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Car Suspension, coupled ODE

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img27.imageshack.us/img27/2193/matlapassig2fig.jpg [Broken]
    http://img132.imageshack.us/img132/7126/matlapassig2.jpg [Broken]

    2. Relevant equations
    My impulse response is entirely different which make me believe that i have messed up my ODE. I have taken the laplace tranform to get it into the from of y to f


    3. The attempt at a solution
    [tex]m_{s}\frac{d^{2}y}{dt^{2}}-c(\frac{dy}{dt} - \frac{dy1}{dt}) + k(y - y1)=0[/tex]
    [tex]m_{u}\frac{d^{2}y1}{dt^{2}}+c(\frac{dy1}{dt} - \frac{dy}{dt}) + k(y1 - y) +k_{t}(y1-f(t))=0[/tex]

    Laplace Transform:

    [tex]Y(S)(m_{s}s^{2}-sc+k)+Y1(S)(sc-k)=0[/tex]
    [tex]Y1(S)(m_{u}s^{2}+sc+k+k_{t})+Y(S)(-sc-k)-F(S)k_{t}=0[/tex]

    Rearranging First Eqn:

    [tex]Y(S)\frac{m_{s}s^{2}-sc+k}{sc-k}=-Y1(S)[/tex]

    Substitution:

    [tex]-Y(S)\frac{m_{s}s^{2}-sc+k}{sc-k}(m_{u}s^{2}+sc+k+k_{t})+Y(S)(-sc-k)=F(S)k_{t}[/tex]



    [tex]-Y(S)(s^{4}(m_{u}m_{s})+s^{3}cm_{s}+s^{2}m_{s}k+s^{2}m_{s}k_{t}-s^{3}cm_{u}-s^{2}c^{2}-sck-sck_{t}+s^{2}m_{u}k+csk+k^{2}+kk_{t}+Y(S)(-c^{2}s^{2}+k^{2}=F(S)(sck_{t}-kk_{t}[/tex]

    [tex]Y(S)(-s^{4}(m_{u}m_{s})-s^{3}(cm_{s}-cm_{u})-s^{2}(m_{s}k+m_{s}k_{t}+m_{u}k+s(ck_{t})-kk_{t})=F(S)(sck_{t}-kk_{t})[/tex]

    So to get the transform from y to f we want this right?
    [tex]\frac{sck_{t}-kk_{t}}{-s^{4}(m_{u}m_{s})-s^{3}(cm_{s}-cm_{u})-s^{2}(m_{s}k+m_{s}k_{t}+m_{u}k+s(ck_{t})-kk_{t}}[/tex]

    When substituting the values in from the question my impulse response done through matlab doesnt look anything like the one given in the question...


    Have i done something wrong?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 1, 2009 #2
    Hey, at first look, I found these equations to be fishy: is the f(t) a force being applied on the sprung mass m_s? If so, then shouldn't the expression f(t) appear in the equation of motion of m_s as opposed to m_u as you have formulated?
     
  4. Oct 1, 2009 #3
    I did think of that, but being a car suspension system, it will be traveling on a road which will have the forces acting on m_u. So i put f(t) as the reaction force due to the road...

    much like this
    mechanics_small.jpg

    Mind you, i am not strong in mechanics, so there could be other flaws in it too
     
  5. Oct 1, 2009 #4
    Even so, the way you have used f(t) in the expressions, it appears to be the road displacement rather than a road reaction force. If that is the case then your formulation seems correct. If however, f(t) is a force on the sprung mass (like you showed in your first illustration), then f(t) should just appear as a force in the first equation.
     
  6. Oct 1, 2009 #5
    So, with that being said

    [tex]m_{s}\frac{d^{2}y}{dt^{2}}-c(\frac{dy}{dt} - \frac{dy1}{dt}) + k(y - y1)=f(t)[/tex]
    [tex]m_{u}\frac{d^{2}y1}{dt^{2}}+c(\frac{dy1}{dt} - \frac{dy}{dt}) + k(y1 - y) +k_{t}(y1)=0[/tex]

    I will see how this spans out in the impulse response latter but just as a check, do those new ODE's seem correct?
     
  7. Oct 1, 2009 #6
    Ok so assuming downward direction as positive and also assuming that the spring damper push the sprung mass upwards(i.e. the spring gets compressed), here are the ODEs:
    [tex]F = ma[/tex]
    [tex]f(t)-k(y - y1)-c(\frac{dy}{dt} - \frac{dy1}{dt}) = m_{s}\frac{d^{2}y}{dt^{2}}[/tex]
    and
    [tex]c(\frac{dy}{dt} - \frac{dy1}{dt}) + k(y - y1) -k_{t}(y1)=m_{u}\frac{d^{2}y1}{dt^{2}}[/tex]

    So your equations are correct except for the sign on the damper force in the first equation.
     
  8. Oct 1, 2009 #7
    doing what i did before i gte the transfer function to be

    [tex]
    \frac{s^{2}m_{u}+sc+k+k_{t}}{s^{4}(m_{u}m_{s})+s^{3}(cm_{s}+cm_{u})+s^{2}(m_{s}k+m_{s}k_{t})+s(ck_{t})+kk_{t}}
    [/tex]

    plugging in the values i get an unstable system and the impulse response is exponentially increasing

    for k = 30 N/mm (and also kt) i put in the value of 30*10^-3 right ?
    I have tried it both with 30 and 30*10^-3 and it is still the same response...

    and i thought this was going to be a simple question...
     
  9. Oct 6, 2009 #8
    k (and kt) are in N/mm so their values in SI units would be 30N/mm or 30x10^3 N/m
     
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