Systematic Calculation of Yield and Failure Curvatures of Reinforced Concrete Cross-Sections

Abstract

This paper examines and provides a robust solution to the problem of yield and failure curvatures of reinforced concrete (RC) cross-sections, taking into account cracking. At the same time, it calculates the corresponding necessary reinforcement or the moment of resistance in both yield and failure limit states. Computationally, the problem of determining the actual curvatures is reduced to the bending design problem of the cross-section in the yield and failure limit states. This study shows the researcher and the designer how to systematically calculate the strains for different concrete and steel grades and for standard or random cross-sections. This complex process is quite necessary to determine the respective curvatures. The main concept is presented with an emphasis on the “solution regions” as well as the critical cases of the “asymptotic regions”, both in yield and failure limit states. Our wide-ranging research on RC element design under biaxial bending with axial force for both yield and failure limit states has been completed and validated via sophisticated algorithms and is available for publication.

1. Introduction

The yield and failure curvatures, as well as the respective stiffnesses, depend on cross-section geometry, the amount and distribution of reinforcement, the reinforcement material properties, the concrete material properties, and the axial forces. The large relative displacements of the column heads in relation to their bases, and therefore those displacements of the associated diaphragms, significantly increase the characteristic periods of the building, thus resulting in low seismic accelerations. This fact is very important for the structural robustness of the building, especially when checking the resistance of an existing building structure to an earthquake.

Many researchers dealt with the specific problem in the past. Chen and Hsu [1] developed a semi-empirical formula for the curvature ductility of doubly reinforced beam sections, which, via performance-based design, takes into account the effect of reinforcement ratios as well as the reinforcement and concrete strengths. Hernández-Montes et al. [2] related the curvature ductility capacity of cross-sections designed with optimal reinforcement to those with symmetric reinforcement, for both unconfined and confined concrete cases, under varying axial loads, gross section area, and concrete strength. Chandrasekaran et al. [3] developed a closed form solution to estimate the curvature ductility of RC elements under service loads, considering the nonlinear characteristics of constitutive materials and the reinforcement ratios as required by Eurocodes. Arslan and Cihanli [4] produced a formula predicting the curvature ductility of reinforced high-strength concrete beams based on the parametric study of experimental results to evaluate the effects of various structural parameters. Lee [5] provided a prediction formula for the curvature ductility factor of doubly reinforced beam sections, taking into account the concrete strength, the tensile yield strength of steel, and the compressive ultimate strength of steel. Laterza et al. [6] performed an efficiency study of codal detailing rules for reinforcement design of primary columns and beams within the critical regions by comparing the codal design results to the measured curvature ductility. By examining the effects of spectral acceleration and a strong column factor, Zhou et al. [7] provided an empirical model in the form of a quantitative relationship between the curvature ductility demands of columns and the global displacement ductility demands of frame structures. Baji and Ronagh [8] developed a probabilistic method used to calculate curvature ductility by means of the central limit theorem, considering the specific behavior of the moment redistribution factor with respect to curvature ductility and plastic hinge length. Research on biaxial bending by Breccolotti et al. [9] produced a formula for the curvature ductility of reinforced short columns of varying section geometry, neutral axis direction, reinforcement ratios, and axial forces. Kollerathu [10] proposed an equation to evaluate and compare the curvature ductilities of reinforced masonry and RC walls, as a result of diagrams of flexural strength versus curvature. Recently, Foroughi and Yuksel [11] developed a predictive formula for the curvature ductility of doubly-reinforced beams by performing a numerical parametric study.

Finding the actual curvatures, both in yield and failure states, requires the calculation of concrete strain and steel strain under axial force and bending moment (or equivalently, reinforcement), an extremely complex computational problem with a wide range of solutions. Calculation tables were also used in the past, but they were available in failure states only and usually for specific materials. Nowadays, due to the variety of available materials and the demand for checking the actual strength and possibly retrofitting existing buildings, the design in limit states also becomes imperative. This is the reason for our extensive research to find a robust theoretical solution to the bending design problem, both in yield and failure limit states, a part of which is presented in this article.

2. Column Limit States

2.1. Column in Yield Limit State

Figure 1 presents an exaggerated model of a column in the yield limit state. Flexural cracks are perpendicular to the axis of the bar, while shear cracks have an inclination of 45° to 60° to the axis of the bar. Here, δ₁ is the displacement due to shear, which is linear and does not affect the curvature, and δ₂ is the displacement due to crack causing bending at the yield limit state.

Considering a differential length ds of the column at the side of its cross-section, the inner fiber compresses and shortens by ds∙ε_c, while the outer fiber stretches and expands by ds∙ε_s1. Then, the resulted differential central angle dθ is as follows:

$$d\theta =ds·\left({\epsilon }_c+{\epsilon }_{s1}\right)/d \mathrm{a}\mathrm{n}\mathrm{d} d\theta =ds/R$$

(1)

Equation (1) yields the following:

$$ds/R=ds·\left({\epsilon }_c+{\epsilon }_{s1}\right)/d\to {\phi }_y=1/R=\left({\epsilon }_c+{\epsilon }_{s1}\right)/d$$

(2)

where φ_y represents the actual yield curvature [12].

Note that in the yield state, it must be ε_c ≤ ε_c2 and ε_s1 ≤ ε_yd, where the yield strain for at least one of the two materials, either ε_c of the concrete or ε_s1 of the steel, has been reached.

2.2. Column in the Failure Limit State

The physical behavior of a column functioning in a failure limit state is represented in Figure 2 through the only possible observational method, which is the experimental one. The experiment, which was part of the “Anti-Seismic Thoraces” tests, took place in 1998 in the NTUA’s Reinforced Concrete Laboratory under the auspices of Professor Theodosios Tassios. It is evident that the column failure takes place in relatively small regions at the ends, while the rest of the column is in a yield state (marginal yielding with cracking) [12,13].

Figure 3 presents an exaggerated model of a column in the failure limit state. Flexural and shear cracks are apparent at the critical end regions of the column, while along the rest of the body, they remain similar to the yield limit state case.

The failure curvatures φ_u are calculated in exactly the same way as the yield limit state, as follows:

$${\phi }_{u,j}=1/R_j=\left({\epsilon }_c+{\epsilon }_{s1}\right)/d, j=\mathrm{1,2} (1=\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{d},2=\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$$

(3)

When the reinforcement of the head and base is the same, as is generally the case, then φ_u,1 = φ_u,2 = φ_u. Note again that in the failure state, it must be ε_c ≤ ε_cu2 and ε_s1 ≤ ε_ud, where the failure strain of at least one of the two materials, either ε_c of the concrete or ε_s1 of the steel, has been reached.

2.3. Example: Calculation of Limit State Curvatures

Let us consider a fixed–fixed support column of height h = 3.0 m under axial force N_d = −800 kN (see Figure 1). The cross-section is 400 mm × 400 mm, f_ck = 30 MPa, γ_c = 1.50, f_yk = 500 MPa, γ_s = 1.15, ε_su = 20‰, and Κ = 1.0 with d₁ = d₂ = 50 mm. The applied reinforcement is 4Φ20 + 4Φ14 (=1860 mm², ρ = 1.16%). It is considered that 50% of the total reinforcements are placed at the corners, while the rest are distributed along the sides (see Figure 4).

Required: the yield curvatures φ = 1/R and the moments of resistance M_Rd at the limit state:

(1)

Yield limit y.

(2)

Failure limit u.

2.3.1. Calculation of Yield Curvature

The uniaxial bending design of the cross-section yields x = 187.5 mm, M_Rd,y = 185 kN∙m, ε_c = 2.0‰, and ε_s1 = 1.734‰. Hence, Equation (2) provides the yield curvature as follows:

$${\phi }_y=\left(2.0+1.734\right)\times {10}^{-3}/(0.40-0.05)=10.67\times {10}^{-3}/\mathrm{m}\to R_y=94 \mathrm{m}$$

The respective elastic curvature (without cracks) at the base of the column is provided by the relation as follows:

$${\phi }_e=1/R=M_{Rd,y}/({\rm E}·{\rm I})$$

According to Eurocode 2 [14], §3.1.2(3), it is f_cm = f_ck + 8 = 38 MPa and E = 22 × (f_cm/10)^0.3 × 10³ = 32.8 GPa, so E∙I = 32.8 × (0.4 × 0.4³/12) = 70.0 × 10³ kN∙m². Substituting, we get:

$${\phi }_{\mathrm{e}}=185/(70.0\times {10}^3)=2.64\times {10}^{-3}/\mathrm{m}\to R_e=379 \mathrm{m}$$

Therefore, it is:

$${\phi }_y/{\phi }_e=10.67\times {10}^{-3}/2.64\times {10}^{-3}=4.04,$$

which is very important for determining the effective stiffness of a column according to Eurocode 8 [15], §4.3.1(6, 7).

2.3.2. Calculation of Failure Curvature

The uniaxial bending design of the cross-section yields x = 156.1 mm, M_Rd,u = 219 kN∙m, ε_c = 3.5‰, and ε_s1 = 4.35‰. Hence, from Equation (3), the failure curvature is as follows:

$${\phi }_u=\left(3.5+4.35\right)\times {10}^{-3}/(0.40-0.05)=22.42\times {10}^{-3}/\mathrm{m}\to R_u=45 \mathrm{m}$$

Similarly to the yield state case, the elastic curvature is as follows:

$${\phi }_{\mathrm{e}}=219/(70.0\times {10}^3)=3.13\times {10}^{-3}/\mathrm{m}\to R_e=320 \mathrm{m}$$

Therefore, it is:

$${\phi }_u/{\phi }_y=22.42\times {10}^{-3}/10.67\times {10}^{-3}=2.10 \mathrm{a}\mathrm{n}\mathrm{d} {\phi }_u/{\phi }_e=22.42\times {10}^{-3}/3.13\times {10}^{-3}=7.16.$$

3. Equilibrium of Internal and External Forces

The following relations are derived from Figure 5:

$$x=d·{\epsilon }_c/\left({\epsilon }_c+{\epsilon }_s\right),{\epsilon }_c={\epsilon }_s·x/\left(d-x\right),{\epsilon }_s={\epsilon }_c·\left(d-x\right)/x,{\epsilon }_{s2}={\epsilon }_c·\left(x-d_2\right)/x$$

(4)

$$k_F={\alpha }_{cc}·f_{cd}·b,F_c=k_F·x·\alpha ,z_c=x·\kappa$$

(5)

$$F_{s1}=A_{s1}·{\sigma }_{s1},F_{s2}=A_{s2}·{\sigma }_{s2}$$

(6)

There are two basic equations balancing the internal forces with the external forces of a cross-section under uniaxial bending with axial force (see Figure 5):

(a)

Equilibrium equation of the forces F_s1, F_c, and F_s2 with the axial force N_d

$$F_{s1}-F_c-F_{s2}=N_d$$

(7)

(b)

Equilibrium equation of the moments of the forces F_c and F_s2 with the bending moment M_d in the cross-section center and the moment of axial force N_d

$$M_d-N_d·z_{s1}=F_c·\left(d-z_c\right)+F_{s2}·\left(d-d_2\right)$$

(8)

Using the Relations (4)–(6), Equations (7) and (8) transform into:

$$A_{s1}·{\sigma }_{s1}={{A_{s2}·{\sigma }_{s2}+F}_c+N}_d$$

(9)

$${M_d=F}_c·\left(d-z_c\right)+A_{s2}·{\sigma }_{s2}·\left(d-d_2\right)+N_d·z_{s1}$$

(10)

If ρ₁ = A_s1/A_c and ρ₂ = A_s2/A_c, so ρ₂/ρ₁ = A_s2/A_s1 and A_s2 = A_s1∙ρ₂/ρ₁, then Equations (9) and (10) can be written as follows:

$$A_{s1}=\left({F_c+N}_d\right)/\left({\sigma }_{s1}-\frac{{\rho }_2}{{\rho }_1}·{\sigma }_{s2}\right)$$

(11)

$$M_d=F_c·\left(d-z_c\right)+A_{s1}·\frac{{\rho }_2}{{\rho }_1}·{\sigma }_{s2}·\left(d-d_2\right)+N_d·z_{s1}$$

(12)

It is emphasized that the axial force N_d always has a given value, independent of the above relations.

The system of Equations (11) and (12) has three unknown variables in the corresponding problem, that is, ε_c, ε_s, and A_s1 or M_d. Therefore, the system solution requires additional conditions to be set, as follows:

• First condition: the reinforcement ratio ρ₂/ρ₁ is provided.
• Second condition: either ε_c or ε_s should be in the limit state.

These two conditions, under certain assumptions, can replace the third equation. Nevertheless, the solution is rather difficult, especially in the failure state, due to numerous and complex combinations. The difficulty could be removed by using the trial solution method. However, this process would require the determination of the solution boundaries, which is also a quite complex problem.

4. Solution Regions in the Yield Limit State

The regions comprising possible solutions in the yield limit state are presented in Figure 6.

Let ρ₂/ρ₁ be the ratio of the compressive to the tensile reinforcement. For any given value of the ratio ρ₂/ρ₁, there is a characteristic case having compression zone depth x₀₁ (see Figure 6), where the denominator of Equation (11) becomes zero. That is

$${\sigma }_{s1}-\frac{{\rho }_2}{{\rho }_1}·{\sigma }_{s2}=0\to \stackrel{{\sigma }_{s1}}{\overbrace{{E_s·\epsilon }_{s1}}}-\frac{{\rho }_2}{{\rho }_1}·\stackrel{{\sigma }_{s2}}{\overbrace{E_s·{\epsilon }_{s2}}}=0\to {\epsilon }_{s1}=\frac{{\rho }_2}{{\rho }_1}·{\epsilon }_{s2}$$

Taking into account Equation (4), the above relation can be written as

$$\stackrel{{\epsilon }_{s1}}{\overbrace{{\epsilon }_c·\left(d-x_{01}\right)/x_{01}}}=\frac{{\rho }_2}{{\rho }_1}·\stackrel{{\epsilon }_{s2}}{\overbrace{{\epsilon }_c·\left(x_{01}-d_2\right)/x_{01}}}\to x_{01}=\left(d+\frac{{\rho }_2}{{\rho }_1}·d_2\right)/\left(1+\frac{{\rho }_2}{{\rho }_1}\right)$$

(13)

Thus, at location 01, corresponding to compression zone depth x₀₁ provided by Equation (13), both tensile reinforcement A_s1 and bending moment M_d will be infinite (see Equations (11) and (12)). Let us name this location existing in the yield limit state “Asymptotic Location”.

5. Solution Regions in the Failure Limit State

The regions comprising possible solutions in the failure limit state are presented in Figure 7. Region 1, where the steel reaches its failure limit [A], and region 2, where the concrete reaches its failure limit [B], are divided to subregions 1a, 1b and 2a, 2b respectively. Along the boundaries 2″A and 1″B, both M_d and A_s1 tend to infinite values.

For ρ₂/ρ₁ = 1, the denominator of Equation (11) is written as dσ = σ_s1 − σ_s2.

By observing the strain distributions in Figure 7, the following conclusions can be drawn:

(1)

The boundary 1″B of the subregions 2a and 2b, has ε_s1 = ε_yd → σ_s1 = f_yd and ε_c = ε_cu2. Since it is usually ε_s2 ≥ ε_yd → σ_s2 = f_yd, we have dσ = σ_s1 − σ_s2 = f_yd − f_yd = 0.

(2)

To the left of location 1″B will continue to be dσ = 0 until the specific location 01 with ε_s1 = ε_s2.

From the last of Equation (4), for ε_s2 = ε_yd and ε_c = ε_cu2, the compression zone depth at location 01 is as follows:

$${\epsilon }_{yd}={\epsilon }_{cu2}·(x_{01}-d_2)/x_{01}\to x_{01}=d_2·{\epsilon }_{cu2}/({\epsilon }_{cu2}-{\epsilon }_{yd})$$

(14)

Thus, for the subregion extended between locations 1″B and 01, corresponding to compression zone depths x₀₁ provided by Equation (14), both the bending moment M_d and the tensile reinforcement A_s1 will be infinite (see Equations (11) and (12)). Let us name this region existing in the failure limit state for ρ₂/ρ₁ = 1 “Asymptotic Region”. Notice that this asymptotic region is independent of the concrete class.

6. Application: The “Typical Rectangular Section” in Limit States

Consider a structural element made of C30/37 concrete and B500 steel with a 300 mm × 550 mm rectangular cross-section, as shown in Figure 8.

Provided:

b = 300 mm, h = 550 mm, d₂ = 50 mm, d₁ = 50 mm,

f_ck = 30 MPa, γ_c = 1.50, a_cc = 0.85, ε_c2 = 2.0‰, ε_cu2 = 3.5‰,

f_yk = 500 MPa, γ_s= 1.15, E_s = 200 GPa (for steel grades B500a,b,c)

Derived:

d = h − d₁ = 500 mm, z_s1 = h/2 − d₁ = 0.225 m

f_cd = f_ck/γ_c = 20.0 MPa, k_F = a_cc·b·f_cd = 0.85·0.30·20.0·10³ = 5100 kN/m,

f_yd = f_yk/γ_s = 500/1.15 = 434.78 MPa, ε_yd = f_yd/E_s=434.78/(200 × 10³) = 2.174‰

For ε_c = ε_c2 = 2‰, it is α = 0.6667 and κ = 0.375, while for ε_c = ε_cu2 = 3.5‰, it is α = 0.8095 and κ = 0.416.

For B500c steel grade, ε_ud = 20‰ with K = 1.0 is used in the simplified stress–strain diagram, while ε_ud = 67.5‰ is used with K = 1.15 in the exact stress–strain diagram.

6.1. Yield Limit State

Using the regions for the yield limit state presented in Figure 6, we form Figure 9 for the “typical rectangular cross-section”, where the strain-based region boundaries and the corresponding compression zones are clearly illustrated. We define the origin of the compression zone as the outermost upper fiber of the cross-section, while x_ij represents the compression zone depth corresponding to the location ‘ij’. For practical representation reasons, we consider the cross-section to lie horizontally.

6.1.1. Strain and Curvature Diagrams in the Yield Limit State

The diagrams of strain ε_c, ε_s and the corresponding yield curvatures φ_y are shown in Figure 10. These values are independent of the axial force N_d and the reinforcement ratio ρ₂/ρ₁.

The compression zone depth is presented on the diagrams in millimeters up to a depth of h = 550 mm, and thereafter in meters on a logarithmic scale. In practice, the values of A_s1 and M_d are required in each characteristic case of the cross-section with respect to x. These values, as obtained from Equations (11) and (12), depend on both the reinforcement ratio ρ₂/ρ₁ and the axial force N_d.

6.1.2. Asymptotic Location in Yield Limit State for ρ₂/ρ₁ = 1

For ρ₂/ρ₁ = 1, Equation (13) yields:

$$x_{01}=(500+1\times 50)/(1+1)=275 \mathrm{m}\mathrm{m} \mathrm{a}\mathrm{n}\mathrm{d} (d-x_{01})/x_{01}=0.818.$$

Since the concrete reaches its critical value in this case (i.e., ε_c = ε_c2 = 2.0‰), Equation (4) yield a tensile strain value for steel ε_s1 = 2.0‰ × 0.818 = 1.636‰. On the other hand, for ε_c = ε_c2 = 2‰, it is α = 0.6667 and κ = 0.375 (see Section 6). Consequently, from Equation (5) the compressive force F_c received by the concrete is $F_c=k_F·x_{01}·\alpha =5100\times 0.275\times 0.6667=935.0 \mathrm{k}\mathrm{N}$.

6.1.3. Solution Nomogram in Yield Limit State for ρ₂/ρ₁ = 1

The method of reinforcement with A_s2 = A_s1 (i.e., ρ₂/ρ₁ = 1) is used in cases where significant axial forces are exerted mainly on columns and/or in cases of beams with special anti-seismic requirements that entail high plasticity requirements [16].

Figure 11 presents a solution nomogram in the yield limit state for ρ₂/ρ₁ = 1, in the form of paired diagrams (M_d, A_s1) corresponding to different compression zone depths x and axial forces N_d. The asymptotic location here stands for x₀₁ = 275.0 mm, and therefore, region II is divided into two subregions (see Figure 9). For this case, it is ε_c = 2.0‰ and F_c = 935.0 kN (see results in Section 6.1.2).

At each position x, there is a specific pair (A_s1, M_d) calculated from Equations (11) and (12). For example, for bending moment M_d = 200 kN∙m and axial force N_d = 0 kN, the steel reaches the yield state first, so the required tensile reinforcement is A_s1 = 1028 mm² (see Figure 11). The respective compression zone depth is found to be x = 153.0 mm, resulting in strains ε_c = 0.958‰ and ε_s = 2.174‰, clearly indicating that the steel has reached its yield point (see Figure 10). Then, Equation (2) provides the yield curvature (also presented in Figure 10).

$${\phi }_y=\left(0.958+2.174\right)\times {10}^{-3}/(0.55-0.05)=6.26\times {10}^{-3}/\mathrm{m}.$$

6.2. Failure Limit State

Using the regions for the failure limit state presented in Figure 7, we form Figure 12 for the “typical rectangular cross-section” where the strain-based region boundaries and the corresponding compression zones are illustrated. In any algorithmic process adopted, the region boundaries should be first determined, because the upper and lower bounds of x and the corresponding values of the non-critical strain of the steel or concrete are needed. In the case of an accurate stress–strain diagram of the steel, for example, for ε_ud = 67.5‰ and K = 1.15, the region boundaries A2” and AB change significantly, but the calculation process remains the same. Furthermore, such differences in reinforcement design values are trivial in practice.

6.2.1. Strain and Curvature Diagrams in the Failure Limit State

The diagrams of strain ε_c, ε_s and the corresponding failure curvatures φ_u state are presented in Figure 13. For comparison reasons, the corresponding yield curvatures φ_y are also shown. Notice that all values are independent of the axial force N_d and the reinforcement ratio ρ₂/ρ₁. The compression zone depth x is given in millimeters up to the total depth of h = 550 mm, and from there on, in meters on a logarithmic scale. In practice, the values of A_s1 and M_d are required at each characteristic location of the cross-section with respect to x. These values, obtained from Equations (11) and (12), depend on both the reinforcement ratio ρ₂/ρ₁ and the axial force N_d.

6.2.2. Asymptotic Region in Failure Limit State for ρ₂/ρ₁ = 1

Equation (14) gives x₀₁ = 50 × 3.5/(3.50 − 2.174) = 132.0 mm, while Equation (5) give the force F_c = 5100 × 0.132 × 0.8095 = 544.9 kN received by the concrete at the location 01.

Equation (4) give x_1″B = 500 × 3.5/(3.5 + 2.174) = 308.4 mm, while Equation (5) give the force F_c = 5100 × 0.3084 × 0.8095 = 1273.2 kN received by the concrete at the location 1″B.

Thus, Equation (11) maintains a zero denominator throughout the interval between x₀₁ = 132.0 mm and x_1″B = 308.4 mm.

6.2.3. Solution Nomogram in Failure Limit State for ρ₂/ρ₁ = 1

Figure 14 presents a solution nomogram for ρ₂/ρ₁ = 1, in the form of paired diagrams (M_d, A_s1) corresponding to different compression zone depths x and axial forces N_d. At each position x, there is a specific pair (A_s1, M_sd) calculated from Equations (11) and (12). The diagram comprises the areas of dominant bending on the left and the areas of dominant compression on the far right. A multiple solution area is also apparent in the middle, theoretically extending to infinity. It should be pointed out that in our case, with ρ₂/ρ₁ = 1, there are two asymptotic boundary locations (in the sense of zeroing the denominator of Equation (11)), that is, 01 and 1″B, as determined in Section 6.2.2. Consequently, region 2a is divided into two subregions (AB, 01) and (01, 1″B) (see Figure 7 and Figure 12).

6.2.4. Indeterminacy or Multiple Solution Region in Failure Limit State

This region extends between the two asymptotic locations 01 and 1″B, corresponding to compression zone depths x₀₁ = 132 mm and x_1″B = 308.4 mm, respectively (see Figure 14).

By relating those to the respective forces, we can say that a cross-section is in the indeterminacy region when stressed by axial forces of 544.9 kN ≤ N_d ≤ 1273.2 kN (see results in Section 6.2.2).

In this region, it is σ_s1 = σ_s2. Thus, Equation (11) gives indeterminacy A_s1 = ∞ for F_c ≠ −N_d and an infinite number of solutions for F_c = −N_d. For each N_d in the region, there is a certain x that gives F_c = −N_d. This location has $x=F_c/{\alpha }_{cc}·f_{cd}·b·\alpha$, and the force F_c is exerted at the position $z_c=x·\kappa$ (where α = 0.8095 and κ = 0.416 because ε_c = ε_cu2 = 3.5‰—see values in Section 6). Furthermore, the tensile strain is ${\epsilon }_{s1}={\epsilon }_c·\left(d-x\right)/x,$while the force is $F_{s2}=A_{s2}{·\sigma }_{s2}=A_{s1}·f_{yd}$.

Since F_c = −N_d, ρ₂/ρ₁ = 1, and σ_s2 = σ_s1 = f_yd, Equation (12) yields:

$$M_d=A_{s1}·f_{yd}·\left(d-d_2\right){-N}_d·\left(d-z_c-z_{s1}\right)$$

(15)

Notice that Equation (15) directly relates A_s1 to M_d. So, when M_d is given in a problem, A_s1 is uniquely calculated, while when A_s1 is given, M_d is uniquely calculated.

Remark: In the multiple solution region, all pairs (M_d, A_s1) having the same N_d correspond to the same x, implying the same strains ε_s1 and ε_c and, hence, constant failure curvatures φ_u.

6.2.5. Application in Failure Limit State for ρ₂/ρ₁ = 1 and N_d = −1000 kN

It is x = 1000/(0.85 × 20 × 10³ × 0.30 × 0.8095) = 0.2422 m, z_c = 0.2422 × 0.416 = 0.101 m, ε_s1 = 3.5 × (0.50 − 0.2422)/0.2422 = 3.73‰, ε_c = 3.5‰, and φ_u = (3.5 + 3.73)/0.50 = 14.46‰/m.

For A_s1 = A_s2 = 0 (when reinforcement is barely required), Equation (15) yields M_d = 0 + 1000 × (0.50 − 0.101 − 0.225) = 174 kN∙m. That is, for axial force N_d = −1000 kN and moment M_d ≤ 174 kN∙m, no reinforcement is required.

For A_s1 = A_s2 = 7000 mm² (selection of maximum reinforcement value A_s1 in Figure 14), Equation (15) yields M_d = 7000 × 10⁻⁶ × 434.78 × 10³ × (0.50 − 0.05) + 1000 × (0.50 − 0.101 − 0.225) = 1544 kN∙m.

For M_d = 800 kN∙m, inverted Equation (15) yields A_s1 = [800–1000 × (0.50 − 0.101 − 0.225)]/[434.78 × 10³ × (0.50 − 0.05)] = 3.200 × 10⁻³ m² = 3200 mm². Hence, the solution is determined as a pair (800, 3200) from the infinite number of pairs of specific solutions on the line N_d = −1000 kN.

7. Conclusions

• The influence of cracking on the curvature of RC elements is significant in relation to the corresponding elastic curvature, even in the yield state, so the corresponding yield stiffness is significantly smaller than the elastic stiffness.
• For specific flexural reinforcement, the yield curvature is much smaller than the failure curvature, while the yield moment of resistance is of the same order of magnitude as the failure moment of resistance.
• During the bending design of a cross-section in the failure limit state, there is an extended region of compressive axial forces where the curvature is practically constant regardless of the acting moment. Then, for any given axial force, the necessary flexural reinforcement is derived from the acting moment based on a first-order equation.
• It is proposed that the design process should take place in the following order:

  (a)

  Calculate reinforcement in the failure state.

  (b)

  Choose and apply reinforcement.

  (c)

  Calculate failure curvatures φ_u, yield curvatures φ_y, and elastic curvatures φ_e.

  (d)

  Estimate curvature ductilities from the ratios φ_u/φ_y and/or φ_u/φ_e.

  (e)

  Determine effective stiffness and resolve.