1/35 Components of Alignment Highway Design Project Horizontal Alignment Vertical Alignment Vertical Alignment Amir Samimi Civil Engineering Department Sharif University of Technology Cross-section /35 3/35 Horizontal Alignment Vertical Alignment
4/35 5/35 Vertical Alignment & Topography Vertical Alignment Determinants Maximum/minimum grade Properties of vertical curves Technical design of vertical curves Compatibility with existing grade Design Speed Sight Distance Vertical Clearance ength of grade Drainage Consideration Cost 6/35 7/35 Vertical Alignment: Tangents and Curves Grade Vertical alignment is made up of tangent and curves: What is MAX and MIX grade that can be allowed on the tangent section? Crest Curve G 3 Sag Curve
8/35 9/35 Grade AASHTO Recommended MAX Grades MIN grade used is typically 0.5% MAX grade is generally a function of the Design Speed Terrain (evel, rolling, Mountainous) High speed facilities: MAX grade is generally kept to 5% where the terrain allows. 3% is desirable since anything larger starts to affect the operations of trucks. 30 mph design speed Acceptable MAX is in the range of 7 to 1%. 10/35 11/35 AASHTO Recommended MAX Grades Grade ength The gradient in combination o with its length will determine e the truck speed reduction on upgrades: For general design purposes, a 10 mph speed reduction should be used. Curves are for grades between 0 and 9%. Source: A Policy on Geometric Design of Highways and Streets, AASHTO, 004. Chapter 3 Elements of Design
1/35 13/35 Truck Climbing anes Curves Vertical curves are employed to: Effect gradual change between road grades, Vertical curves should be: Simple is application, Safe and comfortable in operation, Pleasing in appearance, Adequate for drainage. 14/35 15/35 General Considerations General Considerations Vertical curves are in the shape of a parabola. abo a. Vertical alignment should use a smooth grade line with gradual changes, consistent with the type of highway and character of terrain. Grades with break points and short tangent lengths should be avoided. On long ascending grades, it is preferable to place the steepest grade at the bottom and flatten the grade near the top. It is also preferable to break a sustained grade with short intervals of flatter grades. Maintain moderate grades through intersections s to facilitate starting and turning movements. Roller Coaster type profiles, where the roadway profile closely follows a rolling natural ground line along a relatively straight horizontal alignment, should be avoided. Broken back curvature (short tangent between two curves in same direction) should be avoided. Avoid using sag vertical curves in a cut section unless adequate drainage can be provided.
16/35 17/35 Elevation = y / Change in grade: A = G in % (positive /, negative \) For a crest curve, A is negative For a sag curve, A is positive / Rate of change of curvature: K = / A Which is a gentler curve - small K or large K? Rate of change of grade: r = (g g 1 ) / K and r are both measuring the same characteristic of the curve but in different ways Equation for determining the elevation at any point on the curve: y = y 0 + g 1 x + 1/ rx y 0 = elevation at the x = horizontal distance from g = grade expressed as a ratio r = rate of change of grade 18/35 19/35 Distance to the turning point: x t = -(g 1 /r) Distance to the turning point: This can be derived as follows: y = y 0 + g 1 x + 1/ rx dy/dx = g 1 + rx At the turning point, dy/dx = 0, Therefore x t = - (g 1 / r) = -1% = +% Elevation of = 15.00 m Station of = 500 Station of = 400 ength of curve? / = Sta. Sta. / = 500 m - 400 m = 100 m = 00 m
0/35 1/35 = -1% = +% Elevation of = 15.00 m Station of = 500 Station of = 400 r - value? r = (g - g 1 )/ r = (0.0 - [-0.01])/00 m r = 0.00015 / meter = -1% = +% Elevation of = 15.00 m Station of = 500 Station of = 400 Station of low point? x = -(g 1 /r) x = -([-0.01] / [0.00015/m]) x = 66.67 m Station = [300] + 67.67 m Station 367 /35 3/35 = -1% = +% Elevation of = 15.00 m Station of = 500 Station of = 400 Elevation at low point? y = y 0 + g 1 x + 1/ rx y 0 = Elev. Elev. = Elev. - g 1 / Elev. = 15 m - [-0.01][100 m] Elev. = 16 m = -1% = +% Elevation of = 15.00 m Station of = 500 Station of = 400 Elevation at low point? y = y 0 + g 1 x + 1/ rx y = 16 m + [-0.01][66.67 m] + 1/ [0.00015/m][66.67 m] y = 15.67 m
4/35 5/35 Design of Vertical Curves = -1% = +% Elevation of = 15.00 m Station of = 500 Station of = 400 Elevation at station 350? y = 16 m + [-0.01][50 m] + 1/ [0.00015/m][50 m] y = 15.69 m Elevation at station 450? y = 16 m + [-0.01][150 m] + 1/ [0.00015/m][150 m] y = 16.19 m 6/35 7/35 Design of Vertical Curves The first step in the design is to determine e the minimum length (or minimum K) for a given design speed. K: number of horizontal feet needed for a 1% change in slope. Factors affecting the minimum length include Sufficient sight distance V SSD 1.47Vt 1.075 a t = brake reaction time,.5 sec. V = design speed, mph a = deceleration rate, ft/s - 11. ft/s Driver comfort Appearance Crest Vertical Curves SSD PVI ine of Sight PVC PVT h 1 h For SSD < ASSD 00 h1 h h 100 h 1 SSD For SSD > A
8/35 9/35 Crest Vertical Curves Design Controls for Crest Vertical Curves Assumptions for design h 1 = driver s eye height = 3.5 ft. h = tail light height =.0 ft. Simplified Equations For SSD < ASSD SSD 158 For SSD > 158 A from AASHTO s A Policy on Geometric Design of Highways and Streets 001 30/35 31/35 Sag Vertical Curves Sag Vertical Curves ight Beam Distance (SSD) headlight beam (diverging from OS by β degrees) G Assumptions s for design h 1 = headlight height =.0 ft. β = 1 degree h 1 PVC PVI PVT h =0 Simplified Equations ASSD SSD h S tan 00 1 For SSD < For SSD > 00 h1 SSD A tan For SSD < For SSD > ASSD SSD 400 3.5SSD 400 3.5 A SSD
3/35 33/35 Design Controls for Sag Vertical Curves Example 1 from AASHTO s A Policy on Geometric Design of Highways and Streets 001 A car is traveling at 30 mph in the country at night on a wet road through a 150 ft. long sag vertical curve. The entering grade is -.4 percent and the exiting grade is 4.0 percent. A tree has fallen across the road at approximately the PVT. Assuming the driver cannot see the tree until it is lit by her headlights, is it reasonable to expect the driver to be able to stop before hitting the tree? ANSWER: Assume S>,therefore S = 146.3 ft. which is less than. Must use S< : S=146.17 feet Required SSD = 196.53 ft. Therefore, she s not going to stop in time. 34/35 35/35 Example Example 3 A car is traveling at 30 mph in the ecountry yat night on awet road through a 150 ft. long crest vertical curve. The entering grade is 3.0 percent and the exiting grade is -3.4 percent. A tree has fallen across the road at approximately the PVT. Is it reasonable to expect the driver to be able to stop before hitting the tree? ANSWER: Assume S>, therefore S = 43.59 ft. which is greater than. Required SSD = 196.53 ft. Therefore, she will be able to stop in time. A roadway is being designed ed using a 45 mph design speed. One section of the roadway must go up and over a small hill with an entering grade of 3. percent and an exiting grade of -.0 percent. How long must the vertical curve be? ANSWER: For 45 mph we get K=61, = KA = (61)(5.) = 317. ft.