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Unformatted text preview: The Pennsylvania State University
Department of Civil and Environmental Engineering CE 321: Highway Engineering CE 321: Highway Engineering Class #20 – Vertical Curvature II Spring 2008 Sag Curve Design and Sag Curve Design and Stopping Sight Distance H = height of headlight (ft) β = inclined angle of headlight beam (degrees) L = length of vertical curve (ft) S = sight distance (ft) 200 ( H + S tanβ ) Lm = 2 S − A Sight Distance greater than length of vertical curve: ( S > L )
Headlight distance is the critical length Sag Curve Design Sag Curve Design H = Height of headlight above the roadway (ft) Lm = Minimum length of vertical curve (ft) A = Algebraic difference in grades (%) Algebraic S = Sight Distance (ft) β = the incline angle of the headlight beam relative to the horizontal plane of the car the Sight Distance less than length of vertical curve: ( S < L )
AS 2 Lm = 200 ( H + S tanβ )
Headlight distance is the critical length Sag Curve Design Lm = Minimum length of vertical curve (m) H = Height of headlight above the roadway (m) A = Algebraic difference in grades (%) S = Sight Distance (m) β = the incline angle of the headlight beam relative to the horizontal plane of the car the Sag Curve Design
When you assume AASHTO guidelines:
H = 2.0 ft (headlight height) β = 1° S = SSD SSD > L
Lm = 2 SSD – 400 + 3.5SSD A SSD < L
Lm = A SSD2 400 + 3.5SSD Lm = Minimum length of vertical curve (ft) SSD = StoppingSight Distance (ft)  controlled by headlights A = Algebraic difference in grades (%) Sag Curve Design (Kvalues) Sag Curve Design (Kvalues) Source: AASHTO Green Book (2001) Example Problem 3 Example Problem 3 PVI sta. 375+22; PVI elev. = 1765.03 ft; V = 70 mi/hr; G1 = 2.5%; G2 = +1.2% Determine the minimum length of the curve and if a bridge is placed at sta. 376+00, with a clearance of 14 ft, what elevation would the bridge be set at? Example Problem 3 (con’t) Example Problem 3 (con’t)
376 + 00 14 ft
G1 G2 PVI (375 + 22) elev. 1765.03 ft PVT L/2 L/2 Example Problem 3 (con’t) Example Problem 3 (con’t)
Assume an average running speed = 70 mi/hr (min speed) The corresponding SSD = 730 ft Example Problem 3 (con’t) Example Problem 3 (con’t) Assume: Lm = SSD < L
A SSD2 400 + 3.5SSD Lm = 3.7 (730)2 400 + 3.5(730) = 667.25 ft Since 730 is not < 667.25 , therefore try SSD > L Example Problem 3 (con’t) Example Problem 3 (con’t) Assume: SSD > L
Lm = 2 SSD – 400 + 3.5SSD A Lm = 2 (730) – 400 + 3.5 (730) 3.7 = 661.35 ft Since 730 > 661.35 , therefore use Lm = 661.35 ft Example Problem 3 (con’t) Example Problem 3 (con’t)
376 + 00 Determine sta. and Elevation of PVC with Lmin PVC
(371 + 91) elev. 1773.3 ft G1 PVI (375 + 22) elev. 1765.03 ft 14 ft PVT
G2 L/2 L/2 sta. of PVC = PVI  (1/2)(L) = 375 + 22  3 + 31 = 371 + 91 Elev. of PVC = Elev. of PVI + G1*( L / 2 ) = 1765.03 + (0.025)*(330.7) = 1773.3 ft Example Problem 3 (con’t) Example Problem 3 (con’t)
The Bridge at sta. 376 + 00 (clearance = 14 ft)
376 + 00 x
PVC
(371 + 91) elev. 1773.3 ft G1 PVI (375 + 22) elev. 1765.03 ft G2 14 ft PVT L/2 L/2 To find elevation of Bridge: vertical curve equation yx = a x2 + G1x + Elevation at PVC
x = distance from PVC to point on curve (sta.) yx = elev. of vertical curve at point x L = length of vertical curve (sta.) a = Algebraic difference in grades (%) / 2L G1 = initial grade of curve (%) Example Problem 3 (con’t) Example Problem 3 (con’t)
To find elevation of Bridge: vertical curve equation
376 + 00 PVC
(371 + 91) elev. 1773.3 ft G1 PVI (375 + 22) elev. 1765.03 ft x
14 ft
G2 PVT L/2 L/2 yx = a x2 + G1x + Elevation at PVC
x = 376 + 00  371 + 91 4 + 09 = 4.09 sta. yx = (3.7 / 2(6.62)) 4.092 + (2.5)(4.09) + 1773.3 = 1767.75 ft elevation for bridge = yx + 14 ft = 1781.75 ft Crest Curve Design and Passing Sight Distance
When you assume AASHTO guidelines:
H1 = 3.5 ft (Driver eye height) H2 = 3.5 ft (Opposing vehicle height) PSD > L
Lm = 2 PSD – 2800 A
Lm = Minimum length of vertical curve (ft) PSD = Passing Sight Distance (ft) A = Algebraic difference in grades (%) PSD < L
Lm = A PSD2 2800 Design speed (mph) 20 25 30 35 40 45 50 55 60 65 70 75 80 Crest Curve Design (PSD Kvalues) Crest Curve Design (PSD Kvalues)
Passing Sight Distance (ft) 710 900 1090 1280 1470 1625 1835 1985 2135 2285 2480 2580 2680 180 289 424 585 772 943 1203 1407 1628 1865 2197 2377 2565 Rate of Vertical Curvature, K Underpass Sight Distance Underpass Sight Distance Sag Curve Design and Sag Curve Design and Underpass Sight Distance
S<L,
AS 2 Lm = H + H2 800 H c − 1 2 S>L,
H + H2 800 H c − 1 2 Lm = 2 S − A Lm = minimum length of vertical curve (ft) A = absolute different in vertical grades (%) H1 = height of driver’s eye (ft) H2 = height of object (ft) Hc = clearance height of overpass structure above roadway (ft) S = sight distance (ft) Sag Curve Design and Underpass Sight Distance
When you assume AASHTO guidelines:
H1 = 8 ft (Truck driver eye height) H2 = 2.0 ft (Vehicle taillights) SSD > L
800( H c − 5) Lm = 2 × SSD − A
Lm = Minimum length of vertical curve (ft) SSD = Stopping Sight Distance (ft) A = Algebraic difference in grades (%) Hc = clearance height (ft) SSD < L
A × SSD 2 Lm = 800( H c − 5) Vertical Curve Variables Vertical Curve Variables H = height of headlight (ft) β = inclined angle of headlight beam (degrees) L = length of vertical curve (ft) S = sight distance (ft) Lm = Minimum length of vertical curve (ft) SSD = stopping sight distance (ft) x = distance from PVC to point on curve (sta.) yx = elev. of vertical curve at point x L = length of vertical curve (sta.) a = Algebraic difference in grades (%) / 2L G1 = initial grade of curve (%) Hc = clearance height (ft) PSD = passing sight distance (ft) H1 = height of driver’s eye (ft) H2 = height of object (ft) ...
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This note was uploaded on 06/08/2009 for the course CE 321 taught by Professor Petrucha during the Spring '02 term at Pennsylvania State University, University Park.
 Spring '02
 Petrucha
 Environmental Engineering

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