Lec 9 center line method

Estimation For 02 Room Office
D4’x7’

Footing Details
2’’ 2’’
P.L
G.LG.L
P.L

4) 1-1/2’’ thick brick tiles joined and
pointed in cement sand mortar (1:3)
3) 1’’ thick mud plaster
2) 4’’ thick earth filling
1) Roof insulation comprising of 2
coats of hot bitumen
Roof Details

 SPECIFICATIONS
 P.C.C
 Damp Proof Coarse (D.P.C) = 1:2:4
 Floors = 1:4:8, 1:3:6, 1:2:4
 R.C.C
 Roof Slab, Lintel, and Sun Shades = 1:2:4
 C/S Mortar
 Brickwork in foundation = 1:6
 Brickwork in superstructure = 1:4
 Brick tiles on roof = 1:3
 Plaster Work
 ½” thick 1:3 C/S Plaster

 SPECIFICATIONS
 Clear height of rooms :12’
 Clear height of Verandah : 12’
 Plinth level : 1’-6”
 Thickness of roof slab : 4”
 Thickness of RCC shade : 3”
 Depth of RCC Beams in Verandah : 1’-6” below verandah slab
 Parapet wall : 1’-0”
 Ventilators (4 No.) : 2’-6” x 1’-6”
 RCC lintel : 6” in depth
 Damp proof coarse : 1 ½”thick PCC (1:2:4) + 2 coats of
hot bitumen +polythene sheet
 Internal finishes : 03 coats of white wash/ distemper
paint
 External finishes : 03 coats of Weather shield paint
 Lintels extend : 6” on each side

 This item is described in detail but the price of this item is usually
indicated as lump sum (LS).
 The cost of this item is provided in the estimate by judgment,
according to the description of the item and is indicated as Lump
sum (L.S).
 Estimator must determine
 Depth of soil to be removed
 Equipment to be used
 STEP 1: Clearing and Grubbing

 Topsoil is generally removed from all building, walk, roadway, and
parking areas.
 The volume of topsoil is figured in cubic yards.
 A clearance around the entire basic plan must also be left to allow
for the slope required for the general excavation; usually about 5
feet is allowed on each side of a building and 1 to 2 feet for walks,
roadways, and parking areas.

Plot Area = 28’ x 28‘ = 784 Sq. ft.
Efficiency of Dozer = 437 cu. ft. / hr.
Rate = PRs 4000/ hr. (if more than 12,000 cu. ft.)
Rate = PRs 8000/ hr. (if less than 12,000 cu. ft.)
Let us assume 3 in. depth
Quantity = 784 * (3/12) = 784 * 0.25 = 196 cu. ft.
Equipment hours = 196 cu. ft / 437 cu. ft /hr = 0.45 hours
Equipment cost = 0.45 hours * 8000 per hour = Rs. 3600
Equipment cost = Rs. 3600

 For symmetrical footings, which is the usual case, earthwork in
excavation in foundations, foundation concrete, brickwork in
foundation and plinth, and brickwork in superstructure may be
estimated by either of the two methods:
(1) Separate or Individual Wall Method
(2) Center Line Method
Methods Of Detailed Estimate

 In this method, total length of center lines of walls, long and short,
has to be found out
 Find the total length of center lines of walls of same type, having
same type of foundations and footings and then find the quantities
by multiplying the total center length by the respective breadth and
the height
 In this method, the length will remain the same for excavation in
foundations, for concrete in foundations, for all footings, and for
superstructure (with slight difference when there are cross walls or
number of junctions)
 This method is quicker but requires special attention and
considerations at the junctions, meeting points of partition or cross
walls
Center Line Method

 For rectangular, circular polygonal (hexagonal, octagonal etc.)
buildings having no inter or cross walls, this method is quite simple
 For buildings having cross or partition walls, for every junction, half
breadth of the respective item or footing is to be deducted from the
total center length
 Thus in the case of a building with one partition wall or cross wall
having two junctions, deduct one breadth of the respective item of
work from the total center length
Center Line Method

Centre to Centre Distance
 (4.5” = 0.37’)
 (9” =0.75’)
 H-1= H-2= H-3=10’+12’+0.37’
+0.37’+0.37’
=23.11’
 S-1= S-2= S-5=12’+0.37’+0.37’
= 12.74’
 S-3= S-4= 7.25’+0.37’+0.37’
= 8’
S-1 S-2 S-5
H-1
H-2
H-3
S-3
S-4

STEP 2: Excavation
 Earthwork in excavation for foundation trenches is calculated by
taking the dimensions of each trench as length x breadth x depth.
 It is measured in cubic ft, cubic yard or cubic meter, according to
the prevailing practice.
 The payment for this item is generally done as Rs. per hundred
cubic ft.
 Filling in trenches after the construction of foundation masonry is
ordinary neglected. If the trench filling is, also taken in account, it
may be calculated by deducing the volume of masonry in trenches
from that of the volume of excavation

 To determine the amount of excavation, it is necessary to determine
the following:
1. The size of building (building dimensions).
2. The distance the footing will project beyond the wall.
3. The amount of working space required between the edge of the
footing and the beginning of excavation.
4. The elevation of the existing land, by checking the existing contour
lines on the plot (site) plan.
Continuous footing section

6. The type of soil that will be encountered. This is determined by first
checking the soil borings (on the drawings), but must also be checked
during the site investigation.
7. Whether the excavation will be sloped or shored.
8. The depth of the excavation. This is done by determining the bottom
elevation of the cut to be made. Then take the existing elevation, deduct
any topsoil removed, and subtract the bottom elevation of the cut to
determine the depth of excavation.

When calculating the amount of excavation to be done for a project, the
estimator must be certain that the dimensions used are the
measurements of the outside face of the footings and not those of the
outside of the building. The footings usually project beyond the wall.
Also, an extra 6 inches to 1 foot is added to all sides of the footing to
allow the workers to install and remove forms. The estimator must also
allow for the sloping of the banks to prevent a cave in. The amount of
slope required must be determined by the estimator who considers the
depth of excavation, type of soil, and possible water conditions.

Excavation Plan
Let Sa = S-1 + S-3
Sb = S-4 + S-5
 L1 = H-1 + Sa + H-3 + Sb
 L2 = S-2 – (Overlap)
= S-2 – ½(Width) – ½(Width)
= S-2 – Width
 L3 = H-2 - (Overlap)
= H-2 - ½(Width) – ½(Width)
= H-2 – Width
Using Center Line Method
S-1 S-2 S-5
H-1
H-2
H-3
S-3
S-4

Let Sa = S-1 + S-3
Sb = S-4 + S-5
(As width of excavation is 2’-6”)
 L1 = H-1 + Sa + H-3 + Sb
 L2 = S-2 – ½(Width) – ½(Width)
= S-2 – Width
= S-2 – 2’-6”
 L3 = H-2 - ½(Width) – ½(Width)
= H-2 – Width
= H-2 – 2’-6”
S-2
H-1
H-2
H-3
Sa Sb

S No Description N
Measurement
Quantity
(cu. ft.)
Total
Quantity
(cu. ft.)
Remarks
L’ B’ D’
1 L1 1 87.70 2.5 3.5 767.38 767.38
2 L2 1 10.2 2.1 3.5 75 75 L=12.74-2.5
3 L3 1 20.61 2.5 3.5 180.34 180.34 L=23.11-2.5
Total Quantity 1022.7

Quantity = 1022.7 cu. ft.
Wage rate = Rs. 650/ day
Productivity rate = 0.0134 day / cu ft
Labour Hours = Productivity Rate * Quantity
Labour hours = 0.0134 day / cu ft *1023.88 = 13.65 days
Labour cost = Labour hours * Wage rate per hour
Labour cost = 13.65 days * Rs. 650/day
Labour cost = Rs. 8873

P.C.C
 The type of concrete must be clearly mentioned. The mix proportions and
the type of cement, sand and coarse aggregate must be specified.
 This item is measured in cubic ft and the unit for measurement is,
generally Rs. per 100 cubic ft.
 When the soil is soft or weak, one layer of dry bricks or stone soling is
applied below the foundation concrete. The soling layer is computed in
sq.ft (length x breadth), specifying the thickness in description of item.
 In estimating quantities, the estimator makes no deductions for holes
smaller than 2 sf or for the space that reinforcing bars or other
miscellaneous accessories take up.
 Waste ranges from 5 percent for footings, columns, and beams to 8
percent for slabs.

Procedure for calculating concrete
1. Review the specifications to determine the requirements for each
area in which concrete is used separately (such as footings, floor
slabs, and walkways) and list the following:
(a) Type of concrete
(b) Strength of concrete
(c) Color of concrete
(d) Any special curing or testing
2. Review the drawings to be certain that all concrete items shown on
the drawings are covered in the specifications. If not, a call will have
to be made to the architect-engineer so that an addendum can be
issued.
3. List each of the concrete items required on the project.
4. Determine the quantities required from the working drawings.
Footing sizes are checked on the wall sections and foundation plans.
Watch for different size footings under different walls.

Let Sa = S-1 + S-3
Sb = S-4 + S-5
 L1= H-1 + Sa + H-3 + Sb
 L2 = S-2 – ½(Width) – ½(Width)
= S-2 – Width
 L3 = H-2 - ½(Width) – ½(Width)
= H-2 – Width
S-2
H-1
H-2
H-3
Sa Sb
STEP 3: P.C.C (1:4:8)

S No Description N
Measurement
Quantity
(cu. ft.)
Total
Quantity
(cu. ft.)
Remarks
L’ B’ D’
1 L1 1 87.70 2.5 0.5 109.63 109.63
2 L2 1 10.24 2.125 0.5 10.88 10.88 L=12.74-2.5
3 L3 1 20.61 2.5 0.5 25.76 25.76 L=23.11-2.5
STEP 3: P.C.C (1:4:8)

Concrete in Foundations = 146.27 cu. ft.
Add 5% for waste
Concrete in Foundation = 146.27 + 5/100 * 146.27 = 153.58 cu. ft.
Thumb Rules
Dry material for 100 cu. ft. of cement concrete = 154 cu. ft.
Volume of cement bag = 1.25 cu. ft.
Materials
Concrete = 153.58 cu. ft *1.54 = 236.51 cu. ft.
Cement
= 1/13 * 236.51
= 18.19/1.25
= 18.19 cu. ft.
= 14.55 = 15 Bags
Sand = 4/13*236.51 = 72.77 cu. ft.
Aggregate = 8/13*236.51 = 145.54 cu. ft.
STEP 3: P.C.C (1:4:8)

 Care must be taken, while taking dimensions from the drawings in the
bill of quantities because the walls in this part of the structure are in
the form of steps with changing dimensions.
 This item is calculated in cft and the unit for payment is Rs. per 100
cft.
 In the description of work, the quality of bricks and mortar ratio must
be specified. For example, "Brickwork in foundation and plinth using
first class bricks laid in (1:4) or (1:6) cement-sand (c/s) mortar———
————“
STEP 4: Brickwork up to Plinth Level
with 1:6 (C/S)

 Wastage for masonry = 5%
 Wastage for mortar = 40%
with 1:6 (C/S)

Let Sa = S-1 + S-3
Sb = S-4 + S-5
 L1= H-1 + Sa + H-3 + Sb
 L2 = S-2 – ½(Width) – ½(Width)
= S-2 – Width
 L3 = H-2 - ½(Width) – ½(Width)
= H-2 – Width
S-2
H-1
H-2
H-3
Sa Sb
with (1:6 C/S)

Description N
Measurement Quantity
(cu. ft.)
Total
Quantity
(cu. ft.)
Remarks
L’ B’ D’
L1 = H-1 + Sa + H-3 + Sb
Step-1 1 87.70 1.5 0.5 65.78 65.78
Step-2 1 87.70 1.125 0.5 49.33 49.33
Step-3 up
to plinth 1 87.70 0.75 3.37 221.66 221.66 D = 2+1.5 – 0.13
with 1:6 (C/S)

Description N
(cu. ft.)
Total
Quantity
(cu. ft.)
Remarks
L’ B’ D’
L2 = S-2 – Width
Step-1 1 11.24 1.125 0.5 6.32 6.32 L=12.74-1.5
Step-2 1 11.62 0.75 0.5 4.36 4.36 L=12.74-1.125
Step-3 up
to plinth 1 11.99 0.38 3.37 15.35 15.35
L = 12.74 – 0.75
D = 2+1.5 –0.13
with 1:6 (C/S)

Description N
(cu. ft.)
Total
Quantity
(cu. ft.)
Remarks
L’ B’ D’
L3 = H-2 - Width
Step-1 1 21.61 1.5 0.5 16.20 16.20 L=23.11 –1.5
Step-2 1 21.985 1.125 0.5 12.36 12.36 L=23.11 –1.125
Step-3 up
to plinth 1 22.36 0.75 3.37 56.51 56.51
L=23.11 -0.75
D=2+1.5 –0.13
with 1:6 (C/S)

Total Quantity = 336.77 + 26.03 + 85.07 = 447.87 cu. ft.
 Thumb Rules
Bricks for 100 cu. ft. of brick work = 1350 Nos.
Dry mortar for 100 cu. ft. of brick work = 30 cu. ft.
 Material
Bricks = 1350 / 100 * 447.87 = 6046 Bricks
Add 5% for wastage
Total bricks = 6046 + 5/100 * 6046 = 6349 Bricks
Mortar = 30 / 100 * 447.87 = 134.361
Add 40% for wastage
Total mortar = 134.361 + 134.361 * 40/100 = 188.10
with 1:6 /(C/S)

Cement
= 1 / 7 * 188.10
= 26.87 / 1.25 = 21.5
= 26.87
= 22 Bags
Sand = 6 / 7 * 188.10 = 161.22 cu. ft.
with 1:6 (C/S)

STEP 5: Steps in Front of Verandah with
1:6 (C/S)

Description N
Measurement
Quantity
(cu. ft.)
Total
Quantity
(cu. ft.)
Remarks
L’ B’ D’
Length of steps = 0.75 + 10 + 0.37 + 12 + 0.75 = 23.87’
Step-1 1 23.87 1 0.5 11.94 11.94
Step-2 1 23.87 2 0.5 23.87 23.87
1:6 (C/S)

 Thumb Rules
Mortar for 100 cu. ft. of brick work = 30 cu. ft.
 Materials
Bricks = 1350 / 100 * 35.81 = 483.44 = 484 Bricks
Add 5% for wastage
Total Bricks = 484 * 1.05 = 509 Bricks
Mortar = 30/100 * 35.81 = 10.74 cu. ft
Add 40% for wastage
Total Mortar = 10.74 * 1.40 = 15.04 cu. ft
1:6 (C/S)

Total Mortar = 10.74 * 1.40 = 15.04 cu. ft
Cement = 1 / 7 * 15.04 = 2.14 cu. ft
= 2.14 / 1.25 = 1. 71 = 2 Bag
Sand = 6 / 7 * 15.04 = 12.89 cu. ft
1:6 (C/S)

 Horizontal D.P.C. shall extend the full width of the super structure
walls, however, it shall not be provided across doorways and veranda
openings. It is also provided in roof and floors.
 Vertical D.P.C. is provided in external walls, especially, in the
walls of basements.
 The quantity of D.P.C. is estimated in square ft.(on area basis) and
standard unit for payment is Rs. per 100 sft
STEP 6: 1-1/2” DPC (1:2:4)
& 02 coats of hot bitumen

S-1 S-2 S-5
H-1
H-2
STEP 6: 1-1/2” DPC (1:2:4) & 02 coats of
hot bitumen
 (4.5” = 0.37’)
 (9” =0.75’)
 H-1= H-2= H-3=10’+12’+0.37’
+0.37’+0.37’
=23.11’
 S-1= S-2= S-5=12’+0.37’+0.37’
= 12.74’

TakingArea
Description N Measurement Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
L’ B’ D’
L1 1 71.70 0.75 - 53.78 53.78
L= H-1 + S-1 + H-2
+ S-5
L2 1 12 0.37 - 4.44 4.44
L = 12.74 - 0.37 –
0.37
Columns 3 0.75 0.75 - 0.56 1.69
STEP 6: 1-1/2” DPC (1:2:4) & 02 coats of
hot bitumen

(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
L’ B’ D’
Total
Area - - - - 59.16 59.91
Deductions
Doors 4 0.75 - 3.00 3.00
4 0.37 - 1.50 1.50
STEP 6: 1-1/2” DPC (1:2:4) & 02 coats of
hot bitumen

Concrete Volume = 55.41 x 0.125 = 6.93 cu. ft.
Add 5% for wastage
Total concrete volume = 6.93 * 1.05 = 7.28 cu. ft
 Thumb Rules
Bitumen for 100 sq. ft. of DPC (first coat) = 15 Kg
Bitumen for 100 sq. ft. of DPC (second coat) = 10 Kg
 Materials
Dry Concrete = 7.28 x 1.54 = 11.21 cu. ft.
STEP 6: 1-1/2” DPC (1:2:4) & 02 coats of
hot bitumen

Cement = 1/7 x 11.21 = 1.60 /1.25 = 1.28 = 2 Bags
Sand = 2/7 x 11.21 = 3.20 cu. ft.
Coarse aggregate = 4/7 x 11.21 = 6.4 cu. ft.
Bitumen = 25/100 x 55.41 = 13.85 Kg
STEP 6: 1-1/2” DPC (1:2:4) & 02 coats of
hot bitumen

Important considerations are:
1. Measurements of walls shall be taken in the same order and in the
same manner as for brickwork in foundations and plinth.
2. In the first measurements, all openings such as doors, windows,
veranda openings etc. shall be neglected. However, deductions shall
be made for all openings in the walls, at the end of the item.
3. In the description of the work, the quality of bricks and mortar
ratio have to be specified.
4. Masonry for arches shall be paid separately, at a different rate.
STEP 7: Brick work in super structure

5. The height of super structure is very important. Generally the
quantities are worked out for each storey separately and rates
would be different for different storeis because of additional
labor work, scaffolding and shuttering.
6. The item is worked out in cft and the standard unit for payment is
Rs. Per 100 cft.

S-1 S-2 S-5
H-1
H-2
STEP 7: Brick work in super structure using
first class bricks in (1:4) cement sandmortar
 (4.5” = 0.37’)
 (9” =0.75’)
 H-1= H-2= H-3=10’+12’+0.37’
+0.37’+0.37’
=23.11’
 S-1= S-2= S-5=12’+0.37’+0.37’
= 12.74’

4) 1-1/2’’ thick brick tiles joined and
pointed in cement sand mortar (1:3)
3) 1’’ thick mud plaster
2) 4’’ thick earth filling
1) Roof insulation comprising of 2
coats of hot bitumen

Description N
(cu. ft.)
Total
Quantity
(cu. ft.)
Remarks
L’ B’ D’
L1 1 71.70 0.75 12 645.3 645.3
L= H-1 + S-1 + H-
2
+ S-5
L2 1 12 0.37 12 53.28 53.28
L = 12.74 - 0.37 –
0.37
Columns 3 0.75 0.75 10.5 5.9 17.71
716.29

H-2
 H-1= H-2= H-3=10’+12’+0.37’
+0.37’+0.37’
=23.11’
Sa = Sb = 12’.74 + 8’
= 20.74’
Sa Sb
parapet wall

Description N
Measurement
Quantity
(cu. ft.)
Total
Quantity
(cu. ft.)
Remarks
L’ B’ D’
Total
Volume 716.29
Parapet
Wall 1 87.7 0.75 1.541 101.35 101.35
L= H-1 + Sa + H-2
+ Sb
Total Volume 817.64

Description N
Measurement
Quantity
(cu. ft.)
Total
Quantity
(cu. ft.)
RemarksL’ B’ D’
Doors
1 4 0.75 7 21 21
1 4 0.37 7 10.5 10.5
Windows 3 4 0.75 4 12 36
Ventilators 4 2.5 0.75 1.5 2.81 11.25
Shelves 2 4 0.5 5 10 20
RCC Lintels
Doors
1 5 0.75 0.5 1.87 1.87
1 5 0.37 0.5 0.93 0.93
Windows 3 5 0.75 0.5 1.87 5.62
Ventilators 4 3.5 0.75 0.5 1.31 5.25
Shelves 2 5 0.75 0.5 1.87 3.75
116.18
Net Total = (817.64-116.18) 701.46
(Deductions)

Volume = 701.46 cu. ft.
 Thumbs Rules
Dry mortar for 100 cu. ft. of brick work = 30 cu. ft.
 Materials
Bricks = 1350 / 100 * 701.46 = 9469.71 = 9470 Bricks
Add 5% for wastage
Bricks = 9470 * 1.05 = 9944 Bricks
Mortar = 701.46 * 30/100 = 210.44 cu.ft
Add 40% for wastage
Total Mortar = 210.44 * 1.40 = 294.61 cu.ft

Cement = 1 / 5 * 294.61 = 58.92
= 58.92/ 1.25 = 47.13 = 48 Bags
Sand = 4 / 5 * 294.61 = 235.69 cu. ft.

 For R.C.C. Roof slabs and beams, the total quantities of concrete and
steel are estimated, separately.
 The quantity of plain concrete is estimated in cft and the standard
unit for payment of concrete is Rs. per 100 cft.
 Volume of Reinforcing Steel is not deducted , while estimating the
volume of plain concrete for payment.
 R.C.C. lintels over wall openings such as doors and windows are
also included in R.C.C. work.
STEP 8: Reinforced Cement Concrete

(4.5” = 0.37’)
(9” = 0.75’)
 L = 10’+12’+0.37’+0.75+0.75
= 23.87’
 B = 12’+0.75+0.75+8
= 21.5’
L
B
(1:2:4)(Slab, Lintels & Beams)

(cu. ft.)
Total
Quantity
(cu. ft.)
Remarks
L’ B’ D’
Roof Slab 23.87 21.5 0.33 169.43 169.43
Verandah
Long Beam 1 23.87 0.75 1.5 26.86 26.86
Verandah
Short Beam 2 7.25 0.75 1.5 8.156 16.31

(cu. ft.)
Total
Quantity
(cu. ft.)
Remarks
L’ B’ D’
Volume 212.60
Lintels
Doors
1 5 0.75 0.5 1.87 1.87
1 5 0.37 0.5 0.93 0.93
Windows 3 5 0.75 0.5 1.87 5.62
Ventilators 4 3.5 0.75 0.5 1.31 5.25
Shelves 2 5 0.75 0.5 1.87 3.75
Shades 2 5 1.5 0.25 1.87 3.75

STEP 8: Reinforcement Steel / General
Steel Work
 Steel is provided separately from R.C.C. per ton, per Kg, or per cwt
(standard weight also called Quintal or century weight equal to 112
Ibs = 50Kg).
 Quantity of steel can either be worked out by rules of thumb practice
or by intensive calculations taking the length and diameter of steel
bars from the working drawings showing reinforcement details and
bar-bending schedules. In taking length of bars, due margin of
hooks, bends and overlapping is given.
 As a Rule Of Thumb Practice, for ordinary beams and slabs for
residences, assume 6.75 Ibs of steel per cft of R.C.C. work.
However, for R.C.C. columns, it varies from 8 to 10 Ibs per cft.
normally, we use 2% of steel in columns.
 Percentage of steel means, area of steel divided by total area of the
column multiplied by 100 and 1% of steel in columns corresponds to
a quantity of 4.5 Ibs/cft.
 Wastage for slab is 8% while wastage for other member 10%
(including laps)

Description N Measurement
Quantity
(lbs)
Total
Quantity
(lbs)
Remarks
6.75 Ibs. of steel per cu. ft. of R.C.C. work
Steel 6.75 64.34 434.30 434.30
233.77-169.43
=64.34
Total Weight 434.30
STEP 8: Mild Steel Round Bars as in
Reinforcement

 Thumbs Rules
6.75 Ibs. of steel per cu. ft. of R.C.C. work
 Materials
Concrete for slab = 169.43 cu. ft
Add 8% for wastage in slab
Total Concrete for slab = 160.43 * 1.08 = 183 cu. ft
Concrete for other R.C.C members = 233.77 – 169.43 = 64.34 cu.ft
Add 5% for wastage
Total Concrete for other R.C.C members = 64.34 * 1.05 = 67.55 cu.ft

Total Concrete for R.C.C members = 183 + 67.55 = 250.55 cu.ft
Cement = 1/7 x 385.85 = 55.07/1.25 = 44.06 = 45 Bags
Sand = 2/7 x 385.85 = 110.24 cu. ft.
Coarse aggregate= 4/7 x 385.85 = 220.28 cu.ft
Mild steel round bars= 434.30 lbs
Add 10% for wastage
Total mild steel = 434.30 * 1.1 = 477.72 lbs
Slab steel = 2682.41 lbs (From Lecture 10)
Total steel = 2682.41 + 477.72 = 3160.13 lbs
Steel = 3160.13/ 2000
Steel = 1.58 tons

 Roof consisting of beams, battens, and tiles or wooden planks is
estimated for each part, separately
 Steel beam is estimated by weight, whereas, wooden beam is
measured in cft. Battens are estimated by numbers indicating there
size and lengths. Tiles are also estimated by size and numbers.
 Roof finishing may consist of bitumen coating and/or Polythene
sheets (water proofing) , earth filling (heat proofing) and brick tiles,
etc.
 Dimensions are taken from inner face to inner face of parapet walls.
 This item is estimated in sft and a composite rate for payment is
taken as Rs. per 100 sft of the roof area.
Roofing

 L = 10’+12’+0.37’ = 22.37’
 B = 12’+0.75+8-0.75 = 20’
Description N Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
Taking Area
Area 22.37 20 447.4 447.4
Total Area 447.4
STEP 9: Roof Insulation (02 coats of hot bitumen, 4”
thick earth filling, 1” thick mud plaster and 1-1/2” thick brick tiles
jointed and pointed in cement sand mortar (1:3))

 Thumbs Rules
100 sq. ft. surface area using bricks on bed= 360 Nos.
Dry mortar for 100 sft = 9.00 cft
Bitumen for 100 sq. ft. (first coat) = 15 Kg
Bitumen for 100 sq. ft. (second coat) = 10 Kg
 Materials
Brick Tiles = 360/100 x 447.4 = 1610.6 = 1611 No.
Add 5% for wastage
Total brick tiles = 1161 * 1.05 = 1220 No.
Mortar = 447.4 * 9 / 100 = 40.27 cu.ft
Add 40% for wastage
Total mortar = 40.27 * 1.4 = 56.38 cu.ft

Cement = 1/4 x 56.38 = 14.09 / 1.25 = 11.25 = 12 Bags
Sand = 3/4 x 56.38 = 42.27 cu. ft.
Mud/ Earth Filling = 447.4 x 0.42 = 188 cu. ft.
Bitumen= 25/100 x 447.4 = 111.9 kg
Polythene Sheet = 447.4 sq. ft.

 Cement concrete floors, mosaic floors, and brick floors are most
commonly used.
 Payments are made separately for different layers, like, topping, lean
concrete, sand filling, earth filling, etc.
 Earth filling, sand filling and lean concrete are paid by volume,
whereas, topping is paid on area basis, mentioning thickness in
the description.
 Standard unit for payment of topping is, usually, Rs. per 100 sft.
 The skirting is estimated in running ft.
Flooring

Verandah Length = 10’+12’+0.37’
= 22.37’
B = 8’ – 0.75’ = 7.25’
L
B
STEP 10: Flooring (sand under floor)
(4.5” = 0.37’)
(9” = 0.75’)

Description N
Measurement
Quantity
(cu. ft.)
Total
Quantity
(cu. ft.)
Remarks
L’ B’ D’
Room 1 1 12 10 0.5 60 60
Room 2 1 12 12 0.5 72 72
Verandah 1 22.37 7.25 0.5 81.11 81.11
STEP 10: Flooring (sand under floor)

 4” thick P.C.C = 0.33’
L
B
(4.5” = 0.37’)
(9” = 0.75’)
STEP 10: Flooring (Cement concrete
(1:4:8) as under layer of floors)

Description N
(cu. ft.)
Total
Quantity
(cu. ft.)
Remarks
L’ B’ D’
Room 1 1 12 10 0.33 40 40
Room 2 1 12 12 0.33 48 48
Verandah 1 22.37 7.25 0.33 54.07 54.07

Concrete = 142.07 cu. ft.
Add 5% for wastage
Total concrete =142.07 * 1.05 = 149.17 cu.ft
 Thumbs Rules
 Materials
Cement = 1/13 x 229.73 = 17.67/1.25 = 14.13 = 15 Bags
Sand = 4/13 x 229.73 = 70.69 cu. ft.

TakingArea
Description N
Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Room 1 1 12 10 - 120 120
Room 2 1 12 12 - 144 144
Verandah 1 23.87 8 - 191 191
L = 12+ 10
+0.37+0.75+0.75
Door Sill 1 1 4 0.75 - 3 3
Door Sill 2 1 4 0.37 - 1.5 1.5
Deductions
Columns 3 0.75 0.75 - 0.56 1.68
STEP 10: Flooring (1-1/2” P.C.C (1:2:4) as
top layer of floor, finished smooth)

Concrete Volume = 457.8 x 0.125 = 57.23 cu. ft.
Add 5% for wastage
Total concrete = 57.23 * 1.05 = 60.09 cu.ft
 Thumbs Rules
 Materials
Cement = 1/7 x 92.54 = 13.22/1.25 = 10.576 = 11 Bags
Sand = 2/7 x 92.54 = 26.44 cu. ft.
STEP 10: Flooring (1-1/2” P.C.C (1:2:4) as
top layer of floor, finished smooth)

 Materials
Total Cement bags = 15 + 11 = 26 Bags
Total Sand = 213.11 + 70.69 + 26.44 = 310.24 cu. ft.
Total Coarse aggregate = 141.13 + 52.88 = 194.01 cu. ft.

 The type of plaster, proportioning of materials and minimum
thickness of plaster have to be specified.
 The quantity is calculated for total wall surface without deduction for
openings such as doors windows, ventilators, etc. However, if the
wall is being plastered on both the faces, the deductions for opening
areas are made from one side only.
 Standard unit for payment is Rs. per 100 sft.
 Height is also specified for plastering because, for greater heights,
labor cost increases. The rate varies according to the number of the
storey
Plastering

STEP 11: ½” thick (1:3) C/S plaster to
walls finished smooth
(4.5” = 0.37’)
(9” = 0.75’)
 Inner Height = 12’
 Outer Height =

TakingArea
Description N
Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
L’ B’ D’
INNER SIDE
Room 1
(Long Wall) 2 12 - 12 144 288
Room 1
(Short Wall) 2 10 - 12 120 240
Room 1
(Ceiling) 1 10 12 - 120 120
Room 2
(Long Wall) 2 12 - 12 144 288
Room 2
(Short Wall) 2 12 - 12 144 288

Description N
Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
L’ B’ D’
INNER SIDE
Room 2
(Ceiling) 1 12 12 - 144 288
Verandah
Wall 1 23.87 - 12 286.56 286.56
Verandah
Ceiling 1 22.37 7.25 - 162.18 162.18
Columns 3 3 - 10.5 31.5 94.5
L= 0.75 + 0.75 +
0.75 + 0.75
Long Beam
(Internal) 1 22.37 - 1.5 33.55 33.55

Description N
Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
INNER SIDE
Long Beam
(Soffit) 1 21.62 - 0.75 8.10 8.10
L= 23.87 – 0.75 -
0.75 -0.75
Short Beam
(Internal) 2 7.25 - 1.5 10.875 10.875
Short Beam
(Soffit) 2 7.25 - 0.75 5.43 10.87
OUTER SIDE
Rear Wall 1 23.87 - 15.33 365.92 365.92 L =1.5+12+0.33+1.5

Description N
Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
L’ B’ D’
Left & Right
Side
2 13.5 - 15.33 206.95 413.91
Front Side
(above
verandah)
1 23.87 - 3.33 79.4 79.4 L = 1.5 + 1.5 + 0.33
Left & Right
Side
(above
verandah)
2 8 - 3.33 26.64 53.28
L = 1.5+0.125
+12+0.33+1.5
Verandah
sides
(lower area)
2 8 - 1.5 12 24
Parapet Wall
Inner side 1 84.74 - 1 84.74 84.74
L = 22.37 + 22.37 +
20 + 20
Top side 1 87.7 - 0.75 65.77 65.77 Center to center
Outer side
Already calculated
above

Description N
Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
L’ B’ D’
Verandah Steps
Tread 2 23.87 - 1 23.87 47.74
Riser 3 23.87 - 0.5 11.93 35.80
Sides
2 2 - 0.5 1.00 2.00
2 1 - 0.5 0.5 1.00
Misc
Door Jambs
2 7 0.75 - 5.25 10.50
2 7 0.37 - 2.63 5.25
1 4 0.75 - 3.0 1.50
1 4 0.37 - 1.50 3.00

Description N
Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
L’ B’ D’
Window
Jambs
6 0.75 - 4 3.00 18.00
6 0.75 4 - 3.00 18.00
Ventilator
Jambs
8 0.75 2.5 - 1.88 15.00
8 0.75 - 1.5 1.19 9.00
Shelves 4 0.5 - 5 2.5 10.00
Total Area = 3211.42
DEDUCTION
Doors 2 4 - 7 28.00 56.00
Windows 3 4 - 4 16.00 48.00
Ventilators 4 2.5 - 1.5 3.75 15.00
Net Area = 3211.42 – 119 = 3281.2

 Thumbs Rules
Dry mortar for 100 sq. ft. of ½” thick cement plaster = 6 cu. ft.
or
100 cu. ft. wet mortar = 128 cu. ft. dry mortar
 Materials
Volume = 3294.7 x (0.5/12) = 137.28 cu. ft.
Add 40% for wastage
Total volume = 137.28 * 1.4 = 192.19 cu. ft.
Dry volume = 192.19 * 1.28 = 246 cu. ft
Cement = 1 /4 * 246 = 61.5
Sand
= 61.5/ 1.25
= 3 / 4 * 246
= 49.2 = 50 Bags
= 184.5 cu. ft.

 The type of material used and the quantity of finish required should
be clearly indicated in the description of the item.
 The rate for any type of woodwork includes cutting of timber to
required sizes, joinery work, fittings and fastenings, three coats of oil
paints or varnish, bolts, locks, handles, etc.
 The measurements are taken for the overall area of doors, windows,
etc. If volume of timber required for these items is to be find out, the
computed area is multiplied with the nominal thickness and an
allowance of 25% is made for wastage of timber.
 Rectangular wooden beams, vertical columns, trusses, etc., are
measured in cft.
 Wooden stairs are measured in number of steps and description of the
item includes the riser, tread, and width of the steps.
 Wooden shelves are measured in running ft (RFT).
Wood work / Carpentry

Description N
Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
L’ B’ D’
1 ½” Thick
Wooden
Doors with
Chowkat
2 4 - 7 28 56
Windows
and
Ventilators
3 4 - 4 16 48
4 2.5 - 1.5 3.75 15
Three Coats
of
Painting to
Doors
2 - - - 56 112 Already calculated
STEP 12: Door, Windows & Ventilators
Frames

 Thumbs Rules
Timber for 100 sq. ft. of Panelled Doors = 13 cu. ft.
Timber for 100 sq. ft. of Glazed windows and Ventilators = 8 cu. ft.
 Materials
Timber for doors = 13 / 100 x 56 = 7.28 cu. ft.
Timber for windows and Ventilators = 8 / 100 x 63 = 5.04 cu. ft.
Total = 13.32 cu. ft.
Add 25% for wastage
Total wood required = 13.32 * 1.25 = 16.65 cu. ft
STEP 12: Door, Windows & Ventilators
Frames

Description N
Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
L’ B’ D’
Room No.1
(Long wall) 2 12 - 12 144 288
Room No.1
(Short wall) 2 10 - 12 120 240
Room No.1
(Ceiling) 1 10 12 - 120 120
Room No.2
(Long wall) 2 12 - 12 144 288
Room No.2
(Short wall) 2 12 - 12 144 288
Room No.2
(Ceiling) 1 12 12 - 144 144
Verandah wall 1 23.875 - 12 286.5 286.5
Verandah
Ceiling 1 22.375 - 7.25 162.26 162.26
STEP 13: Three coats of distempering/
white washing to walls (Internal Side)

Description N
Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
L’ B’ D’
Columns 3 3 - 10.5 25.50 76.50
Long beam
(inner side)
1 22.375 - 1.5 33.55 33.55
Long beam
(soffit)
1 21.62 - 0.75 16.21 16.21
Short beam
(sides)
2 7.25 - 1.5 10.87 21.74
Short beam
(soffit)
2 7.25 - 0.75 5.43 10.87
Door Jambs
2 0.75 - 7 5.25 10.50
2 0.375 - 7 2.63 5.25
1 0.75 - 4 3 1.50
1 0.375 - 4 1.50 3.00

Description N
Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
L’ B’ D’
Window
Jambs
6 0.75 - 4 3 18
6 0.75 - 4 3 18
Ventilator
Jambs
8 0.75 - 2.5 1.88 15
8 0.75 - 1.5 1.19 9
Shelves
4 0.5 - 5 2.5 10
4 0.5 - 4 2.0 10
DEDUCTION
Doors 4 4 - 7 28 112
Windows 4 4 - 4 16 64
Ventilators 6 2.5 - 1.5 3 22.50
Net Total Area 1877.38

 Thumbs Rules
Lime for 100 sq. ft. of white wash (one coat) = 1.00 Kg
 Materials
Lime for three coats=1/100 x 3 x 1877.3 = 56.3 Kg

Description N
Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
L’ B’ D’
Rear wall 1 23.875 - 15.375 367.07 367.07
Left & Right
side wall 2 13.5 - 15.375 207.5 415.13
Front side
(above
verandah
roof)
1 23.875 - 3.33 79.50 79.50
Sides (above
verandah
roof)
2 8 - 3.33 26.64 53.28
Left & Right
side wall of
verandah
2 8 - 1.5 12 24
STEP 14: Three coats of Weather shield
paint to walls. (External side)

Description N
Measurement
Quantity
(sq. ft.)
Total
Quantity
(sq. ft.)
Remarks
L’ B’ D’
PARAPET WALL
Top of
parapet
wall
1 87.7 - 0.75 65.775 65.775
Parapet
wall
(inside)
2 20 - 1 20 40
2 22.37 - 1 22.37 44.74
DEDUCTION
Windows 2 4 - 4 16.00 32.00
Ventilators 2 2.5 - 1.5 3.75 7.50
Net Total 1049.99
STEP 14: Three coats of Weather shield
paint to walls. (External side)

Market Rates
Sr. No Item Rate (Rupees)
1 Cement 550 / bag
2 Sand 2000 / 100 cu.ft
3 Aggregate 4500 / 100 cu.ft
4 Bricks 10 / each
5 Steel 92,000 / ton
6 Bitumen 105 / kg
7 Weather Shield 30 / sft
8 Distemper 15 / sft
9 Brick Tile 12 / each
10 Framic tile or face tile 18 / each

Market Rates
Sr. No Item Rate (Rupees)
11 Wood (Deodar) 4500 / cft
12 Lime 10 / kg
13 Earth Filling 3 / cft
14 Polythene sheet 2 / sft

Cost Summary
Sr.
No
Description Item Unit Quantity
Unit Rate
(Rs.)
Amount
(Rs.)
1 Top soil (3’’ depth) 3600
2
Excavate
foundation trenches
(3’6’’ depth)
8873
3
Foundation
concrete (1:4:8)
Cement
Sand
Aggregate
Bags
Cu.ft
Cu.ft
15
73
146
550
20
45
8250
1460
6570
4
Foundation walls
(1:6 C/S)
Bricks
Cement
Sand
No.
Bags
Cu.ft
6349
22
162
10
550
20
63490
12100
3240
5
Brick steps in front
of verandah
(1:6 C/S)
Bricks
Cement
Sand
No.
Bags
Cu.ft
484
2
13
10
550
20
4840
1100
260

Sr.
No
Unit Rate
(Rs.)
Amount
(Rs.)
6
1/2’’ D.P.C (1:2:4)
&
2 coats of bitumen
Cement
Sand
Aggregate
Bitumen
Bags
Cu.ft
Cu.ft
kg
2
3
6
14
550
20
45
105
1100
60
270
1470
7
Brick work in
superstructure
(1:4 C/S)
Bricks
Cement
Sand
No.
Bags
Cu.ft
9944
48
236
10
550
20
99440
26400
4720
8
R.C.C
slab, beams, lintels
(1:2:4)
Cement
Sand
Aggregate
Steel
Bags
Cu.ft
Cu.ft
Tons
45
110
220
1.58
550
20
45
92000
24750
2200
9900
145360
9
Roof insulation
1-1/2’’ tiles
1:3 C/S
Earth filling 5’’
2 coats of hot
bitumen
Polythene sheet
Brick tiles
Cement
Sand
Mud
Bitumen
Sheet
No.
Bags
Cu.ft
Cu.ft
Kg
sft
1220
12
42
188
112
447
12
550
20
3
105
2
14640
6600
840
564
11760
894

Sr.
No
Unit Rate
(Rs.)
Amount
(Rs.)
10
Flooring
Sand cushion 6’’
4’’ P.C.C (1:4:8)
1-1/2’’ P.C.C (1:2:4)
Cement
Sand
Aggregate
Bags
Cu.ft
Cu.ft
26 bags
310 cu.ft
194 cu.ft
550
20
45
14300
6200
8730
11
Plastering
1-1/2’’ thick (1:3)
C/S
Cement
Sand
Bags
Cu.ft
50 bags
185 cu.ft
550
20
27500
3700
12
Wood work
Door, windows,
ventilators & frames
Wood Cu.ft 17 cu.ft 4500 76,500
13
3 coat distempering/
white wash
Lime
Distemper
kg
sft
56 kg
1877 sft
10
15
560
56310
14
3 coat of weather
shield paint
Weather
shield
sft 1050 sft 30 31500
Total Cost 690,051

Cost per square foot
 Total cost = Rs. 690,051
 Plot covered area = 23.87’ x 21.5’ = 513.205 sft
 Cost per sft = Total cost / Covered Area
 Cost per sft = 690,051 / 513.205 = Rs. 1345 / sf t

Instructions for Term Project
 For slab assume suitable bars and calculate steel using new
method.
 Perform quantity take by assuming slab as a single RCC
member of 6’’- 8’’ depth instead of using roof layers.
 Add wastage for each item of work.
 Use market rates provided above for calculating material cost.
 Develop a BoQ/Cost summary.
 Divide total cost with your buildings covered area to calculate
cost per square foot.

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