Doubly reinforced beams are reinforced concrete beams that incorporate both compression steel (in the compression zone) and tension steel (in the tension zone). This design is typically employed when the required moment capacity of a beam exceeds that which can be provided by singly reinforced beams (those with only tension steel).  

In singly reinforced concrete beams without moment redistribution, k = M_Ed/f_ckbd² is limited to k’ = 0.167. Therefore, once the value of k > k’, the beam must be designed as a doubly reinforced beam. In doubly reinforced beams, compression steel reinforcements are required because the concrete section (in compression) by itself cannot develop the required moment of resistance.

However, when there is a moment redistribution in the section, k > k’ where k’ is dependent on the moment redistribution ratio (δ) and is given by:

k’ = 0.6δ – 0.18δ² – 0.21 where δ ≤ 1.0

In all cases, it is recommended that k’ be limited to 0.167 to ensure ductile failure. It is important to understand that the redistribution of moments derived from elastic analysis of a reinforced concrete structure accounts for the plastic behaviour of the material as it approaches its ultimate limit state.

To facilitate this plastic response, concrete sections must be designed to permit the formation of plastic hinges upon the yielding of tensile reinforcement. This approach ensures a ductile structure characterized by a gradual failure at the ultimate limit state, rather than a sudden, catastrophic collapse due to compressive failure of the concrete.

To guarantee the formation of plastic hinges and the necessary section rotation for ductile behaviour, Eurocode 2 imposes limitations on the maximum neutral axis depth, x_bal. These restrictions ensure that the tensile steel experiences high strains and this promotes the desired plastic response.

The permissible value of x_bal is determined based on the desired degree of moment redistribution. According to Eurocode 2, the limit is x_bal ≤ 0.8(δ – 0.44)d for f_ck ≤ C50. However, this limit was modified to x_bal as x_bal ≤ (δ – 0.4)d in the UK Annex to the EC2.

Design of Doubly Reinforced Beam

To ensure a ductile response and prevent sudden compressive failure of the concrete, it is required that the neutral axis depth should not exceed 0.45 times the effective depth (0.45d). This guideline is typically adopted in the design of sections incorporating compression reinforcement (doubly reinforced beam sections).

In the design of reinforced concrete beams, if the design ultimate moment is greater than the ultimate moment of resistance i.e. M_Ed > M_lim, 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 steel in compression, A_s2, is given by:
A_s1 = (M_Ed – M_lim) / (f_sc(d – d₂)) —- (1)
f_sc = E_sε_sc ≤ 0.87f_yk
ε_sc = steel compressive strain = 0.0035(x – d’)/x
E_s = Modulus of elasticity of steel = 200000 N/mm²

Area of tension reinforcement:
A_s1 = M_lim / (0.87f_ykz) + A_s2 (f_sc/0.87f_yk) —- (2)

Where:
M_Ed = Design bending moment of the section
M_lim = 0.167f_ckbd²
z = d[0.5+ √(0.25 – 0.882k’)]
where
k’ = 0.167

Design Example of a Doubly Reinforced Beam

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 600 mm, and the width is 400 mm.
f_ck = 35 N/mm² ; f_yk = 500 N/mm²; concrete cover = 40mm

Top reinforcement (Hogging moment)
Support A (No moment redistribution)
Effective depth (d) = 600 – 40 – 16 – 10 = 534 mm
d₂ = 40 + 8 + 10 = 58 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_ckb_wd²) = (761.24 × 10⁶) / (35 × 400 × 534²) = 0.1906
z = d[0.5+ √((0.25 – 0.882(0.167))] = 0.82d = 0.82 × 534 = 437.88 mm
x = (d − z)/0.4 = (534 − 437.88)/0.4 = 240.3 mm

d₂/x = 58/240.3 = 0.241 < 0.38 (therefore reinforcement bar has yielded and f_sc = 0.87f_yk)

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

M_lim = 0.167f_ckbd² = (0.167 × 35 × 400 × 534²) × 10^-6 = 666.69 kNm

A_S2 = ((761.24 – 666.69) × 10⁶) / (0.87 × 500 × (534 – 58)) = 457 mm²
Provide 3H16 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 × 500 × 0.82 × 534) + 457 mm² = 3957 mm²

Provide 3H32mm + 4H25 mm Top (A_Sprov = 4376 mm²)

Eurocode 2 uses the variable strut inclination method for shear design. It is slightly more complex than the procedure in BS 8110 but can result in savings in the amount of shear reinforcement required.

Like BS 8110, Eurocode 2 models shear behaviour using the truss analogy in which the concrete acts as the diagonal struts, the stirrups act as the vertical ties, the tension reinforcement forms the bottom chord, and the compression steel/ concrete forms the top chord. Whereas in BS 8110 the strut angle has a fixed value of 45°, in EC2 it can vary between 21.8° and 45° and it is this feature which is responsible for possible reductions in the volume of shear reinforcement.

The concrete strut angle changes with respect to the magnitude of the shear force applied. The angle decreases with a reduction in shear stress, with a minimum angle of 21.8° (where cot = 2.5) which happens to be the most frequently used design strut angle.

Shear design in Eurocode 2 often involves comparing the design shear stress (v_Ed) to the concrete shear resistance (v_Rd,c). If v_Ed < v_Rd,c then in the case of one-way spanning slabs no shear reinforcement is required; in the case of beams nominal shear reinforcement is required. v_Rd,c is dependent on the amount of design tensile reinforcement and the design concrete strength.

If v_Ed < v_Rd,max (ie v_{Rd,cot θ = 2.5}), then the shear reinforcement may be calculated as:
A_sw/s ≥ v_Ed/(0.87 × f_yk × z × cot 2.5)

If v_Ed > v_Rd,max (ie v_{Rd,cot θ = 2.5}), then checks against v_{Rd,cot θ = 1.0} and determination of θ are required.
If v_Ed > v_{Rd,cot θ = 1.0} then a larger section or greater concrete strength is required.

Where applied shear stresses are higher, one has to check against the maximum allowable shear stress at the maximum strut angle of θ = 45º (e.g. 5.28 MPa for a C30/37 concrete when cot θ = 1.0) and where necessary determine the actual intermediate strut angle, θ.

Solved Example for Shear Design

For the beam loaded as shown below and disregarding load factors, we are going to obtain the shear reinforcement required for section A. The shear force at the centreline of support has been adopted for this design, and the area of tension steel provided at this section A_s1 = 4825 mm².

Design Data:
Concrete Strength = 35 N/mm²
Grade of steel = 460 N/mm²
Width of beam = 400mm
Depth of beam = 600mm
Effective depth = 543mm
Area of tension steel provided at section A_s1 = 4825 mm²

Shear Design to EC2

Support A; V_Ed = 500.46 kN

V_Rd,c = [C_Rd,c.k.(100ρ₁ f_ck)^(1/3) + k₁.σ_cp]b_w.d ≥ (V_min + k₁.σ_cp) b_w.d

C_Rd,c = 0.18/γ_c = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/543) = 1.606 > 2.0, therefore, k = 1.606
V_min = 0.035k^(3/2) f_ck^0.5
V_min = 0.035 × (1.606)^1.5 × 35^0.5 = 0.421 N/mm²
ρ₁ = A_s/bd = 4825/(400 × 543) = 0.022 > 0.02; Therefore take 0.02

σ_cp = N_Ed/Ac < 0.2f_cd (Where N_Ed is the axial force at the section, A_c = cross-sectional area of the concrete), f_cd = design compressive strength of the concrete.) Take N_Ed = 0

V_Rd,c = [0.12 × 1.606 (100 × 0.02 × 35 )^(1/3)] 400 × 543 = 172511.992 N = 172.511 kN

Since V_Rd,c (172.511 kN) < V_Ed (500.46 kN), shear reinforcement is required.
The compression capacity of the compression strut (V_Rd,max) assuming θ = 21.8° (cot θ = 2.5)

V_Rd,max = (b_w.z.v₁.f_cd) / (cot⁡θ + tanθ)
V₁ = 0.6(1 – f_ck/250) = 0.6(1 – 35/250) = 0.516
f_cd = (α_cc f_ck) / γ_c = (0.85 × 35) / 1.5 = 19.833 N/mm²
Let z = 0.9d

V_Rd,max = [(400 × 0.9 × 543 × 0.516 × 19.833) / (2.5 + 0.4)] × 10^-3 = 689.83 kN

Since V_Rd,c < V_Ed < V_Rd,max

Hence A_sw / S = V_Ed / (0.87.F_yk.z.cot θ) = 500460 / (0.87 × 460 × 0.9 × 543 × 2.5 ) = 1.0235

Minimum shear reinforcement;
A_sw / S = ρ_w,min × b_w × sinα (α = 90° for vertical links)
ρ_w,min = (0.08 × √(F_ck)) / F_yk = (0.08 × √35) / 460 = 0.001028
A_sw/S_min = 0.001028 × 400 × 1 = 0.411
Maximum spacing of shear links = 0.75d = 0.75 × 543 = 407.25

Provide X10mm @ 150mm c/c as shear links (A_sw/S = 1.0467) Ok!!!!

Shear Design to BS 8110-1:1997

Design of support A
Ultimate shear force at the centerline of support
V = 500.46 kN

Using the shear force at the centreline of support;
Shear stress v = V/(bd ) = (500.46 × 10³) / (400 × 543) = 2.304 N/mm²

v(2.304 N/mm² ) < 0.8√f_cu (4.732 N/mm² ). Hence, the dimensions of the cross-section are adequate for shear.

Concrete resistance shear stress
v_c = 0.632 × (100A_s/bd)^1/3 (400/d)^1/4
(100A_s/bd) = (100 × 4825) / (400 × 543) = 2.221 < 3 (See Table 3.8 BS 8110-1;1997)
(400/d)^1/4 = (400/543)^1/4 = 0.926; But for members with shear reinforcement, this value should not be less than 1. Therefore take the value as 1.0

v_c = 0.632 × (2.221)^1/3 × 1.0 = 0.8245 N/mm²
For f_cu = 35 N/mm², v_c = 0.8245 × (35/25)^1/3 = 0.922 N/mm²

Let us check;
(V_c + 0.4)1.322 N/mm² < v(2.304 N/mm²) < 0.8√f_cu (4.732 N/mm²)

Therefore, provide shear reinforcement links.
Let us try 2 legs of Y10mm bars (Area of steel provided = 157 mm²

A_sv/Sv = (0.95 × F_yv) / (b_v (v – v_c)) = [400 × (2.304 – 0.922)] / (0.95 × 460) = 1.264

Maximum spacing = 0.75d = 0.75 × 543 = 407.25 mm
Provide Y10mm @ 100 mm c/c links as shear reinforcement (A_sv/Sv = 1.57)

We can therefore see that disregarding load factor and flexural design requirements, EC2 is more economical than BS 8110 in shear design, and in this case study by about 19%.