with FRP Professor Dr. Björn Täljsten Luleå University of Technology Sto Scandinavia AB 9/12/2013
Agenda Case study Restrained transverse expansion (confinement) Circular and rectangular cross sections A simplified approach for combined normal force and flexure An example for calculation
Case study Parking garage
Case study Parking garage Object: KV Smedby Repair and upgrading of 96 concrete columns due to chloride intrusion and corroded reinforcement - maintaining and increasing load capacity Measure: Repair with StoCrete TK samt StoCrete GM1 (EN1504-3). Sawing grooves for StoFRP Bar E10C (flexural strengthening) Surface preparation and mounting of StoFRP Sheet (confinement strengthening)
Case study Parking garage Mounting CFRP bars (StoFRP Bar E10C) in sawed grooves and bonding with epoxy adhesive (StoBPE Lim 567)
Case study Parking garage Mounting carbon fibre sheets in the longitudinal direction (StoFRP Sheet S300C300) Bonded with an epoxy adhesive (StoBPE Lim 417) - Flexural strengthening
Case study Parking garage Mounting of transverse carbon fibre sheets (StoFRP S300C300) Bonding (StoBPE Lim 417) - Confinement strengthening
Confinement effect Case study Parking garage
End result Case study Parking garage
Restrained transverse expansion - Lateral confinement 1 =f ck,c r r = f l 2 3 2 r E f t f fe E f t f fe E f t f t E f t f t
Confinement effect The lateral confinement pressure, r 1 =f ck,c r 2 Ef t f t r d 2 3 2 r E f t f t E f modulus of elasticity for fibres or FRP t f thickness of fibres or FRP t is the transversal strain E f t f t
Confinement effect The maximum lateral confinement, f l 1 =f ck,c r f l 2E f t f t, rup d 2 3 2 r E f t f t E f modulus of elasticity for fibres or FRP t f thickness of fibres or FRP t,rup is the transversal strain at rupture E f t f t
Confinement effect The maximum transverse strain, t,rup 1 =f ck,c r fu t,rup 2 3 2 r Curvature of the FRP jacket (wrapping) The deformation localisation of the cracked concrete The existence of an overlapping zone (The biaxial stress state that the jacket is subjected to) E f t f t E f t f t
Confinement effect A reasonable and accurate stress-strain model for the FRP confined concrete must be based on the actual transverse rupture strain of the FRP Ascending type Descending type f c f c,c f c,c f c,cu c,c cu,c cu,c c
Confinement effect The most established models for describing the restrained transverse confinement are: ACI 440.2R (2008) [US] Fib bulletin 14 (2001) [international] CSA (2002) [Canadian Standards Association) Eurocode 2, 1992-1-1, (2008) [Europe]
Confinement effect Comparison of different models, confinement effect: 2 = 2 MPa f ck = 30 MPa
Confinement effect Comparison of different models, confinement effect: 2 = 8 MPa f ck = 30 MPa
Centric compression Axial stresses and strain (no eccentricity) 2 Ec E2 2 c Ecc c 0c t 4 f ck f E c ck 2 c t c cu, c Parabolic Linear t E 2 f ck c E 2 c f c,c Intersection E 2 E 2 f ck, c cu, c f ck t cu,c c
Centric compression Maximum confinement pressure f f 33 f cd, c cd f, c. a l Design compressive strength f l 2 E f nt f fe D Maximum confinement pressure a Efficiency factor, considering shape and geometry fe Effective strain (ultimate state for FRP system)
Centric compression Effective strain fe fu Efficiency factor considering FRP premature failure due to the triaxial state of stress (relation between actual failure strain and ultimate coupon failure strain). Correlations to test gives a value of 0.55.
Centric compression Avoiding descending stress strain behaviour f f l ck 0.08 In order to assure that the confinement effects keeps increasing, ascending linearly.
Centric compression The maximum compression strain in confined concrete (f ck <70 MPa) 0.45 f l fe cu, c c2 1.50 12 b 10 fck c2 To avoid excessive crack formation and to secure the integrity of the concrete cross section. Calculated in. b Efficiency factor considering the geometry of the cross section
Circular cross section The confinement effect is largest for circular cross sections a a 1.0 Shape/geometry factors
Non circular cross sections Experimental results indicate that the confinement effect is much less for non circular cross sections (e.g. rectangular cross sections). Larger the cross sections less effect. h b 2 b and h 900mm Limitations
Non circular cross sections The maximum confinement pressure corresponds to an area bound by an equivalent circular cross section. h Diameter 2 2 D b h b D r c
Non circular cross sections The maximum confinement pressure corresponds to an area bound by an equivalent circular cross section. h Geometrical efficiency factors (depending on the effective confinement area and the side ratio b/h) b D r c a Ace b A h c 2 b Ace b A h c 0.5
Non circular cross sections The maximum confinement pressure corresponds to an area bound by an equivalent circular cross section. h Effective confinement area (depending on the rounding of the corners, the side ratio b/h AND the reinforcement ratio) b D r c A A ce c 1 b h h b 3A h2r b 2r c 2 2 c g 1 g g
Interaction between normal forces (N) and flexure (M) Simplified approach N D D Med omslutningseffekt Utan omslutningseffekt Dimensionerande last, N och M Limitations: Slenderness ratio, l lim 50 The effective strain B B fe 0.004 fu A M -N
Interaction between normal forces (N) and flexure (M) cök cu s2,d > sy Simplified approach Nc D s1,d > sy cuk = cu C cök = cu s2,c NL s1,c < sy cök = cu cuk s2,b A B M cök = cu NL s1,b = sy cuk s2,a NL Nt s1,a > sy cuk
Example, confinement Rectangular cross section, increase of normal force by 15% (current capacity 3000 kn). Geometrical properties d2 A s2 0.5d Value Unit Description b = h 400 mm Side of column P d 1 = 40 mm Concrete cover h G d 2 = 40 mm Concrete cover dpg d1 A s1 b h/2 A s1 = 1256 mm 2 Reinforcement area A s2 = 1256 mm 2 Reinforcement area A c = 157487 mm 2 Concrete area (no reinforcement) A g = 160000 mm 2 Gross sectional area
Example, confinement Rectangular cross section, increase of normal force by 15% (current capacity 3000 kn). Geometrical properties d2 A s2 0.5d Value Unit Description b = h 400 mm Side of column P d 1 = 40 mm Concrete cover h G d 2 = 40 mm Concrete cover dpg d1 A s1 b h/2 A s1 = 1256 mm 2 Reinforcement area A s2 = 1256 mm 2 Reinforcement area A c = 157487 mm 2 Concrete area (no reinforcement) A g = 160000 mm 2 Gross sectional area
Example, confinement Rectangular cross section, increase of normal force by 15% (current capacity 3000 kn). Reduction factors in ULS h d2 A s2 P G 0.5d Concrete Steel FRP g c =1.5 g s =1.15 g frp =1.35 cc =0.85 ct =0.85 dpg A s1 h/2 φ ef =2.0 d1 b g ce =1.2
Example, confinement Rectangular cross section, increase of normal force by 20% (current capacity 3300 kn). Material properties h d2 dpg A s2 A s1 P G 0.5d h/2 C o n c r e t e Characteristic values S Characteristic values f ck 25 MPa t e f yk 500 MPa f ctm 2.2 MPa e l E s 210 GPa E cm 31 GPa d1 b C o n c r e t e Design values S Design values f cd 14 MPa t e f yd 435 MPa f ctm 2.2 MPa e l E sd 183 GPa
Example, confinement Estimate the confinement effect, using a carbon fibre sheet with a thickness of 0.17mm (300g/m 2 ) Mechanical properties for sheet d2 A s2 P 0.5d FRP Characteristic values Design values fk 19 f 14.1 E fk 290 GPa E f 214.8 GPa h G dpg A s1 h/2 d1 b
Example, confinement Step 1. Calculate the geometrical factors D b 2 h 2 400 2 400 2 566 mm A ce A c b 1 h h b 3A h 2r b 2r c 2 2 g 1 g c 400 2 400 2 400 230 400 230 360 360 1256 1 3400 400 400 400 0.507 1 1256 400 400 g
Example, confinement Step 1. Calculate the geometrical factors 2 2 Ace b 400 a 0.507 0.507 A h 400 c b A A ce c b h 0.5 0.507 0.55 e 0.55 0.00141 0.0076 fe Normal loading only min 0.004, 0.004 fe e f Considering N-M interaction and shear capacity integrity
Example, confinement Step 2. Calculate the confinement pressure, normal force only Choose the number of layers, n = 4 f l 2E f nt f fe 2214815 40.17 0.0076 3.9 MPa D 566 fcd, c fcd f, c3.3a fl 14 0.953.3 0.507 3.9 20,2 MPa Strengthened normal force capacity: 4347 kn > 3800 kn OK!
Example, confinement Step 2. Calculate the confinement pressure, considering possible N-M interaction Choose the number of layers, n = 6 f l 2E f nt f fe 2214815 60.17 0.004 3.1 MPa D 566 fcd, c fcd f, c3.3a fl 14 0.953.3 0.507 3.1 18.92 MPa Strengthened normal force capacity: 4072 kn > 3800 kn OK!