Originally Posted by
Monoborracho
Why is this true? Force = pressure x area. Wider tires generally run at lower pressure.
Maybe I was not clear that other variables are being held constant. At the same pressure, and the same rim, wider tires put more outward pressure on the rim bead than narrower tires.
The angle at which the tire wall attaches to the rim also effects the outward pressure on the rim. A narrow tire attaches at a nearly vertical angle while a fat tire is angled away from the rim or in an extreme case nearly 90 degrees to the rim wall. While fatter tires are usually run at lower pressures the problem is that the increase in stress due to the change in angle rises at a faster rate than the reduction of stress due to the reduction in pressure. Below is a quote from another thread that explains it in more detail:
Influence of tire pressure and width on the rim stress
The tour Forum discussion about the effects of different tire widths
and Luftdrücke was on the load on the rim. This particular case was
about whether a MTB rim as the a 50mm tire with max. Designed 4bar, is
a 25mm tire to cope with 8 bar pressure can be bent without them too.
Since I has not left in peace, I once counted something: The bending
stress on the rim sidewalls consist primarily of two components, the
direct internal pressure acting on the flanks, and the horizontal
portion of the circumferential force of the tire. The latter one has
to think about this: The tire section is simplified seen a ring. When
this pressure is evenly under tension. Because of the "ring" but the
lower part is open and there is held together only available via the
rim, the rim must take this tension there. Interesting for the bent-up
in the case only the horizontal component of the voltage at the
junction of tire rim. This is the exit angle of the sidewall of the
rim depends. All horizontal forces are not neglected because they
hardly relevant for the question have. The calculation is based only
on one edge of the rim on the other hand, the burden of course
identical. Exposure to peripheral force: The exit angle α the sidewall
can be calculated geometrically: with b = B = inside rim width tire
width
The tangential force is calculated according to the boiler formula as
follows: p = tire pressure (1 bar corresponds to 0.1 N/mm2) D =
Outside diameter of the rim horizontal component FU_X obtained cos α
available via:
Calculation of the resulting force in the middle of the edge: the area
of the inside of the rim edge A is: h = height inner rim flank This
results in the resultant force Fp_R:
Calculation of the moment on the rim: at the foot of the Cross, the
forces lead to a bending moment to the outside: It all looks a mess
out in a formula, you get this:
h = internal height rim flank
p = tire pressure (1 bar corresponds to 0.1 N/mm2)
B = Tyre width
D = Outside diameter of the rim
b = internal width of the rim
means of this formula, you can now compare what you nicely with the appropriate
tires and air pressure of its rim so exacts. For ease, I've also
created a small Excel document to calculate.