Waffle slab structure design

CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN CONTENTS CHAPTER 1: OVERVIEW OF BUILDING ARCHITECTURE 1.1 Demand for construction building: 10 1.2 Location of construction of works: 10 1.3 ARCHITECTURAL SOLUTIONS: 11 1.3.1 Site plan: 11 1.3.2 Front elevation: 12 1.3.3 Traffic system: .13 1.4 TECHNICAL SOLUTION: 13 1.4.1 Electrical system: 13 1.4.2 Water system: 13 1.4.3 Ventilation: 13 1.4.4 Lightening: .13 1.4.5 Fire escape: 14 1.4.6 Anti thunder: 14 1.4.7 Garbage system: .14 CHAPTER 2: OVERVIEW OF HIGH-BUILDING STRUCTURE DESIGN 15 2.1 CHOICE STRUCTURAL SOLUTIONS: 15 2.1.1 Vertical bearing structure system .15 2.1.2 Horizontal bearing structure system 15 2.1.3 Conclusion 18 2.2 MATERIAL CHOICE: 19 2.2.1 Concrete 19 2.2.2 Rebar 19 2.3 SHAFE OF BUILDING .20 2.3.1 Horizontally 20 2.3.2 Vertically 21 2.4 PRELIMINARY SECTION FOR ALL ELEMENT OF BUILDING 21 2.4.1 Preliminary section of slabs and beam .21 2.4.2 Preliminary section of column 22 2.4.3 Preliminary section of shear wall 25 2.4.4 Preliminary section of stair .26 2.5 CALCULATION OF HIGH-RISE BUILDING STRUCTURE 27 2.5.1 Structural diagram 27 STUDENT: VO QUOC TIN ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN 2.5.2 Theories used to calculate high-rise buildings: .27 2.5.3 Calculating method to determine internal force .27 2.5.4 Content calculation 28 CHAPTER 3: ASSIGN LOADS APPLIED ON BUILDING 29 3.1 TYPES OF LOADS ON STRUCTURAL AND BUILDING: 29 3.2 DEAD LOADS (DL) 29 3.3 LIVE LOADS (LL) 31 3.4 WIND LOADS (WL) 31 3.4.1 Static wind loads 31 3.4.2 Dynamic wind loads 32 3.5 EARTHQUAKE LOADS 38 3.5.1 Calculation combination 38 3.5.2 Reaction spectrum method .39 3.6 ASSIGN COMBINATION LOADS 43 3.6.1 Type loads pattern 43 3.6.2 Combination loads 44 CHAPTER 4: VERIFY STABILITY OF BUILDING 47 4.1 VERYFY DISPLACEMENT OF TOP BUILDING 47 4.1.2 Top floor displacement 47 CHAPTER 5: DESIGN SLABS FOR TYPICAL FLOOR 49 5.1 MODELING WAFFLE SLABS IN SAFE .49 5.2 TAKE DATA FROM SAFE SOFTWARE 51 5.3 Design reinforcement for waffle slabs 56 5.3.1 Theoretical calculation 56 5.3.2 Design reinforcement and arrangement rebar in waffle slabs 57 5.4 Verify deflection of waffle slabs 62 CHAPTER 6: DESIGN STAIRCASE FOR TYPICAL FLOOR 63 6.1 THE GEOMETRIC AND SIZE OF STAIR 63 6.2 Materials .64 6.3 Loading 64 6.3.1 Dead load .64 6.3.2 Live loads .65 6.3.3 Chose structural diagram and find internal force by etabs .65 6.3.4 Design main reinforcement for stair 66 STUDENT: VO QUOC TIN ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN 6.3.5 Design beam landing .68 CHAPTER 7: DESIGN FRAME AXIAL C 71 7.1 DESIGN BEAM AXIAL C 71 7.1.2 Theoretical basic 72 7.1.3 Calculation process of C-frame beams: 72 7.1.4 Design main reinforcement for beam 73 7.1.5 Design stirrup for beam 76 7.2 DESIGN COLUMN AXIAL C 77 7.2.1 Theoretical basis: 77 7.2.3 Cases of calculation for column .79 7.2.4 Design main reinforcement for column of frame axial C 83 7.2.5 Design stirrup for column 85 7.3 DESIGN SHEAR WALL OF FRAME AXIAL C 86 7.3.1 Theoretical basic 86 7.3.2 Hypothesis method of moment-bearing area 86 7.3.3 Steps to calculate and design the reinforcement of the wall 87 7.3.4 Design reinforcement for shear wall 90 7.3.5 Design stirrup for shear wall 92 CHAPTER 8: DESIGN FOUDATION FOR BUILDING 93 8.1 THEORETICAL BASIC .93 8.2 CONSTRUCTION GEOGRAPHY: 94 8.3 BORED PILE METHOD: 95 8.4 STRUCTURE AND PILE SIZE 96 8.5 CALCULATE BEARING CAPACITY OF BORED PILE 97 8.5.1 Determination of the pile load capacity according to the material strength 97 8.5.2 Determine the load capacity of the pile according to the physical and mechanical criteria of the ground: 97 8.5.3 Pile load capacity according to SPT standard penetration results 99 8.5.4 Design load capacity 101 8.6 STIFFENER OF PILE .102 8.7 DESIGN FOUNDATION FOR M1 103 8.7.1 Preliminary number of pile 103 8.7.2 Verify bearing capacity of singe pile 104 8.7.3 Check the use of piles considering the group effects: 104 STUDENT: VO QUOC TIN ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN 8.7.4 Check foundation stability and settlement of foundation 105 8.7.5 Verify punching shear 108 8.7.6 Design main reinforcement for foundation cap M1 109 8.8 DESIGN FOUNDATION FOR M2 111 8.8.1 Preliminary number of pile 111 8.8.2 Verify bearing capacity of singe pile 111 8.8.3 Check the use of piles considering the group effects: 112 8.8.4 Check foundation stability and settlement of foundation 112 8.8.5 Verify punching shear 116 8.8.6 Design main reinforcement for foundation cap M1 117 8.9 DESIGN FOUNDATION CORE M3 119 8.9.1 Preliminary number of pile 119 8.9.2 Verify bearing capacity of singe pile 120 8.9.3 Check the use of piles considering the group effects: 120 8.9.4 Check foundation stability and settlement of foundation 121 8.9.5 Verify punching shear 125 8.9.6 Design main reinforcement for foundation cap M1 125 CHAPTER 9: DESIGN WATER TANK 129 9.1 OVERVIEW ABOUT WATER TANK 129 9.2 DATA FOR DESIGN WATER TANK .129 9.2.1 The amount of water needed for the building 129 9.2.2 Preliminary size of water tank .130 9.2.3 Materials 130 9.3 DESIGN TOP SLABS OF WATER TANK 131 9.3.1 Chose structural diagram 131 9.3.2 Loads acting on top slabs of water tank 132 9.3.3 Find internal force 133 9.3.4 Design reinforcement for top slabs 133 9.3.5 Check deflection for top slab of water tank 134 9.3.6 Check cracked for top slab of water tank 137 9.4 CALCULATED WALL OF WATER TANK 137 9.4.1 Chose structural diagram .137 9.4.2 Loads acting on wall of water tank 139 9.4.3 Find internal force 139 STUDENT: VO QUOC TIN ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN 9.4.4 Design reinforcement for wall of water tank .141 9.5 DESIGN BOTTOM SLAB OF WATER TANk .142 9.5.1 Chose structural diagram .142 9.5.2 Load acting on bottom slabs of water tank 143 9.5.3 Find internal force 144 9.5.4 Design reinforcement for bottom slab of water tank 144 9.5.5 Check deflection for bottom slab of water tank 145 9.5.6 Check cracked for bottom slab of water tank 148 9.6 DESIGN BEAM OF WATER TANK .148 9.6.1 Loads acting on beam of water tank 148 9.6.1.5 Design beam DN3 and DD3 : 152 9.6.2 Design reinforcement for beam of water tank .152 9.6.3 Design stirrup for beam of top slab: 153 9.6.4 Calculated stirrup for beam of bottom slab: 155 9.6.5 Design hang reinforcement: 158 9.7 DESIGN COLUMN OF WATER TANK 159 STUDENT: VO QUOC TIN ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN CHAPTER 1: OVERVIEW OF BUILDING ARCHITECTURE 1.1 DEMAND FOR CONSTRUCTION BUILDING: - Today, in the process of integration of the country, the growing economy leads to the life of the people is increasingly improved A large part of the people needs to find a place to live with a fresh environment, many utilities and services to help the unemployed require the birth of many high-end apartments In this trend, many companies build high-class apartments to meet the daily needs of the people Tan Tao Apartment is a construction project of this type - With increasing demand for housing and less available land in the city center, highrise apartment building projects in the suburbs are reasonable and encouraged to invest The above-mentioned projects also contribute to creating an urban face if well organized and in harmony with the surrounding landscape and environment - So, the investment in construction of Tan Tao apartment building is consistent with the policy of encouraging investment in Ho Chi Minh City, meeting the urgent needs of people's houses and promoting economic development, complete the system urban infrastructure system 1.2 LOCATION OF CONSTRUCTION OF WORKS: - Address: Quốc Lộ 1A, Phường Tân Tạo A, Quận Bình Tân, TP Hồ Chí Minh + Tan Tao apartment building, located in Bac Luong Beo residential area, is located in Tan Tao A Ward on the front of National Highway 1A Next to Tan Tao Industrial Park and Pou Yen Industrial Park Convenient transportation, arteries of Binh Tan District and West Saigon New Urban Center such as National Road 1A, Ba Hom Road, Road No 7, Provincial Highway 10, Kinh Duong Vuong Street (extended Hung Vuong) connect together Tan Tao Building with District 6, District 12, Tan Phu District, Binh Tan District and Binh Chanh District - Benefit: + Tan Tao Apartment is near to Ba Hom Market, near Binh Tan Primary School, Star High School, Coopmart Supermarket, Big C An Lac Supermarket, Quoc Anh Hospital, Trieu An Hospital + Ensure 15% of the greenery area and green corridor are isolated from National Highway 1A for shade, fresh air, closed environment and utilities STUDENT: VO QUOC TIN 10 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN 1.3 ARCHITECTURAL SOLUTIONS: 1.3.1 Site plan: -Tan Tao Apartment consists of 15 floors including: basement, 13 floors and roof - The project has an area of 38x40m Construction length is 40m, construction width is 38m - Floor area of building - Designed for: block with 96 apartments - Including elevators and staircases - Parking basement - Ground floor layout of trade - services - Walkways, corridors in the apartment are cool and comfortable STUDENT: VO QUOC TIN 11 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN - The height of elevation is selected at the surface elevation on the basement floor, the height of the completed ground level, the elevation on the basement floor, the height of the building 1.3.2 Front elevation: - The building has a vertical shape Construction height is 48.5m - The facade works in harmony with the surrounding landscape - The works using the main materials are Granite stone, water paint, aluminum bar, decorative stainless-steel frame and soundproof safety glass to create harmonious colors, STUDENT: VO QUOC TIN 12 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN 1.3.3 Traffic system: - The horizontal traffic system in the building is the corridor system - The vertical transportation system is stairs and elevators The stairs include stairs on either side of the building and staircase in the middle of the building The elevator consists of elevators located in the middle of the project - The elevator system is designed to be comfortable, convenient and suitable for use in the project on the basement floor, the height of the building 1.4 TECHNICAL SOLUTION: 1.4.1 Electrical system: - The system receives electricity from the common electricity system of the urban area into the project through the engine room From here electricity is transmitted throughout the building through the internal electrical network In addition, when there is a power failure, you can immediately use a backup generator located in the basement to generate electricity for the project 1.4.2 Water system: - The water source is taken from the regional water supply system and led to the water tank in the basement, the roof water tank, by the automatic pumping system of water pumped to each room through the main genome system near the service room - Waste water is pushed into the common drainage system of the area 1.4.3 Ventilation: - The building is not much restricted by the adjacent buildings so it is convenient to catch the wind, the building uses the wind as the natural wind, and the side still uses artificial wind systems (thanks to the air conditioning system) degree) to help the ventilation system for the work is convenient and better 1.4.4 Lightening: -The lighting solution for the building is calculated separately for each functional area based on the necessary illumination and color requirements - Most areas use white light fluorescent lamps and compact energy saving lamps Minimize the use of incandescent light bulbs Particularly in the outer area, use highpressure halogen or sodium lamps STUDENT: VO QUOC TIN 13 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN 1.4.5 Fire escape: - Reinforced concrete works arranged hollow brick walls with soundproof and heat insulation - Along the corridor, there are fire boxes with CO2 tanks - The floors have stairs to ensure people escape when there is a fire incident Besides, on the top of the roof is a large water tank for fire protection 1.4.6 Anti thunder: - The project is using lightning protection needle in the roof and lightning conduction system to the ground 1.4.7 Garbage system: - On the floor, there is a garbage collection room, the garbage that is transferred from these rooms is gathered and brought to the garbage compartment in the basement, from here there will be a department to take garbage out of the building STUDENT: VO QUOC TIN 14 ID STUDENT: 16149334 CAPSTONE PROJECT m  INSTRUCTOR: Dr DAO DUY KIEN  R bh A M ;     2 m ; As  b b o ;   s  b Rbbho b.ho Rs + With : b  1000 ; ho  h  a  150  20  130(mm) + Concrete B30 ; Rb  170(kg / cm2 ) ; Rbt  12(kg / cm2 ) ;  b  + Rebar AI : Rs  Rsc  2250(kg / cm2 ) ; Rsw  1750(kg / cm2 ) + Conditions  m   R  0, 418 ;    R  0,596 + Percentage of reinforcement: 0,05%  min    As   R 0,596.1.170  max  R b b   4,5% b.ho Rs 2250 Table 9.9: Design reinforcement for bottom slabs Slab S1 S2 M(T.m) m  AS (Cal) Choose AS (chose % M1 0.938444 0.0327 0.0332 326.254  8@150 335.103 choose 0.2578 M2 0.524095 0.0182 -2.124378 0.0739 0.0184 0.0769 180.843  8@ 200 755.325 10 @100 251.327 785.398 0.1933 0.6042 M II -1.194893 0.0416 M 0.938444 0.0327 M 0.524095 0.0182 0.0425 417.378 10 @180 436.332 0.3356 0.0332 326.254  8@150 335.103 0.2578 0.0184 180.843  8@ 200 251.327 0.1933 MI 0.0769 0.0425 755.325 10 @100 417.378 10 @180 785.398 436.332 0.6042 0.3356 MI M II -2.124378 0.0739 -1.194893 0.0416 - Distribution reinforcement  8@ 250 9.5.5 Check deflection for bottom slab of water tank - Select the largest slab in the top slab water tank to check deflection - Choose S1(5500x4000) With span are l1  4(m) ; l2  5,5(m) - Cut the floor plot in each direction with b = 1m; consider the deflection of the slab at this time as the deflection of the beam of size b = 1m; h = 0.1m is fix at both ends - When calculating deflection, we use standard load to calculate - The deflection of the slab in two directions is the same so we just need to calculate the deflection in one direction is enough - Deflection limit according to TCXDVN 356-2005 table Page 15: STUDENT: VO QUOC TIN 145 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN - With slab L   6(m)   f   1 L 400  2(cm) 200 200 - Calculate deflection and crack according to formula of TCXDVN 356-2005 and “Sách Tính tốn thực hành cấu kiện bê tơng cốt thép” of GS.TS.Nguyễn Đình Cống - The deflection is calculated as a beam with two fix ends f  M L 16 B - The standard total short-term gross load is all exerted on the slab S1 S1 : q1  qtc  p  486  1540  2026(kg / m2 ) Calculate the slab plan according to the foursided edge M1  908,3166(kg.m) - The standard total long-term gross load is all exerted on the slab S1 : q2  qtc  p  486  1540  2026(kg / m2 ) Calculate the slab plan according to the foursided edge M  908,3166(kg.m) - The quantities characteristic calculation: + Section b  100(cm); h  15(cm); L  400(cm) a  abv  0.5  1.5  0.5.8  1.9(cm); h0  h  a  15  1.9  13.1(cm) + Concrete B30: Rb.ser  220(kg / cm2 ); Rbt ser  18(kg / cm2 ); Eb  325000(kg / cm2 ) + Rebar AI: Rs.ser  2350(kg / cm2 ); Es  2100000(kg / cm2 ) + Steel layout needs checking AI: 8a150; As  3,351(cm2 ); As'  0(cm2 ) - Modify coefficient between concrete and steel + Es 2100000   6, 462(kg / cm2 ) Eb 325000 + Ared  bh   ( As  As' )  100.15  6, 462(2,351  0)  1521,65(cm2 ) + Sred  + x0  bh 100.152   ( As h0  As' a ' )   6, 462(3,351.13,1  0)  11533, 65(cm3 ) 2 Sred 11533,65   7,58(cm) Ared 1521,65 STUDENT: VO QUOC TIN 146 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN b(h  x0 )3 bx03    As (h0  x0 )  As' ( x0  a ' ) 3 100(15  7,58)3 100.7,583    6, 462.3,351.(13,1  7,58)  3  28794,37(cm ) I red  + + Wred  + r0  I red 28794,37   3880, 48(cm3 ) (h  x0 ) 15  7,58 Wred 3880, 48   2,55(cm) Ared 1521, 65 + rpl  r0  2,55(cm) - Rectangular cross section +   1,75  Wpl   Wred  1,75.3880, 48  6790,84(cm3 ) - Concrete B30  sc  400(kg / cm2 ) + M rp   sc As (h0  x0  rpl )   sc As' ( x0  a '  rpl )  400.3,351.(13,1  7,58  2,55)  10817, 784(kg cm) - Moment resistance cracked M cr  Rbt ser Wpl  M rp  18.6790,84 10817,784  111417,31(kg.cm) - Check the crack resistance of components: M1  18740, 2(kg.cm)  M cr  49239,94(kg.cm) M  18740, 2(kg.cm)  M cr  49239,94(kg.cm)  Top slab is not cracked - Calculate the deflection for the slab without cracks - Slab deflection is equal to the total short-term load and the total long-term load applied - Deflection f1 due to the total short-term live load applied q1  1540(kg / m2 ) Calculate the floor plan according to the four-sided edge M1  609, 4248(kg.m) - Coefficient of heavy concrete b1  0,85; b  - Stiffness due to short-term live load applied + B1  b1Eb I red  0,85.325000.28794,37  7,954.109 (kg.cm2 ) + f1  M1 60942, 48 L  4002  0, 077(cm) 16 B1 16 7,954.10 - Deflection f due to long-term live load applied STUDENT: VO QUOC TIN 147 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN q2  qtc  p  486  1540  2026(kg / m2 ) Calculate the floor plan according to the four- sided edge M  908,3166(kg.m) + Coefficient of heavy concrete b1  0,85; b  + Stiffness due to long-term live load applied B2  b1Eb I red 0,85.325000.28794,37   3,977.109 (kg.cm2 ) b 2 f2  M 2 90831, 66 L  4002  0, 23(cm) 16 B2 16 3,977.10 - Total deflection f  f1  f  0,077  0, 23  0,307(cm)   f   2(cm)  It is satisfy 9.5.6 Check cracked for bottom slab of water tank - Checking cracks for slabs by external moment due to the total short-term load and total long-term load applied the crack-proof moment of the structure - Crack-proof moment M cr  Rbt ser Wpl  M rp  18.6790,84 10817,784  111417,31(kg.cm) -Check the crack resistance of components: M1  18740, 2(kg.cm)  M cr  49239,94(kg.cm) M  18740, 2(kg.cm)  M cr  49239,94(kg.cm)  Top slab is not cracked 9.6 DESIGN BEAM OF WATER TANK 9.6.1 Loads acting on beam of water tank 9.6.1.1 Load applied on beam of top slab - The load applied on the short edge is triangular - Load applied to the long edge is trapezoidal a 493,  986, 6(kg / m) - Maximum load q1   2 q - Self-weight of beam: q2  n bh  1,1.2500.0, 2.(0,3  0,1)  110(kg / m) 9.6.1.2 Load applied on beam of bottom slab - The load applied on the short edge is triangular STUDENT: VO QUOC TIN 148 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN - Load applied to the long edge is trapezoidal a 2093, 2  4186, 4(kg / m) - Maximum load q1   2 q - Self-weight of main beam: q2  n bh  1,1.2500.0,3.(0,  0,15)  371, 25(kg / m) -Self-weight of secondary beam: q3  n bh  1,1.2500.0, 2.(0,5  0,15)  192,5(kg / m) - Self-weight wall of water tank: q4  n bh  1,1.2500.0,12.(1,5  0,3)  396(kg / m) q24  q2  q4  371, 25  396  767, 25(kg / m) Figure 9.12: Deal load and live load Figure 9.12: Self-weight STUDENT: VO QUOC TIN 149 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN Figure 9.13: Shear force diagram Figure 9.14: Moment force diagram 9.6.1.3 Design beam DN1 and DD1 Figure 9.15: Shear force diagram STUDENT: VO QUOC TIN 150 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN Figure 9.16: Moment force diagram 9.6.1.4 Design beam DN2 and DD2 Figure 9.17: Shear force diagram Figure 9.18: Moment force diagram STUDENT: VO QUOC TIN 151 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN 9.6.1.5 Design beam DN3 and DD3 : Figure 9.19: Shear force diagram Figure 9.19: Moment force diagram 9.6.2 Design reinforcement for beam of water tank + m   R bh M ;     2 m ; As  b b o ;  m   R ;    R  b Rbbho Rs + With : b  200 ; ho  h  a  300  40  260(mm) for DN1, DN2, DN3 + With : b  200 ; ho  h  a  500  50  450(mm) for DD3 + With : b  300 ; ho  h  a  600  60  540(mm) for DD1, DN2 + Concrete B30 : Rb  170(kg / cm2 ) ; Rbt  12(kg / cm2 ) ;  b  + Rebar AIII   10  layout for main beam: Rs  Rsc  3650(kg / cm2 ) ; Rsw  2900(kg / cm2 ) ; min  0,05%     R  0,393 ;  R  0,541 As   R 0,541.1.170  max  R b b   2,52% bh0 Rs 3650 + Rebar AI   10  stirrup: STUDENT: VO QUOC TIN 152 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN Rs  Rsc  2250(kg / cm2 ) ; Rsw  1750(kg / cm2 ) ;  R  0, 418 ;  R  0,596 min  0,05%    As   R 0,596.1.170  max  R b b   4,5% bh0 Rs 2250 Beam Position M(T.m) DN1 DN2 DN3 DD1 DD2 DD3 m As As (cal)  Choose mm 218 195.34 1.40259 0.0587 0.0606 149.537 Support -0.00185 Span - - - 6.04396 0.2531 0.2973 733.981 Support -13.4379 0.0825 0.0863 680.985 Span 38.37733 0.2357 0.273 2155.102 6.04984 0.0372 0.0379 299.024 Support -0.00762 Span - - 214 307.876 0.5809 214 307.876 0.5809 214 307.876 0.5809 318 763.407 1.4404 218  116 Support -8.66366 0.0532 0.0547 431.923 Span 508.938 0.9603 8.40007 0.3518 0.4556 1124.641 218  2 20 1137.26 2.1458 Support -1.81467 0.076 0.0791 Span % mm Support -3.62505 0.1518 0.1655 408.595 Span (choose) - 710 0.4189 7 20 2199.11 1.2974 218 508.938 0.3003 214 307.876 0.1816 214 307.876 0.3311 24.58004 0.3344 0.4244 1838.329 618  216 1928.94 2.0741 9.6.3 Design stirrup for beam of top slab: - Design stirrup for beam DN1, DN2, DN3: + Maximum shear force in support: Q  3386, 6(kg ) + Tensile strength of concrete: Rbt  12(kg / cm ) + Stirrup AI Strength of stirrup AI : Rsw  1750(kg / cm ) + Rectangular cross-section beams:  f  ; n  + Heavy concrete: b  ; b  0,6 ; b  1,5 - Bearing capacity of shear force of concrete: Q  b3 (1   f  n ) b Rbt bh0  0,6.(1   0).1.12.20.26  3816(kg )  => No need to calculate the stirrup - The stirrup span is chosen according to the detailed: sct STUDENT: VO QUOC TIN 153 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN h 300   150 sct   sct  150(mm) 150 - The stirrup span is calculated according to the earthquake: sdd  100mm - Selecting the span of stirrup: s  min( sct , sdd )  100mm - Check the condition after choosing the stirrup: + We have: Eb  325000(kg / cm2 ) ; Es  2100000(kg / cm2 ) + Asw   0,8  1, 0053(cm ) + b1   0,001 b Rb   0,0001.1.170  0,83 + w1   + Asw Es 1,0053.2100000  1  1,162 bsEb 20.10.325000 Q  0,3b1w1 b Rbbh0  0,3.0,83.1,162.1.170.20.26  26078, 24(kg )  It is satisfy - Stirrup reinforcement is arranged on two ends of the beam: L is  8a100 - Selected according to detail 3h 3.300   225 sct   sct  200(mm) 500 - Stirrup reinforcement is located in middle of the beam L is  8a200 - Check the skewer reinforcement: + We have: qsw  Rsw Asw  1750.1, 0053  175, 929(kg / cm) s + 10 Q  Qwb  b (1   f ) b Rbt bh02 qsw  2.(1  0).1.12.20.262.175,929  15401,59(kg )  It is satisfied No need to arrange skewer reinforcement STUDENT: VO QUOC TIN 154 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN 9.6.4 Calculated stirrup for beam of bottom slab: - Calculated stirrup for beam DD3: + Maximum shear force in support: Q  7453, 6(kg ) + Tensile strength of concrete: Rbt  12(kg / cm ) + Stirrup AI Strength of stirrup AI : Rsw  1750(kg / cm ) + Rectangular cross-section beams:  f  ; n  + Heavy concrete: b  ; b  0,6 ; b  1,5 - Bearing capacity of shear force of concrete: Q  b3 (1   f  n ) b Rbt bh0  0,6.(1   0).1.12.20.45  6552(kg ) => Need to calculate the stirrup - Selecting stirrup  and two branches n  - Conditions of stirrup span s  min( stt , smax , sct , sdd ) - The stirrup span is calculated according to the detailed: stt stt  nRsw 2b (1   f ) b Rbt bh02 Q2 2.1750. 0,82.2.(1  0).1.12.20.452   125,87(cm) 7453, 62  1258, 7(mm) - Maximum span of stirrup: smax smax  b (1   f ) b Rbt bh02 Q 1,5.(1  0).1.12.20.452   99,991(cm)  999,91(mm) 7453, - The stirrup span is chosen according to the detailed: sct h 500   250 sct   sct  150(mm) 150 - The stirrup span is calculated according to the earthquake: STUDENT: VO QUOC TIN 155 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN sdd  100mm - Selecting the span of stirrup: s  min(stt , smax , sct , sdd )  100mm - Check the condition after choosing the stirrup: 2 + Eb  325000(kg / cm ) ; Es  2100000(kg / cm ) ; + Asw   0,8  1, 0053(cm ) + b1   0,001 b Rb   0,0001.1.170  0,83 + w1   Asw Es 1,0053.2100000  1  1,162 bsEb 20.10.325000 Q  0,3b1w1 b Rbbh0  0,3.0,83.1,162.1.170.20.45  44775,85( kg )  It is satisfied - Stirrup reinforcement is arranged on two ends of the beam L is  8a100 - Selected according to detail 3h 3.500   375 sct   sct  200(mm) 500 - Stirrup reinforcement is located in middle of the beam L is  8a200 - Check the skewer reinforcement: qsw  Rsw Asw 1750.1, 0053   175, 929(kg / cm) s 10 Q  Qwb  b (1   f ) b Rbt bh02 qsw  2.(1  0).1.12.20.452.175,929  26444, 2( kg )  It is satisfied No need to arrange skewer reinforcement - Calculated stirrup for beam DD1, DD2: STUDENT: VO QUOC TIN 156 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN + Maximum shear force in support: Q  15168, 6( kg ) + Tensile strength of concrete: Rbt  12(kg / cm ) + Stirrup AI Strength of stirrup AI: Rsw  1750(kg / cm ) + Rectangular cross-section beams:  f  ; n  + Heavy concrete : b  ; b  0,6 ; b  1,5 - Bearing capacity of shear force of concrete: Q  b3 (1   f  n ) b Rbt bh0  0,6.(1   0).1.12.30.54  11772(kg)  => Need to calculate the stirrup - Selecting stirrup  and two branches n  - Condition span of stirrup s  min( stt , smax , sct , sdd ) - The stirrup span is calculated according to the detailed: stt stt  nRsw 2b (1   f ) b Rbt bh02 Q2 2.1750. 0,82.2.(1  0).1.12.30.542   65, 408(cm) 15168, 62  654, 082(mm) - Maximum span of stirrup: smax b (1   f ) b Rbt bh02 1,5.(1  0).1.12.30.542 smax   Q 15168,6  105,74(cm)  1057, 4( mm) - The stirrup span is chosen according to the detailed: sct h 600   200 sct   sct  200(mm) 300 - The stirrup span is calculated according to the earthquake: sdd  100mm - Selecting the span of stirrup: s  min(stt , smax , sct , sdd )  100mm Check the condition after choosing the stirrup: STUDENT: VO QUOC TIN 157 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN 2 + Eb  325000(kg / cm ) ; Es  2100000(kg / cm ) ; + Asw   0,8  1, 0053(cm ) + b1   0,001 b Rb   0,0001.1.170  0,83 + w1   Asw Es 1,0053.2100000  1  1,108 bsEb 30.10.325000 Q  0,3b1w1 b Rbbh0  0,3.0,83.1,108.1.170.30.54  76702,46(kg )  It is satisfied - Stirrup reinforcement is arranged on two ends of the beam L is  8a100 - Selected according to detail 3h 3.600   450 sct   sct  200(mm) 500 - Stirrup reinforcement is located in middle of the beam L is  8a200 - Check the skewer reinforcement: qsw  Rsw Asw 1750.1, 0053   175, 929(kg / cm) s 10 Q  Qwb  b (1   f ) b Rbt bh02 qsw  2.(1  0).1.12.30.542.175,929  38793,7(kg )  It is satisfy No need to arrange skewer reinforcement 9.6.5 Design hang reinforcement: - Arrange hang reinforcement of beam DD1 DD3 transmission in: - The reaction of the beams DD3 applied on DD1 is F  7453, 6(kg ) - We have: hdc  60(cm); a  6(cm); hdp  50(cm); hs  hdc  a  hdp  60   50  4(cm) - Hang reinforcement area: St  hdp  2hs  20  2.4  28(cm) - Hang reinforcement arranged on both sides of the secondary beam Choose hang reinforcement branches: n  2;8; Asw  1,0053(cm ) STUDENT: VO QUOC TIN 158 ID STUDENT: 16149334 CAPSTONE PROJECT INSTRUCTOR: Dr DAO DUY KIEN - Number of hang bars required: hs ) 7453, 6.(1  ) h0 54  3,92(thanh) x  Asw Rsw 1, 0053.1750 F (1  - Therefore, just choose bars and arrange on both sides, the distance between bars is 50mmm 9.7 DESIGN COLUMN OF WATER TANK Table 9.10: Internal force of column Selecting the maximum internal force of column c1 Story Column Load P V2 V3 T M2 M3 Story1 C1 Comb1 -41.55 -7.98 -20.3 0.006 -5.685 -2.261 Story1 C1 Comb1 -41.08 -7.98 -20.3 0.006 14.617 5.721 Story1 C1 Comb1 -41.08 -7.98 -20.3 0.006 14.617 5.721 Story1 C1 Comb1 -41.08 -7.98 -20.3 0.006 14.617 5.721 Story2 C1 Comb1 -8.53 -5.24 -13.45 0.008 -11.847 -4.659 Story2 C1 Comb1 -8.53 -5.24 -13.45 0.008 -11.847 -4.659 Ast (mm2) 1608.495 1608.495 1608.495 1608.495 1608.495 1608.495 s (%) 1.13 1.13 1.13 1.13 1.13 1.13 Table 9.11: Design reinforcement for column N (T) -41.55 -41.08 -41.08 -41.08 -8.53 -8.53 Mx (T.m) -2.261 5.721 5.721 5.721 -4.659 -4.659 My (T.m) -5.685 14.617 14.617 14.617 -11.85 -11.85 Cx Cy a h0 cm cm cm cm 40 40 35 40 40 35 40 40 35 40 40 35 40 40 35 40 40 35 STUDENT: VO QUOC TIN e (cm) 34.062 63.586 63.586 207.85 207.85 63.586 159 Ast mm2 1420 1420 1420 1420 1420 1420 n 8 8 8 16 16 16 16 16 16 ID STUDENT: 16149334 ... 47 CHAPTER 5: DESIGN SLABS FOR TYPICAL FLOOR 49 5.1 MODELING WAFFLE SLABS IN SAFE .49 5.2 TAKE DATA FROM SAFE SOFTWARE 51 5.3 Design reinforcement for waffle slabs ... b Waffle Slabs - Structural detail: A waffle slab or two-way joist slab is a concrete slab made of reinforced concrete with concrete ribs running in two directions on its underside The name waffle. .. ribs Waffle slabs are preferred for spans greater than 40 feet (12 m), as they are much stronger than flat slabs, flat slabs with drop panels, two-way slabs, one-way slabs, and one-way joist slabs