User:Eml4500.f08.team.Bengtson/teamHW7

November 21st, 2008: Meeting 38
Motivation: The deformed shape of the tress element and the interpolation of the transverse displacements v(s)and s = \tilde{x} elements.

The PVW for Beams is as follows: $$\int_{0}^{L}{w(\tilde{x})}\left[-\frac{\partial ^2}{\partial x^2} \left((EI)\frac{\partial ^2v}{\partial x^2}\right) + f_t - m\ddot{v}\right]dx = 0 $$

This is true for all possible w(x), $$ x=\tilde{x}$$

Integration by parts of the first term.

$$(\alpha) = \int_{0}^{L}{w(\tilde{x})}\frac{\partial^2}{\partial x^2}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}dx$$

To do integration by parts, one must separate the different parts of the integration to do them. $$ r(x) = \left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}$$ $$r'(x) = \frac{\partial}{\partial x}\left(\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}\right)$$ $$ s(x)= w(x) $$ $$s'(x) = \frac{\partial w}{\partial x}$$ Putting these together into an integration by parts, you get:

$$ (\alpha) = \left[w\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}\right]^L_0 - \int_{0}^{L}\frac{dw}{dx}\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}dx$$

Where $$(B_1)=\left[w\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}\right]^L_0$$

Continue Integrating $$= (B_1) - \left[\frac{dw}{dx}\left(EI\right)\frac{\partial ^2v}{\partial x^2}\right]^L_0 + \int_{0}^{L}{\frac{\partial^2w}{\partial x^2}\left(EI\right)\frac{\partial ^2v}{\partial x^2}}dx$$

This gives us: $$(B_2) = \left[\frac{dw}{dx}(EI)\frac{\partial ^2v}{\partial x^2}\right]^L_0$$ and $$v_\gamma =\int_{0}^{L}{\frac{\partial^2w}{\partial x^2}\left(EI\right)\frac{\partial ^2v}{\partial x^2}}dx$$

Please note $$v_\gamma $$ is symmetrical

$$-(B)_1 + (B)_2 - v_\gamma + \int_{0}^{L}{wf_tdx} - \int_{0}^{L}{wm\ddot{v}dx} = 0 $$

For all possible w(x)

Stiffness Term Lets focus on the stiffness term for $$ v_\gamma$$ to derive the beam stiffness matrix and to identify the beam shape function. $$v(\tilde{x}) = N_2(\tilde{x})\tilde{d}_2 + N_3(\tilde{x})\tilde{d}_3 + N_5(\tilde{x})\tilde{d}_5 + N_6(\tilde{x})\tilde{d}_6$$ Recall that $$u(\tilde{x}) = N_1(\tilde{x})\tilde{d}_1 + N_4(\tilde{x})\tilde{d}_4$$ $$N_2(\tilde{x}) = 1 - \frac{3\tilde{x}^2}{L^2} + \frac{2\tilde{x}^3}{L^3}\,\,\,\tilde{d}_2 $$ $$N_3(\tilde{x}) = \tilde{x} - \frac{2\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2}\,\,\,\tilde{d}_3 $$

$$N_5(\tilde{x}) = \frac{3\tilde{x}^2}{L^2} - \frac{2\tilde{x}^3}{L^3}\,\,\,\tilde{d}_5$$ $$N_6(\tilde{x}) = -\frac{\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2}\,\,\,\tilde{d}_6$$