to define [math]\displaystyle{ k=\pm ik_0 }[/math] and write the incident wave as [math]\displaystyle{ e^{\pm ikx} }[/math] so that k is positive real. This will make things much easier. Also, you need to define whether is [math]\displaystyle{ e^{i\omega t} }[/math] or [math]\displaystyle{ e^{-i\omega t} }[/math]. This will clear up a lot of silly errors.

---

for large x

[math]\displaystyle{ \phi = AT e^{-k_0x}\cos{(k_n(z+H))} }[/math] [math]\displaystyle{ \phi_x = -Ak_0T e^{-k_0x}\cos{(k_n(z+H))} }[/math]

where we choose A to normalise and assume k_0 is positive imaginary.

Also note that

[math]\displaystyle{ G_x = -k_0(sgn(x-x^\prime))G }[/math]

---

Can you write down the exact form for G It's in the introduction. Can you write down the exact form taking the limit of large or small x' so that only the n=0 term remains.

---

Now,

[math]\displaystyle{ \int_{-h}^0 \left( G {\partial \phi \over \partial x}|_{x=N} - \phi {\partial G \over \partial x}|_{x=N}\right)\, dz }[/math]

[math]\displaystyle{ = \int_{-h}^0 \left( -Ak_0GT e^{-k_0N}\cos{(k_n(z+H))} + Ak_0GT e^{-k_0N}\cos{(k_n(z+H))} \right)\, dz = 0 }[/math]

and

[math]\displaystyle{ -\int_{-h}^0 \left( G {\partial \phi \over \partial x}|_{x=-N} - \phi {\partial G \over \partial x}|_{x=-N}\right)\, dz }[/math]

[math]\displaystyle{ =-\int_{-h}^0 \left( Ak_0G(e^{-k_0N} + R e^{-k_0N})\cos{(k_n(z+H))} - Ak_0G(e^{-k_0N} + R e^{-k_0N})\cos{(k_n(z+H))} \right)\, dz =0 }[/math]

---

???????????? I have done something wrong here. My incident contribution seems to have cancelled out. ?????????????

---

Also, 0ur governing equations imply [math]\displaystyle{ G {\partial \phi \over \partial z}|_{z=-h} = 0 }[/math] and [math]\displaystyle{ \phi {\partial G \over \partial z}|_{z=-h} = 0 }[/math] so that,

[math]\displaystyle{ \int_{-\infty}^\infty \left( G {\partial \phi \over \partial z}|_{z=-h} - \phi {\partial G \over \partial z}|_{z=-h}\right)\, dx =0 }[/math]

and we are left with

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

At z=0, the z variable disappears to give

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

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

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

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

[math]\displaystyle{ \int_{-\infty}^\infty(\partial_x^4\phi_z)G_z dx = \int_{-\infty}^{x^\prime}(\partial_x^4\phi_z)G_z dx + \int_{x^\prime}^\infty(\partial_x^4\phi_z)G_z dx }[/math]

where

[math]\displaystyle{ \int_{-a}^b(\partial_x^4\phi_z)G_z dx = \int_a^b\phi(\partial_xG)dx - \phi(b)(\partial_x^3G(b) + \phi(a)(\partial_x^3G(a) + (\partial_x\phi(b))(\partial_x^2G(b)) }[/math]

[math]\displaystyle{ - (\partial_x\phi(a))(\partial_x^2G(a)) - (\partial_x^2\phi(b))(\partial_xG(b)) + (\partial_x^2\phi(a))(\partial_xG(a)) + (\partial_x^3\phi(b))G(b) - (\partial_x^3\phi(a))G(a) }[/math]

and obtain

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

[math]\displaystyle{ + \frac{\beta}{\alpha}\left(\partial_{x}^3G_z(x,x^\prime)[\phi_z] - \partial_{x}^2 G_z(x,x^\prime)\partial_x[\phi_z] + \partial_{x} G_z(x,x^\prime)\partial_{x}^2[\phi_z] - G_z(x,x^\prime)\partial_{x}^3[\phi_z] \right) =0 }[/math]

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

The integral can be simplified using the delta function property of the Green function to give us

[math]\displaystyle{ \phi_{z}\left( x\right) = -\beta\left(\partial_{x}^3 G_z [\phi_z] - \partial_{x}^2 G_z [\partial_{x}\phi_z] + \partial_{x} G_z [\partial_{x}^2\phi_z] - G_z [\partial_{x}^3\phi_z]\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_z }[/math] since this represents the displacement.

We include the boundary conditions at infinity, which we omitted earlier, to give the full equation

[math]\displaystyle{ \phi_{z}(x,z) = \phi_z^\mathrm{In} - \beta(\partial_x^3 G_z[\phi_z] - \partial_{x}^2 G_z[\partial_{x}\phi_z] + \partial_{x} G_z [\partial_{x}^2\phi_z] - G_z [\partial_{x}^3\phi_z]) }[/math]

which can be solved by applying the edge conditions at [math]\displaystyle{ x=x^\prime }[/math] and z = 0

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

Solution

We re-express [math]\displaystyle{ \phi_z }[/math] as

[math]\displaystyle{ \phi_{z} = \phi_{z}^{\mathrm{In}} - \psi_a [\phi_z] + \psi_s [\partial_{x}\phi_z] - \chi_a[\partial_{x}^2\phi_z] + \chi_s [\partial_{x}^3\phi_z] }[/math]

where

[math]\displaystyle{ \chi_s = \beta G_z = \frac{i\beta}{2\alpha} \sum_{n=-2}^\infty\frac{k_n\sin^2{(k_n H)}}{C(k_n)}e^{-k_n|x-x^\prime|}, }[/math] [math]\displaystyle{ \chi_a = \beta\partial_x G_z = -sgn(x-x^\prime)\frac{i\beta}{2\alpha} \sum_{n=-2}^\infty\frac{k_n^2\sin^2{(k_n H)}}{C(k_n)}e^{-k_n|x-x^\prime|}, }[/math] [math]\displaystyle{ \psi_s = \beta\partial_x^2 G_z = \frac{i\beta}{2\alpha} \sum_{n=-2}^\infty\frac{k_n^3\sin^2{(k_n H)}}{C(k_n)}e^{-k_n|x-x^\prime|}, }[/math] [math]\displaystyle{ \psi_a =\beta\partial_x^3 G_z = -sgn(x-x^\prime)\frac{i\beta}{2\alpha} \sum_{n=-2}^\infty\frac{k_n^4\sin^2{(k_n H)}}{C(k_n)}e^{-k_n|x-x^\prime|} }[/math]

and

[math]\displaystyle{ \phi_z^{In} = e^{k_0x}\frac{\sin(k_0(z+H))}{\sin(k_0H)} }[/math]

The edge conditions given above imply that [math]\displaystyle{ [\partial_{x}^2\phi_z] }[/math] and [math]\displaystyle{ [\partial_{x}^3\phi_z] }[/math] are zero so that [math]\displaystyle{ \phi_z }[/math] becomes

[math]\displaystyle{ \phi_{z}( x) = \phi_{z}^{\mathrm{In}}(x) -\psi_a [\phi_z] + \psi_s [\partial_{x}\phi_z] }[/math]

We are now left with two unknowns which can be solved using the two edge conditions. To solve, we use

[math]\displaystyle{ \partial_x^2\phi_z = \partial_x^2\phi_z^{In} - \partial_x^2\psi_a [\phi_z] + \partial_x^2\psi_s [\partial_{x}\phi_z] }[/math]

[math]\displaystyle{ = k_0^2e^{k_0x} \frac{\sin(k_0z+H)}{\sin(k_0H)} + \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x-x^\prime|} \left[sgn(x-x^\prime)k_n^6[\phi_z] + k_n^5[\partial_{x}\phi_z] \right] }[/math]

and

[math]\displaystyle{ \partial_x^3\phi_z = \partial_x^3\phi_z^{In} - \partial_x^3\psi_a [\phi_z] + \partial_x^3\psi_s [\partial_{x}\phi_z] }[/math]

[math]\displaystyle{ = k_0^3e^{k_0x} \frac{\sin(k_0z+H)}{\sin(k_0H)} - \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x-x^\prime|} \left[k_n^7[\phi_z] + sgn(x-x^\prime)k_n^6[\partial_{x}\phi_z] \right] }[/math]

At [math]\displaystyle{ x=x^\prime }[/math] and z=0, the first edge conditions gives

[math]\displaystyle{ k_0^2e^{k_0x^\prime} + \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} \left[sgn(x-x^\prime)k_n^6[\phi_z] + k_n^5[\partial_{x}\phi_z] \right] = 0 }[/math]

and the second edge condition gives

[math]\displaystyle{ k_0^3e^{k_0x^\prime} - \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} \left[k_n^7[\phi_z] + sgn(x-x^\prime)k_n^6[\partial_{x}\phi_z] \right] = 0 }[/math]

The jump conditions [math]\displaystyle{ [\phi_z] }[/math] and [math]\displaystyle{ [\partial_{x}\phi_z] }[/math] can be solved by solving the edge conditions simultaneously.

The reflection and transmission coefficients, [math]\displaystyle{ R }[/math] and [math]\displaystyle{ T }[/math] can be found by taking the limit of [math]\displaystyle{ \phi_z }[/math] as [math]\displaystyle{ x\rightarrow\pm\infty }[/math] ie

[math]\displaystyle{ R = \lim\limits_{x\rightarrow\infty} \left( \phi_{z}^{\mathrm{In}}(x) -\psi_a [\phi_z] + \psi_s [\partial_{x}\phi_z]\right) }[/math] [math]\displaystyle{ = - \frac{i\beta\sin^2(k_0h)}{2\alpha C(k_0)}\left[k_0^4[\phi_z] - k_0^3[\partial_{x}\phi_z]\right] }[/math]

and

[math]\displaystyle{ T = \lim\limits_{x\rightarrow-\infty} \left(\phi_{z}^{\mathrm{In}}(x) -\psi_a [\phi_z] + \psi_s [\partial_{x}\phi_z]\right) }[/math] [math]\displaystyle{ =1 + \frac{i\beta\sin^2(k_0h)}{2\alpha C(k_0)}\left[k_0^4[\phi_z] + k_0^3[\partial_{x}\phi_z]\right] }[/math]

---

These are a bit wrong. The reflected is at minus infinity. Also - the reflected wave is the part travelling away from the point - you need to sort out the k stuff first

---

More complicated boundary conditions

More complicated boundary conditions can be treated using this formulation.

Previously, we considered plates with free edges ie with a zero bending moment and zero shear force at each edge. Here, we consider that each plate be connected by a series of flexural rotational springs (stiffness denoted by [math]\displaystyle{ S_r }[/math]) and vertical linear springs (stiffness denoted by [math]\displaystyle{ S_l }[/math]). The edge conditions become:

[math]\displaystyle{ \partial_x^3\phi_z^+(x^\prime) = -\frac{S_l}{\beta}\left( \phi_z^+(x^\prime) - \phi_z^-(x^\prime)\right) }[/math] [math]\displaystyle{ \partial_x^3\phi_z^-(x^\prime) = -\frac{S_l}{\beta}\left( (\phi_z^+(x^\prime) - \phi_z^-(x^\prime)\right) }[/math] [math]\displaystyle{ \partial_x^2\phi_z^+(x^\prime) = \frac{S_r}{\beta} \left( \partial_x\phi_z^+(x^\prime) - \partial_x\phi_z^-(x^\prime)\right) }[/math] [math]\displaystyle{ \partial_x^2\phi_z^-(x^\prime) = \frac{S_r}{\beta}\left( \partial_x\phi_z^+(x^\prime) - \partial_x\phi_z^-(x^\prime) \right) }[/math]

where [math]\displaystyle{ \phi^+ }[/math] is the left edge of the right plate and [math]\displaystyle{ ^- }[/math] is the right edge of the left plate.