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- Homework Statement
- Obtain mass and stiffness matrices given equation of motion
- Relevant Equations
- [tex] A_{11e} \frac{d^2 u_1}{dx^2} + (A_{12e} + A_{66e}) \frac{d^2 v_1}{dx,dy} + A_{66e} \frac{d^2 u_1}{dy^2} + \frac{G_2}{h_2} \left( u_3 - u_1 - d \frac{dw}{dx} \right) = (\rho_p h_p + \rho_s h_s) \frac{d^2 u_1}{dt^2} [/tex]
Hello,
This is not homework but I am trying to replicate some results I found in a paper. In short, the situation is as follows. The following equation is given:
[tex]A_{11e} \frac{d^2 u_1}{dx^2} + (A_{12e} + A_{66e}) \frac{d^2 v_1}{dxdy} + A_{66e} \frac{d^2 u_1}{dy^2} + \frac{G_2}{h_2} \left( u_3 - u_1 - d \frac{dw}{dx} \right) = (\rho_p h_p + \rho_s h_s) \frac{d^2 u_1}{dt^2}[/tex]
where [itex]u_1(x,y)[/itex], [itex]v_1(x,y)[/itex], [itex]u_3(x,y)[/itex], and [itex]w(x,y)[/itex] are given by:
[tex]u_1(x,y) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} U_{1mn}(t) \frac{dX(x)}{dx} Y(y)[/tex]
[tex]u_3(x,y) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} U_{3mn}(t) \frac{dX(x)}{dx} Y(y)[/tex]
[tex]v_1(x,y) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} V_{1mn}(t) X(x) \frac{dY(y)}{dy}[/tex]
[tex]w(x,y) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} W_{mn}(t) X(x) Y(y)[/tex]
with [itex]X(x) = \sin\left( \frac{m \pi x}{a} \right)[/itex] and [itex]Y(y) = \sin\left( \frac{n \pi y}{b} \right)[/itex].
All other quantities are constants.
As far as I understand, the author substitutes the proposed solutions and groups terms according to the temporal variable that accompanies them.
He then rewrites the equation as:
[tex]K_{11} U_{1mn} + K_{12} V_{1mn} + K_{13} U_{3mn} + K_{15} W_{mn} + M_{11} \ddot{U}_{1mn} = 0[/tex]
where (according to him):
[tex]K_{11} = -A_{11e} \left( \frac{m \pi}{a} \right)^2 - A_{66e} \left( \frac{n \pi}{b} \right)^2 - \frac{G_2}{h_2}[/tex]
[tex]K_{12} = -(A_{12e} + A_{66e}) \left( \frac{m \pi}{a} \right) \left( \frac{n \pi}{b} \right)[/tex]
[tex]K_{13} = \frac{G_2}{h_2}[/tex]
[tex]K_{15} = -\frac{G_2 d}{h_2} \left( \frac{m \pi}{a} \right)[/tex]
[tex]M_{11} = -(\rho_p h_p + \rho_s h_s)[/tex]
The problem is that he doesn’t explain how these elements [itex]K_{ij}[/itex] and [itex]M_{ij}[/itex] are obtained.
I assume he multiplied both sides by [itex]\frac{dX(x)}{dx} Y(y)[/itex] and integrated over [itex]x \in [0,a][/itex] and [itex]y \in [0,b][/itex], as if applying orthogonality conditions.
However, when I do that, I only recover some of the terms:
[tex]K_{11} = -A_{11e} \left( \frac{m \pi}{a} \right)^2 - A_{66e} \left( \frac{n \pi}{b} \right)^2 - \frac{G_2}{h_2}[/tex]
[tex]K_{12} = -(A_{12e} + A_{66e}) \left( \frac{n \pi}{b} \right)^2[/tex]
[tex]K_{13} = \frac{G_2}{h_2}[/tex]
[tex]K_{15} = -\frac{G_2 d}{h_2}[/tex]
[tex]M_{11} = -(\rho_p h_p + \rho_s h_s)[/tex]
I cross-checked my results with a Mathematica script, and they seem consistent.
Does anyone have an idea of what the author might be doing differently?
Thanks in advance.
This is not homework but I am trying to replicate some results I found in a paper. In short, the situation is as follows. The following equation is given:
[tex]A_{11e} \frac{d^2 u_1}{dx^2} + (A_{12e} + A_{66e}) \frac{d^2 v_1}{dxdy} + A_{66e} \frac{d^2 u_1}{dy^2} + \frac{G_2}{h_2} \left( u_3 - u_1 - d \frac{dw}{dx} \right) = (\rho_p h_p + \rho_s h_s) \frac{d^2 u_1}{dt^2}[/tex]
where [itex]u_1(x,y)[/itex], [itex]v_1(x,y)[/itex], [itex]u_3(x,y)[/itex], and [itex]w(x,y)[/itex] are given by:
[tex]u_1(x,y) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} U_{1mn}(t) \frac{dX(x)}{dx} Y(y)[/tex]
[tex]u_3(x,y) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} U_{3mn}(t) \frac{dX(x)}{dx} Y(y)[/tex]
[tex]v_1(x,y) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} V_{1mn}(t) X(x) \frac{dY(y)}{dy}[/tex]
[tex]w(x,y) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} W_{mn}(t) X(x) Y(y)[/tex]
with [itex]X(x) = \sin\left( \frac{m \pi x}{a} \right)[/itex] and [itex]Y(y) = \sin\left( \frac{n \pi y}{b} \right)[/itex].
All other quantities are constants.
As far as I understand, the author substitutes the proposed solutions and groups terms according to the temporal variable that accompanies them.
He then rewrites the equation as:
[tex]K_{11} U_{1mn} + K_{12} V_{1mn} + K_{13} U_{3mn} + K_{15} W_{mn} + M_{11} \ddot{U}_{1mn} = 0[/tex]
where (according to him):
[tex]K_{11} = -A_{11e} \left( \frac{m \pi}{a} \right)^2 - A_{66e} \left( \frac{n \pi}{b} \right)^2 - \frac{G_2}{h_2}[/tex]
[tex]K_{12} = -(A_{12e} + A_{66e}) \left( \frac{m \pi}{a} \right) \left( \frac{n \pi}{b} \right)[/tex]
[tex]K_{13} = \frac{G_2}{h_2}[/tex]
[tex]K_{15} = -\frac{G_2 d}{h_2} \left( \frac{m \pi}{a} \right)[/tex]
[tex]M_{11} = -(\rho_p h_p + \rho_s h_s)[/tex]
The problem is that he doesn’t explain how these elements [itex]K_{ij}[/itex] and [itex]M_{ij}[/itex] are obtained.
I assume he multiplied both sides by [itex]\frac{dX(x)}{dx} Y(y)[/itex] and integrated over [itex]x \in [0,a][/itex] and [itex]y \in [0,b][/itex], as if applying orthogonality conditions.
However, when I do that, I only recover some of the terms:
[tex]K_{11} = -A_{11e} \left( \frac{m \pi}{a} \right)^2 - A_{66e} \left( \frac{n \pi}{b} \right)^2 - \frac{G_2}{h_2}[/tex]
[tex]K_{12} = -(A_{12e} + A_{66e}) \left( \frac{n \pi}{b} \right)^2[/tex]
[tex]K_{13} = \frac{G_2}{h_2}[/tex]
[tex]K_{15} = -\frac{G_2 d}{h_2}[/tex]
[tex]M_{11} = -(\rho_p h_p + \rho_s h_s)[/tex]
I cross-checked my results with a Mathematica script, and they seem consistent.
Does anyone have an idea of what the author might be doing differently?
Thanks in advance.