Equations of Motion & Forces of a Motorcycle Suspension

[baranij]

Spring Mass Damper Solution of a Motorcycle Suspension

I am trying to create a model of a motorcycle suspension to figure how road bump forces get transferred to the bike.
I have created a free body diagram of the system and also attached a pdf. If you need the original powerpoint in case there are mistakes, just let me know and I can email it since its not one of the accepted formats on this forum.

The following are the equations of motion that I derived from my Free Body diagram. Since its been a few years since college, I am at a loss as how to proceede to solve for forces and velocities at different points in the diagram. I am looking for equations of force at points b, d and e a function of input velocity and displacement.

Any guidance would be greatly appreciated.

1. \sum FORCES_{a} = F_{R} + (b-a)K_{W} + (\dot{b}-\dot{a})C_{W} = 0

2. \sum FORCES_{b} = F_{b} - \ddot{b}M_{W} + (a-b)K_{W} + (\dot{a}-\dot{b})C_{W} = 0


3a. \sum FORCES_{c} = F_{b} + F_{c} + F_{d} = 0 \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ F_{c} = -\ddot{e}M_{X}

3b. \sum MOMENTS_{c} = (cd)F_{d} - (bc)F_{b} = 0 \ \ \ \ \ \ \ \ so \ \ \ \ \ \ \ \ F_{d} = \frac{bc}{cd}F_{b} \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ F_{b} = \frac{cd}{bc}F_{d}


4. \sum FORCES_{d} = F_{d} + (e-d)K_{X} + (\dot{e}-\dot{d})C_{X} = 0

5. \sum FORCES_{e} = -\ddot{e}M_{X} - eK_{Y} + (d-e)K_{X} + (\dot{d}-\dot{e})C_{X} + F_{c}= 0 \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ F_{c} = -F_{b}-F_{d}

Motion relationship of points b, c, and / d on the lever.
b = \left(\frac{\ \overline{bc}\ }{\ \overline{cd}\ }+1\right) c \ - \ \frac{\ \overline{bc}\ }{\ \overline{cd}\ }d \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ e = c \ \ \ \ \ point \ \ c \ \ and \ point \ \ e \ \ are \ linked \ and \ move \ the \ same \ amount.

 

[baranij]

I managed to combine equations 2 and 4 to get the following.


<br /> K_{W}a \ + \ C_{W}\dot{a} \ + \ R^{2}K_{X}e \ + \ R^{2}C_{X}\dot{e} \ = \ \left(K_{W} + R^{2}K_{X}\right)b \ \ + \ \left(C_{W} + R^{2}C_{X}\right)\dot{b} \ -\ M_{W}\ddot{b}<br />

To keep things simple, I substituted R for the lever motion ratio. R = \frac{\ \overline{cd}\ }{\ \overline{bc}\ }

Now how do I simplify it further?
Should I integrate to get forces...if yes, how do I integrate \ddot{b}
Do I get \dot{b} + C How do I get C?

From combining equations 2 and 4 & 5 I get the following.

<br /> M_{X}\ddot{e} \ + \ K_{Y}e \ - \ K_{W}a \ + \ C_{W}\dot{a} \ = M_{W}\ddot{b} \ - \ C_{W}\dot{b} \ - \ K_{W}b<br />

 

[DannoXYZ]

Hmmm, it's actually much more complicated than that and you need to integrate the force over the amount of actual suspension-travel. Damping values in the dampers will change over this travel distance as the gas compresses and the dampers increase its resistance.

There are also harmonics from the mass of the bike, wheel, spring-rates, damping that causes the overall damping-value to be completely different for a 2nd bump of the exact same size and magnitude as the 1st. This will vary as a function of time.

 

[baranij]

Thank you, that's some very good points. I am trying to keep this as simple as possible and don't want to model a gas damper, just a regular coil spring with a constant damping coefficient damper.

What I'm trying to figure out is the reaction to a specific initial condition...lets say a 2 inch tall bump with a ramp up speed of 4 in/sec

What is confusing me is how to integrate with respect to bump height when the integrals are with respect to Time.

ANY SUGGESTIONS OF HOW I SHOULD PROCEED?

\int s = \int^{bump}_{0} s \ dt = \frac{s^2}{2}+C_{0}

\int v = \int^{bump}_{0} \frac{ds}{dt} \ dt= s+C_{1}

\int a = \int^{bump}_{0} \frac{dv}{dt} \ dt= v+C_{2}

\int (v+C_{2}) = \int^{bump}_{0} \frac{ds}{dt} \ dt\ \ + \int^{bump}_{0} C_{2} \ dt = (s+C_{3}) +(C_{2}s+C_{4})