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Impact force calculation of a Full Suspension Mountain Bike

  1. May 21, 2015 #1
    Hello Folks!
    I want to find the maximum applicable force on a full suspension mountain bike and i will use this force on a FEA software structural analysis for a uni design project.

    This bike has a rear suspension with 400 lbs/in spring rate.Also rear wheel vertical displacement(travel amount) is 120 mm.But i am confused about the assumptions.
    What will be the impact time?What is the weight distribution on wheels?Should i make the calculation solely on the bottom out position of rear suspension because stress will be maximum?
    I think i should make some manipulation on impulse momentum and kinetic energy equations.

    How can i calculate impact forces especially on the rear wheel?
  2. jcsd
  3. May 23, 2015 #2

    Simon Bridge

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    A 400lb force will compress the spring 1in ... given enough time. You can work out that time.
    However - real-life impact times are something you measure through experiments and depend strongly on the specifics of the impact.
    Presumably the rear suspension is also damped - which can affect things.

    You need to decide on a model - start with how much detail you want.
  4. May 24, 2015 #3
    You can use experiment, and this is typical, but another option is to use a multi-body dynamics code such as Simpack or Adams to generate the loads (I am presuming you work for a bike manufacturer or are at a university, as these codes are not inexpensive).
  5. May 26, 2015 #4

    jack action

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    Let's do a thought experiment.

    Imagine an unloaded spring with stiffness [itex]k[/itex] and a weight [itex]W[/itex] over it. The weight is barely touching the top of the spring, such that the spring doesn't support it. We then release the weight. Of course, we know the settling displacement will be [itex]x_0 = \frac{W}{k}[/itex]. But what about the impact load? Well the energy absorbed by the spring must be the same as the potential energy, so:
    [tex]\frac{1}{2}kh_0^2 = Wh_0[/tex]
    [tex]h_0 = 2\frac{W}{k}[/tex]
    This means that the impact force will be:
    [tex]F_i = kh_0 = 2W[/tex]
    So just releasing the weight on the unloaded spring will give an impact force equivalent to twice the weight, no matter the characteristic of the spring!

    Now holding the weight an additional height [itex]h[/itex] above the unloaded spring, you get:
    [tex]\frac{1}{2}kx^2 = W\left(h+h_0\right)[/tex]
    [tex]x = \sqrt{2\frac{W}{k}\left(h+h_0\right)} = \sqrt{2\frac{W}{k}\left(h+2\frac{W}{k}\right)}[/tex]
    And the impact force would be:
    [tex]F_i = kx = \sqrt{2kW\left(h+2\frac{W}{k}\right)} = \sqrt{2kWh+4W^2}[/tex]
    So if [itex]h = 0[/itex], [itex]F_i[/itex] is still equal to [itex]2W[/itex]. But now the stiffer the spring is, the stronger will be the additional impact force.

    The maximum energy the spring can take is [itex]E_{max} = \frac{1}{2}kx_{max}^2[/itex]. At that point - if [itex]F_i > kx_{max}[/itex] - the spring bottoms out and it becomes a solid bar with stiffness [itex]k_{bar} = \frac{EA_0}{L_0}[/itex] (see Wikipedia). The new total energy is found by [itex]E_{total} = W\left(h+x_{max}+2\frac{W}{k_{bar}}\right)[/itex], and the exceeding not absorbed by the spring will be absorbed by this new «bar» spring:
    [tex]\frac{1}{2}k_{bar}x_{bar}^2 = E_{total} - E_{max}[/tex]
    [tex]x_{bar} = \sqrt{2\frac{E_{total} - E_{max}}{k_{bar}}}[/tex]
    Such that the new calculated impact force is:
    [tex]F_i = kx_{max} + k_{bar}x_{bar} = kx_{max} + \sqrt{2k_{bar}\left(E_{total} - E_{max}\right)}[/tex]
    You might have to double check that, but applying this to your bike, [itex]h[/itex] is the maximum drop height you think the bike can fall, [itex]W[/itex] is the weight of the bike and its driver and [itex]k[/itex] is the stiffness of the suspension. You might have to model your suspension design as multiple springs in series to get [itex]k[/itex] or [itex]k_{bar}[/itex].

    Bottom line: The impact force can at least reach twice the total weight, assuming a wheel never leaves the ground and the entire weight is absorbed by that wheel.
  6. Jan 15, 2017 #5
    thanks for this information.

    we are an organizing a Downhil event in a mall- Downmall. please help us justify that MTBikes will not damage the escaltors, in both technical and layman's term what is the impact of a full suspension bike and a rider on a moving escalator? thanks
  7. Jan 16, 2017 #6

    Simon Bridge

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    We cannot do that. That would be a liability/warranty issue... you need hire an engineer. It may just be that the mall's insurance won't cover damage if bikes have been on them (does not matter if bikes caused the damage or not) so it won't matter.
  8. Jan 16, 2017 #7
    Thanks Simon. This type of activity is practiced in Europe http://www.easterngravity.com/downmall , we are trying to have this event in Asia.

    I need your help on the impact on escalators as this is what the engineers are asking from me. I appreciate any help . Thanks
  9. Jan 19, 2017 #8

    Simon Bridge

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    You have engineers (plural) who have advised you to ask for my help in particular?
    I had no idea I was so well thought of.

    Tell them to do their jobs. They are getting paid for it, and, if they work for a comptent firm, they will have some sort of endemnity vs liability if they get it wrong.

    However... it is likely that it is the mall's insurance that is the problem, not physics. What have you done to find out?
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