tutorial 10.pptx
TRANSCRIPT
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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