Friday, June 19, 2020
Report on design of a Car Suspension System - 1650 Words
Report on design of a Car Suspension System (Essay Sample) Content: Report on design of a (Simplified) Car Suspension SystemInstitution AffiliationStudents NameDateDesign of a (Simplified) Car Suspension System A car suspension system is basically aimed at providing a smooth movement as the car moves along the road. The design of a suspension system is more of a control engineering problem. In this study, a à ¼ model (a representation of the 4wheels) is considered. This simplification to a 1D problem makes the analysis simpler. The design requirements for any suspension system is the ability to have a good road holding capacity and at the same providing comfort when riding on bumpy road. In case of any disturbance caused by the road, the chassis of the vehicle should not have large oscillations and these oscillations should be quickly dissipated.. Fig: Illustrating the working of a cars suspension system (Source, Moodle.autolab.uni-pannon.hu)2There is slight compression experienced when moving on a flat surface. Assuming that this slight compression is QUOTE , and applying Hookes law then the spring force acting on the mass upwards is given by QUOTE . Where QUOTE is the spring stiffness. The other force acting downwards in this case, is the weight of the chassis i.e. W= QUOTE . The free body diagram can be illustrated as below ;And from Newtons second law of motion, QUOTE and the chassis in this case is at rest i.e. QUOTE then QUOTE Now assuming that the wheel is displaced upward from the surface of an height QUOTE and which causes the chassis (mass) to move upwards a distance, h . As a result of this movement, the spring will thus be compressed by a length, QUOTE .When Hookes law is applied at the spring, it will produce a reaction force equal to QUOTE . The free body diagram can be represented as in figure below; Applying Newtons second law for this case; QUOTE MgRearranging we have ; QUOTE Considering the typic al road condition as sinusoidal then we get the expression, QUOTE where H is the maximum displacement of the wheel.This differential equation becomes;The frequency responseSinusoidal input linear time invariant gives sinusoidal output as shown below And differentiating with respect to t becomes;Differentiating further QUOTE **We can then substitute the expression in the linear constant coefficient equation** Factoring out the common terms and then rearranging;The frequency response QUOTE is a scalar multiple of the input signal. It thus scales down the amplitude of the input. It can be deduced from the above equation that this is a scaler function of frequency of vibration .Therefore the amplitude depends on the frequency.The answer for this is yes. Since by having a look at the frequency response,Multiplying by QUOTE in both numerator and the denominator QUOTE This is the frequency that will cause a great concern.The customer c omplains of a bumpy ride is simply because the nature of the surface has changed in such a way that there is increase in the output frequency. From the analyses above, this output is still sinusoidal. Yes with this kind of a system I would recommend changing of the systems parameters in order to achieve a system that will cut down the input vibrations to a comfortable level (Dukkipati, 2010)4 a) The new linear constantà ¢Ã¢â ¬Ã coefficient differential equation is as shown, The introduced term results from the shear forces from the fluid that is moving between the fixed and the moving component making up the shock absorber. From Newtonà ¢Ã¢â ¬s law of viscous fluid, we can get the expression of shear stress in the fluid between the moving plate and the fixed plate as; The frequency response can be calculated s below Input y(t)= QUOTE | QUOTE Since QUOTE C) In or der to utilize the frequencies function, we will need to convert the frequency response into a lapse formThis can be done by squaring the denominator and this will give;Then taking components in the denominator,Multiplying by QUOTE all through we have,Taking the square root of -1 w this becomes;Substituting QUOTE ; QUOTE Let H=1 thend)For a unit response.For different damping ratiosFor different value of natural frequency5 a) Cut-off frequency is usually the frequency in which the frequency equals to QUOTE , This cut-oof frequency can be related by natural frequency by the expression;The cut-off frequency is usually controlled by the spring constant and damping constant. Spring constant determines the systems natural frequency and the damping constant governs the damping factor.Cut-off frequency is a corner frequency where amplitude having higher frequency than the input frequency are reduced. When this frequency is reached, then the ride fe els smooth. This cut-off frequency can be reduced by reducing either the natural frequency or the damping factor. In so doing, the step response will become rapid as can also be seen from step-response graph r. For a hard suspension system, it will have a lower cut-off frequency meaning that the time-response of a unit step input will give a... Report on design of a Car Suspension System - 1650 Words Report on design of a Car Suspension System (Essay Sample) Content: Report on design of a (Simplified) Car Suspension SystemInstitution AffiliationStudents NameDateDesign of a (Simplified) Car Suspension System A car suspension system is basically aimed at providing a smooth movement as the car moves along the road. The design of a suspension system is more of a control engineering problem. In this study, a à ¼ model (a representation of the 4wheels) is considered. This simplification to a 1D problem makes the analysis simpler. The design requirements for any suspension system is the ability to have a good road holding capacity and at the same providing comfort when riding on bumpy road. In case of any disturbance caused by the road, the chassis of the vehicle should not have large oscillations and these oscillations should be quickly dissipated.. Fig: Illustrating the working of a cars suspension system (Source, Moodle.autolab.uni-pannon.hu)2There is slight compression experienced when moving on a flat surface. Assuming that this slight compression is QUOTE , and applying Hookes law then the spring force acting on the mass upwards is given by QUOTE . Where QUOTE is the spring stiffness. The other force acting downwards in this case, is the weight of the chassis i.e. W= QUOTE . The free body diagram can be illustrated as below ;And from Newtons second law of motion, QUOTE and the chassis in this case is at rest i.e. QUOTE then QUOTE Now assuming that the wheel is displaced upward from the surface of an height QUOTE and which causes the chassis (mass) to move upwards a distance, h . As a result of this movement, the spring will thus be compressed by a length, QUOTE .When Hookes law is applied at the spring, it will produce a reaction force equal to QUOTE . The free body diagram can be represented as in figure below; Applying Newtons second law for this case; QUOTE MgRearranging we have ; QUOTE Considering the typic al road condition as sinusoidal then we get the expression, QUOTE where H is the maximum displacement of the wheel.This differential equation becomes;The frequency responseSinusoidal input linear time invariant gives sinusoidal output as shown below And differentiating with respect to t becomes;Differentiating further QUOTE **We can then substitute the expression in the linear constant coefficient equation** Factoring out the common terms and then rearranging;The frequency response QUOTE is a scalar multiple of the input signal. It thus scales down the amplitude of the input. It can be deduced from the above equation that this is a scaler function of frequency of vibration .Therefore the amplitude depends on the frequency.The answer for this is yes. Since by having a look at the frequency response,Multiplying by QUOTE in both numerator and the denominator QUOTE This is the frequency that will cause a great concern.The customer c omplains of a bumpy ride is simply because the nature of the surface has changed in such a way that there is increase in the output frequency. From the analyses above, this output is still sinusoidal. Yes with this kind of a system I would recommend changing of the systems parameters in order to achieve a system that will cut down the input vibrations to a comfortable level (Dukkipati, 2010)4 a) The new linear constantà ¢Ã¢â ¬Ã coefficient differential equation is as shown, The introduced term results from the shear forces from the fluid that is moving between the fixed and the moving component making up the shock absorber. From Newtonà ¢Ã¢â ¬s law of viscous fluid, we can get the expression of shear stress in the fluid between the moving plate and the fixed plate as; The frequency response can be calculated s below Input y(t)= QUOTE | QUOTE Since QUOTE C) In or der to utilize the frequencies function, we will need to convert the frequency response into a lapse formThis can be done by squaring the denominator and this will give;Then taking components in the denominator,Multiplying by QUOTE all through we have,Taking the square root of -1 w this becomes;Substituting QUOTE ; QUOTE Let H=1 thend)For a unit response.For different damping ratiosFor different value of natural frequency5 a) Cut-off frequency is usually the frequency in which the frequency equals to QUOTE , This cut-oof frequency can be related by natural frequency by the expression;The cut-off frequency is usually controlled by the spring constant and damping constant. Spring constant determines the systems natural frequency and the damping constant governs the damping factor.Cut-off frequency is a corner frequency where amplitude having higher frequency than the input frequency are reduced. When this frequency is reached, then the ride fe els smooth. This cut-off frequency can be reduced by reducing either the natural frequency or the damping factor. In so doing, the step response will become rapid as can also be seen from step-response graph r. For a hard suspension system, it will have a lower cut-off frequency meaning that the time-response of a unit step input will give a...
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