Design of Doubly-Reinforced Beams According to EC2 | Worked Example

In the design of reinforced concrete beams, if the design ultimate moment is greater than the ultimate moment of resistance i.e. MEd > MRd, then compression reinforcement is required. Provided that d₂/x ≤ 0.38 (i.e. compression steel has yielded) where d₂ is the depth of the compression steel from the compression face and x = (d − z)/0.4

The area of compression reinforcement, A_s2, is given by:

Area of compression reinforcement A_S2 = (M_Ed – M_Rd) / (0.87f_yk (d – d₂)) ————- (1)

Area of tension reinforcement A_s1 = M_Rd / (0.87f_ykz) + A_S2 ———– (2)
Where z = d[0.5+ √(0.25 – 0.882K’)]
where
K’ = 0.167

DESIGN EXAMPLE
To show how this is done, let us consider the flexural design of support A of the beam loaded as shown below. The depth of the beam is 600mm, and the width is 400mm.
f_ck = 35 N/mm² ; f_yk = 460 N/mm²; concrete cover = 40mm

 

Top reinforcement (Hogging moment)
Support A
Effective depth (d) = 600 – 40 – 16 – 10 = 534 mm

M_Ed = 761.24 KNm

But since the flange is in tension, we use the beam width to calculate the value of k (this applies to all support hogging moments)
k = M_Ed / (f_ck b_wd² ) = (761.24 × 10⁶) / (35 × 400 × 534²) = 0.1906

Since k < 0.167, compression is required
Area of compression reinforcement A_S2 = (M_Ed – M_Rd) / (0.87f_yk (d – d₂))

M_Rd = 0.167f_ckbd² = (0.167 × 35 × 400 × 534²) × 10^-6 = 666.69 KNm

d₂ = 40 + 16 + 10 = 66 mm

A_S2 = ((761.24 – 666.69) × 10⁶) / (0.87 × 460 × (534 – 66)) = 504.22 mm2
Provide 3X16 Bottom (A_sprov = 603 mm²)

Area of tension reinforcement A_s1 = M_Rd / (0.87f_yk z) + A_S2
Where z = d[0.5+ √(0.25 – 0.882K’)]
K’ = 0.167
z = d[0.5+ √((0.25 – 0.882(0.167))] = 0.82d

A_s1 = M_Rd / (0.87f_yk z) + A_S2 = ( 666.69 × 10⁶) / (0.87 × 460 × 0.82 × 534) + 504.22 mm² = 4308.66 mm²

Provide 6X32mm TOP (A_Sprov = 4825mm²)

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