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Geometric Design Session 02-06 Matakuliah: S0753 – Teknik Jalan Raya Tahun: 2009

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Bina Nusantara University 3 Contents Concepts Vertical Alignment Fundamentals Crest Vertical Curves Sag Vertical Curves Examples Horizontal Alignment Fundamentals Superelevation

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Bina Nusantara University 4 Introduction Alignment is a 3D problem broken down into two 2D problems –Horizontal Alignment (plan view) –Vertical Alignment (profile view) Stationing –Along horizontal alignment Piilani Highway on Maui

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Bina Nusantara University 5 Stationing Horizontal Alignment Vertical Alignment Introduction

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Bina Nusantara University 6 From Perteet Engineering Introduction

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Bina Nusantara University 7 Geometric Design Elements Sight Distances Superelevation Horizontal Alignment Vertical Alignment

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Bina Nusantara University 8 Vertical Alignment Objective: –Determine elevation to ensure Proper drainage Acceptable level of safety Primary challenge –Transition between two grades –Vertical curves G1G1 G2G2 G1G1 G2G2 Crest Vertical Curve Sag Vertical Curve

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Bina Nusantara University 9 Vertical Curve Fundamentals Parabolic function –Constant rate of change of slope –Implies equal curve tangents y is the roadway elevation x stations (or feet) from the beginning of the curve

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Bina Nusantara University 10 Vertical Curve Fundamentals G1G1 G2G2 PVI PVT PVC L L/2 δ x Choose Either: G 1, G 2 in decimal form, L in feet G 1, G 2 in percent, L in stations

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Bina Nusantara University 11 Relationships Choose Either: G 1, G 2 in decimal form, L in feet G 1, G 2 in percent, L in stations

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Bina Nusantara University 12 Example A 400 ft. equal tangent crest vertical curve has a PVC station of 100+00 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. G 1 =2.0% G 2 = - 4.5% PVI PVT PVC: STA 100+00 EL 59 ft.

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Bina Nusantara University 13 G 1 =2.0% G 2 = -4.5% PVI PVT PVC: STA 100+00 EL 59 ft.

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Bina Nusantara University 14 Other Properties G1G1 G2G2 PVI PVT PVC x YmYm YfYf Y G 1, G 2 in percent L in feet

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Bina Nusantara University 15 Other Properties K-Value (defines vertical curvature) –The number of horizontal feet needed for a 1% change in slope

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Bina Nusantara University 16 Crest Vertical Curves G1G1 G2G2 PVI PVT PVC h2h2 h1h1 L SSD For SSD < LFor SSD > L Line of Sight

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Bina Nusantara University 17 Crest Vertical Curves Assumptions for design –h 1 = driver’s eye height = 3.5 ft. –h 2 = tail light height = 2.0 ft. Simplified Equations For SSD < LFor SSD > L Assuming L > SSD…

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Bina Nusantara University 18 Design Controls for Crest Vertical Curves from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001

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Bina Nusantara University 19 Design Controls for Crest Vertical Curves from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001

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Bina Nusantara University 20 Sag Vertical Curves G1G1 G2G2 PVI PVT PVC h 2 =0 h1h1 L Light Beam Distance (SSD) For SSD < LFor SSD > L headlight beam (diverging from LOS by β degrees)

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Bina Nusantara University 21 Sag Vertical Curves Assumptions for design –h 1 = headlight height = 2.0 ft. –β = 1 degree Simplified Equations For SSD < L For SSD > L Assuming L > SSD…

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Bina Nusantara University 22 Design Controls for Sag Vertical Curves from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001

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Bina Nusantara University 23 Design Controls for Sag Vertical Curves from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001

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Bina Nusantara University 24 Horizontal Alignment Objective: –Geometry of directional transition to ensure: Safety Comfort Primary challenge –Transition between two directions –Horizontal curves Fundamentals –Circular curves –Superelevation Δ

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Bina Nusantara University 25 Horizontal Curve Fundamentals R T PC PT PI M E R Δ Δ/2 L

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Bina Nusantara University 26 Horizontal Curve Fundamentals R T PC PT PI M E R Δ Δ/2 L

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Bina Nusantara University 27 Superelevation α α F cp F cn WpWp WnWn FfFf FfFf α FcFc W 1 ft e ≈ RvRv

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Bina Nusantara University 28 Superelevation

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Bina Nusantara University 29 Selection of e and f s Practical limits on superelevation (e) –Climate –Constructability –Adjacent land use Side friction factor (f s ) variations –Vehicle speed –Pavement texture –Tire condition

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Bina Nusantara University 30 Side Friction Factor from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004

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Bina Nusantara University 31 Minimum Radius Tables

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Bina Nusantara University 32 WSDOT Design Side Friction Factors from the 2005 WSDOT Design Manual, M 22-01 For Open Highways and Ramps

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Bina Nusantara University 33 WSDOT Design Side Friction Factors from the 2005 WSDOT Design Manual, M 22-01 For Low-Speed Urban Managed Access Highways

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Bina Nusantara University 34 Design Superelevation Rates - AASHTO from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004

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Bina Nusantara University 35 Design Superelevation Rates - WSDOT from the 2005 WSDOT Design Manual, M 22-01 e max = 8%

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Bina Nusantara University 36 Circular Curve Geometrics PC = Point of Curvature PT = Point of Tangency PI = Point of Intercept 100/D = L/Δ, so, L = 100 (Δ /D) where: L = arc length(measured in Stations (1 Sta = 100 ft) Δ = internal angle (deflection angle) D = 5729.58/R M = middle ordinate m=R [1 – cos(Δ /2) ] M - is maximum distance from curve to long chord

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Bina Nusantara University 37 Degree of curvature: D = central angle which subtends an arc of 100 feet D=5729.58/R where R – radius of curve For R=1000 ft. D = 5.73 degrees Maximum degree of curve/min radius: D max = 85,660 (e + f)/V 2 or R min = V 2 /[15 (e + f)] Circular Curve Geometrics

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Bina Nusantara University 38 Horizontal Sight Distance 1) Sight line is a chord of the circular curve 2) Applicable Minimum Stopping Sight Distance (MSSD) measured along centerline of inside lane Criterion: no obstruction within middle ordinate Assume: driver eye height = 3.5 ft object height = 2.0 ft. Note: results in line of sight obstruction height at middle ordinate of 2.75 ft

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Bina Nusantara University 39 Horizontal Alignment Basic controlling expression: e + f = V 2 /15R Example: –A horizontal curve has the following characteristics: Δ = 45˚, L = 1200 ft, e = 0.06 ft/ft. What coefficient of side friction would be required by a vehicle traveling at 70 mph?

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Bina Nusantara University 40 Circular Curve Geometrics PC = Point of Curvature PT = Point of Tangency PI = Point of Intercept 100/D = L/Δ, so, L = 100 (Δ /D) where: L = arc length(measured in Stations (1 Sta = 100 ft) Δ = internal angle (deflection angle) D = 5729.58/R M = middle ordinate m=R [1 – cos(Δ /2) ] M - is maximum distance from curve to long chord

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Bina Nusantara University 41 Stopping Sight Distance RvRv ΔsΔs Obstruction MsMs SSD

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Bina Nusantara University 42 Cross Section

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Bina Nusantara University 43 Superelevation Transition from the 2001 Caltrans Highway Design Manual

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Bina Nusantara University 44 Superelevation Transition from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001

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Bina Nusantara University 45 Spiral Curves No Spiral Spiral from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001

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Bina Nusantara University 46 Spiral Curves Involve complex geometry Require more surveying Are somewhat empirical If used, superelevation transition should occur entirely within spiral

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Bina Nusantara University 47 Desirable Spiral Lengths from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001

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