Upload
khawla2789
View
250
Download
2
Embed Size (px)
Citation preview
7/30/2019 CE426 Block2 Roadway Geometric Design 1
1/52
12/28/2
CE 426Highway Design
Dr. Khaled Hamad
Deptartment of Civil Engineering
Block 2:
Roadway Geometric Design
Outline
Geometric Design Concept
Road Design
Horizontal Alignment
Vertical Alignment
2
7/30/2019 CE426 Block2 Roadway Geometric Design 1
2/52
12/28/2
GEOMETRIC DESIGN
3
4
Design Concepts
Geometric alignment is a 3D problem broken
down into two 2D problems:
Horizontal Alignment (plan view)
Vertical Alignment (profile view)
7/30/2019 CE426 Block2 Roadway Geometric Design 1
3/52
12/28/2
5
3D Perspective
2D Plan
(Horizontal)
2D Profile
(Vertical)
6
Stationing: Distance Referencing Along HorizontalAlignment
Horizontal Alignment
Vertical Alignment
Each roadway point is uniquely defined by stationing along the x-axis. E.g., station 345+60or @ the 345th 100 ft-station and 60 ft toward the 346th 100 ft-station)
7/30/2019 CE426 Block2 Roadway Geometric Design 1
4/52
12/28/2
7
Road Plan and Profile
8
Geometric DesignConsists of the following:
1. Horizontal Design Horizontal Alignment (HA)
2. Vertical Design Vertical Alignment (VA)
3. Cross-sectional Design Right-of-way, roadway features, Pavement
4. Junction Design Intersections & Interchanges (Layouts and traffic control)
5. Pavement Design
6. Terminal Facilities Parking Lots, Garages, etc.
8
7/30/2019 CE426 Block2 Roadway Geometric Design 1
5/52
12/28/2
9
HORIZONTAL ALIGNMENT
Roadway Geometric Design
9
10
Horizontal Alignment
Definition:
Series of straight segments of road (tangents)
connected by suitable curves (horizontal curves)
Objective: Establish geometry of directional transition to ensure
safety and comfort
Primary challenges Transition between two directions using tangents and
curves
Relationship between design speeds and curvature
Relationships with superelevation and side friction
7/30/2019 CE426 Block2 Roadway Geometric Design 1
6/52
12/28/2
Horizontal Alignment
Tangents Curves
11
Tangents & Curves
Tangent
Curve
Tangent toCircular Curve
Tangent to
Spiral Curve to
Circular Curve12
7/30/2019 CE426 Block2 Roadway Geometric Design 1
7/52
12/28/2
Motion on Circular Curves
dt
dva
t
Rvan
2
13
coscossin ns amWfW
cos)(cossin2
R
v
g
WWfW s e
tan
cos
sin
gR
vfe s
2
Motion on Circular Curves
gR
vfs
2
tan
W =weight of vehiclefs =coefficient of side frictiong =acceleration of gravityv = vehicle speedR =radius of curve
angle of inclination
e =tan (rate of superelevation)
14
)(
2
sfeg
vR
7/30/2019 CE426 Block2 Roadway Geometric Design 1
8/52
12/28/2
Determine Radius of CurveRadius of curve,R, can be calculated as follows:
15
Where:
v= design speed
e= rate of superelevation
0.04 0.12 depending on type of facility
and local environment
fs = coefficient of side friction
0.20 at low speed to 0.08 at high speed
g= acceleration of gravity = 9.81 m/s2 =32.2 ft/s2
)(
2
sfeg
vR
)(15
2
sfe
vR
For v in mph and Rin ft, you can use:
)(127
2
sfe
vR
For v in kph and Rin m, you can use:
Minimum Radius of a Circular Curve Minimum value of radiusRmin can be calculated as follows:
UsingR > Rmin allows the superelevation rate and side friction
to be smaller than maximum.
See selection ofeand fsin next slides
16
)(maxmax
2
min
sfeg
vR
7/30/2019 CE426 Block2 Roadway Geometric Design 1
9/52
12/28/2
Side friction factor (fs) It is a function of vehicle speed, pavement/roadway surface,
weather condition, tire condition It decreases as speed increases (less tire/pavement contact)
The maximum side friction factor is the point at which the tiresbegin to skid.
See table for range of values:
17
Design
Speed (mph)
Coefficient of
side friction, fs
Design
Speed (km/h)
20 0.20 30
30 0.18 50
40 0.16 65
50 0.14 80
60 0.12 9570 0.10 110
80 0.08 130
Superelevation e Controlled by following factors:
Climate conditions (frequency and amount of ice and snow)
Terrain (flat, rolling, mountainous)
Adjacent land use and type of area (rural or urban)
Frequency of slow moving vehicles who might be influenced byhigh superelevation rates
Constructability
Highest in common use is 10% Never exceed 12% which could be suitable for roads with no
ice and snow on low volume gravel-surfaced roads
8% is logical maximum to minimize slipping by stoppedvehicles, considering snow and ice Typically use 4 or 6% for urban design areas
Can be omitted on low-speed urban streets
Higher values are permitted on freeways as for arterial roads
For consistency use a single rate within a project or on ahighway
18
7/30/2019 CE426 Block2 Roadway Geometric Design 1
10/52
12/28/2
19
AASHTO
Minimum
Radius Table
20
Design Superelevation Rates - AASHTOGreen Book provides charts for 4% to 12% superelevations
7/30/2019 CE426 Block2 Roadway Geometric Design 1
11/52
12/28/2
Simple Curve Fundamentals
PC = Point of curvature PT = Point of tangency
PI = Point of intersection
= Central angle
R = Radius of curve
L = Length of curve
T = Tangent length
C = Chord length
E = External distance
M = Middle ordinate
21
Simple Curve Fundamentals
DDR
5729.6180100
2tan
RT
RL180
LD
100
For US units, degree of curvature, D :2sin2 RC
22
1
2cos
1RE
2cos1 RM
Sta. of PC = Sta. of PI T
Sta. of PT = Sta. of PC + L
7/30/2019 CE426 Block2 Roadway Geometric Design 1
12/52
12/28/2
23
Horizontal Curve Design Procedure
1. Select tangents, PIs, and general curves making
sure you meet minimum radius criteria2. Determine radius of the curve (R)
3. Measure angle () between tangents
4. Determine length of the curve (L) and Tangent (T)
5. Determine station alignment: stations for PC, PI,PT
6. Calculate spiral (transition) lengths (Ls), if needed
7. Check for widening of curves (Check SSDrequirements)
8. Develop edge profiles (superelevation runoffs)
9. Add information to plans
23
Example 1A horizontal curve is designed with a 1,500 ft radius. The
tangent length is 400 ft and the PT station is 20+00. What
are the PI and PC stations?
Solution:
Since we know R and T:
T = R tan(/2) => = 29.86
L = x R x / 180 = (3.14)(1500)(29.86)/180 = 781 ft
Sta. of PC = Sta. of PT L = 2000 781 = 1219.00 = 12+19.00
Sta. of PI = Sta. of PC + T = 1219.0 + 400 = 1619.00 = 16+19.00
24
7/30/2019 CE426 Block2 Roadway Geometric Design 1
13/52
12/28/2
Example 2 A section of a road is being designed as a high-speed divided
highway. The design speed of 60 mph. What is the minimumcurve radius for safe vehicle operation assuming: e is 8% and4%
For 60 mph, use fs = 0.12
Foremax = 8%:R
min= V2/g(fs+e)
= (60)2/15(0.08 + 0.12)
= 1200 ft
Foremax = 4% (urban situation) :R
min= V2/g(fs+e)
= (60)2
/15(0.04 + 0.12)= 1500 ft
25
Example 3Given: Simple horizontal curve with = 50.25, PC = Station (331+38.75)
design speed = 70 mph; and e = 0.07.
Design this horizontal curve (find radius of the curve and Station of PT)
Solution:
For 70 mph, use max f = 0.10
Rmin = v2/g(fs+e)
= (70 x 1.47)2/32.2(0.10 + 0.07) = 1910.33 ft => Use R = 1950 ft
L = 3.14 R /180 = (1950 x 50.25 x 3.14) /180 = 1,710.20 ftSta. of PT = Sta. of PC + L = (331+38.75) + (17+10.20) = 348+48.95
T = R tan(/2) = 1950 tan (50.25 /2) = 914.49 ft
Sta. of PI = Sta. of PC + T = (331+38.75) + (9+14.49) = 340+53.24
26
7/30/2019 CE426 Block2 Roadway Geometric Design 1
14/52
12/28/2
Setting Out Horizontal Curve in Field1. Locate PI in the field
2. Using R and , determine T and L (if not shown on plan)3. Measure out T distance from PI to locate BC and EC4. Determine deflection angles for required stations
5. Set up at PC, sight PI and measure out /2 angle
6. If /2 angle line-of-sight hits the PT mark, layout the curveusing the deflection angles.
7. If the /2 angle line-of-sight does not come acceptably closeto the PT mark, check the computations for T and then re-measure the locations of PC and PT if necessary
27
/2 = l/ 2L
Setting Out Horizontal Curve in Field
FindL = 3.14R /180 = 89.71 m
Then calculate = l/L
Deflection angle = /2
Calculate chord for each station
using
28
Station l /2
0+196.738
0+200 3.262 3.262 0 28' 2" 0 14' 1"
0+220 20 23.262 3 19' 55" 1 39' 58"
0+240 20 43.262 6 11' 49" 3 5' 54"
0+260 20 63.262 9 3' 42" 4 31' 51"
0+280 20 83.262 11 55' 35" 5 57' 48"
0+286.448 6.448 89.71 12 51' 0" 6 25' 30"
Given:
= 12 51'
R = 400 m
PC Sta. = 0+196.738
Determine information to stake out
this curve.
2sin2 RC
7/30/2019 CE426 Block2 Roadway Geometric Design 1
15/52
12/28/2
More Horizontal Curve Types
Compound Curves: multiplecurves connected directly togethergo from large radii to smaller radii(use with caution)
Reverse Curves: two curves inopposite direction (requireseparation typically forsuperelevation attainment)
Broken-Back Curves: two curvessame direction (should avoid)
Simple Curves with Spirals
29
Compound Circular Curve Ratio of flatter (larger) radius to
sharper (smaller) radius < 2:1
Each curve should pass minlength requirement by AASHTO
30
7/30/2019 CE426 Block2 Roadway Geometric Design 1
16/52
12/28/2
Reverse Curves
Seldom used Cause
discomfort andsafety problem
Preferabledesigns: 2 simple curves
w/ sufficienttangents
2 simple curvesseparated byspirals
31
Transition Curves (Spirals) Spiral curves are curves with a continuously changing radius
Design considerations: Safety & Comfort When vehicles enter a curve, the gain or loss of centrifugal forces is not instantaneous
Advantages: Provides natural and easy-to-follow path for drivers (promotes more uniform speeds and
lateral force increases and decreases gradually)
Provides location for attainment of superelevation runoff
Adds flexibility when widening a horizontal curve
Aesthetics, pleasant view
Disadvantages
Involve complex geometry Require more surveying
Are somewhat empirical
If used, superelevation transition should occur entirely within spiral
AASHTO: There is no definitive evidence that transition curves are essentialto safe operation of roadways; therefore, they are not used by many agencies
3232
7/30/2019 CE426 Block2 Roadway Geometric Design 1
17/52
12/28/2
33
Transition Curves (Spirals)
No Spiral
Spiral
Length of Spiral Curves US Customary Units Metric Units
Min spiral length, Ls,min is the larger of:
Max spiral length,
Ls,min = minimum length of spiral
Ls,max = maximum length of spiral V = vehicle speed, mph or km/hr
R = radius of the curve
C = rate of increase of radial acceleration (4 ft/sec3 or 1.2 m/sec3)
pmin = minimum lateral offset between tangent an circular curve (0.66 ft or 0.20 m)
pmax = maximum lateral offset between tangent an circular curve (3.3 ft or 1.0 m)
3434
RC
VL
RpL
s
s
3
min,
minmin,
15.3
)(24
RpLs )(24 maxmax,
RC
VL
RpL
s
s
3
min,
minmin,
0214.0
)(24
7/30/2019 CE426 Block2 Roadway Geometric Design 1
18/52
12/28/2
35
Desirable Spiral Lengths
Length correspond to travel time of 2 seconds at design speed
Insertion of Transition Curves(Spirals)
Euler (clothoid) curves are used
Radius varies from infinity attangent to radius of circularcurve
Spirals are not used when: Speed < 60 km/hr
Radius larger than that selected foremax of 2%
36
7/30/2019 CE426 Block2 Roadway Geometric Design 1
19/52
12/28/2
Transition Curves Fundamentals
37
38
Attainment of Superelevation(Superelevation Transition):
On a tangent, road cross-section is crowned to provideeasy drainage of storm water.
On a curve, cross-section slope (normal crown) shouldchange to that of super-elevation of road, i.e., obtain fullyelevated section.
Change should be gradually achieved over sufficient lengthbecause: Safety and comfort (gentle lateral acceleration and introduction of
centripetal force)
Pleasing appearance (pavement not distorted)
Transition should be achieved along: Spiral curve, if exists
Tangent-to-curve, if spiral does not exist.
38
7/30/2019 CE426 Block2 Roadway Geometric Design 1
20/52
12/28/2
39
Superelevation Transition
Superelevation transition
length is summation of: Tangent Runout: Length of
roadway needed toaccomplish a change inoutside lane cross-sectionfrom normal crown to flat(zero slope), or vice versa.
Superelevation Runoff:Length of roadway neededto accomplish a change inoutside lane cross-section
from flat (zero slope) to fullsuperelevation, or viceversa.
39
40
Superelevation Transition
Normal Crown controlpoints:
OE: Outer Edge
CL: Center Line
IE: Inner Edge
7/30/2019 CE426 Block2 Roadway Geometric Design 1
21/52
12/28/2
Length of Superelevation Runoff, Lr
41
Length of Tangent Runout, Lt
42
7/30/2019 CE426 Block2 Roadway Geometric Design 1
22/52
12/28/2
43
Length of Superelevation Transition
Example For a 4-lane divided highway with cross-section
rotated about centerline, design superelevationrate = 4% and NC = 2%. Design speed is 50mph. What are the minimum length ofsuperelevation runoff (ft) and tangent runout?
.
44
ft144
005.0
75.004.02121
wdr
bewnL
ft7214404.0
02.0 r
d
NCt L
e
eL
7/30/2019 CE426 Block2 Roadway Geometric Design 1
23/52
12/28/2
Superelevation Transition
Superelevation runoffis all attained in spiralcurve if used
If no spiral curve isused: Superelevation runoff
must be attained overa length that includesthe tangent and thecurve
Typically, use 66-70%
on tangent and 33-30% on curve
45
46
Superelevation Transition
7/30/2019 CE426 Block2 Roadway Geometric Design 1
24/52
12/28/2
Methods of Super-
elevation Attainment
1. Pavement revolvedabout center line (CL)
Centerline is point ofcontrol
2. Pavement revolvedaround inner edge (IE)
Inner edge is point ofcontrol
3. Pavement revolvedaround outer edge (OE)
Outer edge is point ofcontrol
47
48
Curve Widening
A traffic lane on a curve mustbe widened because:
Off-tracking: Rear wheels donot follow exactly the samepath/trajectories of front wheels.
Vehicles front overhangrequires an additional lateralspace.
Difficulty of driving on curvesjustifies wider lateral clearance.
Tendency of drivers to steeraway from the pavements edge.
7/30/2019 CE426 Block2 Roadway Geometric Design 1
25/52
12/28/2
49
Curve Widening
Metric
50
ExampleCalculate the widening required for passenger
cars on a curve with radius R =570 ft and
design speed v = 40 mph. The roadway has
two lanes and is 22 ft wide on the tangent
section.
ft3Aft,11Lft,7uft,2.5Cft,22Wn
ft7.11U
115705707U
LRRuU
22
22
ft0.07F
5703)113(2570F
RA)A(2LRF
A
2
A
2
A
ZFC)2(UW Ac
nc WWno widening is needed for
passenger carsft20.11.680.072.5)2(7.11W
ZFC)2(UW
c
Ac
ft1.68570
40Z
R
vZ
7/30/2019 CE426 Block2 Roadway Geometric Design 1
26/52
12/28/2
51
Sight Distance on Horizontal Curve
Sight Distance onhorizontal curve iscompromised bypresence of obstructions: Walls
Cut slopes
Trees
SSD is measured aroundmiddle of inside lane
Line of sight is a chord ofcurve
Ms is sight obstructionoffset
Stopping Sight Distance
where:
v = initial velocity when brakesare applied
tP/R = time to perceive/react
f= coefficient of friction
g = acceleration of gravity = 9.81
m/s2 = 32.2 ft/s2
G = grade (decimal)
Distance to stop vehicle = Distance during P/R + Braking Distance
Gfgv
vtSSD RP
2
2
/
This is distance traveled by a vehicle between the time driver observes
an object in vehicles path and the time the vehicle actually comes to
complete stop.
52
Gfv
vtSSD RP
30
47.12
/
Gfv
vtSSD RP
254
278.02
/
Speed in mph and SSD in ft:
Speed in kph and SSD in m:
7/30/2019 CE426 Block2 Roadway Geometric Design 1
27/52
12/28/2
53
Sight Obstruction on Horizontal Curves
v
sR
SSD
180
DRSSD ssv
100180
v
vsR
SSDRM
90cos1
v
svv
R
MRRSSD 1cos
90
54
Sight Obstruction on Horizontal Curves
7/30/2019 CE426 Block2 Roadway Geometric Design 1
28/52
12/28/2
55
Example
Horizontal curve with 800 ft radius of 35 mph posted speedlimit. Determine horizontal sightline offset that a largebillboard can be placed from the centerline of inside lanewithout reducing required SSD. Assume: e=0%; t=2.5 sec;f=0.35.
Determine SSD:
Determine M:
.
Check value on monograph.
ft9.245
035.030
355.23547.1
3047.1
22
Gf
vvtSSD
ft6.9
80014.3
29.24590cos1800
90cos1
v
vs
R
SSDRM
56
VERTICAL ALIGNMENTRoadway Geometric Design
56
7/30/2019 CE426 Block2 Roadway Geometric Design 1
29/52
12/28/2
57
Vertical Alignment
G1 G2
G1 G2
Crest Vertical Curve
Sag Vertical Curve
Definition:
Straight segments of highway (tangents or grades)
connected by vertical curves.
Purpose: Select suitable grades for tangent sections and
appropriate lengths for vertical curves.
58
Vertical Alignment Objectives:
Provide a gradual change from one tangent grade toanother to allows vehicles to run smoothly on road
Determine elevation to ensure proper drainage andacceptable level of safety
Primary Challenges Topography of area has significant impact (following
existing surfaces to minimize earthwork cut & fill) Grades can significantly impact vehicles
Main Criteria: Minimum stopping distance Adequate drainage Comfort Pleasant appearance Minimize cut and fill
7/30/2019 CE426 Block2 Roadway Geometric Design 1
30/52
12/28/2
Grades
Grade: measure of inclination or slope (rise over the run)
Grades can significantly impact vehicles, especially trucks(performance increase on downgrades & decrease onupgrades) Cars: negotiate 4-5% grades without significant speed reduction
Trucks: significant speed changes
Maximum grade: depends on terrain type, road functionalclass, and design speed Max grade of 5% for design speed of 110 km/hr (70 mph)
Max grade in range of 7% to 12% for design speed of 50 km/hr (30mph)
AASHTO Recommendation (next slide)
Minimum Grade: Typically 0.5%
Flatter grades sometimes are justified
59
60
AASHTOMaximum
Grades
7/30/2019 CE426 Block2 Roadway Geometric Design 1
31/52
12/28/2
Critical Length of Grade
Maximum length which aloaded truck can travel
without unreasonable
speed reduction
Based on speed distance
curve of a typical heavy
truck (with 10 mph speed
reduction as threshold
Critical Length of Grade
General
Design
Speed
Reduction
7/30/2019 CE426 Block2 Roadway Geometric Design 1
32/52
12/28/2
63
Types of Vertical Curves
63
Type I
Type II
Type IIIType IV
Vertical Curve PVI = Point of Vertical
Intersection PVC = Point of Vertical
Curvature PVT = Point of Vertical
Tangency L = Length of vertical curve
Y = Offset of curve from initialgrade line
Ym = mid-curve offset Yf= end-curve offset
G1 = Grade of initial tangent, % G2 = Grade of final tangent, %
64
7/30/2019 CE426 Block2 Roadway Geometric Design 1
33/52
12/28/2
Properties of Vertical Curve
65
Parabolic function:
yis roadway elevation xdistance from beginning ofcurve.
Characteristics: Constant rate of change of slope.
Implies equal curve tangents:Horizontal distance PVC to PVI = Horizontal distance PVI to PVT= Curve length
Length of vertical curve is the horizontal projectionof the curve and not the length along the curve.
cbxaxy 2
Vertical Curve Fundamentals
10:PVCAt the Gbdxdyx
PVCofElev.0:PVCAt the cyx
L
GGa
L
GGa
dx
yd
22:Anywhere 1212
2
2
cbxaxy 2
PVCofElev.2
Therefore, 1212
xGx
L
GGy
66
Choose Either:
G1, G2 in decimal form, L in ft or m G1, G2 in percent, L in stations
(i.e., divide distance by 100)
7/30/2019 CE426 Block2 Roadway Geometric Design 1
34/52
12/28/2
Vertical Curve Fundamentals
2
2x
L
AY
12 GGA Algebraic difference
in grade =
Sta. of PVC = Sta. of PVI - L/2
Offset =
Sta. of PVT = Sta. of PVI + L/2 = Sta. of PVC + L
Elevation of PVC = Elevation of PVI - G1 L/2
Elevation of PVT = Elevation of PVI + G2 L/2
Elevation at any point x = Y +G1 x +Elev. of PVC
67
G1
G2
PVI
PVTPVC
x
Y
8
ALYm
2
ALYf
Mid-curve offset:
Final
offset:
Choose Either: G1, G2 in decimal form, L in ft or m
G1, G2 in percent, L in stations
(i.e., divide distance by 100)
68
Vertical Curve Procedure1. Determine the minimum length of curve to satisfy sight
distance requirements and other criteria.
2. Determine from the layout plans the station andelevation of the PVI (the point where the gradesintersect).
3. Compute the elevations of the PVC (or BVC) and endof vertical curve (EVC or PVT).
4. Compute the offsets, Y, from the tangent to the curve atequal distances or stations.
5. Compute elevations on the curve for each station as:elevation of the tangent offset from tangent, Y. Forcrest curves the offset is (-) and for sag curves theoffset is (+).
6. Compute the location and elevation of highest/lowestpoint on curve.
68
7/30/2019 CE426 Block2 Roadway Geometric Design 1
35/52
12/28/2
69
Criteria for Minimum Length of Curve
Minimum length of crest vertical curve shouldbe based on stopping sight distance.
Other (passing or decision) sight distances couldalso be considered if required.
Minimum length of sag vertical curve shouldbe based on:
1. Night sight distance
2. Comfort
3. Appearance4. Drainage
Stopping Sight Distance
where:
v = initial velocity when brakesare applied
tP/R = time to perceive/react
f= coefficient of friction
g = acceleration of gravity = 9.81
m/s2 = 32.2 ft/s2
G = grade (decimal)
Distance to stop vehicle = Distance during P/R + Braking Distance
Gfgv
vtSSD RP
2
2
/
This is distance traveled by a vehicle between the time driver observes
an object in vehicles path and the time the vehicle actually comes to
complete stop.
70
Gfv
vtSSD RP
30
47.12
/
Gfv
vtSSD RP
254
278.02
/
Speed in mph and SSD in ft:
Speed in kph and SSD in m:
7/30/2019 CE426 Block2 Roadway Geometric Design 1
36/52
12/28/2
71
AASHTO Stopping Sight Distance
Decision Sight Distance (DSD) For unexpected stimulus or when driver has to make
unusual maneuvers AASHTOs recommends using
decision sight distance (DSD).
DSD depends on
Type of maneuver required
Road location (Urban/Rural)
SSD is often inadequate when drivers must make
complex/instantaneous decisions; DSD is longer thanSSD
Exhibit 3.3 from AASHTOs Green Book is used in
design
72
7/30/2019 CE426 Block2 Roadway Geometric Design 1
37/52
12/28/2
Decision Sight Distances for Different Design Speeds and
Avoidance Maneuvers
73
74
Passing Sight DistanceSight distance is determined for a single vehicle passing a single vehicle with the
assumption that cover majority of situations observed in the real-world conditions.
4321 ddddPSD
d1 is the distance traveledduring the P-R time and theinitial acceleration until thepoint where the vehiclebegins to enter the passinglane
d2 is distance traveled whilethe vehicle is passing in thepassing lane
d3 is the distance betweenthe passing vehicle andopposing vehicle
d4 is the distance theopposing vehicle moveswhile the passing vehicle isin the passing lane
7/30/2019 CE426 Block2 Roadway Geometric Design 1
38/52
12/28/2
Passing Sight Distance (PSD)
75
Where
t1 = time of initial maneuver, sec
t2 = time the passing vehicle occupies leftlane, sec
a = average acceleration, mph/sec
v = average speed of passing vehicle,mph
m = difference in speed of passed andpassing vehicle, mph
247.11
11
at
mvtd
22 47.1 vtd
d3 = 110 to 300 ft
d4 = 2d2 /3 3/2
m9030
278.0
2278.0
24
3
22
1
11
dd
d
vtd
atmvtd
a = average acceleration, km/h/s
v = average speed of passing vehicle,
km/h
m = difference in speed between
passing and passed vehicles, km/h
U.S. Customary Metric
Passing Sight Distance on Two-lane Highways
76
7/30/2019 CE426 Block2 Roadway Geometric Design 1
39/52
12/28/2
77
Minimum Length of Crest Curve
G1 G2
PVI
PVTPVC
h2h1 L
S
2
21
2
min
200
hh
SAL
A
hhSL
2
21
min
2002
For S < L For S > L
Line of Sight
Use , where G1 & G2 in percent, not in decimal form12 GGA
78
Minimum Length of Crest Curve AASHTO Values for h1 and h2:
where:SSD (Stopping Sight Distance)
PSD (Passing Sight Distance)
Parameters (US Units) SSD (ft) PSD (ft)
Height of drivers eye (h1) 3.5 3.5
Height of object (h2) 2.0 3.5
Parameters (Metric) SSD (m) PSD (m)
Height of drivers eye (h1) 1.08 1.08
Height of object (h2) 0.60 1.08
7/30/2019 CE426 Block2 Roadway Geometric Design 1
40/52
12/28/2
79
Minimum Length of Crest Curve
The length of crest curve should not be lessthan:
Lmin = 3v(design speed in mph and L in ft)
Lmin = 0.6v(design speed in km/h and L in m)
Generally, minimum lengths of crest curves
are about 100 to 325 ft (30 to 100 m).
80
Minimum Length of Sag Curve
G1 G2
PVI
PVTPVC
h2=0h
L
Headlight Sight Distance (S)
1tan200
2
minSh
SAL
A
ShSL
1tan2002min
For S < L For S > L
headlight beam (diverging by 1 degrees)
h = headlight height
= 2.0 ft (or 0.6 m)
7/30/2019 CE426 Block2 Roadway Geometric Design 1
41/52
12/28/2
81
Other Criteria for Minimum Length of Sag Curve
Comfort Criteria: A comfortable ride will be provided if the radial
acceleration is not greater than 1 ft/sec2
or 0.3 m/sec2
General Appearance Criteria: The rule of thumb for the general
appearance criteria is as follows:
Drainage Criteria(only when curve is curbed): The maximum length inwhich a minimum slope of 0.30% be provided within 50 ft (15 m) oflowest/level point.
mph)inft where(in5.46
2
min vAv
L
ft)(in100min AL m)(in30min AL
km/h)inwherem(in395
2
min vAv
L
82
Example 1For a design speed of 50 mph andgrades shown below, determine theminimum length of a vertical curvethat meets the criteria for safepassing.
Solution:
%32 G1%1 G
%4)1(3GGA 12
:
ft939,1)1tan18002(200
18004 2
L
The answer is correct
because the SL condition
is met.
LS)tan(200
2
Sh
SAL
S = 1,800 ft
7/30/2019 CE426 Block2 Roadway Geometric Design 1
42/52
12/28/2
Example 2 Given:
A sag vertical curve of a highway with -3% grade joining
+3% grade
Design speed = 55 mph
Perception-reaction time = 2.5 sec
Determine the minimum length of the curve?
Solution:
Determine the SSD required for the design conditions.
Because the grade change constantly on a vertical curve,
the worst-case value for G is used to determine the braking
distance:
ft
GfvvtSSD rp
41.51926.31715.202
)03.035.0(30555.25547.1
)(3047.1
22
/
83
Solution:This is a sag curve.
How we determine if S>L or SL:
The condition is not appropriate since 519.41
7/30/2019 CE426 Block2 Roadway Geometric Design 1
43/52
12/28/2
8585
Solution:
Check for other criteria:
Rider comfort:
General Appearance:
Rule of thumb:
So, the minimum length to satisfy all criteria is 731 ft
ftAv
L 3905.46
556
5.46
22
ftAL 6006100100
ftvL 165)55(33min
86
Finding High or Low Point on Vertical Curve
High or low point on a vertical curve is of interest tothe designer to investigate: Drainage conditions
Clearance beneath overhead structures
A
GLy
GKGG
LG
GG
LGx
GxL
GG
dx
dy
xGxL
GGy
LowHigh
LowHigh
2-PVCofElev.
:isPointHigh/LowofElevation
02
2:curveon thepointHigh/LowoflocationfindTo
PVCofElev.2
:usingfoundbecancurveverticalaonpointanyofElevation
21
/
112
1
121/
112
1212
7/30/2019 CE426 Block2 Roadway Geometric Design 1
44/52
12/28/2
Example 3
A 400 ft equal-tangent crest vertical curve has a PVC station of 100+00 at
59.00 ft elevation. The initial grade is 2.0% and the final grade is -4.5%.Determine the elevation and stationing of PVI, PVT, and the highest point of
the curve.
PVI
PVT
PVC: STA 100+00
EL 59 ft.
87
Solution:
PVI STA is 102+00 and PVT STA = 102+00 + 200 = 104+00
Elevation of the PVI = 59 + 0.02(200) = 63 ft.
Elevation of the PVT = 63 0.045(200) = 54 ft.
High point elevation requires figuring out the equation for a vertical curve:
y=ax2+bx+c:
At x = 0, y = c = 59 ft
At x = 0, dy/dx = b = G1 = +2% = 0.02
a = (G2 G1)/2L = (-0.045 0.02)/(2*400) = - 0.00008125
Therefore, y = -0.00008125x2 + 0.02x + 59.
Highest point is where dy/dx = 0. That is, dy/dx = -0.0001625x + 0.02 = 0.
Solve for x, you get x = 123 ft (i.e., 1.23 stations).
Find elevation at x = 123 stations:
y = -0.00008125(123)2 + 0.02(123) + 59 = 60.23 ft
88
7/30/2019 CE426 Block2 Roadway Geometric Design 1
45/52
12/28/2
Another Solution:
You could obtain the same results using grades in percentage and length
of curve in stations:
The equation for the vertical curve is y=ax2+bx+c:
At x = 0, y = c = 59 ft
At x = 0, dy/dx = b = G1 = +2.0%
a = (G2 G1)/2L = (-4.5 2)/(2(4)) = - 0.8125
Therefore, y = -0.8125x2 + 2x + 59.
Highest point is where dy/dx = 0.
That is, dy/dx = -1.625x + 2 = 0. Solve for x, you get x = 1.23 stations.
Find elevation at x = 1.23 stations:
y = -0.8125(1.23)2
+ 2(1.23) + 59 = 60.23 ft
89
90
Length of Crest & Sag Curves Based on KFactors
Rate of Vertical Curvature, K, is the length of thevertical curve per percent algebraic difference inintersecting grades:
Kis a function of design speed (see tables); thus,
it could be viewed as a design control It is a shortcut for computing minimum length of
curves if known for design speed:
ALK
AKL
7/30/2019 CE426 Block2 Roadway Geometric Design 1
46/52
12/28/2
91
Design Controls for Crest Vertical Curves Based on SSD
92
Design Controls for Crest Vertical Curves Based on PSD
7/30/2019 CE426 Block2 Roadway Geometric Design 1
47/52
12/28/2
93
Design Controls for Sag Vertical Curves
94
Design Controls for Crest Vertical Curves (Metric)
7/30/2019 CE426 Block2 Roadway Geometric Design 1
48/52
12/28/2
95
Design Controls for Crest Vertical Curves (US Units)
96
Design Controls for Sag Vertical Curves (Metric)
7/30/2019 CE426 Block2 Roadway Geometric Design 1
49/52
12/28/2
97
Design Controls for Sag Vertical Curves (US Units)
98
Example 4A roadway is being designed using a 45 mph design speed. One section of
the roadway must go up and over a small hill with an entering grade of 3.2%
and an exiting grade of -2.0%. How long must the vertical curve be?
Solution:
A = |-2 3.2| = 5.2%
For 45 mph, we get K=61.
Therefore, L = KA = (61)(5.2) = 317.2 ft
7/30/2019 CE426 Block2 Roadway Geometric Design 1
50/52
12/28/2
99
Example 5A 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 -2.4% and the exiting grade is
4.0%. A tree has fallen across the road at approximately the PVT. Assumingthe 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?
Solution:
Assume that S>L (it usually is not but for example well do it this way):
S = 146.23 ft < L
Must use S
7/30/2019 CE426 Block2 Roadway Geometric Design 1
51/52
12/28/2
101
Sight Distance at Undercrossings
L = Minimum curve length (m or ft)
S =Sight distance (m or ft)
A = |G2-G1|, grade change in %
C= Vertical clearance (m or ft)
h1 = Height of drivers eye (usually 8 ft or 2.4 m)
h2 = Height of object (usually 2 ft or 0.6 m)
2800
21
2
hhC
SAL
A
hhC
SL
2
800
2
21
For S < L For S > L
102
Example 6A +2% grade intersects with a -1% grade at station (535+24.25) at an elevation
of 300 ft. If the design speed is 65 mi/h, determine:
the stations and elevations of the BVC and EVC
the elevation of each 100-ft station
Solution:
A = |2 (-1)| = 3%
For 65 mph, we get K=193
Therefore, L = KA = (193)(3) = 579 ft
7/30/2019 CE426 Block2 Roadway Geometric Design 1
52/52
12/28/2
103
Example 6 Continue
END OF BLOCK 2