User:Eml4507.s13.team2/Report7

Honor Pledge
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Given: A stiffness and displacement matrix
A local element stiffness matrix and displacement matrix for a frame element are presented as $$\tilde{k}_{ij}$$ and $$\tilde{d_j}$$.

Verify: Dimensions after matrix multiplication
Verify the dimensions of: $$\tilde{k}_{ij} \tilde{d_j}$$

for $$i=1,...,6$$ and $$j=1,...,6$$.

Stiffness Matrix
The matrix $$\tilde{k}_{ij}$$ with the given dimensions is as follows.

$$ \tilde{\mathbf{k}}_{ij}=\begin{bmatrix} \frac{EA}{L} & 0 & 0 & -\frac{EA}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & -\frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{2EI}{L} \\ -\frac{EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0 \\ 0 & -\frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{4EI}{L} \end{bmatrix} $$

Individual variable dimensions are given below. Constants are dimensionless and so can be ignored.

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

$$[A]= L^2$$

$$[I]= L^4$$

As a result, combined terms have dimensions as follows.

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

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

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

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

Displacement Matrix
Likewise, the matrix $$\tilde{d_j}$$ is

$$ \tilde{\mathbf{d}_j}= \begin{Bmatrix} \tilde{d_{1}} \\ \tilde{d_{2}} \\ \tilde{d_{3}} \\ \tilde{d_{4}} \\ \tilde{d_{5}} \\ \tilde{d_{6}} \end{Bmatrix} $$

$$[\tilde{d}_j]=L$$ for $$j=1,2,4,5$$

$$[\tilde{d}_j]=1$$ for $$j=3,6$$

Verifying Dimensions
$$ \tilde{\mathbf{k}}_{ij} \tilde{\mathbf{d}_j}=\begin{bmatrix} \frac{EA}{L} & 0 & 0 & -\frac{EA}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & -\frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{2EI}{L} \\ -\frac{EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0 \\ 0 & -\frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{4EI}{L} \end{bmatrix} \begin{Bmatrix} \tilde{d_{1}} \\ \tilde{d_{2}} \\ \tilde{d_{3}} \\ \tilde{d_{4}} \\ \tilde{d_{5}} \\ \tilde{d_{6}} \end{Bmatrix} $$

Since the stiffness matrix contains many zero elements, many products will be dimensionless.

$$ \left [ \tilde{k}_{12}\tilde{d}_2 \right ]=\left [\tilde{k}_{13}\tilde{d}_3 \right ]=\left [\tilde{k}_{15}\tilde{d}_5 \right ]=\left [\tilde{k}_{16}\tilde{d}_6 \right ] = 1 $$

$$ \left [ \tilde{k}_{21}\tilde{d}_1 \right ]=\left [ \tilde{k}_{24}\tilde{d}_4\right ]=1 $$

$$ \left [\tilde{k}_{31}\tilde{d}_1\right ]=\left [\tilde{k}_{34}\tilde{d}_4\right ]=1 $$

$$ \left [\tilde{k}_{42}\tilde{d}_2 \right ]=\left [\tilde{k}_{43}\tilde{d}_4\right ]=\left [\tilde{k}_{45}\tilde{d}_5\right ]=\left [\tilde{k}_{46}\tilde{d}_6\right ]=1 $$

$$ \left [\tilde{k}_{51}\tilde{d}_1\right ]=\left [\tilde{k}_{54}\tilde{d}_4 \right ]= 1$$

$$ \left [\tilde{k}_{61}\tilde{d}_1\right ]=\left [\tilde{k}_{64}\tilde{d}_4 \right ]= 1$$

The remaining terms are:

$$ \left [\tilde{k}_{11}\tilde{d}_1\right ]=\left [\tilde{k}_{14}\tilde{d}_4 \right ]= \frac{F}{L} $$

$$ \left [\tilde{k}_{22}\tilde{d}_2\right ]=\left [\tilde{k}_{25}\tilde{d}_5 \right ]= \frac{F}{L} $$ and $$ \left [\tilde{k}_{23}\tilde{d}_3\right ]=\left [\tilde{k}_{26}\tilde{d}_6 \right ]= F $$

$$ \left [\tilde{k}_{32}\tilde{d}_2\right ]=\left [\tilde{k}_{35}\tilde{d}_5 \right ]= F $$ and $$ \left [\tilde{k}_{33}\tilde{d}_3\right ]=\left [\tilde{k}_{36}\tilde{d}_6 \right ]= FL $$

$$ \left [\tilde{k}_{41}\tilde{d}_1\right ]=\left [\tilde{k}_{44}\tilde{d}_4\right ]= \frac{F}{L}$$

$$ \left [\tilde{k}_{52}\tilde{d}_2\right ]=\left [\tilde{k}_{55}\tilde{d}_5 \right ]= \frac{F}{L} $$ and $$ \left [\tilde{k}_{53}\tilde{d}_3\right ]=\left [\tilde{k}_{56}\tilde{d}_6 \right ]= F $$

$$ \left [\tilde{k}_{62}\tilde{d}_2\right ]=\left [\tilde{k}_{65}\tilde{d}_5 \right ]= F $$ and $$ \left [\tilde{k}_{63}\tilde{d}_3\right ]=\left [\tilde{k}_{66}\tilde{d}_6 \right ]= FL $$

The resulting 6x1 matrix $$\tilde{\mathbf{f}}_j$$ has dimensions of:

$$\left [ \tilde{f}_1 \right ]=\frac{F}{L} $$

$$\left [ \tilde{f}_2 \right ]=\frac{F}{L} + F $$

$$\left [ \tilde{f}_3 \right ]=F+FL $$

$$\left [ \tilde{f}_4 \right ]= \frac{F}{L} $$

$$\left [ \tilde{f}_5 \right ]=\frac{F}{L} + F $$

$$\left [ \tilde{f}_6 \right ]=F+FL $$

Honor Pledge
On our honor this problem was attempted alone using no outside sources other than the class notes located here and our own previous works

Given: A 10-bar Truss System


Consider the truss system shown above under free vibration with an initial displacement of $$F=5$$.

Solve the truss
Using the first 3 lowest modes, solve for the motion of the truss by modal superposition.

Plot the time history
Plot the time history for the vertical displacement of node 2 over 5 periods, i.e. $$0\leq t\leq T=5T$$

Solution
Given: $$ A= 0.5 $$ $$ E= 5 $$ $$ \rho = 2 $$ $$ L_{12}=L_{24}=L_{46}=1 $$ $$ L_{23}=L_{45}=1$$

Solving for all the truss lengths we get the following values:



We first want to project the initial conditions for the displacements into global coordinates: $$d(0)=\sum_{j=1}^{n}z_j(0)\bar\phi _j$$ $$d'(0)=\sum_{j=1}^{n}z'_j(0)\bar\phi _j$$

The initial condition can be obtained from mass-orthogonality: $$\bar\phi_{i}^{T}Md(0)=\sum_{j=1}^{n}z_j(0)\bar\phi_{i}^{T}M\bar\phi_j$$ and $$\bar\phi_{i}^{T}Md(0)=\sum_{j=1}^{n}z_j(0)\delta_{ij}=z_i(0)$$ In terms of the velocity, the initial conditions for i=1,...,n is given by: $$z_i(0)=\bar\phi_{i}^{T}Md(0), z'_i(0)=\bar\phi_{i}^{T}Md'(0)$$