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| [[Image:BendingCircularPlate.png| thumb | 300px | Bending of an edge clamped circular plate under the action of a transverse pressure. The left half of the plate shows the deformed shape while the right half shows the undeformed shape. This calculation was performed using [[Ansys]].]]
| | Hi there! :) My name is Shellie, I'm a student studying Arts and Sciences from Interlaken, United States.<br><br>Feel free to visit my site; [http://www.oyunmatik.net/profile/wisimcox.html Here is your mountain bike sizing.] |
| '''Bending of plates''' or plate bending refers to the [[Deflection (engineering)|deflection]] of a [[plate]] perpendicular to the plane of the plate under the action of external [[force]]s and [[Moment (physics)|moments]]. The amount of deflection can be determined by solving the differential equations of an appropriate [[plate theory]]. The [[stress (physics)|stress]]es in the plate can be calculated from these deflections. Once the stresses are known, [[material failure theory|failure theories]] can be used to determine whether a plate will fail under a given load.
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| == Bending of Kirchhoff-Love plates ==
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| [[Image:PlateForcesMoments upd.png|thumb | 350px | Forces and moments on a flat plate.]]
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| In the [[Kirchhoff–Love plate theory]] for plates the governing equations are<ref name=Reddy>Reddy, J. N., 2007, '''Theory and analysis of elastic plates and shells''', CRC Press, Taylor and Francis.</ref>
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| :<math>
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| N_{\alpha\beta,\alpha} = 0
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| </math>
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| and
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| :<math>
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| M_{\alpha\beta,\alpha\beta} - q = 0
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| </math>
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| In expanded form,
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| :<math>
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| \cfrac{\partial N_{11}}{\partial x_1} + \cfrac{\partial N_{21}}{\partial x_2} = 0 ~;~~
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| \cfrac{\partial N_{12}}{\partial x_1} + \cfrac{\partial N_{22}}{\partial x_2} = 0
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| </math>
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| and
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| :<math>
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| \cfrac{\partial^2 M_{11}}{\partial x_1^2} + 2\cfrac{\partial^2 M_{12}}{\partial x_1 \partial x_2} +
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| \cfrac{\partial^2 M_{22}}{\partial x_2^2} = q
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| </math>
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| where <math>q(x)</math> is an applied transverse [[load]] per unit area, the thickness of the plate is <math>H=2h</math>, the stresses are <math>\sigma_{ij}</math>, and
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| :<math>
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| N_{\alpha\beta} := \int_{-h}^h \sigma_{\alpha\beta}~dx_3 ~;~~
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| M_{\alpha\beta} := \int_{-h}^h x_3~\sigma_{\alpha\beta}~dx_3~.
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| </math>
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| The quantity <math>N</math> has units of [[force]] per unit length. The quantity <math>M</math> has units of [[Moment (physics)|moment]] per unit length.
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| For [[isotropic]], [[homogeneous]], plates with [[Young's modulus]] <math>E</math> and [[Poisson's ratio]] <math>\nu</math> these equations reduce to<ref name=Timo>Timoshenko, S. and Woinowsky-Krieger, S., (1959), '''Theory of plates and shells''', McGraw-Hill New York.</ref>
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| :<math>
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| \nabla^2\nabla^2 w = -\cfrac{q}{D} ~;~~ D := \cfrac{2h^3E}{3(1-\nu^2)} = \cfrac{H^3E}{12(1-\nu^2)}
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| </math>
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| where <math>w(x_1,x_2)</math> is the deflection of the mid-surface of the plate.
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| In rectangular Cartesian coordinates,
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| :<math>
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| \cfrac{\partial^4 w}{\partial x_1^4} + 2\cfrac{\partial^4 w}{\partial x_1^2 \partial x_2^2} +
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| \cfrac{\partial^4 w}{\partial x_2^4} = -\cfrac{q}{D} \,.
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| </math>
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| ==Circular Kirchhoff-Love plates==
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| The bending of circular plates can be examined by solving the governing equation with
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| appropriate boundary conditions. These solutions were first found by Poisson in 1829.
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| Cylindrical coordinates are convenient for such problems.
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| The governing equation in coordinate-free form is
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| :<math>
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| \nabla^2 \nabla^2 w = -\frac{q}{D} \,.
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| </math>
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| In cylindrical coordinates <math>(r, \theta, z)</math>,
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| :<math>
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| \nabla^2 w \equiv \frac{1}{r}\frac{\partial }{\partial r}\left(r \frac{\partial w}{\partial r}\right) +
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| \frac{1}{r^2}\frac{\partial^2 w}{\partial \theta^2} + \frac{\partial^2 w}{\partial z^2} \,.
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| </math>
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| For symmetrically loaded circular plates, <math> w = w(r)</math>, and we have
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| :<math>
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| \nabla^2 w \equiv \frac{1}{r}\cfrac{d }{d r}\left(r \cfrac{d w}{d r}\right) \,.
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| </math>
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| Therefore, the governing equation is
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| :<math>
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| \frac{1}{r}\cfrac{d }{d r}\left[r \cfrac{d }{d r}\left\{\frac{1}{r}\cfrac{d }{d r}\left(r \cfrac{d w}{d r}\right)\right\}\right] = -\frac{q}{D}\,.
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| </math>
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| If <math>q</math> and <math>D</math> are constant, direct integration of the governing equation gives us
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| <blockquote style="border: 1px solid black; padding:10px; width:530px">
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| :<math>
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| w(r) = -\frac{qr^4}{64 D} + C_1\ln r + \cfrac{C_2 r^2}{2} + \cfrac{C_3r^2}{4}(2\ln r - 1) + C_4
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| </math>
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| </blockquote>
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| where <math>C_i</math> are constants. The slope of the deflection surface is
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| :<math>
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| \phi(r) = \cfrac{d w}{d r} = -\frac{qr^3}{16D} + \frac{C_1}{r} + C_2 r + C_3 r \ln r \,.
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| </math>
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| For a circular plate, the requirement that the deflection and the slope of the deflection are finite
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| at <math>r = 0</math> implies that <math>C_1 = C_3 = 0</math>.
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| ===Clamped edges===
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| For a circular plate with clamped edges, we have <math>w(a) = 0</math> and <math>\phi(a) = 0</math> at the edge of
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| the plate (radius <math>a</math>). Using these boundary conditions we get
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| <blockquote style="border: 1px solid black; padding:10px; width:530px">
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| :<math>
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| w(r) = -\frac{q}{64 D} (a^2 -r^2)^2 \quad \text{and} \quad
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| \phi(r) = \frac{qr}{16 D}(a^2-r^2) \,.
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| </math>
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| </blockquote>
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| The in-plane displacements in the plate are
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| :<math>
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| u_r(r) = -z\phi(r) \quad \text{and} \quad u_\theta(r) = 0 \,.
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| </math>
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| The in-plane strains in the plate are
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| :<math>
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| \varepsilon_{rr} = \cfrac{d u_r}{d r} = -\frac{qz}{16D}(a^2-3r^2) ~,~~
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| \varepsilon_{\theta\theta} = \frac{u_r}{r} = -\frac{qz}{16D}(a^2-r^2) ~,~~
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| \varepsilon_{r\theta} = 0 \,.
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| </math>
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| The in-plane stresses in the plate are
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| :<math>
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| \sigma_{rr} = \frac{E}{1-\nu^2}\left[\varepsilon_{rr} + \nu\varepsilon_{\theta\theta}\right] ~;~~
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| \sigma_{\theta\theta} = \frac{E}{1-\nu^2}\left[\varepsilon_{\theta\theta} + \nu\varepsilon_{rr}\right] ~;~~
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| \sigma_{r\theta} = 0 \,.
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| </math>
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| For a plate of thickness <math>2h</math>, the bending stiffness is <math>D = 2Eh^3/[3(1-\nu^2)]</math> and we
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| have
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| <blockquote style="border: 1px solid black; padding:10px; width:430px">
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| :<math>
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| \begin{align}
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| \sigma_{rr} &= -\frac{3qz}{32h^3}\left[(1+\nu)a^2-(3+\nu)r^2\right] \\
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| \sigma_{\theta\theta} &= -\frac{3qz}{32h^3}\left[(1+\nu)a^2-(1+3\nu)r^2\right]\\
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| \sigma_{r\theta} &= 0 \,.
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| \end{align}
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| </math>
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| </blockquote>
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| The moment resultants (bending moments) are
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| :<math>
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| M_{rr} = -\frac{q}{16}\left[(1+\nu)a^2-(3+\nu)r^2\right] ~;~~
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| M_{\theta\theta} = -\frac{q}{16}\left[(1+\nu)a^2-(1+3\nu)r^2\right] ~;~~
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| M_{r\theta} = 0 \,.
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| </math>
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| The maximum radial stress is at <math>z = h</math> and <math>r = a</math>:
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| :<math>
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| \left.\sigma_{rr}\right|_{z=h,r=a} = \frac{3qa^2}{16h^2} = \frac{3qa^2}{4H^2}
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| </math>
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| where <math>H := 2h</math>. The bending moments at the boundary and the center of the plate are
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| :<math>
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| \left.M_{rr}\right|_{r=a} = \frac{qa^2}{8} ~,~~
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| \left.M_{\theta\theta}\right|_{r=a} = \frac{\nu qa^2}{8} ~,~~
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| \left.M_{rr}\right|_{r=0} = \left.M_{\theta\theta}\right|_{r=0} = -\frac{(1+\nu) qa^2}{16} \,.
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| </math>
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| ==Rectangular Kirchhoff-Love plates==
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| [[Image:RectangularPlateBending.svg|thumb | 250px | Bending of a rectangular plate under the action of a distributed force <math>q</math> per unit area.]]
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| For rectangular plates, Navier in 1820 introduced a simple method for finding the displacement and stress when a plate is simply supported. The idea was to express the applied load in terms of Fourier components, find the solution for a sinusoidal load (a single Fourier component), and then superimpose the Fourier components to get the solution for an arbitrary load.
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| ===Sinusoidal load===
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| Let us assume that the load is of the form
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| :<math>
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| q(x,y) = q_0 \sin\frac{\pi x}{a}\sin\frac{\pi y}{b} \,.
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| </math>
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| Here <math>q_0</math> is the amplitude, <math>a</math> is the width of the plate in the <math>x</math>-direction, and
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| <math>b</math> is the width of the plate in the <math>y</math>-direction.
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| Since the plate is simply supported, the displacement <math>w(x,y)</math> along the edges of
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| the plate is zero, the bending moment <math>M_{xx}</math> is zero at <math>x=0</math> and <math>x=a</math>, and
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| <math>M_{yy}</math> is zero at <math>y=0</math> and <math>y=b</math>.
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| If we apply these boundary conditions and solve the plate equation, we get the
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| solution
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| :<math>
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| w(x,y) = \frac{q_0}{\pi^4 D}\,\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^{-2}\,\sin\frac{\pi x}{a}\sin\frac{\pi y}{b} \,.
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| </math>
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| We can calculate the stresses and strains in the plate once we know the displacement.
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| For a more general load of the form
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| :<math>
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| q(x,y) = q_0 \sin\frac{m \pi x}{a}\sin\frac{n \pi y}{b}
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| </math>
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| where <math>m</math> and <math>n</math> are integers, we get the solution
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| <blockquote style="border: 1px solid black; padding:10px; width:530px">
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| :<math> \text{(1)} \qquad
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| w(x,y) = \frac{q_0}{\pi^4 D}\,\left(\frac{m^2}{a^2}+\frac{n^2}{b^2}\right)^{-2}\,\sin\frac{m \pi x}{a}\sin\frac{n \pi y}{b} \,.
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| </math>
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| </blockquote>
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| ===Navier solution===
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| Let us now consider a more general load <math>q(x,y)</math>. We can break this load up into
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| a sum of Fourier components such that
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| :<math>
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| q(x,y) = \sum_{m=1}^{\infty} \sum_{n=1}^\infty a_{mn}\sin\frac{m \pi x}{a}\sin\frac{n \pi y}{b}
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| </math>
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| where <math>a_{mn}</math> is an amplitude. We can use the orthogonality of Fourier components,
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| :<math>
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| \int_0^a \sin\frac{k\pi x}{a}\sin\frac{\ell \pi x}{a}\text{d}x =
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| \begin{cases} 0 & k \ne \ell \\ a/2 & k = \ell \end{cases}
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| </math>
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| to find the amplitudes <math>a_{mn}</math>. Thus we have, by integrating over <math>y</math>,
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| :<math>
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| \int_0^b q(x,y)\sin\frac{\ell\pi y}{b}\,\text{d}y =
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| \sum_{m=1}^{\infty} \sum_{n=1}^\infty a_{mn}\sin\frac{m \pi x}{a}
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| \int_0^b \sin\frac{n \pi y}{b} \sin\frac{\ell\pi y}{b}\,\text{d}y =
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| \frac{b}{2}\sum_{m=1}^{\infty} a_{m\ell}\sin\frac{m \pi x}{a} \,.
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| </math>
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| If we repeat the process by integrating over <math>x</math>, we have
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| :<math>
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| \int_0^b \int_0^a q(x,y)\sin\frac{k\pi x}{a}\sin\frac{\ell\pi y}{b}\,\text{d}x\text{d}y =
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| \frac{b}{2}\sum_{m=1}^{\infty} a_{m\ell}
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| \int_0^a \sin\frac{m \pi x}{a} \sin\frac{k\pi x}{a}\,\text{d}x =
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| \frac{ab}{4} a_{k\ell} \,.
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| </math>
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| Therefore,
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| :<math>
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| a_{mn} = \frac{4}{ab}
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| \int_0^b \int_0^a q(x,y)\sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b}\,\text{d}x\text{d}y \,.
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| </math>
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| Now that we know <math>a_{mn}</math>, we can just superpose solutions of the form given in
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| equation (1) to get the displacement, i.e.,
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| <blockquote style="border: 1px solid black; padding:10px; width:630px">
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| :<math> \text{(2)} \qquad
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| w(x,y) = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{a_{mn}}{\pi^4 D}\,\left(\frac{m^2}{a^2}+\frac{n^2}{b^2}\right)^{-2}\,\sin\frac{m \pi x}{a}\sin\frac{n \pi y}{b} \,.
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| </math>
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| </blockquote>
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| | |
| ====Uniform load====
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| Consider the situation where a uniform load is applied on the plate, i.e.,
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| <math>q(x,y) = q_0</math>. Then
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| :<math>
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| a_{mn} = \frac{4q_0}{ab}
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| \int_0^a \int_0^b \sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b}\,\text{d}x\text{d}y \,.
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| </math>
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| Now
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| :<math>
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| \int_0^a \sin\frac{m\pi x}{a}\,\text{d}x = \frac{a}{m\pi}(1 - \cos m\pi) \quad\text{and}\quad
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| \int_0^b \sin\frac{n\pi y}{b}\,\text{d}y = \frac{b}{n\pi}(1 - \cos n\pi)\,.
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| </math>
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| We can use these relations to get a simpler expression for <math>a_{mn}</math>:
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| :<math>
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| a_{mn} = \frac{4q_0}{mn\pi^2}(1 - \cos m\pi)(1 - \cos n\pi) \,.
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| </math>
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| Since <math>\cos m\pi = \cos n\pi = 1</math> [ so <math>(1 - \cos m\pi) = (1 - \cos n\pi) = 0</math> ] when <math>m</math> and <math>n</math> are even, we can get an even simpler expression for <math>a_{mn}</math> when both <math>m</math> and <math>n</math> are odd:
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| :<math>
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| a_{mn} = \begin{cases}
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| 0 & m~\text{or}~n~\text{even}, \\
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| \cfrac{16q_0}{mn\pi^2} & m~\text{and}~n~\text{odd}\,.
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| \end{cases}
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| </math>
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| Plugging this expression into equation (2) and keeping in mind
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| that only odd terms contribute to the displacement, we have
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| <blockquote style="border: 1px solid black; padding:10px; width:630px">
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| :<math>
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| \begin{align}
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| w(x,y) & = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{16 q_0}{(2m-1)(2n-1)\pi^6 D}\,\left[\frac{(2m-1)^2}{a^2}+\frac{(2n-1)^2}{b^2}\right]^{-2} \,\times\\
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| & \qquad \qquad \quad \sin\frac{(2m-1) \pi x}{a}\sin\frac{(2n-1) \pi y}{b} \,.
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| \end{align}
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| </math>
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| </blockquote>
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| The corresponding moments are given by
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| :<math>
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| \begin{align}
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| M_{xx} & = -D\left(\frac{\partial^2 w}{\partial x^2} + \nu \frac{\partial^2 w}{\partial y^2}\right) \\
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| & = \sum_{m=1}^\infty \sum_{n=1}^\infty\frac{16 q_0}{(2m-1)(2n-1)\pi^4}\,
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| \left[\frac{(2m-1)^2}{a^2}+\nu\frac{(2n-1)^2}{b^2}\right] \,\times\\
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| & \qquad \qquad \left[\frac{(2m-1)^2}{a^2}+\frac{(2n-1)^2}{b^2}\right]^{-2}
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| \sin\frac{(2m-1) \pi x}{a}\sin\frac{(2n-1) \pi y}{b} \\
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| M_{yy} & = -D\left(\frac{\partial^2 w}{\partial y^2} + \nu \frac{\partial^2 w}{\partial x^2}\right) \\
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| & = \sum_{m=1}^\infty \sum_{n=1}^\infty\frac{16 q_0}{(2m-1)(2n-1)\pi^4}\,
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| \left[\frac{(2n-1)^2}{b^2}+\nu\frac{(2m-1)^2}{a^2}\right] \,\times\\
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| & \qquad \qquad \left[\frac{(2m-1)^2}{a^2}+\frac{(2n-1)^2}{b^2}\right]^{-2}
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| \sin\frac{(2m-1) \pi x}{a}\sin\frac{(2n-1) \pi y}{b} \,.
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| \end{align}
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| </math>
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| The stresses in the plate are
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| :<math>
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| \sigma_{xx} = \frac{3z}{2h^3}\,M_{xx} = \frac{12 z}{H^3}\,M_{xx} \quad \text{and} \quad
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| \sigma_{yy} = \frac{3z}{2h^3}\,M_{yy} = \frac{12 z}{H^3}\,M_{yy} \,.
| |
| </math>
| |
| :{{multiple image
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| | width = 400
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| | footer = Displacement and stresses along <math>x=a/2</math> for a rectangular plate with <math>a=20</math> mm, <math>b=40</math> mm, <math>H=2h=0.4</math> mm, <math>E=70</math> GPa, and <math>\nu=0.35</math> under a load <math>q_0 = -10</math> kPa. The red line represents the bottom of the plate, the green line the middle, and the blue line the top of the plate.
| |
| | image1 = wx rectangularPlate.svg
| |
| | caption1 = Displacement (<math>w</math>)
| |
| | image2 = sxx rectangularPlate.svg
| |
| | caption2 = Stress (<math>\sigma_{xx}</math>)
| |
| | image3 = syy rectangularPlate.svg
| |
| | caption3 = Stress (<math>\sigma_{yy}</math>)
| |
| }}
| |
| <!--
| |
| <gallery widths=270px heights=270px perrow=2 caption="Displacement and stresses along <math>x=a/2</math> for a rectangular plate with <math>a=20</math> mm, <math>b=40</math> mm, <math>H=2h=0.4</math> mm, <math>E=70</math> GPa, and <math>\nu=0.35</math> under a load <math>q_0 = -10</math> kPa. The red line represents the bottom of the plate, the green line the middle, and the blue line the top of the plate.">
| |
| file:wx rectangularPlate.svg|Displacement (<math>w</math>)
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| file:sxx rectangularPlate.svg|Stress (<math>\sigma_{xx}</math>)
| |
| file:syy rectangularPlate.svg|Stress (<math>\sigma_{yy}</math>)
| |
| </gallery>
| |
| -->
| |
| | |
| ===Levy solution===
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| Another approach was proposed by Levy in 1899. In this case we start with an
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| assumed form of the displacement and try to fit the parameters so that the
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| governing equation and the boundary conditions are satisfied.
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| | |
| Let us assume that
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| :<math>
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| w(x,y) = \sum_{m=1}^\infty Y_m(y) \sin \frac{m\pi x}{a} \,.
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| </math>
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| For a plate that is simply supported at <math>x=0</math> and <math>x=a</math>, the boundary conditions
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| are <math>w=0</math> and <math>M_{xx} = 0</math>. The moment boundary condition is equivalent to
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| <math>\partial^2 w/\partial x^2 = 0</math> (verify). The goal is to find <math>Y_m(y)</math> such that
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| it satisfies the boundary conditions at <math>y = 0</math> and <math>y = b</math> and, of course, the
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| governing equation <math>\nabla^2 \nabla^2 w = q/D</math>.
| |
| | |
| ====Moments along edges====
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| Let us consider the case of pure moment loading. In that case <math>q = 0</math> and
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| <math>w(x,y)</math> has to satisfy <math>\nabla^2 \nabla^2 w = 0</math>. Since we are working in rectangular
| |
| Cartesian coordinates, the governing equation can be expanded as
| |
| :<math>
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| \frac{\partial^4 w}{\partial x^4} + 2 \frac{\partial^4 w}{\partial x^2\partial y^2}
| |
| + \frac{\partial^4 w}{\partial y^4} = 0 \,.
| |
| </math>
| |
| Plugging the expression for <math>w(x,y)</math> in the governing equation gives us
| |
| :<math>
| |
| \sum_{m=1}^\infty \left[\left(\frac{m\pi}{a}\right)^4 Y_m \sin\frac{m\pi x}{a}
| |
| - 2\left(\frac{m\pi}{a}\right)^2 \cfrac{d^2 Y_m}{d y^2} \sin\frac{m\pi x}{a}
| |
| + \frac{d^4Y_m}{dy^4} \sin\frac{m\pi x}{a}\right] = 0
| |
| </math>
| |
| or
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| :<math>
| |
| \frac{d^4Y_m}{dy^4} - 2 \frac{m^2\pi^2}{a^2} \cfrac{d^2Y_m}{dy^2} + \frac{m^4\pi^4}{a^4} Y_m = 0 \,.
| |
| </math>
| |
| This is an ordinary differential equation which has the general solution
| |
| :<math>
| |
| Y_m = A_m \cosh\frac{m\pi y}{a} + B_m\frac{m\pi y}{a} \cosh\frac{m\pi y}{a} +
| |
| C_m \sinh\frac{m\pi y}{a} + D_m\frac{m\pi y}{a} \sinh\frac{m\pi y}{a}
| |
| </math>
| |
| where <math>A_m, B_m, C_m, D_m</math> are constants that can be determined from the boundary
| |
| conditions. Therefore the displacement solution has the form
| |
| <blockquote style="border: 1px solid black; padding:1px; width:800px">
| |
| :<math>
| |
| w(x,y) = \sum_{m=1}^\infty \left[
| |
| \left(A_m + B_m\frac{m\pi y}{a}\right) \cosh\frac{m\pi y}{a} +
| |
| \left(C_m + D_m\frac{m\pi y}{a}\right) \sinh\frac{m\pi y}{a}
| |
| \right] \sin \frac{m\pi x}{a} \,.
| |
| </math> | |
| </blockquote> | |
| Let us choose the coordinate system such that the boundaries of the plate are
| |
| at <math>x = 0</math> and <math>x = a</math> (same as before) and at <math>y = \pm b/2</math> (and not <math>y=0</math> and
| |
| <math>y=b</math>). Then the moment boundary conditions at the <math>y = \pm b/2</math> boundaries are
| |
| :<math>
| |
| w = 0 \,, -D\frac{\partial^2 w}{\partial y^2}\Bigr|_{y=b/2} = f_1(x) \,,
| |
| -D\frac{\partial^2 w}{\partial y^2}\Bigr|_{y=-b/2} = f_2(x)
| |
| </math>
| |
| where <math>f_1(x), f_2(x)</math> are known functions. The solution can be found by
| |
| applying these boundary conditions. We can show that for the ''symmetrical'' case
| |
| where
| |
| :<math>
| |
| M_{yy}\Bigr|_{y=-b/2} = M_{yy}\Bigr|_{y=b/2}
| |
| </math>
| |
| and
| |
| :<math>
| |
| f_1(x) = f_2(x) = \sum_{m=1}^\infty E_m\sin\frac{m\pi x}{a}
| |
| </math>
| |
| we have
| |
| <blockquote style="border: 1px solid black; padding:1px; width:800px">
| |
| :<math>
| |
| w(x,y) = \frac{a^2}{2\pi^2 D}\sum_{m=1}^\infty \frac{E_m}{m^2\cosh\alpha_m}\,
| |
| \sin\frac{m\pi x}{a}\, \left(\alpha_m \tanh\alpha_m \cosh\frac{m\pi y}{a}
| |
| - \frac{m\pi y}{a}\sinh\frac{m\pi y}{a}\right)
| |
| </math>
| |
| </blockquote>
| |
| where
| |
| :<math>
| |
| \alpha_m = \frac{m\pi b}{2a} \,.
| |
| </math>
| |
| Similarly, for the ''antisymmetrical'' case where
| |
| :<math>
| |
| M_{yy}\Bigr|_{y=-b/2} = -M_{yy}\Bigr|_{y=b/2}
| |
| </math>
| |
| we have
| |
| <blockquote style="border: 1px solid black; padding:1px; width:800px">
| |
| :<math>
| |
| w(x,y) = \frac{a^2}{2\pi^2 D}\sum_{m=1}^\infty \frac{E_m}{m^2\sinh\alpha_m}\,
| |
| \sin\frac{m\pi x}{a}\, \left(\alpha_m \coth\alpha_m \sinh\frac{m\pi y}{a}
| |
| - \frac{m\pi y}{a}\cosh\frac{m\pi y}{a}\right) \,.
| |
| </math>
| |
| </blockquote>
| |
| We can superpose the symmetric and antisymmetric solutions to get more general
| |
| solutions.
| |
| | |
| ==== Uniform and symmetric moment load ====
| |
| For the special case where the loading is symmetric and the moment is uniform, we have at <math>y=\pm b/2</math>,
| |
| :<math>
| |
| M_{yy} = f_1(x) = \frac{4M_0}{\pi}\sum_{m=1}^\infty \frac{1}{2m-1}\,\sin\frac{(2m-1)\pi x}{a} \,.
| |
| </math>
| |
| :{{multiple image
| |
| | width = 400
| |
| | footer = Displacement and stresses for a rectangular plate under uniform bending moment along the edges <math>y=-b/2</math> and <math>y=b/2</math>. The bending stress <math>\sigma_{yy}</math> is along the bottom surface of the plate. The transverse shear stress <math>\sigma_{yz}</math> is along the mid-surface of the plate.
| |
| | image1 = surfRecBMIso_w.png
| |
| | caption1 = Displacement (<math>w</math>)
| |
| | image2 = surfRecBMIso_sy.png
| |
| | caption2 = Bending stress (<math>\sigma_{yy}</math>)
| |
| | image3 = surfRecBMIso_syz.png
| |
| | caption3 = Transverse shear stress (<math>\sigma_{yz}</math>)
| |
| }}
| |
| The resulting displacement is
| |
| <blockquote style="border: 1px solid black; padding:1px; width:800px">
| |
| :<math>
| |
| \begin{align}
| |
| w(x,y) & = \frac{2M_0 a^2}{\pi^3 D}\sum_{m=1}^\infty
| |
| \frac{1}{(2m-1)^3\cosh\alpha_m}\sin\frac{(2m-1)\pi x}{a} \times\\
| |
| & \qquad \left[
| |
| \alpha_m\,\tanh\alpha_m\cosh\frac{(2m-1)\pi y}{a} -\frac{(2m-1)\pi y}{a}
| |
| \sinh\frac{(2m-1)\pi y}{a}\right]
| |
| \end{align}
| |
| </math>
| |
| </blockquote>
| |
| where
| |
| :<math>
| |
| \alpha_m = \frac{\pi (2m-1)b}{2a} \,.
| |
| </math>
| |
| The bending moments and shear forces corresponding to the displacement <math>w</math> are
| |
| :<math>
| |
| \begin{align}
| |
| M_{xx} & = -D\left(\frac{\partial^2 w}{\partial x^2}+\nu\,\frac{\partial^2 w}{\partial y^2}\right) \\
| |
| & = \frac{2M_0(1-\nu)}{\pi}\sum_{m=1}^\infty\frac{1}{(2m-1)\cosh\alpha_m}\,
| |
| \sin\frac{(2m-1)\pi x}{a}
| |
| \left[
| |
| -\frac{(2m-1)\pi y}{a}\sinh\frac{(2m-1)\pi y}{a} + \right. \\
| |
| & \qquad \qquad \qquad \qquad
| |
| \left. \left\{\frac{2\nu}{1-\nu} + \alpha_m\tanh\alpha_m\right\}\cosh\frac{(2m-1)\pi y}{a}
| |
| \right] \\
| |
| M_{xy} & = (1-\nu)D\frac{\partial^2 w}{\partial x \partial y} \\
| |
| & = -\frac{2M_0(1-\nu)}{\pi}\sum_{m=1}^\infty\frac{1}{(2m-1)
| |
| \cosh\alpha_m}\,\cos\frac{(2m-1)\pi x}{a}
| |
| \left[\frac{(2m-1)\pi y}{a}\cosh\frac{(2m-1)\pi y}{a} + \right. \\
| |
| & \qquad \qquad \qquad \qquad
| |
| \left. (1-\alpha_m\tanh\alpha_m)\sinh\frac{(2m-1)\pi y}{a}\right] \\
| |
| Q_{zx} & = \frac{\partial M_{xx}}{\partial x}-\frac{\partial M_{xy}}{\partial y} \\
| |
| & = \frac{4M_0}{a}\sum_{m=1}^\infty \frac{1}{\cosh\alpha_m}\,
| |
| \cos\frac{(2m-1)\pi x}{a}\cosh\frac{(2m-1)\pi y}{a}\,.
| |
| \end{align}
| |
| </math>
| |
| The stresses are
| |
| :<math> | |
| \sigma_{xx} = \frac{12z}{h^3}\,M_{xx} \quad \text{and} \quad
| |
| \sigma_{zx} = \frac{1}{\kappa h}\,Q_{zx}\left(1 - \frac{4z^2}{h^2}\right)\,.
| |
| </math>
| |
| | |
| === Cylindrical plate bending ===
| |
| Cylindrical bending occurs when a rectangular plate that has dimensions <math>a \times b \times h</math>, where <math>a \ll b</math> and the thickness <math>h</math> is small, is subjected to a uniform distributed load perpendicular to the plane of the plate. Such a plate takes the shape of the surface of a cylinder.
| |
| | |
| ==== Simply supported plate with axially fixed ends ====
| |
| For a simply supported plate under cylindrical bending with edges that are free to rotate but have a fixed <math>x_1</math>. Cylindrical bending solutions can be found using the Navier and Levy techniques.
| |
| | |
| ==Bending of thick Mindlin plates==
| |
| For thick plates, we have to consider the effect of through-the-thickness shears on
| |
| the orientation of the normal to the mid-surface after deformation. Mindlin's theory
| |
| provides one approach for find the deformation and stresses in such plates. Solutions
| |
| to Mindlin's theory can be derived from the equivalent Kirchhoff-Love solutions using
| |
| canonical relations.<ref name=lim03>Lim, G. T. and Reddy, J. N., 2003, ''On canonical bending
| |
| relationships for plates'', International Journal of Solids and Structures, vol. 40,
| |
| pp. 3039-3067.</ref>
| |
| | |
| ===Governing equations===
| |
| The canonical governing equation for isotropic thick plates can be expressed as<ref name=lim03/>
| |
| :<math>
| |
| \begin{align}
| |
| & \nabla^2 \left(\mathcal{M} - \frac{\mathcal{B}}{1+\nu}\,q\right) = -q \\
| |
| & \kappa G h\left(\nabla^2 w + \frac{\mathcal{M}}{D}\right) =
| |
| -\left(1 - \cfrac{\mathcal{B} c^2}{1+\nu}\right)q \\
| |
| & \nabla^2 \left(\frac{\partial \varphi_1}{\partial x_2} - \frac{\partial \varphi_2}{\partial x_1}\right)
| |
| = c^2\left(\frac{\partial \varphi_1}{\partial x_2} - \frac{\partial \varphi_2}{\partial x_1}\right)
| |
| \end{align}
| |
| </math>
| |
| where <math>q</math> is the applied transverse load, <math>G</math> is the shear modulus, <math>D = Eh^3/[12(1-\nu^2)]</math>
| |
| is the bending rigidity, <math>h</math> is the plate thickness, <math>c^2 = 2\kappa G h/[D(1-\nu)]</math>,
| |
| <math>\kappa</math> is the shear correction factor, <math>E</math> is the Young's modulus, <math>\nu</math> is the Poisson's
| |
| ratio, and
| |
| :<math>
| |
| \mathcal{M} = D\left[\mathcal{A}\left(\frac{\partial \varphi_1}{\partial x_1} + \frac{\partial \varphi_2}{\partial x_2}\right)
| |
| - (1-\mathcal{A})\nabla^2 w\right] + \frac{2q}{1-\nu^2}\mathcal{B} \,.
| |
| </math>
| |
| In Mindlin's theory, <math>w</math> is the transverse displacement of the mid-surface of the plate
| |
| and the quantities <math>\varphi_1</math> and <math>\varphi_2</math> are the rotations of the mid-surface normal
| |
| about the <math>x_2</math> and <math>x_1</math>-axes, respectively. The canonical parameters for this theory
| |
| are <math>\mathcal{A} = 1</math> and <math>\mathcal{B} = 0</math>. The shear correction factor <math>\kappa</math> usually has the
| |
| value <math>5/6</math>.
| |
| | |
| The solutions to the governing equations can be found if one knows the corresponding
| |
| Kirchhoff-Love solutions by using the relations
| |
| :<math>
| |
| \begin{align}
| |
| w & = w^K + \frac{\mathcal{M}^K}{\kappa G h}\left(1 - \frac{\mathcal{B} c^2}{2}\right)
| |
| - \Phi + \Psi \\
| |
| \varphi_1 & = - \frac{\partial w^K}{\partial x_1}
| |
| - \frac{1}{\kappa G h}\left(1 - \frac{1}{\mathcal{A}} - \frac{\mathcal{B} c^2}{2}\right)Q_1^K
| |
| + \frac{\partial }{\partial x_1}\left(\frac{D}{\kappa G h \mathcal{A}}\nabla^2 \Phi + \Phi - \Psi\right)
| |
| + \frac{1}{c^2}\frac{\partial \Omega}{\partial x_2} \\
| |
| \varphi_2 & = - \frac{\partial w^K}{\partial x_2}
| |
| - \frac{1}{\kappa G h}\left(1 - \frac{1}{\mathcal{A}} - \frac{\mathcal{B} c^2}{2}\right)Q_2^K
| |
| + \frac{\partial }{\partial x_2}\left(\frac{D}{\kappa G h \mathcal{A}}\nabla^2 \Phi + \Phi - \Psi\right)
| |
| + \frac{1}{c^2}\frac{\partial \Omega}{\partial x_1}
| |
| \end{align}
| |
| </math>
| |
| where <math>w^K</math> is the displacement predicted for a Kirchhoff-Love plate, <math>\Phi</math> is a
| |
| biharmonic function such that <math>\nabla^2 \nabla^2 \Phi = 0</math>, <math>\Psi</math> is a function that satisfies the
| |
| Laplace equation, <math>\nabla^2 \Psi = 0</math>, and
| |
| :<math>
| |
| \begin{align}
| |
| \mathcal{M} & = \mathcal{M}^K + \frac{\mathcal{B}}{1+\nu}\,q + D \nabla^2 \Phi ~;~~ \mathcal{M}^K := -D\nabla^2 w^K \\
| |
| Q_1^K & = -D\frac{\partial }{\partial x_1}\left(\nabla^2 w^K\right) ~,~~
| |
| Q_2^K = -D\frac{\partial }{\partial x_2}\left(\nabla^2 w^K\right) \\
| |
| \Omega & = \frac{\partial \varphi_1}{\partial x_2} - \frac{\partial \varphi_2}{\partial x_1} ~,~~ \nabla^2 \Omega = c^2\Omega \,.
| |
| \end{align}
| |
| </math>
| |
| | |
| ===Simply supported rectangular plates===
| |
| For simply supported plates, the ''Marcus moment'' sum vanishes, i.e.,
| |
| :<math>
| |
| \mathcal{M} = \frac{1}{1+\nu}(M_{11}+M_{22}) = D\left(\frac{\partial \varphi_1}{\partial x_2}+\frac{\partial \varphi_2}{\partial x_2}\right) = 0 \,.
| |
| </math>
| |
| In that case the functions <math>\Phi</math>, <math>\Psi</math>, <math>\Omega</math> vanish, and the Mindlin solution is
| |
| related to the corresponding Kirchhoff solution by
| |
| :<math>
| |
| w = w^K + \frac{\mathcal{M}^K}{\kappa G h} \,.
| |
| </math>
| |
| | |
| == Bending of Reissner-Stein cantilever plates ==
| |
| Reissner-Stein theory for cantilever plates<ref name=Reissner51>E. Reissner and M. Stein. Torsion and transverse bending of cantilever plates. Technical Note 2369, National Advisory Committee for Aeronautics,Washington, 1951.</ref> leads to the following coupled ordinary differential equations for a cantilever plate with concentrated end load <math>q_x(y)</math> at <math>x=a</math>.
| |
| :<math>
| |
| \begin{align}
| |
| & bD \frac{\mathrm{d}^4w_x}{\mathrm{d}x^4} = 0 \\
| |
| & \frac{b^3D}{12}\,\frac{\mathrm{d}^4\theta_x}{\mathrm{d}x^4} - 2bD(1-\nu)\cfrac{d^2 \theta_x}{d x^2} = 0
| |
| \end{align}
| |
| </math>
| |
| and the boundary conditions at <math>x=a</math> are
| |
| :<math>
| |
| \begin{align}
| |
| & bD\cfrac{d^3 w_x}{d x^3} + q_{x1} = 0 \quad,\quad
| |
| \frac{b^3D}{12}\cfrac{d^3 \theta_x}{d x^3} -2bD(1-\nu)\cfrac{d \theta_x}{d x} + q_{x2} = 0 \\
| |
| & bD\cfrac{d^2 w_x}{d x^2} = 0 \quad,\quad \frac{b^3D}{12}\cfrac{d^2 \theta_x}{d x^2} = 0 \,.
| |
| \end{align}
| |
| </math>
| |
| Solution of this system of two ODEs gives
| |
| :<math>
| |
| \begin{align}
| |
| w_x(x) & = \frac{q_{x1}}{6bD}\,(3ax^2 -x^3) \\
| |
| \theta_x(x) & = \frac{q_{x2}}{2bD(1-\nu)}\left[x - \frac{1}{\nu_b}\,
| |
| \left(\frac{\sinh(\nu_b a)}{\cosh[\nu_b (x-a)]} + \tanh[\nu_b(x-a)]\right)\right]
| |
| \end{align}
| |
| </math>
| |
| where <math>\nu_b = \sqrt{24(1-\nu)}/b</math>. The bending moments and shear forces corresponding to the displacement
| |
| <math>w = w_x + y\theta_x</math> are
| |
| :<math>
| |
| \begin{align}
| |
| M_{xx} & = -D\left(\frac{\partial^2 w}{\partial x^2}+\nu\,\frac{\partial^2 w}{\partial y^2}\right) \\
| |
| & = q_{x1}\left(\frac{x-a}{b}\right) - \left[\frac{3yq_{x2}}{b^3\nu_b\cosh^3[\nu_b(x-a)]}\right]
| |
| \times \\
| |
| & \quad \left[6\sinh(\nu_b a) - \sinh[\nu_b(2x-a)] +
| |
| \sinh[\nu_b(2x-3a)] + 8\sinh[\nu_b(x-a)]\right] \\
| |
| M_{xy} & = (1-\nu)D\frac{\partial^2 w}{\partial x \partial y} \\
| |
| & = \frac{q_{x2}}{2b}\left[1 -
| |
| \frac{2+\cosh[\nu_b(x-2a)] - \cosh[\nu_b x]}{2\cosh^2[\nu_b(x-a)]}\right] \\
| |
| Q_{zx} & = \frac{\partial M_{xx}}{\partial x}-\frac{\partial M_{xy}}{\partial y} \\
| |
| & = \frac{q_{x1}}{b} - \left(\frac{3yq_{x2}}{2b^3\cosh^4[\nu_b(x-a)]}\right)\times
| |
| \left[32 + \cosh[\nu_b(3x-2a)] - \cosh[\nu_b(3x-4a)]\right. \\
| |
| & \qquad \left. - 16\cosh[2\nu_b(x-a)] +
| |
| 23\cosh[\nu_b(x-2a)] - 23\cosh(\nu_b x)\right]\,.
| |
| \end{align}
| |
| </math>
| |
| The stresses are
| |
| :<math>
| |
| \sigma_{xx} = \frac{12z}{h^3}\,M_{xx} \quad \text{and} \quad
| |
| \sigma_{zx} = \frac{1}{\kappa h}\,Q_{zx}\left(1 - \frac{4z^2}{h^2}\right)\,.
| |
| </math>
| |
| If the applied load at the edge is constant, we recover the solutions for a beam under a
| |
| concentrated end load. If the applied load is a linear function of <math>y</math>, then
| |
| :<math>
| |
| q_{x1} = \int_{-b/2}^{b/2}q_0\left(\frac{1}{2} - \frac{y}{b}\right)\,\text{d}y = \frac{bq_0}{2} ~;~~
| |
| q_{x2} = \int_{-b/2}^{b/2}yq_0\left(\frac{1}{2} - \frac{y}{b}\right)\,\text{d}y = -\frac{b^2q_0}{12} \,.
| |
| </math>
| |
| | |
| == See also ==
| |
| *[[Bending]]
| |
| *[[Infinitesimal strain theory]]
| |
| *[[Kirchhoff–Love plate theory]]
| |
| *[[Linear elasticity]]
| |
| *[[Mindlin–Reissner plate theory]]
| |
| *[[Plate theory]]
| |
| *[[Stress (mechanics)]]
| |
| *[[Stress resultants]]
| |
| *[[Structural acoustics]]
| |
| *[[Vibration of plates]]
| |
| | |
| == References ==
| |
| {{reflist}}
| |
| | |
| {{DEFAULTSORT:Bending Of Plates}}
| |
| [[Category:Continuum mechanics]]
| |