Reinforcement - spacing of bars - BEAMS

According to EN 1992-1-1 SECTION 8: DETAILING OF REINFORCEMENT AND PRESTRESSING TENDONS (clause 8.2)

Clear distance between bars (the spacing of bars) shall be such that the concrete can be placed and compacted satisfactorily for the development of adequate bond.

a - the clear distance (horizontal and vertical) between individual parallel bars or the clear distance between horizontal layers of parallel bars should be not less than the maximum value of:

Where:
a is minimum clear distance between bars [mm],
Φ is bar diameter [mm],
d_g is the maximum size of aggregate [mm],
k₂ is in [mm]

Note: The value of k₁ and k₂ for use in a Country may be found in its National Annex.
The recommended values are:
k₁ = 1
k₂ = 5 mm

 

If bars are positioned in separate horizontal layers, the bars in each layer should be located vertically above each other.
There should be sufficient space between the resulting columns of bars to allow access for vibrators and good compaction of the concrete.
Lapped bars may be allowed to touch one another within the lap length - this is override of the minimum clear distance rule (within the lap length):

Example

Can we put 5 Φ 16 in one row of RC beam?

b = 30 cm (width of RC beam)
c = 3 cm (side cover)
d_g = 32 mm (maximum size of aggregate)
Φ_w = 8 mm (transversal reinforcement diameter)
Φ = 16 mm (longitudinal reinforcement diameter)

(assuming k1 = 1 and k2 = 5 mm)

 

Minimum clear distance between bars shall be: a = 37 mm

Total clearance:
n ⋅ a₁ = b − 2 ⋅ c − 2 ⋅ Φ_w − 5 ⋅ Φ = 30 − 2 ⋅ 3 − 2 ⋅ 0,8 − 5 ⋅ 1,6 = 14,4 cm
Total number of clear spaces between bars:
n = 4
Clearance between bars:
a₁ = 14,4 / 4 = 3,6 cm = 36 mm

a₁ = 36 mm ≈ a = 37 mm   ⇒   we can put 5 Φ 16 in one row.

 

Otherwise, possible solutions are:

• reduce the maximum size of aggregate
• try with smaller bar diameter
• add another layer of bars