Vehicle Ride 1

Given Data:

Front sprung masss natural frequency = 1.15Hz

Wheel base = 2700mm = 2.7m

Speed = 60mph = 26.8m/s

Solution:

Frequency = `1/(2pi)* sqrt((riderate)/(eachmass)`.....................(a)

consider design condition

sprung mass each corner = (1825-90-60)/4 = 418.75kg

using this mass in (a)

1.15= `1/(2pi)* sqrt((riderate)/(418.75)`

ride rate front = sqr(7.23)* 418.75 = 21889.3N/m

We know that

`ride rate (sprung mass) = (wheel rate * tire rate) / (wheel rate + tire rate)` .........(b)

21.89N/mm= (245*WR)/(WR+245)

We get (245-21.9)*WR = 5365.5

WR = 24.04N/mm

Using this wheel rate we can deduce the spring rate:

`Wheel rate = (Sprrate) * (motion ratio)^2` ..............(c)

From this we get Spring rate for front desgin condition = 24.04/sqr(0.97)

Spring rate for front desgin condition= 25.4N/mm ...........(iii)

Time lag between front and rear wheel

time taken = 2.7/26.8= 0.1s

Time at which rear wheel comes to the same position as the front wheel = (1/1.15)-0.1

= 0.769s which gives the rear sprung mass frequency

So the recommended rear sprung mass frequency based on the wheelbase and speed at design condition = 1.3Hz (reciprocal of 0.769s) ...............................(i)

Using the same method previously used in front sprung mass frequency

Using (a)

`1.3 = (1/(2pi)) sqrt((rear ride rate)/418.75)`

rear ride rate`= (8.16)^2*418.75`

Rear ride rate = 27.89N/mm

Using (b)

`27.89 = (WR* 245)/(WR+245)`

`WR = 6833.05/217.11`

Rear wheel rate = 31.47N/mm

Using (c)

Spring rate`=`31.47/(0.85)^2`

Rear spring rate = 43.55N/mm .........................(iv)

Consider curb condition (ride rate doesn't change)

using the same method

Sprung mass at each corner = `1/4(1500-90-60) = 337.5kg`

Front sprung mass natural frequency `= (1/2pi)sqrt(21800/337.5)`

Front sprung mass natural frequency = 1.27Hz .................(v)

Rear sprung mass natural frequency `= (1/2pi)sqrt(27890/337.5)`

Rear sprung mass natural frequency = 1.44Hz ....................(vi)

Consider GVW condition (ride rate doesn't change)

using the same method

sprung mass at each corner = `1/4(2000-90-60) = 462.5kg`

Front sprung mass natural frequency = `(1/2pi)sqrt(21800/462.5)`

Front sprung mass natural frequency = 1.092Hz .................(v)

Rear sprung mass natural frequency = `(1/2pi)sqrt(27890/462.5)`

Rear sprung mass natural frequency = 1.23Hz ....................(vi)

Unsprung mass natural frequency:

Unsprung mass ride rate front = 24.04+25.5 = 49.54 N/mm

Using this Front unsprung mass natural frequency

at design = at curb = at GVW =`(1/2pi)sqrt(49.54/(90/2))`

Front unsprung mass at design= curb= GVW = 10.561Hz.............(vii)

Unsprung mass ride rate rear = 31.47+43.55 = 75.02 N/mm

at design = at curb = at GVW = `(1/2pi)sqrt(75.02/(60/2))`

Rear unsprung mass at design= curb= GVW = 15.917Hz..................(viii)

Ratio of frequencies (rear/front)..............................(ix)

Sprung mass

Curb = 1.44/1.27 = 1.13

Design = 1.3/1.15 = 1.13

GVW = 1.23/1.092 = 1.13

Unsprung mass at design = at curb = at GVW = 15.917/10.561 = 1.5

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2. The plot is attached in the excel file attached here