Maximum Forces on the uprights

[MP4/4]

Hi,

I have calculated the forces in vertical, horizontal and under maximum braking.
It seems that they are a big high. Note that it is a first years team car I am designing.

So here are the calculations. The formulas are taken out of "Rennwagentechnik". A very good german book that gives an overview above all important topics concerning race car design.

Droop (hole in the surface)– maximal vertical force
F_WZ=3〖×F〗_(wz,stat.)
F_WZ=3×321,4 kg ×9,81 m/s^2
F_WZ=9458,80 N


Cornering– outer wheel
F_WZ=2〖×F〗_(wz,stat.)
F_WZ=6305,87 N

F_WY=〖2×μ〗_WY 〖×F〗_(wz,stat.)
F_WY=2×1,7〖×3152,92 N〗_
F_WY= 10719,93 N


Maximum braking
Vorne:
F_WZv=2〖×F〗_(WZv,stat.)

F_WX=〖1,5×μ〗_WX 〖×F〗_(WZv,stat.)
F_WX=1,5×0,7×3152,92 N
F_WX= 3310,57 N

[Unhealthy_FSAE_Addiction]

So I'm seeing three different load cases here:

- Droop (I think you're actually describing bump though)
- Cornering
- Braking

It looks like they're suggesting multiplying a static weight by a "guesstimate" multiplier (3x, 2x, 1.5x ect). I expect this weight might be the static corner weight of the corner of the car you're analyzing and not the entire car's weight.

Here is a great thread on using these contact patch forces to determine forces in the suspension members, which you can use to apply forces to your uprights. You will have to read through the whole thread carefully, as there is some incorrect information in there as well.

http://www.fsae.com/forums/showthread.php?11179-analysis-of-wishbones





Hi,

I have calculated the forces in vertical, horizontal and under maximum braking.
It seems that they are a big high. Note that it is a first years team car I am designing.

So here are the calculations. The formulas are taken out of "Rennwagentechnik". A very good german book that gives an overview above all important topics concerning race car design.

Droop (hole in the surface)– maximal vertical force
F_WZ=3〖×F〗_(wz,stat.)
F_WZ=3×321,4 kg ×9,81 m/s^2
F_WZ=9458,80 N


Cornering– outer wheel
F_WZ=2〖×F〗_(wz,stat.)
F_WZ=6305,87 N

F_WY=〖2×μ〗_WY 〖×F〗_(wz,stat.)
F_WY=2×1,7〖×3152,92 N〗_
F_WY= 10719,93 N


Maximum braking
Vorne:
F_WZv=2〖×F〗_(WZv,stat.)

F_WX=〖1,5×μ〗_WX 〖×F〗_(WZv,stat.)
F_WX=1,5×0,7×3152,92 N
F_WX= 3310,57 N