Cast-in-Place Concrete Connections for Precast Deck Systems (2011)

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B-i Appendix B NCHRP 10-71 Design Examples

B-ii Foreword The NCHRP 10-71 study involved the development of design recommendations and details for Cast-in-Place Concrete Connections for Precast Deck Systems. The project covered two very different systems: (1) the precast composite slab-span system (PCSSS), which is an entire bridge system, and (2) transverse and longitudinal cast-in-place connection concepts between the flanges of precast decked bulb-Ts and full-depth precast deck panels on girders. In the case of the longitudinal connections, flexure and flexure-shear are to be transferred across the joint. In the case of the transverse joints, tension or compression is to be transferred across the joint depending upon the location of the joint along the span. The most critical location for the transverse connection would be at a pier in a continuous system, where the connection would be required to transmit tension, equilibrated by compression in the girder. The five design examples presented in this appendix are separated according to the connection concepts, and the level of detail and complexity in the examples is commensurate with the type of connection concept. The first two examples represent a 50 ft. simply-supported PCSSS bridge (Example 1) and a 40-50-40 ft. three-span continuous PCSSS bridge (Example 2). Because these examples cover the design of complete systems, they have a significant level of detail. The latter three examples provide guidance on the detailing of longitudinal and transverse connection concepts. Example 3 details the design of a longitudinal joint between decked bulb- T members, while the design of a transverse joint over the piers of a continuous bulb-T girder bridge that incorporates full-depth deck panels is presented in Example 4. Finally, Example 5 illustrates the longitudinal connection between two full-depth deck panels. The connection in the latter example is required where the width of the bridge exceeds the practical span of a single full-depth deck panel. Because Examples 3 through 5 cover the longitudinal (flexure or flexure-shear) or transverse joints (tension) that transfer flexure, flexure- shear, or tension, the level of detail In the examples is restricted to the detailing of the connections themselves, rather than the entire bridge systems. Each of the five examples illustrates the use of the recommendations from the design guide provided in Appendix A.

Example Problem 1 1.1 Introduction This example covers the design of the primary load-carrying superstructure elements of a precast composite slab-span system (PCSSS) bridge. The structural system is single span, simple-span beam with a 50' design span. The steps required to design a representative composite panel are illustrated. The design is generally carried out in accordance with the AASHTO LRFD Bridge Design Specifications, 5th Edition, plus interim revisions through 2010. 1.2 Materials, Geometry, Loads and Load Factors Units: kcf kip ft 3 Defined unit: kips per cubic foot ksf kip ft 2  Defined unt: kips per square foot Materials: Concrete: f'c 7.0 ksi Strength of beam concrete at 28 days f'ci 5.5 ksi Strength of beam concrete at transfer of prestressing force wc 0.150 kcf Density of beam concrete f'ct 4.0 ksi Strength of CIP concrete at 28 days wct 0.15 kcf Density of CIP concrete H 70 Average ambient relative humidity α 0.000006 /oF Coefficient of thermal expansion Strand: Astrand 0.217 in 2  Area of one prestressing strand. db 0.6 in Nominal diameter of prestressed strand. fpu 270 ksi Tensile strength of prestressing steel Ep 28500 ksi Modulus of elasticity of prestressing steel fpy 0.9 fpu Yield strength of prestressing steel Pull 0.75 Pull of strands expressed as a fraction of fpu t 18 hr Time from tensioning to detensioning of strands Rebar: fy 60 ksi Yield stress of ordinary rebar Es 29000 ksi Modulus of elasticity of non-prestressed reinforcement Geometry: Beam: Section "IT" Precast section name h 18.0 in Height of precast A 936.0 in 2  Gross area of precast section I 27120 in 4  Gross moment of inertia of precast cross section about centroidal x-x axis yb 8.42 in Center of gravity of gross precast cross section bf 72.0 in Width of bottom flange of precast section tflg 3.00 in Effective thickness of bottom flange B1-1

bv 72.00 in Shear width of precast section (web and longitudinal trough width) Ac 450 in 2  Area of concrete on flexural tension side of member (see LRFD B5.2-3) VSb 5.2 in Volume to surface ratio of precast section VSd 6.0 in Volume to surface ratio of CIP slab Slab: tslab 6.00 in Thickness of CIP slab above precast beam th 15.00 in Thickness of CIP region between precast beams bh 24.00 in Width of CIP region between precast units. tws 0.0 in Thickness of portion of CIP slab assumed to be wear Span: Lovr 50.0 ft Overall length of precast section Ldes 49.0 ft Design span of precast section Lpad 12 in Length of bearing pad Bridge: S 6.00 ft Beam spacing Ng 8 Number of precast sections in bridge cross section Widthoverall 47.5 ft Overall width of bridge Widthctc 44.0 ft Curb to curb width of bridge Nl 2 Number of lanes Loads: Dead: Nbarriers 2 Number of barriers wbarrier 0.300 klf Weight of single barrier wfws 0.023 ksf Weight of future wearing surface allowance Live: HL-93 Notional live load per LRFD Specs wlane 0.64 klf Design lane load wconst 10 psf Construction live load Construction Timing: ttransfer 1.00 day Time from release tensioning of strands to release of prestress tdeck 90 day Time to placement of superimposed dead load (SDL) tfinal 20000 day Time final Load & Resistance Factors: f (variable) Resistance factor for flexure ϕv 0.90 Resistance factor for shear DLA 0.33 Dynamic load allowance (LRFD 3.6.2.1-1) B1-2

1.3 Plan, Elevation, and Typical Section Figure 2: General elevation of bridge. Figure 1: General plan of bridge. Figure 3: Typical cross section of bridge. Strand Pattern: No, Strands Elevation (in) Pat_n 0 0 12 12              Pat_h 16 6 4 2             in B1-3

1.4 Section Properties Note: Flange assumed to be a constant thickness of 3" for computation of section properties. Figure 4: Precast slab dimensions. Non-Composite Section Properties: Sb I yb  Sb 3220.9 in 3  yt h yb St I yt  St 2830.9 in 3  Effective Width: (LRFD 4.6.2.6.1) The effective width of the composite section may be taken as 1/2 the distance to the adjacent beam. beff S 2 S 2  beff 72 in n f'ct f'c  n 0.7559 btran n beff btran 54.4269 in Composite Section Properties: Aslab btran tslab tws  Aslab 326.6 in2 Ah n th bh Ah 272.1 in2 Acomp A Ah Aslab Acomp 1534.7 in 2  ybc A yb Ah tflg th 2         Aslab h tslab tws 2         Acomp  ybc 11.466 in hc h tslab tws hc 24 in ytc hc ybc ytc 12.53 in Islab btran tslab tws 3 12  Islab 979.7 in 4  Ih n bh th 3  12  Ih 5102.5 in 4  Ic I A yb ybc 2 Ih Ah tflg th 2  ybc       2  Islab Aslab h tslab tws 2  ybc       2  Ic 71824 in 4  Sbc Ic ybc  Sbc 6264.3 in 3  Stc Ic ytc n  Stc 7580.3 in 3  B1-4

Composite section modulus at the top of the prestressed beam: ytcb h ybc Stcb Ic ytcb  Stcb 10991.8 in 3  1.5 Strand Pattern Properties Figure 5: Cross section with reinforcement. No_Strands Pat_n No_Strands 24 i 1 last Pat_n( ) ycg i Pat_ni Pat_hi  No_Strands  ycg 3 in ecc yb ycg ecc 5.42 in 1.6 Moments and Shears At Release: Self-weight of beam at release: At transfer, there are two locations along the beam that are of interest: 1. Transfer point of strands: xr1 60 db (LRFD 5.8.2.3) 2. Midspan of beam: xr2 Lovr 2  xr T 3 25( ) ft wsw wc A wsw 0.975 klf i 1 2 Mswri wsw xri  2 Lovr xri    Mswr T 68.7 304.7( ) kip ft B1-5

At Final Conditions: Beam self-weight at final: At final conditions, there are also two points of interest: 1. The critical section for shear is dv from the face of the support, with dv taken as 0.72h (see discussion in Theory section). xf1 0.72 hc Lpad 2  2. Midspan of beam: xf2 0.5 Ldes xf T 1.94 24.5( ) ft j 1 2 Mswfj wsw xfj  2 Ldes xfj    Mswf T 44.5 292.6( ) kip ft Vswfj wsw Ldes 2 xfj         Vswf T 22 0( ) kip Deck Weight: wd tslab tws  S th bh  wct wd 0.825 klf Mdeckj wd xfj  2 Ldes xfj    Mdeck T 37.7 247.6( ) kip ft Vdeckj wd Ldes 2 xfj         Vdeck T 18.6 0( ) kip Barrier Weight (Composite Dead Load): wbarrier 0.3 klf Per barrier: wb Nbarriers wbarrier Ng  wb 0.075 klf Mbarrierj wb xfj  2 Ldes xfj    Mbarrier T 3.4 22.5( ) kip ft Vbarrierj wb Ldes 2 xfj         Vbarrier T 1.7 0( ) kip Future Wearing Surface: wfws 0.023 ksf Per Beam: wf Widthctc wfws Ng  wf 0.1265 klf Mfwsj wf xfj  2 Ldes xfj    Mfws T 5.8 38( ) kip ft Vfwsj wf Ldes 2 xfj         Vfws T 2.9 0( ) kip B1-6

Live Load: Live Load Distribution Factors: Assume superstructure acts like a slab-type bridge. Utilize provisions of LRFD Art. 4.6.2.3 to compute width of equivalent strip to resist lane load. Single-Lane Loading: L1 if Ldes 60 ft 60 ft Ldes  L1 49.00 ft W1_1 if Widthoverall 30 ft 30 ft Widthoverall  W1_1 30.00 ft Estrip1 10 in 5.0 in L1 ft W1_1 ft  Estrip1 202 in Estrip1 16.8 ft Put in terms of fraction of one lane to be distributed to one precast unit: DF1lane bf Estrip1  DF1lane 0.357 Double-Lane Loading: W1_2 if Widthoverall 60 ft 60 ft Widthoverall  W1_2 47.50 ft Estrip2 84 in 1.44 in L1 ft W1_2 ft  Estrip2 153 in Estrip2 12.8 ft DF2lane bf Estrip2  DF2lane 0.4691 Governing Case: DF if DF1lane DF2lane DF1lane DF2lane  DF 0.4691 This distribution factor is applicable to both shear and moment: DFm DF DFv DF Live Load Moments (HL-93): Maximum Moments Due to Design Truck and Design Lane: Due to the Design Truck: A closed-form solution for the maximum moment at any point along a simply-supported beam due to the LRFD design truck is given below. There are two formulae, one which is valid for the region between the support and the L/3 point of the beam, and the other which is valid between L/3 and midspan. These two formulae correspond to different orientations of the truck (i.e., when it faces one way or the other). L Ldes (purely to condense the expression) Mtruckj if xfj L 3  8 kip xfj  L 9 L 9 xfj  84 ft   8 kip L 9 xfj  2  9 xfj  L 42 ft xfj  14 ft L          Mtruck T 107.5 602( ) kip ft B1-7

Due to the Design Lane: Mlanej wlane xfj  2 L xfj    Mlane T 29.2 192.1( ) kip ft Maximum Service Live Load Moments (HL-93): The dynamic load allowance (DLA) is applied to the truck portion only: (LRFD 3.6.2.1-1) MLLj DFm Mlanej 1 DLA( ) Mtruckj    MLL T 80.8 465.7( ) kip ft Live Load Shears: Vtruckj 8 kip L 9 L 9 xfj  84 ft   Vtruck T 55.4 22.3( ) kip Vlanej wlane L xfj   2  2 L  Vlane T 14.5 3.9( ) kip VLLj DFv Vlanej 1 DLA( ) Vtruckj    VLL T 41.4 15.7( ) kip Thermal Gradient: (LRFD 3.12.3) γTG 1.0 (no live live) γTG_L 0.5 (with live load) Effects due to uniform temperature change: Since superstructure is not restrained axially, uniform temperature change causes no internal stress. Effects due to temperature gradient: Fig. 6: Positive temperature gradient (from LRFD 3.12.3-2) Assume AASHTO temperature Zone 1: T1 54 (deg F) T2 14 T3 0 Atemp if hc 16 in 12 in hc 4 in  Atemp 12.00 in A1 4 in A2 Atemp A2 1 ft B1-8

Compute gradient-induced curvature: ψ α l Σ Tai yi Ai ΔTi di Ii       = (LRFD C4.6.6) Area1 A1 bf Area1 288 in 2  I1 bf A1 3  12  I1 384 in 4  Area2 A2 bf Area2 864 in 2  I2 bf A2 3  12  I2 10368 in 4  εgr α A T1 Area1 T2 Area2  εgr 0.000177 Epr 33000 wc 1.5  kcf 1.5  f'c ksi .5  Epr 5072 ksi Fgr Epr Ac εgr Fgr 404.5 kip y1 ytc A1 2  y1 10.53 in y2 ytc A1 A2 2         y2 2.53 in ψ α Ic T1 T2 2       y1 Area1 T1 T2  A1 2 I1 T2 2       y2 Area2 T2  A2 2 I2          ψ 0.000108 ft 1  Only the internal stress component affects the unrestrained simple span: σE E α TG α TuG ψ z = (LRFD C4.6.6-6) Evaluate at top and bottom of precast: fTGt Epr α T2 A1 A2 tslab A2        ψ ytcb        fTGt 0.0564 ksi fTGb Epr ψ ytcb fTGb 0.2987 ksi fTGtt Epr α T1 ψ ytc  fTGtt 1.0705 ksi B1-9

1.7 Flexural Stresses At Release: Beam Self-Weight Stresses: fswrtj Mswrj St  fswrt 0.291 1.292       ksi fswrbj Mswrj Sb  fswrb 0.256 1.135       ksi At Final Conditions: Note: Since for a simple-span structural system of this type, it is unlikely that compression at the top of the deck at a given section would exceed its allowable value, calculation of those stresses will be omitted for simplicity. Only the stresses at the bottom and top of the precast beam itself will be computed. The precast weight and CIP deck weight are carried by the precast section only. Additional loads (i.e. barrier, overlay, live load) are carried by the composite section. Self-Weight: fswtj Mswfj St  fswt 0.189 1.24       ksi fswbj Mswfj Sb  fswb 0.166 1.09       ksi Deck Weight: fdecktj Mdeckj St  fdeckt 0.16 1.05       ksi fdeckbj Mdeckj Sb  fdeckb 0.14 0.922       ksi Barriers: fbarriertj Mbarrierj Stcb  fbarriert 0.004 0.025       ksi fbarrierbj Mbarrierj Sbc  fbarrierb 0.007 0.043       ksi fbarrierttj Mbarrierj Stc  fbarriertt 0.005 0.036       ksi (top of topping) Future Wearing Surface: ffwstj Mfwsj Stcb  ffwst 0.006 0.041       ksi ffwsbj Mfwsj Sbc  ffwsb 0.011 0.073       ksi ffwsttj Mfwsj Stc  ffwstt 0.009 0.06       ksi (top of topping) Live Load: fLLtj MLLj Stcb  fLLt 0.088 0.508       ksi fLLbj MLLj Sbc  fLLb 0.155 0.892       ksi fLLttj MLLj Stc  fLLtt 0.128 0.737       ksi (top of topping) B1-10

Prestress Losses At Release: At release, two components of prestress loss are significant: relaxation of the prestressing steel and elastic shortening. Elastic shortening is the loss of prestress that results when the strands are detensioned and the precast beam shortens in length due to the applied prestress. When the strands are tensioned in the prestress bed and anchored at the abutments, the steel gradually begins to relax as a function of time. By the time the strands are detensioned a small, but measurable, loss due to steel relaxation has occurred. Steel Relaxation (short term): fpj Pull fpu fpj 202.5 ksi fpy 243 ksi ΔfpR1 log t hr       40.0 fpj fpy 0.55        fpj ΔfpR1 1.801 ksi (LRFD 5.9.5.4.4b-2) Elastic Shortening: Eci 33000 wc 1.5  kcf 1.5  f'ci ksi .5  Eci 4496 ksi (LRFD 5.4.2.4-1) Aps No_Strands Astrand Aps 5.208 in 2  (Area of strand at midspan) fpbt fpj ΔfpR1 fpbt 200.7 ksi ΔfpES Aps fpbt I ecc 2 A  ecc Mswr2 A Aps I ecc 2 A  A I Eci Ep   ΔfpES 8.986 ksi Total Prestress Loss at Release: Δfsr ΔfpES ΔfpR1 Δfsr 10.786 ksi %Loss Δfsr Pull fpu 100 %Loss 5.3264 fper fpj ΔfpES ΔfpR1 fper 191.7 ksi Pr fper No_Strands Astrand Pr 998.4 kip At Final Conditions: Total Loss of Prestress: ΔfpT ΔfpES ΔfpLT= (LRFD 5.9.5.1-1) where: fpES = Sum of all losses due to elastic shortening at time of application of prestress load (ksi). fpLT = Total loss due to long-term effects, which include shrinkage and creep of the concrete and relaxation of the prestressing steel (ksi). B1-11

In pretensioned bridge girder design, stresses have traditionally been assessed at two timeframes: at release of prestress (i.e., when the prestress force is applied to the girder) and at final (long-term) conditions. Therefore, loss of prestress has been evaluated at these two periods in the life of the girder. However, with the new prestress loss procedure introduced in the 2005 Interim Revisions, long-term losses are computed in two steps: (a) The time period between prestress transfer and placement of the cast-in-place concrete deck and (b) the time period between deck placement and the end of the service life of the girder. These two periods correspond to the non-composite and composite phases of the structural system. The rate of stress change in the strands can differ significantly in each phase, hence the need to subdivide the long-term losses into two components. Loss at Release: The principal component of prestress loss when the strands are released and the prestress force in the strands is imparted into the girder is elastic shortening. This reduction in prestress (prestress loss) occurs essentially instantaneously. Loss at Final Conditions: The time-dependent loss of prestress consists of three distinct components. Loss due to: 1. Creep of girder concrete, 2. Relaxation of prestressing strands, and 3. Shrinkage of girder concrete. These are computed in two stages: 1. From time prestress force is imparted to the girder to the time the girder is erected and 2. From the time the CIP deck is placed to final time. Mathematically, this is expressed as: fpLT = (fpSR+fpCR+fpR1)id + (fpSD+fpCD+fpR2-fpSS)df (LRFD 5.9.5.4.1-1) Material Properties: The shrinkage and creep properties of the girder and deck need to be computed in preparation for the prestress loss computations. These are addressed in LRFD Article 5.4.2.3. Creep coefficients are computed in accordance with Article 5.4.2.3.2 and shrinkage strains are computed in accordance with Article 5.4.2.3.3. Creep Coefficients Girder creep coefficient at final time due to loading at transfer: ψb tf ti  1.9 kvs khc kf ktd ti 0.118= (LRFD Eq. 5.4.2.3.2-1) where: b = Ratio of creep strain to elastic strain. kvs = Factor for the effect of volume to surface ratio. khc = Humidity factor for creep. kf = Factor for strength of concrete. ktd = Factor for time development. B1-12

kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0. (LRFD 5.4.2.3.2-1) kvs 1.000 khc 1.56 0.008H khc 1.000 (LRFD Eq. 5.4.2.3.2-3) kf 5 1 f'ci ksi   kf 0.769 (LRFD 5.4.2.3.2-4) tf tfinal ti ttransfer t tf ti t 19999 day ktd t day 61 4 f'ci ksi  t day   ktd 0.998 (LRFD 5.4.2.3.2-5) ψbfi 1.9 kvs khc kf ktd ti day       0.118  ψbfi 1.459 Girder creep coefficient at time of CIP placement due to loading at transfer: ti ttransfer td tdeck t td ti t 89 day ktd t day 61 4 f'ci ksi  t day   ktd 0.695 kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0. (LRFD 5.4.2.3.2-1) kvs 1.000 khc 1.56 0.008H khc 1.000 (LRFD Eq. 5.4.2.3.2-3) kf 5 1 f'ci ksi   kf 0.769 (LRFD 5.4.2.3.2-4) ψbdi 1.9 kvs khc kf ktd ti day       0.118  ψbdi 1.016 B1-13

Girder creep coefficient at final time due to loading at CIP placement: ti tdeck t tf ti t 19910 day ktd t day 61 4 f'ci ksi  t day   ktd 0.998 kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0. (LRFD 5.4.2.3.2-1) kvs 1.000 khc 1.56 0.008H khc 1.000 (LRFD Eq. 5.4.2.3.2-3) kf 5 1 f'ci ksi   kf 0.769 (LRFD 5.4.2.3.2-4) ψbfd 1.9 kvs khc kf ktd ti day       0.118  ψbfd 0.858 CIP concrete creep coefficient from time of casting to final: khc 1.56 0.008 H khc 1.000 kf 5 1 0.8 f'ct ksi   kf 1.19 kvs 1.45 0.13 VSd in  kvs 0.670 Note: kvs must be greater than 1.0. kvs 1.000 khc 1.56 0.008H khc 1.000 t tf tdeck t 19910 day ktd t day 61 4 f'ci ksi  t day   ktd 0.998 ψdfd 1.9 kvs khc kf ktd tdeck day       0.118  ψbfd 0.858 B1-14

Shrinkage Strains Girder concrete shrinkage strain between transfer and final time: The concrete shrinkage strain, bid, is computed in accordance with Art. 5.4.2.3.3: εsh kvs khs kf ktd 0.48 10 3 = (LRFD 5.4.2.3.3-1) where: kvs = Factor for the effect of volume to surface ratio. khs = Factor for humidity. kf = Factor for strength of concrete. ktd = Factor for time development. kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0 (LRFD 5.4.2.3.2-1) kvs 1.000 khs 2.00 0.014 H khs 1.020 (LRFD 5.4.2.3.3-2) kf 5 1 f'ci ksi   kf 0.769 (LRFD 5.4.2.3.2-4) tf tfinal ti ttransfer t tf ti t 19999 day ktd t day 61 4 f'ci ksi  t day   ktd 0.998 (LRFD 5.4.2.3.2-5) εbif kvs khs kf ktd 0.48 10 3  εbif 376 10 6  Girder concrete shrinkage strain between transfer and CIP placement: kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0 (LRFD 5.4.2.3.2-1) kvs 1.000 khs 2.00 0.014 H khs 1.020 (LRFD 5.4.2.3.3-2) kf 5 1 f'ci ksi   kf 0.769 (LRFD 5.4.2.3.2-4) t td ti t 89 day B1-15

ktd t day 61 4 f'ci ksi  t day   ktd 0.695 (LRFD 5.4.2.3.2-5) εbid kvs khs kf ktd 0.48 10 3  εbid 262 10 6  CIP concrete shrinkage from deck placement to final: kvs 1.45 0.13 VSd in  kvs 0.670 Note: kvs must be greater than 1.0 (LRFD 5.4.2.3.2-1) kvs 1.000 khs 2.00 0.014 H khs 1.020 (LRFD 5.4.2.3.3-2) kf 5 1 f'ci ksi   kf 0.769 (LRFD 5.4.2.3.2-4) tf tfinal ti tdeck t tf ti t 19910 day ktd t day 61 4 f'ci ksi  t day   ktd 0.998 (LRFD 5.4.2.3.2-5) εddf kvs khs kf ktd 0.48 10 3  εddf 376 10 6  Girder concrete shrinkage strain between CIP placement and final time: The girder concrete shrinkage between deck placement and final time is the difference between the shrinkage at time of deck placement and the total shrinkage at final time. εbdf εbif εbid εbdf 114 10 6  Loss from Transfer to CIP Placement: The prestress loss from transfer of prestress to placement of CIP consists of three loss components: shrinkage of the girder concrete, creep of the girder concrete, and relaxation of the strands. That is, Time-Dependent Loss from Transfer to CIP Placement = fpSR+fpCR+fpR1 B1-16

Shrinkage of Concrete Girder: ΔfpSR εbid Ep Kid= (LRFD5.9.5.4.2a-1) where: bid = Concrete shrinkage strain of girder between transfer and CIP placement. Computed using LRFD Eq. 5.4.2.3.3-1 Ep = Modulus of elasticity of prestressing strand (ksi). Kid = Transformed steel coefficient that accounts for time-dependent interaction between concrete and bonded steel in the section being considered for the time period between transfer and CIP placement. The transformed section coefficient, kid, is computed using: Kid 1 1 Ep Eci Aps A  1 A epg 2  Ig       1 0.7 ψb tf ti   = (LRFD Eq. 5.9.5.4.2a-2) where: epg = Eccentricity of strands with respect to centroid of girder (in). b(tf,ti) = Creep coefficient at final time due to loading introduced at transfer. epg ecc epg 5.42 in I 27120 in 4  A 936 in 2  Kid 1 1 Ep Eci Aps A  1 A epg 2  I       1 0.7 ψbfi   Kid 0.8745 Therefore, the prestress loss due to shrinkage of the girder concrete between time of transfer and CIP placement is: ΔfpSR εbid Ep Kid ΔfpSR 6.526 ksi Creep of Concrete Girder: ΔfpCR Ep Eci fcgp ψb td ti  Kid= (LRFD 5.9.5.4.2b-1) where: fcgp = Concrete stress at cg of prestress pattern due to the prestressing force immediately after transfer and the self-weight of the girder at the section of maximum moment (ksi). Section modulus at cg of strand pattern: epti ecc epti 5.42 in Scgp I epti  Scgp 5004 in 3  B1-17

pInitial prestress force: fpj Pull fpu Pinit fpj Aps Pinit 1054.6 kip fcgp Pinit 1 A epti Scgp         Mswf2 Scgp  fcgp 1.567 ksi ΔfpCR Ep Eci fcgp ψbdi Kid ΔfpCR 8.829 ksi For this example, it will be assumed to be equal to 1.2 ksi (Article 5.9.5.4.2b permits this). ΔfpR1 1.2 ksi Total prestress loss at time of CIP placement: ΔfpLTid ΔfpSR ΔfpCR ΔfpR1 Calculated above: ΔfpSR 6.53 ksi ΔfpCR 8.83 ksi ΔfpR1 1.20 ksi ΔfpLTid 16.555 ksi Loss from CIP Placement to Final: The prestress loss from placement of CIP to final conditions consists of four loss components: shrinkage of the girder concrete, creep of the girder concrete, and relaxation of the strands. That is, Time-Dependent Loss from CIP Placement to Final = fpSD + fpCD + fpR2 - fss Shrinkage of Concrete Girder: ΔfpSD εbdf Ep Kdf= (LRFD5.9.5.4.3a-1) where: bdf = Concrete shrinkage strain of girder between time of CIP placement and final time. Computed using LRFD Eq. 5.4.2.3.3-1 Kdf = Transformed steel coefficient that accounts for time-dependent interaction between concrete and bonded steel in the section being considered for the time period between the time of CIP placement and final time. Compute Kdf: Kdf 1 1 Ep Eci Aps Ac  1 Ac epc 2  Ic       1 0.7ψb tf ti   = (LRFD 5.9.5.4.3a-2) B1-18

where: epc = Eccentricity of strands with respect to centroid of composite section Ac = For composite sections, the gross area of the composite section should be used. However, since this girder is non-composite, the gross area of the non-composite section is used. Ic = Gross area of composite section for composite systems, gross area of bare beam for non-composite systems. b(tf,ti) = Girder creep coefficient epc ecc epc 5.42 in Ac: A 936 in 2  Ic: I 27120 in 4  Kdf 1 1 Ep Eci Aps A  1 A epc 2  I       1 0.7ψbfi   Kdf 0.874 Therefore, prestress loss due to shrinkage of girder concrete between CIP placement and final is: ΔfpSD εbdf Ep Kdf ΔfpSD 2.842 ksi Creep of Concrete Girder: ΔfpCD Ep Eci fcgp ψb tf ti  ψb td ti   Kdf Ep Ec Δfcd ψb tf td  Kdf 0.0= (LRFD5.9.5.4.3b-1) where: fcd = Change in concrete stress at centroid of prestressing strands due to long-term losses between transfer and CIP placement combined with superimposed loads (ksi). b(tf,td) = Girder creep coefficient at final time due to loading at CIP placement per Eq. 5.4.2.3.2-1. Let: ΔfpCD ΔfpCD1 ΔfpCD2= compute fpCD1: ΔfpCD1 Ep Eci fcgp ψbfi ψbdi  Kdf ΔfpCD1 3.844 ksi compute fpCD2: compute fcd: Δfcd ΔP 1 Ag epg 2 Ig         Mfws Mbarrier Mdeck Scgp       = ΔP ΔfpLTid Aps ΔP 86.2 kip epg yb ycg epg 5.42 in B1-19

Δfcd ΔP 1 A epg 2 I       Mfws2 Mbarrier2  Mdeck2  Scgp        Δfcd 0.924 ksi Ec 33000 1.0 0.14 f'c 1000 ksi        1.5  f'c ksi  ksi Ec 4921 ksi ΔfpCD2 Ep Ec Δfcd ψbfd Kdf ΔfpCD2 4.016 ksi Therefore, ΔfpCD ΔfpCD1 ΔfpCD2 ΔfpCD 0.171 ksi Shrinkage of the CIP deck: Ad Aslab n  Ad 432 in 2  Ecd 33000 wc 1.5  kcf 1.5  f'ct ksi .5  Ecd 3834 ksi ed h 2  ed 9 in Δfcdf εddf Ad Ecd 1 0.7 ψdfd  1 Ac epc ed Ic         Δfcdf 0.498 ksi ΔfpSS Ep Ec Δfcdf Kdf 1 0.7 ψbfd  ΔfpSS 4.0364 ksi Relaxation of Prestressing Strands: ΔfpR2 ΔfpR1 ΔfpR2 1.2 ksi (LRFD 5.9.5.4.3c-1) Total prestress loss from CIP placement to final, therefore, is: ΔfpLTdf ΔfpSD ΔfpCD ΔfpR2 ΔfpSS Calculated above: ΔfpSD 2.84 ksi ΔfpCD 0.17 ksi ΔfpR2 1.2 ksi ΔfpSS 4.0364 ksi ΔfpLTdf 7.907 ksi Summary of Time-Dependent Losses Losses from Transfer to CIP Placement Girder shrinkage: ΔfpSR 6.53 ksi Girder creep: ΔfpCR 8.83 ksi Strand relaxation: ΔfpR1 1.2 ksi_______________ Total = ΔfpLTid 16.555 ksi B1-20

Losses from CIP Placement to Final Girder shrinkage: ΔfpSD 2.84 ksi Girder creep: ΔfpCD 0.17 ksi Strand relaxation: ΔfpR2 1.2 ksi Differential Shrinkage:ΔfpSS 4.0364 ksi _______________ Total = ΔfpLTdf 7.907 ksi ΔfpLT ΔfpLTid ΔfpLTdf ΔfpLT 24.46 ksi ΔfPT ΔfpES ΔfpLT ΔfPT 33.45 ksi %Loss ΔfPT Pull fpu 100 %Loss 16.52 Check effective stress after losses: fpe Pull fpu ΔfPT fpe 169.1 ksi fallow 0.80 fpy fallow 194.4 ksi (LRFD 5.9.3-1) Stresses Due to Prestress at End of Transfer Length and Midspan At Release Conditions: j 1 2 fpsrbj Pr 1 A ecc Sb       fpsrb T 2.747 2.747( ) ksi fpsrtj Pr 1 A ecc St       fpsrt T 0.845 0.845( ) ksi At Final Conditions: distj xfj Lovr Ldes 2  dist T 2.44 25( ) ft Lt 60 db Lt 36 in dtj if distj Lt 1.0 distj Lt         dt T 0.8133 1( ) (Fraction strands are transferred.) j 1 2 Pfj fpe dtj No_Strands Astrand Pf T 716.1 880.4( ) kip fpsbj Pfj 1 A ecc Sb       fpsb 1.97 2.422       ksi fpstj Pfj 1 A ecc St       fpst 0.606 0.745       ksi B1-21

Service Stress Check At Release Conditions: Top of precast section (tension): frtj fpsrtj fswrtj  frt T 0.554 0.447( ) ksi fallow_rt 0.24 f'ci ksi fallow_rt 0.563 ksi Status_ServiceLSrtj if frtj fallow_rt "OK" "NG"   Status_ServiceLSrt T "OK" "OK"( ) Bottom of precast section (compression): frbj fpsrbj fswrbj  frb T 2.491 1.612( ) ksi fallow_rc 0.6 f'ci fallow_rc 3.3 ksi Status_ServiceLSrcj if frtj fallow_rc "OK" "NG"   Status_ServiceLSrc T "OK" "OK"( ) Compute required amount of top tension steel at transfer point: xtt h frt1 frt1 frb1           xtt 5.14 in (LRFD C5.9.4.1.2) Ttt frt1 2 bv xtt Ttt 102.49 kip Att Ttt 30 ksi  Att 3.4162 in 2  Assume #8 bars will be used as top tension steel. Required number of bars: Ntt Att 0.79 in 2   Ntt 4 At Final Conditions: Check Service Limit States: Service III Limit State (Tensile Stresses in Bottom of Beam): fIIIbj fpsbj fswbj  fdeckbj  fbarrierbj  ffwsbj  0.8 fLLbj  0.5 fTGb fIIIb T 1.672 0.271( ) ksi fallow_ft 0.19 f'c ksi fallow_ft 0.503 ksi (LRFD 5.9.4.2.2b) Status_ServiceLSftj if fIIIbj fallow_ft "OK" "NG"   Status_ServiceLSft T "OK" "OK"( ) Service I (Compressive Stresses in Top of Beam): Compressive Stress Due to Permanent Loads: fIdtj fpstj fswtj  fdecktj  fbarriertj  ffwstj  fTGt fIdt T 0.1912 1.6673( ) ksi fallow_fcd 0.45 f'c fallow_fcd 3.15 ksi (LRFD 5.9.4.2.1) Status_ServiceLSfcdj if fIdtj fallow_fcd "OK" "NG"   Status_ServiceLSfcd T "OK" "OK"( ) B1-22

Compressive Stress Due to Full Dead Load + Live Load: fIltj fpstj fswtj  fdecktj  fbarriertj  ffwstj  fLLtj  0.5 fTGt fIlt T 0.1312 2.1476( ) ksi fallow_fcl 0.6 f'c fallow_fcl 4.2 ksi (LRFD 5.9.4.2.1) Status_ServiceLSfclj if fIltj fallow_fcl "OK" "NG"   Status_ServiceLSfcl T "OK" "OK"( ) Service I (Compressive Stresses in Top of Topping): Only loads acting on composite section cause stresses in the topping. Since flange slenderness ratio is less than 15, set ϕw = 1. Compressive Stress Due to Permanent Loads: fIdttj fbarrierttj ffwsttj  fTGtt fIdtt T 1.085 1.1662( ) ksi Status_ServiceLSfcdtj if fIdttj fallow_fcd "OK" "NG"   Status_ServiceLSfcdt T "OK" "OK"( ) Compressive Stress Due to Full Dead Load + Live Load: fIlttj fbarrierttj ffwsttj  fLLttj  0.5 fTGtt fIltt T 0.6777 1.3682( ) ksi Status_ServiceLSfcltj if fIlttj fallow_fcl "OK" "NG"   Status_ServiceLSfclt T "OK" "OK"( ) 1.8 Flexural Strength Check Muj 1.25 Mswfj Mdeckj  Mbarrierj    1.5 Mfwsj  1.75 MLLj  Mu T 257 1575( ) kip ft β1 if f'ct 4 ksi( ) 0.85 if f'ct 8 ksi( ) 0.65 0.85 f'ct 4 ksi( ) 1 ksi( ) 0.05                      β1 0.85 (LRFD 5.7.2.2) Preliminary estimate of Ld: Ld 270.0 ksi 2 3 fpe       db ksi 1  Ld 94.38 in (LRFD Eq. 5.11.4.2-1) Kld if h 24 in 1.0 1.6( ) Kld 1 dfj if distj Lt distj Lt fpe fpu  if distj Kld Ld fpe distj Lt Kld Ld Lt       fpu fpe  fpu  1.0                       df T 0.5092 1( ) (fraction strands are developed) Apsj No_Strands dfj Astrand Aps T 2.6522 5.208( ) in 2  b beff dp h th tslab tws ycg k 2 1.04 fpy fpu         k 0.28 (LRFD 5.7.3.1.1-2) hf tslab B1-23

cj Apsj fpu 0.85 f'ct β1 b k Apsj  fpu dp   c T 3.35 6.42( ) in (LRFD 5.7.3.1.1-4) bwj if cj hf b bv  bwT 72 72( ) in cj Apsj fpu 0.85 f'ct b bwj    hf 0.85 f'ct β1 bwj  k Apsj  fpu dp   c T 3.35 6.42( ) in (LRFD 5.7.3.1.1-3) fpsj fpu 1 k cj dp         fps T 263 256.5( ) ksi (LRFD 5.7.3.1.1-1) aj β1 cj a T 2.8489 5.4573( ) in Ld fps1 2 3 fpe       db ksi 1  Ld 90.16 in (LRFD 5.11.4.1-1) distj xfj Lovr Ldes 2  dist T 2.44 25( ) ft dfj if distj Lt distj Lt fpe fpsj  if distj Kld Ld fpe distj Lt Kld Ld Lt       fpsj fpe    fpsj  1.0                       df T 0.5229 1( ) (fraction strands are developed) Apsj No_Strands dfj Astrand Aps T 2.7231 5.208( ) in 2  Mnj Apsj fpsj  dp aj 2         0.85 f'c b bwj    hf aj 2 hf 2         (LRFD 5.7.3.2.2-1) Mn T 2063 3704( ) kip ft Compute phi for each section: ϕfj 0.583 0.25 dp cj 1        ϕf T 3.02 1.73( ) (LRFD Eq. 5.5.4.2.1-1) ϕfj if ϕfj 0.75 0.75 if ϕfj 1.0 1.0 ϕfj      ϕf T 1.00 1.00( ) ϕMnj ϕfj Mnj  Mrj ϕMnj  Mr T 2063 3704( ) kip ft Mu T 257 1575( ) kip ft Status_StrengthLSj if Muj Mrj  "OK" "NG"   Status_StrengthLS T "OK" "OK"( ) B1-24

Maximum Steel Check: Note: The provisions contained in Art. 5.7.3.3.1 to check maximum reinforcement were deleted in 2005. This check is now effectively handled by varying phi, depending upon whether the section is compression or tension controlled. See Art. 5.5.4.2.1. Minimum Steel Check: Compute Cracking Moment at Midspan: fr 0.37 f'c ksi fr 0.979 ksi Mcr Sbc fr fpsb2    Mswf2 Mdeck2   Sbc Sb 1        Mcr if Mcr Sbc fr Sbc fr Mcr  Mcr 1265 kip ft 1.2 Mcr 1518 kip ft Ref: Mr2 3704.1 kip ft 1.33 Mu2  2095.3 kip ft Mmin if 1.2 Mcr 1.33 Mu2  1.2Mcr 1.33 Mu4    Mmin 1518 kip ft Status_MinStl if Mmin Mr2  "OK" "NG"   Status_MinStl "OK" 1.9 Vertical Shear Design At each section the following must be satisfied for shear: Vu Vr (LRFD 5.8.2.1-2) Vr ϕVn= Vn Vc Vs Vp= (LRFD 5.8.3.3-1) Critical Section for Shear: (LRFD 5.8.3.2) The critical section for shear near a support in which the reaction force produces compression in the end of the member is, from the face of support (Fig. 2), the greater of: a. 0.5dvcot(), or b. dv where, dv = Effective shear depth = Distance between resultants of tensile and compressive forces = de - a/2 Compute Aps & dp Note that Aps in the equation used to compute x is the area of the prestressing steel on the flexural tension side only. It is not the total area of strands. The variable Aps_ex is introduced below to handle this. B1-25

NoAps_ft Pat_n Pat_h hc_2( ) j last Pat_n( ) N 0 N N Pat_nj j j 1 break j 0=if Pat_hj hc_2 2 while N  NAps_ft NoAps_ft Pat_n Pat_h hc  NAps_ft 24 CGAps_ft Pat_n Pat_h hc_2( ) j last Pat_n( ) N 0 N_cg 0 N N Pat_nj N_cg N_cg Pat_nj Pat_hj j j 1 break j 0=if Pat_hj hc_2 2 while N_cg N  CGAps_ft CGAps_ft Pat_n Pat_h hc  CGAps_ft 3 in dpv h th tslab tws CGAps_ft dpv 36 in Aps_ex dt1 NAps_ft Astrand Aps_ex 4.236 in 2  dt1 0.8133 Compute "a" based on Aps on Flexural-Tension Side: Aps_a df1 NAps_ft Astrand Aps_a 2.723 in 2  df1 0.5229 cv Aps_a fpu 0.85 f'ct β1 b k Aps_a fps1 dpv   cv 3.44 in bwv if cv hf b bv  bwv 72 in cv Aps_a fpu 0.85 f'ct b bwv  hf 0.85 f'ct β1 bwv k Aps_a fps1 dpv   cv 3.44 in av β1 cv av 2.9252 in Mnv Aps_a fps1  dpv av 2         0.85 f'c b bwv  hf av 2 hf 2         Mnv 2061 kip ft B1-26

Compute dv: dv Mnv Aps_a fps1   dv 34.537 in (LRFD C5.8.2.9-1) But dv need not be taken less than the greater of 0.9de and 0.72h. Thus, 0.9 dp 32.4 in 0.72 hc 17.28 in Min_dv if 0.9 dp 0.72 hc 0.9 dp 0.72 hc  Min_dv 32.4 in dv if dv Min_dv Min_dv dv  dv 34.537 in To compute critical section, assume: θ 20.8 deg 0.5 dv cot θ( ) 45.4602 in Crit_sec if dv 0.5 dv cot θ( ) dv 0.5 dv cot θ( )  (LRFD 5.8.2.7) Crit_sec 45.46 in Crit_sec 3.788 ft Assuming that the distance from the face of support to the centerline of bearing is half the bearing pad length, the critical section for shear is: xf1 Crit_sec Lpad 2  xf1 4.288 ft (Note: Compare this to previous assumption) At the critical section, the factored shear is: Vu 1.25 Vswf1 Vdeck1  Vbarrier1    1.5 Vfws1  1.75 VLL1  Vu 129.6 kip Since the pattern profile for PCSSS beams will always be straight, there will be no vertical component of the prestressing force, Vp: Vp 0.0 kip Compute maximum permissible shear capacity at a section: Vr_max ϕv 0.25 f'c bv dv Vp  Vr_max 3917 kip (LRFD 5.8.3.3-2) Status_Vrmax if Vu Vr_max "OK" "NG"  Status_Vrmax "OK" The shear contribution from the concrete, Vc, is given by: Vc 0.0316 β fc bv dv= (LRFD 5.8.3.3-3) To calculate β, first compute fpo: This can be taken as 0.70fpu per the 2000 Interim Specifications: fpo 0.75 fpu fpo 202.5 ksi (LRFD 5.8.3.4.2) In the 2008 Interim the procedure for the calculation of θ and β was moved to an appendix. The new procedure for the calculation of these two values involves a new value, εs. Check lower bound for Mu: MuLB if Mu1 dv Vu Vp  dv Vu Vp  Mu1  MuLB 372.9 kip ft εs MuLB dv Vu Vp  Aps_ex fpo Ep Aps_ex  εs 0.00495882 (LRFD 5.8.3.4.2-4) B1-27

If εs is less than zero, it can be taken equal to zero: εs if εs 0 0.0 εs  εs 0 β 4.8 1 750 εs  β 4.8 (LRFD 5.8.3.4.2-1) θ 29 3500 εs θ 29 (LRFD 5.8.3.4.2-3) New value for Vc Vc 0.0316 β f'c ksi bv dv Vc 997.9 kip Required Vs is, therefore: Vs Vu ϕv Vc Vp Vs 854 kip Assuming two vertical legs of No. #4 bars: Av Vs fy dv cot θ( )  Av 4.387 in 2 ft  (LRFD C5.8.3.3-1) Spac 2 0.2 in 2  Av  Spac 1.1 in (stirrup spacing) Check minimum transverse reinforcement: Av_min 0.0316 f'c ksi bv fy  Av_min 1.2 in 2 ft  (LRFD 5.8.2.5-1) Check maximum stirrup spacing: (LRFD 5.8.2.7-2) Vspc 0.1 f'c bv dv Vspc 1740.7 kip Ref: Vu 129.6 kip dv 34.54 in Max_spac if Vu Vspc if 0.8 dv 24 in 0.8 dv 24 in  if 0.4 dv 12 in 0.4 dv 12 in   Max_spac 24 in 1.10 Longitudinal Reinforcement Check LRFD requires that the longitudinal steel be checked at all locations along the beam. This requirement is made to ensure that the longitudinal reinforcement is sufficient to develop the required tension tie, which is required for equilibrium. Equation 5.8.3.5-1 is the general equation, applicable at all sections. However, for the special case of the inside edge of bearing at simple-end supports, the longitudinal reinforcement must be able to resist a tensile force of (Vu/ - 0.5Vs - Vp)cot(). Note that when pretensioned strands are used to develop this force, only a portion of the full prestress force may be available near the support due to partial transfer. Additionally, only those strands on the flexural tension side of the member contribute to the tension tie force. B1-28

Required Tension Tie Force: If only the minimum amount of transverse reinforcement that is required by design is provided, the required tension tie force is: FL_reqd Vu ϕv 0.5 Vs Vp       cot θ( ) FL_reqd 643.6 kip Eq. 5.8.3.5-2 However, a greater amount of stirrup reinforcement is typically provided than is required, which increases the actual Vs. Note that by Eq. 5.8.3.5-2, increasing Vs decreases the required tension tie force. Hence, it is helpful to use the computed value of Vs that results from the transverse reinforcement detailed in the design. In this case, the required tension tie force is: Assume 2 legs of No. 4 bars at 12" on center (amount of steel at the critical section for shear): Av_actual 0.4 in 2  Vs_actual Av_actual fy dv cot θ( ) 12 in  Vs_actual 77.9 kip Check the upper limit of Vs: Vs_actual_max Vu ϕv  Vs_actual_max 144 kip LRFD 5.8.3.5 Adopt the lesser of provided Vs and the upper limit of Vs: Vs_actual if Vs_actual Vs_actual_max Vs_actual Vs_actual_max  Vs_actual 77.9 kip The revised value of the required tension tie force is: FL_reqd Vu ϕv       0.5 Vs_actual Vp       cot θ( ) FL_reqd 118.4 kip Provided Tension Tie Force: The longitudinal reinforcement that contributes to the tension tie are strands that are on the flexural tension side of the precast section. Near the ends of the precast section, the strands are typically only partially effective. C5.8.3.5 of the 2006 Interim Revisions permits the strand stress in regions of partial development to be estimated using a bilinear variation, as shown in Fig. 4. Figure 6: Variation in strand stress in relation to distance from beam end. B1-29

The stress in the strands at a given section depends on the location of the section with respect to the end of the precast section. If the section is between the end of the beam and L t (see Fig. 5), a linear interpolation is performed using a stress variation of 0.0 at the end of the beam to f pe at a distance of Lt from the end of the precast section. If the section is to the right of Lt but to the left of Ld, then the stress is interpolated between fpe and fps. If the section is to the right of Ld, then the stress is assumed to be a constant value of f ps. At the face of bearing, the stress in the effective strands is: xFB Lovr Ldes 2 Lpad 2  xFB 1.00 ft (Distance from physical end of beam to face of bearing) Astr Astrand FL_prov if xFB Lt NAps_ft Astr fpe xFB Lt  if xFB Kld Ld NAps_ft Astr fpe xFB Kld Ld Lt Kld Ld Lt       fps fpe         NAps_ft Astr fpe               FL_prov 293.5 kip Status_Vl if FL_prov FL_reqd "OK" "NG"  Status_Vl "OK" Refined Estimate of Provided Tension Tie Force: If it is assumed that the point of intersection of the bearing crack (at angle theta) and c.g. of the strands is where the force in the strands is computed, then additional tensile capacity from the strands can be utilized. Figure 7: Elevation view of end of beam showing location where assumed failure crack crosses the c.g. of that portion of the strand pattern that is effective for resisting tensile forces caused by moment and shear. Distance from end of beam to point of intersection of assumed crack and center of gravity of effective strands: xc Lpad 2       CGAps_ft cot θ( ) xc 0.8 ft (Measured from L face of bearing) xc Lovr Ldes 2       Lpad 2        CGAps_ft cot θ( ) xc 1.3 ft (Measured from L end of beam) FL_prov if xc Lt NAps_ft Astr fpe xc Lt  if xc Kld Ld NAps_ft Astr fpe xc Kld Ld Lt Kld Ld Lt       fps fpe         NAps_ft Astr fpe               FL_prov 376.2 kip Status_Vl "OK"Status_Vl if FL_prov FL_reqd "OK" "NG"  B1-30

1.11 Interface Shear Design The ability to transfer shear across the interface between the top of the precast beam and the cast-in-place deck must be checked. This check falls under the interface shear or shear friction section of LRFD (5.8.4). Recall that under the Standard Specs, this check falls under the horizontal shear section. Little guidance is offered by the LRFD Specs on how to compute the applied shear stress at the strength limit state. The procedure presented here uses the approach recommended by the PCI Bridge Design Manual, which is a strength limit state approach. Applied Factored Shear: Vu 129.6 kip vuh_s Vu dv bv  vuh_s 0.052 ksi vnh_reqd vuh_s ϕv  vnh_reqd 0.0579 ksi Acv bv 1.0 ft Acv 864 in 2  Vnhr vnh_reqd Acv Vnhr 50 kip Nominal Shear Resistance of the Interface (Capacity): Vn cAcv μ Avf fy Pc = (LRFD 5.8.4.1-2) Interface is CIP concrete slab on clean, roughened beam surface, no reinforcement crossing shear plane: (LRFD 5.8.4.3) c 0.135 ksi (cohesion factor) μ 1.000 (friction factor) K1 0.2 (fraction of concrete strength available to resist interface shear) K2 0.8 ksi (limiting interface shear resistance) Since there is no permanent net compressive stress normal to shear plane, Pc = 0. (LRFD 5.8.4.2) Check Maximum Allowable Shear: Vni_max1 K1 f'ct Acv Vni_max1 691 kip (LRFD 5.8.4.1-4) Vni_max2 K2 Acv Vni_max2 691 kip (LRFD 5.8.4.1-5) Vnh_max if Vni_max1 Vni_max2 Vni_max1 Vni_max2  (LRFD 5.8.4.1-2,3) Vnh_max 691 kip Vnh_reqd vnh_reqd Acv Vnh_reqd 50 kip Status_Vuh_max if Vnh_reqd Vnh_max "OK" "NG"  Status_Vuh_max "OK" Assuming no horizontal shear reinforcement crossing the shear plane, provided horizontal shear resistance is: Vnh_prov c Acv Vnh_prov 116.6 kip Status_Vnh_prov if Vnhr Vnh_prov "OK" "NG"  Status_Vnh_prov "OK" B1-31

1.12 Spalling Forces If the maximum spalling stress on the end face of the girder is less than the direct tensile strength of the concrete, then spalling reinforcement is not required when the member depth is less than 22 in. The maximum spalling stress is estimated as: σs P A 0.1206 e 2 h db  0.0256     0= And the direct tensile strength is computed as: fr_dts 0.23 f'c ksi fr_dts 0.609 ksi (LRFD C5.4.2,7) Check reinforcement requirement: Ref: A 936 in2 h 18 in ecc 5.42 in db 0.6 in Pjack Aps2 fpj Pjack 1 10 3  kip σs Pjack A 0.1206 ecc 2 h db  0.0256      σs 0.341 ksi Check whether spalling stress is below threshold and thus is spalling/busting reinforcement is needed: Status_Spalling if σs fr_dts "OK" "NG"  Status_Spalling "OK" 1.13 Transverse Load Distribution The transverse load distribution reinforcement is computed by: Atld kmild Al_mild α kps Al_ps= where: α dcgs dtrans = kps 100 L fpe 60  50%= kmild 100 L 50%= dcgs hc ycg dcgs 21.0 in Compute dtrans: dtrans hc 4in db 2  0.75 in 2  dtrans 19.3 in α dcgs dtrans  α 1.0867 Assume there is no mild longitudinal reinforcement A l_mild in tension at the strength limit state. Al_mild 0.0 in 2  kmild 100 ft Ldes 100  kmild 14.29 % B1-32

kps 100 ft Ldes fpe 60 ksi  100  kps 40.25 % Al_ps Aps2  Al_ps 5.208 in 2  Total amount of transverse load distribution is: Atld kmild Al_mild α kps Al_ps Atld 2.28 in 2  Since the longitudinal reinforcement is per beam width, the area of distribution reinforcement per foot is: Atld_per_ft Atld S  Atld_per_ft 0.38 in 2 ft  Set transverse load distribution reinforcement spacing at 12 in.:Assuming transverse bars are #6, maximum spacing is: Sld_spac_max 0.44 in 2  Atld_per_ft ft ft Sld_spac_max 13.9 in Sld_spac 12in 1.14 Reflective Crack Control Reinforcement Reflective crack control reinforcement is provided from both the transverse load distribution reinforcement as well a drop in cage consisting of vertical stirrups. The total amount of reflective crack control reinforcement required is given as follows: ρcr_req 6 f'ct psi fy  ρcr_req 0.00632 (LRFD 5.14.4.3.3f-1) The crack control reinforcement ratio is defined, per unit length of span, as follows: ρcr Ascr h tflg  1 ft = (LRFD 5.14.4.3.3f-2) The required area of reinforcement of reflective crack control is therefore calculated, per unit length of span, as: Ascr_req ρcr_req h tflg  1 ft Ascr_req 1.1384 in2 The required area of cage reinforcement is subsequently calculated, per unit length of span, as the difference between the total required area of crack control reinforcement and that provided by the reinforcement for transverse load distribution; both transverse bars are effective in providing crack control, however only the lower horizontal legs of the stirrups are considered in the calculation. All calculations are per unit length of span: Ald 2Sld_spac .44 in 2 1ft  Ald 0.88 in 2  Acr_cage_req Ascr_req Ald Acr_cage_req 0.2584 in 2  B1-33

Provide No. 5 stirrups at 12 in. on center: Scage_spac 12in Ascage 0.31in 2  Acr_cage_prov Ascage 1ft Scage_spac        Acr_cage_prov 0.31 in 2  Ascr_prov Ald Acr_cage_prov Status_Ascrack if Ascr_req Ascr_prov "OK" "NG"  Status_Ascrack "OK" Figure 8: Cross section of bridge showing CIP regions. Figure 9: Detail of drop-in cage. B1-34

Figure 10: Plan view of drop-in cage. 1.14 Bottom Flange Reinforcement Determine steel required to resist construction loads on bottom flange: Assume a 1' wide strip: Loads: Self-weight of flange: wflng_sw tflg 12 in wct wflng_sw 0.0375 klf CIP weight: wflng_cip h tflg  12 in wct wflng_cip 0.1875 klf Construction live load (assume 10 psf): wflng_LL wconst 12 in wflng_LL 0.0100 klf Moments: bcant bh 2  bcant 1.00 ft (Length of cantilever) Mflng_sw wflng_sw bcant 2  2  Mflng_sw 0.0187 kip ft Mflng_cip wflng_cip bcant 2  2  Mflng_cip 0.0937 kip ft Mflng_LL wflng_LL bcant 2  2  Mflng_LL 0.005 kip ft Strength Limit State I: Mu_flng 1.25 Mflng_sw Mflng_cip  1.75 Mflng_LL Mu_flng 0.15 kip ft B1-35

Try #3 bars at 12" o.c: As_flng 0.11 in 2  As_flng 0.11 in 2  cflng As_flng fy 0.85 f'c β1 12 in  cflng 0.11 in β1p if f'c 4 ksi( ) 0.85 if f'c 8 ksi( ) 0.65 0.85 f'c 4 ksi( ) 1 ksi( ) 0.05                      β1p 0.70 aflng β1p cflng aflng 0.0761 in ds tflg 1 in 0.5 in 0.5 in 2  ds 1.25 in Mn_flng As_flng fy ds aflng 2         Mn_flng 0.67 kip ft ϕf_flng 0.65 0.15 ds cflng 1        ϕf_flng 2.22 ϕf_flng if ϕf_flng 0.75 0.75 if ϕf_flng 0.9 0.9 ϕf_flng   ϕf_flng 0.9 Mr_flng ϕf_flng Mn_flng Mr_flng 0.60 kip ft Status_StrengthLSflng if Mu_flng Mr_flng "OK" "NG"  Status_StrengthLSflng "OK" Use: Minimum #3 bars @ 12" o.c. in bottom flange. B1-36

Example Problem 2 2.1 Introduction This example covers the design of the multi-span continuous bridge superstructure consisting of precast composite slab span system (PCSSS) elements. The structural system is a three-span, continuous structure with a 40'-50'-40' span layout. The steps required to design representative composite slab of the center span are illustrated. The design is carried out in accordance with the AASHTO LRFD Bridge Design Specifications, 5th Edition (2010). 2.2 Materials, Geometry, Loads and Load Factors Units: kcf kip ft 3 Defined unit: kips per cubic foot Materials: Concrete: f'c 7.0 ksi Strength of beam concrete at 28 days f'ci 5.0 ksi Strength of beam concrete at transfer of prestressing force wc 0.150 kcf Density of beam concrete f'ct 4.0 ksi Strength of CIP concrete at 28 days wct 0.15 kcf Density of CIP concrete H 70 Average ambient relative humidity ϕu 2.1 Ultimate creep coefficient εshp_u 0.00056 Ultimate shrinkage strain in precast concrete εshc_u 0.00069 Ultimate shrinkage strain in CIP concrete α 0.000006 /oF Coefficient of thermal expansion Strand: Astrand 0.217 in 2  Area of one prestressing strand. db 0.6 in Nominal diameter of prestressed strand. fpu 270 ksi Tensile strength of prestressing steel Ep 28500 ksi Modulus of elasticity of prestressing steel fpy 0.9 fpu Yield strength of prestressing steel Pull 0.75 Pull of strands expressed as a fraction of fpu t 18 hr Time from tensioning to detensioning of strands Rebar: fy 60 ksi Yield stress of ordinary rebar Es 29000 ksi Modulus of elasticity of non-prestressed reinforcement Geometry: Beam: Section "IT" Precast section name h 18.0 in Height of precast A 936.0 in 2  Gross area of precast section I 27120 in 4  Gross moment of inertia of precast cross section about cenroidal x-x axis B2-1

yb 8.42 in Center of gravity of gross precast cross section bf 72.0 in Width of bottom flange of precast section tflg 3.00 in Effective thickness of bottom flange bv 72.00 in Shear width of precast section (web and longitudinal trough width) Ac 450 in 2  Area of concrete on flexural tension side of member (see LRFD B5.2-3) VSb 5.2 in Volume to surface ratio of precast section VSd 6.0 in Volume to surface ratio of CIP slab Slab: tslab 6.00 in Thickness of CIP slab above precast beam th 15.00 in Thickness of CIP region between precast beams bh 24.00 in Width of CIP region between precast units.beams tws 0.0 in Thickness of portion of CIP slab assumed to be wear Span: Lp2p 50.0 ft Centerline of pier to centerline of pier dimension Lovr 49.0 ft Overal length of precast section Ldes 48.0 ft Design span of precast section Lpad 12 in Length of bearing pad Pieroffset 0.5 ft Distance from centerline of pier to end of beam Brngoffset 0.5 ft Distance from end of beam to centeriline of bearing Bridge: S 6.00 ft Beam spacing Ng 8 Number of precast sections in bridge cross section Widthoverall 47.5 ft Overall width of bridge Widthctc 44.0 ft Curb to curb width of bridge Nl 2 Number of lanes Loads: Dead: Nbarriers 2 Number of barriers wbarrier 0.300 klf Weight of single barrier wfws 0.023 ksf Weight of future wearing surface allowance Live: HL-93 Notional live load per LRFD Specs wlane 0.64 klf Design lane load Construction Timing: ttransfer 1.00 day Time from release tensioning of strands to release of prestress tdeck 7 day Time when continuity is established tfinal 20000 day Assumed end of service life of bridge (time final) Load & Resistance Factors: f (variable) Resistance factor for flexure ϕv 0.90 Resistance factor for shear DLA 0.33 Dynamic load allowance (LRFD 3.6.2.1-1) B2-2

Figure 1: Plan view of bridge. Figure 2: ELevation view of bridge. Figure 3: Cross section of bridge. B2-3

Loads: Composite Dead Load: Future Wearing Surface: wfws 0.025 ksf wfws wfws Widthctc Ng  wfws 0.1375 klf (per beam) The moments and shears shown below were manually entered into this template. They represent 10th-point values of Span 2 that were generated by a 2-D continuous beam program using the model given in Fig. 5 along with the uniform load given above applied to all spans. All similarly highlighted regions represent manually entered values that were generated in a similar fashion. Span 2: Mfws_c 30.4 13.8 0.8 8.5 14.0 15.9 14.0 8.5 0.8 13.9 30.4                                 kip ft Vfws_c 3.7 3.0 2.2 1.5 0.7 0 0.7 1.5 2.2 3.0 3.7                                 kip Barrier Loads: wbarrier 0.300 klf wbarrier Nbarriers wbarrier Ng  wbarrier 0.075 klf (per beam) Span 2: Mbarrier_c 15.4 7.0 0.4 4.3 7.1 8.0 7.1 4.3 0.4 7.0 15.4                                 kip ft Vbarrier_c 1.9 1.5 1.1 0.8 0.4 0 0.4 0.8 1.1 1.5 1.9                                 kip B2-4

Live Load: HL-93 As with the composite dead load moments and shears, the moments and shears for each component of the HL-93 live load were manually entered below. Design Truck, Span 2, +M: Design Truck, Span 2, -M: Mtruck_pc 57.9 79.3 189.1 299.1 365.7 377.9 365.7 299.9 189.1 79.7 57.6                                 kip ft Vtruck_pc 59.6 52.7 45.2 37.2 29.3 21.0 15.1 9.0 5.3 5.3 5.3                                 kip Mtruck_nc 277.4 181.9 155.3 128.6 121.3 121.3 124.7 128.6 155.3 181.9 277.4                                 kip ft Vtruck_nc 5.3 5.3 5.3 9.0 15.1 21.9 29.3 37.2 45.2 52.7 59.6                                 kip Design Tandem, Span 2, +M: Design Tandem, Span 2, -M: Mtandem_pc 50.8 86.1 206.0 299.9 357.6 373.8 357.6 299.9 206.0 86.1 50.8                                 kip ft Vtandem_pc 48.6 44.6 39.8 34.4 28.6 22.6 16.8 11.3 6.3 4.7 4.7                                 kip Mtandem_nc 216.9 159.4 136.1 112.7 89.4 66.0 89.4 112.7 136.1 159.4 216.9                                 kip ft Vtandem_nc 4.7 4.7 5.4 10.2 15.6 21.4 27.4 33.2 38.7 43.7 47.9                                 kip Design Lane, Span 2, +M Design Lane, Span 2, -M Mlane_pc 17.1 17.9 42.7 81.0 105.0 113.0 105.0 81.0 42.7 17.9 17.1                                 kip ft Vlane_pc 17.6 14.5 11.7 9.1 7.0 5.2 3.8 2.7 2.1 1.7 1.6                                 kip Mlane_nc 148.6 77.3 46.2 44.5 40.1 40.1 40.1 44.5 46.2 77.3 148.6                                 kip ft Vlane_nc 1.7 1.6 2.1 2.7 3.8 5.2 7.0 9.1 11.7 14.5 17.6                                 kip B2-5

Load & Resistance Factors: Load Factors: γpDC 1.25 γpDW 1.5 γLL 1.75 Resistance Factors: Flexure: (variable) ϕfn 1.00 Shear: ϕv 0.90 Strand Pattern: Pat_n 0 0 12 14              Pat_h 16 6 4 2             in As 8 0.79 in 2  As_top 12 0.44 in 2  (longitudinal rebar in slab) No, Strands Elevation (in) 2.3 Section Properties Figure 4: Section dimensions. Non-Composite Section Properties: Sb I yb  Sb 3220.9 in 3  yt h yb St I yt  St 2830.9 in 3  Effective Width: (LRFD 4.6.2.6.1) The effective width of the composite section may be taken as 1/2 the distance to the adjacent beam beff S 2 S 2  beff 72 in n f'ct f'c  n 0.7559 btran n beff btran 54.43 in B2-6

Composite Section Properties: Aslab btran tslab tws  Aslab 326.6 in2 Ah n th bh Ah 272.1 in2 Acomp A Ah Aslab Acomp 1534.7 in 2  ybc A yb Ah tflg th 2         Aslab h tslab tws 2         Acomp  ybc 11.466 in hc h tslab tws hc 24 in ytc hc ybc ytc 12.53 in Islab btran tslab tws 3 12  Islab 979.7 in 4  Ih n bh th 3  12  Ih 5102.5 in 4  Ic I A yb ybc 2 Ih Ah tflg th 2  ybc       2  Islab Aslab h tslab tws 2  ybc       2  Ic 71824 in 4  Sbc Ic ybc  Sbc 6264.3 in 3  Stc Ic ytc n  Stc 7580.3 in 3  Composite section modulus at the top of the prestressed beam: ytcb h ybc Stcb Ic ytcb  Stcb 10991.8 in 3  2.4 Strand Pattern Properties Figure 5: Strand pattern. No_Strands Pat_n No_Strands 26 i 1 last Pat_n( ) ycg i Pat_ni Pat_hi  No_Strands  ycg 2.9231 in ecc yb ycg ecc 5.50 in B2-7

2.5 Moments and Shears At Release: At release, when the prestress force is transferred to the beam, the structural model is a simple-span beam. The assumed length of the beam can be the overall length of the beam, or it can be somewhat less than the overall length to model supports that are located some distance in from the ends of the beam. For this example, the effective beam length at release will be assumed to be the overall length of the precast beam. While the beam is in the prestress yard, there are two locations along the beam that are potentially of interest: the transfer point of the strands and the midspan of the beam. For beams with no debonding, the net stress due to prestress and beam self-weight will achieve maximum and minimum values at one or more of these locations . 1. Transfer point of strands: Lt 60 db Lt 36 in xr1 Lt (LRFD 5.8.2.3) 2. Midspan of beam: xr2 Lovr 2  xr T 3 24.5( ) ft beam self-weight at release: wsw wc A wsw 0.975 klf i 1 2 Mswri wsw xri  2 Lovr xri    Mswr T 67.3 292.6( ) kip ft At Final Conditions: Check Points: At final conditions, there are two structural models that are required to perform the analysis. For a composite system, such as this one, some of the loads act on the bare precast beam and some of the loads act on the continuous, multi-span system. Using the principle of superposition, the effects of the loads that act on the simple-span model can be added to the moments, shears, and stresses caused by loads that act on the continuous system. Two frames of reference are convenient for locating checkpoints at final conditions on a continuous structure. For checking stresses in the beam, for checking the positive flexural capacity, and for computing the required transverse reinforcement (i.e., stirrups), the centerline of the left bearing of the beam will be the reference point and will be designated xf. For computing gross moments and shears acting on the continuous system and for assessing the negative flexural capacity of the system, the centerline of the left pier of the span will be used as the point of reference and will be designated xfc. For this particular structure, for the first frame of reference, there are four points of interest in Span 2 at final conditions: 1. Centerline of left bearing xf1 0.0 ft (Note: All final checkpoints are referenced to L. bearing) 2. Left transfer point of strands: xf2 Lt Lovr Ldes 2  3. Left critical section for shear. The critical section for shear is dv from the face of the support. This can be conservatively assumed to be 0.72h. xf3 0.72 hc Lpad 2  B2-8

4. Midspan of beam: xf4 Ldes 2  xf T 0 2.5 1.94 24( ) ft For the second frame of reference there are eleven points of interest in Span 2 at final conditions, which correspond to tenth points of Span 2, measured from centerline of the left pier (Pier 2) to the centerline of the right pier (Pier 3) : k 1 11 Lp2p 50 ft xfck k 1 10 Lp2p xfc T 1 2 3 4 5 6 7 8 9 10 11 1 0 5 10 15 20 25 30 35 40 45 50 ft Loads Acting on Simple-Span Model beam self-weight at final: j 1 4 Mswfj wsw xfj  2 Ldes xfj    Mswf 0 55.5 43.6 280.8             kip ft Vswfj wsw Ldes 2 xfj         Vswf 23.4 21 21.5 0             kip CIP Weight: wd tslab tws  S th bh  wct wd 0.825 klf Mdeckj wd xfj  2 Ldes xfj    Mdeck 0 46.9 36.9 237.6             kip ft Vdeckj wd Ldes 2 xfj         Vdeck 19.8 17.7 18.2 0             kip Loads Acting on Continuous Model For beam design itself, the results of the continuous structure (10th points, measured from centerline of pier to centerline of pier) must be mapped onto the checkpoints of the beam. The following function maps the results from the continuous beam analysis to the beam check points at final using linear iterpolation. Map PO BO xf xfc MVc  jmax last xfc  j 1 j j 1 break j jmaxif PO BO xfi   xfcj while C PO BO xfi  xfcj 1  xfcj xfcj 1   MVi MVcj 1 C MVcj MVcj 1  i 1 last xf for MV  B2-9

Barrier Weight (composite, continuous structure): Mbarrier Map Pieroffset Brngoffset xf xfc Mbarrier_c  MbarrierT 13.72 9.52 10.4608 8( ) kip ft Vbarrier Map Pieroffset Brngoffset xf xfc Vbarrier_c  VbarrierT 1.82 1.62 1.6648 0( ) kip Future Wearing Surface (composite, continuous structure): Mfws Map Pieroffset Brngoffset xf xfc Mfws_c  MfwsT 27.1 18.8 20.6 15.9( ) kip ft Vfws Map Pieroffset Brngoffset xf xfc Vfws_c  VfwsT 3.6 3.2 3.3 0( ) kip Live Load: Dynamic Load Allowance (impact): DLA 0.33 DLAf 0.15 (LRFD 3.6.2.1-1) Distribution Factors: Assume superstructure acts like a slab-type bridge. Utilize provisions of LRFD Art. 4.6.2.3 to compute width of equivalent strip to resist lane load. Single-Lane Loading: L1 if Ldes 60 ft 60 ft Ldes  L1 48.00 ft W1_1 if Widthoverall 30 ft 30 ft Widthoverall  W1_1 30.00 ft Estrip1 10 in 5.0 in L1 ft W1_1 ft  Estrip1 200 in Estrip1 16.6 ft Put in terms of fraction of one lane to be distributed to one precast unit: DF1lane bf Estrip1  DF1lane 0.3605 Double-Lane Loading: W1_2 if Widthoverall 60 ft 60 ft Widthoverall  W1_2 47.50 ft Estrip2 84 in 1.44 in L1 ft W1_2 ft  Estrip2 153 in Estrip2 12.7 ft DF2lane bf Estrip2  DF2lane 0.4713 Governing Case: DF if DF1lane DF2lane DF1lane DF2lane  DF 0.4713 This distribution factor is applicable to both shear and moment: DFm DF DFv DF B2-10

Live Load (HL-93) Moments & Shears: Positive Moment Envelope: Max_Vehicle Vehicle1 Vehicle2  Vehiclei Vehicle1i  Vehicle1i Vehicle2i if Vehiclei Vehicle2i  otherwise i 1 last Vehicle1 for Vehicle  MVehicle_pc Max_Vehicle Mtruck_pc Mtandem_pc  MVehicle_pc T 1 2 3 4 5 6 7 8 9 10 11 1 58 86 206 300 366 378 366 300 206 86 58 kip ft Mvehicle_p Map Pieroffset Brngoffset xf xfc MVehicle_pc  Mvehicle_pT 64 78 74 378( ) kip ft Mlane_p Map Pieroffset Brngoffset xf xfc Mlane_pc  Mlane_pT 17 18 18 113( ) kip ft MLL_pj DFm Mlane_pj 1.0 DLA( ) Mvehicle_pj    MLL_p T 48 57 55 290( ) kip ft VVehicle_pc Max_Vehicle Vtruck_pc Vtandem_pc  VVehicle_pc T 1 2 3 4 5 6 7 8 9 10 11 1 60 53 45 37 29 23 17 11 6 5 5 kip Vvehicle_p Map Pieroffset Brngoffset xf xfc VVehicle_pc  Vvehicle_pT 58.2 54.8 55.5 22.6( ) kip Vlane_p Map Pieroffset Brngoffset xf xfc Vlane_pc  Vlane_pT 17 15.4 15.8 5.2( ) kip VLL_pj DFv Vlane_pj 1.0 DLA( ) Vvehicle_pj    VLL_p T 44.5 41.6 42.3 16.6( ) kip Negative Moment Envelope: Min_Vehicle Vehicle1 Vehicle2  Vehiclei Vehicle1i  Vehicle1i Vehicle2i if Vehiclei Vehicle2i  otherwise i 1 last Vehicle1 for Vehicle  MVehicle_nc Min_Vehicle Mtruck_nc Mtandem_nc  MVehicle_nc T 1 2 3 4 5 6 7 8 9 10 11 1 -277 -182 -155 -129 -121 -121 -125 -129 -155 -182 -277 kip ft Mvehicle_n Map Pieroffset Brngoffset xf xfc MVehicle_nc  Mvehicle_nT 258 211 221 121( ) kip ft Mlane_n Map Pieroffset Brngoffset xf xfc Mlane_nc  Mlane_nT 134 99 107 40( ) kip ft B2-11

MLL_nj DFm Mlane_nj 1.0 DLA( ) Mvehicle_nj    MLL_n T 225 179 189 95( ) kip ft VVehicle_nc Min_Vehicle Vtruck_nc Vtandem_nc  VVehicle_nc T 1 2 3 4 5 6 7 8 9 10 11 1 -5 -5 -5 -10 -16 -22 -29 -37 -45 -53 -60 kip Vvehicle_n Map Pieroffset Brngoffset xf xfc VVehicle_nc  Vvehicle_nT 5 5 5 22( ) kip Vlane_n Map Pieroffset Brngoffset xf xfc Vlane_nc  Vlane_nT 2 2 2 5( ) kip VLL_nj DFv Vlane_nj 1.0 DLA( ) Vvehicle_nj    VLL_n T 4 4 4 16( ) kip Positive Restraint Moment: (Peterman, 1998) Restraint moments are calculated using the Peterman Method (P-Method) Compute restraint moments at the piers due to shrinkage and creep effects at the end of the service life of the bridge, assumed to be time t final. Continuity is established at time tdeck. Calculate the moment due to eccentric prestressing: Estimate stress in prestressing strands at time of continuity as: fpe 185 ksi Pps No_Strands fpe Astrand Pps 1.0 10 3  kip Mp Pps ecc( ) Mp 478.1 kip ft Dead load moments for Span 2: Mdp2 Mswf4  Mdp2 280.8 kip ft Mdd2 Mdeck4  Mdd2 237.6 kip ft Dead load moments for Span 1 are assumed to be 70% of those of Span 2: Mdp1 0.70 Mdp2 Mdp1 196.6 kip ft Mdd1 0.70 Mdd2 Mdd1 166.32 kip ft Estimate differential shrinkage at end of service life of bridge: εsh kvs khs kf ktd 0.48 10 3 = (LRFD 5.4.2.3.3-1) where: kvs = Factor for the effect of volume to surface ratio. khs = Factor for humidity. kf = Factor for strength of concrete. ktd = Factor for time development. Shrinkage strain in precast at time continuity is established: kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0. (LRFD 5.4.2.3.2-1) kvs 1.000 B2-12

khs 2.00 0.014 H khs 1.020 (LRFD 5.4.2.3.3-2) kf 5 1 f'ci ksi   kf 0.833 (LRFD 5.4.2.3.2-4) td tdeck td 7 day ti ttransfer ti 1 day t td ti t 6 day ktd t day 61 4 f'ci ksi  t day   ktd 0.128 (LRFD 5.4.2.3.2-5) εbid kvs khs kf ktd 0.48 10 3  εbid 52 10 6  Shrinkage strain in precast at time final: kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0. (LRFD 5.4.2.3.2-1) kvs 1.000 khs 2.00 0.014 H khs 1.020 (LRFD 5.4.2.3.3-2) kf 5 1 f'ci ksi   kf 0.833 (LRFD 5.4.2.3.2-4) tf tfinal ti ttransfer t tf ti t 19999 day ktd t day 61 4 f'ci ksi  t day   ktd 0.998 (LRFD 5.4.2.3.2-5) εbif kvs khs kf ktd 0.48 10 3  εbif 407 10 6  Net precast shrinkage strain: εshp εbif εbid εshp 0.000355 Shrinkage strain in CIP at time final: εsh kvs khs kf ktd 0.48 10 3 = (LRFD 5.4.2.3.3-1) where: kvs = Factor for the effect of volume to surface ratio. khs = Factor for humidity. kf = Factor for strength of concrete. ktd = Factor for time development. B2-13

kvs 1.45 0.13 VSd in  kvs 0.670 Note: kvs must be greater than 1.0. (LRFD 5.4.2.3.2-1) kvs 1.000 khs 2.00 0.014 H khs 1.020 (LRFD 5.4.2.3.3-2) kf 5 1 0.8 f'ct ksi   kf 1.190 (LRFD 5.4.2.3.2-4) tf tfinal ti tdeck t tf ti t 19993 day ktd t day 61 4 0.8 f'ct ksi  t day   ktd 0.998 (LRFD 5.4.2.3.2-5) εddf kvs khs kf ktd 0.48 10 3  εddf 581 10 6  Differential shrinkage: εsh εddf εshp εsh 0.000226 Calculate the uniform shrinkage moment: Ed 33000 wc 1.5  kcf 1.5  f'ct ksi .5  Ed 3834 ksi ee ytcb ee 6.5343 in Epr 33000 wc 1.5  kcf 1.5  f'c ksi .5  Epr 5072 ksi Ad Aslab n  Ad 432 in 2  η 1 1 Epr A Ed Ad        1 1 Es As_top Ed Ad               η 0.2368 Ms εsh Ed Ad ee tslab 2         η Ms 70.5 kip ft Creep effects on precast Mp and Md: Creep in precast at continuity: ti ttransfer td tdeck t td ti t 6 day ktd t day 61 4 f'ci ksi  t day   ktd 0.128 B2-14

kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0. (LRFD 5.4.2.3.2-1) kv 1.000 khc 1.56 0.008H khc 1.000 (LRFD Eq. 5.4.2.3.2-3) kf 5 1 f'ci ksi   kf 0.833 (LRFD 5.4.2.3.2-4) ψbdi 1.9 kvs khc kf ktd ti day       0.118  ψbdi 0.156 Creep in precast at time final: kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0. (LRFD 5.4.2.3.2-1) kvs 1.000 khc 1.56 0.008H khc 1.000 (LRFD Eq. 5.4.2.3.2-3) kf 5 1 f'ci ksi   kf 0.833 (LRFD 5.4.2.3.2-4) tf tfinal ti ttransfer t tf ti t 19999 day ktd t day 61 4 f'ci ksi  t day   ktd 0.998 (LRFD 5.4.2.3.2-5) ψbfi 1.9 kvs khc kf ktd ti day       0.118  ψbfi 1.580 Creep effects on CIP Md and Ms: ti tdeck t tf ti t 19993 day ktd t day 61 4 f'ci ksi  t day   ktd 0.998 kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0. (LRFD 5.4.2.3.2-1) kvs 1.000 khc 1.56 0.008H khc 1.000 (LRFD Eq. 5.4.2.3.2-3) kf 5 1 f'ci ksi   kf 0.833 (LRFD 5.4.2.3.2-4) B2-15

ψbfd 1.9 kvs khc kf ktd ti day       0.118  ψbfd 1.256 Compute Mr at interior face of Pier 1: Factors: F1 1 e ψbfi   1 e ψbdi  F1 0.6492 F2 1 e ψbfd  F2 0.7152 Using moment distribution, the following values were obtained: Moment at inside face of Pier 1 due to prestress: Mrp 462 kip ft Moment at inside face of Pier 1 due to differential shrinkage: Mrs 68 kip ft Moments due to superimposed dead load: Mrdp 182 kip ft Mrdd 153 kip ft Calculated restraint moment: Mrm Mrp Mrdp  F1 Mrdd F2 Mrs F2 ψbfd        Mrm 34 kip ft Thermal Gradient: (LRFD 3.12.3) γTG 1.0 (no live live) γTG_L 0.5 (with live load) Effects due to uniform temperature change: Since superstructure is not restrained axially, uniform temperature change causes no internal stress. Effects due to temperature gradient: Fig. 6: Positive temperature gradient (from LRFD 3.12.3-2) Assume AASHTO temperature Zone 1: T1 54 (deg F) T2 14 T3 0 Atemp if hc 16 in 12 in hc 4 in  Atemp 12.00 in A1 4 in B2-16

A2 Atemp A2 1 ft Compute graident-induced curvature: ψ α l Σ Tai yi Ai ΔTi di Ii       = (LRFD C4.6.6-3) Area1 A1 bf Area1 288 in 2  I1 bf A1 3  12  I1 384 in 4  Area2 A2 bf Area2 864 in 2  I2 bf A2 3  12  I2 10368 in 4  εgr α A T1 Area1 T2 Area2  εgr 0.000177 Fgr Epr Ac εgr Fgr 404.5 kip y1 ytc A1 2  y1 10.53 in y2 ytc A1 A2 2         y2 2.53 in ψ α Ic T1 T2 2       y1 Area1 T1 T2  A1 2 I1 T2 2       y2 Area2 T2  A2 2 I2          ψ 0.000108 ft 1  Compute graident-induced fixed-end moment: FEMgr Epr Ic ψ FEMgr 274 kip ft From moment distribution, the moment in Span 2 was calculated: Mgr 265kip ft Compute gradient-induced internal stresses: σE E α TG α TuG ψ z = (LRFD C4.6.6-6) Evaluate at top and bottom of precast: fist Epr α T2 A1 A2 tslab A2        ψ ytcb        fist 0.0564 ksi fisb Epr ψ ytcb fisb 0.2987 ksi fistt Epr α T1 ψ ytc  fistt 1.0705 ksi B2-17

2.6 Flexural Stresses Note: Since for a structural system of this type, it is unlikely that compression at the top of the deck at a given section would exceed its allowable value, calculation of those stresses will be omitted for simplicity. Only the stresses at the bottom and top of the precast beam itself will be computed. The precast weight and CIP deck weight are carried by the precast section only. Additional loads (i.e. barrier, overlay, live load) are carried by the composite section. At Release j 1 2 fswrtj Mswrj St  fswrt 0.285 1.24       ksi fswrbj Mswrj Sb  fswrb 0.251 1.09       ksi At Final Conditions Self-Weight: j 1 4 fswtj Mswfj St  fswt 0 0.235 0.185 1.19             ksi fswbj Mswfj Sb  fswb 0 0.207 0.162 1.046             ksi Deck Weight: fdecktj Mdeckj St  fdeckt 0 0.199 0.156 1.007             ksi fdeckbj Mdeckj Sb  fdeckb 0 0.175 0.137 0.885             ksi Barriers: fbarriertj Mbarrierj Stcb  fbarriert 0.015 0.01 0.011 0.009             ksi fbarrierbj Mbarrierj Sbc  fbarrierb 0.026 0.018 0.02 0.015             ksi fbarrierttj Mbarrierj Stc  fbarriertt 0.022 0.015 0.017 0.013             ksi (top of topping) Future Wearing Surface: ffwstj Mfwsj Stcb  ffwst 0.03 0.021 0.023 0.017             ksi ffwsbj Mfwsj Sbc  ffwsb 0.052 0.036 0.04 0.03             ksi ffwsttj Mfwsj Stc  ffwstt 0.043 0.03 0.033 0.025             ksi (top of topping) B2-18

Live Load: Positive Moment Envelope: fpLLtj MLL_pj Stcb  fpLLt T 0.052 0.062 0.06 0.317( ) ksi fpLLbj MLL_pj Sbc  fpLLb T 0.092 0.109 0.105 0.556( ) ksi fpLLttj MLL_pj Stc  fpLLtt T 0.076 0.09 0.087 0.459( ) ksi (top of topping) Negative Moment Envelope: fnLLtj MLL_nj Stcb  fnLLt T 0.246 0.195 0.206 0.104( ) ksi fnLLbj MLL_nj Sbc  fnLLb T 0.431 0.342 0.362 0.182( ) ksi fnLLttj MLL_nj Stc  fnLLtt T 0.357 0.283 0.299 0.15( ) ksi (top of topping) Restraint Moment: Note: For the center span, restraint moments are constant across the span. For the end spans, restraint moment will vary linearly to zero at the abutments. Also, using the P-method, the restraint moment at the pier on the end span will be larger than the restraint moment at the same pier on the center span. frmt Mrm Stcb  frmt 0.0367 ksi frmb Mrm Sbc  frmb 0.0644 ksi frmtt Mrm Stc  frmtt 0.0532 ksi Thermal Gradient: fgrt Mgr Stcb fist fgrt 0.3457 ksi fgrb Mgr Sbc  fisb fgrb 0.209 ksi fgrtt Mgr Stc fistt fgrtt 1.49 ksi B2-19

Continuity Check: The continuity check from LRFD 5.14.1.4.5 is required for all simply-supported beams made continuous. The sum of stresses due to post-continuity dead load, restraint moment, 50% live load, and 50% thermal gradient at the bottom of the diaphragm must have no net tension: fbarrier frm 0.5 fLL 0.5 fTG 0 (LRFD 5.14.1.4.5) The live load stress used in the continuity check can be taken as that which exists when the effect of continuity is most needed (i.e.- when the maximum positive moment in the continuous live load envelope occurs). Using the vehicle position that creates the maximum positive live load moment at midspan, the negative moment at the pier is evaluated and the corresponding stress at the bottom of the section is determined: fbarrierc fbarrierb1  fbarrierc 0.0263 ksi fpLLc 0.247ksi Generated by the 2D live load model and entered manually frmc frmb frmc 0.0644 ksi fgrc fgrb fgrc 0.209 ksi fc fbarrierc frmc 0.5 fpLLc 0.5fgrc fc 0.0191 ksi There is net tension at the bottom of the diaphragm, and the continuity check fails. The stresses at the diaphragm from Span 1 would also need to be checked for continuity. However, the continuity check fails for Span 2 and will be used to evaluate partial continuity. The continuous live load positive moment envelop cannot be used in design and a partial continuity envelope must be calculated. Partial continuity is a nonlinear load redistribution through an inhomogeneous cross section that requires FE modeling for an exact analysis. An approximate method can be used for a rational design approach. The following is a simplified, linear-elastic method that is reasonable, though there are other ways to model partial continuity: A portion of the live load is required to satisfy the continuity check (i.e.- close the assumed cracks from the bottom tension). This portion is calculated below: LLreq fbarrierc frmc 0.5 fgrc  fpLLc  LLreq 0.5774 Therefore, 57.7% of the live load is used to obtain a continuous system, applied on a simple span. The remaining 42.3% is applied on a continuous span. The two live load envelopes are proportioned as such, and a new partial continuity live load envelope is formed: Continuous system live load positive moment envelope: Simple span live load moment envelope: MLLs 0 90 76 355             kip ft MLL_p 48 57 55 290             kip ft MLLpar LLreq MLLs 1 LLreq  MLL_p MLLparT 20 76 67 328( ) kip ft MLL_p if LLreq 0.5 MLL_p MLLpar  MLL_pT 20 76 67 328( ) kip ft B2-20

Computing partial continuity live load stresses: fpLLtj MLL_pj Stcb  fpLLt T 0.022 0.083 0.073 0.358( ) ksi fpLLbj MLL_pj Sbc  fpLLb T 0.039 0.146 0.129 0.628( ) ksi fpLLttj MLL_pj Stc  fpLLtt T 0.032 0.12 0.106 0.519( ) ksi (top of topping) 2.7 Prestress Losses At Release: At release, two components of prestress loss are significant: relaxation of the prestressing steel and elastic shortening. Elastic shortening is the loss of prestress that results when the strands are detensioned and the precast beam shortens in length due to the applied prestress. When the strands are tensioned in the prestress bed and anchored at the abutments, the steel gradually begins to relax as a function of time. By the time the strands are detensioned a small, but measurable, loss due to steel relaxation has occurred. Steel Relaxation (short term): fpj Pull fpu fpj 202.5 ksi fpy 243 ksi ΔfpR1 log t hr       40.0 fpj fpy 0.55        fpj ΔfpR1 8.149 ksi (LRFD 5.9.5.4.4b-2) Elastic Shortening: Eci 33000 wc 1.5  kcf 1.5  f'ci ksi .5  Eci 4287 ksi (LRFD 5.4.2.4-1) Aps No_Strands Astrand Aps 5.642 in 2  (Area of strand at midspan) fpbt fpj ΔfpR1 fpbt 194.4 ksi ΔfpES Aps fpbt I ecc 2 A  ecc Mswr2 A Aps I ecc 2 A  A I Eci Ep   ΔfpES 10.333 ksi Total Prestress Loss at Release: Δfsr ΔfpES ΔfpR1 Δfsr 18.482 ksi %Loss Δfsr Pull fpu 100 %Loss 9.1268 fper fpj ΔfpES ΔfpR1 fper 184 ksi Pr fper No_Strands Astrand Pr 1038.2 kip B2-21

At Final Conditions: Total Loss of Prestress: ΔfpT ΔfpES ΔfpLT= (LRFD 5.9.5.1-1) where: fpES = Sum of all losses due to elastic shortening at time of application of prestress load (ksi). fpLT = Total loss due to long-term effects, which include shrinkage and creep of the concrete and relaxation of the prestressing steel (ksi). The shrinkage and creep properties of the girder need to be computed in preparation for the prestress loss computations. These are addressed in LRFD Article 5.4.2.3. Creep coefficients are computed in accordance with Article 5.4.2.3.2 and shrinkage strains are computed in accordance with Article 5.4.2.3.3. Creep Coefficients Girder creep coefficients were calculated previously for restraint moments Girder creep coefficient at final time due to loading at transfer: ψbfi 1.580 Girder creep coefficient at time of CIP placement due to loading at transfer: ψbdi 0.156 Girder creep coefficient at final time due to loading at CIP placement: ψbfd 1.256 CIP deck creep coefficient at final time due to loading at CIP placement: tf tfinal ti tdeck t tf ti t 19993 day ktd t day 61 4 0.8 f'ct  ksi  t day   ktd 0.998 kvs 1.45 0.13 VSd in  kvs 0.670 Note: kvs must be greater than 1.0. kvs 1.000 khc 1.56 0.008H khc 1.000 kf 5 1 0.8 f'ct  ksi   kf 1.19 ψdfd 1.9 kvs khc kf ktd ti day       0.118  ψdfd 1.794 B2-22

Shrinkage Strains Shrinkage strains were calculated previously for restraint moment. Girder concrete shrinkage strain between transfer and final time: εbif 407 10 6  Girder concrete shrinkage strain between transfer and CIP placement: εbid 52 10 6  Girder concrete shrinkage strain between CIP placement and final time: The girder concrete shrinkage between deck placement and final time is the difference between the shrinkage at time of deck placement and the total shrinkage at final time. εbdf εbif εbid εbdf 355 10 6  CIP concrete shrinkage strain between CIP placement and final time: εddf 581 10 6  Loss from Transfer to CIP Placement: The prestress loss from transfer of prestress to placement of CIP consists of three loss components: shrinkage of the girder concrete, creep of the girder concrete, and relaxation of the strands. That is, Time-Dependent Loss from Transfer to CIP Placement = fpSR+fpCR+fpR1 Shrinkage of Concrete Girder: ΔfpSR εbid Ep Kid= (LRFD5.9.5.4.2a-1) where: bid = Concrete shrinkage strain of girder between transfer and CIP placement. Computed using LRFD Eq. 5.4.2.3.3-1 Ep = Modulus of elasticity of prestressing strand (ksi). Kid = Transformed steel coefficient that accounts for time-dependent interaction between concrete and bonded steel in the section being considered for the time period between transfer and CIP placement. The transformed section coefficient, kid, is computed using: Kid 1 1 Ep Eci Aps A  1 A epg 2  Ig       1 0.7 ψb tf ti   = (LRFD Eq. 5.9.5.4.2a-2) where: epg = Eccentricity of strands with respect to centroid of girder (in). b(tf,ti) = Creep coefficient at final time due to loading introduced at transfer. Note: The eccentricity of the strand pattern is stored in the vectors ecc_r and ecc_f. Vector B2-23

ecc_r contains values at check points relative to release (i.e., the end of the girder), and vector ecc_r is relative to final check points (i.e., relative to the left bearing of the girder). The eccentricity at the midspan of the girder is the value of interest. epg ecc epg 5.5 in I 27120 in 4  A 936 in 2  Kid 1 1 Ep Eci Aps A  1 A epg 2  I       1 0.7 ψbfi   Kid 0.8529 Therefore, the prestress loss due to shrinkage of the girder concrete between time of transfer and CIP placement is: ΔfpSR εbid Ep Kid ΔfpSR 1.266 ksi Creep of Concrete Girder: ΔfpCR Ep Eci fcgp ψb td ti  Kid= (LRFD 5.9.5.4.2b-1) where: fcgp = Concrete stress at cg of prestress pattern due to the prestressing force immediately after transfer and the self-weight of the girder at the section of maximum moment (ksi). Section modulus at cg of strand pattern: epti yb ycg epti 5.5 in Scgp I epti  Scgp 4934 in 3  Initial prestress force: fpj Pull fpu Pinit fpj Aps Pinit 1142.5 kip fcgp Pinit 1 A epti Scgp         Mswf2 Scgp  fcgp 2.359 ksi ΔfpCR Ep Eci fcgp ψbdi Kid ΔfpCR 2.092 ksi Relaxation of Prestressing Strands: Since, according to LRFD C5.9.5.4.2c, the second equation is the more accurate equation, fpR1 should be computed using the second. However, for this example, it will be assumed to be equal to 1.2 ksi (Article 5.9.5.4.2b permits this). ΔfpR1 1.2 ksi Total prestress loss at time of CIP placement: ΔfpLTid ΔfpSR ΔfpCR ΔfpR1 Calculated above: ΔfpSR 1.27 ksi ΔfpCR 2.09 ksi ΔfpR1 1.20 ksi ΔfpLTid 4.559 ksi B2-24

Loss from CIP Placement to Final: The prestress loss from placement of CIP to final conditions consists of four loss components: shrinkage of the girder concrete, creep of the girder concrete, and relaxation of the strands. That is, Time-Dependent Loss from CIP Placement to Final = fpSD+fpCD+fpR2 Shrinkage of Concrete Girder: ΔfpSD εbdf Ep Kdf= (LRFD5.9.5.4.3a-1) where: bdf = Concrete shrinkage strain of girder between time of CIP placement and final time. Computed using LRFD Eq. 5.4.2.3.3-1 Kdf = Transformed steel coefficient that accounts for time-dependent interaction between concrete and bonded steel in the section being considered for the time period between the time of CIP placement and final time. Compute Kdf: Kdf 1 1 Ep Eci Aps Ac  1 Ac epc 2  Ic         1 0.7ψb tf ti   = (LRFD 5.9.5.4.3a-2) where: epc = Eccentricity of strands with respect to centroid of composite section Ac = For composite sections, the gross area of the composite section should be used. However, since this girder is non-composite, the gross area of the non-composite section is used. Ic = Gross area of composite section for composite systems, gross area of bare beam for non-composite systems. b(tf,ti) = Girder creep coefficient epc yb ycg epc 5.4969 in Ac: A 936 in 2  Ic: I 27120 in 4  Kdf 1 1 Ep Eci Aps A  1 A epc 2  I       1 0.7ψbfi   Kdf 0.853 Therefore, prestress loss due to shrinkage of girder concrete between CIP placement and final is: ΔfpSD εbdf Ep Kdf ΔfpSD 8.632 ksi B2-25

Creep of Concrete Girder: ΔfpCD Ep Eci fcgp ψb tf ti  ψb td ti   Kdf Ep Ec Δfcd ψb tf td  Kdf 0.0= (LRFD5.9.5.4.3b-1) where: fcd = Change in concrete stress at centroid of prestressing strands due to long-term losses between transfer and CIP placement combined with superimposed loads (ksi). b(tf,td) = Girder creep coefficient at final time due to loading at CIP placement per Eq. 5.4.2.3.2-1. Let: ΔfpCD ΔfpCD1 ΔfpCD2= compute fpCD1: ΔfpCD1 Ep Eci fcgp ψbfi ψbdi  Kdf ΔfpCD1 19.041 ksi compute fpCD2: compute fcd: Δfcd ΔP 1 Ag epg 2 Ig         Mfws7 Mbarrier7  Scgp       = ΔP ΔfpLTid Aps ΔP 25.7 kip epg yb ycg epg 5.497 in Δfcd ΔP 1 A epg 2 I       Mfws2 Mbarrier2  Scgp        Δfcd 0.013 ksi Ec 33000 1.0 0.14 f'c 1000 ksi        1.5  f'c ksi  ksi Ec 4921 ksi ΔfpCD2 Ep Ec Δfcd ψbfd Kdf ΔfpCD2 0.079 ksi Therefore, ΔfpCD ΔfpCD1 ΔfpCD2 ΔfpCD 19.12 ksi Relaxation of Prestressing Strands: ΔfpR2 ΔfpR1 ΔfpR2 1.2 ksi (LRFD 5.9.5.4.3c-1) Shrinkage of the CIP deck: Ad Aslab n  Ad 432 in 2  Ecd 33000 wc 1.5  kcf 1.5  f'ct ksi .5  Ecd 3834 ksi ed h 2  ed 9 in B2-26

Δfcdf εddf Ad Ecd 1 0.7 ψdfd  1 Ac epc ed Ic         Δfcdf 0.6548 ksi ΔfpSS Ep Ec Δfcdf Kdf 1 0.7 ψbfd  ΔfpSS 6.0784 ksi Total prestress loss from CIP placement to final, therefore, is: ΔfpLTdf ΔfpSD ΔfpCD ΔfpR2 ΔfpSS Calculated above: ΔfpSD 8.63 ksi ΔfpCD 19.12 ksi ΔfpR2 1.2 ksi ΔfpSS 6.0784 ksi ΔfpLTdf 22.873 ksi Summary of Time-Dependent Losses Losses from Transfer to CIP Placement Girder shrinkage: ΔfpSR 1.27 ksi Girder creep: ΔfpCR 2.09 ksi Strand relaxation: ΔfpR1 1.2 ksi_______________ Total = ΔfpLTid 4.559 ksi Losses from CIP Placement to Final ΔfpSD 8.63 ksiGirder shrinkage: ΔfpCD 19.12 ksiGirder creep: ΔfpR2 1.2 ksiStrand relaxation: ΔfpSS 6.0784 ksiDifferential Shrinkage: _______________ Total = ΔfpLTdf 22.873 ksi ΔfpLT 10.0 fpj Apsm A  γh γst 12.0 ksi γh γst ΔfpR= (LRFD 5.9.5.3-1) γh 1.7 0.01 H γh 1.00 (LRFD 5.9.5.3-2) γst 5 ksi 1 ksi f'ci  γst 0.833 (LRFD 5.9.5.3-3) ΔfpR 2.4 ksi ΔfpLT ΔfpLTid ΔfpLTdf ΔfpLT 27.43 ksi ΔfPT ΔfpES ΔfpLT ΔfPT 37.77 ksi %Loss ΔfPT Pull fpu 100 %Loss 18.65 Check effective stress after losses: fpe Pull fpu ΔfPT fpe 164.7 ksi fallow 0.80 fpy fallow 194.4 ksi (LRFD 5.9.3-1) B2-27

2.8 Stresses Due to Prestress at End of Trnasfer Length and Midspan At Release: Pr fper No_Strands Astrand Pr 1038.2 kip j 1 2 fpsrbj Pr 1 A ecc Sb       fpsrb T 2.881 2.881( ) ksi fpsrtj Pr 1 A ecc St       fpsrt T 0.907 0.907( ) ksi At Final Conditions: j 1 4 distj xfj Lovr Ldes 2  dist T 0.5 3 2.44 24.5( ) ft dtj if distj Lt 1.0 distj Lt         dt T 0.1667 1 0.8133 1( ) (Fraction strands are transferred.) Pfj fpe dtj No_Strands Astrand Pf T 154.9 929.4 755.9 929.4( ) kip ecc 0.4581 ft fpsbj Pfj 1 A ecc Sb       fpsb T 0.43 2.579 2.098 2.579( ) ksi Sb 1.8639 ft 3  A 6.5 ft 2  fpstj Pfj 1 A ecc St       fpst T 0.135 0.812 0.66 0.812( ) ksi 2.9 Service Stress Check At Release: j 1 2 Bottom of beam (compression): frbj fpsrbj fswrbj  frb T 2.63 1.791( ) ksi fallow_rc 0.6 f'ci fallow_rc 3 ksi Status_ServiceLSrcj if frbj fallow_rc "OK" "NG"   Status_ServiceLSrc T "OK" "OK"( ) Top of beam (tension): frtj fpsrtj fswrtj  frt T 0.622 0.334( ) ksi fallow_rt 0.0948 f'ci ksi fallow_rt 0.212 ksi Status_ServiceLSrtj if frtj fallow_rt "OK" "NG"   Status_ServiceLSrt T "NG" "OK"( ) Since top tension exceeds limit, compute required amount of top tension steel at transfer point: xtt h frt1 frt1 frb1           xtt 5.57 in (LRFD C5.9.4.1.2) B2-28

Ttt frt1 2 bv xtt Ttt 124.64 kip Att Ttt 30 ksi  Att 4.1545 in 2  Assume #8 bars will be used as top tension steel: Ntt Att 0.79 in 2   Ntt 5 #8 bars At Final Conditions: Positive Moment Envelope Service III Limit State (Tensile Stresses in Bottom of Beam): j 1 4 fpAllbj fpsbj fswbj  fdeckbj  fbarrierbj  ffwsbj  0.8 fpLLb  j frmb 0.5 fgrb fpAllb T 0.308 1.967 1.586 0.069( ) ksi fallow_ft 0.19 f'c ksi fallow_ft 0.503 ksi (LRFD 5.9.4.2.2b) Status_ServiceLSftj if fpAllbj fallow_ft "OK" "NG"   Status_ServiceLSft T "OK" "OK" "OK" "OK"( ) Service I (Compressive Stresses in Top of Beam): Compressive Stress Due to Permanent Loads: fpPermtj fpstj fswtj  fdecktj  fbarriertj  ffwstj  frmt fgrt fpPermt T 0.203 0.026 0.029 1.794( ) ksi fallow_fcd 0.45 f'c fallow_fcd 3.15 ksi (LRFD 5.9.4.2.1) Status_ServiceLSfcdj if fpPermtj fallow_fcd "OK" "NG"   Status_ServiceLSfcd T "OK" "OK" "OK" "OK"( ) Compressive Stress Due to Full Dead Load + Live Load: fallow_fcl 0.6 f'c fallow_fcl 4.2 ksi (LRFD 5.9.4.2.1) fpAlltj fpstj fswtj  fdecktj  fbarriertj  ffwstj  fpLLtj  frmt 0.5 fgrt fpAllt T 0.052 0.116 0.07 1.979( ) ksi Status_ServiceLSfclj if fpAlltj fallow_fcl "OK" "NG"   Status_ServiceLSfcl T "OK" "OK" "OK" "OK"( ) Negative Moment Envelope Compressive Stress Due to Full Dead Load + Live Load: fnAllbj fpsbj fswbj  fdeckbj  fbarrierbj  ffwsbj  fnLLbj  frmb 0.5 fgrb fnAllb T 0.771 2.425 2.051 0.615( ) ksi fallow_fcn 0.6 f'c fallow_fcn 4.2 ksi (LRFD 5.9.4.2.1) Status_ServiceLSfncj if fnAllbj fallow_fcn "OK" "NG"   Status_ServiceLSfnc T "OK" "OK" "OK" "OK"( ) B2-29

2.10 Flexural Strength Check Positive Moment Envelope Muj 1.25 Mswfj Mdeckj  Mbarrierj    1.5 Mfwsj  1.75 MLL_pj  Mrm Mu T 11 255 208 1289( ) kip ft β1 if f'ct 4 ksi( ) 0.85 if f'ct 8 ksi( ) 0.65 0.85 f'ct 4 ksi( ) 1 ksi( ) 0.05                      β1 0.85 (LRFD 5.7.2.2) Ld 270.0 ksi 2 3 fpe       db ksi 1  Ld 96.11 in (Preliminary estimate of Ld) Kld if h 24 in 1.0 1.6( ) Kld 1 (LRFD Eq. 5.11.4.2-1) df j if distj Lt distj Lt fpe fpu  if distj Kld Ld fpe distj Lt Kld Ld Lt       fpu fpe  fpu  1.0                       df T 0.1017 0.6101 0.4962 1( ) (Preliminary estimate of fraction strands are developed) Apsj No_Strands dfj Astrand Aps T 0.5737 3.4423 2.7998 5.642( ) in 2  b beff b 72 in dp h tslab tws ycg  dp 21.08 in k 2 1.04 fpy fpu         k 0.28 (LRFD 5.7.3.1.1-2) cj Apsj fpu 0.85 f'ct β1 b k Apsj  fpu dp   c T 0.74 4.22 3.47 6.67( ) in (LRFD 5.7.3.1.1-4) hf tslab (Height of flange is slab thickness since this is a composite section) bwj if cj hf b bv  bwT 72 72 72 72( ) in cj Apsj fpu 0.85 f'ct b bwj    hf 0.85 f'ct β1 bwj  k Apsj  fpu dp   c T 0.74 4.22 3.47 6.67( ) in (LRFD 5.7.3.1.1-3) aj β1 cj a T 0.627 3.584 2.946 5.671( ) in fpsj fpu 1 k cj dp         fps T 267.4 254.9 257.6 246.1( ) ksi (LRFD 5.7.3.1.1-1) Ld fps1 2 3 fpe       db ksi 1  Ld 94.52 in (LRFD 5.11.4.1-1) B2-30

dfj if distj Lt distj Lt fpe fpsj  if distj Kld Ld fpe distj Lt Kld Ld Lt       fpsj fpe    fpsj  1.0                       df T 0.1027 0.6463 0.5202 1( ) (fraction strands are developed) Apsj No_Strands dfj Astrand Aps T 0.5794 3.6466 2.9349 5.642( ) in 2  cj Apsj fpu 0.85 f'ct β1 b k Apsj  fpu dp   bwj if cj hf b bv  bwT 72 72 72 72( ) in cj Apsj fpu 0.85 f'ct b bwj    hf 0.85 f'ct β1 bwj  k Apsj  fpu dp   c T 0.74 4.45 3.62 6.67( ) in Mnj dfj Apsj  fpsj  dp aj 2         0.85 f'ct b bwj    hf aj 2 hf 2         (LRFD 5.7.3.2.2-1) Mn T 28 965 642 2110( ) kip ft Compute phi for each section: ϕfj 0.583 0.25 dp cj 1        ϕf T 7.41 1.52 1.79 1.12( ) (LRFD Eq. 5.5.4.2.1-1) ϕfj if ϕfj 0.75 0.75 if ϕfj 1.0 1.0 ϕfj      ϕf T 1.00 1.00 1.00 1.00( ) Mrj ϕfj Mnj  Mr T 28 965 642 2110( ) kip ft Mu T 11 255 208 1289( ) kip ft Status_StrengthLSj if Muj Mrj  "OK" "NG"   Status_StrengthLS T "OK" "OK" "OK" "OK"( ) Maximum Steel Check: Note: The provisions contained in Art. 5.7.3.3.1 to check maximum reinforcement were deleted in 2005. This check is now effectively handled by varying phi, depending upon whether the section is compression or tension controlled. See Art. 5.5.4.2.1. B2-31

Minimum Steel Check: (LRFD 5.7.3.3.2) Compute Cracking Moment at Midspan: fr 0.37 f'c ksi fr 0.979 ksi Mcr Sbc fr fpsb4    Mswf4 Mdeck4   Sbc Sb 1        Mcr if Mcr Sbc fr Sbc fr Mcr  Mcr 1367.6 kip ft 1.2 Mcr 1641.1 kip ft Ref: Mr4 2110.4 kip ft 1.33 Mu4  1714.1 kip ft Mmin if 1.2 Mcr 1.33 Mu4  1.2Mcr 1.33 Mu4    Mmin 1641.1 kip ft Status_MinStl if Mmin Mr4  "OK" "NG"   Status_MinStl "OK" Negative Moment Envelope MLL_nc DFm Mlane_nc1 1.0 DLA( ) MVehicle_nc1    MLL_nc 244 kip ft Mun 1.25 Mbarrier_c1   1.5 Mfws_c1  1.75 MLL_nc Mun 492 kip ft β1n if f'c 4 ksi( ) 0.85 if f'c 8 ksi( ) 0.65 0.85 f'c 4 ksi( ) 1 ksi( ) 0.05                      β1n 0.7 (LRFD 5.7.2.2) cn As fy 0.85 f'c β1n bf  cn 1.26 in As 6.32 in 2  (Reinforcement at the top of the precast) (LRFD 5.7.3.1.1-4) bwn if cn tflg bf bv  6 ft bwn 72 in cn As fy 0.85 f'ct bf bwn  hf 0.85 f'c β1n bwn  cn 1.26 in an β1n cn an 0.885 in (LRFD 5.7.3.1.1-3) Compute phi for each section: ϕfj 0.583 0.25 dp cj 1        ϕf T 7.41 1.52 1.79 1.12( ) (LRFD Eq. 5.5.4.2.1-1) ϕfj if ϕfj 0.75 0.75 if ϕfj 1.0 1.0 ϕfj      ϕf T 1.00 1.00 1.00 1.00( ) ds hc tslab 2  Mrn ϕf4 As fy ds an 2         0.85 f'c bf bwn  bf an 2 tflg 2         (LRFD 5.7.3.2.2-1) Mrn 650 kip ft Status_StrengthLSn if Mun Mrn "OK" "NG""  Status_StrengthLSn "OK" B2-32

2.11 Vertical Shear Design At each section the following must be satisfied for shear: Vu Vr (LRFD 5.8.2.1-2)Note: Evaluation has been disabled for these three equations (as indicated by the small boxes) to enable them to be shown without first evaluating their parameters. Vr ϕVn= Vn Vc Vs Vp= (LRFD 5.8.3.3-1) Critical Section for Shear: (LRFD 5.8.3.2) The critical section for shear near a support in which the reaction force produces compression in the end of the member is, from the face of support (Fig. 2), the greater of: a. 0.5dvcot(), or b. dv where, dv = Effective shear depth = Distance between resultants of tensile and compressive forces = d e - a/2 dv ds an 2  dv 20.557 in an 0.885 in But dv need not be taken less than the greater of 0.9de and 0.72h. Thus, de ds 0.9 de 18.9 in 0.72 hc 17.28 in Min_dv if 0.9 de 0.72 hc 0.9 de 0.72 hc  Min_dv 18.9 in dv if dv Min_dv Min_dv dv  dv 20.557 in To compute critical section, assume: θ 32.3 deg 0.5 dv cot θ( ) 16.2593 in Crit_sec if dv 0.5 dv cot θ( ) dv 0.5 dv cot θ( )  Crit_sec 20.56 in (LRFD 5.8.2.7) Assuming that the distance from the face of support to the centerline of bearing is half the bearing pad length, the critical section for shear is: Crit_sec Crit_sec Lpad 2  Crit_sec 2.213 ft The above calculations for determining the location of the critical section for shear per LRFD are for illustrative purposes. PSBeam uses a conservative approach to estimating the location of the critical section that is consistent with the LRFD provisions, but which requires no assumptions and no iteration. PSBeam assumes the critical section to be located at 0.72h from the face of bearing. The logic of this is as follows: since dv need not be lower than the greater of 0.9de and 0.72h, either of these two values is permissible. And since the greater of the two values may be used, either value may be used since adopting the lesser value is conservative since the design shear is higher closer to the support. Recall that 0.72h from the face of the support was computed earlier as: xf3 1.94 ft At the critical section, the factored shear is: Vu 1.25 Vswf3 Vdeck3  Vbarrier3    1.5 Vfws3  1.75 VLL_p3  Vu 130.6 kip Compute maximum permissible shear capacity at a section: Vr_max ϕv 0.25 f'c bv dv 0.0 kip  Vr_max 2331.2 kip (LRFD 5.8.3.3-2) Status_Vrmax if Vu Vr_max "OK" "NG"  Status_Vrmax "OK" B2-33

The shear contribution from the concrete, Vc, is given by: Muv 1.25 Mswf3 Mdeck3  Mbarrier3    1.5 Mfws3  1.75 MLL_n3  Muv 274 kip ft In the 2008 Interim the procedure for the calculation of θ and β was moved to an appendix. The new procedure for the calculation of these two values involves a new value, εs. Check lower bound for Mu: MuLB if Muv dv Vu  dv Vu  Muv  MuLB 274.2 kip ft εs MuLB dv Vu  Es As  εs 0.00158587 (LRFD 5.8.3.4.2-4) If εs is less than zero, it can be taken equal to zero: εs if εs 0 0.0 εs  εs 0.00158587 β 4.8 1 750 εs  β 2.1924 (LRFD 5.8.3.4.2-1) θ 29 3500 εs θ 34.5505 (LRFD 5.8.3.4.2-3) New value for Vc Vc 0.0316 β f'c ksi bv dv Vc 271.3 kip Required Vs is, therefore: Vs Vu ϕv Vc Vs 126.2 kip Assuming two vertical legs of No. #4 bars: Av Vs fy dv cot θ deg( )  Av 0.845 in 2 ft  (LRFD C5.8.3.3-1) Spac 2 0.2 in 2  Av  Spac 5.7 in (stirrup spacing) Check minimum transverse reinforcement: Av_min 0.0316 f'c ksi bv fy  Av_min 1.204 in 2 ft  (LRFD 5.8.2.5-1) Check maximum stirrup spacing: (LRFD 5.8.2.7-2) Vspc 0.1 f'c bv dv Vspc 1036.1 kip Ref: Vu 130.6 kip dv 20.56 in Max_spac if Vu Vspc if 0.8 dv 24 in 0.8 dv 24 in  if 0.4 dv 12 in 0.4 dv 12 in   Max_spac 16.4 in B2-34

2.12 Longitudinal Reinforcement Check LRFD requires that the longitudinal steel be checked at all locations along the beam. This requirement is made to ensure that the longitudinal reinforcement is sufficient to develop the required tension tie, which is required for equilibrium. Equation 5.8.3.5-1 is the general equation, applicable at all sections. However, for the special case of the inside edge of bearing at simple-end supports, the longitudinal reinforcement must be able to resist a tensile force of (Vu/ - 0.5Vs - Vp)cot(). Note that when pretensioned strands are used to develop this force, only a portion of the full prestress force may be available near the support due to partial transfer. Additionally, only those strands on the flexural tension side of the member contribute to the tension tie force. Required Tension Tie Force: If only the minimum amount of transverse reinforcement that is required by design is provided, the required tension tie force is: Vp 0 kip FL_reqd Vu ϕv 0.5 Vs Vp       cot θ deg( ) FL_reqd 302.4 kip Eq. 5.8.3.5-2 However, a greater amount of stirrup reinforcement is typically provided than is required, which increases the actual Vs. Note that by Eq. 5.8.3.5-2, increasing Vs decreases the required tension tie force. Hence, it is helpful to use the computed value of Vs that results from the transverse reinforcement detailed in the design. In this case, the required tension tie force is: Assume 2 legs of No. 4 bars at 12" on center (amount of steel at the critical section for shear): Av_actual 0.4 in 2  Vs_actual Av_actual fy dv cot θ deg( ) 12 in  Vs_actual 59.7 kip Check the upper limit of Vs: Vs_actual_max Vu ϕv  Vs_actual_max 145.1 kip LRFD 5.8.3.5 Adopt the lesser of provided Vs and the upper limit of Vs: Vs_actual if Vs_actual Vs_actual_max Vs_actual Vs_actual_max  Vs_actual 59.7 kip The revised value of the required tension tie force is: FL_reqd Vu ϕv       0.5 Vs_actual Vp       cot θ( ) FL_reqd 1.7 10 4  kip Provided Tension Tie Force: The longitudinal reinforcement that contributes to the tension tie are strands that are on the flexural tension side of the precast section. Near the ends of the precast section, the strands are typically only partially effective. C5.8.3.5 of the 2006 Interim Revisions permits the strand stress in regions of partial development to be estimated using a bilinear variation, as shown in Fig. 4. B2-35

Figure 7: Variation in strand stress in relation to distance from beam end. The stress in the strands at a given section depends on the location of the section with respect to the end of the precast section. If the section is between the end of the beam and L t (see Fig. 5), a linear interpolation is performed using a stress variation of 0.0 at the end of the beam to f pe at a distance of Lt from the end of the precast section. If the section is to the right of Lt but to the left of Ld, then the stress is interpolated between fpe and fps. If the section is to the right of Ld, then the stress is assumed to be a constant value of f ps. At the face of bearing, the stress in the effective strands is: xFB Lovr Ldes 2 Lpad 2  xFB 1.00 ft (Distance from physical end of beam to face of bearing) Astr No_Strands Astrand Astr 5.642 in 2  FL_prov if xFB Lt Astr fpe xFB Lt  if xFB Kld Ld Astr fpe xFB Kld Ld Lt Kld Ld Lt       fps fpe         Astr fpe               FL_prov 309.8 kip Status_Vl if FL_prov FL_reqd "OK" "NG"  Status_Vl "OK" Refined Estimate of Provided Tension Tie Force: If it is assumed that the point of intersection of the bearing crack (at angle theta) and c.g. of the strands is where the force in the strands is computed, then additional tensile capacity from the strands can be utilized. Figure 8: Elevation view of end of beam showing location where assumed failure crack crosses the c.g. of that portion of the strand pattern that is effective for resisting tensile forces caused by moment and shear. B2-36

Distance from end of beam to point of intersection of assumed crack and center of gravity of effective strands: xc Lpad 2       ecc cot θ deg( ) xc 1.2 ft (Measured from L face of bearing) xc Lovr Ldes 2       Lpad 2        ecc cot θ deg( ) xc 1.7 ft (Measured from L end of beam) FL_prov if xc Lt Astr fpe xc Lt  if xc Kld Ld Astr fpe xc Kld Ld Lt Kld Ld Lt       fps fpe         Astr fpe               FL_prov 515.9 kip Status_Vl "OK"Status_Vl if FL_prov FL_reqd "OK" "NG"  2.13 Interface Shear Design The ability to transfer shear across the interface between the top of the precast beam and the cast-in-place deck must be checked. This check falls under the interface shear or shear friction section of LRFD (5.8.4). Recall that under the Standard Specs, this check falls under the horizontal shear section. Little guidance is offered by the LRFD Specs on how to compute the applied shear stress at the strength limit state. The procedure presented here uses the approach recommended by the PCI Bridge Design Manual, which is a strength limit state approach. Applied Factored Shear: Vu 130.6 kip vuh_s Vu dv bv  vuh_s 0.088 ksi dv 20.56 in xFB 1.00 ft (Distance from physical end of beam to face of bearing) vnh_reqd vuh_s ϕv  vnh_reqd 0.098 ksi Acv bv 1.0 ft Acv 864 in 2  Vnhr vnh_reqd Acv Vnhr 84.7 kip Nominal Shear Resistance of the Interface (Capacity): Status_Vl if FL_prov FL_reqd "OK" "NG" Vn cAc μ Avf fy Pc = (LRFD 5.8.4.1-2) Interface is CIP concrete slab on clean, roughened beam surface, no reinforcement crossing shear plane: (LRFD 5.8.4.3) c 0.135 ksi (cohesion factor) μ 1.000 (friction factor) K1 0.2 (fraction of concrete strength available to resist interface shear) K2 0.8 ksi (limiting interface shear resistance) B2-37

Since there is no permanent net compressive stress normal to shear plane, Pc = 0. (LRFD 5.8.4.2) Check Maximum Allowable Shear: Vni_max1 K1 f'ct Acv Vni_max1 691 kip (LRFD 5.8.4.1-4) Vni_max2 K2 Acv Vni_max2 691 kip (LRFD 5.8.4.1-5) Vnh_max if Vni_max1 Vni_max2 Vni_max1 Vni_max2  (LRFD 5.8.4.1-2,3) Vnh_max 691 kip Vnh_reqd vnh_reqd Acv Vnh_reqd 84.7 kip Status_Vuh_max if Vnh_reqd Vnh_max "OK" "NG"  Assuming no horizontal shear reinforcement crossing the shear plane, provided horizontal shear resistance is: Vnh_prov c Acv Vnh_prov 116.6 kip Status_Vnh_prov if Vnhr Vnh_prov "OK" "NG"  Status_Vnh_prov "OK" 2.14 Spalling Forces If the maximum spalling stress on the end face of the girder is less than the direct tensile strength of the concrete, then spalling reinforcement is not required when the member depth is less than 22 in. The maximum spalling stress is estimated as: σs P A 0.1206 e 2 h db  0.0256     0= And the direct tensile strength is computed as: fr_dts 0.23 f'c ksi fr_dts 0.609 ksi (LRFD C5.4.2,7) Check reinforcement requirement: Ref: A 936 in2 h 18 in ecc 5.4969 in db 0.6 in Pjack Aps4 fpj Pjack 1142.5 kip σs Pjack A 0.1206 ecc 2 h db  0.0256      σs 0.381 ksi Check whether spalling stress is below threshhold and thus is spalling/busting reinforcement is needed: Status_Spalling if σs fr_dts "OK" "NG"  Status_Spalling "OK" B2-38

2.15 Transverse Load Distribution The transverse load distribution reinforcement is computed by: Atld kmild Al_mild α kps Al_ps= where: α dcgs dtrans = kps 100 L fpe 60  50%= kmild 100 L 50%= dcgs hc ycg dcgs 21.1 in Compute dtrans: dtrans hc 4in db 2  0.75 in 2  dtrans 19.3 in α dcgs dtrans  α 1.0907 Assume there is no mild longitudinal reinfocement Al_mild in tension at the strength limit state. Al_mild 0.0 in 2  kmild 100 ft Ldes 100  kmild 14.43 % kps 100 ft Ldes fpe 60 ksi  100  kps 39.63 % Al_ps Aps2  Al_ps 3.6466 in 2  Total amount of transverse load distribution is: Atld kmild Al_mild α kps Al_ps Atld 1.58 in 2  Since the longitudinal reinforcement is per beam width, the area of distribution reinforcement per foot is: Atld_per_ft Atld S  Atld_per_ft 0.26 in 2 ft  Set transverse load distribution reinforcement spacing at 12 in.:Assuming transverse bars are #6, maximum spacing is: Sld_spac_max 0.44 in 2  Atld_per_ft ft ft Sld_spac_max 20.1 in Sld_spac 12in B2-39

2.16 Reflective Crack Control Reinforcement Reflective crack control reinforcement is provided from both the transverse load distribution reinforcment as well as drop in cage consisting of vertical stirrups. The total amount of reflective crack control reinforcement required is given as follows: ρcr_req 6 f'ct psi fy  ρcr_req 0.00632 (LRFD 5.14.4.3.3f-1) The crack control reinforcement ratio is defined, per unit length of span, as follows: ρcr Ascr h tflg  1 ft = (LRFD 5.14.4.3.3f-2) The required area of reinforcement of reflective crack control is therefore calculated, per unit length of span, as: Ascr_req ρcr_req h tflg  1 ft Ascr_req 1.1384 in2 The required area of cage reinforcement is subsequently calculated, per unit length of span, as the difference between the total required area of crack control reinforcement and that provided by the reinforcement for transverse load distribution; both transverse bars are effective in providing crack control, however only the lower horizontal legs of the stirrups are considered in the calculation. All calculations are per unit length of span: Ald 2Sld_spac .44 in 2 1ft  Ald 0.88 in 2  Acr_cage_req Ascr_req Ald Acr_cage_req 0.2584 in 2  Provide No. 5 stirrups at 12 in. on center: Scage_spac 12in Ascage 0.31in 2  Acr_cage_prov Ascage 1ft Scage_spac        Acr_cage_prov 0.31 in 2  Ascr_prov Ald Acr_cage_prov Status_Ascrack if Ascr_req Ascr_prov "OK" "NG"  Status_Ascrack "OK" Figure 9: Cross section of bridge showing CIP regions. B2-40

Figure 10: Detail of drop-in cage. Figure 11: Plan view of drop-in cage. 2.17 Bottom Flange Reinforcement Determine steel required to resist construction loads on bottom flange: Assume a 1' wide strip: Loads: Self-weight of flange: wflng_sw tflg 12 in wct wflng_sw 0.0375 klf CIP weight: wflng_cip h tflg  12 in wct wflng_cip 0.1875 klf Construction live load (assume 10 psf): wconst 10psf wflng_LL wconst 12 in wflng_LL 0.0100 klf Moments: B2-41

bcant bh 2  bcant 1.00 ft (Length of cantilever) Mflng_sw wflng_sw bcant 2  2  Mflng_sw 0.0187 kip ft Mflng_cip wflng_cip bcant 2  2  Mflng_cip 0.0937 kip ft Mflng_LL wflng_LL bcant 2  2  Mflng_LL 0.005 kip ft Strength Limit State I: Mu_flng 1.25 Mflng_sw Mflng_cip  1.75 Mflng_LL Mu_flng 0.15 kip ft Try #3 bars at 12" o.c: As_flng 0.11 in 2  As_flng 0.11 in 2  cflng As_flng fy 0.85 f'c β1 12 in  cflng 0.11 in β1p if f'c 4 ksi( ) 0.85 if f'c 8 ksi( ) 0.65 0.85 f'c 4 ksi( ) 1 ksi( ) 0.05                      β1p 0.70 aflng β1p cflng aflng 0.0761 in ds tflg 1 in 0.5 in 0.5 in 2  ds 1.25 in Mn_flng As_flng fy ds aflng 2         Mn_flng 0.67 kip ft ϕf_flng 0.65 0.15 ds cflng 1        ϕf_flng 2.22 ϕf_flng if ϕf_flng 0.75 0.75 if ϕf_flng 0.9 0.9 ϕf_flng   ϕf_flng 0.9 Mr_flng ϕf_flng Mn_flng Mr_flng 0.60 kip ft Status_StrengthLSflng if Mu_flng Mr_flng "OK" "NG"  Status_StrengthLSflng "OK" Use: Minimum #3 bars @ 12" o.c. in bottom flange. B2-42

2.18 Reinforcement for Positive Restraint Moment at Pier Load factors for each load component are derived from AASHTO (2009) Table 3.4.1-2. Because dead loads induce negative moments at the piers, they were underestimated by 10 percent (i.e.: γp = 0.9). The moments due to creep, shrinkage, and prestress force were designed with a load factor ( γRM) of 1.0, as given by AASHTO (2009) Table 3.4.1-3. γp 0.9 γRM 1.0 From Section 2.5: The design restraint moment due to the time depenent effects of prestress, creep, and shrinkage have been included. Postive restraint moments due to the design thermal gradient have been calculated (Section 2.5) based on AASHTO (2009) Temperature Zone 1, though will not be specifically designed for in this example because of the difference in the relative magnitude of the restraint moment due to thermal gradients and the remaining time dependent effects. Mrm 34 kip ft Mgr 265 ft kip Muprm γp Mbarrier_c1  γp Mfws_c1  Mrm Muprm 8 ft kip All reinforcement provided to resistive positive restraint moments is located in the 24 in. wide trough region between precast panels. The reinforcement is located immediately above the precast flange. Assume No. 4 bars are used for positive continuity reinforcement, and (4) bars are used in each 24 in. trough region: dNo4 0.5in As 0.2in 2  AsRM_prov 4 As AsRM_prov 0.8 in 2  ds_RM h 6in 3.25in 1in dNo4 2  ds_RM 19.5 in The reinforcement required is calculated at the ultimate limit state (i.e.: assuming Whitney's stress block) a AsRM_prov fy 0.8 f'ct 72 in  a 0.21 in AsRM_req Muprm fy ds_RM a 2        AsRM_req 0.08 in 2  Status_AsRM if AsRM_req AsRM_prov "OK" "NG"  Status_AsRM "OK" B2-43

Figure 12: Detail of the positive restraint moment reinforcement Figure 13: Cross section of the positive restraint moment reinforcement References Peterman, R. and Ramirez, J., "Restraint Moments in Bridges with Full-Span Prestressed Concrete Form Panels", PCI Journal, V. 43, No. 2, Jan.-Feb. 1998, pp. 54-73. B2-44

Example Problem 3 3.1 Introduction The design of the longitudinal joint between decked bulb tee members is illustrated in this example. 3.2 Materials, Geometry, Loads and Load Factors Units: kcf kip ft 3  Defined unit: kips per cubic foot ksf kip ft 2  Defined unt: kips per square foot Materials: f'c 7.0 ksi Strength of beam concrete at 28 days f'ci 6.0 ksi Strength of beam concrete at transfer of prestressing force wc 0.150 kcf Density of beam concrete Es 29000 ksi Modulus of elasticity of non-prestressed reinforcement fy 60.0 ksi Yield stress of stainless steel rebar Geometry: Lovr 141.0 ft Overal length of girder Ldes 140.0 ft Design span of girder S 7.00 ft Girder spacing Ng 4 Number of girders in bridge cross section Widthoverall 28.00 ft Overall width of bridge Curb to curb width of bridge Widthctc 25.5 ft Nl 2 Number of lanes tflng 6.25 in Thickness of girder flange Widthbarrier 1.25 ft Assumed width of a typical barrier Loads: Nbarriers 2 Number of barriers (assumed typical weight) wbarrier 0.300 klf Weight of single barrier wfws 0.025 ksf Weight of future wearing surface allowance HL-93 Notional live load per LRFD Specs wlane 0.64 klf Design lane load Factors: f (variable) Resistance factor for flexure ϕv 0.90 Resistance factor for shear DLA 0.33 Dynamic load allowance (LRFD 3.6.2.1-1) B3-1

3.3 Plan, Elevation, and Typical Section Fig. 1: Framing plan of bridge. Fig. 2: Elevation view of girder layout. Fig. 3: Typical section. B3-2

Fig. 4: Girder dimensions. 3.4 Strip Widths Fig. 5: Geometry of exterior girder and wheel load. (LRFD 4.6.2.1)Overhang: Xoverhang Widthoverall Ng 1  S  2 Widthbarrier 1.00 ft 1.25 ft Eoverhang 45.0 in 10.0 Xoverhang 12  Eoverhang 57.5 in B3-3

+Moment: XpM 7.0 ft EpM 26.0 in 6.6 XpM 12  EpM 72.2 in -Moment: XnM 7.0 ft EnM 48.0 in 3.0 XnM 12  EnM 69.0 in 3.4 Analysis Dead Load: Self-weight: Assume dead load acts on simple span, Use 1-ft strip: wsw tflng 12 in wc wsw 0.0781 klf Msw wsw 7 ft( ) 2  8  Msw 0.4785 kip ft Barriers: Use continuous beam model with barriers modeled as point loads at each cantilever. Mbarrier 1.0 kip ft Future wearing surface: Use continuous beam model with FWS modeled as uniform load: MFWS 0.020 kip ft Live Load: Use mulit-span continous model to assess live load moments: Fig. 6: Conitnuous beam model of live load analysis of panel strip. Maximum live load moment: MpLL 21.2 kip ft MLL_perfoot MpLL EpM ft  MLL_perfoot 3.52 kip ft B3-4

3.5 Load Combintations Use Strength Limit State I: Mu 1.25 Msw Mbarrier  1.5 MFWS 1.75 MLL_perfoot Mu 8.04 kip ft 3.6 Flexural Analysis Fig. 7: Girder reinforcement layout. Fig. 8: Joint reinforcement and geometry. Closure pour material should be based on recommended performance criteria from NCHRP 10-71 design recommendations. Fig. 9: Bend diameter of U bars. Note: This is a different than the 6db requirement in AASHTO, but acceptable because of the additional ductility of stainless steel. B3-5

Mn ϕ As fy ds a 2       ϕ 0.90 Check #5 bars at 9" o.c. Areabar 0.31 in 2  Spacingbar 9 in As Areabar 12 in Spacingbar  As 0.41 in 2  a As fy 0.85 f'c 12 in  a 0.35 in ds tflng 1 in 0.625 in 2  ds 4.9375 in Mn ϕ As fy ds a 2       Mn 8.9 kip ft Distribution reinforcement: As_dist_pct 100 S ft  (LRFD 5.14.4.1-1) (< 50%) As_dist_pct 37.8 (percent of primary reinforcement) As_dist As_dist_pct 100 As As_dist 0.16 in 2  B3-6

Example Problem 4 4.1 Introduction The design of the transverse joint over the piers of a girder bridge that incorporates full-depth deck panels is presented. 4.2 Materials, Geometry, Loads and Load Factors Units: kcf kip ft 3 Materials: Concrete: f'c 7.0 ksi Strength of beam concrete at 28 days wc 0.150 kcf Density of beam concrete fy 60 ksi Yield stress of stainless steel rebar Geometry: Beam: h 72.0 in Height of girder bf 26.0 in Width of bottom flange of precast section tf 10.50 in Thickness of bottom flange of girder Sbeam 12.00 ft Beam spacing Deck: tslab 6.00 in Thickness of precast deck panel 4.3 Plan, Elevation, and Typical Section Fig. 1: Plan view of bridge. Fig. 2: Elevation view of bridge. B4-1

Fig. 3: Typical section of bridge. Fig. 4: Girder dimensions. B4-2

Fig. 5: Panel layout. Fig. 6: Panel dimensions. Fig. 7: Transverse panel section. Fig. 8: Detail of panel-to-panel connection. B4-3

Fig. 9: Detail panel-to-panel connection over piers (see Fig. 5). Fig. 10: Section through girders at pier. B4-4

4.3 Analysis Only loads that act on the composite section cause negative moment over the piers. Barrier Weight: Mbarrier 213.1 kip ft Future Wearing Surface: MFWS 248.6 kip ft Live Load: MDesign_Truck 1265 kip ft MDesign_Lane 843 kip ft Fraction of live load moment for one design lane distributed to girder: DFm 0.7404 Dynamic load allowance (applied to truck only): DLA 0.33 Effective Negative Live Load Moment: MLL DFm MDesign_Lane 1 DLA( ) MDesign_Truck  MLL 1869.8 kip ft 4.4 Load Combinations Use Strength Limit State I Mu 1.25 Mbarrier MFWS  1.75 MLL  Mu 3849.4 kip ft Mu_per_ft Mu Sbeam ft  Mu_per_ft 320.8 kip ft (per foot) 4.5 Reinforcement Mn ϕ As fy ds a 2       Try #5 bars at 6" o.c (2 legs of u-bar): Areabar 2 0.31 in 2  Spacingbar 6 in As Areabar 12 in Spacingbar  As 1.24 in 2  c As fy 0.85 f'c 12 in  c 1.04 in (< 10.5 in OK) β1 if f'c 4 ksi( ) 0.85 if f'c 8 ksi( ) 0.65 0.85 f'c 4 ksi( ) 1 ksi( ) 0.05                      β1 0.7 a β1 c a 0.729 in B4-5

Center of gravity of deck reinforcement lies at mid-height of deck. ds h tslab 2  ds 75 in ϕf 0.65 0.15 ds c 1        ϕf if ϕf 0.75 0.75 if ϕf 0.9 0.9 ϕf   ϕf 0.9 Mn ϕf As fy ds a 2       Mn 416.5 kip ft OK B4-6

Example Problem 5 5.1 Introduction The required width of a bridge can exceed the practical length of a full-depth deck panel. To accommodate this, a longitudinal joint must be introduced into the cross section of the bridge. The joint can occur between girders or, preferably, over the centerline of a girder. This example provides typical details for making such connections. Previously presented methods for designing the joints apply. 5.2 Materials, Geometry, and Loads Units: kcf kip ft 3 Materials: Concrete: f'c 7.0 ksi Strength of beam concrete at 28 days wc 0.150 kcf Density of beam concrete fy 60.0 ksi Yield stress of stainless steel rebar Geometry: Beam: h 72.0 in Height of girder bf 26.0 in Width of bottom flange of precast section tf 10.50 in Thickness of bottom flange of girder Sbeam 10.30 ft Beam spacing Deck: tslab 6.00 in Thickness of precast deck panel 5.3 Plan, Elevation, and Typical Section Fig. 1: Plan view of bridge. Fig. 2: Elevation view of bridge. B5-1

Fig. 3: Bridge cross section. Fig. 4: Panel layout with joint between girders. Fig. 5: Girder cross section. B5-2

Fig. 6: Type "A" panel (see Fig. 4). Connection bars in the longitudinal direction should be outside the connection bars in the transverse direction through the section depth. Fig. 7: Section A-A through Type "A" panel. Fig. 8: Type "B" panel (see Fig. 4). Fig. 9: Typical transverse joint between panels. B5-3

Fig. 10: Typical longitudinal joint between panels.This joint is used for cases where the length of a panel exceeds the the limit for shipping or hauling of a panel, which requires that the panel be subdivided into two or more panels of shorter length. Fig. 11: Section through longitudinal joint. B5-4