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

Principle of Virtual Work (PVW) concerning dynamics of the elastic bar
From the PDE equation derived from the previous homework concerning the elastic bar: $$ \frac{\delta }{\delta x}\begin{bmatrix} (EA)\frac{\delta U}{\delta x} \end{bmatrix}+\textit{f}=m\ddot{u} $$  (1) We can then discrete the above equation with the single degree of freedom equation concerning an elastic spring: $$ \mathbf{M\ddot{d}+Kd=F} $$  (2) which then gives $$ \mathbf{-Kd+F=M\ddot{d}} $$ for multiple degrees of freedom in an elastic bar. Deriving equation (2) from equation (1) gives us: $$ \int_{x=0}^{x=L}{W(x)\begin{Bmatrix} \frac{\delta }{\delta x}\begin{bmatrix} EA\frac{\delta U}{\delta x} \end{bmatrix}+\textit{f}-m\ddot{u} \end{Bmatrix}}dx=0 $$  (3) for all possible weighting functions W(x). Deriving equation (1) to (3) is pretty simple a task to complete, but deriving equation (3) from (1) is a whole new different ball game. Equation (3) can be rewritten as: $$ \int W(x)g(x)dx=0 $$ for all W(x) as before. Since equation (3) is equivalent for all weighting functions, we can then select W(x) = g(x). Then equation (3) becomes: $$ \int g^{2}dx=0\Rightarrow g(x)=0 $$ which then ends our derivation of equation (3).

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: 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}=0 \! $$  in order to eliminate equations involving unknown reactions. We are then left with :  $$ 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 $$ w(0)=0 \! $$

Continuous PVW vs Discrete Setting
For the stiffness in the elastic bar, we assume displacement u(x) for $$ x_{i}\leq x\leq x_{i+1} \!$$ For the two-bar truss, we derived the undeformed and deformed shape for each section of the truss as shown below. In the image above, there is an implicit assumption of linear interpolation of displacement between the two nodes. Considering the case when there is only axial displacement comes an expression for $$ U(x) \! $$ in terms of $$ d_{i}=U(x_{i}) \!$$ and $$ d_{i+1}=U(x_{i+1}) \!$$ as a linear function on the x-direction. $$ U(x)=N_{i}(x)d_{i}+N_{i+1}(x)d_{i+1} \!$$ where $$ N_{i}(x) \!$$ and $$ N_{i+1}(x) \!$$ are linear functions in the x-direction.

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} \! $$ Recall the element dof's: $$\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}} \! $$ Using the k value ubtained in the above equation, we can then compare it to the expression given in the textbook. $$ \frac{E}{L^{(i)}}\frac{(A_{1}+A_{2})}{2}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}=\textbf{k}^{(i)} $$ with $$ \frac{(A_{1}+A_{2})}{2} $$ representing the average area of the truss example. Then a comparison was made with both equations above: $$ \frac{(E_{1}+E_{2})(A_{1}+A_{2})}{4L^{(i)}}\begin{bmatrix} 1 &-1 \\ -1 & 1 \end{bmatrix}=\textbf{k}^{(i)}_{ave} $$ with the part of finding $$ \textbf{k}^{(i)}-\textbf{k}^{(i)}_{ave} $$. As a reminder, the Mean Value Theorem in relation to the centroid is called upon. Mean Value Theorem (MVT)   $$ \int_{x=a}^{x=b}{f(x)dx}=f(\tilde{x})[b-a] $$   for $$ \bar{x} \epsilon [a,b] $$, with epsilon representing as "belongs to" $$ a\leq \bar{x}\leq b $$ $$ \int_{A}{xdA}=\bar{x}\int_{A}{dA}=\bar{x}A $$ $$ \int_{x=a}^{x=b}{f(x)g(x)dx}=f(\bar{x})g(\bar{x})[b-a] $$ with $$ a\leq \bar{x}\leq b $$ But $$ f(\bar{x})\neq \frac{1}{b-a}\int_{a}^{b}{f(x)dx} $$ with $$ \frac{1}{b-a}\int_{a}^{b}{f(x)dx} $$ representing the average value of f.  $$ g(\bar{x})\neq \frac{1}{b-a}\int_{a}^{b}{g(x)dx} $$ with $$ \frac{1}{b-a}\int_{a}^{b}{g(x)dx} $$ representing the average value of g.  end

Homework Solutions
This section covers the solutions to in-class assigned homework problems not addressed in the lecture notes above. Therefore, it is recommended that the reader refer to the above notes if a homework problem in this section has been seemingly omitted.

Principal of Virtual Work and the continuous weighting function integral
Equation (1), a partial differential equation (PDE), was given in lecture 29 and used to derive Equation (2). $$ \frac{\partial }{\partial x}\left[(EA)\frac{\partial u}{\partial x} \right] + f = m\ddot{u} $$             (1) $$ \mathbf{M\ddot{d}}+\mathbf{kd} = \mathbf{F} $$             (2) The partial differential equation, Equation (1), represents a summation of forces for infinitesimally thin sections dx over the bounds 0 to length L.  Each of the terms in Equation (1) can be represented discretely by a matrix of individual points. Equation (1) is just the continuous expression of the same thing. Therefore, every $$m\ddot{u}$$ between 0 and L added up continuously will have very similar results as the matrix expression $$\mathbf{M\ddot{d}}$$. The same is true for the other terms as well. Knowing that Equations (1) and (2) are the continuous and discrete versions of the same system we can derive the continuous Equation (3) and use the discrete case to verify that our process is correct. First, we know that $$\int w(x)g(x)dx = 0$$ for all $$w(x)$$. In the case that $$w(x) = g(x)$$, our above equation becomes $$\int g^{2}(x)dx$$. As long as $$g(x) \geq 0$$ then $$g(x) = 0$$. Therefore, since Equation (1) can also be rearranged to equal zero, Equation (3) is true. $$ 0 = \int_{x = 0}^{x = L}{W(x) \left\{ \frac{\partial }{\partial x}\left[ EA\frac{\partial u}{\partial x}\right] + f - m\ddot{u} \right\} dx} $$             (3) To verify this process, we recall a proof for discrete matrices from lecture 24 that was used to justify weighting matrices. For the case where $$\textbf{W}_{6x1}\cdot (\textbf{Kd}-\textbf{F})_{6x1}=0_{1x1}$$ We use the following proof from Team Bottle HW 5.

Proof: Select W at $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}_{1x6}\ $$ plugging W back into the Principle of Virtual Work equation gives: $$ \begin{bmatrix}1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 1 \cdot \bigg[ \sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1} \bigg]+ 0\cdot \bigg[\sum_{j=1}^{6}{K_{2-6j}d_{j}} - F_{2-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{1j}d_{j}=F_{1}}\ $$ Following through as before for the rest of the matrices gives these results: $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1j}d_{j}} - F_{1} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{2j}d_{j}} - F_{2} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{3-6j}d_{j}} - F_{3-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{2j}d_{j}=F_{2}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-2j}d_{j}} - F_{1-2} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{3j}d_{j}} - F_{3} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{4-6j}d_{j}} - F_{4-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{3j}d_{j}=F_{3}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-3j}d_{j}} - F_{1-3} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{4j}d_{j}} - F_{4} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{5-6j}d_{j}} - F_{5-6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{4j}d_{j}=F_{4}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-4j}d_{j}} - F_{1-4} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{5j}d_{j}} - F_{5} \bigg]+ 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{5j}d_{j}=F_{5}}\ $$ $$ \textbf{W}^{T}_{1x6}=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}_{1x6}\ $$ $$ \begin{bmatrix}0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \cdot (\textbf{Kd}-\textbf{F}) = 0 \cdot \bigg[ \sum_{j=1}^{6}{K_{1-5j}d_{j}} - F_{1-5} \bigg]+ 1\cdot \bigg[\sum_{j=1}^{6}{K_{6j}d_{j}} - F_{6} \bigg]=0 $$ Therefore, $$ \textbf{W}\cdot (\textbf{Kd}-\textbf{F})=\sum_{j=1}^{6}{K_{6j}d_{j}=F_{6}}\ $$ Hence, adding up all of these results gives us $$ \textbf{Kd}=\textbf{F} $$.

Finding an expression for k(i)
Given the following equation, find an expression for $$ \mathbf{k^{(i)}} $$. $$ \mathbf{k}^{(i)}_{2x2}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}\mathbf{B^{T}}(\tilde{x})(EA)(\tilde{x})\mathbf{B}(\tilde{x})d\tilde{x}$$ It is already known that $$ \mathbf{B}(\tilde{x}) = \begin{bmatrix}N_{i}(\tilde{x}) & N_{i+1}(\tilde{x})\end{bmatrix} $$ $$ \mathbf{B^{T}}(\tilde{x}) = \begin{bmatrix}N_{i}(\tilde{x})\\ N_{i+1}(\tilde{x})\end{bmatrix} $$ Also, from lecture 33, it is known that $$A(\tilde{x})=N^{(i)}_1(\tilde{x})A_1+N^{(i)}_2(\tilde{x})A_2$$ $$E(\tilde{x})=N^{(i)}_1(\tilde{x})E_1+N^{(i)}_2(\tilde{x})E_2$$ Since Ni is assigned to Node 1 and Ni+1 is assigned to Node 2, the following two equations are true. $$N_{i}(x) = N_{1}^{(i)}(\tilde{x}) N_{i+1}(x) = N_{2}^{(i)}(\tilde{x}) $$ Plugging these expressions back into our original $$ \mathbf{k^{(i)}} $$ equation results in $$ \mathbf{k^{(i)}} = \begin{Bmatrix} N_{1}^{(i)}(L)\\ N_{2}^{(i)}(L) \end{Bmatrix} \begin{bmatrix}N_{1}^{(i)}(L) & N_{2}^{(i)}(L)\end{bmatrix} E(L)A(L) - \begin{Bmatrix} N_{1}^{(i)}(0)\\ N_{2}^{(i)}(0) \end{Bmatrix} \begin{bmatrix}N_{1}^{(i)}(0) & N_{2}^{(i)}(0)\end{bmatrix} E(0)A(0) $$ Multiplying the 2x1 and 1x2 matrices together, the equation above becomes $$ \mathbf{k^{(i)}} = \left[   (N_{1}^{(i)}(L))^{2} + (N_{2}^{(i)}(L))^{2}   \right] E(L)A(L) - \left[   (N_{1}^{(i)}(0))^{2} + (N_{2}^{(i)}(0))^{2}   \right] E(0)A(0) $$ According to the definition of N at the point x = 0, N1 = 1, N2 = 0 At the point x = L, N1 = 0, N2 = 1. Therefore, $$ \mathbf{k^{(i)}} = \left[   (0)^{2} + (1)^{2}   \right] \left[(0)E_{1} + (1)E_{2} \right] \left[ (0)A_{1} + (1)A_{2}  \right] - \left[   (1)^{2} + (0)^{2}   \right] \left[(1)E_{1} + (0)E_{2} \right] \left[ (1)A_{1} + (0)A_{2}  \right] $$ $$ \mathbf{k^{(i)}}_{2x2} = \left[ E_{2}A_{2} - E_{1}A_{1} \right] \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} $$

Formula and MATLAB Code for Linear Interpolation
The conventional formula for linear interpolation is:

$$ N_{i+1}(x)= \frac{x - x_{i}}{x_{i + 1} - x_{i}} $$

This can be programmed in MATLAB as:

Tapered Bar with Constant Modulus of Elasticity
Let $$\displaystyle E_1 = E_2 = E = 70 GPa $$ and let $$\displaystyle A( \tilde{x}) $$ be linear. Also let $$\displaystyle A_1 = 2400 mm^2 $$, $$\displaystyle A_2 = 600 mm^2 $$ and $$\displaystyle L = 300 mm $$.

Given the following equation

$$\displaystyle {E \over L^{(i)}} {(A_1 + A_2) \over 2} \begin{bmatrix} 1 & -1 \\ -1 & 1  \end{bmatrix} = \textbf{k}^{(i)} $$

$$\displaystyle {(70 GPa) \over (300mm)} {(2400mm^2 + 600 mm^2) \over 2} \begin{bmatrix} 1 & -1 \\ -1 & 1  \end{bmatrix} = \textbf{k}^{(i)} $$

$$\displaystyle \textbf{k}^{(i)} = \begin{bmatrix} 714 & -714 \\ -714 & 714  \end{bmatrix} $$



Electric Pylon Analysis
Our team was asked to conduct an analysis of a 91 bar, 60 m high electric pylon. The properties of the pylon are:

Material = 300 M Steel Height = 60 m Young's Modulus = 200 GPa Area = 4 cm2 Density = 7.8 g/cm3

MATLAB was implemented in solving for the displacements, stresses in each bar, the reaction forces, the eigenvectors, and periods of oscillation.

Code
The following MATLAB code was developed and run:

Results
The results array for the axial stress on each bar is: Stresses =

2.1562e+006 -1.3071e+006 1.3468e+006 1.2911e+006 -7.0304e+005 -4.7596e+006 -2.465e+006 1.4236e+006 3.8538e+006 -5.7655e+006 -37882      1755.6  3.8655e+006 2.9395e+006 -3.6988e+006 -5.7661e+006 3.5896e+005 5.242e+006 1.9797e+006 -3.7749e+006 -6.5159e+006 5.5784e+005 6.4388e+005 5.9656e+006 -8.1704e+005 -3.9616e+006 2.32e+006 -8.1307e+005 -5.7556e+006 -28474 -9.9353e+005 -15214 5.9216e+006 55813 -4.8717e+006 1.3915e+006 -3023.3 -5.7588e+006 -4.854e+006 1.3904e+006 -1.8831e+005 2.258e+006 -6.32e+005 -4.0102e+006 1.0133e-007 1.4647e-007 64147 2.7001e+005 2.7149e+005 -2.0778e+006 -4.4271e+006 -5.8777e+006 -7.3549e+006 -8.662e+006 -8.6957e+006 2.5807e-008 -6.1883e-008 -5.397e-008 9.7638e-008 -1.7704e+005 2.2069e+005 -1.9193e+005 1.9887e+005 2.1464e+006 -2.083e+006 2.2895e+006 -2.2643e+006 2.1035e+006 -2.1464e+006 1.4138e+006 -1.3772e+006 1.5835e+006 -3.6928e+006 -97393      -23690        31208  9.0479e+006 9.3927e-008 9.0317e+006 -2.3815e-007 9.0511e+006 -1.3175e+005 -2.9645e+005 1.7966e+006 4.4876e+006 6.6233e+006 7.7755e+006 -1.0741e-008 -2.4017e-008 -1.9214e-008 -1.0741e-008

The highest stresses occur at: MaxComp = -8.6957e+006 CompBar = 55

MaxTens = 9.0511e+006 TensBar = 81

The following is a plot of the deformed (solid black line) vs. undeformed (dashed blue line) pylon structure:

Notice that the bars in which the maximum stresses occur are labeled with arrows.

Effects of Eigenvectors
The lowest eigenvalues and their corresponding oscillation periods are:

lam =  132.16       2468.4       2940.3

T = 0.54654      0.12647      0.11587

Plots of the effects of the three lowest eigenvalues are shown below. The blue dashed line represents the undeformed structure while the solid black line represents the deformed structure.

Statically Determinant?
The final step in this problem was to determine whether the structure can be solved using statics. Since there are only two fixed points and four bars for which reactions must be computed, it can be determined that this problem is statically indeterminate.

Honesty, Innovation, and Ethics, a Brief Commentary
Honesty, Innovation, and Ethics.

These three concepts are fundamental traits of the professional engineer, but easily extend beyond the singular. These are principles which can drive a society and can forge interpersonal relationships without which the world would be subject to a Damoclean verge of chaos.

Without honesty, the integrity of mankind would be shattered, replaced with a nebulous abyss of lies and deceit. By internalizing and reciprocating the concept of honesty, individuals are able to exchange ideas and communications with the knowledge that this information transfer is absolute in self-perceived truth (clearly little "t"). Here, suspicion would only lay not in the wonderment of peril, but in the motivation for the honesty itself. However, this phenomenon is speculated to exist insomuch as the function of integrated human history void of the ideal absolute honesty theorized herein.

Indeed, even Sir Thomas More's Utopia proved that absolute honesty could only be assumed as a satirical tool against those lambasted in metaphor by the book itself. If honesty had this sway, then truth would be self-evident and secrets disavowed as public would be all knowledge if only received upon request. This is within Loc Vu-Quoc's Vision: that is, for the free dissemination of all academic resources via the Wikipedia and Wikiversity interfaces. Here, it is believed that a modern categorical imperative would be enforced (See HW 1) by the willingness of the masses, and hence contributions would be made for the good of mankind to strengthen the collective information pool.

The hitch, however, is that this functions specifically on honesty as a tool to be utilized by all in a Kantian fashion. Reality, though, suggests this is not the universal tendency of human behavior as seen by the necessitation of Wiki admins who tirelessly endeavour to ensure validity of posted material. If the material was always valid, the need for admins would be minimalized significantly; as is, this is unforunately not the case. Henceforth, honesty can only be considered a fluctual tool of mind, something situationally applicable, but never universally so if only due to the fallacies of tabula rasa and the ubiquitous corruptibility of man. More simply, there is no Übermensch.

Innovation, meanwhile, is something which is a limitless pursuit as elusive as Máni to Hati Hróðvitnisson. It is something which inspires and drives, but is ever-presently infinite and inherently unattainable. Innovation is what leads to new ideas and technologies, governments, relationships, and all other forms of human and animal advances. It could be argued that the aspiration of all engineers is to innovate, but it can also be argued that innovation is a function not only on the synthesis of knowledge by the thinker but also of the collective knowledge itself. Ergo, Loc Vu-Quoc's Vision is further realized: through the free and open osmosis of ideas in the international community, especially when governed by the principle of honesty, innovation can occur based upon the synergy intrinsic in the global neural network.

Without innovation, it can be speculated that mankind would have never survived as many generations of procreation as it has managed. Fire, the wheel, and primitive agricultural techniques were innovations to the early mind as modern technology like the iPod, the Nintendo Wii, the iFly VAMP, or Wikipedia are today.

But innovation itself is subject to the concept of ethics, a catchall philosophical term generally finding root in the belief and execution of a "correct" and "proper" behavior of organisms to ensure a "good life". Ethics is exceptionally questionable, as the moralities of good and evil (fundamentally diametric forces defining ethics) can be perceived universally from perspectives arguing favorably to any singular action. For example, if a student is caught plagiarizing from Wikipedia, the common reaction would be to assume the student is exhibiting some form of "evil" as the connotation of the very act is often considered "wrong" by society; conversely, if the action was done in response to the conformities of society as a way of breaking the mold, innovating thought, and pushing the limits of what is convential, how is this different than the "goodly" perceived notions of the American Revolution against King George III? Or further: what if there was a justified reason for the plagiarism, such as the student wasn't able to do the work originally as he had invested time needed to satisfy such criteria in saving a school bus of orphaned children from the cold-hearted wrath of indifferent terrorists threatening their very lives‽

Thus, ethics cannot be objectively considered without understanding the universal. Sometimes this is difficult, and hence open communication that exhibits honesty will aid in a more complete understanding of the situation(s). Yet for the sake of argument based upon the expectations of Loc Vu-Quoc's Vision, ethics should be considered a very heartfelt principle to an engineer, as "proper" moral principles are what would be argued under the Vision as things which will lead to ideal interpersonal communications, and hence refine and improve mankind while leading to innovation through honesty.

Finally, as William Shakespeare once famously noted: "Honesty is the best policy. If I lose mine honor, I lose myself."

Three Bar Truss Problem
This problem is a continuation of a problem introduced in the previous homework assignment where a three bar truss system as seen in the following graphic was to be analyzed for its eigenvectors and eigenvalues using MATLAB. The K matrix for the structure is defined as: where Implying the eigenvectors each correspond to the column vectors composing the V matrix, with the eigenvalues consisting of the diagonal values of the D matrix. Plotting the eigenvectors as deformations to the original structure results in the following series of graphics, each graphic sequentially corresponding the eigenvector column for which the deformation displacements were derived.

MATLAB code for the generation of the plots is as follows:



Effects of Cross-bar to Three Bar System
The addition of a cross-bar to the Three Bar system above results in the following figure:



Where K is defined as:

with eigenvectors and eigenvalues defined similarly to those above as the following:

MATLAB code to model this system can be seen as follows:



Individual Contributions
Eml4500.f08.bottle.vitello 01:16, 10 November 2008 (UTC) Eml4500.f08.bottle.butler 15:55, 17 November 2008 (UTC) Eml4500.f08.bottle.hipps 21:00, 18 November 2008 (UTC) Eml4500.f08.bottle.barnes 22:21, 20 November 2008 (UTC) Eml4500.f08.bottle.loschak 02:47, 21 November 2008 (UTC) Eml4500.f08.bottle.brockmiller 17:34, 21 November 2008 (UTC) Eml4500.f08.bottle.ranto 20:50, 21 November 2008 (UTC)