Worked Example Singly Reinforced Beam Design using EC2

Beam design according to Eurocode 2 is discussed with a worked example. Eurocode 2 worked example for singly reinforced beam section is discussed. The design of the beam for bending moment is explained in this article.

Design calculation of the reinforcement to be provided for a given section based on its properties and loading is done this article. The basic steps that need to be followed are discussed in the article “Design of Singly Reinforced Beam“

Design Data

• Height of the section, h =450mm
• Width of the section, b = 225mm
• Cover to reinforcement= 25mm
• Bending Moment, M = 60 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. K = M / ( b d² f_ck ) = 60×10⁶ / (225×405²x20) = 0.081
3. Therefore, K < K’ = 0.167 : section is singly 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 = 373.6 mm
7. As = M / (0.87 f_yk Z) = 60×10⁶ / (0.87 x 500 x 373.6) = 369.2 mm²
8. Provide 2T16 ( A_{s provided} = 400 mm² )
9. Minimum reinforcements; A_{s provided} > 0.26 (f_ctm / f_yk )bd but not less than 0.0013bd,
10. A_{s provided} > 0.26 (f_ctm / f_yk )bd = 0.26 x (2.2/500)x225x405 = 104.2 mm²
11. 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.
12. Maximum reinforcements; (100 A_{s provided} /A_c) < 4
13. 100 A_{s provided} /A_c = 100 x 400 / (225 x 450) = 0.395. Hence Ok.