Apidays Singapore 2024 - Building Digital Trust in a Digital Economy by Veron...
Quantity notes
1. 1 | P a g e
SAQIBIMRAN 1
Assala mu alykumMyNameissaqibimranandI amthestudentofb.tech (civil) insarhad
univeristyofscience andtechnologypeshawer.
I have writtenthisnotes bydifferentwebsitesandsome byselfandprepare itforthestudentand
alsoforengineerwhoworkonfieldtogetsome knowledge fromit.
I hopeyou allstudentsmaylike it.
Rememberme inyourpray, allah blessme andallofyou friends.
If u have anyconfusioninthisnotescontactme onmygmail id: Saqibimran43@gmail.com
or textmeon0341-7549889.
Saqib imran.
2. 2 | P a g e
SAQIBIMRAN 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
Unitsof
measurement
Unitsof
payment
1 Earthwork
1. Earthwork inexcavation CUM PerCUM
2. Earthwork infillinginfoundationtrenches CUM PerCUM
3. 3 | P a g e
SAQIBIMRAN 3
3. Earth workinfillinginplinth CUM PerCUM
2 Concrete
1. Lime concrete in foundation CUM PerCUM
2. Cementconcrete inlintels CUM PerCUM
3. RCC in slab CUM PerCUM
4. Cementconcrete or RCCin chujja,sunshade CUM PerCUM
5. Lean concrete inroof terracing(thicknessspecified) SQM PerSQM
6. Cementconcrete bed CUM PerCUM
7. Reinforcedconcrete sunshade(specifiedwidthandheight) CUM PerCUM
3 Damp proofcourse (DPC) – thicknessmentioned SQM PerSQM
4 Brick work
1. Brickworkin foundation CUM PerCUM
2. Brickworkin plinth CUM PerCUM
3. Brickworkin superstructure CUM PerCUM
4. Thin partitionwalls SQM PerSQM
5. Brickworkin arches CUM PerSQM
6. Reinforcedbrickwork CUM PerCUM
5 Stone work
4. 4 | P a g e
SAQIBIMRAN 4
1. Stone masonry CUM PerCUM
6 Woodwork
1. Doors and windowsframesorchaukhats,rafters, beams CUM PerCUM
2. Shuttersof doors and windows(thicknessspecified) SQM PerSQM
3. Doors and windowsfittings(like hinges,towerbolts,sliding
bolts,handles)
Each PerEach
7 Steel work
1. Steel reinforcementbarsetc.inRCC and reinforced brickwork Quintal PerQuintal
2. Bending,bindingof steelreinforcement Quintal Perquintal
3. Rivets,boltsandnuts,anchor bolts,lewisbolts,holdingdown
bolts
Quintal Perquintal
4. Iron holdfasts – –
5. Iron railing(heightandtypes specified) – –
6. Iron grills SQM PerSQM
8 Roofing
1. RCC and RB slab roof (excludingsteel) CUM PerCUM
2. Lean concrete roof overand inclusive of tilesorbrickor stone
slabetc. (thicknessspecified)
SQM PerSQM
3. Centeringandshuttering formwork SQM PerSQM
4. AC sheetroofing SQM PerSQM
5. 5 | P a g e
SAQIBIMRAN 5
9 Plastering,pointsand finishing
1. Plastering–cementor lime mortar(thicknessandproportion
specified)
SQM PerSQM
2. Pointing SQM PerSQM
3. White washing,colorwashing,cementwashing, (numberof
coats specified)
SQM PerSQM
4. Distempering(numberof coatsspecified) SQM PerSQM
5. Painting,varnishing(numberof coatsspecified) SQM PerSQM
10 Flooring
1. 25mm cementconcrete over75mm lime concrete floor
(includinglean concrete)
SQM PerSQM
2. 25mm or 40mm cementconcrete floor SQM PerSQM
3. Doors and windowsills(CCorcementmortarplain) SQM PerSQM
11 Rainwater pipe /plainpipe RM PerRM
12 Steel woodentruss Each Pereach
13 Glasspanels(supply) SQM PerSQM
14 Fixingof glasspanelsorcleaning Each PerEach
Note:
SQM = Square meter
CUM = Cubic meter
RM = Running meter
6. 6 | P a g e
SAQIBIMRAN 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.
7. 7 | P a g e
SAQIBIMRAN 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
8. 8 | P a g e
SAQIBIMRAN 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
9. 9 | P a g e
SAQIBIMRAN 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.
10. 10 | P a g e
SAQIBIMRAN 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.
11. 11 | P a g e
SAQIBIMRAN 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
12. 12 | P a g e
SAQIBIMRAN 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.
13. 13 | P a g e
SAQIBIMRAN 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
14. 14 | P a g e
SAQIBIMRAN 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.
15. 15 | P a g e
SAQIBIMRAN 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:
16. 16 | P a g e
SAQIBIMRAN 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:
17. 17 | P a g e
SAQIBIMRAN 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
18. 18 | P a g e
SAQIBIMRAN 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.
19. 19 | P a g e
SAQIBIMRAN 19
Bricks per square foot
Wall thickness Bricks per square foot
4.5 inch 5.0625 bricks
9 inch 10.125 bricks
13.5 inch 15.1875 bricks
Bricks per square meter
Wall thickness Bricks per square meter
4.5 inch 57.16 bricks
9 inch 114.329 bricks
13.5 inch 171.49 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
20. 20 | P a g e
SAQIBIMRAN 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.
21. 21 | P a g e
SAQIBIMRAN 21
Standard conversionfactors
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 FARENHEITX 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 weightof MS Sheetof 1mm thicknessandsize 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 andlength1 metre thenthe weightshall 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 Roundof 20mm diameter andlength1 metre thenthe weightshall be.
3.14 X 0.00000785 X ((20/2) X ( 20/2)) X 1000 mm= 2.46 kgs / metre
SS ROUND
DIA (mm) XDIA (mm) X0.00623 = WEIGHT PER METRE
SS / MS Pipe
22. 22 | P a g e
SAQIBIMRAN 22
OD ( mm) – W.Tthick(mm) XW.Thick(mm) X0.0248 = WeightPerMetre
OD ( mm) – W.Tthick(mm) XW.Thick(mm) X0.00756 = WeightPerFoot
SS / MS CIRCLE
DIA(mm) XDIA (mm) X THICK(mm) 0.0000063 = Kg PerPiece
SS sheet
Length(Mtr) X Width(Mtr) X Thick(mm) X8 = WeightPerPiece
Length(ft) X Width(ft) XThick(inch) X3 /4 = WeightPerPiece
S.S HEXAGONALBAR
DIA (mm) XDIA (mm) X0.00680 = WT. PER Mtr
Dia (mm) XDia (mm) X 0.002072 = Wt. Per foot.
BRASS SHEET
WEIGHT (KGS) = LENGTH (MM) X BREADTH (MM) X 0. 0000085 X THICKNESS
Example – The weightof brasssheetof thickness1 mm,length1220mm andbreadth355mm shall be
1220 X355X 0.0000085 X 1 = 3.68 Kgs/Sheet
Followingtable shows how can we convert various most commonlyused unitsfrom one unitsystem
to another.
Unitsto convert Value
Square footto Square meter 1 ft²= 0.092903 m²
Footper secondsquared to Meterpersecond
squared
1 ft²= 0. 3048 m²
Cubicfootto Cubicmeter 1 ft³= 0.028316 m³
Poundpercubic inchto Kilogrampercubic
meter
1 lb/in³=27679.9 047102 kg/m³
Gallonperminute = Literper second 1 Gallonperminute = 0.0631 Literpersecond
Poundpersquare inch= Kilopascal 1 Psi (PoundPerSquare Inch) = 6.894757 Kpa
(Kilopascal)
Poundforce = Newton 1 Poundforce = 4.448222 Newton
PoundperSquare Foot to Pascal 1 lbf/ft2= 47.88025 Pascal
Acre foot perday = Cubic meterpersecond 1 Acre footper day= 1428 (m3/s)
Acre to square meter 1 acre = 4046.856 m²
Cubicfootper second= Cubicmeterper
second
1 ft³/s= 0.028316847 m³/s
Measurementunitsandstandardsare differentindifferentcountriesbuttomaintainastandard,SI
unitsare mostlyusedwhendealingwithprojectsinvolvingdifferentcountriesorevendifferentstates.
23. 23 | P a g e
SAQIBIMRAN 23
Small projectscanbe done withthe locallyusedunitsystembutwhenthe projectisbig,one standard
unitsystemisto be used.
Two most common system usedin the UnitedStatesare
UnitedStatesCustomarySystem (USCS)
SystemInternational(SI)
But the SI unitsystemismore widelyusedall overthe world.Followingisthe table whichshowshow
youcan convertUSCS measurementsinSImeasurements.( Justmultiplythe USCSamountwiththe
correspondingfigure givenintable below
ConvertUSCS into SI Units
USCS unitX Factor = SI unit SI symbol
Square footX 0.0929 = Square meter M2
Cubicfoot X 0.2831 = Cubic meter M3
Poundpersquare inchX 6.894 = Kilopascal KPa
Poundforce X 4.448 = Newton Nu
Footpoundtorque X 1.356 = Newtonmeter N-m
KipfootX 1.355 = Kilonewtonmeter LN-m
Gallonperminute X0.06309 = Liter persecond L/s
Kippersquare 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 UNITOF 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.
25. 25 | P a g e
SAQIBIMRAN 25
Afterestimatingitisveryimportanttoknow the unitweightof steel barsbecause,we estimate as100
meter20mm ø bar or 100 feet16mm ø bar, etc (ø isthe symbol of diameter).
But steel barsupplierswill notunderstandthisnotation,theymeasure the steel barsinweight.Sowe
have to orderthemin kg or quintal orton.In thisarticle,we will discusshow tocalculate unitweightof
steel barsof differentdiameter.
The formula isW = D²L/162
Where
W = Weightof steel bars.
D = Diameterof steel barsinmm.
L = Lengthof bars inmeter.
Example 1: Calculate the weightof 60 meterslong 12 mm ø bar.
Here,D = 12 mm.
L= 60 m.
We knowthat,W = D²L/162
W = 12² x 60/162 = 53 kg
Weightof 60 m 12mm ø bar is 53 kg.
Let’slookfor anotherexample.
Example 2: Calculate the weightof 100 m 16 mm ø bar.
Here,D = 16 mm.
L = 100 m.
W = 16² x 100/162 = 158 kg.
If we put1 meterlength foreachdiameterof steel barinthe formulathenwe will getthe unitweight.
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 multiplythe lengthof barswiththisunitweightwe will getthe total weightof steel bars.
For example,total weightof 1000 meterlong20mm ø steel baris,
1000 x 2.469 = 2469 kg.
Usingthe same methodwe can calculate the unitweightof differentsteelbars.
Here I have calculatedinmeterbutwe can also calculate infoot.To calculate infootwe have to use the
followingformula:
W= D²L/533
Where D = Diameterof barsin mm.
L = Lengthof bars infoot.
HOW TO CALCULATE STEEL QUANTITY FOR RCC BEAM, COLUMN AND SLAB
Followingare the stepstocalculate the quantityof steel forRCCslab
1. Prepare a bar bendingschedule inordertoclassifydifferentshapesof bars(bentupbar, straight
anchor bar, eosbar, curtail bar etc) and diameters.
2. List downall the shapesof bars from the drawing.
3. Count the numberof bars of each of those shapes.
4. Then calculate the cuttinglengthof eachof those bars.
26. 26 | P a g e
SAQIBIMRAN 26
Cuttinglengthof bar = (lengthof the member–deductionforcoveron bothsides) + development
length.
5. Then calculate the unitweightof eachdiabars bythe followingformula:
W = d^2/162
Where d isthe dia inmm and weight(w) inkg.
6. Then calculate weightof rebar
Weightof rebar= no of bars x cuttinglength x unitweight
7. Add all the weighttogetthe total steel quantity.
NOTE – 1. Minimum% ofsteel as perIndian standard are:
1. Beam (Teinsionreinforcement):
As = 0.85bd/fy of grosscross-sectionarea.
2. Slab – 0.12% of total area
3. Column – 0.8% of cs area
2. Maximum% steel asper Indianstandardare:
Beam– 4 % of cross-sectionarea.
Slab– 4% of cross-sectionarea.
column— 6% of cross-sectionarea.
3. . DevelopmentLength isusuallyspecifiedinthe drawings,butif notthenyoucancalculate itas,
D.L = Depth– 2 timescover
4. Bindingwire = 10 grams perkg of reinforcement.
5. No. of stirrups = (lengthof member–2 x cover) / spacing+ 1
WHATIS BAR BENDING SCHEDULE?
Bar bendingschedule commonlyknownasBBSis one of the mostimportanttermsinCivil Engineering.
Because itplaysa vital role inbuildingconstruction.
Like otherbuildingmaterialsestimationof steel isalsorequiredforconstructingabuildingandhere BBS
comeswithan easysolution.Barbendingschedule providesthe reinforcementcalculationandsome
otherimportantdetails suchasbar mark, bar diameter,barshape,cuttinglength,numberof bars,the
weightof bar,total weightof steel etc.Sothatwe can orderthe requiredamountof steel inadvance.
HISTORY OF BAR BENDING SCHEDULE:
Long yearsback whenthere wasnotoday’sbar bendingschedule,certainrecommendationsgivenby
Prof.BN Dutta were usedforestimatingsteel fordifferentcomponentsof abuilding.
27. 27 | P a g e
SAQIBIMRAN 27
But these are nowbackdated(notwrong) andwe don’tuse themintoday.Because Noaccurate values
were specifiedif we use more barsina single member.Nowadayswe are constructing150+ floors
building.Itisnowpossiblewithourmoderntechniques,equipmentetc.Estimationof steel becomes
easierbecause of BBS.
Civil engineerswhoare workingonsite or fresherswhoare goingtowork at constructionsitesmust
knowhowto prepare bar bendingschedule.Inthisarticle,Iwill discusssome importantthingsusedin
preparingbarbendingschedule.
These are justthe basicthings.Infuture,Iwill come upwithsome otherarticlesrelatedtoBBSsuch as
BBS for beams,columns,slabs,footingsetc.
HOOK LENGTH:
The hook isthe extralengthleftatthe 4th cornerof a stirrupso that the stirrupretainsitsshape.
Generally,hooklengthistakenas9d for one side.
Where d = Diameterof the bar.
The total lengthof stirrups= Total lengthof the bar + 2 x hooklength(fortwohooks)
= L + 2 x 9d
= L + 18d.
Where L = lengthof the bar for stirrup.
BEND LENGTH:
The bar isbentat the columnendtotie withthe footings.Thisextralengthforbendiscalledbend
length.
Bendlengthisgenerallyconsideredas16 d
BendLength= 16d
28. 28 | P a g e
SAQIBIMRAN 28
DEVELOPMENT LENGTH:
Developmentlengthisthe lengthof barrequiredfortransferringthe stressintoconcrete.
In simple words,the quantityof the rebarlengththatisactuallyrequiredtobe embeddedintothe
concrete to create the desiredbondstrengthbetweensteel andconcrete andfurthermoretoproduce
requiredstressforthe steel in thatarea.
The formulafor developmentisgivenbelow:
Developmentlength(Ld) =d x σs/τbd
Where
d = Diameterof the bar.
σs = Stressin the bar at the sectionconsideredasdesignload
τbd= Designbondstress.
29. 29 | P a g e
SAQIBIMRAN 29
LAP LENGTH:
Lap lengthisthe overlappinglengthof twobarsside byside whichgives requireddesignlength.InRCC
structure if the lengthof a bar isnot sufficientlyavailable tomake designlength,lappingisdone.
Suppose we needtobuilda20 m tall building.Butisthere any20 m bar available inthe market?No,the
maximumlengthof rebarisusually12 m, so we needto jointwobars to get20 m bar.
Lap lengthfor tensionmembers= 40d
Lap lengthfor compressionmembers= 50d.
d = Diameterof bars.
30. 30 | P a g e
SAQIBIMRAN 30
Look at the image below.Youmaybe seenthison the terrace (topfloor) of buildings.Thisextrarebaris
leftforfuture constructionpurpose.Ihope youunderstandthis.
CRANK LENGTH:
Generally,barsare bentnearthe supportat an angle of 45°. The angle of bendmay alsobe 30° in
shallowbeams.The purpose of bendnearthe supportisfirstlytoresistthe negative bendingmoment
whichoccurs inthe regionof the supportandsecondlytoresistthe shearforce whichisgreaterat the
support.
31. 31 | P a g e
SAQIBIMRAN 31
Crank bars are mostly providedin slabs.
Crank length= D/sin45°– D/tan45° =1.42D – D = 0.42D
So CrankLength= 0.42D
Where D = Clearheightof the bar = Thicknessof slab – (Topcover+ Bottomcover) – Diameterof the bar
UNIT WEIGHTOF STEEL:
The weightof bar iscalculatedbythe followingformula
W = d²L/162
Where W = Weightof bars.
L = Lengthof bars inmeter.
d = Diameterof the bar.
Example: Calculate the weightof 20 meterslong16 mm ø bar
W = 16² x 20/162 = 32 kg.
CUTTING LENGTH OF BENT UP BARS IN SLAB:
As a site engineer,youneedtocalculate the cuttinglengthof bars accordingto the slabdimensionsand
give instructionstothe barbenders.
For small areaof construction,youcanhand overthe reinforcementdetailingtothe bar benders.They
will take care of cuttinglength.Butbeware,thatmustnotbe accurate.Because theydo notgive
importance tothe bendsand cranks.Theymay give some extrainchestothe bars forthe bendswhich
are totallywrong.Soitis alwaysrecommendedthatasa site engineercalculate the cuttinglength
yourself.Inthisarticle,we will discusshow tothe calculate lengthforreinforcementbarsof slab.Let’s
start withan example.
EXAMPLE:
Where,
32. 32 | P a g e
SAQIBIMRAN 32
Diameterof the bar = 12 mm
ClearCover= 25 mm
ClearSpan(L) = 8000
SlabThickness= 200 mm
DevelopmentLength(Ld) =40d
CALCULATION:
CuttingLength= ClearSpan of Slab+ (2 x DevelopmentLength)+(2 x inclinedlength) –(45° bendx 4) –
(90° bendx 2)
Inclinedlength=D/(sin45°) – dD/(tan 45°) = (D/0.7071) – (D/1)=(1D – 0.7071D)/0.7071= 0.42 D
As youcan see there are four 45°bendsat the innerside (1,2,3 & 4) and two 90° bends( a,b ).
45° = 1d ; 90° = 2d
CuttingLength= ClearSpan of Slab+ (2 X Ld) +(2 x 0.42D) – (1dx 4) – (2d x 2) [BBSShape Codes]
Where,
d = Diameterof the bar.
Ld = Developmentlengthof bar.
D = Heightof the bendbar.
In the above formula,all valuesare knownexcept‘D’.
So we needtofindoutthe value of “D”.
D = SlabThickness – (2 x clearcover) – (diameterof bar)
= 200 – (2 × 25) – 12
= 138 mm
Now,puttingall valuesinthe formula
CuttingLength= ClearSpan 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)
∴ CuttingLength= 8980 mm or 8.98 m.
So forthe above dimension,youneedtocut the mainbars 8.98 m in length.
BAR BENDING SCHEDULE OF LINTEL BEAM:
In thisarticle,Iwill discusshowtoprepare BBSof RCC Lintel Beam.
33. 33 | P a g e
SAQIBIMRAN 33
1. Calculate Total Length Of Main Bars:
Lengthof 1 bar = Lengthof lintel –clearcoverfor bothsides
= 2500 – 2 x 25 [Clearcoverforboth sides]
=2450 mm
= 2.4 m.
Lengthof 4 bars = 2.4 x 4 = 9.6 m
2. Calculate Weight Of Steel For Main Bars:
Weightof steel for12 mm bar = D²L/162 = 12² x 9.6/162 = 8.53 kg.
3. Calculate No Of Stirrups:
No of stirrups= (Total lengthof lintel/c/cdistance between strriups) +1
= (2500/150) + 1 = 18
4. Calculate Total Length Of Stirrups:
Innerdepthdistance =150 -25 -25 -8 =84 mm
Widthdistance = 150 – 25 – 25 -8 = 84 mm.
Cuttinglengthof stirrups =(2x Innerdeothdiatance) +(2xWidthdepth)+HooksLength – Bend
Hookslength= 10d
Bend= 2d
We have 2 hooksand 5 bend
So,
Cuttinglengthof stirrups= (2×84) + (2 x84) +2x10x8 -5x2x8= 418 mm= 0.418 m
Total lengthof stirrups= 0.418 x 18 = 7.54 m
5. Calculate WeightOfSteel For Stirrups:
Weightof steel forstirrups= D²L/162 = 8² x 7.54/162 = 7.61 kg.
Total weightof steel forlintel =8.53 + 7.61 = 16.14 kg.
34. 34 | P a g e
SAQIBIMRAN 34
THE FORMULA D²L/162 FOR CALCULATING WEIGHT OF STEEL BARS FULLY DERIVED
THE FORMULA D²L/162 FULLY DERIVED:
To calculate the weightof steel barsyoumustuse the formulaD²L/162, but doyou know where the
formulaD²L/162 came from? Okay,noproblemif youdon’tknow.Inthisarticle,I will share how to
derive the formulaD²L/162. Solet’sstart fromthe beginning.
The formula isD²L/162
Where,
D = Diameterof steel barinmillimeter
L = Lengthof steel barsinmeter
CalculationOf The Formula D²L/162:
We knowthat,
Weight=Crosssectional areax Length x Density
For steel bar,thisalsoremainssame.
The weightof steel bars=Cross sectional areaof steel barx Lengthof steel barx Densityof steel bar.
That means,
W = A x L x ρ
Where,
W = Weightof steel bars
A = Area = πD²/4
π (pi) = 3.14
D = Diameterof steel barinmillimeter
L = Lengthof steel barinmeter
ρ (Rho) = Densityof steel bar= 7850 kg/m³
Therefore,
W = 3.14 x D²/4 x L x 7850
But there istwoconflictingunitinthe formula.WhichismillimeterforDand meterforρ (Rho).
So we needtoconverteitherD or ρ to the same unit.
Let’schange the unitof D from millimetertometer.
1 millimeter=0.001 meter
Let’sput thisintothe formula,
W= 3.14 x {(D²x 0.001 x 0.001)/4} x L x 7850
= D²L/162
Hope you all understoodthiscalculation.Usingthisformulawe caneasilycalculate the weightof steel
bars.
10 + THUMB RULES FOR CONCRETEMIXDESIGN
35. 35 | P a g e
SAQIBIMRAN 35
FOR ADDING 4 LITERS OF WATER IN 1 CU.M FRESHLY MIXED CONCRETE
1. The slumpvalue will be increasedby25 mm.
2. The compressive strengthof concrete will be decreasedby1.5to 2.0 N/mm2
3. The shrinkage potential will be increasedby10%.
4. 1/4 bag of cementwill be wasted.
IF THE TEMPERATURE OF FRESHLY MIXED CONCRETE IS INCREASED BY 1%, THEN
1. 4 litersof waterpercu.m will give equal slump.
2. The air contentwill be decreasedby1%.
3. The compressive strengthof concrete will be decreasedby 1.0to 1.5 N/mm2.
IF THE AIR CONTENT OF FRESHLY MIXED CONCRETE IS
1. Increasedby1% thenthe compressive strengthwill be decreasedby5 %.
2. Decreasedby1%, yieldwill be decreasedby0.03 cu.mper 1 cu.m.
3. Decreasedby1%, thenthe slumpvalue will be decreasedabout12.5 mm.
4. Decreasedby1%, thenthe durabilityof the concrete will be reducedby10%.
CALCULATIONOF MATERIALS FOR DIFFERENT MIX RATIO:
Quantityestimationof materialsisessentiallyrequiredinanyconstructionworksandquantity of
materialsdependonthe mix proportionsof the concrete.Inourpreviousarticle,we have already
discussed howtocalculate bricksina wall. Todaywe will discusshow tocalculate quantitiesof materials
for differentmix ratioof concrete.(Drymix method)
We will calculate quantitiesof materialsfor1 m3 concrete (By volume).
Let usassume the mix proportionis1 : 2: 4 (cement:sand:stone=a:b:c)
Volume of wetconcrete =1 m3
Volume of dryconcrete = 1 × 1.54 = 1.54 m3
What Is 1.54 GivenBelow:
36. 36 | P a g e
SAQIBIMRAN 36
CALCULATIONFOR CEMENT:
Formula,Cement=(Volume of dryconcrete/a+b+c) ×a
= (1.54/a+b+c) × a = [(1.54/1+2+4)] × 1 = 0.22 cum
Nowdensityof cement=1440 kg/cum
∴ Volume of cement=0.22 × 1440 =316.8 kg.
As we know,1 bag of cementcontains50 kg of cement.
∴ Cementbagsrequired=316.8/50 = 6.33 bags.
CALCULATIONFOR SAND:
Formula, Sand= (Volume of dryconcrete/a+b+c) ×b
= (1.54/a+b+c) × b = (1.54/1+2+4) × 2 = 0.44 cum.
CALCULATIONFOR 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.
CALCULATIONFOR WATER CONTENT:
Let usassume the water/cementratioof the concrete is0.45.
w/c = 0.45
Requiredwaterfor1 bag cement= 0.45 × 0.0353 = 0.0159 cum.
Where volume of 50 kg cement= 0.0353 cum
1 m3 water= 1000 Litre
Requiredwaterfor1 bag cement=0.0159 × 1000 = 15.9 Litre.
∴ Requiredwaterfor6.33 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:Yieldof concrete isconsideredas67% & Wastagesof materials=2%
You can use the same formulafordifferentmix proportionssuchas1:1.5:3 etc.
Here,we have usedcubicmeterunitbut youcan alsocalculate incubic feetunit.
UNIT WEIGHT OF BUILDING MATERIALS
Unit WeightOf BuildingMaterials:
38. 38 | P a g e
SAQIBIMRAN 38
Clinker 750 Kg/m3
Rubber 1300 Kg/ m3
Slag 1500 Kg/ m3
ClaySoil 1900 Kg/ m3
Sand (dry) 1540 to 1600 Kg/ m3
Sand (wet) 11760 to2000 Kg/m3
Steel 7850 Kg/ m3
Chalk 2100 Kg/ m3
Bitumen 1040 Kg/ m3
CALCULATION OF BRICKS IN A WALL
Estimationof buildingmaterialsisessentiallyrequiredbefore startinganew constructionproject.Inour
previousarticle,we have alreadydiscussed how tocalculate cement,sand,andaggregates. Todaywe
will discussthe easiestmethodtocalculate the requirednumberof bricksina wall.Solet’sstartfrom
the beginning.
REQUIRED DATA:
1. Volume of the wall.
2. Volume of a standardbrick.
3. Detailsof openingsinthe wall.
PROCEDURE:
1. First, calculate the volume of the wall tobe built:
Let usassume,
The lengthof the wall (l) = 10 foot.
The heightof the wall (h) = 10 foot.
Thicknessof the wall (b) = 1 foot.
Calculate the volume of the wall by multiplyingthe length,height,and thickness.
39. 39 | P a g e
SAQIBIMRAN 39
∴ Volume of the wall =l×h×b = 10×10×1 =100 Cu.F
2. Calculate the volume ofthe brick:
The standard size of a brick (ISStandard) is190 mm ×90 mm ×90 mm and
withthe mortar joint,itbecomes200mm × 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 findout the total nosof 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. Consider10% wastages ofbricks.
2. If the wall has anyopeningssuchas doors,windowsetc subtractthe volume of the openingsfromthe
volume of wall andthendivide itbythe 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 methodsforcalculatingnumberof blocksinawall.Inthisarticle,Iwill explaintwo
simple methodstofindouttotal no.of blocksfor a wall.
Surface area method.
Volume method.
PROCEDURE:
1. CALCULATION BY SURFACE AREA METHOD:
REQUIRED DATA:
40. 40 | P a g e
SAQIBIMRAN 40
1. Surface area of the wall.
2. Surface area of a standardconcrete block.
3. Surface area of openingsinthe wall.
STEP 1:
Determine The Length&HeightOf The Wall:
Let,the Lengthof the wall = 10 feet.
The heightof the wall = 10 feet.
STEP 2:
Calculate The Surface AreaOf The Wall:
We know,Surface Area= Lengthx Height
∴ Surface areaof the wall = 10 x 10 = 100 sq.feet.
STEP 3:
Calculate The Surface AreaOf Openings:
If the wall hasany openingssuchasdoor,windowsetc,calculate the surface areaof openings.
Let the wall has a windowof 3′ x 3′
∴ Surface areaof the window= 3 x 3 =9 sq.ft
STEP 4:
SubstractThe Surface AreaOf OpeningsFromSurface AreaOf Walls:
∴ Final Surface Areaof wall = 100 – 9 = 91 sq.ft.
STEP 5:
Calculate The Surface AreaOf A Concrete BlockWith Mortar Allowance:
Let the blockis16″ x 8″ x 8″ and mortar allowance is1″
∴ Surface Areaof one blockwithmortar = (16+1) x (8+1)= 153 inch²
= 153/12×12 = 1.0625 sq.ft
STEP 6:
Divide The Total Surface AreaOf The Wall By Surface AreaOf One Block:
∴ Numberof blocks= Surface Areaof wall/SurfaceAreaof a block= 91/1.0625 = 86
41. 41 | P a g e
SAQIBIMRAN 41
Consider5%wastage of concrete blocks.
So the total numbersof blocksrequired=86 + (86 x 5/100) = 86 + 4 = 90.
2. CALCULATION BY VOLUME METHOD:
REQUIRED DATA:
1. Volume of the wall.
2. Volume of a standardconcrete block.
3. Volume of openingsinthe wall.
STEP 1:
Determine The Width,Height,AndThicknessOf The Wall:
Let,the widthof the wall = 10 feet.
The heightof the wall = 10 feet.
The thicknessof the wall = 8″ = 0.67 feet.[ Thicknessof wall willbe same asthe thicknessof one
concrete block]
STEP 2:
Calculate The Volume Of The Wall:
We know,Volume =Widthx Heightx Thickness
∴ Volume of the wall =10 x 10 x 0.67 = 67 cubic feet.
STEP 3:
Calculate The Volume Of Openings:
If the wall hasany openingssuchasdoor,windowsetc,calculate the volume of openings
Let the wall hasa windowof 3′ x 3′ x 0.67′ [ Thicknessof the window will be same asthe thicknessof the
wall]
∴ Volume of the window =3 x 3 x 0.67 =6 cu. ft
STEP 4:
SubstractThe Volume Of OpeningsFromVolumeOf Walls:
∴ Final volume of wall =67 – 6 = 61 cu.ft.
STEP 5:
Calculate The Volume Of A Concrete BlockWithMortar Allowance:
Let the blockis16″ x 8″ x 8″ and mortar allowance is1″
∴ Volume of one blockwithmortar= (16+1) x (8+1) x 8 = 1224 inch3 [The mortal will be givenonthe
upperside andone side of everyconcrete block]
= 1224/12x12x12 = 0.70 cu.ft
STEP 6:
Divide The Total Volume Of The Wall ByVolume Of One Block:
∴ Numberof blocks= Volume of wall/Volumeof ablock= 61/0.70 = 86
Consider5%wastage of concrete blocks.
So the total numbersof blocksrequired=86 + (86 x 5/100) = 86 + 4 = 90.
Note:
If the wall doesnothave anyopening skipstep4, and justdivide the surface areaof wall bythe surface
area of one block.
HOW TO CALCULATE QUANTITY OF MATERIALS IN ARCH
In thisarticle,Iwill discusshowtwocalculate quantityof materialsinarch.
42. 42 | P a g e
SAQIBIMRAN 42
1. SEMI CIRCULAR ARCH:
Given:
Span = 8 ft
Height= 4 ft
Thickness= 1 ft
Breadth= 1ft
To calculate the quantityof materialsinsemi circulararchwe needto calculate the volume of arch.
Volume of arch= Cross-sectionof archx Archlength
We knowCross-sectionarea=Thicknessof arch x Breadth of arch = T x B
From the above image,youcan see the thicknessandbreadthare 1 ft.
Cross-sectionof arch= 1 x 1 = 1 ft²
NowArch length= Circumferenceof half circle (The archissemi circular)
We knowcircumference of circle =2πR
Circumference of half circle =2πR/2 = πR
We don’twhatis R. Sowe needtofindoutthe ‘R’.
R = Half span + Thickness/2= 4 +1/2 = 4.5 ft.
Arch length= πR =3.14 x 4.5 = 14.13 ft.
Quantityof materialsinarch = 1 x 14.13 = 14.13 cft.
2. SEGMENTALARCH:
Given:
Span = 8 ft
Height= 2ft
Thickness= 1 ft
Breadth= 1ft
Central angle = 105°
43. 43 | P a g e
SAQIBIMRAN 43
SOLUTION:
Volume arch= Cross-sectionof archx Arch length
We knowCross-sectionarea=Thicknessof arch x Breadth of arch = T x B
Arch length= θπR/180
NowR = (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
Quantityof materialsinarch = 1 x 7.61 = 7.61 cft.
HOW TO CALCULATE SHUTTERING AREA FOR CONSTRUCTION
Shutteringorformworkisa temporary,rigidstructure inwhichthe cast insituconcrete ispouredfor
castingthe memberstorequiredshape.Differentformworksare usedfordifferentshape members.
44. 44 | P a g e
SAQIBIMRAN 44
In thisarticle,we will discusshowtocalculate the shutteringareabefore construction.
Before calculatingthe shatteringareafirst,we needto know how to calculate the peripheral lengthof
any shape.
PERIPHERAL LENGTH/PERIMETER:
We knowthatperimeteristhe distance arounda2D (twodimensional) shape.
For example,square hasfoursides.Let‘s’be the lengthof one side,then
∴ The peripheral length=s+s+s+s = 4s
Rectangle hasfoursidesalsobutthe lengthandbreadthis different,letone sidelengthis‘l’and
breadthis‘b’
∴ Peripheral length=l+b+l+b= 2l+2b
Nowcomingto the calculationof shutteringarea.
Shtteringarea= Pheripheral length*Depth
Let me give youan example
Pheripheral length=(2 * 1) + (2*0.8) = 2 +1.6 =3.6
∴ Total shutteringarea= 3.6 * 4 = 14.4 sq.m
HOW TO CALCULATE CEMENT BAGS IN 1 CUBIC METER
PROCEDURE TO CALCULATE CEMENT BAGS IN 1 CUBICMETER:
Let usconsiderthe nominal mix is1:2:4
Loss of cementisconsideredas2%
Outputof mix isconsideredas67%
To achieve 1 cum output,we need1/0.67 = 1.49 say 1.50 cum dry mix.
Nowadd 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
45. 45 | P a g e
SAQIBIMRAN 45
As we know,the densityof cementis1440 kg/cumand
Weightof 1 bag cement= 50 kg.
Therefore volumeof 1 bag cement= 50/1440
=0.0347 cum.
∴ No.of cementbags requiredin1cubic meter= 0.2171/0.0347
= 6.25 bags.
Note:You can use the same formulaforcalculatingcementforothernominal mixes.
QUANTITY OF CEMENT, SAND, & AGGREGATES FOR 1000 SQ FTSLAB
To calculate the quantityof cement,sand,andaggregatesrequiredforaflatslabfirstwe needto
calculate the volume.
Areaof flatslab= 1000 sq.ft=93 sq. m.
The minimumthicknessprovidedforslabis150mm .( Accordingto IS 456)
150 mm= 0.15 metres
Hence Wet volume =Areaof slabx Thicknessof slab
Wet volume =93 x 0.15 = 13.95 cu. m.
To calculate the quantitiesof materialswe needdryvolume
Dry volume = Wetvolume x 1.54
Check why 1.54 is multipliedwithwetvolume to get dry volume.
∴ Dry volume =13.95 x 1.54 = 21.48 cu.m (approximately).
Generally,M20 grade concrete ispredominantlyusedforconstructionpurposes.
M20 = 1: 1.5: 3 (Cement:Sand:Aggregates) =(a:b:c)
QUANTITY OF CEMENT:
Quantityof cement= Dry volume x (a/{a+b+c}
Quantityof 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 needthe quantityof cementintermsof bags.
So bymultiplyingthe quantityof cementwiththe densityof cement,we will the getquantityof cement
interms of kilograms.
Unit weightorDensityof cement= 1440 kg/cu.m.
Quantityof cementinkg(s) = 1440 x 3.90 = 5616 kg
We knowthat1 cementbag = 50 kgs
∴ No.of cementbags required=5616/50 kg= 112.32 bags =112 bags.
QUANTITY OF SAND:
Quantityof sand = Dry volume x (b/{a+b+c}
∴ Quantityof 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:
Quantityof aggregates= Dry volume x (c/{a+b+c}
∴ Quantityof 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.
46. 46 | P a g e
SAQIBIMRAN 46
Sand = 5.85 cu.m
Aggregates= 11.71 cu.m
Note:ConsiderYieldof concrete as67% &
Wastagesof materials=2%
You can use the same formulafordifferentmix proportionssuchas1:2:4 etc.
Here, we have calculatedincubicmeterunitbutyou can calculate incubicfeetunit.
HOW TO CALCULATE VOLUME OF CONCRETE FOR STAIRCASE:
Staircase isone of the essential partsof a building.Ithelpspeople toaccessdifferentfloorsinabuilding.
In thisarticle,Iwill explainhowtoestimate concrete volume andquantityof cement,sand,aggregates
for a staircase.Before startingthe estimation,we needtoknow differentcomponentsof astaircase
whichI will use inthiscalculation.
COMPONENTSOF STAIRCASE:
STEPS:
A portionof stairwaycomprisingthe treadandriserwhichpermitsascentanddescentfromone floorto
another.
TREAD:
The horizontal upperpartof a stepon whichfootisplacedinascendingordescendingstairway.
RISER:
The vertical portion of a stepprovidingsupporttothe tread.
FLIGHT:
A seriesof stepswithoutanyplatform,breakorlandingintheirdirection.
LANDING:
A platformorrestingplace providedbetweentwoflights.
Nowcomingto the calculationof concrete volume forthe staircase.
Let’s take an example:
47. 47 | P a g e
SAQIBIMRAN 47
The heightof one floor= 12 feet.
Heightto be acquiredbyone flight= 12/2 = 6 feet.
Risers= 6″ = 0.5 feet.
Numberof risers= Heightof flight/Riser= 6/0.5 = 12
Tread = 10″ = 0.8 feet.
Numberof treads= (Numbersof risers-1) = 12 – 1
Thicknessof waistslab= 6″ = 0.5 feet
Lengthof step= 5 feet
VOLUME OF CONCRETE FOR STEPS:
Volume of one step= 1/2 x riserx treadx lengthof step
= 1/2 x 0.5 x 0.8 x 5 = 1.03 = 1 cubic feet.
As we have 11 numberof stepsina flightthe volume of stepsforfirstflight =11 x 1 = 11 cubicfeet.
CONCRETE VOLUME FOR WAIST SLAB:
To calculate the volume of concrete forwaistslabwe needtoknow the inclinedlengthof the waistslab.
Horizontal lengthof waistslab= Treadx Numberof steps
=0.8333′ x 11 = 9.2 feet.
48. 48 | P a g e
SAQIBIMRAN 48
The heightof the top of the landingfromfloor= Numberof riserx Heightof riser
= 12 x 0.5 = 6 feet.
The volume of concrete forwaistslab= Inclinedlengthof waistslabx Widthof waistslabx Thicknessof
waistslab
= 11′ x 5′ x 0.5′
= 27.5 cubicfeet
∴ Volume of concrete forfirstflight=Volume of waistslab+Volume of steps
= 27.5+11 = 38.5 cubic feet
VOLUME OF CONCRETE FOR LANDING OF STAIRCASE:
Lengthof landing=10.5 feet
Widthof landing=5 feet
Thicknessof landing=0.5 feet
∴ Concrete volume forlanding=10.5′ x 5′ x 0.5′ = 26 cubicfeet.
CONCRETE VOLUME FOR 2ND FLIGHT OF THE STAIRCASE:
As the 1st lightand 2nd flightare same inour staircase sothe volume of concrete will be same.
Concrete volume for2ndflightof the staircase = 38.5 cubicfeet.
∴ Total volume of concrete forstaircase = Volume of firstflight+Volume of secondflight+Volume of
landing
= 38.5 + 38.5 + 26
= 103 cubic feet.
Well,thisisthe wetvolume of concrete.Toconvertwetvolume intodryvolume we needtomultiply
with1.54
∴ Dry volume of concrete = 103 x 1.54 = 159 cubic feet.
QUANTITY OF MATERIALS FOR STAIRCASE:
We will calculate for1:2:4 Concrete mix.
1. Volume of cement= 1/7 x Dry volume = 1/7 x 159 = 22.66 cubic feet.
∴ Numberof Cementbags= volume of cement/Volume of cementbag = 22.66 / 1.25 = 18 No’sof bag.
(The volume of one bagcementof 50 kg = 1.25 cubic feet)
2. Volume of Sand= 2/7 x Dry volume = 2/7 x 159 = 45 cubicfeet.
3. Volume of Aggregate =4/7 x Dry volume =4/7 x 159 = 91 cubic feet.
Note:Here I have usedcubicfeetunitand1:2:4 mix ratio,you can use cubic meter unitandany other
mix ratioof concrete.
WHAT IS 1.54 IN CONCRETE CALCULATION?
49. 49 | P a g e
SAQIBIMRAN 49
In concrete calculation,we alwaysmultiplyWetvolumewith1.54to get dry volume.Butdoyouknow
whatis 1.54 or where this1.54 came from?If You don’t know,no problem. Inthisarticle,Iwill explain
whatis 1.54 while calculatingquantityof cement,sand,aggregatesforconcrete.
letus take a concrete cube.
The lengthof the concrete cube = 1 m
The widthof the concrete cube = 1 m
The heightof the concrete cube = 1 m
Volume of concrete cube = lengthx Widthx Height= 1 x 1 x 1 = 1 m³ (Wetvolume)
Whenwe convertthiswe volume intodryvolume,the volumeisincreasedby54% of wetvolume.
∴ Dry volume =Wet volume +54% of Wet volume
= 1 + (54/100) x 1 =1 + 0.54 = 1.54
To convertwetvolume of 1 m³ concrete intodry volume =1.54
To convertwetvolume of “n” m³ concrete into dryvolume = 1.54 x n
Where n = Wet volume of concrete.
Example:
50. 50 | P a g e
SAQIBIMRAN 50
The lengthof the concrete cube = 4 m
The widthof the concrete cube = 3 m
The heightof the concrete cube = 2 m
Wet volume =4 x 3 x 2 = 24 m³
∴ Dry volume =( Wet volume +54% of wetvolume)
= 24 + {(54/100) x 24}
= 36.96 m³
Note:
Alwaysrememberinconcrete calculation,we getwetvolume first,toconvertwetvolume intodry
volume,multiplywetvolumewith1.54.
HOW TO CALCULATE RECTANGULAR WATER TANK SIZE AND CAPACITY
For highrise building,the dimensionof watertankalongwithdrawingsare providedfromarchitect
office.
But for small residentialbuilding,we needtocalculate byitself.Inthisarticle,we will discusshow to
calculate rectangularwatertanksize andcapacity.
VOLUME OF WATER REQUIREMENT:
As perIS code,a personneeds135 litreswaterperday fordailyuse.
Drinking– 5 Litre
Cooking– 5 litres
Bathingand Toilet–85 litres
WashingclothesandUtensils –30 litres
House Cleaning–10 litres
CALCULATIONOF TANK SIZE AND CAPACITY:
We will calculate foratypical familyhaving4members.
Total waterrequirement=135 x 4 = 540 litres
To calculate the size of watertank we mustknow any one of the following:
51. 51 | P a g e
SAQIBIMRAN 51
Lengthor widthor depthof the tank.
From the formulaof volume of water
1 m3 = 1000 litres
1 litre = 0.001 m3
We are requiring540 litre water
∴ 540 litres= 0.54 m3
Let usassume the depthof water tankis 0.6 m.
∴ Areaof tank= 0.54/0.6 = 0.9 m2
L x B = 0.9 m2
Nowif you knowthe lengthorbreadthyoucan putthe value onthe above formula.
otherwise,take the lengthas2 timesof 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 watertank size
Length(L) = 1.34 m
Breatdh(B) = 0.67 m
Depth(D) = 0.6 m
52. 52 | P a g e
SAQIBIMRAN 52
SEPTIC TANK SHAPE, SIZE & DIMENSIONS WITH TABLE
SEPTIC TANK SHAPE, SIZE & DIMENSIONS:
Septictankis a devise whichdealswiththe sewage fromwaterclosets.Inotherwords,itisa watertight
single-storeyed,undergroundtankinwhichsewage isretainedsufficientlylongtopermitsedimentation
of suspendedsolidsandpractical digestionof settledsludge byanaerobicbacterial action.
SHAPE OF SEPTIC TANK:
The simplestformof a septicformisa single compartment,rectangularorcylindrical inshape.The
shape of septictank influencesthe flowspeedof wastewaterandsludge accumulation.
The dimensionof the septictankshouldnotbe toodeepertooshallow.Because itcancause short-
circuitingof the inletandoutletflow.
Septictankhavinga greatersurface area withsufficientdepthshouldalwaysbe preferred.
Rectangularshapedsingle compartmenttank(withlengththree timesof itswidth) ismore favorable.
53. 53 | P a g e
SAQIBIMRAN 53
In some cases,cylindrical shape withsuitable size isalsosoundtobe a betterchoice.
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
HeavyDuty
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
54. 54 | P a g e
SAQIBIMRAN 54
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:
Septictankis a tank where sewage iskeptalongtime forsedimentationof suspendedsolidby
anaerobicmicroorganismactivity.
Septic Tank Design
Basically,the tankisrectangularinshape consistingof rooftopwithtwochambers,isolatedfromeach
otherby a baffle wall.The firstchamberiscalledgritchamberinwhichthe sewage entersfirstandthe
restone is calledanaerobicchamber.
Inletandoutletpipe shouldbe fittedinsucha mannerthat while enteringorexistingthere shouldnot
be any unsettledinfluenceinthe sewage of anaerobicchamber.Sand,gritetcissettleddowninthe grit
chamberbefore the sewage isdirectedtothe anaerobicchamber.Inthe anaerobicchamber,natural
solidssettle atthe base of the tank where anaerobicbacteriafollowsuponitand changesovercomplex
unsteadymixestomore straightforwardstable mixes.
SEPTIC TANK DESIGN:
The followingmeasure shouldbe keptinview todesignaseptictank:
1. The floorzone of the git chambermustbe adequate todecrease the speedof steamandtoallow
sedimentation.
2. The capacityof a septictankshouldbe sufficienttogive adetainmentperiodfluctuatingfrom12
hoursto 3 days,24 hours time frame isnormallyviewedassatisfactory.
3. The depthbeneathsegmentdivideropeningshouldbe adequatetoallow aggregationforthe
predeterminedperiod.A minimumareaof 0.07 m2 for eachclientingrit chamberandminimum
volumetricsubstance of the gritchamberof 0.02 m3 foreveryclientisrequired.
55. 55 | P a g e
SAQIBIMRAN 55
4. The minimumwidthanddepthof aseptictank shouldbe 0.75 m and1 meterunderneathwaterlevel.
The lengthshouldbe 2 to 4 timesof the width.The minimumcapacityof a septictankisgranted1 m3.
5. Every septictankshouldbe furnishedwithaventilatingchannel minimumdiameterof 5cm.
HOW TO CALCULATE VOLUME OF CONCRETE, SHUTTERING AREA & BITUMEN PAINT AREA
FOR PLAIN FOOTING
In thisarticle Iwill discusshowtocalculate the volume of concrete,shutteringareaandbitumenpaint
area forplain footing(Square footing&Rectangularfooting).
A. SQUARE FOOTING:
Given,
Lengthof footing=0.5 m
Breadthof footing=0.5 m
Depthof footing=0.25 m
Size of column= 0.2 x 0.1
1. Volume ofconcrete = Area of sq. footingx Depth of sq footing
Areaof 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. Shutteringarea = 2 (L + B) x D
= 2 (0.5+0.5) x 0.25
= 0.5m²
3. Bitumenpaint area = Shutteringarea + (Top area offooting – column area)
= 0.5 + [(0.5 x0.5) -(0.2 x 0.1)]
=0.5 + [0.25 – 0.2)]
= 0.73 m².
2. RECTANGULARFOOTING:
56. 56 | P a g e
SAQIBIMRAN 56
Given,
Lengthof footing=1 m
Breadthof footing=0.7 m
Depthof footing=0.25 m
Size of column= 0.4 x 0.3
1. Volume ofconcrete = Area of sq. footingx Depth of sq footing
Areaof sq. footing= L x B = 1 x 0.7 = 0.7 m²
Volume of concrete =0.7 x 0.25 = 0.175 m³.
2. Shutteringarea = 2 (L + B) x D
= 2 (1+0.7) x 0.25
= 0.85m²
3. Bitumenpaint area = Shutteringarea + (Top area offooting – column area)
= 0.85 + [(1x0.7) -(0.4 x 0.3)]
=0.85 + [0.7 – 0.12)]
= 1.43 m².
HOW TO CALCULATECEMENT, SAND QUANTITY FOR PLASTERING?
Followingpointsshouldbe rememberedwhile calculatingthe quantityof cement,sandforplastering
work.
1. For wall plastering,Cement:Sand= 1 : 6
2. For ceilingplastering,Cement:Sand= 1 : 4
3. Thicknessof plastershouldbe inbetween12-15 mm.If an additional coatisrequiredthendonotdo
it at one go.
4. Use good qualityof cement&Sand.
5. Use measuringbox (notheadpan) forsite mix.
We will calculate cementandsandfor100 m2 plasteringareain1:6 ratio and thicknessof 12 mm.
CementMortar Required:
Plasteringthickness=12 mm
= 12/1000 = 0.012m
57. 57 | P a g e
SAQIBIMRAN 57
Volume of cementmortarrequired=( PlasteringAreax thickness)
= 100 m2 x 0.012m = 1.2 m3
(Thisiswetvolume of cementmortar(aftermixingwater) butwe needdryvolume.Togetdry volume
consider35% bulkingof sandand20% wastages
= 1.2 m3 x (1+0.2+0.35) (Ratherthan35% sandbulkage and20% wastage youcan add 1.54 as constant)
= 1.86 m3
Cement:Sand = 1 : 6
Sumof ratio =( 1 + 6) = 7
∴ Cementrequired
= 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 calculatedinSq.mbut youcan also calculate itinSq.ft.
HOW TO CALCULATE QUANTITY OF PAINT FOR BUILDING:
Paintingisone of the most essentialstepstobe done afterconstructionof anybuilding/house.Painting
isdone to protectthe surface of the wallsas well asto increase appearance.
In thisarticle,Iwill discusshowtocalculate the quantityof paintrequiredinabuilding.
Before startingthe calculationletme tell youone importantthingthatquantityof paintiscalculatedin
gallon.
There are two typesof gallons
US gallonand Uk gallon.
1 US gallon= 3.785 litres
1 Uk gallon= 4.546 litres.
In thiscalculationwe will use USgallon.
58. 58 | P a g e
SAQIBIMRAN 58
Let the lengthof the wall = 15 feet
The heightof the wall = 12 feet.
Areaof the wall 15 x 12 = 180 sft.
THUMB RULE FOR PAINT CALCULATION:
Generally,1gallonof paintcan coverup to 350 sftarea of wetwall and
1 gallonof paintcan coverup to 200 sftarea of dry wall.
PAINT QUANTITY OF WET WALL:
Areaof wetwall =180 sft
Paintcoveredarea= 300 sft
Quantityof paint= 180/350 = 0.514 litre.
Let the price of 1 gallonpaint= Rs 1200.
Price of 0.514 gallonpaint= 0.514 x 1200 = 617 Rs.
PAINT QUANTITYOF DRY WALL:
Areaof dry wall = 180 sft.
Paincoveredarea= 200
Quantityof paint= 180/200 =0.9 gallon.
Price of 0.9 gallonpaint= 0.9 x 1200 = 1080 Rs.
Note:
1. If the wall has anyopenings,subtract the areaof openingsfromtotal area& thencalculate paint
quantity.
HOW TO CALCULATE SHUTTERING OIL AND ITS PRICE
Shutteringoil isveryuseful material requiredforeveryconstructionprojects.Soitisveryimportantto
knowthe shutteringoil quantity.Inthisarticle,Iwill discusshow tocalculate shutteringoil anditsprice.
Before startingthe calculationletme tell youfirstwhatisshutteringoil andwhywe use itin
shuttering/formwork.
WHAT IS SHUTTERING OIL?
59. 59 | P a g e
SAQIBIMRAN 59
Shutteringoil are highqualitymineral oil whichisusedonthe shutteringplates/mouldforeasyremoval
of shuttering.
WHYWE USE SHUTTERING OIL INFORMWORK?
The main purposesof usingshutteringoil are asfollows:
1. Shutteringoil worksasa releasingagentbetweenshutteringplatesandconcrete.Ithelpseasy
removal of shutteringatthe time of deshuttering.
2. It givesa goodand betterfinishafterdeshuttering.
3. It keepsthe shutteringplatesandmouldingood conditionand ensuresrepetitive use inmore
projects.
HOW TO CALCULATE SHUTTERING OIL & ITS PRICE:
Let uscalculate the quantityof shutteringoil forthe above slab.
Lengthof slab= 12 m
Breadthof slab= 10 m
Areaof slab= L x B = 12x 10 = 120 m²
There are twotypesof shutteringplatesusedinconstruction.
1. WOODSHUTTERING:
Generally,1litershutteringoil coversupto 30to 40 m² area in woodshuttering.
The quantityof shutteringoil =Total area/1 litercoveringarea
= 120/30 = 4 litre
Price: Letthe price of 1 litershutteringoil =Rs.250.
The price of 4 litershutteringoil =250 x 4 = Rs. 1000.
2. STEEL SHUTTERING:
Generally,1litershutteringoil coversupto 60to 70 m² area in steel shuttering.
The quantityof shutteringoil =Total area/1 litercoveringarea
= 120/60 = 2 litre
Price: Letthe price of 1 litershutteringoil isRs.250.
The price of 2 litershutteringoil =250 x 2 = Rs. 500.
60. 60 | P a g e
SAQIBIMRAN 60
HOW TO CALCULATE CONCRETE VOLUME FOR RETAINING WALL:
Retainingwall isamasonrywall constructedtoresistthe pressure of liquid,earthfilling,sandorother
granularmaterial filledbehindit.
In thisarticle,Iwill discusshowtocalculate the concrete volume forretainingwall.Let’sgetstarted.
EXAMPLE 1:
To calculate the volume of retainingwall we needtocalculate the volumeof retainingwall.
Here I have dividedthe retainingwall intotwoparts,partA isthe base slab andpart B isthe stemof
retainingwall.
So Volume of retainingwall =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 stemis 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 retainingwall=6 + 21 = 27 m³
So the volume of concrete forthe retainingwall = 27 m³
Let’stake anotherexample:
EXAMPLE 2:
61. 61 | P a g e
SAQIBIMRAN 61
Here I have dividedthe retainingwall intothree parts,partA isthe base slab,part B isthe stemandpart
C is the counterfortof the retainingwall.
The volume of retainingwall =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 isalsoa 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 retainingwall there is2counterfortso
Volume of C= 0.75 x 2 = 1.5 m³
The total volume of retainingwall=6 + 9 + 1.5 = 16.5 m³.
HOW TO CALCULATE NUMBER OF TILES IN A ROOM:
Calculationof tilesneededforfloorsandwallsissosimple andeasy.inthispost,Iwill discusshowto
calculate no.of tilesrequiredinaroom.So let’sstart.
STEP 1- CALCULATE AREA OF THE FLOOR:
62. 62 | P a g e
SAQIBIMRAN 62
letthe lengthof the floor= 14 feet.
Breadthof the floor= 12 feet.
∴ Areaof floor= lengthof floorx breadthof floor= 14 x 12 = 168 sft.
STEP 2 – CALCULATE AREA OF 1 TILE:
There are differenttile sizessuchas20cmx20cm, 30cmx30cm, 45×45, 60cmx60cm etc.In this
calculattionwe will use 60×60 cm tiles.
Lengthof 1 tile = 60 cm
Breadthof 1 tile = 60 cm.
∴ Areaof 1 tile = 60 x 60 = 3600 cm² = 3.6 sft.
STEP 3 – DIVIDE FLOOR AREA BY AREA OF 1 TILE:
∴ No.of tilesrequired=Areaof floor/areaof 1 tile = 168/3.6 = 46.6 ≈ 47.
∴ 47 tilesare requiredforthe above floor.
Note:
Consider5%wastages.
Similarly,youcanalsocalculate tilesneededforwalls.
HOW TO CALCULATE LAND AREA OR PLOT AREA:
Sometimeswe mayneedtoknowthe areaof any landor plot forvariousreasonssuchas buying,selling
of land,orconstructinghouse/buildingetc.Inthisarticle,Iwill explainhow tocalculate landare orplot
area fora site.
Area Formula For DifferentShapes:
63. 63 | P a g e
SAQIBIMRAN 63
But for complex shapedlandyoucannoteasilycalculatethe area.Inthat case, we use HERON’sFormula
whichisgivenbelow:
64. 64 | P a g e
SAQIBIMRAN 64
Where S = Perimeterof the triangle
S = (a+ b + c)/2
a = Distance of AB
b = Distance of BC
c = Distance of AC
Let’stake an example,
Firstof all,divide the landareaintominimumpossible no.of triangle.Thenmeasure all the required
distance (like AB,AC,AEetc) byusinga 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.
65. 65 | P a g e
SAQIBIMRAN 65
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 AreaOf Land = 58.78 + 75.89 + 71.55 = 206.22 ft²
66. 66 | P a g e
SAQIBIMRAN 66
NOTE: This methoddoesnot give appropriate resultforcurve shapedarea,forcurve shapedarea we
needtouse difinite integral method.
CENTER LINE METHOD OF ESTIMATION:
There are three differentmethodsof estimation.
1. Centerline method.
2. Long and shortwall method.
3. Crossingmethod.
In thisarticle,Iwill discusscenterlinemethodbrieflyforthe estimationof materials.
WHAT IS CENTER LINE METHOD:
In thismethodof estimation,the total centerline lengthof wallsinabuildingisfirstcalculated,thenthe
centerline lengthis multipliedwiththe breadthanddepthof respectiveitemtogetthe total quantityat
a time.
The centerline lengthfordifferentsections of wallsinabuildingshallbe workedoutseparately.For
verandahwallsorpartitionwallsjoiningthe mainwalls,the centerline lengthshallbe reducedbyhalf of
the breadthof the layerof mainwall that joinswiththe partitionorverandahwall atthe same level.The
numberof such jointsisstudiedfirsttoobtainthe centerline length.
By usingthismethodestimationcanbe finishedmore quickly.Thismethodisasaccurate as other
methods(exceptunsymmetrical wall).Thismethodissuitablyusedforestimatingcircular,rectangular,
hexagonal,octagonal etcshapedbuilding.
Let ustake an example:
FOR ONE ROOM BUILDING:
Firstwe needto calculate total centerline lengthof the building.
67. 67 | P a g e
SAQIBIMRAN 67
Total center line length = 2 x 5.3 + 2 x 4.3 = 19.2 m.
1. Earthwork inexcavation =Total centerline lengthx breadthx 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 foundationfor 1st footing = 19.2 x 0.6 x 0.3 cu.m
b) Brickwork in foundationfor 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 inexcavation = Total centerline lengthx breadthx 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 foundationfor 1st footing = 34.1 x 0.6 x 0.3 = 6.13 cu.m
b) Brickwork in foundationfor 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 youcan estimate quantityof differentitemssuchas
Quantity of D.P.C = Total Centre line lengthx Breadthof foundationx Thicknessof D.P.C
Quantity of plinthbeam = Total Centre line lengthx Breadthof the beamx Depthof beam.
Quantity of wall plasterfor 2 sides = Total centre line lengthx Heightof wall x Thicknessof plaster.
Quantity of Paint for 2 sidesofwall = Total centre line lengthx Heightof wall x 2 = Areaof paint
insq.ft
68. 68 | P a g e
SAQIBIMRAN 68
ONE WAY SLAB DESIGN – HOW TO DESIGN ONE WAY SLAB
In thisarticle,Iwill discusshowtodesignOne wayslabwithexample.
WHAT IS ONE WAYSLAB?
Whena slab issupportedonall foursidesandthe ratio of longspan to shortspan isequal or greater
than two,itwill be consideredasone wayslab.The loadon the slabis carriedbythe shortspan inone
direction.Howevermainreinforcementbaranddistributionbarintransverse direction.
Longerspan (l)/Shorterspan(b) ≥ 2
ACI CODE SPECIFICATIONS FOR ONE WAY SLAB DESIGN:
1. MINIMUMSLAB THICKNESS:
To control deflection,ACICode 9.5.2.1 specifiesminimumthicknessvaluesforone-waysolidslabs.
2. SPAN:
Accordingto ACIcode 8.7.1 If the slabrestsfreelyonitssupports,the spanlengthmaybe takenas equal
to the clear spanplusthe depthof the slab butneednotexceedthe distance betweencentersof
supports.
3. BAR SPACING:
The lateral spacingof the flexural barsshouldnotexceed3timesthe thicknesshor18 inch
accordingto ACI code 7.6.5 The lateral spacingof temperature andshrinkage reinforcementshouldnot
be placedfartherapart than 5 timesthe slabthicknessor18 inch accordingto ACIcode 7.12.2
4. MAXIMUMREINFORCEMENT RATIO:
Reinforcementratioisthe ratioof reinforcementareatogross concrete area basedontotal
depthof slab.One-waysolidslabsare designedasrectangularsectionssubjectedtoshearandmoment.
Thus,the maximumreinforcementratiocorrespondstoa nettensile straininthe
reinforcement,€tof 0.004
5. MINIMUMREINFORCEMENT RATIO:
A) FOR TEMPERATURE AND SHRINKAGE REINFORCEMENT :
According to ACI Code 7.12.2.1
SlabswithGrade 40 or 50 deformedbars –> 0.0020
SlabswithGrade 60 deformedbars –> 0.0018
Slabswhere reinforcementwithyieldstrengthExceeding60000 psi- ->( 0.0018 x 60000/fy)
B) FOR FLEXURAL REINFORCEMENT :
Accordingto ACICode 10.5.4, the minimumflexural reinforcementisnottobe lessthanthe shrinkage
reinforcement,or0.0018
EXAMPLE PROBLEM:
A reinforcedconcrete slabisbuiltintegrallywithitssupportsandconsistsof equal spanof 15 ft. The
service live loadis100 psf and 4000 psi concrete is specifiedforuse withsteel withayieldstressequal
to 60000 psi.Designthe slabfollowingthe provisionsof the ACIcode.
69. 69 | P a g e
SAQIBIMRAN 69
THICKNESS ESTIMATION:
For beingbothendscontinuousminimumslabthickness= L/28 =(15 x 120)/28= 6.43 in.
Let a trial thicknessof 6.50 in.
LOAD CALCULATION:
Consideronlya1 ftwidthof beam.
Deadload = 150 x (6.50/12)) = 81 psf
Live load= 100 psf
FactoredDL andLL ={81+1.2+(100 x 1.6)} =257 psf
DETERMINE MAXIMUMMOMENTS:
Factoredmomentsatcritical sectionsbyACIcode :
At interiorsupport: -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 exteriorsupport: -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,
70. 70 | P a g e
SAQIBIMRAN 70
= 2.64 in
CHECK FOR AVAILABILITYOF THICKNESS:
As ‘d’islessthanthe effective depthof (6.50-1.00) = 5.50 in,the thicknessof 6.50 in
can be adopted.
REINFORCEMENT CALCULATION:
Let,a = 1 inch
At interiorsupport:
Checkingthe assumeddepthof a by
Similarlyat Midspan:
As = (4.13 x 12)/(0.90 x 60 x 5.29) = 0.17 in²
At Exteriorsupport:
As = (2.41 x 12)/(0.90 x 60 x 5.29) = 0.10 in²
MINIMUMREINFORCEMENT:
As = 0.0018 x 12 x 6.50 = 0.14 in²
So we have to provide thisamountof reinforcementwhere Asislessthan0.14 in².
SHRINKAGE REINFORCEMENT:
Minimumreinforcementforshrinkageandtemperatureis
As = 0.0018 x 12 x 6.50 = 0.14 in²
FINAL DESIGN:
71. 71 | P a g e
SAQIBIMRAN 71
LAYOUT OF ONE WAYSLAB:
72. 72 | P a g e
SAQIBIMRAN 72
DESIGN OF COLUMNS AS PER ACI:
1. Maximumand MinimumReinforcementRatio:
The minimumreinforcementratioof 1 % isto be usedintiedor spirallyreinforcedcolumns.This
minimumreinforcementisneededtosafeguardagainstanybending,reduce the effectof shrinkageand
creepand enhance ductilityof columns.
2. MinimumNumberof ReinforcingBars:
Minimumfourbars withinrectangularorcircularsections;orone bar in eachcorner of the crosssection
for othershapesanda minimumof six barsin spirallyreinforcedcolumnsshouldbe used.
3. ClearDistance betweenReinforcingBars:
For tiedor spirallyreinforcedcolumns,cleardistance betweenbarsshouldnot be lessthanthe largerof
150 timesbardiameteror4 cm.
4. Concrete ProtectionCover:
The clear concrete covershouldnotbe lessthan4 cm for columnsnotexposedtoweatherorincontact
withground.It isessential forprotectingthe reinforcementfromcorrosionorfire hazards.
5. MinimumCross-Sectional Dimensions:
For practical considerations,columndimensionscanbe taken asmultiplesof 5cm.
6. Lateral Reinforcement:
Tiesare effective inrestrainingthe longitudinalbarsfrombucklingoutthroughthe surface of the
column,holdingthe reinforcementcage togetherduringthe constructionprocess,confiningthe
concrete core and whencolumnsare subjectedtohorizontal forces,theyserve asshearreinforcement.
73. 73 | P a g e
SAQIBIMRAN 73
Spirals,onthe otherhand,serve inadditiontothese benefitsincompensatingforthe strengthlossdue
to spallingof the outside concrete shell atultimate columnstrength.
7. Ties:
For longitudinal bars,32 mm or smaller,lateral ties10 mmin diametershouldbe used.Inourcountry
and insome neighboringcountries,tiesof 8 mmdia are usedfor columnconstruction
CALCULATION OF EARTH PRESSURE:
The thrust due to the back filling,whichmaybe assumedtobe earth,is generallycalculatedby
Rankine’stheory.The theoryisbasedonthe assumptionthatthe backingmaterial orearthconsistsof
cohesionlessgranularparticles.The formulaederivedfromthistheory underdifferentconditionsof
backfillingare givenbelow:
Case 1: Wallswith earth levelledwiththe top of the wall:
a) Horizontal pressure persq.m(ph) ata depthof (h) meterbelow the levelledtopisgivenbythe
followingformula:
Where
w = Weightof fillinginkg/m3
ϕ = Angle of repose of the soil.
b) Total horizontal pressure (P) atadepthof (h) meterpermeterlengthof the wall isgivenbythe
followingformula:
Acting at h/3 meter from the base.
Case 2: In case of submergedretaining wall orwall retainingearthfilledatslope of a° to the horizontal,
the formulagivinglateral earthpressure (ph)isgivenby:
Acting parallel to the surcharge slope of the filling.
Total pressure (P) atdepthhmeterpermeterlengthof the wall is givenbythe formula:
