2. Benchmark solution

2.1. Calculation method

The densities of the longitudinal steels are calculated in accordance with the 3 Pivots method (A, B, C).

2.2. Reference quantities and results

2.2.1. Calculations at ELU

Details of the calculation will be given with respect to the case by default (Configuration 1 + Configuration 2 for cases where compression steel is required).

2.2.1.1. Charging case 1

The beam is subjected to a compression of \(N=-1000000\text{N}\) and to a shear force of \({V}_{z}=100000\text{N}\).

Longitudinal reinforcement:

The compressive strength of unreinforced concrete is given by:

\({N}_{\mathit{Rd}}={A}_{c}\times {f}_{\mathit{cd}}=\mathrm{0,30}m\times \mathrm{0,50}m\times (\frac{35\mathit{MPa}}{\mathrm{1,5}})=\mathrm{3,5}\text{MN}>1\text{MN}\)

Therefore, the section is in Pivot C (case of uniform deformation) such that the concrete alone can withstand the force ⇒ no longitudinal reinforcement required.

\({A}_{\mathit{SYS}}={A}_{\mathit{SYI}}={A}_{\mathit{SZS}}={A}_{\mathit{SZI}}=0{\mathit{cm}}^{2}\)

Transverse reinforcement:

In Eurocode 2, the calculation of the transverse reinforcement can take into account the normal compression force and it is a question of starting by checking the resistance of the non-reinforced concrete transversely to the shear force, given by the formula below:

\({V}_{\mathit{Rd},c}=[\mathit{max}({C}_{\mathit{Rd},c}k{(100{\mathrm{\rho }}_{l}{f}_{\mathit{ck}})}^{1/3};{v}_{\mathit{min}})+{k}_{1}{\mathrm{\sigma }}_{\mathit{cp}}]\times {b}_{w}d\)

That is, if we take into account the impact of normal effort:

\({V}_{\mathit{Rd},c}=[\mathit{max}(\frac{\mathrm{0,18}}{\mathrm{1,5}}\cdot (1+\sqrt{\frac{200}{460}})\cdot {(100\times 0\times 35)}^{1/3};\frac{\mathrm{0,053}}{\mathrm{1,5}}\cdot (1+\sqrt{\frac{200}{460}})\cdot {35}^{\mathrm{0,5}})+\mathrm{0,15}\times (\frac{\mathrm{1,0}}{\mathrm{0,3}\times \mathrm{0,5}})]\times \mathrm{0,3}\times \mathrm{0,46}\)

\({V}_{\mathit{Rd},c}=185868N\)

This term is greater than the value of the shear force applying to the section, and which is equal to 100,000 N.

Therefore, no cross reinforcement is required.

If we do not take into account the beneficial contribution of normal effort, we will then have:

\({V}_{\mathit{Rd},c}=69884N\)

Which then becomes less than the shear force involving the section, namely \({V}_{\mathit{Ed}}=100000N\); we then go through the Ritter-Mosche lattice modeling given at the Eurocode 2 level, by considering:

• \({\mathrm{\theta }}_{b}=\mathrm{21,8}°\) - angle of inclination of concrete compression rods

• \({V}_{\mathit{Rd},\mathit{max}}=\frac{{\mathrm{\alpha }}_{\mathit{cw}}{b}_{w}z{\mathrm{\nu }}_{1}{f}_{\mathit{cd}}}{\mathrm{tan}({\mathrm{\theta }}_{b})+\mathrm{cot}({\mathrm{\theta }}_{b})}=\frac{\mathrm{1,0}\times \mathrm{0,3}\times (\mathrm{0,9}\times \mathrm{0,46})\times \mathrm{0,6}\times (35/\mathrm{1,5})}{\mathrm{tan}(\mathrm{21,8}°)+\mathrm{cot}(\mathrm{21,8}°)}=564798N>100000N\) - checking that the connecting rods have not been crushed

• \({V}_{\mathit{Rd},s}=(\frac{{A}_{\mathit{sw}}}{s})z{f}_{\mathit{yd}}/\mathrm{tan}({\mathrm{\theta }}_{b})\) - resistance of the transverse reinforcement frames to be expected. In order to verify the stability of the mesh, the dimensioning then leads to the installation of: \({V}_{\mathit{Rd},s}={V}_{\mathit{Ed}}\); or: \(\frac{{A}_{\mathit{sw}}}{s}=\frac{{V}_{\mathit{Ed}}}{z{f}_{\mathit{yd}}/\mathrm{tan}({\mathrm{\theta }}_{b})}=\mathrm{3,93}{\mathit{cm}}^{2}/m\)

2.2.1.2. Charging case2

The beam is subjected to a normal tensile force of \(N=1000000\text{N}\) and to a shear force of \({V}_{z}=-100000\text{N}\).

Longitudinal reinforcement:

It is a section that is fully stretched symmetrically.

The steel section is therefore equal to \({A}_{S}=\frac{N}{{f}_{\mathit{yd}}}=\frac{N}{({f}_{\mathit{yk}}/{\mathrm{\gamma }}_{s})}\) (Pivot A — Case of uniform deformation).

Each frame bed therefore takes up a quarter of the effort, i.e.:

\({A}_{\mathit{SYS}}={A}_{\mathit{SYI}}={A}_{\mathit{SZS}}={A}_{\mathit{SZI}}=\frac{{A}_{S}}{4}=\mathrm{5,75}{\mathit{cm}}^{2}\)

F**transverse reinforcement:**

Same as the previous case, in the absence of a compression force:

\(\frac{{A}_{\mathit{sw}}}{s}=\mathrm{3,93}{\mathit{cm}}^{2}/m\)

2.2.1.3. Charging case 3

The beam is subjected to a normal tensile force of \(N=1000000\text{N}\), a shear force \({V}_{z}=-100000\text{N}\) and a torsional moment \({M}_{T}=-10000\text{Nm}\)

Longitudinal reinforcement:

Same as the previous case: \({A}_{\mathit{SYS}}={A}_{\mathit{SYI}}={A}_{\mathit{SZS}}={A}_{\mathit{SZI}}=\mathrm{5,75}{\mathit{cm}}^{2}\)

F**transverse reinforcement:**

At Eurocode 2, the aim is to start by verifying the resistance of sheared concrete to the coupling of shear force and torsional moment:

• for the shear force:

\({V}_{\mathit{Rd},c}=[\mathit{max}({C}_{\mathit{Rd},c}k{(100{\mathrm{\rho }}_{l}{f}_{\mathit{ck}})}^{1/3};{v}_{\mathit{min}})+{k}_{1}{\mathrm{\sigma }}_{\mathit{cp}}]\times {b}_{w}d\)

Either:

\({V}_{\mathit{Rd},c}=[\mathit{max}(\frac{\mathrm{0,18}}{\mathrm{1,5}}\cdot (1+\sqrt{\frac{200}{460}})\cdot {(100\times \mathrm{8,846}\times {10}^{-3}\times 35)}^{1/3};\frac{\mathrm{0,053}}{\mathrm{1,5}}\cdot (1+\sqrt{\frac{200}{460}})\cdot {35}^{\mathrm{0,5}})]\times \mathrm{0,3}\times \mathrm{0,46}\)

\({V}_{\mathit{Rd},c}=91948N\)

• for twisting:

\({T}_{\mathit{Rd},c}=2{f}_{\mathit{ctd}}{t}_{k}{a}_{k}\)

Either:

\({T}_{\mathit{Rd},c}=2\times (\frac{\mathrm{2,25}}{\mathrm{1,5}})\times (\mathrm{9,375}\times {10}^{-2})\times (\mathrm{8,379}\times {10}^{-2})\)

\({T}_{\mathit{Rd},c}=23534\mathit{Nm}\)

• In other words, the strength of sheared concrete is calculated as follows:

\(\frac{{V}_{\mathit{Ed}}}{{V}_{\mathit{Rd},c}}+\frac{{T}_{\mathit{Ed}}}{{T}_{\mathit{Rd},c}}=\frac{100000}{91948}+\frac{10000}{23534}=\mathrm{1,512}>\mathrm{1,0}\)

Consequently, transverse reinforcement is required; we then use the Ritter-Mosche truss modeling given at the Eurocode 2 level, considering:

• \({\mathrm{\theta }}_{b}=\mathrm{21,8}°\) - angle of inclination of concrete compression rods

• \({V}_{\mathit{Rd},\mathit{max}}=\frac{{\mathrm{\alpha }}_{\mathit{cw}}{b}_{w}z{\mathrm{\nu }}_{1}{f}_{\mathit{cd}}}{\mathrm{tan}({\mathrm{\theta }}_{b})+\mathrm{cot}({\mathrm{\theta }}_{b})}=564798N\) and

\({T}_{\mathit{Rd},\mathit{max}}=2\mathrm{\nu }{\mathrm{\alpha }}_{\mathit{cw}}{f}_{\mathit{cd}}{a}_{k}{t}_{k}\mathrm{sin}({\mathrm{\theta }}_{b})\mathrm{cos}({\mathrm{\theta }}_{b})=65222\mathit{Nm}\)

Yes, \(\frac{{V}_{\mathit{Ed}}}{{V}_{\mathit{Rd},\mathit{max}}}+\frac{{T}_{\mathit{Ed}}}{{T}_{\mathit{Rd},\mathit{max}}}=\frac{100000}{564798}+\frac{10000}{65222}=\mathrm{0,33}<\mathrm{1,0}\) - checking that the connecting rods have not crashed

• \({V}_{\mathit{Rd},s}=(\frac{{A}_{\mathit{sw}}}{s})z{f}_{\mathit{yd}}/\mathrm{tan}({\mathrm{\theta }}_{b})\) - resistance of the transverse reinforcement frames to be expected. In order to verify the stability of the mesh, dimensioning then leads to the installation of: \({V}_{\mathit{Rd},s}={V}_{\mathit{Ed}}+2{V}_{\mathit{EdT}}\), such that \({V}_{\mathit{EdT}}=\frac{{T}_{\mathit{Ed}}}{2{a}_{k}}\times (h-{t}_{k})=12308N\) is:

\(\frac{{A}_{\mathit{sw}}}{s}=\frac{{V}_{\mathit{Ed}}+2{V}_{\mathit{EdT}}}{z{f}_{\mathit{yd}}/\mathrm{tan}({\mathrm{\theta }}_{b})}=\mathrm{4,899}{\mathit{cm}}^{2}/m\)

2.2.1.4. Charging case4

The beam is subjected to a bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=100000\text{Nm}\). This flexural moment corresponds to a stretched lower fiber.

At the ELU:

The ultimate moment reduced \(\mathrm{\mu }=\frac{{M}_{\mathit{FZ}}}{hd\mathrm{²}\mathrm{\eta }{f}_{\mathit{cd}}}=\frac{\mathrm{0,1}}{\mathrm{0,3}\cdot \mathrm{0,46²}\cdot \mathrm{1,0}\cdot (35/\mathrm{1,5})}=\mathrm{0,0675}\).

The relative position of neutral fiber \(\mathrm{\alpha }=1-\sqrt{1-2\mathrm{\mu }}=\mathrm{0,08745}\).

The \(y=d(1-\frac{\mathrm{\alpha }}{2})=\mathrm{0,439}\) reduced lever arm.

The frame section is therefore equal to \({A}_{\mathit{SYI}}=\frac{{M}_{\mathit{FZ}}}{y{\mathrm{\sigma }}_{s}}=\mathrm{5,181}\text{cm²}\) (lower \(\text{Y}\) bed).

2.2.1.5. Charging case5

The beam is subjected to a bending moment \(\text{Y}\) equal to \({M}_{\mathit{FY}}=-100000\text{Nm}\). This bending moment corresponds to a stretched straight fiber.

At the ELU:

The ultimate moment reduces: math: mathrm {mu} =frac {mu} =frac {left| {mathit {FY}}}right|} {{b} _ {w} dmathrm {²}}mathrm {²}}mathrm {²}}mathrm {²}}mathrm {²}mathrm {²}mathrm {²}mathrm {²}mathrm {²}mathrm {²}mathrm {²}mathrm {²}mathrm {²}}mathrm {2}}mathrm {2}}mathrm {eta} {f} | {0.5}cdotmathrm {0.26²}cdotmathrm {1.0}cdot (35/mathrm {1.5})} =mathrm {0.26²}} =mathrm {0.127}.

The relative position of neutral fiber \(\mathrm{\alpha }=1-\sqrt{1-2\mathrm{\mu }}=\mathrm{0,136}\).

The \(z=d(1-\frac{\mathrm{\alpha }}{2})=\mathrm{0,242}\) reduced lever arm.

The frame section is therefore equal to:math: {A} _ {mathit {SZS}} =frac {left| {M} _ {mathit {FY}}right|} {z {mathrm {sigma}} _ {sigma}}} _ {sigma}}} _ {s}} =mathrm {9,492}text {cm²} (lit:math: text {Z}} superior).

2.2.1.6. Charging case 6

The beam is subjected to a following bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-100000\text{Nm}\) and to a normal tensile force equal to \(N=100000\text{N}\).

The section is partially stretched.

At the ELU:

The moment to resume is:math: `M=left| {M} | {M} _ {mathit {FZ}}right|-N (d-frac {{b} _ {w}} {w}} {2}) =79000text {Nm} `

The ultimate moment reduced \(\mathrm{\mu }=\frac{M}{hd\mathrm{²}\mathrm{\eta }{f}_{\mathit{cd}}}=\mathrm{0,0533}\).

The relative position of neutral fiber \(\mathrm{\alpha }=1-\sqrt{1-2\mathrm{\mu }}=\mathrm{0,0548}\).

The \(y=d(1-\frac{\mathrm{\alpha }}{2})=\mathrm{0,447}\) reduced lever arm.

The frame section is therefore equal to \({A}_{\mathit{SYS}}=\frac{M}{y{\mathrm{\sigma }}_{s}}+\frac{N}{{\mathrm{\sigma }}_{s}}=\mathrm{6,361}\text{cm²}\) (upper \(\text{Y}\) bed).

2.2.1.7. Charging case7

The beam is subjected to a following bending moment \(\text{Y}\) equal to \({M}_{\mathit{FY}}=-100000\text{Nm}\) and to a normal tensile force equal to \(N=100000\text{N}\).

The section is partially stretched.

At the ELU:

The moment to resume is:math: `M=left| {M} | {M} _ {mathit {FY}}right|-N (d-frac {h} {2}) =89000text {Nm} `

The ultimate moment reduced \(\mathrm{\mu }=\frac{M}{{b}_{w}d\mathrm{²}\mathrm{\eta }{f}_{\mathit{cd}}}=\mathrm{0,1128}\).

The relative position of neutral fiber \(\mathrm{\alpha }=1-\sqrt{1-2\mathrm{\mu }}=\mathrm{0,120}\).

The \(z=d(1-\frac{\mathrm{\alpha }}{2})=\mathrm{0,244}\) reduced lever arm.

The frame section is therefore equal to \({A}_{\mathit{SZS}}=\frac{M}{z{\mathrm{\sigma }}_{s}}+\frac{N}{{\mathrm{\sigma }}_{s}}=\mathrm{10,676}\text{cm²}\) (upper \(\text{Z}\) bed).

2.2.1.8. Charging case 8

The beam is subjected to a following bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-100000\text{Nm}\) and to a normal tensile force equal to \(N=2000000\text{N}\).

The section is completely extended:math: M=| {M} _ {mathit {FZ}} |-N (d-frac {{b} _ {w}} {2}) =-320000text {Nm} <0

At the ELU:

The reinforcement cross section is therefore equal to:

\({A}_{\mathit{SYS}}=\frac{M}{(d-{b}_{w}){\mathrm{\sigma }}_{s}}+\frac{N}{{\mathrm{\sigma }}_{s}}=\mathrm{17,524}\text{cm²}\) (\(\text{Y}\) upper bed).

\({A}_{\mathit{SYI}}=\frac{-M}{(d-{b}_{w}){\mathrm{\sigma }}_{s}}=\mathrm{28,476}\text{cm²}\) (lower \(\text{Y}\) bed).

2.2.1.9. Charging case 9

The beam is subjected to a following bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=100000\text{Nm}\) and to a following bending moment \(\text{Y}\) equal to \({M}_{\mathit{FY}}=-150000\text{Nm}\). It is therefore in deflected bending, and the resolution is carried out according to the iterative method based on the verification of the Bresler inequality:

• We start by performing a simple bending calculation along the Z axis: we obtain a reinforcement similar to CAS 4, namely \({A}_{\mathit{SYI}}=\mathrm{5,181}\text{cm²}\)

• Idem, we perform a simple bending calculation along the Y axis: we obtain the following reinforcement \({A}_{\mathit{SZS}}=\mathrm{14,85}\text{cm²}\)

• The normal resisting force, as well as the resisting bending moments are calculated using the interaction diagrams constructed from the reinforcement calculated previously; we obtain:

\({N}_{\mathit{Rd}}=4370905N\), \({M}_{\text{Rd,z}}=99888\mathit{Nm}\), and \({M}_{\text{Rd,y}}=-149500\mathit{Nm}\)

• The Bresler equation is then written:

\(\mathit{BRES}={(\frac{{M}_{\mathit{FZ}}}{{M}_{\text{Rd,z}}})}^{a=f(N/{N}_{\mathit{Rd}})}+{(\frac{{M}_{\mathit{FY}}}{{M}_{\text{Rd,y}}})}^{a}={(\frac{100000}{99888})}^{\mathrm{1,0}}+{(\frac{150000}{149500})}^{\mathrm{1,0}}\approx \mathrm{2,0}\) (which is normal, since this reinforcement pair is the result of a decoupled calculation (according to Y and Z) of sizing).

• The iteration algorithm then consists in increasing the reinforcement sections, so that the sum above falls below 1:

• \({A}_{\mathit{SYI}}=\mathrm{5,181}\text{cm²}\) and \({A}_{\mathit{SZS}}=\mathrm{14,85}\text{cm²}\) ⇒ \({M}_{\text{Rd,z}}=99888\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-149500\mathit{Nm}\) ⇒ BRES = 2.0

• \({A}_{\mathit{SYI}}=\mathrm{5,699}\text{cm²}\) and \({A}_{\mathit{SZS}}=\mathrm{16,335}\text{cm²}\) ⇒ \({M}_{\text{Rd,z}}=109467\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-162435\mathit{Nm}\) ⇒ ⇒ BRES = 1,837

• \({A}_{\mathit{SYI}}=\mathrm{6,269}\text{cm²}\) and \({A}_{\mathit{SZS}}=\mathrm{17,968}\text{cm²}\) ⇒ \({M}_{\text{Rd,z}}=119920\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-176233\mathit{Nm}\) ⇒ ⇒ BRES = 1.685

…………

…………

• \({A}_{\mathit{SYI}}=\mathrm{12,217}\text{cm²}\) and \({A}_{\mathit{SZS}}=\mathrm{35,015}\text{cm²}\) ⇒ \({M}_{\text{Rd,z}}=223619\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-293740\mathit{Nm}\) ⇒ ⇒ BRES = 0.958

2.2.1.10. Charging case 10

The beam is subjected to a following bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=100000\text{Nm}\), to a following bending moment \(\text{Y}\) equal to \({M}_{\mathit{FY}}=-150000\text{Nm}\) and to a compression force equal to \(N=-3000000\text{N}\). It is therefore in deflected and composed bending, and the resolution is carried out according to the iterative method based on the verification of the Bresler inequality:

• We start by performing a composite bending calculation along the Z axis:

The moment to resume is:math: `M=left| {M} | {M} _ {mathit {FZ}}right|-N (d-frac {{b} _ {w}} {2}) =730000text {Nm} `

The ultimate moment reduced equivalent is then written \(\mathrm{\mu }=\frac{M}{hd\mathrm{²}\mathrm{\eta }{f}_{\mathit{cd}}}=\mathrm{0,493}>{\mathrm{\mu }}_{\mathit{BC}}=\mathrm{0,48}\).

The section is therefore entirely compressed, and the balance changes to PIVOT C so that the problem is solved by iteration; in the end, we obtain the optimal solution below:

• Extreme fiber deformations: \({\mathrm{\epsilon }}_{\text{c,sup}}=\mathrm{3,33}\cdot {10}^{-3}\), \({\mathrm{\epsilon }}_{\text{c,inf}}=\mathrm{2,23}\cdot {10}^{-4}\)

• PIVOT C depth: \({x}_{c}=(1-\frac{{\mathrm{\epsilon }}_{c2}}{{\mathrm{\epsilon }}_{\mathit{cu}2}}){b}_{w}=(1-\frac{\mathrm{2,0}\cdot {10}^{-3}}{\mathrm{3,5}\cdot {10}^{-3}})\times \mathrm{0,5}=\mathrm{0,2143}m\)

• Resistant compressive force of concrete: \({N}_{\mathit{cc}}={b}_{w}h{f}_{\mathit{cd}}\cdot (1+{m}_{2}{X}^{{n}_{c}})=2973627N\)

• Moment resistant to the center of gravity of the section: \({M}_{\mathit{cc}}=h{b}_{w}^{2}{f}_{\mathit{cd}}\cdot {m}_{1}{X}^{{n}_{c}}=93995\mathit{Nm}\)

Such as:

• \(X=\frac{\mathrm{\Delta }\mathrm{\epsilon }}{{\mathrm{\epsilon }}_{c2}}=\frac{{\mathrm{\epsilon }}_{\text{c,sup}}-{\mathrm{\epsilon }}_{\text{c,inf}}}{{\mathrm{\epsilon }}_{c2}}=\mathrm{1,555}\)

• \({n}_{c}=\mathrm{2,0}\)

• \({m}_{1}=\frac{{(1-{x}_{c}/{b}_{w})}^{{n}_{c}+1}}{2({n}_{c}+1)}\cdot [:ref:`1-2\frac{(1-{x}_{c}/{b}_{w})}{{n}_{c}+2} <1-2\frac{(1-{x}_{c}/{b}_{w})}{{n}_{c}+2}>\)] and :math:`{m}_{2}=frac{-{(1-{x}_{c}/{b}_{w})}^{{n}_{c}+1}}{{n}_{c}+1}

The dimensioning reinforcement is then deduced from the equilibrium equations below:

• \({A}_{\mathit{SYI}}{\mathrm{\sigma }}_{\mathit{SYI}}+{A}_{\mathit{SYS}}{\mathrm{\sigma }}_{\mathit{SYS}}=N-{N}_{\mathit{cc}}\)

• \(-{A}_{\mathit{SYI}}{\mathrm{\sigma }}_{\mathit{SYI}}\cdot ({b}_{w}/2-{c}_{\text{inf}})+{A}_{\mathit{SYS}}{\mathrm{\sigma }}_{\mathit{SYS}}\cdot ({b}_{w}/2-{c}_{\text{sup}})={M}_{\mathit{FZ}}-{M}_{\mathit{cc}}\)

Which results in: \({A}_{\mathit{SYI}}=0\text{cm²}\) and \({A}_{\mathit{SYS}}=\mathrm{0,632}\text{cm²}\) (Compression reinforcement)

• A composite flexure calculation along the Y axis is similarly performed:

The moment to resume is:math: `M=left| {M} | {M} _ {mathit {FY}}right|-N (d-frac {h} {2}) =480000text {Nm} `

The ultimate moment reduced equivalent is then written \(\mathrm{\mu }=\frac{M}{{b}_{w}d\mathrm{²}\mathrm{\eta }{f}_{\mathit{cd}}}=\mathrm{0,609}>{\mathrm{\mu }}_{\mathit{BC}}=\mathrm{0,48}\).

The section is therefore entirely compressed, and the balance changes to PIVOT C so that the problem is solved by iteration; in the end, we obtain the optimal solution below:

• Extreme fiber deformations: \({\mathrm{\epsilon }}_{\text{c,inf}}=\mathrm{3,5}\cdot {10}^{-3}\), \({\mathrm{\epsilon }}_{\text{c,sup}}=0\)

• PIVOT C depth: \({x}_{c}=(1-\frac{{\mathrm{\epsilon }}_{c2}}{{\mathrm{\epsilon }}_{\mathit{cu}2}})h=(1-\frac{\mathrm{2,0}\cdot {10}^{-3}}{\mathrm{3,5}\cdot {10}^{-3}})\times \mathrm{0,3}=\mathrm{0,1286}m\)

• Resistant compressive force of concrete: \({N}_{\mathit{cc}}={b}_{w}h{f}_{\mathit{cd}}\cdot (1+{m}_{2}{X}^{{n}_{c}})=2660934N\)

• Moment resistant to the center of gravity of the section: \({M}_{\mathit{cc}}=-{b}_{w}{h}^{2}{f}_{\mathit{cd}}\cdot {m}_{1}{X}^{{n}_{c}}=-89900\mathit{Nm}\)

Such as:

• \(X=\frac{\mathrm{\Delta }\mathrm{\epsilon }}{{\mathrm{\epsilon }}_{c2}}=\frac{{\mathrm{\epsilon }}_{\text{c,sup}}-{\mathrm{\epsilon }}_{\text{c,inf}}}{{\mathrm{\epsilon }}_{c2}}=-\mathrm{1,75}\)

• \({n}_{c}=\mathrm{2,0}\)

• \({m}_{1}=\frac{{(1-{x}_{c}/h)}^{{n}_{c}+1}}{2({n}_{c}+1)}\cdot [:ref:`1-2\frac{(1-{x}_{c}/h)}{{n}_{c}+2} <1-2\frac{(1-{x}_{c}/h)}{{n}_{c}+2}>\)] and :math:`{m}_{2}=frac{-{(1-{x}_{c}/h)}^{{n}_{c}+1}}{{n}_{c}+1}

The dimensioning reinforcement is then deduced from the equilibrium equations below:

• \({A}_{\mathit{SZI}}{\mathrm{\sigma }}_{\mathit{SZI}}+{A}_{\mathit{SZS}}{\mathrm{\sigma }}_{\mathit{SZS}}=N-{N}_{\mathit{cc}}\)

• \(-{A}_{\mathit{SZI}}{\mathrm{\sigma }}_{\mathit{SZI}}\cdot (h/2-{c}_{\text{inf}})+{A}_{\mathit{SZS}}{\mathrm{\sigma }}_{\mathit{SZS}}\cdot (h/2-{c}_{\text{sup}})={M}_{\mathit{FY}}-{M}_{\mathit{cc}}\)

Which results in: \({A}_{\mathit{SZS}}=0\text{cm²}\) and \({A}_{\mathit{SZI}}=\mathrm{10,902}\text{cm²}\) (Compression reinforcement)

• The normal resisting force, as well as the resistant bending moments, are then calculated using the interaction diagrams constructed from the reinforcement calculated previously; we obtain:

\({N}_{\mathit{Rd}}=4001505N\), \({M}_{\text{Rd,z}}=99965\mathit{Nm}\), and \({M}_{\text{Rd,y}}=-150014\mathit{Nm}\)

• The Bresler equation is then written:

\(\mathit{BRES}={(\frac{{M}_{\mathit{FZ}}}{{M}_{\text{Rd,z}}})}^{a=f(N/{N}_{\mathit{Rd}})}+{(\frac{{M}_{\mathit{FY}}}{{M}_{\text{Rd,y}}})}^{a}={(\frac{100000}{99965})}^{\mathrm{1,58}}+{(\frac{150000}{150014})}^{\mathrm{1,58}}\approx \mathrm{2,0}\) (which is normal, since this reinforcement pair is the result of a decoupled calculation (according to Y and Z) of sizing).

• The iteration algorithm then consists in increasing the reinforcement sections, so that the sum above falls below 1; in the end, we obtain:

• \({A}_{\mathit{SYS}}=\mathrm{2,93}\text{cm²}\) and \({A}_{\mathit{SZI}}=\mathrm{50,09}\text{cm²}\) ⇒

\({N}_{\text{Rd}}=5804403N\), \({M}_{\text{Rd,z}}=138358\mathit{Nm}\), and \({M}_{\text{Rd,y}}=-330836\mathit{Nm}\) ⇒

\(a=f(N/{N}_{\text{Rd}}=\mathrm{0,5168})=\mathrm{1,347}\) ⇒

BRES = 0.9901 ⇒ OK

2.2.1.11. Charging case 11

The beam is subjected to a bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-150000\text{Nm}\).

The section is partially stretched.

The ultimate moment reduces: math: mathrm {mu} =frac {mu} =frac {left| {M} _ {mathit {FZ}}right|} {hdmathrm {²}}mathrm {eta} {mu}} =frac {mathrm {0.15}}} {mathrm {²}}mathrm {²}}mathrm {eta} {f} _ {mathit {cd}}} =frac {mathrm {0.15}}} {mathrm {0.3}}cdotmathrm {0.46²}cdotmathrm {1.0}cdot (35/mathrm {1.5})}} =mathrm {0.1013}.

The relative position of neutral fiber \(\mathrm{\alpha }=1-\sqrt{1-2\mathrm{\mu }}=\mathrm{0,1337}\).

The \(y=d(1-\frac{\mathrm{\alpha }}{2})=\mathrm{0,429}m\) reduced lever arm.

The frame section is therefore equal to:math: {A} _ {mathit {SYS}} =frac {left| {M} _ {mathit {FZ}}right|} {y {mathrm {sigma}} _ {sigma}}} _ {sigma}}} _ {s}} =mathrm {7,923}text {cm²} (lit:math: text {Y}} (lit:math: text {Y}}) higher).

2.2.1.12. Charging case 12

The beam is subjected to a bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-260000\text{Nm}\).

The section is partially stretched.

The ultimate moment reduces: math: mathrm {mu} =frac {mu} =frac {left| {M} _ {mathit {FZ}}right|} {hdmathrm {²}}mathrm {eta} {mu}} =frac {mathrm {0.26}} {mathrm {²}}mathrm {²}}mathrm {eta} {f} _ {mathit {cd}}} =frac {mathrm {0.26}}} {mathrm {0.3}}cdotmathrm {0.46²}cdotmathrm {1.0}cdot (35/mathrm {1.5})}} =mathrm {0.1755}.

The relative position of neutral fiber \(\mathrm{\alpha }=1-\sqrt{1-2\mathrm{\mu }}=\mathrm{0,243}\).

The \(y=d(1-\frac{\mathrm{\alpha }}{2})=\mathrm{0,404}m\) reduced lever arm.

The frame section is therefore equal to:math: {A} _ {mathit {SYS}} =frac {left| {M} _ {mathit {FZ}}right|} {y {mathrm {sigma}} _ {sigma}}} _ {sigma}}} _ {s}} =mathrm {14.4}text {cm²} (lit:math: text {Y}} superior).

2.2.1.13. Charging case 13

The beam is subjected to a bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-380000\text{Nm}\).

The section is partially stretched.

The ultimate moment reduces: math: mathrm {mu} =frac {mu} =frac {left| {M} _ {mathit {FZ}}right|} {hdmathrm {²}}mathrm {eta} {mu}} =frac {mathrm {0.38}}} {mathrm {²}}mathrm {²}}mathrm {eta} {f} _ {mathit {cd}}} =frac {mathrm {0.38}}} {mathrm {0,3}} {mathrm {0,3}}cdotmathrm {0.46²}cdotmathrm {1.0}cdot (35/mathrm {1.5})}} =mathrm {0.2565}.

The relative position of neutral fiber \(\mathrm{\alpha }=1-\sqrt{1-2\mathrm{\mu }}=\mathrm{0,378}\).

The \(y=d(1-\frac{\mathrm{\alpha }}{2})=\mathrm{0,373}m\) reduced lever arm.

The frame section is therefore equal to:math: {A} _ {mathit {SYS}} =frac {left| {M} _ {mathit {FZ}}right|} {y {mathrm {sigma}} _ {sigma}}} _ {sigma}}} _ {s}} =mathrm {22,382}text {cm²} (lit:math: text {Y}} superior).

2.2.1.14. Charging case 14

The beam is subjected to a compression of \(N=-4500000\text{N}\), to a bending moment of \({M}_{\mathit{FZ}}=380000\text{Nm}\) and to a shear force of \({V}_{z}=100000\text{N}\).

Longitudinal reinforcement:

The moment to resume is:math: `M=left| {M} | {M} _ {mathit {FZ}}right|-N (d-frac {{b} _ {w}} {2}) =1325000text {Nm} `

The ultimate moment reduced equivalent is then written \(\mathrm{\mu }=\frac{M}{hd\mathrm{²}\mathrm{\eta }{f}_{\mathit{cd}}}=\mathrm{0,895}>{\mathrm{\mu }}_{\mathit{BC}}=\mathrm{0,48}\).

The section is therefore entirely compressed, and the balance changes to PIVOT C so that the problem is solved by iteration; in the end, we obtain the optimal solution below:

• Extreme fiber deformations: \({\mathrm{\epsilon }}_{\text{c,sup}}=\mathrm{3,213}\cdot {10}^{-3}\), \({\mathrm{\epsilon }}_{\text{c,inf}}=\mathrm{3,829}\cdot {10}^{-4}\)

• PIVOT C depth: \({x}_{c}=(1-\frac{{\mathrm{\epsilon }}_{c2}}{{\mathrm{\epsilon }}_{\mathit{cu}2}}){b}_{w}=(1-\frac{\mathrm{2,0}\cdot {10}^{-3}}{\mathrm{3,5}\cdot {10}^{-3}})\times \mathrm{0,5}=\mathrm{0,2143}m\)

• Resistant compressive force of concrete: \({N}_{\mathit{cc}}={b}_{w}h{f}_{\mathit{cd}}\cdot (1+{m}_{2}{X}^{{n}_{c}})=3064141N\)

• Moment resistant to the center of gravity of the section: \({M}_{\mathit{cc}}=h{b}_{w}^{2}{f}_{\mathit{cd}}\cdot {m}_{1}{X}^{{n}_{c}}=77832\mathit{Nm}\)

Such as:

• \(X=\frac{\mathrm{\Delta }\mathrm{\epsilon }}{{\mathrm{\epsilon }}_{c2}}=\frac{{\mathrm{\epsilon }}_{\text{c,sup}}-{\mathrm{\epsilon }}_{\text{c,inf}}}{{\mathrm{\epsilon }}_{c2}}=\mathrm{1,415}\)

• \({n}_{c}=\mathrm{2,0}\)

• \({m}_{1}=\frac{{(1-{x}_{c}/{b}_{w})}^{{n}_{c}+1}}{2({n}_{c}+1)}\cdot [:ref:`1-2\frac{(1-{x}_{c}/{b}_{w})}{{n}_{c}+2} <1-2\frac{(1-{x}_{c}/{b}_{w})}{{n}_{c}+2}>\)] and :math:`{m}_{2}=frac{-{(1-{x}_{c}/{b}_{w})}^{{n}_{c}+1}}{{n}_{c}+1}

The dimensioning reinforcement is then deduced from the equilibrium equations below:

• \({A}_{\mathit{SYI}}{\mathrm{\sigma }}_{\mathit{SYI}}+{A}_{\mathit{SYS}}{\mathrm{\sigma }}_{\mathit{SYS}}=N-{N}_{\mathit{cc}}\)

• \(-{A}_{\mathit{SYI}}{\mathrm{\sigma }}_{\mathit{SYI}}\cdot ({b}_{w}/2-{c}_{\text{inf}})+{A}_{\mathit{SYS}}{\mathrm{\sigma }}_{\mathit{SYS}}\cdot ({b}_{w}/2-{c}_{\text{sup}})={M}_{\mathit{FZ}}-{M}_{\mathit{cc}}\)

Which results in: \({A}_{\mathit{SYI}}=0\text{cm²}\) and \({A}_{\mathit{SYS}}=\mathrm{33,06}\text{cm²}\) (Compression reinforcement)

F**transverse reinforcement:**

Same as Charging Case 1

2.2.2. Calculations with the ELS Feature

We will give the details of the calculation with respect to the case by default (Configuration 1 + Configuration 2 for the case where compression steel is required), and we will focus only on the calculation of the longitudinal flexural reinforcement.

2.2.2.1. Charging case 1

The beam is subjected to a compression of \(N=-1000000\text{N}\) and to a shear force of \({V}_{z}=100000\text{N}\).

Longitudinal reinforcement:

The compressive strength of unreinforced concrete is given by:

\({N}_{\mathit{Rd}}={A}_{c}\times {\mathrm{\sigma }}_{\text{c,lim}}=\mathrm{0,5}m\times \mathrm{0,3}m\times (21\mathit{MPa})=\mathrm{3,15}\text{MN}>1\text{MN}\)

Therefore, the section is in Pivot C (case of uniform deformation) such that the concrete alone can withstand the force ⇒ no longitudinal reinforcement required.

Transverse reinforcement:

Similar to the calculation in ELU, replacing \({f}_{\mathit{cd}}\) with \({\mathrm{\sigma }}_{\text{c,lim}}\)

2.2.2.2. Charging case2

The beam is subjected to a normal tensile force of \(N=1000000\text{N}\) and to a shear force of \({V}_{z}=-100000\text{N}\).

Longitudinal reinforcement:

I It is a section that is fully extended symmetrically.

The steel section is therefore equal to \({A}_{S}=\frac{N}{{\mathrm{\sigma }}_{\text{s,lim}}}=\frac{1\mathit{MN}}{400\mathit{MPa}}\) (Pivot A — Case of uniform deformation).

Each frame bed therefore takes up a quarter of the effort, i.e.:

\({A}_{\mathit{SYS}}={A}_{\mathit{SYI}}={A}_{\mathit{SZS}}={A}_{\mathit{SZI}}=\frac{{A}_{S}}{4}=\mathrm{6,25}{\mathit{cm}}^{2}\)

Transverse reinforcement:

Similar to the calculation in ELU, replacing \({f}_{\mathit{cd}}\) with \({\mathrm{\sigma }}_{\text{c,lim}}\)

2.2.2.3. Charging case 3

The beam is subjected to a normal tensile force of \(N=1000000\text{N}\), a shear force \({V}_{z}=-100000\text{N}\) and a torsional moment \({M}_{T}=-10000\text{Nm}\)

Longitudinal reinforcement:

Same as the previous case: \({A}_{\mathit{SYS}}={A}_{\mathit{SYI}}={A}_{\mathit{SZS}}={A}_{\mathit{SZI}}=\mathrm{6,25}{\mathit{cm}}^{2}\)

Transverse reinforcement:

Similar to the calculation in ELU, replacing \({f}_{\mathit{cd}}\) with \({\mathrm{\sigma }}_{\text{c,lim}}\)

2.2.2.4. Charging case4

The beam is subjected to a bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=100000\text{Nm}\). This flexural moment corresponds to a stretched lower fiber.

At the ELS:

The resistant moment of concrete is equal to:

\({M}_{\text{lim}}=\frac{1}{2}{\mathrm{\sigma }}_{b}{y}_{\text{lim}}h(d-\frac{{y}_{\text{lim}}}{3})=250527\text{Nm}\)

with \({y}_{\text{lim}}=d\frac{n{\mathrm{\sigma }}_{b}}{n{\mathrm{\sigma }}_{b}+{\mathrm{\sigma }}_{s}}=({b}_{w}-{c}_{\text{y,inf}})\frac{n{\mathrm{\sigma }}_{b}}{n{\mathrm{\sigma }}_{b}+{\mathrm{\sigma }}_{s}}=\mathrm{0,203}\text{m}\)

So we are in the case where \(M={M}_{\mathit{FZ}}⩽{M}_{\text{lim}}\). Thus, only tensile steels are required.

The reduced moment of service is equal to: \(\mathrm{\mu }=\frac{M}{hd\mathrm{²}{\mathrm{\sigma }}_{\text{c,lim}}}=\mathrm{0,075}\)

The coefficient α is the solution of the equation: \(\mathrm{\alpha }\mathrm{³}-3\mathrm{\alpha }\mathrm{²}-6\mathrm{\mu }(1-\mathrm{\alpha })=0\)

By iterative resolution, we get: \(\mathrm{\alpha }=\mathrm{0,299}\)

The required steel cross section is equal to: \({A}_{\mathit{SYI}}=\frac{{M}_{\mathit{FZ}}}{{\mathrm{\sigma }}_{s}d(1-\frac{\mathrm{\alpha }}{3})}=\mathrm{5,89}\text{cm²}\)

2.2.2.5. Charging case5

The beam is subjected to a bending moment \(\text{Y}\) equal to \({M}_{\mathit{FY}}=-100000\text{Nm}\). This bending moment corresponds to a stretched straight fiber.

At the ELS:

The resistant moment of concrete is equal to:

\({M}_{\text{lim}}=\frac{1}{2}{\mathrm{\sigma }}_{b}{z}_{\text{lim}}{b}_{w}(d-\frac{{z}_{\text{lim}}}{3})=133393\text{Nm}\)

with \({z}_{\text{lim}}=d\frac{n{\mathrm{\sigma }}_{b}}{n{\mathrm{\sigma }}_{b}+{\mathrm{\sigma }}_{s}}=(h-{c}_{\text{z,sup}})\frac{n{\mathrm{\sigma }}_{b}}{n{\mathrm{\sigma }}_{b}+{\mathrm{\sigma }}_{s}}=\mathrm{0,115}\text{m}\)

So we are in the case where:math: `M=left| {M} | {mathit {FY}}right|{ M} _ {text {lim}} _ {text {lim}}} `. Thus, only tensile steels are required.

The reduced moment of service is equal to: \(\mathrm{\mu }=\frac{M}{{b}_{w}d\mathrm{²}{\mathrm{\sigma }}_{\text{c,lim}}}=\mathrm{0,141}\)

The coefficient α is the solution of the equation: \(\mathrm{\alpha }\mathrm{³}-3\mathrm{\alpha }\mathrm{²}-6\mathrm{\mu }(1-\mathrm{\alpha })=0\)

By iterative resolution, we get: \(\mathrm{\alpha }=\mathrm{0,396}\)

The required steel section is equal to: :math: {A} _ {mathit {SZS}} =frac {left| {M} _ {mathit {FY}}right|} {{mathrm {sigma}} _ {sigma}}} _ {sigma}}} _ {sigma}}} _ {sigma}}} _ {s} d (1-frac {mathrm {alpha}} {3})} =mathrm {11}} 24}text {cm²}

2.2.2.6. Charging case 6

The beam is subjected to a following bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-100000\text{Nm}\) and to a normal tensile force equal to \(N=100000\text{N}\).

The moment to resume is:math: `M=| {M} _ {mathit {FZ}} |-N (d-frac {{b} _ {w}} {2}) =79000text {Nm} `

The section is partially tense, as we are in the case of \(M⩽{M}_{\text{lim}}=250527\text{Nm}\). Thus, only tensile steels are required.

The reduced moment of service is equal to: \(\mathrm{\mu }=\frac{M}{hd\mathrm{²}{\mathrm{\sigma }}_{\text{c,lim}}}=\mathrm{0,059}\)

The coefficient α is the solution of the equation: \(\mathrm{\alpha }\mathrm{³}-3\mathrm{\alpha }\mathrm{²}-6\mathrm{\mu }(1-\mathrm{\alpha })=0\)

By iterative resolution, we get: \(\mathrm{\alpha }=\mathrm{0,2719}\)

The required steel cross section is equal to:

\({A}_{\mathit{SYS}}=\frac{M}{{\mathrm{\sigma }}_{s}d(1-\frac{\mathrm{\alpha }}{3})}+\frac{N}{{\mathrm{\sigma }}_{s}}=\mathrm{7,171}\text{cm²}\)

2.2.2.7. Charging case7

The beam is subjected to a following bending moment \(\text{Y}\) equal to \({M}_{\mathit{FY}}=-100000\text{Nm}\) and to a normal tensile force equal to \(N=100000\text{N}\).

The moment to resume is:math: `M=| {M} _ {mathit {FY}} |-N (d-frac {h} {2}) =89000text {Nm} `

The section is partially tense, as we are in the case of \(M⩽{M}_{\text{lim}}=133393\text{Nm}\). Thus, only tensile steels are required.

The reduced moment of service is equal to: \(\mathrm{\mu }=\frac{M}{{b}_{w}d\mathrm{²}{\mathrm{\sigma }}_{\text{c,lim}}}=\mathrm{0,125}\)

The coefficient α is the solution of the equation: \(\mathrm{\alpha }\mathrm{³}-3\mathrm{\alpha }\mathrm{²}-6\mathrm{\mu }(1-\mathrm{\alpha })=0\)

By iterative resolution, we get: \(\mathrm{\alpha }=\mathrm{0,375}\)

The required steel cross section is equal to:

\({A}_{\mathit{SZS}}=\frac{M}{{\mathrm{\sigma }}_{s}d(1-\frac{\mathrm{\alpha }}{3})}+\frac{N}{{\mathrm{\sigma }}_{s}}=\mathrm{12,26}\text{cm²}\)

2.2.2.8. Charging case 8

The beam is subjected to a following bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-100000\text{Nm}\) and to a normal tensile force equal to \(N=2000000\text{N}\).

The section is completely extended:math: M=| {M} _ {mathit {FZ}} |-N (d-frac {{b} _ {w}} {2}) =-320000text {Nm} <0

At the ELS:

The reinforcement cross section is therefore equal to:

\({A}_{\mathit{SYS}}=\frac{M}{(d-h){\mathrm{\sigma }}_{s}}+\frac{N}{{\mathrm{\sigma }}_{s}}=\mathrm{30,952}\text{cm²}\) (\(\text{Y}\) superior bed)

\({A}_{\mathit{SYI}}=\frac{-M}{(d-h){\mathrm{\sigma }}_{s}}=\mathrm{19,048}\text{cm²}\) (lower \(\text{Y}\) bed)

2.2.2.9. Charging case 9

The beam is subjected to a following bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=100000\text{Nm}\) and to a following bending moment \(\text{Y}\) equal to \({M}_{\mathit{FY}}=-150000\text{Nm}\). It is therefore in deflected bending, and the resolution is carried out according to the iterative method based on the verification of the Bresler inequality:

• We start by performing a simple bending calculation along the Z axis: we obtain a reinforcement similar to CAS 4, namely \({A}_{\mathit{SYI}}=\mathrm{5,89}\text{cm²}\)

• Idem, we perform a simple bending calculation along the Y axis: we obtain the following reinforcement \({A}_{\mathit{SZS}}=\mathrm{22,868}\text{cm²}\)

• The normal resisting force, as well as the resisting bending moments are calculated using the interaction diagrams constructed from the reinforcement calculated previously; we obtain:

\({N}_{\mathit{Rd}}=4300237N\), \({M}_{\text{Rd,z}}=97519\mathit{Nm}\), and \({M}_{\text{Rd,y}}=-150001\mathit{Nm}\)

• The Bresler equation is then written:

\(\mathit{BRES}={(\frac{{M}_{\mathit{FZ}}}{{M}_{\text{Rd,z}}})}^{a=f(N/{N}_{\mathit{Rd}})}+{(\frac{{M}_{\mathit{FY}}}{{M}_{\text{Rd,y}}})}^{a}={(\frac{100000}{97519})}^{\mathrm{1,0}}+{(\frac{150000}{150001})}^{\mathrm{1,0}}\approx \mathrm{2,0}\) (which is normal, since this reinforcement pair is the result of a decoupled calculation (according to Y and Z) of sizing).

• The iteration algorithm then consists in increasing the reinforcement sections, so that the sum above falls below 1:

• \({A}_{\mathit{SYI}}=\mathrm{5,89}\text{cm²}\) and \({A}_{\mathit{SZS}}=\mathrm{22,868}\text{cm²}\) ⇒ \({M}_{\text{Rd,z}}=97519\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-150501\mathit{Nm}\) ⇒ BRES = 2.0

• \({A}_{\mathit{SYI}}=\mathrm{6,476}\text{cm²}\) and \({A}_{\mathit{SZS}}=\mathrm{25,155}\text{cm²}\) ⇒ \({M}_{\text{Rd,z}}=106795\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-153736\mathit{Nm}\) ⇒ ⇒ BRES = 1.912

• \({A}_{\mathit{SYI}}=\mathrm{7,124}\text{cm²}\) and \({A}_{\mathit{SZS}}=\mathrm{27,671}\text{cm²}\) ⇒ \({M}_{\text{Rd,z}}=116936\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-157433\mathit{Nm}\) ⇒ ⇒ BRES = 1.808

…………

…………

• \({A}_{\mathit{SYI}}=\mathrm{52,719}\text{cm²}\) and \({A}_{\mathit{SZS}}=\mathrm{204,769}\text{cm²}\) ⇒

\({M}_{\text{Rd,z}}=336053\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-215865\mathit{Nm}\) ⇒ BRES = 0.992

2.2.2.10. Charging case 10

The beam is subjected to a following bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=100000\text{Nm}\), to a following bending moment \(\text{Y}\) equal to \({M}_{\mathit{FY}}=-150000\text{Nm}\) and to a compression force equal to \(N=-3000000\text{N}\). It is therefore in deflected and composed bending, and the resolution is carried out according to the iterative method based on the verification of the Bresler inequality:

• We start by performing a composite bending calculation along the Z axis:

The moment to resume is:math: `M=left| {M} | {M} _ {mathit {FZ}}right|-N (d-frac {{b} _ {w}} {2}) =730000text {Nm} `

The section is fully compressed, as we are in the case where \(M>{M}_{\text{lim}}=250527\text{Nm}\) Thus, compression reinforcement is required.

The section is therefore entirely compressed, and the balance changes to PIVOT B (\({\mathrm{\sigma }}_{c}={\mathrm{\sigma }}_{\text{c,lim}}\)) such that the problem is solved by iteration; in the end, we obtain the optimal solution below:

• Extreme fiber stresses: \({\mathrm{\sigma }}_{\text{c,sup}}=21\mathit{MPa}\), \({\mathrm{\sigma }}_{\text{c,inf}}=\mathrm{15,05}\mathit{MPa}\)

• Resistant compressive force of concrete: \({N}_{\mathit{cc}}=\mathrm{0,5}{b}_{w}h\cdot ({\mathrm{\sigma }}_{\text{c,sup}}+{\mathrm{\sigma }}_{\text{c,inf}})=2703750N\)

• Moment resistant to the center of gravity of the section: \({M}_{\mathit{cc}}=(1/12)h{b}_{w}^{2}\cdot ({\mathrm{\sigma }}_{\text{c,sup}}-{\mathrm{\sigma }}_{\text{c,inf}})=37\mathrm{187,5}\mathit{Nm}\)

The dimensioning reinforcement is then deduced from the equilibrium equations below:

• \(n\cdot {A}_{\mathit{SYI}}{\mathrm{\sigma }}_{\mathit{SYI}}+n\cdot {A}_{\mathit{SYS}}{\mathrm{\sigma }}_{\mathit{SYS}}=N-{N}_{\mathit{cc}}\)

• \(-n\cdot {A}_{\mathit{SYI}}{\mathrm{\sigma }}_{\mathit{SYI}}\cdot ({b}_{w}/2-{c}_{\text{inf}})+n\cdot {A}_{\mathit{SYS}}{\mathrm{\sigma }}_{\mathit{SYS}}\cdot ({b}_{w}/2-{c}_{\text{sup}})={M}_{\mathit{FY}}-{M}_{\mathit{cc}}\)

Which results in: \({A}_{\mathit{SYI}}=0\text{cm²}\) and \({A}_{\mathit{SYS}}=\mathrm{9,67}\text{cm²}\) (Compression reinforcement)

• A composite flexure calculation along the Y axis is similarly performed:

The moment to resume is:math: `M=left| {M} | {M} _ {mathit {FY}}right|-N (d-frac {h} {2}) =480000text {Nm} `

The section is fully compressed, as we are in the case where \(M>{M}_{\text{lim}}=133393\text{Nm}\) Thus, compression reinforcement is required.

The section is therefore entirely compressed, and the balance changes to PIVOT B (\({\mathrm{\sigma }}_{c}={\mathrm{\sigma }}_{\text{c,lim}}\)) such that the problem is solved by iteration; in the end, we obtain the optimal solution below:

• Extreme fiber stresses: \({\mathrm{\sigma }}_{\text{c,inf}}=21\mathit{MPa}\), \({\mathrm{\sigma }}_{\text{c,sup}}=\mathrm{7,15}\mathit{MPa}\)

• Resistant compressive force of concrete: \({N}_{\mathit{cc}}=\mathrm{0,5}{b}_{w}h\cdot ({\mathrm{\sigma }}_{\text{c,sup}}+{\mathrm{\sigma }}_{\text{c,inf}})=2111250N\)

• Moment resistant to the center of gravity of the section: \({M}_{\mathit{cc}}=(1/12){b}_{w}{h}^{2}\cdot ({\mathrm{\sigma }}_{\text{c,sup}}-{\mathrm{\sigma }}_{\text{c,inf}})=-51937\mathit{Nm}\)

The dimensioning reinforcement is then deduced from the equilibrium equations below:

• \({A}_{\mathit{SZI}}{\mathrm{\sigma }}_{\mathit{SYI}}+{A}_{\mathit{SZS}}{\mathrm{\sigma }}_{\mathit{SZS}}=N-{N}_{\mathit{cc}}\)

• \(-{A}_{\mathit{SZI}}{\mathrm{\sigma }}_{\mathit{SZI}}\cdot (h/2-{c}_{\text{inf}})+{A}_{\mathit{SZS}}{\mathrm{\sigma }}_{\mathit{SZS}}\cdot (h/2-{c}_{\text{sup}})={M}_{\mathit{FZ}}-{M}_{\mathit{cc}}\)

Which results in: \({A}_{\mathit{SZS}}=0\text{cm²}\) and \({A}_{\mathit{SZI}}=\mathrm{30,98}\text{cm²}\) (Compression reinforcement)

• The normal resisting force, as well as the resistant bending moments, are then calculated using the interaction diagrams constructed from the reinforcement calculated previously; we obtain:

\({N}_{\mathit{Rd}}=4776053N\), \({M}_{\text{Rd,z}}=91\mathrm{703,6}\mathit{Nm}\), and \({M}_{\text{Rd,y}}=-149\mathrm{901,6}\mathit{Nm}\)

• The Bresler equation is then written:

\(\mathit{BRES}={(\frac{{M}_{\mathit{FZ}}}{{M}_{\text{Rd,z}}})}^{a=f(N/{N}_{\mathit{Rd}})}+{(\frac{{M}_{\mathit{FY}}}{{M}_{\text{Rd,y}}})}^{a}={(\frac{100000}{91\mathrm{703,6}})}^{\mathrm{1,44}}+{(\frac{150000}{149\mathrm{901,6}})}^{\mathrm{1,44}}\approx \mathrm{2,0}\) (which is normal, since this reinforcement pair is the result of a decoupled calculation (according to Y and Z) of sizing).

• The iteration algorithm then consists in increasing the reinforcement sections, so that the sum above falls below 1; in the end, we obtain:

• \({A}_{\mathit{SYS}}=\mathrm{18,88}\text{cm²}\) and \({A}_{\mathit{SZI}}=\mathrm{60,42}\text{cm²}\)

2.2.2.11. Charging case 11

The beam is subjected to a bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-150000\text{Nm}\).

At the ELS:

The resistant moment of concrete is equal to:

\({M}_{\text{lim}}=\frac{1}{2}{\mathrm{\sigma }}_{b}{y}_{\text{lim}}h(d-\frac{{y}_{\text{lim}}}{3})=250527\text{Nm}\)

with \({y}_{\text{lim}}=d\frac{n{\mathrm{\sigma }}_{b}}{n{\mathrm{\sigma }}_{b}+{\mathrm{\sigma }}_{s}}=({b}_{w}-{c}_{\text{y,inf}})\frac{n{\mathrm{\sigma }}_{b}}{n{\mathrm{\sigma }}_{b}+{\mathrm{\sigma }}_{s}}=\mathrm{0,203}\text{m}\)

So we are in the case where:math: `M=left| {M} | {mathit {FZ}}right|{ M} _ {text {lim}} _ {text {lim}}} `. Thus, only tensile steels are required.

The reduced moment of service is equal to: \(\mathrm{\mu }=\frac{M}{hd\mathrm{²}{\mathrm{\sigma }}_{\text{c,lim}}}=\mathrm{0,113}\)

The coefficient α is the solution of the equation: \(\mathrm{\alpha }\mathrm{³}-3\mathrm{\alpha }\mathrm{²}-6\mathrm{\mu }(1-\mathrm{\alpha })=0\)

By iterative resolution, we get: \(\mathrm{\alpha }=\mathrm{0,361}\)

The required steel section is equal to: :math: {A} _ {mathit {SYS}} =frac {left| {M} _ {mathit {FZ}}right|} {{mathrm {sigma}} _ {sigma}}} _ {sigma}}} _ {sigma}}} _ {s} d (1-frac {mathrm {alpha}} {3})} =mathrm {9,sigma}}} _ {sigma}}} _ {s} d (1-frac {mathrm {alpha}} {3})} =mathrm {9,sigma}}} 41}text {cm²}

2.2.2.12. Charging case 12

The beam is subjected to a bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-260000\text{Nm}\).

So we are in the case:math: M=left| {M} | {M} _ {mathit {FZ}}right|> {M} _ {text {lim}} `, but:math: `M=left| {M} | {M} _ {mathit {FZ}}} such as:

\({M}_{\text{R}}=\mathrm{\mu }(\mathrm{\alpha }=1)\cdot hd\mathrm{²}{\mathrm{\sigma }}_{\text{c,lim}}=444360\mathit{Nm}\) (maximum moment for a partially compressed section).

Thus, if we restrict ourselves to tensile reinforcement, we then find ourselves in Pivot B (\({\mathrm{\sigma }}_{c}={\mathrm{\sigma }}_{\text{c,lim}}\)), such that the stress experienced at the level of the steel is so \({\mathrm{\sigma }}_{s}<{\mathrm{\sigma }}_{\text{s,lim}}\); and the steel will therefore not be in optimal operation. We will then have:

Reduced moment of service equal to: \(\mathrm{\mu }=\frac{M}{hd\mathrm{²}{\mathrm{\sigma }}_{\text{c,lim}}}=\mathrm{0,195}\)

The coefficient α is the solution of the equation: \(-{\mathrm{\alpha }}^{2}+3\mathrm{\alpha }-6\mathrm{\mu }=0\)

By iterative resolution, we get: \(\mathrm{\alpha }=\mathrm{0,461}>{\mathrm{\alpha }}_{\text{lim}}=\mathrm{0,44}\)

The stress felt at the right of the tension steel is then obtained by the linearity of the stress diagram: \({\mathrm{\sigma }}_{s}=\mathrm{368,48}\mathit{MPa}<{\mathrm{\sigma }}_{\text{s,lim}}=400\mathit{MPa}\)

The required steel section is equal to: :math: {A} _ {mathit {SYS}} =frac {left| {M} _ {mathit {FZ}}right|} {{mathrm {sigma}} _ {sigma}}} _ {sigma}}} _ {sigma}}} _ {sigma}}} _ {s} d (1-frac {mathrm {alpha}} {3})} =mathrm {1818}} ,12}text {cm²}

If we allow ourselves to have compression reinforcement, the steel section can then be optimized by fixing the position of the neutral axis at the level of the limit depth, that is to say \(\mathrm{\alpha }={\mathrm{\alpha }}_{\text{lim}}=\mathrm{0,44}\); we will then have both: \({\mathrm{\sigma }}_{c}={\mathrm{\sigma }}_{\text{c,lim}}\) and \({\mathrm{\sigma }}_{c}={\mathrm{\sigma }}_{\text{c,lim}}\); writing the equilibrium equations then allows us to determine the equilibrium reinforcement below:

\({A}_{\mathit{SYS}}=\mathrm{16,52}\text{cm²}(\mathit{Tendu})\) and \({A}_{\mathit{SYI}}=\mathrm{0,973}\text{cm²}(\mathit{Comprimé})\)

2.2.2.13. Charging case 13

The beam is subjected to a bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-380000\text{Nm}\).

So we are in the case where:math: M=left| {M} | {M} _ {mathit {FZ}}}right|> {M} _ {text {lim}} `, but:math: `M=left| {left| {M} | {M}}}, but:math: `M=left| {M} | {M}}} `(Similar to the previous case).

Thus, if we restrict ourselves to tensile reinforcement, we then find ourselves in Pivot B (\({\mathrm{\sigma }}_{c}={\mathrm{\sigma }}_{\text{c,lim}}\)), such that the stress experienced at the level of the steel is so \({\mathrm{\sigma }}_{s}<{\mathrm{\sigma }}_{\text{s,lim}}\); and the steel will therefore not be in optimal operation. We will then have:

Reduced moment of service equal to: \(\mathrm{\mu }=\frac{M}{hd\mathrm{²}{\mathrm{\sigma }}_{\text{c,lim}}}=\mathrm{0,285}\)

The coefficient α is the solution of the equation: \(-{\mathrm{\alpha }}^{2}+3\mathrm{\alpha }-6\mathrm{\mu }=0\)

By iterative resolution, we get: \(\mathrm{\alpha }=\mathrm{0,765}>{\mathrm{\alpha }}_{\text{lim}}=\mathrm{0,44}\)

The stress felt at the right of the tension steel is then obtained by the linearity of the stress diagram: \({\mathrm{\sigma }}_{s}=\mathrm{96,56}\mathit{MPa}<{\mathrm{\sigma }}_{\text{s,lim}}=400\mathit{MPa}\)

The required steel cross section is equal to: :math: {A} _ {mathit {SYS}} =frac {left| {M} _ {mathit {FZ}}right|} {{mathrm {sigma}} _ {sigma}}} _ {sigma}}} _ {sigma}}} _ {s} d (1-frac {mathrm {alpha}} {3})} =mathrm {sigma}}}} {3})} =mathrm {sigma}}} _ {sigma}}} _ {s} d (1-frac {mathrm {alpha}} {3})} =mathrm {sigma}}} ,85}text {cm²}

If we allow ourselves to have compression reinforcement, the steel section can then be optimized by fixing the position of the neutral axis at the level of the limit depth, that is to say \(\mathrm{\alpha }={\mathrm{\alpha }}_{\text{lim}}=\mathrm{0,44}\); we will then have both: \({\mathrm{\sigma }}_{c}={\mathrm{\sigma }}_{\text{c,lim}}\) and \({\mathrm{\sigma }}_{c}={\mathrm{\sigma }}_{\text{c,lim}}\); writing the equilibrium equations then allows us to determine the equilibrium reinforcement below:

\({A}_{\mathit{SYS}}=\mathrm{23,66}\text{cm²}(\mathit{Tendu})\) and \({A}_{\mathit{SYI}}=\mathrm{12,3}\text{cm²}(\mathit{Comprimé})\)

2.2.2.14. Charging case 14

The beam is subjected to a compression of \(N=-4500000\text{N}\), to a bending moment of \({M}_{\mathit{FZ}}=380000\text{Nm}\) and to a shear force of \({V}_{z}=100000\text{N}\).

Longitudinal reinforcement:

The moment to resume is:math: `M=left| {M} | {M} _ {mathit {FZ}}right|-N (d-frac {{b} _ {w}} {2}) =1325000text {Nm} `

The section is fully compressed, as we are in the case where \(M>{M}_{\text{lim}}=250527\text{Nm}\) Thus, compression reinforcement is required.

The section is therefore entirely compressed, and the balance changes to PIVOT B (\({\mathrm{\sigma }}_{c}={\mathrm{\sigma }}_{\text{c,lim}}\)) such that the problem is solved by iteration; in the end, we obtain the optimal solution below:

• Extreme fiber stresses: \({\mathrm{\sigma }}_{\text{c,sup}}=21\mathit{MPa}\), \({\mathrm{\sigma }}_{\text{c,inf}}=\mathrm{16,65}\mathit{MPa}\)

• Resistant compressive force of concrete: \({N}_{\mathit{cc}}=\mathrm{0,5}{b}_{w}h\cdot ({\mathrm{\sigma }}_{\text{c,sup}}+{\mathrm{\sigma }}_{\text{c,inf}})=2823750N\)

• Moment resistant to the center of gravity of the section: \({M}_{\mathit{cc}}=(1/12)h{b}_{w}^{2}\cdot ({\mathrm{\sigma }}_{\text{c,sup}}-{\mathrm{\sigma }}_{\text{c,inf}})=27\mathrm{187,5}\mathit{Nm}\)

The dimensioning reinforcement is then deduced from the equilibrium equations below:

• \(n\cdot {A}_{\mathit{SYI}}{\mathrm{\sigma }}_{\mathit{SYI}}+n\cdot {A}_{\mathit{SYS}}{\mathrm{\sigma }}_{\mathit{SYS}}=N-{N}_{\mathit{cc}}\)

• \(-n\cdot {A}_{\mathit{SYI}}{\mathrm{\sigma }}_{\mathit{SYI}}\cdot ({b}_{w}/2-{c}_{\text{inf}})+n\cdot {A}_{\mathit{SYS}}{\mathrm{\sigma }}_{\mathit{SYS}}\cdot ({b}_{w}/2-{c}_{\text{sup}})={M}_{\mathit{FZ}}-{M}_{\mathit{cc}}\)

Which results in: \({A}_{\mathit{SYI}}=0\text{cm²}\) and \({A}_{\mathit{SYS}}=\mathrm{54,17}\text{cm²}\) (Compression reinforcement)

F**transverse reinforcement:**

Same as Charging Case 1

2.2.3. Calculations at the ELS Quasi-Permanent

We will give the details of the calculation with respect to the case by default (Configuration 1 + Configuration 2 for the case where compression steel is required), and we will focus only on the calculation of the longitudinal flexural reinforcement.

2.2.3.1. Charging case 1

The beam is subjected to a compression of \(N=-1000000\text{N}\) and to a shear force of \({V}_{z}=100000\text{N}\).

Longitudinal reinforcement:

The compressive strength of unreinforced concrete is given by:

\({N}_{\mathit{Rd}}={A}_{c}\times {\mathrm{\sigma }}_{\text{c,lim,NL}}=\mathrm{0,5}m\times \mathrm{0,3}m\times (\mathrm{15,75}\mathit{MPa})=\mathrm{2,36}\text{MN}>1\text{MN}\)

Therefore, the section is in Pivot C (case of uniform deformation) such that the concrete alone can withstand the force ⇒ no longitudinal reinforcement required.

Transverse reinforcement:

Similar to the calculation in ELU, replacing \({f}_{\mathit{cd}}\) with \({\mathrm{\sigma }}_{\text{c,lim,NL}}\)

2.2.3.2. Charging case2

The beam is subjected to a normal tensile force of \(N=1000000\text{N}\) and to a shear force of \({V}_{z}=-100000\text{N}\).

Longitudinal reinforcement:

It is a section that is fully stretched symmetrically.

The normal force is then distributed equally along the two axes, such that a separate calculation of 0.5 MN is carried out along Y and along Z.

With regard to the sizing of the reinforcement at ELS QP, the calculation is done through an iterative approach that is explained in the following. In fact, it is a question of looking for the most economical reinforcement configuration that makes it possible to verify the criterion for limiting the opening of cracks. Since the equations of Eurocode 2 with respect to the method for verifying the opening of cracks are non-linear equations in relation to the steel section and that they require a pre-established knowledge of the stress field, the design algorithm at ELS QP will iteratively involve the deterministic search algorithm from the dimensioning to the ELS Characteristic (dimensioning of the reinforcement to respect the constraint criterion).

Indicatively, the details of the calculation steps are given for reinforcement along the Z axis:

1. We start the calculation at ELS Characteristic (criterion: stress limitation, with \({\mathrm{\sigma }}_{\mathit{clim}}={{\mathrm{\sigma }}_{b}}^{\mathit{elsqp}}=\mathrm{15,75}\mathit{MPa}\) for the compression pivot of concrete and \({\mathrm{\sigma }}_{\mathit{slim}}={k}_{\mathit{var}}^{A}\times {f}_{\mathit{yk}}=1\times 500=500\mathit{MPa}\) for the traction pivot in line with tension steel).

The algorithm then returns the following data:

• ETAT: « Pure Traction »

• PIVOT: \({\mathrm{\sigma }}_{\mathit{slim}}\)

• Neutral axis depth (AN): \({x}_{\mathit{AN}}=\mathrm{\infty }\)

• Stress on the right side of higher steel: \({\mathrm{\sigma }}_{\mathit{ssup}}=-500\mathit{MPa}(\mathit{TRACTION})\)

• Stress on the right side of the lower steel: \({\mathrm{\sigma }}_{\mathit{sinf}}=-500\mathit{MPa}(\mathit{TRACTION})\)

• Stress at the right of the upper concrete fiber: \({\mathrm{\sigma }}_{\text{csup}}=-\mathrm{33,33}\mathit{MPa}(\mathit{TRACTION})\)

• Stress at the right of the lower concrete fiber: \({\mathrm{\sigma }}_{\text{cinf}}=-\mathrm{33,33}\mathit{MPa}(\mathit{TRACTION})\)

• Calculated steel section: \({A}_{\mathit{ssup}}=\mathrm{5,0}{\mathit{cm}}^{2};{A}_{\mathit{sinf}}=\mathrm{5,0}{\mathit{cm}}^{2}\)

2. The crack opening is then verified in accordance with the equations of Eurocode 2:

• Made of superior fiber:

\({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}}=\frac{{\mathrm{\sigma }}_{s}-{k}_{t}\times \frac{{f}_{\mathit{ctm}}}{{\mathrm{\rho }}_{\mathit{peff}}}\times (1+{\mathrm{\alpha }}_{e}\times {\mathrm{\rho }}_{\mathit{peff}})}{{E}_{s}}=\frac{\mathrm{500,0}-\mathrm{0,4}\times \frac{\mathrm{3,21}}{\mathrm{0,005}}\times (1+\mathrm{15,0}\times \mathrm{0,005})}{210000}=\mathrm{1,429}\times {10}^{-3}\)

With:

\({\mathrm{\rho }}_{\mathit{peff}}={A}_{\mathit{ssup}}/{\mathit{hb}}_{\mathit{ceff}}=\mathrm{0,005}\)

\({b}_{\mathit{ceff}}=\mathit{min}(\mathrm{2,5}\times ({b}_{w}-d);{b}_{w}/2)=\mathrm{100,0}\mathit{cm}\)

\({s}_{\mathit{rmax}}=\mathit{min}({k}_{3}\times {c}_{\text{sup}}+{k}_{1}{k}_{2}{k}_{4}\times ({\mathrm{\varphi }}_{\text{sup}}/{\mathrm{\rho }}_{\mathit{peff}});\mathrm{1,3}\times {b}_{w})\)

\({s}_{\mathit{rmax}}=\mathit{min}(\mathrm{2,485}\times \mathrm{40,0}+\mathrm{0,8}\times \mathrm{1,0}\times \mathrm{0,425}\times (\mathrm{25,0}/\mathrm{0,005});\mathrm{1,3}\times 500)=\mathrm{650,0}\mathit{mm}\)

From where:

\({w}_{\mathit{ksup}}={s}_{\mathit{rmax}}\times ({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}})=\mathrm{650,0}\times \mathrm{1,429}\times {10}^{-3}=\mathrm{0,9286}\mathit{mm}\text{}>\text{}{{w}_{\mathit{max}}}^{s}=\mathrm{0,15}\mathit{mm}\)

• In lower fiber:

We obtain by symmetry \({w}_{\mathit{kinf}}=\mathrm{0,9286}\mathit{mm}\text{}>\text{}{{w}_{\mathit{max}}}^{i}=\mathrm{0,15}\mathit{mm}\)

⇒ Since at least one of the crack openings is not verified, the algorithm will remember \(\mathit{COND}({k}_{\mathit{var}}^{A})=\mathit{FAUX}\)

3. We restart the calculation at ELS Characteristic (criterion: stress limitation, with \({\mathrm{\sigma }}_{\mathit{clim}}={{\mathrm{\sigma }}_{b}}^{\mathit{elsqp}}=\mathrm{15,75}\mathit{MPa}\) for the compression pivot of concrete and \({\mathrm{\sigma }}_{\mathit{slim}}={k}_{\mathit{var}}^{B}\times {f}_{\mathit{yk}}=\mathrm{0,5}\times 500=250\mathit{MPa}\) for the traction pivot in line with tension steel).

The algorithm then returns the following data:

• ETAT: « Pure Traction »

• PIVOT: \({\mathrm{\sigma }}_{\mathit{slim}}\)

• Neutral axis depth (AN): \({x}_{\mathit{AN}}=\mathrm{\infty }\)

• Stress on the right side of higher steel: \({\mathrm{\sigma }}_{\mathit{ssup}}=-250\mathit{MPa}(\mathit{TRACTION})\)

• Stress on the right side of the lower steel: \({\mathrm{\sigma }}_{\mathit{sinf}}=-250\mathit{MPa}(\mathit{TRACTION})\)

• Stress at the right of the upper concrete fiber: \({\mathrm{\sigma }}_{\text{csup}}=-\mathrm{16,67}\mathit{MPa}(\mathit{TRACTION})\)

• Stress at the right of the lower concrete fiber: \({\mathrm{\sigma }}_{\text{cinf}}=-\mathrm{16,67}\mathit{MPa}(\mathit{TRACTION})\)

• Calculated steel section: \({A}_{\mathit{ssup}}=\mathrm{10,0}{\mathit{cm}}^{2};{A}_{\mathit{sinf}}=\mathrm{10,0}{\mathit{cm}}^{2}\)

4. The crack opening is then verified in accordance with the equations of Eurocode 2:

• Made of superior fiber:

\({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}}=\frac{{\mathrm{\sigma }}_{s}-{k}_{t}\times \frac{{f}_{\mathit{ctm}}}{{\mathrm{\rho }}_{\mathit{peff}}}\times (1+{\mathrm{\alpha }}_{e}\times {\mathrm{\rho }}_{\mathit{peff}})}{{E}_{s}}=\frac{\mathrm{250,0}-\mathrm{0,4}\times \frac{\mathrm{3,21}}{\mathrm{0,01}}\times (1+\mathrm{15,0}\times \mathrm{0,01})}{210000}=\mathrm{7,143}\times {10}^{-4}\)

With:

\({\mathrm{\rho }}_{\mathit{peff}}={A}_{\mathit{ssup}}/{\mathit{hb}}_{\mathit{ceff}}=\mathrm{0,01}\)

\({b}_{\mathit{ceff}}=\mathit{min}(\mathrm{2,5}\times ({b}_{w}-d);{b}_{w}/2)=\mathrm{100,0}\mathit{cm}\)

\({s}_{\mathit{rmax}}=\mathit{min}({k}_{3}\times {c}_{\text{sup}}+{k}_{1}{k}_{2}{k}_{4}\times ({\mathrm{\varphi }}_{\text{sup}}/{\mathrm{\rho }}_{\mathit{peff}});\mathrm{1,3}\times {b}_{w})\)

\({s}_{\mathit{rmax}}=\mathit{min}(\mathrm{2,485}\times \mathrm{40,0}+\mathrm{0,8}\times \mathrm{1,0}\times \mathrm{0,425}\times (\mathrm{25,0}/\mathrm{0,01});\mathrm{1,3}\times 500)=\mathrm{650,0}\mathit{mm}\)

From where:

\({w}_{\mathit{ksup}}={s}_{\mathit{rmax}}\times ({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}})=\mathrm{650,0}\times \mathrm{7,143}\times {10}^{-4}=\mathrm{0,464}\mathit{mm}\text{}>\text{}{{w}_{\mathit{max}}}^{s}=\mathrm{0,15}\mathit{mm}\)

• In lower fiber:

We obtain by symmetry \({w}_{\mathit{kinf}}=\mathrm{0,464}\mathit{mm}\text{}>\text{}{{w}_{\mathit{max}}}^{i}=\mathrm{0,15}\mathit{mm}\)

⇒ Since at least one of the crack openings is not verified, the algorithm will remember \(\mathit{COND}({k}_{\mathit{var}}^{B})=\mathit{FAUX}\)

5. The algorithm will restart procedure 3) and 4) by dividing the value of \({k}_{\mathit{var}}^{B}\) by 2, until both crack openings become verified.

So, for \({k}_{\mathit{var}}^{B}=\mathrm{0,125}\), we get \({w}_{\mathit{ksup}/\mathit{inf}}=\mathrm{0,0557}\mathit{mm}\text{}<\text{}{{w}_{\mathit{max}}}^{i/s}=\mathrm{0,15}\mathit{mm}\), and now we have \(\mathit{COND}({k}_{\mathit{var}}^{B})=\mathit{VRAI}\).

6. It is then a question of determining by dichotomy (variation between \({k}_{\mathit{var}}^{A}\) and \({k}_{\mathit{var}}^{B}\)) the value of the coefficient \({k}_{\mathit{var}}\) optimal for the saturation of the dimensioning criterion at the ELS QP (i.e. the largest value of this coefficient making it possible to respect the upper and lower fiber crack openings, while respecting the limitation of the compression stress in concrete):

\({k}_{\mathit{var}}\)	0.5625	0.3438	0.3438	0.234	0.179	0.207	…	0.2209
\({A}_{\mathit{ssup}/\mathit{inf}}({\mathit{cm}}^{2})\)	8.889	14.55	14.55	21.33	27.83	24.15	…	22.62
\({\mathrm{\sigma }}_{c}(\mathit{MPa})\)	-18.75	-11.46	-11.46	-7.81	-5.98	-6.91	…	-7.37
\({w}_{\mathit{ksup}/\mathit{inf}}(\mathit{mm})\)	0.522	0.319	0.319	0.167	0.104	0.133	…	0.149997

So we remember in the end: \({A}_{\mathit{SYS}}={A}_{\mathit{SYI}}=\mathrm{22,62}{\mathit{cm}}^{2}\)

A similar calculation along the Z axis then leads to the following reinforcement: \({A}_{\mathit{SZS}}={A}_{\mathit{SZI}}=\mathrm{18,57}{\mathit{cm}}^{2}\) (by substituting the roles of \({b}_{w}\) and \(h\)).

Transverse reinforcement:

Similar to the calculation in ELU, replacing \({f}_{\mathit{cd}}\) with \({\mathrm{\sigma }}_{\text{c,lim,NL}}\)

2.2.3.3. Charging case 3

The beam is subjected to a normal tensile force of \(N=1000000\text{N}\), a shear force \({V}_{z}=-100000\text{N}\) and a torsional moment \({M}_{T}=-10000\text{Nm}\)

Longitudinal reinforcement:

Same as the previous case: \({A}_{\mathit{SYS}}={A}_{\mathit{SYI}}=\mathrm{22,62}{\mathit{cm}}^{2}\) and \({A}_{\mathit{SZS}}={A}_{\mathit{SZI}}=\mathrm{18,57}{\mathit{cm}}^{2}\)

Transverse reinforcement:

Similar to the calculation in ELU, replacing \({f}_{\mathit{cd}}\) with \({\mathrm{\sigma }}_{\text{c,lim,NL}}\)

2.2.3.4. Charging case4

The beam is subjected to a bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=100000\text{Nm}\). This flexural moment corresponds to a stretched lower fiber.

At the ELS QP :

1. We start the calculation at ELS Characteristic (criterion: stress limitation, with \({\mathrm{\sigma }}_{\mathit{clim}}={{\mathrm{\sigma }}_{b}}^{\mathit{elsqp}}=\mathrm{15,75}\mathit{MPa}\) for the compression pivot of concrete and \({\mathrm{\sigma }}_{\mathit{slim}}={k}_{\mathit{var}}^{A}\times {f}_{\mathit{yk}}=1\times 500=500\mathit{MPa}\) for the traction pivot in line with tension steel).

The algorithm then returns the following data:

• ETAT: « Partially Compressed »

• PIVOT: \({\mathrm{\sigma }}_{\mathit{slim}}\)

• Neutral axis depth (AN): \({x}_{\mathit{AN}}=\mathrm{\alpha }\times d=\mathrm{0,273}\times (50-4)=\mathrm{12,57}\mathit{cm}\)

• Stress on the right side of higher steel: \({\mathrm{\sigma }}_{\mathit{ssup}}=+\mathrm{128,17}\mathit{MPa}(\mathit{COMPRESSION})\)

• Stress on the right side of the lower steel: \({\mathrm{\sigma }}_{\mathit{sinf}}=-500\mathit{MPa}(\mathit{TRACTION})\)

• Stress at the right of the upper concrete fiber: \({\mathrm{\sigma }}_{\text{csup}}=+\mathrm{12,53}\mathit{MPa}(\mathit{COMPRESSION})\)

• Stress at the right of the lower concrete fiber: \({\mathrm{\sigma }}_{\text{cinf}}=-\mathrm{37,32}\mathit{MPa}(\mathit{TRACTION})\)

• Calculated steel section: \({A}_{\mathit{ssup}}=\mathrm{0,0}{\mathit{cm}}^{2};{A}_{\mathit{sinf}}=\mathrm{4,726}{\mathit{cm}}^{2}\)

2. the crack opening is then verified in accordance with the equations of Eurocode 2:

• Made of superior fiber: \({w}_{\mathit{ksup}}=0\mathit{mm}<{{w}_{\mathit{max}}}^{s}\) (no cracking in the absence of traction)

• In lower fiber:

\({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}}=\frac{{\mathrm{\sigma }}_{s}-{k}_{t}\times \frac{{f}_{\mathit{ctm}}}{{\mathrm{\rho }}_{\mathit{peff}}}\times (1+{\mathrm{\alpha }}_{e}\times {\mathrm{\rho }}_{\mathit{peff}})}{{E}_{s}}=\frac{\mathrm{500,0}-\mathrm{0,4}\times \frac{\mathrm{3,21}}{\mathrm{0,00473}}\times (1+\mathrm{15,0}\times \mathrm{0,00473})}{210000}=\mathrm{1,429}\times {10}^{-3}\)

With:

\({\mathrm{\rho }}_{\mathit{peff}}={A}_{\mathit{sinf}}/{\mathit{hb}}_{\mathit{ceff}}=\mathrm{0,0473}\)

\({h}_{\mathit{ceff}}=\mathit{min}(\mathrm{2,5}\times (h-d);(h-{x}_{\mathit{AN}})/3;h/2)=\mathrm{10,0}\mathit{cm}\)

\({s}_{\mathit{rmax}}=\mathit{min}({k}_{3}\times {e}_{\text{sup}}+{k}_{1}{k}_{2}{k}_{4}\times ({\mathrm{\varphi }}_{\text{sup}}/{\mathrm{\rho }}_{\mathit{peff}});\mathrm{1,3}\times ({b}_{w}-{x}_{\mathit{AN}}))\)

\({s}_{\mathit{rmax}}=\mathit{min}(\mathrm{2,485}\times \mathrm{40,0}+\mathrm{0,8}\times \mathrm{0,5}\times \mathrm{0,425}\times (\mathrm{25,0}/\mathrm{0,0473});\mathrm{1,3}\times (500-\mathrm{125,7}))=\mathrm{486,6}\mathit{mm}\)

From where:

\({w}_{\mathit{kinf}}={s}_{\mathit{rmax}}\times ({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}})=\mathrm{486,6}\times \mathrm{1,429}\times {10}^{-3}=\mathrm{0,695}\mathit{mm}\text{}>\text{}{{w}_{\mathit{max}}}^{i}=\mathrm{0,15}\mathit{mm}\)

⇒ Since at least one of the crack openings is not verified, the algorithm will remember \(\mathit{COND}({k}_{\mathit{var}}^{A})=\mathit{FAUX}\)

3. We restart the calculation at ELS Characteristic (criterion: stress limitation, with \({\mathrm{\sigma }}_{\mathit{clim}}={{\mathrm{\sigma }}_{b}}^{\mathit{elsqp}}=\mathrm{15,75}\mathit{MPa}\) for the compression pivot of concrete and \({\mathrm{\sigma }}_{\mathit{slim}}={k}_{\mathit{var}}^{B}\times {f}_{\mathit{yk}}=\mathrm{0,5}\times 500=250\mathit{MPa}\) for the traction pivot in line with tension steel).

The algorithm then returns the following data:

• ETAT: « Partially Compressed »

• PIVOT: \({\mathrm{\sigma }}_{\mathit{slim}}\)

• Neutral axis depth (AN): \({x}_{\mathit{AN}}=\mathrm{\alpha }\times d=\mathrm{0,368}\times (50-4)=\mathrm{16,94}\mathit{cm}\)

• Stress on the right side of higher steel: \({\mathrm{\sigma }}_{\mathit{ssup}}=+\mathrm{111,28}\mathit{MPa}(\mathit{COMPRESSION})\)

• Stress on the right side of the lower steel: \({\mathrm{\sigma }}_{\mathit{sinf}}=-250\mathit{MPa}(\mathit{TRACTION})\)

• Stress at the right of the upper concrete fiber: \({\mathrm{\sigma }}_{\text{csup}}=+\mathrm{9,71}\mathit{MPa}(\mathit{COMPRESSION})\)

• Stress at the right of the lower concrete fiber: \({\mathrm{\sigma }}_{\text{cinf}}=-\mathrm{18,96}\mathit{MPa}(\mathit{TRACTION})\)

• Calculated steel section: \({A}_{\mathit{ssup}}=\mathrm{0,0}{\mathit{cm}}^{2};{A}_{\mathit{sinf}}=\mathrm{9,87}{\mathit{cm}}^{2}\)

4. The crack opening is then verified in accordance with the equations of Eurocode 2:

• Made of superior fiber: \({w}_{\mathit{ksup}}=0\mathit{mm}<{{w}_{\mathit{max}}}^{s}\) (no cracking in the absence of traction)

• In lower fiber:

\({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}}=\frac{{\mathrm{\sigma }}_{s}-{k}_{t}\times \frac{{f}_{\mathit{ctm}}}{{\mathrm{\rho }}_{\mathit{peff}}}\times (1+{\mathrm{\alpha }}_{e}\times {\mathrm{\rho }}_{\mathit{peff}})}{{E}_{s}}=\frac{\mathrm{250,0}-\mathrm{0,4}\times \frac{\mathrm{3,21}}{\mathrm{0,00987}}\times (1+\mathrm{15,0}\times \mathrm{0,00987})}{210000}=\mathrm{7,143}\times {10}^{-4}\)

With:

\({\mathrm{\rho }}_{\mathit{peff}}={A}_{\mathit{sinf}}/{\mathit{hb}}_{\mathit{ceff}}=\mathrm{0,00987}\)

\({b}_{\mathit{ceff}}=\mathit{min}(\mathrm{2,5}\times ({b}_{w}-d);({b}_{w}-{x}_{\mathit{AN}})/3;{b}_{w}/2)=\mathrm{10,0}\mathit{cm}\)

\({s}_{\mathit{rmax}}=\mathit{min}({k}_{3}\times {e}_{\text{sup}}+{k}_{1}{k}_{2}{k}_{4}\times ({\mathrm{\varphi }}_{\text{sup}}/{\mathrm{\rho }}_{\mathit{peff}});\mathrm{1,3}\times ({b}_{w}-{x}_{\mathit{AN}}))\)

\({s}_{\mathit{rmax}}=\mathit{min}(\mathrm{2,485}\times \mathrm{40,0}+\mathrm{0,8}\times \mathrm{0,5}\times \mathrm{0,425}\times (\mathrm{25,0}/\mathrm{0,0473});\mathrm{1,3}\times (500-\mathrm{169,4}))=\mathrm{429,8}\mathit{mm}\)

From where:

\({w}_{\mathit{kinf}}={s}_{\mathit{rmax}}\times ({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}})=\mathrm{429,8}\times \mathrm{7,143}\times {10}^{-4}=\mathrm{0,307}\mathit{mm}\text{}>\text{}{{w}_{\mathit{max}}}^{i}=\mathrm{0,15}\mathit{mm}\)

⇒ Since at least one of the crack openings is not verified, the algorithm will remember \(\mathit{COND}({k}_{\mathit{var}}^{B})=\mathit{FAUX}\)

5. The algorithm will restart procedure 3) and 4) by dividing the value of \({k}_{\mathit{var}}^{B}\) by 2, until both crack openings become verified.

So, for \({k}_{\mathit{var}}^{B}=\mathrm{0,25}\), we get \({w}_{\mathit{kinf}}=\mathrm{0,1033}\mathit{mm}\text{}<\text{}{{w}_{\mathit{max}}}^{i}=\mathrm{0,15}\mathit{mm}\), and now we have \(\mathit{COND}({k}_{\mathit{var}}^{B})=\mathit{VRAI}\).

6. It is then a question of determining by dichotomy (variation between \({k}_{\mathit{var}}^{A}\) and \({k}_{\mathit{var}}^{B}\)) the value of the coefficient \({k}_{\mathit{var}}\) optimal for the saturation of the dimensioning criterion at the ELS QP (i.e. the largest value of this coefficient making it possible to respect the upper and lower fiber crack openings, while respecting the limitation of the compression stress in concrete):

\({k}_{\mathit{var}}\)	0.6250	0.4375	0.4375	0.344	0.297	0.320	…	0.3060
\({A}_{\mathit{ssup}}({\mathit{cm}}^{2})\)	0.0	0.0	0.0	0.0	0.0	0.0	…	0.0
\({A}_{\mathit{sinf}}({\mathit{cm}}^{2})\)	7.647	11.11	11.11	14.89	17.56	16.14	…	17.0
\({\mathrm{\sigma }}_{\text{csup}}(\mathit{MPa})\)	10.4	9.14	9.14	8.14	8.14	8.14	8.63	…	8.36
\({\mathrm{\sigma }}_{\text{cinf}}(\mathit{MPa})\)	-23.55	-16.65	-16.65	-13.21	-11.48	-12.34	…	-11.81
\({w}_{\mathit{ksup}}(\mathit{mm})\)	0.0	0.0	0.0	0.0	0.0	0.0	…	0.0
\({w}_{\mathit{kinf}}(\mathit{mm})\)	0.403	0.262	0.262	0.262	0.189	0.141	0.165	…	0.14999

So in the end we remember: \({A}_{\mathit{SYS}}=0{\mathit{cm}}^{2}\) and \({A}_{\mathit{SYI}}=\mathrm{17,0}{\mathit{cm}}^{2}\)

2.2.3.5. Charging case5

The beam is subjected to a bending moment \(\text{Y}\) equal to \({M}_{\mathit{FY}}=-100000\text{Nm}\). This bending moment corresponds to a stretched straight fiber.

At the ELS QP :

1. We start the calculation at ELS Characteristic (criterion: stress limitation, with \({\mathrm{\sigma }}_{\mathit{clim}}={{\mathrm{\sigma }}_{b}}^{\mathit{elsqp}}=\mathrm{15,75}\mathit{MPa}\) for the compression pivot of concrete and \({\mathrm{\sigma }}_{\mathit{slim}}={k}_{\mathit{var}}^{A}\times {f}_{\mathit{yk}}=1\times 500=500\mathit{MPa}\) for the traction pivot in line with tension steel).

The algorithm then returns the following data:

• ETAT: « Partially Compressed »

• PIVOT: \({\mathrm{\sigma }}_{\mathit{clim}}\)

• Neutral axis depth (AN): \({x}_{\mathit{AN}}=\mathrm{\alpha }\times d=\mathrm{0,44}\times (30-4)=\mathrm{11,448}\mathit{cm}\)

• Stress on the right side of higher steel: \({\mathrm{\sigma }}_{\mathit{ssup}}=-\mathrm{300,29}\mathit{MPa}(\mathit{TRACTION})\)

• Stress on the right side of the lower steel: \({\mathrm{\sigma }}_{\mathit{sinf}}=+\mathrm{153,71}\mathit{MPa}(\mathit{COMPRESSION})\)

• Stress at the right of the upper concrete fiber: \({\mathrm{\sigma }}_{\text{csup}}=-\mathrm{25,52}\mathit{MPa}(\mathit{TRACTION})\)

• Stress at the right of the lower concrete fiber: \({\mathrm{\sigma }}_{\text{cinf}}=+\mathrm{15,75}\mathit{MPa}(\mathit{COMPRESSION})\)

• Calculated steel section: \({A}_{\mathit{ssup}}=\mathrm{15,011}{\mathit{cm}}^{2};{A}_{\mathit{sinf}}=\mathrm{0,0}{\mathit{cm}}^{2}\)

2. the crack opening is then verified in accordance with the equations of Eurocode 2:

• In lower fiber: \({w}_{\mathit{kinf}}=0\mathit{mm}<{{w}_{\mathit{max}}}^{i}\) (no cracking in the absence of traction)

• Made of superior fiber:

\({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}}=\frac{{\mathrm{\sigma }}_{s}-{k}_{t}\times \frac{{f}_{\mathit{ctm}}}{{\mathrm{\rho }}_{\mathit{peff}}}\times (1+{\mathrm{\alpha }}_{e}\times {\mathrm{\rho }}_{\mathit{peff}})}{{E}_{s}}=\frac{\mathrm{300,29}-\mathrm{0,4}\times \frac{\mathrm{3,21}}{\mathrm{0,02427}}\times (1+\mathrm{15,0}\times \mathrm{0,02427})}{210000}=\mathrm{1,086}\times {10}^{-3}\)

With:

\({\mathrm{\rho }}_{\mathit{peff}}={A}_{\mathit{ssup}}/{b}_{w}{h}_{\mathit{ceff}}=\mathrm{0,02427}\)

\({h}_{\mathit{ceff}}=\mathit{min}(\mathrm{2,5}\times (h-d);(h-{x}_{\mathit{AN}})/3;h/2)=\mathrm{6,18}\mathit{cm}\)

\({s}_{\mathit{rmax}}=\mathit{min}({k}_{3}\times {c}_{\text{sup}}+{k}_{1}{k}_{2}{k}_{4}\times ({\mathrm{\varphi }}_{\text{sup}}/{\mathrm{\rho }}_{\mathit{peff}});\mathrm{1,3}\times (h-{x}_{\mathit{AN}}))\)

\({s}_{\mathit{rmax}}=\mathit{min}(\mathrm{2,485}\times \mathrm{40,0}+\mathrm{0,8}\times \mathrm{0,5}\times \mathrm{0,425}\times (\mathrm{25,0}/\mathrm{0,02427});\mathrm{1,3}\times (300-\mathrm{114,48}))=\mathrm{241,17}\mathit{mm}\)

From where:

\({w}_{\mathit{ksup}}={s}_{\mathit{rmax}}\times ({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}})=\mathrm{241,17}\times \mathrm{1,086}\times {10}^{-3}=\mathrm{0,262}\mathit{mm}\text{}>\text{}{{w}_{\mathit{max}}}^{s}=\mathrm{0,15}\mathit{mm}\)

⇒ Since at least one of the crack openings is not verified, the algorithm will remember \(\mathit{COND}({k}_{\mathit{var}}^{A})=\mathit{FAUX}\)

3. We restart the calculation at ELS Characteristic (criterion: stress limitation, with \({\mathrm{\sigma }}_{\mathit{clim}}={{\mathrm{\sigma }}_{b}}^{\mathit{elsqp}}=\mathrm{15,75}\mathit{MPa}\) for the compression pivot of concrete and \({\mathrm{\sigma }}_{\mathit{slim}}={k}_{\mathit{var}}^{B}\times {f}_{\mathit{yk}}=\mathrm{0,5}\times 500=250\mathit{MPa}\) for the traction pivot in line with tension steel).

The algorithm then returns the following data:

• ETAT: « Partially Compressed »

• PIVOT: \({\mathrm{\sigma }}_{\mathit{slim}}\)

• Neutral axis depth (AN): \({x}_{\mathit{AN}}=\mathrm{\alpha }\times d=\mathrm{0,474}\times (30-4)=\mathrm{12,336}\mathit{cm}\)

• Stress on the right side of higher steel: \({\mathrm{\sigma }}_{\mathit{ssup}}=-\mathrm{250,0}\mathit{MPa}(\mathit{TRACTION})\)

• Stress on the right side of the lower steel: \({\mathrm{\sigma }}_{\mathit{sinf}}=+\mathrm{152,53}\mathit{MPa}(\mathit{COMPRESSION})\)

• Stress at the right of the upper concrete fiber: \({\mathrm{\sigma }}_{\text{csup}}=-\mathrm{21,55}\mathit{MPa}(\mathit{TRACTION})\)

• Stress at the right of the lower concrete fiber: \({\mathrm{\sigma }}_{\text{cinf}}=+\mathrm{15,05}\mathit{MPa}(\mathit{COMPRESSION})\)

• Calculated steel section: \({A}_{\mathit{ssup}}=\mathrm{18,56}{\mathit{cm}}^{2};{A}_{\mathit{sinf}}=\mathrm{0,0}{\mathit{cm}}^{2}\)

4. The crack opening is then verified in accordance with the equations of Eurocode 2:

• Made of superior fiber: \({w}_{\mathit{ksup}}=0\mathit{mm}<{{w}_{\mathit{max}}}^{s}\) (no cracking in the absence of traction)

• In lower fiber:

\({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}}=\frac{{\mathrm{\sigma }}_{s}-{k}_{t}\times \frac{{f}_{\mathit{ctm}}}{{\mathrm{\rho }}_{\mathit{peff}}}\times (1+{\mathrm{\alpha }}_{e}\times {\mathrm{\rho }}_{\mathit{peff}})}{{E}_{s}}=\frac{\mathrm{250,0}-\mathrm{0,4}\times \frac{\mathrm{3,21}}{\mathrm{0,0315}}\times (1+\mathrm{15,0}\times \mathrm{0,0315})}{210000}=\mathrm{9,048}\times {10}^{-4}\)

With:

\({\mathrm{\rho }}_{\mathit{peff}}={A}_{\mathit{ssup}}/{b}_{w}{h}_{\mathit{ceff}}=\mathrm{0,0315}\)

\({h}_{\mathit{ceff}}=\mathit{min}(\mathrm{2,5}\times (h-d);(h-{x}_{\mathit{AN}})/3;h/2)=\mathrm{5,89}\mathit{cm}\)

\({s}_{\mathit{rmax}}=\mathit{min}({k}_{3}\times {c}_{\text{sup}}+{k}_{1}{k}_{2}{k}_{4}\times ({\mathrm{\varphi }}_{\text{sup}}/{\mathrm{\rho }}_{\mathit{peff}});\mathrm{1,3}\times (h-{x}_{\mathit{AN}}))\)

\({s}_{\mathit{rmax}}=\mathit{min}(\mathrm{2,485}\times \mathrm{40,0}+\mathrm{0,8}\times \mathrm{0,5}\times \mathrm{0,425}\times (\mathrm{25,0}/\mathrm{0,0315});\mathrm{1,3}\times (300-\mathrm{123,36}))=\mathrm{229,63}\mathit{mm}\)

From where:

\({w}_{\mathit{ksup}}={s}_{\mathit{rmax}}\times ({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}})=\mathrm{229,63}\times \mathrm{9,048}\times {10}^{-4}=\mathrm{0,2078}\mathit{mm}\text{}>\text{}{{w}_{\mathit{max}}}^{s}=\mathrm{0,15}\mathit{mm}\)

⇒ Since at least one of the crack openings is not verified, the algorithm will remember \(\mathit{COND}({k}_{\mathit{var}}^{B})=\mathit{FAUX}\)

5. The algorithm will restart procedure 3) and 4) by dividing the value of \({k}_{\mathit{var}}^{B}\) by 2, until both crack openings become verified.

So, for \({k}_{\mathit{var}}^{B}=\mathrm{0,25}\), we get \({w}_{\mathit{ksup}}=\mathrm{0,065}\mathit{mm}\text{}<\text{}{{w}_{\mathit{max}}}^{s}=\mathrm{0,15}\mathit{mm}\), and now we have \(\mathit{COND}({k}_{\mathit{var}}^{B})=\mathit{VRAI}\).

6. It is then a question of determining by dichotomy (variation between \({k}_{\mathit{var}}^{A}\) and \({k}_{\mathit{var}}^{B}\)) the value of the coefficient \({k}_{\mathit{var}}\) optimal for the saturation of the dimensioning criterion at the ELS QP (i.e. the largest value of this coefficient making it possible to respect the upper and lower fiber crack openings, while respecting the limitation of the compression stress in concrete):

\({k}_{\mathit{var}}\)	0.6250	0.4375	0.4375	0.3438	0.3906	0.4141	…	0.4094
\({A}_{\mathit{ssup}}({\mathit{cm}}^{2})\)	15.011	21.530	21.530	27.175	24.051	22.167	…	22.525
\({A}_{\mathit{sinf}}({\mathit{cm}}^{2})\)	0.0	0.0	0.0	0.0	0.0	0.0	…	0.0
\({\mathrm{\sigma }}_{\text{csup}}(\mathit{MPa})\)	-25.520	-19.060	-19.060	-15.275	-17.172	-18.081	…	-17.9
\({\mathrm{\sigma }}_{\text{cinf}}(\mathit{MPa})\)	15.750	14.520	14.520	13.352	13.965	14.014	…	14.01
\({w}_{\mathit{ksup}}(\mathit{mm})\)	0.262	0.167	0.167	0.167	0.167	0.167	…	0.14999
\({w}_{\mathit{kinf}}(\mathit{mm})\)	0.0	0.0	0.0	0.0	0.0	0.0	…	0.00000

So in the end we remember: \({A}_{\mathit{SZS}}=\mathrm{22,525}{\mathit{cm}}^{2}\) and \({A}_{\mathit{SZI}}=\mathrm{0,0}{\mathit{cm}}^{2}\)

2.2.3.6. Charging case6

The beam is subjected to a following bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-100000\text{Nm}\) and to a normal tensile force equal to \(N=100000\text{N}\).

At the ELSQP:

The algorithm for iterating on the value of \({k}_{\mathit{var}}\) leads to the following results:

Criteria: \({w}_{\mathit{ksup}}<\mathrm{0,15}\mathit{mm}\); \({w}_{\mathit{kinf}}<\mathrm{0,15}\mathit{mm}\); \({\mathrm{\sigma }}_{c,\mathit{compression}}<\mathrm{15,75}\mathit{MPa}\)

\({k}_{\mathit{var}}\)	1.0	0.5	0.5	0.25	0.625	…	0.3225
\({A}_{\mathit{ssup}}({\mathit{cm}}^{2})\)	5.77	11.93	11.93	24.09	9.41	…	18.794
\({A}_{\mathit{sinf}}({\mathit{cm}}^{2})\)	0	0	0	0	…	0
\({x}_{\mathit{AN}}(\mathit{cm})\)	11.42	15.54	15.54	20.33	14.08	…	18.58
\({\mathrm{\sigma }}_{\text{csup}}(\mathit{MPa})\)	-37.19	-18.86	-9.63	-23.44	…	-12.32
\({\mathrm{\sigma }}_{\text{cinf}}(\mathit{MPa})\)	11.01	8.5	8.5	6.6	9.19	…	7.29
\({\mathrm{\sigma }}_{\text{ssup}}(\mathit{MPa})\)	-500	-250	-250	-125	-312.5	…	-121.26
\({\mathrm{\sigma }}_{\text{sinf}}(\mathit{MPa})\)	107.2	94.71	94.71	79.52	98.68	…	85.76
\({w}_{\mathit{ksup}}(\mathit{mm})\)	0.717	0.319	0.98	0.417	…	0.149998
\({w}_{\mathit{kinf}}(\mathit{mm})\)	0.0	0.0	0.0	0.0	…	0.0

So in the end we remember: \({A}_{\mathit{SYS}}=\mathrm{18,794}{\mathit{cm}}^{2}\) and \({A}_{\mathit{SYI}}=\mathrm{0,0}{\mathit{cm}}^{2}\)

2.2.3.7. Charging case7

The beam is subjected to a following bending moment \(\text{Y}\) equal to \({M}_{\mathit{FY}}=-100000\text{Nm}\) and to a normal tensile force equal to \(N=100000\text{N}\).

At the ELSQP:

The algorithm for iterating on the value of \({k}_{\mathit{var}}\) leads to the following results:

Criteria: \({w}_{\mathit{ksup}}<\mathrm{0,15}\mathit{mm}\); \({w}_{\mathit{kinf}}<\mathrm{0,15}\mathit{mm}\); \({\mathrm{\sigma }}_{c,\mathit{compression}}<\mathrm{15,75}\mathit{MPa}\)

\({k}_{\mathit{var}}\)	1,0	0.5	0.5	0.25	0.625	…	0.4156
\({A}_{\mathit{ssup}}({\mathit{cm}}^{2})\)	12.96	20.13	20.13	41.67	15.85	…	24.765
\({A}_{\mathit{sinf}}({\mathit{cm}}^{2})\)	0	0	0	0	…	0
\({x}_{\mathit{AN}}(\mathit{cm})\)	9.97	11.74	11.74	14.94	10.76	…	12.64
\({\mathrm{\sigma }}_{\text{csup}}(\mathit{MPa})\)	-31.66	-21.34	-21.34	-11.35	-26.3	…	-18.01
\({\mathrm{\sigma }}_{\text{cinf}}(\mathit{MPa})\)	15.75	13.73	13.73	11.26	14.7	…	13.12
\({\mathrm{\sigma }}_{\text{ssup}}(\mathit{MPa})\)	-380.02	-250	-250	-125	-312.5	…	-207.78
\({\mathrm{\sigma }}_{\text{sinf}}(\mathit{MPa})\)	141.44	135.81	123.74	138.51	…	134.49
\({w}_{\mathit{ksup}}(\mathit{mm})\)	0.365	0.208	0.208	0.065	0.287	…	0.149996
\({w}_{\mathit{kinf}}(\mathit{mm})\)	0.0	0.0	0.0	0.0	…	0.0

So in the end we remember: \({A}_{\mathit{SZS}}=\mathrm{24,765}{\mathit{cm}}^{2}\) and \({A}_{\mathit{SZI}}=\mathrm{0,0}{\mathit{cm}}^{2}\)

2.2.3.8. Charging case 8

The beam is subjected to a following bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-100000\text{Nm}\) and to a normal tensile force equal to \(N=2000000\text{N}\).

At the ELSQP:

The algorithm for iterating on the value of \({k}_{\mathit{var}}\) leads to the following results:

Criteria: \({w}_{\mathit{ksup}}<\mathrm{0,15}\mathit{mm}\); \({w}_{\mathit{kinf}}<\mathrm{0,15}\mathit{mm}\); \({\mathrm{\sigma }}_{c,\mathit{compression}}<\mathrm{15,75}\mathit{MPa}\)

The required steel cross section is therefore equal to: \({A}_{\mathit{SYS}}=\mathrm{77,545}\text{cm²}\) and \({A}_{\mathit{SYI}}=\mathrm{47,72}\text{cm²}\)

2.2.3.9. Charging case 9

The beam is subjected to a following bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=100000\text{Nm}\) and to a following bending moment \(\text{Y}\) equal to \({M}_{\mathit{FY}}=-150000\text{Nm}\). It is therefore in deflected bending, and the resolution is carried out according to the iterative method based on the verification of the Bresler inequality:

• We start by performing a simple bending calculation along the Z axis: we obtain a reinforcement similar to CAS 4, namely \({A}_{\mathit{SYI}}=\mathrm{17,0}\text{cm²}\)

• Idem, we perform a simple bending calculation along the Y axis: we obtain the following reinforcement \({A}_{\mathit{SZS}}=\mathrm{31,827}\text{cm²}\) and \({A}_{\mathit{SZI}}=\mathrm{9,722}\text{cm²}\)

• The normal resisting force, as well as the resisting bending moments are calculated using the interaction diagrams constructed from the reinforcement calculated previously; we obtain:

\({N}_{\mathit{Rd}}=3467290N\), \({M}_{\text{Rd,z}}=101595\mathit{Nm}\), and \({M}_{\text{Rd,y}}=-97295\mathit{Nm}\)

• The Bresler equation is then written:

\(\mathit{BRES}={(\frac{{M}_{\mathit{FZ}}}{{M}_{\text{Rd,z}}})}^{a=f(N/{N}_{\mathit{Rd}})}+{(\frac{{M}_{\mathit{FY}}}{{M}_{\text{Rd,y}}})}^{a}={(\frac{100000}{101595})}^{\mathrm{1,0}}+{(\frac{150000}{97295})}^{\mathrm{1,0}}\approx \mathrm{2,53}\)

• The iteration algorithm then consists in increasing the reinforcement sections, so that the sum above falls below 1:

• \({A}_{\mathit{SYI}}=\mathrm{17,0}\text{cm²}\), \({A}_{\mathit{SZI}}=\mathrm{9,722}\text{cm²}\) and \({A}_{\mathit{SZS}}=\mathrm{31,827}\text{cm²}\) ⇒ \({M}_{\text{Rd,z}}=101595\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-97295\mathit{Nm}\) ⇒ BRES = 2.53

• \({A}_{\mathit{SYI}}=\mathrm{18,7}\text{cm²}\), \({A}_{\mathit{SZI}}=\mathrm{10,694}\text{cm²}\) and \({A}_{\mathit{SZS}}=\mathrm{35,01}\text{cm²}\) ⇒ \({M}_{\text{Rd,z}}=111079\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-98982\mathit{Nm}\) ⇒ BRES = 2.42

• \({A}_{\mathit{SYI}}=\mathrm{20,57}\text{cm²}\), \({A}_{\mathit{SZI}}=\mathrm{11,76}\text{cm²}\) and \({A}_{\mathit{SZS}}=\mathrm{38,51}\text{cm²}\) ⇒ \({M}_{\text{Rd,z}}=121438\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-100612\mathit{Nm}\) ⇒ BRES = 2.31

…………

…………

• \({A}_{\mathit{SYI}}\mathrm{\to }\mathrm{\infty }\text{cm²}\), \({A}_{\mathit{SZI}}\mathrm{\to }\mathrm{\infty }\text{cm²}\) and \({A}_{\mathit{SZS}}\mathrm{\to }\mathrm{\infty }\text{cm²}\) ⇒ BRES → 1.503 > 1

Thus, it is not possible to calculate a deflected flexural reinforcement density (MFY and MFZ) in the following case. The method currently implemented is an iterative resolution based on verifying the inequality of BRESLER. The algorithm is running out of capacity (a lot of iterations attempted).

The code returns the value of “-1” for all reinforcement sections.

2.2.3.10. Charging case 10

The beam is subjected to a following bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=100000\text{Nm}\), to a following bending moment \(\text{Y}\) equal to \({M}_{\mathit{FY}}=-150000\text{Nm}\) and to a compression force equal to \(N=-3000000\text{N}\). It is therefore in deflected and composed bending, and the resolution is carried out according to the iterative method based on the verification of the Bresler inequality:

• We start by performing a composite bending calculation along the Z axis:

\({k}_{\mathit{var}}\)	1,0
\({A}_{\mathit{ssup}}({\mathit{cm}}^{2})\)	23.57
\({A}_{\mathit{sinf}}({\mathit{cm}}^{2})\)	3.41
\({x}_{\mathit{AN}}(\mathit{cm})\)	399
\({\mathrm{\sigma }}_{\text{csup}}(\mathit{MPa})\)	15.75
\({\mathrm{\sigma }}_{\text{cinf}}(\mathit{MPa})\)	15.75
\({\mathrm{\sigma }}_{\text{ssup}}(\mathit{MPa})\)	236.25
\({\mathrm{\sigma }}_{\text{sinf}}(\mathit{MPa})\)	236.25
\({w}_{\mathit{ksup}}(\mathit{mm})\)	0.0
\({w}_{\mathit{kinf}}(\mathit{mm})\)	0.0

We would then remember in this case: \({A}_{\mathit{SYS}}=\mathrm{23,57}\text{cm²}(\mathit{Comprimé})\) and \({A}_{\mathit{SYI}}=\mathrm{3,41}\text{cm²}(\mathit{Comprimé})\)

• A composite flexure calculation along the Y axis is similarly performed:

\({k}_{\mathit{var}}\)	1,0
\({A}_{\mathit{ssup}}({\mathit{cm}}^{2})\)	0.0
\({A}_{\mathit{sinf}}({\mathit{cm}}^{2})\)	50.95
\({x}_{\mathit{AN}}(\mathit{cm})\)	71.28
\({\mathrm{\sigma }}_{\text{csup}}(\mathit{MPa})\)	15.75
\({\mathrm{\sigma }}_{\text{cinf}}(\mathit{MPa})\)	9.12
\({\mathrm{\sigma }}_{\text{ssup}}(\mathit{MPa})\)	150.07
\({\mathrm{\sigma }}_{\text{sinf}}(\mathit{MPa})\)	223.99
\({w}_{\mathit{ksup}}(\mathit{mm})\)	0.0
\({w}_{\mathit{kinf}}(\mathit{mm})\)	0.0

In this case, we would remember: \({A}_{\mathit{SZI}}=\mathrm{50,95}\text{cm²}(\mathit{Comprimé})\)

• The normal resisting force, as well as the resistant bending moments, are then calculated using the interaction diagrams constructed from the reinforcement calculated previously; we obtain:

\({N}_{\mathit{Rd}}=6648918N\), \({M}_{\text{Rd,z}}=109\mathrm{903,5}\mathit{Nm}\), and \({M}_{\text{Rd,y}}=-165\mathrm{694,7}\mathit{Nm}\)

• The Bresler equation is then written:

\(\mathit{BRES}={(\frac{{M}_{\mathit{FZ}}}{{M}_{\text{Rd,z}}})}^{a=f(N/{N}_{\mathit{Rd}})}+{(\frac{{M}_{\mathit{FY}}}{{M}_{\text{Rd,y}}})}^{a}={(\frac{100000}{109\mathrm{903,5}})}^{\mathrm{1,29}}+{(\frac{150000}{165\mathrm{694,7}})}^{\mathrm{1,29}}=\mathrm{1,8}\approx \mathrm{2,0}\) (which is normal, since this reinforcement pair is the result of a decoupled calculation (according to Y and Z) of sizing).

• The iteration algorithm then consists in increasing the reinforcement sections, so that the sum above falls below 1; in the end, we obtain:

• \({A}_{\mathit{SYI}}=\mathrm{3,41}\text{cm²}\), \({A}_{\mathit{SYS}}=\mathrm{23,57}\text{cm²}\) and \({A}_{\mathit{SZI}}=\mathrm{50,95}\text{cm²}\) ⇒ \({N}_{\mathit{Rd}}=6648918N\), \({M}_{\text{Rd,z}}=109\mathrm{903,5}\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-165\mathrm{694,7}\mathit{Nm}\) ⇒ a = 1.29 and BRES = 1.8

• \({A}_{\mathit{SYI}}=\mathrm{3,76}\text{cm²}\), \({A}_{\mathit{SYS}}=\mathrm{25,93}\text{cm²}\) and \({A}_{\mathit{SZI}}=\mathrm{56,05}\text{cm²}\) ⇒ \({N}_{\mathit{Rd}}=7077559N\), \({M}_{\text{Rd,z}}=120\mathrm{087,9}\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-182\mathrm{662,2}\mathit{Nm}\) ⇒ a = 1.26 and BRES = 1.57

• \({A}_{\mathit{SYI}}=\mathrm{4,13}\text{cm²}\), \({A}_{\mathit{SYS}}=\mathrm{28,52}\text{cm²}\) and \({A}_{\mathit{SZI}}=\mathrm{61,66}\text{cm²}\) ⇒ \({N}_{\mathit{Rd}}=7549065N\), \({M}_{\text{Rd,z}}=130\mathrm{517,1}\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-200\mathrm{830,4}\mathit{Nm}\) ⇒ a = 1.24 and BRES = 1.41

…………

…………

• \({A}_{\mathit{SYI}}=\mathrm{8,85}\text{cm²}\), \({A}_{\mathit{SYS}}=\mathrm{61,13}\text{cm²}\) and \({A}_{\mathit{SZI}}=\mathrm{132,15}\text{cm²}\) ⇒ \({N}_{\mathit{Rd}}=12469648N\), \({M}_{\text{Rd,z}}=206\mathrm{670,3}\mathit{Nm}\) and \({M}_{\text{Rd,y}}=-259\mathrm{125,0}\mathit{Nm}\) ⇒ a = 1.12 and BRES = 0.99

2.2.3.11. Charging case 11

The beam is subjected to a bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-150000\text{Nm}\).

At the ELSQP:

The algorithm for iterating on the value of \({k}_{\mathit{var}}\) leads to the following results:

Criteria: \({w}_{\mathit{ksup}}<\mathrm{0,15}\mathit{mm}\); \({w}_{\mathit{kinf}}<\mathrm{0,15}\mathit{mm}\); \({\mathrm{\sigma }}_{c,\mathit{compression}}<\mathrm{15,75}\mathit{MPa}\)

\({k}_{\mathit{var}}\)	1,0	0.5	0.5	0.25	0.25	0.625	0.438	0.343	0.391	0.391	0.367	…	0.3697
\({A}_{\mathit{ssup}}({\mathit{cm}}^{2})\)	7.83	15.55	15.55	31.72	31.72	12.29	17.85	23.28	19.74	19.74	21.39	…	21.201
\({A}_{\mathit{sinf}}({\mathit{cm}}^{2})\)	0.0	0.0	0.0	0.0	0.0	0.0	0.0	0.0	0.0	…	0.0
\({x}_{\mathit{AN}}(\mathit{cm})\)	15.41	20.08	20.08	25.5	25.5	18.41	21.09	23.09	21.84	22.45	…	22.38
\({\mathrm{\sigma }}_{\text{csup}}(\mathit{MPa})\)	-35.36	-19.24	-19.24	-9.96	-23.85	-16.92	-13.46	-15.18	-15.18	-15.18	-14.32	…	-14.412
\({\mathrm{\sigma }}_{\text{cinf}}(\mathit{MPa})\)	15.75	12.91	12.91	10.37	10.37	13.9	12.34	11.55	11.77	11.66	…	11.676
\({\mathrm{\sigma }}_{\text{ssup}}(\mathit{MPa})\)	-469.13	-250	-250	-250	-125	-312.5	-218.75	-171.88	-195.31	-195.31	-195.31	-183.59	…	-184.87
\({\mathrm{\sigma }}_{\text{sinf}}(\mathit{MPa})\)	174.91	155.09	155.09	131.1	131.1	163.24	143.24	144.18	143.79	…	143.84
\({w}_{\mathit{ksup}}(\mathit{mm})\)	0.612	0.265	0.265	0.075	0.075	0.369	0.129	0.168	0.168	0.148	…	0.149996
\({w}_{\mathit{kinf}}(\mathit{mm})\)	0.0	0.0	0.0	0.0	0.0	0.0	0.0	0.0	0.0	…	0.0

The required steel cross section is therefore equal to: \({A}_{\mathit{SYS}}=\mathrm{21,201}\text{cm²}\) and \({A}_{\mathit{SYI}}=0\text{cm²}\)

2.2.3.12. Charging case 12

The beam is subjected to a bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-260000\text{Nm}\).

At the ELSQP:

The algorithm for iterating on the value of \({k}_{\mathit{var}}\) leads to the following results:

Criteria: \({w}_{\mathit{ksup}}<\mathrm{0,15}\mathit{mm}\); \({w}_{\mathit{kinf}}<\mathrm{0,15}\mathit{mm}\); \({\mathrm{\sigma }}_{c,\mathit{compression}}<\mathrm{15,75}\mathit{MPa}\)

\({k}_{\mathit{var}}\)	1,0	0.5	0.25	0.25	0.625	0.438	0.344	0.391	0.391	0.414	…	0.4117
\({A}_{\mathit{ssup}}({\mathit{cm}}^{2})\)	13.73	26.98	26.98	57.41	57.41	20.73	30.84	40.48	35.29	32.6	…	33.732
\({A}_{\mathit{sinf}}({\mathit{cm}}^{2})\)	14.80	6.75	6.75	1.17	1.17	9.65	5.46	3.09	4.24	4.24	5.42	…	4.642
\({x}_{\mathit{AN}}(\mathit{cm})\)	15.64	22.54	22.54	29.9	29.9	19.78	23.92	26.68	25.3	24.38	…	24.84
\({\mathrm{\sigma }}_{\text{csup}}(\mathit{MPa})\)	-34.61	-19.19	-19.19	-10.4	-24.01	-17.17	-13.77	-15.38	-15.38	-15.38	-16.36	…	-15.953
\({\mathrm{\sigma }}_{\text{cinf}}(\mathit{MPa})\)	15.75	15.75	15.75	15.75	15.75	15.75	15.75	15.75	15.75	15.56	…	15.75
\({\mathrm{\sigma }}_{\text{ssup}}(\mathit{MPa})\)	-458.6	-245.89	-245.89	-125	-312.5	-218.08	-171.08	-193.3	-193.3	-207.03	…	-201.25
\({\mathrm{\sigma }}_{\text{sinf}}(\mathit{MPa})\)	175.83	194.32	194.32	201.09	201.09	188.07	196.74	200.83	198.89	198.89	195.16	…	198.21
\({w}_{\mathit{ksup}}(\mathit{mm})\)	0.673	0.212	0.212	0.064	0.064	0.335	0.169	0.109	0.136	0.136	0.155	…	0.1497
\({w}_{\mathit{kinf}}(\mathit{mm})\)	0.0	0.0	0.0	0.0	0.0	0.0	0.0	0.0	0.0	…	0.0

The required steel section is therefore equal to:

\({A}_{\mathit{SYS}}=\mathrm{33,732}\text{cm²}(\mathit{Tendu})\) and \({A}_{\mathit{SYI}}=\mathrm{4,642}\text{cm²}(\mathit{Comprimé})\)

2.2.3.13. Charging case 13

The beam is subjected to a bending moment \(\text{Z}\) equal to \({M}_{\mathit{FZ}}=-380000\text{Nm}\).

At the ELSQP:

The algorithm for iterating on the value of \({k}_{\mathit{var}}\) leads to the following results:

Criteria: \({w}_{\mathit{ksup}}<\mathrm{0,15}\mathit{mm}\); \({w}_{\mathit{kinf}}<\mathrm{0,15}\mathit{mm}\); \({\mathrm{\sigma }}_{c,\mathit{compression}}<\mathrm{15,75}\mathit{MPa}\)

\({k}_{\mathit{var}}\)	1,0	0.5	0.5	0.25	0.625	0.438	…	0.433305
\({A}_{\mathit{ssup}}({\mathit{cm}}^{2})\)	19.05	38.61	38.61	80.27	29.87	43.94	…	44.212
\({A}_{\mathit{sinf}}({\mathit{cm}}^{2})\)	31.91	21.45	21.45	15.38	24.84	19.98	…	20.283
\({x}_{\mathit{AN}}(\mathit{cm})\)	15.18	22.54	22.54	29.9	19.78	23.92	…	23.93
\({\mathrm{\sigma }}_{\text{csup}}(\mathit{MPa})\)	-36.13	-19.18	-19.18	-10.41	-24.01	-17.17	…	-17.06
\({\mathrm{\sigma }}_{\text{cinf}}(\mathit{MPa})\)	15.75	15.75	15.75	15.75	15.75	…	15.65
\({\mathrm{\sigma }}_{\text{ssup}}(\mathit{MPa})\)	-479.66	-245.89	-245.89	-125	-312.5	-218.08	…	-216.65
\({\mathrm{\sigma }}_{\text{sinf}}(\mathit{MPa})\)	173.99	194.32	194.32	201.09	188.07	196.74	…	195.46
\({w}_{\mathit{ksup}}(\mathit{mm})\)	0.604	0.187	0.187	0.061	0.288	0.152	…	0.149999
\({w}_{\mathit{kinf}}(\mathit{mm})\)	0.0	0.0	0.0	0.0	0.0	0.0	…	0.0

The required steel section is therefore equal to:

\({A}_{\mathit{SYS}}=\mathrm{44,212}\text{cm²}(\mathit{Tendu})\) and \({A}_{\mathit{SYI}}=\mathrm{20,283}\text{cm²}(\mathit{Comprimé})\)

2.2.3.14. Charging case 14

The beam is subjected to a compression of \(N=-4500000\text{N}\), to a bending moment of \({M}_{\mathit{FZ}}=380000\text{Nm}\) and to a shear force of \({V}_{z}=100000\text{N}\).

Longitudinal reinforcement:

At the ELS QP:

We start the calculation at ELS Characteristic (criterion: stress limitation, with \({\mathrm{\sigma }}_{\mathit{clim}}={{\mathrm{\sigma }}_{b}}^{\mathit{elsqp}}=\mathrm{15,75}\mathit{MPa}\) for the compression pivot of concrete and \({\mathrm{\sigma }}_{\mathit{slim}}={k}_{\mathit{var}}^{A}\times {f}_{\mathit{yk}}=1\times 500=500\mathit{MPa}\) for the traction pivot in line with tension steel).

The resistant moment of concrete is equal to:

\({M}_{\text{lim}}=\frac{1}{2}{{\mathrm{\sigma }}_{b}}^{\mathit{elsqp}}h(d-\frac{{y}_{\text{lim}}}{3})=\mathrm{0,5}\times {\mathrm{15,75}}^{\mathit{MPa}}\times {\mathrm{0,3}}^{m}\times ({\mathrm{0,46}}^{m}-{\mathrm{0,1476}}^{m}/3)=143253\text{Nm}\)

with \({y}_{\text{lim}}=d\frac{n{{\mathrm{\sigma }}_{b}}^{\mathit{elsqp}}}{n{{\mathrm{\sigma }}_{b}}^{\mathit{elsqp}}+{\mathrm{\sigma }}_{\mathit{slim}}}=({b}_{w}-{c}_{\text{y,inf}})\frac{n{{\mathrm{\sigma }}_{b}}^{\mathit{elsqp}}}{n{{\mathrm{\sigma }}_{b}}^{\mathit{elsqp}}+{\mathrm{\sigma }}_{\mathit{slim}}}=({\mathrm{0,5}}^{m}-{\mathrm{0,04}}^{m})\times (\frac{15\times {\mathrm{15,75}}^{\mathit{MPa}}}{15\times {\mathrm{15,75}}^{\mathit{MPa}}+{500}^{\mathit{MPa}}})=\mathrm{0,1476}\text{m}\)

The moment to resume is:math: `M=left| {M} | {M} _ {mathit {FZ}}right|-N (d-frac {{b} _ {w}} {2}) =1325000text {Nm} `

The section is fully compressed, as we are in the case where \(M>{M}_{\text{lim}}=250527\text{Nm}\) Thus, compression reinforcement is required.

The section is therefore entirely compressed, and the balance changes to PIVOT B (\({\mathrm{\sigma }}_{c}={{\mathrm{\sigma }}_{b}}^{\mathit{elsqp}}\)) such that the problem is solved by iteration; in the end, we obtain the optimal solution below:

• Extreme fiber stresses: \({\mathrm{\sigma }}_{\text{c,sup}}=\mathrm{15,75}\mathit{MPa}\), \({\mathrm{\sigma }}_{\text{c,inf}}=\mathrm{15,75}\mathit{MPa}\)

• Resistant compressive force of concrete: \({N}_{\mathit{cc}}=\mathrm{0,5}{b}_{w}h\cdot ({\mathrm{\sigma }}_{\text{c,sup}}+{\mathrm{\sigma }}_{\text{c,inf}})=2362500N\)

• Moment resistant to the center of gravity of the section: \({M}_{\mathit{cc}}=(1/12)h{b}_{w}^{2}\cdot ({\mathrm{\sigma }}_{\text{c,sup}}-{\mathrm{\sigma }}_{\text{c,inf}})=\mathrm{0,0}\mathit{Nm}\)

The dimensioning reinforcement is then deduced from the equilibrium equations below:

• \(n\cdot {A}_{\mathit{SYI}}{\mathrm{\sigma }}_{\mathit{SYI}}+n\cdot {A}_{\mathit{SYS}}{\mathrm{\sigma }}_{\mathit{SYS}}=N-{N}_{\mathit{cc}}\)

• \(-n\cdot {A}_{\mathit{SYI}}{\mathrm{\sigma }}_{\mathit{SYI}}\cdot ({b}_{w}/2-{c}_{\text{inf}})+n\cdot {A}_{\mathit{SYS}}{\mathrm{\sigma }}_{\mathit{SYS}}\cdot ({b}_{w}/2-{c}_{\text{sup}})={M}_{\mathit{FZ}}-{M}_{\mathit{cc}}\)

Which results in: \({A}_{\mathit{SYI}}=\mathrm{6,941}\text{cm²}\) and \({A}_{\mathit{SYS}}=\mathrm{83,535}\text{cm²}\) (Compression reinforcement)

With regard then to the verification of the crack openings, the section being entirely compressed, no cracking will take place, namely:

\({w}_{\mathit{ksup}}=0\mathit{mm}<{{w}_{\mathit{max}}}^{s}\) and \({w}_{\mathit{kinf}}=0\mathit{mm}<{{w}_{\mathit{max}}}^{i}\)

The reinforcement obtained through the calculation of stress limitations is therefore retained.

Transverse reinforcement:

Similar to the calculation in ELU, replacing \({f}_{\mathit{cd}}\) with \({\mathrm{\sigma }}_{\text{c,lim,NL}}\)

2.3. Uncertainties about the solution

None.