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Calculations on a set of drumbrakes

  1. May 7, 2006 #1
    I'm doing some calculations on a set of drumbrakes. In a book I have from my school days it says that the normal force on each shoe are not equal (this is the force appearing on the outer left and right sides of the drum when viewing the drum from the side). This apparently leads to different amounts of wear on each shoe. I can't figure out why they shouldn't be equal, so if someone can explain it to me I'd be greatful.

    Also, if someone could explain why a band brake works better in one direction than the other that would be real nice.
     
    Last edited: May 7, 2006
  2. jcsd
  3. May 8, 2006 #2

    Clausius2

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    That's a nice mechanics problem. That's right, they don't need the same force. If you think about it, assuming the wheel is turning clockwise, the rightwards shoe is being pressed by the external force BUT also is being pressed by the friction against the wheel. On the contrary, the leftwards shoe is being expelled because the friction force whereas the external force still presses it towards the wheel.
     
  4. May 8, 2006 #3

    Mech_Engineer

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    A little searching goes a long way...

    http://auto.howstuffworks.com/drum-brake.htm
     
  5. May 8, 2006 #4
    I do not like this explanation. The friction between the wheel and the brake shoe is due to the normal force. What do you mean the right brake shoe is also being pressed by friction against the wheel? This is statement is wrong, or at least appears to be to me.
     
  6. May 8, 2006 #5
    The wheel cylinder in a drum brake system is on the top of the shoes normally. The bottom of the shoes rest against a common fulcrum. Clear it up for you?
     
  7. May 8, 2006 #6

    Clausius2

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    In fact you should take a look at a drum brake system first of all and realise that the shoes are PIVOTING usually at the bottom. The brake system has a horseshoe shape. If you think a little about it and make some draws, you will realise that the friction force is exerting a clockwise torque around the pivot of the rightwards shoe and hence collaborating positively in the braking process. Also, a good mechanical engineer knows that the one of the shoes become degradated sooner than the other, and the first one is just the rightwards one in my example.
     
  8. May 8, 2006 #7
    Right, the friction force cause by both shoes is causing a counter torque to stop the rotation of the wheel. But this is not happening for ONLY the right brake shoe. I still do not see what you are saying :confused:

    http://auto.howstuffworks.com/drum-brake1.htm (play the animation)

    based on this picture, it appears that the left brake shoe has a more even distribution along its length, and the right shoe should wear out near the top, as the bottom has little to no contact.
     
  9. May 8, 2006 #8

    Clausius2

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    :rofl: curious, the drumbrake of that web page is just designed for taking advantage of the self-actuating effect in BOTH shoes. In fact it is an intelligent way of design. Classically, a drum brake has two pivots, one of each shoe, allocated both at the bottom of the drum. Draw an scheme of this set up and see that only the right shoe suffers self-actuating. If you put the left pivot on the top and the right one at the bottom, both shoes suffer self actuating, whereas the hydraulic system for pushing them results to be more complicated (you need two cylinders). The drum of that webpage is mechanically designed for taking advantage of self actuation, and it is complex by the way (I think the op was referring to the classical scheme of a drum, with two pivots at the bottom).
     
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