Hi Guys,

putting the last touches on my bachelor thesis about the design of a FSAE suspension system

for the team that maybe will exist in the future of my former college.

I am going through the last calculations of tire forces experienced in the worst cases.

I assumed braking and cornering as the worst condition in normal circumstances. I excluded hitting the kerb, because my professor is just interested in

the forces under normal conditions.

I will use the formula of the inverse matrix. Ax=B as discussed in an other thread for the forces in the suspension members.

Let's tell you guys how I calculated my final forces for the right front wheel in a left hand turn.

In front view

x Axis: positive to the front of the car

y Axis: positive to the right side

z Axis: positive vertical upwards

The accelerations I am using for braking are 1,5G

and for the lateral acceleration is 1,5 G.

First I calculated the weight transfer between front and rear axle during braking. The car has a weight including driver of 321,4 kg.

With my CG heights and wheelbase, I come to a weight transfer of 90,38 kg.

That gives me a total weight at the front axle of 239,4 kg.

Assuming the car is symmetrical side to side that gives me an initial loading of the front right tires of

119,6 kg.

Next I add the lateral acceleration, which gives me a weight transfer from the left to the right wheel

of 101,4 kg.

That makes a total of 221 kg at the front right wheel. Making it 2168,01 N. At the left tire 18,2 kg. Almost tipping

That`s for the Z component of the B matrix.

For the X component, I made following calculation.

The car is breaking at 1,5G.

321,4 kg * 1,5 G *9,81 m/s² = 4729 N.

Doing the force equilibrium in x direction and assuming a break distribution of 50:50

each wheel takes the same amount of force to equal the decelerating force.

I simplify the tire load vs. grip model quiet a bit here.

This gives me 1182,35 N at the front right wheel in x direction.

Now the Y direction.

The centrifugal force just for the front axle weight to simplify the calculation

at 1,5 G is 2189,6 N.

This centrifugal force has to be put into equilibrium by the tires.

Assuming the load cases for the outer tire with 221 kg = 2168 N and the inner

tire, 18,2 kg = 179 N. Now lets look at the load vs. slip angle vs. lateral force (unfortunatly I had to use a random one I found on the interwebs, so lateral force

with FS tires might be higher)

At an assumed slip angle of 5° the outer tire is capable of roughly 1850 N lateral force.

The inner tire now has to cope with 339 N.

Now I have force in the x, y, z direction and can apply the moments in respect to the cg

to complete my matrix.

X: 1182,35 N

Y: 1850 N

Z : 2168,01 N

Are these sensible numbers to calculate my stresses in the suspension from?