Lesson 5 of 6
Make bars reach with development length and laps
Use grade-dependent development length, account for stock-length splices and include lap steel in the quantity.
This builds on Lesson 4, where we worked out column tie sets and review-worthy ambiguity — go back to Lesson 4.
- 1Development depends on the grades
- 2A splice consumes steel
- 3Check that every piece fits stock
- 4Your turn
Learn · Part 1
Development depends on the grades
Keep the bar diameter fixed and compare the two grade combinations.
M20 / Fe415
752 mm development
47d × 16mm = 752mm for this fully stressed deformed tension-bar example. Use the engineer’s specified anchorage for a real member.
M25 / Fe500
784 mm development
49d × 16mm = 784mm for this fully stressed deformed tension-bar example. Use the engineer’s specified anchorage for a real member.
Learn · Part 2
A splice consumes steel
For the worked grades only, assume the drawing specifies a tension lap equal to Ld. That is an example input, not a universal lap rule.
Required reach
15,000 mm reach
Reach is the distance the joined bar must cover; it excludes the duplicated material within each overlap.
Include the lap
15,752 mm of steel
1 lap × 752mm = 752mm extra steel. Total steel is reach plus all overlaps.
Learn · Part 3
Check that every piece fits stock
The stock limit applies to each fabricated piece.
Piece breakdown
2 pieces
12,000mm + 3,752mm; subtract the 752mm overlap to recover the required reach.
Apply what you learned
Your turn.
Using the 49d factor for M25 / Fe500, what is the development length of a 16mm bar?
See the worked solution
49 × 16 = 784mm.
Answer: 784.0 mm.
Lesson notes & reference
Worked exampleThe worked 16mm bar requires 752mm development for M20 / Fe415.
Your starting point
Main bars do not stop at a support face. Their cutting length must include anchorage, and long runs must include every lap.
Read the complete lesson
1. Development depends on the grades
Keep the bar diameter fixed and compare the two grade combinations.
M20 / Fe415 · 752 mm development
47d × 16mm = 752mm for this fully stressed deformed tension-bar example. Use the engineer’s specified anchorage for a real member.
M25 / Fe500 · 784 mm development
49d × 16mm = 784mm for this fully stressed deformed tension-bar example. Use the engineer’s specified anchorage for a real member.
2. A splice consumes steel
For the worked grades only, assume the drawing specifies a tension lap equal to Ld. That is an example input, not a universal lap rule.
Required reach · 15,000 mm reach
Reach is the distance the joined bar must cover; it excludes the duplicated material within each overlap.
Include the lap · 15,752 mm of steel
1 lap × 752mm = 752mm extra steel. Total steel is reach plus all overlaps.
3. Check that every piece fits stock
The stock limit applies to each fabricated piece.
Piece breakdown · 2 pieces
12,000mm + 3,752mm; subtract the 752mm overlap to recover the required reach.
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.
You have every calculation needed for the beam. Can you place the results into a schedule and catch an implausible total?
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