Highway Design Project

Horizontal Alignment

Highway Design Project

Amir Amir SamimiSamimi

Civil Engineering DepartmentSharif University of Technology

1/42

Horizontal Alignment

2/42

Curve Types

1. Simple curves with spirals

. S p e cu ves w t sp a s2. Broken Back – two curves same direction (avoid) 3. Compound curves: multiple curves connected directly

together (use with caution) go from large radii to smaller radii and have R(large) < 1.5 R(small)

4. Reverse curves – two curves, opposite direction (require separation typically for superelevation attainment)

3/42

Types of Circular Curves

Si l C

Simple Curve Compound Curves Broken-Back Curves

B k B k C h ld b id d if

Reverse CurvesBroken-Back Curves should be avoided if possible. It is better to replace the Curves with a larger radius circular curve.

A tangent should be placed between reverse Curves.

4/42

Typical Configurations of Curves

Spirals are typically placed

Spirals are typically placed between tangents and circular curves to provide a transition from a normal crown section to a superelevated one.

Spirals are typically used at intersections to increase the room for large trucks to make turning movements.

5/42

Horizontal Alignment

Objective:

Objective: Geometry of directional transition to ensure: Safety Comfort

Primary challenge Transition between two directions

H i t l

Δ

Horizontal curves Fundamentals Circular curves Superelevation

6/42

Horizontal Alignment

1. Tangents

. a ge ts2. Curves3. Transitions

Curves require superelevation Retard sliding, g, Allow more uniform speed, Allow use of smaller radii curves (less land)

7/42

Horizontal Curve Fundamentals

2tan

RT

RL

100

RRD

000,18

180100

DRL

180

D = degree of curvature (Delta / L = D / 100)

8/42

Horizontal Curve Fundamentals

11RE

12cos

RE

2cos1RM

9/42

Example

A horizontal curve is designed with a 1500 ft. radius. The

A horizontal curve is designed with a 1500 ft. radius. The tangent length is 400 ft. and the PT station is 20+00. What are the PI and PT stations?

10/42

Superelevation

cpfp FFW ≈Rv

Fc

W 1 fte

cossincossin22

vvs gR

WVgR

WVWfW

α

11/42

Superelevation

cossincossin22 WVWVWfW

cossincossinvv

s gRgRWfW

tan1tan2

sv

s fgRVf

fVf 12

efgR

fe sv

s 1

efgVR

sv

2

12/42

Radius Calculation

Rmin related to max. f and max. e allowed

Rmin related to max. f and max. e allowed Rmin use max e and max f and design speed f is a function of speed, roadway surface, weather condition,

tire condition, and based on comfort AASHTO: 0.5 @ 20 mph with new tires and wet pavement to

0.35 @ 60 mph f decreases as speed increases (less tire/pavement contact)

13/42

Selection of e and fs

Practical limits on superelevation (e)

Practical limits on superelevation (e) Climate Constructability Adjacent land use

Side friction factor (fs) variations Vehicle speed Pavement texture Tire condition

14/42

Maximum e

Controlled by 4 factors:

Co t o ed by acto s: Climate conditions (amount of ice and snow) Terrain (flat, rolling, mountainous) Frequency of slow moving vehicles who might be influenced by

high superelevation rates Highest in common use = 10%, 12% with no ice and snow on low

volume gravel surfaced roadsvolume gravel-surfaced roads 8% is logical maximum to minimized slipping by stopped vehicles

15/42

Side Friction Factor

from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004

16/42

Minimum Radius Tables

17/42

WSDOT Design Side Friction FactorsFor Open Highways and Ramps

18/42

WSDOT Design Side Friction FactorsFor Low-Speed Urban Managed Access Highways

19/42

Design Superelevation Rates - AASHTO

from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004

20/42

Design Superelevation Rates - WSDOT

emax = 8%

from the 2005 WSDOT Design Manual, M 22-01

21/42

Radius Calculation (Example)

Assume a maximum e of 8% and design speed of 60 mph, what

Assume a maximum e of 8% and design speed of 60 mph, what is the minimum radius?

fmax = 0.12 (from Green Book)

.)120080(15

60)(15

22

min

feVR

Rmin = 1200 feet

)12.008.0(15)(15 fe

22/42

Radius Calculation (Example)

Assume a maximum e of 4%

ssu e a a u e o %

fmax = 0.12 (from Green Book)

.)12.004.0(15

60)(15

22

min

feVR

Rmin = 1500 feet

23/42

Stopping Sight Distance

svRSSD 180 SSD

Obstruction

Ms v

s RSSD

180

180

SSDRM 90cos1

Rv

Δs

v

vs RRM

cos1

v

svv

RMRRSSD 1cos

90

24/42

Sight Distance Example

A horizontal curve with R = 800 ft is part of a 2-lane highway

A horizontal curve with R 800 ft is part of a 2 lane highway with a posted speed limit of 35 mph. What is the minimum distance that a large billboard can be placed from the centerline of the inside lane of the curve without reducing required SSD? Assume p/r =2.5 sec and a = 11.2 ft/sec2

VSS 412

Ga

VVtSSD

2.3230

47.1

feetmphmph 2460

2.322.1130

)35(sec)5.2)(35(47.12

25/42

Sight Distance Example

Now estimate the minimum distance that the billboard can be

Now estimate the minimum distance that the billboard can be placed :

f tCOSR

RSSDRm

vv

439)246(65.281

90cos1

in radians not degrees

feetCOSR 43.9)800(

)(1

26/42

Horizontal Curve Example

Deflection angle of a 4º curve is 55º25’, PC at station 238 +

e ect o a g e o a cu ve s 55 5 , C at stat o 3844.75. Find length of curve, T, and station of PT. D = 4º = 55º25’ = 55.417º

. ftRRR

D 4.143258.5729180100

ftftRL 4.1385360

)417.55)(4.1432(2360

2

ftRT 3.7522417.55tan4.1432

2tan

27/42

Horizontal Curve Example

Stationing goes around horizontal curve.

Stat o g goes a ou d o o ta cu ve. What is station of PT?

PC = 238 + 44.75 L = 1385.42 ft = 13 + 85.42 Station at PT = (238 + 44.75) + (13 + 85.42) = 252 + 30.17

28/42

Superelevation Transition

29/42

Superelevation Transition

30/42

Superelevation Runoff/Runout

31/42

Superelevation Runoff - WSDOT

32/42

Purpose of Transition Curves

Provides path for vehicle to move from straight to a circular

Provides path for vehicle to move from straight to a circular curve

Improved appearance of curve to driver Allows introduction of superelevation and pavement widening

33/42

Characteristics of Transition Curve

Should have constant rate of change of radius of curvature

Should have constant rate of change of radius of curvature Transition should be equal to zero at start of straight and equal

to radius of curvature at circular curve Allows passengers to adjust to change in rate of curvature

34/42

Types of Transition Curve

Clothoid

C ot o d Most commonly used Will be examined in more detail

Lemniscate Used for large deflection

angles on high speed roads

Cubic Parabola Cubic Parabola Unsuitable for large

deflection angles

35/42

Geometry of Clothoid

K = Lp.R

p. Lp = Length of plan transition R = Radius of circular curve K = constant

Coordinates can be represented by x = l l5/40(RL )2 x = l – l5/40(RLp)2 - … y = l3/6RLp – l7/336(RLp)3 +…

x and y are measured along the tangent and at right angles from the tangent respectively.

36/42

Shift of Curve for Transition

To accommodate the transition curve the circular curve is

To accommodate the transition curve the circular curve is normally shifted inwards towards the centre of the curve.

37/42

Shift of Curve for Transition

The shift can be calculated by:

The shift can be calculated by: Shift = S =Lp2/24R Lp is the length of transition If S <0.25m then the transition

is usually ignored or not required

38/42

Terminology

39/42

Tangent Length

When a transition curve is used :

W e a t a s t o cu ve s used :

Tangent Length = (R +S) tan (/2)

The distance from the IP to the TS = (R+S) tan (/2) + Lp/2

The circular arc length from SC – CS is reduced by Lp: Arc = R* - Lp

40/42

Transition Length

Radial Acceleration Method:

Radial Acceleration Method: Lp = V3/46.73Ra Lp = length of plan transition V = design speed (km/hr) R = radius of circular curve a = radial acceleration

a varies with design speed and design authority.g p g y Typical values:

41/42

Transition Length

Rate of Rotation of Pavement Method:

Rate of Rotation of Pavement Method: Most calculations for plan transition are done in conjunction with

the superelevation development length (Le). Usually relies on design speed and rate of rotation of pavement:

Lp = Le – 0.4V

Le = (e1 - e2) V / 0.09 Rotations of 2.5%/sec – this is most common, where e1 = 0

Le = (e1 - e2) V / 0.126 Rotation rates of 3.5%/sec

42/42

Tables