UG Geometric Design II

Indian Institute of Technology, Kharagpur

Geometric Design of HighwaysLesson 5

Horizontal Alignment – Part I

Dr. Bhargab MaitraDepartment of Civil Engineering

Indian Institute of Technology KharagpurIndia

Email: [email protected]

mailto:[email protected]


Specific Instructional objectives

• Identify various design elements related to

horizontal alignment

• Appreciate the need and understand the basis for

providing superelevation at horizontal curves

(IRC)


Design elements• Horizontal curve

• Superelevation (IRC & AASHTO approach for design)

• Type and length of transition curve

• Extra widening

• Set-back distance


Horizontal curves

• Should be fluent and blend well with surrounding topography

• Short curves for small deflection angle (kinks) should be avoided

Horizontal curve

Tangent

Tangent


• Curve in same direction separated by short tangents (broken back) should be avoided and replaced by a large single curve.

• Sharp curves should not be introduced at the end of long tangents

• Compound curve should be avoided. When unavoidable limiting value of ratio of flatter curve radius and sharper curve radius is 1.5:1 (IRC)


• When a vehicle traverses a horizontal curve, the centrifugal force acts horizontally outwards through the centre of gravity of the vehicle.

• Centrifugal Force:

• Centrifugal Ratio or Impact Factor:


Overturning effect

• Overturning moment = P * h• Restoring moment = W * b/2• For equilibrium, Ph = Wb/2

or P/W = b/2h

P

W

b/2b/2

C.G

h

Two effects

• Tendency to overturn the vehicle outwards about the outer wheels

• Tendency to skid the vehicle laterally outwards


• Centrifugal ratio P/W = f (Tendency to skid the vehicle laterally outwards)

P

W

C.G

FA=f.RA FB=f.RBBA

RA RB

Skidding effectFor equilibrium P = FA + FB

= f (RA + RB ) = f W


Superelevation• To counteract the effect of centrifugal force and

to reduce the tendency of the vehicle to overturn or skid, the outer edge of the pavement is raised with respect to the inner edge – this transverse inclination to the pavement surface is known as Superelevation or Cant

mv2/R

mg

F

CG

N

• Superelevation e = tanθ

θ

Indian Institute of Technology, KharagpurConsidering equilibrium along F

Pcosθ

= W sinθ + F

= W sinθ + fN

= W sinθ + f (W cosθ+ P sinθ)

P(cosθ - f sinθ) = W sin θ + fW cosθ

P/W = (tanθ + f)/ (1-f tanθ) or, P/W = e+f

e + f = v2/gR = V2/127R [ V in km/h]

P

N

F

CG

θ


Maximum and Minimum Superelevation (IRC)• Maximum allowable superelevation

7 % for plain and rolling terrain

10 % for mountainous terrain not bound by snow

• Minimum superelevation

If calculated superelevation is equal or less than camber, then minimum superelevation equal to camber should be provided from drainage consideration


¾ th Assumption

• Superelevation - fully counteract the centrifugal force or to counteract a fixed proportion of centrifugal force

• In the former case, the superelevation needed would be more than 7 percent on sharp curves causing inconvenience to slow moving vehicles

• When a vehicle negotiates a flat curve, friction would not be developed to the maximum – this is not a balance design


• It is desirable that the superelevation should be such that a moderate amount of friction is developed while negotiating flat curves and friction not exceeding the maximum allowable value be developed at sharp curves

• Indian practice – Superelevation should counteract centrifugal force developed by 3/4th of design speed

e = (0.75V)2/127R = V2/225R


Equilibrium SuperelevationIf coefficient of friction is assumed to be zero then superelevation required to counteract the centrifugal force fully will be given by e = V2/127R

For equilibrium superelevation, the pressures on the outer and inner wheels will be equal, but this will result in a very high value of superelevation


Friction FactorDepends on a number factors• Speed of vehicle

• The type and condition of the roadway surface

• The type and condition of the vehicle tires.

• IRC: 0.15 constant design valueAASHTO recommended values

Design Speed (kmph)

20 30 40 50 60 70 80 90 100 110 120 130

Max. side friction values

0.18 0.17 0.17 0.16 0.15 0.14 0.14 0.13 0.12 0.11 0.09 0.08


Superelevation with/without Transition Curves

• Superelevation should be attained gradually over the full length of the transition curve so that the design superelevation is available at starting point of circular portion

• Where due to some reason transition curves are not provided two-third superelevation may be attained on the straight portion (tangent) and balance one-third on the curve


Ruling and Minimum Radius of Horizontal Curves• For a particular speed the centrifugal force is

dependent on the radius of horizontal curve• To keep the centrifugal ratio within a limit the

radius of the curve should be correspondingly high

e + f = V2/127RRruling= V2/127(e+f) , where V=Ruling design speed

in Km/hrRmin= V’2/127(e+f) , where V’=Minimum design

speed in Km/hr


Example Problem-1Ruling design speed and minimum design speed values 80 km/hr and 60 km/hr respectivelyCalculate the Ruling and minimum radius of horizontal curve

Maximum values of e = 0.07 and f = 0.15

Rruling= V2/127(e+f) = 802/127(0.07+0.15) = 229.1 = 230m (say)

Rmin= V2/127(e+f) = 602/127(0.07+0.15) = 188.84 = 190m (say)


Steps for Superelevation DesignStep-1: The superelevation for 75 percent of design speed is calculated neglecting the friction.ecal = V2/225R

Step-2: If ecal < emax(7%) , ecal is providedIf ecal > emax(7%) , then provide the maximum

superelevation equal to 0.07 & proceed with step (3) & step (4).Step-3: Check the coefficient of friction developed for the maximum value of ‘e’ =0.07 at the full value of design speed.


f = {(V2 / 127 R) - 0.07}

If the value of f thus calculated is less than 0.15, the superelevation of 0.07 is safe for the design speed. If not, calculate the restricted speed as given in step (4)

Step-4: Allowable speed ( Va km/h) at the curve is calculated by considering the design coefficient of lateral friction and the maximum superelevation

e + f = 0.07+ 0.15 = 0.22 = Va2/127 R


Example Problem-2Design the superelevation for a horizontal curve of 400m radius and design speed 100 km/hr

ecal = V2/225R = 1002/225x400 = 0.11>0.07

Take value of e = 0.07Value of developed coefficient of lateral frictionshould be checked against maximum valuef = V2/127R – 0.07= 1002/127x400 – 0.07 = 0.13 < 0.15 (O.K)


Example Problem-3Design the superelevation for a horizontal curve of 1500m and design speed 80 km/hr. Normal camber = 2 %

ecal = V2/225R = 802/225x1500 = 0.0189 < 0.02Normal camber of 2 % may be retained on horizontal curveSafety check should be done along with negative superelevation at the outer half of the pavement dueto normal camberNet transverse skid resistance = - e + f = - 0.02+0.15

= 0.13


Centrifugal ratio = V2/127R= 802/127x1500= 0.033 << 0.13

This Horizontal curve with normal camber is quite safe for design speed of 80 km/hr



Horizontal Alignment – Part II







• Understand AASHTO approach for design of

superelevationDifferent methods for distribution of e and f over a range of curves Maximum superelevationMaximum side friction factorEffect of grade


AASHTO: Distribution of e and f Five methods to distribute e and fMethod-1:e and f are directly proportional to 1/R

• A straight line relation for e and f between 1/R = 0 and 1/R = 1/Rmin for vehicles traveling at design or average running speed

• Considerable merit and logic but simple• Appropriate if each vehicle travels at a constant

speed on tangent and curve (intermediate degree or with minimum radius)

• Some drivers drive faster on tangents and flatter curves than on sharper curves


2

3

4

2

3

3

451

3

5

212

4

1 / R

4

1 / RSupe

r ele

vatio

nSi

de fr

ictio

n fa

ctor

Corresponding f at design speed

Distribution of superelevation


Method-2: f first reaches the maximum value then e starts increasing

• First f - then e are increased in inverse proportion to the radius of curvature

• Possibility of no superelevation on flatter curves• Superelevation, when introduced, increases

rapidly• Particularly advantageous on low speed urban

streets where because of practical constraints, superelevation frequently cannot be provided


2

3

4

2

3

3

451

3

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212

4

1 / R

4

1 / R

Side

frSu

per e

leva

tion

rate

ictio

n fa

ctor




Method-3: e first reaches the maximum value then f starts increasing

• e upto emax for vehicles traveling at design speed • No side friction on flat curves (e < emax at design

speed)• Beyond emax, f increases rapidly as curves

become sharper• Results in negative f on flat curves for vehicles

traveling at average running speed• Marked difference in f for different curves: Not

logical and may result into erratic driving, either at design or average running speed


2

3

4

2

3

3

451

3

5

212

4

1 / R

4

1 / R

Side

frSu

per e

leva

tion

rate

ictio

n fa

ctor




Method-4: Same as method 3 but this is based on average running speed instead of design speed

• Overcomes deficiency of method-3 by using e at speeds lower than design speed

• emax is reached near the middle of the curvature range – at average running speed no f is required upto this curvature

• f, when introduced, increases rapidly and in direct proportion for sharper curves

• Same disadvantage of method-3 but with a smaller degree


2

3

4

2

3

3

451

3

5

212

4

1 / R

4

1 / R

Side

frSu

per e

leva

tion

rate

ictio

n fa

ctor




Method-5: e and f are in a curvilinear relation with 1/R• For overdriving (likely) on flat to intermediate

curves: desirable to have ‘e’ similar to method-4• Very little risk for overdriving on such curves: ‘e’

adequate for average running speed and considerable ‘f’ for greater speed

• Method-1 is also desirable: avoids use of emax for a substantial part of the range of curve radii

• Method-5: distribution of ‘e’ and ‘f’ reasonably retaining the advantages of both Methods 1 and 4

• Curve 5- unsymmetrical parabolic form: a practical distribution of ‘e’ & ‘f’ over the range of curvature


2

3

4

2

3

3

451

3

5

212

4

1 / R

4

1 / R

Side

frSu

per e

leva

tion

rate

ictio

n fa

ctor




Maximum Superelevation (AASHTO)• Depends on Four Factors

Climate condition (frequency & amount of snow and ice)

Terrain condition (flat, rolling, mountainous)

Type of area (urban, rural)

Frequency of slow moving vehicles (affect high ‘e’)

• No single emax is universally applicable

• Desirable to use one emax (uniformity) within a region and similar climate for design consistency


12 %: Practical maximum value where snow and ice do not exist

10 %: Highest superelevation rate for highways in common use

(above 8% only in areas without snow and ice)

8 %: Reasonable max value for low volume gravel surfaced roads – also max practical limit where snowand ice are factors

4-6 %: where traffic congestion and extensive marginal development acts to restrict top speeds

• Five emax values: 4, 6, 8, 10 and 12 percent


• Side friction factor vary with speed • Different curves have been developed• AASHTO recommends a curve for values of ‘f’

(0.17 to 0.08) with reasonable margin of safety at high speeds and lead to somewhat low ‘e’ for low design speed than do some of the other curves

80Side

fric

tion

fact

or

Maximum Side Friction Factor (AASHTO)

Speed (km/h)


• When these f values are used in conjunction with the recommended Method-5, they determine the f distribution curves for the various speeds

• Subtracting these computed f values from the computed value of (e/100 + f) at the design speed, the finalized e distribution is obtained

finalized e/100 distribution

1/Rmin

(e/1

00) o

r f o

r ((e

/100

)+f)

e/100 + f (design speed)

f-distribution

1/R


• Finalized e distribution curves are obtained for various speed and different emax. AASHTO gives different charts and tables for different emax.

emax = 4 %

20

Radius of curve (m)

Supe

r ele

vatio

n ra

te, e

100 km/h

8060

40


Effect of grade• On long or fairly steep grades, drivers tend to

travel faster in downgrade than in upgrade direction

• On divided highway with each roadway independently superelevated or on a one-way ramp this tendency should be recognized

• Design speed should be considered slightly higher in downgrade and slightly lower in upgrade

• Reduction or increase in design speed depends on rate and length of the grade, and the magnitude of the curve radius


• Should similar adjustments be made on two-lane and multilane undivided roadways?Two directions of traffic tend to balance each other: No adjustment necessaryDowngrade speed is the most critical: adjustment for it may be desirableLanes can also be constructed at different cross slope (not common practice) Adjustment for the whole traveled way as determined by the downgrade speed (more practical)In general, it is advisable to follow the common practice: no adjustment on undivided roadways



Horizontal Alignment – Part III







• Understand various stages and alternative approaches for attainment of superelevation

• Appreciate the need and understand the basis for estimating required set-back distance on horizontal curves


Attainment of superelevationSplit-up into two parts::

• Elimination of crown of the cambered section• Rotation of pavement to attain full superelevationElimination of crown of the cambered section1st Method: Outer edge rotated about the crown

C/L


2nd Method: Crown shifted outwards

Disadvantages• Small length of road – cross slope less than

camber • Drainage problem in outer half

Disadvantages• Large negative superelevation on outer half• Drivers have the tendency to run the vehicle along

shifted crown

C/L


Rotation of pavement to attain full superelevation1st Method: Rotation about the C/L (depressing the inner edge and raising the outer edge each by half the total amount of superelevation)

E/2

E/2

C.L

Advantages• Earthwork is balanced • Vertical profile of the C/L

remains unchangedDisadvantages• Drainage problem: depressing

the inner edge below the general level


E

2nd Method: Rotation about the Inner edge (raising both the centre as well as outer edge – outer edge is raised by the total amount of superelevation)

C.L

Advantages • No drainage problem Disadvantages• Additional earth filling• C/L of the pavement is also raised

(vertical alignment of the road is changed)

3rd Method: Rotation about the outer edge


Center Line

Inner edge

Outer edge

Fully Superelevated Circular Curve

Normal Camber

Transition Curve

Typical Superelevation Diagram


Various design elements related to horizontal alignment

Horizontal curveSuperelevationSet-back distanceTransition curveExtra widening


Set-back distance• Adequate sight distance on horizontal curves: An

essential consideration• Obstruction to sight distance: buildings, trees, cut

slopes etc. along the inner side of the horizontal curves

• Set-back distance: to provide adequate sight distance on horizontal curves


• On narrow roads, the SD is measured along the C/L of the road

• On wider roads, the SD is measured along the C/L of the inner-side lane

• Set-back distance depends onRequired sight distance (S)Radius of horizontal curve (R)Length of curve (L)


• Two cases are considered for analysis:Sight distance is less than length of curve (S<L)Sight distance is more than length of curve (S>L)

• Required sight distance could be Stopping Sight DistanceIntermediate Sight DistanceOvertaking Sight Distance


R

S

m

α

Case-I: When S < L

Narrow roads

α/2 = S/2R radians=180S/2πR degree

Set-back distance m = R – R Cos(α/2)


R

m

α

Wide roads

α/2 = S/2(R-d) radians=180S/(2π(R-d)) degree

Set-back distance m = R – (R-d) Cos(α/2)

d= distance between C/L of the road and the C/L of the inner-side lane


Case-II: When S > LNarrow roadsα/2 = L/2R radians

=180L/2πR degreeR

αSet-back distance m = R – R Cos(α/2)

+ ((S-L)/2) Sin (α/2)

Wide roadsα/2 = L/(2(R-d))radians

=180L/2π(R-d) degreeSet-back distancem = R – (R-d)Cos(α/2) + ((S-L)/2) Sin (α/2)


Example Problem-1Calculate required set-back distance considering• Intermediate sight distance• Length of the curve = 300 m• Radius of horizontal curve = 230 m• Design speed = 80 km/h• Coefficient of friction = 0.35• Reaction time = 2.5 sec • Total width of pavement on curve = 7.71 m (two

lane road)


SSD = 0.278 x 80 x 2.5 + 802/(254 x 0.35) = 127.6 mISD = 2 x SSD = 2 x 127.6 m = 255 m < 300 mA case where S < Ld = 7.71/4 m = 1.93 mα/2 = (180 x 255)/ (2 x π x (230-1.93))

= 32 degreeSet-back distance = 230 – (230-1.93) Cos (320)= 36.6 m


Curve Resistance• Automobiles are steered by turning the front

wheels• Vehicles, driven by rear wheels: on horizontal

curves, the direction of rotation of rear and front wheels are different

• Tractive force given by rear wheels: T• Tractive force available in the direction of

movement = T Cos α (less than T)


• Curve Resistance: loss of tractive force due to turning of vehicles on horizontal curve = T-T Cos α = T(1- Cos α) (function of turning angle α)

• Vehicles with front driving wheels: Problem does not exist

• Most of the heavy vehicles have rear driving wheels: more curve resistance on sharper curves

• Problem of curve resistance is acute on hill roads: curves are often sharp and in addition roads have steep gradients



Horizontal Alignment – Part IV







• Understand basic elements/ terminologies related to transition control

• Understand the need for providing transition curve, and basis for using spiral transition curve

• Understand the IRC approach for design of length of transition curve


Tangent runout and superelevation runoff (AASHTO)Tangent runout: Length of roadway needed to accomplish a change in outside-lane cross slope from normal cross slope rate to zero, or vice versaSuperelevation runoff:The length of roadway needed to accomplish a change in outside-lane cross slope from zero to full superelevation, or vice versa


Center Line

Inner edge

Outer edge

Tangent runout Fully Superelevated

Circular Curve

Superelevation runoff

Typical Superelevation Diagram


Horizontal transition control• Tangent-to-Curve Transition: Transition curve is

not used – roadway tangent directly adjoins the main circular curve

• Transition Curve or Spiral Curve Transition: A transition curve is used between roadway tangent and the main circular curve


Horizontal transition curve• Transition curve: radius decreases from infinity

to a designated radius over the length • A transition curve between straight (tangent)

and circular curve: radius changes from infinity (at tangent point) to the radius of the circular curve (at beginning of the circular curve)

• Rate of change of radius of transition curve: shape or equation of the curve


Necessity for transition curve

• Introduce gradually the centrifugal force between the tangent point and the beginning of the circular curve avoiding a sudden jerk on the vehicle

• Enable driver to turn the steering gradually with comfort and safety: easy-to-follow path for drivers

Horizontal curve

Transition Curve

Tangent

Transition Curve

P.I

Tangent


• Minimize encroachment on adjoining traffic lanes and tend to promote uniformity in speed

• Enable gradual introduction of designed superelevation (i.e. superelevation runoff)

• Enable gradual introduction of required extra widening

• Improve aesthetic appearance of the road


Different types of transition curvesa) Cubic parabolab) Lemniscates c) Spiral (Clothoid)

• Up to deflection angle of 9o: No significant difference in these curves

• Radius decreases with an increase in length for all these curves

• Lemniscates and cubic parabola: rate of change of radius and hence the rate of change of centrifugal acceleration is not constant for large deflection angle

Lemniscates

Spiral

Cubic Parabola


• Spiral: Radius is inversely proportional to the length and the rate of change of centrifugal acceleration is uniform throughout the length of curve


Ideal shape of transition curve• Rate of introduction of centrifugal force or rate of

change of centrifugal acceleration should be consistent

• The length should be inversely proportional to radius


• Spiral fulfils the condition of an ideal transition curve

• The geometric property of spiral is such that the calculations and setting out the curve in the field is simple and easy (LR= constant)

• Spiral transition curve simulates the natural turning path of a vehicle

• Spiral Transition Curve: IRC and AASHTO


Length of transition curveIRC method: Length of transition curve is higher of

the following two values.

i)Based on rate of change of centrifugal acceleration

• Time taken to traverse transition length (Ls) in design speed v

t = Ls/v (t- sec, Ls – m, and v – m/sec)• Maximum centrifugal acceleration v2/R is

introduced in time ‘t’ through the transition length Ls


• Rate of change of centrifugal acceleration C =v2/Rt = v2 / R.(Ls/v) =v3/LsR

• if speed is V km/h then,Ls=V3 / (3.6)3 CR = 0.0215V3 / CR

• IRC: C= 80/(75+V) (C- m/sec3, R- m)(0.5<C<0.8)


ii) Based on rate of change of superelevation• Longitudinal grade developed at the pavement

edge compared to through grade along the C/L• Rate of change should not cause discomfort to

travelers or make the road appear unsightly• Rate of change (IRC):

1 in 150 for roads in plain and rolling terrain1 in 60 for mountainous/steep terrain

• Required length depends on the method of attainment of superelevation


• Empirical formula for calculation of length of transition curve on the basis of recommended maximum rate of change of superelevation

For Plain and Rolling Terrain:

Ls = 2.67V2 / Rc

For Mountainous and Steep Terrain:

Ls = 1.0V2 / Rc


Elements of a combined circular and transition curve

RcS P.T

H.I.P

Lc

T.P

Es

cθs

Rc

Ts

LsLs

θs

Ls: Length of transition Lc: length of circular curveEs: Apex distance T.P:Tangent PointS: Shift H.I.P:Horizontal Intersection Pt.

: Total deviation angle: Deviation angle of transition curve: Deviation and central angle of circular arc


RcS P.T

H.I.P

Lc

T.P

Es

cθs

Rc

Ts

LsLs

θs

θs

c

Shift S = Ls2/ (24 Rc)


Example problem-1Design the length of transition curve for a two lane highway having a circular curve of 250m radius. Design speed 70 km/h

Based on rate of change of centrifugal acceleration:Rate of change of centrifugal acceleration,C= 80/(75+V) = 80/(75+70) = 0.55 (0.5<0.55<0.8) 0.K.Ls= 0.0215V3 / CR = 0.0215x703/ (0.55x250)

= 53.6 = 55 m (say)Based on allowable rate of introduction of superelevation:e = V2/225R = 702/225x250 = 0.087

Indian Institute of Technology, KharagpurMaximum allowable value of e = 0.07Side friction factor f = V2/127R – e

= 702/127x250 – 0.07= 0.154 – 0.07 = 0.084<0.15

(O.K)Pavement width B = 7.5mTotal raise of outer edge w.r.t C/L = eB/2 = 0.07x7.5/2

= 0.26mMaximum allowable rate: 1 in 150Ls = 0.26x150 = 39mLength of transition curve = maximum of two cases

i.e. 55m


Minimum Transition lengths for different speeds and curve radii

Plain terrain Rolling terrainDesign Speed (km/hr) Design Speed (km/hr)

100 80 65 50 40 50 40 30 25

600 80

1800 30

20

Curve

radius (m)

Curve radius

(m)

100

200 NA

300 NA 75

360 130 60

400 115

2000 NR


• Where transition curve can not be provided for some reasons, two-third of the superelevation will be attained on the straight section before start of the circular curve and balance one-third on the curve (IRC)

UG Geometric Design II

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