(Đồ án hcmute) tan tao 1 apartmment building

MINISTRY OF EDUCATION AND TRAINING HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY AND EDUCATION FACULTY FOR HIGH QUALITY TRAINING CAPSTONE PROJECT CIVIL ENGINEERING TECHNOLOGY TAN TAO APARTMMENT BUILDING LECTURER: ASSOC PROF DR LE ANH THANG STUDENT: TRAN XUAN PHONG SKL010478 Ho Chi Minh City, August 2022 HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY AND EDUCATION FACULTY FOR HIGH QUALITY TRAINING CAPSTONE PROJECT TAN TAO APARTMMENT BUILDING Student: TRAN XUAN PHONG Student ID: 18149026 Course: 2018 Major: CIVIL ENGINEERING TECHNOLOGY Advisor: ASSOC PROF DR LE ANH THANG Ho Chi Minh City, August 2022 Faculty for High Quality Training – HCMC University of Technology and Education HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY AND EDUCATION FACULTY FOR HIGH QUALITY TRAINING CAPSTONE PROJECT TAN TAO APARTMMENT BUILDING Student: TRAN XUAN PHONG Student ID: 18149026 Course: 2018 Major: CIVIL ENGINEERING TECHNOLOGY Advisor: ASSOC PROF DR LE ANH THANG Ho Chi Minh City, August 2022 Faculty for High Quality Training – HCMC University of Technology and Education THE SOCIALIST REPUBLIC OF VIETNAM Independence – Freedom – Happiness -Ho Chi Minh City, August 26th, 2022 CAPSTONE PROJECT ASSIGNMENT Student name: TRAN XUAN PHONG Student ID: 18149026 Major: Civil Engineering Technology Class: 18149CLA2 Advisor: Assoc Prof Dr LE ANH THANG Phone number: 0938308076 Date of assignment: 24/08/2022 Date of submission: 13/02/2023 Project title: TAN TAO APARTMENT BUILDING Initial materials provided by the advisor: - Architectural profile of the project - Geotechnical profile of the project Content of the project: a Architectural: - Introduction to the architecture of the project - Functions of building architecture - Show architectural drawings b Structural: - Modeling, analyzing, calculating and designing typical slab - Modeling, analyzing, calculating and designing typical stairs - Check the overall stability of the project - Modeling, analyzing, calculating, designing frames, columns, rigid wall, core - Modeling, analysis, calculation and design of bored pile foundation - Modeling, analyzing, calculating and designing water tanks Final product: - 01 explanation - 01 appendix - Drawings (3 architectural drawings, 29 structural drawings) CHAIR OF THE PROGRAM (Sign with full name) ADVISOR (Sign with full name) THE SOCIALIST REPUBLIC OF VIETNAM Independence – Freedom– Happiness -Ho Chi Minh City, August 26th, 2022 ADVISOR’S EVALUATION SHEET Student name: TRAN XUAN PHONG Student ID: 18149026 Major: Civil Engineering Technology Class: 18149CLA2 Project title: TAN TAO APARTMENT BUILDING Advisor: Assoc Prof Dr LE ANH THANG EVALUATION Content of the project: Strengths: Weaknesses: Approval for oral defense? (Approved or denied) Overall evaluation: (Excellent, Good, Fair, Poor) Mark: ……………… (in words: ) Ho Chi Minh City, month day, year ADVISOR (Sign with full name) THE SOCIALIST REPUBLIC OF VIETNAM Independence – Freedom– Happiness -Ho Chi Minh City, January 20, 2020 PRE-DEFENSE EVALUATION SHEET Student name: TRAN XUAN PHONG Student ID: 18149026 Major: Civil Engineering Technology Class: 18149CLA2 Project title: TAN TAO APARTMENT BUILDING Advisor: Assoc Prof Dr LE ANH THANG Name of Reviewer: Assoc Prof Dr HA DUY KHANH EVALUATION Content and workload of the project: Strengths: Weaknesses: Approval for oral defense? (Approved or denied) Overall evaluation: (Excellent, Good, Fair, Poor) Mark: (in words: .) Ho Chi Minh City, month day, year REVIEWER (Sign with full name) ACKNOWLEDGEMENT For each construction student, the Capstone Project is like a big first project in life when preparing to end the study process at the University, and at the same time opening up a new direction in life reality in the future Through the process of doing the Project, I have more opportunities to synthesize and systematize the knowledge I have learned, and at the same time collect and add new knowledge that I still lack, practice my calculation ability and solve problems that may arise in practice During the time of writing my thesis, I have received a lot of guidance and enthusiastic help from Mr Le Anh Thang I would like to express my sincere and deepest gratitude to Mr Le Anh Thang The knowledge and experience that you have imparted to me is the foundation, the golden key for me to be able to complete this Capstone Project Although I have tried my best, but due to limited knowledge and experience, it is difficult to avoid errors in my project, I hope to receive your guidance so that I can try and improve my skills, improve your knowledge Finally, I would like to wish you success and good health so that you can continue your career of imparting knowledge to the next generation Thank you sincerely Ho Chi Minh City, February 12, 2023 Student TRAN XUAN PHONG TABLE OF CONTENTS CHAPTER 1: PROJECT OVERVIEW .19 1.1 Project introduction: 19 1.1.1 Purpose of building construction: 19 1.1.2 Location and project characteristics: .19 1.2 Scale of construction works: 19 1.2.1 Project level: 19 1.2.2 Project height: 19 1.2.3 Construction area: 19 1.2.4 Construction function: 19 1.3 Architectural solutions: 20 1.3.1 Construction ground solutions: .20 1.3.2 Construction traffic solutions: .20 CHAPTER 2: STRUCTURAL OVERVIEW 21 2.1 Analysis and selection of structural: 21 2.1.1 Vertical structural system: .21 2.1.2 Horizontal structural system: 21 2.1.3 Solution for underground structure: 21 2.2 Materials used: 22 2.2.1 Concrete: .22 2.2.2 Reinforcing steel: 22 2.3 Preliminary section: 23 2.3.1 Slab cross-section: 23 2.3.2 Beam cross section: .23 2.3.3 Column cross-section: 24 2.3.4 Section of the wall: 25 CHAPTER 3: LOAD AND IMPACT .26 3.1 Load classification: 26 3.1.1 Regular load: 26 3.1.2 Temporary load: 26 3.2 Dead load: 26 3.2.1 Self-weight of the structure: 26 3.2.2 Weight of retaining wall and fixed wall: .26 3.2.3 Finished load weight: 28 3.3 Live load: 30 3.4 Analysis of building dynamics: .30 3.4.1 Theoretical basis: 30 3.4.2 The volume of participation fluctuates: 31 3.4.3 Specific oscillate survey: .32 3.4.4 Space frame model: .32 3.5 Wind load: .36 3.5.1 Static component of wind load: .36 3.5.2 Dynamic component of wind load: .38 3.5.3 Earthquake load: 42 3.6 Load combination: 44 CHAPTER 4: STAIRS .47 4.1 General concept: 47 4.2 Stair structure: 47 4.2.1 Floor plan and section of stairs: 47 4.2.2 Choose the preliminary size of the stairs: .48 4.3 Active load: 50 4.3.1 Load acting on the landing: 50 4.3.2 Load acting on the skirtboard: .51 4.3.3 Load acting on beam: 52 4.4 Calculation diagram: .53 4.5 Calculation of reinforcement: 56 4.5.1 Skirtboard: .56 4.5.2 Check the shear resistance of the skirtboard: 57 4.5.3 Beam: 57 CHAPTER 5: TYPICAL FLOOR DESIGN 60 5.1 General concept: 60 5.2 Computational hypothesis: 60 5.3 Dead loads acting on the floor: 60 5.4 Live load acting on the floor: 60 5.5 Set up calculation scheme: 61 5.6 Floor test by limit state I: .65 5.6.1 Sequence of calculation of reinforcement for floor: .65 5.6.2 Calculation and arregement rebar for floor: 66 5.7 Floor test by limit state II – section 8.2, TCVN 5574-2018: 67 5.7.1 Calculation cases: 67 5.7.2 Load combination: 68 5.7.3 Material specification: 68 5.7.4 Check for crack formation according to section 8.2.2.1.1 TCVN 5574-2018: .69 5.7.5 Check the crack width according to section 8.2.2.1.1 TCVN 5574-2018: 71 5.7.6 Check the deflection of the floor: 76 CHAPTER 6: FRAME DESIGN .78 6.1 Shape and size: 78 6.2 Section size: .78 6.3 Evaluation of frame internal force results: 78 6.3.1 Internal force diagram: 78 6.3.2 Displacement of the top of the building: .85 6.3.3 Check the relative horizontal displacement between floors: 86 6.4 Calculation of reinforcement: 90 6.4.1 Theory of calculation of beam reinforcement: 90 6.4.2 Theory of calculation of stirrup for beams: 91 6.4.3 Apply calculation for reinforcement: 92 6.4.4 Apply calculation for sitirrup: .93 6.4.5 Anchor rebar for beam: 95 6.4.6 Reinforced stirrup for beam: 96 6.4.7 Calculation of column rebar: 97 Step 7: Checking the crack formation conditions reinforced concrete structures: Cracks in reinforced concrete structures form when conditions occur: 𝑀 = 30.21 𝑘𝑁/𝑚 ≥ 𝑀𝑐𝑟𝑐 = 9.44 𝑘𝑁/𝑚 → The crack is occurred 9.6.2 Crack expansion: According to section 8.2.2.1.2 TCVN 5574:2018, when cracks appear in the members, it is necessary to calculate the crack width It is necessary to calculate reinforced concrete members according to short-term and long-term cracks Short-term crack width is determined by the simultaneous action of permanent and temporary loads (long-term and short-term), while long-term crack width is determined only by permanent and temporary loads Calculation of crack expansion: Step 1: Determination of coefficients taking into account the uneven distribution of relative strain of tensile reinforcement between cracks 𝚿𝒔 : Allow to take Ψ𝑠 = when no cracks appear according to 8.2.2.3.1 TCVN 5574:2018 If crack appear (M>Mcrc) then the value should be determined according to the formula for flexural members according to section 8.2.2.3.4 TCVN 5574:2018 Ψ𝑠 = − 0.8 × 𝑀𝑐𝑟𝑐 9.44 = − 0.8 × = 0.723 𝑀1 27.228 Step 2: Determination of clearance between adjacent orthogonal cracks: Base clearance between adjacent perpendicular cracks 𝐿𝑠 , determined according to formula 174, section 8.2.2.3.3 TCVN 5574:2018, but not less than 10𝑑𝑠 , and 100mm and not more than 40𝑑𝑠 and 400mm: 𝐿𝑠 = 0.5 𝐴𝑏𝑡 75000 𝑑𝑠 = 0.5 × × 12 = 61.224 𝑚𝑚 → 𝑐ℎ𝑜𝑜𝑠𝑒 𝐿𝑠 = 110 𝑚𝑚 𝐴𝑠 735 𝐴𝑏𝑡 : Tensile concrete cross-sectional area 𝐴𝑏𝑡 = 𝑏 × (ℎ − 𝑦𝑐 ); 2𝑎 ≤ ℎ − 𝑦𝑐 ≤ 0.5ℎ 𝐴𝑏𝑡 = 1000 × (150 − 41.45) = 108550.1 > 0.5ℎ = 1000 × 150 = 75000 𝑚𝑚2 → 𝐶ℎ𝑜𝑜𝑠𝑒 𝐴𝑏𝑡 = 75000 𝑚𝑚2 244 𝑦𝑐 : the height of the compression zone of the converted cross-section of the member is determined according to 8.2.3.3.6 TCVN 5574: 2018 𝑦𝑐 = ℎ0 [√(𝜇𝑠 𝛼𝑠2 )2 + 2(𝜇𝑠 𝛼𝑠2 ) − (𝜇𝑠 𝛼𝑠2 )] = 125 × √(0.006 × 13.64)2 + × (0.006 × 13.64) − (0.006 × 13.64) = 41.45 𝑚𝑚 𝛼𝑠2 = 𝛼𝑠1 : Conversion coefficient of reinforcement to concrete, according to 8.2.2.3.2 TCVN 5574: 2018 𝛼𝑠1 = 𝐸𝑠 𝐸𝑏,𝑟𝑒𝑑 = 200000 = 13.64 14666.667 𝐸𝑏,𝑟𝑒𝑑 : Converted strain modulus of compressive concrete, including elastic deformation of compressive concrete 𝐸𝑏,𝑟𝑒𝑑 = 𝑅𝑏,𝑛 𝜀𝑏1,𝑟𝑒𝑑 = 22 == 14666.667 𝑀𝑃𝑎 0.0015 𝜇𝑠 : Tensile reinforcement content 𝜇𝑠 = 𝐴𝑠 735 = = 0.006 𝑏ℎ0 1000 × 125 𝐴𝑠 : Area of tensile reinforcement 𝐴𝑠 = 735 𝑚𝑚 𝑑𝑠 : Reinforcement diameter 𝑑𝑠 = 12 𝑚𝑚 Step 3: Determine the moment of inertia of the converted cross-section: Ired According to section 8.2.3.3.5 TCVN 5574:2018 ′ 𝐼𝑟𝑒𝑑 = 𝐼𝑏 + 𝐼𝑠 𝛼𝑠2 + 𝐼 𝑠 𝛼𝑠1 𝑏𝑦𝑐 = + 𝐴𝑠 (ℎ0 − 𝑦𝑐 )2 𝛼𝑠2 + 𝐴′ 𝑠 (𝑦𝑐 − 𝑎′ )2 𝛼𝑠1 1000 × 41.453 = + 735 × (125 − 41.45)2 × 13.64 = 9.370 × 107 𝑚𝑚4 245 Step 4: Determination of stress in tensile longitudinal reinforcement at perpendicular section with crack caused by external force: According to section 8.2.2.3.2 TCVN 5574:2018 𝑀1 (ℎ0 − 𝑦𝑐 ) 27.228 × (125 − 41.45) × 106 𝜎𝑠 = 𝛼𝑠1 = × 13.64 = 331.1 𝑁/𝑚𝑚2 𝐼𝑟𝑒𝑑 9.370 × 107 Step 5: Determination of crack width perpendicular to member axis 𝒂𝒄𝒓𝒄,𝒊 : 𝑎𝑐𝑟𝑐,𝑖 = 𝜑1 𝜑2 𝜑3 Ψ𝑠 𝜎𝑠,𝑖 𝐿 𝐸𝑠 𝑠 𝜎𝑠 : Stress in the longitudinal reinforcement of the member at the perpendicular section with cracks 𝐿𝑠 : Base distance between adjacent orthogonal cracks Ψ𝑠 : Coeficient taking into account the unequal distribution of relative strain of the crack tensile reinforcement 𝜑1 : Coeficient taking into account the time of application of the load 𝜑1 = 1: 𝑊ℎ𝑒𝑛 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑎 𝑠ℎ𝑜𝑟𝑡 − 𝑡𝑒𝑟𝑚 𝑒𝑓𝑓𝑒𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑜𝑎𝑑 𝜑1 = 1.4: 𝑊ℎ𝑒𝑛 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑎 𝑙𝑜𝑛𝑔 − 𝑡𝑒𝑟𝑚 𝑒𝑓𝑓𝑒𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑜𝑎𝑑 𝜑2 : Coefficient taking into account the type of surface shape of longitudinal reinforcement 𝜑2 = 0.5: 𝐹𝑜𝑟 𝑟𝑖𝑏𝑏𝑒𝑑 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑛𝑑 𝑐𝑎𝑏𝑙𝑒𝑠 𝜑2 = 0.8: 𝐹𝑜𝑟 𝑝𝑙𝑎𝑖𝑛 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝜑3 : Factor taking into account the bearing characteristics, taken as for members subjected to eccentric bending and compression Step 6: Determination of long-term and short-term crack widths of perpendicular and oblique cracks: According to section 8.2.2.1.4 TCVN 5574:2018: The long-term crack width is determined by the formula 𝑎𝑐𝑟𝑐 = 𝑎𝑐𝑟𝑐,1 = 0.2107 𝑚𝑚 246 Where: 𝑎𝑐𝑟𝑐,1 : Crack width due to long-term effects of permanent and temporary loads 𝑎𝑐𝑟𝑐,1 = 𝜑1 𝜑2 𝜑3 Ψ𝑠 𝜎𝑠 331.1 𝐿𝑠 = 1.4 × 0.5 × × 0.723 × × 110 = 0.0921 𝑚𝑚 𝐸𝑠 200000 According to section 8.2.2.1.3 TCVN 5574:2018: Checking the crack expansion condition of reinforced concrete members: 𝑎𝑐𝑟𝑐 𝑑ℎ = 0.0921 𝑚𝑚 ≤ 𝑎𝑐𝑟𝑐,𝑢 = 0.3 𝑚𝑚 Where: 𝑎𝑐𝑟𝑐 : Crack width due to external force 𝑎𝑐𝑟𝑐,𝑢 : Permissible limit crack width (Table 17-TCVN 5574:2018) Table 9.14 Permissible crack limit check 𝒂𝒄𝒓𝒄 value of crack (mm) Reinforcement type Standard Long-term Short-term According to the condition to ensure the integrity of the reinforcement CB240T, CB300T TCVN 1651-1:2008 CB300V, CB400V TCVN 1651-2:2018 CB500V,CB600V 0.3 0.4 TCVN 6288:1997 Cold drawn steel wire (ISO 10544:1992) High-strength bar reinforcement TCVN 6284-5:1997 (with conventional yield strength (ISO 6934-5:1991) 835, 930 and 080 MPa) TCVN 6284-2:1997 High strength cold drawn steel wire (ISO 6394-2:1991) 0.2 0.3 7-strand cable 12.4mm or more in TCVN 6284-4:1997 diameter (ISO 6934-4:1991) 19 fiber cable 7-strand cable with diameter less TCVN 6284-4:1997 0.1 0.2 than 12.4 mm (ISO 6934-4:1991) According to the condition of limiting seepage for the structure 0.2 0.3 247 9.7 Check for the deflecton: 9.7.1 Check for the deflection for top plate: Figure 9.1 Deflection for top plate 𝑓 = 𝑚𝑚 < [𝑓] = 40 𝑚𝑚 → Deflection condition is sastified 248 9.7.2 Check for the deflection for bottom plate: Figure 9.2 Deflection for bottom plate 𝑓 = 15.6 𝑚𝑚 < [𝑓] = 40 𝑚𝑚 → Deflection condition is sastified 249 9.8 Calculate reinforcement of beam: 9.8.1 Calculation rebar for beam: Table 9.15 Material Concrete B30 Density γ 25 Calculated compressive strength 𝑅𝑏 17 × 103 Elastic modulus 𝐸𝑏 3.25 × 107 Steel CB400-V (𝜙 ≥ 10𝑚𝑚) Calculated tensile strength 𝑅𝑠 350 × 103 Calculated compressive strength 𝑅𝑠𝑐 350 × 103 Elastic modulus 𝐸𝑏 20 × 107 𝑊𝑖𝑡ℎ: { 𝑘𝑁/𝑚2 𝑘𝑁/𝑚2 𝑘𝑁/𝑚2 𝑘𝑁/𝑚2 𝑘𝑁/𝑚2 𝑘𝑁/𝑚2 𝑏 × ℎ = 300 × 700 𝑚𝑚 → ℎ0 = 675 𝑚𝑚 𝑎 = 25 𝑚𝑚 Table 9.16 Calculation rebar for beam Name Main bottom beam Main top beam Secondary bottom beam Secondary top beam Location Left Sup Span Right Sup Left Sup Span Right Sup Left Sup Span Right Sup Left Sup Span Right Sup M 𝒉𝟎 b kN/m 177.371 321.058 177.371 177.371 72.615 177.371 111.080 122.959 111.080 44.621 122.959 44.621 mm 675 675 675 675 675 675 675 675 675 675 675 675 mm 300 300 300 300 300 300 300 300 300 300 300 300 𝑹𝒃 17 17 17 17 17 17 17 17 17 17 17 17 𝜶𝒎 0.076 0.138 0.076 0.105 0.043 0.105 0.145 0.160 0.145 0.058 0.160 0.058 𝝃 𝑨𝒔 0.079 0.149 0.079 0.111 0.044 0.111 0.157 0.176 0.157 0.060 0.176 0.060 𝑚𝑚2 781.9 1468.6 781.9 933.3 368.9 933.3 725.1 810.8 725.1 276.7 810.8 276.7 Choose rebar 𝜙 n mm 25 25 25 25 22 25 22 25 22 22 25 22 Reinforced 𝜙 n mm 25 𝑨𝒔 𝝁 𝑚𝑚2 982 1473 982 982 760 982 760 982 760 760 982 760 % 0.48 0.73 0.48 0.57 0.44 0.57 0.80 1.03 0.80 0.80 1.03 0.80 250 9.8.2 Calculation stirrup for beam: 𝑄𝑚𝑎𝑥 = 160.358 𝑘𝑁 > 0.5𝑅𝑏𝑡 ℎ0 𝑏 = 0.5 × 1.15 × 675 × 300 × 10−3 = 116.44 𝑘𝑁 → Need to calculate and arrange stirrup Stirrup structure according to TCVN 198-1997: Distance of belt reinforcement according to seismic resistance condition Within the critical zones of the main seismic girder, reinforcement should be arranged to satisfy the following: The diameter 𝑑𝑏𝑤 of the reinforcing bars should not be less than 6mm The distance s of the reinforcement rings should not exceed 𝑠 = 𝑚𝑖𝑛 { ℎ𝑤 ; 24𝑑𝑏𝑤 ; 225; 8𝑑𝑏𝑙 } Where: 𝑑𝑏𝑙 : Minimum diameter of longitudinal steel bar (mm) ℎ𝑤 : Height of beam section (mm) → 𝑠 = 𝑚𝑖𝑛{175; 600; 225; 550} = 175 𝑚𝑚 The first reinforcement must be placed no more than 50 (mm) from the end cross section of the beam Figure 9.3 Transverse reinforcement in the critical region of the beam 251 Stirrup structure according to TCVN 198-1997: Within the length of 3hd (hd is the height of the beam cross section) of the beam from the edge of the column, reinforcement must be placed thicker than the area between the beams The distance between the reinforcing bars is not greater than the calculated value according to the shear force requirements but at the same time must be ≤0.25hd and not greater than times the diameter of the longitudinal reinforcement In no case should this distance exceed 150 mm In the area between the beams (outside the range mentioned above), the distance between the reinforcements should be ≤ 0.5hd and not greater than 12 times the diameter of the longitudinal reinforcement and not exceed 300mm Conclusion Choose the diameter of the stirrup to be 8mm So, the first part of the beam is L/4 (mm) from the joint frame arrangement stirrup ϕ8, branches with distance s = 150 (mm), the rest ϕ8, branches with distance s = 250 (mm) Check condition again: according to TCVN 9386:2012 and TCVN 5574:2018: 𝑄𝑏 = 𝜑𝑏2 𝑅𝑏𝑡 𝑏ℎ𝑜2 1.5 × 1.15 × 300 × 6752 = = 174.656 𝑘𝑁 𝐶 × 675 × 1000 𝑄𝑠𝑤 = 𝜑𝑠𝑤 𝑞𝑠𝑤 𝐶 = 𝑞𝑠𝑤 = 0.75 × 1350 × 141.4 = 143.168 𝑘𝑁 1000 𝑅𝑠𝑤 𝐴𝑠𝑤 210 × 101 = = 141.4 𝑘𝑁𝑚 𝑆𝑤 150 Where: 𝐴𝑠𝑤 = 101 𝑚𝑚2 (𝜙8 − 𝑡𝑤𝑜 𝑏𝑟𝑎𝑛𝑐ℎ𝑒𝑠) 𝑅𝑠𝑤 = 210 𝑀𝑃𝑎 𝑆𝑤 = 150 𝑚𝑚 (𝑎𝑡 𝐿/4 𝑠𝑝𝑎𝑛) → 𝑄𝑚𝑎𝑥 = 160.358 𝑘𝑁 ≤ 𝑄 = 𝑄𝑏 + 𝑄𝑠𝑤 = 174.656 + 143.168 = 317.823 𝑘𝑁 → Satisfied condition 252 9.8.3 Anchor rebar for beam: According to section 10.3.5 TCVN 5574:2018, basis length anchor rebar: 𝐿0,𝑎𝑛 = 𝑅𝑠 𝐴𝑠 𝑅𝑏𝑜𝑛𝑑 𝑢𝑠 We have: 𝜙12 → 𝐴𝑠 = 113.1 𝑚𝑚2 ; 𝜙10 → 𝐴𝑠 = 78.5 𝑚𝑚2 𝑅𝑏𝑜𝑛𝑑 is the calculated adhesion strength of reinforcement to concrete, with the assumption that this adhesion is evenly distributed along the anchor length, and is determined by the formula: 𝑅𝑏𝑜𝑛𝑑 = 𝜂1 𝜂2 𝑅𝑏𝑡 𝜂1 = 2: with ribbed cold rolled steel 𝛼 = 1: with the tensile state 𝜂2 = 1: when the diameter of reinforcement is less than or equal to 32 → 𝑅𝑏𝑜𝑛𝑑 = × × 1.15 = 2.3 𝑀𝑃𝑎 𝐿0,𝑎𝑛 = 𝑅𝑠 𝐴𝑠 350 × 113.1 = = 456.532 𝑚𝑚 𝑅𝑏𝑜𝑛𝑑 𝑢𝑠 2.3 × 𝜋 × 12 → 𝑙0 = 𝐿0,𝑎𝑛 = 𝐿0,𝑎𝑛 456.532 = = 38.044 × 𝐷 𝑑𝑠 12 𝑅𝑠 𝐴𝑠 350 × 78.5 = = 380.242 𝑚𝑚 𝑅𝑏𝑜𝑛𝑑 𝑢𝑠 2.3 × 𝜋 × 10 → 𝑙0 = 𝐿0,𝑎𝑛 380.242 = = 38.024 × 𝐷 𝑑𝑠 10 Required calculated anchor length of rebar: 𝐿𝑎𝑛 = 𝛼 × 𝐿0,𝑎𝑛 𝐴𝑠,𝑐𝑎𝑙 𝐴𝑠,𝑒𝑓 Where: 𝐿0,𝑎𝑛 : is basis length anchor rebar 𝐴𝑠,𝑐𝑎𝑙 , 𝐴𝑠,𝑒𝑓 : is the cross-sectional area of the reinforcement according to calculation and actuality, respectively 253 𝛼: is the coefficient, taking into account the influence of the stress state of the concrete and the reinforcement and the influence of the solution for the structure of the anchor area of the member on the anchorage length Tensile rebar: 𝛼 = and 𝐴𝑠,𝑐𝑎𝑙 /𝐴𝑠,𝑒𝑓 = →{ 𝐿𝑎𝑛 = 38.030 × 𝐷 → 𝐶ℎ𝑜𝑜𝑠𝑒 50𝐷 𝐿𝑎𝑛 = 32.473 × 𝐷 → 𝐶ℎ𝑜𝑜𝑠𝑒 50𝐷 So, the length of anchoring rebar are: 𝜙12 = 1100 𝑚𝑚 𝜙10 = 1255 𝑚𝑚 9.8.4 Reinforced stirrup for beam: Theory basis: According to TCVN 5574:2018, section 10.4.12 and “Tài liệu hướng dẫn đồ án môn học bê tông cốt thép” of Dr Nguyễn Văn Hiệp Figure 9.4 Arrangement of reinforcement in the area where beams intersect Calculation: Choose beam B40 to calculate reinforced stirrup, based on the diagram of shear force, reaction at the position of the secondary beam at the intersection of the main beam, we have: 𝑃 = 105.111 𝑘𝑁 Required stirrup area: 𝐴= 𝑃 105.111 × 10 = = 4.0427 𝑐𝑚2 𝑅𝑠 260 254 Number of stirrup: Choose 𝜙8, branches: 𝑛= 𝐴 4.0427 = = 4.018 → 𝐶ℎ𝑜𝑜𝑠𝑒 𝑛 = × 0.503 × 0.503 Distance between stirrups: 𝑠= 700 − 500 = 6.667 → 𝑐ℎ𝑜𝑜𝑠𝑒 𝑠 = 7𝑐𝑚 × 10 255 REFERENCE DOCUMENTS Vietnamese: [1] Thầy Võ Bá Tầm (2012), “Kết Cấu Bê Tông Cốt Thép – Tập 1”, Nhà xuất ĐHQG TP Hồ Chí Minh, 392 trang [2] Thầy Võ Bá Tầm (2007), “Kết Cấu Bê Tông Cốt Thép – Tập 2”, Nhà xuất ĐHQG TP Hồ Chí Minh, 470 trang [3] Thầy Võ Bá Tầm (2005), “Kết Cấu Bê Tông Cốt Thép – Tập 3”, Nhà xuất ĐHQG TP Hồ Chí Minh, 328 trang [4] GS Nguyễn Đình Cống (2006), “Kết Cấu Bê Tông Cốt Thép – Tập 2”, Nhà xuất xây dựng Hà Nội, 199 trang [5] Thầy Phan Quang Minh, Thầy Ngơ Thế Phong, Thầy Nguyễn Đình Cống (2006), “Kết Cấu Bê Tông Cốt Thép”, Nhà xuất Khoa học Kỹ thuật Hà Nội, 395 trang [6] Thầy Võ Bá Tầm (2012), “Nhà Cao Tầng Bê Tông Cốt Thép”, Nhà xuất ĐHQG TP Hồ Chí Minh, 247 trang [7] TCVN 2737-1995, “Tải Trọng Tác Động”, 56 trang [8] TCVN 5574-2018, “Thiết Kế Kết Cấu Bê Tông Bê Tông Cốt Thép”, Viện Khoa học Công nghệ Xây dựng biên soạn, Bộ Xây dựng đề nghị, Tổng cục Tiêu chuẩn Đo lường Chất lượng thẩm định, Bộ Khoa học Công nghệ công bố, 195 trang [9] TCVN 33-2006, “Cấp Nước-Mạng Lưới Đường Ống Cơng Trình”, Bộ Xây Dựng, 158 trang [10] TCXD 198-1997, Nhà Cao Tầng – Kết Cấu Bê Tông Cốt Thép Tồn Khối, 17 trang [11] TCVN 9386-2012, “Thiết Kế Cơng Trình Chịu Động Đất”, Viện Khoa học Cơng nghệ Xây dựng - Bộ Xây dựng biên soạn, Bộ Xây dựng đề nghị, Tổng cục Tiêu chuẩn Đo lường Chất lượng thẩm định, Bộ Khoa học Công nghệ công bố, 230 trang [12] TCVN 10304-2014, “Móng Cọc – Tiêu Chuẩn Thiết Kế”, trường Đại học Xây dựng biên soạn, Bộ Xây dựng đề nghị, Tổng cục Tiêu chuẩn Đo lường Chất lượng thẩm định, Bộ Khoa học Công nghệ công bố, 86 trang [13] TCVN 9362-2012, “Tiêu Chuẩn Thiết Nền Nhà Cơng Trình”, Viện Khoa học Công nghệ Xây dựng - Bộ Xây dựng biên soạn, Bộ Xây dựng đề nghị, Tổng cục Tiêu chuẩn Đo Iường Chất ượng thẩm định, Bộ Khoa học Cơng nghệ cơng bố, 58 trang [14] TCXD 205-1998, “Móng Cọc – Tiêu Chuẩn Thiết Kế”, 66 trang 256 [15] TCXDVN 375-2006, “Thiết Kế Cơng Trình Chịu Động Đất”, Bộ Xây Dựng, 340 trang English: [16] ACI Standard (318-19), “Building Code Requirements for Structural Concrete”, 628 pages 257 S K L 0