lec 05 horizontal alignment ( transportation engineering dr.lina shbeeb )

Dr. Lina Shbeeb

Horizontal alignment

Transportation Engineering

Dr. Lina Shbeeb

Geometric design

• Consists of the following:

• Horizontal Design of roadways (Horizontal

design)

• Vertical Design of roadways (Vertical design)

• Cross-sectional Design (pavement design, no of

lane, etc)

• Terminal Facilities (Parking Lots, Garages, etc)

• Intersections & Interchanges.

Dr. Lina Shbeeb

Horizontal alignment

Vertical alignment

Dr. Lina Shbeeb

Horizontal Alignment (HA)

• Definition: Straight segments of roadways

(tangents) connected by suitable curves (horizontal curves) there might be a need to provide transitions.

• Curves require superelevation to provide banking of curve, retard sliding, allow more uniform speed.

• The length of the facility is measured along the horizontal of control line like the center line and is usually expressed in terms of 100-ft stations from a reference point (i.e. sta 14 is located at 1400 ft from a reference point.

Dr. Lina Shbeeb

Types of curves in horizontal

alignment • simple curves: it will be discussed in the

following section.

• Compound Curves consist of 2 or more curves in succession turning in the same direction with a common tangent point

• Reverse Curves consist of two simple curves with equal radii turning opposite directions with a common tangent

• Transition or Spiral curves are placed between tangents and circular curves or between two adjacent circular curves with substantially different radii.

Dr. Lina Shbeeb

Types of curves

Compound curve Reverse curve

Dr. Lina Shbeeb

2sin2:

100:

2tantan:

)2

cos1(tan:

)12

(sectan:

:

RcordLongLC

DcurveofLengthL

RgentofLengthT

RcedisordinateMiddleM

RcedisExternalE

curveofDegreeD

2cos

1

2sec

/2 /2

Simple curve:

simple circular

curve connects

two tangents

Point of intersection (PI): The point that resembles intersection of the two tangents

Point of curvature (PC): the point where the curve begins

Point of tangency (PT): the point where the curve ends

3602 RL

Dr. Lina Shbeeb

Radius Calculation

Rmin = ___V2______

15(e + fs)

Where:

V = velocity (mph)

e = superelevation

fs =side friction coefficient

15 = gravity and unit conversion

fs is a function of

• speed,

• Roadway surface,

• weather condition,

• tire condition, ect

• AASHTO: 0.5 if the

speed is 20 mph with

new tires and wet

pavement to 0.35 if

speed becomes 60

mph e-value as function of road

and surface conditions

0.1: commonly used,

highway

0.08: snow and ice,

highway

0.12: low-volume, gravel-

surfaced, rural (drainage)

0.4-0.6: urban, traffic

congestion

Dr. Lina Shbeeb

Degree of curve

• As two tangents intersect they can be connected with infinite

number of curves that differ by D: Degree of curve they

have two definitions one for highway and one for railway

– The arc definition (highway) is equal to the central angle in degree

subtended by an arc of 100 ft; the relation between D and R is

estimated by

– The Chord definition (Railway) is equal to the central angle in degree

subtended by a cord of 100 ft; the relation between D and R is

estimated by

RD

D

R

58.5729

3602

100

R

D 50

2sin

Dr. Lina Shbeeb

Degree of curve

100ft

Arc Definition Chord Definition

Dr. Lina Shbeeb

Radius of Curve Design Chart

Design speed

(mph)

emax

fmax Rmin

(ft)

Recommended

(ft)

30 0.10 0.16 231 300

40 0.10 0.15 432 500

50 0.10 0.14 694 750

60 0.10 0.12 1091 1200

70 0.10 0.10 1637 1800

Dr. Lina Shbeeb

Radius Calculation (Example)

Design radius example: assume a maximum e

of 8% and design speed of 60 mph, what is

the minimum radius?

fmax = 0.12 (from ASSHTO for speed 60)

Rmin = 602/ 15(0.08 + 0.12)=1200 ft

Dr. Lina Shbeeb

Horizontal Curve Example

• Deflection angle of a 5º curve is 55º30’, PC at

station 238 + 44.75. Find length of curve,T, and

station of PC.

• D = 4º

• = 55º30’ = 55.5º

• D = 5729.58 R = 5729.58 = 1,145.9 ft

R 5

Dr. Lina Shbeeb


• D = 5º

• = 55.5º

• R = 1,145.9 ft

• L = 2R = 2(1,145.9 ft)(55.5º) = 1109.4ft

360 360

Dr. Lina Shbeeb


• D = 4º

• = 55.417º

• R = 1,145.9 ft

• L = 1109.4 ft

• T = R tan = 1,145.9 ft tan (55.5) = 583.97 ft

2 2

Dr. Lina Shbeeb

Sight Distance for Horizontal

Curves

• Location of object along chord length that blocks line of sight around the curve

• m = R(1 – cos [28.65 S])

R

Where:

m = line of sight

S = stopping sight distance

R = radius

Dr. Lina Shbeeb

Sight Distance Example

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 and a = 11.2 ft/sec2

SSD = 1.47vt + _________v2____

30(__a___ G)

32.2

Dr. Lina Shbeeb


SSD = 1.47(35 mph)(2.5 sec) +

_____(35 mph)2____ = 246 feet

30(__11.2___ 0)

32.2

Dr. Lina Shbeeb


m = R(1 – cos [28.65 S])

R

m = 800 (1 – cos [28.65 {246}]) = 9.43 feet

800

Dr. Lina Shbeeb

Development of superelevation Distance AB defined as the tangent runout

Distance BE defined as the superelevation runoff

Dr. Lina Shbeeb

Methods of Attaining Superelevation

1. Pavement revolved about center line

Centerline is point of control

2. Pavement revolved around inner edge

Inner edge is point of control

3. Pavement revolved around outer edge

Outer edge is point of control