UNIT
1
Introduction
Structure
1.1 Definition - Purpose - Data required for estimation
1.2 Types of estimates
1.3 Units of measurement of various items of work as per IS code
1200
Learning Objectives
After studying this unit, the student will be able to
• Understand the purpose for estimation
• Understand the various types of estimates
•Learn the various items of works as per IS code 1200
1.1 Definition - Purpose - Data required for estimation
1.1.1 Estimation
Estimating is the technique of calculating or computing the various
quantities and the expected expenditure to be incurred on a particular works or
project.
1.1.2 Purpose of Estimating
Sanction or approval of any project or work, its estimated cost is worked
out and necessary funds are sanctioned by the competent authority. In case of
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funds available are less than the estimated cost, the work is done in part or by
reducing it or specifications are altered so that the work can be completed
within available funds.
1.1.3 Data Required For Estimation
(a) Drawing : The drawing is the basis from which quantities of various
items for a work are calculated so fully dimensioned drawing must be prepared
showing plans, different sections and other relevant details or the work. For
plans, sections and elevations: 1 cm=0.5m to 1 cm = 2m and for details drawings
1cm= 1 cm to 1 cm are the scales normally used.
(b) Specifications
(i) General Specifications : In General Specifications the nature and
class of work and the names of materials that should be used are described.
This forms a general idea for the project.
(ii) Details Specifications : Detailed Specifications describe every
item of work in the estimate. These specify the qualities, quantities and proportions
of materials, workmanship, the method of preparation and execution for different
items of works in a project. Thus specification of a work serves as a guide to the
supervising staff of the contractor as well as to the owner to execute the work to
their satisfaction.
(c) Rates : Quantities of different items of works are estimated from
the drawing and these are multiplied by the rates. So, rates for different items of
works are vital factors to determine the estimated cost. Normally the Engineering
departments provide with the current schedule of rates per unit of work, materials,
wages of labour, transport etc. In case when such rate is not found in the schedule,
this is worked out by analysis.
(d) Up dated mode of measurement : For standard deductions or
additions are also necessary to determine the correct quantities of works.
(e) Standing circulars : For taxes and insurance etc, are required to
fix up rates those items which are not in the schedule of rates.
1.2 Types of Estimates
1.2.1 Detailed Estimate
Quantities of all items of work are calculated from their respective
dimensions on the drawings on the drawings on a measurement sheet. Multiplying
these quantities by their respective rates in a separate sheet, the cost of items of
work are worked out individually and then summarized, i.e., abstracted which is
the detailed actual estimated cost work. All other expenses required for
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satisfactory completion of the project are added to the above cost to frame the
total of a detailed estimate.
Item No Description of item No Length Breadth Height or Depth Quantity
Abstract Estimate
The cost of each and every individual item of work is calculated by
multiplying the quantity computed is called Abstract Estimate.
Sl.No Description of item Quantity Unit Rate RS. Unit of rate Amount Rs.
1.2.2 Types of Estimates
(a) Approximate Estimate
(b) Plinth Area Estimate
(c) Cubic Rate Estimate
(a) Approximate Estimate : The other name of Approximate estimate
is preliminary or rough estimate. This is made to find out an approximate cost in
a short time and thus enable the responsible authority concerned to consider the
financial asp.ect of the scheme for according sanction to the same. Such an
estimate is prepared adopting different methods for difference types of works.
During preparation of the estimate detailed surveying, design, drawings, etc, are
not required. This estimate is prepared after preliminary investigation, preliminary
surveying and where required sub-soil investigations and tests to determine the
safe bearing capacity may be conducted. A line sketch of the project according
to its requirements may be required. Rates are determined either from practical
knowledge or from records for similar works.
(b) Plinth Area Estimate : The Plinth area is the built up covered area
measured at the floor level of the basement or of any storey of a building. Plinth
area can be calculated by taking the external dimensions of the building excluding
plinth offsets.
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Plinth area shall be calculated including the following,
(i) Area of the floor level excluding plinth offsets if any, when the building
consists of columns projecting beyond cladding the plinth area shall be taken up
to the external face of cladding in case of corrugated sheet cladding outer edge
or corrugation shall be considered,
(ii) Stair cover,
(iii) Internal shaft for sanitary installations and garbage chute provided
these do not exceed 2 sqm in area,
(iv) Lift well including landing,
(v) Machine room and
(vi) Area of porch other than cantilevered.
(c) Cubic Rate Estimate : The method of estimating building cost by
cubic metre volume is more accurate in general, than the method of estimating
cost by plinth area, because the cost of building depends not only on its plinth
area but also on the volume of the building. By this method the volume or cubic
content of the proposed building is worked out and multiplied by the rate per
cubic volume of similar buildings in that locality, constructed recently.
So the preparation of such a estimate depends on
(i) Determination of total volume, in cubic metres, of the proposed
building and
(ii) Determination of the present rate per cubic metre of similar buildings
constructed recently in that locality.
1.3 Units of measurements
The units of measurements are mainly categorized for their nature, shape
and size and for making payments to the contractor also. The principle of units
of measurements normally consists the following:
(a) Single units work like doors, windows, trusses etc., are expressed
in numbers
(b) Works consists linear measurements involve length like fencing hand
rail, bands of specified width etc., are expressed in running meter.
(c) Works consists surface area measurements involve area like
plastering, whitewashing etc., area expressed in square meters (m2)
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(d) Works consists cubical contents which involve volume like
earthwork, cement concrete, masonry etc., expressed in cubic
metres (m3).
S. No
Description of work
Unit
1
Earth work excavation
2
Concreting for foundations
1cum
3
R.R/ Brick masonry fir foundations,
basement and superstructure
1cum
4
Filling in basement with sand
1cum
5
Steel reinforcement in R.C.C.
1 KN
6
R.C.C. for beams, slabs, lintels
1cum
7
Plastering, whitewashing, flooring, painting
10 Sq. m
8
Weather proof course
1 Sq. m
9
A.C. sheet tiled roofing
1 Sq. m
10
R.C. sunshade
1 RM
11
D.P.C.
1 RM
12
Rain water pipe /Drain pipe
1 RM
13
Steel/Wooden trusses
1 No.
14
Door, Windows, Ventilators
1 No.
10 cum
Short Answer Type Questions
1. What is estimating ?
2. What is the purpose of estimating ?
3. What are the methods of estimating?
4. Write the units of the following;
(a) Brick work (b) Plasting (c) Flooring (d) Masonry
Long Answer Type Questions
1. Explain the methods of Estimate?
2. Explain the types of Estimate in detail ?
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UNIT
2
Methods of Building Estimates
Structure
2.1 Preparation of detailed estimates of building
Learning Objectives
After studying this unit, the student will be able to
• Learn how to prepare the detailed estimates for buildings
2.1 Preparation of detailed estimates of building
2.1.1 Building Estimating Methods
The quantities like earthwork, foundation, concrete and brickwork plinth
and superstructure, etc can be worked out by the following two methods’
1. Long wall – Short wall method
2. Centre line method
Long Wall – Short Wall Method : In this method, the wall along the
length of room is considered to be long wall while the wall perpendicular to long
was is said to be short wall. To get the length to long wall or short wall, calculate
first the centre line lengths of individual walls. Then the length of long wall, (out
to out) may be calculated after adding half breadth at each end to its centre line
length.
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Thus the length of short wall measured in to in and may be found by
deducting half breadth from its centre length at each end. These lengths are
multiplied by breadth and depths to get quantities.
Centre Line Method : This method is suitable for walls of similar
cross sections. Here the total centre loine length is multiplied by breadth and
depth of respective item to get the total quantity at time. When cross walls or
partitions or verandah walls join with main wall, the centre line length gets reduced
by half of breadth for each junction. Such junction or joints are studied carefully
while calculating total centre line length.
2.1.2 Measurement of Building
The rules for measurement of each item are invariably described in IS1200. However some of the general rules are listed below.
1. Measurement shall be made for finished item of work and description
of each item shall included materials, transport, labour, fabrication, tools and
plant 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 or thickness.
3. All works shall be measured subject to the following tolerances.
(i) Linear measurement shall be meaured to the nearest 0.01m
(ii) Areas shall be measured to the nearest 0.01 Sq.m
(iii) Cubic contents shall be worked-out to the nearest 0.01 cum.
4. Same type of work under differenct conditions and natures shall be
measured separately under separate items.
5. The bill of quantities shall fully describe the materials, proportions,
workmanship and accurately represent the work to be executed.
6. In case of Masonry (stone or brick) or structural concrete, the
categories shall be measured separately and the heights shall be described.
(a) From foundation to plinth level
(b) From plinth level to first floor level
(c) From first floor to second floor level and so on.
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Example : Estimate the quantities of single room building by using the
following methods shown in Fig.1
(a) Longwall – Short wall method ,
(b) Centrelinemethod
Fig 2.1
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Long wall – Short wall Method
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332
Centre Line Method
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2.1a - Estimate of Single Room Building
Description of items
2.
Nos
3.
Earth work excavation and
depositing on banks with an initial
lead and lift in all soil, like loomy,
red earth mixed with boulders,
black cotton soil etc., complete.
Measurements
L
4.
B
5.
D
6.
Area or
Contents
7.
1x1
16.06
0.80
0.75
9.64 cum
1x1
16.06
0.80
0.15
1.92 cum
For footing
1x1
16.06
0.60
0.45
4.33 cum
For basement
1x1
16.06
0.45
0.45
3.25 cum
For steps
1x1
1.00
0.25
0.25
0.06 cum
Around the building
Cement Concrete (1: 5 : 10) bed
using 40 mm, size H.T. Metal
from approved quarry including
cost and conveyance of all
materials to site of work and
labour charges etc., complete.
Around the building
R.R.Masonry in CM (1 : 8)
including cost and conveyance of
all materials to site of work and
labour charges etc., completely
around the building.
7.64 cum
Refilling foundation trench with
excavated soil including labor
charges etc., complete.
Qty. for earthwork for alround the
building
Qty. for C.C (1: 5 : 10 ) for
alround the building
Qty. for R.R. Masonry for
foundation
9.64
-1.81
-6.26
1.57 cum
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Filling basement with excavated soils
including labour charges etc., complete
1x1 4.33
2.80
0.45
5.46 cum
1x1 16.06 0.45
0.12
0.87 cum
R.C.C.(1 : 2 : 4 ) using 20 mm. HBG metal
including cost and conveyance of all
materials and labour charges etc., complete
for plinth beam.
Around the building
Suppling, cutting, bending, and fixing 8mm.
steel for rods and 6 mm.steel for stirrups
for plinth beam.
Long wall 8 mm.
2x4 5.18
41.44
Short wall 8 mm.
2x4 3.65
29.00
Stirrups 6 mm.
1x50 1.00
50.00
Steel 8 mm.=70.44 m * 0.39 Kg./R.M
27.55 kgs.
Steel 6 mm.=50.00 m * 0.22 Kg./R.M
11.00 kgs.
0.45 kgs
39.00 kgs
Wastage
Filling with sand including watering,
1x1 4.33
tampering etc., for plinth level.
2.80
0.12
1.45 cum
1x1 16.06 0.20
0.60
8.35 cum
Doors
1x1 1.00
0.20
2.00
-0.40 cum
Windows
1x2 0.60
0.20
0.75
-0.18 cum
Lintels over doors
1x1 1.30
0.20
0.05 -0.013 cum
Lintels over window
1x2 0.91
0.20
0.05 -0.020 cum
Lafts
1x1 3.45
0.20
0.05 -0.035 cum
Solid cement block masonry in CM (1 : 6)
using cement blocks including cost and
conveyance of all materials to site of work
and labour charges and curing etc.,
Around the building
Deductions
7.62 cum
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120 mm TH R.C Slab
2.60 m
0.3 m
Sand Filling
0.20 mm Cement Solid
Block masonry
Cuddapah Slab
flooring
0.20 mm th Plinth beam
R.R. Masonry
150 m m TH C.C (1:5:10) Bed
Sand Filling
Hall 3.05 x 4.58 m
D = 1.00 x 2.00 M
W = 1.00 x 1.20 M
PLAN
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2c. Estimation of Two Room Building
Description of items
2.
Earth work excavation and
depositing on banks with an initial
lead and lift in all soil, like loomy,
red earth mixed with boulders,
black cotton soil etc., complete.
Measurements
Nos
3.
L
B
D
4.
5.
6.
Area or
Contents
7.
14.22 cum
Around the building
1x1
31.60 m 0.60 m 0.75 m
1x1
4.40m
Cross wall
0.60 m 0.75 m
1.98 cum
16.30 cum
Sand Filling in foundation trench
including cost and conveyance of
all materials and labor charges
etc., complete.
2.34 cum
Around the building
1x1
31.60 m 0.60 m 0.15 m
Cross wall
1x1
4.40 m
0.60 m 0.15 m
0.396 cum
2.73 cum
Cement Concrete (1: 5 : 10) bed
using 40 mm, size H.T. Metal
from approved quarry including
cost and conveyance of all
materials to site of work and
labour charges etc., complete.
2.34 cum
Around the building
1x1
31.60 m 0.60 m 0.15 m
1x1
4.40 m
Cross Wall
R.R.Masonry in CM (1 : 8)
including cost and conveyance of
all materials to site of work and
labour charges etc., completely
around the building.
0.60 m 0.15 m
0.396 cum
2.73 cum
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0.30 mm Height Solid
Block Masonry
120 mm TH R.C Slab
200 mm TH Solid Block
Masonry
Cuddapah Slab Flooring
350 mm TH Basement in
R.R. Masonry & M (1:8)
160 m TH C.C Bed
150 mm TH SAND Fill
0.60 m WIDE
0.70 m
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2.1(d) Estimate of Two Room Building
Esitmate the following quantities from the given drawing below.
(a) Earth work excavation
(b) C.C Bed (1:5:10)
(c) R.R Masonry in C.M (1:8)
(d) Brick Masonry in C.M (1:8)
T.C.L for Brick Work
= 2 (8.70 + 4.35) + 4.35 - 2 x 1/2 x 0.25
= (30.45 - 0.35)
= 30.10 m
Plan at Super Structure Level
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(a)Estimation of Single Bed Room Building
Fig 2.2 Social elevation at ABBC
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Fig 2.3 Line plan of a building
Measurement details and bill of quantities of various items of work of
example
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2.1 F Estimation of Double Bed Room Residential building
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2.1 G Estimation of Basement Steps (Two Way)
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2.1G 2 - Estimation of Stair Case (Three Way)
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Estimation of basement steps (Three way steps)
Long Answer Type Questions
1. Prepare an estimate for single room building using the Centre Line
Method ?
2. Prepare an estimate for two room building by using any method?
3. Prepare an estimate for single bed room building by using Longwall –
Short wall method?
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UNIT
3
Analysis of Rates
Structure
31. Definition - Data required
3.2 Factors effecting Rate Analysis
3.3 Cost of material at source and at asite
3.4 Standard Schedule of Rates (SSR) of different materials
3.5 Types of Labour - Wages as per SSR
3.6 Lead and lift - Preparation of lead statement
3.7 Preparation of unit rates for finished items of work
3.8 Methods of calculating quantities of ingredients of varous proportions
3.1 Definition - Data Required
Rate analysis is the study of principal role played by various constituents,
elements of construction such as equipments, cost of labor, number of equipments
etc.
Rate of an item = Cost of material (A) + Cost of Labor (B) + Cost of
scaffolding (C)+ Cost of water charges (D) + Cost of sundries (E)*
Sundries mean cost of all small items which cannot be accounted
separately.
In order to determine the rate of a particular item, the factors affecting
the rate of that item are studied carefully and then finally a rate is decided for that
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item. This process of determining the rates of an item is termed as analysis of
rates or rate analysis.
The rate of particular item of work depends on the following
1. Specifications of works and material about their quality, proportion
and constructional 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
conveyance charges.
5. Overhead and establishment charges
6. Profit
3.2 Factors affecting Rate Analysis
• Locality and situation.
• Size and extent of work.
• Nature of project.
• Height/Level of work at which it is being executed.
• Environmental and climatic conditions.
3.3 Cost of material at source and at site
The costs of materials are taken as delivered at site inclusive of the
transport 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
completing the 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
and labour or due to change in technique.
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Description of
material
Conveyance
Lead in
K.M Initial Cost Charges Total Amount
Sand for motor
50
470/-
9.30/km
935.00
Sand for concrete
50
375/-
9.30/km
840.00
Sand for filling
20
288/-
9.30/km
474.00
Rough Stone
30
293/-
11.20/km
629.00
40 mm HBG
20
588/-
11.20/km
812.00
20 mm HBG
20
1024/-
11.20/km
629.00
12mm HBG
20
1076
11.20/km
1248.00
6 mm HBG
20
788/-
11.20/km
1300.00
Cement
Local
5100/ Tonne
—-
4800.00
Steel
Local 48000/ Tonne
—-
1012.00
17.80/km
812.00
Bricks
20
3412
*The lead & lift charges inclusive of 14% contractors profit
Table showing the cost of material at source and at site
3.4 Standard Schedule of Rates (SSR) of Different materials
Unit
S.Rate for
2012-13
Coarse aggregate 10-4.75mm (including
seigniorage charges)
Cum
788
Coarse aggregate 20-10mm (including
seigniorage charges)
Cum
1076
Coarse aggregate 40 - 20mm (including
seigniorage charges)
Cum
1024
Coarse aggregate 80-40mm (including
seigniorage charges)
Cum
588
S.No Description of Material
1.
2.
3.
4.
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5.
(a)Fine aggregate / sand (unscreened) for
concrete items and filter items (including
seigniorage charges)
357
Cum
375
Cum
288
Fine aggregate / sand (screened) for
mortar, plastering items and sand blasting
items (including seigniorage charges)
Cum
490
7.
G.I pipe 100 mm dia B Class
Rm
650
8.
G.I pipe 15 mm dia A Class
Rm
95
9.
G.I pipe 25 mm dia A Class
Rm
145
10. G.I pipe 40 mm dia B Class
Rm
265
11. G.I pipe 50 mm dia A Class
Rm
290
12. G.I pipe 80 mm dia B Class
Rm
450
13. G.I bolts / nuts and washers
Kg
97
14. G.I Sheet (Plain)
Tonne
15. M.S Bolts / Nuts / Washers
Kg
76
16. PVC Sealing Strip
Rm
39
17. PVC water stopper 310 mm wide
Rm
435
18. PVC Pipe of 100 mm dia
Rm
157
19. Reinforcement steel (including loading
charges) ( HYSD / TMT)Note: As per
monthly rate fixed by sub-committee
Tonne
20. Rough stone 20x20x75cms (including
seigniorage charges)
Each
21
21. Stone chips (at dump yard / spoil)
(including seigniorage charges and loading
charges)
Cum
195
Cum
337
(b) Sand for filling ((including seigniorage
charges)
6.
22. Stone chips (at quarry) (including
seigniorage charges and loading charges
57,375
48000 Rate
for 6/2012)
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Structural steel angle by channel /
beam / bars (including loading
charges)Note: as per monthly rate
fixed by sub-committee
To n ne
Structural steel plate by flats
(including loading charges)Note: as
per monthly rate fixed by subcommittee
Tonne
25.
Synthetic enamel Paint first quantity
Ltr
26.
Through stones 20x20x30 to 45 cms
long (including seigniorage charges
and loading charges)
Each
21
Through stones 25x25x45 to 60 cms
long (including seigniorage charges
and loading charges)
Each
35
Cum
178
Un-coursed rubble stones (at
Quarry) (including seigniorage
charges and loading charges)
Cum
293
Water Proofing compound
KG
66
23.
24.
27.
28.
29.
30.
Un-coursed rubble stones (at dump
yard / spoil) (including seigniorage
charges and loading charges)
48,500 Rate
for 6/2012
46,000 Rate
for 6/2012
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3.5 Types of Labour
The labour can be classified in to
1. Skilled – 1st class
2. Skilled – 2d Class
3. Unskilled
The labour charges can be obtained from the standard schedule of rates
30% of the skilled labour provided in the data may be taken as Ist class, remaining
70% as II class. The rates of materials for Government works are fixed by the
superintendent Engineer for his circle every year and approved by the Board of
Chief Engineers. These rates are incorporated in the standard schedule of rates.
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Fig. 4.1
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3.6 Lead and Lift - Preparation of Lead Statement
Lead : Lead is the average horizontal distance between the center of excavation
to the center of deposition.
Lift : Lift is the average height through which the earth has to be lifted from
source to the place of spreading or heaping.
3.6.1 Preparation of Lead Statement
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3.6.2 Data - Cement Mortar
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3.7 Preparation of Unit Rates for finished items of work
using standard data & SSR
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3.8 Methods calculating quantities of ingredients of various
proportions of cement concrete
A: Analyse Cement Concrete (1:2:4) with graded stone chips from 20
mm down to 6 mm from RC works excluding shuttering and reinforcement –
1m3.
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B:Prepare the data of CC for foundations (1:4:8) using 40mm HBG
– 1 cu.ml
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C:Analyse Cement Concrete (1:5:10) using 40mm HBG metal
from approval quarry for foundation laid in layers not more than 15cms
thick well compacted
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Short Answer Type Questions
1. What is analysis of rates?
2. Write the rates of following materials as per SSR ?
(a) Bricks (b) Sand (c) Cement (g) Coarse aggregate
3. What is full form of SSR?
4. Write the types of labour?
5. What is Lead and Lift ?
Long Answer Type Questions
1. Analyse the quantities of brick work with C.M (1:6)?
2. Write the rates of the following;
(a) Mason Class – I
(b) Man Mazdoor
(c) Bhisti
UNIT
4
Estimation of open Drains
and Roads
Structure
4.1 Estimation of open drain in rural area
4.2 Estimation of earthwork by trapeziodal rural
4.3 Estimation of earthwork by prismoidal method
4.4 Estimation of roads (abstract estimates)
4.1 Estimation of Open Drain in rural area
Estimate the construction of surface drain for length of 3meters. The
figure below shows cross section of surface drain .
150
100
90
100
150
150
300
524
All dimensions are in mm
Fig. 4.1
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Leads
Lead is the average horizontal distance between the centre the excavation
to the centre of deposition and not necessarily the route actually taken.
Lift
It is the average height through which the earth has to be lifted from
source to the place of spreading or heaping.
Fig. 4.2
Calculation of Earth Work
It is not always to lay the roads on fairly level ground. And they may
pass through undulating grounds, hills, the rivers etc. The formation of road shall
follow the natural slope of gro7und to achieve the economy. In other words the
earth work is greatly reduced. The average depth at each cross section either in
failing or in cutting is determined from L.S and C.S of ground. For which levels
have already been taken for this purpose.
Fig 4.3
Building Construction & Maintenance Technician
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Volume = Area of C/S x Length
= (bd+2x ½ x nd x d) L
= (bd + nd2) L
4.2 Estimation of earthwork by Trapeziodal rural
This method is also called as average cross sectional area method. The
average sectional area is to be found by two end sections.
Let A1 and A2 be two end sections.
Am = Mean area of end cross section.
Am = A1 + A2
2
Where A1 = bd1 + nd12
A2 = bd2 + nd22
Sections Depth ‘d’ Area of Area of Total Sec Mean
Quantity = L
Length
or
(bd + nd2)
Central ends nd2 tional area sectional between
chainage
portion
bd+nd2
area Am sections L Em Cutting
bd
bank
ment
Volume of earth work (V) = L x Am
The volume of earth work for series of embankments are computed
from table 4.2.
4.2.1 Tapezoidal Formula for a series of Cross-Sectional Areas at Equal
Intervals:
Let A1 + A2 + A3 ———— An be the cross sectional areas along L.S.
of Road.
Paper - III Estimating & Costing
379
‘ L’ be the distance between two cross sections.
Then,
V=L
A1 + An
+ A2 + A3 ———— An-1
2
OR
V = L/2
(A1 + An ) + 2 (A2 + A3 + A4———— An-1
= Length [ Sum of first and last areas plus two time the remaining areas]
2
4.3 Prismoidal Formula Method
In this, the earth work is given by that the sum of end sections and four
times the mean sectional area together multiplie by one sixth of length between
end sections Mathematically,
V = L/6 (A1 + A2 + 4Am)
Where
Am= A1 + A2
2
A1 = bd1 + nd12
A2 = bd2 + nd22
4.3.1 Prismoidal Formula for a series of Cross-Sectional Areas at Equal
Intervals
This method is adopted when there is odd number of cross sections.
Volume of earth (V) = L/3 (A1 + An ) + 4 (A2 + A4 + A6+—— An-1)
+ 2 (A3 + A5 +—— An-2)
= Length [ Sum of first and last areas +4 times even areas + 2 times odd
areas]
2
Example 1 : Find the volume of work in an embank of length 15.0m,
top width 7.0m and depth 3.5m. The side slopes are 1 ½ : 1
Top Width ‘b’ = 7.0 m
Depth
d = 3.50 m
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Side Slopes
=1½:1
Bottom width = (b + 2nd)
= [7.0+2x 1 ½ x 3.5] = 17.5 m.
Volume of earh work = (bd + nd2) L
= (7 x 3.5 + 1.5 x 3.5)2 x 15
= (24.5 + 18.375 ) 15
= 643.125 m3
(or)
V = ½ (7.0 +17.50) 3.5 x 15.0 = 643.125 m3
Example 2 : Find volume of earth work in a canal of depth 2.50 m and
bottom width 1.50 m for a length 50 m. The side slopes are 1 : 1
Bottom width of canal = 1.50m
Top wisdth of canal
= b + 2nd
= 1.5 + 2 x 1x 2.5 = 6.50m
Depth = 2.50m
Paper - III Estimating & Costing
381
Volume of earthwork (V) = (bd + nd2) L
= (1.5 x 2.5 + 1 x 2.52) 50
= 500.0 m3
(or)
V = ½ (1.5 + 6.50) x 2.5 x 50 = 500.0 m3
Example 3 : The depths at two ends of an embankment of road of
length 80.0 m are 2.5 m and 3.40 m. The formation width and side slopes are
12.0 m and 2 : 1 respectively. Estimate the quantity of earth work by
(a) Prismodial formula
Volume of earth work (V) by Prismodial formula
A1 = 42.50 m2
A2 = 63.92 m2
Am = 52.805 (for mid section depth 2.95m)
Using prismoidal rule
V = L/6 (A1 + 4 Am + A2]
= 80/6 ( 40.5+4x52.805+63.92)
= 4208.53 m3
Example 4 : The formation width of a road embankment is 10.0m .
The side slopes are 2:1. The depths along the centre line of road at 50.0 m
intervals are
1.20, 1.10, 1.40, 1.20, 0.90, 1.5 and 1.0
Calcualte the quantity of earth work by
(a) Trapezodial rule and
(b) Prismoidal rule
(a) Volume of earthwork (v) by Trapezoidal formula sectional areas at
50m, intervals
A1 = bd1 + nd12
= 10 x 1.20 + 2 x (1.20) = 14.88 m2
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A2 = 10 x 1.10 + 2 (1.10)2 = 13.42 m2
A3 = 10 x 1.40 + 2 x (1.40)2 = 17.92 m2
A4 = 14.88 m2
A5 = 10 x 0.90 + 2 x (0.90)2 = 17.92 m2
A6 = 10 x 1.5 + 2 (1.5)2 = 19.50 m2
A7 = 10 x 1.0 + 2 x (1.0)2 = 12.0 m2
Then, using Trapezoidal rule
V = L/2 (A1 + A7 +2 (A2 + A3 + A4 +A5 +A6)]
= 50/2 [14.88+12.0+2(13.42+17.92+14.88+10.62+19.50]
= 4489.0 m3
(c) Volume of earth work by prismodial Rule.
V = L/3 (A1 + A7 +4 (A2 + A4 + A6) +2 (A3 +A5)]
= 50/2 (14.88 +12.0+4(13.42+14.88+19.5) + 2(17.92+10.62)]
= 4586.0 m3
Example 5 : The area within contour lines at the site of a reservoir and
the face of proposed dam are as follows.
Contour in metres
Area in Sqm
350
370.0
352
12350.0
354
74500.0
356
1,50,000.0
358
2,75,000.0
360
4,21,000.0
362
4,70,900.0
364
5,95,050.0
366
6,40,700.0
Paper - III Estimating & Costing
383
Taking 350.0 as the bottom level of the reservoir and 366 as the
maximum water level of reservoir find the volume of water in cubic metres by
using
(a) Trapezoidal formula and
(b) Prismoidal formula
Solution
Given data
Contour interval (h) = 2.0m
A1 = 370 m2
A2 = 12350 m2
A3 = 74500 m2
A4 = 150000 m2
A5 = 275500 m2
A6 = 4,21,000 m2
A7 = 4,70,900 m2
A8 = 595050 m2
A7 = 6,40,700 m2
(a) Volume of water by Trapezodial rule
V = h/2 (A1 + A9 +2 (A2 + A3 + A4 +A5 +A6+A7 +A8)]
=2.0/2[370 + 640700 + 2 (12350 + 74500 + 1500000 + 275500
+ 421000 + 470900+595050+640700)]
= 4639670.0 m3
(b) Volume of water by prismoidal rule.
V = h/3 (A1 + A9 +4 (A2 + A4 + A6+A8) + 2 (A3 + A5 + A7) ]
= 2.0/3[370+640700+ 4(12350+150000+421000+595050)
+2(74500+275500+470900)]
= 46,64,313.33 m3
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Example 6 : The ground levels along the ridge of proposed canal are
as shown below. The bed of the canal is 4.0 m wide and sloped at 1 in 100 in
longitudinal direction.
The side slopes are 1 ½ : 1.
Determine the volume of earth work in cutting by
(a) Trapezoidal method and
(b) Prismoidal method
Also find the area for canal lining.
Ground level, Bed level and Depth of cutting are in metres. The cross
sectional areas at 25.0 m intervals are
A1 = bd1 + nd12
= 4 x 2 + 1.5 x 22 = 14.0 m2
A2 = 4 x 2.4 + 1.5 x 2.42 = 18.24 m2
A3 = 4 x 2.2 + 1.5 x 2.22 = 16.06 m2
A4 = 4 x 2.5 + 1.5 x 2.52 = 19.375 m2
A5 = 4 x 2.95 + 1.5 x 2.952 = 24.853 m2
A6 = 4 x3.1 + 1.5 x 3.12 = 26.815 m2
A7 = 4 x3.5 + 1.5 x 3.52 = 32.375 m2
(a) Volume of earth work in cutting by Trapezoidal formula
V = L/2 (A1 + A7 +2 (A2 + A3 + A4 +A5 +A6)]
= 25/2 [14.0+32.375+2(18.24+16.06+19.375+24.852+26.815)]
= 3213.26 m3
(b) Volume of earthwork (V) in cutting by prismoidal formula
V = L/3 (A1 + A7 +4 (A2 + A4 + A6) +2 (A3 +A5)]
= 25/3[14.0+32.375+4(18.24+19.375+26.815)+2(16.06+24.853)]
= 3216.0 m3
Sloped area for canal lining (least C/S)
= 2 x d n2 + 1 x L
= 2 x 2 1.52+1 x 150 = 1081.66 m2
Paper - III Estimating & Costing
385
Example 7 : The area enclosed by contour has lines of boil heap are
as follows
Contour in metres
Area in Sqm
200
1.0
190
4.0
198
15.0
197
47.0
196
120.0
195
180.0
194
260.0
193
340.0
192
430.0
Taking 192.0 as the general ground level and 200 level as the crest
point of heap find the volume of earth work by using
(a) Trapezoidal rule
(b) Prismoidal rule
Solution:
Data given:
Contour interval (h) = 1.0m
A1 = 1.0 m2
A2 = 4.00 m2
A3 = 15.0 m2
A4 = 47.0 m2
A5 =120.0 m2
A6 =180.0 m2
A7 = 260.00 m2
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A8 = 340.0 m2
A7 =430.0 m2
(a) Volume of earth work by Trapezodial rule
V = h/2 (A1 + A9 +2 (A2+ A3 + A4 +A5 +A6+A7 +A8)]
= 1.0/2 [1+430.0+2(4+15+47+120+180+260+340))]
= 1181.5 m3
(b) Volume of eathwork by prismoidal rule.
V = h/3 (A1 + A9 +4 (A2+ A4 + A6+A8) + 2 (A3 + A5 + A7) ]
= 1.0/3[(1+430+4(4+47+180+340) + 2(15+120+260)]
=1168.33 m3
4.4 Estimation of Roads
Cement Concrete Roads : The cement concrete roads can be constructed
by the following methods :
(i) Premixed Concrete Roads : These roads are constructed by mixing
cement, sand and aggregates in the ratio of 1:2:4 with required quantity of water
in the concrete mixers. The mixed concrete is laid over the base of W.B.M in
the designed thickness depending on the density of the traffic.
(ii) In these roads, the road metal is laid on the prepared bed of subgrade
or W.B.M base and rolled to the desired thickness. Over this rolled surface thin
cement mortar or cement slurry is spread and the surfaces is finished. After
laying the cement mortar, the road is cured for atleast 10-15 days.
Fig. 4.4
Paper - III Estimating & Costing
387
Water Bound Macadam RSuch types of roads are being constructed
since very olden days. The W.B.M road is used as village road or as base for
bituminous roads. In most of the road projects water bound macadam road is
constructed in the first phase. When more founds are made available the surfacing
with premix carpet, bituminous macadam or cement concreting is done.
Fig. 4.5
Short Answer Type Questions
1. What is meant by drain ?
2. Write the Trapezoidal Rule for calculating earth work?
3. Write the Prismoidal Rule for calculating earth work?
4. Write about the types of roads ?
5. What is C.C. Road ?
Long Answer Type Questions
1. Prepare an estimate for WBM Road. Assume data?
2. Prepare an estimate for CC Road. Assume data?
UNIT
5
Estimation of Public health
Engineering works
Structure
5.1 Preparation of detailed estimates of septic tank
5.2 Estimation of quantity of sanitary pipes and pipe specials and fittings
Learning Objectives
After studying this unit, the student will be able to
•Know how to prepare the detailed estimates for septic tank
• Learn how to prepare estimation of quantity of sanitary pipes
• Learn how to prepare estimation of quantity of pipe specials and fitting
for building
2.1 Septic Tank
The Plan and Sectional elevation of a house hold setic tank for 20
useres prepare detailed estamate two construct the tank. as shown in the figure.
100 mm dia
Ventilation Pipe
45 cm Manhole
Cover
Inspection
Chamber
10cmx10cm
Vent hole
8 cm
10 cm x 10 cm
Vent hole
10 cm
2.25 m
20 mm cement
plaster (1:2)
18 cm
83 cm. Thick live
conc.
Section on A-B
10 cm
45 CM dia Man
hole Cover
Plan
Fig. 5.1 Plan and Sectional Elevation of Septic Tank
100 mm
324
Building Construction & Maintenance Technician
Paper - III Estimating & Costing
325
Building Construction & Maintenance Technician
326
Traditional Size for house hold Septic Tanks
Size of the Tank
No. of
Users
L.
m
B.
m
H.
m
Cubical
content
cu m
10
1.08
0.60
0.95
1.03
20
2.30
0.60
1.30
1.79
30
2.80
0.60
1.40
2.35
50
3.40
0.70
1.60
3.81
100
4.30
0.75
1.95
6.29
H - Liquid depth. Consider 45 cm free board
5.2 Estimation of Quantity of Sanitary Pipes and pipes & Fittings
The Material used in the construction of pipes required in house drainage.
In house drainage works stone wear, asbestos cement, lead and iron pipes are
used for jointing laying and fixing of soil waste, rain water and vent pipes of
various types of fittings are required.
Fig. 5.2 Construction of pipes required in house drainage
Paper - III Estimating & Costing
327
Fitments
For Male personnel
For Female Personnel
Water closets
1 for every 25 persons.
1 for every 15 persons.
Ablution taps
1 in each W.C.
1 in each W.C
Drinking fountain
1 For every 100 persons. 1 For every 100 persons.
Urinals
Nil up to 6 persons.
Same as for male personnel
1 for 7-20 persons.
Same as for male personnel
2 for 21-45 persons.
Same as for male personnel
3 for 46-70 persons.
Same as for male personnel
4 for 71-100 persons.
Same as for male personnel
From 101 to 200 persons. Same as for male personnel
add @2.5%
1 for over 200 person
and @3%
Same as for male personnel
Wash basins
1 for every 25 persons
Same as males.
Baths. Perferably
1 on each floor
Same as males
Cleaner’s Sinks
1 per floor
Same as males
Rain Water Pipe
13
13
Inspection chambers 13
13
Short Answer Type Questions
1. Mention the sanitary fittings.
2. Write the types of joints in plumbing work.
3. Write the parts of Septic Tank.
Long Answer Type Questions
1. Prepare an estimate of Septic Tank for the following;
(a) Earth work
(b) C.C
(c) Brick Work
(d) R.C.C Slab