Banking of Roads

Banking of roads, also known as road superelevation, is a crucial engineering technique used to enhance vehicle safety and stability on curved pathways. By elevating the outer edge of the road, this design helps counteract the outward centrifugal force that vehicles experience during turns, reducing the risk of skidding and improving overall driving comfort. Understanding the concept, purpose, and benefits of road banking is essential for anyone studying transportation engineering, road design, or vehicle dynamics.

1.0Definition of Banking of Roads

• Banking of roads, also known as superelevation, is the practice of raising the outer edge of a curved road above the inner edge to create a sloped surface. This inclination helps provide the necessary centripetal force for vehicles to safely negotiate curves by reducing reliance on friction and minimizing the risk of skidding.

When vehicles go through turnings, they travel along a nearly circular arc. There must be some force which provides the required centripetal acceleration. If the vehicles travel in a horizontal circular path, this resultant force is also horizontal. The necessary centripetal force is being provided to the vehicles by the following three ways :

1. By friction only.

2.By banking of roads only.

3.By friction and banking of roads both.

In real life the necessary centripetal force is provided by friction and banking of roads both.

2.0Centripetal Force Provided by Friction Only

$$\begin{gathered} \text{Suppose a car of mass m is moving with a speed v in a horizontal circular arc of radius r. In this case, the necessary centripetal force will be provided to the car by the force of friction f acting towards centre of the circular path.}f=\frac{m{v}^2}{r} \\ {f}_L=\mu N=\mu mg \\ \text{Therefore, for a safe turn without skidding,} \\ \frac{m{v}^2}{r}\le f_L\Rightarrow \frac{m{v}^2}{r}\le \mu mg\Rightarrow v\le \sqrt{\mu rg} \end{gathered}$$

3.0Centripetal Force Provided by Banking of Roads Only

• Friction is not always reliable at turns particularly when high speeds and sharp turns are involved. To avoid dependence on friction, the roads are banked at the turn in the sense that the outer part of the road is somewhat lifted compared to the inner part.

$$\begin{gathered} N\sin \theta =\frac{m{v}^2}{r}\text{and}N\cos \theta =mg \\ \tan \theta =\frac{v^2}{rg}\Rightarrow v=\sqrt{rg\tan \theta }\therefore \tan \theta =\frac{h}{b} \\ \tan \theta =\frac{v^2}{rg}=\frac{h}{b} \end{gathered}$$

4.0Centripetal Force Provided by Both Friction and Banking of Roads

• If a vehicle is moving on a circular road which is rough and banked also, then three forces may act on the vehicle.
• Of these the first force, i.e., weight (mg) is fixed both in magnitude and direction.
• The direction of the second force, i.e., normal reaction N is perpendicular to the road.
• The direction of the third force, i.e., friction f can be either inwards or outwards, while its magnitude can be varied up to a maximum limit $\left(f_L=\mu N\right)$
• So, direction and the magnitude of friction f are so adjusted that the resultant of the three forces mentioned above is  $\frac{m{v}^2}{r}$ towards the Centre.

$$\begin{gathered} \text{(A) If v <}\sqrt{rg\tan \theta },\text{then friction acts outwards} \\ N\cos \theta +f\sin \theta =mg\dots \dots .(1) \\ N\sin \theta -f\cos \theta =\frac{m{v}^2}{r}(2) \\ \text{For minimum speed,}f=\mu N \end{gathered}$$       

$$\begin{gathered} \text{So, by dividing equation (1) by equation (2),} \\ \frac{N\cos \theta +\mu N\sin \theta }{N\sin \theta -\mu N\cos \theta }=\frac{mg}{\frac{m{v}_{\min }^2}{r}} \\ Therefore,v_{\min }=\sqrt{rg\left(\frac{\tan \theta -\mu }{1+\mu \tan \theta }\right)} \\ \text{If we assume}\mu =\tan \varphi , \\ {v}_{\min }=\sqrt{rg\left(\frac{\tan \theta -\tan \varphi }{1+\tan \varphi \tan \theta }\right)}=\sqrt{rg\tan (\theta -\varphi )} \end{gathered}$$

(B) If the speed of the vehicle is high then friction acts inwards.

$$\begin{gathered} \text{In this case for maximum speed,} \\ N\cos \theta -\mu N\sin \theta =mg \\ N\sin \theta +\mu N\cos \theta =\frac{m{v}_{\max }^2}{r} \\ \text{Which gives} \\ {v}_{\max }=\sqrt{rg\left(\frac{\tan \theta +\mu }{1-\mu \tan \theta }\right)} \\ \text{If we assume}\mu =\tan \varphi , \\ {v}_{\max }=\sqrt{rg\left(\frac{\tan \theta +\tan \varphi }{1-\tan \varphi \tan \theta }\right)}=\sqrt{rg\tan (\theta +\varphi )} \end{gathered}$$

Note: The expression $\tan \theta =\frac{v^2}{rg}$ also gives the angle of banking for an aircraft, i.e., the angle through which it should tilt while negotiating a curve, to avoid deviation from the circular path. The expression $\tan \theta =\frac{v^2}{rg}$ also gives the angle at which a cyclist should lean inward, when rounding a corner. In this case, is the angle which the cyclist must make with the vertical.