User:Eml4507.s13.team4.vcb

User:Eml4507.s13.team4.vcb/report

=Table of Assignments R7=

=Problem R7.1a: Verify the dim of the $$ \underline{k}^{(e)}_{6x6}\underline{d}^{(e)}_{6x1} $$ matrices (fead.f08.mtgs.[37-41] pg. 2)=

On our honor, we did this assignment on our own.

Given: The desired dimensions for k and d with index of 6
$$ \underline{k}^{(e)}_{6x6}\underline{d}^{(e)}_{6x1} $$

1.Use an index of 6 to verify the dimensions of k and d
Obtain:

$$ \underline{k}^{(e)}_{6x6}\underline{d}^{(e)}_{6x1} = \underline{f}^{(e)}_{6x1} $$

With supporting: $$ \underline{k}^{(e)}_{6x6} = \underline{\tilde{T}}^{(e)T}_{6x6} \underline{\tilde{k}}^{(e)}_{6x6} \underline{\tilde{T}}^{(e)}_{6x6} $$

From:

$$ \underline{\tilde{k}}^{(e)}_{6x6} \underline{\tilde{d}}^{(e)}_{6x1} = \underline{\tilde{f}}^{(e)}_{6x1} $$

Solution: Verify the dimensions of k and d
The constructed 6x6 matrix shown below represents $$ \underline{\tilde{T}} $$

$$ \begin{Bmatrix} \tilde{d}_{1}\\ \tilde{d}_{2}\\ \tilde{d}_{3}\\ \tilde{d}_{4}\\ \tilde{d}_{5}\\ \tilde{d}_{6}\end{Bmatrix}= \begin{Bmatrix} R_{11} & R_{12} & 0 & 0  & 0  & 0\\ R_{21} & R_{22} & 0 & 0  & 0  & 0\\ 0 & 0 & 1 & 0  & 0  & 0\\ 0 & 0 & 0 & R_{11} & R_{12}  & 0\\ 0 & 0 & 0 & R_{21} & R_{22} & 0\\ 0 & 0 & 0 & 0  & 0  & 1\end{Bmatrix} \begin{Bmatrix}	d_{1}\\ d_{2}\\ d_{3}\\ d_{4}\\ d_{5}\\ d_{6}\end{Bmatrix} $$

The following matrix represents the element stiffness matrix in local coordinates.

$$\underline{\tilde{k}}^{(e)}_{6x6} = \begin{bmatrix} \frac{EA}{L} & 0 & 0 & \frac{-EA}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{2EI}{L} \\ \frac{-EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0 \\ 0 & \frac{-12EI}{L^{3}} & \frac{-6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{-6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{4EI}{L}\end{bmatrix}$$

Using the above two matricies and the transpose of T, you can construct the following equation:

$$ \underline{k}^{(e)}_{6x6} = \underline{\tilde{T}}^{(e)T}_{6x6} \underline{\tilde{k}}^{(e)}_{6x6} \underline{\tilde{T}}^{(e)}_{6x6} $$

Multiplying $$ \underline{\tilde{k}}^{(e)}_{6x6} $$ by $$ \underline{\tilde{d}}^{(e)}_{6x1} $$ will yield the required $$ \underline{\tilde{f}}^{(e)}_{6x1} $$

=Problem R7.1b: Solve 2 element frame system (fead.f08.mtgs.[37-41] pg. 3)=

On our honor, we did this assignment on our own.

Given: Information on the two element truss system
Assume a square cross section. Also use same data from fea.f08.mtgs.p5-4.



Element length: (1) =4

Element length: (2) =2

Young's modulus: (1) =3

Young's modulus: (2) =5

Cross section area: (1) =1

Cross section area: (2) =2

Inclination angle: (1) = 30 deg

Inclination angle: (2) = -45 deg

1. Plot Undeformed Shape
Undeformed Shape.



View of element with single force member view.



2. Plot Deformed Shape 2-bar Truss
Plotting the deformed shape 2-bar truss for the given problem from p5-2 of the fead08 lecture notes. The deformation can be seen the truss using the forces given from p5-2.



3. Plot Deformed Shape 2-bar Frame
Plotting the deformed shape 2-bar frame for the given problem from p5-2 of the fead08 lecture notes. The constraints are preserved during the analysis in this mode to keep the frame.



ref
= Problem R7.2 =

On my honor, I have neither given nor recieved unauthorized aid in doing this assignment.

Description
We are to resolve problem 5.7. We are to solve for motion of the truss using modal superposition using the three lowest eigenvalues.

Solution
The following code was used for problem 5.7 to obtain the K and M matrices and to obtain the eigenvalues and the eigenvectors:

From this, we obtain M and K matrices which are:

K =

Columns 1 through 9

3.3839   0.8839   -2.5000         0   -0.8839   -0.8839         0         0         0    0.8839    0.8839         0         0   -0.8839   -0.8839         0         0         0   -2.5000         0    5.8839    0.8839         0         0   -2.5000         0   -0.8839         0         0    0.8839    3.3839         0   -2.5000         0         0   -0.8839   -0.8839   -0.8839         0         0    4.2678         0   -0.8839    0.8839   -2.5000   -0.8839   -0.8839         0   -2.5000         0    4.2678    0.8839   -0.8839         0         0         0   -2.5000         0   -0.8839    0.8839    5.8839   -0.8839         0         0         0         0         0    0.8839   -0.8839   -0.8839    3.3839         0         0         0   -0.8839   -0.8839   -2.5000         0         0         0    4.2678         0         0   -0.8839   -0.8839         0         0         0   -2.5000         0         0         0         0         0         0         0   -2.5000         0   -0.8839         0         0         0         0         0         0         0         0    0.8839

Columns 10 through 12

0        0         0         0         0         0   -0.8839         0         0   -0.8839         0         0         0         0         0         0         0         0         0   -2.5000         0   -2.5000         0         0         0   -0.8839    0.8839    4.2678    0.8839   -0.8839    0.8839    3.3839   -0.8839   -0.8839   -0.8839    0.8839

M =

Columns 1 through 9

1.2071        0         0         0         0         0         0         0         0         0    1.2071         0         0         0         0         0         0         0         0         0    2.2071         0         0         0         0         0         0         0         0         0    2.2071         0         0         0         0         0         0         0         0         0    2.4142         0         0         0         0         0         0         0         0         0    2.4142         0         0         0         0         0         0         0         0         0    2.2071         0         0         0         0         0         0         0         0         0    2.2071         0         0         0         0         0         0         0         0         0    2.4142         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0

Columns 10 through 12

0        0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0         0    2.4142         0         0         0    1.2071         0         0         0    1.2071

This will also output the lowest eigenpairs:

$$ \gamma_{1}, \phi_{1} $$

$$ \gamma_{2}, \phi_{2} $$

$$ \gamma_{3}, \phi_{3} $$

Then, with these, we will find what the modal equations are according to the equation:

$$ z^{''} + \gamma_{i}z = \phi_{i}^{T} F(t) $$

This will yield 3 unique differential equations. We solve for the complete solution to these differential equation using the boundary conditions from:

$$ z_{i} (0) = \bar{\phi}_{i}^{T} M d(0) $$ and

$$ z_{i}^{'} (0) = \bar{\phi}_{i}^{T} M d^{'}(0) $$

With these solutions for z, we can then find and plot the actual displacements from modal superposition using:

$$ d(t) = \Sigma^{3}_{j=1} z_{j} \phi_{j} $$

=Problem 6.4 Deformation in 3D Space= On my honor, I have neither given nor received unauthorized aid in doing this assignment.

Problem Statement
There is a 3D truss system with applied forces. We need to optimize each beam to meet a factor of safety of 1.5.

Conclusion
The final area of each member was able to be reduced to its minimum of 0.1 inches and stay above a factor of safety of 1.5. Final stresses are in psi:
 * 1) 	13.94
 * 2) 	-8487.15
 * 3) 	-8487.15
 * 4) 	8611.63
 * 5) 	8611.63
 * 6) 	-12871.54
 * 7) 	13045.99
 * 8) 	-12871.54
 * 9) 	13045.99
 * 10) 	63.42
 * 11) 	63.42
 * 12) 	788.95
 * 13) 	-778.48
 * 14) 	-5171.02
 * 15) 	5174.49
 * 16) 	-5171.02
 * 17) 	5174.49
 * 18) 	-7787.74
 * 19) 	-7787.74
 * 20) 	7792.39
 * 21) 	7792.39
 * 22) 	14680.46
 * 23) 	-14679.76
 * 24) 	-14679.76
 * 25) 	14680.46

CALFEM Verification
The Matlab calculations were confirmed using the CALFEM toolbox in Matlab.

The following presents the constants used and options for different designs

This Edof matrix uses column 1 as the as the element number. The next three columns are the DOFs for the first node and the next three columns are for the second node since this is a three dimensional problem.

Coord contains the x, y, and z coordinates for each node.

The bar3e is used to make the stiffness matrix for a 3D system. The bar3s is used to find the member forces in a 3D system. One can see that the 3 denotes the dimension of the system being observed. The stresses can be found by dividing the forces by the areas.

=Problem R5.7: Consider the free vibration of truss system p.53-22b (fead.f13.sec53b.[21])= On our honor, we did this assignment on our own.

Given: Truss system under free vibration with applied force and zero initial velocity
Consider the truss system. The initial force is applied at node four and the force equals five.

Use standard node and element naming conventions.



Solution: Solve for the truss motion
K =

Columns 1 through 5

3.3839   0.8839   -2.5000         0   -0.8839    0.8839    0.8839         0         0   -0.8839   -2.5000         0    5.8839    0.8839         0         0         0    0.8839    3.3839         0   -0.8839   -0.8839         0         0    4.2678   -0.8839   -0.8839         0   -2.5000         0         0         0   -2.5000         0   -0.8839         0         0         0         0    0.8839         0         0   -0.8839   -0.8839   -2.5000         0         0   -0.8839   -0.8839         0         0         0         0         0         0         0         0         0         0         0

Columns 6 through 10

-0.8839        0         0         0         0   -0.8839         0         0         0         0         0   -2.5000         0   -0.8839   -0.8839   -2.5000         0         0   -0.8839   -0.8839         0   -0.8839    0.8839   -2.5000         0    4.2678    0.8839   -0.8839         0         0    0.8839    5.8839   -0.8839         0         0   -0.8839   -0.8839    3.3839         0   -2.5000         0         0         0    4.2678         0         0         0   -2.5000         0    4.2678         0   -2.5000         0   -0.8839    0.8839         0         0         0    0.8839   -0.8839

Columns 11 through 12

0        0         0         0         0         0         0         0         0         0         0         0   -2.5000         0         0         0   -0.8839    0.8839    0.8839   -0.8839    3.3839   -0.8839   -0.8839    0.8839

=Table of Assignments R5=

=Problem R5.5a: Eigen vector plot for zero evals of the 2 bar truss system p.21-2 (fead.f08.mtgs.[21])= On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Given: Two member zero evals
Consider a 2 member system. Plot the eigen vectors corresponding to zero evals and interpret the results.

Use standard node and element naming conventions.



Solution: Plot the eigen values and interpret
This is the general stiffness matrix for a two bar system as given.

$$ \vec{K}=\begin{bmatrix} &1&2&3&4&5&6\\ 1&(l^{(1)} )^{2}*K^{(1)} & l^{(1)}m^{(1)}*K^{(1)} & -(l^{(1)} )^{2}*K^{(1)} & -l^{(1)}m^{(1)}*K^{(1)} & 0 & 0\\ 2&l^{(1)}m^{(1)}*K^{(1)} & (m^{(1)} )^{2}*K^{(1)} & -l^{(1)}m^{(1)}*K^{(1)} & -(m^{(1)} )^{2}*K^{(1)} & 0 & 0\\ 3& -(l^{(1)} )^{2}*K^{(1)}& -l^{(1)}m^{(1)}*K^{(1)} & (l^{(1)} )^{2}*K^{(1)}+(l^{(2)} )^{2}*K^{(2)} & l^{(1)}m^{(1)}*K^{(1)}+l^{(2)}m^{(2)}*K^{(2)} & -(l^{(2)} )^{2}*K^{(2)} & -l^{(2)}m^{(2)}*K^{(2)}\\ 4&-l^{(1)}m^{(1)}*K^{(1)} & -(m^{(1)} )^{2}*K^{(1)} & l^{(1)}m^{(1)}*K^{(1)}+l^{(2)}m^{(2)}*K^{(2)} & (m^{(1)} )^{2}*K^{(1)}+(m^{(2)} )^{2}*K^{(2)}& -(l^{(2)} )^{2}*K^{(2)} & -l^{(2)}m^{(2)}*K^{(2)}\\ 5&0 & 0& -(l^{(2)} )^{2}*K^{(2)} & -l^{(2)}m^{(2)}*K^{(2)} & (l^{(2)} )^{2}*K^{(2)} & l^{(2)}m^{(2)}*K^{(2)}\\ 6&0 & 0 &-l^{(2)}m^{(2)}*K^{(2)} & -(m^{(2)} )^{2}*K^{(2)} & l^{(2)}m^{(2)}*K^{(2)} & (m^{(2)} )^{2}*K^{(2)}\\ \end{bmatrix} $$

The global stiffness matrix in numerical form match to this specific problem is as follows.

$$ \vec{K}=\begin{bmatrix} 0.5625 & 0.3248 & -0.5625 & -0.3248 & 0 & 0\\ 0.3248 & 0.1875 & -0.3248 & -0.1875 & 0 & 0\\ -0.5625 & -0.3248 & 3.0625 & -2.1752 & -2.5000 & 2.5000\\ -0.3248 & -0.1875 & -2.1752 & 2.6875 & 2.5000 & -2.5000\\ 0 & 0 & -2.5000 & 2.5000 & 2.5000 & -2.5000\\ 0 & 0 & 2.5000 & -2.5000 & -2.5000 & 2.5000\\ \end{bmatrix} $$

The eigen vector matrix is as follows.

$$ \vec{V}=\begin{bmatrix} -0.1118 & 0.5043 & -0.0000 & -0.5931 & 0.6174 & -0.0139\\ -0.0814 & -0.8634 & 0.0000 & -0.3476 & 0.3565 & -0.0080\\ -0.4628 & 0.0089 & 0.0000 & -0.4803 & -0.5409 & 0.5123\\ 0.5266 & -0.0053 & -0.0000 & -0.5429 & -0.4330 & -0.4904\\ -0.4947 & 0.0071 & -0.7071 & 0.0313 & -0.0765 & -0.4984\\ 0.4947 & -0.0071 & -0.7071 & -0.0313 & 0.0765 & 0.4984\\ \end{bmatrix} $$

$$ \vec{D}=\begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1.4705 & 0\\ 0 & 0 & 0 & 0 & 0 & 10.0295\\ \end{bmatrix} $$

The eigen vectors correspond to the columns of the eigen vector matrix shown above. The eigen value mode shapes are a linear combination of the pure mode shapes (pure rigid body and pure mechanism).



As a thought experiment, we plotted the position plus the deformation derived from the eigen vector matrix. This will supply 4 shapes but they are not constrained to the boundary conditions. Applying the boundary conditions to the K matrix reduces the number of non zero eigen vector outputs to two. The results are shown below. The figure plots came from the eigen vectors in the fifth and sixth columns of the constrained eigen vector matrix. Using the other columns produced a figure that was not constrained to the fixed supports.



=Problem R5.5b: Eigen vector plot for zero evals of the square 3 bar truss system p.21-3 (figure a) (fead.f08.mtgs.[21])= On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Given: Square 3 bar truss system
Consider a 3 member system. Plot the eigen vectors corresponding to zero evals and interpret the results.

a=b=1 E=2 A=3



1.Plot the eigen values
Use standard node and element naming conventions.

Solution: Plot the eigen values
$$ \vec{V}=\begin{bmatrix} -0.7071  &      0    &     0  & -0.7071\\         0  &  1.0000    &     0     &    0\\   -0.7071    &     0     &    0  &  0.7071\\         0    &     0  &  1.0000    &     0\\ \end{bmatrix} $$

$$ \vec{D}=\begin{bmatrix} 0 &   0  &   0   &  0\\     0  &   6  &   0  &   0\\     0   &  0 &    6   &  0\\     0   &  0  &   0  &  12\\ \end{bmatrix} $$



=Table of Assignments=

=Problem R4.3: Post processing of member forces and reactions p.11-3 (fead.f08.mtgs.[10-18])= On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Given: Two member system with initial force P
Consider a 2 member system with an applied force. The displacements are assumed to be known. This allows for the post processing for the reactions both globally and for the members.



Solution: Post process for member forces and reactions
Both methods first require the displacements which are assumed as given since we are post processing for the member forces.

This is the general stiffness matrix for a two bar system as given.

$$ \vec{K}=\begin{bmatrix} &1&2&3&4&5&6\\ 1&(l^{(1)} )^{2}*K^{(1)} & l^{(1)}m^{(1)}*K^{(1)} & -(l^{(1)} )^{2}*K^{(1)} & -l^{(1)}m^{(1)}*K^{(1)} & 0 & 0\\ 2&l^{(1)}m^{(1)}*K^{(1)} & (m^{(1)} )^{2}*K^{(1)} & -l^{(1)}m^{(1)}*K^{(1)} & -(m^{(1)} )^{2}*K^{(1)} & 0 & 0\\ 3& -(l^{(1)} )^{2}*K^{(1)}& -l^{(1)}m^{(1)}*K^{(1)} & (l^{(1)} )^{2}*K^{(1)}+(l^{(2)} )^{2}*K^{(2)} & l^{(1)}m^{(1)}*K^{(1)}+l^{(2)}m^{(2)}*K^{(2)} & -(l^{(2)} )^{2}*K^{(2)} & -l^{(2)}m^{(2)}*K^{(2)}\\ 4&-l^{(1)}m^{(1)}*K^{(1)} & -(m^{(1)} )^{2}*K^{(1)} & l^{(1)}m^{(1)}*K^{(1)}+l^{(2)}m^{(2)}*K^{(2)} & (m^{(1)} )^{2}*K^{(1)}+(m^{(2)} )^{2}*K^{(2)}& -(l^{(2)} )^{2}*K^{(2)} & -l^{(2)}m^{(2)}*K^{(2)}\\ 5&0 & 0& -(l^{(2)} )^{2}*K^{(2)} & -l^{(2)}m^{(2)}*K^{(2)} & (l^{(2)} )^{2}*K^{(2)} & l^{(2)}m^{(2)}*K^{(2)}\\ 6&0 & 0 &-l^{(2)}m^{(2)}*K^{(2)} & -(m^{(2)} )^{2}*K^{(2)} & l^{(2)}m^{(2)}*K^{(2)} & (m^{(2)} )^{2}*K^{(2)}\\ \end{bmatrix} $$

The first two columns and the last two columns are struck out due to fixed positions of nodes 1 and 3. Since we are post processing for the reactions forces and member forces, we know the displacements at node 2. These displacements are used along with the global stiffness matrix to solve for the reaction forces. The reaction forces will be parallel to the members since the truss elements are two force members.

$$ \begin{Bmatrix}F_{1x}\\ F_{1y}\\ F_{3x}\\ F_{3y}\end{Bmatrix}= \begin{Bmatrix} -(l^{(1)})^{2}*K^{(1)} & -l^{(1)}m^{(1)}*K^{(1)}\\ -(l^{(1)})m^{(1)}*K^{(1)} & -(m^{(1)})^{2}*K^{(1)}\\ -(l^{(2)})^{2}*K^{(2)} & -l^{(2)}m^{(2)}*K^{(2)}\\ -(l^{(2)})m^{(2)}*K^{(2)} & -(m^{(2)})^{2}*K^{(2)}\end{Bmatrix} \begin{Bmatrix} u2\\ v2\end{Bmatrix} $$

Method 1 (square root of sum of squares)
Once the x and y components of the global forces on node 1 have been found, the square root of the sum of squares of the component can be used to find the member force magnitude. This is the member force since we are observing a two force member.

$$ \mathbf{Element 1 force}=\sqrt{F_{1x}^2+F_{1y}^2} $$

$$ \mathbf{Element 2 force}=\sqrt{F_{3x}^2+F_{3y}^2} $$

Method 2 (transformation matrix)
The local items are denoted by the overhead bar for each term. The global items do not have this accent.

This first step describes how you obtain the local displacements from the global system.

$$ \begin{bmatrix}\mathbf{T}\end{bmatrix}\begin{Bmatrix}\mathbf{q}\end{Bmatrix}=\begin{Bmatrix}\mathbf{\overline{q}} \end{Bmatrix} $$

$$ \begin{bmatrix}\overline{u_1}\\ \overline{v_1} \\ \overline{u_2} \\ \overline{v_2} \end{bmatrix}=\begin{bmatrix} l & m & 0 & 0\\ -l & m & 0 & 0\\ 0 & 0 & l & m\\ 0 & 0 & -l & m \end{bmatrix} \begin{bmatrix}u_1\\ v_1 \\ u_2 \\ v_2 \end{bmatrix} $$

$$ \begin{bmatrix}\mathbf{T}\end{bmatrix}\begin{Bmatrix}\mathbf{f}\end{Bmatrix}=\begin{Bmatrix}\mathbf{\overline{f}} \end{Bmatrix} $$

$$ \begin{bmatrix}f_{1\overline{x}}\\ f_{1\overline{y}} \\ f_{2\overline{x}} \\ f_{2\overline{y}} \end{bmatrix}= \frac{EA}{L} \begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix}u_1\\ v_1 \\ u_2 \\ v_2 \end{bmatrix} $$

Discussion of Methods
As you could see the method 1 is most simple for easy systems. However, getting into larger systems, the second method will be more useful for automating with an application such as Matlab. The systematic approach of method two allows for large sets of data to be systematically crunched based on the global system matrix. It can be seen that the computational efficiency could vary based on the application. I would choose method two for most approaches since it is more generally applied to large systems.

=Problem Pb-53.5: Draw the FBDs and derive the dif. eq. of motion from p.53-13 (2 DOFs spring-mass-damper system)= On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Given a Linearly Elastic Spring-Mass_Damper System With 2 DOFs
Consider a system of three linearly elastic springs and dampers with two masses.



2.Derive the differential equation of motion and the coefficient matrices
$$ \mathbf M \ddot{\mathbf d} + \mathbf C \dot{\mathbf d} + \mathbf K {\mathbf d} = \mathbf F(t) $$

$$ \mathbf d=\begin{Bmatrix}d_1\\d_2\end{Bmatrix}_{2 \times 1} $$

$$ \mathbf F=\begin{Bmatrix}F_1\\F_2\end{Bmatrix}_{2 \times 1} $$

$$ \mathbf M=\begin{bmatrix}m_1 & 0\\0& m_2\end{bmatrix}_{2 \times 2} $$

$$ \mathbf C= \begin{bmatrix} (c_1+c_2) & -c_2\\ -c_2 & (c_2+c_3) \end{bmatrix}_{2 \times 2} $$

$$ \mathbf K= \begin{bmatrix} (k_1+k_2) & -k_2\\ -k_2 & (k_2+k_3) \end{bmatrix}_{2 \times 2} $$

Construct the first FBD (mass 1)
This FBD is of the left mass (mass 1).



Construct the second FBD (mass 2)
This FBD is of the right mass (mass 2).



Construct the third FBD (left wall support)
This FBD is of the left wall support (item 3).



Construct the fourth FBD (right wall support)
This FBD is of the right wall support (item 4).



Approach
The position of the masses in the system are described by d1 and d2. The walls are fixed. This means that the values for d3 and d4, for the left and right walls respectively, are zero.

Force Derivations
The springs are assumed to be at their unstretched length (equilibrium) in the figure. A positive d1 causes tension in spring 1 and compression in spring 2. This observation is made from holding m2 stationary. Viewing the last spring from equilibrium, if the mass 2 moves to the right and d2 becomes positive, the third spring will compress.

The differential equations of motion will show that the system is coupled.

$$ F_{s1} = k_1d_1 $$

$$ F_{d1} = c_1 \dot d_1 $$

$$ F_{s2} = k_2(d_2-d_1) $$

$$ F_{d2} = c_2(\dot d_2-\dot d_1) $$

$$ F_{s3} = k_3d_2 $$

$$ F_{d3} = c_3 \dot d_2 $$

Summing the Forces
$$ m_1 \ddot d_1 + (c_1+c_2) \dot d_1 - c_2 \dot d_2 + (k_1+k_2)d_1-k_2d_2=F_1 $$

$$ m_2 \ddot d_2 - c_2 \dot d_1 + (c_2+c_3) \dot d_2 - k_2 d_1 + (k_2+k_3) d_2 =F_2 $$

Writing the Two Differential Equations in Matrix Form
Grouping the dependent equations and writing the coefficients in matrix form yields the results shown below.

$$ \mathbf M \ddot{\mathbf d} + \mathbf C \dot{\mathbf d} + \mathbf K {\mathbf d} = \mathbf F(t) $$

$$ \mathbf d=\begin{Bmatrix}d_1\\d_2\end{Bmatrix}_{2 \times 1} $$

$$ \mathbf F=\begin{Bmatrix}F_1\\F_2\end{Bmatrix}_{2 \times 1} $$

$$ \mathbf M=\begin{bmatrix}m_1 & 0\\0& m_2\end{bmatrix}_{2 \times 2} $$

$$ \mathbf C= \begin{bmatrix} (c_1+c_2) & -c_2\\ -c_2 & (c_2+c_3) \end{bmatrix}_{2 \times 2} $$

$$ \mathbf K= \begin{bmatrix} (k_1+k_2) & -k_2\\ -k_2 & (k_2+k_3) \end{bmatrix}_{2 \times 2} $$

Contribution
=Problem CALFEM 3.4 Manual exs1 on a Spring System and Verification=

Given a Linearly Elastic Spring System
Consider a system of 3 linearly elastic springs

Use the following diagram for the node locations

Construct the topology matrix
This matrix contains the node numbers and degrees of freedom.

$$ \begin{bmatrix} 1 & 1 & 2 \\ 2 & 2 & 3 \\ 3 & 2 & 3 \end{bmatrix} $$

Construct the global stiffness matrix and load vector f
$$ \mathbf{K}= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$

The load vector is in position 2 with F=100

$$ \mathbf{f}= \begin{bmatrix} 0 \\ 100 \\ 0 \end{bmatrix}

$$

Generating the element stiffness matrices
The element stiffness matrices are generated using the CALFEM function spring1e. The matrices are generated using k and 2k where k=1500

$$ \mathbf{Ke1}= \begin{bmatrix} 1500 & -1500 \\ -1500 & 1500 \end{bmatrix} $$

$$ \mathbf{Ke2}= \begin{bmatrix} 3000 & -3000 \\ -3000& 3000 \end{bmatrix} $$

Assembling the element stiffness matrices into the global stiffness matrix
The assembly is done using the assem fucntion from CALFEM.

Assembling the second element at the first row:

$$ \mathbf{K}= \begin{bmatrix} 3000 & -3000 & 0 \\ -3000 & 3000 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$

Assembling the first element at the second row:

$$ \mathbf{K}= \begin{bmatrix} 3000 & -3000 & 0 \\ -3000 & 4500 & -1500 \\ 0 & -1500 & 1500 \end{bmatrix} $$

Assembling the second element at the third row:

$$ \mathbf{K}= \begin{bmatrix} 3000 & -3000 & 0 \\ -3000 & 7500 & -4500 \\ 0 & -4500 & 4500 \end{bmatrix} $$

Solve the system of equations using boundary conditions
The CALFEM fucntion solveq is used to solve the system of equations subject to certain boundary conditions.

$$ \mathbf{bc}= \begin{bmatrix} 1 & 0 \\ 3 & 0 \end{bmatrix} $$

$$ \mathbf{a}= \begin{bmatrix} 0 \\ 0.0133 \\ 0 \end{bmatrix} $$

$$ \mathbf{a}= \begin{bmatrix} -40 \\ 0 \\ -60 \end{bmatrix} $$

Evaluate the element forces from the element displacements
$$ \mathbf{ed1}= \begin{bmatrix} 0 & 0.0133 \end{bmatrix} $$

$$ \mathbf{ed2}= \begin{bmatrix} 0.0133 & 0 \end{bmatrix} $$

$$ \mathbf{ed3}= \begin{bmatrix} 0.0133 & 0 \end{bmatrix} $$

Evaluate the spring forces from function spring1s
$$ \mathbf{es1}= \begin{bmatrix} 40 \end{bmatrix} $$

$$ \mathbf{es2}= \begin{bmatrix} -20 \end{bmatrix} $$

$$ \mathbf{es3}= \begin{bmatrix} -40 \end{bmatrix} $$

Solution: Results of verification
This section verifies the CALFEM code by manual calculations

Nodal equilibrium equations
$$f_1^{(1)}=F_1 = R_1$$

$$f_2^{(1)}+ f_2^{(2)}+ f_2^{(3)}=F = 100 $$

$$f_3^{(2)}+f_3^{(3)}=F_2=R_2$$

Setting up the equation to solve
$$ \begin{bmatrix}K_s\end{bmatrix}\begin{Bmatrix}Q_s\end{Bmatrix}=\begin{Bmatrix}F_s \end{Bmatrix} $$

$$ 1000\begin{bmatrix}3 &-3 &-2  &0 \\  -3&7.5  &-4.5 \\  0&-4.5  &4.5 \end{bmatrix}\begin{Bmatrix}u_1\\ u_2\\ u_3\end{Bmatrix}=\begin{Bmatrix}R_1\\ 100\\ R_2\end{Bmatrix}$$

Implementing the boundary conditions
$$ 1000\begin{bmatrix}7.5 \end{bmatrix}\begin{Bmatrix} u_2 \end{Bmatrix}=\begin{Bmatrix}100 \end{Bmatrix} $$

Inverting the global stiffness matrix to find displacements
$$ u_2=0.01333 $$

Displacements 1 and 3 are zero due to boundary conditions.

Calculate forces in springs from equation
The forces in the springs are found by the following equations. Negative forces display compressions and positive tensions.

$$P=k(u_j-u_i)$$

$$ P^{(1)}=3000(0.01333-0)=40$$

$$P^{(2)}=1500(0-0.01333)=-20$$

$$P^{(3)}=3000(0-0.01333)=-40$$

Discuss if the solution is verified
The solution by CALFEM is verified since both the displacements and the forces are the same by both methods.

=R.1.2=

R.1.2a
$$

\begin{array}{lcr} \mbox{Inertial Frame fixed at t=0 shown on diagram}\\ \underline{E}_{y}: to\ the\ right \\ \underline{E}_{x}:upward \\ \underline{E}_{z}:\underline{E}_{y} \times \underline{E}_{x} \\ \\ \mbox{Kinematics}\\ \mbox{Observing rectilinear motion}\\ _{}^{F}\textrm{\underline{r}}=y\underline{E}_{y}\\ _{}^{F}\textrm{\underline{v}}=\dot{y}\underline{E}_{y}=\mbox{v}\underline{E}_{y}\\ \\ \mbox{Kinetics}\\ \underline{F}_{s}=-k(l-l_{o})\underline{u}_{s}\\ l=l_{o}+y\\ \underline{F}_{s}=-k(y)\underline{E}_{y}\\ \underline{F}_{d}=-c\frac{d\underline{r}}{d\underline{t}}=-c\mbox{v}\underline{E}_{y}\\ \underline{f}(t)=f(t)\underline{E}_{y}\\ \underline{F}=-k(y)\underline{E}_{y}-c\mbox{v}\underline{E}_{y}+f(t)\underline{E}_{y}=m\ddot{y}\underline{E}_{y}\\ \mbox{The components in the y-direction yield a differential equation}\\ m\ddot{y}+c\dot{y}+ky=\underline{f}(t) \end{array} $$\\

R.1.2b
$$ 	\begin{array}{lcr}

\mbox{Kinematics}\\ y=y_{k}+y_{c} (1)\\ \\ \mbox{Kinetics}\\ m{y}''+f_{1}=f(t) (2)\\ f_{1}=f_{k}=f_{c} (3)\\ \\ \mbox{Constitutive Equations}\\ f_{k}=ky_{k} (4)\\ f_{c}=c{y_c}' (5)\\ (1): y={y_k}+{y_c}'' (6)\\ (3): f_{k}=f_{c}\Rightarrow k{y_k}=c{y_c}' (7)\\ \Rightarrow{y_c}'=\frac{k}{c}{y_k} (8)\\ \mbox{Using}(1) and (8)\\ {y}={y_k}+{({y_c}')}'={y_k}+{(\frac{k}{c}{y_k})}'={y_k}+\frac{k}{c}{{y_k}'} (9)\\ \mbox{Using} (9) and (2)-(3)\\ m({y_k}''+\frac{k}{c}{{y_k}'})+k{y_k}=f(t) (10)

\end{array} $$\\

=R.1.5=