03 Horizontal Alignment & Superelevation PDF

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HORIZONTAL ALIGNMENT

Learning Objectives 

Be able to calculate minimum curve radius.



Be able to analyse and design a simple circular curve.



Explain and identify the axis of rotation of superelevation.



Determine the superelevation development length



Calculate the chainages related to superelevation development

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What is Road Horizontal Alignment? 

The horizontal alignment of a road usually consists of a series of straights or tangents and circular curves which are joined to each other by transition curves.



Straights may be omitted on continuously curving (curvilinear) alignment (Appropriate for divided roads).

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How is a horizontal curve developed??? Angle

Radius Angle

Directionoftravel– lefttoright ALWAYS! CVE20005 ROAD ENGINEERING

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Types of Horizontal Curves 

(Simple) circular curve



Transition curve



Reverse curve



Compound curve



Broken back curve

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Angle Changes (I) NOT Requiring Curves I

Operating speed (km/h) 40 50 60 70 80 90 100 110 120

Maximum deflection angle below which no curve is required 2 lane pavement 4 lane pavement 1.5 N/A 1.5 N/A 1 0.5 1 0.5 1 0.5 1 0.5 1 0.5 0.5 0.25 0.5 0.25

Minimum horizontal curve lengths (TS to ST) 45 70 100 140 180 230 280 340 400 (Austroads 2010, page 146)

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Determining Deflection Angle from Bearings 



Example: A tangent with bearing 125 14’ 20” meets another tangent N(0) with a bearing of 75 28’ 30”. 

Draw the tangent lines.



Determine the deflection angle.

W

E

Solution: S(180)

Deflectionangle(I)



Deflection angle (I) = 125 14’ 20” - 75 28’ 30” = 49 455’ 50”

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Circular Curve Details – Simple HC Notation:

R = radius of circular curve, (m) IP = intersection point, the point where the two straights join TP = tangent point, the point where a straight and a circular curve join I = intersection angle, the angle between the two straights (degrees) T = Tangent distance (m) T= Tangent distance (m) SD= Secant distance (m) Arc= length of circular arc (m), length between the two TPs along the curve

I T  R tan   2

Arc  RI c 

RI 180

Degree of Curvature = CVE20005 ROAD ENGINEERING

Chainages measured along the curve

Chainage TP1 = CH IP – T Chainage TP2 = CH TP1 + Arc

??? 5488 (anglesubtendedby30.5mlengthofcurve) R 8

Selection of Curves Radii 

Criterion 1: Lateral forces on the car – must be safe and comfortable (MV2/Rg) = radial force, M = vehicle mass (kg), g = gravitational acceleration (m/s2), R = radius (m), v = speed (m/s2), FN = Normal reaction force, Ff = Frictional force, FN*Ff = lateral friction coefficient)



Criterion 2: Adequate sight distance



Determine Rmin to satisfy Criterion 1, then check suitability for Criterion 2.

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If it does not satisfy Criterion 2, then must provide suitable offset. If this is not possible then change radius. 9

Criterion 1: Lateral Forces on the Car Forces Acting on Car on A Right Hand Curve with Superelevation

Forcescontributingtothe centripetalforce:Fg +Ff (Mv2/R)cosӨ =Fg +Ff

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Fg = Mg sinӨ Ff =f * N N ~ Mgcos Ө Ff = f * Mg cosӨ

M=vehiclemass f=lateralfrictioncoefficient N=normalforce g=gravitationalacceleration R=radiusofcurve v=speed,m/s Fg=gravityforce,parallelto surface Ff =lateralfrictionforce, paralleltosurface

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Forces Acting on Car on a Right Hand Curve Without Superelevation 

Normal crossfall along the curve



No crossfall or superelevation along the curve

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Minimum Curve Radius 

     

where; Rmin V E f

= minimum curve radius (metres) = operating speed in km/hr = superelevation (+ve towards the centre of the curve, -ve fall away from centre of curve) = coefficient of lateral friction

In practice, it is usual to adopt radii greater than those calculated from the above formula, to reduce superelevation and side friction below their maximum values. Use of maximum superelevation will need to be considered where the radius of a curve is approaching the minimum for section operating speed (usually in steep terrain). CVE20005 ROAD ENGINEERING

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Recommended side friction factors for cars and trucks f Operating speed (km/h)

Cars Des max Abs max

Trucks Des max Abs max

40

0.30

0.35

-

-

50

0.30

0.35

0.21

0.25

60

0.24

0.33

0.17

0.24

70

0.19

0.31

0.14

0.23

80

0.16

0.26

0.13

0.20

90

0.13

0.20

0.11

0.15

100

0.12

0.16

0.12

0.12

110

0.12

0.12

0.12

0.12

120

0.11

0.11

130

0.11

0.11

0.11 -

0.11 -

f is the coefficient of sideways friction. Its limit occurs when the vehicle is about to slide on the road. However, this is not the value for which road curves are designed. Drivers adjust their speed round a curve to produce f values lower than the maximum. Drivers choose a "safe and comfortable limit". CVE20005 ROAD ENGINEERING

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Typical ranges for E value Operating speed

Greater than 90

Between 70 and 89

Less than 70

0.05 - 0.06 (typical 0.06)

0.05-0.07 (typical 0.07)

0.05-0.10 (typical 0.07)

(km/h) Superelevation E (m/m)

0.1m/m used in mountainous terrain

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Selecting Suitable R value 

When using R > Rmin, E value needs to be checked using the graph in next slide.



E value obtained from graph should be rounded up. 



E.g. 4.1 becomes 4.5 and 4.7 becomes 5.

Use the new E and R values to check that ‘f’ is below the max for the operating speed using this formula.

V2 f  E 127R

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Criterion 2 – Adequate Sight Distance •

Calculate STSD required for the operating speed

•

Compare STSD with Curve length (ARC), •

If STSD > ARC, accept design Radius

•

If STSD < ARC, check offset distance requirement

•

ARC measured along CL (centreline) of road

•

STSD measured along CL of the inside lane

•

Offset measured from CL of the inside lane to obstruction

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When STSD < circular curve ARC ARC

Insidelane

STSD

offset TP

TP

ForV=80km/hr,STSD=115m Rmin =160m,I=50o ARC=140m STSD1000.



Not recommended as drivers may not anticipate the changes caused by different radii, which may cause: 

Drivers may not stay in the proper lane where the change of radius is not obvious



Motorcyclists loosing stability of the motorcycle

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Broken Back Curves  

Comprise two or more adjacent unidirectional curves joined by short tangent or transitional curve Avoid if possible, impossible to provide correct amount of superelevation throughout.

Driver's View of Broken-Back Curve CVE20005 ROAD ENGINEERING

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Broken Back Curves 

Poor appearance due to short tangents 

Recommended  3V - 4V (V m), V = Operating Speed



Minimum distance is V m. e.g. V = 80km/h, minimum length of straight would be 80m.



This length will result in visual satisfaction and would be enough for the normal crossfall to be regained and maintained for a distance corresponding to about 4 seconds travel time.



If this minimum length cannot be obtained, the alignment should be altered by use of a compound curve or a transitional curve between the two constant radius curves.

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SUPERELEVATION

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Superelevation Axis of Rotation 

Superelevation is a slope on a curved pavement selected so as to enhance forces assisting a vehicle to maintain a circular path.



Superelevation is developed by rotating each half of the crosssection (pavement and shoulders) about an axis: 

On undivided rural roads, rotation about the carriageway centreline



On undivided urban roads, the centreline is used, unless some constraints or obstructions may dictate the kerb line is used.



Divided rural roads, about the median edge of each carriageway

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Superelevation Development Undivided Road

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Superelevation Development Divided Road

Axis of Rotation Axis of Rotation

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Length of Superelevation Development (Lsd) 



The length over which superelevation is developed should be adequate to provide safe and comfortable riding quality and good appearance. It can be determined using the following criteria: 

The rate of rotation of the pavement



The relative grade (RG). RG is the relative change of grade of the carriageway edges with respect to the longitudinal grade of the road (axis of rotation).

The selected superelevation development length is the higher value resulting from any of these two criteria. The higher the Operating Speed or wider the carriageway, the longer the superelevation development length will need to be to meet appearance and comfort requirements.

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Superelevation Development Diagram Changes in Road Cross-Section Along a Curve With a Spiral Two-Way Two-Lanes Road

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Superelevation development without a transition curve 

Superelevation development occurs along the curve from TP up to 30%-40% of the curve length, both sides.



This leave 40-20% of curve length with full superelevation. Lsd

TP

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Circular Curve

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Simulation of Superelevation Development http://techalive.mtu.edu/modules/module0003/Superelevation.htm http://www.cee.mtu.edu/~balkire/Superelevation.ppt

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Length of Superelevation Development (Lsd) Criterion 1 = Rate of Rotation of the Pavement



Generally, road length (L) that corresponds to a change in crossfall from E1 and E2 is:

E  E1 V L 2 3.6r

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Eis+veforfalltowardsthecentreofcurve Eis–veforfallawayfromthecentreofcurve

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Length of Superelevation Development (Lsd) Criterion 1 = Rate of Rotation of the Pavement



For a change in crossfall from normal to full superelevation, L is referred to as Superelevation Development Length (Lsd, in metres)  

  3.6



E = superelevation, n = normal crossfall



V = operating speed, km/h



r = rate of rotation, m/m/sec r = 0.025m/m/s, V80km/h r = 0.035m/m/s, V 80 km/hr r = 0.035 m/m/s for V < 80 km/hr (which would occur in steeperterrain)



These values refer to a change of crossfall of 'r' m/m for each second of travel time. 

eg. for r = 0.025 m/m/s and speed V 100 km/hr i.e. 28 m/s. in travel time of 1 sec, a change in crossfall of 0.025m/m occurs in 28 m.

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Example – Lsd at 100km/h 

r = 0.025 m/m/sec for V = 100km/h, E = 0.06, normal crossfall n = 0.03m/m.



 



At 100km/h, you travel 27.78m per second



Every second of travel, the change in crossfall is 0.025m/m and occurs over 27.78m



You travel 27.78m in 1s

. . . .

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 100  100



Remaining distance of the Lsd=100m is 100m-27.78m x 3 = 16.67m



16.67m needs 0.6sec



The change in crossfall is 0.6sec = 0.6 x 0.025 = 0.015m/m



From CH 1083.33 to 1100m, you travel 16.67m in 0.6m

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Length of Superelevation Development (Lsd) Criterion 2 = Relative Grade (RG)



RG is the relative change in grade (reduced level or elevation) of the carriageway edges with respect to the longitudinal grade of the road axis of rotation



From SSD (normal crossfall) to SC (full superelevation)  

    

 100%

RG = Relative change in grade W = width of carriageway from centreline (axis of rotation) to edge of pavement (does not include shoulder)

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Relative Grade Diagram- 2 way 2 lane road Outer edge Reduced levels (RL) relative to road centreline (axis of rotation) Outer edge

Lsd

Inner edge

RL +ve

SSD

SC

TS

EW

0

CL

nW -ve 0 n

nW

nW

W

n

nW

E EW

E EW

W

SSD Centreline (axis of rotation)

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n

W

W

W

TS Outer pavement edge

W

SC Inner pavement edge

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The inner pavement edge starts rotating when the outer edge is at positive normal crossfall i.e.+ n, m/m (between TS and SC)

IP

The length of road that corresponds to a change in crossfall from 0 (Ch.TS) to + n = Linn. Linn = (n-0)*V/3.6r Linn = Tro Tro (from –n to 0) = Lsd-Lsp Ch inner = Ch TS + Linn

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Relative Grade Diagram-2 way 2 lane road Inner edge levels relative to road centreline RL SSD

+ve

0

SC

TS

nW

CL

nW

EW

-ve

n

n nW

nW

E

n

EW

nW

E

n

EW

nW W

W W

SSD &TS Centreline (axis of rotation)

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W

W

Point where inner edge starts to rotate Outer pavement edge

W

SC

Inner pavement edge

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Relative Grade Diagram – 2-way 2-lane road Lsd

SSD

RL

SC

TS

+ve

0

Outer edge

nW

nw

EW

nw

nW

EW+(-nW)

-ve

RG % = slope of this line

EW EW

CL

Inner edge

󰇛󰇜 

 CVE20005 ROAD ENGINEERING

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Relative Grade Diagram for the whole curve (circular and two transition curves) TS

SC

CS

Elevation,m

SSD

ST

ESD

Chainage,m

Inner edge – the side where the curve centre is

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Outer edge

Inner edge

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Relative Grade Diagram for the whole curve (circular and two transition curves) SSD

TS

SC

CS

ST

ESD

SSD

TS

SC

CS

ST

ESD

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Maximum Relative Grade Between Edge Carriageway and Control Line (axis of rotation) in Superelevation Development Operating Speed, km/hr One lane (1)

Relative Grade % Two lanes (2)

More than two lanes (3)

(WR = 3.5)

(WR = 7.0)

(WR = 10.5)

0.9 0.75

1.3 1.15

1.7 1.5

60

0.6

1.0

1.3

70

0.55

0.9

1.15

80

0.5

0.8

1.0

 40 50

1.

90

0.45

0.75

0.95

100

0.4

0.7

0.9

110

0.4

0.65

0.85

0.6

0.8

> 120 0.4 Applies to normal two-lane road with the axis of rotation on the centerline

2.

Applies to two-lane roadway with control along one edge; four-lane roadway with control on centerline; and two-lane road with climbing lane and control on centerline of basic two lanes

3.

Applies to multilane road with more than two lanes between the control and the edge of running lanes

RG = 12.6*W/V RG = 9.0*W/V

V 80km/hr

Relative Grade should be limited for appearance reasons. Note: if Operating Speed V is greater than 80 km/hr, the required superelevation length will generally result in an acceptable relative grade and appearance. If RG is greater than those in the table, use max RG to determine acceptable Lsd. Maximum relative grade between edge of carriageway and control line (axis of rotation) in super-elevation development (Austroads, 2002) CVE20005 ROAD ENGINEERING

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RG Diagram for a divided road Cwy 2

Cwy 1 Lsd

Cwy 1

sc SSD TS EW

nW

Median ed...