Estimation of the spacers’ quantity
If there is no other rule to follow when calculating the spacers of the formwork, one could use the empirical rule presented below:
RULES FOR ESTIMATING THE QUANTITTIES OF SPACERS AND REBAR CHAIRS
Superstructure and foundation beams: (2 m of ‘linear spacers + 2 ‘pieces of point support’) in every meter of ‘clear beam length’
Columns: (8 ‘pieces of point support’) per column
Shear walls: 2 m of ‘linear spacers’ in every meter of the shear wall’s length
Foundation slabs and footings: (1.25 m of ‘linear spacers’) in every square meter of ‘clear slab area’
Slabs’ free edges: (1 m of ‘linear spacers’ + 1m of ‘rebar chairs’ + 1 ‘piece of point support’) in every edge meter
Slabs’ supports: 1 m of ‘rebar chairs’ per meter of ‘every slab support’
• 1 m of ‘linear spacers’ might be 1 m of a plastic spacer, or 5 special formed spacers, or any other number of local spacers, etc.
• 1m of ‘rebar chairs’ might be 3 fourlegged point spacers, or 1 m of a folded wire mesh, or 2 pieces of impromptu steel rebar chairs, etc.
• 1 ‘piece of point support’ might be a properly formed plastic or concrete point support, or a peg, etc.
Footings of the strip and the spread footing foundation:
linear spacers with thickness equal to 40 mm:
[4,40 x 1,00 +5,00 x 0,85 +2 x 0,15 x 1,00] x 1,25 = 11,20 m
linear spacers of 40 mm: 2 x 4,20 x 2,00 = 16,80 m
point spacers of 40 mm: 2 x 4,20 x 2,00 ≈16 pieces
Columns: 8 x 4 = 32 pcs. of point support
Shear wall: 3,60 x 2,50 x 2 = 18,00 m of linear spacers with thickness equal to 25 mm
Beams: (2 x 4,20 +3,60) x 2,0 = 24,00 m of linear spacers equal to 25 mm
(2 x 4,20 +3,60) x 2,0 = 24 pieces of point spacers equal to 25 mm
Slab: 3,90 x 4,50 x 1,25 ≈ 22,00 m of linear spacers with thickness equal to 25 mm
Columns: 8 x 4 = 32 pcs. of point support
Beams: 2 x 4,20 x 2,0 = 16,80 m of linear spacers equal to 25 mm
2 x 4,20 x 2 ≈ 16 pieces of point spacers equal to 25 mm
Slab: 3,80 x 5,00 x 1,25 = 23,80 m of linear spacers with thickness equal to 25 mm
3,60 x 1,00 x 2 = 7,20 m of linear spacers with thickness equal to 25 mm
3,60 x 1,00 x 2 = 7,20 m rebar chairs with height equal to 110 mm
3,60 x 1,00 x 2 ≈ 8 pieces of point spacers equal to 25 mm
Quantities sum: 

m 
Linear spacers of 40 mm: 
11,20 +16,80 
= 
28,00 
Linear spacers of 25 mm: 
18,00 + 24,00 + 22,00 + 7,20 +16,80 + 23,80 
≈ 
112,00 
Rebar chairs of 110 mm: 

= 
7,20 



pieces 
Point spacers of 40 mm: 

= 
16 
Point spacers of 25 mm: 
32 + 24 + 8 + 32 + 24 
= 
120 



