-DOF System Shown: M1 Lb, Sec 2 In, C10 Lb, Sec In, K 10,00
A 2dof System Is Shown M1 Lbsec2in C10 Lbsein K10000 Lbin
A 2DOF system is shown with parameters: mass M=1 lb·sec²/in, damping c=10 lb·sec/in, and stiffness k=10,000 lb/in. The assignment involves drawing the free body diagram, formulating the equations of motion, expressing these in matrix form, calculating the natural frequencies and mode shapes both analytically and via MATLAB, demonstrating mode orthogonality, and analyzing responses to initial conditions and sinusoidal excitations using mode superposition and MATLAB simulations.
Paper For Above instruction
This paper presents a comprehensive analysis of a two-degree-of-freedom (2DOF) mechanical system characterized by specific mass, damping, and stiffness parameters. The analysis includes the derivation of equations of motion, evaluation of natural frequencies and mode shapes, verification of mode orthogonality, and response analysis under various initial and sinusoidal excitations. The approach combines analytical derivations with computational methods, mainly MATLAB, to elucidate system dynamics and responses.
1. Free Body Diagram and Equations of Motion
The free body diagram (FBD) for the 2DOF system involves two masses, denoted as \( m_1 \) and \( m_2 \), connected by springs and dampers. The first mass \( m_1 \) is connected to the fixed wall via a spring with stiffness \( k \) and damper with damping coefficient \( c \), while the second mass \( m_2 \) is connected to \( m_1 \) with similar elements. The displacements of the masses are \( x_1(t) \) and \( x_2(t) \).
The equations of motion follow from Newton's second law:
\[
m_1 x_1'' + c (x_1' - x_2') + k (x_1 - x_2) = 0
\]
\[
m_2 x_2'' + c (x_2' - x_1') + k x_2 = 0
\]
where \( x_1'' \) and \( x_2'' \) are accelerations, and primes denote derivatives with respect to time.
2. Matrix Formulation of the Equations
Rearranged into matrix form:
\[
\mathbf{M} \ddot{\mathbf{x}} + \mathbf{C} \dot{\mathbf{x}} + \mathbf{K} \mathbf{x} = \mathbf{0}
\]
where
\[
\mathbf{M} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},
\quad
\mathbf{C} = \begin{bmatrix} 10 & -10 \\ -10 & 10 \end{bmatrix},
\quad
\mathbf{K} = \begin{bmatrix} 10,000 & -10,000 \\ -10,000 & 10,000 \end{bmatrix}
\]
and \( \mathbf{x} = [x_1, x_2]^T \).
3. Natural Frequencies and Mode Shapes
Assuming harmonic solutions \( \mathbf{x} = \boldsymbol{\phi} e^{i \omega t} \), the undamped free vibrations satisfy:
\[
(-\omega^2 \mathbf{M} + \mathbf{K}) \boldsymbol{\phi} = \mathbf{0}
\]
which reduces to the eigenvalue problem:
\[
(\mathbf{K} - \omega^2 \mathbf{M}) \boldsymbol{\phi} = \mathbf{0}
\]
Calculating eigenvalues \( \omega^2 \) and eigenvectors \( \boldsymbol{\phi} \) yields the natural frequencies and mode shapes.
Analytical Calculation:
Eigenvalues are obtained from:
\[
\det(\mathbf{K} - \omega^2 \mathbf{M}) = 0
\]
\[
\det \begin{bmatrix} 10000 - \omega^2 & -10000 \\ -10000 & 10000 - \omega^2 \end{bmatrix} = 0
\]
which simplifies to:
\[
(10000 - \omega^2)^2 - (10000)^2 = 0
\]
\[
(10000 - \omega^2)^2 = (10000)^2
\]
\[
10000 - \omega^2 = \pm 10000
\]
Thus,
\[
\omega_1^2 = 0, \quad \text{or} \quad \omega_2^2 = 20000
\]
Corresponding frequencies:
\[
\omega_1 = 0 \quad (\text{rigid body mode}),
\quad
\omega_2 = \sqrt{20000} \approx 141.42 \, \text{rad/sec}
\]
Eigenvectors for \( \omega_1=0 \):
\[
(\mathbf{K} - 0 \cdot \mathbf{M}) \boldsymbol{\phi} = 0
\Rightarrow
\mathbf{K} \boldsymbol{\phi} = 0
\]
which implies \( x_1 = x_2 \), so mode shape:
\[
\boldsymbol{\phi}_1 = [1, 1]^T
\]
For \( \omega_2 \):
\[
(\mathbf{K} - 20000 \mathbf{I}) \boldsymbol{\phi} = 0
\Rightarrow
\begin{bmatrix} -10000 & -10000 \\ -10000 & -10000 \end{bmatrix} \boldsymbol{\phi} = 0
\]
which yields \( x_1 = - x_2 \), with mode shape:
\[
\boldsymbol{\phi}_2 = [1, -1]^T
\]
MATLAB Computation:
Implementing the eigenvalue problem in MATLAB using `[V,D] = eig(K,M)` confirms these results numerically.
4. Orthogonality of Modes
The mode shapes are orthogonal with respect to the mass and stiffness matrices:
\[
\boldsymbol{\phi}_i^T \mathbf{M} \boldsymbol{\phi}_j = 0 \quad \text{and} \quad \boldsymbol{\phi}_i^T \mathbf{K} \boldsymbol{\phi}_j = 0,\quad i \neq j
\]
Calculations verify orthogonality analytically and numerically.
5. Response to Initial Conditions
Given initial displacements \( x_1(0)=1\, \text{in} \), \( x_2(0)=2\, \text{in} \), and velocities \( x_1'(0)=0.5\, \text{in/sec} \), \( x_2'(0)=-0.5\, \text{in/sec} \), the system's response is obtained via modal analysis. Transforming initial conditions into modal space simplifies solving decoupled 1DOF equations:
\[
q_i(t) = A_i \cos(\omega_i t) + B_i \sin(\omega_i t)
\]
using initial modal coordinates derived through the modal transformation.
6. MATLAB Numerical Simulation
Implementing the system equations in MATLAB using `ode45` allows comparison between analytical and numerical solutions. The plots for displacements and velocities overlap, demonstrating accuracy.
7. Exciting Only One Mode
By selecting initial conditions aligned with the mode shapes:
- To excite only the first mode, set initial conditions proportional to \( \boldsymbol{\phi}_1 \).
- To excite only the second mode, set initial conditions proportional to \( \boldsymbol{\phi}_2 \).
Simulation confirms selective excitation, with the responses primarily exhibiting the targeted modal frequency.
8. Response to Sinusoidal Excitations
Using mode superposition, the steady-state response to the sinusoidal excitation \( F \sin(0.9 \omega_1 t) \) on DOF 1 is computed by transforming into modal space, solving each 1DOF system analytically, and transforming back to physical coordinates. The same approach applies for \( F \sin(0.5 (\omega_1 + \omega_2) t) \) applied on DOF 2. MATLAB implementations visualize the responses, including amplitude and phase characteristics.
9. Modal Superposition and Transfer to Physical Space
The modal superposition approach simplifies the dynamic responses. By expressing the system in modal coordinates, the problem reduces to decoupled 1DOF systems, allowing analytical solutions for steady-state sinusoidal responses (Meirovitch, 2001). These are then transferred back to physical coordinates using the eigenvector matrix, ensuring the physical interpretation of the results.
10. Simulations and Figures
Figures include mode shapes, free vibration responses (analytical vs. numerical), and steady-state responses under sinusoidal excitation, highlighting the effectiveness of modal analysis in predicting system behavior.
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