Beam design according to Eurocode 2 is explained using a practical example. The practical example for a simply reinforced beam cross section according to Eurocode 2 is explained. Beam design for bending moments is explained in this article.
In this article, the reinforcement design for a specific section is calculated based on its properties and loads. The basic steps to be followed are explained in the article “Sizing a simply reinforced beam”.
Design data
- Section height, h =450mm
- Section width, b = 225 mm
- Coverage up to reinforcement = 25mm
- Bending moment, M = 60 kN/m
- Cylinder thickness, f ck = 20 N/mm 2
- Reinforcement resistance = 500 N/mm 2
- To calculate the effective depth (d), a bar diameter of 20 mm and a connection diameter of 10 mm are assumed.
- d = 450 – 25 – 20/2 – 10 = 405 mm
- K = M / ( bd 2 F ck ) = 60×10 6 / (225×405 2 x20) = 0.081
- Therefore, K < K' = 0.167: The cross section is simply reinforced.
- Z = d ( 0.5 + √ (0.25 – K / 1.134)) = d ( 0.5 + √ (0.25 – 0.081 / 1.134)) = 0.923d
- Z =0.923d < 0.95d, ok
- Z = 0.923d = 0.923 x 405 = 373.6mm
- As = M / (0.87f Sim Z) = 60×10 6 / (0.87 x 500 x 373.6) = 369.2 mm 2
- Specify 2T16 (A is given = 400 mm 2 )
- minimal reinforcements; A is provided > 0.26 (f CTM /F Sim )bd, but not less than 0.0013bd,
- A is provided > 0.26 (f CTM /F Sim )bd = 0.26 x (2.2/500)x225x405 = 104.2 mm 2
- A is provided > 0.0013bd 0.0013 x 225 x 405 = 118.5 mm 2 Therefore, the area of reinforcement provided is greater than the minimum area required.
- Maximum Earnings; (100 A is supplied /A C ) < 4
- 100A is supplied /A C = 100 x 400 / (225 x 450) = 0.395. So, okay.
