Design of FRP Systems for Strengthening Concrete Girders in Shear (2011)

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59 These proposed guidelines are the recommendations of the NCHRP Project 12-75 Research Team at the Missouri University of Science and Technology. These guidelines have not been approved by NCHRP or any AASHTO committee nor formally accepted for the AASHTO specifications. A T T A C H M E N T B Recommended Design Guidelines for Concrete Girders Strengthened in Shear with FRP

60 SPECIFICATIONS COMMENTARY B1 GENERAL This attachment presents recommended design guidelines for concrete girders strengthened in shear using externally bonded fiber reinforced polymers (FRPs). Design examples developed using these guidelines are presented in the appendix. B1.1 Design Philosophy n uV V (B1-1) where: Vn : Nominal shear resistance Vu : Required shear strength : Strength reduction factor (0.9) Careful consideration for all possible failure modes and subsequent strains and stresses should be considered in determining the nominal shear strength of a member. B1.2 Scope These design guidelines focus on presenting design procedures including design equations. Specific limits of applying the proposed design guidelines are also presented in the relevant sections throughout this document. B2 EVALUATION AND REPAIR OF EXISTING RC BEAMS FRP strengthening is usually performed on structurally deficient or damaged RC beams. Before a strengthening procedure is implemented, the extent of deficiency and suitability of FRP strengthening should be evaluated. The necessary evaluation criteria for repair of existing concrete structures and post repair evaluation criteria are well established in the following documents. CB2 The proposed design guidelines were based on the traditional reinforced concrete (RC) design principles adopted by the current AASHTO LRFD Bridge Design Specifications and the knowledge on the mechanical behavior of FRP obtained from work performed under the NCHRP Project 12-75. As such, the factored shear resistance, φV n , of a concrete member should meet or exceed the factored shear force applied to the member, V u . The applied factored shear force and the factored shear resistance should be computed based on the load and resistance factors specified in the AASHTO LRFD Bridge Design Specifications. The factored shear resistance shall be determined as: Information, such as evaluation and repair of existing RC beams as well as proper application of FRP, is available; an attempt was made to provide references to other publications where additional details can be found.

61 SPECIFICATIONS ACI 201.1R: Guide for Making a Condition Survey of Concrete in Service ACI 224.1R: Causes, Evaluation, and Repair of Cracks in Concrete ACI 364.1R-94: Guide for Evaluation of Concrete Struct ures Prior to Rehabilita ti on ACI 440.2R-08: Guide for the Design and Construction of Externally Bonded FRP Systems for Strengthening Concrete Structures ACI 503R: Use of Epoxy Compounds with Concrete ACI 546R: Concrete Repair Guide International Concrete Repair Institute (ICRI) ICRI 03730: Guide for Surface Preparation for the Repair of Deteriorated Concrete Resulting from Reinforcing Steel Corrosion International Concrete Repair Institute (ICRI) ICRI 03733: Guide for Selecting and Specifying Materials for Repairs of Concrete Surfaces NCHRP Report 609: Recommended Construction Specifications Process Control Manual for Repair and Retrofit of Concrete Structures Using Bonded FRP Composites Relevant specifications and guidelines provided by FRP manufacturers should also be carefully reviewed prior to th e design of any strengthening syste m. B3 STRENGTHENING SCHEMES FRP shear reinforcement is commonly attached to a beam , as shown in Figure B3.1 with (a) side bonding, in which the FRP is only bonded to the sides, (b) U-wrap, in which FRP U-jackets are bonded to both th e sides and soffit , and (c) complete wrapping, in which th e FRP is wrapped around the entire cross section. (a) (b) (c) Figure B3.1 Strengthening Scheme: Cross-Sectional View (a) Side bonding, (b) U-wrap, and (c) Complete wrap For all wrapping schem es, the FRP can be applied continuously along the portion of th e me mb er length to be strengthened or as discrete strips. The fibers of th e FRP may also be oriented at various angles to meet a range of strengthening requirem ents as shown in Figure B3.2 CB3 Com plete wrapping of th e cross section is th e mo st effective schem e and is comm only used in strengthening columns where there is sufficient access for such application. Beams are typically limited to U-wrap and side bonding applications since the integral slab ma kes it im practical to co mp letely wrap such me mb ers. U-wrapping has been experi me ntally shown to be mo re effective in im proving the shear resistance of a me mb er than side bonding.

62 SPECIFICATIONS COMMENTARY (a) Width of FRP Strips (w )f fCenter-to-Center Spacing of FRP Strip (s ) (b) Width of FRP Strips (w )f Center-to-Center Spacing of FRP Strip (s )f Figure B3.2 Strengthening Scheme: Side View — (a) Fibers at 90° Direction, and (b) Fibers at Inclined Direction B4 APPLICATION OF FRP B4.1 General In general, procedures for the installation of FRP systems are developed by the manufacturer and can vary between different systems. Procedures may also vary depending on the type and condition of the structure to be strengthened. The application of FRP systems will not stop the ongoing corrosion of existing steel reinforcements. The cause of corrosion to internal steel reinforcements should be addressed and corrosion-related deterioration should be repaired prior to application of any FRP system. B4.2 Surface Preparation The concrete surface should be prepared to a minimum concrete surface profile (CSP) 3 as defined by the ICRI- surface-profile chips (ICRI 03732, NCHRP Report 609). Localized out-of-plane variations, including form lines, should not exceed 1/32 inch or the tolerances recommended by the FRP system manufacturer, whichever is smaller. Bug holes and voids should be filled with epoxy putty. It is recommended that surface preparation be accomplished using abrasive or water-blasting techniques. All laitance, dust, dirt, oil, curing compound, existing coatings, and any other matter that could interfere with the bond between the FRP system and concrete substrate should be removed. When fibers are wrapped around corners, the corners should be rounded to a minimum 1/2 inch radius to prevent stress concentrations in the FRP system and voids between the FRP system and the concrete. Rough edges should also be smoothed by grinding or with putty prior to FRP application. CB4.1 It is recommended that FRP applications be performed by a contractor trained in accordance with the installation procedures specified by the manufacturer. Comprehensive guidelines in this regard are provided in NCHRP Report 609, Recommended Construction Specifications and Process Control Manual for Repair and Retrofit of Concrete Structures Using Bonded FRP Composites CB4.2 Bond behavior of the FRP system is highly dependent on a sound concrete substrate and can significantly influence the integrity of the FRP strengthening system. Proper preparation and profiling of the concrete substrate is necessary to achieve optimum bond strength. Improper surface preparation can lead to premature debonding or delamination.

63 SPECIFICATIONS COMMENTARY B4.3 Inspection, Evaluation, and Acceptance Application of FRP systems should be inspected by a licensed engineer or qualified inspector knowledgeable in FRP systems and installation procedures. The following should be recorded at the time of installation: Date and time of installation Ambient temperature, relative humidity, and general weather observations and surface temperature of concrete Surface dryness, surface preparation methods and resulting profile using the ICRC-surface-profile- chips Qualitative description of surface cleanliness Type of auxiliary heat source, if applicable Widths of cracks not injected with epoxy Fiber or pre-cured laminate batch number(s) and approximate locations in structure Batch numbers, mixture ratios, mixing times, and qualitative descriptions of the appearance of all mixed resins, including primers, putties, saturants, adhesives, and coatings mixed for the day Observations of progress of cure of resins Conformance with installation procedures Location and size of any delaminations or air voids General progress of work Level of curing of resin in accordance with ASTM D3418. Adhesion strength B5 MATERIAL PROPERTIES OF FRP The following mechanical properties should be obtained from manufacturers or coupon tests in accordance with ASTM D3039. fE : the modulus of elasticity of FRP fu : the ultimate strain of FRP. Then, the nominal resistance, fuf , can be determined assuming linear behavior of FRP stress-strain relationship up to failure as: fu f fuf E (B5-1) CB4.3 When concrete and atmospheric temperatures exceed 90°F, difficulties may be experienced in application of the epoxy compound owing to acceleration of the reaction and hardening rates. If ambient temperatures above 90°F are anticipated, work should be scheduled when the temperature is lower, such as in the early morning hours. If it is necessary to apply epoxy compounds at temperatures exceeding 90 °F, the work should be supervised by a person experienced in applying epoxy at high temperatures. Epoxy systems formulated for elevated temperature are available (ACI 530R-93). At temperatures below 40°F, difficulties may occur due to deceleration of the reaction rates. The presence of frost or ice crystals may also be detrimental to the bond between the FRP and the concrete. Evaluate moisture content or outgassing of the concrete by determining if moisture will collect at bond lines between old concrete and epoxy adhesive before epoxy has cured. This may be accomplished by taping a 4 x 4 ft (1 x 1 m) polyethylene sheet to concrete surface. If moisture collects on underside of polyethylene sheet before epoxy would cure, then allow concrete to dry sufficiently to prevent the possibility of a moisture barrier between old concrete and new epoxy (ACI 530R-93). During installation, sample cups of mixed resin should be prepared according to a predetermined sampling plan and retained for testing to determine level of curing in accordance with ASTM D3418. The relative cure of the resin can also be evaluated on the project site by physical observation of resin tackiness and hardness of work surfaces or hardness of retained resin samples. For bond-critical applications, tension adhesion testing of cored samples should be conducted using the methods in ACI 530R or ASTM D 4541 or the method described by ISIS (1998). The sampling frequency should be specified. Tension adhesion strengths should exceed 200 psi and exhibit failure of the concrete substrate before failure of the adhesive (ACI 440.2R-08).

64 SPECIFICATIONS COMMENTARY B6 NOMINAL SHEAR RESISTANCE An interaction is known to exist between the shear contributions of concrete, transverse steel reinforcement, and FRP. However, this interaction mechanism is not yet fully understood and thus is not reflected in the design procedures. Therefore, following the current reinforced concrete design principals, the nominal shear resistance ( nV ) is determined by adding the contribution of the FRP reinforcement to the contributions from concrete and internal transverse steel reinforcement: n c s fV V V V (B6-1) where, cV is the contributions of concrete, sV is the contribution of transverse steel reinforcement (stirrups), and fV is the contribution of FRP. The contributions from the concrete ( cV ) and transverse steel reinforcement ( sV ) can be computed based on the current AASHTO LRFD Bridge Design Specifications. Calculation of the FRP contribution ( fV ) is presented in the following sections. B7 SHEAR CONTRIBUTION OF FRP B7.1 Calculation of Contribution of FRP The contribution of FRP ( fV ) can be computed using the 45° truss model as: sin cos sin cos sin cos f f f ffe f f f ff fe f f f ff fe v f ff ( + ) f dA V s E ( + ) dA s E b ( + )d (B7-1) where, fA is the area of FRP covering two sides of the beam and can be determined by 2 f f fn t w ( fn is number of FRP plies, ft is the FRP reinforcement thickness, fw is the width of the strip), fef is the effective stress of FRP, fd is the effective depth of FRP measured from the top of FRP reinforcement to the centroid of the longitudinal reinforcement, fs is the center-to-center spacing of FRP, f is the angle of inclination of FRP with respect to the longitudinal axis of the member as shown in Figure B3.2, fE is the modulus of elasticity of FRP, fe is the effective strain of FRP, f is the reinforcement ratio of FRP, and vb is the effective web width taken as the minimum web width within the effective depth ( fd ) CB7.1

65 SPECIFICATIONS COMMENTARY The FRP shear reinforcement ratio, f , is determined as: For discrete strips 2 f f f f v f n t w b s (B7-2) For continuous sheets 2 f f f v n t b (B7-3) The effective strain ( fe ) represents the average strain experienced by the FRP at shear failure of the strengthened member and can be expressed as: For Full Anchorage (Rupture Failures Expected): Complete Wrap or U-Wrap with Anchors fe f fuR (B7-4) where .670.088 4( ) 1.0f f fR E For Other Anchorage (Non-Rupture Failures more likely): Side bonding or U-Wrap 0.012fe f fuR (B7-5) where .670.066 3( ) 1.0f f fR E The effective strain, fe , is largely dependent on the failure modes as discussed in Appendix A - Sections A3 and A4. Therefore, the experimental database collected in this project was grouped by the failure mode of the test specimens, i.e., either as debonding or rupture of the FRP and then regression analyses were performed to obtain Eqn. B7-4 and B7-5. The upper bound for the quantity fEf in Eqs. B7-4 and B7-5 is 300 ksi. Substituting this value in these two equations results in the lower bound value of Rf shown in the two equations.

66 SPECIFICATIONS COMMENTARY B7.2 Limitations B7.2.1 Shear span-to-depth ratio The reduction factors (Rf) were developed from tests in which the loading was at a distance from the support sufficient to assume plane sections before deformation remain plane after deformation, i.e. shallow beam behavior. Thus, these provisions are only applicable to beams with a shear span-to-depth ratio greater than 2.5. B7.2.2 Maximum Amount of FRP Shear Reinforcement The amount of FRP should be determined so that the nominal shear strength calculated by Eq. B 6-1 should not exceed the nominal shear strength calculated by 0.25n c v v pV f b d V (AASHTO 5.8.3.3-2) B.7.2.3 Maximum Spacing of FRP Shear Reinforcement The clear spacing between externally bonded FRP shear reinforcement shall not exceed the maximum permitted spacing ( maxs ) in accordance with the current AASHTO LRFD Bridge Design Specifications, expressed as: If '0.125 thenu cv f max 0.8 24in.vs d (AASHTO 5.8.2.7-1) If '0.125 thenu cv f max 0.4 12in.vs d (AASHTO 5.8.2.7-2) where uv = the shear stress calculated in accordance with AASHTO LRFD – Article 5.8.2.9 (ksi) and vd =effective shear depth as defined in AASHTO LRFD — Article 5.8.2.9 (in.) B7.3 Use of Anchorage Systems Different types of anchorage systems are available for shear strengthening with FRP. Examples of mechanical anchorage systems consisting of FRP composite plates and concrete anchor bolts are available in the literature [NCHRP Report 12-75]. However, it should be noted that additional horizontal FRP strips cannot ensure FRP rupture failure. Thus, it is recommended that Equation B7-5 be used to calculate the FRP contribution, realizing that such approach will result in conservative estimates. CB7.2.2 This provision is required to avoid web crushing failure of FRP strengthened beams due to excessive transverse shear reinforcement (both FRP and steel stirrups).

67 The following six design examples are presented to illustrate use of the recommended guidelines: Example 1-1: RC T-beam without internal transverse steel reinforcement strengthened with FRP in U-wrap configuration without anchorage systems Example 1-2: RC T-beam without internal transverse steel reinforcement strengthened with FRP in U-wrap configuration with an anchorage system Example 2-1: RC T-beam with internal transverse steel reinforcement strengthened with FRP in U-wrap configuration without anchorage systems Example 2-2: RC T-Beam with internal transverse steel reinforcement strengthened with FRP in U-wrap configuration with an anchorage system Example 3-1: PC I-Beam with internal transverse steel reinforcement strengthened with FRP in U-wrap configuration without anchorage systems Example 3-2: PC I-Beam with internal transverse steel reinforcement strengthened with FRP in U-wrap configuration with an anchorage system A P P E N D I X Design Examples

68 Reinforcement Strengthened with FRP in U-wrap Configuration without Anchorage Systems DESIGN EXAMPLE 1-1: RC T-Beam without Internal Transverse Steel 1. INTRODUCTION This example demonstrates the design procedures for externally bonded FRP shear reinforcement of an older reinforced concrete (RC) bridge using a U-wrap configuration without anchorage. The bridge consists of simply supported T-beams spanning 42 feet and spaced at 4.5 feet on center. The T-beams contain no transverse steel reinforcement. Additional details of the T-beam are provided in Figures 1 and 2. 42 ft 114 ft 4.5 ft 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 Figure 1. Bridge plan and transverse section. 2. MATERIAL PROPERTIES The following material properties have been chosen to represent those anticipated in an older bridge for which shear deficiencies might be expected. 2.1. Concrete Compressive strength f' c := 3.0 ksi Modulus of elasticity ksi E c = 3321 ksi E c := 33 1.5( )1.5⋅ f' c 1000⋅

69 β1 0.85 f'c 4≤if 0.65 f' c 8≥if 0.85 0.05 f' c 4−( )⋅− otherwise := β1 0.85= 2.2. Longitudinal Reinforcement Yield strength fy := 60 ksi Modulus of elasticity E s := 29000 ksi 2.3. FRP Reinforcement Carbon Fiber Sheets are used in this example. Thickness tf := 0.0065 in. Failure strength ffu := 550 ksi Modulus of elasticity Ef := 33000 ksi Failure strain εfu ffu Ef := εfu = 0.017 in/in 3. GEOMETRICAL PROPERTIES Total Height hT := 37 in. Flange Thickness hf := 7 in. Width of the web b v := 18 in. Effective Width of the Flange b eff := 54 in. Tensile reinforcement = 12#11 A s := 18.72 in2 Internal shear reinforcement = Not provided A v := 0.0 in2 Distance from the extreme compression fiber to the center of the steel at the section d := 32.7 in.

70 Figure 2. Cross-Section of an Intermediate Beam 4. CALCULATION OF THE FACTORED SHEAR FORCE AND NOMINAL SHEAR RESISTANCE 4.1 Factored Shear Force at the Critical Section V u_crit := 100 kips 4.2. Calculation of Nominal Shear Resistance For this example, the simplified approach is followed. θ := 45 deg β := 2 The nominal shear resistance provided by the concrete, V c , is calculated in accordance with LRFD Eqn.5.8.3.3-3 as: Assuming rectangular section behavior with no compression steel, the distance from the extreme compression fiber to the neutral axis, cc, may be calculated as: c c1 := A s ⋅fy 0.85⋅f' c ⋅b eff⋅β1 c c1 = 9.6 in.

71 a c1 := β1⋅cc1 c c2 := A s ⋅fy – 0.85⋅f'c⋅(beff – b1)⋅hf 0.85⋅f' c ⋅b v ⋅β1 a c1 = 8.16 in. check_a c1 := "Assumption is correct" if (ac1 ≤ hf) "Not behave as rectangular" otherwise check_a c1 = "Not behave as rectangular" a c2 := β1⋅cc2 ac2 = 10.47 in. c c2 = 12.32 in. check_a c2 := "Assumption is correct" if (ac2 ≥ hf) "Not behave as rectangular" otherwise check_a c2 = "Assumption is correct" Assuming T-beam section behavior with no compression steel, the distance from the extreme compression fiber to the neutral axis, c c , may be calculated as: Therefore c c = 12.32 in. a c = 10.47 in. c c := c c2 a c := a c2 The effective shear depth dv is taken as the distance, measured perpendicular to the neutral axis, between the resultants of the tensile and compressive forces due to flexure; it need not be taken less than the greater of 0.9d e or 0.72h (LRFD Article5.8.2.9) d v1 := d a c 2 − (dv2 := 0.9⋅d) (dv3 := 0.72⋅hT) (dv := max(dv1, dv2, dv3)) (dv) = 29.4 in. The nominal shear resistance provided by the concrete is: (LRFD Eqn. 5.8.3.3-3) v c := 0.0316⋅β⋅ f' c ⋅b v ⋅d v (vc) = 58 kips v s := 0 The nominal shear resistance provided by the internal steel reinforcement is:

72 v n := v c + v s + vp v n = 58 kips (vp := 0) The nominal shear resistance provided by the vertical component of prestressing strands is: The nominal shear resistance of the member is: (LRFD Eqn. 5.8.3.3-1) 5. DESIGN OF FRP SHEAR REINFORCEMENT 5.1 Check if FRP Reinforcement is Necessary or Not Strength reduction factor for shear (φ := 0.9) Check_FRP_Needed := "NOT need shear reinforcement" if φ V n ≥ V u_crit⋅ "NEED shear reinforcement" otherwise Check_FRP_Needed = "NEED shear reinforcement" 5.2 Computation of Required Vf Vf_req := V u_crit φ − Vn Vf_req = 53.1 kips 5.3 Selection of FRP Strengthening Scheme U-wrap configuration is used without anchorage systems at the end of the sheets. The FRP sheets will be applied at 90 degree with respect to the longitudinal axis of the girder as shown in Figure 3. First, the spacing of FRP strips is chosen to meet the maximum spacing requirement. Then, the width of the FRP strips is selected to adjust the amount of FRP strips. Figure 3 FRP strengthening scheme.

73 Use number of plies of FRP sheets nf := 1 Use the width of FRP sheets wf := 8 in. df = 25.7 in. Use the center-to-center spacing of FRP sheets sf := 15 in. αf := 90 degOrientation of FRP sheets Effective depth of FRP sheets df := d – hf Check if the selected spacing is acceptable or not Shear stress on concrete is: (LRFD Eqn. 5.8.2.9-1) v u := V u_crit − φ⋅Vp φ⋅b v ⋅d v (vu) = 0.21 ksi The maximum spacing of the transverse reinforcement is: (LRFD Eqn. 5.8.2.7-1) (LRFD Eqn. 5.8.2.7-2) s max := min(0.8⋅dv, 24) if vu < 0.125⋅f'c min(0.4⋅dv, 12) otherwise s max = 23.5 Check_Spacing := "Acceptable" if sf ≤ smax "NOT_Acceptable_Change_the_Spacing" otherwise Check_Spacing = "Acceptable" 5.4 Calculation of Shear Resistance of FRP, Vf The FRP reinforcement ratio is: ρf := 2⋅nf⋅wf⋅tf b v ⋅sf (Attachment A Eqn. 5.8.3.3-10) ρf( ) = 3.852× 10−4

74 The FRP strain reduction factor is: Rf min 3 ρf Ef⋅( )⋅ −0.67, 1.0 := (Attachment A Eqn. 5.8.3.3-9) The effective strain of FRP is: (Attachment A Eqn. 5.8.3.3-7) The effective stress of FRP is: (Attachment A Eqn. 5.8.3.3-6) ffe( ) = 300.4 ksi The shear contribution of the FRP can be then calculated. Vf := ρf⋅Ef⋅εfe⋅bv⋅df⋅(sin(αf) + cos(αf)) ffe := εfe⋅Ef εfe = 9.103× 10–3 in./in. Rf = 0.546 εfe := min (Rf⋅εfu, 0.012) (Attachment A Eqn. 5.8.3.3-5) Vf( ) 53.5 kips= Vf_check1 := "Change FRP Strengthening Scheme" if (Vf < Vf_req) "Provided FRP Strength Large Enough" otherwise Vf_check1 = "Provided FRP Strength Large Enough" Vf_check2 := "Provided FRP amount is adequate" if (Vf_req ≤ Vf < 1.1⋅Vf_req) "Change the FRP amount slightly" otherwise (Vf_check2) = "Provided FRP amount is adequate" 5.5 Calculation of Design Shear Resistance of the Member The design strength of the member is: (Attachment A Eqn. 5.8.3.3-1) φV n_total := φ⋅(Vc + Vp + Vs + Vf) (φVn_total) = 100.4 kips

75 Web_crushing_limit := 0.25⋅f' c ⋅b v ⋅d v + Vp (LRFD Eqn. 5.8.3.3-2) Web_crushing_limit = 397.3 kips Check_web_crushing_limit := "OK" if (Vc + Vs + Vf + Vp) ≤ Web_crushing_limit "No Good" otherwise Check_web_crushing_limit = "OK" 6. SUMMARY Externally bonded FRP sheets were designed in this example. The FRP sheets are applied at 90 degrees with respect to the longitudinal axis of the member with the U-wrap configuration and without anchorage systems as shown in Figure 4. The final design is summarized as: Use number of plies of FRP sheets Use the width of FRP sheets Use the center-to-center spacing of FRP sheets sf = 15 in. nf = 1 wf = 8 in. Figure 4. Final design of FRP strengthening.

76 DESIGN EXAMPLE 1-2: RC T-Beam without Internal Transverse Steel Reinforcement Strengthened with FRP in U-wrap Configuration with an Anchorage System 1. INTRODUCTION This example demonstrates the design procedures for externally bonded FRP shear reinforcement of an older reinforced concrete (RC) bridge using a U-wrap configuration with anchorage. The bridge consists of simply supported T-beams spanning 42 feet and spaced at 4.5 feet on center. The T-beams contain no transverse steel reinforcement. Additional details of the T-beam are provided in Figures 1 and 2. 42 ft 114 ft 4.5 ft 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 Figure 1. Bridge plan and transverse section. 2. MATERIAL PROPERTIES The following material properties have been chosen to represent those anticipated in an older bridge for which shear deficiencies might be expected. 2.1. Concrete Compressive strength (f' c := 3.0) ksi Modulus of elasticity E c 33 1.5( )1.5⋅ f'c 1000⋅ := (E c ) = 3321 ksi ksi

77 ( ) β1 := 0.85 f'c ≤ 4if 0.65 f' c ≥ 8if 0.85 0.05 f' c − 4⋅− otherwise β1 = 0.85 2.2. Longitudinal Reinforcement Yield strength fy := 60 ksi Modulus of elasticity E s := 29000 ksi 2.3. FRP Reinforcement Carbon Fiber Sheets are used in this example. Thickness tf := 0.0065 in. Failure strength ffu := 550 ksi Modulus of elasticity Ef := 33000 ksi Failure strain εfu ffu Ef := εfu = 0.017 in/in 3. GEOMETRICAL PROPERTIES Total Height hT := 37 in. Flange Thickness hf := 7 in. Width of the web b v := 18 in. Effective Width of the Flange b eff := 54 in. Tensile reinforcement = 12#11 A s := 18.72 in2 Internal shear reinforcement = Not provided (A v := 0.0) in2 Distance from the extreme compression fiber to the center of the steel at the section d := 32.7 in.

78 Figure 2. Cross-section of an intermediate beam. 4. CALCULATION OF THE FACTORED SHEAR FORCE AND NOMINAL SHEAR RESISTANCE 4.1 Factored Shear Force at the Critical Section V u_crit := 100 kips 4.2. Calculation of Nominal Shear Resistance For this example, the simplified approach is followed. (θ := 45 deg) (β := 2) The nominal shear resistance provided by the concrete, V c , is calculated in accordance with LRFD Eqn.5.8.3.3-3 as: Assuming rectangular section behavior with no compression steel, the distance from the extreme compression fiber to the neutral axis, c c , may be calculated as: c c1 := A s ⋅fy 0.85⋅f' c ⋅b eff⋅β1 (cc1) = 9.6 in. (a c1 := β1⋅cc1) (ac1) = 8.16 in.

79 (a c2 := β1⋅cc2) ( ) check_a c1 "Assumption is correct" if (ac1 ≤ hf) "Not behave as rectangular" otherwise := (check_a c1) = "Not behave as rectangular" Assuming T-beam section behavior with no compression steel, the distance from the extreme compression fiber to the neutral axis, c c , may be calculated as: c c2 A s fy⋅ 0.85 f'c⋅ beff bv−⋅ hf⋅− 0.85 f' c ⋅ b v ⋅ β1⋅ := (c c2) = 12.32 in. (a c2) = 10.47 in. check_a c2 "Assumption is correct" if (ac2 ≥ hf) "Not behave as rectangular" otherwise := (check_a c2) = "Assumption is correct" (c c := c c2) (cc) = 12.32 in. (a c := a c2) (ac) = 10.47 in. Therefore The effective shear depth d v is taken as the distance, measured perpendicular to the neutral axis, between the resultants of the tensile and compressive forces due to flexure; it need not be taken less than the greater of 0.9d e or 0.72h (LRFD Article5.8.2.9) d v1 d a c 2 − := (d v2 := 0.9 ⋅ d) (dv3 := 0.72 ⋅ hT) (d v := max (d v1, dv2, dv3)) (d v ) = 29.4 in. The nominal shear resistance provided by the concrete is: V c := 0.0316⋅β⋅ f' c ⋅b v ⋅d v (LRFD Eqn. 5.8.3.3-3) (V c ) = 58 kips The nominal shear resistance provided by the internal steel reinforcement is: V s := 0

80 The nominal shear resistance provided by the vertical component of prestressing strands is: (Vp := 0) The nominal shear resistance of the member is: (LRFD Eqn. 5.8.3.3-1) V n := V c + V s + Vp V n = 58 kips 5. DESIGN OF FRP SHEAR REINFORCEMENT 5.1 Check if FRP Reinforcement is Necessary or Not Strength reduction factor for shear (φ := 0.9) Check_FRP_Needed := "NOT need shear reinforcement" if φ⋅V n ≥ V u_crit "NEED shear reinforcement" otherwise Check_FRP_Needed = "NEED shear reinforcement" 5.2 Computation of Required Vf Vf_req := V u_crit φ Vn− Vf_req = 53.1 kips 5.3 Selection of FRP Strengthening Scheme U-wrap configuration is used with anchorage systems at the end of the sheets. The FRP sheets will be applied at 90 degrees with respect to the longitudinal axis of the girder as shown in the Figure 3 below. Anchorage systems will be installed at the top end portion of the FRP sheets to increase the effectiveness of FRP shear strengthening. First, the spacing of FRP strips is chosen to meet the maximum spacing requirement. Then, the width of the FRP strips is selected to adjust the amount of FRP strips. Figure 3. FRP strengthening scheme.

81 Use number of plies of FRP sheets nf := 1 Use the width of FRP sheets wf := 5.5 in. Use the center-to-center spacing of FRP sheets sf := 18 in. Orientation of FRP sheets αf := 90 deg Effective depth of FRP sheets df := d − hf df = 25.7 in. Check if the selected spacing is acceptable or not Shear stress on concrete is: v u V u_crit − φ⋅Vp φ⋅b v ⋅d v := (LRFD Eqn. 5.8.2.9-1) (v u ) = 0.21 ksi The maximum spacing of the transverse reinforcement is: s max := min(0.8 ⋅d v , 24) if v u < 0.125⋅f' c min(0.4⋅d v , 12) otherwise (LRFD Eqn. 5.8.2.7-1) (LRFD Eqn. 5.8.2.7-2) s max = 23.5 Check_Spacing := "Acceptable" if sf ≤ smax "NOT_Acceptable_Change_the_Spacing" otherwise Check_Spacing = "Acceptable" 5.4 Calculation of Shear Resistance of FRP, Vf The FRP reinforcement ratio is: ρf := 2⋅nf⋅wf⋅tf b v ⋅sf (Attachment A Eqn. 5.8.3.3-10) ρf( ) = 2.207 × 10−4

82 ( )−0.67 The FRP strain reduction factor is: Rf := min 4 ρf Ef⋅⋅ 1.0, (Attachment A Eqn. 5.8.3.3-8) Rf = 1 The effective strain of FRP is: εfe := Rf ⋅ εfu (Attachment A Eqn. 5.8.3.3-7) εfe = 0.017 in./in. The effective stress of FRP is: ffe := εfe ⋅ Ef (Attachment A Eqn. 5.8.3.3-6) (ffe) = 550 ksi The shear contribution of the FRP can be then calculated. (Vf := ρf ⋅ Ef ⋅ εfe ⋅ bv ⋅ df ⋅ (sin (αf) + cos (αf))) (Attachment A Eqn. 5.8.3.3-5) (Vf) = 56.1 kips Vf_check1 := "Change FRP Strengthening Scheme" if (Vf < Vf_req) "Provided FRP Strength Large Enough" otherwise Vf_check1 = "Provided FRP Strength Large Enough" Vf_check2 := "Provided FRP amount is adequate" if (Vf_req ≤ Vf < 1.1 ⋅ Vf_req) "Change the FRP amount slightly" otherwise (Vf_check2) = "Provided FRP amount is adequate" 5.5 Calculation of Design Shear Resistance of the Member The design strength of the member is: φV n_total := φ ⋅ (Vc + Vp + Vs + Vf) (Attachment A Eqn. 5.8.3.3-1) (φV n_total) = 102.722 kips

83 φV n_check := "Not Good" if Vu_crit > φVn_total "OK" otherwise (φV n_check) = "OK" Web_crushing_limit := 0.25 ⋅ f' c ⋅ b v ⋅ d v + Vp (LRFD Eqn 5.8.3.3-2) Web_crushing_limit = 397.3 kips Check_web_crushing_limit := "OK" if (V c + V s + Vf + Vp) ≤ Web_crushing_limit "No Good" otherwise Check_web_crushing_limit = "OK" 6. SUMMARY Externally bonded FRP sheets were designed in this example. The FRP sheets are applied at 90 degrees with respect to the longitudinal axis of the member with the U-wrap configuration as shown in Figure 4. In addition, an anchorage system is installed. The final design is summarized as: Use number of plies of FRP sheets Use the width of FRP sheets wf = 5.5 in. Use the center-to-center spacing of FRP sheets sf = 18 in. nf = 1 Figure 4. Final design of FRP strengthening.

84 DESIGN EXAMPLE 2-1: RC T-Beam with Internal Transverse Steel Reinforcement Strengthened with FRP in U-wrap Configuration without Anchorage Systems 1. INTRODUCTION This example demonstrates the design procedures for externally bonded FRP shear reinforcement of an older reinforced concrete (RC) bridge using a U-wrap configuration without anchorage. The bridge consists of simply supported T-beams spanning 42 feet and spaced at 4.5 feet on center. The T-beams contain transverse steel reinforcement spaced at 12 inches on center. Additional details of the T-beam are provided in Figures 1 and 2. 42 ft 114 ft 4.5 ft 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 Figure 1. Bridge plan and transverse section. 2. MATERIAL PROPERTIES The following material properties have been chosen to represent those anticipated in an older bridge for which shear deficiencies might be expected. 2.1. Concrete Compressive strength f' c := 3.0 ksi Modulus of elasticity ksiE c := 33 1.5( )1.5⋅ f' c 1000⋅ E c( ) = 3321 ksi

85 β1 := 0.85 if f 'c ≤ 4 0.65 if f ' c ≥ 8 0.85 − 0.05 f' c 4−( )⋅ otherwise β1 = 0.85 2.2. Longitudinal Reinforcement Yield strength fy := 60 ksi Modulus of elasticity E s := 29000 ksi 2.3. Internal Steel Shear Reinforcement Yield strength fyt := 60 ksi 2.4. FRP Reinforcement Carbon Fiber Sheets are used in this example. Thickness tf := 0.0065 in. Failure strength ffu := 550 ksi Modulus of elasticity Ef := 33000 ksi Failure strain εfu := ffu Ef εfu = 0.017 in./in. 3. GEOMETRICAL PROPERTIES Total Height hT := 37 in. Flange Thickness hf := 7 in. Width of the web b v := 18 in. Effective width of the flange b eff := 54 in. Tensile reinforcement = 12#11 A s := 18.72 in.2 Internal shear reinforcement = #3 at 12 in. spacing A v := 0.22 in.2 s v := 12 in. α := 90⋅deg Distance from the extreme compression fiber to the center of the steel at the section d := 32.7 in.

86 7 in. 30 in. 18 in. 1.5 in. 1.5 in. 1.5 in.32.7 in. 54 in. Figure 2. Cross-section of an intermediate beam. 4. CALCULATION OF THE FACTORED SHEAR FORCE AND NOMINAL SHEAR RESISTANCE 4.1 Factored Shear Force at the Critical Section V u_crit := 120 kips 4.2. Calculation of Nominal Shear Resistance For this example, the simplified approach is followed. (θ := 45 deg) (β := 2) The nominal shear resistance provided by the concrete, V c , is calculated in accordance with LRFD Eqn.5.8.3.3-3 as: Assuming rectangular section behavior with no compression steel, the distance from the extreme compression fiber to the neutral axis, c c , may be calculated as: c c1 := c c1( ) = 9.6 in. A s ⋅fy 0.85⋅f' c ⋅b eff⋅β1 a c1( ) = 8.16 in.ac1 := β1⋅cc1( )

87 Assuming T-beam section behavior with no compression steel, the distance from the extreme compression fiber to the neutral axis, cc , may be calculated as: A s ⋅fy−0.85⋅f'c⋅(beff−bv)⋅hf 0.85⋅f' c ⋅b v ⋅β1 c c2cc2 := ( ) = 12.32 in. a c2( ) = 10.47 in.ac2 := β1⋅cc2( ) check_a c2 "Assumption is correct" if ac2 hf≥( ) "Not behave as rectangular" otherwise := check_a c2( ) = "Assumption is correct" Therefore c c := c c2( ) cc( ) a c := a c2( ) ac( ) = 12.32 in. = 10.47 in. check_a c1 "Assumption is correct" if ac1 hf≤( ) "Not behave as rectangular" otherwise := check_a c1( ) = "Not behave as rectangular" The effective shear depth d v is taken as the distance, measured perpendicular to the neutral axis, between the resultants of the tensile and compressive forces due to flexure; it need not be taken less than the greater of 0.9de or 0.72h (LRFD Article5.8.2.9) d v1 := d − a c 2 The nominal shear resistance provided by the concrete is: V c := 0.0316⋅β⋅ f' c ⋅b v ⋅d v (LRFD Eqn. 5.8.3.3-3) (Vc) = 57.988 kips (dv) = 29.43 in. (dv := max (dv1, dv2, dv3)) (dv2 := 0.9⋅d) (dv3 := 0.72⋅hT) The nominal shear resistance provided by the internal steel reinforcement is: V s := A v ⋅fyt⋅dv⋅(cot(θ) + cot(α)) sin(α) s v (LRFD Eqn. 5.8.3.3-4) V s = 32.373 kips

88 The nominal shear resistance provided by the vertical component of prestressing strands is: Vp := 0( ) The nominal shear resistance of the member is: (LRFD Eqn. 5.8.3.3-1) V n = 90.36 kips V n := V c + V s + Vp 5. DESIGN OF FRP SHEAR REINFORCEMENT 5.1 Check if FRP Reinforcement is Necessary or Not Strength reduction factor for shear (φ := 0.9) Check_FRP_Needed := "NOT need shear reinforcement" if φ V n ≥ V u_crit⋅ "NEED shear reinforcement" otherwise Check_FRP_Needed = "NEED shear reinforcement" 5.2 Computation of Required Vf Vf_req := V u_crit φ − Vn Vf_req = 43 kips 5.3 Selection of FRP Strengthening Scheme U-wrap configuration is used with anchorage systems at the end of the sheets. The FRP sheets will be applied at 90 degrees with respect to the longitudinal axis of the girder as shown in the Figure 3 below. First, the spacing of FRP strips is chosen to meet the maximum spacing requirement. Then, the width of the FRP strips is selected to adjust the amount of FRP strips. Figure 3. FRP strengthening scheme.

89 Use number of plies of FRP sheets nf := 1 Use the width of FRP sheets wf := 4 in. df = 25.7 in. Use the center-to-center spacing of FRP sheets sf := 12 in. αf := 90 degOrientation of FRP sheets Effective depth of FRP sheets df := d – hf Check if the selected spacing is acceptable Shear stress on concrete is: (LRFD Eqn. 5.8.2.9-1) v u := V u_crit − φ⋅Vp φ⋅b v ⋅d v (vu) = 0.252 ksi The maximum spacing of the transverse reinforcement is: (LRFD Eqn. 5.8.2.7-1) (LRFD Eqn. 5.8.2.7-2) s max := min(0.8⋅dv, 24) if vu < 0.125⋅f'c min(0.4⋅dv, 12) otherwise s max = 23.5 Check_Spacing := "Acceptable" if sf ≤ smax "NOT_Acceptable_Change_the_Spacing" otherwise Check_Spacing = "Acceptable" 5.4 Calculation of Shear Resistance of FRP, Vf The FRP reinforcement ratio is: ρf := 2⋅nf⋅wf⋅tf b v ⋅sf (Attachment A Eqn. 5.8.3.3-10) ρf( ) = 2.407 × 10−4

90 The FRP strain reduction factor is: Rf min 3 ρf Ef⋅( )⋅ −0.67, 1.0 := (Attachment A Eqn. 5.8.3.3-9) The effective strain of FRP is: (Attachment A Eqn. 5.8.3.3-7) The effective stress of FRP is: (Attachment A Eqn. 5.8.3.3-6) ffe( ) = 396 ksi The shear contribution of the FRP can be then calculated. Vf := ρf⋅Ef⋅εfe⋅bv⋅df⋅(sin(αf) + cos(αf)) ffe := εfe⋅Ef εfe = 0.012 in./in. Rf = 0.748 εfe := min (Rf⋅εfu, 0.012) (Attachment A Eqn. 5.8.3.3-5) Vf( ) 44.1 kips= Vf_check1 := "Change FRP Strengthening Scheme" if (Vf < Vf_req) "Provided FRP Strength Large Enough" otherwise Vf_check1 = "Provided FRP Strength Large Enough" Vf_check2 := "Provided FRP amount is adequate" if (Vf_req ≤ Vf < 1.1⋅Vf_req) "Change the FRP amount slightly" otherwise (Vf_check2) = "Provided FRP amount is adequate" 5.5 Calculation of Design Shear Resistance of the Member The design strength of the member is: (Attachment A Eqn. 5.8.3.3-1) φV n_total := φ⋅(Vc + Vp + Vs + Vf) (φVn_total) = 121.02 kips

91 "Not Good" if V u_crit > φVn_total "OK" otherwise Web_crushing_limit := 0.25⋅f' c ⋅b v ⋅d v + Vp (LRFD Eqn. 5.8.3.3-2) Web_crushing_limit = 397.3 kips Check_web_crushing_limit := "OK" if (Vc + Vs + Vf + Vp) ≤ Web_crushing_limit φV n_check := (φVn_total) = "OK" "No Good" otherwise Check_web_crushing_limit = "OK" 6. SUMMARY Externally bonded FRP sheets were designed in this example. The FRP sheets are applied at 90 degrees with respect to the longitudinal axis of the member with the U-wrap configuration and without anchorage systems as shown in Figure 4. The final design is summarized as: Figure 4. Final design of FRP strengthening. Use number of plies of FRP sheets Use the width of FRP sheets Use the center-to-center spacing of FRP sheets sf = 12 in. nf = 1 wf = 4 in.

92 DESIGN EXAMPLE 2-2: RC T-Beam with Internal Transverse Steel Reinforcement Strengthened with FRP in U-wrap Configuration with an Anchorage System 1. INTRODUCTION This example demonstrates the design procedures for externally bonded FRP shear reinforcement of an older reinforced concrete (RC) bridge using a U-wrap configuration with anchorage. The bridge consists of simply supported T-beams spanning 42 feet and spaced at 4.5 feet on center. The T-beams contain transverse steel reinforcement spaced at 12 inches on center. Additional details of the T-beam are provided in Figures 1 and 2. 42 ft 114 ft 4.5 ft 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 Figure 1. Bridge plan and transverse section. (E c ) = 3321 ksi 2. MATERIAL PROPERTIES The following material properties have been chosen to represent those anticipated in an older bridge for which shear deficiencies might be expected. 2.1. Concrete Compressive strength f' c := 3.0 ksi Modulus of elasticity E c := 33⋅(1.5)1.5 f' c ⋅1000 ksi

93 β1 := 0.85 if f'c ≤ 4 0.65 if f' c ≥ 8 0.85 − 0.05⋅(f' c − 4) otherwise β1 = 0.85 2.2. Longitudinal Reinforcement Yield strength fy := 60 ksi Modulus of elasticity E s := 29000 ksi 2.3. Internal Steel Shear Reinforcement Yield strength fyt := 60 ksi 2.4. FRP Reinforcement Carbon Fiber Sheets are used in this example. Thickness tf := 0.0065 in. Failure strength ffu := 550 ksi Modulus of elasticity Ef := 33000 ksi Failure strain εfu := ffu Ef εfu = 0.017 in./in. 3. GEOMETRICAL PROPERTIES Total Height hT := 37 in. Flange Thickness hf := 7 in. Width of the web b v := 18 in. Effective Width of the Flange b eff := 54 in. Tensile reinforcement = 12#11 A s := 18.72 in.2 Internal shear reinforcement = #3 at 12 in. spacing A v := 0.22 in2 s v := 12 in. α := 90 deg Distance from the extreme compression fiber to the center of the steel at the section d := 32.7 in.

94 7 in. 30 in. 18 in. 1.5 in. 1.5 in. 1.5 in.32.7 in. 54 in. Figure 2. Cross-section of an intermediate beam. 4. CALCULATION OF THE FACTORED SHEAR FORCE AND NOMINAL SHEAR RESISTANCE 4.1 Factored Shear Force at the Critical Section V u_crit := 120 kips 4.2. Calculation of Nominal Shear Resistance For this example, the simplified approach is followed. (θ := 45 deg) (β := 2) The nominal shear resistance provided by the concrete, V c , is calculated in accordance with LRFD Eqn.5.8.3.3-3 as: Assuming rectangular section behavior with no compression steel, the distance from the extreme compression fiber to the neutral axis, c c , may be calculated as: c c1 := c c1( ) = 9.6 in. A s ⋅fy 0.85⋅f' c ⋅b eff⋅β1 a c1( ) = 8.16 in.ac1 := β1⋅cc1( )

95 Assuming T-beam section behavior with no compression steel, the distance from the extreme compression fiber to the neutral axis, c c , may be calculated as: A s ⋅fy−0.85⋅f'c⋅(beff−bv)⋅hf 0.85⋅f' c ⋅b v ⋅β1 c c2cc2 := ( ) = 12.32 in. a c2( ) = 10.47 in.ac2 := β1⋅cc2( ) check_a c2 "Assumption is correct" if ac2 hf≥( ) "Not behave as rectangular" otherwise := check_a c2( ) = "Assumption is correct" Therefore c c := c c2( ) cc( ) a c := a c2( ) ac( ) = 12.32 in. = 10.47 in. check_a c1 "Assumption is correct" if ac1 hf≤( ) "Not behave as rectangular" otherwise := check_a c1( ) = "Not behave as rectangular" The effective shear depth d v is taken as the distance, measured perpendicular to the neutral axis, between the resultants of the tensile and compressive forces due to flexure; it need not be taken less than the greater of 0.9d e or 0.72h (LRFD Article5.8.2.9) d v1 := d − a c 2 The nominal shear resistance provided by the concrete is: V c := 0.0316⋅β⋅ f' c ⋅b v ⋅d v (LRFD Eqn. 5.8.3.3-3) (Vc) = 58 kips (dv) = 29.4 in. (dv := max (dv1, dv2, dv3)) (dv2 := 0.9⋅d) (dv3 := 0.72⋅hT) The nominal shear resistance provided by the internal steel reinforcement is: V s := A v ⋅fyt⋅dv⋅(cot(θ) + cot(α)) sin(α) s v (LRFD Eqn. 5.8.3.3-4) V s = 32.4 kips

96 The nominal shear resistance provided by the vertical component of prestressing strands is: Vp := 0( ) The nominal shear resistance of the member is: (LRFD Eqn. 5.8.3.3-1) V n = 90.4 kips V n := V c + V s + Vp 5. DESIGN OF FRP SHEAR REINFORCEMENT 5.1 Check if FRP Reinforcement is Necessary or Not Strength reduction factor for shear (φ := 0.9) Check_FRP_Needed := "NOT need shear reinforcement" if φ V n ≥ V u_crit⋅ "NEED shear reinforcement" otherwise Check_FRP_Needed = "NEED shear reinforcement" 5.2 Computation of Required Vf Vf_req := V u_crit φ − Vn Vf_req = 43 kips 5.3 Selection of FRP Strengthening Scheme U-wrap configuration is used with anchorage systems at the end of the sheets. The FRP sheets will be applied at 90 degrees with respect to the longitudinal axis of the girder as shown in the Figure 3 below. Anchorage systems will be installed at the top end portion of the FRP sheets to increase the effectiveness of FRP shear strengthening. First, the spacing of FRP strips is chosen to meet the maximum spacing requirement. Then, the width of the FRP strips is selected to adjust the amount of FRP strips. Figure 3. FRP strengthening scheme.

97 Use number of plies of FRP sheets nf := 1 Use the width of FRP sheets wf := 4 in. df = 25.7 in. Use the center-to-center spacing of FRP sheets sf := 16 in. αf := 90 degOrientation of FRP sheets Effective depth of FRP sheets df := d – hf Check if the selected spacing is acceptable or not Shear stress on concrete is: (LRFD Eqn. 5.8.2.9-1) v u := V u_crit − φ⋅Vp φ⋅b v ⋅d v (vu) = 0.252 ksi The maximum spacing of the transverse reinforcement is: (LRFD Eqn. 5.8.2.7-1) (LRFD Eqn. 5.8.2.7-2) s max := min(0.8⋅dv, 24) if vu < 0.125⋅f'c min(0.4⋅dv, 12) otherwise s max = 23.5 Check_Spacing := "Acceptable" if sf ≤ smax "NOT_Acceptable_Change_the_Spacing" otherwise Check_Spacing = "Acceptable" 5.4 Calculation of Shear Resistance of FRP, Vf The FRP reinforcement ratio is: ρf := 2⋅nf⋅wf⋅tf b v ⋅sf (Attachment A Eqn. 5.8.3.3-10) ρf( ) = 1.806 × 10−4

98 The FRP strain reduction factor is: Rf min 4 ρf Ef⋅( )⋅ −0.67, 1.0 := (Attachment A Eqn. 5.8.3.3-8) The effective strain of FRP is: (Attachment A Eqn. 5.8.3.3-7) The effective stress of FRP is: (Attachment A Eqn. 5.8.3.3-6) ffe( ) = 550 ksi The shear contribution of the FRP can be then calculated. (Vf := ρf⋅Ef⋅εfe⋅bv⋅df⋅(sin(αf) + cos(αf))) ffe := εfe⋅Ef εfe = 0.017 in./in. Rf = 1 εfe := Rf⋅εfu (Attachment A Eqn. 5.8.3.3-5) Vf( ) 45.9 kips= Vf_check1 := "Change FRP Strengthening Scheme" if (Vf < Vf_req) "Provided FRP Strength Large Enough" otherwise Vf_check1 = "Provided FRP Strength Large Enough" Vf_check2 := "Provided FRP amount is adequate" if (Vf_req ≤ Vf < 1.1⋅Vf_req) "Change the FRP amount slightly" otherwise (Vf_check2) = "Provided FRP amount is adequate" 5.5 Calculation of Design Shear Resistance of the Member The design strength of the member is: (Attachment A Eqn. 5.8.3.3-1) φV n_total := φ⋅(Vc + Vp + Vs + Vf) "Not Good" if V u_crit > φVn_total "OK" otherwise φV n_check := (φVn_total) = 122.7 kips (φVn_check) = "OK"

99 Web_crushing_limit := 0.25⋅f' c ⋅b v ⋅d v + Vp (LRFD Eqn. 5.8.3.3-2) Web_crushing_limit = 397.3 kips Check_web_crushing_limit := "OK" if (Vc + Vs + Vf + Vp) ≤ Web_crushing_limit "No Good" otherwise Check_web_crushing_limit = "OK" 6. SUMMARY Externally bonded FRP sheets were designed in this example. The FRP sheets are applied at 90 degrees with respect to the longitudinal axis of the member with the U-wrap configuration as shown in Figure 4. In addition, an anchorage system is installed. The final design is summarized as: Figure 4. Final design of FRP strengthening. Use number of plies of FRP sheets Use the width of FRP sheets Use the center-to-center spacing of FRP sheets sf = 16 in. nf = 1 wf = 4 in.

100 DESIGN EXAMPLE 3-1: PC I-Beam with Internal Transverse Steel Reinforcement Strengthened with FRP in U-wrap Configuration without Anchorage Systems 1. INTRODUCTION This example demonstrates the design procedures for externally bonded FRP shear reinforcement of a prestressed I-beam bridge using a U-wrap configuration without anchorage. The bridge consists of five simply supported pretensioned I-beams spanning 42 feet and spaced at 7.5 feet on center. The I-beams are lightly reinforced with transverse steel reinforcement. Additional details of the bulb-tees are provided in Figures 1 and 2. 17.2 ft 23.0 in. phi 8 Strands 6 Strands Beam Length = 43 ft Center Line 4.3 ft (a) Prestressed I-Beam Bridge Deck Cross-Section (b) Beam Tendon Geometry Figure 1. AASHTO bulb-tee bridge deck bridge (Ref. PCI Bridge Design Manual). 6 in. Thick Uniform Deck

101 2. MATERIAL PROPERTIES 2.1. Concrete 2.1.1 Deck 2.1.1 I-Beam E cd = 3834 ksi Compressive strength f' cd := 4.0 ksi Modulus of elasticity E cd := 33⋅(1.5)1.5 f'cd⋅1000 ksi β1d := 0.85 if f'cd ≤ 4 0.65 if f' cd ≥ 8 0.85 − 0.05⋅(f'cd − 4) otherwise β1d = 0.85 E cb = 5072 ksi Compressive strength f' cb := 7.0 ksi Modulus of elasticity E cb := 33⋅(1.5)1.5 f'cb⋅1000 ksi β1b := 0.85 if f'cb ≤ 4 0.65 if f' cb ≥ 8 0.85 − 0.05⋅(f'cb − 4) otherwise β1b = 0.7

102 2.2. Prestressing Strands Specified tensile strength fpu := 270 ksi Diameter = 0.5 in. Total Area of the 14 strands Aps := 2.142 in.2 k := 0.28 for low-relaxation steel Yield strength fpy := 243 ksi Modulus of elasticity Eps := 28500 ksi 2.3. Internal Steel Shear Reinforcement Yield strength fyt := 60 ksi 2.4. FRP Reinforcement Carbon Fiber Sheets are used in this example. Thickness tf := 0.0065 in. Failure strength ffu := 550 ksi Modulus of elasticity Ef := 33000 ksi Failure strain εfu := ffu Ef εfu = 0.017 in./in. 3. GEOMETRICAL PROPERTIES Total height including deck slab hT := 38 in. Flange thickness hf := 6 in. Width of the web b v := 7 in. Effective width of the Flange b eff := 79.0 in. Internal shear reinforcement = #3 at 12 in. spacing A v := 0.22 in.2 s v := 12 in. α := 90⋅deg

103 (a) I-Beam Prestressing Pattern. (b) Cross-Section of an Intermediate Beam Figure 2. Cross-section of an intermediate beam.

104 4. CALCULATION OF THE FACTORED SHEAR FORCE AND NOMINAL SHEAR RESISTANCE 4.1 Factored Shear Force at the Critical Section 4.2. Calculation of Nominal Shear Resistance For this example, the simplified approach is followed. The nominal shear resistance provided by the concrete, V c , is calculated in accordance with LRFD Eqn.5.8.3.3-3 as: The distance from the extreme compression fiber to the center of gravity of the strands at the midspan: Assuming rectangular section behavior with no compression steel, the distance from the extreme compression fiber to the neutral axis, cc , may be calculated as: a c := β1d⋅cc V u_crit := 100 kips dp := 34.6 in. θ := 45⋅deg β := 2 c c := Aps⋅fpu dp fpu0.85⋅f' cd⋅beff⋅β1d + k⋅Aps⋅ check_a c "Assumption is correct" if a c hf≥( ) "Not behave as rectangular" otherwise := check_a c( ) = "Assumption is correct" c c( ) = 2.482 in. a c( ) = 2.11 in. The effective shear depth dv is taken as the distance, measured perpendicular to the neutral axis, between the resultants of the tensile and compressive forces due to flexure; it need not be taken to be less than the greater of 0.9de or 0.72h (LRFD Article5.8.2.9).

105 Since some of the strands are harped, the effective depth varies point-to-point. However, the effective depth must be calculated at the critical section in shear, which is not yet determined; therefore, an iterative procedure is required. For this example, only the final cycle of the iteration is shown. Assume d v d v_trial := 27.36 in. Calculate the distance from the extreme compression face to the center of gravity of the strand, de at the location, d v away from the centerline of the support. etr := 2⋅4 + 4⋅4 + 23 206.4 206.4 d v_trial− + 2 ⋅2 + 23 206.4 206.4 d v_trial− + 4 ⋅2 + 23 206.4 206.4 d v_trial−( )( )( ) + 6 ⋅2 6 + 6 + 2 d e := hT − etr d e = 26.021 in. Determine d v d v1 := de − a c 2 d v2 := 0.9⋅de dv3 := 0.72⋅hT( ) (dv_max := max (dv1, dv2, dv3)) (dv) = 27.36 in. (dv := max (dv_max, 0.5⋅dv_max⋅cot(θ))) Final d v Check_dv1 := "OK" 0.995 d v_trial d v ≤ 1.005≤if "Try Again" otherwise Check_dv1 = "OK" Check_dv2 := "OK" d v b v 4≤if "NOT GOOD" otherwise Check_dv2 = "OK"

106 Harped tendon force = 6 x 0.153 x 149.0 = 136.8 kips (assuming fpe = 149 ksi) slope of the tendons ψ := 0.111 The nominal shear resistance provided by the concrete is: V c := 0.0316⋅β⋅ f' c ⋅b v ⋅d v (LRFD Eqn. 5.8.3.3-3) The nominal shear resistance provided by the internal steel reinforcement is: V s := A v ⋅fyt⋅dv⋅(cot(θ) + cot(α)) sin(α) s v (LRFD Eqn. 5.8.3.3-4) (Vc) = 32 kips The nominal shear resistance of the member is: (LRFD Eqn. 5.8.3.3-1) V n = 77.3 kips V s = 30.1 kips V n := V c + V s + Vp Vp := 136.8⋅ψ Vp = 15.2 kips 5. DESIGN OF FRP SHEAR REINFORCEMENT 5.1 Check if FRP Reinforcement is Necessary or Not Strength reduction factor for shear (φ := 0.9) Check_FRP_Needed := "NOT need shear reinforcement" if φ V n ≥ V u_crit⋅ "NEED shear reinforcement" otherwise Check_FRP_Needed = "NEED shear reinforcement" 5.2 Computation of Required Vf Vf_req := V u_crit φ − Vn Vf_req = 33.8 kips

107 5.3 Selection of FRP Strengthening Scheme U-wrap configuration is used with anchorage systems at the end of the sheets. The FRP sheets will be applied at 90 degrees with respect to the longitudinal axis of the girder as shown in the Figure 3 below. First, the spacing of FRP strips is chosen to meet the maximum spacing requirement. Then, the width of the FRP strips is selected to adjust the amount of FRP strips. Figure 3. FRP strengthening scheme. Use number of plies of FRP sheets nf := 1 Use the width of FRP sheets wf := 8 in. df = 28.6 in. Use the center-to-center spacing of FRP sheets sf := 12 in. αf := 90 degOrientation of FRP sheets Effective depth of FRP sheets df := dp – hf Check if the selected spacing is acceptable or not Shear stress on concrete is: (LRFD Eqn. 5.8.2.9-1) v u := V u_crit − φ⋅Vp φ⋅b v ⋅d v (vu) = 0.501 ksi The maximum spacing of the transverse reinforcement is: (LRFD Eqn. 5.8.2.7-1) (LRFD Eqn. 5.8.2.7-2) s max := min(0.8⋅dv, 24) if vu < 0.125⋅f'cb min(0.4⋅dv, 12) otherwise s max = 21.9

108 5.4 Calculation of Shear Resistance of FRP, Vf The FRP reinforcement ratio is: ρf := 2⋅nf⋅wf⋅tf b v ⋅sf (Attachment A Eqn. 5.8.3.3-10) The FRP strain reduction factor is: Rf min 3 ρf Ef⋅( )⋅ −0.67, 1.0 := (Attachment A Eqn. 5.8.3.3-9) The effective strain of FRP is: (Attachment A Eqn. 5.8.3.3-7) The effective stress of FRP is: (Attachment A Eqn. 5.8.3.3-6) ffe( ) = 137.4 ksi The shear contribution of the FRP can be then calculated. (Vf := ρf⋅Ef⋅εfe⋅bv⋅df⋅(sin(αf) + cos(αf))) ffe := εfe⋅Ef εfe = 4.163 × 10−3 in./in. Rf = 0.25 εfe := min (Rf⋅εfu, 0.012) (Attachment A Eqn. 5.8.3.3-5) Vf( ) 34.1 kips= ρf( ) = 1.238 × 10−3 Vf_check1 := "Change FRP Strengthening Scheme" if (Vf < Vf_req) "Provided FRP Strength Large Enough" otherwise Vf_check1 = "Provided FRP Strength Large Enough" Check_Spacing := "Acceptable" if sf ≤ smax "NOT_Acceptable_Change_the_Spacing" otherwise Check_Spacing = "Acceptable"

109 5.5 Calculation of Design Shear Resistance of the Member The design strength of the member is: (Attachment A Eqn. 5.8.3.3-1) "Not Good" if V u_crit > φVn_total "OK" otherwise Web_crushing_limit := 0.25⋅f' cb⋅bv⋅dv⋅Vp (LRFD Eqn. 5.8.3.3-2) Web_crushing_limit = 350.3 kips Check_web_crushing_limit := "OK" if (Vc + Vs + Vf + Vp) ≤ Web_crushing_limit φV n_total := φ⋅(Vc + Vp + Vs + Vf) φV n_check := (φVn_total) = 100.2 kips (φVn_check) = "OK" "No Good" otherwise Check_web_crushing_limit = "OK" Vf_check2 := "Provided FRP amount is adequate" if (Vf_req ≤ Vf < 1.1⋅Vf_req) "Change the FRP amount slightly" otherwise (Vf_check2) = "Provided FRP amount is adequate" 6. SUMMARY Externally bonded FRP sheets were designed in this example. The FRP sheets are applied at 90 degrees with respect to the longitudinal axis of the member with the U-wrap configuration and without anchorage systems as shown in Figure 4. The final design is summarized as: Use number of plies of FRP sheets Use the width of FRP sheets Use the center-to-center spacing of FRP sheets sf = 12 in. nf = 1 wf = 8 in.

110 Figure 4. Final design of FRP strengthening.

111 DESIGN EXAMPLE 3-2: PC I-Beam with Internal Transverse Steel Reinforcement Strengthened with FRP in U-wrap Configuration with an Anchorage System 1. INTRODUCTION This example demonstrates the design procedures for externally bonded FRP shear reinforcement of a prestressed I-beam bridge using a U-wrap configuration with anchorage. The bridge consists of five simply supported prestensioned I-beams spanning 42 feet and spaced at 7.5 feet on center. The I-beams are lightly reinforced with transverse steel reinforcement. Additional details of the bulb-tees are provided in Figures 1 and 2. 17.2 ft 23.0 in. phi 8 Strands 6 Strands Beam Length = 43 ft Center Line 4.3 ft (a) Prestressed I-Beam Bridge Deck Cross-Section (b) Beam Tendon Geometry Figure 1. AASHTO bulb-tee bridge deck bridge (Ref. PCI Bridge Design Manual).

112 (f' cd − 4) 2. MATERIAL PROPERTIES 2.1. Concrete 2.1.1 Deck Compressive strength f' cd := 4.0 ksi Modulus of elasticity E cd := 33 ⋅ (1.5)1.5 f'cd ⋅ 1000 ksi E cd = 3834 ksi Modulus of elasticity E cb := 33 ⋅ (1.5)1.5 f'cb ⋅ 1000 ksi E cb = 5072 ksi β1d := 0.85 if f'cd ≤ 4 0.65 if f' cd ≥ 8 0.85 − 0.05⋅ otherwise β1d = 0.85 (f' cb − 4) β1b := 0.85 if f'cb ≤ 4 0.65 if f' cb ≥ 8 0.85 − 0.05⋅ otherwise β1b = 0.7 2.1.1 I-Beam Compressive strength f' cb := 7.0 ksi 2.2. Prestressing Strands Specified tensile strength fpu := 270 ksi Yield strength fpy := 243 ksi Modulus of elasticity Eps := 28500 ksi Diameter = 0.5 in. Total Area of the 14 strands Aps := 2.142 in2 k := 0.28 for low-relaxation steel

113 2.3. Internal Steel Shear Reinforcement Yield strength fyt := 60 ksi 2.4. FRP Reinforcement Carbon-Fiber Sheets are used in this example. Thickness tf := 0.0065 in. Failure strength ffu := 550 ksi Modulus of elasticity Ef := 33000 ksi Failure strain εfu := ffu Ef εfu = 0.017 in./in. 3. GEOMETRICAL PROPERTIES Total height including deck slab hT := 38 in. Flange thickness hf := 6 in. Width of the web b v := 7 in. Effective width of the flange b eff := 79.0 in. Internal shear reinforcement = #3 at 12 in. spacing A v := 0.22 in2 s v := 12 in. α := 90 ⋅ deg (a) I-Beam Prestressing Pattern

114 (b) Cross-Section of an Intermediate Beam Figure 2. Cross-section of an intermediate beam. 4. CALCULATION OF THE FACTORED SHEAR FORCE AND NOMINAL SHEAR RESISTANCE 4.1 Factored Shear Force at the Critical Section 4.2. Calculation of Nominal Shear Resistance For this example, the simplified approach is followed. The nominal shear resistance provided by the concrete, V c , is calculated in accordance with LRFD Eqn.5.8.3.3-3 as: The distance from the extreme compression fiber to the center of gravity of the strands at the midspan: V u_crit := 100 kips dp := 34.6 in. θ := 45⋅deg β := 2

115 Assuming rectangular section behavior with no compression steel, the distance from the extreme compression fiber to the neutral axis, c c , may be calculated as: a c := β1d⋅cc c c := Aps⋅fpu dp fpu0.85⋅f' cd⋅beff⋅β1d + k⋅Aps⋅ check_a c "Assumption is correct" if a c hf≥( ) "Not behave as rectangular" otherwise := check_a c( ) = "Assumption is correct" c c( ) = 2.482 in. a c( ) = 2.11 in. The effective shear depth d v is taken as the distance, measured perpendicular to the neutral axis, between the resultants of the tensile and compressive forces due to flexure; it need not be taken to be less than the greater of 0.9d e or 0.72h (LRFD Article5.8.2.9). Since some of the strands are harped, the effective depth varies point-to-point. However, the effective depth must be calculated at the critical section in shear, which is not yet determined; therefore, an iterative procedure is required. For this example, only the final cycle of the iteration is shown. Assume d v d v_trial := 27.36 in. Calculate the distance from the extreme compression face to the center of gravity of the strand, d e at the location, d v away from the centerline of the support. etr := 2⋅4 + 4⋅4 + 23 206.4 206.4 d v_trial− + 2 ⋅2 + 23 206.4 206.4 d v_trial− + 4 ⋅2 + 23 206.4 206.4 d v_trial−( )( )( ) + 6 ⋅2 6 + 6 + 2 d e := hT − etr d e = 26.021 in.

116 Determine d v d v1 := de − a c 2 d v2 := 0.9⋅de dv3 := 0.72⋅hT( ) (dv_max := max (dv1, dv2, dv3)) (dv) = 27.36 in. (dv := max (dv_max, 0.5⋅dv_max⋅cot(θ))) Final d v Check_dv1 := "OK" 0.995 d v_trial d v ≤ 1.005≤if "Try Again" otherwise Check_dv1 = "OK" Check_dv2 := "OK" d v b v 4≤if "NOT GOOD" otherwise Check_dv2 = "OK" Harped tendon force = 6 x 0.153 x 149.0 = 136.8 kips (assuming fpe = 149 ksi) slope of the tendons ψ := 0.111 The nominal shear resistance provided by the concrete is: V c := 0.0316⋅β⋅ f' cb⋅bv⋅dv (LRFD Eqn. 5.8.3.3-3) The nominal shear resistance provided by the internal steel reinforcement is: V s := A v ⋅fyt⋅dv⋅(cot(θ) + cot(α)) sin(α) s v (LRFD Eqn. 5.8.3.3-4) (Vc) = 32 kips V s = 30.1 kips Vp := 136.8⋅ψ Vp = 15.2 kips

117 The nominal shear resistance of the member is: (LRFD Eqn. 5.8.3.3-1) V n = 77.3 kips V n := V c + V s + Vp 5. DESIGN OF FRP SHEAR REINFORCEMENT 5.1 Check if FRP Reinforcement is Necessary Strength reduction factor for shear (φ := 0.9) Check_FRP_Needed := "NOT need shear reinforcement" if φ V n ≥ V u_crit⋅ "NEED shear reinforcement" otherwise Check_FRP_Needed = "NEED shear reinforcement" 5.2 Computation of Required Vf Vf_req := V u_crit φ − Vn Vf_req = 33.8 kips 5.3 Selection of FRP Strengthening Scheme U-wrap configuration is used with anchorage systems at the end of the sheets. The FRP sheets will be applied at 90 degrees with respect to the longitudinal axis of the girder as shown in the Figure 3 below. First, the spacing of FRP strips is chosen to meet the maximum spacing requirement. Then, the width of the FRP strips is selected to adjust the amount of FRP strips. Figure 3. FRP strengthening scheme. Use number of plies of FRP sheets nf := 1 Use the width of FRP sheets wf := 4 in. Use the center-to-center spacing of FRP sheets sf := 12 in.

118 df = 28.6 in. αf := 90 degOrientation of FRP sheets Effective depth of FRP sheets df := dp – hf Check if the selected spacing is acceptable or not Shear stress on concrete is: (LRFD Eqn. 5.8.2.9-1) v u := V u_crit − φ⋅Vp φ⋅b v ⋅d v (vu) = 0.501 ksi The maximum spacing of the transverse reinforcement is: (LRFD Eqn. 5.8.2.7-1) (LRFD Eqn. 5.8.2.7-2) s max := min(0.8⋅dv, 24) if vu < 0.125⋅f'cb min(0.4⋅dv, 12) otherwise s max = 21.9 Check_Spacing := "Acceptable" if sf ≤ smax "NOT_Acceptable_Change_the_Spacing" otherwise Check_Spacing = "Acceptable" 5.4 Calculation of Shear Resistance of FRP, Vf The FRP reinforcement ratio is: ρf := 2⋅nf⋅wf⋅tf b v ⋅sf (Attachment A Eqn. 5.8.3.3-10) The FRP strain reduction factor is: Rf min 4 ρf Ef⋅( )⋅ −0.67, 1.0 := (Attachment A Eqn. 5.8.3.3-8) Rf = 0.53 ρf( ) = 6.19 × 10−4 The effective strain of FRP is: (Attachment A Eqn. 5.8.3.3-7) εfe = 8.832 × 10−3 in./in. εfe := Rf⋅εfu

119 The effective stress of FRP is: (Attachment A Eqn. 5.8.3.3-6) ffe( ) = 291.4 ksi The shear contribution of the FRP can be then calculated. (Vf := ρf⋅Ef⋅εfe⋅bv⋅df⋅(sin(αf) + cos(αf))) ffe := εfe⋅Ef (Attachment A Eqn. 5.8.3.3-5) Vf( ) 36.1 kips= Vf_check1 := "Change FRP Strengthening Scheme" if (Vf < Vf_req) "Provided FRP Strength Large Enough" otherwise Vf_check1 = "Provided FRP Strength Large Enough" Vf_check2 := "Provided FRP amount is adequate" if (Vf_req ≤ Vf < 1.1⋅Vf_req) "Change the FRP amount slightly" otherwise Vf_check2 = "Provided FRP amount is adequate" 5.5 Calculation of Design Shear Resistance of the Member The design strength of the member is: (Attachment A Eqn. 5.8.3.3-1) "Not Good" if V u_crit > φVn_total "OK" otherwise Web_crushing_limit := 0.25⋅f' cb⋅bv⋅dv⋅Vp (LRFD Eqn. 5.8.3.3-2) Web_crushing_limit = 350.3 kips Check_web_crushing_limit := "OK" if (Vc + Vs + Vf + Vp) ≤ Web_crushing_limit φV n_total := φ⋅(Vc + Vp + Vs + Vf) φV n_check := (φVn_total) = 102.1 kips (φVn_check) = "OK" "No Good" otherwise Check_web_crushing_limit = "OK"

120 6. SUMMARY Externally bonded FRP sheets were designed in this example. The FRP sheets are applied at 90 degrees with respect to the longitudinal axis of the member with the U-wrap configuration and without anchorage systems as shown in Figure 4. The final design is summarized as: Figure 4. Final design of FRP strengthening. Use number of plies of FRP sheets Use the width of FRP sheets Use the center-to-center spacing of FRP sheets sf = 12 in. nf = 1 wf = 4 in.