Estimating & Costing

UNIT

1

Introduction 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..

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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 for completion. 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.

UNIT

2

Measurement of Materials and Works 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 following principle. 1. Massive or volumetric items of work such as earth work, concerete for foundations, R.R Masonry , Brick Masonry etc. The measurements of length, breadth , height or depth shall be taken to compute the volume or cubical contents. 2. Shallow, thin and surface work shall be taken in square unit or in area. The measurements of length and breadth or height shall be taken to compute the area, Ex. Plastering, white washing etc.

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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 Civil Engineering Works Units of measurements

Unit of payment

(a) Earth work excavation in all types of soils except rock requiring blastering.

10.00cum

10.00cum

(b) Earth work excavation in the soils hard rock requirng blastering.

1.00cum

1.00cum

(c) Excavation of pipe line through of specified width and depth inall types of soils

1.00 rmt

1.00rmt

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

10.00cum

10.00cum

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

10.00cum

10.00cum

Plain cement concrete foundation.

for

1.00cum

1.00cum

R.R.masonry or brick masonry for foundation basement, super strucrture, parapet wall etc.

1.00cum

1.00cum

Filling the basement with sand.

1.00cum

1.00cum

(a) RCC 1:2: 4 with normal reinforcement for plinth beam , columns, lintels, verandah beam- T beam etc.

1.00cum

Sl.No Particulars of items 1.

2. 3.

4. 5.

1.00cum

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

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10.00sqm

10.00sqm

6.

Plastering pointing, flooring, weather proof coarse, white washing, colour washing, painting.

1.0.00sqm1.00sqm

10.00sqm

7.

Roofing with A.C sheets, tiled roofing, Kurnool trerrace, Madras terrace etc.

10.00sq m.

10.00sq m.

8.

D.P.C specified width and thick-

1.00Rmt

1.00Rmt

ness

9.

Wooden and steel trusses

1.00No

1.00No

10

Doors, windows, ventilators.

1.00 No

1.00No

11.

Ornamentel border of specified width and thickness.

1.00Rmt

1.00Rmt

12.

R.C.C pipes, A.C pipes GI or C.I pipes, stone ware pipes etc.

1.00Rmt

1.00Rmt

13.

Steel reinforcement in R.C.C.

Kg/unit

kg/unit

14.

Rough stone pitching revetment and soiling of specified thickness.

1.00cum

1.00cum

15.

(a) Roads works : Metal collections , gravel collections, solving stones, pitching any stones, revetment stones etc.

1.00cum

1.00cum

10.00sqm

1.00sqm

10.00cum

1.00cum

(b) Road works : Spreading metal gravel and consolidation with roller of specified thickness. (c) Cement concrete payments of specified thickness.

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2.2 Rules For Measurement Measurement of works occupies a very important place in the planning and 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 or in 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 into-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 in excavation, 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 at the 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.

Short Answer Type Questions 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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UNIT

3

Types of Estimates 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 the completion 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.

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3.1 Detailed estimate The 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, 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 class of 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 S.no Description of work No Length Breadth Height/Depth Quantity Remarks

Abstract estimate S.No. Description of work Quantity Rate Per Amount

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.

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3.3. Problems in preliminary estimate 1. 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 student s = 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.

Short Answer Type Questions 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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UNIT

4

Detailed and Abstract Estimate of Buildings 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 After studying this unit student will be able to Prepare detailed estimates of single roomed, Building roomed, Double roomed buildings, for load bearing walls and Framed structures. Detailed Estimate of Primary School Building, Compound walls and steps. Detailed estimate Dog legged and Open Well STair case. Preparational estimate for ground and first floor.

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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 + 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

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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 – ½ width x number of junctions.

3.0 0.5 0.7 0.9 1.2

E L E V A T I O N

0.3 0.6 0.3

S E C T I O N

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.

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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 wall and 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. 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.

Sl. No. Description No. L m of work

B m

H m

Quantity m3

1

Earth work in 1 excavation

21.2

1.2

1.2

30.528

2

C.C. bed in 1 foundation

21.2

1.2

0.3

7.63

3

R.R. masonry in foundation and plinth 1

21.2

0.9

0.6

11.45

Second footing 1

21.2

0.7

0.3

4.45

Basement

21.2

0.5

1.2

12.72

First footing

1

28.62 4

Brick work in 1 super structure

Detailed Estimate

21.2

0.3

3

19.08

Remarks

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Sl. Description of work No. L No.

B

H

Quantity Remarks

1

m

m

m3

Earth work in excavation in foundation

m

Long Walls

2

7.5 1.2 1.2

21.6

L=6.3+1.2=7.5

Short Walls

2

3.1 1.2 1.2

8.93

L=4.31.2=3.1

Total 30.53 2

Plain cement concrete in foundation (1:5:10) Long Walls

2

7.5 1.2 0.3

5.4

L=6.3+1.2=7.5

Short Walls

2

3.1 1.2 0.3

2.68

L=4.31.2=3.1

Total 8.08 3

R.R. Masonry in foundation & basement c.m (1:8) First footing Long Walls

2

7.2 0.9 0.6

7.78

L=6.3+0.9=7.2

Short WaLLS

2

3.4 0.9 0.6

3.67

L=4.30.9=3.4

11.45 Second footing Long Walls

2

7

0.7 0.3

2.94

L=6.3+0.7=7.0

Short WaLLS

2

3.6 0.7 0.3

1.51

L=4.30.7=3.6

4.45 Basement Long Walls

2

6.8 0.5 1.2

8.16

L=6.3+0.5=6.8

Short Walls

2

3.8 0.5 1.2

4.56

L=4.30.5=3.8

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12.72 Total R.R. masonry 28.62 of 4

Brick work in super structure c.m. ( 1:8) Long Walls

2

6.6

0.3 3

11.88

L=6.3+0.3=6.6

Short Walls

2

4

0.3 3

7.2

L=4.30.3=4.0

19.08

4.2 Two roomed building( load bearing type structure) Detailed Estimate Of A Double Roomed Building By Long Wall And Short Wall Method Centre to centre length of long wall = 5.0 + 0.3 + 5.0 + 2x0.3/2 = 10.6 m. 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 Total centre to centre line lenght = 10.6 x 2 + 5.3x3 = 37.1 m Sl. No. Description of work 1

No. L

B

H

Quantity Remarks

m

m

m

m3

Earthwork in excavation Long Walls

2

11.8 1.2

1.2

33.98

L= 10.6 + 1.2 = 11.8

Short Walls

3

4.1

1.2

17.71

L = 5.3 1.2 = 4.1

1.2

Total 51.69 2

C.C. bed in foundation Long Walls

2

11.8 1.2

0.3

8.5

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Short Walls

3

4.1

1.2

0.3

4.43 12.93

3

R.R. masonry in foundation & plinth First footing Long Walls

2

11.5 0.9

0.6

312.42

L = 10.6 + 0.9 = 11.5

Short walls

3

4.4

0.6

7.13

L = 5.3 - 0.9 = 4.4

0.9

19.55 Second footing Long Walls

2

11.3 0.7

0.3

4.75

L = 10.6 + 0.7 = 11.3

Short Walls

3

4.6

0.3

2.9

L = 5.3 -0.7 = 4.6

0.7

7.65 Third footing & plinth Long Walls

2

11.1 0.5

1.2

13.32

L = 10.6 + 0.5 = 11.1

Short walls

3

4.8

1.2

8.64

L = 5.3 - 0.5 = 4.8

0.5

21.96 R.R. masonry Total

4

49.16

Brick work in super structure Long Walls

2

10.9

0.3

3

19.62

Short Walls

3

5

0.3

3

13.5 33.12

L = 10.6 + 0.3 = 10.9 L = 5.3 - 0.3 = 5.0

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Centre line method Sl. No. Description of work Earthwork in 1 excavation C.C. bed in 2 foundation R.R. masonry in 3 foundation First footing

4

No. L

B

H

1

35.9 1.2

1

35.9 1.2

1

36.2 0.9

Second footing

1

36.4 0.7

Basement

1

36.6 0.5

1 1

36.8 0.3

Brickwork in superstructure

Quantity

Remarks

1.2 51.69 m3 L= 37.1 2x1/2x1.2 0.3 12.93 m3

0.6 19.55 me L = 37.1 2x1/2x0.9 0.3 7.65 m3 L = 37.1 2x1/2x0.7 1.2 21.96 m3 L = 37.1 2x1/2x0.5 Total 49.165 m3 3 33.12 m3 L = 37.1 2x1/2x0.3

3.0 0.5 0.7 0.9 1.2

E L E V A T I O N

5m x 5 m

S E

5m x 5m

Fig 4.2 Double Room

C T I O N

0.3 0.6 0.3

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

D

D

6.0 x 5.0 m

5.0 x 5.0 m

D

D

5.0 x 4.0 m

D

3.0m

5.0 x 4.0 m

D

0.3 0.6

0.9 0.9

0.3

1.2m

Fig 4.3 Plan Section

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m

m

m

m3

5m 4m

5m 4m

5m 4m

5m 4m Basement

56.43

Basement

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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. 100 mm thick RCC slab

0.23 x 0.23 R.C.C Column

3.0 m

1.2 m G.L

G.L 6.0 x 5.0 m

0.9 m

5.0 x 5.0 m R.C.C. Footing 5.0 x 4.0 m

6.0 x 4.0 m

0.3 m 0.3 m

R.C.C. 1.2 m

P L A N

S E C T I O N

Fig 4.4 Residential Building Framed Structure

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S. Description of work No. 1

175

No. L

B

H

Quantity Remarks

m

m

m

m3

Earthwork in excavation Columns

9

1.2

1.2 1.8 23.33

In between columns

1

63

0.9 0.9 51.03

Deduct for columns

9

0.6

0.6 0.9 -2.92

L=12x3+ 9x3=63

71.44 2

C.C. bed in foundation Columns

9

1.2

1.2 0.3 3.89

In between columns

1

63

0.9 0.3 17.01

Deduct for columns

9

0.6

0.6 0.3 -0.972 19.93

3

R.R. masonry in foundation First footing

1

63

0.7 0.6 26.46

Second footing

1

63

0.45 1.2 34.02 60.48

4

Brickwork in superstructure

1

63

0.23 3

43.47

Deductions Doors

6

1

0.23 2

-2.76

Windows

8

1.2

0.23 1.2 -2.65

Net Brickwork in super structure 38.06

5

R.C.C. column footing

9

1.2

1.2 0.3 3.89

9 Trapezoidal section Stem

(1.44+4x0.985+0.053)/6 0.3 2.44

9

0.23 0.23 5.1 2.43 8.76

H=0.9+1.2+ 3.0=5.1

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Construction Technology

6

R.C.C. Plinth beam

1

63

0.23 0.3 4.35

7

R.C.C. in lintels&sunshades Lintels

1

63

0.23 0.1 1.45

Sunshades

1

45.6 0.7

0.07 2.23

L=2(12.9+ 9.9)=45.6

3.68 8

R.C.C. slab and beams Beams under slab

9

63

0.23 0.3 4.35

1m. Projection from 9 slab

1

0.23 0.3 0.62

R.C.C. Slab.

14.9 11.9 0.12 21.28

L=12.9+1.0+ 1.0=14.9

26.25

B=9.9+1.0+ 1.0=11.9

1

1

L=2(12.9+9.9) =45.6

External plastering 20 mm Thick

1

45.6

3.12 142.27

Doors

6

1

2

Windows

8

1.2

1.2 -11.52

H=3.0+0.12

Deductions -12

Net External plastering area 118.75

10 Internal Plastering 12 mm thick Rooms 6mx5m

2

22

3

132

L=2(6+5)=22

Rooms5mx4m

2

18

3

108

L=2(5+4)=18

240 11 Sand filling in rooms Rooms 6mx5m

2

6

5

1.2 72

Paper - II Estimating and Costing

2

Rooms 5mx4m

177

5

4

1.2

48 120

12 C.C. bed in rooms Rooms 6mx5m

2

6

5

0.1

6

Rooms5mx4m

2

5

4

0.1

4 10

13 Flooring in rooms Rooms 6mx5m

2

6

5

60

Rooms5mx4m

2

5

4

40 100

14 Fabrication & placement of steel

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

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.

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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

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PRIMARY SCHOOL BUILDING WITH SLOPING ROOF

les Ti

1.5 m

s Tile

2.0 m

0.9 m

ELEVATION

0.9 m 0.6 m

0.9 m 0.6 m

0.9 m 0.3 m

0.9 m 0.3 m

1.2 m

References W

W D - Door 1.00 m x 2.00 m W - Window 1.2 m x 1.2 m

Room 3.0 x 3.0 m

D

Room 3.0 x 3.0 m

D

P L A N

Width of 1st footing : 0.9 m Second footing : 0.7 m Basement : 0.5 m S E C T I O N

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Construction Technology

4.6 RCC Dog legged – open well stairs

1650

250 1650 150 Floor

E L E V A T I O N

S E C T I O N

2500

1000

P L A N

Fig 4.5 Dog Legged Stair case

- A A

Paper - II Estimating and Costing

Sloping side

22

181

0.28 Tota l

0.4

2.464 11.264

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Construction Technology

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

300 152 8 C

7

300 152 3 4 B

300 152 8 8

SECTION AT ‘AA’

A

Flight No. No. of Risers

No. of Treads

Each Riser

Each Tread

4.6.1 Open Well Staircase

Note : 1. All dimensions are in Milli meters 2. Follow the written dimensions only OPEN WELL TYPE STAIRCASE Scale 1:50 DRG. No. 18 Fig 4.6 Open well Stair case

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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. Height of risers = 0.15x4=0.6 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. Height of risers = 0.15x8=1.2 m. Sloping length of flight = Square root of (2.1x2.1+1.2x1.2)=2.42 m.

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Construction Technology

4.7 Two storied residential building (framed structure type) Parpet wall Weathering course Lintel & sunshade Brick masonry

3.05 m

Roof slab 3.05 m C.C. flooring

0.902 Elevation

E L E V A T I O N Fig 4.7 Two storied residential building

R.C.C Mix 1:4:1 Sand filling C.C. floring 1:4:8

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185

Fig 4.8 Ground Floor & First Floor

Ground floor Number of columns = 15 Height of columns in ground flo or & 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 – 3 Nos., 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.

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Construction Technology

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

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188

Construction Technology

Paper - II Estimating and Costing

189

190

Construction Technology

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.

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191

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. 0.23m 0.15 0.15 0.15

1.5 m

0.45

1.0 m

0.6

0.6 m

Front View

Side View

0.3 m 0.9 m

0.3

0.23

0.3 0.3

1.0 m Top view

4.0 m

Fig. 4.9 Plan and Section of a compound Wall

0.23

6.0m

0.23

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Construction Technology

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 – width In 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 not applicable. 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 sloping side.) Number of risers = Height of the flight/ rise. Number of treads = Number of risers – 1. Treads length = Number of treads x Tread.

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193

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 ½.

Short Answer Type Questions 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.

Long Answer Type Questions 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.

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UNIT

5

Specifications and Analysis of Rates 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 After studying this unit student will be able to • 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

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195

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.

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Construction Technology

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 first class 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.

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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

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Construction Technology

123 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.

Paper - II Estimating and Costing

Category of worker S. Rate for 2012-13 123 • 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

199

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Construction Technology

• 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 2012-13 123 • 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

Paper - II Estimating and Costing

• 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 2012-13 123 • 32 Helper Vibratory plain/ pad foot roller 285 • 33 Helper Wagon drill/ Drifter 285

201

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Construction Technology

• 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

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283 Sl No. Category of worker S. Rate for 2012-13 123 • 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

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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 source to 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. Mat- Source Unit Cost at Lead Equi Blas Seign Cess Cru Deduc Net Re source inKm. valent ting orage charges shing tions rate mar No erial char if any at ks metal char char ges ges ges site led road

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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 Mix Quantity of Quantity Cost of propor cement in of sand cement -tion bags in m3

Cost of sand

Total cost Mix -ing charges

1:2

9.5 bags

0.66

Rs.2422. Rs. 323. 50 40

Rs.85

2831.50

1:3

7.2

0.75

Rs. 1836. Rs. 367. 50 00

Rs.85

2288.50

1:4

5.76

0.8

Rs. 1469. Rs. 392. 00 00

Rs.85

1946.00

1:5

4.79

0.83

Rs. 1221. Rs. 406. 50 70

Rs.85

1713.20

1:6

4.11

0.857

Rs.1048. Rs.419.95 Rs.85 05

1553.00

1:8

3.19

0.89

Rs.813. 45

Rs. 436.10 Rs.85

1334.55

1:10

2.62

0.91

Rs.668. 10

Rs. 445.9 Rs.85

1199.00

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.

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Construction Technology

Particulars

Quantity

Rate

Cost

Materials Cement

2.74 bags

Rs. 255/bag

Rs. 698.70

Sand

0.475 m3

Rs. 375/m3.

Rs. 178.15

Coarse aggregate

0.95 m3

Rs. 588/bag

Rs. 558.60

Labor: Head mason

0.05 No.

Rs. 350/No.

Rs. 17.50

Mason

0.15 No.

Rs. 315/No.

Rs. 47.25

Men mazdoor

1.2 NO.

Rs. 250/No.

Rs. 300

Women mazdoor

1.8 No.

Rs. 250/No.

Rs. 450

Waterman

0.4 No.

Rs. 250/No.

Rs. 100

Add 20% for labor

Rs.182.95 Total

Rs.2533.15

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

Amount

Quantity

Rate

6.25 bags

Rs. 255/bag Rs. 1593.75

R.C.C(1:2:4) including cost of materials, labour charges, centering charges but excluding cost of steel and its fabrication. Materials Cement

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207

Sand

0.434 m3

Rs. 375/m3

Rs. 162.75

Coarse aggregate

0.868 m3

Rs. 1161.80/ Rs. 1003.80 m3 Total

Rs. 2760.30

Labour Head mason

0.05

Rs. 350/No.

Rs. 17.50

Mason

0.3

Rs. 315/No.

Rs. 94.50

Men mazdoor

1.2 No.

Rs. 250/No.

Rs. 300.00

Women mazdoor

2.0 NO.

Rs. 250/No.

Rs. 500.00

Waterman

0.6 No.

Rs. 250/No.

Rs. 150.00 Rs. 1062.00

20% local allowance

Rs. 212.40 Rs. 1274.40

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

Centering Cost of Total Cost charges materials and including labour materials and labour

Lintel

Rs.1215

Rs. 4034.70

Rs. 5249.70

Slab

Rs. 1533.33 Rs. 4034.70

Rs. 5568.00

Beam

Rs. 1637

Rs. 4034.70

Rs. 5671.70

Column

Rs.929

Rs. 4034.70

Rs. 4963.70

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Construction Technology

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 Particulars

Quantity

Rate

Bricks

500 Nos.

Rs. 5021.30per Rs.2510.65 1000 Nos.

Cement

1.32 bags

Rs. 255 per bag Rs. 336.60

Sand

0.274 m3.

Rs. 490/m3.

Amount

Brick masonry in superstructure including cost of materials and labour Materials

Materials cost Total

Rs. 134.30 Rs. 2981.55

Labour Head mason

0.05 No.

Rs. 350/No.

Rs. 17.50

Mason

1.0 No.

Rs. 315/No.

Rs. 315.00

Men mazdoor

0.7 NO.

Rs. 250/No.

Rs. 175.00

Women mazdoor

1.0 No.

Rs. 250/No.

Rs. 250.00

Waterman

0.2 No.

Rs. 250/No.

Rs. 50.00

Paper - II Estimating and Costing

209

Total

Rs. 807.50

Add 20%

Rs.161.50 Rs. 969.00

Materials and Labour Total Cost

Rs. 3950.55

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. Particulars

Quantity or No. Rate

Amount

Materials Stone including bond 1.25 m3. stone and wastage

Rs.535.60/m3

Rs. 669.5

Cement

1.8 bags

Rs. 255/ bag

Rs. 459

Sand

0.36 m3.

Rs. 490/m3.

Rs. 176.40 Rs. 1304.90

Labour Head mason

0.05 No.

Rs. 350/No.

Rs. 17.50

Mason

1.6 No.

Rs. 315/No.

Rs. 504.00

Men mazdoor

1.6 No.

Rs. 250/No.

Rs. 400.00

Women mazdoor

0.8 No.

Rs. 250/No.

Rs. 200.00

Waterman

0.15 No.

Rs. 250/No.

Rs. 37.50 Rs.1159.00

Add 20% allowance

Rs. 231.80 Rs. 1390.80

Total cost of materials and labour

Rs. 2695.70

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Construction Technology

Quantit y 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= 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. Total cost of labour = Rs. 7204.20+1440.80=Rs. 8645.00 Total cost of external plastering=Rs.4659.00+ Rs. 8645.00=Rs. 13304.00 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 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

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211

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=6 bags. Cost of cement= Rs. 255/ bag; Cost of sand= Rs. 490/m3.; Cost of aggregate = Rs.1161.80/m3.

212

Particulars

Construction Technology

Quantity or No. Rate per

Amount

Materials Stone aggregate

2.40 m3.

Rs. 1161.80/m3.

Rs. 2788.40

Sand (coarse)

1.20 m3.

Rs. 490/m3.

Rs. 588.00

Cement

18 bags

Rs. 255/ bag

Rs. 4590.00

Rs. 255/ bag

Rs.1530.00

Cement for surface 6 bags finishing

Rs. 9496.40 Labour etc. Head mason

¾ no.

Rs. 350/day

Rs. 262.50

Mason

10 Nos.

Rs. 315/day

Rs. 3150.00

Men mazdoor

5 Nos.

Rs. 250/day

Rs. 1250.00

Women mazdoor

5 Nos.

Rs. 250/day

Rs.1250.00

Waterman

2 Nos.

Rs. 250/day

Rs.500.00 Rs. 6412.50

Add 20% extra

Rs.1282.50 Rs. 7695.00

Side forms

Lump sum

Rs. 300.00

Total cost of materials

Rs. 9496.40

Total cost of labour

Rs. 7695.00

Side forms for finishing

Lump sum

Rs. 300.00

Total cost

Rs.17491.00

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.

Paper - II Estimating and Costing

213

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 = 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

To tal 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. No. L

B

Stiles

4

2.00

0.10 0.04

0.032

Top rail

1

1.00

0.10 0.04

0.004

Frieze rail

1

1.00

0.10 0.04

0.004

Lock rail

1

1.00

0.15 0.04

0.006

Bottom rail

1

1.00

0.10 0.04

0.004

Particulars

Thick ness

Quantity/ Rate Nos.

Materials:timber

Amount

214

Construction Technology

Planks for 1 panels

1.55

0.6

0.025

0.023

0.073 Ad d 5%

for

wastage 0.00365 0.0767 m3 Rs.1054 Rs.8090.80 86.00/m3

Brass accessories Tower bolt 1No. 30 cm.

1 No.

Rs.248 Rs. 248.00 .00/No.

Tower bolt 1No. 15 cm.

1 No.

Rs.121.00/ Rs.121.00 No.

Handle 10 2.no cm.

2 Nos.

Rs.337.00/ Rs.674.00 No..

Hinges

6 Nos.

Rs.112.00/ Rs. 672.00 No.

6.no

Door stopper

Rs. Rs. 729.00 729.00/ No. Rs.146.00 1 No. Rs. 146.00/ No. Rs.2590.00 accessories 1 No.

Aldrop 30 1 No. cm. 1 No.

Cost Of Labour Head 1/15 carpenter No.

1/15 No.

Rs. 350/ Rs. 23.35 day

Carpenter 4 Nos.

4 Nos.

Rs. 315/ Rs. day 1260.00

Helpers

2 Nos.

Rs. 250/ Rs.500.00 day Rs. 1783.35

Paper - II Estimating and Costing

215

Add 20% Rs. 356.65 extra Total

Rs. 2140.00

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.

216

Construction Technology

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 required for 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.

Short Answer Type Questions 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.

Long Answer Type Questions 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).

Paper - II Estimating and Costing

217

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.

218

Construction Technology

UNIT

6

Earthwork Calculations 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 After studying this unit student will be able to • 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.

Paper - II Estimating and Costing

219

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 Quantity Stations Depth Mean Central Area of Total Length or depth or area Bd sides Sd2 sec between (Bd+Sd2)xL tional stations L Embank Height Height ment cutt area ing Bd+Sd2 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 Station Height Area of Area of or depth cent ral s i d e s portion Sd2 Bd

To t a l M e a n sectional sectional a r e a area Bd+Sd2

Length between stations L

Quantity =(Bd+Sd2) x LBanking Cutting

220

Construction Technology

1:S

d1

Sd1

Sd2

1:S

Sd1

B 1:S

1:S

B

d2

Sd2

B d1 L Fig 6.1

Trapezoidal-Prismoidal Formula : In the prismoidal formula the areas at the ends and the mid sectional area are also taken into consideration. If the area 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 cross sections : When the series of cross sections A0,A1, A2,A3, …………An are at 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 odd areas)+4(Sum of even areas))xD/3 Example 1 : Calculate the quantity of earthwork for 200 metre length for a portion of a road in an uniform ground. The heights of the banks at the two ends are 1.00 and 1.60 m. The formation width is 10 metre and side slopes are 2:1. Assume that there is no transverse slope.

Paper - II Estimating and Costing

221

Mid sectional area method : Height d1 = 1.00m. Height d2 = 1.60 m. 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 bank for a length of 200 m. The heights of the banks at the two ends are 2.50 m and 3.50 m. and the ratio of side slope 2:1. If the side slopes are to be provided with 15 cm. thick stone pitching, calculate the cost of pitching at the rate of Rs. 200 per cu m.

222

Construction Technology

• 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 a proposed road from chainage 10 to chainage 20 are given below. The formation level at the 10th chainage is 107 m. and the road is in downward gradient of 1 in 150 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, 104.30 105.00 , 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 Height Mean Central or Depth height area or depth Bd

Side area Sd2

Length in Quantity = Total area between [(Bd+S(d)2]xL Bd+Sd2 chainage Banking Cutting

10

2.00

-

11

1.20

1.60

16.00

5.12

21.12

30

633.6

-

12

1.16

1.18

11.80

2.78

14.58

30

437.4

-

13

0.50

0.83

8.30

1.38

9.68

30

290.4

-

14

0.78

0.64

6.40

0.82

7.22

30

216.6

-

15

1.60

1.19

11.90

2.83

14.73

30

441.9

-

16

0.60

1.10

11.00

2.42

13.42

30

402.6

-

Paper - II Estimating and Costing

223

17

1.20

0.90

9.00

1.62

10.62

30

318.6

-

18

0.38

0.79

7.90

1.25

9.15

30

274.5

-

19

0.70

0.54

5.40

0.58

5.98

30

179.4

-

20

1.10

0.90

9.00

1.62

10.62

30

318.6

-

Total

3513.6 cu m.

A railway embankment is 10 m. wide with side slopes 11/2 to1. Assume the ground to be level in direction transverse to the centre line, calculate the volume contained in a length of 120 metres, the centre heights at 20 m. intervals being 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 height is banking of height 0.3 m. and at 31st chainage, it is cutting at a depth of 0.40 m.

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Find the volume of banking and cutting if the formation width is 10 m. and the side slopes are 2:1 in banking and 11/2 : 1 in cutting. Chainage distance = 40 m. Let the height of embankment be zero at a distance of x mts. • Length of cutting =( 40-x) . (x/0.3) =[(40-x)/0.4] 0.3x 0.7x= 12 x=17.14 say 17.0 m.

0.4x=12-

• 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. 0.3 0.4 x

(40-x) 40 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.

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225

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

Short Answer Type Questions 1. 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 Questions 1. 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. Questions 1. Prepare a detailed estimate for earthwork for a portion of road from the following data.

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Construction Technology

Distance in metres R.L. of ground

R.L. of formation

0

114.50

115.000

100

114.75

Upward grad. 1 in 200

200

115.25

300

115.20

400

116.10

500

116.85

600

118.00

700

118.25

800

118.10

900

117.80

1000

117.75

1100

117.90

1200

117.50

Downward grad 1 in 400

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

UNIT

7

Detailed Estimates Structure 7.0 Introduction 7.1 Estimate of gravel roads 7.2 Cement concrete road 7.3 Septic tank with soak pit

Learning Objectives After studying this unit student will be able to • Calculate the quantities of material required for gravel and cement concrete roads. Calculate the quantities of Septic Tank.

7.0 Introduction A road consists of sub base, base course and wearing course. The sub base consists of earthwork prepared as per the height of formation. Over this sub base a base course of stone ballast or brick ballast of 12 cm. Thickness compacted 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. Depending upon the wearing course provided the roads are classified as cement concrete roads, bituminous roads and gravel roads. Depending upon the cost involved the appropriate road required is decided. In order to estimate the cost of the

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Construction Technology

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

7.1. Estimate of gravel roads In 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 gravel consists of thickness of their respective layers multiplied by its thickness. Calculate the quantity of metal required for a 3.70 m. wide road for one kilometer length for one layer of 8 cm. compacted thickness. Metal of 12 cm. is required for compact thickness of 8 cm. as volume of 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 length W.B.M. road. The formation width of the road is 10.0 m. and the average height of the bank is 1.0 m. and the side slopes are 2:1. The metal led width is 3.7 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, inter coat and top coat of 12 cm. compacted to 8 cm. A gravel coat of 5 cm. thick is laid 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 10.0m 3.70m

3.15m

3.15m 1.0 m

1.0 m Saeing coat Inter coat Top coat Fig 7.1 Cross section road

Gravel

Paper - II Estimating and Costing

S.No. Particulars of work

229

B Hor D Quantity m3 m m2

No. L m

Metal ling

1

Preparation of sub (a)

grade Soling coat

1

1000 4

0.15

600

(b)

Inter coat

1

1000 3.7 0.12

444

(c)

Top coat

1

1000 3.7 0.12

444

2

Layer of gravel

1

1000 3.7 0.05

185

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 rammed kankar. No Length m Cement concrete 2 1000 1:2:4in tracks including laying.

Breadth Thickness Quantity m m2 m3 0.1 0.6 120

Kankar metal loose 2 under c.c.tracks in between c.c.tracks. 1

1000

0.9

0.2

360

1000

0.9

0.133

120

S.no Particular 1

2

480 For consolidating kankar an allowance of 1/3 is to be provided while taking 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.

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Construction Technology

10 cm Thick cc .Track

Rammed kankar 60 cm

60 cm

10 cm cc 15 cm kankar 90 cm

90 cm Rammed kankar

Fig 7.2 C.C. Track

7.3. Septic tank with soak pit Septic 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 shall be finished with 12 mm cement plaster and floor shall be finished with 20 mm cement plaster with 1:3 cement mortar. Upper and lower portions of soak pit shall be of second class brick work in 1:6 cement mortar and middle portion shall be of dry brickwork. Roof covering slabs and baffle wall shall be of precast R.C.C. Details of Measurement & Calculation Of Quantities S.No Particulars of items 1

2

Earthwork in excavation septic tank Soak pit upto 3.0 m Soakpit Lowerportion

No. Length m

Breadth Height Quantity or m3 m Depth m

1 1

1.7 2.8 (22/28)x(2.0)2

1.95 3

9.28 9.42

1

(22/28)x(1.4)2

0.2

0.3 19

Floor& Foundation

1

2.8

1.7

0.2

0.95

Sloping floor

1

2

0.9

0.05

0.09

Cementconcrete 1:3:6

Paper - II Estimating and Costing

3

231

First class brickwork in 1:4 c.m. in septic tank First step Long walls

2

2.6

0.3

0.6

0.94

Short wall

2

0.9

0.3

0.6

0.32

2nd step Long wall

2

2.4

0.2

1.15

1.1

2

0.9

0.2

1.15

0.42

Short wall

2.78 4

2nd class brickwork in 1:6 cement mortar in soak pit Upper portion Lower portion

5

2nd class dry brickwork in soak pit

6

Precast R.C.C. work Coverslab septictank Coverslab Soak pit Baffle wall septictank

7

12 mm cement plaster 1:3 in septic tank

1

(22/7) x 1.20

1

(22/7) x 1.20

1

(22/7) x 1.20

1

2.4

1

(22/ 28)x(1.40)2

1

1

0.2 0.2 0.2

1.3

0.04

0.5

0.38

0.2

0.15 0.53

2.5

1.88

0.075

0.234

0.075

0.115

0.45

0.018 0.367

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Construction Technology

Long walls

2

2

1.7

6.8

Short walls

2

0.9

1.7

3.06 9.86 sq m.

8

20 mm c e m e n t plaster 1:3 in floor of septic tank

1.80 sq m.

2

1

Baffle wall

Section

0.4 m 0.9

In let

Out let

2.0 m

Plan Fig. 7.3 Septic Tank

Summary Structure 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 cm thick, inter coat and top coat 8 cm to 10 cm thick and wearing course of gravel 5 cm thick.

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233

Structure of a cement concrete road : Plain cement concrete is provided over rammed earth. Component parts of a septic tank : A septic tank consists of Plain cement 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 the top, inlet and outlet pipes. A soak pit is connected to the septic tank to collect the discharge effluent. A soak pit consists of hollow circular brickwork constructed with cement mortar. Dry brickwork is placed in the hollow section.

Short Answer Type Questions 1. 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 Questions 1. Prepare a detailed estimate for the construction of one kilometer length over a formation of an embankment. The formation width is 10.0 m. and side slope 2:1. The metal led width is 4.0 m. and three coats of metal ling are to be provided. Soling coat of 15 cm. boulders, inter coat and top coats of 12 cm loose compacted to 8 cm thick. Wearing coat of gravel 5 cm thick. 2. Prepare a detailed estimate for one kilometer length cement concrete road 4.0 m wide and 15 cm thick. It is laid over rammed earth 6.0 m. wide and 20 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 cast slabs , baffle wall, inlets and oulets.

O.J.T. Questions 1. Calculate the materials required for proposed construction of gravel road and cement concrete road over an existing formation.