Quantity notes pdf

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SAQIB IMRAN 1
Assala mu alykum My Name is saqib imran and I am the
student of b.tech (civil) in sarhad univeristy of
science and technology peshawer.
I have written this notes by different websites and
some by self and prepare it for the student and also
for engineer who work on field to get some knowledge
from it.
I hope you all students may like it.
Remember me in your pray, allah bless me and all of
you friends.
If u have any confusion in this notes contact me on my
gmail id: Saqibimran43@gmail.com
or text me on 0341-7549889.
Saqib imran.

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SAQIB IMRAN 2
Methods of Measurements and
Units of Civil Construction Works
The methods and units of measurements for civil construction works are mainly
categorised for their nature, shape and size and for making payments to the
contractor. 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 cornice, fencing,
hand rail, bands of specified width etc., are expressed in running metres (RM)
c) Works consists areal surface measurements involve area like plastering,
white washing, partitions of specified thickness etc., and are expressed in
square meters (m2)
d) Works consists cubical contents which involve volume like earth work,
cement concrete, Masonry etc are expressed in Cubic metres.
Table below shows units of measurement of various items of civil
engineering works based on IS 1200.
Sl.
No.
Particulars of item
Units of
measurement
Units of
payment
1 Earthwork
1. Earthwork in excavation CUM Per CUM
2. Earthwork in filling in foundation trenches CUM Per CUM

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SAQIB IMRAN 3
3. Earth work in filling in plinth CUM Per CUM
2 Concrete
1. Lime concrete in foundation CUM Per CUM
2. Cement concrete in lintels CUM Per CUM
3. RCC in slab CUM Per CUM
4. Cement concrete or RCC in chujja, sunshade CUM Per CUM
5. Lean concrete in roof terracing (thickness specified) SQM Per SQM
6. Cement concrete bed CUM Per CUM
7. Reinforced concrete sunshade (specified width and height) CUM Per CUM
3 Damp proof course (DPC) – thickness mentioned SQM Per SQM
4 Brick work
1. Brickwork in foundation CUM Per CUM
2. Brickwork in plinth CUM Per CUM
3. Brickwork in super structure CUM Per CUM
4. Thin partition walls SQM Per SQM
5. Brickwork in arches CUM Per SQM
6. Reinforced brickwork CUM Per CUM
5 Stone work

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SAQIB IMRAN 4
1. Stone masonry CUM Per CUM
6 Wood work
1. Doors and windows frames or chaukhats, rafters, beams CUM Per CUM
2. Shutters of doors and windows (thickness specified) SQM Per SQM
3. Doors and windows fittings (like hinges, tower bolts, sliding
bolts, handles)
Each Per Each
7 Steel work
1. Steel reinforcement bars etc. in RCC and reinforced brick work Quintal Per Quintal
2. Bending, binding of steel reinforcement Quintal Per quintal
3. Rivets, bolts and nuts, anchor bolts, lewis bolts, holding down
bolts
Quintal Per quintal
4. Iron hold fasts – –
5. Iron railing (height and types specified) – –
6. Iron grills SQM Per SQM
8 Roofing
1. RCC and RB slab roof (excluding steel) CUM Per CUM
2. Lean concrete roof over and inclusive of tiles or brick or stone
slab etc. (thickness specified)
SQM Per SQM
3. Centering and shuttering formwork SQM Per SQM
4. AC sheet roofing SQM Per SQM

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SAQIB IMRAN 5
9 Plastering, points and finishing
1. Plastering – cement or lime mortar (thickness and proportion
specified)
SQM Per SQM
2. Pointing SQM Per SQM
3. White washing, color washing, cement washing, (number of
coats specified)
SQM Per SQM
4. Distempering (number of coats specified) SQM Per SQM
5. Painting, varnishing (number of coats specified) SQM Per SQM
10 Flooring
1. 25mm cement concrete over 75mm lime concrete floor
(including lean concrete)
SQM Per SQM
2. 25mm or 40mm cement concrete floor SQM Per SQM
3. Doors and window sills (CC or cement mortar plain) SQM Per SQM
11 Rain water pipe / plain pipe RM Per RM
12 Steel wooden truss Each Per each
13 Glass panels (supply) SQM Per SQM
14 Fixing of glass panels or cleaning Each Per Each
Note:
 SQM = Square meter
 CUM = Cubic meter
 RM = Running meter

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SAQIB IMRAN 6
Methods for Measurement of civil engineering
works:
The rules for measurement of each item are invariably described in IS – 1200.
However some of the general rules are listed below:
1. Measurement shall be made for finished item of work and description of each
item shall include materials, transport, labour, fabrication tools and plant and all
types of overheads for finished 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.
 Linear measurement shall be measured to the nearest 0.01m.
 Areas shall be measured to the nearest 0.01 SQM
 Cubic contents shall be worked out to the nearest 0.01 cum.
4. Same type of work under different conditions and nature shall be measured
separately under separate items.
5. The bill of quantities shall fully describe the materials, proportions,
workmanships and accurately represent the work to be executed.
6. In case of masonry (stone or brick) or structural concrete, the categories
shall be measured separately and the heights shall be described as:
 From foundation to plinth level
 From plinth to first floor level
 From first floor to second floor level and so on.

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SAQIB IMRAN 7
Units of Measurement for Payments of Civil
Construction Works
Units of Measurements for payment of construction works in civil
engineering projects are given in table below:
Sl.
No.
Civil Construction Works
Units of
measurement
Unit of
payment
1
Earthwork: Excavation, filling, cutting,
banking
m3
100 m3
2 Surface dressing m2
m2
3 Cutting of trees Number Per number
4 Stones: quarrying, blasting m3
m3
5 Concrete: PCC, RCC, Precast m3
m3
6 Jail works m2
m2
7 Damp proof course m2
m2
8 Brick work of any description m3
m3
9 Thin partition wall m2
m2
10
String course, drip course, water course
coping etc.
m m
11 Stone work of any description m3
m3
12 Stone work in wall facing (thickness specified) m2
m2

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SAQIB IMRAN 8
13 Wood work: truss, rafter, beam etc. m3
m3
14 Door, window shutters m2
m2
15 Door, window fittings Number Per number
16
Timbering, boarding, sawing of timber,
timbering of trenches, partition, etc.
m2
m2
17 Steel work Quintal Per quintal
18
Expanded metal, fabric reinforcement, cutting
angles, plates, tees
cm2
cm2
19 Threading; welding; solder of sheets cm Per cm
20 Iron gate, grill collapsible gate, rolling shutter m2
m2
21 Iron railing m m
22
Roofing: tiled, corrugated iron, caves board
(thickness specified)
m2
m2
23 Centering, shuttering m2
m2
24 Ridges; valleys; gutters (girth given) m m
25 Expansion and contraction joints m m
26 Ceiling timber, A.C. sheet, board, etc. m2
m2
27
Plastering; pointing; white washing;
distempering; painting; varnishing; polishing;
coal tarring; removing of paints
m2
m2
28 Flooring of any kind m2
m2

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SAQIB IMRAN 9
29 Pipes, laying of pipes m m
30 Dismantling of brick masonry m3
m3
31 Grouting m2
m2
32 Grouting of cracks, joints m m
33
Supply of sand; brick ballast; aggregates;
timber
m3
m3
34 Supply of cement Bag Per bag
35 Supply of steel, G.I. sheet, bare electric line Quintal Per quintal
36 Supply of GI sheet m2
m2
37 Supply of sanitary items Number Per number
38 Supply of paint, varnishes Liter Per liter
39 Supply of explosives, stiff paint Kg Per kg
Measurement of Masonry Brick
Works in Construction including
Deductions
Measurement of masonry works in construction is required for calculation of
quantities of materials in masonry and to measure completed work. Masonry
works are those where cement and fine aggregates without any coarse
aggregates are used for construction purpose.

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SAQIB IMRAN 10
There different types of masonry work:
1. Brick Masonry
2. Concrete block / brick masonry
3. Stone / rubble masonry
4. Clay tiles, etc.
Measurement of Masonry Works:
Masonry works are such as brick masonry, concrete block, stone or rubble
masonry are generally measured in volume, but are also measured in terms of
area where the thickness of masonry is limited to single brick or single block.
Tiling is measured in area.
While measuring the quantities of masonry work, the wastage of materials such
as bricks, cement or sand is not taken into account. Also, the types and classes
of bricks, blocks or tiles are not considering while measuring the quantity of
work. This is taken care in pricing of the different masonry works. Each type of
masonry works are measured separately into categories to calculate exact cost
of construction based on its price.

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SAQIB IMRAN 11
Rules for Measurement of Masonry Works:
Following are the general rules for measuring masonry works:
1. Masonry is measured as “net in place” with deductions for openings such as
doors, windows, ventilation etc.
2. Different shapes of masonry units are measured separately such rectangular,
circular etc.
3. Masonry at different heights are measured separately because masonry at
higher elevation from ground may require scaffolding and hoisting.
4. Masonry work is measured separately in the different categories such as:
a) Facings
b) Backing to facings

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SAQIB IMRAN 12
c) Walls and partitions
d) Furring to walls
e) Fire protection
5. Masonry work which requires cleaning of surface are measured in area such
as square meter or square feet.
6. Any special treatment to masonry surfaces are measured in area such as
square feet or square meter, with the count of number of coats applied.
7. Any joints in masonry structure such as expansion joints or control joints are
measured in length such as meter or feet with description of type of joint. Also
the type of joint filler material for the joint used is indicated in description.
8. For different types of mortar, or mortar with different types of admixtures are
measured separately. These are measured in volume such as cubic meter or
cubic feet.
9. If any reinforcement is used in masonry, then it is measured separately.
10. Extra items in masonry construction such as anchor bolts, sleeves, brackets,
and similar items that are built into masonry are described in the measurement
and measured separately.
11. Enumerate weep holes where they are required to be formed using plastic
inserts and such like.
12. Measure rigid insulation to masonry work in square feet or square meters,
describing the type and thickness of material.

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SAQIB IMRAN 13
Measurement of Brick Masonry
For measurement of brick masonry, general rules specified above are used
along with following points:
1. Brick masonry is measured in volume for thickness more than single
bricks. For masonry with single bricks, it is measured in square meters.
2. Facing bricks are measured separately.
3. Different types or class of brick masonry are measured separately.
Measurement of Concrete Block Masonry
Concrete block masonry are measured as general rules described above along
with following points:
1. Different types of concrete block masonry such as type and size of
concrete blocks are measured separately.
2. Special units required at corners, jambs, heads, sills, and other similar
locations are measured separately.
3. Measure concrete to core fills and bond beams in cubic feet or cubic
meters, stating the strength and type of concrete to be used. These
are measured separately for different types of concrete block based on
strength.
4. Measure in linear feet or meters reinforcing steel to core fills and bond
beams, stating the size and type of rebar to be used.
Deduction for Openings and Bearings in Masonry
Works Measurements
No deduction is made for following:
(1) Opening each up to 1000sq cm 0.1 square meter

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SAQIB IMRAN 14
(2) Ends of beams, post, rafters, etc. up to 500 sq.cm or 0.05 sq.m in section.
(3) Bed plate, wall plate, bearing of balcony (chajja) and the like up to 10 cm
depth. bearing of floor and roof slabs are not deducted from masonry.
For other openings deduction are made in following manner:
Deductions for Rectangular Opening:
Full deduction is made for rectangular openings in masonry walls
Deductions = L x H x thickness of wall
Deductions for Reinforced Concrete and Reinforced Brickwork:
Reinforced concrete and reinforced brickwork may be in roof or floor slab, in
beam lintel, column, foundation etc and the quantities are calculated in cubic
meter. Length, breadth and thickness are taken correctly from the plan,
elevation and section or from other detailed drawings.
The quantities are calculated in cubic meter exclusive of steel reinforcement and
its bending but inclusive of centering and shuttering and fixing and binding
reinforcement in position. The reinforcement including its bending is taken up
separately under steel work in as per weight.

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SAQIB IMRAN 15
For this purpose, 0.6% to 1% (usually 1%) of R.C.C. or reinforced brickwork by
volume may be taken for steel if other details are not available. The volume of
steel is not required to deducted from R.C.C. or reinforced brick work.
Reinforced concrete and reinforced brickwork may also be estimated inclusive of
steel and centering and shuttering for the complete work, if specified. Centering
and shuttering are usually included in the R.C.C or R.B. work but may also be
taken separately as per square meter of surface in contact with concrete.
Deductions for Flooring and Roofing Works:
Ground floor:
The base line concrete and floor finishing of cement concrete or stone or marble
or mosaic etc. are usually taken as one job or one item and the quantity is
calculated in square meter multiplying the length by the breadth.
The length and breadth are measured as inside dimensions from wall to wall of
superstructure. both the work of base concrete and floor finishing are paid
under one item.
Roof:
Supporting structure is taken separately in cubic meter and the lime concrete
terracing is computed in square meter with thickness specified under a separate
item including surface rendering smooth. The compacted thickness of lime
concrete terracing is 7.5 cm to 12 cm average. Lime concrete terracing may
also be calculated in cubic meter with average thickness.
Plastering:

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SAQIB IMRAN 16
Plastering usually 12mm thick is calculated in square meter. for wall the
measurement are taken for whole face of the wall for both side as solid, and
deducting for opening are made in following manner.
(1) No deduction is made for the end of beams, posts, rafters etc.
(2) For small opening up to 0.5 square meter no deduction is made, and at the
time no additions are made for jambs, soffits and sills of these opening.
(3) For opening exceeding 0.5 square meter but not exceeding 3 square meter
deduction is made for one side and other face is allowed for jambs soffits and
sills which are not taken into account separately
(4) For opening above 3 square meter deduction is made for both faces of the
opening and the jambs, soffits and sills are into account and added.
Pointing:
Pointing in wall calculated in square meter for whole surface and deduction
similar to plastering is made.
Cornice:
Ornamental and large cornice is measured in running meter or running foots for
the complete work which includes masonry, plastering, moldings etc. and paid
for in running meter.
Similarly, string course, corbelling, coping etc are measured and paid for in
running meters for the complete work.
Pillars:

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SAQIB IMRAN 17
Pillars are taken separately in cubic meter foe their net volume and quantities
are calculated.
Wood Works:
Wooden beams, posts, wooden roof trusses, doorpost etc. come under this
items and the quantities are computed in cubic meter. The dimension of finished
work shall be taken.
Iron Works:
This computed weight in kg or quintal and the quantities are calculated correctly
by multiplying the weights per running meters by length. the weight per running
meter can be obtained from the steel section book. For steel joint the length is
equal to the clear span plus two bearing the bearing may taken third forth
thickness of wall or 20 to30 cm.
White Wash or Color Washing or Distempering:
The quantities are computed in square meter and are usually same as for
plastering. The inside is usually white washed or distempered and this item will
be same as for outside plaster. These item need not be calculated separately
but simply written as same as for outside plaster and inside plaster.
Brick calculator | Brickwork calculation and
brick estimation
November 22, 2017 by admin 2 Comments

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SAQIB IMRAN 18
In this article i will tell you how to calculate bricks in a wall. By the explanation describe
below, you can make your own MS Excel Brick calculator program to calculate bricks in
a wall.
1. We know that standard dimensions of brick is 9 inch x 3 inch x 4.5 inch.
2. Convert these dimensions into feet.
3. Now multiply these dimensions to get volume of a brick.
4. 0.75 feet x 0.25 feet x 0.375 feet = 0.0703125 cubic feet.
5. Bricks required for 1 cubic feet brickwork will be (1/0.070) = 14.285 bricks
6. 10% space of brickwork is covered by mortar.
7. Subtract 10% bricks from 14.285, 14.285-1.4285=12.85 bricks.
8. Add 5% wastage of bricks.
9. 5% of 12.85 is 0.64 bricks.
10. Add 0.64 and 12.85 to get number of bricks in 1 cubic feet, we get 0.64 + 12.85
=13.492 bricks or we can say that 13.5 bricks.
Number of bricks in 100 Cubic feet brickwork
As we know that there are 13.5 bricks in 1 cubic feet. So in 100 cubic feet there will be
1350 bricks.

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SAQIB IMRAN 19
Bricks per square foot
Wall thickness Bricks per square foot
4.5 inch 5.0625 bricks
9 inch 10.125 bricks
Bricks per square meter
Wall thickness Bricks per square meter
9 inch 114.329 bricks
How to calculate bricks in a wall ?
To calculate bricks in a wall, we need to know the dimensions of the wall. For instance,
if a wall is 10 feet long with 10 feet height and 9 inch thickness. Then we can find its
bricks in a following way. Multiply dimensions of wall. 10 feet x 10 feet x 0.75 feet. We
will get 75 cubic feet. We have seen above that there are 13.5 bricks in 1 cubic feet
brickwork. So there will be 75 x 13.5 =1012.5 bricks in 75 cubic feet. In this way we
calculate bricks in any wall with the known dimensions.
Bricks calculation formula
Bricks calculation formula is written below.
In feet
 Length of wall in feet x height of wall in feet x thickness of wall in feet x 13.5 =
number of bricks

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SAQIB IMRAN 20
In meter
 length of wall in meter x height of wall in meter x thickness of wall in meter x 500 =
number of bricks
Number of bricks in 1 Cubic meter brickwork
Standard dimensions of brick in metric units are 225 x 112.5 x 75 mm.
1. multiply these dimensions to get volume of a brick, 0.225 m x 0.1125 m x 0.075
m=0.00189 cubic meter.
2. In one cubic meter, number of bricks will be (1/0.00189)=529.1 bricks.
3. 10% of brickwork will be covered by mortar.
4. Subtract 10% bricks from the 529.1, we will get 529.1-52.91=476.20 bricks.
5. Add 5% wastage of bricks.
6. 5% of 476.20 is 23.80 bricks.
7. Add 23.80 and 476.20 to get number of bricks in one cubic meter brickwork.
8. 23.80+476.20=500 bricks.

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SAQIB IMRAN 21
Standard conversion factors
INCH = 25.4 MILLIMETRE
FOOT = 0.3048 METRE
YARD = 0.9144 METRE
MILE = 1.6093 KILOMETER
ACRE = 0.4047 HECTARE
POUND = 0.4536 KILOGRAM
DEGREE FARENHEIT X 5/9 – 32 = DEGREE CELSIUS
MILLIMETRE= 0.0394 INCH
METRE = 3.2808FOOT
METRE = 1.0936YARD
1) MILD STEEL (MS)
SHEET
WEIGHT (KGS) = LENGTH (MM) X WIDTH (MM) X 0. 00000785 X THICKNESS
example – The weight of MS Sheet of 1mm thickness and size 1250 MM X 2500 MM shall be
2500MM X 1250 MM X 0.00000785 X 1 = 24.53 KGS/ SHEET
ROLLED STEEL CHANNELS
MS SQUARE
WEIGHT (KGS ) = WIDTH X WIDTH X 0.00000785 X LENGTH.
Example : A Square of size 25mm and length 1 metre then the weight shall be.
25x25X 0.00000785 X 1000mm = 4.90 kgs/metre
MS ROUND
WEIGHT (KGS ) = 3.14 X 0.00000785 X ((diameter / 2)X( diameter / 2)) X LENGTH.
Example : A Round of 20mm diameter and length 1 metre then the weight shall be.
3.14 X 0.00000785 X ((20/2) X ( 20/2)) X 1000 mm = 2.46 kgs / metre
SS ROUND
DIA (mm) X DIA (mm) X 0.00623 = WEIGHT PER METRE
SS / MS Pipe

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SAQIB IMRAN 22
OD ( mm) – W.Tthick(mm) X W.Thick (mm) X 0.0248 = Weight Per Metre
OD ( mm) – W.Tthick(mm) X W.Thick (mm) X 0.00756 = Weight Per Foot
SS / MS CIRCLE
DIA(mm) X DIA (mm) X THICK(mm) 0.0000063 = Kg Per Piece
SS sheet
Length (Mtr) X Width (Mtr) X Thick(mm) X 8 = Weight Per Piece
Length (ft) X Width (ft) X Thick(inch) X 3 /4 = Weight Per Piece
S.S HEXAGONAL BAR
DIA (mm) X DIA (mm) X 0.00680 = WT. PER Mtr
Dia (mm) X Dia (mm) X 0.002072 = Wt. Per foot.
BRASS SHEET
WEIGHT (KGS) = LENGTH (MM) X BREADTH (MM) X 0. 0000085 X THICKNESS
Example – The weight of brass sheet of thickness 1 mm, length 1220mm and breadth 355mm shall be
1220 X355X 0.0000085 X 1 = 3.68 Kgs/Sheet
Following table shows how can we convert various most commonly used units from one unit system
to another.
Units to convert Value
Square foot to Square meter 1 ft² = 0.092903 m²
Foot per second squared to Meter per second
squared
1 ft² = 0. 3048 m²
Cubic foot to Cubic meter 1 ft³ = 0.028316 m³
Pound per cubic inch to Kilogram per cubic
meter
1 lb/in³ = 27679.9 047102 kg/m³
Gallon per minute = Liter per second 1 Gallon per minute = 0.0631 Liter per second
Pound per square inch = Kilopascal 1 Psi (Pound Per Square Inch) = 6.894757 Kpa
(Kilopascal)
Pound force = Newton 1 Pound force = 4.448222 Newton
Pound per Square Foot to Pascal 1 lbf/ft2 = 47.88025 Pascal
Acre foot per day = Cubic meter per second 1 Acre foot per day= 1428 (m3/s)
Acre to square meter 1 acre = 4046.856 m²
Cubic foot per second = Cubic meter per
second
1 ft³/s = 0.028316847 m³/s
Measurement units and standards are different in different countries but to maintain a standard, SI
units are mostly used when dealing with projects involving different countries or even different states.

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Small projects can be done with the locally used unit system but when the project is big, one standard
unit system is to be used.
Two most common system used in the United States are
 United States Customary System (USCS)
System International (SI)
But the SI unit system is more widely used all over the world. Following is the table which shows how
you can convert USCS measurements in SI measurements. ( Just multiply the USCS amount with the
corresponding figure given in table below
Convert USCS into SI Units
USCS unit X Factor = SI unit SI symbol
Square foot X 0.0929 = Square meter M2
Cubic foot X 0.2831 = Cubic meter M3
Pound per square inch X 6.894 = Kilopascal KPa
Pound force X 4.448 = Newton Nu
Foot pound torque X 1.356 = Newton meter N-m
Kip foot X 1.355 = Kilonewton meter LN-m
Gallon per minute X 0.06309 = Liter per second L/s
Kip per square inch X 6.89 = Megapascal MPa
BASIC STANDARDS:
1 inch = 25.4 millimeters = 2.54cm.
1 meter = 39.37 inches =1.09 yards.
1 liter = 0.22 galls (imp.)
1 gallon (imp.) = 4.546 liters.
1 gallon (US) = 3.785 liters.
1 Kilogram (kg) = 2.2046 pounds (lb).
METRIC UNIT OF WEIGHT/MASS:
1 tonne = 1000 kilograms = 1,000,000 grams.
1 quintal = 100 kilograms = 100,000 grams.
1 Slug = 14.606 kg
1 Slug = 32.2 lb
MEASUREMENTS OF LENGTH:
1 foot = 12 inches.
1 yard = 3 feet.
1 furlong = 220 yards.

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1 mile = 8 fulongs.
1 Kilometer (Km) = 1000 meters.
1 meter (m) = 100 centimeters (cm).
1 cm = 10 millimeter (mm).
METRIC UNITS FOR LIQUID MEASUREMENTS:
1 liter = 1000ml.
1 liter = 1kg.
1 kiloliter (1000 liters) = 1 cubic meter/ 1 cu.m/ 1 m3.
CONVERSION FACTORS:
1 cu. ft. = 28.317 liters.
1 cu. ft. = 0.028 cu. meters.
1 cu. ft. = 6.24 galls (imperial).
1 cu. ft. = 7.48 galls (US).
1 imp. gall = 1.20 galls (US), liquid.
1 imp. gall = 1.03 galls (US), dry.
MEASUREMENTS OF AREA:
1 Acre = 43560 sq. ft
1 Acre = 4046.46 sq. m
1 Acre = 8 Kanals.
1 Kanal = 20 Marlas.
1 Marla = 225 sq. ft (* in some regions 272 sq. ft)
1 Marla = 15.50 sq.m
MISCELLENIUS CONVERSION FACTORS:
1 cu.m = 35.32 cu.ft.
1 Pound = 4.448 Newton (Force).
1 klb = 4.448 kN.
1 Psi (lb/sq.in) = 6.689 Pascal (N/sq.m)
1 (lb/sq. ft) = 0.048 (kN/sq.m)
CALCULATION OF UNIT WEIGHT OF STEEL BARS:

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SAQIB IMRAN 25
After estimating it is very important to know the unit weight of steel bars because, we estimate as 100
meter 20mm ø bar or 100 feet 16mm ø bar, etc (ø is the symbol of diameter).
But steel bar suppliers will not understand this notation, they measure the steel bars in weight. So we
have to order them in kg or quintal or ton. In this article, we will discuss how to calculate unit weight of
steel bars of different diameter.
The formula is W = D²L/162
Where
W = Weight of steel bars.
D = Diameter of steel bars in mm.
L = Length of bars in meter.
Example 1: Calculate the weight of 60 meters long 12 mm ø bar.
Here, D = 12 mm.
L= 60 m.
We know that, W = D²L/162
W = 12² x 60/162 = 53 kg
Weight of 60 m 12mm ø bar is 53 kg.
Let’s look for another example.
Example 2: Calculate the weight of 100 m 16 mm ø bar.
Here, D = 16 mm.
L = 100 m.
W = 16² x 100/162 = 158 kg.
If we put 1 meter length for each diameter of steel bar in the formula then we will get the unit weight.
 10mm ø bar = 10² x 1/162 = 0.617 kg/m
 12mm ø bar = 12² x 1/162 = 0.888 kg/m
 16mm ø bar = 16² x 1/162 = 1.580 kg/m
 20mm ø bar = 20² x 1/162 = 2.469 kg/m
If we multiply the length of bars with this unit weight we will get the total weight of steel bars.
For example, total weight of 1000 meter long 20mm ø steel bar is,
1000 x 2.469 = 2469 kg.
Using the same method we can calculate the unit weight of different steel bars.
Here I have calculated in meter but we can also calculate in foot. To calculate in foot we have to use the
following formula:
W= D²L/533
Where D = Diameter of bars in mm.
L = Length of bars in foot.
HOW TO CALCULATE STEEL QUANTITY FOR RCC BEAM, COLUMN AND SLAB
Following are the steps to calculate the quantity of steel for RCC slab
1. Prepare a bar bending schedule in order to classify different shapes of bars (bent up bar, straight
anchor bar, eos bar, curtail bar etc) and diameters.
2. List down all the shapes of bars from the drawing.
3. Count the number of bars of each of those shapes.
4. Then calculate the cutting length of each of those bars.

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Cutting length of bar = (length of the member – deduction for cover on both sides) + development
length.
5. Then calculate the unit weight of each dia bars by the following formula:
W = d^2/162
Where d is the dia in mm and weight (w) in kg.
6. Then calculate weight of rebar
Weight of rebar= no of bars x cutting length x unit weight
7. Add all the weight to get the total steel quantity.
NOTE – 1. Minimum % of steel as per Indian standard are:
1. Beam (Teinsion reinforcement):
As = 0.85bd/fy of gross cross-section area.
2. Slab – 0.12% of total area
3. Column – 0.8% of cs area
2. Maximum % steel as per Indian standard are:
Beam – 4 % of cross-section area.
Slab – 4% of cross-section area.
column — 6% of cross-section area.
3. . Development Length is usually specified in the drawings, but if not thenyou can calculate it as,
D.L = Depth – 2 times cover
4. Binding wire = 10 grams per kg of reinforcement.
5. No. of stirrups = (length of member – 2 x cover ) / spacing + 1
WHAT IS BAR BENDING SCHEDULE?
Bar bending schedule commonly known as BBS is one of the most important terms in Civil Engineering.
Because it plays a vital role in building construction.
Like other building materials estimation of steel is also required for constructing a building and here BBS
comes with an easy solution. Bar bending schedule provides the reinforcement calculation and some
other important details such as bar mark, bar diameter, bar shape, cutting length, number of bars, the
weight of bar, total weight of steel etc. So that we can order the required amount of steel in advance.
HISTORY OF BAR BENDING SCHEDULE:
Long years back when there was no today’s bar bending schedule, certain recommendations given by
Prof. BN Dutta were used for estimating steel for different components of a building.

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But these are now backdated (not wrong) and we don’t use them in today. Because No accurate values
were specified if we use more bars in a single member. Nowadays we are constructing 150+ floors
building. It is now possible with our modern techniques, equipment etc. Estimation of steel becomes
easier because of BBS.
Civil engineers who are working on site or freshers who are going to work at construction sites must
know how to prepare bar bending schedule. In this article, I will discuss some important things used in
preparing bar bending schedule.
These are just the basic things. In future, I will come up with some other articles related to BBS such as
BBS for beams, columns, slabs, footings etc.
HOOK LENGTH:
The hook is the extra length left at the 4th corner of a stirrup so that the stirrup retains its shape.
Generally, hook length is taken as 9d for one side.
Where d = Diameter of the bar.
The total length of stirrups = Total length of the bar + 2 x hook length (for two hooks)
= L + 2 x 9d
= L + 18d.
Where L = length of the bar for stirrup.
BEND LENGTH:
The bar is bent at the column end to tie with the footings. This extra length for bend is called bend
length.
Bend length is generally considered as 16 d
Bend Length = 16d

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DEVELOPMENT LENGTH:
Development length is the length of bar required for transferring the stress into concrete.
In simple words, the quantity of the rebar length that is actually required to be embedded into the
concrete to create the desired bond strength between steel and concrete and furthermore to produce
required stress for the steel in that area.
The formula for development is given below:
Development length (Ld) = d x σs/τbd
Where
d = Diameter of the bar.
σs = Stress in the bar at the section considered as design load
τbd = Design bond stress.

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LAP LENGTH:
Lap length is the overlapping length of two bars side by side which gives required design length. In RCC
structure if the length of a bar is not sufficiently available to make design length, lapping is done.
Suppose we need to build a 20 m tall building. But is there any 20 m bar available in the market? No, the
maximum length of rebar is usually 12 m, so we need to join two bars to get 20 m bar.
Lap length for tension members = 40d
Lap length for compression members = 50d.
d = Diameter of bars.

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Look at the image below. You may be seen this on the terrace (top floor) of buildings. This extra rebar is
left for future construction purpose. I hope you understand this.
CRANK LENGTH:
Generally, bars are bent near the support at an angle of 45°. The angle of bend may also be 30° in
shallow beams. The purpose of bend near the support is firstly to resist the negative bending moment
which occurs in the region of the support and secondly to resist the shear force which is greater at the
support.

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Crank bars are mostly provided in slabs.
Crank length = D/sin45° – D/tan45° =1.42D – D = 0.42D
So Crank Length = 0.42D
Where D = Clear height of the bar = Thickness of slab – (Top cover + Bottom cover) – Diameter of the bar
UNIT WEIGHT OF STEEL:
The weight of bar is calculated by the following formula
W = d²L/162
Where W = Weight of bars.
L = Length of bars in meter.
Example: Calculate the weight of 20 meters long 16 mm ø bar
W = 16² x 20/162 = 32 kg.
CUTTING LENGTH OF BENT UP BARS IN SLAB:
As a site engineer, you need to calculate the cutting length of bars according to the slab dimensions and
give instructions to the bar benders.
For small area of construction, you can hand over the reinforcement detailing to the bar benders. They
will take care of cutting length. But beware, that must not be accurate. Because they do not give
importance to the bends and cranks. They may give some extra inches to the bars for the bends which
are totally wrong. So it is always recommended that as a site engineer calculate the cutting length
yourself. In this article, we will discuss how to the calculate length for reinforcement bars of slab. Let’s
start with an example.
EXAMPLE:
Where,

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Diameter of the bar = 12 mm
Clear Cover = 25 mm
Clear Span (L) = 8000
Slab Thickness = 200 mm
Development Length(Ld) = 40d
CALCULATION:
Cutting Length = Clear Span of Slab + (2 x Development Length) + (2 x inclined length) – (45° bend x 4) –
(90° bend x 2)
Inclined length = D/(sin 45°) – dD/ (tan 45°) = (D/0.7071) – (D/1)= (1D – 0.7071D)/0.7071= 0.42 D
As you can see there are four 45°bends at the inner side (1,2,3 & 4) and two 90° bends ( a,b ).
45° = 1d ; 90° = 2d
Cutting Length = Clear Span of Slab + (2 X Ld) +(2 x 0.42D) – (1d x 4) – (2d x 2) [BBS Shape Codes]
Where,
Ld = Development length of bar.
D = Height of the bend bar.
In the above formula, all values are known except ‘D’.
So we need to find out the value of “D”.
D = Slab Thickness – (2 x clear cover) – (diameter of bar)
= 200 – (2 × 25) – 12
= 138 mm
Now, putting all values in the formula
Cutting Length = Clear Span of Slab + (2 x Ld) +(2 x 0.42D) – (1d x 4) – (2d x 2)
= 8000 + (2 x 40 x 12) +(2 x 0.42 x 138) – (1 x 12 x 4) – (2 x 12 x 2)
∴ Cutting Length = 8980 mm or 8.98 m.
So for the above dimension, you need to cut the main bars 8.98 m in length.
BAR BENDING SCHEDULE OF LINTEL BEAM:
In this article, I will discuss how to prepare BBS of RCC Lintel Beam.

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1. Calculate Total Length Of Main Bars:
Length of 1 bar = Length of lintel – clear cover for both sides
= 2500 – 2 x 25 [Clear cover for both sides]
=2450 mm
= 2.4 m.
Length of 4 bars = 2.4 x 4 = 9.6 m
2. Calculate Weight Of Steel For Main Bars:
Weight of steel for 12 mm bar = D²L/162 = 12² x 9.6/162 = 8.53 kg.
3. Calculate No Of Stirrups:
No of stirrups = (Total length of lintel/c/c distance between strriups) + 1
= (2500/150) + 1 = 18
4. Calculate Total Length Of Stirrups:
Inner depth distance = 150 -25 -25 -8 =84 mm
Width distance = 150 – 25 – 25 -8 = 84 mm.
Cutting length of stirrups =(2x Inner deoth diatance) +(2xWidth depth) + Hooks Length – Bend
Hooks length = 10d
Bend = 2d
We have 2 hooks and 5 bend
So,
Cutting length of stirrups = (2×84) + (2 x84) +2x10x8 -5x2x8 = 418 mm = 0.418 m
Total length of stirrups = 0.418 x 18 = 7.54 m
5. Calculate Weight Of Steel For Stirrups:
Weight of steel for stirrups = D²L/162 = 8² x 7.54/162 = 7.61 kg.
Total weight of steel for lintel = 8.53 + 7.61 = 16.14 kg.

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THE FORMULA D²L/162 FOR CALCULATING WEIGHT OF STEEL BARS FULLY DERIVED
THE FORMULA D²L/162 FULLY DERIVED:
To calculate the weight of steel bars you must use the formula D²L/162, but do you know where the
formula D²L/162 came from? Okay, no problem if you don’t know. In this article, I will share how to
derive the formula D²L/162. So let’s start from the beginning.
The formula is D²L/162
Where,
D = Diameter of steel bar in millimeter
L = Length of steel bars in meter
Calculation Of The Formula D²L/162:
We know that,
Weight= Cross sectional area x Length x Density
For steel bar, this also remains same.
The weight of steel bars= Cross sectional area of steel bar x Length of steel bar x Density of steel bar.
That means,
W = A x L x ρ
Where,
W = Weight of steel bars
A = Area = πD²/4
π (pi) = 3.14
D = Diameter of steel bar in millimeter
L = Length of steel bar in meter
ρ (Rho) = Density of steel bar = 7850 kg/m³
Therefore,
W = 3.14 x D²/4 x L x 7850
But there is two conflicting unit in the formula. Which is millimeter for D and meter for ρ (Rho).
So we need to convert either D or ρ to the same unit.
Let’s change the unit of D from millimeter to meter.
1 millimeter = 0.001 meter
Let’s put this into the formula,
W= 3.14 x {(D² x 0.001 x 0.001)/4} x L x 7850
= D²L/162
Hope you all understood this calculation. Using this formula we can easily calculate the weight of steel
bars.
10 + THUMB RULES FOR CONCRETE MIX DESIGN

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FOR ADDING 4 LITERS OF WATER IN 1 CU.M FRESHLY MIXED CONCRETE
1. The slump value will be increased by 25 mm.
2. The compressive strength of concrete will be decreased by 1.5 to 2.0 N/mm2
3. The shrinkage potential will be increased by 10%.
4. 1/4 bag of cement will be wasted.
IF THE TEMPERATURE OF FRESHLY MIXED CONCRETE IS INCREASED BY 1%, THEN
1. 4 liters of water per cu.m will give equal slump.
2. The air content will be decreased by 1%.
3. The compressive strength of concrete will be decreased by 1.0 to 1.5 N/mm2.
IF THE AIR CONTENT OF FRESHLY MIXED CONCRETE IS
1. Increased by 1% then the compressive strength will be decreased by 5 %.
2. Decreased by 1%, yield will be decreased by 0.03 cu.m per 1 cu.m.
3. Decreased by 1%, then the slump value will be decreased about 12.5 mm.
4. Decreased by 1%, then the durability of the concrete will be reduced by 10%.
CALCULATION OF MATERIALS FOR DIFFERENT MIX RATIO:
Quantity estimation of materials is essentially required in any construction works and quantity of
materials depend on the mix proportions of the concrete. In our previous article, we have already
discussed how to calculate bricks in a wall. Today we will discuss how to calculate quantities of materials
for different mix ratio of concrete. (Dry mix method)
We will calculate quantities of materials for 1 m3 concrete (By volume).
Let us assume the mix proportion is 1 : 2: 4 (cement:sand:stone =a:b:c)
Volume of wet concrete = 1 m3
Volume of dry concrete = 1 × 1.54 = 1.54 m3
What Is 1.54 Given Below:

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SAQIB IMRAN 36
CALCULATION FOR CEMENT:
Formula, Cement = (Volume of dry concrete/a+b+c) × a
= (1.54/a+b+c) × a = [(1.54/1+2+4)] × 1 = 0.22 cum
Now density of cement = 1440 kg/cum
∴ Volume of cement = 0.22 × 1440 =316.8 kg.
As we know, 1 bag of cement contains 50 kg of cement .
∴ Cement bags required = 316.8/50 = 6.33 bags.
CALCULATION FOR SAND:
Formula, Sand = (Volume of dry concrete/a+b+c) × b
= (1.54/a+b+c) × b = (1.54/1+2+4) × 2 = 0.44 cum.
CALCULATION FOR STONE CHIPS:
Formula, Stone Chips = (Volume of dry concrete/a+b+c) × c
= (1.54/a+b+c) ×c = (1.54/1+2+4) × 4 = 0.88 cum.
CALCULATION FOR WATER CONTENT:
Let us assume the water/cement ratio of the concrete is 0.45.
w/c = 0.45
Required water for 1 bag cement = 0.45 × 0.0353 = 0.0159 cum.
Where volume of 50 kg cement = 0.0353 cum
1 m3 water = 1000 Litre
Required water for 1 bag cement=0.0159 × 1000 = 15.9 Litre.
∴ Required water for 6.33 bags cement = 6.33 × 15.9 = 101 Litre.
SUMMARY:
Cement = 6.33 bags.
Sand = 0.44 cum
Stone chips = 0.88 cum
Water = 101 litre.
Note: Yield of concrete is considered as 67% & Wastages of materials = 2%
You can use the same formula for different mix proportions such as 1:1.5:3 etc.
Here, we have used cubic meter unit but you can also calculate in cubic feet unit.
UNIT WEIGHT OF BUILDING MATERIALS
Unit Weight Of Building Materials:

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There are many building materials used in construction. In the below table we have tried to cover the
unit weight of building materials which are most commonly used at the construction site.
Material Unit Weight
Water 1000 Kg/ m3
Bricks (broken) 1420 Kg/ m3
Bricks(common) 1600 Kg/ m3
Cement(ordinary) 1440 Kg/ m3
Cement (rapid hardening) 1250 Kg/ m3
Cement Mortar 2000 Kg/ m3
Cement Concrete (Plain) 2400 Kg/ m3
Cement Concrete (Reinforced) 2500 Kg/ m3
Glass 2500 Kg/ m3
Lime Concrete 1900 Kg/ m3
Cement Plaster 2000 Kg/ m3
Lime Plaster 1700 Kg/ m3
Stones (Ballast) 1720 Kg/ m3
Stones (Aggregates) 1750 Kg/ m3
Stones (Basalt) 2850 Kg/ m3
Stones (Granite) 2450 Kg/ m3
Stones (Marble) 2650 Kg/ m3
Timber (Oak, Sal) 510 Kg/ m3
Timber (Mango) 650 Kg/ m3
Timber (Teak) 625 Kg/ m3
Coal 600 Kg/ m3
Plastics 1250 Kg/ m3
Oils 800 Kg/ m3
Ashes 650 Kg/ m3

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SAQIB IMRAN 38
Clinker 750 Kg/ m3
Rubber 1300 Kg/ m3
Slag 1500 Kg/ m3
Clay Soil 1900 Kg/ m3
Sand (dry) 1540 to 1600 Kg/ m3
Sand (wet) 11760 to 2000 Kg/ m3
Steel 7850 Kg/ m3
Chalk 2100 Kg/ m3
Bitumen 1040 Kg/ m3
CALCULATION OF BRICKS IN A WALL
Estimation of building materials is essentially required before starting a new construction project. In our
previous article, we have already discussed how to calculate cement, sand, and aggregates. Today we
will discuss the easiest method to calculate the required number of bricks in a wall. So let’s start from
the beginning.
REQUIRED DATA:
1. Volume of the wall.
2. Volume of a standard brick.
3. Details of openings in the wall.
PROCEDURE:
1. First, calculate the volume of the wall to be built:
Let us assume,
The length of the wall (l) = 10 foot.
The height of the wall (h) = 10 foot.
Thickness of the wall (b) = 1 foot.
Calculate the volume of the wall by multiplying the length, height, and thickness.

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SAQIB IMRAN 39
∴ Volume of the wall = l×h×b = 10×10×1 =100 Cu.F
2. Calculate the volume of the brick:
The standard size of a brick (IS Standard) is 190 mm ×90 mm ×90 mm and
with the mortar joint, it becomes 200mm × 100 mm× 100 mm.
l = 200 mm= 0.656168 ft.
b = 100 mm =0.328084 ft.
h = 100 mm = 0.328084 ft.
∴ Volume of the brick = l× b×h = 0.656168× 0.328084× 0.328084 = 0.0706 Cu.F
3. To find out the total nos of brick divide the volume of wall by the volume of the brick.
∴ No. of bricks required = 100/0.0706 = 1416 no. of bricks.
Note: 1. Consider 10% wastages of bricks.
2. If the wall has any openings such as doors, windows etc subtract the volume of the openings from the
volume of wall and then divide it by the volume of brick.
HOW TO CALCULATE NUMBER OF CONCRETE BLOCKS IN A WALL
Concrete Block Wall
NUMBER OF CONCRETE BLOCKS IN A WALL:
There are several methods for calculating number of blocks in a wall. In this article, I will explain two
simple methods to find out total no. of blocks for a wall.
 Surface area method.
 Volume method.
PROCEDURE:
1. CALCULATION BY SURFACE AREA METHOD:
REQUIRED DATA:

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SAQIB IMRAN 40
1. Surface area of the wall.
2. Surface area of a standard concrete block.
3. Surface area of openings in the wall.
STEP 1:
Determine The Length & Height Of The Wall:
Let, the Length of the wall = 10 feet.
The height of the wall = 10 feet.
STEP 2:
Calculate The Surface Area Of The Wall:
We know, Surface Area = Length x Height
∴ Surface area of the wall = 10 x 10 = 100 sq. feet.
STEP 3:
Calculate The Surface Area Of Openings:
If the wall has any openings such as door, windows etc, calculate the surface area of openings.
Let the wall has a window of 3′ x 3′
∴ Surface area of the window = 3 x 3 =9 sq. ft
STEP 4:
Substract The Surface Area Of Openings From Surface Area Of Walls:
∴ Final Surface Area of wall = 100 – 9 = 91 sq.ft.
STEP 5:
Calculate The Surface Area Of A Concrete Block With Mortar Allowance:
Let the block is 16″ x 8″ x 8″ and mortar allowance is 1″
∴ Surface Area of one block with mortar = (16+1) x (8+1)= 153 inch²
= 153/12×12 = 1.0625 sq.ft
STEP 6:
Divide The Total Surface Area Of The Wall By Surface Area Of One Block:
∴ Number of blocks = Surface Area of wall/Surface Area of a block = 91/1.0625 = 86

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SAQIB IMRAN 41
Consider 5% wastage of concrete blocks.
So the total numbers of blocks required = 86 + (86 x 5/100) = 86 + 4 = 90.
2. CALCULATION BY VOLUME METHOD:
REQUIRED DATA:
1. Volume of the wall.
2. Volume of a standard concrete block.
3. Volume of openings in the wall.
STEP 1:
Determine The Width, Height, And Thickness Of The Wall:
Let, the width of the wall = 10 feet.
The thickness of the wall = 8″ = 0.67 feet. [ Thickness of wall will be same as the thickness of one
concrete block]
STEP 2:
Calculate The Volume Of The Wall:
We know, Volume = Width x Height x Thickness
∴ Volume of the wall = 10 x 10 x 0.67 = 67 cubic feet.
STEP 3:
Calculate The Volume Of Openings:
If the wall has any openings such as door, windows etc, calculate the volume of openings
Let the wall has a window of 3′ x 3′ x 0.67′ [ Thickness of the window will be same as the thickness of the
wall]
∴ Volume of the window = 3 x 3 x 0.67 =6 cu. ft
STEP 4:
Substract The Volume Of Openings From Volume Of Walls:
∴ Final volume of wall = 67 – 6 = 61 cu.ft.
STEP 5:
Calculate The Volume Of A Concrete Block With Mortar Allowance:
Let the block is 16″ x 8″ x 8″ and mortar allowance is 1″
∴ Volume of one block with mortar = (16+1) x (8+1) x 8 = 1224 inch3 [The mortal will be given on the
upper side and one side of every concrete block]
= 1224/12x12x12 = 0.70 cu.ft
STEP 6:
Divide The Total Volume Of The Wall By Volume Of One Block:
∴ Number of blocks = Volume of wall/Volume of a block = 61/0.70 = 86
Consider 5% wastage of concrete blocks.
So the total numbers of blocks required = 86 + (86 x 5/100) = 86 + 4 = 90.
Note:
If the wall does not have any opening skip step 4, and just divide the surface area of wall by the surface
area of one block.
HOW TO CALCULATE QUANTITY OF MATERIALS IN ARCH
In this article, I will discuss how two calculate quantity of materials in arch.

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SAQIB IMRAN 42
1. SEMI CIRCULAR ARCH:
Given:
Span = 8 ft
Height = 4 ft
Thickness = 1 ft
Breadth = 1ft
To calculate the quantity of materials in semi circular arch we need to calculate the volume of arch.
Volume of arch = Cross-section of arch x Arch length
We know Cross-section area = Thickness of arch x Breadth of arch = T x B
From the above image, you can see the thickness and breadth are 1 ft.
Cross-section of arch = 1 x 1 = 1 ft²
Now Arch length = Circumference of half circle (The arch is semi circular)
We know circumference of circle = 2πR
Circumference of half circle = 2πR/2 = πR
We don’t what is R. So we need to find out the ‘R’.
R = Half span + Thickness/2 = 4 +1/2 = 4.5 ft.
Arch length = πR =3.14 x 4.5 = 14.13 ft.
Quantity of materials in arch = 1 x 14.13 = 14.13 cft.
2. SEGMENTAL ARCH:
Given:
Span = 8 ft
Height = 2ft
Thickness = 1 ft
Breadth = 1ft
Central angle = 105°

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SAQIB IMRAN 43
SOLUTION:
Volume arch = Cross-section of arch x Arch length
We know Cross-section area = Thickness of arch x Breadth of arch = T x B
Arch length = θπR/180
Now R = (a² + h²)/2h [a =half span = 8/2 =4]
=(4² + 3²)/2 x 3 = 25/6 =4.16 ft.
Arch length = (105 x 3.14 x 4.16)/180 = 7.61
Quantity of materials in arch = 1 x 7.61 = 7.61 cft.
HOW TO CALCULATE SHUTTERING AREA FOR CONSTRUCTION
Shuttering or formwork is a temporary, rigid structure in which the cast in situ concrete is poured for
casting the members to required shape. Different formworks are used for different shape members.

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In this article, we will discuss how to calculate the shuttering area before construction.
Before calculating the shattering area first, we need to know how to calculate the peripheral length of
any shape.
PERIPHERAL LENGTH/PERIMETER:
We know that perimeter is the distance around a 2D (two dimensional) shape.
For example, square has four sides. Let ‘s’ be the length of one side, then
∴ The peripheral length = s+s+s+s = 4s
Rectangle has four sides also but the length and breadth is different, let one side length is ‘l’ and
breadth is ‘b’
∴ Peripheral length = l+b+l+b = 2l+2b
Now coming to the calculation of shuttering area.
Shttering area = Pheripheral length * Depth
Let me give you an example
Pheripheral length = (2 * 1) + (2*0.8) = 2 +1.6 =3.6
∴ Total shuttering area = 3.6 * 4 = 14.4 sq.m
HOW TO CALCULATE CEMENT BAGS IN 1 CUBIC METER
PROCEDURE TO CALCULATE CEMENT BAGS IN 1 CUBIC METER:
Let us consider the nominal mix is 1:2:4
Loss of cement is considered as 2%
Output of mix is considered as 67%
To achieve 1 cum output, we need 1/0.67 = 1.49 say 1.50 cum dry mix.
Now add the wastage of 2%, i.e (1.50 + 0.02) = 1.52 cum.
Volume of cement = (cement/cement+sand+aggregate) × Total material
=(1/1+2+4) × 1.52
=0.2171 cum

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SAQIB IMRAN 45
As we know, the density of cement is 1440 kg/cum and
Weight of 1 bag cement = 50 kg.
Therefore volume of 1 bag cement = 50/1440
=0.0347 cum.
∴ No. of cement bags required in 1 cubic meter = 0.2171/0.0347
= 6.25 bags.
Note: You can use the same formula for calculating cement for other nominal mixes.
QUANTITY OF CEMENT, SAND, & AGGREGATES FOR 1000 SQ FT SLAB
To calculate the quantity of cement, sand, and aggregates required for a flat slab first we need to
calculate the volume.
Area of flat slab = 1000 sq.ft =93 sq. m.
The minimum thickness provided for slab is 150mm .( According to IS 456)
150 mm = 0.15 metres
Hence Wet volume = Area of slab x Thickness of slab
Wet volume = 93 x 0.15 = 13.95 cu. m.
To calculate the quantities of materials we need dry volume
Dry volume = Wet volume x 1.54
Check why 1.54 is multiplied with wet volume to get dry volume.
∴ Dry volume = 13.95 x 1.54 = 21.48 cu.m (approximately).
Generally, M20 grade concrete is predominantly used for construction purposes.
M20 = 1: 1.5: 3 (Cement:Sand:Aggregates) = (a:b:c)
QUANTITY OF CEMENT:
Quantity of cement = Dry volume x (a/{a+b+c}
Quantity of cement = Dry volume x (1/{1+1.5+3}) for M20 grade concrete.
= 21.48 x (1/5.5}) = 3.90 cu.m = 3.90 cu.m
We need the quantity of cement in terms of bags.
So by multiplying the quantity of cement with the density of cement, we will the get quantity of cement
in terms of kilograms.
Unit weight or Density of cement = 1440 kg/cu.m.
Quantity of cement in kg(s) = 1440 x 3.90 = 5616 kg
We know that 1 cement bag = 50 kgs
∴ No. of cement bags required = 5616/50 kg = 112.32 bags =112 bags.
QUANTITY OF SAND:
Quantity of sand = Dry volume x (b/{a+b+c}
∴ Quantity of sand = Dry volume x (1.5/{1+1.5+3}) for M20 grade concrete.
= 21.48 x (1.5/5.5}) = 5.85 cu.m
QUANTITY OF AGGREGATES:
Quantity of aggregates = Dry volume x (c/{a+b+c}
∴ Quantity of aggregates = Dry volume x (3/{1+1.5+3}) for M20 grade concrete.
= 21.48 x (3/5.5}) = 11.71 cu.m
SUMMARY:
Cement = 112 bags.

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Sand = 5.85 cu.m
Aggregates = 11.71 cu.m
Note: ConsiderYield of concrete as 67% &
Wastages of materials = 2%
You can use the same formula for different mix proportions such as 1:2:4 etc.
Here, we have calculated in cubic meter unit but you can calculate in cubic feet unit.
HOW TO CALCULATE VOLUME OF CONCRETE FOR STAIRCASE:
Staircase is one of the essential parts of a building. It helps people to access different floors in a building.
In this article, I will explain how to estimate concrete volume and quantity of cement, sand, aggregates
for a staircase. Before starting the estimation, we need to know different components of a staircase
which I will use in this calculation.
COMPONENTS OF STAIRCASE:
STEPS:
A portion of stairway comprising the tread and riser which permits ascent and descent from one floor to
another.
TREAD:
The horizontal upper part of a step on which foot is placed in ascending or descending stairway.
RISER:
The vertical portion of a step providing support to the tread.
FLIGHT:
A series of steps without any platform, break or landing in their direction.
LANDING:
A platform or resting place provided between two flights.
Now coming to the calculation of concrete volume for the staircase.
Let’s take an example:

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The height of one floor = 12 feet.
Height to be acquired by one flight = 12/2 = 6 feet.
Risers = 6″ = 0.5 feet.
Number of risers = Height of flight / Riser = 6/0.5 = 12
Tread = 10″ = 0.8 feet.
Number of treads = (Numbers of risers -1) = 12 – 1
Thickness of waist slab = 6″ = 0.5 feet
Length of step = 5 feet
VOLUME OF CONCRETE FOR STEPS:
Volume of one step = 1/2 x riser x tread x length of step
= 1/2 x 0.5 x 0.8 x 5 = 1.03 = 1 cubic feet.
As we have 11 number of steps in a flight the volume of steps for first flight = 11 x 1 = 11 cubic feet.
CONCRETE VOLUME FOR WAIST SLAB:
To calculate the volume of concrete for waist slab we need to know the inclined length of the waist slab.
Horizontal length of waist slab = Tread x Number of steps
=0.8333′ x 11 = 9.2 feet.

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The height of the top of the landing from floor = Number of riser x Height of riser
= 12 x 0.5 = 6 feet.
The volume of concrete for waist slab = Inclined length of waist slab x Width of waist slab x Thickness of
waist slab
= 11′ x 5′ x 0.5′
= 27.5 cubic feet
∴ Volume of concrete for first flight = Volume of waist slab + Volume of steps
= 27.5+11 = 38.5 cubic feet
VOLUME OF CONCRETE FOR LANDING OF STAIRCASE:
Length of landing = 10.5 feet
Width of landing = 5 feet
Thickness of landing = 0.5 feet
∴ Concrete volume for landing = 10.5′ x 5′ x 0.5′ = 26 cubic feet.
CONCRETE VOLUME FOR 2ND FLIGHT OF THE STAIRCASE:
As the 1st light and 2nd flight are same in our staircase so the volume of concrete will be same.
Concrete volume for 2nd flight of the staircase = 38.5 cubic feet.
∴ Total volume of concrete for staircase = Volume of first flight + Volume of second flight + Volume of
landing
= 38.5 + 38.5 + 26
= 103 cubic feet.
Well, this is the wet volume of concrete. To convert wet volume into dry volume we need to multiply
with 1.54
∴ Dry volume of concrete = 103 x 1.54 = 159 cubic feet.
QUANTITY OF MATERIALS FOR STAIRCASE:
We will calculate for 1:2:4 Concrete mix.
1. Volume of cement = 1/7 x Dry volume = 1/7 x 159 = 22.66 cubic feet.
∴ Number of Cement bags = volume of cement / Volume of cement bag = 22.66 / 1.25 = 18 No’s of bag.
(The volume of one bag cement of 50 kg = 1.25 cubic feet)
2. Volume of Sand = 2/7 x Dry volume = 2/7 x 159 = 45 cubic feet.
3. Volume of Aggregate = 4/7 x Dry volume = 4/7 x 159 = 91 cubic feet.
Note: Here I have used cubic feet unit and 1:2:4 mix ratio, you can use cubic meter unit and any other
mix ratio of concrete.
WHAT IS 1.54 IN CONCRETE CALCULATION?

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In concrete calculation, we always multiply Wet volume with 1.54 to get dry volume. But do you know
what is 1.54 or where this 1.54 came from? If You don’t know, no problem. In this article, I will explain
what is 1.54 while calculating quantity of cement, sand, aggregates for concrete.
let us take a concrete cube.
The length of the concrete cube = 1 m
The width of the concrete cube = 1 m
The height of the concrete cube = 1 m
Volume of concrete cube = length x Width x Height = 1 x 1 x 1 = 1 m³ (Wet volume)
When we convert this we volume into dry volume, the volume is increased by 54% of wet volume.
∴ Dry volume = Wet volume + 54% of Wet volume
= 1 + (54/100) x 1 =1 + 0.54 = 1.54
To convert wet volume of 1 m³ concrete into dry volume = 1.54
To convert wet volume of “n” m³ concrete into dry volume = 1.54 x n
Where n = Wet volume of concrete.
Example:

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The length of the concrete cube = 4 m
The width of the concrete cube = 3 m
The height of the concrete cube = 2 m
Wet volume = 4 x 3 x 2 = 24 m³
∴ Dry volume = ( Wet volume + 54% of wet volume)
= 24 + {(54/100) x 24}
= 36.96 m³
Note:
Always remember in concrete calculation, we get wet volume first, to convert wet volume into dry
volume, multiply wet volume with 1.54.
HOW TO CALCULATE RECTANGULAR WATER TANK SIZE AND CAPACITY
For high rise building, the dimension of water tank along with drawings are provided from architect
office.
But for small residential building, we need to calculate by itself. In this article, we will discuss how to
calculate rectangular water tank size and capacity.
VOLUME OF WATER REQUIREMENT:
As per IS code, a person needs 135 litres water per day for daily use.
Drinking – 5 Litre
Cooking – 5 litres
Bathing and Toilet – 85 litres
Washing clothes and Utensils – 30 litres
House Cleaning – 10 litres
CALCULATION OF TANK SIZE AND CAPACITY:
We will calculate for a typical family having 4 members.
Total water requirement = 135 x 4 = 540 litres
To calculate the size of water tank we must know any one of the following:

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Length or width or depth of the tank.
From the formula of volume of water
1 m3 = 1000 litres
1 litre = 0.001 m3
We are requiring 540 litre water
∴ 540 litres = 0.54 m3
Let us assume the depth of water tank is 0.6 m.
∴ Area of tank = 0.54/0.6 = 0.9 m2
L x B = 0.9 m2
Now if you know the length or breadth you can put the value on the above formula.
otherwise, take the length as 2 times of B.
∴ L = 2B
2B x B = 0.9 m2
B2 = 0.45
B = 0.67
∴ L = 2 x 0.67 = 1.34 m
So For 540 litre water tank size
Length (L) = 1.34 m
Breatdh (B) = 0.67 m
Depth (D) = 0.6 m

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SEPTIC TANK SHAPE, SIZE & DIMENSIONS WITH TABLE
SEPTIC TANK SHAPE, SIZE & DIMENSIONS:
Septic tank is a devise which deals with the sewage from water closets. In other words, it is a watertight
single-storeyed, underground tank in which sewage is retained sufficiently long to permit sedimentation
of suspended solids and practical digestion of settled sludge by anaerobic bacterial action.
SHAPE OF SEPTIC TANK:
The simplest form of a septic form is a single compartment, rectangular or cylindrical in shape. The
shape of septic tank influences the flow speed of wastewater and sludge accumulation.
The dimension of the septic tank should not be too deeper too shallow. Because it can cause short-
circuiting of the inlet and outlet flow.
Septic tank having a greater surface area with sufficient depth should always be preferred.
Rectangular shaped single compartment tank (with length three times of its width) is more favorable.

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In some cases, cylindrical shape with suitable size is also sound to be a better choice.
SEPTIC TANK SIZES, DIMENSIONS & WEIGHTS:
SEPTIC TANK
SIZE
DIMENSIONS
WEIGHT
(LBS)
ANCHOR WEIGHT
(LBS)
SOIL COVER
(IN)
5000 gal
204″L x 96″W
x93″H
41,400 30,850 24
3000 gal
165″L x 92″W x
76″H
20,300 23,320 23
2600 gal
147″L x 90″W x
73″H
18,100 20,625 24
2000 gal
162″L x 78″W x
64″H
16,100 15,675 19
1600 gal
145″L x 78″W x
61″H
14,000 11,270 16
1200 gal
111″L x 78″W x
61″H
11,400 9,532 17
1000 gal
Low Profile
120″L x 67″W x
57″H
9,500 8,705 17
1000 gal
Heavy Duty
96″L x 78″W x
61″H
9,200 8,945 18
800 gal
96″L x 67″W x
57″H
8,000 6,560 16
600 gal
78″L x 56″W x
60″H
6,600 3,810 14

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1600/1400 gal
174″L x 90″W x
73″H
23,000 22,410 22
1250/750 gal
162″L x 78″W x
64″H
16,400 15,725 19
1000/600 gal
145″L x 78″W x
61″H
14,700 12,705 17
DESIGN CRITERIA OF SEPTIC TANK
SEPTIC TANK:
Septic tank is a tank where sewage is kept a long time for sedimentation of suspended solid by
anaerobic microorganism activity.
Septic Tank Design
Basically, the tank is rectangular in shape consisting of rooftop with two chambers, isolated from each
other by a baffle wall. The first chamber is called grit chamber in which the sewage enters first and the
rest one is called anaerobic chamber.
Inlet and outlet pipe should be fitted in such a manner that while entering or existing there should not
be any unsettled influence in the sewage of anaerobic chamber. Sand, grit etc is settled down in the grit
chamber before the sewage is directed to the anaerobic chamber. In the anaerobic chamber, natural
solids settle at the base of the tank where anaerobic bacteria follows up on it and changes over complex
unsteady mixes to more straightforward stable mixes.
SEPTIC TANK DESIGN:
The following measure should be kept in view to design a septic tank:
1. The floor zone of the git chamber must be adequate to decrease the speed of steam and to allow
sedimentation.
2. The capacity of a septic tank should be sufficient to give a detainment period fluctuating from 12
hours to 3 days, 24 hours time frame is normally viewed as satisfactory.
3. The depth beneath segment divider opening should be adequate to allow aggregation for the
predetermined period. A minimum area of 0.07 m2 for each client in grit chamber and minimum
volumetric substance of the grit chamber of 0.02 m3 for every client is required.

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4. The minimum width and depth of a septic tank should be 0.75 m and 1 meter underneath water level.
The length should be 2 to 4 times of the width. The minimum capacity of a septic tank is granted 1 m3.
5. Every septic tank should be furnished with a ventilating channel minimum diameter of 5 cm.
HOW TO CALCULATE VOLUME OF CONCRETE, SHUTTERING AREA & BITUMEN PAINT AREA
FOR PLAIN FOOTING
In this article I will discuss how to calculate the volume of concrete, shuttering area and bitumen paint
area for plain footing (Square footing & Rectangular footing).
A. SQUARE FOOTING:
Given,
Length of footing = 0.5 m
Breadth of footing = 0.5 m
Depth of footing = 0.25 m
Size of column = 0.2 x 0.1
1. Volume of concrete = Area of sq. footing x Depth of sq footing
Area of sq. footing = L x B = 0.5 x 0.5 = 0.25 m²
Volume of concrete = 0.25 x 0.25 = 0.0625 m³.
2. Shuttering area = 2 (L + B) x D
= 2 (0.5+0.5) x 0.25
= 0.5m²
3. Bitumen paint area = Shuttering area + (Top area of footing – column area)
= 0.5 + [(0.5 x0.5) -(0.2 x 0.1)]
=0.5 + [0.25 – 0.2)]
= 0.73 m².
2. RECTANGULAR FOOTING:

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Given,
Length of footing = 1 m
Breadth of footing = 0.7 m
Depth of footing = 0.25 m
Size of column = 0.4 x 0.3
1. Volume of concrete = Area of sq. footing x Depth of sq footing
Area of sq. footing = L x B = 1 x 0.7 = 0.7 m²
Volume of concrete = 0.7 x 0.25 = 0.175 m³.
2. Shuttering area = 2 (L + B) x D
= 2 (1+0.7) x 0.25
= 0.85m²
3. Bitumen paint area = Shuttering area + (Top area of footing – column area)
= 0.85 + [(1 x0.7) -(0.4 x 0.3)]
=0.85 + [0.7 – 0.12)]
= 1.43 m².
HOW TO CALCULATE CEMENT, SAND QUANTITY FOR PLASTERING?
Following points should be remembered while calculating the quantity of cement, sand for plastering
work.
1. For wall plastering, Cement : Sand = 1 : 6
2. For ceiling plastering, Cement : Sand = 1 : 4
3. Thickness of plaster should be in between 12-15 mm. If an additional coat is required then do not do
it at one go.
4. Use good quality of cement & Sand.
5. Use measuring box (not head pan) for site mix.
We will calculate cement and sand for 100 m2 plastering area in 1:6 ratio and thickness of 12 mm.
Cement Mortar Required:
Plastering thickness = 12 mm
= 12/1000 = 0.012m

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Volume of cement mortar required = ( Plastering Area x thickness )
= 100 m2 x 0.012m = 1.2 m3
(This is wet volume of cement mortar (after mixing water) but we need dry volume. To get dry volume
consider 35% bulking of sand and 20% wastages
= 1.2 m3 x (1+0.2+0.35) (Rather than 35% sand bulkage and 20% wastage you can add 1.54 as constant)
= 1.86 m3
Cement : Sand = 1 : 6
Sum of ratio =( 1 + 6) = 7
∴ Cement required
= 1.86 x 1/7
= 0.265 m3
= 0.265/0.0347 ( 0.0347 m3 = 1 bag = 50 kg cement)
= 7.66 bags (≈ 8 Bags)
∴ Sand required
= 1.86 x 6/7
= 1.59 m3
Here we have calculated in Sq.m but you can also calculate it in Sq.ft.
HOW TO CALCULATE QUANTITY OF PAINT FOR BUILDING:
Painting is one of the most essential steps to be done after construction of any building/house. Painting
is done to protect the surface of the walls as well as to increase appearance.
In this article, I will discuss how to calculate the quantity of paint required in a building.
Before starting the calculation let me tell you one important thing that quantity of paint is calculated in
gallon.
There are two types of gallons
US gallon and Uk gallon.
1 US gallon = 3.785 litres
1 Uk gallon = 4.546 litres.
In this calculation we will use US gallon.

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Let the length of the wall = 15 feet
Area of the wall 15 x 12 = 180 sft.
THUMB RULE FOR PAINT CALCULATION:
Generally, 1 gallon of paint can cover up to 350 sft area of wet wall and
1 gallon of paint can cover up to 200 sft area of dry wall.
PAINT QUANTITY OF WET WALL:
Area of wet wall =180 sft
Paint covered area = 300 sft
Quantity of paint = 180/350 = 0.514 litre.
Let the price of 1 gallon paint = Rs 1200.
Price of 0.514 gallon paint = 0.514 x 1200 = 617 Rs.
PAINT QUANTITY OF DRY WALL:
Area of dry wall = 180 sft.
Pain covered area = 200
Quantity of paint = 180/200 =0.9 gallon.
Price of 0.9 gallon paint = 0.9 x 1200 = 1080 Rs.
Note:
1. If the wall has any openings, subtract the area of openings from total area & then calculate paint
quantity.
HOW TO CALCULATE SHUTTERING OIL AND ITS PRICE
Shuttering oil is very useful material required for every construction projects. So it is very important to
know the shuttering oil quantity. In this article, I will discuss how to calculate shuttering oil and its price.
Before starting the calculation let me tell you first what is shuttering oil and why we use it in
shuttering/formwork.
WHAT IS SHUTTERING OIL?

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Shuttering oil are high quality mineral oil which is used on the shuttering plates/mould for easy removal
of shuttering.
WHY WE USE SHUTTERING OIL IN FORMWORK?
The main purposes of using shuttering oil are as follows:
1. Shuttering oil works as a releasing agent between shuttering plates and concrete. It helps easy
removal of shuttering at the time of deshuttering.
2. It gives a good and better finish after deshuttering.
3. It keeps the shuttering plates and mould in good condition and ensures repetitive use in more
projects.
HOW TO CALCULATE SHUTTERING OIL & ITS PRICE:
Let us calculate the quantity of shuttering oil for the above slab.
Length of slab = 12 m
Breadth of slab = 10 m
Area of slab = L x B = 12x 10 = 120 m²
There are two types of shuttering plates used in construction.
1. WOOD SHUTTERING:
Generally, 1 liter shuttering oil covers upto 30 to 40 m² area in wood shuttering.
The quantity of shuttering oil = Total area/1 liter covering area
= 120/30 = 4 litre
Price: Let the price of 1 liter shuttering oil = Rs. 250.
The price of 4 liter shuttering oil = 250 x 4 = Rs. 1000.
2. STEEL SHUTTERING:
Generally, 1 liter shuttering oil covers upto 60 to 70 m² area in steel shuttering.
The quantity of shuttering oil = Total area/1 liter covering area
= 120/60 = 2 litre
Price: Let the price of 1 liter shuttering oil is Rs. 250.
The price of 2 liter shuttering oil = 250 x 2 = Rs. 500.

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HOW TO CALCULATE CONCRETE VOLUME FOR RETAINING WALL:
Retaining wall is a masonry wall constructed to resist the pressure of liquid, earth filling, sand or other
granular material filled behind it.
In this article, I will discuss how to calculate the concrete volume for retaining wall. Let’s get started.
EXAMPLE 1:
To calculate the volume of retaining wall we need to calculate the volume of retaining wall.
Here I have divided the retaining wall into two parts, part A is the base slab and part B is the stem of
retaining wall.
So Volume of retaining wall = Volume of base slab + Volume of stem.
Volume of base slab = l x b x h = 10 x 3 x 0.2 = 6 m³
The stem is a trapezoid.
So Volume of stem = [{(a + b)/2} x h] x l
= [{(0.5 + 0.2)/2} x 3] x 10
= 21 m³
The total volume of retaining wall = 6 + 21 = 27 m³
So the volume of concrete for the retaining wall = 27 m³
Let’s take another example:
EXAMPLE 2:

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Here I have divided the retaining wall into three parts, part A is the base slab, part B is the stem and part
C is the counterfort of the retaining wall.
The volume of retaining wall = Volume of base slab + Volume of stem + Volume of counterfort
= Volume of A + Volume of B + Volume of C
Volume of A = l x b x h = 12 x 2.5 x 0.2 = 6 m³
Part B is a trapezooid.
So Volume of B = [{(a + b)/2} x h] x l = [{(0.2 + 0.3)/2} x 3] x 12
= 9 m³
As part C is also a trapezoid
So
Volume of C = [{(a + b)/2} x h] x l = [{(0.5 + 2)/2} x 3] x 0.2 = 0.75 m³
In the above retaining wall there is 2 counterfort so
Volume of C = 0.75 x 2 = 1.5 m³
The total volume of retaining wall = 6 + 9 + 1.5 = 16.5 m³.
HOW TO CALCULATE NUMBER OF TILES IN A ROOM:
Calculation of tiles needed for floors and walls is so simple and easy. in this post, I will discuss how to
calculate no. of tiles required in a room. So let’s start.
STEP 1- CALCULATE AREA OF THE FLOOR:

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let the length of the floor = 14 feet.
Breadth of the floor = 12 feet.
∴ Area of floor = length of floor x breadth of floor = 14 x 12 = 168 sft.
STEP 2 – CALCULATE AREA OF 1 TILE:
There are different tile sizes such as 20cmx20cm, 30cmx30cm, 45×45, 60cmx60cm etc. In this
calculattion we will use 60×60 cm tiles.
Length of 1 tile = 60 cm
Breadth of 1 tile = 60 cm.
∴ Area of 1 tile = 60 x 60 = 3600 cm² = 3.6 sft.
STEP 3 – DIVIDE FLOOR AREA BY AREA OF 1 TILE:
∴ No. of tiles required = Area of floor/ area of 1 tile = 168/3.6 = 46.6 ≈ 47.
∴ 47 tiles are required for the above floor.
Note:
Consider 5% wastages.
Similarly, you can also calculate tiles needed for walls.
HOW TO CALCULATE LAND AREA OR PLOT AREA:
Sometimes we may need to know the area of any land or plot for various reasons such as buying, selling
of land, or constructing house/building etc. In this article, I will explain how to calculate land are or plot
area for a site.
Area Formula For Different Shapes:

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But for complex shaped land you cannot easily calculate the area. In that case, we use HERON’s Formula
which is given below:

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Where S = Perimeter of the triangle
S = (a + b + c)/2
a = Distance of AB
b = Distance of BC
c = Distance of AC
Let’s take an example,
First of all, divide the land area into minimum possible no. of triangle. Then measure all the required
distance (like AB, AC, AE etc) by using a tape or chain.
1. FOR △ ACD:
AC = 10 feet.
CD = 12 feet.
AD = 14 feet.
S = (a + b + c)/2 = (10 + 12 + 14)/2 = 18 feet.

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2. FOR △AED:
AD = 14 feet.
AE = 14feet.
DE = 12 feet.
S = (a + b + c)/2 = (14+14+12)/2 = 20 feet.
3. FOR △AEB:
AB = 16 feet.
BE = 12 feet.
AE = 12 feet.
S = (a + b + c)/2 = (16 + 12 + 12)/2 = 20 feet.
Total Area Of Land = 58.78 + 75.89 + 71.55 = 206.22 ft²

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NOTE: This method does not give appropriate result for curve shaped area, for curve shaped area we
need to use difinite integral method.
CENTER LINE METHOD OF ESTIMATION:
There are three different methods of estimation.
1. Center line method.
2. Long and short wall method.
3. Crossing method.
In this article, I will discuss center line method briefly for the estimation of materials.
WHAT IS CENTER LINE METHOD:
In this method of estimation, the total center line length of walls in a building is first calculated, then the
center line length is multiplied with the breadth and depth of respective item to get the total quantity at
a time.
The center line length for different sections of walls in a building shall be worked out separately. For
verandah walls or partition walls joining the main walls, the center line length shall be reduced by half of
the breadth of the layer of main wall that joins with the partition or verandah wall at the same level. The
number of such joints is studied first to obtain the center line length.
By using this method estimation can be finished more quickly. This method is as accurate as other
methods (except unsymmetrical wall). This method is suitably used for estimating circular, rectangular,
hexagonal, octagonal etc shaped building.
Let us take an example:
FOR ONE ROOM BUILDING:
First we need to calculate total center line length of the building.

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Total center line length = 2 x 5.3 + 2 x 4.3 = 19.2 m.
1. Earthwork in excavation = Total center line length x breadth x depth
= 19.2 x 0.9 x (0.3+0.3+0.3)
= 19.2 x 0.9 x 0.9
= 15.52 cu.m
2. Concrete in foundation = 19.2 x 0.9 x 0.3 = 5.18 cu.m
3. a) Brickwork in foundation for 1st footing = 19.2 x 0.6 x 0.3 cu.m
b) Brickwork in foundation for 2nd footing = 19.2 x 0.5 x 0.3 = 2.88 cu.m
4. Brickwork in superstructure = 19.2 x 3.5 x 0.3 = 20.16 cu.m
FOR TWO ROOM BUILDING:
Total center line length = 2 x (5.3+5.3) + 3 x 4.3 = 34.1 m.
1. Earthwork in excavation = Total center line length x breadth x depth
= 34.1 x 0.9 x (0.3+0.3+0.3)
= 34.1 x 0.9 x 0.9
= 27.62 cu.m
2. Concrete in foundation = 34.1 x 0.9 x 0.3 = 9.20 cu.m
3. a) Brickwork in foundation for 1st footing = 34.1 x 0.6 x 0.3 = 6.13 cu.m
b) Brickwork in foundation for 2nd footing = 34.1 x 0.5 x 0.3 = 5.11 cu.m
4. Brickwork in superstructure = 34.1 x 3.5 x 0.3 = 35.8 cu.m
Thus you can estimate quantity of different items such as
Quantity of D.P.C = Total Centre line length x Breadth of foundation x Thickness of D.P.C
Quantity of plinth beam = Total Centre line length x Breadth of the beam x Depth of beam.
Quantity of wall plaster for 2 sides = Total centre line length x Height of wall x Thickness of plaster.
Quantity of Paint for 2 sides of wall = Total centre line length x Height of wall x 2 = Area of paint
in sq.ft

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ONE WAY SLAB DESIGN – HOW TO DESIGN ONE WAY SLAB
In this article, I will discuss how to design One way slab with example.
WHAT IS ONE WAY SLAB?
When a slab is supported on all four sides and the ratio of long span to short span is equal or greater
than two, it will be considered as one way slab. The load on the slab is carried by the short span in one
direction. However main reinforcement bar and distribution bar in transverse direction.
Longer span (l)/Shorter span (b) ≥ 2
ACI CODE SPECIFICATIONS FOR ONE WAY SLAB DESIGN:
1. MINIMUM SLAB THICKNESS:
To control deflection, ACI Code 9.5.2.1 specifies minimum thickness values for one-way solid slabs.
2. SPAN:
According to ACI code 8.7.1 If the slab rests freely on its supports, the span length may be taken as equal
to the clear span plus the depth of the slab but need not exceed the distance between centers of
supports.
3. BAR SPACING:
The lateral spacing of the flexural bars should not exceed 3 times the thickness h or 18 inch
according to ACI code 7.6.5 The lateral spacing of temperature and shrinkage reinforcement should not
be placed farther apart than 5 times the slab thickness or 18 inch according to ACI code 7.12.2
4. MAXIMUM REINFORCEMENT RATIO:
Reinforcement ratio is the ratio of reinforcement area to gross concrete area based on total
depth of slab. One-way solid slabs are designed as rectangular sections subjected to shear and moment.
Thus, the maximum reinforcement ratio corresponds to a net tensile strain in the
reinforcement, €t of 0.004
5. MINIMUM REINFORCEMENT RATIO:
A) FOR TEMPERATURE AND SHRINKAGE REINFORCEMENT :
According to ACI Code 7.12.2.1
Slabs with Grade 40 or 50 deformed bars –> 0.0020
Slabs with Grade 60 deformed bars –> 0.0018
Slabs where reinforcement with yield strength Exceeding 60000 psi- ->( 0.0018 x 60000/fy)
B) FOR FLEXURAL REINFORCEMENT :
According to ACI Code 10.5.4, the minimum flexural reinforcement is not to be less than the shrinkage
reinforcement, or 0.0018
EXAMPLE PROBLEM:
A reinforced concrete slab is built integrally with its supports and consists of equal span of 15 ft. The
service live load is 100 psf and 4000 psi concrete is specified for use with steel with a yield stress equal
to 60000 psi. Design the slab following the provisions of the ACI code.

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THICKNESS ESTIMATION:
For being both ends continuous minimum slab thickness = L/28 =(15 x 120)/28= 6.43 in.
Let a trial thickness of 6.50 in.
LOAD CALCULATION:
Consider only a 1 ft width of beam.
Dead load = 150 x (6.50/12)) = 81 psf
Live load = 100 psf
Factored DL and LL ={81+1.2+(100 x 1.6)} =257 psf
DETERMINE MAXIMUM MOMENTS:
Factored moments at critical sections by ACI code :
At interior support : -M=1/9 x 0.257 x 152 = 6.43 k-ft
At midspan : +M=1/14 x 0.257 x 152 = 4.13 k-ft
At exterior support : -M=1/24 x 0.257 x 152 = 2.41 k-ft
Mmax = 6.43 k-ft
=0.85 x 0.85 x 4/60 x 0.003/(0.003+0.004)
= 0.021
Now,

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SAQIB IMRAN 70
= 2.64 in
CHECK FOR AVAILABILITY OF THICKNESS:
As ‘d’ is less than the effective depth of (6.50-1.00) = 5.50 in, the thickness of 6.50 in
can be adopted.
REINFORCEMENT CALCULATION:
Let, a = 1 inch
At interior support:
Checking the assumed depth of a by
Similarly at Midspan:
As = (4.13 x 12)/(0.90 x 60 x 5.29) = 0.17 in²
At Exterior support:
As = (2.41 x 12)/(0.90 x 60 x 5.29) = 0.10 in²
MINIMUM REINFORCEMENT:
As = 0.0018 x 12 x 6.50 = 0.14 in²
So we have to provide this amount of reinforcement where As is less than 0.14 in².
SHRINKAGE REINFORCEMENT:
Minimum reinforcement for shrinkage and temperature is
As = 0.0018 x 12 x 6.50 = 0.14 in²
FINAL DESIGN:

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SAQIB IMRAN 71
LAYOUT OF ONE WAY SLAB:

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SAQIB IMRAN 72
DESIGN OF COLUMNS AS PER ACI:
1. Maximum and Minimum Reinforcement Ratio:
The minimum reinforcement ratio of 1 % is to be used in tied or spirally reinforced columns. This
minimum reinforcement is needed to safeguard against any bending, reduce the effect of shrinkage and
creep and enhance ductility of columns.
2. Minimum Number of Reinforcing Bars:
Minimum four bars within rectangular or circular sections; or one bar in each corner of the cross section
for other shapes and a minimum of six bars in spirally reinforced columns should be used.
3. Clear Distance between Reinforcing Bars:
For tied or spirally reinforced columns, clear distance between bars should not be less than the larger of
150 times bar diameter or 4 cm.
4. Concrete Protection Cover:
The clear concrete cover should not be less than 4 cm for columns not exposed to weather or in contact
with ground. It is essential for protecting the reinforcement from corrosion or fire hazards.
5. Minimum Cross-Sectional Dimensions:
For practical considerations, column dimensions can be taken as multiples of 5 cm.
6. Lateral Reinforcement:
Ties are effective in restraining the longitudinal bars from buckling out through the surface of the
column, holding the reinforcement cage together during the construction process, confining the
concrete core and when columns are subjected to horizontal forces, they serve as shear reinforcement.

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SAQIB IMRAN 73
Spirals, on the other hand, serve in addition to these benefits in compensating for the strength loss due
to spalling of the outside concrete shell at ultimate column strength.
7. Ties:
For longitudinal bars, 32 mm or smaller, lateral ties 10 mm in diameter should be used. In our country
and in some neighboring countries, ties of 8 mm dia are used for column construction
CALCULATION OF EARTH PRESSURE:
The thrust due to the back filling, which may be assumed to be earth, is generally calculated by
Rankine’s theory. The theory is based on the assumption that the backing material or earth consists of
cohesionless granular particles. The formulae derived from this theory under different conditions of
backfilling are given below:
Case 1: Walls with earth levelled with the top of the wall:
a) Horizontal pressure per sq.m (ph) at a depth of (h) meter below the levelled top is given by the
following formula:
Where
w = Weight of filling in kg/m3
ϕ = Angle of repose of the soil.
b) Total horizontal pressure (P) at a depth of (h) meter per meter length of the wall is given by the
following formula:
Acting at h/3 meter from the base.
Case 2: In case of submerged retaining wall or wall retaining earth filled at slope of a° to the horizontal,
the formula giving lateral earth pressure (ph) is given by:
Acting parallel to the surcharge slope of the filling.
Total pressure (P) at depth h meter per meter length of the wall is given by the formula:

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