StudentShare
Contact Us
Sign In / Sign Up for FREE
Search
Go to advanced search...
Free

Suspension Systems - Article Example

Cite this document
Summary
The paper  "Suspension Systems" tells that designing a car suspension system is regarded as a control engineering menace. This study considers a quarter model which is an explication of the four wheels. The analysis is simplified by the simplification of the 1D problem…
Download full paper File format: .doc, available for editing
GRAB THE BEST PAPER96% of users find it useful

Extract of sample "Suspension Systems"

A Simple Car Suspension System Design Institution Affiliation Student’s Name Date Overview A suspension systems is pivotal in a car. It helps them to move smoothly along the road. Designing car suspension system is regarded as a control engineering menace. This study considers a quarter models which is an explication of the four wheels. The analysis is simplified by the simplification to the 1D problem. For any car suspension system to be effectively designed, it must contain proper road holding capacity and offer comfort when moving on rutted road. In case of any interference instigated by the path, there should not exist large oscillations on the chassis of the car and such oscillations need to be debauched quickly. Figugure 1.0 : Demonstrating the functioning of the suspension system (Retrieved From: (Moodle.autolab.uni-pannon.hu) 2 When moving on flat surface, light compression is often experienced. Supposing the small firmness measure as, using Hooke’s law, the spring power exerted on the upward quantity can be estimated using  , of which  is the roughness of the spring. Another Downward potency exerted in this instance is chassis mass, for instance W=. Figure below indicates free body diagram; According to the second principle of motion by Newton,  . In this context, the chassis does not move i.e.  then k∂ - Mg= 0 i.e k∂- Mg Therefore, supposing the wheel moves ascends at a height h, making the weight of the chassis to ascend at a distance of h. This movement will compel the spring to be compressed by a given length, (l=y). Applying Hook’s law to the spring makes it to produce a reaction force which is equivalent to . The figure below indicates a diagram of a free body; Applying Newton’s second law for this case; M dy/dt = (h - y) k + ∂k) - Mg But k∂ -Mg = 0 Readjusting the equation; M dy/dt + ky= kh Bearing in mind the distinctive lane situation as sinusoidal results to;  H stands for the highest dislocation of the wheel. This equation of differentiation becomes; Ky + M dy/dt = kH sin (ῳt) However Kh = F0 ky +M dy/dt = F0 sin (ῳt) a) The occurrence reaction Sinusoidal involvement linear time invariant offers sinusoidal output as indicated Y (t) = G (jῳ) X (jῳ) X (jῳ) = sin (ῳt) Thus |G (jῳ)| is the frequency response Therefore, differentiating with respect to t, the equation becomes y1 = | G (jῳ)| x (jῳ) x cos (ῳt) Further differentiation of the equation; Y11 = - | G (jῳ)| x ῳ x ῳ x sin (ῳt) The expression can be substituted in the linear constant coefficient equation. -M | G (jῳ) | x ῳ2 x sin (ῳt) + K| G (jῳ) | (sin (ῳt)) = F0 sin (ῳt) Common terms are factored out and rearranging the equation -M |G (ῳ) | x ῳ2 + k | G (ῳ) | = F0 |G (ῳ) | = F0/k-Mῳ2 b) The rate of recurrence  can be rated as a scalar multiple of the integrated signal. Consequently, lowering the input level. As noted in the calculation a scalar function of vibration rate of occurrence .Thus; the amplitude relies on the rate of recurrence. This rate of recurrence causes the great concern because multiplying the original frequency by 1/k in the numerator and denominator. | G (ῳ) | = [ (F0/k)/ (1- Mῳ2/k)] = [ (H) / (1- ῳ2/ῳ2n)]  3) When the clients’ keeps on complaining of rutted ride, then it means that the characteristics of the surface is transformed and there is a rise in output rate of recurrence. It can be concluded that this yield is still sinusoidal. I agree with the approach. Additionally, I can indorse the need to implement modification of the parameters of the system to obtain an arrangement that cuts lowers the vibrations input to a rate that is substantial (Adhwarjee, 2007). 4 a) An equation of depicting constant‐coefficient differential is presented as follows My11+ cy1+ky = F0 sin (ῳt) The introduction of the term cy1 originates from the fluid flowing between the moving and fixed element thus a shock absorber emerges. According to Newton’s principle of viscous – Manifestation of shear stress in the fluid moving between fixed and moving plate as; τ= µdu/dy =µv/p µ is the vibrant viscosity v = dy/dt (velocity of the moving plate) p = is the detachment between the moving and fixed plates Shearing force is obtained by multiplying shearing force by interface area. For instance F = µAv/p = cv = cy1 c = µA/P = Damping constant b) The rate of recurrence response can be deliberates as: Input y(t)= | G( jῳ) | sin (ῳt + Ǿ) |G (jῳ) | sin (ῳt + Ǿ) y1 = | G(jῳ) | ῳ cos ( (ῳt + Ǿ) ) y11 = - |G (jῳ) | ῳ2sin (ῳt + Ǿ) Substituting in My11 + cy1 + ky = F0 sin (ῳt) -M |G ((jῳ) | ῳ2 sin (ῳt + Ǿ) + c| G (jῳ) ῳ cos ((ῳt + Ǿ)) + k |G (jῳ)|sin(ῳt + Ǿ) = CF0 ῳ cos ((ῳt + Ǿ)). |G (jῳ) | (k -Mῳ2) sin (ῳt + Ǿ) + c|G (jῳ)|ῳ cos ((ῳt+ Ǿ) = F0sin (ῳt) Applying trigonometric relations Sin (ῳt + Ǿ) = sin ῳtcosǾ + sin Ǿcosῳt Cos ((ῳt + Ǿ)) = cosῳtcosǾ - sinǾsinῳt Substituting in the equation we obtain; |G (jῳ) | (k- Mῳ2) sinῳtcosǾ + | G (jῳ) | (k - Mῳ2) sinǾcosῳt + c|G (jῳ) |ῳcosῳtcosǾ) -c|G (jῳ)|ῳsinǾsinῳt = F0sin(ῳt) By equating the coefficient of cosῳt and sinῳt on both sides of equation |G (jῳ)| (k-Mῳ2) cosǾ- c|G (jῳ)| ῳsinǾ=F0 c| G (jῳ)|ῳcosǾ + |G (jῳ)| (k-Mῳ2) sinǾ= 0 Solving |G (jῳ)|= [(F0)/((k-Mῳ2)2+ c2ῳ2)1/2)] Since ῳn=√k/m ζ= c/2mῳn H= F0/k Dividing by k G (jῳ) = [(H)/{(1-(ῳ/ῳn)2)2 + ( 2 ζ ῳ/ῳn)2}1/2 Ǿ= tan-1 (2 ζ ῳ/ῳn/1-(ῳ/ῳn)2 C) To use the function of the frequencies, changes are to be made on the frequency response to a lapse form. The process is performed by aligning the denominator as shown; Then taking components in the denominator, = (H)/1- (ῳ/ῳn)2 + 2 ζ ῳ/ῳn Multiplying both sides by ῳn2 (H ῳn2)/ (ῳn2- ῳ2 +2 ζ ῳnῳ) Taking the square root of -1 w this becomes; (Hῳn2)/ (ῳn2+ (ῳj)2+2 ζ ῳnῳj) Substituting ῳj= s G(s)= (Hῳn2)/(s2+2 ζ ῳns+ῳn Let H=1 thus, G (s)= (ῳn2)/(s2 +2 ζ ῳns + ῳn) d) - Unit response -Different damping ratios c) Diverse values of the natural frequency d) 5 a) The rate of recurrence is often the frequency represented by. The frequency is associated to natural frequency as follows: ῳc = ῳn [(1-2ζ2) +/- √4ζ4+ 4ζ2- 2]1/2 The spring and damping constant is often used to control cut-off frequency. Spring constant usually defines the natural frequency, while damping constant manages the damping factor. Additionally, cut-off frequency is seen to be a turning frequency at which amplitude with a greater frequency is generated. Upon reaching this frequency, the journey becomes smooth. Cut- off frequency can be lowered by lowering damping factor of natural frequency. After reducing damping factor and natural frequency, step response is likely to become quick as indicated in the step-response graph r. in case of hard suspension system, the system is likely to have low frequency cut-off implying that time- response of the unit step input is likely to give rapid response lethargic. (b) Decreasing the damping factor leads to decreased frequency thus making the travel smooth because the cut-off frequency is already reduced, thus it is vital to decrease damping factor to decrease the cut-off frequency. This set-up renders the response slow as indicated in the step response of damping factors (Fraden, 2015). 6. It is evident from the findings that high natural frequency and lower values of damping factors need to be attained in the procedure of designing. This can be attained by using a spring of high constant and a damper that has a high level of damping constant. Tooley and Dingle, (2013) proposes that there should be much care to prevent the frequency parameter from equaling the natural frequency. When this occurs, it can result in damaging resonating frequency. Summary and conclusion From the study, it is evident that car suspension system acts as a filter and creates stability when the car is on motion. Additionally, the suspension often eradicates high vibrations that occur. The study considered two different systems. The first system provided that the swift amplitude is lowered by introducing the same vibration but with scaled amplitude. However, this system retained induced vibrations. Clients were still complaining of poor journey. Introduction of a damper was conducted with the objective of controlling the scale of these amplitudes. Nevertheless, damper offered the cut-off frequency that provided grounds for identifying the systems functional point. References Adhwarjee, D, K. (2007). Theory and Applications of Mechanical Vibration. Firewall Media. Fraden, J. (2015). Handbook of Modern Sensors: Physics, Designs, and Applications. Springer. Tooley, M and Dingle, L. (2013). Engineering Science: For Foundation Degree and Higher National. Routledge. Moodle.autolab.uni-pannon.hu,. 'Chapter 1.  Introduction'. N.p., 2015. Web. 21 Oct. 2015. . Read More
Cite this document
  • APA
  • MLA
  • CHICAGO
(Paraphrasing Article Example | Topics and Well Written Essays - 2500 words, n.d.)
Paraphrasing Article Example | Topics and Well Written Essays - 2500 words. https://studentshare.org/engineering-and-construction/2094668-suspension-system
(Paraphrasing Article Example | Topics and Well Written Essays - 2500 Words)
Paraphrasing Article Example | Topics and Well Written Essays - 2500 Words. https://studentshare.org/engineering-and-construction/2094668-suspension-system.
“Paraphrasing Article Example | Topics and Well Written Essays - 2500 Words”. https://studentshare.org/engineering-and-construction/2094668-suspension-system.
  • Cited: 0 times
sponsored ads
We use cookies to create the best experience for you. Keep on browsing if you are OK with that, or find out how to manage cookies.
Contact Us