User:Eml4500.f08.jamama.adam/homework6

Meeting 30
Integration by parts:$$ r(x), s(x) $$

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

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

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

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

We can recall cont. PVW (Principal of Virtual Work) Eq. (3) p. 29-1

First term: $$ r(x)=(EA)\frac{\partial{u}}{\partial{x}}  ;   s(x) = W(x)$$

By intergrating by parts we cen get:

$$ \int_{x=0}^{x=L}{\underbrace{W(x)}_{s}\frac{\partial}{\partial{x}}\underbrace{\left[(EA)\frac{\partial{u}}{\partial{x}}\right]}_{r} dx} = \left[W(EA)\frac{\partial{u}}{\partial{x}}\right]_{x=0}^{x=L} - \int_{x=0}^{x=L}{\frac{dw}{dx}(EA)\frac{\partial{u}}{\partial{x}}dx}$$

$$ = W(L)(EA)(L)\frac{\partial{u}}{\partial{x}}(L,t) - W(0)(EA)(0)\frac{\partial{u}}{\partial{x}}(0,t) - \int_{0}^{L}{\frac{dw}{dx}(EA)\frac{\partial{u}}{\partial{x}}dx}$$

Now lets consider the following model problem: p.28-3  2 b.c.'s



At x=o, select W(x) so that W(0)=0 (i.e. Kinematically admissible)

Motivation : Discrete PVW applied to Equation on  p. 10-1

$$\mathbf{W}_{6x1}. \underbrace{\left(\left[      \right]_{6x2}\begin{Bmatrix}d3\\d4\end{Bmatrix}_{2x1} - \mathbf{F}_{6x1}\right)}_{6x1} = 0 $$     for all W

$$ \mathbf{F}^T = \begin{bmatrix} F_1&F_2&F_3&F_4&F_5&F_6\end{bmatrix}$$  with $$F_1, F_2, F_5, F_6$$: unknown reactions

Since W can be selectedarbitrarily, select W so that:

$$ W_1 = W_2 = W_5 = W_6 = 0$$; so to eliminate equations involving unknown reactions => eliminate rows 1,2,5,6;

Let equation (1) be: $$ \mathbf{K}_{2x2} \mathbf{d}_{2x1} = \mathbf{F}_{2x1}$$

with: $$ \mathbf{d}_{2x1} = \begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix}$$ and: $$ \mathbf{F}_{2x1} = \begin{Bmatrix} F_3 \\ F_4 \end{Bmatrix}$$

Let us note that for the step before eq.(1)


 * $$ \mathbf{W}.\left(\mathbf{K}\mathbf{d}-\mathbf{F}\right) = 0$$ with $$ \mathbf{W} = \begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix}$$

Now back to the PVW:

unknown reaction $$ N(o,t) = (EA)(0)\frac{\partial{u}}{\partial{x}}(0,t)$$

Cont. PVW :

$$W(L)f(t)- \int_0^L{\frac{dw}{dx}(EA)\frac{\partial{u}}{\partial{x}}dx} + \int_0^L{W(x)\left[f - m\ddot{u}\right]dx} = 0$$

So for all W(x) so that W(0) = 0,

$$=> \int_0^L{W(m\ddot{u})}dx + \int_0^L{\frac{dW}{dx}(EA)\frac{\partial{u}}{\partial{x}}}dx = W(L)f(t) + \int_0^L{Wf}dx$$
 * for all W so that W(0) = 0.