EXAMPLE 1: THREESPAN CONTINUOUS STRAIGHT COMPOSITE I GIRDER Load and Resistance Factor Design (Third Edition  Customary U.S.


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1 EXAMPLE 1: THREESPAN CONTINUOUS STRAIGHT COMPOSITE I GIRDER Load and Resistance Factor Design (Third Edition  Customary U.S. Units) by Michael A. Grubb, P.E. Bridge Software Development International, Ltd. Cranberry Township, PA and Robert E. Schmidt, E.I.T. SITEBlauvelt Engineers Pittsburgh, PA DESIGN PARAMETERS SPECIFICATIONS: LRFD Third Edition (004) STRUCTURAL STEEL:  ASTM A 709 Grade HPS 70W for flanges in negativeflexure regions  ASTM A 709 Grade 50W for all other girder and crossframe steel CONCRETE: f' c 4.0 ksi REINFORCING STEEL: F y 60 ksi ADTT:,000 trucks per day Design Example 31
2 BRIDGE CROSSSECTION CROSSFRAMES (Article ) The need for diaphragms or crossframes shall be investigated for all stages of assumed construction procedures and the final condition. The investigation should include, but not be limited to the following: Transfer of lateral wind loads from the bottom of the girder to the deck & from the deck to the bearings, Stability of the bottom flange for all loads when it is in compression, Stability of the top flange in compression prior to curing of the deck, Consideration of any flange lateral bending effects, and Distribution of vertical dead & live loads applied to the structure. Design Example 3
3 BRIDGE FRAMING PLAN CROSSSECTION PROPORTIONS Web Depth SpantoDepth Ratios (Table ) 0.03L 0.03(175.0) 5.6 ft 67. in. Use 69.0 in. Web Thickness (Article ) D ( ) 0.46 in. t w min. Design Example 33
4 CROSSSECTION PROPORTIONS (continued) Flange Width (Article ) ( ) D/6 69.0/ in. b f min. L (1) 85 ( ) 14.1in. b fc min. Flange Thickness (Article ) ( t ) 1.1t 1.1( 0.565) 0.6 in. f min w Design Example 34
5 CROSSSECTION PROPORTIONS (continued) Flange WidthtoThickness (Article ) b t f 18 ( 0.875) 10.3 f < 1.0 ok Flange Moments of Inertia (Article ) I I yc yt ( ) ( 18) < 0.51 < 10 ok 1 DEAD LOADS (Article 3.5.1) Component Dead Load (DC 1 ) DC 1 component dead load acting on the noncomposite section  Concrete deck k/ft (incl. integral w.s.)  Overhang tapers 0.14 k/ft  Deck haunches k/ft  SIP forms k/ft  Crossframes 0.10 k/ft & details TOTAL k/ft 4 girders k/ft + girder weight Design Example 35
6 DEAD LOADS (continued) Component Dead Load (DC ) DC component dead load acting on the composite section  Barriers 0.50/ 0.60 k/ft Note: Distributed equally to exterior girder & adjacent interior girder Wearing Surface Load (DW)  Wearing surface [0.05 x 40.0]/4 girders 0.50 k/ft Note: Distributed equally to each girder Basic LRFD Design Live Load HL93  (Article ) Design Truck: or Design Tandem: Pair of 5.0 KIP axles spaced 4.0 FT apart superimposed on Design Lane Load 0.64 KLF uniformly distributed load Design Example 36
7 LRFD Negative Moment Loading (Article ) For negative moment between points of permanentload contraflexure & interiorpier reactions, check an additional load case: Add a second design truck to the design lane load, with a minimum headway between the front and rear axles of the two trucks equal to 50 feet. Fix the rearaxle spacing of both design trucks at 14 feet, and Reduce all loads by 10 percent. LRFD Fatigue Load (Article ) Design Truck only > w/ fixed 30ft rearaxle spacing placed in a single lane Design Example 37
8 LOAD for OPTIONAL LIVELOAD DEFLECTION EVALUATION Refer to Article : Deflection is taken as the larger of:  That resulting from the design truck by itself.  That resulting from 5% of the design truck together with the design lane load. WIND LOADS (Article 3.8) DZ D V P PB Ł VB ł VDZ PB 10,000 P B base wind pressure ksf for beams V DZ design wind velocity at elevation Z V B base wind velocity at 30 ft height 100 mph Eq. ( ) For this example, assume the bridge is 35 ft above low ground & located in open country. Design Example 38
9 WIND LOADS (continued) V DZ.5V o V Ł V 30 B ln ł Ł Z Z o ł Eq. ( ) V o friction velocity 8. mph for open country V 30 wind velocity at 30 ft above low ground V B 100 mph in absence of better information Z height of structure above low ground (> 30 ft) Z o friction length of upstream fetch 0.3 ft for open country WIND LOADS (continued) w V DZ.5 Ł100 ł Ł0.3 ł Ø( 103.0) ø PD 0.050Œ œ ksf º 10,000 ß ( 8.0) ln mph PD hexp 0.053(10.41) 0.55 kips / ft > 0.3 kips / ft ok Design Example 39
10 Basic LRFD Design Equation S? i? i Q i fr n R r Eq. ( ) where:? i? D? R? I? i 0.95 for maximum g s? i h h h 0.95 for minimum g s D R I? i Load factor f Resistance factor Q i Nominal force effect R n Nominal resistance R r Factored resistance fr n 1 Load Combinations and Load Factors Load Combination Limit State DC DD DW EH EV ES LL IM CE BR PL LS WA WS WL FR TU CR SH TG SE Use One of These at a Time EQ IC CT CV STRENGTHI? p /1.0? TG? SE STRENGTHII? p /1.0? TG? SE STRENGTHIII? p /1.0? TG? SE STRENGTHIV EH, EV, ES, DW DC ONLY? p / STRENGTHV? p /1.0? TG? SE EXTREMEI? p? EQ EXTREMEII? p SERVICEI /1.0? TG? SE SERVICEII / SERVICEIII /1.0? TG? SE FATIGUELL, IM & CE ONLY Design Example 310
11 Load Factors for Permanent Loads,? p Load Factor Type of Load Maximum Minimum DC: Component and Attachments DD: Downdrag DW: Wearing Surfaces and Utilities EH: Horizontal Earth Pressure Active AtRest EV: Vertical Earth Pressure Overall Stability Retaining Structure Rigid Buried Structure Rigid Frames N/A LRFD LOAD COMBINATIONS (continued) Construction Loads (Article 3.4.): STRENGTH I  Construction loads > Load factor DW > Load factor 1.5 STRENGTH III  Construction dead loads > Load factor Wind loads > Load factor DW > Load factor 1.5 STRENGTH V  Construction dead loads > Load factor DW > Load factor 1.5 Design Example 311
12 STRUCTURAL ANALYSIS Summary  LiveLoad Distribution Factors: Strength Limit State Interior Girder Exterior Girder Bending Moment lanes lanes Shear 1.08 lanes lanes Fatigue Limit State Interior Girder Exterior Girder Bending Moment lanes lanes Shear lanes lanes STRUCTURAL ANALYSIS (continued) Distribution Factor for LiveLoad Deflection: NL DF m3 ŁNb ł lanes Ł 4 ł Design Example 31
13 STRUCTURAL ANALYSIS (continued) Dynamic Load Allowance Impact (IM) COMPONENT Deck Joints All Limit States All Other Components  Fatigue & Fracture Limit State  All Other Limit States (applied to design truck only not to design lane load) IM 75% 15% 33% Design Example 313
14 Design Example 314
15 STRUCTURAL ANALYSIS (continued) Live Load Deflection Design Truck + IM (SERVICE I): (D LL+IM ) end span 0.91 in. (governs) (D LL+IM ) center span 1.3 in. (governs) 100% Design Lane + 5% Design Truck + IM (SERVICE I): (D LL+IM ) end span (0.91) 0.83 in. (D LL+IM ) center span (1.3) 1.16 in. Design Example 315
16 LRFD LIMIT STATES The LRFD Specifications require examination of the following limit states: SERVICE LIMIT STATE FATIGUE & FRACTURE LIMIT STATE STRENGTH LIMIT STATE  (CONSTRUCTIBILITY) EXTREME EVENT LIMIT STATE SECTION PROPERTIES Section L 1 ) Effective Flange Width (Article ): Interior Girder or or 1.0t s L 4 b x 1 4 tf 1.0 ( 9.0) in in. (governs) average spacing of girders in. + Design Example 316
17 SECTION PROPERTIES (continued) Section L 1 ) or or Effective Flange Width (Article ): Exterior Girder L x in t s btf ( 9.0) in. 4 + widthof the overhang in in. (governs) SECTION PROPERTIES (continued) Section L 1 ) Plastic Moment (Article D Appendix D): P + P t w + P 3,060 kips < 3,763 kips \ PNA is in the top flange,use Case II P c s M p A F steel y 0.85f ' b c 75.5(50) 3,763 kips eff t s 0.85(4.0)(100.0)(9.0) 3,060 kips tc ØPw + Pt  Ps ø y 1 Œ + P œ º c ß 0.44 in. from the top of the top flange [ y + ( tc  y) ] + [ Psds + Pw dw + Ptdt ] Pc tc M p 170,38 kip in. 14,199 kip ft Design Example 317
18 SECTION PROPERTIES (continued) Section L 1 ) Yield Moment (Article D Appendix D): M M M M + M MD1 MD M F y + + S S S NC LT AD (,0)( 1) 1.5( 335)( 1) ( 3)( 1) Ø Œ + º 1,973 M 78,06 kip in. 6,517 kipft AD y y y D1 [ (,0) + 1.5( 335) ( 3) + 6,517] D + M 10,171kipft AD (M p / M y 1.4) ST,483 MAD ø +,706œ ß SECTION PROPERTIES (continued) Section  Interior Pier) Effective Flange Width (Art ): Exterior Girder in. Min. Concrete Deck Reinforcement (Article ): 9.0 Ø ø A 1 Œ º 1 Ł łł 1 łß œ ( 43.0) ft 4,776 in deck. 0.01(4,776) in in. ft in. in (100.5) in. from bot. of the deck Design Example 318
19 Constructibility DECKPLACEMENT SEQUENCE Design Example 319
20 Table 1: Moments from DeckPlacement Analysis Span > 1 Unfactored DeadLoad Moments (kipft) Length (ft) Steel Weight SIP Forms (SIP) Cast Sum of Casts + SIP Max. +M DC + DW Deck, haunches + SIP M 35 +,537,889 kipft Table : Vertical Deflections from DeckPlacement Analysis Span >1 Unfactored Vertical DeadLoad Deflections (In.) Length (ft) Steel Weight SIP Forms (SIP) Cast Sum of Casts + SIP DC + DW Total Deck, haunches + SIP Design Example 30
21 Table 3: Unfactored Vertical DeadLoad Reactions from DeckPlacement Analysis (kips) Abut. 1 Pier 1 Pier Abut. Steel Weight sum SIP Forms (SIP) sum Cast sum Cast sum Cast sum Sum of Casts + SIP DC + DW Total Deck, haunches SIP DECKPLACEMENT ANALYSIS (continued) Calculate f bu : (at Section 11 > 560 from abut.) For STRENGTH I: Top flange: Bot. flange: 1.0(1.5)(,889)(1) fbu 7.41ksi 1, (1.5)(,889)(1) f bu 1.96 ksi 1,973 For STRENGTH IV: Top flange: Bot. flange: 1.0(1.5)(,889)(1) fbu ksi 1, (1.5)(,889)(1) f bu 6.36 ksi 1,973 Design Example 31
22 DECKOVERHANG LOADS F P tan a 3.5 ft a tan 1 Ł 5.75 ft ł o 31.3 DECK OVERHANG LOADS (continued) Deck overhang weight: P 55 lbs/ft Construction loads: Overhang deck forms: Screed rail: Walkway: Railing: Finishing machine: P 40 lbs/ft P 85 lbs/ft P 15 lbs/ft P 5 lbs/ft P 3000 lbs Design Example 3
23 DECK OVERHANG LOADS (continued) Determine if amplification of firstorder compressionflange f l is required: L b 40 If: L 1.L b p CbR f F bm b yc then, no amplification f 0.85 l fl1 fbm Otherwise: Eq. ( ) 1 Ł F cr ł f l1 Or: f l (AF)fl f 1 l1 DECK OVERHANG LOADS (continued) R b 1.0 L 1.L b bm C b 1.0 f bm f bu ksi (STRENGTH IV) p CbR f F b yc Eq. ( ) L 1.0r p yc Eq. ( ) t E F where: r t b fc 1 Dct 1 1+ Ł 3bfct w fc ł Eq. ( ) Design Example 33
24 DECK OVERHANG LOADS (continued) For the steel section at Section 11, D c in. 16 r t 3.90 in (0.5) 1 1 Ł (1) ł 1.0(3.90) 9,000 L p 7.83 ft (1.0) 1. b ( 7.83) ft < L 4.0 ft DECK OVERHANG LOADS (continued) Therefore, amplification of the firstorder compressionflange f l is required: CbRbp E Fcr Calculate F cr : L Eq. ( ) b Ł rt ł 1.0(1.0) p (9,000) 4(1) Ł 3.90 ł Fcr 5.49 ksi Note: F cr may exceed R b R h F yc in this calculation. Note: assumes K 1.0 (see Appendix A of example) Design Example 34
25 DECK OVERHANG LOADS (continued) The amplification factor is determined as: For STRENGTH I: 0.85 AF 1.78 > Ł 5.49 ł ok For STRENGTH IV: 0.85 AF.8 > Ł 5.49 ł ok DECK OVERHANG LOADS (continued) For STRENGTH I: Dead loads: [ + 1.5( ) ] lbs / ft P (55) F F P tana 731.3tan( 31.3 o l ) F L ( 4) M b l l 1.34 kip  ft 1 1 M 1.34(1) Top flange: f l l 6.00 ksi Sl 1(16) lbs / ft M 1.34(1) Bot. flange: f l l 3.45 ksi Sl 1.375(18) 6 Design Example 35
26 DECK OVERHANG LOADS (continued) For STRENGTH I: Finishing machine: [ ] 4,500 lbs P (3000) F P Ptana 4,500 tan( 31.3 o l ),736 lbs P.736( 4) M Lb l l 8.1 kip  ft 8 8 M 8.1(1) Top flange: f l l.31ksi Sl 1(16) 6 M 8.1(1) Bot. flange: f l l 1.33 ksi Sl 1.375(18) 6 DECK OVERHANG LOADS (continued) For STRENGTH I: Top flange: f l total ksi * AF (8.31)(1.78) ksi < 0.6F yf 30 ksi ok Bot. flange: f l total ksi * AF (4.78)(1.0) 4.78 ksi < 0.6F yf 30 ksi ok Design Example 36
27 DECK OVERHANG LOADS (continued) For STRENGTH IV: Dead loads: P 1.0[ 1.5( )] 795 lbs / ft F F P tan a 795 tan(31.3 o l ) lbs / ft F L ( 4) M b l l 3.0 kip  ft 1 1 M 3.0(1) Top flange: f l l 6.5 ksi Sl 1(16) 6 M 3.0(1) Bot. flange: f l l 3.75 ksi Sl 1.375(18) 6 Finishing machine: Not considered DECK OVERHANG LOADS (continued) For STRENGTH IV: Top flange: f l total 6.5 ksi * AF 6.5(.8) ksi ksi < 0.6F yf 30 ksi ok Bot. flange: f l total 3.75 ksi * AF 3.75(1.0) 3.75 ksi 3.75 ksi < 0.6F yf 30 ksi ok Design Example 37
28 CONSTRUCTIBILITY  FLEXURE (Article ) Determine if the section is a slenderweb section: 5.7 E F yc D t D t w w c E F yc (38.63) 0.5 c Eq. ( ) , < Therefore, the section is a slenderweb section. Go to Article to compute F nc. CONSTRUCTIBILITY  FLEXURE (Article ) For discretely braced compression flanges: f f f R F Eq. ( ) bu + l f h yc f 1 3 bu + fl f bu f F f F f f crw nc Eq. ( ) Eq. ( ) For discretely braced tension flanges: f f f R F Eq. ( ) bu + l f h yt Design Example 38
29 LOCAL BUCKLING RESISTANCE Top Flange (Article ) Determine the slenderness ratio of the top flange: l 0.38 l b t 16 1 fc f fc ( ) E F , pf yc 9. Since l f <l pf : F R R F Eq. ( ) F FLB nc b nc 1.0(1.0)(50) 50.0 ksi h yc Flexural Resistance  Composite Sections in Negative Flexure & Noncomposite Sections F n or M n F max or M M max max Basic Form of All FLB & LTB Eqs Fyr λf λpf Fnc 1 1 RbR hfyc nc b hf yc R F Anchor point 1 h yc λrf λpf nc b hf yc Fyr Lb L p Fnc Cb 1 1 RbRhFyc RbRhFyc RhF yc Lr L p F RR F RR Anchor point F r or M r r compact noncompact (inelastic buckling) nonslender F F RRF nc cr slender (elastic buckling) b h yc b bπ Lb C R r t E L p or λ λ p pf L r λor r λ rf L b or b fc /t fc Design Example 39
30 LAT. TORSIONAL BUCKLING RESISTANCE (Article ) Determine the limiting unbraced length, L r : L pr r t E F yr Eq. ( ) where: F 0.7F F yr yc yw F yr 0.7(50) 35.0 ksi < 50 ksi ( 0.5F yc 5 ksi ok) Therefore: p(3.90) 9,000 L r 9.39 ft LAT. TORSIONAL BUCKLING RESISTANCE (Article ) Since L p 7.83 ft < L b 4.0 ft < L r 9.39 ft: Ø F  L yr b p F nc Cb Œ11 RbRhFyc Ł RhFyc łł Lr Lp ł F nc º Ø 1.0Œ1 1 º  Ł  L Eq. ( ) Therefore: F ncltb ksi (< F ncflb 50.0 ksi) œ ß ø ø ( 1.0) (1.0)(50) ksi 1.0(50) œ łł  łß < 1.0(1.0)(50) 50 ksi R b R h F yc \ F nc F ncltb ksi Design Example 330
31 Flexural Resistance  Composite Sections in Negative Flexure & Noncomposite Sections F n or M n F max or M M max max Basic Form of All FLB & LTB Eqs Fyr λf λpf Fnc 1 1 RbR hfyc nc b hf yc R F Anchor point 1 h yc λrf λpf nc b hf yc Fyr Lb L p Fnc Cb 1 1 RbRhFyc RbRhFyc RhF yc Lr L p F RR F RR Anchor point F r or M r r compact noncompact (inelastic buckling) nonslender F F RRF nc cr slender (elastic buckling) b h yc b bπ Lb C R r t E L p or λ λ p pf L r λor r λ rf L b or b fc /t fc CONSTRUCTIBILITY  FLEXURE Top Flange For STRENGTH I: fbu 4.0 ksi < 50.0 ksi + fl ffrhfyc Eq. ( ) f bu + f l ksi ksi 4.0 ksi ffrhfyc 1.0(1.0)(50) 50.0 ksi f bu + l f F f nc f bu 3.34 ksi < ksi ok ( Ratio ) 1 + fl fffnc Eq. ( ) f ksi + ksi 3.34 ksi (38.75) ksi ok (Ratio 0.835) Design Example 331
32 WEB BENDBUCKLING RESISTANCE (Article ) F 0.9Ek D Ł t w ł crw k k min(r F h 9 ( D ) c D 9 ( ) yc,f yw ) Eq. ( ) 0.9(9,000)(8.7) Fcrw ksi < RhFyc 1.0(50) 50 ksi 69.0 Ł 0.5 ł ok CONSTRUCTIBILITY  FLEXURE Web & Top Flange (continued) fbu fffcrw Eq. ( ) fffcrw 1.0(39.33) ksi ksi < ksi ok (Ratio 0.697) For STRENGTH IV: fbu + fl ffrhfyc Eq. ( ) f bu + f l ksi ksi ksi ffrhfyc 1.0(1.0)(50) 50.0 ksi ksi < 50.0 ksi ok (Ratio 0.955) Design Example 33
33 CONSTRUCTIBILITY  FLEXURE Top Flange (continued) & Web 1 fbu + fl fffnc Eq. ( ) fbu + fl ksi + ksi ksi 3 3 fffnc 1.0(38.75) ksi 37.85ksi < ksi ok (Ratio 0.977) fbu fffcrw Eq. ( ) fffcrw 1.0(39.33) ksi ksi < ksi ok (Ratio 0.836) CONSTRUCTIBILITY Wind Load  Section 11 Calculate f bu due to the steel weight within the unbraced length containing Section 11: For STRENGTH III: Top Flange: Bottom Flange: 1.0(1.5)(35)(1) f bu ksi 1, (1.5)(35)(1) f bu.68 ksi 1,973 Calculate the factored wind force on the steel section: ( 1.5) 1.0 (0.053)( ) W kips / ft 1 Design Example 333
34 BRIDGE FRAMING PLAN CONSTRUCTIBILITY Wind Load  Section 11 (continued) Assume Span 1 of the structure resists the lateral wind force as a propped cantilever with an effective span length of 100 (i.e. assume top lateral bracing provides an effective line of fixity 00 from the pier): 9 9 M11 WLe (0.403)(10.0) 408.0kip  ft (Note: refined 3D analysis > kipft) Design Example 334
35 CONSTRUCTIBILITY Wind Load  Section 11 (continued) Proportion total lateral moment to top & bottom flanges according to relative lateral stiffness of each flange. Then, divide total lateral moment equally to each girder: 1(16) Top Flg: I l in Bot Flg: 1.375(18) I l in Top Flange: Bottom Flange: 408.0(341.3) Ml kip  ft ( ) (668.3) Ml 67.5 kip  ft ( )4 CONSTRUCTIBILITY Wind Load  Section 11 (continued) Separate calculations indicate that lateral bending stresses in the top (compression) flange may be determined from a firstorder analysis (i.e. no amplification is required). Top Flange: f l 34.48(1) 9.70 ksi < 0.6Fyf 30.0 ksi 1(16) 6 ok Bottom Flange: f l 67.5(1) ksi < 0.6Fyf 30.0 ksi 1.375(18) 6 ok Design Example 335
36 BRIDGE FRAMING PLAN CONSTRUCTIBILITY Wind Load  Section 11 (continued) Calculate the shear in the propped cantilever at the assume effective line of fixity: V 5 WL 8 5 (0.403)(10.0) 8 f f e 30.3 kips Resolve the shear into a compressive force in the diagonal of the top bracing: P 30.3 Ł (0.0) + (1.0) 1.0 ł kips Design Example 336
37 CONSTRUCTIBILITY Wind Load  Section 11 (continued) Separate calculations (see example) indicate a compressive force of kips in the diagonal to the selfweight of the steel. Therefore, the total compressive force in the bracing diagonal is: ( kips) + ( kips) kips (Note: refined 3D analysis > kips) CONSTRUCTIBILITY Wind Load  Section 11 (continued) Estimate the maximum lateral deflection of Span 1 of the structure (i.e. the propped cantilever) due to the factored wind load using the total lateral moments of inertia of the top & bottom flanges of all four girders at Section 11: 4 4 D l max. WL 0.403(10.0) (1,78) e 185EI 185(9,000)( )4 (Note: refined 3D analysis > 7.0 inches) 6.7 in. If the top lateral bracing were not present: L e > D l max. 1.3 inches Design Example 337
38 CONSTRUCTIBILITY Performance Ratios POSITIVEMOMENT REGION, SPAN 1 (Section 11) Constructibility (Slenderweb section) Flexure (STRENGTH I) Eq. ( ) Top flange Eq. ( ) Top flange Eq. ( ) Web bend buckling Eq. ( ) Bottom flange Flexure (STRENGTH III Wind load on noncomposite structure) Eq. ( ) Top flange 0.61 Eq. ( ) Top flange Eq. ( ) Web bend buckling Eq. ( ) Bottom flange 0.7 Flexure (STRENGTH IV) Eq. ( ) Top flange Eq. ( ) Top flange Eq. ( ) Web bend buckling Eq. ( ) Bottom flange 0.60 Shear (960 from the abutment) (STRENGTH IV) CONSTRUCTIBILITY Shear (Article ) Interior panels of stiffened webs must satisfy: V V u fvvcr Eq. ( ) ( V ) 1.0(1.5)( 79) 119 kips u DC  1 at 960 from the abutment V n Vcr CVp Eq. ( ) Vp 0.58F yw Dtw 1,001kips C 0.66 (for 07inch stiffener spacing) 0.66(1,001) 66 kips > V 119 kips (Ratio 0.447) cr u  Design Example 338
39 CONSTRUCTIBILITY Section  (Interior Pier) In regions of negative flexure, the constructibility checks for flexure generally do not control because the sizes of the flanges in these regions are normally governed by the sum of the factored dead and live load stresses at the strength limit state. Also, the maximum accumulated negative moments during the deck placement in these regions typically do not differ significantly from the calculated DC 1 negative moments. Deck overhang brackets and wind loads do induce lateral bending into the flanges, which can be considered using the flexural design equations. Web bendbuckling and shear should always be checked in these regions for critical stages of construction (refer to the design example). CONSTRUCTIBILITY Concrete Deck (Article ) Unless longitudinal reinforcement is provided according to the provisions of Article , f deck ff r 0.9f r ' fr 0.4 fc ksi ff r 0.90(0.480) 0.43 ksi Design Example 339
40 Table 1: Moments from DeckPlacement Analysis Span > 1 Unfactored DeadLoad Moments (kipft) Length (ft) Steel Weight SIP Forms (SIP) Cast Sum of Casts + SIP Max. +M DC + DW Deck, haunches + SIP M 35 +,537,889 kipft CONSTRUCTIBILITY Concrete Deck (continued) Calculate the longitudinal concrete deck tensile stress at the end of Cast 1 (use n 8): 1.0(1.5)( 1,403)(3.0)(1) f deck ksi > 0.43 ksi 161,518(8) Therefore, provide onepercent longitudinal reinforcement (No. 6 bars or 1 ). Extend to 95.0 feet from the abutment. Tensile force (0.453)(100.0)(9.0) 408 kips Design Example 340
41 Service Limit State SERVICE LIMIT STATE Elastic Deformations (Article ) Use suggested minimum spantodepth ratios (optional  Article ) Check liveload deflections (optional  Article.5..6.): 140.0(1) End Spans: DALLOW.10 in. > 0.91in (1) Center Span: DALLOW.63 in. > 1.3 in. 800 ok ok Design Example 341
42 SERVICE LIMIT STATE Permanent Deformations (Article ) Under the SERVICE II load combination: 1.0DC + 1.0DW + 1.3(LL+IM) Top steel flange of composite sections: f f 0.95RhFyf Eq. ( ) Bottom steel flange of composite sections: f f f + l 0.95RhFyf Eq. ( ) Web bendbuckling: f c Fcrw Eq. ( ) SERVICE LIMIT STATE Permanent Deformations (continued) Check top flange (Section 11): 0.95Rh F 0.95(1.0)(50) ksi yf f f 0.95RhFyf Ø1.0(,0) 1.0(335+ 3) 1.3(3,510) ø f f 1.0Œ ksi 1,581 4,863 13,805 œ º ß .30 ksi < ksi (Ratio 0.469) ok Design Example 34
43 SERVICE LIMIT STATE Permanent Deformations (continued) Check bottom flange (Section 11): f f f + l 0.95RhFyf f f Ø1.0(,0) 1.0Œ º 1, ( ) 1.3(3,510) ø ksi,483,706 œ ß (Ratio 0.775) For composite sections in positive flexure with D/t w 150, web bendbuckling need not be checked at the service limit state ksi + 0 < ksi ok SERVICE LIMIT STATE Permanent Deformations (continued) Check Section  (interior pier): Article for members with shear connectors provided throughout their entire length that also satisfy the provisions of Article (i.e. one percent longitudinal reinforcement is provided in the deck wherever the tensile stress in the deck due to the factored construction loads or the SERVICE II load combination exceeds the modulus of rupture), flexural stresses caused by SERVICE II loads applied to the composite section may be computed using the shortterm or longterm composite section, as appropriate, assuming the concrete deck is effective for both positive and negative flexure. Design Example 343
44 SERVICE LIMIT STATE Permanent Deformations (continued) Check Section  (interior pier): Flange majoraxis bending stresses at Section  and at the first flange transition located 150 from the interior pier are checked under the SERVICE II load combination and do not control. Stresses acting on the composite section are computed assuming the concrete is effective for negative flexure, as permitted in Article Web bendbuckling must be checked for composite sections in negative flexure under the SERVICE II load combination: f c F crw Eq. ( ) WEB BENDBUCKLING RESISTANCE (Article ) F 0.9Ek D Ł t w ł crw min(r F h yc 0.7) Eq. ( ) 9 where: k D Eq. ( ) According to Article D6.3.1 (Appendix D), for composite sections in negative flexure at the service limit state where the concrete is considered effective in tension for computing flexural stresses on the composite section, as permitted in Article , D c is to be computed as:  fc Eq. (D ) D c d  t fc 0 Ł fc + ft ł,f ( ) c D yw Design Example 344
45 WEB BENDBUCKLING RESISTANCE (Article ) Check the bottomflange transition (controls): Ø1.0( ,656 ) 1.0( ) 1.3( ,709 ) ø f f 1.0 Œ ksi º 1,789,46,463 œ ß  ( ) D c in. > 0 Ł ł 9 k.9 ( ) ok 0.9(9,000)(.9) 39.7ksi < 39.7 ksi 69.0 ok Ratio (0.979) Ł 0.565ł Fcrw SERVICE LIMIT STATE Performance Ratios POSITIVEMOMENT REGION, SPAN 1 (Section 11) Service Limit State Liveload deflection Permanent deformations (SERVICE II) Eq. ( ) Top flange Eq. ( ) Bottom flange INTERIORPIER SECTION (Section ) Permanent deformations (SERVICE II) Eq. ( ) Top Section Eq. ( ) Top Flange transition Eq. ( ) Bottom Section Eq. ( ) Bottom Flange transition Eq. ( ) Web bend Section Eq. ( ) Web bend Flange transition Design Example 345
46 SERVICE LIMIT STATE Concrete Deck (Article ) Calculate the longitudinal concrete deck tensile stress in Span 1 at 95.0 ft from the abutment (use n 8): 1.0[ 1.0(87) + 1.0(83) + 1.3( 1,701)](3.0)(1) fdeck ksi 161,518(8) Therefore, extend the onepercent longitudinal reinforcement (No. 6 bars or 1 ) to 94.0 feet from the abutment. f deck ksi < 0.43 ksi ok > 0.90fr 0.43 ksi Fatigue & Fracture Limit State Design Example 346
47 FATIGUE RESISTANCE FIRST PRINCIPAL For lower traffic volumes, fatigue resistance is inversely proportional to the cube of the effective stress range. Design Example 347
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