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# Chassis Design

Updated: Jan 28, 2022

### "Adding power makes you faster on the straights, subtracting weight makes you faster everywhere."

##### - Anthony Colin Bruce Chapman

### Basic requirements:

###### The body should be light.

###### It should have minimum number of components.

###### It should provide sufficient space for passengers and luggage.

###### It should withstand vibrations while in motion.

###### It should offer minimum resistance to air.

###### It should be cheap and easy in manufacturing.

###### It should have uniformly distributed load.

###### It should have long fatigue life.

###### It should provide good vision and ventilation.

### Load cases:

##### The vehicle designer should know the load conditions in order to ensure the structure doesn’t fail due to some instantaneous load.

#### Some Basic Load Cases-

#### Bending-

##### In static condition (vehicle at rest or moving with constant velocity) :

##### 1. Loading (due to the weight of components distributed along the vehicle frame) in vertical plane (let, x-z plane) cause bending about the y-axis.

##### 2. Unsprung mass do not impose loads in static condition.

##### 3. Determination and Result:

###### Static condition vehicle structure can be treated as 2D beam because vehicle is approximate symmetry at x-y plane.

###### Can be solved using static equilibrium balance.

###### Results in set of algebraic equation.

##### In dynamic condition (vehicle moving at variable velocity or vehicle moving at bumpy road even at constant velocity) :

##### 1. Inertia of the structure contributes in total loading

##### 2. Always higher than static loading

##### (For road vehicles: 3 times than static loads; For off-road vehicles: 4 times than static load)

##### 3. Determination and result:

###### Can be solved using dynamic equilibrium balance.

###### Generally results in differential equation.

#### Torsion-

##### Vehicle body is subjected to a moment applied at the axle centerlines by applying upward and downward loads at each axle. These loads result in twisting action or torsion moment about the longitudinal x-axis.

##### 1. Cause: When vehicle traverse on an uneven road and front and rear axles experiences a moment.

##### 2. For pure simple torsion torque is applied to one axle and reacted by the other axle.

##### 3. Determination and Result:

###### Maximum torsion moment is based on the loads at the lighter loaded axle.

###### Load at front axle 𝑅 and at rear axle 𝑟 can be determined by;

##### 𝒓×𝒕=𝑹×𝒇 (Where 𝑡 and 𝑓 are track width of rear and front axle)

##### 4. Dynamic Factor:

###### On road: 1.3

###### Off road: 1.5 to 1.8

#### Combined bending and torsion-

##### 1. Bending and torsional loads are super imposed. (assumed to be linear)

##### 2. All loads of lighter axle is applied to one bending torsion wheel

##### 3. Determination: Loads can be determined by moment balance.

#### Lateral Loading-

##### 1. Generated at the tire to ground contact patch

##### 2. Cause: Cornering

##### 3. Result of lateral loading: When inside wheel reaction becomes zero, vehicle rollover.

##### 4. Determination:

##### i) Practical:

###### It can either be open loop or close loop

###### Open loop or directional response: In this case, driver gives some input through steering and corrects it. We study both response and feedback.

###### Close loop: In this case, driver gives some input and we study vehicle response.

##### ii) Theoretical:

###### These loads are balanced by centrifugal force.

###### Centrifugal acceleration: 𝑉^2/𝑅=𝑔𝑡/2ℎ

###### Force at CG, at the moment of roll over: 𝑀𝑉2𝑅=𝑀𝑔𝑡2ℎ

###### Taking ΣF = 0 and ΣM = 0, it can be shown that;

###### Lateral force at front wheel 𝑌𝑓= (MV^2/R)* [b/(a+b)]

###### Lateral force at rear wheel 𝑌𝑓= (MV^2/R)* [a/(a+b)]

##### Width of car and reinforcement provide sufficient bending stiffness to withstand lateral force.

##### Longitudinal Loading-

##### 1. Cause: When vehicles accelerate and decelerate, longitudinal loads generated

##### 2. Determination:

##### During acceleration:

###### Weight transferred from front to back.

###### Taking ΣM = 0;

###### Reaction force on the front wheel-

###### 𝑅f={𝑀𝑔(𝐿−𝑎)−𝑀ℎ(𝑑𝑣/𝑑𝑡)}/𝐿

###### Reaction force on rear wheel-

###### 𝑅r={𝑀𝑔(𝑎)+𝑀ℎ(𝑑𝑣/𝑑𝑡)}/𝐿

##### During deceleration:

###### Weight transferred from back to front.

###### Taking ΣM = 0;

###### Reaction force on the front wheel-

###### 𝑅f={𝑀𝑔(𝐿−𝑎)+𝑀ℎ(𝑑𝑣/𝑑𝑡)}/𝐿

###### Reaction force on rear wheel-

###### 𝑅r={𝑀𝑔(𝑎)-𝑀ℎ(𝑑𝑣/𝑑𝑡)}/𝐿

##### 3. Tractive and braking forces add to bending through suspension.

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