User:Eml4500.f08.delta 6.castrillon/HW 6

Integration by Parts Recalling the method of integration by pats for integration, it is described below. We have two functions, $$r(x)$$ and $$s(x)$$. When we want to determine the derivative of the product of these two functions, we have: $$(rs)'=r's+rs'$$ where, $$r'=\frac{dr}{dx}$$ and $$s'=\frac{ds}{dx}$$ Then, $$(rs)=\int (rs)'dx=\int rs'dx+\int r'sdx$$ $$\int r'sdx=rs-\int rs'dx$$

Recalling the Principle of Virtual Work, we can apply this method of integration to this principle.

Let $$r(x)=EA \frac{\partial u}{\partial x}$$ and $$s(x)=w(x)$$

Now applying Integration by Parts,


 * $$\int _{x=0}^{L}w(x) \frac{\partial }{\partial x} \left[ EA \frac {\partial u}{\partial x} \right] dx$$
 * $$\left[w(EA)\frac {\partial u}{\partial x} \right]_{0}^{L}-\int_{0}^{L}\frac{dw}{dx}EA\frac{\partial u}{\partial x}dx$$
 * $$w(L)EA(L) \frac{\partial u(L,t)}{\partial x}-w(0)EA(0) \frac{\partial u(0,t)}{\partial x}- \int_{0}^{L} \frac{dw}{dx}EA\frac{\partial u}{\partial x}dx$$

It is noticeable that
 * $$N(L,t)=EA(L) \frac{\partial u(L,t)}{\partial x}$$

and
 * $$N(0,t)=EA(0) \frac{\partial u(0,t)}{\partial x}$$

Now we need to account for the boundary conditions (BCs). We use the BC defined in the diagram below:



At the wall, (x=0), we have to select w(x) such that w(0)=0. This requirement would be kinematically admissible as it would adhere to the constraints.

Descrete PVW applied to the reduction of the global FD rel

 * $$w$$ $$_{6x1}$$•$$\left( \begin{bmatrix}

K \end{bmatrix}_{6x2}

\begin{pmatrix} d \end{pmatrix}_{2x1}-F_{6x1} \right)=0_{1x1}$$; for all $$w$$

We have the transpose of F:


 * $$F^T$$$$=\begin{bmatrix}

F_1 & F_2 & F_5 & F_6 \end{bmatrix}$$

where F_1, F_2, F_5, and F_6 are all unknown reactions.

Since $$w$$ can be selected arbitrarily, we say that $$w$$ is such that:
 * $$w_1=w_2=w_5=w_6=0$$

Therefore, to eliminate the equation involving unknown reactions (element rows 1, 2, 5, and 6), we say that:
 * $$\bar{w}_{2x1}$$•$$\left( \mathbf{\bar{K} \bar{d} \bar{F}} \right)=0$$; for all $$w$$.

This yields to,
 * $$\bar{K}_{2x2} \bar{d}_{2x1}$$ = $$\bar{F}_{2x1}$$

where,
 * $$\bar{d}_{2x1}$$=$$\begin{pmatrix}

d_3\\ d_4 \end{pmatrix}$$ and
 * $$\bar{F}_{2x1}$$=$$

\begin{pmatrix} F_3\\ F_4 \end{pmatrix}$$

Turning our attention back to the Principle of Virtual Work, and recalling the reaction terms,


 * $$N(L,t)=EA(L) \frac{\partial u(L,t)}{\partial x}$$

and
 * $$N(0,t)=EA(0) \frac{\partial u(0,t)}{\partial x}$$,

we can build the following expression:


 * $$w(L)F(t)-\int_{0}^{L}\frac{dw}{dx}EA\frac{\partial d}{\partial x}dx+\int_{0}^{L} w(x)\left[f-m\ddot{u}\right]dx=0$$ for all w(x) such that w(0)=0.

Rewriting in row form,


 * $$\int_{0}^{L}w \left(m\ddot {u}\right)dx+\int_{0}^{L} \frac{dw}{dx}EA\frac{\partial u}{\partial x}dx=w(L)F(t)+\int_{0}^{L}wfdx$$ for all w(x) such that w(0)=0.