Steering Force Calculation

Vehicle Specs

Mass of vehicle:

Total mass = 140 kg

Mass of the vehicle with driver = 200 kg

Mass on the front = 0.45 x 200= 90 kg

Mass on the rear =110kg

Corner mass front = 45kg
Corner mass rear = 55 kg

COEFFICIENT OF FRICTION taking as 0.8

Steering Force Calculations:

The basic concept is that torque required to turn the wheel should be more than resisting torque by friction

Force of friction (on one wheel) = µ x g x corner mass

= 0.8 x 9.81 x 45

= 353.16 N

As we know that steer happens about the kingpin axis of the car

Input torque from the ground (on one wheel) = force of friction x perpendicular distance from contact patch to kingpin axis

we can take it approx to scrub radius

The scrub radius is the distance in front view between the kingpin axis and the center of the contact patch of the wheel, where both would theoretically touch the road.

Torque due to friction force = 353.16 x .12831

= 45.31 Nm

This torque will be equal to the lateral push from the tie rod.

The torque due to lateral push from tie rod = force on tie rod x perpendicular distance between the outer tie rod point end to kingpin axis

Torque due to lateral push = force on tie rod (f_t) x 0.0945m

As I already define that both the torque would be equal

Therefore,

f_t =479.47N

As the tie rod pushing the tire from one side & pulling it from another side

So,

Total force on rack = 479..47x 2 = 958 ~

Radius of pinion = 26.11mm= .0261 m

Torque on pinion = 1000 x .0261 =26.1Nm

Torque on steering wheel = Torque on pinion = 26.1Nm

Radius of steering wheel = 5 inch = 0.127 m

Force applied by the driver to steer the wheels = torque on steering wheel/radius of the steering wheel

= 26.11/.127

= 205.59N