User:EML4500.f08.JAMAMA/Homework6

Meeting 29
Initial Conditions  (2 conditions) at t = 0, prescribe:

Condition 1: $$ u(x, t=0) = \bar{u}(x) $$
 * $$ \uparrow $$
 * known function displacement

Condition 2: $$ \frac{\partial u}{\partial t}(x, t=0) = \dot {u}(x, t=0) = \bar{v}(x)$$
 * $$ \uparrow $$
 * known function velocity

PVW (continuous) of dynamics of elastic bar

PDE (Partial Differential Equation):

$$ \frac{\partial}{\partial x}[(EA)\frac{\partial u}{\partial x}] + f = m \ddot u \; \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (1)$$
 * $$ \uparrow $$

Discrete EOM (Equation of Motion)
 * $$\downarrow $$

$$ -\mathbf{M\ddot d} + \mathbf{F} = \mathbf{Kd} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad (2)$$


 * $$\Downarrow $$

$$ \mathbf{M\ddot d} + \mathbf{Kd} = \mathbf{F} $$
 * $$ \nwarrow $$
 * Multiple Degree of Freedom System (MDOF)

Single Degree of Freedom System (SDOF)

$$ \Rightarrow \ \mathbf{m\ddot d} + \mathbf{kd} = \mathbf{f} $$

Derive (2) from (1)

$$ \int \limits_0^L ( \frac{\partial}{\partial x}[EA \frac{\partial u}{\partial x}] + f - m \ddot {u} ) dx = 0,$$ $$ \qquad $$   for all possible  $$\mathbf{w(x)} \qquad \qquad \qquad (3)$$
 * $$ \uparrow $$
 * weighting function

$$ (1) \Rightarrow (3) $$  >>  trivial

$$ (3) \Rightarrow (1) $$  >>  not trivial

(3) rewritten as : $$ \int w(x) g(x) dx, \qquad $$ for all $$ \mathbf{w(x)}$$

Since (3) holds for all $$ \mathbf{w(x)}$$, select $$ \mathbf{w(x)} = \mathbf{g(x)}$$, then (3) becomes $$\int \underbrace{g^2(x)}_{\ge 0} dx = 0 \Rightarrow g(x) = 0 $$

Meeting 30
Integration by parts:$$ r(x), s(x) $$

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

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

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

$$ => \underline{\int{r's} = rs - \int{rs'}}$$

We can recall cont. PVW (Principal of Virtual Work) Eq. (3) p. 29-1

First term: $$ r(x)=(EA)\frac{\partial{u}}{\partial{x}}  ;   s(x) = W(x)$$

By integrating by parts we can get:

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

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

Now lets consider the following model problem: p.28-3  2 b.c.'s



At x=o, select W(x) so that W(0)=0 (i.e. Kinematically admissible)

Motivation : Discrete PVW applied to Equation on  p. 10-1

$$\mathbf{W}_{6x1}. \underbrace{\left(\left[      \right]_{6x2}\begin{Bmatrix}d3\\d4\end{Bmatrix}_{2x1} - \mathbf{F}_{6x1}\right)}_{6x1} = 0 $$     for all W

$$ \mathbf{F}^T = \begin{bmatrix} F_1&F_2&F_3&F_4&F_5&F_6\end{bmatrix}$$  with $$F_1, F_2, F_5, F_6$$: unknown reactions

Since W can be selected arbitrarily, select W so that:

$$ W_1 = W_2 = W_5 = W_6 = 0$$; so to eliminate equations involving unknown reactions => eliminate rows 1,2,5,6;

Let equation (1) be: $$ \mathbf{K}_{2x2} \mathbf{d}_{2x1} = \mathbf{F}_{2x1}$$

with: $$ \mathbf{d}_{2x1} = \begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix}$$ and: $$ \mathbf{F}_{2x1} = \begin{Bmatrix} F_3 \\ F_4 \end{Bmatrix}$$

Let us note that for the step before eq.(1)


 * $$ \mathbf{W}.\left(\mathbf{K}\mathbf{d}-\mathbf{F}\right) = 0$$ with $$ \mathbf{W} = \begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix}$$

Now back to the PVW:

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

Cont. PVW :

$$W(L)f(t)- \int_0^L{\frac{dw}{dx}(EA)\frac{\partial{u}}{\partial{x}}dx} + \int_0^L{W(x)\left[f - m\ddot{u}\right]dx} = 0$$

So for all W(x) so that W(0) = 0,

$$=> \int_0^L{W(m\ddot{u})}dx + \int_0^L{\frac{dW}{dx}(EA)\frac{\partial{u}}{\partial{x}}}dx = W(L)f(t) + \int_0^L{Wf}dx$$
 * for all W so that W(0) = 0.

Meeting 31






Assumption: displacement within element $$i$$ is linear displacement $$u(x)$$ for $$x_{i}\le x\le x_{i+1}$$, (i.e., $$x\in \left[ x_{i},x_{i+1} \right]$$)

Motivation for linear interpolation of $$u(x)$$

2-bar truss:



Deformed shape is a straight line (linear interpolation) of the displacement of 2-nodes.

Consider the case where there are only axial displacements:

Q: Express $$x$$ in terms of $$u_{i}$$ $$(u(x_{i})\text{ and }d_{i+1}=u(x_{i+1}))$$ as a linear function of $$x$$ (i.e. linear interpolation)

$$u(x)=N_{i}(x)d_{i}+N_{i+1}(x)d_{i+1}$$

express $$N_{i}(x)$$ and $$N_{i+1}(x)$$ as linear functions of $$x$$.

$$N_{i}(x)=N_{1}^{(i)}(\tilde{x})$$



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

This linear interpolation utilizes the geometry properties of similar triangles in order to provide the unknown value, which lies in between two known values.

Meeting 32
Rube Goldberg Device: This is a device that accomplishes a simple task through many unnecessary steps, and in a very indirect fashion.

An example of this type of device can be seen in the following youtube link: []

NOTE: Honesty, imagination, ethics. It is always very important to do your own work, and not plagiarize other people's work. Honesty is the best policy.

Cont PVW to discrete PVW.

Lagrangian Interpolation.


 * Motivation for form of Ni(x) and Ni+1(x)

1) Mi(x) and Mi+1(x) are linear (straight lines), thus any linear combination of Ni and Ni+1 are also linear and in the particular expression for U(x) on Page 31-3.

$$N_{i}(x)=\alpha _{i}+\beta _{i}x $$ with $$(\alpha _{i},\beta _{i})$$ being numbers.

$$N_{i+1}(x)=\alpha _{i+1}+\beta _{i+1}x$$ Linear combo of Ni and $$(\alpha _{i},\beta _{i})$$ numbers.

$$N_{i+1}$$: $$N_{i}d_{i}+N_{i+1}d_{i+1}=(\alpha _{i}+\beta _{i}x)d_{i}+(\alpha _{i+1}+\beta _{i+1}x)d_{i+1}$$ $$=(\alpha _{i}d_{i}+\alpha _{i+1}d_{i+1})+(\beta _{i}d_{i}+\beta _{i+1}d_{i+1})x$$ is clearly a function of x.

2) Recall equation for U(x) (interp of U(x)) on Page 31-3:

$$U(x_{i})=N_{i}(x_{i})d_{i}+N_{i+1}(x_{i})d_{i+1}=d_{i}$$

=1 =0

Meeting 33
FEM (Finite Element Method) via (Principles of Virtual Work) Continued

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

Applying the same interpolation for W(x) as for u(x)

$$\displaystyle W(x)=N_{i}(x)W_{i}+N_{i+1}(x)W_{i+1}$$

The element stiffness matrix then becomes

$$\beta =\int_{x_{i}}^{x_{i+1}}{\left[\frac{dN_{i}}{dx}(x)W_{i}+\frac{dN_{i+1}}{dx}(x)W_{i+1} \right]\left[EA \right]\left[\frac{dN_{i}}{dx}(x)d_{i}+\frac{dN_{i+1}}{dx}(x)d_{i+1} \right]}dx$$

$$u(x)=\begin{bmatrix} N_{i}(x) & N_{i+1}(x) \end{bmatrix} \begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix} \Rightarrow u(x)=\textbf{N}(x) \begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix}$$

$$\frac{du(x)}{dx}=\begin{bmatrix} N^{'}_{i}(x) & N^{'}_{i+1}(x) \end{bmatrix} \begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix} \Rightarrow \frac{du(x)}{dx}=\textbf{B}(x) \begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix}$$

Just like u(x), W(x) is represented the same way

$$W(x)=\textbf{N}(x) \begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix}$$

$$\frac{dW(x)}{dx}=\textbf{B}(x) \begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix}$$

Now recall the element degrees of freedom as shown in the picture below



$$\begin{bmatrix} d_{i}\\ d_{i+1} \end{bmatrix}= \begin{bmatrix} d_{1}^{(i)}\\ d_{2}^{(i)} \end{bmatrix}=\textbf{d}^{(i)}$$

$$\begin{bmatrix} W_{i}\\ W_{i+1} \end{bmatrix}= \begin{bmatrix} W_{1}^{(i)}\\ W_{2}^{(i)} \end{bmatrix}=\textbf{W}^{(i)}$$

Now the element stiffness matrix becomes

$$\beta =\int_{x_{i}}^{x_{i+1}}{\left[\textbf{BW}^{(i)}\right]\left[EA \right]\left[\textbf{Bd}^{(i)}\right]}dx=\textbf{W}^{(i)}\bullet \left(\textbf{k}^{(i)} \textbf{d}^{(i)}\right)$$

Each set of matrices and scalars multiplied inside the brackets when evaluated come out to be scalars so the bracketed items can be multiplied in any order to come out to the same result.

$$\beta =\int_{x_{i}}^{x_{i+1}}{\left[EA \right]\left[\textbf{BW}^{(i)}\right]\left[\textbf{Bd}^{(i)}\right]}dx$$

Since the the quantity of $$\left[\textbf{BW}^{(i)}\right]$$ is a scalar value the transpose of the value is the same. The transpose of the scalar value $$\left[\textbf{BW}^{(i)}\right]^{T}$$ is also equal to $$\left[\textbf{W}^{(i)T} \textbf{B}^{(i)T}\right]$$

$$\beta =\textbf{W}^{(i)}\left( \int_{x_{i}}^{x_{i+1}}{\textbf{B}^{T}\left[EA \right]\textbf{B}dx}\right) \textbf{d}^{(i)}$$

So the stiffness matrix is equal to

$$\textbf{k}^{(i)}=\int_{x_{i}}^{x_{i+1}}{\textbf{B}^{T}\left(EA \right)\textbf{B}dx}$$

$$\textbf{B}=\begin{bmatrix} \frac{-1}{L^{(i)}} & \frac{1}{L^{(i)}} \end{bmatrix}$$

$$\displaystyle L^{(i)}=x_{i+1}-x_{i}$$



Transformation of variable from $$\displaystyle x$$ to $$\displaystyle \tilde{x}$$

$$\tilde{x}=x-x_{i}$$

$$d\tilde{x}=dx$$

$$\textbf{k}^{(i)}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}{\textbf{B}(\tilde{x})^{T}\left(EA \right)(\tilde{x})\textbf{B}(\tilde{x})d\tilde{x}}$$

Find expression for $$\textbf{k}^{(i)}$$ using the equation above.

$$A(\tilde{x})=N^{(i)}_{1}(\tilde{x})A_{1}+N^{(i)}_{2}(\tilde{x})A_{2}$$

$$E(\tilde{x})=N^{(i)}_{1}(\tilde{x})E_{1}+N^{(i)}_{2}(\tilde{x})E_{2}$$

$$\textbf{k}^{(i)}=\int_{\tilde{x}=0}^{\tilde{x}=L^{(i)}}{\textbf{B}(\tilde{x})^{T}\left(\left(N^{(i)}_{1}(\tilde{x})E_{1}+N^{(i)}_{2}(\tilde{x})E_{2}\right)\left(N^{(i)}_{1}(\tilde{x})A_{1}+N^{(i)}_{2}(\tilde{x})A_{2}\right) \right)\textbf{B}(\tilde{x})d\tilde{x}}$$

Meeting 34
$$N_{1}^{(i)}(\tilde{x})=\tilde{x}=\left\{ \begin{align} & 1\text{ at }\tilde{x}=0 \\ & 1\text{ at }\tilde{x}=L^{(i)} \\ \end{align} \right.$$

$$N_{2}^{(i)}=\frac{L^{(i)}}=\left\{ \begin{align} & 0\text{ at }\tilde{x}=0 \\ & 1\text{ at }\tilde{x}=L^{(i)} \\ \end{align} \right.$$

Shape functions (basis): $$N_{1}^{(i)},N_{2}^{(i)}$$

$$K^{(i)}=\frac{(E_{1}+\frac{x}{L}(E_{2}-E_{1}))(A_{1}+\frac{x}{L}(A_{2}-A_{1}))}{L}\begin{bmatrix} 1 &-1 \\ -1 &1 \end{bmatrix}$$
 * HW 6: Book p. 159:

$$K^{(i)}=\frac{(E)(A_{1}+\frac{x}{L}(A_{2}-A_{1}))}{L}\begin{bmatrix} 1 &-1 \\ -1 &1 \end{bmatrix}$$

$$\int_{0}^{L}{(A_{1}+\frac{x}{L}(A_{2}-A_{1}))dx}=(A_{1}x+\frac{x^{2}}{2L}(A_{2}-A_{1}))\mid ^{L}_{0}$$

$$A_{1}L+\frac{L^{2}}{2L}(A_{2}-A_{1})=A_{1}(L-\frac{L}{2})+\frac{L}{2}A_{2}=\frac{A_{1}+A_{2}}{2}$$

$$A_{1}=A_{L}$$...$$A_{2}=A_{R}$$

so:

$$K^{(i)}=\frac{E(A_{1}+A_{2})}{2L}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}=K^{(i)}=\frac{E(A_{L}+A_{R})}{2L}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

Set $$E_{1}=E_{2}=E$$ and let $$A(\tilde{x})$$ be linear as on p. 33-5. Obtain $$k^{(i)}$$ from the previous problem (p. 33-5) and compare it to the expression given in the book.

$$\frac{E(A_{1}+A_{2})}{L^{(i)}2}\left[ \begin{matrix} 1 & -1 \\   -1 & 1  \\ \end{matrix} \right]=K^{(i)}$$



Next, compare the general $$k^{(i)}$$ on p. 33-5 to the stiffness matrix obtained by using $$\frac{1}{2}(A_{1}+A_{2})$$ and $$\frac{1}{2}(E_{1}+E_{2})$$.

NOTE: $$E_{1}\ne E_{2}$$

$$\frac{(E_{1}+E_{2})(A_{1}+A_{2})}{4L^{(i)}}\left[ \begin{matrix} 1 & -1 \\   -1 & 1  \\ \end{matrix} \right]=k_{ave}^{(i)}$$

find $$k^{(i)}-k_{ave}^{(i)}$$

Recall the mean value theorem (MVT) and its relation to the centroid.

MVT: $$\int_{x=a}^{x=b}{f(x)dx=f(\bar{x})\left[ b-a \right]}$$ for $$\bar{x}\in [a,b]\text{ or }a\le \bar{x}\le b$$



$$Centroid\int_{A}{xdA}=\bar{x}\int_{A}{dA}=\bar{x}A$$

$$\int_{x=a}^{x=b}{f(x)g(x)dx=f(\bar{x})g(\bar{x})[b-a]\text{ for }a\le \bar{x}\le b}$$

but: $$f(\bar{x})\ne \underbrace{\frac{1}{b-a}\int_{a}^{b}{f(x)dx}}_{average\text{ }f}$$

$$g(\bar{x})\ne \underbrace{\frac{1}{b-a}\int_{a}^{b}{g(x)dx}}_{average\text{ g}}$$

The following electric pylon is statically indeterminate because it has nodes with multiple (more than two) elements running to them. Because of this fact, there are too many unknowns for the amount of equations that can be used to solve it statically, therefore it is statically indeterminate.

Electric Pylon
{| class="toccolours collapsible collapsed" style="width:100%" ! Matlab Code For Electric Pylon 

Matlab Code


results=[]; for i=1:n_elem lm=lmm(i,:); location = [x(node_connect(1,i)),y(node_connect(1,i)); x(node_connect(2,i)),y(node_connect(2,i))]; results = [results; PlaneTrussResults(E,A,location,d(lm))]; end results










 * }

Contributing Members
William Mueller Eml4500.f08.jamama.mueller 22:19, 16 November 2008 (UTC)

Justin McIntire --Eml4500.f08.jamama.justin 18:38, 20 November 2008 (UTC)

Cedric Adam--Eml4500.f08.jamama.adam 08:35, 21 November 2008 (UTC)

Arunas Janulevicius --Eml4500.f08.jamama.jan 10:13, 21 November 2008 (UTC)

Megan Alvarez EML4500.f08.jamama.megan 17:10, 21 November 2008 (UTC)