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)