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  • Crank Torque/Piston Force Question

    Discussion in 'Calculations' started by Idiotic, Jun 29, 2012.

    1. Idiotic

      Idiotic New Member

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      Hi all,

      Before I throw myself out of the office window could someone please help me out a bit here...I'm sure I used to handle these sort of calcs years ago but now I'm not sure of anything!

      Got a pneumatic press module here and I'd like to calc the force on the piston at 1mm from TDC, I've drawn it out here as clear as possible and it's like this:
      [​IMG]
      A formula I came across in many of the books I've been looking through said Fp = T/OM
      I worked out OM as OM=OC(Sin(17.13+4.34)/Sin(90-4.34); then transposed and came up with Fp=3935N

      Could some one please let me know if I'm along the right lines or am I going mad?

      Many thanks

      I've just previewed the post and it looks a lot like a school/college homework question...I can assure you it's not, although that doesn't make me feel any better about my maths!
       
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    3. Erich

      Erich Well-Known Member

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      I approached the problem a different way, but I got the same answer. Go with it.
       
    4. Idiotic

      Idiotic New Member

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      Phew, that's good to know. Thanks Erich

      I'd be interested to know the way you approached it - I originally did it a different way, got a crazy answer which completely threw me and then it went downhill from there!
       
    5. Erich

      Erich Well-Known Member

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      I just broke the problem down and drew Free body diagrams.

      Calculated the Force in the conrod as a function of the piston force.

      The I went up to the crank and applied that force to the crank. I broke that force into a component normal to the crank and a component in line with the crank.

      Came up with equation relating Torque to Conrod force.

      substituted and solve for Fp vs T
       
    6. Idiotic

      Idiotic New Member

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      Got it...finally! Thanks for explaining your methods and also thanks very much for taking the time and trouble to help me out, it is very much appreciated.
       
    7. Auto Engineer

      Auto Engineer Active Member

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      Looking at the diagram but applogies in advance if anyone thinks I am slightly off the mark, but the obtuse triangle appears to me that the length I would not equate to 70mm?
       
    8. Auto Engineer

      Auto Engineer Active Member

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      Sorry my mistake should have worked it out first
       
    9. Auto Engineer

      Auto Engineer Active Member

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      Yep I was thinking about something should be done with the value of I, but I found this method works as well;

      Working with trig, POC triangle OC = 18mm, theta = 17.13 degress, Phi = 4.34 degress.

      Using the sine rule we have; PC = 18(sin 17.13/sin4.34) = 70.06 which is I.

      OCM triangle = angle C = Phi + theta = 21.47 degress
      POM = 90 degress

      Angle m = 85.66 degress

      Using the sine rule we get;

      OM = 18(sin 21.47/sin 85.66) = 6.61

      So Fp = 26 / 6.61 x 10^3 = 3.94 x 10^3 N
       
    10. BobInOH

      BobInOH New Member

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      A simpler approach to solve is: the applied moment is constructed of vectors, which are the applied force, Fp, and the moment arm, r, or, T = r X F = r * F * sin(angl OCP).
      Since the sum of angles in a triangle is 180 degrees, angle OCP = 180 - 4.34 - 17.13 = 158.53 degrees.
      So Fp = T / r sin(158.53) = 26 Nm / [18mm * sin(158.53)] = 26000 Nmm / (18mm * 0.3660) = 3946.42 N. QED!
       
    11. Auto Engineer

      Auto Engineer Active Member

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      I would'nt of thought that was a simpler approach. Idiotic gave a result of 3935N and asked for the force Fp acting on the piston, other posters said he was close to it and should go with it, however my worked out example showed how he did it, but the method showed how to calculate the compoent force, and not the force acting on the piston, which is the force 3946.4N, and is a different force acting.

      ;)
       

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