b10. Design method for tensile reinforcement of a rectangular section in bending at the SLS with known compressive reinforcement, A_{s}
Eurocode 2  Design of concrete sections
Assumptions
1. Plane sections remain plane after straining, so that there is a linear distribution of strains across the section.
2. Reinforcing steels have the same deformation as the nearby concrete.
3. The tensile strength of concrete is ignored.
4. A triangular distribution of the compressive stress in the concrete is assumed.
5. The serviceability limit state occurs when the tensile stress in the reinforcement reaches the limit σ_{s,ser} = k_{3} f_{yk} (Pivot A) and/or the compressive stress in the concrete reaches the limit σ_{c,ser} = k_{1} f_{ck} (Pivot B). The parameters k_{1} and k_{3} are chosen by National Annex, see § 7.2 (2) and § 7.2 (5) respectively.
Considering the depth of the neutral axis/the effective depth of the crosssection:
α_{ser} = x/d  (b10.1) 
The linear distribution of strains and the stress diagram gives:
α_{ser} = ε_{c}/(ε_{c} + ε_{s}) = n_{e} σ_{c}/(n_{e} σ_{c} + σ_{s})  (b10.2) 
where:
 n_{e}

is the effective modular ratio
n_{e} = E_{s} / E_{c,eff}with:
 E_{s}
 the design value of the modulus of elasticity of the reinforcing steel, see § 3.2.7 (4)
 E_{c,eff}
 the effective modulus of elasticity for concrete.
Balanced section AB
We consider a balanced section for which the Pivot A and Pivot B are reached at the same time: σ_{s} = σ_{s,ser} and σ_{c} = σ_{c,ser}.
For the balanced section, calculating:
α_{AB} = n_{e} σ_{c,ser}/(n_{e} σ_{c,ser} + σ_{s,ser})  (b10.3) 
If α_{ser} > α_{AB} ⇔ the Pivot B is reached first. Otherwise, the Pivot A is reached first.
Serviceability limit state at Pivot B
For equilibrium, the ultimate design moment must be balanced by the moment of resistance of the section:
M_{ser} = F_{c} z = 0,5 b σ_{c,ser} x (d  x/3) = 0,5 b d^{2} σ_{c,ser} α_{c,ser} (1  α_{c,ser}/3)  (b10.4) 
Substituting the following in (b10.4):
μ_{c,ser} = M_{ser} /(b d^{2} σ_{c,ser})  (b10.5) 
(b10.4) ⇒ μ_{c,ser} = 0,5 α_{c,ser} ( 1  α_{c,ser}/3)
⇔ α_{c,ser}^{2}  3 α_{c,ser} + 6 μ_{c,ser} = 0  (b10.6) 
If μ_{c,ser} ≤ 0,375, this quadratic equation has a solution:
α_{c,ser} = 1,5 [1  (1  8 μ_{c,ser}/3)^{0.5}]  (b10.7) 
Serviceability limit state at Pivot A
For equilibrium, the ultimate design moment must be balanced by the moment of resistance of the section:
M_{ser} = F_{c} z = 0,5 b σ_{c} x (d  x/3) = 0,5 b d^{2} σ_{c} α_{s,ser} (1  α_{s,ser}/3)  (b10.8) 
The stress σ_{c} is deducted from the stress diagram:
σ_{c} = σ_{s,ser} α_{s,ser} /[n_{e} (1  α_{s,ser})]  (b10.9) 
Calculating:
μ_{s,ser} = M_{ser} /(b d^{2} σ_{s,ser})  (b10.10) 
Substituting (b10.9) and (b10.10) in (b10.8),
(b10.8) ⇒ μ_{s,ser} = 0,5 n_{e} α_{s,ser}^{2} ( 1  α_{s,ser}/3)
⇔ α_{s,ser}^{3}  3 α_{s,ser}^{2}  6 n_{e} μ_{s,ser} α_{s,ser} + 6 n_{e} μ_{s,ser} = 0  (b10.11) 
By solving this cubic equation, we get a root α_{s,ser} ∈ (0, 1).
Two fictive sections
Considering that the section is an overlap of two fictive sections (see Figure b102):
 One concrete section with only a tensile reinforcement A_{s1};
 One section with only two reinforcements A_{s2} and A_{sc}.
Iterative calculation of the stress σ_{sc}
Assumimg the stress in the compressive reinforcement σ_{sc} = f_{yd} = f_{yk}/γ_{s}.
The bending moment M_{s} supported by the section without compression reinforcement:
M_{s} = M_{ser}  A_{sc} σ_{sc} (d  d')  (b10.12) 
Hence:
μ_{c,ser} = M_{s} /(b d^{2} σ_{c,ser}) 
α_{c,ser} = 1,5 [1  (1  8 μ_{c,ser}/3)^{0.5}] 
• α_{c,ser} ≤ α_{AB} ⇒ Pivot A
μ_{s,ser} = M_{s} /(b d^{2} σ_{s,ser}) 
(b10.11) ⇒ α_{s,ser} 
Considering α_{ser} = α_{s,ser}.
The stress diagram ⇒ the compressive stress:
σ_{sc*} = σ_{s,ser} (α_{ser}  d'/d)/(1  α_{ser})  (b10.13) 
• α_{u} > α_{AB} ⇒ Pivot B
Considering α_{ser} = α_{c,ser}.
The stress diagram ⇒ the compressive stress:
σ_{sc*} = n_{e} σ_{c,ser} (α_{ser}  d'/d)/α_{ser}  (b10.14) 
This calculated value of σ_{sc*} must be compared with the initial value σ_{sc}. If the difference is greater than 5 %, we start over with a lower value of σ_{sc}. If the difference is lower than or equal to 5 %, we accept the initial value σ_{sc} as the stress in the compressive reinforcement.
Reinforcement A_{s}
For the first fictive section without compression steel, the tensile stress in the reinforcement σ_{s} is equal to σ_{s,ser} in case of Pivot A, is equal to n_{e} σ_{c,ser} (1  α_{ser})/α_{ser} in case of Pivot B.
The tensile reinforcement A_{s1}:
A_{s1} = M_{s} /[(1  α_{ser} /3) d σ_{s}]  (b10.15) 
The second fictive section at equilibrium gives the tensile reinforcement A_{s2}:
A_{s2} = A_{sc} σ_{sc} /σ_{s}  (b10.16) 
Finally, the total tensile reinforcement is:
A_{s} = A_{s1} + A_{s2}  (b10.17) 
Flowchart
Figure b103 resumes this design method for a rectangular reinforced concrete section in bending at the SLS with known compressive reinforcement: