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baranij
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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 [tex]b, d[/tex] and [tex]e[/tex] a function of input velocity and displacement.
Any guidance would be greatly appreciated.
1. [tex]\sum FORCES_{a} = F_{R} + (b-a)K_{W} + (\dot{b}-\dot{a})C_{W} = 0 [/tex]
2. [tex]\sum FORCES_{b} = F_{b} - \ddot{b}M_{W} + (a-b)K_{W} + (\dot{a}-\dot{b})C_{W} = 0 [/tex]
3a. [tex]\sum FORCES_{c} = F_{b} + F_{c} + F_{d} = 0 \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ F_{c} = -\ddot{e}M_{X}[/tex]
3b. [tex]\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}[/tex]
4. [tex]\sum FORCES_{d} = F_{d} + (e-d)K_{X} + (\dot{e}-\dot{d})C_{X} = 0 [/tex]
5. [tex]\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}[/tex]
Motion relationship of points [tex]b, c, and / d[/tex] on the lever.
[tex] 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.[/tex]
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 [tex]b, d[/tex] and [tex]e[/tex] a function of input velocity and displacement.
Any guidance would be greatly appreciated.
1. [tex]\sum FORCES_{a} = F_{R} + (b-a)K_{W} + (\dot{b}-\dot{a})C_{W} = 0 [/tex]
2. [tex]\sum FORCES_{b} = F_{b} - \ddot{b}M_{W} + (a-b)K_{W} + (\dot{a}-\dot{b})C_{W} = 0 [/tex]
3a. [tex]\sum FORCES_{c} = F_{b} + F_{c} + F_{d} = 0 \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ F_{c} = -\ddot{e}M_{X}[/tex]
3b. [tex]\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}[/tex]
4. [tex]\sum FORCES_{d} = F_{d} + (e-d)K_{X} + (\dot{e}-\dot{d})C_{X} = 0 [/tex]
5. [tex]\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}[/tex]
Motion relationship of points [tex]b, c, and / d[/tex] on the lever.
[tex] 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.[/tex]
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