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Force generated at caliper bracket while braking

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  1. Jul 3, 2013 #1
    Hi to all,

    I'm new to the forum.

    I'm designing a bracket for a front caliper of a motorcycle. I'm learning to use solidworks just to design a bracket that will be safe to use in the real world :-).. but I'm stuck on calculating the force that will be generated on the bracket (or better the connecting bolts) while braking.
    I've been looking all over the net to find an example/guidance, and this seems the best place to ask for help.

    I'm interested in the worst case scenario, so I consider the total mass of the motorcycle+rider to be 300Kg,with the total weight being only of the front tire.
    The front rim is ~19 inch and the front disk has a diameter of 320mm. As a limiting factor I consider standard tires which will slip at ~1.1g.

    I would like to know what is the force generated at the bracket while braking, at 1.1g.
    In my understanding the friction between the pads and the disk, and the force generated by the pistons in the caliper, shouldn't be relevant, as they can be any value as long as they provide a 1,1g deceleration.

    Could someone point me in the right direction?

    Thanks :)
    Riccardo
     
  2. jcsd
  3. Jul 4, 2013 #2
    no replies :-(

    Let me know if I'm asking the wrong question..
    Basically I'm interested in the rotational forces on the brake bracket while braking.

    Any help??
     
  4. Jul 7, 2013 #3
    hey Badrock
    welcome

    Do you about torques? F = T x r
    and angular velocity and acceleration, ω and α
    and mass moment of inertia

    simply,
    You have a torque at the wheel-ground contact, and a torque at the brake-disk contact.

    1. wheel not rotating and sliding -
    Disk-pad --> static friction
    wheel-ground --> kinetic friction
    Torque at ground = torque at disk

    2. Wheel rotating at constant angular velocity -
    Disk-pad --> kinetic friction ( can be less than static friction of disk pad )
    wheel ground --? static friction ( can be more than kinetic friction wheel ground )
    wheel rotates at same speed
    Ground torque = Disk torque ( cinsider braking down a hill )

    3. Wheel decelerating
    Disk-pad --> kinetic friction
    Wheel-ground --> static friction
    wheel slows down
    Ground torque < disk torque
    Calculate using change in angular acceleration of wheel and mass moment of inertia.
    http://en.wikipedia.org/wiki/Angular_acceleration
    Note: putting all the mass into the wheel is going to skew your result, perhaps to the worst case scenario - you will have to find that out.

    4. dynanmics and kinematics
    The force from the ground on the wheel when braking is what slows the whole bike down.
    Equations of motion
    http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
    ( plod through this site to find out a lot of stuff and it is easy to follow )

    Attempt to understand and try out some of the equations, and if you have difficulty don't hesitate to come back and ask some more questions, on their application.

    Cherios!
     
  5. Jul 9, 2013 #4
    To generate 1.1g braking the front tyre will need to be transmitting 330kgf of 3237N to the tarmac.

    The rolling radius acts as a lever, if we assume 1" of tyre added to the 9.5" radius of the rim we have a 266.7mm rolling radius. This means we need to be producing 3237N / .2667m = 12137Nm at the axle.

    Multiply this back out by the disc radius to get the force required at the outer radius of the disc 12137Nm * .160m = 1941N. You will obviously need to be more accurate with your mesurements (the force is not applied at the outer edge of the disc etc) and don't forget to add on your (significant) FoS.
     
  6. Jul 9, 2013 #5

    rcgldr

    User Avatar
    Homework Helper

    The grip of the front tire under maximum braking is probably greater than 1.1 g. Take a look at the actual stopping distances for sport type bikes.

    Anther issue is how much lateral (squeezing) force will be generated at the pads. You'd need to know the worst case expected coefficient of friction, size of the pads, in addition to all the other data you need to calculate the required torque on the brake rotors.
     
  7. Jul 9, 2013 #6
    Thank you all you very much for the answers and help, really appreciated!! :smile: :smile:
    @1Byte, thanks also for the link to hyperphysics.. a nice resource to check/learn basic concepts.

    There is something that still confuses me quite a bit, shouldn't the Torque = Force * Radius, instead of F =T x r?
    I thought that 1Byte had a typo, but Kozy used the same formula in his example calculation... which makes me believe that I'm missing something. :confused:

    I did a bit of research and rcgldr is right, the static coefficient of friction of a motorcycle tire is a bit higher than what I thought. Studies show that on a dry surface it can peak at 1.2, and the dynamic around 0.7-0.9 (http://papers.sae.org/2010-01-0054/)

    I did some calculations and this is what I came up with.
    Please correct me if I'm wrong or imprecise...

    Assumptions:
    The limiting factor while braking is the traction of the tires.
    I only consider the worst case scenario where all the weight is transferred to the front tire.
    Weight of the bike + rider = 300Kg
    Radius of wheel + tire = 254mm
    Radius of brake disk = 160mm. Pads are 40mm wide => the force is applied at 158mm from the axle.
    The force that is applied to decelerate the bike is:
    F = m * a​
    Which is limited by the traction of the tires:
    F = μ * W = μ * m * g​
    Hence:
    m * a = μ * m * g => a = μ * g => a = 1.2 * 9.8 = 11.76 m*s^-2
    This removes the mass of the bike from the equation. I checked and it seams to be correct except for big vehicles like trucks, where the limiting factor are the brakes, or race cars which can reach very high speeds (that's not the case for me, it's a bike from the 60s :smile:).
    So, the biggest deceleration I can archive is 11.76 m*s^-2, 1.2g.
    This deceleration creates a force tangential to the wheel which is:
    F = m * a = 300kg * 11.75 = 3528 N

    The torque at the axle is in turn:
    T = F * r = 3528 * 0.254 = 896.112 Nm
    Which means that the force at the center of the brake pads is:
    F = T / r = 896.112 / 0.158 = 5671.595 N


    This looks like a huge force to me, but it's also considering the worst case scenario.
    In reality the μ of the tires is closer to 1, and the weight on the front tire is ~176Kg (80% of 220Kg, the real expected weight) which gives me a more reasonable 2772.8 N force at the caliper.
    Still much higher than I thought! :uhh:

    Any feedback, comment or suggestion is very welcome.

    Thanks again for the input so far!!
     
    Last edited: Jul 9, 2013
  8. Jul 9, 2013 #7
    You've got the tangential force correct, but the axle torque is required to generate the tangential force, so the torque is F/R, then back to the pad is F*R. End result is the tangential force is 3525N, the axle torque should be 13,877Nm and the force at the caliper should be 2137N.
     
  9. Jul 10, 2013 #8
    Yes, you are correct, sorry about that.
     
  10. Jul 10, 2013 #9
    Don't worry, I imagined it was a typo :smile:

    Thank you Kozy for going through my calculation and giving me feedback.
    Sorry for maybe asking dumb questions.. but I can't get my head around the facts that the formula for calcuating the torque changes depending on its origin.
    I'm trying to understand what is the difference between a torque which is generates a force (t = f/r) and a torque which is generated by a force (t = f*r). Am I missing something? :confused:

    Could someone explain it to me or point me to some online resources?

    @Kozy: not that I don't trust you, I just want to understand what is going on :smile:.

    Thanks
     
  11. Jul 10, 2013 #10
    You are right not to trust me, the summer heat must have gone to my head. I had a massive brain fart. You are in fact correct.
     
  12. Jul 10, 2013 #11
    Hi Kozy, thanks for coming back to me.
    So, my caclutions look more or less correct right? :smile:
    When designing the bracket I'll confider a FoS of at least 3, just to be on the safe side.

    Another question on the same line. I'm considering the material to use, either aluminium (6082 T6 / 7075 T5) or steel.
    In my understanding the braking torque is progagated from the caliper to the forks through the friction of the surfaces in contant, rather than the bolts as I thought before.
    Since the fork mounting brackets and the caliper mount are made of cast aluminium, I was thinking of using aluminium instead of steel, as alu-alu has a higher μ than alu-steel.
    I would prefer steel, as it will be able to spot a failure, or aluminium with steel inserts.

    Any idea or thought?

    Thanks again!
     
  13. Jul 13, 2013 #12

    Bandit127

    User Avatar
    Gold Member

    On a sports bike (and some street bikes) the CoG is high enough that the bike will "stoppie" (i.e. raise the rear wheel under braking). In this situation the front wheel takes the full weight. 300 Kg in this case.
     
  14. Feb 18, 2015 #13
    Hi, I'm designing a brake caliper for an FSAE vehicle and maybe someone can help me. I don't know exactly which forces are acting when we are braking. I know how to calculate the Torque required and so on but, for a simulation, where should I apply the forces or torques and why?.
    Thank you.
     
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