The area of one bar, the area you need, and the number of bars that gets you there
← Back to the calculatorThere are two different numbers on every drawing and they are easy to swap. Area of one bar is a property of the bar and never changes: #4 is 0.20 in² today and 0.20 in² in ten years. Area required is a property of the member and comes from the design.
The arithmetic that joins them is one division, and the part that actually goes wrong is not the division: it is whether the bars you counted fit in the width you have, at the clear spacing the code allows. That is the second half of this page.
This is the whole table. Everything else on the page is arithmetic on these six numbers per row.
| Bar | Metric | Dia (in) | Area (in²) | Area (mm²) | Weight (lb/ft) |
|---|---|---|---|---|---|
| #3 | #10 | 0.375 | 0.11 | 71 | 0.376 |
| #4 | #13 | 0.500 | 0.20 | 129 | 0.668 |
| #5 | #16 | 0.625 | 0.31 | 200 | 1.043 |
| #6 | #19 | 0.750 | 0.44 | 284 | 1.502 |
| #7 | #22 | 0.875 | 0.60 | 387 | 2.044 |
| #8 | #25 | 1.000 | 0.79 | 509 | 2.670 |
| #9 | #29 | 1.128 | 1.00 | 645 | 3.400 |
| #10 | #32 | 1.270 | 1.27 | 819 | 4.303 |
| #11 | #36 | 1.410 | 1.56 | 1006 | 5.313 |
| #14 | #43 | 1.693 | 2.25 | 1452 | 7.650 |
| #18 | #57 | 2.257 | 4.00 | 2581 | 13.600 |
Two things to notice. The size number is eighths of an inch of diameter, not area: #4 is 4/8 = 0.500 in across. And area grows with the square of the diameter, so the step from #4 to #8 is 4× the steel while the diameter only doubles.
The row in blue is #4 because that is the bar most residential and light commercial work uses, and because it is the one people quote from memory and get wrong. It is 0.20 in², not 0.25.
One division, rounded up, then a check that the answer is physically placeable.
N = As,required ÷ Ab, rounded up
Where As,required is the steel area the design calls for, in square inches, and Ab is the area of one bar from the table above. A 24 in wide footing strip that needs 1.24 in² takes 1.24 ÷ 0.31 = 4.0 #5 bars, so four. If the division gives 4.03 you take five, because bars come in whole pieces.
Where the required area comes from, if you do not have a design in front of you:
As,min = 0.0018 × Ag for Grade 60 deformed bars, ACI 318-19 24.4.3.2. A 5 in slab is 60 in² gross per foot of width, so 0.108 in² per foot.f'c is guessing.Symbols and units, once, so the rest of the page reads cleanly:
| Symbol | Meaning | Unit | Where it comes from |
|---|---|---|---|
| As,req | Steel area required | in² | Design, or 24.4.3.2 for slabs |
| Ab | Area of one bar | in² | Table above, ASTM A615 |
| N | Number of bars | count | Calculated |
| db | Nominal bar diameter | in | Table above |
| b | Member width | in | Drawing |
| c | Clear cover | in | ACI 318-19 Table 20.5.1.3.1 |
| s | Minimum clear spacing | in | ACI 318-19 25.2.1 |
| dagg | Nominal maximum aggregate | in | Mix design |
The count is easy. The count that fits is the one that matters, and it comes from the clear spacing in ACI 318-19 25.2.1: the greater of one bar diameter, 1 in, and 4/3 of the nominal maximum aggregate size.
s = max(db, 1 in, 4/3 × dagg) N per layer = (b − 2c + s) ÷ (db + s), rounded down
The + s in both places is what makes it a count rather than a length: n bars have n−1 gaps between them, so you add one gap, divide, and drop the remainder. Then:
layers = N ÷ N per layer, rounded up As,provided = N × Ab
Minimum clear spacing by bar and by mix, since the aggregate term is the one people forget:
| Bar | db (in) | s, 3/4 in agg | s, 1 in agg | s, 1½ in agg |
|---|---|---|---|---|
| #3 | 0.375 | 1.00 | 1.33 | 2.00 |
| #4 | 0.500 | 1.00 | 1.33 | 2.00 |
| #5 | 0.625 | 1.00 | 1.33 | 2.00 |
| #6 | 0.750 | 1.00 | 1.33 | 2.00 |
| #7 | 0.875 | 1.00 | 1.33 | 2.00 |
| #8 | 1.000 | 1.00 | 1.33 | 2.00 |
| #9 | 1.128 | 1.13 | 1.33 | 2.00 |
| #11 | 1.410 | 1.41 | 1.41 | 2.00 |
Below #9 the aggregate governs at 1 in and above, and above #9 the bar diameter takes over. That is why a 1 in aggregate mix costs you bar capacity in a narrow member: the gaps grow before the bars do.
For the times you just want the count. Read across the row for the area you need, and the bracket is the area you actually get.
| As needed (in²) | #3 | #4 | #5 | #6 | #7 | #8 |
|---|---|---|---|---|---|---|
| 0.50 | 5 (0.55) | 3 (0.60) | 2 (0.62) | 2 (0.88) | 1 (0.60) | 1 (0.79) |
| 0.75 | 7 (0.77) | 4 (0.80) | 3 (0.93) | 2 (0.88) | 2 (1.20) | 1 (0.79) |
| 1.00 | 10 (1.10) | 5 (1.00) | 4 (1.24) | 3 (1.32) | 2 (1.20) | 2 (1.58) |
| 1.50 | 14 (1.54) | 8 (1.60) | 5 (1.55) | 4 (1.76) | 3 (1.80) | 2 (1.58) |
| 2.00 | 19 (2.09) | 10 (2.00) | 7 (2.17) | 5 (2.20) | 4 (2.40) | 3 (2.37) |
| 3.00 | 28 (3.08) | 15 (3.00) | 10 (3.10) | 7 (3.08) | 5 (3.00) | 4 (3.16) |
| 4.00 | 37 (4.07) | 20 (4.00) | 13 (4.03) | 10 (4.40) | 7 (4.20) | 6 (4.74) |
Everything above, in one place. The count, the fit, the layers and the weight per foot all move together.
Try #5. N = 1.24 ÷ 0.31 = 4.0, so four bars, providing 1.24 in². Clear spacing with a 1 in aggregate mix is max(0.625, 1, 1.33) = 1.33 in. One layer holds (24 − 6 + 1.33) ÷ (0.625 + 1.33) = 19.33 ÷ 1.955 = 9.9, so nine bars. Four fit with five places to spare, and the answer is four #5.
Now try #4 for the same area: N = 1.24 ÷ 0.20 = 6.2, so seven bars, providing 1.40 in². One layer holds (24 − 6 + 1.33) ÷ (0.5 + 1.33) = 19.33 ÷ 1.83 = 10.5, so ten. Seven #4 fit as easily as four #5, weigh 7 × 0.668 = 4.68 lb/ft against 4.17 lb/ft, and develop over 14.2 in against 17.8 in. That trade — more bars, less development, slightly more steel — is the whole reason bar size is a choice rather than a calculation.
#6 gives 3.00 ÷ 0.44 = 6.8, so seven bars at 3.08 in². One layer holds (12 − 3 + 1.33) ÷ (0.75 + 1.33) = 10.33 ÷ 2.08 = 4.9, so four. Seven bars do not fit in one layer: the answer is two layers, four and three, and the stem wall now needs its layers spaced per ACI 318-19 25.2.2 and its effective depth recalculated. Seven #6 in a 12 in wall is the classic "the area checks, the detail does not" failure.
As,required has to come from a moment, a shear or a minimum, and the only minimum this page gives is the slab shrinkage case.0.20 in², on a 0.500 in nominal diameter. The series runs #3 0.11, #4 0.20, #5 0.31, #6 0.44, #7 0.60, #8 0.79, #9 1.00, #10 1.27 and #11 1.56 in². These are nominal areas from ASTM A615/A615M, not measured areas, and the difference between nominal and actual is a mill tolerance, not something you design around.
Five, exactly: 5 × 0.20 = 1.00 in². Whether five fit is a separate question. In a 12 in wide member at 2 in of cover the clear width is 8 in, and five #4 bars plus four gaps at the 1 in minimum need 6.5 in, so they fit with 1.5 in left over. In an 8 in wide member the clear width is 4 in, and five bars would need 6.5 in, so the count goes up a layer or the bar size goes up.
No, and the number is not an area. A #4 bar is 4/8 in in diameter, so the number is eighths of an inch of diameter. Its area is 0.20 in². A #8 bar is 1.000 in across and 0.79 in² of steel. Doubling the size number roughly quadruples the area, which is why the step from #4 to #8 is 4 times the steel and not twice.
Multiply by 645.16. A #5 bar at 0.31 in² is 200 mm², and that 200 is the number printed on a metric drawing for a 16 mm bar. The conversion is exact, but the metric bar is not: a 16 mm bar has 201 mm², a soft-metric #5 has 200, and the 1 mm² difference is inside the mill tolerance.
For shrinkage and temperature in a slab, ACI 318-19 24.4.3.2 sets 0.0018 of the gross section for Grade 60 deformed bars. A 5 in slab is 60 in² per foot of width, so the minimum is 0.108 in² per foot, which #4 at 18 in satisfies at 0.133. That is a floor on the area, not a design: the slab still has to be checked for the moment it actually carries.
Because development and spacing both work per bar. ACI 318-19 25.4.2.3 develops each bar separately, so eight #4 bars carry the same area as four #6 bars but need twice the development length total and twice the clear gaps. Crowding small bars to save depth is how a section that checks on paper fails to place in the field.
No. Nominal area is the area of the plain round bar of the same nominal diameter, and the deformations add a few per cent of surface but are not counted. Bond relies on them, section capacity does not. This is also why a #4 bar weighs 0.668 lb/ft and not the 0.667 lb/ft that a plain 0.5 in round would give: the mill weight includes the deformations.