How To Make A Steel Beam

Transcription

1 Resistenza a taglio di travi in calcestruzzo fibrorinforzato Università degli Studi di Brescia [email protected] Milano June 17 th, 2015

2 Outlines Shear Action Factor affecting the shear strength Experimental tests on PC and FRC beams Wide-shallow beams in FRC Shear Design of FRC beams 2

3 Optimized reinforcement: definition Place the best performing reinforcement (fibers and/or rebars) where required by tensile stresses in the structural elements 3

4 Reinforcement use in structural elements In structural elements both distributed and localized stresses are generally present Conventional rebars represent the best reinforcement for localized stresses Fibers represent the best reinforcement for diffused stresses Structural optimization generally requires the use of a combination of rebars and fibers Structural ductility is generally enhanced 4

5 Mechanisms of Shear Transfer Uncracked concrete zone Unterface shear transfer Dowel action Residual tensile stress across cracks 5

6 Shear in beams without stirrups In FRC elements there is an additional contribution to shear resistance provided by fiber reinforcement: V = V c + V f V c represents the concrete contribution. V f represents the fiber contribution (post cracking strength). 6

7 Shear in beams with stirrups V = V c + V w + V f 7

8 Factor affecting the shear strength Tensile strength of concrete Longitudinal reinforcement ratio Shear span-to-depth ratio (a/d) Axial forces Coarse aggregate size Size effect 8

9 Factors affecting the Shear Strength Distribution of longitudinal reinforcement along beam height (fiber?) 9

10 Size effect 10

11 Experimental campaign on PC and FRC beams d a V 2Ø24 Bars, L=4550 mm mm V Steel Plate 200x90x30 mm 45 2Ø24 Deformed Bars 480 mm Campaign I 8 beams NSC 3 beams HSC a/d=2.5 Longitudinal reinforcement ratio 1% No stirrups 11

12 Ø 0.18 Ø 1.0 Ø 0.6 Ø 0.18 Ø 0.62 Ø 0.38 Experimental campaign on PC and FRC beams 400 Normal Strength Concrete, f' c = 24.8 MPa Load [kn] kg/m 3, 30/0.6 V 30 kg/m 3, 30/ kg/m 3, 12/0.18 V NSC1-PC NSC1-FRC1 NSC1-FRC Displacement [mm] Crack Width [mm] Normal Strength Concrete, f' c = 24.8 MPa Average First Cracking CPT TPT V V V 30 kg/m 3, 30/0.6 NSC1-PC NSC1-FRC1 NSC1-FRC Load [kn] V 30 kg/m 3, 30/ kg/m 3, 12/0.18 f c = 24.8 MPa Fibers: % macro-fibres, 30 mm long with an aspect ratio of % micro-fibres, 12 mm long with an aspect ratio of

13 Ø 0.18 Ø 0.18 Ø 0.6 Ø 0.62 Ø 0.38 Ø 0.6 Ø 0.62 Ø 0.38 Experimental campaign on PC and FRC beams Load [kn] Crack Width [mm] High Strength Concrete 50 kg/m 3, 45/30 V 50 kg/m 3, 80/30 HSC-PC HSC-FRC1 HSC-FRC Displacement [mm] High Strength Concrete Average First Cracking HSC-PC HSC-FRC1 HSC-FRC2 V CPT TPT Load [kn] V V V 50 kg/m 3, 45/30 V 50 kg/m 3, 80/30 f c = 60 MPa. Fibers: 0.6% macro-fibers, Normal strength fibers (45/30) High strength fibers (80/30) A relative small amount of 12 fibers signifiantly enhances both the shear capacity and the ductility, both in case of normal strength concretes and high strength concretes

14 Experimental campaign on PC and FRC beams No fibre Fibre 50/1.0 PC beams: very brittle and sudden failure FRC: prior warning of impending collapse, more distributed crack pattern Fibre 80/30 14

15 Final crack patterns PC Fibres 45/30 90 kn 150 kn Failure Fibres control and stabilize the shear crack propagation The combination of macrofibres with micro-fibres lead to a better crack control at small and high values of crack opening. 90 kn 150 kn Failure Fibres 80/30 15

16 Size effect Beams without web reinforcement fail when inclined cracking occurs or shortly afterwards. The inclined cracking load of a beam is affected by many principal variables. One of these is the size of beam (size effect). An increase in the effective depth of a beam results in a decrease of the shear bearing capacity. d V u v u = b w d Ohio shear collapse (1955) 16

17 Experimental campaign II d a V 2Ø24 Bars, L=4550 mm beams 4350 mm 45 V Steel Plate 200x90x30 mm 2Ø24 Deformed Bars 480 mm d d a a P 6Ø20 Bars, L=4600 mm P 6Ø20 Bars, L=4600 mm 4550 mm 4600 mm 4550 mm 4600 mm Steel Plate 200x90x30 mm 45 Steel Plate 200x90x30 mm 6Ø20 Deformed Bars 1000 mm 6Ø20 Deformed Bars 1000 mm a/d= Longitudinal reinforcement ratio 1% NSC e HSC (f c =25.7 and 55 MPa) H=500 mm and H=1000 mm 17

18 Experimental campaign II Load Displacement Curve Small Size Specimens H = 500 mm; f c = 25.7 MPa Load-Displacement Curve Large Size Specimens H = 1000 mm; f c = 25.7 MPa Load [kn] Displacement [mm] P PC-50 FRC kg/cm Test 1 FRC kg/cm Test 2 MSR-50 Test 1 MSR-50 Test 2 d Load [kn] P d PC-100 MSR-100 FRC kg/cm Displacement [mm] Small and Large Size Specimens, f c =25.7 MPa. FRC beams: Steel fibres 0.25% l=50 mm Aspect ratio= 50 Minimum shear reinforcement: MSR-50 with 2f8@300mm, 2f8@650mm MSR-100 with 18

19 Experimental campaign III 1000 Load [kn] 800 P Large Size Specimens H = 1000 mm; f c = 55 MPa 5 Crack Width [mm] 4 P Large Size Specimens H = 1000 mm; f c = 55 MPa d TPT4 TPT5 TPT2 TPT P TPT6 TPT H-PC-100 H-MSR-100 H-FRC H-PC-100 H-MSR-100 H-FRC Displacement [mm] Large Size Specimens, f c =55.0 MPa. The same amount of fibers Stable propagation of cracks, higher ductility and shear capacity, prior warning of impeding collapse. 0 Load [kn]

20 Experimental campaign IV SPECIMEN P u v u v u /(f c ) 1/2 u V u /V u,fl [kn] [MPa] [-] [mm] [-] PC MSR MSR FRC FRC PC MSR-100 NSC FRC PC MSR-100 HSC FRC Fibres can completely substitute the minimum amount of shear reinforcement kg/m 3 is the amount required in many structural applications PC H100 MSR H100 Fibres resulted less effective than the minimum amount of shear reinforcement on the size effect. FRC-20 H100 20

21 Numerical Analyses NFEA v/v max,flex [-] Size Effect on PC and FRC Members g c = 1, r s = 1%, NSC f' c = 22 MPa, HSC f' c = 58.3 MPa Effective Depth [mm] Skin Reinforcement Recommended f eq(0.6-3) HSC-FRC3 NSC-FRC2 NSC-FRC1 NSC-PC exp PC exp FRC Vu,num v= b d w M v = u,flex max,flex 2.5 b d 2 w d from 250 to 2000 mm. Increasing the fiber toughness, the size effect influence decreases 21

22 Experimental Program on DEEP BEAMS H=500 mm H=1000 mm H=1500 mm 9 full scale beams a/d=3.0 ρ= 1% NSC (f cm =35 MPa) H=500 mm H=1000 mm H=1500 mm 0, 50 or 75 kg/m 3 of steel fibers (SFRC) No stirrups anywhere 22

23 Deep beams: Experimental results Fibers enhance the ductility and the shear bearing capacity (50 kg/m 3 of fibers doubled the shear strength). ULS Fibers increase the load at which the shear crack becomes unstable. PC - max shear crack = mm; FRC - max shear crack = 3-4 mm Beams H

24 Deep beams: Experimental results Fibers enhance the behavior at Serviceability Limit State as well. A significant enhanced post-cracking stiffness is observed in FRC beams, which is mainly due to the bridging effect of fibers. SLS Fibers improve the durability of structural elements Beams H

25 Deep beams: Crack patterns In FRC elements: MULTIPLE SHEAR CRACKS STABLE SHEAR CRACKS Progressive formation of multiple cracks, with a stable propagation. Fibers are highly effective in controlling development and propagation of cracking Fibers more distributed crack pattern, with more closely spaced and smaller cracks H1500 FRC50 H=1500 mm 25

26 Deep beams in PFRC: Experimental results Unlike MSR samples (r min =0.17%), shear failure was observed in PFRC beams. PP fibres enhanced the ultimate shear strength and ductility of deep beams, which are much more critical in shear than WSBs. 13 kg/m 3 of PP fibres was able to double both the shear strength and the ductility, as compared to control samples PP fibres were found to be effective also in presence of prestressing Specimen Failure mode V u v u v u /(f cm ) 1/2 [kn] [MPa] [-] PC 300x800-1 Shear MSR 300x800-1 Flexure MSR 300x800-2 Flexure PFRC 300x800-1 Shear PFRC 300x800-2 Shear PC 150x800-1 Shear PC 150x800-2 Shear MSR 150x800-1 Flexure MSR 150x800-2 Flexure PFRC 150x800-1 Shear PFRC 150x800-2 Shear PC 150x800 PT-1 Shear PFRC 150x800 PT-1 Shear* PC 150x600-1 Shear PC 150x600-2 Shear MSR 150x600-1 Flexure MSR 150x600-2 Flexure PFRC 150x600-1 Shear PFRC 150x600-2 Shear

27 Deep beams in PFRC: Experimental results PP fibres and MSR increase the load at which shear crack becomes instable PP fibres have been able to slightly increase the post-cracking stiffness Flexural crack spacing was 20% smaller in PFRC deep beams 1000 Load 900 [kn] Deep Beams 300x800 d=761 mm PC 300x800-1 MSR 300x800-1 MSR 300x800-2 PFRC 300x800-1 PFRC 300x800-2 P/2 P/2 8 Crack Width 7 [mm] Deep Beams 300x800 d=761 mm PC 300x800-1 MSR 300x800-1 MSR 300x800-2 PFRC 300x800-1 PFRC 300x800-2 V P/ Mid-span deflection [mm] Load [kn]

28 Prestressed Double tees 6 full-scale double tees H=500mm; d=390 mm; L=6000mm b w =120 mm 3 tendons 0.6 ρ l =0.89% 2 double tees WR1 DT 1 double tees WR2+SCPFRC DT 28

29 Prestressed Double Tees 3 double tees SCPFRC-DT 10kg/m 3 of PP fibres (V f =1.10%) 29

30 Tests in zone with uniform prestressing 3 experimental tests a/d 3 WR1 DT-1 (RC) SCPFRC DT-1 (only PP fibres) SCPFRC DT-2 (only PP fibres) 30

31 Tests in zone with uniform prestressing All specimens showed a shear failure % Shear crack % Both PP fibres and MSR led to: - An increment of the shear capacity of about 10-15% - An increment of the ductility at least of 70%. PP fibres can completely substitute the minimum amount of shear reinforcement 31

32 Tests in the end zone 6 experimental tests a/d 3 WR1 DT-2 a/b (RC) SCPFRC DT-3 a/b (only PP fibres) WR2+SCPFRC DT a/b (RC + PP fibers) 32

33 Tests in the end zone +15% The more effective solution in the end zones seems to be the combination of PP fibres and conventional reinforcement (steel wire mesh Ø5 200x300), in which after shear cracking a stable behaviour with an increment (+20%) of the ultimate bearing capacity was observed. 33

34 Wide-Shallow Beams (WSBs) Typical structure typology of residential building in: Southern Europe, Australia, Middle-East, Central and South-America. Topping concrete layer Beam RC spandrel wide-shallow beam d b d b Lightweight ribbed one-way reinforced concrete slab RC central wide-shallow beam d b Wide-shallow beam 2d b d b Benefits: floors completely flat, more architectural flexibility, easier formwork Disadvantages: deformability, small net span, transfer of load to column. 34

35 Experimental campaigns on WSBs FIRST EXP. CAMPAIGN SECOND EXP. CAMPAIGN 16 full scale WSBs Main purposes: 1) Study both shear and flexure behavior of WSBs 2) Evaluate the possibility of completely substitute the minimum amount of shear reinforcement by steel fibers (SFRC) 14 full scale WSBs Main purposes: 1) Study the shear behavior of PC wide-shallow beams 2) Evaluate structural polymer fibers in full scale tests 3) Replace the minimum shear reinforcement by polymer fibers (PFRC) 35

36 SFRC and PFRC WSBs: Experimental results Either minimum shear reinforcement or fibers greatly influences the shear behavior of wide-shallow beams (by altering the collapse from shear to flexure, with enhanced bearing capacity and ductility). Both STEEL FIBERS and POLYPROPYLENE FIBERS have been able to completely substitute the minimum amount of transverse reinforcement. 13 kg/m 3 SFRC PFRC 36

37 FRC WSBs: Flexure behavior Steel fibers even in small amount, provides a greater flexure bearing capacity and increases the overall ductility under flexure, due to the positive effect in increasing the compression softening. In the case of PFRC WSBs this behavior is less pronounced. μ δ =4.3 SFRC (+6%) +40% (+10%) μ δ =6.1 (+7%) (+10%) 37

38 Shear design of FRC elements Shear strength without shear reinforcement V Rd k g c (100 r 1 f ck ) 0.15 CP b W d Modified in MC2010 in order to consider fibres contribution - Fibres contribution included in the shear concrete strength - Fibres contribution is taken into account considering the residual strengths according to EN14651 (3PBT) 38

39 Shear design of FRC elements f Ftuk VRd,F k 100 r f ck cp b d [tensio W g c fctk f Ftuk characteristic value of the residual strength (SLU), considering w u = 1.5 mm, in MPa; f Ftu f 3 R3 This model has been adopted by Model Code 2010 (MC2010) 39

40 Shear Strength vs. Toughness Range used in structural application V u,frc /V u,ec2 [-] New model for FRC w/o stirrups f Ftu MPa f eq(0.6-3) [MPa] fctm = 2.0 MPa fctm = 2.5 MPa fctm = 3.0 MPa fctm = 3.5 MPa fctm = 4.0 MPa fctm = 5.0 MPa The shear strength significantly increases using fibres 40

41 MC2010 model vs experimental results fib vs. RILEM V u,exp /V u,model Conservative RILEM DATA BRESCIA DATA Unconservative fib RILEM Test # MC2010 model seems to be consistent 41

42 Minimum shear reinforcement Crack control Ductility Warning of impending collapse V f r Rd, ct, FIBERS eq,min w,min f yk V Rd, ct f ck f ck b w d f eq(0.6-3) [MPa] 3 f f ct ck k 1/3 100 r f ck Minimum FRC Transverse Reinforcement f eq,min f 0.7 eq,min f ck 4.5 Calculation Design Formulation Best Fitting Formulation 3 f f ct ck 1 1 1/ k 100 r f ck Compressive Strength [MPa] f Ftuk f ck 20 42

43 MC2010 Shear Workshop Salò Ottobre

44 Example of Application for Shear: I p = 35 kn/m 200 d 2Ø24 Bars 6 m 500 mm 500 2Ø24 Deformed Bars p u 35 kn / m ( ULS) h 500 mm; d 460mm f 30 MPa; f 500 MPa ck c s yk g 1.5; g fcd 20 MPa; f yk 435 MPa f 2 MPa ( EC2) ctk M max p l kn m Vmax p l kn As 904 mm rl 0.98% b d 200 mm 460 mm M u w 161 kn m 44

45 Example of Application for Shear: I VRd, ct k (100 r1 fck ) 0.15CP b d 49 kn W g c Minimum Shear Reinforcement meters requiring design shear reinforcement; 2.8 meters requiring minimum shear reinforcement. Minimum Shear Reinforcement: Design Shear Reinforcement: s 0.75d 345 mm Asw VR, ds z fyd VRd VRd, ct 56 kn f s ck rw,min f s 321 mm yk mm 28@ 300 mm 45

46 Example of Application for Shear: I Assume 30 kg/m 3 of steel fibers having l/f =67 and fftk,u=0.90 MPa (tested at the University of Brescia) V Rd, F 0.18 Ftk, u 1 3 k (100 r1 (1 7.5 ) f ck ) g c f ctk f 0.15 CP b W d V Rd, F Minimum shear reinforcement f ck f Ftuk MPa Design Shear Reinforcement 1 1 ( (1 7.5 ) 20) OK Asw VR, ds z fyd VRd VRd, ct 24 kn s 2 6@ 300 mm s 420 mm kn Minimum Shear Reinforcement 46

47 Example of Application for Shear: I 2Ø8@300mm 2Ø6@300mm Plain concrete 2Ø6@300mm FRC 47

48 Example of Application for Shear: WSBs kn/m 500 cm 500 cm d=26cm g = 40 kn/m WSBs q = 10 kn/m p Concrete C30/37 B450C ϒ u 35 kn / m ( ULS) h 500 mm; d 460mm f 30 MPa; f 500 MPa ck c s yk g 1.5; g fcd 20 MPa; f yk 435 MPa f 2 MPa ( EC2) ctk H = 30 cm d = 26 cm b = 80 cm ϒ 48

49 Example of Application for Shear: WSBs Shear design according to Level I of Model Code 2010 (inclination of the compressive stress field of 30 ). MSR MSR Steel incidence on concrete volume: 145 kg/m 3 49

50 Example of Application for Shear: WSBs FRC classified as 3c (f R1k =3MPa; f R3k =3MPa, i.e. 35 kg/m 3 of steel fibres 60 mm long with diameter of 0.75 mm) MSR by fibres MSR by fibres Steel incidence on concrete volume: 115 kg/m 3 50

51 Thank you for your kind attention University of Brescia 51

52 Examples of Shear Design: II p (UDL) L=10 m 1- Given material properties (NSC) and reinf. details, calculation of M rd; 2- Corresponding ultimate load (UDL) p u =8M rd /l 2 ; 3- Design ultimate shear V rd =1.1 p u l (-p u 1m) where 1.1 is a incremental safety factor to guard against brittle collapse phenomena; 4- Concrete contribution to shear without fibers, fiber contribution, concrete+fiber (V cd,fibers ) contribution; 5- A sw /s ratio necessary to fully reach the required shear capacity V rd, being V wd =A sw /s 0.9 d f ywd = V rd V cd,fibers ; 6- Transverse reinforcement saving ratio due to fibers as: Asw / sfibres S. R. 1 A / s sw NO FIBRES 52/20 52

53 Fibre Contribution [kn] Examples of Shear Design: II Fibre Contribution to Shear Prestressed and not-prestressed Member CNR rho = 0.8% CNR rho = 1% CNR rho = 1.2% CNR rho = 1.4% CNR rho = 1.6% RILEM all rho RILE M Model f eq(0.6-3) [MPa] r f R3 =0.90f eq(0.6-3) Parameters: -prestressing; -Fiber toughness; -Reinforcement ratio. V f CNR model is equal to MC2010 one Model RILEM does not depends on the reinforcement ratio. 0.7 k k b d f Linear trend vs. f eq l fd w 53/20 53

54 Concrete + Fibre Contribution [kn] Concrete + Fibre Contribution [kn] Examples of Shear Design: II Concrete + Fibre Shear Contribution Prestressed Member Concrete + Fibre Shear Contribution Not-Prestressed Member 250 r 140 r CNR 80 CNR 50 RILEM 60 RILEM f eq(0.6-3) [MPa] f eq(0.6-3) [MPa] Good agreement up to f eq(0.6-3) <3 MPa, which covers most of practical applications (for strain-softening materials). 54/20 54

55 1-(A sw /s, FIBRES )/(A sw /s,no FIBERS ) [-] 1-(A sw /s, FIBRES )/(A sw /s,no FIBRES ) [-] Examples of Shear Design: II Stirrup Saving ratio S. R. 1 Asw / sfibres Asw / sno FIBRES Transverse Shear Reinforcement Saving Ratio Prestressed Member Transverse Shear Reinforcement Saving Ratio Not-Prestressed Member 0.30 CNR 0.25 CNR RILEM 0.80 RILEM r 0.20 r f eq(0.6-3) [MPa] f eq(0.6-3) [MPa] 55/20 55

56 Pictures at failure W770 PC AND W770 FRC-2 PC 15x60 A and FRC 15x80 B 56

57 L'/L [-] L'/L [-] Examples of Shear Design: II L'/L CNR Not-Prestressed Member L'/L RILEM Not-Prestressed Member CNR rho = 0.8% 0.40 RILEM rho = 0.8% 0.30 CNR rho = 1% CNR rho = 1.2% 0.30 RILEM rho = 1% RILEM rho = 1.2% 0.20 CNR rho = 1.4% 0.20 RILEM rho = 1.4% 0.10 CNR rho = 1.6% 0.10 RILEM rho = 1.6% f eq(0.6-3) [MPa] f eq(0.6-3) [MPa] L : Length of the central portion of a beam where no traditional reinforcement is necessary. L: Total length. 57/20 57

58 Examples of Shear Design: II Plain concrete FRC The introduction of fibers is industrially convenient, cost-effective and time-saving when the design process can be suitably optimized. 58/20 58