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External contracting block brake mechanism

  1. Nov 8, 2008 #1
    In the brake shown in the pic, what ll be the point of application of force to disengage the brake(it is holding type brake). It is an assignment problem in machine drawing class, I have calculated the shoe dimensions, but I have no idea where ll the force be applied to disengage the brake. And how ll the force be transmitted to the second shoe?

    Force Fs shown doesnt mean anything, i was trying to work it out.

    Attached Files:

  2. jcsd
  3. Nov 8, 2008 #2


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    This is probably just a personal shortcoming on my part, but I really can't make out what's what on that diagram. Do you have other views of it?
  4. Nov 8, 2008 #3
    yups, thats exactly the problem. it looks messy.

    lowermost part is the plate on which everything is mounted. Both links are mounted on this plate, brake shoes are mounted symmetrically on these links. On top of the brake drum is some plate on whic some mechanism to transfer force is there. This is what i am not getting.

    this problem is from the book 'Machine Design - Robert Norton'. I couldn't find this book in the library. If anyone has it, could you please explain it to me?

    I have MD by shigley, it is nice. Is norton worth buying??
  5. Nov 8, 2008 #4


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    It still seems to be missing something, but I think that I've got the basic idea now. If I'm reading it correctly, a downward force on the vertical rod (left side of pic) should disengage the shoes.
  6. Nov 9, 2008 #5


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    I'm with Danger. Crap picture, but pushing/pulling the rod (top left of picture) downwards would disengage the brake.
  7. Nov 9, 2008 #6
    i know the pic is too bad.

    Anyways i got it, the top left lever is a bell crank, when pushed down, the far end of the bell crank pushes away the push rod for the right shoe, & the left shoe moves leftward because the pivot of the moves left.
  8. Nov 9, 2008 #7


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    That's what it looks like to me. What bothers me a bit is that I don't see any sort of equalization mechanism to make sure that both shoes retract fully.
  9. Nov 9, 2008 #8
    But there is an adjustment screw mounted to the actuator that would limit the open position of the left shoe.
    Last edited: Nov 9, 2008
  10. Nov 9, 2008 #9


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    Oh... okay. I can't make that out, but it explains a bit.
  11. Nov 9, 2008 #10


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    It appears that the brake is released by retracting the actuating ram into the cylinder. That compresses the coil spring, pushing the opposing shoe away from the shaft that is being braked.

    If such a brake is to be used in a fail-closed application pressure would need to be applied to the top of the cylinder to open the brake. Failure of actuating pressure would result in the brake being applied.
  12. Nov 9, 2008 #11
    Yeah, that's what I get out of it, Turbo. The bell crank appears to press against an adjustable nut on a threaded shaft to open the jaws.

    Danger's point is still standing though: what keeps the brakes from rubbing in the open position? If that's really a stop screw mounted to the cylinder, as it appears to be, what forces tend to move the pair of jaws to the left stop to prevent rubbing?--or maybe it doesn't matter in the application.
  13. Nov 10, 2008 #12
    I don't understand this part, why would the shoe rub against the drum? Once it loses contact, its enough. That screw on the left must set the maximum retraction possible.

    Anyways, thanks a lot everyone:cool:
  14. Nov 11, 2008 #13
    hi ankit i still didnt understand how it works. plz help me out.
  15. Nov 12, 2008 #14
    On second glance, there's something obvious missed. Retracting the actuator will move would move the left shoe slightly into the drum as the bell crank pivots become horizontally displaced from one another. So I don't get it.
  16. Nov 12, 2008 #15
    I am not sure if I understand you correctly phrak, but I assume you are talking about what happens if the left arm of bell crank starts above the pivot, then it ll get horizontal first as it goes down & push into the drum.

    But I don't see any problem in that, we can start with the lever arm below the pivot.
  17. Nov 12, 2008 #16
    I ll tell you in class on 19th:biggrin:
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