ece 2404 highway engineering 1 - 2016
TRANSCRIPT
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ECE 2404 HIGHWAY ENGINEERING 1
Course outline
Principles of highway design
Geometric design; Horizontal and vertical curves,
Factors and elements of geometric designs.
Intersection design: at-grade junctions, roundabouts, conflict points, kerbed and Ghost
islands, lane separation, acceleration and deceleration lanes, intersection site triangle.
Speed, sight distance and capacity
References
1. Highway and traffic Vol.1 by C.A oflaherty
2. Highway and traffic Engineering in developing countries by Bant thageson
3.
Principles of highway Eng and traffic analysis by Fred L. Mannering and Walter P.
4.
Highway traffic analysis and design by R. J salter.
5. MOR. Design manual.
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CHAPTER ONE
PRINCIPLES OF HIGHWAY DESIGN
INTRODUCTION TO HIGHWAY ENGINEERING
Highway engineering is the process of design and construction of efficient and safe highways
and roads. Concepts such as grade, surface texture, sight distance and radii of horizontal
bends and vertical slopes in relation to design speed and in addition to road junction design
(intersections and interchanges) are all important elements of highway engineering
Design is the process of selecting the elements that once combined will make up an end
product. In engineering, these elements are primarily features, dimensions and materials.
Highway geometric design is selection of a road's visible features and dimensions (lane or
shoulder width, for example). These have important bearing on how the road will function, its
capacity, driver behaviour and safety.
HIGHWAY DESIGN
OBJECTIVES;
*To provide facilities which are safe, efficient, comfortable, of adequate capacity
economical (in the long run) and structurally sound.
FACTORS CONTROLLING DESIGN:
1. Topography and land use.
- A fundamental consideration in geometric design, as it influences such design elements as
alignment, sight distances, design speed etc.
- Man made features e.g. Agricultural and other centres of economics importance influence the
final location of highway.
2. Environmental conditions.
- Preservation of natural beauty of the country sight.
- Preservation of areas of some particular value e.g. national parks, common
Monuments etc
-
Preservation of soil erosion.
- Reduction of environmental pollution in terms of noise and air pollution.
http://en.wikipedia.org/wiki/Designhttp://en.wikipedia.org/wiki/Road_safetyhttp://en.wikipedia.org/wiki/Highwayhttp://en.wikipedia.org/wiki/Roadhttp://en.wikipedia.org/wiki/Grade_separationhttp://en.wikipedia.org/wiki/Radiushttp://en.wikipedia.org/wiki/Design_speedhttp://en.wikipedia.org/wiki/Junction_(road)http://en.wikipedia.org/wiki/Intersection_(road)http://en.wikipedia.org/wiki/Interchange_(road)http://en.wikipedia.org/wiki/Interchange_(road)http://en.wikipedia.org/wiki/Intersection_(road)http://en.wikipedia.org/wiki/Junction_(road)http://en.wikipedia.org/wiki/Design_speedhttp://en.wikipedia.org/wiki/Radiushttp://en.wikipedia.org/wiki/Grade_separationhttp://en.wikipedia.org/wiki/Roadhttp://en.wikipedia.org/wiki/Highwayhttp://en.wikipedia.org/wiki/Road_safetyhttp://en.wikipedia.org/wiki/Design
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3. Road safety consideration.
- Reduction of number of accidents.
- Reduction of severity of accidents.
4.
Road function.
- The functional classification of roads must be considered in the determination of
geographical design standard.
5. Traffic considerations, in terms of
- Traffic volume
- Directional distribution.
- Composition of the traffic
- Capacity and level of service.
6. Design speed.
- Is the maximum safe speed that can be maintained over a section of road when design
or ideal or favourable operation condition exists.
- Selection of design speed may be influenced by a number of factors:
a) Functions of the road.
b)
Nature of terrain.
c) Traffic volume.
d) Land use (Accessibility)
e) Economic considerations.
Terrain Road class
A&B C D&E
Flat
Rolling
Mountainous
100-120
90-100
50-70
90-100
60-90
40-60
80
50-80
20-50
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CLASSIFICATION OF ROADS
Roads can be classified according to:
- Basic roads system; Basic urban classification and Basic rural classification.
- Structural classification; murram, earth, asphalt material.
- Functional classification.
FUNCTIONAL CLASSIFICATION
There are 5 classes according to the major functions in road network.
Class A. International trunk roads.
These are roads linking centres of international importance and crossing international
boundaries or terminating in international points.
Class B. National trunk roads
These are roads linking nationally important centres (Principle towns) or urban centres.
Class C. Primary roads
Are roads linking provincially important centres to each other or to a higher classes roads
(Urban or rural centres)
Class D. Secondary roads
Are roads linking locally important centres to each other, to a more important centres or to a
higher class road (Rural or market centres)
Class E. Minor roads
These are any roads linking minor centres. (Market or local centre)
Roads of the highest classes are A&B,have there major function to provide mobility, while
the function of class E road is to provide access. The roads class C and D have for all
practical purpose to provide both mobility and access with emphasis on mobility for primary
roads and access for secondary roads.
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7. Design vehicle;
This is a selected motor vehicle whose physical characteristics in terms of its weight and
dimensions (used in determination of highway design element) will accommodate all vehicles
within a designated class. Vehicle dimensions that influence design includes:
i. Minimum turning radius.
ii. Path of the inner run wheel or tyre.
iii.
Thread width.
iv. Wheel bar-dimension between the back and front wheel.
Design elements influenced by design vehicle characteristics are:
I. Road cross-section.
II.
Junction layout.
III. Road widening and horizontal alignment.
Categories of design vehicles include:
a) Passenger vehicles cars.
b) Single unit trucks.
c) Semi trailer combinations.
8. Economic considerations.
The geometrical design features of a road, in relation to traffic expected on it and type of
terrain should be evaluated and justified in terms of cost and benefit that will come about as a
result of providing the road.
Asst One
Using relevant examples discuss how Roads have been classified in Kenya according to:
a)
Basic road system
b) Structural classification
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CHAPTER TWO
GEOMETRIC DESIGN
Geometric design elements are
1. Sight distance.
2. Horizontal alignment.
3. Vertical alignment.
4. Cross-section
SIGHT DISTANCE
There are two types
Stopping sight distance (SSD)
Passing sight distance (PSD)
a) STOPPING SIGHT DISTANCE (SSD)
This is defined as minimum distance required by a driver travelling under the designed speed
of road to bring a vehicle to a safe stop on sighting on an unexpected object on the road ahead
of him. This includes
i. Perception reaction time (lag time)
Time between sighting of the hazard and the first application of brakes
Dependent on drivers alertness, distance of obstruction, speed, object colour etc
ii. Braking distance
Distance the vehicle moves from first application of brakes to stop.
Usually dependent on the condition of the tyre, roads, vehicle speed etc.
For horizontal roads braking distance f
V Vt d
254
2
For sloppy roads braking distance
G f
V Vt d
254
2
i.e Down slope is – ve while Upslope is +ve
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Where V = Vehicle speed in km/hr
f =Longitudinal coefficient of friction.
= Brake efficiency.
d =Braking distance in meters.
G = Slope of the road
Example
Calculate the braking distance for a vehicle moving at 80km/h on a road with coefficient of
friction of 0.4 and grade of 1:40 for a perception time of 2.5 sec.
Soln
G f V
Vt d
254
2
Given
V = 80 km/hr
f =0.4.
= 1.
G = 1:40
So for Down slope,
1
40
14.0254
80
6.3
5.280 2
x
d =55.56+67.19=122.75=123m
For Up slope
140
14.0254
80
6.3
5.280 2
x
d =55.56+59.27=114.85=115m
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b) PASSING SIGHT DISTANCE (PSD)
This is a requirement for a two lane, two way, and carriage way.
Def: Minimum distance on a two lane two way carriage way that a driver should have
ahead of him to safely overtake a slow moving vehicle in front of him without interfering
with the speed and safety of an oncoming vehicle.
Key
d1= Preliminary delay distance.
It`s the distance taken or moved by the fast vehicle while trailing the slower vehicle while
waiting opportunity for safe overtaking, after which it hurriedly moves to the opposite lane.
d2= Overtaking distance.
This is the distance that the fast vehicle travels as it moves on the opposite lane, passes the
slower vehicle and safely returns to its lane.
d3= Safety distance.
Clearance distance between overtaking and oncoming vehicles after overtaking
d4= Distance moved by an oncoming vehicle as overtaking vehicle overtakes.
It is usually ⅔ d2 if the speeds of overtaking and oncoming vehicles are the same (design
speed)
Safe PSD= d1+ d2+ d3+d4
Reduced PSD= 2/3d2+d3+d4.
d1 d2 d3 d4
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ROAD GRADIENT
The grade of the road is the slope in the longitudinal direction. The descending and ascending
roads are said to have – ve and +ve grades respectively.
The grade to be used in a highway is governed by
i. Topography of the country or terrain.
ii. Type of traffic e.g. automobile, animal drawn etc.
iii.
Drainage.
iv. Access to adjoining property.
v.
Obligatory points and roads or railway or canal crossing.
vi. Appearance.
RESISTANCE TO TRACTION
The factor that determines the traction force are various resistances which come into place
on moving traffics and includes:
I. Wind resistance.
II.
Axle resistance.
III. Grade resistance.
IV. Rolling resistance.
WIND RESISTANCE
May be calculated from the relation, Ra= CAV2
Where: Ra is Kg/m2
C is Constant normally =0.01
V is Speed in Km/h
A is projected area of vehicle in m2
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Rolling resistance
Varies
1)
Inversely with diameter of the wheel2) Increases with decreasing width of the wheel as the lower width increases pressure on
the surface resulting in increase in resistance.
3) Nature of the road surface. Hard surface offer less resistance e.g driving on sand
Axle resistance
For well designed vehicle, this resistance may be neglected
Grade resistance
On a level ground
When up an incline
Resistance to traction consists of gravity (Weight component) and friction component
i.e Traction force T= Psinα +fPcosα
Traction force T
P=W
fP=fW
Traction force T
Psinα
Psinα
P
Pcosα
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For small α
sinα = tanα and cosα=1
Therefore T= Ptanα +fP
tanα =Slope of grade =G
Hence T=PG+fP
When down an inclination G is – ve
Therefore T=fP-PG
Example
A vehicle moving at 30km/h on a level road suddenly joins a graded section and continues
moving without changing the tractive force until it stops after moving 23.6m in 5.7sec.
- Estimate the grade of the road if the coefficient of friction is 0.02
- What would be the tractive force in terms of vehicle weight and in terms of initial tractive
force to maintain the speed of the vehicle
Soln
Given d=23.6m t=5.7sec and u=30km/h
Recall Deceleration
d= ut at 2
2
1=23.6m= 7.56.3
307.52
1 2
aX
a=-1.47m/s
On level ground, Tractive force T =fW
On inclination, Tractive force T= WG+fP=W(G+f)
T’-T=a=mass Force =
g
W WG = gG= 47.1
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Then G = g
47.1=
81.9
47.1=0.15
Therefore T’= W(G+f)=W(0.15+0.02)=0.17W
But T=Wf, hence W= f
T
Therefore
T’=0.17W=0.17 f
T =0.17
02.0
T =8.5T
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HORIZONTAL ALIGNMENT
It is an important feature in design, influencing the efficiency and safety of a highway. The
alignment consists of a series of intersecting tangents which are connected by the use of
simple circular curves and transition curves. The design is dependent on the radius of the
curve and super elevation of the carriage way. Proper design leads to.
i. Improved speed
ii. Higher highway capacity
iii.
Lower accidents
SUPER ELEVATION
This is the provision of a transverse inclination of the carriageway i.e outer edge being raised
with respect to the inner edge to overcome the effect of centrifugal force due to curvature.
This has the advantage of
a. Lower danger of overturning
b. Higher traffic volume
c. Lower maintenance on outer wheel and lower wear of its line on carriageway
d.
Drainage of water on one side only.
gR
WV 2
N
W
P
a. Non super elevated
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Fig (a) illustrates the forces acting on a vehicle as it moves about a horizontal curve on a flat
carriage way (non super elevated).
N
W
P gR
WV 2
α
Wcosα Wsinα
gR
WV 2cosα
gR
WV 2sinα
b. Super elevated
gR
WV 2Centrifugal force
W (Weight)
Friction
C) PLAN
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With the forces in equilibrium
gR
WV P
2
where P = lateral frictional force resisting the centrifugal
V = Speed of the vehicle
R = Radius of the curve
g = Acceleration due to gravity
But P μN=μW= gR
WV P
2
Hence μ= gR
V 2=W
P
μ= transverse coefficient of friction
W
P =Centrifugal ratio for velocity in km/h
g=9.81m/s2
V=kph
μ=
gR x
V 2
22
)6060(
1000=
R
V
127
2
Therefore for non limiting value of μ, the minimum curve radius can be calculated for any
given design speed. The centrifugal force causes an overturning moment through the centre
of gravity of the vehicle and the carriage way surface. The moment is resisted by a righting
moment caused by the weight of the vehicle acting thro’ its centre of gravity.
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For equilibrium
2
2 d W xh
gR
WV
gR
V
d W h
2
2
= 2
d
Where
d=Lateral width btn the wheels
h=Height of centre of gravity above the carriageway.
When the carriage way is super elevated, the forces acting on the vehicle are as in Fig b.
At equilibrium
N=Wcosα+ sin2
gR
WV
μN=P= cos2
gR
WV -Wsinα
Dividing equation ii and i
μ=
sincos
sincos
2
2
gRV
gR
V
Since R is big, the effect of sinα (which is very small almost equal to zero) on sin2
gR
V
makes it =0
Since R is big, the effect of cosα (which is almost equal to one) on cos2
gR
V makes it remain
the same.
Eqn i
Eqn ii
Eqn iii
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Hence
μ= gR
V 2-tanα
For an angle of super elevation where α is usually small, the term tanα is called super
elevation and denoted by e.
Therefore
μ= gR
V 2-e
Since V=kph and g=9.81m/s2
μ= R
V
127
2
-e
μ+e= R
V
127
2
for a case of no lateral friction and the centrifugal force is counteracted by the super elevation
i.e μ=0, we have a phenomenon of “hands off speed” (self steering).
Hence e= R
V
127
2
In a case where the super elevation takes 40% of the centrifugal force
e= R
V
1274.0
2
Example
If super elevation for R is equal to 450m is e and it balances 45% centrifugal force estimated
the value of e for a velocity of 85km/h. If the above e and R balances the hand off speed,
estimate the design speed.
e= 057.0450127
8545.0 2
x
x
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V2=127Re
V=57.075km/h
Maximum super elevation emax depends on
i.
Climatic conditions
ii. Terrain conditions
iii. Type of area, whether urban or rural – influences vehicle speed.
iv. Road surface texture
Recommended super elevation emax
AASHTO
– Urban - 6%
– Rural - frequent snow - 8%
– Rural - normal condition - 10%
UK
– Urban - 4%
– Rural - 7%
Kenya
– Urban - 6%
– Rural - 6%
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CURVES
Circular Curves
A circular curve joining two road tangents is described either by its radius or its degree of
curvature. The degree of curvature , is defined as the central angle subtended by 100m arc
of the curve.
2tan
RT
2sin2
R Lc
D L
100
4tan1
2sec
T R E
2cos1 Rm
T
PI
E
m
PTPC 2
2
2
Lc
RR
PI=Point of intersection
PC=Point of Curvature
PT=Point of tangency
=External Angle=
m=Middle ordinate, offset from line of sight of road
L=Length of Curvature
Lc=Long chord length
D=Degree of curvature
T=Length of tangent
R=Radius of Curvature
E=External distance
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R D
5730
2
sin
50
R
222 m R X R
2222 2 mmR R X R
but 22
2
2m
S X
222
2
2 22
mmR RmS
R
mRS 22
2
R
S m
8
2
If an immovable object limits the sight
line AB for safe stopping sight distance ,
S, along the circular curve ACB, then the
middle ordinate distance, M, may be
estimated by considering that the track of
the vehicle is along the chord AM and
MB, rather than the arc of the curve. By
geometry
X
C
m
BA
RR
O
R-m
X M
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In the above situation where the required sight distance is greater than the available length of
curve, L, and the sight distance overlaps the tangent to the curve for a distance, , on either
side of the curve , then S =L + 2
2
LS
By geometry 222
2m X
S
And 222 m Rd X while 22
2
2 R
LS d
Hence 222
2m X
S
= 22 m Rd +m2= 2
2
2
2 R
LS d
2m R +m2
222222
2
224
1
4 mm Rm R R L LS S S
Rm L LS S S 224
1
4
222
R
LS Lm
8
2
DE
M
X
C
m
BA
dd
O
R-m
X
L
RR
DCE=L
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WIDENING WIDTH OF CARRIAGE WAY ON CURVES
The rear wheels of the vehicle follow a path of a shorter radius than front wheels and have an
effect of increasing effective width of carriage way to maintain the safety clearance between
opposing vehicles on curves sections. This is more so necessitated due to tendency of drivers
moving away from the edge of carriage way hence increasing the potential of accident on the
curved sections.
Where: R = radius of path of outer front wheel.
R = radius of the path of inner rear wheel.
d = lateral width between wheels.
= length between front and rear axles.
TRANSITION CURVES
Enable vehicles moving at a high speed to make a change from tangent sections to the curve
sections and to the next tangent sections of a road in a safe and comfortable fashion (way).
Therefore a transition curve provides:
i. A gradual introduction of radial acceleration from zero in the tangent to a maximum
of R
V 2 on the circular curve. for spiral (or Clothoid)
espiral Lengthofth R
1
ii. Enough length of roads on which the full super elevation required on the circular
curve may be applied.
Direction
wd
r R
w = Distance to be increased.
The extra width W may be estimated from
w= R – (r+d)
22 Rd r
Hence 22 R Rw
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iii. Smooth aesthetic that matches or harmonises with terrain and removes any kinky
appearance that would otherwise occur as intersection point.
Transition curves are generated from lemniscates spiral or clothoid or cubic parabola curves.
The spiral transition curve is widely accepted due to the ease in which it can be set out in thefield.
R
Ls
2 Radians =
R
Ls
2
3.57 degrees
R P R E
2
sec
R
L P s
24
2
10
12
sc L X
423
3 c y
= Spiral angle
Ls= Spiral length=Transition curve length
R = Radius of the circular curve
Xc and yc = Coordinates of SC
P = Shift
2
Ls K Approximately
E
M
X
C
m
STTS
dd
O
R-m
s s X
Ls
RR
DCE=L
SC CS
pXc
yc
K
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LENGTH OF TRANSITION CURVES
1. Shortt’s method
Uses the rate of change of radial acceleration in relation to safety and comfort. Radial
acceleration changes from 0 to R
V 2over the length of the transition curve.
Time taken to travel the transition length is given by.
V
Lt s
For rate of gain of radial acceleration C:
s
s
RL
V
V
L
R
V C 32
m/sec3
CR
V L s
3
when V is in m/sec orCR
V L s 3
3
6.3 when V is in km/h
2. Rate of attainment of super elevation.
%5.22
11 WxY
max1
2
1WxeY
Slope
Y Y L s
21
Where
W=Width of carriage way
2.5% is the transverse slope
Y1
Y2
Ls
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Example
Speed Max Slope (Longitudinal)
50 0.66%
85 0.50%
100 0.40%140 0.36%
For a two lane single carriage with lane width of 5.65m, cross fall on tangent of 2.5 percent
and super elevation on circular curve of 6 percent, determine Ls an R, V=85km/h and C=
0.45m/s3.
Solution
14125.02
025.065.52%5.2
2
11
x xWxY
339.02
06.065.52%6
2
12
x xWxY
m x
Slope
Y Y L s 05.96
5
100048025.0
%5.0
339.014125.021
CR
V L s 3
3
6.3 hence
m
CL
V R
s
54.30405.9645.06.3
85
6.3 3
3
3
3
VERTICAL ALIGNMENT
Vertical curves are provided whenever there is change in gradient. The curves contributes to
safety, visibility and comfort .There are two types of vertical curves, namely
- Crest or summit vertical curves
- Sag vertical curves.
1. CREST VERTICAL CURVES
₵
Eq%p%
y
xBVC EVC
L
a800
GL E for X
2
L and G=p-q
At the location of crest or sag
G
Lp
q p
Lp x
G Lp
q p
Lp y
200200
22
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The forms generally adopted for vertical curves are a circle or a simple parabola. The latter is
however preferred and adopted in profile alignment design because of ease at which can be
set out as well as enabling the comfortable transition from one grade to another.
The above equations have been derived from parabolic curves.
The rate of change of slope in a simple parabola is constant i.e k xd
yd
2
2
1ckxdx
dy
But
pdx
dy at x=0
And qdxdy at x=L
1c pdx
dy
pkLc Lk qdx
dy 1
pkLq
k L
pq
Then
p x L
pq
dx
dy
2
2
2c px
x
L
pq y
At X=0, y=0, hence c2=0
Therefore px x
L
pq y
2
2
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LENGTH OF VERTICAL CURVES
The length of vertical curve may be determined on the bases of
1. Safety considerations - sight distance e.g SSD
2.
Driver comfort consideration depending on the rate of change of radial acceleration
3. Aesthetic appearance.
4. Drainage requirement i.e systems should be designed such that effective drainage system is
provided.
A. SAFETY CONSIDERATIONS
The decision on whether to provide condition for L > S or L < S depends on the slope of
the two straight gradients as well as the driver ’s eye height, h1, and object height, h2, (where
S = SSD)
Generally, if
800
S q p >h1, then design for
L > S where S depends on design speed.
If L > S
221
2
min
200 hh
GS L
Where G is in percentage
h1 =1.05 m
h2 = 0.26m
If L < S i.e
800
S q p
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Radial acceleration, C=v R
V 2 where R v=
G
L100
R v=Equivalent Radius of the vertical curves
Substituting this value in C, C= L
GV
100
2
for V in m/s or C= L
GV
1300
2
for V in km/h
Therefore given values of C, G and V, L may be calculated
C. AESTHETIC CONDITIONS
L≥0.5V for V in km/h
D. DRAINAGE REQUIREMENTS
Slope taken at 50fts on either side of the road from the crest should be greater than 0.5%
GRADIENTS
Long steep up hill grades have considerable effects on speeds of heavy commercial vehicles.
They reduce their speed causing difficulty in overtaking by small vehicles and subsequently
affect operating conditions along a given road section.
Climbing lanes/Creeper lanes
Where longitudinal gradients are long enough and/or steep enough to cause significant
increases in the speed differences between cars and heavy commercial vehicles, both traffic
safety and road capacity may be adversely affected. Climbing lanes are then introduced at
points where speeds fall by certain levels for a given road class. The lane is terminated when
>0.5% >0.5%
50fts 50fts
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the gain in speed reaches the original level at point of introduction. Determination should
ensure that no traffic hazard is created i.e passing sight distance should be adequate.
The introduction and termination of a climbing lane should be affected by tapers of length
60m and should not be considered as part of the climbing lane. The width of the climbinglane shall be equal to that of the adjacent reduced single lane so as to give three traffic lanes
of equal width.
Example
A -3% grade is being joined to a -5% grade by means of a parabolic curve of length 1200m.
Calculate the vertical offset at the point of intersection of the tangent.
2
L x
pX x
L
pq X
2
2
y
pL
E 2
m x
x
x E 3600
100
3
2
600
1200100
35
100
6003
-3%
-5%
E
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VERTICAL SAG CURVES
The widely accepted criteria for determining the minimum length of vertical sag culves
includes
i. Safety criteria – the minimum vertical clearance e.g when passing under a bridge
ii.
Comfort criteria/Consideration
iii. Head light sight distance – mainly for safety at night
iv. Drainage control
v. General aesthetics
i. Comfort criteria
This is a faction of vertical radial acceleration where C is given by
v R
V C
2
and100
G R L v
C
V R L v
1300
2
for V in km/h
pq
G
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ii. Vertical clearance
The absolute minimum stopping sight distance is the factor controlling the criteria
When L > S i.e
800
S q p >h1, then
2800 21
2
minhh
GS L
Where =Vertical clearance to edge of structure of obstruction eg bridge, underpass
h1 =1.05 m (driver ’s eye height)
h2 = 0.26m (object height)
When L < S i.e
800
S q p
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Example
Determine the length of the sag curve for the absolute and desirable cases
Solution
Absolute case
When L > S i.e 800
S q p >h1, then
2800 21
2
minhh
GS L
4.1
800
16043
>1.05 Then
2800
21
2
minhh
GS L =
m24.58
2
26.005.1
5.4800
16043 2
Desirable case
2800 21
2
minhh
GS L =
m19.105
2
26.005.15.4800
21543 2
p
q
G
P=-3%
q=+4%
=4.5m
S=160m (Absolute Min SSD for VD=85)
S=215m (Desirable Min SSD for VD=85)