Shaft Support

Discussion in 'Boat Design' started by Wayne Kendall, Apr 28, 2013.

  1. Wayne Kendall
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    Wayne Kendall Junior Member

    I have a 1 1/2 inch stainless shaft in my boat. Wondering how much distance should there be between supports. Is 4 feet too much with a steady bearing
     
  2. JSL
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    JSL Senior Member

    A lot depends on material, rpm, hp, etc. but a very good approximation is 40D max. and 20D min.
    1.5" shaft = 60" to 30". You are at about 26D so make sure the alignment is very good
     
  3. Wayne Kendall
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    Wayne Kendall Junior Member

    My question in the previous post should have read. Is a 4 foot distance too far without some type of a steady bearing? The hp of the engine is 135 and I run around 2000 Rpms

    Thanks
    Wayne
     
  4. JSL
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    JSL Senior Member

    you should be okay. One source is Gerr's 'Propeller Hanbook', pg 93. Or, check with your shaft or prop people, they should be able to help
    You can also use ABYC Part P6.6
     
    Last edited: Apr 28, 2013
  5. tom kane
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    tom kane Senior Member

    The twenty forty rule applies. No closer than 20 times the diameter of the shaft and no further than forty times the shaft diameter. Of course you may have to compromise.
     
  6. DCockey
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    DCockey Senior Member

    What is the reason for a minimum distance between bearings?

    How does maximum shaft speed enter into the "40/20" rule?
     
  7. Ad Hoc
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    Ad Hoc Naval Architect

    To prevent excessive bending and whirling. Whirling calc's are standard on Class vessels.

    Luckily MikeJohns has answered this one previously. He also kindly gives you a simple rule of thumb formula. In your case 1.60m or approx 5 feet. Viz:
    http://www.boatdesign.net/forums/me...bearings-ideas-please-27981-2.html#post282610
     
  8. DCockey
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    DCockey Senior Member

    Maximum distance, not minimum distance, between bearings would be to prevent excessive bending and whirling.

    Anyone know the rationale for the minimum distance recommedations.
     
  9. daiquiri
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    daiquiri Engineering and Design

    Ad Hoc probably wanted to say "Max. bending stress". If shaft bearings are not properly aligned, both shaft and the bearings will be subject to a mechanical stress induced by the misalignment.

    The rationale goes like this:

    Consider a cantilever beam of length L, with a load F applied at the free end. It can be a simple model of a shaft connected to a gearbox at the fixed end, and supported by a bearing at the free end.
    The bending moment induced by the load F will create a displacement Y of the free end, and the well-known relationship is valid:
    Y = 1/3 (F L^3) / (E J)​
    But the same situation can be created also by displacing the free end by a quantity Y. The load necessary for this will again be equal to F. By solving the above equation for F, you get:
    F = 3YEJ / L^3​
    The bending moment at the root of the beam is M = FL, hence:
    M = 3YEJ / L^2​
    As you can see, the smaller the length L, the higher the load acting on a beam (or a shaft) for the same displacement (misalignment) - and hence the higher the bending moment-induced stress at the root of the beam. Hence, there will be a minimum distance (for a given misalignment) below which the stress will exceed the maximum allowable for the shaft material.

    It should be noted that the same loads will be transmitted to the gearbox or to the coupling, and to the bearing(s) - and from these transmitted to the hull. Each of these components have maximum allowable radial, axial forces and bending moment which they can safely handle. Exceeding them can cost money and nerves. ;)

    Cheers
     
  10. DCockey
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    DCockey Senior Member

    Or he simply mis-read "minimum" as "maximum" and gave a respone which is valid for "maximum". That would be consistent with Occam's razor.

    My guess also was that that the generic recommendations for minimum bearing spacing are connected with potential bearing mis-alignment and resulting radial loads on the bearings and bending stress in the shaft increasing as you showed inversely proportional to square of the distance between bearings. The radial load and stress are also proportional to the magnitude of the bearing mis-alignmet (Y above), the section modulus (J above) and modulus (E above). A generic guideline for minimum bearing spacing of a multiple of shaft diameter (such as 20D suggested previously in this thread) would need to be linked to an assumption or guideline for bearing alignment accuracy.

    Also, could radial load on the bearings become a concern before maximum stress in the shaft become a concern?
     
  11. daiquiri
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    daiquiri Engineering and Design

    Yes it can, and is practically nearly always so. As you probably know, radially loaded rotating components are subject to increased wear and to fatigue. Hence, their lifespan (measured in hours or in number of cycles) will decrease as radial loads increase. An example of how life rating of a ball-bearing is calculated in function of radial loads and rpm can be seen here: http://www.astbearings.com/radial-ball-bearings-life-and-load-ratings.html
    The same principle is valid for any type of bearings or rotating equipment, including the plain bearings and gearboxes. But it is a type of calculations performed for industrial, automotive and similar money-aware applications. In boat and ship design there is an additional problem of hull sagging and hogging in seaway, which introduce an additional and highly variable factor, difficult to translate into the above equations. So we are often satisfied with generic rules (at least in small-boat design - don't know how it works for big projects), like 20-40 shaft diameters - which allow us to consider the shaft as a slender elastic beam and keep radial loads to levels which are statistically proven to give a lifespan sufficient for the average (whatever it means) use.

    Cheers
     
  12. Ad Hoc
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    Ad Hoc Naval Architect

    Correct. I am actually slightly dyslexic and when I am tired, or its late or early or a combo of these, I miss read incorrectly more than I read correctly. So, yes, I miss read min for max. Sorry about that.

    It is not a cantilever beam D. It is a continuous beam which is simply supported. If you were to consider a uniform beam between supports that is vibrating, we can assume this to displacement to be “y”. If a small element of the shaft say “dx” in length it therefore has a mass of (m/l)dx then it is simply Newtons 2nd law F = ma, the force coming from the shaft.

    The mass = (m/l)dx
    The acceleration = d2y/dt2

    From the general Eqn of bending of beams = EI.d2/dy2 = M

    From this, as usual, we can integrate to get the slope or once more to get the deflection, y, or differentiating to get the shear force and once more to get the ‘rate of loading’, which leaves a forth order differential equation, viz:

    EI d4/dx4 = M = (m/l)d2y/dt2
     
  13. daiquiri
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    daiquiri Engineering and Design

    Hi AH,

    Your considerations on vibrations of rotating masses imo add to my observation about static radial loads introduced by misaligned bearings. The dynamic vibrational analysis define the higher end of the allowable supported length, the static radial loads define the lower end.

    the cantilever beam in my post is a simple model which serve simply to show the importance of supported length L in the generation of radial loads, in presence of a misalignment. If bearings allow an angular dis-alignment of the shaft (like rolling-element bearings), then it is surely not a cantilever beam - as you rightly said. But if they do not allow it, or if they partially allow it, then we cannot consider a shaft as a simply supported one, because reaction moments will be created by bearings. For example, depending on design, plain or sleeve bearings might allow just a very limited amount of misalignment and can exert a significant radial load on the misaligned shaft, while at the same time suffering a significant wear created by the shaft.

    In any case, the general formula for beam deflection under static load (or static load due to beam deflection) should remain the same in case of cantilever or otherwise supported beam - the only difference being the factor which multiplies the right-hand term:
    M = K (yEJ / L^2)​
    where K will depend on the type of shaft bearings.

    I think this is a great reading (generic, not a boating-related one) for those interested in misalignment-induced wear and failures in mechanical transmissions: http://www.unitechinc.com/pdf/Why_Misalignment_Continues_to_Rank1_Article.pdf

    Cheers
     
  14. Ad Hoc
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    Ad Hoc Naval Architect

    It is still simply supported. If the shaft is not rotating, i.e. it is static, what is holding or supporting the shaft weight?..the bearings. Whether these are “fixed” or “loose”, the shaft is being supported. Having a misalignment does not make the change the nature of the support, only the loads imparted to it. Since if the support is fixed or “built-in” in beam nomenclature, this requires a bearing of sufficient length, along the shaft to “fix” it. Since to be “built-in” implies zero slope and deflection at the ends, or in this case the bearing. Which clearly it can’t be the case. Thus it is simply supported.

    Since if the bearing allows some degree of angular movement, it does not move in one axis in-plane at the left and right hand side of the bearing as a rigid body, as that would imply the shaft is no longer horizontal at that support, since what about the other end? The other end, the bearing, if this also allows angular movement it must be equal and opposite to the other to remain horizontal., and thus has a defection somewhere along the span between the supports. Otherwise the shaft is simply rotating about a single bearing like a handle. (Having non uniform misalignment at each support can occur but is rather more complex.)

    So the bearing is a “node”. When the shaft is static (not rotating), the bearing quite simply supports the weight. When the shaft is rotating it still supports the weight. But it also allows, depending upon the type of bearing, longitudinal, angular or radial movement, or all 3. The amount of contact is determined by the bearing and the design of the shaft.

    So when the shaft has angular deflection it is the first integral i.e. dy/dx of the bending moment eqn. But when rotating the restoring force, or ‘spring’ if you like, to bring the shaft back to equilibrium is the not the slope (dy/dx) but comes from the shear force (noted above), which is the differential of the bending moment, M, dM/dx or simply d3y/dx3.

    You can treat it though as if the shaft has a central mass, and the shaft is “flexible”. Then the stiffness of the shaft (like a spring) is directly related to the deflection.
     

  15. daiquiri
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    daiquiri Engineering and Design

    AH, frankly it looks like we are talking about the same thing, but not quite about the same thing here... :p

    Might be a matter of different terminology used. Forget about shaft rotation for a moment, and consider just statics. The aim is to consider 2D elastic line of the shaft. Hence, under these conditions, in my part of the world, a simply supported beam has compliant bearings which allow non-axial angular movements. A fixed support does not allow such movements. A semi-compliant support would be the one which allows rotation, but with some reaction moment by the bearings.

    If I understood you properly, you are talking about a shaft with compliant bearings. If that is the case, the question is: if the shaft is simply supported (compliant bearings) why should there be a recommended minimum (!) distance between supports?

    You could have two supports at 1cm distance, for example, if they were compliant. The shaft would remain straight and simply change it's rigid-body in-plane angular position to pass through both of them...
     
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