User:Eml4507.s13.team2/Report3

Honor Pledge
On our honor, we consulted the solution of FEAD F08 Team "Ramrod" HW4 located here. The following MATLAB codes are based upon those used to solve and plot a different truss system. The codes were rewritten to suit the particular plane structure and conditions of this problem and produce the desired results. Dependencies in the MATLAB code were obtained from a finite element companion site located here.

Given
A three-bar plane truss with identical members is presented, and is shown in the image below.

Element 2 is vertical, and all members make equal angles with each other. A force is applied at Node 1, while Nodes 2 and 3 have non-zero prescribed displacement degrees of freedom.



The following member properties are given:

$$ E = 206 \ GPa $$

$$ A = 1*10^{-4} \ m^2 $$

$$ L = 1 \ m $$

The following force and displacement DOFs are given:

$$ F = 20,000 \ N $$

$$ d_3= 2 \ cm $$

$$ d_4= -1 \ cm $$

$$ d_5= -3 \ cm $$

$$ d_6= 5 \ cm $$

$$ d_7= 0 \ cm $$

$$ d_8= 0 \ cm $$

Find
Perform the following:


 * Using MATLAB, solve for:
 * All unknown global displacement DOFs
 * All unknown force components (reactions)
 * All member forces


 * Plot the deformed truss shape superposed on the undeformed truss shape.


 * Verify the results with CALFEM.

Solving with MATLAB
In order to solve for the desired values, the following MATLAB code was used:

Output of MATLAB Code
The above code returns the following output:

k = 1.545e+007 -8.9201e+006 -1.545e+007  8.9201e+006 -8.9201e+006   5.15e+006  8.9201e+006   -5.15e+006 -1.545e+007 8.9201e+006   1.545e+007 -8.9201e+006 8.9201e+006  -5.15e+006 -8.9201e+006    5.15e+006 k = 7.7238e-026 1.2614e-009 -7.7238e-026 -1.2614e-009 1.2614e-009   2.06e+007 -1.2614e-009   -2.06e+007 -7.7238e-026 -1.2614e-009 7.7238e-026  1.2614e-009 -1.2614e-009  -2.06e+007  1.2614e-009    2.06e+007 k = 1.545e+007 8.9201e+006  -1.545e+007 -8.9201e+006 8.9201e+006   5.15e+006 -8.9201e+006   -5.15e+006 -1.545e+007 -8.9201e+006  1.545e+007  8.9201e+006 -8.9201e+006  -5.15e+006  8.9201e+006    5.15e+006 K = Columns 1 through 5 3.09e+007 1.8626e-009 -7.7238e-026 -1.2614e-009  -1.545e+007 1.8626e-009   3.09e+007 -1.2614e-009   -2.06e+007  8.9201e+006 -7.7238e-026 -1.2614e-009 7.7238e-026  1.2614e-009            0 -1.2614e-009  -2.06e+007  1.2614e-009    2.06e+007            0 -1.545e+007 8.9201e+006            0            0   1.545e+007 8.9201e+006  -5.15e+006            0            0 -8.9201e+006 -1.545e+007 -8.9201e+006           0            0            0 -8.9201e+006  -5.15e+006            0            0            0 Columns 6 through 8 8.9201e+006 -1.545e+007 -8.9201e+006 -5.15e+006 -8.9201e+006  -5.15e+006 0           0            0            0            0            0 -8.9201e+006            0            0 5.15e+006           0            0 0  1.545e+007  8.9201e+006 0 8.9201e+006    5.15e+006 R = 14142       14142            0            0            0            0            0            0 d = -0.028976    0.010785         0.02        -0.01        -0.03         0.05            0            0 rf = -2.6217e-011 -4.2816e+005 -3.6562e+005 2.1109e+005 3.5148e+005 2.0293e+005 forces = -4.2219e+005 -4.2816e+005 -4.0586e+005

The first, second, and third outputs k give the element stiffness matrices for element 1, 2, and 3, respectively.

K gives the global stiffness matrix.

R gives the RHS external force matrix.

d gives the global displacement DOFs.

rf gives the reaction force components.

forces gives the member forces.

Plot of Deformed and Undeformed Shape
To plot the figure, the following MATLAB code was used.

The resulting figure is shown below. Blue dashed lines represent the undeformed truss member shape, while red lines indicate the deformed shape. Axis units are in meters.



Honor Pledge
On our honor, this problem was solved on our own using knowledge from vibrations class and referring to Dr. Vu-Quoc's lecture notes located here.

Problem Statement
Image taken from Dr. Vu-Quoc's FEAD.s13.sec53b notes. (1) draw the free body diagrams for each of the bodies (2) derive the mass, spring, and dashpot matrices (3) derive the force equation

Solution
Setting up a force balance for each free body diagram we get:
 * {| style="width:100%" border="0"

$$  \displaystyle f_1 (t)=m_1 a_1= m_1 \ddot{x_1}+(k_1 + k_2)x_1 - k_2 x_2 +(c_1 + c_2) \dot{x_1} -c_2 \dot{x_1}
 * style="width:95%" |
 * style="width:95%" |

$$      (2.0)
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle f_2 (t)=m_2 a_2= m_2 \ddot{x_2}+(k_2 + k_3)x_2 - k_3 x_2 +(c_2 + c_3) \dot{x_2} -c_2 \dot{x_1}
 * style="width:95%" |
 * style="width:95%" |

$$      (2.1) given the standard for of the equation:
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle F(t)=M \ddot(d) + C \dot(d) + Kd
 * style="width:95%" |
 * style="width:95%" |

$$      (2.2) To form the individual matrices we group the like terms from equation 2.0 and 2.1:
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle M=\begin{bmatrix} m_1 & 0\\ 0 & m_2 \end{bmatrix} $$      (2.3)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle C=\begin{bmatrix} c_1 + c_2 & -c_2\\ -c_2 & c_2 + c_3 \end{bmatrix} $$      (2.4)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle K=\begin{bmatrix} k_1 + k_2 & -k_2\\ -k_2 & k_2 + k_3 \end{bmatrix} $$      (2.5) As a quick check, we know that each matrix should be symmetric, which in the case above they are. Looking at the diagrams we can also see:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
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$$  \displaystyle F=\begin{bmatrix} f_1\\ f_2 \end{bmatrix} $$      (2.6)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle d=\begin{bmatrix} d_1 = 0\\ d_2 = 0 \end{bmatrix} $$      (2.7) Plugging in our matrices into the standard form of the equation shown in 2.2 we get:
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="1"

$$  \displaystyle F(t)=\begin{bmatrix} m_1 & 0\\ 0 & m_2 \end{bmatrix} \ddot(d) + \begin{bmatrix} c_1 + c_2 & -c_2\\ -c_2 & c_2 + c_3 \end{bmatrix} \dot(d) + \displaystyle \begin{bmatrix} k_1 + k_2 & -k_2\\ -k_2 & k_2 + k_3 \end{bmatrix}d
 * style="width:95%" |
 * style="width:95%" |

$$      (2.8)
 * 
 * }

Solved By:--Eml4507.s13.team2.rosenberg (discuss • contribs) 16:11, 20 February 2013 (UTC)

Honor Pledge
On our honor, we consulted the solution of Fead.s13.sec53b and the help of CALFEM Manual

Problem Statement
Description: We have a two degrees of freedom spring-mass-damper system as shown in the following figure. The displacement (d), force (F), mass (M), damping (C), and stiffness (K) matrices are given to be:


 * {| style="width:100%" border="0"

$$  \displaystyle d = \left[\begin{array}{cc} d_{1} \\ d_{2} \\ \end{array} \right] $$      (3.0)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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$$  \displaystyle F = \left[\begin{array}{cc} F_{1} \\ F_{2} \\ \end{array} \right] $$      (3.1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle d = \left[\begin{array}{cc} m_{1} & 0 \\ 0 & m_{2} \\ \end{array} \right] $$      (3.2)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle C = \left[\begin{array}{cc} (c_{1}+c_{2}) & -c_{2} \\ -c_{2} & (c_{2}+c_{3}) \\ \end{array} \right] $$      (3.3)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle K = \left[\begin{array}{cc} (k_{1}+k_{2}) & -k_{2} \\ -k_{2} & (k_{2}+k_{3}) \\ \end{array} \right] $$      (3.4)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

We are given the following numerical values:
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$$  \displaystyle m_{1}=3, m_{2}=2 $$      (3.5)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle c_{1}=\frac{1}{2}, c_{2}=\frac{1}{4}, c_{3}=\frac{1}{3} $$      (3.6)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle k_{1}=10, k_{2}=20, k_{3}=15 $$      (3.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle F_{1}(t)=0, F_{2}(t)=0 $$      (3.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle d_{1}(0)=-1, d_{2}(0)=2, d'_{1}(0)=0, d'_{2}(0)=0 $$      (3.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Also, the time step size, defined as dt, is given as:


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$$  \displaystyle dt=\frac{T_{1}}{10} $$      (3.10) where T1 is equal to the fundamental period of the system.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Objective: To integrate the system numerically in order to find the time histories for the displacement degrees of freedom. In other words, given the previous information, we need to find the matrix d.

Solution
First, we input all of the given numerical values and any of the given matrices we can create with these values in MATLAB with the code shown below:

We then use the CALFEM command "eigen" in MATLAB in order to solve the generalized eigenvalue problem:


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$$  \displaystyle Kx=\lambda Mx $$ (3.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Done in MATLAB, with L being the matrix of eigenvalues, this comes out to be:

We then take the smallest eigenvalue, which will be denoted as λ1, and this eigenvalue can then be equated to:


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$$  \displaystyle \lambda_{1}=(\omega_{1})^{2} $$     (3.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where ω1 is defined as the system's "fundamental circular frequency." The next step is to find the fundamental period of the system, which is related to the fundamental circular frequency by the relation:


 * {| style="width:100%" border="0"

$$  \displaystyle \omega_{1} = \frac{2\pi }{T_{1}}=2{\pi}f_{1} $$     (3.13) T1 can be easily found using this relationship in MATLAB, with T1 being the fundamental period, as shown:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Referring to the given relation between time step size and fundamental period, we then use the CALFEM command "step2" to numerically integrate with respect to time t along the interval:
 * {| style="width:100%" border="0"

$$  \displaystyle 0\leq t\leq T = 5T_{1} $$     (3.14) The MATLAB code is as follows:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Finally, we then plot the time histories of the displacement degrees of freedom along the time interval, or the functions d1(t) and d2(t), which are the values in the matrix d.

Honor Pledge
On our honor, we consulted the equations from the course book "Introduction to Finite Element Analysis and Design."

Solution
As seen above there are 4 elements and 5 global nodes. Which have x,y and z coordinates of

x=[0 0.3 0.5 0.7 1];

y=[0 0 0 0 0];

z=[0 0 0 0 0];

The stiffness matrices for each element can be found using

$$ k_{1}= \frac{EA_1}{L_1} \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}

k_{2}= \frac{EA_2}{L_2} \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}

k_{3}= \frac{EA_2}{L_2} \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}

k_{4}= \frac{EA_3}{L_3} \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}

$$

k_2 and k_3 are the same because they are part of the same beam and have the same length

Honor Pledge
On our honor, we consulted the equations from the course book "Introduction to Finite Element Analysis and Design."

Problem Statement
Develop MATLAB code to solve a simple Plane Truss system using finite element methods.

Solution
MATLAB Output

U1 is the displacement at node 1. The first row is the displacement in the y direction and the second is the displacement in the x direction. Fn2n3 is the matrix of forces at nodes 2 and 3. The k matrix is formed using equation 2.39 in "Introduction to Finite Element Analysis and Design" by Nam-Ho Kim and Bhavani V. Sankar.

Displacement graph zoomed in to show displacements. Blue graph is original and red after applied force.

Honor Pledge
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. The MATLAB code is based on that used in Problem 3.1 above, which contains references in its own honor pledge. The code was expanded to fit the plane truss and conditions of this particular problem.

Given
A plane truss structure is shown in the figure below with element numbers and encircled global node numbers.

The horizontal and vertical members have length $$ L $$, while inclined members have length $$ \sqrt{2}L $$.

Assume a Young's modulus of $$ E=100 \ GPa $$, cross sectional area $$ A = 1.0 \ cm^2 $$, and $$ L = 0.3 \ m $$.



Find

 * Develop a MATLAB program to determine the deflections and element forces for the following three load cases:
 * Case A: $$ F_{x13}=F_{x14}=10,000 \ N $$
 * Case B: $$ F_{y13}=F_{y14}=10,000 \ N $$
 * Case C: $$ F_{x13}=10,000 \ and \ F_{x14}=-10,000 \ N $$


 * Plot the deformed shape


 * Use the average tip deflections obtained to compute the equivalent section properties
 * $$ (EA)_{eq} $$
 * $$ (EI)_{eq} $$
 * $$ (GA)_{eq} $$

Load Analysis with MATLAB
In order to solve for the desired values, the following MATLAB code was used. Lines 47-50 were altered as needed to fit each load condition.

Load Case A
For Case A, the MATLAB code returns the following deflections in d and member forces in forces:

d = 0 -4.1321e-019 0           0       0.0003      -0.0003       0.0003      -0.0003       0.0006      -0.0006       0.0006      -0.0006       0.0009      -0.0009       0.0009      -0.0009       0.0012      -0.0012       0.0012      -0.0012       0.0015      -0.0015       0.0015      -0.0015       0.0018      -0.0018       0.0018      -0.0018 forces = 10000       10000        10000        10000        10000        10000        10000        10000        10000        10000        10000        10000  1.3774e-011 1.4456e-011 1.807e-011 2.1684e-011 2.1684e-011 7.228e-012 0 -2.0585e-011 -2.2999e-011 -2.9388e-011 -4.0888e-011 -1.5333e-011 5.111e-012

Load Case B
For Case B, the MATLAB code returns the following deflections in d and member forces in forces:

d = 0      0.0006            0            0        0.003    0.0064971      -0.0036    0.0058971       0.0054     0.018394      -0.0066     0.017794       0.0072     0.035091       -0.009     0.034491       0.0084     0.055388      -0.0108     0.054788        0.009     0.078085       -0.012     0.077485        0.009      0.10168      -0.0126      0.10138 forces = 1e+005 80000       60000        40000        20000 -1.1565e-010 -1.2e+005 -1e+005 -80000      -60000       -40000       -20000       -20000       -20000       -20000       -20000       -20000       -20000       -10000        28284        28284        28284        28284        28284        28284

Load Case C
For Case C, the MATLAB code returns the following deflections in d and member forces in forces:

d = 0  2.628e-018 0           0       0.0003       0.0003      -0.0003       0.0003       0.0006       0.0012      -0.0006       0.0012       0.0009       0.0027      -0.0009       0.0027       0.0012       0.0048      -0.0012       0.0048       0.0015       0.0075      -0.0015       0.0075       0.0018       0.0108      -0.0018       0.0108 forces = 10000       10000        10000        10000        10000        10000       -10000       -10000       -10000       -10000       -10000       -10000 -8.7601e-011 -8.8543e-011 -1.0119e-010 -1.1565e-010 -1.1565e-010 -2.8912e-011 0 1.2295e-010 1.2777e-010 1.38e-010 1.6355e-010 8.1776e-011 -6.1332e-011

Equivalent Cross-Sectional Properties
The desired cross sectional values are included in the following relations:

Axial rigidity from the axial deflection of a beam due to an axial force F:
 * {| style="width:100%" border="0"

$$  \displaystyle u_{tip}=\frac{Fl}{(EA)_{eq}} $$     (6.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Flexural and shear rigidity from the transverse deflection due to a transverse force F at the tip:
 * {| style="width:100%" border="0"

$$  \displaystyle v_{tip}=\frac{Fl^3}{3(EI)_{eq}}+\frac{Fl}{(GA)_{eq}} $$     (6.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Flexural rigidity from the transverse deflection due to an end couple C:
 * {| style="width:100%" border="0"

$$  \displaystyle v_{tip}=\frac{Cl^2}{2(EI)_{eq}} $$     (6.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

First, the results from Load Case A are used to solve for $$ (EA)_{eq} $$ using Eq. 6.1. The average axial deflection at the tip was 0.0018 meters. The beam length is 1.8 meters, and the total applied axial load was 20,000 newtons.


 * {| style="width:100%" border="0"

$$  \displaystyle 0.0018=\frac{20000*1.8}{(EA)_{eq}} $$     (6.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="1"

$$  \displaystyle (EA)_{eq}=20*10^6 \ N $$ (6.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Next, the results from Load Case C are used to solve for $$ (EI)_{eq} $$ using Eq. 6.3. The average transverse deflection at the tip was 0.0108 meters. The beam length is 1.8 meters, and the magnitude of the end couple C was 10,000 newtons.


 * {| style="width:100%" border="0"

$$  \displaystyle .0108=\frac{10000*(1.8)^2}{2(EI)_{eq}} $$     (6.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border=""

$$  \displaystyle (EI)_{eq}=1500 \ kN $$ (6.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Finally, the results from Load Case B and the flexural rigidity above are used to solve for $$ (GA)_{eq} $$ using Eq. 6.2. The average transverse deflection at the tip was 0.10153 meters. The beam length is 1.8 meters, and the total applied transverse load was 20,000 newtons.


 * {| style="width:100%" border="0"

$$  \displaystyle 0.10153=\frac{20000(1.8)^3}{3(1500000)}+\frac{20000*1.8}{(GA)_{eq}} $$     (6.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border=""

$$  \displaystyle (GA)_{eq} \approx 476 \ kN $$ (6.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Honor Pledge
On our honor, we consulted the solution of EML4500.f08.2 and Kim.2008.chap2.pdf

Problem Statement
a) Verify that the bar element stiffness matrix in global coordinates can be obtained by
 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf k^{(e)} = {\mathbf T}^{{(e)}T} \, \widehat{\mathbf k} ^ \,{\mathbf T}^{(e)}
 * style="width:95%" |
 * style="width:95%" |

$$     (7.0)
 * <p style="text-align:right">
 * }

b) Verify that you also get the same expression for the bar element stiffness matrix in global coordinates
 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf k^{(e)} = \widetilde{\mathbf T}^{{(e)}T} \, \widetilde{\mathbf k}^{(e)} \, \widetilde{\mathbf T}^{(e)}
 * style="width:95%" |
 * style="width:95%" |

$$     (7.1)
 * <p style="text-align:right">
 * }

Solution
In order to verify both parts of the solution we are going to use the transformation matrix and its special property.
 * {| style="width:100%" border="0"

$$  \displaystyle {\mathbf T}^{T} = ^{-1}
 * style="width:95%" |
 * style="width:95%" |

$$     (7.2)
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \widetilde{\mathbf T}^{T} = \widetilde^{-1}
 * style="width:95%" |
 * style="width:95%" |

$$     (7.3)
 * <p style="text-align:right">
 * }

a) We will start out in local coordinates and use the property shown in Eq(1.2). The bar above the symbol means "local". The axial definitions will be used
 * {| style="width:100%" border="0"

$$  \displaystyle \bar{\mathbf p} = \hat{\mathbf k} \, \bar
 * style="width:95%" |
 * style="width:95%" |

$$     (7.4)
 * <p style="text-align:right">
 * }

The transformation matrix relates the "local" to the "global" coordinates.
 * {| style="width:100%" border="0"

$$  \displaystyle \bar{\mathbf q} = \,
 * style="width:95%" |
 * style="width:95%" |

$$     (7.5)
 * <p style="text-align:right">
 * }


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$$  \displaystyle \bar{\mathbf p} = \,
 * style="width:95%" |
 * style="width:95%" |

$$     (7.6)
 * <p style="text-align:right">
 * }

Using Eq(7.5) and Eq(7.6) we arrive at


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$$  \displaystyle \hat{\mathbf k} \,{\mathbf T} \, {\mathbf d} = \, {\mathbf f}
 * style="width:95%" |
 * style="width:95%" |

$$     (7.7)
 * <p style="text-align:right">
 * }

We want to take Eq(7.7) and compare it to the force-displacement equation in the axial direction


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$$  \displaystyle \bar{\mathbf p} = \hat \, \bar{\mathbf q}
 * style="width:95%" |
 * style="width:95%" |

$$     (7.8)
 * <p style="text-align:right">
 * }

Because of the Transformation matrix property, we can change Eq(7.8) to what we want


 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf f = {\mathbf T}^{-1} \, \hat{\mathbf k} \,\mathbf T \,\mathbf d
 * style="width:95%" |
 * style="width:95%" |

$$     (7.9) However, because the transformation matrix is not square, it is a (2x4) rectangular matrix, we cannot take its inverse. This is where the Transformation matrix property comes into play
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf f = {\mathbf T}^{T} \, \hat{\mathbf k} \,\mathbf T \,\mathbf d $$ (7.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Looking at Eq(7.10) we can see that the first three terms on the right hand side of the equation relate to the bar element stiffness matrix


 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf k = {\mathbf T}^{T} \, \hat{\mathbf k} \,\mathbf T $$ (7.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Looking at the right side matrices, we get a (4x2) * (2x2) * (2x4) which equals to a (4x4) matrix by using simple matrix multiplication

b) We use start out with Eq(7.3)
 * {| style="width:100%" border="0"

$$  \displaystyle \bar{\mathbf f} = \widetilde \, \bar
 * style="width:95%" |
 * style="width:95%" |

$$     (7.12)
 * <p style="text-align:right">
 * }

The transformation matrix relates the "local" to the "global" coordinates.
 * {| style="width:100%" border="0"

$$  \displaystyle \bar{\mathbf q} = \widetilde{\mathbf T} \,
 * style="width:95%" |
 * style="width:95%" |

$$     (7.13)
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \bar{\mathbf f} = \widetilde{\mathbf T} \,
 * style="width:95%" |
 * style="width:95%" |

$$     (7.14)
 * <p style="text-align:right">
 * }

Plugging in Eq(7.13) and Eq(7.14) into Eq(7.12) we get


 * {| style="width:100%" border="0"

$$  \displaystyle \widetilde{\mathbf T} \,{\mathbf f} = \widetilde{\mathbf k} \, \widetilde{\mathbf T} \, {\mathbf q}
 * style="width:95%" |
 * style="width:95%" |

$$     (7.15)
 * <p style="text-align:right">
 * }

We want to take Eq(7.15) and compare it to the force-displacement equation


 * {| style="width:100%" border="0"

$$  \displaystyle {\mathbf f} = \widetilde \, {\mathbf q}
 * style="width:95%" |
 * style="width:95%" |

$$     (7.16)
 * <p style="text-align:right">
 * }

Because of the Transformation matrix property, we can change Eq(7.16) to what we want


 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf f = \widetilde{\mathbf T}^{-1} \, \widetilde{\mathbf k} \,\widetilde{\mathbf T} \,\mathbf q
 * style="width:95%" |
 * style="width:95%" |

$$     (7.17)
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf f = \widetilde{\mathbf T}^{T} \, \widetilde{\mathbf k} \,\widetilde{\mathbf T} \,\mathbf q $$ (7.18)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Looking at Eq(7.18) we can see that the first three terms on the right hand side of the equation relate to the bar element stiffness matrix


 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf k = \widetilde{\mathbf T}^{T} \, \widetilde{\mathbf k} \,\widetilde{\mathbf T} $$ (7.19)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="1"

$$  \displaystyle \mathbf k = {\mathbf T}^{T} \, \hat{\mathbf k} \,\mathbf T $$ (7.20)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


 * {| style="width:100%" border="1"

$$  \displaystyle \mathbf k = \widetilde{\mathbf T}^{T} \, \widetilde{\mathbf k} \,\widetilde{\mathbf T} $$ (7.21)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Honor Pledge
On our honor, we consulted the solution of EML4507.s13.team2/Report2 and FEAD.s13.sec53a

Problem Statement
a) Consider the linear 2nd order differential equation with constant coefficients described by $$ \lambda _{1,2}=-0.5 \pm 2i $$


 * Find the system natural frequency.

b) Let the system be excited by $$ f\left ( t \right )=f_{0}cos\left ( \bar{\omega }t \right )$$
 * Find the solution to the particular equation.

c) Consider the following information about the excitation $$ \frac{f_{0}}{m}=r_{0}=1 \bar{\omega }=.9\omega $$
 * Find $$ y_{p}\left ( t \right ) and A $$.

d) Find the complete solution that satisfies the initial conditions $$ y\left ( 0 \right )=1 $$ and $${y}'\left ( 0 \right )=0 $$

Solution
a) Using the characteristic equation $$\lambda^{2}+a\lambda+b=0$$ $$\lambda= \frac{-a\pm \sqrt{a^{2}-4b}}{2}$$

$$-.5=\frac{-a}{2}$$

$$ a=1 $$

$$ 2i=\frac{\sqrt{a^{2}-4b}}{2} $$

$$ -16=a^{2}-4b $$

$$b=\frac{17}{4} $$

$$ b=\omega ^{2}$$

$$ \omega = 2.0615$$

b) $$y_{p}\left ( t \right )=de^{i\left ( \bar{\omega } t-\phi \right )}$$

$$ d=\frac{\frac{f_{0}}{m}}{\sqrt{\left ( \omega ^{2}-\bar{\omega } ^{2}\right )^{2}+\left ( 2\zeta \omega \bar{\omega } \right )^{2}}} $$

$$ \zeta = \frac{a}{2\omega } $$

$$ y_{p}\left ( t \right )=\frac{\frac{f_{0}}{m}}{\sqrt{\left ( \omega ^{2}-\bar{\omega } ^{2}\right )^{2}+\left ( a \bar{\omega } \right )^{2}}}e^{i\left ( \bar{\omega } t-\phi \right )} $$

c) $$ y_{p}\left ( t \right )=\frac{1}{\sqrt{\left ( 2.06^{2}-1.86^{2} \right )^{2}+\left ( 1.86 \right )^{2}}}e^{i\left ( 1.86t-1.17 \right )} $$

$$ \rho =\frac{\bar{\omega }}{\omega } $$

$$ \rho =\frac{.9\omega }{\omega } $$

$$ \rho =.9 $$

$$ A=\frac{1}{\left [ \left ( 1-\rho ^{2} \right ) +\left ( 2\zeta \rho \right )^{2}\right ]^{\frac{1}{2}}} $$

$$ A=2.09 $$

d) $$ y_{h}\left ( t \right )=Ae^{\left ( \frac{-at}{2} \right )}cos\left ( \tilde{\omega }t \right )+Be^{\left ( \frac{-at}{2} \right )}sin\left ( \tilde{\omega }t \right ) $$

$$ \tilde{\omega }^{2}=b-\frac{a^{2}}{4} $$

$$ \tilde{\omega }=2 $$

$$ y\left ( t \right )=2.02e^{i\left ( 1.86t-1.17 \right )}+Ae^{\left ( \frac{-at}{2} \right )}cos\left ( \tilde{\omega }t \right ) $$

$$ y\left ( t \right )=2.02cos\left ( 1.86t-1.17 \right )+Ae^{\left ( -.5t \right )}cos\left (2t \right )+Be^{\left ( -.5t \right )}sin\left ( 2t \right ) $$

$$ y\left ( t \right )=2.02cos\left ( 1.86t-1.17 \right )+.21e^{\left ( -.5t \right )}cos\left (2t \right )-1.78e^{\left ( -.5t \right )}sin\left ( 2t \right ) $$

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