Bike porn

Ok;

Assume four point bend deflection (but works for all small scale deflection cases), non-dynamic loading case. Where major span is 100mm, and minor span is 50mm. Again, as this is constant this does not need to be exactly the same as real BB's.

Assume square axle of width (a) 17.68 mm, (b) 21.2mm. Where (a) is the maximum width for a square axle on a standard 25mm BB, and the same for (b) where 21.2mm is the maximum for the BB30 standard.

Then assume an exposed axle length of 25mm (not that this is used in the calculation, but needs to be non-zero to allow for over roller rotation). Apply symmetric loading on the central loading rollers.

As in;

     .......↓.........          ↓.........

---

---

..↑...................... ↑..

Then calculate area moment of inertia, I. Where;
I=bh^3/12 where b=width, h; height

From above we say that b=h, thus I=h^4/3

Thus for,
BB25 I=(0.01768)^4/12=8.14e-9m^4
BB30 I=(0.02122)^4/12=1.68e-8m^4

Let stiffness be, say.... 163GPa, (medium stiffness UD CFRP - In this case thats Hex IM7/8552)

Then using;

E[flex]=F(L^3)/4Id;

where d; deflection. Re-arrange for d

d=F(l^3)/E4I

Letting F (quasi again!) =1kN

Thus;

d|(a)=(1000)(0.1^3)/(163,000,000,000)(4)(8.14e-9)
d|(b)=(1000)(0.1^3)/(163,000,000,000)(4)(1.68e-8)

Thus;
Deflection for
(a)[25mmBB]=1/5308=1.88e-4m, or 0.188mm
(b)[BB30}=1/11016=9.08e--5m, or 0.0908mm

Which is pretty much exactly 2. So for the same load, the deflection is half. Not a third.

I think that will do. And there was me thinking that I finished work at 5pm :-)

This just might be one of th greatest posts in LFGSS history