Problem set 3 -week 3 challenge: Anti-roll bar optimization

Answer:

Given Data:

Formula used:

1. Roll moment (ft-lb/g): sprung mass * sprung mass roll moment arm * (`pi/180`)

2. Roll gradient (deg/g): sprung mass * sprung mass roll moment arm / [Front roll stiffness + Rear roll stiffness - sprung mass * sprung mass roll moment arm * (`pi/180`)]

3. Load transfer due to roll or elastic component (lb/g): Roll moment / Track width

4. Front roll stiffness (ft-lb/deg) at wheel center level: Tire rate contribution * Total roll front stiffness / (Tire rate contribution - Total roll front stiffness)

5. Tire rate contribution: [Tire stiffness * track width^2 / 2] * `180/pi`

6. Front anti-roll spring rate (lb/in) for one wheel jounce: `(500000*(OD^4-ID^4)*MR^2)/(0.4422A^2B+0.2264C^3`

7. Front anti-roll spring rate (lb/in) for roll: 2*Front anti roll spring rate for one wheel jounce

8. Front anti-roll bar stiffness (ft-lb/deg): [Front anti roll spring rate for roll * track width^2 / 2] / `180/pi`

9. Auxiliary roll stiffness (%): (Front roll bar contribution / Front roll stiffness at wheel centre level)*100

10. Front roll bar contribution: Front roll stiffness at wheel centre level - wheel rate contribution

11. Wheel rate contribution: [Wheel rate * track width^2 / 2] / `180/pi`

RESULTS:

(i) Roll moment (ft-lb/g) = 7641.51

(ii) Roll gradient (deg/g) = 3.1849

(iii) Load transfer due to roll or elastic component (lb/g) = 1563.21

(iv) Front roll stiffness (ft-lb/deg) at wheel center level = 1988.66

(v) Front anti-roll spring rate (lb/in) for one wheel jounce = 149.10

(vi) Front anti-roll spring rate (lb/in) for roll = 298.20

( vii) Front anti-roll bar stiffness (ft-lb/deg) = 1491.67

(viii) Auxiliary roll stiffness (%) = 70.95

The Excel sheet where the calculation are done is attached below: