solved question papers unit 1 1. june july 09, dec 12 · pdf filetransportation engineering 1...

TRANSPORTATION ENGINEERING 1 10CV56

Dept of Civil Engineering, SJBIT

SOLVED QUESTION PAPERS

UNIT 1

1. Discuss the role of transportation in National development.

June July 09, Dec 12

Transportation is vital for the economic development of any region since every

Commodity produced whether it is food, clothing, industrial products or medicine

needs transport 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 centres viz; farms and factories to the marketing centres and later to

the retailers and the consumers for distribution

These above are some of the necessities which make us to use

transportation. The importance and adequacy of transportation system of a

country indicates its economic and social development.

Economic Activity: Two important factors well known in economic activity are:

Production or supply and

Consumption for human wants or demand.

Social Effects: The various social effects of transportation may be further

classified into:

Sectionalism and transportation

Concentration of population into urban area

Aspect of safety, law and order.



2. Compare road transportation with other modes of transportation.

June July 09, June July 11, Dec 12

Three basic modes of transport are by land, water and air. Land has given

development of road and rail transport. Water and air have developed waterways

and airways respectively. Apart from these major modes of transportation, other

modes include pipelines, elevators, belt conveyors, cable cars, aerial ropeways

and monorails. Pipe lines are used for the transportation of water, other fluids and

even solid particles

The four major modes of transportation are:

Roadways or highways

Railways

Airways

Waterways.

Airways:

The transportation by air is the fastest among the four modes.

Air also provides more comfort apart from saving in transportation time

for the

Passengers and the goods between the airports.

Waterways:

Transportation by water is the slowest among the four modes.

This mode needs minimum energy to haul load through unit distance

Railways:

The transportation along the railway track could be advantageous by

railways between the stations both for the passengers and goods,

particularly for longer distances.



The energy requirement to haul unit load through unit distance by the

railway is only a fraction (one fourth to one sixth) of the required by road.

Hence, full advantage of this mode of transportation should be taken for the

transportation of bulk goods along land where the railway facilities are

available.

Roadways:

The transportation by road is the only mode which could give maximum

service tone and all.

The road or highways not only include the modern highway system but

also the city streets, feeder roads and village roads, catering for a wide-

range of road vehicles and the pedestrians.

This mode has also maximum flexibility for travel with reference to route,

direction, time and speed of travel etc. through any mode of road vehicle.

3. What is Jayakar committee and what are its recommendations How is

in implemented (June July 09, June July 11, Dec2010 May June 2010,

Dec 12)

RECOMMENDATIONS: Over a period after the First World War, motor

vehicles using the roads increased and this demanded a better road network

which can carry mixed traffic conditions. The existing roads when not capable to

withstand the mixed traffic conditions. For the improvement of roads in India

government of India appointed Mr. Jayakar Committee to study the situations and

to recommend suitable measures for road improvement in 1927 and a report was

submitted in 1928 with following recommendations:



Road development in the country should be considered as a national

interest. As the provincial and local government do not have the financial

and technical capacity for road development.

Extra tax to be levied from the road users as fund to develop road.

A Semi-official technical body has to be formed to collect and pool

technical knowhow from various parts of the country and to act as an

advisory body on various aspects of the roads.

A research organization should be instituted at National level to carry out

research and development work and should be available for consultation

IMPLEMENTATIONS:

Majority of the recommendations were accepted by the government implemented

by Jayakar Committee. Some of the technical bodies were formed such as,

• Central Road Fund (CRF) in 1929

• Indian Road Congress (IRC) in 1934

• Central Road Research Institute (CRRI) in 1950.

4. List the characteristics of road transport ( June July 11, Dec 12)

It is accepted that the fact road transport is the nearest to the people. The

passengers and goods have to be first transported by road before reaching a

railway station or an airport. It is seen that road network alone could serve the

remotest villages of the vast country like occurs. The various characteristics

(advantages) and disadvantages of different mode of transport are briefly listed

here:



Roadways: Advantages:

Flexibility: It offers complete freedom to the road users.

It requires relatively smaller investments and cheaper in construction with

respect to other modes.

Disadvantages:

Speed is related to accidents and more accidents results due to higher

speed.

Not suitable for long distance travel

Railways

Advantages:

Can transport heavy loads of goods at higher speed

Power required per tone is less compared to roadways

Disadvantages

Entry and exist points are fixed

Requires controlling system and no freedom of movement

Waterways:

Advantages:

Cheapest: Cost per tonne is lowest

Possess highest load carrying capacity



Disadvantages:

Slow in operation and consumes more time

Depends on weather condition

5. Describe the scope and functions of transportation Dec 2010

The purpose of a transportation system is to coordinate the movement of

people, goods and vehicles in order to utilize routes most efficiently. When

implemented, Transportation systems seek to reduce transport costs and improve

delivery times through effective timetabling and route management. Periodic re-

evaluations and the development of alternative routes allow for timely changes to

the transportation system in order to maintain efficiency.

standard transportation system will usually feature multiple timetables designed

to inform the user of where each vehicle in the fleet is expected to be at any given

point in time. These timetables are developed alongside an array of route plans

designed to coordinate vehicle movements in a way that prevents bottlenecks in

any one location

6. Explain briefly the contribution of the following in road development in

India. Indian roads congress, Central road fund. Dec 09, Dec 12

Central Research Fund (CRF):

Central Research Fund (CRF) was formed on 1st March 1929

The consumers of petrol were charged an extra levy of 2.64 paisa/liter of

petrol to buildup this road development fund.

From the fund collected 20 percent of the annual revenue is to be retained

as meeting expenses on the administration of the road fund, road



experiments and research on road and bridge projects of special

importance.

The balance 80 percent of the fund to be allotted by the Central

Government to the various states based on actual petrol consumption or

revenue collected

The accounts of the CRF are maintained by the Accountant General of

Central Revenues.

Indian Road Congress (IRC):

It is a semi-official technical body formed in 1934. It was formed to

recommend standard specifications.

It was constituted to provide a forum of regular technical pooling of

experience and ideas on all matters affecting the planning, construction and

maintenance of roads in India.

IRC has played an important role in the formulation of the 20-year road

development plans in India.

Now, it has become an active body of national importance controlling

specifications, guidelines and other special publications on various aspect

of Highway Engineering.



UNIT 2

1. With neat sketches, explain different road patterns

June july 09, Dec 10

The various road patterns may be classified as follows:

Rectangular or block pattern: In this, entire area is divided into

rectangular segments having a common central business and marketing

area. This area has all the services located in the central place. This pattern

is not convenient or safe from traffic operation point of view and it results

into more number of accidents at intersections. Eg: Chandigarh city.

Radial or star and block pattern: In this, roads radially emerge from the

central business area in all directions and between two built-up area will be

there. The main advantage in this, central place is easy accessible from all

the directions. Eg: Nagpur

Radial or star and circular pattern: In this roads radiate in all the

directions and also circular ring roads are provided. Advantages: Traffic

will not touch the heart of the city and it flows radially and reaches the

other radial road and thereby reducing the congestion in the centre of the

city. This ring road system is well suited for big cities where traffic

problems are more in the heart of the city. Eg: Connaught place in New

Delhi.

Radial or star and grid pattern: It is very much similar to star and the

circular pattern expects the radial roads are connected by grids. In this

pattern a grid is formed around the central point which is a business centre.

Eg: Nagpur road plan.



Hexagonal pattern: In this entire zone of planning is divided into

hexagonal zones having separate marketing zone and central services

surrounded by hexagonal pattern of roads. Each hexagonal element is

independent. At each corner of hexagon three roads meet

2. What is saturation system of road planning? How is it used to decide the

best proposed among several alternative proposals? June July 09

In this system optimum road length is calculated for an area based on the concept

of attaining maximum utility per unit length of the road. This is also called as

maximum utility system.

Factors to attain maximum utility per unit length are:

Population served by the road network

Productivity (industrial and agricultural) served by the road network.

The various steps to be taken to obtain maximum utility per unit length are:

Population factors or units: Since, the area under consideration consists of

villages and towns with different population these are grouped into some

convenient population range and some reasoning values of utility units to

each range of population serve are assigned.

Population less than 500, utility unit = 0.25

501 to 1001, utility unit = 0.50

Productivity Factors or units: The total agricultural and industrial products

served by each road system are worked out and the productivity served may be

assigned appropriate values of utility units per unit weight.



Optimum Road length: Based on the master plan the targeted road length is

fixed for the country on the basis of area or population and production or both.

And the same may be taken as a guide to decide the total length of the road

system in each proposal.

3 The area of certain backward district in India is 1 8400 KM2 and there

are 12 towns as per 1981 census. Determine the length of different categories

of roads by third road development plan

June July 09, June July 11

Soln:

1. NH in km = 18400/50 = 368km

2. SH in km

a. By area 18400/25 = 736km

b. By area and no of towns = 62.5*12 – 18400/50 = 382km

Adopt 736km

3. MDR in km

a. By area 18400/12.5 = 1472km

b. By area and no of towns = 90*12 = 382km

Adopt 1472km

4. Total length of road network as per 2001 per 100sqm is

(NH+SH+MDR+ODR+VR)= 18400*(82/100)

= 15088km

Length of NH+SH+MDR = 638+736+1472 = 2846km

Length of ODR & VR = 15088- 2846 = 12242km



4. What are the objectives? How do you determine road lengths by 3rd

road development plan formulae? June July 11

The third twenty year road plan was prepared by the Road Wing of the

Ministry of Shipping and Transport with the active co-operation from a number

of organizations and the experts in the field of Highway Engineering and

Transportation. This document was released during the 45th Annual Session and

the Golden Jubilee celebrations of the Indian Road Congress in February 1985 at

Lucknow. Therefore, this plan for 1981-2001 is also called as ‘Lucknow Road

Plan’.

Salient Features:

Road development is based on the primary, secondary and tertiary road

system

All the villages with the population above 500 is to be connected with the

metalled road

Aim is 82km/100 sq km

NH will form the main guideline

The length of SH and MDR are calculated based on the area of population

Town >1500 –MDR

1000-1500 –ODR

Roads should also be built in less industrialized area

Long term master plans should be prepared at taluk, district and state

levels

Existing roads should be improved

There should be improved environmental quality and road safety



5. Explain briefly various types of classification of roads June July 11

Roads are classified based on various aspects namely,

1) Based on the carriage way,

Paved Roads: These roads are provided with a hard pavement course

which should be at least a water bound macadam (WBM) layer.

Unpaved Roads: These roads are not provided with a hard pavement

course of at least a WBM layer. Thus earth roads and gravel roads may be

called as unpaved roads.

2) Based on Surface pavement provided

Surface Roads: These roads are provided with a bituminous or cement

concrete surfacing.

Unsurfaced Roads: These are not provided with bituminous or cement

concrete surfacing.

Roads which are provided with bituminous surfacing are called as

black toped roads and that of concrete are referred to as concrete roads

respectively.

3) Based on Traffic Volume:

Heavy

Medium

Light traffic roads.

4) Based on Load transported or tonnage:



Class-I or Class-A

Class-II or Class-B.

5) Based on location and Function:

National Highways (NH)

State Highways (SH)

Major District Roads (MDR):

Other District Roads (ODR):

Village Roads (VR)

6. There are three alternate proposals for a backward district shown

below. Suggest the order of priority for phasing based on the utility

units of 0.5, 1, 2, 4 and 8 for the five population ranges and 1 and 5 per

1000 t of agricultural and industrial products. Dec12

Proposal Total road

length

km

Number of villages served with

population range

Productivity in

(1000 t)

<200

0

2001-

5000 5001

-

1

0

0

0

0

10001

-

2

0

0

0

0

> 20000 Agricultur

e

Industrial

P 500 100 150 40 20 10

-

150 20

Q 600 200 250 68 28 3 220 25

R 700 270 350 82 36 4 300 35

Soln:

Road P= 100*05+150*1+40*2+20*4+10*8+150*1+20*5 = 690 / 500 = 1.38 3rd

Road Q= 200*.5+250*1+68*2+28*4+3*8+220*1+25*5 = 967/600 = 1.61 2nd

Road R = 270*.5+350*1+82*2+36*4+4*8+300*1+35*5 = 2515/700= 3.59 1st



7. Define Master plan and Saturation system. Dec09, Dec 12

Master plan is referred to as road development plan of a city; district or a street or for

whole country. It is an ideal plan showing full development of the area at some future date. It

serves as the guide for the plan to improve some of the existing roads and to plan the network of

new roads. It helps in controlling the industrial, commercial and agricultural and habitat growth

in a systematic way of that area. It gives a perceptive picture of a fully developed area in a plan

and scientific way.

UNIT 3

1.. Explain briefly any two important surface characteristics influencing

highway geometric design. June July 09

Friction between the wheel and the pavement surface is a crucial factor in the

design of horizontal curves and thus the safe operating speed. Further, it also

affects the acceleration and deceleration ability of vehicles. Lack of adequate

friction can cause skidding or slipping of vehicles.

Various factors that affect friction are:

Type of the pavement (like bituminous, concrete, or gravel),

Condition of the pavement (dry or wet, hot or cold, etc),

Condition of the tire (new or old), and

Speed and load of the vehicle

Unevenness

It affects the vehicle operating cost, speed, riding comfort, safety, fuel

consumption and wear and tear of tires. Unevenness index is a measure of



unevenness which is the cumulative measure of vertical undulation of the

pavement surface recorded per unit Horizontal length of the road.

2. Define right of way. List the factors affecting right of way with

dimensions draw a typical cross section of divided highway in urban

area

June July 09, June July 11, Dec 10,

Right of way

Right of way (ROW) or land width is the width of land acquired for the road,

along its alignment. It should be adequate to accommodate all the cross-sectional

elements of the highway and may reasonably provide for future development. To

prevent ribbon

The right of way width is governed by:



Width of formation: It depends on the category of the highway and width

of roadway and road margins.

Height of embankment or depth of cutting: It is governed by the

topography and the vertical alignment.

Side slopes of embankment or cutting: It depends on the height of the

slope, soil type etc.

Drainage system and their size which depends on rainfall, topography etc.

Sight distance considerations : On curves etc. there is restriction to the

visibility on the inner side of the curve due to the presence of some

obstructions like building structures etc.

Reserve land for future widening: Some land has to be acquired in advance

3. Explain briefly the various factors governing the highway alignment

June July 11

The factors governing the highway alignment are as follows

Traffic: The alignment should suit the traffic requirements. Based on the

origin destination data of the area, the desire lines should be drawn. The

new alignment should be drawn keeping in view the desire lines, traffic

flow pattern etc.

Geometric design: Geometric design factors such as gradient, radius of

curve, sight distance etc. also governs the alignment of the highway. To

keep the radius of curve minimum, it may be required to change the

alignment of the highway. The alignments should be finalized such that the

obstructions to visibility do not restrict the minimum requirements of sight



distance. The design standards vary with the class of road and the terrain

and accordingly the highway should be aligned.

Economy: The alignment finalized should be economical. All the three

costs i.e. construction, maintenance, and operating cost should be

minimum. The construction cost can be decreased much if it is possible to

maintain a balance between cutting and filling. Also try to avoid very high

embankments and very deep cuttings as the construction cost will be very

higher in these cases.

Other considerations: various other factors that govern the alignment are

drainage considerations, political factors and monotony.

4. Explain atleast two objectives of providing: i) Camber, ii) Shoulders

Camber

Camber or cant is the cross slope provided to raise middle of the road surface in

the transverse direction to drain off rain water from road surface.

The objectives of providing camber are:

Surface protection especially for gravel and bituminous roads

Sub-grade protection by proper drainage

Quick drying of pavement which in turn increases safety Too steep slope

is undesirable for it will erode the surface. Camber is measured in 1 in n or

n% (Eg. 1 in 50 or2%) and the value depends on the type of pavement surface.



Shoulders

A shoulder are provided along the road edge and is intended for

accommodation of stopped vehicles, serve as an emergency lane for vehicles and

provide lateral support for base and surface courses. The shoulder should be

strong enough to bear the weight of a fully loaded truck even in wet conditions

5. What is an ideal alignment? Explain with neat sketches, how you will

align through a Hill pass, a bridge site and Marshy land

Dec 10, Dec 09, Dec 12

The requirements of an ideal alignment are

The 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.

The alignment should be easy to construct and maintain. It should be easy

for the operation of vehicles. So to the maximum extend easy gradients and

curves should be provided.

It should be safe both from the construction and operating point of view

especially at slopes, embankments, and cutting. It should have safe

geometric features.

The alignment should be economical and it can be considered so only

when the initial cost, maintenance cost, and operating cost is minimum.

Bridge site: The bridge can be located only where the river has straight and

Permanent path and also where the abutment and pier can be strongly founded.

The road approach to the bridge should not be curved and skew crossing should

be avoided as possible. Thus to locate a bridge the highway alignment may be

changed.



Mountain: While the alignment passes through a mountain, the various

alternatives are to either construct a tunnel or to go round the hills. The suitability

of the alternative depends on factors like topography, site conditions and

construction and operation cost.

Intermediate town: The alignment may be slightly deviated to connect an

intermediate town or village nearby. These were some of the obligatory points

through which the alignment should pass. Coming to the second category, that is

the points through which the alignment should not pass are:

6. In a region with light rainfall, a two lane bituminous road is to be

designed. Determine the height of the crown of parabolic camber.

Draw the sketch of the chamber board Dec 12

Soln

Provide a camber rate of 1 in 33 as the rainfall is heavy rise of the crown

w.r.t edges

7.0/2 * 1/33 = 1.06m



UNIT 4

1 List the factors affecting SSD and OSD. June July 09, June July 11, Dec 12

The factors affecting are as follows

Reaction time of the driver

Reaction time of a driver is the time taken from the instant the object is visible to

the driver to the instant when the brakes are applied. The total reaction time may

be split up into four components based on PIEV theory. In practice, all these

times are usually combined into a total perception- reaction time suitable for

design purposes as well as for easy measurement.

Speed of the vehicle

The speed of the vehicle very much affects the sight distance. Higher the speed,

more time will be required to stop the vehicle. Hence it is evident that, as the

speed increases, sight distance also increases.

Efficiency of brakes

The efficiency of the brakes depends upon the age of the vehicle, vehicle

characteristics etc. If the brake efficiency is 100%, the vehicle will stop the

moment the brakes are applied. But practically, it is not possible to achieve

100% brake efficiency

Frictional resistance between the tire and the road The frictional

resistance between the tire and road plays an important role to bring the

vehicle to stop. When the frictional resistance is more, the vehicles stop

immediately. Thus sight required will be less. No separate provision for

brake efficiency is provided while computing the sight distance

Gradient of the road



Gradient of the road also affects the sight distance. While climbing up a

gradient, the vehicle can stop immediately. Therefore sight distance

required is less.

2. The design speed of overtaking vehicle is 60 Kmph. The rate of accelerate

of the observed vehicle is 3.6 Kmph/sec. The difference in speed between

overtaking and overtaken vehicle is 20Kmph. Calculate OSD as per IRC

guidelines for a lane with two way traffic. June July 09

Soln:

OSD = d1 + d2 + d3

d1 = vb*t

= 44*2

= 88m

s = 0.7vb + 6

= .7*44+6 = 36.8m

T= sqrt 14.4*S/A

T= 12.13secs

d2 = 0.278*44*12.13 + 2*36.8

d2= 74.8m

d3= 0.28*60*12.13 = 203.17m

therefore OSD= d1 + d2 + d3

=365.17m



3. Calculate the length of transition curve for a plan, the Design speed - 80

kmph Radius of curve 250m. Road width = 7.5 m Maxim allowable

rate of super elevation I in 150, Super elevation max introduced to 0.07

Assume pavement is rotated with respect to center line. June July 09

Soln:

1.The length of the transition curve Ls1 in m is

Ls1 = 0.0215v3/ CR = 84.67m

C= 80/(75+V) = 0.52

2. Rate of introduction of super-elevation

Ls2 = Ne (W +We)

= 150*.07*7.616

= 79.9m

3. By empirical formula

IRC suggest the length of the transition curve is minimum for a plain and rolling terrain

Ls3 = 2.7V2/R

= 69.12m

Adopt 84.67m as transition curve length



4. List the objects of providing:

i) Super elevation ii) Extra widening of pavement on horizontal curves.

June July 11, Dec10

Super-elevation or cant or banking is the transverse slope provided at

horizontal curve to counteract the centrifugal force, by raising the outer edge of

the pavement with respect to the inner edge, throughout the length of the

horizontal curve. Forces acting on a vehicle on horizontal curve of radius R m at a

speed of v m=sec2are:

Extra widening of pavement on horizontal curves

Mechanical widening

The reasons for the mechanical widening are: When a vehicle negotiates a

horizontal curve, the rear wheels follow a path of shorter radius than the front

wheels as shown in figure. This phenomenon is called off tracking, and has the

effect of increasing the effective width of a road space required by the vehicle.



Psychological widening

Widening of pavements has to be done for some psychological reasons also.

There is a tendency for the drivers to drive close to the edges of the pavement on

curves. Some extra space is to be provided for more clearance for the crossing

and overtaking operations on curves

Widening = Mechanical widening + Psychological widening

5. Determine the SSD on a : i) Level road, ii) Down grade of 1 in 120. Take

t = 2.5 secs, f = 0.35, V = 50 kmph. June July 11

Soln

a. Level ground

SSD = v*t + v2

= 13.88*2.5 + 13.882/2*9.81*.37

2gf

= 61.23m

b. Down grade of 1 in 120.

= 13.88*2.5 + 13.882/(2*9.81*0.37-0.12)

= 75.61m

6. A valley curve is formed when a descending gradient of 1 in 40 meets an

ascending gradient of 1 in 30. Design the length of valley curve to fulfill

both comfort and headlight sight distance conditions. Take design speed

as 100 kmph. Also find the position of the lowest point of valley curve to

locate a culvert June July 11, Dec 10



Soln

Sight distance

SSD = v*t + v2

= 27.77*2.5 + 27.772/2*9.81*.35

2gf

= 181.73m

N= -.058

Comfort condition

Ls = NV3

* 2 =91.28m

C

Case 1: Length of valley curve greater than stopping sight distance (L > S)

= 323.44m

Assumption is correct Length of valley curve is 323.44m

Lowest pt of culvert lies at a distance

X0 = L sqrt (n1*2N)

X0 = 149.7m

7. A NH passing through a flat terrain has a horizontal curve of radius

equal to the ruling minimum radius. If the design speed is 100 kmph,

calculate the: i) Design super elevation, ii) Extra widening and iii)

Length of transition. Make suitable assumptions Dec 10



Soln:

= 357.9m

Super elevation:

e = v2/ 225R = 0.12 take e value of 0.07

Extra widening:

= 0.65m

Transition curve

1. The length of the transition curve Ls1 in m is

Ls1 = 0.0215v3/ CR = 130.6m

C= 80/(75+V) = 0.46

2. Rate of introduction of super-elevation

Ls2 = Ne (W +We)

= 150*.07*7.66

= 80.43m

3. By empirical formula

IRC suggest the length of the transition curve is minimum for a plain and rolling

terrain

Ls3 = 2.7V2/R



= 75.44m

Adopt 130.6 m as transition curve length

8. Explain, with the help of a sketch, the effect of centrifugal force on a vehicle

negotiating horizontal curve Dec 12



9. Explain briefly the attainment of designed super elevation in practice

Dec12, May June 2010

Attainment of super-elevation

Elimination of the crown of the cambered section by:

Rotating the outer edge about the crown: The outer half of the cross slope is

rotated about the crown at a desired rate such that this surface falls on the

same plane as the inner half.

Rotation of the pavement cross section to attain full super elevation

There are two methods of attaining super elevation by rotating the pavement

Rotation about the center line: The pavement is rotated such that the inner edge

is depressed and the outer edge is raised both by half the total amount of super

elevation, i.e., by E=2 with respect to the centre

Rotation about the inner edge: Here the pavement is rotated raising the outer

edge as well as the centre such that the outer edge is raised by the full amount of



Super elevation with respect to the inner edge.

10 A summit curve is formed when an ascending gradient of 1 in 25 meets

another ascending gradient of 1 in 100. Find the length or summit curve

to provide the required SSD for design speed of 80 kmph May June 2010

Soln:

SSD = v*t + v2

= 22.22*2.5 + 22.222/2*9.81*.35

2gf

=127.44m

N= 0.03

L= 2S- 4.4/N = 108.21m > 127.44m

UNIT 5

1. Distinguish between bitumen end tar June July 09, June July 11

Bitumen

It has black to dark brown color

It is natural petroleum product

It is soluble in carbon disulphide & in carbon tetrachloride

It has better weather resisting property

Tar

It also has black to dark brown in color

Tar is produced by the destructive distillation of coal or wool



Tar is soluble only in toluene

It has inferior weather resisting proper

2. Explain the desirable properties of good subgrade soil.

June July 09, Dec 10, Dec 12, May June 2010

The desirable properties of soil as a highway material are

Stability

Incompressibility

Permanency of strength

Minimum changes in volume and stability under adverse conditions of

weather and ground water

Good drainage, and

Ease of compaction.

The soil should possess adequate stability or resistance to permanent

deformation under loads, and should possess resistance to weathering, thus

retaining the desired subgrade support. Minimum variation in volume will

ensure minimum variation in differential.

3. A plate load-test was conducted on a soaked subgrade using a. Plate

diameter of 30 cm. The load values corresponding to the mean

settlement dial readings are given below. Determine the modulus of

subgrade reaction for the standard plate.

June July 11, June July 09, Dec 10

Modulus of subgrade reaction of soil is found as follows



Plate bearing test

The plate bearing test has been devised to evaluate the supporting power of

subgrade or any other pavement layer by using plates of larger diameter.

Plate bearing test was originally meant to find the modulus of subgrade

reaction in the westergards’s analysis for wheel load stresses in cement

concrete pavement.

In the plate bearing test a compressive stress is applied to the soil or

pavement layer through rigid plates of relatively large size & the deflection

are measurement for various stress values.

The deflection level is generally limited to a low value of 1.25mm to 5mm.

Modulus subgrade reaction (k)

K may be defined as the pressure sustained per unit deformation of

subgrade at specified pressure level using specified plate size.

The standard plate size for finding K value is 75cm dia in same test a

smaller plate of 30cm dia is also used (75, 60, 45, 30 & 22.5 cm dia).

4. List the soil classification systems and explain briefly the HRB soil

classification system June July 11

Soil classification is done as follows

1. Grain size analysis

2. Unified soil classification

3. IS Soil classification

4. Highway Research Board (HRB) classification of soils



Highway Research Board (HRB) classification of soils

This is also called American Association of State Highway Officials (AASHO)

Classification of Revised Public Roads Administration (PRA) soil classification

system. Soils are divided into seven groups A-I to A-7. A-I, A-2 and A-3 soils are

granular soils, percentage fines passing 0.074 mm sieve being less than 35. A-4,

A-5, A-6 and A-7, soils are fine grained or silt-clay soils, passing 0.074 mm sieve

being greater than 35 percent.

A-1 soils are well graded mixture of stone fragments, gravel coarse sand, fine

sand and non-plastic or slightly plastic soil binder. The soils of this group are

subdivided into two subgroups, A- 1-a, consisting predominantly of stone

fragments or gravel and A-I-b consisting predominantly of coarse sand.

A-2 group of soils include a wide range of granular soils ranging from A- 1 to A-

3 groups, consisting of granular soils and upto 35% fines of A-4, A-5, A-6 or A-7

groups. Based on the fines content, the soils of A-2 groups are subdivided into

subgroups A-2-4, A-2- 5, A-2-6 and A-2-7.

A-3 soils consist mainly, uniformly graded medium or fine sand similar to beach

sand or desert blown sand. Stream-deposited mixtures of poorly graded fine sand

with some coarse sand and gravel are also included in this group.

A-4 soils are generally silty soils, non-plastic or moderately plastic in nature with

liquid limit and plasticity index values less than 40 and 10 respectively

A-5 soils are also silty soils with plasticity index less than 10%, but with liquid

limit values exceeding 40%. These include highly elastic or compressible, soils,

usually of diatomaceous of micaceous character.



A-6 group of soils are plastic clays, having high values of plasticity index

exceeding 10% and low values of liquid limit below 40%; they have high volume

change properties with variation in moisture content.

A-7 soils are also clayey soils as A-6 soils, but with high values of both liquid

limit and plasticity index, (LL greater than 40% and P1 greater than 10%). These

soils have low permeability and high volume change properties with changes in

moisture content

5. Describe how the quality of toughness and hardness of aggregates is

evaluated in the lab Dec 10

Abrasion test

Abrasion test is carried out to test the hardness property of aggregates and to

decide whether they are suitable for different pavement construction works. Los

Angeles abrasion test is a preferred one for carrying out the hardness property and

has been standardized in India (IS:2386 part-IV). The principle of Los Angeles

abrasion test is to find the percentage wear due to relative rubbing action between

the aggregate and steel balls used as abrasive charge. Los Angeles machine

consists of circular drum of internal diameter 700 mm and length 520 mm

mounted on horizontal axis enabling it to be rotated (see Figure 2). An abrasive

charge consisting of cast iron spherical balls of 48 mm diameters and weight 340-

445 g is placed in the cylinder along with the aggregates. The number of the

abrasive spheres varies according to the grading of the sample. The quantity of

aggregates to be used depends upon the gradation and usually ranges from 5-10

kg. The cylinder is then locked and rotated at the speed of 30-33 rpm for a total of

500 -1000 revolutions depending upon the gradation of aggregates.

After specified revolutions, the material is sieved through 1.7 mm sieve and

passed fraction is expressed as percentage total weight of the sample. This value



is called Los Angeles abrasion value. A maximum value of 40 percent is allowed

for WBM base course in Indian conditions. For bituminous concrete, a maximum

value of 35 is specified.

Impact test

The aggregate impact test is carried out to evaluate the resistance to impact

of aggregates. Aggregates passing 12.5 mm sieve and retained on 10 mm sieve is

filled in a cylindrical steel cup of internal dia 10.2 mm and depth 5 cm which is

attached to a metal base of impact testing machine. The material is filled in 3

layers where each layer is tamped for 25 numbers of blows. Metal hammer of

weight 13.5 to 14 Kg is arranged to drop with a free fall of 38.0 cm by vertical

guides and the test specimen is subjected to 15 number of blows. The crushed

aggregate is allowed to pass through 2.36 mm IS sieve. And the impact value is

measured as percentage of aggregates passing sieve (W2) to the total weight of

the sample (W1).

Aggregates to be used for wearing course, the impact value shouldn't exceed 30

percent. For bituminous macadam the maximum permissible value is 35 percent.

For Water bound macadam base courses the maximum permissible value defined

by IRC is 40 percent



6. Explain briefly, how CBR value of given soil is determined

May June 2010

This is a penetration test developed by the California division of highway.

Forevaluating the stability of soil subgrade and other pavement materials. The test

results have been correlated with flexible pavement thickness requirement for

highway and airfield. CBR test may be conducted in the laboratory on a prepared

specimen in a mould or in situ in the field.

Procedure

CBR test may be performed on undisturbed soil specimens.

About 5kgs of soil is taken passing though 20mm IS sieve and retained on

4.75mm IS sieve

The soil is mixed with water up to OMC.

The spacer disc is placed at the bottom of the mould over the base plate & a

coarse filter paper is placed over the spacer disc.



Then the moist soil sample is to be compacted over this in the mould by

adopting either IS light compaction or IS heavy compaction.

For IS heavy compaction 3 equal layers of compacted thickness about

44mm by applying 56 evenly distributed blows from 2.6kgs rammer.

For IS heavy compaction 5 equal layers of compacted thickness about

26.5mm by applying 56 evenly distributed blows from 4.89 kg rammer.

After compacting the last layer, The collar is removed and the excess soil

above the top of the mould is evenly trimmed off by means of straight edge

(of 5mm thickness).

Clamps are removed ant the mould with compacted soil is lifted leaving

below the perforated base plate & the spacer disc which is removed.

Then the mould with compacted soil is weighed

Filter paper is placed on the perforated base plate &* the mould with

compacted soil is inverted & placed in position over the base plate.

Now the clamps of the base is tightened

Another filter paper is placed on the placed on the top surface of the sample

& the perforated plate with adjustable stem is placed over it.

Now surcharge weights of 2.5 or 5kgs are placed over the perforated plate

& the whole mould with the weights is placed in a water tank for soaking

such that water can enter the specimen both from the top & bottom.

The initial dial gauge readings is recorded & the test set up is kept

undisturbed in the water tank to allow soaking of the soil specimen for full

4 days or 96 hrs.

The final dial gauge reading is noted to measure the expansion & swelling

of the specimen due to soaking.

Then the specimen is clamped over base plate surcharge wt’s are placed on

specimens

centrally such that the penetration test could be conducted.



The mould with base plate is placed under the penetration plunger of

loading machine.

The penetration plunger is seated +at the centre of the specimen & is

brought in contact with the top surface of the soil sample by applying a

seating load of 4kgs.

The dial gauge for measuring the penetration values of the plunger is fitted

in position

The dial gauge of proving ring & the penetration dial gauge are set to 0.

The load is applied though the penetration plunger at a uniform rate of

1.5mm/min

The load reading are recorded at penetration reading 0, 0.5, 1.0, 1.5, 2, 2.5,

3, 4, 5, 7.5, 10 & 12.5mm.

In case the load reading starts decreasing before 12.5mm penetration, the

max load & the corresponding penetration values are recorded.

After the final reading the load is released & the mould from loading

machine.

The proving ring calibration factor is noted so that load dial gauge value

can be converted into the load in kg

CBR at 2.5mm = P1 (kg) x 100%

1370

CBR at 5mm = P2 (kg) x 100%

1370



7. Explain briefly the desirable properties of road aggregates. Mention the

laboratory tests conducted on aggregates May June 2010

Desirable Properties of aggregates are as follows

Strength

The aggregates to be used in road construction should be sufficiently

strong to withstand the stresses due to traffic wheel load. The aggregates

which are to be used in top layers of the pavements, particularly in the

wearing course have to be capable of with4jnhighs1cssesinaddItion to -

wear and tear; hence they should possess sufficient strength resistance to

crushing.

Toughness

Aggregates in the pavements are also subjected to impact due to moving

wheel loads. Sever impact like hammering is quite move on water bound

macadam roads where stones protrude out especially after the monsoons.

Durability

Durability

The stone used in pavement construction should be durable and should

resist disintegration due to the action of weather. The property of the

stones to withstand the adverse action of weather may called soundness

Shape of Aggregates

The size of the aggregates is first qualified by the size of

square sieve opening through which an aggregate may pass, and not by the

shape. Aggregates which happen to fall in a particular size range may have

rounded cubical, angular flaky or elongated shape of particles. It is e and donated

particles will have less strength and durability when compared



with cubical angular or rounded articles of the same Stone. Hence too

flaky and too much elongated aggregates should be avoided as far as possible.

Tests for Road Aggregate

In order to decide the suitability of the road stones for use in construction, the

Following tests are carried out:

(a) Crushing test

(b) Abrasion test

(c) Impact test

(d) Soundness

(e) Shape test

(f) Specific gravity and water absorption test

8. Enumerate the steps for the determination of modulus of subgrade

reaction and for making corrections for plate size and subsequent soaking

Dec10

Procedure

At the test site, about 20 cm top soils is removed & the site is leveled & the

plate is properly seated on the prepared surface.

The stiffening plates of decreasing dia are placed & the jack & proving ring

assembly are fitted to provide reaction against the frame.

3 or 4 dial gauges are fixed on the periphery of the palte from the

independent datum frame foe measuring settlement.

A seating load of 0.07 kg/cm2 (320kgs for 75 dia) is applied & released

after a few sec.

The settlement dial gauges reading are now noted corresponding to zero

load.

A load is applied by means of jack sufficient to cause an average settlement

of about 0.25mm.



When there is no perception increase in settlement or when the rate of

settlement is less than 0.025mm/min (case of clayey soil or wet soil), the

reading of the settlement dial gauge are noted & the avg settlement is found

& the load is noted from the proving ring dial reading.

The load is then increased till settlement increases to a further amount of

about 0.25mm & the avg settlement & load are found.

The procedure is repeated till the settlement reaches 0.175cm.

A graph is plotted with mean settlement versus mean bearing pressure

(load/unit area) as shown in fig.

Bearing pressure settlement curve.

The pressure p (kg/cm2) corresponding to a settlement delta = 0.125cm

(obtaines from the graph shown above)

The modulus of subgrade reaction k is calculated from the relation.

K = P kg/cm2

0.125

Correction for smaller plate size

In some cases the load capacity may not be adequate to cause 75cm dia

plate to settle 0.175cm.

n such a case a plate of smaller dia (say 30cm) may be used.

Then K value should be found by applying a suitable correction for plate

size.

Assuming the subgrade to be an elastic medium with modulus of elasticity

E (kg/cm2), the theoretical relationship of deformation (cm) under a rigid

plate of radius a (cm) is



given by

Delta = 1.18Pa

E

But, K = P

D

Substitute the value of D in K

Therefore K = P x E

2.18 Pa

K

1.18a

If the value of E is taken as constant for a soil, Then k x a = constant

i.e. Ka = ka or K = ka

9. The properties of a subgrade soil are given below:

Passing 0.074 mm sieve = 55% Liquid limit = 50% plastic limit =

41%

Find group index and classify the soil by HRB classification and discuss

the suitability of soil as subgrade material. Dec10

Soln:

LL = 50%

PI = LL- PL = 50-41 = 9

GI = 0.2a+0.005ac+.01bd

= 0.2*20+0.005*20*10+.01*40*0

= 5.0

Sincle value is 5 soil classification is A- 5(5)



UNIT 6

1. Explain the following: Modulus of subgrade reaction, Radius of resisting

section Radius of relative stiffness June July 11, Dec 12

Modulus of subgrade reaction

The modulus of subgrade reaction (k) is used as a primary input for rigid

pavement design. It estimates the support of the layers below a rigid pavement

surface course (the PCC slab). The k-value can be determined by field tests or by

correlation with other tests. There is no direct laboratory procedure for

determining k-value.

The modulus of subgrade reaction came about because work done by

Westergaard during the 1920s developed the k-value as a spring constant to

model the support beneath the slab

Radius of resisting section Radius of relative stiffness

When load is applied on the pavement surafce certain portion of pavement

section resits or regains its original shape after removal load and this portion of

slab which resists the deformatiom is called as Radius of resisting section

Radius of relative stiffness

2. Explain the significance of ESWL in pavement design June July 11

Equivalent single wheel load

To carry maximum load within the specified limit and to carry greater load, dual

wheel or dual tandem assembly is often used. Equivalent single wheel load

(ESWL) is the single wheel load having the same contact pressure, which



produces same value of maximum stress, deflection, tensile stress or contact

pressure at the desired depth. The procedure of finding the ESWL for equal stress

criteria is provided below. This is a semi-rational method, known as Boyd and

Foster method, based on the following assumptions:

equalancy concept is based on equal stress;

contact area is circular;

influence angle is 45o; and

The main significance of ESWL is used to determine the pavement thickness and

to analysis the stress distribution behavior of wheel under load. It also provided

the intensity of stress with respect to depth of the pavement

3. Design the flexible pavement for construction of new highway for the

following data, as per

IRC: 37-2001: Number of commercial vehicles as per last count = 1000

commercial vehicles, Period of construction = 3 years, Annual growth rate

- 8.0%, Design CBR of soil = 4%, Category of road = national highway,

two lane single carriageway, Design life = 10 years, VDF = 2.5, LDF = 0.75

(75%). Calculate the overall thickness and individual layer thickness using

pavement design chart. June July 11, Dec10, Dec 12

Soln:

1. Distribution factor = 0.75



Total pavement thickness for CBR 4% and traffic 7.2 msa from IRC:37 2001

chart1 = 660 mm

Pavement composition can be obtained by interpolation from Pavement Design

Catalogue (IRC:37 2001).

1. Bituminous surfacing = 25 mm SDBC + 70 mm DBM

2. Road-base = 250 mm WBM

3. sub-base = 315 mm granular material of CBR not less than 30 %

4.Distinguish between flexible pavement and rigid pavement. Dec10, Dec 09

Flexible Pavement Rigid Pavement

1. Deformation in the sub grade is

transferred to the upper layers

2. Design is based on load

distributing characteristics of the

component layers

3. Have low flexural strength

4. Load is transferred by grain to

grain contact

5. Have low completion cost but

repairing cost is high

1. Deformation in the sub grade is not

transferred to subsequent layers

2. Design is based on flexural strength or slab

action

3. Have high flexural strength

4. No such phenomenon of grain to grain

load transfer exists

5. Have low repairing cost but completion

cost is high

6. Life span is more as compare to flexible



5. Determine the warping stresses at the interior, edge and corner in a 20

cm thick C.C. pavement with transverse joints at 8.5 m interval and

longitudinal joints at 3.6 m intervals. The modulus of sub-grade reaction k =

6.9 kg/cm3. The temperature differential is 0.6°C/cm of the slab thickness.

Take a = 15 cm, e = 10 x 10"6/°C, E = 3 x I05 kg/cm2, i = 0.15. Dec10, June

July 09, May June 2010

Soln

Radius of relative stiffness = 73.79cm

Length of the slab = 8.5m = 850cm

Cx = Lx/l = 11.51 = 1.03

= 3 x I05*10x10

6*0.6 (1.03+.15*.55/1-.15

2)

= 2

= 25.61kg/cm2

= 23.18kg/cm2

= 6.3618kg/cm2



6. Briefly explain the design factors to be considered in pavement design

May June 2010

Factors affecting pavement design are as follows

Traffic and loading

Traffic is the most important factor in the pavement design. The key factors

include contact pressure, wheel load, axle configuration, moving loads, load, and

load repetitions.

Contact pressure

The tire pressure is an important factor, as it determines the contact area and the

contact pressure between the wheel and the pavement surface. Even though the

shape of the contact area is elliptical, for sake of simplicity in analysis, a circular

area is often considered.

Wheel load

The next important factor is the wheel load which determines the depth of the

Pavement required to ensure that the subgrade soil is not failed. Wheel

configuration affects the stress distribution and deflection within a pavement.

Many commercial vehicles have dual rear wheels which ensure that the contact

pressure is within the limits. The normal practice is to convert dual wheel into an

equivalent single wheel load so that the analysis is made simpler.

Repetition of Loads

The influence of traffic on pavement not only depends on the magnitude of the

wheel load, but also on the frequency of the load applications. Each load

application causes some deformation and the total deformation is the summation

of all these



Environmental factors

Environmental factors affect the performance of the pavement materials and

cause various damages. Environmental factors that affect pavement are of two

types, temperature and precipitation.

7. Explain the design steps for the design of flexible pavement using CSA

method May June 2010

Design procedure

Based on the performance of existing designs and using analytical approach,

simple design charts and a catalogue of pavement designs are added in the code.

The pavement designs are given for subgrade CBR values ranging from 2% to

10% and design traffic ranging from 1 msa to 150 msa for an average annual

pavement temperature of 35 C. The

later thicknesses obtained from the analysis have been slightly modified to adapt

the designs to stage construction. Using the following simple input parameters,

appropriate designs could be chosen for the given traffic and soil strength:

Design traffic in terms of cumulative number of standard axles; and

CBR value of subgrade.

Design traffic

The method considers traffic in terms of the cumulative number of standard axles

(8160 kg) to be carried by the pavement during the design life. This requires the

following information:

1. Initial traffic in terms of CVPD

2. Traffic growth rate during the design life

3. Design life in number of years



4. Vehicle damage factor (VDF)

5. Distribution of commercial traffic over the carriage way.

8. Explain the meaning of ESWL. How is it determined for a dual wheel

assembly and what are its applications? Dec 09

To carry maximum load within the specified limit and to carry greater load,

dualwheel or dual tandem assembly is often used. Equivalent single wheel load

(ESWL) is the single wheel load having the same contact pressure, which

produces same value of maximum stress, deflection, tensile stress or contact

pressure at the desired depth. The procedure of finding the ESWL for equal stress

criteria is provided below. This is a semi-rational method, known as Boyd and

Foster method, based on the following assumptions:

equalancy concept is based on equal stress;

contact area is circular;

influence angle is 45o; and



The main application of ESWL is used to determine the pavement thickness and

to analysis the stress distribution behavior of wheel under load. It also provided

the intensity of stress with respect to depth of the pavement

9. Calculate the load and temperature stresses at edge regions of a cement

concrete pavement using 1RC:58 - 2002 stress equations Dec 09

Wheel load : 51 kN

Modulus of elasticity of concrete : 3x10 5 N/mm2

Modulus of subgrade reaction -: 0.06 N/mm2

Radius of contact area : 150 mm

C: 0.820

Poisson's ratio of concrete : 0.15

L/l: 5.5

Thickness of pavement: 200 mm

Temperature differential of the given location : 18°C

Soln:

Edge load stress from the chart is given for h=20cm, k= 6kg/cm2

Se= 27.5kg/cm2

Radius of relative stiffness = 60.8cm

Length of the slab = 4.5 m = 850cm

Cx = Lx/l = 5.5 =0 .75



Ste = ( E*e*t/2)

= 22.55kg/cm2

UNIT 7

1. Explain the construction steps of 1) Water bound Macadam; ii) Cement,

concrete pavement June July 09, May June 2010, Dec12

Preparation of Foundation for Receiving the WBM course

The foundation for receiving the new layer of WBM may be either the subgrade

orsub-base or base course. This foundation layer is prepared to the required grade

and camberand the dust and either loose materials are cleaned. On existing road

surface, the depressionsand pot-holes are filled and the corrugations are removed

by scarifying and reshaping thesurface to the required grade and camber as

necessary. If the existing surface is a bituminous

surfacing, ftirrows of depth 50 mm and width 50 mm cut at 1.0 m intervals and at

45 degreesto the centre line of the carriageway before laying the Coarse

aggregate.

Provision of Lateral confinement

Lateral confinement is to be provided before starting WBM construction. This

may be done by constructing the shoulders to advance, to a thickness equal to that

of the compactedWBM layer and by trimming the inner sides vertically

Spreading of Coarse Aggregates

The coarse aggregates are spread uniformly to proper profile to even thickness

upon the prepared foundation and checked by templates. The WBM course is



normally constructed to compacted thickness of 7.5 cm except in the case of

WBM sub-base course using coarse aggregate grading no.1 which is of 10.0 cm

compacted thickness.

Rolling

After spreading the coarse aggregates properly, compaction is done by a three

wheeled power roller of capacity 6 to 10 tons or alternatively by an equivalent

vibratoryroI1q the weight of the roller depends on the type of coarse aggregates.

Application of Screenings.

After the coarse aggregates are rolled adequately, the dry screenings are gradually

over the surface to fill the interstices in three or more applications. Dry rolling is

continued as the screenings are being spread and brooming carried out.

Sprinkling and Grouting

After the application of screenings, the surface is sprinkled with water, swept

rolled. Wet screenings are swept into the voids using hand brooms. Ad•

screenings are applied and rolled till the coarse aggregates are well bonded and

firmly set.

Application of Binding Material

After the application of screening and rolling, binding material is applied at a

uniform and slow rate at two or more successive thin layers. After each

application of binding material, the surface is copiously sprinkled with water and

wet slurry swept with brooms to fill the voids.



Setting and Drying

After final compaction, the WBM course is allowed to set over-night. On the next

day the ‘hungry’ spots are located and are filled with screenings or binding

material, lightly sprinkled with water if necessary and rolled. No traffic is allowed

till the WBM layer sets and dries out.

Cement concrete pavement construction procedure

(i) Preparation of Subgrade and Sub-base

The subgrade or sub-base for laying of the concrete slabs should comply with the

wing requirement; that no soft spots are present in the subgrade or sub-base; that

the uniformly compacted subgrade or sub-base extends atleast 30 cm on either

side of the width to be concreted; that the subgrade is properly drained; that the

minimum modulus subgrade reaction obtained with a plate bearing test is 5.54

kg/cm2. over the soil subgrade. In such a case, the moistening of the subgrade

prior to placing of the concrete is not required.

ii) Placing of Forms

The steel or wooden forms are used for the purpose. The steel forms are of M.S.

channel sections and their depth is equal to the thickness the pavements. The

sections have a length of at least 3 m except on curves of less than 45.0 m

radius, where shorter sections are used.

(ii) Batching of Material and Mixing

After determining the proportion of ingredients for the field mix, the fine

aggregates and coarse aggregates are proportioned by weight in a weight-

batching plant and placed into the hopper along with the necessary quantity of



cement. Cement is measured by the bag. All batching of material is done on the

basis of one or more whole bags of cement.

(iii) Transporting and Placing of Concrete

The cement concrete is mixed in quantities required for immediate use and is

deposited on the soil subgrade or sub-base to the required depth and width of the

pavement section within the form work in continuous operation.

(iv) Compaction and Finishing

The surface of pavement is compacted either by means of a power-driven

finishing machine or by a vibrating hand screed. For areas where the width of the

slab is very small as at the corner of road junctions, etc., hand consolidation and

finishing may be adopted:

2. Explain the important design step- for longitudinal drains of a road to

drain off surface water. June July 11, Dec12, June July 09

Simplified steps for the design of longitudinal drains of a road to drain off the

surface water given below:

The frequency of return period such as 10 years, 25 years etc. is decided

based on finances available and desired margin of safety, for the design of

the drainage system.

The values of coefficients of run-off C1, C2, C3 etc. from drainage areas

A1, A2, A3 etc. are found and the weighted value of C is computed.

Inlet time for the flow of storm water from the farthest point in the drainage

area to the drain inlet along the steepest path of flow is estimated from the

distance, slope of the ground and type of the cover. Figure 11.3 may be

used for this purpose.



Time of flow along the longitudinal drain T2 is. determined for the

estimated length of longitudinal drain L upto the nearest cross drainage or a

water course and for the allowable velocity of flow V in the drain i.e., T2 =

L.

The total time T for inlet flow and flow along the drain is taken as the time

of concentration or the design value of rain fall duration, T = T1 + T2.

The required depth of flow in the drain is calculated for a convenient

bottom width and side slop of the drain. The actual depth of the open

channel drain may be increased slightly to give a free board. The hydraulic

mean radius of flow R is determined.

The required longitudinal slope S of the drain is calculated using

Manning’s formula adopting suitable value of roughness coefficient n.

3. Explain the construction steps for Bituminous concrete. June July 11

Constructions Steps

Preparation of existing layer: The existing layer is prepared to a proper

profile. Pot holes are patched and irregularities are made even. The surface

is properly cleaned.

Tack coat or prime coat application: A track coat is applied of thin layer

of bitumen binder on the existing layer either using the sprayer or a pouring

can. the quantity of application is 40 to 7.5 kg per 10 m2 for black top layer

and 7.5 to 10kg per 10 m2 for untreated WBM layer.

Premix preparation: The bitumen binder and aggregates as per

recommended gradings are separately heated to the specified temperatures

and are then placed in the mixer chosen for the job. The mixing



temperature for each grading and the bitumen binder is also specified based

on. the laboratory results. A tolerance of ± 10°C is allowed. The mixing is

done till a homogeneous mixture is obtained. The mixture is then carried to

the site for its placement through a transporter or a wheel barrow.

Placement. The bituminous paving mixture is then immediately placed on

the desired location and is spread with rakes to a pre-determined thickness.

The camber profile is checked with a template. It may be stated here that a

compacting temperature also influences the strength characteristic of the

resulting pavement structure. It is therefore required that the minimum time

is spent between the placement of the mix and the rolling operations.

Rolling and finishing The paving mix. The rolling is done with 8 to 10

tones tandem roller. The rolling is commenced from the edges of the

pavement construction towards the centre, and uniform overlapping is

provided. The finished surface should not show separate lines of markings

due to defective or improper rolling. The roller wheels are kept damp,

otherwise the paving mix may partly stick to the wheels and the finishing

may not be good. A variation of 6 mm over 3 m length is allowed in the

cross profile. The number of undulations exceeding 10 nun should be less

than 30 in 300 m length of pavement

5 Differentiate between seal coat and prime coat June July 11, Dec10

Prime coat:

Bituminous prime coat is the first application of a low viscosity liquid

bituminous material over an existing porous or absorbent pavement surface like

the WBM base course.



Tack coat.

Bituminous tack coat is the application of bituminous material over existing

Pavement surface which is relatively impervious like an existing bituminous

surface or a cement concrete pavement or a pervious surface like the WBM

which has already been

treated by a prime coat.

6. What is penetration macadam? Describe the method of constructing

penetration macadam road, stepwise. Dec10

Penetration Macadam

Bituminous Penetration Macadam or Grouted Macadam is used as a base or

binder course. The coarse aggregates are first spread and compacted well in dry

state and after that hot bituminous binder of relatively high viscosity is sprayed in

fairly large quantity at the top.

Construction steps is as follows

Preparation of Foundation for Receiving the WBM course

Spreading of Coarse Aggregates

Rolling

Bitumen application

Application of key aggregates

Seal coat

Finishing

Setting and Drying

Open to traffi



7. Explain the methods of sub-surface drainage to control the seepage flow,

capillary rise and water table Dec10, May June 2010, Dec09

Lowering of Water Table

The Highest level of water table should be fairly below the level of subgrade, in

order that the subgrade and pavement layers are not subjected to excessive

moisture. From practical considerations it is suggested that the water table should

be kept atleast 1.0 to 1.2 m the subgrade. In places where water table is high

(almost at ground level at times) the best remedy is to take the road formation on

embankment of height not less than 1.0 to 1.2 meter. When the formation is to be

at or below the general ground level, it would be necessary to lower the water

table.

Control of capillary rise

A layer of granular material of suitable thickness is provided during the

construction of embankment between the subgrade and the highest level of

subsurface water table. another method of providing capillary cutoff is by

inserting an impermeable bituminous layer in place of granular blanket

Control of seapage flow

The Highest level of water table should be fairly below the level of subgrade, in

order that the subgrade and pavement layers are not subjected to excessive

moisture. From practical considerations it is suggested that the water table should

be kept atleast 0.6 to 0.9 m the subgrade



UNIT 8

1. Explain the terms : i) BOOT and BOT, Mud pumping, Alligator

cracking Bump integrator June July 09, Dec09, Dec10, June July 11

Alligator cracking: This is the most common type of failure that occurs due the

the relative movement of pavement materials. this is caused due to the repeated

application moving loads resulting in fatigue failure or due to the moisture

variation resulting in swelling shrinkage of subgrade and other pavement

materials

Build–operate–transfer (BOT) or build–own–operate–transfer (BOOT) is a

form of project financing, wherein a private entity receives a concession from the

private or public sector to finance, design, construct, and operate a facility stated

in the concession contract. This enables the project proponent to recover its

investment, operating and maintenance expenses in the project.

Due to the long-term nature of the arrangement, the fees are usually raised during

the concession period. The rate of increase is often tied to a combination of

internal and external variables, allowing the proponent to reach a satisfactory

internal rate of return for its investment.

Bump integrator

An Indigenous device developed by CRRI It comprises of a standard

pneumatic wheel mounted within a rectangular frame with single leaf spring on

either side. Spring dashpots mounted on the leaf spring provide damping for the

suspension Integrating unit, mounted on one side of the frame integrates the

unevenness in cm. For the measurement It is towed by a jeep at a constant speed

of 32 kmph under standard tyre pressure of 2.1 kg/cm2 along the designated

http://en.wikipedia.org/wiki/Concession_%28contract%29

http://en.wikipedia.org/wiki/Public_sector

http://en.wikipedia.org/wiki/Internal_rate_of_return



wheel path Bumps in cm and corresponding road length in terms of wheel revolution pulses

are displayed / recorded on a panel board

2. Explain briefly various highway user benefits. June July 11

The various benefits due to highway improvement may be classified

into two categories: (i) quantifiable or tangible benefits in terms of market values

and (ii) nonquantifiable or intangible benefits.

Quantifilab1e Benefits

Various benefits which can be quantified include benefits to road user such as

reduction in vehicle operation cost, time cost and accident cost. The other benefits

include enhancement in land value. These are briefly explained below:

Saving in vehicle operation cost is due to reduction in fuel and oil consumption

and reduction in wear and tear of tyres and other maintenance costs. A road

withsharp curves and steep grades require frequent speed changes; presence of

intersections require stopping idling and accelerating; vehicle operation on road

stretches with high traffic volume or congestion necessitates speed changes and

stopping and increased travel time.

Non-quantable Benefits

The non-quantifiable benefits due to improvements in highway facilities include

reduction in fatigue and discomfort during travel, increase in comfort and

conveniences and improvement in general amenities, social and educational

aspects, development of recreational and medical services, improved mobility of

essential services and defence forces, aesthetic values, etc..

3. It is proposed to widen a stretch of a single lane road of length 40km to two

lanes at a total cost of Rs.7.5 lakhs per km and the rate of interest is 7.0%



per year. The annual cost of maintenance of the existing single lane road is

Rs. 10000 per km and that of the improved two lane road is Rs. 14000 per

km. The average vehicle operation cost on the existing road is Rs.150 per

vehicle-km and that on the improved road is estimated to be Rs.1.20 per

vehicle-km. If the present traffic is 2000 motor vehicles per day and by the

end of 15 years design period, the traffic is estimated tc be doubles,

determine whether the investment on the "improvement of the road is

economically viable during the 15 years period June July 11, 09

Soln:

Average traffic during the design period = ( 2000+4000)/2 3000mv/day

Average road user cost on existing road per year = 365*40*3000*1.5

= 657laks

Average road user cost on improved road per year = 365*40*3000*1.2

= 525.6Laks

Total benefit = 657-525.6 = 131.4Lakhs

CRF for n=15 and i=7% 0.10

Improved cost Cr= 7.5*40 = 300Lakhs

Present annual cost Cr = 300*.010 = 30Lakhs

Additional maintaince cost= 14000-1000*40 = 1.6Lakhs

BCR= 131.4/30 = 4.38 > 1 hence project is economical.

4. Explain the various benefits that a road user gets by the improvement

of a road May June 2010, Dec 12

The benefits for road user is as follows

Quantifilab1e Benefits

Various benefits which can be quantified include benefits to road user such as



reduction in vehicle operation cost, time cost and accident cost. The other benefits

include enhancement in land value. These are briefly explained below:

Saving in vehicle operation cost is due to reduction in fuel and oil consumption

and reduction in wear and tear of tyres and other maintenance costs. A road

withsharp curves and steep grades require frequent speed changes; presence of

intersections require stopping idling and accelerating; vehicle operation on road

stretches with high traffic volume or congestion necessitates speed changes and

stopping and increased travel time.

Non-quantable Benefits

The non-quantifiable benefits due to improvements in highway facilities include

reduction in fatigue and discomfort during travel, increase in comfort and

conveniences and improvement in general amenities, social and educational

aspects, development of recreational and medical services, improved mobility of

essential services and defence forces, aesthetic values, etc..

5. Mention the types and causes of failures in flexible pavement

May June 2010

The various types of failures and causes of flexible pavement are as follows

Alligator cracking: This is the most common type of failure that occurs

due the the relative movement of pavement materials. this is caused due to

the repeated application moving loads resulting in fatigue failure or due to

the moisture variation resulting in swelling shrinkage of subgrade and other

pavement materials

Consolidation of pavement layers: this type of failure occurs due to the

repeated application of load on the wheel path



Shear failure and cracking: This failure causes upheaval of pavement

materials by forming a fracture or cracking

Longitudinal cracking: this is due to frost action and volume changes in

subgrade

Reflection cracking: this type of failure is seen during overlays when the

bottom layer cracks reflects to the top surface

6. Compare the annual costs of two types of pavement structures) WBM

with thin bituminous surface at total cost of Rs. 2.2 lakhs per km, life of

5 years, interest at 10%, salvage value of Rs. 0.9 lakhs after 5 years,

annual average maintenance cost of Rs. 0.35 lakhs per km and

Bituminous macadam base and bituminous concrete surface, total cost

of Rs. 4.2 lakhs per km, life of 15 years, interest at 8%, salvage value of

Rs. 2.0 lakhs at the end of 15 years ; annual average maintenance cost

of Rs. 0.25 lakhs per Ion.

May June 2010, Dec 12

Soln:

Annual cost of pavement with WBM base and thin bituminous surface

= (2.2 -0.9 CRF) ( i= 0.08, n= 5) + 0.1*0.9+0.35

= 1.3*0.2638+.09+0.35 0.78Lakhs

Annual cost of bituminous pavement

= (4.2 -2.0) CRF ( i= 0.08, n= 5) + 0.08*2+0.25

= 2.2*0.1168+0.16+0.25 0.66 Lakhs

Annual cost of bituminous is lower compared to the WBM pavement



7. Write short notes on: Annual cost method, Benefit - cost ratio method

Dec09

Annual cost method

The annual cost of each element of capital improvement is found by multiplying

by the appropriate CRF value calculated for the assume life span. The annual cost

Cr may be found using the relation

C1 = P. i(1+i)n = P(CRF)

(1+i)n-1

Benefit Cost ratio Method

Principle of this method is to assess the merit of a particular scheme by

comparing the annual benefits with the increase in annual cost

Benefit cost ration = Annual benefits from improvement

Annual cost of the improvement

= R – R1

H1 - H

Where R = total annual road user cost for axisting highway

The benefit-cost ratios are determined between alternate proposals and those

plans dub are not attractive are discarded. Then the benefit cost ratios for various

increments of added investment are computed to arrive at the best proposal. hi

order to justify the proposed improvement, the ratio should be greater than 1.0.

However, the choice of interest rate would affect the results of the benefit-cost

solutions.

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