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User:Eas4200c.f08.ZYX.reger/Week3,4

Problem 1.1:

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User:Jtneill student2

Jtneill student2 02:38, 26 September 2008 (UTC)
 * A second test account of User:Jtneill (see also User:Jtneill student).
 * I'm trying to learn just a bit about Help:Blocking
 * It seems I've had no problem creating this new user account, despite I thought blocking of User:Jtneill student and account creation from this IP address.

title: User:Jtneill student2


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User:Eml4500.f08.gravy.sms/HW2

HW: Verify equilibrium of element 1

Let $$P_{1}^{(1)} = f_{1}^{(1)} + f_{2}^{(1)} = -4.4378 i\hat{} -2.5622j\hat{}$$ and let $$P_{2}^{(1)} = f_{3}^{(1)} + f_{4}^{(1)} = 4.4378 i\hat{} + 2.5622j\hat{}$$

Note that $$P_{1}^{(1)} + P_{2}^{(1)} = 0$$ which illustrates that $$\sum{}F=0$$.

Also, note that $$P_{1}^{(1)} \times P_{2}^{(1)} = \left( -4.4378 i\hat{} -2.5622j\hat{} \right) \times \left( 4.4378 i\hat{} +2.5622j\hat{} \right) = \left( \left( -4.4378 \times 2.5622 \right) - \left( 4.4378 \times -2.5622 \right) \right) k\hat{} = 0$$.

This shows that $$P_{1}^{(1)}$$ and $$P_{2}^{(1)}$$ are in line with each other. Since all forces are in line with each other, $$\sum{}M_{anypoint} = 0$$.

HW: Verify equilibrium of node 2

Note that $$\sum{}F = P_{1}^{(1)} + P_{2}^{(2)} + P = \left( -4.4378 i\hat{} -2.5622j\hat{} \right) + \left( 4.4378 i\hat{} -4.4378j\hat{} \right) + 7j\hat{} = 0$$.

Also, since all forces are acting on the same point, $$\sum{}M_{2} = 0$$.

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User:Eml4500.f08.group/Homework2/Homework Problems

=SOLVE USING FEA=

For the first part of our homework, the 2-truss system shown in the figure below had to be analyzed and solved using the F.E.A. procedure. Where P is an applied downward force of 7, R1x, R1y, R3x, and R3y are the reactions in the x and y directions at nodes 1 and 3 respectively.

The element 1 is inclined at an angle of 30 degrees relative to the x axis while element 2 is declined at an angle of 45 degrees.

L1 = 4 L2 = 2 A1 = 1

A2 = 2 E1 = 3

E2 = 5

The next step is to draw the Free Body Diagram for the 1st element and determine its stiffness matrix. This Free Body Diagram is shown in the following figure and the stiffness matrix is defined as follows.

where k(e) = k(1) = E(1)*A(1)/L(e), l(e) = cos(θ(e)), m(e) = sin(θ(e))

In the case of element 1,

k(1) = 3*1/4 = 3/4 = 0.75

l(1) = cos(30) = 0.866

m(1) = sin(30) = 0.500

The resulting stiffness matrix for element 1 is, k(1) $$ = \begin{bmatrix} 0.5625&0.32476&-0.5625&-0.32476 \\ 0.32476&0.1875&-0.32476&-0.1875\\ -0.5625&-0.32476 & 0.5625&0.32476\\ -0.32476&-0.1875&0.32476&0.1875 \end{bmatrix} $$

Next it is necessary to calculate the stiffness matrix for the second element.

k(2) = 5*2/2 = 5

l(2) = cos(-45) = 0.500

m(2) = sin(-45) = -0.500

The resulting stiffness matrix for element 2 is, k(2) $$ = \begin{bmatrix} 2.5&-2.5&-2.5&2.5 \\ -2.5&2.5&2.5&-2.5\\ -2.5&2.5&2.5&-2.5\\ 2.5&-2.5&-2.5&2.5 \end{bmatrix} $$

Given that each element stiffness matrix is a 4x4 (because there are 4 unknown forces), it is next necessary to assemble a global (6x6) force-displacement matrix. This is done by recognizing that

F(1)1 = R(1)x

F(1)2 = R(1)y

F(2)3 = R(3)x

F(2)4 = R(3)y

F(1)3 = F(2)1 = Px

F(1)4 = F(2)2 = Py

Applying the above relationships to the global (6x6) matrix yields the following

K = $$ \begin{bmatrix} 0.5625&0.32476&-0.5625&-0.32476&0&0 \\ 0.32476&0.1875&-0.32476&-0.1875&0&0 \\ -0.5625&-0.32476&3.0625&-2.1752&-2.5&2.5 \\ -0.32476&-0.1875&-2.1752&2.6875&2.5&-2.5 \\ 0&0&-2.5&2.5&2.5&-2.5 \\ 0&0&2.5&-2.5&-2.5&2.5 \end{bmatrix} $$

Next, we observe that since node 1 and node 3 are pinned and fixed. This means that no displacement can occur at these points and thus d11, d12, d23, and d24 = 0. We have also been given that Py = 7 and is oriented vertically Px = 0. Applying these observations into the equation F = KD yields the following system.

$$ \begin{bmatrix} 0.5625&0.32476&-0.5625&-0.32476&0&0 \\ 0.32476&0.1875&-0.32476&-0.1875&0&0 \\ -0.5625&-0.32476&3.0625&-2.1752&-2.5&2.5 \\ -0.32476&-0.1875&-2.1752&2.6875&2.5&-2.5 \\ 0&0&-2.5&2.5&2.5&-2.5 \\ 0&0&2.5&-2.5&-2.5&2.5 \end{bmatrix} $$ * $$\begin{bmatrix} 0\\ 0\\ D_3\\ D_4\\ 0\\ 0 \end{bmatrix} $$ = $$ \begin{bmatrix} R_{x1}\\ R_{y1}\\ 0\\ 7\\ R_{x3}\\ R_{y3} \end{bmatrix}$$

These equations can be used to solve for D3 and D4 (the x and y displacement of node 2).

$$D_3 = 4.353$$

$$D_4 = 6.1271$$

We can now plug these values back into the matrix equation and solve for the unknown reactions at nodes 1 and 2. When we do this, we find:

$$R_{x1} = -4.4378$$

$$R_{y1} = -2.5622$$

$$R_{x3} = 4.4378$$

$$R_{y3} = -4.4378$$

=SOLVE USING STATICS=

To verify these results, it is possible to solve the same problem using statics. Thus can been done by realize that in a truss, any reaction is directed along the axis of the truss. When the reactions at nodes 1 and 3 are combined into a single reaction forces (R1 and R3)we can solve the problem as follows.

ΣFy = R1sin(30) + R2sin(45) = 7

ΣFx = -R1cos(30) + R2cos(45) =

$$\begin{bmatrix} sin(30)&sin(45) \\ -cos(30)&cos(45) \end{bmatrix}$$ * $$\begin{bmatrix} R1\\ R3 \end{bmatrix}$$ = $$\begin{bmatrix} 7\\ 0 \end{bmatrix}$$

R1 = 5.12436

R3 = 6.27603

Plugging these values into the above equations, we find that

Rx1 = -5.12436*cos(30) = -4.4378

Ry1 = -5.12436*sin(30) = -2.562

Rx3 = 6.27603*cos(45) = 4.4378

Ry3 = -6.27603*sin(45) = -4.4378

These values agree perfectly with the ones attained using the Finite Element method thus confirming the results.

=PROVE THE EXAMPLE IS A 2-FORCE BODY=

The above statics solutions assumes a 2-force body and that the solutions attained via this assumption match those of the solutions attained using Finite Element Analysis. Given that the two solutions sets match, this proves that the trusses in the example problem are indeed 2-force bodies.

=NODE 2 IN EQUILIBRIUM=

In order for node 2 to be in equilibrium, the sum of the forces in the y direction have to equal to 0 as to the sum of the forces in the x direction. That is

R3sin(45) + R1sin(30) = 7

R3cos(45) - R1 cos(30) = 0

(6.27603)(0.707107) + (5.12436)(0.5) = 7

(6.27603)(0.707107) - (5.12436)(0.866025) = 0

Thus the sum of the forces in both the x and y directions equal 0, i.e. the node does not move and is in equilibrium.


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User:Eml4500.f08.qwiki.gatch

Not sure if you're gonna get this (still trying to figure out wikipedia), but I was wondering why you changed the diagrams on our page. I noticed a couple arrows were going the opposite way that Vu Quoc drew them in class. Was there a reason for this?

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User:Eas4200c.f08.ZYX/HW Week 3-4

=Problem 1.1=

= Problem 1.7 =

=Stringers=

=Shear=

=HW Problem Based on Figure 1.2= {| class="toccolours collapsible collapsed" style="width:100%" ! Moment of Inertia for Solid Circle and equal-area Channel

Moment of Inertia: Case 2 (equal-area Channel) - Figure 1.2

 * }

=Proof that (ab)_max with constant perimeter L is a square=

= Contributing Members=

title: User:Eas4200c.f08.ZYX/HW Week 3-4


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User:Eml4500.f08.ateam/hw 3

A-Team Teampage

Homework Report #3

This page is intended for Homework Report #3 of the A-Team.

Due date: Wednesday, 8 October 2008 5 pm EDT = 17:00 EDT (or 21:00 UTC).

Derivation of Element Force Displacement Relationship
The element force displacement relation can be derived with respect to the global coordinate system from the axial displacement and the axial forces. The internal forces and internal displacements of any element can be represented by the member forces in the local coordinate system instead of by two variables in the global coordinates. The elemental forces and elemental displacements can be shown with an axial displacement of Element e at Local Node i, $$ {q_i^{(e)}}$$, and an axial force of Element e at Local Node i, $$ {p_i^{(e)}}$$.

Our goal is to derive Equation 1 from Equation 2.

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

 * $$ \mathbf{k^{(e)}} \mathbf{q^{(e)}} = \mathbf{P^{(e)}} $$ (2)

The relationship between $$q^{(e)} $$ and $$d^{(e)} $$, and $$p^{(e)} $$ and $$f^{(e)} $$ also needs to be determined. We know that $$ {T^{(e)}} {d^{(e)}} = {q^{(e)}} $$. Now we consider the displacement vector of Local Node 1, $$ {d_1^{(e)}} $$

Node 1

The vector d1(e) can be broken down in to two unit vectors.

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

The vector $$ {\vec d_1^{(e)}} $$ of Local Node 1 is on axis $$ \tilde{x} $$. It can be shown that $$ q_1^{(e)} $$ is the axial displacement of Local Node 1 and is the orthogonal projection of the displacement. $$ q_1^{(e)} $$ can now be related to $$ d_1^{(e)} $$.

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

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

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

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

These equations can be shortened because, as proven in Homework 1,

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

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

$$ q_1^{(e)} $$ can now be rewritten again:

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

 * $$ = \left[ \begin{matrix} l^{(e)} & m^{(e)} \end{matrix} \right] \left[ \begin{matrix} d_1^{(e)} \\ d_2^{(e)} \end{matrix} \right] $$

Node 2
The reactions and relationships are very similar for Local Node 2.

 * $$ {d_2^{(e)}} = d_3^{(e)} \vec i + d_4^{(e)} \vec j $$.

Therefore,

 * $$ q_2^{(e)} = \vec d_3^{(e)} \cdot \vec \tilde{j} $$

 * $$ q_2^{(e)} = (d_2^{(e)} \tilde{i} + d_4^{(e)} \tilde{j}) \cdot \vec \tilde{j} $$

 * $$ q_2^{(e)} = d_3^{(e)} (\vec i \cdot \vec \tilde{j}) + d_4^{(e)} (\vec j \cdot \vec \tilde{j}) $$

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

Similarly,

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

 * $$ q_2^{(e)} = m^{(e)} d_3^{(e)} +  l^{(e)} d_4^{(e)} $$

 * $$ q_2^{(e)} = \left[ \begin{matrix} l^{(e)} & m^{(e)} \end{matrix} \right] \left[ \begin{matrix} d_3^{(e)} \\ d_4^{(e)} \end{matrix} \right] $$

After solving for the axial displacement in both nodes, an overall reaction can be formed.

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

Derivation of Eigenvalues
From the relationship betweeen qi(e) and di(e):

$$ \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} l^{(e)} & m^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \end{matrix} \right] \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} $$  where  q(e) = T(e)×d(e)

Use the same argument to solve for the axial displacements, qi(e), and the axial forces, pi(e).

$$ \begin{Bmatrix} P_1^{(e)} \\ P_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} T^{(e)} \end{matrix} \right] \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} $$         where  P(e) = T(e)×f(e)

Now recall the force-displacement relationship:

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

The goal here is to have the original force-displacement relationship: $$ k^{(e)}d^{(e)} = f^{(e)} $$. To accomplish this we would need to move matrix $$ T^{(e)} $$ from the right hand side to the left hand side. This could be achieved by multiplying the equation by the inverse $$ T^{-1} $$ of the matrix $$ T^{(e)} $$. As seen above though, the matrix $$ T^{(e)} $$ is a rectangular matrix and therefore cannot be inverted.

To overcome this problem, we will use the transpose of $$ T^{(e)} $$ and replace $$ k^{(e)} $$ with $$ [T^{(e)T}\hat{k}^{(e)}T^{(e)}] $$. The force-displacement relationship now becomes:

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

We can recall from previous lectures that $$ \theta^{(1)} = 30^{o} $$, $$ \theta^{(2)} = 45^{o} $$ and $$ l^{(e)} = cos(\theta^{(e)}) $$, $$ m^{(e)} = sin(\theta^{(e)}) $$. From these values and equations we can determine the matrix $$ T^{(e)} $$.

$$ \begin{Bmatrix} T^{(1)} \end{Bmatrix} = \left[ \begin{matrix} l^{(1)} & m^{(1)} & 0 & 0 \\ 0 & 0 & l^{(1)} & m^{(1)} \end{matrix} \right] = \left[ \begin{matrix} 0.866 & 0.5 & 0 & 0 \\ 0 & 0 & 0.866 & 0.5 \end{matrix} \right] $$

And the transpose of $$ T^{(1)} $$ is:

$$\left[ \begin{matrix} 0.866 & 0 \\ 0.5 & 0 \\ 0 & 0.866 \\ 0 & 0.5 \end{matrix} \right]$$

From lecture 12 we know that, $$ \hat{k}^{(1)} = \left[ \begin{matrix} 1 & -1 \\ -1 & 1 \end{matrix} \right] $$

So therefore,

$$ [T^{(1)T}\hat{k}^{(1)}T^{(1)}] $$ = $$\left[ \begin{matrix} 0.866 & 0 \\ 0.5 & 0 \\ 0 & 0.866 \\ 0 & 0.5 \end{matrix} \right]$$ $$ \left[ \  \begin{matrix} 1 & -1 \\ 1 & -1 \end{matrix} \right] $$ $$\left[ \begin{matrix} 0.866 & 0.5 & 0 & 0 \\ 0 & 0 & 0.866 & 0.5 \end{matrix} \right] $$ = $$\left[ \begin{matrix} 0.7499 & 0.433 & -0.7499 & -0.433 \\ 0.433 & 0.25 & -0.433 & -0.25 \\ -0.7499 & -0.433 & 0.7499 & 0.433 \\ -0.433 & -0.25 & 0.433 & 0.25 \end{matrix} \right] $$

And if we recall from previous lectures, we will see that these are the coefficients of the 4x4 matrix $$ \underline k^{(1)} $$.

Calculating the eigenvalues of this matrix leads to values of:

$$ \lambda = \left[ \begin{matrix} 0 & 2 & 0 & 0 \end{matrix} \right] $$

The three zero eigenvalues in this matrix correspond to the stored elastic energy associated with the three possible rigid body motions. Two of these being translational and the other rotational.

Closing the loop between the FEM and Statics
To close the loop between the finite element method and the method of statics, we use the principle of virtual displacement to solve for the reactions and thus the member forces in the 2 force body.

Two Bar Truss System
To solve the two bar truss system and thus close the loop here is the progression of steps:

FEM -> Compute Displacement -> Compute Reactions Statics -> Compute Reactions -> Compute Displacement When solved, both of these solution steps should yeild the same values for the reactions and displacement. This is how we can check our work.

We are given the 2-Force bodies for each element(1,2). By statics, the reactions are known and thus the member forces are known. Member Forces: P1 and P2 Next, we compute the axial displacement degrees of freedom. (This gives us the amount of extension in the bars) q2 = P1/K1 = AC  q1 = 0  (fixed node 1) q1 = -P2/K2 = -AB q2 = 0  (fixed node 3)

Question: How do we back out from the above results, the displacement degrees of freedom of node 2?

The dotted lines represent the deformed object and the solid lines represent the original object. The distances AB and CD are the displacements of node 2. Given the forces P1 and P2 we can back out the displacement of node 2.

Below is an explanation of the displacement in the figure above. This is why we can approximate the displacement in 1 direction as being zero and being the length of the bar multiplied by the displacement angle.

Matlab: Two-Bar Truss System Deformation
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 $$\theta^{(1)}=30^\circ$$ and $$\theta^{(2)}=135^\circ$$, we find that $$(x_B,y_B)=(-0.88756,0.88756)$$ and $$(x_C,y_C)=(5.91703,3.4162)$$. Finally, it is important to find $$(x_D,y_D)$$ to verify that it is equal to displacements calculated with the FEM method. We can calculate $$(x_D,y_D)$$ 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 $$(x_D,y_D)=(4.135,6.127)$$. Finally, by the definition of AD, we obtain the following relation from the FEM and the loop is closed.

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

Three-Bar Truss System Introduction
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.

Review of Moment Analysis Techniques
How to sum the moment about point B:

The moment is the same as any point along the line of action of the force. In this example, the point $$A'$$ is used to simplify the cross product since the vector $$ \vec {BA'}$$ > is perpendicular to $$\vec F$$ >.

$$\sum M_B = \vec {AB} x  \vec F$$

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

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

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

$$ M_B = \vec {BA} x \vec F + \vec {AA'} x  \vec F  = 0$$

Three-Bar Truss System Problem

In this problem, there are three bars instead of two in the last example. The numbering of the elements and nodes is important to make the subsequent matrix manipulations as simple as possible. As a rule of thumb, one should use the overlapping numbering system. Put Local Node 1 on Local Node 2 of the previous element, if possible. This is shown in the following image.

The angle ʘ is determined by the vector going from Local Node 1 to 2 in each element. Therefore the angle from Local Node 1 to 2 in Element 3 is pointing down and the corresponding angle is negative. By following these conventions when the Global Stiffness Matrix is made, the three elements add into the same place in order to solve for the unknown forces. In this case, they all add into the columns and rows 3 and 4.

The finished matrix without numerical values is given in the following figure.

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

Contributing Team Members
The following students contributed to this report:

Daniel DiDomenico Eml4500.f08.ateam.didomenico 03:35, 7 October 2008 (UTC)

Matthew Philbin Eml4500.f08.Ateam.philbin 17:57, 7 October 2008 (UTC)

Mark Barry Eml4500.f08.Ateam.barry 19:58, 7 October 2008 (UTC)

Sean Miller EML4500.f08.Ateam.Miller 01:21, 8 October 2008 (UTC)

Jil Paquette Eml4500.f08.Ateam.paquette 23:00, 7 October 2008 (UTC)

Ian Slevinski Eml4500.f08.Ateam.Slevinski 23:45, 7 October 2008 (UTC)

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User:Eas4200c.f08.radsam/Lecture2

= Lecture Notes=

title: User:Eas4200c.f08.radsam/Lecture2


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User:Eml4500.f08.delta 6.guzman/Lecture Notes 09/12

Relationships between Global Diagrams and Element Diagrams
It is necessary to understand the relationships between the global free diagrams and the element free diagrams in order to solve the truss system of matrices going from the element matrices to the global matrices.

Global Free Diagram Relationship
The global free diagram relationship is given by the equation: $$\begin{bmatrix}K_{ij}^{}\end{bmatrix}_{6x6} \left\{d_{ij}\right\}_{6x1} = \left\{F_{i}\right\}_{6x1}$$ where the d and F  elements are composed by 6 x 6 matrices as following:

$$\sum_{j=1}^{6} K_{ij}d_{j}= F_{i}$$ where i goes from 1 through 6 since there are six equations
 * Global Stiffness Matrix:

$$K_{nxn} = \begin{bmatrix}K_{ij}^{}\end{bmatrix}_{nxn}$$
 * Global Displacement Matrix:

$$d_{nx1} = \left\{d_{j}^{}\right\}_{nx1}$$
 * Global Force Matrix:

$$F_{nx1} = \left\{F_{i}^{}\right\}_{nx1}$$

Element Free Diagram Relationship
The element free diagrams relationship is given by the equation: $$k_{4x4}^{(e)} d_{4x4}^{(e)} = f_{4x1}^{(e)}$$ where e indicates the element number. The d and f elements are composed by 4 x 1 matrices as following:

$$d_{4x1}^{(e)} = \left\{d_{j}^{(e)}\right\}_{4x1}$$ $$d_{4x1}^{(e)}=\begin{Bmatrix} d_{1}^{(e)} \\ d_{2}^{(e)} \\ d_{3}^{(e)} \\ d_{4}^{(e)} \end{Bmatrix}$$ $$f_{4x1}^{(e)} = \left\{f_{i}^{(e)}\right\}_{4x1}$$ $$f_{4x1}^{(e)}=\begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \\ f_{3}^{(e)} \\ f_{4}^{(e)} \end{Bmatrix}$$ $$k_{4x4}^{(e)} = \begin{bmatrix}k_{ij}^{(e)}\end{bmatrix}_{4x4}$$ $$k_{4x4}^{(e)}=\mathbf \begin{bmatrix} K_{11}^{(e)} & K_{12}^{(e)} & K_{13}^{(e)} & K_{14}^{(e)}\\K_{21}^{(e)} & K_{22}^{(e)} & K_{23}^{(e)} & K_{24}^{(e)}\\K_{31}^{(e)} & K_{32}^{(e)} & K_{33}^{(e)} & K_{34}^{(e)}\\K_{41}^{(e)} & K_{42}^{(e)} & K_{43}^{(e)} & K_{44}^{(e)}\end{bmatrix}$$
 * Element Displacement Matrix:
 * Element Force Matrix:
 * Element Stiffness Matrix:

Since the matrix k is symmetric in general the upper triangular part of the matrix should transpose to the lower triangular part, i.e. $$k_{ij}^{(e)} = k_{ji}^{(e)}$$

Assembly Process
From the free body general relationships it is possible to generate new associations in order to solve the stiffness, displacement and forces matrices:

 * Global level: $$\mathbf\begin{Bmatrix} d_{1} & d_{2} & d_{3} & d_{4} & d_{5} & d_{6}\end{Bmatrix}$$

 * Element level:
 * Element 1: $$\mathbf\begin{Bmatrix} d_{1}^{(1)} & d_{2}^{(1)} & d_{3}^{(1)} & d_{4}^{(1)} & d_{5}^{(1)} & d_{6}^{(1)}\end{Bmatrix}$$
 * Element 2: $$\mathbf\begin{Bmatrix} d_{1}^{(2)} & d_{2}^{(2)} & d_{3}^{(2)} & d_{4}^{(2)} & d_{5}^{(2)} & d_{6}^{(2)}\end{Bmatrix}$$

 * Nodes:
 * Node 1: $$d_{2} = d_{1}^{(1)}$$
 * Node 2: $$d_{3} = d_{2}^{(1)}$$
 * $$d_{3} = d_{3}^{(1)} = d_{1}^{(2)}$$
 * $$d_{4} = d_{4}^{(1)} = d_{2}^{(2)}$$

Note: These points are from the same node therefore must move in the same direction =>> (same global number = same direction)
 * Node 3: $$d_{5} = d_{3}^{(2)}$$
 * $$d_{6} = d_{4}^{(2)}$$


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Theory of Equations

''"Art of mathematics consists finding the special case which contains all germs of generality." -David Hilbert''

In mathematics, the Theory of Equations comprises a major part of traditional algebra. Topics include polynomials, algebraic equations, separation of roots including Sturm's theorem, approximation of roots, and the application of matrices and determinants to the solving of equations.

From the point of view of abstract algebra, the material is divided between symmetric function theory, field theory, Galois theory, and computational considerations including numerical analysis.

The first chapter Systems of Equations is basically about Linear Algebra, that is the study of multivariable equation systems through matrices and vector spaces. It extends to complete study Matrices and related topics of Determinants, Eigenvalues,etc. along with their applications in equation systems.

The second chapter Polynomial Functions is about the study of polynomials and their properties such as Roots and Determinants, Coefficients and Symmetric Function of Roots, Galois Groups, Derivatives, Maxima and Minima, etc.

The third chapter Functional Equations

Chapters

 * Systems of Equations
 * Polynomial Functions
 * Functional Equations

==Textbooks=

Practice Questions
1. Let a, b be the roots of the equation  and  be the roots of the equation ew. Then, the value of r is

2. If a, b, c, are the sides of a triangle ABC such that has real roots. then

3. If one root is square of the other root of the equation, then the relation between p and q is

4. For all ‘x’, x2 + 2ax + (10–3a) > 0, then the interval in which ‘a’ lies is

(a)    a < – 5

(b)    –5 < a < 2

(c)    a > 5

(d)    2 < a < 5

[IIT JEE 2004]

5. The set of all real numbers x for which  is

6. For the equation, if one of the root is square of the other, the n p is equal to

(a)    1/3

(b)    1

(c)    3

(d)    2/3

[IIT JEE 2000]

7. If b > a, then the equation (x – a) (x – b) – 1 = 0 has

(a)    Both roots in (a,b)

(b)    Both roots in (– ¥ ,a)

(c)    Both roots in (b, + ¥)

(d)    One root in (– ¥, a) and the other in (b, ¥)

[IIT JEE 2000]

8. If a and b (a < b) are the roots of the equation, where c < 0 < b, then

9. If the roots of the equation  are real and less than 3, then

(a)    a < 2

(b)    2£a£3

(c)    3 < a < 4

(d)    a > 4

[IIT JEE 1999]

10. The equation  has

(a)    No solution

(b)    One solution

(c)    Two solutions

(d)    More then two solutions

[IIT JEE 1997]

11. Let a ,b be the roots of the equation  Then the roots of the equation

(a)    a,c

(b)    b,c

(c)    a,b

(d)    a + c, b + c

[IIT JEE 1992]

12. Let f(x) be quadratic expression which is positive for all real values of x. If g(x) = f(x)+ f'(x)+ f” (x), then for any real x

(a)    g(x) < 0

(b)    g(x) > 0

(c)    g(x) = 0

(d)    g(x) ³ 0

[IIT JEE 1990]

13. Let a, b, c be real number, a ¹ 0. If a is a root of a2x2 + bx + c = 0, b is the root of a2x2 – bx – c = 0 and 0 < a < b, then the equation a2x2 + 2bx + 2c = 0 has a root g that always satisfies

14. If a and b are the roots of  and  are roots of  then the equation  has always

(a)    Two real roots

(b)    Two positive roots

(c)    Two negative roots

(d)    One positive and one negative root

[IIT JEE 1989]

15. If a, b and c are distinct positive numbers, then the expression  is

(a)    Positive

(b)    Negative

(c)    Non-positives

(d)    Non-negative

[IIT JEE 1986]

16. The equation  has

(a)    No root

(b)    One root

(c)    Two equal roots

(d)    Infinitely many roots

[IIT JEE 1984]

17. a + b + c = 0, then the quadratic equation  has

(a)    At least one root in (0, 1)

(b)    One root in (2, 3) and the other in (–2, –1)

(c)    Imaginary roots

(d)    None of the above

[IIT JEE 1983]

18. The largest interval for which  is

(a)    – 4 < x £ 0

(b)    0 < x < 1

(c)    – 100 < x < 100

(d)    – ¥ < x < ¥

[IIT JEE 1982]

19. If x1, x2, ……, xn are any real numbers and n is any positive integer, the

20. The number of real solutions of the equation |x|2 –3 |x| + 2 = 0 is

(a)    4

(b)    1

(c)    3

(d)    2

[IIT JEE 1982]

Answer keys:
1.    d    2. a    3. a    4. b    5. b

6.    c    7. d    8. b    9. a    10. a

11.    c    12. b    13. d    14. d    15. b

16.    a    17. a    18. d    19. d    20. a

Related WebApps

 * Function Calculator

title: Theory of Equations


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User:Eml4500.f08.delta 6.guzman/MATLAB Two Bar Truss Example

Two Bar Truss Example (MATLAB)
It is important to understand how to solve a bar truss system employing a finite element MATLAB code in order to compare the resulting matrices with those obtained from hand calculations.

Take this truss example with the following: Data

MATLAB Code Note: this MATLAB code was obtained from Dr. L. Vu-Quoc website http://clesm.mae.ufl.edu/~vql/courses/fead/2008.fall/codes/twoBarTrussEx.txt

Results:


text:
Trigonometry

Trigonometry is the study of angles. It is a branch of mathematics that is used for algebraic solving in geometry. With the knowledge of trigonometry anything from calculus to vector algebra can be solved.

Trigonometry is used in finding out the heights of tall objects like mountains, trees, etc. The tool used in trigonometry for this purpose is called clinometer.

''"There is perhaps nothing which so occupies the middle position of mathematics as trigonometry."-J. F. Herbart''

Lessons

 * /Functions/
 * /Identities/

 * Divisions of trigonometry
 * /Plane/
 * /Spherical/
 * /Analytic/
 * Coordinate
 * /Polar/
 * /Transformation/
 * /Conic/
 * /Barycentric/

 * Triangle Geometry/
 * /Trigonometric Analysis/

title: Trigonometry


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User:Eml4500.f08.Lulz.js/HW1

Part 11: Submatrices and colon notation
This section of the primer discusses more advanced matrix manipulation techniques.

 * The first example shows the use of non-integral increments and boundaries in matrix declaration:
 * The next example employs a table of  values to demonstrate matrix generation from a set of vectors:
 * The following set of examples display the various uses of the colon operator in MATLAB:
 * The final set of examples for this section explores reordering of vectors using both colon notation and the  and   commands.

Part 12: M-files: script files, function files
This section of the primer discusses MATLAB's M-file system for executing command sequences. MATLAB handles two types of M-files: scripts and functions.

 First, the MATLAB code saved as :
 * The first example shows the use of a simple script for creating a matrix:

And then the execution of the script to create the matrix:

 First, the MATLAB code saved as the function :
 * The next example employs a function to generate a n-by-m matrix of random numbers from 0 to 9:

And the calling of the function to generate a 4-by-5 matrix:

 First, the MATLAB code saved as : Next, two function calls are made to. The first displays the aforementioned multiple outputs: And the second displays a single output (no square bracket required): Lastly, the  command is used to display the header for  :
 * The next function highlights the capability to have multiple outputs:
 * The last function examples highlight the use of the  command to pass functions to other functions:

First, the MATLAB code for the function, which finds the zero of a passed function via bisection method:

And the function call to find the zero of the sin function near x = 2:
 * Finally, the primer notes the usefulness of the  command to identify the nature of functions:

Part 13: Text strings, error messages, input
This section of the primer discusses the various methods of handling text in MATLAB.

  Note that when the  command is used, the resulting output is displayed in red.
 * The first example shows the necessity of single quotes for defining a simple text string:
 * The next example features use of the  and   commands:
 * The final example features text-based input:

title: User:Eml4500.f08.Lulz.js/HW1


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User:Eml4500.f08.Lulz.js/HW2

Derivation of $$\vec{\tilde{i}}\cdot\vec{j}$$

The definition of $$\vec{\tilde{i}}$$ given in the lecture notes is: $$\vec{\tilde{i}}=\cos\theta^{(e)}\vec{i}+\sin\theta^{(e)}\vec{j}$$

Taking the dot product of $$\vec{\tilde{i}}$$ and $$\vec{j}$$ yields: $$\vec{\tilde{i}}\cdot\vec{j}=\cos\theta^{(e)}\vec{i}\cdot\vec{j}+\sin\theta^{(e)}\vec{j}\cdot\vec{j}$$

By definition, $$\vec{i}\cdot\vec{j}$$ is equal to 0, and $$\vec{j}\cdot\vec{j}$$ is equal to 1. Using these relationships reduces the above equation to: $$\vec{\tilde{i}}\cdot\vec{j}=\sin\theta^{(e)}$$

Previously, the direction cosine $$m^{\left(e\right)}$$ was defined as $$ \cos \left(\frac{\pi }{2}-\theta ^{(e)}\right)$$, which is equivalent to $$\sin\theta^{\left(e\right)}$$. Thus the derived expression for $$\vec{\tilde{i}}\cdot\vec{j}$$ is: $$\vec{\tilde{i}}\cdot\vec{j}=m^{(e)}$$

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Talk:Syphilis

The page on std's has general background information on all of the common std's with the exception of syphilis. Syphilis is an std that has deadly consequences, if untreated, and is easily misdiagnosed. With respect to all of the other std's described on the std page, this page contains very pertinent information to this site and should not be disregarded due to a fear that other encyclopedias have similar information. The majority of wikiversity information can be found in alternate sources. Unless the information has an error or false statement in it this page can only add to the plethera of information contained in wikiversity.

title: Talk:Syphilis


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User:Eml4500.f08.jamama.adam/MatlabHomework2

Solving a Two-Bar Truss System using MatLab
The same system can be solved using MatLab.

The following code solves for the parameters: "Stiffness matrices, displacements, and reactions."

Here are the calculaled results from MatLab:

It can be observed that the values that were calculated previously are similar to the ones calculated using Matlab.

The two "k_local" that MatLab calculated are our k(1) and k(2) respectively.

The two lowercase k's (k) from MatLab are our local stiffness matrices k(1) and k(2).

The Uppercase k (K) found in the Matlab Results is equivalent to our global stiffness matrix K.

The "R" matrix is the know forces matrix.

The "d" matrix is the displacement matrix.

The "reaction" matrix is for the reaction forces.

And the "results" matrix is to present the


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User:EML4500.f08.JAMAMA/MATLABTWOBAR

Solving a Two-Bar Truss System using MatLab
The same system can be solved using MatLab.

The following code solves for the parameters: "Stiffness matrices, displacements, and reactions."

Here are the calculaled results from MatLab:

It can be observed that the values that were calculated previously are similar to the ones calculated using Matlab.

The two "k_local" that MatLab calculated are our k(1) and k(2) respectively.

The two lowercase k's (k) from MatLab are our local stiffness matrices k(1) and k(2).

The Uppercase k (K) found in the Matlab Results is equivalent to our global stiffness matrix K.

The "R" matrix is the know forces matrix.

The "d" matrix is the displacement matrix.

The "reaction" matrix is for the reaction forces.

And the "results" matrix is to present the

--Eml4500.f08.jamama.justin 18:27, 26 September 2008 (UTC)

title: User:EML4500.f08.JAMAMA/MATLABTWOBAR


text:
Standard solutions

'Preparing a Primary Standard Solution'

A primary standard is one that:

 * Is available in a highly pure state.
 * Is stable in air.
 * Is readiily soluble in water.
 * Has a high relative formula mass.
 * Will react rapidly when in solution during volumetric analysis.

The recipe:

 * 1) Work out how much you want (volume) and select the appropriate glassware!
 * 2) Wash out the glassware carefully.
 * 3) Work out how much solid you require.
 * 4) Weigh a clean and empty weighing bottle.
 * 5) Weigh out your solid using a clean spatula on a balance which will weight to 0.01g.
 * 6) Transfer the solid from your weighing bottle into a clean beaker.
 * 7) Either wash out any remaining solids into the beaker, using distilled water, or re-weigh the weighing bottle and adjust the mass used in your laboratory book for the solid that was not transferred.
 * 8) Use a glass stirring rod to gently break up any crystals to a smaller size.
 * 9) Dissolve the solid in a mixture of about 80% of your final required volume.
 * 10) Decant the liquid into a volumetric flask.
 * 11) Rinse out the beaker with a little more solvent and add the washings to the flask (REPEAT).
 * 12) Top up the volume with distilled water from a dropping pipette until the bottom of the meniscus is level with the graduation.
 * 13) Place the stopper in the volumetric and invert several times to ensure complete mixing and a homogeneous solution.
 * 14) Calculate the molarity (concentration) of the solution.
 * 15) Transfer your solution to a clean bottle.
 * 16) Label the solution with your name, date, name of solution and molarity.

title: Standard solutions


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User:Eml4500.f08.RAMROD.A/MyContributionHW2

Statically Indeterminate 3Truss System
A look at the system with global nodes identifited:

Element Picture
We want to look at the elemental picture with $$\displaystyle {\color{Green} Global }$$ and $$\displaystyle {\color{Blue} Local }$$ coordinate systems:

While the $$\displaystyle {\color{Blue} Local }$$ Coordinates are denoated by $$\displaystyle {\color{Blue} (\tilde{x}, \tilde{y}) } $$.

 * $$\displaystyle \tilde{x} $$. is defined as along the element from local node 1 to 2.

Vector Notation

 * A common occurrence when dealing with forces such as P or the lengths of elements is to use vector notation: $$\displaystyle ({i}, {j})$$ a long with the angle between $$\displaystyle x $$ and $$\displaystyle \tilde{x}$$; denoted as $$\displaystyle \theta $$.

Next Step of Finite Element Method
It is important to go back to the equation described in the recipe from [earlier articles] on Finite Element Method:

K(e) * d(e) = f(e)

Where:

 * K is the element stiffness matrix

 * d is the element displacement matrix

 * f is the element force matrix

It is now possible to determine ke using the aforementioned $$\displaystyle \theta $$ and what we call the direction cosines

Direction Cosines
$$\displaystyle l^e $$ and $$\displaystyle m^e $$.

 * $$\displaystyle l^e = \cos (\theta) $$

 * $$\displaystyle m^e = \cos (90-\theta) = \sin (\theta) $$

Element Stiffness Matrix, k
$$\displaystyle \mathbf{k^e}=\frac{E^e*A^e}{L^e}\left[Direction Cosine Matrix \right] $$

Where: "$\displaystyle L = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} $"
 * E = Young's Modulus of element
 * A = Cross sectional Area of element
 * L = Length of element

When the cosine matrix is added in, we get: $$\displaystyle \mathbf{k^e}=\frac{E^e*A^e}{L^e} \begin{bmatrix} l^2 & (l\cdot m) & -l^2  & -(l\cdot m) \\ \\(l\cdot m) & m^2 & -(l\cdot m) & -m^2 \\ \\-l^2 & -(l\cdot m) & l^2 & (l\cdot m) \\ \\-(l\cdot m) & -m^2 & (l\cdot m) & m^2 \end{bmatrix}$$


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Talk:Paleontology/Anatomy

--80.87.90.102 10:09, 26 September 2008 (UTC)I am a student of KNUST specialising Wildlife and Range management. I want information on evaluation to assist scientific authorities in making non-detriment findings.80.87.90.102 10:09, 26 September 2008 (UTC)sammy

title: Talk:Paleontology/Anatomy


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User:EML4500.f08.RAMROD.E/HW2

Step 4) Elimination of known degrees of freedom in order to reduce the global Force-Displacement relationship.

For this problem, the reduction is due to the fact that global nodes 1 and 3 are fixed and therefore have zero displacement, or:

$$d_{1} = d_{2} = d_{5} = d_{6} = 0$$

Setting these displacements to zero is the equivalent of eliminating the corresponding columns (1,2,5 & 6) of the global stiffness matrix, K. This elimination is shown below.

$$K = \begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24}\\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix}$$

In addition to removing these columns because of the fixed boundary conditions, rows 1, 2, 5, & 6 can also be eliminated due to the principle of virtual work (PVW). After deletion of these rows, the resulting K becomes:

$$K = \begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44}\\ \end{bmatrix}$$

And, the resulting global Force-Displacement relationship is:

$$\begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44}\\ \end{bmatrix} * \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix} = \begin{Bmatrix} F_{3}\\ F_{4} \end{Bmatrix}$$

We have already solved for each element in K previously such that

$$K = \begin{bmatrix} 3.0625 & -2.175\\ -2.175 & 2.6875\\ \end{bmatrix}$$

Looking at the problem drawing, we can see that F3 is zero and that F4 is equal to -7. Plugging these values and the values of K into the above global F-D relationship gives the relationship:

$$ \begin{bmatrix} 3.0625 & -2.175\\ -2.175 & 2.6875\\ \end{bmatrix} * \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix} = \begin{Bmatrix} 0\\ -7 \end{Bmatrix}$$

Now, to solve for the unknown displacements, we can multiply both sides by K-1, where:

$$K^{-1} = \frac{1}{det(K)}\begin{bmatrix} K_{33} & -K_{34}\\ -K_{43} & K_{44} \end{bmatrix} = \begin{bmatrix} 0.875 & 0.621\\ 0.621 & 0.768 \end{bmatrix}$$

It is important to not that K-1 is not the same as the transpose of K(KT).

Therefore, the equation to solve for the unknown displacements are:

$$\begin{bmatrix} 0.875 & 0.621\\ 0.621 & 0.768 \end{bmatrix}*\begin{Bmatrix} 0\\ -7 \end{Bmatrix} = \begin{Bmatrix} d_{3}\\ d_{4} \end{Bmatrix}$$

Solving for these displacements gives

$$d_{3} = -4.347 $$ $$d_{4} = -5.376$$

 Step 5: Finding Reactions  The final step for this problem is to solve for the unknown reactions at global nodes 1 and 3. For this we use the element F-D relationships:

$$K^{(1)}* d^{(1)} = F^{(1)}$$ $$K^{(2)}* d^{(2)} = F^{(2)}$$

In the above equation, K(e), d(e), and  half of f(e) are known for each element, e. For this situation, the F-D relationships with all known values substituted is:

$$\begin{Bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16}\\ \frac{3\sqrt{3}}{16}\ & \frac{3}{16} \end{Bmatrix} \begin{Bmatrix} -4.347\\ -5.376 \end{Bmatrix}= \begin{Bmatrix} f^{(1)}_1\\ f^{(1)}_2\end{Bmatrix}$$

$$\begin{Bmatrix} \frac{5}{2} & \frac{-5}{2}\\ \frac{-5}{2} & \frac{5}{2} \end{Bmatrix} \begin{Bmatrix} -4.347\\ -5.376 \end{Bmatrix}= \begin{Bmatrix} f^{(2)}_3\\ f^{(2)}_4 \end{Bmatrix}$$

Using this equation, we can now solve for the f(1) matrix and thus we find the values for the reactions: $$f^{(1)}_{1} = F_{1} = -4.191$$ $$ f^{(1)}_{2} = F_{2} = -2.420$$

Using this same procedure applied to element 2, we can solve for the values of the other two unknown reactions giving us: $$f^{(2)}_{3} = F_{5} = 2.564$$ $$f^{(2)}_{4} = F_{6} = -2.564$$

title: User:EML4500.f08.RAMROD.E/HW2


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User:Eml4500.f08.RAMROD.F/HW1

Chris Fontana's Part
= Notes Summary for Friday, September 5 =

Intro
We discussed the ordered steps to slove a the truss problem from Septmeber 3. We discussed a generalized format to begin with the diagram of a model system and to work our way down to computing reaction forces.

Step 1: Global Picture
The global picture is the overall structural picture of the system. The picture includes the following:
 * 1) Global Forces
 * 2) Global Degrees of Freedom (aka DOF)

The displacement definition are partioned into:
 * 1) Known part, eq., fixed DOF, constraints
 * 2) Unknown part solved by using Finite Elemental Analysis (see the truss example above)

Similarly, the global forces are partioned into:
 * 1) Known part: Applied forces
 * 2) Unknown part: Reactions

Step 2: Element Picture

 * 1) Element definitions
 * 2) Element forces

These can be found in either global coordinate systems or in local coordinate systems

Step 3: Global Force Diagram relationship

 * 1) Element stiffness matrices in global coordinates
 * 2) Element force matrices in global coordinates
 * 3) assembly of element stiffness matrix and element force matrix into global force displacement relation.

K · d = F

(NxN) (Nx1) = (Nx1)

Step 4: Elimination of DOF
Eliminate degrees of known degrees of freedom to reduce the golbal force displacement relationship. (The Stiffness Matrix is non-singular-invertible)
 * 1) M = the number of unkwown DOF
 * 2) N = the number of both known and unknown DOF

K · d = F

(MxN) (Mx1) = (Mx1)                       M>N

Stiffness Matrix K nonsingular => k-1 exists (K is invertible)

K-1 · d = F

Step 5: Compute Element Forces
Compute element forces from known displacements => element stresses

Step 6: Compute Reactions
Compute reaction forces or the unknown forces.

title: User:Eml4500.f08.RAMROD.F/HW1


text:
User:Eml4500.f08.RAMROD.F/HW2

=Friday September 19 2008=

Element 1:

K(1) · d(1) = F(1)

$$= \begin{bmatrix} 0.5625 & 0.032476 & -0.5625 & -0.032476 \\ 0.032476 &

0.1875 & -0.032476 & -0.1875 \\ -0.5625 & -0.032476 & 0.5625 & 0.032476 \\ -0.032476 & -0.1875 & 0.032476 & 0.1875 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}$$

You may simplify the matrix because of the boundary conditions e.g. the first two numbers in the displacement matrix are zero.

$$= \begin{bmatrix} -0.5625 & -0.032476 \\ -0.032476 & -0.1875 \\ 0.5625 & 0.032476 \\ 0.032476 & 0.1875 \end{bmatrix} \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

$$=\begin{Bmatrix} -4.4378 \\ -2.5622 \\ +4.4378 \\ +2.5622 \end{Bmatrix} = \begin{Bmatrix} F_1^1 \\ F_2^1 \\ F_3^1 \\ F_4^1 \end{Bmatrix}$$

Reactions forces are actually $$F_1^1$$ & $$F_2^1$$

Internal forces are actually $$F_3^1$$ & $$F_4^1$$

Element 1:

$$P_1^1 = \sqrt((F_1^1)^2 + (F_2^1)^2)$$

$$P_1^1 = \sqrt((-4.4378)^2 + (-2.5622)^2$$

$$ P_1^1 = 5.69064                     $$

Element 2:

$$P_1^2 = \sqrt((F_1^2)^2 + (F_2^2)^2)$$

$$P_1^2 = \sqrt((-4.4378)^2 + (-2.5622)^2$$

$$P_1^2 = 5.69064  $$

Method Two: Statics
Solving the two bar truss system using Euler Cut Principle.

Back to FE Solution: How to bring P back into the big picture?

Ans: Using the Euler cut method to isolate node two, equilibrium can be established at global node 2. Statics can be applied to the system by using summation of the forces in the x and y directions. This provides two equations with two unknowns that can be solved by using a system of equations.

$$\sum{F_x}= P_1cos30 - P_2cos45=0$$

$$\sum{F_y}=7- P_1cos30 -P_2cos45=0$$

$$\sum{F_x}= P_1 \frac{\sqrt{3}}{2} - P_2\frac{\sqrt{2}}{2}=0$$

$$\sum{F_y}=7- \frac{P_1}{2}-P_2\frac{\sqrt{2}}{2}=0$$

$$P_1=5.122436\;$$

$$P_2 = 6.27603\;$$

title: User:Eml4500.f08.RAMROD.F/HW2


title: File:Truss15.JPG


text: File:Truss Variables15.JPG

title: File:Truss Variables15.JPG


text:
User talk:Trevorl314

 Hello Trevorl314, 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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User:Eas4200c.f08.vqcrew.c/Homework 2/Contributing Team Members

HW2 Contributing Team Members
The following students contributed to this report:

John Saxon Eas4200c.f08.vqcrew.c 20:46, 26 September 2008 (UTC)

Adam Edstrand Eas4200c.f08.vqcrew.a 20:25, 26 September 2008 (UTC)

Javier Stober Eas4200c.f08.vqcrew.b 20:26, 26 September 2008 (UTC)

Darin Toscano Eas4200c.f08.vqcrew.d 20:08, 26 September 2008 (UTC)

Kevin Klauk Eas4200c.f08.vqcrew.E 16:22, 26 September 2008 (UTC)


text:
Talk:Elasticity/Williams asymptotic solution

Please check formulae 27 to 29, since in the skew-symmetric situation the sigma should be null and only the tau should exist


text: File:Hwtruss.jpg

title: File:Hwtruss.jpg


title: File:Hwpic2.JPG


text:
User:Eas4200c.f08.carbon.clausen

Aaron J Clausen

Is also User:Eml4500.f08.group.clausen

Team Carbon - EAS4200c

Team Group - EML4500

title: User:Eas4200c.f08.carbon.clausen


text:
User:Eml4500.f08.group.clausen

Aaron J Clausen

Is also User:Eas4200c.f08.carbon.clausen

Team Carbon - EAS4200c

Team Group - EML4500

title: User:Eml4500.f08.group.clausen


text:
User talk:Oscarmarian

 Hello Oscarmarian, 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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User:Eml4500.f08.echo.mott/ECHO.HW2

This page contains material required for EML4500 Homework Report No.2. The previous version of this page was submitted to the class.

Statically Determinate Problem
Here is an example of a statically determinate problem. It can be solved using the traditional methods learned in statics.

2 Force body example:

 * Drawing and Labeling the Global Picture:

 * Breaking it down into two elements:

$$\sum F_x = 0; R_1cos( \theta_1) - R_2cos( \theta_2) + Pcos( \theta_P) = 0$$

$$\sum F_y = 0; R_1sin( \theta_1) - R_2sin( \theta_2) + Psin( \theta_P) = 0$$

$$\sum M_1 = 0; R_2sin( \theta_2)*(l_2cos( \theta_2) + l_1cos( \theta_1)) + Psin( \theta_P)*(l_1cos( \theta_1)) = 0$$

Three equations and three unknowns, therefore it is statically determinate.

Statically Indeterminate
Some problems cannot be solved using the 3 equations from statics. That is, if the problem contains more than 3 unknowns it cannot be solved using traditional statics methods.

The example on the left can be solved because each member is a 2-force member. Hence, there are only 3 unknowns. The example on the right is a statically indeterminate structure because there are 6 unknowns.

The Element Stiffness Matrix
After the diagrams have been drawn for each element in the structure, the element stiffness matrix should be computed for each element. The stiffness matrix k looks like this:

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

The scalar $$k^{(e)}$$ can be found from the properties of the structure where E is the Elastic Modulus, A is the cross-sectional area of the element, and L is the length of the element.

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

$$l^{(e)}$$ is the x-length of the element in reference to global coordinates and $$m^{(e)}$$ is the y-length of the element from a global perspective. They can be thought of as the length of the element projected onto global coordinates from local coordinates.

The x-length can be found by using: $$l_{s}= \frac{x_{2}-x_{1}}{L}$$

The y-length of the element can be found by using: $$l_{s}= \frac{y_{2}-y_{1}}{L}$$

With the inclined angle referenced to the horizontal, the horizontal and vertical lengths can be found by:

$${l_s}^x = {cos \theta}$$

$${l_s}^y = sin \theta$$

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

$$m^{(e)}= \mathbf{\tilde{i}} \cdot \mathbf{j}=cos( \frac{ \pi}{2} - \theta^{(e)})=sin( \theta^{(e)})$$

$$\mathbf{\tilde{i}}=cos( \theta^{(e)})\mathbf{i}+sin( \theta^{(e)})\mathbf{j}=l^{(e)}+m^{(e)}$$

$$\mathbf{\tilde{i}}=cos( \theta^{(e)})\mathbf{i}+sin( \theta^{(e)})\mathbf{j}$$

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

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

After reviewing the stiffness matrix two observations can be made:

1) Of all the numbers in the matrix, only three need to be computed. The other coefficients have the same absolute values.
 * $$l_s^{}m_s $$
 * $${l_s}^2$$
 * $${m_s}^{2}$$

2) Matrix k is symmetric. That is $$k_{i j}^{1}=k_{j i}^{1}$$.

The Global Stiffness Matrix
Consider the Global Force-Displacement Relationship for a Truss with 2 elastic bars. There are 6 degrees of freedom in the global picture.

Written in compact notation we have the following:

$$\begin{bmatrix}K_{ij}^{}\end{bmatrix}_{6x6} \left\{d_{ij}\right\}_{6x1} = \left\{F_{i}\right\}_{6x1}$$

In the more general form $${nxn}$$ :

$$\sum_{j=1}^{6} K_{ij}d_{j}= F_{i}$$; i = 1, 2, ..., 6

$$K_{nxn} = \begin{bmatrix}K_{ij}^{}\end{bmatrix}_{nxn}$$ = global stiffness matrix

$$d_{nx1} = \left\{d_{j}^{}\right\}_{nx1}$$ = global displacement matrix

$$F_{nx1} = \left\{F_{i}^{}\right\}_{nx1}$$ = global force matrix

Now recall the Element Force-Displacement relationship:

$$k_{4x4}^{(e)} d_{4x4}^{(e)} = f_{4x1}^{(e)}$$

$$k_{4x4}^{(e)} = \begin{bmatrix}k_{ij}^{(e)}\end{bmatrix}_{4x4}$$ = element stiffness matrix

$$d_{4x1}^{(e)} = \left\{d_{j}^{(e)}\right\}_{4x1}$$ = element displacement matrix

$$f_{4x1}^{(e)} = \left\{f_{i}^{(e)}\right\}_{4x1}$$ = element force matrix

Use an assembly process to go from element matrices(stiffness, displacement, force) to global matrices.

 * First, Identify the correspondence between the element displacement degrees of freedom's and the global displacement degrees of freedom's.

Global Level: $$\left\{d_1, d_2, \dots, d_6\right\}$$

Element level:
 * $$\left\{d_1^{(1)}, d_2^{(1)}, d_3^{(1)}, d_4^{(1)}\right\}$$
 * $$\left\{d_1^{(2)}, d_2^{(2)}, d_3^{(2)}, d_4^{(2)}\right\}$$

The relationship between global-local(element) dof's are as follows:

$$\text{Global Node 1} = \begin{cases} d_1 = d_1^{(1)} \\ d_2 = d_2^{(1)}\end{cases}$$

$$\text{Global Node 2} = \begin{cases} d_3 = d_3^{(1)} = d_1^{(2)} \\ d_4 = d_4^{(1)} = d_2^{(2)}\end{cases}$$

$$\text{Global Node 3} = \begin{cases} d_5 = d_3^{(2)} \\ d_6 = d_4^{(2)}\end{cases}$$

Conceptual step of the assembly:

Taking the global stiffness matrix and substituting in the appropriate values:

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

Plugging in values,

$$K_{11}=k_{11}^{(1)}=\frac{9}{16}=0.5625$$

$$K_{12}=K_{21}=k_{12}^{(1)}=\frac{3\sqrt{3}}{16}=0.3248$$

$$K_{13}=K_{31}=k_{13}^{(1)}=-k_{11}^{(1)}=-0.5625$$

$$K_{14}=K_{23}=K_{32}=K_{41}=k_{14}^{(1)}=-k_{12}^{(1)}=-0.3248$$

$$K_{22}=k_{22}^{(1)}=\frac{3}{16}=0.1875$$

$$K_{24}=K_{42}=k_{24}^{(1)}=-k_{22}^{(1)}=-0.1875$$

$$K_{33}=k_{33}^{(1)}+k_{11}^{(2)}=	\begin{matrix} \frac{9}{16} \end{matrix}+\begin{matrix} \frac{5}{2} \end{matrix}=3.0625$$

$$K_{34}=k_{34}^{(1)}+k_{12}^{(2)}=	\begin{matrix} \frac{3\sqrt{3}}{16} \end{matrix}+\begin{matrix} \frac{-5}{2} \end{matrix}=-2.1752$$

$$K_{35}=K_{53}=k_{13}^{(2)}=-k_{11}^{(2)}=\frac{-5}{2}=-2.5$$

$$K_{36}=K_{45}=K_{54}=K_{63}=k_{14}^{(2)}=-k_{12}^{2}=2.5$$

$$K_{43}=K_{34}=k_{43}^{(1)}+k_{21}^{(2)}=k_{34}^{(1)}+k_{12}^{(2)}=-2.1752$$

$$K_{44}=k_{44}^{(1)}+k_{22}^{(2)}=\frac{3}{16}+\frac{5}{2}=2.6875$$

$$K_{46}=K_{64}=k_{24}^{(2)}=\frac{-5}{2}=-2.5$$

$$K_{55}=k_{33}^{(2)}=\frac{5}{2}=2.5$$

$$K_{56}=K_{65}=k_{34}^{(2)}=\frac{-5}{2}=-2.5$$

$$K_{66}=k_{44}^{(2)}=\frac{5}{2}=2.5$$

$$K_{15}^{}=K_{16}=K_{25}=K_{26}=K_{51}=K_{52}=K_{61}=K_{62}=0$$

$$\mathbf{K}=\begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0\\0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0\\-0.5625 & -0.3248 & 3.0625 & -2.1752 & -2.5 & 2.5\\-0.3248 & -0.1875 & -2.1752 & 2.6875 & 2.5 & -2.5\\0 & 0 & -2.5 & 2.5 & 2.5 & -2.5\\0 & 0 & 2.5 & -2.5 & -2.5 & 2.5\end{bmatrix}$$

Solving for Unknown Displacements
Next, the elimination of known degree's of freedom can create the following matrices.

$$ d_1^{} = d_2^{} = d_5^{} = d_6^{}=0 $$

$$	\mathbf{K}\begin{Bmatrix} d_1=0\\d_2=0\\d_3\\d_4\\d_5=0\\d_6=0 \end{Bmatrix}=\begin{vmatrix} k_{13} & k_{14} \\ k_{23} & k_{24}\\k_{33} & k_{34}\\k_{43} & k_{44}\\k_{53} & k_{54}\\k_{63} & k_{64}\end{vmatrix}\begin{Bmatrix} d_3\\d_4\end{Bmatrix}=\mathbf{F}$$

Applying fixed boundary conditions allows us to delete the corresponding columns in the global stiffness matrix K

By the Principal of Virtual Work (PVW) we're able to delete the corresponding rows (in this case - 1, 2, 5, 6). Corresponding rows of F are also deleted.

The resulting FD relation is: $$ \begin{bmatrix} k_{33} & k_{34} \\ k_{43} & k_{44} \end{bmatrix} \begin{Bmatrix} d_3\\d_4\end{Bmatrix}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}$$

$$\mathbf{F}=\begin{Bmatrix} F_3\\F_4\end{Bmatrix}=\begin{Bmatrix} 0\\P\end{Bmatrix}$$

And P =>

What is the inverse of K without using a calculator?

Taking the determinant of K:

$$det \mathbf{K}=k_{33}k_{44}-k_{k34}k_{43}$$

$$\mathbf{K}^{-1}=\frac{1}{det\mathbf{K}}\begin{bmatrix} k_{44} & -k_{34} \\ -k_{43} & k_{33} \end{bmatrix}$$

Verifying:

$$\mathbf{K}\mathbf{K}^{-1}=\mathbf{K}^{-1}\mathbf{K}=\mathbf{I}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

This is essentially:

$$\frac{1}{det(\mathbf{K})}\begin{bmatrix} det(\mathbf{K}) & 0 \\ 0 & det(\mathbf{K}) \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Note: $$\mathbf{K}^T=\begin{bmatrix} k_{33} & k_{43} \\ k_{34} & k_{44} \end{bmatrix}$$

Therefore: $$\mathbf{K}^{-1}\ne \ \frac{1}{det(\mathbf{K})}\mathbf{K}^T$$

Going back to the 2 bar truss problem...

$$\begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix}=\mathbf{K}^{-1}\begin{Bmatrix} 0 \\ P \end{Bmatrix}=\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

Solving for Reactions
Now we must compute the reactions:

 * Method 1:

Use element FD relations.

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

Taking a look at element 1 again and applying the knowns:

$$d^{(1)}=\begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}$$

Taking a look at element 2:

$$\mathbf{k}^{(2)}\mathbf{d}^{(2)}=\mathbf{f}^{(2)}$$

$$d^{(1)}=\begin{Bmatrix} 4.352 \\ 6.1271 \\ 0 \\ 0 \end{Bmatrix}=\begin{Bmatrix} d_1^{(2)} \\ d_2^{(2)} \\ d_3^{(2)} \\ d_4^{(2)} \end{Bmatrix}$$

Code

 * The 2-bar truss example outlined above can be solved using MATLAB. The code is shown here:

clear all;

e = [3 5]; A = [1 2]; P = 7; L=[4 2]; alpha = pi/3; beta = pi/4;

nodes = [0, 0]; L(1)*cos(pi/2-alpha), L(1)*sin(pi/2-alpha); L(1)*cos(pi/2-alpha)+L(2)*sin(beta),L(1)*sin(pi/2-alpha)-L(2)*cos(beta)];

dof = 2*length(nodes);

conn = [1,2; 2,3];

lmm = [1, 2, 3, 4; 3, 4, 5, 6];

elems = size(lmm,1);

K=zeros(dof); R = zeros(dof,1); debc = [1, 2, 5, 6];

ebcVals = zeros(length(debc),1);

 * The load vector

R = zeros(dof,1); R(4) = P;

 * Assembly of global stiffness matrix

K=zeros(dof);

for i=1:elems lm=lmm(i,:); con=conn(i,:); k_local=e(i)*A(i)/L(i)*[1 -1; -1 1] k=PlaneTrussElement(e(i), A(i), nodes(con,:)) K(lm, lm) = K(lm, lm) + k;     end

K

R

 * The nodal solution and reactions

[d, reactions] = NodalSoln(K, R, debc, ebcVals)

results=[];

for i=1:elems results = [results; PlaneTrussResults(e, A, ...                       nodes(conn(i,:),:), d(lmm(i,:)))]; end

format short g

results

Results

 * The code shown above generates the following results:

k_local = 0.75       -0.75        -0.75         0.75

k = 0.5625     0.32476      -0.5625     -0.32476      0.32476       0.1875     -0.32476      -0.1875      -0.5625     -0.32476       0.5625      0.32476     -0.32476      -0.1875      0.32476       0.1875

k_local = 5   -5           -5     5

k = 2.5        -2.5         -2.5          2.5         -2.5          2.5          2.5         -2.5         -2.5          2.5          2.5         -2.5          2.5         -2.5         -2.5          2.5

K = 0.5625     0.32476      -0.5625     -0.32476            0            0      0.32476       0.1875     -0.32476      -0.1875            0            0      -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5     -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5            0            0         -2.5          2.5          2.5         -2.5            0            0          2.5         -2.5         -2.5          2.5

R = 0    0     0     7     0     0

d = 0           0        4.352       6.1271            0            0

reactions = -4.4378     -2.5622       4.4378      -4.4378

results = 1.7081      5.1244       8.5406       5.1244       17.081       0.6276       1.8828        3.138       1.8828        6.276

 * The previous code was taken from the EML4500 course website. A text version of the data can be found                                                                                here

Contributing Members HW2
Chris Mott 18:40, 26 September 2008 (UTC)

Charles 18:41, 26 September 2008 (UTC)

Bradley LaCroix Eml4500.f08.echo.lacroix 18:54, 26 September 2008 (UTC)

C. Sell 18:58, 26 September 2008 (UTC)

Brandon Sell Eml4500.f08.echo.sell.br 19:10, 26 September 2008 (UTC)

Jason Bruce 18:40, 26 September 2008 (UTC)

title: User:Eml4500.f08.echo.mott/ECHO.HW2


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User:Davichito

This is my page.

Projects

 * Collaboration
 * Collaborative writing
 * Game programming
 * Fiction writing
 * Wiki Campus Radio
 * Music performances
 * Parqués
 * Simulated reality

title: User:Davichito


text:
User talk:Davichito

Feel free to post messages here

 Hello Davichito, 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! --mikeu talk 16:12, 26 September 2008 (UTC)

Thucydides
Hi Davichito,

The Thycydides course is inactive. I will make a sign on it. You could, of course, read the text on your own. Sorry.Daanschr 07:00, 2 October 2008 (UTC)

IRC
I haven't been on IRC since I last talked to you on #fiction-wika and I don't think I'll be using it much at all since I am not comfortable with it. Perhaps it is due to my infamiliarity with the VisualIRC client. How do I use the Wiki interface? Maybe that would be better... KirbyPuckettFan 01:42, 5 November 2008 (UTC)

new audio

This new audio file includes most of the newly recorded versions of your questions and your answer. It does not include question #1 because there was some low frequency noise in that part. All of the new audio was very low volume. Particularly on the "answer" part, the volume was not even, as if the microphone was moving or something. Control panel: in the advanced settings for your sound input, is there anything like "microphone boost" as an option? --JWSchmidt 14:37, 7 November 2008 (UTC)

fiction
Long time, no talk! I've just updated my story page on Ficciones. Check it out and let's start discussing the story in more detail. I've also written to John and told him that I'm interested in the three of us collaborating on this. KirbyPuckettFan 01:58, 25 November 2008 (UTC)
 * Que pasa? It's been too long, I know! I apologize for leaving you and John hanging with the story...school and everything else got in the way of writing, and then my PC crashed and I lost all the plot notes I had written!  Hopefully, the three of us can piece it back together. KirbyPuckettFan 05:47, 30 August 2009 (UTC)

Scoreing request
Any chance of scoring this with chords for MIDI?

http://en.wikisource.org/wiki/Image:Eternal_Father,_Strong_to_Save.jpg

Sfan00 IMG 14:33, 2 December 2008 (UTC)

title: User talk:Davichito


title: File:Element2final.JPG


text:
User:Eml4500.f08.Lulz.Layton.Eric/Hw2/Hw2 contributing team members

Contributing Team Members
Aaron Fisher - Eml4500.f08.Lulz.fisher 18:12, 26 September 2008 (UTC)

Eric Layton - Eml4500.f08.Lulz.Layton.Eric 17:55, 26 September 2008 (UTC)

Sam Miorelli - Eml4500.f08.lulz.abcd 18:01, 26 September 2008 (UTC)

Benjamin Mitchell - Eml4500.f08.Lulz.mitchell.bm 17:58, 26 September 2008 (UTC)

John Saxon - Eml4500.f08.Lulz.js 16:27, 26 September 2008 (UTC)

Andrew Strack - Eml4500.f08.lulz.strack 18:50, 26 September 2008 (UTC)


text: File:Mohrsss.PNG

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text:
User:EML4500.f08.A-team.kirley

Loops in MatLab
A loop is a programming tool in Matlab that is used to do repetitive work. It saves the user time. In the tutorial, matrix a represents the probabilities that two island populations will move or stay put on their side of the island. x is a column vector representing the initial populations.

a=    0.8  0.1 0.2 0.9

x=    1 Subscript text             0

Successive population states can be predicted by multiplying the two matrices. Using a loop, the operation can be repeated a defined number of times using the following form:

>> a = [ 0.8 0.1; 0.2 0.9 ] >> x = [ 1; 0 ] >> for i = 1:20, x = a*x, end

This operation will be repeated 20 times, and the populations of the island can be predicted 20 time units away.

Meeting 8, September 12, 2008
In compact notation:

[Kij]6x6[dj]6x1=[Fi]6x1 (more generally n x n)

ΣKijdj = Fi     i = 1, 2, ... , 6

Kn x n = [Kij} = global stiffness matrix

dn x 1 = {dj}n x 1 = global displacement matrix

Fn x 1 = {Fi}n x 1 = global force matrix

Recall elem FD relation:

k(l)4x4 d(l)4x1 = f(l)4x1

Where k, d, and f are the element stiffness, displacement, and force matrices, respectively.

Assembly of global matrices from element matrices:

Identify the correspondence between element displacement degrees of freedom and global displacement degrees of freedom

Global level: {d1, -d2, ... , d6}

Elem level:

*Elem 1: {d(1)1, d(1)2, d(1)3, d(1)3, d(1)4

*Elem 2: {d(2)1, d(2)2, d(2)3, d(2)3, d(2)4

d1 = d(1)1

d2 = d(1)2

d3 = d(1)3 = d(2)1

d4 = d(1)4 = d(2)2

d5 = d(2)3

d6 = d(2)4

Visualization of assembly:

Meeting 9, September 15, 2008

Kundefinedk(1)11 = 9/16

Kundefinedk(1)12 = .3247595

K33 = k(1)33 + k(2)11 = 9/16 + 5/2 = 3.0625

K34 = K34 = k(1)34 + k(2)12 = .3247595 + 5/2

The rest of the stiffness matrix can be computed similarly

Elimination of known dofs:

reduce global FD rel.

d1 = d2 = d5 = d6 = 0

EML4500.f08.A-team.kirley 20:25, 26 September 2008 (UTC)

Infinitesimal displacement

2 unknowns (XD YD)

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

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

Therefore

X - X_{P} = PQ\cdot \cos{\theta} $$

Y - Y_{P} = PQ\cdot \sin{\theta} $$

\frac {Y-Y_{P}}{X-X_{P}} = \tan{\theta} $$

Y-Y{P} = (\tan{\theta})\cdot(X-X_{P}) $$

Equation for a line perpendicular to above line, passing P:

Y-Y_{P} = \tan{(\theta + \frac{\pi}{2})}(X-X{P}) $$

 Three-bar Truss System

Local node numbering by convenience in assembly of K:

EML4500.f08.A-team.kirley 20:54, 8 October 2008 (UTC)

Connectivity Array "conn"
Consider the two-bar truss system (5-6):

conn(e,j) = global node number of local node j of element e

Location Matrix Master Array "lmm"

lmm(i,j) = eq. number (global dof #) for the element stiffness coefficient corresponding to the jth local dof #

Meeting 20 10/10/08
Note: consider case: $$\tilde{d}^{(e)}_4 \neq 0$$

$$ \tilde{d}^{(e)}_1 {=} \tilde{d}^{(e)}_2 {=} \tilde{d}^{(e)}_3 {=} 0$$

$$ \underline{\tilde{f}}^{(e)}_{4x1} {=} \underline{\tilde{k}}^{(e)}_{4x4} \cdot \underline{\tilde{d}}^{(e)}_{4x1} {=} \underline{0}_{4x1} \leftarrow 4^{th}$$ column of $$\underline{\tilde{k}}^{(e)}$$

Interpretation of trans. dofs

p. 19-3: $$ \underline{\tilde{d}}^{(e)} {=} \underline{\tilde{T}}^{(e)} \cdot \underline{d}^{(e)}$$

Similarly  $$ \underline{\tilde{f}}^{(e)} {=} \underline{\tilde{T}}^{(e)} \cdot \underline{f}^{(e)}$$

Also:  $$ \underline{\tilde{k}}^{(e)} \cdot \underline{\tilde{d}}^{(e)} {=} \underline{\tilde{f}}^{(e)} $$

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

If $$\underline{\tilde{T}}$$ is invertible, then:

$$ [\underline{\tilde{T}}^{(e)-1} \cdot \underline{\tilde{k}}^{(e)} \cdot \underline{\tilde{T}}^{(e)}] \cdot \underline{d}^{(e)} {=} \underline{f}^{(e)} $$

$$ \underline{\tilde{T}}^{(e)}$$ block diag. matrix (p. 19-3)

Consider a general block-diag. matrix:

Question: What is A-1

Simpler example: Diag. matrix:

$$ {=} diag[d_{11}, d_{22} , d_{33} , ... , d_{nn}] $$

$$ \underline{B}^{-1} = Diag[\frac{1}{d_{11}}, \frac{1}{d_{22}} , ... , \frac{1}{d_{nn}}]$$

Assuming $$ d_{ii} \neq 0 $$ for $$ i {=} 1, ... , n$$

For a block-diag matrix A:

$$ \underline{A} {=} Diag[\underline{D}_1, ... , \underline{D}_s]$$

$$ \underline{A}^{-1} {=} Diag[\underline{D}_1^{-1}, ... , \underline{D}^{-1}_{s}]$$

$$ \underline{\tilde{T}}^{(e)-1} {=} Diag[\underline{R}^{(e) -1}, ... , \underline{R}^{(e) -1}] $$

p. 19-2: $$ \underline{R}^{(e) T} {=} \begin{vmatrix} l^{(e)} & -m^{(e)} \\ m^{(e)} & l^{(e)} \end{vmatrix} $$

$$ \underline{R}^{(e)T}_{2x2} \cdot \underline{R}^{(e)}_{2x2} {=} \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}_{2x2} {=} \underline{I}_{2x2} $$

$$ \rightarrow \underline{R}^{(e)-1} = \underline{R}^{(e)T} $$

$$ \rightarrow \underline{\tilde{T}}^{(e)-1} = Diag[\underline{R}^{(e)T}, \underline{R}^{(e)T}] $$

$$ {=} \underbrace{ (Diag[ \underline{R}^{(e)}, \underline{R}^{(e)}]) }_{\underline{\tilde{T}}^{(e)}} $$

$$ \underline{\tilde{T}}^{(e)-1} {=} \underline{\tilde{T}}^{(e)T} $$

p. 20-1: $$ \underbrace{ [\underline{\tilde{T}}^{(e)T} \cdot \underline{\tilde{k}}^{(e)} \cdot \underline{\tilde{T}}^{(e)}]}_{\underline{k}^{(e)}} \cdot \underline{d}^{(e)} {=} \underline{f}^{(e)} $$

HW: Verify: $$ \underline{k}^{(e)} {=} \underline{\tilde{T}}^{(e)T} \cdot \underline{\tilde{k}}^{(e)} \cdot \underline{\tilde{T}}^{(e)}$$

EML4500.f08.A-team.kirley 01:19, 30 October 2008 (UTC)

Meeting 27 10/29/08
$$ \hat{\underline{w}}_{2x1} {=}$$ virtual axial displacement, corresponding to $$ q^{(e)}_{2x1} $$

$$ \underline{w}_{4x1} {=} $$ virtual displacement in global coordinate system, corresponding to $$ d^{(e)}_{4x1} $$

Replace Equations (3) and (4) into (2):

$$ \hat{\underline{w}}_{2x1} \cdot (\hat{\underline{k}}^{(e)} \underline{q}^{(e)} - \underline{p}^{(e)}) {=} 0_{1x1} $$ (2)

$$ \Rightarrow (\underline{T}^{(e)}\underline{w}) \cdot [\hat{\underline{k}}^{(e)}(\underline{T}^{(e)}\underline{d}^{(e)}) - \underline{p}^{(e)}] {=} 0 $$ for all $$ w_{4x1} $$ (5)

Recall: $$ (\underline{A}\underline{B})^T {=} \underline{B}^T\underline{A}^T $$ (6)

Recall: $$ \underline{a}_{nx1} \cdot \underline{b}_{nx1} {=} \underbrace{ \underline{a}^T_{1xn}\underline{b}_{nx1} }_{1x1 (scalar)} $$ (7)

Apply (7) and (6) in (5):

$$ \underbrace{(\underline{T}^{(e)} \underline{w})^T}_{(7)}[\underline{\hat{k}}^{(e)}(\underline{T}^{(e)}\underline{d}^{(e)}) - \underline{p}^{(e)}] {=} 0_{1x1} $$ for all $$ \underline{w}_{4x1} $$

$$ \Rightarrow \underbrace{\underline{w}^T\underline{T}^{(e)T}}_{(6)}[-] {=} 0 $$

$$ \Rightarrow \underbrace{\underline{w}}_{(7)} \cdot [(\underbrace{\underline{T}^{(e)T} \underline{\hat{k}}^{(e)} \underline{T}^{(e)} \underline{d}^{(e)}}_{\underline{k^{(e)}}}) - \underbrace{\underline{T}^{(e)T} \underline{p}^{(e)}}_{\underline{f}^{(e)}}] {=} 0 $$ for all $$ \underline{w}_{4x1} $$

$$ \Rightarrow \underline{w} \cdot [\underline{k}^{(e)}\underline{d}^{(e)} - \underline{f}^{(e)}] {=} 0 $$ for all $$ \underline{w} $$

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

So far, only discrete cases have been discussed. Now continuous cases will be discussed.

Motivational model problem: Elastic bar with varying A(x), E(x), subjected to varying axial load (distributed) and concentrated load and inertia force (dynamics). Concentrated load and inertia force are time dependent.

Meeting 33: Wednesday, 12 November 2008. EML4500
FEM vs. PVW (cont.)

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

p. 31-1: interpolation for u(x)

Apply same interpolation for w(x), i.e.

$$ w(x) {=} N_i(x)w_i + N_{i+1}(x)w_{i+1} $$

Element stiffness matrix for element i:

$$ \int_{x_i}^{x_{i+1}} [N_i^'w_i + N_{i+1}^'w_{i+1}](EA)[N_i^'d_i + N_{i+1}^'d_{i+1}]\, dx $$

$$ N_i^' :{=} \frac{dN_i(x)}{dx} $$

Likewise for $$ N_{i+1}^' $$:

$$ N_{i+1}^' :{=} \frac{dN_{i+1}(x)}{dx} $$

Note: $$ u(x) {=} \underbrace{ \begin{vmatrix} N_i(x) & N_{i+1}(x) \end{vmatrix}}_{N(x)_{1x2}} \begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix}_{2x1} $$

$$ \frac{du(x)}{dx} = \underbrace{ \begin{vmatrix} N_i^'(x) & N_{i+1}^'(x) \end{vmatrix}}_{\underline{B}(x)_{1x2}} \begin{Bmatrix} d_i \\ d_{i+1} \end{Bmatrix}_{2x1} $$

Similarly: $$ w(x) {=} \underline{N}(x) \begin{Bmatrix} W_i \\ W_{i+1} \end{Bmatrix} $$

$$ \frac{dW(x)}{dx} {=} \underline{B}(x) \begin{Bmatrix} W_i \\ W_{i+1} \end{Bmatrix} $$

Recall the element degree of freedom conventions:

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

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

$$ \beta {=} \int_{x^i}^{x^{i+1}} \underbrace{(\underline{B} \underline{w}^{(i)})}_{1x1} \underbrace{ (EA)}_{1x1} \underbrace{(\underline{B}\underline{d}^{(i)})}_{1x1}, dx (scalar) $$

$$ {=} \underline{w}^i \cdot (\underline{k}^{(i)} \underline{d}^{(i)}) $$

$$ \beta {=} \int_{x^i}^{x^{i+1}} (EA) \underbrace{(\underline{B} \underline{w}^{(i)})}_{1x1} \cdot \underbrace{(\underline{B}\underline{d}^{(i)})}_{1x1} dx $$

$$ (\underline{B}\underline{w}^{(i)})^T(\underline{B}\underline{d}^{(i)}) $$

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

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

$$ \underline{k}^{(i)}_{2x2} {=} \int_{x_i}^{x_{i+1}} \underbrace{\underline{B}^T(x)}_{2x1} \underbrace{(EA)}_{\underbrace{(x)}_{1x1}} \underbrace{\underline{B}(x)}_{1x2} dx $$

$$ B(x) {=} \begin{vmatrix} HW & \frac{1}{L^{(i)}} \end{vmatrix} $$

$$ L^{(i)} {=} x_{i+1} - x_i $$ (Length of element i)

HW 6: Consider EA = const.

$$ k^{(i)} {=} \frac{EA}{L^{(i)}} \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} $$

Transfer of variable coordinates from x to $$ \tilde{x} $$

$$ \tilde{x} :{=} x - x_i $$

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

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

HW6: Find expression for $$ \underline{k}^{(i)} $$ using above:

$$ A(\tilde{x}) {=} N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2 $$

$$ E(\tilde{x}) {=} N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2 $$

$$ \underline{k}^{(i)} {=} ? $$

Meeting 40: Friday, 5 December 2008. EML4500
Note: Eq. (2) (39-2), dimension analysis:

$$ [u] {=} L $$

(31-3): $$ [N_1] {=} [N_4] {=} 1 $$

$$ Eq. (2) (39-2) \underbrace{[N_1]}_{1} \underbrace{[\tilde{d}_1]}_{L} + \underbrace{[N_4]}_{1} \underbrace{[\tilde{d}_4]}_{L} $$

$$ [V] {=} L $$

$$ \underbrace{[N_2]}_{1} \underbrace{[\tilde{d}_2]}_{L} {=} L $$ (displ. transv.)

$$ \underbrace{[N_2]}_{1} \underbrace{[\tilde{d}_2]}_{L} {=} L $$ (rot.)

$$ \underbrace{[N_2]}_{1} \underbrace{[\tilde{d}_2]}_{L} and \underbrace{[N_2]}_{1} \underbrace{[\tilde{d}_2]}_{L} {=} L $$ are to be done for homework

Derivation of the beam shape functions

$$ N_2, N_3, N_5, N_6 $$ (38-4)

p. (38-3) plots

Recall: Governing PDE for beams

p. (37-4), Eq. (1), without $$ f_t $$ (distributed transverse load) and without inertia force $$ m\ddot{v} $$ (static case):

$$ \frac{\delta^2}{\delta x^2} \left \{(EI) \frac{\delta^2v}{\delta x^2}\right \} {=} 0 $$

Further, consider constant EI:

$$ \frac{\delta^4}{\delta x^4} v {=} 0 $$ Integrate 4 times to get 4 constants.

$$ \Rightarrow V(x) {=} C_0 + C_1x^1 + C_2x^2 + C_3x^3 $$

To obtain $$ N_2(x) $$ ($$ \tilde{x} $$ {=} x for simplicity)

V(0) = 1,  V(L) = 0

V'(0)   =   V'(L) = 0

Use above boundary conditions to solve for $$ L_0, ... , L_3 $$

V(0) = 1 = C0

(1) $$ V(L) {=} 1 + C_1(L) + C_2(L)^2 + C_3(L)^3 {=} 0 $$

$$ V^{\prime}(x) {=} C_1 + 2C_2x + 3C_3x^2 $$

$$ V^{\prime}(0) {=} C_1 {=} 0 $$

(2) $$ V^{\prime}(L) {=} 2C_2(L) + 3C_3(L)^2 {=} 0 $$

(2) $$ \rightarrow C_3 {=} - \frac{2}{3} \frac{C_2}{L} $$

(1) $$ \rightarrow 0 {=} 1 + C_2L^2 + (\frac{-2}{3} \frac{C_2}{L})L^3 {=} 1 + C_2L^2[1 - \frac{2}{3}] $$

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

$$ C_3 {=} - \frac{2}{3} \frac{1}{L}(- \frac{3}{L^2}) {=} \frac{2}{L^3} $$

Compare with expression for N2 on p. (38-4)

For N3: boundary conditions         $$ \tilde{d}_3 $$ rotation

V(0) = V(L) = 0

V'(0) = 1, V'(L) = 0

For N5: V(0) = 0, V(L) = 1    $$ \tilde{d}_5 $$ displacement

V'(0) = 0, V'(L) = 0

For N6: $$ d_6 $$ rotation

V(0) = V(L) = 0

V'(0) = 0, V'(L) = 1

See p. (39-1) for plots of N5, N6

Derive coefficient in $$ \underline{\tilde{k}} $$ (elem. stiffness matrix)

Coefficient for EA: Done

Coefficient for EI: To Be Done

\tilde{k}_{22} {=} \frac{12EI}{L^3} \int^L_0 \frac{d^2N_2}{dx^2}(EI)\frac{d^2N_2}{dx^2}dx

title: User:EML4500.f08.A-team.kirley


text:
User:Eml4500.f08.lulz.abcd/HW2

We are asked to prove that P1 = P2. From previous calculations we have a full FD calculation for element 1:

$$\begin{bmatrix} -0.5625 & -0.32476 \\ -0.32476 & -0.1875 \\ 0.5625 & 0.32476 \\ 0.32476 & 0.1875 \end{bmatrix}\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix} = \begin{Bmatrix} -4.4378 \\ -2.5622 \\ 4.4378 \\ 2.5622 \end{Bmatrix} = \begin{Bmatrix} f_1^{(1)} \\ f_2^{(1)} \\ f_3^{(1)} \\ f_4^{(1)} \end{Bmatrix}$$

The root sum squared method can be used for the proof required. Generally the root sum squared equation is:

$$P_1^{(e)} = \sqrt{\left(f_1^{(e)}\right)^2 + \left(f_2^{(e)}\right)^2}$$

We can then expand that to our condition and create a system of equations for P1 and P2.

$$P_1^{(1)} = \sqrt{\left(f_1^{(1)}\right)^2 + \left(f_2^{(1)}\right)^2}$$

$$P_2^{(1)} = \sqrt{\left(f_3^{(1)}\right)^2 + \left(f_3^{(1)}\right)^2}$$

From the FD matrix we have values for all of the local forces used in this system so P1 and P2 can be calculated directly and their equality proven.

$$P_1^{(1)} = \sqrt{(-4.4378)^2 + (-2.5622)^2} = 5.1243$$

$$P_2^{(1)} = \sqrt{(4.4378)^2 + (2.5622)^2} = 5.1243$$

title: User:Eml4500.f08.lulz.abcd/HW2


text:
User:Eml4500.f08.FEABBQ/HW3

Homework 3

 * Lecture Notes (current version).

People who contributed to HW3
J.Jayma 21:07, 5 October 2008 (UTC)

Eml4500.f08.FEABBQ.vitt 21:40, 5 October 2008 (UTC)

Stark 22:20, 5 October 2008 (UTC)

Esmail Hadman 03:27, 8 October 2008 (UTC)

William Kurth 14:39, 8 October 2008 (UTC)

Andy Koby 16:12, 8 October 2008 (UTC)

title: User:Eml4500.f08.FEABBQ/HW3


text:
User:Eml4500.f08.FEABBQ/HW3/Notes

Class Notes: September 22, 2008
Derivation of element FD cont. global coordinate system

Equation [1]

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

Equation [2]

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

$$ k^{(e)} \left[ \begin{matrix} 1 & -1 \\ -1 & 1 \end{matrix} \right] \begin{Bmatrix} q_{1}^{(e)} \\ q_{2}^{(e)} \end{Bmatrix} = \begin{Bmatrix} P_{1}^{(e)} \\ P_{2}^{(e)} \end{Bmatrix}$$

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

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

Goal : Derive [1] from [2]
The relationship can be expressed in form of,

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

Now consider the displacement vector of the local node 1, which is denoted by $$ \mathbf{d_1^{(e)}} $$

$$ \mathbf{d_1^{(e)}} = d_i^{(e)} \vec i + d_2^{(e)} \vec j $$

Now $$ q_1^{(e)} $$ equals the axial displacement of node i which is the orthogonal projection of displacement vector of $$ \mathbf{d_1^{(e)}} $$ of node 1 on the axis $$ \tilde{x} $$ of element (e)

Node 1:

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

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

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

REMEMBER: from lecture Sept. 8th that

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

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

Therefore,

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

$$ q_1^{(e)} = \left[ \begin{matrix} l^{(e)} & m^{(e)} \end{matrix} \right] \left[ \begin{matrix} d_1^{(e)} \\ d_2^{(e)} \end{matrix} \right] $$

Node 2:

$$ \Rightarrow q_2^{(e)} = \vec d_3^{(e)} \cdot \vec \tilde{j} $$

$$ q_2^{(e)} = (d_3^{(e)} \tilde{i} + d_4^{(e)} \tilde{j}) \cdot \vec \tilde{j} $$

$$ q_2^{(e)} = d_3^{(e)} (\vec i \cdot \vec \tilde{j}) + d_4^{(e)} (\vec j \cdot \vec \tilde{j}) $$

$$ q_2^{(e)} = m^{(e)} d_3^{(e)} +  l^{(e)} d_4^{(e)} $$

$$ q_2^{(e)} = \left[ \begin{matrix} l^{(e)} & m^{(e)} \end{matrix} \right] \left[ \begin{matrix} d_3^{(e)} \\ d_4^{(e)} \end{matrix} \right] $$

Now you can combine $$ q_1^{(e)} $$ and $$ q_2^{(e)} $$ into one matrix,

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

Class Notes: September 26, 2008
Equation [2]

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

Fallowing the same argument above we get the relation

$$ \begin{Bmatrix} p_1^{(e)} \\ p_2^{(e)} \end{Bmatrix} = \left[ \begin{matrix} l^{(e)} & m^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \end{matrix} \right] \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} $$

Equation [3]

$$ \mathbf{p_{2x1}^{(e)}} = \mathbf{T_{2x4}^{(e)}} \mathbf{f_{4x1}^{(e)}} $$

recalling element axial FD relation

Equation [4]

$$ \mathbf {k_{2x4}^{(e)}} \mathbf{q_{4x1}^{(e)}} =\mathbf{p_{2x1}^{(e)}}  $$

and Equation [5]

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

Substituting equations 3 and 5 into equation 4 yields to

$$ \mathbf {k_{2x4}^{(e)}} \mathbf{T_{2x4}^{(e)}} \mathbf{d_{4x1}^{(e)}} =\mathbf{T_{2x4}^{(e)}} \mathbf{f_{4x1}^{(e)}}  $$

The Goal is to have $$ \mathbf{k^{(e)}} \mathbf{d^{(e)}} = \mathbf{f^{(e)}} $$ so move $$ \mathbf{T^{(e)}} $$ from R.H.S. to L.H.S. by premultiplying the equation by $$ \mathbf{{(T^{(e)})}^{(-1)}} $$        ( inverse $$ \mathbf{T^{(e)}} $$)

But unfortunately $$ \mathbf{T^{(e)}} $$ is a rectangular matrix [2x4], can't inverse $$ \mathbf{T^{(e)}} $$

to solve this problem the transpose property was used as follows

$$ \mathbf{{T_{4x2}^{(e)}}^{T}} \mathbf {k_{2x4}^{(e)}} \mathbf{T_{2x4}^{(e)}} \mathbf{d_{4x1}^{(e)}} = \mathbf{f_{4x1}^{(e)}} $$

$$ \mathbf{{T_{4x2}^{(e)}}^{T}} \mathbf {k_{2x4}^{(e)}} \mathbf{T_{2x4}^{(e)}} \mathbf{d_{4x1}^{(e)}} = \mathbf{f_{4x1}^{(e)}} $$

$$ \mathbf {k_{2x4}^{(e)}} \mathbf{d_{4x1}^{(e)}} = \mathbf{f_{4x1}^{(e)}}  $$

Homework

Verify that

$$ \mathbf {I_{4x4}} = \mathbf{{T_{4x2}^{(e)}}^{T}} \mathbf{T_{2x4}^{(e)}}   $$

$$ \mathbf {I_{4x4}} =  \left[ \begin{matrix} l^{(e)} & 0 \\ m^{(e)} & 0 \\ 0 & l^{(e)} \\ 0 & m^{(e)}  \end{matrix} \right]  \left[ \begin{matrix} l^{(e)} & m^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \end{matrix} \right]   $$

Applying the principle of virtual work will result reduction of global FD relation:

$$ \mathbf {k_{2x4}^{(e)}} = \mathbf{{T_{4x2}^{(e)}}^{T}}  \mathbf{\tilde{k}_{4x1}^{(e)}} \mathbf{T_{2x4}^{(e)}}   $$

Remark

why not solve as follows $$ \mathbf{d_{4x1}^{(e)}} = \mathbf{{(T^{(e)})}^{(-1)}}  \mathbf{f_{4x1}^{(e)}}  $$  ?

The answer is that the used methods to compute $$ \mathbf{{(K^{(e)})}^{(-1)}} $$ could not be used since K is singular. this will result a determinants value of zero and thus K is not inverted.

Recall

for an unconstrained structure system there are three possible body motion in 2-D (plane) two translation and one rotation.

Homework

Compute the eigenvalues of K and make observation about the number of zero eigenvalues.

Stiffness matrix K is defined as

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

with values from sample problem in homework-2 K becomes

$$\textbf{K}= \begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0 \\ -0.5625 & -0.3248 & 3.0625 & -2.1752 & -2.5 & 2.5 \\ -0.3248 & -0.1875 & -2.1752 & 2.6875 & 2.5 & -2.5 \\ 0 & 0 & -2.5 & 2.5 & 2.5 & -2.5 \\ 0 & 0 & 2.5 & -2.5 & -2.5 & 2.5  \end{bmatrix}\,$$

Solving for the eigenvalues of K using MATLAB results in

Matlab: Computing eigenvalues
The following is a matlab script and output for the two-bar K example.

Zero Eigenvalues correspond to zero stored elastic energy, rigid body modes- (shape, eigenvictors)

Class Notes: September 29, 2008
Using the Global Force-Displacement Relation

Use the global force-displacement relationship matrix from lecture 10:

$$ \left[ \begin{array}{cc} k_{13}^{(1)} & k_{14}^{(1)} \\ k_{23}^{(1)} & k_{24}^{(1)} \\ (k_{33}^{(1)} + k_{11}^{(2)}) & (k_{34}^{(1)} + k_{12}^{(2)}) \\  (k_{43}^{(1)} + k_{21}^{(2)}) & (k_{44}^{(1)} + k_{22}^{(2)})  \\  k_{31}^{(2)} & k_{32}^{(2)} \\ k_{41}^{(2)} & k_{42}^{(2)}  \\ \end{array} \right] $$ $$\left[ \begin{array}{l} d_1 \\ d_2  \\ d_3  \\ d_4 \\ d_5 \\ d_6 \end{array} \right] = \left[ \begin{array}{l} F_1 \\ F_2  \\ F_3  \\ F_4 \\ F_5 \\ F_6 \end{array} \right]$$ (6x2) x (2x1)  =  (6x1)

Only do computations for rows 1,2,5,6 to get F1,F2,F5,F6. ....

Closing the loop between the Finite Element Method and statics: Virtual Displacement

For the two bar truss system:

Going from computing reactions to computing the displacements in statics "closes the loop".

by statics, reactions are known and therefor the member forces $$P_1^{(1)}$$ and $$P_2^{(1)}$$

Compute the axial displacement of the degrees of freedom (extension of bars):

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

$$ q_1^{(1)} = 0$$, fixed global node 1

$$q_1^{(2)} = \frac{-P_2^{(2)}}{k^{(2)}} = -AB$$

$$q_2^{(2)} = 0 $$, fixed global node 3

Question: How to get the displacement DOF's from the above results for node 2?

Class Notes: Wednesday, October 1, 2008

In order to find the coordinates of point D, one must first know the coordinates of points B and C. The magnitudes of AC and AB are solved for below.

 * $$AC=\frac{\vert P_2^{(1)}\vert}{k^{(1)}}=\frac{5.1243}{\frac{3}{4}}=6.8324$$

 * $$AB=\frac{\vert P_1^{(2)}\vert}{k^{(2)}}=\frac{6.2760}{5}=1.2552$$

Now that the distances of points B and C are known from the reference point, point A, we must solve for the coordinates of points B and C by constructing lines parallel to initial elements and measuring that distance along the line to find $$(x_B,y_B)$$ and $$(x_C,y_c)$$.

To do this we need the equations for the line AB and line AC. The figure below shows a line being constructed that runs through P and some arbitrary point Q at $$(x,y)$$. This line has the same slope as $$\vec \tilde{i}$$.

Constructing this line is as follows

 * $$\overrightarrow{P Q} = \vert PQ \vert \vec \tilde{i}=(PQ)[\cos\theta \vec i + \sin\theta \vec j]$$

 * $$=(x-x_p)\vec i + (y-y_p)\vec j$$

From this one can see that
 * $$(x-x_p)=(PQ)\cos\theta$$

 * $$(y-y_p)=(PQ)\sin\theta$$

This is sufficient to solve for the coordinates of points B and C, however to make one further useful extension, these two expressions can be combined to create the equation of a line.

 * $$\frac{(y-y_p)}{(x-x_p)}=\frac{\sin\theta}{\cos\theta}=\tan\theta \qquad \Rightarrow \qquad y-y_p = (\tan\theta)(x-x_p)$$

Using these expressions the coordinates for B and C are now to be solved for. Chosing A as the origin will make calculations convenient, so for these instances that is what was chosen.

 * $$(x_B,y_B)\Rightarrow(x_B - x_A, y_B - y_A)=((AB)\cos\theta^{(2)},(AB)\sin\theta^{(2)})$$
 * $$\quad =((1.2552)\cos 135^\circ,(1.2552)\sin 135^\circ)$$
 * $$(x_B,y_B) = (-0.8876,0.8876)$$

 * $$(x_C,y_C)\Rightarrow(x_C - x_A, y_C - y_A)=((AC)\cos\theta^{(1)},(AC)\sin\theta^{(1)})$$
 * $$\quad =((6.8324)\cos 30^\circ,(6.8324)\sin 30^\circ)$$
 * $$(x_C,y_C) = (5.9170,3.4162)$$

Once the equation of a line is known in the form shown earlier, one can easily derive the equation of a line perpendicular to the original, passing through a given point by adding a multiple of $$\frac {\pi}{2}$$ to the argument of the tangent term. The equation of the line perpendicular to the equation of the line derived above, passing through point P is given below.

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

The last unknown position is that of point D. This can either be calculated via the intersection of perpendicular lines passing through B and C, or via the results of the FEM solution. Choosing point A to be the origin once again for the same convenience, the coordinates of point D can be expressed as follows:

 * $$\overrightarrow{AD} = (x_D-x_A)\vec i +(y_D-y_a)\vec j = x_D \vec i + y_D \vec j$$

By definition $$\overrightarrow{AD}$$ is the displacement vector which is defined as

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

where $$d_3$$ and $$d_4$$ are the displacement DoFs previously solved for.

 * $$\therefore (x_D,y_D) = (d_3,d_4)$$

 * $$ \Rightarrow (x_D,y_D) = (4.3502,6.1254)$$

3-Bar Truss System

When numbering local nodes be sure to consider convenience in assembling the global stiffness matrix, $$\mathbf{\underline{K}}$$. Careful selection of local nodes can greatly ease the process and structure of the global stiffness matrix. The figure below shows the most convenient local node numbering.

Class Notes: Friday, October 3, 2008
There is a clarification/correction to the angle for element 3, $$\theta^{(3)}$$, as shown by the following image and equation...

$$ \theta^{(3)}=-90^{\circ}-45^{\circ}=-135^{\circ} $$

Now that the angle for element 3 is clear, we can examine the three-bar truss system with respect to the reaction forces and the external force P...

We know the following equations to be true for this three-bar truss...

$$\Sigma{F}_{x}=0\, $$

$$\Sigma{F}_{y}=0\, $$

$$\Sigma{M}_{A}=0\, $$

where $$\Sigma{M}_{A}=0$$ is trivial, due to the fact that all force vectors pass through this point.

But what about point B ~ $$\Sigma{M}_{B}\, $$?

We will use the following image and subsequent equations to explain the moment about point B. Note: Points A and B are unrelated to the original three-bar truss figure...

The following is true for all $$A'\,$$ on the line of action of $$\vec{F}$$... $$ \Sigma{\vec{M}}_{B}=\vec{BA}\times\vec{F}=\vec{BA'}\times\vec{F}$$ $$ \vec{BA'}=\vec{BA}+\vec{AA'}$$ $$ \Sigma{\vec{M}}_{B}=(\vec{BA}+\vec{AA'})\times\vec{F}=\vec{BA}\times\vec{F}+\vec{AA'}\times\vec{F}$$

where we know that $$\vec{AA'}\times\vec{F}=\vec{0}\, $$. Now we go back to the three-bar truss: Node $$A$$ is in equilibrium: $$ \sum_{i=0}^3\vec{F}=\vec{0}\, $$

And we know that $$ \sum_{i}\vec{M}_{B,i}=\sum_{i}\vec{BA'_{i}}\times\vec{F_{i}}\, $$

$$A'_{i}\, $$ is any point on the line of action of $$\vec{F_{i}}\, $$

$$ \sum_{i}\vec{M}_{B,i}=\sum_{i}\vec{BA'_{i}}\times\vec{F_{i}}=\vec{BA}\times\sum_{i}\vec{F_{i}}=\vec{0}\, $$, therefore proving $$ \Sigma{M}_{B}=\vec{0}\, $$.

Next, we move on to the global stiffness matrix for the three-bar truss. A general outline of the overlapping nature of the three individual element stiffness matrices is shown below...

Next, we fill this matrix...

HW: Complete the global stiffness matrix for the three-bar truss system.
Using the method described in HW2 and the values $$ k_{(1)}=1.2 $$, $$ k_{(2)}=0.8 $$ and $$ k_{(3)}=0.6 $$, we can find each individual element stiffness matrix...

$$ k^{(1)} = \left[ \begin{array}{llll} 0.9 & 0.52 & -0.9 & -0.52\\ 0.52 & 0.3 & -0.52 & -0.3\\ -0.9 &-0.52 & 0.9 & 0.52\\ -0.52 & -0.3 & 0.52 & 0.3 \end{array} \right] $$

$$ k^{(2)} = \left[ \begin{array}{llll} 0.6 & -0.346 & -0.6 & 0.346\\ -0.346 & 0.2 & 0.346 & -0.2\\ -0.6 & 0.346 & 0.6 & -0.346\\ 0.346 & -0.2 & -0.346 & 0.2 \end{array} \right] $$

$$ k^{(3)} = \left[ \begin{array}{llll} 0.3 & 0.3 & -0.3 & -0.3\\ 0.3 & 0.3 & -0.3 & -0.3\\ -0.3 & -0.3 & 0.3 & 0.3\\ -0.3 & -0.3 & 0.3 & 0.3 \end{array} \right] $$

$$ \underline{K} = \left[ \begin{array}{llllllll} 0.9 & 0.52 & -0.9 & -0.52 & 0 & 0 & 0 & 0\\ 0.52 & 0.3 & -0.52 & -0.3 & 0 & 0 & 0 & 0\\ -0.9 & -0.52 & 1.8 & 0.474 & -0.6 & 0.346 & -0.3 & -0.3\\ -0.52 & -0.3 & 0.474 & 0.8 & 0.346 & -0.2 & -0.3 & -0.3\\ 0 & 0 & -0.6 & 0.346 & 0.6 & -0.346 & 0 & 0\\ 0 & 0 & 0.346 & -0.2 & -0.346 & 0.2 & 0 & 0\\ 0 & 0 & -0.3 & -0.3 & 0 & 0 & 0.3 & 0.3\\ 0 & 0 & -0.3 & -0.3 & 0 & 0 & 0.3 & 0.3\end{array} \right] $$

Matlab Homework - Truss Example
The following image is the modified output of the above code...

People who contributed to HW3
Jose Jayma 21:07, 5 October 2008 (UTC)

Eml4500.f08.FEABBQ.vitt 21:40, 5 October 2008 (UTC)

Stark 22:20, 5 October 2008 (UTC)

Esmail Hadman 03:27, 8 October 2008 (UTC)

William Kurth 14:39, 8 October 2008 (UTC)

Andy Koby 16:12, 8 October 2008 (UTC)

title: User:Eml4500.f08.FEABBQ/HW3/Notes


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