Transportation Engineering-I

Transportation Engineering I
Faruque Abdullah
Lecturer
Dept. of Civil Engineering
Dhaka International University

Assignment
1. Discuss about the role of transportation.
2. Discuss role of transportation in rural development.
3. Write down the different modes of transportation.
4. Write down the advantages and disadvantages of road
transportation.
5. Discuss about the characteristics of road transport.

Introduction
 Role of Transportation:
 Transportation contributes to the economic, industrial, social and cultural
development of any country.
 Transportation is vital for the economic development of any region, since
every commodity produced whether it is food, clothing, industrial
products or medicine needs transportation at all stages from production to
distribution.
 In the production stage, transportation is required for carrying raw
materials like seeds, manure, coal, steel, etc.

 In the distribution stage, transportation is required from the production
centers, farms and factories to the marketing centers and later to the
retailers and the consumers for distribution.
 The inadequate transportation facilities retard the process of socio-
economic development of the country.
 The adequacy of transportation system of a country indicates its economic
and social development.

 Role of Transportation in rural development:
 Over 75% of the population of the country living in the villages.
 The development in urban countries alone don’t indicate the overall
development of the country.
 Only with the improvement in transportation facilities in rural areas, there
could be faster development of the rural centers.
 The fertilizer and other inputs for agriculture and cottage industries could
reach the rural population easily.
 Products can be sold at the nearest marketing centers for more
remunerative price.

 Faster economic growth and decreased wastage.
 With improved facilities for education, health care and other social needs
in the villages, the urge for the migration to urban centers decreases.
 Balance development of the country as a whole.

 Modes of transportation:
 Land transportation
 Roadways or Highways
 Railways
 Water transportation
 Air transportation
 Other modes of transportation
 Pipe lines
 Elevators
 Cable cars
 Monorails
 Belt conveyors

 Advantages & disadvantages of road transport:
Less investment.
Door to door service.
Service in rural areas.
Flexible service.
Suitable for short distance.
Lesser risk of damage in
transit.
Saving packing cost.
Seasonal nature.
Accident & breakdown.
Unsuitable for long distance
and bulking traffic.
Slow speed.
Lack of organization.
Advantages Disadvantages

 Characteristics of road transport:
⁕ Roads are used by various types of road vehicles, but railway tracks are
used only by rail locomotives and wagons, waterways are used by only
ships and boats.
⁕ Road transports requires a relatively small investment for the government.
⁕ Motor vehicles are much cheaper than other carriers.
⁕ Road transport offer a complete freedom to road users to transfer vehicle
from one lane to another and from one road to another according to the
need and convenience.
⁕ In particular, for short distance travel, road transport saves time.
⁕ Road transport is the only means of transport that offers itself to the whole
community alike.

Highway Development & Planning
Historical development of road construction:
 History of highway engineering gives us an idea abut roads of
ancient times.
 Roads in Rome were constructed in a large scale and it radiated
in many directions helping them in military operations.
 Thus they are considered to be pioneer in road construction.

Ancient roads:
 Most primitive mode of transport was by foot.
 These human pathways would have been developed for
specific purposes leading to camp sites, food, streams for
drinking water etc.
 Invention of wheel in Mesopotamian civilization lead to
development of animal drawn vehicles.
 To provide adequate strength to carry wheels, new ways tended
to folly sunny drier side of path.

 After invention of wheel animal drawn vehicle were developed
and need for hard surface road emerged.
 Traces of such hard roads were developed from carious ancient
civilization dated as old as 3500 BC.
 Earliest authentic record of road was found from Assyrian
empire constructed about 1900 BC.

Necessity of highway planning:
 In present era planning is considered as pre-requisite before
attempting any development program.
 Particularly true for any engineering work as planning is basic
requirement for any new project or an expansive program.
 Highway planning is also a basic need for highway
development.
 Planning is of great importance when funds available are
limited where as total requirement is much higher.
 Planning helps best utilization of available funds.

 Objects of highway planning:
 To plan a road network for efficient and safe traffic operation but at
minimum cost.
 To achieve at road system and length of different categories of roads
which could provide maximum utility and could be constructed with in
the available resources during plan period under consideration.
 To fix up date wise priorities for development of each road link based on
utility as main criteria for phasing road development program.
 To plan for future requirements and improvements of road in view of
anticipated development.
 To workout financing system.

 Classification of roads based on location and function:
 Expressways:
Separate class of highways with superior facilities, design standards and
high speed ( 120 Km/hr)
Should permit only fast moving vehicle.
Parking, loading and unloading of goods and pedestrian crossing is not
allowed.
 National Highways (NH):
=> Main highways running through a country connecting major ports, foreign
highways, capitals of large states and large industrial area.

 State Highway(SH):
Arterial roads of a state, connecting up with the national highways of
adjacent states, district and headquarters with in the state.
Also serve as main arteries for traffic to and from adjacent roads.
 Major District Roads (MDR):
Important roads with in a district serving areas of production and
connecting these with each other or with the main highways of a district.
Has lower speed and geometric design specifications than NH/SH.
 Other District Roads (ODR):
Roads serving rural areas of production and providing them with outlet to
market and centers.
 Village Roads (VR)

Problem-1: Calculate the additional length of metaled roads for the
following road system.
i. Total area = 9600 𝐾𝑚2
ii. Agricultural area = 3200 𝐾𝑚2
iii. Existing Railway track = 105 Km
iv. Existing length of metaled road = 322 Km
Population > 5000 2001-5000
Number of Villages and towns 8 40
Solution:
A = 3200 𝐾𝑚2
B= 9600-3200 = 6400 𝐾𝑚2
N = 40, T = 8, D = 15% of RL, R = 105 Km

Total length of metaled road = [
𝐴
8
+
𝐵
32
+ 1.6 N + 8T] + D –R
= [
3200
8
+
6400
32
+ 1.6 x 40 + 8 x 8] + D -105
= 728 + 15 % of 728 – 105
= 732.2 Km.
Additional metaled road needed = 732.2 – 322 = 410.2 Km.

Road Pattern
 Rectangular or Block pattern
 Whole area is divided into rectangular blocks of plots with streets
intersecting at right angles.
 Main roads which passing through centre of area should be sufficiently
wide and other branch roads may be comparatively narrow.
 May be further divided into small rectangular blocks for construction of
building placed back to back adding road on their front.
 Construction and maintenance is easier.
 But from traffic point of view this pattern is not very much convenient
because of number of intersections present here.

 Hexagonal pattern
 Radial/ Star/Block pattern

 Radial/ Star/ Circular pattern
 Radial/ Star/ Grid pattern

Transportation System
 Traffic: Traffic means movement of persons (people) and vehicles (on
road)
 Transport: Transport means movement of persons and goods from one
place to another place.
 Transportation: Transportation means the entire activity involving traffic
and transport.

The land use and transport feedback cycle
Transport
system
Activities
Accessibility
Land use

Traffic Characteristics
 Road user characteristics:
1. Physical characteristics.
2. Mental characteristics.
3. Psychological characteristics.
4. Environmental characteristics.
 Vehicular characteristics:
1. Width 6. Speed
2. Length 7. Power
3. Height 8. Braking system
4. Weight 8. Lighting system

 Road characteristics:
1. Friction
2. Light reflection characteristics.
3. Unevenness.
4. Surface slope.

Traffic Volume Study
 Traffic volume: Traffic volume is the number of vehicles that pass a given
line of road per unit time at any selected period. It is expressed in vehicles
per hour.
 PCU: A passenger car is equal to 1 PCU then if 100 cars can play on a
road then it is 100 PCU. If a bus is introduced on the same road, then the
number of cars may reduce to 94, so 100-94 = 6 PCU. Thus 6 PCU is
equivalent to 1 bus. The number of cars a vehicle can substitute is known
as PCU of that vehicle.

Traffic flow Characteristics
 At road intersection, traffic maneuvers are,
1. Diverging:
2. Merging:

Values of PCU
Type of Vehicle PCU Values
Car 1
Cycle/Motor Cycle 0.5
Auto Rickshaw 1
Tempo/CNG 1
Truck 4.5
Bus 3
Bullock Cart 6

Problem-2: Calculate the PCU/hr for a road using following data:
Solution:
PCU/hr = 8 x 0.5 + 6 x 4.5 + 19 x 1 + 26 x 1 + 3 x 6 + 9 x 1 = 103 PCU/hr
Vehicles Type Cycle Truck Car CNG BUS Bullock Cart Auto rickshaw
Number of Vehicle 8 6 19 26 17 3 9
Type of Vehicle PCU Values
Car 1
Cycle/Motor Cycle 0.5
Auto Rickshaw 1
Tempo/CNG 1
Truck 4.5
Bus 3
Bullock Cart 6

Problem-3: Calculate the PHF and also the PCU/hr for the PHF for a
stream on certain road using following data:
Time Cycle Truck Car CNG BUS
8:00 8 8 19 23 11
8:10 9 7 18 24 16
8:20 12 5 22 26 17
8:30 7 3 21 28 19
8:40 5 2 20 20 18
8:50 6 2 17 16 15
9:00 7 3 15 18 14
9:10 5 1 18 18 15
9:20 9 3 22 19 19
9:30 13 1 24 22 22
9:40 8 2 27 21 11

Solution:
Time Cycle
(A)
Truck
(B)
Car
(C)
CNG
(D)
BUS
(E)
PCU =
0.5A+4.5B+C+D+3E
8:00 8 8 19 23 11 115
8:10 9 7 18 24 16 126
8:20 12 5 22 26 17 127.5
8:30 7 3 21 28 19 123
8:40 5 2 20 20 18 105.5
8:50 6 2 17 16 15 90
9:00 7 3 15 18 14 92
9:10 5 1 18 18 15 88
9:20 9 3 22 19 19 116
9:30 13 1 24 22 22 123
9:40 8 2 27 21 11 94

Total six maximum 10 min interval PCU summation = 127.5 + 126 + 123 +
123 + 116 + 115
= 730.5 PCU
Peak 10 min interval PCU = 127.5 PCU
 PHF =
730.5
6 𝑋 127.5
= 0.95
 PCU/hr = 730.5 / 0.95 = 769 PCU/hr

TMS (Time Mean Speed):
Time mean speed (TMS) is defined as the average speed of all vehicles passing
a point on a highway over a specified time period.
 It is the simple average of spot speed.
 Expression for 𝑣 𝑡 =
1
𝑛
 𝑣𝑖 is the spot speed of ith vehicle.
 n is the number of observation.
𝑖=0
𝑛
𝑣𝑖

SMS (Space Mean Speed):
Space mean speed (SMS) is defined as the average speed of all vehicles
occupying a given section of a highway over a specified time period.
 Consider unit length of a road and 𝑣𝑖 is the spot speed of ith vehicle.
 Let 𝑡𝑖 is the time taken to complete unit distance, 𝑡𝑖 =
1
𝑣 𝑖
 If there are n such vehicles, then the average travel time 𝑡 𝑠 is given by,
𝑡 𝑠 =
∑𝑡 𝑖
𝑛
=
1
𝑛
∑
1
𝑣 𝑖
 If average travel time is 𝑡 𝑠 then average speed 𝑣𝑠 is 1
𝑡 𝑠
.
𝑣𝑠 =
1
𝑡 𝑠
=
𝑛
∑ 𝑖=0
𝑛 1
𝑣 𝑖

Problem-4: If the spot speeds are 50, 40, 60, 54 and 45 m/s then
calculate the SMS (Space Mean Speed) & TMS (Time Mean Speed).
Solution: TMS (𝑣 𝑡) is the average of spot speed. Therefore,
𝑣 𝑡 =
∑𝑣 𝑖
𝑛
=
50+40+60+54+45
5
= 49.8 m/s
SMS (𝑣𝑠) is the harmonic mean of spot speed. Therefore,
𝑣𝑠 =
𝑛
∑
1
𝑣 𝑖
=
5
1
50
+
1
40
+
1
60
+
1
54
+
1
45
=
5
0.12
= 48.82 m/s

Problem-5: If SMS is 20.38 m/s and 𝜎2 is 138.727 for a stream in a
certain road of 88 vehicle. Then find out the TMS and density of the
stream.
Solution: TMS, 𝑣 𝑡 = 𝑣𝑠 +
𝜎2
𝑣 𝑠
= 20.38 +
138.727
20.38
= 27.184 m/s
Density, k =
𝑞
𝑣 𝑠
=
88
20.38
= 4.3 vehicle/km

Parking Study
Out of 8760 hours in a year, the car runs on an average for only 400 hours,
remaining 8360 hours when it is parked. As the city grows, the use of land
increases considerably creating problem for space for parking purposes. The
demand of parking space is one of the major problems in big cities. Vehicles
if improperly parked cause lot of problems like-traffic congestions, accidents,
obstruction to fire-fighting operation, environmental pollution, etc.

Parking System
On street (curb) parking: Parked along the curb which is very convenient for
peoples but causes of traffic jams.
Common modes of on-street parking are (i) Parallel parking (ii) Angle
parking (iii) Right angle parking.
To ensure safety and convenience, it is desirable to prohibit parking at
following locations:
(i) Near intersections (ii) Near level crossing (iii) Pedestrian crossing (iv)
Narrow streets (v) Structures, etc.

Off-street parking: When parking places are provided away from the road
curb.
Methods of off-street parking are (i) Surface cur parks (ii) Multi-storey car
parks (iii) Roof parks (iv) Mechanical car parks (v) Underground car parks.
Peripheral parking schemes: Parking facilities are provided at the periphery
of the town. They are: (i) Park and walk (ii) Park and ride (iii) Good-bye
(kiss) and ride.

Road Accident
An accident which occurred or originated on a road open to public traffic
resulting in either injury or loss of life or damage to property, in which at
least one moving vehicle was involved.

Causes of road accidents:
 Road factors:
1. Design speed
2. Control of access
3. Pavement width
4. Pavement surface
5. Shoulders
6. Horizontal curve
7. Vertical alignment
8. Median width
9. Sight distance
10. Intersections, etc.
 Environment factors: Snow, ice, rain water, fog, smoke, heavy rainfall, etc.
Road Accident Causes

 Vehicle factors:
1. Vehicle body and it’s features
2. Tyers
3. Braking System
4. Lighting system
5. Vehicle inspection and maintenance
6. Mixed traffic
7. Overloading, etc.
 Road user factors:
1. Age of driver
2. Gender of driver
3. Violation of traffic laws
4. Competition with other vehicle or motor cycle.
5. Talking with pillion rider.
 Others factors: Ribbon development, advertisement boards, trees,
attractive views, material on road, etc.

Motor vehicle moving ahead
Motor vehicle backing
Pedestrian
Parked vehicle
Fixed object
Rear end collision
Side swipe
Out of control
Fatal accident
Personal injury
Property damage only
Head on collision
Right-angle collision
Collision Diagram & Symbol

Traffic Signs
 Warning Signs (Danger signs)
 Regulatory Signs (Mandatory signs)
 Informatory Signs.

Traffic Signals
 Traffic control signal
1. Manually operated signals
2. Fixed-time signals
3. Traffic actuated signals
 Pedestrian signals

Pavement Marking
 Centre lines: These broken line are meant to separate the opposite streams of
traffic on undivided two-way roads.
 Route direction arrows: It is used to guide effectively the
traffic in the correct direction.

Pavement Marking
 Stop lines: These are solid white lines provided transversely to the carriageway
and used to indicate the point behind which vehicles are required to stop.
 Cross walk line: Pedestrian crossing (2m wide) are very important for their
safety.

Pavement Marking
 Bus stops: Kerb length reserved for buses to stop are marked by continuous
yellow line on kerb indicating “ parking prohibited” or marked by the word
“BUS STOP”.

Traffic Island
 Divisional Island
 Channelizing Island
 Central Island

 Unchanneled Intersection: There is no provision of any direction island or
central island.
Tee Box
Road Intersections

 Intersection at grade or level: These include all roads which meat at more or less
the same level.
Tee Wye

 Channelized Intersection: It is achieved by introducing island into the
intersection area, thus reducing the total conflict area available in the
unchannelezed intersection.
Advantages:
1. Vehicles can be confined to definite paths.
2. Points of conflicts can be separated.
3. Refuse island can be provided for pedestrians.
4. Angle of merging kept minimum.

 Rotary Intersection: By provision of rotary, necessity of stopping at road
intersection is eliminated. Traffic from all converging roads keeps on moving
round the central island in clockwise direction and adopts its road of desire to
move out of rotary without stopping any where. On, rotary crossing conflicts are
completely eliminated.

1. Traffic holding capacity is the
highest.
2. It is more safe.
3. All vehicle get equal preference.
4. Operational cost of vehicle is less.
5. No need of traffic police or signal
6. Number of accidents are low.
1. Requires large area of land and
hence it is costly in build up areas.
2. If large number of pedestrians and
cyclist are involved, design and
operation of rotary becomes
complicated.
3. If the vehicular traffic have to stop
to allow pedestrian to cross, the
main purpose of rotary is defeated.

 Grade Separated Intersection: The intersecting roads are separated by
passing one road over or below the other and eliminating the crossing
manoeuvers.

1. Maximum facility is given to the
crossing traffic.
2. There is increased safety for turning
traffic.
3. They provide comfort and
convenience to the driver and
saving the time & operational cost.
4. It can be designed for any angle of
intersection.
1. It involves very large areas.
2. It involves lot of expenditure in
providing bridges under passes and
interchange ramps.
3. Unnecessary rising grades and sags
are introduced in vertical alignment.

 Over pass: When the major highway is taken above by raising its profile above
the general ground level by embankment and an over-bridge across another
highway.
 Under pass: When the highway is taken by depressing it below the ground level
to cross another road by means of an under-bridge.

 By-pass: It is an arrangement for diverting a traffic flow to avoid unnecessary
hazards.
 Fly-over: It is a bridge over a large road intersection for safe traffic flow
avoiding conflict.

 Some well-known traffic management measures are:
1. Restriction on right-turning.
2. One-way streets.
3. Tidal-flow operations.
4. Exclusive Bus-lanes.
5. Closing side-streets (Frontage road).
6. Traffic Aids:
a) Roadway Delineators: These are intended at turning point to provide
visual aids to drivers at night.
b) Hazard Markers: These are intended to define obstructions like guardrails
& abutment at road sides.

c) Object markers: These are used to indicate hazards and obstructions.
d) Speed breaker
e) Rumple strip
f) Guard rails
g) Safety barrier
h) Barrier and Channelizers
i) Traffic Attenuators, etc.

Highway Alignment & Surveys
 Alignment:
 Marking position or layout of central line on
ground and giving direction to highway is
called alignment.
1. Horizontal alignment
2. Vertical alignment
 Once an alignment is fixed and constructed,
it is not easy to change it due to increase in
cost adjoining land and construction of costly
structures by the roadside.

Requirements of an ideal alignment
 Short:
 Alignment between two terminal stations should be short and as far as
possible be straight.
 But due to some practical considerations deviations may be needed.
 Easy:
 Alignment should be easy to construct and maintain.
 It should be easy for operation of vehicles.
 Maximum extend easy gradient and curves should be provided.

 Safe:
 It should be safe both from construction and operating point of view
specially at slopes, embankments and cutting.
 It should have safe geometric features.
 Economical:
 Alignment should be economical.
 It can be considered so only when initial cost, maintenance cost and
operating cost is minimum,

Factors Controlling Alignment
 Obligatory points:
 These are control points governing highway alignment.
 These points are classified into two categories.
Points through which it should pass:
1. Bridge site:
 Bridge can be located only where river has straight and permanent path
and also where abutment and pier can be strongly funded.
 Road approach to bridge should not be curved and skew crossing should
be avoided or possible.
 Thus to locate a bridge highway alignment may be changed.

2. Mountain:
 While alignment passes through a mountain, various alternatives are to
either construct a tunnel or to go round hills.
 Suitability of alternative depends on factors like topography, site
conditions and construction and operation cost.
3. Intermediate town:
 Alignment may be slightly deviated to connect an intermediate town or
village nearby.

Points through which it should not pass:
1. Religious places:
 These have been protected by law from being acquired for any purpose.
 Therefore, these points should be avoided while aligning.
2. Very costly structure:
 Acquiring such structures means heavy compensation which would result
in an increase in initial cost.
 So, alignment may be deviated not to pass through that point.
3. Lakes/ ponds:
 Presence of lake or pond on alignment path would be also necessitate
deviation of alignment.

 Traffic:
 Alignment should suit traffic requirement.
 New alignment should be drawn keeping in view desire lines, traffic
pattern, etc.
 Geometric design:
 Geometric design factors such as gradient, radius of curve, sight distance,
etc. also governs alignment of highway.
 To keep radius of minimum, it may be required to change alignment of
highway.

 Economics:
 Alignment finalized based on above factors should be economical.
 Need to avoid high embankment and deep cutting to decrease initial cost.
 Alignment is chosen in a manner to balance cutting and filling.
 Other consideration:
 Drainage considerations, hydrological factors, political considerations and
monotony govern alignment.
 Subsurface water level, seepage flow and high flood level are factors to
be kept in view.

Elements of geometric design:
1. Cross-sectional elements – includes cross slope, various width of road
and features in road margins.
2. Sight distance consideration – includes cross slope, various width and
features in the road margins.
3. Horizontal alignment detail’s – includes features like super elevation,
transition curve, extra widening and set back distance, etc.
4. Vertical alignment distance and design of length of curve.
5. Intersection features – includes layout, capacity, etc.

Factors Affecting Geometric Design
 Design speed.
 Topography.
 Traffic factors.
 Vehicles.
 Human.
 Design hourly volume and capacity.
 Environment.
 Economy.
 Others factors.

Skid & Slip ( )
 Skid:
 Occurs when wheels slide without revolving or when wheels partially
revolves.
 Skidding happens when path travelled along road surface is more than
circumferential movement of wheel due to their rotation.
 When brakes are applied, wheels are locked partially or fully and if
vehicles moves forward, longitudinal skid takes places.
 May vary from 0% to 100%
*****

 Slip:
 Takes place in a horizontal curve, if centrifugal force is greater than
counteracting force.
 Considered dangerous as vehicle goes out of control leading to an
accident.
 Vehicles travel less distance than the wheel revolves.

 Camber:
 Slope provided to road surface in transverse direction to drain off rain water from road
surface.
 Usually camber is provided on straight roads by raising centre of carriage way with
respect to edges, forming a crown or highest point on center line.
 It is also expressed in %.
 Steep camber is provided in areas of heavy rainfall.
 Types: 1. Parabolic; 2. Straight; 3. Combination of straight & parabolic.
 Carriageway:
 Metaled portion of road over which vehicles move.
 Different types of carriageway according to material which used for construction.

 Traffic Lane:
 Carriage intended for one line of traffic movement is called a traffic lane.
 Kerb:
 Indicate boundary between carriageway and shoulder island or footpath.
 Different types of kerb:
1. Low or mountable kerb (10 cm height)..
2. Semi barrier type kerb (15 cm height)..
3. Barrier type kerb (20 cm height).
4. Submerged kerb (Rural areas to provide lateral stability).

 Shoulder:
 Provided along road edge and are intended for accommodation of stooped
vehicles.
 Serve as an emergency lane for vehicles.
 Provide lateral support for base and surface course.
 It is desirable to have a width of 4.6 m for shoulder.
 A minimum width of 2.5 m is recommended for 2 lane rural highways.

 Right of Way (ROW):
 ROW or land width is width of land acquired for road, along its alignment.
 It should be adequate to accommodate all cross-sectional elements of highway
and many reasonably provided for future developments.
 To prevent ribbon developments along highways, control lies and building
lines may be provided.
 Control line is a line which represents nearest limits of future uncontrolled
building activity in relation to a road.
 Building line represents a line on either side of the road between which and
road no building is permitted at all.

 Sight distance:
 Safe and efficient operation of vehicle on roads depends among factors or
road length at which an obstruction, if any becomes visible to driver
indirection of travel.
 In other words feasibility to see ahead or visibility is very important for safe
vehicle operation on a highway.
 Sight distance available from a point is actual distance along road surface,
which a driver from specified height above carriageway has visibility of
stationary or moving objects.
 In other words, sight distance is length of road visible ahead to driver at any
distance.

 Stopping Sight Distance (SSD):
 SSD is defined as distance needed for drivers to see an object on roadway
ahead and bring their vehicles to safe stop before colliding with an object.
 Stopping Distance is the summation of lag distance (distance travelled by
vehicle during total reaction time) and braking distance (distance travelled by
vehicle after application of brake to a dead stop position).
 SD = Vt +
𝑉2
2𝑔𝑓
[Here, V in m/sec & g = 9.81 m/𝑠𝑒𝑐2
]
 SD = 0.278 Vt +
𝑉2
254𝑓
[Here, V in Km/hr]
Friction, f = 0.4~0.35 depending upon speed from 30 Kmph to 80 Kmph

 Stopping Sight Distance (SSD) in slope:
 SD for ascending gradient,
SD = vt +
𝑉2
2𝑔(𝑓+0.01𝑛)
[Here, V in m/sec & n in %]
SD = 0.278 Vt +
𝑉2
254(𝑓+0.01𝑛)
[Here, V in Km/hr & n in %]
 SD for descending gradient,
SD = vt +
𝑉2
2𝑔(𝑓−0.01𝑛)
[Here, V in m/sec & n in %]
SD = 0.278 Vt +
𝑉2
254(𝑓−0.01𝑛)
[Here, V in Km/hr & n in %]
 Minimum SSD for one-way traffic lane is equal to SD and for two-way traffic
lanes it is double of SD.

Problem-6: Calculate the safe SSD for design speed of 50 Kmph for (a) Two-way
traffic on a two lane road (b) Two-way traffic on a single plane road. Assume
coefficient of friction as 0.37 and reaction time of driver as 2.5 sec.
Solution: v = 50 Kmph =
50
3.6
m/sec = 13.9 m/sec
Stopping Distance, SD = vt +
𝑣2
2𝑔𝑓
= 13.9 x 2.5 +
13.92
2 𝑥 9.81 𝑥 0.37
= 61.4 m
(a) SSD for two-way traffic on a two lane road = SD = 61.4 m
(b) SSD for two-way traffic on a single lane = 2 x SD = 2 x 61.4 = 122.8 m

Problem-7: Calculate the safe SSD for design speed of 80 Kmph on a highway at a
descending gradient of 2% for f = 0.35.
80
3.6
m/sec = 22.22 m/sec
Assume, t = 2.5
𝑣2
2𝑔(𝑓−0.01𝑛)
= 22.22 x 2.5 +
22.222
2 𝑥 9.81 𝑥 (0.35−0.01 𝑥 2)
= 131.8 m.

Exercise-1: Calculate the safe SSD for design speed of 70 Kmph on a highway at a
ascending gradient of 1% for f = 0.37.
70
3.6
m/sec = 19.44 m/sec
Assume, t = 2.5
𝑣2
2𝑔(𝑓+0.01𝑛)
= 19.44 x 2.5 +
19.442
2 𝑥 9.81 𝑥 (0.37+0.01 𝑥 1)
= 99.29 m.

Exercise-2: Calculate the safe SSD for design speed of 65 Kmph on a highway at a
descending gradient of 3% for f = 0.4.
65
3.6
m/sec = 18.06 m/sec
Assume, t = 2.5
𝑣2
2𝑔(𝑓−0.01𝑛)
= 18.06 x 2.5 +
18.062
2 𝑥 9.81 𝑥 (0.4−0.01 𝑥 3)
= 90.08 m.

 PIEV Theory:
According to this theory total reaction time of driver is split into four parts-
 Perception:
 It is the time required for sensation received by eyes or ears to be transmitted
to brain through nervous system and spinal chord.
 In other words, it is time required to perceive an object or situation.
 Intellection:
 It is the time required for understanding situation.
 It is also time required for comparing different thoughts, regrouping and
registering new sensation.

 Emotion:
 It is time elapsed during emotional sensation and disturbance such as fear,
anger or other emotional feelings such as superstition, etc. to situation.
 Emotional time of a driver is likely to vary considerably during upon
problems involved.
 Volition:
 It is the time taken for final action.
 It is also possible that driver may apply brakes or take any avoiding action by
reflux action, even without thinking.

 Overtaking Sight Distance (OSD):
 OSD is the minimum distance open to vision of driver of a vehicle intending
to overtake slow vehicle ahead safely against traffic in the opposite direction.
 OSD is measured along the centre line.

 OSD = 𝑑1 + 𝑑2 + 𝑑3 = 𝑑1 + (S + b + S) + 𝑑3
 𝑑1 = 𝑣 𝑏t [t is the reaction time & 𝑣 𝑏is the velocity of overtaken vehicle (m/s)]
 S = (0.7 𝑣 𝑏 + 6) m
 b = 𝑣 𝑏T [ T =
4𝑆
𝑎
[a is the acceleration in m/𝑠𝑒𝑐2]
 𝑑3 = vT [v is the design velocity of vehicle moving from opposite direction (m/s)]
𝑣 𝑏
v

 OSD = 𝑑1 + 𝑑2 + 𝑑3 = 𝑑1 + (S + b + S) + 𝑑3
= 𝑣 𝑏t + 2 (0.7 𝑣 𝑏 + 6) + 𝑣 𝑏T + vT [𝑣 𝑏 & v in m/sec]
=0.278𝑉𝑏t + 2 (0.7 𝑉𝑏 + 6) + 0.278𝑉𝑏T + 0.278VT [𝑉𝑏 & V in Km/hr]
In case of speed of overtaken vehicle 𝑉𝑏 is not given, then assume (V-16) Kmph.
Where, V is design speed in Kmph. Assume, 𝑣 𝑏 = (v - 4.5) m/sec and v is design
speed in m/sec.
 For one-way traffic, OSD = 𝑑1 + 𝑑2
 For two-way traffic, OSD = 𝑑1 + 𝑑2 + 𝑑3
 Over taking zone = 3 OSD to 5 OSD
 𝑆𝑃1 = Sign Post “ Overtaking zone ahead”
 𝑆𝑃2 = Sign Post “ End of overtaking zone”

Problem-8: The speed of overtaking and overtaken vehicles are 70 & 40 Kmph
respectively on a two-way traffic road. If the acceleration of overtaking vehicle is
0.99 m/𝑠𝑒𝑐2, calculate and draw the overtaking zone.
Solution: v =
70
3.6
= 19.4 m/sec; 𝑣 𝑏 =
40
3.6
= 11.11 m/sec; a = 0.99 m/𝑠𝑒𝑐2
Assume, t = 2 sec.
𝑑1 = 𝑣 𝑏t = 11.1 x 2 = 22.2 m
S = (0.7 𝑣 𝑏 + 6) m = 0.7 x 11.11 + 6 = 13.8 m
T =
4𝑆
𝑎
=
4 𝑥 13.8
0.99
= 7.47 sec
b = 𝑣 𝑏T = 11.1 x 7.47 = 82.9 m
𝑑2 = 𝑣 𝑏T + 2S = 82.9 + 2 x 13.8 = 110.5 m
𝑑3 = vT = 19.4 x 7.47 = 144.9 m
OSD = 𝑑1 + 𝑑2 + 𝑑3
= 22.2 + 110.5 + 144.9
= 277.6 ≈ 278 m

Minimum length of overtaking zone = 3 OSD = 3 x 278 m = 834 m
Desirable length of overtaking zone = 5 OSD = 5 x 278 m = 1390 m
𝑆𝑃1
𝑆𝑃2
𝑆𝑃1
𝑆𝑃2
278 m
278 m
278 m
278 m
834 m to 1390 m
Fig. Overtaking Zone
𝑆𝑃1 = Sign Post “ Overtaking zone ahead”
𝑆𝑃2 = Sign Post “ End of overtaking zone”

 Superelevation:
 Superelevation or cant or banking is transverse slope provided at horizontal
curve to counteract centrifugal force by raising the outer edge, throughout
length of horizontal curve.
 When outer edge is raised, a component of curve weight will be
complemented in counteracting effect of centrifugal force.

 e + f =
𝑣2
𝑔𝑅
…………….(1)
e = Rate of superelevation
f = Co-efficient of lateral friction 0.15
v = Speed of vehicle, m/sec
R = Radius of curve, m and
g = 9.8 m/𝑠𝑒𝑐2
 If there is no friction due to some practical reasons, then f = 0 and eq (1) becomes,
e =
𝑣2
𝑔𝑅
=
𝑣2
127𝑅
This results in situation where pressure on outer and inner wheels are same, requiring
very high superelevation e.

 If there is no superelevation provided due to some practical reasons, then e = 0 and
eq (1) becomes,
f =
𝑣2
𝑔𝑅
=
𝑣2
127𝑅
This results in a very high co-efficient of friction.
 If e = 0 and f = 0.15 then for safe travelling speed from eq (1) is given by,
𝑣 𝑎 = 𝑓𝑔𝑅
Where, 𝑣 𝑎 is restricted speed.

 Design of Superelevation:
 Step-1: Find e for 75% of design speed neglecting f. i.e,
𝑒1 =
(0.75𝑣)2
𝑔𝑅
 Step-2: If 𝑒1 ≤ 0.07, then e = 𝑒1 =
(0.75𝑣)2
𝑔𝑅
else if, 𝑒1 > 0.07. Then e = 0.07 and go to step 3.
 Step-3: Find 𝑓1 for design speed and max e i.e,
𝑓1 =
𝑣2
𝑔𝑅
- e
If 𝑓1 < 0.15, then maximum e = 0.07 is safe, else go to step-4.

 Step-4: Find allowable speed 𝑣 𝑎 for maximum e = 0.07 and f = 0.15,
𝑣 𝑎 = 0.22𝑔𝑅 = 27.94 𝑅 (Kmph)
If 𝑣 𝑎 ≥ v, then design is adequate. Otherwise, use speed adopt control
measures or look for speed control measures.

Problem-9: Design the rate of superelevation for a horizontal highway curve of
radius 500 m and speed 100 Kmph.
Solution:
v =
100
3.6
= 27.78 m/sec
Step-1: 𝑒1 =
(0.75𝑣)2
𝑔𝑅
=
(0.75 𝑥 27.78)2
9.81 𝑥 500
= 0.089
Step-2: As 𝑒1 > 0.07 , then provide maximum superelevation, e = 0.07
Step-3: 𝑓1 =
𝑣2
𝑔𝑅
- e =
27.782
9.81 𝑥 500
- 0.07 = 0.087
As 𝑓1 < 0.15.
So, Design is safe with a superelevation of 0.07.

Problem-10: Design the rate of superelevation for a horizontal highway curve of
radius 200 m and speed 80 Kmph.
Solution:
v =
80
3.6
= 22.22 m/sec
Step-1: 𝑒1 =
(0.75𝑣)2
𝑔𝑅
=
(0.75 𝑥 22.22)2
9.81 𝑥 200
= 0.14
Step-2: As 𝑒1 > 0.07 , then provide maximum superelevation, e = 0.07
Step-3: 𝑓1 =
𝑣2
𝑔𝑅
- e =
22.222
9.81 𝑥 200
- 0.07 = 0.18
As 𝑓1 > 0.15, then provide maximum friction co-efficient, f = 0.15 and
speed should be restricted.

 Step-4: Allowable speed 𝑣 𝑎 for maximum e = 0.07 and f = 0.15,
𝑣 𝑎 = 27.94 𝑅 (Kmph)
= 27.94 𝑥 200 = 74.75 Kmph ≈ 70 Kmph
Hence, the speed may be restricted to 70 Kmph at this curve.

 Attainment of Superelevation:
May be split into two categories:
i ) Elimination of crown of cambered section.
1. Rotating outer edge about crown: Outer half of cross slope is rotated about
crown at a desired rate such that this surface falls on same plane as inner half.
Fig. Outer edge rotated about the crown
Levels of outer edge

2. Shifting position of crown: Also known as diagonal crown method. Position of
crown is progressively shifted outwards. Thus increasing width of inner half of
cross section progressively.
Positions of crown
Fig. Crown shifted outwards

ii) Rotation of pavement cross section to attain full superelevation.
1. Rotation about centre line: Pavement is rotated such that inner edge is
depressed and outer edge is raised both by half total amount of super elevation.
i.e. by
𝐸
2
w. r. to centre.
Inner edge
Outer edge
𝐸
2
𝐸
2
Fig. Rotating about centers

2. Rotation about the inner edge: Pavement rotated raising outer edge as well as
centre such that outer edge is raised by full amount of super elevation, E w. r. to
inner edge.
Inner edge Outer edge
𝐸
2
E
Fig. Rotating about the inner edge

Problem-11: An engineer checking the speed characteristics confidence level 95%,
acceptable limit 1.0 kmph and 130 spot variance 25 Kmph. Has the engineer meet
the entire requirement?
Solution:
We know,
Sample, N = (
𝑍𝜎
𝑑
)2
For 95% confidence level, Z = 1.96 and 𝜎2 = 25. So, σ = 5 and d = 1 Kmph.
N = (
1.96 𝑥 5
1
)2 = 96.04 > 95
∴ Yes, the engineer has meet the requirement.

Transportation Engineering-I

More Related Content

What's hot

Similar to Transportation Engineering-I

More from Rajshahi University of Engineering and Technology

Recently uploaded

Transportation Engineering-I