Center Line Method Of Estimation

In the Center Line method, sum total length of the center lines of walls, long and short has to be found out. Find the total length of the center lines of walls of the same type.

Long and short have the same type of foundation and footing and they find the quantities by multiplying the total center length by the respective breadth and height.

In this method, the length will remain the same for the excavation in a foundation for concrete in the foundation. For all footings and for the superstructure. (With slight difference when there are cross walls or the number of junctions).

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Center Line Method

In center line method is quick but requires special attention and consideration at the junction, meeting points of cross walls, etc.

One Room Estimation

Centerline length of one wall = Inner dimension of room + Wall thickness

= Inner dimension of room + ½ x Wall thickness of one side + ½ x wall thickness of other.

Centre line length of horizontal wall

= 4 + 0.45 + 0.45

= 4.90 m

Centre line length of vertical wall

= 3 + 0.45 + 0.45

= 3.90 m

Total Length of excavation for a foundation

= 2(4.90 + 3.90)

= 17.6 m

Quantity of excavation for a foundation

= Total length of centre line of walls x breath of trench x Depth of trench

= 17.6 x 0.9 x 1.2

= 19.01 m3

For rectangular, circular, or polygonal buildings having no inter or cross walls, this method is quite simple but for buildings having a cross or partition walls, for every junction of partition or cross walls with main walls, a half breath of the respective item or footing is to be deducted from the total center length.

Net Centre line length – (½ x breath of wall x number of the junction)

The following points in relation to the center–line method should be noted.

The term junction is used to indicate the meeting points of cross walls or partition walls with main walls.

For instance, if there is one cross wall there will be two junctions. At the corner of a building, no junction is formed.

For rectangular buildings with No cross wall, there will not be any junction and hence no deduction from the center line length will be required.

It is fact that a certain portion taken twice is counterbalanced by the same amount of portion left out at the corner.

The breadth indicates the width of an item whose center line is being worked, for instance, it will be the width of a trench of excavation and foundation concrete.

But for brick masonry up to the plinth or for the superstructure will be equal to the corresponding thickness of the wall.

Two-Room Estimation By Center Line Method

For a two-room building with an inner cross wall, two junctions will be formed shown in fig.

Total center line length

= (2 x AB + 2 X BC + EF)

= (2 X 9.50 + 2 X 5.5 + 5.5)

If this center line length is multiplied by the width and depth, the quantity at junctions P and Q will come twice, the quantity of each being equal to ( ½ x width x Depth )

Hence the quantity of these two junctions must be deducted to get the net centerline length.

Net center line length

=( Total Centre line length – ( width/ 2) x No. of the junction)

= 35. 50 – (½ x 0.90) x 2

= 34. 60 m

centerline method of estimation example pdf

House Estimation by center line method,

Item No.	Item Description (Work Details)	No	Length (L)	Breadth (B)	Height (H)	Quantity
1.	Earthwork in excavation for the foundation	1	44.4	0.9	1.10	43.96
	Total Centre Line Length = 48.9 m
	No. of Junction = 10
	L = 48.9 – (1/2) x 0.90 x 0.90 x 10
	L = 44.4 m
OR
1.	Brickbat cement concrete (1:4:8)for the foundation	1	44.4	0.9	0.2	7.99 m²
2.	Brick masonry up to plinth in C.M.1:6
	First Step: L= 48.0-1/2X0.5X10=46.40m	1	46.4	0.5	0.3	6.96
	Second Step:L= 48.9-1/2X0.4X10= 46.9m	1	46.9	0.4	0.3	5.63
	Third Step:L= 48.9-1/2X0.3X10= 47.4m	1	47.4	0.3	0.85	12.08
	Steps:
	First Step:	1	1.1	0.9	0.15	0.15
	Second Step:	1	1.1	0.6	0.15	0.10
	Third Step:	1	1.1	0.3	0.15	0.05
	For steps L =D1= 1.1m					24.97 m²
3.	Brick masonry above plinth up to slab level in C.M. 1:6
	L= 48.9=1/2X0.2X10= 47.9m	1	47.9	0.2	3.0	28.74 m²

	Deduction For Doors & Windows
	D1	3	1.10	0.2	2.1	1.39
	D2	2	0.90	0.2	2.1	0.76
	G1	1	1.2	0.2	2.1	0.50
	W1	4	1.8	0.2	1.4	2.02
	W2	1	1.2	0.2	1.4	0.34
	W3	1	1.5	0.2	1.4	0.42
	V	2	0.6	0.2	0.6	0.14
	Deduction for door/ windows					(-)5.57 m²

	Deduction For Linter Above Door & Windows with 15 cm Bearing at each End
	D1	3	1.4	0.2	0.15	0.126
	D2	2	1.2	0.2	0.15	0.072
	G1	1	1.5	0.2	0.15	0.045
	W1	4	2.1	0.2	0.15	0.25
	W2	1	1.5	0.2	0.15	0.045
	W3	1	1.8	0.20	0.15	0.054
	V	2	0.9	0.2	0.15	.054
	Deduction for lintel					(-)0.646 m²
	(Same as that obtained in the long wall-short wall method)				Net quantity	22.52 m²
OR
3.	Smooth plaster inside the rooms and ceilings in C.M. 1:3					238.39 m²
	Plaster For Wall
	Drawing Room	4	4.0		3.0	48.0
	Bed Room	2	4.0		3.0	24.0
		2	3.0		3.0	18.0
	Kitchen	2	4.0		3.0	24.0
		2	3.0		3.0	18.0
	Bath	2	3.0		3.0	18.0
		2	1.7		3.0	10.2
	W. C.	2	1.3		3.0	7.8
		2	1.1		3.0	6.6
	In Front Of W. C.	2	1.5		3.0	9.0
		2	1.1		3.0	6.6
	Ceiling Plaster:
	Drawing Room	1	4.0	4.0		16.0
	Bedroom	1	4.0	3.0		12.0
	Kitchen	1	3.0	4.0		12.0
	Bath	1	3.0	1.7		5.1
	W. C.	1	1.3	1.1		1.43
	In front of W. C.	1	1.5	1.1		1.65
						238.39 m²
	Deduction For Plaster:
	D1	5/2	1.1		2.1	5.78
	D2	4/2	0.9		2.1	3.78
	G1	2/2	1.2		2.1	2.52
	W1	4/2	1.8		1.4	5.04
	W2	1/2	1.2		1.4	0.84
	W3	1/2	1.5		1.4	1.05
						(-)9.01 m²
					Net quantity	219.37 m²

Important Notes

The area of ventilation is less than 0.5 sq m, therefore no deduction is made.
For Doors, Half the number of faces is deducted.
For calculation of internal plaster, inner doors have two faces and the external door has one inner face
For calculation of internal plaster, windows have one internal face.
for windows, half the number of faces is deducted.

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