Presentation on theme: "Geometric Design It deals with visible elements of a highway."— Presentation transcript:
1 Geometric Design It deals with visible elements of a highway. It is influenced by:Nature of terrain.TypeComposition and hourly volume / capacity of trafficTraffic FactorsOperating speed (Design Speed)Landuse characteristics (Topography)Environmental Factors (Aesthetics).
2 TERRAIN CLASSIFICATION Terrain typePercentage cross slopeof the countryPlain0-10Rolling10-25Mountainous25-60Steep>60
3 goals of geometric design Maximize the comfortSafety,Economy of facilitiesSustainable Transportation Planning.
11 ARTERIALNo frontage access, no standing vehicle, very little cross traffic.Design Speed : 80km/hrLand width : 50 – 60mSpacing 1.5km in CBD & 8km or more in sparsely developed areas.Divided roads with full or partial parkingPedestrian allowed to walk only at intersection
12 SUB ARTERIAL Bus stops but no standing vehicle. Less mobility than arterial.Spacing for CBD : 0.5kmSub-urban fringes : 3.5kmDesign speed : 60 km/hrLand width : 30 – 40 m
13 Collector Street Collects and distributes traffic from local streets Provides access to arterial roadsLocated in residential, business and industrial areas.Full access allowed.Parking permitted.Design speed : 50km/hrLand Width : 20-30m
14 Local Street Design Speed : 30km/hr. Land Width : 10 – 20m. Primary access to residence, business or other abutting propertyLess volume of traffic at slow speedOrigin and termination of trips.Unrestricted parking, pedestrian movements. (with frontage access, parked vehicle, bus stops and no waiting restrictions)
15 CUL–DE- SACDead End Street with only one entry access for entry and exit.Recommended in Residential areas
16 HIGHWAY CROSS SECTIONAL ELEMENTS 1.Carriage way (Pavement width)2.Camber3.Kerb4.Traffic Separators5.Width of road way or formation width6.Right of way (Land Width)7.Road margins8.Pavement Surface(Ref: IRC 86 – 1983)
17 GEOMETRIC CROSS SECTION The primary consideration in the design of cross sections is drainage.Highway cross sections consist of traveled way, shoulders (or parking lanes), and drainage channels.Shoulders are intended primarily as a safety feature.Shoulders provide:accommodation of stopped vehiclesemergency use,and lateral support of the pavement.Shoulders may be either paved or unpaved.Drainage channels may consist of ditches (usually grassed swales) or of paved shoulders with berms of curbs and gutters.
19 Divided highway cross section, depressed median, with ditches.
20 Geometric cross section cont.. Standard lane widths are 3.6 m (12 ft).Shoulders or parking lanes for heavily traveled roads are 2.4 to 3.6 m (8 to 12 ft) in width.narrower shoulders used on lightly traveled road.
21 CARRIAGE WAY (IRC RECOMMENDATIONS) Single lane without Kerbs = 3.50mTwo lane without kerbs = 7mTwo lane with kerbs = 7.5m3 lane with or without kerbs = 10.5 /11.04 lane with or without kerbs = 14.0m6 lane with or without kerbs = 21.0 mIntermediate carriage way = 5.5mMultilane pavement = 3.5m/lane
22 Footpath (Side walk) No of Persons/Hr Required Width of footpath (m) All in one directionIn both direction12008001.5240016002.036002.5480032003.0600040004.0
23 Cycle Track Minimum = 2m Each addln lane = 1m Separate Cycle Track for peak hour cycle traffic more than 400 with motor vehicle of traffic 100 – 200 vehicles/Hr.Motor Vehicles > 200; separate cycle track for cycle traafic of 100 is sufficient.
24 Median Width of Median Depends on: Mim Width of Median: Available ROW TerrainTurn LanesDrainage.Mim Width of Median:Pedestrian Refuge =1.2mTo protect vehicle making Right turn = 4.0m (Recc – 7.0m)To protect vehicle crossing at grade = 9 – 12m.For Urban area 1.2 to 5m
25 KERBS Road kerbs serve a number of purposes: - retaining the carriageway edge to prevent 'spreading' and loss of structural integrity- acting as a barrier or demarcation between road traffic and pedestrians or verges- providing physical 'check' to prevent vehicles leaving the carriageway- forming a channel along which surface water can be drained
26 KERBSLow or mountable kerbs : height = 10 cm provided at medians and channelization schemes and also helps in longitudinal drainage.Semi-barrier type kerbs : When the pedestrian traffic is high.Height is 15 cm above the pavement edge.Prevents encroachment of parking vehicles, but at acute emergency it is possible to drive over this kerb with some difficulty.Barrier type kerbs : Designed to discourage vehicles from leaving the pavement. They are provided when there is considerable amount of pedestrian traffic.Height of 20 cm above the pavement edge with a steep batter.Submerged kerbs :They are used in rural roads.The kerbs are provided at pavement edges between the pavement edge and shoulders.They provide lateral confinement and stability to the pavement.
27 CAMBER (OR) CROSS FALL S. No Type of Surface % of camber in rainfall range Heavy to light1Gravelled or WBM surface2.5 % - 3 %( 1 in 40 to 1 in 33)2Thin bituminous Surface2.0 % %( 1 in 50 to 1 in 40)3Bituminous Surfacing or Cement Concrete surfacing1.7 % %4Earth4 % - 3 %
28 Types of Camber Parabolic or Elliptic Straight Line Straight and Parabolic
29 Sight DistancesThe actual distance along the road surface up to which the driver of a vehicle sitting at a specified height has visibility of any obstacle.The visibility ahead of the driver at any instance.29
30 SIGHT DISTANCETHE SIGHT DISTANCE AVAILABLE ON A ROAD TO A DRIVER DEPENDS ONFEATURE OF ROAD AHEADHEIGHT OF THE DRIVER’S EYE ABOVE THE ROAD SURFACE
32 Sight Distance in Design Stopping Sight Distance (SSD) – object in roadwayPassing Sight Distance (PSD) – pass slow vehicle32
33 Stopping Sight Distance (SSD) THE DISTANCE WITHIN WHICH A MOTOR VEHICLE CAN BE STOPPED DEPENDS ONTotal reaction time of driverSpeed of vehiclesEfficiency of brakesGradient of roadFrictional resistance
34 TOTAL REACTION TIMEPERCEPTION TIMEBRAKE REACTION TIME
35 TOTAL REACTION TIME DEPENDS ON PIEV THEORY PERCEPTIONINTELLECTIONEMOTIONVOLIATION
36 Perception-Reaction Process IdentificationEmotionReaction (volition)PIEVUsed for Signal Design and Braking Distance3636
37 Perception-Reaction Process Sees or hears situation (sees deer)IdentificationIdentify situation (realizes deer is in road)EmotionDecides on course of action (swerve, stop, change lanes, etc)Reaction (volition)Acts (time to start events in motion but not actually do action)Foot begins to hit brake, not actual deceleration3737
38 0.5 to 7 seconds Affected by a number of factors. Typical Perception-Reactiontime range0.5 to 7 secondsAffected by a number of factors.3838
39 Perception-Reaction Time Factors EnvironmentUrban vs. RuralNight vs. DayWet vs. DryAgePhysical ConditionFatigueDrugs/AlcoholDistractions3939
40 AgeOlder driversMay perceive something as a hazard but not act quickly enoughMore difficulty seeing, hearing, reactingDrive slowerLess flexible4040
41 AgeYounger driversQuick Response but not have experience to recognize things as a hazard or be able to decide what to doDrive fasterAre unfamiliar with driving experienceAre less apt to drive safely after a few drinksAre easily distracted by conversation and others inside the vehicleMay be more likely to operate faulty equipment.Poorly developed risk perceptionFeel invincible, the "Superman Syndrome”4141
42 Alcohol Affects each person differently Slows reaction time Increases risk takingDulls judgmentSlows decision-makingPresents peripheral vision difficulties4242
43 Stopping Sight Distance (SSD) Required for every point along alignment (horizontal and vertical) – Design for it, or sign for lower, safe speed.Available SSD = f(roadway alignment, objects off the alignment, object on road)SSD = LD + BDLag distanceBraking Distance43
44 Lag Distance Speed of the vehicle = v m/sec Reaction Time of Driver = t sec ; (2.5 sec)Lag Distance = v t mIf the design speed is V kmph,Lag Distance = V x x t60 x 60= V t m
45 Braking DistanceKinetic Energy at the design speed of v m/sec= ½ m v2= W v2 ; m = W/g2gW = weight of the VehicleG = acceleration due to gravity (9.9 m/sec2)Work done in stopping the vehicle = F x lF = Frictional forceL = braking distanceF = coeff of friction = 0.35Wv2 = fWl ; l = v22g fg
46 SSD Equation SSD,m = 0.278V t + _____V2_____ 254f SSD in meter V = speed in kmphT = perception/reaction time (in seconds)f = design coefficient of friction46
47 STOPPING SIGHT DISTANCE FOR ASCENDING GRADIENT AND DESCENDING GRADIENT SSD = 0.278vt v22g(f+ (n/100))(or)SSD = 0.278Vt V2254(f - n/100)
48 Passing Distance Applied to rural two-lane roads The distance required for a vehicle to safely overtake another vehicle on a two lane, two-way roadway and return to the original lane without interference with opposing vehiclesDesigners assume single vehicle passingSeveral assumptions are considered (vehicle being passed s traveling at a uniform speed, and others)Normally use car passing carPassing distance increased by type of vehicleMinimum passing distance currently used are conservative
55 Stationing Horizontal Alignment Vertical Alignment Therefore, roads will almost always be a bit longer than their stationing because of the vertical alignmentDraw in stationing on each of these curves and explain it
56 Alignment Design Definition of alignment: Definitions from a dictionaryIn a highway design manual: a series of straight lines called tangents connected by circular curves or transition or spiral curves in modern practiceDefinition of alignment design: also geometric design, the configuration of horizontal, vertical and cross-sectional elements (first treated separately and finally coordinated to form a continuous whole facility)Horizontal alignment designComponents of horizontal alignmentTangents (segments of straight lines)Circular/simple curvesSpiral or transition curves
57 Alignment Design Horizontal curves Simple curves This consists of a single arc of uniform radius connecting two tangentsCompound curvesA compound curve is formed by joining a series of two or more simple curves of different radius which turn in same direction..
61 Alignment Design Horizontal curves TRANSITION CURVE Reverse curves A curve having its radius varying gradually from a radius equal to infinity to a finite value equal to that of a circular curveReverse curvesA circular curve consistings of two simple curves of same or different radii and turn in the opposite direction is called reverse curveSaturday, March 25, 2017
63 VERTICAL ALIGNMENTThe vertical alignment of a transportation facility consists oftangent grades (straight line in the vertical plane)vertical curves. Vertical alignment is documented by the profile.
71 Concepts Alignment is a 3D problem broken down into two 2D problems Horizontal Alignment (plan view)Vertical Alignment (profile view)StationingAlong horizontal alignment12+00 = 1,200 ft.Piilani Highway on Maui
72 Stationing Horizontal Alignment Vertical Alignment Therefore, roads will almost always be a bit longer than their stationing because of the vertical alignmentDraw in stationing on each of these curves and explain it
73 From Perteet Engineering Typical set of road plans – one page onlyFrom Perteet Engineering
75 Vertical Alignment Objective: Primary challenge Determine elevation to ensureProper drainageAcceptable level of safetyPrimary challengeTransition between two gradesVertical curvesSag Vertical CurveG1G2G1G2Crest Vertical Curve
76 Vertical Curve Fundamentals Parabolic functionConstant rate of change of slopeImplies equal curve tangentsy is the roadway elevation x stations (or feet) from the beginning of the curve
77 Vertical Curve Fundamentals PVIG1δPVCG2PVTL/2LxChoose Either:G1, G2 in decimal form, L in feetG1, G2 in percent, L in stations
78 Relationships Choose Either: G1, G2 in decimal form, L in feet G1, G2 in percent, L in stationsRelationships
79 ExampleA 400 ft. equal tangent crest vertical curve has a PVC station of at 59 ft. elevation. The initial grade is 2.0 percent and the final grade is -4.5 percent. Determine the elevation and stationing of PVI, PVT, and the high point of the curve.PVIPVTG1=2.0%G2= - 4.5%PVC: STAEL 59 ft.
80 PVI G1=2.0% PVT G2= -4.5% PVC: STA 100+00 EL 59 ft. 400 ft. vertical curve, therefore:PVI is at STA and PVT is at STAElevation of the PVI is 59’ (200) = 63 ft.Elevation of the PVT is 63’ – 0.045(200) = 54 ft.High point elevation requires figuring out the equation for a vertical curveAt x = 0, y = c => c=59 ft.At x = 0, dY/dx = b = G1 = +2.0%a = (G2 – G1)/2L = (-4.5 – 2)/(2(4)) =y = x2 + 2x + 59High point is where dy/dx = 0dy/dx = x + 2 = 0x = 1.23 stationsFind elevation at x = 1.23 stationsy = (1.23)2 + 2(1.23) + 59y = ft
81 Other Properties G1, G2 in percent L in feet G1 x PVT PVC Y Ym G2 PVI YfLast slide we found x = 1.23 stations
82 Other Properties K-Value (defines vertical curvature) The number of horizontal feet needed for a 1% change in slopeG is in percent, x is in feetG is in decimal, x is in stations
83 Crest Vertical Curves For SSD < L For SSD > L SSD h2 h1 L PVILine of SightPVCG1PVTG2h2h1LFor SSD < LFor SSD > L
84 Crest Vertical Curves Assumptions for design Simplified Equations h1 = driver’s eye height = 3.5 ft.h2 = tail light height = 2.0 ft.Simplified EquationsMinimum lengths are about 100 to 300 ft.Another way to get min length is 3 x (design speed in mph)For SSD < LFor SSD > L
86 Design Controls for Crest Vertical Curves from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001
87 Design Controls for Crest Vertical Curves from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001
88 Light Beam Distance (SSD) Sag Vertical CurvesLight Beam Distance (SSD)G1headlight beam (diverging from LOS by β degrees)G2PVCPVTh1PVIh2=0LFor SSD < LFor SSD > L
89 Sag Vertical Curves Assumptions for design Simplified Equations h1 = headlight height = 2.0 ft.β = 1 degreeSimplified EquationsWhat can you do if you need a shorter sag vertical curve than calculated? Provide fixed-source street lightingMinimum lengths are about 100 to 300 ft.Another way to get min length is 3 x design speed in mphFor SSD < LFor SSD > L
91 Design Controls for Sag Vertical Curves from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001
92 Design Controls for Sag Vertical Curves from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001
93 Example 1A car is traveling at 30 mph in the country at night on a wet road through a 150 ft. long sag vertical curve. The entering grade is -2.4 percent and the exiting grade is 4.0 percent. A tree has fallen across the road at approximately the PVT. Assuming the driver cannot see the tree until it is lit by her headlights, is it reasonable to expect the driver to be able to stop before hitting the tree?Assume that S>L (it usually is not but for example we’ll do it this way), therefore S = ft. which is less than LMust use S<L equation, it’s a quadratic with roots of ft and ft.The driver will see the tree when it is feet in front of her.Available SSD is ft.Required SSD = (1.47 x 30)2/2(32.2)( ) + 2.5(1.47 x 30) = ft.Therefore, she’s not going to stop in time.ORL/A = K = 150/6.4 = 23.43, which is less than the required K of 37 for a 30 mph design speedStopping sight distance on level ground at 30 mph is approximately 200 ft.
94 Example 2 Similar to Example 1 but for a crest curve. A car is traveling at 30 mph in the country at night on a wet road through a 150 ft. long crest vertical curve. The entering grade is 3.0 percent and the exiting grade is -3.4 percent. A tree has fallen across the road at approximately the PVT. Is it reasonable to expect the driver to be able to stop before hitting the tree?Assume that S>L (it usually is), therefore SSD = ft. which is greater than LThe driver will see the tree when it is feet in front of her.Available SSD = ft.Required SSD = (1.47 x 30)2/2(32.2)( ) + 2.5(1.47 x 30) = ft.Therefore, she will be able to stop in time.ORL/A = K = 150/6.4 = 23.43, which is greater than the required K of 19 for a 30 mph design speed on a crest vertical curveStopping sight distance on level ground at 30 mph is approximately 200 ft.
95 Example 3A roadway is being designed using a 45 mph design speed. One section of the roadway must go up and over a small hill with an entering grade of 3.2 percent and an exiting grade of -2.0 percent. How long must the vertical curve be?For 45 mph we get K=61, therefore L = KA = (61)(5.2) = ft.
96 Trinity Road between Sonoma and Napa valleys Horizontal Alignment
97 Horizontal Alignment Objective: Primary challenge Fundamentals Geometry of directional transition to ensure:SafetyComfortPrimary challengeTransition between two directionsHorizontal curvesFundamentalsCircular curvesSuperelevationΔ
98 Horizontal Curve Fundamentals PITΔEMLPCΔ/2PTD = degree of curvature (angle subtended by a 100’ arc)RRΔ/2Δ/2
100 Example 4A horizontal curve is designed with a 1500 ft. radius. The tangent length is 400 ft. and the PT station is What are the PI and PT stations?Since we know R and T we can use T = Rtan(delta/2) to get delta = degreesD = /R. Therefore D = 3.82L = 100(delta)/D = 100(29.86)/3.82 = 781 ft.PC = PT – PI = 2000 – 781 =PI = PC +T = =Note: cannot find PI by subtracting T from PT!
102 Superelevation Divide both sides by Wcos(α) Assume fse is small and can be neglected – it is the normal component of centripetal acceleration
103 Selection of e and fs Practical limits on superelevation (e) ClimateConstructabilityAdjacent land useSide friction factor (fs) variationsVehicle speedPavement textureTire conditionThe maximum side friction factor is the point at which the tires begin to skidDesign values of fs are chosen somewhat below this maximum value so there is a margin of safety
104 Side Friction Factor New Graph from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004