User:Eml4500.f08.jamama.adam/homework7

Meeting 38
To get motivated lets consider the deformed shape of a truss element interp. of transverse displacement $$ V(s) with s = \tilde{x}$$.

Principal of Virtual Work (PVW): Eq. (1)

$$\int_{0}^{1}{w(x)}\left[-\frac{\partial{}^{2}}{\partial{x^{2}}}(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}} + f_{T} - m\ddot{v}\right]dx = 0$$ for all possible $$w(x) $$

For the first order term we will integrate by parts:

Let us recall the integration by parts method:

$$ r(x), s(x) $$

$$ (rs)' = r's + rs'$$

$$ r' = \frac{dr}{dx}  ;   s' = \frac{ds}{dx}$$

$$ \underbrace{\int {(rs)'}}_{(rs)}=\int{r's} + \int{rs'}$$

$$ =>\int{r's} = rs - \int{rs'}$$

Therefore if we let (α) = $$\int_{0}^{1}{\underbrace{w(x)}_{s(x)}}\frac{\partial{}^{2}}{\partial{x^{2}}}\left[(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]dx $$

then we can write that $$\underbrace{\frac{\partial{}}{\partial{x}}\left(\underbrace{\frac{\partial{}}{\partial{x}}\left[(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]}_{r(x)}\right)}_{r^'(x)}$$

$$= \left[\underbrace{W\frac{\partial{}}{\partial{x}}\left[(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]}_{\beta^1}\right]_{0}^{L} - \int_{0}^{L}\underbrace{\frac{dw}{dx}}_{s'(x)}\underbrace{\frac{\partial{}}{\partial{x}}\left[(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]dx}_{r(x)}$$

$$= \beta^{1} - \underbrace{\left[\frac{dw}{dx}(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}\right]_{0}^{L}}_{\beta^{2}} + \underbrace{\int_{0}^{L}\frac{d^{2}w}{dx^{2}}(EI)\frac{\partial^{2}{v}}{\partial{x^{2}}}dx}_{\gamma}$$ The symmetry should be noted!

Then Eq. (1) becomes:

$$- \beta^{1} + \beta^{2} - \gamma + \int_{0}^{L}{wf_{t}}dx - \int_{0}^{L}{wm\ddot{v}}dx = 0$$ for all possible $$w(x)$$

We will now focus on the stiffness term $$\gamma$$ to derive the beam stiffness matrix and to identify the beam shape functions.



$$V(\tilde{x}) = N_{2}(\tilde{x})\tilde{d}_{2} + N_{3}(\tilde{x})\tilde{d}_{3} + N_{5}(\tilde{x})\tilde{d}_{5} + N_{6}(\tilde{x})\tilde{d}_{6}$$

Let us recall: $$U(\tilde{x}) = N_{1}(\tilde{x})\tilde{d}_{1} + N_{4}(\tilde{x})\tilde{d}_{4}$$











From the book p.246, the following functions can be found:

$$N_{2}(\tilde{x}) = 1 - \frac{3\tilde{x}^{2}}{L^{2}} + \frac{2\tilde{x}^{3}}{L^{3}}$$ corresponding to $$ \tilde{d}_{2}$$

$$N_{3}(\tilde{x}) = \tilde{x} - \frac{2\tilde{x}^{2}}{L} + \frac{\tilde{x}^{3}}{L^{2}}$$ corresponding to $$ \tilde{d}_{3}$$

$$N_{5}(\tilde{x}) = \frac{3\tilde{x}^{2}}{L^{2}} - \frac{2\tilde{x}^{3}}{L^{3}}$$ corresponding to $$ \tilde{d}_{5}$$

$$N_{6}(\tilde{x}) = -\frac{\tilde{x}^{2}}{L} + \frac{\tilde{x}^{3}}{L^{2}}$$ corresponding to $$ \tilde{d}_{6}$$