Cutting Length of Stirrups (Beam & Column)

Cutting length of a rectangular stirrup = 2(a + b) + 2 × hook length − bend deductions, where a and b are the stirrup's internal dimensions. With standard 135° hooks of 10d and five bends, the practical formula is: 2(a+b) + 20d − 13d ≈ 2(a+b) + 7d (many sites conservatively use 2(a+b) + 24d − 13d with 12d hooks). Worked examples below.

a and b are measured to the STIRRUP, not the concrete — that's the deduction people forget. The dashed outline is the concrete face; the gap is the 25 mm cover.

Step-by-step (rectangular stirrup)

For a beam/column section B × D with clear cover c and stirrup diameter d:

1. Stirrup outer dimensions: a = B − 2c, b = D − 2c (measured out-to-out of stirrup).
2. Perimeter: 2 × (a + b).
3. Add hooks: two 135° hooks, each 10d (IS 2502 minimum; use 75 mm floor for small d).
4. Deduct bends: three 90° bends × 2d each, two 135° bends × 3d each → 6d + 6d ≈ 13d (rounding practice varies 12–13d).

Cutting length = 2(a + b) + 20d − 13d = 2(a + b) + 7d (with 10d hooks)

Worked example — 230 × 450 beam

8 mm stirrups, 25 mm cover:

Number of stirrups in a 4 m beam @ 150 c/c = (4000 ÷ 150) + 1 = 28 nos → 28 × 1.22 m × 0.395 kg/m ≈ 13.5 kg of 8 mm steel.

230 × 300

916

230 × 450

1,216

230 × 600

1,516

300 × 450

1,356

300 × 600

1,656

Straight from the formula above: 2(a+b) + 7d. The 230 × 450 bar is the worked example on this page — 1,216 mm. Every millimetre here multiplies by the stirrup count in the beam, which is why the 13d bend deduction is not optional book-keeping: skip it across 28 stirrups and you have over-ordered by nearly 3 metres of steel.

Other stirrup shapes

Site rules that protect the work

• Hooks must be 135° (seismic requirement), never 90° — 90° hooks open up in an earthquake.
• Hook length ≥ 10d and ≥ 75 mm, whichever is more.
• First stirrup within 50 mm of the support face; closer spacing (d/4 or 100 mm) within 2×D of supports in seismic detailing.
• Bend stirrups on a proper mandrel — sharp bends crack the bar at the corner.

Why cut length is never the perimeter

The arithmetic on this page exists because of one physical fact: a bar bent around a corner does not follow the corner. It follows an arc, and the arc is shorter than the two straight lines it replaces.

Picture it. A bar bent 90° around a pin: the outside of the bend travels further than the inside, and the bar's centreline — which is what its length actually is — cuts the corner. If you calculate the stirrup as the sum of its sides, you have measured the outside of the shape, and the steel you cut will be too long.

Hence bend deductions. For each bend, you subtract an allowance that depends on the angle and the bar diameter. IS 2502 codifies these. Roughly:

• 90° bend: deduct about 2d
• 135° bend: deduct about 3d
• 180° bend: deduct about 4d

And hooks add length. A stirrup must be anchored — the ends cannot simply stop. IS 2502 sets a minimum hook of 9d (with a practical floor around 75 mm for small bars), and seismic detailing uses 10d hooks at 135°.

So the formula assembles from three parts:

Cut length = perimeter of the bent shape + hook allowances − bend deductions

For a rectangular stirrup with four 90° corners and two 135° hooks, working with 10d hooks: 2(a + b) + 2(10d) − [3 × 2d + 2 × 3d] = 2(a + b) + 20d − 12d ≈ 2(a + b) + 7d, with rounding practice varying between 12d and 13d for the total deduction.

Why it matters at scale. A single stirrup being 50 mm long is nothing. A beam at 150 mm centres over 4 m has 28 stirrups; a house has hundreds of beams and columns. Systematically over-cutting by 50 mm across 3,000 stirrups is 150 m of 8 mm bar — around 60 kg, and more importantly it is 3,000 stirrups that do not quite close properly because the bar bender is forcing the last bend.

And under-cutting is worse than over-cutting. A stirrup cut short cannot close: the hooks do not reach the core, or they get bent flatter than 135° to make them fit. That is the exact detail the hooks existed to provide, quietly removed.

Cover, and where a and b come from

The formula uses a and b — the stirrup's own dimensions — and getting them from the member size is where the second class of error lives.

a = B − 2c and b = D − 2c, where B and D are the member's overall dimensions and c is the clear cover.

The trap: cover is measured to the outside of the stirrup. So subtracting 2c from the member size gives you the stirrup's outside dimensions directly. If someone instead works to the main bar and subtracts cover plus a bar diameter, the stirrup comes out undersized and the actual cover ends up larger than specified — which sounds harmless and is not, because it reduces the effective depth of the member.

The cover to use:

• Beams: 25 mm (mild exposure). This is the one people get wrong by importing the column figure.
• Columns: 40 mm.
• Slabs: 20 mm.
• Footings: 50 mm (cast against earth, with a blinding layer).
• Severe/marine exposure: more, per IS 456, with a higher concrete grade alongside.

A worked check, 230 × 450 beam, 8 mm stirrup, 25 mm cover:

• a = 230 − 50 = 180 mm
• b = 450 − 50 = 400 mm
• Perimeter = 2(180 + 400) = 1,160 mm
• Hooks = 2 × 10 × 8 = 160 mm
• Deduction ≈ 13 × 8 = 104 mm
• Cut length = 1,160 + 160 − 104 = 1,216 mm

Use 40 mm cover here by mistake — the column figure — and you get 1,096 mm: a stirrup 120 mm short, which will not close, on every stirrup in every beam in the house.

Other shapes, and the ones that catch people

Circular stirrups (spirals/helical). Length per turn is π × (D − 2c − d) plus the lap. In a circular column the helical reinforcement is continuous, and its pitch is the spacing.

Diamond and rhombus ties in columns with many bars — the perimeter is longer than it looks because the diagonal legs are longer than the sides.

Two-legged vs four-legged. A wide beam needs cross-ties or a four-legged stirrup so that intermediate bars are restrained too. The cutting length of a cross-tie is a different, simpler shape — a straight bar with a 135° hook each end — and it is routinely forgotten in the schedule.

Open (U) stirrups in some slab and beam details.

And the one that matters most: the 135° hook is not optional. In seismic detailing it is the whole point. A stirrup with 90° hooks opens up when the cover spalls — which is exactly the moment the confinement was needed. It is faster to bend, it looks nearly identical in a photograph, and it is one of the most consequential substitutions made on Indian sites. Check the hooks; do not just count the stirrups.

Turning cut length into a steel order

Cut length is only useful once it becomes weight, and that is a two-step conversion nobody does in their head correctly.

Step 1 — how many stirrups? Number = (length of the member ÷ spacing) + 1. The +1 is real and routinely dropped: a 4 m beam at 150 mm centres is (4,000 ÷ 150) + 1 = 28 stirrups, not 27. Across a house, that dropped +1 is dozens of missing stirrups.

But spacing is not uniform. Under IS 13920, the confinement zones at each end are far tighter than mid-span. So the count is really: (zone length ÷ tight spacing) for each end, plus (middle length ÷ normal spacing), plus one. A beam with 100 mm spacing over 900 mm at each end and 150 mm in the middle has substantially more stirrups than the uniform calculation suggests — and this is exactly where a steel estimate comes up short.

Step 2 — weight. Total length in metres × unit weight from D²/162. For 8 mm that is 0.395 kg/m.

Worked, for that 230 × 450 beam, 4 m long:

• 28 stirrups × 1.216 m = 34.05 m of 8 mm bar
• 34.05 × 0.395 = 13.45 kg for one beam's stirrups

Scale it: a modest house has perhaps 40 beams and 20 columns. At roughly 13 kg per beam and similar per column, that is around 780 kg of 8 mm — 13% of a 6-tonne house steel order, all of it in the smallest, most expensive-per-kg diameter.

Which is the practical point of this page. Stirrups are not a rounding error. They are an eighth of the steel bill, they are in the diameter that costs the most per kilogram (thin bars need more rolling passes), and they are the bars whose length is most often calculated wrong.

And they are what offcuts are for. Short lengths left from cutting main bars are exactly the right size for stirrups, chairs and dowels — so a yard that scraps them and then orders fresh 8 mm has charged you twice for the same steel. Ask whether offcuts are being nested back into the schedule.

Bending: hand, machine, and what to watch

Hand bending around a pin is universal on small Indian sites. It works, and it has two failure modes: the bend radius is whatever the pin happens to be, and consistency across 3,000 stirrups depends on one person's attention.

The bend radius matters. Too tight and you damage the bar — TMT gets its strength from a quenched, tempered structure, and a sharp bend cracks that hard outer rim. IS 2502 sets minimum mandrel diameters; for mild steel and TMT in ordinary sizes the working figure is a pin of at least 4d for bars up to 20 mm.

Never heat a bar to bend it. Heating undoes exactly the tempering that makes TMT what it is, and the bar becomes brittle at the bend. It is quick, it is common, and it is a genuine defect.

Never re-bend. A bar bent, straightened and bent again is work-hardened and may have cracked. This is what happens to column starter bars that get flattened so people can walk on a slab, then straightened for the next lift.

Machine bending on a bar bending machine gives consistent radii and consistent lengths, and on any quantity it pays for itself in reduced wastage alone.

What to check at the bar bending yard:

• Measure a finished stirrup. Against the schedule. Pick a few, from different batches.
• Check the hooks are 135°, not 90°, and that they turn inward.
• Check the corners aren't cracked. Look at the outside of each bend for fine transverse cracks — that is a bar that was bent too tight or was cold and brittle.
• Check the shape closes. A stirrup that has to be forced around the cage is the wrong size, and forcing it opens the hooks.

The scrap conversation. Bar bending is charged per tonne or per kg, and the offcuts have value. Settle in the contract whether the scrap money is yours or the contractor's — on a six-tonne house it is not trivial, and it is never mentioned unless you raise it.

The three numbers to carry

Cut length = 2(a + b) + 7d, with a = B − 2c and b = D − 2c, using 10d hooks and a 13d total bend deduction.

Cover for beams is 25 mm. Not 40 — that is columns, and importing it shortens every stirrup by 120 mm.

Count = (length ÷ spacing) + 1, calculated separately for the tight confinement zones at each end and the wider middle. The +1 gets dropped, and the tight zones get forgotten, and between them that is where a stirrup order comes up short.

Why this page is worth the arithmetic

Stirrups look like the trivial part of a steel order and they are not. They are roughly an eighth of the tonnage, in the diameter that costs the most per kilogram, calculated with the formula most likely to be got wrong, bent by the process with the least supervision, and carrying the detail (the 135° hook) that matters most in an earthquake.

Get the cover right, get the +1 right, get the confinement zones counted, and check the hooks in the yard. That is the whole job, and it is twenty minutes of attention against a line item worth tens of thousands of rupees and a structural property you cannot inspect once the concrete is in.

A note on rounding practice

You will see the bend deduction written as 12d in some references and 13d in others, and the resulting formula as 2(a+b)+8d or 2(a+b)+7d. Both are in use, both are defensible, and the difference on an 8 mm stirrup is 8 mm of bar.

What matters is being consistent within one schedule, and rounding the final cut length up to the nearest 5 or 10 mm. A stirrup 8 mm long is trimmed in a second. A stirrup 8 mm short cannot close, and the bar bender's solution to that is to flatten the hooks — which removes the detail the hooks were there for.

Frequently asked questions

What is the formula for stirrup cutting length? 2(a+b) + 2 hook lengths (10d each) − bend deductions (≈13d total) for a rectangular stirrup, where a and b are the stirrup dimensions after deducting cover.

Why deduct for bends? Steel stretches along the outer face at each bend, so the flat cutting length is shorter than the sum of sides — 2d per 90° bend and 3d per 135° bend is deducted.

What hook length is required on stirrups? Minimum 10 times the stirrup diameter (and not less than 75 mm), bent at 135° into the core.

How many stirrups in a beam? (Clear span ÷ spacing) + 1, with tighter spacing near supports per the structural drawing.

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