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An Introduction to FRP- Strengthening of Concrete Structures ISIS Educational Module 4: Updated October 2010 for ISIS Canada ISIS EC Module 4.

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Presentation on theme: "An Introduction to FRP- Strengthening of Concrete Structures ISIS Educational Module 4: Updated October 2010 for ISIS Canada ISIS EC Module 4."— Presentation transcript:

1 An Introduction to FRP- Strengthening of Concrete Structures ISIS Educational Module 4: Updated October 2010 for ISIS Canada ISIS EC Module 4

2 Module Objectives ISIS EC Module 4 The use of FRPs in civil infrastructure is steadily increasing in Canada and around the world. This module is directed mainly to students to: Provide a background and general awareness of FRP materials, their properties, their behaviour and their potential uses Introduce the philosophies and procedures for strengthening structures with FRPs Familiarize the students with designing using the Canadian Highway Bridge Design Code (CHBDC)

3 Overview ISIS EC Module 4 1.Introduction 2.FRP Materials 3.Evaluation of Existing Structures 4.Flexural Strengthening 5.Shear Strengthening 6.Column Strengthening 7.Installation of FRP strengthening systems 8.Quality control and quality assurance 9.Additional applications 10.Field applications

4 1 - Introduction The worlds population depends on an extensive infrastructure system Roads, sewers, highways, buildings The system has suffered in past years Neglect, deterioration, lack of funding Global Infrastructure Crisis ISIS EC Module 4

5 Why is strengthening needed? Many structures, including bridges and parking garages, have become structurally deficient due to deterioration. In Canada, more than 40% of the bridges currently in use were built more than 40 years ago. Many structures are also becoming functionally obsolete due to increased loading. 1 - Introduction ISIS EC Module 4

6 2 - FRP Materials Why repair with the same materials? Why repeat the cycle? FRP Materials Light weight Easy to install High Strength 5x steel Corrosion resistant Durable structures Highly versatile Suits many projects ISIS EC Module 4

7 FRP is a composite consisting of fibres and matrix. Fibres: – Provide strength and stiffness –Their quality, orientation and shape affect the final product Matrix (resin): –Coats the fibres –Protects the fibres from abrasion –Transfers stresses between the fibres Strain [%] 0.5-4.82-8 50-90 2400-4300 Stress [MPa] Fibres Matrix FRP 2 - FRP Materials

8 ISIS EC Module 4 FRP material properties are a function of: Fibre quality, orientation and shape Fibre volumetric ratio Adhesion to the matrix Manufacturing process (additives and fillers) 2 - FRP Materials

9 Wide range of FRP products are available: Plates (Rigid strips) Sheets (Flexible fabric) Rods The fibres could be: Carbon Glass Aramid ISIS EC Module 4 FRP sheet

10 2 - FRP Materials ISIS EC Module 4 FRP advantages: Does not corrode High strength to weight ratio Reduced installation time and cost Non-conductive and non-metallic Low maintenance requirements Disadvantage: High temperature is a serious concern

11 3 - FRP Materials ISIS EC Module 4 FRP properties versus steel: Linear elastic behaviour to failure No yielding Higher ultimate strength Lower strain at failure Strain [%] 123 500 1000 1500 2000 2500 Stress [MPa] Steel CFRP GFRP

12 ISIS EC Module 4 2 - FRP Materials FRP SystemFiber Type Weight [g/m 2 ] Thickness [mm] Tensile Strength [MPa] Tensile Elastic Modulus [GPa] Strain at Failure [%] Fyfe Co. LLC [www.fyfeco.com] Tyfo SEH51AGlass9151.346020.91.8 Tyfo SEH25AGlass5050.541720.91.8 Tyfo SCH41Carbon6441.0834820.9 Hughes Brothers Inc. [www.hughesbros.com] Aslan 200Carbon--6.4-12.72068-17241240.017-0.015 Aslan 500 #2Carbon--2.020681241.7 Aslan 500 #3Carbon--4.519651241.5 Aslan 400 CFRP LaminatesCarbon--1.424001311.9 Sika Canada Inc. [www.sika.ca] SikaWrap 430GGlass4300.550424.61.9 SikaWrap 100GGlass9151.055824.42.2 SikaWrap 230CCarbon2300.471561.01.1 SikaWrap 103CCarbon6101.071765.11.0 CarboDur SCarbon--1.2-1.428001651.7 CarboDur MCarbon--1.424002101.2 CarboDur HCarbon--1.413003000.5 BASF Building Systems Inc. [www.BASFBuildingSystems.com] MBrace S&P LaminateCarbon--1.427001591.7 MBrace EG 900Glass9000.37151772.42.1 MBrace CF 130Carbon3000.1738002271.7 MBrace CF 160Carbon6000.3338002271.7 MBrace CF 530Carbon3000.1735003730.94 MBrace AK 60Aramid6000.2820001201.6

13 ISIS EC Module 4 1) Wet lay-up system: - Used with flexible sheets - Multiple layers can be used - Saturate sheets with epoxy adhesive, then place on the concrete surface and press with a roller Epoxy Roller Installation: - The system is installed on the surface of the concrete element while the resin matrix is still wet, and the polymerization occurs on site 2 - FRP Materials

14 - Used with FRP plates or laminates - Used for surface bonded plates or near surface mounted reinforcements - The pre-cured laminates should be placed on or into the wet adhesive - Place on the concrete surface - Multiple layers can not be used - Not as flexible for variable structural shapes 2) Pre-cured system: ISIS EC Module 4

15 3- Near Surface Mounted (NSM) It is a newer class of FRP strengthening technique. Un-strengthened concrete T-beam Longitudinal grooves cut into soffit FRP strips placed in grooves Grooves filled with epoxy grout Research indicates that NSM reinforcement is effective and efficient for strengthening. 2 - FRP Materials

16 ISIS EC Module 4 Type Application Schematic FRP-Strengthening Applications : Fibre Dir. Confinement Around the column Circumferential Section Shear Side face of the beam (U or closed wrap) Perpendicular to long. axis of the beam Section Flexural face of the beam Tension and/or side axis of the beam Along long. Section

17 3 - Evaluation of Existing Structures ISIS EC Module 4 Repair process includes : 1.Evaluation of the existing structure Understanding the cause and the effect of the deterioration 2.Determining if repair is required and its extent (quantify) 3.Analysis and design 4.Introducing a repair strategy

18 3 - Evaluation of Existing Structures ISIS EC Module 4 Fixing the effect without understanding the cause is likely to result in premature failure of repair. Proper repair requires an understanding of the cause to eliminate the effect. Evaluation is important to: Determine concrete condition Identify the cause of the deficiency Establish the current load capacity Evaluate the feasibility of FRP strengthening

19 3 - Evaluation of Existing Structures ISIS EC Module 4 Evaluation should include: All past modifications Actual size of elements Actual material properties Location, size and cause of cracks and spalling Location and extent of corrosion Quantity and location of rebar

20 Problems in a structure could be due to: Defects: In design or material or during construction Damage: Due to overloading, earthquake or fire Deterioration: Due to corrosion or sulphate attack 3 - Evaluation of Existing Structures ISIS EC Module 4

21 3 - Evaluation of Existing Structures ISIS EC Module 4 Examples of some deficiencies: 1. Environmental effects Freeze-Thaw Chloride Ingress Wet-Dry

22 3 - Evaluation of Existing Structures A primary factor leading to extensive degradation….. 2. Corrosion Moisture, oxygen and chlorides penetrate Concrete Reinforcing Steel Through concrete Through cracks Corrosion products form Volume expansion occurs More cracking Corrosion propagation End result ISIS EC Module 4

23 3 - Evaluation of Existing Structures ISIS EC Module 4 Deficiencies could be due to: 3.Updated design loads 4.Updated design code procedures 5.Increase in traffic loads Then Now

24 3 - Evaluation of Existing Structures ISIS EC Module 4 Deficiencies could be due to: 6.Fire damage 7.Earthquakes

25 ISIS EC Module 4 Material resistance factors (CHBDC): - Concrete c = 0.75 - Steel reinforcement: - Reinforcing bars s = 0.90 - Prestressing strand s p = 0.95 - Base FRP for pultruded FRP: - AFRP FRP = 0.55 (for externally bonded applications) - AFRP FRP = 0.65 (for NSMR) - CFRP FRP = 0.80 (for externally bonded applications and NSMR) - GFRP FRP = 0.70 (for externally bonded applications and NSMR) - Non-pultruded FRP made by wet lay-up: 0.75 times base FRP 4 - Flexural Strengthening

26 ISIS EC Module 4 Failure Modes The analysis of the flexural strength of FRP strengthened elements is based on the following assumptions: 1) The internal stresses are in equilibrium with the applied loads. 2) Plane sections remain plane. 3) Strain compatibility exists between adjacent materials. (ie. Perfect bond between: concrete and steel, concrete and FRP) 4) The maximum tensile strain of the FRP ( FRPt ) is 0.006. 5) The maximum compressive strain in the concrete ( cu ) is 0.0035. 6) The contributions of FRPs in compression and of the concrete in tension are neglected. 4 - Flexural Strengthening

27 ISIS EC Module 4 Failure Modes The potential modes of failure are: 1) Concrete crushing before steel yielding or rupture of the FRP. 2) Steel yielding followed by concrete crushing before rupture of the FRP. 3) Steel yielding followed by rupture of the FRP. 4) Peeling, debonding, delamination or anchorage failure of the FRP (considered premature tension failures to avoid). 4 - Flexural Strengthening

28 ISIS EC Module 4 Rectangular section without compression steel: b d Cross Section AsAs Strain Distribution FRP c h b FRP s c Stress Distribution fsfs f FRP Equiv. Stress Distribution a = 1 c 1 Φ c f c TsTs T FRP CcCc C c = c 1 f c 1 bcT FRP = FRP A FRP E FRP FRP T s = s A s f s 4 - Flexural Strengthening

29 The equilibrium equations are: 1) Force equilibrium in the section : Compression forces =Tension forces 2) Moment equilibrium in the section: External applied moment= Internal moment ISIS EC Module 4 C c = T s + T FRP M applied = T s d - a 2 + T FRP h - a 2 4 - Flexural Strengthening

30 ISIS EC Module 4 b d Cross Section AsAs Rectangular section with compression steel : h b FRP Strain Distribution FRP s c Stress Distribution fsfs f FRP Equiv. Stress Distribution a = 1 c 1 Φ c f c TsTs T FRP CcCc AsAs cu s fsfs CsCs Add a compressive stress resultant C s = s f s A s d 4 - Flexural Strengthening M r = T s d-d- a 2 +T FRP h- a 2 +C s a 2 - d

31 Iterative design procedure: 1.Assume initial strains 2.Select FRP material and A FRP 3.Assume a failure mode If concrete crushing before steel yields, then: cu = 0.0035 and steel < yield If concrete crushing after steel yields, then: cu = 0.0035 and steel > yield If FRP ruptures after steel yields, then: FRP = FRPt and steel > yield b d AsAs FRP c h b FRP s c ISIS EC Module 4 4 - Flexural Strengthening

32 4.Determine the compressive stress block factors ( 1, 1 ) 1 = 0.85 – 0.0015 f c > 0.67 1 = 0.97 – 0.0025 f c > 0.67 d AsAs b a = 1 c 1 Φ c f c TsTs CcCc h b FRP T FRP ISIS EC Module 4 4 - Flexural Strengthening

33 5.Calculate c (neutral axis position) Using equilibrium equation the following equations can be derived and used: - Concrete crushing before steel yields ( s and s < y ) -Steel yielding followed by concrete crushing ( s and s > y ) - Steel yielding followed by FRP rupture ( s and s > y ) 1 c f c 1 bc 2 + s E s cu (A s +A s )+ FRP E FRP ( cu + fi )A FRP s E s cu (A s ' d ' +A s d)+ FRP E FRP cu A FRP h = 0 c - s f y (A s -A s )+ FRP E FRP ( cu + fi ) A FRP c- FRP E FRP cu A FRP h = 0 ISIS EC Module 4 c = s f y (A s -A s )+ FRP E FRP FRPt A FRP 1 c f c 1 b 4 - Flexural Strengthening 1 c f c 1 bc 2 +

34 6.Check failure mode assumption with the material strains If failure is initiated by: Concrete crushing: FRP rupture: - If the assumption is proven to be false, go back to step 3 and make another assumption - If the assumption is correct, proceed to the next step ISIS EC Module 4 cu = 0.0035 s = cu d - c c FRP = cu h - c c fi s ' = cu c - d ' c FRP = FRPu FRPt fi + FRPu ) s ' = fi + FRPt ) fi + FRPu ) s = fi + FRPt ) fi + FRPu ) c = fi + FRPt ) c-d ' h-c c-d ' h-c d-c h-c d-c h-c c c 4 - Flexural Strengthening

35 7.Compute internal forces 8.Calculate the section moment resistance ( M r ) 9.Compare M r to the applied moment ( M applied ) –If M r < M applied, go to step 2 and change A FRP –If M r > M applied, then the design is safe C s = s f s A' s T FRP = FRP A FRP E FRP FRP T s = s A s f s ISIS EC Module 4 M r = T s d-d- a 2 +T FRP h- a 2 4 - Flexural Strengthening +C s a 2 - d

36 Optimized determination of A FRP Assuming: All the steel has yielded and combining the equilibrium equations: 1.Determine c using the following equation 2.Determine strain in FRP ISIS EC Module 4 1 c f c b w 1 2 c 2 2 1 c f c b w h 1 c C s (h-d)+T s (d s -h) - M f FRP = cu h - c c - fi 0.006 FRPu 4 - Flexural Strengthening

37 3.Determine successively: 4.Optimize value of AFRP Use this A FRP as an input for the iterative design method ISIS EC Module 4 A FRP = T FRP FRP E FRP FRP C c = c 1 f c 1 bc T FRP = C c + C s - T s 4 - Flexural Strengthening

38 Geometric parameters If the neutral axis lies in the web (c < h f ), then treat it as a rectangular section with a compression zone width = b e. If the neutral axis lies outside the web (c > h f ), then treat it as a T-section. ISIS EC Module 4 T-section: 4 - Flexural Strengthening

39 Factored moment subdivisionGeometric parameters =+ =+ The section is treated as the summation of: flange (M rf ) and web (M rw ). - Flange (M rf ) - Web (M rw ) ISIS EC Module 4 s f y 1 c f c (b e -b w )h f A sf = A sw = A s – A sf M rw = s f y A sw d - a 2 +T FRP h - a 2 M rf = s f y A sf d - hfhf 2 4 - Flexural Strengthening

40 Design procedure: 1.Select FRP material and A FRP 2.Determine behaviour of the section ( Rect. or T) If Then, it is rectangular behaviour Else, it is a T-section 3.Determine A sf and M rf 4.Determine A FRP to obtain required M rw ISIS EC Module 4 h f s f y A s + FRP E FRP FRP A FRP 1 c f c 1 b e 4 - Flexural Strengthening

41 ISIS EC Module 4 Example: Calculate the moment resistance (M r ) for an FRP-strengthened rectangular concrete section Section information: f c = 40 MPa FRPu = 1.26 % A FRP = 110 mm 2 f y = 400 MPa E s = 200 GPa E FRP = 210 GPa b = 125 mm h = 360 mm 2-15M barsd = 320 mm CFRP 4 - Flexural Strengthening

42 ISIS EC Module 4 Solution: Step 1: Assume failure mode Assume failure of beam due to crushing of concrete in compression after yielding of internal steel reinforcement Thus, cu = 0.0035 and steel > yield 4 - Flexural Strengthening

43 ISIS EC Module 4 Step 2: Calculate concrete stress block factors 1 = 0.85 – 0.0015 f c > 0.67 1 = 0.85 – 0.0015 (40) = 0.79 1 = 0.97 – 0.0025 f c > 0.67 1 = 0.97 – 0.0025 (40) = 0.87 4 - Flexural Strengthening

44 ISIS EC Module 4 Step 3: Find depth of neutral axis, c Steel yielding followed by concrete crushing 1 c f c bc 2 + s f y (A s -A s ) + FRP E FRP ( cu + fi ) A FRP c - FRP E FRP cu A FRP h = 0 0.79(0.75)40(0.87) 125(c 2 ) 0.9(400) (0-400)+ 0.80(210000)(0.0035+0)110 c - 0.80(210000) (0.0035)(110)360 = 0 2577.375(c 2 )-79320(c)-23284800 = 0 c = 111.7 mm or c = -80.9 mm (rejected) 4 - Flexural Strengthening

45 ISIS EC Module 4 Step 4: Check mode of failure Steel yielding followed by concrete crushing cu = 0.0035 s = cu = 0.0035 Trial 2, assume the steel yields and the strain in the FRP is 0.006 d s -c c 320-111.7 111.7 = 0.0065 > 0.002( y ) FRP = cu h-c c = 0.0035 360-111.7 111.7 = 0.0078 > 0.006 not O.K. 4 - Flexural Strengthening

46 ISIS EC Module 4 Step 5: Trial 2 s f y A s + FRP E FRP FRP A FRP Compression in concrete = Tension in ( Steel + FRP) 1 c f c b c = 0.79(0.75)40(0.87)125(c) 0.9(400)400 + 0.80(210000)0.006(110) = c = 98.9 mm 4 - Flexural Strengthening

47 ISIS EC Module 4 Step 6: Check mode of failure FRP = 0.006 s = FRP = 0.006 The assumed mode of failure is correct d s -c 320-98.9 = 0.0051 > 0.002( y ) cu = FRP c h-c = 0.006 98.9 360-98.9 = 0.0023 < 0.0035 O.K. h-c 360-98.9 4 - Flexural Strengthening

48 ISIS EC Module 4 Step 7: Moment Resistance s f y A s + FRP E FRP FRP A FRP = 0.9(400)400 +0.80(210000)0.006(110) = M r = Tsds-ds- 1 c 2 +T FRP h- 1 c 2 ds-ds- 2 h- 1 c 2 320- 0.87 (98.9) 2 360- 2 M r = 75 10 6 N.mm = 75 kN.m 4 - Flexural Strengthening

49 ISIS EC Module 4 Example: The T-beam requires strengthening to upgrade its moment capacity to 600 kN-m. Calculate the required area of FRP (A FRP ). Section information: f c = 30 MPa FRPu = 1.26 % A s = 8 x 300 mm 2 f y = 400 MPa E s = 200 GPa E FRP = 155 GPa 4 - Flexural Strengthening

50 ISIS EC Module 4 Step 1: Calculate concrete stress block factors 1 = 0.85 – 0.0015 f c > 0.67 1 = 0.85 – 0.0015 (30) = 0.805 1 = 0.97 – 0.0025 f c > 0.67 1 = 0.97 – 0.0025 (30) = 0.895 4 - Flexural Strengthening

51 ISIS EC Module 4 Step 2: Evaluating the moment capacity of the existing section s f y A s Compression in concrete = Tension in steel 1 c f c b 1 c = 0.805(0.75)30(0.895)650(c) 0.9(400)(300× 8) = c = 82mm > h f Assume neutral axis is inside the flange (c < h f ) The assumption (c < h f ) is wrong. 4 - Flexural Strengthening

52 ISIS EC Module 4 Step 2: Evaluating the moment capacity of the existing section s f y 1 c f c (b e -b w )h f =1409 mm 2 Assume neutral axis is outside the flange (c > h f ) A sf = 0.9(400) 0.805(0.75)30(650-250)70 A sf = A sw = A s –A sf = 2400 -1409 = 991 mm 2 M rf = 0.9 (400)1409 (510 - 70 2 ) = 240.939 ×10 6 N.mm = 240.939 kN.m 4 - Flexural Strengthening

53 ISIS EC Module 4 Step 2: Evaluating the moment capacity of the existing section s f y A sw Compression in concrete = Tension in steel (A sw ) 1 c f c b 1 c = 0.805(0.75)30(0.895)250(c) 0.9(400)(991) = c = 88.03mm > h f M rw = 0.9 (400)991(510 - 0.895 (88.03) 2 ) = 167.89 ×10 6 N.mm = 167.89 kN.m M r = 167.89 + 240.939 =408.8 kN.mMoment resistance of the section 4 - Flexural Strengthening

54 ISIS EC Module 4 Step 3: Optimized determination of A FRP 1) Determine c u sing the following equation: 1 c f c b w 1 2 c 2 c= 1153 (rejected) or 187mm (accepted) 2 1 c f c b w h 1 c C s (h-d)+ s f y A sw (d s -h)-(M f -M rf ) = 0 0.805(0.75)30(250)(0.895) 2 c 2 2 - 0.805(0.75)30(250)600(0.895)c - 0.9(400)991(510-600) + (600-240.9)x10 6 = 0 1813.57c 2 - 2431603.1c + 391208400 = 0 4 - Flexural Strengthening

55 ISIS EC Module 4 Step 3: Optimized determination of A FRP 2) Strain in FRP Select: 2 layers b = 250mm and t FRP = 1.5mm, A FRP = 750mm 2 FRP = cu h-c c = 0.0035 600-187 187 = 0.0077 > 0.006 FRP = 0.006 T FRP = 1 c f c b w 1 c - s f y A sw =0.805(0.75)30(250)0.895(187) - 400(0.9)991=401 089.6 N A FRP = T FRP FRP E FRP FRP = 401 089.6 0.75x0.8×155×10 3 ×0.006 = 718.8 mm 2 4 - Flexural Strengthening

56 ISIS EC Module 4 Step 4: Check the design Assume tension failure of the FRP and yielding of steel = 191.3 mm >h f …………O.K c = T FRP + s f y A sw 1 c f c b 1 = 0.75(0.8)(155000)0.006(750)+ 0.9(400)991 0.805(250)0.75(30)0.895 1) Neutral axis location 2) Check strains c = FRP c h-c = 0.006 191.3 600-191.3 = 0.0028 < 0.0035 s = FRP d s -c h-c = 0.006 510-191.3 600-191.3 = 0.00467 > y ………….O.K 4 - Flexural Strengthening

57 ISIS EC Module 4 Step 4: Check the design M rw = = 0.75(155000)0.006(600) 3) Moment resistance of the section M rw = Ts d s - 1 c 2 +T FRP h - 1 c 2 0.9(400)991 600 - 0.895(191.3) 2 510 - 0.895(191.3) 2 × 10 -6 + = 366.68 kN.m × 10 -6 M rf = 0.9(400)1409(510 - 70 2 ) = 240.939 ×10 6 N.mm = 240.939 kN.m M rt = M rw + M rf = 366.68 + 240.94 = 607.62 kN.m 4 - Flexural Strengthening

58 ISIS EC Module 4 5 - Shear Strengthening The shear resistance of the concrete element depends on the interaction between the concrete and the reinforcement. FRP sheets can be applied to increase shear resistance. The sheets are placed perpendicular or at an angle to the beams longitudinal axis. The shear capacity from the FRP stirrups is related to the angle of the cracks in the concrete, the direction and the effective strain of the FRP.

59 ISIS EC Module 4 d FRP is the effective shear depth for FRP s FRP is the spacing of the FRP stirrups w FRP is the width of the FRP stirrup 5 - Shear Strengthening is the angle of inclination of diagonal cracks in the concrete. is the angle of the FRP stirrups

60 ISIS EC Module 4 Many different possible configurations: 1) Continuous wraps or finite width sheets (width and spacing) 2) Angle between the sheet and the beams axis 3) Wrap configuration with respect to the cross section U-Wrap ContinuousFinite = 90 5 - Shear Strengthening Fully Wrapped 90

61 ISIS EC Module 4 Shear resistance of a beam (V r ): 1) Existing capacity - Resistance from concrete (V c ) - Resistance from steel (V s ) 2) Additional capacity - Resistance from FRP wraps (V FRP ) V r =VcVc VsVs V FRP ++ 5 - Shear Strengthening

62 ISIS EC Module 4 Shear resistance of a beam (V r ): 1) Resistance provided by concrete (V c ) d v (0.72h, 0.9d) 2) Resistance provided by steel (V s ) V s = s f y A v d v (cot +cot )sin s V c = 2.5 v c f cr b v d v 5 - Shear Strengthening

63 ISIS EC Module 4 3) Resistance provided by FRP: V FRP = FRP A FRP E FRP FRPe d FRP (cot + cot ) sin s FRP A FRP = 2 t FRP w FRP FRPe is the effective strain in the FRP stirrups d FRP is the effective depth s FRP is the spacing of the FRP stirrups 5 - Shear Strengthening

64 ISIS EC Module 4 Effective depth of FRP, d FRP : Closed wrap shear FRP No flexural FRP Closed wrap shear FRP Tension FRP for flexure d FRP (0.9d, 0.72h)d FRP 0.9h d 5 - Shear Strengthening

65 ISIS EC Module 4 Effective depth of FRP, d FRP : U-Shaped FRP stirrup No flexural FRP U-Shaped FRP stirrup Tension FRP for flexure d FRP (0.9h FRP, 0.72h)d FRP (0.72h,h FRP ) h frp 5-Shear Strengthening

66 ISIS EC Module 4 frpe = 0.004 0.75 frpu (For completely wrapped sections) frpe = K v frpu 0.004 (For other configurations) where: K v = K 1 = Effective strain in FRP, FRPe : K1K2LeK1K2Le 11900 FRPu 0.75 f c 27 2/3 K 2 = d FRP -L e d FRP Le=Le= 23300 (t FRP E FRP ) 0.58 5 - Shear Strengthening

67 ISIS EC Module 4 Checks: - Spacing of strips, s FRP : s FRP w FRP + d FRP 4 - Maximum allowable shear strengthening, V FRP : V c + V s + V FRP 0.25 c f c b v d v 5 - Shear Strengthening

68 Shear Strengthening ISIS EC Module 4 Example Example: Calculate the shear capacity (V r ) for an FRP-strengthened concrete section Section information Section b = 150 mm h FRP = 450 mm d s =550mm CFRP wrap Section Elevation f c = 45 MPa FRPu = 1.5% f y = 400 MPa (re-bar & stirrup) Steel used is 10M E FRP = 230GPa s = 225 mm c/c t FRP = 1.02 mm w FRP = 100 mm s FRP = 200 mm 150mm h=600 mm 5 - Shear Strengthening

69 ISIS EC Module 4 Solution: 1) Resistance provided by concrete (V c ) V c = 2.5 v c f cr b v d v f cr = 0.4* f c = 0.4* 45=2.68 d v (0.72h and 0.9d) (0.72*600 and 0.9*550) (432 and 495) = 495mm V c = 2.5*0.18*0.75*2.68*150*495*10 -3 = 67.24 kN 5 - Shear Strengthening

70 ISIS EC Module 4 2) Resistance provided by steel (V s ) V s = s f y A v d v (cot + cot )sin s V s = (0.9)400(200)495(cot42 + cot90)sin90 225 V s = 175,921 N = 175.9 kN 5 - Shear Strengthening

71 ISIS EC Module 4 3) Resistance provided by GFRP (V FRP ) d FRP (0.9 h FRP,0.72h) (0.9 × 450, 0.72 × 600) (405,432) = 432mm A FRP = 2 t FRP w FRP = 2(1.02)(100) = 204 mm 2 V FRP = FRP A FRP E FRP FRPe d FRP (cot + cot ) sin s FRP 5 - Shear Strengthening

72 ISIS EC Module 4 3) Resistance provided by FRP: 27 f c 2/3 = 27 45 2/3 = 1.406 K2=K2= d FRP -L e d FRP Le=Le= 23300 (t FRP E FRP ) 0.58 = 23300 (1.02 x 230 000) 0.58 = 17.888mm = 432 - 17.88 432 = 0.959 Kv=Kv= K1=K1= K1K2LeK1K2Le 11900 FRPu = (1.406)(0.959)(17.888) 11900 (1.5)(10 -2 ) = 0.135 < 0.75 =0.135 5 - Shear Strengthening

73 ISIS EC Module 4 FRPe 0.004 FRPe K v FRPu = 0.135 (1.5)(10 -2 )= 0.002025 FRPe = 0.002025 Effective strain in FRP, frpe : V FRP = FRP A FRP E FRP FRPe d FRP (cot + cot ) sin s FRP V FRP = 0.6(204)(230000)(0.002025)(432)(cot42) 200(1000) =136.8 kN 5 - Shear Strengthening

74 ISIS EC Module 4 Total resistance of the section (V r ): V r =VcVc VsVs V FRP ++ V r = 67.24 + 175.9 + 136.8 = 379.9 kN 5 - Shear Strengthening

75 ISIS EC Module 4 Checks: 1) Maximum allowable shear strengthening, V FRP : V c + V s + V FRP 0.25 c f c b v d v 379.9 0.25(0.75)(45)(150)(495)(10 -3 ) 379.9 626.48 kN…………………………O.K. 5 - Shear Strengthening

76 ISIS EC Module 4 Checks: 2) Spacing of strips, s FRP : s FRP w FRP + d FRP 4 200 100 + 432 4 200 208 mm…………………….O.K 5 - Shear Strengthening

77 ISIS EC Module 4 6 - Column Strengthening FRP sheets can be wrapped around concrete columns to increase strength How it works: Concrete shortens… …and dilates… …FRP confines the concrete… f l FRP …and places it in triaxial stress… Internal reinforcing steel Concrete FRP wrap

78 ISIS EC Module 4 The result: Increased load capacity Increased deformation capability 6 - Column Strengthening

79 ISIS EC Module 4 Confinement efficiency –Best: circular cross-section –Worst: rectangular section Areas of concrete unconfined by the small bending stiffness of FRP system Stress concentration at corners confined unconfined Stress distribution in rectangular section Uniform stress distribution in circular section 6 - Column Strengthening

80 ISIS EC Module 4 Slenderness of the column If the column is not slender, then the column is designed and analyzed for axial load only (short column). If the column is slender, then the column is designed and analyzed for combined axial load and bending moment. 6 - Column Strengthening

81 ISIS EC Module 4 Slenderness of the column Slenderness could be ignored if: kluklu r < 34 - 12 M1M1 M2M2 Where: k is the effective length factor for the column l u is the unsupported length of the column r is the radius of gyration of the section M 1 is the smaller end moment at ULS due to factored loads M 2 is the larger end moment at ULS due to factored loads Braced columns kluklu r < 22 Un-braced columns 6 - Column Strengthening

82 ISIS EC Module 4 1 - Short column (axial load only) 6 - Column Strengthening

83 ISIS EC Module 4 1) Confinement Pressure (f l FRP ) : f l FRP = Where: f l FRP is the confinement pressure t FRP is the thickness of the FRP confining system D g is the external diameter of the circular section or the diagonal of the rectangular section 2t FRP FRP f FRPu DgDg …………………………Eq 6-2 6 - Column Strengthening

84 ISIS EC Module 4 Minimum confinement pressure Maximum confinement pressure Why? To ensure adequate ductility of column Limit f l FRP 0.1 f c To prevent excessive deformations of column Limit Why? f l frp 0.33 f c Confinement Limits: 6 - Column Strengthening

85 ISIS EC Module 4 2) Confined concrete strength (f cc ) : Where : f c is the unconfined specified concrete strength f cc = f c + 2 f l FRP The benefit of the confining pressure is to increase the confined compressive concrete strength, f cc …………………………Eq 6-3 6 - Column Strengthening

86 ISIS EC Module 4 3) Axial Load capacity (P r ) : Where : A g is the gross area of the cross section A s is the total cross- sectional area of the longitudinal steel reinforcing bars P r = 0.8 c f cc (A g -A s )+ s f y A s The factored axial load resistance for an FRP-confined reinforced concrete column, P r is given by: …………………………Eq 6-5 6 - Column Strengthening

87 ISIS EC Module 4 Design steps for short column (axial load only): 1) Determine the required confined concrete ( f cc ) strength according to Equation 6-5. 2) Determine the required confinement pressure ( f l FRP ) from Equation 6-3. 3) Using the properties of the selected FRP system, determine a minimal thickness for the FRP (t FRP ) from Equation 6-2. 4) Check for the confinement limits. 6 - Column Strengthening

88 ISIS EC Module 4 2 - Slender Column (axial load + moment) 6 - Column Strengthening

89 ISIS EC Module 4 Section analysis is based on stress and strain compatibility Equivalent stress distribution confined concrete unconfined concrete Internal forces side FRP tension face FRP Axial strain distribution Cross section FRP Steel bars c = = s c-d = sj d sj -c = s d-c = FRP h-c …………………Eq 6-6 C cc + C c + C s – F sj – T s - T FRP,side - T FRP,face = P r P f 6 - Column Strengthening = 5 cc c f cc fcfc +1

90 ISIS EC Module 4 1) Assuming concrete crushing Internal force f c +f cc cc C cc 2 b c- c c cc CcCc c c CsCs cc F sj c T s s f y A s or s s E s A s if s < y T FRP,side FRP c E FRP (h-c) t FRP (h-c) cc T FRP,face FRP c E FRP bt FRP (h-c) cc 6 - Column Strengthening c c 1 f c b 1 s f y A s s s f y A sj E s A sj (d sj -c) cc - c - c c 2 h 2f c +f cc 3f c +3f cc Distance from the centre of the section c cc c 2 h - c + 1- 2 1 h /2 -d d sj - 2 h 2 h (h-c) 2 h 3 - d – h/2

91 ISIS EC Module 4 2) Assuming maximum FRP tension ( FRP = FRPt ): f c + f c FRPt C cc c 2 bc- h-c c FRPt C c c 1 f c b 1 h-c c C s s f y A s or f s s E s A s if s < y FRPt F sj s s f y A sj h-c E s A sj (d sj -c) T s s f y A s T FRP,side FRP f FRPu (h-c) t FRP FRP E FRP FRPt (h-c) t FRP f c = (f cc -f c ) cc - c c - c T FRP,face FRP f FRPu b t FRP FRP E FRP FRPt b t FRP 6 - Column Strengthening Internal force Distance from the centre of the section 2 h - d d-h/2 2 h d sj - 2 h (h-c) 2 h 3 - c FRPt h-c 2 h - c + 1- 2 1 FRPt c - h-c c 2 h 3f c + f c 6f c +3 f c

92 ISIS EC Module 4 Design steps for slender column: 1) Assuming a linear distribution of strain, identify the relationship of strain in the various materials as a function of the assumed failure strain. 2) Determine the resultant force for each material. 3) Calculate the position of the neutral axis using equilibrium of forces. 4) Check the validity of the assumptions of strains and stresses for all materials. 6 - Column Strengthening

93 ISIS EC Module 4 Design steps for slender column: 5) Determine P r as the sum of the resultant force from each material. 6) Determine M r as the sum of the internal resultant forces multiplied by their respective distances to the centroid of the section. 6 - Column Strengthening

94 ISIS EC Module 4 Rectangular Columns External FRP wrapping may be used with rectangular columns. However, strengthening is not as effective and is more complex. Confinement all around Confinement only in some areas 6 - Column Strengthening

95 ISIS EC Module 4 Some geometrical limitations are imposed: Sharp edge concrete should be rounded to promote an intimate and continuous contact of the FRP with the concrete. - minimum radius is 35 mm The aspect ratio of the section (h/b) 1.5 The smaller cross section dimension (b) 600 mm The equations used are the same. D g is taken as the diagonal of the cross section. 6 - Column Strengthening f l FRP = 2t FRP FRP F FRPu DgDg = h 2 +b 2

96 ISIS EC Module 4 Additional Considerations External FRP wrapping may also be used with circular and rectangular RC columns to strengthen for shear. Particularly useful in seismic upgrade situations where increased lateral loads are a concern. 6 - Column Strengthening

97 ISIS EC Module 4 The confining effects of FRP wraps are not activated until significant radial expansion of concrete occurs. Therefore, ensure service loads are kept low enough to prevent failure by creep and fatigue To avoid creep failure: 6 - Column Strengthening P D 0.85 0.8 1 c f` c (A g -A s )+ f s A s f s 0.0015 Es 0.8f y Where: P D is the dead load f s is the stress in the axial steel reinforcement

98 Example ISIS EC Module 4 Example: Determine the number of layers of GFRP wrap that are required to increase the factored axial load capacity of the column to 3450 kN. Information RC column factored axial resistance (after strengthening) = 3450 kN l u = 2500 mm D g = 450 mm A g = 159000 mm 2 A s = 2500 mm 2 f y = 400 MPa f c = 30 MPa f FRPu = 600 MPa t FRP = 1 mm FRP = 0.70*0.75 6 - Column Strengthening

99 ISIS EC Module 4 Solution: Step 1: Check for the slenderness effect kluklu r < 34 - 12 M1M1 M2M2 k =1.0, M 1 =0 and M 2 =0 2500 112.5 = 22.2 < 34 Thus, the slenderness effect can be ignored 6 - Column Strengthening

100 ISIS EC Module 4 Step 2: Determine the required confined concrete strength, f cc P r = 0.8 1 c f cc (A g -A s )+ s f y A s 1 = 0.85 – 0.0015 f c > 0.67 1 = 0.85 – 0.0015 (30) = 0.81 3450 000 = 0.80.81(0.75) f cc (159000-2500)+ 0.9(400)2500 f cc = 35.9 MPa 6 - Column Strengthening

101 ISIS EC Module 4 Step 3: Determine the required confinement pressure ( f l FRP ) 35.9 = 30+ 2 f l FRP f cc = f c + 2 f l FRP f l FRP = 2.95 MPa Step 4: Check for the confinement limits f l FRP 0.1 f c =0.1(30) = 3 MPa f l FRP 0.33 f c =0.33(30) = 9.9 MPa f l FRP = 3 MPa 6 - Column Strengthening

102 ISIS EC Module 4 Step 5: Determine the minimal thickness for the FRP (t FRP ) and number of layers f l FRP = 2t FRP FRP F FRPu DgDg 3 = 2t FRP (0.70×0.75)600 450 t FRP = 2.14 mm Since t GFRP = 1.0 mm, 3 layers of GFRP are required. 6 - Column Strengthening

103 Example ISIS EC Module 4 Example: Check the design of the following column. It is required to resist a factored axial load of 6000 kN and a factored moment of 1600 kN.m. Information A st = 4000 mm 2 f y = 400 MPa f c = 30 MPa f FRPu = 3450MPa t FRP = 0.167 mm FRPu = 0.015 75 325 75 800 600 Axial FRP 2 layers Hoop FRP 6 layers Steel bars 6 - Column Strengthening

104 ISIS EC Module 4 Step 1: Determine the properties of the confined concrete f l FRP = b D g = b 2 +h 2 = 600 2 +800 2 = 1000 mm = 1.25 1.5 Confinement limits: h b < 800 Equivalent diameter: Confining pressure: 2t FRP FRP f FRPu DgDg 2(6 × 0.167)(0.75 ×0.70)3450 1000 == 3.63 MPa 0.33 f c f l FRP 0.1 f c 10 f l FRP 3 ………………….O.K 6 - Column Strengthening

105 ISIS EC Module 4 Step 1: Determine the properties of the confined concrete Confined concrete strength: Concrete strain: f cc = f c + 2 f l FRP f cc = 30+2×3.63 = 37.26 MPa = 5 cc c f cc +1 fcfc cc = 0.0035( 5 -1 +1) = 0.0077 37.26 30 6 - Column Strengthening

106 ISIS EC Module 4 Step 2: Determine the equations of the resultant forces The following assumptions were made: - Compression failure (concrete crushing) - f c varies linearly from f c to f cc - Yielding of both tension and compression steel - Intermediate steel in elastic domain 1 = 0.85 – 0.0015 f c > 0.67 1 = 0.85 – 0.0015 (30) = 0.805 1 = 0.97 – 0.0025 f c > 0.67 1 = 0.97 – 0.0025 (30) = 0.895 6 - Column Strengthening

107 ISIS EC Module 4 Assuming concrete crushing: f c +f cc cc C cc = c 2 b c- c c cc C c = c 1 f c b 1 c c C s = s f y A s = 0.75 30+37.26 2 600 c - c × 0.0035 0.0077 = 0.75(0.805)30(600)0.895 c × 0.0035 0.0077 =15133.5 c - c × 0.0035 0.0077 = 9726.4 c × 0.0035 0.0077 = 0.9(400)1500 = 540 000 N 6 - Column Strengthening

108 ISIS EC Module 4 cc F sj = s c E s A sj (d sj -c) T s = s f y A s = 0.9 (400) 1500 = 540 000 N T FRP,side = FRP c E FRP (h-c) t FRP = 0.75×0.70 (h-c) cc T FRP,face = FRP c E FRP bt FRP = 0.75×0.70 (h-c) cc = 0.9 c (400-c) 0.0077 200 000×1000 c (400-c) 0.0077 =180 × 10 6 230 000 × 0.334 c (800-c) 2 0.0077 = 40330.5 c (800-c) 2 0.0077 c (800-c) 0.0077 230 000 ×600 ×0.334 = 24198300 c (800-c) 0.0077 6 - Column Strengthening

109 ISIS EC Module 4 C cc +C c +C s -F sj -T s -T FRP,side -T FRP,face = P r 15133.5 c - c × 0.0035 0.0077 + 9726.4 c × 0.0035 0.0077 + 540 000 c (400-c) 0.0077 -180 × 10 6 -540 000- 40330.5 c (800-c) 2 0.0077 - 24198300 c (800-c) 0.0077 = 6000 × 10 3 c = 472 mm Step 3: Determine the position of the neutral axis, c: 6 - Column Strengthening

110 ISIS EC Module 4 Step 4: Check the assumptions for strains: = cc c s = 0.0077 ×= 0.0065 > 0.002 OK sj (c-d ) 472-75 472 cc c = (d sj -c) = 0.0077 × 400-472 472 = -0.0012 < ± 0.002 OK = cc c s = 0.0077 ×= 0.0041 > 0.002 OK (d-c ) 725-472 472 = cc c FRP = 0.0077 ×= 0.0054 < 0.006 OK (h-c ) 800-472 472 = c cc c = 0.0035 ×= 214.5 mm 472 0.0077 6 - Column Strengthening

111 ISIS EC Module 4 P r = C cc +C c +C s -F sj -T s -T FRP,side -T FRP,face =15133.5 472-214.5 + 9726.4 214.5 + 540 000 -0.0012 -180 × 10 6 - 540 000- 40330.5 472 (800-472) 2 0.0077 - 24198300 472 (800-472) 0.0077 = 5997 × 10 3 N Step 5: Determine P r : 6 - Column Strengthening

112 ISIS EC Module 4 Step 6: Determine M r : cc C cc - c - c c CcCc c CsCs cc c 2 h 2f c +f cc 3f c +3f cc 2 h - d 2 h - c + 1- 2 1 - 472-214.5 2 800 2 X 30+37.3 3 X 30+3 X 37.3 = 3897000 = 1075 X 10 6 N.mm =2086000 2 800 - 472 + 1- 2 0.895 X 214.5= 97 X 10 6 N.mm = 540 000 2 800 -75 = 176 X 10 6 N.mm 6 - Column Strengthening

113 ISIS EC Module 4 Step 6 : Determine M r : F sj d sj - T FRP,side (h-c) T FRP,face 2 h 2 h 3 - 2 h = 176 X 10 6 N.mm = 0 N.mm T s d - 2 h = 540 000 725 - 2 800 = 71400 (800-472) 2 800 3 - = 21 X 10 6 N.mm =131 000 2 800 = 52 X 10 6 N.mm Total = 1597 X 10 6 N.mm The flexural resistance is adequate M r = 1597 kN.m 1600 kN.m 6 - Column Strengthening

114 ISIS EC Module 4 Includes: 2) Handling and storage of FRP materials 3) Staff and contractor qualifications 4) Concrete surface preparation 5) Installation of FRP systems 7) Protection and finishing for FRP system 7 - Installation of FRP Strengthening Systems 6) Curing the FRP system 1) Approval of FRP materials

115 ISIS EC Module 4 2) Handling and storage of FRP materials: - Must be carried out in accordance with manufacturer specifications. - Contractor and supplier must ensure that FRP materials are shipped in adequate conditions. Do not use opened or damaged containers. - FRP components must be stored in clean & dry area, sheltered from sun rays. - Do not use material that has exceeded its shelf life. - Material safety data sheet for all FRP materials and components should be obtained from the manufacturer and should be accessible at the job site. 1) Approval of FRP materials: The use of certified FRP materials is recommended. Qualification testing can be used for the approval of the FRP materials. 7 - Installation of FRP Strengthening Systems

116 ISIS EC Module 4 3) Staff and contractor qualifications: The workers must have a basic knowledge of all stages of the installation of the FRP systems. The minimum required knowledge includes: - An understanding of the security instructions - Mixing proportions of resins - Application rates - Pot life and curing times - Installation techniques 7 - Installation of FRP Strengthening Systems

117 ISIS EC Module 4 4) Concrete surface preparation: -Repair of existing substrate: - The concrete surfaces must be free of particles and pieces that no longer adhere to the structure. - The surface must be cleaned from oil residuals or contaminants. - Rough surface should be smoothed. - Sections with sharp edges must be rounded. - Surface preparation for contact critical applications - A continuous contact between the concrete and the FRP confinement system should be guaranteed. - Rounding of corners, filling holes and elimination of depression are of prime importance. 7 - Installation of FRP Strengthening Systems

118 ISIS EC Module 4 5) Installation of FRP systems : - Primer, putty, saturating resin and fibres should be a part of the same system. - All equipment should be clean and in good operating condition - Ambient air and concrete surface temperature should be 10°C or more - The mixing of resins should be done in accordance with the FRP system manufacturer recommended procedure. All components should be mixed at a proper temperature and in the correct ratio until there is a uniform mix, free from trapped air. - The installation of FRP is either hand wet applied system or precured system. 7 - Installation of FRP Strengthening Systems

119 ISIS EC Module 4 6) Curing the FRP system - FRP materials should be cured according to the recommendations of the manufacturer unless the curing process is accelerated by heating, chemical reactant or other external supply. - The curing time should not be less than 24 hours before further work is done on the repaired surface. - Chemical contamination from gases, dust or liquid must be prevented during the cure of all materials. 7 - Installation of FRP Strengthening Systems

120 ISIS EC Module 4 7) Protection and finishing for FRP system - When the surface of the FRP materials is sufficiently dry or hard, a protection system and/or paint compatible with the installed reinforcement can be added. - The coating must dry for a minimum of 24 hours. - A certificate of compatibility of the protection system with the selected type of FRP reinforcement must be obtained from the manufacturer of the FRP materials. 7 - Installation of FRP Strengthening Systems

121 ISIS EC Module 4 8 - Quality Control and Quality Assurance The FRP material suppliers, the FRP installation contractors and all others associated with the FRP strengthening project should maintain a comprehensive quality assurance and quality control program. 1) Material qualification and acceptance : The FRP manufacturer, distributor or their agent should provide information demonstrating that the proposed FRP meets all mechanical, physical and chemical design requirements. Tensile strength, type of fibres, resins, durability, etc. 2) Qualification of contractor personnel: The selection of contractors should be based on evidence regarding their qualifications and experience for FRP strengthening projects.

122 ISIS EC Module 4 8 - Quality Control and Quality Assurance 3) Inspection of concrete substrate: - The concrete surface should be inspected and tested before application of FRP. The inspection should include: - Smoothness or roughness of the surface - Holes and cracks - Corners radius - Cleanliness - Pull-off tests should be performed to determine the tensile strength of the concrete for bond-critical applications.

123 ISIS EC Module 4 8 - Quality Control and Quality Assurance 4) FRP material inspection: Inspection of the FRP materials shall be conducted before, during and after their installation. - Before Construction The FRP supplier should submit certification & identification of all the FRP materials to be used. The installation procedure should be submitted as well. - During Construction Keep records for: - Quantity and mixture proportions of resin - The date and time of mixing - Ambient temperature & humidity - All other useful information Visual inspection of fibres orientation and waviness should be carried out.

124 ISIS EC Module 4 8 - Quality Control and Quality Assurance 4) FRP material inspection: - At completion of the project: A record of all final inspection and test results related to the FRP material should be retained. Samples of the cured FRP materials should be retained as well. 5)Testing: - Qualification testing: It is a specification for the product certification of FRPs used for rehabilitation. It includes some guidelines as: - FRP systems whose properties have not been fully established should not be considered - Constituent materials, fibres, matrices and adhesives, should be acceptable by the applicable code and known for their good performance.

125 ISIS EC Module 4 8 - Quality Control and Quality Assurance 5) Testing: - Field testing: Confirmatory test samples of the FRP material systems should be prepared at the construction site and tested at an approved laboratory. In-place load testing can be used to confirm the behaviour of the FRP strengthened member.

126 ISIS EC Module 4 9 - Additional Applications Prestressed FRP Sheets One way to improve FRP effectiveness is to apply prestress to the sheet prior to bonding This allows the FRP to contribute to both service and ultimate load- bearing situations It can also help close existing cracks, and delay the formation of new cracks Prestressing FRP sheets is a promising technique, but is still under development

127 ISIS EC Module 4 10 - Field Applications Maryland Bridge - Winnipeg, Manitoba - Constructed in 1969 - Twin five-span continuous precast prestressed girders - CFRP sheets to upgrade shear capacity

128 ISIS EC Module 4 John Hart Bridge - Prince George, BC - 84 girder ends were shear strengthened with CFRP - Increase in shear capacity of 15-20% - Upgrade completed in 6 weeks Locations for FRP shear reinforcement 10 - Field Applications

129 ISIS EC Module 4 Country Hills Boulevard Bridge - Calgary, AB - Deck strengthened in negative bending with CFRP strips - New wearing surface placed on top of FRP strips 10 - Field Applications

130 ISIS EC Module 4 A Canadian code exists for the design of FRP-strengthened concrete members CAN/CSA-S806-02: Design and Construction of Building Components with Fibre Reinforced Polymers (Currently under revision) Design Guidance CAN/CSA-S6-10: Canadian Highway Bridge Design Code

131 Additional Information ISIS EC Module 4 Available from www.isiscanada.com ISIS EC Module 1 : Mechanics Examples Incorporating FRP Materials ISIS EC Module 2: An Introduction to FRP Composites for Construction ISIS EC Module 3: An Introduction to FRP-Reinforced Concrete ISIS EC Module 5: Introduction to Structural Health Monitoring ISIS EC Module 6: Application & Handling of FRP Reinforcements for Concrete ISIS EC Module 7: Introduction to Life Cycle Engineering & Costing for Innovative Infrastructure ISIS EC Module 8: Durability of FRP Composites for Construction ISIS EC Module 9: Prestressing Concrete Structures with Fibre Reinforced Polymers


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