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Rebar Calculator
Estimate rebar quantity, length, weight & material list for slabs, walls & footings
Enter the slab, wall or footing size and it returns bar count, total length, lap allowance, weight and a cost estimate. A 20 ft × 20 ft slab in #4 at 12 in on centre both ways is 40 bars, 800 ft, 534.4 lb.
0.668 lb/ft#4, the default bar
40bars in 20 × 20 ft at 12 in
534.4 lbthat slab, both ways
1.3Class B lap multiplier
No account and no upload — the arithmetic runs in your browser. Cover, lap class and waste are inputs, not hidden assumptions.
Inputs
ft
ft
in
in
in
$ / bar
Results
Where to Buy Rebar
Once you have your bar sizes, lengths and piece counts, the next step is buying. eBay lists rebar, rebar ties, wire mesh and bar cutters from many sellers, so you can compare price per piece before you order.
Affiliate disclosure: the links below are affiliate links. If you buy through them, we may earn a commission at no extra cost to you, and it does not change the price you pay.
Total Length = Σ(Bars × Bar Length) + Lap Allowance
Lap Allowance = Lap Length × (Pieces per Bar − 1) × Bars (when bar > standard length)
Weight (lb/ft) = (π × d² × 490) / 576 d = diameter in inches, 490 = steel density (lb/ft³)
Total Weight = Total Length × Unit Weight
Weight to Order = Pieces × Standard Bar Length × Unit Weight
Pieces = Σ⌈ Bar Length / Standard Bar Length ⌉ × Bars
Lap length defaults to the Class B splice of ACI 318-19 Table 25.4.2.3 at 4,000 psi — 19 in for a #3 up to 62 in for a #8, in the row where clear spacing and cover both reach db. It is not 40 × db, which is short on every bar size. Override with your own value if needed.
Total Weight is the steel in the finished work, laps included.
Weight to Order counts whole stock bars, so it includes offcut waste — use that figure when buying by weight.
Results are for material estimation only. Always follow your project's engineering specifications.
Not sure which bar size to pick? See the rebar size & weight chart. Weighing bar against welded wire mesh? See rebar vs wire mesh.
Circular Slabs
A 10 ft diameter slab, #4 at 12 in on centre each way, single layer, takes 9 bars in each direction and 69.80 ft of bar per direction — 139.6 ft and 93.3 lb for the whole slab. Double layer is 186.5 lb.
That number rests on three things, and none of them are printed on a round slab:
Single layer or double layer. A slab on grade is one mat, both ways. A structural slab is a top and a bottom mat — multiply every figure below by 2.
Spacing s. The tables are at s = 12 in. Change s and the bar count changes with it, because every bar is re-cut to the circle at its new offset. You cannot scale 69.80 ft by 12/s.
What R is. R = D/2 − cover − db/2 — the radius to the centre line of the bar, not to the face of the slab. A 10 ft slab with 3 in cover and #4 gives R = 60 − 3 − 0.25 = 56.75 in.
A square formula has no length and no width to work with here. The square that just encloses a 10 ft circle is 100 sq ft; the circle is 78.5 sq ft. Order off the square and a quarter of the steel is still in the yard when the pour is done.
The cut grid (this is the answer above)
Two mats at right angles, spacing s, every bar cut where it leaves the circle, so the bars shorten as they move off centre.
R = D / 2 − cover − db / 2
Bars per direction = 2k + 1, where k = largest i with i · s < R
Bar at offset i · s: length = 2 √(R² − (i · s)²)
Cut-grid length per direction = 2R + Σ 4 √(R² − (i · s)²), i = 1 … k
Both ways, single layer = 2 × that length · double layer = 4 × that length
The radial and hoop mat (comparison only)
Bars run out from the centre and hoops ring them at the same spacing. Different mat, different arithmetic.
Radial bars m = ⌈ 2πR / s ⌉, each one R long
Hoops n = ⌈ R / s ⌉, the k-th hoop at radius R − (k−1)s
Hoop length total = 2π [ nR − s · n(n−1)/2 ]
The comparison table below counts every radial bar at full length R. Stopping alternate bars at R/2 is a placing option, and it is not applied to these figures.
Three numbers, three names
Same 10 ft slab, s = 12 in, R = 56.75 in. They sit within 0.7% of each other, which is exactly why they get mixed up. Only one of them is the answer.
Name
How it is produced
Per direction
Use it for
Exact
Chords added one at a time, R includes db/2
69.80 ft
The answer on this page
Grid total, added by hand
Same chords with R rounded to 57, nine bars summed on paper
70.29 ft
Re-checking the page on a phone, +0.7%
Area estimate
πR² / s
70.26 ft
A 30-second sanity check, +0.7%
The hand sum behind 70.29: half-bars of 114.00, 111.45, 103.40, 88.39 and 61.48 in, so 114.00 + 2 × (111.45 + 103.40 + 88.39 + 61.48) = 843.43 in = 70.29 ft.
The area estimate walks away when the spacing is coarse. An 8 ft slab at 9 in comes to 56.48 ft exactly and 58.25 ft by area — +3.1%. Keep it for checking, never for ordering.
Diameter against both methods
#4 bar, 3 in cover, s = 12 in, single layer unless stated. R already carries the half bar diameter.
Diameter
R (in)
Bars each way
Cut grid, ft/dir
Cut grid, single, ft
Cut grid, single, lb
Cut grid, double, lb
Radial bars
Hoop, ft
Radial + hoop, ft
Radial + hoop, lb
Ratio to cut grid
6 ft
32.75
5
23.0
46.1
30.8
61.6
18
32.6
81.7
55
1.77
8 ft
44.75
7
43.3
86.6
57.8
115.6
24
56.0
145.5
97
1.68
10 ft
56.75
9
69.8
139.6
93.3
186.5
30
85.7
227.6
152
1.63
12 ft
68.75
11
102.6
205.2
137.1
274.2
36
121.7
328.0
219
1.60
15 ft
86.75
15
165.5
331.0
221.1
442.3
46
187.4
520.0
347
1.57
Cut grid columns: bars = 2k + 1, ft/dir = 2R + Σ4√(R² − (i·s)²), lb = ft × 0.668. Radial and hoop columns: radial = m × R, hoop = 2π[nR − s·n(n−1)/2]. Ratio = radial and hoop total ÷ cut grid single layer.
Why the two mats differ by 1.6 to 1.8 times
On the 10 ft slab the same bar at the same spacing runs 93.3 lb cut square and 152 lb radial and hoop — 1.63 times. At 6 ft the gap is 1.77; at 15 ft it closes to 1.57. The difference sits at the centre, where radial bars converge: at a spacing that works at the edge, they are already too tight to place and to consolidate near the middle.
Slabs on grade are placed as a cut grid. Radial and hoop mats belong to tank bases and to slabs designed for radial reinforcement. Take 1.7 as a default and roughly half the order is still banded when the job is finished.
Laps are not counted anywhere in this section. They follow bar length and splice class, not the fact that the slab is round, so they are added after this length, not inside it.
Before the bar schedule, the slab has to have a thickness — see concrete slab thickness for the minimums by use and by cover.
L-Shaped and Odd Slabs
A 12 ft × 10 ft L-shaped slab with 4 ft legs, #4 at 12 in on centre, single layer, takes 133.0 ft of bar and 88.8 lb in the mat — and 139.0 ft / 92.9 lb once the single bar at the inside corner is added.
That number stands on three things, and an L-shaped slab has no drawing to read them off:
Single layer or double layer. A slab on grade is one mat, both ways. A structural slab is a top mat and a bottom mat — double every figure below.
Spacing s. The tables are at s = 12 in. Change s and the count changes with it, because every bar is re-cut to the span it happens to cross. 133.0 ft does not scale by 12/s: the same slab at 18 in is 94.5 ft, not 88.7.
Where the first bar starts, c′. c′ = cover + db/2 = 3.25 in for #4 at 3 in cover, and every bar is cut to its own span less 2 × cover. So each bar loses 6 in off the span. "Cut to the slab width" is not an instruction — it is the source of the error.
Why an L is the easiest slab to over-order
An L has no single length and no single width, so both shortcuts are wrong, and they are wrong in opposite directions.
Bounding box treated as the slab. The 12 × 10 L with 4 ft legs is 72 sq ft of slab sitting inside a 120 sq ft rectangle. That is +66.7% on area (120 ÷ 72 − 1) and +72.2% on steel (229.0 ÷ 133.0 − 1). Two different numbers with two different names, and the steel one is the bigger: every bar in the box runs a full 138 in or 114 in, while the bars that only reach into the vertical leg run 42 in.
Split into two rectangles and added up. The bars that cross the joint get counted in both pieces — +10.5% (147.0 ft against 133.0 ft). It looks close enough to be right, which is what makes it dangerous.
The correct rule fits in one line: each bar runs the full span of the slab along its own line. A bar that stops at the vertical leg is 4 ft long, not 12 ft.
One bar at a time (this is the answer above)
Put the origin at the outside corner. x runs along the 12 ft side, y along the 10 ft side, and both legs are 4 ft wide.
Bars per direction n = ⌊ (outline − 2c′) / s ⌋ + 1 · c′ = cover + db/2
Bar parallel to x, at height y: length = (y < leg) ? L − 2·cover : leg − 2·cover
Bar parallel to y, at position x: length = (x < leg) ? W − 2·cover : leg − 2·cover
Mat length = Σ (every bar × its own length), both directions
Inside corner bar = one #4, 3 ft each way = 6.0 ft, carried as its own line, not inside the mat
Parallel to x: the 4 bars below y = 48 in run the full 138 in (4 × 138 = 552 in); the 6 bars above it run 42 in (6 × 42 = 252 in). Total 804 in = 67.0 ft.
Parallel to y: the 4 bars below x = 48 in run the full 114 in (4 × 114 = 456 in); the 8 bars beyond it run 42 in (8 × 42 = 336 in). Total 792 in = 66.0 ft.
67.0 + 66.0 = 133.0 ft, and 133.0 × 0.668 = 88.8 lb. Add the corner bar and the order is 139.0 ft / 92.9 lb.
The 10.5% that the two-rectangle method adds is exactly the 4 bars at x < 48 in counted twice: 4 × 42 = 168 in = 14 ft, and 133.0 + 14 = 147.0 ft.
Three numbers, three names
Same 12 × 10 slab, same bar and spacing. Only the first row is the answer.
Name
How it is produced
Length
Steel
Use it for
Exact, bar by bar
Every bar cut to the span on its own line, count started at c′
133.0 ft
88.8 lb
The answer on this page
Two rectangles, added
Two pieces sized separately, joint bars counted in both
147.0 ft
98.2 lb
A counterexample, +10.5%
Bounding box
12 × 10 filled as one plain rectangle
229.0 ft
153.0 lb
The expensive mistake, +72.2%
Steel column is length × 0.668 lb/ft throughout. The 72.2% is a steel deviation; the area deviation for the same box is 66.7%. They are not the same number and must not be quoted for each other.
Size against all three methods
#4 bar, 3 in cover, 4 ft legs, s = 12 in except the last row, single layer. Both deviation columns are measured against the exact mat length, not against area.
Outline
Leg width
Mat, ft
Mat, lb
Two rectangles, ft
Deviation (steel)
Bounding box, ft
Deviation (steel)
Bounding box deviation (area)
12 × 10 ft
4 ft
133.0
88.8
147.0
+10.5%
229.0
+72.2%
+66.7% (120 vs 72 sq ft)
10 × 8 ft
4 ft
103.0
68.8
117.0
+13.6%
151.0
+46.6%
+42.9% (80 vs 56 sq ft)
16 × 12 ft
4 ft
178.0
118.9
192.0
+7.9%
370.0
+107.9%
+100.0% (192 vs 96 sq ft)
12 × 10 ft @ s=18
4 ft
94.5
63.1
105.0
+11.1%
156.5
+65.6%
+66.7% (120 vs 72 sq ft)
The steel deviation and the area deviation are not one number, and they do not always point the same way. At 16 × 12 the area is exactly double (+100.0%) and the steel is more than double (+107.9%). At 12 × 10 on 18 in centres it flips: steel +65.6% is slightly less than area +66.7%, because the wider spacing drops a row out of the box. Quote the column you mean.
Narrower legs on a bigger slab push the bounding box further out of reach. The two-rectangle error follows the number of bars crossing the joint, which is why it stays small and looks acceptable.
Working a two-way mat by hand, including the T-junctions and corners that an L shares with a plain rectangle — see rebar grid calculator.
Columns
A 16 × 16 in column with eight #6 verticals, #3 ties at 12 in on centre, a 10 ft floor-to-floor height and 1.5 in cover takes 120.2 lb of vertical bar and 20.0 lb of tie with 135° seismic hooks. The 140.2 lb total is a total, not an order: the two are bought as separate line items.
That answer stands on four things, and a column drawing usually shows only two of them:
How the ties are counted. N = ⌈ height / spacing ⌉ + 1, one at the top and one at the bottom. Drop the +1 and it is 10 ties instead of 11 — 9.1% light.
How many covers come out of one tie. 2(b + h) − 8 × cover. Eight covers, because the tie turns two directions at each of four corners. Take out two instead of eight and the 16 × 16 #3 tie goes from 58.0 in to 67.0 in — 15.5% heavy.
The hook, with its db and its angle. 135° seismic hook = 6db, not less than 3 in. 90° hook = 6db for #3–#5 (not less than 2.5 in), or 12db for #6–#8. The same #6 tie is 61.0 in at 135° and 70.0 in at 90° — 14.7% apart. "Two hooks" is not a dimension.
No laps. These are one storey, 10 ft, spliced nowhere. A lap at every floor adds 8 × lap length on the verticals, added outside these figures, not inside them.
Two sets of numbers that look alike
A column is a vertical set and a horizontal set, and they are not interchangeable. One vertical bar here is 120 in long; one tie is 58 in. Read the tie off the same line as the vertical and the tie comes out longer than the bar it wraps.
The mistakes are all quiet ones. The tie count loses a bar at the top of the lift. The cover is deducted twice instead of eight times, which leaves 6 in of steel in the cage. The hook is left as a word, and the reader supplies whichever default their bar bender uses. None of these throw an error; each one moves the number by 9% to 15%.
The two sets, worked separately
Vertical bars. The count is set by the tie, not by the steel area: every vertical has to land on a corner or a leg of the tie. Where verticals sit more than 6 in apart along a face, a crosstie goes in, otherwise the bar between them is not restrained and does not count as braced.
Vertical count = corners + one per leg · one vertical = height
Vertical weight = count × height × unit weight (laps outside this: 8 × lap length per floor)
Tie count N = ⌈ height / spacing ⌉ + 1
One tie L = 2(b + h) − 8 × cover + 2 × extension
135°: extension = max(6db, 3 in) · 90°: max(6db, 2.5 in) for #3–#5, 12db for #6–#8
Tie weight = N × L × unit weight
Unit weights used below: #3 0.376, #4 0.668, #5 1.043, #6 1.502 lb/ft.
The 16 × 16 worked by hand (this is the answer above)
Verticals. 8 × #6 at 10 ft = 80.0 ft × 1.502 = 120.2 lb.
Ties. ⌈ 120 / 12 ⌉ + 1 = 10 + 1 = 11. One tie: 2(16 + 16) − 8 × 1.5 + 2 × 3.0 = 64 − 12 + 6 = 58.0 in. The 6db of a #3 is 2.25 in, short of the 3 in floor, so the floor sets the extension here, not the bar.
11 × 58.0 = 638 in = 53.17 ft × 0.376 = 20.0 lb.
120.2 + 20.0 = 140.2 lb for the storey with both sets together.
Checking it without the formula
The tie is a rectangle sitting inside the cover: 16 − 2 × 1.5 = 13 in a side, one turn is 4 × 13 = 52 in — the same 52 that 2(16 + 16) − 8 × 1.5 gives. Add the two hooks: 52 + 2 × 3.0 = 58.0 in. Eleven of them is 638 in, 53.17 ft, 20.0 lb. The verticals are the easy half: 80 ft at 1.502 lb/ft.
Column against column
10 ft height, ties at 12 in, 1.5 in cover, 135° hooks. Verticals and ties are kept in separate columns on purpose.
Column (section · bars)
Verticals
Vertical ft
Vertical lb
Tie bar
Ties
One tie, in
Tie lb
Total lb
16 × 16 · 8×#6 + #3 (the answer above)
8×#6
80.0
120.2
#3
11
58.0
20.0
140.2
12 × 12 · 4×#4 + #3
4×#4
40.0
26.7
#3
11
42.0
14.5
41.2
16 × 16 · 8×#5 + #4
8×#5
80.0
83.4
#4
11
58.0
35.5
118.9
18 × 18 · 8×#6 + #4
8×#6
80.0
120.2
#4
11
66.0
40.4
160.6
Ties run 14.3% / 35.2% / 29.9% / 25.2% of the total, in row order. The two are never folded into one figure for ordering: #6 comes by the bar, #3 by the coil, and a single total has to be taken apart again before anyone can place the order — which is where the tie gets read as a vertical.
Same 16 × 16 section, two different answers: 20.0 lb of #3 against 35.5 lb of #4, 78% apart. A section size does not name a quantity on its own. Section and bar sizes have to be written together.
All four rows sit at 12 in on a 10 ft height, so all four have 11 ties. The count follows the height and the spacing only. A bigger section changes the length of one tie, not how many there are.
Hook type, same tie
This one changes the hook and nothing else. It deliberately uses a #6 tie, which is not one of the rows above — its 84.0 lb does not belong beside them.
Hook
Extension
One tie, in
Tie lb
Against 135°
135° seismic
6db = 4.5 in
61.0
84.0
—
90°
12db = 9.0 in
70.0
96.4
+14.7%
Same bar, same 12 in spacing, same 11 ties, 14.7% apart on the hook alone. 6 × 0.75 = 4.5 in clears the 3 in floor, so the 6db governs at 135°; 12 × 0.75 = 9.0 in at 90°. Seismic hoops are 135°. A page that writes "two hooks" has handed this choice to the reader's default, and the default is where the 14.7% goes.
The same tie arithmetic read against beam stirrups, and the cover that sets both — see concrete beam reinforcement.
Metric Slabs
A 9.0 × 12.0 m slab with 12 mm bar at 300 mm on centre, one layer each way and 40 mm cover takes 40 + 30 bars, 750.0 m of bar and 666.7 kg — 6.17 kg/m². Worked as integer bars, not as an area estimate.
Three things stand next to that answer, and a metric drawing usually shows none of them:
One layer or two. Everything here is a single mat: one layer, one direction each way. A second layer doubles it. It is not ×1.1, and it is not a rounding difference.
Centre to centre, not clear. 300 mm is centre to centre. Read it as clear spacing and the same slab picks up a large block of extra bars, because every gap grows by one bar diameter.
12 mm is a nominal diameter, and it is not #4. #4 is 12.7 mm. Count #4 as 12 mm and the order comes up 10.6% short — the same 11.8% that separates 666.7 kg from 745.6 kg on this slab.
Why the metric slab is the easiest one to get wrong
None of these four produce an error message. Each one moves the number by a few per cent, which is exactly the size that reads as correct.
12 mm is not #4. 12 mm has a geometric area of 113.1 mm². #4 has a nominal diameter of 12.7 mm and a nominal area of 0.20 in² = 129.0 mm² (geometric at 12.7 mm is 126.7, which is 1.8% off the nominal — both numbers are right, they are two definitions). Unit weight is 0.8889 against 0.994 kg/m, and on this slab that is 666.7 kg against 745.6 kg, +11.8%.
300 mm is not 12 in. 12 in is 304.8 mm. That is 1.6%, small enough to ignore until the count lands on a rounding boundary, and then a whole bar appears or disappears.
The count only rounds one way: floor + 1. ⌊ clear span ÷ s ⌋ + 1. Round up instead, or round to nearest, and this slab goes from 40 + 30 bars to 41 + 31, 750 m to 762 m, 1.6% heavy.
Unit weight has two spellings. D²/162 and 0.006165 D². At 12 mm they give 0.8889 and 0.8878 kg/m, 0.13% apart. This section uses D²/162 throughout and names the other in the table notes.
The four lines this section is built on
Bars each way n = ⌊ (clear span mm ÷ spacing mm) ⌋ + 1 clear span = span − 2 × cover
Total length L = n_long × long side + n_short × short side
Unit weight w = D² / 162 (kg/m, D in mm)
Total weight W = L × w · per area = W ÷ slab area
Estimate, comparison only: 2 × (1000 ÷ s) × w
The 9 × 12 m slab, worked by hand (this is the answer above)
Clear spans. Long side 12000 − 80 = 11920 mm. Short side 9000 − 80 = 8920 mm. Two covers of 40 mm come off each direction, not one.
Along the 12 m direction. ⌊ 11920 ÷ 300 ⌋ + 1 = 39 + 1 = 40 bars, each 12 m long → 480.0 m.
Along the 9 m direction. ⌊ 8920 ÷ 300 ⌋ + 1 = 29 + 1 = 30 bars, each 9 m long → 270.0 m.
Total. 480.0 + 270.0 = 750.0 m. 12 mm unit weight is 144 ÷ 162 = 0.8889 kg/m. 750.0 × 0.8889 = 666.7 kg, and 666.7 ÷ 108 = 6.17 kg/m².
Three numbers for one slab, three names
Every column below is a real result on this slab. Only the first one can be ordered from.
What this column is called
How it is built
Result
Can it be ordered from
Integer bars, added up (the answer)
40 + 30 bars, 750.0 m × 0.8889
666.7 kg / 6.17 kg/m²
Yes. Quote the quantity and the weight off this one.
Area method estimate
2 × (1000 ÷ 300) × 0.8889 = 5.93 kg/m² × 108
640.0 kg (−4.0%)
No. For a sense of scale only.
Bought length
40 bars of 12 m plus 30 bars of 9 m cut from 12 m stock = 70 bars = 840 m
746.7 kg (+12.0%)
Not as a quantity. It decides how many bars you buy.
745.6 and 746.7 are two different things, 0.15% apart, and they are the pair most likely to end up under one name. 745.6 is this same layout rebarred as #4 at 12 in — a diameter change. 746.7 is 12 mm bar but bought as 12 m stock — a stock length change. Nothing connects them. Anyone who checks one against the other will never get them to agree, and will be right both times.
Five ways to reinforce the same slab
Bar & spacing
Bars (12 m way + 9 m way)
Total m
Total kg
kg/m²
10 mm @ 200
60 + 45
1125.0
694.4
6.43
10 mm @ 150
80 + 60
1500.0
925.9
8.57
12 mm @ 300 (the answer)
40 + 30
750.0
666.7
6.17
12 mm @ 200
60 + 45
1125.0
1000.0
9.26
16 mm @ 200
60 + 45
1125.0
1777.8
16.46
Every row uses the same clear spans, 11920 and 8920 mm, and the same 40 mm cover. The count moves with the spacing alone; the weight moves with the count and the diameter together. 12 mm at 200 and 10 mm at 200 both give 60 + 45 bars and both take 1125.0 m — the 44% difference between 1000.0 and 694.4 kg is all diameter.
The same slab in imperial bar
Bar & spacing
Bars
Total m
Total kg
kg/m²
#3 @ 12 in
40 + 30
750.0
419.7
3.89
#4 @ 16 in
30 + 22
558.0
554.7
5.14
#4 @ 12 in
40 + 30
750.0
745.6
6.90
12 mm at 300 and #4 at 12 in come out with the same 40 + 30 bars and the same 750.0 m, and 11.8% apart in weight. All of that difference is in the diameter, none of it in the layout. So when a drawing says "12", the first question is millimetres or bar number.
D²/162 gives 0.8889 kg/m, 0.006165 D² gives 0.8878 kg/m, 0.13% apart, which is 0.8 kg over this 750 m. Both formulas are correct. Mixing them inside one page is what makes someone else's re-check fail, so a section uses one and says in the note which one.
10 mm at 200 and 12 mm at 300 are only 4.2% apart in kg/m² (6.43 against 6.17), but the first is 105 bars and the second is 70. Same weight of steel, half the tying labour. That column never appears on a bar schedule, and on small-diameter metric mats it is the one that decides the crew's day.
4 / 5 / 6 / 8 inch against 100 / 125 / 150 / 200 mm
Thickness converted to metric is not taken to the exact millimetre on site. It is rounded to the nearest 25 mm, and the rounding always goes down, so every row is short by the same 1.57%:
Imperial thickness
Exact mm
As written in metric
Difference
4 in
101.6
100 mm
−1.57%
5 in
127.0
125 mm
−1.57%
6 in
152.4
150 mm
−1.57%
8 in
203.2
200 mm
−1.57%
Spacing behaves the same way, and it is the spacing that sets the bar count:
Imperial spacing
Exact mm
As written in metric
Difference
12 in
304.8
300 mm
−1.57%
16 in
406.4
400 mm
−1.57%
18 in
457.2
450 mm
−1.57%
24 in
609.6
600 mm
−1.57%
Thickness down 1.57% and spacing down 1.57% is steel down 3.13% (0.98425² = 0.96875). This is not scatter, it is a systematic bias from rounding in one direction every time. A 4 in slab and a 100 mm slab are not the same slab: 3.1% of the steel separates them, and no drawing ever writes that 3.1% down.
One 10 × 10 ft, 4 in slab at 3 in cover with #4 at 12 in is 10 + 10 bars, 200.0 ft, 133.6 lb. The same slab as 3.048 × 3.048 m with 12 mm at 300 and 76 mm cover is also 10 + 10 bars, but 60.96 m, 54.2 kg (= 119.5 lb). Same bar count, 10.6% apart in weight — the #4 against 12 mm gap again, this time arriving through the 4 inch to 100 mm conversion.
Three ways this section goes wrong, with the numbers
1. #4 counted as 12 mm. 745.6 kg written down as 666.7 kg is 10.6% short. The other way round, 12 mm counted as #4, is 11.8% over. One name, two diameters, and the direction of the error depends on which way the reader guessed.
2. 12 m stock cut for 9 m bars. The 40 bars of the long direction come off 12 m stock one for one, no waste. The 30 bars of the short direction each leave 3 m behind, 90 m of offcut in total. Bought length is 70 × 12 = 840 m against 750 m used, +12.0%. Whether that offcut is usable depends entirely on whether the job has a short span to put it in — and that is a site question, not a calculator question.
3. Area method taken as the answer. 2 × (1000 ÷ 300) × 0.8889 × 108 = 640.0 kg, 4.0% under the integer count. The gap is stable on a big slab and ragged on a small one, which is precisely why it can size a budget and cannot place an order.
The diameter table behind the 0.8889, including the metric sizes that have no imperial equivalent — see rebar size chart.
Stairs
A 9 ft floor-to-floor flight, 15 risers at 7.2 in, 11 in treads, 42 in wide, 6 in waist — takes 4 #4 bars down the slope and 16 #3 across, 113.70 ft = 60.77 lb of steel in the concrete. You buy it as 7 sticks: 4 × 20 ft #4 and 3 × 20 ft #3, 76.00 lb — 25.1% more than ends up poured. That gap is the order, not an error.
Three things stand next to that answer, and a stair drawing usually shows none of them:
The steps carry no bars. All the steel runs in the waist slab, under the risers. Riser faces and treads are shapes sitting on the slab, not members.
The waist thickness is perpendicular to the slope. The 6 in on the drawing reads 7.33 in with a plumb rule — 22.1% apart. Same slab, same steel, two rulers.
Both spacings are centre to centre. 12 in across the width for the main bars, 12 in down the slope for the distribution. Clear spacing adds a bar.
The flight geometry: riser count sets the slope
Total rise sets the riser count, and the count lands on a whole number: 108 ÷ 7 = 15.43, and 0.43 of a riser does not exist. Take 15 risers at 7.2 in or 16 at 6.75 — the height per step moves, the count does not. One flight has n risers and n − 1 treads, so the run is (n − 1) × tread, and the slope length is √(run² + rise²). IRC checks — riser ≤ 7¾ in, tread ≥ 10 in — and every row below clears both.
The count and the lengths
Main bars (laid down the slope, spaced across the width) = ⌊ (width − 2×cover) / s ⌋ + 1
One main bar = slope length − 2×cover · both ends land on stair-beam faces
Distribution bars (laid across, spaced down the slope) = ⌊ main bar length / s ⌋ + 1
One distribution bar = width − 2×cover
Weight = count × bar length × unit weight (#4 0.668, #3 0.376 lb/ft)
No laps on a single flight: the longest main bar here is 16.18 ft, one 20 ft stick carries it whole. The ends hook into the stair beams — a 90° hook on a #4 is 6db = 3.0 in (2.5 in floor) — and a hook is bent, not spliced, so it never enters a lap account.
The answer flight, worked by hand
Slope. Run = 14 × 11 = 154 in, rise 108 in → slope = √(154² + 108²) = 188.1 in = 15.67 ft, at 35.0°. The 6 in waist reads 6 ÷ 0.8187 = 7.33 in plumb.
Main #4 @ 12 across. ⌊ 39 / 12 ⌋ + 1 = 4 bars, each 188.1 − 3 = 185.1 in = 15.43 ft → 4 × 15.43 = 61.70 ft × 0.668 = 41.22 lb.
Distribution #3 @ 12 down. ⌊ 185.1 / 12 ⌋ + 1 = 16 bars, each 39 in = 3.25 ft → 16 × 3.25 = 52.00 ft × 0.376 = 19.55 lb.
61.70 + 52.00 = 113.70 ft = 60.77 lb in the concrete.
Three numbers for the same flight
Name of the column
Where it comes from
Number
Safe to order from?
Whole-bar total (the answer)
4 #4 down the slope + 16 #3 across, 113.70 ft
60.77 lb
Yes — report quantity and weight from this one
Area-method estimate
—
—
No. A sloped strip has no flat area; a stair psf figure is a flat-slab habit borrowed where it does not exist
What you actually buy
4 #4 sticks (one bar each) + 16 #3 cuts off 3 sticks (6 cuts per 20 ft stick)
7 sticks = 140 ft = 76.00 lb (+25.1%)
Not usage. This column decides how many sticks leave the yard
The +25.1% is the saw's geometry, not waste to apologize for: each #4 stick leaves 4.58 ft, the three #3 sticks leave 8.0 ft together. Offcuts live or die by whatever short work the site has next.
Same stairs, three riser heights
Total rise 108 in, 42 in wide, 11 in treads, #4 @ 12 down the slope, #3 @ 12 across, 6 in waist, 1.5 in cover.
Risers
Riser, in
Slope, in
Angle
Plumb waist, in
#4 bars
#4 ft
#3 bars
#3 ft
In the concrete, lb
16
6.75
197.2
33.2°
7.17
4
64.73
17
55.25
64.02
15 (the answer)
7.20
188.1
35.0°
7.33
4
61.70
16
52.00
60.77
14
7.71
179.2
37.1°
7.52
4
58.73
15
48.75
57.56
Risers 16 → 14 and the counts barely move: #4 stays at 4, #3 drops two. The count belongs to the width and the slope, not to the step count — what moves is the length of every bar, and the steel only falls 10.1% (64.02 → 57.56 lb).
All three flights order the identical bundle: 4 × 20 ft #4 + 3 × 20 ft #3. The longest main bar (16.18 ft) fits one 20 ft stick at every riser count, so no flight ever laps. The stairs change, the order does not; only the offcuts move — 15.3 / 18.3 / 21.3 ft of #4.
Plumb waist reads 7.17 / 7.33 / 7.52 in as the flight steepens, while the perpendicular 6 in never moves. The drawing writes one, the site measures the other, and 22% lives in the gap.
Three quiet ways stairs go wrong
Plumb drawn as perpendicular. Write 6 in plumb and the true waist is 6 × 0.8187 = 4.91 in — 18% thin before the first bar is tied. The steel did not shrink; the slab did.
Barring the steps. A bar in every riser face turns one flight into 30 short bars off the load path. The four #4 in the waist are the flight's steel; the steps are the concrete's own work.
Borrowing the flat-slab psf. There is no flat area on a slope. A stair psf neither sets the count nor the order — on this page it does not even qualify as an estimate.
Slabs with an opening
A 20 × 16 ft slab in #4 at 12 in o.c. both ways, with a 4 × 4 ft opening in the middle, takes 663.00 ft = 442.88 lb of steel in the concrete — 21.38 lb more than the same slab with no hole. The opening cuts 28 ft of main steel out and forces 60 ft of trim steel back in. You buy it as 40 sticks: 800 ft, 534.40 lb — 20.7% more than ends up poured.
Three things stand next to that answer, and an opening detail usually shows none of them:
The bars the hole cuts are not scrapped. They are cut at the opening face and hooked where they stop — a 90° hook on a #4 is 6db = 3.0 in (2.5 in floor) — and a hook is bent, not spliced, so it never enters a lap account.
Every opening buys steel back. Four edge trimmers run along the faces (opening width + ld on each side) and two diagonal bars cross each corner at 45°, 2ld apiece.
One number sizes all of it: ld = 22 in for a #4 in 3,000 psi concrete (ACI 318-19 25.4.2.3, C = 25; 21.9 rounded up to the next whole inch). That is why a 2 ft hole already needs 68 in trim bars — 2.83× the hole it serves.
What the opening actually costs
Bars cut, each way = bars whose centre line falls inside the opening span
One cut segment = edge-to-face distance + the 90° hook (6db = 3.0 in for #4, 2.5 in floor)
Edge trimmer = opening width + 2 × ld · one bar along each of the four faces
Corner diagonal = 2 × ld, run at 45° · two bars per corner
Weight = total ft × unit weight (#4 = 0.668 lb/ft)
A trimmer that stops at the opening edge has nothing to anchor into — ld each side is the minimum, and the anchor length is charged to the hole, not to the slab.
The answer slab, worked by hand
x-way. 16 bars run the 19.75 ft length; the hole (72–120 in) catches 4 centrelines (73.5 / 85.5 / 97.5 / 109.5). Each cut bar becomes two 97.5 in segments (94.5 to the face + 3.0 hook) = 16.25 ft. 12 full × 19.75 = 237.00 ft + 4 × 16.25 = 65.00 ft → 302.00 ft.
y-way. 20 bars; 4 caught (97.5 / 109.5 / 121.5 / 133.5), two 73.5 in segments (70.5 + 3.0) = 12.25 ft each. 16 × 15.75 = 252.00 ft + 4 × 12.25 = 49.00 ft → 301.00 ft.
Trim steel. 4 edge trimmers × 92 in (48 + 2 × 22) = 30.67 ft, plus 8 corner diagonals × 44 in = 29.33 ft → 60.00 ft.
302.00 + 301.00 + 60.00 = 663.00 ft = 442.88 lb in the concrete. The solid slab is 631.00 ft = 421.51 lb: the hole removed 28.00 ft of main steel and the trim added 60.00 ft — net +32.00 ft, +21.38 lb.
Three numbers for the same slab
Name of the column
Where it comes from
Number
Safe to order from?
Whole-bar total (the answer)
302.00 x-way + 301.00 y-way + 60.00 trim = 663.00 ft
442.88 lb
Yes — report quantity and weight from this one
Area-method estimate
304 sq ft (hole subtracted) × 1.336 lb/sq ft
406.14 lb (−8.3%)
No. It sees the hole and never the trim steel — the bigger the hole, the further under it reads
What you actually buy
40 sticks: 28 full bars + 16 hooked segments off 8 sticks + trimmers and diagonals off 4
800 ft = 534.40 lb (+20.7%)
Not usage. This column decides how many sticks leave the yard
800 ft bought, 663.00 ft poured — 137 ft comes home as offcut.
Same slab, three holes
20 × 16 ft, #4 @ 12 both ways, single layer, openings centred, ld = 22 in (#4, 3,000 psi), 90° hooks on cut ends, trim per opening = 4 edge trimmers + 8 corner diagonals.
Opening
Bars cut, each way
Main steel, ft
Trim steel, ft
In the concrete, lb
Area method, lb
Sticks
Order, lb
None (solid reference)
—
631.00
—
421.51
427.52
—
—
2 × 2 ft
2 + 2
625.00
52.00
452.24
422.18
40
534.40
4 × 4 ft (the answer)
4 + 4
603.00
60.00
442.88
406.14
40
534.40
6 × 6 ft
6 + 6
565.00
68.00
422.84
379.42
38
507.68
Bigger hole, less steel in the slab: 452.24 → 422.84 lb (−6.5%) from the 2 ft hole to the 6 ft hole — yet every hole ends heavier than the solid 421.51: +30.73 / +21.38 / +1.34 lb. The 6 ft hole cuts 66 ft of main steel and puts 68 ft of trim back; 2 ft apart, still positive. An opening never pays for itself.
The order barely moves: 40 / 40 / 38 sticks. The hole changes what sits on the sticks — full bars become hooked segments plus trim steel — far more than it changes how many sticks leave the yard (534.40 → 507.68 lb).
The area method moves the wrong way: it falls 10.1% from the 2 ft hole to the 6 ft hole while the true steel falls 6.5%, and its gap widens from −6.6% to −10.3%. It tracks the hole; the trim steel it never counts.
Three quiet ways openings go wrong
“The hole saves steel.” A 4 × 4 ft opening removes 28 ft of main steel and forces 60 ft of trim back in — net +32.00 ft, +21.38 lb. What shrank was the area, never the steel.
“Trimmers stop at the opening edge.” A bar that ends at the edge has nothing to anchor into. ld = 22 in each side is the minimum, so even a 2 ft hole needs 68 in = 5.67 ft trimmers — 2.83× the hole they serve.
“The estimate shrinks with the hole.” The area method runs 422.18 → 379.42 lb (−10.1%) against a true 452.24 → 422.84 (−6.5%). The bigger the hole, the further under it reports — and the trim steel is missing from every row.
Where the 16-and-20 bar counts and the both-way grid behind this slab come from — the rebar grid calculator.
Cantilever slabs
A 10 ft wide canopy with a 4 ft cantilever, #4 top bars at 12 in o.c. plus #3 distribution at 12, takes 118.08 ft = 61.80 lb of steel in the concrete. Every main bar is 71.5 in long, and 22 in of each lies inside the back-span — past the support face, off the canopy footprint. You buy 7 sticks: 140 ft = 76.00 lb, 23.0% more than ends up poured.
Three things stand next to that answer, and a canopy detail usually shows none of them:
The tension is on top. A cantilever is a negative-moment member: the working steel is the top mat. The bottom bars of the back-span do not continue into the cantilever as tension steel — that is the definition of a cantilever, not an oversight.
22 in of every top bar lives in the span. ld = 22 in for a #4 in 3,000 psi concrete (ACI 318-19 25.4.2.3, C = 25; 21.9 rounded up to the next whole inch). It is charged to the cantilever but lies inside the back-span: on the answer canopy, 10 bars × 22 in = 18.33 ft = 12.25 lb sits entirely outside the canopy footprint.
The tip hook is 90°, not “a hook”. 6db = 3.0 in for a #4 (2.5 in floor for #3–#5). It is bent at the free edge, not spliced, so it never enters a lap account.
What a cantilever bar is made of
Main bar count = floor((width − 2 × cover) ÷ spacing) + 1 · spaced across the width
Main bar length = (cantilever − cover) + 90° hook + ld · ld measured from the support face into the back-span
Distribution zone = cantilever − cover + ld · count = floor(zone ÷ spacing) + 1
A top bar that stops at the support face has nothing developed past it — ld into the span is the minimum, and that length is part of the cantilever's steel even though it lies over the back-span.
The answer canopy, worked by hand
Main top bars, #4 @ 12. floor((120 − 3) ÷ 12) + 1 = 10 bars. Each runs 48 − 1.5 = 46.5 in to the tip, bends down 3.0 in, and reaches 22 in past the support face: 71.5 in = 5.96 ft. 10 × 5.96 ft = 59.58 ft = 39.80 lb.
Distribution bars, #3 @ 12. The zone runs 48 − 1.5 + 22 = 68.5 in from the free edge; floor(68.5 ÷ 12) + 1 = 6 bars, each 9.75 ft across the width. 6 × 9.75 = 58.50 ft = 22.00 lb. The last bar lands 13.5 in past the support face — inside ld, because the distribution steel follows the top mat into the span.
59.58 + 58.50 = 118.08 ft = 61.80 lb in the concrete.
Three numbers for the same canopy
Name of the column
Where it comes from
Number
Safe to order from?
Whole-bar total (the answer)
59.58 ft main + 58.50 ft distribution = 118.08 ft
61.80 lb
Yes — report quantity and weight from this one
Area-method estimate
40 sq ft canopy footprint × 1.336 lb/sq ft
53.44 lb (−13.5%)
No. It measures the footprint and never the 22 in that lies past the support face
What you actually buy
4 sticks of #4 (3 cuts each) + 3 sticks of #3 (2 cuts each)
140 ft = 76.00 lb (+23.0%)
Not usage. This column decides how many sticks leave the yard
140 ft bought, 118.08 ft poured — 21.92 ft comes home as offcut.
Same canopy, three cantilever lengths
10 ft wide, #4 @ 12 top bars + #3 @ 12 distribution, cover 1.5 in, ld = 22 in (#4, 3,000 psi), 90° tip hooks.
Cantilever
Main bar
Main total
Distribution bars
Distribution total
In the concrete, lb
Area method, lb
Sticks
Order, lb
3 ft
10 × 4.96 ft
49.58 ft = 33.12 lb
5
48.75 ft
51.45
40.08 (−22.1%)
6
62.64 (+21.7%)
4 ft (the answer)
10 × 5.96 ft
59.58 ft = 39.80 lb
6
58.50 ft
61.80
53.44 (−13.5%)
7
76.00 (+23.0%)
5 ft
10 × 6.96 ft
69.58 ft = 46.48 lb
7
68.25 ft
72.14
66.80 (−7.4%)
9
96.88 (+34.3%)
The count never moves: 10 main bars for any length — the width sets the count, the cantilever sets the length. Each bar grows exactly 1.00 ft per extra foot of cantilever, so steel goes 51.45 → 72.14 lb (+40.2%) while the cantilever grows 66.7%.
The area method's gap widens as the cantilever shrinks: −7.4% at 5 ft, −22.1% at 3 ft. The shorter the cantilever, the bigger a share the 22 in extension is of every bar — 26.3% / 30.8% / 37.0% across the three rows — and the footprint never sees any of it.
The order jumps 6 → 7 → 9 sticks. At 5 ft each main bar is 6.96 ft and a 20 ft stick yields only 2 cuts, so the #4 column jumps from 4 sticks to 5 — saw geometry, not a 34.3% leap in steel.
Three quiet ways cantilevers go wrong
“Continue the bottom bars through.” Over a cantilever the bottom steel is not tension steel. Splicing back-span bottom bars into the canopy pays lap for a bar that carries nothing there — the bar that must reach into the span is the top one, by ld = 22 in.
“The top bar can stop at the support face.” Without the 22 in extension the #4 has no development past the face. On the answer canopy that is 12.25 lb of steel hanging on one detailing line — cutting it removes the cantilever's own anchorage.
“Quote the canopy by psf.” 40 sq ft × 1.336 = 53.44 lb against a true 61.80 lb (−13.5%) — and −22.1% at 3 ft. The psf rate only sees the footprint; half of every top bar's anchorage stands outside it.
A 3-pile cap, 5 × 5 ft plan, #4 bottom bars at 12 in o.c. both ways plus a #3 top mat at 12, takes 97.50 ft = 51.26 lb of steel in the concrete. Every bottom bar is 5.00 ft long with a hook bent at each end. You buy 6 sticks: 120 ft = 62.64 lb, 22.2% more than ends up poured. The slab psf rate reports 33.40 lb — 34.8% under.
The sizes here are a worked example, stated as assumptions: 12-in piles at 36-in centers (3D), the cap overhanging each pile edge by 6 in, a 36-in-thick cap, #4 @ 12 bottom mat, #3 @ 12 top mat. A real cap's thickness is set by punching shear on each pile head, which this section does not compute — what follows is countable arithmetic, not a construction prescription.
Three things stand next to that answer, and a pile-cap detail usually shows all three:
The piles set where the bars go. Bottom bars bear directly on the pile heads, so the grid is shifted onto the pile center lines. On the 2-pile cap a straight 12-in grid lands at 3 / 15 / 27 / 39 / 51 in from the edge while the pile centers sit at 12 and 48 — not one bar on a head. Shift the grid, keep the count: still 5 bars.
The pile steel anchors above the head. ld for a #5 in 3,000 psi concrete (ACI 318-19 25.4.2.3, C = 25) is 27.39 rounded up to 28 in of projection above each pile top. A 36-in cap leaves 31.5 in after both covers — OK by 3.5 in. A 30-in cap leaves 25.5 in and comes up 2.5 in short. That #5 belongs to the pile order, not the cap mat.
Every edge gets a hook. Bottom bars carry a 90° hook at both ends — 6db = 3.0 in for a #4 (2.5 in floor) — and both directions are hooked, so all four cap edges get bent ends: 10 bars × 2 × 3.0 in = 5.00 ft = 3.34 lb of pure bends on the answer cap.
What a pile-cap bar is made of
Cap plan = pile centers + pile diameter + 2 × overhang · 2-pile: 36 + 12 + 12 = 60 in × 24 in
3-pile triangle (side 36) bounds 60 × 55.18 in · rounded UP to a 60 × 60 in square cap
Top bar count = same formula at 1.5 in cover · length = dim − 2 × cover, no hooks
Anchorage ld = 60000 ÷ (C × √f'c) × db · #5, 3,000 psi, C = 25 → 28 in above each pile head
Weight = count × length × unit weight (#4 = 0.668, #3 = 0.376 lb/ft), summed before rounding.
The answer cap, worked by hand
Bottom mat, #4 @ 12 both ways. floor((60 − 6) ÷ 12) + 1 = 5 bars each way. Each runs 54 in plus two 3.0-in hooks: 60 in = 5.00 ft. 10 × 5.00 ft = 50.00 ft = 33.40 lb.
Top mat, #3 @ 12 both ways. floor((60 − 3) ÷ 12) + 1 = 5 bars each way, each 57 in = 4.75 ft. 10 × 4.75 = 47.50 ft = 17.86 lb.
50.00 + 47.50 = 97.50 ft = 51.26 lb in the concrete.
Three numbers for the same cap
Name of the column
Where it comes from
Number
Safe to order from?
Whole-bar total (the answer)
50.00 ft bottom + 47.50 ft top = 97.50 ft
51.26 lb
Yes — report quantity and weight from this one
Area-method estimate
25 sq ft plan × 1.336 lb/sq ft — a slab rate that sees one mat, not the cap's two
33.40 lb (−34.8%)
No. It never sees the second mat, the hooks, or the piles
What you actually buy
3 sticks of #4 + 3 sticks of #3 (one cut length per family)
120 ft = 62.64 lb (+22.2%)
Not usage. This column decides how many sticks leave the yard
120 ft bought, 97.50 ft poured — 22.50 ft comes home as offcut.
Two, three, and four piles under the same rules
12-in piles at 36-in centers, 6-in overhang, 36-in-thick cap, #4 @ 12 bottom (hooked both ends) + #3 @ 12 top, cover 3.0 / 1.5 in.
Piles
Plan
Bottom mat
Top mat
In the concrete, lb
Area method, lb
Sticks
Order, lb
2
5 × 2 ft
29.00 ft = 19.37 lb
27.25 ft = 10.25 lb
29.62
13.36 (−54.9%)
6
62.64 (+111.5%)
3 (the answer)
5 × 5 ft
50.00 ft = 33.40 lb
47.50 ft = 17.86 lb
51.26
33.40 (−34.8%)
6
62.64 (+22.2%)
4
5 × 5 ft
50.00 ft = 33.40 lb
47.50 ft = 17.86 lb
51.26
33.40 (−34.8%)
6
62.64 (+22.2%)
The 3- and 4-pile rows are identical steel: the same 36-in pile spacing produces the same 60 × 60 plan, and the pile count changes the layout inside (triangle vs 2 × 2) and the number of heads to anchor — 3 × 28 in vs 4 × 28 in of #5 projection — not the mat math.
The 2-pile order nearly doubles (+111.5%): a 5 × 2 ft footprint still consumes six 20-ft sticks when each family keeps one cut length. Small plans pay for stock length, not for steel.
Real 3-pile caps are often triangular or chamfered in plan; this section keeps rectangles so every bar length stays countable. Chamfers only shorten the corner bars.
Three quiet ways pile caps go wrong
“The psf rate works for caps too.” 25 sq ft × 1.336 = 33.40 lb against a true 51.26 lb (−34.8%) — and −54.9% on the 2-pile cap. The slab rate is calibrated to one mat; a cap stacks two over a footprint the psf method treats as the whole story, and neither mat knows where the piles are.
“30 inches of cap is deep enough.” 30 − 3 − 1.5 = 25.5 in of room above the pile head against a 28-in ld for the #5 — 2.5 in short, inside a cap that looks plenty deep. Depth is measured from the pile top to the top mat, and 36 in is what clears it with 3.5 to spare.
“12-in spacing, done.” A straight grid lands at 3 / 15 / 27 / 39 / 51 in while the pile centers sit at 12 and 48 — no bar over a head. In a cap the bottom bars bear on the piles, so the grid shifts onto the center lines; the count formula never moves (still 5), the detail does.