Vehicle on Curve Track

1. Vehicle Dynamics
P e r t e m u a n 6 :
V E H I C L E O N A C U R V E D T R A C K
D R . B U D I W A L U Y O , M T
F U N G K Y D I Y A N P E R T I W I , S T , M T

2. When a vehicle is moving round a curve, such as a corner on the road, a number
of factors affect the speed at which the vehicle can proceed. The two factors that
are considered here are:
(1) Maximum speed before overturning occurs;
(2) Maximum speed before skidding occurs

3. O v e r t u r n i n g s p e e d
The forces acting on the vehicle are:
mg –> the weight of the vehicle pressing down
through the center of gravity;
RA and RB – the normal reactions at the Centre of a
wheel’s point of contact on the road;
Centrifugal force= mv2 /r
v = velocity of vehicle in m/s
r = radius of turn in meters.

4. Take moments about the contact point of the outer wheel:
Overturning moment = Righting moment
 -
𝑚.𝑣2
𝑟
. h + m.g.d/2 = 0 ⇒
𝑚.𝑣2
𝑟
. h = m.g.d/2
 v2 = g.r.d/2.h
 v = ⎷(g.r.d/2h) ( Maks Speed to avoid overturning)
m = mass of vehicle in kg;
g = gravitational constant;
h = height of center of gravity;
d = track width in meters.

5. Example
In the vehicle shown in Figure the track width d = 144 m and the center
of gravity is 0.8 m above ground level. Determine The overturning speed
on a curve of 80 m radius. Take g = 9.81m/s2

6. Solution
Known:
h = 0.8 m;
d = 144 m;
r = 80 m;
g = 981 m/s2
So….
v = √(grd)/2.h = √ (981×80 ×144)/2×0.8
= √706.3 = 26.58 m/s
= 26.58 m/s x1 km/1000 m x 3600 s/h
= 95.7 km/h

7. Skidding speed
The friction coefficient the tires & the track = µ.
The friction force = µ.m.g Newton
The force pushing :
the vehicle sideways = centrifugal force = m.v2.r
Newton
At the point of skidding the centrifugal force is equal and opposite to the friction force.
m.v2. r = µ.m.g, where ever, v2 = µ.g.r
Skidding velocity v = √ µ.g.r (Maks speed to avoid skidding)

8. Example
Determine the skidding speed of a vehicle on a curve of 80 m radius if the coefficient
of friction between the tire and the road is 0.7. Take g = 9.81m/s2.

9. Solution
Known:
g = 9.81 m/s2;
µ = 0.7;
r = 80 m
So……
v = √ µ.g.r
v = √0.7 × 9.81 × 80
= √5494
v = 23.43 m/s
= 84.4 km/h

10. Thank You for your attention………………… !!!!!
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