SUPER_ELEVATION.pdf

Highway Engineering 1

1. Highway Geometric Design
Analysis and Design of Superelevation

2. •WHAT IS SUPERELEVATION IN HIGHWAY ENGINEERING?
Superelevationisthetransverseslopeprovidedtocounteracttheeffect
ofcentrifugalforceandreducethetendencyofvehicletooverturnand
toskidlaterallyoutwardsbyraisingthepavementouteredgewith
respecttoinneredge.superelevationisrepresentedby“e”.
Superelevationorcantorbankingisthetransverseslopeprovidedat
horizontalcurvetocounteractthecentrifugalforce,byraisingtheouter
edgeofthepavementwithrespecttotheinneredge,throughoutthe
lengthofthehorizontalcurve.

3. •Whenvehiclesapproachingahorizontalcurve,therewillbeforceresulted
centripetalaccelerationtryingtopushthisvehicleoutsidethecurve.
•Thisforceisnormallybalancedbytheforceresultedfromthefriction
betweenvehicles’tiresandroadsurface.
•Athighspeedsand/orlowradius,thefrictionalforceisnotgenerally
sufficienttobalancethecentrifugalforce.
•Forthisreason,thecarriagewayshouldbesuper-elevatedtoincreasethe
resistanceasshownbelow.

4. Superelevation

6. Carriage way showing superelevation

7. •Objective of providing super elevations:
i.Tocounteracttheeffectofcentrifugalforceactingonthemoving
vehicletopullthesameoutwardonahorizontalcurve.
ii.Tohelpafastmovingvehicletonegotiateacurvedpathwithout
overturningandskidding.
iii.Toensuresafetyofthefastmovingvehicle.
iv.Topreventdamagingeffectontheroadsurfaceduetoimproper
distributionofload.

8. Figure: Superelevation

11. SuperelevationinHighwayEngineering

12. •Analysis of superelevation
Let us say Design speed = V m/s Radius = R m

13. •Various forces acting on the vehicle:

15. Where;
•e=rateofsuperelevationin%
•f=lateralfrictionfactor=0.15
•V=velocityofvehicleinm/s
•g=accelerationduetogravity=9.81??????/??????
2
•R=radiusofcircularcurveinmeters.

16. In order to find out how much this raising should be, the following analysis may
be done as presented
Figure below shows: Vehicle on curves, acting forces

18. W sin ??????+ w cos ??????. f = P cos ??????
Dividing above EQUATION by w. cos ??????, we get

20. •Limits for minimum superelevation
•Minimum superelevation = camber or cross slope Camber: Slope provide in
the transverse direction to drain off rain water quickly is known as Camber
or Cross slope. This will also prevents slipping and skidding of vehicles.

21. Design Of Superelevation
•There are four steps involved in the design of superelevation. And
they are;
•
Step 1:

22. •Step 1:
Calculatethesuperelevationnecessaryfor75%designspeedand
assumeNolateralfrictionisdeveloped
If e value is less than �
�????????????= 0.07,
provide calculated e value.
Otherwise proceed to next step

23. •Step 2:
•When �
????????????�>�
�????????????,
Provide e = �
�????????????= 0.07 in this step and go to next step.

24. •Step 3:
•From the above step we have the value of e. so, check for lateral friction
factor is applied in this step for the known value of e.
If�
????????????&#3627408473;<&#3627408467;
&#3627408474;????????????(0.15),thene=0.07issafe.
Butif&#3627408467;
????????????&#3627408473;>0.15.
Thenrestrictthevaluestof=0.15,e=0.07
Andgotolaststep.

25. •Step 4:
•In this step we will find out the value of restricted speed. Let V = Va
If ??????
??????> V, then e = 0.07, f = 0.15 if
??????
??????< V, then also e = 0.07, f = 0.15
but,
speed restriction board is provided
which consists the value of ??????
??????As
shown in the figure.

29. HORIZONTAL ALIGNMENT
•DEFINITION AND TYPES OF HORIZONTAL CURVES
•Horizontalalignmentisoneofthemostimportantfeaturesofahighway
design.Itsproperdesigncanresultinhighperformanceregardingspeed,
safety,efficiency,andcomfort.
•Inaddition,itmayresultinthesavingofeconomyandincreasethehighway
capacity.

30. •Thedesignofhorizontalalignmentsrequirestheunderstandingofdesign
speedandhorizontalcurves.
•Horizontalalignmentincludesofaroadcompriseaseriesofstraightlines
knownastangentswiththeprovisionofcurvestochangedirection.It
alsoincludesthedesignofsuperelevation,extrawidening,setback
distance,transitioncurvedesign,etc.

31. •Typicallyhorizontalcurvesconsistofacircularcurvewithaconstant
radius.
•Forfasterdesignspeedscircularcurvesarejoinedtothetangentsusing
transitioncurveswhichhavevaryingradii.
•Thesecurvesimprovetheoccupantsafetyandcomfortbyprovidinga
gradualincreaseofthesidewaysforcefeltbythevehicleduetothe
introductionofthecurve.
•Acircularcurvemayalsobeaccompaniedbytheintroductionof
superelevation.
•Thisisweretheroadistiltedintothecurvetoreducethelikelihoodof
vehiclesrunningofftheroadatthecurve.

33. •PC is the Point of Curvature. This is the point where the straight tangent
ends and the curve starts.
•IP is the point of intersection between the 2 tangent lines. It can be used
to help setting out of the road curve on site.
•PT is the Point of Tangency. This is the point where the curve ends and the
second straight starts.
•R is the radius of the curve determined as detailed below.
•Δ is the deflection angle.

34. Horizontal Curve
•Horizontalcurvesareprovidedineachandeverypointofintersection
oftwostraightalignmentsofhighwaysinordertochangethe
direction.
•Thedirectionchangeshouldbegradualtoensuresafetyandcomfort
tothepassengers.
•Thenecessityofcurvearisesduetothefollowingreasons:

35. •Topographyoftheterrain
•Restrictionsimposedbyproperty
•Providingaccesstocertainlocality
•Restrictionsbysomeunavoidablereasonsofland
•Restrictionsbycertainreligious,monumentalorsomeotherstructures
•Makinguseofexistingrightofway
•Minimizingearthworkquantity
•Preservationofexistingamenities
•Maintainingconsistencywiththetopographicalfeaturesoftheterrain.

36. •Theapplicationofhorizontalcurvesenhancescomforttothe
passengersbyavoidingthesuddenchangeindirectionandreduces
mentalstrainbytravellingmonotonouslyalongthestraightroute.
•Inaddition,itmakesthedrivermorealertwhiletravellingalongthe
curvedpathwhichhelpstoreduceroadaccidents.
•Thisalsoplaysanimportantroleinspeedcontrolandcompelsthe
drivertomaintainthespeedofvehiclewithinareasonablelimit.

37. Types Of Curves
•The simplest type of highway curve is the circular curve. It is a curve
used in highway having the constant radius.
i.Simple circular curves: It consists of a single arc connecting two
straight lines.
ii.Reverse circular curves: It consists of two or more arcs of one or
different circles turning in two opposite directions that join at the
common tangent point.
iii.Compound circular curves: It consists of series of simple circular
curves of one or different radius that turns in the same direction
and meet at the common tangent point.

38. Fundamental Horizontal Curve Properties
Asidefrommomentum,whenavehiclemakesaturn,twoforcesare
actinguponit;
-Thefirstisgravity,whichpullsthevehicletowardtheground.
-Thesecondiscentrifugalforce,forwhichitsopposite,centripetal
accelerationisrequiredtokeepthevehicleonacurvedpath.
-Foranygivenvelocity,thecentripetalforceneedstobegreaterfora
tighterturn(onewithasmallerradius)thanabroaderone(onewith
alargerradius).
-Onalevelsurface,sidefriction&#3627408467;
??????servesasacounteringforcetothe
centrifugalforce,butitgenerallyprovidesverylittleresistance/force.
Thus,avehiclehastomakeaverywidecircleinordertomakea
turnonthelevel.

39. •Giventhatroaddesignsusuallyarelimitedbyverynarrowdesignareas,
wideturnsaregenerallydiscouraged.
•Todealwiththisissue,designersofhorizontalcurvesincorporateroadsthat
aretiltedataslightangle.
•Thistiltisdefinedassuperelevation,ore,whichistheamountofriseseen
onanangledcross-sectionofaroadgivenacertainrun,otherwiseknownas
slope.
•Thepresenceofsuperelevationonacurveallowssomeofthecentripetal
forcetobecounteredbytheground,thusallowingtheturntobeexecuted
atafasterratethanwouldbeallowedonaflatsurface.

40. •Superelevationalsoplaysanotherimportantrolebyaidingindrainage
duringprecipitationevents,aswaterrunsofftheroadratherthan
collectingonit.
•Generally,superelevationislimitedtobeinglessthan14percent,as
engineersneedtoaccountforstoppedvehiclesonthecurve,where
centripetalforceisnotpresent.
•TheallowableradiusR
•forahorizontalcurvecanthenbedeterminedbyknowingtheintended
designvelocityV
•,thecoefficientoffriction,andtheallowedsuperelevationonthecurve.

41. •TheallowableradiusRforahorizontalcurvecanthenbedeterminedby
knowingtheintendeddesignvelocityV,thecoefficientoffriction,andthe
allowedsuperelevationonthecurve.
Withthisradius,practitionerscandeterminethedegreeofcurvetoseeifit
fallswithinacceptablestandards.

42. •How to Calculate the Degree of a Curve

43. •Theratiobetweenthedegreeofcurvature(D)and360isthesameas
theratiobetween100feetofarcandthecircumference(C)ofacircle
havingthesameradius.Thatmaybeexpressedasfollows:
Da=Degreeofcurve[anglesubtendedbya30.5-m(100ft)arcalong
thehorizontalcurve

44. 1

Superelevation or Banking of Road: -
When a vehicle travels in a circular path or curved path, it is subjected to an outward force which
makes a vehicle to overturn and skid due to Centrifugal force. To overcome this force and for safe
travel of a vehicle, the outer edge of the road is raised above the inner edge. This is known as
superelevation or banking of road.


Superelevation/Banking of road reduces the effect of centrifugal force on the running wheels. If
superelevation is not provided with the entire centripetal force is produced by the friction
between the vehicle’s tires and the roadway, thus results in reducing the speed of a vehicle.

Advantages of providing Super elevation: -
i. Super elevation is provided to achieve the higher speed of vehicles.
ii. It increases the stability of fast-moving vehicles when they pass through a horizontal
curve, and it also decreases the stresses on the foundation.
iii. In the absence of super elevation on the road along curves, potholes are likely to occur at
the outer edge of the road.
iv. The max value of Super Elevation is 1 in 15.

45. 2

Derivation of Super Elevation:
As per the figure, the below forces are acting on a car

In order to find out the angle of elevation (Super Elevation) the “tan” formula is used;
From above fig, tan??????=
&#3627408476;&#3627408477;&#3627408477;&#3627408476;&#3627408480;&#3627408470;&#3627408481;&#3627408466; &#3627408480;&#3627408470;&#3627408465;&#3627408466;
??????&#3627408465;&#3627408471;??????&#3627408464;&#3627408466;&#3627408475;&#3627408481; &#3627408480;&#3627408470;&#3627408465;&#3627408466;


Therefore, tan??????=
??????
??????

The below forces are acting on the vehicle as mentioned in figure:
Weight of the vehicle = W kg (↓);
Centripetal force = P (→);
Frictional forces = F1 & F2 (←);
You can check out the below figure more idea.

Hence, P. Cosθ = W. Sinθ + F1 + F2

Where, F = fR

P. Cosθ = W. Sin θ + fR1 + fR2

= W. Sin θ + f (R1 + R2)
= W. Sin θ + f (P Sin θ + W Cos θ)

46. 3

P. Cos θ - f. P Sin θ = W. Sin θ + f. W Cos θ
Divide with “W. Cos θ”;
(??????.??????&#3627408424;&#3627408428;??????−&#3627408519;.??????.&#3627408428;??????&#3627408423;??????)
&#3627408510;.??????&#3627408424;&#3627408428;??????
=
(&#3627408510;.&#3627408428;??????&#3627408423;??????+&#3627408519;.&#3627408510;.??????&#3627408424;&#3627408428;??????)
&#3627408510;.??????&#3627408424;&#3627408428;??????

??????
&#3627408510;
−
&#3627408519;.??????
&#3627408510;
&#3627408429;??????&#3627408423;??????= &#3627408429;??????&#3627408423;??????+&#3627408519;
??????
&#3627408510;
{&#3627409359;−&#3627408519;.&#3627408429;??????&#3627408423;??????}= &#3627408429;??????&#3627408423;??????+&#3627408519;
??????
&#3627408510;
=
(&#3627408429;??????&#3627408423;??????+&#3627408519;)
(&#3627409359;−&#3627408519;.????????????????????????)

??????
&#3627408510;
= &#3627408518;+&#3627408519;
But,
??????
&#3627408510;
=
&#3627408509;
&#3627409360;
&#3627408520;??????

Therefore, e + f =
&#3627408509;
&#3627409360;
&#3627408520;??????

Where,
e = rate of Super elevation in %
V = velocity of vehicle in m/s
f = lateral friction factor = 0.15
g = acceleration due to gravity = 9.81 m/s
2
R = radius of circular curve in meters.
If velocity is in KMPH then; e + f = V
2/ 127R
Super Elevation formula: -