Stability of vehicles in curved paths

In the analysis below the vehicle is considered as a rigid body, and the effects of the suspension system are neglected.

Lateral sliding

A vehicle will skid or slide outwards (see figure below) when the centrifugal force (CF) exceeds the frictional forces (Fwi and Fwo) or force µ (Fi + Fo) which are equal to µw or μmg. Sliding will be about the commence when the force CF is equal to µmg.  therefore and sliding velocity m/s

The transferred weight (wt) during sliding equals N

but if sliding ceases, the transferred weight wt will be equal N

* Lateral sliding and impact:

A vehicle sliding or drifting sideways may hit a curbstone or grass verge which brings the slide to zero in a very short period of time. A sudden lateral deceleration produces an inertia force (ma) acting through the center of gravity (CG) in addition to the centrifugal force, and together they provide a large overturning couple where h is the height of the CG above the road surface, r the radius of the curved path (both in meters), m the mass of the vehicle (kg), and the lateral deceleration (m/s2).

Before any sliding takes place the value of CF must become greater than the frictional forces between the tires and road surface μ (Fi + Fo) or μmg. Under sliding conditions the overturning moment is equal to μmgh, but on impact the full centrifugal force will become active as an overturning force acting with the impact or inertia force, and overturning moment becomes A vehicle sliding sideways or drifting does not necessarily have to hit an obstruction to provide a dangerous overturning couple. For example, a car sliding sideways on a wet road surface suddenly transferring to a dry surface will undergo a considerable lateral deceleration sufficiently large to cause overturning, or at best, give steering and stability problems. When possible, all lateral sliding should be corrected and brought under control ‘slowly’ for safety.

Overturning

A vehicle will commence to overturn when all the vertical loading carried by the inner wheels has been transferred. If the center of gravity CG is considered as acting in the center of vehicles track (t), when the value of ½  weight or ½ mg has been transferred, the vehicle will commence to overturn. Taking moments about the outer wheels (Fo),in the above figure gives: thus and overturning velocity m/s

Another method is to use the resultant of the two forces CF and mg. When the resultant of these forces passes through the center of the outer wheels’ contact area, the vehicle will be on the point of overturning. Consider CG is in the center of the track. Taking moments about Ro, as shown in the figure below.  thus the angle θ can be determined. thus Note: in the first method In the second method Thus both formulae are virtually the same.

If the CG is not in the center of the track, instead of tan θ = t / 2h the actual distance between the CG and the center of the outer wheels is substituted for ½ t. For example, if x was 0.9 m and y (0.7 m the tan θ = y/h =0.7/h.

VEHICLE ON CURVED BANKED TRACK

No tendency to slide

A vehicle travelling at a certain velocity in a curved path on a banked road will have three forces acting upon it, in the figure below

- the centrifugal force CF,

- the value of the vehicle’s weight (W= mg),

- and the friction forces (Fwi and Fwo).   * If the resultant force R of the forces CF and w, acts normal or at right angles to the road surface then no weight or loading has been transferred from the inner wheels, and there is no tendency for the vehicle to slide at this velocity and road radii.

When the resultant force is acting normal to the road surface the vehicle should behave as if it was traveling along level road surface. Good design and engineering paly an important part towards safety on the road thus m/s Vehicle sliding on banking

In the case of the resultant mg cos θ passing normal to the road surface, if the vehicle speed was reduced there would be a tendency for the vehicle to slide down the banking due to the force mg sin θ. This is resisted by the frictional force µ (Fwi+Fwo) or µmg cos θ + CF sin θ. In each case the tire construction will play an important role in preventing sliding and providing stability and safety. Two methods of solving problems concerning curved banked road surfaces are given.

Taking the forces acting parallel to the road surface and those perpendicular: therefore divide each side by m cos θ and sliding velocity when θ = 0 Worked examples

1-A car is travelling in a curved path at a radius of 42 m which is banked for safety at an angle of 16o. If the coefficient of friction is 0.404, at what speed would this vehicle commence to slide up the banking?  2- Vehicle sliding outwards: level road

A vehicle weighing 13.4 kN has its centre of gravity in the centre of the 1.4 m track, at a height of 70 cm above the road surface. If lateral adhesion is 0.6:

(a) At what velocity will the vehicle begin to slide on a level surface, and what weight remains on the inner wheels when this vehicle is travelling in a curve of 36.6 m radius?

(b) If the road becomes banked at 20o for the same radius,

i- what velocity the vehicle begin to slide on a level surface?

ii-weight remains on the inner wheels? And

iii- what will be the velocity at which no weight transferred?  where wt is the transferred weight, h the height of CG and t the width of track. weight on inner wheels (Fi) (b) i- The sliding velocity, banked road ii- Σ Fy = 0 Σ Mo = 0 iii- the velocity at no weight transferred Vehicle overturning on curved banked road

A vehicle negotiating a banked curved road will be on the point of overturning when all the weight carried by the inner wheels has been transferred to the outer wheels. For a given set of vehicle dimensions, such as track, height of centre of gravity and its position relative to the nearside and offside wheels, the overturning velocity will depend upon the value of the centrifugal force, the road curvature or radius, and the angle of the road banking.

Incorrect road banking, such as banking in the wrong plane, could lead to a serious accident. Examples are given of both sliding or overturning on a reversed banking. Both sliding and overturning speeds should be studied for correct and incorrect banking.

At the figure below:

Consider that the centre of gravity lies in the center of the track. Taking forces parallel and normal to the road surface:   divide each side by m cosq therefore when q = 0 worked examples

A car is traveling on a left-hand curved surface of 34 m radius, banked at an angle of 22o to improve the safety factor. The car’s track is 1.6 m, and its centre of gravity lies 0.8 m above the road surface, but in the center of the track. Determine the minimum overturning velocity.  sliding and overturning

1- A vehicle travelling at 24.4 m/s on a level road surface which is curved to a radius of 40.2 m is about to overturn. The centre of gravity is at the centre of the 1.4 m track.

(a) At what height is the center of gravity?

(b) At what velocity could this vehicle negotiate the same curve, but banked to an angle of 20o, without any weight leaving the inside wheels?

(c) What velocity could be attained on a level road curved to the same radius before sliding commences if the coefficient of adhesion is 0.65?

(a) (b) thus (c) Vehicle will slide when force therefore 2-Determine the minimum speed at which a vehicle would overturn on a level road when negotiating a curve of 20 m radius. The vehicle's track is 1.3 m wide and the center of gravity is o.6m above the road.

Overturning velocity (v) Or where 