tutorial 10.pptx

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Introduction toTransportation Engineering

Instructor Dr. Norman GarrickTutorial Instructor Hamed Ahangari

March 2016

Geometric Design of Highways

The engineering aspects of alignment design is usually referred to as geometric design

Highway alignment is in reality a three-dimensional problem

Vertical Alignment

Profile View

Components of The Alignment

Horizontal Alignment

Plan View

Cross-section

www.atlantaeng.com

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safety.fhwa.dot.govwikipedia

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Horizontal Curve

• Objective: • Safety• Comfort

• Primary challenge :– The critical design feature of horizontal alignment

is the horizontal curve that transitions the roadway between two straight (tangent) sections.

Horizontal Alignment

Δ

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Key Elements Δ– Intersection angle R - Radius

R

PC PT

PI

R

L

Δ

Key Elements

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R

PC PT

PI

R

L

ΔT

Δ

Δ/2

Length L = R. Δ L=. R. Δ /180

Tangent (tan(Δ /2)= T/R)

T= R.tan(Δ /2)

T

RΔ/2

Stationing Reference System

Horizontal Alignment0+00

1+00

2+003+00 4+00

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ΔΔ = Deflection Angle

R = Radius of Circular Curve

L = Length of Curvature

BC = Beginning of Curve (PC) EC = End of Curve (PT)

PI = Point of Intersection

T = Tangent Length

C = Chord Length

M = Middle Ordinate

E = External Distance

R R

BC(PC)

EC(PT)

PI

T TL

M

E

C

ΔΔ/2

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Basic Definitions

• BC/PC: Point of Curvature• BC = PI – T

– PI = Point of Intersection – T = Tangent

• EC/PT: Point of Tangency• EC = BC + L

– L = Length12

R

PC(BC)

PT(EC)

PI

R

L

ΔT

Δ

Δ/2

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Degree of Curvature• D used to describe curves

• D defines Radius

• Arc Method: – D/ Δ = 100/L (1) (360/D)=100/(2R)

R = 5730/D (2)

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Curve Calculations

Length L = 100.Δ/D (3)

Tangent T = R.tan(Δ /2) (4)

Chord C = 2R.sin(Δ /2) (5)

External Distance E = = R sec(Δ/2) – R (6)

Mid Ordinate M = R-R.cos(Δ /2) (7)

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Example 1• A horizontal curve is designed with a 2000 ft.

radius. The tangent length is 500 ft. and the EC station is 30+00. What are the BC and PI stations?

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• Since we know R and T we can use T = R.tan(Δ /2) so Δ = 28.07 degrees

• D = 5730/R. Therefore D = 2.86

• L = 100(Δ)/D = 100(28.07)/2.86 = 980 ft.

• BC = EC – L = 3000 – 980 =2020~(20+20)

• PI = BC +T = 2020 + 500 = 2520~(25+20)

Solution

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Example 2A curve has external angle of 22.30’ degrees, a degree of curvature is 2°30’ and the PI is at 175+00. Calculate:

RadiusLength of Curve BC and EC ChordExternal Distance Mid Ordinate

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• Given: D = 2°30’, Δ=22.30’

• Part i) Radius:

• Part ii ) Length of Curve

• Part iii ) BC and EC

83.22915.2

5730

R

'87.45525.22tan38.2291

T

)13.44171()87.554()00175( BC

00.9005.25.22100

L

)13.44180()009()13.44171( EC

Solution

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• Part iv ) Chord

• Part v ) External Distance

• Part vi ) Mid Ordinate

'23.89425.22sin)83.2291(2

C

'90.44125.22sec(83.2291

E

'04.4425.22cos183.2291

M

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Example 3In a horizontal curve the intersection angle is 60 degree. We also know that the minimum internal sight distance (Mid Ordinate) for this road is 100 ft. If the BC of the curve is in 10+50 Calculate:

RadiusLength of Curve PI and EC

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• Given: Δ= 60 , M= 100 ft and BC= 10+50

• Part i) Radius: M = R-R.cos(Δ /2)

R = 747.15 ft

• Part ii ) Length of Curve L=. R. Δ /180

35.782180/14.3)60(747 L

Solution

)2/60cos(1*100 R

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• Given: Δ= 60 , M= 100 ft and BC= (10+50)

• Part iii ) PI and EC

)10.8114()10.314()5010( TBCPI

)35.3218()35.827()5010( LBCEC

Solution

'10.431)260tan(747

T