Worked example for the design of a double reinforced section according to EC2

The basic steps to be followed were explained in another article and should also be consulted for more information. There are different methods for calculating the area of ​​reinforcement with different names. However, all methods produce the same result, even if they look different.

Design data

• Section height, h =450mm
• Section width, b = 225 mm
• Coverage up to reinforcement = 25mm
• Bending moment, M = 150 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. d' = 25 + 10 + 20/2 = 45 mm
3. K = M / (bd ² F _ck ) = 150×10 ⁶ / (225×405 ² x20) = 0.203
4. Therefore K > K' = 0.167: the cross section is double reinforced
5. d'/d = 45/405 = 0.111 < 0.171. Therefore compression gives R/F
6. Compressive reinforcement area As' =(KK') f _ck bd ² / (0.87f _Yes (dd'))
7. As' =(0.203-0.167) x 20 x 225 x 405 ² / (0.87 x 500 x (405-45)) = 170mm ²
8. Z = 0.82d <0.95d, OK, Design procedure for Z equation
9. Z = 0.82d = 0.82 x 405 = 332.1 mm
10. area of ​​tensile reinforcement; How = (K'f _ck bd ² / 0.87f _Sim Z) + How'
11. As = (0.167 x 20 x 225 x 405 ² / 0.87 x 500 x 332.1) + 170 = 1024 mm ²
12. Specify 2T25+1T16 ( A _{is supplied} = 1180 mm ² )
13. minimal reinforcements; A _{is provided} > 0.26 (f _CTM /F _Sim )bd, but not less than 0.0013bd,
14. A _{is provided} > 0.26 (f _CTM /F _Sim )bd = 0.26 x (2.2/500)x225x405 = 104.2 mm ²
15. 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.
16. Maximum Earnings; (100 A _{is supplied} /A _C ) < 4
17. 100A _{is supplied} /A _C = 100 x 400 / (225 x 450) = 0.395. So, okay.