# User:EML4500.f08.JAMAMA/LectureNotesHW2

## 9/8/08

The Global Forces are determined by the following matrix

$\begin{Bmatrix} f1\\ f2\\ f3\\ f4\\ f5\\ f6\\ \end{Bmatrix} = \begin{bmatrix} & & & & & \\ & & & & & \\ & & K& & & \\ & & & & & \\ & & & & & \\ & & & & & \end{bmatrix} \times \begin{Bmatrix} d1\\ d2\\ d3\\ d4\\ d5\\ d6\\ \end{Bmatrix}$

6X1          6X6          6X1

Global Force   Global Stiffness   Global Displacement

Column Matrix     Matrix        Column Matrix

Before the global matrix can be constructed, you first need to find the displacements in the elements themselves. To do this for each element, the following equation may be used:

$K^{\left(e \right) } \times D^{\left(e \right)} = F^{\left(e \right)}$

4X4      4X1      4X1

where: $K^{\left(e \right) }$=Element stiffness matrix for element e

$D^{\left(e \right)}$=Element displacement matrix of element e

$F^{\left(e \right)}$=Element force matrix

Element Stiffness Matrix:

The element stiffness matrix $K^{\left(e \right)}$ for the equation above is calculated with the matrix shown (below).

$k^{\left(e \right)}\times \begin{bmatrix} \left(l^{\left(e \right)} \right)^2 & l^{\left(e \right)}m^{\left(e \right)} & -\left(l^{\left(e \right)} \right)^2 & -l^{\left(e \right)}m^{\left(e \right)}\\ l^{\left(e \right)}m^{\left(e \right)} & \left(m^{\left(e \right)} \right)^2 & -l^{\left(e \right)}m^{\left(e \right)} & -\left(m^{\left(e \right)} \right)^2 \\ -\left(l^{\left(e \right)} \right)^2& -l^{\left(e \right)}m^{\left(e \right)} & \left(l^{\left(e \right)} \right)^2 & l^{\left(e \right)}m^{\left(e \right)} \\ -l^{\left(e \right)}m^{\left(e \right)}& -\left(m^{\left(e \right)} \right)^2 & l^{\left(e \right)}m^{\left(e \right)} & \left(m^{\left(e \right)} \right)^2 \end{bmatrix}$

where: $k^{\left(e \right)}= \frac{E^{\left(e \right)}A^{\left(e \right)}}{L^{\left(e \right)}}$ Axial stiffness of bar element "e"

and $l^{\left(e \right)}, m^{\left(e \right)}$=Director cosines of the $\tilde{x}$ axis with respect to the (x,y) coordinates for element e.

Director Cosines

Director cosines relate the angle at which the element is on to the traditional x,y axis. These values are needed to calculate the stiffness matrix k.

$l^{\left(e \right)}=\vec{\tilde{i}}\bullet \vec{i}=\cos \theta^{\left(e \right)}$

$m^{\left(e \right)}=\vec{\tilde{i}}\bullet \vec{j}=\cos \left(\frac{\pi }{2} - \theta^{\left(e \right)}\right)=\sin \theta ^{\left(e \right)}$

$\vec{\tilde{i}}=\cos \theta ^{\left(e \right)} \vec{i} + \sin \theta ^{\left(e \right)}\vec{j}$

$\vec{\tilde{i}}\bullet \vec{i}=\cos \theta ^{\left(e \right)} \vec{i}\bullet \vec{i} + \sin \theta ^{\left(e \right)}\vec{j}\bullet \vec{i}$

$\vec{\tilde{i}}\bullet \vec{j}=\cos \theta ^{\left(e \right)} \vec{i}\bullet \vec{j} + \sin \theta ^{\left(e \right)}\vec{j}\bullet \vec{j}$

## 9/10/08

Model of the bar turss system (continued)

### Elem. 1

Elem. 1 FBD + angle of incline

θ(1) = 30°

l(1) = cos(1)(θ) = cos(1)(30°) = $\frac{\sqrt{3}}{2}$

m(1) = sin(1)(θ) = sin(1)(30°) = $\frac{1}{2}$

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

k(1)= $\begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} \\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} \\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)} & k_{34}^{(1)} \\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)} & k_{44}^{(1)} \end{bmatrix}$ = [k$_{ij}^{(e)}]$4 × 4, where $_{j = 1,2,3,4\ \Rightarrow \ column\ index}^{i = 1,2,3,4\ \Rightarrow \ row\ index}$ and $e$ = number of element

$k_{11}^{(1)} = k^{(1)} [l^{(1)}]^2 = \frac {3}{4} * [\frac{\sqrt{3}}{2}]^2 = \frac {9}{16}$

$k_{12}^{(1)} = k^{(1)} [l^{(1)}m^{(1)}] = \frac {3}{4} * \frac{\sqrt{3}}{2}]*\frac{1}{2} = \frac {3\sqrt{3}}{16}$

$k_{42}^{(1)} = k^{(1)} (-[m^{(1)}]^2) = \frac {3}{4} * (-\frac{1}{4}) = -\frac {3}{16}$

Note: this matrix is symetircal with respect to the diagnol $k_{xx}^{(1)}, where\; x= 1,2,3... \to k_{13}^{(1)}=k_{31}^{(1)}.$

In genral, $k_{ij}^{(e)}= k_{ji}^{(e)}\ or\ [{\mathbf{k}^{(e)}}]^T = \mathbf{k}^{(e)}$

### Elem. 2

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

θ(2) $= -45^\circ \Rightarrow _{m^{(2)} =\ sin(-45^\circ)\ =-\frac{\sqrt{2}}{2}}^{l^{(2)} =\ cos(-45^\circ)\ =\frac{\sqrt{2}}{2}}$

$\mathbf{k}^{(e)}$ = [k$_{ij}^{(e)}]$4 × 4$,\qquad \begin{matrix} k_{11}^{(2)} = k^{(2)} [l^{(2)}]^2 = 5 * [\frac{\sqrt{2}}{2}]^2 = \frac {5}{2}\\ k_{12}^{(2)} = k^{(2)} l^{(2)}m^{(2)} = 5 * \frac{\sqrt{2}}{2}* (-\frac{\sqrt{2}}{2}) = -\frac {5}{2}\\ k_{42}^{(2)} = k^{(2)} (-[m^{(2)}]^2) = 5 * (-[\frac{\sqrt{2}}{2}]^2) = -\frac {5}{2} \end{matrix}$

Observe: Absolute values for all the coeficients are the same [k$_{ij}^{(e)}]$4 × 4, where $e =\ 2,and\ _{j = 1,2,3,4}^{i = 1,2,3,4}$. All that differs are the positive/negative signs in front of the magnitudes of the coeficients.

Element F-D relation: $\mathbf{k}_{4 \times 4}^{e}*\mathbf{d}_{4\times1}^{e} = \mathbf{f}_{4\times1}^{e}$

$\mathbf{d}^{e}\ ={ \begin{Bmatrix} d_1^{(e)}\\ \vdots\\ d_4^{(e)} \end{Bmatrix}}_{4 \times 1};\quad \mathbf{f}^{e}\ = {\begin{Bmatrix} f_1^{(e)}\\ \vdots\\ f_4^{(e)} \end{Bmatrix}}_{4 \times 1}$

### Global F-D Relation("free-free" structure)

$\mathbf{K}_{n \times n}*\mathbf{d}_{n\times1} = \mathbf{F}_{n\times1} \Longrightarrow$ there are $\mathbf{6}$ degrees of freedom (6 forces are acting upon the sturcture) $\quad \Rightarrow \mathbf{n=6}$

## 9/12/08

Global Force Displacement Relationship

• Note: The following matrix formulas show that there is a difference between the notation for the stiffness and force matrices $k$, $K$, and $f$, $F$. For example, $K$ is the global stiffness matrix, and $k$ is the element stiffness matrix.

The matrix form for the example problem in progress is as follows:

$\begin{bmatrix} k_{11} & k_{12} & k_{13} & k_{14} & k_{15} & k_{16}\\ k_{21} & k_{22} & k_{23} & k_{24} & k_{25} & k_{26}\\ k_{31} & k_{32} & k_{33} & k_{34} & k_{35} & k_{36}\\ k_{41} & k_{42} & k_{43} & k_{44} & k_{45} & k_{46}\\ k_{51} & k_{52} & k_{53} & k_{54} & k_{55} & k_{56}\\ k_{61} & k_{62} & k_{63} & k_{64} & k_{65} & k_{66} \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix} = \begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix}$

The matrix formula for the element matrices is written as follows:

In order to express the element matrices in global matrices to solve for the displacements, there is an assembly process.

You must identify the correspondence between the element displacement degrees of freedom and the global displacement degrees of freedom.

For the current example, the global displacement degrees of freedom are: $\{d_1,d_2,d_3,d_4,d_4,d_5,d_6\}$

At the element level:

    Element 1: $\left\{d_1^{(1)},d_2^{(1)},d_3^{(1)},d_4^{(1)},d_5^{(1)},d_6^{(1)}\right\}$

    Element 2: $\left\{d_1^{(2)},d_2^{(2)},d_3^{(2)},d_4^{(2)},d_5^{(2)},d_6^{(2)}\right\}$


Then it is necessary to identify the global - element degrees of freedom. In other words, determine the element displacement degrees of freedom that are actually equivalent to one another if the separate elements were to be put back together, as well as which are in turn actually describing a global displacement degree of freedom:

$\left. \begin{array}{c} d_1 = d^{(1)}_1 \\ d_2 = d^{(1)}_2 \end{array} \right\}$ at Global Node 1

$\left. \begin{array}{c} d_3 = d^{(1)}_3 = d^{(2)}_1\\ d_4 = d^{(1)}_4 = d^{(2)}_2 \end{array} \right\}$ at Global Node 2

$\left. \begin{array}{c} d_5 = d^{(2)}_3\\ d_6 = d^{(2)}_4 \end{array} \right\}$ at Global Node 3


Conceptual Step of Assembly: Topology of K

After calculation of the local stiffness matrices K1 and K2, they can be combined following the conceptual step of assembly method called: The Topology of K as follows:

Within the intersection of the two matrices, the terms that would overlap there are simply added together. (See 9/15/08 for the matrix completed for the current example.)

## 9/15/08

The resulting global stiffness matrix is:

$\begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\[1ex] k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\[1ex] k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)}+k_{11}^{(2)}) & (k_{34}^{(1)}+k_{12}^{(2)}) & k_{13}^{(2)} & k_{14}^{(2)}\\[1ex] k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)}+k_{21}^{(2)}) & (k_{44}^{(1)}+k_{22}^{(2)}) & k_{23}^{(2)} & k_{24}^{(2)}\\[1ex] 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)}\\[1ex] 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)}\\ \end{bmatrix}$

A sample calculation of the values of K:

$K_{11} = k_{11}^{(1)}=\dfrac{9}{16}$
$K_{12} = k_{12}^{(1)}=\dfrac{3\sqrt 3}{16}$
$K_{33} = k_{33}^{(1)}+k_{11}^{(2)}=\dfrac{9}{16}+\dfrac{5}{2}=3.0625$
$K_{34} = k_{34}^{(1)}+k_{12}^{(2)}=\dfrac{3\sqrt 3}{16}+\left(-\dfrac{5}{2}\right)=-2.175$
$K_{43} = K_{34}=k_{43}^{(1)}+k_{21}^{(2)}=k_{34}^{(1)}+k_{12}^{(2)}=-2.175$
$K_{44} = k_{44}^{(1)}+k_{22}^{(2)}=\dfrac{3}{16}+\dfrac{5}{2}=2.6875$

## 9/17/08

Step 4: Introduction of Boundary Equations

In this step eliminating known degrees of freedom will reduce the global Force Displacement Relationship. By knowing the displacements for global nodes 1 and 3 are zero because of the boundary conditions expressed in the problem statement. The boundary conditions have been applied to the global stiffness matrix below.

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

The global stiffness matrix reduces to a 6X2 matrix when applying the boundary equations. The top and bottom two rows are also eliminated due to the principle of virtual work. The Force Displacement relation then becomes the equation shown below.

$\begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44} \end{bmatrix} \begin{bmatrix} d_3 \\ d_4 \end{bmatrix} = \begin{bmatrix} F_3 \\ F_4 \end{bmatrix}$

To solve for the displacement of global node 2. The equation above needs to be multiplied on each side by K-1, and the values for the stiffness matrix and the force matrix inserted. This solves d3 and d4 to be 4.352 and 6.1271 respectively.

## 9/19/08

In order to solve for the displacements of the example, each element is considered:

The displacements are found by solving the 4x4 matrix, after the eliminations discussed previously. This method provides the solutions:

$\underline{k}^{(1)} \underline{d}^{(1)} = \underline{f}^{(1)}$

$\underline{k}^{(2)} \underline{d}^{(2)} = \underline{f}^{(2)}$

The internal forces for nodes 1 and 3 will be zero because they are fixed points. (See "Boundary Conditions") Therefore, the matrix will reduce to:

$\begin{bmatrix} 3.0625 & 2.7152\\ -2.7152 & 2.6875 \end{bmatrix} \begin{Bmatrix} d_3 \\ d_4 \end{Bmatrix} = \begin{Bmatrix} F_3 \\ F_4 \end{Bmatrix}$

Where the known external forces will be plugged in.

To continue, for example, using specifically element 1, the now known displacements from the previous step, can be plugged back in for the variable labels, and the reactions at the supports can be found. This is the last step.

$\begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0\\ 0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0\\ 0 & 0 & -2.5 & -2.5 &2.5&2.5\\ 0 & 0 & -2.5 & -2.5 &2.5&2.5 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix} = \begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2 \\ f^{(1)}_3 \\ f^{(1)}_4 \end{Bmatrix}$

Where the subscripts 1 and 2 denote the reactions at the supports , and the subscripts 3 and 4 denote the internal forces. (As labeled on the free body diagrams for element 1 above.)

The solutions of the reactions and internal forces for element 1 are as follows:

$\begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2 \\ f^{(1)}_3 \\ f^{(1)}_4 \end{Bmatrix} = \begin{Bmatrix} -4.4378 \\ -2.5622 \\ 4.4378 \\ -2.5622 \end{Bmatrix}$

Take note of the fact that each of the internal forces and reactions for the x and y directions will sum to zero, providing an algebraic proof that the element is in equilibrium.

$\begin{array}{l} \sum f_1^{(1)}+f_2^{(1)}=0\\ \sum f_3^{(1)}+f_4^{(1)}=0 \end{array}$

## Solving a Two-Bar Truss System using MatLab

The same system can be solved using MatLab.

The following code solves for the parameters: "Stiffness matrices, displacements, and reactions."

Here are the calculaled results from MatLab:

It can be observed that the values that were calculated previously are similar to the ones calculated using Matlab.

The two "k_local" that MatLab calculated are our k(1) and k(2) respectively.

The two lowercase k's (k) from MatLab are our local stiffness matrices k(1) and k(2).

The Uppercase k (K) found in the Matlab Results is equivalent to our global stiffness matrix K.

The "R" matrix is the known forces matrix.

The "d" matrix is the displacement matrix.

The "reaction" matrix is for the reaction forces.

## Contributing Team Members

--Eml4500.f08.jamama.justin 18:26, 26 September 2008 (UTC)