QUANTITY SURVEY For Residential Building By MD USOOF ALAM

 Estimation is the scientific way of working out
the approximate cost of an engineering project
before execution of the work.
 In estimation and costing we will estimate all the
quantities like concrete, shuttering, brickwork,
plastering, reinforcement, etc.
 Then cost is applied.
 Estimation and costing is also known as quantity
survey.
 Quantity survey can be done manually or with
help of Microsoft excel.
2Dept of Civil Engg, A.H.C.E.T, Damaragidda,Chevella,T.S.

 Types of footings.
 Types of columns.
 Types of beams.
 Types of Slabs.
 Types ofWalls.
 Length calculation.
 Area calculation.
 Volume calculation.

Module-i~ Sub-structure calculation
<below ground level>.
Module-ii~ Super-structure calculation
<above ground level>.
Module-iii~ Reinforcement calculation
for RCC & Steel structure.
Module-iv~ Computer application < MS
excel sheet>.
Module-v~ Project work.

 PCC under footings.
 Footings.
 Neck columns.
 Tie beams.
 PCC under tie beams.
 Plinth beams.
 PCC under plinth beams.
 Grade slabs.
 Septic tank.
 Etc.

o Volume of concrete in m3
o Area of shuttering in m2
o Area of bitumen paint in m2
o Area of termite control in m2
o Area of polythene sheet in m2
o Reinforcement in Kgs. Or tons.

 Floor columns.
 Floor beams.
 Floor slabs.
 Stair case.
 Brick work.
 Door.
 Windows
 Over head water tank
 Flooring calculations

This is the process of digging earth on site where the foundation is to be laid

After excavation, compaction is done on the earth’s surface and then it is treated
with termite control (Anti fungus spray).

Polythene sheet is laid on the surface of ground afterTermite
Control Spray is spread. Generally they are 2mm to 6mm thick.

This is also called as blinding. It is just a mixture of cement, sand and
aggregate without reinforcement. Generally the Grade of concrete
for PCC is used as M7.5 with ratio or proportion of concrete as 1:4:8.

This is a Concrete with cement, sand and aggregate with
reinforcement. Generally the Grade of concrete is used as M15 with
ratio 1:2:4,

The part of Column which comes between the top of footing to
bottom of Ground beam is called as Neck column.

It’s a black colour paint, painted to the structure which is under ground before
back filling of Soil.The purpose of using the Bitumen paint is to
(i) Increase the life of the structure
(ii) Give strength to it
(iii)To acts as water-proofing agent
(iv)And to safe the structure from Corrosion.
(iv) Liquid or sheet form.

Formwork
Masonry (Stone Masonry and Brick Masonry)
Flooring
Plastering
Pointing
Doors and Windows
Whitewashing, Colour washing and Distempering
Painting
Lump sum Item
Electrification
Nomenclature if Item
Rates

Dept of Civil Engg, A.H.C.E.T, Damaragidda,Chevella,T.S. 31

 PCC CONCRETE = 7.5565m3
 GRADE OF CONCRETE USED FOR PCC M7.5
 RATIO = 1:4:8
 SAND: FOR 1m3 OF CONCRETE = 0.46m3
 FOR 7.5565m3 = 7.5565 X 0.46 =3.47599m3
 AGGRIGATE OF SIZE 40mm
 For 1 m3 of concrete = 0.92 m3
 For 7.5565 m3 = 7.5565 x 0.92 =6.95198 m3
 CEMENT FOR 1m3 OF CONCRETE = 165.60KGS
 FOR 7.5565 m3
 7.5565 X 165.60 =1251.3564KGS
 RCC CONCRETE = 101 m3
 GRADE OF CONCRETE USED FOR PCC M15
 RATIO = 1:2:4
 SAND: FOR 1m3 OF CONCRETE = 0.46m3
 FOR ,101m3 = 101 X 0.46 =46.46m3
 AGGRIGATE OF SIZE 20mm
 For 1 m3 of concrete = 0.92 m3
 For 101 m3 = 101 x 0.92 =92.92 m3

 CEMENT FOR 1m3 OF CONCRETE = 331.20KGS
 FOR 101 m3
 101 X 331.20 =33451.2KGS
 TOTAL SAND = 3.47599 + 46.46 = 49.95 m3
 AGGRIGATE 40mm = 6.95198 m3
 AGGRIGATE 20mm = 92.92 m3
 TOTALCEMENT = 1251.3564 + 33451.2 =34702.5564KGS
 = 695 BAGS
 TOTALQUANTITYOF SAND = 49.95 m3
 VOLUMEOF SAND EACHTRUCK ORTIPPER = AREA X DEPTH
 = 4.29 X 2.5 X 1.4 = 15.01 m3
 NOOF SAND IN EACHTRUCK = 15.01 m3
 NOOF SANDTRUCK REQUIRED =VOL OF SAND /VOL OFTRUCK
 = 49.95 / 15.01 = 3.765TRUCKSAPPROX 4
TRUCKS
 TOTALVOLUMEOF AGGRIGATE 40mm & 20mm = 100 m3
 NOOF AGGRIGATE EACHTRUCK = 9 m3
 NOOF AGGRIGATETRUCK REQUIRED =VOL OF AGGRIGATE /VOL OF
TRUCK
= 100 / 15.01 = 6.668TRUCKSAPPROX 7TRUCKS

 BRICK SIZE: 0.2m x 0.1m x 0.1m
 WALLTHICKNESS: 0.2m
 ROLLER SHUTTER: 2.43m x 2.43m
 WALLTHICKNESS: 0.23m
 BRICKWORK CALCULATION:
 BRICKWORK AREA:
 1.WALL: length=21.73m, depth=4m.
 Area=21.73 x 4= 86.92m2
 2.WALL: length=21.73m, depth=4m.
 Area=21.73 x 4= 86.92m2
 3.WALL: length=13m, depth=4m.
 Area=13 x 4=52m2
 Area=13 x 4=52m2
 WALL 4 Length same
 52 +52 +52 + 52 + 52 + 52 =312m2
 Area=21 x 4=84m2
 WALL 5 Length same
 84 + 84 +84 =252m2

 BRICKWORK AREA =86.92+86.92+52+312+252 = 789.84M2
 DEDUCTIONS:
 1. ROLLER SHUTTER: length=2.43m, depth=2.43m.
 Area=2.43 x 2.43= 5.91m2
 ROLLER SHUTTERTOTAL = 18
 PER ROLLER SHUTTER AREA = 5.91 X 18 = 107m2
 2. MAIN GATE AREA: length=3m, depth=4m.
 Area=3 x 4= 12m2
 MAIN GATE AREATOTAL = 6
 PER MAIN GATE AREA: 12 X 6 =72m2
 3. STAIR CASE AREA: length=3.35m, depth=4m.
 Area=3.35 x 4= 13.4m2
 STAIR CASE AREATOTAL = 3
 PER STAIR CASE AREA: 13.4 X 3 = 40.2m2
 TOTAL DEDUCTION AREA = 107 + 72 + 40.2 = 219.2m2
 TOTAL BRICKWORK AREA = BRICKWORK AREA – DEDUCTION AREA
 = 789.84 – 219.2
 = 570.64M2
 VOLUME OF BRICK WORK =TOTAL BRICKWORK AREA XWALLTHICKNESS
 = 570.64 X 0.23
 = 131.2472M2
 VOLUME OF SINGLE BRICK = LENGTH X BREADTH X DEPTH
 = 0.2 X 0.1 X 0.1
 = 0.002m3
 NUMBEROF BRICKS =VOLUME OF BRICKWORK /VOLUME OF SINGLE BRICK
 = 131.2472/0.002m3
 =65623.6 BRICKS
 = 65624 BRICKS 46Dept of Civil Engg, A.H.C.E.T, Damaragidda,Chevella,T.S.

 ROLLER SHUTTER: 2.43m x 2.43m
 PLASTERINGTHICKNESS: 0.02m
 PLASTERINGCALCULATION:
 PLASTERINGAREA: INTERNALAREA 1
 INTERNAL DIMENSION 1: LENGTH = 3.83m, BREADTH = 3.83m,
DEPTH = 4m.
 INTERNALAREA 1= (LENGTH + BREADTH) X 2 X DEPTH m2
 = (3.83 + 3.83) x 2 x4
 = 61.28m2
 TOTAL INTERNALAREA 1: =61.28 * 15
 = 919.2m2
 INTERNAL DEDUCTIONS 1:
 SHUTTER: LENGTH = 2.43m, DEPTH = 2.43m.
 SHUTTERAREA = 2.43 x 2.43=5.9049m2
 TOTAL DEDUCTIONAREA: 5.9049 X 18 =106.2882m2
 INTERNAL DIMENSIONS 2:
 STAIR CASE : LENGTH = 3.83, DEPTH = 4m.
 INTERNALAREA 2= (LENGTH + BREADTH + LENGTH) X DEPTH m2
 = (3.83 + 3.83 = 3.83) X 4
 = 11.49 X 4 = 45.96m2
 TOTAL DIMENSION = 45.96 X 2 =91.92m2 49Dept of Civil Engg, A.H.C.E.T, Damaragidda,Chevella,T.S.

 TOTAL INTERNAL AREA = INTERNAL AREA – DEDUCTION AREA
 = (91.92 + 919.2)–106.1882
 = 904.9318m2
 EXTERNAL AREA 1: LENGTH = 4.05, BREADTH = 4.05, DEPTH = 4M
 = (L + B +L)X 4
 = (4.05+4.05+4.05)X 4=48.6m2
 TOTAL EXTERNAL AREA 1: 48.6X3=145.8m2
 = (L + B)X 4
 = (4.05+4.05)X4 =32.4m2
 = (L)X 4
 = (4.05)X4 =16.2m2
 TOTAL EXTERNAL AREA: 145.8 + 226.8 +132.0m2
 = 504.6m2
 EXTERNAL DEDUCTIONS 1:
 SHUTTER: LENGTH = 2.43m, DEPTH = 2.43m.
 SHUTTERAREA = 2.43 x 2.43=5.9049m2
 TOTAL EXTERNAL DEDUCTION AREA: 5.9049 X 18 = 106.2882
 TOTAL EXTERNAL AREA = EXTERNAL AREA – DEDUCTION AREA
 = 504.6–106.1882
 = 398.3118m2
 TOTAL PLASTERING AREA: INTERNAL + EXTERNAL PLASTERING AREA
 = 904.9318m2 + 398.3118m2
 = 1303.2436m2

 1. DIAMETERS:
 6mm,8mm,10mm,12mm,16mm,20mm,25mm,28mm
 2. STANDARD LENGTH OF BAR = 12m.
 3.WEIGHT PER METER = D2/162 kg/m. D=DIAMETER
 FOR EXAMPLE SAY 8mm BAR.
 WEIGHT PER METER = 82/162 = 0.39 kg/m.
 4.WEIGHT PER BAR = (D2/162)*12 KGS/BAR.
 FOR EXAMPLE SAY 8mm BAR.
 WEIGHT PER METER = (82/162)*12 = 4.74 KGS/BAR
 5. HOOK :
 LENGTH OF HOOK =9d, d = diameter.
 LENGTH OF BAR = L + 9d +9d m
 6. BEND :
 LENGTH OF BEND =10d – 16d, d = diameter.
 LENGTH OF BAR = L + 12d +12d m. (BEND = 12d).

 NECK COLUMNTO FOOTING OVERLAPPING = 40d – 50d, or 16D
 COLUMNTO COLUMN = 50d,
 GENERALOVERLAPPINGWHEN 12m BAR FINISHES = 50d.
D=DIAMETER
 8. CRANK BAR :
 CRANK = 0.42d,
 LENGTHOFTHE CRANK BAR = L + 0.42d + 0.42d,
 D= SLAB OR BEAMTHICKNESS –TOP N BOTTOM COVER.
 LENGTHOF EXTRA BAR = L/4.
 9. COVER: SPACE BETWEEN REINFORCEMENTAND SHUTTERING.
COVERING BLOCKSARE USED FOR PROVIDINGCOVER.
 FOOTING = 75mm
 COLUMN = 25mm - 50mm
 BEAM = 25mm - 50mm
 SLAB = 25mm
 10. SPACING: DISTANCE BETWEENTWO REINFORCEMENT BARS.
 11. NUMBER OF BARS = (OPPOSITE LENGTH /SPACING) + 1.
 12. NUMBER OF STIRRUPS = (ACTUAL LENGTH OF COLUMN OR
BEAM /SPACING) + 1.

 FOOTING REINFORCEMENT CALCULATIONS:
 SHORTWAY BARS:
 1. LENGTH OF S/W BAR = L - 0.1 – 0.1 + D + D M
 LENGTH OF S/W BAR = 1.4 – 0. 1 – 0.1 +0.3 + 0.3 M
 LENGTH OF S/W BAR = 1.80 M

 2. NUMBER OF SHORTWAY BAR = (OPPOSITE LENGTH / SPACING) + 1
 NUMBER OF SHORTWAY BAR = {(1.8 – 0.1 – 0.1)/0.1} + 1
 NUMBER OF SHORTWAY BAR = {1.6/0.1} + 1
 NUMBER OF SHORTWAY BAR = 17 BARS
 3.TOTAL LENGTH OF SHORT WAY BARS = LENGTH OF BAR * NUMBER OF BARS M
 TOTAL LENGTH OF SHORTWAY BARS = 1.80 *17 M
 TOTAL LENGTH OF SHORTWAY BARS = 30.60 M
 4. DIAMETER OF SHORT WAY BAR = 10 MM (GIVEN IN DRAWING)
 5.WEIGHT PER METER OF SHORTWAY BAR K= D2/162 GS/M
 WEIGHT PER METER OF SHORT WAY BAR = 10*10/162 KGS/M
 WEIGHT PER METER OF SHORT WAY BAR = 0.62 KGS/M
 6.TOTAL WEIGHT OF SHORTWAY BARS =TOTAL LENGTH * WEIGHT PER METER * NO OF
MEMBER KGS
 TOTALWEIGHT OF SHORT WAY BARS = 30.60 * 0.62 * 7 KGS
 TOTALWEIGHT OF SHORT WAY BARS = 132.804 KGS

 LONGWAY BARS:
 1. LENGTH OF L/W BAR = L - 0.1 – 0.1 + D + D M
 LENGTH OF L/W BAR = 1.8 – 0. 1 – 0.1 +0.3 + 0.3 M
 LENGTH OF L/W BAR = 2.20 M

 2. NUMBER OF LONGWAY BAR = (OPPOSITE LENGTH / SPACING) + 1
 NUMBER OF LONGWAY BAR = {(1.4 – 0.1 – 0.1)/0.1} + 1
 NUMBER OF LONGWAY BAR = {1.2/0.1} + 1
 NUMBER OF LONGWAY BAR = 13 BARS

 3.TOTAL LENGTH OF LONGWAY BARS = LENGTH OF BAR * NUMBER OF BARS M
 TOTAL LENGTH OF LONGWAY BARS = 2.2 *13 M
 TOTAL LENGTH OF LONGWAY BARS = 28.60 M

 4. DIAMETER OF LONGWAY BAR = 10 MM (GIVEN IN DRAWING).

 5.WEIGHT PER METER OF LONGWAY BAR = D2/162 KGS/M
 WEIGHT PER METER OF LONGWAY BAR = 10*10/162 KGS/M
 WEIGHT PER METER OF LONGWAY BAR = 0.62 KGS/M

 6.TOTAL WEIGHT OF LONGWAY BARS =TOTAL LENGTH * WEIGHT PER METER * NO OF
MEMBER KGS
 TOTAL WEIGHT OF LONGWAY BARS = 28.60 * 0.62 * 7 KGS
 TOTAL WEIGHT OF LONGWAY BARS = 124.124 KGS

 NECK COLUMN (MAIN BAR)
 L = NECK COLUMN + FOOTING BOTTOM MESH
 L = 1.2 + 0.3
 L = 1.5 DIA = 12MM
 LENGTHOF EACH MAIN BAR = L + 50D + 16D
 = 1.5 + 66D
 = 1.5 + 66 (0.012)
 = 2.30 M
 NO OF MAIN BAR = 8
 TOTAL LENGTH OF MAIN BAR = LENGTH OF EACH BAR X NO OF MAIN
BAR
 = 2.30 X 8
 = 18.4 M
 WEIGHT OF STEEL BAR IN KGS / M = D2/162 KGS/M
 = 12*12/162
 = 0.88 KGS/ M
 REQUIREDWEIGHT OF STEEL FOR MAIN BAR = WEIGHT/M XTOTAL
LENGTHOF MAIN BAR X NO OF MEMBER
 = 0.88 X 18.4 X 7
 = 113.344 KGS

 NECK COLUMN (STIRRUPS)
 25 MM (0.025 M) CONCRETE COVER HASTO BE DEDUCTED FROM EACH
SIDE OF COLUMN
 L = LENGTH OF EACH STIRRUPS
 L = (0.40 + 0.18) X 2
 L = 1.16 M
 LENGTH OF EACH STIRRUPSWITH HOOK = L +18D
 = 1.16 + 18(0.008)
 = 1.304 M
 NO OF STIRRUPS = (HEIGHT OF NECK COLUMN / SPACING) + 1
 = (1.5 / 0.1) + 1
 = 16 STIRRUPS
 TOTAL LENGTH OF STIRRUPS = LENGTH OF EACH STIRRUPS X NO OF
STIRRUPS
 = 1.304 X 16
 = 20.864 M
 WEIGHT OF STEEL BAR IN KGS / M = D2/162 KGS/M
 = 8*8 / 162
 = 0.39 KGS / M
 REQUIREDWEIGHT OF BAR FOR STIRRUPS = WEIGHT OF STEEL BAR IN
KGS/M XTOTAL LENGTH X NO OF ITEM
 = 0.39 X 20.864 X 7 = 56.959 KGS

QUANTITY SURVEY For Residential Building By MD USOOF ALAM

Recommended

Recommended

More Related Content

Similar to QUANTITY SURVEY For Residential Building By MD USOOF ALAM

Similar to QUANTITY SURVEY For Residential Building By MD USOOF ALAM (20)

Recently uploaded

Recently uploaded (20)

QUANTITY SURVEY For Residential Building By MD USOOF ALAM