2. Benchmark solution#

2.1. Calculation method#

The densities of the longitudinal steels are calculated according to the Capra and Maury method. Given the directions of the efforts, the « dimensioning » facet is obvious. The analytical calculation therefore boils down to a cross-section calculation making it possible to determine the forces to which the 2 steel beds (upper and lower) are subjected.

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 the case where compression steel is required).

2.2.1.1. Charging case 1#

The plate is subjected to a compression of \({N}_{\mathit{yy}}=-1000000\text{N}\) and to a shear force of \({Q}_{y}=100000\text{N}\).

Longitudinal reinforcement:

The compressive strength of unreinforced concrete is given by:

\({N}_{\mathit{Rd}}={A}_{c}\times {f}_{\mathit{cd}}=\mathrm{0,20}m\times \mathrm{1,0}m\times (\frac{35\mathit{MPa}}{\mathrm{1,5}})=\mathrm{4,67}\text{MN}>1\text{MN}\)

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

F**transverse reinforcement:**

For the calculation in BAEL, only the shear force is taken into account:

\({A}_{\mathit{ST}}=\frac{\sqrt{{Q}_{x}^{2}+{Q}_{y}^{2}}}{z(\mathrm{cot}\mathrm{\theta }+\mathrm{cot}\mathrm{\alpha })\mathrm{sin}\mathrm{\alpha }{\mathrm{\sigma }}_{s}}=\frac{\sqrt{{Q}_{x}^{2}+{Q}_{y}^{2}}}{z{\mathrm{\sigma }}_{s}}\) where \(\mathrm{\alpha }=90°\text{et}\mathrm{\theta }=45°\) and \(z=\mathrm{0,9}(h-c)\)

Be \({A}_{\mathit{ST}}=\mathrm{15,967}\mathit{cm}\mathrm{²}/m\mathrm{²}\)

In Eurocode 2, the calculation of the transverse reinforcement is different, it takes into account the normal compression force and the aim is to start by checking the resistance of non-reinforced concrete transversely to the shear force, given by the formula below:

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

Either:

\({V}_{\mathit{Rd},c}=[\mathit{max}(\frac{\mathrm{0,18}}{\mathrm{1,5}}\cdot (1+\sqrt{\frac{200}{160}})\cdot {(100\times 0\times 35)}^{1/3};\frac{\mathrm{0,35}}{\mathrm{1,5}}\cdot {35}^{\mathrm{0,5}})+\mathrm{0,15}\times (\frac{\mathrm{1,0}}{\mathrm{0,20}})]\times \mathrm{0,16}=341000N\)

This term is greater than the maximum value of the shear forces of “facets”, obtained for the facet oriented perpendicular to the Y axis of the plate (\(\mathrm{\theta }=\frac{\mathrm{\pi }}{2}\)), and which is equal to 100,000 N.

Therefore, no cross reinforcement is required.

2.2.1.2. Charging case2#

The plate is subjected to a tensile force of \({N}_{\mathit{xx}}=1000000\text{N}\) along the \(\text{X}\) axis and to a shear force \({Q}_{x}=-600000\text{N}\).

Longitudinal reinforcement:

I It is a section that is fully extended symmetrically.

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

Each frame bed therefore takes up half of the effort i.e.: \({A}_{\mathit{SXS}}={A}_{\mathit{SXI}}=\frac{{A}_{S}}{2}=\mathrm{11,5}{\mathit{cm}}^{2}\)

F**transverse reinforcement:**

For BAEL, only the cutting effort is taken into account:

For the « user » code \({A}_{\mathit{ST}}=\mathrm{95,785}\mathit{cm}\mathrm{²}/m\mathrm{²}\), for the code BAEL \({A}_{\mathit{ST}}=\mathrm{95,833}\mathit{cm}\mathrm{²}/m\mathrm{²}\)

In Eurocode 2, the calculation of transverse reinforcement is different, it takes into account the normal compressive force (in this case, it is a tensile force, no impact) and the aim is to start by checking the resistance of the non-reinforced concrete transversely to the shear force, given by the formula below:

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

Either:

\({V}_{\mathit{Rd},c}=[\mathit{max}(\frac{\mathrm{0,18}}{\mathrm{1,5}}\cdot (1+\sqrt{\frac{200}{160}})\cdot {(100\times (\frac{\mathrm{11,5}\cdot {10}^{-4}}{\mathrm{0,16}})\times 35)}^{1/3};\frac{\mathrm{0,35}}{\mathrm{1,5}}\cdot {35}^{\mathrm{0,5}})+\mathrm{0,15}\times (\frac{0}{\mathrm{0,20}})]\times \mathrm{0,16}=220867N\)

This term is less than the maximum value of the shear forces of “facets”, obtained for the facet oriented perpendicular to the X axis of the plate (\(\mathrm{\theta }=0\)), and which is equal to 600,000 N.

Therefore, transverse reinforcement is required, and is calculated according to the Ritter-Mosche truss model; in the end, we obtain:

\({A}_{\mathit{SXT}}=\mathrm{40,11}\mathit{cm}\mathrm{²}/m\mathrm{²}\) and \({A}_{\mathit{SYT}}=\mathrm{14,09}\mathit{cm}\mathrm{²}/m\mathrm{²}\)

2.2.1.3. Charging case 3#

The plate is subjected to a tensile force of \({N}_{\mathit{yy}}=1000000\text{N}\) along the \(\text{Y}\) axis and to a shear force \({Q}_{x}=-20000\text{N}\) and \({Q}_{y}=80000\text{N}\).

Longitudinal reinforcement:

The theoretical results are symmetric to those of configuration 2: \({A}_{\mathit{SYS}}={A}_{\mathit{SYI}}=\mathrm{11,5}{\mathit{cm}}^{2}\)

F**transverse reinforcement:**

For BAEL, only the cutting effort is taken into account:

where \(\alpha =90°\text{et}\theta =45°\) with \(z=h-2c\) (BAEL)

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

This term is greater than the maximum value of the shear forces of “facets”.

Therefore, no cross reinforcement is required.

2.2.1.4. Charging case4#

The plate is subjected to a bending moment \(\text{Y}\) equal to \({M}_{\mathit{yy}}=100000\text{Nm}\). This bending moment corresponds to a stretched upper fiber.

At the ELU:

The ultimate moment reduced \(\mathrm{\mu }=\frac{{M}_{\mathit{yy}}}{d\mathrm{²}{\mathrm{\sigma }}_{b}}=0.167\).

The relative position of neutral fiber \(\mathrm{\alpha }=1-\sqrt{1-2\mathrm{\mu }}=0.184\).

The \(z=d(1-\frac{\mathrm{\alpha }}{2})=0.145\) reduced lever arm.

The frame section is therefore equal to \({A}_{\mathit{SYS}}=\frac{{M}_{\mathit{yy}}}{z{\mathrm{\sigma }}_{s}}=15.835\text{cm²/m}\) (upper \(\text{Y}\) bed).

2.2.1.5. Charging case5#

The plate is subjected to a bending moment \(\text{X}\) equal to \({M}_{\mathit{xx}}=100000\text{Nm}\). This bending moment corresponds to a stretched upper fiber.

The theoretical results are symmetric to those of configuration 4.

The frame section is therefore equal to \({A}_{\mathit{SXS}}=15.835\text{cm²/m}\) (bed \(\text{X}\) greater) than ELU.

2.2.1.6. Charging case6#

The plate is subjected to a bending moment \(\text{X}\) equal to \({M}_{\mathit{xx}}=100000\text{Nm}\) and to a compression force following \(\text{X}\) equal to \({N}_{\mathit{xx}}=-100000\text{N}\).

The section is partially stretched.

At the ELU:

The time to start again is \(M={M}_{\mathit{xx}}-{N}_{\mathit{xx}}(d-\frac{h}{2})=106000\text{Nm}\)

The ultimate moment reduced \(\mathrm{\mu }=\frac{M}{d\mathrm{²}{\mathrm{\sigma }}_{b}}=0.177\).

The relative position of neutral fiber \(\mathrm{\alpha }=1-\sqrt{1-2\mathrm{\mu }}=0.197\).

The \(z=d(1-\frac{\mathrm{\alpha }}{2})=0.144\) reduced lever arm.

The frame section is therefore equal to \({A}_{\mathit{SXS}}=\frac{{M}_{\mathit{xx}}}{z{\mathrm{\sigma }}_{s}}+\frac{{N}_{\mathit{xx}}}{{\mathrm{\sigma }}_{s}}=14.601\text{cm²/m}\) (upper \(\text{X}\) bed).

2.2.1.7. Charging case7#

The plate is subjected to a following bending moment \(\text{X}\) equal to \({M}_{\mathit{xx}}=100000\text{Nm}\) and to a tensile force equal to \({N}_{\mathit{xx}}=100000\text{N}\).

The moment to resume is:math: `M=| {M} _ {mathit {xx}} |- {N} _ {mathit {xx}} (d-frac {h} {2}) =94000text {Nm}} `

The section is therefore partially stretched.

At the ELU:

The ultimate moment reduced \(\mathrm{\mu }=\frac{M}{d\mathrm{²}{\mathrm{\sigma }}_{b}}=0.157\).

The relative position of neutral fiber \(\mathrm{\alpha }=1-\sqrt{1-2\mathrm{\mu }}=0.172\).

The \(z=d(1-\frac{\mathrm{\alpha }}{2})=0.146\) reduced lever arm.

2.2.1.8. Charging case 8#

The plate is subjected to a bending moment \(\text{X}\) equal to \({M}_{\mathit{xx}}=100000\text{Nm}\) and to a tensile force equal to \({N}_{\mathit{xx}}=2000000\text{N}\) along \(\text{X}\).

The section is completely extended:math: M=| {M} _ {mathit {xx}} |- {N} _ {mathit {xx}} (d-frac {h} {2}) =-20000text {Nm} <0

At the ELU:

The reinforcement cross section is therefore equal to:

\({A}_{\mathit{SXS}}=\frac{M}{(d-e){\mathrm{\sigma }}_{s}}+\frac{{N}_{\mathit{xx}}}{{\mathrm{\sigma }}_{s}}=42.167\text{cm²/m}\) (\(\text{X}\) upper bed).

\({A}_{\mathit{SXI}}=\frac{-M}{(d-e){\mathrm{\sigma }}_{s}}=3.833\text{cm²/m}\) (lower \(\text{X}\) bed).

2.2.1.9. Charging case 9#

The plate is subjected to a following bending moment \(\text{X}\) equal to \({M}_{\mathit{xx}}=100000\text{Nm}\) and to a following bending moment \(\text{Y}\) equal to \({M}_{\mathit{yy}}=-75000\text{Nm}\).

The section is partially stretched.

At the ELU:

The frame section following \(\text{X}\) is the same as configuration 5.

Let’s say \({A}_{\mathit{SXS}}=15.835\text{cm²/m}\) (lower \(\text{X}\) bed).

Next \(\text{Y}\):

The ultimate moment reduced \(\mathrm{\mu }=\frac{{M}_{\mathit{yy}}}{d\mathrm{²}{\mathrm{\sigma }}_{b}}=0.125\).

The relative position of neutral fiber \(\mathrm{\alpha }=1-\sqrt{1-2\mathrm{\mu }}=0.135\).

The \(z=d(1-\frac{\mathrm{\alpha }}{2})=0.149\) reduced lever arm.

The frame section is therefore equal to \({A}_{\mathit{SYI}}=\frac{{M}_{\mathit{yy}}}{z{\mathrm{\sigma }}_{s}}=11.559\text{cm²/m}\) (lower \(\text{Y}\) bed).

2.2.1.10. Charging case 10#

The plate is subjected to a bending moment \(\text{Y}\) equal to \({M}_{\mathit{yy}}=-150000\text{Nm}\).

The section is partially stretched.

At the ELU:

The ultimate moment reduced \(\mathrm{\mu }=\frac{{M}_{\mathit{fx}}}{d\mathrm{²}{\mathrm{\sigma }}_{b}}=0.251\).

The relative position of neutral fiber \(\mathrm{\alpha }=1-\sqrt{1-2\mathrm{\mu }}=0.294\).

The \(z=d(1-\frac{\mathrm{\alpha }}{2})=0.136\) reduced lever arm.

The frame section is therefore equal to \({A}_{\mathit{SYI}}=\frac{{M}_{\mathit{yy}}}{z{\mathrm{\sigma }}_{s}}=25.285\text{cm²/m}\) (lower \(\text{Y}\) bed).

2.2.1.11. Charging case 11#

The plate is subjected to a bending moment \(\text{Y}\) equal to \({M}_{\mathit{yy}}=-260000\text{Nm}\).

The section is partially stretched.

The ultimate moment reduced \(\mathrm{\mu }=\frac{{M}_{\mathit{yy}}}{d\mathrm{²}{\mathrm{\sigma }}_{b}}=0.435\).

The relative position of neutral fiber \(\mathrm{\alpha }=1-\sqrt{1-2\mathrm{\mu }}=0.642\).

The \(z=d(1-\frac{\mathrm{\alpha }}{2})=0.109\) reduced lever arm.

The frame section is therefore equal to \({A}_{\mathit{SYI}}=\frac{{M}_{\mathit{yy}}}{z{E}_{A}{\mathrm{\epsilon }}_{A}}=130.26\text{cm²/m}\) (lower \(\text{Y}\) bed).

In the case where compression reinforcement is possible (CONFIG 2 - FERR_COMP = “OUI”):

In order to prevent the lower steel from falling into the “elastic” domain of its law of behavior (which goes against optimization reasoning), the position of the neutral fiber is then blocked at the following depth: \(\mathrm{\alpha }={\mathrm{\alpha }}_{R}=\frac{1}{1+\frac{{\mathrm{\epsilon }}_{\mathit{sy}}}{{\mathrm{\epsilon }}_{\mathit{cu}2}}}=\frac{1}{1+0.207/0.35}=0.628\) (Pivot B “locked”)

The balance of forces and moments will then lead to the determination of the two unknowns of the problem, namely the upper steel sections (compressed steel) and lower steel sections (tension steel):

\({A}_{\mathit{SYI}}\times {\mathrm{\sigma }}_{\mathit{SYI}}+{A}_{\mathit{SYS}}\times {\mathrm{\sigma }}_{\mathit{SYS}}+{F}_{c}=0\)

\(-{A}_{\mathit{SYI}}\times {\mathrm{\sigma }}_{\mathit{SYI}}\cdot (h-{c}_{\text{inf}})+{A}_{\mathit{SYS}}\times {\mathrm{\sigma }}_{\mathit{SYS}}\cdot (h-{c}_{\text{sup}})+{M}_{c}=0\)

Such as: \({\mathrm{\sigma }}_{\mathit{SYI}}<0(\mathit{Traction})\); \({\mathrm{\sigma }}_{\mathit{SYS}}>0(\mathit{Compression})\)

\({F}_{c}=\mathrm{\lambda }\cdot \mathrm{\alpha }\cdot d\times \mathrm{\eta }\cdot {f}_{\mathit{cd}}\); \({M}_{c}=\mathrm{\lambda }\cdot \mathrm{\alpha }\cdot d\times \mathrm{0,5}\cdot (h-\mathrm{\lambda }\cdot \mathrm{\alpha }\cdot d)\times \mathrm{\eta }\cdot {f}_{\mathit{cd}}\)

We will then find:

\({A}_{\mathit{SYI}}=50.57\text{cm²/m}\) and \({A}_{\mathit{SYS}}=6.83\text{cm²/m}\)

2.2.1.12. Charging case 12#

The plate is subjected to a bending moment \(\text{Y}\) equal to \({M}_{\mathit{yy}}=-380000\text{Nm}\).

The ultimate reduced moment is therefore:

\(\mathrm{\mu }=\frac{{M}_{\mathit{yy}}}{d\mathrm{²}{\mathrm{\sigma }}_{b}}=0.63>{\mathrm{\mu }}_{\mathit{BC}}=\mathrm{\lambda }\cdot {\mathrm{\alpha }}_{\mathit{BC}}\cdot (1-\mathrm{0,5}\cdot \mathrm{\lambda }\cdot {\mathrm{\alpha }}_{\mathit{BC}})=0.8\times 1.0\times (1-0.5\times 0.8\times 1.0)=0.48\).

So we are:

Or in PIVOT C, such that the upper and lower steels are completely compressed.
Or in PIVOT B, with compression steel (the lower steel remains tense).

The resolution is carried out iteratively, and the algorithm contains the configuration leading to optimal reinforcement (i.e. to a minimum total reinforcement cross section); in this case, the optimal configuration is obtained below:

Balance: \(\mathit{PIVOT}B\) \(\mathrm{\alpha }={\mathrm{\alpha }}_{R}=0.628\)
Reinforcement obtained: \({A}_{\mathit{SYI}}=73.81\text{cm²/m}\) and \({A}_{\mathit{SYS}}=29.83\text{cm²/m}\)

2.2.1.13. Charging case 13#

The plate is subjected to a compression of \({N}_{\mathit{yy}}=-1500000\text{N}\) and to a shear force of \({Q}_{y}=800000\text{N}\).

Longitudinal reinforcement:

The compressive strength of unreinforced concrete is given by:

\({N}_{\mathit{Rd}}={A}_{c}\times {f}_{\mathit{cd}}=\mathrm{0,20}m\times \mathrm{1,0}m\times (\frac{35\mathit{MPa}}{\mathrm{1,5}})=\mathrm{4,67}\text{MN}>1\mathrm{,5}\text{MN}\)

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

F**transverse reinforcement:**

For BAEL, only the cutting effort is taken into account:

where \(\mathrm{\alpha }=90°\text{et}\mathrm{\theta }=45°\) and \(z=h-2c\) (BAEL)

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

Either:

This term is less than the maximum value of the “facet” shear forces, obtained for the facet oriented perpendicular to the Y axis of the plate (\(\mathrm{\theta }=\mathrm{\pi }/2\)), and which is equal to 800,000 N.

Therefore, transverse reinforcement is required, and is calculated according to the Ritter-Mosche truss model; in the end, we obtain:

\({A}_{\mathit{SXT}}=\mathrm{13,02}\mathit{cm}\mathrm{²}/m\mathrm{²}\) and \({A}_{\mathit{SYT}}=\mathrm{63,15}\mathit{cm}\mathrm{²}/m\mathrm{²}\)

2.2.1.14. Charging case 14#

The plate is subjected to a compression of \({N}_{\mathit{xx}}=-4500000\text{N}\), to a bending moment of \({M}_{\mathit{xx}}=380000\text{N}\) and to a shear force of \({Q}_{y}=100000\text{N}\).

Longitudinal reinforcement:

The time to start again is \(M={M}_{\mathit{xx}}-{N}_{\mathit{xx}}(d-\frac{h}{2})=650000\text{Nm}\)

The ultimate reduced moment is then:

\(\mathrm{\mu }=\frac{M}{d\mathrm{²}{\mathrm{\sigma }}_{b}}=\frac{0.65}{{0.16}^{2}\times 35/1.5}=1.088>{\mathrm{\mu }}_{\mathit{BC}}=\mathrm{\lambda }\cdot {\mathrm{\alpha }}_{\mathit{BC}}\cdot (1-\mathrm{0,5}\cdot \mathrm{\lambda }\cdot {\mathrm{\alpha }}_{\mathit{BC}})=0.8\times 1.0\times (1-0.5\times 0.8\times 1.0)=0.48\).

So we are:

Or in PIVOT C, such that the upper and lower steels are completely compressed.
Or in PIVOT B, with compression steel (the lower steel remains tense).

Balance: \(\mathit{PIVOT}B\) \(\mathrm{\alpha }={\mathrm{\alpha }}_{R}=0.628\)
Reinforcement obtained: \({A}_{\mathit{SXI}}=81.58\text{cm²/m}\) and \({A}_{\mathit{SXS}}=21.74\text{cm²/m}\)

F**transverse reinforcement:**

In Eurocode 2, the calculation of the transverse reinforcement takes into account the normal compression force and the aim is to start by verifying the resistance of non-reinforced concrete transversely to the shear force, given by the formula below:

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

Either:

\({V}_{\mathit{Rd},c}=[\mathit{max}(\frac{\mathrm{0,18}}{\mathrm{1,5}}\cdot (1+\sqrt{\frac{200}{160}})\cdot {(100\times (\frac{\mathrm{21,74}\times {10}^{-4}}{\mathrm{0,16}})\times 35)}^{1/3};\frac{\mathrm{0,35}}{\mathrm{1,5}}\cdot {35}^{\mathrm{0,5}})+\mathrm{0,15}\times (\frac{\mathrm{4,5}}{\mathrm{0,20}})]\times \mathrm{0,16}\)

\({V}_{\mathit{Rd},c}=760867N\)

Therefore, no cross reinforcement is required.

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 plate is subjected to a compression of \({N}_{\mathit{yy}}=-1000000\text{N}\) and to a shear force of \({Q}_{y}=100000\text{N}\).

Longitudinal reinforcement:

The compressive strength of unreinforced concrete is given by:

\({N}_{\mathit{Rd}}={A}_{c}\times {\mathrm{\sigma }}_{\text{c,lim}}=\mathrm{0,20}m\times \mathrm{1,0}m\times (21\mathit{MPa})=\mathrm{4,2}\text{MN}>1\text{MN}\)

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 plate is subjected to a tensile force of \({N}_{\mathit{xx}}=1000000\text{N}\) along the \(\text{X}\) axis and to a shear force \({Q}_{x}=-600000\text{N}\).

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 half of the effort i.e.: \({A}_{\mathit{SXS}}={A}_{\mathit{SXI}}=\frac{{A}_{S}}{2}=\mathrm{12,5}{\mathit{cm}}^{2}/m\)

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 plate is subjected to a tensile force of \({N}_{\mathit{yy}}=1000000\text{N}\) along the \(\text{Y}\) axis and to a shear force \({Q}_{x}=-20000\text{N}\) and \({Q}_{y}=80000\text{N}\).

Longitudinal reinforcement:

The theoretical results are symmetric to those of configuration 2: \({A}_{\mathit{SYS}}={A}_{\mathit{SYI}}=\mathrm{12,5}{\mathit{cm}}^{2}/m\)

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 plate is subjected to a bending moment \(\text{Y}\) equal to \({M}_{\mathit{yy}}=100000\text{Nm}\). This bending moment corresponds to a stretched upper fiber.

At the ELS:

The resistant moment of concrete is equal to:

\({M}_{\text{lim}}=\frac{1}{2}{\mathrm{\sigma }}_{b}{y}_{\text{lim}}(d-\frac{{y}_{\text{lim}}}{3})=116315\text{Nm}\)

with \({y}_{\text{lim}}=d\frac{n{\mathrm{\sigma }}_{b}}{n{\mathrm{\sigma }}_{b}+{\mathrm{\sigma }}_{s}}=0.0839\text{m}\)

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

The reduced moment of service is equal to: \(\mathrm{\mu }=n\frac{M}{d\mathrm{²}{\mathrm{\sigma }}_{s}}=0.205\)

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: \(\mathrm{\alpha }=0.497\)

The required steel cross section is equal to: \({A}_{\mathit{SYS}}=\frac{{M}_{\mathit{yy}}}{{\mathrm{\sigma }}_{s}d(1-\frac{\mathrm{\alpha }}{3})}=19.14\text{cm²/m}\)

2.2.2.5. Charging case5#

The plate is subjected to a bending moment \(\text{X}\) equal to \({M}_{\mathit{xx}}=100000\text{Nm}\). This bending moment corresponds to a stretched upper fiber.

The theoretical results are symmetric to those of configuration 4.

The frame section is therefore equal to \({A}_{\mathit{SXS}}=19.14\text{cm²/m}\) (bed \(\text{X}\) greater) than ELS.

2.2.2.6. Charging case6#

The plate is subjected to a bending moment \(\text{X}\) equal to \({M}_{\mathit{xx}}=300000\text{Nm}\) and to a compression force following \(\text{X}\) equal to \({N}_{\mathit{xx}}=-20000\text{N}\).

The moment to take into account is: \(M={M}_{\mathit{xx}}-{N}_{\mathit{xx}}(d-\frac{h}{2})=301200\text{Nm}\)

So we are in the case where \(M>{M}_{\text{lim}}\). Thus, compression reinforcement is required, and the resolution is carried out iteratively, so as to seek the equilibrium configuration leading to optimal reinforcement; in this case, we obtain:

Balance configuration: \(y={y}_{\text{lim}}=0.0839m\); \(\mathrm{\sigma }={\mathrm{\sigma }}_{\text{s,lim}}=400\mathit{MPa}\); \(\mathrm{\sigma }={\mathrm{\sigma }}_{\text{c,lim}}=21\mathit{MPa}\)
Reinforcement selected: \({A}_{\mathit{SXI}}=65.78\text{cm²/m}(\mathit{Comprimé})\) and \({A}_{\mathit{SXS}}=61.44\text{cm²/m}(\mathit{Tendu})\)

2.2.2.7. Charging case7#

The plate is subjected to a following bending moment \(\text{X}\) equal to \({M}_{\mathit{xx}}=100000\text{Nm}\) and to a tensile force equal to \({N}_{\mathit{xx}}=100000\text{N}\).

The moment to resume is:math: `M=| {M} _ {mathit {xx}} |- {N} _ {mathit {xx}} (d-frac {h} {2}) =94000text {Nm}} `

The limit values are the same as calculated in configuration 4.

We are in the case where \(M⩽{M}_{\text{lim}}\). Thus, only tensile steels are required.

The reduced moment of service is equal to: \(\mathrm{\mu }=n\frac{M}{d\mathrm{²}{\mathrm{\sigma }}_{s}}=0.192\)

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

By iterative resolution, we get: \(\mathrm{\alpha }=0.134\)

The required steel cross section is equal to:

\({A}_{\mathit{SXS}}=\frac{M}{{\mathrm{\sigma }}_{s}d(1-\frac{\mathrm{\alpha }}{3})}+\frac{{N}_{\mathit{xx}}}{{\mathrm{\sigma }}_{s}}=20.301\text{cm²/m}\)

2.2.2.8. Charging case 8#

The plate is subjected to a bending moment \(\text{X}\) equal to \({M}_{\mathit{xx}}=100000\text{Nm}\) and to a tensile force equal to \({N}_{\mathit{xx}}=2000000\text{N}\) along \(\text{X}\).

The section is completely extended:math: M=| {M} _ {mathit {xx}} |- {N} _ {mathit {xx}} (d-frac {h} {2}) =-20000text {Nm} <0

At the ELS:

The reinforcement cross section is therefore equal to:

\({A}_{\mathit{SXS}}=\frac{M}{(d-e){\mathrm{\sigma }}_{s}}+\frac{{N}_{\mathit{xx}}}{{\mathrm{\sigma }}_{s}}=45.833\text{cm²/m}\) (\(\text{X}\) superior bed)

\({A}_{\mathit{SXI}}=\frac{-M}{(d-e){\sigma }_{s}}=4.167\text{cm²/m}\) (lower \(\text{X}\) bed)

2.2.2.9. Charging case 9#

The section is partially stretched.

At the ELS:

The frame section following \(\text{X}\) is the same as configuration 5.

Let’s be \({A}_{\mathit{SXS}}=18.70\text{cm²/m}\).

Next \(\text{Y}\):

The limit values are the same as calculated in configuration 4.

We are in the case where \(M⩽{M}_{\text{lim}}\). Thus, only tensile steels are required.

The reduced moment of service is equal to: \(\mathrm{\mu }=n\frac{M}{d\mathrm{²}{\mathrm{\sigma }}_{s}}=0.154\)

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

By iterative resolution, we get: \(\mathrm{\alpha }=0.447\)

The required steel cross section is equal to:

\({A}_{\mathit{SYS}}=\frac{M}{{\mathrm{\sigma }}_{s}d(1-\frac{\mathrm{\alpha }}{3})}+\frac{{N}_{\mathit{xx}}}{{\mathrm{\sigma }}_{s}}=13.83\text{cm²/m}\)

2.2.2.10. Charging case 10#

The plate is subjected to a bending moment \(\text{Y}\) equal to \({M}_{\mathit{yy}}=-300000\text{Nm}\).

Balance configuration: \(y={y}_{\text{lim}}=0.0848m\);

\(\mathrm{\sigma }=-391\mathit{MPa}(\mathit{Traction})\); \(\mathrm{\sigma }={\mathrm{\sigma }}_{\text{c,lim}}=21\mathit{MPa}(\mathit{Compression})\)

Reinforcement selected: \({A}_{\mathit{SXI}}=65.35\text{cm²/m}(\mathit{Tendu})\) and \({A}_{\mathit{SXS}}=61.7\text{cm²/m}(\mathit{Comprimé})\)

2.2.3. Calculations at the ELS Quasi-Permanent#

2.2.3.1. Charging case 1#

The plate is subjected to a compression of \({N}_{\mathit{yy}}=-1000000\text{N}\) and to a shear force of \({Q}_{y}=100000\text{N}\).

Longitudinal reinforcement:

The compressive strength of unreinforced concrete is given by:

\({N}_{\mathit{Rd}}={A}_{c}\times {\mathrm{\sigma }}_{\text{c,lim,NL}}=\mathrm{0,20}m\times \mathrm{1,0}m\times (\mathrm{15,75}\mathit{MPa})=\mathrm{3,15}\text{MN}>1\text{MN}\)

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 \({\mathrm{\sigma }}_{\text{c,lim}}\)

2.2.3.2. Charging case2#

The plate is subjected to a tensile force of \({N}_{\mathit{xx}}=1000000\text{N}\) along the \(\text{X}\) axis and to a shear force \({Q}_{x}=-600000\text{N}\).

Longitudinal reinforcement:

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 for modeling (C) are given:

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 \({\mathrm{\sigma }}_{\mathit{slim}}={k}_{\mathit{var}}^{A}\times {f}_{e}=1\times 500=500\mathit{MPa}\) 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: \({\mathrm{\sigma }}_{\text{csup}}=-\mathrm{23,81}\mathit{MPa}(\mathit{TRACTION})\)
Stress at the right of the lower concrete fiber: \({\mathrm{\sigma }}_{\text{cinf}}=-\mathrm{23,81}\mathit{MPa}(\mathit{TRACTION})\)
Calculated steel section: \({A}_{\mathit{ssup}}=\mathrm{10,0}{\mathit{cm}}^{2}/m;{A}_{\mathit{sinf}}=\mathrm{10,0}{\mathit{cm}}^{2}/m\)

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

On the upper side:

\({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}}=\frac{{\mathrm{\sigma }}_{s}-{k}_{t}\times \frac{{f}_{\mathit{ctm}}}{{\mathrm{\rho }}_{\mathit{peff}}}\times (1+{\mathrm{\alpha }}_{e}\times {\mathrm{\rho }}_{\mathit{peff}})}{{E}_{s}}=\frac{\mathrm{500,0}-\mathrm{0,6}\times \frac{\mathrm{3,21}}{\mathrm{0,01}}\times (1+\mathrm{21,0}\times \mathrm{0,01})}{210000}=\mathrm{1,429}\times {10}^{-3}\)

With:

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

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

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

\({s}_{\mathit{rmax}}=\mathit{min}(\mathrm{2,485}\times \mathrm{40,0}+\mathrm{0,8}\times \mathrm{1,0}\times \mathrm{0,425}\times (\mathrm{25,0}/\mathrm{0,01});\mathrm{1,3}\times 200)=\mathrm{260,0}\mathit{mm}\)

From where:

\({w}_{\mathit{ksup}}={s}_{\mathit{rmax}}\times ({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}})=\mathrm{260,0}\times \mathrm{1,429}\times {10}^{-3}=\mathrm{0,371}\mathit{mm}\text{}>\text{}{{w}_{\mathit{max}}}^{s}=\mathrm{0,15}\mathit{mm}\)

On the lower side:

We obtain by symmetry \({w}_{\mathit{kinf}}=\mathrm{0,371}\mathit{mm}\text{}>\text{}{{w}_{\mathit{max}}}^{i}=\mathrm{0,15}\mathit{mm}\)

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

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 \({\mathrm{\sigma }}_{\mathit{slim}}={k}_{\mathit{var}}^{B}\times {f}_{e}=\mathrm{0,5}\times 500=250\mathit{MPa}\) 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: \({\mathrm{\sigma }}_{\mathit{ssup}}=-500\mathit{MPa}(\mathit{TRACTION})\)
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: \({\mathrm{\sigma }}_{\text{csup}}=-\mathrm{11,9}\mathit{MPa}(\mathit{TRACTION})\)
Stress at the right of the lower concrete fiber: \({\mathrm{\sigma }}_{\text{cinf}}=-\mathrm{11,9}\mathit{MPa}(\mathit{TRACTION})\)
Calculated steel section: \({A}_{\mathit{ssup}}=\mathrm{20,0}{\mathit{cm}}^{2}/m;{A}_{\mathit{sinf}}=\mathrm{20,0}{\mathit{cm}}^{2}/m\)

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

On the upper side:

\({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}}=\frac{{\mathrm{\sigma }}_{s}-{k}_{t}\times \frac{{f}_{\mathit{ctm}}}{{\mathrm{\rho }}_{\mathit{peff}}}\times (1+{\mathrm{\alpha }}_{e}\times {\mathrm{\rho }}_{\mathit{peff}})}{{E}_{s}}=\frac{\mathrm{250,0}-\mathrm{0,6}\times \frac{\mathrm{3,21}}{\mathrm{0,02}}\times (1+\mathrm{21,0}\times \mathrm{0,02})}{210000}=\mathrm{7,143}\times {10}^{-4}\)

With:

\({\mathrm{\rho }}_{\mathit{peff}}={A}_{\mathit{ssup}}/{h}_{\mathit{ceff}}=\mathrm{0,02}\)

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

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

From where:

\({w}_{\mathit{ksup}}={s}_{\mathit{rmax}}\times ({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}})=\mathrm{260,0}\times \mathrm{1,429}\times {10}^{-3}=\mathrm{0,186}\mathit{mm}\text{}>\text{}{{w}_{\mathit{max}}}^{s}=\mathrm{0,15}\mathit{mm}\)

On the lower side:

We obtain by symmetry \({w}_{\mathit{kinf}}=\mathrm{0,186}\mathit{mm}\text{}>\text{}{{w}_{\mathit{max}}}^{i}=\mathrm{0,15}\mathit{mm}\)

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

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 \({w}_{\mathit{ksup}/\mathit{inf}}=\mathrm{0,093}\mathit{mm}\text{}<\text{}{{w}_{\mathit{max}}}^{i/s}=\mathrm{0,15}\mathit{mm}\), and now we have \(\mathit{COND}({k}_{\mathit{var}}^{B})=\mathit{VRAI}\).

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):

\({k}_{\mathit{var}}\)	0.625	0.438	0.438	0.344	0.344	0.391	0.402	0.408	0.408	0.405	…	0.4038
\({A}_{\mathit{ssup}/\mathit{inf}}({\mathit{cm}}^{2}/m)\)	16.0	22.86	22.86	29.91	29.91	25.60	24.60	24.85	24.50	24.50	24.67	…	24,762
\({\mathrm{\sigma }}_{c}(\mathit{MPa})\)	-14.88	-10.42	-10.42	-8.18	-9.18	-9.86	-9.58	-9.72	-9.72	-9.65	…	-9.62
\({w}_{\mathit{ksup}/\mathit{inf}}(\mathit{mm})\)	0.232	0.163	0.163	0.163	0.128	0.145	0.149	0.152	0.152	0.151	…	0.1499988

So in the end, we remember: \({A}_{\mathit{SXS}}={A}_{\mathit{SXI}}=\mathrm{24,762}{\mathit{cm}}^{2}/m\)

Transverse reinforcement:

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

2.2.3.3. Charging case 3#

The plate is subjected to a tensile force of \({N}_{\mathit{yy}}=1000000\text{N}\) along the \(\text{Y}\) axis and to a shear force \({Q}_{x}=-20000\text{N}\) and \({Q}_{y}=80000\text{N}\).

Longitudinal reinforcement:

The theoretical results are symmetric to those of configuration 2: \({A}_{\mathit{SYS}}={A}_{\mathit{SYI}}=\mathrm{24,762}{\mathit{cm}}^{2}/m\)

Transverse reinforcement:

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

2.2.3.4. Charging case4#

The plate is subjected to a bending moment \(\text{Y}\) equal to \({M}_{\mathit{yy}}=100000\text{Nm}\). This bending moment corresponds to a stretched upper fiber.

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 \({\mathrm{\sigma }}_{\mathit{slim}}={k}_{\mathit{var}}^{A}\times {f}_{e}=1\times 500=500\mathit{MPa}\) 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): \({x}_{\mathit{AN}}=\mathrm{\alpha }\times d=\mathrm{0,627}\times (20-4)=\mathrm{10,034}\mathit{cm}\)
Stress on the right side of higher steel: \({\mathrm{\sigma }}_{\mathit{ssup}}=-\mathrm{196,65}\mathit{MPa}(\mathit{TRACTION})\)
Stress on the right side of the lower steel: \({\mathrm{\sigma }}_{\mathit{sinf}}=+\mathrm{198,89}\mathit{MPa}(\mathit{COMPRESSION})\)
Stress at the right of the upper concrete fiber: \({\mathrm{\sigma }}_{\text{csup}}=-\mathrm{15,643}\mathit{MPa}(\mathit{TRACTION})\)
Stress at the right of the lower concrete fiber: \({\mathrm{\sigma }}_{\text{cinf}}=+\mathrm{15,75}\mathit{MPa}(\mathit{COMPRESSION})\)
Calculated steel section: \({A}_{\mathit{ssup}}=\mathrm{40,18}{\mathit{cm}}^{2}/m;{A}_{\mathit{sinf}}=\mathrm{0,0}{\mathit{cm}}^{2}/m\)

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

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

\({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}}=\frac{{\mathrm{\sigma }}_{s}-{k}_{t}\times \frac{{f}_{\mathit{ctm}}}{{\mathrm{\rho }}_{\mathit{peff}}}\times (1+{\mathrm{\alpha }}_{e}\times {\mathrm{\rho }}_{\mathit{peff}})}{{E}_{s}}=\frac{\mathrm{196,65}-\mathrm{0,6}\times \frac{\mathrm{3,21}}{\mathrm{0,121}}\times (1+\mathrm{21,0}\times \mathrm{0,121})}{210000}=\mathrm{6,68}\times {10}^{-4}\)

With:

\({\mathrm{\rho }}_{\mathit{peff}}={A}_{\mathit{ssup}}/{h}_{\mathit{ceff}}=\mathrm{0,121}\)

\({h}_{\mathit{ceff}}=\mathit{min}(\mathrm{2,5}\times (h-d);(h-{x}_{\mathit{AN}})/3;h/2)=\mathrm{3,32}\mathit{cm}\)

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

\({s}_{\mathit{rmax}}=\mathit{min}(\mathrm{2,485}\times \mathrm{40,0}+\mathrm{0,8}\times \mathrm{0,5}\times \mathrm{0,425}\times (\mathrm{25,0}/\mathrm{0,121});\mathrm{1,3}\times (200-\mathrm{100,34}))=\mathrm{129,6}\mathit{mm}\)

From where:

\({w}_{\mathit{ksup}}={s}_{\mathit{rmax}}\times ({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}})=\mathrm{129,6}\times \mathrm{6,68}\times {10}^{-4}=\mathrm{0,087}\mathit{mm}\text{}<\text{}{{w}_{\mathit{max}}}^{s}=\mathrm{0,15}\mathit{mm}\)

Since both crack openings are verified, the reinforcement sizing algorithm at ELS QP will stop at this point. In the constrained case, it would have been a question of iterating on the value of \({k}_{\mathit{var}}\) in order to determine the maximum value of this pivot coefficient making it possible to ensure compliance with the two criteria for opening cracks.

So in the end, we remember: \({A}_{\mathit{SYS}}=\mathrm{40,1813}\text{cm²/m}\)

2.2.3.5. Charging case5#

The plate is subjected to a bending moment \(\text{X}\) equal to \({M}_{\mathit{xx}}=100000\text{Nm}\). This bending moment corresponds to a stretched upper fiber.

The theoretical results are symmetric to those of configuration 4.

The frame section is therefore equal to \({A}_{\mathit{SXS}}=\mathrm{40,1813}\text{cm²/m}\) (bed \(\text{X}\) greater than) to the ELS QP.

2.2.3.6. Charging case 6#

The plate is subjected to a bending moment \(\text{X}\) equal to \({M}_{\mathit{xx}}=300000\text{Nm}\) and to a compression force following \(\text{X}\) equal to \({N}_{\mathit{xx}}=-15000\text{N}\).

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	0.75	0.625	…	0.6339	0.6338
\({A}_{\mathit{ssup}}({\mathit{cm}}^{2}/m)\)	73.24	97.59	73.24	79.45	…	76.48	79.450
\({A}_{\mathit{sinf}}({\mathit{cm}}^{2}/m)\)	109.3	93.27	93.27	109.23	103.96	…	106.92	103.960
\({x}_{\mathit{AN}}(\mathit{cm})\)	8.00	9.12	8.00	8.32	…	8.16	8.32
\({\mathrm{\sigma }}_{\text{csup}}(\mathit{MPa})\)	-23.63	-18.79	-18.79	-23.63	-22.11	…	-22.79	-22.11
\({\mathrm{\sigma }}_{\text{cinf}}(\mathit{MPa})\)	15.75	15.75	15.75	15.75	…	15.75	15.75
\({\mathrm{\sigma }}_{\text{ssup}}(\mathit{MPa})\)	-330.75	-249.51	-330.75	-305.31	…	-316.96	-305.31
\({\mathrm{\sigma }}_{\text{sinf}}(\mathit{MPa})\)	165.37	185.68	185.68	165.37	171.74	…	168.18	171.74
\({w}_{\mathit{ksup}}(\mathit{mm})\)	0.163	0.111	0.111	0.163	0.146	…	0.1541	0.146
\({w}_{\mathit{kinf}}(\mathit{mm})\)	0.0	0.0	0.0	0.0	…	0	0.0

The steel section required for the final includes compression reinforcement in the lower layer:

\({A}_{\mathit{SXI}}=\mathrm{103,96}\text{cm²/m}\)

In the upper layer, the reinforcement is normally tense; on the other hand, the iteration algorithm on the Capra-Maury facets results in an optimum reinforcement including reinforcement along the “Y” axis (due to the non-linearity of the problem at ELS QP), such as:

\({A}_{\mathit{SXS}}=\mathrm{80,61}\text{cm²/m}\) and \({A}_{\mathit{SYS}}=\mathrm{4,22}\text{cm²/m}\)

2.2.3.7. Charging case7#

The plate is subjected to a following bending moment \(\text{X}\) equal to \({M}_{\mathit{xx}}=100000\text{Nm}\) and to a tensile force equal to \({N}_{\mathit{xx}}=100000\text{N}\).

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 \({\mathrm{\sigma }}_{\mathit{slim}}={k}_{\mathit{var}}^{A}\times {f}_{e}=1\times 500=500\mathit{MPa}\) 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): \({x}_{\mathit{AN}}=\mathrm{\alpha }\times d=\mathrm{0,537}\times (20-4)=\mathrm{8,598}\mathit{cm}\)
Stress on the right side of higher steel: \({\mathrm{\sigma }}_{\mathit{ssup}}=-\mathrm{242,08}\mathit{MPa}(\mathit{TRACTION})\)
Stress on the right side of the lower steel: \({\mathrm{\sigma }}_{\mathit{sinf}}=+\mathrm{187,54}\mathit{MPa}(\mathit{COMPRESSION})\)
Stress at the right of the upper concrete fiber: \({\mathrm{\sigma }}_{\text{csup}}=-\mathrm{18,347}\mathit{MPa}(\mathit{TRACTION})\)
Stress at the right of the lower concrete fiber: \({\mathrm{\sigma }}_{\text{cinf}}=+\mathrm{15,75}\mathit{MPa}(\mathit{COMPRESSION})\)
Calculated steel section: \({A}_{\mathit{ssup}}=\mathrm{34,18}{\mathit{cm}}^{2}/m;{A}_{\mathit{sinf}}=\mathrm{0,0}{\mathit{cm}}^{2}/m\)

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

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

\({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}}=\frac{{\mathrm{\sigma }}_{s}-{k}_{t}\times \frac{{f}_{\mathit{ctm}}}{{\mathrm{\rho }}_{\mathit{peff}}}\times (1+{\mathrm{\alpha }}_{e}\times {\mathrm{\rho }}_{\mathit{peff}})}{{E}_{s}}=\frac{\mathrm{242,08}-\mathrm{0,6}\times \frac{\mathrm{3,21}}{\mathrm{0,0899}}\times (1+\mathrm{21,0}\times \mathrm{0,0899})}{210000}=\mathrm{8,5818}\times {10}^{-4}\)

With:

\({\mathrm{\rho }}_{\mathit{peff}}={A}_{\mathit{ssup}}/{h}_{\mathit{ceff}}=\mathrm{0,0899}\)

\({h}_{\mathit{ceff}}=\mathit{min}(\mathrm{2,5}\times (h-d);(h-{x}_{\mathit{AN}})/3;h/2)=\mathrm{3,80}\mathit{cm}\)

\({s}_{\mathit{rmax}}=\mathit{min}(\mathrm{2,485}\times \mathrm{40,0}+\mathrm{0,8}\times \mathrm{0,5}\times \mathrm{0,425}\times (\mathrm{25,0}/\mathrm{0,0899});\mathrm{1,3}\times (200-\mathrm{85,98}))=\mathrm{146,7}\mathit{mm}\)

From where:

\({w}_{\mathit{ksup}}={s}_{\mathit{rmax}}\times ({\mathrm{\epsilon }}_{\mathit{sm}}-{\mathrm{\epsilon }}_{\mathit{cm}})=\mathrm{146,7}\times \mathrm{8,5818}\times {10}^{-4}=\mathrm{0,126}\mathit{mm}\text{}<\text{}{{w}_{\mathit{max}}}^{s}=\mathrm{0,15}\mathit{mm}\)

Since both crack openings are verified, the reinforcement sizing algorithm at ELS QP will stop at this point. In the constrained case, it would have been a question of iterating on the value of \({k}_{\mathit{var}}\) in order to determine the maximum value of this pivot coefficient making it possible to ensure compliance with the two criteria for opening cracks.

So in the end, we remember: \({A}_{\mathit{SXS}}=\mathrm{34,18}\text{cm²/m}\)

2.2.3.8. Charging case 8#

The plate is subjected to a bending moment \(\text{X}\) equal to \({M}_{\mathit{xx}}=100000\text{Nm}\) and to a tensile force equal to \({N}_{\mathit{xx}}=2000000\text{N}\) along \(\text{X}\).

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 section is therefore equal to: \({A}_{\mathit{SXS}}=\mathrm{92,20}\text{cm²/m}\) and \({A}_{\mathit{SXI}}=\mathrm{8,38}\text{cm²/m}\)

2.2.3.9. Charging case 9#

At the ELS QP :

The frame section following \(\text{X}\) is the same as configuration 5.

Let’s be \({A}_{\mathit{SXS}}=\mathrm{40,1813}\text{cm²/m}\).

Following \(\text{Y}\), the iteration algorithm 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	0.75	0.75	0.625	0.5625	0.59375	0.578125	…	0.57
\({A}_{\mathit{ssup}}({\mathit{cm}}^{2}/m)\)	0.0	0.0	0.0	0.0	0.0	0.0	0.0	0.0	…	0.0
\({A}_{\mathit{sinf}}({\mathit{cm}}^{2}/m)\)	12.769	22.697	22.697	14.713	14.713	17.512	20.328	18.839	19.562	…	20.184
\({x}_{\mathit{AN}}(\mathit{cm})\)	0.000	8.472	8.472	7.472	7.327	7.777	8.174	7.971	8.071	…	8.155
\({\mathrm{\sigma }}_{\text{csup}}(\mathit{MPa})\)	15.75	13.39	13.39	15.07	15.07	14.07	13.98	14.03	14.01	…	15.75
\({\mathrm{\sigma }}_{\text{cinf}}(\mathit{MPa})\)	-29.49	-18.23	-18.23	-26.09	-22.12	-20.24	-21.18	-21.18	-20.71	…	-30.48
\({\mathrm{\sigma }}_{\text{ssup}}(\mathit{MPa})\)	140.72	148.49	148.49	143.59	143.59	143.54	150.01	146.81	148.42	…	149.72
\({\mathrm{\sigma }}_{\text{sinf}}(\mathit{MPa})\)	-429.35	-250.0	-250.0	-375.0	-312.5	-281.3	-296.9	-296.9	-289.1	…	-282.7
\({w}_{\mathit{ksup}}(\mathit{mm})\)	0	0	0	0	0	0	0	0	…	0
\({w}_{\mathit{kinf}}(\mathit{mm})\)	0.357	0.126	0.126	0.126	0.219	0.172	0.149	0.160	0.155	…	0.15000

We therefore remember:

\({A}_{\mathit{SYI}}=\mathrm{20,184}\text{cm²/m}\)

2.2.3.10. Charging case 10#

The plate is subjected to a bending moment \(\text{Y}\) equal to \({M}_{\mathit{yy}}=-125000\text{Nm}\).

At the ELS QP :

By default, if we restrict ourselves to a configuration without compression steel, the algorithm for iterating on the value of \({k}_{\mathit{var}}\) converges from the first step for the maximum value of the coefficient (equal to 1):

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
\({A}_{\mathit{ssup}}({\mathit{cm}}^{2}/m)\)	0
\({A}_{\mathit{sinf}}({\mathit{cm}}^{2}/m)\)	234.77
\({x}_{\mathit{AN}}(\mathit{cm})\)	14.01
\({\mathrm{\sigma }}_{\text{csup}}(\mathit{MPa})\)	15.75
\({\mathrm{\sigma }}_{\text{cinf}}(\mathit{MPa})\)	-6.73
\({\mathrm{\sigma }}_{\text{ssup}}(\mathit{MPa})\)	236.31
\({\mathrm{\sigma }}_{\text{sinf}}(\mathit{MPa})\)	-46.99
\({w}_{\mathit{ksup}}(\mathit{mm})\)	0.00
\({w}_{\mathit{kinf}}(\mathit{mm})\)	0.1046

So we would remember: \({A}_{\mathit{SYI}}=\mathrm{234,77}\text{cm²/m}(\mathit{Tendu})\)

Moreover, if compression reinforcement is authorized, the algorithm then optimizes the total quantity of reinforcement (so as to make the tension reinforcement work as much as possible in the selected equilibrium configuration); in this case, we will then obtain:

\({k}_{\mathit{var}}\)	1,0	0.5	0.750	…	0.5597
\({A}_{\mathit{ssup}}({\mathit{cm}}^{2}/m)\)	21.785	14.326	21.785	…	15.9120
\({A}_{\mathit{sinf}}({\mathit{cm}}^{2}/m)\)	28.187	39.445	28.187	…	36.216
\({x}_{\mathit{AN}}(\mathit{cm})\)	7.839	9.119	7.839	…	8.8
\({\mathrm{\sigma }}_{\text{csup}}(\mathit{MPa})\)	15.75	15.75	15.75	…	15.75
\({\mathrm{\sigma }}_{\text{cinf}}(\mathit{MPa})\)	-24.429	-18.789	-24.429	…	-20.045
\({\mathrm{\sigma }}_{\text{ssup}}(\mathit{MPa})\)	162.000	185.684	162.000	…	180.409
\({\mathrm{\sigma }}_{\text{sinf}}(\mathit{MPa})\)	-344.250	-249.513	-344.250	…	-270.613
\({w}_{\mathit{ksup}}(\mathit{mm})\)	0.0	0.0	0.0	…	0.0
\({w}_{\mathit{kinf}}(\mathit{mm})\)	0.208	0.126	0.208	…	0.143

We would then remember in this case: \({A}_{\mathit{SYS}}=\mathrm{15,91}\text{cm²/m}(\mathit{Comprimé})\) and \({A}_{\mathit{SYI}}=\mathrm{36,22}\text{cm²/m}(\mathit{Tendu})\)

2.3. Uncertainties about the solution#

None.