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Chapter 2 of 10

Estimation and Bar Bending Schedule

In the AAI Manager (Civil) syllabus under Estimation, Costing, Contracts & Project Management · 2 parts

Building Estimates — Long Wall-Short Wall & Centre Line Methods · Bar Bending Schedule

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📑 Contents (12 sections)

Part 1 of 2

Building Estimates — Long Wall-Short Wall & Centre Line Methods

Last reviewed 16 Sept 2026 · 7 min read

Methods of estimating walls

In a building, walls and their foundations have the same cross-section along their length, so quantities are found by multiplying length × cross-sectional dimensions for each layer (excavation, concrete bed, footings, plinth and superstructure). The difficulty lies in corners and junctions, where quantities could be double counted or omitted.

1. Long wall–short wall method

Walls in one direction are treated as long walls and those at right angles as short walls.

FormulaLong wall–short wall lengths (for each layer of width b)

Long wall length (out-to-out) $=$ centre-to-centre length $+$ one breadth $b$
Short wall length (in-to-in) $=$ centre-to-centre length $-$ one breadth $b$

The breadth $b$ is the width of that particular layer (trench width, footing width, wall thickness), so the lengths change from layer to layer.

Quantities for each layer are computed separately for long walls and short walls.
Simple and less error-prone for rectangular buildings; laborious for buildings with many walls.

2. Centre line method

The total length of the centre lines of all walls is multiplied by the cross-sectional dimensions of each layer.

FormulaCentre line method

Quantity of a layer = (total centre line length - junction deduction) \times b \times h

At each T-junction (where a cross wall meets a main wall), deduct half the breadth of that layer ( $b /2$ ) from the total centre line length for each junction — because the overlap would otherwise be counted twice.
At corners (L-junctions), no deduction is needed — the extra length on the outer side balances the shortage on the inner side.
At a cross (+) junction of walls, deduct one full breadth (two halves).

Quick and suitable for buildings with walls of uniform section and many junctions (after applying deductions carefully).

3. Partly centre line and partly cross-wall method

Main walls are computed by the centre line method, and cross walls by in-to-in lengths — used where walls have different sections.

Sequence of items in a building estimate

No.	Item	Unit
1	Earthwork in excavation in foundation trenches	m³
2	Lean concrete (e.g. 1 : 4 : 8 PCC) in foundation bed	m³
3	Brick/stone masonry in foundation and plinth (footing steps)	m³
4	Damp-proof course at plinth level	m²
5	Filling in plinth (sand/earth) and floor base	m³
6	Masonry in superstructure (with deductions for openings, lintels, embedded RCC)	m³
7	RCC work — lintels, chajjas, beams, columns, slabs	m³
8	Formwork for RCC items	m²
9	Reinforcement (from BBS)	kg
10	Flooring — base concrete and floor finish; skirting	m², m³
11	Plastering — internal walls, external walls, ceiling	m²
12	Doors, windows, ventilators — frames and shutters, fittings	m², no.
13	Roof treatment / waterproofing, parapet, coping	m²
14	Whitewashing, distempering, painting	m²
15	Steps, sunshades, rainwater pipes, miscellaneous	as applicable
16	Water supply, sanitary and electrical installations	lump sum/%
17	Contingencies, work-charged establishment	%

Worked estimate — single-room building

Worked ExampleEstimate of a single-room building

Data:

Internal room size 5.0 m × 4.0 m; walls 300 mm thick in superstructure; height of superstructure (floor to roof slab bottom) 3.0 m.
Foundation: trench 0.9 m wide, 1.0 m deep below ground level; PCC 1 : 4 : 8 bed 0.9 m wide × 0.2 m thick.
Brick masonry footings: first step 0.6 m wide × 0.3 m; second step 0.45 m wide × 0.3 m; plinth wall 0.3 m wide × 0.6 m (up to plinth level).
Openings: one door 1.0 m × 2.1 m; two windows 1.2 m × 1.2 m; RCC lintels 0.3 m wide × 0.15 m deep with 150 mm bearing on each side.

Centre line lengths: long walls c/c $= 5.0 + 0.3 = 5.3$ m; short walls c/c $= 4.0 + 0.3 = 4.3$ m Total centre line length $= 2 (5.3 + 4.3) = 19.2 m$ (four L-corners, no T-junctions → no deduction)

1. Earthwork in excavation $19.2 \times 0.9 \times 1.0 = 17.28 m^{3}$ Check by long wall–short wall: long walls out-to-out $= 5.3 + 0.9 = 6.2$ m (×2 = 12.4 m); short walls in-to-in $= 4.3 - 0.9 = 3.4$ m (×2 = 6.8 m); total 19.2 m ✓

2. PCC 1 : 4 : 8 in foundation $19.2 \times 0.9 \times 0.2 = 3.46 m^{3}$

3. Brick masonry in foundation and plinth

First step: $19.2 \times 0.6 \times 0.3 = 3.456$ m³
Second step: $19.2 \times 0.45 \times 0.3 = 2.592$ m³
Plinth wall: $19.2 \times 0.3 \times 0.6 = 3.456$ m³ Total $= 9.50 m^{3}$

4. Damp-proof course (at plinth level) $19.2 \times 0.3 = 5.76$ m², less door opening $0.3 \times 1.0 = 0.30$ m² → 5.46 m²

5. Brick masonry in superstructure

Gross: $19.2 \times 0.3 \times 3.0 = 17.28$ m³
Deduct door: $1.0 \times 2.1 \times 0.3 = 0.630$ m³
Deduct windows: $2 \times 1.2 \times 1.2 \times 0.3 = 0.864$ m³
Deduct lintels: length $= (1.0 + 0.3) + 2 (1.2 + 0.3) = 4.3$ m → $4.3 \times 0.3 \times 0.15 = 0.194$ m³ Net $= 17.28 - 0.630 - 0.864 - 0.194 = 15.59 m^{3}$

6. RCC lintels $4.3 \times 0.3 \times 0.15 = 0.194 m^{3}$

7. RCC roof slab (120 mm thick, resting over walls, no projection) $(5.0 + 0.6) \times (4.0 + 0.6) \times 0.12 = 5.6 \times 4.6 \times 0.12 = 3.09 m^{3}$

8. Internal plaster (12 mm) on walls Internal perimeter $= 2 (5.0 + 4.0) = 18.0$ m → gross $= 18.0 \times 3.0 = 54.0$ m² Door (2.1 m²) and windows (1.44 m² each) are between 0.5 and 3 m² → deduct on one face (taken on the inside face here): $2.1 + 2.88 = 4.98$ m² Net internal plaster $= 49.02 m^{2}$

9. External plaster (on superstructure) External perimeter $= 2 (5.6 + 4.6) = 20.4$ m → $20.4 \times 3.0 = 61.20 m^{2}$ (no deductions on this face under the one-face rule for these openings)

10. Ceiling plaster $5.0 \times 4.0 = 20.00 m^{2}$

11. Flooring Floor area $= 5.0 \times 4.0 = 20.00 m^{2}$ (base concrete and finish computed from thicknesses)

The abstract of cost then multiplies each quantity by its rate (from the schedule of rates) and adds contingencies and other provisions.

Worked ExampleDeduction at T-junctions (centre line method)

If an internal cross wall (same foundation section) is added across the room of the example (c/c length 4.3 m), find the centre line length for the excavation layer.

Solution. The cross wall meets the long walls at two T-junctions. For the 0.9 m wide trench, deduct $2 \times \frac{0.9}{2} = 0.9$ m. Centre line length $= 19.2 + 4.3 - 0.9 = 22.6 m$ (a different deduction applies to each layer according to its breadth)

Part 2 of 2

Bar Bending Schedule

Last reviewed 16 Sept 2026 · 7 min read

Bar bending schedule (BBS)

A bar bending schedule is a detailed list of all reinforcement bars in a structural element (or project), giving their mark, shape, diameter, number, cutting length and weight.

Purposes

Accurate quantity of steel for estimation, procurement and billing.
Instructions for cutting and bending bars in the yard — reduces wastage and errors.
Better quality control — verification of reinforcement against drawings before concreting.
Planning of bar lengths (standard 12 m lengths) to minimise offcuts.
Basis for measurement of reinforcement (kg/tonne) in the Measurement Book.

Format of a BBS

Column	Content
Member / location	e.g. beam B1, slab S2, column C3
Bar mark	Identification of each bar type
Shape (sketch)	Straight, hooked, cranked, stirrup etc. with dimensions
Diameter (mm)
Number of members
Number of bars per member
Total number of bars
Cutting length (m)
Total length (m)
Unit weight (kg/m)	$d^{2} /162$
Total weight (kg)

Relevant references include IS 2502 (code of practice for bending and fixing of bars) and SP 34 (handbook on concrete reinforcement and detailing).

Unit weight of bars

FormulaWeight of reinforcement

w = \frac{d ^{2}}{162} kg/m

( $d$ in mm)

Diameter (mm)	6	8	10	12	16	20	25	32
Weight (kg/m)	0.222	0.395	0.617	0.889	1.580	2.469	3.858	6.321

Hooks, bends and allowances

Code requirements (IS 456)

Anchorage value of a standard U-type (180°) hook is taken as 16Ø and of a 90° bend as 8Ø (with limits on bend radius) in calculating development length.
Minimum cover and bend radii follow code provisions; stirrups in seismic detailing (IS 13920) use 135° hooks with specified extensions.

Typical site practice for cutting lengths

When a bar is bent, its measured outer dimensions exceed its centre-line length, so an allowance (bend deduction) is subtracted from the sum of the outer dimensions. Commonly used practical values are:

Bend angle	Deduction (approx.)
45°	1Ø
90°	2Ø
135°	3Ø
180°	4Ø

Hook length additions commonly adopted: about 9Ø for a 180° hook on main bars; about 10Ø (or a minimum length) for 135° stirrup hooks. These are practical approximations; exact values depend on bend radius and project specifications.

Development and lap lengths

FormulaDevelopment length (IS 456)

L_{d} = \frac{ϕ σ _{s}}{4 τ _{b d}}

$ϕ$ = bar diameter; $σ_{s}$ = stress in bar at the section (0.87 f_y at limit state); $τ_{b d}$ = design bond stress.

Design bond stress for plain bars in tension: M20 — 1.2, M25 — 1.4, M30 — 1.5, M35 — 1.7, M40 and above — 1.9 N/mm².
For deformed bars, increase by 60%; for bars in compression, increase by 25%.

Lap lengths (IS 456, typical):

Tension laps: not less than $L_{d}$ or 30Ø, whichever is greater (more at critical locations).
Compression laps: not less than the development length in compression or 24Ø, whichever is greater.
Laps are staggered and avoided at points of maximum stress.

Cutting length formulas

FormulaCommon bars

Straight bar (no hooks): $L =$ member length $-$ 2 × end cover

Straight bar with two 180° hooks: $L =$ member length $-$ 2 × end cover $+ 2 \times 9 ϕ$ (practical hook allowance) $-$ bend deductions as applicable

Cranked (bent-up) bar with 45° cranks: Extra length per crank $= 1.414 d^{'} - d^{'} = 0.414 d^{'}$ (commonly taken as $0.42 d^{'}$ ) $d^{'}$ = vertical distance between the centres of the horizontal portions $= D - 2 \times cover - ϕ$ (approximately) $L =$ straight length $+$ (number of cranks × $0.42 d^{'}$ ) $-$ bend deductions (if applied)

Rectangular stirrup (two 135° hooks): Centre-line dimensions: $a = b - 2 c - ϕ$ , $h = D - 2 c - ϕ$ ( $c$ = clear cover to stirrup) $L = 2 (a + h) + 2 \times (hook length) - (bend deductions: 3 \times 2 ϕ + 2 \times 3 ϕ)$

Circular tie/ring: $L = π \times$ centre-line diameter + overlap/hooks

Number of stirrups: $n = \frac{length over which provided}{spacing} + 1$ (rounded up)

Number of bars in a slab: $n = \frac{clear span in the perpendicular direction - 2 \times cover}{spacing} + 1$

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