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User:Eml4500.f08.Lulz.Layton.Eric/Hw3

Class Notes
Derivation of element force displacement with respect to the global coordinate system (p. 6-1)

$$\textbf{k}^{(e)}\textbf{d}^{(e)}=\textbf{f}^{(e)}$$ (1)

Where k(e) is the 4 x 4 element stiffness for element e, d(e) is the 4 x 1 element displacement of element e, and f(e) is the element force matrix.

A two force body argument example is given below:

The matrix equation for solving for the axial force can be written below:

$$k^{(e)}\hat{\textbf{k}}^{(e)}\textbf{q}^{(e)}=\textbf{p}^{(e)}$$ (2)

$$k^{(e)}\left[\begin{array}{c c} 1 & -1\\ -1 & 1 \end{array}\right] \begin{Bmatrix} q_1^{(e)}\\ q_2^{(e)} \end{Bmatrix}= \begin{Bmatrix} P_1^{(e)}\\ P_2^{(e)} \end{Bmatrix}$$

where:

$$q_i^{(e)}$$ = axial displacement of element e at local node i

$$P_i^{(e)}$$ = axial force of element e at local node i

The goal is to derive Equation (1) from Equation (2). We would like to find a relationship between q(e) and d(e) and between P(e) and f(e)

This relationship can be expressed in the form:

$$\textbf{q}^{(e)}=\textbf{T}^{(e)}\textbf{f}^{(e)}$$

Considering the displacement vector of local node 1:

$$d_1^{(e)}=d_1^{(e)}\overrightarrow{i}+d_2^{(e)}\overrightarrow{j}$$

$$q_1^{(e)}$$ = axial disp of node 1 is the orthogonal projection of the displacement. Vector $$\overrightarrow{d}_1^{(e)}$$ of node 1 on the axis x of elem e.

$$q_1^{(e)}=\vec{d_1^{(e)}}\cdot\vec{\tilde{i}}$$

$$q_1^{(e)}=(d_1^{(e)}\vec{i}+d_2^{(e)}\vec{j})\cdot\vec{\tilde{i}}$$

$$q_1^{(e)}=d_1^{(e)}(\vec{i}\cdot\vec{\tilde{i}})+d_2^{(e)}(\vec{j}\cdot\vec{\tilde{i}})$$

$$(\vec{i}\cdot\vec{\tilde{i}})$$ and $$(\vec{j}\cdot\vec{\tilde{i}})$$ are director cosines

$$q_1^{(e)}=l^{(e)}d_1^{(e)}+m^{(e)}d_2^{(e)}$$

$$q_1^{(e)}=\lfloor\begin{array}{c c} l^{(e)} & m^{(e)} \end{array}\rfloor \begin{Bmatrix} d_1^{(e)}\\ d_2^{(e)} \end{Bmatrix}$$

where $$q_1^{(e)}$$ is a 1 X 1 scalar matrix. similarly for local node 2:

$$q_2^{(e)}=\lfloor\begin{array}{c c} l^{(e)} & m^{(e)} \end{array}\rfloor \begin{Bmatrix} d_3^{(e)}\\ d_4^{(e)} \end{Bmatrix}$$

Therefor:

$$\begin{Bmatrix} q_1^{(e)}\\ q_2^{(e)} \end{Bmatrix}= \left[\begin{array}{c c c c} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{array}\right] \left\{\begin{array}{c} d_1^{(e)}\\ d_2^{(e)}\\ d_3^{(e)}\\ d_4^{(e)} \end{array}\right\}$$

Similarly:

$$\begin{Bmatrix} P_1^{(e)}\\ P_2^{(e)} \end{Bmatrix}=\textbf{T}^{(e)} \left\{\begin{array}{c} f_1^{(e)}\\ f_2^{(e)}\\ f_3^{(e)}\\ f_4^{(e)} \end{array}\right\}$$

Recall the element axial Force-Displacement relationship:

$$\hat{\textbf{k}}^{(e)}\textbf{q}^{(e)}=\textbf{P}^{(e)}$$

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

Now we want to have:

$$\hat{\textbf{k}}^{(e)}\textbf{d}^{(e)}=\textbf{f}^{(e)}$$

So we move $$\textbf{T}^{(e)}$$ to left hand side. Unfortunately T(e) is a rectangular matrix (2x4) so we can't invert.

Therefore we get:

$$[\textbf{T}^{(e)T}\hat{\textbf{k}}^{(e)}\textbf{T}^{(e)}]\text{d}^{(e)}=\textbf{f}^{(e)}$$

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

This equation is given without proof, so a homework problem was given to prove that the left hand side is equal to the right hand side of the equation. The justification for pursuing the above equation is due to the Principle of Virtual Work (PVW). Page 10-1 has the 1st applied use of PVW reduction of global force displacement.

Why not solve like so?

$$\textbf{d}=\textbf{K}^{-1}\textbf{F}$$

This method cannot be pursued because the determinant of K is zero, which causes a singularity when moving to the right hand side.

$$\frac{1}{det\textbf{K}}=\frac10$$

$$\frac{1}{det\textbf{K}}$$ is needed to find the inverse K-1. In an unconstrained structural system, there are 3 possible rigid body motions in 3-D. This includes 2 translational and 1 rotational reactions.

Closing the loop between the FEM and statics is virtual displacement.

The two-bar truss system

Closing the Loop FEM Computational Displacement Computational Reactions Statics

From Page 11-2, the Free Body Diagram of the 2 force bodies. (elem 1,2). Using statics, the reactions are known and the number of forces are then known.

$$P_1^{(1)}$$ and $$P_2^{(2)}$$ are known by statics. In order to compute the axial displacement (amount of extension in the bars) dof, $$q_2^{(1)} $$ must be found.

$$q_2^{(1)} = \frac{|P_1^{(1)}|}{k^{(1)}} = \frac{|P_2^{(2)}|}{k^{(1)}}$$

$$q_2^{(1)} = 0 $$. This is fixed at node 1.

The following picture reveals an undeformed truss system and deformed truss system.

How can we be certain that the above picture represents the displacements correctly?

From this picture we see that:

$$u_y=Rsin(\alpha)$$

$$u_x=R[1-cos(\alpha)]$$

Assuming that alpha is small allows us to assume:

$$u_y=R\alpha$$

$$u_x=R$$

There are two unknowns $$(x_D, y_D)$$

We need equations for line AB and line BC

The following is a general problem to further explain this method:

Where Q can slide anywhere along the line.

$$\textbf{PQ}=(PQ)\overrightarrow{\tilde{i}}=(PQ)[cos\theta\overrightarrow{i}+sin\theta\overrightarrow{j}]$$

$$\textbf{PQ}=(x-x_P)\overrightarrow{i}+(y-y_P)\overrightarrow{j}$$

Therefore:

$$x-x_P=(PQ)cos\theta$$

$$y-y_P=(PQ)sin\theta$$

Further:

$$\frac{y-y_P}{x-x_P}=tan\theta$$

Finally:

$$y-y_P=(tan\theta)(x-x_P)$$

So, an equation for a line perpendicular to the above line, passing through point P is:

$$y-y_P=tan(\theta+\frac{\pi}{2}(x-x_P)$$

3-bar Truss System

For element 1:

$$E^{(1)}=2$$ $$A^{(1}=3$$ $$L^{(1)}=5$$ $$\theta^{(1)}=30$$(deg)

For element 2:

$$E^{(2)}=4$$ $$A^{(2}=1$$ $$L^{(2)}=5$$ $$\theta^{(2)}=30$$(deg)

For element 3:

$$E^{(3)}=3$$ $$A^{(3}=2$$ $$L^{(3)}=10$$ $$\theta^{(3)}=45$$(deg)

If we break down each element into its free-body diagram we get the following 3 pictures:

If we put these 3 pictures together we get:

We cannot solve this problem by statics because we only have 2 useful equations.

$$\Sigma F_x=0$$

$$\Sigma F_y=0$$

The moment equation $$\Sigma M_A=0$$ is trivial and does not help us solve the problem.

But what about taking the moment about an arbitrary point B?

Following the math for this we get:

$$\Sigma\textbf{M}_B=\textbf{BA}\times\textbf{F}$$

$$\Sigma\textbf{M}_B=\textbf{BA}^'\times\textbf{F}$$

$$\textbf{BA}^'=\textbf{BA}\times\textbf{AA}^'$$

$$\Sigma\textbf{M}_B=(\textbf{BA}+\textbf{AA}^')\times\textbf{F}$$

$$\Sigma\textbf{M}_B=\textbf{BA}\times\textbf{F} + \textbf{AA}^'\times\textbf{F}$$

Where: $$\textbf{AA}^'\times\textbf{F}=0$$ which is trivial and doesn't help us.

Now if we look back at the three-bar truss example we see that node A is in equilibrium.

$$\Sigma_{i=0}^3\textbf{F}_i=\overrightarrow{0}$$

So, where A' is any point on the line of action of Fi:

$$\Sigma\textbf{M}_{Bi}=\Sigma\textbf{BA}_i^' x \textbf{F}_i$$

Where $$\Sigma_i\textbf{F}_i=0$$ and is trivial.

Now we need to find the global stiffness matrix. This can be found in the following manner:

So, for example we see:

$$K_{33}=k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)}$$

$$K_{34}=k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)}$$

Axial displacement calculation for local node 2
During meeting 12, a calculation for the axial displacement of local node 1 for an arbitrary element e was considered. What follows is a similar calculation for local node 2 of the same element.

First, we define the displacement vector at local node 2 (Please note the [#] subscript was used in place of the "box" subscript used in class):

$$ \vec{\mathbf{d}^{(e)}_{[2]}} = d_{3}^{(e)} \vec{i} + d_{4}^{(e)} \vec{j} $$

Next, the axial displacement at node 2 is defined: $$ q_{2}^{(e)} = \vec{\mathbf{d}^{(e)}_{[2]}}\cdot\vec{\tilde{i}} $$

$$ q_{2}^{(e)} = (d_{3}^{(e)}\vec{i} + d_{4}^{(e)}\vec{j})\vec{\tilde{i}} $$

$$ q_{2}^{(e)} = d_{3}^{(e)}(\vec{i}\cdot\vec{\tilde{i}}) + d_{4}^{(e)}(\vec{j}\cdot\vec{\tilde{i}}) $$

From a previous homework question and the class notes, the following relations are valid:

$$\vec{\tilde{i}}\cdot\vec{i}=\cos\theta^{(e)}$$ $$\vec{\tilde{i}}\cdot\vec{j}=\sin\theta^{(e)}$$

Substituting,

$$ q_{2}^{(e)} = d_{3}^{(e)}\cos \theta^{(e)} + d_{4}^{(e)}\sin\theta^{(e)} $$

From the notes, the following relations are valid:

$$\cos\theta^{\left(e\right)} = l^{(e)}$$ $$\sin\theta^{\left(e\right)} = m^{(e)}$$

Substituting,

$$ q_{2}^{(e)} = d_{3}^{(e)}l^{(e)} + d_{4}^{(e)}m^{(e)} $$

Putting the above equation yields the following result, consistent with that of the axial displacement of local node 1:

$$ q_{2}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}\begin{Bmatrix} d_{3}^{(e)}\\ d_{4}^{(e)} \end{Bmatrix} $$

Proof of $$\mathbf{k}^{(e)} = \mathbf{T}^{(e)^{T}} \mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}$$
In this question we are asked to prove that the above matrix product generates the generic element stiffness matrix $$\mathbf{k}^{(e)}$$.

First, the three matrices used to generate the product:

$$\mathbf{T}^{(e)} = \begin{bmatrix} l^{(e)}&m^{(e)}&0&0 \\ 0&0&l^{(e)}&m^{(e)} \end{bmatrix}$$

$$\mathbf{T}^{(e)^{T}} = \begin{bmatrix} l^{(e)}&0 \\ m^{(e)}&0 \\ 0&l^{(e)} \\ 0&m^{(e)} \\ \end{bmatrix}$$

$$\mathbf{\hat{k}}^{(e)} = \begin{bmatrix} 1&-1 \\ -1&1 \\ \end{bmatrix}$$

The product of the first two matrices:

$$\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)} = \begin{bmatrix} 1&-1 \\ -1&1 \\ \end{bmatrix} \begin{bmatrix} l^{(e)}&m^{(e)}&0&0 \\ 0&0&l^{(e)}&m^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)}&m^{(e)}&-l^{(e)}&-m^{(e)} \\ -l^{(e)}&-m^{(e)}&l^{(e)}&m^{(e)} \\ \end{bmatrix}$$

The entire product multiplied out:

$$\mathbf{T}^{(e)^{T}}\mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)} = \begin{bmatrix} l^{(e)}&0 \\ m^{(e)}&0 \\ 0&l^{(e)} \\ 0&m^{(e)} \\ \end{bmatrix} \begin{bmatrix} l^{(e)}&m^{(e)}&-l^{(e)}&-m^{(e)} \\ -l^{(e)}&-m^{(e)}&l^{(e)}&m^{(e)} \\ \end{bmatrix} $$ $$ = \begin{bmatrix} l^{(e)^{2}} & m^{(e)}l^{(e)} & -l^{(e)^{2}} & -m^{(e)}l^{(e)} \\ m^{(e)}l^{(e)} & m^{(e)^{2}} & -m^{(e)}l^{(e)} & -m^{(e)^{2}}\\ -l^{(e)^{2}} & -m^{(e)}l^{(e)} & l^{(e)^{2}} & m^{(e)}l^{(e)}\\ -m^{(e)}l^{(e)} & -m^{(e)^{2}} & m^{(e)}l^{(e)} & m^{(e)^{2}} \\ \end{bmatrix} $$

As shown in previous homework question, the element stiffness matrix:

$$\mathbf{k}^{(e)} = \begin{bmatrix} l^{(e)^{2}} & m^{(e)}l^{(e)} & -l^{(e)^{2}} & -m^{(e)}l^{(e)} \\ m^{(e)}l^{(e)} & m^{(e)^{2}} & -m^{(e)}l^{(e)} & -m^{(e)^{2}}\\ -l^{(e)^{2}} & -m^{(e)}l^{(e)} & l^{(e)^{2}} & m^{(e)}l^{(e)}\\ -m^{(e)}l^{(e)} & -m^{(e)^{2}} & m^{(e)}l^{(e)} & m^{(e)^{2}} \\ \end{bmatrix}$$

Thus, it is easily shown that the following relation is indeed true:

$$\mathbf{k}^{(e)} = \mathbf{T}^{(e)^{T}} \mathbf{\hat{k}}^{(e)}\mathbf{T}^{(e)}$$

Eigenvalues of the Global Stiffness Matrix
From Homework 2, the MATLAB code for the two bar truss system produced the global stiffness matrix (K) verified by manual calculations. This matrix was used to determine the eigenvalues. After running the M-file "TwoBarTrussSystem" and the command "format", the following output was produced.

>> K

K =

0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.1875   -0.3248   -0.1875         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

>> eig(K)

ans =

-0.0000  -0.0000    0.0000    0.0000    1.4705   10.0295 The negative sign in front of the first two zeros implies that all four "zeros" are actually very small values. Due to the precision in MATLAB, it is assumed that the "zeros" are essentially zero.

Under this assumption, the result of four zeros as eigenvalues represents the four degrees of freedom of the two bar truss system that are fixed. Those four degrees of freedom are the x and y directional displacements at global nodes 1 and 3.

Comparison of Global F-d relationships and Elemental F-d relationships
From Homework 2, the elemental F-d relationships produced reactions at global nodes 1 and 3 as shown below.

$$ \begin{bmatrix} F_{1} \\ F_{2} \\ F_{5} \\ F_{6} \end{bmatrix}= \begin{bmatrix} -4.4378 \\ -2.5622 \\ 4.4378 \\ -4.4378 \end{bmatrix} $$

For the global F-d relationships method, the following calculations were performed to produce the reactions at all the nodes. Columns 1, 2, 5, and 6 were removed from the global stiffness matrix, and rows of the same rank were removed from the displacement matrix. This is because the four directional displacements of Nodes 1 and 3 are all zero because the nodes are fixed.

$$ d_1 = d_2 = d_5 = d_6 = 0 $$

$$ \left[ \begin{array}{cc} K_{13} & K_{14} \\ K_{23} & K_{24} \\ K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{array} \right]\begin{bmatrix} d_{3} \\ d_{4} \end{bmatrix}= \begin{bmatrix} F_{1} \\ F_{2} \\ F_{3} \\ F_{4} \\ F_{5} \\ F_{6} \end{bmatrix} $$

After running the M-file "TwoBarTrussSystem" and the command "format", the following output was produced. The global stiffness matrix (K) was the same one from the previous section.

>> d

d =

0        0    4.3520    6.1271         0         0

>> K34 = K(1:6,3:4)

K34 =

-0.5625  -0.3248   -0.3248   -0.1875    3.0625   -2.1752   -2.1752    2.6875   -2.5000    2.5000    2.5000   -2.5000

>> d34 = d(3:4)

d34 =

4.3520   6.1271

>> F = K34*d34

F =

-4.4378  -2.5622    0.0000    7.0000    4.4378   -4.4378

The reactions produced using the global F-d relationships method are identical to the results from the elemental F-d relationships method. The additional two values (rows 3 and 4) of the global method verify the load to the system.

Solve for AC and AB
As given in class, AC and AB are dependent on their elemental P's and K's. In HW 2 we solved for both elemental K values and one of the P values: that work is reproduced as appropriate in the proof below

$$AC = \frac{|P^{(1)}_2|}{K^{(1)}} = \frac{\sqrt{\left(f_3^{(1)}\right)^2 + \left(f_4^{(1)}\right)^2}}{\left(\frac{EA}L\right)} = \frac{\sqrt{(4.4378)^2 + (2.5622)^2}} = \frac{5.1243}{0.75} = 6.8324$$

$$AB = \frac{|P^{(2)}_1|}{K^{(2)}} = \frac{\sqrt{\left(f_1^{(2)}\right)^2 + \left(f_2^{(2)}\right)^2}}{\left(\frac{EA}L\right)} = \frac{\sqrt{(-4.438)^2 + (4.438)^2}} = \frac{6.2763}5 = 1.2553$$

Find (x,y) coordinates for $$(x_B, y_B), (x_C, y_C)$$, and $$(x_D,y_D)$$
From the previous homework problem we know that:

AC = 6.8324

AB = 1.255

We also know from the notes that the two following equations are true:

$$x_C=AC\cos(\theta)$$

$$y_C=AC\sin(\theta)$$

$$x_B=AB\cos(\theta)$$

$$y_B=AB\sin(\theta)$$

Knowing that for $$(x_b,y_b)$$ $$\theta$$ equals 135 and for $$(x_c,y_c)$$ $$\theta$$ equals 30 the two coordinates are found to be:

$$(x_B,y_B) = (-.887,.887)$$

$$(x_C,y_C) = (5.917,3.416)$$

The coordinates for $$(x_d,y_d)$$ are obtained using the following equation:

$$AD = (x_D - x_A)i + (y_D - y_A)j$$

Which reduces to:

$$AD = (x_D)i + (y_D)j$$

due to point a being at the origin. AD also equals:

$$AD = d_3i +d_4j$$

Knowing that $$d_3 = 4.35$$ and that $$d_4 = 6.127$$ it then can be determined that:

$$(x_D,y_D) = (4.35,6.127)$$

Global Stiffness Matrix for a Three-bar Truss System
For this portion of the homework we were asked to fill in the global stiffness matrix for the three-bar truss system. The global K matrix follows:

$$\left[\begin{array}{c c c c c c c c} k_{11}^{(1)}&k_{12}^{(1)}&k_{13}^{(1)}&k_{14}^{(1)}&0&0&0&0\\ k_{21}^{(1)}&k_{22}^{(1)}&k_{23}^{(1)}&k_{24}^{(1)}&0&0&0&0\\ k_{31}^{(1)}&k_{32}^{(1)}&(k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)})& (k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)})&k_{13}^{(2)}&k_{14}^{(2)}& k_{13}^{(3)}&k_{14}^{(3)}\\ k_{41}^{(1)}&k_{42}^{(1)}&(k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)})& (k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)})&k_{23}^{(2)}&k_{24}^{(2)}& k_{23}^{(3)}&k_{24}^{(3)}\\ 0&0&k_{31}^{(2)}&k_{32}^{(2)}&k_{33}^{(2)}&k_{34}^{(2)}&0&0\\ 0&0&k_{41}^{(2)}&k_{42}^{(2)}&k_{43}^{(2)}&k_{44}^{(2)}&0&0\\ 0&0&k_{31}^{(3)}&k_{32}^{(3)}&0&0&k_{33}^{(3)}&k_{34}^{(3)}\\ 0&0&k_{41}^{(3)}&k_{42}^{(3)}&0&0&k_{43}^{(3)}&k_{44}^{(3)} \end{array}\right]$$

Matlab Code
The following Matlab code was written to plot the shape of the undeformed and deformed two-bar truss system that was discussed in class. The first part of the code clears all previous data and closes any open figures. The next part is the definition of the inputs. First we define the displacements as obtained from running the Matlab code from HW2, the number of nodes and the number of elements. Next we input the material properties of element 1 and element 2 which have already been defined. Using the lengths and angels of each element, and assuming that global node 1 is at the origin, we populate a position array pos with the x and y coordinates of the local nodes of each element, where row 1 is element 1 and row 2 is element 2. Then, using a simple for loop, we transfer this data two separate column arrays, x and y, that hold the x and y coordinates of each node respectively. The following part sets up the connectivity arrays for element 1 and element 2 using the following formula:

node_conn(local node number, elemnt number) = global node number

Following this we have a for loop which plots the undeformed shape of the two-bar truss system and a couple lines to add text to the figure to label the nodes and elements. To be able to plot the deformed shape, we first need to re-define the position matrix to account for the displacements of each node. We then re-populate the x and y arrays with the new values. Now we can plot the deformed shape using the same basic for loop and the same connectivity arrays.

The Matlab code and the resulting plot follow:

% Matlab Script %***************************************************** % % Purpose: %  Sketch the undeformed and deformed %  two-bar truss system from the clas %  example problem. % % Author: Benjamin Mitchell % % Created: 10-07-2008 00:25:38 EST % %***************************************************** clear; close; d1 = 0;        %disp of node 1 in the x direction d2 = 0;        %disp of node 1 in the y direction d3 = 4.3520;   %disp of node 2 in the x direction d4 = 6.1271;   %disp of node 2 in the y direction d5 = 0;        %disp of node 3 in the x direction d6 = 0;        %disp of node 3 in the y direction %input nodal coordinates num_nodes = 3;              %total number of nodes num_elems = 2;              %total number of elements tot_dofs = 2 * num_nodes;   %total dofs of the system %element 1 details L1 = 4; E1 = 3; A1 = 1; theta1 = 30; %deg thetar1 = (pi * theta1)/180; %element 2 details L2 = 2; E2 = 5; A2 = 2; theta2 = -45; %deg thetar2 = (pi * theta2)/180; %undeformed positions pos(1,:)=[0,L1*cos(thetar1),L1*cos(thetar1) + L2*cos(thetar2)]; pos(2,:)=[0,L1*sin(thetar1),0]; %undeformed nodal coordinate arrays for i=1:num_nodes x(i)=pos(1,i); y(i)=pos(2,i); end %element connectivity array %follows: %node_conn(local node number, elemnt number) = global node number node_conn(1,1)=1;    %element 1 node_conn(2,1)=2; node_conn(1,2)=2;    %element 2 node_conn(2,2)=3; %plot the undeformed 2-bar truss system for i=1:num_elems node_1=node_conn(1,i); node_2=node_conn(2,i); xx=[x(node_1),x(node_2)]; yy=[y(node_1),y(node_2)]; axis([-4 10 -4 10]) plot(xx,yy,'--rs') hold on end text(x(node_conn(1,1)),y(node_conn(1,1))-.5,'Global Node 1','HorizontalAlignment','center') text(x(node_conn(2,1)),y(node_conn(2,1))+.5,'Global Node 2','HorizontalAlignment','center') text(x(node_conn(2,2)),y(node_conn(2,2))-.5,'Global Node 3','HorizontalAlignment','center') text(x(node_conn(2,1))/2,y(node_conn(2,1))/2,'Element 1','HorizontalAlignment','center') text(x(node_conn(2,1))+(x(node_conn(2,2))-x(node_conn(2,1)))/2,(y(node_conn(1,2)))/2,... 'Element 2','HorizontalAlignment','center') %deformed positions pos(1,:)=[0 + d1,L1*cos(thetar1) + d3,L1*cos(thetar1) + L2*cos(thetar2) + d5]; pos(2,:)=[0 + d2,L1*sin(thetar1) + d4,0 + d6]; %deformed nodal coordinate arrays for i=1:num_nodes x(i)=pos(1,i); y(i)=pos(2,i); end %plot the deformed 2-bar truss system for i=1:num_elems hold on    node_1=node_conn(1,i); node_2=node_conn(2,i); xx=[x(node_1),x(node_2)]; yy=[y(node_1),y(node_2)]; plot(xx,yy,'-s') end text(x(node_conn(2,1)),y(node_conn(2,1))-.5,'Global Node 2','HorizontalAlignment','center') text(x(node_conn(2,1))/2,y(node_conn(2,1))/2,'Element 1','HorizontalAlignment','center') text(x(node_conn(2,1))+(x(node_conn(2,2))-x(node_conn(2,1)))/2,(y(node_conn(1,2)))/2,... 'Element 2','HorizontalAlignment','center') title('Deformed and Undeformed Two-bar Truss System') xlabel('X') ylabel('Y')

Contributing Members
Ben Mitchell - Eml4500.f08.Lulz.mitchell.bm 16:46, 8 October 2008 (UTC)

Sam Miorelli - Eml4500.f08.lulz.abcd 19:09, 8 October 2008 (UTC)

Aaron Fisher - Eml4500.f08.Lulz.fisher 19:11, 8 October 2008 (UTC)

Andrew Strack - Eml4500.f08.lulz.strack 19:19, 8 October 2008 (UTC)

Eric Layton - Eml4500.f08.Lulz.Layton.Eric 19:23, 8 October 2008 (UTC)

John Saxon - Eml4500.f08.Lulz.js 19:31, 8 October 2008 (UTC)

title: User:Eml4500.f08.Lulz.Layton.Eric/Hw3


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User:EML4500.f08.Ateam.Miller/homework3

Closing the Loop between FEM and Statics - continued
We continue to back out the displacement DOF's of global node 2. From the previous figure, we can calculate AB and AC.

$$ AB=\frac{|P_1^{(2)}|}{K^{(2)}}=\frac{6.276}{5}=1.2552 $$

$$ AC=\frac{|P_2^{(1)}|}{K^{(1)}}=\frac{5.1243}{0.75}=6.8324 $$

Using these results and the following figure, the location, or (x,y) coordinates of each point can be calculated with trigonometry.

$$ \overrightarrow{PQ}=(\underline{P}Q)\hat{\tilde{i}}=(PQ)[cos\theta\hat{i}+sin\theta\hat{j}] =(x-x_{\underline{p}})\hat{i}+(y-y_{\underline{p}})\hat{j} $$

$$\Rightarrow x-x_{\underline{p}}=(\underline{P}Q)cos\theta $$

$$ y-y_{\underline{p}}=(\underline{P}Q)sin\theta $$

$$\Rightarrow y-y_{\underline{p}}$$

$$\Rightarrow \frac{y-y_{\underline{p}}}{x-x_{\underline{p}}}=tan\theta$$

$$y-y_{\underline{p}}=(tan\theta)(x-x_{\underline{p}})$$

Therefore, the equation for the line perpendicular to the above line passing through P is:

$$y-y_{\underline{p}}=(tan(\theta+\frac{\pi }{2}))(x-x_{\underline{p}})$$

Now, using θ(1}=30° and θ(2}=135° (xB,yB)=(-0.88756,0.88756) and (xC,yC)=(5.91703,3.4162). Finally, it is important to find (xD,yD) to verify that it is equal to displacements calculated with the FEM method. We can calculate (xD,yD) with the equation below.

$$\overrightarrow{AD}=(x_{D}-x_{A})\hat{i}+(y_{D}-y_{A})\hat{j}$$

By choosing the origin to be at point A, xA=yA=0 and the above equation reduces to

$$\overrightarrow{AD}=(x_D)\hat{i}+(y_D)\hat{j}$$

and (xD,yD)=(4.135,6.127).Finally, by the definition of AD,

$$\overrightarrow{AD}=d_3\hat{i}+d_4\hat{j}=4.135\hat{i}+6.127\hat{j}$$

from the FEM method and the loop is closed.

3-Bar Truss System
We now extend the FEM method to a three-bar truss system as shown below.

The properties of the truss system are listed below.

E(1) = 2 ;  A(1) = 3  ;  L(1) = 5

E(2) = 4 ;  A(2) = 1  ;  L(2) = 5

E(3) = 3 ;  A(3) = 2  ;  L(3) = 10

P =30

In order simplify the assembly of the K matrix, a convenient local numbering system must be assigned. A convenient local node numbering system is shown in the figure below.

title: User:EML4500.f08.Ateam.Miller/homework3


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User:Eml4500.f08.delta 6.castrillon/P & K Values Problem

=Problem: Get the p and k values for the stretching of each truss element=

Determining p values
We know that $$p^{(e)}$$ = $$T^{(e)}$$ $$f^{(e)}$$, where

 * $$T^{(e)}$$ = $$\begin{bmatrix}

l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ \end{bmatrix}$$

and, from past exercises,

 * $$f^{(e)}$$ = $$\begin{bmatrix}

-4.4378\\ -2.5622\\ 4.4378\\ 2.5622\end{bmatrix}$$

Therefore, we can say that the p values become:

 * $$p^{(1)}_{2}$$ = |-4.4378cos(30°) + (-2.5622)sin(30°)|= 5.1243
 * $$p^{(2)}_{1}$$ = |4.4378cos(30°) + 2.5622sin(30°)|= 5.1243

Determining k values
From past exercises it is known that the axial stiffness for both bar elements are as follow:

 * $$k^{(e)}$$ = $$\frac{E^{e}A^{e}}{L^{e}}$$

Therefore,

 * $$k^{(1)}$$ = $$\frac{(3)(1)}{4}$$ = 0.75

 * $$k^{(2)}$$ = $$\frac{(5)(2)}{2}$$ = 5

Determining the displacements
Following the same idea of $$d = \frac{f}{k}$$, we have

 * AC = $$\frac{p^{(1)}_{2}}{k^{(1)}}$$ = $$\frac{5.1243}{0.75}$$ = 6.8324

 * AB = $$\frac{p^{(2)}_{1}}{k^{(2)}}$$ = $$\frac{5.1243}{5}$$ = 1.0249


text:
Topic talk:Historiography/Philosophy of history

I propose that this page be moved and put as a subsection of Topic:Historiography. It appears to not be in development, and its focus is very similar to that of the Department of Historiography. There's no point in dividing labor on this one. The Jade Knight (d'viser) 04:15, 7 October 2008 (UTC)
 * I second the proposal. This department can go. Geo.plrd 04:50, 7 October 2008 (UTC)
 * As it has been six days without further discussion, I am going to build a course Introduction to History out of this page. At the conclusion of this process, I am going to redirect this page to Topic:Historiography. Please post here if you have objections ASAP. Geo.plrd 05:29, 13 October 2008 (UTC)
 * Please be aware that here at Wikiversity, it can sometimes take individuals quite a while to respond. That said, I don't want to discourage you from being bold.  The Jade Knight (d'viser) 05:41, 13 October 2008 (UTC)
 * This page will be archived in its entirety at the end of the process (about a week?). Geo.plrd 06:15, 16 October 2008 (UTC)


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Cohen's d

Cohen's d is an effect size used to indicate the standardised difference between two means. It can be used, for example, to accompany reporting of t-test and ANOVA results. It is also widely used in meta-analysis.

Cohen's d is an appropriate effect size for the comparison between two means. APA style strongly recommends use of Eta-Squared. Eta-squared covers how much variance in a dependent variable (DV) is explained by an independent variable (IV), but that IV possibly has multiple levels and hence partial eta-squared doesn't explain the size of difference between each of the pairwise mean differences.

Cohen's d can be calculated as the difference between the means divided by the pooled SD: $$\frac {\text{mean difference}} {\text{standard deviation}} $$ or $$\frac {\text{pooled standard deviation}} $$

Cohen's d, etc. is available in SPSS in the t test procedure and a number of other places.

In an ANOVA, you need to be clear about which two means you are interested in knowing about the size of difference between. This could most likely mean that you are interested in several ds, e.g., to compare marginal totals (for main effects) or cells (for interactions). In general, it is recommended to report all relevant Cohen's d values unless you've got a particular reason to just focus on a one or some of the possible values. From a descriptive statistics table, calculating Cohen's d is relatively straightforward.

Calculating Cohen's d provides useful information for discussion (e.g., allows ready comparison with meta-analyses and the size of effects reported in other studies). Where you are reporting about differences between two means, then a standardised mean effect size (such as d) would be an appropriate accompaniment to inferential testing.

title: Cohen's d


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User:Inon

See My hebrew user page

title: User:Inon


text:
User:Rano~enwikiversity

Hi, My name's Vikki Rano. I'm a Ugandan, African. My goal's to be a prominent Human Resource manager for the biggest company in the world!

title: User:Rano~enwikiversity


text:
Talk:First Line Management/Team building

Copyright violation
I've rolled back this recent anonymous contribution: because a google search for the content reveals copyright violation. Some of the material might be able to added more carefully, with appropriate attribution. -- Jtneill - Talk - c 12:49, 7 October 2008 (UTC)


text:
User talk:Rano~enwikiversity

 Hello Rano, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 12:50, 7 October 2008 (UTC)

 Hello , we are the Baah and Bahh twins and we'd both like to welcome you to Wikiversity. While you sit back and explore Wikiversity, enjoy some of our fresh goat milk. You can help us learn by creating new learning resources or help us by improving the guided tours and other exploration tools. We look forward to seeing you around. Baaaahbye for now.

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Rano. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Rano~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:41, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:18, 19 April 2015 (UTC)

title: User talk:Rano~enwikiversity


text:
User:Laurion

I say. There will not be Wikiversity on Russian soon. "Learn free" is our (Russian) national feature, not yours, folks. So, it's new for you all, but not new for us ;-)

title: User:Laurion


text:
User talk:Laurion

 Hello Laurion, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random random project.
 * Browse everything on Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 13:14, 7 October 2008 (UTC)

 Hi , I am Percy. Nice to meet you. I love the beach, swimming, fishing, and I am currently doing some research at Wikiversity on oceanography. I am also learning to play piano. What do you like? Can I be your mascot?

Mascot? For me? As you wish. I'm not really bothered :-) I'm a Russian industrial, graphic and web designer and musician, I like arts, music, things. I like lots of things, I just can't tell about every one. Laurion 11:43, 8 October 2008 (UTC)

title: User talk:Laurion


text:
User talk:Eduard.orasanu

 Hello Eduard.orasanu, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 13:30, 7 October 2008 (UTC)

 Hi! Our name is Vicky-Verity. We are twins and are Wikiversity mascots. We think Wikiversity is a great place to learn and make friends (when we aren't sleeping). Leave us a message!

title: User talk:Eduard.orasanu


text:
Transactional analysis

 (TA) is a psychodynamic way of understanding human interpersonal relationships and interactions. Basically, it builds on the Freudian notion of Id, Ego, and Superego as three competing parts of our "self" to consider relationships in terms of what happens when people play each of these roles -- in the form of Child (Id), Adult (Ego), and Parent (Superego). TA involves the notion of "hooking" whereby if I play a Child role, it will put you at risk of being hooked into playing Parent. When in a Child-Parent game, someone needs to risk playing Adult, and thereby hook the other person back into an Adult-Adult dynamic. However, being a Child is not to be discouraged - this is about being playful, creative, spontaneous, loving, etc. And sometimes we need to be a Parent - this is about being nurturing, responsible, guiding, supportive, etc. But the "worst" of this dynamic can be, for example, the screaming child and the overbearing parent. Other combinations have other dominant dynamics. The common dynamics are described as "games". There is much more to TA, but this introduction gives the essence of it. Enough, really, for anyone interested to start asking questions, reading, and making observations and analysis, etc.

Key concepts

 * Ego state model (PAC model): ,
 * Transactions, Strokes (Warm Fuzzies & Cold Pricklies), Time structuring
 * Life script
 * Discounting, redefining, symbiosis
 * Rackets, stamps, and games
 * Autonomy

Philosophical assumptions

 * People are OK.
 * Everyone has the capacity to think.
 * People decide their own destiny, and these decisions can be changed.

From these assumptions follow two basic principles of transactional analysis practice:
 * 1) The contractual method: Emphasizes that the transactional analysis practitioner and the client take joint responsibility for achieving whatever change the client wants to make, and
 * 2) Open communication: Means that the client as well as the practitioner should have full information about what is going on in their work together.

Games people play
There are some common dynamics that occur between people. TA describes many of these - such as in Berne's book "Games People Play".

An example dynamic is the Karpman drama triangle.

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User:Eas4200c.f08.blue.a/Lecture 12

Assigned Reading from Mechanics of Aircraft Structures (C. T. Sun): pg. 451 - Shear Panel pg 85-86 - Shear Flow (Eq. 3.49, 3.50) pg. 20 - Fig. 2.2 pg. 64 - Fig, 3.2 pg 115 - Fig. 4.1

The lecture began with a discussion of shear in non-uniform objects such as the one in Figure 1 below.

Figure 1

Then, curved panels were discussed and reference was made to equations 1.4 and 1.5 (Sun, 5). To illustrate the concept of shear in a curved panel, problem 1.5 from the text book (Sun, 17) was discussed.

Problem 1.5
Problem statement: Derive the relations given by equations 1.4 $$\left( V_{x}=\tau ta\right)$$ and 1.5 $$\left( V_{y}=\tau tb\right)$$ in a curved panel as in Figure 2 below.

Figure 2

A Closer look is taken at the z,y plane and the shear flow on the edge of the curve as seen below:

Figure 3

Now let q be equal to the shear flow traveling from point A to point B as shown by the direction of the arrows above in Figure 3. $$q= \left( \tau \right) \left( t \right)$$

The infinitesimal change of shear force $$\left( d\vec{F}\right)$$ is shown below as

$$d\vec{F}=qd\vec{l}=q(dl_{y}\hat{j}+dl_{z}\hat{k})=q(dlcos\theta \hat{j}+dlsin\theta \hat{k})=qdydz$$

A closer look at $$d\vec{F}$$is shown in Figure 4 below.

Figure 4

The Resultant shear force vector is found as

$$\vec{F}=\int_{B}^{A}{d\vec{F}}=q\left(\left(\int_{B}^{A}{dy}\right)\hat{j}+\left(\int_{A}^{B}{dz} \right) \hat{k}\right)$$

Substituting equations 1.4 and 1.5,

$$\vec{F}=q\left(a\hat{j} +b\hat{k}\right)=F_{y}\hat{j}+F_{z}\hat{k}$$

where

$$F_{y}=\left( q\right)\left(a\right)$$ as in equation 3.49b

and

$$F_{z}=\left( q\right)\left(b\right)$$ as in equation 3.49c

yielding

$$\frac{F_{y}}{F_{z}}=\frac{a}{b}$$

The Resultant magnitude is

$$\left|\left|\vec{F} \right| \right|=\sqrt{\left[\left(F_{y} \right)^{2} +\left(F_{z} \right)^{2}\right]}=q\left[a^{2} +b^{2}\right]^{1/2}$$

$$R=\left|\left|\vec{F} \right| \right|$$

$$T=2\left( q\right)\left( A\right)$$

Now, a thin walled cross section can be analyzed in the same way as the curved panel. The thin walled cross section is seen in the z,y plane below.

Figure 5

a close up of the section in question:

Figure 6

It is assumed that the Torque (T) is in the x direction ie. $$\vec{T}=T\hat{i}$$

$$\vec{T}=\vec{r}\bowtie d\vec{F}$$

It follows that

$$dT = \rho dF=\rho \left(qdl \right)$$

and therefore

$$T=\oint_{}^{}{dT}=q\oint_{}^{}{\rho dl}$$

From Figure 6 above, it can be seen that $$ dA=\frac{1}{2}\rho dl$$

Therefore

$$T=\oint_{}^{}{dT}=q\oint_{}^{}{2dA}=2q\oint_{}^{}{dA}=2q\bar{A}$$

Where $$\bar{A}$$ is equal to the average area as measured from the middle of the thickness dl.

title: User:Eas4200c.f08.blue.a/Lecture 12


text: Social psychology (psychology)/Lectures/Subpages


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User:Eas4200c.f08.ZYX/HW Week 5-6

NACA Airfoil: Coordinates and Area in Matlab

Class Notes

Lecture 13 09/24: exam - no lecture

Proving the Area of a Triangle

Polar Moment of Inertia of Circular Cross-Section

title: User:Eas4200c.f08.ZYX/HW Week 5-6


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User:Eml4500.f08.gravy.jad/Notes3.g

FEA Notes and Homework 3

Force Displacement
Derivation of element Force Displacement with respect to global coordinate system.

$$\ k^{(e)}_{4x4} d^{(e)}_{4x1}=f^{(e)}_{4x1} \,$$ equation 1

The figure is labeled with the correct conventions to aid with the solving method. The four forces and displacements are collapsed to two forces and displacements by combining the x and y directions into a single vector.

Force Displacement matrices calculations
The force displacement can be written as

$$ \ K^{(e)} * \begin{bmatrix} 1 & -1 \\ -1 &1 \end{bmatrix}_{\hat{K}_{2x2}^{(e)}}* \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix}_{q_{2x1}^{(e)}} = \begin{Bmatrix} p_1^{(e)} \\ p_2^{(e)} \end{Bmatrix}_{p_{2x1}^{(e)}} \,$$ equation 2

$$\ q_i^{(e)} = \,$$ axial disp.of element at local node i

$$\ p_i^{(e)} = \,$$ axial disp.of element at local node i

The force displacement formula simplifies to the above and the unknowns can be solved for.

Goal: Derive Equation 1 from Equation 2

Want the relationship between $$\underline {q^{(e)}}$$ and $$\underline {d^{(e)}}$$ also $$\underline {p^{(e)}}$$ and $$\underline {f^{(e)}}$$

These relationships can be expressed in the form:

$$\ \underline {q^{(e)}_{2x1}}= \underline { T^{(e)}_{2x4}} * \underline {d^{(e)}_{4x1}} \,$$

To relate the q and d matrices and the p and f matrices a new matrix must be introduced. This matrix is the T matrix.

Consider the displacement vector of local node 1 denoted by

$$ \vec{d^{(e)}_{1}} $$

$$ \vec{d^{(e)}_{1}}= \vec{d^{(e)}_{1}}* \vec{i} + \vec{d^{(e)}_{2}} *\vec{j} $$

$$\ q_i^{(e)} = \,$$ axial displacement of node 1 is the orthogonal projection of the displacement.

Vector $$ \vec{d^{(e)}_{1}} $$ of node 1 on the axis $$\tilde{x}$$ of element $$e^{(1)}$$

$$\ = ( d_1^{(e)}*\vec{i} + d_2^{(e)}*\vec{j}) \cdot \vec{\tilde{i}} \,$$

$$\ = d_1^{(e)}* (\vec{i} \cdot \vec{\tilde{i}}) + d_2^{(e)}*(\vec{j} \cdot \vec{\tilde{i}} ) \,$$

$$\ (\vec{i} \cdot \vec{\tilde{i}} )= cos \theta^{(e)} = l^{(e)} \,$$

$$\ (\vec{j} \cdot \vec{\tilde{i}} )=sin \theta^{(e)} = m^{(e)} \,$$

$$\ q_1^{(e)}=l^{(e)}*d_1^{(e)} + m^{(e)}* d_2^{(e)} \,$$

$$\ = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}_{1x2} * \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix}_{2x1} \,$$

Similarly for node 2 : HW.
$$\ q_2^{(e)}=l^{(e)}*d_1^{(e)} + m^{(e)}* d_2^{(e)}= \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}_{1x2} * \begin{Bmatrix} d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} \,$$

$$ \begin{Bmatrix}q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix}_{\underline{q_{2x1}^{(e)}}} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0& 0\\0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}_{\underline{T_{2x4}^{(e)}}}* \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix}_{\underline{d_{4x1}^{(e)}}}$$

$$\ \underline{ q_{2x1}^{(e)} }= \underline{ T_{2x4}^{(e)} }* \underline{ d_{4x1}^{(e)} } \,$$

Similarly (same argument) :

$$\ \begin{Bmatrix} P_1^{(e)}\\P_2^{(e)} \end{Bmatrix}_{\underline{ P_{2x1}^{(e)} } } = \underline{ T_{2x4}^{(e)} } * \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} _{\underline{ f_{4x1}^{(e)} } } \,$$

$$\underline{ P^{(e)} } = \underline{ T^{(e)}* f^{(e)} }$$

Recall element axial Force Displacement relation:

$$\ \underline{\hat{k}_{2x2}^{(e)}} * \underline{q_{2x1}^{(e)}}= \underline{P_{2x1}^{(e)}} \,$$

$$\underline{\hat{k}^{(e)}} *( \underline{ T^{(e)} } * \underline{ d^{(e)} })_{\underline{q^{(e)}}} = ( \underline{ T^{(e)} } * \underline{ f^{(e)} })_{\underline{P^{(e)}}}$$

Goal: Want to have $$\ \underline{k^{(e)}} * \underline{d^{(e)}}= \underline{f^{(e)}} \,$$

So "move " $$\underline{ T^{(e)}}$$ from the right hand side to the left hand side by multiplying the equation by $$ \underline{ T^{(e) -1}}$$ (inverse of $$\underline{ T^{(e)}} $$)

Unfortunately, $$\underline{ T^{(e)}}$$ is a rectangular matrix (2x4) so you can not invert it.

Ans: $$[\underline{ T_{4x2}^{(e)T} } * \underline{\hat{k}_{2x2}^{(e)}} * \underline{ T_{2x4}^{(e)}} ]_{is the matrix we are looking for} * d_{4x1}^{(e)}= f_{4x1}^{(e)}$$

$$\ \underline{k^{(e)} } * \underline{d^{(e)} }= \underline{f^{(e)} } \,$$

1) $$\ \underline{k^{(e)} } = \underline{T^{(e)T}} * \underline{\hat{k}^{(e)} }* \underline{T^{(e)} } \,$$

HW: verify above.
Verify the following equation:

$$\ \underline{k}^{(e)} = \underline{T}^{(e)T} * \underline{k}^{(e)} * \underline{T}^{(e)}$$

Where: $$\ \underline{T}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}$$ and $$\ \underline{k}^{e)} = k^{(e)} * \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

Since $$k^{(e)}$$ is a scaler, we can move it within the equation without changing the problem. So after moving the scaler and inserting in the matrices, the following equation is found:

$$\ \underline{k}^{(e)} = k^{(e)} * \begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)}\end{bmatrix} * \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} * \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}$$

Solving the first set of matrices gives:

$$\ \underline{k}^{(e)} = k^{(e)} * \begin{bmatrix} l^{(e)} & -l^{(e)}\\ m^{(e)} & -m^{(e)}\\ -l^{(e)} & l^{(e)}\\ -m^{(e)} & m^{(e)}\end{bmatrix} * \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}$$

Solving the second set of matrices gives:

$$\ \underline{k}^{(e)} = k^{(e)}\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -(l^{(e)}m^{(e)})  \\  l^{(e)}m^{(e)} & (m^{(e)})^2 & -(l^{(e)}m^{(e)}) & -(m^{(e)})^2  \\ -(l^{(e)})^2 & -(l^{(e)}m^{(e)}) & (l^{(e)})^2 & l^{(e)}m^{(e)}\\ -(l^{(e)}m^{(e)}) & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2 \end{bmatrix} $$

Which is the definition of the stiffness matrix, therefore the equation is true.

Justification for 1) : PVW (later see 10-1 for fist approach of PVW Reduction of Global Force Displacement relation)

$$ \underline{K_{6x6}}* \underline{d_{6x1}}= \underline{F_{6x1}} which reduces to \underline{K_{2x2}}* \underline{d_{2x2}}= \underline{F_{2x2}}$$

Remark: Why not solve as follows

$$ \underline{d} = \underline{K^{-1}}* \underline{F} $$ ?

The reason is that K can't be unversed because it is singular

Singular matrix has a determinate of 0 so it is not invertible

Recall: need to compute $$\frac{1}{det(K)}$$ to find K-1

Why? For an unconstrained structure system there are three possible rigid body motion in 2-D (2 translation, and 1 rotation)

HW: Eigenvalues of the Global Stiffness Matrix K
For an unconstrained structure system, there are three possible rigid body motionsin 2-D (2 translation and 1 rotation).

Dynamic Evaluation Problem

 K v = λ Mv

$$\ \underline{k} = \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}_{6x6}\,$$

$$\ \ = \begin{bmatrix} \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{49}{16} & \frac{3\sqrt{3} - 40}{16} & -\frac{5}{2} & \frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3} - 40}{16} & \frac{43}{16} & \frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2} \\ 0 & 0 & \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$

MATLAB Code

Values of the stiffness matrix are assigned to matrix K >> K= [9/16 (3*sqrt(3))/16 -9/16 -(3*sqrt(3))/16 0 0; (3*sqrt(3))/16 3/16 -(3*sqrt(3))/16 -3/16 0 0; -9/16 -(3*sqrt(3))/16 49/16 ((3*sqrt(3)-40)/16) -5/2 5/2; -(3*sqrt(3))/16 -3/16 ((3*sqrt(3)-40)/16) 43/16 5/2 -5/2; 0 0 -5/2 5/2 5/2 -5/2; 0 0 5/2 -5/2 -5/2 5/2]

Values of K in decimal form K = 0.5625   0.3248   -0.5625   -0.3248         0         0    0.3248    0.1875   -0.3248   -0.1875         0         0   -0.5625   -0.3248    3.0625   -2.1752   -2.5000    2.5000   -0.3248   -0.1875   -2.1752    2.6875    2.5000   -2.5000         0         0   -2.5000    2.5000    2.5000   -2.5000         0         0    2.5000   -2.5000   -2.5000    2.5000

Matlab function to calculate eigenvalues >> d = eig(K)

List of the eigenvalues for global stiffness matrix K d = -0.0000  -0.0000    0.0000    0.0000    1.4705   10.0295

λ = 0, 0, 0, 0, 1.4705, 10.0295

It is important to note that four of the six eigenvalues are zero. These values represent zero stored elastic energy (rigid body at the modes).

//(end Remark)

HW: FD Rel. on p. 10-1
$$(\underline{k}_{6x2}) (\underline{d}_{2x1}) = \underline{F}_{6x1} $$

$$\ \begin{bmatrix} K_{13}^{(1)} & K_{14}^{(1)} \\ K_{23}^{(1)} & K_{24}^{(1)} \\ (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) \\ (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) \\ K_{31}^{(2)} & K_{32}^{(2)} \\ K_{41}^{(2)} & K_{42}^{(2)} \end{bmatrix}_{6x2}\,$$ $$\ \begin{bmatrix} d_3 \\ d_4 \end{bmatrix}_{2x1}\,$$ $$\ \ = \begin{bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{bmatrix}_{6x1}\,$$

$$\ \begin{bmatrix} -\frac{9}{16} & -\frac{3\sqrt{3}}{16} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} \\ \frac{49}{16} & \frac{3\sqrt{3} - 40}{16}\\ \frac{3\sqrt{3} - 40}{16} & \frac{43}{16} \\ -\frac{5}{2} & \frac{5}{2} \\ \frac{5}{2} & -\frac{5}{2} \end{bmatrix}_{6X2}\,$$ $$\ \begin{bmatrix} 4.3527 \\ 6.1271 \end{bmatrix}_{2x1}\,$$ $$\ \ = \begin{bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_3 \\ F_4 \end{bmatrix}_{6x1}\,$$

$$F_1 = (-\frac{9}{16})(4.3527) + (-\frac{3\sqrt{3}}{16})(6.1271) = -4.4382$$

$$F_2 = (-\frac{3\sqrt{3}}{16})(4.3527) + (-\frac{3}{16})(6.1271) = -2.5624$$

$$F_5 = (-\frac{5}{2})(4.3527) + (\frac{5}{2})(6.1271) = 4.436$$

$$F_5 = (\frac{5}{2})(4.3527) + (-\frac{5}{2})(6.1271) = -4.436$$

Note: F3 and F4 were calculated in previous section to be 0 and 7 respectively

Closing the loop between the FEM and statics: Virtual Displacement
There are two different ways on solving these types of problems: using FEM, or using statics.

Element 1:

Element 2:

These pictures are FBD's of 2 force bodies of element 1 & 2. By Statics reactions are known and therefore the member forces $$\ [P_{1}^{(1)}, P_{2}^{(2)}] \,$$ are equal and opposite in direction. We need to compute the axial displacement dofs ( amount of extension of bars):

$$\ q_{2}^{(1)}= P^{(1)}/k^{(1)}= \underline{AC} \,$$ . ( $$\ q_{1}^{(1)}=0 \,$$ because it's fixed at global node 1.

$$\ q_{1}^{(2)}= -P^{(2)}/k^{(2)}= -\underline{AC} \,$$ . ( $$\ q_{2}^{(2)}=0 \,$$ because it's fixed at global node 3.

Q : How to back out from above results of displacement of dofs of global node 2?

MATLAB code for truss displacement

The above picture is the output graph from MATLab after the above code was run. The solid line represents the deformed truss, while the dashed lines show the initial, or un-deformed, truss.

$$\ u_{y}=R*sin(\alpha) \simeq R*\alpha \,$$

$$\ u_{x}=R*(1-cos(\alpha)) \simeq 0 \,$$ $$\ (cos(\alpha) \simeq 1) \,$$

hence

Here we are assuming that alpha is very small. When we make that assumption sin(alpha) becomes alpha, and cos(alpha) becomes one. The picture shown above is exaggerated, and alpha is very small.

Infinitesimal Displacement (Related to Virtual Displacement)
We can solve for the displacement vectors AC and AB by first referencing all displacement from point A which will be (x,y)=(0,0) on the coordinate system. The actual displacement vector can be defined by the x and y displacement vectors in the form of d(3)(x-coord) and d(4)(y-coord). Points B and C are the axial displacement vectors in regards to element 1 and element 2.

$$\ AC= \frac{|P_{2}^{(1)}|}{k^{(1)}} = \frac{5.1243}{3/4} = 6.8324 \,$$

HW: Solving AC and AB
From previous lecture

Element Length

$$\ L^{(1)}=4 \,$$

$$\ L^{(2)}=2 \,$$

Young's Modules

$$\ E^{(1)}=3 \,$$

$$\ E^{(2)}=5 \,$$

Cross-sectional Area

$$\ A^{(1)}=1 \,$$

$$\ A^{(2)}=2 \,$$

Inclination angle

 * $$\ \theta^{(1)} = 30^o \,$$

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

 * $$\ \theta^{(1)} = 30^o \,$$

Since $$ \ P_{(e)} = T_{(e)} * f_{(e)}\,$$

And the T matrix is

$$\ T^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)}& 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)}\end{bmatrix} \,$$

we can solve for P

The f matrix was solved in a previous lecture and is:

$$\ F= \begin{bmatrix} -4.4378 \\ -2.5622 \\ +4.4378 \\ +2.5622 \end{bmatrix}_{4X1}\,$$

l m and k were also solved in a previous lecture and are:

$$\ l^{(1)}=cos\theta^{(1)}= cos 30^o= \frac{\sqrt{3}}{2} \,$$

$$\ m^{(1)}=sin\theta^{(1)}= sin 30^o= \frac{1}{2} \,$$

$$\ k^{(1)}= \frac{E^{(1)}A^1}{L^1} \frac{\left(3\right ) \left(1\right )}{4}= \frac {3}{4}\,$$

$$\ l^{(2)}=cos\theta^{(2)}= cos -45^o= .707 \,$$

$$\ m^{(2)}=sin\theta^{(2)}= sin -45^o= -.707 \,$$

$$\ k^{(1)}= \frac{E^{(5)}A^1}{L^1} \frac{\left(2\right ) \left(1\right )}{2}= 5\,$$

$$T^{(1)}$$ and $$T^{(2)}$$ are solved as the following:

$$\ T^{(1)} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix} \,$$

$$\ T^{(2)} = \begin{bmatrix} .707 & -.707 & 0 & 0 \\ 0 & 0 & .707 & -.707\end{bmatrix} \,$$

With T and f the P s were found as being,

$$P_{2}^{(1)}$$= $$P_{1}^{(2)}$$=5.1243

so

$$\ AC= \frac{|P_{2}^{(1)}|}{k^{(1)}} = \frac{5.1243}{3/4} = 6.8324 \,$$

$$\ AC= \frac{|P_{1}^{(2)}|}{k^{(2)}} = \frac{5.1243}{5} = 1.0249 \,$$

HW sove (x,y) for points B and C
Knowing that the angle to find x_b and y_b is 30 degrees, and to find x_c and y_c is 135 degrees, we can use simple trigonometry to solve for the values:

$$\ x_i = L*cos(\theta), yi = L*sin(\theta)\,$$

$$\ x_b = AC*cos(30)= 6.8324 * cos(30) = 5.9170, y_b = AC*sin(30) = 6.8324 * sin(30) = 3.4162\,$$

$$\ x_c = AB*cos(135) = 1.0249 * cos(135) = -.7247, y_c = AB*sin(135) = 1.0249 * sin(135) = .7247 \,$$

(xb,yb) = (5.9170, 3.4162) (xc,yc) = (-0.7247, 0.7247)

- P is fixed, Q is "sliding" on the line

- the two lines are parallel $$\ \bar{PQ} =(PQ)*\vec {\tilde i} = (PQ)[cos\theta \vec {i} + sin\theta  \vec {j}] = (x-x_p)\vec i + (y-y_p)\vec j \,$$

$$\ \Rightarrow x-x_p=PQ cos\theta \,$$

$$\ \Rightarrow y-y_p=PQ sin\theta \,$$

$$\ \frac{y-y_p}{x-x_p}= tan\theta \,$$

$$\ (y-y_p)= (tan\theta)(x-x_p) \,$$

eq for a line perpendicular to the line, passing P: $$\ (y-y_p)= \tan(\theta+\frac{\pi}{2})(x-x_p) \,$$

HW solve (x,y) for point D
To find x_d and y_d, the following equation is used:

$$\vec{AD} = x_d\vec{i}+ y_d\vec{j}$$

From HW2, we know that:

$$\vec{AD} = d_3\vec{i} + d_4\vec{j}$$

Also from HW2, we know that d3=4.352 and d4=6.1271. This means that:

$$\vec{AD} = 4.352\vec{i} + 6.1271\vec{j}$$

hence, (xd,yd) = (4.352, 6.1271).

since $$\ x_A=0 and y_A=0 \,$$ then $$\ \vec {AD} =(x_D-x_A)\vec i + (y_D - y_A)\vec j = x_D-\vec i + y_D\vec j \,$$

By definition: $$\ \vec{AD} = d_3 \vec i + d_4 \vec j \,$$ (these are displacement vectors of A from FEM

///end closing loop

3-Bar Truss Syst:

The FEA method of solving for a 3-Bar Truss system is done similarly to the 2-Bar Truss system except with an extra element. A 3-Bar Truss system cannot be solved though statics due to the fact that there are too many unknowns.

$$\ E^{(1)} =2, E^{(2)} =4, E^{(3)} =3 \,$$

$$\ A^{(1)} =3, A^{(2)} =1, A^{(3)} =2 \,$$

$$\ L^{(1)} =5, L^{(2)} =5, L^{(3)} =10 \,$$

$$\ P=30, \theta^{(1)}=30^\circ = -\theta^{(2)}, \theta^{(3)}=45^\circ \,$$

Local node numbering is equivalent to the convenience in assembly k.

Convenient local node numbering

$$ \sum{F_x} =0$$

$$ \sum{F_y} =0$$

$$ \sum{M_A} =0$$ trivial

The sum of the moments about A equals 0 so it is trivial. This means that there are three unknowns and only two equations. So the problem can not be solved using statics and must be solved using FEA.

Question: How about $$\sum{M_B}?$$ It still equals 0

$$ \sum{M_B} = \vec{BA'}\times \vec{F}$$

for A' on line of action of $$\vec{F}$$

$$\vec{BA'}= \vec{BA} + \vec{AA'}$$

$$ \sum{M_B}=(\vec{BA}+ \vec{AA'})\times \vec{F}$$ $$ = \vec{BA}\times F + \vec{AA'}\times \vec{F}= 0 $$

Back to 3-bar truss

Node A is in equilibrium : $$\sum_{i=0}{\vec{F_i}} =\vec{0}$$

$$ \sum_i{M_{B,i}}=\sum_i{\vec{BA'_i}}\times \vec{F_i}$$

A'i= any point on line of action of $$\vec{F_i}$$

$$ \sum_i{M_{B,i}}=\sum_i{\vec{BA}}\times \vec{F_i} \vec{BA} \times \sum_i{\vec{F_i}} = 0 $$

Converting from k matrices to K matrix
The calculations above prove that it doesn't matter what point you choose the moment will still be 0. So no matter what you only have two equations which means the problem is still statically indeterminate.

This figure is a representation of the k matrices forming the K matrix. The third element k matrix is split so that the ks corresponding to node 2 are added to the ks corresponding to node 2 for both element one and two.

Where $$K_{33}= k^{(1)}_{33}+ k^{(2)}_{11}+k^{(3)}_{11} $$ and

$$K_{34}= k^{(1)}_{34}+ k^{(2)}_{13}+k^{(3)}_{12} $$

HW: fill out the rest
$$ \begin{bmatrix} k^{(1)}_{11} & k^{(1)}_{12} & k^{(1)}_{13} & k^{(1)}_{14} & 0 & 0 & 0 & 0\\ k^{(1)}_{21} & k^{(1)}_{22} & k^{(1)}_{23} & k^{(1)}_{24} & 0 & 0 & 0 & 0\\ k^{(1)}_{31} & k^{(1)}_{32} & k^{(1)}_{33}+k^{(2)}_{11}+k^{(3)}_{11} & k^{(1)}_{34}+k^{(2)}_{12}+k^{(3)}_{12} & k^{(2)}_{13} & k^{(2)}_{14} & k^{(3)}_{13} & k^{(3)}_{14}\\ k^{(1)}_{41} & k^{(1)}_{42} & k^{(1)}_{43}+k^{(2)}_{21}+k^{(3)}_{21} & k^{(1)}_{44}+k^{(2)}_{22}+k^{(3)}_{22} & k^{(2)}_{23} & k^{(2)}_{24} & k^{(3)}_{23} & k^{(3)}_{24}\\ 0 & 0 & k^{(2)}_{31} & k^{(2)}_{32} & k^{(2)}_{33} & k^{(2)}_{34} & 0 & 0\\ 0 & 0 & k^{(2)}_{41} & k^{(2)}_{42} & k^{(2)}_{43} & k^{(2)}_{44} & 0 & 0\\ 0 & 0 & k^{(3)}_{31} & k^{(3)}_{32} & 0 & 0 & k^{(3)}_{33} & k^{(3)}_{34}\\ 0 & 0 & k^{(3)}_{41} & k^{(3)}_{42} & 0 & 0 & k^{(3)}_{43} & k^{(3)}_{44} \end{bmatrix} $$

Contributing Team Members
The following students contributed to this report:

Jessica Dyer Eml4500.f08.gravy.jad 01:45, 8 October 2008 (UTC)

Mikal McFarlane Eml4500.f08.gravy.mmm 15:37, 7 October 2008 (UTC)

Patrick Clayton Eml4500.f08.gravy.psc 22:49, 7 October 2008 (UTC)

Sean Sullivan Eml4500.f08.gravy.sms 01:01, 8 October 2008 (UTC)

Brian Eggeman Eml4500.f08.gravy.bje 01:43, 8 October 2008 (UTC)

Justin Tennant Eml4500.f08.gravy.JMT 01:47, 8 October 2008 (UTC)

title: User:Eml4500.f08.gravy.jad/Notes3.g


text:
User:Eas4200c.f08.blue.a/Homework3

Aerospace Structures Team Blue Homework Report # 3

Lecture Notes
Assigned Reading from Mechanics of Aircraft Structures (C. T. Sun): pg. 451 - Shear Panel pg 85-86 - Shear Flow (Eq. 3.49, 3.50) pg. 20 - Fig. 2.2 pg. 64 - Fig, 3.2 pg 115 - Fig. 4.1

The lecture began with a discussion of shear in non-uniform objects such as the one in Figure 1 below.

Figure 1

Then, curved panels were discussed and reference was made to equations 1.4 and 1.5 (Sun, 5). To illustrate the concept of shear in a curved panel, problem 1.5 from the text book (Sun, 17) was discussed.

Problem 1.5
Problem statement: Derive the relations given by equations 1.4 $$\left( V_{x}=\tau ta\right)$$ and 1.5 $$\left( V_{y}=\tau tb\right)$$ in a curved panel as in Figure 2 below.

Figure 2

A Closer look is taken at the z,y plane and the shear flow on the edge of the curve as seen below:

Figure 3

Now let q be equal to the shear flow traveling from point A to point B as shown by the direction of the arrows above in Figure 3.

$$q= \left( \tau \right) \left( t \right)$$

The infinitesimal change of shear force $$\left( d\vec{F}\right)$$ is shown below as

$$d\vec{F}=qd\vec{l}=q(dl_{y}\hat{j}+dl_{z}\hat{k})=q(dlcos\theta \hat{j}+dlsin\theta \hat{k})=qdydz$$

A closer look at $$d\vec{F}$$is shown in Figure 4 below.

Figure 4

The Resultant shear force vector is found as

$$\vec{F}=\int_{B}^{A}{d\vec{F}}=q\left(\left(\int_{B}^{A}{dy}\right)\hat{j}+\left(\int_{A}^{B}{dz} \right) \hat{k}   \right)$$

Substituting equations 1.4 and 1.5,

$$\vec{F}=q\left(a\hat{j} +b\hat{k}\right)=F_{y}\hat{j}+F_{z}\hat{k}$$

where

$$F_{y}=\left( q\right)\left(a\right)$$ as in equation 3.49b

and

$$F_{z}=\left( q\right)\left(b\right)$$ as in equation 3.49c

yielding

$$\frac{F_{y}}{F_{z}}=\frac{a}{b}$$

The Resultant magnitude is

$$\left|\left|\vec{F} \right| \right|=\sqrt{\left[\left(F_{y} \right)^{2} +\left(F_{z} \right)^{2}\right]}=q\left[a^ {2} +b^{2}\right]^{1/2}$$

$$R=\left|\left|\vec{F} \right| \right|$$

$$T=2\left( q\right)\left( A\right)$$

Now, a thin walled cross section can be analyzed in the same way as the curved panel. The thin walled cross section is seen in the z,y plane below.

Figure 5

a close up of the section in question:

Figure 6

It is assumed that the Torque (T) is in the x direction ie. $$\vec{T}=T\hat{i}$$

$$\vec{T}=\vec{r}\bowtie d\vec{F}$$

It follows that:

$$dT = \rho dF=\rho \left(qdl \right)$$

and therefore:

$$T=\oint_{}^{}{dT}=q\oint_{}^{}{\rho dl}$$

From Figure 6 above, it can be seen that $$   dA=\frac{1}{2}\rho dl$$

Therefore:

$$T=\oint_{}^{}{dT}=q\oint_{}^{}{2dA}=2q\oint_{}^{}{dA}=2q\bar{A}$$

Where $$\bar{A}$$ is equal to the average area as measured from the middle of the thickness dl.

 * {| class="toccolours collapsible collapsed" width="100%" style="text-align:left"

! Homework Solution: Triangle Area Proof

Problem: prove that $$A_{BEC}=\frac{1}{2}(ED)(BC)$$

Solution: The area of the triangle BED is calculated from the formula for the area of a right triangle: $$A=\frac{1}{2}bh$$ b= lenght of horizontal base h= length of vertical height Thus, the area of triangle BED is:

$$A_{BED}=\frac{1}{2}(ED)(BD)$$

and the area of the interior right triangle CED is:

$$A_{CED}=\frac{1}{2}(ED)(CD)$$

Therefore, the area of triangle BEC can be found by subtracting these two areas:

$$A_{BEC}=\frac{1}{2}(ED)(BD)-frac{1}{2}(ED)(CD)=\frac{1}{2}(ED)(BC)$$ |}

For the open thin walled cross section:

This means that: $$T=2q\bar{A}$$ and that the reation: J=\frac{1}{2}\pi a^{4}

Uniform Bar with Circular Cross Section
Nonwarping case: Cross Section behaves as rigid disks

For this case, Torque is calculated thus:

$$T=\int_{A}^{}{\int_{}^{}{}r\tau dA}$$  with   $$\tau =G\gamma $$

and define:

$$\gamma =\frac{rd\alpha }{dx}$$

Therefore: $$T=\int_{A}^{}{rG(r\theta )dA}$$ where G and $$\theta$$ are independent of variables y and z.

This means that the above formula can be written as: $$T=G\theta (\int_{A}^{}{r^{2}dA})$$.

The integral enclosed in brackets is known as the polar moment of inertia, J. if the radius of a circular cross section is defined as a, then J can written as:

$$J=\frac{1}{2}\pi a^{4}$$

Now consider the case of a hollow circular cross section with thin walls (t<<a)

Let: $$r_{inner}=a$$ and $$r_{outer}=b$$

This yields a polar moment of inertia of:

$$J=\frac{1}{2}\pi (b^{4}-a^{4})$$

and with some manipulation:

$$J=\frac{1}{2}\pi(b-a)(b+a) (b^{2}-a^{2})$$

Where (b-a) can be approximated as t and (b+a) is approximately $$2r\bar{}$$.

in addition we assume:

$$b^{2}\cong r\bar{}^{2}$$ $$a^{2}\cong r\bar{}^{2}$$

Substituting yields the eqution:

$$J=2\pi tr\bar{}^{3}$$

rearranging gives the equation:

$$J=2\pi ^{-\frac{1}{2}}t(\pi r\bar{}^{2})^{\frac{3}{2}}$$

where $$\pi r\bar{}^{2}=A\bar{}$$. Therefore, J is proportional to $$A\bar{}^{\frac{3}{2}}$$ with $$(2\pi ^{-\frac{1}{2}}t)$$ as the proportionality factor.

Open Thin Walled Cross Sections
Imagine a uniform bar with circular cross section that is said to be non-warping:

The cross section behaves as a rigid disk

The relation we will take as the connection between Torque that is applied and the resulting twist produced with respect to the strength of the material is called Hooke's Law. Hooke's Law can be derived from the equation of torque by setting certain variables as constants.

where

and

which can then be represented as

or

where a=the radius of the circular cross section

You will notice this equation bears a striking resemblance to the equation for the 2nd area moment of inertia of a body. In this case, though, it is in polar coordinates, and is considered the 2nd polar area moment of inertia. The 2nd polar area moment of inertia is also denoted as J.

Hollow Thin Wall Cross Section
Ri = inner radius | Ro = outer radius

are approximations which give us

or

by arranging the equation this way, we can now see that J is proportional to the area to the 3/2 power.

$$\theta = \frac{\alpha}{x} $$ rate of twist    (1)

$$u_{y}=-\theta \times z $$ is the horizontal portion and

$$u_{z}=+(PP' )\cos \beta$$ is the vertical portion

$$=\alpha y_{p}$$ where $$\alpha = \theta x$$

$$u_{z}=+\theta \simeq y$$    (2)

z is the displacement of a point on the cross section

For the warping displacement along the x-axis is

$$ u_{x} = \theta \psi(y, z)$$    (3)

The kinematic assumptions are equations (1), (2), and (3)

After the following road map, this derivation will continue.

Road map for torsional analysis of an aircraft wing
A. Kinematic Assumption

B. Strain Displacement Relationship

C. Equilibrium Equations

D. Prandth Stress Function $$/phi$$

E. Strain Compatibility Equations

F. Equation for $$\phi$$

G. Boundary Condition for $$\phi$$

H. $$T = 2\int \int \phi dA $$

T = GJ$$\theta$$ J = moment of inertia for a circular cross section Moment of inertia for a non circular cross section is    $$J = \frac{-4}{\bigtriangledown ^{2}\phi}\int \int \phi dA$$

I. Thin walled cross section

Ad-Hoc Assumption Formal Derivation $$T = 2q\bar{A}$$

J. Twist angle $$\theta$$: Method 1

$$\theta = \frac{1}{2G\bar{A}} \oint_{}^{}{}\frac{q}{t}ds$$ s = curvilinear coordinate along this wall

K. Sec. 3.6 on multi cell thin walled cross-section

Continuing the discussion on multicell torsion: The two important equations for single cell torsion are $$T=2*q*A$$ and $$\theta=\frac{1}{2(G)(A)}\oint\frac{q}{t}dl$$

Which is the torque and angle of twist respectively of the cell.

When moving to a multicell view in which:

cell $$i=1,....,n_{cell}$$

Torque on the multicell system now becomes equal to the sum of all the individual cell torques.

$$T=2*\sum{q_i*A_i}$$

in which:

$$q_i$$ is the shear flow in cell i

$$A_i$$ is the "average" area in cell i

so therefore:

$$T_i=2*q_i*A_i$$

so the torque on the for the entire multicell system would then be the sum of all the $$T_i$$

$$T=\sum_{n=1}^{n_c}T_i$$

the same approach is used for the angle of twist

$$\theta$$

in which:

$$\theta= \theta_1,.....,\theta_i$$

Therefore:

$$\theta_i=\frac{1}{2*G_i*A_i}\oint\frac{q_i}{t_i}dl$$ in which $$G_i$$ is the shear modulus of cell i   and $$t_i$$ is the thickness of cell i

So therefore the angl$$e$$ of twist for the entire multicell system would then be the sum of all the

$$\theta_i$$ $$\theta=\sum_{n=1}^{n_c}\theta_i$$

Example

$$t_1=.008m$$ $$t_2=t_3=.01m$$ $$a=4m$$ $$b=2m$$

Therefore the area will equal the sum of the semicircle and triangle which make up the overall airfoil shape.

$$A=\frac{\pi*(\frac{b}{a})^2}{2}+\frac{1}{2}*b*a=5.5708m $$ and $$T=2*q*A$$

So therefore:

$$q=\frac{T}{2*A}$$

in which T is the variable, so the twist angle will equal:

$$\theta=\frac{1}{2*G*A}\sum_{j=1}^{3}\frac{q_j*l_j}{t_j}$$

in which l is the length of the path.

NACA Airfoil Matlab Code
functi on NACA_Airfoil_Generator(m,p,t)

% This function takes the 4 digit NACA airfoil digits and plots the % airfoil. It then plots the centroid of the airfoil and calculates the % Area of the airfoil.

% m = maximum camber in percentage of chord length. % p = position of maximum camber in tenths of the chord length. % t = maximum airfoil thickness in percentage of the chord.

% Chord Length = 1

% Converting the NACA numbers into useable values for calculations.

m = 0.01*m; p = 0.1*p; t = 0.01*t;

% Creating the x input arrays.

x = [0:0.01:1];

x_c1 = [0:0.01:p]; x_c2 = [p+0.01:0.01:1];

% Calculating the y coordinates of the mean camber line.

y_c1 = (m/(p^2)).*(2*p.*x_c1-(x_c1.^2)); y_c2 = (m/((1-p)^2)).*((1-2*p)+(2*p.*x_c2)-(x_c2.^2));

y_c = cat(2, y_c1, y_c2)

% Computing the thickness distribution of the airfoil

y_t = 5*t*((0.2969.*sqrt(x))-0.1260.*x-(0.3516.*(x.^2))+(0.2843.*(x.^3))-(0.1015.*(x.^4)))

% Taking the derivative of the camber line. dy_c1 = (m/(p^2)).*(2*p-(2.*x_c1)); dy_c2 = (m/((1-p)^2)).*((1-2*p)+(2*p)-(2.*x_c2));

dy_c = cat(2,dy_c1, dy_c2)

% Computing the angle between the x-axis to the tangent of the mean camber % line.

theta = atan(dy_c)

% Computing the upper and lower coordinates of the airfoil

x_u = x - y_t.*sin(theta); y_u = y_c + y_t.*cos(theta);

x_l = x + y_t.*sin(theta); y_l = y_c - y_t.*cos(theta);

% Plotting the mean camber line, upper coordinates, and lower coordinates % of the airfoil. Calculating and plotting the centroid as a cross-hair.

% Calculating the centroid ns = 100; x_centroid = 0; y_centroid = 0; total = 0;

% Upper airfoil section for i = 1:ns dt = (x_u(i+1)-x_u(i))*(y_u(i+1)+y_u(i))*0.5; d_centroid_x = (x_u(i+1)+x_u(i))*0.5; d_centroid_y = (y_u(i+1)+y_u(i))*0.5; x_centroid = x_centroid + (dt*d_centroid_x); y_centroid = y_centroid + (dt*d_centroid_y); total = total + dt; end

% Lower Airfoil Section for i = 1:ns dt = (x_l(i+1)-x_l(i))*(y_l(i+1)+y_l(i))*0.5; d_centroid_x = (x_l(i+1)+x_l(i))*0.5; d_centroid_y = (y_l(i+1)+y_l(i))*0.5; x_centroid = x_centroid + (dt*d_centroid_x); y_centroid = y_centroid + (dt*d_centroid_y); total = total + dt; end

centroid_x = x_centroid/total centroid_y = y_centroid/total

% Plotting

plot(x,y_c) hold on plot(x_u,y_u) plot(x_l,y_l) plot(centroid_x, centroid_y, '+') axis on hold off

% Calculating the value of A-bar.

ns = 100; x_cent = 0; y_cent = 0; a_bar = 0;

% Upper airfoil section for i = 1:ns r = [x_u(i) - x_cent, y_u(i) - y_cent, 0]; dl = [x_u(i+1) - x_u(i), y_u(i+1) - y_u(i), 0]; d_abar = 0.5.*cross(dl,r); a_bar = a_bar+d_abar; end

% Now for the lower airfoil section for i = 1:ns r = [x_l(i) - x_cent, y_l(i) - y_cent, 0]; dl = [x_l(i+1) - x_l(i), y_l(i+1) - y_l(i), 0]; d_abar = 0.5.*cross(dl,r); a_bar = a_bar+d_abar; end

From the code, the calculated A-bar for the NACA 2415 airfoil is 0.1020.

Team Members that Contributed to this Report
Mark Barry Eas4200c.f08.blue.a 04:09, 8 October 2008 (UTC)

Jeffrey Shea Eas4200c.f08.Blue.E 17:38, 7 October 2008

Christian Garabaya Eas4200c.f08.blue.d 12:58, 8 October 2008 (UTC)

Thomas Sexton EAS4200c.f08.blue.c 13:14, 8 October 2008 (UTC)

Marc Pelletier EAS4200c.f08.blue.b 16:16, 8 October 2008 (UTC)

Clay Robertson eas4200c.f08.blue.f 23:25, 8 October 2008 (UTC)

title: User:Eas4200c.f08.blue.a/Homework3


text:
User:Stevenarntson/English Composition/Vispo

"VisPo" is short for "Visual Poetry." While many people think about poetry as something that is read aloud, poems are also often just looked at on the page. For this reason, some poets have become interested in creating poetry as part of visible, instead of audible, culture. The worlds of typography, visual art, and poetry all intersect here. Below are some links that will take you to some interesting examples of VisPo on the internet. We'll look at some others in class as well.

Origins of Written Language

 * Mysteries of the Alphabet by Mark Alain-Ouaknin

Visual Poetry

 * "Swan and Shadow" by John Hollander, and an anagrammatic version by Richard Grantham
 * l(a by ee cummings
 * Calligrammes by Guillaume Apollinaire
 * "Constellations" by Eugen Gomringer
 * "The Sorrows of Young Werther (Goth Phase)" by Donato Mancini.
 * Poems That Go
 * Evidence that Everything Is Exploding
 * Fossil Sky by David Hinton.
 * Primiti Too Ta, an excerpt from Ursonate by Kurt Schwitters. This is a great visualization of Schwitters' experimental sound poetry.


text:
User:Eml4500.f08.ateam.mcnally/Week 5

Meeting 16 on Wednesday October 1st 2008 was a continuation of the “closing the loop” infinitesimal displacement lecture given during meeting 15 on September 29 2008. The image below is the infinitesimal displacement of the two bar truss system outlined in previous lectures. . The infinitesimal displacements are very closely related to virtual displacement from the principal of virtual work.

It is important to note in the image the coordinate system (in red) is taken to be at point A which correlates to the Global node 2 of the two bar truss system. The black vector ending at point D is the virtual displacement vector.

The length of the displacements AC and AB of the deformed shape are then solved for using the following equations.

$$

AC = \frac{|P_{1}^{(1)}|}{k^{(1)}} =\frac{5.1243}{3/4} = 6.8324 $$

$$ AB = \frac{|P_{1}^{(2)}|}{k^{(2)}} =\frac{6.276}{5} = 1.2552

$$

The next step is to solve for the (x,y) coordinates of point B and Point C using the following equations. $$ X_{B} = ABcos(135) = 1.2552cos(135) = -.88756 $$

$$ Y_{B} = ABsin(135) = 1.2552sin(135) = .88756 $$

$$ X_{C} = ACcos(30) = 6.8324cos(30) = 5.91703 $$

$$ Y_{C}=ACsin(30) = 6.8324sin(30) = 3.4162 $$

The final two unknowns are XD and YD which can be solved for with the following image and equations.

The two unknowns are (XD, YD) so we need an equation for the line AB and the line BC

Note:Θ=a

$$

PQ = (PQ)i^{~} = (X-X_{P})i + (Y-Y_{P})j $$

$$

(PQ)i^{~} = PQ[ cos(a)i + sin(a)j ] $$

Therefore

$$ X-X_{P} = PQcos(a) $$

$$ Y-Y_{P} = PQsin(a) $$

$$ \frac {Y-Y_{P}}{X-X_{P}} = tan(a) $$

$$ Y-Y{P} = (tan(a))(X-X_{P}) $$

and the equation for a line perpendicular to the above line passing through P

$$ Y-Y_{P} = tan(a + pi/2)(X-X{P}) $$

And by definition and from FEM in which we solved for d3 and d4

$$ AD = d_{3}i + d_{4}j $$

Therefore $$ X_{D} = 4.35 $$ and $$ Y_{D} = 6.125 $$

The lecture concluded with the start of the Finite Element Method ‘’’FEM’’’ for a three bar truss. The following data corresponds the Global Picture of the three bar truss.

$$ P=30 $$

$$ \begin{matrix} E^{(1)}=2 & E^{(2)}=4 & E^{(3)}=3\\ A^{(1)}=3 & A^{(2)}=1 & A^{(3)}=2\\ L^{(1)}=5 & L^{(2)}=5 & L^{(3)}=10 \end{matrix}

$$

The corresponding element picture is a follows.

--Eml4500.f08.ateam.mcnally 17:33, 7 October 2008 (UTC)

title: User:Eml4500.f08.ateam.mcnally/Week 5


text:
User:Eml4500.f08.wiki1.aguilar/hw3

Derivation of Elemental Free-body Diagram with respect to the Global Coordinate System

 * $$ \textbf{k}_{4x4}^{(e)}\textbf{d}_{4x1}^{(e)}=\textbf{f}_{4x1}^{(e)}

$$

$$ {k}^{(e)} \begin{bmatrix} 1&-1\\ -1&1 \end{bmatrix} \begin{Bmatrix} {q}_{1}^{(e)}\\ {q}_{2}^{(e)} \end{Bmatrix} = \begin{Bmatrix} {P}_{1}^{(e)}\\ {P}_{2}^{(e)} \end{Bmatrix} $$

$${q}_{i}^{(e)}=$$ axial displacement of element 'e' at local node 'i' $${P}_{i}^{(e)}=$$ axial force of element 'e' at local node 'i'

We want to find the relationship between $$\textbf{q}^{(e)}$$ and $$\textbf{d}^{(e)}$$, and $$\textbf{P}^{(e)}$$ and $$\textbf{F}^{(e)}$$.

These relationships can be expressed in the form: $$ \textbf{q}_{2x1}^{(e)}\textbf{T}_{2x4}^{(e)}=\textbf{d}_{4x1}^{(e)} $$

Consider the displacement vector of the local node 1 denoted by $$\vec{d}_{1}^{(e)}$$.

$$\vec{d}_{[1]}^{(e)}={d}_{1}^{(e)}\vec{i}+{d}_{1}^{(e)}\vec{j}$$

$${q}_{1}^{(e)}$$ = axial displacement of node 1 is the prthagonal projection of the displacement vector $$\vec{d}_{1}^{(e)}$$ of node 1 on the $$\tilde{x}$$ axis of element 'e'.

$${q}_{1}^{(e)}=\vec{d}_{1}^{(e)}+\vec{\tilde{i}}$$

$${q}_{1}^{(e)}=({d}_{1}^{(e)}\vec{i}+{d}_{2}^{(e)}\vec{j})+\vec{\tilde{i}}$$

$${q}_{1}^{(e)}={d}_{1}^{(e)}(\vec{i}\cdot\vec{\tilde{i}})+({d}_{2}^{(e)}(\vec{j}\cdot\vec{\tilde{i}})$$

$$(\vec{i}\cdot\vec{\tilde{i}})=\cos\theta^{(e)}=l^{(e)}$$

$$(\vec{i}\cdot\vec{\tilde{i}})=\sin\theta^{(e)}=m^{(e)}$$

$${q}_{1}^{(e)}=l^{(e)}{d}_{1}^{(e)}+m^{(e)}{d}_{2}^{(e)}$$

$${q}_{1}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)}\end{bmatrix}_{1x2}\begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}  \end{Bmatrix}_{2x1}$$

Here we can see that $${q}_{1}^{(e)}$$ is a 1x1 scalar.

We do this for node 2 as well. $${q}_{2}^{(e)}= \begin{bmatrix}l^{(e)} & m^{(e)}\end{bmatrix}_{1x2}\begin{Bmatrix}{d}_{3}^{(e)}\\{d}_{4}^{(e)}  \end{Bmatrix}_{2x1}$$

which leads us to:

$$\begin{Bmatrix} {q}_{1}^{(e)}\\{q}_{2}^{(e)}\end{Bmatrix}= \begin{bmatrix} l^{(e)}& m^{(e)}&0&0\\0&0& l^{(e)}& m^{(e)} \end{bmatrix} \begin{Bmatrix}{d}_{1}^{(e)}\\{d}_{2}^{(e)}\\{d}_{3}^{(e)}\\{d}_{4}^{(e)} \end{Bmatrix}$$

where this is the equation we set out to derive, $$ \textbf{q}_{2x1}^{(e)}\textbf{T}_{2x4}^{(e)}=\textbf{d}_{4x1}^{(e)} $$

Using the Global Free-body Diagram Relationship
Using d3, and d4 in $$ \textbf{K}\textbf{d}=\textbf{F} $$ and using boundary conditions to eliminate rows in the global displacement matrix we get:

$$ \begin{bmatrix} K_{1,3} & K_{1,4} \\ K_{2,3} & K_{2,4} \\ K_{3,3} & K_{3,4} \\ K_{4,3} & K_{4,4} \\ K_{5,3} & K_{5,4} \\ K_{6,3} & K_{6,4} \end{bmatrix}_{6x2} \begin{Bmatrix}d_3\\d_4\end{Bmatrix}_{2x1}=\begin{Bmatrix}F_1\\F_2\\F_3\\F_4\\F_5\\F_6\end{Bmatrix}_{6x1}$$

Note: We really only need to do the computations for rows 1,2,5, and 6 to get F1,F2,F5,and F6 because computation from rows 3 and 4 gives the applied load which is already known.

Closing the Loop between FEM and Statics: Virtual Displacement
Two-bar truss system:

Since our two-bar truss system is statically determinant because we can view element 1 and 2 as two force bodies. By statics, the reactions are known and therefore, so are the member forces $$P^{(1)}_1$$, and $$P^{(1)}_1$$.

We then compute axial displacement degrees of freedom. This is the amount of extension within the bars.

$${q}_{2}^{(1)} = \frac{{P}_{1}^{(1)}}{{k}^{(1)}}= \frac{{P}_{2}^{(1)}}{{k}^{(1)}}=AC$$

$${q}_{1}^{(1)} = 0$$    (Node 1 is fixed)

$${q}_{2}^{(1)} = \frac{{-P}_{2}^{(2)}}{{k}^{(2)}}=AB$$

$${q}_{2}^{(2)} = 0$$    (Node 2 is fixed)

How do we back out the displacement DOFs of node 2 from above results?

Note: The displacement of node 2 is in the direction of vector D.

title: User:Eml4500.f08.wiki1.aguilar/hw3


text:
User:EML4500.f08.RAMROD.E/HW3

Lecture Notes: Meeting 14 - 9/26/08
In order to relate the axial coordinate system's forces, f(e), to the element coordinate system's axial forces, p(e), it is necessary to employ a transformation matrix, T(e) such that:

$$\begin{Bmatrix}

P^{(e)}_{1}\\ P^{(e)}_{2} \end{Bmatrix} = T^{(e)}*

\begin{Bmatrix}

f^{(e)}_{1}\\ f^{(e)}_{2}\\ f^{(e)}_{3}\\ f^{(e)}_{4} \end{Bmatrix}$$

Looking at this equation, we can tell that T(e) must be a 2X4 matrix in order for the dimensions to work out.

Knowing this, and using the element F-D relationship,

$$\hat{k}^{(e)}*q^{(e)} = p^{(e)}$$

we can substitute in the Transformation Matrix relationship above to get an equation relating the cartesion forces and displacements via the transformation matrix and \hat{k}. The resulting equation is shown below.

$$\hat{k}^{(e)}*(T^{(e)}*d^{(e)}) = T^{(e)}$$

Rearranging this equation requires moving the T(e) term to the left side to allow us to solve for the internal forces in f(e). Since T(e) is rectangular, the transpose, rather than the inverse, is multiplied by each side resulting in:

$$[T^{(e)T}*\hat{k}^{(e)}*(T^{(e)}]*d^{(e)} = f^{(e)}$$

In order for this to be true, it must be proven that,

$$k^{(e)}= [T^{(e)T}*\hat{k}^{(e)}*(T^{(e)}]$$

This proof is given below as the assigned homework.

Now that we have proven the derivation of k^e, we can continue to solving for the global displacement degrees of freedom.

Using the principle of virtual work (i.e. d1 = d2 = d5 = d6 = 0) we can reduce the global equation to two unknown DOF's, d3 and d4.

The next homework assignment was to solve for the eigenvalues of the global stiffness matrix, K.

title: User:EML4500.f08.RAMROD.E/HW3


text:
User:Eml4500.f08.jamama.adam/homework3

Meeting 15
The previous simplification that was applied to the Stiffness Matrix was done by application of the boundary conditions as follows:

$$\left[\begin{array}{cccccc} K_{11}&K_{12}&K_{13}&K_{14}&K_{15}&K_{16}\\ K_{21}&K_{22}&K_{23}&K_{24}&K_{25}&K_{26}\\ K_{31}&K_{32}&K_{33}&K_{34}&K_{35}&K_{36}\\ K_{41}&K_{42}&K_{43}&K_{44}&K_{45}&K_{46}\\ K_{51}&K_{52}&K_{53}&K_{54}&K_{55}&K_{56}\\ K_{61}&K_{62}&K_{63}&K_{64}&K_{65}&K_{66}\end{array}\right] \left[\begin{array}{c}0\\0\\d_3\\d_4\\0\\0\end{array}\right]= \left[\begin{array}{cc} K_{13}&K_{14}\\K_{23}&K_{24}\\K_{33}&K_{34}\\K_{43}&K_{44}\\K_{53}&K_{54}\\K_{63}&K_{64} \end{array}\right] \left[\begin{array}{c}d_3\\d_4\end{array}\right]= \left[\begin{array}{c}F_1\\F_2\\F_3\\F_4\\F_5\\F_6\end{array}\right]$$

F3 and F4 being the known applied loads, the computation for rows 3 and 4 is no longer needed for it would only return F3 and F4. The only needed computations would be for rows 1, 2, 5, 6 to get F1, F2, F5, and F6

To close the loop between Statics and Finite Element Methods, Virtual Displacement can be used as follows:

Two bar Truss syst:

By statics, the reactions forces are known leading to the member forces P1(1), P2(1), component axial displacement degrees of freedom:

$$\begin{array}{l} q_1^{(1)}=\dfrac{P_1^{(1)}}{k^{(1)}}=\dfrac{P_2^{(1)}}{k^{(1)}}=AC\\ q_1^{(1)}=0,\, fixed\, node\, (1)\\ q_1^{(2)}=-\dfrac{P_2^{(2)}}{k^{(2)}}\\ q_2^{(2)}=0,\, fixed\, node\, (3)\\ \end{array}$$

Such a situation makes us wonder about how it would possible to back out the displacement degrees of freedom of node (2) from the results above?

Uy = R sin(α) ≈ Rα (if α small)

Ux = R(1 - cos(α)) ≈ 0 (first Order)

title: User:Eml4500.f08.jamama.adam/homework3


text:
User:Eml4500.f08.delta 6.lopez/HW3

Three Bar Truss System
The three bar truss system is shown in the image below. The characteristics of each element is shown in the table that follows the image.

Convenient methods of numbering the local nodes are shown in the images below for each element. Element 1

Element 2

Element 3

$$ \sum{F_{x}}=0 $$ $$ \sum{F_{y}}=0 $$ $$ \sum{M_{A}}=0 $$

Question: What if we take the moment about a point B?

$$ \sum{M_{B}} = \vec{BA}\; X\; \vec{ F} = \vec{BA'}\; X\; \vec{ F} $$ The above equation stands only when A' lies on the line of action. We can derive this moment by defining the BA' vector, then substituting into the moment equation.

$$ \vec{BA'} = \vec{BA}+\vec{AA'} $$

$$ \vec{M_B} = {(\vec{BA}+\vec{AA'})}\; X\; \vec{F}= \vec{BA}\; X\; \vec{F} + \vec{AA'}\; X\; \vec{F} $$

where $$ \vec{AA'}\; X\; \vec{F} = 0 $$

The derivation of the moment about point B is quite simple:
 * NOTE

$$ \sum{\vec{F}_{i}} = 0 $$ $$ \sum{\vec{M}_{B_{i}}} = \sum{\vec{BA'}_{i}}\; X\; \vec{F}_{i} = \sum{\vec{BA}_{i}}\; X\; \sum{\vec{F}_{i}} $$ where $$ \sum{\vec{F}_{i}} = 0 $$ Therefore, the moment about point B is also equal to zero.

The 8x8 global stiffness matrix is found below, where the local stiffness matrices for each element correspond to the color-coded areas. The k values for the local stiffness matrix for element 3 can be found as shown below. Only two examples are shown, yet the values can be found under the homework section of this page. $$ k_{33} = k^{(1)}_{33} + k^{(2)}_{11} + k^{(3)}_{11} $$ $$ k_{34} = k^{(1)}_{34} + k^{(2)}_{12} + k^{(3)}_{12} $$

=HOMEWORK= The displacement vector for node 2 can be found by breaking it down into vector components. $$q^{(e)}_2= (d^{(e)}_3\vec{i}+d^{(e)}_4\vec{j}) \cdot \vec{\tilde{i}}$$ This is equivalent to $$q^{(e)}_2= d^{(e)}_{3}(\vec{i}\cdot \vec{\tilde{i}})+d^{(e)}_4(\vec{j} \cdot \vec{\tilde{i}})$$

q2(e) is shown below, after substituting in the director cosines.

$$q_2^{(e)}=l^{(e)}d_3^{(e)}+m^{(e)}d_4^{(e)}$$ where $$ l^{(e)} = cos\theta $$ $$ m^{(e)} = sin\theta $$

$$q_2^{(e)}=\begin{bmatrix}l^{(e)} & m^{(e)} \end{bmatrix}\begin{bmatrix}d_3^{(e)}\\ d_4^{(e)}\end{bmatrix}$$

title: User:Eml4500.f08.delta 6.lopez/HW3


text:
User:Eml4500.f08.gravy.sms/HW3

Element 1:

Element 2:

Solving for AC and AB:

Since $$ d = \frac{f}{k} $$,

$$\ AC= \frac{|P_{2}^{(1)}|}{k^{(1)}} = \frac{5.1243}{3/4} = 6.8324 \,$$

$$\ AC= \frac{|P_{1}^{(2)}|}{k^{(2)}} = \frac{5.1243}{5} = 1.0249 \,$$

Filling in K matrix for 3-bar truss:

$$ \begin{bmatrix} k^{(1)}_{11} & k^{(1)}_{12} & k^{(1)}_{13} & k^{(1)}_{14} & 0 & 0 & 0 & 0\\ k^{(1)}_{21} & k^{(1)}_{22} & k^{(1)}_{23} & k^{(1)}_{24} & 0 & 0 & 0 & 0\\ k^{(1)}_{31} & k^{(1)}_{32} & k^{(1)}_{33}+k^{(2)}_{11}+k^{(3)}_{11} & k^{(1)}_{34}+k^{(2)}_{12}+k^{(3)}_{12} & k^{(2)}_{13} & k^{(2)}_{14} & k^{(3)}_{13} & k^{(3)}_{14}\\ k^{(1)}_{41} & k^{(1)}_{42} & k^{(1)}_{43}+k^{(2)}_{21}+k^{(3)}_{21} & k^{(1)}_{44}+k^{(2)}_{22}+k^{(3)}_{22} & k^{(2)}_{23} & k^{(2)}_{24} & k^{(3)}_{23} & k^{(3)}_{24}\\ 0 & 0 & k^{(2)}_{31} & k^{(2)}_{32} & k^{(2)}_{33} & k^{(2)}_{34} & 0 & 0\\ 0 & 0 & k^{(2)}_{41} & k^{(2)}_{42} & k^{(2)}_{43} & k^{(2)}_{44} & 0 & 0\\ 0 & 0 & k^{(3)}_{31} & k^{(3)}_{32} & 0 & 0 & k^{(3)}_{33} & k^{(3)}_{34}\\ 0 & 0 & k^{(3)}_{41} & k^{(3)}_{42} & 0 & 0 & k^{(3)}_{43} & k^{(3)}_{44} \end{bmatrix} $$

title: User:Eml4500.f08.gravy.sms/HW3


text:
User:Eml4500.f08.jamama.jan/Meeting12

Mtg.12:

$$ \mathbf{q}_{2 \times 1}^{(e)} = \mathbf{T}_{2 \times 4}^{(e)} * \mathbf{d}_{4 \times 1}^{(e)} \qquad \qquad \qquad \qquad \qquad \qquad (1) $$

Derivation of element F-D with respect to global coordinate sstem (p. 6-1)

$$ \mathbf{k}_{4 \times 4}^{(e)} * \mathbf{d}_{4 \times 1}^{(e)} = \mathbf{f}_{4 \times 1}^{(e)} $$

(p. 4-5) $$ \ \ k^{(e)} \begin{bmatrix}

1 & -1 \\ -1 & 1

\end{bmatrix}

\begin{Bmatrix}

q_1^{(e)} \\ q_2^{(e)}

\end{Bmatrix}=

\begin{Bmatrix}

P_1^{(e)} \\ P_2^{(e)}

\end{Bmatrix} \qquad \qquad (2)$$

$$ q_1^{(e)}\ $$ = axial displacement of element $$ \mathbf{e}\ $$ at local node

$$ P_1^{(e)}\ $$ = axial force of element $$ \mathbf{e}\ $$ at local node

Global: Derive eq. $$ (1) $$ from eq. $$(2)$$ $$ \rightarrow $$ equation $$(2)$$ was already derived in Mtg.4

Want to find relation between $$ \begin{matrix} \mathbf{q}_{2 \times 1}^{(e)} & and & \mathbf{d}_{4 \times 1}^{(e)} \\ \mathbf{P}_{2 \times 1}^{(e)} & and & \mathbf{f}_{4 \times 1}^{(e)} \end{matrix} $$

These relations can be expressed in the form: $$ \mathbf{q}_{2 \times 1}^{(e)} = \mathbf{T}_{2 \times 4}^{(e)} * \mathbf{d}_{4 \times 1}^{(e)} $$

Consider the displacement vector of local node, denoted by $$ \mathbf{d}_{(1)}^{(e)} $$

$$ \Rightarrow $$ $$ \vec {\mathbf{d}}_{(1)}^{(e)} =  \mathbf{d}_1^{(e)}\vec {i} +  \mathbf{d}_2^{(e)} \vec {j}$$

$$\Rightarrow \ q_{1}^{(e)} $$ = axial displacement of node is the orthogonal projection of the dispalcement. Vector $$ \vec {\mathbf{d}}_{1}^{(e)}$$ of node on the axis of $$ \tilde{i} $$ of element $$ \mathbf{e} $$

$$ q_{1}^{(e)}= \vec {\mathbf{d}}_{(1)}^{(e)}* \vec{\tilde{i}} \Rightarrow (\mathbf{d}_1^{(e)} \vec {i} +  \mathbf{d}_2^{(e)} \vec {j})* \vec{\tilde{i}}= \mathbf{d}_1^{(e)}( \vec{i} * \vec{\tilde{i}} ) + \mathbf{d}_2^{(e)} (\vec {j}* \vec{\tilde{i}}) \Rightarrow $$

$$ \Rightarrow \begin{matrix}

\vec{i} * \vec{\tilde{i}} & = & cos(\theta^{(e)}) & = & l^{(e)} \\ \vec{j} * \vec{\tilde{i}} & = & sin(\theta^{(e)}) & = & m^{(e)}

\end{matrix} \Rightarrow \ $$ $$q_{1}^{(e)}= l^{(e)} \mathbf{d}_1^{(e)} + m^{(e)}\mathbf{d}_2^{(e)} = {\llcorner l^{(e)} m^{(e)} \lrcorner}_{1 \times 2}

{\begin{Bmatrix}

d_{1}^{(e)} \\ d_{2}^{(e)}

\end{Bmatrix}}_{2 \times 1}$$

Similarly for node, $$ q_{2}^{(e)}= {\llcorner l^{(e)} m^{(e)} \lrcorner}_{1 \times 2} {\begin{Bmatrix}

d_{3}^{(e)} \\ d_{4}^{(e)}

\end{Bmatrix}}_{2 \times 1}$$

$$ \begin{matrix}

\begin{Bmatrix}

q_{1}^{(e)} \\ q_{2}^{(e)} \end{Bmatrix} & = &

\begin{bmatrix}

l^{(e)} & m^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix} &

\begin{Bmatrix}

d_{1}^{(e)} \\ d_{2}^{(e)} \\ d_{3}^{(e)} \\ d_{4}^{(e)} \end{Bmatrix} \\

& & & \\

\mathbf{q}_{2 \times 1}^{(e)} & = & \mathbf{T}_{2 \times 4}^{(e)} & \mathbf{d}_{4 \times 1}^{(e)} \end{matrix}

$$

title: User:Eml4500.f08.jamama.jan/Meeting12


text:
User:Eas4200c.f08.spars.lee/Homework 3

22 Sep notes
A shear panel (fig 15) is a thin sheet of material that is intended to withstand in-plane shear forces. The shear force in the panel is distributed across the depth of the panel, but because the panel is assumed to be very thin this force may be viewed as being constant across the depth. With this assumption the internal forces can be seen as a shear flow (q) represented by the equation
 * $$ q=\tau \!t$$

Where $$\tau \!$$ is the uniform shear stress and $$\!t$$ is the thickness of the panel.

With this in mind the panel may be viewed as a line in the (Z,Y) plane (fig 16). The shear force acting in the panel can therefore be broken up into an infintessimal force ($$\overrightarrow{dF}$$) acting on an infintessimal length ($$\overrightarrow{dF}$$) represented by the following equation.

 * $$ \overrightarrow{dF} = q\overrightarrow{dL} = q(dL_y\overrightarrow{j}+dL_z\overrightarrow{k})$$

Zooming in on an infintessimal length (fig. 18) allows this equation to be taken further.

 * $$\overrightarrow{dF} = q(dL cos \theta \overrightarrow{j} + dL sin\theta\overrightarrow{k}) = q(dy\overrightarrow{j} + dz\overrightarrow{k})$$

The resultant shear force vector can then be determined by integrating this infintessimal force along the line representing the panel as follows.

 * $$\overrightarrow{F}=\int_A^B\overrightarrow{dF}=q[(\int_A^Bdy)\overrightarrow{j} + (\int_A^Bdz)\overrightarrow{k}]=q(a\overrightarrow{j} + b\overrightarrow{k})$$

Breaking this force into components

 * $$\overrightarrow{F}=F_y\overrightarrow{j} + F_z\overrightarrow{k}$$ → $$F_y=qa\!$$ $$F_z=qb\!$$

Recalling that $$ q=\tau \!t$$ this equation is the same as equations (1.4) and (1.5) (Sun, pg.5). Calculating the resultant shear force magnitude

 * ||$$\overrightarrow{F}$$|| $$=[(F_y)^2 + (F_z)^2]^{1/2}= q[a^2+b^2]^{1/2} = qd \!$$

Where d is the length of a straight line from point A to point B and does not depend of the curve. This equation is the same as equation (3.49a) (sun, pg.85)

Relating the force (||$$\overrightarrow{F}$$||) to torque (T)

 * $$T=2q\overline{A}$$

Which is proven by considering a thin walled cross-section with a shear flow along that wall.

 * $$T=\oint \rho q ds = \int \int_{\overline{A}} 2q dA = 2q\overline{A}$$ (equation 3.48 Sun, pg 85)

comparing J
The area of the two cylinders are determined.

case a

 * $$\pi \,\! r^2 = \pi \,\! (0.01\mathbf{m})^2 = (0.0001)\pi \,\!$$ $$ \mathbf{m^2}$$

case b

 * $$\pi \,\! r_o^2 - r_i^2 =\pi \,\! [(0.051\mathbf{m})^2 - (0.05\mathbf{m})^2] =(0.000101)\pi \,\!$$ $$ \mathbf{m^2}$$

As the difference in area is negligible the two areas can be approximated as being equal which means that in both cases the same amount of material is used such that the shape resulting in the more efficient torsion member can be determined.

The torsional constant for the two cylinders are determined as follows,

case a

 * $$J=\frac{1}{2}\pi \,\!r^4 = \frac{1}{2}\pi \,\!(0.01\mathbf{m})^4 = (0.000000005)\pi \,\!$$ $$\mathbf{m^4}$$

case b

$$J = \frac{1}{2}\pi \,\!(r_o^4 - r_i^4) = \frac{1}{2}\pi \,\!(r_o - r_i)(r_o - r_i)(r_o^2 + r_i^2) = \frac{1}{2}\pi \,\!(0.001\mathbf{m})(0.101\mathbf{m})(0.05101\mathbf{m}) = (0.0000002576)\pi \,\!$$ $$\mathbf{m^4}$$

Comparing the two,

 * $$\frac{J^{(a)}}{J^{(b)}} = 0.0194099002$$

In other words the torsional constant of the shape in case a is less than two percent of the torsional constant of the shape in case b despite both shapes having the same area. To illustrate this consider a hollow cylinder of interior radius $$r_i$$ and thickness $$ t = 0.02 r_i$$ where the torsional constant is equal to the torsional constant of case a.

 * $$J = \frac{1}{2}\pi \,\![r_i(1+0.02)-r_i][r_i(1+0.02)+r_i][[r_i(1+0.02)]^2+r_i^2] = (0.000000005)\pi \,\! = \frac{1}{2}\pi \,\!(0.02r_i)(2.02r_i)(2.0404r_i^2) = 0.04121608r_i^4\pi \,\!$$

Which leads to

 * $$\!r_i = 0.0186627582$$ $$\mathbf{m}$$

 * $$\!t =  0.000373255$$ $$\mathbf{m}$$

Where the area is determined by

 * $$A=\pi \,\!(0.0190360134^2-0.0186627582^2)=(0.00001407126)\pi \,\!$$ $$\mathbf{m^2}$$
 * $$A=\pi \,\!(0.0190360134^2-0.0186627582^2)=(0.00001407126)\pi \,\!$$ $$\mathbf{m^2}$$

Comparing this area to the area of case a

 * $$\frac{A^{(a)}}{A^{(c)}} = 7.10668352$$

Which indicates that although both cases have the same torsional constant the area of this third case is over seven times less than the area of case a.

Big Picture Road Map
In order to build up to the torsional analysis of multicell structures Dr. Vu-Quoc has suggested the following set of steps.

 * Kinematic assumptions
 * Strain displacement relationship
 * Equilbrium equation for stresses
 * Prondtl stress function $$\phi \,\! (x,y)$$
 * Strain compatability equation
 * Equation fo $$\phi \,\!$$
 * Boundary conditions for $$\phi \,\!$$
 * Relationship between $$t$$ and $$\phi \,\!$$
 * Thin-walled cross section
 * Twist angle $$\theta \,\!$$: method one
 * '''Section 3.6 multicell thin-walled cross section

title: User:Eas4200c.f08.spars.lee/Homework 3


text:
User:EML4500.F08.RAMROD.B/Homework 3

 Referring back to the picture of the truss system, it can be seen that the displacement of point $$\displaystyle D$$ is simply the displacement of the global node $$\displaystyle 2$$ at point $$\displaystyle A$$. In other words,

$$\displaystyle Y_{D} = d_{4} $$ and

$$\displaystyle X_{D} = d_{3} $$

We can now proceed to calculate the length of lines $$\displaystyle AC $$ and $$\displaystyle AB $$ be computed which as we already know are defined to be the axial forces. In other words it is the amount the bars extend or elongate due to the applied force. As previously stated the formula to find line $$\displaystyle AC $$ is as follows,

$$\displaystyle AC = \frac{\left|P_{2}^1 \right|}{k^1}$$

where

$$\displaystyle \left|P_{2}^1 \right| = 5.1243 $$ $$\displaystyle k^1 = \frac{3}{4}$$

Therefore,

$$\displaystyle AC = 6.8324$$

Remember that $$\displaystyle \left|P_{2}^1 \right|$$ is the internal force $$\displaystyle 2$$ at element $$\displaystyle 1 $$. Remember also that in order to obtain the value of $$\displaystyle \left|P_{2}^1 \right|$$, it is necessary to take the Root Sum Squared, RSS for short, of the two forces making acting at that node. The formula for the RSS of line AC extending from element 1 is,

$$\displaystyle \sqrt{(f_3)^2 + (f_4)^2}$$

The same can be done for $$\displaystyle AB $$ with the axial force from element two.

$$\displaystyle AB = \frac{\left|P_{1}^2 \right|}{k^2}$$ $$\displaystyle \left|P_{1}^2 \right| = 6.276 $$

$$\displaystyle k^2 = 5$$

Therefore

$$\displaystyle AB = 1.2552$$

From this point it is now possible to find the coordinates of points $$\displaystyle B $$ and $$\displaystyle C $$. Simple trigonometry will be used to carry out these calculations. First we will find the coordinates for point C, $$\displaystyle (x_C,y_C) $$. If you recall from previous lessons

$$\displaystyle \theta^1 = 30^{\circ} $$

and we already know the length of $$\displaystyle AC $$ so therefore the coordinates of $$\displaystyle C $$ can be expressed as

$$\displaystyle x_C = AC \cos (30^{\circ}) = 6.8324 *\frac{\sqrt{3}}{2} = 5.917$$

$$\displaystyle y_C = AC \sin (30^{\circ}) = 6.8324 *\frac{1}{2} = 3.4162$$

The coordinates of point $$\displaystyle B $$ can be found in the same way. the angle used is the angle the line $$\displaystyle AB $$ makes with the negative x-axis.

$$\displaystyle x_B = AB \cos (45^{\circ}) = 1.2552 *\frac{\sqrt{2}}{2} = -.88756 $$

$$\displaystyle y_B = AB \sin (30^{\circ}) = 1.2552*\frac{\sqrt{2}}{2} = .886756$$

We next need to find the coordinates of the point $$\displaystyle D $$. There are two unknowns at point $$\displaystyle D $$, the x and the y coordinate $$\displaystyle (x_D,y_D) $$. Since there are two unknowns, we need to find two equations in which these two variables are being utilized. The equation for the lines $$\displaystyle AB $$ and $$\displaystyle AC $$ are the two equations we will utliize. In order to do so we will work with the following figure.

The line $$\displaystyle PQ $$ can be represented in the following way;

$$\displaystyle \vec{PQ} = (PQ) \tilde{i} = PQ [\cos(\theta) \vec{i} + \sin(\theta) \vec{j}] $$

The cosine and the sine of the angle can then be expressed as;

$$\displaystyle \vec{PQ} = (x-x_p) \vec{i} + (y-y_p) \vec{j} $$

Dividing the previous two equations will yield the following;

$$\displaystyle \frac{(y-y_p)}{(x-x_p)} = \tan(\theta) $$

Multiplying through by $$\displaystyle (x-x_p)$$ and the equation of the line $$\displaystyle \vec{PQ} $$ is given by the following;

$$\displaystyle y-y_p= \tan(\theta)(x-x_p) $$

This equation for line $$\displaystyle \vec{PQ} $$ is equivalent to the equation for line $$\displaystyle \vec{AC} $$. The equation for a line that is perpendicular to line $$\displaystyle \vec{AC} $$ and passing throught the point $$\displaystyle A $$ is given by the following;

$$\displaystyle y-y_p= \tan(\theta +\frac{\pi}{2})(x-x_p) $$

This equation is equivalent to the equation for line $$\displaystyle \vec{AC} $$. When the coordinates for lines $$\displaystyle \vec{BD} $$ and $$\displaystyle \vec{CD} $$ are put into the above equation then it is possible to solve for the coordinates of point $$\displaystyle D $$.

$$\displaystyle y_D-y_B= \tan(\theta +\frac{\pi}{2})(x_D-x_B) $$

 $$\displaystyle y_D-y_C= \tan(\theta +\frac{\pi}{2})(x_D-x_C)$$

 $$\displaystyle y_D-.886756= \tan(\frac{\pi}{4})(x_D+.886756) $$

 $$\displaystyle y_D-3.4162= \tan (\frac{2\pi}{3}) - (x_D-5.917)$$

Solving for $$\displaystyle y_D $$ for each equation will allow you to then solve for $$\displaystyle x_D$$. Plug this value back into one of the equations will allow you to get the value of $$\displaystyle y_D$$. When done, the coordinate of point $$\displaystyle D $$ becomes;

$$\displaystyle D = (4.352, 6.1271) $$

Using the Finite Element Method (FEM), we have already found the displacement of point $$\displaystyle D $$. In other words;

$$\displaystyle \vec{AD} = (x_D-x_A) \vec{i} + (y_D-y_A) \vec{j} $$

In this equation the variables $$\displaystyle x_A $$ and $$\displaystyle y_A $$ are zero and the only variables that remain are the coordinates of point $$\displaystyle D $$. By definition;

 $$\displaystyle x_D = d_3 $$ and $$\displaystyle y_D = d_4 $$

And therefore,

$$\displaystyle \vec{AD} = d_3 \vec{i} + d_4 \vec{j} $$

where the coefficients are the global displacements of point $$\displaystyle A $$ to point $$\displaystyle D $$.

=Three Truss Bar System=

We will now look at the process of analyzing a three bar truss system as shown below.

It is important to note the helpfulness in numbering the local nodes of each element. Doing this in a certain manner make the assembly of the global stiffness matrix more convenient. The following numbering is the best and most convenient system.

title: User:EML4500.F08.RAMROD.B/Homework 3