User:Eml4507.s13.team3.chen/R3

=Problem 7 (fead.f08 p.14-3, HW)= On my honor, I have neither given nor received unauthorized aid in doing this assignment.

Given
Bar element stiffness matrix: $$\mathbf k^{(e)} = k^{(e)} \begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)} \\l^{(e)}m^{(e)} & (m^{(e)})^2 & -l^{(e)}m^{(e)} & -(m^{(e)})^2 \\-(l^{(e)})^2 & -l^{(e)}m^{(e)} & (l^{(e)})^2 & l^{(e)}m^{(e)} \\-l^{(e)}m^{(e)} & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2 \end{bmatrix}$$,

Where $$k^{(e)} = \frac{EA}{L}$$, $$l^{(e)} = \cos \theta^{(e)}$$, and $$m^{(e)} = \sin \theta^{(e)}$$

Find

 * Verify that the bar element stiffness matrix $$\mathbf k^{(e)}$$ in global coordinates can be obtained by $$\mathbf k^{(e)} = {\mathbf T}^{{(e)}T} \, \hat{\mathbf k}^{(e)} \, {\mathbf T}^{(e)}$$.
 * Verify that you also get the same expression for the bar element stiffness matrix in global coordinates with $$\mathbf k^{(e)} = \widetilde{\mathbf T}^{{(e)}T} \, \widetilde{\mathbf k}^{(e)} \, \widetilde{\mathbf T}^{(e)}$$, where $$\widetilde{\mathbf T}^{(e)}$$ is the 4x4 transformation matrix in K.2008.2 p.32, and $$\widetilde{\mathbf k}^{(e)}$$ is the 4x4 bar element stiffness matrix in local coordinates in K.2008.2 p.31.

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

Check that given matrix dimensions correspond
$$\mathbf k^{(e)} = {\mathbf T}^{{(e)}T} \, \hat{\mathbf k}^{(e)} \, {\mathbf T}^{(e)} \Rightarrow \begin{bmatrix} 4 \times 4 \end{bmatrix} = \begin{bmatrix} 4 \times 2 \end{bmatrix} \begin{bmatrix} 2 \times 2 \end{bmatrix} \begin{bmatrix} 2 \times 4 \end{bmatrix} $$

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


 * $$= k^{(e)} \begin{bmatrix} l^{(e)} & -l^{(e)} \\m^{(e)} & -m^{(e)} \\-l^{(e)} & l^{(e)} \\-m^{(e)} & m^{(e)} \end{bmatrix}$$, where $$k^{(e)}$$ can be moved in front because it is a scalar.

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


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

Given
4x4 bar element stiffness matrix in local coordinates: $$\widetilde{\mathbf k}^{(e)} = k^{(e)} \begin{bmatrix} 1 & 0 & -1 & 0 \\0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

Transformation matrix: $$\widetilde{\mathbf T}^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 \\-m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)} \\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix}$$

Transposed transformation matrix: $$\widetilde{\mathbf T}^{(e)T} = \begin{bmatrix} l^{(e)} & -m^{(e)} & 0 & 0 \\m^{(e)} & l^{(e)} & 0 & 0 \\ 0 & 0 & l^{(e)} & -m^{(e)} \\ 0 & 0 & m^{(e)} & l^{(e)} \end{bmatrix}$$

Check that given matrix dimensions correspond
$$\mathbf k^{(e)} = \widetilde{\mathbf T}^{{(e)}T} \, \widetilde{\mathbf k}^{(e)} \, \widetilde{\mathbf T}^{(e)} \Rightarrow \begin{bmatrix} 4 \times 4 \end{bmatrix} = \begin{bmatrix} 4 \times 4 \end{bmatrix} \begin{bmatrix} 4 \times 4 \end{bmatrix} \begin{bmatrix} 4 \times 4 \end{bmatrix} $$

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


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

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


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

=Problem 8 (sec.53a, p.53-2b, Pb-53.4)= On my honor, I have neither given nor received unauthorized aid in doing this assignment.

Given

 * Consider an L2-ODE-CC with the following roots: $$\lambda_1,_2 = -0.5+i2$$. Find the system natural frequency $$\omega$$.


 * Let the system be excited by a periodic force of $$f(t) = f_0cos(\bar \omega t)$$. Find the expression for the particular solution $$y_p(t)$$.


 * Consider:
 * $$f_0/m = r_0 = 1$$


 * $$\bar \omega = 0.9\omega$$


 * Initial Conditions
 * $$y(0) = 1$$
 * $$y'(0) = 0$$

Find

 * Find the particular solution $$y_p(t)$$ and the amplification factor $$\mathbb A$$ for the given input.
 * Find the complete solution $$y(t)$$ satisfying the given initial conditions.
 * Plot in separate figures: $$y_h(t), y_p(t), y(t)$$
 * Plot in the same figure: $$y_h(t), y_p(t), y(t)$$

System natural frequency $$\omega$$
From R1.5 (Case 3) :


 * $$ (\lambda-\lambda_1)(\lambda-\lambda_2)$$
 * $$ (\lambda+.5+i2)(\lambda+.5-i2)$$
 * $$ \lambda^2 + .5\lambda - i2\lambda + .5\lambda + .25 + .5i2 +i2\lambda - .5i2 + 4$$
 * $$ \lambda^2 + \lambda + 4.25 = 0$$

The standard L2-ODE-CC is then
 * $$y'' + y' + 4.25y = 0 $$

Relate L2-ODE-CC to standard form of equation of motion:


 * $$y + y' + 4.25y = y + ay' + by$$
 * $$y'' + 2\zeta \omega y' + \omega^2 y$$
 * $$a = 2\zeta \omega$$
 * $$b = \omega^2$$
 * $$\omega = \sqrt{4.25}$$
 * $$\zeta = \frac1{2\sqrt{4.25}}$$

The system natural frequency: $$\omega = \sqrt{4.25}$$

Expression for particular solution $$y_p(t)$$
To find $$y_p(t)$$, use the formula for trial particular solution:


 * $$y_p(t) = de^{i(\bar \omega t - \phi)}$$,

Where $$d$$ is defined as:
 * $$d=\frac{f_0/m}{\sqrt{(\omega^2 - \bar\omega^2)^2 + (2 \zeta \omega \bar\omega)^2}}$$

And $$\phi$$ is defined by:
 * $$\tan\phi=\frac{2\zeta\omega\bar\omega}{\omega^2 - \bar\omega^2}$$

And the damped circular frequency $$\bar \omega$$
 * $$\bar \omega = \omega \sqrt{1 - \zeta^2}$$

Combining the previous equations, the entire expression is written as:
 * $$y_p(t) = \frac{f_0/m}{\sqrt{(\omega^2 - \bar\omega^2)^2 + (2 \zeta \omega \bar\omega)^2}}e^{i(\bar \omega t - \phi)}$$

But since this gives a complex solution, $$e^{i(\bar \omega t - \phi)}$$ is redefined with Euler's formula:
 * $$e^{i(\bar \omega t - \phi)} = \cos(\bar \omega t - \phi) + i \sin(\bar \omega t - \phi)$$

Since $$f(t) = f_0 \cos \bar \omega t$$ is a real forcing function, $$i \sin(\bar \omega t - \phi)$$ can be removed.

So the final expression for the particular solution $$y_p(t)$$ is:
 * $$y_p(t) = \frac{f_0/m}{\sqrt{(\omega^2 - \bar\omega^2)^2 + (2 \zeta \omega \bar\omega)^2}}\cos(\bar \omega t - \phi)$$

Which can be simplified to:
 * $$y_p(t) = d \cos(\bar \omega t - \phi)$$

Particular solution $$y_p(t)$$ for the case and amplification factor $$\mathbb A$$
Consider the case:
 * $$f_0/m = r_0 = 1$$


 * $$\bar \omega = 0.9\omega$$

The amplification factor $$\mathbb A$$ is given as:
 * $$\mathbb A := \frac1{\left[(1 - \rho^2)^2+(2 \zeta \rho)^2 \right]^{1/2}}$$,

Where $$\rho = \frac {\bar \omega}{\omega} = 0.9$$
 * $$\mathbb A = 2.10032$$

The particular solution $$y_p(t)$$ for this case is:
 * $$y_p(t) = d \cos(\bar \omega t - \phi)$$

Where $$d$$ is redefined in terms of $$\mathbb A$$ as:
 * $$d = \mathbb A \frac{f_0}{k}$$, where $$k = bm = \omega^2 m$$


 * $$d = \mathbb A \frac{f_0}{\omega^2 m}$$

Since $$f_0/m = r_0 = 1$$,


 * $$d = \frac{\mathbb A }{\omega^2}$$


 * $$y_p(t) = \frac{\mathbb A }{\omega^2} \cos(\bar \omega t - \phi)$$

Plugging in the values and simplifying results in the particular solution $$y_p(t)$$:
 * $$y_p(t) = 0.494193 \cos(1.8554 t - 1.1603)$$

Complete solution $$y(t)$$ satisfying $$y(0) = 1, y'(0) = 0$$
By superposition: $$y(t)=y_h(t)+y_p(t)$$
 * $$y_h(t)=C_1 e^{(-0.5 t)} sin(2 t)+C_2 e^{(-0.5 t)} cos(2 t)$$


 * $$y_p(t) = 0.494193 \cos(1.8554 t - 1.1603)$$


 * $$y(t)=C_1 e^{(-0.5 t)} sin(2 t)+C_2 e^{(-0.5 t)} cos(2 t)+ 0.494193 \cos(1.8554 t - 1.1603)$$

Using the conditions $$y(0) = 1, y'(0) = 0$$ with the complete solution and the derivative of the complete solution:


 * $$y(t)=C_1 e^{(-0.5 t)} sin(2 t)+C_2 e^{(-0.5 t)} cos(2 t)+ 0.494193 \cos(1.8554 t - 1.1603)$$
 * $$y'(t)=e^{-0.5t}[(-0.5 C_1 - 0.5 C_2) \sin(2t) + (2 C_1 + 2 C_2) \cos(2t)] - 0.916929 \sin(1.8554t - 1.1603)$$


 * $$y(0)=1=0+C_2+\cos(1.1603)$$
 * $$y(0)=1=0+C_2+0.197213$$
 * $$C_2 = 0.802787$$


 * $$y'(0)=0=-0.916926 \sin(1.1603) + 2 C_1 + 2 C_2$$


 * $$y'(0)=0=0.840752 + 2 C_1 + 1.60557$$


 * $$C_1 = -1.22316$$

The complete solution is:
 * $$y(t)=-1.22316 e^{(-0.5 t)} sin(2 t)+0.802787 e^{(-0.5 t)} cos(2 t)+ 0.494193 \cos(1.8554 t - 1.1603)$$

Plots








=References=