Masonry Horizontal Joint-Reinforcement Takeoff (IRC R606.12.2)
Ladder or truss wire laid in the bed joints of a masonry wall.
Example
You enter
- Wall length (ft) 40
- Wall height (ft) 12
- Vertical spacing (in, code cap 16) 16
- Wire piece length (ft) 10
You get
- Reinforced courses 9
- Total wire pieces 36
Details, formula, and sources
Reinforced courses = ceil(height in / vertical spacing), pieces per course = ceil(length / piece length), total = courses x pieces. IRC R606.12.2 / TMS 402 cap the vertical spacing at 16 in (every other 8 in course); some specs tighten it to 8 in or add wire at bond beams and above and below openings. A 40 x 12 ft CMU wall at 16 in and 10 ft pieces is 9 courses x 4 = 36 pieces; a 30 x 10 ft wall is 24. Wire laps at least 6 in (the lap is not added here). A material count; the spacing, the lap, and the extra wire at openings come from the structural spec and the adopted code.
reinforced_courses = ceil(wall_height_ft x 12 / vertical_spacing_in); pieces_per_course = ceil(wall_length_ft / piece_length_ft); total_pieces = reinforced_courses x pieces_per_course.
IRC R606.12.2 and TMS 402 masonry horizontal joint-reinforcement provisions, by name; the spacing and lap come from the adopted code and the structural spec.
The count arithmetic is public first-principles; the 16 in maximum vertical spacing and 6 in minimum lap are in IRC R606.12.2 (many jurisdictions post the adopted IRC).
Estimate. AHJ and licensed professional govern.
Field names used by the API: wall_length_ft, wall_height_ft, vertical_spacing_in, piece_length_ft, reinforced_courses, total_pieces
- Vertical spacing IRC R606.12.2 / TMS 402 max 16 in o.c. (every other 8 in course); some specs tighten to 8 inIRC R606.12.2
- Lap not counted wire laps >= 6 in and gets extra runs at openings/corners; not added into this countmasonry practice