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  • Calculate clamp force on bike handlebar

    Discussion in 'Calculations' started by FLM, Jan 22, 2015.

    1. FLM

      FLM Member

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      Hi Folks

      I am looking for some help in a calculation if possible.

      I am looking at a bike design where the handlebars are clamped in place in the stem using only 1 bolt - see the picture below for an example.

      http://i.ebayimg.com/00/s/MTE5OFgxNjAw/z/rdgAAOxyrYFR2wbP/$(KGrHqF,!k8FHQwpchTzBR2wbP!Now~~60_35.JPG

      There has been issues with the handle bars slipping forward when in use. I think that this is a manufacturing problem and not a design issue as this type of stem with one bolt is quite common.

      However I would like to prove this mathematically. How do I calculate the force that the bolt and clamp have on the handlebars? I would like to compare a calculation of 1 bolt versus 2 and 4.

      Any help greatly appreciated.

      Thanks
       
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    3. Lochnagar

      Lochnagar Well-Known Member

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      Just so that we are all on the same page about what we are talking about here "in general terms" – please refer to the pictures in the hyperlink below of two fundamentally different types of coupling here:-

      There is a one-piece clamping collar – which has been “weakened” by cutting partially through the wall on the opposite side – however, as you will see in the pictures in the link below – they can have one or more bolts to clamp them.

      There is a two-piece clamping collar – which has been made by cutting completely through the wall on both sides – however, as you will see in the pictures in the link below – they can have two or more bolts to clamp them.


      http://www.ondrives.com/collars-and-clamps


      The two-piece clamping collar has two tangential screws. The main advantage of the two-piece collar is that it can be installed without having to remove other components on the shaft. The two-piece collar also provides more “holding power” than the one-piece clamping collar. With a one-piece collar some of torque force applied to the screw serves to close the collar to the bore. In the two-piece collar all torqueing effort is available because of two screws equally spaced on both sides of the collar.

      To compute the “torque resistance” of both systems – you would realistically need to do an FEA analysis – as the tolerance of the hole and bore are of great significance in determining this torque resistance – and what you need to determine is the pressure distribution around the handlebar – which as I said will be very much affected by the bore and hole diameters – and also the wall thickness of the collar – since it has to allow distortion of the clamp diameter to the shaft diameter.

      I would take a guess – that in the case of your bike – you have probably got a very loose fit between the collar and the handlebar – to avoid scoring the handlebars when assembling – which then means in “simple terms” that you probably don’t have a uniform pressure distribution around the handlebars when they are clamped – since the clamp is being distorted from a “large circle” to a “smaller circle” – and as a consequence the “large circle” is probably becoming a little oval in shape as a consequence.

      Hope this helps.
       
    4. FLM

      FLM Member

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      Hi lochnagar

      Sorry to be a bit more clearer I have attached a proper picture of the design. I understand exactly what you are saying and I agree - the problem is most likely a manufacturing/assembly issue. What I am trying to do is find and equation to compare the force of 1 bolt versus 2 or 4. [​IMG]
       
    5. teja

      teja New Member

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      sir i am having doubt on loadings on shaft.which is of slag transfer car which moves on a rail
      now there is a shaft which is having two wheels and two bearings(axle box) at ends and gear box at middle
      and my doubt is where the entire body weight act and and scrap weight i should add and where reaction forces act?
      i will be glad if you reply
      thanks
       
    6. Lochnagar

      Lochnagar Well-Known Member

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      Hi FLM,

      You say the bolt is 7mm - which notionally appears an unusual size to me - because the "standard sizes" are M5, M6, M8, M10. I am not saying it is not an M7 - all I am saying is that is not a "standard size" - therefore it might be worth double checking the bolt shank/thread diameter.


      If the bolt is a stainless bolt M6 x 1.0Pitch (Grade A4-70) - this will have a proof strength of 450N/mm^2 - so we would typically tighten it to 75% of the proof strength - which roughly computes to 8.5/9Nm. (It varies a little depending on bolt washer diameter). So at 8.5Nm tightening torque - this equates to a clamping force of 6730N.

      So if you look at the clamping bracket now - to get equilibrium - we have 6730 x 51.7 - P x 28.67 = 0 (Where P is the clamp force on the handle bar). P = 12136N.

      You haven't stipulated the material of the handlebar - or the clamp - but lets assume it is aluminium. The coefficient of friction for aluminium to aluminium - is roughly 0.3 - though higher figures are sometimes quoted - but you need to remember that the surfaces here are going to be smooth.

      So the friction force on the handle bar - resisting rotation of the handlebar in the clamp is 0.3 x 12136 = 3640N. So we have torsional resistance of 3640 x 2 x 11/1000 = 80Nm. As your handlebars look to have your hand grips about 200mm above the pivot point - then if you apply a force of 400N (40Kg) - they will rotate.

      However, there are a few points to note.

      1) You appear to indicate that the bolt does not have a nut in your picture - but instead goes into a tapped hole in the lower "U" shaped bracket. If this is so - then you will not tighten the bolt up to 8.5Nm as you would strip the threads in the aluminium bracket - so you would probably tighten it to something closer to 5Nm - if you are using 6082-T6 aluminium. (I personally don't like a bolt that does not use a nut).
      2) I don't like the design of the lower "U" shaped bracket - because on the right hand side you have made it completely inflexible by welding the gusset to it - therefore it can't deform to clamp the handlebar.

      Hope this helps.
       
    7. FLM

      FLM Member

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      Ok this is appears to be on the lines of what I am looking for.

      I am confirming the bolt details but yes as far as I know it is an M7 bolt. I dont as of yet know the pitch or thread quantity.

      The materials used are steel - i am awaiting confirmation of the grades of steel. The CoF for steel is 0.2.

      Your proof strength I assume is from a table? I will get this once I get the bolt specs. I am not sure where you get 8.5Nm from (something simple I am missing). You then work out clamping force - what eqn are you using there? F=T/cD?

      If you could explain this paragraph below and where you get the equations and figures from that would be great:

      So the friction force on the handle bar -resisting rotation of the handlebar in the clamp is 0.3 x 12136 = 3640N. So wehave torsional resistance of 3640 x 2 x 11/1000 = 80Nm. As your handlebars lookto have your hand grips about 200mm above the pivot point - then if you apply aforce of 400N (40Kg) - they will rotate.

      And also to clarify there is a nut used - the pic I posted does not show this. Have a look at the picture below:[​IMG]
       
    8. Lochnagar

      Lochnagar Well-Known Member

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      Hi FLM,

      The equation to compute the 8.5Nm - I use is from Shigley - which is the "Engineers Bible" - see the link below - you will see the equation I use is at the bottom of page 437. There are a few other expressions you need from Chapter 8 of this book - and so I have brought everything together and made my own Excel Spreadsheet.

      https://books2all.files.wordpress.com/2013/12/shigley-machine-design-9th-edition.pdf

      The bolt clamps the tube - which exerts a radial force of 12136N. The tangential force is the friction force resisting rotation - and that is the friction coefficient times the normal (or radial force). (If you think about a block you are trying to push across a desk - the resistance to motion is the coefficient of friction between the two surfaces times the weight of the block - so the friction force is perpendicular to the normal or "weight force"). Hence we have 12136 x 0.3 = 3640N.

      So the force on the top of the handlebar is 12136N - and there must be an equal and opposite force on the bottom of the handlebar of 12136N. So from the centre of the handle bar - we have a radius of 11mm (=0.011m). So the total torque resistance is the "top friction force" x radius + "bottom friction force" x radius. Hence we have 3640 x 2 x 11/1000 = 80Nm.

      Your handle grips (where your brake levers are) - look to be 200mm above the centre of the 22mm diameter. So the torque is the force x perpendicular distance. So if you exert a force of 400N (40Kg) - horizontally - then I would expect the handlebar to rotate.

      You should remember - that there are a number of approximations/assumptions in any calculations - and I guess the "assumed" coefficient of friction is open to some "debate". Ideally, you would seek to quantify this value more precisely with experiment on the surfaces that you are doing the calculations on.

      Hope this helps.
       
    9. FLM

      FLM Member

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      Hey sorry but I am still not seeing where you are getting 8.5 from? You are using T=KFd?
       
    10. Lochnagar

      Lochnagar Well-Known Member

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      Have you looked at page 437 of the book I posted a link to - see above? As I also said in my earlier posting - you need to preload the bolt to about 75% of the proof strength. So the axial load on the bolt needs to equivalent to 75% of the proof strength. So for a M6 x 1P the cross sectional area is defined (not as pi D^2 / 4) - but as defined on page 412 - see the equations at the foot of the table. So you will see that the XSA = 20.1mm^2. So 6730/20.1 = 335N/mm^2. And 335/450 = 75%.

      If you don't like the more complex equation I have detailed on page 437 - then you can use T=0.2 x F x D - though I have to say I prefer the full equation on page 437 - since it deals with washer diameter too. However, if you use T = 0.2 x 6730 x (6/1000) = 8Nm. If you use the full equation on page 437 - you get 8.5Nm.

      Hope this helps - and by the way I will have a pint of Guiness:)
       
    11. FLM

      FLM Member

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      Yep I was looking at that equation and went on to page 438 where they have the simplified eqn which I had been using anyway. I thought surely he can't mean that huge eqn!! Ok leave this wit me as I don't have all the required info but hopefully will early next week..... And yes after all your help I will send you a 6 pack of the black stuff!!
       

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