Tuesday, February 24, 2009

Quick Release Force

Have you ever wondered about  quick-release clamp force? (that which holds your wheel in the frame) ?   Way back, this came up in a forum, and it was mentioned that Bicycling Magazine actually tested/measured it! (imagine the time wasted!) At the time,  I commented that, that was pretty silly because it is easily calculated. Which I didn't bother to do until now and yes it is easy calculation!   

From BLOG Pictures

It is a statically indeterminate calculation, meaning it is not solvable with force equations alone, deflection/stiffness of components  is also considered.  Since Metal (steel ti or alum in most cases for bicycles), under elastic conditions, behaves linearly (simply meaning that if the force doubles, so does the deflection, and so on), this is not an issue.  The components can be thought of as Springs, where Force=SpringConstant*Deflection  or  (F=k*x).

In the case of a bicycle hub, the main components are the quick release rod elongating, and the hub's axles, compressing.  Also involved are the nuts, and frame dropouts. 

For items under axial loading the equation to calculate the spring rate is ElasticModulus*CrossSectArea / Length    or   (k=EA/L).

The amount of deflection introduced to the system is the "throw" of the quick release lever, which is typically reccomended to be 90 degrees (so we will use that).  With most quick releases this equates to 1 mm of deflection.   

In the table below (done with MS Excel) the measurements came from a typical shimano rear hub setup.  Everything is assumed to be steel (thus E=30e6psi).   Most of the measurement are not exact and for these purposes really don't need to be, with maybe exception of the quick release rod.  Assumptions also have to be made as to where in the threads does the force path change from nut to bolt. (good assumption is 3 threads in...)
ItemModulusLengthOuter DiaInner DiaOD of clamped volX-Sect AreaSpring Rateamount of deflection
ELdodid3AAk (EA/L)
Axle Nut30E638179.5156.10.2421910150.00091
Axle Nut30E638179.5156.10.2421910150.00091
QR Nut30E66154164.10.25412721510.00014
QR Nut30E65160201.10.31218698800.00009
QR Rod30E61565019.60.03058530.02958
Overall Spring Rate [lbf/in]4398
amount of throw [in]0.039
Amount of Clamping Force: [lbf]173
Amount of Force if consider only QR rod:  [lbf]230

Only 170 lb !  Not much.  Mostly dependent on the Quick Release rod, as expected since it is considerably more flexible.  Other notes:
-Compare to a clamp force of bolting: ~8000 lbs vs 170 lbs.  Thus the need for locknuts that "bite" into and deform dropouts.  Thus  vertical dropouts, Lawyer Lips, disc brakes causing pullout, etc. 
-If you going to use a Ti-rod, be wary.  With a modulus of about half of steel, you will only get 1/2 the clamp force (<100lbf!)>
-The calculation is a lot easier and can be almost as accurate if you consider only the qr rod.  It =   E*A/L*AmtThrow.  If the axle is titanium or aluminum with small x-sect area then you will want to include that as well.  
-Obviously if you are able to apply more Throw/deflection into the quick release, the amount of force goes up linearly.  Ie in this particular case, 2mm would give 340 lbf.  
-If "things" plastically deform during the clamping, then all bets are off on clamp force. 
-How about quick releases with plastic (actual plastic) in the force path ??   Sucks to be you, Good luck !!    No actually the plastic is so thin, ie the "L" is small thus the spring rate is still high. But yeah, personally I do not use them.  
-Note the approx 7 mil of axle compression (for a typical 10mmx1 steel axle..).  Thus the reason to have a little play in the bearings under no-load. 
-Note the nice mix of American and Metric Units.  Ametrican!  Like a lot of other American Engineers, we tend to use what we are use to.  Another quirk,  7 mil = 7 thou (thousandths) = 0.007 inch (mil is not millimeters, but rather is short for a milli-inch, crazy huh?)