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Machine parts, conic clutch two teachers, two different results!

  1. May 14, 2013 #1

    Femme_physics

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    Machine parts, conic clutch...two teachers, two different results! :(

    So a friend of mine who studies for the same test scanned me his teacher's solution to a problem they solved in class. We solved the same problem with a different teacher. I could really use your help to figure out who out of the two teachers went wrong (or maybe they both did?). This is important.


    1. The problem statement, all variables and given/known data

    The Question:
    A conic clutch, shown in the drawing, rotates at 700 [RPM].
    Calculate the power that can be conveyed (calculate by uniform surface stress).


    [Broken]




    Uploaded with ImageShack.us
    2. Relevant equations

    Since there is a conflict between the used formulas, I won't list them here. I will write the variables...

    Rf = Friction's Radius
    P = Power
    n = rotational speed [RPM]
    Fa = Required axial force for the conveyance of the moment
    Dm = Needed average radius
    α= Conical Angle
    [P] = Max allowable surface contact stress
    [τ]= Max allowable torsion

    THE SOLUTIONS


    My friend's teacher:

    [Broken]

    My teacher:

    [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 16, 2013 #2

    Femme_physics

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    *BUMP* Anyone?
     
  4. May 21, 2013 #3

    I like Serena

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    What happened to your second scan? :surprised


    Anyway, according to this PDF (google rules!):

    http://www.freestudy.co.uk/dynamics/clutches.pdf
    the formula for a conic cludge with constant pressure is:
    $$M_{eff} = \frac {\pi p \mu}{12 \sin \alpha} ({D_o}^3 - {D_i}^3)$$
    with ##D_i = D_o - 2b\sin \alpha##.

    So:
    $$M_{eff} = \frac {\pi \cdot 0.3 \cdot 0.3}{12 \sin 15^\circ} (300^3 - (300 - 2 \cdot 60 \cdot \sin 15^\circ)^3) = 687100.7 \text{ Nmm}$$
    This appears to come close to your friend's teacher.



    According to the same article:
    $$P=\frac {2\pi n}{60} M_{eff}$$
    So:
    $$P=\frac {2\pi \cdot 700}{60} \cdot 687100.7 = 50367.1 \text{ W}$$
    The difference with your friend's teacher is mostly the division by an extra factor 2, of which I do not know where it is coming from.
    There seems to be some assumption that only half of the torque can actually be transmitted.


    Now what did your teacher write again?
    Your picture link appears to be broken.
     
  5. May 23, 2013 #4

    Femme_physics

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    [Broken]

    There's my teacher's solution, reuploaded!

    Wow, I can't believe my teacher's solution is wrong :( This is very disconcerting.

    Thank you so much for the input!!
     
    Last edited by a moderator: May 6, 2017
  6. May 24, 2013 #5

    Femme_physics

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    Printing this to show it to my teacher...see what he has to say!
     
  7. May 24, 2013 #6

    I like Serena

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    Well... it remains curious that your teacher is effectively dividing by sin(15) two times.
    That doesn't seem right.
     
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