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Unformatted text preview: Highway Geometric Design Highway
Horizontal Alignment of Highways Horizontal Alignment Horizontal
The critical aspect of horizontal alignment is the horizontal curve that focuses on the design of the directional transition of the roadway in a horizontal plane. Vehicle Cornering α = angle of inclined W = weight of the vehicle (lb) Wn = Weight normal to the roadway surface Wp = Weight parallel to the roadway surface Ff = Side friction force (lb) Fc = Centrifugal force Fcn = Normal force Fcp = Parallel force Rv = Radius defined to the vehicle’s traveled path (ft). Rv is measured from the innermost vehicle path, which is assumed to be the midpoint of the innermost lane (worst case). Wp + FF = Fcp FF = Fs (Wn + Fcn) Where: Fs is the coefficient of side friction
w p + F s (W n + Fcn ) = F cp w sin( α ) + F s ( w cos( α ) + Fc sin( α ) = Fc cos( α ) wv 2 wv 2 sin( α )) = cos( α ) w sin( α ) + F s ( w cos( α ) + gR v gR v g is the gravitational constant v vehicle speed in ft/sec Dividing both sides of the equation by w cos(α)
v2 tan(α ) + Fs = (1 − f s tan α ) gR v
fs and tan α are very small tan( α ) is the superelevation denoted as e e is the number of vertical feet rise per 100 feet horizontal V2 V2 Rv = R min = e 15 ( f s + e ) g fs + 100 e max = 0 . 08 Example A roadway is being designed for a speed of 70 mph. At one horizontal curve, it is known that the superelevation is 0.08 and the coefficient of side friction is 0.10 Determine the minimum radius of curve that will provide for safe vehicle operation. Solution (70 * 1 . 47 ) = 1826 . 85 ft v Rv = = g ( f s + e ) 32 . 2 ( 0 . 08 + 0 . 1)
2 2 This is the minimum radius, larger radii will generate centrifugal forces lower than those safely supported by the superelevation and the side frictional force. AASHTO provides general guidance for the selection of e & fs for Horz. Curve design, Because high rates of superelevation can cause steering problems, ice on roadways can reduce fs and vehicles travelling less than design speed could be forced inwardly off the curve by gravitational forces. AASHTO table 314 p.145 Values in table are grouped by e’s e.g. higher max. e’s are permitted on freeways relative to arterial and local roads. Example A roadway is being designed for a speed of 120 km/hr. At one horizontal curve, it is known that the superelevation is 8 percent and the coefficient of side friction is 0.09 Determine the minimum radius of curve (measured to the traveled path) that will provide for safe vehicle operation. Solution
1000 120 * 2 v 60 * 60 = = 666 . 457 m Rv = 8 e ) g fs + 9 . 807 ( 0 . 09 + 100 100 2 This is the minimum radius, large radii will generate centrifugal forces lower than those safely supported by the superelevation and the side frictional force. Horizontal Curve Fundamentals Horizontal
Simple Curve: A standard curve with a simple standard radius Compound Curve: Two or more simple curves in succession Spiral Curve: Curves with a continuously changing radius. Simple Curve: R = radius measured to the centerline of th road PC = beginning point of the horizontal curve (point of curvature) T = tangent length PI = tangent intersection (Point of intersection) ∆ = central angle of the curve PT = ending point of the curve (point of tangency) M = middle ordinate E = external distance T = tangent length L = length of the curve C = Chord length D = degree of the curve (in degrees) = angle opposite to a 100 ft arc along the curve D is a measure of the sharpness of the curve and is frequently used instead of the radius in the actual construction D = 5729.6 /R From curve geometric
T = R tan (∆ / 2 ) C = 2 (R sin (∆ / 2 )) R 1 E= − R = R cos (∆ / 2 ) − 1 cos (∆ / 2 ) M = R − R cos (∆ / 2 ) = R (1 − cos (∆ / 2 )) L ∆ = 2π R 360 L ∆ = 100 D ∆R L= 57 . 3 ∆ L = 100 D Example A horizontal curve is designed with a 2000 ft. radius. The curve has a tangent of 400 ft and the PI is at station 103+00. Determine the stationing of the PT. Solution
R= 2000 ft T = R tan(∆/2) tan(∆ /2) = T/R = 400/2000 = 0.2 ∆ /2 = 11.3 ∆ = 22.6 ∆ / 360 = L / (2 * 3.14 R ) L = 2x3.14x2000x22.6 / 360 = 788.89 ft Stationing of PT = 103+00  4+00 + 7+88.89 = 106+88.89 T = 400 ft Example
Given ∆ = 16 38 R = 1000 ft = 330 m
o PI at station 6+26.57 Calculate the station of the PC and PT; also calculate lengths C,M, and E. Solution 16 o 38 − T = R tan( ) = 146 . 18 ft = 48 . 24 m 2 1 E = R( − 1 ) = 10 . 628 ft = 3 . 507 m ∆ cos 2 ∆ M = R (1 − cos ) = 10 . 516 ft = 3 . 470 m 2 ∆ ) = 289 . 29 ft = 95 . 465 m C = 2 ( R sin 2 L ∆ = 360 2π R L = 290 . 31 ft , 95 . 801 m Solution
Station PC = 6+26.57  1+46.18 = 4+80.39 Station PT = 4+80.39 + 2+90.31 = 7+70.7 Station PC = 6+026.56  1+048.24 = 5+978.32 Station PT = 5+978.32 + 0+095.801 = 6+074.121 Stopping Sight Distance and Horizontal Curve Design
Sight distance restrictions occur when obstructions are present. Such obstructions are frequently encountered on Hwys due to the cost of rightofway acquisition and/or cost of moving earthen materials (e.g. rock) SSD is measured along the horizontal curve from the center of the traveled lane. Stopping Sight Distance and Horizontal Curve Design For a specified SSD, some distance Ms, the middle ordinate must be visually cleared , so that the line of sight is such that sufficient SD is available. Remember;
∆R L= 57 . 3 ∆ SRv SSD = 57 . 3 57 . 3 SSD ∆S = Rv Rv = radius to the vehicle’s path ∆ s = angle subtended by an arc equal to the SSD in length ∆ Q M = R (1 − cos ) 2 ∴M ∴M
S ∆s ) = R v (1 − cos = R v (1 − cos 2 28 . 65 SSD = R v (1 − cos ) Rv 57 . 3 SSD Rv 2) S Ms is the middle ordinate necessary to provide adequate SSD Rv SSD = (cos 28 . 65 −1 Rv − M S ) Rv Example
A horizontal curve on a twolane Hwy is designed with a 2000 ft radius, 12ft lanes, and a 60 mph design speed. Determine the distance that must be cleared from the inside edge of the the inside lane to provide sufficient sight distance for SSD. Solution R v = 2000 − 6 = 1994 From table III1 ASSHTO SSD = 570 ft MS MS 28.65 SSD = R v (1 − cos ) Rv 28.65 x 570 = 1994(1 − cos ) = 20.34 ft 1994 20.3 ft must be cleared from the center of the inside lane. From the inside edge 14.3ft. Highway Geometric Design Highway
Vertical Alignment of Highways Vertical Alignment Vertical
The objective of vertical alignment is The to determine the elevation of highway points to ensure proper roadway drainage and an acceptable level of safety. The primary challenge of vertical The alignment lies in the transition of roadway elevation between two grades. This transition is achieved by means of a vertical curve. Crest Vertical Curves Crest
G1 = initial roadway grade G2 = final roadway grade PVC = initial point of the curve PVT = final point of the curve L = length of the curve PVI = the point of vertical intersection A = absolute value of the difference in grade Sag Vertical Curves Sag
Equal tangent vertical curve G1 = initial roadway grade PVC = initial point of the curve PVI = the point of vertical intersection G2 = final roadway grade PVT = final point of the curve L = length of the curve Highway Positioning Highway
– Highway positioning and length is defined as the actual distance along the highway (usually measured along the centerline of the highway, on a horizontal plane) from some specified point. – This distance is generally measured in terms of stations, with each station constituting 100 ft of highway alignment distance. Vertical Curve Fundamentals Vertical
– A parabolic function defines roadway elevations at every point. It provides a constant rate of change of slope and implies equal curve tangents. y = ax + bx + c
2
Where y = roadway elevation Where X = stations from the beginning of the curve stations C = elevation of the PVC elevation – The first derivative gives the slope and is dy dx = 2 ax + b
At the PVC x=0 dy b= dx =G 1 – The rate of change of slope is dy = 2a dx
Average rate of change of slope is 2 G 2 − G1 ∴a = 2L d y G2 − G1 = dx L
2 Example 1 Example – A 600 ft equal tangent sag vertical curve has the PVC at station 170+00 and elevation 1000 ft. The initial grade is –3.5 percent and the final grade is 0.5 percent. Determine the elevation and stationing of the PVI, PVT, and the lowest point on the curve. – Solution Since the curve is equal tangent, PVI will Since be 600/2 (3 stations) from PVC; PVT will be 600 (6 stations) from PVC Stationing of PVI = 170+00 + 3+00 = Stationing 173+00 Stationing of PVT =170+00 + 6+00 = Stationing 176+00 Elevation of PVI = 1000 – 3.5/100 * 300 = Elevation 989.5 ft Elevation of PVT = 989.5 + 0.5/100 * 300 = Elevation 991.0 ft Regarding the lowest point on Regarding the curve, it will occur when the first derivative of the parabolic function vanishes dy = 2 ax + b = 0 dx Q b = G 1 = − 3 .5 0 .5 − ( − 3 .5 ) Qa= = 0 . 33 2 6 dy ∴ = 2(0.33) x − 3.5 = 0 dx x = 5.3stations The stationing of the lowest The point: –170+00 + 5+30 = 175+30 Y = ax2 + bx + c ax = 0.33(5.3)2 + (3.5)(5.3) + 1000 = 990.72 ft Example 2 Example – A equal tangent vertical curve is to be constructed between grades of –2.0% (initial) and 1.0% (final). The PVI is a station 110+00 and elevation 420 ft. Due to a street crossing the roadway, the elevation of the roadway at station 112+00 must be at 424.5 ft. Design the curve. Solution Solution – The design problem is one of the determining the length of the curve required to ensure that station 112+00 is at elevation 424.5 ft. G2 − G1 1.0 − (−2.0) 1.5 = a= = 2L 2L L y = ax + bx + c
2 b =G = −2.0 1 c = 420 + 2.0(0.5L) = 420 + L X = 0.5L + 2
424.5 = (1.5 / L)(0.5L + 2) + (−2.0)(0.5L + 2) + (420 + L)
2 4 .5 = 0 .375 L + 3 + 6 / L − 4 2 0 = 0 . 375 L + 5 . 5 L − 6
Gives: L = 13.466 stations Gives: Using the value of L, Elevation of PVC = c = 420 + 13.466 = 433.47 ft Stationing of PVC = 110+00 – (13+46.6)/2 =103+26.7 Elevation of PVT = elevation of PVI + (0.5L)G2 =420 + [0.5(13.466)](1.0) = 426.73 ft Stationing of PVT = 110+00 + (13+46.6)/2 =116+73.3 And X =0.5L+2.0 =8.733 stations from the PVC Offsets: vertical distance from Offsets: the initial tangent to the curve. A 2 y= x 200 L Where Y: offset at any distance x A: absolute value of the difference in grades L: length of vertical curve in feet X: distance from PVC in feet A 2 y= x 200 L AL ym = 800 AL yf = 200 Define a value K, a horizontal Define distance, in feet, required to effect a percent change in slope. (measure of curvature) K x = = K L A G
1 x: distance from PVC to the high/low point. (distance to slope=0, occurs at a distance from PVC=K times the approach grade G1.) Example: Example:
– A vertical curve crosses a 4ft diameter pipe at right angles. The pipe is located at station 110+85 and its centerline is at elevation 1091.60 ft. The PVI of the vertical curve is at station 110+00 and elevation 1098.4 ft. The vertical curve is equal tangent, 600 ft long, and connects an initial grade of +1.20 percent and a final grade of –1.08 percent. – Using offsets, determine the depth, below the surface of the curve, to the top of the pipe, and determine the station of the highest point on the curve. Solution: Solution: – The PVC is at station 107 (110+00 – 3+00, which is half of the curve length), so the pipe is 385 ft (110+85 – 107+00) from the beginning of the curve (PVC). – The elevation of the PVC: 1098.4  (3 stations x 1.2 ft/station) = 1098.4 1094.8 ft – The elevation of the initial tangent: 1094.8 + (3.85 stations x 1.2 ft/station) = 1094.8 1099.42 ft Determine the offset above the pipe at x Determine = 385 ft A 2 y= x 200 L
y= 1.2 − (−1.08) 200(600) (385) = 2.82 ft
2 Thus the elevation of the curve above the pipe is 1096.6 ft (1099.422.82). The elevation of the top of the pipe is 1093.60 ft (elevation of the centerline plus one half of the pipe’s diameter), so the pipe is 3 ft below the surface of the curve (1096.61093.6) Find K to determine the location of the Find highest point on the curve: 600 K= = 263 .16 1.2 − ( −1.08)
And the distance from the PVC to the highest point is: x = K × G1 = 263.16×1.2 = 315.79 ft
So the station of the highest point: (107+00) + (3+15.79) = 110+15.79 ...
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This note was uploaded on 01/13/2011 for the course CECS Transporta taught by Professor Essamradwan during the Spring '10 term at University of Central Florida.
 Spring '10
 EssamRadwan

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