SSLS134 - Calculation of reinforcement CAPRA and MAURY: []

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Titre : SSLS134 - Calcul de ferraillage CAPRA et MAURY : [...] Date : 11/07/2018 Page : 1/28Responsable : HAELEWYN Jessica Clé : V3.03.134 Révision :

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SSLS134 - Reinforcement according to the method of Capra and Maury: analytical calculation

Summary:

This test relates to the analytical checking of the densities of reinforcement calculated using the operatorCALC_FERRAILLAGE. Several cases of loadings are studied to the ELS and with the ELECTED OFFICIAL by using 3 codes theBAEL, the eurocode 2 and one code “user”.

Warning : The translation process used on this website is a "Machine Translation". It may be imprecise and inaccurate in whole or in part and isprovided as a convenience.Copyright 2021 EDF R&D - Licensed under the terms of the GNU FDL (http://www.gnu.org/copyleft/fdl.html)



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1 Problem of reference

1.1 Geometry

One considers a square plate of with dimensions 1m and thickness 0,2 m .

1.2 Properties of material

Nothing.

1.3 Boundary conditions and loadings

There is no really mechanical calculation.One arranges oneself so that the field of efforts in the plate corresponds to the one of the 10 followingconfigurations:1. compressive force of 1000000 N exerted along the axis Y , and a shearing action of

100000 N according to Y2. force of traction of 1000000 N exerted along the axis X , and a shearing action of

−600 000 N according to X3. force of traction of 1000000 N exerted along the axis Y , and a shearing action of

−20000 N according to X and of 80000 N according to Y4. bending moment of 100 000 Nm around the axis X .5. bending moment of 100000 Nm around the axis Y .6. bending moment of 100000 Nm around the axis Y and compression forces of 100000 N

(with the ELECTED OFFICIAL) and 20 000 N with the ELS exerted along the axis X 7. bending moment of 100000 Nm around the axis Y and traction forces of 100000 N exerted

along the axis X 8. bending moment of 100000 Nm around the axis Y and traction forces of 2 000000 N

exerted along the axis X 9. bending moment of 100 000 Nm around the axis Y and bending moment of −75000 Nm

around the axis X 10. bending moment of −150000 Nm (with the ELECTED OFFICIAL) and −100000 Nm with

the ELS around the axis X .11. bending moment of −260000 Nm (with the ELECTED OFFICIAL).12. bending moment of −380000 Nm (with the ELECTED OFFICIAL).13. compressive force of 1500 000 N exerted along the axis Y , and a shearing action of

800 000 N according to Y (with the ELECTED OFFICIAL)14. compressive force of 4500000 N exerted along the axis X , and a shearing action of

100000 N according to Y (with the ELECTED OFFICIAL)




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1.4 Other parameters of calculation

Coatings (higher and lower) are fixed at e=4cm . Calculations are carried out with the State Ultimatelimit ( ELU ) and with the State Limit of Service ( ELS ).

Modeling With B C D

Coding USER BAEL EC2 EC2

Lower coating4 cm

4 cm

Higher coating 2 cm

f e : limitcharacteristic ofelasticity of thereinforcing steel

500 MPa 500MPa 500MPa

γs 1.15 1.15 1.15

σ s : elastic limit ofsteel to the

ELECTED OFFICIAL435 MPa σ s=

f eγs

=434,78 MPa

σ sels : elastic limit of

steel with the ELS400 MPa

f cj : resistancecharacteristic in

compression of theconcrete

35 MPa 35 MPa 55 MPa

γc 1.5 1.5 1.5αcc 0.85 1 1

σb : resistance incompression of the

concrete to theELECTED OFFICIAL

23.3 MPaαcc f cj

γc

19,33MPa

αcc f cjγc

23,33 MPa

αcc f cjγc

36,37MPa

σbels : resistance in

compression of theconcrete with the

ELS

21 MPa 33 MPa

Ea : Young modulusof steel

210 GPa

PIV A 10‰ By default:10‰

Cweary steel b:εuk=5 %

PIV A=0,9εuk

Cweary steel a:εuk=2,5 %

PIV A=0,9εuk

PIV B 3,5‰ By default: 3,5‰

NR or CEQUI orALPHA_E: value of

the coefficient ofequivalence with the

ELS

15.1 15.1 21




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2 Reference solution

2.1 Method of calculating

The densities of longitudinal steels are calculated according to the method of Capra and Maury. Takinginto account the directions of the efforts, the “dimensioning” facet is obvious. Analytical calculation isthus summarized with a calculation of section making it possible to determine the efforts to which the2 steel beds are subjected (higher and lower). Three codes are used but the reference solutions aregenerally rather close. The efforts and moments are given in convention “civil engineer”, compressionis positive.

2.2 Sizes and results of reference 2.2.1 Configuration 1

The plate is subjected to a compression of N yy=1000 000 N and with a shearing action

Q y=100 000 N .

Longitudinal reinforcement:ELECTED OFFICIAL: 0 in all the directions.ELS : 0 in all the directions.

Ferraillage transverse:For the codes “user” and BAEL, only the shearing action are taken into account:

AST =√ Q x

2+Q y

2

z (cot θ+cot α)sinασ s

=√ Q x

2+Q y

2

z σ s

where α=90 ° et θ=45 ° and z=0,9(h−c )

For the code “user” AST =15,964cm ² /m ² , for code BAEL AST =15,967cm ²/ m ²

In Eurocode 2, the calculation of transverse reinforcement is different, it takes account of the normaleffort of compression : Qmax=100000 N

Normal effort corresponding to this section ( θ=−π2 ) is of N=N yysin (θ) ²=1000000 N .

σ cp=Nh

>0 One thus has σcpσb

=0,214<0,25 thus αcc=1+σ cpσb

=1,214

v1=0.6(1−f ck

250)=0.516 and thus X =

Qmax

zv1αcc

=0.057 and like X =0.3448 : cotθ=2.5

AST =Qmax

zcotθσ s

=6,3888888cm²/m² with z=0,9 (h−c )

2.2.2 Configuration 2

The plate is subjected to a tractive effort of N xx=−1000000N according to the axis X and of a

shearing action Q x=−600000 N .

Longitudinal reinforcement: IL acts of an entirely tended section in a symmetrical way.

The steel section is thus equal to AS=Nσ s

.

Each bed of reinforcements thus takes again half of the effort is: ASXS=ASXI=AS

2




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USER BAEL Eurocode 2(C and D)

ELECTED

OFFICIAL

11,494 11.5 11.5

ELS 12.5 12.5 12.5

Ldensity D has‘braces theoretical along the axis Y is worthless taking into account the absence ofeffort in this direction.


AST =√ Q x

2+Q y

2


=√ Q x

2+Q y

2

z σ s

where α=90 ° et θ=45°

For the code “user” AST =95,785cm ²/ m ² , for code BAEL AST =95,833cm ² /m ²

Normal effort corresponding to this section ( θ=0 ) is of N=N xx cos(θ) ²=1000000N in traction(thus negative with the direction civil engineer).

In Eurocode 2 (modeling C and D), the calculation of transverse reinforcement is different, it takesaccount of the normal effort of compression : Qmax=600000 N

σ cp=Nh

≤0 One thus has αcc=1

v1=0.6(1−f ck

250)=0.516(C )et 0,468(D) and thus X =

Qmax

zv1 αcc

=0.346 (C )et 0,216 (D)

in (C) X≥0.3448 : cotθ=2,488 from where AST =Qmax

zcotθσ s

=38,524cm²/m² with

z=0,9 (h−c )

in (D) X <0.3448 : cotθ=2.5 AST =Qmax

zcotθσ s

=34,074cm²/m² with z=0,9 (h−csup)


The plate is subjected to a tractive effort of N yy=−1000000 N according to the axis X and of a

shearing action Q x=−20 000 N and Q y=80 000 N .

Longitudinal reinforcement:The theoretical results are symmetrical those of configuration 2: ASYS=ASYI

USER BAEL Eurocode 2

ELECTED

OFFICIAL

11,494 11.5 11.5

ELS 12.5 12.5 12.5





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AST=√Q x

2+Q y2


=√Q x

2+Q y2

z σ s

=13.164/15.805cm²/m²

where α=90 ° et θ=45° with z=0.9(h−c ) (USER) and z=h−2c (BAEL)

In Eurocode 2, the calculation of transverse reinforcement is different, it takes account of the normaleffort of compression : Qmax=82450,44 N

Normal effort corresponding to this section ( θ= π12 ) is of N=N yysin (θ) ²=66987,3 N in traction

(thus negative with the direction civil engineer).

σ cp=Nh

≤0 One thus has αcc=1

v1=0.6(1−f ck

250)=0.516 and thus X =

Qmax

0.9 dv1αcc

=0.057 and like X⩽0.3448 : cotθ=2.5

AST=Qmax

zcot θσ s

=6.32cm²/m² with z=h−2c


The plate is subjected to one bending moment according to Y (around the axis X ) equal toM fx=−100 000 Nm . This bending moment corresponds to a tended top fibre.

With the ELECTED OFFICIAL: (the values are given in the order “user”/BAEL/Eurocode 2)

Reduced ultimate moment μ=M fx

d ²σb

=0.167/0.196 /0.167 .

The relative position of neutral fibre α=1−√1−2μ=0.184 /0.221 /0.184 .

The arm of reduced lever z=d (1−α2)=0.145/0.142/0.145 .

The section of reinforcement is thus equal to ASYS=M fx

z σ s

=15.83/16.165/15.835cm²/m (bed Y

superior).

With the ELS: (the values are given in the order “user”/BAEL/Eurocode 2)The moment of resistance of the concrete is equal to:

M lim=12σb y lim(d−

y lim

3)=101342/101342/116315 Nm

with y lim=dnσb

nσb+σ s

=0.0707 /0.0707/0.0839 m

We are thus if M =M fx⩽M lim . Thus, only of tended steels are necessary.

The reduced moment of service is equal to: μ=nM

d ²σ s

=0.148/0.148/0.205

The coefficient α is solution of the equation: α ³−3α ²−6μ(1−α)=0By iterative resolution, one obtains: α=0.440 /0.440/0.497

The steel section necessary is equal to: ASYS=M fx

σ s d (1−α3

)=18.31/18.31/18.729cm²/m


The theoretical results are symmetrical those of configuration 4.




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The section of reinforcement is thus equal to ASXS=15.83 /16.165 /15.835cm²/m (bed Xsuperior) with the ELECTED OFFICIAL.The section of reinforcement is thus equal to ASXS=18.31 /18.31 /18.729cm²/m (bed X superior)with the ELS.For modeling D, d =h−csup because M fy<0 :

Reduced ultimate moment μ=M fy

d ²σb

=0.0863 .

The relative position of neutral fibre α=1−√1−2μ=0.0904 .

The arm of reduced lever z=d (1−α2)=0.172 .

The section of reinforcement is thus equal to ASXS=M fy

z σ s

=13.383cm²/m (bed Y superior).

With the ELS, LE moment of resistance of the concrete is equal to:

M lim=12

σb y lim(d−y lim

3)=267318 Nm

with y lim=dnσb

n σb+σ s

=0.114 m

We are thus if M =M fx⩽M lim . Thus, only of tended steels are necessary.


d ² σ s

=0.162

The coefficient α is solution of the equation: α ³−3α ²−6μ(1−α)=0By iterative resolution, one obtains: α=0.456

The steel section necessary is equal to: ASYS=M fx

σ s d (1−α3

)=16.378 cm²/m


The plate is subjected to one bending moment according to X (around the axis Y ) equal toM fy=−100 000 Nm and with a compressive force according to X equal to N xx=100 000 N with

the ELECTED OFFICIAL and equal to N xx=20 000N with the ELS.

The section is partially tended.

With the ELECTED OFFICIAL:(the values are given in the order “user”/BAEL/Eurocode 2)

The moment to begin again is M =M fy−N xx (d −h2)=106000 Nm

Reduced ultimate moment μ=M

d ²σb

=0.178/0.209 /0.177 .

The relative position of neutral fibre α=1−√1−2μ=0.197/0.237 /0.197 .

The arm of reduced lever z=d (1−α2)=0.144/0.141 /0.144 .

The section of reinforcement is thus equal to ASXS=Mz σ s

+N xxσ s

=14.596/14.984/14.601cm²/m

(bed X superior).

With the ELS:(the values are given in the order “user”/BAEL/Eurocode 2)

The normal effort of compression is reduced because one would exceed the working stress ofcompression of the concrete.




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The limiting values are the same ones as calculated in configuration 4.

The moment to take into account is: M =M fy−N xx (d −h2)=101200 Nm

We are thus if M ⩽M lim . Thus, only of tended steels are necessary.


d ²σ s

=0.149/0.149 /0.207

The coefficient α is solution of the equation: α ³−3α ²−6μ(1−α)=0By iterative resolution, one obtains: α=0.442 /0.442/0.499

The steel section necessary is equal to: ASXS=M

σ s d (1−α3)+

N xxσ s

=18.044/18.044 /18.47cm²/m


The plate is subjected to one bending moment according to X (around the axis Y ) equal toM fy=−100 000 Nm and with a tractive effort equal to N xx=−100 000 N according to X .

The moment to begin again is M =|M fy|−N xx(d−h2)=94 000 Nm for modelings A/B/C

The moment to begin again is M =|M fy|−N xx (d−h2)=92 000 Nm for modelings D with

d =h−csup because M fy<0 :The section thus is partially tended.

With the ELECTED OFFICIAL: (the values are given in the order “user”/BAEL/Eurocode2(C)/Eurocode 2(D))

Reduced ultimate moment μ=M

d ²σb

=0.157/0.185/0.157 /0.079 .

The relative position of neutral fibre α=1−√1−2μ=0.172 /0.206/0.172/0.0828 .

The arm of reduced lever z=d (1−α2)=0.146 /0.143/0.146/0.173 .

The section of reinforcement is thus equal to

ASXS=Mz σ s

+N xxσ s

=17.079/17.368/17.085/14.563 cm²/m (bed X superior).

With the ELS: (the values are given in the order “user”/BAEL/Eurocode 2(C)/Eurocode 2(D))

The limiting values are the same ones as calculated in configuration 4.We are if M ⩽M lim . Thus, only of tended steels are necessary.


d ² σ s

=0.139/0.139/0.193 /0.149

The coefficient α is solution of the equation: α ³−3α ²−6μ(1−α)=0By iterative resolution, one obtains: α=0.430/0.430 /0.486/0.442The steel section necessary is equal to:

ASXS=M

σ s d (1−α3)+

N xxσ s

=19.642/19.642/ 20.028/17.484 cm²/m





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The plate is subjected to one bending moment according to X (around the axis Y ) equal toM fy=−100 000 Nm and with a tractive effort equal to N xx=−2000000 N according to X .

The section is completely tended M =|M fy|−N xx(d−h2)=−20000 Nm<0 for modelings A/B/C

The moment to begin again is M =|M fy|−N xx(d−h2)=−60 000 Nm for modeling D with

d =h−csup and e=csup because M fy<0With the ELECTED OFFICIAL: (the values are given in the order “user”/BAEL/Eurocode 2)The section of reinforcement is thus equal to:

ASXS=M

(d−e )σ s

+N xxσ s

=42.146/ 42.167/42.167 /37.375cm²/m (bed X superior).

ASXI =−M

(d −e)σ s

=3.831/3.833 /3.833 /8.625 cm²/m (bed X inferior).

With the ELS: the values are identical for the 3 codes.The section of reinforcement is thus equal to:

ASXS=M

(d−e)σ s

+N xxσ s

=45.833cm²/m (bed X superior) for modelings A/B/C

ASXS=M

(d−e )σ s

+N xxσ s

=40.625 cm²/m (bed X superior) for modeling D

ASXI=−M

(d−e)σ s

=4.167 cm²/m (bed X inferior) for modelings A/B/C

ASXI=−M

(d −e)σ s

=4.062 cm²/m (bed X inferior) for modeling D


The plate is subjected to one bending moment according to X (around the axis Y ) equal toM fy=−100000 Nm and at one bending moment according to Y (around the axis X ) equal to

M fx=75000 Nm .


With the ELECTED OFFICIAL: (the values are given in the order “user”/BAEL/Eurocode 2(C)/Eurocode 2 (D))The section of reinforcement according to X is the same one as configuration 5.

That is to say ASXS=15.83/16.165/15.835/13.383cm²/m .

According to Y :

for modeling D with d =h−c inf because M fx>0


d ²σb

=0.125/0.148/0.125 /0.082 .

The relative position of neutral fibre α=1−√1−2μ=0.134 /0.161/0.135/0.086 .

The arm of reduced lever z=d (1−α2)=0.149 /0.147/0.149/0.153 .

The section of reinforcement is thus equal to ASYS =M fx

z σ s

=11.555/11.723 /11.559 /11.263cm²/m

(bed Y superior).




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With the ELS: (the values are given in the order “user”/BAEL/Eurocode 2 (C)/Eurocode 2 (D))The section of reinforcement according to X is the same one as configuration 5.

That is to say ASXS=18.31/18.31/18.729 /16.378 cm²/m .

According to Y :The limiting values are the same ones as calculated in configuration 4.We are if M ⩽M lim . Thus, only of tended steels are necessary.


d ² σ s

=0.111/0.111/0.154/0.162

The coefficient α is solution of the equation: α ³−3α ²−6μ(1−α)=0By iterative resolution, one obtains: α=0.393/0.393/0.447 /0.447The steel section necessary is equal to:

ASYS=M

σ s d (1−α3

)+

N xxσ s

=13.486 /13.486/13.771/13.771cm²/m


The plate is subjected to one bending moment according to Y (around the axis X ) equal toM fx=150 000 Nm with the ELECTED OFFICIAL and with M fx=100000 Nm with the ELS.


With the ELECTED OFFICIAL: (the values are given in the order “user”/BAEL/Eurocode 2)


d ²σb

=0.251/0.295 /0.251 .



The section of reinforcement is thus equal to ASYI=M fx

z σ s

=25.28/26.302 /25.285cm²/m (bed Y

inferior).With the ELS: (the values are given in the order “user”/BAEL/Eurocode 2)The bending moment is reduced because one would exceed the working stress of compression of theconcrete.The limiting values are the same ones as calculated in configuration 4.We are if M ⩽M lim . Thus, only of tended steels are necessary.The configuration is the same one as configuration 5 but for lower steels.The steel section necessary is thus equal to: ASYI=18.31 /18.31 /18.729cm²/m


The plate is subjected to one bending moment according to Y (around the axis X ) equal toM fx=260 000 Nm for calculation with the eurocode 2 and “user” and M fx=220 000 Nm for

calculation with the BAEL. Is calculated only to the ELECTED OFFICIAL. (the values are given in theorder “user”/BAEL/Eurocode 2).



d ²σ b

=0.435/0.433 /0.435 .





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The section of reinforcement is thus equal to ASYI=M fx

z E AεA

=55.014 /46.329/54.971 cm²/m (bed

Y inferior).


The plate is subjected to one bending moment according to Y (around the axis X ) equal toM fx=380000 Nm . Is calculated only to the ELECTED OFFICIAL.


d ²σb

=0.63>0.48 . Steels are completely compressed, the steel

sections are put at -1. Attention, in this case, certain facets of Capra-Maury are partially tended (seeR7.04.05).


The plate is subjected to a compression of N yy=1500000 N and with a shearing action

Q y=8 00000 N .

Longitudinal reinforcement:ELECTED OFFICIAL: -1 (Pivot C not calculated, all the facets are compressed)


AST=√Q x

2+Q y2


=√Q x

2+Q y2

z σ s

=127.714/153.333cm²/m²

where α=90 ° et θ=45 ° and z=0.9(h−c ) (USER) and z=h−2c (BAEL)

In Eurocode 2, the calculation of transverse reinforcement is different, it takes account of the normaleffort of compression :Qmax=800000 N


σ cp=Nh


=0,321>0,25 thus αcc=1,25

v1=0.6(1−f ck

250)=0.516 and thus X =

Qmax

zv1αcc

=0.443 and like X≥0.3448 : cotθ=1.652

AST=Qmax

zcot θσ s

=92.801 cm²/m² with z=h−2c


The plate is subjected to a compression of N yy=2500000 N and with a shearing action

Q y=100000 N .

Longitudinal reinforcement:ELECTED OFFICIAL: -1 (Pivot C not calculated, all the facets are compressed)





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AST=√Q x

2+Q y2

z (cot θ+cot α)sinα σ s

=√Q x

2+Q y2

z σ s

=15.964/19.167 cm²/m²

where α=90 ° et θ=45 ° and z=0.9(h−c ) (USER) and z=h−2c (BAEL)

In Eurocode 2, the calculation of transverse reinforcement is different, it takes account of the normaleffort of compression :Qmax=800000 N


σ cp=Nh


=0,535>0,5 thus αcc=2,5(1−σ cpσb

)=1,161

v1=0.6(1−f ck

250)=0.516 and thus X =

Qmax

zv1αcc

=0.059 and like X≤0.3448 : cotθ=2.5

AST=Qmax

zcot θσ s

=7.667 cm²/m² with z=h−2c

2.3 Uncertainties on the solution

None.




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3 Modeling A

3.1 Characteristics of modeling

A modeling is used DKT. One uses coding “USER”.

3.2 Characteristics of the grid

The grid contains 1 element of the type QUAD4.

3.3 Sizes tested and results

Configuration Absolutelimit

considered

Identification Type ofreference

Value ofreference

cm²/m

Relative error

1

ELECTEDOFFICIAL

DNSXI ‘ANALYTICAL’ 0 0

ELECTEDOFFICIAL

DNSXS ‘ANALYTICAL’ 0 0

ELECTEDOFFICIAL

DNSYI ‘ANALYTICAL’ 0 0

ELECTEDOFFICIAL

DNSYS ‘ANALYTICAL’ 0 0

ELECTEDOFFICIAL

DNST ‘ANALYTICAL’ 15,964 0,001

ELS DNSXI ‘ANALYTICAL’ 0 0

ELS DNSXS ‘ANALYTICAL’ 0 0

ELS DNSYI ‘ANALYTICAL’ 0 0

ELS DNSYS ‘ANALYTICAL’ 0 0

2

ELECTEDOFFICIAL

DNSXI ‘ANALYTICAL’ 11,494 0,002

ELECTEDOFFICIAL

DNSXS ‘ANALYTICAL’ 11,494 0,002

ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL

DNST ‘ANALYTICAL’ 95.785 0,004

ELS DNSXI ‘ANALYTICAL’ 12.5 0

ELS DNSXS ‘ANALYTICAL’ 12.5 0



3 ELECTEDOFFICIAL


ELECTED DNSXS ‘ANALYTICAL’ 0 0




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OFFICIAL

ELECTEDOFFICIAL

DNSYI ‘ANALYTICAL’ 11,494 0,002

ELECTEDOFFICIAL

DNSYS ‘ANALYTICAL’ 11,494 0,002

ELECTEDOFFICIAL




ELS DNSYI ‘ANALYTICAL’ 12.5 0

ELS DNSYS ‘ANALYTICAL’ 12.5 0

4

ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL

DNSYS ‘ANALYTICAL’ 15.83 0,001




ELS DNSYS ‘ANALYTICAL’ 18.31 0,001

5

ELECTEDOFFICIAL


ELECTEDOFFICIAL

DNSXS ‘ANALYTICAL’ 15.83 0,001

ELECTEDOFFICIAL


ELECTEDOFFICIAL



ELS DNSXS ‘ANALYTICAL’ 18.31 0,001



6 ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL



ELS DNSXS ‘ANALYTICAL’ 18,044 0,002




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7

ELECTEDOFFICIAL


ELECTEDOFFICIAL

DNSXS ‘ANALYTICAL’ 17,079 0.0004

ELECTEDOFFICIAL


ELECTEDOFFICIAL



ELS DNSXS ‘ANALYTICAL’ 19,642 0.0001



8

ELECTEDOFFICIAL


ELECTEDOFFICIAL

DNSXS ‘ANALYTICAL’ 42,146 0.0009

ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELS DNSXI ‘ANALYTICAL’ 4,167 0,008




9

ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL





ELS DNSYS ‘ANALYTICAL’ 13,486 0,003

10 ELECTEDOFFICIAL

DNSXI ‘ANALYTICAL’ 25.28 0,001

ELECTEDOFFICIAL


ELECTEDOFFICIAL





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ELECTEDOFFICIAL


ELS DNSXI ‘ANALYTICAL’ 18.31 0,001




11

ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL


12

ELECTEDOFFICIAL

DNSXI ‘ANALYTICAL’ -1 -1

ELECTEDOFFICIAL

DNSXS ‘ANALYTICAL’ -1 -1

ELECTEDOFFICIAL

DNSYI ‘ANALYTICAL’ -1 -1

ELECTEDOFFICIAL

DNSYS ‘ANALYTICAL’ -1 -1

13

ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL


14

ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL





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4 Modeling B


A modeling is used DKT. One uses coding “BAEL”.





considered


Value ofreference

cm²/m

Relative error

1

ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL


ELECTEDOFFICIAL






2

ELECTEDOFFICIAL


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3

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DNSYI ‘ANALYTICAL’ 11.5 0,002

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ELS DNSXS ‘ANALYTICAL’ 18,044 0,002



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DNSXS ‘ANALYTICAL’ 17.368 0.0004

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ELS DNSYS ‘ANALYTICAL’ 13,486 0,003

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DNSYI ‘ANALYTICAL’ 26,302 0,001

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ELS DNSYI ‘ANALYTICAL’ 18.31 0,001


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5 Modeling C


A modeling is used DKT. One uses coding “EC2”.





considered


Value ofreference

cm²/m

Relative error

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ELECTED DNSXS ‘ANALYTICAL’ 0 0




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DNSYI ‘ANALYTICAL’ 11.5 0,002

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7

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6 Modeling D


A modeling is used DKT. One uses coding “EC2”.





considered


Value ofreference

cm²/m

Relative error

1

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DNST ‘ANALYTICAL’ 0 0





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ELECTED DNSXS ‘ANALYTICAL’ 16.588 0,001




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DNST ‘ANALYTICAL’ 0 0





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7 Summary of the results

This test makes it possible to highlight the validity of calculations of density of reinforcement on simple cases.The results got with the model are indeed in conformity with the given values in an analytical way.


Documents

SSLS134 - Calculation of reinforcement CAPRA and MAURY: []