Stirrup Calculator
Calculate rebar stirrups (ties) for a beam or column: cut length with 135° or 90° hooks, how many stirrups at closer end spacing, total metres, weight in kg and how many 12 m bars to buy.
How it works
a = width − 2·cover − d · b = depth − 2·cover − d (centre-line) · cut length = 2(a + b) + 2 × hook extension, hook = 6d (≥ 75 mm for 135° hooks, ACI 318 Table 25.3.2) · stirrups per member = 2 × (zone ÷ s₁ + 1) + ((L − 2·zone) ÷ s₂ − 1) · weight = length × d² ÷ 162 kg/m
A stirrup (tie) wraps the main bars of a beam or column to resist shear and confine the concrete. Its size is the section minus the cover on both sides, measured to the centre of the stirrup bar, and its cut length adds two hook extensions — 6 bar diameters, and at least 75 mm for the 135° seismic hooks required in earthquake zones by ACI 318 and most Latin American codes. Near supports, where shear is highest, stirrups are placed closer together over a confinement zone, then opened up in the middle. Bar weight follows from steel density: d² ÷ 162 kg per metre with d in millimetres. Bending deductions are small and ignored here; your bar-bending schedule may trim 1–2 cm.
Worked example
A 30 × 45 cm beam with 4 cm cover and 3/8″ (9.5 mm) stirrups: centre-line size 21.1 × 36.1 cm; hooks max(6 × 9.5, 75) = 7.5 cm each; cut length = 2 × (21.05 + 36.05) + 15 = 129.2 cm. Over a 5 m span with 90 cm end zones at 10 cm and 20 cm in the middle: 10 + 15 + 10 = 35 stirrups, 45.22 m of bar weighing 25.2 kg — 4 bars of 12 m (9 stirrups from each).
Sources: ACI 318-19 Building Code Requirements for Structural Concrete · CRSI — Concrete Reinforcing Steel Institute
Frequently asked questions
How do I calculate the cut length of a stirrup?
Take the beam width and depth, subtract twice the cover and one bar diameter to get the centre-line sides, add them and double (the perimeter), then add two hook extensions. Many bar-bending schedules also subtract 2d for each 90° bend; the difference is a centimetre or two.
Why are stirrups closer near the columns?
Shear force and earthquake demands are highest near supports, so codes require tighter spacing there — typically d/4 or 10 cm over a length of twice the beam depth — and allow wider spacing (d/2) in the middle.
What is a 135° hook and when is it required?
It is a hook bent back into the core of the member so it cannot open when the cover spalls. Seismic codes (ACI 318 Chapter 18, NTC Mexico, NSR-10, E.060) require 135° hooks with 6d extensions (min 75 mm) on hoops in frames.
How much does a stirrup bar weigh?
Weight per metre ≈ d² ÷ 162 kg with d in mm: 1/4″ ≈ 0.25 kg/m, 3/8″ ≈ 0.56 kg/m, 1/2″ ≈ 1.0 kg/m.
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This calculator is for educational purposes. Double-check important results before acting on them.