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Torque in ft/lbs from engine

  1. Mar 29, 2014 #1
    If a gas engine makes 20 hp at 3000 rpm I figure the ft/lbs of torque equals about 460.
    Negating friction if the drive axle the engine turns goes 10 rpm with 20 hp the ft/lbs of torque equals about 10,000.

    I was using the formula hp x 5252 divided by rpm of shaft

    What I need to know if 10,000 ft/lbs is right. The drive axle has a sprocket hub with a 3/8th inch grade 8 bolt going through the hub and axle . It seems the 10,000 ft/lbs is right out of wack given the engine size.

    The axle dia. is 1.25 inches so I figure the un-threaded part of the bolt would take up to 460ft/lbs of torque before shearing.
    The shear on the bolt according to the chart is 80,000 pounds/inch square

    Thanks ED
     
  2. jcsd
  3. Mar 29, 2014 #2

    SteamKing

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    You might want to check your initial calculation: 20 HP @ 3000 RPM is not equivalent to 460 lb-ft of torque. If that were true, a John Deere lawn tractor would require a Chevy 454 motor to get around.
     
  4. Mar 29, 2014 #3
    20hp@3000rpm=35ft/lbs
    20hp@10rpm=10,000 ft/lbs
    using the formula hp x 5252 divided by rpm of shaft
    Can that be right
     
  5. Mar 29, 2014 #4

    SteamKing

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    Assuming you have some sort of reduction gear between the gas engine and your drive axle, the gear ratio would be 3000/10 = 300:1. The gas engine torque required to produce 20 HP @ 3000 RPM = 35 lb-ft, therefore, the torque acting on the drive axle after the speed reduction will be 35*300 = 10,500 lb-ft torque. An axle diameter of 1.25 in seems a bit undersized, so lets check:

    http://www.engineeringtoolbox.com/torsion-shafts-d_947.html

    Max. Torque = (π/16)*σ*D[itex]^{3}[/itex]

    where D is the diameter of a solid shaft,
    σ is the max. shear stress of the shaft material.

    Use AISI 4140 HTS steel: yield stress = approx. 700 MPa = 101.5 ksi
    use 0.4*yield for max shear stress = 40.61 ksi

    So, for a 1.25" OD drive axle of AISI 4140 material

    Max. Torque = (π/16)*40.61*1000*1.25[itex]^{3}[/itex]
    Max. Torque = 15,574 in-lbs = 1297 lb-ft

    Drive torque = 10,500 lb-ft

    Drive axle needs to be much bigger than 1.25 in. OD
     
  6. Mar 30, 2014 #5
    Thanks SteamKing for your response
    I have been thinking about the problem for a time now and concluded the senario is unlikely.
    This contraption I want to build have 6 wheels driven on 6 separate axles (3 on one side and 3 on the other side) so it is a skid steer arrangement.
    Would the drive axles only get that torque load if tires suddenly stopped rotating and could not slip while the engine was making 20hp and the axles was turning 10rpm. Thanks ED
     
  7. Mar 30, 2014 #6

    SteamKing

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    Without knowing more about the arrangement of your drivetrain, I can't say. With wheels turning at only 10 RPM, it doesn't seem like this vehicle will be going anyplace quickly.
     
  8. Mar 30, 2014 #7
    No it doesn't need to go fast. I am using it to pull a boat on rough narrow trails around Hearst Ontario Canada. Last thing I need is a collision with a tree when months could go by before seeing another person.
    About 10 miles per hour is max.The drive train is like an Argo ATV which has a skid steer setup. A drive chain comes from the one side of the transmission for the left side front axle and the right side front axle is connected to the trans. again with roller chain. In other words the transmission has 2 output shafts one on the right and the other on the left.When the right side is locked up with the right brake the contraption turns to the right.
    The front axle driven from the trans.,is connected to the second axle with a roller chain and the last (third axle is connected to the front axle again with roller chain.Sounds confusing but that's how it works. Thanks ED
     
  9. Mar 31, 2014 #8

    jack action

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    That's your problem right there. If you want to go 10 mph with a wheel turning 10 rpm, that would mean your tire is 336" in diameter !

    Use these equations to find the appropriate numbers.

    You will most likely need something closer to 200 rpm (at the wheel, i.e. gear ratio = 1) - which would mean a 525 lb.ft torque instead of 10 000 lb.ft.
     
  10. Apr 1, 2014 #9
    Thank You Jack Action for your valuable input.
    So given the wheel is driven with a 525ft/lb torque I need to select a grade 8 bolt to use as a pin through the drive sprocket hub and 1.25 dia drive axle.
    Would a 3/8th dia.grade 8 bolt be ok if the unthreaded shank part of the bolt is on the shear planes. . It would be a double shear because the bolt goes through both sides of the hub.
    I figured the tensile of the bolt 150,000psi x 60% =90,000psi shear
    bolt area=3.14 x .188 x .188 x90,000=10,000lbs to shear the bolt
    1.25 inch axle dia.divided by 2 = .625inches to feet =.052 feet x 10,000lbs = 520 ft/lbs to shear the bolt
    but because it is double shear one bolt should do it. Is that right???
     
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