Design of a Car Suspension System - Report Example

Design of a Car Suspension System Student University Design of a Car Suspension System Introduction Car suspension system is a combination of springs, tires and its air, linkages and shock absorbers. The shock absorbers connect the wheels to the vehicles in order to allow relative motion to take place between them. This suspension system serves the roles of road holding and braking and that of creating a comfort atmosphere to the passengers through isolation of roads noise and vibration (Haykin, 2002). All these to be achieved, right trade-offs are required in the suspension system in order to serve the purpose. In this report, a focus on the design of a simplified car suspension system was undertaken in which certain parameter questions were answered in a report structured format as follows; Since the car suspension system helps the car to mitigate the force at which the boarders experience variations in frequency of the moving car, then, the car suspension system is perceived to like a filter that filters high frequency variations beyond which the comfort of boarders is affected. Therefore, car suspension system is used to act like a low pass filter in order to filter out unwanted frequencies beyond certain level. In this case, the car suspension system to be designed in the first case is that of undammed spring mass system. Because of this, the free body diagram of the system is as shown in figure 1. In this figure the forces acting on the mass, chassis, is also shown where represents the force due to the spring while represent the force due gravitational pull. In addition to that, the coordinates of the wheel and the masses have been shown as h(t) and y(t) respectively Figure 1: A free Body Diagram of the Chassis Showing the Co-ordinates and the Direction of Forces acting on it From this system, differential equation representing the behavior of the chassis can be generated by use of Newton is Second Law of motion as shown in equation (1) (1) Where; And are as However, at equilibrium, acceleration is equal to zero. Because of that, . Therefore, By considering the change in the position of the mass with respect to y direction, the force experienced by the spring is given by; Therefore, At this point , the mass experiences an acceleration (Haykin, 2002). This acceleration of the mass can be derived from Newton’s second law of motion as; By rearing this equation, Where, Based on the above differential equation, the frequency response of this linear transfer invariant system is obtained by letting the particular solution of the equation to be; Therefore, the complementary function is obtained to be; At t=0, Also, at t=0 Therefore, Therefore On the other hand, the particular function is solved as follows Therefore, Also; Therefore, For the frequency response, By differentiation the above equation twice and substituting into the net force experienced by the chassis, then; Also; Therefore, Therefore, Based on the result of the frequency response of the system, it is observed that the system amplitude of the response depends on the frequency due to the bump spacing. In this case, when the frequency response of the system will be at its minimum level of change from the neutral point. This implies that the value of the spring constant is not enough to sustain the system at its minimum point of variation from the neutral point. With respect to that, the frequency of the system depends on the mass of the chassis and the spring constant as shown below By changing the parameters of the system, parameters k and M, the system will not help because changing these parameters only changes the natural frequencies of the system which does not have any much effect on the frequency response of the chassis. Adding of a shock absorber reduces the impact of the variation of the amplitudes experienced by the passengers. This absorber acts in a directly proportional to the velocity of the chassis and as a result, it mitigates the impact of the change in amplitude with respect to the frequency of the moving car. Therefore, an introduction of the shock absorber into the car suspension system will change the LCCDE equation to like the one shown below; Considering the system at time zero, the change in the direction of the height of the wheels is equal to zero. Therefore, writing the above equation, it becomes; Based on the above differential equation, the frequency response of this linear transfer invariant system is obtained by; Therefore; Hence, differentiating the above and substituting it into the mass spring damper equation, By substituting the above into the equation, By assuming the The following Matlab command was used in plotting the frequency response of the system % Frequency response of the mass-damper spring system M=2000; %Mass of the Chassis in kg w=2; %Natural Frequency of the system eps=0.1 % The value of Zeta a=[0 0 w.^2]; b=[1 (2*eps*w) w.^2]; fre=[0:0.1:2*pi]; figure; freqs(a,b,fre); Grid on As a result of this code, the following was the plot output. Figure 2: The Frequency Response Plot using the Freqs command For the case when the natural frequency is 3, below is the code used and figure 3 shows its output. % Frequency response of the mass-damper spring system M=2000; %Mass of the Chassis in kg w=3; %Natural Frequency of the system eps=0.1 % The value of Zeta a=[0 0 w.^2]; b=[1 (2*eps*w) w.^2]; fre=[0:0.1:2*pi]; figure; freqs(a,b,fre); Grid on Figure 3: frequency Response when natural frequency is 3 and damping ratio=0.1 When natural frequency is 10 and damping ratio=0.1 % Frequency response of the mass-damper spring system M=2000; %Mass of the Chassis in kg w=10; %Natural Frequency of the system eps=0.1 % The value of Zeta a=[0 0 w.^2]; b=[1 (2*eps*w) w.^2]; fre=[0:0.1:2*pi]; figure; freqs(a,b,fre); Grid on On the other hand, the frequency response was plotted with constant natural frequency while varying the damping ratio as shown in the figures below. % Frequency response of the mass-damper spring system M=2000; %Mass of the Chassis in kg w=3; %Natural Frequency of the system eps=0.5 % The value of Zeta a=[0 0 w.^2]; b=[1 (2*eps*w) w.^2]; fre=[0:0.1:2*pi]; figure; freqs(a,b,fre); Grid on When When In addition to this, the step response of the system was solved by rewriting the second order differential equation to first order differential equation and then converted to state space for purpose of analysis. In this case, the equation was solved as follows; by rewriting the above into the form of two first order differ tai; equations as shown below, Let Then it implies that; Also; Rewriting equation 2 and 3 into matrix form In addition that, the response of the system with respect to the output only can be represented as follows; Defining these matrices into and putting them into the Matlab, the step repose can be obtained. But, And By using the following code, the following was the plot of the graph. %Solution of the sytem using Step response figure; for eps=0.2:0.4:1; for w=2:4:10; s=-(w)^2 d=(2*eps*w) A=[0 1; s d]; B=[0; -s]; C=[0 1]; D=[0]; sys=ss(A, B, C,D); if eps Read More

With respect to that, the frequency of the system depends on the mass of the chassis and the spring constant as shown below By changing the parameters of the system, parameters k and M, the system will not help because changing these parameters only changes the natural frequencies of the system which does not have any much effect on the frequency response of the chassis. Adding of a shock absorber reduces the impact of the variation of the amplitudes experienced by the passengers. This absorber acts in a directly proportional to the velocity of the chassis and as a result, it mitigates the impact of the change in amplitude with respect to the frequency of the moving car.

Therefore, an introduction of the shock absorber into the car suspension system will change the LCCDE equation to like the one shown below; Considering the system at time zero, the change in the direction of the height of the wheels is equal to zero. Therefore, writing the above equation, it becomes; Based on the above differential equation, the frequency response of this linear transfer invariant system is obtained by; Therefore; Hence, differentiating the above and substituting it into the mass spring damper equation, By substituting the above into the equation, By assuming the The following Matlab command was used in plotting the frequency response of the system % Frequency response of the mass-damper spring system M=2000; %Mass of the Chassis in kg w=2; %Natural Frequency of the system eps=0.

1 % The value of Zeta a=[0 0 w.^2]; b=[1 (2*eps*w) w.^2]; fre=[0:0.1:2*pi]; figure; freqs(a,b,fre); Grid on As a result of this code, the following was the plot output. Figure 2: The Frequency Response Plot using the Freqs command For the case when the natural frequency is 3, below is the code used and figure 3 shows its output. % Frequency response of the mass-damper spring system M=2000; %Mass of the Chassis in kg w=3; %Natural Frequency of the system eps=0.1 % The value of Zeta a=[0 0 w.^2]; b=[1 (2*eps*w) w.

^2]; fre=[0:0.1:2*pi]; figure; freqs(a,b,fre); Grid on Figure 3: frequency Response when natural frequency is 3 and damping ratio=0.1 When natural frequency is 10 and damping ratio=0.1 % Frequency response of the mass-damper spring system M=2000; %Mass of the Chassis in kg w=10; %Natural Frequency of the system eps=0.1 % The value of Zeta a=[0 0 w.^2]; b=[1 (2*eps*w) w.^2]; fre=[0:0.1:2*pi]; figure; freqs(a,b,fre); Grid on On the other hand, the frequency response was plotted with constant natural frequency while varying the damping ratio as shown in the figures below.

% Frequency response of the mass-damper spring system M=2000; %Mass of the Chassis in kg w=3; %Natural Frequency of the system eps=0.5 % The value of Zeta a=[0 0 w.^2]; b=[1 (2*eps*w) w.^2]; fre=[0:0.1:2*pi]; figure; freqs(a,b,fre); Grid on When When In addition to this, the step response of the system was solved by rewriting the second order differential equation to first order differential equation and then converted to state space for purpose of analysis. In this case, the equation was solved as follows; by rewriting the above into the form of two first order differ tai; equations as shown below, Let Then it implies that; Also; Rewriting equation 2 and 3 into matrix form In addition that, the response of the system with respect to the output only can be represented as follows; Defining these matrices into and putting them into the Matlab, the step repose can be obtained.

But, And By using the following code, the following was the plot of the graph. %Solution of the sytem using Step response figure; for eps=0.2:0.4:1; for w=2:4:10; s=-(w)^2 d=(2*eps*w) A=[0 1; s d]; B=[0; -s]; C=[0 1]; D=[0]; sys=ss(A, B, C,D); if eps

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