User:Eml4500.f08.gravy.jad/NotesRedo3+Oldversion

FEA Notes and Homework 3

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

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



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

Force Displacement matrices calculations
The force displacement can be written as

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

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

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

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

Goal: Derive Equation 1 from Equation 2

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

These relationships can be expressed in the form:

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

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

Consider the displacement vector of local node 1 denoted by

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



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

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

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

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

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

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

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

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

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

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

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

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

Similarly (same argument) :

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

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

Recall element axial Force Displacement relation:

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

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

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

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

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

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

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

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

HW: verify above.
Verify the following equation:

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

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

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

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

Solving the first set of matrices gives:

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

Solving the second set of matrices gives:

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

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

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

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

Remark: Why not solve as follows

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

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

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

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

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

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

Dynamic Evaluation Problem

 K v = λ Mv 

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

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

MATLAB Code

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

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

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

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

λ = 0, 0, 0, 0, 1.4705, 10.0295

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

//(end Remark)

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

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

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

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

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

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

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

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

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



Element 1:



Element 2:



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

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

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

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



MATLAB code for truss displacement


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



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

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

hence

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

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



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

HW: Solving AC and AB
From previous lecture

Element Length

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

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

Young's Modules

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

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

Cross-sectional Area

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

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

Inclination angle


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


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


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

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

And the T matrix is

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

we can solve for P

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

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

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

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

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

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

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

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

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

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

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

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

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

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

so

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

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

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

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

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

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

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



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

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

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

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

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

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

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

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

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

From HW2, we know that:

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

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

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

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

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

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

///end closing loop

3-Bar Truss Syst:


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

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

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

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

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

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

Convenient local node numbering





$$ \sum{F_x} =0$$

$$ \sum{F_y} =0$$

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

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

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



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

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

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

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

Back to 3-bar truss

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

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

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

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

Converting from k matrices to K matrix


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

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

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

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

Contributing Team Members
The following students contributed to this report:

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

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

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

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

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

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