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User:Eml4500.f08.bottle.butler/HW 7

Homework Problem
Resuming the 2-bar truss system from lecture 34, we assign the elements the following properties: Element 1: E^{(1)}_{1}=2, A^{(1)}_{1}=0.5 \! E^{(1)}_{2}=4, A^{(1)}_{2}=1.5 \! Element 2: E^{(2)}_{1}=3, A^{(2)}_{1}=1 \! E^{(2)}_{2}=7, A^{(2)}_{2}=3 \!

We can then compute the solution for the 2-bar truss system with tapered elements. The solution and plot of the deformed shapes can be found below.

Intro to Frame Elements
A frame element is composed of a truss (bar) element and a beam element. The truss element undergoes axial deformation while the beam element undergoes transverse deformation. A model problem can be found below. Note the rigid connection where the force P \! is applied. This implies that the angle between the two elements will remain constant after deformation.

Frame FBD’s
Due to the axial and transverse deformation, we now draw the free body diagrams for each element as seen below.

In general, we assume d^{(e)}_{i} \! correlates to f^{(e)}_{i} \! (This means that we generalize the forces, understanding that the rotational forces are the bending moments about that node), where $$ e = 1, 2 \! $$, $$ i = 1,2,3,4,5,6 \! $$

We can then draw the global dof as follows:

We are then able to construct 2 element stiffness matrices: $$ k^{(e)}_{6x6}, e = 1, 2 \! $$ $$ K_{9x9}=Ak^{(e)}_{6X6} \! $$

We can then display K as:

Meeting 36
For ease of computation, we then chose a different coordinate system. The free body diagram for element 1 is

Where $$ \tilde{k}^{(e)}_{6x6}\tilde{d}^{(e)}_{6x6}=\tilde{f}^{(e)}_{6x6} \! $$ and $$\tilde{d}^{(e)}=\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \cdot \cdot \cdot \\ \tilde{d}_{6}^{(e)} \end{Bmatrix} \! $$ and $$ \tilde{f}^{(e)}=\begin{Bmatrix} \tilde{f}_{1}^{(e)}\\ \cdot \cdot \cdot \\ \tilde{f}_{6}^{(e)}

\end{Bmatrix} \! $$

Note: Because of the moments about $$ \tilde{x} \! $$, $$ \tilde{f}^{(e)}_{3}=f^{(e)}_{3} \! $$ and $$ \tilde{f}^{(e)}_{6}=f^{(e)}_{6} \! $$

Constructing $$ \tilde{k} \! $$
We find

$$\tilde{k} \! $$ $$\begin{bmatrix} \frac{EA}{L} & 0 & 0 & \frac{-EA}{L} &0 &0 \\ \\ & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{-12EI}{L^{3}} &\frac{-6EI}{L^{2}} \\ \\ & &  \frac{4EI}{L}&0  & \frac{-6EI}{L^{2}} & \frac{2EI}{L}\\ \\ & &  & \frac{EA}{L} &0  & 0\\ \\ & &  &  &\frac{12EI}{L^{3}}  &\frac{6EI}{L^{2}} \\ \\ symmetric & &  &  &  & \frac{4EI}{L} \end{bmatrix} \! $$

Dimensional Analysis
Note: For our dimensional analysis we will use the following symbol $$ [ ] \! $$ to denote "dimension of...".

$$ [\tilde{d}_{1}] = L = [\tilde{d}_{i}], i=1,2,4,5 \! $$ $$ [\tilde{d}_{3}]= 1 (dimensionless) = [\tilde{d}_{6}] \! $$

We use the following proof for $$ [\theta ] \! $$

Meeting 37
$$ [\theta ] = \frac{[\bar{AB}]}{[R]}= \frac{L}{L} = 1 \! $$ $$ \sigma =E\epsilon \Rightarrow [\sigma ]=[E]\underbrace{[\epsilon ]}_{=1} \! $$  [\epsilon ] = \frac{du}{dx}=\frac{L}{L}=1 \! $$ [\sigma ]=[E]=\frac{F}{L^{2}} \! $$ $$ [A] = L^{2}, [I] = L^{4} \! $$ $$ [\frac{EA}{L}]=[\tilde{k}_{11}= \frac{\frac{F}{L^{2}}L^{2}}{L}= \frac{F}{L} \! $$ $$ [\tilde{k}_{11}\tilde{d_{1}}]=F \! $$ $$ [\tilde{k}_{23}\tilde{d}_{3}]=[\tilde{k}_{23}]1=\frac{[6][E][I]}{L^{2}}= \frac{1(\frac{F}{L^{2}})L^{4}}{L^{2}}=F \! $$

From Local to Global Coordinates
We now to the elemental FD relationship in global coordinates from the elemental FD relationship in local coordinates.

$$ k^{(e)}_{6x6}d^{(e)}_{6x6}=f^{(e)}_{6x6} \! $$ $$ k^{(e)}_{6x6}= \!$$$$\tilde{T}^{(e)} \!$$$$_{6x6}^{T} \!$$ $$\tilde{k}^{(e)}_{6x6}\tilde{T}^{(e)}_{6x6} \! $$ We can then construct the following matrices

$$\begin{Bmatrix} \tilde{d}_{1}\\ \tilde{d}_{2}\\ \tilde{d}_{3}\\ \tilde{d}_{4}\\ \tilde{d}_{5}\\ \tilde{d}_{6} \end{Bmatrix}_{6x1}= \underbrace{\begin{bmatrix} l^{(e)} & m^{(e)} & 0 &0 &0  &0 \\ -m^{(e)}&l^{(e)} &0  & 0 & 0 & 0\\ 0& 0 & 1 & 0 &0 &0 \\ 0 &0  &0  & l^{(e)} &m^{(e)}  & 0\\ 0 &0 &0  & -m^{(e)} & l^{(e)} &0 \\ 0& 0 &0 &0  & 0 & 1 \end{bmatrix}_{6x6}}_{\tilde{T}} \begin{Bmatrix} d_{1}\\ d_{2}\\ d_{3}\\ d_{4}\\ d_{5}\\ d_{6} \end{Bmatrix}_{6x1} \! $$

where $$\begin{bmatrix} l^{(e)} & m^{(e)}\\ -m^{(e)}& l^{(e)} \end{bmatrix} = R \! $$

The 2-element frame system is solved below. The same data used for the 2-bar truss system is used to solve the current system. The deformation of the two systems are then plotted on the same graph to give the reader a better understanding of a truss system vs a frame system.

PVW for Beams
$$ \int_{0}^{L}{W(x)[\frac{-\partial^{2} }{\partial x^{2}}((EI)\frac{\partial ^{2}v}{\partial x^{2}})+f_{t}-m\ddot{v}]dx}=0 |! $$ for all $$ W(x)\; equation 1 \! $$ We can use integration by parts on the first term:

$$ \alpha =\; \int_{0}^{L}{\underbrace{W(x)}_{s(x)}\frac{\partial ^{2}}{\partial x^{2}}((EI)\frac{\partial ^{2}v}{\partial x^{2}})dx} \! $$ where $$\underbrace{\frac{\partial }{\partial x}\underbrace{(\frac{\partial }{\partial x}(EI)\frac{\partial ^{2}v}{\partial x^{2}})}_{r(x)}}_{r'(x)} \! $$

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

By integrating the second part we achieve: $$ [\beta _{1}]^{L}_{0}-\underbrace{[\frac{dW}{dx}(EI)\frac{\partial ^{2}v}{\partial x^{2}}]^{L}_{0}}_{\beta _{2}}+\underbrace{\int_{0}^{L}{\frac{d^{2}W}{dx^{2}}((EI)\frac{\partial ^{2}v}{\partial x^{2}})dx}}_{\gamma } \! $$

Therefore, $$\alpha =\beta _{1}+\beta _{2}-\gamma \! $$ >br> and equation 1 can be written as $$ -\beta _{1}+\beta _{2}-\gamma +\int_{0}^{L}{wf_{t}dx}-\int_{0}^{L}{w\, m\, \ddot{v}dx}=0 \! $$

Shape Functions
Now we'll focus on the stiffness term $$ \gamma \! $$ to derive the beam stiffness matrix and to identify the shape functions. Equation 2$$ = \; v(\tilde{x})=N_{2}(\tilde{x})\tilde{d}_{2}+ \! $$$$N_{3}(\tilde{x})\! $$$$\tilde{d}_{3}+ \! $$$$N_{5}(\tilde{x})\tilde{d}_{5}+N_{6}(\tilde{x})\tilde{d}_{6} \! $$ Recall equation 1 $$=\; u(\tilde{x})=\ N_{1}(\tilde{x})\tilde{d}_{1}+N_{4}(\tilde{x})\tilde{d}_{4} \! $$ The shape functions can be written as $$N_{2}(\tilde{x})=1-\frac{3\tilde{x}^{2}}{L^{2}}+\frac{2\tilde{x}^{3}}{L^{3}} \! $$ $$N_{3}(\tilde{x})=\tilde{x}-\frac{2\tilde{x}^{2}}{L}+\frac{\tilde{x}^{3}}{L^{2}} \! $$ $$N_{5}(\tilde{x})=\frac{3\tilde{x}^{2}}{L^{2}}-\frac{2\tilde{x}^{3}}{L^{3}} \! $$ $$N_{6}(\tilde{x})=-\frac{\tilde{x}^{2}}{L}+\frac{\tilde{x}^{3}}{L^{2}} \! $$ $$\tilde{d}^{(e)}_{6x1}=\tilde{T}^{(e)}_{6x6} d^{(e)}_{6x1} \! $$ Where $$d^{(e)}_{6x1} \! $$ is known after solving the FE system. Compute $$ u(\tilde{x}), v(\tilde{x})\! $$:  $$u(\tilde{x})=u(\tilde{x})\vec{\tilde{i}}+v(\tilde{x})\vec{\tilde{j}} \!$$ $$ = u_{x}(\tilde{x})=u(\tilde{x})\vec{\tilde{i}}+u_{y}(\tilde{x})\vec{\tilde{j}} \!$$ We can compute $$ u(\tilde{x}), v(\tilde{x})\! $$ by using equation 1 and 3. $$\begin{Bmatrix} u_{x}(\tilde{x})\\ u_{y}(\tilde{x}) \end{Bmatrix}= R^{T}\begin{Bmatrix} u(\tilde{x})\\ v(\tilde{x}) \end{Bmatrix} \! $$  $$\begin{Bmatrix} u(\tilde{x})\\ v(\tilde{x}) \end{Bmatrix}=\underbrace{\begin{bmatrix} N_{1} &0 &0  & N_{4} & 0 & 0\\ 0 & N_{2} &N_{3} & 0 & N_{5} & N_{6} \end{bmatrix}}_{\mathbb{N}(\tilde{x})}\begin{Bmatrix} \tilde{d}_{1}^{(e)}\\ \tilde{d}_{2}^{(e)}\\ \tilde{d}_{3}^{(e)}\\ \tilde{d}_{4}^{(e)}\\ \tilde{d}_{5}^{(e)}\\ \tilde{d}_{6}^{(e)} \end{Bmatrix} \! $$  $$ \begin{Bmatrix} u_{x}(\tilde{x})\\ u_{v}(\tilde{x}) \end{Bmatrix}=R^{T}\mathbb{N}(\tilde{x})\tilde{T}^{(e)}d^{(e)} \! $$

title: User:Eml4500.f08.bottle.butler/HW 7


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title: User:Stevenarntson/Student Hub/archive