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

Presentation on theme: "Geometric Design Session 02-06 Matakuliah: S0753 – Teknik Jalan Raya Tahun: 2009."— Presentation transcript:

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

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

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

Bina Nusantara University 5 Stationing Horizontal Alignment Vertical Alignment Introduction

Bina Nusantara University 6 From Perteet Engineering Introduction

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

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

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

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

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

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.

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

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

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

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

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…

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

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

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)

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…

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

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

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

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

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

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

Bina Nusantara University 28 Superelevation

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

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

Bina Nusantara University 31 Minimum Radius Tables

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

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

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

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

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

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

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

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?

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

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

Bina Nusantara University 42 Cross Section

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

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

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

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

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

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