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ATTACHMENT B Illustrative Examples The following examples are presented to illustrate calculations associated with a number of commonly used FRP strengthening techniques in accordance with the recommended Guide Specification for the Design of Bonded FRP Reinforcement Systems for Repair and Strengthening of Concrete Bridge Elements. These examples should illustrate how to approach bridge strengthening projects in practice. These examples cover the five sections of the proposed Guide Specification. Example 1: Calculation of the characteristic value of the strength of an FRP reinforcement system Example 2: Flexural strengthening of a T-beam in an unstressed condition Example 3: Flexural strengthening of a T-beam in a stressed condition Example 4: Shear strengthening of a T-beam using U-jacket FRP reinforcement Example 5: Shear strengthening of a rectangular beam using complete wrapping FRP reinforcing system Example 6: Strengthening of an axially loaded circular column.
B-1 Example 1 It is required to calculate the characteristic value of the strength of a field-manufactured FRP system to be bonded externally to strengthen an existing bridge member. Tensile tests were conducted on coupons in accordance with ASTM D3039. Results are given in Table B1. It is also required to establish the linear load-strain relationship for use in the design following the recommended Guide Specifications. Table B1: Summary of test data Coupon ID Strength (kips/in) at 1% strain 1 2.00 2 2.17 3 2.01 4 2.10 5 1.87 6 2.14 7 2.08 8 2.12 9 2.10 10 2.09 11 2.10 12 2.14 13 2.12 14 2.21 15 1.95 16 2.22
B-2 Step 1: Examine the data set to find out if there is any outlier. To do so, the Maximum Normed Residual outlined in ASTM D7290 Standard Practice will be used. To do so 1.1 Sort the data in ascending order as shown in Table B2. Table B2- Strength data sorted in ascending order Coupon ID Strength (kips/in) at 1% strain 5 15 1 3 7 10 4 9 11 8 13 6 12 2 14 16 1.87 1.95 2.00 2.01 2.08 2.09 2.1 2.1 2.1 2.12 2.12 2.14 2.14 2.17 2.21 2.22 1.2 Calculate the arithmetic mean, x , and the standard deviation, s, of the sample population from the equations n x x n i i 1 1 1 2 n xx s n i i
B-3 inkipsx /09.2 16 22.221.2....95.187.1 inkips x s i i /092.0)116( )09.2( 16 1 2 Step 3 1.2 Calculate the Maximum Normed Residual (MNR) as follows: For each data value, compute the term s xxi representing the deviation from the sample mean divided by the sample standard deviation. For the first data value, coupon ID 5: 388.2 092.0 09.287.1 s xxi For the second data value, coupon ID38: 52.1 092.0 09.295.1 s xxi For the remaining data, computed s xxi values are given in Table B3. Compute the critical value, Cr, from the equation: 56.2 165 82 5 82 22 n Cr The Maximum Normed Residual (MNR) is 39.2max s xx MNR i
B-4 Because 56.239.2 rCMNR , the sample population is outlier-free, and one can proceed with the statistical evaluation. Table B3 Strength deviation from the sample mean Coupon ID Strength (kips/in) at 1% strain s xxi 5 15 1 3 7 10 4 9 11 8 13 6 12 2 14 16 1.87 1.95 2.00 2.01 2.08 2.09 2.1 2.1 2.1 2.12 2.12 2.14 2.14 2.17 2.21 2.22 2.39 1.52 0.98 0.87 0.11 0.00 0.11 0.11 0.11 0.33 0.33 0.54 0.54 0.87 1.30 1.41 Step 4 The sample size is greater than 10 and the coefficient of variation 15.0044.0 09.2 092.0COV , the composite material strength data meet the requirements of Article 1.4.3 of the Guide Specifications. Step 5 : Estimate the parameters of the two-parameter Weibull distribution from the following equations (See Commentary C1.4 of the Guide Specifications)
B-5 13.2)09.2(02.1)09.2()044.0)(8/3(18/31 xCOVu 3.27 044.0 2.12.1 COV Step 6: Compute the characteristic value of the strength from the equation: 1/ 1/27.3 0.10 0.1054 (2.13) 0.1054 1.96 /x u kips in Step 7: Establish the FRP reinforcement strength-strain design relationship as shown in B-1. Strain Lo ad /u n it w id th 0.01 1.96 kip/in. Figure B-1 Load-strain relationship of the FRP reinforcement
B-6 Example 2 Flexural strengthening of a simply supported cast in-place reinforced concrete girder This example illustrates the flexural strengthening of a reinforced concrete T-beam with an externally bonded carbon fiber-reinforced polymeric reinforcement system to accommodate higher loading. Bridge Data Span: 39 ft Type: Cast in-place reinforced concrete Year built: 1957 Location: State of Georgia Concrete compression Strength: ' 3.9cf ksi (from in-situ testing) Reinforcing steel yield strength: 40yf ksi Girder dimensions and Steel Reinforcement: See Figure B.2 FRP reinforcement: Shop-fabricated carbon fiber/Epoxy composite plates Plate thickness, 0.039t Glass Transition Temperature: 165ogT F Tensile strain in the FRP reinforcement at failure: 0.013tufrp Tensile strength in the FRP reinforcement at 1% strain: 9.3 /frpP kips in Shear modulus of the adhesive = 185 ksi Structural Analysis Results under the New Loading For Strength I Load Combination: ftkipM D 239 and ftkipM IL 615 . For Fatigue Limit State: ftkipM IL 308 Special Notes Hydraulic jacking procedure of the bridge will be used so that strengthening is carried out in an unstressed condition.
B-7 24.5" 6" 7'-2" 7'-2" 18" Detail A # 4 8#11 Bundled 2" cover 18" Figure B-2 Bridge cross section at mid-span
B-8 SOLUTION Step 1: Determine if the FRP reinforcement material is in compliance with Section 2 of the Guide Specification and be sure that the glass transition temperature is higher than the maximum design temperature plus 40oF. The maximum design temperature, MaxDesignT , determined from Article 3.12.2.2 of AASHTO LRFD Bridge Design Specifications for the location of the bridge (State of Georgia) 110oMaxDesignT F 40 110 40 150 165o o o o oMaxDesign gT F F F F T F . Thus, Article 2.2.4.1 of the Guide Specification is satisfied. Step 2: Establish the linear stress-strain relationship of the FRP reinforcement based on the design assumptions specified in Article 3.2 of the Guide and compute the tensile strength corresponding to a strain value of 0.005. Results are presented in Figure B.3 ./65.4)3.9( 01.0 005.0 inkipNb 9.3 kip/in. 1.3% Strain Load Tensile failure of the carbon composite 1%0.5% Nb = 4.65 kip/in. Figure B-3 FRP reinforcement stress-strain diagram for design purposes
B-9 Step 3: Calculate the flexural strength of the T-beam Effective depth .59.2641.15.025.30 ind Effective Flange Width As per Article 4.6.2.6.1 of AASHTO LRFD Bridge Design Specifications, the effective flange width is taken as the minimum of One-quarter of the effective span length; Twelve times the average depth of the slab, plus the greater of web thickness or one-half the width of the top flange of the girder; or The average spacing of adjacent beams. .86 .9018)6(1212 .117 4 )12)(39( 4 ins inbt inl Minimumb ws e e .86 inbe Assumptions: A rectangular stress block to represent the distribution of concrete compression stresses (Article 5.7.2.2 of AASHTO LRFD Bridge Design Specifications), No contribution of the steel in the compression zone to the flexural strength, The strain in the tension steel is greater than the yield strain, and The neutral axis is located in the flange of the section Thus, the compression and tension forces are '0.85c c eC f b a and s yT A f , respectively, as illustrated in Figure B-5. From the condition of equilibrium of forces: # 4 8#11 Bundeled 2" cover 18" be = 86" 30 . 5" Figure B-4 Reinforced Concrete T-Beam
B-10 '0.85 c e s yf b a A f Thus, .75.1 869.385.0 4048.12 85.0 ' in bf fA a ec ys sA be = 86" d= 26 .5 9" h= 30 .5 " a '0.85 c ef b a ys fA Figure B-5 Force equilibrium on a reinforced concrete T-beam The depth of the neutral axis: .06.2 85.0 75.1 1 inac Since .6.06.2 intinc s , the assumption that depth of the neutral axis fall within the flange is appropriate. Referring to Figure B-5, the strain in the tension steel can be computed as follows: 0.003 s d c c 036.0003.0 06.2 06.259.26 s Since 400.036 0.00138 29,000 y s s f E , the assumption that the tension steel yielded is correct. The nominal flexural strength of the girder can then be computed from .837,12 2 75.159.264048.12 2 inkipadfAM ysn
B-11 .553,11837,129.0 inkipM n Check compliance with Article 1.4.4 of the proposed Guide Specifications .248,10854615239.553,11 inkipftkipMMinkipM ILDn Proceed with the design of an externally bonded FRP reinforcement system. Step 4: Estimate the amount of FRP reinforcement required to accommodate the increase in flexural strength. The factored moment for Strength I limit state is .500,16375,1)615(75.123925.175.125.1 inkipftkipMMM ILDu For a preliminary estimate of the amount of FRP reinforcement necessary to resist 1,375 k-ft of moment, the following approximate design equation can be used: infunre orced u n FRP M MT h kipsTfrp 1625.30 12963375,1 frpbfrp bnNT Where n is the number of FRP reinforcement plates. Use a reinforcement width of 41frpb , the number of required layers is: 162 2.5(4.65)(14) frp b frp T n N b Try 3 layers of the FRP reinforcement, for which kipsTfrp 3.1951465.43 Step 5: Compute the factored flexural resistance of the strengthened T-beam Location of the neutral axis The depth of the neutral axis can be determined from both strain compatibility and force equilibrium conditions as follows:
B-12 Figure B-6 Reinforced concrete T-beam externally strengthened with FRP reinforcement Assume .6 inc 6 (0.005) 0.00122 30.5 6c FRP c h c '1,820 1,820 3.9 3,594c cE f ksi ' 3.91.71 1.71 0.00186 3,594 c o c f E 0.00122 0.66 0.00186 c o 2 2 2 1 1 (0.66) 0.548(0.66) c o c o Ln Ln Compression force in the concrete: ' 20.9 0.9(3.9)(0.548)(6)(86) 992.5c c eC f cb kips Tension Force in the tension steel: Strain in the steel: 001379.0 000,29 4000418.000122.0 6 659.26 E f c cd y ycs Figure B-7 Strain and stress diagrams for the reinforced concrete T-beam externally reinforced with bonded carbon fiber FRP reinforcement
B-13 Thus, (12.48)(40) 499.2s s yT A f kips Tension Force in the FRP reinforcement: kipsTfrp 3.1951465.43 Total Tension Force kipsTTT sfrp 5.6942.4993.195 Clearly equilibrium of the forces is not satisfied 992.5 694.5 298cC T kips , and the assumed depth for the neutral axis ( 6 .c in ) is incorrect. By trial and error, one can find that by assuming a depth of the neutral axis, 4.96 .c in , and repeating the above calculations, the following values are computed: For 4.97 .c in 0.00097c , 0.0042s y , 0.53c o , 2 0.46 , 695.2cC kips , 499.2sT kips , kipsT frp 3.195 , kipsTTT sfrp 5.6942.4993.195 , and 695.2 694.5 0.7cC T kips , close enough to zero. The factored flexural resistance 2 20.9r s s s frp FRPM A f d k c T h k c With 2 2 2 2 2 arctan 2 0.53 arctan 0.53 1 1 0.35 0.46 0.53 c c o o c o k and 0.85frp 0.9 12.48 40 26.59 0.35 4.97 0.85 195.3 30.5 0.35 4.97 15,939 . r M kips in 15,939 . 16,500 . r M kip in kip in Increase the width of the FRP reinforcement to 17 .frpb in and re-compute the flexural resistance rM . By doing so, we can find 5.1 .c in and 16,930 16,500 . r M kips in kip in .
B-14 Thus, AASHTO Strength I Load Combination limit is satisfied. Step 6: Check ductility requirements (Article 3.4.2 of the Guide) When reinforcing steel first yields at 40 0.00138 29,000 y s y s f E . For such a case, the strain and stress diagrams are shown in Figure B-8. cy y c y frp s y ys fA k2c frpT cC d ' 2 (0.9 )av cf f Figure B-8 Strain and stress distribution in the T-beam when tension steel reinforcement yield By satisfying the conditions of force equilibrium and strain compatibility, the strain in the FRP reinforcement when the steel tensile reinforcement yields can be found numerically to be 0.0016yfrp . Thus, the ductility requirement of Article 3.4.2 of the guide specification is: 0.005 3.1 2.5 0.0016 u frp y frp . OK Step 7: Development length int 237.15 109 . 9.1 0.065 3.9(17) frp d frp T L in ft b
B-15 B-9 FRP Reinforcement location Distance of FRP reinforcement end termination from the girder centerline = 9.10 + 2.33 =11.43 ft. Use 12 ft and reinforce symmetrically as shown in Figure B-10. B-10 FRP reinforcement Detail Step 8: Check fatigue load combination limit state For the fatigue load combination: .772,2231)308(75.075.0 inkipftkipM IL Determine the cracking moment: t g rcr y IfM with ksiff cr 474.09.324.024.0 ' Section Properties: 4096,78 inI g inyt 4.20 78,096(0.474) 1,815 . 2,772 . 20.4cr M kip in kip in
B-16 Neglect the concrete part in tension and calculate the moment of inertia of an equivalent transformed FRP section: From the FRP reinforcement load-strain data: / 4.65 / (0.039) 23,850 0.005 frp b frp frp frp frp f N t E ksi Modular ratio for the concrete: 3,594 0.15 23,850 c c frp E n E Modular ratio of the steel: 29,000 1.2 23,850 s s frp E n E Based on the modular ratios for the concrete and for the steel, an equivalent FRP transformed section is constructed as shown below with the neutral axis assumed to lie in the flange. nc Ac =nc be z z ns As Afrp N.A. 11 dh Figure B11-Equivalent FRP transformed section By summing the moment of areas about reference line 1-1: 2 2 frp frp s s c c frp s s c c t zA h n A d n A A n A n A z 2 2 frp frp s s c e frp s s c e t zA h n A d n b z A n A n b z z 2 2 2 2 0 frp frp s s frp s s c e c e t A h n A d A n A z z n b n b
B-17 2 2 (3)(17)(0.039) 1.2(12.48) 2.6 .(0.15)(86) frp s s c e A n A in n b 2 0.1172 (3)(17)(0.039) 30.5 (1.2)(12.48) 26.632 2 71.22 .(0.15)(86) frp s s c e A h n A d in n b The equation ( 2 2.6 71.22 0z z ) has the solutions of 7.24 .z in or 9.84 .z in and only the positive solution 7.24 .z in is valid. Because 7.24 . 6 .z in in , the assumption that the neutral axis fall in the flange was incorrect. Assume that the neutral axis is located at a distance 6 .z in By summing the moment of areas about reference line 1-1: ( ) ( ) 2 2 2 frp s frp s s c e w s c w frp s s c e w s c w nt t zA h n A d n b b t n b z A n A n b b t n b z z 2 2 ( )2 ( 2 2 0 frp s frp s s c e w s frp s s c e w s c w c w nt tA h n A d n b b tA n A n b b t z z n b n b By substituting all parameters into the above equation, the following equation is obtained 2 57.9 476.4 0z z Which has a positive solution 7.31 .z in The moment of inertia of the equivalent transformed FRP section can be computed to be 48,345 .TI in Strain in the concrete, steel reinforcement, and FRP reinforcement, respectively, due to the fatigue load combination: '(231)(12)(7.31) 3.90.00010 0.36 0.36 0.00039(8,345)(23,850) 3,595 f c c T frp c M z f I E E ( ) (231)(12)(26.63 7.31) (40)0.0003 0.8 0.8 0.0011(8,345)(23,850) (29,000) f s y T frp M d z I E
B-18 ( ) (231)(12) 30.50 3(0.039) 7.31 0.00032 0.8(0.013) 0.0104(8,345)(23,850) f frp u frp frp T frp M h t z I E Step 9: Check reinforcement end termination peeling The reinforcement end terminates at a distance of 19.5-12 = 7.5 ft from each of the end supports. It is required to calculate the moment and shear at 7.5 ft from the end support. From analysis, we will use the following combinations: 1.25 1.75 503u D L IM M M kip ft 1.25 1.75 112u D L IV V V kips Calculate the peel stress from the equation: 1 4 3 a FRP peel av FRP a E tf E t 2 1a a aE G 1/2 FRPa av u u frp FRP a T t h zGV M E t t I 1/2 185 (0.117)(30.5 7.31)112 (503)(12) 1.5(23,850)(0.117)(0.125) 8,345av ksi 1 4 3(500) 0.117(1.5) 0.740 0.065 3.9 0.128 23,850 0.125peel f ksi ksi Provide mechanical anchors at the FRP reinforcement ends.
B-19 Example 3 Flexural strengthening of a simply supported cast in-place reinforced concrete girder This example illustrates the flexural strengthening of a reinforced concrete T-beam with an externally bonded carbon fiber-reinforced polymeric reinforcement system to accommodate higher loading. This example is identical to Example 2 except that strengthening of the bridge will be carried out under the effect of the bridge dead load (stressed condition). Bridge Data Use the same data provided in Example 2. SOLUTION Step 1: Determine if the FRP reinforcement material is in compliance with Section 2 of the Guide Specification and be sure that the glass transition temperature is higher than the maximum design temperature plus 40oF. Based upon the location of the bridge (State of Georgia), the maximum design temperature determined from Article 3.12.2.2 of AASHTO LRFD Bridge Design Specifications is: 110oMaxDesignT F Because 40 110 40 150 165o o o o oMaxDesign gT F F F F T F , Article 2.2.4.1 of the Guide is satisfied. Step 2: Determine the cracking moment for the T-beam Determine the cracking moment: t g rcr y IfM ksiff cr 474.09.324.024.0 ' 4096,78 inI g
B-20 inyt 4.20 78,096(0.474) 1,815 151 20.4cr M kip in kip ft Step 3: Determine the initial strain at the time of strengthening Because ftkipMftkipM crD 2.152239 , the cracked moment of inertia of the section will be used to compute the initial strain resulting from the dead load. 29,000 8 3,594 s c c E n E Assume that the neutral axis lies inside the flange of the T-section. In such a case the depth of the neutral axis can be computed from: 2 (8)(12.48) 2(26.59)(86)1 1 1 1 6.79 . 6 . 86 (8)(12.48) s e N s e s nA dby in t in b nA The assumption associated with the location of the neutral axis is not correct. Assume that the neutral axis falls in the web at a distance Ny ,from the top of the flange, which can be determined by considering a cracked transformed section as follows: 6(86)(6)(3) (18)( 6) 6 (8)(12.48)(26.59) (86)(6) (18)( 6) (8)(12.48) 2 N N N N yy y y From which 6.82 .Ny in 3 3 2 2 4(86)(6) (6.82 6)(86)(6)(6.82 3) (18) (8)(12.48)(26.59 6.82) 48,104 . 12 3cr I in The initial tensile stress at the bottom concrete surface: ( ) (239)(12)(30.5 6.82) 1.41 48,104 D N bo cr M h y ksi I At the time of installing the externally bonded FRP reinforcement, the dead load initial strain at the bottom surface is: 1.41 0.00039 3,594 bo bo cE
B-21 Step 4: Determine the maximum strain in the FRP reinforcement With a maximum useable strain of 0.005 at the FRP reinforcement/concrete interface, the maximum strain in the FRP reinforcement that can be developed is: 0.005 0.005 0.00039 0.0046bo The force per unit width in the FRP reinforcement corresponding to a strain of 0.0047 is: 0.0047 0.0046 (9.3) 4.28 / . 0.01 N kip in To estimate the amount of FRP reinforcement necessary to resist 1,375 k-ft of moment, the following approximate design equation can be used: infunre orced u n frp M MT h kipsTfrp 1625.30 )12(963375,1 frp b frpT nN b Where n is the number of FRP reinforcement plates. Use a reinforcement width of 17 .frpb in , the number of required layers is: 0.0047 162 2.23(4.28)(17) frp frp T n N b Try 3 layers of the FRP reinforcement, for which 3(4.28)(17) 218.3frpT kips Step 5: Compute the factored flexural resistance of the strengthened T-beam The computation procedure is similar to that of Example 2. By iteration, we find 5.24 .c in , and 16,475 . 1,373 1,375 r M kip in kip ft kip ft The remaining steps can be followed as presented in Example 2.
B-22 Example 4 U-Jacket Shear strengthening of a reinforced concrete bridge This example illustrates the shear strengthening of a reinforced concrete T-beam with an externally bonded carbon fiber-reinforced polymeric reinforcement U-jacket system to accommodate higher loading. Bridge Data Span: 39 ft Type: Cast in-place reinforced concrete Year built: 1957 Location: State of Georgia Concrete compression Strength: ' 3.9cf ksi (from in-situ testing) Reinforcing steel yield strength: 40yf ksi Girder dimensions and Steel Reinforcement: See Figure B-12 FRP reinforcement: Shop-fabricated carbon fiber/Epoxy composite plates Plate thickness, 0.039t Glass Transition Temperature: 165ogT F Tensile strain in the FRP reinforcement at failure: 0.013tufrp Tensile strength in the FRP reinforcement at 1% strain: 9.3 /frpP kips in kipsVD 24 , kipsV IL 61
B-23 Figure B-12 Beam geometry and reinforcement SOLUTION Step 1: Calculate Nominal Shear Strength of Reinforced Concrete T-beam In accordance with Article 5.8.2.9 of the AASHTO LRFD Bridge Design Specifications (2007) The effective web width: 18 .vb in The effective shear depth: h d D Maximumd e CT v c 72.0 9.0 & cCTD & = Distance between the resultants of the tensile and compressive forces due to flexure ed = Effective depth from extreme compression fiber to the centroid of the tensile force in the tensile reinforcement--Article 5.7.3.3.1 of AASHTO (2007) h = Overall depth of the member
B-24 From Example 2, & 1.7526.59 25.72 . 2cT C D in , 26.59 .ed in , 30.5 .h in 25.72 . 0.9 0.9 26.59 23.93 . 0.72 0.72 30.5 21.96 . v e in d Maximum d in h in 25.72 .vd in Check if the transverse reinforcement of the reinforced concrete girder meets the minimum transverse shear reinforcement specified in Article 5.8.2.5 of AASHTO (2007) yv v cv f SbfA '0316.0 For 2#4 steel stirrups, 20.4 .vA in 2 2(18)(12)0.4 0.0316 ' 0.0316 3.9 0.35 . 40 v v c yv b SA in f inf O.K Nominal Shear ResistanceâArticle 5.8.3.3 of AASHTO (2007) pvvc psc n Vdbf VVV MinimumV '25.0 Where vvcc dbfV '0316.0 , S dfA V vyvs sin)cot(cot , and 0pV (non-prestressed girder) Because the minimum transverse reinforcement requirement of Article 5.8.2.5 of AASHTO (2007) is met and the girder is neither prestressed nor axially loaded, the values of and can be determined by the simplified procedures of Article 5.8.3.4.1 of the AASHTO LRFD Bridge Design Specifications (AASHTO, 2007). Therefore: = 2.0 = 45° 0.0316(2) 3.9(18)(25.72) 57.78cV kips
B-25 (0.4)(40)(25.72) cot(45) cot(90) sin(90) 34.3 12s V kips 57.78 34.3 0 92.08 0.25 ' 0.25(3.9)(18)(25.72) 0 451.4 c s p n c v v p V V V kips V Minimum f b d V kips 92.08nV kips Step 2: Estimate the amount of FRP reinforcement needed to increase the shear strength to 156 kips kipsVVV ILDu 1376175.12425.175.125.1 137 (0.9)(92.08) 98.41 0.55 u n frp frp V VV kips From the linear stress-strain relationship of the FRP reinforcement compute the tensile strength corresponding to a strain value of 0.004. 0.004 (9.3) 3.72 / . 0.01s N kip in For a U-jacket type of shear reinforcement without mechanical anchors: 3.72 / .efrp sN N kip in , with 30.5 6 24.5 .frpd in If intermittent transverse shear reinforcement is used, FRP shear reinforcement shall be provided symmetrically on both sides of the member with spacing not to exceed the smaller value of 0.4 dv or 12 inches (Article 4.3, Guide). 0.4 0.4(25.56) 10.2 . ( ) 12 . v v d in Governs s Minimum in Try 2-in wide FRP plates spaced at 10 inches apart. vvv frpfrp e frp frp sss dwN V 6.3645.24)90cos(90sin272.32)cos(sin
B-26 Table B4 presents the FRP reinforcement shear strength for different values of vs Table B4-FRP Reinforcement Shear Strength for various values of vs vs (in.) frpV (kips) 10 36.5 4 91.2 3 121.5 2.75 132 Use 2-in wide FRP plates with 3 in. center-to-center. This will leave 1- in.- gaps between the plates to facilitate future inspection, using currently employed inspection techniques, of the bridge girder. If continuous FRP shear reinforcement is provided, then sin cos 2(3.72)(1 0)(24.5) 182.3efrp frp frpV N d kips , and future inspection will require thermography techniques. In either case, intermittent or continuous reinforcement, the nominal shear strength provided by the externally bonded FRP shear reinforcement shall satisfy Article 4.3.5 stipulating, frpwcfrps dbfVV '8 kipsdbfV frpwcfrp 967,6)5.24(189.38834 ' kipsV frp 933,634967,6 In both cases, the provision of Article 4.3.5 of the Guide is satisfied.
B-27 Example 5 Shear strengthening Using Complete wrapping Reinforcing System This example illustrates the shear strengthening of a reinforced concrete member with an externally bonded carbon fiber-reinforced polymeric reinforcement completely wrapped around the section that requires strengthening. The shear forces obtained from structural analysis are: kipsVD 80 and kipsV IL 100 Figure B-13 Reinforcement details Span: 150 in. Type: Reinforced concrete Concrete compression Strength: ksifc 9.3' Reinforcing steel yield strength: 60yf ksi P
B-28 Steel Reinforcement: See Figure B-13. FRP reinforcement: Field-fabricated carbon fiber/Epoxy composites Tensile strain in the FRP reinforcement at failure: 018.0tufrp Tensile strength in the FRP reinforcement is 0.94 kip/in. at a strain of 0.01. SOLUTION Step 1: Calculate nominal shear strength of the reinforced concrete member and Check compliance with Article 1.4.4 of the proposed Guide Specifications In accordance with Article 5.8.2.9 of the AASHTO LRFD Bridge Design Specifications (2007) The effective web width: 11.8 .vb in The effective shear depth: h d D Maximumd e CT v c 72.0 9.0 & cCTD & = Distance between the resultants of the tensile and compressive forces due to flexure ed = Effective depth from extreme compression fiber to the centroid of the tensile force in the tensile reinforcement--Article 5.7.3.3.1 of AASHTO (2007) h = Overall depth of the member Calculation of cCTD & : Assume a rectangular concrete compression stress block for the section and establish the section force equilibrium by a trial-and-error procedure: 22.3 . sec 0.9 0.9 26.61 23.9 . 0.72 0.72 30.5 20.4 . v e in from flexural analysis of the tion d Maximum d in h in
B-29 23.9 .vd in Check if the transverse reinforcement of the reinforced concrete girder meets the minimum transverse shear reinforcement specified in Article 5.8.2.5 of AASHTO (2007) yv v cv f SbfA '0316.0 For 2 double leg #5 steel stirrups, 21.24 .vA in 2 2(11.8)(9.5)1.24 . 0.0316 ' 0.0316 3.9 0.11 . 60 v v c yv b SA in f inf O.K Nominal Shear ResistanceâArticle 5.8.3.3 of AASHTO (2007) pvvc psc n Vdbf VVV MinimumV '25.0 Where vvcc dbfV '0316.0 , S dfA V vyvs sin)cot(cot , and 0pV (non-prestressed girder) Because the minimum transverse reinforcement requirement of Article 5.8.2.5 of AASHTO (2007) is met and the girder is neither prestressed nor axially loaded, the values of and can be determined by the simplified procedures of Article 5.8.3.4.1 of the AASHTO LRFD Bridge Design Specifications (AASHTO, 2007). Therefore: = 2.0 = 45° kipsVc 2.35)9.23)(8.11(9.3)2(0316.0 kipsVs 2.1875.9 )90sin()90cot()45cot()9.23)(60)(24.1( kipsVdbf kipsVVV MinimumV pvvc psc n 0.2750)9.23)(8.11)(9.3(25.0'25.0 4.22202.1872.35 kipsVn 4.222
B-30 kipsVVkipsVV ILDnr 18010080200)4.222(9.0 The member meets the requirement of Article 1.4.4. Thus, proceed with strengthening using externally bonded FRP shear reinforcement wrapped around the member. Step 2: Estimate the amount of FRP reinforcement needed to increase the shear strength to 156 kips kipsVVV ILDu 275)100(75.1)80(25.175.125.1 275 200 115 0.65 u r frp required frp V VV kips From the linear stress-strain relationship of the FRP reinforcement compute the tensile strength corresponding to a strain value of 0.004. kipsN s 376.0)94.0(01.0 004.0 Check if Sutwfrp NNN 5.0, , 0.5 (0.5)(1.69) 0.85 / . 0.376 / .frp w utN N kip in kip in , 1 10.376 0.85 0.376 0.613 / . 2 2 e frp s frp w sN N N N kip in (sin cos ) 2 0.613 sin 90 cos(90) 23.9 29.3 efrp frp frpV N d kips Number of required layers: e 115 3.9 29.3 frp r quired frp V n V Use 4 layers for which the provided 4(29.3) 117.2frpV kips The nominal shear strength provided by the externally bonded FRP shear reinforcement shall satisfy Article 4.3.5 stipulating, frpwcfrps dbfVV '8 kipsVdbfV sfrpwcfrp 268,42.187)9.23(8.119.388 '
B-31 117.2 4,268frpV kips kips The provision of Article 4.3.5 of the Guide is satisfied. Calculate the factored shear resistance 0.9 35.2 187.2 0.65(117.2) 276 r c s frp frpV V V V kips O.K.
B-32 Example 6 Axial Strength of a confined circular column It is required to strengthen the column shown below so that the axial compression strength is 4,000 kips. Column Data Column height: 24 ft Column diameter: 42 inches Compression Concrete Strength: 4 ksi Vertical reinforcement: 16 #8 Type of transverse reinforcement: #3 ties Tie spacing: 12 in. FRP reinforcement: Field-fabricated carbon fiber/Epoxy composites Dry fabric weight: 9 oz/yd2 Glass Transition Temperature: 165ogT F Thickness of a single layer after curing: 0.0397 .in Tensile strength of a single layer FRP reinforcement at 1% strain: 3.8 / .frpP kips in 42 " 16 # 8 Bars # 3 ties at 12 in. spacing 2" Figure B-14 Reinforcement details of a circular column
B-33 SOLUTION Step 1: Compute the axial strength of the column stystgcn AfAAfP '85.080.0 nnr PPP 75.0 2 2 2(42) 1385 . 4 4g DA in 264.12)79.0(16 inAst 0.80 0.85(4) 1385 12.64 (60)(12.64) 4,340nP kips 0.75(4,340) 3,255 r nP P kips Step 2: Compute the FRP reinforcement strength at a strain of 0.004. 0.004 3.8 1.52 / . 0.01frpo N kip in Step 3: Compute the required confined concrete strength Using Eq. 5.3.1-1 of the Guide Specifications: ustystgccr PAfAAfP '85.080.0 From which ksi AA AfP f stg sty u cc 07.546.12138585.0 )46.12)(60()75.0)(8.0( 4000 85.0 80.0 ' ' ' 2 ' 1 5.07lcc c c ff f ksif
B-34 24 1 5.07 4 lf ksif l 535.0 As per Article 5.3.2.2, the confinement pressure shall be greater or equal to 600 psi but less than that specified in Eq. 5.3.3.3-2 as follows ' 1 4 10.6 1 1 1.33 2 2 (0.8)(0.75) c l e ff ksi ksi k O.K. D Nf frpfrpl 2 (0.6)(42) 19.38 / . 2 2(0.65) l frp frp f DN kip in Required number of layers 19.38 12.75 1.52 frp frpo N n N Use 13 layers that will have a thickness of 0.507 .in . Accordingly, the column axial strength is computed as follows: ksi k fksi D Nf e cfrp frpl 33.11 1 2 611.0 42 )52.1)(13(265.02 ' ksif fff c l ccc 22.54 )611.0(21)4(21'' ' kipsAfAAfP stystgccn 5470)46.12)(60()46.121385)(22.5(85.08.0'85.080.0 kipskipsPP nr 000,4102,4)470,5(75.0
