OConnor_SimulationProject
- 1. Final Simulation Vibrations Project Kevin O'Connor April 16, 2016 Professor Ma
- 2. Simulation 1 Case A : Diagram of Setup: Simulation Variables Mass (kg) 0.25 Damper (Ns/m) 10 Spring Constant (N/m) 2500 F_o (N) 26 Tau (seconds) 0.001 Length of Step (seconds) 0.001 Equation of Motion: 𝑘𝑘 𝑚𝑚𝑥𝑥̈ + 𝑐𝑐𝑥𝑥̇ + 𝑘𝑘𝑥𝑥 = 𝐹𝐹(𝑡𝑡) 0.25𝑥𝑥̈ + 10𝑥𝑥̇ + 2500𝑥𝑥 = 28𝛿𝛿(0)
- 3. Simulation 1 Case A Setup:
- 4. Simulation 1 Case A Signal Builder:
- 5. Simulation 1 Case A Scope Output: Peak Displacement (m) Peak Force (N) 0.000845 26
- 6. Simulation 1 Case b: Parameters: Simulation Variables Mass (kg) 0.25 Damper (Ns/m) 10 Spring Constant (N/m) 2500 Fo (N) 96 Signal Builder Step Function:
- 7. Simulation 1 Case b Setup:
- 8. Simulation 1 Case B Scope Output: Peak Displacement (m) Peak Force (N) 0.054189 96
- 9. Part C: MATLAB Code:
- 11. 𝑠𝑠𝑠𝑠 Conclusion: The MATLAB theoretical solution presented in the graph above uses the convolution integral taken with respect to tau to solve the step function in part a. I have used MATLAB to solve the complex integral needed to integrate a damped step function and then have plotted it. It should be noted that the blue wave form is the displacement of the system from time 0 < 𝑡𝑡 < 0.001 and the yellow wave form is the displacement of the system from time 𝑡𝑡 > 0.001. The theoretical simulation using the convolution integral results in a maximum value of approximately 0.0175 meters and the Simulink simulation results in a maximum value of 0.00845 meters. However it should be noted that when the time increment of the simulation is decreased the accuracy of the simulation increases as well.
- 12. Simulation Part2: Parameters: Simulation Inputs First Name Initial, K 11 Last Name Initial, O 15 Spring Constant, k1 (N/m) 452000 Spring Constant, k2 (N/m) 500000 Mass, m1 (kg) 1026 Mass, m2 (kg) 300 Amplitude, Y(m) 0.1 Period, L(m) 6 Velocity, v(kph) 80 Frequency, w(rad/s) 23.27106 Equation of Motion: 𝑚𝑚1 𝑥𝑥1̈ + 𝑘𝑘1(𝑥𝑥2 − 𝑥𝑥1) = 0 1028𝑥𝑥1̈ + 456000(𝑥𝑥2 − 𝑥𝑥1) = 0 𝑒𝑒𝑒𝑒: (1) 𝑚𝑚2 𝑥𝑥2̈ + 𝑘𝑘2(𝑥𝑥2 − 𝑦𝑦) − 𝑘𝑘1(𝑥𝑥2 − 𝑥𝑥1) = 0 300𝑥𝑥2̈ + 44000𝑥𝑥2 + 456000𝑥𝑥1 = 500000𝑦𝑦 𝑒𝑒𝑒𝑒: (2)
- 13. Simulink Setup for Part 1:
- 14. Initial Response of the Vehicle with no Damping: Results: Max Displacement (m) Peak 1 0.17587 Peak 2 0.172002 Peak 3 0.162139
- 15. Displacement Adjustments: To minimize the maximum displacement for this 2 degree of freedom system there are a few adjustments that could be made. First the stiffness of the springs could be increased. However if they are increased too much, this could have an adverse effect on the displacement and cause the bus to have an even greater displacement. The easiest way to reduce the buses displacement would be to add dampers. Adding a damper does not run the risk of increasing the displacement like increasing the spring constant does, which makes it the optimal choice for our application. For this design we will use a model of a car spring-damper system where the ideal damping ratio, ζ = 0.2 to 0.4. Too calculate the damping ratio we will use the following equation: ζ = 𝐶𝐶 𝑚𝑚 It is important to note that for this example we will assume the tire will have no damper (𝑐𝑐2) and the set up will resemble to following: 0.3 = 2�𝑚𝑚1 𝑘𝑘1 We will design the damping ratio to be 0.3 and we will receive the following value for 𝑐𝑐 1: 𝑐𝑐 1 , 𝑐𝑐1 = 12990 𝑁𝑁𝑁𝑁/𝑚𝑚
- 16. Parameters: Simulation Inputs First Name Initial, K 11 Last Name Initial, O 15 Spring Constant, k1 (N/m) 452000 Spring Constant, k2 (N/m) 500000 Damper, c1 (Ns/m) 12990 Mass, m1 (kg) 1028 Mass, m2 (kg) 300 Amplitude, Y(m) 0.1 Period, L(m) 6 Velocity, v(kph) 80 Frequency, w(rad/s) 23.27106 Equations of Motion: 𝑚𝑚1 𝑥𝑥1̈ + 𝑐𝑐1(𝑥𝑥̇2 − 𝑥𝑥̇1) + 𝑘𝑘1(𝑥𝑥2 − 𝑥𝑥1) = 0 1028𝑥𝑥1̈ + 1300(𝑥𝑥̇2 − 𝑥𝑥̇1) + 456000(𝑥𝑥2 − 𝑥𝑥1) = 0 𝑒𝑒𝑒𝑒: (1) 𝑚𝑚2 𝑥𝑥2̈ − 𝑐𝑐1(𝑥𝑥2̇ − 𝑥𝑥̇1) + 𝑘𝑘2(𝑥𝑥2 − 𝑦𝑦) − 𝑘𝑘1(𝑥𝑥2 − 𝑥𝑥1) = 0 300𝑥𝑥2̈ − 1300(𝑥𝑥̇2 − 𝑥𝑥̇1) + 44000𝑥𝑥2 + 456000𝑥𝑥1 = 500000𝑦𝑦 𝑒𝑒𝑒𝑒: (2)
- 17. Simulink Setup for 2 Degree of Freedom System with Damping:
- 18. Response of the system:
- 19. Results: When damping is added there is a significant decrease in the displacement of the bus, which can be seen below in the table. Before Damping Displacement (m) Damping Added Displacement (m) % Reduction in Displacement (%) Peak 1 0.17587 0.120211 30.36 Peak 2 0.172002 0.100011 40.75 Peak 3 0.162139 0.089022 37.69 When we compare the scope without damping to the one with damping there is also a significant change in the shape of the “Total Displacement” plot. Without damping, the plot had many peaks, and the total displacement never came close to the input displacement of 0.1 meters. When damping is added however there is a peak in the “Total Displacement” initially but this is quickly compensated for by the damping and within 2 seconds, the “Total Displacement” plot is only displacing the input displacement of 0.1 meters. This is the important take away from this simulation, that when dampers are added the car does not continue to have peak displacements across the entire simulation, which is what was observed in the initial simulation. The dampers are working to absorb the input of the road and rebound of the springs, which is clearly evident in the shape of the scope plots.