User:Eml4500.f08.FEABBQ/HW6

 See my comments below. Please don't remove these comment boxes; just add your comment to these comment boxes in case you fix the problems. Eml4500.f08 19:52, 26 November 2008 (UTC)

Class Notes November 03, 2008
Notes from November-03-08

HW: Refine the Mode-Shape Plots
We revisit the 2-bar truss example to refine our mode-shape plot. First, we output the global stiffness matrix, K, and the corresponding eigenvectors and eigenvalues...

Next, we plot each column of the eigenvector matrix as a mode-shape for only the columns which correspond to a zero eigenvector (the first four columns)...





Next, we study the unstable 3 bar truss (Case A in the image below)...



The following global stiffness matrix and eigenvectors and eigenvalues were solved for using Matlab (Note: The matlab code is available upon request, but it is omitted here due to similarity to previous solving and plotting code provided.)

The following are the five mode shapes associated with the five zero-value eigenvectors for the three bar truss (Case A). Note: some magnification has been applied (3x).











END HW

Continuing the derivation:

Initial conditions: At t=0, prescribed

$$\left(x,t=0 \right) =\bar{U}\left(x \right)$$ Known Function Displacement $$\frac{\partial }{\partial t}\left(x,t=0 \right)=\dot{U }\left(x,t=0 \right)=\bar{V}\left(x \right)$$Known Function Velocity

$$\frac{\partial }{\partial x}\left[\left(EA \right)\frac{\partial }{\partial x} \right]+f=m\ddot{U}$$ Eq.(1)
 * PVW (continuous) of the dynamics of elastic bar
 * PDE

Discrete EOM $$\Leftarrow $$ Equation of motion. $$-Kd+F=M\ddot{d}$$

$$\frac{\partial }{\partial x}\left[\left(EA \right)\frac{\partial }{\partial x} \right]\Rightarrow -Kd$$

$$f \Rightarrow F$$

$$m\ddot{U} \Rightarrow M\ddot{d} $$

$$M\ddot{d}+Kd=F$$    Eq.(2)

Goal is to derive Eq.(2) from Eq.(1):
 * MDOF $$\Rightarrow$$ Multiple Degree of Freedom system
 * SDOF $$\Rightarrow$$ Single Degree of Freedom system

Motivation a SDOF spring mass system shown below where : $$M\ddot{d}+Kd=F$$



$$\int_{0}^{x=L}{W\left(x \right)\left\{\frac{\partial }{\partial x}\left[EA\frac{\partial U}{\partial x} \right]+f-m\ddot{U} \right\}}dx=0$$ Eq.(3)    For all posible $$W(x)$$

As seen
 * Eq.(1)$$\Rightarrow$$ Eq.(3) is trivial


 * Eq.(3)$$\Rightarrow$$ Eq.(1) is not trivial

Eq.(3) can be rewritten as $$\int W(x)g(x)dx=0$$

Since Eq.(3) holds for all $$W(x)$$, select $$W(x)=g(x)$$, then Eq.(3) becomes:

$$\int g^{2}dx=0 $$ where $$g^2\geq 0$$  \Rightarrow $$g(x)=0$$

Class Notes November 05, 2008
Continuing with the discussion from November 03, 2008, we revisit the concept of Integration by Parts:

Given $$r(x)$$ and $$s(x)$$, $$(rs)'=r's+rs'$$ where $$r'=\frac{dr}{dx}$$ and $$s'=\frac{ds}{dx}$$. The integral is therefore $$\int{(rs)'}=\int{r's}+\int{rs'}$$. By parts we have $$\int{(r's)}=rs-\int{rs'}$$.

Recall the continuous Principal of Virtual Work. For the first term, think of $$r(x)=(EA)\frac{\partial{u}}{\partial{x}}$$ and $$s(x)=W(x)$$.

Then By Parts: $$\int_{x=0}^{x=L} W(x)\frac{\partial}{\partial{x}}[EA\frac{\partial{u}}{\partial{x}}]dx=[W(EA)\frac{\partial{u}}{\partial{x}}]-\int_0^L \frac{\partial{W}}{\partial{x}}(EA)\frac{\partial{u}}{\partial{x}}dx$$.

This equals $$=W(L)EA(L)\frac{\partial{u}(L,t)}{\partial{x}}-W(0)EA(0)\frac{\partial{u}(0,t)}{\partial{x}}-\int_0^L \frac{dW}{dx}(EA)\frac{\partial{u}}{\partial{x}}dx$$.

Now we can examine the following model problem...



There are two boundary conditions: At $$x=0$$, select $$W(x)$$ at $$ W(0)=0$$, i.e. kindematically admissible. Recall $$ \underline{W}\cdot([]\left\{\begin{array}{c} d_{3} \\ d_{4} \end{array}\right\}-\underline{F})=0$$.

$$\underline{F}^{T}=\left[\begin{array}{cccccc} F_{1} & F_{2} & F_{3} & F_{4} & F_{5} & F_{6} \end{array}\right] $$, where $$ F_{3}$$ & $$F_{4}$$ are known.

Select $$w_{1}=w_{2}=w_{5}=w_{6}=0$$.

Then we return to the familiar $$ \underline{\overline{K}}\underline{\overline{d}}=\underline{\overline{F}} $$, where $$\underline{\overline{d}}=\left\{\begin{array}{c} d_{3} \\ d_{4} \end{array}\right\} $$ and $$\underline{\overline{F}}=\left\{\begin{array}{c} F_{3} \\ F_{4} \end{array}\right\}$$.

Note: $$ \underline{\overline{W}}\cdot(\underline{\overline{K}}\underline{\overline{d}}-\underline{\overline{F}})=0$$ for all $$\underline{\overline{W}}$$, where $$\underline{\overline{W}}=\left\{\begin{array}{c} W_{3} \\ W_{4} \end{array}\right\} $$

Now we have an unkown reaction: $$ N(0,t)=(EA)(0)\frac{\partial{u}(0,t)}{\partial{x}}$$

We use the Principal of Virtual Work (continuous): $$ W(L)F(t)-\int_0^L \frac{dW}{dx}(EA)\frac{\partial{u}}{\partial{x}}dx+\int_0^L W(x)[f-m\ddot{u}]dx=0$$ for all $$W(x)$$ such that $$W(0)=0$$.

This leads to $$ \int_0^L W(m\ddot{u})dx+\int_0^L \frac{dW}{dx}(EA)\frac{\partial{u}}{\partial{x}}dx=W(L)F(t)+\int_0^L Wfdx$$ for all $$W(x)$$ such that $$W(0)=0$$.

Interpolation




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



Deformed shape is a straight line. i.e. there was an implicit assumption of linear interpolation of displacement between 2 nodes.

Consider the case where there are only asial displacements (i.e. zero transverse displacement)

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

$$ u(x) = N_{i}(x)d_{i} + N_{i+1}(x)d_{i+1} $$ where $$ N_{i}(x) $$ and $$ N_{i+1}(x) $$ are linear functions in x.

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



HW: Equation and Picture for $$ N_{i}(x) $$
$$ N_{i}(x)= \frac{x-x_{i+1}}{x_{i+1} - x_{i}} $$

Continuous Principle of Virtual Work to Discrete Principle of Virtual Work (Continued)
Lagrangian Interpolation

What is the motivation for the form of $$N_i(x)\,\!$$ and $$N_{i+1}(x)\,\!$$? 1) $$N_i(x)\,\!$$ and $$N_{i+1}(x)\,\!$$ are both linear, thus any linear combination of $$N_i\,\!$$ and $$N_{i+1}\,\!$$ is also linear.  In particular, the expression for $$u(x)=N_i(x)d_i+N_{i+1}(x)d_{i+1}\,\!$$ must be linear.  Below is a demonstration:

Let: $$N_i(x)=\alpha_i + \beta_ix\,\!$$ -$$\alpha_i, \beta_i\,\!$$ are real numbers

$$N_{i+1}(x)=\alpha_{i+1} + \beta_{i+1}x\,\!$$ -$$\alpha_{i+1}, \beta_{i+1}\,\!$$ are real numbers

Now we can combine $$N_i(x)\,\!$$ and $$N_{i+1}(x)\,\!$$ :

$$u(x)=N_i(x)d_i+N_{i+1}(x)d_{i+1}\,\!$$

$$=(\alpha_i + \beta_ix)d_i+(\alpha_{i+1}, \beta_{i+1})d_{i+1}\,\!$$

$$=(\alpha_i+\alpha_{i+1}+(\beta_i+\beta_{i+1})x\,\!$$ < is a linear function

2) Recall the equation for $$u(x)\,\!$$, an interpolation of x:

$$u(x_i)=\underbrace{N_i(x_i)d_i}_{1}+N_{i+1}(x_i)d_{i+1} = d_i\,\!$$

FEM via PVW:(continued)
Continuing with the discussion of $$u(x_{i})$$ we can find that

$$u(x_{i+1})=N_{i}(x_{i+1})d_{i}+N_{i+1}(x_{i+1})d_{i+1}=d_{i+1}$$ simply because the terms $$N_{i}(x_{i+1})=0$$ and $$N_{i+1}(x_{i+1})=1$$.

Applying the 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

$$\Beta=\int_{x_{i}}^{x_{i+1}}{[N_{i}'w_{i}+N_{i+1}'w_{i+1}] (EA)[N_{i}'d_{i}+N_{i+1}'d_{i+1}]}dx $$

where the term $$ [N_{i}'w_{i}+N_{i+1}'w_{i+1}] = w'(x) $$ and the term $$ [N_{i}'d_{i}+N_{i+1}'d_{i+1}] = u'(x) $$, along with these terms below.

$$N_{i}'=\frac{d}{dx}N_{i}(x)$$

$$N_{i+1}'=\frac{d}{dx}N_{i+1}(x)$$

Note:

$$ u(x) = \left[ \begin{array}{cc} N_{i}(x) & N_{i+1}(x)\end{array}\right] \left\{\begin{matrix} d_{i}\\ d_{i 1} \end{matrix} \right\} $$

$$\frac{\partial }{\partial x}u(x) = \left[ \begin{array}{cc} N_{i}'(x) & N_{i+1}'(x)\end{array}\right]\left\{\begin{matrix} d_{i}\\ d_{i 1}

\end{matrix} \right\}$$

where the terms $$\left[ \begin{array}{cc} N_{i}(x) & N_{i+1}(x)\end{array}\right]=\underline{N}(x) $$ and $$\left[ \begin{array}{cc} N_{i}'(x) & N_{i+1}'(x)\end{array}\right]=B(x)$$

Similarly:

$$w(x)=\underline{N}(x)\left\{\begin{array}{c} w_{i} \\ w_{i+1} \end{array}\right\} $$

$$\frac{dw(x)}{dx}=\underline{B}(x)\left\{\begin{array}{c} w_{i} \\ w_{i+1} \end{array}\right\} $$

Recall the element dofs



$$\left\{\begin{array}{c} d_{i}\\d_{i+1}\end{array}\right\}=\left\{\begin{array}{c} d^{(i)}_{1}\\d^{(i)}_{2}\end{array}\right\}=\underline{d}^{(i)}$$

$$\left\{\begin{array}{c} w_{i}\\w_{i+1}\end{array}\right\}=\left\{\begin{array}{c} w^{(i)}_{1}\\w^{(i)}_{2}\end{array}\right\}=\underline{w}^{(i)}$$

$$\Beta=\int_{x_{i}}^{x_{i+1}}(\underline{B}\underline{w}^{(i)})(EA)(\underline{B}\underline{d}^{(i)})dx=\underline{w}^{(i)}\cdot(\underline{k}^{(i)}\underline{d}^{(i)})$$

$$\Beta=\int_{x_{i}}^{x_{i+1}}(EA)(\underline{B}\underline{w}^{(i)})\cdot(\underline{B}\underline{d}^{(i)})dx\Rightarrow(\underline{B}\underline{w}^{(i)})^{T}(\underline{B}\underline{d}^{(i)})\Rightarrow\underline{w}^{(i),T}\underline{B}^{T}\Rightarrow\underline{w}^{(i)}\cdot\underline{B}^{T}$$

$$\Beta=\underline{w}^{(i)}\cdot(\int\underline{B}^{T}(EA)\underline{B}dx)d^{(i)}$$

$$\underline{k}^{(i)}=\int_{x_{i}}^{x_{i+1}}\underline{B}^{T}(x)(EA)\underline{B}(x)dx$$

$$\underline{B}(x)=\left[ \begin{array}{cc} -\frac{1}{L^{(i)}} & \frac{1}{L^{(i)}} \end{array} \right] $$

HW: Find an expression for $$\underline{k}^{(i)}$$
The following problem refers to the figure below



where

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

and

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

Consider the case where $$EA$$ is a constant...

$$ \underline{k}^{(i)}=\frac{EA}{L^{(i)}} \left[ \begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array} \right] $$

The transformation of variables (coordinates) from $$x$$ to $$\tilde{x}$$ is as follows...

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

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

The task is to find an expression for $$\underline{k}^{(i)}$$ using the equation below...

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

Due to the above equation's similarity to the equation used to find the stiffness matrix for EA=Constant, we assume that the stiffness matrix with varying cross-sectional area and modulus is

$$\underline{k}^{(i)}=\frac{[N^{(i)}_{1}(\tilde{x})A_{1}+N^{(i)}_{2}(\tilde{x})A_{2}][N^{(i)}_{1}(\tilde{x})E_{1}+N^{(i)}_{2}(\tilde{x})E_{2}]}{L^{(i)}} \left[ \begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array} \right] $$

Although this may not be correct...

 Error: You are right; it was not right! See my comments in Team Bike (with annotations). Eml4500.f08 19:52, 26 November 2008 (UTC)

Class Notes November 14, 2008
$$ N_{i}(x)=N_{1}^{(i)}(\tilde{x})=1-\frac{\tilde{x}}{L^{(i)}}= \begin{cases} 1 & \text{ at } \tilde{x}=0 \\ 0 & \text{ at } \tilde{x}=L^{(i)} \end{cases}$$

$$ N_{i+1}(x)=N_{2}^{(i)}(\tilde{x})=\frac{\tilde{x}}{L^{(i)}}= \begin{cases} 0 & \text{ at } \tilde{x}=0 \\ 1 & \text{ at } \tilde{x}=L^{(i)} \end{cases} $$

===HW: Set $$ E_{1}=E_{2}=E_{3}$$, let $$A(\tilde{x})$$ be linear as before, and obtain $$\underline{K}^{(i)}$$ and compare to expression in the book on p.159===

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

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

Compare general $$\underline{K}^{(i)}$$ from before to $$\underline{K}_{avg}^{(i)}$$ where,

$$ \frac{(E_{1}+E_{2})}{2}\frac{(A_{1}+A_{2})}{2}\frac{1}{L^{(i)}}\begin{bmatrix} 1 &-1 \\ -1 & 1 \end{bmatrix}=\underline{K}_{avg}^{(i)}$$

end HW

HW: Find, $$\underline{K}^{(i)}-\underline{K}_{avg}^{(i)}$$
Solution

end HW

Remark: recall the Mean value Theorem (MVT) and its relations to centroid:

MVT: for $$\bar{x}\epsilon [a,b]$$,     $$a\leq \bar{x}\leq b$$,

$$ \int_{x=a}^{x=b}{f(x)dx}=f(\tilde{x})[b-a] $$

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

$$\int_{x=a}^{x=b}{f(x)g(x)dx} = f(\bar{x})g(\bar{x})[b-a]$$ where  $$ a\leq \bar{x}\leq b$$

But in general,

$$f(\bar{x})\neq \frac{1}{b-a} \int_{a}^{b}{f(x)dx}$$

$$g(\bar{x})\neq \frac{1}{b-a} \int_{a}^{b}{g(x)dx} $$

Matlab HW: Electric Pylon Analysis
The following is the Matlab code which solves the Electric Pylon problem and plots the deformed shape, amongst other things. Please use the comments embedded within the code to clarify the analysis.

The output for the deformed shape overlayed on the undeformed shape is shown below...



The output of the code given above displays many things, but most importantly are the following


 * The axial stress in each element; the highest tensile and compressive strengths and their associated elements

Which can be interpretted as  being the highest tensile stress associated with element 81 and   being the highest compressive stress associated with element 55.


 * The plot of the three lowest eigenpairs, which were found to be 132.34 associated with eigenvector column 79, 2471.7 associated with eigenvector column 82 and 2944.2 associated with eigenvector column 83. The plots of the three eigenvectors associated with the three lowest eigenpairs is shown below...








 * The three lowest vibrational periods of the electric pylon were also calculated and the combined output of the section of the code which finds the three lowest eigenpairs and the corresponding three lowest vibrational periods is shown below...


 * As for the electric pylon being statically determinate, we believe it is not so. The two lowest nodes (the nodes set to 0 for boundary conditions) have two elements attached to them.  Cutting the rest of the pylon away and analyzing only those two nodes and the -1000N external force, it is clear that there are too many unknowns because of the added elements connected to the nodes.


 * The general results for the electric pylon code are shown below, placed last due to similarity to the output of previous problems (2-bar truss, 3-bar truss, etc.)... Note: the K matrix is omitted due to its enormity...

Contributing Members:

 * GrantStark 21:05, 21 November 2008 (UTC)
 * J.Jayma 23:25, 17 November 2008 (UTC)
 * Esmail Hamdan 14:55, 21 November 2008 (UTC)
 * Nik Vitt 05:10, 21 November 2008 (UTC)
 * Andy Koby 15:07, 21 November 2008 (UTC)