How many ties, how much wire, and the one number nobody publishes: pounds per ton
← Back to the calculatorTwo numbers decide the wire: how many intersections there are per ton, and how much wire one tie eats. The first comes from the grid and the bar size, the second from the bar diameter and how long you leave the twist tail. Everything on this page is those two, multiplied.
The part that goes wrong on site is not the arithmetic, it is that the tie count is treated as a habit. Tying every intersection on a #4 mat doubles the wire to 13 lb per ton and buys nothing that CRSI asks for.
Sixteen gauge black annealed, ties at every other intersection (the checkerboard that most flat work uses), wire cut at eight bar diameters plus a 6 in tail. Pounds of wire per ton of rebar:
| Bar | 12 in grid | 18 in grid | 24 in grid |
|---|---|---|---|
| #3 | 10.4 | 6.9 | 5.2 |
| #4 | 6.5 | 4.3 | 3.3 |
| #5 | 4.6 | 3.1 | 2.3 |
| #6 | 3.5 | 2.3 | 1.7 |
| #7 | 2.8 | 1.8 | 1.4 |
| #8 | 2.3 | 1.5 | 1.1 |
| #9 | 1.9 | 1.3 | 1.0 |
| #11 | 1.4 | 0.9 | 0.7 |
Read the top left and bottom right corners together: 10.4 lb per ton for #3 at 12 in, 0.7 lb per ton for #11 at 24 in. That is a 15 to 1 spread inside one material, which is why "a coil per ten tons" is either generous or short depending on the job.
References. Wire gauge and the tie frequency: CRSI Manual of Standard Practice, which names 16 gauge black annealed and states that a tie at every fourth or fifth intersection is sufficient. Wire properties: ASTM A1064/A1064M. Clear spacing that the ties must not violate: ACI 318-19 25.2.1. There is no ACI 318 clause for tie wire quantity, because tie wire is a placing aid and not reinforcement.
The frequency is a floor, not a rule for every point. What is tied more densely than the floor:
| Location | Tie density | Why |
|---|---|---|
| Interior of a flat mat | Every 4th or 5th intersection | CRSI minimum; the mat cannot shift once the bars interlock |
| First and last row of a mat | Every intersection | Free ends lift and kick out under a screed |
| Perimeter bars | Every intersection | First thing a boot or a chute hits |
| Lap splices | Both ends of every lap | A lap that opens loses its length, and length is the whole splice |
| At chairs and dobies | Every support point | The tie is what holds the bar down onto the support |
| Wall and column cages | Every intersection, or as drawn | Vertical bars slide down horizontals without it |
| Beam stirrups to main bars | Every intersection | Stirrups float up during vibration |
Those extra points are why a real job always beats the table. The table counts interior intersections only; add roughly 10 to 15 per cent for perimeter, laps and supports on a mat, more on a wall cage where every intersection is tied.
Six steps, each one a quantity you could stop at. Symbols and units first, because the whole chain is one unit conversion deep.
| Symbol | Meaning | Unit | Source |
|---|---|---|---|
| w | Weight of the bar | lb/ft | ASTM A615, 0.668 for #4 |
| s | Bar spacing, both ways | in | Drawing |
| w_sf | Steel weight per area | lb/ft² | Step 1 |
| I | Intersections per area | count/ft² | Step 2 |
| k | Tie density factor | — | 1.0, 0.5 or 0.25 |
| n | Ties per ton | count/ton | Step 4 |
| L_tie | Wire per tie | in | Step 5 |
| ρ | Wire weight per length | lb/in | 0.00087 for 16 ga |
1. w_sf = 2 × (12 ÷ s) × w 2. I = (12 ÷ s)² 3. T = I × k 4. n = T ÷ (w_sf ÷ 2000) 5. L_tie = 8 × db + 6 6. W = n × L_tie × ρ
Step 1 is two directions of bar in one square foot. Step 2 is the same geometry turned 90 degrees: spacing in feet is s ÷ 12, so intersections per square foot is the inverse squared. Step 4 converts ties per square foot into ties per ton by dividing by the tonnage that same square foot holds — which is why heavier bar always means less wire per ton.
The constants, once:
db each, plus a 6 in tail to twist and bend down. For #4 that is 10 in, for #8 it is 14 in.n grows seven times faster than the numerator.Area 1,200 ft². Step 1: w_sf = 2 × 1 × 0.668 = 1.336 lb/ft². Step 2: I = 1² = 1 intersection per square foot, so 1,200 intersections. Step 3: T = 1,200 × 0.5 = 600 ties. Step 4: total steel is 1,200 × 1.336 = 1,603 lb = 0.80 ton, so n = 600 ÷ 0.80 = 748 ties per ton. Step 5: L_tie = 8 × 0.5 + 6 = 10 in. Step 6: W = 748 × 10 × 0.00087 = 6.5 lb per ton, and 0.80 ton needs 5.2 lb of wire — one small coil, not a 50 lb one.
Step 1: w_sf = 2 × 0.667 × 1.043 = 1.391 lb/ft². Step 2: I = 0.667² = 0.444. Step 3: T = 0.222 per square foot. Step 4: n = 0.222 ÷ (1.391 ÷ 2000) = 319 ties per ton. Step 5: L_tie = 8 × 0.625 + 6 = 11 in. Step 6: W = 319 × 11 × 0.00087 = 3.1 lb per ton, so 20 tons needs 61 lb — two 50 lb coils. The habit of one coil per ten tons would have sent four.
Between about 1 and 10 lb per ton, and the spread is bar size and grid, not luck. A #3 mat at 12 in on centre tied every other intersection takes 10.4 lb per ton; a #8 mat at 24 in takes 1.1 lb. The number people quote, around 10 lb per ton, is the worst corner of that table. For a typical #4 or #5 mat at 12 to 18 in it is 4 to 7 lb per ton, which means one 50 lb coil ties 7 to 12 tons.
About 115 ties at 10 in of wire each. 16 gauge is 0.0625 in across, which is 0.00307 in² of section and 0.00087 lb per inch, so a pound is 1,149 in of wire. A 50 lb coil is roughly 5,700 ties, and a 100 lb coil roughly 11,500.
No. CRSI's Manual of Standard Practice states that a tie at every fourth or fifth intersection is sufficient, and that any tying beyond normal practice has to be specified. What is normally tied more densely is the perimeter of a mat, the first and last row, lap splices at both ends, and every point where the steel meets a chair or a dobie.
Because ties are counted at intersections and tonnage is counted in weight. A #8 mat at 12 in has the same number of intersections per square foot as a #4 mat, but each square foot holds 2.67 lb of bar instead of 0.67 lb, so one ton of #8 covers four times the area, and the same area needs four times fewer ties per ton. The wire per tie grows a little with bar size, nowhere near enough to close that gap.
16 gauge black annealed is the default and is what CRSI's Manual of Standard Practice names unless the engineer says otherwise. 16.5 gauge is softer and twists faster on high-volume work, 15 gauge is for #8 and larger, and 14 gauge for very heavy cages. Galvanised and stainless are for exposed or corrosive work and cost several times as much.
At #5 in a 12 in grid tied every other intersection, 4.6 lb per ton, so 20 tons is 92 lb: two 50 lb coils with 8 lb left. The same 20 tons of #3 at 12 in is 208 lb, or five coils, which is the reason the estimate is worth doing rather than rounding to one coil per ten tons.
No. Ties hold bars in position until the concrete is placed and set; they carry nothing after that. A bar that is tied but at the wrong cover is still at the wrong cover, and a mat that is tied at every intersection but sitting on the ground has no bottom cover at all. Tying is a placing activity, not a structural one.