b1. Design method for longitudinal reinforcements of a rectangular section in bending at the ULS, A_{sc} and 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 rectangular distribution of the compressive stress in the concrete is assumed, see 3.1.7 (3).
5. The ultimate limit state occurs when the strain in the reinforcing steel reaches the limit ε_{ud} (Pivot A) and/or the strain in the concrete reaches the limit ε_{cu3} (Pivot B).
Considering the depth of the neutral axis/the effective depth of the cross-section:
α_{u} = x/d | (b1.1) |
The linear distribution of strains gives:
α_{u} = ε_{c}/(ε_{c} + ε_{s}) | (b1.2) |
Balanced section AB
We consider a balanced section for which the Pivot A and Pivot B are reached at the same time: ε_{c} = ε_{cu3} and ε_{s} = ε_{ud}.
For the balanced section, calculating:
α_{AB} = ε_{cu3}/(ε_{cu3} + ε_{ud}) | (b1.3) |
If α_{u} > α_{AB} ⇔ the Pivot B is reached first. Otherwise, the Pivot A is reached first.
Yield tensile steel section
At the beginning of the top branch of the design stress-strain diagram for reinforcing steel (see Figure 3.8), the strain of steel is equal to ε_{se} = f_{yd}/E_{s}.
Calculating:
α_{se} = ε_{cu3}/(ε_{cu3} + ε_{se}) | (b1.4) |
If α_{u} > α_{se} ⇔ the steel does not reach yield. In this case, the design is not economical. Therefore, a compressive reinforcement should be designed.
If α_{u} ≤ α_{se}, no compressive reinforcement is needed.
Singly reinforced section
For equilibrium, the ultimate design moment must be balanced by the moment of resistance of the section:
M_{Ed} = F_{c} z = b η f_{cd} λ x (d - λ x/2) | (b1.5) |
M_{Ed} = F_{s} z = A_{s} σ_{s} (1 - λ α_{u} /2) d ⇔ A_{s} = M_{Ed} /[(1 - λ α_{u} /2) d σ_{s}] | (b1.6) |
Substituting the following in (b1.5):
μ_{u} = M_{Ed} /(b d^{2} η f_{cd}) | (b1.7) |
(b1.5) ⇒ 2 μ_{u} = 2 λ (x/d) - λ^{2} (x/d)^{2}
⇔ 2 μ_{u} = 2 λ (α_{u}) - λ^{2} (α_{u})^{2} |
(b1.8) |
If μ_{u} ≤ 0,5, this quadratic equation has a solution:
α_{u} = [1 - (1 - 2μ_{u})^{0.5}] /λ | (b1.9) |
• α_{u} ≤ α_{AB} ⇒ Pivot A, no compressive reinforcement is needed
The strain in the tensile reinforcement ε_{s} = ε_{ud}.
By reading the design stress-strain diagram for reinforcing steel (see more in 3.2.7 (2))
⇒ the stress in the tensile reinforcement σ_{s}.
(b1.6) ⇒ the tensile reinforcement A_{s}.
• α_{AB} < α_{u} ≤ α_{se} ⇒ Pivot B, the tensile reinforcement has reached yield, no compressive reinforcement is needed
The strain in the concrete ε_{c} = ε_{cu3}.
The strain in the tensile reinforcement:
ε_{s} = ε_{cu3}⋅(d - x)/x = ε_{cu3}⋅(1 - α_{u})/α_{u} | (b1.10) |
By reading the design stress-strain diagram for reinforcing steel
⇒ the stress in the tensile reinforcement σ_{s}.
(b1.6) ⇒ the tensile reinforcement A_{s}.
If μ_{u} > 0,5, the quadratic equation (b1.8) does not have a solution.
In this case, we consider that the singly reinforced section is not suitable because the bending moment is too big. A compressive reinforcement should be calculated.
Doubly reinforced section
• α_{u} > α_{se} ⇒ Pivot B, tensile reinforcement does not reach yield, compressive reinforcement is needed;
• Or μ_{u} > 0,5
Considering that the section is an overlap of two fictive sections (see Figure b1-2):
- One concrete section with only a tensile reinforcement A_{s1} that has a strain ε_{s} = ε_{se} = f_{yd}/E_{s};
- One section with only two reinforcements A_{s2} and A_{sc}.
For the first section:
(b1.4) ⇒ α_{se}; (b1.8) ⇒ μ_{se}.
(b1.7) ⇒ the bending moment supported by this section:
M_{se} = μ_{se} (b d^{2} η f_{cd}) | (b1.11) |
(b1.6) ⇒ the tensile reinforcement A_{s1} = M_{se} /[(1 - λ α_{se} /2) d f_{yd}].
For the second section, at equilibrium:
M_{Ed} - M_{se} = A_{sc} (d - d') σ_{sc} ⇔ A_{sc} = (M_{Ed} - M_{se}) /[(d - d') σ_{sc}] | (b1.12) |
F_{s2} = F_{sc} ⇔ A_{s2} = A_{sc} σ_{sc} /σ_{s2} | (b1.13) |
The strain diagram ⇒ the strain in the compression reinforcement
ε_{sc} = ε_{cu3}⋅(α_{se} - d/d')/α_{se}.
By reading the design stress-strain diagram for reinforcing steel
⇒ the stress in the compressive reinforcement σ_{sc}.
(b1.12) ⇒ the compressive reinforcement A_{sc}.
σ_{s2} = σ_{s1} = f_{yd}
(b1.13) ⇒ the tensile reinforcement A_{s2}.
Finally, the total tensile reinforcement is:
A_{s} = A_{s1} + A_{s2} | (b1.14) |
Flowchart
Figure b1-3 resumes this design method for a rectangular reinforced concrete section in bending: