Highway Engineering Test 1

Geometric design of highways are done  in such a way that from every point on highway the length of view available is sufficient so that the vehicle could be stopped in that visible distance or operation like overtaking can be performed safely.

We have various types of sight distances and their factor affecting them are listed inn below table:

Concept:

The extra widening on the road is given by:

\({E_w} = \frac{{n{L^2}}}{{2R}} + \frac{V}{{9.5\sqrt R }}\)

Where,

n = number of lane

L = wheel base

R = Radius of the curve

V = Design speed of the road in kmph

E_w = Extra widening

The extra width is not required for radius greater than 300 meters.

Concept:

The shift of the transition curve:

\(S = \frac{{L_S^2}}{{24R}}\)

Where,

L_s = Length of transition curve

R = Radius of circular curve

L_s will be taken as maximum of:

(i) Based on allowable rate of centrifugal acceleration.

(ii) By allowable rate of introduction of superelevation (e)

(iii) According to IRC

Calculation:

L_S = Maximum (50 m, 45m, 48 m)

L_S = 50 m

R = 240 m

\(S = \frac{{{{\left( {50} \right)}^2}}}{{24 \times 240}}\)

S = 0.434 m

Concept:

Braking Efficiency:

• Braking efficiency is expressed by the ratio of the actual coefficient of friction mobilized during braking to the coefficient of friction should be developed in ideal case and expressed in percentage. An efficiency of 100% implies that the vehicle will stop at the moment the brake is applied. Obtaining 100% is not practicable. This is an ideal condition of the vehicle.
• Skid length is given by, \({\rm{L}} = \frac{{{{\rm{u}}^2}}}{{2{\rm{gf}}}}\), where u is the velocity of the vehicle, f is the actual coefficient of friction mobilized and g is the acceleration due to gravity.

Calculation:

Given, L = 14 m; u = 36 kmph = 10 m/s; g =10 m/s²

\(\therefore {\rm{f}} = {\rm{\;}}\frac{{{{\rm{u}}^2}}}{{2{\rm{gL}}}} = \frac{{{{10}^2}}}{{2 \times 10 \times 14}} = 0.357\)

Given, for the ideal application of brakes the friction coefficient that should be developed is 0.45.

∴ Braking Efficiency = 0.357/0.45 = 0.7936 = 79.4% 

Concept:

Design rate of super-elevation, e = \(\frac{{{V^2}}}{{127 \times R}}\)

where

V = speed of the vehicle, in km/hr

R = radius of the curve

Calculation:

R = 480 m, V = 70 kmph

\(\text{e}=\frac{{{70}^{2}}}{127\;\times \;480}\)

∴ e = 0.08

Note:

On a highway in plain terrain, the maximum design rate of super-elevation allowed is 7%

On a highway in hilly terrain (Not bound to snow), the maximum design rate of super-elevation allowed is 10%

Given data

Radius of the curve(R) = 350 m

Stopping sight distance (SSD)  = 122 m.

Length of the curve (Ls) = 300 m

∵ L_s > SSD

∴ Setback distance (m) for 4 lanes from the center line of the road =

R – (R – 3d) cos (∝/2)

Where, R = Radius of the curve

\(d = \frac{{W + W_e}}{{2 \times n}}\)

where n = number of the lane.

∝ = angle b/w the segment of the curve.

Also \(\frac{ \propto }{2} = \frac{{SSD}}{{2\left( {R - 3d} \right)}} \times \frac{{180^\circ }}{\pi }\)

\(\begin{array}{l} \therefore d = \frac{{14 + 0}}{8} = 1.75\\ \frac{ \propto }{2} = \frac{{122}}{{2\left( {350 - 3 \times 1.75} \right)}} \times \frac{{180^\circ }}{\pi } = 10.14^\circ \end{array}\)

\(\therefore M = R\;-\left( {R\;-3d} \right)\cos \left( {\frac{ \propto }{2}} \right)\)

= 350 – (350 – 3 × 1.75) cos (10.14°)

Concept:

Transition Curve: Transition curve is provided to change the horizontal alignment from straight to circular curve gradually and has a radius which decreases from infinity at the straight end (tangent point) to the desired radius of the circular curve at the other end (curve point). There are 3 criteria to determine the length of transition curve on an Indian road and length provided will be maximum of those.

1. Rate of change of centrifugal acceleration:

\({{\rm{L}}_1} = \frac{{{{\rm{v}}^3}}}{{{\rm{CR}}}}\) where v is design speed in m/sec; R is radius of horizontal curve and C is rate of change in centrifugal acceleration and IRC suggests that \({\rm{C}} = \frac{{80}}{{75 + 3.6{\rm{v}}}}\)

2. Rate of change of super elevation:

 where e’ is effective super elevation; N is rate of change in super-elevation and B is total width of pavement including the widening if any. While providing super-elevation, if the pavement is rotated about the inner edge then e’ = e and if the pavement is rotated about the centre line then e’ = 0.5e where e is the actual super-elevation provided. IRC suggests value for N = 150, 100 and 60 for rolling terrain, built-up area and hilly terrain respectively.

3. IRC Empirical Formula:

\({{\rm{L}}_3} = \left\{ {\begin{array}{*{20}{c}}{\frac{{2.7{{\rm{V}}^2}}}{{\rm{R}}},\;For\;plain\;and\;rolling\;terrain}\\{\frac{{{{\rm{V}}^2}}}{{\rm{R}}},\;For\;hilly\;and\;steep\;terrain}\end{array}} \right.\), where V is speed in km/hr and R is radius of horizontal curve

Calculation:

Given, design speed in km/hr, V = 75 km/hr.        ∴ design speed in m/sec, v = 20.83 m/sec.

Radius of horizontal curve, R = 200 m; Total width of pavement including widening, B = 7.75 m.

Rate of change in super-elevation for hilly terrain, N = 60.

Rate of change in centrifugal acceleration, \({\rm{C}} = \frac{{80}}{{75 + 3.6 \times 20.83}} = 0.533{\rm{\;}}{{\rm{m}}^3}/{\rm{sec}}\)

∴ \({{\rm{L}}_1} = \frac{{{{20.83}^3}}}{{0.533 \times 200}} = 84.77{\rm{\;m}}\)

Super-elevation, \({\rm{e}} = \frac{{{{\rm{V}}^2}}}{{225{\rm{R}}}} = \frac{{{{75}^2}}}{{225 \times 200}} = {\rm{\;}}0.125\) 

But as per IRC guidelines the maximum super-elevation that should be provided on a hilly terrain is 0.1.

So, e = 0.1.

As, the pavement is rotated by the centre line, e’ = 0.5e = 0.5 × 0.1 = 0.05.

\(\therefore {{\rm{L}}_2} = 0.05 \times 60 \times 7.75 = 23.25{\rm{\;m}}\)

\(\therefore {{\rm{L}}_3} = \frac{{{{75}^2}}}{{200}} = 28.125{\rm{\;m\;}}\)

So, length of transition curve should be, \({\rm{L}} = \max \left\{ {{{\rm{L}}_1},{{\rm{L}}_2},{{\rm{L}}_3}} \right\} = 84.77{\rm{\;m}}\)

Concept:

Stopping Sight Distance (SSD): It is the absolute minimum sight distance available in a highway at any spot which should be of sufficient length to stop a vehicle traveling as design speed, safely without collision with any other obstruction.

Head on Collision Distance: It is the minimum distance required to avoid head-on collision between two vehicles on a highway. Head-on collision distance for a,

1. Single lane two-way road = 2 × SSD.
2. Double lane (or more) two-way road = SSD

Now, the stopping distance of a vehicle is the sum of:

1. The distance traveled by the vehicle during the total reaction time known as lag distance = vt and

2. The distance traveled by the vehicle after the application of the brakes, to a dead stop position which is known as the braking distance = \(\frac{{{{\rm{v}}^2}}}{{2{\rm{g }}\left( {{\rm{f\eta}} \pm {\rm{S}}} \right)}}\)

\(\therefore {\rm{SSD}} = {\rm{vt}} + \frac{{{{\rm{v}}^2}}}{{2{\rm{g}}\left( {{\rm{f\eta }} \pm {\rm{S}}} \right)}}\) where, v = Speed in m/sec; t = Reaction time in sec; f = Coefficient of friction; η is braking efficiency; S is gradient + ve sign for ascending gradient & -ve sign for descending gradient.

Calculation:

Given, design speed = 60 km/hr ∴ v = 60/3.6 = 16.67 m/sec

If nothing is mentioned reaction time t should be assumed as 2.5 sec; coefficient of friction, f = 0.36; braking efficiency, η = 60% =0.6; g = 9.81 m/s² and gradient is given as 3% upwards. ∴ S = +0.03

\(\therefore {\rm{SSD}} = 16.67 \times 2.5 + \frac{{{{16.67}^2}}}{{2 \times 9.81 \times \left( {0.36\times 0.6 + 0.03} \right)}} = 41.67 + 57.58 = 99.25{\rm{m}}\)

Now, as the road is a single lane two-way road, so the head-on collision distance = 2 × SSD.

∴ Head on collision distance = 2 × 99.25 = 198.5 m.

Concept:

Valley Curve:

• Valley curve or sag curve are the vertical curves with convexity downwards. These curves are provided either when a descending gradient meets an ascending gradient or a flat gradient or a descending gradient or when an ascending gradient meets another ascending gradient. Valley curves are designed for two criteria viz. comfort criteria and headlight sight distance.
• The length of the valley curve (L), when provided as per comfort criteria, is given by, \({\bf{L}} = 2{\left[ {\frac{{{\bf{N}}{{\bf{v}}^3}}}{{\bf{C}}}} \right]^{0.5}}\) where v is the design speed in m/sec; N is deviation angle which is the algebraic difference in gradients and C is the rate of change in centrifugal acceleration in m/sec³.
• If the valley curve is provided in form of a cubic parabola, then the location of the lowest point on the valley curve measured from the first tangent point on the first gradient is, \( = {\bf{L}}\sqrt {\frac{{{{\bf{n}}_1}}}{{2{\bf{N}}}}} \) , where n₁ is the first gradient and L is the length of valley curve.

Calculation:

Given, design speed is 70 km/hr ∴ v = 70/3.6 = 19.44 m/sec.

\({{\rm{n}}_1} = - 1{\rm{\;in\;}}45 = - 0.022{\rm{\;and\;}}{{\rm{n}}_2} = + 1{\rm{\;in\;}}65 = 0.0154.\)

\(\therefore {\rm{N}} = {\rm{\;}}\left| {{{\rm{n}}_1} - {{\rm{n}}_2}} \right| = 0.0376\) and C = 0.6 m/sec³