Worked example of constructing a beam simply reinforced with EC2

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.

1. d = 450 – 25 – 20/2 – 10 = 405 mm
2. K = M / ( bd ² F _ck ) = 60×10 ⁶ / (225×405 ² x20) = 0.081
3. Therefore, K < K' = 0.167: The cross section is simply reinforced.
4. Z = d ( 0.5 + √ (0.25 – K / 1.134)) = d ( 0.5 + √ (0.25 – 0.081 / 1.134)) = 0.923d
5. Z =0.923d < 0.95d, ok
6. Z = 0.923d = 0.923 x 405 = 373.6mm
7. As = M / (0.87f _Sim Z) = 60×10 ⁶ / (0.87 x 500 x 373.6) = 369.2 mm ²
8. Specify 2T16 (A _{is given} = 400 mm ² )
9. minimal reinforcements; A _{is provided} > 0.26 (f _CTM /F _Sim )bd, but not less than 0.0013bd,
10. A _{is provided} > 0.26 (f _CTM /F _Sim )bd = 0.26 x (2.2/500)x225x405 = 104.2 mm ²
11. 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.
12. Maximum Earnings; (100 A _{is supplied} /A _C ) < 4
13. 100A _{is supplied} /A _C = 100 x 400 / (225 x 450) = 0.395. So, okay.