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Titrate:
SSLL111 - Multifibre Eléments of beam (straight lines)


Date:
05/11/02
Author (S):
S. MILL, L. DAVENNE, F.GATUINGT Key
:
V3.01.111-A Page:
1/8

Organization (S): EDF-R & D/AMA, LMT Cachan

Handbook of Validation
V3.01 booklet: Linear statics of the linear structures
Document: V3.01.111

SSLL111 - Static Réponse of a beam concrete
armed (section in T) with linear behavior

Summary:

The problem consists in analyzing the response of a concrete beam reinforced via a modeling
multifibre beam. This test corresponds to a static analysis of a beam having a linear behavior.
Three successive loading cases are tested: a specific force, the actual weight and a rise in
temperature. For the first loading case, two grids of the section, one coarse and the other finer are
tested.

Handbook of Validation
V3.01 booklet: Linear statics of the linear structures
HT-66/02/001/A

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Titrate:
SSLL111 - Multifibre Eléments of beam (straight lines)


Date:
05/11/02
Author (S):
S. MILL, L. DAVENNE, F.GATUINGT Key
:
V3.01.111-A Page:
2/8

1 Characteristics
general

1.1 Geometry

Beam in inflection three points, defined by:
y
X
With
B
5 m


With a section in double T:
30 cm
10 C m
10 C m
5 cm
y
12, 5 cm
O
20 C m
Z
10 cm
12,5 cm
8 cm
8 cm
5 cm
20 cm

On this diagram, O is located at middle height of the section.

The total section of higher steels is 3.10­4 m2 and that of lower steels is 4.10­4 m2.

1.2
Material properties

· concrete: E = 2. 1010 Pa; = 0.2; = 2400 kg.m3; = 10-5 K1
· steel: E = 2,1. 1011 Pa; = 0.33; = 7800 kg.m3; = 10-5 K1

1.3
Boundary conditions

Simple support in b: Dy = 0
Support “doubles” in a: dx = Dy = dz = 0 just as X-ray = ry = 0.

1.4 Loadings

Three loading cases are tested successively:
Loading 1: effort concentrated in the mediums of the beam, F = 10000 NR
Loading 2: actual weight of the beam, G = 9,8 Mr. s2
Loading 3: homogeneous heating of the beam T = 100 K
Handbook of Validation
V3.01 booklet: Linear statics of the linear structures
HT-66/02/001/A

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Version
6.2
Titrate:
SSLL111 - Multifibre Eléments of beam (straight lines)


Date:
05/11/02
Author (S):
S. MILL, L. DAVENNE, F.GATUINGT Key
:
V3.01.111-A Page:
3/8

2
Reference solution

Calculations of reference are carried out starting from a simple elastic design in RdM.

2.1 Center
rubber band

In pure bending, for an elastic behavior, the neutral axis passes by the elastic center
(barycentre of the sections balanced by the modules of materials):
C such as E CM dS = 0
S
One determines initially the position of the centers of gravity of the concrete only G and steel only G by
B
has
report/ratio at the point O.

y = 0,125 × 0,3 ×0,05 - 0,125 × 0,2 ×0,05 =1,38888.10-2 m
G B
0,2 × 0,05 + 0,1×0,2 + 0,3× 0,05
y = 0,125 × 3 - 0,125 × 4 = - 1,78571.10-2 m

G has
3+ 4
Z = Z = 0 m
Ga
GB

One can then determine the position compared to O of the elastic center C.

EaSa OGa + EbSb
B
OG
OC =

EaSa + EbSb

The concrete S section is 0,045 m2 and the section of steel S is 7.10­4 m2. The Young modulus of
B
has
concrete is 2.1010 MPa and that of steel 21.1010 MPa. One thus has

y = 2 × 0,045 ×1,38888 - 21× 7.10-4 ×1,78571 = 0,94317.10-2 m
C
2 ×0,045 + 21× 7.10-4

Z = 0 m
C

2.2 Moments
quadratic

The quadratic moments of the rectangular concrete sections are calculated by the formula
following:
3
bh
2
+ B × H × D
12

Where, B represents the width, H the height and D the distance from the center of gravity of the section by report/ratio
with the axis for which one calculates the moment.
One then obtains the quadratic moment of the concrete section compared to axis Z passing by
center elastic:

3
3
,
0 × 05
,
0
×
I
-
-
concrete =
+ (3
,
0 × 05
,
0
) (125
,
0
-
10
.
94317
,
0
)
3
2
2
1
,
0
,
0 2
+
+ (1,
0 ×,
0 2) (
.
94317
,
0
10) 2
2
12
12
3
,
0 2 × 05
,
0
+
+ (,
0 2 × 05
,
0
) (125
,
0
+ 94317
,
0
10
. -) 2
2
3
-
4
=,
0
.
4547 10
m
12

Handbook of Validation
V3.01 booklet: Linear statics of the linear structures
HT-66/02/001/A

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Titrate:
SSLL111 - Multifibre Eléments of beam (straight lines)


Date:
05/11/02
Author (S):
S. MILL, L. DAVENNE, F.GATUINGT Key
:
V3.01.111-A Page:
4/8

Inertias of steels are calculated by the following formula:

4 + S ×d2 S×d2
64

Where, represents the diameter of steel, S the steel section and D the distance from the center of gravity from
section compared to the axis for which one calculates the moment. The diameter of steels being small, one
neglect the first term.
One then obtains the quadratic moment of the steel sections compared to axis Z passing by the center
rubber band:

3.10-4 × (0,125 - 0,94317.10-2) 2 + 4.10-4 × (0,125 + 0,94317.10-2) 2 = 0,1124.10-4 m4

For the complete section of the beam, the quadratic moment balanced by the Young moduli of
materials is:

I.E.(internal excitation) = 2.1010 × 0,4547.10-3 + 21.1010 × 0,1124.10-4 = 11,4544.106 Pa.m4

2.3
Loading case 1

In the case of load 1 (loading concentrated in the middle of the beam), the arrow is calculated by
formulate following RDM:

F L 3
×
F =

48EI

What gives the arrow:
F =
10000 × 53
= 2,2735.10-3 m
48 ×11,4544.106

One can also calculate the following generalized efforts:
F
· the shearing action at the beginning of the beam (left left) is worth
= 5000 NR,
2
F × L
· the bending moment in the middle of the beam is worth:
=1,25.104 N.m.
4

2.4
Loading case 2

In the case of load 2 (actual weight of the beam), the arrow is calculated by the formula of RDM
following:
F = 5 × p × l4
384 I.E.(internal excitation)

where p is the linear load due to the weight of materials:

p = G (S + S) = 9,8 × (2800 × 7.10-4 + 2400 × 0,045) =1111,9 NR .m-1
has has
B B

What gives the arrow:
F = 5 ×1111,9 × 54 = 7,9.10-4 m
384 ×11,4544.106
Handbook of Validation
V3.01 booklet: Linear statics of the linear structures
HT-66/02/001/A

Code_Aster ®
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Titrate:
SSLL111 - Multifibre Eléments of beam (straight lines)


Date:
05/11/02
Author (S):
S. MILL, L. DAVENNE, F.GATUINGT Key
:
V3.01.111-A Page:
5/8

2.5
Loading case 3

In the case of load 3 (homogeneous rise in temperature), the beam being isostatic and them
dilation coefficients of the concrete and steel being identical, the solution is simple:
The generalized constraints and efforts are null.
The lengthening of the beam is: L = × L ×T

What gives with the values of our case:

L =10-5 ×5 ×100 = 5.10-3 m

3 Modeling
3.1
Characteristics of modeling

Longitudinal grid of the beam:
We have 3 nodes and two elements (POU_D_EM).

With
C
B


The concrete part of the cross section of the beam is with a grid (AFFE_SECT) while steels
are given directly in the form of 4 specific fibers in AFFE_CARA_ELEM (AFFE_PONCT).

Two grids of the concrete part are tested in the case of load 1. The fine grid consists of
120 fibers and the coarse grid consists of 16 fibers:




Note:

The problem being 2D, only one fiber in the width could seem sufficient
(multi-layer), but that would result in having null terms in the matrix of rigidity
(the own inertia of fibers not being taken into account) and with an error at the time of the resolution of
system of equations.
Handbook of Validation
V3.01 booklet: Linear statics of the linear structures
HT-66/02/001/A

Code_Aster ®
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Titrate:
SSLL111 - Multifibre Eléments of beam (straight lines)


Date:
05/11/02
Author (S):
S. MILL, L. DAVENNE, F.GATUINGT Key
:
V3.01.111-A Page:
6/8

3.2 Functionalities
tested

Commands



CREA_MAILLAGE
CREA_GROUP_MA

AFFE_MODELE
MAILLAGE
“MECANIQUE”
“POU_D_EM”

DEFI_MATERIAU
“ELAS”

AFFE_MATERIAU
GROUP_MA

MATER
AFFE_CARA_ELEM
POUTRE
GROUP_MA

SECTION

ORIENTATION
GROUP_MA

CARA
“ANGL_VRIL”

AFFE_SECT
GROUP_MA

MAILLAGE_SECT

TOUT_SECT
“OUI”

COOR_AXE_POUTRE

NOM

AFFE_FIBER
GROUP_MA
“SURFACE”

CARA

VALE

COOR_AXE_POUTRE

NOM

AFFE_CHAR_MECA
MODELE

DDL_IMPO
GROUP_NO

FORCE_NODALE
GROUP_NO

PESANTEUR

TEMP_CALCULEE

MECA_STATIQUE
MODELE

CHAM_MATER

CARA_ELEM

EXCIT
CHARGE

CALC_ELEM
REUSE

RESULTAT

MODELE

CHAM_MATER

CARA_ELEM

OPTION
EFGE_ELNO_DEPL

EXCIT

CALC_NO
REUSE

RESULTAT

OPTION
EFGE_NOEU_DEPL


Handbook of Validation
V3.01 booklet: Linear statics of the linear structures
HT-66/02/001/A

Code_Aster ®
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Titrate:
SSLL111 - Multifibre Eléments of beam (straight lines)


Date:
05/11/02
Author (S):
S. MILL, L. DAVENNE, F.GATUINGT Key
:
V3.01.111-A Page:
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4 Results

4.1
Loading case 1


Reference
Modeling Ar
Relative error %
Arrow
(fine grid)
2,2735 10­3 2,2740
10­3 0,02
Arrow


(coarse grid)
2,2735 10­3 2,2956
10­3 1,0
(1)
Sharp effort



(supports A)
5000
2500
0,0 (2)
Bending moment



(Medium) 1,25
104 6,25
103 0,0
(2)

1) Calculations are carried out without taking into account the own inertia of each fiber. Results
show that it is not nevertheless very useful to hold account of it because the difference between one
coarse grid and a fine grid is not obvious.
The grid of the section does not need to be very fine to have precise results (in
elasticity).
2) Option EFGE_NOEU_DEPL used to calculate the efforts generalized with the nodes does one
average of the generalized efforts of all the elements connected to the node. In our case,
we have 2 superimposed elements of beam (for the concrete, for steel), the efforts
calculated are thus divided by 2.
If one adds the values with efforts by element (EFGE_ELNO_DEPL) of the element concrete and with
the element steel, one finds the theoretical values well.

Note:

If one makes a calculation of arrow by taking O (middle height) like reference axis to the place
elastic center (COOR_AXE_POUTRE), the relative error on the arrow is 0,2% here (bus it
center elastic is practically with middle height (see 1.2.1).

4.2
Loading case 2


Reference
Modeling Ar
Relative error %
Arrow
(fine grid)
7,900 10­4 7,902
10­4 0,02

4.3
Loading case 3


Reference
Modeling Ar
Relative error %
Lengthening 5,00
10­3 5,00
10­3 0,0
Efforts 0,00
0,00
0,0

Handbook of Validation
V3.01 booklet: Linear statics of the linear structures
HT-66/02/001/A

Code_Aster ®
Version
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Titrate:
SSLL111 - Multifibre Eléments of beam (straight lines)


Date:
05/11/02
Author (S):
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:
V3.01.111-A Page:
8/8

5
Summary of the results

The results obtained are in concord with the results of reference.

Handbook of Validation
V3.01 booklet: Linear statics of the linear structures
HT-66/02/001/A

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