Originally Posted by
MikeX
In a fit of geekishness I decided to calculate the amount of deflection one might expect from 22.2 mm versus 23.8 mm diameter aluminum handlebars. It’s one way to unwind on a Saturday morning. I’m assuming a 23 inch wide straight “flat” bar with 5 inch bar ends, 2mm tube thickness, a pull force of 100 pounds at the end of the bar end, and ignoring any flex in the bar end or elsewhere. I’m also using 10,000,000 psi for the elastic modulus (E) and 3,760,000 psi for the torsion modulus (G) of aluminum.
Using the I = π R cubed t formula for thin wall tubes to calculated the moment of inertia I get .0207 inches to the forth power for the 22.2 mm bar and .0255 for the 23.8 mm bar.
Using the deflection = W L cubed / 3EI formula for a simple cantilever beam with the load concentrated at the end and using the 11 inches of bar sticking out from the stem as the length (L) I calculated there will be .214” deflection for the 22.2 mm bar and .174” for the 23.8 mm bar.
For torsion the twist in radians is = torque L / J G where J = 2I and I get .0353 radians for the 22.2 mm bar and .0287 radians for the 23.8 mm bar which is equivalent to about .177 deflection for the 22.2 mm bar and .144” for the 23.8 mm bar.
Adding the bending to the twist gets you .391” deflection at the end of the bar end for the 22.2 mm bar and .318" deflection for the 23.8 mm bar. I guess you could feel that .073” difference particularly if you consider that the total deflection would double as you push and pull on the bars and that any curves in the bar would increase the deflection.
For steel you’d get about a third of those deflections (the elastic and torsional moduli of steel are about 3X that of aluminum) IF the wall thickness stays at 2 mm. For less than 2mm wall thickness the deflection would increase proportionately: ½ the wall thickness equals double the deflection.
Thanks for this very informative analysis, but I think you're seriously underestimating by not counting the flex of the bar end. No one is complaining about the flex of just a 22" flat bar, it's the combined flex of the flat section and the bar end which is the problem.
If curving increases deflection then we can consider the deflection in a cantilever 11+5=16" long as a
lower bound for the swept/curved bar. That means the bending should be multiplied by
at least (16/11)^3 = 3 which is at least 0.642" or the 22.2mm bar and 0.522" for the 23.8mm bar for a difference of 0.12" plus the torsion.
Moreover most of the bars cited here have ends which are 7 inches or more. So when combined with the push/pull, the curvature, and the torsion it appears we're looking at at least a 1/4" to 1/2" difference in deflection between the 22.2 and 23.8mm bar. That's huge.
Originally Posted by
DannoXYZ
The steel bar is probably 1/16". Can you try calculating with that?
1/16" = 1.59mm
So for 1/16" thick steel the deflections would be divided by 3*1.59/2 = 2.39
The cost is that bar weight would also be multiplied by 2.39 (ouch) vs about 1.15 for 23.8mm aluminum.