User:Eml4500.f08.ateam/hw 2

A-Team Teampage

Homework Report #2

This page is intended for Homework Report #2 of the A-Team.

Due date: Friday, 26 September 08, 5 pm EDT = 17:00 EDT (or 21:00 UTC).

Statically Determinate versus Statically Indeterminate
A statically determinate problem is one that can be solved using simple statics relationships. There are three main equations, setting the sum of forces in the x equal to zero and setting the sum of the forces in the y direction equal and setting the sum of the moments about a point equal to zero. With these three equations, any problem with up to three unknowns can be solved. Problems with multiple beams or more than three reaction forces are statically indeterminate since there are insufficient equations to solve for all the unknowns.

An exception to the indeterminate rules is shown in the following example.

Step 1: Global Picture


Here there are four reaction forces, two at each support and a known applied force at the top. It can be noted that the beams in this problem are both two-force members. By looking at each element separately and summing the forces in any direction it is found that they must be both equal and opposite. Summing the moment about either end shows that both forces' line of action must go through each end. Otherwise the moments would not sum to zero.



Therefore the two forces are constrained in one direction. With these simple straight beams, the line of action is along the beam itself. If the beam were curved or had an angle in it, the same logic would apply and the line of action of the two forces will be along the invisible line going through the two attachment points.

This simplification of the problem lowers the degrees of freedom (unknowns) from four to three. Since the direction of the reaction forces are now known, only the magnitude is left to be found.

Step 2: Element Picture
Here the element degrees of freedom and element forces are shown in either the global system or in a local coordinate system. The element number is in the parentheses and the forces (and displacements) are numbered from node 1 to 3, each starting with the x and then y axes.



Step 3: Global Force-Displacement Relationship
The next step is to create a relationship between the forces and displacements in the global coordinate system. The matrices will be in the following form.

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

Where k (e) is a 4x4 matrix and d (e) and f (e) are 4x1 matrices. In order to find the values of k (e), the formula found on page 225 of Fundamental Finite Element Analysys and Applications must be used. It states that the scalar, $$K = \frac{EA}{L}$$, is multiplied by a matrix composed of functions of the director of cosines.



The director of cosines is a function of the angle of the element and given as, $$l^{(e)}= \vec \tilde{i} \centerdot \vec i = \cos \ \Theta^{(e)}$$ and  $$m^{(e)}= \vec \tilde{i} \centerdot \vec j = \sin \ \Theta^{(e)}$$. The completed matrix for K (e) is as follows.

$$ \underline K^{(e)}= K*\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)} \\ l^{(e)}m^{(e)} & (m^{(e)})^2 & -l^{(e)}m^{(e)} & -(m^{(e)})^2 \\ -(l^{(e)})^2 & -l^{(e)}m^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)} \\ -l^{(e)}m^{(e)} & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2 \end{bmatrix} $$

The k (e) matrices for elements 1 and 2 were calculated based on the given data and are as follows.

$$ \underline K^{(1)}= \begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & \frac{-9}{16} & \frac{-3\sqrt{3}}{16} \\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & \frac{-3\sqrt{3}}{16} & \frac{-3}{16} \\ \frac{-9}{16} & \frac{-3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16} \\ \frac{-3\sqrt{3}}{16} & \frac{-3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix}=\begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 \\ 0.3248 & 0.1875 & -0.3248 & -0.1875 \\ -0.5625 & -0.3248 & 0.5625 & 0.3248 \\ -0.3248 & -0.1875 & 0.3248 & 0.1875 \end{bmatrix} $$

$$ \underline K^{(2)}= \begin{bmatrix} \frac{5}{2} & \frac{-5}{2} & \frac{-5}{2} & \frac{5}{2} \\ \frac{-5}{2} & \frac{5}{2} & \frac{5}{2} & \frac{-5}{2} \\ \frac{-5}{2} & \frac{5}{2} & \frac{5}{2} & \frac{-5}{2} \\ \frac{5}{2} & \frac{-5}{2} & \frac{-5}{2} & \frac{5}{2} \end{bmatrix}=\begin{bmatrix} 2.5 & -2.5 & -2.5 & 2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ -2.5 & 2.5 & 2.5 & -2.5 \\ 2.5 & -2.5 & -2.5 & 2.5 \end{bmatrix} $$

We can observe that K is symmetric, meaning $$k^{(e)T} = K ^(e)$$.

$$K^(e)= $$$$\begin{bmatrix} K_{11}&K_{12}&K_{13}&K_{14} \\ &K_{22}&K_{23}&K_{24} \\ & &K_{33}&K_{34} \\ &sym. & &K_{44} \\ \end{bmatrix} $$

Element 1:
From the FBD of element 1, we see that the angle from the x-axis is 30°.

θ(1) = 30ο l(1) = cosθ(1) = cos30ο = $$\frac{\sqrt3}{2}$$ m(1) = sinθ(1) = sin30° = $$\frac{1}{2}$$

For element 1, the stiffness is given as: $$K^{(1)} = \frac{E^{(1)}A^{(1)}}{L^{(1)}}$$ = $$\frac{3}{4}$$

And the matrix of k(1) is given as:

$$ \underline K^{(1)}= K*\begin{bmatrix} k^{(1)}(l^{(1)})^2 & k^{(1)}l^{(1)}m^{(1)} & -k^{(1)}(l^{(1)})^2 & -k^{(1)}l^{(1)}m^{(1)} \\ k^{(1)}l^{(1)}m^{(1)} & k^{(1)}(m^{(1)})^2 & -k^{(1)}l^{(1)}m^{(1)} & -k^{(1)}(m^{(1)})^2 \\ -k^{(1)}(l^{(1)})^2 & -k^{(1)}l^{(1)}m^{(1)} & k^{(1)}(l^{(1)})^2 & k^{(1)}l^{(1)}m^{(1)} \\ -k^{(1)}l^{(1)}m^{(1)} & -k^{(1)}(m^{(1)})^2 & k^{(1)}l^{(1)}m^{(1)} & k^{(1)}(m^{(1)})^2 \end{bmatrix} $$

In this matrix, only three values need to be calulated. The rest only differ by being positive or negative. As you can see this matrix is symetric. Therefore,

 kij(1) = kji(1)

And in general:

 kij(e) = kji(e) and  k(e)T = k(e)

Element 2:
From the FBD of element 2, we see that the angle from the x-axis is -45°.

θ(2) = -45°

l(2) = cosθ(2) = cos(-45°) = $$\frac{\sqrt2}{2}$$

m(2) = sinθ(2) = sin(-45°) = $$\frac{-\sqrt2}{2}$$

For element 2, the stiffness is given as: $$K^{(2)} = \frac{E^{(2)}A^{(2)}}{L^{(2)}}$$ = 2.5

And the matrix of k(2) is given as:

$$ \underline K^{(2)}= K*\begin{bmatrix} k^{(2)}(l^{(2)})^2 & k^{(2)}l^{(2)}m^{(2)} & -k^{(2)}(l^{(2)})^2 & -k^{(2)}l^{(2)}m^{(2)} \\ k^{(2)}l^{(2)}m^{(2)} & k^{(2)}(m^{(2)})^2 & -k^{(2)}l^{(2)}m^{(2)} & -k^{(2)}(m^{(2)})^2 \\ -k^{(2)}(l^{(2)})^2 & -k^{(2)}l^{(2)}m^{(2)} & k^{(2)}(l^{(2)})^2 & k^{(2)}l^{(2)}m^{(2)} \\ -k^{(2)}l^{(2)}m^{(2)} & -k^{(2)}(m^{(2)})^2 & k^{(2)}l^{(2)}m^{(2)} & k^{(2)}(m^{(2)})^2 \end{bmatrix} $$

Global Force Displacement Relation
The Global Force Displacement Relation is known as K d = F. Matrix K is the element stiffness matrix in global coordinates. In our example, the global stiffness matrix is a 6 x 6 grid, the global displacement matrix is 6 x 1, and the global force matrix is 6 x 1, and is shown below:


 * $$\begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix} = \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}$$

Since the stiffness matrix is 6 x 6, there are six degrees of freedom in the problem. The Global Force Displacement Relation can be shown in a more compact notation:
 * $$\begin{bmatrix} K_{ij} \end{bmatrix} \begin{Bmatrix} d_{j} \end{Bmatrix} = \begin{Bmatrix} F_{i} \end{Bmatrix}$$

or    $$\sum_{j=1}^6 $$ Kij dj  = Fi , i=1,2,3,4,5,6

K or $$\begin{bmatrix} K_{ij} \end{bmatrix}$$n x n is the global stiffness matrix. d or $$\begin{Bmatrix} d_{j} \end{Bmatrix}$$n x 1 is the global displacement matrix. F or $$\begin{Bmatrix} F_{i} \end{Bmatrix}$$n x 1 is the global force matrix.

These global matrices are different than the element force displacement relation k(e) d(e) = f(e). In this relationship, recall the k(e) is a 4 x 4 matrix, called the element stiffness matrix where (e) is the element number. d(e) and f(e) are the element displacement and force matrices, both of which are 4 x 1. These element matrices are smaller than the global ones because each one pertains to an element rather than the entire picture. To go from an element matrix to a global matrix, the correspondence between the element displacement degrees of freedom and the global displacement degrees of freedom need to be identified.



At the global level, the degrees of freedom are $$d_1,\ d_2,\ d_3,\ d_4,\ d_5,\ d_6$$. At the element level, there is a set of degrees of for each element.

In Element 1, the degrees of freedom are $$d_1^{(1)},\ d_2^{(1)},\ d_3^{(1)},\ d_4^{(1)}$$. Element 2 is $$d_1^{(2)},\ d_2^{(2)},\ d_3^{(2)},\ d_4^{(2)}$$.

The schematic above shows that some of the global degrees of freedom are the same as the element. At node 1:
 * d1 = d1(1)
 * d2 = d2(1)

Node 2 on the global schematic has the same displacement degrees of freedom as on both the first and second element; therefore:
 * d3 = d3(1) = d1(2)


 * d4 = d4(1) = d2(2)

Node 3 is only:
 * d5 = d3(2)
 * d6 = d4(2)

The conceptual step of assembly has a small matrix for each element stiffness matrix and two zero matrices all within the global force displacement relation. The basic matrix is shown below:

$$\begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix} = \begin{Bmatrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ F_5 \\ F_6 \end{Bmatrix}$$

Assembly of Force Displacement Matrix
The global force displacement matrix is assembled from the 2 element displacement matrices.

The Global Stiffness Matrix with symbolic representation:
 * $$\begin{bmatrix}

K^1_{11}&K^1_{12}&K^1_{13}&K^1_{14}&0&0\\ K^1_{21}&K^1_{22}&K^1_{23}&K^1_{24}&0&0\\ K^1_{31}&K^1_{32}&K^1_{33}+K^2_{11}&K^1_{34}+K^2_{12}&K^2_{13}&K^2_{14}\\ K^1_{41}&K^1_{42}&K^1_{43}+K^2_{21}&K^1_{44}+K^2_{22}&K^2_{23}&K^2_{24}\\ 0&0&K^2_{31}&K^2_{32}&K^2_{33}&K^2_{34}\\ 0&0&K^2_{41}&K^2_{42}&K^2_{43}&K^2_{44} \\ \end{bmatrix} $$

Note: The overlap in the 2 matrices comes from the fact that global node 2 is shared between both elements.

Calculating the Matrix Values:
K11 = k111 = 0.5625 K12 = k112 = 0.3247 K33 = k133 + k211 = 3.0625 K34 = k134 + k212 = -2.1752 K43 = k143 + k221 = -2.1752 K44 = k144 + k222 = 2.6875

Elimination of the known degrees of freedom reduces the global force displacement relationship. d1 = d2 = d5 = d6 = 0.

Two Bar Truss Matlab Example
Here is the Matlab code run for this example:

% 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

Here are the results for the code:

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

Step 4: Elimination of Known Degrees of Freedom
The next step in the Finite Element Method is to eliminate the known degrees of freedom by reducing the global force-displacement relationship. The displacement matrix from the previous two-truss system can be reduced by applying the fixed boundary conditions:

$$d_1=d_2=d_5=d_6=0$$

The resulting displacement matrix is:

$$d= \begin{bmatrix} 0 \\ 0 \\ d_3  \\ d_4 \\ 0 \\ 0 \end{bmatrix} $$

After applying these conditions, the corresponding columns in the global stiffness matrix can be deleted and the resulting matrix is shown below.

$$F=K \begin{bmatrix} 0 \\ 0 \\ d_3  \\ d_4 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} k_{13} & k_{14} \\ k_{23} & k_{24} \\ k_{33} & k_{34} \\ k_{43} & k_{44} \\ k_{53} & k_{54} \\ k_{63} & k_{64} \\ \end{bmatrix} \begin{bmatrix} d_3 \\ d_4 \\ \end{bmatrix} $$

The next step in reducing the force-displacement relationship is to apply the Principle of Virtual Work. This allows the corresponding rows of the global stiffness matrix and the global force matrix to be deleted. The resulting force-displacement relation is

$$ \begin{bmatrix} k_{33} & k_{34} \\ k_{43} & k_{44} \\ \end{bmatrix} \begin{bmatrix} d_3 \\ d_4 \\ \end{bmatrix}= \begin{bmatrix} F_3 \\ F_4 \\ \end{bmatrix} $$ or $$\mathbf{K}\cdot\mathbf{d} = \mathbf{F}$$

Based on the given information, the reduced force matrix is a known, as shown below.

$$\underline{\overline{F}}= \begin{bmatrix} F_3 \\ F_4 \\ \end{bmatrix}= \begin{bmatrix} 0 \\ P \\ \end{bmatrix}= \begin{bmatrix} 0 \\ 7 \\ \end{bmatrix} $$

Finally, the reduced global displacement matrix can be solved by multiplying the reduced force matrix by the inverse of the reduced global stiffness matrix.

$$ \begin{bmatrix} d_3 \\ d_4 \\ \end{bmatrix} =\begin{bmatrix} k_{33} & k_{34} \\ k_{43} & k_{44} \\ \end{bmatrix}^{-1} \begin{bmatrix} 0 \\ P \\ \end{bmatrix}= \begin{bmatrix} 4.352 \\ 6.127 \\ \end{bmatrix} $$

Step 5: Compute Reactions
After the displacements of Node 2 are computed, then the next step is to compute the reactions. There are two methods that can be used to compute the reactions. The first method is to use the element force-displacement relations

$$k^{(e)}d^{(e)}=f^{(e)}$$

for elements 1 and 2. The element displacement matrices can be determined from the global displacement matrix.

$$\underline{d}^{(1)}= \begin{bmatrix} 0 \\ 0 \\ 4.352 \\ 6.127 \\ \end{bmatrix}

\underline{d}^{(2)}= \begin{bmatrix} 4.352 \\ 6.127 \\ 0 \\ 0 \\ \end{bmatrix} $$

Solving for element forces of Element 1
$$k^{(1)}d^{(1)} = f^{(1)}$$

$$\begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2\\ f^{(1)}_3\\ f^{(1)}_4 \end{Bmatrix}$$ = $$\begin{bmatrix} 0.5625 & 0.32776 & -0.5625 & -0.32476 \\ 0.32476 & 0.1875 & -0.32476 & -0.1875 \\ -0.5625 & -0.32476 & 0.5625 & 0.32476 \\ -0.32476 & -0.1875 & 0.32476 & 0.1875 \end{bmatrix} $$ $$\begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}$$

This relation can be reduced to the following by the observation that the first two terms in the element displacement matrix are zero. This is due to matrix algebra causing the first two columns of the element stiffness matrix will completely drop out.

 $$\begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2\\ f^{(1)}_3\\ f^{(1)}_4 \end{Bmatrix}$$ = $$\begin{bmatrix} -0.5625 & -0.32476 \\ -0.32476 & -0.1875 \\ 0.5625 & 0.32476 \\ 0.32476 & 0.1875 \end{bmatrix} $$ $$\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

Carrying out this relation reveals the element force matrix.

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

In which, $$f^{(1)}_1$$ and $$f^{(1)}_2$$ are reaction forces while $$f^{(1)}_3$$ and $$f^{(1)}_4$$ are internal forces.

Element 1 is in equilibrium because the reaction forces and internal forces offset each other and the sum of the moments about any point on Element 1 is zero.



$$\sum F_x = f^{(1)}_1 + f^{(1)}_3 = 0$$

$$\sum F_x = f^{(1)}_2 + f^{(1)}_4 = 0$$

$$\sum M_{any} = 0$$

Solving for element forces of Element 2


$$P^{(1)}_1 = \sqrt{f^{(1)}_1 + f^{(1)}_2}$$

Accounting for the applied force
In order to proceed, the applied force P must be accounted for in the free-body diagram and the equilibrium of all nodes must be verified. The equilibrium of global nodes 1 and 3 were previously verified.



Statics method
There is a second method to solve for the reactions and internal forces of the 2-bar truss depicted here. This method uses the principles of statics, but also must employ the Euler Cut Principle. The Euler Cut Principle must be utilized because the truss is considered statically indeterminate.



Method 1



Method 2

There are multiple ways of applying the Euler Cut Principle to the truss. Each way is determined simply by where the cut is made in the truss system. Depending on this choice, there are two or three resulting free-body diagrams from which the forces are derived.

Contributing Team Members
The following students contributed to this report:

Matthew Philbin Eml4500.f08.Ateam.philbin 22:28, 24 September 2008 (UTC)

Mark Barry Eml4500.f08.Ateam.barry 13:41, 25 September 2008 (UTC)

Jil Paquette Eml4500.f08.Ateam.paquette 17:29, 25 September 2008 (UTC)

Sean Miller EML4500.f08.Ateam.Miller 04:00, 26 September 2008 (UTC)

Daniel DiDomenico Eml4500.f08.ateam.didomenico 04:02, 26 September 2008 (UTC)

Ian Slevinski Eml4500.f08.Ateam.Slevinski 18:02, 26 September 2008 (UTC)