User:Eml4500.f08.RAMROD.A/HW3

=HW 3 Ramrod=

Derivation of elemental force displacement with respect to the global system.
$$ K^{(e)}_{4x4} \cdot d^{(e)}_{4x1} = F^{(e)}_{4x1}$$



Statics dictates that this can be defined as a two-force member. Therefore, it can be drawn like this.



The force displacement relationship is as follows for P and q.

$$K^{(e)}_{2x2} \cdot q^{(e)}_{2x1} = P^{(e)}_{2x1}$$

$$K^{(e)}\begin{vmatrix} 1 & -1\\ -1 & 1 \end{vmatrix} \begin{Bmatrix} q^{(e)}_1\\ q^{(e)}_2 \end{Bmatrix}=\begin{Bmatrix} P^{(e)}_1\\ P^{(e)}_2 \end{Bmatrix}$$

$$q^{(e)}_i$$ = the axial displacement of element e at local node i.

$$ P^{(e)}_i$$ = the axial force of element e at local node i.

Goal : Derive equation $$ K^{(e)}_{4x4} \cdot d^{(e)}_{4x1} = F^{(e)}_{4x1}$$ from $$K^{(e)}_{2x2} \cdot q^{(e)}_{2x1} = P^{(e)}_{2x1}$$ . Find the relationship between matrix $$q^{(e)}$$ and $$d^{(e)}$$ and between matrix $$P^{(e)}$$ and $$F^{(e)}$$.

These relations can be expressed in the form:

$$q^{(e)}_{2x1}= T^{(e)}_{2x1}\cdot d^{(e)}_{4x1}$$

== Consider the displacement vector of local node 1, denoted by $$d^{(e)}_1 $$ == center text



$$ d^{(e)}_1 =  d^{(e)}_1\vec{i} \cdot d^{(e)}_2\vec{j}$$

$$q^{(e)}_1 $$ = the axial displacement of node 1 is the orthogonal projection of the displacement vector of node one. $$d^{(e)}_1 $$ on the zxis of the bar $$\tilde{x}$$ of element e.

$$q^{(e)}_1=\vec{d}^{(e)}_1\cdot \vec{\tilde{i}}$$

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

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

$$q^{(e)}_1=l^{(e)}d^{(e)}_1+ m^{(e)}d^{(e)}_2$$

$$q^{(e)}_1=\begin{vmatrix} l^{(e)} & m^{(e)} \end{vmatrix}\begin{Bmatrix} d^{(e)}_1 \\d^{(e)}_2

\end{Bmatrix}$$

Consider the displacement vector of local node 2, denoted by $$d^{(e)}_2$$


$$ d^{(e)}_2 =  d^{(e)}_3\vec{i} \cdot d^{(e)}_4\vec{j}$$

$$q^{(e)}_2 $$ = the axial displacement of node 2 is the orthogonal projection of the displacement vector of node two. $$d^{(e)}_2 $$ on the zxis of the bar $$\tilde{x}$$ of element e.

$$q^{(e)}_2=\vec{d}^{(e)}_2\cdot \vec{\tilde{i}}$$

$$q^{(e)}_2=(d^{(e)}_3\vec{i}+d^{(e)}_4\vec{j})\cdot \vec{\tilde{i}}$$

$$q^{(e)}_2=d^{(e)}_3(\vec{i}\cdot \vec{\tilde{i}})+d^{(e)}_4(\vec{j}\cdot \vec{\tilde{i}})$$

$$q^{(e)}_2=l^{(e)}d^{(e)}_3+ m^{(e)}d^{(e)}_4$$

$$q^{(e)}_2=\begin{vmatrix} l^{(e)} & m^{(e)} \end{vmatrix}\begin{Bmatrix} d^{(e)}_3 \\d^{(e)}_4

\end{Bmatrix}$$

The result of the nodal analysis is combined to form this matrix.

$$\begin{Bmatrix} q^{(e)}_1\\ q^{(e)}_2 \end{Bmatrix}=\begin{vmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0& 0 & -l^{(e)} & -m^{(e)} \end{vmatrix}\begin{Bmatrix} d^{(e)}_1\\ d^{(e)}_2 \\ d^{(e)}_3 \\ d^{(e)}_4

\end{Bmatrix} $$

The Transformation Matrix
In order to relate the axial coordinate system's forces, f(e), to the element coordinate system's axial forces, p(e), it is necessary to employ a transformation matrix, T(e) such that:

$$\begin{Bmatrix}

P^{(e)}_{1}\\ P^{(e)}_{2} \end{Bmatrix} = T^{(e)}*

\begin{Bmatrix}

f^{(e)}_{1}\\ f^{(e)}_{2}\\ f^{(e)}_{3}\\ f^{(e)}_{4} \end{Bmatrix}$$

Looking at this equation, we can tell that T(e) must be a 2X4 matrix in order for the dimensions to work out.

Knowing this, and using the element F-D relationship,

$$\hat{k}^{(e)}*q^{(e)} = p^{(e)}$$

we can substitute in the Transformation Matrix relationship above to get an equation relating the cartesion forces and displacements via the transformation matrix and \hat{k}. The resulting equation is shown below.

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

Rearranging this equation requires moving the T(e) term to the left side to allow us to solve for the internal forces in f(e). Since T(e) is rectangular, the transpose, rather than the inverse, is multiplied by each side resulting in:

$$[T^{(e)T}*\hat{k}^{(e)}*(T^{(e)}]*d^{(e)} = f^{(e)}$$

In order for this to be true, it must be proven that,

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

This proof is given below as the assigned homework.

Now that we have proven the derivation of k^e, we can continue to solving for the global displacement degrees of freedom.

Using the principle of virtual work (i.e. d1 = d2 = d5 = d6 = 0) we can reduce the global equation to two unknown DOF's, d3 and d4.

The next homework assignment was to solve for the eigenvalues of the global stiffness matrix, K.

2 Bar Truss System


Closing the loop refers to the two upward pointing arrows on the right side of the figure, making your way back to computing the reaction forces.

Using Statics
Taking a look back at the statics method, the reaction forces known and unknown therefore the member forces are P1(1), P2(2)
 * Note: Both members are in tension

How to Compute the axial displacement of dofs (degrees of freedom) or the amount of extension of the bars:
$$ q_2^{(1)} = \frac {P_1^{(1)}}{d^(1)} = \frac {P_2^{(1)}}{k^{(1)}} = AC $$
 * P1(1) = P2(1)
 * q1(1) = 0, Fixed global node 1

$$ q_1^{(2)} = \frac {-P_2^{(2)}}{k^{(2)}} = -AB $$


 * q2(2) = 0, Fixed global node 3

Finding The Displacements


Referring back to the picture of the truss system, it can be seen that the displacement of point $$\displaystyle D$$ is simply the displacement of the global node $$\displaystyle 2$$ at point $$\displaystyle A$$. In other words,

$$\displaystyle Y_{D} = d_{4} $$ and

$$\displaystyle X_{D} = d_{3} $$

We can now proceed to calculate the length of lines $$\displaystyle AC $$ and $$\displaystyle AB $$ be computed which as we already know are defined to be the axial forces. In other words it is the amount the bars extend or elongate due to the applied force. As previously stated the formula to find line $$\displaystyle AC $$ is as follows,

$$\displaystyle AC = \frac{\left|P_{2}^1 \right|}{k^1}$$

where

$$\displaystyle \left|P_{2}^1 \right| = 5.1243 $$ $$\displaystyle k^1 = \frac{3}{4}$$

Therefore,

$$\displaystyle AC = 6.8324$$

Remember that $$\displaystyle \left|P_{2}^1 \right|$$ is the internal force $$\displaystyle 2$$ at element $$\displaystyle 1 $$. Remember also that in order to obtain the value of $$\displaystyle \left|P_{2}^1 \right|$$, it is necessary to take the Root Sum Squared, RSS for short, of the two forces making acting at that node. The formula for the RSS of line AC extending from element 1 is,

$$\displaystyle \sqrt{(f_3)^2 + (f_4)^2}$$

The same can be done for $$\displaystyle AB $$ with the axial force from element two.

$$\displaystyle AB = \frac{\left|P_{1}^2 \right|}{k^2}$$ $$\displaystyle \left|P_{1}^2 \right| = 6.276 $$

$$\displaystyle k^2 = 5$$

Therefore

$$\displaystyle AB = 1.2552$$

From this point it is now possible to find the coordinates of points $$\displaystyle B $$ and $$\displaystyle C $$. Simple trigonometry will be used to carry out these calculations. First we will find the coordinates for point C, $$\displaystyle (x_C,y_C) $$. If you recall from previous lessons

$$\displaystyle \theta^1 = 30^{\circ} $$

and we already know the length of $$\displaystyle AC $$ so therefore the coordinates of $$\displaystyle C $$ can be expressed as

$$\displaystyle x_C = AC \cos (30^{\circ}) = 6.8324 *\frac{\sqrt{3}}{2} = 5.917$$

$$\displaystyle y_C = AC \sin (30^{\circ}) = 6.8324 *\frac{1}{2} = 3.4162$$

The coordinates of point $$\displaystyle B $$ can be found in the same way. the angle used is the angle the line $$\displaystyle AB $$ makes with the negative x-axis.

$$\displaystyle x_B = AB \cos (45^{\circ}) = 1.2552 *\frac{\sqrt{2}}{2} = -.88756 $$

$$\displaystyle y_B = AB \sin (30^{\circ}) = 1.2552*\frac{\sqrt{2}}{2} = .886756$$

We next need to find the coordinates of the point $$\displaystyle D $$. There are two unknowns at point $$\displaystyle D $$, the x and the y coordinate $$\displaystyle (x_D,y_D) $$. Since there are two unknowns, we need to find two equations in which these two variables are being utilized. The equation for the lines $$\displaystyle AB $$ and $$\displaystyle AC $$ are the two equations we will utliize. In order to do so we will work with the following figure.



The line $$\displaystyle PQ $$ can be represented in the following way;

$$\displaystyle \vec{PQ} = (PQ) \tilde{i} = PQ [\cos(\theta) \vec{i} + \sin(\theta) \vec{j}] $$

The cosine and the sine of the angle can then be expressed as;

$$\displaystyle \vec{PQ} = (x-x_p) \vec{i} + (y-y_p) \vec{j} $$

Dividing the previous two equations will yield the following;

$$\displaystyle \frac{(y-y_p)}{(x-x_p)} = \tan(\theta) $$

Multiplying through by $$\displaystyle (x-x_p)$$ and the equation of the line $$\displaystyle \vec{PQ} $$ is given by the following;

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

This equation for line $$\displaystyle \vec{PQ} $$ is equivalent to the equation for line $$\displaystyle \vec{AC} $$. The equation for a line that is perpendicular to line $$\displaystyle \vec{AC} $$ and passing throught the point $$\displaystyle A $$ is given by the following;

<p style="text-align:center;">$$\displaystyle y-y_p= \tan(\theta +\frac{\pi}{2})(x-x_p) $$

This equation is equivalent to the equation for line $$\displaystyle \vec{AC} $$. When the coordinates for lines $$\displaystyle \vec{BD} $$ and $$\displaystyle \vec{CD} $$ are put into the above equation then it is possible to solve for the coordinates of point $$\displaystyle D $$.

<p style="text-align:center;">$$\displaystyle y_D-y_B= \tan(\theta +\frac{\pi}{2})(x_D-x_B) $$ <br \> <br \> $$\displaystyle y_D-y_C= \tan(\theta +\frac{\pi}{2})(x_D-x_C)$$ <br \> <br \> $$\displaystyle y_D-.886756= \tan(\frac{\pi}{4})(x_D+.886756) $$ <br \> <br \> $$\displaystyle y_D-3.4162= \tan (\frac{2\pi}{3}) - (x_D-5.917)$$

Solving for $$\displaystyle y_D $$ for each equation will allow you to then solve for $$\displaystyle x_D$$. Plug this value back into one of the equations will allow you to get the value of $$\displaystyle y_D$$. When done, the coordinate of point $$\displaystyle D $$ becomes;

<p style="text-align:center;">$$\displaystyle D = (4.352, 6.1271) $$

Using the Finite Element Method (FEM), we have already found the displacement of point $$\displaystyle D $$. In other words;

<p style="text-align:center;">$$\displaystyle \vec{AD} = (x_D-x_A) \vec{i} + (y_D-y_A) \vec{j} $$

In this equation the variables $$\displaystyle x_A $$ and $$\displaystyle y_A $$ are zero and the only variables that remain are the coordinates of point $$\displaystyle D $$. By definition;

<p style="text-align:center;"> $$\displaystyle x_D = d_3 $$ and $$\displaystyle y_D = d_4 $$

And therefore, <p style="text-align:center;">$$\displaystyle \vec{AD} = d_3 \vec{i} + d_4 \vec{j} $$

where the coefficients are the global displacements of point $$\displaystyle A $$ to point $$\displaystyle D $$.

Three Truss Bar System
We will now look at the process of analyzing a three bar truss system as shown below.



It is important to note the helpfulness in numbering the local nodes of each element. Doing this in a certain manner make the assembly of the global stiffness matrix more convenient. The following numbering is the best and most convenient system.







The 3 Bar Truss and the Moment Equation
A look at the 3 Bar Truss Setup



From Statics, We know that there are typically three useful equations:

$$\displaystyle \sum{Fx}=0 $$

$$\displaystyle \sum{Fy}=0 $$

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

Statically Indeterminate
For this truss setup, even the moment equation renders no useful information because all reactions pass through the same point, A, and the setup is statically indeterminate

Now, we ask the question: what if we tried to sum the moments about some arbitrary point, B?

We will need to look at another picture and vector math to find out answers:

Can we get Information Summing Moments about B?


A look at the vector algebra:

$$\displaystyle \sum{\overline{M_B}}= \overline{BA} \times \overline{F} $$

Thus, $$\displaystyle \sum{\overline{M_B}}= \overline{BA'} \times \overline{F} $$ for all A' on line of action of F (see image right)

From Vector Algebra: $$\displaystyle \overline{BA'} = \overline{BA}+ \overline{AA'} $$

$$\displaystyle \overline{M_B} = (\overline{BA}+ \overline{AA'}) \times \overline{F} $$

$$\displaystyle \overline{M_B} = \overline{BA} \times \overline{F} + \overline{AA'} \times \overline{F} $$,

With The latter cross product in the previous equation equal to zero, we are left with: $$\displaystyle \overline{M_B} = \overline{BA} \times \overline{F} + 0 $$

More Generally:

$$\displaystyle \sum_{i}^{}{B_i}=\sum_{i}^{}{\overline{BA_i} \times \overline}$$ Where Ai is any point on the line of action of F.

Which Indicates: $$\displaystyle \sum{\overline{M_B}} = \overline{BA} \times \sum{\overline{F}} $$

We are already aware that: $$\displaystyle \sum{\overline{F}} = 0 $$

We can conclude that: $$\displaystyle \sum{\overline{M_B}} = \overline{BA} \times 0 = 0 $$

So, even if we take the moment from any arbitrary point, it will ultimately not provide any helpful information to solve the problem. The problem truly is statically indeterminate.

Three Truss Stiffness Matrix
Now we can solve for overall K by combining K(1), K(2) and K(3)

Matrix K $$\displaystyle K = \begin{bmatrix} k_{33}^{(1)}+k_{11}^{(2)}+k_{11}^{(3)} & k_{34}^{(1)}+k_{12}^{(2)}+k_{12}^{(3)}\\k_{43}^{(1)}+k_{21}^{(2)}+k_{21}^{(3)} & k_{44}^{(1)}+k_{22}^{(2)}+k_{22}^{(3)}\\k_{31}^{(2)}+k_{31}^{(3)} & k_{32}^{(2)}+k_{32}^{(3)}\\k_{41}^{(2)}+k_{41}^{(3)} & k_{42}^{(2)}+k_{42}^{(3)}\\k_{51}^{(3)} & k_{52}^{(3)}\\k_{61}^{(3)} & k_{62}^{(3)} \end{bmatrix} $$  =   $$\displaystyle K = \begin{bmatrix} k_{33} & k_{34}\\k_{43} & k_{44}\\k_{53} & k_{54}\\k_{63} & k_{64}\\k_{73} & k_{74}\\k_{83} & k_{84} \end{bmatrix} $$

MATLAB Code to Plot Undeformed and Deformed Two Bar Truss System
The code below plots the two bar truss system that was solved in class with the deformed system superimposed on top. The plot of the system is below the code.

% Matlab Script %********************************************************************* % filename: two_bar_truss.m  % % PURPOSE: %  Plot the two bar truss system discussed in class and the deformed %  system with matlab % % AUTHOR: %   Daniel Olivero %   Team Ramrod %   EML 4500 FEAD, Fall 2008. % % Modified on: vql, Fri, 03 Oct 2003, 08:37:46 EDT % Created on : Sun, 05 Oct 2008, 16:22:00 EDT % % DEPENDENCIES: %  call: % % REMARKS: % %*********************************************************************  %Plot Two beam truss system % clear; close;

%model with 2-D beam elements dof = 2; %dof per node: axial disp x, 2= disp y %obtain the coordinatates of all nodes % n_node = 3;             %number of nodes n_elem = 2;           %number of elements total_dof = 2 * n_node; %total dof of system

position(:, 1) = [0; 0]; position(:, 2) = [3.46; 2]; position(:, 3) = [4.9; .59];

% print the node coord. for i = 1 : n_node; x(i) = position (1,i); y(i) = position (2,i); end

node_connect (1, 1) = 1;   %element 1 node_connect (2, 1) = 2;

node_connect (1, 2) = 2;   %element 2 node_connect (2, 2) = 3;

% connect all beam elements by connectivity array for i = 1 : n_elem; node_1 = node_connect(1,i); node_2 = node_connect(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; axis([ -2 5 -2 5]) plot(xx,yy,'--') hold on end

text(x(node_connect(1,1)),y(node_connect(1,1)),'Global Node 1', 'HorizontalAlignment', 'center') text(x(node_connect(2,1)),y(node_connect(2,1)),'Global Node 2', 'HorizontalAlignment', 'center') text(x(node_connect(2,2)),y(node_connect(2,2)),'Global Node 3', 'HorizontalAlignment', 'center') text(x(node_connect(2,1))/2,y(node_connect(2,1))/2,'Element 1', 'HorizontalAlignment', 'center') text( x(node_connect(2,1)) + (x(node_connect(2,2))-x(node_connect(2,1)))/2, y(node_connect(2,1)) +(y(node_connect(2,2))-y(node_connect(2,1)))/2,'Element 2', 'HorizontalAlignment', 'center')

hold on

position_disp(:,1) = [0;0]; position_disp(:,2) = [7.812; 8.1271]; position_disp(:,3) = [4.9; .59];

% print the node coord. for i = 1 : n_node; x(i) = position_disp (1,i); y(i) = position_disp (2,i); end

node_connect_disp (1, 1) = 1;   %element 1 node_connect_disp (2, 1) = 2;

node_connect_disp (1, 2) = 2;   %element 2 node_connect_disp (2, 2) = 3;

% connect all beam elements by connectivity array for i = 1 : n_elem; node_1 = node_connect_disp(1,i); node_2 = node_connect_disp(2,i); xx = [x(node_1),x(node_2)]; yy = [y(node_1),y(node_2)]; axis([ -2 10 -2 10]) plot(xx,yy,'-') hold on end

text(x(node_connect_disp(2,1)),y(node_connect_disp(2,1)),'Global Node 2', 'HorizontalAlignment', 'center') text(x(node_connect_disp(2,1))/2,y(node_connect_disp(2,1))/2,'Element 1', 'HorizontalAlignment', 'center') text( x(node_connect_disp(2,1)) + (x(node_connect_disp(2,2))-x(node_connect_disp(2,1)))/2, y(node_connect_disp(2,1)) +(y(node_connect_disp(2,2))-y(node_connect_disp(2,1)))/2,'Element 2', 'HorizontalAlignment', 'center')

title ('Two Barr Truss System') xlabel('x') ylabel('y')

Contributing Team Members
Christopher Pergola EML4500.F08.RAMROD.B 19:32, 8 October 2008 (UTC)

EML4500.f08.RAMROD.E 19:19, 7 October 2008 (UTC)

Eml4500.f08.ramrod.D 19:24, 7 October 2008 (UTC)

Eml4500.f08.ramrod.c 03:14, 8 October 2008 (UTC

Chris Fontana Eml4500.f08.RAMROD.F 13:34, 8 October 2008 (UTC)

Eml4500.f08.RAMROD.A 16:24, 8 October 2008 (UTC)