Difference between revisions of "Floating Elastic Plates of Identical Properties"
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We now present the solution for the case of a set of cracks with waves incident from normal. | We now present the solution for the case of a set of cracks with waves incident from normal. | ||
− | The solution is similiar to the case of one crack except that we solve for <math> x^\prime = {a_1, a_2 a_3 . . .} </math> where <math>a_n</math> is the | + | The solution is similiar to the case of one crack except that we solve for <math> x^\prime = {a_1, a_2, a_3 . . .} </math> where <math>a_n</math> is the position of the <math>n^th</math> crack. |
Revision as of 22:21, 27 May 2007
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
The Free-Surface Green Function for a Floating Elastic Plate satisfies the following equations (plus the Sommerfeld Radiation Condition far from the body)
where
and [math]\displaystyle{ k_n }[/math] are the solutions of the Dispersion Relation for a Floating Elastic Plate,
with [math]\displaystyle{ e^{i\omega t} }[/math] so that [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 negative 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
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
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]
where we take the limit as N goes to infinity.
We now introduce the incident potential so that for x small
for large x
where we choose A to normalise and assume k_0 is positive imaginary.
and for both small and large x
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]
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]
???????????? Hmmm is this right? I know A normalises, but should it be this complex. ?????????????
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,
and we are left with
At z=0, the z variable disappears to give
We then substitute [math]\displaystyle{ G = \phi }[/math] to remove [math]\displaystyle{ \phi }[/math] and obtain
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
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]
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]
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
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
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
where
and
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
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]
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]
At [math]\displaystyle{ x=x^\prime }[/math] and z=0, the first edge conditions gives
and the second edge condition gives
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
and
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:
where [math]\displaystyle{ \phi^+ }[/math] is the left edge of the right plate and [math]\displaystyle{ \phi^- }[/math] is the right edge of the left plate.
These edge conditions imply [math]\displaystyle{ \frac{\partial^2\phi_n^+(x^\prime)}{\partial x^2} = \frac{\partial^2\phi_n^-(x^\prime)}{\partial x^2} }[/math] and [math]\displaystyle{ \frac{\partial^3\phi_n^+(x^\prime)}{\partial x^3} = \frac{\partial^3\phi_n^-(x^\prime)}{\partial x^3} }[/math] which imply [math]\displaystyle{ [\partial_x^2\phi_n] = 0 }[/math] and [math]\displaystyle{ [\partial_x^3\phi_n] = 0 }[/math].
[math]\displaystyle{ \phi_n }[/math] can now be re-expressed by
We now have two unknowns which can be solved simultaneously using the following two edge conditions
and
where [math]\displaystyle{ s_l = \beta/S_l }[/math] and [math]\displaystyle{ s_r = \beta/S_r }[/math].
A Set of Floating Elastic Plates of Identical Properties
We now present the solution for the case of a set of cracks with waves incident from normal.
The solution is similiar to the case of one crack except that we solve for [math]\displaystyle{ x^\prime = {a_1, a_2, a_3 . . .} }[/math] where [math]\displaystyle{ a_n }[/math] is the position of the [math]\displaystyle{ n^th }[/math] crack.