Doubly Reinforced Beam Design to BS 8110
According to BS 8110, the concrete strain is limited to 0.0035 and the yielding of steel is considered as 0.002 for x/d value of 0.5. When the reinforcement strain reaches 0.002, it starts to yield, at that pint strain of the concrete is 0.002. After reaching concrete strain to a value of 0.002, we start providing compression reinforcement for the section as if the strain of the concrete reaches to 0.0035 concrete will fail and it will be a brittle failure which does not give sufficient warning at collapse.
Design example
Design Data
Bending Moment = 250 kNm
Shear Force = 200 kN
Beam width = 225 mm
Beam height = 500 mm
Grade of the concrete = 25 N/mm2
Strength of the steel = 460 N/mm2
Cover to R/F = 30 mm
Shear links diameter = 10 mm
No redistribution of the moment is assumed.
Calculations
effective depth = 500-30-10-20/2
= 450 mm
K = M/bd2fcu
= 250×106/(225×3502x25)
= 0.219
K > K’ = 0.156
Hence, section is doubly reinforced.
find d’ which is distance to center of compression reinforcement form the compression face
Assume T16 as compression reinforcements.
d’ = 30+10+16/2
= 48 mm
Area of compression reinforcement As’
As’ = (K-K’)fcubd2 /{0.95fy(d-d’)}
= (0.219-0.156)25x225x4502 /{0.95×460(450-48)}
= 277 mm2
Provide 2T16 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’fcubd2 / (0.95fyZ)} +As’
= 0.156x25x225x4502 / (0.95x460x349.599) + 277
= 1592 mm2
Provide three 25 mm bars in one layer and two 12 mm bars in a other layer.
Bottom reinforcements 3T25 + 2T12
Compression reinforcements 2T12
