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Structure1.0 Introduction

1.1 Definition

1.2 Need for estimation and costing

Learning ObjecyivesAfter studing this unit, student will be able to

• Have an idea of the introduction to estimating and costing.

1.0 IntroductionIn the civil engineering field, the construction activity contains the

following three steps.

1. Plans : Preparation of drawings plan, section, elevation, with fulldimension and detailed, specifications meeting the requirements of the proposedstructure.

2. Estimation : Preparation of an estimate is for arriving the cost ofthe structure to verify the available funds or to procure the required funds forcompletion 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

Construction Technology152

The plans contains size of room and dimensions of the work and the estimatecontains the quantity and quality aspects of the structure.

1.1 DefinitionEstimation 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 theprocess 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 Questions1. What is meant by Estimating and Costing ?

2. State need for Estimation and Costing.


2.1 Units of measurements

2.2 Rules For Measurement

2.3 Different methods of tasking out quantities

Learning ObjectivesAfter the studying this unit student will be able to

• To measure various quantities as per rules.

2.0 IntroductionThe 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


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 ofpayment

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 alltypes of soils except rock re-quiring blastering.

(b) Earth work excavation inthe soils hard rock requirngblastering.

(c) Excavation of pipe linethrough of specified width anddepth inall types of soils

(d) Earthwork for road for-mation ,bund formation etc.cutting , embankment.

(e) Refilling of foundations ,basements, pipe lines, trencheswith excavated soils.

Plain cement concrete forfoundation.

R.R.masonry or brick ma-sonry for foundation base-ment, super strucrture, parapetwall etc.

Filling the basement withsand.

(a) RCC 1:2: 4 with normalreinforcement for plinth beam ,columns, lintels, verandahbeam- 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

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 ofspecified thickness .

Plastering pointing, flooring,weather proof coarse, whitewashing, colour washing, paint-ing.

Roofing with A.C sheets, tiledroofing, Kurnool trerrace, Ma-dras terrace etc.

D.P.C specified width and thick-nessWooden and steel trusses

Doors, windows, ventilators.

Ornamentel border of speci-fied width and thickness.

R.C.C pipes, A.C pipes GI orC.I pipes, stone ware pipesetc.

Steel reinforcement in R.C.C.

Rough stone pitching revet-ment and soiling of specifiedthickness.

(a) Roads works : Metal col-lections , gravel collections,solving stones, pitching anystones, revetment stones etc.

(b) Road works : Spreadingmetal gravel and consolidationwith 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


2.2 Rules For MeasurementMeasurement of works occupies a very important place in the planning

and execution of any work or project, from the time of the first estimate aremade until the completion and settlement of payments. The methods followedfor the measurement are not uiform and the practices or prevalent differconsiderably in between the states. Even in the same state different departmentsfollow different methods. For convernience a uniform method should be followedthroughout the country. The uniform methods of measurement to be followedwhich is applicable to the preparation of the estimates and bill of quantitiesand 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


(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 thestory 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 thefollowing principle.

(i) Mass, voluminious and thick works shall be taken in cabic unit orvolumne. The measurement of length, breadth, and height or depth shall betaken 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 tocompute the area (sq.m).

(iii) Long and thin work shall be taken in linear or running unit andlinear measurement shall be taken(running meter).

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

2.3 Different methods of taking out quantitiesThe items of work like earth work in excavation in foundation, foundation

concrete stone masonry in foundation and basement, stone or brick masonry insuper 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 runningin the direction generally the long walls out-to-out and the internal length ofwalls 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 ofwall. The same rule applicable to the excavation in foundation, to concrete infoundaiuon and to masonry.

The simple mehtod is to take the long walls of short or erros wallsseparately and to find out the centre to centre lengths of long wall anf shortwalls from the plan. For symmetrical footing on either sides, the centre lineremians same for suepr structure and for foundation and plinth.


For long walls add to the center length one breadth of wall, whichgives the length of the wall out-to-out ,multiplying this length by the breadth andheight 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 onebreadth of foundaiton. Adopt the same process for foudation conceret and foreacth footing. It should be noted that each footing is to be taken separately andthe 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 onone 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 centrelength one breadth of wall, which gives the length in-to-in, and repeat the sameprocess 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 wallsubstract one breadth from the centre length to get the corresponding lengths.

It will be noticed that by taking dimensions in this ways, the long wallsare gradually decreasing in length from foundation to superstructure, while theshort 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 generalmethod.

2.3.2 Centre line method

In this method known as centre line method. This method is easy andquick 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 ofsimilar cross sections. In this method the total centre line multiplied by breadthand 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 arecross walls or number of junctions). It requires special attention and considerationat the junctions, meeting points of partition or cross walls, etc.

For rectangular, circular polygonal (hexagonal, octagonal etc) buildinghaving no inter or cross walls, this method is quite simple. For each junction halfbreadth of the respective items or footings is to be deducted from the totalcentre length. Thus in the case of a building with one partition wall or cross wallhaving two junctions, for earthwork in foundation trench and foundation concretededuct one breadth of trench or concrete from the total centre length (half breadth


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 centrelength 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 typewalls shall be taken jointly first , and then all B type walls should be takentogether separately. In such cases no deductions of any kind need be made forA type walls, but when B type walls are taken, for each junction deducting ofhalf breadth of A type wall (main wall) shall have to be made from the totalcentre length of walls.

It may be noted that at corners of the building where two walls aremeeting no substraction or addition is required.

Note : Student should practice method I first and when they have becomesufficiently acquainted with method I, then only they should take up the methodII.

Short Answer Type Questions1. 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.



3.1 Detailed estimate

3.2 Preliminary or approximate estimate

3.3 Problems in preliminary estimate

Learning ObjectivesAfter studying this unit student will be able to

• Understand the definition of detailed estimate, stages of preparationof estimate, Data required for an estimate and types of estimate.

3.0 IntroductionAn 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 knowbeforehand 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 therewill not be much difference in the estimated cost and actual cost. The estimatorshould be fully acquainted with the methods of construction, skilled andexperienced for accurate estimating.

3UNIT

Types of Estimates

161Paper - II Estimating and Costing

3.1 Detailed estimateThe estimate may be approximate or preliminary estimate or accurate

estimate. 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, takingdetailed measurements of each item of work and calculating their quantities isknown as detailed estimate.

3.1.2. Stages of preparation

To prepare the complete estimation of the project, besides the estimatedcost 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 andthe cost of external services are to be provided. Provision of supervision chargesand contractors profit are to be provided in the estimate.

Data required for preparing an estimate : To prepare an estimatefor a work the following data are necessary.

Drawings : The detailed drawings of plan, elevation and section, drawnto a scale are necessary to take the details of measurements of various items ofwork.

Specifications : The specifications gives the nature, quality and classof materials, their proportion, method of execution and workmanship and theclass of labour required. The cost of the work varies with its specifications. Thecement 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 materialsto be used in construction, the wages of different categories of labour should beavailable for preparing an estimate. The location of the work and its distance ofsource of materials and cost of transport should be known. These rates may beobtained from the Standard Schedule of Rates prepared by the engineeringdepartments.

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 ofvarious items of work and an abstract estimate of the quantities and their unitrates are required.


Detailed Estimate

Abstract estimate

3.2. Preliminary or approximate estimatePreliminary or approximate estimate is required for preliminary studies

of various items of work or project , to decide the financial position and policyfor administrative sanction by the competent authority. The preliminary estimateis prepared by different methods for different types of works. The variousmethods of preparing the preliminary estimate are plinth area estimate, cubicalrate 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 buildingand multiplying by the plinth area rate. The plinth area rate is obtained bycomparing the cost of the cost of similar building having similar specifications inthe locality.

3.2.2. Cubic area estimate

The cubic rate estimate is prepared on the basis of the cubical contentsof the building. The cubic rate is obtained from the cost of the similar building inthe locality having similar specifications. The cost of the building is estimated bymultiplying the volume of the building with the cubic area rate. Cubic rate estimateis 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 andworks 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 andfor irrigation works, the preliminary estimate is made per kilometer. For bridgesand culverts per running meter. For sewerage and water supply projects on thebasis 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


3.3. Problems in preliminary estimate1. If the cost of school building per student is Rs. 25000. Calculate the

cost 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

SummaryDetailed 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 rateestimate and estimate per unit base.

Short Answer Type Questions1. 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 Questions1. Describe the various types of preliminary estimates.



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


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.

4UNIT

Detailed and AbstractEstimate of Buildings


4.0 IntroductionTo estimate the cost of any building or a structure, drawings,

specifications and rates are required. Regarding the detailed estimate by longwall and short wall method and centre line method, the drawings consisting ofplan elevation and section are sufficient. The estimator should be able to take allthe dimensions from the drawings. The length and breadth are taken from theplan, while the height or depth are taken from the section and elevations. In longwall and short wall method the walls are taken separately, while in the centre linemethod, the centre line lengths of all the walls are combined. The accuracy ofestimate depends upon the skill of the estimator in studying the drawings. Thelong wall and short wall method is useful for load bearing type structure, but itcannot 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 asseparate 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 + ½ width of the wall.

6. Centre to centre length of short wall = internal length of short wall + ½ 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


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 thecentre line lengths of long walls and short walls are added to get the total centreline length. At the junctions of two walls, the length is present in both of thewalls. Hence half of the length of that width is to be subtracted from the totalcentre line length.

Length = Total centre line length – ½ 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


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 are7.0 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.

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

Descriptionof work

Earth work inexcavation

C.C. bed infoundation

R.R. masonryin foundationand plinth

First footing

Second footing

Basement

Brick work insuper 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


Sl.No.

1

2

3

Description of work

Earth work inexcavation infoundation

Long Walls

Short Walls

Plain cement concretein foundation (1:5:10)

Long Walls

Short Walls

R.R. Masonry infoundation& 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


4.2 Two roomed building( load bearing type structure)Detailed Estimate Of A Double Roomed Building By Long Wall And

Short Wall MethodCentre to centre length of long wall = 5.0 + 0.3 + 5.0 + 2x0.3/2 = 10.6m.Centre to centre length of short wall = 5.0 + 2x0.3/2 = 5.3 m.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 superstructure 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 ofwork

Earthwork inexcavation

Long Walls

Short Walls

C.C. bed infoundation

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


3

4

Short Walls

R.R. masonry infoundation &plinth

First footing

Long Walls

Short walls

Second footing

Long Walls

Short Walls

Third footing &plinth

Long Walls

Short walls

Brick work insuper 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 =10.9L = 5.3 - 0.3 =5.0

R.R. masonry Total


Centre line method

Fig 4.2 Double Room

No.

1

1

1

1

1

11

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

Total3

Sl. No.

1

2

3

4

Description ofworkEarthwork inexcavationC.C. bed infoundationR.R. masonry infoundationFirst footing

Second footing

Basement

Brickwork insuperstructure

Remarks

L= 37.1 -2x1/2x1.2

L = 37.1 -2x1/2x0.9L = 37.1 -2x1/2x0.7L = 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 m333.12 m3

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

5m x 5 m 5 m x 5 m

1.20.90.70.5

0.30.60.3

3.0


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

0.90.9

1.2m

3.0m


56.43

m m mm3

5m4m

5m4m

5m4m

5m4m

Basement

Basement


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 beamsunder 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 ofthe 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 andbreadth 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 0.23R.C.C Column

3.0 m

1.2 m

0.9 m

0.3 m

0.3 m

R.C.C.FootingR.C.C.

G.L G.L

100 mmthickRCC slab

1.2 m


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 infoundation

First footing

Second footing

Brickwork insuperstructure

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


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 fromslab

R.C.C. Slab.

External plastering20 mm

Thick

Deductions

Doors

Windows

Internal Plastering12 mm thick

Rooms 6mx5m

Rooms5mx4m

Sand filling in rooms

Rooms 6mx5m

1

1

1

1

9

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


4.5 Primary school building with sloped roofWall 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 ofsteel

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

40

100

(8.76+4.35+3.68+26.25)x1.25x87.5/100x1000 78.5x100/100x1000tonnes 4.22 t


Length of the sloping roof = square root of (1.5mx1.5m + 1.0m.x1.0m.) = 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 cmseach = (4.6/0.1)+1=47


ELEVATION

W W

DD

Room3.0 x 3.0 m

Room3.0 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

TilesTiles

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


4.6 RCC Dog legged – open well stairs

Fig 4.5 Dog Legged Stair case

1650

250

150Floor

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


Sloping side 2 2 0.28 2.4640.411.264Total


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 m3Landing 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 = 11Height of the first flight = 11x0.15=1.65 m.Number of treads = 10Length 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

ht N

o. N

o. o

f R

isers

N

o. o

f Tr

eads

E

ach

Rise

r

Each

Tre

ad

A

8

8

152

3

00

B

4

3

152

30

0

C

8

7

152

300

SECTION AT ‘AA’

Note :1. All dimensions are in Milli meters2. Follow the written dimensions only

OPEN WELL TYPE STAIRCASE

Scale 1:50

DRG. No. 18


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.


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

Weatheringcourse

Lintel &sunshadeBrickmasonry

Roof slab

C.C. flooring

R.C.C Mix1:4:1

Sand fillingC.C. floring1:4:8

Elevation0.902

3.05 m

3.05 m


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.


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


4.8 Detailed estimate of compound wall and stepsLength 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.


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

0.150.15

0.3

0.3

0.3

1.0 mTop view

Front View

Side View

Fig. 4.9 Plan and Section of a compound Wall


SummaryTo 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 methodand 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 – width

In centre line method, the length = Total centre line length – (number ofjunctions)xwidth/2

For a double room building, the total centre line length = sum of thecentre line lengths of two long walls and three short walls. The numberof junctions = 2.

For a building with number of rooms, the total centre line length = sumof the centre to centre lengths of three long walls, three short walls of length 5.3m. 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, Plinthbeam, lintels and beams under slab are obtained by adding the internal dimensionsof the rooms.

The roof for the primary school building is a gable roof, having its slopein two directions. The roof under consideration is the roof having its width = 3.0m. 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.


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

Length of the sloping side = Square root of [(Treads length)2 + (Heightof flight)2].

Area of brickwork in each step = (Rise x Tread) x ½.

Short Answer Type Questions1. 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.

Long Answer Type Questions1. Find the earthwork in excavation, C.C. bed in foundation, R.R.

masonry in foundation, Brick work in superstructure and plastering for singleroom building and double room building by long wall short wall method andcentre 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.



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


• Prepare the unit ratio of various items of works. Find the cost of materials, specifications of various of various items of works.

5.0 IntroductionTo 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 workare calculated from the specifications of the various types of materials. The ratesare calculated as per the rates in the standard schedule of rates. The unit rates ofvarious items of work increase considerably with the specifications. The

5UNIT

Specifications and Analysisof Rates


specifications indicate the quality of the work while the drawings are used forthe quality of the work.

5.1 Prepare specifications for different items of workSpecifications specifies or describes the nature and the class of work,

materials to be used in the work, workmanship etc. From the study of thespecifications one can easily understand the nature of the work and what thework shall be.

Detailed specifications : Detailed specifications are written to expressthe 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 inexcavations 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 infoundation.

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.


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 rubblemasonry

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.


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 thecost 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 areunder 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


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.


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


• 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


• 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


• 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


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


given. If transport is required on a cart track or a sand track, to reachthe 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 givenvolume in the case of materials like sand and coarse aggregate. Bricks are stackedfor a given number. Sometimes are stacking charges are included in loading andunloading. Loading and unloading charges are fixed for a given volume or weightfor different materials.

The cost of labor wages for each category of labor are given above asper 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 laborare prepared. It is prepared by the board of chief engineers and approve it forthat year.

5.4 Concept of lead and lift- leads statementThe distance between the source of material to the worksite is known

as the lead. This lead distance changes from one project to another projectdepending upon the location. The vertical height through which the material is tobe 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 forMetalled 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 atsite. It includes basic rate, plus conveyance, blasting charges, seignorage chargesetc.

Lead Statement

S.No

Mat-erial

Source Unit Cost atsource

LeadinKm.

Equivalentmetalledroad

Blastingcharges

Seignoragecharges

Cesscharges

Crushingcharges

Deductionsif any

Netrateatsite

Remarks


5.5 Preparation of unit rates for finished items of worksCost 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=2.74bags.

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 ofCoarse aggregate=Rs.588/m3.

Cost ofsand

Rs. 323.40

Rs. 367.50

Rs. 392.00

Rs. 406.70

Rs.419.95

Rs. 436.10

Rs. 445.9

Mix-ingcharges

Rs.85

Rs.85

Rs.85

Rs.85

Rs.85

Rs.85

Rs.85

Mixpropor-tion

1:2

1:3

1:4

1:5

1:6

1:8

1:10

Quantity ofcement inbags

9.5 bags

7.2

5.76

4.79

4.11

3.19

2.62

Quantityof sandin m3

0.66

0.75

0.8

0.83

0.857

0.89

0.91

Cost ofcement

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


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 ofmaterials, labour charges,centering charges but excludingcost of steel and its fabrication.

Materials

Cement

Quantity

6.25 bags

Rate

Rs. 255/bag

Amount

Rs. 1593.75


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 withCasuarinas 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.

Centeringchargesincludingmaterials andlabour

Rs.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% localallowance

Cost ofmaterials andlabour

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


1 m3 of R.C.C. work requires approximately 90 kgs. of steel. The costof 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 volumeof 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 insuperstructure including costof materials and labour

Materials

Bricks

Cement

Sand

Labour

Head mason

Mason

Men mazdoor

Women mazdoor

Waterman

Rate

Rs. 5021.30per1000 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


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 LabourTotal Cost

Particulars

Materials

Stone including bondstone and wastage

Cement

Sand

Labour

Head mason

Mason

Men mazdoor

Women mazdoor

Waterman

Total cost of materials andlabour

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


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% dryvolume=2.0+0.4+0.6=3.0 m3.

Cost of 1:6 cement mortar = Rs. 1553.00/m3. Cost of 3.0 m3 cementmortar=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= ¾ 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. Totalcost of labour = Rs. 7204.20+1440.80=Rs. 8645.00

Total cost of external plastering=Rs.4659.00+ Rs. 8645.00=Rs.13304.00

Cost 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 unevensurfaces 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 m3mortar = 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


5.5.6 Pointing in cement mortar

For pointing in brickwork the total dry volume of materials is taken as0.60 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% forunevenness of concrete

Quantity of concrete = 2.5+0.25=2.75 m3. Add 50% for dry volumeof 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=18bags.

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 ofaggregate = Rs.1161.80/m3.


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 thedoor is 1.00 m. x 2.00 m.

Particulars

Materials

Stone aggregate

Sand (coarse)

Cement

Cement for surfacefinishing

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

¾ 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


Materials : Teakwood of cross section 8 cmx12 cm.

Length of the frame = 2x( 2.14+1.2)=6.68 m. Quantity oftimber=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 = ½ No. Cost =250x1/2= Rs.125.00

Cost 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. (widthof 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 rail

Frieze rail

Lock rail

Bottomrail

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

Thickness

0.04

0.04

0.04

0.04

0.04

Quantity/Nos.

0.032

0.004

0.004

0.006

0.004

Rate


Planks forpanels

Brassaccessories

Tower bolt30 cm.

Tower bolt15 cm.

Handle 10cm.

Hinges

Aldrop 30cm.

Doorstopper

Labour

Headcarpenter

Carpenter

Helpers

1

1No.

1No.

2.no

6.no

1 No.

1 No.

1/15No.

4 Nos.

1.55

Add5%

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.105486.00/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


Cost of materials = Rs. 8090.80

Cost of brass accessories=Rs.2590.00

Cost of labour = Rs. 2140.00

Total cost = Rs.12820.80

SummarySpecification 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 purchasethe 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 ratesconsists 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 thematerial to the work site is known as the lead. The vertical height through whichthe material is lifted is known as the lift.

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

Total

Rs. 356.65

Rs. 2140.00


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

Short Answer Type Questions1. 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.

Long Answer Type Questions1. 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).


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. Questions1. 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.



6.1 Trapezoidal, Prismoidal, Mid ordinate

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


• 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 overearthwork. 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 ofearthwork may be calculated by the following methods.

6UNIT

Earthwork Calculations


6.1. Trapezoidal, Prismoidal, Mid ordinateSectional and mean sectional area methods for calculating earthwork.

Mid sectional area method : In the mid sectional area method, theaverage height of the two ends is taken as the mean depth. L is the length of thesection. B is the formation width, and S:1 is the side slope and d1 and d2 are theheight of the embankment at the two ends

Mean height dm = (d1+d2)/2

Area of midsection = Area of rectangular portion+ area of two triangularportions=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 endsof 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 orHeight

Meandepth orHeight

Centralarea Bd

Area ofsides Sd2

TotalsectionalareaBd+Sd2

Lengthbetweenstations L

Quantity(Bd+Sd2)xLEmbankment cutting

Station Heightor depth

Area ofcentralportionBd

Area ofs i d e sSd2

T o t a lsectionala r e aBd+Sd2

M e a nsectionalarea

LengthbetweenstationsL

Quantity=(Bd+Sd2)x LBankingCutting


Fig 6.1

Trapezoidal-Prismoidal Formula : In the prismoidal formula the areasat the ends and the mid sectional area are also taken into consideration. If thearea at the ends are A1 and A2 respectively and Am is the mid sectional area,Quantity or volume = (A1+A2+4Am)xL/6

• Cross sectional area at one end A1 = Bd1+S(d1)2

• Cross sectional area at the other end = A2 = Bd2+ S(d2)2

• Depth at the mid section = dm = (d1+d2)/2

• Area at the mid section = Bdm+S(dm)2 = Am

• Quantity = (A1+A2+4Am)xL/6

Trapezoidal formula and prismoidal formula for a series of crosssections : When the series of cross sections A0,A1, A2,A3, …………An areat equal distances D, then the volume by the trapezoidal formula is given by V =((A0+An)/2+A1+A2+A3+ ………..+An-1 +An)

Volume by Prismoidal formula : V=((A0+An)+2(Sum of the oddareas)+4(Sum of even areas))xD/3

Example 1 : Calculate the quantity of earthwork for 200 metre lengthfor a portion of a road in an uniform ground. The heights of the banks at the twoends are 1.00 and 1.60 m. The formation width is 10 metre and side slopes are2:1. Assume that there is no transverse slope.

B

Sd1 Sd1

1:S

1:S

d1

B

Sd2 Sd2

1:S

1:Sd2

d1

B

L


Mid sectional area method : Height d1 = 1.00m. Height d2 = 1.60m. Formation width = B = 10 m.

• Height at the mid section dm = (d1+d2)/2 = (1.00+1.60)/2=1.3 m. Side slopes S = 2.

• Area at the mid section = Bdm + S(dm)2 =10x1.3 + 2(1.3)2 = 16.38 sq. m. Length = L = 200 m.

• Quantity = Area x length = ((Bdm+S(dm)2)xL=16.38x200 = 3276 cu m.

• Mean sectional area method : Quantity = Mean sectional area x length

• A1 = Sectional area at one end = Bd1 + S(d1)2 =10x1+2(1.0)2 = 12 sq m.

• A2 = Sectional area at another end = Bd2+S(d2)2 =10x1.6+2(1.6)2= 21.12 sq m.

• Mean sectional area = Am = (A1+A2)/2 =(12+21.12)/2 = 16.56 sq m.

• Quantity = Mean sectional area x length = 16.56x200=3312 cu m.

• Prismoidal formula : Quantity = (A1+A2+4Am)xL/6

• A1 = sectional area at one end = Bd1+S(d1)2 = 10x1.0+2(1)2 = 12 sq m.

• A2 = Sectional area at another end = Bd2+S(d2)2 = 10x1.6+2(1.6)2= 21.2 sq m.

• Am = Mid sectional area = Bdm+S(dm)2 dm = (d1+d2)/2= (1.0+1.6)/2 = 1.3 m.

• Am = Bdm+S(dm)2 = 10x1.3+2(1.3)2 = 16.38 sq m.

• Quantity = (12+21.12+4x16.38)x200/6 = 98.64x200/6= 3288 cu m.

• Area of side sloping surface : Area of side slopes = Lxdx(square root of (S2+1))

Example 2 : Calculate the area of the side slopes of a portion of a bankfor a length of 200 m. The heights of the banks at the two ends are 2.50 m and3.50 m. and the ratio of side slope 2:1. If the side slopes are to be provided with15 cm. thick stone pitching, calculate the cost of pitching at the rate of Rs. 200per cu m.


• Mean height = (2.5+3.5)/2 = 3.0 m.

• Sloping breadth at the mid section = d(square root of s2+1)=3[Square root of( 2x2)+1] = 6.71 m.

• Area of the two side slopes = 2x200x6.71 = 2684 sq m.

• Quantity of pitching = Area x thickness =2684x0.15 = 402.6 cu m.

• Cost of stone pitching = 402.6 x 400=Rs. 161040.

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

Example : Reduced level (R.L.) of ground along the centre line of aproposed road from chainage 10 to chainage 20 are given below. The formationlevel at the 10th chainage is 107 m. and the road is in downward gradient of 1 in150 up to the chainage 14 and then the gradient changes to 1 in 100 downward.Formation width of the road is 10 metre and side slopes of banking are 2:1.Length of the chain is 30 metre. Calculate the quantity of earthwork.

Chainage : 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20

R.L. of ground : 105.00, 105.60, 105.44, 105.90, 105.42, 104.30105.00 , 104.10, 104.62, 104.00, 103.30

R.L. formation : 107.00, 106.80, 106.60, 106.40, 106.20,105.90.105.60 105.30 105.00 104.70 104.40

Height of bank : 2.00, 1.20, 1.16, 0.50, 0.78, 1.60, 0.60, 1.20,0.38, 0.70, 1.10

Chainage

10

11

12

13

14

15

16

Heightor Depth

2.00

1.20

1.16

0.50

0.78

1.60

0.60

Meanheightor depth

-

1.60

1.18

0.83

0.64

1.19

1.10

CentralareaBd

16.00

11.80

8.30

6.40

11.90

11.00

SideareaSd2

5.12

2.78

1.38

0.82

2.83

2.42

TotalareaBd+Sd2

21.12

14.58

9.68

7.22

14.73

13.42

Length inbetweenchainage

30

30

30

30

30

30

Quantity =[(Bd+S(d)2]xLBankingCutting

633.6 -

437.4 -

290.4 -

216.6 -

441.9 -

402.6 -


A railway embankment is 10 m. wide with side slopes 11/2 to1. Assumethe ground to be level in direction transverse to the centre line, calculate thevolume contained in a length of 120 metres, the centre heights at 20 m. intervalsbeing 2.2, 3.7, 3.8, 4.0, 3.8, 2.8, 2.5 m.

For a level section, the area is given by A=(b+nh)h

• Slope is 11/2:1. Hence n=1.5

The areas at different sections will be as under

• A1 = (10+1.5x2.2)2.2=29.26 m2.

• A2 = (10+1.5x3.7)3.7=57.54 m2.

• A3 = (10+1.5x3.8)3.8=59.66 m2.

• A4 = (10+1.5x4.0)4.0=64.00 m2.

• A5 = (10+1.5x3.8)3.8=59.66 m2.

• A6 = (10+1.5x2.8)2.8=39.76 m2.

• A7 = (10+1.5x2.5)2.5=34.37 m2.

• Volume by trapezoidal rule : V = d[(A1+An)/2 +A2+A3+A4+. . . . +An-1 ]

• V = 20[( 29.26+34.37)/2 +57.54+59.66+64.00+59.66+39.76] = 6258.9 m3.

• Volume by prismoidal rule : V=d/3[(A1+An)+2(Sum of odd areas)+4(sum of even areas)]

•V=20/ 3[(29.26+34.37)+2(59.66+59.66)+4(57.54+64.00+39.76)]=6316.5 m3.

Problems involving banking and cutting : At the 30th chainage the heightis banking of height 0.3 m. and at 31st chainage, it is cutting at a depth of 0.40 m.

17

18

19

20

1.20

0.38

0.70

1.10

0.90

0.79

0.54

0.90

9.00

7.90

5.40

9.00

1.62

1.25

0.58

1.62

10.62

9.15

5.98

10.62

30

30

30

30

Total

318.6 -

274.5 -

179.4 -

318.6 -

3513.6 cu m.


Find the volume of banking and cutting if the formation width is 10 m. and theside slopes are 2:1 in banking and 11/2 : 1 in cutting.

Chainage distance = 40 m. Let the height of embankment be zero at adistance of x mts.

• Length of cutting =( 40-x) . (x/0.3) =[(40-x)/0.4] 0.4x=12- 0.3x 0.7x= 12 x=17.14 say 17.0 m.

• Volume of banking : Mean height = (0.3+0.0)/2=0.15 m. Central area = 10x0.15 = 1.5 sq m.

• Side area = 2x(0.15x0.15)=0.05 sq m. Total area = 1.5+0.05=1.55 sq m.

• Volume of banking = Area x length = 1.55x17=26.35 m3.

• Volume of cutting : Mean depth = (0.0+0.4)/2 = 0.2 m. Central area = 10x0.2 =2.0 sq m.

• Side areas = 1.5(0.2x0.2) = 0.06 sq m. Total area = 2.0+0.06 = 2.06 sq m.

• Volume of cutting = Area x length = 2.06 x 23 = 47.38 m3.

Fig 6.2

Summary• Earthwork calculations are required for various engineering works as roads, railways, irrigation and water supply and sanitary works.

• The various methods of calculation of earthworks are Mid sectional area method, mean sectional area method, trapezoidal rule and prismoidal rule.

• Prismoidal formula is not applicable for even number of areas.

• Banking : If the earthwork is above the ground level it is banking.

0.3

0.4

40

(40-x)x


• Cutting : If the earthwork is below the ground level, it is cutting.

Short Answer Type Questions1. List out the various types of engineering works involving earthwork.

2. What are the various methods of calculating earthwork?

3. Define banking and cutting

4. Mention the relationship between the Reduced level of formation and the ground line

5. What is the formula for calculating the side slope area.?

Long Answer Type Questions1. The areas within the contour line at the site of reservoir and the proposed face of the dam are as follows

Contour Area

101 1,000 m2

102 12,800m2

103 95,200 m2

104 147,600 m2

105 872,500 m2

106 1350,000 m2

107 1985,000 m2

108 2286,000 m2

109 2512,000 m2

Taking 101 as the bottom level of the reservoir and 109 as the top level,calculate the capacity of the reservoir.

O.J.T. Questions1. Prepare a detailed estimate for earthwork for a portion of road from

the following data.


Formation width of road is 10 m. wide. The side slopes are 2:1 in bankingand 11/2:1 in cutting.

Distance in metres

0

100

200

300

400

500

600

700

800

900

1000

1100

1200

R.L. of ground

114.50

114.75

115.25

115.20

116.10

116.85

118.00

118.25

118.10

117.80

117.75

117.90

117.50

R.L. of formation

115.000

Upward grad. 1 in 200

Downward grad 1 in 400


7.1 Estimate of gravel roads

7.2 Cement concrete road

7.3 Septic tank with soak pit


• Calculate the quantities of material required for gravel and cement concrete roads. Calculate the quantities of Septic Tank.

7.0 IntroductionA road consists of sub base, base course and wearing course. The sub

base consists of earthwork prepared as per the height of formation. Over thissub base a base course of stone ballast or brick ballast of 12 cm. Thicknesscompacted to 8 cm. is laid. Finally a wearing coat is laid over this base course.The wearing course may be of cement concrete, bitumen or gravel. Dependingupon the wearing course provided the roads are classified as cement concreteroads, bituminous roads and gravel roads. Depending upon the cost involvedthe appropriate road required is decided. In order to estimate the cost of the

7UNIT

Detailed Estimates


road, we should be able to prepare the detailed estimate of the various types ofroads and calculate the materials required. In the sixth unit we studied aboutcalculation of earthwork involved in the formation of roads. In this unit we shallfind the quantities of the base course and wearing course.

7.1. Estimate of gravel roadsIn a gravel roads, the gravel is generally laid over stone ballast. It is laid

over the entire width of the road. The quantity of stone boulders and gravelconsists of thickness of their respective layers multiplied by its thickness.

Calculate the quantity of metal required for a 3.70 m. wide road for onekilometer length for one layer of 8 cm. compacted thickness.

Metal of 12 cm. is required for compact thickness of 8 cm. as volumeof loose metal gets reduced on half compaction.

Quantity of metal = 1000 x 3.70 x 0.12 = 444 cu m.

Prepare a detailed estimate for the construction of one kilometer lengthW.B.M. road. The formation width of the road is 10.0 m. and the averageheight of the bank is 1.0 m. and the side slopes are 2:1. The metal led width is3.7 m. m. and three coats of metal are to be provided as per cross section.Soiling coat of 15 cm. thick boulders at the base. Over this soiling coat, intercoat and top coat of 12 cm. compacted to 8 cm. A gravel coat of 5 cm. thick islaid over these metal led surface.

Quantity of earth work = [Bd+S(d)2] x L = [10 x 1.0 + 2(1)2] x 1000=12000 cum.

Length of the soling coat = 3.7 +0.15 + 0.15 = 4.0 m.

Detailed estimate of wbm road with gravel

Fig 7.1 Cross section road

Top coat

Inter coat

Saeing coat

Gravel

1.0 m

3.15 m

1.0 m

3.70 m 3.15 m

10.0 m


7.2 Cement concrete road Prepare an estimate for one kilometer length of a cement concrete track

way with 60 cm wide tracks 1.50 meter centre to centre over 15 cm rammedkankar.

For consolidating kankar an allowance of 1/3 is to be provided whiletaking loose thickness of kankar.

Eg. For 0.10 m. thickness loose kankar taken = 0.1 + 0.1 x 1/3 = 0.133 m.

Similarly for 0.15 m thickness loose kankar = 0.15 x 1.33 = 0.20 m.

S.No.

1

(a)

(b)

(c)

2

Particulars ofwork

Metal ling

Preparation of sub

grade Soling coat

Inter coat

Top coat

Layer of gravel

No.

1

1

1

1

L

1000

1000

1000

1000

B

4

3.7

3.7

3.7

Hor D

0.15

0.12

0.12

0.05

Quantity

600

444

444

185

S.no

1

2

Particular

Cement concrete1:2:4in tracks includ-ing laying.

Kankar metal looseunder c.c.tracks in be-tween c.c.tracks.

No

2

2

1

Length

1000

1000

1000

Breadth

0.6

0.9

0.9

Thickness

0.1

0.2

0.133

Quantity

120

360

120

480

m m m2

m m m2 m3

m3


Fig 7.2 C.C. Track

7.3. Septic tank with soak pitSeptic tank shall be of first class brickwork in 1:4 cement mortar, the

foundation and floor shal be of 1:3:6 cement concrete. Inside septic tank shallbe finished with 12 mm cement plaster and floor shall be finished with 20 mmcement plaster with 1:3 cement mortar. Upper and lower portions of soak pitshall be of second class brick work in 1:6 cement mortar and middle portionshall be of dry brickwork. Roof covering slabs and baffle wall shall be of precastR.C.C.

Details of Measurement & Calculation Of Quantities

Rammed kankar

Thick cc .Track Rammed kankar10 cm

60 cm 60 cm10 cm cc15 cm kankar

90 cm 90 cm

S.No

1

2

Particulars ofitems

Earthwork inexcavationseptic tankSoak pit upto3.0 mSoakpitLowerportion

Cementconcrete1:3:6

Floor&Foundation

Sloping floor

No.

11

1

1

1

Length

2.8(22/28)x(2.0)2

(22/28)x(1.4)2

2.8

2

Breadth

1.7

1.7

0.9

HeightorDepth

1.953

0.2

0.2

0.05

Quantity

9.289.42

0.319

0.95

0.09

m3m m

m


2

2

2

2

1

1

1

1

1

1

0.3

0.3

0.2

0.2

0.2

0.2

0.2

1.3

0.04

0.94

0.32

1.1

0.42

2.78

0.38

0.15

0.53

1.88

0.234

0.115

0.018

0.367

3

4

5

6

7

First classbrickwork in1:4 c.m. inseptic tank

First stepLong walls

Short wall

2nd step Longwall

Short wall

2nd classbrickwork in1:6 cementmortar in

soak pitUpper portion

Lower portion

2nd class drybrickworkin soak pit

PrecastR.C.C. work

Coverslabseptictank

CoverslabSoak pitBaffle wallseptictank

12 mmcement plaster1:3 in septictank

2.6

0.9

2.4

0.9

(22/7) x 1.20

(22/7) x 1.20

(22/7) x 1.20

2.4

(22/28)x(1.40)2

1

0.6

0.6

1.15

1.15

0.5

0.2

2.5

0.075

0.075

0.45


Fig. 7.3 Septic Tank

SummaryStructure of a road : The structure of a road from base to the top is as

follows. Earthwork formation , sub base, base course and wearing course.

Types of roads : Gravel road, cement concrete road, bituminous road.

Structure of a gravel road : Soling coat of boulders about 15 cmthick, inter coat and top coat 8 cm to 10 cm thick and wearing course of gravel5 cm thick.

8

Long walls

Short walls

20 mmc e m e n tplaster1:3 in floor ofseptic tank

2

2

1

2

0.9

2

1.7

1.7

6.8

3.06

9.86 sq m.

1.80 sq m.

Baffle wall

Out let

Section

In let

Plan

2.0 m

0.4 m0.9


Structure of a cement concrete road : Plain cement concrete isprovided over rammed earth.

Component parts of a septic tank : A septic tank consists of Plaincement concrete at its base, Walls on all the four sides in brickwork or R.R.masonry, baffle wall, sum board for large tanks, Precast R.C.C. slabs at thetop, inlet and outlet pipes. A soak pit is connected to the septic tank to collectthe discharge effluent. A soak pit consists of hollow circular brickwork constructedwith cement mortar. Dry brickwork is placed in the hollow section.

Short Answer Type Questions1. What is the structure of a road ?

2. List out the various types of roads.

3. Mention the various parts of a gravel road.

4. What are the various parts of a septic tank?

Long Answer Type Questions1. Prepare a detailed estimate for the construction of one kilometer

length over a formation of an embankment. The formation width is 10.0 m. andside slope 2:1. The metal led width is 4.0 m. and three coats of metal ling are tobe provided. Soling coat of 15 cm. boulders, inter coat and top coats of 12 cmloose compacted to 8 cm thick. Wearing coat of gravel 5 cm thick.

2. Prepare a detailed estimate for one kilometer length cement concreteroad 4.0 m wide and 15 cm thick. It is laid over rammed earth 6.0 m. wide and20 cm thick.

3. Prepare a detailed estimate for a septic tank 2.0 m. long and 1.0 m.wide. The height of the septic tank is 2.0 m. Assume suitable data for pre castslabs , baffle wall, inlets and oulets.

O.J.T. Questions1. Calculate the materials required for proposed construction of gravel

road and cement concrete road over an existing formation.

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