Estimation & costing guide

Structure
1.0Introduction
1.1Definition
1.2 Need for estimation and costing
Learning Objecyives
After studingthis unit, student willbe able to
• Have anidea ofthe introductionto estimating and costing.
1.0 Introduction
In the civilengineering field, the construction activity contains the
following three steps.
1. Plans : Preparation ofdrawings plan, section, elevation, with full
dimensionand detailed, specificationsmeeting therequirementsoftheproposed
structure.
2. Estimation : Preparation ofan estimate is for arriving the cost of
the structure to verify the available funds or to procure the required funds for
completion of the proposed structure.
3. Execution(construction) :It is a grounding theproposedstructure,
for constructionas per the provision contained in drawings and estimation..
Introduction
1
UNIT

Construction Technology
152
The plans contains size of roomand dimensions ofthe work and the estimate
contains the quantity and quality aspects ofthe structure.
1.1 Definition
Estimation and costing there are two basic points involved in
constructionofstructures are :
1. Quantity: Thequantity aspectsiswith referencetothemeasurement
in thedrawings (plan, elevation, section)
2. Quality: Thequalityaspectsis with reference to the specifications,
i.e properties ofmaterials, workmanship etc.
Note : The estimation and costing ofanystructure is defined as the
process of determination of quantities of items of work, and its cost for
completion.
2. Estimate ofa project is therefore, a forecast ofits probable cost.
1.2 Need for Estimation and Costing
The object ofpreparing the estimatefor anycivilengineering structure is
1. To know the quantities ofvarious items of work, a materialand
labour and theirsource ofidentification.
2. To decide whether the proposalcanmatch the available funds to
complete the structure.
3. To obtainthe administrative and technicalsanction ofestimate
from the competent authoritiesto release thefundsforconstruction.
4. To invite tenders or quotations based on the estimate quantities for
entrust of works to the execution.
Short Answer Type Questions
1. What is meant byEstimating and Costing ?
2. Stateneed for Estimationand Costing.

Structure
2.0 Introduction
2.1 Units ofmeasurements
2.2 Rules For Measurement
2.3 Different methods oftasking out quantities
Learning Objectives
After the studyingthis unit student willbe able to
• To measure various quantities as per rules.
2.0 Introduction
The units ofdifferents works depends on their nature, size and shape.
.In general, the units of different items of works are based on the following
principle.
1. Massive or volumetric items ofwork such asearth work, concerete
for foundations, R.R Masonry , Brick Masonry etc. The measurements of
length, breadth, height ordepthshallbe takentocomputethe volume or cubical
contents.
2. Shallow, thin and surface work shallbe taken in square unit or in
area. The measurements of length and breadth or height shall be taken to
compute the area, Ex. Plastering, white washing etc.
Measurement of Materials
and Works
2
UNIT

154
3. Long and Thin work shall be taken inlinear or running units and
linear measurement shall, be taken. Ex : Fencing, Rainwater pipes,
ornamental borders etc.
4. Single units ofwork are expressedinnumbers.Ex. Doors,Windows,
Rafters, Trusses etc.
2.1 Units of measurement for various items of Civil
Engineering Works
Sl.No
1.
2.
3.
4.
5.
Unit of
payment
10.00cum
1.00cum
1.00rmt
10.00cum
10.00cum
1.00cum
1.00cum
1.00cum
1.00cum
Particulars of items
(a)Earthworkexcavationinall
types of soils except rock re-
quiringblastering.
(b) Earth work excavation in
the soils hard rock requirng
blastering.
(c) Excavation of pipe line
throughofspecified widthand
depth inalltypes ofsoils
(d) Earthwork for road for-
mation ,bund formation etc.
cutting , embankment.
(e) Refilling of foundations ,
basements, pipelines, trenches
withexcavated soils.
Plain cement concrete for
foundation.
R.R.masonry or brick ma-
sonry for foundation base-
ment, superstrucrture, parapet
walletc.
Filling the basement with
sand.
(a) RCC 1:2: 4 with normal
reinforcement forplinthbeam,
columns, lintels, verandah
beam- T beam etc.
Units of measure-
ments
10.00cum
1.00cum
1.00 rmt
10.00cum
10.00cum
1.00cum
1.00cum
1.00cum
1.00cum

Paper - II Estimating and Costing 155
10.00sqm
10.00sqm
10.00sq m.
1.00Rmt
1.00No
1.00No
1.00Rmt
1.00Rmt
kg/unit
1.00cum
1.00cum
1.00sqm
1.00cum
6.
7.
8.
9.
10
11.
12.
13.
14.
15.
(b) R.C.C 1: 2: 4 for slabs of
specified thickness .
Plastering pointing, flooring,
weather proof coarse, white
washing,colourwashing,paint-
ing.
Roofing withA.C sheets, tiled
roofing, Kurnooltrerrace, Ma-
dras terrace etc.
D.P.C specified width and thick-
ness
Wooden and steel trusses
Doors, windows, ventilators.
Ornamentelborder of speci-
fied widthand thickness.
R.C.C pipes, A.C pipes GI or
C.I pipes, stone ware pipes
etc.
Steelreinforcement inR.C.C.
Rough stone pitching revet-
ment and soiling of specified
thickness.
(a) Roads works : Metal col-
lections , gravel collections,
solving stones, pitching any
stones, revetment stones etc.
(b) Road works : Spreading
metalgraveland consolidation
with roller ofspecified thick-
ness.
(c) Cement concrete pay-
ments ofspecified thickness.
10.00sqm
1.0.00sqm1.00sqm
10.00sq m.
1.00Rmt
1.00No
1.00 No
1.00Rmt
1.00Rmt
Kg/unit
1.00cum
1.00cum
10.00sqm
10.00cum

156
2.2 Rules For Measurement
Measurement ofworks occupiesaveryimportant placeintheplanning
and execution ofanywork or project, fromthe time ofthe first estimate are
madeuntilthe completion and settlement ofpayments. The methods followed
for the measurement are not uiform and the practices or prevalent differ
considerablyinbetweenthe states. Eveninthe samestatedifferent departments
followdifferent methods. Forconvernienceauniformmethodshould befollowed
throughout the country. The uniform methods ofmeasurement to be followed
which is applicable to the preparation of the estimates and billofquantities
and to theside measurement ofcompleted works have beendescribed below.
General Rules
1. Measuremet shallbe itemwise forthe finished items ofwork and the
descriptionofeachitemsshall be held toinculde materials, transport,
labour, fabrication, hoisting, tools and plants, over hands and other
incidentalcharges forfinishing the work to therequired shape, size,
designand specifications.
2. In booking dimensions the order shallbe in the sequence oflength,
breadth and height or depth or thickness.
3.Allworksshallbemeasured not subject to following tolerances unless
otherwise stated.
(a) Dimensions shallbe measured to the nearest 0.01 meter i.e 1cm(1/
211
).
(b)Areas shall eb measured to the nearest 0.01 sq.m(0.1 sqft).
(c) Cubiccontents shallbe workedup to the nearest 0.01 cum(0.1cuft)
4. Same type of work under different condition and nature shallbe
measured separatelyunder separate items.
5. The billofquantities shallfullydescribe thematerials proportions
andwork-manships andaccuratelyrepresent theworktobe executed.
Work which byits nature cannot be accuratelytaken offor which
requires site measuremets shallbe described as provisional.
6. In case of structurealconcrete, brick work or stone masonry, the
work under the following categories shall be measured separately
and the heights shallbe described.
(a) Fromfirst floor level

(b) Fromplinth levelto first floor level.
(c) Fromfirst levelto second floor level and so on.
The parapet shall be measured with the corresponding items of the
storynext below.
Principle ofunits:Theunitsofdifferent worksdependontheir nature,
size and shape. In generalthe units ofdifferent itemofwork are based onthe
following principle.
(i) Mass, voluminious and thick works shallbe taken in cabic unit or
volumne. The measurement of length, breadth, and height or depth shall be
taken to compute the volume cubic contents(cum).
(ii) Shallow, thin and surface work shallbe taken in separate units or
in area. The measurement of length and breadth or height shall be taken to
compute the area (sq.m).
(iii) Long and thin work shall be taken in linear or running unit and
linear measurement shallbe taken(running meter).
(iv) Piece work, job work etc taken in number
2.3 Different methods of taking out quantities
Theitems ofworklikeearthworkinexcavation infoundation,foundation
concrete stonemasonryinfoundationand basement, stone or brick masonryin
super stucrture maybyestimated bu either ofthe following methods.
1. Long wall and short wallmethod (or) Generalmethod
2. Centreline method
2.3.1 Long wall and short wall method
Inthis methodmeasure orfind out the external lengthsofwalls running
in the directiongenerallythe long walls out-to-out and the internal length of
walls runningin the transverse direction in-to-in i.e. ofcross or short wallin-
to-inand calculatequantities multiplying the lengthbythe breadthand height of
wall. The same rule applicable to the excavation in foundation, to concrete in
foundaiuonandto masonry.
The simple mehtod is to take the long walls of short or erros walls
separately and to find out the centre to centre lengths of long wall anf short
walls from the plan. For symmetrical footing on either sides, the centre line
remians same for suepr structure and for foundation and plinth.

158
For long walls add to the center length one breadth of wall, which
givesthelengthofthe wallout-to-out ,multiplying thislengthbythebreadthand
height and get the quantities,. Thusfor findingthe quantities ofearthwork in
excavation, for the length of trench out-to-out add to the centre length one
breadthoffoundaiton.Adopt the sameprocess for foudationconceret and for
eacth footing. It should be notedthat eachfooting isto be taken separately and
the breadthofthe particular footing is to be added to the centre length.
Long walllengthout-to-out = centreto centre length+ halfbreadth on
oneside+halfbreadthon the other side =centre to centrelength+ onebreadth.
For short or cross walls subtract ( instead ofadding) from the centre
lengthonebreadthofwall, which givesthe lengthin-to-in, and repeat the same
process as for the long walls, subtracting one breadthinstead ofadding.
Short walllengthin-to-in= Centre to centre length- one breadth.
That is, in case of long wall add one breadth and in case ofshort wall
substract one breadth fromthe centre lengthto get the corresponding lengths.
It willbe noticed that bytaking dimensions in this ways, the long walls
are graduallydecreasingin length fromfoundationto superstructure, while the
short walls areincreasing inlength.
Thismethodissimpleandaccurateandthereisno chanceofanymistake.
This method may be named as long wall and short wall method, or general
method.
2.3.2 Centre line method
In this method known as centre line method. This method is easy and
quick incalculations. Inthis method sumtotallengthofcentre lines ofallwalls,
long and short has to be found out. This method is well suitable for walls of
similar cross sections. Inthis method the totalcentre line multiplied bybreadth
and depthofconcerned itemgivesthetotalquantityofeachitem.Inthismethod,
thelengthwillremainsameforexcavationinfoundationforconcreteinfoundation,
for allfootings and for super structure (with slight difference where there are
crosswallsornumberofjunctions).It requiresspecialattentionandconsideration
at the junctions, meeting points ofpartition or cross walls, etc.
For rectangular, circular polygonal(hexagonal, octagonaletc) building
havingno interor cross walls, thismethod isquite simple. Foreachjunctionhalf
breadth of the respective items or footings is to be deducted from the total
centre length. Thus inthe caseofa building withone partitionwallorcross wall
havingtwo junctions,forearthworkinfoundationtrenchandfoundationconcrete
deduct onebreadthoftrenchorconcretefromthetotalcentrelength(halfbreadth

for onejunctionand thebreadth( 2 x1/2= one) for twojunctions. Forfootings,
similarlydeduct one breadth offootingfor two junctions fromthe totalcentre
length andso on. Iftwo walls come fromoppositedirections and meet awallat
the samepoint, than there willbe two junctions.
Inthecaseofabuildinghavingdifferent type ofwalls,suppose the other
(main) walls are ofAtype and inter cross walls are ofB type, then allA type
walls shall be taken jointly first , and then all B type walls should be taken
together separately. Insuch cases no deductions ofanykind need be made for
A type walls, but when B type walls are taken, for eachjunction deducting of
half breadth ofAtype wall (main wall) shall have to be made from the total
centre lengthofwalls.
It may be noted that at corners of the building where two walls are
meeting no substractionor addition is required.
Note:Student shouldpracticemethodIfirstandwhentheyhavebecome
sufficientlyacquaintedwithmethod I, thenonlytheyshould takeupthe method
II.
1.Writetheunit ofmeasurements. Earthwork,P.C.C, R.C.C, Masonary,
Plastering, Flooring, Fencing, Ornamentalborder, Door, Windows,
Trusses etc.
2. Write generalrules for measurement.
3. Write different methods oftaking out quantities and describe.

160
Structure
3.0 Introduction
3.1 Detailed estimate
3.2 Preliminaryor approximate estimate
3.3 Problems in preliminaryestimate
Learning Objectives
After studying this unit student willbe able to
• Understand the definition ofdetailed estimate, stages ofpreparation
ofestimate, Data required for an estimate and types ofestimate.
3.0 Introduction
An estimate is a probable cost ofa work. It is usuallyprepared before
the construction is taken up. The primary object of an estimate is to know
beforehand the cost ofthe work. The actualcost ofthe work is knownafter the
completionofthe work. Iftheestimate is prepared carefullyand correctlythere
willnot be muchdifference intheestimated cost and actualcost. The estimator
should be fully acquainted with the methods of construction, skilled and
experienced foraccurate estimating.
3
UNIT
Types of Estimates

161
Paper - II Estimating and Costing
3.1 Detailed estimate
The estimate maybe approximate or preliminaryestimate or accurate
estimate. Inapproximate estimate theapproximatecost ofthe workisestimated.
In the accurate estimate the detailsofvarious items are taken and calculated.
3.1.1 Definition
The estimate preparedbydividing the work into different items, taking
detailed measurements ofeach itemofwork and calculating their quantities is
known as detailed estimate.
3.1.2. Stages of preparation
To prepare thecompleteestimationofthe project, besidesthe estimated
cost ofdifferentmainitemsofwork,Thecost ofpreliminaryworksandsurveying,
cost ofland and its acquisition, cost ofleveling and preparationofground and
thecostofexternalservices areto be provided. Provisionofsupervisioncharges
and contractors profit are to be provided in the estimate.
Data required for preparing an estimate : To prepare an estimate
for a work the following data are necessary.
Drawings: Thedetaileddrawings ofplan, elevationandsection, drawn
to a scaleare necessaryto take the details ofmeasurements ofvarious items of
work.
Specifications : The specifications gives the nature, qualityand class
ofmaterials, their proportion, method ofexecution and workmanship and the
class oflabourrequired. The cost ofthework varies withits specifications. The
cement mortarwith1:3 is morecostlier thancement mortar with 1:6.
Rates:Theratesforvariousitemsofwork, theratesofvariousmaterials
to be usedinconstruction, the wagesofdifferent categories oflabour should be
available for preparinganestimate. The locationofthe work andits distance of
source ofmaterials and cost oftransport should be known. Theserates maybe
obtained from the Standard Schedule of Rates prepared by the engineering
departments.
3.1.3 Detailsof measurements and calculation ofquantities and abstract
of estimated cost
To prepare an accurate estimate, a detailed estimate ofquantities of
various items ofwork and an abstract estimate ofthe quantities and their unit
rates are required.

162
Detailed Estimate
Abstract estimate
3.2. Preliminary or approximate estimate
Preliminaryor approximate estimate is requiredfor preliminarystudies
ofvarious items ofwork or project , to decide the financialposition and policy
foradministrative sanctionbythecompetent authority.Thepreliminaryestimate
is prepared by different methods for different types of works. The various
methods ofpreparing the preliminaryestimate are plintharea estimate, cubical
rate estimate and estimate per unit base.
3.2.1 Plinth area estimate
The plintharearateis calculated byfinding the plinthareaofthe building
and multiplying by the plinth area rate. The plinth area rate is obtained by
comparing thecost ofthecost ofsimilar building havingsimilar specifications in
the locality.
3.2.2. Cubic area estimate
The cubic rate estimate is prepared on the basis ofthe cubicalcontents
ofthebuilding. Thecubic rateis obtained fromthecost ofthe similarbuilding in
the localityhaving similar specifications. Thecost ofthe buildingis estimated by
multiplyingthevolumeofthebuildingwiththecubicarearate.Cubicrateestimate
is more accurate as compared to the plinth area estimate.
3.2.3 Estimate per unit base
Thepreliminaryestimate maybe prepared for different structures and
works byvarious ways. For schools and hostels, per class rooms for schools,
per bed forhospitals, per seat fortheater halls, etc. For roadsand highways and
for irrigationworks, the preliminaryestimateismade per kilometer.For bridges
and culvertsper running meter. For sewerage and water supplyprojects on the
basis ofper head of population served.
S.no Description of work No Length Breadth Height/Depth Quantity Remarks
S.No. Description of work Quantity Rate Per Amount

163
3.3. Problems in preliminary estimate
1. Ifthecost ofschoolbuildingper student is Rs. 25000. Calculate the
cost ofschoolbuilding for 100 students.
Cost of the school building for 100 students = Rs.
25000x100=Rs.2500000.
2. Ifthe cost ofconstruction of1 km. length ofa highwayis Rs.
10000000. Find the cost ofconstructionfor 20 km.
Cost ofconstruction for 20 km= Rs. 10000000x20=Rs.200000000.
3. Ifthe plinth area rate ofa residentialbuilding is Rs.10000/sq m.
Calculate the cost ofconstruction ofa residentialbuilding of100 sq. m.
Cost of construction of 100 sq. m.= plinth area rate x area =
10000x100=Rs.1000000
Summary
Detailedestimateconsistsoftakingthedetailedmeasurements oflength,
breadth, height andcalculating thequantities.
Data required for estimate : Drawings, specifications and rates.
Types of preliminary estimates : Plinth area estimate, cubic rate
estimate and estimate per unit base.
1. Define detailed estimate.
2. What are stages for preparation ofan estimate?
3. List out the data required for preparation ofan estimate.
4. Write the tabular formfor the detailed estimate.
5. Write the tabular formfor preparation ofanabstract estimate.
Long Answer Type Questions
1. Describe the various typesofpreliminaryestimates.

164
Structure
4.0 Introduction
4.1 Single roomed building (loadbearing structure)
4.2 Two roomed building( load bearing type structure)
4.3 Singlestoriedresidentialbuildingwithnumberofrooms(loadbearing
type structure)
4.4 Singlestoried residentialbuilding withnumber ofrooms (framed
structure type)
4.5 Primaryschoolbuilding withsloped roof
4.6 RCC Dog legged – open wellstairs
4.7 Two storied residentialbuilding (framed structure type)
4.8 Detailed estimate ofcompound walland steps
Learning Objectives
Prepare detailedestimates ofsingle roomed, Building roomed, Double
roomed buildings, forload bearing walls and Framedstructures. Detailed Esti-
mateofPrimarySchoolBuilding, Compound walls andsteps. Detailed estimate
Dog legged and OpenWellSTair case. Preparationalestimate for ground and
first floor.
4
UNIT
Detailed and Abstract
Estimate of Buildings

165
4.0 Introduction
To estimate the cost of any building or a structure, drawings,
specifications and rates are required. Regarding the detailed estimate bylong
wallandshort wallmethod and centre line method, the drawings consisting of
planelevationandsectionare sufficient. Theestimator should be ableto take all
the dimensions from the drawings. The length and breadth are taken fromthe
plan, while theheight or deptharetakenfromthe sectionandelevations. Inlong
wallandshortwallmethod thewallsaretakenseparately, while inthe centre line
method, the centre line lengths ofallthe walls are combined. The accuracyof
estimate depends upon the skillofthe estimator in studying the drawings. The
long walland short wallmethod is usefulfor load bearing type structure, but it
cannot be applied for framed structure.
4.1 Single roomed building (load bearing structure)
There are two steps inestimating the cost ofa building or a structure.
1.Takingoutquantities andcalculationofquantitiesindetailedestimate.
2. Determiningthe cost fromthe abstract estimate.
Long wall and short wall method : This method is also called as
separate orindividualwallmethod. Thisis simple and it gives accurate values.
The following procedure is adopted.
1.Thedimensionsoflongwallandshort wallshouldbetakenseparately.
2. Irrespective ofitslengths, thewallwhichistakenfirst islong walland
the wallwhichis taken next is the short wall.
3. Thecentrelineofthewallofthebuildingisconsideredfordetermining
the centre to centre line length oflong walls and short walls.
4. The centre to centreto centre lengthoflong walls or short walls is
obtained byadding halfthe width ofthe wallto theinternallength of
either long wallor short wall.
5. Centreto centre lengthoflong wall=internallengthoflongwall+ ½
width ofthe wall.
6. Centre to centre lengthofshort wall=internallengthofshort wall+
½ widthof the wall.
7. To determinethe lengths ofdifferent quantities such as earthwork,
c.c. bedinfoundation, R.R. masonryetc, lengthoflongwall= centre

166
to centre lengthoflong wall+ width, the widthis the respective
width ofthe itemin consideration.
8. Similarlylengthoftheshort wall=centreto centrelengthofthe short
wall – width, where the widthis the respective widthofthe item
such as earthwork, c.c. bed etc.
Centre line method : In the centre line method, the sum of all the
centre linelengths oflong wallsand short walls areadded to get thetotalcentre
line length. At the junctions of two walls, the length is present in both of the
walls. Hence halfofthe lengthofthat width is to be subtracted fromthe total
centrelinelength.
Length =Totalcentre line length– ½ widthxnumber ofjunctions.
Fig 4.1 Plan Single Room
Centre to centre length oflong wall= 6.0 + 2x0.3/2 = 6.3 m.
Centre to centre length ofshort wall= 4.0 + 2x0.3/2 = 4.3 m.
Length ofLongWall= Centre to centre Length ofLongWall+Width
Length ofShort Wall= Centre to centre Length ofShort Wall– width
For earthwork inexcavation LengthofLongWall= 6.3 + 1.2 =7.5 m.
E L E V A T I O N S E C T I O N
1.2
0.9
0.7
0.5
0.3
0.6
0.3
3.0

167
For earthwork inexcavationLengthofShort Wall=4.3 – 1.2 =3.1 m.
In cement concrete in foundation the length and widthofthe long wall
and shortwallare thesame, but theheight isdifferent fromthat ofthe foundation
For R.R. masonryFirst footing Lengthoflong wall=6.3 + 0.9 =7.2 m.
Length of Short Wall= 4.3 -0.9 = 3.4 m.
Similarlyfor secondfooting &Thirdfooting,LengthofLongWalls are
7.0 and 6.8 and for short walls are 3.6 m and 3.8 m respectively.
Detailed estimate ofa single roomed building bycentre line method
Centre to centre length oflong wall= 6.0 + 2x0.3/2 = 6.3 m.
Totalcentre line length = 2(6.3 + 4.3) = 21.2 m.
Detailed Estimate
Quantity
30.528
7.63
11.45
4.45
12.72
28.62
19.08
Remarks
Sl. No.
1
2
3
4
Description
of work
Earth work in
excavation
C.C. bed in
foundation
R.R. masonry
infoundation
and plinth
First footing
Secondfooting
Basement
Brick work in
superstructure
No.
1
1
1
1
1
1
L
21.2
21.2
21.2
21.2
21.2
21.2
B
1.2
1.2
0.9
0.7
0.5
0.3
H
1.2
0.3
0.6
0.3
1.2
3
m m m m3

168
Sl.
No.
1
2
3
Description of work
Earthwork in
excavationin
foundation
Long Walls
Short Walls
Plaincement concrete
infoundation(1:5:10)
Long Walls
Short Walls
R.R. Masonryin
foundation
& basement c.m(1:8)
First footing
Long Walls
Short WaLLS
Secondfooting
Long Walls
Short WaLLS
Basement
Long Walls
Short Walls
No.
2
2
2
2
2
2
2
2
2
2
L
m
7.5
3.1
7.5
3.1
7.2
3.4
7
3.6
6.8
3.8
B
m
1.2
1.2
1.2
1.2
0.9
0.9
0.7
0.7
0.5
0.5
H
m
1.2
1.2
Total
0.3
0.3
Total
0.6
0.6
0.3
0.3
1.2
1.2
Quantity
m3
21.6
8.93
30.53
5.4
2.68
8.08
7.78
3.67
11.45
2.94
1.51
4.45
8.16
4.56
Remarks
L=6.3+1.2=7.5
L = 4 . 3 -
1.2=3.1
L=6.3+1.2=7.5
L = 4 . 3 -
1.2=3.1
L=6.3+0.9=7.2
L = 4 . 3 -
0.9=3.4
L=6.3+0.7=7.0
L = 4 . 3 -
0.7=3.6
L=6.3+0.5=6.8
L = 4 . 3 -
0.5=3.8

169
4.2 Two roomed building( load bearing type structure)
Detailed Estimate OfADouble Roomed Building ByLong WallAnd
Short WallMethod
Centre to centre length oflong wall= 5.0 + 0.3 + 5.0 + 2x0.3/2 = 10.6
m.
Number oflong walls = 2. Number of short walls = 3.
Lengthoflongwall= centre to centre lengthoflong walls + width
Length ofshort wall= centre to centre lengthofshort wall- width
4 Brick work insuper
structure c.m. ( 1:8)
Long Walls
Short Walls
2
2
Total
of
6.6
4
R.R.
0.3
0.3
masonry
3
3
12.72
28.62
11.88
7.2
19.08
L=6.3+0.3=6.6
L = 4 . 3 -
0.3=4.0
Sl. No.
1
2
Description of
work
Earthworkin
excavation
Long Walls
Short Walls
C.C. bed in
foundation
Long Walls
No.
2
3
2
L
m
11.8
4.1
11.8
B
m
1.2
1.2
1.2
H
m
1.2
1.2
Total
0.3
Quantity
m3
33.98
17.71
51.69
8.5
Remarks
L= 10.6 +
1.2 = 11.8
L = 5.3 -
1.2 = 4.1
Totalcentre to centre line lenght = 10.6 x 2 + 5.3x3 = 37.1 m

170
3
4
Short Walls
R.R. masonryin
foundation&
plinth
First footing
Long Walls
Short walls
Secondfooting
Long Walls
Short Walls
Thirdfooting&
plinth
Long Walls
Short walls
Brick work in
super structure
Long Walls
Short Walls
3
2
3
2
3
2
3
2
3
4.1
11.5
4.4
11.3
4.6
11.1
4.8
10.9
5
1.2
0.9
0.9
0.7
0.7
0.5
0.5
0.3
0.3
0.3
0.6
0.6
0.3
0.3
1.2
1.2
3
3
4.43
12.93
312.42
7.13
19.55
4.75
2.9
7.65
13.32
8.64
21.96
49.16
19.62
13.5
33.12
L = 10.6 +
0.9 = 11.5
L = 5.3 - 0.9
= 4.4
L = 10.6 +
0.7 = 11.3
L = 5.3 -0.7
= 4.6
L = 10.6 +
0.5 = 11.1
L = 5.3 - 0.5
= 4.8
L= 10.6+ 0.3=
10.9
L = 5.3 - 0.3 =
5.0
R.R. masonryTotal

171
Centre line method
Fig 4.2 Double Room
No.
1
1
1
1
1
1
1
L
35.9
35.9
36.2
36.4
36.6
36.8
B
1.2
1.2
0.9
0.7
0.5
0.3
H
1.2
0.3
0.6
0.3
1.2
Total
3
Sl. No.
1
2
3
4
Description of
work
Earthworkin
excavation
C.C. bed in
foundation
R.R. masonryin
foundation
First footing
Secondfooting
Basement
Brickwork in
superstructure
Remarks
L= 37.1 -
2x1/2x1.2
L = 37.1 -
2x1/2x0.9
L = 37.1 -
2x1/2x0.7
L = 37.1 -
2x1/2x0.5
L = 37.1 -
2x1/2x0.3
Quantity
51.69 m3
12.93 m3
19.55 me
7.65 m3
21.96 m3
49.165 m3
33.12 m3
E L E V A T I O N S E C T I O N
5m x 5 m 5 m x 5 m
1.2
0.9
0.7
0.5
0.3
0.6
0.3
3.0

172
4.3 Single storied residential building with number of rooms
(load bearingtype structure)
Length oflong walls = 6.0+0.3+5.0+2x0.3/2=11.6 m.
Number oflong walls = 3
Length ofshort wallof5.0 m. length= 5.0+2x0.3/2=5.3 m.
Number of5.0 mshortwalls =3
Length of4.0 m. length short walls = 4.0+2x0.3/2=4.3m.
Number of4.0 m. length short walls = 3
Totalcentre line length= 11.6x3+5.3x3+4.3x3=63.6m.
Fig 4.3 Plan Section
6.0 x 5.0 m 5.0 x 5.0 m
5.0 x 4.0 m
5.0 x 4.0 m
D D
D
D
D D
0.3
0.6
0.3
0.9
0.9
1.2m
3.0m

173
56.43
m m m
m3
5m
4m
5m
4m
5m
4m
5m
4m
Basement
Basement

174
4.4 Single storied residential building with number of rooms
(framed structure type)
Number ofcolumns in a framed structure = 9
Size ofthe columns = 230mmx230 mm
Length of R.R. masonry, Brickwork, lintels, plinth beam and beams
under slab = (6+6)x3+(5+4)x3=63 m.
Length ofsunshades and externalplastering = (12.9+9.9)x2= 45.6 m.
Length ofslab with 1 m. extension on both sides = 1.0+1.0=2.0 m.
ExternalPlastering :Area ofexternalplastering =LengthxHeight
Length ofPlastering = 2x(12.9+9.9)=45.6 m.
Height ofexternal plastering = 3.0+0.12, where 3.0mis the height of
the room and 0.12 m. is the thickness ofthe slab.
Internalplastering: Area ofinternalplastering = Lengthx Height
Length of plastering = 2(L+B) , Where L and B are the length and
breadthofthe roomrespectively.
For 6mx5mroom, length = 2(6+5)=22m. Similarlyfor 5mx4mroom,
length=2(5+4)=18 m.
Fig 4.4 Residential Building Framed Structure
6.0x5.0m 5.0x5.0m
5.0x4.0m
6.0x4.0m
P L A N S E C T I O N
0.23x0.23
R.C.CColumn
3.0m
1.2m
0.9m
0.3m
0.3m
R.C.C.
Footing
R.C.C.
G
.L G
.L
100 mm
thick
RCC slab
1.2m

175
B
m
1.2
0.9
0.6
1.2
0.9
0.6
0.7
0.45
0.23
0.23
0.23
1.2
0.23
S.
No.
1
2
3
4
5
Description of work
Earthworkinexcavation
Columns
Inbetweencolumns
Deduct for columns
C.C. bed in foundation
Columns
Inbetweencolumns
Deduct for columns
R.R. masonryin
foundation
First footing
Secondfooting
Brickwork in
superstructure
Deductions Doors
Windows
R.C.C. column footing
Trapezoidalsection
Stem
No.
9
1
9
9
1
9
1
1
1
6
8
9
9
9
L
m
1.2
63
0.6
1.2
63
0.6
63
63
63
1
1.2
1.2
0.23
H
m
1.8
0.9
0.9
0.3
0.3
0.3
0.6
1.2
3
2
1.2
0.3
0.3
5.1
Quantity
m3
23.33
51.03
-2.92
71.44
3.89
17.01
-0.972
19.93
26.46
34.02
60.48
43.47
-2.76
-2.65
38.06
3.89
2.44
2.43
8.76
Remarks
L=12x3+
9x3=63
H=0.9+1.2+
3.0=5.1
Net Brickwork in super structure
(1.44+4x0.985+0.053)/6

176
6
7
8
9
10
11
R.C.C. Plinth beam
R.C.C. in
lintels&sunshades
Lintels
Sunshades
R.C.C. slaband beams
Beams under slab
1m. Projection from
slab
R.C.C. Slab.
Externalplastering
20 mm
Thick
Deductions
Doors
Windows
InternalPlastering
12 mmthick
Rooms6mx5m
Rooms5mx4m
Sandfillinginrooms
Rooms6mx5m
1
1
1
1
9
1
1
6
8
2
2
2
63
63
45.6
63
1
14.9
45.6
1
1.2
22
18
6
0.23
0.23
0.7
0.23
0.23
11.9
5
0.3
0.1
0.07
0.3
0.3
0.12
3.12
2
1.2
3
3
1.2
4.35
1.45
2.23
3.68
4.35
0.62
21.28
26.25
142.27
-12
-11.52
118.75
132
108
240
72
L=2(12.9+
9.9)=45.6
L=12.9+1.0+
1.0=14.9
B=9.9+1.0+
1.0=11.9
L=2(12.9+9.9)
=45.6
H=3.0+0.12
L=2(6+5)=22
L=2(5+4)=18
Net External plastering area

177
4.5 Primary school building with sloped roof
Wallthickness = 0.3 m. inbrick masonry.
Width offoundation = 1.2 m. Depth offoundation= 1.8 m.
Widthoffirst footing = 0.9 m. Depthoffirst footing = 0.9 m.
Second footing width= 0.7 m. Depth = 0.6 m.
Width ofthird footing and plinth = 0.5 m. Height = 0.9 m.
Centre to centre length oflong walls = 3.0+0.3+3.0+2x0.3/2=6.6 m.
Centre to centre length ofshort walls = 3.0+2x0.3/2=3.3 m.
Totalcentre line length= 6.6x2+3.3x3=23.1 m.
Number ofjunctions = 2.
Height ofthesloping roof=1.0 m.
12
13
14
Rooms5mx4m
C.C. bed in rooms
Rooms6mx5m
Rooms5mx4m
Flooringinrooms
Rooms6mx5m
Rooms5mx4m
Fabrication &
placement of
steel
2
2
2
2
2
5
6
5
6
5
4
5
4
5
4
1.2
0.1
0.1
48
120
6
4
10
60
40
100
(8.76+4.35+3.68+26.25)x1.25x87.5/100x1000 78.5x100/100x1000
tonnes 4.22 t

178
Length ofthe sloping roof = square root of (1.5mx1.5m+ 1.0m.x1.0
m.) = 1.8 m.
Number of gable rafters at a spacing of 30 cms. Centre to centre =(
6.0/0.3)+1=21
Length ofthe gable rafters = 1.8+1.8+0.5+0.5=4.6 m.
Number ofreapers along a length of6.05 mts.At a spacing of10 cms
each = (4.6/0.1)+1=47

179
ELEVATION
W W
D
D
Room
3.0 x 3.0 m
Room
3.0 x 3.0 m
References
D - Door 1.00 m x 2.00 m
W - Window 1.2 m x 1.2 m
Width of1st footing : 0.9 m
Second footing : 0.7 m
Basement : 0.5 m
P L A N
S E C T I O N
Tiles
T
i
l
e
s
1.2 m
0.9 m
0.6 m
0.3 m
0.9 m
0.9 m
0.6 m
0.3 m
0.9 m
2.0 m
1.5 m
0.9 m
PRIMARYSCHOOLBUILDINGWITHSLOPINGROOF

180
4.6 RCC Dog legged – open well stairs
Fig 4.5 Dog Legged Stair case
1650
250
150
Floor
1650
2500 1000
E L E V A T I O N S E C T I O N - A A
P L A N

181
Sloping side 22 0.28 2.464
0.4
11.264
Total

182
Lengthoftheinclined flight = Squareroot of(1.65x1.65+2.5x2.5)=3.0
m.
Size ofbase offlight = 1.0x0.5x0.25 m3
Landing at the middle and topfloor =2.0mx1.0mx0.15m.
Length ofthe hand rail = (2x3.0+0.40)=6.8 m.
Number ofrisers = 11
Height ofthe first flight = 11x0.15=1.65 m.
Number oftreads = 10
Length oftreads in each flight = 10x0.25=2.5 m.
Triangular portionofthe brick has abase of0.25 m. and height 0.15 m.
Area ofthe brickwork = 1/2x(0.25x0.15) m2.
4.6.1 Open Well Staircase
Fig 4.6 Open well Stair case
Flight
No.
No.
of
Risers
No.
of
Treads
Each
Riser
Each
Tread
A
8
8
152
300
B
4
3
152
300
C
8
7
152
300
SECTION AT ‘AA’
Note :
1.All dimensionsare in Milli meters
2. Follow the written dimensions only
OPENWELLTYPESTAIRCASE
Scale 1:50
DRG. No. 18

183
Flight No.A
Horizontaldistance oftreads = 0.3x8=2.4 m.
Height of risers = 0.15x9=1.35 m.
Sloping lengthofflight = Square root of(2.4x2.4+1.35x1.35)=2.75 m.
Flight No. B
Horizontallengthoftreads = 0.3x3=0.9 m.
Height ofrisers = 0.15x4=0.6 m.
Sloping lengthofflight= Square root of(0.9x0.9+0.6x0.6)=1.08 m.
Flight No. C
Horizontallengthoftreads = 0.3x7=2.1 m.
Height ofrisers = 0.15x8=1.2 m.
Sloping length offlight = Square root of(2.1x2.1+1.2x1.2)=2.42 m.

184
4.7 Two storied residential building (framed structure type)
Fig 4.7 Twostoried residential building
E L E V A T I O N
Parpet wall
Weathering
course
Lintel &
sunshade
Brick
masonry
Roof slab
C.C. flooring
R.C.CMix
1:4:1
Sandfilling
C.C. floring
1:4:8
Elevation
0.902
3.05m
3.05m

185
Fig 4.8 Ground Floor & First Floor
Ground floor
Number ofcolumns = 15
Height of columns in ground floor & first floor =
0.90+0.9+3.05+0.1+3.05+0.1+0.8=8.9 m.
Height ofcolumn in ground floor = 0.9+9+3.05+0.1=4.95 m.
Height ofcolumn in first floor = 3.05+0.1+0.8=3.95 m.
Length of brickwork, lintels and beams =
4.21x4+4.20x4+3.05x2+3.00x2+2.00x2+4.00x2+3.34x2 = 64.42 m.
Openings – Main door – 1.00mx2.1m -1 No., Door – D 0.9x2.1 – 3
Nos., Door D1 – 0.76x2.1 – 2 Nos.
Windows - W – 1.8mx1.2m – 5 Nos., W1 – 1.2mx1.2m – 2 Nos.
Length of wall100 mm. thick = 4.21+3.79+1.5= 9.5 m.
Length ofsunshade = 2.1x5+1.5x2+1.1x1+1.3x1 = 15.9 m.

186
Length ofslab = 12.68 m., Width of slab = 9.10 m.
Length ofexternalplastering = 2(12.68+9.10)=43.56 m.
Trapezoidal section of the column foundation : Area of base A1 =
1.0x1.0=1.0 m2.
Area ofthe column stem= 0.23x0.23=0.0529 m2=A2

187

189

190
4.8 Detailed estimate of compound wall and steps
Length of the compound wall between the brick columns 230 mm x
230 mm = 6.0 + 4.0 = 10.0 m.
Height ofthe compound wall= 1.5 m.
Depthofexcavation below ground level= 0.9 m.
Width ofthe foundation = 0.9 m.
Thickness ofthe C.C. bed = 0.3 m.
Size ofthe first footing = 0.6 m. x 0.6 m.
Size ofthe plinth = 0.45 x 1.0m2.
Size ofthe brickwork in columns = 0.23 x 0.23 x 1.5 m.
Number ofbrick columns = 3
Lengthoftheearthwork inexcavation=
6.0+0.23+0.23+4.0+0.23=10.69
Quantityofearthwork in excavation = 10.69x0.9x0.9=8.66 m3.
QuantityofC.C. bed in foundation = 10.69x0.9x0.3=2.89 m3.
R.R. masonryfirst footing = 10.69x0.6x0.6= 3.85 m3.
R.R. masonryin plinth = 10.69x0.45x1.0= 4.81 m3.
R.R. masonry total = 3.85+4.81= 8.66 m.
Brick masonryincolumns = 3x0.23x0.23x1.5=0.24 m3.
Brickwork inbetween columns = 10.0x0.10x1.5= 1.5 m3.
Totalbrick masonry= 0.24+1.5=1.74 m.
Deduction for gate 2.0mx1.5m= 2.0x0.1x1.5=0.3 m3.
Net brickwork in superstructure = 1.74-0.3 = 1.44 m.
Plastering in columns= 4x0.23x1.5x3=4.14 m2.
Plastering in betweencolumns = 10x1.5x2=30 m2.

191
Total area ofplastering = 4.14+30=34.14 m2.
Estimate of steps
Quantityoffirst step = 1.0x0.9x0.3=0.27 m3.
Quantityof second step = 1.0x0.6x0.3=0.18 m3.
Quantityofthird step = 1.0x0.3x0.3=0.09 m3.
Total quantityofbrickwork in steps = 0.27+0.18+0.09=0.54 m3.
1.5 m
1.0 m
0.45
0.6
0.9m
0.3 m
0.6 m
0.23m
0.23
4.0 m
0.23
6.0m
0.23
0.15
0.15
0.15
0.3
0.3
0.3
1.0 m
Topview
FrontView
Side View
Fig. 4.9 Plan and Section of a compoundWall

192
Summary
To estimate the cost ofa buildingor a structure the steps involved are
1. Taking out the measurement ofvarious items and calculate the
quantities as per the detailed estimate.
2. Determining the cost ofthe calculated quantities as perAbstract
estimate.
ThemethodsofcalculatingquantitiesareLongwallandshortwallmethod
and Centreline method.
Length ofLongwall= Centre to centre length ofthe longwall+ width
Lengthofshort wall= Centre to centre lengthofthe short wall– width
Incentre linemethod, the length=Totalcentre line length– (number of
junctions)xwidth/2
For a double roombuilding, the totalcentre line length =sumofthe
centre linelengths oftwo longwalls and three short walls. The number
ofjunctions = 2.
For a building with number ofrooms, the totalcentre linelength= sum
ofthecentre to centre lengthsofthree long walls, three short walls oflength5.3
m. and three short walls oflength4.3 m. Number ofjunctions = 6.
The long wall short wall method and the centre line method are not
applicable. The lengthsofthe R.R. masonry, Brickworkinsuperstructure, Plinth
beam, lintelsandbeamsunderslabareobtainedbyaddingtheinternaldimensions
ofthe rooms.
The roofforthe primaryschoolbuildingis a gable roof, having its slope
intwo directions. The roofunder considerationisthe roofhaving itswidth=3.0
m. and its length = 6.0 m.
Length ofthe gable rafter = square root of[(width/2)2 + (Rise)2]
Number ofgable rafters = Lengthofthe roof/ spacingofthe rafters.
Area ofthetiledsurface = 2x(Lengthofthe roof)xWidthofthesloping
side.)
Number ofrisers = Height ofthe flight/ rise.
Number oftreads = Number ofrisers – 1.
Treads length= Number oftreads xTread.

193
Horizontallengthofthe stairs =Treads length +Widthofthe landing
Lengthofthe sloping side =Squareroot of[(Treadslength)2 + (Height
offlight)2].
Area ofbrickwork in each step = (Rise x Tread) x ½.
1. What are the steps involved in finding the cost ofthe building?
2. What are the methodsinvolved intaking measurementsina detailed
estimate.
3. Write the tabular formula ofa detailed estimate.
4. Calculatethe numberofrisers ina flightofheight 1.50 m. andthe rise
of 15 cms.
5. Ifthe number of risers = 10, find the number oftreads.
6. Find the length ofthe gable rafter for a roomofwidth 6.0 m. and
length 12.0 m and the rise is 1.5 m.
1. Find the earthwork in excavation, C.C. bed in foundation, R.R.
masonryin foundation, Brick work in superstructure and plastering for single
roombuilding and double roombuilding by long wall short wallmethod and
centre line method.
2. Detailed estimate ofa dog legged stair case.
3. Detailed estimate ofcompound walland steps.
O.J.T. Type Questions
1. Detailed estimate ofa number ofrooms.
2. Detailed estimate ofa framed structure.
3. Detailed estimate ofa Primaryschoolbuilding.
4. Detailed estimate ofan openwellstair case.
5. Detailedestimate ofa doublestoried building.

194
Structure
5.0 Introduction
5.1 Prepare specifications for different items ofwork.
5.2 Find the cost ofmaterials at source and at site.
5.3 Studyofthe cost oflabor types oflabor using standard schedule
of rates
5.4 Concept oflead and lift- leads statement
5.5 Preparationofunit rates for finished items ofworks
Learning Objectives
• Prepare the unit ratio ofvarious items ofworks. Find the cost of
materials, specifications ofvarious ofvarious itemsofworks.
5.0 Introduction
To estimate the cost ofthe building, the quantities ofvarious items of
work are calculatedfromthe drawings. Theunit rates ofvariousitems ofwork
arecalculatedfromthespecifications ofthe varioustypesofmaterials. The rates
are calculatedas per the ratesinthe standard scheduleofrates. The unit rates of
various items of work increase considerably with the specifications. The
5
UNIT
Specifications andAnalysis
of Rates

195
specifications indicate the qualityofthe work while the drawings are used for
the qualityofthe work.
5.1 Prepare specifications for different items of work
Specifications specifies or describes the nature and the class of work,
materials to be used in the work, workmanship etc. From the study of the
specifications one can easily understand the nature ofthe work and what the
work shallbe.
Detailed specifications : Detailed specificationsare writtento express
the requirements clearlyin a concise form avoiding repetition and ambiguity.
The detailed specificationsfor various items ofwork are as follows.
Earthwork excavationoffoundation
The following specifications shall be followed in the earthwork in
excavations infoundations.
1. Foundationtrench shallbe dug to the exact widthand depth of
foundation.
2. Excavated earth shallnot be placed within 1 m. ofthe edge ofthe
foundation.
3. The bottomofthe trenches shallbe perfectlyleveled both
longitudinallyand transversely.
4. Ifwater accumulates in the trench, it should be pumped out. Care
should be taken to prevent water fromentering the trench.
5. Ifrocks and boulders are found during excavation, theyshould be
removed and the bed ofthe trench should be leveled and
consolidated.
6. Foundation concrete should be laid onlyafter the inspection and
approval by the Engineer in charge.
Cement concrete in foundation (1:5:10)
The following specifications should be followed incement concrete in
foundation.
1. Course aggregate should be ofhard broken stone, free fromdust,
dirt and foreign matter.
2. Fineaggregate shallbe ofcoarse sand, consisting ofhard, sharp and
angular grains and shallpassthrough screen of5 mm. square mesh.

196
3. Sand should be free fromdust, dirt and organic matters.
4. Water shallbe cleanand free from alkaline and acid matter.
5. Mixing should be done onmasonryplatformor sheet irontrayin
hand mixing.
6. Coarse aggregate and sand should be mixed byvolume and cement
by weight.
Random rubble masonry
The following specifications should be followed in random rubble
masonry
1. The stones should besound, hard and durable. Stones withrounded
surface shallnot be used.
2. No stone shall be less than 15 cm. in size.
3. Bond stones should be provided at every1 m. length.
4. Cement mortar 1:3 to 1:6 shallbe provided.
5. The joints in the stone masonryshallnot be thicker than 2 cm.
6. The masonryshall be watered for at least 10 days.
Brick masonry
The following specifications should be followed inbrick masonryfirst
class
1. Bricks ofstandard size, copper red color, regular in shape, having
sharp square edges should be used.
2. Thebricksshouldnotabsorbmorethan20%ofwaterwhenimmersed
in water for 24 hours.
3. The mortar used inbrick masonryshallbe 1:3 to 1:6.
4. The bricks shallbe wellbonded and laid in English bond unless
otherwise specified.
5. Mortarjoints shallnot exceed6 mm. inthickness andthe joints shall
be fullyflushed with mortar.
6. The bricks should be soaked in water before use in masonry.
7. The brick masonry shallbe watered for at least 10 days.

197
Plastering
The followingspecifications should be followedinplastering
1. The materials ofmortar, cement and sand used in plastering should
be as per specifications.
2. The joints ofthebrickwork shallbe raked for a depth of18 mm. on
the surface.
3. Ceiling plastering should be completed before the start ofwall
plastering.
4. The thickness ofthe plastering should not be less than12 mm. for
internalplastering and 20 mm. for externalplastering.
5. The plastering work shallbe checked forhorizontalitywitha straight
edge and for verticality with a plumb bob.
6. Anydefectiveplasteringshallbecut inrectangularshapeandreplaced.
7. The plastering should be watered for at least 10 days.
5.2 Find the cost of materials at source and at site.
The amount required to purchase the material at the source of its
production is the cost ofmaterials at the source.
Cost of materials at site : The cost of materials at site includes the
cost ofmaterials at source along withthe cost ofseignories, taxes, royalties,
transport, stacking, loadingand unloading etc.
Seignories are collected for materials like sand, stones etc., which are
under the controlofrespective localagencies under government control.
5.3 Study of the cost of labor types of labor using standard
schedule of rates
Labour rates
Si
No.
Category of worker
S. Rate
For
2012-13

198
1 2 3
Skilled catregory
• 1 Bar bender 330
• 2 Black smith / Tin smith/ Rivetor 315
• 3 Blaster ( Licensed ) 355
• 4 Carpenter Cl- I 315
• 5 Electrician ( Licensed ) 355
• 6 Fitter Cl- I 315
• 7 Floor Polisher /Tile Layer 315
• 8 Foreman 355
• 9 Gauge reader 300
• 10
• Maistry/ Work Inspector with Non-technicalQualification
• SSLC/SSC/HSC
• 300
• 11 Mason Cl- I / Brick layer Cl- I 315
• 12 Mechanic Cl- I 315
• 13 OperatorAir compressor / DG set 315
• 14 Operator Batching plant 355
• 15 Operator Bus/Ambulance/ Lorry/Tanker 315
• 16 Operator Concrete /Asphalt mixer 315
• 17 Operator Concrete /Asphalt paver 315
• 18 Operator Concrete pump / Placer/ ice plant 315
Common SoR 2012 : 13
280
Sl
No.

199
Category of worker
S. Rate
for
2012-13
1 2 3
• 19 Operator Core drilling machine 355
• 20 Operator Crane/ Tower crane/ Cable way 355
• 21 Operator Drilling jumbo / Loco / Winch 315
• 22 Operator Grouting/ Guniting/Shotcreting 315
• 23 Operator Jackhammer/Pneumatic tamper 315
• 24 Operator Pump /Ventilation fan 315
• 25Operator Lathe/Drilling/Shearing machine 355
• 26 Operator Bending / Planing machine 315
• 27 Operator Road roller 315
• 28 Operator Shovel / Scraper / Dozer 355
• 29 Operator Spillway/ Sluice gate 315
• 30 Operator Crusher / Conveyor / Mucker 315
• 31 Operator Tipper / Dumper /Transit mixer 355
• 32 Operator Concrete vibrator 315
• 33 Operator Vibratory plain / pad foot roller 315
• 34 Operator Wagon drill/ Drifter 355
• 35 Painter Cl- I 350
• 36 Plumber / Pipe fitter 350
• 37 Sarang / Khalasi 315
• 38 Spun pipe moulder 315
• 39 Stone chiseller CI- I / Stone cutter Cl- l 315
• 40 Struct. steelFabricator / Marker / Erector 355

200
• 41 Welder / Gas Cutter 315
• 42 Welder (X-ray quality) 355
II. Semi skilled category
1. Asphalt Sprayer / Boiler attendant 285
2. Bhisti 285
3. Boatmanwith boat 300
Common SoR 2012:13
281
Sl
No.
Category of worker
S. Rate
for
2012-13
1 2 3
• 4 Carpenter Cl- II / Erector shuttering 285
• 5 Cartman with double bullock cart 330
• 6 Cartman with single bullock cart 310
• 7 Chavali/ Navagani 285
• 8 Crowbarman / Jumper man 285
• 9 Fitter Cl- II 285
• 10 Gang man / Head / Survey mazdoor 285
• 11 Gardener / Trained mali285
• 12 HelperAir compressor / DG set 285
• 13 Helper Batching plant 285
• 14 Helper Blasting 285
• 15 Helper Bus/Ambulance/ Lorry/ Tanker 285

201
• 16Helper Bending/Shearing/Planing machine 285
• 17 Helper Carpenter 285
• 18 Helper Concrete /Asphalt mixer 285
• 19 Helper Concrete /Asphalt paver 285
• 20 Helper Core drilling machine 285
• 21 Helper Crane/ Tower crane/ Cable way 285
• 22 Helper Drilling jumbo / Loco / Winch 285
• 23 Helper Fitter / Fabrication/Electrician 285
• 24 Helper Grouting/ Guniting/ Shotcreting 285
• 25 Helper Jack hammer / Pneumatic tamper 285
• 26 Helper Laboratory/ Instrumentation 285
• 27 Helper Road roller 285
• 28 Helper Shovel/ Scraper / Dozer 285
• 29 Helper Crusher / Conveyor / Mucker 285
• 30 Helper Tipper / Dumper/ Transit mixer 285
• 31 Helper Vibrator 285
• Common SoR 2012:13
• 282
Sl
No.
Category of worker
S. Rate
for
2012-13
1 2 3
• 32 Helper Vibratory plain/ pad foot roller 285
• 33 Helper Wagon drill/ Drifter 285

202
• 34 Lineman Electric / Telephone 285
• 35 Mason Cl- ll / Brick layer Cl-II 285
• 36 Mechanic Cl- II 285
• 37 Painter Cl- II 300
• 38 Patkari / Neeraganti/ Sowdy 285
• 39 Stone Chiseller Cl- II 285
• 40 Stone breaker / Hammer man 285
• 41 Valve man / Canalsluice operator 285
III. Un-skilled category
• 1 Cement /Asphalt handling mazdoor 250
• 2 Civic worker 250
• 3 Heavy mazdoor 250
• 4 Light mazdoor 250
• 5 Watchman 250
IV. Other category
• 1 Care-taker / conductor / LiftAttender 300
• 2 Cook / Mess man 300
• 3 Dhobi 300
• 4 Diploma Engineer / Surveyor 450
• 5 Diver with headgear 365
• 6 Graduate / LaboratoryAssistant 350
• 7 Graduate Engineer/ Geologist 600
• 8 HorticultureAssistant /Photographer 300
• 9 ITI certificate holder / Tracer / Printer 350
• 10 Literate mazdoor 285
• 11 Stenographer / Computer Operator 400
• Common SoR 2012:13

203
283
Sl
No.
Category of worker
S. Rate
for
2012-13
1 2 3
• 12 Telephone / Wireless Operator 350
• 13 Typist / Job Typist 350
• 14
• CADoperator withDiploma inEngineering/Generaldegree with
• CAD certificate
• 500
• 15 Jeep Driver 355
• 16 Data Processing Operator 500
• Note : 1. The wage should not be less than the minimumwages of
schedule ofemployment,
• Subject to out turn. 2. 25% extra over the corresponding labour
rates in respect of the work to be
• Done during night time subject to issue ofcertificate accordinglyby
the concerned estimate.
• Sanctioning authorityfor providing in the data andbyconcerned
Executive Engineerincharge oftheworkfor payment. Thenight time
allowance is applicable only to the works done under Greater
• HyderabadMunicipalCorporation, GreaterVisakhapatnamMunicipal
Corporationand Vijayawada MunicipalCorporationlimits only.of
various government agencies.
• Transport cost includes cost oftransporting the materialfromsource
to the site. In S.S.R., the cost oftransporting on a mettaled road is

204
given. Iftransport is required on a cart track or a sand track, to reach
the site, that distance is converted to equivalent metalled road.
Distance on cart track = Distance on metalled road x 1.1
Distance on sand track = Distance on metalled road x 1.4
Stacking includes placing the material in a specified heap for a given
volumeinthecaseofmaterialslikesandandcoarseaggregate.Bricksarestacked
for agivennumber. Sometimes arestacking charges are includedinloading and
unloading. Loadingandunloadingcharges arefixedforagivenvolumeorweight
fordifferent materials.
The cost oflabor wages for eachcategoryoflabor are given above as
per Standard schedule ofrates 2012-13.
Standard schedule of rates : In standard schedule ofrates (S.S.R.) ,
the rates ofvarious materials, machineryand hiring charges andwages oflabor
are prepared. It is prepared bythe board ofchiefengineers and approve it for
that year.
5.4 Concept of lead and lift- leads statement
The distance between the source ofmaterialto the worksite is known
as the lead. This lead distance changes from one project to another project
depending uponthelocation. The verticalheight throughwhichthematerialisto
be disposed is known as the lift.
Lead charges : The conveyance charges ofthe materials fromsource
to thesite ofwork iscalled lead charge. InS.S.R. the lead chargesare given for
Metalledroads.Theequivalent distanceofmetalledroadforcarttrack=1.1xlead,
while for sandytrack = 1.4xlead.
Lead statement : Lead statement gives the cost ofvarious materials at
site. It includesbasic rate, plusconveyance, blastingcharges, seignorage charges
etc.
Lead Statement
S.
No
Mat-
erial
Source Unit Cost at
source
Lead
inKm.
Equi
valent
metal
led
road
Blas
ting
char
ges
Seign
orage
char
ges
Cess
charges
Cru
shing
char
ges
Deduc
tions
if any
Net
rate
at
site
Re
mar
ks

205
5.5 Preparation of unit rates for finished items of works
Cost of sand as per S.S.R. : For concrete = Rs. 375., For filling =
Rs. 288., For plastering = Rs.490.
Cost ofcement = Rs. 5100/ton., = Rs. 255 per bag.
Mixing charges for mixing 1 m3 ofmortar = Rs. 85.
Cost of preparation of 1 m3 mortarfor different proportions
5.5.1. Cement concrete in foundation (1:5:10)1
Quantity of cement =(1.52/16)x1=0.095 m3=0.095x1440/50=2.74
bags.
Quantityofsand = (1.52/16)x5=0.475 m3
Quantityofaggregate= (1.52/16)x10=0.95 m3.
Cost ofcement = Rs.255 per bag., Cost ofsand=Rs. 375/m3., Cost of
Coarse aggregate=Rs.588/m3.
Cost of
sand
Rs. 323.
40
Rs. 367.
50
Rs. 392.
00
Rs. 406.
70
Rs.419.95
Rs. 436.10
Rs. 445.9
Mix
-ing
charges
Rs.85
Rs.85
Rs.85
Rs.85
Rs.85
Rs.85
Rs.85
Mix
propor
-tion
1:2
1:3
1:4
1:5
1:6
1:8
1:10
Quantity of
cement in
bags
9.5 bags
7.2
5.76
4.79
4.11
3.19
2.62
Quantity
of sand
inm3
0.66
0.75
0.8
0.83
0.857
0.89
0.91
Cost of
cement
Rs.2422.
50
Rs. 1836.
00
Rs. 1469.
00
Rs. 1221.
50
Rs.1048.
05
Rs.813.
45
Rs.668.
10
Total cost
2831.50
2288.50
1946.00
1713.20
1553.00
1334.55
1199.00

206
R.C.C. (1:2:4) works in Beams, slab, columns etc
Quantityofcement = 1.52x1/7=0.217 m3=0.217x1440/50=6.25 bags.
Quantity of sand = 1.52x2/7=0.434 m3.
Quantityofcoarse aggregate = 1.52x4/7=0.869 m3.
Quantityofsteel=1.1x78.5/100=0.86quintals=86.35 kgs.
Centering and scaffolding charges with casurina ballies, bamboos,
wooden reapers, poles etc.
Lintel=Rs. 1215/m3; Sunshades = Rs. 214/m2., Columns = Rs. 929/
m2., Beams = Rs. 1637/m2.
Slabs up to 150 mm. = Rs. 184/m2.
Particulars
Materials Cement
Sand
Coarse aggregate
Labor: Head mason
Mason
Menmazdoor
Womenmazdoor
Waterman
Add 20% for labor
Quantity
2.74 bags
0.475 m3
0.95 m3
0.05 No.
0.15 No.
1.2 NO.
1.8 No.
0.4 No.
Rate
Rs. 255/bag
Rs. 375/m3.
Rs. 588/bag
Rs. 350/No.
Rs. 315/No.
Rs. 250/No.
Rs. 250/No.
Rs. 250/No.
Total
Cost
Rs. 698.70
Rs. 178.15
Rs. 558.60
Rs. 17.50
Rs. 47.25
Rs. 300
Rs. 450
Rs. 100
Rs.182.95
Rs.2533.15
Particulars
R.C.C(1:2:4) including cost of
materials, labour charges,
centering chargesbut excluding
cost ofsteeland its fabrication.
Materials
Cement
Quantity
6.25 bags
Rate
Rs. 255/bag
Amount
Rs. 1593.75

207
Sand
Coarse aggregate
Labour
Head mason
Mason
Menmazdoor
Womenmazdoor
Waterman
Totalcost ofmaterials & labour
= Rs.2760.30+1274.40=
Rs.4034.70
R.C.C. works in lintel, slab,
beams and columns
Centering chargeswith
Casuarinas baileys, bamboos,
poles, wallplates etc.
Item
Lintel
Slab
Beam
Column
0.434 m3
0.868 m3
0.05
0.3
1.2 No.
2.0 NO.
0.6 No.
Centering
charges
including
materials and
labour
Rs.1215
Rs. 1533.33
Rs. 1637
Rs.929
Rs. 375/m3
Rs. 1161.80/
m3
Total
Rs. 350/No.
Rs. 315/No.
Rs. 250/No.
Rs. 250/No.
Rs. 250/No.
20% local
allowance
Cost of
materials and
labour
Rs. 4034.70
Rs. 4034.70
Rs. 4034.70
Rs. 4034.70
Rs. 162.75
Rs. 1003.80
Rs. 2760.30
Rs. 17.50
Rs. 94.50
Rs. 300.00
Rs. 500.00
Rs. 150.00
Rs. 1062.00
Rs. 212.40
Rs. 1274.40
TotalCost
Rs. 5249.70
Rs. 5568.00
Rs. 5671.70
Rs. 4963.70

208
1 m3ofR.C.C. work requiresapproximately90 kgs. ofsteel. The cost
offabricationofsteelincluding bending and placement in positionis Rs. 6.00/
Kg.
5.5.3 Brick masonry in cement mortar
The size ofthe bricks considered are 19 cmx9 cmx9 cm. The volume
ofmortar is 0.32 m3. Cost ofbrick masonryfor 1.0 m3 is considered.
Number of bricks required = 500
Mortar witha proportionof1:6 is considered.
Quantityofcement = 0.32/7=0.0457 m3=0.0457x1440/50=1.32 bags
QuantityofSand = 0.32x6/7=0.274 m3
Cost of1000 no. ofbricks 19cmx9cmx9cm as per S.S.R. =Rs. 4687,
Loading and unloading charges=Rs.37.30, Conveyance charges
=118.65+17.80x10=Rs. 297.( for 15 K.M.)
Totalcost ofbricks = Rs.4687+Rs.37.30+297=Rs.5021.30
Quantity
500 Nos.
1.32 bags
0.274 m3.
Materials cost
0.05 No.
1.0 No.
0.7 NO.
1.0 No.
0.2 No.
Particulars
Brick masonryin
superstructure includingcost
ofmaterials and labour
Materials
Bricks
Cement
Sand
Labour
Head mason
Mason
Menmazdoor
Womenmazdoor
Waterman
Rate
Rs. 5021.30per
1000 Nos.
Rs. 255 per bag
Rs. 490/m3.
Total
Rs. 350/No.
Rs. 315/No.
Rs. 250/No.
Rs. 250/No.
Rs. 250/No.
Amount
Rs.2510.65
Rs. 336.60
Rs. 134.30
Rs. 2981.55
Rs. 17.50
Rs. 315.00
Rs. 175.00
Rs. 250.00
Rs. 50.00

209
5.5.4 Course rubble stone masonry(CRS) in cement mortar
Quantity of stone required = 1.25 m3. Volume of mortar required
=40%=0.4.
Quantityofcement requiredfor C.M. 1:6= 0.4/7=0.06m3=0.06x1440/
50=1.8 bags.
Total
Add 20%
Rs. 807.50
Rs.161.50
Rs. 969.00
Rs. 3950.55
Materials and Labour
TotalCost
Particulars
Materials
Stone including bond
stone and wastage
Cement
Sand
Labour
Head mason
Mason
Menmazdoor
Womenmazdoor
Waterman
Totalcostofmaterialsand
labour
Quantity or No.
1.25 m3.
1.8 bags
0.36 m3.
0.05 No.
1.6 No.
1.6 No.
0.8 No.
0.15 No.
Rate
Rs.535.60/m3
Rs. 255/ bag
Rs. 490/m3.
Rs. 350/No.
Rs. 315/No.
Rs. 250/No.
Rs. 250/No.
Rs. 250/No.
Add 20%
allowance
Amount
Rs. 669.5
Rs. 459
Rs. 176.40
Rs. 1304.90
Rs. 17.50
Rs. 504.00
Rs. 400.00
Rs. 200.00
Rs. 37.50
Rs.1159.00
Rs. 231.80
Rs. 1390.80
Rs. 2695.70

210
Quantity of sand= 0.36 m3. Cost of rubble stone =
Rs.293+Rs.74.60+11.20x15 = Rs. 535.60 for a conveyance of 20 K.M.
5.5.5 Plastering
Externalplastering 20 mm. thickand Internalplastering 12mm. thick.
Materials for 20 mm. thick plastering in a wallof100 sq. m.
Volume ofplastering = 100x20/1000=2.0 m3.
Add 20% for wet volume and increasing 25% dry
volume=2.0+0.4+0.6=3.0 m3.
Cost of1:6 cement mortar = Rs. 1553.00/m3. Cost of3.0 m3 cement
mortar=1553.00x3=Rs.4659.00
Labourcharges : Head mason=1/3 no. Cost=(1/3)x350=Rs. 116.70
Mason=12 Nos. Cost=10x315=Rs. 3150.00 Men mazdoor=15 Nos.
= 15x250= Rs. 3750.00
Waterman= ¾ No. Cost = (3/4)x250=Rs. 187.50.
Cost oflabour = Rs.116.70+Rs. 3150+Rs.3750.00+Rs. 187.50= Rs.
7204.20 Add 20% allowance =Rs. 1440.80. Total
cost of labour = Rs. 7204.20+1440.80=Rs. 8645.00
Total cost of external plastering=Rs.4659.00+ Rs. 8645.00=Rs.
13304.00
Cost of20 mm. thick plastering/m2 = 13304.00/100= Rs.133.04
Materials for internalplastering 12 mm. thick for 100 m2.
Volume of plastering= 100x12/1000=1.2 m3. Add 30% for uneven
surfaces and 25% for dryvolume.
Total volume ofplastering = 1.2+0.36+0.29=1.95 m3. say2.0 m3.
Cost of 1:6 cement mortar for 1 m3= Rs. 1553.00 Cost of 2.0 m3
mortar = 2x1553.00= Rs.3106.00
Labour charges = Rs. 8645.00.
Totalcost ofplastering 12 mm. thick = Rs. 3106.00+ Rs.8645.00=Rs.
11751.00
Cost ofplastering 12 mm thick per m2= 11751/100=Rs. 117.51

211
5.5.6 Pointing in cement mortar
For pointing in brickwork the totaldryvolume ofmaterialsis taken as
0.60 m3 for 100 m2.
Pointing withcement mortar ofproportion1:2 : Dryvolume ofmortar
= 0.60 m3
Cost of mortar 1:2 for 1 m3=Rs. 2831.50. Cost of 0.6 m3 mortar =
0.6x2831.50=Rs. 1699.00
Labour: Head mason (1/3)x350=Rs. 116.70
Mason = 10x315=Rs.3150.00; Men mazdoor=10x250=Rs.2500.00;
Waterman=0.5x250=Rs. 125.00
l Cost oflabour = 116.70+3150+2500+125.00=Rs. 5891.70
Add 20% allowance=Rs.1178.30; Total cost = 5891.70+1178.30=
Rs.7070.00
Totalcost ofmaterials and labour = 1699.00+7070.00=Rs.8769.00
Cost ofpointing per m2= 8769.00/100=Rs. 87.70
5.5.7. Cement concrete flooring
Considering 2.5 cm. thick concrete for an area of floor = 100 m2.
Volume of concrete floor = 100x2.5/100=2.5 m3. Add 10% for
unevenness ofconcrete
Quantityof concrete = 2.5+0.25=2.75 m3. Add 50% for dry volume
ofconcrete=1.375 m3.
Totalquantityofconcrete= 2.75+1.375=4.125 m3.
Quantity of cement required = 4.125/7=0.60 m3.=0.6x1440/50=18
bags.
Quantityofsand= 0.6x2=1.2 m3. Quantityofstone aggregate= 0.6x4
= 2.4 m3.
Cement forsurface finishing = 100x2/1000=0.2 m3. = 0.2x1440/50=6
bags.
Cost of cement= Rs. 255/ bag; Cost of sand= Rs. 490/m3.; Cost of
aggregate = Rs.1161.80/m3.

212
Cost of cement concrete flooring per sq. meter = 17491.00/
100=Rs.174.91/sq m.
5.5.8. Doors and windows – paneled and glazed
Consider preparation of door frame with Sal wood . The size of the
door is 1.00 m. x 2.00 m.
Particulars
Materials
Stone aggregate
Sand (coarse)
Cement
Cement for surface
finishing
Labour etc.
Head mason
Mason
Menmazdoor
Womenmazdoor
Waterman
Totalcost ofmaterials
Totalcost oflabour
Side formsforfinishing
Quantity orNo.
2.40 m3.
1.20 m3.
18 bags
6 bags
¾ no.
10 Nos.
5 Nos.
5 Nos.
2 Nos.
Add 20% extra
Side forms
Rate per
Rs. 1161.80/m3.
Rs. 490/m3.
Rs. 255/ bag
Rs. 255/ bag
Rs. 350/day
Rs. 315/day
Rs. 250/day
Rs. 250/day
Rs. 250/day
Lumpsum
Lumpsum
Totalcost
Amount
Rs. 2788.40
Rs. 588.00
Rs. 4590.00
Rs.1530.00
Rs. 9496.40
Rs. 262.50
Rs. 3150.00
Rs. 1250.00
Rs.1250.00
Rs.500.00
Rs. 6412.50
Rs.1282.50
Rs. 7695.00
Rs. 300.00
Rs. 9496.40
Rs. 7695.00
Rs. 300.00
Rs.17491.00

213
Materials : Teakwood of cross section 8 cmx12 cm.
Length of the frame = 2x( 2.14+1.2)=6.68 m. Quantity of
timber=6.68x0.08x0.12=0.064 m3.
Add 5% for wastage = 0.0032 m3. Total quantity of timber
=0.064+0.0032=0.0672 m3.
Rate of salwood = Rs. 40012.00/m3.
Cost of timber = 0.0672x40012.00= Rs. 2688.80
Labour : Head carpenter =1/16 No. Cost =350x1/16= Rs.21.90
Carpenter =1/4 No. Cost =315x1/4= Rs.78.75
Men mazdoor = ½ No. Cost =250x1/2= Rs.125.00
Cost oflabour Rs.225.65
Add 20% allowance Rs.45.20
Totalcost oflabour Rs.270.85
Total cost of materials and labour = Rs. 2688.80+Rs.
270.85=Rs.2959.65 say Rs. 2960.00
Width ofthe plank=1.0-0.10-0.10-0.10=0.6 m. (Widthofthe stiles)
Length ofthe plank = 2.0-0.10-0.10-0.10-0.15-0.10=1.55 m. (width
oftop, frieze, lock and bottomrails
Unit rate of 40 mm. thick paneled door shutter ofsize 1.0x2.0 sq m.
double door in teak wood.
Amount
Particulars
Materials:-
timber
Stiles
Toprail
Frieze rail
Lock rail
Bottom
rail
No.
4
1
1
1
1
L
2.00
1.00
1.00
1.00
1.00
B
0.10
0.10
0.10
0.15
0.10
Thick
ness
0.04
0.04
0.04
0.04
0.04
Quantity/
Nos.
0.032
0.004
0.004
0.006
0.004
Rate

214
Planks for
panels
Brass
accessories
Towerbolt
30 cm.
Towerbolt
15 cm.
Handle 10
cm.
Hinges
Aldrop 30
cm.
Door
stopper
Labour
Head
carpenter
Carpenter
Helpers
1
1No.
1No.
2.no
6.no
1 No.
1 No.
1/15
No.
4 Nos.
1.55
Add
5%
0.6
for
Cost
0.025
wastage
Of
0.023
0.073
0.00365
0.0767 m3
1 No.
1 No.
2 Nos.
6 Nos.
1 No.
1 No.
accessories
1/15 No.
4 Nos.
2 Nos.
Rs.1054
86.00/m3
Rs.248
.00/No.
Rs.121.00/
No.
Rs.337.00/
No..
Rs.112.00/
No.
Rs.
729.00/
No.
Rs.
146.00/
No.
Rs. 350/
day
Rs. 315/
day
Rs. 250/
day
Rs.8090.80
Rs. 248.00
Rs.121.00
Rs.674.00
Rs. 672.00
Rs. 729.00
Rs.146.00
Rs.2590.00
Rs. 23.35
Rs.
1260.00
Rs.500.00
Rs.
1783.35

215
Cost of materials = Rs. 8090.80
Cost ofbrass accessories=Rs.2590.00
Cost oflabour = Rs. 2140.00
Totalcost = Rs.12820.80
Summary
Specification definesthe nature and classofwork, materials to be used
in the work, workmanship etc.
Cost of materials at the source : The amount required to purchase
the materialsat the source ofits productionis thecost ofmaterialsat the source.
Cost ofmaterialsat thesite=Cost ofmaterialsat thesource+Seignories
+ Taxes +Royalties + Transport + Loading+ unloading etc.
Cost oftransport on metalled road is givenin the S.S.R.
Distance oncart track = 1.1 x Distance on metalled road
Distance onsand track = 1.4 x Distance on metalled road
Standard Schedule of Rates (S.S.R.) : Standard schedule of rates
consists ofthe ratesofmaterials, machinery, hiringchargesandwages oflabour.
It is prepared by the board ofchiefengineers and approved for that year.
Lead and Lift : The horizontal distance between the source of the
materialto thework siteis knownas the lead. The verticalheight throughwhich
the materialis lifted is knownas the lift.
Lead Statement :Thestatement indetailofthe cost ofmaterialsat the
site is knownas the lead statement.
Quantityofmaterials in Plaincement concrete (1:5:10) :
Quantity ofcement = 1.52 x 1/16 = 0.095 cu m. = 0.095 x 1440/50 =
2.74 bags
Quantity ofsand = 1.52 x 5/16 = 0.475 cu m.
Add 20%
extra
Total
Rs. 356.65
Rs. 2140.00

216
Quantity ofcoarse aggregate = 1.52 x 10/16 = 0.95 cu m.
Brick masonryin cement mortar for 1.0 cu m.
Number of bricks of size 19 cm. x 9 cm. x 9 cm. = 500
Volume ofmortar = 0.32 cu m.
Course rubble masonry:
Quantityofstone = 1.25 cu m.
Volume ofmortar = 0.40 cu m.
Plastering20 mm. thick: Thevolumeofcement sandmortarrequired
for an area of 100 sq m. and a thickness of 20 mm. is 3.0 cu m.
Plastering 12mm. thick: Thevolume ofcement sandmortarrequired
for an area of 100 sq m. and a thickness of 12 mm. is 2.0 cu m.
Pointing : The volume ofcement sand mortar requiredfor pointing of
an area of100 sq m. with a mix proportion 1:2 is 0.60 cu m.
1. Define specification.
2. What is cost ofmaterials at the source.?
3. What is the cost ofmaterials at the site?
4. Write a tabular formfor an abstract estimate.
5. List out the various types oflabour.
6. Define standard schedule ofrates.
7. What is lead and lift?
8. What is a lead statement.
1. Prepare specifications for the following
(a) Earthwork inexcavation, (b) Cement concrete in foundation, (c)
R.R. masonry, (d) Brick work in cement mortar.
2. Find the unit rate for Plaincement concrete (1:6:12)
3. Find the unit rate for course rubble masonryofcement mortar (1:6).

217
4. Findtheunit rateforbrickworkincement mortar(1:6)usingstandard
size of bricks.
5. Find the unit rate of plastering 12 mm. and 20 mm. thick with a
proportion of(1:5) cement mortar.
O.J.T. Questions
1. Prepare a unit rate of brickwork in cement mortar for 1.0 cu m.
using modular bricks.
2. Prepare a unit rate of R.C.C. (1:2:4) for 1.0 cu m. in slabs, beams
and columns.
3. Find the cost of a door (1.00m. x 2.00 m.) in country wood
4. Find the cost ofa window (1.2 m x 1.2 m) in Sal wood.

218
Structure
6.0 Introduction
6.1 Trapezoidal, Prismoidal, Mid ordinate
6.2 Taking out quantitiesfromL.S. andC.S. incuttingandembankment
Learning Objectives
• Calcualate the quantities ofearthwork inbanking and cutting by
Trapezoidaland PrismoidalRule
6.0 Introduction
Alltypes ofroads, railways and irrigation works are constructed over
earthwork.Tounderstandthecalculationofearthworkinvolvedinthesestructures,
these methods ofcalculationhave to be studied in detail.
Crosssectionofearthwork isintheformofa trapezium. The quantityof
earthwork maybe calculated bythefollowing methods.
6
UNIT
EarthworkCalculations

219
6.1. Trapezoidal, Prismoidal, Mid ordinate
Sectionalandmeansectionalarea methods for calculating earthwork.
Mid sectional area method : In the mid sectional area method, the
average height ofthe two ends is takenas the meandepth. L is thelengthofthe
section. B isthe formationwidth, andS:1 is the sideslope and d1 andd2 are the
height ofthe embankment at the two ends
Mean height dm= (d1+d2)/2
Areaofmidsection=Areaofrectangularportion+areaoftwo triangular
portions=Bdm+1/2sdm2+1/2sdm2=Bdm+2dm2.
Quantityofearthwork = (Bdm+sdm2)xL
The quantitiesofearthworkmay becalculated in a tabularform as below
Mean SectionalArea Method : In this method, the area at the ends
ofdepthd1 and d2 are calculated and the meanarea of the section is found.
Sectionalare at one endA1 = Bd1+S(d1)2
Sectional area at the other end = Bd2+S(d2)2=A2
The meansectionalareaA=(A1+A2)/2
QuantityQ=((A1+A2)/2)xL
Thequantitiesofearthworkmay becalculated ina tabularformasfollows
Stations Depth
or
Height
Mean
depth or
Height
Central
area Bd
Area of
sides Sd2
Total
sec
tional
area
Bd+Sd2
Length
between
stations L
Quantity
(Bd+Sd2)xL
Embank
ment cutt
ing
Station Height
or depth
Area of
central
portion
Bd
Area of
s i d e s
Sd2
T o t a l
sectional
a r e a
Bd+Sd2
M e a n
sectional
area
Length
between
stations
L
Quantity
=(Bd+Sd2)
xLBanking
Cutting

220
Fig 6.1
Trapezoidal-Prismoidal Formula:Intheprismoidalformulatheareas
at the ends and the mid sectionalarea are also taken into consideration. If the
area at the ends areA1 andA2 respectively andAm is the mid sectional area,
Quantityorvolume = (A1+A2+4Am)xL/6
• Cross sectional area at one endA1 = Bd1+S(d1)2
• Cross sectionalarea at the other end =A2 = Bd2+ S(d2)2
• Depth at the mid section = dm = (d1+d2)/2
• Area at the mid section = Bdm+S(dm)2 =Am
• Quantity= (A1+A2+4Am)xL/6
Trapezoidal formula and prismoidal formula fora series of cross
sections : When the series ofcross sectionsA0,A1,A2,A3, …………An are
at equaldistances D, thenthevolume bythe trapezoidalformula is givenbyV =
((A0+An)/2+A1+A2+A3+ ………..+An-1 +An)
Volume by Prismoidal formula : V=((A0+An)+2(Sum of the odd
areas)+4(Sumofevenareas))xD/3
Example 1 : Calculatethe quantityofearthwork for200 metre length
for a portionofa road inanuniformground. Theheights ofthe banksat the two
ends are 1.00 and 1.60 m. The formation widthis 10 metre and side slopes are
2:1.Assume that there is no transverse slope.
B
Sd1
Sd1
1
:
S
1
:
S
d1
B
Sd2
Sd2
1
:
S
1
:
S
d2
d1
B
L

221
Mid sectional area method : Height d1 = 1.00m. Height d2 = 1.60
m. Formationwidth = B = 10 m.
• Height at the mid section dm= (d1+d2)/2 = (1.00+1.60)/2=1.3 m.
Side slopes S = 2.
• Area at the mid section = Bdm+ S(dm)2 =10x1.3 + 2(1.3)2 = 16.38
sq. m. Length = L = 200 m.
• Quantity=Area x length = ((Bdm+S(dm)2)xL=16.38x200 = 3276
cu m.
• Mean sectional area method : Quantity= Mean sectional area x
length
• A1 = Sectional area at one end = Bd1 + S(d1)2 =10x1+2(1.0)2 =
12 sq m.
• A2 = Sectionalareaat another end= Bd2+S(d2)2 =10x1.6+2(1.6)2=
21.12 sq m.
• Mean sectionalarea =Am = (A1+A2)/2 =(12+21.12)/2 = 16.56 sq
m.
• Quantity= Mean sectionalarea x length = 16.56x200=3312 cu m.
• Prismoidal formula : Quantity= (A1+A2+4Am)xL/6
• A1 = sectionalarea at one end = Bd1+S(d1)2 = 10x1.0+2(1)2 = 12
sq m.
• A2 = Sectional area at another end = Bd2+S(d2)2 =
10x1.6+2(1.6)2= 21.2 sq m.
• Am = Mid sectional area = Bdm+S(dm)2 dm = (d1+d2)/2=
(1.0+1.6)/2 = 1.3 m.
• Am = Bdm+S(dm)2 = 10x1.3+2(1.3)2 = 16.38 sq m.
• Quantity = (12+21.12+4x16.38)x200/6 = 98.64x200/6= 3288 cu
m.
• Areaofside sloping surface :Area ofside slopes= Lxdx(square root
of (S2+1))
Example 2 : Calculate the area oftheside slopes ofa portionofa bank
for a lengthof200 m. The heights ofthe banks at the two ends are 2.50 mand
3.50 m. andthe ratio ofsideslope 2:1. Iftheside slopes are to be provided with
15 cm. thick stone pitching, calculate the cost ofpitching at the rate ofRs. 200
per cu m.

222
• Mean height = (2.5+3.5)/2 = 3.0 m.
• Sloping breadthat themid section=d(squareroot ofs2+1)=3[Square
root of( 2x2)+1] = 6.71 m.
• Area of the two side slopes = 2x200x6.71 = 2684 sq m.
• Quantityof pitching =Area x thickness =2684x0.15 = 402.6 cu m.
• Cost of stone pitching = 402.6 x 400=Rs. 161040.
6.2. Taking out quantities from L.S. and C.S. in cutting and
embankment
Example : Reduced level (R.L.) of ground along the centre line of a
proposed roadfromchainage10 to chainage 20 aregivenbelow. The formation
levelat the 10th
chainageis 107 m. andthe road isin downward gradient of1 in
150 upto the chainage 14and thenthe gradient changes to 1 in100 downward.
Formation width of the road is 10 metre and side slopes of banking are 2:1.
Length ofthe chainis 30 metre. Calculate the quantityofearthwork.
Chainage : 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
R.L. of ground : 105.00, 105.60, 105.44, 105.90, 105.42, 104.30
105.00 , 104.10, 104.62, 104.00, 103.30
R.L. formation : 107.00, 106.80, 106.60, 106.40, 106.20,105.90.
105.60 105.30 105.00 104.70 104.40
Height of bank : 2.00, 1.20, 1.16, 0.50, 0.78, 1.60, 0.60, 1.20,
0.38, 0.70, 1.10
Chainage
10
11
12
13
14
15
16
Height
or Depth
2.00
1.20
1.16
0.50
0.78
1.60
0.60
Mean
height
or depth
-
1.60
1.18
0.83
0.64
1.19
1.10
Central
area
Bd
16.00
11.80
8.30
6.40
11.90
11.00
Side
area
Sd2
5.12
2.78
1.38
0.82
2.83
2.42
Total
area
Bd+Sd2
21.12
14.58
9.68
7.22
14.73
13.42
Length in
between
chainage
30
30
30
30
30
30
Quantity=
[(Bd+S(d)2]xL
Banking
Cutting
633.6 -
437.4 -
290.4 -
216.6 -
441.9 -
402.6 -

223
Arailwayembankment is 10 m. wide withside slopes11/2 to1.Assume
the ground to be level in direction transverse to the centre line, calculate the
volume contained ina lengthof120metres, the centre heights at 20 m. intervals
being 2.2, 3.7, 3.8, 4.0, 3.8, 2.8, 2.5 m.
For a levelsection, the area is given byA=(b+nh)h
• Slope is 11/2:1. Hence n=1.5
The areas at different sections will be as under
• A1 = (10+1.5x2.2)2.2=29.26 m2.
• A2 = (10+1.5x3.7)3.7=57.54 m2.
• A3 = (10+1.5x3.8)3.8=59.66 m2.
• A4 = (10+1.5x4.0)4.0=64.00 m2.
• A5 = (10+1.5x3.8)3.8=59.66 m2.
• A6 = (10+1.5x2.8)2.8=39.76 m2.
• A7 = (10+1.5x2.5)2.5=34.37 m2.
• Volume by trapezoidalrule : V = d[(A1+An)/2 +A2+A3+A4+. . .
. +An-1 ]
• V = 20[( 29.26+34.37)/2 +57.54+59.66+64.00+59.66+39.76] =
6258.9 m3.
• Volume by prismoidal rule : V=d/3[(A1+An)+2(Sumof odd
areas)+4(sum ofeven areas)]
•V=20/
3[(29.26+34.37)+2(59.66+59.66)+4(57.54+64.00+39.76)]=6316.5 m3.
Problems involvingbanking and cutting :At the30th
chainagethe height
is banking ofheight 0.3 m. and at 31st
chainage, it is cuttingat a depthof0.40 m.
17
18
19
20
1.20
0.38
0.70
1.10
0.90
0.79
0.54
0.90
9.00
7.90
5.40
9.00
1.62
1.25
0.58
1.62
10.62
9.15
5.98
10.62
30
30
30
30
Total
318.6 -
274.5 -
179.4 -
318.6 -
3513.6cum.

224
Find the volume ofbanking and cutting ifthe formationwidth is 10 m. and the
side slopes are 2:1 in banking and 11/2 : 1 in cutting.
Chainage distance = 40 m. Let the height ofembankment be zero at a
distance ofx mts.
• Length ofcutting =( 40-x) . (x/0.3) =[(40-x)/0.4] 0.4x=12-
0.3x 0.7x= 12 x=17.14 say 17.0 m.
• Volume of banking : Mean height = (0.3+0.0)/2=0.15 m. Central
area = 10x0.15 = 1.5 sq m.
• Side area = 2x(0.15x0.15)=0.05 sq m. Totalarea = 1.5+0.05=1.55
sq m.
• Volume of banking =Area x length = 1.55x17=26.35 m3.
• Volume of cutting : Mean depth = (0.0+0.4)/2 = 0.2 m. Central
area = 10x0.2 =2.0 sq m.
• Side areas = 1.5(0.2x0.2) = 0.06 sq m. Total area = 2.0+0.06 =
2.06 sq m.
• Volume of cutting =Area x length = 2.06 x 23 = 47.38 m3.
Fig 6.2
Summary
• Earthworkcalculations are required for variousengineering works as
roads, railways, irrigation and water supplyand sanitaryworks.
• The various methods ofcalculation ofearthworks are Mid sectional
area method, mean sectionalarea method, trapezoidalrule and
prismoidalrule.
• Prismoidalformula is not applicable for even number ofareas.
• Banking : Ifthe earthwork is above the ground levelit is banking.
0.3
0.4
40
(40-x)
x

225
• Cutting : Ifthe earthwork is below the ground level, it is cutting.
1. List out the varioustypes ofengineering worksinvolving earthwork.
2. What are the various methods ofcalculating earthwork?
3. Definebanking and cutting
4. Mentionthe relationship between the Reduced levelofformation
and the ground line
5. What is the formula for calculating the side slope area.?
1. Theareas within the contour line at the site ofreservoir and the
proposed face of the dam are as follows
Contour Area
101 1,000 m2
102 12,800m2
103 95,200 m2
104 147,600 m2
105 872,500 m2
106 1350,000 m2
107 1985,000 m2
108 2286,000 m2
109 2512,000 m2
Taking 101 asthe bottomlevelofthe reservoir and 109 asthe top level,
calculate the capacityofthe reservoir.
O.J.T. Questions
1. Prepare a detailed estimate for earthwork for a portionofroad from
thefollowing data.

226
Formationwidthofroadis10m.wide. Thesideslopesare2:1inbanking
and 11/2:1incutting.
Distance in metres
0
100
200
300
400
500
600
700
800
900
1000
1100
1200
R.L. of ground
114.50
114.75
115.25
115.20
116.10
116.85
118.00
118.25
118.10
117.80
117.75
117.90
117.50
R.L. offormation
115.000
Upward grad. 1 in 200
Downward grad1 in400

Structure
7.0 Introduction
7.1 Estimate ofgravelroads
7.2 Cement concrete road
7.3 Septic tank with soak pit
Learning Objectives
• Calculate the quantities ofmaterialrequired for graveland cement
concrete roads. Calculate the quantities ofSeptic Tank.
7.0 Introduction
A road consists ofsub base, base course and wearing course. The sub
base consists ofearthwork prepared as per the height offormation. Over this
sub base a base course of stone ballast or brick ballast of 12 cm. Thickness
compacted to 8 cm. is laid. Finallya wearing coat is laid over this base course.
The wearingcourse maybe ofcement concrete, bitumenorgravel. Depending
upon the wearing course provided the roads are classified as cement concrete
roads, bituminousroads and gravelroads. Depending upon thecost involved
the appropriate road required is decided. Inorder to estimate the cost of the
7
UNIT
Detailed Estimates

228
road, we shouldbe able to preparethe detailed estimate ofthe various types of
roads and calculate the materials required. In the sixth unit we studied about
calculationofearthworkinvolved inthe formationofroads. Inthisunit we shall
find the quantities ofthe base course and wearing course.
7.1. Estimate of gravel roads
Ina gravelroads, the gravelis generallylaid over stone ballast. It is laid
over the entire width of the road. The quantity of stone boulders and gravel
consists ofthicknessoftheir respective layers multipliedbyits thickness.
Calculate the quantityofmetalrequired fora 3.70 m. wideroad for one
kilometer length for one layer of8cm. compacted thickness.
Metalof 12 cm. is required for compact thickness of8 cm. as volume
ofloose metalgets reduced onhalfcompaction.
Quantity ofmetal = 1000 x 3.70 x 0.12 = 444 cu m.
Prepare a detailedestimate for the constructionofone kilometer length
W.B.M. road. The formation width of the road is 10.0 m. and the average
height of the bank is 1.0 m. and the side slopes are 2:1. The metalled width is
3.7 m. m. and three coats of metal are to be provided as per cross section.
Soiling coat of15 cm. thick boulders at the base. Over this soiling coat, inter
coat and top coat of12 cm. compacted to 8 cm.Agravelcoat of5 cm. thick is
laid over thesemetalled surface.
Quantityofearthwork =[Bd+S(d)2] xL = [10 x1.0 +2(1)2] x 1000
=12000 cum.
Length ofthe soling coat = 3.7 +0.15 + 0.15 = 4.0 m.
Detailed estimate ofwbm road with gravel
Fig 7.1 Cross section road
Top coat
Inter coat
Saeing coat
Gravel
1.0m
3.15m
1.0m
3.70m 3.15m
10.0m

229
7.2 Cement concrete road
Prepareanestimate for onekilometer lengthofacement concrete track
way with60 cmwide tracks 1.50 meter centre to centre over 15 cm rammed
kankar.
For consolidating kankar an allowance of1/3 is to be provided while
taking loose thickness ofkankar.
Eg. For 0.10 m. thickness loose kankar taken = 0.1 + 0.1 x 1/3 = 0.
133 m.
Similarly for 0.15 m thickness loose kankar = 0.15 x 1.33 = 0.20 m.
S.No.
1
(a)
(b)
(c)
2
Particulars of
work
Metal ling
Preparation ofsub
grade Soling coat
Inter coat
Top coat
Layer ofgravel
No.
1
1
1
1
L
1000
1000
1000
1000
B
4
3.7
3.7
3.7
Hor D
0.15
0.12
0.12
0.05
Quantity
600
444
444
185
S.no
1
2
Particular
Cement concrete
1:2:4in tracks includ-
inglaying.
Kankar metal loose
under c.c.tracksinbe-
tween c.c.tracks.
No
2
2
1
Length
1000
1000
1000
Breadth
0.6
0.9
0.9
Thickness
0.1
0.2
0.133
Quantity
120
360
120
480
m m m2
m m m2 m3
m3

230
Fig 7.2 C.C. Track
7.3. Septic tank with soak pit
Septic tank shall be offirst class brickwork in 1:4 cement mortar, the
foundation and floor shalbe of1:3:6 cement concrete. Inside septic tank shall
be finished with12 mm cement plaster and floor shall be finished with 20 mm
cement plaster with 1:3 cement mortar. Upper and lower portions ofsoak pit
shall be ofsecond class brick work in 1:6 cement mortar and middle portion
shallbe ofdrybrickwork. Roofcoveringslabs and baffle wallshallbe ofprecast
R.C.C.
Details of Measurement & Calculation Of Quantities
Rammedkankar
Thick cc .Track Rammedkankar
10cm
60cm 60cm
10 cm cc
15cm kankar
90cm 90cm
S.No
1
2
Particulars of
items
Earthworkin
excavation
septic tank
Soak pit upto
3.0 m
Soakpit
Lowerportion
Cementconcrete
1:3:6
Floor&
Foundation
Sloping floor
No.
1
1
1
1
1
Length
2.8
(22/28)x(2.0)2
(22/28)x(1.4)2
2.8
2
Breadth
1.7
1.7
0.9
Height
or
Depth
1.95
3
0.2
0.2
0.05
Quantity
9.28
9.42
0.3
19
0.95
0.09
m3
m m
m

231
2
2
2
2
1
1
1
1
1
1
0.3
0.3
0.2
0.2
0.2
0.2
0.2
1.3
0.04
0.94
0.32
1.1
0.42
2.78
0.38
0.15
0.53
1.88
0.234
0.115
0.018
0.367
3
4
5
6
7
First class
brickwork in
1:4 c.m. in
septic tank
First step
Longwalls
Short wall
2ndstepLong
wall
Short wall
2nd class
brickwork in
1:6 cement
mortarin
soak pit
Upperportion
Lowerportion
2nd class dry
brickwork
in soak pit
Precast
R.C.C. work
Coverslab
septictank
Coverslab
Soak pit
Bafflewall
septictank
12 mm
cementplaster
1:3 in septic
tank
2.6
0.9
2.4
0.9
(22/7) x 1.20
(22/7) x 1.20
(22/7) x 1.20
2.4
(22/
28)x(1.40)2
1
0.6
0.6
1.15
1.15
0.5
0.2
2.5
0.075
0.075
0.45

232
Fig. 7.3 Septic Tank
Summary
Structure of aroad : The structure ofa road frombase to the topis as
follows. Earthwork formation , sub base, base course and wearing course.
Types of roads : Gravelroad, cement concrete road, bituminous road.
Structure of a gravel road : Soling coat of boulders about 15 cm
thick, intercoat and top coat 8 cmto 10cmthick and wearingcourse ofgravel
5 cmthick.
8
Longwalls
Short walls
20 mm
c e m e n t
plaster
1:3 infloor of
septic tank
2
2
1
2
0.9
2
1.7
1.7
6.8
3.06
9.86 sq m.
1.80 sq m.
Bafflewall
Out let
Section
In let
Plan
2.0m
0.4m
0.9

233
Structure of a cement concrete road : Plain cement concrete is
provided over rammed earth.
Component parts of a septic tank : Aseptic tank consists of Plain
cement concrete at its base, Walls on all the four sides in brickwork or R.R.
masonry, baffle wall, sum board for large tanks, Precast R.C.C. slabs at the
top, inlet and outlet pipes.Asoak pit is connected to the septic tank to collect
thedischargeeffluent.Asoakpitconsistsofhollowcircularbrickworkconstructed
withcement mortar. Drybrickwork is placed inthe hollow section.
1. What is the structure ofa road ?
2. List out the various typesofroads.
3. Mentionthe various parts ofa gravelroad.
4. What are the various parts ofa septic tank?
1. Prepare a detailed estimate for the construction of one kilometer
lengthover aformationofanembankment. The formationwidthis10.0 m. and
side slope 2:1. The metalled widthis 4.0 m. andthree coats ofmetalling are to
be provided. Soling coat of15 cm. boulders, intercoat and top coats of12 cm
loose compacted to 8 cmthick. Wearing coat ofgravel5 cmthick.
2. Prepare adetailed estimate for one kilometerlengthcement concrete
road 4.0 mwide and 15 cmthick. It is laid over rammed earth 6.0 m. wide and
20 cmthick.
3. Prepare a detailed estimate for a septic tank 2.0 m. long and 1.0 m.
wide. The height ofthe septic tank is 2.0 m.Assume suitable data for pre cast
slabs , baffle wall, inlets and oulets.
O.J.T. Questions
1. Calculate thematerials required for proposedconstruction ofgravel
road and cement concrete road over an existing formation.

Estimation & costing guide

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