User:Eml 4500.f08.delta 6.specht/HW3

Determination of Element FD (wrt global coordinates)
We have the FD for springs:

k$$^{(e)}$$ d$$^{(e)}$$ = f$$^{(e)}$$   (1) Where each represents a 4x4, 4x1, and 4x1 matrix, respectively.

This is represented by the following diagram.



Also, we can treat this element as a spring element turned 45° with the following analysis.



The reaction forces have been taken along the direction of the element so that we can analyze the stretching or shrinkage of the bar element as a force is applied on it. The p vectors translate to be the force on the node and the q vectors are the displacement of the node as the bar stretches or shrinks. These pertain to the FD relationship mentioned above as:

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

In long-hand form:

$$k^{(e)} \begin{bmatrix} 1 & -1 \\ -1 & 1\\ \end{bmatrix} \begin{pmatrix} q^{(e)}_{1}\\ q^{(e)}_{2} \end{pmatrix} = \begin{pmatrix} p^{(e)}_{1}\\ p^{(e)}_{2} \end{pmatrix}$$

Where $$q^{(e)}_i$$ is the axial displacement of element e at local node i, and $$p^{(e)}_i$$ is the axial force of element e at local node i. Also, these matrices are sized as 2x2, 2x1, and 2x1, respectively.

Derivation of eqn (1) from eqn (2)
We want to find the relationship between $$q^{(e)}_{2x1}$$ and $$d^{(e)}_{4x1}$$, and $$p^{(e)}_{2x1}$$ and $$f^{(e)}_{4x1}$$. It can be expressed as follows:


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

Consider the displacement of node 1, denoted as $$\vec{d}^{(e)}$$.




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

We say that $$q^{(e)}_{1}$$, the axial displacement of local node 1, is the orthogonal projection of the displacement. It is defined as:

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

where
 * $$\vec{i}$$•$$\vec{\tilde{i}}$$ = cosθ^{(e)} = l$$^{(e)}$$
 * $$\vec{j}$$•$$\vec{\tilde{i}}$$ = sinθ^{(e)} = m$$^{(e)}$$

Which leads to the construction of the matrix equation that relates $$d$$ and $$q$$:

$$q^{(e)}_{1}$$ = l$$^{(e)}$$ $$d^{e}_{1}$$ + m$$^{(e)}$$ $$d^{e}_{2}$$
 * = $$\begin{bmatrix}

l^{(e)} & m^{(e)}\\ \end{bmatrix}_{1x2}$$ $$\begin{pmatrix} d^{e}_{1}\\ d^{e}_{2}\\ \end{pmatrix}_{2x1}$$

Similarily for node 2:

$$q^{(e)}_{2}$$ = l$$^{(e)}$$ $$d^{e}_{2}$$ + m$$^{(e)}$$ $$d^{e}_{2}$$
 * = $$\begin{bmatrix}

l^{(e)} & m^{(e)}\\ \end{bmatrix}_{1x2}$$ $$\begin{pmatrix} d^{e}_{3}\\ d^{e}_{4}\\ \end{pmatrix}_{2x1}$$

These two matrix equations combine to yield:


 * $$\begin{pmatrix}

q^{e}_{1}\\ q^{e}_{2}\\ \end{pmatrix}_{2x1}$$ = $$\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ \end{bmatrix}_{2x4} \begin{pmatrix} d^{e}_{1}\\ d^{e}_{2}\\ d^{e}_{3}\\ d^{e}_{4}\\ \end{pmatrix}_{4x1}$$

and in short-hand form:


 * $$q^{(e)}$$ = $$T^{(e)}$$ $$d^{(e)}$$

Similarly: $$\begin{Bmatrix}P_{1}^{(e)}\\P_{2}^{(e)}\end{Bmatrix}_{2x1}= T^{(e)}\begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \\ f_{3}^{(e)} \\ f_{4}^{(e)} \end{Bmatrix}_{4x4} => P^{(e)}=T^{(e)}*F^{(e)}$$ $$\hat{K}^{(e)}_{2x2}*q^{(e)}_{2x1} = P^{(e)}_{2x1}$$ where $$ q^{(e)}_{2x1}=T^{(e)}*d^{(e)}$$ and $$ P^{(e)}_{2x1}= T^{(e)}*F^{(e)} $$ therefore $$\Rightarrow \hat{K}^{(e)}_{2x2}*T^{(e)}*d^{(e)}= T^{(e)}*F^{(e)}$$

The goal with these matrices is to obtain $$K^{(e)}*d^{(e)}=F^{(e)}$$ in order to do that we need to move $$T^{(e)}$$ from the right hand side to the left hand side and this usually can be done by multiplying by the inverse $$T^{(e)-1}$$ but since $$T^{(e)}$$ is a rectangular matrix with (2x4) dimensions it can not be inverted therefore what needs to be done is the transpose $$T^{(e)^(T)}$$

$$[T^{(e)^(T)}_{4x2}*\hat{K}^{(e)}_{2x2}*T^{(e)}_{2x4}]$$$$ * d^{(e)}_{4x1}$$ = $$F^{(e)}_{4x1}$$ where $$K^{(e)}_{4x4}=[T^{(e)^(T)}_{4x2}*\hat{K}^{(e)}_{2x2}*T^{(e)}_{2x4}]\Rightarrow $$ $$K^{(e)}*d^{(e)}=F^{(e)}\Rightarrow K^{(e)}= T^{(e)^(T)} * \hat{K}^{(e)} * T^{(e)}$$

By the principle of virtual work (PVW) the reduction of the global force displacement relation is $$\Rightarrow$$

$$K_{6x6}*d_{6x1}=F_{6x1}\Rightarrow K_{2x2}*d_{2x1}=F_{2x1}$$

Then why not solve these matrices at this stage as follows $$d=K^{-1}*F$$ ? The reason why this can not be done is due to the matrix singularity of $$ K $$ Since the determinant of matrix $$ K $$ is equal to zero its inverse will be $$ \frac{1}{det [K]} $$ which will be undefined since the denominator is equal to zero.

The Eigenvalue and Stiffness Matrix K Relationship
$$K= \ \begin{bmatrix} K_{11}^{(1)} & K_{12}^{(1)} & K_{13}^{(1)} & K_{14}^{(1)} & 0 & 0  \\ K_{21}^{(1)}& K_{22}^{(1)} & K_{23}^{(1)} & K_{24}^{(1)} & 0 & 0 \\ K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)} \\ 0 & 0 & K_{31}^{(2)} & K_{32}^{(2)} & K_{33}^{(2)} & K_{34}^{(2)} \\ 0 & 0 & K_{41}^{(2)} & K_{42}^{(2)} & K_{43}^{(2)} & K_{44}^{(2)}  \end{bmatrix}_{6X6}\,$$ $$\ \ = \begin{bmatrix}  \frac{9}{16}& \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & 0 & 0 \\  \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & 0 & 0 \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & (\frac{9}{16} + \frac{5}{2}) & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & (\frac{3\sqrt{3}}{16} + \frac{5}{2}) & (\frac{3}{16} + \frac{5}{2}) & -\frac{5}{2} & -\frac{5}{2} \\  0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} \\ 0 & 0 & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} & \frac{5}{2}  \end{bmatrix}_{6X6}\,$$

For an unconstrained structure system there are 3 possible rigid body motions in 2-D (2 in translation plus 1 in rotation) The eigenvalues of the stiffness matrix are really important for design because they are related to the vibrations of the structure. A zero eigenvalue is related to a rigid body movement since it lacks the strain energy according to the equation $$K*v=\lambda *M*v $$ where $$K$$ is the stiffness matrix, λ is the eigenvalue related to vibrational frequency, and $$M$$ is the mass matrix. A zero eigenvalue means there is no elastic energy therefore there is no energy stored in the structure.

Closing the Loop Between FEA and Statics
Use global force displacement relation. $$ \begin{bmatrix} K_{13} & K_{14}\\ K_{23} & K_{24}\\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \textbf{F} $$ This matrix is a 6x2 multiplied by a 2x1 matrix and equals a 6x1 matrix. The forces $$F_{3}$$ and $$F_{4}$$are the known applied loads. Therefore, there is no need to compute them. The computations that need to be done are for rows 1,2,5 and 6. This will give The reactions $$F_{1}$$, $$F_{2}$$, $$F_{5}$$ and $$F_{6}$$. Closing the loop between The finite element method (FEM) and statics using Virtual Displacement Solving the two bar truss system uses a process shown in the figure below. Where side 1 is the finite element method and side two is the statics approach. The statics approach is used to confirm the finite element method and is known as closing the loop. The process of determinig the displacements using statics was discussed above in "Determination of Element FD (wrt global Coordinates). In the statics process the reactions are known. Therefore, the member forces $$P^{(1)}_{1}$$ and $$P^{(2)}_{2}$$ are known. To compute the axial displacement dof's ($$q$$) or the amount of extension of the bars the member forces and the element stiffiness's need to be used. The following formulas show the relationship between the member forces, element stiffness's and the axial displacements. $$ P=Kq$$ so, $$ q=P/K$$ $$q^{(1)}_{2}= \frac{P^{(1)}_{1}}{K^{(1)}}= \frac{P^{(1)}_{2}}{K^{(1)}}= AC$$ $$q^{(2)}_{1}= \frac{-P^{(2)}_{2}}{K^{(2)}}= \frac{-P^{(2)}_{2}}{K^{(2)}}= -AB$$ The axial displacements dof's at node 1 and 3 are equal to zero.  $$q^{(1)}_{2}=q^{(2)}_{2}= 0$$  Question: How does one back out from the displacement dof's of node 2?  Where: A = node 2, AC is the extension of 1A and AB is the extension of 3A. First an extensions must be drawn perpendicular to AC and AB. Then a line can be drawn between the point D, where the two extensions meet, and node 1. A line is also drawn between node 3 and point D. These two lines are the dotted line and denote the deformed shape of the two-bar truss system. The vector R is the displacement of A or node 2. It is found by taking the square root of the sum of the squared magnitudes of AC and AB. $$R= (AC^{2} + AB^{2})^{\frac{1}{2}}$$ Review of the small angle approximation:

Where A is the angle between the two bars. $$U_{y}= Rsin(A)$$ $$U_{x}= R(1-cos(A))$$ If the angle is small then $$sin(A)$$ is approximately equal to A and $$cos(A)$$ is approximately equal to 1. The above equations become: $$U_{y}= R*A$$ $$U_{x}= 0$$ The image above presents the two bar truss system that is being analyzed. The vector AD is the displacement vector of the whole system, which has an unknown magnitude. The length of the vectors AB and AC can be determined using the method shown below. The computation shows the length of the component AC. The magnitude of AB is determined in the homework section of this report. $$ AC = \begin{vmatrix} \frac{P^{(1)}_{2}}{k^{(1)}} \end{vmatrix}=\frac{5.1243}{\frac{3}{4}}=6.8234 $$ $$ AB=\begin{vmatrix} \frac{P_{2}^{1}}{k^{2}} \end{vmatrix} $$ The x and y coordinates of points B, C, and D are denoted as: $$ \begin{pmatrix} x_{B}, y_{B} \end{pmatrix}    , \begin{pmatrix} x_{C}, y_{C} \end{pmatrix}    and \begin{pmatrix} x_{D}, y_{D} \end{pmatrix} $$ The values for these points are computed in the homework. The slopes of the lines BD and CD are given below. $$ Slope of CD = \theta ^{(1)}+ 90^{\circ} $$ $$ Slope of BC = \theta ^{(2)}- 90^{\circ} $$ If the magnitudes of the vectors BD and CD are known, the point D can be found using the equations given below. Points B and C must first be found.



$$ \vec{PQ}=(PQ)\vec{\tilde{i}}=PQ(cos\vec{i}+sin\vec{j}) $$ $$ \vec{PQ}=(x-x_{P})\vec{i}+(y-y_{P})\vec{j} $$ $$ x-x_{P}=(PQ)cos\theta $$ $$ y-y_{P}=(PQ)sin\theta $$ $$ \frac{y-y_{P}}{x-x_{P}}=tan\theta $$ $$ y-y_{P}=(tan\theta )(x-x_{P}) $$ Once the points B and C have been found, the equation shown below can be used to find the point D. The equation below is the equation of a line, passing through a point P.  If the values for B and C are input into these equations, the point D can be found geometrically. $$ y-y_{P}=(tan\theta + \frac{\pi }{2})(x-x_{P}) $$ The vector AD can be expressed as shown below, where the x and y values of the point A are zero, since the origin is defined at A.  $$ \vec{AD}= (x_{D}-x_{A})\vec{i}+(y_{D}-y_{A})\vec{j} $$ The vector AD can also be expressed in terms of the global displacement vectors, as shown in the diagram earlier (d3 and d4). $$ \vec{AD}=d_{3}\vec{i}+d_{4}\vec{j} $$

Three Bar Truss System
The three bar truss system is shown in the image below. The characteristics of each element is shown in the table that follows the image.





Convenient methods of numbering the local nodes are shown in the images below for each element. Element 1



Element 2



Element 3



$$ \sum{F_{x}}=0 $$ $$ \sum{F_{y}}=0 $$ $$ \sum{M_{A}}=0 $$

Question: What if we take the moment about a point B?



$$ \sum{M_{B}} = \vec{BA}\; X\; \vec{ F} = \vec{BA'}\; X\; \vec{ F} $$ The above equation stands only when A' lies on the line of action. We can derive this moment by defining the BA' vector, then substituting into the moment equation.

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

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

where $$ \vec{AA'}\; X\; \vec{F} = 0 $$

The derivation of the moment about point B is quite simple:
 * NOTE

$$ \sum{\vec{F}_{i}} = 0 $$ $$ \sum{\vec{M}_{B_{i}}} = \sum{\vec{BA'}_{i}}\; X\; \vec{F}_{i} = \sum{\vec{BA}_{i}}\; X\; \sum{\vec{F}_{i}} $$ where $$ \sum{\vec{F}_{i}} = 0 $$ Therefore, the moment about point B is also equal to zero.

The 8x8 global stiffness matrix is found below, where the local stiffness matrices for each element correspond to the color-coded areas. The k values for the local stiffness matrix for element 3 can be found as shown below. Only two examples are shown, yet the values can be found under the homework section of this page. $$ k_{33} = k^{(1)}_{33} + k^{(2)}_{11} + k^{(3)}_{11} $$ $$ k_{34} = k^{(1)}_{34} + k^{(2)}_{12} + k^{(3)}_{12} $$

Determination of Point Coordinates
The following are the unknown variables that will be determined:

(XD, YD) (XB, YB) (XC, YC)

We will begin by finding the coordinates of point D. Note: Point A has been designated as the origin.

Using the definition of line AD, (XD, YD) can be found.

$$\vec{AD} = (X_{D} - X_{A}) \vec{i} + (Y_{D} - Y_{A}) \vec{j}$$

Since point A is referenced as the origin, XA and YA are equal to zero.

By definition of the displacement vector of A,

$$\vec{AD} = d_{3}\vec{i} + d_{4}\vec{j}$$

Thus,

(XD, YD) = (d3, d4) = (4.352, 6.1271)

Next, using the relationship of two points on the same line gives,

$$y - y _{C} = tan(\theta) (x - x_{C})$$

where θ = 30°, (XC, YC) = (5.92, 3.42)

The relationship between point B and point D gives,

$$y - y _{B} = tan(\theta + \frac{\pi }{2}) (x - x_{B})$$

where θ = 135°,(XB, YB) = (-0.88, 0.88)

Results:

Point D: (4.35, 6.13) Point C: (5.92, 3.42) Point B: (-0.88, 0.88)

Global K Matrix for Three Bar System
The following variables are known: E(1) = 2 A(1) = 3 L(1) = 5 θ(1) = 30° E(2) = 4 A(2) = 1 L(2) = 5 θ(2) = -30° E(3) = 3 A(3) = 2 L(3) = 10 θ(3) = -135° The following variables can be determined:

$$k_{(1)} = \frac{E^{(1)}A^{(1)}}{L^{(1)}} = \frac{6}{5} = 1.2$$ $$k_{(2)} = \frac{E^{(2)}A^{(2)}}{L^{(2)}} = \frac{4}{5} = 0.8$$ $$k_{(3)} = \frac{E^{(3)}A^{(3)}}{L^{(3)}} = \frac{6}{10} = 0.6$$

$$l^{(1)} = cos\theta ^{(1)} = 0.866$$ $$m^{(1)} = sin\theta ^{(1)} = 0.5$$ $$l^{(2)} = cos\theta ^{(2)} = 0.866$$ $$m^{(2)} = sin\theta ^{(2)} = -0.5$$ $$l^{(3)} = cos\theta ^{(3)} = -0.707$$ $$m^{(3)} = sin\theta ^{(3)} = -0.707$$ Note: l(3) = m(3) The following are the element stiffness matrices, which are needed to compute the global stiffness matrix.

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

$$k^{(1)} = \begin{bmatrix} 0.9 & 0.52 & -0.9 & -0.52\\ 0.52 & 0.3 & -0.52 & -0.3 \\ -0.9 & -0.52 & 0.9 & 0.52\\ -0.52 & -0.3 & 0.52 & 0.3 \end{bmatrix}$$

$$k^{(2)} = \begin{bmatrix} 0.6 & -0.346 & -0.6 & 0.346\\ -0.346 & 0.2 & 0.346 & -0.2 \\ -0.6 & 0.346 & 0.6 & -0.346\\ 0.346 & -0.2 & -0.346 & 0.2 \end{bmatrix}$$

$$k^{(3)} = \begin{bmatrix} 0.3 & 0.3 & -0.3 & -0.3\\ 0.3 & 0.3 & -0.3 & -0.3\\ -0.3 & -0.3 & 0.3 & 0.3\\ -0.3 & -0.3 & 0.3 & 0.3 \end{bmatrix}$$

Now the global stiffness matrix can be computed.

$$K = \begin{bmatrix} 0.9 & 0.52 & -0.9 & -0.52 & 0 & 0 & 0 & 0\\ 0.52 & 0.3 & -0.52 & -0.3 & 0 & 0 & 0 & 0\\ -0.9 & -0.52 & 1.8 & 0.474 & -0.6 & 0.346 & -0.3 & -0.3\\ -0.52 & -0.3 & 0.474 & 0.8 & 0.346 & -0.2 & -0.3 & -0.3\\ 0 & 0 & -0.6 & 0.346 & 0.6 & -0.346 & 0 & 0\\ 0 & 0 & 0.346 & -0.2 & -0.346 & 0.2 & 0 & 0\\ 0 & 0 & -0.3 & -0.3 & 0 & 0 & 0.3 & 0.3\\ 0 & 0 & -0.3 & -0.3 & 0 & 0 & 0.3 & 0.3 \end{bmatrix}$$

Matrix Proof
Note: Since the inverse of T(e) cannot be used the transpose is utilized. The following demonstrates that both sides are equal.

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

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

$$k_{(e)} = \frac{E^{(e)}A^{(e)}}{L^{(e)}}$$

thus,

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

To prove that both sides are equal, the matrices must be multiplied in the correct order. The matrices are given as:

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

$$\hat{k}^{(e)} = k_{(e)}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

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

Multiplying out the first two matrices gives:

$$T^{(e)T}\hat{k}^{(e)} = \begin{bmatrix} l^{(e)} & 0 \\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)} \end{bmatrix} k_{(e)}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} = k_{(e)} \begin{bmatrix} l^{(e)} & -l^{(e)}\\ m^{(e)} & -m^{(e)}\\ -l^{(e)} & l^{(e)}\\ -m^{(e)} & m^{(e)} \end{bmatrix}$$

Taking the result from this and now multiplying it by the third matrix T(e) gives:

$$k_{(e)} \begin{bmatrix} l^{(e)} & -l^{(e)}\\ m^{(e)} & -m^{(e)}\\ -l^{(e)} & l^{(e)}\\ -m^{(e)} & m^{(e)} \end{bmatrix} T^{(e)} = k_{(e)} \begin{bmatrix} l^{(e)} & -l^{(e)}\\ m^{(e)} & -m^{(e)}\\ -l^{(e)} & l^{(e)}\\ -m^{(e)} & m^{(e)} \end{bmatrix} \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix} = k_{(e)}\begin{bmatrix} l^{(e)2} & l^{(e)}m^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & m^{(e)2} & -l^{(e)}m^{(e)} & -m^{(e)2} \\ -l^{(e)2} & -l^{(e)}m^{(e)} & l^{(e)2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -m^{(e)2} & l^{(e)}m^{(e)} & m^{(e)2} \end{bmatrix} $$

Observation: both sides are equal.

Determination of Reaction Forces using 3 Different Methods
Three methods of solving the same problem are shown below. Each method returns the same result upon careful consideration of the geometry of the situation, and the properties of the elements. The only difference between each method of solving this problem is the amount of effort put into the calculations. It can be seen that the Global Reduced Matrix Method can take significantly less time to find a solution than for each of the other methods.

Global Reduced Matrix Method
This method uses the global K matrix (which has been reduced to only the 3rd and 4th columns of the 6 x 6 matrix), and the global displacement vector to calculate the resultant forces. These forces are given as F1, F2, F5, and F6 in the global force matrix. $$ F=\begin{bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}\\ \end{bmatrix}=\begin{bmatrix} K_{13} &K_{14} \\ K_{23} &K_{24} \\ K_{33} &K_{34} \\ K_{43} &K_{44} \\ K_{53} &K_{54} \\ K_{63} &K_{64} \end{bmatrix}*\begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} $$ $$ F=\begin{bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}\\ \end{bmatrix}=\begin{bmatrix} -0.5625 &-0.3248 \\ -0.3248 &-0.1875  \\  (2.5+0.5625) &(-2.5+0.3248)  \\  (-2.5 +0.3248) &(2.5+0.1875)  \\  -2.5 &2.5  \\  2.5 &-2.5  \end{bmatrix}*\begin{Bmatrix} 4.352\\ 6.1271 \end{Bmatrix}=\begin{Bmatrix} -4.44\\ -2.56\\ 0.00\\ 7.00\\ 4.44\\ -4.44\\ \end{Bmatrix} $$

Element Method (Section 5.1)
The second method computes the reaction forces at the fixed nodes by considering each element indivudually. First, the reaction forces for element 1 are computed using the local k matrix and the local displacement vector. Second, the reaction forces for element 2 are computed using the same method. The general equation is presented first, followed by the specific equation for each element, with values for the k matrices and the displacement vectors inserted. $$ f^{(e)}=k^{(e)}*d^{(e)} $$ $$ f^{(1)}=\begin{bmatrix} 0.5625 &0.3248 &-0.5625  &-0.3248 \\ 0.3248 &0.1875  &-0.3248  &-0.1875 \\ -0.5625 &-0.3248  &0.5625  &0.3248 \\ -0.3248 &-0.1875  &0.3248  &0.1875 \end{bmatrix}*\begin{Bmatrix} 0\\ 0\\ 4.352\\ 6.1271 \end{Bmatrix}=\begin{Bmatrix} -4.44\\ -2.56\\ 4.44\\ 2.56 \end{Bmatrix} $$ $$ f^{(2)}=\begin{bmatrix} 2.5 &-2.5 &-2.5  &2.5 \\ -2.5 &2.5  &2.5  &-2.5 \\ -2.5 &2.5  &2.5  &-2.5 \\ 2.5 &-2.5  &-2.5  &2.5 \end{bmatrix}*\begin{Bmatrix} 4.352\\ 6.1271\\ 0\\ 0 \end{Bmatrix}=\begin{Bmatrix} -4.44\\ 4.44\\ 4.44\\ -4.44 \end{Bmatrix} $$

Statics Method
The last method computes the reaction forces by using equations from statics considerations. The equations used are presented below, along with the values obtained for the reaction forces.



P = 7 L(1) = 4 L(2) = 2 θ(1) = 30° θ(2) = 45°



Observation:

By using trigonometry and the angles given the unknown lengths can be determined. These lengths will be needed when analyzing the summation of the moments of each node.

$$\sin {30^{\circ} } = \frac{L_c}{4} \Rightarrow L_c = 4 \sin {30^{\circ} } = 2$$

$$\cos {30^{\circ} } = \frac{L_x}{4} \Rightarrow L_x = 4 \cos {30^{\circ} } = 3.46$$

$$\sin {45^{\circ} } = \frac{L_b}{2} \Rightarrow L_b = 2 \sin {45^{\circ} } = 1.4$$

$$\cos {45^{\circ} } = \frac{L_y}{2} \Rightarrow L_y = 2 \cos {45^{\circ} } = 1.4$$

Lz = Lx + Ly = 3.46 + 1.4 = 4.86

La = Lc - Lb = 2 - 1.4 = 0.6

Statics equations can now be analyzed since all lengths are known. In order to do this, two moment equations and the equations for the sum of forces about the x and y axes will be used, and are presented below.

$$ \sum{M_{2}^{(1)}}=(-F_{1}*L_{C})-(F_{2}*L_{X})=0 $$

$$ \sum{M_{2}^{(2)}}=(F_{3}*L_{B})-(F_{4}*L_{Y})=0 $$

$$ \sum{F_{X}}=F_{1}+F_{3}=0 $$

$$ \sum{F_{Y}}=P+F_{2}+F_{4}=0 $$

Solving these equations simultaneously, with the value of P=7 will yield these results:

$$ F_{1}=-4.44 $$

$$ F_{2}=-2.56 $$ $$ F_{3}=4.44 $$ $$ F_{4}=-4.44 $$

Determination of q for Node 2
The displacement vector for node 2 can be found by breaking it down into vector components. $$q^{(e)}_2= (d^{(e)}_3\vec{i}+d^{(e)}_4\vec{j}) \cdot \vec{\tilde{i}}$$ This is equivalent to $$q^{(e)}_2= d^{(e)}_{3}(\vec{i}\cdot \vec{\tilde{i}})+d^{(e)}_4(\vec{j} \cdot \vec{\tilde{i}})$$

q2(e) is shown below, after substituting in the director cosines.

$$q_2^{(e)}=l^{(e)}d_3^{(e)}+m^{(e)}d_4^{(e)}$$ where $$ l^{(e)} = cos\theta $$ $$ m^{(e)} = sin\theta $$

$$q_2^{(e)}=\begin{bmatrix}l^{(e)} & m^{(e)} \end{bmatrix}\begin{bmatrix}d_3^{(e)}\\ d_4^{(e)}\end{bmatrix}$$

Determining p values
We know that $$p^{(e)}$$ = $$T^{(e)}$$ $$f^{(e)}$$, where


 * $$T^{(e)}$$ = $$\begin{bmatrix}

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

and, from past exercises,


 * $$f^{(e)}$$ = $$\begin{bmatrix}

-4.4378\\ -2.5622\\ 4.4378\\ 2.5622\end{bmatrix}$$

Therefore, we can say that the p values become:


 * $$p^{(1)}_{2}$$ = |-4.4378cos(30°) + (-2.5622)sin(30°)|= 5.1243
 * $$p^{(2)}_{1}$$ = |4.4378cos(30°) + 2.5622sin(30°)|= 5.1243

Determining k values
From past exercises it is known that the axial stiffness for both bar elements are as follow:


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

Therefore,


 * $$k^{(1)}$$ = $$\frac{(3)(1)}{4}$$ = 0.75


 * $$k^{(2)}$$ = $$\frac{(5)(2)}{2}$$ = 5

Determining the displacements
Following the same idea of $$d = \frac{f}{k}$$, we have


 * AC = $$\frac{p^{(1)}_{2}}{k^{(1)}}$$ = $$\frac{5.1243}{0.75}$$ = 6.8324


 * AB = $$\frac{p^{(2)}_{1}}{k^{(2)}}$$ = $$\frac{5.1243}{5}$$ = 1.0249

MATLAB Plot of a Two-bar Truss Structure (undeformed and deformed)
The following code plots the analysis of an undeformed and deformed two-bar truss system using MATLAB MATLAB Code

Note: this MATLAB code was created using the principles obtained from:


 * The MATLAB script for the analysis of the Two Bar Truss System obtained from Dr. L. Vu-Quoc website http://clesm.mae.ufl.edu/~vql/courses/fead/2008.fall/codes/twoBarTrussEx.txt


 * The MATLAB script for the plot of a crab-leg structure created by X.G. Tan (2003) and obtained from Dr. L. Vu-Quoc website http://apollo.mae.ufl.edu/~vql/courses/fead/2006.fall/codes/matlab_plot_example2.m


 * The MATLAB script for the plot of an electric pylon (undeformed five-bar truss structure) created by Team MeshWorks (2003) and obtained from Dr. L. Vu-Quoc website http://apollo.mae.ufl.edu/~vql/courses/fead/2006.fall/code.web.pages/electric_pylon.m

Results



Contributing Team Members
Iain Specht--Eml 4500.f08.delta 6.specht 18:13, 5 October 2008 (UTC) Nicolas Castrillon --Eml4500.f08.delta 6.castrillon 02:37, 6 October 2008 (UTC) Diana Guzman--Eml4500.f08.delta 6.guzman 21:34, 6 October 2008 (UTC) Sean Lopez--Eml4500.f08.delta 6.lopez 16:02, 7 October 2008 (UTC) Micheal Krueger--Eml4500.f08.delta 6.krueger 20:29, 7 October 2008 (UTC) Eric Ramirez--Eml4500.f08.delta 6.ramirez 01:49, 8 October 2008 (UTC)