Calculating Stiffness of Springs/Suspensions for Inclined Suspension

[Vincent Mazzo]

Hello guys

How can I calculate the following stiffness of the springs/suspensions in the case of inclined suspension (angle between suspension and axle)?

I know that the lateral separation of the springs causes them to develop a roll resisting moment proporcional to the difference roll angle between the body and the axle.

And in the case of independent suspension? How substitute the rate at the wheel for K_s and use the tread as the separation distance?

 

[jack action]

Derivation of the roll stiffness:

The reaction torque ($T_{\phi}$) can be found with the spring force ($F_{s}$) times the lever arm length ($s/2$) for both springs:

$T_{\phi} = 2 \times F_{s} \frac{s}{2} = F_{s} s$

The spring force is equal to the spring rate times the spring compression (or extension) based on the roll angle ($\phi$) and lever arm length ($s/2$):

$F_{s} = K_{s} \phi \frac{s}{2}$

Which leads to:

$T_{\phi} = 0.5 K_{s} s^{2} \phi$

Since roll stiffness is defined as the amount of torque per roll angle:

$K_{\phi} = \frac{T_{\phi}}{\phi} = 0.5 K_{s} s^{2}$

So for independent suspensions, the same procedure applies. All you have to do is to replace $K_{s}$ with the http://enderw88.wordpress.com/automotive-theory/spring-rate-theory/" and $s$ with the track width (where the wheel rates are acting).

For the straight axle with inclined springs, multiply the roll rate by the cosine of the inclination angle (see the wheel rate link above, same concept).