Thiết kế dầm thép theo TCVN 11823 2017 Thiết kế dựa trên phần mềm MIDAS và 1 số tài liệu tham khảo kèm theo.
MINISTRY OF EDUCATION AND TRAINING UNIVERSITY OF TRANSPORT AND COMMUNICATIONS PARTIAL PROJECT OF STEEL BRIDGE DESIGN Student : ID : Class : Nguyen Thien Long Lecturer : Hanoi, 6/2021 Steel Bridge Design Nguyen Thien Long - 172603217 Contents CHAPTER 1: INTRODUCTION I INPUT DATA II DESIGNED DATA Designed method and loading Materials 2.1 Concrete 2.2 Girder steel 2.3 Designed cross section CHAPTER 2: INTERNAL FORCE CALCULATION AND COMBINATION FOR MAIN GIRDER 10 I GEOMETRICAL CHARACTERISTICS 10 1.1 Calculation stages 10 1.2 Calculate effective flange width 10 1.2.1 Interior beam 10 1.2.2 Exterior beam 10 1.3 Geometrical characteristic of stage 10 1.4 Transverse distribution coefficient calculation 12 1.4.1 Longitudinal stiffness coefficient calculation 12 1.4.2 Calculate moment distribution coefficient 12 1.4.3 Calculate shear force distribution coefficient 14 1.4.4 Calculate pedestrian path distribution coefficient 14 1.4.5 Calculate distribution coefficient in fatigue limit state 15 1.4.6 Calculate moment distribution coefficient 15 1.4.7 Calculate shear force distribution coefficient 15 1.4.8 Calculate defection distribution coefficient 16 1.4.9 Table distribution coefficient 16 1.5 Geometrical characteristic of stage 16 II 1.5.1 Short term loading(n) 16 1.5.2 Long term loading(3n) 17 LOADING 18 Steel Bridge Design Nguyen Thien Long - 172603217 2.1 Dead load 18 2.1.1 DC 19 2.1.2 DW 19 2.1.3 Total dead load 19 2.1.4 Dead load in limit state 20 2.2 Live load 21 2.2.1 Loading combination and coefficient 22 2.2.2 Strength limit state I 23 2.2.3 Service limit state II 23 2.2.4 Fatigue limit state II 24 CHAPTER 3: CALCULATED FOR DESIGN GIRDER 25 I CHARACTERISTICS PROPRERTIES OF CROSS SECTION 25 1.1 Plastic moment 25 1.1.1 Determine plastic force 25 1.1.2 Determine plastic neutral axis(PNA) 25 1.1.3 Determine plastic moment 26 1.2 Yielding moment 26 II 1.2.1 Determine for components moment 26 1.2.2 Determine yielding moment 27 CHECKING IN LIMIT STATE I 27 2.1 Bending moment 27 III 2.1.1 Checking the plastic properties 27 2.1.2 Checking the ability to apply Part 6, appendix A 27 2.1.3 Checking bending resistance for composite cross section 27 2.1.4 Checking for shearing resistance 28 CHECKING IN SERVICE LIMIT STATE II 29 3.1 Elastic deformation 29 3.2 Unrestorable deformation related to bending 29 IV FATIGUE LIMIT STATE AND CRACKING 29 4.1 Fatigue related to loading 30 Steel Bridge Design Nguyen Thien Long - 172603217 4.2 Fatigue stress calculation 30 V 4.2.1 Stress in bottom edge of top flange 30 4.2.2 Stress in top edge of bottom flange 30 ULTIMATE FATIGUE REQUIREMENT FOR WEB 31 CHAPTER 4: DESIGN FOR STIFFENERS AND SHEAR CONNECTORS 32 I SHEAR CONNECTORS 32 1.1 Composition 32 1.2 Fatigue Limit State(FLS) 33 1.3 Strength limit state 34 Steel Bridge Design Nguyen Thien Long - 172603217 CHAPTER 1: INTRODUCTION I II - INPUT DATA Length of span, L: 36(m) Carriageway width, Bc: 7(m) Pedestrian path width, Bp: 1.5(m) Design load: HL – 93, PL 3.10-3 (MPa) Superstructure: Composite steel I girder bridge Type of connection: Bolted Connections Concrete type: 30(MPa) Steel type: G40, G60 or the same Other data: DESIGNED DATA Designed method and loading This project was designed based on specification TCVN 11823 – 2017 and some reference books - Design load: HL – 93; PL 3.10-3 MPa - Span structure: x 36(m) - Bridge width: 11(m) Materials 2.1 Concrete According to C3.5.1.B6: - Specific weight of wearing coat: 𝛾𝐷𝑊 = 2250(𝑘𝑔⁄𝑚3 ) = 2250 × 9.81 × 10−3 = 22.0725(𝑘𝑁⁄𝑚3 ) - Specific weight of bridge slab’s concrete: 𝛾𝐶 = 2450(𝑘𝑔⁄𝑚3 ) = 2450 × 9.81 × 10−3 = 24.0345(𝑘𝑁⁄𝑚3 ) - Compressive strength of concrete at 28 days: 𝑓𝑐′ = 30(𝑀𝑃𝑎) - Elastic modulus of slab concrete( C5.4.2.4.E8): 0.33 𝐸𝑐 = 0.0017 × 𝐾1 × 𝑊𝑐2 × 𝑓 ′ 𝑐 𝐸𝑐 = 0.0017 × × 24502 × 300.33 = 31349.535(𝑀𝑃𝑎) 2.2 Girder steel - Cross section: I girder - Specification: ASTM A709M, Grade 50(345MPa) - Specific weight of steel(C3.5.1): 𝛾𝑆 = 7850(𝑘𝑔⁄𝑚3 ) = 77.0085(𝑘𝑁⁄𝑚3 ) - Elastic modulus of steel(C5.4.3.2): 𝐸𝑆 = 200000(𝑀𝑃𝑎) - Minimum tensile strength of steel(C6.4.1): 𝑓𝑢 = 450(𝑀𝑃𝑎) Steel Bridge Design Nguyen Thien Long - 172603217 - Minimum plastic strength of steel(C6.4.1): 𝐹𝑦 = 𝐹𝑦𝑏 = 𝐹𝑦𝑤 = 𝐹𝑦𝑡 = 345(𝑀𝑃𝑎) Where: Fyt, Fyw, Fyc is minimum tensile strength of tensile flange, web and compressive flange, respectively - Convertible coefficient for elastic modulus, approximately(C6.10.1.1.1.2): +, For short – term loading: n = 8, +, For long – term loading: n = 24 2.3 Designed cross section 2.3.1 Deck slab - Bases on C9.7.1.1, minimum thickness of deck slab is 175(mm) - According to C2.5.6.3.B2, minimum thickness of deck slab is: 1.2 × (𝑆 + 3000) 1.2 × (3000 + 3000) 𝑡𝑆 ≥ = = 240(𝑚𝑚) 30 30 Where: +, S: Spacing between main girders, S = 3000(mm) - So, choose thickness of deck slab ts = 250(mm) 2.3.2 Girder cross section Because the girder was bended on the whole length by positive moment, so choose one of cross section for all location Note Name Dimension Width(mm) bv Top haunch width bt Top flange width 360 tw Web thickness 15 bb Bottom flange width 420 Height(mm) tv Haunch thickness 50 tt Top flange thickness 20 D = hw Web height 1550 tb Bottom flange thickness 30 h1 Total steel girder height 1600 Total concrete slab h2 300 height Steel Bridge Design Nguyen Thien Long - 172603217 Figure I steel cross section 2.3.3 Wearing coat - Resistance water layer: 5(mm) thickness - Asphalt concrete pavement layer: 70(mm) thickness - Total wearing coat thickness: 75(mm) 2.3.4 Parapet Parapet’s dimensions would be indicated on the figure, total area of parapet: Apar = 387500 mm2 Figure Parapet cross section Steel Bridge Design Nguyen Thien Long - 172603217 2.3.5 Span cross section Span cross section include main girders, dimensions as the table: Note Name Dimension Unit S Girders spacing m Se Overhang m Total width of W 11 m bridge Carriageway Wcar m width Pedestrian path Wpes 1.5 m width Wpar Parapet width 0.5 m Figure Span cross section 2.3.6 Checking limit dimension for main girder 2.3.6.1 Web height - Based on C2.5.2.6.3.B2, minimum height for I girder is 0.033L (Simple beam) - So, 𝐿𝑚𝑖𝑛 = 0.033 × 36 = 1.188(𝑚) = 1188(𝑚𝑚) - Check: ℎ𝑤 = 1550(𝑚𝑚) > 𝐿𝑚𝑖𝑛 = 1188(𝑚𝑚) → TRUE 2.3.6.2 Web thickness 𝐷 - Based on C6.10.2.1.1 without using longitudinal stiffeners: ≤ 150(𝑚𝑚) 𝑡𝑤 - Check: 𝐷 𝑡𝑤 = 1550 15 = 103.333(𝑚𝑚) < 150(𝑚𝑚) → TRUE 2.3.6.3 Flange dimensions - Based on C6.10.2.2, conditions should be checked - C6.10.2.2.E95: 𝑏𝑓 2×𝑡𝑓 ≤ 12 Steel Bridge Design Nguyen Thien Long - 172603217 +, For top flange, 𝑏𝑓 = 2×𝑡𝑓 𝑏𝑓 +, For bottom flanger, - C6.10.2.2.E96: 𝑏𝑓 ≥ 360 2×20 2×𝑡𝑓 = = ≤ 12 → TRUE 420 2×30 = ≤ 12 → TRUE 𝐷 +, For top flange, 𝑏𝑓 ≥ 𝐷 ↔ 360 ≥ +, For bottom flanger, 𝑏𝑓 ≥ 𝐷 1550 ↔ 420 ≥ = 258.33 → TRUE 1550 = 258.33 → TRUE - C6.10.2.2.E97: 𝑡𝑓 ≥ 1.1 × 𝑡𝑤 +, For top flange, 𝑡𝑓 ≥ 1.1 × 𝑡𝑤 ↔ 20 ≥ 1.1 × 15 = 16.5 → TRUE +, For bottom flanger, 𝑡𝑓 ≥ 1.1 × 𝑡𝑤 ↔ 30 ≥ 1.1 × 15 = 16.5 → TRUE - C6.10.2.2.E98: 0.1 ≤ - Check: 0.1 ≤ 𝐼𝑦𝑐 𝐼𝑦𝑡 𝐼𝑦𝑐 𝐼𝑦𝑡 ≤ 10 ≤ 10 ↔ 0.1 ≤ 240000 945000 = 0.254 ≤ 10 → TRUE Steel Bridge Design Nguyen Thien Long - 172603217 CHAPTER 2: INTERNAL FORCE CALCULATION AND COMBINATION FOR MAIN GIRDER I GEOMETRICAL CHARACTERISTICS 1.1 Calculation stages - Stage 1: Non – composite steel girder cross section - Stage 2: +, Composite girder cross section under short – term loading(Modulus coefficient n) +, Composite girder cross section under long – term loading(Modulus coefficient 3n) Calculate geometrical characteristic for stage in this chapter About stage 2, calculate after determined transverse distribution coefficient 1.2 Calculate effective flange width Effective span length: 𝐿𝑒𝑓𝑓 = 35(𝑚) 1.2.1 Interior beam 1 effective span length: 𝐵𝑒1 = × 35 = 8.75(𝑚) = 8750(𝑚𝑚) - 12 times of deck slab thickness plus maximum of (Web thickness; top flange width): 𝐵𝑒2 = 12 × 300 + max (15; × 360) = 3780(𝑚𝑚) Spacing of girders: 𝐵𝑒3 = 𝑆 = 3000(𝑚𝑚) Takes the minimum value to set for effective flange width: 𝐵𝑒𝑖 = 𝐵𝑒3 = 3000(𝑚𝑚) 1.2.2 Exterior beam 1 effective span length: 𝐵𝑒1 = × 35 = 4.375(𝑚) = 4375(𝑚𝑚) - - 1 times of deck slab thickness plus maximum of ( Web thickness; top flange 1 width): 𝐵𝑒2 = × 300 + max ( × 15; × 360) = 1890(𝑚𝑚) Overhang: 𝐵𝑒3 = 𝑆𝑒 = 2500(𝑚𝑚) Takes the minimum value to set for effective flange width: 𝐵𝑒𝑒 = 𝐵𝑒3 = 2500(𝑚𝑚) 1.3 Geometrical characteristic of stage Set up coordination system: - X axis: Coincide with neutral axis of web - Y axis: Symmetrical axis - 10 Steel Bridge Design Cross section 10 11 Nguyen Thien Long - 172603217 DC1 Shear force -430.4 -344.32 -258.24 -172.16 -86.08 86.08 172.16 258.24 344.32 430.4 Moment 1355.74 2410.21 3163.4 3615.32 3765.96 3615.32 3163.4 2410.21 1355.74 DC2+DW Shear force Moment -157.54 -126.03 496.25 -94.52 882.22 -63.02 1157.91 -31.51 1323.32 1378.46 31.51 1323.32 63.02 1157.91 94.52 882.22 126.03 496.25 157.54 2.2 Live load - Using Midas Civil 2019 to calculate the internal force of main girder +, Based on HL – 93 loading combination, put it on moment/shear force influence line and calculate +, HL – 93 loading combination: • Designed truck (35kN – 145kN – 145kN) • Designed tandem (110kN – 110kN) • Designed lane (9.3kN/m) - Figure HL – 93 loading combination Moment influence lines of some characteristic cross section: +, Shear force: • Middle of span: 21 Steel Bridge Design ã ã 3ì Nguyen Thien Long - 172603217 of span: of span: +, Moment: ã Middle of span: ã ã 3ì of span: of span: 2.2.1 Loading combination and coefficient 22 Steel Bridge Design Nguyen Thien Long - 172603217 Loading DC DW combination LL according to Maximum Minimum Maximum Minimum limit state Strength Limit 1.25 0.9 1.5 0.65 1.75 State I Service Limit 1 1.3 State II Fatigue Limit 0 0 0.75 State II 2.2.2 Strength limit state I 2.2.2.1 Combination loading: HL – 93K Shear Force Moment Cross section Maximum Minimum Maximum Minimum -753.93 -1720.13 0 -572.46 -1430.23 5401.44 2374.88 -361.92 -1148.69 9564.86 4222 -133.49 -875.52 12490.27 5541.38 106.21 -610.72 14234.2 6333 354.28 -354.28 14768.38 6596.88 610.72 -106.21 14234.2 6333 875.52 133.49 12490.27 5541.38 1148.69 361.92 9564.86 4222 10 1430.23 572.46 5401.44 2374.88 11 1720.13 753.93 0 2.2.2.2 Combination loading: HL – 93M Shear Force Moment Cross section Maximum Minimum Maximum Minimum -753.93 -1579.17 0 -564.76 -1307.49 4971.86 2374.88 -360.01 -1044.18 8833.26 4222 -146.9 -789.23 11584.18 5541.38 74.59 -542.65 13224.64 6333 304.43 -304.43 13754.63 6596.88 542.65 -74.59 13224.64 6333 789.23 146.9 11584.18 5541.38 1044.18 360.01 8833.26 4222 10 1307.49 564.76 4971.86 2374.88 11 1579.17 753.93 0 2.2.3 Service limit state II 2.2.3.1 Combination loading: HL – 93K 23 Steel Bridge Design Shear Force Cross section Maximum Minimum -587.93 -1305.68 -447.55 -1084.75 -285.58 -870.04 -110.31 -661.54 73.33 -459.25 263.18 -263.18 459.25 -73.33 661.54 110.31 870.04 285.58 10 1084.75 447.55 11 1305.68 587.93 2.2.3.2 Combination loading: HL – 93M Shear Force Cross section Maximum Minimum -587.93 -1200.97 -441.83 -993.58 -284.16 -792.4 -120.27 -597.43 49.83 -408.68 226.15 -226.15 408.68 -49.83 597.43 120.27 792.4 284.16 10 993.58 441.83 11 1200.97 587.93 2.2.4 Fatigue limit state II Shear Force Cross section Maximum Minimum -277.4 10.28 -240.57 29.59 -205.38 53.94 -171.82 80.96 -139.9 109.61 -109.61 139.9 -80.96 171.82 -53.94 205.38 -29.59 10 240.57 -10.28 11 277.4 Nguyen Thien Long - 172603217 Moment Maximum Minimum 0 4100.3 1851.99 7261.41 3292.43 9483.34 4321.31 10808.1 4938.64 11214.68 5144.42 10808.1 4938.64 9483.34 4321.31 7261.41 3292.43 4100.3 1851.99 0 Moment Maximum Minimum 0 3781.18 1851.99 6717.93 3292.43 8810.25 4321.31 10058.14 4938.64 10461.6 5144.42 10058.14 4938.64 8810.25 4321.31 6717.93 3292.43 3781.18 1851.99 0 Moment Maximum Minimum 0 867.74 1529.18 1984.33 2253.38 2326.23 2253.38 1984.33 1529.18 867.74 0 24 Steel Bridge Design Nguyen Thien Long - 172603217 CHAPTER 3: CALCULATED FOR DESIGN GIRDER I 1.1 CHARACTERISTICS PROPRERTIES OF CROSS SECTION Plastic moment Plastic moment, Mp, calculated by moment according plastic force rotate along plastic axis Plastic force in steel part of cross section have to calculate with limited plastic strength, respectively With compressive concrete, calculate plastic force according to rectangular stress diagram, 𝑃 = 0.85 × 𝑓𝑐′ = 0.85 × 30 = 25.5(𝑀𝑃𝑎) Based on C6.D1, ignore tensile concrete Ignore steel in slab deck when determine plastic moment if consider more safety according to C6.D1: 1.1.1 Determine plastic force Using C6.D1.BD1: 1.1.1.1 Deck slab 𝑃𝑠 = 0.85 × 𝑓𝑐′ × 𝑡𝑠 × 𝑏𝑒 = 0.85 × 30 × 250 × 2500 × 10−3 = 15937.5(𝑘𝑁) 1.1.1.2 Compressive flange area 𝑃𝑐 = 𝐹𝑦𝑡 × 𝑏𝑡 × 𝑡𝑡 = 345 × 360 × 20 × 10−3 = 2484(𝑘𝑁) 1.1.1.3 Web area 𝑃𝑤 = 𝐹𝑦𝑤 × ℎ𝑤 × 𝑡𝑤 = 345 × 1550 × 15 = 8021.25(𝑘𝑁) 1.1.1.4 Tensile flange area 𝑃𝑡 = 𝐹𝑦𝑏 × 𝑏𝑏 × 𝑡𝑏 = 345 × 420 × 30 = 4347(𝑘𝑁) 1.1.2 Determine plastic neutral axis(PNA) Figure Plastic neutral axis 25 Steel Bridge Design Nguyen Thien Long - 172603217 Using equilibrium plastic force condition to determine Checking conditions: - Condition 1: 𝑃𝑤 + 𝑃𝑡 < 𝑃𝑐 + 𝑃𝑠 ↔ 8021.25 + 4347 < 2484 + 15937.5 → TRUE So, PNA not go through web - Condition 1: 𝑃𝑤 + 𝑃𝑡 + 𝑃𝑐 < 𝑃𝑠 ↔ 8021.25 + 4347 + 2484 < 15937.5 → TRUE So, PNA not go through top flange Conclusion: PNA goes through deck slab - Calculate for compressive area height based on C6.D1.BD1: 𝑃𝑐 + 𝑃𝑤 + 𝑃𝑡 2484 + 8021.25 + 4347 ] = 250 × [ ] = 232.9765(𝑚𝑚) 𝑌̅ = 𝑡𝑠 × [ 𝑃𝑠 15937.5 1.1.3 Determine plastic moment 𝑌̅ × 𝑃𝑠 𝑀𝑝 = ( ) + (𝑃𝑐 × 𝑑𝑐 + 𝑃𝑤 × 𝑑𝑤 + 𝑃𝑡 × 𝑑𝑡 ) × 𝑡𝑠 Where: 𝑑𝑖 : Distance from PNA to components centre: 𝑡 250 - Deck slab: 𝑑𝑠 = 𝑌̅ − 𝑠 = 232.9765 − = 107.9765(𝑚𝑚) - Top flange: 𝑑𝑐 = 𝑡𝑠 − 𝑌̅ + 𝑡𝑣 + 𝑡𝑡 2 = 250 − 232.9765 + 50 + 20 = 77.023(𝑚𝑚) ℎ 1550 Web: 𝑑𝑤 = 𝑡𝑠 − 𝑌̅ + 𝑡𝑣 + 𝑡𝑡 + 𝑤 = 250 − 232.9765 + 50 + 20 + = 2 862.0235(𝑚𝑚) 𝑡 30 Bottom flange: 𝑑𝑡 = ℎ1 + ℎ2 − 𝑏 − 𝑌̅ = 1600 + 300 − − 232.9765 = 2 1652.024(𝑚𝑚) The value of plastic moment: 232.97652 × 15937.5 𝑀𝑝 = ( ) + (15937.5 × 107.9765 × 250 + 8021.25 × 862.0235 + 4347 × 1652.024 = 16017.29136(𝑘𝑁 𝑚) 1.2 Yielding moment Yielding moment, My, is moment that firstly causes yielding phenomenon in steel flange Based on C6.D2.2, yielding moment of composite cross section equals total acted moment on each stage of cross section, ignore transverse bending moment 1.2.1 Determine for components moment 𝑀𝐷1 𝑀𝐷2 𝑀𝐴𝐷 𝐹𝑦𝑡 = + + 𝑆𝑁𝐶 𝑆𝐿𝑇 𝑆𝑆𝑇 Where: - MD1: Moment according to DC1, acted on steel cross section 𝑀𝐷1 = 4707.45(𝑘𝑁 𝑚) 26 Steel Bridge Design Nguyen Thien Long - 172603217 - MD2: Moment according to DC2+DW, acted on long term composite cross section 𝑀𝐷2 = 1889.43(𝑘𝑁 𝑚) - MAD: Addition moment, according to addition live load on short term composite cross section 𝑀𝐷1 𝑀𝐷2 𝑀𝐴𝐷 = 𝑆𝑆𝑇 × (𝐹𝑦𝑡 − − ) 𝑆𝑁𝐶 𝑆𝐿𝑇 4704.45 × 106 1889.43 × 106 = 33658341.28 × (345 − − ) 23439681.01 30567478.71 = 2771956848(𝑁 𝑚𝑚) = 2771.957(𝑘𝑁 𝑚) 1.2.2 Determine yielding moment The value of plastic moment: 𝑀𝑦 = 𝑀𝐷1 + 𝑀𝐷2 + 𝑀𝐴𝐷 = 4707.45 + 1889.43 + 2771.957 = 9368.837(𝑘𝑁 𝑚) II CHECKING IN LIMIT STATE I Checking based on C6.10.6 2.1 Bending moment 2.1.1 Checking the plastic properties Based on C6.10.7.3.E124, checking to prevent concrete deck slab not crack: 𝐷𝑝 ≤ 0.42𝐷𝑡 Where: Dp: Distance from top of deck slab to centre axis of composite cross section that includes plastic moment, 𝐷𝑝 = 𝑌 = 232.976(𝑚𝑚) Dt: Total height of composite cross section, 𝐷𝑡 = ℎ1 + ℎ2 = 1600 + 300 = 1900(𝑚𝑚) So, check the condition: 232.976 ≤ 0.42 × 1900 → TRUE 2.1.2 Checking the ability to apply Part 6, appendix A Using C10.6.2.2 to check requirement - Minumum yielding strength required is less than 485 MPa: 𝐹𝑦 = 345 < 485 → TRUE - Web required is satisfied with C10.2.1.1: TRUE - Slenderness of web according to C6.10.6.2.2.E113: 2×𝐷𝑐𝑝 𝑡𝑤 ≤ 3.76 × √ 𝐸 𝐹𝑦𝑐 Where: +, Dcp: compressive height of web according to plastic moment, based on D3.2, appendix D, Dcp = 0(mm) So, slenderness of web: 2×0 15 ≤ 3.76 × √ 200000 345 → TRUE All of conditions are satisfied, this cross section is compact 2.1.3 Checking bending resistance for composite cross section 27 Steel Bridge Design Nguyen Thien Long - 172603217 2.1.3.1 Compact cross section Checking based on C6.10.7.1.1.E116: 𝑀𝑢 + × 𝑓𝜆 × 𝑆𝑥𝑡 ≤ 𝜙𝑓 × 𝑀𝑛 Where: - Mu: Maximum moment in strength limit state I - 𝜙𝑓 : Resistance bending coefficient, based on C6.5.4.2, 𝜙𝑓 = - 𝑓𝜆 : Transverse bending stress, 𝑓𝜆 = 0( Straight girder, continuous braces) - 𝑆𝑥𝑡 : Static moment of elastic cross section along main axis according to tensile flange Equation: 𝑀𝑢 ≤ 𝜙𝑓 × 𝑀𝑛 2.1.3.2 Nominal resistance bending strength Checking condition: 𝐷𝑝 ≤ 0.1𝐷𝑡 ↔ 232.9765 ≤ 0.1 × 1900 → FALSE Because the condition is not satisfied, nominal resistance bending strength would be calculated as: 𝐷𝑝 𝑀𝑛 = 𝑀𝑝 × (1.07 − 0.7 × ) 𝐷𝑡 232.9765 = 16017.29136 × (1.07 − 0.7 × ) = 15763.6826(𝑘𝑁 𝑚) 1900 - Checking equation in 2.1.3.1: 𝑀𝑢 ≤ 𝜙𝑓 × 𝑀𝑛 ↔ 14768.38 ≤ × 15763.6826 → TRUE - Checking for transverse bending stress: 𝑓𝜆 ≤ 0.6 × 𝐹𝑦𝑡 ↔ ≤ 0.6 × 345 → TRUE 2.1.4 Checking for shearing resistance Without stiffeners based on C6.10.9.2.E146, calculate shearing resistance as below: 𝑉𝑛 = 𝑉𝑐𝑟 = 𝐶 × 𝑉𝑝 Where: - Vp: Plastic force, 𝑉𝑝 = 0.58 × 𝐹𝑦𝑤 × 𝐷 × 𝑡𝑤 = 0.58 × 345 × 1550 × 15 = 4652325(𝑁) = 4652.325(𝑘𝑁) - C: Buckling resistance ratio for shearing based on C6.10.9.2.E151,152,153, with buckling coefficient k = - Checking conditions to calculate C: +, Condition 1: 𝐷 𝑡𝑤 𝐸×𝑘 < 1.12 × √ 𝐸×𝑘 +, Condition 2: 1.12 × √ 200000×5 1.4 × √ 345 +, Condition 3: 𝐹𝑦𝑤 ≤ 𝐹𝑦𝑤 𝐷 𝑡𝑤 ↔ 1550 15 200000×5 < 1.12 × √ < 1.4 × √ 𝐸×𝑘 𝐹𝑦𝑤 345 ↔ 1.12 × √ →FALSE 200000×5 345 ≤ 1550 15 < → FALSE 𝐷 𝑡𝑤 𝐸×𝑘 > 1.4 × √ 𝐹𝑦𝑤 ↔ 1550 15 200000×5 > 1.4 × √ 345 → TRUE 28 Steel Bridge Design Nguyen Thien Long - 172603217 So, C would be calculated follows condition 𝐸×𝑘 1.57 200000 × √ = = 0.4262 2× 𝐹𝑦𝑤 345 𝐷 1550 ( ) ( ) 𝑡𝑤 15 Resistance shearing: 𝑉𝑛 = 0.4262 × 4652.325 = 1982.758(𝑘𝑁) - Consider shear force in exterior beam at strength limit state I, 𝑉𝑢 = 1720.13(𝑘𝑁) - With 𝜙 = 1, check the condition: 𝑉𝑢 < 𝜙 × 𝑉𝑛 ↔ 1720.13 < × 1982.758 → TRUE III CHECKING IN SERVICE LIMIT STATE II Checking based on C6.10.4 3.1 Elastic deformation 𝐿 36000 - Based on C2.5.2.6.2, maximum deflection according to live load is = = 𝐶= 1.57 ×√ 1000 1000 36(𝑚𝑚) - Actual deflection related to live load, 𝛿, calculated based on C2.5.2.6.2 with distributed coefficient gmV = 0.5 𝐷𝑒𝑠𝑖𝑔𝑛𝑒𝑑 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑜𝑓 𝑑𝑒𝑠𝑖𝑔𝑛𝑒𝑑 𝑡𝑟𝑢𝑐𝑘 𝛿 = 𝑚𝑎𝑥 { 𝐷𝑒𝑠𝑖𝑔𝑛𝑒𝑑 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡 𝑜𝑓 25% 𝐻𝐿 − 93𝐾 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 ↔ 𝛿 = 0.25 × 0,09556 = 0.02389(𝑚) = 23.89(𝑚𝑚) Check the condition: 𝛿 = 23.89(𝑚𝑚) < 36(𝑚𝑚) → TRUE 3.2 Unrestorable deformation related to bending Based on C6.10.4.2.2.E108, top flange need to check: 𝑓𝑓 ≤ 0.95 × 𝑅ℎ × 𝐹𝑦𝑓 Where: - 𝑆𝑒𝑆𝐿 𝑀𝐷𝐶1 𝑓𝑓 : Stress in top flange, 𝑓𝑓 = ( 1378.46 58771606.86 - + 11214.68 165870476 𝑡 𝑆𝑁𝐶 + 𝑆𝑒𝑆𝐿 𝑀𝐷𝐶2+𝐷𝑊 𝑡 𝑆𝐿𝑇 + 𝑆𝑒𝑆𝐿 𝑀𝐻𝐿−93𝐾 𝑡 𝑆𝑆𝑇 )=( 3765.96 18466922.27 + ) × 106 = 294.9956(𝑀𝑃𝑎) 𝑅ℎ : Hybrid coefficient, determined by C6.10.1.10.1, 𝑅ℎ = Checking the condition: 294.9956 ≤ 0.95 × × 345 → TRUE Similarly, bottom flange need to check: 𝑓𝑓 + Checking the condition: 𝑓𝑓 = ( 1378.46 30567478.71 + 11214.68 33658341.28 𝑆𝑒𝑆𝐿 𝑀𝐷𝐶1 𝑐 𝑆𝑁𝐶 + 𝑓𝜆 𝑆𝑒𝑆𝐿 𝑀𝐷𝐶2+𝐷𝑊 𝑐 𝑆𝐿𝑇 ≤ 0.95 × 𝑅ℎ × 𝐹𝑦𝑓 + 𝑆𝑒𝑆𝐿 𝑀𝐻𝐿−93𝐾 𝑐 𝑆𝑆𝑇 )=( 3765.96 23439681.01 + ) × 106 = 326.8821189(𝑀𝑃𝑎) Checking the condition: 326.8821189 ≤ 0.95 × × 345 → TRUE So, all conditions were satisfied in service limit state Based on C6.10.2.1.1, composite cross section not have to check 𝑓𝑐 ≤ 𝐹𝑐𝑟𝑤 IV FATIGUE LIMIT STATE AND CRACKING Checking conditions related to FLS in C6.10.5.1, cracking in C6.10.5.2, ultimate FLS in C6.10.5.3 29 Steel Bridge Design Nguyen Thien Long - 172603217 4.1 Fatigue related to loading Based on C6.6.1, stress related to loading is less than nominal fatigue resistance: 𝛾 × (∆𝑓) ≤ (∆𝐹 )𝑛 Where: - 𝛾: Loading coefficient according to fatigue loading combination - ∆𝑓: Stress amplitude related to fatigue loading, Mpa - ∆𝐹: Nominal fatigue resistance − In this project, consider that bridge’s lifetime is eternity, so calculate as below: (∆𝐹 )𝑛 = (∆𝐹 ) 𝑇𝐻 +, Nominal fatigue resistance depends on mold materials and transverse connections Based on table 3, obtains that fatigue parts belong to A type, fatigue amplitude (∆𝐹 ) 𝑇𝐻 = 165(𝑀𝑃𝑎) So, minimum allowable stress (∆𝐹 )𝑛 = 165(𝑀𝑃𝑎) 4.2 Fatigue stress calculation Using fatigue loading combination with IM = 15% 4.2.1 Stress in bottom edge of top flange 𝑀𝑓1 𝛾 × (∆𝑓) = 𝐼𝑆𝑇 𝑦𝑡 Where: - 𝑀𝑓1 : Maximum moment in FLS I(Using FLS II value multiply with 1.5, divide with 0.75) - 𝐼𝑆𝑇 : Inertia moment of short term composite steel cross section - 𝑦 𝑡 : Distance from neutral axis of cross section to bottom edge of top flange So, stress in bottom edge of top flange: 2326.23 × 1.5 0.75 𝛾 × (∆𝑓) = × 106 = 25.97031773(𝑀𝑃𝑎) 44768872311 269.9026 − 20 Checking the condition: 𝛾 × (∆𝑓) < (∆𝐹 )𝑛 ↔ 25.97031773 < 165 → TRUE 4.2.2 Stress in top edge of bottom flange 𝑀𝑓1 𝛾 × (∆𝑓) = 𝐼𝑆𝑇 𝑦𝑐 Where: - 𝑦 𝑐 : Distance from neutral axis of cross section to top edge of bottom flange So, stress in top edge of bottom flange: 2326.23 × 1.5 0.75 𝛾 × (∆𝑓) = × 106 = 133.03(𝑀𝑃𝑎) 44768872311 1330.0974 − 50 30 Steel Bridge Design Nguyen Thien Long - 172603217 Checking the condition: 𝛾 × (∆𝑓) < (∆𝐹 )𝑛 ↔ 133.03 < 165 → TRUE V ULTIMATE FATIGUE REQUIREMENT FOR WEB Based on C6.10.5.3, shear force have to check in FLS: 𝑉𝑢 ≤ 𝑉𝑐𝑟 Because not use stiffeners in web, so above condition certainly satisfied 31 Steel Bridge Design Nguyen Thien Long - 172603217 CHAPTER 4: DESIGN FOR STIFFENERS AND SHEAR CONNECTORS I 1.1 SHEAR CONNECTORS Composition In composite cross section, set up shear connectors at contacted plane between deck slab and steel girder to resist shear force In this project, choose mushroom stud with below properties: Properties of stud Note Value Unit Height h 150 mm Diameter d 20 mm Setup follows horizontal n studs direction Minimum tensile Fu 420 MPa strength Yielding strength Fy 345 MPa Checking condition: ℎ ≥4↔ 150 ≥ → TRUE - Based on C6.10.10.1.1, - Stud spacing have to determined in FLS, based on C6.10.10.2 and C6.10.10.3 Quantities of studs are more than required quantities studs, based on C6.10.10.4 𝑑 20 Figure Set up shear connectors 32 Steel Bridge Design 1.2 - Nguyen Thien Long - 172603217 Fatigue Limit State(FLS) Spacing of stud have to satisfied with C6.10.10.1.2.E158 𝑛 × 𝑍𝑟 𝑝≤ 𝑉𝑠𝑟 Where: 𝑉𝑠𝑟 : Amplitude of transverse shear force for each length unit, 𝑁⁄𝑚𝑚, 𝑉𝑠𝑟 = √(𝑉𝑓𝑎𝑡 ) + (𝐹𝑓𝑎𝑡 ) +, 𝑉𝑓𝑎𝑡 : Amplitude of longitudinal shear force for each length unit, , = ì ã In positive moment area, the maximum fatigue shear force 𝑉𝑓 at abutment, 𝑉𝑓 = × (277.4 − 0) = 554.8(𝑘𝑁) • I: Inertia moment of short term composite cross section, 𝐼 = 𝐼𝑆𝑇 = 44768872311(𝑚𝑚4 ) • Q: Static moment of short term equivalent area of concrete slab( effective width) with neutral axis of short term composite cross section, 𝑄= So, 𝑉𝑓𝑎𝑡 = - - 𝑉𝑓 ×𝑄 𝐼 𝑡𝑠 𝑡 × 𝑡𝑠 × 𝐵𝑒𝑓𝑓 × (𝑦𝑆𝑇 + 𝑡𝑣 + ) 250 = × 250 × 2500 × (269.9025974 + 50 + ) = 34758015.42(𝑚𝑚3 ) = 554.8×34758015.42 44768872311 × 103 = 430.7400647(𝑘𝑁⁄𝑚) +, 𝐹𝑓𝑎𝑡 : Amplitude of , 𝑁⁄𝑚𝑚, 𝐹𝑓𝑎𝑡 = 0(Straight bridge) So, 𝑉𝑠𝑟 = 𝑉𝑓𝑎𝑡 = 430.7400647(𝑘𝑁⁄𝑚) Assume that AADT is more than 960(veh), using fatigue combination I and unlimited fatigue resistance based on C6.10.10.2 to calculate Shearing resistance on each stud: 𝑍𝑟 = 38 × 𝑑 = 38 × 202 × 10−3 = 15.2(𝑘𝑁) Check the stud’s spacing: 𝑛 × 𝑍𝑟 × 15.2 𝑝≤ = = 176.44(𝑚𝑚) 𝑉𝑠𝑟 430.7400647 Choose equal stud’s spacing 𝑝 = 120(𝑚𝑚) for on the whole length of girder based on C6.10.10.1.3 Check the additional condition for stud: +, Spacing of stud 𝑝 ≤ 600(𝑚𝑚) ↔ 120 ≤ 600(𝑚𝑚) → TRUE +, Spacing of stud 𝑝 ≥ × 𝑑 ↔ 120 ≥ × 20 → TRUE +, Follows horizontal direction, perpendicular spacing of stud 33 Steel Bridge Design Nguyen Thien Long - 172603217 𝑝 ≥ × 𝑑 ↔ 120 ≥ × 20 → TRUE +, Distance from edge of top flange and closest edge of stud is more than 25 mm→ TRUE 1.3 Strength limit state Based on C6.10.10.4.1.E170, designed resistance of simple stud, 𝑄𝑟 , calculate as below: 𝑄𝑟 = 𝜙𝑠𝑐 × 𝑄𝑛 Where: - 𝜙𝑠𝑐 : Resistance coefficient according to stud, based on C5.4.2, 𝜙𝑠𝑐 = 0.85 - 𝑄𝑛 : Nominal shearing resistance of simple stud, based on C6.10.10.4.3, calculate as below: 𝑄𝑛 = 0.5 × 𝐴𝑠𝑐 × √𝑓𝑐′ × 𝐸𝑐 ≤ 𝐴𝑠𝑐 × 𝐹𝑢 +, 𝐴𝑠𝑐 : Area of simple stud, 𝐴𝑠𝑐 = 𝜋×𝑑2 = 𝜋×202 = 314.1592654(𝑚𝑚2 ) So, 𝑄𝑛 = 0.5 × 𝐴𝑠𝑐 × √𝑓𝑐′ × 𝐸𝑐 = 0.5 × 314.1592654 × √30 × 31349.535 = 152333.7231(𝑁) Check condition: 𝑄𝑛 ≤ 𝐴𝑠𝑐 × 𝐹𝑢 ↔ 152333.7231 ≤ 314.1592654 × 420 → FALSE Recalculate Qn: 𝑄𝑛 = 𝐴𝑠𝑐 × 𝐹𝑢 = 314.1592654 × 420 = 131946.8915(𝑁) = 131.9468915(𝑘𝑁) So, 𝑄𝑟 = 𝜙𝑠𝑐 × 𝑄𝑛 = 0.85 × 131.9468915 = 112.1548577(𝑘𝑁) In strength limit state, based on C6.10.10.4.1.E171, minimum quantities of studs 𝑃 𝑛𝑠 = 𝑄𝑟 - Where: P: Total nominal shear force determined by C6.10.10.4.2 +, In simple span, P calculated as below: 𝑃 = √𝑃𝑝2 + 𝐹𝑝2 Where: • 𝑃𝑝 : Total vertical shear force in concrete slab at maximum positive moment 𝑃𝑝1 point, accoring to live load with IM Value of 𝑃𝑝 : 𝑃𝑝 = 𝑚𝑖𝑛 { 𝑃𝑝2 ′ 𝑃𝑝1 = 0.85 × 𝑓𝑐 × 𝑏𝑠 × 𝑡𝑠 = 0.85 × 30 × 2500 × 250 × 10−3 = 15937.5(𝑘𝑁) 𝑃𝑝2 = 𝐹𝑦𝑤 × 𝐷 × 𝑡𝑤 + 𝐹𝑦𝑡 × 𝑏𝑓𝑡 × 𝑡𝑓𝑡 + 𝐹𝑦𝑐 × 𝑏𝑓𝑐 × 𝑡𝑓𝑐 = 345 × (1550 × 15 + 360 × 20 + 420 × 30) ì 103 = 14852.25() So, = 14852.25() ã 𝐹𝑝 : Total centripetal shear force in concrete slab at maximum positive moment point, accoring to live load with IM Value of 𝐹𝑝 : 𝐹𝑝 = 0(Straight span) The value of P: 𝑃 = √𝑃𝑝2 + 𝐹𝑝2 = √14852.252 = 14852.25(𝑘𝑁) 34 Steel Bridge Design Nguyen Thien Long - 172603217 Minimum quantities of studs 𝑛𝑠 = 𝑃 𝑄𝑟 = 14852.25 112.1548577 = 132.43(𝑠𝑡𝑢𝑑𝑠) Consider the part which calculates from bearing to maximum positive moment point: 𝐿1 = 𝐿𝑒𝑓𝑓 = 35 = 17.5(𝑚) = 17500(𝑚𝑚) With above length L1, use quantities of studs with spacing p = 120mm : 𝑛𝐿1 = 𝑛×𝐿1 5×17500 = = 729.16(𝑠𝑡𝑢𝑑𝑠) 𝑝 120 Choose 730(studs) Check the condition: 𝑛𝐿1 ≥ 𝑛𝑠 ↔ 730 ≥ 132.43 → TRUE With remaining parts, because of symmetrical properties, also use 730(studs) Conclusion: Using stud’s spacing 𝑝 = 120(𝑚𝑚) for each line On cross section, set up stud lines along whole girder’s length→ Satisfied SLS and FLS conditions 35 ... DESIGNED DATA Designed method and loading This project was designed based on specification TCVN 11823 – 2017 and some reference books - Design load: HL – 93; PL 3.10-3 MPa - Span structure: x 36(m)