connection degree in composite continuous beams - New Page 1

For steel-concrete continuous beams of bridges, a full connection between the ... case of composite continuous beam in buildings especially connected with ...
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EUROSTEEL 2008, 3-5 September 2008, Graz, Austria

CONNECTION DEGREE IN COMPOSITE CONTINUOUS BEAMS Influence on the bending moment capacity Samy Guezouli a, Mohammed Hjiaja and Quang Huy Nguyena a

Natianal Institure of Applied Science (INSA), Rennes, France

INTRODUCTION For steel-concrete continuous beams of bridges, a full connection between the girder and the slab is generally set contrary to what it is sometimes available for buildings : a partial connection. The Eurocode classification recommends for beams of class 1 in hogging zone (the stronger crosssection) to make a « rigid-plastic » global analysis that leads to an important economy of material while for the beams of classes 2, 3 or 4, an elastic global analysis (EGA) is generally required. It is noted that the EGA is also allowed for cross-sections of class 1 and this work concerns this specific case of composite continuous beam in buildings especially connected with ductile studs. In this case, the plastic design of the connection is possible leading to a full connection and the influence of a partial connection on the bending resistance of the beam could be investigated. On other side, the material non linearities, the concrete creep and shrinkage, the tension stiffening and the temperature difference effects cause a redistribution of the bending moment along the beam. The percentage of this redistribution is clearly influenced by the degree of connection and the Eurocodes predictions [1] appear to be, for some cases, questionable. These predictions represent the maximum moment redistribution from intermediate supports to mid-spans depending on the type of EGA. For “cracked” EGA, the concrete slab is supposed cracked in hogging zones and the maximum percentage of redistribution is 25% at Ultimate Limit States. The user friendly program “Pontmixte” is used to model a symmetrical 3-span beam including different material characteristics mentioned above. The behaviour of the studs must be previously identified by calibrating a non-linear model with experimental results from “push-out” tests. The characteristic curve “Q-γ" (Shear force – Slip) of the stud must be in accordance with a ductile behaviour. The continuous beams under investigation (with cross-sections of class 1 all along the beams), are calculated twice. The first calculation consists in a non linear calculation that should stop when the elastic resistant moment is reached at one of the intermediate supports. This limit correspond to a load level noted F% . The second one is a linear elastic calculation with the same load level F% . The percentage of moment redistribution can be deduced from these computations. 1

PUSH-OUT TEST – DUCTILE STUDS

Several push-out test were carried out with the aim to calibrate a nonlinear behaviour model. During an EGA, the studs are supposed all working at PRd, this hypothesis suppose a perfect adherence between the girder and the slab. Nevertheless, experimental testing revealed the non linear behaviour of the stud is given by :

(

Q = Qu 1 − e −C1 γ

)

C2

with Qu = 1.25PRd = fu (πd2/4)

(1)

where Q is the force at the top of the stud and γ the stud slip. For fixed value of PRd = 121.6 kN (Qu = 152 kN). The stiffness seems to increase (from stud 1 to stud 7) if the parameter C1 increases and the parameter C2 decreases. This influence is not negligible because the slip could double from stud 1 to stud 7 for a load equal to 120 kN for example. Moreover, for fixed values (C1 = 0.7 and C2 = 0.8) for example, the limit force Qu has also an influence on the stud behaviour (Fig. 1). A set of three push-out tests (Fig. 2) were performed in order to get the previous parameters (C1, C2 and Qu)

using a non linear regression with 3 parameters ; the results show that the stiffness increases from stud A to stud C. Fig.3 shows that the values C1 = 0.7 and C2 = 0.8, commonly used, correspond to ductile studs. 160

160 Qu = 152 kN . .

120 100

Stud 1 (C1=0.7 - C2=0.8) Stud 2 (C1=0.7 - C2=0.7) Stud 3 (C1=0.7 - C2=0.6) Stud 4 (C1=0.7 - C2=0.5) Stud 5 (C1=0.8 - C2=0.5) Stud 6 (C1=0.9 - C2=0.5) Stud 7 (C1=1.0 - C2=0.5)

80 60 40 20

Force Q (kN)

Force Q (kN) .

140

100 80

Stud 1a (Qu = 100 kN) Stud 1b (Qu = 110 kN) Stud 1c (Qu = 120 kN) Stud 1d (Qu = 130 kN) Stud 1e (Qu = 140 kN) Stud 1f (Qu = 150 kN)

60 40

0

0

1

2 3 4 Stud slip (mm)

Max mm 5 =66

Fig. 1. Influence of C1 and C2 on Eq. (1). 160

A B C

140 .

120

20

0

Force Q (kN)

140

120 100

0

1

2

3 4 Stud slip (mm)

Max = 66mm 5

Fig. 2. Influence of Qu on Eq. (1)

1% reinforcing bars (2500×140)

Q = Qu(1 − e−C1IγI)C2

80

(200×14.5) Stud A : C1 = 0.85, C2 = 0.51 and Qu = 136.2 kN Stud B : C1 = 1.01, C2 = 0.50 and Qu = 144.8 kN Stud C : C1 = 0.81, C2 = 0.38 and Qu = 155.7 kN Dotted curve : C1 = 0.7, C2 = 0.8 and Qu = 152 kN

60 40 20

8.2

(497×8.4) (200×14.5)

0 0

1

2

3 4 Stud slip (mm)

5

6

Fig. 3. Parameters for ductile studs. 2

Fig. 4. The composite cross-section.

BEAMS UNDER INVESTIGATION

The finite element investigation concerns a symmetrical 3-span beam with various ratios of adjacent span lengths. Keeping always the mid-span length equal to 12 m, the end-spans are varied : 3 m – 6 m – 9 m and then 12 m. The composite cross section remains always the same (Fig. 4). The load is distributed uniformly in accordance with the most critical loading cases : “A” for the maximum moment in sagging zone and “B” for the maximum one in hogging zone (Fig. 5).The non linear material behaviours are shown on Fig. 6. Self-weight

3m

Increasing load

12 m

Fig. 5.a. Loading case “A”.

Self-weight

3m

3m

Increasing load

12 m

Fig. 5.b. Loading case “B”

3m

σ(a)

σ(c)

fu(a) fy(a)

fcm

Arctg(αE(a))

Arctg(E(a))

fck=0.8 fcm

ε(a) εe(a) µ1εe(a) µ2εe(a) Tension

0.4 fcm

Arctg(Ecm)

Tension

εm

fctm

σ(s)

ε(c)

εr

fu(s) fy(s)

Compression

kη − η 2 σ(c ) ε (c) ; η= = f cm 1 + ( k − 2 ) η εm

; k = 1.1E cm

Arctg(E(s))

εm f cm

ε(s) εe(s)

µ2εe(s) Tension

Fig. 6.a. Concrete behaviour. 3

Fig. 6.b. Steel girder and bars behaviours.

FULL AND PARTIAL CONNECTION

The partial connection has no sense if firstly an elastic calculation is carried out, this means that the connection of the beams under investigation must be fully connected using a rigid-plastic method. This calculation is possible in our case because the composite cross-section (Fig. 4) is of class 1 (the compressed flange as well as the web) according to Eurocodes classification. For the end-span (Fig. 7), the shear force between the upper flange of the girder and the slab depends on the critical length :  Aa f y( a )  Aa f y( a ) (s) and On BC : Fcf = Ask f y + Min  On AB : Fcf = Min  0.85 f ck beff hc 0.85 f ck beff hc

(2)

where Aa is the girder area, Ask the reinforcing bars area, beff the effective width and hc the height of the slab. fya is the yield stress of the girder, fsk the yield stress of reinforcing bars and fck the compression strength of the concrete slab. The resistance of the stud is given by :  πd2 PRd = Min  0.8 fu   4  A

B

 2  ,0.29α d Ecm f ck 

  

(3)

M pl− .Rd

15

20

20

C A

M pl+ .Rd

B

C

D

15% AC

Fig. 7. Plastic bending moments (sagging and hogging zones).

Fig. 8. Number Nf of connected crosssections (Full connection).

The mechanical characteristics are given in Table 1. The longitudinal mesh of the beam is independent of the stud’s longitudinal distribution; this facility avoids the errors that could be generated by the difference of the mesh density. Each connected cross-section will always have 2 studs. The numerical simulations concern the beams given in Fig. 8. The distance between the

connected cross-sections is calculated according to the span lengths of the beams (see Fig. 8). For each critical length the distribution remains regular (we recall that CD = 12/2 = 6 m and AC = [3, 6, 9, 12] m). The degree of partial connection is represented by a coefficient κ defined by :

κ=

N = [1 / 3, 2 / 3,1] Nf

(4)

Table 1. Material mechanical characteristics Material

Parameters

Steel girder

Bars

Aa = 10110 mm 2

As = 3500 mm 2

f y( a ) = 355 MPa

f y( s ) = 400 MPa

f u( a ) = 510 MPa

f u( s ) = 432 MPa

Concrete

Ac = beff hc = 350000 mm 2

E ( a ) = 210 GPa E ( s ) = 200 GPa µ1 = 10 and µ 2 = 25 µ1 = 1 and µ 2 = 25 4

f ck = 40 MPa f tcm = 3.5 MPa Ecm = 35000 MPa

Stud h = 125 mm d = 19 mm →α = 1 f u = 500 MPa C1 = 0.8, C2 = 0.7 Qu = 152 kN

MOMENT REDISTRIBUTION PERCENTAGE

The moment redistribution percentage is calculated using Eq. (5) :

P

sup port

(% ) =

−( cr ,comp ) num,comp ) M Ed − M el−(,Rd −( cr ,comp ) M Ed

× 100

(5)

where the superscript (cr,comp) is related to the hogging bending moment obtained from the “virtual” cracked elastic global analysis, (num,comp) is the one related to the numerical value obtained from the non linear calculation and both of them concern a composite cross-section. In this case, the maximum of P, given by the Eurocodes, is equal to 25 % and if the beams are prefabricated, the composite action is supposed to be effective from the very beginning of loading. The bending moment is plotted for each value of κ and each loading case (Figs. 9, 10, 11 and 12). The moment redistribution remains available since the sagging bending does not exceed the plastic resistance M pl+ .Rd (1345 MPa). Using Eq. (5), Table 2 gives the values of P. Nevertheless, it is pointed out that when the span lengths ratio becomes greater than 0.6, the loading case of type A (Figs. 11.a and 12.a), needs a relative high load level to reach the hogging elastic resistance M el− .Rd (the criterion for stopping calculation). For these cases and for a low degree of connexion, this criterion can not be reached because the slip at the interface becomes important exceeding the limit value fixed to 6 mm. The failure of the connexion occurs for the case of κ = 1/3 (the corresponding curves can not be plotted). 5

SUMMARY

The F.E. program “Pontmixte” was used for these numerical simulations, references [2] and [3] gives sufficient details about the specific element of composite beam considered. For loading case A, the redistribution increases by increasing κ, the prediction of 25% remains available while the span-lengths ratio is lower than 0.6 and appears questionable if it is greater than 0.6. The bending moment capacity increases by increasing κ and also by increasing the span-lengths ratio (Figs. 13.a and 14.a). A low degree of connection (κ = 1/3) could be insufficient to resist to the shear flow at the interface girder-slab. For loading case B, the redistribution increases by increasing κ, the prediction of 25% remains available while κ > 2/3. The bending moment capacity increases by increasing κ but it decreases when the span-lengths ratio is higher than 0.6 (Figs. 13.b and 14.b).

1500

1500 M+pl,Rd = 1345 kNm

M+pl,Rd = 1345 kNm 1000 Self-weight (g)

500 Loading case A (q)

0 0

3

6

9 12 Distance X (m)

-500

15

18

-1000

Bending moment (kNm)

Bending moment (kNm)

1000

Loading case B (q)

0 0

3

6

-500

9 12 Distance X (m)

15

18

-1000

Numerical calculation Virtual elastic calculation

Numerical calculation Virtual elastic calculation

-1500

-1500

Fig. 9.a. Redistribution of bending moments – Beam 3-12-3 – Loading case A

Fig. 9.b. Redistribution of bending moments – Beam 3-12-3 – Loading case B

1500

1500

M+pl,Rd = 1345 kNm

M+pl,Rd = 1345 kNm

1000

1000 Self-weight (g)

500 Loading case A (q)

0 0

3

6

9 12 15 Distance X (m)

-500

18

21

24

-1000

Bending moment (kNm)

Bending moment (kNm)

Self-weight (g)

500

Self-weight (g)

500 Loading case B (q)

0 0

3

6

9 12 15 Distance X (m)

-500

18

21

24

-1000 Numerical calculation Virtual elastic calculation

Numerical calculation Virtual elastic calculation -1500

-1500

Fig. 10.a. Redistribution of bending moments – Fig. 10.b. Redistribution of bending moments – Beam 6-12-6 – Loading case A Beam 6-12-6 – Loading case B 1500

1500

M+pl,Rd = 1345 kNm

M+pl,Rd = 1345 kNm

1000 Self-weight (g)

500 Loading case A (q)

0 0

3

6

9

-500

12 15 18 Distance X (m)

21

24

27

30

-1000

Bending moment (kNm)

Bending moment (kNm)

1000

Self-weight (g)

500 Loading case B (q)

0 0

3

6

9

-500

12 15 18 Distance X (m)

21

24

27

30

-1000 Numerical calculation Virtual elastic calculation

-1500

-1500

Fig. 11.a. Redistribution of bending moments – Fig. 11.b. Redistribution of bending moments – Beam 9-12-9 – Loading case A Beam 9-12-9 – Loading case B 1500

1500

M+pl,Rd = 1345 kNm

M+pl,Rd = 1345 kNm 1000 Self-weight (g)

500 Loading case A (q)

0 0 -500

3

6

9

12

15 18 21 24 Distance X (m)

27

30

33

36

Bending moment (kNm)

Bending moment (kNm)

1000

Self-weight (g)

500 Loading case B (q)

0 0 -500

-1000

-1000

-1500

-1500

3

6

9

12 15 18 Distance X (m)

21

24

27

30

Numerical calculation Virtual elastic calculation

Fig. 12.a. Redistribution of bending moments – Fig. 12.b. Redistribution of bending moments – Beam 12-12-12 – Loading case A Beam 12-12-12 – Loading case B

Table 2. Bending moment redistribution − ( cr ,comp ) M Ed (kNm)

κ

Case A

1/3 2/3 1

842 886 921

num ,comp ) M el−(,Rd

P support (% )

Load (g + q)

num ,comp ) M el+ (,Rd

Case B Case A Case B Case A Case B Case A Case B Case A Case B Beam : 3-12-3 827 885 901

627 639 651

664 652 650

26% 28% 29%

20% 26% 28%

81 88 92

80 86 88

834 953 1004

815 921 950

93 99 101

80 84 86

1014 1145 1177

872 950 990

− 102 105

68

− 1205 1234

717

− 104 106

62

− 1215 1250

588

Beam : 6-12-6 1/3 2/3 1

824 883 901

828 865 892

659 646 649

660 651 658

20% 27% 28%

20% 25% 26%

Beam : 9-12-9 1/3

820

− 836 853

2/3 1

654

− 645 655

858 877

− 23% 24%

651 655

20% 24% 25%

71 72

777 802

Beam : 12-12-12 1/3

976

− 799 813

752

− 653 657

977 988

721 710

23% 26% 28%

62 62

608 615

.

− 18% 19%

30

25

25 Eurocode 20 prediction

Redistribution P(%)

30 Eurocode 20 prediction Stud failure

15 10

κ = 1/3 κ = 2/3 κ=1

5 0 0

0,2

15 10

κ = 1/3 κ = 2/3 κ=1

5 0

0,4 0,6 Span-lengths ratio

0,8

0

1

0,4 0,6 Span-lengths ratio

0,8

1

Fig. 13.b. Redistribution P – Case B

.

Fig. 13.a. Redistribution P – Case A

0,2

.

Redistribution P(%)

.

2/3 1

+

M

pl.Rd

1500

= 1345 kNm

Bending Moment (kNm)

Bending Moment (kNm)

1500 1200

+

M

pl.Rd

= 1345 kNm

1200

Stud failure

900 600

κ = 1/3 κ = 2/3 κ=1

300

900 600

κ = 1/3 κ = 2/3 κ=1

300 0

0 0

0,2

0,4 0,6 0,8 Span-lengths ratio

1

Fig. 14.a. Bending moment capacity – Case A

0

0,2

0,4 0,6 0,8 Span-lengths ratio

1

Fig. 14.b. Bending moment capacity – Case B

REFERENCES [1] CEN (European Committee for Standardisation), prEN 1994 – Part 2 – Design of composite steel and concrete structures [2] Guezouli S. & Aribert J.M., Numerical Investigation of moment redistribution in continuous beams of composite bridges, Composite Construction V : 18 to 23 July, 2004 [3] Guezouli S. & Yabuki T., “Pontmixte” a User Friendly Program for Continuous Beams of Composite Bridges, SDSS’06, Lisbon, September 6-8-2006