User:Uf.team.wiki/HW3

= Recall: Element Force-Displacement Relationship with respect to the Global Coordinate System =

The Traditional Element Force-Displacement Relationship:

$$ \mathbf{\underset{4 \times 4}k^{(e)}}\mathbf{\underset{4 \times 1}d^{(e)}}=\mathbf{\underset{4 \times 1}f^{(e)}} $$ (Eqn. 1)

Bar Element 1:



= Element Force-Displacement Relationship: Axial Forces =



Element Force-Displacement Relationship: Axial Forces

$$ \mathbf{\underset{2 \times 2}k^{(e)}}\mathbf{\underset{2 \times 1}q^{(e)}}=\mathbf{\underset{2 \times 1}p^{(e)}} $$  (Eqn. 2)

$$ \Rightarrow 	\underbrace{k^{(e)} \begin{bmatrix}

1 &-1 \\

-1 & 1 \\ \end{bmatrix}}_{\hat k^{(e)}} \underbrace{\begin{pmatrix}

q^{(e)}_1 \\

q^{(e)}_2 \end{pmatrix}}_{\underset{2 \times 1}q^{(e)}} = \underbrace{\begin{pmatrix}

p^{(e)}_1 \\

p^{(e)}_2 \end{pmatrix}}_{\underset{2 \times 1}p^{(e)}} $$

Where:


 * q(e)i is the axial displacement of element "e" at local node "i"
 * p(e)i is the axial force of element "e" at local node "i"

GOAL:  Derive equation (1) from equation (2) in order to find:
 * Relationship between q (e) and d (e)
 * Relationship between p (e) and f (e)

These relationships can be expressed as:

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

Consider the displacement vector of local node [1], denoted by $$\mathbf d_{[1]}^{(e)}$$:



Where:


 * $$\overrightarrow{d}^{(e)}_{[1]} = d^{(e)}_1\overrightarrow{i}+{d}^{(e)}_2\overrightarrow{j}$$


 * $$\overrightarrow{q}^{(e)}_{[1]}$$ = axial displacement of node [1] is the orthogonal projection of the displacement vector d(e)[1] of node [1] on the axis $$\overrightarrow{\tilde{x}} $$ of element "e".

$$ \mathbf q_1^{(e)} = \overrightarrow{d}^{(e)}_{[1]}\overrightarrow{i} $$
 * NOTE: Very important relationship***

$$ = (d^{(e)}_1 \overrightarrow{i}+d^{(e)}_2 \overrightarrow{j}) \overrightarrow{\tilde{i}} $$

$$ = d^{(e)}_1 (\overrightarrow{i} \cdot \overrightarrow{\tilde{i}}) + d^{(e)}_2 (\overrightarrow{j} \cdot \overrightarrow{\tilde{i}})$$

Recall:
 * $$(\overrightarrow{i} \cdot \overrightarrow{\tilde{i}}) = \cos \Theta\,\! = l^{(e)}$$
 * $$(\overrightarrow{j} \cdot \overrightarrow{\tilde{i}}) = \sin \Theta\,\! = m^{(e)}$$

$$ q_1^{(e)}= l^{(e)} d_1^{(e)} + m^{(e)} d_2^{(e)} \Rightarrow \begin{bmatrix}

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

d^{(e)}_1 \\

d^{(e)}_2 \end{pmatrix} $$

Similarly for node [2]:

$$ q_2^{(e)}= \begin{bmatrix}

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

d^{(e)}_3 \\

d^{(e)}_4 \end{pmatrix} $$

This leads to the following relationship:

$$	\underbrace{\begin{pmatrix}

q^{(e)}_1 \\

q^{(e)}_2 \end{pmatrix}}_{\underset{2 \times 1}q^{(e)}} = \underbrace{\begin{bmatrix}

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

0 & 0 & l^{(e)} & m^{(e)} \\ \end{bmatrix}}_{\underset{2 \times 4}T^{(e)}}\underbrace{\begin{pmatrix}

d^{(e)}_1 \\ d^{(e)}_2 \\ d^{(e)}_3 \\ d^{(e)}_4 \end{pmatrix}}_{\underset{4 \times 1}d^{(e)}}$$

Similarly, the following represents the same argument:

$$	\underbrace{\begin{pmatrix}

p^{(e)}_1 \\

p^{(e)}_2 \end{pmatrix}}_{\underset{2 \times 1}p^{(e)}} = \mathbf\underset{2 \times 4}T^{(e)}\underbrace{\begin{pmatrix}

f^{(e)}_1 \\ f^{(e)}_2 \\ f^{(e)}_3 \\ f^{(e)}_4 \end{pmatrix}}_{\underset{4 \times 1}f^{(e)}}$$

Recall the element axial force-displacement relationship:

$$ \mathbf{\underset{2 \times 2} \hat k^{(e)}}\mathbf{\underset{2 \times 1}q^{(e)}}=\mathbf{\underset{2 \times 1}p^{(e)}} $$

$$ \Rightarrow \mathbf \hat k^{(e)} (\underbrace{\mathbf T^{(e)} \mathbf d^{(e)}}_{\mathbf q^{(e)}}) = (\underbrace{\mathbf T^{(e)} \mathbf f^{(e)}}_{\mathbf p^{(e)}}) $$

GOAL: To get into the form $$\mathbf{k^{(e)}d^{(e)}=f^{(e)}}$$.


 * Must "move" $$ \mathbf T^{(e)} $$ from left-hand side to right-hand side by premultiplying the equation by $$ \mathbf T^{(e)^{-1}} $$ {the inverse of T}.
 * However, $$ \mathbf T^{(e)} $$ is a rectangular matrix, it cannot be inverted!

Solution: Divide left-hand side by the transpose of T:

$$ \underbrace{\mathbf{ [T^{(e)^T}\hat k^{(e)} T^{(e)} }]}_{4 \times 4} \mathbf {\underset{4 \times 1}d^{(e)}}\mathbf {\underset{4 \times 1}f^{(e)}} \Rightarrow \mathbf{k^{(e)}d^{(e)}=f^{(e)}}$$

$$\mathbf k^{e} = \mathbf {T^{(e)^T} \hat k^{(e)} T^{(e)}}$$ Equation (3)

Further justification for (3) : Principal of Virtual Work

Example:

$$ \underset{6 \times 6}\mathbf K \underset{6 \times 1}\mathbf d = \underset{6 \times 1}\mathbf F \longrightarrow \bar \underset{2 \times 2}\mathbf K \bar \underset{2 \times 1}\mathbf d = \bar \underset{2 \times 1}\mathbf F$$

For an unconstrained structural system, there are 3 possible rigid body motions in two-dimensional space (2 translational and 1 rotational).

Example: Dynamics evaluation problem:

$$ \mathbf {Kv} = \lambda \mathbf {Mv} $$

Where:
 * K : stiffness matrix
 * λ : eigenvalue (related to the vibrational frequency)
 * M : mass matrix

Use example from HW2 to compute eigenvalues of $$ \underset{6 \times 6}\mathbf K$$:

Recall:

$$ \underset{6 \times 6}\mathbf K = \begin{bmatrix} k_{11}^{(1)} &  k_{12}^{(1)}  &  k_{13}^{(1)}  &  k_{14}^{(1)}  &  0  &  0 \\ k_{21}^{(1)} &  k_{22}^{(1)}  &  k_{23}^{(1)}  &  k_{24}^{(1)}  &  0  &  0 \\ k_{31}^{(1)} &  k_{32}^{(1)}  &  k_{33}^{(1)}+k_{11}^{(2)}  &  k_{34}^{(1)}+k_{12}^{(2)}  &  k_{13}^{(2)}  &  k_{14}^{(2)} \\ k_{41}^{(1)} &  k_{42}^{(1)}  &  k_{43}^{(1)}+k_{21}^{(2)}  &  k_{44}^{(1)}+k_{22}^{(2)}  &  k_{23}^{(2)}  &  k_{24}^{(2)} \\ 0 &  0  &  k_{31}^{(2)}  &  k_{32}^{(2)}  &  k_{33}^{(2)}  &  k_{34}^{(2)} \\ 0 &  0  &  k_{41}^{(2)}  &  k_{42}^{(2)}  &  k_{43}^{(2)}  &  k_{44}^{(2)} \\ \end{bmatrix} = \begin{bmatrix} \frac{9}{16} &  \frac{3\sqrt{3}}{16}  &  -\frac{9}{16}  &  -\frac{3\sqrt{3}}{16}  &  0  &  0 \\ \frac{3\sqrt{3}}{16} &  \frac{3}{16}  &  -\frac{3\sqrt{3}}{16}  &  -\frac{3}{16}  &  0  &  0 \\ -\frac{9}{16} &  -\frac{3\sqrt{3}}{16}  &  \frac{49}{16}  &  \frac{3\sqrt{3}-40}{16}  &  -\frac{5}{2}  &  \frac{5}{2} \\ -\frac{3\sqrt{3}}{16} &  -\frac{3}{16}  &  \frac{3\sqrt{3}-40}{16}  &  \frac{43}{16}  &  \frac{5}{2}  &  -\frac{5}{2} \\ 0 &  0  &  -\frac{5}{2}  &  \frac{5}{2}  &  \frac{5}{2}  &  -\frac{5}{2} \\ 0 &  0  &  \frac{5}{2}  &  -\frac{5}{2}  &  -\frac{5}{2}  &  \frac{5}{2} \\ \end{bmatrix}$$

$$ \lambda \Rightarrow \begin{pmatrix} 10.03 &  1.47  &  0  &  0  &  0  &  -1.75 \\ \end{pmatrix} $$

NOTE: THREE of the eigenvalues are 0 → zero eigenvalue corresponds to zero stored elastic energy (in rigid body modes).

= Two-Bar Truss System MATLAB Plot = The following MATLAB Code plots the two-bar truss system solved in HW2 in the initial and deformed state.

Output:

= Closing the loop between FEM and Statics =

Use Force Displacment relation on (p10-1)


 * (6x2 matrix) x (2x1 matrix) = (6x1 matrix)
 * Closing the Loop between the Finite Element Method (FEM) and Statics is virtual displacement




 * How do we find the displacement to close the loop?
 * Statics tells us the reactions known and therefore the member forces P1(1), P2(2) are equivalent and in opposing directions.


 * Computing axial displacement DOF's, the amount of extension of the bars themselves:

$${\mathbf{q}_2^{(1)}}= \frac{\mathbf{K}^{(1)}}= \frac{|{\underset{2}\mathbf{P}^{(1)}}|}{\mathbf{K}^{(1)}} $$ $${\mathbf{q}_1^{(2)}}= \frac{-{\underset{2}\mathbf{P}^{(2)}}}{\mathbf{K}^{(2)}} $$
 * where q1(1) = 0 is the fixed node 1
 * where q2(2) = 0 is the fixed node 3

= Infinitesimal Displacement =

Figure 5 to the right is a display of the 2 bar truss system we solved previously.

It also displays the displaced element bar system and the location ff points $$\mathbf{B}$$, $$\mathbf{C}$$, and $$\mathbf{D}$$


 * Line $$\mathbf{A}\mathbf{B}$$ is $$\perp$$ to $$\mathbf{B}\mathbf{D}$$


 * Line $$\mathbf{A}\mathbf{C}$$ is $$\perp$$ to $$\mathbf{C}\mathbf{D}$$

$$\mathbf{A}\mathbf{C}= \frac{|{\underset{2}\mathbf{P}^{(1)}}|}{\mathbf{K}^{(1)}}= \frac{|5.124|}{.75}=6.832 $$

$$\mathbf{A}\mathbf{B}= \frac{|{\underset{1}\mathbf{P}^{(2)}}|}{\mathbf{K}^{(2)}}= \frac{|-6.276|}{5}=1.255 $$



The equations for $${\mathbf(U)_y}$$ and $${\mathbf(U)_x}$$ are given as follows

$${\mathbf(U)_y} = \mathbf(R)\sin\alpha$$


 * for small $$\alpha$$

$${\mathbf(U)_x} = \mathbf(R)({1} - \cos\alpha)\cong {0}$$


 * for first order


 * note that $$({1} - \cos\alpha)\simeq {1}$$



Figure 7 displays the trigonometric ratios given by the components above.

We take note that for very small $$\boldsymbol{\alpha}$$,


 * $$\sin\boldsymbol{\alpha}\simeq{\boldsymbol{\alpha}}$$



We use these following equations to fin the x and y coordinates of point B and C

$$\mathbf{X} - {\mathbf{X}_p} = (\mathbf{P}\mathbf{Q})\cos\theta$$

$$\mathbf{Y} - {\mathbf{Y}_p} = (\mathbf{P}\mathbf{Q})\sin\theta$$

We also know that $$\frac{\mathbf{Y} - {\mathbf{Y}_p}}{\mathbf{X} - {\mathbf{X}_p}} = \tan\theta$$

We also deduce that $$\mathbf{Y} - {\mathbf{Y}_p} = \tan\theta(\mathbf{X} - {\mathbf{X}_p}) $$

For a line that is perpendicular to the line above the equation is

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

To determine $$\mathbf{A}\mathbf{D}$$ and point $$\mathbf{D}$$'s components.

$$\overrightarrow{\mathbf{A} \mathbf{D}}= ({\mathbf{X}_d} - {\mathbf{X}_a})\tilde{i} + ({\mathbf{Y}_d} - {\mathbf{Y}_a})\tilde{j} $$

From the definition of the displacement vectors we know

$$\overrightarrow{\mathbf{A} \mathbf{D}} = {\mathbf{d}_3}\tilde{i} + {\mathbf{d}_4}\tilde{j}$$

$${\mathbf{X}_d} = {\mathbf{d}_3} = {4.35}$$

$${\mathbf{Y}_d} = {\mathbf{d}_4} = {6.13}$$

We can now solve for the x and y coordinates of point B and C

For point B we used the following formula to find thew coordinates

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

and our results were

$${\mathbf{X}_b} = {-0.88}$$

$${\mathbf{Y}_b} = {0.88}$$

For point C we used the following formula to find thew coordinates

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

and our results were

$${\mathbf{X}_c} = {5.92}$$

$${\mathbf{Y}_c} = {3.42}$$

= 3 Bar Truss System =

As shown in our previous Wikipedia page, HW2, the 3 bar truss system below is statically indeterminate. The moment equation for this system is proven useless in the section below.

We simply have to many unknowns and not enough equations.

To solve this system we must turn to FEM (Finite Element Method) to compute the reactions and displacement vectors.

Figure 10 clearly shows the 3 bar truss system configuration.

Clearly defined in the figure for each element is:


 * The Young's Modulus, $$\mathbf{E}$$


 * The cross sectional area, $$\mathbf{A}$$


 * The Length of the element, $$\mathbf{L}$$


 * The angle associated with each element, $$\boldsymbol{\theta}$$

We now have what we need to begin forming elemnt stiffness matrices for each element.



Figure 10 shows the 3 bar truss system in its element form. It displays the corresponding $$\theta$$ and the numbering system used.

All of the nodes are labeled accordingly and the elements are numbered. It shows the proper numbering system to allow for the most overlap. This simplifies the compilation of the global stiffness matrix.

This assembly process is shown below in the following section.


 * $$\theta^{(1)} = {30^0}$$


 * $$\theta^{(1)} = {-120^0}$$


 * $$\theta^{(1)} = {-30^0}$$

Going back to the 3 bar truss system: We can begin to start solving the problem by writing the moment equation, with sigma notation and for moment compenents, i. Therefore we take the sum, $$\sum_{i}{\overrightarrow{\mathbf{M_{B,i}}}}=\sum_{i}{(\overrightarrow{\mathbf{BA'_{i}}}\times\overrightarrow{\mathbf{F_{i}}})}$$ $$\mathbf{A'_{i}}$$= any point on line of action of $$\overrightarrow{\mathbf{F_{i}}}$$ $$\sum_{i}{\overrightarrow{\mathbf{M_{B,i}}}}=\sum_{i}{\overrightarrow{\mathbf{BA_{i}}}\times\overrightarrow{\mathbf{F_{i}}}}$$ $$=\overrightarrow{\mathbf{BA}}\times\overrightarrow{\mathbf{F_{i}}}=\overrightarrow{\mathbf{0}}$$ $$\sum_{i=0}^{3}{} = \overrightarrow{\mathbf{0}}$$
 * Note: A is in equilibrium

Next, we continue the problem by looking to begin to setup the entire global stiffness matrix from all 3 element stiffness matricies. Below is the assembly block diagram of the global stiffness matrix. It displays the proper method to combine the 3 element stiffness matrices to allow for proper calculation.



This is the global stiffness matrix fully labeled. It displays the addition used from all of the element stiffness matrices.

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

= References =

In this section we will list some of our references and sources to help readers improve and further there understanding of the material covered in this assignment.

For matrices, the transpose was used to help evaluate an expression. [Source: Wikipedia]

The eigenvalues of a matrix were also mentioned. [Source: Wikipedia]

It is helpful for readers to review methods in ][Source: Wikipedia] to successfully navigate harder problems.

= Contributing Members = The following Team Wiki members contributed to this report.

Rodney Dagulo --Eml4500.f08.wiki.dagulo.r 16:52, 7 October 2008 (UTC)

Thomas Kehoe --EML4500.f08.Wiki.Kehoe 22:36, 7 October 2008 (UTC)

Ricardo Lopez --EML4500.f08.Wiki.Lopez 15:13, 8 October 2008 (UTC)

Cortland Glowacki --EML4500.f08.Wiki.cort 09:46, 8 October 2008 (UTC)

Renee Hood --EML4500.f08.Wiki.hood 16:19, 8 October 2008 (UTC)

Gonzalo Barcia --EML4500.f08.Wiki.barcia 14:21, 8 October 2008 (UTC)