User:Eml4500.f08.team.figuerrez/hw6

Meeting 30 - Principle of Virtual Work using Integration by Parts
Integration by Parts:

The integration by parts formula is used to rewrite an integral of a product of a derivative function as follows:

Give two functions $${\boldsymbol{r(x)}}$$, $${\boldsymbol{s(x)}}$$

and taking the derivatives of each of the two functions yields: $${\boldsymbol{r'}}=\frac{dr}{dx},{\boldsymbol{s'}}=\frac{ds}{dx}$$

From the equation $${\boldsymbol{(rs)'}}={\boldsymbol{r's}}+{\boldsymbol{rs'}}$$, we integrate both sides to get:

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

Since $$\int{\boldsymbol{(rs)'}}={\boldsymbol{(rs)}}$$ and solving for $$\int{\boldsymbol{r's}}$$ the resulting integrand is:

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

Recall the Cont. Principle of Virtual Work (PVW):

1st term: $$r(x)=(EA)\frac{du}{dx},s(x)=w(x)$$

Using integration by parts we get:

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

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

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

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

Now consider the 2 boundary condition problem shown below:

At $${\boldsymbol{x=0}}$$ select $${\boldsymbol{w(x)}}$$ such that $${\boldsymbol{w(0)=0}}$$. (i.e. kinematically admissible)

Motivation: Discrete PVW applied to the equation shown below:

$$ \underline{W}_{6x1}\left(\left[\underline K\right]_{6x2}\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}_{2x1}-\underline{F}_{6x1}\right)=0_{1x1}$$, for all $${\boldsymbol{\underline{W}}}$$

where the transpose of $${\boldsymbol{\underline{F}}}$$ is equal to $$\underline{F}^{T}=\begin{Bmatrix}F_{1} & F_{2} & F_{3} & F_{4} & F_{5} & F_{6}\end{Bmatrix}$$

$${\boldsymbol{F_{3}, F_{4}}}$$ are known reactions and $${\boldsymbol{F_{1}, F_{2}, F_{5}, F_{6}}}$$ are unknown reactions.

Since $${\boldsymbol{\underline{w}}}$$ can be selected arbitrarily, select $${\boldsymbol{\underline{w}}}$$ such that $${\boldsymbol{w_{1} = w_{2} = w_{5} = w_{6} = 0}}$$ in order to eliminate equations involving unknown reactions; rows $${\boldsymbol{1, 2, 5,}}$$ and $${\boldsymbol{6}}$$ are eliminated.

The resulting equation is:

$$\underline{K}_{2x2}*\underline{d}_{2x1}=\underline{F}_{2x1}$$   $${\boldsymbol{(1)}}$$

$$\underline{K}_{2x2}* \begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix} = \begin{Bmatrix} F_{3} \\ F_{4}\end{Bmatrix}$$

Back to the Cont. Principle of Virtual Work (PVW), the unknown reaction is:

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

The equation then becomes:

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

The final equation then becomes:

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