Why the cut length is not the perimeter
Take a beam 16 inches wide and 24 deep, with an inch and a half of cover to the stirrup. The stirrup is 13 by 21 inches out to out. Measured to the bar centreline, which is where length is developed, it is one bar diameter smaller in each direction: 12.625 by 20.625 for a number 3 stirrup. Multiply that rectangle out and the sharp-cornered perimeter is 66.5 inches.
But the bar does not turn a sharp corner. It goes round a pin, and around that pin it follows an arc that is shorter than the two straight tangents it replaces. For a 90 degree bend at a centreline radius R, the two tangents are 2R and the arc is 1.5708R, so each corner saves 0.4292R of bar. Four corners on a stirrup, and then the hooks add their own tails back and take their own deduction. The result is a cut length that is a definite number and is never the number you measured off the drawing.
None of that is a code question. It is the same tangent-against-arc geometry that governs a sheet metal bend allowance, applied to a round bar instead of a flat sheet. What is a code and schedule question is the bend diameter, the hook length and the hook angle, and all three are fields on this page for that reason.
One limit of the method is worth stating. The deduction is the difference between two tangent lengths and an arc, and as the bend angle approaches 180 degrees the tangent length grows without limit, so the arithmetic stops describing the shape. That is why this page offers 90 and 135 degree hooks and not 180. A bar turned right back on itself is a semicircle plus two straight legs, and the developed length for one belongs on the fabricator schedule rather than in this deduction.
Deriving the weight instead of looking it up
Rebar weight per foot is usually copied off a table, and the table is right, but it is worth knowing where it comes from because it makes the whole cut list checkable. A number 4 bar is half an inch nominal diameter, so its cross-sectional area is pi times 0.25 squared, which is 0.19635 square inches. Twelve inches of that is 2.3562 cubic inches. Steel runs about 0.2836 pounds per cubic inch. Multiply and you get 0.6682 pounds per foot, which is the 0.668 in every table.
| Bar | Nominal diameter | Area (sq in) | Derived lb/ft |
|---|---|---|---|
| #3 | 0.375 in | 0.1104 | 0.376 |
| #4 | 0.500 in | 0.1963 | 0.668 |
| #5 | 0.625 in | 0.3068 | 1.044 |
| #6 | 0.750 in | 0.4418 | 1.503 |
| #8 | 1.000 in | 0.7854 | 2.673 |
The bar number is the diameter in eighths of an inch, which is the piece of rebar trivia that actually gets used on site. This calculator computes the weight from the diameter every time rather than reading a stored table, so the number in the results and the number in this paragraph come from the same arithmetic.
Stirrups per member, and the plus one again
Stirrups at 12 inch centres over a 24 foot beam with 3 inches off each end have 282 inches to fill, which is 23 full spaces and 24 stirrups. The last one closes the final space. This is the same plus-one that catches people on a slab mat and on a stud layout, and being one stirrup short at each end of a run of beams is a trip back to the yard.
Where the spacing tightens near supports — and it usually does — run the calculator once for each spacing zone and add the results. A single spacing across a whole beam is a simplification that this page makes because you told it to, not one it believes.
Laps on the long bars
A 24 foot member with 3 inches off each end wants bars 23 feet 6 inches long, and the yard sells 20 foot sticks. That is not one and a bit sticks: the two pieces have to overlap by the lap length, which is dead steel doing no work at either end, so the pair covers less than 40 feet. At 40 bar diameters on a number 5, the lap is 25 inches, two sticks lapped cover 37 feet 11 inches, and 23 feet 6 fits inside that with room to spare — but push the member to 42 feet and you need three sticks per bar rather than two, and the third is mostly offcut.
The lap figure itself is a field because it belongs to the drawing. Lap and development lengths depend on concrete strength, bar coating, bar spacing, cover, whether the bar is top steel and what the designer specified, and no page should be handing you one.
The line this page does not cross
Everything here is a cut list. Nothing here says whether the cage you described is right for the member it goes in, whether the cover is enough, whether the bend diameter is permitted for that bar size, or whether the hooks are the right ones. Those are structural and code decisions belonging to the engineer of record and to the adopted code, and the whole schedule arrives at this page from them.
Two hazards worth naming: bar projecting from a pour is an impalement hazard with requirements governed elsewhere, and bending bar by hand stores energy in a piece of steel that straightens violently if the tool slips. Neither is a procedure here.
Questions people ask
How do I work out the cut length of a stirrup?
Take the member size, subtract the cover from each face to get the out-to-out of the stirrup, then subtract one bar diameter in each direction to get the bar centreline rectangle. Add the perimeter of that rectangle to the hook extensions, then subtract a deduction for every bend. For a 90 degree bend the deduction is about 0.43 times the centreline bend radius, because the arc the bar follows is shorter than the two tangents it replaces.
Why is the cut length shorter than the outside dimensions?
Because the bar takes the inside of every corner rather than going out to the corner and coming back. Around a 90 degree bend the two tangent lengths total twice the centreline radius while the arc is only about 1.571 times it, so each corner saves roughly 0.43 radii of steel. Four corners on a rectangular stirrup, and the total saving is enough to matter on a hundred of them.
What bend diameter and hook length should I use?
The ones on your bar bending schedule, or from the fabricator, or from the code your jurisdiction has adopted. This page supplies no minimum bend diameter and no hook length, because both depend on bar size, grade and where the hook is being used, and getting them from a calculator would be the wrong way round. They are inputs here so the geometry can be worked for whatever your schedule says.
How is the weight per foot calculated?
From the nominal diameter and the density of steel rather than from a stored table. A half inch bar has an area of 0.19635 square inches, a foot of it is 2.3562 cubic inches, and at about 0.2836 pounds per cubic inch that is 0.668 pounds per foot — which is the standard table figure. Doing it this way means the weight in the results and the arithmetic in the guide cannot drift apart.
Does this check whether my reinforcement is adequate?
No. It is a cut list. Bar sizes, spacing, cover, bend diameters, hook angles and lap lengths all come into it as inputs from your drawings, and it turns them into lengths, counts and weights. Whether the cage suits the member is a structural question for the engineer of record, and a reinforced element that carries load is engineered rather than estimated.