CON-130 · Lesson 9 of 12 · 34 min
Quantity takeoff arithmetic
A quantity takeoff is the systematic measurement of every material and item of work shown on the drawings and described in the specifications. It is the foundation of the estimate, the source of purchase orders and a baseline for tracking progress and changes. Lessons 3 to 5 covered areas, volumes and earthwork; this lesson applies that geometry to the materials that are counted rather than poured: framing lumber, sheet goods such as drywall, reinforcing steel, masonry units and tile. Each has its own units of sale, standard sizes and counting rules, and each rewards a careful, tabulated method. The full estimating process, pricing and software are covered in the construction estimating course; here the focus is on the arithmetic that makes quantities right.
Principles of a good takeoff
- Work in an organized order, usually by specification division and then by floor, area or system, so that nothing is missed and nothing is counted twice.
- Measure net quantities from the drawings, then apply waste, laps and allowances separately and visibly, so that others can check them.
- Use the unit of measure in which the item is bought or priced: m³ or yd³ for concrete, t or tons for rebar and aggregates, m² or ft² for drywall, pieces for doors and fixtures.
- Record dimensions, not just results, in a takeoff sheet or software, with references to drawing numbers and details.
- Mark up the drawings as items are measured, so that what has been counted is visible.
- Round only at the end, and round up to the units in which materials are sold.
Counting rules matter as much as formulas. When items are spaced along a length, the number of items is the number of spaces plus one if there is an item at each end. Forgetting the extra one, called the fence-post error, is one of the most frequent small errors in takeoffs of studs, joists, bars and anchors.
Here L is the length over which items are spaced, s is the spacing, and the brackets mean 'round up to the next whole number'. A wall 3.0 m long with studs at 400 mm needs ⌈3,000 ÷ 400⌉ + 1 = ⌈7.5⌉ + 1 = 9 studs, before any extras for corners, openings and connections.
Framing lumber
Canadian lumber is graded and sized to the same standards as US lumber and sold under imperial names such as 2×6, with metric equivalents (38 × 140 mm) on drawings. Studs and plates are usually priced per piece or per linear foot or metre, and larger volumes may be priced per thousand board feet (MBF), a unit still common in Canadian lumber trading. Precut studs in standard lengths, such as 92 5/8 in (2,353 mm) for 8 ft walls, reduce cutting.
A board foot is a volume of wood equal to a board 1 inch thick, 12 inches wide and 1 foot long, or 144 cubic inches. Board feet are calculated from nominal, not actual, dimensions. The metric equivalent volume is the cubic metre; one cubic metre of nominal dimensions is about 424 board feet.
Worked problem
Board feet in a joist order
A floor needs 24 joists of 2×10, each 16 ft long, plus 3 rim boards of 2×10 × 16 ft. (a) How many board feet is that? (b) At $1,150 per MBF, what does the lumber cost? (c) What is the actual volume of wood in cubic metres?
Show solution
- (a) One 2×10 × 16 ft = (2 × 10 × 16) ÷ 12 = 26.67 BF. Pieces = 24 + 3 = 27; total = 27 × 26.67 = 720 BF.
- (b) Cost = 720 ÷ 1,000 × $1,150 = $828.
- (c) Actual size 38 × 235 mm and 16 ft = 4.877 m: one piece = 0.038 × 0.235 × 4.877 = 0.0436 m³; 27 pieces = 1.18 m³. The nominal volume, 720 BF ÷ 424, is about 1.70 m³; the difference is the surfacing and drying allowance built into nominal sizes.
Answer: 720 BF costing about $828; the actual wood volume is about 1.18 m³.
Worked problem
Studs and plates for a wall
An exterior wall is 12.0 m long and 2.4 m high, framed with 38 × 140 mm studs at 400 mm on centre, with a single bottom plate and a double top plate. It has two window openings, each needing 2 extra studs on each side (king and jack studs), and two corners each needing 2 extra studs. Count the studs and the linear metres of plate, and convert the plate length to 3.66 m (12 ft) pieces.
Show solution
- Common studs: ⌈12,000 ÷ 400⌉ + 1 = 30 + 1 = 31.
- Extras: openings 2 × 2 × 2 = 8; corners 2 × 2 = 4; total = 31 + 8 + 4 = 43 studs. (Studs that fall inside openings could be deducted; many estimators keep them to cover cripples and blocking.)
- Plates: 3 × 12.0 = 36.0 m; in 3.66 m pieces, 36.0 ÷ 3.66 = 9.8, so 10 pieces, plus allowance for splices staggered over studs.
- With a 10 % waste allowance on studs, order 43 × 1.10 = 47.3, so 48 studs.
Answer: 43 studs net (order 48) and 36 m of plate (about 10 pieces of 3.66 m).
Sheet goods: drywall and sheathing
Drywall, plywood, OSB and rigid insulation are sold in sheets. The simplest takeoff divides the net area by the sheet area and adds waste, then rounds up to whole sheets. More careful takeoffs lay out sheets wall by wall, choosing lengths that minimize joints and waste: a 2.7 m high wall may be boarded horizontally with sheets 1.22 m wide, and longer sheets (3.66 m or 12 ft) reduce butt joints on long walls.
| Sheet size | Area | Common uses |
|---|---|---|
| 1220 × 2440 mm (4 × 8 ft) | 2.977 m² (32 ft²) | Plywood, OSB, drywall |
| 1220 × 3050 mm (4 × 10 ft) | 3.721 m² (40 ft²) | Drywall on taller walls |
| 1220 × 3660 mm (4 × 12 ft) | 4.465 m² (48 ft²) | Drywall on long walls and ceilings |
| 1220 × 4880 mm (4 × 16 ft) | 5.954 m² (64 ft²) | Drywall in commercial work |
Worked problem
Drywall for a room
The room in Lesson 3 (5.4 m × 4.2 m, 2.7 m ceiling) has 46.35 m² of net wall area. Board the walls and the ceiling with 1220 × 3660 mm sheets with a 10 % waste allowance. How many sheets are needed?
Show solution
- Ceiling area = 5.4 × 4.2 = 22.68 m².
- Total = 46.35 + 22.68 = 69.03 m²; with waste 69.03 × 1.10 = 75.93 m².
- Sheet area = 1.22 × 3.66 = 4.465 m²; sheets = 75.93 ÷ 4.465 = 17.0, so 17 sheets, or 18 to give a margin for damage, since the result is almost exactly a whole number.
Answer: About 17 to 18 sheets.
Drywall takeoffs also generate quantities for screws, tape, joint compound and corner bead, usually by applying rates per square metre or square foot from manufacturers' data, and for the labour of hanging and finishing, which is often priced per unit area at a specified level of finish.
Reinforcing steel
Reinforcing bars are bought by mass and placed by count and length. The takeoff counts the bars of each size and length from the drawings and bar schedules, adds laps where bars are spliced, and converts total length to mass using the mass per unit length of each bar size. Detailing and lap lengths are set by the structural engineer and the governing concrete standard; the construction manager's role is to check quantities and understand how they were produced.
Canadian bars are metric sizes under CSA G30.18, named by their approximate diameter in millimetres with an M suffix. Common stock lengths are 6, 12 and 18 m. Bar masses are fixed by the standard:
| Bar | Nominal diameter | Mass per metre |
|---|---|---|
| 10M | 11.3 mm | 0.785 kg/m |
| 15M | 16.0 mm | 1.570 kg/m |
| 20M | 19.5 mm | 2.355 kg/m |
| 25M | 25.2 mm | 3.925 kg/m |
| 30M | 29.9 mm | 5.495 kg/m |
| 35M | 35.7 mm | 7.850 kg/m |
To count bars in a mat, find the length available for spacing (the slab dimension less the cover at each edge), divide by the spacing, round up and add one. Each bar's length is the other slab dimension less the cover at both ends. If a bar is longer than the stock length, it must be spliced, and each lap adds the lap length to the total.
Worked problem
Reinforcing for a slab
The 12.35 m × 8.60 m slab from Lesson 1 is reinforced with 15M bars at 300 mm each way, with 75 mm edge cover. Stock bars are 12 m long and the engineer specifies a 650 mm lap where bars are spliced. Find the number and length of bars in each direction, the total length and the mass.
Show solution
- Long bars run 12.35 m and are spaced across 8.60 m: spacing length = 8,600 − 2 × 75 = 8,450 mm; ⌈8,450 ÷ 300⌉ + 1 = ⌈28.2⌉ + 1 = 30 bars.
- Each long bar = 12.35 − 0.15 = 12.20 m, longer than 12 m stock, so each needs one lap: 12.20 + 0.65 = 12.85 m of bar.
- Short bars run 8.60 m and are spaced along 12.35 m: spacing length = 12,350 − 150 = 12,200 mm; ⌈12,200 ÷ 300⌉ + 1 = ⌈40.7⌉ + 1 = 42 bars, each 8.60 − 0.15 = 8.45 m.
- Total length = 30 × 12.85 + 42 × 8.45 = 385.5 + 354.9 = 740.4 m.
- Mass = 740.4 × 1.570 = 1,162 kg, about 1.16 t before waste and accessories such as chairs.
Answer: 72 bars totalling about 740 m and 1.16 t of 15M.
Masonry units
Masonry is taken off by wall area, net of openings, and converted to units using the number of units per square metre or square foot. That number depends on the unit size plus one mortar joint in each direction, because each unit effectively occupies its own face plus half a joint all round.
Canadian masonry uses metric modular units. A metric modular brick (about 190 × 57 mm face) with 10 mm joints occupies 200 × 66.7 mm, so 75 bricks cover a square metre. A standard concrete block (390 × 190 mm face) with 10 mm joints occupies 400 × 200 mm, so 12.5 blocks cover a square metre.
Worked problem
Concrete block for a wall
A block wall is 30.0 m long and 3.0 m high with two window openings of 1.2 m × 1.2 m. Using 12.5 blocks per m² and 5 % waste, how many blocks are needed?
Show solution
- Gross area = 30.0 × 3.0 = 90.0 m²; openings = 2 × 1.44 = 2.88 m²; net = 87.12 m².
- Blocks = 87.12 × 12.5 = 1,089; with waste 1,089 × 1.05 = 1,143.5, so 1,144 blocks.
- Special units, such as bond beam, lintel and corner blocks, are counted separately from the drawings and deducted from the standard count.
Answer: 1,144 blocks.
Tile and other boxed materials
Tile, flooring planks, shingles and many finishes are sold in boxes or bundles that cover a stated area. Divide the area including waste by the coverage per box and round up to whole boxes. Owners often require attic stock, extra material from the same production lot kept for future repairs, usually stated as a percentage or a number of boxes in the specification. Because colour and shade vary between production lots, all the tile for one area, including attic stock, should be ordered at the same time.
Worked problem
Ordering tile by the box
The tile floor in Lesson 2 needs 211.4 m² (2,275 ft²) including waste. Boxes cover 1.44 m² (15.5 ft²). The specification requires attic stock of 2 % of the installed quantity. How many boxes should be ordered?
Show solution
- Installed (net) quantity = 186 m²; attic stock = 0.02 × 186 = 3.7 m².
- Total = 211.4 + 3.7 = 215.1 m²; boxes = 215.1 ÷ 1.44 = 149.4, so 150 boxes.
- In imperial: net 2,002 ft²; attic stock 40 ft²; total 2,275 + 40 = 2,315 ft²; 2,315 ÷ 15.5 = 149.4, so 150 boxes.
Answer: 150 boxes.
Units of measure
| Item | Typical takeoff unit | Typical purchase unit |
|---|---|---|
| Excavation and fill | Bank m³ or BCY | Truckloads, tonnes or tons |
| Concrete | m³ or yd³ | m³ or yd³ delivered |
| Formwork | m² or ft² of contact area | Rental or material by piece |
| Reinforcing steel | Length by bar size | Tonnes or tons, or by bar list |
| Masonry | m² or ft² of wall | Units, pallets |
| Framing lumber | Pieces and linear m or ft | Pieces or MBF |
| Drywall and sheathing | m² or ft² | Sheets |
| Roofing | m² or squares | Bundles, rolls |
| Doors, windows, fixtures | Each | Each |
Common mistakes
MistakeForgetting to add one when counting spaced items.
InsteadNumber of items = spaces (rounded up) + 1 when there is an item at each end.
MistakeUsing actual sizes to calculate board feet.
InsteadBoard feet use nominal dimensions; use actual sizes only for fit and real volume.
MistakeIgnoring laps when bars exceed stock length.
InsteadAdd one lap length for each splice before converting to mass.
MistakeCounting masonry units from the unit face without joints.
InsteadInclude one joint in each direction: units per area = 1 ÷ ((ℓ + j)(h + j)).
MistakeOrdering tile in several lots.
InsteadOrder all tile, including attic stock, from one production lot.
Summary and key terms
- A takeoff measures net quantities, then applies waste and allowances visibly.
- Spaced items number one more than the spaces between them.
- Board feet = T × W (inches, nominal) × L (feet) ÷ 12.
- Sheet goods: area plus waste divided by sheet area, rounded up.
- Rebar: count bars in each direction, add laps, convert length to mass.
- Masonry: net wall area times units per area, including joints.
- Boxed materials are rounded up to whole boxes, with attic stock where required.
- Quantity takeoff
- Systematic measurement of quantities from drawings.
- Fence-post error
- Miscounting items by forgetting the last one.
- Board foot
- Volume 1 in × 12 in × 1 ft of nominal lumber.
- MBF
- One thousand board feet.
- Stock length
- Standard length in which a material is supplied.
- Lap splice
- Overlap joining two reinforcing bars.
- Cover
- Concrete between reinforcing and the concrete surface.
- Modular unit
- Masonry unit sized to a grid including joints.
- Attic stock
- Extra material left with the owner for repairs.
- Bar list
- Schedule of reinforcing bars by size, length and shape.
Practice problems
Work each problem on paper first, then reveal the answer to check.
How many joists at 400 mm on centre over 6.0 m?
Show answer
⌈15⌉ + 1 = 16.
How many studs at 16 in on centre over 22 ft?
Show answer
⌈16.5⌉ + 1 = 18.
Board feet in 40 pieces of 2×6 × 12 ft?
Show answer
40 × 12 = 480 BF.
Sheets of 4 × 8 for 1,150 ft² plus 12 % waste?
Show answer
1,288 ÷ 32 = 40.3, so 41 sheets.
Mass of 250 m of 20M bar?
Show answer
250 × 2.355 = 588.75 kg.
Weight of 600 ft of #5 bar?
Show answer
600 × 1.043 = 625.8 lb.
Bars at 250 mm in a 5.0 m wide slab with 75 mm cover each side?
Show answer
⌈4,850 ÷ 250⌉ + 1 = 21.
Bricks for 40 m² at 75 per m² with 5 % waste?
Show answer
3,150.
Blocks for 520 ft² at 1.125 per ft²?
Show answer
585.
Boxes of tile covering 1.2 m² each for 50 m²?
Show answer
41.7, so 42 boxes.
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