Stirrup cutting length = the bar's centreline around the four sides, three 90° corners and two 135° hooks, plus the hook extensions. For a 230 × 450 mm beam with 25 mm cover, an 8 mm stirrup is 180 × 400 mm outside and its cutting length is 1,306 mm with 75 mm hooks.
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Cutting length per bar
1,306mm
Total length
52.3 m
Total weight
20.6 kg
90° bend deduction
17 mm (2.07 db)
Dimensions are outside (out-to-out) as on a BS 8666 / ACI 315 bar schedule. Check your fabricator’s bending conventions.
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Cutting length = 2 × (A − 2(r + d)) + 2 × (B − 2(r + d)) + 3 × 90° arcs + 2 × 135° arcs + 2 × hook extension, where A and B are the outside sizes, d is the bar size and r the bend radius. The arcs are measured on the bar's centreline, radius r + d/2.
How do I find the stirrup size from the beam size?
Take the clear cover off each face. A 230 × 450 mm beam with 25 mm cover to the stirrup has an 8 mm stirrup 230 − 50 = 180 mm wide and 450 − 50 = 400 mm deep, measured outside.
What is the quick site formula for stirrups?
Many sites use 2 × (A + B) + 2 × 10d for the hooks − 3 × 2d for the 90° bends − 2 × 3d for the 135° bends, which comes to 2(A + B) + 8d. It ignores the real bend radius, so it can be several centimetres short: for the 180 × 400 mm example it gives 1,224 mm, against 1,316 mm exact with the same 80 mm hooks. Check which method your bar bending schedule uses.
How long should a stirrup hook be?
ACI 318 asks for a 135° hook with an extension of 6 bar diameters, at least 75 mm (3 in). In India, IS 13920 asks for 6d but at least 65 mm in earthquake-resistant members, and many sites simply use 10d (80 mm for an 8 mm bar).
References
ACI 318-19 §25.3.2: stirrup hooks, 135° with an extension of 6db and at least 75 mm (3 in) for #3 to #5
IS 13920:2016: 135° hooks with a 6d extension, at least 65 mm, in earthquake-resistant members
BS 8666:2020: minimum bend (former) diameter 4d for bars up to 16 mm