ME 3143 System Dynamics & Modeling

Transcription

1 ME 3143 System Dynamics & Modeling Project II Car Suspension System Analysis Using Bond-Graph Approach Group # 1 Group members Cooperation (%) Lauren Baxter /100 Michael Bryant /100 Ross Hedges /100 Aida Hendrickson /100 (Group leader) Nick Miller /100 Leah Troskot /100 Cullen Walsh /100 Project Advisor Milad Khaledyan Semester/Year Fall

2 Table of Contents Introduction... 3 Executive Summary... 4 Design Method Discussion... 4 Results and Analysis... 5 Bond Graph... 5 Constitutive Laws... 6 Linear Constitutive Laws... 6 Nonlinear Constitutive Laws Junction and 0-Junction Relationships... 6 Compliances, Intertances, and Resistances... 6 Given Parameters... 7 Differential Equations... 7 Simulink Model: Nonlinear System... 8 Plots: Nonlinear System Response for U=0.9 m/s... 9 Plots: Nonlinear System Response for U=13.5 m/s Simulink Model: Linearized Model Plots: Linearized System Response for U=0.9 m/s System Response Linearized System Conclusion References Appendix MATLAB Code Nonlinear System Linear System Linearization of the System about Equilibrium Points

3 Introduction The objective of the quarter car model system is to not only determine the correlating bond graph, power strokes, causalities, input and state variables, and determining the differential equations that govern the system, but to linearize the non-linear spring, damper, and tire spring elements in the quarter car as well. Both non-linearized and linearized systems were analyzed via Simulink models and the spring and damper forces and input velocities were graphed with respect to displacement or time. Once the system was completely solved, the stability of the linear system was analyzed. Any elements that were not defined to be non-linear were assumed to be linear. Simulink, MATLAB, and 20-Sim were used to aid in the visual and simulation aspects of the project. 3

4 Executive Summary The bottom quarter of a car was modeled via a non-linear unsolved system represented by un-stretched spring, damper, mass, and stretched spring, restricted only to vertical motion. The system was subjected to an uneven surface, which was represented by a sine curve. The objective of the quarter car problem was to develop a bond graph to model the system, assign appropriate power strokes, numberings, and causalities, and define the state and input variables along with the constitutive laws for each of the non-linear and linear elements. From these representations of the quarter car, differential equations were built and were linearized to solve the system. Under different input conditions, the system was tested via Simulink models and the different forces from the springs, damper, and masses along with input velocities were analyzed versus either displacement or time. In conclusion, the stability of the linearized system was analyzed. Design Method Discussion In the problem, the team was given the task to model one quarter of a car using the knowledge and methods taught in ME To better visualize and simplify the system, the quarter-car had to be broken up into familiar elements that could be used to model different aspects of the car. In this particular case, the car was modeled using a sprung mass to symbolize the mass of ¼ of a car, and an unsprung mass to symbolize the mass of the tire, wheel, and parts of the brakes and suspension. In between these two elements, a suspension spring (ks) and damper (bs) were used to model the suspension system. The tire must also be modeled as a spring with spring constant (kt). It should be noted that the suspension spring and damper, along with the tire spring are to be treated as nonlinear elements. The system is constrained to only move vertically, and is in the presence of gravity. The construction of the bond graph began with an effort source consisting of the car mass in the presence of gravity. The quarter mass of the car was also modeled as an inertance connected to the same one-junction. A zero junction separates the quarter car body and the tire. The suspension system that separates the two masses was modeled by a spring and damper that move with a common velocity, therefore the two nonlinear elements were connected to a 1-Junction. The tire was modeled with an inertance and an effort source. The spring like properties of the tire were modeled as a compliance connected to the 0 junction that connected the final flow source to the bond graph. After constructing the bond graph causality can be assigned. Starting with the sources and following the 1 and 0 junction rules the system was found to be normally causal. After causalities were assigned, the system was found to be fourth order with state variables q2, P5, q9, and P11. State variables occur when inertance and compliance elements have causal strokes that allow the use of the integral form of each governing equation. The integral forms are preferred because they reduce the overall noise in the system. These integral forms were later used to write the constitutive laws for each element. Since the system stays in one energy domain (Mechanical Translation) there was no need for any 4

5 transformers are gyrators, therefore all constitutive laws were derived from causal strokes and 1 and 0-junction rules. However, the damper and two compliances were both written in their nonlinear forms. Before linearization, the state equations were split up into 2 scenarios depending on which q9 equation was satisfied. The Phi equations were then set equal to zero, in order to solve for the equilibrium points of the state variables. Then each phi equation was differentiated in respect to each of the state variables. These partial derivatives make up each row of the Jacobian matrix used to linearize the system. Once the Jacobian matrix was formed, the equilibrium points of each variable were plugged in to get the matrix in strictly numerical terms. The Eigenvalues were then found by solving for the determinant of (λi- A). The stability of a system can be indicated by the real parts of its eigenvalues. If the eigenvalues have negative real parts the system is stable. If the eigenvalues have positive real parts the system is unstable, and if the eigenvalues have 0 real parts, the system is marginally stable. The linearized eigenvalues we received for scenario 1 (q 9 q s0 ) were (0+/ j, 0+/ j). These eigenvalues were then used to determine that the system was marginally stable at this point. The eigenvalues for scenario 2 (q 9 q s0 ) were (0+/ j, 0+/ ). These numbers indicate that the system was marginally stable in this particular scenario. Results and Analysis Bond Graph 5

6 Constitutive Laws Linear Constitutive Laws f 5 = 1 I 5 p 5 f 11 = 1 I 11 p 11 e 12 = m s g e 4 = m us g Nonlinear Constitutive Laws f 1 = { (h ) πucos (πut d d ) 0 for 0 < ( U d ) t 1 otherwise e 8 = B(f 8 ) 3 e 2 = { k tq 2 0 q 2 0 q 2 < 0 k e 9 = { s1 q 9 k s1 q s0 + k s2 (q 9 q s0 ) q 9 q s2 q 9 q s2 1-Junction and 0-Junction Relationships e 3 = e 2 = e 1 f 3 + f 2 = f 1 e 3 = e 6 + e 4 + e 5 f 3 = f 4 = f 5 = f 6 e 7 = e 8 + e 9 f 7 = f 8 = f 9 e 6 = e 7 = e 10 f 6 = f 7 + f 10 e 10 = e 11 + e 12 f 10 = f 11 = f 12 Compliances, Intertances, and Resistances C 2 = 1 k t I 5 = m us C 9 = 1 k s1 or 1 k s2 R 8 = b s I 11 = m s 6

7 Given Parameters q s ini = m sg k s1 q us ini = (m s + m us )g k t U = 0.9 m for part 2 s U = 13.5 m for part 3 s Differential Equations φ 1 = q 2 = 1 p m 5 + { (h ) πucos (πut d d ) us 0 φ 2 = p 5 = B [ 1 p m p us m 11 ] s φ 3 = q 9 = 1 m s p m us p 5 φ 4 = p 11 = m s g + B [ 1 p m p us m 11 ] s for 0 < ( U d ) t 1 otherwise k { s1 q 9 k s1 q s0 + k s2 (q 9 q s0 ) k + { s1 q 9 k s1 q s0 + k s2 (q 9 q s0 ) q 9 q s0 q 9 q s0 } m us g + q 2 k t q 9 q s0 q 9 q s0 7

8 Simulink Model: Nonlinear System 8

9 Plots: Nonlinear System Response for U=0.9 m/s Displacement [m] Displacement [m] 9

10 Plots: Nonlinear System Response for U=13.5 m/s Displacement [m] Displacement [m] 10

11 Simulink Model: Linearized Model 11

12 Plots: Linearized System Response for U=0.9 m/s Displacement [m] Displacement [m] 12

13 System Response The response of the nonlinear system was modeled using Simulink. A simulation time 2 seconds was chosen because of the highly oscillatory response of the system. Simulations were made with the car hitting the bump at a low speed, 0.9 m/s, and a high speed, 13.5 m/s so a comparison could be made of the effects of the two speeds on the suspension. A model was also linearized about the equilibrium points of the system in order to express a general representation of the behavior of the system. The linearized model produced similar plots to those of the two velocity inputs. At low speed, the input velocity rose and fell to 0.5 m/s and -0.5 m/s in one second before reaching zero velocity. At high speed, the velocity input dropped from 10 m/s to -10 m/s before reaching zero velocity in less than a quarter of a second. In both cases, the Damper Force vs Velocity plots followed a cubic model despite the oscillations in velocity. However, the rapid and drastic velocity change of the 13.5 m/s system produced a more forceful and abrupt response from the damper than its 0.9 m/s counterpart over time. The repeated abuse of the damper by hitting bumps at high speed would wear down the materials more quickly than if bumps were hit at low speed. The response of the suspension and tire, which were modeled as springs, of the 13.5 m/s system also demonstrated actions that could result in increased material abuse. The maximum force response of the suspension systems was very similar, however, the time both systems took to reach the maximum was very different. The 13.5 m/s system reached its maximum, and also its minimum, in less than a quarter of a second. On the other hand, the 0.9 m/s reached its maximum after one second after responding with a much lower frequency of oscillation. The tires responded similarly to the suspension: the maximum and minimum forces were equal in value in each system, but the time it took the 13.5 m/s system to reach the values was much shorter than the 0.9 m/s system. 13

14 Linearized System Calculations can be found in the Appendix Equilibrium Points Case 1: q 9 q s0 q 2e = p 5e = 0 q 9e = p 11e = 0 Case 2: q 9 q s0 q 2e = p 5e = 0 q 9e = p 11e = 0 Characteristic Equation λ λ λ λ Eigenvalues λ = 0 ± j λ = 0 ± j λ = 0 ± j λ = 0 ± j Stability Marginally Stable because the real part of the eigenvalues are 0 Marginally Stable because the real part of the eigenvalues are 0 14

15 Conclusion This project demonstrated how engineers may model complex systems using nonlinear ideal elements. The simplified system was used to develop a bond graph and then differential equations. Because the differential equations included multiple variables, it was evident that once one aspect of the system changes, the other variables were affected as well. There was only one input, but due to varying road bumps, it created many different cosinusoidal input levels. The model portrays how these nonlinear elements react under all ranges of inputs. The responses in machine components further from the input than their counterparts was diminished in scale because of the absorption and dissipation of energy in compliance and inertance components. Having completed the project, the team feels more confident in interpreting nonlinear complex systems. 15

16 References [1] Brown, Forbes T. "Engineering System Dynamics: A Unified Graph-Centered Approach, Second Edition 2nd Edition." Engineering System Dynamics: A Unified Graph-Centered Approach, Second Edition: Forbes T. Brown: : Books. N.p., n.d. Web. 23 Oct [2] Murphy, Michael. "ME 3143 Lecture 5-6." Moodle. N.p., 5 Sept Web. 16

17 Appendix MATLAB Code Nonlinear System U=0.9; Ms=320; %Kg Mus=(Ms/6); Ks1= ; %N/m^2 Ks2=10*Ks1; Kt=10*Ks1; g=9.81; Qsi=(Ms*g)/Ks1; Qusi=(Ms+Mus)*g/Kt; Qso=1.3*Qsi; B=1500; %N/(m/s)^3 d=1; %m h=.25; %m Qus=((Ms+Mus)*g)/Kt; clf grid on; figure(1); subplot(2,4,1); plot(vin_t(:,1),vin_t(:,2)); title('input Velocity vs Time'); xlabel('time [s]'); ylabel('input Velocity [m/s]'); Fd = B*(((1/Mus).*p5(:,2)- (1/Ms).*p11(:,2)).^3); subplot(2,4,2); plot(q9dot(:,2),fd); title('dampening Force vs Velocity'); xlabel('velocity [m/s]'); ylabel('dampning Force [N]'); subplot(2,4,3); plot(time(:,2),fd); title('dampening Force vs Time'); xlabel('time [s]'); ylabel('dampening Force [N]'); subplot(2,4,4); plot(time(:,2),e9(:,2)); title('suspension Force vs Time'); xlabel('time [s]'); ylabel('force [N]'); Ft = Kt*q2(:,2); subplot(2,4,5.25); plot(time(:,2),ft); title('tire Spring Force vs Time'); 17

18 xlabel('time [s]'); ylabel('force [N]'); subplot(2,4,6.5); plot(q9(:,2),e9(:,2)); title('suspension Force vs Suspension Displacement'); xlabel('time [s]'); ylabel('force [N]'); subplot(2,4,7.75); plot(q2(:,2),ft); title('tire Spring vs Tire Displacement'); xlabel('time [s]'); ylabel('force [N]'); function [y,vint] = fcn(p5_mus,t) %#codegen U=13.5; Ms=320; %Kg Mus=(Ms/6); Ks1= ; %N/m^2 Ks2=10*Ks1; Kt=10*Ks1; g=9.81; Qsi=(Ms*g)/Ks1; Qso=1.3*Qsi; B=1500; %N/(m/s)^3 d=1; %m h=.25; %m Qus=((Ms+Mus)*g)/Kt; if (((U/d)*t) >= 0) && (((U/d)*t) <= 1) Vint=(h/d)*pi*U*cos((pi*U*t)/d); elseif (((U/d)*t) > 1) Vint=0; else Vint=0; end y = -p5_mus+vint; function [y,e9] = fcn(p5_mus,q2_kt,q9,p11_ms) %#codegen U=13.5; Ms=320; %Kg Mus=(Ms/6); Ks1= ; %N/m^2 Ks2=10*Ks1; Kt=10*Ks1; g=9.81; Qsi=(Ms*g)/Ks1; Qso=1.3*Qsi; B=1500; %N/(m/s)^3 d=1; %m h=.25; %m Qus=((Ms+Mus)*g)/Kt; 18

19 if (q9 <= Qso) e9=ks1*q9; else e9=(ks1*qso)+ks2*(q9-qso); end y = (-B*(p5_mus-p11_ms)^3)-e9-(Mus*g)+q2_kt; function y = fcn(p5_mus,q9,p11_ms) %#codegen U=13.5; Ms=320; %Kg Mus=(Ms/6); Ks1= ; %N/m^2 Ks2=10*Ks1; Kt=10*Ks1; g=9.81; Qsi=(Ms*g)/Ks1; Qso=1.3*Qsi; B=1500; %N/(m/s)^3 d=1; %m h=.25; %m Qus=((Ms+Mus)*g)/Kt; if (q9 <= Qso) e9=ks1*q9; else e9=(ks1*qso)+ks2*(q9-qso); end y = (-Ms*g)+(B*(p5_mus-p11_ms)^3)+e9; Linear System U=0.9; Ms=320; %Kg Mus=(Ms/6); Ks1= ; %N/m^2 Ks2=10*Ks1; Kt=10*Ks1; g=9.81; Qsi=(Ms*g)/Ks1; Qusi=(Ms+Mus)*g/Kt; Qso=1.3*Qsi; B=1500; %N/(m/s)^3 d=1; %m h=.25; %m Qus=((Ms+Mus)*g)/Kt; clf grid on; figure(1); subplot(2,4,1); plot(vin_t(:,1),vin_t(:,2)); title('input Velocity vs Time'); xlabel('time [s]'); 19

20 ylabel('input Velocity [m/s]'); Fd = B*(((1/Mus).*p5(:,2)- (1/Ms).*p11(:,2)).^3); subplot(2,4,2); plot(q9dot(:,2),fd); title('dampening Force vs Velocity'); xlabel('velocity [m/s]'); ylabel('dampning Force [N]'); subplot(2,4,3); plot(time(:,2),fd); title('dampening Force vs Time'); xlabel('time [s]'); ylabel('dampening Force [N]'); subplot(2,4,4); plot(time(:,2),e9(:,2)); title('suspension Force vs Time'); xlabel('time [s]'); ylabel('force [N]'); Ft = Kt*q2(:,2); subplot(2,4,5.25); plot(time(:,2),ft); title('tire Spring Force vs Time'); xlabel('time [s]'); ylabel('force [N]'); subplot(2,4,6.5); plot(q9(:,2),e9(:,2)); title('suspension Force vs Suspension Displacement'); xlabel('time [s]'); ylabel('force [N]'); subplot(2,4,7.75); plot(q2(:,2),ft); title('tire Spring vs Tire Displacement'); xlabel('time [s]'); ylabel('force [N]'); function [y,vint] = fcn(p5_mus,t) %#codegen U=13.5; Ms=320; %Kg Mus=(Ms/6); Ks1= ; %N/m^2 Ks2=10*Ks1; Kt=10*Ks1; g=9.81; Qsi=(Ms*g)/Ks1; Qso=1.3*Qsi; B=1500; %N/(m/s)^3 d=1; %m 20

21 h=.25; %m Qus=((Ms+Mus)*g)/Kt; if (((U/d)*t) >= 0) && (((U/d)*t) <= 1) Vint=(h/d)*pi*U*cos((pi*U*t)/d); elseif (((U/d)*t) > 1) Vint=0; else Vint=0; end y = -p5_mus+vint; function [y,e9] = fcn(p5_mus,q2_kt,q9,p11_ms) %#codegen U=13.5; Ms=320; %Kg Mus=(Ms/6); Ks1= ; %N/m^2 Ks2=10*Ks1; Kt=10*Ks1; g=9.81; Qsi=(Ms*g)/Ks1; Qso=1.3*Qsi; B=1500; %N/(m/s)^3 d=1; %m h=.25; %m Qus=((Ms+Mus)*g)/Kt; if (q9 <= Qso) e9=ks1*q9; else e9=ks2*(q9); end y = -e9+q2_kt; function y = fcn(p5,q9,p11) %#codegen U=13.5; Ms=320; %Kg Mus=(Ms/6); Ks1= ; %N/m^2 Ks2=10*Ks1; Kt=10*Ks1; g=9.81; Qsi=(Ms*g)/Ks1; Qso=1.3*Qsi; B=1500; %N/(m/s)^3 d=1; %m h=.25; %m Qus=((Ms+Mus)*g)/Kt; if (q9 <= Qso) e9=ks1*q9; else e9=ks2*q9; end y = e9; 21

22 Linearization of the System about Equilibrium Points 22

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