Simplified Shear Design of Structural Concrete Members: Appendixes (2005)

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Appendix J: Examples of Shear Design The purpose of these design examples is to demonstrate the application of the shear design methods developed in Tasks 4 through 6, i.e., Proposal 1 (Modified STD Approach) and Proposal 2 (Modified CSA Approach). These design examples are based on real bridges that have been constructed, or on published design examples such as those in the PCI Bridge Design Manual. These design examples show the shear design of three types of superstructure: simple-span non-composite precast pretensioned box beams with straight debonded strands, pretensioned I- beams with draped strands made continuous with unstressed reinforcement, and continuous cast- in-place post-tensioned box beams. The design examples also include the shear design of three types of non-prestressed substructure elements: cap beams, columns, and footings. Section 5.8.3.2 of the AASHTO LRFD Specifications allows the location of the critical section for shear to be taken as the larger of 0.5dvcot(θ) or dv from the internal face of the support. However, the 0.5dvcot(θ) value requires an initial assumption as to the value of theta and is therefore another drawback to use of the LRFD Sectional Design Model. In the following examples, the critical sections used are those used by the designers who provided the case studies on which these examples are based. The moments and shears acting at the specified critical sections are those calculated by the original designers. J-1

J.1 Example 1: Precast, Pretensioned Non-Composite Box Beam J.1.1 Example Description This example demonstrates the shear design at a specific section of a 95-ft single-span AASHTO Type BIII-48 box beam bridge with no skew. The bridge is that of Example 9.2 of the PCI Bridge Design Manual. This non-composite pretensioned beam is 39-inch deep and simply supported. All strands are straight and some are debonded for a distance of 5ft from the end of the beam. The shear design is accomplished in accordance with the Proposal 1 (Modified STD Approach) and Proposal 2 (Modified CSA Approach). J.1.2 Geometry and Loading The beam is simply supported on a 95-ft single-span and is part of a bridge for which the superstructure consists of seven beams abutted as shown in Fig. J-1. The individual beams are transversely post-tensioned together to form the bridge through 8-in.-thick full-depth diaphragms located at quarter-points. The design live load is HL-93. As shown in Fig. J-2, the section at a distance of 42.74 in. from support is considered for shear design in this example. Typical interior beam for design Figure J-1 Bridge Cross-Section Location of design section Figure J-2 Location of design section J-2

J.1.3 Material Properties The material properties are given in Table J-1. Table J-1 Material Properties CONCRETE PROPERTIES Concrete strength at 28 days, f’c 5 ksi Concrete unit weight, wc 0.150 kcf Modulus of elasticity of concrete, Ec = '5.1)(000,33 cc fw [LRFD Eq. 5.4.2.4-1] 4,287ksi PRESTRESSING STRANDS Type 0.5 in. dia., seven-wire, low-relaxation Area of a strand 0.153 in2 Ultimate strength, fpu 270.0 ksi Yield strength, fpy (=0.9 fpu) [LRFD Table 5.4.4.1-1] 243.0 ksi A parameter for prestressing, fpo = 0.7 fpu 189 ksi Effective prestress after all loses, fse 171.6 ksi Modulus of elasticity, Ep [LRFD Art. 5.4.4.2] 28,500 ksi REINFORCING BARS Yield strength, fy #3 Grade 60 ksi Modulus of elasticity, Es [LRFD Art. 5.4.3.2] 29,000 ksi J.1.4 Sectional Properties and Forces The sectional properties and forces are given in Table J-2. Fig. J-3 provides detailed dimensions of the cross-section of AASHTO Box Beam Type BIII-48. The strand patterns at midspan and at the design section are shown in Figs. J-4 and J-5, respectively. Other basic calculations are also provided. Table J-2 Sectional Properties and Forces OVERALL GEOMETRY AND SECTIONAL PROPERTIES Span length, L 95 ft Overall depth of girder, h 39 in. Width of Web, bv 10 in. (Width of each leg is 5 in.) Area of cross-section of girder, Ag 813 in.2 Moment of inertia, Ig 168,367 in.4 Distance from centroid to extreme bottom fiber, yb 19.29 in. Distance from centroid to extreme top fiber, yt 19.71 in. J-3

Section modulus for the extreme bottom fiber, Sb 8,728 in3 Section modulus for the extreme top fiber, St 8,542 in3 Area of non-prestressed tension reinforcement, As 0 in2 Distance from extreme compression fiber to centroid of longitudinal tension reinforcement, ds 0 in Area of prestressed tension reinforcement, Ap =29(0.153)=4.437 in2 Distance from the top fiber to the centroid of prestressed tendons, dp 36.59 in Weight of beam 0.847 kip/ft SECTIONAL FORCES AT DESIGN SECTION Unfactored shear force caused by dead load, Vd 47.6 kips Factored shear force, Vu 146.5 kips Unfactored moment caused by dead load, Md 176.0 ft-kips Factored moment, Mu 424.6 ft-kips Figure J-3 Cross-Section of AASHTO Box Beam Type BIII-48 J-4

Figure J-4 Strand Pattern at Midspan Figure J-5 Strand Pattern at the Design Section Calculation of effective depth, : vd Note that only 22 strands (16@2 in. and 6@4 in.) are effective at the design section for shear, because 7 strands are debonded and the top level of strands is ignored. See Fig. J-5. The equivalent compressive block depth, , is calculated by flexural analysis ( =9.03 in.). a a The center of gravity of strand in tension at the design section is: 16(2) 6(4) 2.55 22bs y i+= = n n n n n Therefore, 39 2.55 36.45ed i= − = vd is the greater of: / 2 [36.45 0.5(9.03)] 31.94ed a in− = − = (Controls) 0.9 0.9(36.45) 32.81ed i= = 0.72 0.72(39) 28.08h i= = Therefore, . 32.81vd i= J-5

The design shear stress is: 146.5 0 0.496 (0.9)(10)(32.81) u p u v v V V v k b d si φ φ − −= = = Thus, '/ 0.496 / 5 0.099u cv f = = J.1.5 Shear Design by Proposal 1: Modified STD Approach Shear design procedures in accordance with the Modified STD Approach are used to determine the amount and spacing of the shear reinforcement required at a section located 42.74 in. from the support. a) Evaluation of Web-Shear Cracking Strength Compute web-shear cracking strength, : cwV '(0.06 0.3 )cw c pc v v pV f f b d= + V+ ips The effective prestress force is: 24(0.153)(171.6) 630.1 seP k= = Compressive stress in concrete at the centroid of cross section due to prestress is: 630.1 0.775 813 se pc g Pf ksi A = = = Therefore, [0.06 5.0 0.3(0.775)](10)(32.81) 0 120.3cwV k= + + = ips b) Evaluation of Flexure-Shear Cracking Strength Compute flexure-shear cracking strength, : ciV ' max 0.02 crci c v v d i MV f b d V V M = + + '0.06 c v vf b d≥ , 146.5 47.6 98.9i u dV V V kips= − = − = max 424.6 176.0 248.6u dM M M ft kips= − = − = ⋅ . Moment causing flexural cracking at the design section due to externally applied loads: '(0.2 )gcr c pe d b I M f f f y = + − The center of gravity of strand pattern at the design section is: 16(2) 6(4) 2(36) 5.33 24bs y i+ += = n The eccentricity of the strands at the design section is: 19.29 5.33 13.96 .b bse y y in= − = − = Compressive stress in concrete due to effective prestress forces only: 630.1 630.1(13.96)(19.29) =1.783 813 168,367 se se b pe g g P P eyf ksi A I = + = + J-6

Stress due to service dead load: 176.0(12)(19.29) 0.242 168,367 d b d g M yf ksi I = = = Therefore, 168,367 (0.2 5.0 1.783 0.242) /12 1,446.1 19.29cr M ft kips= + − = ⋅ The flexure-shear cracking strength, , is: ciV 98.9(1,446.1)0.02 5.0(10)(32.81) 47.6 637.6 248.6ci V k= + + = ips '0.06 0.06 5.0(10)(32.81) 44.0 c v vf b d kips> = = c) Evaluation of Concrete Contribution The nominal shear strength provided by concrete is the lesser of and . ciV cwV Web-shear cracking strength (Governs), 120.3cwV = kips ips ips Flexure-shear cracking strength 637.6 ciV k= Thus, the nominal shear strength provided by concrete is: 120.3 cV k= d) Evaluation of Required Transverse Reinforcement Check if 0.5u cV Vφ> 146.5 0.5 0.5(0.9)(120.3) 54.1u cV kips V kipsφ= > = = Therefore, transverse reinforcement must be provided. The shear force required is: ( / ) (146.5 / 0.9) 120.3 42.5s u cV V V kipsφ= − = − = The shear strength provided by transverse reinforcement is: s dfA V vyvs ααθ sin)cot(cot += [LRFD Eq. 5.8.3.3-4] When vertical stirrups are used, α =90. Since governs and cwV 424.6 1,446.1u crM ft kips M ft kips= ⋅ < = ⋅ , the angle of compressive strut is obtained as 'cot 1 3 / 1.80pc cf fθ = + ≤ cot 1 3(0.775) / 5.0 2.04 1.80θ = + = > (Governs) Therefore, cot 1.80θ = . The area of transverse reinforcement (in2) within a spacing (s) is: (use fy = 60 ksi) 2/ /( cot ) 42.5 /[(60)(32.81)(1.80)] 0.0120 . / .v s y vA s V f d in inθ= = = J-7

Therefore, use #3 single leg stirrups in each side of web at 18 in. spacing provided > required /vA s 20.22 /18 0.0122 . / .in in= = /vA s 20.0120 . / .in in= Then, provided 0.22 60 32.81 1.80 43.3 18s V k× × ×= = ips e) Checks Maximum Spacing Limit of Transverse Reinforcement Maximum spacing of transverse reinforcement shall not exceed the following: When , [LRFD Eqs. 5.8.2.7-1] '/ 0.099 0.125u cv f = < max 0.8 24.0 .vs d= ≤ in in n n in max 24 .s ≤ (controls) 0.8 (0.8)(32.81) 26.25 .vd i≤ = = Since < O.K. 18 .s i= max 24 .s ≤ Minimum Reinforcement Requirement The area of transverse reinforcement shall not be less than: ' 2,min (10)(18)0.0316 0.0316 5.0 0.212 60 v v c y b sA f f ≥ = = in < provided 20.22vA in= O.K. Maximum Nominal Shear Resistance In order to ensure that the concrete in the web of the girder will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications specify an upper limit of as follows: nV V f ( 0V in this example) '0.25n c v v pb d V≤ + p = = + = 120.3 43.3 163.6nV kips '0.25 0.25(5.0)(10)(32.81) 410.1c v vf b d kips< = = O.K. J.1.6 Shear Design by Proposal 2: Modified CSA Approach Shear design procedures in accordance with the Modified CSA Approach are used to determine the amount and spacing of the shear reinforcement required at a section located at a distance of 42.74 in. from the support. a) Evaluation of xε Calculate the strain in the reinforcement on the flexural tension side, xε : / 0.5 0.002 2( ) u v u u p ps po x s s p ps M d N V V A f E A E A ε + + − −= ≤+ 3424.6(12) / 32.81 0 146.5 0 3.366(189) 1.743 10 2[(0 28,500(3.366)]x ε −+ + − −= =+ − × Since the value of xε is negative, a different equation must be used: / 0.5 0.002 2( ) u v u u p ps po x c c s s p ps M d N V V A f E A E A E A ε + + − −= ≤+ + J-8

where Ac = area of concrete on the flexural tension side as shown in Fig. J-6. = 2(0.5)(813) 406.5 in= Figure J-6 Illustration of Ac 5424.6(12) / 32.81 0 146.5 0 3.366(189) 9.1 10 2[(4,287)(406.5) (28,500)(3.366)]x ε −+ + − −= =+ − × b) Evaluation of β and θ Calculate θ from the longitudinal strain, xε . 29 7,000 xθ ε= + 529 7,000( 9.1 10 ) 28.4−= + − × = ° Assume that at least minimum shear reinforcement is provided. Then, the coefficient, β , is: 4.8 (1 1500 )x β ε= + 5 4.8 5.56 [1 1,500( 9.1 10 )]− = =+ − × c) Evaluation of Concrete Contribution The contribution of the concrete to the nominal shear resistance is: vvcc dbfV '0316.0 β= 0.0316(5.56) 5.0(10)(32.81) 128.9 kips= = d) Evaluation of Required Transverse Reinforcement Check if 0.5 ( )u cV V pVφ> + 146.5 0.5 ( ) (0.5)(0.9)(128.9 0) 58.0u c pV kips V V kipsφ= > + = + = Therefore, transverse shear reinforcement must be provided. The shear resistance to be provided by transverse reinforcement is: ( / ) (146.5 / 0.9) 128.9 0 33.9s u c pV V V V kipsφ= − − = − − = J-9

The shear strength provided by transverse reinforcement is: s dfA V vyvs ααθ sin)cot(cot += [LRFD Eq. 5.8.3.3-4] When vertical stirrups are used, 90α = ° . Then, the required area of transverse reinforcement within a spacing (s) is: (use yf = 60 ksi) 2/ /( cot ) 33.9 /[(60)(32.81)cot 28.4 ] 0.0093 . / .v s y vA s V f d in inθ= = ° = Therefore, use #3 single leg stirrups in each side of the web at 24 in. spacing provided /vA s 20.22 / 24 0.0092 . / .in in= = ≈ / required vA s 20.0093 . / .in in= Then, 0.22 60 32.81 cot 28.4 33.4 24s V k× × × °= = ips e) Checks Maximum Spacing Limit of Transverse Reinforcement Maximum spacing of transverse reinforcement may not exceed the following: When , [LRFD Eqs. 5.8.2.7-1] '/ 0.099 0.125u cv f = < max 0.8 24.0 .vs d= ≤ in in n n in (controls) max 24 .s ≤ 0.8 (0.8)(32.81) 26.25 .vd i≤ = = Since and O.K. 24 .s i= max 24 .s ≤ Minimum Transverse Reinforcement Requirement The area of transverse reinforcement should not be less than: ' 2 ,min (10)(24)0.0316 0.0316 5.0 0.283 60 v v c y b sA f f ≥ = = in > provided 20.22vA in= Therefore minimum transverse reinforcement governs. Use #3 single leg stirrups in each web at 18 in. spacing. Then, provided > required /vA s 20.22 /18 0.0122 . / .in in= = /vA s 20.283/ 24 0.0118 . / .in in= = And 0.22 60 32.81 cot 28.4 44.5 18s V k× × × °= = ips Maximum Nominal Shear Resistance In order to ensure that the concrete in the web of the girder will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications specify an upper limit on Vn as follows: V f ( 0V in this example) '0.25n c v v pb d V≤ + p = = + + = 128.9 44.5 0 173.4nV kips '0.25 0.25(5.0)(10)(32.81) 410.1c v vf b d kips< = = O.K. J-10

J.1.7 Summary and Conclusions Shear design procedures in accordance with the Modified STD Approach and the Modified CSA Approach were used to determine the amount and spacing of the transverse reinforcement required at a section 42.74 in. from the support of a 95-ft single-span AASHTO Type BIII-48 box beam with straight strands and of which seven strands are debonded at the design section.. The shear design results are summarized in Table J-3. In this design example, the amounts of transverse reinforcement required by the two approaches are very similar. Table J-3 Summary of Results Required or calculated Proposal 1: Modified STD Approach Proposal 2: Modified CSA Approach cV , kips 120.3 128.9 sV , kips 42.5 33.9 θ , deg. 29.0 28.4 Reinforcement Provided single leg (each web) #3 bars @18 inches single leg (each web) #3 bars @18 inches J-11

J.2 Example 2: Three-Span Continuous Precast, Pretensioned Girders J.2.1 Example Description This example is based on Example 9.6 of the PCI Bridge Design Manual. The bridge uses 72-inch bulb -tee beams with harped (draped) pretensioned strands on 110-feet end spans and a 120-feet interior span. The beams are made continuous for live load by the addition of non- prestressed reinforcement in the deck in the negative moment region. This example illustrates the shear design of the negative moment region of a beam made continuous with non-prestressed reinforcement. J.2.2 Geometry and Loading As shown in Fig. J-7, the bridge is a three-span continuous structure and is designed to act compositely to resist all superimposed dead loads, live loads and impact. Design live load is AASHTO LRFD HL-93. The superstructure consists of four beams spaced at 12 ft centers as shown in Fig. J-8. The section at a distance of 7.10 ft from support, as indicated in Fig. J-7, is designed for shear in this example. Location of design section (7.10 ft from support) Figure J-7 Span Geometry Typical interior beam for design Figure J-8 Bridge Cross-Section J.2.3 Material Properties The material properties are given in Table J-4. J-12

Table J-4 Material Properties and Basic Information CONCRETE PROPERTIES Concrete strength of girder at 28 days, f’c 7.0 ksi Concrete strength of deck at 28 days, f’c 4.0 ksi Concrete unit weight, wc 0.150 kcf Modulus of elasticity of concrete for girder, Ec,beam = '5.1)(000,33 cc fw [LRFD Eq. 5.4.2.4-1] 5,072 ksi Modulus of elasticity of concrete for deck, Ec,slab = '5.1)(000,33 cc fw [LRFD Eq. 5.4.2.4-1] 3,834 ksi Modular ratio between slab and beam concrete, n = Ec,slab / Ec,beam = 0.7559 PRESTRESSING STRANDS Type 0.5 in. dia., seven-wire, low-relaxation Area of a strand 0.153 in2 Ultimate strength, fpu 270.0 ksi Yield strength, fpy (=0.9 fpu) [LRFD Table 5.4.4.1-1] 243 ksi A parameter for prestressing, fpo = 0.7 fpu 189 ksi Effective prestress after all loses, fse 152.9 ksi Modulus of elasticity, Ep [LRFD Art. 5.4.4.2] 28,500 ksi REINFORCING BARS Yield strength, fy #5 Grade 60 ksi Modulus of elasticity, Es [LRFD Art. 5.4.3.2] 29,000 ksi J.2.4 Sectional Properties and Forces The sectional properties and forces are given in Table J-5. Fig. J-9 provides details of the dimensions of the cross-section of a AASHTO Type IV beam and Fig. J-10 provides details for the composite section. The typical strand patterns at midspan and at the end are shown in Fig. J- 11. Note that 12 strands are draped as shown in Fig. J-11. Some basic calculations are also provided in this sub-section. Table J-5 Sectional Properties and Forces OVERALL GEOMETRY AND SECTIONAL PROPERTIES Non-Composite Section Span length, L 120 ft Overall depth of girder, h 72 in. Width of web, bv 6 in. Area of cross-section of girder, Ag 767 in.2 Moment of inertia, Ig 545,894 in.4 Distance from centroid to extreme bottom fiber, yb 36.60 in. J-13

Distance from centroid to extreme top fiber, yt 35.40 in. Section modulus for the extreme bottom fiber, Sb 14,915 in3 Section modulus for the extreme top fiber, St 15,421 in3 Weight of beam 0.799 kip/ft Composite Section Overall depth of the composite section, hc 80 in. Slab thickness, ts 8.0 in. Total transformed area of the composite section, Ac 1,412 in.2 Moment of inertia of the composite section, Ic 1,097,252 in.4 Distance from centroid of the composite section to extreme bottom fiber, ybc 54.67 in. Distance from centroid of the composite section to extreme top fiber of beam, ytg 17.33 in. Distance from centroid of the composite section to extreme top fiber of slab, ytc 25.33 in. Composite section modulus for the extreme bottom fiber of beam, Sbc 20,070 in3 Composite section modulus for the extreme top fiber of beam, Stg 63,315 in3 Composite section modulus for the extreme top fiber of slab, Stc 57,307 in3 Area of non-prestressed tension reinforcement, As 0 in2 Distance from extreme compression fiber to centroid of longitudinal tension reinforcement, ds 0 in Area of prestressed tension reinforcement, Ap =44(0.153)=6.732 in2 Distance from the top fiber to the centroid of prestressed tendons, dp 74.18 in SECTIONAL FORCES AT DESIGN SECTION Unfactored shear force due to beam weight, Vdg 42.3 kips Unfactored shear force due to deck slab, Vds 64.6 kips Unfactored shear force due to superimposed dead load, Vdw 22.0 kips Unfactored shear force due to total dead load, Vd 128.9 kips Factored shear force, Vu 405.0 kips Unfactored moment due to beam weight, Mdg 272.7 ft-kips Unfactored moment due to deck slab, Mds 417.1 ft-kips Unfactored moment due to superimposed dead load, Mdw -384.0 ft-kips Unfactored moment due to total dead load, Md 305.8 ft-kips Factored moment, Mu -2,877.6 ft-kips J-14

Figure J-9 Cross-Section of AASHTO-PCI, BT-72 Bulb-Tee Beam Figure J-10 Composite Section J-15

Figure J-11 Strand Pattern at Midspan and at Ends Calculation of effective depth, : vd Note that the design section (7.10 ft from the interior support) is located in the negative moment zone. Thus, only the nonprestressed reinforcement in the slab is considered as the main reinforcement for flexure and the prestress reinforcement is neglected. The area of nonprestressed steel on the flexural tension side of the member is and the effective depth is . The equivalent compressive block depth, , is calculated by flexural analysis ( a =6.02 in.). 215.52sA = in in n n n ips= 76.25ed = a vd is the greater of: (Controls) / 2 [76.25 0.5(6.02)] 73.24ed a in− = − = 0.9 0.9(76.25) 68.63ed i= = 0.72 0.72(80.0) 57.60h i= = Therefore, . 73.24vd i= The vertical component of the effective prestressing force, , due to the 12 draped strands (with an angle of ) is: pV 7.2° 12(0.153)(152.9)sin 7.2 35.2pV k= ° The design shear stress is: 405.0 0.9(35.2) 0.944 (0.9)(6)(73.24) u p u v v V V v k b d si φ φ − −= = = Thus, . '/ 0.944 / 7.0 0.135u cv f = = J.2.5 Shear Design by Proposal 1: Modified STD Approach J-16

Shear design procedures in accordance with the Modified STD Approach are used to determine the amount and spacing of the shear reinforcement required at a distance of 7.10 ft from the interior support. a) Evaluation of Web-Shear Cracking Strength Compute web-shear cracking strength, : cwV '(0.06 0.3 )cw c pc v v pV f f b d= + V+ ips The effective prestress force is: 44(0.153)(152.9) 1,029.3 seP k= = The eccentricity of the strands at the design section is 18.79 in. The compressive stress in the concrete at the centroid of the cross section due to both pretensioning and moments and resisted by precast member alone is: ( )(( ) dg ds bc bse se bc b pc g g g )M M y yP P e y yf A I I + −−= − + 1,029.3 1,029.3(18.79)(54.67 36.60) (272.7 417.1)(12)(54.67 36.60) 767 545,894 545,894 0.976 pcf ksi − + −= − + = Therefore, [0.06 7.0 0.3(0.976)](6)(73.24) 35.2 233.6cwV k= + + = ips b) Evaluation of Flexure-Shear Cracking Strength Compute flexure-shear cracking strength, : ciV ' max 0.02 crci c v v d i MV f b d V V M = + + '0.06 c v vf b d≥ , 405.0 128.9 276.1i u dV V V kips= − = − = max 2,877.6 305.8 3,183.4u dM M M ft kip= − = − − = − ⋅ s . Moment causing flexural cracking at the design section due to externally applied loads: '(0.2 )ccr c pe d tc IM f f f y = + − Because the beam at this section is under net negative moment, the compressive stress in the concrete due to effective prestress forces only, at the extreme fiber of the section where tensile stress is caused by externally applied loads, pef , should be evaluated at the top of the deck slab. Pretension has no effect on the deck slab, therefore, pef = 0. Stress due to service dead load: 384.0(12)(25.33) 0.106 1,097,252 dw tc d c M yf ksi I −= = = Note that because this section is under net negative moment dwM was evaluated conservatively by considering the dead load negative moment component as that resulting from the dead load acting on a continuous span. Therefore, J-17

1,097,252 (0.2 4.0 0 0.106) /12 1,061.3 25.33cr M ft kips= + − = ⋅ The flexure-shear cracking strength, , is: ciV 276.1(1,061.3)0.02 7.0(6)(73.24) 128.9 244.2 3,183.4ci V k= + + = ips '0.06 0.06 7.0(6)(73.24) 69.76 c v vf b d kips> = = c) Evaluation of Concrete Contribution The nominal shear strength provided by the concrete is the lesser of and . ciV cwV Web-shear cracking strength 233.6cwV kips= (Governs) Flexure-shear cracking strength 244.2 ciV kips= Thus, the nominal shear strength provided by concrete is: 233.6 cV k= ips d) Evaluation of Required Transverse Reinforcement Check if 0.5u cV Vφ> 405.0 0.5 0.5(0.9)(233.6) 105.1u cV kips V kipsφ= > = = Therefore, transverse reinforcement must be provided. The shear force required is: ( / ) (405.0 / 0.9) 233.6 216.4s u cV V V kipsφ= − = − = The shear strength provided by transverse reinforcement is: s dfA V vyvs ααθ sin)cot(cot += [LRFD Eq. 5.8.3.3-4] When vertical stirrups are used, α =90. Since 2,877.6 1,061.3u crM ft kips M ft kips= ⋅ > = ⋅ , the angle of compressive strut is 45 . ° Therefore, the area of transverse reinforcement (in2) within a spacing (s) is: (use fy = 60 ksi) 2/ /( cot ) 216.4 /[(60)(73.24)(cot 45 )] 0.0492 . / .v s y vA s V f d in inθ= = ° = Therefore, use #5 double leg stirrups at 12 in. spacing provided > required /vA s 22(0.31) /12 0.0517 . / .in in= = /vA s 20.0492 . / .in in= Then, provided 0.62(60)(73.24)cot 45.0 227.0 12s V k°= = ips e) Checks Maximum Spacing Limit of Transverse Reinforcement Maximum spacing of transverse reinforcement shall not exceed the following: J-18

When , [LRFD Eqs. 5.8.2.7-1] '/ 0.135 0.125u cv f = > max 0.4 12.0 .vs d= ≤ in in n in n (controls) max 12 .s ≤ 0.4 (0.4)(73.24) 29.30 .vd i≤ = = Therefore, max 12 .s ≤ Actual spacing, O.K. 12 .s i= Minimum Reinforcement Requirement The area of transverse reinforcement shall not be less than: ' 2,min (6)(12)0.0316 0.0316 7.0 0.1003 60 v v c y b sA f f ≥ = = in < provided 20.62vA in= O.K. Maximum Nominal Shear Resistance In order to ensure that the concrete in the web of the girder will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications specify an upper limit of Vn as follows: '0.25n c v vV f b d≤ + pV ips233.6 227.0 460.6nV k= + = '0.25 0.25(7.0)(6)(73.24) 35.2 804.2c v v pf b d V kips< + = + = O.K. J.2.6 Shear Design by Proposal 2: Modified CSA Approach Shear design procedures in accordance with the Modified CSA Approach are used to determine the amount and spacing of the shear reinforcement required at a distance of 7.10 ft from the interior support. a) Evaluation of xε Calculate the strain in the reinforcement on the flexural tension side, xε : / 0.5 0.002 2( ) u v u u p ps po x s s p ps M d N V V A f E A E A ε + + − −= ≤+ 32,877.6(12) / 73.24 0 405.0 35.2 12(0.153)(189) 0.492 10 2[29,000(15.52) 28,500(12)(0.153)] −+ + − −= =+ × n Note that there are 12 draped strands on the flexural tension side of the member which is taken as the half-depth of the member (LRFD Figure 5.8.3.4.2-3). Thus, 212(0.153) 1.836psA i= = . b) Evaluation of β and θ Calculate θ from the longitudinal strain, xε . 29 7,000 xθ ε= + 329 7,000(0.492 10 ) 32.4−= + × = ° Assume that at least minimum amount of shear reinforcement is provided. Then, the coefficient, β , is: J-19

4.8 (1 1500 )x β ε= + 3 4.8 2.76 [1 1,500(0.492 10 )]− = =+ × c) Evaluation of Concrete Contribution The contribution of the concrete to the nominal shear resistance is: vvcc dbfV '0316.0 β= 0.0316(2.76) 7.0(6.0)(73.24) 101.4 kips= = d) Evaluation of Required Transverse Reinforcement Check if 0.5 ( )u cV V pVφ> + 405.0 0.5 ( ) 0.5(0.9)(101.4 35.2) 61.5u c pV kips V V kipsφ= > + = + = Therefore, transverse shear reinforcement must be provided. The shear resistance to be provided by transverse reinforcement is: ( / ) (405.0 / 0.9) 101.4 35.2 313.4s u c pV V V V kipsφ= − − = − − = The shear strength provided by transverse reinforcement is: s dfA V vyvs ααθ sin)cot(cot += [LRFD Eq. 5.8.3.3-4] When vertical stirrups are used, 90α = ° . Then, the area of transverse reinforcement within a spacing (s) is: (use yf = 60 ksi) 2/ /( cot ) 313.4 /[(60)(73.24)cot 32.4 ] 0.0453 . / .v s y vA s V f d in inθ= = ° = Therefore, use #5 double legs stirrups at 12 in. spacing / provided > required vA s 22(0.31) /12 0.0517 . / .in in= = /vA s 20.0453 . / .in in= Then, provided 0.62(60)(73.24)cot 32.4 357.8 12s V k°= = ips e) Checks Maximum Spacing Limit of Transverse Reinforcement Maximum spacing of transverse reinforcement shall not exceed the following: J-20

When , [LRFD Eqs. 5.8.2.7-1] '/ 0.135 0.125u cv f = > max 0.4 12.0 .vs d= ≤ in in n in n (controls) max 12 .s ≤ 0.4 (0.4)(73.24) 29.30 .vd i≤ = = Therefore, max 12 .s ≤ Actual spacing, O.K. 12 .s i= Minimum Reinforcement Requirement The area of transverse reinforcement shall not be less than: ' 2,min (6)(12)0.0316 0.0316 7.0 0.100 60 v v c y b sA f f ≥ = = in < provided 20.62vA in= O.K. Maximum Nominal Shear Resistance In order to ensure that the concrete in the web of the girder will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications specify an upper limit on Vn as follows: '0.25n c v vV f b d≤ + pV ips 101.4 357.8 35.2 494.4nV k= + + = '0.25 0.25(7.0)(6)(73.24) 35.2 804.2c v v pf b d V kips< + = + = O.K. J.2.7 Summary and Conclusions Shear design procedures in accordance with the Modified STD Approach and the Modified CSA Approach are used to determine the amount and spacing of the transverse reinforcement required at a given section of a 120-ft span 72-inch bulb-tee beam with harped (draped) pretensioned strands and that was made continuous for live load. Shear design results are summarized in Table J-6. While the concrete contribution, (including ), calculated by the Modified STD Approach was about 70% greater than that by the Modified CSA Approach, the difference in the amount of transverse reinforcement required by these two approaches is very small. pV Table J-6 Summary of Results Required or calculated Proposal 1: Modified STD Approach Proposal 2: Modified CSA Approach ( + ), kips cV pV 233.6 136.6 sV , kips 216.4 313.4 θ , deg. 45.0 32.4 Reinforcement Provided double leg #5 bars @12 inches double leg #5 bars @12 inches J-21

J.3 Example 3: Reinforced Concrete Cap Beam J.3.1 Example Description This design example demonstrates the shear design of a section of a 15-ft span cap beam sitting on three circular columns of 3-ft diameter.. The cap beam supports a 3-lane superstructure consisting of six AASHTO Type IV beams. J.3.2 Geometry and Loading Each span of the two-span continuous beam is 15 feet long as shown in Fig. J-12. The section is 39-inches wide and 39-inches deep. The beam rests on three 3-ft diameter and 14-ft long columns. It is designed for HL20 Loading. As shown in the figure, the selected section AA, at a distance of 4.5 ft from the exterior support, is to be designed for shear.. Figure J-12 Elevation View (PBS&J) J.3.3 Material Properties The material properties are given in Table J-7. Table J-7 Material Properties CONCRETE PROPERTIES Concrete strength at 28 days, f’c 3.6 ksi Concrete unit weight, wc 0.150 kcf Modulus of elasticity of concrete, Ec = '5.1)(000,33 cc fw [LRFD Eq. 5.4.2.4-1] 3,834 ksi REINFORCING BARS Yield strength, fy 60 ksi Modulus of elasticity, Es [LRFD Art. 5.4.3.2] 29,000 ksi J.3.4 Sectional Properties and Forces The sectional properties and forces are summarized in Table J-8. Fig. J-13 provides cross- section details. Other basic calculations are also provided. Table J-8 Sectional Properties and Forces OVERALL GEOMETRY AND SECTIONAL PROPERTIES Span length, L 15 ft J-22

Overall depth of beam, h 39 in. Width of web, bv 39 in. Area of cross-section of beam, Ag 1,521 in.2 Moment of inertia, Ig 192,787 in.4 Distance from centroid to extreme bottom fiber, yb 19.5 in. Distance from centroid to extreme top fiber, yt 19.5 in. Section modulus for the extreme bottom fiber, Sb 9,987 in3 Section modulus for the extreme top fiber, St 9,987 in3 Area of non-prestressed tension reinforcement, As 9.36 in2 (6-#11 bars) Distance from extreme compression fiber to centroid of longitudinal tension reinforcement, ds 35.7 in SECTIONAL FORCES AT DESIGN SECTION Unfactored shear force due to dead load, Vd 149.0 kips Factored shear force, Vu 321.7 kips Unfactored moment due to dead load, Md 103.9 ft-kips Factored moment, Mu 300.2 ft-kips Figure J-13 Cross-Section (PBS&J) Calculation of effective shear depth, : vd 35.7e sd d in= = vd is the greater of: (Controls) 0.9 0.9(35.7) 32.1ed i= = n n n 0.72 0.72(39) 28.1h i= = Therefore, . 32.1vd i= J-23

The design shear stress is: 321.7 0.286 (0.9)(39)(32.1) u u v v Vv k b dφ= = = si Thus, . '/ 0.286 / 3.6 0.079u cv f = = J.3.5 Shear Design by Proposal 1: Modified STD Approach Shear design procedures in accordance with the Modified STD Approach are used to determine the amount and spacing of the shear reinforcement required at a distance of 42.74 in. from the support. a) Evaluation of Concrete Contribution Compute nominal shear strength provided by concrete, : cV '0.06 0.06 3.6(39)(32.1) 143c c v vV f b d= = = kips d) Evaluation of Required Transverse Reinforcement Check if 0.5u cV Vφ> 321.7 0.5 0.5(0.9)(143) 64.4u cV kips V kipsφ= > = = Therefore, transverse reinforcement must be provided. The shear force required is ( / ) (321.7 / 0.9) 143 214s u cV V V kipsφ= − = − = The shear strength provided by the transverse reinforcement is: s dfA V vyvs ααθ sin)cot(cot += [LRFD Eq. 5.8.3.3-4] When vertical stirrups are used, α =90. For RC members, the angle of the compressive strut is 45 degrees; i.e., cot 1θ = Therefore, area of transverse reinforcement (in2) required within a spacing (s) is: (use fy = 60 ksi) 2/ /( cot ) 214 /[(60)(32.1)(1.0)] 0.11 . / .v s y vA s V f d in inθ= = = Use #5 four-leg stirrups at 11-in. spacing provided ≥ / required /vA s 20.31 4 /11 0.11 . / .in in= × = vA s 20.11 . / .in in= Therefore, provided 0.31 4 60 32.1 1.0 217 11s V k× × × ×= = ips e) Checks Maximum Spacing Limitation for Transverse Reinforcement Maximum spacing of transverse reinforcement shall not exceed the following: J-24

When , [LRFD Eqs. 5.8.2.7-1] '/ 0.099 0.125u cv f = < max 0.8 24.0 .vs d= ≤ in in n = 24 .in≤ max 24 .s ≤ (controls) 0.8 (0.8)(32.1) 25.7 .vd i≤ = = Since O.K. 11 .s in Minimum Reinforcement Requirement The area of transverse reinforcement shall not be less than: ' 2,min (39)(12)0.0316 0.0316 3.6 0.468 60 v v c y b sA f f = = = in < provided 21.24vA in= O.K. Maximum Nominal Shear Resistance In order to ensure that the concrete in the web of the beam will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications speicfy an upper limit on as follows: nV '0.25n cV f b≤ v vd = + = 143 217 360nV kips '0.25 0.25(3.6)(39)(32.1) 1126.7c v vf b d kips< = = O.K. J.3.6 Shear Design by Proposal 2: Modified CSA Approach Shear design procedures in accordance with the Modified CSA Approach are used to determine the amount and spacing of the shear reinforcement required at a distance of 54 in. from the support. a) Evaluation of xε The design moment is : 300.2uM kips ft= which shall not be less than the following value. /12 321.7(32.1) /12 861u vV d kips ft⋅ = = Therefore, substitute Vu dv for Mu in the xε calculation. Calculate the strain in the reinforcement on the flexural tension side, xε : / 0.5 2( ) u v u u x s s p ps M d N E A E A ε V+ += + 3861(12) / 32.1 0 321.7 1.2 10 0.002 2[29,000(9.36) 0] −+ += = ×+ ≤ OK b) Evaluation of β and θ Calculate θ from the longitudinal strain, xε . xεθ 700029 += J-25

329 7,000(1.2 10 ) 37.4−= + × = ° Assume that at least minimum required amount of shear reinforcement is provided. Then, the coefficient, β , is: 4.8 (1 1500 )x β ε= + 3 4.8 1.71 [1 1,500(1.2 10 )]− = =+ × c) Evaluation of Concrete Contribution The contribution of the concrete to the nominal shear resistance is: vvcc dbfV '0316.0 β= 0.0316(1.71) 3.6(39)(32.1) 128kips= = d) Evaluation of Required Transverse Reinforcement Check if 0.5 ( )u cV V pVφ> + , 321.7 0.5 ( ) 0.5(0.9)(128 0) 57.5u c pV kips V V kipsφ= > + = + = Therefore, transverse shear reinforcement must be provided. The shear resistance required for the transverse reinforcement is: ( / ) (321.7 / 0.9) 128 229s u cV V V kipsφ= − = − = The shear strength provided by transverse reinforcement is: s dfA V vyvs ααθ sin)cot(cot += [LRFD Eq. 5.8.3.3-4] When vertical stirrups are used, 90α = ° . Then, the area of transverse reinforcement within a spacing (s) is: (use yf = 60 ksi) 2/ /( cot ) 229 /[(60)(32.1)cot 37.4 ] 0.091 . / .v s y vA s V f d in inθ= = ° = Use #5 four-leg stirrups at 12-in. spacing provided > required /vA s 20.31 4 /12 0.103 . / .in in= × = /vA s 20.091 . / .in in= e) Checks Maximum Spacing Limitation for Transverse Reinforcement Maximum spacing of transverse reinforcement shall not exceed the following J-26

When , [LRFD Eqs. 5.8.2.7-1] '/ 0.079 0.125u cv f = < max 0.8 24.0 .vs d= ≤ in in n = 24 .in≤ (controls) max 24 .s ≤ 0.8 (0.8)(32.1) 25.7 .vd i≤ = = Since O.K. 12 .s in Minimum Reinforcement Requirement The area of transverse reinforcement shall not be less than: ' 2,min (39)(12)0.0316 0.0316 3.6 0.468 60 v v c y b sA f f = = = in < provided 21.24vA in= O.K. Maximum Nominal Shear Resistance In order to ensure that the concrete in the web of the beam will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications specify an upper limit on Vn as follows: '0.25n c v vV f b d≤ + pV (cot cot )sin 1.24 60 32.1 cot(37.4) 260 12 v y v s A f d V k s ips θ α α+ ⋅ ⋅ ⋅= = = = + + = 128 260 0 388nV kips '0.25 0 0.25(3.6)(39)(32.1) 0 1126.7c v vf b d kips< + = + = O.K. J.3.7 Summary and Conclusions The required amounts of transverse reinforcement were four-leg #5 bar stirrups @12 inches and four-leg #5 bar stirrups @11 inches for Proposal 1 and 2, respectively. Table J-9 summarizes the results. Table J-9 Summary of Results Required or Calculated Proposal 1: Modified STD Approach Proposal 2: Modified CSA Approach ( + ), kips cV pV 143 128 sV , kips 214 229 θ , deg. 45 37.4 Reinforcement Provided four-leg #5 bars @11 inches four-leg #5 bars @12 inches J-27

J.4 Example 4: Reinforced Concrete Column and Footing J.4.1 Example Description This design example demonstrates shear design for two sections of a reinforced concrete column and footing, which are part of a pier designed by Modjeski and Masters, Inc.. The shear design is accomplished in accordance with the Proposal 1 (Modified STD Approach) and Proposal 2 (Modified CSA Approach). In the shear design of the footing, only one-way action is considered for a demonstration of proposals. J.4.2 Geometry and Loading The pier consists of 4 circular columns spaced 14’-1” apart on spread footings founded on sandy soil. Fig. J-14 shows an elevation of the pier. Each column is 39-inches in diameter and 18 ft tall. The footing size is 12 . The bridge supported by the pier was designed for the HS-25 live loading of the AASHTO Standard Specifications. Maximum shear occurs on column 1 at 0.0 ft from the bottom (top face of footing) and equals . The critical section for the footing is shown in Fig. J-15 and the design shear force is 21.9 k/ft. ' 12 ' 3'× × tV =44.8 kips Figure J-14 Elevation (Modjeski and Masters, Inc.) Figure J-15 Critical Section for Footing (Modjeski and Masters, Inc.) J-28

J.4.3 Material Properties The material properties are given in Table J-10. Table J-10 Material Properties CONCRETE PROPERTIES Concrete strength at 28 days, f’c 3.0 ksi Concrete unit weight, wc 0.150 kcf Modulus of elasticity of concrete, Ec = '5.1)(000,33 cc fw [LRFD Eq. 5.4.2.4-1] 3,321ksi REINFORCING BARS Yield strength, fy 60 ksi Modulus of elasticity, Es [LRFD Art. 5.4.3.2] 29,000 ksi J.4.4 Sectional Properties and Forces The sectional properties and forces are summarized in Table J-11. Figures J-16 and J-17 provide cross-section details. Other basic calculations are also provided. Table J-11 Sectional Properties and Forces OVERALL GEOMETRY AND SECTIONAL PROPERTIES COLUMN FOOTING Length (net), L 18 ft - Overall depth (or diameter) of member, h (or dc) 39 in. 36 in. Width of web, bv 1,370 37 in.= 12 in./ft Area of cross-section of member, Ag 1,370 in.2 432 in.2/ft Moment of inertia, Ig 149,292 in.4 46,656 in.4/ft Distance from centroid to extreme bottom fiber, yb 21 in. 18 in. Distance from centroid to extreme top fiber, yt 21 in. 18 in. Section modulus for the extreme bottom fiber, Sb 7,109 in.3 2,592 in.3/ft Section modulus for the extreme top fiber, St 7,109 in.3 2,592 in.3/ft Area of non-prestressed tension reinforcement, As 5.53 in.2 (7-#8 bars) 1.08 in.2/ft (#9 bars @ 11.4 in. spacing) Distance from extreme compression fiber to centroid of longitudinal tension reinforcement, ds 34.1 in. 31.3 in. J-29

SECTIONAL FORCES AT DESIGN SECTION Unfactored shear force due to dead load, Vd 0.0 kips* 0.0 kips* Factored shear force, Vu 44.8 kips 21.9 kips/ft Unfactored moment due to dead load, Md 0.0 ft-kips* 0.0 kips/ft* Factored moment, Mu 491 ft-kips 20.7 ft-kips/ft Factored axial force, Nu 1,062 kips (compression) 0.0 kips/ft * These data are not available. For more accurate calculations, however, shear force and moment at a section due to unfactored dead load must be evaluated. Figure J-16 Cross Section of Columns (Modjeski and Masters, Inc.) Figure J-17 Cross Section of Footings (Modjeski and Masters, Inc.) Calculation of effective shear depth, : vd Column: 34.1 .e sd d in= = n n n vd is the greater of: (Controls) 0.9 0.9(34.1) 30.7 .ed i= = 0.72 0.72(39) 28.1 .h i= = Therefore, 30.7 .vd i= J-30

The design shear stress is: 44.8 0.044 (0.9)(37)(30.7) u u v v Vv k b dφ= = = si Thus, . '/ 0.044 / 3.0 0.015u cv f = = Footing: 31.3 .e sd d in= = n n n vd is the greater of: (Controls) 0.9 0.9(31.3) 28.2 .ed i= = 0.72 0.72(36) 25.9 .h i= = Therefore, 28.2 .vd i= The design shear stress is: 21.9 0.072 (0.9)(12)(28.2) u u v v Vv k b dφ= = = si Thus, . '/ 0.044 / 3.0 0.015u cv f = = J.4.5 Shear Design by Proposal 1: Modified STD Approach Column: Shear design procedures in accordance with the Modified STD Approach are used to determine the amount and spacing of the shear reinforcement required at a section 0.0 in. from the top face of column. a) Evaluation of Concrete Contribution Compute nominal shear strength provided by concrete, : cV '0.06 0.06 3.0(37)(30.7) 118c c v vV f b d= = = kips d) Evaluation of Required Transverse Reinforcement Check if 0.5u cV Vφ> 44.8 0.5 0.5(0.9)(118) 53.1u cV kips V kipsφ= > = = Therefore, transverse reinforcement is not required. However, ties (or spirals) should be spaced at least at 12-in. center-to-center for confinement of the column. Footing: Shear design procedures in accordance with the Modified STD Approach are used to determine the amount and spacing of the shear reinforcement required at a section located 23.1 in. from the edge of footing. a) Evaluation of Concrete Contribution Compute nominal shear strength provided by concrete, : cV '0.06 0.06 3.0(12)(36) 44.9c c v vV f b d= = = kips d) Evaluation of Required Transverse Reinforcement Check if u cV Vφ> J-31

21.9 0.9(44.9) 40.4u cV kips V kipsφ= < = = Therefore, shear reinforcement is not required. J.4.6 Shear Design by Proposal 2: Modified CSA Approach Column: Shear design procedures in accordance with the Modified CSA Approach are used to determine the amount and spacing of the shear reinforcement required at a distance of 0.0 in. from the top face of column. a) Evaluation of xε The design moment is 491uM kips ft= which shall not be less than the following value. /12 44.8(30.7) /12 115u vV d kips ft⋅ = = < Mu OK Calculate the strain in the reinforcement on the flexural tension side, xε : / 0.5 2( ) u v u u x s s p ps M d N E A E A ε V+ += + 4491(12) / 30.7 0.5(1062) 44.8 9.2 10 2[29,000(5.53) 0] −− += =+ − × Since the value of xε is negative, the following equation must be used: / 0.5 2( ) u v u u x c c s s M d N E A E A ε V+ += + 5491(12) / 30.7 0.5(1062) 44.8 6.0 10 2[3,321(1,370 / 2) 29,000(5.53)] −− += =+ − × where Ac = area of concrete on the flexural tension side. b) Evaluation of β and θ Calculate θ from the longitudinal strain, xε . xεθ 700029 += 529 7,000( 6.0 10 ) 28.6−= + − × = ° Assume that at least the minimum amount of shear reinforcement required is provided. Then, the coefficient, β , is: 4.8 (1 1500 )x β ε= + 5 4.8 5.28 [1 1,500( 6.0 10 )]− = =+ − × c) Evaluation of Concrete Contribution J-32

The contribution of the concrete to the nominal shear resistance is: vvcc dbfV '0316.0 β= 0.0316(5.28) 3.0(37)(30.7) 328.3kips= = d) Evaluation of Required Transverse Reinforcement Check if 0.5 ( )u cV V pVφ> + 44.8 0.5 ( ) 0.5(0.9)(328.3 0) 147.7u c pV kips V V kipsφ= < + = + = Therefore, transverse shear reinforcement is not required. If the section contains no transverse reinforcement, β should be calculated by the following equation. 5 4.8 51 4.8 51 6.7 (1 1500 ) (39 ) (1+1500( 6.0 10 )) (39 1.15)x zes β ε −= ⋅ = ⋅+ + − × + = Where the crack spacing parameter, sz, shall be taken as dv or as the maximum distance between layers of distributed longitudinal reinforcement, whichever is less. From Fig. J-16, sz = 5.9 in. and g 1.38 1.38(5.9) 1.15 0.63 a 0.63 0.75 z ze ss = = =+ + . Again, the contribution of the concrete to the nominal shear resistance is: vvcc dbfV '0316.0 β= 0.0316(6.7) 3.0(37)(30.7) 416.5kips= = a half of which is larger than / 50.0uV kipsφ = . Therefore, shear reinforcement is not required. However, ties (or spirals) should be spaced at least at 12-in. center-to-center for confinement of the column. Footing: Shear design procedures in accordance with the Modified CSA Approach are used to determine the amount and spacing of the shear reinforcement at a section 23.1 in. from the edge of footing. a) Evaluation of xε The design moment is 20.7uM kips ft= , which shall not be less than the following value. /12 21.9(28.2) /12 51.5u vV d kips ft⋅ = = Therefore, substitute Vu dv for Mu in the xε calculation. Calculate the strain in the reinforcement on the flexural tension side, xε : / 0.5 2( ) u v u u x s s p ps M d N E A E A ε V+ += + 451.5(12) / 28.2 0 21.9 7.0 10 2[29,000(1.083) 0] −+ += =+ × b) Evaluation of β and θ Calculate θ from the longitudinal strain, xε . xεθ 700029 += J-33

429 7,000(7.0 10 ) 33.9−= + × = ° Assume that at least the minimum amount of shear reinforcement required is provided. Then, the coefficient, β , is: 4.8 (1 1500 )x β ε= + 4 4.8 2.34 [1 1,500(7.0 10 )]− = =+ × c) Evaluation of Concrete Contribution The contribution of the concrete to the nominal shear resistance is: vvcc dbfV '0316.0 β= 0.0316(2.34) 3.0(12)(28.2) 43.3kips= = d) Evaluation of Required Transverse Reinforcement Check if 0.5 ( )u cV V pVφ> + 21.9 0.5 ( ) 0.5(0.9)(43.3 0) 19.5u c pV kips V V kipsφ= > + = + = Therefore, shear reinforcement is not required. J.4.7 Summary and Conclusions Table J-12 summarizes the results. In the column, the shear reinforcement is not required for either approach. However, Vc values are very different. Vc increases as the axial compressive force increases in Proposal 2 while axial force effects are not reflected in Proposal 1. For the design of the footing, both approaches yield similar results. Table J-12 (a) Summary of Results (Column) Required or Calculated Proposal 1: Modified STD Approach Proposal 2: Modified CSA Approach ( + ), kips cV pV 118 328.3 sV , kips 0 0 θ , deg. 45 28.6 Reinforcement Provided #3 bars @ 12 in. (for confinement) #3 bars @ 12 in. (for confinement) Table J-12 (b) Summary of Results (Footing) Required or Calculated Proposal 1: Modified STD Approach Proposal 2: Modified CSA Approach ( + ), kips cV pV 44.9 43.3 J-34

Reinforcement Provided Not Required Not Required J-35

J.5 Example 5: Two-Span Continuous Post-Tensioned Box Bridge in Nevada J.5.1 Example Description This design example demonstrates shear design in the vicinity of the inflection point, (0.9L from the exterior support), of two-span, cast-in-place, post-tensioned box girder bridge. BERGER/ABAM provided the initial design data. This non-composite box girder is 60-inch deep and two-span continuous. The shear design is accomplished in accordance with the Proposal 1 (Modified STD Approach) and Proposal 2 (Modified CSA Approach). J.5.2 Geometry and Loading The beams are post-tensioned with the tendon profiles illustrated in Fig. J-18. The design girder is 120 ft long as shown on the left side of the figure. The design section is located 12 ft from the center of the mid-support, which corresponds to the inflection point. Figure J-18 Tendon Profiles (BERGER/ABAM) J.5.3 Material Properties The material properties are given in Table J-13. Table J-13 Material Properties CONCRETE PROPERTIES Concrete strength at 28 days, f’c 4.0 ksi Concrete unit weight, wc 0.150 kcf Modulus of elasticity of concrete, Ec = '5.1)(000,33 cc fw [LRFD Eq. 5.4.2.4-1] 3,834 ksi POST-TENSIONING STRANDS Type 0.5 in. dia., seven-wire, low-relaxation Area of a strand 0.153 in2 Ultimate strength, fpu 270.0 ksi Yield strength, fpy (=0.9 fpu) [LRFD Table 5.4.4.1-1] 243 ksi A parameter for prestressing, fpo = 0.7 fpu 189 ksi Modulus of elasticity, Ep [LRFD Art. 5.4.4.2] 28,500 ksi J-36

REINFORCING BARS Yield strength, fy 60 ksi Modulus of elasticity, Es [LRFD Art. 5.4.3.2] 29,000 ksi J.5.4 Sectional Properties and Forces The sectional properties and forces are given in Table J-14. Fig. J-19 provides detailed dimensions of cross-section. Other basic calculations are also provided. Table J-14 Sectional Properties and Forces OVERALL GEOMETRY AND SECTIONAL PROPERTIES Span length, L 120 ft Overall depth of girder, h 60 in. Width of web, bv 84 in. Area of cross-section of girder, Ag 14,210 in.2 Moment of inertia, Ig 7,699,484 in.4 Distance from centroid to extreme bottom fiber, yb 33.12 in. Distance from centroid to extreme top fiber, yt 26.88 in. Section modulus for the extreme bottom fiber, Sb 232,472 in3 Section modulus for the extreme top fiber, St 286,439 in3 Area of non-prestressed tension reinforcement, As 18.6 in2 Distance from extreme compression fiber to centroid of longitudinal tension reinforcement, ds 56.0 in Area of prestressed tension reinforcement, Ap 53.09 in2 Distance from the bottom fiber to the centroid of prestressed tendons, dp (negative moment region) 40.8 in SECTIONAL FORCES AT DESIGN SECTION Unfactored shear force due to dead load, Vd 1,114 kips Factored shear force, Vu 2,387 kips Vertical component of the effective prestress force at the section considered, Vp 757 kips Unfactored moment due to dead load, Md 15,612 ft-kips Factored moment, Mu 18,694 ft-kips J-37

Figure J-19 Girder Cross-Section (BERGER/ABAM) Effective depth or distance from the top fiber to the centroid of the tension steel including the prestressed steel is calculated as: 42 .p ps p s y se p ps s y A f d A f d d A f A f += + in= n n n [LRFD Eqs. 5.7.3.3.1-2]. where fps is the average stress in the prestressing steel, which can be calculated from [LRFD Eqs. 5.7.3.1.1-1]. The effective depth, is the greater of: vd 0.9 0.9(42) 37.8 .ed i= = (Controls) 0.72 0.72(60) 43.2 .h i= = Therefore, . 43.2 .vd i= The design shear stress is: 2,387 0.9(757) 0.522 (0.9)(84)(43.2) u p u v v V V v k b d si φ φ − −= = = Thus, '/ 0.522 / 4 0.131u cv f = = J.5.5 Shear Design by Proposal 1: Modified STD Approach Shear design procedures in accordance with the Modified STD Approach are used to determine the amount and spacing of the shear reinforcement at a section located 11.0 ft. from the center of the central support. a) Evaluation of Web-Shear Cracking Strength Compute web-shear cracking strength, : cwV '(0.06 0.3 )cw c pc v v pV f f b d= + V+ ips The effective prestress forces are: (53.09)(178) 9,450 seP k= = Compressive stress in concrete at the centroid of cross section due to prestress is: 9, 450 0.67 14, 210 se pc g Pf ksi A = = = Therefore, [0.06 4.0 0.3(0.67)](84)(43.2) 757 1,922cwV k= + + = ips J-38

b) Evaluation of Flexure-Shear Cracking Strength Compute flexure-shear cracking strength, : ciV ' max 0.02 crci c v v d i MV f b d V V M = + + '0.06 c v vf b d≥ , 2,387 1,114 1,273i u dV V V kips= − = − = max 18,694 15,612 3,082u dM M M ft kip= − = − = ⋅ s . Moment causing flexural cracking at the design section due to externally applied loads: '(0.2 )gcr c pe d t I M f f f y = + − The eccentricity of the strands at the design section is: ( ) 40.8 (60.0 26.88) 7.7 .p te d h y in= − − = − − = Compressive stress in concrete due to effective prestress forces only: 9, 450(7.7)(26.88)0.67 =0.92 7,699,484 se se t pe g g P P eyf ksi A I = + = + Stress due to service dead load: (15,612)(12)(26.88) 0.65 7,699, 484 d t d g M yf ksi I = = = Therefore, 7,699,484 (0.2 4.0 0.92 0.65) /12 15,992 26.88cr M ft kips= + − = ⋅ The flexure-shear cracking strength, , is: ciV 1, 273(15,992)0.02 4.0(84)(43.2) 1,114 7,733 3,082ci V k= + + = ips '0.06 0.06 4.0(84)(43.2) 435 c v vf b d kips> = = c) Evaluation of Concrete Contribution The nominal shear strength provided by the concrete is the lesser of and . ciV cwV Web-shear cracking strength 1,922cwV kips= (Governs) Flexure-shear cracking strength 7,733 ciV kips= Thus, the nominal shear strength provided by the concrete is: 1,922 cV k= ips d) Evaluation of Required Transverse Reinforcement Check if 0.5u cV Vφ> 2,387 0.5 0.5(0.9)(1,922) 865u cV kips V kipsφ= > = = J-39

Therefore, transverse reinforcement must be provided. The shear force required is ( / ) (2,387 / 0.9) 1,922 730s u cV V V kipsφ= − = − = The shear strength provided by the transverse reinforcement is: s dfA V vyvs ααθ sin)cot(cot += [LRFD Eq. 5.8.3.3-4] When vertical stirrups are used, α =90. Since governs but cwV 18,694 15,992u crM ft kips M ft kips= ⋅ > = ⋅ , the angle of compressive strut is 45 degrees; i.e., cot 1θ = . Therefore, area of transverse reinforcement (in2) within a spacing (s) is:(use fy = 60 ksi) 2/ /( cot ) 730 /[(60)(43.2)(1.0)] 0.28 . / .v s y vA s V f d in inθ= = = Therefore, use double-leg #4 bars in each web at 10 in. spacing / provided > required vA s 20.2 2 7 /10 0.28 . / .in in= × × = /vA s 20.28 . / .in in= Then, provided 0.2 2 7 60 43.2 1.0 726 10s V k× × × × ×= = ips Maximum Nominal Shear Resistance In order to ensure that the concrete in the web of the girder will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications specify an upper limit on as follows: nV '0.25n c v vV f b d≤ + pV ips 1,922 726 2,648nV k= + = '0.25 0.25(4.0)(84)(43.2) 757 4,386c v v pf b d V kips< + = + = O.K. J.5.6 Shear Design by Proposal 2: Modified CSA Approach Shear design procedures in accordance with the Modified CSA Approach are used to determine the amount and spacing of the shear reinforcement required at a section located 11.0 ft . from the centerline of the center support. a) Evaluation of xε Calculate the strain in the reinforcement on the flexural tension side, xε : / 0.5 2( ) u v u u p ps po x s s p ps M d N V V A f E A E A ε + + − −= + 418,694(12) / 43.2 0 2,387 757 53.09(189) 7.82 10 2[29,000(18.6) 28,500(53.09)] −+ + − −= =+ − × Since the value of xε is negative, a different equation must be used: J-40

/ 0.5 2( ) u v u u p ps po x c c s s p ps M d N V V A f E A E A E A ε + + − −= + + Where the area of concrete on the flexural tension side is the area above mid-height for the negative moment region; i.e., Ac = 8,126 in.2 . 518,694(12) / 43.2 0 2,387 757 53.09(189) 4.8 10 2[(3,834)(8,126) (29,000)(18.6) (28,500)(53.09)]x ε −+ + − −= =+ + − × b) Evaluation of β and θ Calculate θ from the longitudinal strain, xε . xεθ 700029 += 529 7,000( 4.8 10 ) 28.7−= + − × = ° Assume that at least the minimum amount of shear reinforcement required is provided. Then, the coefficient, β , is: 4.8 (1 1500 )x β ε= + 5 4.8 5.18 [1 1,500( 4.8 10 )]− = =+ − × c) Evaluation of Concrete Contribution The contribution of the concrete to the nominal shear resistance is: vvcc dbfV '0316.0 β= 0.0316(5.18) 4.0(84)(43.2) 1,187 kips= = d) Evaluation of Required Transverse Reinforcement Check if 0.5 ( )u cV V pVφ> + 2387 0.5 ( ) 0.5(0.9)(1187 757) 874.8u c pV kips V V kipsφ= > + = + = Therefore, transverse shear reinforcement must be provided. The shear resistance required for the transverse reinforcement is: ( / ) (2,387 / 0.9) 1,187 757 708.3s u c pV V V V kipsφ= − − = − − = The shear strength provided by the transverse reinforcement is: s dfA V vyvs ααθ sin)cot(cot += [LRFD Eq. 5.8.3.3-4] When vertical stirrups are used, 90α = ° . Then, the area of transverse reinforcement within a spacing (s) is: (use yf = 60 ksi) 2/ /( cot ) 708.3/[(60)(43.2)cot 28.7 ] 0.15 . / .v s y vA s V f d in inθ= = ° = J-41

Therefore, use #3 double-leg stirrups in each web at 10 in. spacing / provided > required vA s 20.11 2 7 /10 0.154 . / .in in= × × = /vA s 20.15 . / .in in= Then, provided 0.11 2 7 60 43.2 cot 28.7 729 10s V k× × × × × °= = ips e) Checks Maximum Spacing Limitation of Transverse Reinforcement Maximum spacing of transverse reinforcement may not exceed the following: When , [LRFD Eqs. 5.8.2.7-1] '/ 0.131 0.125u cv f = > max 0.4 12.0 .vs d= ≤ in in n n (controls) max 12 .s ≤ 0.4 (0.4)(43.2) 17.3 .vd i≤ = = Since < 12 in. O.K. 12 .s i= Maximum Nominal Shear Resistance In order to ensure that the concrete in the web of the girder will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications specify an upper limit on Vn as follows: '0.25n c v vV f b d≤ + pV ips 1, 209 732 757 2,698nV k= + + = '0.25 0.25(4.0)(84)(43.2) 757 4,386c v v pf b d V kips< + = + = O.K. J.5.7 Summary and Conclusions The shear reinforcement required by Proposal 1 is almost twice as much as that required by Proposal 2. This result is mainly due to the discrepancy in the angle assumed for the compressive strut. In Proposal 1, that angle is conservatively taken as 45 degrees when the factored moment is greater than the cracking moment. Table J-15 Summary of Results Required or Calculated Proposal 1: Modified STD Approach Proposal 2: Modified CSA Approach ( + ), kips cV pV 1,922 1,944 sV , kips 730 708 θ , deg. 45 28.7 Reinforcement Provided double-leg #4 bars @ 10 in. double-leg #3 bars @ 10 in. J-42

J.6 Example 6: Shear Design Example of a Multi-Post Bent Cap J.6.1 Example Description This design example is from Tennessee DOT and is for a multi-post bent cap beam that is 86 feet wide. The beam is supported on four columns distributed at 22 ft centers below the beam. Fig. J-20 shows the elevation of the multi-post bent. The design section is taken at the internal face of the first pier in the first bay. Shear design is accomplished in accordance with the Proposal 1 (Modified CSA Approach) and Proposal 2 (Modified STD Approach). Since the design section is in the D-region, the Strut-and-Tie method is also used for shear design. Design Section Figure J-20 Elevation of a Multi-Post Bent J.6.2 Geometry and Loading Fig. J-21shows the load pattern and the design section dimensions. The design section was taken at the interior face of the first pier. Fig. J-22 gives the reinforcement layout for typical sections. 48” 64 .8 ” Design Section Figure J-21 Elevation of a Multi-Post Bent J-43

Figure J-22 Reinforcement Layout for sections through cap J.6.3 Material Properties The material properties are given in Table J-16. Table J-16 Material Properties CONCRETE PROPERTIES Concrete strength at 28 days, f’c 3.0 ksi Concrete unit weight, wc 0.150 kcf Modulus of elasticity of concrete, Ec = '5.1)(000,33 cc fw [LRFD Eq. 5.4.2.4-1] 3,321ksi REINFORCING BARS Yield strength, fy 60 ksi(#10, #6) Modulus of elasticity, Es [LRFD Art. 5.4.3.2] 29,000 ksi J.6.4. Sectional Properties and Forces The sectional properties and forces are given in Table J-17. Table J-17 Sectional Properties and Forces OVERALL GEOMETRY AND SECTIONAL PROPERTIES Span length, L 22 ft Overall depth of girder, h 64.8 in. Width of web, bv 48 in. Area of cross-section of girder, Ag 3,110.4 in.2 Moment of inertia, Ig 1,088,390.8 in.4 Distance from centroid to extreme bottom fiber, yb 32.4in. Distance from centroid to extreme top fiber, yt 32.4 in. Section modulus for the extreme bottom fiber, Sb 33,592 in3 J-44

Section modulus for the extreme top fiber, St 33,592 in3 Area of non-prestressed tension reinforcement, As 30.48 in2 Distance from extreme compression fiber to centroid of longitudinal tension reinforcement, ds 58.8 in Area of prestressed tension reinforcement, Ap Distance from the bottom fiber to the centroid of prestressed tendons, dp Weight of beam Varies with sections SECTIONAL FORCES AT DESIGN SECTION* Unfactored shear force due to dead load, Vd Factored shear force, Vu 1222.4 kips Unfactored moment due to dead load, Md Factored moment, Mu 3477.1 ft-kips *Negative moment, bottom is in compression For variable depth members, LRFD requires components of inclined flexural compression shall be considered when calculating shear resistance. The flexural compression at bottom side is: cC 3477.1 12 855 / 2 58.8 0.5(20) u c e MC k d a ×= = =− − ips Therefore the shear resistance for the flexural compression is: 3tan( ) 855 256.5 10comp c V C kipθ= = × = s in Calculation of effective depth, : vd The compressive block depth, a = 20 in 58.8e sd d in= = / 2 58.8 0.5(20) 48.8 max 0.9 max 0.9(58.8) 52.92 52.92 0.72 0.72(64.8) 46.66 e v e d a d d h − − =⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪= = = =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪=⎩ ⎭ ⎩ ⎭ The design shear stress is: 1222.4 256.5 0 0.423 (0.9)(48)(52.92) u comp p u v v V V V v k b d si φ φ − − − −= = = Thus, < 0.18 '/ 0.423/ 3.0 0.141u cv f = = J.6.5. Shear Design by Proposal 1: Modified STD Approach Shear design procedures in accordance with the Modified STD Approach are used to determine the required amount and spacing of the shear reinforcement. J-45

a) Evaluation of concrete contribution Vc '0.06c cV f= v vb d 0.06 3.0(48)(52.92) 264.0 kips= = d) Evaluation of Required Transverse Reinforcement Check if 0.5u cV Vφ> 965.9 0.5 0.5(0.9)(264.0) 118.8u comp cV V kips V kipsφ− = > = = Therefore, transverse reinforcement must be provided. The shear force required is ( ) / (965.9 / 0.9) 264 809.2s u comp cV V V V kipsφ= − − = − = The shear strength provided by the transverse reinforcement is: v y vs A f d V s = [LRFD Eq. 5.8.3.3-4] 2/ /( ) 809.2 /[(60)(52.92)] 0.255 . / .v s y vA s V f d in in= = = Therefore, use 4-#6 bars at a spacing of 4.5 in. and 24(0.44) 1.76 .vA in= = 4.5 .s in= 2( ) 1.76 / 4.5 0.39 . / .vA provided in in s = = > 2( ) 0.255 . / .vA required in in s = Then, provided 1.76 60 52.92 1242 4.5s V k× ×= = ips e) Checks Maximum Spacing Limitation on Transverse Reinforcement Maximum spacing of the transverse reinforcement shall not exceed the following: When , [LRFD Eqs. 5.8.2.7-1] '/ 0.178 0.125u cv f = ≥ max 0.4 12.0 .vs d= ≤ in in n = ≤ in (controls) max 12 .s ≤ 0.4 (0.4)(52.92) 21 .vd i≤ = = Since O.K. 4.5 .s in max 12 .s = Minimum Reinforcement Requirement The area of transverse reinforcement shall not be less than: ' 2,min (48)(4.5)0.0316 0.0316 3.0 0.197 60 v v c y b sA f f ≥ = = in < provided 21.76vA in= O.K. Maximum Nominal Shear Resistance J-46

In order to ensure that the concrete in the web of the girder will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications specify an upper limit on Vn as follows: '0.25n c v vV f b d≤ + pV 264 1242 1506c sV V kips+ = + = '0.25 0.25(3.0)(48)(52.92) 1905c v vf b d kips< = = Then 1506 ( ) / 1073n u compV kips V V kipsφ= > − = O.K. J.6.6 Shear Design by Proposal 2: Modified CSA Approach Shear design procedures in accordance with the Modified CSA Approach are used to determine the required amount and spacing of the shear reinforcement. a) Evaluation of xε Calculate the strain in the reinforcement on the flexural tension side, xε : / 0.5 0.002 2( ) u v u u p ps po x s s p ps M d N V V A f E A E A ε + + − −= ≤+ 3 3477.1(12) / 52.92 0 1222.4 256.5 0 0 0.992 10 2[(29,000)(30.48)] −+ + − − −= = × b) Evaluation of β and θ Calculate θ from the longitudinal strain, xε . xεθ 700029 += 329 7,000(0.992 10 ) 35.9−= + × = ° cot 1.381θ = Assume that at least the minimum amount of shear reinforcement required is provided. Then, the coefficient, β , is: 3 4.8 4.8 1.929 (1 1500 ) (1 (1500)(0.992 10 ))x β ε −= = =+ + × c) Evaluation of Concrete Contribution The contribution of the concrete to the nominal shear resistance is: vvcc dbfV '0316.0 β= 0.0316(1.929) 3.0(48)(52.92) 268.2kips= = d) Evaluation of Required Transverse Reinforcement Check if 0.5 ( )u cV V pVφ> + 965.9 0.5 ( ) 0.5(0.9)(268.2 0) 120.7u comp c pV V kips V V kipsφ− = > + = + = Therefore, transverse shear reinforcement must be provided. The required shear resistance for the transverse reinforcement is: ( ) / (965.9 / 0.9) 268.2 0 805s u comp c pV V V V V kipsφ= − − − = − − = J-47

The shear strength provided by the transverse reinforcement is: s dfA V vyvs ααθ sin)cot(cot += [LRFD Eq. 5.8.3.3-4] When vertical stirrups are used, 90α = ° . Then, the area of transverse reinforcement within a spacing of (s) is: (use yf = 60 ksi, cot 1.381θ = ) 2/ /( cot ) 805 /[(60)(52.92)(1.381)] 0.184 . / .v s y vA s V f d in inθ= = = Therefore, use 4-#6 bars at a spacing of 4.5 in. and 24(0.44) 1.76 .vA in= = 4.5 .s in= 2( ) 1.76 / 4.5 0.39 . / .vA provided in in s = = > 2( ) 0.184 . / .vA required in in s = Then, provided 1.76 60 52.92 1.381 1715 4.5s V k× × ×= = ips e) Checks Maximum Spacing Limitation on Transverse Reinforcement Maximum spacing of transverse reinforcement shall not exceed the following: When , [LRFD Eqs. 5.8.2.7-1] '/ 0.178 0.125u cv f = ≥ max 0.4 12.0 .vs d= ≤ in in n = ≤ in max 12 .s ≤ (controls) 0.4 (0.4)(52.92) 21 .vd i≤ = = Since O.K. 4.5 .s in max 12 .s = Minimum Reinforcement Requirement The area of transverse reinforcement should not be less than: ' 2 ,min (48)(4.5)0.0316 0.0316 3.0 0.197 60 v v c y b sA f f ≥ = = in < provided 21.76vA in= O.K. Maximum Nominal Shear Resistance In order to ensure that the concrete in the web of the girder will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications specify an upper limit on Vn as follows: V f , ( V'0.25n c v v pb d V≤ + 0p = in this example) 268.2 1715 1983c sV V kips+ = + = '0.25 0.25(3.0)(48)(52.92) 1905c v vf b d kips> = = Then 1905 ( ) / 1073n u compV kips V V kipsφ= > − = O.K. J.6.7 Shear Design (STRUT AND TIE) J-48

Figure J-23 Strut-and-Tie Model h =16.62”a h =48.18”b 44.4” w = 24” θ b P=1212.3kips θ = 47.6o b = 28.9” a P=1212.3kips T(tie) =1118.7kips P(strut)=1649.6kips (Factored load) (Factored load) Figure J-24 Strut-and-Tie Model a) Geometry Properties The bearing width is : 24 .aw i= n in in 3 3 3 6 9 6 1.27 16.62 .a bh d= + + + = + × = 64.8 64.8 16.62 48.18 .b ah h= − = − = Therefore the strut angle is: 48.18arctan( ) 47.3 44.4 oθ = = The strut width can be given as : sin( ) cos( ) 28.9 .b ab w h inθ θ= + = b) Interior forces Compressive force in strut: / sin( ) 1212.3 / sin(47.3) 1649.6strut bearingP P kipsθ= = = Tensile force in ties: J-49

/ tan( ) 1212.3 / tan(47.3) 1118.7tie bearingT P kipsθ= = = c) Strut The tensile strain in the concrete in the direction of the tension tie is: 60 0.00207 29000 y s s f E ε = = = The principal tensile strain of concrete is : 2 2 1 ( 0.002)cot 0.00207 (0.00207 0.002)(0.923) 0.00554s sε ε ε θ= + + = + + = The limiting compressive stress cuf is: ' ' 1 3 1.72 0.85 2.55 0.8 170 0.8 (170)(0.00554) c cu c ff ksi f ksiε= = = ≤ =+ + 1.72cuf ksi= Then (1.72)(48)(28.9) 2386 1649.6n cu cs strutP f A kips P kip= = = > = s s O.K d) Tie (60)(30.48) 1828.8 1118.7n y s tieP f A kips T kip= = = > = O.K f) Node Region Check The limiting compressive stress in the node region is: 'lim 0.85 (0.85)(0.70)(3.0) 1.785it cf f kφ= = = si Compressive stress in the strut is: 1649.6 ((48)(28.9)) 1.19 1.785strut strut csf P A ksi ks= = = ≤ i Compressive stress under bearing is: 1212.3 ((48)(24)) 1.05 1.785bearing bearing cf P A ksi k= = = ≤ si O.K g) Crack control reinforcement The required area of crack control reinforcement with a spacing of 12 inches is : 20.003 (36 ) (12 ) 1.296 1required in in inA ft f × ×= = t Try #6 bars spaced @ 6 inches vertically and at 12 inches horizontally: 2 20.44 2 0.44 1.32 1.296 1s in inA ft ft × += = ≥ ft , O.K. J.6.8. Summary and Conclusions The design results by the Modified STD Approach and the Modified CSA Approach are summarized in Table J-18. The required amounts of transverse reinforcement by those two approaches are very similar. Those two methods require large amounts of shear reinforcement while only minimum reinforcement is required by the strut-and-tie method. The reason is that the design section is within the D-region and strut-and-tie method is more appropriate for shear design in this region even though the LRFD Specifications allow the use of the sectional design method. J-50

calculated Approach Approach ( + ), kips cV pV 264 268.2 sV , kips 809.2 805 θ , deg. 45.0 35.9 Reinforcement Provided 4- #6 bars @4.5 inches 4- #6 bars @4.5 inches J-51 Table J-18 Summary of Results Required or Proposal 1: Modified STD Proposal 2: Modified CSA

J.7 Example 7: Type IV Beam J.7.1 Example Description This example demonstrates the shear design of a section of a 100-ft span AASHTO Type IV beam bridge. Bridge details were provided by Tim Bradberry of the Texas Department of Transportation. The bridge consists of 3 spans with each span simply supported. The composite pretensioned beams are 54-inch deep and have an 8 in. thick deck. Shear design is accomplished in accordance with Proposal 1 (Modified STD Approach) and Proposal 2 (Modified CSA Approach). J.7.2 Geometry and Loading This bridge has 3 spans, each span is simply supported, and the span lengths are 100-ft, 120-ft, and 100-ft as shown in Fig. J-25. Design live load is HS-20, and the section at a distance of 4.93 ft from the support, as marked in Fig. J-25, is designed for shear. As shown in Fig. J-26, the typical interior beam among five beams is designed in this example. . Location of designed section (4.93 ft from support) Figure J-25 Bridge Span Geometry The interior beam for design Figure J-26 Bridge Cross-Section J.7.3 Material Properties J-52

The material properties are given in Table J-19. Table J-19 Material Properties CONCRETE PROPERTIES Concrete strength of girder at 28 days, f’c 6.5 ksi Concrete strength of deck at 28 days, f’c 4.0 ksi Concrete unit weight, wc 0.150 kcf Modulus of elasticity of concrete for girder, Ec,beam = '5.1)(000,33 cc fw [LRFD Eq. 5.4.2.4-1] 4,888 ksi Modulus of elasticity of concrete for deck, Ec,slab = '5.1)(000,33 cc fw [LRFD Eq. 5.4.2.4-1] 3,834 ksi Modular ratio between slab and beam concrete, n = Ec,slab / Ec,beam = 0.784 PRESTRESSING STRANDS Type 0.5 in. dia., seven-wire, low-relaxation Area of a strand 0.153 in2 Ultimate strength, fpu 270.0 ksi Yield strength, fpy (=0.9 fpu) [LRFD Table 5.4.4.1-1] 243 ksi A parameter for prestressing, fpo = 0.7 fpu 189 ksi Effective prestressing stress after all losses, fse 148.2 ksi Modulus of elasticity, Ep [LRFD Art. 5.4.4.2] 28,500 ksi REINFORCING BARS Yield strength, fy #3, #4 :60 ksi Modulus of elasticity, Es [LRFD Art. 5.4.3.2] 29,000 ksi J.7.4 Sectional Properties and Forces The sectional properties and forces are given in Table J-20. Fig. J-27 provides detailed dimensions for the cross-section of an AASHTO Type IV beam. The typical strand pattern at midspan is shown in Fig. J-28, and the strand pattern at the design section is given in Table J-21. Note that 8 strands are draped as shown in Table J-21. Some basic calculations are also provided in this sub-section. Table J-20 Sectional Properties and Forces OVERALL GEOMETRY AND SECTIONAL PROPERTIES Non-Composite Section Span length, L 98.58 ft Overall depth of girder, h 54.0 in. Width of web, bv 8.0 in. Area of cross-section of girder, Ag 789 in.2 Moment of inertia, Ig 260,403 in.4 Distance from centroid to extreme bottom fiber, yb 24.75 in. J-53

Distance from centroid to extreme top fiber, yt 29.25 in. Section modulus for the extreme bottom fiber, Sb 10,521 in3 Section modulus for the extreme top fiber, St 8,902 in3 Weight of beam 0.821 kip/ft Composite Section Overall depth of the composite section, hc 62 in. Slab thickness, ts 8.0 in. Total transformed area of the composite section, Ac 1,604 in.2 Moment of inertia of the composite section, Ic 708,041 in.4 Distance from centroid of the composite section to extreme bottom fiber, ybc 41.66 in. Distance from centroid of the composite section to extreme top fiber of beam, ytg 12.34 in. Distance from centroid of the composite section to extreme top fiber of slab, ytc 20.34 in. Composite section modulus for the extreme bottom fiber of beam, Sbc 16,996 in3 Composite section modulus for the extreme top fiber of beam, Stg 57,378 in3 Composite section modulus for the extreme top fiber of slab, Stc 34,810 in3 Area of non-prestressed tension reinforcement, As 0 in2 Distance from extreme compression fiber to centroid of longitudinal tension reinforcement, ds 0 in Area of prestressed tension reinforcement, Ap =40(0.153)=6.120 in2 Distance from the top fiber to the centroid of prestressed tendons, dp 55.93 in SECTIONAL FORCES AT DESIGN SECTION Unfactored shear force due to beam weight, Vdg 36.4 kips Unfactored shear force due to deck slab, Vds 37.7 kips Unfactored shear force due to superimposed dead load, Vdw 8.3 kips Unfactored shear force due to total dead load, Vd 82.4 kips Factored shear force, Vu 265.0 kips Unfactored moment due to beam weight, Mdg 189.6 ft-kips Unfactored moment due to deck slab, Mds 196.2 ft-kips Unfactored moment due to superimposed dead load, Mdw 43.4 ft-kips Unfactored moment due to total dead load, Md 429.2 ft-kips Factored moment, Mu 1,125.0 ft-kips J-54

Figure J-27 Cross-Section of AASHTO Box Beam Type BIII-48 Figure J-28 Typical Strand Pattern at Midspan Table J-21 Strand Pattern at the Design Section (a total of 40 strands, 8 draped strands) Dist. from bottom (in.) 2.0 4.0 6.0 8.0 46.0 48.0 50.0 52.0 Strands/Row at End 10 10 10 2 2 2 2 2 Strands/Row at CL 12 12 12 4 0 0 0 0 Calculation of effective depth, : vd Note that 8 strands are draped, and 49.40ed in= at the design section for shear. The equivalent compressive block depth, , is calculated by flexural analysis ( =8.40 in.). a a Then, is the greater of: vd J-55

/ 2 [49.4 0.5(8.4)] 45.20ed a in− = − = (Controls) 0.9 0.9(49.40) 44.46ed i= = n n n 0.72 0.72(62) 43.40h i= = Therefore, . 45.20vd i= The vertical component of the effective prestressing force, , due to the 8 draped strands is 17.17 kips. Then, the design shear stress is: pV 265.0 0.9(17.17) 0.767 (0.9)(8.0)(45.20) u p u v v V V v k b d si φ φ − −= = = Thus, '/ 0.767 / 6.5 0.118u cv f = = J.7.5 Shear Design by Proposal 1: Modified STD Approach Shear design in accordance with the Modified STD Approach is used to determine the amount and spacing of the shear reinforcement required at a section located 4.93 ft from the support. a) Evaluation of Web-Shear Cracking Strength Compute web-shear cracking strength, : cwV '(0.06 0.3 )cw c pc v v pV f f b d= + V+ ips The effective prestress force is: 40(0.153)(148.2) 907.0 seP k= = The eccentricity of the strands at the design section is 12.15 in. Then, the compressive stress in the concrete at the centroid of cross section due to both pretensioning and moments and resisted by the precast member alone is: ( )(( ) dg ds bc bse se bc b pc g g g )M M y yP P e y yf A I I + −−= − + 907.0 907.0(12.15)(41.66 24.75) (189.6 196.2)(12)(41.66 24.75) 0.735 789 260,403 260,403pc f kips− + −= − + = Therefore, [0.06 6.5 0.3(0.735)](8)(45.20) 17.17 152.2cwV k= + + = ips b) Evaluation of Flexure-Shear Cracking Strength Compute flexure-shear cracking strength, : ciV ' max 0.02 crci c v v d i MV f b d V V M = + + '0.06 c v vf b d≥ J-56

, 265.0 82.4 182.6i u dV V V kips= − = − = max 1,125.0 429.2 695.8u dM M M ft kip= − = − = ⋅ s . Moment causing flexural cracking at the design section due to externally applied loads: '(0.2 )ccr c pe d bc IM f f f y = + − Compressive stress in concrete due to effective prestress forces only: 907.0 907.0(12.15)(24.75) =2.197 789 260,403 se se b pe g g P P eyf ksi A I = + = + Stress due to service dead load: 429.2(12)(24.75) 0.490 260, 403 d b d g M yf ksi I = = = Therefore, 708,041(0.2 6.5 2.197 0.490) /12 3,139.8 41.66cr M ft kips= + − = ⋅ The flexure-shear cracking strength, , is: ciV 182.6(3,139.8)0.02 6.5(8.0)(45.20) 82.4 924.8 695.8ci V k= + + = ips '0.06 0.06 6.5(8.0)(45.20) 55.3 c v vf b d kips> = = c) Evaluation of Concrete Contribution The nominal shear strength provided by concrete is the lesser of and . ciV cwV Web-shear cracking strength (Governs) 152.2cwV = kips ips ips Flexure-shear cracking strength 924.8 ciV k= Thus, the nominal shear strength provided by concrete is: 152.2 cV k= d) Evaluation of Required Transverse Reinforcement Check if 0.5u cV Vφ> 265.0 0.5 0.5(0.9)(152.2) 68.5u cV kips V kipsφ= > = = Therefore, transverse reinforcement must be provided. The shear force required is ( / ) (265.0 / 0.9) 152.2 142.2s u cV V V kipsφ= − = − = The shear strength provided by the transverse reinforcement is: s dfA V vyvs ααθ sin)cot(cot += [LRFD Eq. 5.8.3.3-4] When vertical stirrups are used, α =90. Since governs and cwV 1,125.0 3,139.8u crM ft kips M ft kips= ⋅ < = ⋅ , the angle of the compressive strut is obtained as: J-57

'cot 1 3 / 1.80pc cf fθ = + ≤ cot 1 3(0.735) / 6.5 1.86 1.80θ = + = > (Governs) Therefore, cot 1.80θ = . Therefore, area of transverse reinforcement (in2) within a spacing (s) is: (use fy = 60 ksi) 2/ /( cot ) 142.2 /[(60)(45.20)(1.8)] 0.0291 . / .v s y vA s V f d in inθ= = = Therefore, use #4 double leg stirrups at 12 in. spacing provided > required /vA s 22(0.20) /12 0.0333 . / .in in= = /vA s 20.0291 . / .in in= Then, provided 0.40(60)(45.20)(1.80) 162.7 12s V k= = ips . e) Checks Maximum Spacing Limitation of Transverse Reinforcement Maximum spacing of transverse reinforcement shall not exceed the following: When , [LRFD Eqs. 5.8.2.7-1] '/ 0.118 0.125u cv f = < max 0.8 24.0 .vs d= ≤ in in n n in (controls) max 24 .s ≤ 0.8 (0.8)(45.20) 36.16 .vd i≤ = = Since < O.K. 12 .s i= max 24 .s ≤ Minimum Reinforcement Requirement The area of transverse reinforcement shall not be less than: ' 2,min (8)(12)0.0316 0.0316 6.5 0.129 60 v v c y b sA f f ≥ = = in < provided 20.40vA in= O.K. Maximum Nominal Shear Resistance In order to ensure that the concrete in the web of the girder will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications specify an upper limit on Vn as follows: '0.25n c v vV f b d≤ + pV ips 152.2 162.7 314.9nV k= + = '0.25 0.25(6.5)(8)(45.20) 17.17 604.8c v v pf b d V kips< + = + = O.K. J.7.6 Shear Design by Proposal 2: Modified CSA Approach Shear design in accordance with the Modified CSA Approach is used to determine the amount and spacing of the shear reinforcement required at a section located 4.93 ft from the support. a) Evaluation of xε Calculate the strain in the reinforcement on the flexural tension side, xε : J-58

/ 0.5 0.002 2( ) u v u u p ps po x s s p ps M d N V V A f E A E A ε + + − −= ≤+ 31,125.0(12) / 45.20 0 265.0 17.17 (40)(0.153)(189) 1.749 10 2[0 28,500(40)(0.153)] −+ + − −= =+ − × Since the value of xε is negative, a different equation must be used: / 0.5 0.002 2( ) u v u u p ps po x c c s s p ps M d N V V A f E A E A E A ε + + − −= ≤+ + where Ac = area of concrete on the flexural tension side . 2473 in= 31,125.0(12) / 45.20 0 265.0 17.17 (40)(0.153)(189) 0.123 10 2[4,888(473) 28,500(40)(0.153)]x ε −+ + − −= =+ − × b) Evaluation of β and θ Calculate θ from the longitudinal strain, xε . 29 7,000 xθ ε= + 329 7,000( 0.123 10 ) 28.1−= + − × = ° Assume that at least the minimum amount of shear reinforcement required is provided. Then, the coefficient, β , is: 4.8 (1 1,500 )x β ε= + 3 4.8 5.88 [1 1,500( 0.123 10 )]− = =+ − × c) Evaluation of Concrete Contribution The contribution of the concrete to the nominal shear resistance is: vvcc dbfV '0316.0 β= 0.0316(5.88) 6.5(8.0)(45.20) 171.4 kips= = d) Evaluation of Required Transverse Reinforcement Check if 0.5 ( )u cV V pVφ> + 265.0 0.5 ( ) 0.5(0.9)(171.4 17.17) 84.8u c pV kips V V kipsφ= > + = + = Therefore, transverse shear reinforcement must be provided. The shear resistance required for the transverse reinforcement is: ( / ) (265.0 / 0.9) 171.4 17.17 105.9s u c pV V V V kipsφ= − − = − − = The shear strength provided by the transverse reinforcement is: s dfA V vyvs ααθ sin)cot(cot += [LRFD Eq. 5.8.3.3-4] J-59

When vertical stirrups are used, 90α = ° . Then, the required area of transverse reinforcement within a spacing (s) is: (use yf = 60 ksi) 2/ /( cot ) 105.9 /[(60)(45.20)cot 28.1 ] 0.021 . / .v s y vA s V f d in inθ= = ° = Therefore, use #3 double leg stirrups at 11 in. spacing provided > required /vA s 22(0.11) /11 0.0200 . / .in in= = /vA s 20.021 . / .in in= Then, provided 0.22(60)(45.20)cot 28.1 101.6 11s V k°= = ips e) Checks Maximum Spacing Limitation of Transverse Reinforcement Maximum spacing of transverse reinforcement shall not exceed the following: When , [LRFD Eqs. 5.8.2.7-1] '/ 0.118 0.125u cv f = < max 0.8 24.0 .vs d= ≤ in in n n in (controls) max 24 .s ≤ 0.8 (0.8)(45.20) 36.16 .vd i≤ = = Since < O.K. 11 .s i= max 24 .s ≤ Minimum Reinforcement Requirement The area of transverse reinforcement shall not be less than: ' 2,min (8)(11)0.0316 0.0316 6.5 0.118 60 v v c y b sA f f ≥ = = in < provided 20.22vA in= O.K. Maximum Nominal Shear Resistance In order to ensure that the concrete in the web of the girder will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications specify an upper limit on Vn as follows: '0.25n c v vV f b d≤ + pV ips 171.4 101.6 17.17 290.1nV k= + + = '0.25 0.25(6.5)(8)(45.20) 17.17 604.8c v v pf b d V kips< + = + = O.K. J.7.7 Summary and Conclusions Shear design in accordance with the Modified STD Approach and the Modified CSA Approach are used to determine the required amount and spacing of the transverse reinforcement at a section of a 100-ft span AASHTO Type IV beam with harped (draped) pretensioned strands. The shear design results are summarized in Table J-22. While the calculated compression strut angle by both approaches is almost same, the required amount of transverse reinforcement is very different, i.e., the Modified STD Approach required about 50% more transverse reinforcement than the Modified CSA Approach due to the differences in the concrete contribution calculated by the two approaches. J-60

Table J-22 Summary of Results Required or calculated Proposal 1: Modified STD Approach Proposal 2: Modified CSA Approach ( + ), kips cV pV 152.2 188.5 sV , kips 142.2 105.9 θ , deg. 29.0 28.1 Reinforcement Provided double leg #4 bars @12 inches double leg #3 bars @11 inches J-61

J.8 Example 8: Precast Balanced Cantilever Construction Using the AASHTO-PCI- ASBI Segmental Box Girder Standards J.8.1 Example Description This example gives the shear design calculations for a 5-span Precast Balanced Cantilever Bridge constructed using AASHTO-PCI-ASBI segmental box girders. The design section is taken from the second bay near the support. Shear design is accomplished in accordance with Proposal 1 (Modified STD Approach) and Proposal 2 (Modified CSA Approach). J.8.2 Geometry and Loading The whole bridge is 270m long and consists of 5 spans. Fig. J-29 gives the plan and elevation of the bridge. Design Section Figure J-29 Plan and Elevation of Design live load is HL-93. Fig. J-30 provides detailed dimensions of the segmental box cross-section. . J-62 Figure J-30 Cross section of AASHTO-PCI-ASBI Segmental Box

Fig. J-31 illustrates the Post-Tensioning anchorages and reinforcement layout. . Figure J-31 Post-Tensioning anchorages J.8.3 Material Properties The material properties are given in Table J-23. Table J-23 Material Properties CONCRETE PROPERTIES Concrete strength at 28 days, f’c 6.1 ksi (42MPa) Concrete unit weight, wc 0.150 kcf Modulus of elasticity of concrete, Ec = '5.1)(000,33 cc fw [LRFD Eq. 5.4.2.4-1] 4,510ksi (31,100MPa) PRESTRESSING STRANDS Type 0.6 in. dia., seven-wire, low-relaxation Area of a strand 0.217 in2 (140mm2) Ultimate strength, fpu 270.0 ksi(1860MPa) Yield strength, fpy (=0.9 fpu) [LRFD Table 5.4.4.1-1] 243 ksi (1674MPa) Initial prestressing stress, fpi 202.5 ksi A parameter for prestressing, fpo 198.8ksi (1370.6MPa) J-63

Modulus of elasticity, Ep [LRFD Art. 5.4.4.2] 28,500 ksi (197,000MPa) REINFORCING BARS Yield strength, fy 25M ,60 ksi (400MPa) Modulus of elasticity, Es [LRFD Art. 5.4.3.2] 29,000 ksi (200,000MPa) J.8.4 Sectional Properties and Forces The design section was located near the interface of second pier as marked in Fig J-29. The sectional properties and forces are given in Table J-24. Table J-24 Sectional Properties and Forces OVERALL GEOMETRY AND SECTIONAL PROPERTIES Span length, L 196.85 ft(60m) Overall depth of girder, h 106.3 in.(2700mm) Width of web, bv 31.5 in. (800mm) Area of cross-section of girder, Ag 12,268 in.2 Moment of inertia, Ig 20,368,314 in.4 Distance from centroid to extreme bottom fiber, yb 63.82in. Distance from centroid to extreme top fiber, yt 42.48 in. Section modulus for the extreme bottom fiber, Sb 319,154 in3 Section modulus for the extreme top fiber, St 480,562 in3 Area of non-prestressed tension reinforcement, As 0 in2 Distance from extreme compression fiber to centroid of longitudinal tension reinforcement, ds 0 in Area of prestressed tension reinforcement, Ap 51.87 in2 (33464mm2) Distance from the bottom fiber to the centroid of prestressed tendons, dp 101.65in Weight of beam 11.88kip/ft (520KN/3m) SECTIONAL FORCES AT DESIGN SECTION* Unfactored shear force due to dead load, Vd 998 kips (4440 KN) Factored shear force, Vu 1803 kips (8022KN) Unfactored moment due to dead load, Md 22793 ft-kips (30903KN-m) Factored moment, Mu 39504 ft-kips (53560KN-m) *Negative moment, bottom slab is in compression Calculation of effective depth, : vd The compressive block depth, a = 15.9 in (403mm) 101.65e pd d i= = n J-64

/ 2 101.65 0.5(15.9) 93.7 max 0.9 max 0.9(101.65) 91.5 93.7 0.72 0.72(106.3) 76.5 e v e d a d d h − − =⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪= = = =⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪=⎩ ⎭ ⎩ ⎭ in The design shear stress is: 1803 0 0.679 (0.9)(31.5)(93.7) u p u v v V V v k b d si φ φ − −= = = Thus, < 0.18 '/ 0.679 / 6.1 0.111u cv f = = J.8.5 Shear Design by Proposal 1: Modified STD Approach Shear design in accordance with the Modified STD Approach is used to determine the amount and spacing of the required shear reinforcement. a) Evaluation of Web-Shear Cracking Strength Compute web-shear cracking strength, : cwV '(0.06 0.3 )cw c pc v v pV f f b d= + V+ ips The effective prestress force is: (51.87)(198.8) 10311.8 seP k= = Compressive stress in concrete at the centroid of cross section due to prestress is: 10311.8 0.841 12268 se pc g Pf kips A = = = Therefore, [0.06 6.1 0.3(0.841)](31.5)(93.7) 0 1182cwV k= + + = ips b) Evaluation of Flexure-Shear Cracking Strength Compute flexure-shear cracking strength, : ciV ' max 0.02 crci c v v d i MV f b d V V M = + + '0.06 c v vf b d≥ , 1803 998 805i u dV V V kips= − = − = max 39504 22793 16711u dM M M ft kip= − = − = ⋅ s . Moment causing flexural cracking at the design section due to externally applied loads: '(0.2 )gcr c pe d t I M f f f y = + − in (Negative moment, bottom in compression) The center of gravity of the strand pattern at the design section is: (from bottom fiber) 101.65pd = The eccentricity of the strands at the design section is: 101.65 63.82 37.83 .p be d y in= − = − = Compressive stress in concrete due to effective prestress forces only: 10311.8 10311.8(37.83) =1.652 12268 480562 se se pe g t P P ef ksi A S = + = + J-65

Stress due to service dead load: 22793 12 0.569 480562 d d t Mf ksi S ×= = = Therefore, 480562(0.2 6.1 1.652 0.569) 63152crM ft kips= + − = ⋅ The flexure-shear cracking strength, , is: ciV 805(63152)0.02 6.1(31.5)(93.7) 998 4186 16711ci V k= + + = ips '0.06 0.06 6.1(31.5)(93.7) 437.4 c v vf b d kips> = = c) Evaluation of Concrete Contribution The nominal shear strength provided by concrete is the lesser of and . ciV cwV Web-shear cracking strength (Governs) 1182cwV = kips ips ips Flexure-shear cracking strength 4186 ciV k= Thus, the nominal shear strength provided by concrete is: 1182 cV k= d) Evaluation of Required Transverse Reinforcement Check if 0.5u cV Vφ> 1803 0.5 0.5(0.9)(1182) 532u cV kips V kipsφ= > = = Therefore, transverse reinforcement must be provided. The shear force required is ( / ) (1803/ 0.9) 1182 821.3s u cV V V kipsφ= − = − = The shear strength provided by the transverse reinforcement is: s dfA V vyvs ααθ sin)cot(cot += [LRFD Eq. 5.8.3.3-4] When vertical stirrups are used, α =90. Since governs and cwV 39,504 63,152u crM ft kips M ft kips= ⋅ < = ⋅ , the angle of the compressive strut is obtained as: 'cot 1 3 / 1.80pc cf fθ = + ≤ cot 1 3(0.841) / 6.1 2.02 1.80θ = + = > (Governs), so cot 1.80θ = . Therefore, the area of the required transverse reinforcement (in2) within a spacing (s) is: (use fy = 60 ksi) 2/ /( cot ) 821.3 /[(60)(93.7)(1.80)] 0.0812 . / .v s y vA s V f d in inθ= = = J-66

Therefore, use 2-25M bars in each web at a spacing of 24 in. and 24(0.775) 3.1 .vA in= = 24 .s in= 2( ) 3.1/ 24 0.129 . / .vA provided in in s = = > 2( ) 0.0812 . / .vA required in in s = Then, provided 3.1 60 93.7 1.8 1307 24s V k× × ×= = ips e) Checks Maximum Spacing Limitation of Transverse Reinforcement Maximum spacing of transverse reinforcement must not exceed the following: When , [LRFD Eqs. 5.8.2.7-1] '/ 0.111 0.125u cv f = < max 0.8 24.0 .vs d= ≤ in n = ≤ in max 24 24 min min 24 0.8 0.8(03.7) 75v in in s i d in ⎧ ⎫ ⎧ ⎫= = =⎨ ⎬ ⎨ ⎬=⎩ ⎭⎩ ⎭ Since O.K. 24 .s in max 24 .s = Minimum Reinforcement Requirement The area of transverse reinforcement shall not be less than: ' 2,min (31.5)(24)0.0316 0.0316 6.1 0.983 60 v v c y b sA f f ≥ = = in < provided 23.1vA in= O.K. Maximum Nominal Shear Resistance In order to ensure that the concrete in the web of the girder will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications specify an upper limit on as follows: nV '0.18n c v vV f b d≤ + pV = + = 1182 1307 2489nV kips '0.18 0 0.18(6.1)(31.5)(93.7) 3241c v vf b d kips< + = = O.K. J.8.6 Shear Design by Proposal 2: Modified CSA Approach Shear design in accordance with the Modified CSA Approach is used to determine the amount and spacing of the shear reinforcement required at the selected section. a) Evaluation of xε Calculate the strain in the reinforcement on the flexural tension side, xε : / 0.5 0.002 2( ) u v u u p ps po x s s p ps M d N V V A f E A E A ε + + − −= ≤+ 339504(12) / 93.7 0 1803 0 51.87(198.8) 1.167 10 2[(0 28,500(51.87)] −+ + − −= =+ − × J-67

Since the value of xε is negative, a different equation must be used: / 0.5 0.003 2( ) u v u u p ps po x c c s s p ps M d N V V A f E A E A E A ε + + − −= ≤+ + where Ac = area of concrete on the flexural tension side(top half of beam) = 26170in 339504(12) / 93.7 0 1803 0 51.87(198.8) 0.059 10 2[(4510)(6172) 28,500(51.87)]x ε −+ + − −= =+ − × b) Evaluation of β and θ Calculate θ from the longitudinal strain, xε . xεθ 700029 += 329 7,000( 0.059 10 ) 28.6−= + − × = ° cot 1.834θ = Assume that at least the minimum amount of shear reinforcement required is provided. Then, the coefficient, β , is: 3 4.8 4.8 5.27 (1 1500 ) (1 (1500)( 0.059 10 ))x β ε −= = =+ + − × c) Evaluation of Concrete Contribution The contribution of the concrete to the nominal shear resistance is: vvcc dbfV '0316.0 β= 0.0316(5.27) 6.1(31.5)(93.7) 1214 kips= = d) Evaluation of Required Transverse Reinforcement Check if 0.5 ( )u cV V pVφ> + 1803 0.5 ( ) 0.5(0.9)(1214 0) 546.3u c pV kips V V kipsφ= > + = + = Therefore, transverse shear reinforcement must be provided. The shear resistance to be provided by the transverse reinforcement is: ( / ) (1803/ 0.9) 1214 0 789s u c pV V V V kipsφ= − − = − − = The shear strength provided by the transverse reinforcement is: s dfA V vyvs ααθ sin)cot(cot += [LRFD Eq. 5.8.3.3-4] By using yf = 60 ksi, 90α = ° , cot 1.834θ = , and: 2/ /( cot ) 789 /[(60)(93.7)cot 28.6 ] 0.0765 . / .v s y vA s V f d in inθ= = ° = J-68

Therefore, use 2-25M bars in each web at a spacing of 24 in. and 24(0.775) 3.1 .vA in= = 24 .s in= 2( ) 3.1/ 24 0.129 . / .vA provided in in s = = > 2( ) 0.0765 . / .vA required in in s = Then, provided 3.1 60 93.7 1.834 1332 24s V k× × ×= = ips e) Checks Maximum Spacing Limitation of Transverse Reinforcement Maximum spacing of transverse reinforcement shall not exceed the following: When , [LRFD Eqs. 5.8.2.7-1] '/ 0.111 0.125u cv f = < max 0.8 24.0 .vs d= ≤ in n = ≤ in max 24 24 min min 24 0.8 0.8(03.7) 75v in in s i d in ⎧ ⎫ ⎧ ⎫= = =⎨ ⎬ ⎨ ⎬=⎩ ⎭⎩ ⎭ Since O.K. 24 .s in max 24 .s = Minimum Reinforcement Requirement The area of transverse reinforcement shall not be less than: ' 2,min (31.5)(24)0.0316 0.0316 6.1 0.983 60 v v c y b sA f f ≥ = = in < provided 23.1vA in= O.K. Maximum Nominal Shear Resistance In order to ensure that the concrete in the web of the girder will not crush prior to yielding of the transverse reinforcement, the LRFD Specifications specify an upper limit on Vn as follows: '0.18n c v vV f b d≤ + pV 1214 1332 2546c sV V kips+ = + = '0.18 0.18(6.1)(31.5)(93.7) 3241c v vf b d kips< = = O.K. J.8.7 Summary and Conclusions Shear design in accordance with the Modified STD Approach and the Modified CSA Approach are used to determine the amount and spacing of the required transverse reinforcement at a section of a 5-span Precast Balanced Cantilever Bridge constructed using AASHTO-PCI- ASBI Segmental Box Girders. The design results are summarized in Table J-25. The required amount of transverse reinforcement by those two approaches is very similar. Table J-25 Summary of Results Required or calculated Proposal 1: Modified STD Approach Proposal 2: Modified CSA Approach ( + ), kips cV pV 1182 1214 J-69

sV , kips 821 789 θ , deg. 29.0 28.6 Reinforcement Provided 4- 25M bars @24 inches 4- 25M bars @24 inches J-70