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Static and dynamic analysis of a crank with pedal

  1. Nov 19, 2012 #1

    me and my friend have a problem that we cant really agree upon.

    we are doing an analysis study on a bike's crank and pedal.

    the first picture shows how we put the force of 210 N.
    picture 2 shows how we used "fixed geometry" to lock the part in place.

    now, this is the static analyses, where we get a stress consentration of 121.3 MPa in the pin connecting the crank and the pedal.

    our unagreement comes here:

    she belives that the stress consentration in the same spot in a dynamic "real" event could be reduced.
    because instead of the part beeing fixed and locked where the gear is, it would rater have a force working against the pedal force.

    i belive the stresses in the pin would be the same nomatter if its static or dynamic in that excact position. The force used on the pedal will still be 210 N and therefor the max stress in that area would remain the same.
    i cant imagine how the stresses at that location can be reduced due to the change from static to dynamic, when all paramters are the same except for resitance insted of it beeing locked.

    (im know the stresses would change due to the change in position in dynamic, but that isnt what we are asking for)

    im not sure if i made my self understandable here. please ask if anything is unclare:)

    Attached Files:

  2. jcsd
  3. Nov 19, 2012 #2


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    I agree, unless you have a significant acceleration of your pedal. Force on it has to cancel if its mass*acceleration is negligible in the setup.
  4. Nov 19, 2012 #3


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    I agree there would not be any significant change between a static and dynamic stress levels, for a realistic acceleration rate of the bike.

    However you might want to think a bit more about how you restrained the crank against rotating, so you get the correct reaction forces on it. You seem to have fixed the chain wheel at three points around its circumference, which doesn't correspond to the real situation.

    The force propelling the bike comes from the tension in the top part of the chain, which is applied tangentially to the top of the gear wheel. You can probablly ignore the tension in the bottom part of the chain which should be much smaller. The other forces on the crank are all through its bearings.

    To model the fact that the crank can turn in the bearing, you can restrain it in the X Y and Z directions all the way along the axis of rotation. You then need to stop it spinning freely, by restraining it tangentially at the top, where the chain starts to wrap around it.

    Without making a model I don't know if that will actually make any difference, but these details can be important if they create the wrong load paths through the structure. Restraining the model the right way is no harder than doing it wrong, except for a bit more thought to decide what is "right".
  5. Nov 20, 2012 #4
    it is true what you are saying. we shouldnt restrain it at all the way around the gear.
    thanks for the heads up;)

    So can someone explain why the stresses at the given point will be equal at that excact position in the two different scenarios?

    i need to know why, (given my theory is right) so i can explain it to her:) or vice versa
  6. Nov 20, 2012 #5


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    Which two scenarios are you asking about? Accelerating and not accelerating? Or two different ways of restraining the model?
  7. Nov 20, 2012 #6
    Accelerating and not accelerating.
  8. Nov 20, 2012 #7


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    The mass of the assembly is very small compared to mass of the wheel and the bicycle, so acceleration will not make any difference here. Nearly all of the stress will come from action between chain, axis on which sprocket is mounted, and the crank/pedal itself.

    As far as stress from the chain, I see two extreme cases that should be considered. First, it's possible that spacing between teeth of the sprocket is slightly shorter than spacing between links of the chain. In that case, chain will mostly pull on the top of the sprocket. On the other hand, if the spacing between teeth is slightly longer, the stress will be distributed more evenly, with more tangential force at the bottom of the sprocket, and radial force through half of the revolution in between. Of course, rounded grooves will help reduce the significance of this effect, so it might be ok to just assume even pull.
  9. Nov 20, 2012 #8


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    The way the chain fits onto the sprocket teeth will affect the local stresses around the rim of the sprocket, but not the way the chain "pulls" on the sprocket as a whole.

    Think about a free body diagram of the sprocket and pedal crank, plus a short piece of the chain from the sprocket towards the rear wheel. You have three statically determinate forces acting (ignoring the weight of the object, which will be small compared with the other forces): the pull in the direction of the chain, the force on the pedal, and the reaction force at the bearing. Those forces are not affected by the details of how the chain interacts with the sprocket.
  10. Nov 20, 2012 #9


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    The acceleration of the bike is limited by the friction force between the tires and the ground, and the maximum will be of the order of 1g (10m/s). So the maximum force required for the linear accleration of the crank is the same order of magnitude as the weight of the crank. That is much smaller than the force applied to the pedal, etc.

    There is also the rotational acceleration to consider, and you need more numbers about the size of the bike wheels and the gear ratios to estimate that, but you would get to the same conclusion - the effect would be small. That's not really surprising, since these parts were designed to be as light as possible to minimize the amount of "wasted" energy propelling the bike.
  11. Nov 20, 2012 #10


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    True. I was thinking about net moment of force around a point other than center of the sprocket, but yeah, that still works out the same. So if you are not interested in stress on the sprocket, but only the pedal and the crank, then it doesn't matter how the chain interacts with the sprocket.
  12. Nov 20, 2012 #11
    Thank you very much, both of you. That helped alot:D
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