Lesson 2 of 6
Calculate a complete stirrup cutting length
Move from concrete size to covered perimeter, bend deduction, hook allowance and the final rounded cutting length.
This builds on Lesson 1, where we worked out how to read the beam and reinforcement inputs — go back to Lesson 1.
- 1Bring the rectangle inside the concrete
- 2Account for all five bends
- 3Add two hook allowances
- 4Finish the whole chain
- 5Your turn
Learn · Part 1
Bring the rectangle inside the concrete
Change cover and watch the working perimeter change.
20mm cover
1,200 mm perimeter
A = 230 − 2 × 20 = 190mm; B = 450 − 2 × 20 = 410mm. Perimeter = 2(A + B).
25mm cover
1,160 mm perimeter
A = 230 − 2 × 25 = 180mm; B = 450 − 2 × 25 = 400mm. Perimeter = 2(A + B).
40mm cover
1,040 mm perimeter
A = 230 − 2 × 40 = 150mm; B = 450 − 2 × 40 = 370mm. Perimeter = 2(A + B).
Why a bend is a deduction
The curve cuts inside the imaginary sharp corner. The outside measurements count more length than the bar needs.
90° bend: deduct 2d = 16 mm
1 corners + 0 hook bends counted. Green marks show the bends included so far.
Schematic geometry. Deductions follow the course convention; they are not measured from this picture.
Why a bend is a deduction
The curve cuts inside the imaginary sharp corner. The outside measurements count more length than the bar needs.
90° bend: deduct 2d = 16 mm
2 corners + 0 hook bends counted. Green marks show the bends included so far.
Schematic geometry. Deductions follow the course convention; they are not measured from this picture.
Why a bend is a deduction
The curve cuts inside the imaginary sharp corner. The outside measurements count more length than the bar needs.
90° bend: deduct 2d = 16 mm
3 corners + 0 hook bends counted. Green marks show the bends included so far.
Schematic geometry. Deductions follow the course convention; they are not measured from this picture.
Why a bend is a deduction
The curve cuts inside the imaginary sharp corner. The outside measurements count more length than the bar needs.
135° bend: deduct 3d = 24 mm
3 corners + 1 hook bends counted. Green marks show the bends included so far.
Schematic geometry. Deductions follow the course convention; they are not measured from this picture.
Why a bend is a deduction
The curve cuts inside the imaginary sharp corner. The outside measurements count more length than the bar needs.
135° bend: deduct 3d = 24 mm
3 corners + 2 hook bends counted. Green marks show the bends included so far.
Schematic geometry. Deductions follow the course convention; they are not measured from this picture.
Learn · Part 2
Account for all five bends
Tap through three corners and two hook bends. Start from the worked 25mm cover case.
1 bend
1,144 mm intermediate
Outside leg dimensions meet at imaginary sharp corners. The steel follows a rounded centreline inside them, so adding the outside legs overcounts the cutting length. Deduct 2d for each 90° corner and 3d for each 135° hook bend under this course’s convention. 1 of 5 bends counted: subtract 16mm in total. This intermediate value is not a cutting length.
2 bends
1,128 mm intermediate
Outside leg dimensions meet at imaginary sharp corners. The steel follows a rounded centreline inside them, so adding the outside legs overcounts the cutting length. Deduct 2d for each 90° corner and 3d for each 135° hook bend under this course’s convention. 2 of 5 bends counted: subtract 32mm in total. This intermediate value is not a cutting length.
3 bends
1,112 mm intermediate
Outside leg dimensions meet at imaginary sharp corners. The steel follows a rounded centreline inside them, so adding the outside legs overcounts the cutting length. Deduct 2d for each 90° corner and 3d for each 135° hook bend under this course’s convention. 3 of 5 bends counted: subtract 48mm in total. This intermediate value is not a cutting length.
4 bends
1,088 mm intermediate
Outside leg dimensions meet at imaginary sharp corners. The steel follows a rounded centreline inside them, so adding the outside legs overcounts the cutting length. Deduct 2d for each 90° corner and 3d for each 135° hook bend under this course’s convention. 4 of 5 bends counted: subtract 72mm in total. This intermediate value is not a cutting length.
5 bends
1,064 mm intermediate
Outside leg dimensions meet at imaginary sharp corners. The steel follows a rounded centreline inside them, so adding the outside legs overcounts the cutting length. Deduct 2d for each 90° corner and 3d for each 135° hook bend under this course’s convention. 5 of 5 bends counted: subtract 96mm in total. This intermediate value is not a cutting length.
One hook, ten bar diameters
d = 6 mm. The two green ends are the hooks. A larger bar needs a longer hook allowance.
10 × 6 = 60 mm · Use 75 mm at each end
60 mm is below the 75 mm minimum, so the 6 mm bar uses 75 mm at each end. The dashed extension shows the extra allowance.
One hook, ten bar diameters
d = 8 mm. The two green ends are the hooks. A larger bar needs a longer hook allowance.
10 × 8 = 80 mm · Use 80 mm at each end
10d exceeds the 75 mm minimum here. Both hooks use this allowance before bend deductions.
One hook, ten bar diameters
d = 10 mm. The two green ends are the hooks. A larger bar needs a longer hook allowance.
10 × 10 = 100 mm · Use 100 mm at each end
10d exceeds the 75 mm minimum here. Both hooks use this allowance before bend deductions.
Learn · Part 3
Add two hook allowances
Compare diameters. The minimum keeps a small bar’s hook from becoming too short.
6mm bar
150 mm for two hooks
Each 135° hook uses 10d, with a 75mm minimum: 75mm per end. This is the allowance before bend deductions.
8mm bar
160 mm for two hooks
Each 135° hook uses 10d, with a 75mm minimum: 80mm per end. This is the allowance before bend deductions.
10mm bar
200 mm for two hooks
Each 135° hook uses 10d, with a 75mm minimum: 100mm per end. This is the allowance before bend deductions.
Learn · Part 4
Finish the whole chain
Return to the 8mm worked example. Round once, after both hooks are included.
Before rounding
1,224 mm before rounding
1160 − 96 + 160 = 1224mm. The arithmetic is complete; rounding remains.
Final cutting length
1,220 mm cutting length
Round to the nearest 10mm, with halfway values rounded up. Only the final cutting length goes into the schedule.
Apply what you learned
Your turn.
Your turn: 300 × 500mm beam, 30mm cover, 10mm bar, two 135° hooks. What is the final cutting length?
See the worked solution
Perimeter 1360 − deductions 120 + hooks 200 = 1440mm → 1440mm.
Answer: 1440 mm.
Lesson notes & reference
Worked exampleThe worked 230 × 450mm beam needs a 1220mm stirrup cutting length.
Your starting point
Use the drawing inputs from Lesson 1: a 230 × 450mm beam, 25mm cover, 8mm stirrup and two 135° hooks.
Read the complete lesson
1. Bring the rectangle inside the concrete
Change cover and watch the working perimeter change.
20mm cover · 1,200 mm perimeter
A = 230 − 2 × 20 = 190mm; B = 450 − 2 × 20 = 410mm. Perimeter = 2(A + B).
25mm cover · 1,160 mm perimeter
A = 230 − 2 × 25 = 180mm; B = 450 − 2 × 25 = 400mm. Perimeter = 2(A + B).
40mm cover · 1,040 mm perimeter
A = 230 − 2 × 40 = 150mm; B = 450 − 2 × 40 = 370mm. Perimeter = 2(A + B).
2. Account for all five bends
Tap through three corners and two hook bends. Start from the worked 25mm cover case.
1 bend · 1,144 mm intermediate
Outside leg dimensions meet at imaginary sharp corners. The steel follows a rounded centreline inside them, so adding the outside legs overcounts the cutting length. Deduct 2d for each 90° corner and 3d for each 135° hook bend under this course’s convention. 1 of 5 bends counted: subtract 16mm in total. This intermediate value is not a cutting length.
2 bends · 1,128 mm intermediate
Outside leg dimensions meet at imaginary sharp corners. The steel follows a rounded centreline inside them, so adding the outside legs overcounts the cutting length. Deduct 2d for each 90° corner and 3d for each 135° hook bend under this course’s convention. 2 of 5 bends counted: subtract 32mm in total. This intermediate value is not a cutting length.
3 bends · 1,112 mm intermediate
Outside leg dimensions meet at imaginary sharp corners. The steel follows a rounded centreline inside them, so adding the outside legs overcounts the cutting length. Deduct 2d for each 90° corner and 3d for each 135° hook bend under this course’s convention. 3 of 5 bends counted: subtract 48mm in total. This intermediate value is not a cutting length.
4 bends · 1,088 mm intermediate
Outside leg dimensions meet at imaginary sharp corners. The steel follows a rounded centreline inside them, so adding the outside legs overcounts the cutting length. Deduct 2d for each 90° corner and 3d for each 135° hook bend under this course’s convention. 4 of 5 bends counted: subtract 72mm in total. This intermediate value is not a cutting length.
5 bends · 1,064 mm intermediate
Outside leg dimensions meet at imaginary sharp corners. The steel follows a rounded centreline inside them, so adding the outside legs overcounts the cutting length. Deduct 2d for each 90° corner and 3d for each 135° hook bend under this course’s convention. 5 of 5 bends counted: subtract 96mm in total. This intermediate value is not a cutting length.
3. Add two hook allowances
Compare diameters. The minimum keeps a small bar’s hook from becoming too short.
6mm bar · 150 mm for two hooks
Each 135° hook uses 10d, with a 75mm minimum: 75mm per end. This is the allowance before bend deductions.
8mm bar · 160 mm for two hooks
Each 135° hook uses 10d, with a 75mm minimum: 80mm per end. This is the allowance before bend deductions.
10mm bar · 200 mm for two hooks
Each 135° hook uses 10d, with a 75mm minimum: 100mm per end. This is the allowance before bend deductions.
4. Finish the whole chain
Return to the 8mm worked example. Round once, after both hooks are included.
Before rounding · 1,224 mm before rounding
1160 − 96 + 160 = 1224mm. The arithmetic is complete; rounding remains.
Final cutting length · 1,220 mm cutting length
Round to the nearest 10mm, with halfway values rounded up. Only the final cutting length goes into the schedule.
The schedule you’re building toward
The same worked example throughout the course: three identical beams, with three bottom and three top main bars in each.
| Bar | Shape | Dia | Cutting length | Bars per member | Total bars | Total length | Weight | Notes |
|---|---|---|---|---|---|---|---|---|
| S1 | Stirrup · rectangular, 135° hooks | 8mm | 1220mm | 27 | 81 | 98.82m | 39.04kg | Uniform centres; no zones |
| B1 | Bottom main · straight | 16mm | 5500mm | 3 | 9 | 49.50m | 78.22kg | Straight anchorage at both ends |
| T1 | Top main · straight | 16mm | 5500mm | 3 | 9 | 49.50m | 78.22kg | Straight anchorage at both ends |
Total steel: 195.48kg. Worked quantities exclude wastage. Main bars assume straight anchorage fits each support; use the actual structural detail for a project.
One stirrup is useful. How do spacing and repetition turn it into a steel quantity for the job?
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