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  • Hydraulic wine press - Beam calculations

    Discussion in 'Calculations' started by Tomer1, Jul 16, 2011.

    1. Tomer1

      Tomer1 Member

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      Hi folks,
      Im a home winemaker from israel DIYing an hydraulic wine press but its been a few years since I finished college and most of the theoretical stuff is.... missing :D
      The design is a very common one among DIYers which borrows from tranditional and modern designs.

      The force applicator is a 5ton bottle jack (cheaper then mounting an hydraulic piston and using a compressor).
      The frame will be RHS galvinized steel which may or may not be powder coated,depending on the final cost.

      Here is how some commerically sold presses with about the same design look like,

      [​IMG]
      [​IMG]
      [​IMG]
      [​IMG]

      Now for the math part...
      I'l upload my calculations in an additional reply
       
    2.  
    3. Tomer1

      Tomer1 Member

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      [​IMG]

      Now,
      I want to use angular support welded to the top beam at L\4 (0.15m and 0.45m) 45 degrees.
      Which means Im now looking at 4 supports rather then two and max torque is reduced from 7500 kNm to 2.3 kNm.

      [​IMG]

      [​IMG]
       
    4. Tomer1

      Tomer1 Member

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      So for St37 I need Zx> 2340/161*E^6=1.454 E^-5 m3
      meaning the RHS size needs to be 44.3X44.3 so by looking at my catalog
      I can choose 50X50X2.5 (Inertia of 17.70 cm^4)

      And for St42 I need Zx> 2340/220*E^6=1.064 E^-5 m3
      meaning RHS needs to be 0.40X0.40 so I can choose 40X40X4 (12.1 cm^4)

      Now to calculate the 45 degres support, I need to use the reaction force of 34400N

      So 34400*cos45=24324 N
      Now the support beam needs to be calculated to withstand bending torque so I'l calculate its length.

      a/L = sin 45 , a=0.6m/4=0.15m

      0.15/sin45=L
      L=~0.213m

      So max torque = 0.213*24324=5181 Nm.
      Say I want to use the same RHS profile instead of round bar and have it cut at 45 degres for welding...

      Zx37> 5181/161^6=3.21*E^-5 profile= square root of 3 (6*3.21*E^-5) =57.7 so I can choose 60X60X3 (36.60)
      Zx42> 5181/220^6=2.35*E^-5 profile= square root of 3 (6*2.35*E^-5) =52 so I can choose 50X50X4 (25.50)

      So are my calcs correct?
      Will it hold?

      Thanks for the help.
       
    5. maniacal_engineer

      maniacal_engineer Well-Known Member

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      your design is statically indeterminate, which means that the loading you show assumes everything is zero before it is loaded. but normal tolerances mean that will probably not be the case.
       
    6. Tomer1

      Tomer1 Member

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      I just noticed a huge mistake misreading the tech sheet,the translation to hebrew got me.
      The profile required is much more substantial, 70X70X3 given ST37.

      Thanks for the reply , so what alse do I need to take into consideration?
      The self wight of the beam as distributed load?

      Can you comment about the 45o support?
      These are critical to my design and If I miscalculate and they end up deflecting the whole frame could possibly distort.
       
    7. hmck57

      hmck57 Member

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      If I understand your intended geometry correctly, then your analysis isn't quite correct but it is conservative for the member analysis. It is unconservative for the joint analysis however. Maniacal is correct: Your design is statically indeterminate, but that means more than residual stresses will be present. It also means that the internal loads are dependent on the relative stiffnesses of the members. So saying what the internal loads/stresses/deflections are isn't possible without more complete geometry.

      Look at the attached geometry (http://twitpic.com/5smh15) - is this something like what you are designing? This is a 1/2 model based on symmetry. Note this is only the left side of the frame, and 1/2 of the load. Each square in the grid is 0.05m. The overall length of the cross member plus the 45deg member is 0.6m, and the side members I used are 0.3m tall.

      If the geometry is near correct, then the deflected shape will look something like this: http://twitpic.com/5smhgz, and the internal bending moments will look like this: http://twitpic.com/5smhoe. I'm not familiar with ST37 - that is probably a European spec, maybe an older one.
       
    8. Tomer1

      Tomer1 Member

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      No I dont think thats the same design,
      Ive set up a quick drawing.

      [​IMG]
       

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