User:EML4500.f08.JAMAMA/Homework3

September 22, 2008
$$ \mathbf{q}_{2 \times 1}^{(e)} = \mathbf{T}_{2 \times 4}^{(e)} * \mathbf{d}_{4 \times 1}^{(e)} \qquad \qquad \qquad \qquad \qquad \qquad (1) $$

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

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



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

1 & -1 \\ -1 & 1

\end{bmatrix}

\begin{Bmatrix}

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

\end{Bmatrix}=

\begin{Bmatrix}

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

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

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

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

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

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

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

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



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



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

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

$$ \Rightarrow \begin{matrix} \vec{i} * \vec{\tilde{i}} & = & \cos(\theta^{(e)}) & = & l^{(e)} \\ \vec{j} * \vec{\tilde{i}} & = & \sin(\theta^{(e)}) & = & m^{(e)} \end{matrix} \Rightarrow \ $$ $$q_{1}^{(e)}= l^{(e)} \mathbf{d}_1^{(e)} + m^{(e)}\mathbf{d}_2^{(e)} = {\llcorner l^{(e)} m^{(e)} \lrcorner}_{1 \times 2}

{\begin{Bmatrix}

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

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

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

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

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

$$ \begin{matrix}

\begin{Bmatrix}

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

\begin{bmatrix}

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

\begin{Bmatrix}

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

& & & \\

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

$$

September 26, 2008
Similarly (same argument)

$$\begin{Bmatrix} P{_{1}}^{(e)}\\ P{_{2}}^{(e)}\\ \end{Bmatrix} =  T^{(e)} \begin{Bmatrix} f{_{1}}^{(e)}\\ f{_{2}}^{(e)}\\ f{_{3}}^{(e)}\\ f{_{4}}^{(e)} \end{Bmatrix} $$  2X1 2X4 4X1

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

Recall element axial force displacement relationship

$$\hat{k}^{(e)}q^{(e)}=P^{(e)}$$ 2X2 2X12X1

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

Goal: Want to have $$k^{(e)}d^{(e)}=f^{(e)}$$ so "move" $$T^{(e)}$$ from right to left hand side by pre-multiplying equation by $$T^{(e)-1}$$ (inverse of $$T^{(e)}$$ ). Unfortunately, $$T^{(e)}$$ is a rectangular matrix (2X4) so it cannot be inverted.

So the actual answer is: $$\begin{bmatrix} T{^{(e)T}}\bullet \hat{k}^{(e)}T^{(e)} \end{bmatrix} d^{(e)}=f^{(e)}$$ [4X22X22X4] 4X14X1 [ equal to a 4X4 ]

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

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

Justification for the above equation: Principle Virtual Work (PVW)

(See notes from the September1 17th for the first application of PVW. This is shown through Reduction of Global FD relationship.)

$$k d = F \Leftrightarrow  \bar{k} \bar{d} = \bar{F}$$ 6X6 6X1   6X6    2X2 2X1   2X1

Question: Why would you not solve as follows? $$d = k^{-1} F$$ ?

It is because the matrix k is a singularity, and the det k=0. Therefore $$k^{-1}$$ is not invertible. This is because

to find $$k^{-1}$$ you need to compute $$\frac{1}{\det k}$$

why? For an unconstrained structural system, there are three possible rigid body motions in 2-D. (2-translational and 1-rotational)

Dynamic eigenvalue problems: $$K v=\lambda Mv $$

K=Stiffness matrix

$$\lambda$$=Eigen value (related to the vibrational frequency)

M=Mass matrix

Zero eigenvalue corresponds to zero stored elastic energy, so rigid body modes. Modes, also called mode shapes, are used to describe the vibrations in an object. These correspond to the number of half wave vibrations in a body.

September 29, 2008
The previous simplification that was applied to the Stiffness Matrix was done by application of the boundary conditions as follows:

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

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

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

Two bar Truss syst:



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

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

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





Uy = R sin(α) ≈ Rα (if α small)

Ux = R(1 - cos(α)) ≈ 0 (first Order)



October 1, 2008
Closing the Loop (Continued)

Infinitesimal displacement (Relative to virtual displacement)



$$AC = \frac{P_2^1}{K^1} = \frac{5.1243}{\frac{3}{4}} = 6.8324$$

$$ AB = \frac{P_1^2}{K^2}$$

The process for finding the coordinates of $$( X_B, Y_B)$$ and $$( X_C , Y_C)$$ is as follows:

$$X_C$$ = 6.8324 cosθ$$ ^1$$

$$Y_C$$ = 6.8324 sinθ$$ ^1$$

By adding the values obtained for $$(X_C, Y_C)$$ and $$(X_B , Y_B)$$ to the values for the coordinate of A, you will obtain the actual values for points C & B.

Once B & C are known you have two unkowns remaining $$(X_D, Y_D)$$. To get these unkowns the equations for lines AB and BC are needed.



$$\overrightarrow {PQ}\ = (PQ)\hat i\ = (PQ) [cos\theta \hat i\ + \sin\theta \hat j] = (x - x_p) \hat i\ + (y - y_p) \hat j$$

$$X - X_p = (PQ) \cos \theta$$ $$Y - Y_p = (PQ) \sin \theta$$

$$\frac {Y - Y_p}{X - X_p} = \tan \theta \Rrightarrow Y - Y_p = (\tan \theta)(X - X_p) $$

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

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

$$ \overrightarrow {AD}\ = (X_d - X_a)\hat i\ + (Y_d - Y_a)\hat j$$

Due to A being at the origin A(0,0). $$\Rrightarrow X_a = 0 ,  Y_a = 0$$

By definition: $$\overrightarrow {AD} = d_3 \hat i\ + d_4 \hat j$$

$$\overrightarrow AD$$ is the displacement vector of A

$$d_3, d_4$$ are acquired from FEM $$\Rrightarrow d_3 = X_d , d_4 = Y_d$$

3 Bar Truss System

Convenient Local Node Numbering



$$\underline {P} = 30$$

$$\begin{matrix}E^1 = 2  &   E^2 = 4   &   E^3 = 3 \\ A^1 = 3   &   A^2 = 1   &   A^3 = 2 \\ L^1 = 5   &   L^2 = 5   &   L^3 = 10 \\ \theta^1 = 30^\circ   &   \theta^1 = -30^\circ   &   \theta^3 = 45^\circ \end{matrix}$$

October 3, 2008
The following schematic shows 2 equations, but 3 unknowns.



For the summation of the moment about $$A$$, the following free body diagram is used:



$$\begin{align} & \sum{F_{x}}=0 \\ & \sum{F_{y}=0} \\ & \sum{M_{A}=0} \\ \end{align}$$

The summation of the moment about the point A provides $$0=0$$, which makes it a trivial answer.

The same goes for the $$\sum{M_{B}}$$, as it also reveals a solution of $$0=0$$. Because the perpendicular distances are equivalent, and all of the forces act on a common line of action, the sum of the moments about B will also equal zero.


 * Note: This goes for both 2D and 3D schematics.

For just the bar with the Reaction Force labeled R1, the moment about the point B will become:

$$\sum{M_{B}=\overrightarrow{BA}\times \overrightarrow{F}}$$

For $${A}'$$ on the line of action labeled $$\vec{F}$$, $$\sum{M_{B}=\overrightarrow{B{A}'}\times \overrightarrow{F}}$$.

Where, $$\overrightarrow{B{A}'}=\overrightarrow{BA}+\overrightarrow{A{A}'}$$, and $$\overrightarrow{M_{B}}=(\overrightarrow{BA}+\overrightarrow{A{A}'})\times \vec{F}=\overrightarrow{BA}\times \vec{F}+\overrightarrow{A{A}'}\times \vec{F}$$


 * Note: $$\overrightarrow{A{A}'}\times \vec{F}=\vec{0}$$

$$\Rightarrow $$ Back to the 3-bar truss...

For the truss, Node A is in equilibrium: $$\sum\limits_{i=0}^{3}{\vec{F}_{i}=\vec{0}}$$

$$\sum\limits_{i}{\overrightarrow{M_{B,i}}=\sum\limits_{i}{\overrightarrow{B{A}'_{i}}}\times \overrightarrow{F_{i}}}$$

Where, $$A_{i}^{\prime }$$ is equal to any point on the line of action of $$\overrightarrow{F_{i}}$$.

Therefore, by combining the knowledge of $$A_{i}^{\prime }$$ with the knoledge of $$\sum\limits_{i}{\overrightarrow{M_{B,i}}=\sum\limits_{i}{\overrightarrow{B{A}'_{i}}}\times \overrightarrow{F_{i}}}$$ provides: $$\sum\limits_{i}{\overrightarrow{M_{B,i}}=\sum\limits_{i}{BA\times \overrightarrow{F_{i}}=\overrightarrow{BA}\times \sum\limits_{i}{\overrightarrow{F_{i}}}}}$$


 * Note that $$\sum\limits_{i}{\overrightarrow{F_{i}}=\vec{0}}$$.

Thus, the schematic above is statically indeterminant, because it has 2 equations and 3 unknowns.



Using the above diagram of the global matrix $$K$$, which combines the 3 element stiffness matrices: $$k_{ij}^{(1)}$$, $$k_{ij}^{(2)}$$, $$k_{ij}^{(3)}$$, the entries into the global matrix will be as follows:

$$\underline{K}=$$

Contributing Team Members
Arunas Janulevicius --Eml4500.f08.jamama.jan 20:55, 7 October 2008 (UTC)

William Mueller -- Eml4500.f08.jamama.mueller 00:02, 8 October 2008 (UTC)

Cedric Adam --Eml4500.f08.jamama.adam 05:46, 8 October 2008 (UTC)

Justin McIntire--Eml4500.f08.jamama.justin 16:04, 8 October 2008 (UTC)

Megan Alvarez --EML4500.f08.jamama.megan 18:34, 8 October 2008 (UTC)

Chris Alford --Eml4500.f08.jamama.chris 19:40, 8 October 2008 (UTC)