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: (2m 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: 2m of ‘linear spacers’ in every meter of the shear wall’s length
Foundation slabs and footings: (1.25m of ‘linear spacers’) in every square meter of ‘clear
slab area’
Slabs’ free edges: (1m of ‘linear spacers’ + 1m of ‘rebar chairs’ + 1 ‘piece of point support’) in every edge meter
Slabs’ supports: 1m 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 four-legged point spacers, or 1m 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.
Foundation
Footings of the strip and the spread footing foundation: linear spacers with thickness equal to 4.0cm:
[4,40 x 1,00 +5,00 x 0,85 +2 x 0,15 x 1,00] x 1,25 = 11,20 m
Connecting beams:
linear spacers of 4,00 cm: 2 x 4,20 x 2,00 = 16,80 m
point spacers of 4,00 cm: 2 x 4,20 x 2,00 ≈16 pieces
Basement
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 2,50 Beams: (2 x 4,20 +3,60) x 2,0 = 24,00 m of linear spacers equal to 2,50 cm
(2 x 4,20 +3,60) x 2,0 = 24 pieces of point spacers equal to 2,50 cm
Slab: 3,90 x 4,50 x 1,25 ≈ 22,00 m of linear spacers with thickness equal to 2,50
Ground floor
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 2,50 cm
2 x 4,20 x 2 ≈ 16 pieces of point spacers equal to 2,50 cm
Slab: 3,80 x 5,00 x 1,25 = 23,80 m of linear spacers with thickness equal to 2,50 cm Slab’s free edges:
3,60 x 1,00 x 2 = 7,20 m of linear spacers with thickness equal to 2,50 cm 3,60 x 1,00 x 2 = 7,20 m rebar chairs with height equal to 11,0 cm
3,60 x 1,00 x 2 ≈ 8 pieces of point spacers equal to 2,50 cm
Quantities sum: m
Linear spacers of 4.00 cm: |
11,20 +16,80 |
= |
28,00 |
Linear spacers of 2,50 cm: |
18,00 + 24,00 + 22,00 + 7,20 +16,80 + 23,80 |
≈ |
112,00 |
Rebar chairs of 11,00 cm: |
= |
7,20 |
pieces
Point spacers of 4,00 cm: |
= 16 |
|
Point spacers of 2,50 cm: |
32 + 24 + 8 + 32 + 24 |
= 120 |