Horizontal Alignment

7/17/2019 Horizontal Alignment

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The various factors to be considered in the

horizontal alignment are:Design speed

Radius of circular curves.

Type and length of transition curves Super elevation.

Widening of pavement on curves.

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Most important factor in the design of

horizontal alignment.Depends on the type of the road as well as

type of terrain.

For instance, Indian Road Congress (IRC) hasclassified the terrains into four categories,

namely plain, rolling, mountainous, and

steep.

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RoadClassificatio

n

Design speed in kmph for various terrains

Plain RollingMountaniou

sSteep

Rulin

gMin

Rulin

gMin

Rulin

gMin

Rulin

gMin

NS&SH 100 80 80 65 50 40 40 30

MDR 80 65 65 50 40 30 30 20

ODR 65 50 50 40 30 25 25 20

VR 50 40 40 35 25 20 25 20

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The presence of horizontal curve imparts

centrifugal force which is a reactive forceacting outward on a vehicle negotiating it.

Centrifugal force depends on speed and

radius of the horizontal curve It counteracted to a certain extent by

transverse friction between the tyre and

pavement surface. Centrifugal force is given by the equation

P=Wv2

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Where P = centrifugal force, kg.

W = weight of the vehicle, kgR = radius of the circular curve, m

v= speed of vehicle, m/sec

g = acceleration due to gravity

= 9.8 m/sec2

The ratio of the centrifugal force to the weight

of the vehicle (P/W) is known as thecentrifugal ratio and is given by:

=

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The centrifugal force acting on a vehicle

negotiating a horizontal curve has twoeffects:

Tendency to overturn the vehicle outwards

about the outer wheels. Tendency to skid the vehicle laterally out

wards.

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Overturning effect

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Taking moments of the forces with respect to

the other when the vehicle is just about tooverride:

Ph = W

or

=

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For safety the following condition must

satisfy:

b2h

>v2

gR

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Transverse skidding effect

The equilibrium condition of transverse skidresistance is given by:

P= FA + FB = f(RA + RB) = fw

Equating fw equal to counteractingcentrifugal force

P = fW or

= f

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Super-elevation or cant or banking is the

transverse slope provided at horizontal curveto counteract the centrifugal force, by

raising the outer edge of the pavement with

respect to the inner edge, throughout the

length of the horizontal curve.

In order to find out how much this raising

should be, the following analysis may be

done.

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P the centrifugal force acting horizontally

out-wards through the center of gravity.W the weight of the vehicle acting down-

wards through the center of gravity, and

F the friction force between the wheels andthe pavement, along the surface inward.

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At equilibrium resolving the forces

P cosθ = W sinθ + FA + FB= W sinθ + f(RA + RB)

= W sinθ + f(W cosθ + P sinθ )

Dividing by Wcosθ

(1 – ftanθ) = tanθ + f

=

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=

This is an exact expression for super elevation.

But normally, f = 0.15 andθ< 4

o

, 1 - f tanθ~1.

= tanθ + f

= e + f, e= tanθ = rate of super elevation

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e + f =

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Steps for super elevation design

i)The super elevation for 75% of design speed,neglecting f, i.e e =

(0:75v)2

ii)If the calculated value is less than .07 the

value so obtained is provided. If the value ofsuper elevation exceeds .07 then proceed withsteps (iii) or (iv)

iii)Check the coefficient of friction developedfor maximum value of e = .07 using equation

e + f =

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If the calculated value of f is less than 0.15,

the super elevation of 0.07 is safe for designspeed.

If not proceed to step (iv)

iv)Find the allowable speed va for themaximum e = 0.07 and f = 0.15,

ie 0.07 + 0.15 =

If the allowable speed is higher than the

design speed, then design is adequate and

provides a super elevation of e equal to 0.07

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If the allowable speed is less than the design

speed, the speed is limited to the allowablespeed va.

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Horizontal Alignment

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