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From page 535... ... B-i Appendix B NCHRP 10-71 Design Examples Read the entire page →
From page 536... ... 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) Read the entire page →
From page 537... ... 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. Read the entire page →
From page 538... ... 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) Read the entire page →
From page 539... ... 1.3 Plan, Elevation, and Typical Section Figure 2: General elevation of bridge. Figure 1: General plan of bridge. Read the entire page →
From page 540... ... 1.4 Section Properties Note: Flange assumed to be a constant thickness of 3" for computation of section properties. Figure 4: Precast slab dimensions. Read the entire page →
From page 541... ... 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( ) Read the entire page →
From page 542... ... 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) Read the entire page →
From page 543... ... Live Load: Live Load Distribution Factors: Assume superstructure acts like a slab-type bridge. Utilize provisions of LRFD Art. Read the entire page →
From page 544... ... 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) Read the entire page →
From page 545... ... 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) Read the entire page →
From page 546... ... 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. Read the entire page →
From page 547... ... 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. Read the entire page →
From page 548... ... 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) Read the entire page →
From page 549... ... kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0. Read the entire page →
From page 550... ... 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. Read the entire page →
From page 551... ... Shrinkage Strains Girder concrete shrinkage strain between transfer and final time: The concrete shrinkage strain, bid, is computed in accordance with Art. Read the entire page →
From page 552... ... 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) Read the entire page →
From page 553... ... 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. Read the entire page →
From page 554... ... 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) Read the entire page →
From page 555... ... 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. Read the entire page →
From page 556... ... 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 Read the entire page →
From page 557... ... 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( ) Read the entire page →
From page 558... ... Service Stress Check At Release Conditions: Top of precast section (tension) : frtj fpsrtj fswrtj  frt T 0.554 0.447( ) Read the entire page →
From page 559... ... 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) Read the entire page →
From page 560... ... cj Apsj fpu 0.85 f'ct β1 b k Apsj  fpu dp   c T 3.35 6.42( ) Read the entire page →
From page 561... ... Maximum Steel Check: Note: The provisions contained in Art. 5.7.3.3.1 to check maximum reinforcement were deleted in 2005. Read the entire page →
From page 562... ... NoAps_ft Pat_n Pat_h hc_2( ) j last Pat_n( ) Read the entire page →
From page 563... ... 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. Read the entire page →
From page 564... ... 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) Read the entire page →
From page 565... ... 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. Read the entire page →
From page 566... ... 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. Read the entire page →
From page 567... ... 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) Read the entire page →
From page 568... ... 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) Read the entire page →
From page 569... ... 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) Read the entire page →
From page 570... ... Provide No. 5 stirrups at 12 in. Read the entire page →
From page 571... ... 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) Read the entire page →
From page 572... ... 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( ) Read the entire page →
From page 573... ... 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. Read the entire page →
From page 574... ... 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) Read the entire page →
From page 575... ... Figure 1: Plan view of bridge. Figure 2: ELevation view of bridge. Read the entire page →
From page 576... ... 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. Read the entire page →
From page 577... ... 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 Read the entire page →
From page 578... ... 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) Read the entire page →
From page 579... ... 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( ) Read the entire page →
From page 580... ... 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. Read the entire page →
From page 581... ... 4. Midspan of beam: xf4 Ldes 2  xf T 0 2.5 1.94 24( ) Read the entire page →
From page 582... ... Barrier Weight (composite, continuous structure) : Mbarrier Map Pieroffset Brngoffset xf xfc Mbarrier_c  MbarrierT 13.72 9.52 10.4608 8( ) Read the entire page →
From page 583... ... 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( ) Read the entire page →
From page 584... ... MLL_nj DFm Mlane_nj 1.0 DLA( ) Mvehicle_nj    MLL_n T 225 179 189 95( ) Read the entire page →
From page 585... ... khs 2.00 0.014 H khs 1.020 (LRFD 5.4.2.3.3-2) Read the entire page →
From page 586... ... kvs 1.45 0.13 VSd in  kvs 0.670 Note: kvs must be greater than 1.0. Read the entire page →
From page 587... ... kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0. Read the entire page →
From page 588... ... 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) Read the entire page →
From page 589... ... 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) Read the entire page →
From page 590... ... 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. Read the entire page →
From page 591... ... Live Load: Positive Moment Envelope: fpLLtj MLL_pj Stcb  fpLLt T 0.052 0.062 0.06 0.317( ) Read the entire page →
From page 592... ... 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) Read the entire page →
From page 593... ... Computing partial continuity live load stresses: fpLLtj MLL_pj Stcb  fpLLt T 0.022 0.083 0.073 0.358( ) Read the entire page →
From page 594... ... 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) Read the entire page →
From page 595... ... 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. Read the entire page →
From page 596... ... 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) Read the entire page →
From page 597... ... 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) Read the entire page →
From page 598... ... 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) Read the entire page →
From page 599... ... 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) Read the entire page →
From page 600... ... 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( ) Read the entire page →
From page 601... ... 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( ) Read the entire page →
From page 602... ... 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( ) Read the entire page →
From page 603... ... 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) Read the entire page →
From page 604... ... 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( ) Read the entire page →
From page 605... ... 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) Read the entire page →
From page 606... ... 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. Read the entire page →
From page 607... ... 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. Read the entire page →
From page 608... ... 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. Read the entire page →
From page 609... ... 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) Read the entire page →
From page 610... ... Since there is no permanent net compressive stress normal to shear plane, Pc = 0. Read the entire page →
From page 611... ... 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 Read the entire page →
From page 612... ... 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) Read the entire page →
From page 613... ... Figure 10: Detail of drop-in cage. Figure 11: Plan view of drop-in cage. Read the entire page →
From page 614... ... 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( ) Read the entire page →
From page 615... ... 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. Read the entire page →
From page 616... ... 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 Read the entire page →
From page 617... ... 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) Read the entire page →
From page 618... ... 3.3 Plan, Elevation, and Typical Section Fig. 1: Framing plan of bridge. Read the entire page →
From page 619... ... Fig. 4: Girder dimensions. Read the entire page →
From page 620... ... +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. Read the entire page →
From page 621... ... 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. Read the entire page →
From page 622... ... 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) Read the entire page →
From page 623... ... 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. Read the entire page →
From page 624... ... Fig. 3: Typical section of bridge. Read the entire page →
From page 625... ... Fig. 5: Panel layout. Read the entire page →
From page 626... ... Fig. 9: Detail panel-to-panel connection over piers (see Fig. Read the entire page →
From page 627... ... 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) Read the entire page →
From page 628... ... 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 Read the entire page →
From page 629... ... 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. Read the entire page →
From page 630... ... Fig. 3: Bridge cross section. Read the entire page →
From page 631... ... Fig. 6: Type "A" panel (see Fig. Read the entire page →
From page 632... ... 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. Read the entire page →

From page 535...

... B-i Appendix B NCHRP 10-71 Design Examples

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From page 536...

... 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)

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From page 537...

... 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.

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From page 538...

... 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)

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From page 539...

... 1.3 Plan, Elevation, and Typical Section Figure 2: General elevation of bridge. Figure 1: General plan of bridge.

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From page 540...

... 1.4 Section Properties Note: Flange assumed to be a constant thickness of 3" for computation of section properties. Figure 4: Precast slab dimensions.

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From page 541...

... 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( )

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From page 542...

... 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)

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From page 543...

... Live Load: Live Load Distribution Factors: Assume superstructure acts like a slab-type bridge. Utilize provisions of LRFD Art.

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From page 544...

... 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)

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From page 545...

... 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)

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From page 546...

... 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.

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From page 547...

... 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.

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From page 548...

... 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)

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From page 549...

... kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0.

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From page 550...

... 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.

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From page 551...

... Shrinkage Strains Girder concrete shrinkage strain between transfer and final time: The concrete shrinkage strain, bid, is computed in accordance with Art.

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From page 552...

... 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)

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From page 553...

... 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.

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From page 554...

... 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)

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From page 555...

... 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.

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From page 556...

... 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

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From page 557...

... 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( )

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From page 558...

... Service Stress Check At Release Conditions: Top of precast section (tension) : frtj fpsrtj fswrtj  frt T 0.554 0.447( )

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From page 559...

... 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)

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From page 560...

... cj Apsj fpu 0.85 f'ct β1 b k Apsj  fpu dp   c T 3.35 6.42( )

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From page 561...

... Maximum Steel Check: Note: The provisions contained in Art. 5.7.3.3.1 to check maximum reinforcement were deleted in 2005.

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From page 562...

... NoAps_ft Pat_n Pat_h hc_2( ) j last Pat_n( )

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From page 563...

... 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.

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From page 564...

... 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)

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From page 565...

... 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.

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From page 566...

... 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.

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From page 567...

... 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)

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From page 568...

... 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)

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From page 569...

... 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)

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From page 570...

... Provide No. 5 stirrups at 12 in.

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From page 571...

... 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)

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From page 572...

... 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( )

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From page 573...

... 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.

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From page 574...

... 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)

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From page 575...

... Figure 1: Plan view of bridge. Figure 2: ELevation view of bridge.

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From page 576...

... 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.

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From page 577...

... 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

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From page 578...

... 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)

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From page 579...

... 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( )

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From page 580...

... 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.

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From page 581...

... 4. Midspan of beam: xf4 Ldes 2  xf T 0 2.5 1.94 24( )

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From page 582...

... Barrier Weight (composite, continuous structure) : Mbarrier Map Pieroffset Brngoffset xf xfc Mbarrier_c  MbarrierT 13.72 9.52 10.4608 8( )

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From page 583...

... 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( )

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From page 584...

... MLL_nj DFm Mlane_nj 1.0 DLA( ) Mvehicle_nj    MLL_n T 225 179 189 95( )

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From page 585...

... khs 2.00 0.014 H khs 1.020 (LRFD 5.4.2.3.3-2)

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From page 586...

... kvs 1.45 0.13 VSd in  kvs 0.670 Note: kvs must be greater than 1.0.

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From page 587...

... kvs 1.45 0.13 VSb in  kvs 0.774 Note: kvs must be greater than 1.0.

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From page 588...

... 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)

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From page 589...

... 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)

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From page 590...

... 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.

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From page 591...

... Live Load: Positive Moment Envelope: fpLLtj MLL_pj Stcb  fpLLt T 0.052 0.062 0.06 0.317( )

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From page 592...

... 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)

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From page 593...

... Computing partial continuity live load stresses: fpLLtj MLL_pj Stcb  fpLLt T 0.022 0.083 0.073 0.358( )

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From page 594...

... 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)

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From page 595...

... 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.

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From page 596...

... 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)

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From page 597...

... 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)

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From page 598...

... 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)

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From page 599...

... 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)

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From page 600...

... 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( )

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From page 601...

... 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( )

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From page 602...

... 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( )

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From page 603...

... 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)

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From page 604...

... 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( )

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From page 605...

... 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)

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From page 606...

... 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.

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From page 607...

... 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.

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From page 608...

... 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.

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From page 609...

... 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)

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From page 610...

... Since there is no permanent net compressive stress normal to shear plane, Pc = 0.

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From page 611...

... 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

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From page 612...

... 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)

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From page 613...

... Figure 10: Detail of drop-in cage. Figure 11: Plan view of drop-in cage.

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From page 614...

... 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( )

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From page 615...

... 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.

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From page 616...

... 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

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From page 617...

... 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)

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From page 618...

... 3.3 Plan, Elevation, and Typical Section Fig. 1: Framing plan of bridge.

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From page 619...

... Fig. 4: Girder dimensions.

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From page 620...

... +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.

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From page 621...

... 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.

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From page 622...

... 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)

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From page 623...

... 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.

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From page 624...

... Fig. 3: Typical section of bridge.

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From page 625...

... Fig. 5: Panel layout.

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From page 626...

... Fig. 9: Detail panel-to-panel connection over piers (see Fig.

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From page 627...

... 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)

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From page 628...

... 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

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From page 629...

... 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.

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From page 630...

... Fig. 3: Bridge cross section.

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From page 631...

... Fig. 6: Type "A" panel (see Fig.

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From page 632...

... 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.

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