estimating & costing

7/23/2019 Estimating & Costing

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Structure

1.0 Introduction

1.1 Definition

1.2 Need for estimation and costing

Learning Objecyives

After studing this unit, student will be able to

Have an idea of the introduction to estimating and costing.

1.0 Introduction

In the civil engineering field, the construction activity contains the

following three steps.

1. Plans : Preparation of drawings plan, section, elevation, with full

dimension and detailed, specifications meeting the requirements of the proposed

structure.2. Estimation : Preparation of an estimate is for arriving the cost of

the structure to verify the available funds or to procure the required funds for

completion of the proposed structure.

3. Execution (construction) : It is a grounding the proposed structure,

for construction as per the provision contained in drawings and estimation..

Introduction

1UNIT


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Construction Technology152

The plans contains size of room and dimensions of the work and the estimate

contains the quantity and quality aspects of the structure.

1.1 Definition

Estimation and costing there are two basic points involved in

construction of structures are :

1. Quantity : The quantity aspects is with reference to the measurement

in the drawings (plan, elevation, section)

2. Quality :The quality aspects is with reference to the specifications,

i.e properties of materials, workmanship etc.

Note : The estimation and costing of any structure is defined as the

process of determination of quantities of items of work, and its cost forcompletion.

2. Estimate of a project is therefore, a forecast of its probable cost.

1.2 Need for Estimation and Costing

The object of preparing the estimate for any civil engineering structure is

1. To know the quantities of various items of work, a material and

labour and their source of identification.

2. To decide whether the proposal can match the available funds to

complete the structure.

3. To obtain the administrative and technical sanction of estimate

from the competent authorities to release the funds for construction.

4. To invite tenders or quotations based on the estimate quantities for

entrust of works to the execution.

Short Answer Type Questions

1. What is meant by Estimating and Costing ?

2. State need for Estimation and Costing.


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Structure

2.0 Introduction

2.1 Units of measurements

2.2 Rules For Measurement

2.3 Different methods of tasking out quantities

Learning Objectives

After the studying this unit student will be able to

To measure various quantities as per rules.

2.0 Introduction

The units of differents works depends on their nature, size and shape..In general, the units of different items of works are based on the followingprinciple.

1. Massive or volumetric items of work such as earth work, conceretefor foundations, R.R Masonry , Brick Masonry etc. The measurements oflength, breadth , height or depth shall be taken to compute the volume or cubicalcontents.

2. Shallow, thin and surface work shall be taken in square unit or inarea. The measurements of length and breadth or height shall be taken tocompute the area, Ex. Plastering, white washing etc.

Measurement of Materialsand Works

2UNIT


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3. Long and Thin work shall be taken in linear or running units and

linear measurement shall, be taken. Ex : Fencing, Rainwater pipes,

ornamental borders etc.

4. Single units of work are expressed in numbers. Ex. Doors, Windows,

Rafters, Trusses etc.

2.1 Units of measurement for various items of CivilEngineering Works

Sl.No

1.

2.

3.

4.

5.

Unit of

payment

10.00cum

1.00cum

1.00rmt

10.00cum

10.00cum

1.00cum

1.00cum

1.00cum

1.00cum

Particulars of items

(a) Earth work excavation in all

types of soils except rock re-

quiring blastering.

(b) Earth work excavation in

the soils hard rock requirng

blastering.

(c) Excavation of pipe line

through of specified width and

depth inall types of soils

(d) Earthwork for road for-

mation ,bund formation etc.

cutting , embankment.(e) Refilling of foundations ,

basements, pipe lines, trenches

with excavated soils.

Plain cement concrete for

foundation.

R.R.masonry or brick ma-

sonry for foundation base-

ment, super strucrture, parapet

wall etc.

Filling the basement with

sand.

(a) RCC 1:2: 4 with normal

reinforcement for plinth beam ,

columns, lintels, verandah

beam- T beam etc.

Units of measure-

ments

10.00cum

1.00cum

1.00 rmt

10.00cum

10.00cum

1.00cum

1.00cum

1.00cum

1.00cum


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Paper - II Estimating and Costing 155

10.00sqm

10.00sqm

10.00sq m.

1.00Rmt

1.00No

1.00No

1.00Rmt

1.00Rmt

kg/unit

1.00cum

1.00cum

1.00sqm

1.00cum

6.

7.

8.

9.

10

11.

12.

13.

14.

15.

(b) R.C.C 1: 2: 4 for slabs of

specified thickness .Plastering pointing, flooring,

weather proof coarse, white

washing, colour washing, paint-

ing.

Roofing with A.C sheets, tiled

roofing, Kurnool trerrace, Ma-

dras terrace etc.

D.P.C specified width and thick-

ness

Wooden and steel trusses

Doors, windows, ventilators.

Ornamentel border of speci-

fied width and thickness.

R.C.C pipes, A.C pipes GI or

C.I pipes, stone ware pipes

etc.

Steel reinforcement in R.C.C.

Rough stone pitching revet-

ment and soiling of specified

thickness.

(a) Roads works : Metal col-

lections , gravel collections,

solving stones, pitching any

stones, revetment stones etc.

(b) Road works : Spreadingmetal gravel and consolidation

with roller of specified thick-

ness.

(c) Cement concrete pay-

ments of specified thickness.

10.00sqm

1.0.00sqm1.00sqm

10.00sq m.

1.00Rmt

1.00No

1.00 No

1.00Rmt

1.00Rmt

Kg/unit

1.00cum

1.00cum

10.00sqm

10.00cum


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2.2 Rules For Measurement

Measurement of works occupies a very important place in the planningand execution of any work or project, from the time of the first estimate are

made until the completion and settlement of payments. The methods followed

for the measurement are not uiform and the practices or prevalent differ

considerably in between the states. Even in the same state different departments

follow different methods. For convernience a uniform method should be followed

throughout the country. The uniform methods of measurement to be followed

which is applicable to the preparation of the estimates and bill of quantities

and to the side measurement of completed works have been described below.

General Rules

1. Measuremet shall be item wise for the finished items of work and the description of each items shall be held to inculde materials, transport,

labour, fabrication, hoisting, tools and plants, over hands and other

incidental charges for finishing the work to the required shape, size,

design and specifications.

2. In booking dimensions the order shall be in the sequence of length,

breadth and height or depth or thickness.

3. All works shallbe measured not subject to following tolerances unless

otherwise stated.

(a) Dimensions shall be measured to the nearest 0.01 meter i.e 1cm(1/ 211).

(b) Areas shall eb measured to the nearest 0.01 sq.m (0.1 sqft).

(c) Cubic contents shall be worked up to the nearest 0.01 cum(0.1cuft)

4. Same type of work under different condition and nature shall be

measured separately under separate items.

5. The bill of quantities shall fully describe the materials proportions

and work-manships and accurately represent the work to be executed.

Work which by its nature cannot be accurately taken off or which

requires site measuremets shall be described as provisional.

6. In case of structureal concrete, brick work or stone masonry, the

work under the following categories shall be measured separately

and the heights shall be described.

(a) From first floor level


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(b) From plinth level to first floor level.

(c) From first level to second floor level and so on.

The parapet shall be measured with the corresponding items of the

story next below.

Principle of units :The units of different works depend on their nature,

size and shape. In general the units of different item of work are based on the

following principle.

(i) Mass, voluminious and thick works shall be taken in cabic unit or

volumne. The measurement of length, breadth, and height or depth shall be

taken to compute the volume cubic contents(cum).

(ii) Shallow, thin and surface work shall be taken in separate units orin area. The measurement of length and breadth or height shall be taken to

compute the area (sq.m).

(iii) Long and thin work shall be taken in linear or running unit and

linear measurement shall be taken(running meter).

(iv) Piece work, job work etc taken in number

2.3 Different methods of taking out quantities

The items of work like earth work in excavation in foundation, foundation

concrete stone masonry in foundation and basement, stone or brick masonry in

super stucrture may by estimated bu either of the following methods.1. Long wall and short wall method (or) General method

2. Centre line method

2.3.1 Long wall and short wall method

In this method measure or find out the external lengths of walls running

in the direction generally the long walls out-to-out and the internal length of

walls running in the transverse direction in-to-in i.e. of cross or short wall in-

to-in and calculate quantities multiplying the length by the breadth and height of

wall. The same rule applicable to the excavation in foundation, to concrete in

foundaiuon and to masonry.

The simple mehtod is to take the long walls of short or erros walls

separately and to find out the centre to centre lengths of long wall anf short

walls from the plan. For symmetrical footing on either sides, the centre line

remians same for suepr structure and for foundation and plinth.


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For long walls add to the center length one breadth of wall, which

gives the length of the wall out-to-out ,multiplying this length by the breadth and

height and get the quantities,. Thus for finding the quantities of earth work inexcavation, for the length of trench out-to-out add to the centre length one

breadth of foundaiton. Adopt the same process for foudation conceret and for

eacth footing. It should be noted that each footing is to be taken separately and

the breadth of the particular footing is to be added to the centre length.

Long wall length out-to-out = centre to centre length + half breadth on

one side + half breadth on the other side = centre to centre length + one breadth.

For short or cross walls sub tract ( instead of adding) from the centre

length one breadth of wall, which gives the length in-to-in, and repeat the same

process as for the long walls, subtracting one breadth instead of adding.

Short wall length in-to-in= Centre to centre length - one breadth.

That is, in case of long wall add one breadth and in case of short wall

substract one breadth from the centre length to get the corresponding lengths.

It will be noticed that by taking dimensions in this ways, the long walls

are gradually decreasing in length from foundation to superstructure, while the

short walls are increasing in length.

This method is simple and accurate and there is no chance of any mistake.

This method may be named as long wall and short wall method, or general

method.

2.3.2 Centre line method

In this method known as centre line method. This method is easy and

quick in calculations. In this method sum total length of centre lines of all walls,

long and short has to be found out. This method is well suitable for walls of

similar cross sections. In this method the total centre line multiplied by breadth

and depth of concerned item gives the total quantity of each item. In this method,

the length will remain same for excavation in foundation for concrete in foundation,

for all footings and for super structure (with slight difference where there are

cross walls or number of junctions). It requires special attention and consideration

at the junctions, meeting points of partition or cross walls, etc.

For rectangular, circular polygonal (hexagonal, octagonal etc) building

having no inter or cross walls, this method is quite simple. For each junction half

breadth of the respective items or footings is to be deducted from the total

centre length. Thus in the case of a building with one partition wall or cross wall

having two junctions, for earthwork in foundation trench and foundation concrete

deduct one breadth of trench or concrete from the total centre length (half breadth


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for one junction and the breadth ( 2 x 1/2 = one) for two junctions. For footings,

similarly deduct one breadth of footing for two junctions from the total centre

length and so on. If two walls come from opposite directions and meet a wall atthe same point, than there will be two junctions.

In the case of a building having different type of walls, suppose the other

(main) walls are of A type and inter cross walls are of B type, then all A type

walls shall be taken jointly first , and then all B type walls should be taken

together separately. In such cases no deductions of any kind need be made for

A type walls, but when B type walls are taken, for each junction deducting of

half breadth of A type wall (main wall) shall have to be made from the total

centre length of walls.

It may be noted that at corners of the building where two walls are

meeting no substraction or addition is required.

Note : Student should practice method I first and when they have become

sufficiently acquainted with method I, then only they should take up the method

II.


1. Write the unit of measurements. Earthwork, P.C.C, R.C.C, Masonary,

Plastering, Flooring, Fencing, Ornamental border, Door, Windows,

Trusses etc.

2. Write general rules for measurement.

3. Write different methods of taking out quantities and describe.


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Structure

3.0 Introduction

3.1 Detailed estimate

3.2 Preliminary or approximate estimate

3.3 Problems in preliminary estimate

Learning Objectives

After studying this unit student will be able to

Understand the definition of detailed estimate, stages of preparation

of estimate, Data required for an estimate and types of estimate.

3.0 Introduction

An estimate is a probable cost of a work. It is usually prepared before

the construction is taken up. The primary object of an estimate is to know

beforehand the cost of the work. The actual cost of the work is known after thecompletion of the work. If the estimate is prepared carefully and correctly there

will not be much difference in the estimated cost and actual cost. The estimator

should be fully acquainted with the methods of construction, skilled and

experienced for accurate estimating.

3UNIT

Types of Estimates


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161Paper - II Estimating and Costing

3.1 Detailed estimate

The estimate may be approximate or preliminary estimate or accurateestimate. In approximate estimate the approximate cost of the work is estimated.

In the accurate estimate the details of various items are taken and calculated.

3.1.1 Definition

The estimate prepared by dividing the work into different items, taking

detailed measurements of each item of work and calculating their quantities is

known as detailed estimate.

3.1.2. Stages of preparation

To prepare the complete estimation of the project, besides the estimated

cost of different main items of work, The cost of preliminary works and surveying,cost of land and its acquisition, cost of leveling and preparation of ground and

the cost of external services are to be provided. Provision of supervision charges

and contractors profit are to be provided in the estimate.

Data required for preparing an estimate : To prepare an estimate

for a work the following data are necessary.

Drawings : The detailed drawings of plan, elevation and section, drawn

to a scale are necessary to take the details of measurements of various items of

work.

Specifications : The specifications gives the nature, quality and classof materials, their proportion, method of execution and workmanship and the

class of labour required. The cost of the work varies with its specifications. The

cement mortar with 1:3 is more costlier than cement mortar with 1:6.

Rates :The rates for various items of work, the rates of various materials

to be used in construction, the wages of different categories of labour should be

available for preparing an estimate. The location of the work and its distance of

source of materials and cost of transport should be known. These rates may be

obtained from the Standard Schedule of Rates prepared by the engineering

departments.

3.1.3 Details of measurements and calculation of quantities and abstract of estimated cost

To prepare an accurate estimate, a detailed estimate of quantities of

various items of work and an abstract estimate of the quantities and their unit

rates are required.


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Detailed Estimate

Abstract estimate

3.2. Preliminary or approximate estimate

Preliminary or approximate estimate is required for preliminary studies

of various items of work or project , to decide the financial position and policy

for administrative sanction by the competent authority. The preliminary estimate

is prepared by different methods for different types of works. The various

methods of preparing the preliminary estimate are plinth area estimate, cubical

rate estimate and estimate per unit base.

3.2.1 Plinth area estimate

The plinth area rate is calculated by finding the plinth area of the building

and multiplying by the plinth area rate. The plinth area rate is obtained by

comparing the cost of the cost of similar building having similar specifications in

the locality.

3.2.2. Cubic area estimate

The cubic rate estimate is prepared on the basis of the cubical contents

of the building. The cubic rate is obtained from the cost of the similar building in

the locality having similar specifications. The cost of the building is estimated by

multiplying the volume of the building with the cubic area rate. Cubic rate estimate

is more accurate as compared to the plinth area estimate.

3.2.3 Estimate per unit base

The preliminary estimate may be prepared for different structures and

works by various ways. For schools and hostels, per class rooms for schools,per bed for hospitals, per seat for theater halls, etc. For roads and highways and

for irrigation works, the preliminary estimate is made per kilometer. For bridges

and culverts per running meter. For sewerage and water supply projects on the

basis of per head of population served.

S.no Description of work No Length Breadth Height/Depth Quantity Remarks

S.No. Description of work Quantity Rate Per Amount


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3.3. Problems in preliminary estimate

1. If the cost of school building per student is Rs. 25000. Calculate thecost of school building for 100 students.

Cost of the school building for 100 students = Rs.

25000x100=Rs.2500000.

2. If the cost of construction of 1 km. length of a highway is Rs.

10000000. Find the cost of construction for 20 km.

Cost of construction for 20 km = Rs. 10000000x20=Rs.200000000.

3. If the plinth area rate of a residential building is Rs.10000/sq m.

Calculate the cost of construction of a residential building of 100 sq. m.

Cost of construction of 100 sq. m.= plinth area rate x area =

10000x100=Rs.1000000

Summary

Detailed estimate consists of taking the detailed measurements of length,

breadth, height and calculating the quantities.

Data required for estimate :Drawings, specifications and rates.

Types of preliminary estimates : Plinth area estimate, cubic rate

estimate and estimate per unit base.


1. Define detailed estimate.

2. What are stages for preparation of an estimate?

3. List out the data required for preparation of an estimate.

4. Write the tabular form for the detailed estimate.

5. Write the tabular form for preparation of an abstract estimate.

Long Answer Type Questions

1. Describe the various types of preliminary estimates.


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Structure

4.0 Introduction

4.1 Single roomed building (load bearing structure)

4.2 Two roomed building( load bearing type structure)

4.3 Single storied residential building with number of rooms (load bearing

type structure)

4.4 Single storied residential building with number of rooms (framed

structure type)

4.5 Primary school building with sloped roof

4.6 RCC Dog legged open well stairs

4.7 Two storied residential building (framed structure type)

4.8 Detailed estimate of compound wall and steps

Learning Objectives


Prepare detailed estimates of single roomed, Building roomed, Doubleroomed buildings, for load bearing walls and Framed structures. Detailed Esti-mate of Primary School Building, Compound walls and steps. Detailed estimateDog legged and Open Well STair case. Preparational estimate for ground andfirst floor.

4UNITDetailed and AbstractEstimate of Buildings


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4.0 Introduction

To estimate the cost of any building or a structure, drawings,specifications and rates are required. Regarding the detailed estimate by long

wall and short wall method and centre line method, the drawings consisting of

plan elevation and section are sufficient. The estimator should be able to take all

the dimensions from the drawings. The length and breadth are taken from the

plan, while the height or depth are taken from the section and elevations. In long

wall and short wall method the walls are taken separately, while in the centre line

method, the centre line lengths of all the walls are combined. The accuracy of

estimate depends upon the skill of the estimator in studying the drawings. The

long wall and short wall method is useful for load bearing type structure, but it

cannot be applied for framed structure.

4.1 Single roomed building (load bearing structure)

There are two steps in estimating the cost of a building or a structure.

1. Taking out quantities and calculation of quantities in detailed estimate.

2. Determining the cost from the abstract estimate.

Long wall and short wall method : This method is also called as

separate or individual wall method. This is simple and it gives accurate values.

The following procedure is adopted.

1. The dimensions of long wall and short wall should be taken separately.2. Irrespective of its lengths, the wall which is taken first is long wall and

the wall which is taken next is the short wall.

3. The centre line of the wall of the building is considered for determining

the centre to centre line length of long walls and short walls.

4. The centre to centre to centre length of long walls or short walls is

obtained by adding half the width of the wall to the internal length of

either long wall or short wall.

5. Centre to centre length of long wall = internal length of long wall +

width of the wall.

6. Centre to centre length of short wall = internal length of short wall +

width of the wall.

7. To determine the lengths of different quantities such as earthwork,

c.c. bed in foundation, R.R. masonry etc, length of long wall = centre


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to centre length of long wall + width, the width is the respective

width of the item in consideration.

8. Similarly length of the short wall = centre to centre length of the short

wall width, where the width is the respective width of the item

such as earthwork, c.c. bed etc.

Centre line method : In the centre line method, the sum of all the

centre line lengths of long walls and short walls are added to get the total centre

line length. At the junctions of two walls, the length is present in both of the

walls. Hence half of the length of that width is to be subtracted from the total

centre line length.

Length = Total centre line length width x number of junctions.

Fig 4.1 Plan Single Room

Centre to centre length of long wall = 6.0 + 2x0.3/2 = 6.3 m.

Centre to centre length of short wall = 4.0 + 2x0.3/2 = 4.3 m.

Length of Long Wall = Centre to centre Length of Long Wall + Width

Length of Short Wall = Centre to centre Length of Short Wall width

For earth work in excavation Length of Long Wall = 6.3 + 1.2 = 7.5 m.

E L E V A T I O N S E C T I O N

1. 20. 90. 70. 5

0.30.60.3

3.0


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For earth work in excavation Length of Short Wall = 4.3 1.2 = 3.1 m.

In cement concrete in foundation the length and width of the long walland short wall are the same, but the height is different from that of the foundation

For R.R. masonry First footing Length of long wall = 6.3 + 0.9 = 7.2 m.

Length of Short Wall = 4.3 -0.9 = 3.4 m.

Similarly for second footing & Third footing, Length of Long Walls are

7.0 and 6.8 and for short walls are 3.6 m and 3.8 m respectively.

Detailed estimate of a single roomed building by centre line method

Centre to centre length of long wall = 6.0 + 2x0.3/2 = 6.3 m.


Total centre line length = 2(6.3 + 4.3) = 21.2 m.

Detailed Estimate

Quantity

30.528

7.63

11.45

4.45

12.72

28.62

19.08

RemarksSl. No.

1

2

3

4

Description

of work

Earth work in

excavation

C.C. bed infoundation

R.R. masonry

in foundation

and plinth

First footing

Second footing

Basement

Brick work in

super structure

No.

1

1

1

1

1

1

L

21.2

21.2

21.2

21.2

21.2

21.2

B

1.2

1.2

0.9

0.7

0.5

0.3

H

1.2

0.3

0.6

0.3

1.2

3

m m m m3


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Sl.

No.

1

2

3

Description of work

Earth work in

excavation in

foundation

Long Walls

Short Walls

Plain cement concrete

in foundation (1:5:10)

Long Walls

Short Walls

R.R. Masonry in

foundation

& basement c.m (1:8)

First footing

Long Walls

Short WaLLS

Second footing

Long Walls

Short WaLLS

Basement

Long Walls

Short Walls

No.

2

2

2

2

2

2

2

2

2

2

L

m

7.5

3.1

7.5

3.1

7.2

3.4

7

3.6

6.8

3.8

B

m

1.2

1.2

1.2

1.2

0.9

0.9

0.7

0.7

0.5

0.5

H

m

1.2

1.2

Total

0.3

0.3

Total

0.6

0.6

0.3

0.3

1.2

1.2

Quantity

m3

21.6

8.93

30.53

5.4

2.68

8.08

7.78

3.67

11.45

2.94

1.51

4.45

8.16

4.56

Remarks

L=6.3+1.2=7.5

L = 4 . 3 -

1.2=3.1

L=6.3+1.2=7.5

L = 4 . 3 -

1.2=3.1

L=6.3+0.9=7.2

L = 4 . 3 -

0.9=3.4

L=6.3+0.7=7.0

L = 4 . 3 -

0.7=3.6

L=6.3+0.5=6.8

L = 4 . 3 -

0.5=3.8


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4.2 Two roomed building( load bearing type structure)

Detailed Estimate Of A Double Roomed Building By Long Wall AndShort Wall Method

Centre to centre length of long wall = 5.0 + 0.3 + 5.0 + 2x0.3/2 = 10.6m.


Number of long walls = 2. Number of short walls = 3.

Length of long wall = centre to centre length of long walls + width

Length of short wall = centre to centre length of short wall - width

4 Brick work in super

structure c.m. ( 1:8)

Long Walls

Short Walls

2

2

Totalof

6.6

4

R.R.

0.3

0.3

masonry

3

3

12.72

28.62

11.88

7.2

19.08

L=6.3+0.3=6.6

L = 4 . 3 -

0.3=4.0

Sl. No.

1

2

Description of

work

Earthwork in

excavation

Long Walls

Short Walls

C.C. bed in

foundation

Long Walls

No.

2

3

2

L

m

11.8

4.1

11.8

B

m

1.2

1.2

1.2

H

m

1.2

1.2

Total

0.3

Quantity

m3

33.98

17.71

51.69

8.5

Remarks

L= 10.6 +

1.2 = 11.8

L = 5.3 -

1.2 = 4.1

Total centre to centre line lenght = 10.6 x 2 + 5.3x3 = 37.1 m


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3

4

Short Walls

R.R. masonry in

foundation &

plinth

First footing

Long Walls

Short walls

Second footing

Long Walls

Short Walls

Third footing &plinth

Long Walls

Short walls

Brick work in

super structure

Long Walls

Short Walls

3

2

3

2

3

2

3

2

3

4.1

11.5

4.4

11.3

4.6

11.1

4.8

10.9

5

1.2

0.9

0.9

0.7

0.7

0.5

0.5

0.3

0.3

0.3

0.6

0.6

0.3

0.3

1.2

1.2

3

3

4.43

12.93

312.42

7.13

19.55

4.75

2.9

7.65

13.32

8.64

21.96

49.16

19.62

13.5

33.12

L = 10.6 +

0.9 = 11.5

L = 5.3 - 0.9

= 4.4

L = 10.6 +

0.7 = 11.3

L = 5.3 -0.7

= 4.6

L = 10.6 +

0.5 = 11.1

L = 5.3 - 0.5

= 4.8

L = 10.6 + 0.3 =

10.9

L = 5.3 - 0.3 =

5.0

R.R. masonry Total


21/83


Centre line method

Fig 4.2 Double Room

No.

1

1

1

1

1

1

1

L

35.9

35.9

36.2

36.4

36.6

36.8

B

1.2

1.2

0.9

0.7

0.5

0.3

H

1.2

0.3

0.6

0.3

1.2

Total

3

Sl. No.

1

2

3

4

Description of

work

Earthwork in

excavation

C.C. bed in

foundation

R.R. masonry in

foundation

First footing

Second footing

Basement

Brickwork in

superstructure

Remarks

L= 37.1 -

2x1/2x1.2

L = 37.1 -

2x1/2x0.9

L = 37.1 -

2x1/2x0.7

L = 37.1 -

2x1/2x0.5

L = 37.1 -

2x1/2x0.3

Quantity

51.69 m3

12.93 m3

19.55 me

7.65 m3

21.96 m3

49.165 m3

33.12 m3

E L E V A T I O N S E C T I O N

5m x 5 m5 m x 5 m

1. 20. 90. 70. 5

0.30.60.3

3.0


22/83


4.3 Single storied residential building with number of rooms (load bearingtype structure)

Length of long walls = 6.0+0.3+5.0+2x0.3/2=11.6 m.

Number of long walls = 3

Length of short wall of 5.0 m. length = 5.0+2x0.3/2=5.3 m.

Number of 5.0 m shortwalls =3

Length of 4.0 m. length short walls = 4.0+2x0.3/2=4.3m.

Number of 4.0 m. length short walls = 3

Total centre line length = 11.6x3+5.3x3+4.3x3=63.6m.

Fig 4.3 Plan Section

6.0 x 5.0 m 5.0 x 5.0 m

5.0 x 4.0 m5.0 x 4.0 m

D D

DD

D D

0.30.6

0.3

0.9

0.9

1.2m

3.0m


23/83


56.43

m mm m3

5m4m

5m4m

5m4m

5m4m

Basement

Basement


24/83


4.4 Single storied residential building with number of rooms (framed structure type)

Number of columns in a framed structure = 9

Size of the columns = 230 mmx230 mm

Length of R.R. masonry, Brickwork, lintels, plinth beam and beams

under slab = (6+6)x3+(5+4)x3=63 m.

Length of sunshades and external plastering = (12.9+9.9)x2= 45.6 m.

Length of slab with 1 m. extension on both sides = 1.0+1.0=2.0 m.

External Plastering : Area of external plastering = Length x Height

Length of Plastering = 2x(12.9+9.9)=45.6 m.

Height of external plastering = 3.0+0.12, where 3.0m is the height of

the room and 0.12 m. is the thickness of the slab.

Internal plastering : Area of internal plastering = Length x Height

Length of plastering = 2(L+B) , Where L and B are the length and

breadth of the room respectively.

For 6mx5m room, length = 2(6+5)=22m. Similarly for 5mx4m room,

length =2(5+4)=18 m.

Fig 4.4 Residential Building Framed Structure

6.0 x 5.0 m 5.0 x 5.0 m

5.0 x 4.0 m6.0 x 4.0 m

P L A N S E C T I O N

0.23 x 0.23

R.C.C Column3.0 m

1.2 m

0.9 m

0.3 m

0.3 m

R.C.C.

Footing

R.C.C.

G.L G.L

100 mm

thick

RCC slab

1.2 m


25/83


B

m

1.2

0.9

0.6

1.2

0.9

0.6

0.7

0.45

0.23

0.23

0.23

1.2

0.23

S.

No.

1

2

3

4

5

Description of work

Earthwork in excavation

Columns

In between columns

Deduct for columns

C.C. bed in foundation

Columns

In between columns

Deduct for columns

R.R. masonry in

foundation

First footing

Second footing

Brickwork in

superstructure

Deductions Doors

Windows

R.C.C. column footing

Trapezoidal section

Stem

No.

9

1

9

9

1

9

1

1

1

6

8

9

9

9

L

m

1.2

63

0.6

1.2

63

0.6

63

63

63

1

1.2

1.2

0.23

H

m

1.8

0.9

0.9

0.3

0.3

0.3

0.6

1.2

3

2

1.2

0.3

0.3

5.1

Quantity

m3

23.33

51.03

-2.92

71.44

3.89

17.01

-0.972

19.93

26.46

34.02

60.48

43.47

-2.76

-2.65

38.06

3.89

2.44

2.43

8.76

Remarks

L=12x3+

9x3=63

H=0.9+1.2+

3.0=5.1

Net Brickwork in super structure

(1.44+4x0.985+0.053)/6


26/83


6

7

8

9

10

11

R.C.C. Plinth beam

R.C.C. inlintels&sunshades

Lintels

Sunshades

R.C.C. slab and beams

Beams under slab

1m. Projection from

slab

R.C.C. Slab.

External plastering

20 mm

Thick

Deductions

Doors

Windows

Internal Plastering

12 mm thick

Rooms 6mx5m

Rooms5mx4m

Sand filling in rooms

Rooms 6mx5m

1

1

1

19

1

1

6

8

2

2

2

63

63

45.6

63

1

14.9

45.6

1

1.2

22

18

6

0.23

0.23

0.7

0.23

0.23

11.9

5

0.3

0.1

0.07

0.3

0.3

0.12

3.12

2

1.2

3

3

1.2

4.35

1.45

2.23

3.68

4.35

0.62

21.28

26.25

142.27

-12

-11.52

118.75

132

108

240

72

L=2(12.9+

9.9)=45.6

L=12.9+1.0+

1.0=14.9

B=9.9+1.0+

1.0=11.9

L=2(12.9+9.9)

=45.6

H=3.0+0.12

L=2(6+5)=22

L=2(5+4)=18

Net External plastering area


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4.5 Primary school building with sloped roof

Wall thickness = 0.3 m. in brick masonry.

Width of foundation = 1.2 m. Depth of foundation = 1.8 m.

Width of first footing = 0.9 m. Depth of first footing = 0.9 m.

Second footing width = 0.7 m. Depth = 0.6 m.

Width of third footing and plinth = 0.5 m. Height = 0.9 m.

Centre to centre length of long walls = 3.0+0.3+3.0+2x0.3/2=6.6 m.

Centre to centre length of short walls = 3.0+2x0.3/2=3.3 m.

Total centre line length = 6.6x2+3.3x3=23.1 m.

Number of junctions = 2.

Height of the sloping roof =1.0 m.

12

13

14

Rooms 5mx4m

C.C. bed in rooms

Rooms 6mx5m

Rooms5mx4m

Flooring in rooms

Rooms 6mx5m

Rooms5mx4m

Fabrication &

placement of

steel

2

2

2

2

2

5

6

5

6

5

4

5

4

5

4

1.2

0.1

0.1

48

120

6

4

10

60

40100

(8.76+4.35+3.68+26.25)x1.25x87.5/100x1000 78.5x100/100x1000

tonnes 4.22 t


28/83


Length of the sloping roof = square root of (1.5mx1.5m + 1.0m.x1.0

m.) = 1.8 m.

Number of gable rafters at a spacing of 30 cms. Centre to centre =(

6.0/0.3)+1=21

Length of the gable rafters = 1.8+1.8+0.5+0.5=4.6 m.

Number of reapers along a length of 6.05 mts. At a spacing of 10 cms

each = (4.6/0.1)+1=47


29/83


ELEVATION

W W

DD

Room

3.0 x 3.0 m

Room

3.0 x 3.0 m

References

D - Door 1.00 m x 2.00 m

W - Window 1.2 m x 1.2 m

Width of 1st footing : 0.9 m

Second footing : 0.7 m

Basement : 0.5 m

P L A N

S E C T I O N

Tiles

Tiles

1.2 m

0.9 m

0.6 m

0.3 m

0.9 m

0.9 m

0.6 m

0.3 m

0.9 m

2.0 m

1.5 m

0.9 m

PRIMARY SCHOOL BUILDING WITH SLOPING ROOF


30/83


4.6 RCC Dog legged open well stairs

Fig 4.5 Dog Legged Stair case

1650

250

150

Floor

1650

2500 1000

E L E V A T I O N S E C T I O N - A A

P L A N


31/83


Sloping side 2 2 0.28 2.4640. 411.264Total


32/83


Length of the inclined flight = Square root of (1.65x1.65+2.5x2.5)=3.0m.

Size of base of flight = 1.0x0.5x0.25 m3

Landing at the middle and top floor =2.0mx1.0mx0.15m.

Length of the hand rail = (2x3.0+0.40)=6.8 m.

Number of risers = 11

Height of the first flight = 11x0.15=1.65 m.

Number of treads = 10

Length of treads in each flight = 10x0.25=2.5 m.

Triangular portion of the brick has a base of 0.25 m. and height 0.15 m.

Area of the brickwork = 1/2x(0.25x0.15) m2.

4.6.1 Open Well Staircase

Fig 4.6 Open well Stair case

Flig

htNo.

No.

ofRisers

No.o

fTrea

ds

Eac

hRiser

Eac

hTrea

d

A

8

8

152

300

B

4

3

152

300

C

8

7

152

300

SECTION AT AA

Note :

1. All dimensions are in Milli meters

2. Follow the written dimensions only

OPEN WELL TYPE STAIRCASE

Scale 1:50

DRG. No. 18


33/83


Flight No. A

Horizontal distance of treads = 0.3x8=2.4 m.

Height of risers = 0.15x9=1.35 m.

Sloping length of flight = Square root of(2.4x2.4+1.35x1.35)=2.75 m.

Flight No. B

Horizontal length of treads = 0.3x3=0.9 m.


Sloping length of flight= Square root of (0.9x0.9+0.6x0.6)=1.08 m.

Flight No. C

Horizontal length of treads = 0.3x7=2.1 m.


Sloping length of flight = Square root of (2.1x2.1+1.2x1.2)=2.42 m.


34/83


4.7 Two storied residential building (framed structure type)

Fig 4.7 Two storied residential building

E L E V A T I O N

Parpet wall

Weathering

course

Lintel &

sunshade

Brick

masonry

Roof slab

C.C. flooring

R.C.C Mix

1:4:1Sand filling

C.C. floring

1:4:8Elevation

0.902

3.05 m

3.05 m


35/83


Fig 4.8 Ground Floor & First Floor

Ground floor

Number of columns = 15

Height of columns in ground floor & first floor =

0.90+0.9+3.05+0.1+3.05+0.1+0.8=8.9 m.

Height of column in ground floor = 0.9+9+3.05+0.1=4.95 m.

Height of column in first floor = 3.05+0.1+0.8=3.95 m.

Length of brickwork, lintels and beams =

4.21x4+4.20x4+3.05x2+3.00x2+2.00x2+4.00x2+3.34x2 = 64.42 m.

Openings Main door 1.00mx2.1m -1 No., Door D 0.9x2.1 3Nos., Door D1 0.76x2.1 2 Nos.

Windows - W 1.8mx1.2m 5 Nos., W1 1.2mx1.2m 2 Nos.

Length of wall 100 mm. thick = 4.21+3.79+1.5= 9.5 m.

Length of sunshade = 2.1x5+1.5x2+1.1x1+1.3x1 = 15.9 m.


36/83


Length of slab = 12.68 m., Width of slab = 9.10 m.

Length of external plastering = 2(12.68+9.10)=43.56 m.

Trapezoidal section of the column foundation : Area of base A1 =

1.0x1.0=1.0 m2.

Area of the column stem = 0.23x0.23=0.0529 m2=A2


37/83



38/83



39/83



40/83


4.8 Detailed estimate of compound wall and steps

Length of the compound wall between the brick columns 230 mm x

230 mm = 6.0 + 4.0 = 10.0 m.

Height of the compound wall = 1.5 m.

Depth of excavation below ground level = 0.9 m.

Width of the foundation = 0.9 m.

Thickness of the C.C. bed = 0.3 m.

Size of the first footing = 0.6 m. x 0.6 m.

Size of the plinth = 0.45 x 1.0m2.

Size of the brickwork in columns = 0.23 x 0.23 x 1.5 m.

Number of brick columns = 3

Length of the earthwork in excavation =

6.0+0.23+0.23+4.0+0.23=10.69

Quantity of earthwork in excavation = 10.69x0.9x0.9=8.66 m3.

Quantity of C.C. bed in foundation = 10.69x0.9x0.3=2.89 m3.

R.R. masonry first footing = 10.69x0.6x0.6= 3.85 m3.

R.R. masonry in plinth = 10.69x0.45x1.0= 4.81 m3.

R.R. masonry total = 3.85+4.81= 8.66 m.

Brick masonry in columns = 3x0.23x0.23x1.5=0.24 m3.

Brickwork in between columns = 10.0x0.10x1.5= 1.5 m3.

Total brick masonry = 0.24+1.5=1.74 m.

Deduction for gate 2.0mx1.5m = 2.0x0.1x1.5=0.3 m3.

Net brickwork in superstructure = 1.74-0.3 = 1.44 m.

Plastering in columns = 4x0.23x1.5x3=4.14 m2.

Plastering in between columns = 10x1.5x2=30 m2.


41/83


Total area of plastering = 4.14+30=34.14 m2.

Estimate of steps

Quantity of first step = 1.0x0.9x0.3=0.27 m3.

Quantity of second step = 1.0x0.6x0.3=0.18 m3.

Quantity of third step = 1.0x0.3x0.3=0.09 m3.

Total quantity of brickwork in steps = 0.27+0.18+0.09=0.54 m3.

1.5 m

1.0 m0.45

0.6

0.9 m

0.3 m

0.6 m

0.23m

0.23

4.0 m

0.23

6.0m0.23

0.15

0.15

0.15

0.3

0.3

0.3

1.0 m

Top view

Front View

Side View

Fig. 4.9 Plan and Section of a compound Wall


42/83


Summary

To estimate the cost of a building or a structure the steps involved are

1. Taking out the measurement of various items and calculate the

quantities as per the detailed estimate.

2. Determining the cost of the calculated quantities as per Abstract

estimate.

The methods of calculating quantities are Long wall and short wall method

and Centre line method.

Length of Long wall = Centre to centre length of the long wall + width

Length of short wall = Centre to centre length of the short wall widthIn centre line method, the length = Total centre line length (number of

junctions)xwidth/2

For a double room building, the total centre line length = sum of the

centre line lengths of two long walls and three short walls. The number

of junctions = 2.

For a building with number of rooms, the total centre line length = sum

of the centre to centre lengths of three long walls, three short walls of length 5.3

m. and three short walls of length 4.3 m. Number of junctions = 6.

The long wall short wall method and the centre line method are notapplicable. The lengths of the R.R. masonry, Brickwork in superstructure, Plinth

beam, lintels and beams under slab are obtained by adding the internal dimensions

of the rooms.

The roof for the primary school building is a gable roof, having its slope

in two directions. The roof under consideration is the roof having its width = 3.0

m. and its length = 6.0 m.

Length of the gable rafter = square root of [(width/2)2 + (Rise)2]

Number of gable rafters = Length of the roof/ spacing of the rafters.

Area of the tiled surface = 2x(Length of the roof )x Width of the slopingside.)

Number of risers = Height of the flight/ rise.

Number of treads = Number of risers 1.

Treads length = Number of treads x Tread.


43/83


Horizontal length of the stairs = Treads length + Width of the landing

Length of the sloping side = Square root of [(Treads length)2 + (Height

of flight)2].

Area of brickwork in each step = (Rise x Tread) x .


1. What are the steps involved in finding the cost of the building?

2. What are the methods involved in taking measurements in a detailed

estimate.

3. Write the tabular formula of a detailed estimate.

4. Calculate the number of risers in a flight of height 1.50 m. and the rise of 15 cms.

5. If the number of risers = 10, find the number of treads.

6. Find the length of the gable rafter for a room of width 6.0 m. and

length 12.0 m and the rise is 1.5 m.


1. Find the earthwork in excavation, C.C. bed in foundation, R.R.

masonry in foundation, Brick work in superstructure and plastering for single

room building and double room building by long wall short wall method and

centre line method.2. Detailed estimate of a dog legged stair case.

3. Detailed estimate of compound wall and steps.

O.J.T. Type Questions

1. Detailed estimate of a number of rooms.

2. Detailed estimate of a framed structure.

3. Detailed estimate of a Primary school building.

4. Detailed estimate of an open well stair case.

5. Detailed estimate of a double storied building.


44/83


Structure

5.0 Introduction

5.1 Prepare specifications for different items of work.

5.2 Find the cost of materials at source and at site.

5.3 Study of the cost of labor types of labor using standard schedule

of rates

5.4 Concept of lead and lift- leads statement

5.5 Preparation of unit rates for finished items of works

Learning Objectives


Prepare the unit ratio of various items of works. Find the cost of

materials, specifications of various of various items of works.

5.0 Introduction

To estimate the cost of the building, the quantities of various items of

work are calculated from the drawings. The unit rates of various items of work

are calculated from the specifications of the various types of materials. The rates

are calculated as per the rates in the standard schedule of rates. The unit rates of

various items of work increase considerably with the specifications. The

5UNIT

Specifications and Analysis

of Rates


45/83


specifications indicate the quality of the work while the drawings are used for

the quality of the work.

5.1 Prepare specifications for different items of work

Specifications specifies or describes the nature and the class of work,

materials to be used in the work, workmanship etc. From the study of the

specifications one can easily understand the nature of the work and what the

work shall be.

Detailed specifications : Detailed specifications are written to express

the requirements clearly in a concise form avoiding repetition and ambiguity.

The detailed specifications for various items of work are as follows.

Earthwork excavation of foundation

The following specifications shall be followed in the earthwork in

excavations in foundations.

1. Foundation trench shall be dug to the exact width and depth of

foundation.

2. Excavated earth shall not be placed within 1 m. of the edge of the

foundation.

3. The bottom of the trenches shall be perfectly leveled both

longitudinally and transversely.

4. If water accumulates in the trench, it should be pumped out. Care should be taken to prevent water from entering the trench.

5. If rocks and boulders are found during excavation, they should be

removed and the bed of the trench should be leveled and

consolidated.

6. Foundation concrete should be laid only after the inspection and

approval by the Engineer in charge.

Cement concrete in foundation (1:5:10)

The following specifications should be followed in cement concrete in

foundation.

1. Course aggregate should be of hard broken stone, free from dust,

dirt and foreign matter.

2. Fine aggregate shall be of coarse sand, consisting of hard, sharp and

angular grains and shall pass through screen of 5 mm. square mesh.


46/83


3. Sand should be free from dust, dirt and organic matters.

4. Water shall be clean and free from alkaline and acid matter.

5. Mixing should be done on masonry platform or sheet iron tray in

hand mixing.

6. Coarse aggregate and sand should be mixed by volume and cement

by weight.

Random rubble masonry

The following specifications should be followed in random rubble

masonry

1. The stones should be sound, hard and durable. Stones with rounded

surface shall not be used.

2. No stone shall be less than 15 cm. in size.

3. Bond stones should be provided at every 1 m. length.

4. Cement mortar 1:3 to 1:6 shall be provided.

5. The joints in the stone masonry shall not be thicker than 2 cm.

6. The masonry shall be watered for at least 10 days.

Brick masonry

The following specifications should be followed in brick masonry firstclass

1. Bricks of standard size, copper red color, regular in shape, having

sharp square edges should be used.

2. The bricks should not absorb more than 20% of water when immersed

in water for 24 hours.

3. The mortar used in brick masonry shall be 1:3 to 1:6.

4. The bricks shall be well bonded and laid in English bond unless

otherwise specified.

5. Mortar joints shall not exceed 6 mm. in thickness and the joints shall

be fully flushed with mortar.

6. The bricks should be soaked in water before use in masonry.

7. The brick masonry shall be watered for at least 10 days.


47/83


Plastering

The following specifications should be followed in plastering

1. The materials of mortar, cement and sand used in plastering should

be as per specifications.

2. The joints of the brickwork shall be raked for a depth of 18 mm. on

the surface.

3. Ceiling plastering should be completed before the start of wall

plastering.

4. The thickness of the plastering should not be less than 12 mm. for

internal plastering and 20 mm. for external plastering.

5. The plastering work shall be checked for horizontality with a straight edge and for verticality with a plumb bob.

6. Any defective plastering shall be cut in rectangular shape and replaced.

7. The plastering should be watered for at least 10 days.

5.2 Find the cost of materials at source and at site.

The amount required to purchase the material at the source of its

production is the cost of materials at the source.

Cost of materials at site : The cost of materials at site includes the

cost of materials at source along with the cost of seignories, taxes, royalties,transport, stacking, loading and unloading etc.

Seignories are collected for materials like sand, stones etc., which are

under the control of respective local agencies under government control.

5.3 Study of the cost of labor types of labor using standard schedule of rates

Labour rates

Si

No.

Category of worker

S. Rate

For

2012-13


48/83


1 2 3

Skilled catregory

1 Bar bender 330

2 Black smith / Tin smith / Rivetor 315

3 Blaster ( Licensed ) 355

4 Carpenter Cl- I 315

5 Electrician ( Licensed ) 355

6 Fitter Cl- I 315

7 Floor Polisher / Tile Layer 315

8 Foreman 355

9 Gauge reader 300

10

Maistry / Work Inspector with Non-technical Qualification

SSLC/SSC/HSC

300

11 Mason Cl- I / Brick layer Cl- I 315

12 Mechanic Cl- I 315

13 Operator Air compressor / DG set 315

14 Operator Batching plant 355

15 Operator Bus/Ambulance/ Lorry/ Tanker 315

16 Operator Concrete / Asphalt mixer 315

17 Operator Concrete / Asphalt paver 315

18 Operator Concrete pump / Placer/ ice plant 315

Common SoR 2012 : 13

280

Sl

No.


49/83


Category of worker

S. Rate

for

2012-13

1 2 3

19 Operator Core drilling machine 355

20 Operator Crane/ Tower crane/ Cable way 355

21 Operator Drilling jumbo / Loco / Winch 315

22 Operator Grouting/ Guniting/ Shotcreting 315

23 Operator Jackhammer/Pneumatic tamper 315

24 Operator Pump / Ventilation fan 315

25 Operator Lathe/Drilling/Shearing machine 355

26 Operator Bending / Planing machine 315

27 Operator Road roller 315

28 Operator Shovel / Scraper / Dozer 355

29 Operator Spillway / Sluice gate 315

30 Operator Crusher / Conveyor / Mucker 315

31 Operator Tipper / Dumper / Transit mixer 355

32 Operator Concrete vibrator 315

33 Operator Vibratory plain / pad foot roller 315

34 Operator Wagon drill / Drifter 355

35 Painter Cl- I 350

36 Plumber / Pipe fitter 350

37 Sarang / Khalasi 315

38 Spun pipe moulder 315

39 Stone chiseller CI- I / Stone cutter Cl- l 315

40 Struct. steel Fabricator / Marker / Erector 355


50/83


41 Welder / Gas Cutter 315

42 Welder (X-ray quality) 355

II. Semi skilled category

1. Asphalt Sprayer / Boiler attendant 285

2. Bhisti 285

3. Boatman with boat 300

Common SoR 2012:13

281

Sl

No.

Category of worker

S. Rate

for

2012-13

1 2 3

4 Carpenter Cl- II / Erector shuttering 285

5 Cartman with double bullock cart 330

6 Cartman with single bullock cart 310

7 Chavali / Navagani 285

8 Crowbarman / Jumper man 285

9 Fitter Cl- II 285

10 Gang man / Head / Survey mazdoor 285

11 Gardener / Trained mali 285

12 Helper Air compressor / DG set 285

13 Helper Batching plant 285

14 Helper Blasting 285

15 Helper Bus/ Ambulance/ Lorry/ Tanker 285


51/83


16 Helper Bending/Shearing/Planing machine 285

17 Helper Carpenter 285

18 Helper Concrete / Asphalt mixer 285

19 Helper Concrete / Asphalt paver 285

20 Helper Core drilling machine 285

21 Helper Crane/ Tower crane/ Cable way 285

22 Helper Drilling jumbo / Loco / Winch 285

23 Helper Fitter / Fabrication/Electrician 285

24 Helper Grouting/ Guniting/ Shotcreting 285

25 Helper Jack hammer / Pneumatic tamper 285

26 Helper Laboratory / Instrumentation 285

27 Helper Road roller 285

28 Helper Shovel / Scraper / Dozer 285

29 Helper Crusher / Conveyor / Mucker 285

30 Helper Tipper / Dumper/ Transit mixer 285

31 Helper Vibrator 285

Common SoR 2012:13

282

Sl

No.

Category of worker

S. Rate

for

2012-13

1 2 3

32 Helper Vibratory plain/ pad foot roller 285

33 Helper Wagon drill/ Drifter 285


52/83


34 Lineman Electric / Telephone 285

35 Mason Cl- ll / Brick layer Cl-II 285

36 Mechanic Cl- II 285

37 Painter Cl- II 300

38 Patkari / Neeraganti / Sowdy 285

39 Stone Chiseller Cl- II 285

40 Stone breaker / Hammer man 285

41 Valve man / Canal sluice operator 285

III. Un-skilled category

1 Cement / Asphalt handling mazdoor 250

2 Civic worker 250

3 Heavy mazdoor 250

4 Light mazdoor 250

5 Watchman 250

IV. Other category

1 Care-taker / conductor / Lift Attender 300

2 Cook / Mess man 300

3 Dhobi 300

4 Diploma Engineer / Surveyor 450

5 Diver with headgear 365

6 Graduate / Laboratory Assistant 350

7 Graduate Engineer/ Geologist 600

8 Horticulture Assistant / Photographer 300

9 ITI certificate holder / Tracer / Printer 350

10 Literate mazdoor 285

11 Stenographer / Computer Operator 400

Common SoR 2012:13


53/83


283

Sl

No.

Category of worker

S. Rate

for

2012-13

1 2 3

12 Telephone / Wireless Operator 350

13 Typist / Job Typist 350

14

CAD operator with Diploma in Engineering/General degree with

CAD certificate

500

15 Jeep Driver 355

16 Data Processing Operator 500

Note : 1. The wage should not be less than the minimum wages of

schedule of employment,

Subject to out turn. 2. 25% extra over the corresponding labour

rates in respect of the work to be

Done during night time subject to issue of certificate accordingly by

the concerned estimate.

Sanctioning authority for providing in the data and by concerned

Executive Engineer in charge of the work for payment. The night time

allowance is applicable only to the works done under Greater

Hyderabad Municipal Corporation, Greater Visakhapatnam Municipal

Corporation and Vijayawada Municipal Corporation limits only.of

various government agencies.

Transport cost includes cost of transporting the material from source

to the site. In S.S.R., the cost of transporting on a mettaled road is


54/83


given. If transport is required on a cart track or a sand track, to reach

the site, that distance is converted to equivalent metalled road.

Distance on cart track = Distance on metalled road x 1.1

Distance on sand track = Distance on metalled road x 1.4

Stacking includes placing the material in a specified heap for a given

volume in the case of materials like sand and coarse aggregate. Bricks are stacked

for a given number. Sometimes are stacking charges are included in loading and

unloading. Loading and unloading charges are fixed for a given volume or weight

for different materials.

The cost of labor wages for each category of labor are given above as

per Standard schedule of rates 2012-13.

Standard schedule of rates : In standard schedule of rates (S.S.R.) ,

the rates of various materials, machinery and hiring charges and wages of labor

are prepared. It is prepared by the board of chief engineers and approve it for

that year.

5.4 Concept of lead and lift- leads statement

The distance between the source of material to the worksite is known

as the lead. This lead distance changes from one project to another project

depending upon the location. The vertical height through which the material is to

be disposed is known as the lift.

Lead charges : The conveyance charges of the materials from sourceto the site of work is called lead charge. In S.S.R. the lead charges are given for

Metalled roads. The equivalent distance of metalled road for cart track = 1.1xlead,

while for sandy track = 1.4xlead.

Lead statement :Lead statement gives the cost of various materials at

site. It includes basic rate, plus conveyance, blasting charges, seignorage charges

etc.

Lead Statement

S.

No

Mat-

erial

Source Unit Cost at

source

Lead

inKm.

Equi

valentmetal

led

road

Blas

tingchar

ges

Seign

oragechar

ges

Cess

charges

Cru

shingchar

ges

Deduc

tionsif any

Net

rateat

site

Re

marks


55/83


5.5 Preparation of unit rates for finished items of works

Cost of sand as per S.S.R. : For concrete = Rs. 375., For filling =Rs. 288., For plastering = Rs.490.

Cost of cement = Rs. 5100/ton., = Rs. 255 per bag.

Mixing charges for mixing 1 m3 of mortar = Rs. 85.

Cost of preparation of 1 m3 mortar for different proportions

5.5.1. Cement concrete in foundation (1:5:10)1

Quantity of cement =(1.52/16)x1=0.095 m3=0.095x1440/50=2.74

bags.

Quantity of sand = (1.52/16)x5=0.475 m3

Quantity of aggregate = (1.52/16)x10=0.95 m3.

Cost of cement = Rs.255 per bag., Cost of sand=Rs. 375/m3., Cost of

Coarse aggregate=Rs.588/m3.

Cost of

sand

Rs. 323.40

Rs. 367.

50

Rs. 392.

00

Rs. 406.

70

Rs.419.95

Rs. 436.10

Rs. 445.9

Mix

-ing

charges

Rs.85

Rs.85

Rs.85

Rs.85

Rs.85

Rs.85

Rs.85

Mix

propor

-tion

1:2

1:3

1:4

1:5

1:6

1:8

1:10

Quantity of

cement in

bags

9.5 bags

7.2

5.76

4.79

4.11

3.19

2.62

Quantity

of sand

in m3

0.66

0.75

0.8

0.83

0.857

0.89

0.91

Cost of

cement

Rs.2422.50

Rs. 1836.

00

Rs. 1469.

00

Rs. 1221.

50

Rs.1048.

05

Rs.813.

45

Rs.668.

10

Total cost

2831.50

2288.50

1946.00

1713.20

1553.00

1334.55

1199.00


56/83


R.C.C. (1:2:4) works in Beams, slab, columns etc

Quantity of cement = 1.52x1/7=0.217 m3 =0.217x1440/50=6.25 bags.

Quantity of sand = 1.52x2/7=0.434 m3.

Quantity of coarse aggregate = 1.52x4/7=0.869 m3.

Quantity of steel =1.1x78.5/100=0.86quintals=86.35 kgs.

Centering and scaffolding charges with casurina ballies, bamboos,

wooden reapers, poles etc.

Lintel = Rs. 1215/m3; Sunshades = Rs. 214/m2., Columns = Rs. 929/

m2., Beams = Rs. 1637/m2.

Slabs up to 150 mm. = Rs. 184/m2.

Particulars

Materials Cement

Sand

Coarse aggregate

Labor: Head mason

Mason

Men mazdoor

Women mazdoor

Waterman

Add 20% for labor

Quantity

2.74 bags

0.475 m3

0.95 m3

0.05 No.

0.15 No.

1.2 NO.

1.8 No.

0.4 No.

Rate

Rs. 255/bag

Rs. 375/m3.

Rs. 588/bag

Rs. 350/No.

Rs. 315/No.

Rs. 250/No.

Rs. 250/No.

Rs. 250/No.

Total

Cost

Rs. 698.70

Rs. 178.15

Rs. 558.60

Rs. 17.50

Rs. 47.25

Rs. 300

Rs. 450

Rs. 100

Rs.182.95

Rs.2533.15

Particulars

R.C.C(1:2:4) including cost of

materials, labour charges,centering charges but excluding

cost of steel and its fabrication.

Materials

Cement

Quantity

6.25 bags

Rate

Rs. 255/bag

Amount

Rs. 1593.75


57/83


Sand

Coarse aggregate

Labour

Head mason

Mason

Men mazdoor

Women mazdoor

Waterman

Total cost of materials & labour

= Rs.2760.30+1274.40=

Rs.4034.70

R.C.C. works in lintel, slab,

beams and columns

Centering charges with

Casuarinas baileys, bamboos,

poles, wall plates etc.

Item

Lintel

Slab

Beam

Column

0.434 m3

0.868 m3

0.05

0.3

1.2 No.

2.0 NO.

0.6 No.

Centering

charges

including

materials and

labourRs.1215

Rs. 1533.33

Rs. 1637

Rs.929

Rs. 375/m3

Rs. 1161.80/m3

Total

Rs. 350/No.

Rs. 315/No.

Rs. 250/No.

Rs. 250/No.

Rs. 250/No.

20% local

allowance

Cost of

materials and

labour

Rs. 4034.70

Rs. 4034.70

Rs. 4034.70

Rs. 4034.70

Rs. 162.75

Rs. 1003.80

Rs. 2760.30

Rs. 17.50

Rs. 94.50

Rs. 300.00

Rs. 500.00

Rs. 150.00

Rs. 1062.00

Rs. 212.40

Rs. 1274.40

Total Cost

Rs. 5249.70

Rs. 5568.00

Rs. 5671.70

Rs. 4963.70


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1 m3 of R.C.C. work requires approximately 90 kgs. of steel. The cost

of fabrication of steel including bending and placement in position is Rs. 6.00/

Kg.

5.5.3 Brick masonry in cement mortar

The size of the bricks considered are 19 cmx9 cmx9 cm. The volume

of mortar is 0.32 m3. Cost of brick masonry for 1.0 m3 is considered.

Number of bricks required = 500

Mortar with a proportion of 1:6 is considered.

Quantity of cement = 0.32/7=0.0457 m3=0.0457x1440/50=1.32 bags

Quantity of Sand = 0.32x6/7=0.274 m3

Cost of 1000 no. of bricks 19cmx9cmx9cm as per S.S.R. =Rs. 4687,

Loading and unloading charges=Rs.37.30, Conveyance charges

=118.65+17.80x10=Rs. 297.( for 15 K.M.)

Total cost of bricks = Rs.4687+Rs.37.30+297=Rs.5021.30

Quantity

500 Nos.

1.32 bags

0.274 m3.

Materials cost

0.05 No.

1.0 No.

0.7 NO.

1.0 No.

0.2 No.

Particulars

Brick masonry in

superstructure including cost

of materials and labour

Materials

Bricks

Cement

Sand

Labour

Head mason

Mason

Men mazdoor

Women mazdoor

Waterman

Rate

Rs. 5021.30per

1000 Nos.

Rs. 255 per bag

Rs. 490/m3.

Total

Rs. 350/No.

Rs. 315/No.

Rs. 250/No.

Rs. 250/No.

Rs. 250/No.

Amount

Rs.2510.65

Rs. 336.60

Rs. 134.30

Rs. 2981.55

Rs. 17.50

Rs. 315.00

Rs. 175.00

Rs. 250.00

Rs. 50.00


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5.5.4 Course rubble stone masonry(CRS) in cement mortar

Quantity of stone required = 1.25 m3. Volume of mortar required=40%=0.4.

Quantity of cement required for C.M. 1:6 = 0.4/7=0.06 m3=0.06x1440/

50=1.8 bags.

Total

Add 20%

Rs. 807.50

Rs.161.50

Rs. 969.00

Rs. 3950.55Materials and Labour

Total Cost

Particulars

Materials

Stone including bond

stone and wastage

Cement

Sand

Labour

Head mason

Mason

Men mazdoor

Women mazdoor

Waterman

Total cost of materials and

labour

Quantity or No.

1.25 m3.

1.8 bags

0.36 m3.

0.05 No.

1.6 No.

1.6 No.

0.8 No.

0.15 No.

Rate

Rs.535.60/m3

Rs. 255/ bag

Rs. 490/m3.

Rs. 350/No.

Rs. 315/No.

Rs. 250/No.

Rs. 250/No.

Rs. 250/No.

Add 20%

allowance

Amount

Rs. 669.5

Rs. 459

Rs. 176.40

Rs. 1304.90

Rs. 17.50

Rs. 504.00

Rs. 400.00

Rs. 200.00

Rs. 37.50

Rs.1159.00

Rs. 231.80

Rs. 1390.80

Rs. 2695.70


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Quantity of sand= 0.36 m3. Cost of rubble stone =

Rs.293+Rs.74.60+11.20x15 = Rs. 535.60 for a conveyance of 20 K.M.

5.5.5 Plastering

External plastering 20 mm. thick and Internal plastering 12 mm. thick.

Materials for 20 mm. thick plastering in a wall of 100 sq. m.

Volume of plastering = 100x20/1000=2.0 m3.

Add 20% for wet volume and increasing 25% dry

volume=2.0+0.4+0.6=3.0 m3.

Cost of 1:6 cement mortar = Rs. 1553.00/m3. Cost of 3.0 m3 cement

mortar=1553.00x3=Rs.4659.00

Labour charges : Head mason =1/3 no. Cost=(1/3)x350=Rs. 116.70

Mason=12 Nos. Cost=10x315=Rs. 3150.00 Men mazdoor=15 Nos.

= 15x250= Rs. 3750.00

Waterman= No. Cost = (3/4)x250=Rs. 187.50.

Cost of labour = Rs.116.70+Rs. 3150+Rs.3750.00+Rs. 187.50= Rs.

7204.20 Add 20% allowance =Rs. 1440.80. Total

cost of labour = Rs. 7204.20+1440.80=Rs. 8645.00

Total cost of external plastering=Rs.4659.00+ Rs. 8645.00=Rs.

13304.00Cost of 20 mm. thick plastering/m2 = 13304.00/100= Rs.133.04

Materials for internal plastering 12 mm. thick for 100 m2.

Volume of plastering= 100x12/1000=1.2 m3. Add 30% for uneven

surfaces and 25% for dry volume.

Total volume of plastering = 1.2+0.36+0.29=1.95 m3. say 2.0 m3.

Cost of 1:6 cement mortar for 1 m3= Rs. 1553.00 Cost of 2.0 m3

mortar = 2x1553.00= Rs.3106.00

Labour charges = Rs. 8645.00.

Total cost of plastering 12 mm. thick = Rs. 3106.00+ Rs.8645.00=Rs.

11751.00

Cost of plastering 12 mm thick per m2= 11751/100=Rs. 117.51


61/83


5.5.6 Pointing in cement mortar

For pointing in brickwork the total dry volume of materials is taken as

0.60 m3 for 100 m2.

Pointing with cement mortar of proportion 1:2 : Dry volume of mortar

= 0.60 m3

Cost of mortar 1:2 for 1 m3=Rs. 2831.50. Cost of 0.6 m3 mortar =

0.6x2831.50=Rs. 1699.00

Labour :Head mason (1/3)x350=Rs. 116.70

Mason = 10x315=Rs.3150.00; Men mazdoor=10x250=Rs.2500.00;

Waterman=0.5x250=Rs. 125.00

l Cost of labour = 116.70+3150+2500+125.00=Rs. 5891.70

Add 20% allowance=Rs.1178.30; Total cost = 5891.70+1178.30=

Rs.7070.00

Total cost of materials and labour = 1699.00+7070.00=Rs.8769.00

Cost of pointing per m2= 8769.00/100=Rs. 87.70

5.5.7. Cement concrete flooring

Considering 2.5 cm. thick concrete for an area of floor = 100 m2.

Volume of concrete floor = 100x2.5/100=2.5 m3. Add 10% for

unevenness of concrete

Quantity of concrete = 2.5+0.25=2.75 m3. Add 50% for dry volume

of concrete=1.375 m3.

Total quantity of concrete= 2.75+1.375=4.125 m3.

Quantity of cement required = 4.125/7=0.60 m3.=0.6x1440/50=18

bags.

Quantity of sand= 0.6x2=1.2 m3. Quantity of stone aggregate = 0.6x4

= 2.4 m3.

Cement for surface finishing = 100x2/1000=0.2 m3. = 0.2x1440/50=6bags.

Cost of cement= Rs. 255/ bag; Cost of sand= Rs. 490/m3.; Cost of

aggregate = Rs.1161.80/m3.


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Cost of cement concrete flooring per sq. meter = 17491.00/100=Rs.174.91/sq m.

5.5.8. Doors and windows paneled and glazed

Consider preparation of door frame with Sal wood . The size of the

door is 1.00 m. x 2.00 m.

Particulars

Materials

Stone aggregate

Sand (coarse)

Cement

Cement for surface

finishing

Labour etc.

Head mason

Mason

Men mazdoor

Women mazdoor

Waterman

Total cost of materials

Total cost of labour

Side forms for finishing

Quantity or No.

2.40 m3.

1.20 m3.

18 bags

6 bags

no.

10 Nos.

5 Nos.

5 Nos.

2 Nos.

Add 20% extra

Side forms

Rate per

Rs. 1161.80/m3.

Rs. 490/m3.

Rs. 255/ bag

Rs. 255/ bag

Rs. 350/day

Rs. 315/day

Rs. 250/day

Rs. 250/day

Rs. 250/day

Lump sum

Lump sum

Total cost

Amount

Rs. 2788.40

Rs. 588.00

Rs. 4590.00

Rs.1530.00

Rs. 9496.40

Rs. 262.50

Rs. 3150.00

Rs. 1250.00

Rs.1250.00

Rs.500.00

Rs. 6412.50

Rs.1282.50

Rs. 7695.00

Rs. 300.00

Rs. 9496.40

Rs. 7695.00

Rs. 300.00

Rs.17491.00


63/83


Materials :Teakwood of cross section 8 cmx12 cm.

Length of the frame = 2x( 2.14+1.2)=6.68 m. Quantity of

timber=6.68x0.08x0.12=0.064 m3.

Add 5% for wastage = 0.0032 m3. Total quantity of timber

=0.064+0.0032=0.0672 m3.

Rate of sal wood = Rs. 40012.00/m3.

Cost of timber = 0.0672x40012.00= Rs. 2688.80

Labour : Head carpenter =1/16 No. Cost =350x1/16= Rs.21.90

Carpenter =1/4 No. Cost =315x1/4= Rs.78.75

Men mazdoor = No. Cost =250x1/2= Rs.125.00Cost of labour Rs.225.65

Add 20% allowance Rs.45.20

Total cost of labour Rs.270.85

Total cost of materials and labour = Rs. 2688.80+Rs.

270.85=Rs.2959.65 say Rs. 2960.00

Width of the plank=1.0-0.10-0.10-0.10=0.6 m. (Width of the stiles)

Length of the plank = 2.0-0.10-0.10-0.10-0.15-0.10=1.55 m. (width

of top, frieze, lock and bottom rails

Unit rate of 40 mm. thick paneled door shutter of size 1.0x2.0 sq m.

double door in teak wood.

AmountParticulars

Materials:-

timber

Stiles

Top railFrieze rail

Lock rail

Bottom

rail

No.

4

1

1

1

1

L

2.00

1.00

1.00

1.00

1.00

B

0.10

0.10

0.10

0.15

0.10

Thick

ness

0.04

0.04

0.04

0.04

0.04

Quantity/

Nos.

0.032

0.004

0.004

0.006

0.004

Rate


64/83


Planks for

panels

Brass

accessories

Tower bolt

30 cm.

Tower bolt

15 cm.

Handle 10

cm.

Hinges

Aldrop 30

cm.

Door

stopper

Labour

Head

carpenter

Carpenter

Helpers

1

1No.

1No.

2.no

6.no

1 No.

1 No.

1/15

No.

4 Nos.

1.55

Add

5%

0.6

for

Cost

0.025

wastage

Of

0.023

0.073

0.00365

0.0767 m3

1 No.

1 No.

2 Nos.

6 Nos.

1 No.

1 No.

accessories

1/15 No.

4 Nos.

2 Nos.

Rs.1054

86.00/m3

Rs.248

.00/No.

Rs.121.00/

No.

Rs.337.00/

No..

Rs.112.00/

No.

Rs.

729.00/No.

Rs.

146.00/

No.

Rs. 350/

day

Rs. 315/day

Rs. 250/

day

Rs.8090.80

Rs. 248.00

Rs.121.00

Rs.674.00

Rs. 672.00

Rs. 729.00

Rs.146.00

Rs.2590.00

Rs. 23.35

Rs.1260.00

Rs.500.00

Rs.

1783.35


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Cost of materials = Rs. 8090.80

Cost of brass accessories=Rs.2590.00

Cost of labour = Rs. 2140.00

Total cost = Rs.12820.80

Summary

Specification defines the nature and class of work, materials to be used

in the work, workmanship etc.

Cost of materials at the source : The amount required to purchase

the materials at the source of its production is the cost of materials at the source.

Cost of materials at the site = Cost of materials at the source + Seignories

+ Taxes + Royalties + Transport + Loading + unloading etc.

Cost of transport on metal led road is given in the S.S.R.

Distance on cart track = 1.1 x Distance on metal led road

Distance on sand track = 1.4 x Distance on metal led road

Standard Schedule of Rates (S.S.R.) :Standard schedule of rates

consists of the rates of materials, machinery, hiring charges and wages of labour.

It is prepared by the board of chief engineers and approved for that year.

Lead and Lift : The horizontal distance between the source of the

material to the work site is known as the lead. The vertical height through which

the material is lifted is known as the lift.

Lead Statement :The statement in detail of the cost of materials at the

site is known as the lead statement.

Quantity of materials in Plain cement concrete (1:5:10) :

Quantity of cement = 1.52 x 1/16 = 0.095 cu m. = 0.095 x 1440/50 =

2.74 bags

Quantity of sand = 1.52 x 5/16 = 0.475 cu m.

Add 20%

extraTotal

Rs. 356.65

Rs. 2140.00


66/83


Quantity of coarse aggregate = 1.52 x 10/16 = 0.95 cu m.

Brick masonry in cement mortar for 1.0 cu m.

Number of bricks of size 19 cm. x 9 cm. x 9 cm. = 500

Volume of mortar = 0.32 cu m.

Course rubble masonry :

Quantity of stone = 1.25 cu m.

Volume of mortar = 0.40 cu m.

Plastering 20 mm. thick : The volume of cement sand mortar required

for an area of 100 sq m. and a thickness of 20 mm. is 3.0 cu m.

Plastering 12 mm. thick :The volume of cement sand mortar requiredfor an area of 100 sq m. and a thickness of 12 mm. is 2.0 cu m.

Pointing : The volume of cement sand mortar required for pointing of

an area of 100 sq m. with a mix proportion 1:2 is 0.60 cu m.


1. Define specification.

2. What is cost of materials at the source.?

3. What is the cost of materials at the site?

4. Write a tabular form for an abstract estimate.

5. List out the various types of labour.

6. Define standard schedule of rates.

7. What is lead and lift?

8. What is a lead statement.


1. Prepare specifications for the following

(a) Earthwork in excavation, (b) Cement concrete in foundation, (c) R.R. masonry, (d) Brick work in cement mortar.

2. Find the unit rate for Plain cement concrete (1:6:12)

3. Find the unit rate for course rubble masonry of cement mortar (1:6).


67/83


4. Find the unit rate for brick work in cement mortar (1:6) using standard

size of bricks.

5. Find the unit rate of plastering 12 mm. and 20 mm. thick with a

proportion of (1:5) cement mortar.

O.J.T. Questions

1. Prepare a unit rate of brickwork in cement mortar for 1.0 cu m.

using modular bricks.

2. Prepare a unit rate of R.C.C. (1:2:4) for 1.0 cu m. in slabs, beams

and columns.

3. Find the cost of a door (1.00m. x 2.00 m.) in country wood

4. Find the cost of a window (1.2 m x 1.2 m) in Sal wood.


68/83


Structure

6.0 Introduction

6.1 Trapezoidal, Prismoidal, Mid ordinate

6.2 Taking out quantities from L.S. and C.S. in cutting and embankment

Learning Objectives


Calcualate the quantities of earth work in banking and cutting by

Trapezoidal and Prismoidal Rule

6.0 Introduction

All types of roads, railways and irrigation works are constructed over

earthwork. To understand the calculation of earthwork involved in these structures,

these methods of calculation have to be studied in detail.

Cross section of earthwork is in the form of a trapezium. The quantity of

earthwork may be calculated by the following methods.

6UNIT

Earthwork Calculations


69/83


6.1. Trapezoidal, Prismoidal, Mid ordinate

Sectional and mean sectional area methods for calculating earthwork.

Mid sectional area method : In the mid sectional area method, the

average height of the two ends is taken as the mean depth. L is the length of the

section. B is the formation width, and S:1 is the side slope and d1 and d2 are the

height of the embankment at the two ends

Mean height dm = (d1+d2)/2

Area of midsection = Area of rectangular portion+ area of two triangular

portions=Bdm+1/2sdm2+1/2sdm2=Bdm+2dm2.

Quantity of earthwork = (Bdm+sdm2)xL

The quantities of earthwork may be calculated in a tabular form as below

Mean Sectional Area Method : In this method, the area at the ends

of depth d1 and d2 are calculated and the mean area of the section is found.

Sectional are at one end A1 = Bd1+S(d1)2

Sectional area at the other end = Bd2+S(d2)2=A2

The mean sectional area A=(A1+A2)/2

Quantity Q=((A1+A2)/2)xL

The quantities of earthwork may be calculated in a tabular form as follows

Stations Depth

or

Height

Mean

depth or

Height

Central

area Bd

Area of

sides Sd2

Total

sec

tional

area

Bd+Sd2

Length

between

stations L

Quantity

(Bd+Sd2)xL

Embank

ment cutt

ing

Station Height

or depth

Area of

central

port ion

Bd

Area of

s i d e s

Sd2

T o t a l

sectional

a r e a

Bd+Sd2

M e a n

sectional

area

Length

between

stations

L

Quantity

=(Bd+Sd2)

x LBankingCutting


70/83


Fig 6.1

Trapezoidal-Prismoidal Formula : In the prismoidal formula the areas

at the ends and the mid sectional area are als

estimating & costing

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