User:Eml4500.f08.gravy.psc

Verify the dimensions of all the terms in the stiffness matrix multiplied by the displacement matrix
Note: The square brackets, in dimensional analysis, indicate that the dimension of the enclosed are being found. Note: The unit 'F' stands for force, such as newtons or pounds-force. Note: The unit 'L' stands for length, such as inches, meters, or something similar.

To find the dimensionals, first we need the equations for the values, which is given by $$\tilde{k_{ij}}$$ and $$\tilde{d_{j}}$$.

$$\tilde{k_{ij}}\tilde{d_{j}}=\begin{bmatrix} \frac{EA}{L} & 0 & 0 & -\frac{EA}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & -\frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{2EI}{L} \\ -\frac{EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0 \\ 0 & -\frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} \\ 0 & -\frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & \frac{6EI}{L^{2}} & -\frac{4EI}{L} \end{bmatrix} \begin{Bmatrix} \tilde{d_{1}} \\ \tilde{d_{2}} \\ \tilde{d_{3}} \\ \tilde{d_{4}} \\ \tilde{d_{5}} \\ \tilde{d_{6}} \end{Bmatrix}$$

The dimensions of the displacements $$\tilde{d_{j}}$$ are split into two different groups. The first group, where j = 1, 2, 4 and 5, are transverse displacements. The dimensional analysis is shown below:

$$[\tilde{d_{1}}] = [\tilde{d_{2}}] = [\tilde{d_{4}}] = [\tilde{d_{5}}] = L$$

The displacements $$\tilde{d_{j}}$$ where j = 3 and 6 are rotational displacements. The dimensions of rotational displacements are unit less, which will be shown as '1'. Shown below is the dimensional analysis.

$$[\tilde{d_{3}}] = [\tilde{d_{6}}] = 1$$

Other important units are:

$$[E] = \frac{F}{L^{2}}$$

$$[A] = L^{2}$$

$$[I] = L^{4}$$

Using the unit analysis listed above, and applying it to the different values used in the stiffness matrix multiplied by the displacement matrix are:

$$[\frac{EA}{L}] = \frac{F}{L}$$

$$[\frac{EI}{L}] = FL$$

$$[\frac{EI}{L^{2}}] = F$$

$$[\frac{EI}{L^{3}}] = \frac{F}{L}$$

The only other value used by the the stiffness matrix multiplied by the displacement matrix are numbers, including 'zero' and negitives. All numbers, like the rotational displacements, are dimensionless, therefore their units are equal to '1'.

Listed below are all of the non-zero values and their associated units from the top half of the resultant matrix.

$$ [\tilde{k_{11}}\tilde{d_{1}}] = [\frac{EA}{L} * L] = F $$ $$ [\tilde{k_{14}}\tilde{d_{4}}] = [\frac{EA}{L} * L] = F$$ $$ [\tilde{k_{22}}\tilde{d_{2}}] = [\frac{EI}{L^2} * 1] = FL$$ $$ [\tilde{k_{23}}\tilde{d_{3}}] = [\frac{EI}{L^3} * L] = F$$ $$ [\tilde{k_{25}}\tilde{d_{5}}] = [\frac{EI}{L^3} * L] = F$$ $$ [\tilde{k_{26}}\tilde{d_{6}}] = [\frac{EI}{L^2} * 1] = FL$$ $$ [\tilde{k_{32}}\tilde{d_{2}}] = [\frac{EI}{L^2} * L] = FL$$ $$ [\tilde{k_{33}}\tilde{d_{3}}] = [\frac{EI}{L} * 1] = FL$$ $$ [\tilde{k_{35}}\tilde{d_{5}}] = [\frac{EI}{L^2} * L] = FL$$ $$ [\tilde{k_{36}}\tilde{d_{6}}] = [\frac{EI}{L} * 1] = FL$$ $$ [\tilde{k_{41}}\tilde{d_{1}}] = [\frac{EA}{L} * L] = F$$ $$ [\tilde{k_{44}}\tilde{d_{4}}] = [\frac{EA}{L} * L] = F$$ $$ [\tilde{k_{52}}\tilde{d_{2}}] = [\frac{EI}{L^3} * L] = F$$ $$ [\tilde{k_{53}}\tilde{d_{3}}] = [\frac{EI}{L^3} * L] = F$$ $$ [\tilde{k_{55}}\tilde{d_{5}}] = [\frac{EI}{L^3} * L] = F$$ $$ [\tilde{k_{56}}\tilde{d_{6}}] = [\frac{EI}{L^2} * 1] = FL$$ $$ [\tilde{k_{62}}\tilde{d_{2}}] = [\frac{EI}{L^2} * L] = FL$$ $$ [\tilde{k_{63}}\tilde{d_{3}}] = [\frac{EI}{L} * 1] = FL$$ $$ [\tilde{k_{65}}\tilde{d_{5}}] = [\frac{EI}{L^2} * L] = FL$$ $$ [\tilde{k_{66}}\tilde{d_{6}}] = [\frac{EI}{L} * 1] = FL$$

N(i)
In order to solve the system, we need to express $$ u\bigl(\begin{matrix}x\end{matrix}\bigr)$$ in terms of $$d_{i}=u\bigl(\begin{matrix}x_{i}\end{matrix}\bigr)$$ and $$d_{i+1}=u\bigl(\begin{matrix}x_{i+1}\end{matrix}\bigr)$$ as a linear function in x (i.e. linear interpolation)

$$u\bigl(\begin{matrix}x_{i}\end{matrix}\bigr)=N_{i}(x)d_{i}+N_{i+1}(x)d_{i+1}$$

where $$N_{i}\bigl(\begin{matrix}x\end{matrix}\bigr)$$ and $$N_{i+1}\bigl(\begin{matrix}x\end{matrix}\bigr)$$ are linear functions of x.



From this plot it can be easily observed that:

$$ N_{i}(x)=\frac{x-x_{i}}{x_{i+1}-x_{i}} $$

and

$$ N_{i+1}(x)=\frac{x-x_{i+1}}{x_{i}-x_{i+1}} $$

Consider EA = Const
HW 6: Consider EA = constant, then the equation for k simplifies to:

$$ k^{(i)} {=} \frac{EA}{L^{(i)}} \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} $$

27.1 - $$(AB)^T = B^T * A^T$$
Prove: $$ (A_{i x j} B_{j x k})^T = B^T A^T$$

Although this will work for any i, j or k selected, for ease of proving, a low value for each was selected and filled in randomly.

The results were $$ A_{2x3} $$ and $$ B_{3x3} $$with values of:

$$ A = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} $$

$$ B = \begin{bmatrix} 7 & 8 & 9\\ 1 & 2 & 3\\  4 & 5 & 6 \end{bmatrix} $$

Step 1: Multiply A and B which gives:

$$ AB = \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 7 & 8 & 9\\ 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} 21 & 27 & 33\\ 57 & 72 & 87 \end{bmatrix} $$

Step 2: Take the Transpose of the resultant matrix:

$$ (AB)^T = \begin{bmatrix} 21 & 57\\ 27 & 72\\ 33 & 87 \end{bmatrix} $$

Step 3: Solve the right side of the equation:

Step 3.a: Solve $$B^T$$

$$ B^T = \begin{bmatrix} 7 & 1 & 4\\ 8 & 2 & 5\\ 9 & 3 & 6 \end{bmatrix}$$

Step 3.b: Solve $$A^T$$

$$ A^T = \begin{bmatrix} 1 & 4\\ 2 & 5\\ 3 & 6 \end{bmatrix}$$

Step 4: Multiply the two resultant matrices together:

$$ B^T A^T = \begin{bmatrix} 7 & 1 & 4\\ 8 & 2 & 5\\ 9 & 3 & 6 \end{bmatrix}\begin{bmatrix}1 & 4\\ 2 & 5\\ 3 & 6 \end{bmatrix} = \begin{bmatrix} 21 & 57\\ 27 & 72\\ 33 & 87 \end{bmatrix}$$

Since both sides give the same answer, the equation $$ (A_{i x j} B_{j x k})^T = B^T A^T$$ is valid for all i, j and k.

27.4
Show that the following equation is true:

$$( T^{(e)}W)\cdot [\hat{k}^{(e)}q^{(e)} - p^{(e)}] = 0 $$

$$( T^{(e)}W)\cdot [\hat{k}^{(e)}(T^{(e)}d^{(e)}) - p^{(e)}]=0 $$

$$(T^{(e)}W)^T [\hat{k}^{(e)}(T^{(e)}d^{(e)})-p^{(e)}] = 0 $$

By applying the relation shown above in Homework 27.1, the equation above can be simplified to:

$$ W^{T}T^{(e)T}[\hat{k}^{(e)}(T^{(e)}d^{(e)}) - p^{(e)}] = 0 $$

$$ W^{T}[T^{(e)T}\hat{k}^{(e)}(T^{(e)}d^{(e)}) - T^{(e)T}p^{(e)}] = 0 $$

$$ W^{T}[(T^{(e)T}\hat{k}^{(e)}T^{(e)})d^{(e)} - T^{(e)T}p^{(e)}] = 0 $$

The following proof is then used: []

$$\ W^{T}[ k^{(e)}d^{(e)} - T^{(e)T}p^{(e)}] = 0 \,$$

$$\ W^{T}[k^{(e)}d^{(e)} - f^{(e)}] = 0 \,$$

Then, apply the dot product property shown above in reverse, and the follow equation comes forth:

$$ W \cdot [k^{(e)}d^{(e)} - f^{(e)}] = 0 $$

Which was proved to be true earlier in this report.

HW 4 -20.4
Verify that the equation: $$ \mathbf k^{(e)} = \tilde \mathbf T^{(e)^{T}}\tilde \mathbf k^{(e)}\tilde \mathbf T^{(e)}$$ is true.

$$\mathbf k^{(e)} = \tilde \mathbf T^{(e)^{T}}\tilde \mathbf k^{(e)}\tilde \mathbf T^{(e)}$$

$$\mathbf k^{(e)} = k^{(e)}\begin{bmatrix} l^{(e)} & -m^{(e)} & 0 & 0 \\ m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & -m^{(e)} \\ 0 & 0 & m^{(e)} & l^{(e)} \end{bmatrix}\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\ -m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix}$$

$$\mathbf k^{(e)} = \underset{}{k^{(e)}} \begin{bmatrix} (l^{(e)})^{2} & l^{(e)}m^{(e)}  &  -(l^{(e)})^{2}  &  -l^{(e)}m^{(e)}   \\ l^{(e)}m^{(e)} &  (m^{(e)})^{2}  &  -l^{(e)}m^{(e)}  &  -(m^{(e)})^{2}   \\ -(l^{(e)})^{2} & -l^{(e)}m^{(e)}  &  (l^{(e)})^{2}  &  l^{(e)}m^{(e)} \\ -l^{(e)}m^{(e)} &  -(m^{(e)})^{2}  &  l^{(e)}m^{(e)}  &  (m^{(e)})^{2} \\ \end{bmatrix} $$

The above is the definition of k(e), therefore the equation is verified and true.

HW 4 - 23.2
Derive Equation 2 :

$$f_4^{(1)}+f_2^{(2)}=P $$  (2)

Writing the forces in terms of displacements and element stiffness:

$$f_4^{(1)}=\begin{bmatrix} k_{41}^{(1)}d_{1}^{(1)} + k_{42}^{(1)}d_{2}^{(1)} + k_{43}^{(1)}d_{3}^{(1)} + k_{44}^{(1)}d_{4}^{(1)} \end{bmatrix}$$

$$ f_2^{(2)}=\begin{bmatrix} k_{21}^{(2)}d_{1}^{(2)} + k_{22}^{(2)}d_{2}^{(2)} + k_{23}^{(2)}d_{3}^{(2)} + k_{24}^{(2)}d_{4}^{(2)} \end{bmatrix}$$

Substituting the previous two equations into Equation 2: $$ \begin{bmatrix} k_{41}^{(1)}d_{1}^{(1)} + k_{42}^{(1)}d_{2}^{(1)} + k_{43}^{(1)}d_{3}^{(1)} + k_{44}^{(1)}d_{4}^{(1)} \end{bmatrix}+\begin{bmatrix} k_{21}^{(2)}d_{1}^{(2)} + k_{22}^{(2)}d_{2}^{(2)} + k_{23}^{(2)}d_{3}^{(2)} + k_{24}^{(2)}d_{4}^{(2)} \end{bmatrix}= P $$

Then, replace the elemental d with the global equivalent:

$$ \begin{bmatrix} k_{41}^{(1)}d_1 + k_{42}^{(1)}d_2 + k_{43}^{(1)}d_3 + k_{44}^{(1)}d_4\end{bmatrix} + \begin{bmatrix} k_{21}^{(2)}d_3 + k_{22}^{(2)}d_4 + k_{23}^{(2)}d_5 + k_{24}^{(2)}d_6 \end{bmatrix}= P$$

Combing terms of the same d value and then turn that equation into a multiplication of matrices to form:

$$\ \begin{bmatrix} K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)}  \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix} = P \,$$

Which is the fourth row of the $$\ \underline{K} \underline{d} = \underline{F} \,$$ equation.

Combining the Results of Eq 1 and Eq 2, and plugging them into a single equation gives the following:

$$\ \begin{bmatrix} K_{31}^{(1)} & K_{32}^{(1)} & (K_{33}^{(1)} + K_{11}^{(2)}) & (K_{34}^{(1)} + K_{12}^{(2)}) & K_{13}^{(2)} & K_{14}^{(2)} \\ K_{41}^{(1)}& K_{42}^{(1)} & (K_{43}^{(1)} + K_{21}^{(2)}) & (K_{44}^{(1)} + K_{22}^{(2)}) & K_{23}^{(2)} & K_{24}^{(2)}  \end{bmatrix} \begin{Bmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix} = \begin{Bmatrix} 0 \\ P \end{Bmatrix}\,$$

HW2 MATLab Redo HW problem
To solve the system above using an automated FEM process, we've turned to MATLab. Below is the code that was used:

In order to run this code, there are a few functions that are needed. All these functions were downloaded from the book's Website. They are listed below.

After running the preceding code, this was the output results: (Note: code was changed from MATLab laungue to a generic matrix in order to be easier to read.)

14-3 HW problem
Verify the following equation:

$$\ \underline{k}^{(e)} = \underline{T}^{(e)T} * \underline{k}^{carrot(e)} * \underline{T}^{(e)}$$

Where: $$\ \underline{T}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}$$ and $$\ \underline{k}^{carrot(e)} = k^{(e)} * \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$$

Since $$k^{(e)}$$ is a scaler, we can move it within the equation without changing the problem. So after moving the scaler and inserting in the matrices, the following equation is found:

$$\ \underline{k}^{(e)} = k^{(e)} * \begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)}\end{bmatrix} * \begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix} * \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}$$

Solving the first set of matrices gives:

$$\ \underline{k}^{(e)} = k^{(e)} * \begin{bmatrix} l^{(e)} & -l^{(e)}\\ m^{(e)} & -m^{(e)}\\ -l^{(e)} & l^{(e)}\\ -m^{(e)} & m^{(e)}\end{bmatrix} * \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}$$

Solving the second set of matrices gives:

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

Which is the definition of the stiffness matrix, therefore the equation is true.

MATLAB code 15-3


The above picture is the output graph from MATLab after the above code was run. The solid line represents the deformed truss, while the dashed lines show the initial, or un-deformed, truss.

HW#1 - M-Files
M-files M-files are are stored files usually ending with the “.m” extension. They allow MATLAB to run an entire preprogrammed series of commands. They effectivly make up the backbone of MATLAB, and most work is usually done to create or edit these files. There are two main types of m-files: script files and function files.

Script Files Script files are a type of m-file that contains a sequence or normal MATLAB commands. They are normally used for the creation and mataince of large variables, such as matrices. The varibable created in a script file are global. To call a script file, simply enter a command into MATLAB that is the name of the script file with the “.m” extension left off. For example, if the file is named “example1.m” than typing in the command “example1” will cause the script file to run.

Function Files Function files are the true brains behind MATLAB. They allow for the creation of any type of function programmable, which greatly extends the cabalities of MATLAB. Function files, unlike script files, default to creating local variables. However they can be declared global if the need arises. A function files starts with the line “function variable=execute(a,b)”. The word “function” tells MATLAb that the following is a function. “variable” is the output argument, and the name can be changed to be more descriptive. “execute(a,b)” is the name of the function (execute) and the variables that need to be passed into the function (a,b). The list of variables (a,b), called the input arguments, can be expanded or deleted to fit the need of the function. Function files, like all m-files, are able to use and call other files within them, allowing for a greater range of programming options.

An example of a function file is shown below. Remember that the “%” symbol allows for comments without affecting any code. MATLAB ignores all statements after the % symbol.

function a=randint(m,n) %RANDINT Randomly generated integral matrix. %       randint(m,n) returns an m-by-n matrix with entries %      between 0 and 9 a=floor(10*rand(m,n));

Putting the command “z=randint(3,6)” into MATLAB will return a 3-by-6 matrix variable named “z”. The great advantage to function files is the ability to expand and create huge programs. This allows for the creation of many smaller function, similar to the randint(a,b) created before, to be combined into a large file that can be programmed to do thousands of different things. Also, there is the ability to combine self-made functions with some of MATLAB’s built-in functions, much like the built in function “rand(x,y)” was used above.

A useful command to find out a list of all the functions that MATLAB has access to, a combination of built-in and user programmed functions, can be seen by typing the command “type functionname”.

HW#2 - Lectures 8 and 10
Lecture 8

$$\underline{Global F-D Relationship}$$

Noting the fact that the equations for the elemental $$\underline{d^{(e)}}$$ and global $$\underline{d}$$ overlap, as seen in the below picture, the following equations can be derived.



$$d_1 = d_1^{(1)}$$ $$d_2 = d_2^{(1)}$$ $$d_3 = d_3^{(1)} = d_1^{(2)}$$ $$d_4 = d_4^{(1)} = d_2^{(2)}$$ $$d_5 = d_3^{(2)}$$ $$d_6 = d_4^{(2)}$$

Below is a representation of the global $$\underline{K}$$ matrix, which is the stiffness matrix for the system. The two colored boxes represent the contributions of the elemental $$\underline{k^{(e)}}$$. The $$\underline{k^{(1)}}$$ matrix is shown in blue, while the red box represents $$\underline{k^{(2)}}$$.

Lecture 10 $$\underline{Elimination of Known Degrees of Freedom}$$

Knowing that global nodes 1 and 3 are actually fixed points, it is seen that there is no possibility of them moving. Therefore, the displacement at those points is zero.

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

$$\ \underline{d}=\begin{Bmatrix} d_1 \\ d_2  \\  d_3 \\ d_4 \\ d_5 \\ d_6 \end{Bmatrix}= \begin{Bmatrix}  0 \\ 0 \\  d_3 \\ d_4 \\ 0 \\ 0 \end{Bmatrix}$$

Since 4 of the values for $$\underline{d}$$ are zero, we can reduce the system. Turning $$\underline{d}$$ into a 2x2 matrix, would mean that you would be able to reduce $$\underline{K}$$ to a 6x2 matrix. This is down by removing the columns that are associated with the rows within $$\underline{d}$$ that are zero. The new $$\underline{K}$$ is shown below.

$$\ \underline{K}=\begin{bmatrix} K_{13} & K_{14} \\ K_{23} & K_{24} \\  K_{33} & K_{34} \\ K_{43} & K_{44} \\ K_{53} & K_{54} \\ K_{63} & K_{64} \end{bmatrix}$$

This makes the F-D equation: $$\ \underline{K}_{6x2} \underline{d}_{2x1} = \underline{F}_{6x1} \,$$

The next step is to see that since global nodes 1 and 3 are fixed, we can able the concept of "Virtual Work" which will allow us to ignore their effects for the moment and let us turn $$\underline{F}$$ into a 2x1 by removing the components from those fixed nodes. However, by doing this, we need to adjust $$\underline{K}$$ to keep the equation true. We can further reduce $$\underline{K}$$ down to a 2x2, and the following equation results:

$$\ \underline{K}_{2x2} * \underline{d}_{2x1} = \underline{F}_{2x1} \,$$

Where $$\underline{K}$$ is equal to:

$$\ \underline{K}=\begin{bmatrix} K_{33} & K_{34} \\ K_{43} & K_{44} \end{bmatrix}$$ Since $$\underline{F}$$ is now only made up of the forces from global node 2, we now know the value of $$\underline{F}$$. $$\ \underline{F} = \begin{Bmatrix} F_3 \\ F_4 \end{Bmatrix}= \begin{Bmatrix}  0 \\ P \end{Bmatrix}$$ We also can solve for $$\underline{K}$$: $$\ \underline{K}=\begin{bmatrix} 3.0625 & 2.1752 \\ 2.1752 & 2.6875 \end{bmatrix}$$

At this point, the only term not known in the global F-D equation is $$\underline{d}$$. In order to solve the system, we must first solve for $$\underline{d}$$, which gives the following equation: $$\ \underline{d}_{2x1} = \underline{K}^{-1}_{2x2} * \underline{F}_{2x1} \,$$

Where $$\underline{K^{-1}}$$ is equal to: $$\ \underline{K^{-1}} = \frac{1}{det(\underline{K})}*\begin{bmatrix} K_{44} & -K_{43} \\ -K_{34} & K_{33} \end{bmatrix}$$

$$\ \underline{d} = \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$