User:Eml4500.f08.bottle.vitello/HW 2

Homework 2 describes the methods involved in solving truss analysis problems using methods of FEA.

Lecture Notes
The following is a diagram for the truss problem which will motivate the notes hereafter.

Note that all constants and physical constraints are specified in the subsequent notes.

When faced with a problem as presented above in Statics, the boundary conditions are removed and replaced with forces. When there are more unknowns then equations of equilibrium for a specific problem, the problem is considered statically indeterminate. The following steps are used to solve a statically indeterminate problem using matrices.

First, draw a global free body diagram. The axis for the FBD are chosen based on the direction of the reaction forces. The FBD is then labeled using the following notation:

1.    Nodes are numbered within circles 2.	Elements are then numbered within triangles 3.	Unknown reaction forces are labeled 4.	The three equations of quality are written.

An example of global free body diagram labeling is shown below.

Then, a FBD for each element is drawn. The following notation is used:

1.    Each node is labeled within a box 2.	Element is numbered within a triangle 3.	The reaction at each node is labeled in the following way: f (e) dof, d (e) dof begins with 1 on X axis and then moves to Y axis, then proceeds to next node, where (e) is the element number and dof is the degree of freedom number.

An example of local free body diagram labeling is shown below.



Once the elements are labeled the force displacement relationship is written. This relationship takes the form:
 * $$\textbf{k}^{(e)}\textbf{d}^{(e)}=\textbf{f}^{(e)}$$

where k(e) is the element's stiffness matrix. The stiffness matrix for element 1 would have the form:


 * $$ \textbf{k}^{(1)}= \left[ \begin{array}{cccc} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} \\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} \\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)} & k_{34}^{(1)} \\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)} & k_{44}^{(1)} \end{array} \right]$$

The big picture: Steps to solve a simple truss problem: 1.	Global Picture (Description) a.	At structure level i.	Global degrees of freedom (displacement, dof’s) ii. Global forces 2.	Displacement degrees of freedom are partitioned into a.	A known part (e.g. fixed dof, constraint/applied force) b.	An unknown part (solve using FEM/resultant forces) 3.	Element picture a.	Element DOF’s b.	Element faces 4.	Global Force Displacement relationship a.	Element stiffness matrices in global coordinates b.	Element force matrices in global c.	Assembly of element stiffness matrix and element force matrix in global FD rel d.	Kd=F 5.	Elimination of known dof’s to reduce global relationship a.	Stiffness matrix is non singular 6.	Compute forces from known d= element stress 7.	Compute reactions

Specific example:

Element Length L(1)= 4 L(2)=2 Youngs modulus E(1)= 3 E(2)=5

θ(1) = 30° θ(2) = -45°

A(1) = 1 A(2) = 2

Where  K = (E A) / L

Matrices: {F}=[k]{d}

(Remember to use convenient coordinate system and apply the given load at only one node)

$$ \textbf{k}^{(1)}= \left[ \begin{array}{cccc} l_{s}^{2} & l_{s}m_{s} & -l_{s}^{2} & -l_{s}m_{s} \\ l_{s}m_{s} & m_{s}^{2} & -l_{s}m_{s} & -m_{s}^{2} \\ -l_{s}^{2} & -l_{s}m_{s} & l_{s}^{2} & l_{s}m_{s} \\ -l_{s}m_{s} & -m_{s}^{2} & l_{s}m_{s} & m_{s}^{2} \end{array} \right]$$

where l(e)= cosθ(e) and m(e)= sinθ(e)

We know that θ(1) = 30°. Therefore, l(1)= cos(θ(1))= cos 30° and m(1) = sinθ(1) = sin30°. Finally, K(1) = (3 * 1) / 4 = 0.75

Therefore, we can compute the stiffness matrix k(1), which is calculated in the following form:


 * $$ \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} $$

When looking at the stiffness matrix we are able to observe that only 3 values need to be calculated: l(e)m(e), m(e), and l(e). The absolute values of these numbers form the stiffness matrix with only the sign changing. We are also able to observe that the matrix is symmetric, or the transpose of k(e) is equal to k(e).

We can use the same method described above to find the stiffness matrix for element 2. l(2)= cos(–45°)

m(2)= sin(-45°)

A brief description of the force-displacement relationship:

Elemental FD: k(e)d(e)= f(e)
 * $$ f^{(e)} = \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix}$$
 * $$ d^{(e)} = \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\ d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix}$$

Global FD:
 * $$ \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 global notation can be simplified into the form [Kij]{dj} = {Fi} where K= [Kij] is the global stiffness matrix, d= {dj} is the global displacement matrix and F= {Fi} is the global force matrix. The elemental notation can be simplified into the form k(e)d(e)=f (e), where k= [kij(e)] is the element stiffness matrix, d= {dj(e)} is the element displacement matrix and f= {fi(e)} is the element force matrix.

One can move from the element to the global matrices through an assembly process. The correspondence between the element degrees of freedom and the global displacement degrees of freedom can be seen below for element 1 and 2:

For node 1: d1 = d(1)1 and d2 = d(1)2

For node 2: d3 = d(1)3= d(2)1 and d4= d(1)4= d(2)2

For node 3: d5 = d(3)4 and d6 = d(2)4 When fully completed we can see that the topography of K resembles the figure below.


 * [[Image:K_Matrix.JPG]]

Therefore, for example 1, K6x6 =
 * $$ \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} $$

Where K11= k(1)11 = ;  K33= k(1)33 + k(2)11 = 3.0625; etc. At this step we can eliminate the known DOF’s. Because the truss example we are calculating is static, we know d1=d2=d5=d6= 0. When we apply these boundary conditions we can delete the corresponding columns in the stiffness matrix and by using the Principle of Virtual Work we can also eliminate the corresponding rows. We are left with the following equation:


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

Once we have used the boundary conditions (global nodes 1 and 3 are fixed) to eliminate the 1st, 2nd, 5th, and 6th columns of the global K matrix, we are left with the matrix below.


 * $$\begin{bmatrix} K_{13}&K_{14}\\ K_{23}&K_{24}\\ K_{33}&K_{34}\\ K_{43}&K_{44}\\ K_{53}&K_{54}\\ K_{63}&K_{64} \end{bmatrix} = \begin{Bmatrix} d_3\\ d_4 \end{Bmatrix} =\overline{F}_{6x1}$$

We can also delete the 1st, 2nd, 5th, and 6th rows in the global K matrix and in the global F (force) matrix. The global F matrix will be a 2x1 matrix containing the values F3, F4. If we recall from the original problem statement that the force P acting on the truss system is directed upwards in the y-direction, we can say that F3=0 and F4=P.

Here are the resulting force displacement reactions.

Our problem is now reduced to a much more manageable matrix. Since we already know the values of F, we want to solve for the global displacement d3 and d4 terms. We will take the inverse of the 2x2 K matrix in order to do this.

The resulting global displacements d3=4.352 and d4=6.1271 are the displacements in the x- and y-direction of global node 2.

To compute the reaction forces in each individual element we solve the matrices k(e)d(e) = f(e) for e = 1,2 Recall that in element 1 the displacement terms d3 and d4 are in the 3rd and 4th rows. It is important here to note that the displacements terms are in different rows for each element. The local displacement matrices are below. Due to the fact that the local displacement matrix for element (1) has zeros in the first two rows, the first two columns of the local k matrix can be deleted. Plugging in numbers for the matrix k(1)d(1) = f(1) We get the local reaction forces for element (1) below. Since we know that element (1) is in equilibrium, the following three equations are true Element (2) can be represented by the following diagram with these forces acting. Taking the root sum squares of the local force components will help solve for the magnitude of forces P for both elements. There are two methods of statics to solve for the 2-bar truss system. These two principles describe the Euler Cut Principle.

Two-Bar Truss System
The following is a MATLAB example for a Two-Bar Truss System

% Two bar truss example clear all; e = [3 5]; A = [1 2]; P = 7; L=[4 2]; alpha = pi/3; beta = pi/4;

nodes = [0, 0; L(1)*cos(pi/2-alpha), L(1)*sin(pi/2-alpha); L(1)*cos(pi/2-alpha)+L(2)*sin(beta),L(1)*sin(pi/2-alpha)-L(2)*cos(beta)];

dof=2*length(nodes);

conn=[1,2; 2,3]; lmm = [1, 2, 3, 4; 3, 4, 5, 6]; elems=size(lmm,1); K=zeros(dof); R = zeros(dof,1); debc = [1, 2, 5, 6]; ebcVals = zeros(length(debc),1);

%load vector R = zeros(dof,1); R(4) = P;

% Assemble global stiffness matrix K=zeros(dof); for i=1:elems lm=lmm(i,:); con=conn(i,:); k_local=e(i)*A(i)/L(i)*[1 -1; -1 1] k=PlaneTrussElement(e(i), A(i), nodes(con,:)) K(lm, lm) = K(lm, lm) + k; end K R % Nodal solution and reactions [d, reactions] = NodalSoln(K, R, debc, ebcVals) results=[]; for i=1:elems results = [results; PlaneTrussResults(e, A, ... nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g results

k_local =

0.75       -0.75        -0.75         0.75

k =

0.5625     0.32476      -0.5625     -0.32476      0.32476       0.1875     -0.32476      -0.1875      -0.5625     -0.32476       0.5625      0.32476     -0.32476      -0.1875      0.32476       0.1875

k_local =

5   -5    -5     5

k =

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

K =

0.5625     0.32476      -0.5625     -0.32476            0            0      0.32476       0.1875     -0.32476      -0.1875            0            0      -0.5625     -0.32476       3.0625      -2.1752         -2.5          2.5     -0.32476      -0.1875      -2.1752       2.6875          2.5         -2.5            0            0         -2.5          2.5          2.5         -2.5            0            0          2.5         -2.5         -2.5          2.5

R =

0    0     0     7     0     0

d =

0           0        4.352       6.1271            0            0

reactions =

-4.4378     -2.5622       4.4378      -4.4378

results =

1.7081      5.1244       8.5406       5.1244       17.081       0.6276       1.8828        3.138       1.8828        6.276

Homework Solutions
These problems supplement the material in the notes, much of which overlaps with or directly relates to the following assigned problems. As the class notes and the assigned problems are very much mutually integrated, it is essential to consider the class notes as a supplement and reference to the following homework solutions.

Derivation of $$\tilde{\textbf{i}}\cdot\textbf{j}$$: Given that m(e) = sin(θ(e)) for a particular element e it follows from the definition of director cosines that:

$$ m^{(e)}=\mathbf{\vec{i}}\cdot \vec{j}=\vec{j}\cdot (\cos \theta ^{(e)}\vec{i} + \sin \theta ^{(e)}\vec{j}=\cos \theta ^{(e)}\vec{j}\cdot \vec{i } + \sin \theta ^{(e)}\vec{j}\cdot \vec{j}=\cos \theta ^{(e)}\ (0) + \sin \theta ^{(e)}\ (1)$$

Such that $$m^{(e)}=0 +\sin \theta ^{(e)}$$

and hence $$m^{(e)}=\sin \theta ^{(e)}$$

Ergo, $$\tilde{\textbf{i}}\cdot\textbf{j}$$ is thus derived.

Find values for k(e) matrices from example truss problem

The example truss problem can be found in the lecture notes, above.

Given that l(1)=cos(30o), m(1)=sin(30o), and k(1)=0.75, the k(1) matrix is then formed using the methods described above, resulting in:

$$\textbf{k}^{(1)}= k^{(1)} \left[ \begin{array}{cccc} l_{s}^{2} & l_{s}m_{s} & -l_{s}^{2} & -l_{s}m_{s} \\ l_{s}m_{s} & m_{s}^{2} & -l_{s}m_{s} & -m_{s}^{2} \\ -l_{s}^{2} & -l_{s}m_{s} & l_{s}^{2} & l_{s}m_{s} \\ -l_{s}m_{s} & -m_{s}^{2} & l_{s}m_{s} & m_{s}^{2} \end{array} \right]=\left[\begin{array}{cccc} 0.5625 & 0.3248 & -0.5625 & -0.3248 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 \\ -0.5625 & -0.3248 & 0.5625 & 0.3248 \\ -0.3248 & -0.1875 & 0.3248 & 0.1875 \end{array} \right]$$

Also, given that l(2)=cos(-45o), m(2)=sin(-45o), and k(2)=5, the k(2) matrix is then formed using the methods described above, resulting in:

$$\textbf{k}^{(2)}= k^{(2)} \left[ \begin{array}{cccc} l_{s}^{2} & l_{s}m_{s} & -l_{s}^{2} & -l_{s}m_{s} \\ l_{s}m_{s} & m_{s}^{2} & -l_{s}m_{s} & -m_{s}^{2} \\ -l_{s}^{2} & -l_{s}m_{s} & l_{s}^{2} & l_{s}m_{s} \\ -l_{s}m_{s} & -m_{s}^{2} & l_{s}m_{s} & m_{s}^{2} \end{array} \right]=\left[\begin{array}{cccc} 2.5 & -2.5 & -2.5 & 2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ 2.5 & -2.5 & -2.5 & 2.5 \end{array} \right]$$

Prove that -P1(1) is equal to P1(2)

Knowing that and given the root sum square method, it follows that the definitions of -P1(1) and P1(2) are

$$-P^{(1)}_{1}=\sqrt{(f_{1}^{(1)})^{2}+(f_{2}^{(1)})^{2}}$$

and

$$P_{1}^{(2)}=\sqrt{(f_{3}^{(1)})^{2}+(f_{4}^{(1)})^{2}}$$

where

$$(f_{1}^{(1)})=-4.4378$$, $$(f_{2}^{(1)})=-2.5622$$, $$(f_{3}^{(1)})= 4.4378$$, and $$(f_{4}^{(1)})= 2.5622$$

such that

$$-P_{1}^{(1)}=\sqrt{(-4.4378)^{2}+(-2.5622)^{2}}=5.1243$$

and

$$P_{1}^{(2)}=\sqrt{(4.4378)^{2}+(2.5622)^{2}}=5.1243$$

Thus -P1(1) is equal to P1(2)

Prove Equilibrium at Node 2 of Example Truss Problem

Using the value calculated above for P1(1) of -5.1243 and calculating P2(2) as 6.2760, it follows that:

$$\sum{F_{y}} = 7-5.1243\sin(30^\circ)-6.2760\sin(45^\circ)=0$$

and

$$\sum{F_{x}} = 5.1243\cos(30^\circ)-6.2760\cos(45^\circ)=0 $$

Hence equilibrium is satisfied.

Individual Contributions
Eml4500.f08.bottle.vitello 21:57, 24 September 2008 (UTC) User:Eml4500.f08.bottle.butler/HW 2 22:00, 24 September 2008 (UTC) Eml4500.f08.bottle.barnes 19:35, 25 September 2008 (UTC) Eml4500.f08.bottle.ranto 21:13, 25 September 2008 (UTC) Eml4500.f08.bottle.hipps 21:21, 25 September 2008 (UTC) Eml4500.f08.bottle.loschak 08:13, 26 September 2008 (UTC)