Intermediate Vocational Course, 2nd Year : ESTIMATING & CONSTING(FOR THE COURSE OF CONSTRUCTION TECHNOLOGY)Authors : B.N. Suresh, Editor : B. Harnath Reddy,First Edition : 2006State Institute of Vocational EducationAndhra Pradesh, Hyderabad.Printed and Published bythe Telugu Akademi, Hyderabad on behalf of theState Institution of Vocational EducationDirectorate of Intermediate EducationGovt. of Andhra Pradesh, Hyderabad.First Edition : 2006Copies : .All rights whatsoever in this book are strictly reserved and noportion of it may be reproduced by any process for any purposewithout the written permission of the copyright owners.Price : Rs. /-Printed in IndiaLaser Typeset Reformatted by Chinmai D.T.P. Works, ChittorTExt Printed at ........................................Andhra Pradesh.AUTHORB.N. SURESH B.E.J.L. in Construction Technology (Vocational)Govt. Junior College for GirlsGuntur.EDITORB. HARNATH REDDY B.Tech.J.L. in Construction Technology (Vocational)New Govt. Junior CollegeMalakpet, Hyderabad.INDEXS.No. Topic Page No.01. Introduction to the subject 1 - 302. Measurement of Materials & Works 4 - 1203. Types of Estimates 13 - 2104. Detail & Abstract Estimates of Buildings 22 - 5505. Analysis of Rates 56 - 7306. Estimation of Quantities of Steel &RCC Elements 74 - 8007. Earth Work Calculations 81 - 9508. Detailed Estimates 96 - 10809. Appendex 109 - 1151 Estimation and CostingINTRODUCTION TO THE SUBJECT1.1 DEFINITION OF ESTIMATING AND COSTINGEstimating is the technique of calculating or Computing the variousquantities and the expected Expenditure to be incurred on a particular work orproject.In case the funds avilable are less than the estimated cost the work isdone in part or by reducing it or specifications are altered, the following require-ment are necessary for preparing an estimate.a ) Drawings like plan, elevation and sections of important points.b) Detailed specifications about workmenship & properties of materials etc.c) Standard schedule of rates of the current year.1.2 NEED FOR ESTIMATION AND COSTING

1. Estimate give an idea of the cost of the work and hence its feasibility canbe determined i..e whether the project could be taken up with in the fundsavailable or not.2. Estimate gives an idea of time required for the completion of the work.3. Estimate is required to invite the tenders and Quotations and to arangecontract.4. Estimate is also required to control the expenditure during the executionof work.5. Estimate decides whether the proposed plan matches the funds availableor not.1.3 PROCEDURE OF ESTIMATING OR METHOD OF ESTIMATING.Estimating involves the following operations1. Preparing detailed Estimate.2. Calculating the rate of each unit of work3. Preparing abstract of estimate1.4 DATA REQUIRED TO PREPARE AN ESTIMATE1. Drawings i.e.plans, elevations, sections etc.2. Specifications.3. Rates.Chapter121.4.1 DRAWINGSIf the drawings are not clear and without complete dimensions the prepa-ration of estimation become very difficult. So, It is very essential before prepar-ing an estimate.1.4.2. SPECIFICATIONSa) General Specifications: This gives the nature, quality, class and work andmaterials in general terms to be used in various parts of wok. It helps noform a general idea of building.b) Detailed Specifications: These gives the detailed description of the vari-ous items of work laying down the Quantities and qualities of materials,their proportions, the method of preparation workmanship and executionof work.1.4.3. RATES:For preparing the estimate the unit rates of each item of work are re-quired.1. For arriving at the unit rates of each item.2. The rates of various materials to be used in the construction.3. The cost of transport materials.4. The wages of labour, skilled or unskilled of masons, carpenters, Mazdoor,etc.,1.5 COMPLETE ESTIMATE:Most of people think that the estimate of a structure includes cost of land,cost of materials and labour, But many other direct and indirect costs includedand is shown below.The Complete EstimateCost of land P.s.andcontingencies at 5%Legal expenses between ownerand contractorCost of StructureActualcost of landCost of Surveying

Cost of Verificationof deeds andexecution of deedsBrochorageif anyCost of labourPermit fees forconstrution water, electricity fromconcerned autoritiescost of materialsConsultingEngineersfeescost forpreparationof plan,estimate anddesignCost ofsupervisionIntroduction to the Subject3 Estimation and Costing1.6 LUMPSUM:While preparing an estimate, it is not possible to workout in detail in caseof petty items. Items other than civil engineering such items are called lumpsumitems or simply L.S.Items.The following are some of L.S. Items in the estimate.1. Water supply and sanitary arrangements.2. Electrical installations like meter, motor, etc.,3. Architectural features.4. Contingencies and unforeseen items.Ingeneral, certain percentage on the cost of estimation is alloted for theabove L.S.ItemsEven if subestimates prepared or at the end of execution of work, theactual cost should not exceed the L.S.amounts provided in the main estimate.1.7 WORK CHARGED ESTABLISHMENT:During the construction of a project considerable number of skilled su-pervisors, work assistance, watch men etc., are employed on temporary basis.The salaries of these persons are drawn from the L.S. amount alloted towardsthe work charged establishment. that is, establishment which is charged directlyto work. an L.S.amount of 1½ to 2% of the estimated cost is provided towardsthe work charged establishment.EXERCISEShort Answer Questions1. State the requirements of an estimate?2. Briefly Explain need for estimation?3. What is work charged establishment?42.1 UNITS OF MEASUREMENTS:The units of measurements are mainly categorised for their nature, shapeand size and for making payments to the contractor and also. The principle ofunits of measurements normally consists the following:a) Single units work like doors, windows, trusses etc., are expressedin numbers.b) Works consists linear measurements involve length like cornice,

fencing, hand rail, bands of specified width etc., are expressed inrunning metres (RM)c) Works consists areal surface measurements involve area likeplastering, white washing, partitions of specified thickness etc., areexpressed in square meters (m2)d) Works consists cubical contents which involve volume like earthwork, cement concrete, Masonry etc are expressed in Cubic metres.[BASED ON IS 1200 REVISED]Particulas of itemEarth work:1. Earth work in Excavation2. Earthwork in fillingin founda-tion trenches3. Earth work in filling in plinthConcrete:1. Lime concretre in foundation2. Cement concrete in Lintels3. R.C.C.in slab4. C.C. or R.C.C. Chujja, Sun-shade5. L.C. in roof terracing(thickness specified)Sl.No.IIIUnits ofMeasurementcumcumcumcumcumcumcumsqmUnits ofpaymentPer%cumPer%cumPer%cumpercumpercumpercumpercumpersqmMEASUREMENT OF MA MEASUREMENT OF MA MEASUREMENT OF MA MEASUREMENT OF MA MEASUREMENT OF MATERIALS TERIALS TERIALS TERIALS TERIALSAND WORKS AND WORKS AND WORKS AND WORKS AND WORKSChapter25 Estimation and Costing6. Cement concrete bed7. R.C. Sunshade (SpecifiedWidth & HightDamp ProofCource (D.P.C)(Thickness should be men-tioned)

Brick work:1. Brickwork in foundation2. Brick work in plinth3. Brick work in super struc-ture4. Thin partition walls5. Brick work in arches6. Reinforced brick work(R.B.Work)Stone Work:Stone masonryWood work:1. Door sand windows framesor chowkhats, raftersbeams2. Shutters of doors and win-dows (thickness specified)3. Doors and windows fittings(like hinges, tower bolts,sliding bolts, handles)Steel work1. Steel reinforcement barsetc in R.C.C. andR.B.work. quintal2. Bending, binding of steelReinforcement3. Rivets, bolts, & nuts, An-chor bolts, Lewis bolts,Holding down bolts.4. Iron hold fasts5. Iron railing (height andtypes specified)6. Iron grillsIIIIVVVIVIIcumcumsqmcumcumcumsqmcumcumcumcumsqmNumberQuintalQuintalQuintalQuintalQuintalsqmper cum1rmpersqm

percumpercumpercumpercumpercumpercumpercumpercumpersqmper numberper quintalper quintalper quintalper quintalper quintalper sqm6Roofing1. R.C.C. and R.B.Slab roof(excluding steel)2. L.C. roof over and inclusiveof tiles or brick or stone slabetc (thickness specified)3. Centering and shutteringform work4. A.C.Sheet roofingPlastering, points&finishing1. Plastering-Cement or LimeMortar (thickness and pro-portion specified)2. Pointing3. White washing, colourwashing, cement wash(number of coats specified)4. Distempering (number ofcoats specified)5. Painting, varnishing (numberof coats specified)Flooring1. 25mm cement concreteover 75mm lime concretefloor (including L.C.)2. 25mm or 40mm C.C. floor3. Doors and window sills(C.C. or cement mortarplain)Rain water pipe /Plain pipeSteel wooden trussesGlass pannels(supply)Fixing of glass panels orcleaning VIIIIXXXIXIIXIIIXIVcumsqm

sqmsqmsqmsqmsqmsqmsqmsqmsqmsqm1RM1NosqmNoper cumper sqmper sqmper sqmper sqmper sqmper sqmper sqmper sqmper sqmper sqmper sqmper RMper 1Noper sqmper no.Measurement of Materials and Works7 Estimation and Costing2.2 RULES FOR MEASUREMENT :The rules for measurement of each item are invaribly described in IS-1200. However some of the general rules are listed below.1. Measurement shall be made for finished item of work and description ofeach item shall include materials, transport, labour, fabrication tools andplant and all types of overheads for finishing the work in required shape,size and specification.2. In booking, the order shall be in sequence of length, breadth and height orthickness.3. All works shall be measured subject to the following tolerances.i) Linear measurement shall be measured to the nearest 0.01m.ii) Areas shall be measured to the nearest 0.01 sq.miii) Cubic contents shall be worked-out to the nearest 0.01 cum4. Same type of work under different conditions and nature shall be measuredseparately under separate items.5. The bill of quantities shall fully describe the materials, proportions,workmanships and accurately represent the work to be executed.6. In case of masonary (stone or brick) or structural concrete, the categoriesshall be measured separately and the heights shall be described:a) from foundation to plinth levelb) from plinth level to First floor levelc) from Fist floor to Second floor level and so on.2.3 METHODS OF TAKING OUT QUANTITIES:The quantities like earth work, foundation concrete, brickwork in plinthand super structure etc., canbe workout by any of following two methods:a) Long wall - short wall methodb) Centre line method.c) Partly centre line and short wall method.

a) Long wall-short wall method:In this method, the wall along the length of room is considered to be longwall while the wall perpendicular to long wall is said to be short wall. To get the8length of long wall or short wall, calculate first the centre line lengths of individualwalls. Then the length of long wall, (out to out) may be calculated after addinghalf breadth at each end to its centre line length. Thus the length of short wallmeasured into in and may be found by deducting half breadth from its centre linelength at each end. The length of long wall usually decreases from earth work tobrick work in super structure while the short wall increases. These lengths aremultiplied by breadth and depth to get quantities.b) Centre line method:This method is suitable for walls of similar cross sections. Here the totalcentre line length is multiplied by breadth and depth of respective item to get thetotal quantity at a time. When cross walls or partitions or verandah walls joinwith mainall, the centre line length gets reduced by half of breadth for eachjunction. such junction or joints are studied caefully while calculating total centreline length.The estimates prepared by this method are most accurate and quick.c) Partly centre line and partly cross wall method:This method is adopted when external (i.e., alround the building) wall isof one thickness and the internal walls having different thicknesses. In such cases,centre line method is applied to external walls and long wall-short wall method isused to internal walls. This method suits for different thicknesses walls and diffeentlevel of foundations. Because of this reason, all Engineering departments arepracticing this method.Measurement of Materials and Works9 Estimation and CostingP.B.-1: From the Drawing given below determine (a) Earth work exca-vation (b) CC (1:5:10) Bed (c) R.R.Masonry in C.M. (1:6) (d)Brick Work in C.M.(1:6).Single Roomed Building (Load Bearing type structure)Note: All Dimensions are in 'M'101.2.3.4.Earth Work excavationfor foundationa) Long wallsb) Short wallsC.C.(1:4:8) bed forfoundationa) Long wallsb) Short wallsR.R.Masonry in CM(1:6) fora) Footings i) Long walls ii) Short wallsb) Basement i) Long walls

ii) Short wallsBrick masonary with CM(1:6) for super structurea) Long Wallb) Short walls22222222226.23.46.23.40.90.90.90.90.60.60.450.450.300.301.41.4Total15.2648.56824.1923.3481.8365.1843.542.225.763.1052.0795.18410.087.2017.28 L=5.3+.45+.45 =6.2 D= 0.3+0.5+0.6 = 1.4 L= 4.3-0.45-0.45= 3.4L= 5.3+0.3+0.3=5.9L=4.3-0.3-0.3 = 3.7L= 5.3+0.225+0.225= 5.75L= 4.3-0.225-0.225 =3.85L=5.3+0.15+0.15=5.6L=4.3-0.15-0.15=4.0S.No. Particulars of Items No. L B H Q ExplanationTotal R.R. Masonry for footings and Basement = 5.76+5.184 = 10.94 m3

0.30.3Total0.50.5Total0.60.6Total3.003.00Total5.93.75.753.855.64.0m3m3m3m3m3Long wall - Short wall MethodMeasurement of Materials and Works11 Estimation and Costing1.2.3.4.Earth Work excavationfor foundationC.C.(1:4:8) bed forfoundationR.R.Masonry in CM(1:6) fora) Footingsb) BasementBrick masany withCM (1:6) for super structure1111119.219.219.219.219.20.90.90.60.450.3

1.40.30.50.6Total0.324.1925.1845.765.18410.94417.28L=2(5.3+4.3)=19.2S.No. Particulars of Items No. L B H Q ExplanationCentre Line Method5.34.3m3m3m3m312EXERCISEI. Short Answer Questions1. List the difference between centre line method and long wall-short wallmethod of taking out measurements.2. What are the rules to be followed while taking the mesurements?3. Mension the units for the following items.a) flooring b) R.R.Masonry c) Plastering for pointing d) Damp proofcourse e) R.C. sunshade (Sepcified width and thickness)II. Essay type questions1. From the Drawing given below determine (a) Earth work excavation (b)CC (1:5:10) Bed (c) R.R.Masonry in C.M. (1:6) (d) Brick Work inC.M.(1:6). by(a) longwall - short wall method(b) Centre line MethodMeasurement of Materials and Works13 Estimation and CostingTYPES OF ESTIMATESChapter33.1 DETAILED ESTIMATE:The preparation of detailed estimate consists of working out quantities ofvarious items of work and then determine the cost of each item. This is preparedin two stages.i) Details of measurements and calculation of quantities:The complete work is divided into various items of work such as earthwork concreting, brick work, R.C.C. Plastering etc., The details of measure-ments are taken from drawings and entered in respective columns of prescribedproforma. the quantities are calculated by multiplying the values that are in num-bers column to Depth column as shown below:Details of measurements formii) Abstract of Estimated Cost :The cost of each item of work is worked out from the quantities thatalready computed in the detals measurement form at workable rate. But the

total cost is worked out in the prescribed form is known as abstract of estimatedform. 4%of estimated Cost is allowed for Petty Supervision, contingencies andUnforeseen items.S.No.Descriptionof ItemNoBreadth(B)mDepth/Height(D/H)mQuantityExplanatoryNotesLength(L)m14ABSTRACT OF ESTIMATE FORMThe detailed estimate should accompained withi) Reportii) Specificationiii) Drawings (plans, elevation, sections)iv) Design charts and calculationsv) Standard schedule of rates.3.1.1.Factors to be consisdered While Preparing Detailed Esti-mate:i) Quantity and transportation of materials: For bigger project, the re-quirement of materials is more. such bulk volume of mateials will be pur-chased and transported definitely at cheaper rate.ii) Location of site: The site of work is selected, such that it should reducedamage or in transit during loading, unloading, stocking of mateirals.iii) Local labour charges: The skill, suitability and wages of local labouresare consideed while preparing the detailed estimate.3.2 DATA:The process of working out the cost or rate per unit of each item is calledas Data. In preparation of Data, the rates of materials and labour are obtainedfrom current standard scheduled of rates and while the quantities of materialsand labour required for one unit of item are taken from Standard Data Book(S.D.B)Item No.Description/ParticularsQuantity Unit Rate Per(Unit)AmountTypes of Estimates15 Estimation and Costing3.2.1 Fixing of Rate per Unit of an Item:The rate per unit of an item includes the following:1) Quantity of materials & cost: The requirement of mateials are takenstrictly in accordance with standard data book(S.D.B). The cost of theseincludes first cost, freight, insurance and transportation charges.ii) Cost of labour: The exact number of labourers required for unit of workand the multiplied by the wages/ day to get of labour for unit item work.iii) Cost of equipment (T&P): Some works need special type of equip-ment, tools and plant. In such case, an amount of 1 to 2% of estimated

cost is provided.iv) Overhead charges: To meet expenses of office rent, depreciation ofequipment salaries of staff postage, lighting an amount of 4% of estimatecost is allocated.3.3 METHODS OF PREPARATION OF APPROXIMATE ESTIMATE:Preliminary or approximate estimate is required for studies of various as-pects of work of project and for its administrative approval. It can decide, incase of commercial projects, whether the net income earned justifies the amountinvested or not. The approximate estimate is prepared from the practical knowl-edge and cost of similar works. The estimate is accompanied by a report duelyexplaining necessity and utility of the project and with a site or layout plan. Apercentage 5 to 10% is allowed for contingencies. The following are the meth-ods used for preparation of approximate estimates.a) Plinth area methodb) Cubical contents methodsc) Unit base method.a) Plinth area method: The cost of construction is determined by multiplyingplinth area with plinth area rate. The area is obtained by multiplying length andbreadth (outer dimensions of building). In fixing the plinth area rate, carefullobservation and necessary enquiries are made in respect of quality and quantityaspect of materials and labour, type of foundation, hight of building, roof, woodwork, fixtures, number of storeys etc.,As per IS 3861-1966, the following areas include while calculating theplinth area of building.16a) Area of walls at floor level.b) Internal shafts of sanitary installations not exceeding 2.0m2, lifts,airconditionsing ducts etc.,c) Area of barsati at terrace level:Barsati means any covered space open on one side constructed on oneside constructed on terraced roof which is used as shelter during rainyseason.d) Porches of non cantilever type.Areas which are not to includea) Area of lofts.b) Unenclosed balconies.c) Architectural bands, cornices etc.,d) Domes, towers projecting above terrace level.e) Box louvers and vertical sunbreakers.b) Cubical Contents Method: This method is generally used for multistoreyedbuildings. It is more accurate that the other two methods viz., plinth area methodand unit base method. The cost of a structure is calculated approximately as thetotal cubical contents (Volume of buildings) multiplied by Local Cubic Rate. Thevolume of building is obtained by Length x breadth x depth or height. The lengthand breadth are measured out to out of walls excluding the plinth off set.The cost of string course, cornice, carbelling etc., is neglected.The cost of building= volume of buildings x rate/ unit volume.c) Unit Base Method: According to this method the cost of structure is deter-mined by multiplying the total number of units with unit rate of each item. In caseschools and colleges, the unit considered to be as 'one student' and in case ofhospital, the unit is 'one bed'. the unit rate is calculated by dividing the actualexpenditure incured or cost of similar building in the nearby locality by the nu

m-ber of units.Types of Estimates17 Estimation and CostingProblems on Plinth Area MethodExample 3.1: Prepare an approximate estimate of building project with totalplinth area of all building is 800 sqm. and from following data.i) Plinth area rate Rs. 4500 per sqmii) Cost of water supply @7½%of cost of building.iii) Cost of Sanitary andElectrical installations each @ 7½% of cost ofbuilding.iv) Cost of architectural features @1% of building cost.v) Cost of roads and lawns @5% of building cost.vi) Cost of P.S. and contingencies @4% of building cost.Determine the total cost of building project.Solution :Data given:Plinth area = 800m2.Plinth area rate = Rs. 4500 per Sqm.∴ Cost of building = 800 x 4500 = Rs. 36,00,000=00Add the cost of the water supply charges @7½%=00 000 , 70 , 21005 . 7 000 , 00 , 36= =×Add the Cost of Sanitary and electrical installation @ 15%= 00 000 , 40 , 510015 000 , 00 , 36= =×Add the cost of archetectural features @1%= 00 000 , 36 1001 000 , 00 , 36= =×Add the cost of Roads Lawns @ 5%= 00 000 , 80 , 11005 000 , 00 , 36= =×Add the Cost of P.S. and contingencies @ 4% = 00 000 , 44 , 11004 000 , 00 , 36= =×Total Rs. 47,70,000=00Assume Add supervision charges 8% on overall cost = 00 600 , 81 , 3

1008000 , 70 , 47 = = ×51,51,600=00 Grand Total Rs.18Example 3.2 : The plinth area of an appartment is 500 sqm. Detemine the totalcost of building from the following data:a) Rate of construction = Rs.1230/--per m3.b) The height of appartment = 16.25 mc) Water Supply, Sanitary and Electrical installations each at 6% ofbuilding cost.d) Architectural appearance @ 1% of building cost.e) Unforeseen item @2% of Building cost.f) P.S. and contingencies @4% of building.Solution :a) The Cost of building = cubic content x cubic rate= 500 ×16.25 ×1230 = Rs. 99,93,750/-b) Provision for water supply, sanitary and Electrical installations water supply and sanitation each @ 6% = − =×/ 875 , 98 , 17 Rs.10018 750 , 93 , 99 i.e total percent = 3×6 = 18% building costc) Architectural appearance @1%=1001 750 , 93 , 99 × = Rs. 99,937/�d) Unforeseen items @2% = Rs. 1,99,875/�e) P.S. and contingenies @4% = Rs. 3,99,750/�Total = Rs.1,24,92,187/�Sundries 7,813/�Total cost of the building project = Grand Total = Rs.1,25,00,000/�Types of Estimates19 Estimation and CostingExample 3.3: The plinth area and plinth area rate of a residential building are100 sqm and Rs. 5000/� respectively. Determine the total cost of building as�suming suitable provisions.Solution :Cost of building = 100 x 5000 = Rs.5,00,000Cost of water supply andsanitary fittings @15% = 10015 000 , 00 , 5 ×= Rs. 75,000Cost of Electrification @7½% =1005 . 7 000 , 00 , 5 ×= Rs. 37,500Cost of Roads & Lawns @5%=1005 000 , 00 , 5 ×= Rs. 25,000Cost of P.S.& contingencies@4%=1004 000 , 00 , 5 ×

= Rs. 20,000Total Cost Rs. 6,57,500/�Example 3.4 : Prepare an approximate Extimate of a proposed building fromthe follwoing?Plinth area of the building = 226 sqm.Cost of the structure = 2500 per sqm.Water supply and sanitary arangements = 12½%Electrification =7%Fluctuation of rates = 5%petty supervision charges = 3%sol: Cost of Building = 226x 2500 = Rs.5,65,000Water supply & Sanitory arrangements @ 12½ %=1005 . 12 000 , 65 , 5 ×= Rs. 70,000Electrification @7% =1007 000 , 65 , 5 ×= Rs. 39,55020Fluctuation of rates 5% =1005 000 , 65 , 5 ×= Rs. 28,250Pettysupervision charges 3%=1003 000 , 65 , 5 ×= Rs.16,950Total Cost Rs. = 7,19,750.00Problem on Cubical content Method:Example 3.5 : Prepare the rough estimate for a proposed commertial complesfor a municipal corporation for the following data.Plinth Area = 500m2/floorHt of each storey = 3.5mNo.of storeys = G+2Cubical content rate = Rs. 1000/m3Provided for a following as a pecentage of structured costa) water supply & Sanitary arrangement �8%b) Electrification �6%c) Fluctuation of rates � 5%d) Contractors profit � 10%e) Petty supervision & contingencies � 3%Sol : Cubical content = No.of storeys (Plinth Area x height of each storey)= 3(500x3.5) = 5250m3Structural cost = Cubical content x cubical content rate= 5250 x 1000 = 52.5 Lakhsother provisons:�a) Water supply and sanitation = 52.5x8/100 = Rs.4.2 Lakhsb) Electrification = 52.5 x 6/100 = Rs.3.15 lakhsc) fluctuation of rates = 52.5 x 5/100 = Rs.2.625Total = Rs. 9.975 LakhsStructural cost = Rs. 52.500 LakhsTotal = Rs.62.475 Lakhsd) P.S./& contingencies = 62.475 x 3/100 = Rs.1.874 Lakhse) Contractors Profit = 62.475 x 10/100 = Rs.6.247 Lakhs

Total Cost = Rs.70.596 LakhsTypes of Estimates21 Estimation and CostingProblems on Unit Base Method:Example 3.6: Prepare an approximate estimate or rough cost estimate of ahospital building for 50 beds. The cost of construction altogether for each bed isRs. 60,000/�. Determine the total cost of hospital building.Solution:No. of beds = 50Cost of construction = Rs. 60,000/�Total Cost of Hospital building = 50x 60,000= Rs. 30,00,000/� Example 3.7: To prepare the rough cost estimate of a hostel building whichaccommodate 150 students. The cost of construction including all provisions isRs. 15,000/� per student. Determine total cost of building.Solution :No.of students= 150Cost of construction including all L.S. provisions = Rs. 15,000/�Total Cost of hostel building =150 x 15000 = Rs. 22,50,000/�(Rupees twenty two lakhs, fifty thousands only)EXERCISEI. SHORT ANSWER QUESTIONS:1. List the factors to be consider while preparing detailed estimate andexplain breifly?2. What are the differences between plinth area method and Unit base method?3. List the requirements of data preparation.II ESSAY TYPE QUESTIONS :1. Prepare the approximate cost of building project (group HOuseing)i) No.of houses = 150ii) Plinth area of each dwelling = 600m2iii) Plinth area rate = Rs. 5,000/�per m2iv) Cost of water supply & sanitary arrangements @12½%v) Electrification at 7½% of cost of builing.vi Cost of roads & Lawns @5%vii) Cost of P.S.& contingencies @4%2. Prepare a rough cost estimate of a cinema theatre which accommodate 1700seats. The cost of construction including all provisions is Rs.6000/� per seat.3. What are the methods of preparation of approximate estimates and explainbriefly.22DET DET DET DET DETAIL & AIL & AIL & AIL & AIL & ABSTRA ABSTRA ABSTRA ABSTRA ABSTRACT ESTIMA CT ESTIMA CT ESTIMA CT ESTIMA CT ESTIMATES TES TES TES TESOF B OF B OF B OF B OF BUILDINGS UILDINGS UILDINGS UILDINGS UILDINGSChapter4Single Roomed Building(Load Bearing type structure)Example 1: From the given figure below calculate the detailed andabstract estimate for the single roomed building (Loadbearing type structure) bya) long wall & short wall method (b) Centre Line MethodNote: All Dimensions are in 'M'23 Estimation and Costing1.2.3.4.

Earth Work excavationfor foundationa) Long wallsb) Short wallsC.C.(1:4:8) bed forfoundationa) Long wallsb) Short wallsR.R.Masonry in CM(1:6) fora) Footings i) Long walls ii) Short wallsb) Basement i) Long walls ii) Short wallsBrick masonary with CM(1:6) for super structurea) Long Wallsb) Short wallsc) for parapetwall a) Long Wallsb) Short walls2222222222226.23.46.23.40.90.90.90.90.60.60.450.450.300.300.20.21.41.4Total15.2648.56824.1923.3481.8365.184

3.542.225.763.1052.0795.18410.087.201.681.3220.28 L=5.3+.45+.45 =6.2 D= 0.3+0.5+0.6 = 1.4 L= 4.3�0.45�0.45= 3.4L= 5.3+0.3+0.3=5.9L=4.3�0.3�0.3 = 3.7L= 5.3+0.225+0.225= 5.75L= 4.3�0.225�0.225 =3.85L=5.3+0.15+0.15=5.6L=4.3�0.15�0.15=4.0S.No. Particulars of Items No. L B H Q ExplanationTotal R.R. Masonry for footings and Basement = 5.76+5.184 = 10.94 m30.30.3Total0.50.5Total0.60.6Total3.003.000.750.75Total5.93.75.753.855.64.05.64.4m3m3m3a) Long wall � Short Method5.64.60.2m3m3

245.6.789Deductions for openingsa)Doorsb) WindowsR.C.C. (1:2:4) fora) Roof slabb) Lintels overi) Doorsii) Windowsc) Beamsi) Long beamsii) short beamsSandfilling forbasementC.C.(1:4:8) forflooringFlooring with MosaictilesPlastering with CM(1:6)for super structureInsideFor walls Out sideFor wallsBasement outsideParapet walla) Insideb) topDeductions for opeiningsDoorsWindows1311322111111111x23x21.01.55.61.21.55.64.0

4.854.855.018.020.421.618.819.61.01.50.30.34.60.30.30.30.33.853.854.0� �� �� �� �0.2� �� �2.11.2Total0.120.150.150.30.3Total0.480.1� �3.03.870.60.75���

Total2.11.20.631.62(�)2.253.0900.0540.2021.0080.7205.0748.961.8620.054.0

61.212.9614.13.92146.184.210.815.0L=5.0�0.075�0.075=4.85B= 4.0�0.075�0.075=3.85 L= 2(5.0+4.0) = 18 .0L=2(5.6+4.6)=20.4H=3.0+0.12+0.75=3.87 (upto parapet wall)S.No. Particulars of Items No. L B H Q Explanationm2m3m2Net Brick Masonry = 20.28 � 2.25 = 18.03m3Net Plastering = 146.18 � 15.0 = 131.18m2m2m3Detail & Abstract Estimates of Buildings25 Estimation and Costing5.01.01.5101112.131415Plastering for Ceilingwith CM(1:5)White Washing with twocoats with Janatha cementSame as quantity ofplastering for walls andceilingColour washing with twocoatsSame as quantity ofplastering for walls andceilingSupply & Fixing of bestcountry wood fora) Doorsb) WindowsPainting with ready mixedsynthetic enamil paits withtwo coats over primary coat

for new wood fora) Doorsb) WindowsPetty supervision andcontingencies at 4% androunding off.1132¼x12¼x34.0���

���

� �2.11.2Total20.0151.18151.181 No.3No.4.72512.1516.875(= 131.18+20= 151.18)(=131.18+20)151.18)S.No. Particulars of Items No. L B H Q Explanationm2m2261.2.3.4.Earth Work exevationfor foundationC.C.(1:4:8) bed forfoundationR.R.Masonry in CM(1:6) fora) Footingsb) BasementBrick masonry withCM (1:6) for super structure11111119.219.219.219.219.220.0

0.90.90.60.450.30.21.40.30.50.6Total3.00.7524.1925.1845.765.18410.94417.283.00L=2(5.3+4.3)=19.2S.No. Particulars of Items No. L B H Q Explanationb) Centre Line Method5.34.35.6. 7Deductions for openingsa)Doorsb) WindowsR.C.C. (1:2:4) fora) roof slabb) Lintels overi) Doorsii) Windowsc) beamsSandfilling forbasementC.C.(1:4:8) forflooring131131111.01.55.61.21.519.24.854.850.30.34.6

0.30.31.33.853.852.11.2Total0.120.150.150.3Total0.480.10.631.62(�)2.253.0900.0540.2021.7285.0748.961.86L=5.0�0.075�0.075=4.85B= 4.0�0.075�0.075=3.85m3m3m3m3m3For parapet wallNet Brick Masony = 17.28 +3.0�2.25 = 18.03m3Detail & Abstract Estimates of Buildings27 Estimation and Costing8.9101112.13flooring with MosaictilesPlastering with CM(1:6)for super structureInsideFor walls Out sideFor wallsBasement outsideParapet walla) Inside

b) topDeductions for opeiningsDoorsWindowsPlastering for Ceilingwith CM(1:5)White Washing with twocoats with Janatha cementSame as quantity ofplastering for walls andceilingColour washing with twocoatsSame as quantity ofplastering for walls andceilingSupply & Fixing of bestcountry wood fora) Doorsb) Windows1111111x23x21135.018.020.421.618.819.61.01.55.04.0� �� �� �� �0.2� �� �4.0� �3.03.870.60.75���

Total2.11.2� �20.0

54.061.212.9614.13.92146.184.210.815.020.0151.18151.181 No.3No.L=5.0�0.075�0.075=4.85B= 4.0�0.075�0.075=3.85(131.18+20=151.18)m2m2Net Plastering = 146.18�15 = 131.18 m2m2m2m2281415Painting with ready mixedsynthetic enamil paints withtwo coats over primary coatfor new wood fora) Doorsb) WindowsPetty supervision andcontingencies at 4% androunding off.2¼x12¼x31.01.5���

���

2.11.2Total4.72512.1516.875S.No. Particulars of Items No. L B H Q Explanationm2Detail & Abstract Estimates of Buildings29 Estimation and CostingAbstract estimate of single roomed building (load bearing structure)S.No. Description of item Quantity Unit Rate Per Amount

1.2.3.4.5.6.7.8.9.10.111213141516171819Earth work excaationCement concrete(1:4:8)RR.masonry in C.M.(1:5)Sand filling in basementBrick masonry in countrybricks of standard size inCM(1:8)R.C.C. (1:2:4) for lintels,beams etc.R.C.C.(1:2:4) for slabs,Cement concrete (1:5:10)for flooringSupplying and fixing ofcountry wood for doors.Supplying and fixing ofcountry wood for windowsand ventilators.Plastering to all exposedsurfaces of brick work andbasement with C.M (1:5)White washing with bestshell limeFlooring with spartek tilesset in C.M (1:3)Painting with ready mixedenamel paintPovision for water supplyand sanitary arangements@12.5%Provision for electrification@7.5%Povision for architecturalappearance @2%Provision for unforeseenitems 2%Provision for P.s.andcontingencies @4%24.1925.18410.948.96

18.031.9843.091.862.15.4151.18151.182016.875m3m3m3m3m3m3m3m3m2m2m2m2m2m246545451391195.20229160306030145216502300582116423033510m31m3m310m3

m3m3m3m3m2m210m210m210m210m21125.008009.3015217.50175.0041306.7311963.5218633.002700.723465.0012420.008798.701753.688460.00565.31134593.4616824.1810094.502691.862691.865383.73172279.65Grand Total Rs. Total30Example :2 :�From the given figure below calculate the details andabstract estimate for the double roomed building (Loadbearing type structure) by a) long wall & short wall method(b) Centre Line MethodRoom14x6mRoom23x6mDetail & Abstract Estimates of Buildings31 Estimation and Costing1.2.3.Earth Work excavationfor foundationa) Long walls

b) Short wallsC.C.(1:4:8) bed forfoundationa) Long wallsb) Short wallsBrick masanory forfootings with CM (1:4) first footinga) Longwallsb) Short walls2nd fooringa) Long wallsb) short wallsii) for base ment long walls short wallsiii) for super structure long walls short wallsiv) Parapet walla) long wallsb) Shot wallsDeductions for openingsDoorsWindowsLintels over doorswindows2323232323232233338.65.38.65.38.455.458.205.708.005.907.906.007.906.201.0

1.51.201.701.01.01.01.00.850.850.60.60.40.40.30.30.20.20.30.30.30.31.0511.05Total0.20.2Total0.40.40.450.450.40.43.03.00.700.70Total2.11.20.100.10Total18.0516.7034.753.443.186.625.7465.5604.4284.6172.5602.83214.2216.202.2121.73660.11

1.891.620.1080.1533.771L=7.6+0.5+0.5=86L=6.3�0.5�0.5=5.3L=7.6+0.425+0.425=8.45L=6.3�0.425�0.425=5.45L=7.6+0.3+0.3=8.2L=6.3�0.3�0.3=5.7L=7.6+0.2+0.0=8.0L=6.3�0.2�0.2= 5.9L=7.6+0.15+0.15=7.9L=6.3�0.15�0.15=6.0Net B.M.=60.11�377=56.34m3S.No. Particulars of Items No. L B H QExplanation7.96.60.2m3m332RCC(1:2:4)for a) roof slabb) for lintles over doorsWindowsc) beamsPlastering for wallsa) Inside room1 room2b) out sideParapet wall(Sides)Deductionsa) doorsb) windowsflooring with cuddapahslab in cm (1:3)Room1Room213311111×21×13×23×2117.91.21.7

33.820.018.029.028.228.21.01.54.03.06.60.30.30.3� ����

���

���

0.20���

6.06.00.120.10.10.3Total3.03.03.00.70� �Total2.101.20Total���

���

Total6.2560.1080.1533.0429.29860.0054.0087.0039.485.64246.1212.610.823.424184242L=2(4.0+6.0)=20L=2(7.9+6.6)=29L=2(7.7+6.4)=28.2

45.6.78910111213 Net Plastering = 246.12� 23.4 = 222.72 m2Plastering for ceiling =same as flooringWhite washing = same as plastering for walls & Ceiling = 222.72 +42 = 264.72Colour washing with two coatsSame as quantity of plastering for walls and ceiling 264.72Supply & Fixing of best country wood fora) Doorsb) WindowsPainting with ready mixed synthetic enamil paints two coatsover primary coat for new wood fora) Doorsb) Windows2% unforeseen items4% P.S& contingenciesand round off.333Nos.3 Nos2¼x32¼x31.01.5��

��

14.17511.1325.305m3m2m2m2m2m2m2Detail & Abstract Estimates of Buildings33 Estimation and Costingb) Centre Line Method1.2.3.

4.Total centre line length=(4.3+3.3)2+6.3x3=34.1mEarth work excavationC.C.(1:4:8) bed forfoundationBrick masonry withCM(1:4)a) for foundationi) first footingii) 2nd footingb) for basementc) for super structured) for parapet wallTotal centre line length= 2(7.7+6.4) =28.2Deductions forOpenings DoorswindowsLintels DoorsWindows1111111333333.133.133.2533.5033.733.8028.21.01.51.21.71.01.00.850.600.400.300.20.30.30.30.31.050.200.400.450.403.0

0.70Total2.11.20.10.1Total34.756.6211.309.0455.39230.423.94860.101.891.620.1081.1533.771L=34.1�2x1/2=33.1L=34.1�0.85 =33.25L=34.1�0.6 x2/2L=34.1�0.4 x2/2L=34.1�0.3x2/2Net B.M.=60.11�3.771=56.34m3S.No. Particulars of Items No. L B H Q Explanation4.3 3.36.3Quantity of R.C.C.Roof, Plastering for walls and cealing andflooring, White washing is same as Longwall &Short wallmethod.7.96.60.2776.4m3m3m334Abstract estimate of two roomed building (Load bearing type structure)S.No. Description of item Quantity Unit Rate Per Amount1.2.3.4.5.6.7.8.9.10.111213

1415161718Earth work excavationCement concrete(1:4:8)Sand filling in basementBrick masonry in countryBricks of standard size inCM(1:8)R.C.C. (1:2:4) for lintels,beams etc.R.C.C.(1:2:4) for slabs,Cement concrete (1:5:10)for flooringSupplying and fixing ofcountry wood for doors.Supplying and fixing ofcountry wood for windowsand ventilators.Plastering to all exposedsurfaces of brick work andbasement with C.M (1:5)White washing with bestshell limeFlooring with spartek tilesset in C.M (1:3)Painting with ready mixedenamel paintProvision for water supplyand sanitary arrangements@12.5%Provision for electrification@7.5%Provision for architecturalappearance @2%Provision for unforeseenitems 2%Provision for P.S.andcontingencies @4%34.756.6212.03656.343.3036.264.26.35.4222.72264.724225.305m3m3m3

m3m3m3m3m3m2m2m2m2m24651545195.20229160306030145216502300582116423033510m31m310m3m3m3m3m3m2m210m210m210m210m21615.90

10228.00235.00129075.0019918.0037748.006098.4010395.0012420.0012962.303070.7517766.008477.17128090.0016011.259606.752561.802561.805123.60163955.23Grand TotalDetail & Abstract Estimates of Buildings35 Estimation and CostingExample 3 :� From the given figure below calculate the details and ab�stract estimate for the single Storeyed residential buildingwith no of rooms (Load bearing type structure) by CentreLine MethodCentre line diagramTotal centre line length =(3.3+3.8)3+3.8×3+4.3×2=41.3mno of T Junctions = 43.33.83.84.37.1361.2.3.4.5.6.Earth work ExcavationC.C. bed (1:5:10)R.R. Masornary in CM1:61st FootingIInd FootingBasementDamp proof courseover basement alroundthe building with CC(1:2:4)Deduct for Door sillsNet QuantityFirst class brick workin wall ina) superstructure withCM 1:6b) Parapet wallDeductions:

DoorsWindowsLintel opening overDoorsWindowsNet QuantityPlastering with 12mmthin CM 1:5Deductions for openings11111131138381x239.539.540.140.340.540.51.040.730.41.01.21.21.440.10.90.90.60.50.40.60.30.30.30.30.30.30.3���

1.00.30.30.40.6Total���

���

3.0

0.6Total2.01.50.10.1Total3.035.5510.6657.2188.069.7225.0016.2� 0.936.635.47242.1021.804.320.1080.3366.564240.641.3�4x0.9/2=39.541.3�4x0.6/2=40.141.3�4x0.5/2=40.341.3�4x0.4/2=40.5L =41.3�4x0.3/2L=2(7.1+8.1)Asue 100mmprojection on eithersideL=41.3�4x0.3=40.1of BM = 42.102�6.564 = 35.538m3=16.2�0.9=15.3sq.mS.No. Particulars of Items No. L B H Q Explanation7.48.40.37.18.1m3m3m2m2m3Detail & Abstract Estimates of Buildings37 Estimation and CostingS.No. Particulars of Items No. L B H Q Explanation7.8.9.

10.11121314DoorswindowsPlastering for parapetwall (sides)TopFlooring with 25mmthCC(1:2:4)KitchenBedHallSills of DoorsCeiling = Same asFlooringRCC(1:2:4) fora) Slabb) lintels over DoorsWindowsc) beams3x28x21x21111313811.01.230.430.43.03.56.81.07.401.21.440.7���

���

���

0.33.53.54.00.38.400.30.30.32.0

1.5Total0.6���

Total� �� �� �� �Total1.50.10.10.3Total12.028.840.836.489.1245.6010.512.2527.200.9050.8550.859.3240.1080.3363.66313.431Net Plastering = 240.6�40.8+45.6 =245.4 m2white washing = Same as Plastering for wallsand ceiling 245.4+50.85 = 296.25 m2Supply & Fixing of best country wood fora) Doorsb) WindowsPainting with ready mixed synthetic enamil paints two coatsover primary coat for new wood fora) Doorsb) Windows2% unforeseen items4% P.S& contingenciesand round off.383Nos.8 Nos2¼x32¼x81.01.2��

��

13.5032.4045.90

2.01.5m2m2m2m2m3m238Abstract estimate of single storeyed residential building with no of rooms (leadbeary type)S.No. Description of item Quantity Unit Rate Per Amount1.2.3.4.5.6.7.8.9.10.111213141516171819Earth work excavationCement concrete(1:4:8)RR.masonry in C.M.(1:5)Sand filling in basementBrick masonry in countrybricks of standard size inCM(1:8)R.C.C. (1:2:4) for lintels,beams etc.R.C.C.(1:2:4) for slabs,Cement concrete (1:5:10)for flooringSupplying and fixing ofcountry wood for doors.Supplying and fixing ofcountry wood for windowsand ventilators.Plastering to all exposedsurfaces of brick work andbasement with C.M (1:5)White washing with bestshell limeFlooring with spartek tiles

set in C.M (1:3)Painting with ready mixedenamel paintProvision for water supplyand sanitary arrangements@12.5%Provision for electrification@7.5%Provision for architecturalappearance @2%Provision for unforeseenitems 2%Provision for P.S.andcontingencies @4%35.5510.66525.0023.77535.5354.1079.3245.0856.0014.40245.40296.2550.8545.90m3m3m3m3m3m3m3m3m2m2m2m2m2m246515451391195.20

229160306030145216502300582116423033510m31m3m310m3m3m3m3m3m2m210m210m210m210m21653.00164.77.5034775.00464.0081417.6024765.2056223.707383.409900.0033120.0014282.303436.5021509.501537.65306945.3538368.2023020.906138.906138.9012277.80392890.00Detail & Abstract Estimates of Buildings

39 Estimation and CostingExample 4:� From the given figure below calculate the details and ab�stract estimate for the single storeid residential building withno.of rooms (Framed Structured type) by Centre LineMethod40S.No. Particulars of Items No. L B H Q Explanation12.3.4.5.Earth work excavationfor foundation fora) pillarsb) around the buildingand cross wallsC.C. (1:4:8) fora) pillarsb) around the buildingand cross wallsBrick Masonry with C.M.(1:6) fora) first footingb) Second Footingc) Superstructured) Parapet wallDeduction for openinga) Doorsb) WindowsR.C.C.(1:1.5:3) forcolumnsa) Rectangular portionb) Trepezoidal portionc) Square portion upto GLd) Squareporiton above GLPlastering with 12mmthin CM 1:5Deductions for openings818111113888881x21.526.31.538.338.338.3

38.330.41.01.21.50.90.30.340.11.50.751.50.750.450.350.30.30.30.31.50.90.30.3���

1.800.85Total0.150.15Total0.350.303.00.6Total2.01.5Total0.30.450.93.0Total3.032.427.960.32.74.37.06.034.694.025.4720.211.84.326.125.402.92

0.652.1611.13240.6L= 5.6+2.8x2+2.3x3+(1.8+2.3)2L= 3.5x3+3x2+3.5x2+4x2+6.8=38.3L=(7.1+8.1)x2=30.4L=41.3�4x0.3=40.16.87.80.37.18.1Net Brick Masonry = 20.21 � 6.12 = 14.09m3m3m3m3m3Detail & Abstract Estimates of Buildings41 Estimation and CostingS.No. Particulars of Items No. L B H Q Explanation6.7.8.9.10111213DoorswindowsPlastering for parapetwall (sides)TopFlooring with 25mmthCC(1:2:4)KitchenBedHallSills of DoorsCeiling = Same asFlooringRCC(1:2:4) fora) Slabb) lintels over DoorsWindowsc) beams3x28x21x211

11313811.01.230.430.43.03.56.81.07.401.21.440.7���

���

���

0.33.53.54.00.38.400.30.30.32.01.5Total0.6���

Total� �� �� �� �Total1.50.10.10.3Total12.028.840.836.489.1245.6010.512.2527.200.9050.8550.859.324

0.1080.3363.66313.431Net Plastring = 240.6�40.8+45.6 =245.4 m2white Washing = Same as Plastering for wallsand ceiling 245.4+50.85 = 296.25 m2Supply & Fixing of best county wood fora) Doorsb) WindowsPainting with ready mixed synthetic enamil paints two coatsover primary coat for new wood fora) Doorsb) Windows2% unforeseen items4% P.S& contingenciesand round off.383Nos.8 Nos2¼x32¼x81.01.2��

��

13.5032.4045.902.01.5m2m2m2m2m342Abstract estimate of single storeyed residential building (framed structuretype)S.No. Description of item Quantity Unit Rate Per Amount1.2.3.3.4.5.6.7.8.9.1011

12131415161718Earth work excavationCement concrete(1:4:8)Sand filling in basementBrick masonry in countrybricks of standard size inCM(1:5) Reefs columnsR.C.C. (1:2:4) for lintels,beams, columns etc.R.C.C.(1:2:4) for slabs,Cement concrete (1:5:10)for flooringSupplying and fixing ofcountry wood for doors.Supplying and fixing ofcountry wood for windowsand ventilators.Plastering to all exposedsurfaces of brick work andbasement with C.M (1:5)White washing with bestshell limeFlooring with spartek tilesset in C.M (1:3)Painting with ready mixedenamel paintProvision for staircaseProvision for water supplyand sanitary arrangements@12.5%Provision for electrification@7.5%Provision for architecturalappearance @2%Provision for unforeseenitems 2%Provision for P.s.andcontingencies @4%60.307.0023.77514.0915.2379.3245.0856.0014.40245.40296.2550.8551.00LSm3

m3m3m3m3m3m3m3m2m2m2m2m2m24651545195.20229174056030145216502300582116423033510m31m3m310m3m3m3m3m2m210m210m

210m210m22804.0010815.00464.0032250.20112830.0056223.707383.409900.0033120.0014282.303436.5021509.501708.5050000.00354584.6044323.0026593.807091.707091.7014183.40453868.00 Total Rs.Detail & Abstract Estimates of Buildings43 Estimation and CostingExample 5 :� From the given figure below calculate the details and ab�stract estimate for the two storeoied residential building withno.of rooms (Framed Structured type) by Centre LineMethodGround Floor PlanFirst Floor Plan44S.No. Particulars of Items No. L B H Q Explanation12.3.4.5.6.7.First FloorR.C.C. (1:1.5:3) fora) Columnsb) Slabsc) beamsd) lintels over doors windowsB.M. with CM(1:8) in thefirst floorParapet wallDeductions for openingsDoorsWindowsPlastering with CM (1:4)for wallsfor parapetwall sidesParapet wall Top

DeductionsDoorsWindowsFlooring with CM(1:3)8111611161x21x211610.37.4040.71.21.428.630.41.01.230.430.430.41.01.26.80.308.40.30.30.30.30.30.30.3� �� �0.3���

� �7.83.00.150.30.10.1Total3.00.62.01.53.0

0.6� �2.01.5Total���

2.169.3243.6630.0360.25215.43525.745.47�0.6�3.24182.436.489.12�2.0�10.8215.253.0453.04The quantities of various items of the building for the Ground floor is same as previousproblem. Here the quantities of various items of the building for the First floor is men�tioned here.Net BM = 25.74+5.47�0.6�3.24 = 27.372Plastering for ceiling with CM(1:3)= Same as FlooringWhite washing or colour washing = same as ceiling & BM = 53.04 + 215.2= 268.24The estimation of a staircase is mentioned sepa�rately in the next problemm3m3m2m2m2m2Detail & Abstract Estimates of Buildings45 Estimation and CostingExample 6: � Estimate the Quantities of the pictured roof shown in figurea) Size of common rafter = 80x40mmb) Size of ridege piece = 120x 200mmc) Size of eaves board = 20 x 300mm230mm thick brick wallCommon rafters at 450mm c/cRise = 1/3Span5005.46

a) Length of Common rafter = 222 2346 . 573 . 23 2 |.|

\|+ = |.|

\|+ |.|

\| Span length= 3.28mb) Length of ridge piece = 7.0+0.23x2 +0.5x2 = 8.46 mc) Length of Eaves board = 2( 8.46+5.46) = 27.84mS.No Description No L B H Qty Remarks 1 Ridge piece 1 8.46 0.12 0.20 0.20 2 Eaves Board 1 27.84 — 0.30 8.35 Unit of eavesBoard in m2 3 Common rafters 40 3.28 0.08 0.04 0.4246Example� 7: � Calculate the quantities of items of the stair case of thefigure shown in below.Detail & Abstract Estimates of Buildings47 Estimation and CostingS.No. Particulars of Items No. L B H Q Explanation12.3.4.5.6.7.8.R.C.C. (1:2:4) excludingsteel and its fabricationbut including centeringand shultering andbinding wire.a) Toe wallb) Waist slab for 1 and IIflightsc) Landing Middle andfirst floorIst class brick work in

C.M. (1:4) for steps20mm. thick cementplastering (1:5) for stepsfinished neata) Treads & Risesb) ends of steps2.5cm No sing in steps2.5cm. C.C. flooringfinished neat cementfloating in middle andfirst floor landing.Supplying and fixing ofbest teak wood hand railfinished smoothsupply and fixing of bestteak wood newel posts &finished smoothCap of Newel post1x11x21x22x112x112x112x121x21x11x21x23.153.212.851.21.21.22.556.671.0���

0.31.21.65� �1.2� �0.1� �0.40.170.17TotalTotal� �� �� �0.1���

0.381.311.60

3.290.4950.4110.9728.8RM6.126.67RM0.022Nos.R.C.C. Stair Case0.3½x( 0.25+1.5) L= (1.2+0.15+1.2+2x0.3) L= m 21 . 3 65 . 1 75 . 222= + L= (1.2+0.15+1.2+2x0.15)x (0.25 +0.15) 10.56½x( 0.25+1.5)m3m2m2m3m348Example 8:� From the given figure below calculate the details esti�mate for the Compound Wall2304503001800230500800150G LNote: 1) Brick Pillers of size 230x 230 size are built every 3 meters2) The expansion joints are provided for every 6m lengthPlot15x 12Cross Section of the compound wallDetail & Abstract Estimates of Buildings49 Estimation and CostingS.No. Particulars of Items No. L B H Q Explanation12.3.4.5.6.7.8.Earth work excavation

for foundationTotal Centerline length =2(15.15+12.15)= 54.6C.C.(1:4:8) for founda�tionFirst class brick work inCM (1:6) in foundationa) footingb) BasementD.P.C.with C.C.(1:1½:3)25mmtha) First Class B.M. inCM(1:6) for wall insuper structureb) Brick piller @3cm c/cDeduction 150mm thwallPlastering with CM(1:5)a) Outer surface &inner surface(0.3+0.04+1.8)b) Top of wallc) Piller Projection fromthe face of the wallWhite washing/coloursame as item(6)11111114141x21x114x254.654.654.654.654.654.60.230.1554.654.60.040.800.800.50.230.230.150.230.23���

0.15���

0.68

0.230.450.3Total���

1.81.81.8Total2.14� �1.8Total29.710.0412.283.7616.0412.5614.741.33�0.87 15.2233.698.192.016243.89243.8915.1512.15m3m3m3m3m2m2m250Example 9:� Estimation of basement steps (one way)S.No. Particulars of Items No.L BH QExplanation12.3.4.5.Earth work excavation forfoundationC.C.(1:4:8) bed forfoundationIst class BM in CM (1:4)

a) 1st stepb) 2nd Stepc) 3rd Stepd) 4th stepPlastering with CM(1:3)a) Threadsb) Risersc) endsa) Ist stepb) 2nd Stepc) 3rd Stepd) 4th Stepwhite washing /colourwashing1111114422221.81.81.51.51.51.51.51.51.20.90.60.31.351.351.200.900.600.30���

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0.150.150.150.150.150.15Total���

0.150.15

0.150.150.15Total0.3600.3600.270.270.130.060.731.80.90.360.270.180.093.603.60 = Same as item (4)1.51.20.50.9 0.90.60.60.30.150.150.51.5 Note: All dimensions are in metres150mm offset allround0.150.15m3m3m3m2m2Detail & Abstract Estimates of Buildings51 Estimation and CostingEXERCISEShort Answer Questions1. The internal dimensions of a single roomed building are 5.75x3.75m.Find the Centre line length of room and parapet. If the wall thickness ofroom and parapet are 300mm and 250mm respectively.2. The internal dimensions of a room are 6.25 x 4.25m. find the quantity ofsand filling in basemet. the height and thickness of basement are 750mmand 450mm respectively the wall thickness of room is 230mm.Essay Type Questions:1. The plan and section of one roomed buildingCalculate the following quantities by a) central line method b) Long wall& shortwall method.i) Earth work excavation .ii) Cement Concrete for foundation.

iii) Brick in CM 1:6 for footing.iv) Brick in CM 1:6 for walls excluding openings522) For a building drawing shown in figure calculatea) Brickwork in CM(1:6)in foundation footing.b) 12mm thick plastering the wall surfaces with CM (1:6) for all superstructure walls by central line method.c) Earth work excavation for the foundation.3) Repare the detailed estimate for the following items of work forthe building shown in figure.a) R.C.C. (1:1.5:3) in columns upto ground level only.b) R.C.C. (1:2:4) in plinth Bleamsc) R.C.C. (1:2:4) in slab.Detail & Abstract Estimates of Buildings53 Estimation and Costing4) Prepare the detailed estimate for the following items of work for abuilding shown .a) R.R. masonry in CM 1:6 for footings and basement.b) Brick work in CM 1:6 for super structure.c) Plastering for ceiling with CM 1:35) From the Hipped roof shown in sketch, calculatea) Length of Hip rafterb) Ridge Piece546) For an R.C.C. Stair case shown in fig. Calculate the followingcontents.a) R.C.C. (1:2:4) for base beam, waist slab, Top and intermediatelandings.b) Brick work in CM(1:4) for steps.Detail & Abstract Estimates of Buildings55 Estimation and Costing7) The section of steps at the front of a residential building is shown infig. Calculatea) Volume of BM in CM (1:5) for all three steps. the length of steps is2.1mb) Plastering with CM (1:4) for all three steps.56Definition : In order to determine the rate of a particular item, the factorsaffecting the rate of that item are studied carefully and then finally a rate is de�cided for that item. This process of determining the rates of an item is termed asanalysis of rates or rate analysis.The rates of particular item of work depends on the following.1. Specifications of works and material about their quality, proportion and con�structional operation method.2. Quantity of materials and their costs.3. Cost of labours and their wages.4. Location of site of work and the distances from source and conveyancecharges.5. Overhead and establishment charges6. ProfitCost of materials at source and at site of construction.The costs of materials are taken as delivered at site inclusive of thetransport local taxes and other charges.Purpose of Analysis of rates:1. To work out the actual cost of per unit of the items.2. To work out the economical use of materials and processes in completingthe particulars item.3. To work out the cost of extra items which are not provided in the contract

bond, but are to be done as per the directions of the department.4. To revise the schedule of rates due to increase in the cost of material andlabour or due to change in technique.Cost of labour �types of labour, standard schedule of ratesThe labour can be classified in to1) Skilled 1st class2) Skilled IInd Class3) un skilledThe labour charges can be obtained from the standard schedule of rates30% of the skilled labour provided in the data may be taken as Ist class, remain�

ing 70% as II class. The rates of materials for Government works are fixed byAN AN AN AN ANAL AL AL AL ALY YY YYSIS OF RA SIS OF RA SIS OF RA SIS OF RA SIS OF RATES TES TES TES TESChapter557 Estimation and Costingthe superintendent Engineer for his circle every year and approved by the Boardof Chief Engineers. These rates are incorporated in the standard schedule ofrates.Lead statement: The distance between the source of availability of materialand construction site is known as "Lead " and is expected in Km. The cost ofconvenayce of material depends on lead.This statement will give the total cost of materials per unit item. It in�cludes first cost, convenayce loading, unloading stacking, charges etc.The rate shown in the lead statement are for mettalled road and includeloading and staking charges . The environment lead on the metalled roads arearrived by multiplying by a factora) for metal tracks � lead x 1.0b) For cartze tracks � Lead x 1.1c) For Sandy tracks � lead x 1.4 Note: For 1m3 wet concrete = 1.52m3 dry concrete approximatelySP.Wt of concrete= 1440 kg/m3 (or) 1.44 t/m3 1 bag of cement = 50 KgExample 1:� Calculate the Quantity of material for the following items.a) R.C.C. (1:2:4) for 20m3 of workb) R.C.C. (1:3:6) for 15m3 of worka) Quantity of cement required = ) 4 2 1 (1+ + x 1.52 ×20 = 4.14m3 x 501440=119.26 bagsQuantity of Sand required =) 4 2 1 (

2+ +×1.52x20=8.28m3Quantity of cource aggreate = 74x1.52x20 = 16.56m3b) Quantity of cement required =101 ×1.52 x 1.5 = 2.28m3 x 501440=65.66Quantity of sand required = 103 x 1.52x 15 = 6.84m3Quantity of CA required = 106×1.52x15= 13.68m3Bags58Example 2:� Calculate the quantity of materials for the following items.a) C.M. (1:4) for 1m3 of workb) CM (1:6) for 1m3 of workHint: Cement will go to fill up the volds in sand. So total volume was be 4 instead of 1+4=5a) Quantity of Cement required = 41×1 = 0.25m3=0.25x501440=7.2 bagsQuantity of Sand required = 44×1=1m3b) Quantity of cement required =61×1=0.16m3=0.16 x 50

1440=4.8bagsQuantity of sand required =66×1 =1m3Example 3:�Calculate the Quantity of Cement required in bags for the follow�ing items.a) B.M. in CM(1:3) for 15 cum of work using 0.2m3 of CM required for1m3 of Brick workb) RCC (1:2:4) for 20m3 of workSol : a) 1m3 of Brick work � 0.2m3 of CM(1:3)15 m3 of Brick work = 15×0.2=3m3Quantity of cement required in bags = 31×3 × 501440=28.8bagsb) Quantity of Cement required in bags=71x 1.52×20×501440=125 bagsExample 4:�Calculate the quantity of Cement required in bags for the followingitems ofwork.a) C.C. (1:4:8) usy 40mm HBG metals for 30m3of workb) RR masanry in CM(1:5) very 0.34m3 of CM for 1m3 of masanry for20m of worksol : a) Quantity of Cement required = 131×1.52×30×501440=101bags

b) 1m3 of RR masanry = 0.34m3 of CM (1:5)20 m3 of RR masanry required = ? 20x 0.34=6.8m3Quantity of cement required =51×6.8×501440=39.2bagsAnalysis of Rates59 Estimation and CostingExample 5:�

Prepare the lead statement for

the following materialsS.No.MaterialRate at SourceLead in KM

Conveyance Charge per kmMTCTST1.2.3.40mm HBG MetalRiver Sand

CementRs.120/m3Rs.15/m3Rs. 135/bags�

�

�

3 25 2�

�

�

7 6 4Rs.5.00/m3Rs.3.5

0/m3Rs. 4.00 per 4km/bag1.2.3.S.No.Mateial40mm HBG MetalRive

r SandCementRs.120/m3Rs.15/m3Rs. 135/bagsRate ofSource�

�

3 25 2�

�

�

Lead in KMMTCTST7 6 4Equalantlead in kmConveyanceChargeTotal co

nvey�

ance ChargeTotal cost5×1.1+7×1.4=15.33x1+2x1.1+6x1.4=13

.62x1+4x1.4=7.65.00/m33.50/m34.00per4km/bag15.3x5=76.513.6x3

.5=47.60.46.7x 4.0=7.6120+76.5=196.5/m315+47.6=62.6/m3135+7.

6=142.6/bagCost of cement at site = 142.6/bag1 bag of cement

= 50kgsp.wt of cement = 1440 kg/m3 = 1.44t/m3Cost of Cement

= 142.6x501440 = 4106.88/m360Rs.2100/10KNRs.850/1000nosRs. 1

5m3Rs. 250/m31.2.3.4.S.No.MaterialCementBricksSand40mmHBGMet

alRate ofSourceEqualantlead in kmConve�

yanceChargeRs.Totalco

nveyanceChargeRs.Total costRs.5x1.4+2x1.1+3x1=11.25x1.4+3x1=

101x1.4+2x1.1+2x1=5.63x1.4+2x1.1+2x1=8.41.50309.00/m36.5/m31

6.80300.0050.4054.62116.8/10KN1198/1000nos107.4/m3354.6/m35 5 1 32

�

�

2 2Lead in KMSTCTMT3 3 2 2SeinerageChargeRs.�

�

353035CessChargeRs.�

�

131215Example 6:�

Prepare the lead statement for the follow

ing materialsS.No.MaterialRate of SourceLead in KMConveyance

Chargeper kmSTCTMT1.2.3.4.CementBricksSand40mm HBG Metal5 5 4 32�

�

2 23 3 5 2Rs.1.5/m3Rs.30/1000Nos/KmRs.9.00 / km/cumRs.6.50/Km/m3Rs.

2100/10 KN (tonn)Rs.850/100nosRs. 15/m3Rs. 250/m3SeinarageCh

argesCessCharges�

�

�

353035�

�

�

131215Analysis of Rates61 Estimation and CostingExample 7:� Prepare a data sheet &calculate the cost of the following itemsof works:a) Plastering with cement mortar (1:4), 20 mm thick unit�10m20.21m3 C.M. (1:4)0.66 Nos. Brick layer I class1.54 Nos. Brick layer II Class0.5 No.s Men Mazdoors3.2 Nos. Women mazdoorsL.S. Sundries.b) R.R. Masonry in C.M. (1:6) �1m31.1 m3Rough stones0.34 m3C.M. (1:6)0.54 No.s Mason I Class1.26 Nos. Mason II Class1.40 Nos. Men mazdoors1.40 Nos. Women mazdoors

LS. Sundries.Lead Statement of materials:Labour Charges :1. Mason / Brick layer I Class Rs.100=00 per day.2. Mason /Brick layer II class Rs. 80=00 per day3. Men mazdoor Rs. 60=00 per day4. Women mazdoor Rs. 60=00 per day5. Mixing charges of cement mortar Rs. 16=00perm3S.No.MaterialsCost atSourceRs. �� Ps.PerConveyanceChargesper kmRough stoneSandCementLead inKm123260=0012=002100=00m3m310knor1tonne1825Local5=00/m34=00/m3�

62Lead Statement :PerS.No. Material Cost atSourceLead inKMConveyanceChargeRs.TotalconvenyanceCharge Rs.TotalcostRs.12

3Rough StonesandCement260.0012.002100m3m310KNor1tonne1825Local500/m34.00/m3���

90.00100.00���

350.00112.002100/tonnea) Plaster with CM (1:4), 20mm thick, unit�10m2Cost of CM (1:4) for 0.21m3cost of Cement = |.|

\| × 44 . 1 x 21 . 041×2100= 158.76Cost of Sand = |.|

\| × 21 . 044 ×112 = 23.52Total Cost Rs. 182.28Unit182.2866.00123.2030.00

192.0028.163.36625.00Total Rs.S.No. Description Quantity Rate per Amount1234567.CM(1:4)Brick layer I classBrick layer II ClassMen mazdoorsWomen mazdoorsMixing ChargesSundrys0.210.661.540.53.20.21L.S.m3NosNosNosNosm3182.28100806060160.21m3daydaydaydaym3b) RR Masanry in CM (1:6) �1m3Cost of CM (1:6) for 0.34m3Cost of Cement = |.|

\|

× 44 . 1 x 34 . 061 ×2100 = 171.36Cost of Sand = |.|

\| × 34 . 066 ×112 = 38.08Total Cost Rs. 209.44Analysis of Rates63 Estimation and Costing385.00209.4454.0010.0884.0084.005.4418.04850.00/m3Total Rs.Example 8:�Prepare a data sheet and calculate the cost of the items givenbelow:a) Brick masonry in C.M. (1:6) with country bricks�unit Icum.600Nos. country bricks.0.38m3 C.M.(1:6)1.40Nos. Mason0.7 Nos. Man Mazdoor2.1 Nos. Woman MazdoorL.S. Sundries.b) C.C.(1:5:10) using 40mm HBG metal unit 1cum.0.92m3....... 40mm size HBG metal0.46m3....... Sand0.092m3..... Cement0.2 Nos ...... Mason1.8 Nos ...... Man Mazdoor1.4 Nos. ...... Woman MazdoorL.S. ............ Sundries.Lead Statement of materials:S.No. Description Quantity Rate per Amount1234

5678Rough StoneCM(1:6)Mason IClassMason II ClassMen MazdoorsWomen MazdoorsMixing ChargesSundries1.10.340.541.261.401.400.34L.S.m3m3NosNosNosNosm3350209.44100.008.0060.0060.0016.00m30.34m3daydaydaydaym3Quantity UnitS.No. Material Cost at SourceRs. Ps.Per Lead inKmConveyanceCharges per Km123440mmHBG metalSand

Bricks countryCement210=0016=00780=002600=00m3m31000Nos10KNor1tonne1618at siteat siteRs.6=00/m3Rs.3=00/m3��

��

64Labour charges:i) Mason� Rs. 90 per day.ii) Man Mazdoor � Rs. 70 per dayiii) Woman Mazdoor � Rs. 70 per day.iv) Mixing Charges of C.M. Rs. 20=00 per m3.Lead Statement :Sl.No.Material Cost atSourceLead inKMConveyanceChargeRs.TotalconveyanceCharge Rs.TotalcostRs.123440mm HBG metalsandCountry bricksCement210.0016.00780.002600

m3m31000nos10knor1tonne1618at SiteAt siteRs.6/m3Rs.3/m3���

���

96.0054.00���

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306.0070.00780.002600/tPera) B.M. CM(1:6) with country bricks � 1m3CM (1:6) � 0.38m3Cost of Cement = |.|

\| × × 44 . 1 38 . 061 ×2600= 237.12Cost of Sand = |.|

\| × 38 . 066 × 70 = 26.60Total Cost Rs. 263.72769.23263.72126.00147.007.6086.44

1400.00 Total Rs.S.No.Description Quantity Rate perAmountRs.123456Country BricksCM (1:6)MasonMan mazdoorsMixing ChargesSundries6000.381.42.10.38L.SNosm3NosNosm3 780263.72907020 10000.38m3daydaym3UnitAnalysis of Rates65 Estimation and CostingS.No. Description Quantity Rate per Amount1234567840mm HBG metalSandCementMasonMan mazdoorwomen Mazdoor

Mixing chargesSun dries0.920.460.0920.21.801.41.0L.Sm3m3m3NosNosNosm330670260090707020m3m3tNosNosNosm3Unitb) CC (1:5:10) using 40mm HBG metal �1m3281.5232.20344.4518.00126.0098.0020.004.83925.00 /m3Total Rs.66Lead St

atementTown data for Guntur Town Buildings as per 2004�

05�

"S

.S.R.Sl.No.Description ofMaterialSource ofsupplyUnitIntial c

ostof materialDeductStackingchargesConvey�

anceChargeBlasting

ChargesSeinorageChargesCrushingchargesTotalCostCementSand fo

r MortarSand for FillingGravel40mm HBG metal20mm HBG Metal12

mm HBG metal10mm HBG metal6mm HBG metalR.T.SteelBricksR.R.St

one formasonry worksLead inKmLocalKrishna RiverT.lapalem Vag

uPerecherlaLamLamlamLamLamLocalKollurPerecherla1 2 3 4 5 6 7 8 9 101112MtCumCu

mCumCumCumCumCumCumMt1000Cums0.0034KM28km12km11km11km11km11k

m11km�

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48km11km2700.0071.0028.0038.50269.00469.00375.00313.

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275.001668.80317.00Analysis of Rates67 Estimation and CostingPreparation of Unit rates for finished items of wordsI a) Cement Concrete in foundation (1:5:10)S.No. Description of Item Quantity Unit Rate Per Amount40mm HBG MetalSandCementMason Ist ClassMason 2nd ClassMan mazdoorWomen MazdoorAdd Extra 15%on M.LAdd T.O.T. @4%Sundries0.920.460.0920.060.141.801.40CumcumCumNoNoNoNo547.75284.802700.00150.00131.00101.00101.00CumCumMTNosNos

NosNos503.93131.00357.709.0018.34181.80141.4052.581395.7555.830.421452.001.2.3.4.5.6.7.8.910b). Cement Concrete in foundation (1:4:8)S.No. Description of Item Quantity Unit Rate Per Amount40mm HBG MetalSandCementMason Ist ClassMason 2nd ClassMan mazdoorWomen MazdoorAdd Extra 15%on M.LAdd T.O.T. @4%Sundries0.920.460.1150.060.141.801.40CumCumCumNoNoNoNo547.75284.802700.00150.00131.00101.00101.00CumCumMT

NosNosNosNos503.93131.00447.129.0018.34181.80141.4052.581485.1759.400.431545.001.2.3.4.5.6.7.8.9102) R.C.C.WorksV.R.C.C.(1:2:4) Nominal mix using 20mm Normal size hard brokengranite metal approved quarry with necessary reinforcement including casting,curing cost & conveyance of all materials. Total Rs.Total Rs.68S.No. Description of Item Quantity Unit Rate Per AmountV.P.C.C (1:2:4)Centering ChargesSteel @0.5% = 0.5/100=0.005m3(0.005x7.85t/m3 =0.04tAdd T.O.T. @4%Sundries1.001.000.04CumCumMT2372.40430.0034580.00CumCumMT2372.40430.001383.204185.60

167.400.004353.001.2.3.4.c) V.R.C.C (1:2:4) for bed blocks, column footings including form workcentering chargesS.No. Description of Item Quantity Unit Rate Per Amount20mm HBG MetalSandCementMason Ist ClassMan mazdoorWomen MazdoorVibrating chargesMachiny mixing concreteAdd Extra 15%on M.L0.920.460.230.21.81.41.01.0CumcumCumNoNoNoCumCum797.75284.802700.00180.00131.00101.00101.0028.80CumCumMTNosNosNosNoscum733.93131.00894.2430.00235.80.141.40101.0028.8076.23

2372.401.2.3.4.5.6.7.8.92 a) P.C.C.(1:2:4)S.No. Description of Item Quantity Unit Rate Per Amountcost of steelFabrication chargesAdd 15% on M.L.Add T.O.T. @4%Sundries1.001.00MTMT275005.00MTKg27500.005000.00750.0033250.001330.000.0034580.001.2.3.4.5.b) For steel reinforcementTotal Rs.Total Rs.Total Rs.Analysis of Rates69 Estimation and CostingS.No. Descrtiption of Item Quantity Unit Rate Per AmountV.P.C.C. (1:2:4)Centering ChargesSteel for columns, beams@1.5% =1.5/100x7.85=0.117tAdd T.O.T. @4%Sundries1.001.000.117CumCumMT2372.40675.0034580.00

CumCumMT2372.40675.004072.007119.40284.770.837405.001.2.3.4.5.d) V.R.C.C (1:2:4) for columns rectangular beams, pedastals includingform work at centering charges.S.No. Descrtiption of Item Quantity Unit Rate Per AmountV.P.C.C (1:2:4)Centering ChargesSteel for slabs@1% =1/100 x 7.85 =0.0785 tAdd T.O.T. @4%Sundries1.0010.000.0785CumCumMT2372.40710.0034580.00CumCumMT2372.40710.002714.535796.63231.871.206030.001.2.3.e) V.R.C.C (1:2:4) for slabs, lintels including form work at centeringcharges upto 100mm, thick3. Pointing to R.R.Masonary in CM(1:4) mix using cost & conveyanceof Cement, sand and all materials from approved sources to siteand labour charges for point neatly etc.Total Rs.Total Rs.70S.No. Description of Item Quantity Unit Rate Per AmountCost of CM(1:4)Cement =09 . 0 x 44 . 14

1×Sand = 09 . 041×Mining Chargesmason Ist Class2nd ClassMan mazdoorWomen MazdoorAdd 15% on MLAdd TOT @ 4%Sundries0.090.0320.091.00.481.120.501.10CumtCumCumNos.NosNosNos2700.00284.8032.50150.00131.00101.00101.00MtCumCumNosNosNosNos87.4825.6332.5072.00146.7255.00111.1057.72588.1523.530.32612.001.2.3.

4.5.6.7.8.9.10.4. Cement concrete flooring (1:2:4) using 12mm HBG machine crushedchips from approved quarry to site of work including curing cost andconveyance of all materials completed.S.No. Description of Item Quantity Unit Rate Per Amount1.2.3.4.5.6.7.8.9.1011.12mm HBG metalcrushed chipsSandCement(0.23m3x1.44=0.33tMason ISt class 2nd ClassMan mazdoorWomen MazdoorAdd 15% Extra on MLAdd TOT @4%Sundries0.920.460.23(or)0.3310.060.141.801.40CumcumcumMTNosnosnosnos680.25284.802700150.00131.00101.00101.00cum

cummtnosnosnosnos625.83131.00894.249.0018.34181.80141.4052.582054.1982.170.642137.00Total Rs. Total Rs.Analysis of Rates71 Estimation and Costing5 a) Supply and fixing teak wood fully panneled with 10x 4 cm styles, and10x4cm rails and 3.5CM TH panels with teak wood fram of 6.25x 10cm sizeincluding cost of hold fasts, but hinges and labour charges for fixing door inposition and fixing furniture etc., complete for one door of size 1.100 x 2.00 ofarea 2.2 sqm.Requirements :i) Verticals = 2x 2.0 x 0.10 x 0.0625 = 0.0250ii) Horizontals = 1x 1.10 x 0.10 x 0.0625 = 0.0068iii) Styles = 4x 1.937 x 0.10 x 0.04 = 0.0300iv) Rails = 2x 5x 0.5075 x 0.10 x 0.04 = 0.0020v) Planks = 2x 4x 0.364 x 0.3475 x .035= 0.03540.0090m3S.No. Description of Item Quantity Unit Rate Per Amount1.2.3.4.wood CostButt HingesZ�hold fastsCost of labour0.009662.2CumNosNossqm250002010800cumeacheach

sqmTotal2470.00120.0060.001760.004410.00Cost of door per 1m2 = 4410/ 2.2 = 2004.54 say Rs.2010/�725 b) Supply and fixing teak wood fully panneled with 10x 4 cm styles, and10x4cm rails and 3.5CM TH panels with teak wood fram of 6.25x 10cm sizeincluding cost of hold fasts, but hinges and labour charges for fixing window inposition and fixing furniture etc., complete for one window of size 1.0x1.2 ofarea 1.2 sqm.Requirements :i) Verticals = 3x1.2 x 0.10 x 0.0625 = 0.0225ii) Horizontals = 3x 1.00 x 0.10 x 0.0625 = 0.0188iii) Styles = 4x 2 x 0.10 x 0.04 = 0.0160iv) Rails = 4x 2x 0.4062 x 0.10 x 0.04 = 0.0012v) Planks = 4x 0.3102x 0.2102 x0.03 = 0.00700.0076m3S.No. Description of Item Quantity Unit Rate Per Amount1.2.3.4.wood CostButt HingesZ�hold fastsCost of labour0.0076641.2CumNosNossqm2500020101000cumeacheachsqmTotal1900.00120.0040.001200.003260.00Cost of door per 1m2 = 3260/ 1.2 = 2716.67 say Rs.2720/�Analysis of Rates73 Estimation and Costing

EXERCISEShort Answer Questions1. Calculate the Cement contents for the followinga) C.C.(1:510) using 40mm H.B.G.Metal for 25m3 workb) Brick work in CM (1:6) using country Bricks for 15m3 of work if 0.38 m3 of CM(1:6) is required for 1m3 of Brick work.2. Calculate the Rates of following materials by using the lead statementgiven below.No.Material Rate of SourceLead in KMConveyanceCharge per ST CT MT1.2.3.4.CementBricksSand40mm HBGMetal314221313552Rs.2.5/m3Rs.40/1000Nos/KmRs.12.00 / km/cumRs.7.50/Km/m3Rs.2100/10 KN (tonn)Rs.850/100nosRs. 15/m3Rs. 250/m3Essay type Questions1 Prepare a data sheet and calculate the cost of the items given below:a) Brick masonry in C.M. (1:6) with country bricks�unit Icum.600Nos. country bricks.0.38m3

C.M.(1:6)1.40Nos. Masons0.7 Nos. Man Mazdoor2.1 Nos. Woman MazdoorL.S. Sundries.b) C.C.(1:5:10) using 40mm HBG metal unit 1cum.0.92m3....... 40mm size HBG metal0.46m3....... Sand0.092m3..... Cement0.2 Nos ...... Mason1.8 Nos ...... Man Mazdoor1.4 Nos. ...... Woman MazdoorL.S. ............ Sundries.Lead Statement of materials:S.No. MaterialCost at SourceRs. Ps.PerLead inKmConveyanceCharges per Km123440mmHBG metalSandBricks countryCement210=0016=00780=002600=00m3m31000Nos10KN or1tonne1618at siteat siteRs.6=00/m3Rs.3=00/m3��

��

Labour charges:i) Mason� Rs. 90 per day.ii)Man Mazdoor � Rs. 70 per day

iii)Woman Mazdoor � Rs. 70 per day.iv)Mixing Charges of C.M. Rs. 20=00 per m3.74ESTIMA ESTIMA ESTIMA ESTIMA ESTIMATION OF Q TION OF Q TION OF Q TION OF Q TION OF QU UU UUANTITIES ANTITIES ANTITIES ANTITIES ANTITIESOF STEEL & R.C.C. ELEMENTS OF STEEL & R.C.C. ELEMENTS OF STEEL & R.C.C. ELEMENTS OF STEEL & R.C.C. ELEMENTS OF STEEL & R.C.C. ELEMENTSChapter6Example 1: Prepare the bar bending schedule of the given figure for R.C.C.beam.75 Estimation and CostingName.ShapeDia.No.Length in mTotal Lengthin mS

elf weight in kg /mTotal Weight in KgB E AM 4000+2x230�

2x32=439

6main bars16 4000+2x230�

2x32=4396Anchor bars1224396+2x(9x12)

= 4612mm= 4.612m4.612 x 2= 9.224m786010001242×| . | \ |×π= 0.890.89

x 9.224= 8.224396+2x(9x16)= 4684mm= 4.684m4.684 x 2= 9.368m7

86010001642×| . | \ |×π= 1.581.58x 9.368= 14.8Cranked bars798280079

862500-2x25-16= 4344396+2x(9x16)+2(0.414x434)= 5043mm= 5.043

mAdditional length foreach crank = 0.414d5.04 x 2= 10.087860

10001642×| . | \ |×π= 1.581.58x 10.08= 15.92450180Height = 500-2x25

=450Width =230-2x25 =18016172(450+180) +2x9x6= 1368mm= 1.368

m1.368x17= 23.25678601000642×| . | \ |×π= 0.220.22x23.256= 5.16No.

of stiru�

s = ((798/210)+1)x2+(2800/400) = 17 Nos76Exam�

le 2:

Pre�

are the bar bending schedule of the given figure for R.

C.C. LintelEstimation of Quantities of Steel of R.C.C. Elements77 Estimation and CostingName.Sha�

eDia.No.Length in mTotal Lengthin mSel

f weight in kg /mTotal Weight in KgLINT E L 1500+2x230-2x25=191

0main bars12 1500+2x230-2x25=1910Anchor bars1022.09 x 2= 4.1

8m786010001042×| . | \ |×π= 0.620.62x 4.18= 2.5921910+2x(9x12)= 212

6mm= 2.1264m2.126 x 2= 4.252m0.89x 4.252= 3.78Cranked bars45

5100045562120-2x25-12= 581910+2x(9x12)+2(0.414x58)= 2174mm=

2.174mAdditional length foreach crank = 0.414d2.174 x 2= 4.3

48786010001242×| . | \ |×π= 0.890.89x 4.348= 1.8770180Height = 120-

2x25=70Width =230-2x25 =18012142(70+180) +2x9x6= 608mm= 0.60

8m0.608x14= 8.51278601000642×| . | \ |×π= 0.220.22x8.512= 1.87No. o

f stiru�

s = ((1910/150)+1) = 14 Nos786010001242×| . | \ |×π= 0.8919

10+2x(9x10)= 2090mm= 2.090m78300230Internal room dimension = 4000

x 2000100Exam�

le 3: Pre�

are the bar bending schedule of the g

iven figure for R.C.C. Lintel40002 0 0 03 0 01 0 0300Estimation of Quantities of Steel of R.C.C. Elements79 Estimation and CostingName.Sha�

eDia.No.Le

ngth in mTotal Lengthin mSelf weight in kg /mTotal Weight i

n KgS L A BCranked bars611704410+=27100-2x13-8= 662410+2x(9x8)+(0.

414x66)= 2581.3mm= 2.581mAdditional length foreach crank = 0

.414d2.581 x 27= 69.778601000842×| . | \ |×π= 0.390.39x 69.7= 27.53

84.41m4.41x15= 66.1578601000642×| . | \ |×π= 0.220.22x66.15= 14.553

2000+2x230-2x25=2410 4000+2x230-2x25=441011802410+=1580EXERCISE1) Pre�are the Bar bending schedule for the beam shown below.2) Pre�are the Bar bending schedule of a sim�ly su��orted R.C.C. Lintels fromthe following s�ecification:Size of lintel 300mm widex 200mm de�th.Main bars in tension zone of Fe

250(grade I) 3 bars of 16mm dia., one bar is cranked through 450 at 170mm from each end2 No. anchor bars at to� 8mm dia.Two legged stirru�s@150mm c/c of 6mm dia. through out.Clear s�an of the lintel is 1150mm.Bearing on either side is 150mm.Estimation of Quantities of Steel of R.C.C. Elements81 Estimation and CostingEARTH WORK EARTH WORK EARTH WORK EARTH WORK EARTH WORKCAL CAL CAL CAL CALCULA CULA CULA CULA CULATIONS TIONS TIONS TIONS TIONSCha�ter77.1 Introduction:-Generally all the Civil Engineering �rojects like roads, railways, earth dams,canal bunds, buildings etc. involves the earth work.This earth work may beeither earth excavation or earth filling or Some times both will get according tothe desired sha�e and level. Basically the volume of earthwork is com�utedfrom length, breadth, and de�th of excavation or filling.In this cha�ter the various methods of calculating the earth work quantitiesshall be discussed.7.2 Lead and Lift:Lead:It is the average horizontal distance between the centre of excavation to thecentre of de�osition. The unit of lead is 50m.Lift :It is the average height through which the earth has to be lifted from sourceto the �lace of s�reading or hea�ing. The unit of lift is 2.00m for first lift and oneextra lift for every 1.0m. for exam�le when earth is to be lifted for 4.5m, Fourlifts are to be �aid to the contractor.i.e. U�to2.0- 1 lift1.0 - 1 Lift1.0 - 1 lift Total 04 lifts0.5 - 1 lift7.3 Calculation of earth work for Roads:7.3.1 case 1) volume of earth work in banking or in cutting having "no longitudi-nal slo�e".}82Volume = Crosectional area x lengthV = (bd+2x1/2x ndx d)LV = (bd+nd2)LCase 2:When the ground is in longitudinal slo�e or the formation has uniform gradi-ent for a length the earth work may be calculated by the following methods.1. By Mid Section or Mid ordinate method.Where d1, d2 = de�th of banks at two endsEarth work Calculations83 Estimation and CostingMid ordinate (or) Average de�th (d

m) = 2d d2 1 +Area of mid section (Am) = ) nd bd (2m m +volume of earth work (v) = Am x L = L ) nd bd (2m m × +ii) Tre�ezoidal formula: (for two sections)In this method also called mean sectional area methodLet A1 &A2 be two areas at two ends.A= ) nd bd (21 1 + , A2 = ) nd bd (22 2 +Am = 2A A2 1 +Volume of earth work (v) = Am ×Liii) Tre�ezoidal formula for a series of c/s areas at equal intervals.Let A1,A2,A3.......An are the cross sectional areas along L.S of Road 'L" isthe distance between two cross sectionsThe volume of earth workV=

+ + + + |.

|

\| +−) A ..... A A (2A AL1 n 3 2n 1 (or)= | | ) A ..... A A ( 2 ) A A (2L1 n 3 2 n 1 −+ + + + += 2length[ (sum of first and last areas ) + 2(remaing Areas) ]iv) Prismoidal formula for a series of cross sectional areas at equal intervals.Note : This method is adopted when there is odd number of cross sections.Volume of earth workV= | | ) A ...... A A ( 2 ) A ..... A A A ( 4 ) A A (3L2 n 5 3 1 n 6 4 2 n 1 − − + + + + + + + + + + = 3length(Sum of first and last areas)+4(even areas)+2(odd Areas)]84Example 7.1 : Find the volume of earth work in embankment of length 12m.Top width is 5.5m and depth is 2.5m the side slopes ara 1½:1Sol : Top width b=5.5mDepth d= 2.5mside slopes =1½:1 i.e. n=1.5length L=12mVolume of earth work V = (bd+nd2)L= (5.5 ×2.5+1.5×2.52)12= 77.5m3Example 7.2 : The depths at two ends of an embankment of road of length70m are 2m and 2.5m. The formation width and side slopes are 8m and 2:1respectively. Estimate the Quantity of earth work bya) Mid Sectional Area (ii)Mean sectional Area method.Sol: a) b=8m, d1=2m, d2=2.5m, l=70m, n=2Mean depth dm = 2d d

2 1+= 25 . 2 2+=2.25mMid sectional Area = Am = bdm+ndm2 = (8x2.25+2x2.252)2=28.125m2Volume of earth work (V)= AmxL = 28.125x70=1968.75m3.b) Area of c/s at one end A1 = bd1 +nd12 = 8x2+2x22=24m2Area of C/s at other end A2=bd2+nd22 =8×2.5+2×2.52 =32.5m2Mean Sectional Area (Am) =2A A2 1 +=25 . 32 24+=28.25m2Volume of earth work (V)= AmxL=28.25x70=1977.5m3.Example 7.3The following width of road embank ment is 10m. The side slopes are 2:1The depth along the centre line road at 50m intervals are 1.25, 1.10, 1.50, 1.20,1.0,1.10, 1.15m calculate the Quantity of earth work bya) Mid sectional ruleb) Trepezoidal rulec) Prismoidal rulea) Mid Sectional rule : b=10m, n=2.5.51½:12.5

Earth work Calculations85 Estimation and CostingChainage DepthsMeandepth (dm)Area of(bdm+ndm2)Quantity (m3)Am×L0501001502002503001.251.101.151.201.001.101.151.1751.1251.1751.101.021.12514.5113.7814.5113.412.7013.78505050505050725.56689.06725.56671.00635.25689.064135.49m3

TotalLength b/wChainagesb) Trepezoidal ruleA= bd +nd2A1 = bd1+nd12= 10x 1.25+2x 1.252= 15.625 m2A2 = bd2+nd22= 10x 1.10+2x 1.102=13.42m2A3 = 10x 1.15+2.1.152= 14.145m2A4 = 10x 1.2+2x1.22=14.88m2A5=10x 1.0+2x12=12.0m2,A6 = 10 x 1.1 +2x1.12 = 13.42m2A7 = 10x1.15+2x1.152= 14.145 m2Volume of earth work by Trepezoidal rulev = L

+ + + |.

|

\| +−) A .... A A (2A A1 n 3 2n 1 = 50

+ + + + + |.|

\| +) 42 . 13 0 . 12 818 . 14 145 . 14 42 . 13 (2145 . 14 625 . 15 = 4137.50 m3}86c) By Prismoidal rule v = | | Areas) 2(Odd Areas) 4(even ) A (A3Ln 1 + + + = | | ) A 2(A ) A A 4(A ) A (A3L5 3 6 4 2 7 1 + + + + + + = | | ) 2 1 2(14.145 ) 42 . 3 1 88 . 4 1 4(13.42 ) 145 . 4 1 (15.625350+ + + + + + = 4149 m3Example 7.4:� Estimate the Quantity of earth work for a portion of road fromthe following data Chainage 0 1 2 3 4 5 6 7 8 9RL 7.50 7.70 7.50 7.25 6.85 6.95 6.70 6.45 6.30 5.95The formation level at Chainage 0 is 8.0 and having falling gradient of 1 in100. The top width is 12m and side slopes 1½ horizontal to 1 vertical assumingthe transverse direction is in level calculate the quantity of earth work Take1 chain = 20m by using trepezoidol & Prismoidol formula. C/S.OF ROAD LEVEL's7.5 7.7

7.2 7.256.85 6.956.76.45 6.35.9587.8 7.67.47.27 6.86.66.46.25.05.56.06.57.07.58.08.50 20 40 60 80 100 120 140 160 180 200DistanceF.L&R.L'sG.L'sF.L'sEarth work Calculations87 Estimation and CostingSol : �b=12m n=5Trepezoidal formula :V=

+ + + + |.|

\| +−) A .... A A (2A A

L1 n 3 2n 1 =

+ + + + + + + + |.|

\| +215 . 1 837 . 1 215 . 1 63 . 0 38 . 4 837 . 1 215 . 1 215 . 1 (209 . 3 375 . 620= 365.53m3Prismoidal formula :V= | | areas) ( 2 ) ( 4 ) (31 Odd areas even A ALn + + + = | | ) A A A A ( 2 ) A A A A ( 4 ) A A (3L9 7 5 3 8 6 4 2 10 1 + + + + + + + + += 320 [ (6.375+3.09+4( 1.215+1.837+0.63+1.837) +2(1.215+4.38+1.815+1.215)]= 317.27 m3ChainageDistanceReducedlevelFormationLevelEmbank�mentCuttingArea of01234567

890204060801001201401601807.507.707.507.256.856.956.706.456.305.958.07.87.67.47.27.06.86.66.46.20.500.100.100.150.350.050.100150.100.256.3751.2751.2151.8394.380.631.2151.8371.2153.09Depth(d) ofEmbank�mentCuttingbd+nd288Example 7.5:� The road has the following data

Chainage 0 20 40 60 80 100 120 RL of 20.6 21.0 21.5 22.1 22.7 22.9 23.0 GroundThe formation level at chainage zero is 22.0 and having a rising gradient of1 in 100 the top width is 12.0m and side slopes are1½ :1 Assuming the trans�verse direction is in level. calculate the quantity of earth work bya) Trepezoidal formula b) Prismoldal formulaChainageDistanceReducedlevelFormationLevelEmbark�mentCut�tingArea of020406080100120Embark�mentCuttingDepth (d)of20.621.021.522.122.722.923.022.022.222.422.622.823.023.21.401.200.900.500.100.100.2019.7416.5612.016.3751.2151.2152.460 C/S.OF ROAD LEVEL's20.6 21

21.5 22.1 22.7 22.9 2322 22.2 22.4 22.6 22.8 23 23.25.010.015.020.025.00 20 40 60 80 100 120 140DistanceF.L&R.L'sSeries1Series2Earth work Calculations89 Estimation and Costinga) Trepezoidal formula:Vol of earth work in embankmentV= L

+ + + + |.|

\| +−) A ........ A A (2A A1 n 3 2n 1 =

+ + + + + |.|

\| +) 215 . 1 215 . 1 375 . 6 01 . 12 56 . 16 (

246 . 2 74 . 1920= 969.5 m3b) Prismoidal formulaV = 3L[( A1+An)+4(even Areas ) +2(Odd Areas) ]= 320[ (19.74+2.46)+4(16.56+6.325+1.2+5)+2(12.01+1.215)]= 968.33m390Example 7.6:�From the above problem if the formation level at 0th chainage in20m. Calculate the volume of earth work by using the formulas?Chainage020406080100120ReducedlevelFormationLevelEmbank�mentCuttingArea ofEmbank�mentCuttingDepth (d)of20.6021.0021.5022.1022.7022.9023.0020.0020.2020.4020.6020.8021.0021.20��

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7.74010.5615.01521.37528.21528.21526.4600.600.801.101.501.901.901.80��

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bd+nd2 C/S.OF ROAD LEVEL's20.6 21 21.5 22.1 22.7 22.9 2320 20.2 20.4 20.6 20.8 21 21.25.010.015.020.025.00 20 40 60 80 100 120 140DistanceF.L&R.L'sG.L'sF.L'SEarth work Calculations91 Estimation and Costinga) Trepezoidal formula:Vol.of earth work in cuttingV= L

+ + + + |.|

\| +−) A ........ A A (2A A1 n 3 2n 1 =

+ + + + + |.|

\| +) 215 . 28 215 . 28 375 . 21 015 . 15 56 . 10 (246 . 26 74 . 720= 2409.6 m3b) Prismoidal formulae :V= | | areas) ( 2 ) ( 4 ) (31 Odd areas even A ALn + + + = | | ) A A ( 2 ) A A A ( 4 ) A A (3L5 3 6 4 2 7 1 + + + + + += 320 [ (7.74+26.46) + 4( 10.56+21.375+28.215) +2(15.015+28.215)]= 2408.4 m3Example 7.7:�From the same above problem 7.6 if the gradient is in 100falling calculate the quantity of earth work by using the formulas

Chainage020406080100120ReducedlevelFormationLevelEmbank�mentCut�tingArea ofEmbank�mentCuttingDepth (d)of20.6021.0021.5022.1022.7022.9023.0020.0019.819.619.419.2019.018.80��

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7.7416.5628.21543.33560.37569.61576.860.601.201.902.703.503.904.20��

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92a) Trepezoidol formulae:Vol.of earth work in cuttingV= L

+ + + + |.|

\| +−) A ........ A A (2A A1 n 3 2n 1 =

+ + + + + |.|

\| +) 615 . 69 375 . 60 335 . 43 215 . 28 56 . 16 (286 . 76 74 . 720= 5208 m3b) Prismoidal formulae :V= | | areas) Odd ( 2 ) areas even ( 4 ) An 1 A (3L+ + + = | | ) A A ( 2 ) A A A ( 4 ) A A (3L5 3 6 4 2 7 1 + + + + + += 320 [ (7.74+76.86) + 4( 16.56+43.335+69.615) +2(28.215+60.375)]= 5198.8 m

3 C/S.OF ROAD LEVEL's20.6 21 21.5 22.1 22.7 22.9 2320 19.8 19.6 19.4 19.2 19 18.85.010.015.020.025.00 20 40 60 80 100 120 140DistanceF.L&R.L'sG.L'sF.L'sEarth work Calculations93 Estimation and CostingChainage(m)R.L. F.L.Depth (d)of .Embank�mentCuttingArea of .EmbankmentCutting020406062.58010012020.621.021.522.122.722.923.022.022.2022.40

22.2022.0021.8021.601.401.200.900.100.00 0.000.701.101.400.0009.13515.01519.74Example 7.8:� From the problem 7.5 if the gradient is 1 in 100 raising upto40th chainage and 1 in 100 falling ragient from 40th Chainage to 120th chainage.Calculate the vol of earth work by using the formulas.bd+nd2 bd+nd219.7416.5612.011.2150.000From similer triangel properties7 . 0x 200.1x −=0.7x = (20�x)0.10.7x = 2�0.1x0.7x+0.1x = 20.8x = 2x=0.82 = 820 = 2.5C/S. of ROAD LEVELS20.6 21 21.522.122.7 22.9 2322 22.2 22.422.2 2221.8 21.6

2020.52121.52222.52323.50 20 40 60 80 100 120 140ChainageR.L&F.L'sG.L'sF.L's0.160 x20�x800.72094Chainage 0 20 40 60 62.5Area 19.74 16.56 12.01 1.215 0.00vol of earth work in embankmenthere the intervals are not equal so we have to take the seperate volumes fromoth chainage to 60th chainage and 60th chainage to 62.5 chainageV = 62.5) � vol(60 60) � (0 Vol +=

++

+ + |.|

\| +20.00 1.2152.5 ) 01 . 12 56 . 16 (2215 . 1 74 . 1920

= 782.46m3By Prismoidal V = | | | | ) 00 . 0 215 . 1 (35 . 201 . 12 2 56 . 16 4 ) 215 . 1 74 . 19 (320+ + × + × + += 742.44 m3Vol of earth work in cuttingChainage 62.5 80 100 120Area 0.00 9.135 15.015 19.74Volume (v) = vol (62.5�80)+Vol (80�120)By Tripezoidal formulaV =

+ |.|

\| ++

+015 . 15274 . 19 135 . 9202135 . 9 05 . 17 = 668.98m3By Prismoidalv = | | ( ) | | 015 . 15 4 74 . 19 135 . 9320135 9 . 035 . 17× + + + + = 646.18 m3Earth work Calculations95 Estimation and CostingEXERCISE

Short Answer Questions1. State the following formulae with usual notationa) Prismoidal formulab) Trepezoidal formula2. For an embankment 90m long of uniform gradient when the height of bank is2.4m at one end and 1.8m at the other end the width of embankment at topis 8m and its side slopes 2 vertical to 1 Horizontal calculate the quantity ofearth work by a) Mid Sectional area method b) Mean sectional area method.3. Find the earthwork in embankment between 5/2km to 5/5km of the pro�posed road whose c/s is given below.Essay type questions1. The road has the following dataChainage in m 0 30 60 90 120G.L. in m 25.8 26.5 27.2 28.1 28.5The Formation level at chinage zero is 28 and having the rising gradient of1 in 100 the top width is 10m and the side slopes are 1½ horizontal to 1vertical Assuming transverse slope is level calculate the volume of earthwork.2. The reduced level of ground along the centre line of a proposed road fromchaiage 0 to 6 are given below. The formation level at '0' chainage is 10.00and the road is in down ward gradient of 1in 100 formation width of roadis 10m and side slopes are 2:1 for both banking and cutting. Length ofchain is 20m calculate the quantity of earth work required by a) Trepezoidalrule b) Prismoidal rule.Chainage 0 1 2 3 4 5 6R L of ground 8.0 7.8 7.6 7.3 6.9 6.2 6.53.51.852:196DETAILED ESTIMATESChapter8A) Gravel RoadA gravel road comprising of a gravel of thickness 100mm compactedthickness and compacted by hand roller. A gravel is placed over an earthernformation which is compacted by a 2 tonne roller. The estimate of gravel road consists of determining the folloiwng quantities.i) Earth work excavation and depositing on bank and compactionii) collection of graveliii) spreading compacting gravel to OMCExample 8.1:� Find the estamation of a gravel road for the fig shown below. fora proposed road from 0km to 12km.S.No. Particulars of Items No.L BH QExplanation1a) Earth work excavation and depositing on bark with anintial lead and lift of soil for formation and filing of pits, potholes etc.Area of C/s at O km (A) = 10x1.2+2x1.22 = 14.88m2Area of C/s at 6 Km (A2) = 10x0.8+2x0.82 = 9.28m2

Area of C/s. at 12 km (A3) = 10x 0.6+2x0.62 = 6.72m2Vol of earth work =

+ |.|

\| +28 . 9272 . 6 88 . 14600= 12048m3b) Add extra for pits & pot holes LS = 52m3Total 12100 m3Deduct for gravel = 1 x 1200 x 5 x 0.1 = 600 m3Net Earth work = 12100�600=11,500m310m1.22:110m0.62:110m0.82:1C/s at 0km C/s at 6km C/s at 12km10m1.2 2:1 5m100mm97 Estimation and CostingCement concrete roadC.C. road is laid over an existing W.B.M road, In certain cases. It islaid over a prepared sub grade and a base course is provided. The concreteused for roads is M15 grade using 20mm H.B.G. metal while for base course aconcrete of 1:4:8 using 40mm HBGmetal the stages of Estimations of a C.C.roadisa) Earth work excavation and deposting on the bankb) Cement concrete (1:4:8) for base coursec) Cement concrete (1:2:8) for wearing course.Example 8.2:� Calculation for the estimation of a C.C.road for a length of 100mand width of C.C.road is 3.50m with 100mm thickness of earh layer.S.No. Particulars of Items No.L BH QExplanation1 C.C.(1:4:8) for base courseincluding cost and convey�ance of all materials at sitemachine mixing, laying cur�

ing etc.1 100 3.5 0.1 35. cum2.34.5.6.Collection of gravel in�cluding cost & convey�ance etc complete 50%allowance is given forOMC compaction.Spreading of gravel andwateringUn forcean items @2%Tools and plant @1%P.S. and continsecis @4%11120012005.005.000.15���

900m36000m2L.S.L.SL.S2345C.C.(1:2:4) for pavementProvision for mastic padsUnforcean items @2%Petty supervision @4%1 100 3.5 0.1 35cumL.S.L.S.L.S98Example 8.3 :� Prepare an estimate for 1 Km length of C.C. track or the figshown below.S.No. Particulars of Items No.L BH QExplanation12.C.C.(1:2:4) in tracksincluding layinglaying of kankar(for loose thicknessincrease with 333

1%)a) in between C.Ctracksb) under C.C.tracks2121000100010000.60.90.90.10.1330.20120m3120360480 m3900900600 6001500150100 C.C(1:4:8)KankarDetailed Estimates99 Estimation and CostingExample 8.4:� Calculate the quantities of different items of the figure shownin below SEPTIC TANK10045.Long wall short wallmethodLong wallShortwalls(or)centre line methodtotal centre line length(3400+1400)2=9600R.C.C. (1:2:4) using20mm HBG metala) R.C.C slabb) Baffle wallc) Scum boardPlastering with CM(1:4)with 20mm tha) Inner surface of septictankb) flooringc) Sides of Scum boardd) Top and bottome) sides of baffle wallf) top of baffle wallDeduct for Pipe openings

Total (net) Plastering113.71.19.63.701.401.408.403.11.11.11.01.0 2) 1 . 0 (4 ×π0.30.30.31.700.10.1---1.1- -0.1- -0.11.21.2Total1.20.10.750.75Total1.2- -0.75- -0.75---Total2.6640.7923.4563.4560.6290.1050.1050.83910.083.411.650.22

1.650.10.015717.1012.3.S.No. Particulars of Items No.L BH QEx�lanationEarth work excation u�toG.L.C.C. (1:4:8)bedBrick masonary in CM1:4 for side walls4.04.02.02.01.90.315.2m32.4m33.71.70.32211x21x21x21x123.41.4(3.1+1.1)2=8.4Detailed Estimates101 Estimation and Costinga) Earth filling with excavated soilaround the brick wallcentre line methodTotal Centre line length =(1.85+3.85)2 = 11.4b) over R.C.C. �annels(neglecting the s�ace forventi �i�e footing)su��ly fixing of steel grillsincluding labour for fabrica-tion @ 750N/m3Provision of 100mm dia inletand out let teesProvision of A.C.ventilatingshaft 3m hight duly embed-ded in b/w at bottomProvision for A.C.cowl for

ventilating �i�eUnforcean itsm @2xP.S.& contingencies @4%S.No. Particulars of Items No.L BH QEx�lanation6.789.1011121111x21x11x111.43.700.839---0.151.70x750=- -1.300.30Total629.25N- -1 No1nosL.SL.S2.2231.18874.1162.92Kgs2Nos1 No1 NoL.SL.S4.02.00.153.851.85102Exam�le 8.5:- Calculate the quantities of different items of the figure shownin belowSEPTIK TANKDetailed Estimates103 Estimation and CostingS.No. Particulars of Items No.L B

H QEx�lanation1.2.3.4.11111114.604.610.6010.204.101.201.202.102.100.400.31.600.100.103.10.301.201.200.11.802.10Total29.952.8985.0883.6720.6560.2160.2521.124Earth work excavation u�toG.L.C.C.(1:4:8) bed forfoundationBrick masonary in CM 1:4for side wallsa) U�to first ste� (400th)centre line methodtotal centre line length= (3900+1400)2=10600b) from Ist to II ste� (300th)Centre line methodTotal centre line length(3800+1300)2=10200R.C.C. (1:2:4) using 20mmHBG metala) RCC roof slab

b) Baffle wallc) 8cum ward41001600300380013004300180040039001400Total Brick Masonry = 5.088+3.672 = 8.760(Assure �rojection100mm inside thewall)104S.No. Particulars of Items No.L BH QEx�lanation5.6.111x211x21212.8019.03.51.01.01.01.02) 1 . 0 (4 ×π- -1.0- -0.1---0.11.602.4- -2.1---1.8- -1.6021.63.15

4.20.13.60.10-0.01533.082.3044.96Plastering with CM(1:4)with 20mm thicka) Inner surface of se�tictankb) flooringc) sides of scum boardd) Bottom of scum boarde) sides of baffle wallf) To� of baffle wallg) deduction for Pi�eo�eningEarth filling with excavatedsoil around the brick worka) u�to first ste�Total length = (4.45+1.95)2= 12.8b) from 1st ste� to u� toGround LevelTotal Centre Line length=2(4.35+1.85) = 12.4L =2(3.5+1.0)=9.0Net Plastering =m247502250150445019500.1512.401.21x21x11x10.254600210025043501850Detailed Estimates105 Estimation and CostingS.No. Particulars of Items No.L BH QEx�lanation7891011

1211x21x11x1---- -- -L.S2Nos1No1 NoL.SL.SSu��ly & fixing of steelgrills including labour forfabrication @750 N/m3Provision of 100mm diainlet & outlet TeesProvision of A.C. cowl forventilating shaft 3 mtheight duly embeded belowat bottomProvision of A.C. cowl forventilating �i�eUnforceen items @2xR.S.& Contingeties @4%---------------Exam�le 8.6:- Calculate the quantities of different items of the figure shownin belowSOAK PIT1600600180070050016001001601000106S.No. Particulars of Items No.L BH QEx�lanation1.2.3.4.5.Earth work excavationin non cohesive soilslike sandy soils with anintial lead & lift

a) Soak �itb) side brick wallBrick work in CM (1:5)with country bricksincluding cost andconveyance etccom�lete alround the�itcentre line methodsu��ly & �ackingincluding cost &con-veyancea) Brick batsb) 80mm brick jellyc)40mm brick jellyd) gravel brick jellyR.C.C.(1:2:4) slab�anels (�recast) using20mm HBG metalinlcuidng cost &conveyanceFilling with clay soil onto� of �it u�to G.L.111126 . 1 x4π) 6 . 1 06 . 2 (42 2−π) 6 . 1 06 . 2 (42 2−π ) 83 . 1 ( π 0.237.761.539.291.193.86 1.16Total0.9 230230 160018300.9 1.19126 . 14 ×π2

6 . 14 ×π26 . 14×π0.6 1.2126 . 14 ×π1.8 3.62126 . 14 ×π0.7 1.4126 . 14 ×π0.5 1.007.22 Total1206 . 24 ×π0.1 0.331206 . 24 ×π0.16 0.53Detailed Estimates107 Estimation and CostingS.No. Particulars of Items No.L BH QEx�lanation7.89Laying of joining 100mm�o�ies including earthworkEncavation, sand filling�acking joints etc

com�letsL=12+0.23+1.6/2Unforcean items ofwork @2%Petty su�ervision andcontingencies @4%11113.03-------------13.03LSLSRMEXERCISE1. Calculate the quantities of various elements of the figure shown in below.1082. Pre�are a detailed estimate for following items of work of "SOAKPIT"from the given figurea) 800mm size brick jelly.b) 40mm size brick jelly.c) Gravel,d) Brick masonry in C.M. (1:6):Detailed Estimates109 Estimation and CostingAPPENDEXQuantities of Materials and their Costs:The includes the quantities of various materials for unit quantity of anitem followed by the s�ecification and costs of various materilas. the cost in-cludes first cost, freight, trans�ortation and insurance charges.Labour and Cost:This includes the number and wages of different categeries of labourers.Skilled, unskilled etc.,Cost of Equi�ment:For big �rojects it is necessary to use s�ecial ty�e of tools and �lantslike s�ecial ty�e of mixed concrete trans�ort vehicle called tri�ing wagons, cranesetc. in order to �urchase such tools and �lants and amount of 2 to 3% of esti-mated cost is �rovided in the estimate.Over head Charges:This includes office rent, de�reciation of equi�ments, salaies of officestaff, �ostage, lighting travelling allowances, tele�hone bills. the contractor may�rovide small tooks like ladders, trowels, ro�es etc., fo his workmen.Here anamount of 5% of estimated cost is �rovided towards overhead charges.Profit :Generally 105 of estimated cost is considered for contractor's �rofitafter allowing the charges of equi�ments and establishments. For small job s15% and large works 8% �rofit is considered.Standard Data Book:This book gives the quantities of materials and labour required for unititem of work.Standard scheduled of rates:The rates of materials and wages of laboures are fixed by su�erintend-ing Engineer for this cicle for evey year.And these rates ae a��roved by board

of enginees. The S.S.R. for 2002-2003 is �resented in the last �ages.Water Charges:For drinking and for work,s the arrangement of water is done either bysinking tube well or by giving connection to the work site from cor�oration bya �i�e line. Centrally 1% of estimated.110Task or out-turn work:This is the quantity of work which can be done by an atisan for tradeworking of 8 hours. Although the task is different from �erson to �erson ac-cording to their �hysical and mental abilities, the average task or out turn workis taken into consideration for �re�aring rate �er unit item. Task does not meanthat the quantity of work done by one oner labour. But other laboureso hel�ersalso be engaged to com�lete the given task.For exam�le a manson can �re�are 2.0m3 of cement concrete �er day�rovided he is hel�ed by two mazdoors to carry and mix the ingredients.The following may be taken as a��roximate quantity of work out-turnwork or task for an average artisan �er day.Sundries:A lum�sum amount is generally �rovided in the analysis of rates, to-wards �urchase of certain tools and other �retty items which cannot be ac-counted in detail. an amount of 2½ to 3% of labour cost is �rovided for this�ur�ose.TABLEEarth work excavation in foundation, trenchesin ordinary soils, lead 50m and lift u� to 1.5mEarth work in excavation in foundation trenchesin hard soils, lead 50m and lift u�to 1.5mEarth work in soft or decom�osed rock byblasting lead u� to 50m and lift u�to 1.5mSand filling in �linth, consolidation and dressendSingle layer brick flat soling including ramming,dresing etc.Lime concrete in foundationC.C.R.C.C. (1:2:4)Brick work in foundation and �linthBrick work in su�er structure (G.F)Half brick work in �artition wall2.75 m3/Mazd2.10m3/ Mazd0.55m3/ MaZd4.0m3/ Mazd9.0Sqm/ Mazd10m3/Mason4.0m3/ Mason3.25m3

/ Mason1.40m3/Mason1.25m3/Mason7.00Sqm/ Mason1.2.3.4.5.6.7.8.9.1011No.Descri�tion of workQuantity of work �erday (8 hours of day)A��endex111 Estimation and CostingBricks in �lain archesReinforced brick work in slabs2.5 cm C.c.D. P.C.2.0cm D.P.C. with C.M.R.R.Masonry foundation & PlinthR.R.Masonry in su�erstructureAshlar masonry in su�erstructureC.R.S. Masonry in su�erstructureBrick on 1st floor with C.M.7.5 cm floor with (1:4:8)Teraced flooring -7.5cm TH2.5cm THC.C. flooringTerrazzo flooring 6mm TH mosaic work ove 2cmthick C.C.(1:2:4)Pre cast Terrazzo tiles 2mm TH, laying on bed of25mm thick L.M.Ranigang Tile roofingMangaloe tile roofing including wooden battens, tilesset in C.M.Corrugated G.I. sheet roofing12mmTH current �laster on new brick workRule �ointing on brick workSingle coat white washing over old white washingWhite washing over one coat �rintingLime �inning over interior surfaces(Plaster)Water �roofing cement �aint to new cement �lasterSnow cem washing on �laster surface two coatsPriming coat with ready mined �rimer on wood orsteelPainting two coats with ready mined �aint for woodworkBreaking of over burnt brick to ballast 40mm downBreaking of over burnt brick to ballast 25mm1.0m3

/ Mason1.00m3/ Mason12.5 m2/Mason20Sqm/ Mason1.00cm/Mason0.9m3/ Mason0.40m20.67m21.0 Sqm/ Mason10.0Sqm/mason20Sqn/mason12.50 Sqm/mason5.0 Sqm/m25.0 Sq/m26.7 Sqm20 m210Sqm10Sqm10Sqm133 Sqm33.70 sqm5.00sqm20.m3/Paints20 m3/sqm40m318m20.75m3/Mazd0.55m3121314151617181920212223

24252627282930313233343536373839112PREAMBLE1. AREA ALLOWANCES:A. MUNICIPALITIESi) Allow 15% extra over basic rates on labour com�onents works (u�toa belt of 12k.m from the Munici�al limits in all District Head Quartersfor all s�ecial class, first class and the remaining Munici�alities.ii) For works at Tirumala Hills 30% extra over the S.S.Rates and 30%extra for Hoarsely Hills over the S.S.Rates of (R&B) circle,CHITTOOR is allowed on labour com�onent works.iii) For works located inside Tirumala Tem�le allow 20% extra over therate for Tirumala Hills.Note: For Items (i) above works within a belt of 12 Kilometers from all theMunici�al limits shall be taken into account for �ur�ose of allowingthe extra �ercentage.B. INDUSTRIAL AREA10% extra over the basic rates on Labour com�onent shall be al-lowed (u�to a belt of 10km from the Munici�al limits).C. RURAL AREAAllow 15% extra on skilled and semi skilled workmen in rural areaswhere no other allowances including im�ortation of labour and ameni-ties are admissibleD. AGENCY/TRIBAL AREANot a��licable to this circle.E. GHAT ROADSFor the Ghat roads stee�er than 1 in 20 gradient, the length of theroad may be taken as 1.50 times of the existing length of the road forthe �ur�ose of leads only for the conveyance of materials based onthe certificate for the Ghat Road given by the Su�erintending Engineerconcerned.NOTE : Under the com�elling circumstances the concerned Chief Engineercan ado�t the equivalent length of the road at 2.5 times of the actuallength.F.JAIL COMPOUNDS15% extra is allowed over labour rates for the works in the Jailscom�ounds, only equivalent number of men mazdoors shall be�rovided for works in jail Premises as no women and Children areallowed inside.113 Estimation and CostingNOTE: If more than one area allowance such as those for (a) Munici�alities(b) Industrial area (c) Ghat Roads are a��licable for a �articularsituation only the maximum out of the allowable �ercentage is to beallowed.II. IMPORTATION OF LABOUR AND LABOUR AMENITIES:

Maximum of 13% towards labour im�ortation and amenities to labourbutting etc., of the total labour com�onent is allowed only in case ofworks where the labour com�onent (i.e., ) excluding the cost of ma-terials such as cement and steel works out to more than Rs. 1.00lakhs vide G.O. Ms. No. 270 T R&B(c-I0 De�artment dated: 20-5-1978 onthe basis of certificate of the Executive Engineer that the locallabour available is not adequate and that labour has to be im�ortedfor executing the work subject to the a��roval of the Chief EngineerConcerned.NOTE:1. Extra �ercentage towards Labour im�ortation and labour amenitieswhere ever necessary is admissible in addition to other �ercentagesallowable.2. The above �ercentages may be allowed where ever necessary on thefollowing item.1. Labour Rates.2. Materials like Sand, Metal Kankar, Quarry rubbish and clay forfoundation or filling etc., bricks and tiles.3. Jungle Clearance.4. Dismantling5. Earth work including leads and lifts.6. Purely labour involving items like grinding, mixing, binding, steeland feeding ingredients into mixer etc.,7. Blasting, Drilling holes etc.,8. Stacking metal, Sand, Gravel, Stone, Picking, metalled, grav-elled surface s�reading metal etc.,9. Loading and unloading materials excluding that �arts of work inconveyance of materials by carts and lories.10. Labour com�onents to be included in the data for items likemasonry, mortar etc.,114III. WATER LEADThe following labour is allowed for conveyance of water for everyhalf kilometer lead or �art there over the initial lead or �art there ofover the initial lead of half Kilometer.a) Cement Concrete 1.50 Woman Mazdoor / cum.b) Masonry 1.60 Woman Mazdoor /cum.c) Plastering 0.50 Woman Mazdoor / 10sqm.IV. EXCAVATION OF TRIAL TRENCHES, TRIAL PITS AND EXCA-VATION IN RESTRICTED PLACES.a) Trial trenches not more than 2 Metres in width and de�th not less thantwice the Width -20% extra.b) 1. Trial �its u�to 2 M de�th 125% extra2. Over 2M de�th and u�to 4M de�th 200% extra3. Over 4M de�th and u�to 6M de�th 300% extra4. Over 6M de�th and u�to 8M de�th 400% extra5. Over 8M de�th and u�to 9M de�th 400% extra6. Over 9M de�th 550% extrac) Excavation in Restricted �laces:i) Foundation of building, excavation of road boundary drains,model sections for canals, excavation of field channels excava-tion of narrow trenches of similar nature not more than 2M inwidth and de�th not less than twice the width.50% Extraii) For �i�e lines where the de�th is less than 1.5times 75% Extrathe widthiii) For �i�e lines where the de�th is 1.5 times or morethan the width 150%Extraiv) Silt removal in restricted area such as channels ofunder tunnels, culverts and sy�hons. 150% Extra

NOTE :i) The extra �ercentage allowed is over S.s., 301 rates for the cor-res�onding soil, it includes the charges of alllifts and initial leadbut do not include dewatering charges if any in res�ect of all theitems under (a) & (b) above.ii) The above extra �ercentage in res�ect of excavation in restricted�laces are not to be allowed in res�ect of items involving blastingcom�onent which is to be taken as 1/3 of the cost.Preamble115 Estimation and CostingV. PROVISIONS OF 1st CLASS AND 2nd CLASS WORKMEN UNDER SKILLED LABOUR30% of the skilled labour �rovided in the data may be taken as 1stClass and emaining 70% as 2nd class.Where the nature of work is same no distinction need be made in caseor men and women workers.VI. CEMENT CONCRETE PROPORTION AND REQUIREMENTS TOCOARSE AGGREGATES ETC.,(UNIT=1cum)OF FINISHED WORKi) For Cement Concrete �ro�ortions (1:4:8) (1:5:10) etc. 0.92 cum ofcoarse aggregate shall be ado�ted and the quantity of mortar requiredcalculated �ro�ortionately in each case.ii) For Cement concrete �ro�ortions (1:5:8) (1:6:10) etc., 0.90 cum ofcoarse aggregate shall be ado�ted and the quantity of mortar requiredcalculated �ro�ortionately in each case.VII. REQUIREMENTS OF CEMENT MORTAR FOR STONE MASONRYPer unit (1cum) of finished work:a) CR. Masonry first sort - 0.28 cum of Cement mortarb) CR.Masonry second sort - 0.32 cum of Cement mortarc) R.R.Masonry - 0.34 cum of Cement mortarNOTE: In massive walls above 3M thick, 0.40cum of cement mortar shall be allowed.VIII. REVETMENT AND APRON WORKSi) The size of stone for the volume range 0.0515 to 0.030 cum shall notbe less than 0.30 x 0.30 x 0.15M to 0.30x 0.225 x 0.225M.ii) The rate of labour com�onents as �er the standard Data book is to beado�ted for revetment work only. However for a�ron work Rs. 2.50�er cum should be deducted.iii) Labour charges for rock to be ado�t two thirds of the labour chargesof revetment item.IX. SEIGNIORAGE CHARGESi) The seigniorage charges as existing actually may be added in the Datarates in the estimates subject to the conditions that the concernedExecutive Engineer who �re�are the estimates should certify in writ-ing the rates of seigniorage charges in all cases where the seignioragecharges are actually �ayable.ii) The revised seigniorage charges as fixed by Government in G.O.M.S.No.154 (Industries and commerce(M-I) De�artment Dt. 23-07-96may be ado�ted as follows.