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

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

BRIDGE CROSS-SECTION CROSS-FRAMES (Article 6.7.4.1) 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-

BRIDGE FRAMING PLAN CROSS-SECTION PROPORTIONS Web Depth Span-to-Depth Ratios (Table.5..6.3-1) 0.03L 0.03(175.0) 5.6 ft 67. in. Use 69.0 in. Web Thickness (Article 6.10..1.1) D 150 69 150 ( ) 0.46 in. t w min. Design Example 3-3

CROSS-SECTION PROPORTIONS (continued) Flange Width (Article 6.10..) ( ) D/6 69.0/6 11.5 in. b f min. L 85 100(1) 85 ( ) 14.1in. b fc min. Flange Thickness (Article 6.10..) ( t ) 1.1t 1.1( 0.565) 0.6 in. f min w Design Example 3-4

CROSS-SECTION PROPORTIONS (continued) Flange Width-to-Thickness (Article 6.10..) b t f 18 ( 0.875) 10.3 f < 1.0 ok Flange Moments of Inertia (Article 6.10..) I I yc yt ( ) 3 1 16 1 3 1.375( 18) 0.51 0.1 < 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 5.106 k/ft (incl. integral w.s.) - Overhang tapers 0.14 k/ft - Deck haunches 0.183 k/ft - SIP forms 0.480 k/ft - Cross-frames 0.10 k/ft & details TOTAL 6.031 k/ft 4 girders 1.508 k/ft + girder weight Design Example 3-5

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 3.6.1..1) 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

LRFD Negative Moment Loading (Article 3.6.1.3.1) 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 3.6.1.4.1) Design Truck only > w/ fixed 30-ft rearaxle spacing placed in a single lane Design Example 3-7

LOAD for OPTIONAL LIVE-LOAD DEFLECTION EVALUATION Refer to Article 3.6.1.3.: 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 0.050 ksf for beams V DZ design wind velocity at elevation Z V B base wind velocity at 30 ft height 100 mph Eq. (3.8.1..1-1) For this example, assume the bridge is 35 ft above low ground & located in open country. Design Example 3-8

WIND LOADS (continued) V DZ.5V o V Ł V 30 B ln ł Ł Z Z o ł Eq. (3.8.1.1-1) 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. 100 35 V DZ.5 Ł100 ł Ł0.3 ł Ø( 103.0) ø PD 0.050Œ œ 0.053 ksf º 10,000 ß ( 8.0) ln 103.0 mph PD hexp 0.053(10.41) 0.55 kips / ft > 0.3 kips / ft ok Design Example 3-9

Basic LRFD Design Equation S? i? i Q i fr n R r Eq. (1.3..1-1) 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 1.75 1.00 - - 1.00 0.50/1.0? TG? SE - - - - STRENGTH-II? p 1.35 1.00 - - 1.00 0.50/1.0? TG? SE - - - - STRENGTH-III? p - 1.00 1.40-1.00 0.50/1.0? TG? SE - - - - STRENGTH-IV EH, EV, ES, DW DC ONLY? p 1.5-1.00 - - 1.00 0.50/1.0 - - - - - - STRENGTH-V? p 1.35 1.00 0.40 0.40 1.00 0.50/1.0? TG? SE - - - - EXTREME-I? p? EQ 1.00 - - 1.00 - - - 1.00 - - - EXTREME-II? p 0.50 1.00 - - 1.00 - - - - 1.00 1.00 1.00 SERVICE-I 1.00 1.00 1.00 0.30 0.30 1.00 1.00/1.0? TG? SE - - - - SERVICE-II 1.00 1.30 1.00 - - 1.00 1.00/1.0 - - - - - - SERVICE-III 1.00 0.80 1.00 - - 1.00 1.00/1.0? TG? SE - - - - FATIGUE-LL, IM & CE ONLY - 0.75 - - - - - - - - - - - Design Example 3-10

Load Factors for Permanent Loads,? p Load Factor Type of Load Maximum Minimum DC: Component and Attachments 1.5 0.90 DD: Downdrag 1.80 0.45 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 1.50 0.65 1.50 1.35 1.35 1.35 1.30 1.35 1.95 0.90 0.90 N/A 1.00 0.90 0.90 0.90 LRFD LOAD COMBINATIONS (continued) Construction Loads (Article 3.4.): STRENGTH I - Construction loads -> Load factor 1.5 - DW -> Load factor 1.5 STRENGTH III - Construction dead loads -> Load factor 1.5 - Wind loads -> Load factor 1.5 - DW -> Load factor 1.5 STRENGTH V - Construction dead loads -> Load factor 1.5 - DW -> Load factor 1.5 Design Example 3-11

STRUCTURAL ANALYSIS Summary -- Live-Load Distribution Factors: Strength Limit State Interior Girder Exterior Girder Bending Moment 0.807 lanes 0.950 lanes Shear 1.08 lanes 0.950 lanes Fatigue Limit State Interior Girder Exterior Girder Bending Moment 0.440 lanes 0.750 lanes Shear 0.700 lanes 0.750 lanes STRUCTURAL ANALYSIS (continued) Distribution Factor for Live-Load Deflection: NL DF m3 ŁNb ł 3 0.85 0.638 lanes Ł 4 ł Design Example 3-1

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

Design Example 3-14

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 0.60 + 0.5(0.91) 0.83 in. (D LL+IM ) center span 0.85 + 0.5(1.3) 1.16 in. Design Example 3-15

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 4.6..6): Interior Girder or or 1.0t s L 4 b + 100.0 x 1 4 tf 1.0 ( 9.0) 300.0 in. 16.0 116.0 in. (governs) average spacing of girders 144.0 in. + Design Example 3-16

SECTION PROPERTIES (continued) Section 1-1 (@ 0.4L 1 ) or or Effective Flange Width (Article 4.6..6): Exterior Girder 116.0 L 100.0 x 1 + 58.0 + 08.0 in. 8 8 116.0 + 6.0t 116.0 s btf + 4 58.0 + 6.0 ( 9.0) 16.0 + 116.0 in. 4 + widthof the overhang 58.0 + 4.0 in. 100.0 in. (governs) SECTION PROPERTIES (continued) Section 1-1 (@ 0.4L 1 ) Plastic Moment (Article D6.1 -- 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

SECTION PROPERTIES (continued) Section 1-1 (@ 0.4L 1 ) Yield Moment (Article D6.. -- Appendix D): M M M M + M MD1 MD M F y + + S S S NC LT AD (,0)( 1) 1.5( 335)( 1) + 1.50( 3)( 1) Ø1.5 50 1.0 Œ + º 1,973 M 78,06 kip- in. 6,517 kip-ft AD y y y D1 [ (,0) + 1.5( 335) + 1.50( 3) + 6,517] 1.0 1.5 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. 4.6..6): Exterior Girder 100.5 in. Min. Concrete Deck Reinforcement (Article 6.10.1.7): 9.0 Ø 1 3.0 18 ø A 1 Œ º 1 Ł łł 1 łß œ ( 43.0) + + 0.5 3.5-33.17 ft 4,776 in deck. 0.01(4,776) 47.76 in. 47.76 1.11in. ft 0.096 in. in. 43.0 0.096(100.5) 9.30 in. @ 4.63 in. from bot. of the deck Design Example 3-18

Constructibility DECK-PLACEMENT SEQUENCE Design Example 3-19

Table 1: Moments from Deck-Placement Analysis Span -> 1 Unfactored Dead-Load Moments (kip-ft) Length (ft) 1.00 4.00 4.00 48.00 56.00 7.00 84.00 96.00 100.00 Steel Weight 143 50 341 353 35 96 06 74 1 --------------------------------------------------------------------------------------------------------------------- SIP Forms (SIP) 63 110 147 151 150 14 84 7 4 Cast 1 870 1544 189 306 387 86 1983 1484 175-168 -336-589 -673-786 -1010-1179 -1347-1403 3 14 8 50 57 67 86 101 115 10 --------------------------------------------------------------------------------------------------------------------- Sum of Casts + SIP 779 1346 1797 1841 1818 1486 989 79-4 Max. +M 933 1654 336 457 537 410 067 1511 179 --------------------------------------------------------------------------------------------------------------------- DC + DW 75 477 643 661 657 551 386 148 5 Deck, haunches + SIP 786 1360 18 1870 1850 158 1038 335 53 M 35 +,537,889 kip-ft Table : Vertical Deflections from Deck-Placement Analysis Span ->1 Unfactored Vertical Dead-Load Deflections (In.) Length (ft) 1.00 4.00 4.00 48.00 56.00 7.00 84.00 96.00 100.00 Steel Weight -.17 -.3 -.47 -.50 -.51 -.47 -.39 -.9 -.5 --------------------------------------------------------------------------------------------------------------------- SIP Forms (SIP) -.07 -.14 -.0 -.1 -.1 -.0 -.16 -.1 -.10 Cast 1-1.3 -.50-3.78-4.04-4.7-4.30-3.95-3.33-3.08.7.5.86.96 1.08 1.5 1.3 1.3 1.31 3 -.01 -.03 -.04 -.04 -.05 -.05 -.05 -.04 -.03 --------------------------------------------------------------------------------------------------------------------- Sum of Casts + SIP -1.14 -.14-3.16-3.34-3.46-3.30 -.84 -.17-1.91 --------------------------------------------------------------------------------------------------------------------- DC + DW -.17 -.3 -.46 -.48 -.49 -.45 -.38 -.8 -.4 Total -1.48 -.78-4.09-4.3-4.46-4. -3.61 -.74 -.40 Deck, haunches + SIP-.9-1.71 -.47 -.59 -.64 -.43 -.0-1.47-1.7 Design Example 3-0

Table 3: Unfactored Vertical Dead-Load Reactions from Deck-Placement Analysis (kips) Abut. 1 Pier 1 Pier Abut. Steel Weight -13. -53. -53. -13. sum -13. -53. -53. -13. ----------------------------------------------------------------------------------------- SIP Forms (SIP) -6. -1. -1. -6. sum -19. -74. -74. -19. Cast 1-80. -55. -55. -80. sum -99. -19. -19. -99. Cast 13. -75. -75. 14. sum -85. -04. -04. -85. Cast 3-1. -110. -110. -1. sum -86. -314. -314. -86. ----------------------------------------------------------------------------------------- Sum of Casts + SIP -73. -61. -61. -73. ----------------------------------------------------------------------------------------- DC + DW -6. -90. -90. -6. Total -11. -404. -404. -11. Deck, haunches -74. -61. -61. -74. + 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 1,581 1.0(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 -3.89 ksi 1,581 1.0(1.5)(,889)(1) f bu 6.36 ksi 1,973 Design Example 3-1

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-

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. (6.10.1.6-4) 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 -3.89 ksi (STRENGTH IV) p CbR f F b yc Eq. (6.10.1.6-) L 1.0r p yc Eq. (6.10.8..3-4) t E F where: r t b fc 1 Dct 1 1+ Ł 3bfct w fc ł Eq. (6.10.8..3-9) Design Example 3-3

DECK OVERHANG LOADS (continued) For the steel section at Section 1-1, D c 38.63 in. 16 r t 3.90 in. 1 38.63(0.5) 1 1 Ł + 3 16(1) ł 1.0(3.90) 9,000 L p 7.83 ft 1 50 1.0(1.0) 1. b - 3.89 50 ( 7.83) 11.59 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. (6.10.8..3-8) 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

DECK OVERHANG LOADS (continued) The amplification factor is determined as: For STRENGTH I: 0.85 AF 1.78 > 1.0-7.41 1- Ł 5.49 ł ok For STRENGTH IV: 0.85 AF.8 > 1.0-3.89 1- Ł 5.49 ł ok DECK OVERHANG LOADS (continued) For STRENGTH I: Dead loads: [ + 1.5(40 + 85 + 5 + 15) ] 731.3 lbs / ft P 1.0 1.5(55) F F P tana 731.3tan( 31.3 o l ) F L 0.4446( 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) 6 444.6 lbs / ft M 1.34(1) Bot. flange: f l l 3.45 ksi Sl 1.375(18) 6 Design Example 3-5

DECK OVERHANG LOADS (continued) For STRENGTH I: Finishing machine: [ ] 4,500 lbs P 1.0 1.5(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 6.00 +.31 8.31 ksi * AF (8.31)(1.78) 14.79 ksi < 0.6F yf 30 ksi ok Bot. flange: f l total 3.45 + 1.33 4.78 ksi * AF (4.78)(1.0) 4.78 ksi < 0.6F yf 30 ksi ok Design Example 3-6

DECK OVERHANG LOADS (continued) For STRENGTH IV: Dead loads: P 1.0[ 1.5(55 + 40 + 85 + 5 + 15 )] 795 lbs / ft F F P tan a 795 tan(31.3 o l ) 483.4 lbs / ft F L 0.4834 ( 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) 14.87 ksi 14.87 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

CONSTRUCTIBILITY - FLEXURE (Article 6.10.3.) Determine if the section is a slender-web section: 5.7 E F yc D t D t w w c 5.7 5.7 E F yc (38.63) 0.5 c Eq. (6.10.6..3-1) 154.5 9,000 137.3 < 154.5 50 Therefore, the section is a slender-web section. Go to Article 6.10.8 to compute F nc. CONSTRUCTIBILITY - FLEXURE (Article 6.10.3.) For discretely braced compression flanges: f f f R F Eq. (6.10.3..1-1) bu + l f h yc f 1 3 bu + fl f bu f F f F f f crw nc Eq. (6.10.3..1-) Eq. (6.10.3..1-3) For discretely braced tension flanges: f f f R F Eq. (6.10.3..-1) bu + l f h yt Design Example 3-8

LOCAL BUCKLING RESISTANCE Top Flange (Article 6.10.8..) Determine the slenderness ratio of the top flange: l 0.38 l b t 16 1 fc f fc ( ) E F 0.38 8.0 9,000 50 pf yc 9. Since l f <l pf : F R R F Eq. (6.10.8..-1) F FLB nc b nc 1.0(1.0)(50) 50.0 ksi h yc 6.10.8 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

LAT. TORSIONAL BUCKLING RESISTANCE (Article 6.10.8..3) Determine the limiting unbraced length, L r : L pr r t E F yr Eq. (6.10.8..3-5) 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 1 35.0 LAT. TORSIONAL BUCKLING RESISTANCE (Article 6.10.8..3) 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. (6.10.8..3-) Therefore: F ncltb 38.75 ksi (< F ncflb 50.0 ksi) œ ß ø 35.0 4.0-7.83 ø ( 1.0) (1.0)(50) 38.75 ksi 1.0(50) 9.39 7.83 œ łł - łß < 1.0(1.0)(50) 50 ksi R b R h F yc \ F nc F ncltb 38.75 ksi Design Example 3-30

6.10.8 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. (6.10.3..1-1) f bu + f l - 7.41ksi + 14.79 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 < 38.75 ksi ok ( Ratio 0.844 ) 1 + fl fffnc Eq. (6.10.3..1-) 3 1 14.79 f - 7.41 ksi + ksi 3.34 ksi 3 3 1.0(38.75) 38.75 ksi ok (Ratio 0.835) Design Example 3-31

WEB BEND-BUCKLING RESISTANCE (Article 6.10.1.9) F 0.9Ek D Ł t w ł crw k k min(r F h 9 ( D ) c D 9 ( 38.63 69.0) yc,f yw 8.7 0.7) Eq. (6.10.1.9.1-1) 0.9(9,000)(8.7) Fcrw 39.33 ksi < RhFyc 1.0(50) 50 ksi 69.0 Ł 0.5 ł ok CONSTRUCTIBILITY - FLEXURE Web & Top Flange (continued) fbu fffcrw Eq. (6.10.3..1-3) fffcrw 1.0(39.33) 39.33 ksi - 7.41ksi < 39.33 ksi ok (Ratio 0.697) For STRENGTH IV: fbu + fl ffrhfyc Eq. (6.10.3..1-1) f bu + f l - 3.89 ksi + 14.87 ksi 47.76 ksi ffrhfyc 1.0(1.0)(50) 50.0 ksi 47.76 ksi < 50.0 ksi ok (Ratio 0.955) Design Example 3-3

CONSTRUCTIBILITY - FLEXURE Top Flange (continued) & Web 1 fbu + fl fffnc Eq. (6.10.3..1-) 3 1 14.87 fbu + fl - 3.89 ksi + ksi 37.85 ksi 3 3 fffnc 1.0(38.75) 38.75 ksi 37.85ksi < 38.75 ksi ok (Ratio 0.977) fbu fffcrw Eq. (6.10.3..1-3) fffcrw 1.0(39.33) 39.33 ksi - 3.89 ksi < 39.33 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 -3.34 ksi 1,581 1.0(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)(.0 + 69.0 +.0) W 0.403 kips / ft 1 Design Example 3-33

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 18 18 (Note: refined 3D analysis > 405.0 kip-ft) Design Example 3-34

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 341.3 in. 1 3 4 Bot Flg: 1.375(18) I l 668.3 in. 1 3 4 Top Flange: Bottom Flange: 408.0(341.3) Ml 34.48 kip - ft (341.3 + 668.3)4 408.0(668.3) Ml 67.5 kip - ft (341.3 + 668.3)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) 10.91 ksi < 0.6Fyf 30.0 ksi 1.375(18) 6 ok Design Example 3-35

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 -58.76 kips Design Example 3-36

CONSTRUCTIBILITY Wind Load - Section 1-1 (continued) Separate calculations (see example) indicate a compressive force of -19.04 kips in the diagonal to the self-weight of the steel. Therefore, the total compressive force in the bracing diagonal is: (-58.76 kips) + (-19.04 kips) -77.80 kips (Note: refined 3D analysis > -67.0 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)(341.3 + 668.3)4 (Note: refined 3D analysis > 7.0 inches) 6.7 in. If the top lateral bracing were not present: L e 140-0 > D l max. 1.3 inches Design Example 3-37

CONSTRUCTIBILITY Performance Ratios POSITIVE-MOMENT REGION, SPAN 1 (Section 1-1) Constructibility (Slender-web section) Flexure (STRENGTH I) Eq. (6.10.3..1-1) Top flange 0.844 Eq. (6.10.3..1-) Top flange 0.835 Eq. (6.10.3..1-3) Web bend buckling 0.697 Eq. (6.10.3..-1) Bottom flange 0.535 Flexure (STRENGTH III Wind load on noncomposite structure) Eq. (6.10.3..1-1) Top flange 0.61 Eq. (6.10.3..1-) Top flange 0.170 Eq. (6.10.3..1-3) Web bend buckling 0.085 Eq. (6.10.3..-1) Bottom flange 0.7 Flexure (STRENGTH IV) Eq. (6.10.3..1-1) Top flange 0.955 Eq. (6.10.3..1-) Top flange 0.977 Eq. (6.10.3..1-3) Web bend buckling 0.836 Eq. (6.10.3..-1) Bottom flange 0.60 Shear (96-0 from the abutment) (STRENGTH IV) 0.447 CONSTRUCTIBILITY Shear (Article 6.10.3.3) Interior panels of stiffened webs must satisfy: V V u fvvcr Eq. (6.10.3.3-1) ( V ) 1.0(1.5)( -79) 119 kips u DC - 1 at 96-0 from the abutment V n Vcr CVp Eq. (6.10.9.3.3-1) 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

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 6.10.3..4) Unless longitudinal reinforcement is provided according to the provisions of Article 6.10.1.7, f deck ff r 0.9f r ' fr 0.4 fc 0.4 4.0 0.480 ksi ff r 0.90(0.480) 0.43 ksi Design Example 3-39

Table 1: Moments from Deck-Placement Analysis Span -> 1 Unfactored Dead-Load Moments (kip-ft) Length (ft) 1.00 4.00 4.00 48.00 56.00 7.00 84.00 96.00 100.00 Steel Weight 143 50 341 353 35 96 06 74 1 --------------------------------------------------------------------------------------------------------------------- SIP Forms (SIP) 63 110 147 151 150 14 84 7 4 Cast 1 870 1544 189 306 387 86 1983 1484 175-168 -336-589 -673-786 -1010-1179 -1347-1403 3 14 8 50 57 67 86 101 115 10 --------------------------------------------------------------------------------------------------------------------- Sum of Casts + SIP 779 1346 1797 1841 1818 1486 989 79-4 Max. +M 933 1654 336 457 537 410 067 1511 179 --------------------------------------------------------------------------------------------------------------------- DC + DW 75 477 643 661 657 551 386 148 5 Deck, haunches + SIP 786 1360 18 1870 1850 158 1038 335 53 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 0.453 ksi > 0.43 ksi 161,518(8) Therefore, provide one-percent longitudinal reinforcement (No. 6 bars or smaller @ 1 ). Extend to 95.0 feet from the abutment. Tensile force (0.453)(100.0)(9.0) 408 kips Design Example 3-40

Service Limit State SERVICE LIMIT STATE Elastic Deformations (Article 6.10.4.1) Use suggested minimum span-to-depth ratios (optional - Article.5..6.3) Check live-load deflections (optional - Article.5..6.): 140.0(1) End Spans: DALLOW.10 in. > 0.91in. 800 175.0(1) Center Span: DALLOW.63 in. > 1.3 in. 800 ok ok Design Example 3-41

SERVICE LIMIT STATE Permanent Deformations (Article 6.10.4.) 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. (6.10.4..-1) Bottom steel flange of composite sections: f f f + l 0.95RhFyf Eq. (6.10.4..-) Web bend-buckling: f c Fcrw Eq. (6.10.4..-4) SERVICE LIMIT STATE Permanent Deformations (continued) Check top flange (Section 1-1): 0.95Rh F 0.95(1.0)(50) 47.50 ksi yf f f 0.95RhFyf Ø1.0(,0) 1.0(335+ 3) 1.3(3,510) ø f f 1.0Œ + + 1 -.30 ksi 1,581 4,863 13,805 œ º ß -.30 ksi < 47.50 ksi (Ratio 0.469) ok Design Example 3-4

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 1,973 1.0(335 + 3) 1.3(3,510) ø + + 1 36.80 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. 36.80 ksi + 0 < 47.50 ksi ok SERVICE LIMIT STATE Permanent Deformations (continued) Check Section - (interior pier): Article 6.10.4..1 -- for members with shear connectors provided throughout their entire length that also satisfy the provisions of Article 6.10.1.7 (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

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 6.10.4..1. Web bend-buckling must be checked for composite sections in negative flexure under the SERVICE II load combination: f c F crw Eq. (6.10.4..-4) WEB BEND-BUCKLING RESISTANCE (Article 6.10.1.9) F 0.9Ek D Ł t w ł crw min(r F h yc 0.7) Eq. (6.10.1.9.1-1) 9 where: k D Eq. (6.10.1.9.1-) 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 6.10.4..1, D c is to be computed as: - fc Eq. (D6.3.1-1) D c d - t fc 0 Ł fc + ft ł,f ( ) c D yw Design Example 3-44

WEB BEND-BUCKLING RESISTANCE (Article 6.10.1.9) Check the bottom-flange transition (controls): Ø1.0( -,656 ) 1.0( -373 + -358) 1.3( -,709 ) ø f f 1.0 Œ + + 1-38.88 ksi º 1,789,46,463 œ ß - (-38.88 ) D c 71.0-1.0 43.5 in. > 0 Ł - 38.88 + 3.51 ł 9 k.9 ( 43.5 69.0) ok 0.9(9,000)(.9) 39.7ksi - 38.88 < 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 0.433 Permanent deformations (SERVICE II) Eq. (6.10.4..-1) Top flange 0.469 Eq. (6.10.4..-) Bottom flange 0.775 INTERIOR-PIER SECTION (Section -) Permanent deformations (SERVICE II) Eq. (6.10.4..-1) Top flange @ Section - 0.415 Eq. (6.10.4..-1) Top flange @ Flange transition 0.368 Eq. (6.10.4..-) Bottom flange @ Section - 0.604 Eq. (6.10.4..-) Bottom flange @ Flange transition 0.609 Eq. (6.10.4..-4) Web bend buckling @ Section - 0.898 Eq. (6.10.4..-4) Web bend buckling @ Flange transition 0.979 Design Example 3-45

SERVICE LIMIT STATE Concrete Deck (Article 6.10.1.7) 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 0.440 ksi 161,518(8) Therefore, extend the one-percent longitudinal reinforcement (No. 6 bars or smaller @ 1 ) to 94.0 feet from the abutment. f deck 0.430 ksi < 0.43 ksi ok > 0.90fr 0.43 ksi Fatigue & Fracture Limit State Design Example 3-46

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

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 3.6.1.4) The specified load condition for fatigue is a single truck; the current HS0 truck with a fixed rearaxle spacing of 30-0. 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

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 1.75 1.00 - - 1.00 0.50/1.0? TG? SE - - - - STRENGTH-II? p 1.35 1.00 - - 1.00 0.50/1.0? TG? SE - - - - STRENGTH-III? p - 1.00 1.40-1.00 0.50/1.0? TG? SE - - - - STRENGTH-IV EH, EV, ES, DW DC ONLY? p 1.5-1.00 - - 1.00 0.50/1.0 - - - - - - STRENGTH-V? p 1.35 1.00 0.40 0.40 1.00 0.50/1.0? TG? SE - - - - EXTREME-I? p? EQ 1.00 - - 1.00 - - - 1.00 - - - EXTREME-II? p 0.50 1.00 - - 1.00 - - - - 1.00 1.00 1.00 SERVICE-I 1.00 1.00 1.00 0.30 0.30 1.00 1.00/1.0? TG? SE - - - - SERVICE-II 1.00 1.30 1.00 - - 1.00 1.00/1.0 - - - - - - SERVICE-III 1.00 0.80 1.00 - - 1.00 1.00/1.0? TG? SE - - - - FATIGUE-LL, IM & CE ONLY - 0.75 - - - - - - - - - - - FATIGUE LOAD - (continued) In the LRFD Specification, 75% of the stress range due to the fatigue load is considered to be representative of the effective stress range; that is, the stress range due to an HS-15 truck weighing 54.0 kips with a fixed rear-axle spacing of 30-0 (FATIGUE load combination -- load factor 0.75) The maximum stress range is assumed to be twice the effective stress range (i.e. stress range due to a 108-kip truck) Design Example 3-49

NOMINAL FATIGUE RESISTANCE (Article 6.6.1..5) In the LRFD Specifications, the nominal fatigue resistance is specified as follows: A ŁNł 1 3 ( DF) n ( DF) TH Eq. (6.6.1..5-1) Eq. (6.6.1..5-) A detail category constant (Table 6.6.1..5-1) n cycles per truck passage (Table 6.6.1..5-) (DF) TH constant-amplitude fatigue threshold (Table 6.6.1..5-3) 1 N (365 )(75 )n(adtt ) SL Design Example 3-50

NOMINAL FATIGUE RESISTANCE In the LRFD Specifications, the nominal fatigue resistance is specified as follows: A ŁNł 1 3 1 ( DF) n ( DF) TH N (365 )(75 )n(adtt ) SL Eq. (6.6.1..5-1) Eq. (6.6.1..5-) A detail category constant (Table 6.6.1..5-1) n cycles per truck passage (Table 6.6.1..5-) (DF) TH constant-amplitude fatigue threshold (Table 6.6.1..5-3) Design Example 3-51

NOMINAL FATIGUE RESISTANCE (continued) First and Second Principles of Fatigue Resistance 1 3 ( Df) e max ---------------------------------------------------------------------- Ł A N ł ( Df ) ( DF) TH ( Df) max.0( Df) e 1 ( f) e ( DF) TH \ D NOMINAL FATIGUE RESISTANCE (continued) These LRFD Specification principals can be used for design or evaluation : Design Example 3-5

FATIGUE & FRACTURE LIMIT STATE Load Induced Fatigue (Article 6.6.1.) (ADTT) SL p x ADTT Eq. (3.6.1.4.-1) where ADTT number of trucks per day in one direction averaged over the design life (assumed to be,000 for this example) For a 3-lane bridge: p 0.80 (Table 3.6.1.4.-1) \ (ADTT) SL 0.80(,000) 1,600 trucks/day Design Example 3-53

FATIGUE & FRACTURE LIMIT STATE Load Induced Fatigue (continued) Article 6.6.1..1 -- for flexural members with shear connectors provided throughout their entire length, and with concrete deck reinforcement satisfying the provisions of Article 6.10.1.7 (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), the liveload stress range may be computed using the shortterm composite section assuming the concrete deck is effective for both positive and negative flexure. FATIGUE & FRACTURE LIMIT STATE Load Induced Fatigue (continued) Check the top-flange connection-plate weld at 7-0 from the abutment in Span 1: Fatigue need not be checked at this detail. ( 1 )( 38.63 ) 1,84 fdc - 13.49 ksi 1 6,658 81( 1 )( 3.13 ) fdc - 0.665 ksi 117,341-14.16 ksi ( ) ( )( ) 0.75-496 1 fll + IM 161,518 10.70-14.16 ksi > 0.591 ksi 0.591 ksi Design Example 3-54

FATIGUE & FRACTURE LIMIT STATE Load Induced Fatigue (continued) Check the bottom-flange connection-plate weld at 7-0 from the abutment in Span 1 -- use the n-composite stiffness to compute the stress range as permitted in Art. 6.6.1..1:? (?f) 0.75 5.96 ksi ( 1,337 )( 1 )( 58.31) 0.75-496 ( 1 )( 58.31) 161,518 + ( F) n ( DF) TH 161,518 1 A 3 1 D Eq. (6.6.1..5-1) Ł Nł Design Example 3-55

FATIGUE & FRACTURE LIMIT STATE Load Induced Fatigue (continued) (ADTT) SL 1,600 > 745 trucks per day (Category C ) DF 1 DF \ ( ) n ( ) TH For a Category C detail, (DF)TH 1.0 ksi (Table 6.6.1..5-3). Therefore: 1 ( DF) n ( 1.0) 6.00 ksi ( Df) ( DF) n g Eq. (6.6.1..-1) 5.96 ksi < 6.00 ksi ok (Ratio 0.993) FATIGUE & FRACTURE LIMIT STATE Performance Ratios POSITIVE-MOMENT REGION, SPAN 1 (Section 1-1) Fatigue and Fracture Limit State Base metal at connection plate weld to bottom flange 0.993 (7-0 from the abutment) Stud shear connector weld to top flange 0.08 (100-0 from the abutment) Special fatigue requirement for webs 0.618 (7-3 from the abutment) INTERIOR-PIER SECTION (Section -) Fatigue and Fracture Limit State Base metal at connection plate weld to top flange 0.113 (0-0 to the left of the interior pier) Special fatigue requirement for webs (shear at interior pier) 0.636 Design Example 3-56

FATIGUE & FRACTURE LIMIT STATE Special Fatigue Req. for Webs (Article 6.10.5.3) Interior panels of stiffened webs must satisfy: V u V cr Eq. (6.10.5.3-1) V u 73.0 + 11 + 10 + (0.75)(47) 165 kips at 7-3 from the abutment V n Vcr CVp Eq. (6.10.9.3.3-1) Vp 0.58F yw Dtw 1,001kips C 0.67 (for 01-inch stiffener spacing) Vcr 0.67(1,001) 67 kips > Vu 165.0 kips ok (Ratio 0.618) Strength Limit State Design Example 3-57

STRENGTH LIMIT STATE Section 1-1 (@0.4L 1 ) Section 1-1 qualifies as a compact section if (Art. 6.10.6..): F y 70 ksi ok D/t w 150 ok D cp E ok 3.76 tw Fyc N.A. of composite section at the plastic moment is in the top flange \ D cp 0 > Go to Art. 6.10.7.1.1 (Compact sections) STRENGTH LIMIT STATE Section 1-1 (continued) Compact composite sections in positive flexure must satisfy the following ductility requirement: D p 0.4D t Eq. (6.10.7.3-1) D p distance from top of deck to N.A. at plastic moment D t total depth of the composite section D p 9.0 + 3.5-1.0 + 0.44 11.94 in. D t 1.375+ 69.0 + 3.5 + 9.0 8.88 in. 0.4Dt 0.4(8.88) 34.81in. > 11.94 in. ok Design Example 3-58

STRENGTH LIMIT STATE Section 1-1 (continued) Compact sections in positive flexure must satisfy the following relationship at the strength limit state (Article 6.10.7.1.1): 1 Mu + f l Sxt f Mn Eq. (6.10.7.1.1-1) 3 where: S xt M yt /F yt STRENGTH LIMIT STATE Section 1-1 (continued) Determine lateral bending stress,f l, in the bottom flange due to the factored wind load: hgpdd W M w WL 10 b Eq. (C4.6..7.1-1) Eq. (C4.6..7.1-) h 1.0 L b 4-0 WL 0 AF 1.0 for tension flanges Design Example 3-59

STRENGTH LIMIT STATE Section 1-1 (continued) For STRENGTH I: Wind not considered For STRENGTH III: 1.0(1.4)(0.053)(1.375 + 69.0 + 1.0)/ 1 W M f w l S l 0.1(4.0) Mw 1.73 kip - ft 10 1.73(1).06 ksi 1.375 6 ( 18) < 0.6F yf * AF 30.0 ksi ok 0.1kips / ft.06(1.0).06 ksi STRENGTH LIMIT STATE Section 1-1 (continued) For STRENGTH IV: Wind not considered For STRENGTH V: 1.0(0.4)(0.053)(1.375 + 69.0 + 1.0) /1 W M f w l S l 0.063(4.0) Mw 3.63 kip - ft 10 3.63(1) 0.587 ksi 1.375 18 6 ( ) < 0.6F yf * AF 30.0 ksi ok 0.063 kips / ft 0.587(1.0) 0.587 ksi Design Example 3-60

STRENGTH LIMIT STATE Section 1-1 (continued) Calculate M u : For STRENGTH I: 1.0[ 1.5(,0 + 335) + 1.5(3) + 1.75(3,510) ] 9,797 kip ft M u - For STRENGTH III: 1.0[ 1.5(,0 + 335) + 1.5(3) ] 3,654 kip ft M u - For STRENGTH IV: 1.0[ 1.5(,0 + 335 + 3) ] 4,89 kip ft For STRENGTH V: 1.0[ 1.5(,0 + 335) + 1.5(3) + 1.35(3,510) ] 8,393 kip ft M u - M u - STRENGTH LIMIT STATE Section 1-1 (continued) Determine the nominal flexural resistance (Article 6.10.7.1.) If D p 0.1D t, then: M n Mp Eq. (6.10.7.1.-1) Dp Otherwise: M n Mp 1.07-0.7 Eq. (6.10.7.1.-) Ł Dt ł In a continuous span, M n is limited to: M n 1.3RhMy Eq. (6.10.7.1.-3) Design Example 3-61

STRENGTH LIMIT STATE Section 1-1 (continued) 0.1D 0.1(8.88) 8.9 in. < Dp t 11.94 in. Dp \ Mn Mp 1.07-0.7 Eq.(6.10.7.1. - ) Ł Dt ł Ø 11.94 ø M n 14,199 Œ 1.07-0.7 13,761kip - ft º Ł8.88łß œ ( 6.10.7.1. 3) In a continuous span : Mn 1.3R hmy Eq. - M n 1.3(1.0)(10,171) 13, kip - ft ( governs) \ M n 13, kip - ft Myt 10,171(1) S xt,441in F 50 yt 3 For STRENGTH I: 1 Mu + flsxt ffmn Eq. (6.10.7.1.1-1) 3 1 Mu + flsxt 9,797 kip - ft + 0 9,797 kip - ft 3 φ M 1.0(13,) 13, kip - ft f n 9,797 kip - ft < 13, kip - ft 3,794 kip - ft < 13, kip - ft (.06) ok ok (Ratio 0.741) 1 1 (,441) Mu + flsxt 3,654 kip - ft + 3,794 kip - ft 3 3 1 φ M 1.0(13,) 13, kip - ft f STRENGTH LIMIT STATE Section 1-1 (continued) For STRENGTH III: n (Ratio 0.87) Design Example 3-6

For STRENGTH IV: For STRENGTH V: 1 Mu + flsxt ffmn Eq. (6.10.7.1.1-1) 3 1 Mu + flsxt 4,89 kip - ft + 0 4,89 kip - ft 3 φ M 1.0(13,) 13, kip - ft f n 4,89 kip - ft < 13, kip - ft 8,433 kip - ft < 13, kip - ft ok ( 0.587) 1 1 Mu + flsxt 8,393 kip - ft + 3 3 φ M 1.0(13,) 13, kip - ft f STRENGTH LIMIT STATE Section 1-1 (continued) n ok (Ratio 0.34) (,441) 8,433 kip - ft 1 (Ratio 0.638) STRENGTH LIMIT STATE Performance Ratios POSITIVE-MOMENT REGION, SPAN 1 (Section 1-1) Strength Limit State (Compact Section) Ductility requirement 0.343 Flexure Eq. (6.10.7.1.1-1) (STRENGTH I) 0.741 Flexure Eq. (6.10.7.1.1-1) (STRENGTH III) 0.87 Flexure Eq. (6.10.7.1.1-1) (STRENGTH IV) 0.34 Flexure Eq. (6.10.7.1.1-1) (STRENGTH V) 0.638 Shear (End panel) (STRENGTH I) 0.995 Design Example 3-63

SHEAR (Article 6.10.9) Highlights: Tension-field action extended to interior panels of hybrid sections. Moment-shear interaction equation is eliminated. No handling requirement since D/t w is limited to 150 for transversely stiffened girders. Shear connector design moved to separate Article 6.10.10. SHEAR End Panels (Article 6.10.9.3.3) End panels must satisfy: V u fvvn Eq. (6.10.9.1-1) [ + 13) + 1.5(13) + 1.75(139) ] V u 1.0 1.5(87 388 kips n Vcr CVp Eq. (6.10.9.3.3-1) V int. girder at the abutment Vp 0.58 F Dt w yw 1,001 kips C 0.390 (for 87-inch stiffener spacing < 1.5D 103.5 in.) Vcr 0.390(1,001) 390 kips > Vu 388 kips (Ratio 0.995) Design Example 3-64

SHEAR Interior Panels (Article 6.10.9.3.) For interior panels of hybrid & nonhybrid members with the section along the entire panel proportioned such that: Dt ( b t + b t ) fc fc w ft.5 Eq. (6.10.9.3.-1) the nominal shear resistance is to be taken as: V n V p ft Ø ø Œ œ Œ 0.87(1- C) C + œ Œ d œ Œ o 1+ œ Œº Ł D ł œß Eq. (6.10.9.3.-) SHEAR Interior Panels (continued) For interior panels of hybrid and nonhybrid members with the panel proportioned such that: ( b t + b t ) fc Dt fc w > Eq. (6.10.9.3.-1) the nominal shear resistance is to be taken as: ft ft.5 V n V p Ø Œ Œ ŒC + Œ Œ º Ł 0.87(1- C) do 1+ Ł D ł do + D ø œ œ œ œ œ łß Eq. (6.10.9.3.-8) Design Example 3-65

SHEAR Interior Panels (continued) d o 16.75 ft 01.0 in. < 3D 07.0 in. For an unstiffened web: [ + 11) + 1.5(11) + 1.75(17) ] V u 1.0 1.5(74 345 kips int. girder at 87 in. from abutment V V CV n cr C 0.39 (calculated w/ k 5) p Eq. (6.10.9.-1) V p 0.58(50)(69.0)(0.5) 1,001kips Vn Vcr 0.39(1,001) 39 kips f v V n 1.0(39) 39 kips < 345 kips\ stiffeners are required Design Example 3-66

SHEAR Interior Panels (continued) Calculate V n for the stiffened panel: (69.0)(0.5) [ 16(1.0) + 18(0.875) ].17 <.5 5 k 5 + 5.59 01.0 Ł 69.0 ł 1.40 Ek F yw 1.40 9,000(5.59) 50 79.7 < D t w 69.0 138.0 0.5 SHEAR Interior Panels (continued) Calculate V n for the stiffened panel (cont d): C Ł D t 1.57 w ł Ł V Ek F yw ł 1.57 9,000(5.59) C 0.67 Ł 50 ł > ( 138.0) V p 0.58(50)(69.0)(0.5) 1,001kips Ø ø Œ œ 0.87(1 0.67) 1,001Œ - 0.67 + œ Œ 01.0 œ Œ 1+ œ º Œ Ł 69.0 ł œß n 475 kips fv Vn 1.0(475) 475 kips > Vu 345 kips ok Design Example 3-67

STRENGTH LIMIT STATE Section - (Interior Pier) Determine if the section is a slender-web section (Article 6.10.6..3): Dc E 5.7 t F 9,000 5.7 116.0 70 (36.55) 130.0 > 116.0 0.565 w Eq. (6.10.6..3-1) Note: Use D c of the steel section + long. reinforcement (D6.3.1) Therefore, the section is a slender-web section. > Go to Article 6.10.8 yc STRENGTH LIMIT STATE Section - (continued) Use the provisions of Article 6.10.8. For discretely braced compression flanges: f bu + l 1 f fff 3 nc Eq. (6.10.8.1.1-1) For continuously braced flanges: f bu f R F f h yf Eq. (6.10.8.1.3-1) Design Example 3-68

STRENGTH LIMIT STATE Section - (continued) Top flange: Calculate the maximum flange major-axis flexural stresses at Section - due to the factored loads under the STRENGTH I load combination: Ø1.5( -4,840) 1.5( -690) 1.5( -664) 1.75( -4,040) ø f 1.0Œ + + + 1 54.57 ksi,94 3,194 3,194 3,703 œ º ß Bottom flange: Ø1.5( -4,840) 1.5( -690) 1.5( -664) 1.75( -4,040) ø f 1.0Œ + + + 1-55.64 ksi 3,149 3,08 3,08 3,310 œ º ß STRENGTH LIMIT STATE Section - (continued) Calculate the hybrid factor, R h (Article 6.10.1.10.1): R h 3 ( 3r -r ) 1+b 1+ b D t b A n w fn Eq. (6.10.1.10.1-1) Eq. (6.10.1.10.1-) F r f yw n 1.0 Design Example 3-69

STRENGTH LIMIT STATE Section - (continued) D ntw Calculate > A (36.55)(0.565) b b 1. 08 40.0 fn D n larger of the distances from the elastic N.A. of the cross-section to the inside face of either flange At Section -: D n D c 36.55 in. (steel + reinf.) A fn sum of the flange area & the area of any cover plates on the side of the neutral axis corresponding to D n. For the top flange, can include reinforcement. At Section -: A fn A bot flg. 0() 40.0 in STRENGTH LIMIT STATE Section - (continued) F Calculate yw 1.0 > f 50.0 r r 0. 714 70.0 n f n for sections where yielding occurs first in the flange, a cover plate or the longitudinal reinforcement on the side of the N.A. corresponding to D n, the largest of the specified minimum yield strengths of each component included in the calculation of A fn. At Section -: f n F yc 70.0 ksi Otherwise, f n is equal to the largest elastic stress in the component on the side of the N.A. corresponding to D n at first yield on the opposite side of the N.A. Design Example 3-70

STRENGTH LIMIT STATE Section - (continued) Calculate R h : R h 3 ( 3r -r ) 1+b 1+ b R 3 [ - (0.714) ] 0. 984 1 + 1.08 3(0.714) 1 + (1.08) h STRENGTH LIMIT STATE Section - (continued) R Calculate web load-shedding factor, R b (Article 6.10.1.10.): D t a wc 1- Ł 100 + 300a E F c 130.0 > lrw 5.7 w yc D łł t w c b - lrw wc fc fc ł 1.0 116.0 D ctw (36.55)(0.565) a wc 1. 08 b t 0() Eq. (6.10.1.10.-3) 1.08 R b 1- (130.0-116.0) 0.990 Ł 100 + 300(1.08) ł Design Example 3-71

LOCAL BUCKLING RESISTANCE Bottom Flange - Section - (Article 6.10.8..) Determine the slenderness ratio of the bottom flange: bfc 0 lf 5.0 t fc Since l f < l pf : F R R F Eq. (6.10.8..-1) ( ) E 9,000 lpf 0.38 0.38 7.73 F 70 yc nc b h ( 0.990 )(0.984)(70.0) 68.19ksi Fnc FLB yc 6.10.8 Flexural Resistance - Composite Sections in Negative Flexure & Noncomposite Sections Basic Form of All FLB & LTB Eqs F n or M n F max or M M max max F RR F nc b h yc F RR F nc b h yc Fyr λf λpf Fnc 1 1 RbR hfyc Anchor point 1 R F h yc λrf λpf Fyr Lb Lp Fnc Cb 1 1 RbRhFyc RbRhFyc RhF yc Lr L p Anchor point F r or M r r compact noncompact (inelastic buckling) nonslender slender F F RRF nc cr b h yc b bπ Lb C R r t E (elastic buckling) L p or λλ p pf L r λor r λ rf L b or b fc /t fc Design Example 3-7

LATERAL TORSIONAL BUCKLING RESISTANCE (Article 6.10.8..3) From Article 6.10.8..3: For unbraced lengths containing a transition to a smaller section at a distance less than or equal to 0 percent of the unbraced length from the brace point with the smaller moment, the lateral torsional buckling resistance may be determined assuming the transition to the smaller section does not exist. Assume L b 17-0 ( 17.0-15.0) 17.0 1% < 0% Design Example 3-73

LATERAL TORSIONAL BUCKLING RESISTANCE (continued) Determine the limiting unbraced length, L p : L 1.0r p Therefore: yc Eq. (6.10.8..3-4) t E F where: 0 r t 5.33 in. 1 36.55(0.565) 1 1 Ł + 3 0() ł 1.0(5.33) 9,000 L p 9.04 ft 1 70 r t b fc Ł + 1 Dct 1 1 3 bfct w fc ł LATERAL TORSIONAL BUCKLING RESISTANCE (continued) Determine the limiting unbraced length, L r : L pr r t E F yr Eq. (6.10.8..3-5) where: F 0.7F F yr yc yw F yr 0.7(70) 49.0 ksi < 50 ksi ( 0.5F yc 35 ksi ok) Therefore: p(5.33) 9,000 L r 33.95 ft 1 49.0 Design Example 3-74

LATERAL TORSIONAL BUCKLING RESISTANCE (continued) Determine the moment gradient modifier, C b : For unbraced cantilevers & members where f mid /f > 1.0 or f 0: C b 1.0 Eq. (6.10.8..3-6) Otherwise: f f C 1.75-1.05 1 + 0.3 1 Ł f ł Ł f ł b.3 Eq. (6.10.8..3-7) f 1 fmid -f fo Eq. (6.10.8..3-10) Examples: f 1 fmid -f fo f /f 0.875 mid f f f mid 1 f0 f 1 /f 0.75 L b C b 1.13 When f mid < (f o +f )/, f 1 f o f 1 f 0 f0 f mid f f 1 /f 0.375 C b 1.40 f 0 f mid f C b 1 f mid > f f 0 < 0 f mid f 0 C b 1 f mid /f 0.75 f mid f f 1 /f 0.5 f 1 f 0 0 C b 1.3 Design Example 3-75

LATERAL TORSIONAL BUCKLING RESISTANCE (continued) Since the moment diagram in this region is concave in shape, f 1 f o (therefore, do not need to compute f mid ). For STRENGTH I: Bottom flange: Ø1.5( -,390) 1.5( -334) 1.5( -31) 1.75( -,615) ø f1 fo 1.0Œ + + + 1 31.34 ksi 3,149 3,08 3,08 3,310 œ º ß 31.34 31.34 Cb 1.75-1.05 + 0.3 1.5 <.3 Ł 55.64 ł Ł 55.64 ł Note: f o is positive for compression. f is always positive. LATERAL TORSIONAL BUCKLING RESISTANCE (continued) F nc Since L p 9.04 ft < L b 17.0 ft < L r 33.95 ft: Ø F - L yr b p F nc Cb Œ1-1- RbRhFyc Ł RhFyc łł Lr -Lp ł º Ø 1.5Œ1 1 º - Ł - L œ ß ø R b R h F yc Eq. (6.10.8..3-) 49.0 17.0-9.04 ø ( 0.990) (0.984)(70) 77.38 ksi 0.984(70) 33.95 9.04 œ łł - łß > 0.990(0.984)(70) 68.19 ksi Therefore: F ncltb 68.19 ksi ( F ncflb 68.19 ksi) \ F nc 68.19 ksi Design Example 3-76

6.10.8 Flexural Resistance - Composite Sections in Negative Flexure & Noncomposite Sections Basic Form of All FLB & LTB Eqs F n or M n F max or M M max max F RR F nc b h yc F RR F nc b h yc Fyr λf λpf Fnc 1 1 RbR hfyc Anchor point 1 R F h yc λrf λpf Fyr Lb Lp Fnc Cb 1 1 RbRhFyc RbRhFyc RhF yc Lr L p Anchor point F r or M r r compact noncompact (inelastic buckling) nonslender slender F F RRF nc cr b h yc b bπ Lb C R r t E (elastic buckling) L p or λλ p pf L r λor r λ rf L b or b fc /t fc LATERAL TORSIONAL BUCKLING RESISTANCE (continued) ( 0.0-15.0) Assume L b 0-0 0.0 From Article C6.10.8..3: For unbraced lengths containing a transition to a smaller section at a distance greater than 0 percent of the unbraced length from the brace point with the smaller moment, the lateral torsional buckling resistance should be taken as the smallest resistance, F nc, within the unbraced length under consideration. This resistance is to be compared to the largest value of the compressive stress, f bu, throughout the unbraced length calculated using the actual properties of the section. The moment gradient modifier, C b, should be taken equal to 1.0 in this case... 5% > 0% Design Example 3-77

LATERAL TORSIONAL BUCKLING RESISTANCE (continued) For the smaller section at the flange transition in the first unbraced length adjacent to the interior pier: r t 4.95 in. L p 8.40 ft L r 31.53 ft R h 0.971 R b 0.977 Set C b 1.0. Design Example 3-78

LATERAL TORSIONAL BUCKLING RESISTANCE (continued) F nc Since L p 8.40 ft < L b 0.0 ft < L r 31.53 ft: Ø F - L yr b p F nc Cb Œ1-1- RbRhFyc Ł RhFyc łł Lr -Lp º Ø 1.0Œ1 1 º - Ł - L ø œ ł ß R b R h F yc Eq. (6.10.8..3-) 49.0 0.0-8.40 ø ( 0.977) (0.971)(70) 57.11ksi 0.971(70) 31.53 8.40 œ łł - łß < 0.977(0.971)(70) 66.41ksi Therefore: F ncltb 57.11 ksi LATERAL TORSIONAL BUCKLING RESISTANCE (continued) For L b 17-0 : Flange transition < 0% of unbraced length from the brace point with the smaller moment For L b 0-0 : F ncltb 68.19 ksi Flange transition > 0% of unbraced length from the brace point with the smaller moment F ncltb 57.11 ksi From Appendix C to the Design Example: F ncltb- 61.14 ksi F ncltbtr 60.48 ksi Design Example 3-79

LOCAL BUCKLING RESISTANCE Bottom-Flange Transition (Article 6.10.8..) Determine the slenderness ratio of the bottom flange: bfc 0 E l f 10.0 > lpf 0.38 7.73 t ( 1) F fc yr l rf 0.56 yc E F yr F 0.7F F yw yc Eq. (6.10.8..-5) F yr 0.7(70) 49.0 ksi < 50 ksi ( 0.5F yc 35 ksi ok) Therefore: 9,000 lrf 0.56 13.6 49.0 LOCAL BUCKLING RESISTANCE Bottom-Flange Transition (continued) Since l pf 7.73 < l f 10.0 < l rf 13.6: F nc Ø Fyr Œ1-1- Œº Ł RhF yc l łł l f rf - l - l pf pf ø œrbrhf łœß Eq. (6.10.8..-) 49.0 10.0 7.73 F Ø ø nc Œ1 1 (0.977)(0.971)(70.0) 59.6 ksi 0.971(70.0) 13.6 7.73 œ º - - Ł - łł - łß yc Therefore: F ncflb 59.6 ksi (> F ncltb 57.11 ksi) \ F nc F ncltb 57.11 ksi Design Example 3-80

6.10.8 Flexural Resistance - Composite Sections in Negative Flexure & Noncomposite Sections Basic Form of All FLB & LTB Eqs F n or M n F max or M M max max F RR F nc b h yc F RR F nc b h yc Fyr λf λpf Fnc 1 1 RbR hfyc Anchor point 1 R F h yc λrf λpf Fyr Lb Lp Fnc Cb 1 1 RbRhFyc RbRhFyc RhF yc Lr L p Anchor point F r or M r r compact noncompact (inelastic buckling) nonslender slender F F RRF nc cr b h yc b bπ Lb C R r t E (elastic buckling) L p or λλ p pf L r λor r λ rf L b or b fc /t fc STRENGTH LIMIT STATE Section - - (continued) For STRENGTH I: Section -: Top Flange: f bu 54.57 ksi Bottom Flange: f -55.64 ksi Flange Transition (Span 1): Top Flange: Ø1.5( -,656 ) 1.5 (-373 ) 1.5( -358 ) 1.75( -,709 ) ø f bu 1.0 Œ + + + 1 5.86 ksi 1,700 1,945 1,945,448 œ º ß Bottom Flange: Ø1.5( -,656 ) 1.5( -373 ) 1.5( -358 ) 1.75( -,709 ) ø f 1.0Œ + + + 1-57.54 ksi 1,789 1,86 1,86 1,975 œ º ß \ f bu (bot. flange) -57.54 ksi Design Example 3-81

STRENGTH LIMIT STATE Section - - Bottom Flange For STRENGTH I: f 1 3 bu + fl f F f nc Eq. (6.10.8.1.1-1) f bu + l f F f nc 1 f - 57.54ksi+ 0 57.54ksi 3 1.0(57.11) 57.11ksi 57.54 ksi > 57.11ksi sayok (Ratio 1.008) STRENGTH LIMIT STATE Section - - Top Flange For STRENGTH I: f f R bu f h F yf Eq. (6.10.8.1.3-1) Section -: f bu 54.57 ksi f R F f h yf 1.0(0.984)(70.00) 68.88 ksi 54.57 ksi < 68.88 ksi ok (Ratio 0.79) Flange transition: f bu 5.86 ksi f R F f h yf 1.0(0.971)(70.00) 67.97 ksi 5.86 ksi < 67.97 ksi ok (Ratio 0.778) Design Example 3-8

STRENGTH LIMIT STATE Performance Ratios (continued) INTERIOR-PIER SECTION (Section -) Strength Limit State (Slender-web section) Flexure (STRENGTH I) Eq. (6.10.8.1.1-1) Bottom flange 1.008 Eq. (6.10.8.1.3-1) Top flange @ Section - 0.79 Eq. (6.10.8.1.3-1) Top flange @ Flange transition 0.778 Flexure (STRENGTH III) Eq. (6.10.8.1.1-1) Bottom flange 0.536 Eq. (6.10.8.1.3-1) Top flange @ Section - 0.460 Eq. (6.10.8.1.3-1) Top flange @ Flange transition 0.445 Flexure (STRENGTH IV) Eq. (6.10.8.1.1-1) Bottom flange 0.617 Eq. (6.10.8.1.3-1) Top flange @ Section - 0.541 Eq. (6.10.8.1.3-1) Top flange @ Flange transition 0.54 Flexure (STRENGTH V) Eq. (6.10.8.1.1-1) Bottom flange 0.896 Eq. (6.10.8.1.3-1) Top flange @ Section - 0.716 Eq. (6.10.8.1.3-1) Top flange @ Flange transition 0.700 OTHER TOPICS COVERED Spacing of Transverse Stiffeners Shear Connector Design Transverse Intermediate Stiffener Design Bearing Stiffener Design Appendix A - Elastic Effective Length Factor for Lateral Torsional Bucking Appendix B - Moment Gradient Modifier, C b (Example cases) Appendix C - Lateral Torsional Buckling Resistance of Stepped Flanges Design Example 3-83

QUESTIONS Michael A. Grubb, P.E. BSDI, Ltd. 74-709-8349 m.grubb@comcast.net Design Example 3-84