User:Eml 4500.f08.delta 6.specht/HW5

 Note: Derivation of 3-D bar element missing.

Cheating: Modified the "results" array for the 2-bar truss after I pointed out the problem in my lecture, without any change in the code and without putting a comment box to alert the readers; here is the comparison between the official version before the deadline and the cheated version. See my comments below.

Please don't remove this comment in case you amend your work; just add a comment box to say what you did. Eml4500.f08 12:20, 11 November 2008 (UTC)

=Homework 5=

Justification
We can justify the elimination of rows 1,2,5,6 to obtain $$\bar{\mathbf{K}}$$ in a 2-bar truss.


 * We begin with the FD relationship, writing it as $$K$$$$d$$ = $$F$$. Based on the model 2-bar system, we can generalize this method for any kind of application.


 * $$K_{6x6}$$$$d_{6x1}$$ = $$F_{6x1}$$

Therefore,
 * $$K$$$$d$$-$$F$$=$$0_{6x1}$$


 * Principle of Virtual Work:


 * $$w$$$$_{6x1}$$•($$K$$$$d$$-$$F$$)$$_{6x1}$$ = $$0_{1x1}$$; for all $$w$$$$_{6x1}$$

Where w is the weighting matrix, and the equation is equalized to zero as a sacalar (or 1x1 matrix). In other words, the weighting matix is dotted with the FD relationship matrix to obtain a scalar. Also, according to the wise Prof. Vu-Quoc, "$$w$$ was not selected out of frivolity," as it is called the weighting coefficient or matrix.We say that the FD relationship is equivalent to the PVW.

Proof of FD relationship and PVW equality

 * 1) FD relationship to PVW is trivial because if FD relationship is true, then $$w$$(0)=0, this being obvious
 * 2) We now want to show how to go from PVW to FD relationship.


 * We have the PVW: $$w$$$$_{6x1}$$•($$K$$$$d$$-$$F$$)$$_{6x1}$$ = $$0_{1x1}$$; for all $$w$$$$_{6x1}$$


 * Choice 1: Select $$w$$ such that $$w_{1}$$=1, $$w_{2}$$=...=$$w_{6}$$=0


 * Therefore,
 * $$w^{T}_{6x1}$$=$$\begin{bmatrix}

1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$
 * $$w$$•($$K$$$$d$$-$$F$$) = 1•$$\left[{\sum_{j=1}^{6}K_{1j}d_{j}-F_1} \right]$$+0•$$\left[{\sum_{j=1}^{6}K_{2j}d_{j}-F_2} \right]$$+...+0•$$\left[{\sum_{j=1}^{6}K_{6j}d_{j}-F_6} \right]$$=0
 * $$\sum_{j=1}^{6}{K_{1j}d_j}=F_1$$ (1$$^{st}$$equation)


 * Choice 2: Select $$w$$ such that $$w_{1}$$=0, $$w_{2}$$=1, $$w_{3}$$=...=$$w_{6}$$=0


 * Therefore,
 * $$w^{T}_{6x1}$$=$$\begin{bmatrix}

0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}$$
 * $$w$$•($$K$$$$d$$-$$F$$) = 0•$$\left[{\sum_{j=1}^{6}K_{1j}d_{j}-F_1} \right]$$+1•$$\left[{\sum_{j=1}^{6}K_{2j}d_{j}-F_2} \right]$$+0•$$\left[{\sum_{j=1}^{6}K_{3j}d_{j}-F_3} \right]$$+...=0
 * $$\sum_{j=1}^{6}{K_{2j}d_j}=F_2$$ (2$$^{nd}$$equation)


 * Choices 3 through 6 repeat this pattern.

Ultimately, we will have achieved the FD relationship.

PVW: Furthermore
According to the boundary conditions (BCs) of the 2-bar truss, $$d_1=d_2=d_5=d_6=0$$.

The weighting cofficient must be kinematically admissible. It cannot violate the BCs: $$w_1=w_2=w_5=w_6=0$$. Therefore, weighting coefficients go hand-in-hand with virtual displacements, as explained by calculus of variations.

For the 2-bar truss at hand, it is evident that $$w$$•($$K$$$$d$$-$$F$$) can be reduced to a rectangular matrix multiplied by a 2x1 matrix with a 6x1 matrix as a result.


 * $$w$$•($$K$$$$d$$-$$F$$)=$$\begin{bmatrix}

w_3\\ w_2 \end{bmatrix}$$•($$\bar{K}_{2x2}$$$$\bar{d}_{2x1}$$-$$\bar{F}_{2x1}$$)=0

Where,
 * $$\bar{K}$$=$$\begin{bmatrix}

K_{33} & K_{34}\\ K_{43} & K_{44} \end{bmatrix}$$
 * $$\bar{d}$$=$$\begin{Bmatrix}

d_3\\ d_4 \end{Bmatrix}$$
 * $$\bar{F}$$=$$\begin{Bmatrix}

F_3\\ F_4 \end{Bmatrix}$$

PVW: The Weighting Coefficient
First the weighting coefficient needs to be defined, or reinforced, and an example needs to be given to emphasize its use and importance. We will let w equal the weighting coefficient.

w = weighting coefficient

Now an Example:

Lets look at the calculation of a final course grade. Where: Exam 1 = 82 and is worth 15 % of final grade Exam 2 = 74 and is worth 20 % of final grade Exam 3 = 92 and is worth 25 % of final grade Hw grade = 87 and is worth 40 % of final grade

The formula to calculate the final grade is given as:

$$ finalgrade = w_0(Hw grade) + \sum_{i=1}^{3}w_iExam_i$$

Where $$ w_i$$ and $$ w_0$$ are the exam weighting coefficients and the homework(Hw) weighting coefficient. Now plugging into the above formula:

$$ finalgrade = .4(87) + .15(82) + .2(74) + .25(92) = 84.9$$

Now to emphasize how the weighting coefficients will change the final value, we'll let the Hw grade and all the exam grades remain the same but change the worth to the following: Exam 1 worth 30 % Exam 2 worth 30 % Exam 3 worth 30 % Hw grade worth 10 %

The final grade now becomes:

$$ finalgrade = .1(87) + .3(82) + .3(74) + .3(92) = 83.1$$

This shows the importance of the weighting coefficients. The importance of a value, within an equation, can be changed by changing the weighting coefficient of that value. The final value will also be changed, as shown above, by changing the weighting coefficients.

Now we will return to the Principle of Virtual Work. Recall: $$K^{(e)}= T^{(e)^(T)} * \hat{K}^{(e)} * T^{(e)}$$

Where $$K^{(e)}$$ is a 4x4 matrix, $$T^{(e)^(T)}$$ is a 4x2 matrix, $$\hat{K}^{(e)}$$ is a 2x2 matrix and $$T^{(e)}$$ is a 2x4 matrix.

Question: How do we use the Principle of Virtual Work?

First Recall the Force Displacement Relationship (FD Rel) with respect to the axial degrees of freedom ($$q^{(e)}$$):

Eqn (1)$$\hat{K}^{(e)}q^{(e)} = p^{(e)}$$

Take note that the matrix p is lower case. This implies it is in reference to the element not the global coordinate system.

We then move the p matrix to the left hand side of the equation. Equation (1) is now in the same form as the FD Rel $$Kd-F=0$$.

$$\hat{K}^{(e)}q^{(e)} - p^{(e)}= 0$$

The equation is now multiplied by the weighting coefficient, which is a 2x2 matrix, to give the Principle of Virtual Work:

Eqn (2)$$w\cdot(\hat{K}^{(e)}q^{(e)} - p^{(e)})= 0$$

This holds true for all weighting coefficient matrices. In previous notes and home work we have shown that $$ Eqn(1) \Leftrightarrow Eqn(2)$$. We can ow exploit the Principle of Virtual Work in order to get expressions in the global coordinate system. Recall: Eqn (3)$$q^{(e)}=T^{(e)}d^{(e)}$$ Where $$q^{(e)}$$ is the real displacement and is a 2x1 matrix. $$T^{(e)}$$ is a 2x4 matrix and $$d^{(e)}$$ is a 4x1 matrix. $$w$$ corresponds to the axial degrees of freedom. Since we are using virtual work the axial weighting coefficient needs to be converted into the virtual weighting coefficient. This is done using Eqn (4) below: Eqn (4)$$\hat{w}=T^{(e)}w$$ Where $$\hat{w}$$ is a 2x1 matrix, $$T^{(e)}$$ is a 2x4 matrix and $$w$$ is a 4x1 matrix.

Proving the Coordinate Systems Relationship
For this proof the following variables should be defined. $$q_{2x1}^{(e)}$$ is given as the real axial displacement. The variable $$\hat{W}_{2x1}$$ is the virtual axial displacement that corresponds to the axial displacement $$q_{2x1}^{(e)}$$. $$W_{4x1}d^{(e)}$$ is the virtual axial displacement in the global coordinate system that corresponds to the real displacement in global coordinates ($$d^{(e)}$$).

The equations for the real axial displacement and the virtual axial displacement are as follows:

$$ q_{2x1}^{(e)}=T_{2x4}^{(e)}d_{4x1}^{(e)} $$

$$ \hat{W}_{2x1}=T_{2x4}^{(e)}W_{4x1} $$ The equation for the axial force can be rearranged and multiplied by the weighting factor, which is the virtual axial displacement vector, in order to obtain:

$$ \hat{W}_{2x1}\cdot \left(\hat{k}_{2x2}^{(e)}q_{2x1}^{(e)}-p_{2x1}^{(e)} \right)=0_{1x1} $$

Once the equations for the real axial displacement and the virtual axial displacement are substituted into the equation given above, the following relation is obtained.

$$ \left( T^{(e)}W\right)\cdot \left[\hat{k}_{2x2}^{(e)}\left(T^{(e)}d^{(e)} \right)-p_{2x1}^{(e)} \right]=0 $$ This relation is true for all $$W_{4x1}$$. This equation can be further manipulated in order to eliminate the dot product that is present. The form of the equation obtained is below.

$$ \left( T^{(e)}W\right)^{T}\left[\hat{k}_{2x2}^{(e)}\left(T^{(e)}d^{(e)} \right)-p_{2x1}^{(e)} \right]=0_{1x1} W_{4x1} $$

By applying the relation below to the equation that is listed above, the equation above can be simplified to:

$$ W^{T}T^{(e)T}\left[\hat{k}_{2x2}^{(e)}\left(T^{(e)}d^{(e)} \right)-p_{2x1}^{(e)} \right]=0_{1x1} $$

This relation is obtained by using the following substitution for the dot product.

$$ a_{nx1}\cdot b_{nx1}=a_{1xn}^{T}b_{nx1} $$

This same relation can be applied again in reverse once the transpose of the T matrix is distributed through the matrix containing the axial force-displacement relationship.

$$ W \cdot \left[\left(T^{(e)T}\hat{k}^{(e)}T^{(e)} \right)d^{(e)} - \left(T^{(e)T}p^{(e)} \right) \right]= 0$$ for all $$ W_{4x1} $$

From this recognize that the value $$T^{(e)T}\hat{k}^{(e)}T^{(e)} $$ is equal to the element stiffness matrix, $$k^{(e)}$$, and that the value $$T^{(e)T}p^{(e)} $$ is equal to the element force vector, $$f^{(e)}$$. This relation can then be simplified to the force-displacement relation in global coordinates.

$$ W \cdot \left[k^{(e)}d^{(e)}-f^{(e)}\right]= 0$$ for all $$ W_{4x1} $$

Since this is true for all $$ W_{4x1}$$, the original force-displacement,$$ k^{(e)}d^{(e)}= f^{(e)} $$, relation can be obtained. This is the second and more powerful method to prove the following relation that is used during this proof.

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

FEA and the Mechanics of Materials


$$ \sum{F_{x}} = 0 = -N(x, t) + N(x + dx, t) + f(x, t)dx - m(x)\ddot{u} $$

$$ = \frac{dN}{dx}(x, t)dx + h.o.t. + f(x, t)dx + m(x)\ddot{u} $$

In the above equation, we can neglect the h.o.t. (higher order terms). If we recall Taylor Series Expansion:

$$ f(x+dx) = f(x) + \frac{df(x)}{dx}dx + \frac{1}{2}\frac{d^{2}f(x)}{dx^{2}} + ... $$

The term with the 1/2 coefficient and everything after that term are know as higher order terms. Equation (1) now turns into:

$$ \frac{dN}{dx} + f = m\ddot{u}\; \; \; \; \; \;\; \; \;\; \; \;Eq. (2) $$

This equation is also know as the equation of motion.

$$ N(x, t) = A(x)\sigma (x, t)\; \; \; \; \; \; \; \; \; \; \; \; Eq. (3) $$

$$ \sigma (x, t) = E(x)\varepsilon (x, t) $$

$$ \varepsilon (x, t) = \frac{du}{dx}(x, t) $$

The equation above is known as the Constitutive Relation. Now, if we substitute Equation (3) into Equation (2):

$$ \frac{d}{dx}[N(x, t) = A(x)\sigma (x, t)] + f(x, t) = m(x)\ddot{u} $$

This is know as the Partial Differential Equation of Motion. We need two boundary conditions, since this is a second order derivative with respect to x. These two conditions are the initial displacements and initial velocities.

where $$ u(0, t) = 0 = u(L, t) $$



$$ N(L, t) = A(L)\sigma (L, t) $$

$$ \sigma (L, t) = E(L)\varepsilon (L, t) $$

$$ \varepsilon (L, t) = \frac{du}{dx}(L, t) $$

$$ \frac{du}{dx}(L, t) = \frac{F(t)}{A(L)E(L)} $$

Chapter 2: Elastic Bar Subject to Variable Properties and Loads
Up to this point the case of a discrete, non-continuous partial differential equation (PDE) has been discussed. The new material is concerning continuous PDEs. The problem that will be used to examine this type of analysis is an elastic bar that has a varying cross sectional area,$$A(x)$$, and modulus of elasticity, $$E(x)$$. This bar will be subject to varying axial loads (distributed loads), time dependent concentrated loads ($$f(x,t)$$), and dynamic inertial forces. An image of this case is shown below with the distributed load shown on the axis of the elastic bar.



Axial FD Relation
$$ k^{(e)}_{6x6}d^{(e)}_{6x1}=f^{(e)}_{6x1} $$

Element Displacements
$$ d^{(e)}=\begin{bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ d_{5}^{(e)}\\ d_{6}^{(e)} \end{bmatrix} $$

Element Forces
$$ f^{(e)}=\begin{bmatrix} f_{1}^{(e)}\\ f_{2}^{(e)}\\ f_{3}^{(e)}\\ f_{4}^{(e)}\\ f_{5}^{(e)}\\ f_{6}^{(e)} \end{bmatrix} $$

Relation Between Axial DOF's and Element DOF's in Global Coordinates
$$ q^{(e)}=\begin{pmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{pmatrix}= \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)}  & n^{(e)} \end{bmatrix}\begin{pmatrix} d_{1}^{(e)}\\ d_{2}^{(e)}\\ d_{3}^{(e)}\\ d_{4}^{(e)}\\ d_{5}^{(e)}\\ d_{6}^{(e)} \end{pmatrix} =T_{2x6}^{(e)} d_{6x1}^{(e)} $$

Relation Between Axial Forces and Element Forces in Global Coordinates
$$ P^{(e)}=\begin{pmatrix} P_{1}^{(e)}\\ P_{2}^{(e)} \end{pmatrix}= \begin{bmatrix} l^{(e)} & m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)}  & n^{(e)} \end{bmatrix}\begin{pmatrix} f_{1}^{(e)}\\ f_{2}^{(e)}\\ f_{3}^{(e)}\\ f_{4}^{(e)}\\ f_{5}^{(e)}\\ f_{6}^{(e)} \end{pmatrix} =T_{2x6}^{(e)} f_{6x1}^{(e)} $$

Element Stiffness Matrix in Global Coordinates in Terms of Axial DOF's
$$ k^{(e)}= \frac{EA}{L} $$

$$ \hat{k}^{(e)}_{2x2}=k^{(e)}\begin{bmatrix} 1& -1 \\ -1 & 1 \end{bmatrix}=\frac{EA}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} $$

$$ k^{(e)}_{6x6}=T^{(e)T}_{6x2}\hat{k}^{(e)}_{2x2}T^{(e)}_{2x6}=\frac{EA}{L}\begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ n^{(e)} & 0\\ 0 & l^{(e)}\\ 0& m^{(e)}\\ 0 & n^{(e)} \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} l^{(e)} &m^{(e)} & n^{(e)} & 0 & 0 & 0\\ 0& 0 & 0 & l^{(e)} & m^{(e)} & n^{(e)} \end{bmatrix} $$

$$ k^{(e)}_{6x6}=\frac{EA}{L}\begin{bmatrix} l^{(e)2} & l^{(e)}m^{(e)} & l^{(e)}n^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)} & -l^{(e)}n^{(e)}\\ m^{(e)}l^{(e)} & m^{(e)2} &m^{(e)}n^{(e)} &-m^{(e)}l^{(e)} & -m^{(e)2} &-m^{(e)}n^{(e)}  \\ n^{(e)}l^{(e)} & n^{(e)}m^{(e)} & n^{(e)2} & -n^{(e)}l^{(e)} & -n^{(e)}m^{(e)} & -n^{(e)2}\\ -l^{(e)2} & -l^{(e)}m^{(e)} & -l^{(e)}n^{(e)} & l^{(e)2} & l^{(e)}m^{(e)} & l^{(e)}n^{(e)}\\ -m^{(e)}l^{(e)} & -m^{(e)2} &-m^{(e)}n^{(e)} & m^{(e)}l^{(e)} & m^{(e)2} &m^{(e)}n^{(e)}  \\ -n^{(e)}l^{(e)} & -n^{(e)}m^{(e)} & -n^{(e)2}& n^{(e)}l^{(e)} & n^{(e)}m^{(e)} & n^{(e)2} \end{bmatrix} $$

Element Force Matrix in Global Coordinates in Terms of Axial Forces
$$ f^{(e)}_{6x1}=\begin{pmatrix} f^{(e)}_1\\ f^{(e)}_2\\ f^{(e)}_3\\ f^{(e)}_4\\ f^{(e)}_5\\ f^{(e)}_6 \end{pmatrix}=T^{(e)T}_{6x2}P^{(e)}_{2x1}=

\begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ n^{(e)} & 0\\ 0& l^{(e)}\\ 0 & m^{(e)}\\ 0 & n^{(e)} \end{bmatrix}\begin{pmatrix} P^{(e)}_1\\ P^{(e)}_2 \end{pmatrix} $$

3-Dimensional Space Truss (MATLAB)
A three-bar space truss is discussed in the course textbook by Asghar Bhatti on page 230. This truss system contains elements with these properties:

a1 = 200; a2 = 600; e = 200000; P = 20000;
 * E$$^{(1)}$$=200,000
 * E$$^{(2)}$$=200,000
 * E$$^{(3)}$$=200,000


 * A$$^{(1)}$$=200 mm
 * A$$^{(2)}$$=200 mm
 * A$$^{(3)}$$=600 mm

The truss system has four nodes, three coplanar and supported by a wall, and belong the the three elements, which meet at the free-moving fourth node. Also, a 20-kN force acts upon the fourth global node in the negative y-direction.

The following MATLAB code shows the colaculation of the system reactions and nodal dofs. It is easy to note that only the fourth global node will showacse any displacement due to the force P.

First defining some functions:

These functions serve to set up each element's stiffness matrix and display the results. The bulk of the code is below, which constructs the stiffness matrix, determines the dof values, and the reactions. The results are also displayed below.

The deformation of this space truss due to the force P and its code is shown below:

This is the truss output from the above code.

x-axis perspective:



y-axis perspective:



z-axis perspective:



(-2,-2,3) perspective:



(1.5,2,2) perspective:



Statics Approach
The three-bar space truss explored above proves to be statically determinate because its reactions were readily able to be determined. Determining these results was possible from the fact that there were enough constraints on the truss, i.e. the fixed nodes on the wall. Since this truss system is statically determinate, then a statics approach can be developed.



This truss system is loaded by -P$$\hat{j}$$.

Defining the vectors from global node 4 to each of the other three global nodes:


 * 41=0.96$$\hat{i}$$+1.92$$\hat{j}$$-2$$\hat{k}$$


 * 42=-.144$$\hat{i}$$+1.44$$\hat{j}$$-2$$\hat{k}$$


 * 43=-2$$\hat{k}$$

And their magnitudes:


 * $$|41|=\sqrt{0.96^2+1.92^2+(-2)^2}=2.9339$$


 * $$|42|=\sqrt{(-1.44)^2+1.44^2+(-2)^2}=2.8543$$


 * $$|43|=\sqrt{(-2)^2}=2$$

Defining the forces along each element (negative for compression) in vector notation:


 * T$$_{41}$$=$$T_{41}$$ λ$$_{41}$$=$$T_{41}\frac{\mathbf{T_{41}}}{|41|}$$=$$(0.3272\mathbf{\hat{i}}+0.6544\mathbf{\hat{j}}-0.6817\mathbf{\hat{k}})T_{41}$$


 * T$$_{42}$$=$$T_{42}$$ λ$$_{42}$$=$$T_{42}\frac{\mathbf{T_{42}}}{|42|}$$=$$(-0.5045\mathbf{\hat{i}}+0.5045\mathbf{\hat{j}}-0.7007\mathbf{\hat{k}})T_{42}$$


 * T$$_{43}$$=$$T_{43}$$ λ$$_{43}$$=$$T_{43}\frac{\mathbf{T_{43}}}{|43|}$$=$$(-\mathbf{\hat{k}})T_{43}$$

Applying the knowledge that the forces will all balance out in all directions, we can solve for T$$_{41}$$, T$$_{42}$$, and T$$_{43}$$:

ΣF$$_x$$:


 * 0.3272$$T_{41}$$-0.5045$$T_{42}$$=0

ΣF$$_y$$:


 * 0.6544$$T_{42}$$+0.5045$$T_{42}$$-P=0

ΣF$$_z$$:


 * -$$T_{43}$$-0.6817$$T_{41}$$-0.7007$$T_{42}$$=0

Three unknowns and three equations allows us to find the reactions along the axis of the elements:


 * $$T_{41}$$=-2,314.533 N (compression)


 * $$T_{42}$$=20,374.691 N (tension)


 * $$T_{43}$$=13,214.269 N (tension)

Notice that these exact reaction values are equal to the values attained with the FEA method above.

2 Bar Truss (MATLAB)
MATLAB Code

The code that was provided by the website for the book for the function PlaneTrussResults was given incorrectly. The code that was used to obtain the results in a 2X6 matrix for the aforementioned function is provided below.

function [results] = PlaneTrussResults(e, A, coord, disps) % [results] = PlaneTrussResults(e, A, coord, disps) % Compute plane truss element results % e = modulus of elasticity % A = Area of cross-section % coord = coordinates at the element ends % disps = displacements at element ends % The output quantities are eps = axial strain % sigma = axial stress and force = axial force. results=[]; x1=coord(1,1); y1=coord(1,2); x2=coord(2,1); y2=coord(2,2); L=sqrt((x2-x1)^2+(y2-y1)^2); ls=(x2-x1)/L; ms=(y2-y1)/L; T=[ls,ms,0,0; 0,0,ls,ms]; d = T*disps; eps= (d(2)-d(1))/L; sigma = e.*eps; force = sigma.*A; results=[eps, sigma, force];

The difference between this function and the old function can be seen in the declaration of the function. The declaration line of the old function is shown below. The variable results in the main program was being filled by the first variable in the matrix, which was given as eps. In the new PlaneTrussResults code the array results is declared within the body of the function, and this variable is returned to the results variable in the main body of the Two Bar Truss program. Because this array contains all three of the values (eps, sigma, and force), and not just the first value for the strain, the whole matrix results, which is a 2x6 matrix, is output correctly. function [eps, sigma, force] = PlaneTrussResults(e, A, coord, disps)

 Cheating: Added two columns for array "results" with the code remaining the same and without adding a comment box to warn readers. See my comments at the top.

Also incomplete listing of the codes used.

Eml4500.f08 12:20, 11 November 2008 (UTC)

6 Bar Truss (MATLAB)
The following describes the reactions and displacements of a 6-bar truss system shown on pg 226 of the textbook.

Same cross-sectional areas and Moduli
Code and results of the reaction forces and displacements:

The following displays the code for the visual representation of the truss system and its deformation.



Different cross-sectional areas and Moduli
Code and results of the reaction forces and displacements of the six-bar truss system, each element with a different cross-sectional area and Modulus value combination, chosen without much thought:

>

The following displays the code for the visual representation of the truss system and its deformation.



Global and Local Weighting Coefficients
Justification of eliminating rows 1,2,5,6 to obtain the K2x2 in 2 bar truss system

Force Displacement Relationship gives,

$$ Kd - F = 0 $$ (1)

Equation 2 is given as,

$$ w(Kd - F) = 0 $$ (2)

for all w.

Choice 1: Select w so that w1 = 1, w2 = 0, ..., w6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 1 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} + 0 [ \sum_{j = 1}^{6}{k_{2j}d_j - F_2]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{1j}d_j - F_1} = 0 $$

Choice 2: Select w so that w2 = 1, w1,3,4,5,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} + 1 [ \sum_{j = 1}^{6}{k_{2j}d_j - F_2]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{2j}d_j - F_2} = 0 $$

Choice 3: Select w so that w3 = 1, w1,2,4,5,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} +...+ 1 [ \sum_{j = 1}^{6}{k_{3j}d_j - F_3]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{3j}d_j - F_3} = 0 $$

Choice 4: Select w so that w4 = 1, w1,2,3,5,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} +...+ 1 [ \sum_{j = 1}^{6}{k_{4j}d_j - F_4]} +...+ 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{4j}d_j - F_4} = 0 $$

Choice 5: Select w so that w5 = 1, w1,2,3,4,6 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} +...+ 1 [ \sum_{j = 1}^{6}{k_{5j}d_j - F_5]} + 0 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{5j}d_j - F_5} = 0 $$

Choice 6: Select w so that w6 = 1, w1,2,3,4,5 = 0

Thus,

$$ W_{1x6}^T = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$

Therefore,

$$ w (Kd - F) = 0 [ \sum_{j = 1}^{6}{k_{1j}d_j - F_1]} + 0 [ \sum_{j = 1}^{6}{k_{2j}d_j - F_2]} +...+ 1 [ \sum_{j = 1}^{6}{k_{6j}d_j - F_6]} = 0 $$

which simplifies to equation (1),

$$ w (Kd - F) = \sum_{j = 1}^{6}{k_{6j}d_j - F_6} = 0 $$

Proof of Axial Displacements
If we set w1 = 1, then w2...w6 = 0

$$ \mathbf{w}^{T} = \begin{bmatrix} 1 & 0& 0 &  0& 0 & 0 \end{bmatrix} $$

$$ \mathbf{w}*(\mathbf{k}\mathbf{q}- \mathbf{p}) $$

$$ = (1)[\sum{k_{1j}}q_j- p_1] + (0)[\sum{k_{2j}}q_j- p_2]+ (0)[\sum{k_{3j}}q_j- p_3]+ (0)[\sum{k_{4j}}q_j- p_4]+ (0)[\sum{k_{5j}}q_j- p_5]+ (0)[\sum{k_{6j}}q_j- p_6] $$

$$ =\sum{k_{1j}}q_j- p_1 $$

For all other values of w (w2...w6), the trend remains the same, and the answers can be found below.

w2 = 1, w1,3,4,5,6 = 0 $$ =\sum{k_{2j}}q_j- p_2 $$

w3 = 1, w1,2,4,5,6 = 0 $$ =\sum{k_{3j}}q_j- p_3 $$

w4 = 1, w1,2,3,5,6 = 0 $$ =\sum{k_{4j}}q_j- p_4 $$

w5 = 1, w1,2,3,4,6 = 0 $$ =\sum{k_{5j}}q_j- p_5 $$

w6 = 1, w1,2,3,4,5 = 0 $$ =\sum{k_{6j}}q_j- p_6 $$

Proof of Transpose
The purpose of this proof is to show that the transpose of two matrices that are multiplied together is equal to the transpose of each individual matrix multiplied together. This relation is shown as follows:

$$ \left(A_{ixj}B_{jxm} \right)^{T}=B_{mxj}^{T}A_{jxi}^{T} $$ For this proof, the values for$$ A_{2x3} $$ and $$ B_{3x3} $$are given as:

$$ A=\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} $$

$$ B=\begin{bmatrix} 7 & 8 & 9\\ 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} $$

In order to find the result of the left hand side of the equation at the top of this section the matrices A and B are multiplied together. Then the transpose of this matrix can be taken to get result of the left hand side.

$$ AB=\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 7 & 8 & 9\\ 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix}=\begin{bmatrix} 21 & 27 & 33\\ 57 & 72 & 87 \end{bmatrix} $$

$$ \left( AB\right)^{T}=\begin{bmatrix} 21 & 27 & 33\\ 57 & 72 & 87 \end{bmatrix}^{T}=\begin{bmatrix} 21 & 57\\ 27 & 72\\ 33 & 87 \end{bmatrix} $$

The right hand side of the equation can be found by first finding the transpose of each matrix individually. Then these two matrices can be multiplied together to obtain the result.

$$ B^{T}A^{T}=\begin{bmatrix} 7 & 1 & 4\\ 8 & 2 & 5\\ 9 & 3 & 6 \end{bmatrix}\begin{bmatrix} 1 & 4\\ 2 & 5\\ 3 & 6 \end{bmatrix}=\begin{bmatrix} 21 & 57\\ 27 & 72\\ 33 & 87 \end{bmatrix} $$

Since each result is the same, the relation established at the top of this section is valid.

Composite materials
Refer to engineered materials composed of usually two constituent materials with considerably different mechanical properties.

Types
In order to have the composite at least one fraction of each type is required.


 * Matrix: by maintaining a given position the matrix materials surround and support the reinforcement materials.


 * Reinforcement: by providing mechanical and electrical properties reinforcement materials enhance the matrix materials.

Examples
Examples of composite materials can be dated to primitive times. One of the most elementary would be the composite for bricks which were made out of mud and straw. More modern examples can be encountered from spacecraft to roadways applications such as asphalt concrete.

Some examples of composite materials where the modulus of elasticity is a function of length are:


 * Hardmetal: is a composite of metal and carbide. The modulus of metal is significantly different than that of carbide in order to increase the composite’s strength.
 * Bone: is a composite of hydroxyapatite with collagen fibers. The modulus of bone is altered by the collagen fibers embedded inside of the dydroxyapatite to lend it the force necessary to support the body movements.
 * Concrete walls (reinforced): the modulus of concrete is changed by using reinforced steel in order to reinforce the concrete in walls.
 * Composite panel: the modulus of two aluminum sheets is changed by inserting in between a non-aluminum material.

For additional information please visit:


 * http://www.mech.utah.edu/~rusmeeha/labNotes/composites.html


 * http://www.societyofrobots.com/materials_carbonfiber.shtml


 * http://www.science.org.au/nova/059/059key.htm


 * http://www.matter.org.uk/matscicdrom/manual/co.html#_Toc360435310

For more in depth research about composite materials please visit the University of Florida engineering databases http://libguides.uflib.ufl.edu/content.php?pid=7641&sid=48706 Other resources are available by engineering subjects guides http://www.uflib.ufl.edu/msl/subjects/engin.html or by consulting the University of Florida engineering subject specialists:
 * Denise Bennett
 * Amy Buhler
 * Vernon Kisling
 * Carrie Newsom

Contributing Team Members
Iain Specht --Eml 4500.f08.delta 6.specht 18:50, 6 November 2008 (UTC) Sean Lopez--Eml4500.f08.delta 6.lopez 20:13, 6 November 2008 (UTC) Eric Ramirez--Eml4500.f08.delta 6.ramirez 21:39, 6 November 2008 (UTC) Nicolas Castrillon--Eml4500.f08.delta 6.castrillon 01:47, 7 November 2008 (UTC) Micheal Krueger--Eml4500.f08.delta 6.krueger 17:32, 7 November 2008 (UTC) Diana Guzman--Eml4500.f08.delta 6.guzman 19:49, 7 November 2008 (UTC)