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

Transcription

1 EXAMPLE 1: THREE-SPAN 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. SITE-Blauvelt Engineers Pittsburgh, PA DESIGN PARAMETERS SPECIFICATIONS: LRFD Third Edition (004) STRUCTURAL STEEL: - ASTM A 709 Grade HPS 70W for flanges in negative-flexure regions - ASTM A 709 Grade 50W for all other girder and cross-frame steel CONCRETE: f' c 4.0 ksi REINFORCING STEEL: F y 60 ksi ADTT:,000 trucks per day Design Example 3-1

2 BRIDGE CROSS-SECTION CROSS-FRAMES (Article ) The need for diaphragms or cross-frames 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 CROSS-SECTION PROPORTIONS Web Depth Span-to-Depth 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 3-3

4 CROSS-SECTION 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 3-4

5 CROSS-SECTION PROPORTIONS (continued) Flange Width-to-Thickness (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 - Cross-frames 0.10 k/ft & details TOTAL k/ft 4 girders k/ft + girder weight Design Example 3-5

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 HL-93 - (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 3-6

7 LRFD Negative Moment Loading (Article ) For negative moment between points of permanent-load contraflexure & interior-pier 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 rear-axle spacing of both design trucks at 14 feet, and Reduce all loads by 10 percent. LRFD Fatigue Load (Article ) Design Truck only > w/ fixed 30-ft rearaxle spacing placed in a single lane Design Example 3-7

8 LOAD for OPTIONAL LIVE-LOAD 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 3-8

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 3-9

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 STRENGTH-I? p /1.0? TG? SE STRENGTH-II? p /1.0? TG? SE STRENGTH-III? p /1.0? TG? SE STRENGTH-IV EH, EV, ES, DW DC ONLY? p / STRENGTH-V? p /1.0? TG? SE EXTREME-I? p? EQ EXTREME-II? p SERVICE-I /1.0? TG? SE SERVICE-II / SERVICE-III /1.0? TG? SE FATIGUE-LL, IM & CE ONLY Design Example 3-10

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 At-Rest 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 3-11

12 STRUCTURAL ANALYSIS Summary -- Live-Load 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 Live-Load Deflection: NL DF m3 ŁNb ł lanes Ł 4 ł Design Example 3-1

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 3-13

14 Design Example 3-14

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 3-15

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 1-1 (@ 0.4L 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 3-16

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 1-1 (@ 0.4L 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 3-17

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 kip-ft AD y y y D1 [ (,0) + 1.5( 335) ( 3) + 6,517] D + M 10,171kip-ft 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 3-18

19 Constructibility DECK-PLACEMENT SEQUENCE Design Example 3-19

20 Table 1: Moments from Deck-Placement Analysis Span -> 1 Unfactored Dead-Load Moments (kip-ft) Length (ft) Steel Weight SIP Forms (SIP) Cast Sum of Casts + SIP Max. +M DC + DW Deck, haunches + SIP M 35 +,537,889 kip-ft Table : Vertical Deflections from Deck-Placement Analysis Span ->1 Unfactored Vertical Dead-Load Deflections (In.) Length (ft) Steel Weight SIP Forms (SIP) Cast Sum of Casts + SIP DC + DW Total Deck, haunches + SIP Design Example 3-0

21 Table 3: Unfactored Vertical Dead-Load Reactions from Deck-Placement 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 DECK-PLACEMENT ANALYSIS (continued) Calculate f bu : (at Section 1-1 -> 56-0 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 3-1

22 DECK-OVERHANG 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 first-order compression-flange f l is required: L b 4-0 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 3-3

24 DECK OVERHANG LOADS (continued) For the steel section at Section 1-1, 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 first-order compression-flange 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 3-4

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 3-5

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 3-6

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 3-7

28 CONSTRUCTIBILITY - FLEXURE (Article ) Determine if the section is a slender-web 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 slender-web 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 3-8

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 3-9

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 Œ1-1- 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 3-30

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 3-31

32 WEB BEND-BUCKLING 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 3-3

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 1-1 Calculate f bu due to the steel weight within the unbraced length containing Section 1-1: 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 3-33

34 BRIDGE FRAMING PLAN CONSTRUCTIBILITY Wind Load - Section 1-1 (continued) Assume Span 1 of the structure resists the lateral wind force as a propped cantilever with an effective span length of 10-0 (i.e. assume top lateral bracing provides an effective line of fixity 0-0 from the pier): 9 9 M1-1 WLe (0.403)(10.0) 408.0kip - ft (Note: refined 3D analysis > kip-ft) Design Example 3-34

35 CONSTRUCTIBILITY Wind Load - Section 1-1 (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 1-1 (continued) Separate calculations indicate that lateral bending stresses in the top (compression) flange may be determined from a first-order 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 3-35

36 BRIDGE FRAMING PLAN CONSTRUCTIBILITY Wind Load - Section 1-1 (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 3-36

37 CONSTRUCTIBILITY Wind Load - Section 1-1 (continued) Separate calculations (see example) indicate a compressive force of kips in the diagonal to the self-weight 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 1-1 (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 1-1: 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 3-37

38 CONSTRUCTIBILITY Performance Ratios POSITIVE-MOMENT REGION, SPAN 1 (Section 1-1) Constructibility (Slender-web 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 (96-0 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 96-0 from the abutment V n Vcr CVp Eq. ( ) Vp 0.58F yw Dtw 1,001kips C 0.66 (for 07-inch stiffener spacing) 0.66(1,001) 66 kips > V 119 kips (Ratio 0.447) cr u - Design Example 3-38

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 bend-buckling 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 3-39

40 Table 1: Moments from Deck-Placement Analysis Span -> 1 Unfactored Dead-Load Moments (kip-ft) Length (ft) Steel Weight SIP Forms (SIP) Cast Sum of Casts + SIP Max. +M DC + DW Deck, haunches + SIP M 35 +,537,889 kip-ft 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 one-percent 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 3-40

41 Service Limit State SERVICE LIMIT STATE Elastic Deformations (Article ) Use suggested minimum span-to-depth ratios (optional - Article ) Check live-load 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 3-41

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 bend-buckling: f c Fcrw Eq. ( ) SERVICE LIMIT STATE Permanent Deformations (continued) Check top flange (Section 1-1): 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 3-4

43 SERVICE LIMIT STATE Permanent Deformations (continued) Check bottom flange (Section 1-1): 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 bend-buckling 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 long-term composite section, as appropriate, assuming the concrete deck is effective for both positive and negative flexure. Design Example 3-43

44 SERVICE LIMIT STATE Permanent Deformations (continued) Check Section - (interior pier): Flange major-axis bending stresses at Section - and at the first flange transition located 15-0 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 bend-buckling must be checked for composite sections in negative flexure under the SERVICE II load combination: f c F crw Eq. ( ) WEB BEND-BUCKLING 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 3-44

45 WEB BEND-BUCKLING RESISTANCE (Article ) Check the bottom-flange 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 POSITIVE-MOMENT REGION, SPAN 1 (Section 1-1) Service Limit State Live-load deflection Permanent deformations (SERVICE II) Eq. ( ) Top flange Eq. ( ) Bottom flange INTERIOR-PIER 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 3-45

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 one-percent 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 3-46

47 FATIGUE RESISTANCE FIRST PRINCIPAL For lower traffic volumes, fatigue resistance is inversely proportional to the cube of the effective stress range. Design Example 3-47

48 FATIGUE RESISTANCE SECOND PRINCIPAL For higher traffic volumes, fatigue resistance is infinite if the maximum stress range is less than the constant-amplitude fatigue threshold. FATIGUE LOAD (Article ) The specified load condition for fatigue is a single truck; the current HS0 truck with a fixed rearaxle spacing of The truck occupies a single lane on the bridge -- not multiple lanes. The fatigue load produces a lower calculated stress range than the Standard Specifications. Design Example 3-48