According to BS 8110, the elongation of concrete is limited to 0.0035 and the yield strength of steel is assumed to be 0.002 at an x/d value of 0.5. When the elongation of the reinforcement reaches 0.002, it begins to fail, as the elongation of the concrete at this value is 0.002. After the concrete elongation reaches a value of 0.002, we start providing compression reinforcement for the section, because when the concrete elongation reaches 0.0035, the concrete ruptures and it is a brittle rupture that does not give adequate warning in case of collapse.
Project example
Design data
Bending moment = 250 kNm
Shear force = 200 kN
Beam width = 225 mm
Beam height = 500 mm
Concrete quality = 25 N/mm2
Steel strength = 460 N/mm2
Coverage up to R/F = 30 mm
Diameter of thrust links = 10 mm
A redistribution of momentum is not assumed.
Calculations
effective depth = 500-30-10-20/2
= 450mm
K = M/bd 2 fcu
= 250 × 10 6 /(225 × 350 2 x25)
= 0.219
K > K' = 0.156
Therefore the section is double reinforced.
Find d', which is the distance to the center of the compression reinforcement from the compression surface
T16 is assumed as the pressure gain.
d' = 30+10+16/2
= 48mm
Compression reinforcement area As'
As' = (KK')fcubd 2 /{0.95fy(tt')}
= (0.219-0.156)25x225x450 2 /{0.95×460(450-48)}
= 277mm2
2T16 must be used as compression reinforcement.
Z = d(0.5+{0.25-K'/0.9} 1/2 )
= 450(0.5+{0.25-0.156/0.9} 1/2 )
= 349.599 mm
As = {(K'fcubd 2 / (0.95fyZ)} +As'
= 0.156 x 25 x 225 x 450 2 / (0.95 x 460 x 349,599) + 277
= 1592 mm 2
Provide three 25mm rods in one layer and two 12mm rods in the other layer.
Floor reinforcements 3Q25 + 2Q12
Compression reinforcements 2Q12
