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Use of Fiber-Reinforced Polymers in Highway Infrastructure (2017)

Chapter: Appendix A - Design Examples

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Suggested Citation:"Appendix A - Design Examples." National Academies of Sciences, Engineering, and Medicine. 2017. Use of Fiber-Reinforced Polymers in Highway Infrastructure. Washington, DC: The National Academies Press. doi: 10.17226/24888.
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Suggested Citation:"Appendix A - Design Examples." National Academies of Sciences, Engineering, and Medicine. 2017. Use of Fiber-Reinforced Polymers in Highway Infrastructure. Washington, DC: The National Academies Press. doi: 10.17226/24888.
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Suggested Citation:"Appendix A - Design Examples." National Academies of Sciences, Engineering, and Medicine. 2017. Use of Fiber-Reinforced Polymers in Highway Infrastructure. Washington, DC: The National Academies Press. doi: 10.17226/24888.
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Suggested Citation:"Appendix A - Design Examples." National Academies of Sciences, Engineering, and Medicine. 2017. Use of Fiber-Reinforced Polymers in Highway Infrastructure. Washington, DC: The National Academies Press. doi: 10.17226/24888.
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Suggested Citation:"Appendix A - Design Examples." National Academies of Sciences, Engineering, and Medicine. 2017. Use of Fiber-Reinforced Polymers in Highway Infrastructure. Washington, DC: The National Academies Press. doi: 10.17226/24888.
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Suggested Citation:"Appendix A - Design Examples." National Academies of Sciences, Engineering, and Medicine. 2017. Use of Fiber-Reinforced Polymers in Highway Infrastructure. Washington, DC: The National Academies Press. doi: 10.17226/24888.
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Suggested Citation:"Appendix A - Design Examples." National Academies of Sciences, Engineering, and Medicine. 2017. Use of Fiber-Reinforced Polymers in Highway Infrastructure. Washington, DC: The National Academies Press. doi: 10.17226/24888.
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Suggested Citation:"Appendix A - Design Examples." National Academies of Sciences, Engineering, and Medicine. 2017. Use of Fiber-Reinforced Polymers in Highway Infrastructure. Washington, DC: The National Academies Press. doi: 10.17226/24888.
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Suggested Citation:"Appendix A - Design Examples." National Academies of Sciences, Engineering, and Medicine. 2017. Use of Fiber-Reinforced Polymers in Highway Infrastructure. Washington, DC: The National Academies Press. doi: 10.17226/24888.
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153 Appendix A design examples Five design examples are provided to help transportation agencies understand fiber-reinforced polymer (FRP) applications. AASHTO guide specifications are referenced, unless otherwise stated. It is important to note that the design approaches shown in the guide specifications are different from those of other design documents, such as ACI 440.2R-08 (ACI 2008a). Because most department of transportation (DOT) engi- neers are familiar with the design of bridge members, essential information related to FRP is emphasized, rather than describing the typical design details. For the convenience of DOT engineers, U.S. customary units are employed. FRp-ReinFoRced concRete BRidge deck Design a typical 8-in. bridge deck reinforced with GFRP bars. The strength of concrete is fc′ = 3,000 psi along with the following GFRP properties: ffu (ultimate strength) = 100 ksi, Ef (modulus) = 6,000 ksi, and efu (rupture strain) = 0.016. The deck is subjected to a factored moment of Mu = 15 ft-kip/ft. 8-in. deck reinforced with GFRP bars Solution: AASHTO guide specifications for GFRP reinforcement (AASHTO 2009) are used • Design strength with the environmental reduction factor of CE = 0.7 (Table 2.6.1.2-1): 0.7 100 70 ksi 0.7 0.016 0.011 ( ) ( ) = = = ε = ε = = f C f C fd E fu fd E fu • Determine the effective depth of the bottom bars (positive moment): cover bar diameter 8 0.75 0.625 6.94 in. 1 2 1 2 ( ) ( )= − − = − − = d h Cover = 0.75 (Table 5.6.2.3-1) Bar diameter = 0.625 in. for assumed #5 bar (Table 4.5.4-1) • Calculate the GFRP reinforcement ratio producing balanced strain condition: f f E E ffb c fd f cu f cu fd ( ) ( ) ( )( ) ρ = β ′ ε ε + = + = 0.85 Eq. 2.7.4.2-2 0.85 0.85 370 6,000 0.003 6,000 0.003 70 0.0063 1 • Select the number of bars: 1.4 for compression-controlled section Figure 2.7.4.2-1 1.4 0.0063 0.0088 0.0088 12 6.94 0.73 in. /ft2 ( ) ( ) ( )( ) ρ ρ = = = ρ = =A bd fb f f f Use three #5 bars: Af = 3(0.31) = 0.93 in.2/ft

154 • Determine the GFRP stress at ultimate condition: f E f E E f ksi f f cu c f f cu f cu fd ( ) ( ) ( ) ( )( )( )( ) ( )( ) = ε + β ′ ρ ε − ε ≤ = × + − = 4 0.85 0.5 Eq. 2.9.3.1-1 6,000 0.003 4 0.85 0.85 3 0.0088 6,000 0.003 0.5 6,000 0.003 58.2 2 1 2 • Determine the capacity of the deck: 2 Eq. 2.9.3.2.2-1 0.65 0.93 58.2 6.94 1.772 17.8 ft-kip ft 0.85 Eq. 2.9.3.2.2-2 0.93 58.2 0.85 3 12 1.77 in. OK ( ) ( )( ) ( ) ( ) ( ) ( )( ) Φ = Φ −  = −   = = ′ = = Φ > M A f d a a A f f b M M n f f f f c n u FRp-StRengthened ReinFoRced concRete BRidge giRdeR in FlexuRe: exteRnAlly Bonded cFRp SheetS An existing reinforced concrete bridge girder is subjected to a factored moment of 750 ft-kip. The T-girder has an effective flange width of 7 ft with a depth of h = 40 in. (deck = 8 in. plus girder = 32 in.) and a girder width of 15 in. The girder has the following properties (compression steel is ignored for design convenience): fc′ (compressive concrete strength) = 3,000 psi, fy (steel yield strength) = 60 ksi, As (tension steel area) = 7 in.2, and d (effective depth) = 28 in. The girder is damaged and a loss of 20% in the steel occurs (As_damaged = 5.6 in.2). It is decided that the girder is strengthened with CFRP sheets having the following properties: ffu (ultimate strength) = 550 ksi, Ef (modulus) = 33,000 ksi, efu (rupture strain) = 0.016, and tf (thickness) = 0.006 in. Deter- mine the amount of CFRP sheets to recover the flexural strength of the damaged girder. 7 ft 8 in. 32 in.40 in. 15 in. As = 7 in.2 EB CFRP 7 ft (typ.) Solution: AASHTO Guide Specifications for Design of Bonded FRP Systems for Repair and Strengthening of Concrete Bridge Elements (AASHTO 2012a) are used, unless otherwise stated. • Calculate the capacity of the undamaged girder (under-reinforced): M A f d an s s s ( ) ( )( ) φ = φ −  = −   = 2 AASHTO LRFD BDS Eq. 5.7.3.2.2-1 0.9 7 60 28 1.962 851 ft-kip

155 a A f f b a h s s c f( )( )( ) ( ) = ′ = × = < =0.85 7 60 0.85 3 7 12 1.96 in. 8 in. • Calculate the capacity of the damaged girder (under-reinforced): 2 AASHTO LRFD BDS Eq. 5.7.3.2.2-1 0.9 5.6 60 28 1.572 685 ft-kip 0.85 5.6 60 0.85 3 7 12 1.57 in. 8 in. _ _ _ ( ) ( ) ( )( ) ( ) ( )( ) φ = φ −  = −   = = ′ = × = < = M A f d a a A f f b a h n damaged s damaged s s s damaged s c f • Calculate the amount of CFRP necessary to recover the strength: 851 685 12 0.85 40 58.6 kips Eq. 3.4.1.1-2 58.6 1.03 15 3.79, say 4 layers _ ( )( ) ( ) ( ) ( ) = φ − φ φ = − = = = = = T M Mh T nN b n T N b f n n damaged f f b f f b f (Nb = FRP tensile strength per unit width at a strain of 0.005 = 171 ksi × 0.006 in. = 1.03 kip/in.; bf = FRP bond width = 15 in.) According to similar triangles in conjunction with force equilibrium, the concrete strain ec and the section’s neutral axis depth (c) at ef = 0.005 are found to be ec = 0.000655 and c = 4.63 in. Cc = 0.9 fc′ b2cb [Cc = resultant compression force, NCHRP Report 655 (Zureick et al. 2010)] T = Ts + Tf (T = total tension force, NCHRP Report 655) 1 Eq. 3.4.1.1-4 1 0.0006550.0016 0.000655 0.0016 0.38 1.71 Eq. 3.2-2 1.71 3 1820 3 0.0016 2 0 2 0 2 0 0 ( ) ( ) β = + ε ε     ε ε   = +      = ε = ′ = ε = = Ln Ln f E c c c c • Calculate the flexural capacity of the strengthened section: Eq. 3.4.1.1-1 0.9 5.6 60 28 0.34 4.63 0.85 4 1.03 15 40 0.34 4.63 1 2 arctan Eq. 3.4.1.1-3 1 2 0.0006550.0016 arctan 0.000655 0.0016 0.378 0.0006550.0016 0.34 834 ft-kip (if 5 CFRP layers are used, 876 ft-kip) OK _ _ 2 2 2 0 0 2 0 2 2 _ _ { } { } ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) φ = φ − + φ − = − + × × − = − ε ε   − εε      β ε ε   = −   −       = = φ = φ > M A f d k c T h k c k M M M n strengthened s damaged s s f f c c c n strengthened n strengthened u • Calculate the tensile development length of the CFRP sheet: L T f b Ld frp c f d( ) ( )( )( )≥ ′ = = = =0.065 Eq. 3.4.3.1-1 4 1.03 15 0.065 3 15 36.6 in. 3.05 ft, use 3.5 ft

156 FRp-StRengthened ReinFoRced concRete BRidge giRdeR in FlexuRe: neAR-SuRFAce-Mounted cFRp StRipS An existing reinforced concrete bridge girder is subjected to a factored moment of 750 ft-kip. The T-girder has an effective flange width of 7 ft with a depth of h = 40 in. (deck = 8 in. plus girder = 32 in.) and a girder width of 15 in. The girder has the following properties (compression steel is ignored for design convenience): fc′ (compressive concrete strength) = 3,000 psi, fy (steel yield strength) = 60 ksi, As (tension steel area) = 7 in.2, and d (effective depth) = 28 in. In accordance with new code requirements, the girder’s flexural strength needs to be increased by 30%. It is decided that the girder is strengthened with near-surface-mounted (NSM) CFRP strips having the following properties: ffu (ultimate strength) = 285 ksi, Ef (modulus) = 18,000 ksi, efu (rupture strain) = 0.015, wf (width) = 0.63 in., and tf (thickness) = 0.177 in. Determine the amount of CFRP strips to satisfy the flexural strength of the retrofitted girder. As = 7 in.2 40 in. 15 in. 32 in. NSM CFRP 8 in. 7 ft (typ.) 7 ft Solution: AASHTO Guide Specifications for Design of Bonded FRP Systems for Repair (AASHTO 2012a) are used. Design factors that are not stated in the AASHTO guide specifications (e.g., environmental reductions factors and bond-dependent coefficients) are not included in this design example. • Calculate the capacity of the original girder (under-reinforced): 2 AASHTO LRFD BDS Eq. 5.7.3.2.2-1 0.9 7 60 28 1.962 851 ft-kip 0.85 7 60 0.85 3 7 12 1.96 in. 8 in.( ) ( ) ( )( ) ( ) ( )( ) φ = φ −  = −   = = ′ = × = < = M A f d a a A f f b a h n s s s s s c f • Determine the capacity required to carry the increased load effect: M M Mn CFRP u new n exist ( )φ = − φ = − =− 1.3 750 851 124 ft-kip_ _ • Calculate the amount of CFRP strips necessary to upgrade the girder: T M h T n w t f f n CFRP f f f f fe( ) ( ) ( )= φ φ = = = 124 12 0.85 40 43.7 kips _ n T w t f f f f fe( ) ( )= = × = 43.7 0.63 0.177 95 4.13, use 6 strips and corresponding Tf = 6(0.63 × 0.177)95 = 63.6 kips (Following the limit specified in the specifications and the method used in NCHRP Report 655, the effective stress of the CFRP, ffe, is assumed to be the stress at a strain of ef = 0.005: ffe = (0.005/0.015) (285) = 95 ksi) According to similar triangles in conjunction with force equilibrium, the concrete strain ec and the section’s neutral axis depth (c) at ef = 0.005 are found to be ec = 0.000739 and c = 5.15 in.

157 Cc = 0.9 fc′ b2cb (Cc = resultant compression force, NCHRP Report 655) T = Ts + Tf (T = total tension force, NCHRP Report 655) 1 Eq. 3.4.1.1-4 1 0.0007390.0016 0.000739 0.0016 0.42 1.71 Eq. 3.2-2 1.71 3 1820 3 0.0016 2 0 2 0 2 0 0 ( ) ( ) β = + ε ε     ε ε   = +      = ε = ′ = ε = = Ln Ln f E c c c c • Calculate the flexural capacity of the strengthened section: Eq. 3.4.1.1-1 0.9 7 60 28 0.35 5.15 0.85 63.6 40 0.35 5.15 1 2 arctan Eq. 3.4.1.1-3 1 2 0.0007390.0016 arctan 0.000739 0.0016 0.42 0.0007390.0016 0.35 997 ft-kip OK _ 2 2 2 0 0 2 0 2 2 _ { } { } ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) φ = φ − + φ − = − + − = − ε ε   − εε      β ε ε   = −   −       = = φ > M A f d k c T h k c k M M n new s s s f f c c c n new u • Calculate the development length of the NSM CFRP strips: L a b a b fd b b b b b fd( ) ( ) ( ) ( ) ( )= + τ = + =2 Eq.13-4 of ACI440.2R-08 0.177 0.63 2 0.177 0.63 1000 95000 6.6 in. where ab and bb are the thickness and width of the CFRP strip, respectively; tb is the bond strength of the strip (tb = 1,000 psi as recommended by ACI440.2R-08); and ffd is the design stress of the FRP. Note that this expression is taken from ACI440.2R-08 because the AASHTO guide specifications do not include NSM CFRP. FRp-StRengthened ReinFoRced concRete BRidge giRdeR in SheAR Owing to the demand for increased traffic, the T-girder shown below requires CFRP strengthening in shear. The centroid of the longitudinal bars is 5 in. from the girder’s bottom. The factored shear used for the initial design is 85 kips, whereas the required shear is 100 kips. The original girder has vertical shear stirrups (Av = 0.4 in.2, fy = 60 ksi) at a spacing of s = 12 in. Determine the amount of CFRP U-wraps to upgrade the shear capacity of the girder with the following properties: ffu (strength) = 550 ksi, Ef (modulus) = 33,000 ksi, efu (rupture strain) = 0.016, and tf (thickness) = 0.006 in. Av = 0.4 in.2 CFRP sheet32 in.40 in. 15 in. 8 in. 7 ft 7 ft (typ.) @ 12 in. C.T.C

158 Solution: AASHTO Guide Specifications for Design of Bonded FRP Systems for Repair (AASHTO 2012a) are used, unless otherwise stated. • Calculate the shear capacity of the original girder: AASHTO LRFD BDS Eq. 5.8.3.3-1 0.9 47.3 56 0 93 kips 0.0316 AASHTO LRFD BDS Eq. 5.8.3.3-3 0.0316 2 3 15 28.8 47.3 kips resistance by concrete section max 0.9 , 0.72 AASHTO LRFD BDS Art. 5.8.2.9 max 0.9 28 , 0.72 40 28.8 in. cot cot sin AASHTO LRFD BDS Eq. 5.8.3.3-4 0.4 60 28 cot 45 cot90 sin90 12 56 kips resistance by stirrups 0 resistance by prestressing steel ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) φ = φ + + = + + = = β ′ = = = = = = θ + α α = + = = V V V V V f b d d d h V A f d s V n c s p c c v v v s v y v p • Calculate the amount of CFRP needed to upgrade the girder: According to Eq. 4.3.1-1 , 100 0.9 47.3 56 0 0.85 8.3 kips From sin cos Eq. 4.3.2-1 , sin cos 8.3 10 34.8 27 sin90 cos90 0.088 in. 0.004 for U-wraps, Eq. 4.3.2-5 33000 0.066 0.016 34.8 ksi min 0.4 ,12in Eq. 5.8.2.7-2 for 10015 28.8 0.23 0.125 11.52 in. controls 2 ( ) ( ) ( ) ( ) [ ] ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) φ = φ + + + φ = − φ + + φ = − + + = = α + α = α + α = + = = ε ε = ε ≤ = = = = = = ≥ = V V V V V V V V V V V A f d s A V sf d f E R s d v Vb d n c s p f f f u c s p f f f fe f f f f f f f fe f f f fe f fe fe f fu f v u u v v Use CFRP vertical U-wraps spaced at a center-to-center spacing of 10 in. df = effective depth of FRP measured from the top of FRP reinforcement to the centroid of the longitu- dinal reinforcement = 32 in. (girder depth) - 5 in. (bar location from girder bottom) = 27 in. Because the thickness of the CFRP is tf = 0.006 in. and Af = 2tfwf (where wf is the width of the CFRP U-wrap), w A tf f f ( )= = =2 0.088 2 0.006 7.3 in. Use 8-in.-wide U-wraps at 10 in. on center FRp-StRengthened ReinFoRced concRete BRidge coluMn A circular short column was designed to carry a factored axial concentric load of 2,500 kips. The column has a diameter of 45 in. and is reinforced by 15 #7 longitudinal bars (As = 9.0 in.2) and #3 ties at 10 in. center- to-center. Because of a superstructure renovation project, the column is subjected to an increased load of 3,000 kips (factored). CFRP strengthening is planned to accommodate the increased load demand. Deter- mine the amount of CFRP sheets to upgrade the axial capacity of the column with the following properties: ffu (ultimate strength) = 550 ksi, Ef (FRP modulus) = 33,000 ksi, efu (FRP-rupture strain) = 0.016, and tf (FRP thickness) = 0.006 in.

159 45 in. No. 7 No. 3 @ 10 in. C.T.C Section A - A CFRP-wrapping AA Solution: AASHTO Guide Specifications for Design of Bonded FRP Systems for Repair (AASHTO 2012a) are used, unless otherwise stated. • Calculate the capacity of the original column: 0.80 0.85 AASHTO LRFD BDS Eq. 5.7.4.4-3 0.75 0.80 0.85 3 1590 9 0 60 9 0 2,743 kips { } { } ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) φ = φ ′ − − + − − ε = − − + × − = P f A A A f A A f En c g st ps y st ps pe p cu • Calculate the compressive strength of the confined concrete ( f ′cc): From 0.80 0.85 Eq. 5.3.1-2 , 0.80 0.85 3000 0.75 0.80 60 9 0.85 1590 9 3.32 ksi { }( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) φ = φ ′ − + ′ = φ φ − − = − × − = P f A A f A f P f A A A n cc g st y st cc n y st g st Rearranging f f ffcc c l c ′ = ′ + ′  1 2 (Eq. 5.3.2.2-1) gives the required confining pressure ( fl), f ff fl cc c c= ′ ′ −   ′ = −  =12 1 12 3.323 1 3 0.16 ksi • Calculate the amount of CFRP ( fl): 2 2 1 Eq. 5.3.2.2-2 gives 2 0.16 45 2 0.65 5.54 kip/in. FRP strength per width corresponding to a strain of 0.004 550 0.004 0.016 0.006 0.825 kip/in. per CFRP layer Number of CFRP layers required 5.540.825 6.72, use 7 CFRP layers 0 0 ( ) ( ) ( ) ( )( ) ( ) ( ) = φ ≤ ′ − φ  = φ = = = = = = = f ND f k N f D N N N N l f f c e c f l f f f f f • Calculate the capacity of the strengthened column: 2 0.652 7 0.82545 0.167 2 1 3 2 1 0.80 0.75 0.75 OK 0.80 for tied column 0.75 concrete material resistance factor 1 2 3 1 2 0.1673 3.334 ksi 0.80 0.85 0.75 0.80 0.85 3.334 1590 9 60 9 3,012 kips 3,000 kips OK { } { } ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )= φ = × = ≤ ′ − φ  = −  = = φ = ′ = ′ + ′   = +  = φ = φ ′ − + = − + × = > = f ND f k k f f ff P f A A f A P l f f c e c e c cc c l c n cc g st y st u

Abbreviations and acronyms used without definitions in TRB publications: A4A Airlines for America AAAE American Association of Airport Executives AASHO American Association of State Highway Officials AASHTO American Association of State Highway and Transportation Officials ACI–NA Airports Council International–North America ACRP Airport Cooperative Research Program ADA Americans with Disabilities Act APTA American Public Transportation Association ASCE American Society of Civil Engineers ASME American Society of Mechanical Engineers ASTM American Society for Testing and Materials ATA American Trucking Associations CTAA Community Transportation Association of America CTBSSP Commercial Truck and Bus Safety Synthesis Program DHS Department of Homeland Security DOE Department of Energy EPA Environmental Protection Agency FAA Federal Aviation Administration FAST Fixing America’s Surface Transportation Act (2015) FHWA Federal Highway Administration FMCSA Federal Motor Carrier Safety Administration FRA Federal Railroad Administration FTA Federal Transit Administration HMCRP Hazardous Materials Cooperative Research Program IEEE Institute of Electrical and Electronics Engineers ISTEA Intermodal Surface Transportation Efficiency Act of 1991 ITE Institute of Transportation Engineers MAP-21 Moving Ahead for Progress in the 21st Century Act (2012) NASA National Aeronautics and Space Administration NASAO National Association of State Aviation Officials NCFRP National Cooperative Freight Research Program NCHRP National Cooperative Highway Research Program NHTSA National Highway Traffic Safety Administration NTSB National Transportation Safety Board PHMSA Pipeline and Hazardous Materials Safety Administration RITA Research and Innovative Technology Administration SAE Society of Automotive Engineers SAFETEA-LU Safe, Accountable, Flexible, Efficient Transportation Equity Act: A Legacy for Users (2005) TCRP Transit Cooperative Research Program TDC Transit Development Corporation TEA-21 Transportation Equity Act for the 21st Century (1998) TRB Transportation Research Board TSA Transportation Security Administration U.S.DOT United States Department of Transportation

TRANSPORTATION RESEARCH BOARD 5 0 0 F ifth S tre e t, N W W a s h in g to n , D C 2 0 0 0 1 A D D R ESS SER VICE R EQ UESTED NO N-PRO FIT O RG . U.S. PO STAG E PA ID CO LUM BIA, M D PER M IT NO . 88 ISBN 978-0-309-39004-0 9 7 8 0 3 0 9 3 9 0 0 4 0 9 0 0 0 0 Use of Fiber-Reinforced Polym ers in Highw ay Infrastructure NCHRP Synthesis 512 TRB

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TRB's National Cooperative Highway Research Program (NCHRP) Synthesis 512: Use of Fiber-Reinforced Polymers in Highway Infrastructure documents the current state of the practice in the use of fiber-reinforced polymers (FRPs) in highway infrastructure. The synthesis identifies FRP applications, current research, barriers to more widespread use, and research needs. The objectives of the study are to synthesize published literature on FRP materials in highway infrastructure and to establish the state of current practice of FRP applications in transportation agencies.

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