User:Eml4500.f08.bottle.butler/HW 6

Integration by Parts (IBP)
Assume we have the following functions: $$r(x)\! $$ and $$ s(x)\! $$ $$ (rs)' = r's + rs' \! $$ where $$ r'=\frac{dr}{dx} \! $$ and $$ s'=\frac{ds}{dx} \! $$ Therefore, $$ \underbrace{\int (rs)'}_{rs}=\int r's + \int rs' \! $$  $$ \int r's = rs-\int rs' \! $$

Applying IBP to PVW
Returning to the continuous PVW, $$ (3)\; \int_{x=0}^{x=L}{W(x)}(\frac{\partial}{\partial x} [EA\frac{\partial u}{\partial x}]+f-m\ddot{u})dx =0 \! $$ for all possible $$ W(x) \! $$ If r(x) = EA\frac{\partial u}{\partial x}\! and $$s(x)=W(x) \!$$ and we integrate by parts we find the following: $$ \int_{x=0}^{x=L}\underbrace_{s(x)} \frac{\partial}{\partial x} \underbrace{[EA\frac{\partial u}{\partial x}]}_{r(x)}dx = [W(EA)\frac{\partial u}{\partial x}]^{x=L}_{x=0}-\int_{x=0}^{x=L}{\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx} \! $$  $$ = W(L)\underbrace{(EA(L))\frac{\partial u (L,x)}{\partial x}}_{N(L,t)} - \!$$$$\underbrace{W(0)(EA(0))\frac{\partial u (0,t)}{\partial x}}_{N(0,t)} \!$$$$ - \int_{0}^{L}{\frac{dw}{dx}(EA)\frac{\partial u}{\partial x}}dx \! $$

Boundary Conditions
We now account for boundary conditions. We consider the model problem: image At $$ x=0 \! $$ select $$ W(x) \! $$ such that $$ W(0)=0 \! $$ (i.e. kinematically adminisble).

Motivation
Let's reexamine the discrete PVW as applied to the equation on page 10-1. $$w_{6X1}\circ ([]_{6X2}\begin{Bmatrix} d_{3}\\ d_{4}

\end{Bmatrix}_{2X1}-F_{6X1})=0_{1X1} \! $$ $$F^{T}=\begin{bmatrix} F_{1} & F_{2} & F_{3} & F_{4} & F_{5} & F_{6} \end{bmatrix} \! $$  where $$F_{2} \! $$ and $$ F_{3} \! $$ are known and $$ F_{1}, F_{2},  F_{5}, F_{6} \! $$ are unknown.  Since $$ W \! $$ can be selected arbitrary, then we select $$ W \! $$ such that $$ W_{1}=W_{2}=W_{5}=W_{6}=) \! $$ in order to eliminate equations involving unknown reactions. We are then left with : $$ K_{2X2}d_{2X1}=F_{2X1} = K\begin{Bmatrix} d_{3}\\ d_{4}

\end{Bmatrix}=\begin{Bmatrix} F_{3}\\ F_{4}

\end{Bmatrix} \! $$

Continuous case PVW cont.
The unknown reaction $$N(0,t)- EA(0)\frac{\partial u (0,t)}{\partial x} \! $$ where $$ N= \! $$ the normal force. $$ W(L)F(t)-\int_{0}^{L}{\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}}dx+\int_{0}^{L}{W(x)[f-m\ddot{u}]dx}=0 \! $$ for all $$ W(x) \! $$ such that $$ W(0)=0 \! $$ $$\Rightarrow \int_{0}^{L}{W(m\ddot{u})dx}+\int_{0}^{L}{\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx} \! $$ $$W(L)F(t)+\int_{0}^{L}{wfdx} \! $$ for all $$ w(x) \! $$ such that $$ w(0)=0 \! $$

Connecting continuous PVW to the discrete PVW
To connect the two theories we use a Lagrangian interpretation. We use the form of $$ N_{i}(x) \! $$ and $$N_{i+1}(x) \! $$ 1) $$ N_{i}(x) \! $$ and $$N_{i+1}(x) \! $$ are linear, thus any linear combination of $$ N_{i}(x) \! $$ and $$N_{i+1}(x) \! $$ are also linear, specifically $$u(x)= N_{i}(x)d_{i}+ N_{i+1}(x) d_{i+1} \! $$ $$N_{i}(x)=\alpha _{i}+\beta _{i} \! $$ $$N_{i+1}(x)=\alpha _{i+1}+\beta _{i+1} \! $$ where $$ \alpha \! $$ and $$\beta \! $$ are real numbers. Therefore we get the linear combination: $$ N_{i}(x)d_{i}+ N_{i+1}(x) d_{i+1} = (\alpha _{i}+\beta _{i})d_{i} + (\alpha _{i+1}+\beta _{i+1})_{i+1} \! $$   $$ = (\alpha _{i}d_{i} +\alpha _{i+1}d_{i+1})+(\beta _{i}d_{i}+\beta _{i+1}d_{i+1})x \! $$  2) Recall the equation  for $$ u(x)\! $$ (the interpolation of $$ u(x) \! $$ consider: $$ \underbrace {N_{i}(x)d_{i}}_{1}+ \underbrace {N_{i+1}(x) d_{i+1}}_{0} \! $$  therefore $$ u(x_{i})=d_{i} \! $$  and $$u(x_{i+1})=d_{i+1} \! $$  We apply the same theory to  W(x) \!  $$ W(x)=N_{i}(x)W_{i}+N_{i+1}W_{i+1} \! $$

Element Stiffness Matrix for Element i
$$ (\beta ) \! \;$$ $$\int_{x_{i}}^{x_{i+1}}{\underbrace{[N_{i}'W_{i}+N_{i+1}'W_{i+1}]}_{W'(x)}}(EA(x))\underbrace{[N_{i}'d_{i}+N_{i+1}'d_{i+1}]}_{u'(x)}dx \! $$ where $$ N'_{i}=\frac{dN_{i}(x)}{dx} \! $$ and $$N'_{i+1}=\frac{dN_{i+1}(x)}{dx} \! $$ Note: $$ u(x) \! $$ can be written as $$ \underbrace{[N_{i}'N_{i+1}']}_{N(x)_{1X2}} \! $$ Therefore,  \frac{du(x)}{dx}=\underbrace{[N_{i}'(x)N_{i+1}'(x)]}_{B(x)_{2X1}}\begin{Bmatrix} d_{1}\\d_{i+1}

\end{Bmatrix}_{2X1} \! Similarly: $$ W(x) = N(x)\begin{Bmatrix} W_{i}\\W_{i+1}

\end{Bmatrix} \! $$  \frac{dW(x)}{dx}=B(x)_{2X1}\begin{Bmatrix} W_{1}\\W_{i+1}

\end{Bmatrix}_{2X1} \! Recall the element dof's: image $$\begin{Bmatrix} d_{i}\\d_{i+1}

\end{Bmatrix}= \begin{Bmatrix} d^{i}_{1}\\d^{1}_{2}

\end{Bmatrix}=d^{i} \! $$ and $$\begin{Bmatrix} W_{i}\\W_{i+1}

\end{Bmatrix}= \begin{Bmatrix} W^{i}_{1}\\W^{1}_{2}

\end{Bmatrix}=W^{i} \! $$  $$ (\beta ) = \int_{x_{i}}^{x_{i+1}}{\underbrace{(B(x)W^{(i)})}_{1X1}}\underbrace{(EA)}_{1X1} \underbrace{(\beta d^{(i)})}_{1X1}dx \! $$ $$ = W^{(i)}\circ (k^{(i)}d^{(i)}) \! $$ where our goal is to find $$ k \! $$ We can rewrite $$ (\beta )=\int_{x_{i}}^{x_{i+1}}{(EA) \underbrace{(B(x)W^{(i)})}_{1X1}} \circ \underbrace{(B d^{(i)})}_{1X1}dx  \! $$ as: $$(BW^{(i)})^{T}(Bd^{(i)}) \! $$ Therefore $$ W^{(i)}\circ (\underbrace{\int B^{T}(EA)Bdx)}_{k^{(i)}}d^{(i)} \! $$  $$ k^{(i)}_{2X2}=\int_{x_{i}}^{x_{i+1}}{\underbrace{B(x)^{T}}}_{2X1}\underbrace{(EA)}_{1X1}\underbrace{B(x)}_{1X2}dx  \! $$ We can then transform the coordinantes from $$ x \! $$ to $$ \tilde{x} \! $$ where $$ \tilde{x}= x-x_{i} \! $$ and $$ d\tilde{x} = dx \! $$  $$ k^{(i)}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}{B^{T}(\tilde{x})(EA(\tilde{x})B(\tilde{x}) d\tilde{x}} \! $$