Composite structure design

Capstone project Instructor: Dao Duy Kien CONTENTS CHAPTER 1: OVERVIEW OF CONSTRUCTION ARCHITECTURE 21 I Demand for construction of work: 21 II Location of construction of works: 21 III Architectural solutions: 22 Site plan: 22 Front elevation: 23 Traffic system: 24 IV Technical solution: .24 Electrical system: 24 Water system: 24 Ventilation: 24 Lightening: 24 Fire escape: 24 Anti thunder: 25 Garbage system: 25 CHAPTER 2: INTRODUCTION TO THE COMPOSITE STRUCTURE 26 I Introduction: 26 II Advantages of composite structures: 27 Architecture: 27 Economy: 27 Temperature resistant: 27 Construction: .28 III Compare with other structural options: 28 IV Connection shear resistance: 30 V Structural solutions for constructions: 32 CHAPTER 3: MATERIAL USED FOR STEEL- CONCRETE COMPOSITE STRUCTURES 33 I Standard: 33 II Concrete: 33 Rules of Eurocode and Eurocode 4: 33 STUDENT: TRAN MINH TIEN - 16149333 Capstone project Instructor: Dao Duy Kien Calculated strength of concrete: 34 Elastic module Ecm: 35 Shrinkage of concrete: 36 Coefficient of thermal expansion: 36 Vietnamese regulations - TCVN 5574: 2012: 37 Instant compression strength: 37 Standard compression strength: 38 Calculated compression strength: .38 10 Elastic module: 38 11 Comparison of mechanical properties of concrete according to Eurocode and Vietnamese standards TCCVN 5574: 2012: 38 III Steel: 40 Steel bars: 40 Structure steel: 42 IV Forming the steel structure of composite slab: 44 V Connection: 45 Connection shear resistance: 45 Bolts: 45 CHAPTER 4: PRELIMINARY SECTION 46 I Preliminary select the dimension of composite slab: .46 Design parameters of steel sheet: 46 Characteristics of the composite slab: 46 Material characteristics: 47 Concrete B20: 47 Reinforcement: 47 Safety factor: .47 II Preliminary select the dimension of the hard wall ( vách): 48 CHATER 5: LOAD AND IMPACT LOAD 50 I Applied standard: 50 II Load: 50 Calculation method: 50 STUDENT: TRAN MINH TIEN - 16149333 Capstone project Instructor: Dao Duy Kien Dead load: 51 a Weight of walls on the slab: 51 b Dead load effects on the frame: 51 Live load: 52 a Live load construction 52 b Live load used 53 Wind load: 54 a Static components of wind load: 54 b Dynamic components of wind load: 56 Earthquake load: 62 a Introduction 62 b Calculation method: 63 Combination load: .67 Chapter -Design steel-concrete composite slab 68 I Introduction: 68 Figure 6.3 - Minimum length of slab 70 II The working of the composite slab: 70 Three working types of composite slab: 70 There are three types of connection between steel and concrete: 71 Types of destructive: 72 Brittle and plastic destruction: 72 III Checking the composite floor at the construction stage: .73 Calculation diagram 73 Internal force analysis 73 Ultimate limit state 73 a Load: 73 b Checking bearing capacity: 74 Service limit state 77 IV Checking slab in the composite stage 78 Calculation diagram 78 Load 78 STUDENT: TRAN MINH TIEN - 16149333 Capstone project Instructor: Dao Duy Kien Internal force analysis 78 a Ultimate limit state 79 b Service limit state 92 V Design structure composite steel-concrete slab: 97 Calculated composite steel-concrete slab, floor in axis C-D/3-4 97 Parameter of the composite slab: 97 Calculated load: 99 a Construction stage: 99 b Composite stage 100 Determine external force: 101 Checking slab in construction stage 104 Ultimate limit state 104 Service limit state 105 Check the slab at the composite stage 105 a Ultimate limit state 106 It is satisfied 108 b Service limit state 109 Calculated composite steel-concrete slab, Ground floor in axis C-D/3-4 111 111 Parameter of the composite slab: 111 Steel sheet is supplied from Tata steel with the code is Comflor 46 / 1.2 with the parameters shown below: 111 Calculated load: 113 Determine external force: 115 Checking slab in construction stage 117 It is satisfied 122 c Service limit state 122 VI Calculated reinforcement for composite slab 124 CHAPTER 7: DESIGN OF STAIRCASE 125 I Preliminary select the dimension of the staircase: .125 II Material: 126 STUDENT: TRAN MINH TIEN - 16149333 10 Capstone project Instructor: Dao Duy Kien III Loading: 126 Dead load: 127 Live load .128 Diagram calculation 128 Analysis model by 2D 128 Calculated reinforcement .131 IV Calculated landing beam (dầm chiếu tới): 132 loading 132 Calculate vertical reinforcement: 133 Chapter 8-INTERNAL FORCE AND DEFLECTION 134 I Bearing structure of the building 134 II Internal force of frame 134 III Considering: 144 Axial force: 144 Moment: 144 Shear force: .144 Select the internal force to check the design 144 IV Deflection of structure 145 Figure 8.11-Maximum displacement of building 145 CHAPTER 9- DESIGN OF COMPOSITE BEAM 146 I Applied standard: 146 II Introduction: 146 General: 146 Effective working width of section 146 Classification of cross section 148 Analytical method 150 a Methods of linear elastic analysis 151 b Methods of analysis of hardness - plasticity (elastic - flexible) 152 Check beams in construction stage 152 a Ultimate limit state .152 b Service limit state 153 STUDENT: TRAN MINH TIEN - 16149333 11 Capstone project Instructor: Dao Duy Kien Check beams in construction stage 153 a Ultimate limit state .153 Moment of plastic resistance of cross section under positive moment 154 Moment of plastic resistance of section under negative moment 158 Cross section durability when bearing capacity of shear and moment forces 159 III Connection between steel beam and composite slab 161 Introduction 161 Welded stud has a cap in the composite slab 162 Welding stud arrangement is required 163 Shear resistant connection design for composite secondary beam 165 Shear resistant connection design for composite main beam 167 IV Horizontal reinforcement 167 Service limit state 169 Check deflection 169 The formation of cracks in concrete in negative moment areas 170 Oscillate 171 V The result calculated of secondary beam, span E-F axis 172 The width work on secondary beam 175 Classify section of secondary beam .176 Load 176 Construction stage: 176 Composite stage 177 Checking in construction stage 178 Checking in composite beam .179 Design of connection resistance .180 VI The result calculated of main beam, span 4-5 axis C 185 The width work on main beam 188 Classification of cross section beam 189 Loading 189 Construction stage .189 Composite stage 190 STUDENT: TRAN MINH TIEN - 16149333 12 Capstone project Instructor: Dao Duy Kien Checking in construction stage 191 a Ultimate limit state 191 b Service limit state 192 Checking in composite stage 192 a Ultimate limit state: 192 Connection resistance design .194 Checking layout connection 195 b Service limit state 197 Deflection 197 Cracked concrete 197 Oscillate 198 CHAPTER 10-COMPOSITE COLUMN STEEL-CONCRETE AND WALL 199 I Standard 199 II Introduction .199 Calculation method 200 General method: 200 Simple method: 200 III Check column stability 200 Composite column under axial compression 201 a Bearing capacity of axial compression according to strength conditions 201 b Composite column stability condition 201 Modify slenderness .201 Ultimate force 202 For short-term loads: 202 For long-term load: .202 Bearing capacity of columns under stable conditions 204 Simple conditions test method 205 Check the column for bending resistance 206 a Interactive axis of the main axis: 207 b Interactive axis of the secondary axis 210 Bearing capacity of one way compression bending 213 STUDENT: TRAN MINH TIEN - 16149333 13 Capstone project Instructor: Dao Duy Kien Bearing capacity of two ways compression bending 214 According to standard EN 1994 – – 214 Bearing capacity of shear force .215 IV Calculation result of column C4 group G1 – axis C 216 Material characteristics 216 Internal force of column 216 Geometric characteristics of columns 217 Checking local buckling 218 Checking composite column under axial force 218 Stable conditions .218 Condition of slenderness 218 Bearing capacity under stable conditions of columns 221 Check the column for bending .221 Interaction curve y-y axis 223 Shear strength: 225 a Bearing capacity of bending compression one direction 225 b Bending compression resistance two direction: 226 V Calculated wall in frame axis 227 Theory 227 Steps to calculate vertical reinforcement for walls 228 Calculate the stirrup for hard walls 231 Calculate reinforcement for a specific case 232 CHAPTER 11: DESIGN CONNECTION BETWEEN MAIN BEAM AND SECONDARY BEAM 235 I Standard 235 II Theory of connection hinge calculation (fin plate) 235 Connection structure 235 a General 235 b Check on structure 236 Check the bolt group 238 Check base plate 241 STUDENT: TRAN MINH TIEN - 16149333 14 Capstone project Instructor: Dao Duy Kien Check on the web beam 245 a Bearing capacity of shear force 246 b Bearing capacity of bending and shear force at the same time 247 c Bearing capacity of bending and shear force at the same time of not beveled beams249 Check bearing capacity of the beveled end of the beam .251 Check the stability of the beveled of the beam 253 Check welded line connection base plate and column flange (column web, beam web) 254 Check on stable of support beam 255 a Local shear force 255 b Puncture 256 I Check on structure 256 II Checking on bolts group: 258 Checking on bearing capacity of resistance shear force 258 Checking pressure face resistance on steel plate 259 Checking pressure face resistance on web beam 260 Checking on steel plate .261 Bearing capacity of resistance shear force 261 Checking bearing capacity of bending .261 Checking stability of steel plate 262 Check local buckling of the web 262 Bearing capacity of resistance shear force 262 Checking bearing capacity of the top beam is beveled 263 Checking the welding line connection between the steel plate and main beam 264 10 Checking the stability of support beam 264 Local shear force 264 Shear punctured 265 CHAPTER 12: DESIGN CONNECTION BETWEEN MAIN BEAM AND COLUMN266 I Standard 266 II Introduction about half-hard composite node 266 III Connections characteristics 268 STUDENT: TRAN MINH TIEN - 16149333 15 Capstone project Instructor: Dao Duy Kien General 268 Arrange springs 270 Slenderness of node 272 Hardness modify coefficient 272 Process of connections calculation 273 Connections the frame nodes according to - The strong axis of the inner column and the boundary column 274 Connections the framing node according to - The weak axis of the inner column and the boundary column 286 Requirements on connection beams- columns 294 Regulations on link design 294 10 Principles of half-hard connection composite design 295 IV Result calculation: 296 Result of the composite connection design of C-axis frame .297 Connection between beam B65 and hard wall 310 CHAPTER 13 : DESIGN connection between COLUMN AND BASE column 315 I Standard: 315 Introduction to column connection: 315 Process of connecting columns: 316 Check the bearing capacity of the tensile zone: 317 Bearing capacity of shear force 319 The case of the joint plate is under compression 319 II Design results connecting columns 319 Parameter: 320 Calculating bearing areas .321 Check the section of the base plate 322 Check the thickness of the base plate 322 Check bolts 323 Check welded 323 III Introducing column base connection: 324 Featured connection: 326 STUDENT: TRAN MINH TIEN - 16149333 16 Capstone project Instructor: Dao Duy Kien - Rectangular cross-section beams:  f  ;  n  - Heavy concrete: b  ; b  0, ; b  1,5 - Bearing capacity of shear force of concrete: Q  b (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 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 / cm ) ; Es  2100000(kg / cm ) ; Asw   0,82  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 3h 3.300   225  sct  200(mm) - Selected according to detail sct  4 500 -Stirrup reinforcement is located in middle of the beam STUDENT: TRAN MINH TIEN - 16149333 L is  8a 200 400 Capstone project Instructor: Dao Duy Kien 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 bh02qsw  2.(1  0).1.12.20.26 2.175,929  15401,59(kg )  It is satisfy No need to arrange skewer reinforcement 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, ; b  1,5 - Bearing capacity of shear force of concrete: Q  b (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 branch n  - Conditions of stirrup span s  min(stt , smax , sct , s dd ) - The stirrup span is calculated according to the detailed: stt stt   nRsw 2b (1   f ) b Rbt bh02 Q2 2.1750. 0,8 2.2.(1  0).1.12.20.45  125,87( cm)  1258, 7( mm) 7453, - Maximum span of stirrup: smax STUDENT: TRAN MINH TIEN - 16149333 401 Capstone project smax   Instructor: Dao Duy Kien b (1   f ) b Rbt bh02 Q 1,5.(1  0).1.12.20.45  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: sdd  100mm - Selecting the span of stirrup: s  min(stt , smax , sct , s dd )  100mm  Check the condition after choosing the stirrup: We have: Eb  325000(kg / cm ) ; Es  2100000(kg / cm ) ; Asw   0,82  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 satisfy -Stirrup reinforcement is arranged on two ends of the beam L is  8a100 3h 3.500   375  sct  200(mm) - Selected according to detail sct  4 500 -Stirrup reinforcement is located in middle of the beam L is  8a 200 Check the skewer reinforcement: We have: qsw  Rsw Asw 1750.1, 0053   175,929(kg / cm ) s 10 STUDENT: TRAN MINH TIEN - 16149333 402 Capstone project - Instructor: Dao Duy Kien Q  Qwb  b (1   f ) b Rbt bh02qsw  2.(1  0).1.12.20.45 2.175,929  26444, 2(kg )  It is satisfy No need to arrange skewer reinforcement  Calculated stirrup for beam DD1, DD2 : - 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, ; b  1,5 - Bearing capacity of shear force of concrete: Q  b (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 branch n  - Condition span of stirrup s  min(stt , smax , sct , s dd ) - The stirrup span is calculated according to the detailed: stt stt   nRsw 2b (1   f ) b Rbt bh02 Q2 2.1750. 0,8 2.2.(1  0).1.12.30.54  65, 408( cm)  654, 082( mm) 15168, - Maximum span of stirrup: smax smax b (1   f ) b Rbt bh02 1,5.(1  0).1.12.30.54    105, 74(cm )  1057, 4(mm ) Q 15168, - The stirrup span is chosen according to the detailed: sct STUDENT: TRAN MINH TIEN - 16149333 403 Capstone project Instructor: Dao Duy Kien 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 , s dd )  100mm  Check the condition after choosing the stirrup: We have: Eb  325000(kg / cm ) ; Es  2100000(kg / cm ) ; Asw   0,82  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 satisfy -Stirrup reinforcement is arranged on two ends of the beam L is  8a100 3h 3.600   450  sct  200(mm) - Selected according to detail sct  4 500 -Stirrup reinforcement is located in middle of the beam L is  8a 200 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 bh02qsw  2.(1  0).1.12.30.54 2.175,929  38793, 7(kg )  It is satisfy No need to arrange skewer reinforcement STUDENT: TRAN MINH TIEN - 16149333 404 Capstone project Instructor: Dao Duy Kien Calculated 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 ) hs ) 7453, 6.(1  ) h0 54  3,92(thanh )  Number of hang bars required: x  Asw Rsw 1, 0053.1750 F (1  Therefore, just choose bars and arrange on both sides, the distance between 50mmm VI Calculated column of water tank Internal force SELECTING THE MAXIMUM INTERNAL FORCE OF COLUMN C1 Story Column Load P V2 V3 T M2 STORY1 C1 COMB1 -41.55 -7.98 -20.3 0.006 -5.685 STORY1 C1 COMB1 -41.08 -7.98 -20.3 0.006 14.617 STORY1 C1 COMB1 -41.08 -7.98 -20.3 0.006 14.617 STORY1 C1 COMB1 -41.08 -7.98 -20.3 0.006 14.617 STORY2 C1 COMB1 -8.53 -5.24 -13.45 0.008 -11.847 STORY2 C1 COMB1 -8.53 -5.24 -13.45 0.008 -11.847 bars is M3 -2.261 5.721 5.721 5.721 -4.659 -4.659 Calculation result 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 cm 40 40 40 40 40 40 Cy cm 40 40 40 40 40 40 STUDENT: TRAN MINH TIEN - 16149333 a cm 5 5 5 h0 cm 35 35 35 35 35 35 e (cm) 34.062 63.586 63.586 207.85 207.85 63.586 Ast mm2 1420 1420 1420 1420 1420 1420 n 8 8 8 16 16 16 16 16 16 Ast (mm2) 1608.495 1608.495 1608.495 1608.495 1608.495 1608.495 405 s (%) 1.13 1.13 1.13 1.13 1.13 1.13 Capstone project Instructor: Dao Duy Kien CHAPTER 16: DESIGN WAFFLE SLAB I Parameter: Preliminary of section - Column mesh dimension L1 x L2 (2000x2000)(mm) - Based on the "Handbook for practicing structural engineering" we have: 1 h L   400(mm) 20 20 + Beam: 1 2 1 2  8 b     h           200( mm) 3 3 3 3  3 + Selecting span of beam L=8 m - Preliminary height of waffle slab hb  70( mm ) Material: - Concrete B30 and steel AIII, AI - Concrete B30 : Rb  17( Mpa ); Rbt  1.2(Mpa);  b  - Steel AIII   10 : Rs  Rsc  365( Mpa ); Rsw  290( Mpa ) - Steel AI   10 : Rs  Rsc  225( Mpa ); Rsw  175( Mpa ) II Calculated waffle slab CALCULATION METHOD - Total load applied on slab include dead load and live load - Dead load include: Self-weight of slab,finishing layer and wall on slab Calculation diagram a Calculation - The slab is calculated as a single plot according to the elastic diagram L + If the slab is of the type:  (slab work in short edge direction): L1  Cut in the short direction a strip with a width of 1m, analyze the connection of the two ends to give a diagram of the beam L + If the slab is of the type:  (slab work in two direction also as a four-sided slab) L1  Connection are considered single support when: + Slab located on the wall, assembly h + Slab located on beam d  hb  Connection are considered fix when: STUDENT: TRAN MINH TIEN - 16149333 406 Capstone project Instructor: Dao Duy Kien + Slab located on beam hd 3 hb b Calculation diagram + Dimension of slab (2x2m) => L2 1 L1 + Height of slab : hb  70( mm) + Height of beam: hd  400(mm) hd 400   5,72  => fix connection hb 70  Slab belong to four-side plate and calculation according to connection fix diagram Figure 16.1-Slab diagram - Cut the plate in each direction with width b = 1m, solve with distributed load to find the moment of span and support Determine internal force Figure 16.2-Internal force of plate four-side Calculated reinforcement m   R bh M ;     2 m ; As  b b o ;  m   R ;    R  b Rb bho Rs - Concrete cover: a = 10(mm)  h0  70 10  60  mm - Concrete B30 : Rb  17( Mpa ); Rbt  1.2(Mpa);  b  STUDENT: TRAN MINH TIEN - 16149333 407 Capstone project Instructor: Dao Duy Kien - Steel AIII: Rs  Rsc  365( Mpa ); Rsw  290( Mpa ); R  0,393;  R  0,541 min  0, 05%    As   R 0,541117  max  R b b   2,52% bh0 Rs 365 - Steel AI: Rs  Rsc  225( Mpa ); Rsw  175( Mpa ); R  0, 418;  R  0,596   0, 05%    As   R 0,596   17   max  R b b   4,5% bh0 Rs 225 Figure 16.3-Plan of waffle slab STUDENT: TRAN MINH TIEN - 16149333 408 Capstone project Instructor: Dao Duy Kien Figure 16.4-Trip series in the X direction Table 16.1-The slab moment value in the X direction Moment M(+) M(-)mũ cột M(-) nhịp Value 1.1494 -3.0088 -0.3981 STUDENT: TRAN MINH TIEN - 16149333 Unit kN.m kN.m kN.m Position SA14 SA14 SA14 409 Capstone project Instructor: Dao Duy Kien M = 1.1494(kN.m) m  M 1.1494  106   0.1878  b Rbbho2 17  100  60     2m     0.1878  0.21 As   b Rbbho 0.21 17  100  60  58.68( mm2 ) Rs 365  Choose Ø6a200( As  142(mm ) ) min  0, 05%    As   R 0,541117 58.68   0.978  max  R b b   2,52% bh0 100  60 Rs 365 With M = -3.0088 (kN.m) m  M 3.0088 10   0.49  b Rbbho2 17 100  602     2 m     0.49  0.86 As   b Rbbho 0.86 17 100  60   240.32(mm ) Rs 365  Choose Ø8a200( As  252(mm ) )   0, 05%    As   R 0,541 1 17 240.32   0.4   max  R b b   2,52% bh0 100  60 Rs 365 With M = -0.3981 (kN.m) M 0.3981 106 m    0.065  b Rbbho2 17  100  60     2m     0.065  0.067 As   b Rbbho 0.067  17  100  60  18.7( mm2 ) Rs 365  Chose Ø6a200( As  142(mm ) ) A   R 0,541117 18.7  0.31  max  R b b   2,52%  min  0, 05%    s  bh0 100  60 Rs 365 STUDENT: TRAN MINH TIEN - 16149333 410 Capstone project Instructor: Dao Duy Kien Figure 16.5-Trip series in the Y direction Table 16.2-The slab moment value in the Y direction Moment Value Unit Position M(+) M(-)mũ cột M(-) nhịp 1.7708 -2.3963 -1.4797 kN.m kN.m kN.m SA14 SA14 SA14 STUDENT: TRAN MINH TIEN - 16149333 411 Capstone project Instructor: Dao Duy Kien With M = 1.7708 (kN.m) m  M 1.7708 10   0.29  b Rbbho2 17 100  602     2 m     0.29  0.35 As   b Rbbho 0.35 17 100  60   97.80(mm ) Rs 365  Choose Ø6a200( As  142(mm ) ) min  0, 05%    As   R 0,541117 97.80   1.63  max  R b b   2,52% bh0 100  60 Rs 365 With M = -2.3963 (kN.m) m  M 2.3963  106   0.39  b Rbbho2 17  100  60     2 m     0.39  0.53 As   b Rbbho 0.53  17  100  60   148.11( mm ) Rs 365  Choose Ø8a200( As  252(mm ) )   0, 05%    As   R 0,541 1 17 148.11   2.46   max  R b b   2,52% bh0 100  60 Rs 365 With M = -1.4797 (kN.m) M 1.4797 10 m    0.24  b Rbbho2 17 100  602     2 m     0.24  0.28 As   b Rbbho 0.28 17 100  60   78.25(mm ) Rs 365  Chose Ø6a200( As  142(mm ) A   R 0,541117 78.25 min  0, 05%    s   1.31  max  R b b   2,52%  bh0 100  60 Rs 365 STUDENT: TRAN MINH TIEN - 16149333 412 Capstone project Instructor: Dao Duy Kien Check deflection Figure 16.5-Deflection of slab - Maximum defection of slab: f  0.011( m ) - The allowed displacement is calculated according to Appendix M in table M.1 TCVN 5574-2018 L   0.032(m ) f  250 250 L   0.032(m )  f  0.011(m)  f  250 250  It is satisfy STUDENT: TRAN MINH TIEN - 16149333 413 Capstone project Instructor: Dao Duy Kien REFERENCES - [1] Standard: “EN 1990:2002: Basic of structural design” - [2] Standard: “EN 1991 – – 1: “Action on structures – General actions – Denities, Self-weight, imposed loads for buildings” - [3] Standard: “EN 1992 – 1– 1: “Design of concrete structure – General rules and rules for buildings” - [4] Standard: “EN 1993 – 1– 1: Design of composite steel and concrete structure – General rules and rules for buildings” - [5] Standard: “EN 1994 – 1– 1: “Design of composite steel and concrete structure – General rules and rules for buildings” - [6] Standard: “EN 1998 – 1: Design of structure for earthquake resistance – General rules, seismic actions and rules for buildings” - [7] Standard: “TCVN 5574 : 2012 – Kết cấu bê tông bê tông cốt thép” and standard “TCVN 5575 :2012 – kết cấu thép” - [8] Standard: “TCVN 2737:1995, tải trọng tác động” - [9] Standard: “TCVN 9386:2012 Thiết kế cơng trình chịu động đất” - [10] Standard: “TCVN 7888:2014 Cọc khoan nhồi” SOFTWARE - [11] ETABS VERSION 2017 - [12] AUTOCAD 2019 - [13] EXCELL - [14] WORD STUDENT: TRAN MINH TIEN - 16149333 414 ... Instructor: Dao Duy Kien STRUCTURE CHAPTER 2: INTRODUCTION TO THE COMPOSITE STRUCTURE I Introduction: Structural steel - concrete composite (referred to as composite structure) is a combination... composite structures: Designing a project must meet requirements on bearing capacity, ductility of structure, architecture, economy, construction and heat resistance Architecture: Composite structures... the composite structure in the calculation, students choose to design and apply the steel - concrete combined structure for the project: Include: _Floor: the floor is designed according to the composite