text:
User:Eml 4500.f08.delta 6.specht/HW7

 See my comments below.

This report is an example of a parallel non-cooperative work, in which each student would take turn to take notes and write up on a lecture; the result is a mosaic of sometimes incoherent pieces that are put together. Several teams work in this non-cooperative fashion, resulting in rampant absenteism. (I noticed a couple of students from this team attended lectures regularly.) Here is an example of a highly parallel non-cooperative report. By contrast, here is an example of a coherent cooperative report (Team Bike HW7).

This team did NOT do the important section on recommended software to improve productivity for future students; the team was too focused on commenting on Mediawiki vs. WebCT in a parallel non-cooperative fashion, instead working cooperatively to provide the team's consensus, as requested. I added my original e-mail about this comparison to set the proper context for this section in The best of HW7.

Eml4500.f08 11:36, 10 December 2008 (UTC)

=Homework 7=

The Two-Bar Frame
Modifying the familiar two-bar truss system, we accomodate it as a frame and varying cross-sectional area and modulus. This two-bar frame can be found below.

This arrangement is effective to accomodate the general $$\mathbf{k}^{(i)}$$ equation:

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

Frame elements can be seen as a combination between a truss and a beam element.

Model Frame Problem with 2 Elements
We have the following 2-bar fram system:

This frame system difers from a typical truss system since the only two pivoting nodes are the ones located at the ends of the global picture. In other words, frame systems have rigid elbows at element joints. The elbow angle will not change after deformation takes place.

The free-body diagrams of the elements are found below.

Notice the additional dof at each node. This dof comes from the ability of the node to rotate, therefore we say that the $$d$$ labels in the elements fbds above not only represent displacement but also rotation. We still refer to this label as a dof, regardless of whether it is a displacement or a rotation.

We can also relate these dofs to forces and moments at each node. As a general rule,

 * $$d_i^{(e)}$$→$$f_i^{(e)}$$

Where $$d_i^{(e)}$$ is the displacement or rotation of node i of element e, and $$f_i^{(e)}$$ is the force or moment at node i of element e.

Furthermore, the rotational dofs:

 * $$d_3^{(e)}, d_6^{(e)}$$

correspond to the bending moments:

 * $$f_3^{(e)}, f_6^{(e)}$$

Assembling The Global Stiffness Matrix
The frame's global dofs are shown below.

There are two element stiffness matrices:

 * $$\mathbf{k}_{6x6}^{(e)}$$, where $$e$$= 1 and 2.

The global stiffness matrix can be assembled by following the same procedure as for the truss system:

 * $$\mathbf{K}_{(9x9)}=A_{e=1}^{e=2}\mathbf{k}_{6x6}^{(e)}$$

The topography of the global stiffness matrix of the above 2-bar frame is:

 * $$\begin{bmatrix}

K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & K_{15}^{(1)} & K_{16}^{(1)} & 0 & 0 & 0 \\ K_{21}^{(1)} & K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & K_{25}^{(1)} & K_{26}^{(1)} & 0 & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & K_{33}^{(1)} & K_{34}^{(1)} & K_{35}^{(1)} & K_{36}^{(1)} & 0 & 0 & 0 \\ K_{41}^{(1)} & K_{42}^{(1)} & K_{43}^{(1)} & (K_{44}^{(1)}+K_{11}^{(2)}) & (K_{45}^{(1)}+K_{12}^{(2)})& (K_{46}^{(1)}+K_{13}^{(2)}) & K_{14}^{(1)} & K_{15}^{(1)} & K_{16}^{(1)}\\ K_{51}^{(1)} & K_{52}^{(1)} & K_{53}^{(1)} & (K_{54}^{(1)}+K_{21}^{(2)}) & (K_{55}^{(1)}+K_{22}^{(2)})& (K_{56}^{(1)}+K_{23}^{(2)}) & K_{24}^{(1)} & K_{25}^{(1)} & K_{26}^{(1)}\\ K_{61}^{(1)} & K_{62}^{(1)} & K_{63}^{(1)} & (K_{64}^{(1)}+K_{31}^{(2)}) & (K_{65}^{(1)}+K_{32}^{(2)})& (K_{66}^{(1)}+K_{33}^{(2)}) & K_{34}^{(1)} & K_{35}^{(1)} & K_{36}^{(1)}\\ 0 & 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)} & K_{45}^{(2)} & K_{46}^{(2)} \\ 0 & 0 & 0 & K_{51}^{(2)} & K_{52}^{(2)} & K_{53}^{(2)} & K_{54}^{(2)} & K_{55}^{(2)} & K_{56}^{(2)} \\ 0 & 0 & 0 & K_{61}^{(2)} & K_{62}^{(2)} & K_{63}^{(2)} & K_{64}^{(2)} & K_{65}^{(2)} & K_{66}^{(2)} \end{bmatrix}_{9x9}$$

Wednesday 11/19
It is important to recognize what elements constitute the local stiffness matrix for each element in the system, the following diagrams show the local element degrees of freedom of element 1 and element 2

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: $$ \tilde{f}^{(e)}_{3}=f^{(e)}_{3} \! $$ and $$ \tilde{f}^{(e)}_{6}=f^{(e)}_{6} \! $$ since these are the moments about the $$ \tilde{z} \! $$ axis.

By using the elements of the diagrams above for element 1 and element 2, the local stiffness matrix $$ \tilde{k}\! $$ would be: $$ \begin{bmatrix}\frac{EA}{L} & 0 & 0 & \frac{-EA}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{2EI}{L} \\ \frac{-EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0 \\ 0 & \frac{-12EI}{L^{3}} & \frac{-6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{-6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{4EI}{L} \end{bmatrix} $$

Note: All the matrix coefficients would not have the same dimensions but when they are multiplied with rotations or displacements depending on location they will end up with the right units.

Dimensional Analysis
$$ \left [ \tilde {d}_1 \right ] = L = \left [ \tilde {d}_i \right ] \qquad i=1, 2, 4, 5$$

where: $$ \left [ \tilde {d}_1 \right ] $$ is a dimension of length and $$ \left [ \tilde {d}_i \right ] $$ is displacement.

Note: For the dimensional analysis the $$ [ ] \! $$ symbolizes "dimension of" instead of the $$ [ ] \! $$ commonly used for matrices.

$$ \left [ \tilde {d}_3 \right ] = 1 = \left [ \tilde {d}_6 \right ] $$

where: $$ 1 $$ is dimensionless and $$ \left [ \tilde {d}_6 \right ] $$ is the rotational displacement

From the graph it can be observed that $$ \bar{AB} = R * \theta $$ where $$ \theta $$ is measured in radians and $$\bar{AB}$$ is the arc length from point A to point B. Another way to express $$ \theta $$ is $$\frac{\bar{AB}}{R}$$ where both have units of length making $$ \theta $$  dimensionless: $$ \left [ \theta \right ] = \frac {[\bar{AB}]}{[R]} = \frac {L}{L} = 1 $$

Applying the same non-dimensional analysis principle: $$ \sigma = \Epsilon \varepsilon \rightarrow \left [ \sigma \right ] = \left [ \Epsilon \right ] \left [ \varepsilon \right ] = 1 $$

Friday 11/21
Continued examination of dimensions for various variables.

$$ [\varepsilon ] = \frac{[d_4]}{[dx]} = \frac{L}{L} = 1 $$

Further,

$$ [\sigma] = [E] = \frac{F}{L^2} $$

$$ [A] = L^2 $$

$$ [I] = \int x^2dx = L^4 $$

$$ [\frac{EA}{L}] = [\tilde{k}_{11}]= \frac{L^2\frac{F}{L^2}}{L} = \frac{F}{L} $$

$$ [\tilde{k}_{11}\tilde{d}_1] = [\tilde{k}_{11}][\tilde{d}_1] = \frac{F}{L}L = F $$

$$ [\tilde{k}_{23}\tilde{d}_3] = [\tilde{k}_{23}][\tilde{d}_3] = \frac{[6][E][I]}{[L^2]} = \frac{1 \frac{F}{L^2} L^4}{L^2} 1 = F $$

Note: $$ [\tilde{d}_3] $$ = 1

Goal: Derive element force displacement relationship in global coordinate system from element force displacement relationship in local coordinates.

In elaboration, obtaining

$$ k^{(e)}_{6x6} d^{(e)}_{6x1} = f^{(e)}_{6x1} $$

and

from

$$ \tilde{k}^{(e)}_{6x6} \tilde{d}^{(e)}_{6x1} = \tilde{f}^{(e)}_{6x1} $$

In order to move on $$ \tilde{T}^{(e)}_{6x6} $$ must be analyzed, which has a block diagonal structure.

Note: It is easy to expand the $$ \tilde{T}^{(e)}_{6x6} $$ matrix to account for the frame problem since the rotational axis $$ z = \tilde{z} $$

Examining the relationship between $$ \tilde{d}_i $$ and $$ d_i $$ gives,

Note: $$ \tilde{d}_3, \tilde{d}_6, d_3, d_6 $$ represent rotational degrees of freedom.

Furthermore, we will now derive $$ \hat{k}^{(e)} $$ from PVW, focusing only on bending effect.

$$ \frac{\partial^2}{\partial x^2}((EI) \frac{\partial^2v}{\partial x^2}) - f_t(x) = m(x) \ddot{v} $$

Note: EI is a function of x, $$ f_t(x) $$ represents the distributive transverse load, and v represents the transverse displacement.

Monday 11/24
Motivation: deformed shape of truss element and Interpolation of transverse displacement v(s), where s=$$tilde{x}$$. Recall the governing partial differential equation for the bar element is: $$ +\frac{d^{2}}{dx^{2}}[ (EA)\frac{\partial u}{\partial x} ] + \mathit{f_{a}(x,t)} = m(x)\ddot{u} $$ The positive sign in front of the equation is to emphasize the difference between the above equation and the governing equation for a distributed transversal load on a bar element. The first term, $$\frac{d}{dx^{2}}[ (EA)\frac{\partial u}{\partial x} ]$$, is a second order differential equation with respect to x. The second term ,$$\mathit{f_{a}(x,t)}$$, refers to an axial distributed load. Where the subscript a is to distinguish the axial load equation from the transversal load equation. The load may be changing with time. Therefore, the load may be a function of time as well as distance (x).

Principle of Virtual Work for Beams:

The derivation is the same as the derivation of the principle of virtual work for the dynamics of elastic bar (refer to homework 6, PVW of dynamics of elastic bar) with one difference. The difference is that integration by parts has to be performed twice instead of once. Eqn (1)

$$ \int_{0}^{L}{w(\tilde{x})\left\{\frac{d^{2}}{dx^{2}} [ (EI) \frac{\partial^{2} u}{\partial x^{2}}] + f_{t} - m\ddot{v}\right\}dx = 0} $$

For all possible w(x), where $$f_{t}$$ is the transversal distributed load. First the first term will be integrated by parts. For simplicity the minus sign will be dropped and the first term will be defined as $$\alpha$$:

$$\alpha:= \int_{0}^{L}{w(\tilde{x})\frac{d^{2}}{dx^{2}} [ (EI) \frac{\partial^{2} u}{\partial x^{2}}]dx} $$ Where $$\tilde{x}=s(\tilde{x})$$. The position (r) is: $$r = \frac{d}{dx} [ (EI) \frac{\partial^{2} v}{\partial x^{2}}] $$ and the velocity ($$ \dot{r}$$) is: $$\dot{r} = \frac{d}{dx}(\frac{d}{dx} [ (EI) \frac{\partial^{2} v}{\partial x^{2}}]) $$

Now the argument x is omitted for simplicity and the equation is integrated.

$$=\left[w\frac{\partial }{\partial x}\left\{(EI) \frac{\partial^{2} v}{\partial x^{2}}\right\}\right]_{0}^{L} - \int_{0}^{L}\frac{dw}{dx}\frac{\partial }{\partial x}\left\{(EI)\frac{\partial^{2} v}{\partial x^{2}}\right\}dx$$

Now again for simplicity $$\beta_{1}$$ is defined as:

$$\beta_{1} = w\frac{\partial }{\partial x}\left\{(EI) \frac{\partial^{2} v}{\partial x^{2}}\right\}$$

Now Integration by parts once more.

$$=\beta_{1} - \left[w\frac{\partial }{\partial x}\left\{(EI) \frac{\partial^{2} v}{\partial x^{2}}\right\}\right]_{0}^{L} + \int_{0}^{L}\frac{d^{2}w}{dx^{2}}\frac{\partial }{\partial x}\left\{(EI)\frac{\partial^{2} v}{\partial x^{2}}\right\}dx$$

Now $$\beta_{2}$$ and $$\gamma$$ are defined as:

$$\beta_{2} = \left[w\frac{\partial }{\partial x}\left\{(EI) \frac{\partial^{2} v}{\partial x^{2}}\right\}\right]_{0}^{L}$$ and  $$\gamma = \int_{0}^{L}\frac{d^{2}w}{dx^{2}}\frac{\partial }{\partial x}\left\{(EI)\frac{\partial^{2} v}{\partial x^{2}}\right\}dx$$

Eqn (1) now becomes:

$$-\beta_{1}+\beta_{2}-\gamma+\int_{0}^{L}wf_{t}dx-\int_{0}^{L}wm\ddot{v}dx=0$$

For all possible $$w(\tilde{x})$$

Now look at the stiffness term $$\gamma$$ to derive the beam stiffness matrix and to identify the beam shape functions.

The axial displacements $$\tilde{d_{1}}$$ and $$\tilde{d_{4}}$$ are ignored because the focus is on the beam.

The transverse displacement ($$v(\tilde{x})$$) is: $$v(\tilde{x})=N_{2}\tilde{d_{2}} + N_{3}\tilde{d_{3}} + N_{5}\tilde{d_{5}} + N_{6}\tilde{d_{6}}$$. Recall: $$u(\tilde{x})= N_{1}(\tilde{x})\tilde{d_{1}} + N_{4}(tilde{x})\tilde{d_{4}}$$

The rotation of $$N_{2}(\tilde{x})$$ is equal to zero. The slope is therefore equal to zero. This is true at the ends of the beam. The expressions for the shape functions are:

$$N_{2}(\tilde{x})=1-\frac{3\tilde{x^{2}}}{L^{2}}+\frac{2\tilde{x^{3}}}{L^{3}}$$

$$N_{3}(\tilde{x})=\tilde{x}-\frac{2\tilde{x^{2}}}{L}+\frac{\tilde{x^{3}}}{L^{2}}$$

$$N_{5}(\tilde{x})=\frac{3\tilde{x^{2}}}{L^{2}}-\frac{2\tilde{x^{3}}}{L^{3}}$$

$$N_{2}(\tilde{x})=-\frac{\tilde{x^{2}}}{L}+\frac{\tilde{x^{3}}}{L^{2}}$$

Observe that when the shape function is multiplied by the displacement (d) it must give a displacement.

Monday 12/1
The applied internal forces

$$ N_{5}\left(\tilde{x}\right) $$ $$ N_{6}\left(\tilde{x}\right) $$

are shown below.

Recall the equation of the displacement matrix below, where

$$ \mathbf{d}_{6x1}^{(e)} $$

is known after solving the system.

$$ \mathbf{\tilde{d}}_{6x1}^{(e)} = \mathbf{\tilde{T}}_{6x6}^{(e)}\mathbf{d}_{6x1}^{(e)} $$

Next, we will look at the computation of

$$ u\left(\tilde{x}\right) $$

and

$$ v\left(\tilde{x}\right) $$

$$ \mathbf{u}\left(\tilde{x}\right) = u(\tilde{x})\tilde{i} + v(\tilde{x})\tilde{j} = u_{x}(\tilde{x})\tilde{i} + u_{y}(\tilde{x})\tilde{j} $$

The computation of

$$ u\left(\tilde{x}\right) $$ and $$ v\left(\tilde{x}\right) $$

required using the following equations:

$$ u(\tilde{x})=N_1(\tilde{x})\tilde{d}_1+N_4(\tilde{x})\tilde{d}_4 $$

$$ v(\tilde{x})=N_2(\tilde{x})\tilde{d}_2+N_3(\tilde{x})\tilde{d}_3+N_5(\tilde{x})\tilde{d}_5+N_6(\tilde{x})\tilde{d}_6 $$

The computation of

$$ u_{x}\left(\tilde{x}\right) $$ and $$ u_{y}\left(\tilde{x}\right) $$ from $$ u\left(\tilde{x}\right) $$ and $$ v\left(\tilde{x}\right) $$

is shown below.

$$ \begin{Bmatrix} u_{x}\left(\tilde{x}\right)\\u_{y}\left(\tilde{x}\right) \end{Bmatrix} = \mathbf{R}^{T}\begin{Bmatrix} u\left(\tilde{x}\right)\\v\left(\tilde{x}\right) \end{Bmatrix} $$

$$ \begin{Bmatrix} u\left(\tilde{x}\right)\\v\left(\tilde{x}\right) \end{Bmatrix} = \begin{bmatrix} N_{1} & 0 & 0 & N_{4} & 0 & 0 \\ 0 & N_{2} & N_{3} & 0 & N_{5} & N_{6} \end{bmatrix}\begin{Bmatrix} \tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)} \end{Bmatrix} $$

where

$$ \begin{bmatrix} N_{1} & 0 & 0 & N_{4} & 0 & 0 \\ 0 & N_{2} & N_{3} & 0 & N_{5} & N_{6} \end{bmatrix} $$

is represented by

$$ \mathbf{N}\left(\tilde{x}\right) $$ and

$$ \begin{Bmatrix} \tilde{d}_{1}^{(e)} \\ \tilde{d}_{2}^{(e)} \\ \tilde{d}_{3}^{(e)} \\ \tilde{d}_{4}^{(e)} \\ \tilde{d}_{5}^{(e)} \\ \tilde{d}_{6}^{(e)} \end{Bmatrix} $$

is denoted

$$ \mathbf{\tilde{d}}^{(e)} $$

$$ \begin{Bmatrix} u_{x}\left(\tilde{x}\right)\\u_{y}\left(\tilde{x}\right) \end{Bmatrix} = \mathbf{R}^{T}\mathbf{N}\left(\tilde{x}\right)\mathbf{\tilde{T}}^{(e)}\mathbf{d}^{(e)} $$

Dimensional Analysis
The dimensional analysis of the axial components has already been performed, but is shown again here for comparison. The shape functions$$N_1$$ through $$N_6$$ correspond to the variables in the matrix given above, N.

Axial Displacements
$$ [u]= L $$ $$ [N_1]=[N_4]=1 $$ $$ [N_1][\tilde{d}^{(e)}_1]=[N_4][\tilde{d}^{(e)}_4]= L $$

Transverse Displacements
$$ [N_2]=[N_3]=[N_5]=[N_6]=1 $$ The dimensions of displacements 2 and 5 correspond to the dimensions of displacements in the $$\tilde{y}$$ direction, which correspond to the following values: $$ [\tilde{d}_2]=[\tilde{d}_5]= L $$ This means that the dimensions of each component of displacement based on the shape functions 2 and 5 are: $$ [N_2][\tilde{d}_2]=[N_5][\tilde{d}_5]= L $$ The dimensions of displacements 3 and 6 correspond to the dimension of the rotation, which is given as: $$ [\tilde{d}_3]=[\tilde{d}_6]=1 $$ This means that the dimensions of each component of displacement based on the shape functions 3 and 6 are: $$ [N_3][\tilde{d}_3]=[N_6][\tilde{d}_6]=1 $$

Derivation of Beam Shape Functions
Governing PDE for Beams:

$$ -\frac{\delta ^2}{\delta x^2}\left[\left(EI \right)\frac{\delta ^2v}{\delta x^2} \right]+f_t(x)=m(x)\ddot{v} $$

Governing PDE for Elastic Bar: (insert equation for bar here) When the transverse loads and the inertial forces are ignored, the first equation reduces to the first term.

$$ -\frac{\delta ^2}{\delta x^2}\left[\left(EI \right)\frac{\delta ^2v}{\delta x^2} \right]=0 $$ If the modulus of elasticity, E, and the moment of inertia, I, are constant across the length of the beam the equation can be further reduced.

$$ -EI\frac{\delta ^4v}{\delta x^4}=0 $$

After integrating four times, four constants of integration are developed. The equation that represents V can be written as the following:

$$ v(x)=C_0+C_1x+C_2x^2+C_3x^3 $$ The shape functions can be obtained from this equation by applying the boundary conditions for each shape function individually. The first function $$N_2$$ can be obtained by the following procedure.

Boundary Conditions
$$ v(0)=1 $$

$$ v(L)=0 $$

$$ v'(0)=v'(L)=0 $$

Solution to Four Simultaneous Equations
The first boundary condition leads to the following equation:

$$ v(x)=C_0+C_1(0)+C_2(0)^2+C_3(0)^3=1 $$

$$ C_0=1 $$ The third boundary condition leads to another coefficient that can be eliminated from the descriptive equation.

$$ v'(0)=C_1+2C_2(0)+3C_3(0)^2=0 $$

$$ C_1=0 $$

The remaining two equations can be solved for the other two coefficients.

$$ v(L)=C_2(L)^2+C_3(L)^3=1 $$

$$ v'(L)=2C_2(L)+3C_3(L)^2 $$ Solving these two equations simultaneously yields the following results for the coefficients, which can then be combined into the equation for $$v(x)$$ to yield the same result as $$N_2$$ that has been used already in class.

$$ C_2=-\frac{3}{L^2} $$

$$ C_3=\frac{2}{L^3} $$

$$ N_2=1-\frac{3\tilde{x}^2}{L^2}+\frac{2\tilde{x}^3}{L^3} $$

Boundary Conditions
$$ v(0)=v(L)=0 $$

$$ v'(0)=1 $$

$$ v'(L)=0 $$

Solution to Four Simultaneous Equations
Using the same methods as described for the shape function $$N_2$$ the coefficients can be determined.

$$ v(0)=C_0+C_1(0)+C_2(0)^2+C_3(0)^3=0 $$

$$ C_0=0 $$

$$ v'(0)=C_1+2C_2(0)+3C_3(0)^2=1 $$

$$ C_1=1 $$

$$ v(L)=C_1(L)+C_2(L)^2+C_3(L)^3=0 $$

$$ v'(L)=C_1+2C_2(L)+3C_3(L)^2=0 $$

$$ C_2=-\frac{2}{L} $$

$$ C_3=-\frac{1}{L^2} $$

$$ N_3=\tilde{x}-\frac{2\tilde{x}^2}{L}-\frac{2\tilde{x}^3}{L^2} $$

Boundary Conditions
$$ v(0)=0 $$

$$ v(L)=1 $$

$$ v'(0)=v'(L)=0 $$

Solution to Four Simultaneous Equations
The four simultaneous equations can be solved for the coefficients in the same manner as for $$N_2$$.

$$ v(0)=C_0+C_1(0)+C_2(0)^2+C_3(0)^3=0 $$

$$ C_0=0 $$

$$ v'(0)=C_1+2C_2(0)+3C_3(0)^2=0 $$

$$ C_1=0 $$

The last two equations can be solved simultaneously in order to yield that last two coefficients.

$$ v(L)=C_2(L)^2+C_3(L)^3=1 $$

$$ v'(L)=2C_2(L)+3C_3(L)^2=0 $$

$$ C_2=\frac{3}{L^2} $$

$$ C_3=-\frac{2}{L^3} $$

$$ N_5=\frac{3\tilde{x}^2}{L^2}-\frac{2\tilde{x}^3}{L^3} $$

Boundary Conditions
$$ v(0)=v(L)=0 $$

$$ v'(0)=0 $$

$$ v'(L)=1 $$

Solution to Four Simultaneous Equations
The coefficients for this shape function can be obtained in the same manner as before.

$$ v(0)=C_0+C_1(0)+C_2(0)^2+C_3(0)^3=0 $$

$$ C_0=0 $$

$$ v'(0)=C_1+2C_2(0)+3C_3(0)^2=0 $$

$$ C_1=0 $$

The last two equations can be solved simultaneously to find the last two coefficients.

$$ v(L)=C_2(L)^2+C_3(L)^3=0 $$

$$ v'(L)=2C_2(L)+3C_3(L)^2=1 $$

$$ C_2=-\frac{1}{L} $$

$$ C_3=\frac{1}{L^2} $$

$$ N_6=-\frac{\tilde{x}^2}{L}+\frac{\tilde{x}^3}{L^2} $$

Derivation of Coefficients in $$\tilde{k}$$
The derivation of the coefficients corresponding to elastic bars (terms involving EA) has already been accomplished. The terms corresponding to elastic beams have yet to be derived. The derivation of the term $$\tilde{k}_{22}$$ is shown below.

$$ \tilde{k}_{22}=\frac{12EI}{L^3}=\frac{12EI}{L^3}=\int_{0}^{L}{\frac{\delta ^2N_2}{\delta x^2}(EI)\frac{\delta ^2N_2}{\delta x^2}dx} $$

Monday 12/8
$$ \tilde{k}_{23} = \frac{6EI}{L^2} = \int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2}dx} $$

In general,

$$ \tilde{k}_{ij} = \int_{0}^{L}{\frac{d^2N_i}{dx^2}(EI)\frac{d^2N_j}{dx^2}dx} $$

With i,j = 2,3,5,6

Elastodynamics (trusses, 2-D frames, 3-D elasticity)
Analyzing the Modal problem using the discrete Principle of Virtual Work is as follows:

$$ \mathbf{\bar{w}} = [\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{K}}\mathbf{\bar{d}} - \mathbf{\bar{F}}] = 0 $$

Above, the boundary conditions are already applied and the K matrix represents the reduced stiffness matrix.

This equation is true for all $$\mathbf{\bar{w}}$$.

$$\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{k}}\mathbf{\bar{d}} = \mathbf{\bar{F}}(t)$$

$$\mathbf{\bar{d}}(0) = \mathbf{\bar{d}}_0$$

$$\mathbf{\dot{\bar{d}}}(0) = \mathbf{\bar{V}}_0$$

The above is labeled as Equation (1).

These are the complete ordinary differential equations (ODE's, second order in time) with initial conditions governing the elastodynamics of the discretized continuous problem with multiple degrees of freedom.

Now, solving the method for solving equation (1) is as follows:

1) Consider the unforced vibrations problem:

$$\mathbf{\bar{M}}_{nxn}\mathbf{\ddot{v}}_{nx1} + \mathbf{\bar{K}}_{nxn}\mathbf{v}_{nx1} = \mathbf{0}_{nx1} $$

The above is labeled as Equation (2).

Assume:

$$\mathbf{v}(t)_{nx1} = (sinwt)\mathbf{\phi _{nx1}} $$

Where the phi matrix is not time dependent.

Thus:

$$\mathbf{\ddot{v}} = -\omega ^2 sin\omega t\mathbf{\phi } $$

This means equation (2) becomes:

$$-\omega ^2 sin\omega t\mathbf{\bar{M}}\mathbf{\phi } + \omega ^2 sin\omega t\mathbf{\bar{K}}\mathbf{\phi } = \mathbf{0} $$

Therefore,

$$\mathbf{\bar{k}}\mathbf{\phi } = \omega ^2\mathbf{\bar{M}}\mathbf{\phi } $$

Which is the generalized eigenvalue problem, as it is of the form:

$$\mathbf{A}\mathbf{x} = \lambda \mathbf{B}\mathbf{x} $$

Where lambda is an eigenvalue.

A standard eigenvalue problem is given when the B matrix is equal to the identity matrix and the equation becomes:

$$\mathbf{A}\mathbf{x} = \lambda \mathbf{x} $$

This means that

$$\lambda =\omega ^2 $$

is an eigenvalue.

Also the eigenpairs can be represented as:

$$(\lambda _i,\mathbf{\phi _i}) $$

with i = 1 through n

Now the 'ith' mode is:

$$\mathbf{v}_i(t) = (sinw_it)\mathbf{\phi _i} $$

for i = 1 through n

Now we perform step 2:

2) Modal superposition method:

By using the orthogonal properties of the eigenpairs we can equate:

$$\mathbf{\phi _i}^T_{1xn}\mathbf{\bar{M}}_{nxn}\mathbf{\phi }_{nx1} = \delta _{ij} = \begin{cases} & \text{1 if } i=j \\ & \text{0 if } i\neq j \end{cases}$$

This delta is known as the Kronecker delta.

This is possible due to the mass orthogonality of the eigenvector.

Now, applying this to Equations (1) and (2) gives:

$$\mathbf{\bar{M}}\mathbf{\phi }_j = \lambda _j \cdot \mathbf{\bar{k}}\mathbf{\phi }_j $$

$$ \mathbf{\phi }_i^T\mathbf{\bar{M}}\mathbf{\phi }_j = \lambda _j\phi _i^T\mathbf{\bar{k}}\mathbf{\phi }_j $$

where the left hand side is equal to the Kronecker delta. Therefore,

$$\phi _i^T\mathbf{\bar{k}}\mathbf{\phi }_j = \frac{1}{\lambda _j}\delta _{ij} $$

$$\mathbf{\bar{d}}_{nx1}(t) = \sum{\zeta _i(t)\mathbf{\phi }_{inx1}} $$

Equation (1) can be written as:

$$\mathbf{\bar{M}}(\sum_{j}^{}{\ddot{\zeta }_j\mathbf{\phi }_j}) + \mathbf{\bar{K}}(\sum_{j}^{}{\zeta _j\mathbf{\phi }_j}) = \mathbf{F} $$

The first term is equal to the second derivative of the d matrix, while the second term is equal to the d matrix itself. Also,

$$\sum_{j}^{}{\ddot{\zeta }_j(\mathbf{\phi }_i^T\mathbf{\bar{M}}\mathbf{\phi }_j} + \sum_{j}^{}{\zeta _j(\mathbf{\phi }_i^T\mathbf{\bar{K}}\mathbf{\phi }_j} = \mathbf{\phi }_i^T\mathbf{F}$$

The first term in this equation is equal to the Kronecker delta, while the second term is equal to an eigenvalue times the Kronecker delta. This means that this equation can be written as an ordinary differential equation as follows:

$$\ddot{\zeta } + \lambda _i\zeta _i = \mathbf{\phi }_i^T\mathbf{F} $$

with i = 1 through n

2-Element Frame and Truss System
Here are the properties of the system we will be solving using MATLAB.

$$

E^{(1)}=3$$

$$

A^{(1)}=1$$

$$

L^{(1)}=4$$

$$

\theta^{(1)}=30$$

$$

I^{(1)}=.0833$$

$$

E^{(2)}=5$$

$$

A^{(2)}=2$$

$$

L^{(2)}=2$$

$$

\theta^{(2)}=-45$$

$$

P=7$$

Electric Pylon Frame MATLAB Code
% Matlab Script %********************************************************************* % filename: electric_pylon.m % % PURPOSE: %  Plot an electric pylon ("superman") structure with matlab %  Perform required FEA analysis % % AUTHOR: %   Team Delta_6 %   EML 4500 Finite Element Analysis and Design  Fall 2008. % % Modified on: vql, Fri, 21 Nov 2008, 08:37:46 EDT % Created on : Spring 2000 % % DEPENDENCIES: %  call: % % REMARKS: % %*********************************************************************

% Plot the "superman" structure %

clear; close;

% model with 3-D beam elements dof = 3;         %  dof per node:  1= disp x,  2= disp y A=.0004;          %  Area of the elements in (m^2) e=200000000000;  %  Young's modulus of the elements in (Pa) p=7800;          %  Density of the elements in (kg/m^3) P=-1000;         %  Downward Vertical Force applied at node #33 h=sqrt(A);

% obtain the coordinates of all nodes %  n_node = 46;             % number of nodes n_elem = 91;            % number of elements total_dof = 2 * n_node; % total dof of system

position(:, 1) = (60/7.43)*[ 1.5; 0; 0]; position(:, 2) = (60/7.43)*[ 4.53; 0; 0];

position(:, 3) = (60/7.43)*[ 1.88; 1.18; 0]; position(:, 4) = (60/7.43)*[ 3.03; 1.18; 0]; position(:, 5) = (60/7.43)*[ 4.17; 1.18; 0];

position(:, 6) = (60/7.43)*[ 2.2; 2.181; 0]; position(:, 7) = (60/7.43)*[ 3.03; 2.181; 0]; position(:, 8) = (60/7.43)*[ 3.85; 2.181; 0];

position(:, 9) = (60/7.43)*[ 2.53; 3.181; 0]; position(:, 10) = (60/7.43)*[ 3.53; 3.181; 0];

position(:, 11) = (60/7.43)*[ 2.34; 3.681; 0]; position(:, 12) = (60/7.43)*[ 3.03; 3.681; 0]; position(:, 13) = (60/7.43)*[ 3.71; 3.681; 0];

position(:, 14) = (60/7.43)*[ 2.16; 4.181; 0]; position(:, 15) = (60/7.43)*[ 2.62; 4.181; 0]; position(:, 16) = (60/7.43)*[ 3.44; 4.181; 0]; position(:, 17) = (60/7.43)*[ 3.89; 4.181; 0];

position(:, 18) = (60/7.43)*[ 2.36; 4.5; 0]; position(:, 19) = (60/7.43)*[ 3.7; 4.5; 0];

position(:, 20) = (60/7.43)*[ 1.8; 5.172; 0]; position(:, 21) = (60/7.43)*[ 4.25; 5.172; 0];

position(:, 22) = (60/7.43)*[ 0; 6; 0]; position(:, 23) = (60/7.43)*[ 1.06; 6; 0]; position(:, 24) = (60/7.43)*[ 1.5; 6; 0]; position(:, 25) = (60/7.43)*[ 1.82; 6; 0]; position(:, 26) = (60/7.43)*[ 2.22; 6; 0]; position(:, 27) = (60/7.43)*[ 2.71; 6; 0]; position(:, 28) = (60/7.43)*[ 3.34; 6; 0]; position(:, 29) = (60/7.43)*[ 3.84; 6; 0]; position(:, 30) = (60/7.43)*[ 4.23; 6; 0]; position(:, 31) = (60/7.43)*[ 4.55; 6; 0]; position(:, 32) = (60/7.43)*[ 5; 6; 0]; position(:, 33) = (60/7.43)*[ 6.05; 6; 0];

position(:, 34) = (60/7.43)*[ .8; 6.23; 0]; position(:, 35) = (60/7.43)*[ 5.25; 6.23; 0];

position(:, 36) = (60/7.43)*[ 1.24; 6.355; 0]; position(:, 37) = (60/7.43)*[ 4.82; 6.355; 0];

position(:, 38) = (60/7.43)*[ 1.5; 6.43; 0]; position(:, 39) = (60/7.43)*[ 2; 6.43; 0]; position(:, 40) = (60/7.43)*[ 2.48; 6.43; 0]; position(:, 41) = (60/7.43)*[ 3.03; 6.43; 0]; position(:, 42) = (60/7.43)*[ 3.58; 6.43; 0]; position(:, 43) = (60/7.43)*[ 4.05; 6.43; 0]; position(:, 44) = (60/7.43)*[ 4.55; 6.43; 0];

position(:, 45) = (60/7.43)*[ 1.5; 7.43; 0]; position(:, 46) = (60/7.43)*[ 4.55; 7.43; 0];

% print the node coord. for i = 1 : n_node; x(i) = position(1,i); y(i) = position(2,i); z(i) = position(3,i);

end

node_connect(1, 1) = 1;  % element 1 node_connect(2, 1) = 3;

node_connect(1, 2) = 1;  % element 2 node_connect(2, 2) = 4;

node_connect(1, 3) = 3;  % element 5 node_connect(2, 3) = 4;

node_connect(1, 4) = 4;  % element 3 node_connect(2, 4) = 2;

node_connect(1, 5) = 4;  % element 6 node_connect(2, 5) = 5;

node_connect(1, 6) = 2;  % element 4 node_connect(2, 6) = 5;

node_connect(1, 7) = 3;  % element 8 node_connect(2, 7) = 7;

node_connect(1, 8) = 5;  % element 9 node_connect(2, 8) = 7;

node_connect(1, 9) = 3;  % element 7 node_connect(2, 9) = 6;

node_connect(1, 10) = 5;  % element 10 node_connect(2, 10) = 8;

node_connect(1, 11) = 6;  % element 11 node_connect(2, 11) = 7;

node_connect(1, 12) = 7;  % element 12 node_connect(2, 12) = 8;

node_connect(1, 13) = 6;  % element 13 node_connect(2, 13) = 9;

node_connect(1, 14) = 7;  % element 14 node_connect(2, 14) = 9;

node_connect(1, 15) = 7;  % element 15 node_connect(2, 15) = 10;

node_connect(1, 16) = 8;  % element 16 node_connect(2, 16) = 10;

node_connect(1, 17) = 9;  % element 17 node_connect(2, 17) = 10;

node_connect(1, 18) = 9;  % element 18 node_connect(2, 18) = 11;

node_connect(1, 19) = 9;  % element 19 node_connect(2, 19) = 12;

node_connect(1, 20) = 10;  % element 20 node_connect(2, 20) = 12;

node_connect(1, 21) = 10;  % element 21 node_connect(2, 21) = 13;

node_connect(1, 22) = 11;  % element 22 node_connect(2, 22) = 12;

node_connect(1, 23) = 12;  % element 23 node_connect(2, 23) = 13;

node_connect(1, 24) = 11;  % element 24 node_connect(2, 24) = 14;

node_connect(1, 25) = 11;  % element 25 node_connect(2, 25) = 15;

node_connect(1, 26) = 12;  % element 26 node_connect(2, 26) = 15;

node_connect(1, 27) = 12;  % element 27 node_connect(2, 27) = 16;

node_connect(1, 28) = 13;  % element 28 node_connect(2, 28) = 16;

node_connect(1, 29) = 13;  % element 29 node_connect(2, 29) = 17;

node_connect(1, 30) = 14;  % element 30 node_connect(2, 30) = 15;

node_connect(1, 31) = 15;  % element 31 node_connect(2, 31) = 16;

node_connect(1, 32) = 16;  % element 32 node_connect(2, 32) = 17;

node_connect(1, 33) = 14;  % element 33 node_connect(2, 33) = 20;

node_connect(1, 34) = 14;  % element 34 node_connect(2, 34) = 18;

node_connect(1, 35) = 15;  % element 35 node_connect(2, 35) = 18;

node_connect(1, 36) = 16;  % element 36 node_connect(2, 36) = 19;

node_connect(1, 37) = 17;  % element 37 node_connect(2, 37) = 19;

node_connect(1, 38) = 17;  % element 38 node_connect(2, 38) = 21;

node_connect(1, 39) = 18;  % element 39 node_connect(2, 39) = 20;

node_connect(1, 40) = 19;  % element 40 node_connect(2, 40) = 21;

node_connect(1, 41) = 20;  % element 41 node_connect(2, 41) = 24;

node_connect(1, 42) = 20;  % element 42 node_connect(2, 42) = 26;

node_connect(1, 43) = 21;  % element 43 node_connect(2, 43) = 29;

node_connect(1, 44) = 21;  % element 44 node_connect(2, 44) = 31;

node_connect(1, 45) = 22;  % element 45 node_connect(2, 45) = 23;

node_connect(1, 46) = 23;  % element 46 node_connect(2, 46) = 24;

node_connect(1, 47) = 24;  % element 47 node_connect(2, 47) = 25;

node_connect(1, 48) = 25;  % element 48 node_connect(2, 48) = 26;

node_connect(1, 49) = 26;  % element 49 node_connect(2, 49) = 27;

node_connect(1, 50) = 27;  % element 50 node_connect(2, 50) = 28;

node_connect(1, 51) = 28;  % element 51 node_connect(2, 51) = 29;

node_connect(1, 52) = 29;  % element 52 node_connect(2, 52) = 30;

node_connect(1, 53) = 30;  % element 53 node_connect(2, 53) = 31;

node_connect(1, 54) = 31;  % element 54 node_connect(2, 54) = 32;

node_connect(1, 55) = 32;  % element 55 node_connect(2, 55) = 33;

node_connect(1, 56) = 22;  % element 56 node_connect(2, 56) = 34;

node_connect(1, 57) = 23;  % element 57 node_connect(2, 57) = 34;

node_connect(1, 58) = 23;  % element 58 node_connect(2, 58) = 36;

node_connect(1, 59) = 24;  % element 59 node_connect(2, 59) = 36;

node_connect(1, 60) = 24;  % element 60 node_connect(2, 60) = 38;

node_connect(1, 61) = 25;  % element 61 node_connect(2, 61) = 38;

node_connect(1, 62) = 25;  % element 62 node_connect(2, 62) = 39;

node_connect(1, 63) = 26;  % element 63 node_connect(2, 63) = 39;

node_connect(1, 64) = 26;  % element 64 node_connect(2, 64) = 40;

node_connect(1, 65) = 27;  % element 65 node_connect(2, 65) = 40;

node_connect(1, 66) = 27;  % element 66 node_connect(2, 66) = 41;

node_connect(1, 67) = 28;  % element 67 node_connect(2, 67) = 41;

node_connect(1, 68) = 28;  % element 68 node_connect(2, 68) = 42;

node_connect(1, 69) = 29;  % element 69 node_connect(2, 69) = 42;

node_connect(1, 70) = 29;  % element 70 node_connect(2, 70) = 43;

node_connect(1, 71) = 30;  % element 71 node_connect(2, 71) = 43;

node_connect(1, 72) = 30;  % element 72 node_connect(2, 72) = 44;

node_connect(1, 73) = 31;  % element 73 node_connect(2, 73) = 44;

node_connect(1, 74) = 31;  % element 74 node_connect(2, 74) = 37;

node_connect(1, 75) = 32;  % element 75 node_connect(2, 75) = 37;

node_connect(1, 76) = 32;  % element 76 node_connect(2, 76) = 35;

node_connect(1, 77) = 33;  % element 77 node_connect(2, 77) = 35;

node_connect(1, 78) = 34;  % element 78 node_connect(2, 78) = 36;

node_connect(1, 79) = 35;  % element 79 node_connect(2, 79) = 37;

node_connect(1, 80) = 36;  % element 80 node_connect(2, 80) = 38;

node_connect(1, 81) = 37;  % element 81 node_connect(2, 81) = 44;

node_connect(1, 82) = 38;  % element 82 node_connect(2, 82) = 39;

node_connect(1, 83) = 39;  % element 83 node_connect(2, 83) = 40;

node_connect(1, 84) = 40;  % element 84 node_connect(2, 84) = 41;

node_connect(1, 85) = 41;  % element 85 node_connect(2, 85) = 42;

node_connect(1, 86) = 42;  % element 86 node_connect(2, 86) = 43;

node_connect(1, 87) = 43;  % element 87 node_connect(2, 87) = 44;

node_connect(1, 88) = 38;  % element 88 node_connect(2, 88) = 45;

node_connect(1, 89) = 39;  % element 89 node_connect(2, 89) = 45;

node_connect(1, 90) = 43;  % element 90 node_connect(2, 90) = 46;

node_connect(1, 91) = 44;  % element 91 node_connect(2, 91) = 46;

% connect all beam elements by connectivity array for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; zz = [z(node_1),z(node_2)]; plot3(xx,yy,zz,'--') hold on  end

title('Electric Pylon') xlabel('x') ylabel('y') zlabel('z')

view([0 0 1])     % xy plane view % view([0 1 0])    % xz plane view %view([1 0 0])    % yz plane view

% produce postscript file for later printing % print -dps electric_pylon.ps  print -deps electric_pylon.ps

R=zeros(dof*n_node,1); R(98)=P; R; %Location Master Matrix Array

for i = 1:n_elem lmm(i,1)= (3*node_connect(1,i)-2); lmm(i,2)= (3*node_connect(1,i)-1); lmm(i,3)= (3*node_connect(1,i)); lmm(i,4)= (3*node_connect(2,i)-2); lmm(i,5)= (3*node_connect(2,i)-1); lmm(i,6)= (3*node_connect(2,i)); end

%Element Stiffness Matrix K=zeros(138); for i=1:n_elem node1=node_connect(1,i); node2=node_connect(2,i); coordinates=[x(node1), y(node1);x(node2), y(node2)];

lm=lmm(i,:); con=node_connect(:,i);

I=(h^4)/12;

EI=e*I; EA = e*A; x1=coordinates(1,1); y1=coordinates(1,2); x2=coordinates(2,1); y2=coordinates(2,2); L=sqrt((x2-x1)^2+(y2-y1)^2); ls=(x2-x1)/L; ms=(y2-y1)/L;

[ke,rq]=PlaneFrameElement(e,I,A,0,0,coordinates); R(lm)=R(lm)+ rq; K(lm, lm) = K(lm, lm) + ke; end

%Computing reactions debc=[1,2,3,4,5,6]; ebcVals=zeros(length(debc),1); [d, reactions] = NodalSoln(K, R, debc, ebcVals); results=[]; f_total=[]; bm_total=[]; V_total=[]; for i=1:n_elem node1=node_connect(1,i); node2=node_connect(2,i); coordinates=[x(node1), y(node1);x(node2), y(node2)]; lm=lmm(i,:); con=node_connect(:,i); [f,bm,V] = PlaneFrameResults(e, I,A,0,0,coordinates,d(lm)); f_total=[f_total;f]; bm_total=[bm_total;bm]; V_total=[V_total;V]; end format short g V_total; bm_total;

%Finding the Max Moment and Shear

max_moment = 0; max_shear=0; max_moment_elem=0; max_shear_elem=0; total_num_forces=2*n_elem; for i=1:total_num_forces if(abs(bm_total(i,3))>abs(max_moment)) max_moment=bm_total(i,3); max_moment_elem=ceil(i/2); end if (abs(V_total(i,3))>abs(max_shear)) max_shear=V_total(i,3); max_shear_elem=ceil(i/2); end end

max_shear; max_moment; max_shear_elem; max_moment_elem;

%Lumped Mass Matrix

M=zeros(3*n_node); m=zeros(n_node); for i=1:n_node for j=1:n_elem if(node_connect(1,j)==i || node_connect(2,j)==i) node1=node_connect(1,j); node2=node_connect(2,j); coordinates=[x(node1), y(node1);x(node2), y(node2)]; x1=coordinates(1,1); y1=coordinates(1,2); x2=coordinates(2,1); y2=coordinates(2,2); L=sqrt((x2-x1)^2+(y2-y1)^2); mass=A*L*p; m(i,i)=m(i,i)+(mass/2);

end end

end

for i=1:n_node M((3*i)-2,(3*i)-2)=m(i,i); M((3*i)-1,(3*i)-1)=m(i,i); % M((3*i),(3*i))=m(i,i); end

%Computation of K_bar and M_bar K_bar=zeros(132); M_bar=zeros(132); z=1;                   %row number for the K_bar and M_bar matrix for i=7:138             %row number for addition n=1;               %column number for the K_bar and M_bar matrix for j=7:138         %column number for addition M_bar(z,n)=M(i,j); K_bar(z,n)=K(i,j); n=n+1; end z=z+1;             %row number for the K_bar matrix end

%Find the 3 Lowest Eigenvalues Eig_sol=pinv(M_bar)*K_bar; [V,D]=eig(Eig_sol); lambda1=1e30; lambda2=1e30; lambda3=1e30; for i=1:132 if (D(i,i)lambda1) lambda2=D(i,i); v2=V(:,i); vect_col2=i; else if(D(i,i)lambda2) lambda3=D(i,i); v3=V(:,i); vect_col3=i; end end end end

lambda1; evect1=transpose(v1); lambda2; evect2=transpose(v2); lambda3; evect3=transpose(v3);

%Computation of the 3 Vibrational Periods Corresponding to the 3 Lowest %Eigenvalues omega1=sqrt(lambda1); f1=omega1/(2*pi); T1=1/f1;

omega2=sqrt(lambda2); f2=omega2/(2*pi); T2=1/f2;

omega3=sqrt(lambda3); f3=omega3/(2*pi); T3=1/f3;

%Finding the 3 Lowest Vibrational Periods vp1=1e100; vp2=1e100; vp3=1e100; Vib_Per=[]; for i=1:88 omega=sqrt(D(i,i)); f=omega/(2*pi); T=1/f; Vib_Per(i,1)=T; if (Tvp1) vp2=T; %vect_col2=i; else if (Tvp2) vp3=T; %vect_col3=i; end end end end vp1; vp2; vp3;

%Printing of Eigenvectors over the original Structure % Original Structure for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)];

%plot(xx,yy,'--') hold on  end % produce postscript file for later printing % print -dps electric_pylon.ps  print -deps electric_pylon.ps

%First Eigenvector FirstRows=zeros(6,132); V_plot=20*[FirstRows;V];

%new positions for the nodes for the first (smallest) eigenvector for i=1:n_node position_vect1(:,i)=[position(1,i)-V_plot((3*i)-2,vect_col1), position(2,i)-V_plot((3*i)-1,vect_col1)]; end

%Nodal Coordinates for i=1:n_node x_vect1(i)=position_vect1(1,i); y_vect1(i)=position_vect1(2,i); end for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x_vect1(node_1),x_vect1(node_2)]; yy = [y_vect1(node_1),y_vect1(node_2)];

%plot(xx,yy,'r-') hold on  end

%title('Electric Pylon (Undeformed and Eigenvector Associated with Smallest Eigenvalue')  %xlabel('x')   %ylabel('y')

%Second Eigenvector FirstRows=zeros(6,132); V_plot=20*[FirstRows;V];

%new positions for the nodes for the second smallest eigenvector for i=1:n_node position_vect2(:,i)=[position(1,i)+V_plot((3*i)-2,vect_col2), position(2,i)+V_plot((3*i)-1,vect_col2)]; end

%Nodal Coordinates for i=1:n_node x_vect2(i)=position_vect2(1,i); y_vect2(i)=position_vect2(2,i); end for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x_vect2(node_1),x_vect2(node_2)]; yy = [y_vect2(node_1),y_vect2(node_2)];

%plot(xx,yy,'r-') hold on  end

%title('Electric Pylon (Undeformed and Eigenvector Associated with Second Smallest Eigenvalue')  %xlabel('x')   %ylabel('y')

%Third Eigenvector FirstRows=zeros(6,132); V_plot=20*[FirstRows;V];

%new positions for the nodes for the second smallest eigenvector for i=1:n_node position_vect3(:,i)=[position(1,i)-V_plot((3*i)-2,vect_col3), position(2,i)-V_plot((3*i)-1,vect_col3)]; end

%Nodal Coordinates for i=1:n_node x_vect3(i)=position_vect3(1,i); y_vect3(i)=position_vect3(2,i); end for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x_vect3(node_1),x_vect3(node_2)]; yy = [y_vect3(node_1),y_vect3(node_2)];

plot(xx,yy,'r-') hold on  end

%title('Electric Pylon (Undeformed and Eigenvector Associated with Third Smallest Eigenvalue')  %xlabel('x')   %ylabel('y')

%Displaced Structure for i=1:n_node %d_def=zeros(46,2); disp1=d((3*i)-2,1); disp2=d((3*i)-1,1);

position(1, i) = (position(1,i))+(150*disp1); position(2, i) = (position(2,i))+(150*disp2);

end

% print the node coord. for i = 1 : n_node; x(i) = position(1,i); y(i) = position(2,i); z(i) = position(3,i);

end % connect all beam elements by connectivity array for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; zz = [z(node_1),z(node_2)]; %plot3(xx,yy,zz,'r-') hold on  end

title('Electric Pylon') xlabel('x') ylabel('y') zlabel('z')

view([0 0 1])     % xy plane view % view([0 1 0])    % xz plane view %view([1 0 0])    % yz plane view

% produce postscript file for later printing % print -dps electric_pylon.ps  print -deps electric_pylon.ps

Additional Functions
PlaneFrameElement function [ke, rq] = PlaneFrameElement(modulus, inertia, A, qs, qt, coord) % [ke, rq] = PlaneFrameElement(modulus, inertia, A, qs, qt, coord) % Generates equations for a plane frame element % modulus = modulus of elasticity % inertia = moment of inertia % A = area of cross-section % qs = distributed load along the element axis % qt = distributed load normal to the element axis % coord = coordinates at the element ends

EI=modulus*inertia; EA = modulus*A; x1=coord(1,1); y1=coord(1,2); x2=coord(2,1); y2=coord(2,2); L=sqrt((x2-x1)^2+(y2-y1)^2); ls=(x2-x1)/L; ms=(y2-y1)/L;

ke = [(EA*L^2*ls^2 + 12*EI*ms^2)/L^3, ((-12*EI + EA*L^2)*ls*ms)/L^3, ...       (-6*EI*ms)/L^2, -((EA*L^2*ls^2 + 12*EI*ms^2)/L^3), ... ((12*EI - EA*L^2)*ls*ms)/L^3, (-6*EI*ms)/L^2; ((-12*EI + EA*L^2)*ls*ms)/L^3, (12*EI*ls^2 + EA*L^2*ms^2)/L^3, ...       (6*EI*ls)/L^2, ((12*EI - EA*L^2)*ls*ms)/L^3, ... -((12*EI*ls^2 + EA*L^2*ms^2)/L^3), (6*EI*ls)/L^2; (-6*EI*ms)/L^2, (6*EI*ls)/L^2, (4*EI)/L, ... (6*EI*ms)/L^2, (-6*EI*ls)/L^2,(2*EI)/L; -((EA*L^2*ls^2 + 12*EI*ms^2)/L^3), ((12*EI -EA*L^2)*ls*ms)/L^3, ...       (6*EI*ms)/L^2, (EA*L^2*ls^2 + 12*EI*ms^2)/L^3, ... ((-12*EI + EA*L^2)*ls*ms)/L^3, (6*EI*ms)/L^2; ((12*EI - EA*L^2)*ls*ms)/L^3, -((12*EI*ls^2 + EA*L^2*ms^2)/L^3),...       (-6*EI*ls)/L^2, ((-12*EI + EA*L^2)*ls*ms)/L^3, ... (12*EI*ls^2 + EA*L^2*ms^2)/L^3, (-6*EI*ls)/L^2; (-6*EI*ms)/L^2, (6*EI*ls)/L^2, (2*EI)/L, (6*EI*ms)/L^2, ... (-6*EI*ls)/L^2, (4*EI)/L]; rq = [(L*(ls*qs - ms*qt))/2; (L*(ms*qs + ls*qt))/2;(L^2*qt)/12; (L*(ls*qs - ms*qt))/2; (L*(ms*qs + ls*qt))/2; -(L^2*qt)/12]; PlaneFrameResults function [f, bm, V] = PlaneFrameResults(modulus, inertia, A, ...   qs, qt, coord, dn) % [f, bm, V] = PlaneFrameResults(modulus, inertia, A, ... %   qs, qt, coord, dn) % Generates frame element results % modulus = modulus of elasticity % inertia = moment of inertia % A = area of cross-section % qs = distributed load along the element axis % qt = distributed load normal to the element axis % coord = coordinates at the element ends % dn = nodal solution % [f, bm, V] = [axial force, bending moment, shear]

EI=modulus*inertia; EA = modulus*A; x1=coord(1,1); y1=coord(1,2); x2=coord(2,1); y2=coord(2,2); L=sqrt((x2-x1)^2+(y2-y1)^2); ls=(x2-x1)/L; ms=(y2-y1)/L;

u = dn([1,4]); v = dn([2,3,5,6]);

f=[]; bm=[]; V=[]; % Change increment to get results at more points for s=0:L:L x = x1 + s*ls; y = y1+ s*ms; f = [f; [x,y,EA*(u(2)-u(1))/L]]; dn2 = [(12*s)/L^3 - 6/L^2, (6*s)/L^2 - 4/L, 6/L^2 - (12*s)/L^3, ... (6*s)/L^2 - 2/L]; bm = [bm; [x, y, EI*dn2*v+(qt*(L^2 - 6*s*L + 6*s^2))/(12)]]; dn3 = [12/L^3, 6/L^2, -(12/L^3), 6/L^2]; V = [V; [x, y, EI*dn3*v+((qt*(12*s - 6*L))/(12))]]; end NodalSoln function [d, rf] = NodalSoln(K, R, debc, ebcVals) % [nd, rf] = NodalSoln(K, R, debc, ebcVals) % Computes nodal solution and reactions % K = global coefficient matrix % R = global right hand side vector % debc = list of degrees of freedom with specified values % ebcVals = specified values dof = length(R); df = setdiff(1:dof, debc); Kf = K(df, df); Rf = R(df) - K(df, debc)*ebcVals; dfVals = Kf\Rf; d = zeros(dof,1); d(debc) = ebcVals; d(df) = dfVals; rf = K(debc,:)*d - R(debc);

Comparison of Nodal Displacements
It can be seen from the table below that the displacements that are calculated when treating the electric pylon as a truss structure are very similar to the values acquired using a frame system analysis. The displacements calculated using a frame analysis differ on the order of $$10^{-8}$$ from the values obtained using a truss analysis.

Frame Element with Highest Nodal Bending Moment
The nodal bending moments were calculated with the function PlaneTrussResults, which is discussed earlier. In order to store the value of the bending moments measured in this function, the following line of code was written to increase the size of the matrix to accomodate every element.

bm_total=[bm_total;bm];

The largest bending moment, and the element in which it occurs can be easily acquired. The code shown below was created to accomplish this task. %Finding the Max Moment max_moment = 0; max_moment_elem=0; total_num_forces=2*n_elem; for i=1:total_num_forces if(bm_total(i,3)>max_moment) max_moment=bm_total(i,3); max_moment_elem=ceil(i/2); end end

max_moment max_moment_elem Using these two sets of code, the maximum bending moment and the element in which it occurs were found to be: max_moment =

6.2874

max_moment_elem =

81

Frame Element with Highest Transverse Shear Force
The transverse shear forces were also calculated using the function PlaneFrameResults that was given earlier. These values were then compiled into a large matrix in order to ease the calculation of the maximum transverse shear force. The following line of code was created to accomplish this task. V_total=[V_total;V];

This matrix was then used to calculate the maximum transverse shear force, and the element in which it occurs, when used in conjuntion with the following code.

%Finding the Max Shear

max_shear=0; max_shear_elem=0; total_num_forces=2*n_elem; for i=1:total_num_forces if (V_total(i,3)>max_shear) max_shear=V_total(i,3); max_shear_elem=ceil(i/2); end end

max_shear max_shear_elem Using this code, the following values for the maximum transverse shear force, and the element in which it occurs were found to be: max_shear =

-5.4827

max_shear_elem =

81

Plot of Undeformed and Deformed Shapes
The plot of the deformed and undeformed electric pylon structures are shown in the figure below. The dashed lines in blue are the original structure. The red lines are the deformed truss structure, while the green lines represent the deformed frame structure.

The difference between the two deformed structures is not directly visible using this method. This image is enlarged and focused on the node at which the load is applied (Node 33) in order to show the small difference obtained using the different methods of analysis. The dotted line in this picture is the frame analysis, and the solid line shown is the truss analysis.

Determination of Static Condition
The electric pylon frame, much like the truss structure, is statically indetermine. This is caused by the lack of available equations to solve for the number of unknowns. Using a static analysis, with the two nodes clamped at the bottom of the structure, there are four unknown reactions (two for each node). Two of these unknowns can be found by summing the forces in the x and y directions. The other two forces could only be eliminated by equations created by the sum of the moments about two separate points. Since the nodes are located at the same vertical position, it is impossible to create an equation based on the moments that would suffice, as these variables would tend to cancel each other. Thus, only two of the four unknowns can be found, making the frame analysis statically indeterminate.

Comparison of Reaction Forces
It can be seen from the table below that the reaction forces calculated using the truss method and the frame method are very similar. It can also be noted that the sum of the forces in the x-direction for each case is equal to zero. The sum of the forces in the y-direction for each method is approximately equal to the force that is applied at the rightmost node on the electric pylon.

Three Lowest Eigenpairs
The code that was used to find the eigenvalues, and the eigenvectors corresponding to these eigenvalues is shown above in the code used for the frame analysis. The values obtained are in the sections below. The eigenvector is displayed as the transpose of the original eigenvector in order to save space. The generalized eigenvalue problem that was used to analyze this frame is given as:

$$ \bar{K}v=\lambda \bar{M}v $$

Smallest Eigenpair
lambda1 =

132.86

evect1 =

Columns 1 through 7

0.002065  -0.0071344            0    0.0014718  1.6264e-005            0    0.0019696

Columns 8 through 14

0.0069748           0   -0.0023269      -0.0119            0   -0.0022818 -3.1083e-006

Columns 15 through 21

0  -0.0022827     0.011793            0     -0.02167    -0.011731            0

Columns 22 through 28

-0.021724    0.011739            0    -0.037126    -0.020181            0    -0.037036

Columns 29 through 35

1.9661e-005           0    -0.037091     0.019808            0    -0.054073    -0.028787

Columns 36 through 42

0   -0.054044    -0.010729            0     -0.05401     0.010722            0

Columns 43 through 49

-0.054025    0.028395            0    -0.067992    -0.020088            0    -0.067966

Columns 50 through 56

0.020103           0      -0.1168    -0.056482            0     -0.11733     0.056306

Columns 57 through 63

0    -0.16187     -0.15094            0     -0.16199    -0.095213            0

Columns 64 through 70

-0.16203   -0.072597            0     -0.16207    -0.055567            0     -0.16242

Columns 71 through 77

-0.035048           0     -0.16304    -0.013353            0     -0.16307     0.013567

Columns 78 through 84

0    -0.16249     0.035595            0     -0.16216     0.055432            0

Columns 85 through 91

-0.16212     0.07229            0     -0.16209     0.095224            0     -0.16197

Columns 92 through 98

0.14995           0     -0.17388     -0.10866            0     -0.17387      0.10804

Columns 99 through 105

0    -0.18022    -0.085947            0     -0.18016     0.086037            0

Columns 106 through 112

-0.18402   -0.072451            0     -0.18357    -0.046196            0     -0.18279

Columns 113 through 119

-0.023355           0     -0.18242   0.00033099            0     -0.18274     0.023969

Columns 120 through 126

0    -0.18348      0.04615            0     -0.18392     0.072153            0

Columns 127 through 132

-0.23694   -0.072646            0     -0.23635     0.072347            0

Second Smallest Eigenpair
lambda2 =

2471.2

evect2 =

Columns 1 through 7

0.0052161    0.026474            0  -0.00058811 -6.2468e-006            0   -0.0067635

Columns 8 through 14

0.025585           0  -0.00031537     0.044409            0  -0.00032035     0.054787

Columns 15 through 21

0 -0.00022037     0.043697            0   -0.0034941     0.061278            0

Columns 22 through 28

0.0036747    0.060487            0   -0.0028717     0.072169            0   0.00075563

Columns 29 through 35

0.058547           0    0.0046154     0.070598            0    0.0055912     0.089114

Columns 36 through 42

0   0.0053101     0.062862            0    -0.002698     0.061897            0

Columns 43 through 49

-0.0029237    0.087342            0     0.026785     0.076037            0    -0.024135

Columns 50 through 56

0.074873           0      0.14696      0.16682            0     -0.14826      0.16685

Columns 57 through 63

0    0.028062    -0.038691            0     0.027044     0.088149            0

Columns 64 through 70

0.026844     0.13645            0     0.025352       0.1861            0     0.018514

Columns 71 through 77

0.23267           0     0.007504      0.26522            0   -0.0080013      0.26512

Columns 78 through 84

0   -0.019169      0.23153            0    -0.025826      0.18579            0

Columns 85 through 91

-0.02732     0.13614            0    -0.027539     0.086607            0    -0.028563

Columns 92 through 98

-0.039489           0   0.00079157     0.058654            0   -0.0012143     0.058164

Columns 99 through 105

0   -0.012969      0.10801            0      0.01258      0.10653            0

Columns 106 through 112

-0.022728     0.14121            0    -0.018876      0.21005            0    -0.011164

Columns 113 through 119

0.25356           0  -7.883e-005       0.2716            0     0.010992      0.25275

Columns 120 through 126

0    0.018464      0.20976            0     0.022308      0.14088            0

Columns 127 through 132

-0.16524     0.13994            0      0.16486      0.13962            0

Third Smallest Eigenpair
lambda3 =

2942.6

evect3 =

Columns 1 through 7

0.045112    0.018917            0      0.03338   0.00039065            0      0.04554

Columns 8 through 14

-0.019648           0     0.098727     0.016366            0     0.097768  -0.00091162

Columns 15 through 21

0    0.098605    -0.017531            0      0.17331    0.0062518            0

Columns 22 through 28

0.17322  -0.0082697            0      0.19259    0.0038836            0      0.18943

Columns 29 through 35

-0.0010322           0      0.19274   -0.0059396            0      0.19753   -0.0059569

Columns 36 through 42

0     0.19634    0.0044044            0      0.19636   -0.0065704            0

Columns 43 through 49

0.19754   0.0038358            0      0.18842   0.00015649            0      0.18816

Columns 50 through 56

-0.0021636           0      0.10485    -0.063574            0      0.10201     0.062351

Columns 57 through 63

0     0.06724     -0.37216            0     0.057031     -0.15138            0

Columns 64 through 70

0.051631   -0.094548            0      0.04841    -0.068008            0     0.045051

Columns 71 through 77

-0.037274           0     0.041954    -0.012713            0     0.041802     0.013652

Columns 78 through 84

0    0.044729      0.03798            0     0.047824     0.066849            0

Columns 85 through 91

0.050927    0.092296            0     0.056192       0.1484            0     0.065903

Columns 92 through 98

0.35895           0     0.023218     -0.19034            0     0.023641      0.18442

Columns 99 through 105

0    0.010042     -0.12645            0     0.010825      0.12435            0

Columns 106 through 112

0.0056352   -0.098291            0     0.011252    -0.053232            0     0.016088

Columns 113 through 119

-0.023365           0     0.018321   0.00073381            0     0.016463     0.024508

Columns 120 through 126

0    0.011981     0.052667            0    0.0066177     0.095907            0

Columns 127 through 132

-0.087066    -0.10046            0    -0.082297     0.097986            0

Plots of Eigenmodes
Plot of the Smallest Eigenvector Plot of the Second Smallest Eigenvector Plot of the Third Smallest Eigenvector

Three Lowest Vibrational Periods
The table below shows the vibrational periods for the truss analysis and the frame analysis of the electric pylon system. All of the values in the table that have a T in the variable name are the vibrational periods corresponding to the three smallest eigenvalues. The variables that contain the letters vp are the smallest vibrational periods for each system. It can be seen from this table that the vibrational periods obtained for each form of analysis are very similar.

Animations
The first animation is the vibration resulting from the smallest eigenvector of the electric pylon truss problem. This calculation was performed in Homework 6.

The second animation is the vibration resulting from the second smallest eigenvector of the electric pylon truss problem. This calculation was performed in Homework 6. The third animation is the vibration resulting from the third smallest eigenvector of the electric pylon truss problem. This calculation was performed in Homework 6.

Two Bar Truss MATLAB Analysis
Data

MATLAB Code

In order to run this MATLAB code the following functions are needed:

Note: this MATLAB functions were originally obtained from the MATLAB.zip folder provided by the Student Companion Site Bhatti: Advanced Topics in Finite Element Analysis of Structures: With Mathematica and MATLAB Computations website http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=2256&itemId=0471648086&resourceId=4981 and later modified in order to applied to the MATLAB code presented in this section.

New Results

Previous Results (HW#2)

Plot

Prove FD Relationship
The only difference with this proof and the proof in earliear notes is that the matrices are 6x6 and 6x1 instead of 4x4 and 4x1. Start with: Eqn 1$$\tilde{f}^{(e)}_{6x1} = \tilde{k}_{6x6} ^{(e)} \tilde{d}_{6x1}^{(e)}$$

Then plug the two following equations into Eqn 1:

$$\tilde{f}^{(e)} = \tilde{T}^{(e)} f^{(e)}   $$ and $$ \tilde{d}^{(e)} = \tilde{T}^{(e)}  d^{(e)} $$

The following equation is found:

$$\tilde{k}^{(e)} \tilde{T}^{(e)} d^{(e)}= \tilde{T}^{(e)}  f^{(e)}$$

The transpose of the $$\tilde{T}$$ is taken to move the matrix from the right side to the left and the following equation is found: Eqn 2$$f^{(e)} = \begin{bmatrix} \tilde{T}^{(e)^{T}} \tilde{k}^{(e)}  \tilde{T}^{(e)}*d^{(e)} \end{bmatrix}$$

Now recall the following equation:

$$k_{6x6}^{(e)}=\tilde{T}_{6x6}^{(e)T}\tilde{k}_{6x6}^{(e)}\tilde{T}_{6x6}^{(e)}$$

This equation is substituted into Eqn 2 to get:

$$f_{6x1}^{(e)}=k_{6x6}^{(e)}d_{6x1}^{(e)}$$

Verify the dimensions of all the terms in the stiffness matrix multiplied by the displacement matrix
Start with the matrices of $$\tilde{k_{ij}}$$ and $$\tilde{d_{j}}$$.

$$\tilde{k_{ij}}\tilde{d_{j}}=\begin{bmatrix} \frac{EA}{L} & 0 & 0 & -\frac{EA}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & -\frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{2EI}{L} \\ -\frac{EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0 \\ 0 & -\frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} \\ 0 & -\frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & \frac{6EI}{L^{2}} & -\frac{4EI}{L} \end{bmatrix} \begin{pmatrix} \tilde{d_{1}} \\ \tilde{d_{2}} \\ \tilde{d_{3}} \\ \tilde{d_{4}} \\ \tilde{d_{5}} \\ \tilde{d_{6}} \end{pmatrix}$$

The displacements $$\tilde{d_{j}}$$ where i=1,2,4 and 5 are transverse displacements. The dimensions of transverse displacements are length. Shown below is the dimensional analysis. $$\left[\tilde{d_{i}}\right] = \left[\tilde{d_{1}}\right] = \left[\tilde{d_{2}}\right] = \left[\tilde{d_{4}}\right] = \left[\tilde{d_{5}}\right] = L$$ The square brackets, in dimensional analysis, indicate that the dimension of the enclosed are being found. The displacements $$\tilde{d_{j}}$$ where i=3 and 6 are rotational displacements. The dimensions of rotational displacements are 1. Shown below is the dimensional analysis. $$\left[\tilde{d_{3}}\right] = \left[\tilde{d_{6}}\right] = 1$$ From the class notes, it is known that: $$\left[E\right] = \frac{F}{L^{2}}$$ and $$\left[A\right] = L^{2}$$ and $$\left[I\right] = L^{4}$$ Therefore: $$\left[\frac{EA}{L}\right] = \frac{F}{L}$$ and $$\left[\frac{EI}{L}\right] = FL$$ and $$\left[\frac{EI}{L^{2}}\right] = F$$ and $$\left[\frac{EI}{L^{3}}\right] = \frac{F}{L}$$ Since zero is dimensionless, it is known that the dimensions of all $$\tilde{k_{ij}}\tilde{d_{j}}$$ that equal to zero have dimensions of 1. Zero is obviously dimensionless, and a dimensionless parameter is 1. Below are all $$\tilde{k_{ij}}\tilde{d_{i}}$$ that have a dimension of 1.

$$\left[\tilde{k_{12}}\tilde{d_{2}}\right] = \left[\tilde{k_{13}}\tilde{d_{3}}\right] = \left[\tilde{k_{15}}\tilde{d_{5}}\right] = \left[\tilde{k_{16}}\tilde{d_{6}}\right] = \left[\tilde{k_{21}}\tilde{d_{1}}\right] = \left[\tilde{k_{24}}\tilde{d_{4}}\right] = \left[\tilde{k_{31}}\tilde{d_{1}}\right] = \left[\tilde{k_{34}}\tilde{d_{4}}\right] = \left[\tilde{k_{42}}\tilde{d_{2}}\right] = \left[\tilde{k_{43}}\tilde{d_{3}}\right] = \left[\tilde{k_{45}}\tilde{d_{5}}\right] = \left[\tilde{k_{46}}\tilde{d_{6}}\right]$$ $$ = \left[\tilde{k_{51}}\tilde{d_{1}}\right] = \left[\tilde{k_{54}}\tilde{d_{4}}\right] = \left[\tilde{k_{61}}\tilde{d_{1}}\right] =  \left[\tilde{k_{64}}\tilde{d_{4}}\right] = 1 $$

When the rest of the stiffness parameters are multiplied out with the corresponding displacements the following dimensional parameters are found:

$$\left[\tilde{k_{11}}\tilde{d_{1}}\right] = \left[\tilde{k_{14}}\tilde{d_{4}}\right] = \left[\tilde{k_{22}}\tilde{d_{2}}\right] = \left[\tilde{k_{25}}\tilde{d_{5}}\right] = \left[\tilde{k_{41}}\tilde{d_{1}}\right] = \left[\tilde{k_{44}}\tilde{d_{4}}\right] = \left[\tilde{k_{52}}\tilde{d_{2}}\right] = \left[\tilde{k_{55}}\tilde{d_{5}}\right] = F$$

$$\left[\tilde{k_{23}}\tilde{d_{3}}\right] = \left[\tilde{k_{26}}\tilde{d_{6}}\right] = \left[\tilde{k_{32}}\tilde{d_{2}}\right] = \left[\tilde{k_{33}}\tilde{d_{3}}\right] = \left[\tilde{k_{35}}\tilde{d_{5}}\right] = \left[\tilde{k_{36}}\tilde{d_{6}}\right] = \left[\tilde{k_{53}}\tilde{d_{3}}\right] = \left[\tilde{k_{56}}\tilde{d_{6}}\right] = \left[\tilde{k_{62}}\tilde{d_{2}}\right] = \left[\tilde{k_{63}}\tilde{d_{3}}\right] = \left[\tilde{k_{65}}\tilde{d_{5}}\right] = \left[\tilde{k_{66}}\tilde{d_{6}}\right] = FL$$

Iain
After using Wikiversity for a semester in order to post homework I feel I am able to give a justified opinion on its usefulness in the classroom. It can be seen as a good learning tool, as it can make acquisition of class notes easier in order to study for a test. It does take copious amounts of time in order to produce a functional reference, which can be detrimental to the amount of time that I, as a student, have to study the class material. While the aim may have been to have the student review and revise these notes, it became a tedious task of simply reproducing what was already written. More effort was put into writing the code to represent equations and pictures than actually studying the material. Correct answers to the proofs were seldom given, and this may have resulted in errors in the learning process for the student. Furthermore, students stopped coming to class after realizing that they could simply read the notes online earlier. This led to larger gaps in the learning process, and subsequently proved detrimental to the student's overall knowledge of the subject.

E-Learning is a website run by WebCT that is commonly used in classes at UF. It allows for the posting of files, including .pdf, .ppt, .doc, and any other common file types that are used by programs in the class. The method of submitting homework can be easier, as the files that were used for analysis, or the documents to be submitted, need only be attached to the submission. Comments can be written by the student to designate some details of the assignment. Large amounts of detail can be written in a program such as Microsoft Word, and also attached to the submission. Grades in E-Learning are easy to post and manage, as there is a tab built into the structure of the website. This would allow the TA's to simply post the grades instead of having to respond to questioning about someone's grades. Overall, this system of submission would give the professor, the TA's, and the students more time to learn the material. This is justified by the amount of time that goes into writing code for equations and making diagrams for Wikiversity. Any images made in E-Learning can simply be scanned and submitted, instead of having to upload the file to Wiki Commons, and then write the code to post it on the webpage.

Nicolas
There are two main differences between both applications, each having a deep impact on the experience of the user, i.e. the student. Wikiversity has proved to be based on very sturdy programming code that ensures its reliability, while E-Learning has showcased bugs and other obstacles. As past experience dictates, E-Learning has caused problems with navigation and file downloads. This problem can be very frustrating as it can lead to crashing of the internet browser, Internet Explorer in my case. Also, the E-Learning site produces various useless pop-up windows that ask for checks and permissions to perform orders from the user. On the other hand, Wikiversity is a very rigid program that is very well maintained and easy to use. It works with any internet browser. You can tell that the site is monitored by site managers constantly to ensure safe, serious, relevant, and clean, information is posted. As far as using E-Learning vs. Wikiversity for a classroom purpose, then they both perform different functions. For once, E-Learning is able to store documents for the students to download and use. This method requires a course that relies on homework. The student downloads the homework problems from the site and solves the problems. However, Wikiversity is geared towards posting information online. This method is geared towards a class that requires assignments with the written word such as essays, poems, editorials, etc. It great advantage that Wikiversity has is that the users can edit content and view changes real time. E-Learning does not feature the possibility to edit documents real time for other contributors to witness, as it acts as a way to post different versions of documents, not edit the original. As far as the EML4500 class goes, I feel that the idea of posting notes online is very advantageous only for the student that did NOT attend class. I think that attending class and having the notes on paper is sufficient, as I learned the material anyways by being in class. Furthermore, having the same set of notes 8 different times seemed superfluous. Having to learn the Wiki code for re-copying notes was not an efficient use of my time, even more since I am a graduating senior who was looking for a job. The aim was to learn about the advantages of the Wiki applications (like being able to immortalize the information), but I think that reading or hearing about these applications and advantages would have been enough to make me understand. Despite it all, I did get a firm grasp of the subject and the Wiki code. The EML4500 class is an engineering course that should be based on complete homework problems for practice, which could be turned in through Wikiversity. Seeing the homework problems online was very useful since they were clear and required minimal time to post. Posting notes was very tedious since learning had occurred in the classroom.

Sean
There are many pros and cons when talking about using Wikiversity versus using E-learning for keeping up a class website.

First, I will discuss the pros and cons of using E-learning. This website it available only to students enrolled in certain classes. The layout and organization of the site is highly efficient and easy to learn, with all sections of a class website on the left hand side of the webpage, in neat tabs. The information posted to one's e-learning page is safe from inaccuracies, since only the person in charge of the account can alter the page in any way. Also, the e-learning system is frequently slow and sometimes shut down for maintenance. These are just some of the traits of the ELS website.

Now, I will evaluate the Wikiversity website. At first, the entering of lecture notes was a tiring and tedious task. However, after a few weeks, submitting my notes for the day became a simple, quick assignment. Images and videos could be entered in the middle of a lecture, creating a fluid easy to read article. These articles became a great way to study for exams, since the notes were accurate and easy to follow. A table of contents allows the user to quickly skip to a section of the notes, allowing for easy reviewing. Also, Wikiversity allows strangers to access your work, and correct or add their own thoughts, alsoallowing groups of people to work together on articles, each entering their own opinion or facts.

Overall, I enjoy using both Wikiversity and E-learning, however, for the normal class, where lectures and all other downloadable content is posted by the teacher, I prefer to use the E-learning system.

Eric Ramirez
The question we are given to evaluate is E-learning vs. Wikiversity.

Well I personally believe that the comparing of the two is like comparing apple and oranges.

First, I have used E-learning for a couple semesters now and dont have any major complaints about the system. However, at times the system does run very slow and the log-in page has frequent errors in successfully logging in a user. The format or layout of the system is fairly easy to comprehend and navigate. The system allows students to message each other, submit homework assignments online, download content, etc. Thus, the E-learning system covers a lot of angles that are beneficial towards students.

Second, Wikiversity from personal experience has been a great tool in uploading class content not only for personal use but for the benefit of others to study or learn from notes posted. Wikiversity also consists of a fairly simple layout, easy to navigate, and one can easily post information on the website. Wikiversity also goes hand in hand with Wikipedia which offers tons of information, hence; being known as the online encyclopedia.

Finally, as stated before comparing the two systems is like mixing apples and oranges. The two systems are used differently and I believe offer different approaches to classroom involvment and curriculum organization. The E-learning system is more of a tool that teachers can use to organize a certain class and keep students involved online. Where as the Wikiversity system can be used more by the students themselves. Moreover, the E-learning system allows teachers to post student grades that can only be viewed by each individual student. Up until this point, I am unaware if Wikiversity provides any way for teachers to offer confidential information to individual students on the site without someone being able to clear or sabatoge the Wikiversity page.

In essence, I believe the two systems are different and should not be compared to each other. They each have their flaws and their advantages over the other but do not carry the same type of intent. E-learning is great for teachers to use to organize their class and keep information online while Wikiversity is optimal in sharing information between students as well as providing vasts amount of information posted on Wikipedia.

Diana
Wikiversity: Pros

 * Reports can be easily edited by any group member.
 * Dr. VuQuoc insists in using it because he strongly believes Wikiversity would be greatly accepted in the future.

Cons

 * Professionals often discourage the use of Wikipedia as a valid source of information.
 * The quality of the work done does not equal the amount of time spent on Wikiversity due to the unfamiliarity with the system.
 * The grade for each team member is given in proportion to their Wiki contributions instead of the equal distribution that would be given to any other group effort.
 * Wikiversity content can be changed by any person with a valid IP address.
 * The quality of work decreases since any other member from another group can observe the Wiki report and just copy its contents as its own.

E-learning: Pros

 * System provided just to the University of Florida students.
 * Secured system that can only be accessed with a valid Gator username and password.
 * Members from other groups can not observe/modify the reports. Only the group members, professors and TA are able to.
 * The system provides a feature that compares reports/essays to any available information online making cheating and copy right infringements almost impossible.
 * The system provides the chat feature which allows students to contribute to other group members without meetings in person facilitating and improving the quality of the reports.
 * Students are familiar with E-learning making the time spent in projects truly reflect the quality of the work.
 * Program usage of Word, Excel and Power Point are encouraged by E-learning since these programs are greatly accepted in the industry.
 * The grade is equally distributed since there was just one report presented by the group instead of having a grade count by the number of collaborations made in the report.
 * E-learning motivates confidentiality since all the other groups’ work is kept private just like in real life project.
 * Students actually learn the material since they do not have to spent time to figure out unfamiliar commands in E-learning.

Cons

 * Editing one report is usually done by one group member instead of all the collaborating members.

Conclusion

I honestly believe that Wikiversity and E-learning can not be compared to each other. The two systems have different purposes. While Wikiversity is intended for easy learning the E-learning system is intended for students who are part of an institution and this system reflects the principles of confidentiality and integrity that the University of Florida stands for. The online editing feature of Wikiversity is the only disadvantage that I believe is holding back the E-learning system. Once this feature is available there won’t be any reason to use Wikiversity. I do enjoy E-learning more because it reflects many of the aspects that I will experience when I get out of college. The use of programs widely accepted is encourage same as the privacy and ethics principles I have learnt while completing my degree. Also the amount of time spent on the reports is truly reflected when using the E-learning system. For these reasons I believe that the E-learning system is a better approach for the college classroom.

Trent
Wiki: Pros: 1) The multiple variations of notes and homeworks made studying more efficient and decreased errors on homeworks that were already graded. 2) The Compilation of the best of the notes can be used by later classes. This is put in the pros, but could also be a con. If the homework was wrong and not corrected it would influence the reader negatively. 3) The wiki network always seems to be available. There hasn't been a time when I could not log on. This is not the case with E-learning. 4) Now that the code for wiki is known the students can use wiki for other projects. Cons: 1) Having homework on a continuously available website with no way of restricting access makes cheating extremely easy. 2) Learning the code added extra work to an already work intense degree program. 3) The time to grade the class notes and the homeworks was slow. The response was much longer than with regular courses. 4) The materials sometimes felt as though it was spread all over the place. I don't feel like it was a compact user friendly way of relating the course material. 5) The pages can be changed by others. This could also lead to inaccuracies in studying. 6) The errors in the homeworks were not always well defined and could lead to inaccuracies in studying.

E-learning: Pros: 1) The format is compact and easy to navigate. 2) Cheating is very difficult. Cons: 1) This format cannot be used to study. 2) E-learning is not really a learning tool. It is more of a line of information between the teacher and the students. 3) The system is often slow and many times the student is unable to log on.

My Opinion I feel that wiki can be a good learning tool, but this takes an extremely large amount of time and diligence by the teacher and the T.A.'s. The homeworks would need to be graded in a shorter amount of time and the errors would need to be well marked and the correct answer given. Also the open site could lead to errors in studying. The pages could be changed and this could introduce errors. Without these problems wiki would be a good tool for courses. I feel that E-learning is a better way to deliver and receive information and reduces errors. Therefore, I feel that E-learning is a better system for the classroom.

Contributing Team Members
Iain Specht--Eml 4500.f08.delta 6.specht 17:35, 24 November 2008 (UTC) Micheal Trent Krueger--Eml4500.f08.delta 6.krueger 01:44, 28 November 2008 (UTC) Nicolas Castrillon--Eml4500.f08.delta 6.castrillon 06:39, 3 December 2008 (UTC) Sean Lopez--Eml4500.f08.delta 6.lopez 03:45, 9 December 2008 (UTC) Eric Ramirez--Eml4500.f08.delta 6.ramirez 17:13, 9 December 2008 (UTC) Diana Guzman --Eml4500.f08.delta 6.guzman 21:11, 9 December 2008 (UTC)

title: User:Eml 4500.f08.delta 6.specht/HW7


text:
User:Stevenarntson/English Composition/assignments

 *  Your grade in this class is divided into thirds: participation, workshop, and project.

Statement of Intent
You are responsible for deciding what you will produce this quarter, with the one caveat that I will vet your project idea(s) based on appropriateness of scope. Your statement of intent will be 300 words, free of typos, and posted to your Wikiversity userspace. Here is an example.

 * overview. What you intend to accomplish, stated in quantitative terms.
 * rationale. What do you hope to get out of this?
 * publication/presentation. How will the public access the work?
 * schedule. Build a realistic order of events.
 * anticipated problems. What might go wrong? How will you compensate?

Workshop
You will have a piece of writing critiqued by the class, at least 1,500 words (or less with my permission if you wrote a villanelle). The workshop is explained fully here.

Reader's Report
You will select and read something of some significance to what you are trying to accomplish this quarter, and you will write a short (200-word) review of it. Not what it was about, but what you got out of it.

Publication/Presentation
Your project will be presented in two ways.
 * On your wikiversity userpage.
 * In one of the following formats:
 * 1) A physical zine of at least ten copies.
 * 2) A performance/recording that is available to the public.
 * 3) Attempts toward publication.

Participation

 * attendance, including field trips
 * active in discussion and during in-class assignments
 * self assessment

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
User:Stevenarntson/group spaces

Winter 2010
<>
 * English Composition
 * Analytical Writing

Fall 2009

 * Analytical Writing

Summer 2009

 * Analytical Writing

Spring 2009

 * Analytical Writing
 * Humanities II

Winter 2009

 * English Composition
 * Analytical Writing

title: User:Stevenarntson/group spaces


text:
User:Stevenarntson/group spaces/winter 2008

 * English Composition
 * Analytical Writing


text:
User:Eas4200c.f08.aero.lee/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.

Warning: User Eas4200c.f08.lee was not registered; please register to create this user account, since otherwise the page may be deleted.

Also, don't put the figures on the right of the collapsible tables, since these figures would block the link to expand/collapse these tables; see your figures to the right of your matlab codes in collapsible tables below. Instead, you may want to put these figures to the left of the collapsible tables. Another way is to shorten the width of the collapsible tables.

Eml4500.f08 13:27, 23 November 2008 (UTC)

-

I reposted under the correct username. It was a typo that led '0' to become 'o' and was overlooked.

Also, it doesn't appear that any of the collapsible table expand/collapse links are blocked at all by any figures. I'm not sure if it looks different when viewed from a different computer.

--Eas4200c.f08.aero.lee 18:08, 24 November 2008 (UTC)

=Plate Buckling= A plate under uniform stress in one dimension (see Figure 1) undergoes deformation according to the following equation:
 * $$u_z(x,y)= \sum_{m=1}^{ \infty }{}\sum_{n=1}^{\infty}\psi_{mn} (x,y) $$ where $$\psi$$ is expressed as
 * $$\psi_z(x,y)=csin(\frac{m\pi x}{a})sin(\frac{n\pi x}{b})$$

and $$u$$ is the out of plane deformation. (Can be thought of as deformation in z direction.) Please note this is for a simply supported plate.

The resulting shape when looking at the plate normal to the xy plane is called the buckling mode shape.

The critical loading, $$P_x$$ is given by the following equation:
 * $$(P_x)=k_c\frac{\pi ^{2}D}{b}$$

 * $$k_c(m,\frac{a}{b})=(\frac{mb}{a}+\frac{a}{mb})^2$$

and

 * $$D=\frac{Eh^3}{12(1-\upsilon ^2)}$$

Where h is the thickness of the plate in the z direction.

The critical buckling stress is:
 * $$(\sigma_{xx})_{cr}=\frac{(P_x)_{cr}}{bh}=k_c \frac{\pi^2D}{b^2h}$$

Components related to the minimum critical value are denoted with a star, e.g. $$\sigma^\star_{cr}$$ The minimum critical buckling stress is dependent on $$k^\star_c$$, which in turn is dependent on $$m^\star$$ by the following equation:
 * $$k_c^\star(a/b)=\min_{m = 1, 2, \cdots}\ k_c(m, a/b)=\min_{m = 1, 2, \cdots}\left( \frac {m b}{a} +\frac{a}{m b}\right)^2 =

\left( \frac{m^\star b}{a}+\frac{a}{m^\star b}\right)^2$$ where
 * $$ m^\star(a/b)=\underset{m=1,2, \cdots}{\rm argmin}\ k_c(m, a/b) =\underset{m=1,2, \cdots}{\rm argmin}\left(\frac{m b}{a}

+ \frac {a}{m b}\right)^2$$ The transition from m to m+1 waves occurs when:
 * $$\frac{a}{b}=\sqrt{m(m+1)}$$

For a square plate with all 4 edges clamped, the buckling mode shape approximates to:
 * $$ \psi(x,y)=\frac{c}{4}\left(1 - \cos \frac{2 \pi x}{a}\right)\left(1 - \cos \frac{2 \pi y}{b} \right)$$

This gives a “bubble” like shape, as seen in the matlab figure below. It makes logical sense since the slope of the buckled shape must approach zero (dz/dx and dz/dy) at all clamped edges.

Shear Load, Simply Supported
Load is denoted as $$N_{xy}$$, which is force per unit length, or stress times h, or plate thickness.

Stored strain energy is equated to the work done by shear load U=W.

The Nxy term is in the equation for strain energy, and the critical Nxy is expressed as:
 * $$\displaystyle

(N_{xy})_{cr} =-\frac{ab D}{32}\frac{\displaystyle\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}C_{mn}^2\left(\frac{m^2 \pi^2}{a^2}+       \frac{n^2 \pi^2}{b^2}\right)^2}{\displaystyle\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\sum_{p=1}^{\infty}\sum_{q=1}^{\infty} C_{mn}C_{pq}\frac{mnpq}{(m^2 - p^2)(n^2 - q^2)}} $$ After further analysis and derivation, crit. buckling stress is:
 * $$(\sigma_{xy})^{\star,5}_{cr}=\pm k^{\star, 5}_c \frac{\pi^2D}{b^2h}, k^{\star,5}_c = \frac{\pi^2}{32}\frac{1}{\lambda\vartheta }$$

=A note on dimensional analysis=

A bracket notation (ex. [m]) is used to indicate the dimensions associated with a variable. Here are some examples:

 * $$[f]=\frac{F}{L}$$

this means the dimensions of "f" are force per length.

Since

 * $$[A]=L^2$$

then

 * $$\left[\frac{f}{A} \right]= \frac{F}{L^{3}}$$

meaning the dimensions of "f" per unit area are force per volume, or length cubed. Note, L is only a unit of measure, not a specific dimension.

The following describes a body force:

 * $$\frac{d\sigma }{dx}+\frac{f(z)}{A}=0\Rightarrow [\sigma ]=\frac{F}{L^2}\Rightarrow \left[\frac{d \sigma}{dx} \right]= \frac{[d \sigma ]}{[dx]}=\frac{F/L^2}{L} = \frac{F}{L^3}$$

To show that axial strain is nondimensional:

 * $$[\epsilon ] = \frac{du}{dx} = \frac{L}{L} = 1...nondimensional...$$

Shear strain is also nondimensional:

 * $$[\gamma ]=\frac{[\epsilon _{yy}]}{[\epsilon _{xx}]}=1...nondimensional...$$

Also note
 * $$\left[\frac{d \sigma_{ij}}{dx_i} \right]=\frac{F}{L^3}=\frac{Force}{Volume}$$

Intro
We introduce a function, $$\phi$$, which plays the role of a potential function. We call this the Prandtl stress function. $$\sigma_{yx}$$ and $$\sigma_{zx}$$ are components of the gradient of $$\phi$$ with respect to y and z.

In wikipedia, a gradient is defined as is a vector field which points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change. In notation, it is defined as
 * $$\bigtriangledown f(x,y,z)= \frac{\partial f}{\partial x}i + \frac{\partial f}{\partial y}j + \frac{\partial f}{\partial z}k$$

where f is the potential.

For our case,
 * $$\sigma_{yx}=\frac{\partial \phi}{\partial z}$$ and
 * $$\sigma_{zx}=-\frac{\partial \phi}{\partial y}$$

We can plug these values into the following equation:
 * $$\frac{\partial \sigma_{yx}}{\partial y}+ \frac{\partial \sigma_{zx}}{\partial z}=0$$

to obtain
 * $$\frac{\partial }{\partial y}\frac{\partial \phi}{\partial z}+\frac{\partial }{\partial z} \frac{(-\partial \phi)}{\partial y}=\frac{\partial^2 \phi}{\partial y \partial z}- \frac{\partial^2 \phi}{\partial z \partial y}=0$$

Phi is continuous and smooth, making the 2nd mixed derivative interchangeable, i.e.
 * $$\frac{\partial^2 \phi}{\partial y \partial z}= \frac{\partial^2 \phi}{\partial z \partial y}$$

These equations can be compared with Mechanics of Aircraft Structures(Sun) Ch. 3 eqs 3.15-3.19. Using the author's notation:
 * $$\frac{\partial \tau_{yz}}{\partial x}-\frac{\partial \tau_{xz}}{\partial y}=2G\theta$$

and
 * $$\frac{\partial^2 \phi}{\partial x^2}+ \frac{\partial^2 \phi}{\partial y^2}=-2G\theta$$

The remainder of section 3.2 brings us to our roadmap equation for J (Roadmap eq 7) and notes that once the Prandtl stress function is solved, GJ, the torsional rigidity (the bar's ability to resist deformation due to torsion) is also found. Torsional analysis of structures fall into two cases - that of structures with a thin walled cross section (also referred to as closed cross section) and those of structures with a solid cross section.

Case 1: Thin Walled Cross Section
A prime example of this case is that of the NACA airfoils we have been studying throughout the semester. For thin walled cross section the boundary conditions for Prandtl stress function, $$ \phi $$, are that $$\phi$$ is a constant on inner and outer surfaces of the cross section. The figure below depicts the inner and outer surfaces, $$ S_1 $$ and $$ S_0 $$, respectively.

The boundary conditions for this case are $$\phi=C_0$$ on $$S_0$$ and $$\phi=C_1$$ on $$S_1$$.

Case 2: Solid Cross Section
This case consists of any bar or beam that has a solid cross section such as the one depicted below.

The constant value of $$\phi$$ at the boundary solid cross sections is arbitrary and so it can be set to zero on that surface. Thus the boundary in this case is $$\phi = 0$$ on $$S_0$$

Uniform Bar With Solid Circular Cross Section
As an example, we will analyze the following bar with a circular cross section as shown below.

For this case, $$\phi$$ can be defined as

 * $$\phi(y,z) = C\left(\frac{y^2}{a^2}+\frac{z^2}{b^2}-1\right)$$

It might also be noted that for the case of a circular cross section a=b. As mentioned before for the case of a solid cross section $$\phi=0$$ on $$S_0$$. Recalling the relation:

 * $$\nabla^2\phi=-2G\Theta$$

an expression for C can be obtained by solving for the laplacian of $$\phi$$.

 * $$\nabla^2\phi = \frac{4C}{a^2} = -2G\Theta$$
 * $$\Rightarrow C = -\frac{1}{2}a^2G\Theta$$

To analyze the Prandtl stress function, $$ \phi $$, as a function of the radius of a circular cross section of diameter $$ 2\alpha $$, consider the definition of the radius as

 * $$ r = \left(y^2 + z^2 \right)^{1/2} $$

Then, a plot of $$ \phi(r) $$ vs. r would yield the figure below

Now, recall from previous lectures the following equations

 * $$ T = 2\int_{A}^{}{\phi dA} = 2C\left(\frac{J}{a^2}-A\right) $$

 * $$ J = \int_{A}^{}{r^2 dA} = \frac{1}{2} \pi a^4 $$

 * $$ \tau = \frac{Tr}{J} $$

Combine these equations with the following

 * $$ \sigma_{yx} = \frac{\partial \phi}{\partial z} = -G \theta z $$

 * $$ \sigma_{zx} = -\frac{\partial \phi}{\partial y} = G \theta y $$

to obtain

 * $$ u_x\left(y,z\right) = 0 $$

This implies that there is no axial displacement at any point along a bar with uniform, solid, circular cross-section. This was assumed to be true during ad hoc derivations of twist; however, we see now that the Theory of Elasticity proves it to be so. A rigorous proof is presented below.

 Error: See my comments in Team VQCrew (with annotation). Eml4500.f08 14:04, 23 November 2008 (UTC)

= Traction Force Equation: Derivation =

Geometric Manipulation Using an Infinitesimal Element
Goal: Using the traction force equation to derive the equation:

$$\displaystyle[t_i]_{3x1}=[\sigma_{ij}]_{3x3}[n_i]_{3x1} $$

Consider the following figure below of an infinitesimal element to be able to be able to derive geometrically, the traction force from the stress tensor and normal vector at the surface of the element.

From the figure the vector $$\boldsymbol\vec{n}$$ has the vector equation: $$\displaystyle \boldsymbol \vec{n}=n_y \boldsymbol \vec{j}+n_z \boldsymbol \vec{k}$$

From this equation, $$\boldsymbol\vec{n}$$ is the unit normal vector and is given as $$\displaystyle ||\boldsymbol \vec{n}||=1$$

Looking at the figure, the elemental geometry helps in the establishing the following relationship between dy and dz:

$$\displaystyle dy=ds*sin(\theta)$$ $$\displaystyle dz=ds*cos(\theta)$$

Additionally, the normal vector equation can be used to rewrite the vector in terms of angle $$\theta$$ and conclude also that:

$$\displaystyle n_y=cos(\theta)$$ $$\displaystyle n_z=sin(\theta)$$

Looking at Figure and continuing through this derivation, the next logical choice would be to take the sum the forces in the y and z directions using the equilibrium equations. It is important to note that the depth into the page is taken as the x-direction and is neglected for this case since the depth is considered to be unity, or 1.

As given in class, the derivation for the equilibrium in the y direction of the sum of forces is given as:

$$\sum {F}_y=0=-\sigma_{yy}*(dz*1)-\sigma_{yz}*(dy*1) +t_y*(d_s*1)$$

Substituting values for dy and dz into the equilibrium equation above, the following equation is produced:

$$\sum {F}_y=0=-\sigma_{yy}ds n_y-\sigma_{yz}ds n_z +t_yd_s$$

Further simplification of the equilibrium equation above and solving for $$\displaystyle t_y : $$

$$\displaystyle t_y= \sigma_{yy}n_y+\sigma_{yz}n_z(1)$$

As part of HW 6, the equilibrium equation for the sum of forces in the z-direction are given below:

Using equations (1) and (2) for tractional force components, we can arrange the equations in the form of a matrix, reflected below to produce equation 3:

$$\begin{Bmatrix} t_y\\ t_z\\ \end{Bmatrix} = \begin{bmatrix} \sigma_{yy} & \sigma_{yz}\\ \sigma_{zy} & \sigma_{zz}\\ \end{bmatrix} \begin{Bmatrix} n_y\\ n_z\\ \end{Bmatrix}(3) $$ Similarly for a general 3D case, a similar matrix may be constructed using indicial notation, shown below:

$$\begin{Bmatrix} t_1\\ t_2\\ t_3\\ \end{Bmatrix} = \begin{bmatrix} \sigma_{11} & \sigma_{12}& \sigma_{13}\\ \sigma_{21} & \sigma_{22}& \sigma_{23}\\ \sigma_{31} & \sigma_{32}& \sigma_{33}\\ \end{bmatrix} \begin{Bmatrix} n_1\\ n_2\\ n_3\\ \end{Bmatrix}(4) $$

An advantage of using this indicial notation allows us to rewrite the entire matrix equation in the form of a single equation that is more compact. This compact notation is shown below:

$$t_i=\sum_{j=1}^3{\sigma_{ij}n_j},i=1,2,3$$

In this equation looking at the stress value, where the subscript i, and j designates the row, and column of the matrix respectively.

Generalizing this formula, produces our end goal and solves for the desired equation: $$\displaystyle [t_i]_{3x1}=[\sigma_{ij}]_{3x3}[n_i]_{3x1} $$

Roadmap Incorporation
Looking at the roadmap on lecture page 16-2, we recall the following three equations:

$$\displaystyle T=2\int\int_{A}^{}{}\phi dA(5)$$ $$\displaystyle T = GJ\theta~(6) $$ $$\displaystyle J=\frac{-4}{\nabla^{2}\phi }\int \int _{A}\phi dA(7)$$

These three equations can be combined, and the equation (8) shown below can be derived by solving the three equations shown above for the Lagrangian of phi.

$$\frac{\partial^2{\phi}}{\partial{y}^2}+\frac{\partial^2{\phi}}{\partial{z}^2}=-2G\theta(8)$$

This becomes important because $$\phi$$ is taken to be constant on the lateral surface of the bar.

= Flexural Shear in Thin-walled Sections =

Motivation
It is known that for a generic loading condition of a beam, there exists not only a bending moment but also a transverse shear load. Consider the beam in the figure below. Some generic loading condition is applied to the straight beam on the left. A cut can be made in the beam at a distance x from the wall. The bending moment and shear loads are drawn for each side of the exposed section, where M(x) represents the moment as a function of distance x and V(x) represents the shear load as a function of distance x.

= Unsymmetrical Thin-walled Cross Section =

General Equation for unsymmetrical cross sections $$\displaystyle \int_{A_s} \frac{\sigma_{xx}}{dx} dA = -q_s $$ Eq. 5.1

Symmetric Cross Section (about y- axis)
$$ \displaystyle \sigma_{xx} = \frac{M_y z}{I_y}$$

$$ \displaystyle q(s) = \frac{-V_z Q_y}{I_y}$$ Eq. 5.2

$$ \displaystyle Q_y = \int_{A_s} z dA = z_c A_s$$

Unsymmetric Cross Section
$$ \displaystyle \sigma_{xx} = (k_y M_z - k_{yz} M_y)y + (k_z M_y - k_{yz} M_z)z$$ → Eq. 5.3

This is true for the following k values from p.26-2 of notes: $$ \displaystyle k_y = \frac{I_y}{D}$$ $$k_{yz} = \frac{I_{yz}}{D}$$ $$k_z = \frac{I_z}{D}$$

The expression for $$\displaystyle \sigma_{xx}$$ can also be put into matrix form:

$$\displaystyle \sigma_{xx} = \begin{bmatrix} z & y\\ \end{bmatrix} \begin{bmatrix} k_z & -k_{yz}\\ -k_{yz} & k_y\\ \end{bmatrix} \begin{Bmatrix} M_y\\ M_z\\ \end{Bmatrix} $$

$$ \displaystyle q(s) = -(k_y V_y - k_{yz} V_z) Q_z - (k_z V_z - k_{yz} V_y) Q_y$$ → Eq. 5.5

$$ \displaystyle => Q_z = \int_{A_s} y dA$$ and $$ \displaystyle Q_y = \int_{A_s} z dA $$

= Matlab Analysis =

The main purpose of this weeks homework problem is to examine the bending stress in the skin of the NACA airfoil. In order to determine this, we will use exact approximations as well as approximations for simply supported and clamped boundary conditions since the airfoil's spars and stringers cause the skin to fall somewhere in between the two.

The first part involves using the previous homework's inertia matrix, centroid, and applied moments to determine the normal bending stress in the skin between the two spars as a function of the y-coordinate using the equations:

 * $$\sigma_{xx}=E\epsilon_{xx}=\frac{I_yM_z-I_{yz}M_y}{I_yI_z-(I_{yz})^2}y+\frac{I_zM_y-I_{yz}M_z}{I_yI_z-(I_{yz})^2}z$$

\begin{bmatrix} I_{22} \\ I_{33} \\ I_{23} \\ \end{bmatrix}=

\begin{bmatrix} 1.298\ x\ 10^{-5}\ m^4 \\ 6.346\ x\ 10^{-7}\ m^4 \\ -9.494\ x\ 10^{-8}\ m^4 \\ \end{bmatrix}$$

 * $$Centroid(y,z)=(0.210\ m,\ 0.210\ m)$$

 The figure does not look right: The normal stress $$\displaystyle \sigma_{xx}$$ is expected to rise and fall with the coordinate $$\displaystyle z$$ of the airfoil skin, i.e., the function would look curved similar to the curve of the airfoil skin, instead of a straight line. See, e.g., Team VQCrew (with annotations) Eml4500.f08 11:36, 26 November 2008 (UTC)

From this we find that the average stresses are $$\sigma_{xx,FB}=26.5\ MPa$$ and $$\sigma_{xx,EH}=32.1\ MPa$$.

Next we wish to explore what shapes the skin would undergo due to these bending stress. For an aspect ratio of 1.5, we can see in Figures XXX and XXX that depending on m and n, different modes of buckling occur. The variables m and n determine the number of half wave lengths in the x and y direction respectively, which is readily apparent in the figures.

We are interested in finding the period T of this function that determines the shape of the buckled plate to prove the m is indeed the number of half waves. This involves finding the period of $$ \sin ( \frac{m \pi x}{a} )$$  by considering   $$\sin ( \frac{m \pi (x + T)}{a} ) = \sin ( \frac{m \pi x}{a} )$$. We know that:

 * $$\sin( \theta ) = \sin( \theta + 2\pi k)=\sin(\frac{m \pi x}{a}+\frac{m \pi T}{a})$$

 * $$Period=T=\frac{2\pi}{\frac{m\pi}{a}}=\frac{2a}{m}$$

For constant a=1, we see that the period is 2/m. For m=1, we see the period is 2 and therefore there is one half of the wave at a=1. For m=2, the period is 1 and two half waves at a=1, so on and so forth.

 I moved this figure to the left, so the figure does not block the link to show the collapsible table containing your matlab code, and added a few blank lines, so this collapsible table does not get pushed too far to the right. Eml4500.f08 11:36, 26 November 2008 (UTC)

In the determining of critical stresses, a variable of large importance is k_c. This number is a function purely of m, the number of half wave lengths, and the aspect ratio of the plate. The figure to the right shows how k_c varies with these variables. This factor is used in critical stress calculations for simply supported plates.

 I moved this figure to the left, so the figure does not block the link to show the collapsible table containing your matlab code, and added a few blank lines, so this collapsible table does not get pushed too far to the right. Eml4500.f08 11:36, 26 November 2008 (UTC)

As we discussed earlier, there are stringers and spars connect to the skin so it is not just simply supported. The other approximation for bending of a flat plate is clamped boundary condition. Here the two ends are not allowed to rotate, making the slope of plate zero at the boundary. The figure to the right shows how this sort of boundary condition affects the buckling mode of the plate.

The final section is interested in modeling the skin section in between the two spars as flat plates and determining the critical stresses in this plate. We will compare the value reached earlier for average stress to each of the two boundary conditions. In this figure, we vary the aspect ratio of plate while holding the width constant to that of the actual skin section. The two cases shown are for the clamped and simply supported boundaries mentioned earlier. We can see that at small aspect ratios, our average stress for the top plate, which the figure was based upon, falls in between the two curves. At higher aspect ratio's however (which is what we would expect for an aircraft wing), the stress is higher the than the critical stress, implying that the skin does not work well to resist bending, which is what we expected.

= Contributing team members =

 * Gonzalo Barcia--Eas4200c.f08.aero.barcia 03:05, 21 November 2008 (UTC)
 * William Kurth--Eas4200c.f08.aero.kurth 14:56, 21 November 2008 (UTC)
 * Melisa Gaar--Eas4200c.f08.aero.gaar 19:21, 21 November 2008 (UTC)
 * Thomas McGilvray--Eas4200c.f08.aero.mcgilvray 19:38, 21 November 2008 (UTC)
 * Ray Strods--Eas4300c.f08.aero.strods 19:19, 21 November 2008 (UTC)
 * Jared Lee--Eas4200c.f08.aero.lee 19:43, 21 November 2008 (UTC)
 * Oliver Oyama--Eas4300c.f08.aero.oyama 19:54, 21 November 2008 (UTC)

title: User:Eas4200c.f08.aero.lee/HW6