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Max Force on a single rototiller blade

  1. Aug 2, 2016 #1
    I am looking to find the max force on a single rototiller blade. The blade is being rotated at 180 RPM with 60Hp. The blade is bolted to a Flange that is welded to center shaft. The shaft has a 3.5" diameter. The diameter of the cutting circle is 24". The blade is a 10.25"x3"x.25" rectangle. If the blade was spinning at full speed and hit a rock what would the max force of that impact be?
     
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  3. Aug 3, 2016 #2

    Mech_Engineer

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    You need to know something about the blade's rotational moment of inertia which you can use to calculate the blade's angular momentum and estimate the force it would take to decelerate the blade in a very short distance.

    Are you trying to find the impact torque that would be applied to the shaft?
     
  4. Aug 3, 2016 #3
    I am looking to see if the blade will brake off if the tiller hit a rock and how big the bolts need to be that hold the blade in place. This will also give me an idea on how much force will be applied to the shaft
     
  5. Aug 3, 2016 #4

    jack action

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    You are looking at this the wrong way.

    Don't try to identify the maximum force you can encounter in an unknown environment which, in theory, could be infinite.

    Instead, design the «weak spot» of your machine, i.e. where you wish it would break such that your design is safe for the user. For example, you could decide that the weak spot is the upper end of the shaft such that the blade and shaft stay in place within the bearings while slowing down, then you know that the bolts holding the blade and the blade itself should resist the torque needed to break the weak spot of your shaft. Therefore, the weak spot will always break before anything happen to the blade or the bolts.
     
  6. Aug 3, 2016 #5
    Jack action, the final design will have a slip clutch or shear bolt in line to be the "weak spot" i just was wanting to know my limits.
     
  7. Aug 3, 2016 #6
    And that is what you were given.
    The slip clutch or shear bolt will allow a maximum torque to be delivered to the blade(s).
    The bolts holding the blade to the shaft will have to withstand that torque, and not break before, when abruptly stopped from rotating.
     
  8. Aug 4, 2016 #7
    The shear bolt will be rated at 60 hp. I just cannot remember how to so solve for when the blade stops abruptly.

    Finding the angular moment of inertia I just use L=Iw. After that what equation do I use to solve for the abrupt stop.
     
  9. Aug 4, 2016 #8

    jack action

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    You're looking for the angular impulse:
    [tex]\int_{t1}^{t_2} Tdt = T_{avg}(t_2 - t_1) = I(\omega_2 - \omega_1)[/tex]
    The problem for you will be to identify ##t_2 - t_1##, i.e. the time the blade takes to stop. So it won't help you at all.

    You don't seem to grasp the fact the shear bolt will break at a predetermined torque so there is no need for your blade (or bolts holding it) to resist more than that torque. Or are we missing what you are looking for?
     
  10. Aug 4, 2016 #9

    Nidum

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    (1) The blade as a whole won't stop instantaneously . The blade and the rotor hub will continue to rotate after blade impact . For a brief period after initial impact the blade will be bending and the motor torque will be rising . The bending force on the blade comes from the max torque achieved at the hub root , the geometry of the blade and the location of the point of the impact on the blade .

    (2) There will be some effect of shock loading but it is unlikely to be as big as is commonly assumed . Shock loading in this type of situation usually gets dissipated by local distortion around the point of impact .
     
  11. Aug 5, 2016 #10
    I am think I am just confusing myself. So if I have 180 rpm and 60 Hp, I have a torque value of T=1750 lb-ft. Is this the max amount of force I will see at the end of the blade if it were to hit a rock (disregard the shear bolt)? Or is there a shock force/ impact force that will be greater? If so what will this force be?
     
  12. Aug 6, 2016 #11

    jack action

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    The torque you will see at any point ##x## on the rotating components when the blade hits a rock will be:
    [tex]T_x = T_{in} - I_x\alpha[/tex]
    Where ##I_x## is the moment of inertia of the components from the shear bolt up to point ##x## and ##T_{in}## is the maximum torque handled by the shear bolt.

    When hitting a rock, the blade will decelerate, so ##\alpha## will be negative thus - you are right - the torque at ##x## is greater than ##T_{in}##.

    How big is ##\alpha##? That is the million dollar question which can only be answered by proper testing. And the results are valid for only the conditions tested. The value of ##\alpha## depends on the amount of deformation of both your machine and the object that was hit, which depends the elasticity of the materials and the shapes of the machine components and the object.

    Are all this trouble worth it? Let's take the bolts holding the blade to the shaft. ##I_x## will be basically the shaft, so probably relatively small compared to the blade itself. Even then, estimating the time (##\alpha = \frac{\Delta\omega}{\Delta t}##) would be a wild guess. Certainly not more precise than estimating a safety factor.

    Estimating deceleration

    Another way to look at it (that doesn't involve knowing the time taken to stop) is to estimate the work done to absorb the energy of the hit. The energy equation when completely stopping the blade from angular velocity ##\omega## will be:
    [tex]\frac{1}{2}I\omega^2 = T\theta[/tex]
    Where ##T## is the average torque (may reach some higher peaks) and ##\theta## is the combined angular deformation of the rotating components. ##\theta## can be found based on the torsion definition (##\theta = \frac{Tl}{JG}##; which do not include the deformation of the hit object, so it gives a safety margin), such that:
    [tex]T = \sqrt{\frac{I\omega^2JG}{2l}}[/tex]
    That would be the torque needed to decelerate the rotating components. So it should be equal to ##I\alpha##, therefore:
    [tex]\alpha = \omega\sqrt{\frac{JG}{2Il}}[/tex]
    Which could be used in the previous equation for any component ##x##.

    I did not test that, but the math looks good.
     
  13. Aug 6, 2016 #12

    Baluncore

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    I would attach each individual blade with two bolts, one large that will act like a hinge, the other small that will act like a shear pin. When a blade hits a rock it will shear the smaller bold and fold out of the way, being retained by the larger bolt.

    This is an agricultural design problem. You will need to do some stepwise improvements based on trial and error. Any attempt to design agricultural equipment to withstand maximum forces will result in heavy expensive equipment, unsuited to the market. You are better to build a prototype “conveniently” that “looks right”, then re-design the bits that break during prototype testing.
     
  14. Aug 6, 2016 #13

    anorlunda

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    If the blade and bolt are too strong, the whole machine could lurch up in the air when a rock is struck.

    If rocks are hit frequently, shear pins would be very annoying.

    The ductility of the materials is a critical parameter.

    I agree with others that testing rather than calculation sounds most promising. For initial guesses, I would reverse engineer a competing rototiller.

    Good luck.
     
  15. Aug 6, 2016 #14

    Nidum

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    +1
     
  16. Aug 6, 2016 #15
    I have a lot of firsthand experience with agricultural equipment designed by engineers (Ugggg)

    What's probably going to happen first is the tiller will lift off the ground to get over the rock..
    First question.. is the tiller blade curved? the 10x3x.25 inch dimension.. is that the raw plate before bending it (so that it has about 4" of cutting edge) leaving it at about 6" from the shaft OD.. so you have a total radius of about 8".. I have no idea what this tiller will weigh.. if it's 1000 lbs, it will take somewhere between 600-1000 lb ft of torque to lift it, factoring some acceleration.

    Baluncore +1 on the shear bolt.. just don't make it out of unobtainium... just a standard 8.8 or Gr 5 so a farmer can get it locally

    Trust me, the whole machine WILL come up.. I prefer that to replacing shear bolts..

    If you want to look at a well built machine, look up the Howard Rotovator.. Ours is a 5ft model from about 1960 and will keep going long after any of these Rinieri and Maschio machines
     
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