Worked Example Design of Doubly Reinforced Section to EC2

Basic steps that need to be followed have discussed in a different article and it shall also be referred for more information. There are different methods of calculating the reinforcement area with used different notations. However, all the methods provide the same answer though the methods look different.

Design Data

• Height of the section, h =450mm
• Width of the section, b = 225mm
• Cover to reinforcement= 25mm
• Bending Moment, M = 150 kN/m
• Cylinder Strength, f_ck = 20 N/mm²
• Reinforcement strength= 500 N/mm²
• Assume bar diameter as 20mm and link diameter as 10mm for calculating the effective depth (d)

1. d = 450 – 25 – 20 / 2 – 10 = 405 mm
2. d’ = 25 + 10 + 20/2 = 45 mm
3. K = M / ( b d² f_ck ) = 150×10⁶ / (225×405²x20) = 0.203
4. Therefore, K > K’ = 0.167 : section is doubly reinforced
5. d’/d = 45/405 = 0.111 < 0.171. Therefore, commpression R/F yielding
6. Compression reinforcemen area As’ =(K-K’) f_ck b d² / [ 0.87 f_yk (d-d’)]
7. As’ =(0.203-0.167) x 20 x 225 x 405² / [ 0.87 x 500 x (405-45)] = 170 mm²
8. Z = 0.82d < 0.95d, Ok, refere design procedure for equation of the Z
9. Z =0.82d = 0.82 x 405 = 332.1 mm
10. Tension reinforcement area; As = [K’ f_ck b d² / 0.87 f_yk Z] + As’
11. As = [0.167 x 20 x 225 x 405² / 0.87 x 500 x 332.1 ] + 170 =1024 mm²
12. Provide 2T25+1T16 ( A_{s provided} = 1180 mm² )
13. Minimum reinforcements; A_{s provided} > 0.26 (f_ctm / f_yk )bd but not less than 0.0013bd,
14. A_{s provided} > 0.26 (f_ctm / f_yk )bd = 0.26 x (2.2/500)x225x405 = 104.2 mm²
15. A_{s provided} > 0.0013bd 0.0013 x 225 x 405 = 118.5 mm² . Therefore, Provided reinforcement area is grater than the minimum required area.
16. Maximum reinforcements; (100 A_{s provided} /A_c) < 4
17. 100 A_{s provided} /A_c = 100 x 400 / (225 x 450) = 0.395. Hence Ok.