User:Eml4500.f08.wiki1.aguilar/TeamHW6

= Team Homework #6=

Rube Goldberg Machine
A Rube Goldberg machine is an overly complicated system designed to accomplish a relatively simple task. An prime example of this is if you have ever played the board game "MouseTrap" as a child. For more information, click the link, [Rube Goldberg Machine].

Honesty, Imagination, and Ethics
[Honesty],[Imagination], and [Ethics]

[The University of Florida Honor Code] We, the members of the University of Florida community, pledge to hold ourselves and our peers to the highest standards of honesty and integrity.

The University of Florida Honor Pledge

"On my honor, I have neither given nor received unauthorized aid in doing this assignment.”

PVW (Continuous) of Dynamics of Elastic Bar
PDE: $$\frac{\partial }{\partial x}\left[(EA)\frac{\partial u}{\partial x} \right]+f=m\ddot{u}\; (1)$$ Discrete EOM (Equation of Motion) $$\Rightarrow \mathbf{-Kd+F=M\ddot{d}}$$ $$\Rightarrow \mathbf{M\ddot{d}+Kd=F}\; (2)$$ (for Multiple Degree of Freedom System (MDOF))

SDOF=Single Degree of Freedom:



Derive (2) from (1): $$\int_{x=0}^{x=L}{W(x)\left\{\frac{\partial }{\partial x}\left[EA\frac{\partial u}{\partial x} \right]+f-m\ddot{u} \right\}dx}=0\; (3)$$ for all possible W(x) (Weighting function)

$$(1)\Rightarrow (3)$$ Trivial $$(3)\Rightarrow (1)$$ NOT Trivial

(3) rewritten as: $$\int W(x)g(x)dx=0$$ for all W(x)

Since (3) holds for all W(x), select W(x)=g(x), then (3) becomes: $$\int g^2dx=0\Rightarrow g(x)=0$$ $$(g^2\geq 0)$$

Integration by Parts: r(x), s(x)

$$(rs)'=r's+rs'$$

$$r'=\frac{dr}{dx}$$ and $$s'=\frac{ds}{dx}$$

$$\int (rs)'=\int r's+rs'$$

$$\Rightarrow (rs)=\int r's+rs'$$

$$\Rightarrow \int r's=rs-\int rs'$$

Recall continuous PVW $$\Rightarrow Eq. (3)$$ from Meeting 29

1st term: $$r(x)=(EA)\frac{\partial u}{\partial x}$$ and $$s(x)= W(x)$$

Through integration by parts:  $$\int_{x=0}^{x=L}{W(x)\frac{\partial }{\partial x}\left[(EA)\frac{\partial u}{\partial x} \right]dx}$$

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

<p style="text-align:center;">$$=W(L)(EA)(L)\frac{\partial u}{\partial x}(L,t)-W(0)EA(0)\frac{\partial u}{\partial x}(0,t)-\int_{0}^{L}{\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx}$$

Now consider a model problem from Meeting 28:

At x=0, select W(x) such that W(0)=0 and kinematically admissible

<p style="text-align:center;">

Motivation: Discrete PVW applied to Equation from Meeting 10 <p style="text-align:center;">$$\mathbf{W}_{6\times1}\cdot(\mathbf{[K]}_{6\times2}\begin{Bmatrix}d_3\\d_4\end{Bmatrix}_{2\times1}-\mathbf{F}_{6\times1})=0_{1\times1}$$

<p style="text-align:center;">$$\mathbf{F^T=\begin{bmatrix} F_1 & F_2 & F_3 & F_4 & F_5 & F_6 \end{bmatrix} }$$

F3 and F4 are known reactions F1, F2, F5, F6 are unknown reactions

Since W can be selected arbitrarily, select W such that W1 = W2 = W5 = W6 = 0 so to eliminate equations involving unknown reactions →Eliminate rows 1,2,5,6

<p style="text-align:center;">$$\mathbf{\bar{K}_{2x2}\bar{d}_{2x1}=\bar{F}_{2x1}}\; (1)$$

<p style="text-align:center;">$$\mathbf{\bar{d}_{2x1}}=\begin{Bmatrix} d_3\\d_4\end{Bmatrix}$$ and $$\mathbf{\bar{F}_{2x1}}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}$$

Note:

<p style="text-align:center;">$$\mathbf{\bar{W}\cdot }(\mathbf{\bar{K}\bar{d}-\bar{F})=0}$$ <p style="text-align:center;">for all $$\mathbf{\bar{W}}$$ <p style="text-align:center;">where $$\mathbf{\bar{W}}=\begin{Bmatrix} W_3\\W_4\end{Bmatrix}$$

Back to Continuous PVW:

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

Continuous PVW gives: <p style="text-align:center;">$$W(L)F(t)-\int_{0}^{L}{\frac{dw}{dx}(EA)\frac{\partial u}{\partial x}dx}+\int_{0}^{L}{W(x)[f-m\ddot{u}]dx=0}$$

<p style="text-align:center;">for all W(x) such that W(0)=0

<p style="text-align:center;">$$\Rightarrow \int_{0}^{L}{W(m\ddot{u})dx+\int_{0}^{L}{\frac{dW}{dx}}(EA)\frac{\partial u}{\partial x}dx}$$

<p style="text-align:center;">$$=W(L)F(t)+\int_{0}^{L}{Wfdx}$$

<p style="text-align:center;">for all W(x) such that W(0) = 0

<p style="text-align:center;">Bar Broken Up Into Elements

<p style="text-align:center;">Single Element Function Interpolation

Assume $$u(x) \,$$ for $$x_i \le x \le x_{i+1} \,$$

Motivation for Linear Interpolation of $$u(x) \,$$
<p style="text-align:center;">2-bar Truss

<p style="text-align:center;">2-Bar Truss Deformation

Continuous Principle of Virtual Work to Discrete Case (continued)
Lagrangian Interpolation

Motivation for Form of $$N_i(x) \,$$ and $$N_{i+1}(x) \,$$:

$$N_i(x) \,$$ and $$N_{i+1}(x) \,$$ are linear (straight lines), thus any linear combination of $$N_i(x) \,$$ and $$N_{i+1} \,$$ is also linear, and in particular, the expression for $$u(x) \,$$.

$$N_i(x)=\alpha_i + \beta_i x \,$$, with $$\alpha_i \,$$, $$\beta_i \,$$ numbers

$$N_{i+1}(x)=\alpha_{i+1}+\beta_{i+1}x \,$$, with $$\alpha_{i+1} \,$$, $$\beta_{i+1} \,$$ numbers

Linear Combination of $$N_i \,$$ and $$N_{i+1} \,$$:

$$N_id_i+N_{i+1}d_{i+1}=(\alpha_i+\beta_ix)d_i+(\alpha_{i+1}+\beta_{i+1}x)d_{i+1} \,$$ $$\quad=(\alpha_id_i+\alpha_{i+1}d_{i+1})+(\beta_id_i+\beta_{i+1}d_{i+1})x \,$$ is clearly linear.

Recall equation for $$u(x) \,$$ (interpolation of $$u(x) \,$$):

$$u(x_i)=N_i(x_i)d_i+N_{i+1}(x_i)d_{i+1} \,$$ $$u(x_i)={x_{i+1}-x_i \over x_{i+1}-x_i}d_i+{x_i-x_i \over x_{i+1}-x_i}d_{i+1} \,$$ $$N_i(x_i)=1 \,$$ $$N_{i+1}(x_i)=0 \,$$ $$u(x_i)=d_i \,$$

Using the same logic:

$$u(x_{i+1})=d_{i+1} \,$$

We have just showed how we got to the relation:

$$ u(x_{i+1})=d_{i+1} $$

Page 31.3 of our notes shows the interpolation for u(x). We can apply this same interpolation for w(x), i.e.,

$$ w(x)=N_i(x)\; W_i\; +\; N_{i+1}(x)\; W_{i+1}$$

Element Stiffness Matrix for Element i
In order to make the leap from the discrete version of the Principle of Virtual Work to the continuous version, the stiffness matrix must involve an integral.

The form of this integral is in the form below:

$$\int_{x_i}^{x_{i+1}}[N'_i\; W_i\; +\; N'_{i+1}\; W_{i+1}](EA)[N'_i\;  d_i\; +\; N_{i+1}\; W_{i+1}]dx        \;\;\;\;\;\;\;\;\;   (eqn.\; \beta) $$

with    $$N'_i:= \frac {dN'_i(x)}{dx}$$      and     $$N'_{i+1}:= \frac {dN'_{i+1}(x)}{dx} $$.

Note:

$$u(x)= \underbrace{ \begin{bmatrix}N_i(x)&N_{i+1}(x)\end{bmatrix}}_{\mathbf{N}(x)_{1\times 2}}  \begin{Bmatrix}d_i\\d_{i+1}\end{Bmatrix}_{2\times 1}$$

and

$$\frac{du(x)}{dx} = \begin{bmatrix}N'_i(x)&N'_{i+1}(x)\end{bmatrix}_{2\times 1}  \begin{Bmatrix}d_i\\d_{i+1}\end{Bmatrix}_{2\times 1} $$

Similarly:

$$w(x)= \mathbf{N}(x) \begin{Bmatrix}w_i\\w_{i+1}\end{Bmatrix}$$

and

$$\frac{dw(x)}{dx} = \mathbf{B}(x) \begin{Bmatrix}w_i\\w_{i+1}\end{Bmatrix} $$

Recall the element degrees of freedom:

<p style="text-align:center;">

$$\begin{Bmatrix}d_i\\d_{i+1}\end{Bmatrix} = \begin{Bmatrix}d_1^{(i)}\\d_2^{(i)}\end{Bmatrix} = \mathbf{d}^{(i)}$$

$$\begin{Bmatrix}w_i\\w_{i+1}\end{Bmatrix} = \begin{Bmatrix}w_1^{(i)}\\w_2^{(i)}\end{Bmatrix} = \mathbf{w}^{(i)}$$

We're looking to take $$ eqn. \beta$$ and move some terms around so that it corresponds to the relation: $$\mathbf{w}^{(i)} \cdot (\mathbf{k}^{(i)}\mathbf{d}^{(i)})$$.

So how do we get: $$\int_{x_i}^{x_{i+1}} (\mathbf{B}\mathbf{w}^{(i)})_{1\times1}(EA)_{1\times1}(\mathbf{B}\mathbf{d}^{(i)})_{1\times1} dx = \mathbf{w}^{(i)} \cdot (\mathbf{k}^{(i)}\mathbf{d}^{(i)}) $$ ?

Well,

$$\int_{x_i}^{x_{i+1}} (EA)(\mathbf{B}\mathbf{w}^{(i)})(\mathbf{B}\mathbf{d}^{(i)})dx - (\mathbf{B}\mathbf{w}^{(i)})^T(\mathbf{B}\mathbf{d}^{(i)}) $$

As a 1x1 matrix, $$(\mathbf{Bw}^{(i)})$$ is equal to its transpose, $$(\mathbf{Bw}^{(i)}))^T$$.

$$(\mathbf{Bw}^{(i)})^T = (\mathbf{w}^{(i)})^T \cdot (\mathbf{B})^T = (\mathbf{w}^{(i)}) \cdot (\mathbf{B})^T$$

So we now have: $$\int (\mathbf{w}^{(i)}\mathbf{B}^T)(EA)(\mathbf{B}\mathbf{d}^{(i)}) dx $$

We can also move parts out of the integral if they are not dependent on 'x', ($$\mathbf{w}^{(i)},\mathbf{d}^{(i)}$$):

Therefore we get: $$ \mathbf{w}^{(i)} \cdot ( \int \mathbf{B}^T (EA) \mathbf{B} dx) \;\mathbf{d}^{(i)}$$

We can now see that:

$$\mathbf{k}^{(i)}_{2\times2}=\int_{x_i}^{x_{i+1}} \mathbf{B}_{2\times1}^T (EA)_{1\times1} \mathbf{B}_{1\times2} dx$$

where $$ \mathbf{B}^T$$, $$(EA)$$, and $$\mathbf{B}$$ are all functions in terms of 'x'.

$$\mathbf{B}= \begin{bmatrix}\frac{1}{x_{i}-x_{i+1}} & \frac{1}{x_{i+1}-x_{i}}\end{bmatrix} = \begin{bmatrix}\frac{1}{-L^{i}} & \frac{1}{L^{i}}\end{bmatrix}$$

The length of element i: $$L^{(i)}=x_{i+1}-x_{i}$$

Transfer of Coordinates
We now do a basic transfer of $$x$$ to $$\tilde x$$.

$$\tilde x := x - x_i$$ and $$d\tilde x = dx$$

We get a new equation based on our old for the element stiffness matrix as: $$\mathbf{k}^{(i)}=\int_{\tilde x=0}^{\tilde x=L^{(i)}} \mathbf{B}(\tilde x) (EA)(\tilde x) \mathbf{B}(\tilde x) d\tilde x$$

$$$$ $$$$ $$$$

Modifying the Two Bar Truss
Using this general formula for stiffness matrix that works with non-constant E and A.

$$\mathbf{k}^{(i)}=\frac{1}{6L} \left( {2E_1A_1}+{(E_1A_2+E_2A_1)}+ {2E_2A_2}\right) \begin{bmatrix} 1 & -1\\ -1 & 1\end{bmatrix}$$

The matlab function of the axial forces and the function that calls it were modified to accomodate teh variances in E and A.

Boundary conditions:

Element 1

$$E_1^{(1)}=2$$ $$E_2^{(1)}=4$$ $$A_1^{(1)}=0.5$$ $$A_2^{(1)}=1.5$$

Element 2

$$E_1^{(2)}=3$$ $$E_2^{(2)}=7$$ $$A_1^{(2)}=1$$ $$A_2^{(2)}=3$$

Averages from before:

$$E_{avg}^{(1)}=3$$ $$A_{avg}^{(1)}=1$$

$$E_{avg}^{(2)}=5$$ $$A_{avg}^{(2)}=2$$

The plot of the undeformed beam, the deformed beam using the new equation, and the deformed beam using just the averages of E and A is as follows. The undeformed beam is the blue dotted line, the deformed beam using the averages is the solid red line, and the deformed beam using the varied E and A values is the solid green line. The matlab code that plots this image is as follows: Two-Bar Plot

The matlab code used to attain the deformed beam using the varied E and A is as follows: 2 Bar Truss with varied E and A

The altered m-file, DeltaPlaneTrussElement.m, is as follows: DeltaPlaneTrussElement.m Note:  The deformed beam using the average values of E and A can be found in previous hw.

The Electric Pylon Problem
The highest axial stresses are 1376.8 lbs in element 1 and -1370.1 lbs in element 4.

The figure below shows the undeformed and deformed pylon. The dashed blue line represents the undeformed pylon while the solid red line shows the deformed pylon with a 20,000 weighting factor applied.



Contributors
Eml4500.f08.wiki1.brannon 17:42, 21 November 2008 (UTC) Eml4500.f08.wiki1.ambrosio 20:34, 21 November 2008 (UTC) Eml4500.f08.wiki1.oatley 20:52, 21 November 2008 (UTC) Eml4500.f08.wiki1.aguilar 21:44, 21 November 2008 (UTC)<BR>