What force does a quick release skewer create?

[Manc33]

Shimano says to use 5.0Nm - 7.5Nm of torque on their quick-release skewers.
Source: Third column at the top here:
https://si.shimano.com/pdfs/um/UM-3000H-006-00-ENG.pdf

When using a quick release, surely your wheel isn't only being held on with 5.0Nm - 7.5Nm of force?

Solid axles with nuts (according to Park Tool) should be tightened to 29.4Nm-44Nm!
Source: Click #4 at the right side then it's under the section "WHEEL, HUB, REAR COG AREA" here:
https://www.parktool.com/blog/repair-he ... d-concepts

So how come this is? It would seem the end result holds the wheel on about the same amount?

I noticed that even Park Tool doesn't specify a torque for QR.

I'm asking because I want to tighten a solid axle on a carbon frame (that at least, does have thin metal inserts on the dropouts) but the axle nuts need to be at around 35Nm-40Nm and I don't know if that's way more force than a quick release would be.

I suppose the crux of the issue is, what's the compressive strength of steel, 6061 aluminium and carbon fiber?

According to Google those strengths are as follows:

Construction Steel: 250 MPa
Carbon Steel: 550 MPa
Aluminium (6061): 290 MPa
Carbon Fiber: 869 MPa

This makes clamping axles at 40Nm to 6061 aluminium look dangerous

I know when I filed down part of my (6061) frame a while back I was like "Oh, it's done already, OK" whereas filing down 5mm thick stainless steel, I was there for ages. I also know that it will blunt three junior hacksaw blades when cutting carbon fork steerers down to size in the past - so yeah that stuff is incredibly hard.

Right now I have axle nuts tightened to 40Nm on my 6061 aluminium dropouts and have done over 250 miles without issue.

[Cyckelgalen]

QR skewers can exert a force of several hundred, up to 500 Kg.
One MPa is roughly 10 kg per square cm. If we assume that the contact surface between the skewer and the dropout is 1 sq cm, then, those 500 kg (at the very most) translate into 50 MPa, nowhere near the compressive strength of the softest material, aluminium.
In real life, it would be much lower than 50 MPa, because most skewers won't go up to 500 kg, and the actual contact surface between skewer and dropout is likely to be more than one sq cm, spreading the force exerted on a larger area.

[Cyckelgalen]

Interesting experiment on SQ clamping forces:
https://m.youtube.com/watch?v=du742eTDgUE

[Jdsk]

A metrologist writes:

I'm completely confused by the units in the discussion above.

When using a quick release, surely your wheel isn't only being held on with 5.0Nm - 7.5Nm of force?

You can't measure force in Nm. You can measure torque and energy. The SI unit of force is the newton.

One MPa is roughly 10 kg per square cm.

Is that kgf rather then kg? And if so why measure a clamping force in gravitational units? The SI unit of force is the newton.

Thanks

Jonathan

[Manc33]

The flange on the axle nuts (about 25mm) is a lot bigger than the surface area of a QR so there is that.

If it cracks my dropout then it cracks it, the frame has done thousands of miles - if not tens of thousands - and I'm probably at least its 3rd or 4th owner. It wasn't expensive, I'd just rather not break it!

[Jamesh]

I've never thought too much about it.

Just tightened them up tight!

Never damaged a. Carbon, alu or steel frame.

Cheers James

[MikeF]

I'm not entirely clear what the torque measuerment refers to. Is it the torque applied to the lever, or the torque applied to the nut with the lever open?.
The lever operates a cam so the force on the skewer will be magnified compared with that applied to the lever

[jb]

Nm is the torque required to turn the nut, it's not the force clamping the axle to the frame.
Skewers are different and the torque on the lever is not easily measured, but again it is not the clamping force on the frame.
As has been said some types of skewers can create quite large compression forces (in Newton's) and can shorten the axle slightly . Which is why Brucey always suggested setting up hub bearings with a little play when not compressed.

[Manc33]

Shimano and Park Tool don't give figures regarding the amount of pressure, they just say tighten bolts to x "Nm".

If I was confusing pressing forces with torque forces then I apologize.

[andrew_s]

Back of an envelope calculations...

A QR lever is 6 cm long. To apply 7.5 Nm with a lever that long takes 7.5/(.06×9.81) = 12.5 kgf on the end of the lever.
From the other direction, a QR skewer tightens by 2 mm when the lever is moved through an arc of 100 degrees, so the movement ratio is (2×.06×pi×100)/(.002×360) = 52.

The clamping force, ignoring friction, is therefore 52×12.5 = 650 kgf.
So, with a guess for friction, a clamping force of 500 kg is about right.

[Jdsk]

A QR lever is 6 cm long. To apply 7.5 Nm with a lever that long takes 7.5/(.06×9.81) = 12.5 kgf on the end of the lever.
From the other direction, a QR skewer tightens by 2 mm when the lever is moved through an arc of 100 degrees, so the movement ratio is (2×.06×pi×100)/(.002×360) = 52.

The clamping force, ignoring friction, is therefore 52×12.5 = 650 kgf.
So, with a guess for friction, a clamping force of 500 kg is about right.

That's interesting, but why use gravitational units of force? The calculation starts in newton-metres, and the pressure considerations for the material, as upthread, end in pascals. I'd express the forces, including the clamping force, in newtons.

Thanks

Jonathan

[jb]

A QR lever is 6 cm long. To apply 7.5 Nm with a lever that long takes 7.5/(.06×9.81) = 12.5 kgf on the end of the lever.
From the other direction, a QR skewer tightens by 2 mm when the lever is moved through an arc of 100 degrees, so the movement ratio is (2×.06×pi×100)/(.002×360) = 52.

The clamping force, ignoring friction, is therefore 52×12.5 = 650 kgf.
So, with a guess for friction, a clamping force of 500 kg is about right.

That's interesting, but why use gravitational units of force? The calculation starts in newton-metres, and the pressure considerations for the material, as upthread, end in pascals. I'd express the forces, including the clamping force, in newtons.

Thanks

Jonathan

Half a tonne is easier to visualise than 4.9KN.

[nirakaro]

why use gravitational units of force?

Perhaps because some of us non-metrologists find it easier to get an intuitive feeling for gravitational units? Probably sacrilegious, but I think of a newton as being about four ounces (not an SI unit, I realise). Wouldn't work on the space station, I know, but hey.

[Jdsk]

why use gravitational units of force?

Perhaps because some of us non-metrologists find it easier to get an intuitive feeling for gravitational units? Probably sacrilegious, but I think of a newton as being about four ounces (not an SI unit, I realise). Wouldn't work on the space station, I know, but hey.

Thanks

There are many situations in which gravitational measures of force work well, mostly of course where weight is important. I wouldn't choose to use them but I can see why others would. But weight doesn't play any rôle in this problem...

Jonathan

PS: It's about the weight of an apple! : -)

[Jdsk]

A QR lever is 6 cm long. To apply 7.5 Nm with a lever that long takes 7.5/(.06×9.81) = 12.5 kgf on the end of the lever.
From the other direction, a QR skewer tightens by 2 mm when the lever is moved through an arc of 100 degrees, so the movement ratio is (2×.06×pi×100)/(.002×360) = 52.

The clamping force, ignoring friction, is therefore 52×12.5 = 650 kgf.
So, with a guess for friction, a clamping force of 500 kg is about right.

That's interesting, but why use gravitational units of force? The calculation starts in newton-metres, and the pressure considerations for the material, as upthread, end in pascals. I'd express the forces, including the clamping force, in newtons.

Half a tonne is easier to visualise than 4.9KN.

And very conveniently a tonne is roughly a ton! Or at least roughly ones of the types of ton!

: - )

Jonathan