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Đồ án tốt nghiệp ngành kỹ thuật xây dựng dân dụng, được tính toán và trình bày bằng ngôn ngữ: tiếng Anh, có file Auto Cad, Etab, Sap2000 đính kèm. Dành cho sinh viên ngành xây dựng có đề tài tham khảo và hỗ trợ đồ án. chia làm 5 part
GRADUATION THESIS PAGE CHAPTER INSTRUCTOR DESIGN STAIRS FOR FLOORS - 15 1.1 PLAN VIEW AND SECTION VIEW OF STAIR 1.1.1.1 PLAN VIEW AND SECTION VIEW We have plan view and section view of stair floor 2-3 are shown in Table 3.1 Table 3.1 Plan view, section view of stair floor - 1.1.2 Preliminary size The stairs type from floor to floor 15 of this building is 3-step stair case form: Stair case has steps - Stair case has steps - Stair case has steps GRADUATION THESIS PAGE INSTRUCTOR Totally we have 22 steps Size of steps: h 3400 150 mm 22 ; l = 300 mm Preliminary steps thickness by this fomulation: Lo 4700 hb (156 �120) mm 30 �40 30 �40 With L is the design bar in Table 3.1: L o 1300 �2 2100 4700 (mm) o So we choose h b 130 mm HeigLL of beam: hd Lo 4700 470 �365 mm 10 �13 10 �13 Choose hd = 400 mm hd 400 200 �100 mm �4 �4 Choose bd = 200 mm Width of beam: bd 1.2 TYPES OF LOADS : 1.2.1 Load by layers of staircase break: We have load by layers of staircase break is shown in Table 3.2 Granite, thickness-20mm Đáhoa cương, dà y 20 mm Lớ p vữ a ló t, dà y 20mm Reinforcement concrete, thickness-130mm Bả n sà n bêtô ng, dà y 130 mm Plaster, -thickness-30mm Lớ p vữ a xi mă ng, dà y 15mm Mortar lining , thickness-30mm Table 3.2 Staircase break’s layers GRADUATION THESIS PAGE INSTRUCTOR 1.2.1.1 Dead load by layers of staircase break n g s1 � i h i n i We definite dead load by this fomulation : With : i : is the weigLL of i layer ; n i : is the load factor i layer; th th h i : is the thickness of i layer th We have dead load by layers is shown in Table 3.4: Thicknes s (kN/m3 (m) ) Granite 0.02 24 1.2 0.58 Mortar lining 0.03 18 1.3 0.70 Reinforcement 0.13 25 1.1 3.58 0.025 18 1.3 0.59 Load type Layers Dead load Design n load (kN/m2) concrete Plaster Total 5.72 Table 3.4 Dead load by layers of staircase break 1.2.1.2 Live load As “TCVN 2737-2012”: ptc = 3kPa, n = 1.2 Live load: ps n �p tc 1.2 �3 3.6 kN / m 1.2.1.3 Load on the slant slab Dead load is calculate by this famulation: n g � i h tdi n i ' s2 With : kN / m i : is the weigLL of i layer ; n i : is the load factor i layer ; th th h tdi : is the thickness of i layer as slant side th GRADUATION THESIS PAGE INSTRUCTOR Calculation the equivalent thickness of layers: cos α We have : lb l2b h 2b 0.3 0.32 0.152 0.894 Granite and Mortar lining ( use the same method for Granite ) l h b i cos (300 150) �25 �0.894 33.52(mm) h td b lb 300 Calculation the equivalent thickness of steps: h td h b cos 150 �0.894 67.05 (mm) 2 We have dead load by layers on slant slab is shown in Table 3.5: Load Type Layers Thicknes Thickness s equivalent (m) (mm) Design (kN/m3) n load (kN/m2) Dead Granite 0.02 26.82 24 1.2 0.77 load Mortar lining 0.025 33.52 18 1.3 0.96 Steps 0.15 67.05 18 1.2 1.45 Reinforcement 0.13 0.13 25 1.1 3.58 0.025 0.025 18 1.3 0.58 concre Plaster Total Table 3.5 Dead load on slant slab Vertical live load on slab: g s2 ' g s2 7.61 8.51 kN / m cos α 0.894 Note: With weigLL of handrails is 0.30 kN/m 1.2.1.4 Total load Staircase break: 7.61 GRADUATION THESIS PAGE INSTRUCTOR q s1 gs1 ps 5.72 3.6 9.32 (kN/m2) Stair case 1, 2, 3: q s2 g s2 ps Stair case : g lc 0.3 8.51 3.6 12.34 A1 1.3 (kN/m2) q s2 cos 12.34 �0.894 11.03 kN / m 1.3 USING “SAP2000” FOR CALCULATION STAIR CASE , We cut a strip wide 1m , then calculate for the stairs which is shown in Table 3.6, Table 3.7 and Table 3.8: Table 3.6 Distributed load on stairs Table 3.7 Moment chart for stairs GRADUATION THESIS PAGE INSTRUCTOR Table 3.8 Joint reaction forces chart for stairs kN.m / m M sup port 0.4M max 0.4 �6.93 2.78 kN.m / m M span 0.7M max 0.7 �6.63 4.64 Correction load : 1.4 REINFORCEMEN T CALCULATION FOR STAIRS Assume that the distance from the reinforced concrete edge to the center of tensile reinforcement group is: a 25 mm Steel ratio : 0.05% max R b R b 0.604 �0.9 �14.5 �100 2.16% Rs 365 for reinforcement group AIII max R b R b 0.651 �0.9 �14.5 �100 3.78% Rs 225 for reinforcement group AI 1.4.1 Reinforcement calculation for supports in slant slab (Stair case 1, 2) Working heigLL of section : h o h a 130 25 105 (mm) Definite coefficient m : m M 2.78 0.02 b R b bh o 0.9 �14.5 �103 �� 0.1052 � 2 m �0.02 0.02 GRADUATION THESIS PAGE INSTRUCTOR Reinforcement : b R b bh o 0.02 �0.9 �14.5 �1000 �105 122 (mm / m) Rs 225 Astt c Choose: 8a200 , A s 250 (mm / m) Checking steel ratio : Asch 250 �100 0.22% bh o 1000 �115 0.05% 0.22% max 3.78% Eligible 1.4.2 Reinforcement calculation for span in slant slab (Stair case 1, 2) Working heigLL of section : h o h a 130 25 105 (mm) M 4.64 m 0.027 R bh 0.9 � 14.5 � 10 �� 0.105 m b b o Definite coefficient : � 2 m �0.027 0.027 Reinforcement : Astt b R b bh o 0.027 �0.9 �14.5 �1000 �115 180(mm / m) Rs 225 c Choose 8a200 , A s 250 (mm / m) Checking ratio : Asch 250 �100 0.22% bh o 1000 �115 0.05% 0.22% max 3.78% Eligible Horizontal reinforcement we choose 6a250 for supports, 6a200 for span 1.4.3 Calculation reinforcement for stair case 3: �d � � 2.1 � ;A � � 2.35m ; 1.3m � � �= �0.894 � We have stair case is a slab with size : �cos GRADUATION THESIS PAGE INSTRUCTOR d � 2A 2.6 m � cos � �� h d 400 3.1 � h s 130 � � We have: Two-way slab; link between segment and staircase break’s beam is fixed supports, edges which ink with staircase break is q cos 12.34 �0.894 11.03 kN / m joint , others is free joint , have load s2 Follow “sơ đồ – Phụ lục 13 ( Sách BTCT tập – Võ Bá Tầm)” d L2 2.35 m cos We have ; L = A = 1.3 (m) x � � y L1 1.30 � 0.553 � � L 2.35 x � � ry � 0.0021 0.0149 0.2785 0.0366 And : d A q s2dA 12.34 �2.1 �1.3 33.68 kN.m / m cos � M x q x 33.68 �0.0021 0.07 kN.m / m � M y q y 33.68 �0.0149 0.501 kN.m / m � �� M x q x 33.68 �0.2785 9.379 kN.m / m � � r M x q r x 33.68 �0.0366 1.232 kN.m / m � Reinforcement calculation: b �h s We calculate reinforcement as bending component has section with b = 1m, q q s2 cos hs = 0.13 (m) Choose a = 25 (mm) � h 0.13 0.025 0.105(m) Definite : m Checking: M b R b bh 02 GRADUATION THESIS PAGE INSTRUCTOR m � R � 2 m Diện tích cốt thép yêu cầu phạm vi bề rộng b = 1m: b R b bh Rs We have reinforement calculation is shown in Table 3.9: As b Positio M h0 (m (mm2/m m n (kN.m/m) (m) ) ) Span L1 0.07 0.10 0.0005 0.0005 Span L2 0.501 0.10 0.003 0.003 18 Fixed 9.379 0.10 0.07 0.073 444 support Consol 1.232 0.10 0.009 0.009 55 e Table 3.9: Reinforcement calculation for staircase As Achs 6a200 6a200 10a170 6a200 1.5 REINFORCEMEN T CALCULATION FOR STAIRCASE BREAK 1.5.1 Load on beam of staircase break ( using “sap2000”) WeigLL of beam it self: Horizontal segment : g d1 bd (h d h s )n btct 0.2 �(0.4 0.13) �1.1 �25 1.485 (kN / m) Slant segment : g d2 bd (h d h s )n btct 1.485 1.66 (kN / m) cos 0.894 WeigLL of wall : Horizontal segment ( staircase break 1): g t1 b t h t1n t 0.2 �(3.4 �0.15) �1.1 �16.5 8.53(kN / m) Horizontal segment ( staircase break 2) g t3 b t h t2 n t 0.2 �(3.4 (7 �0.15 �0.15) 0.4) �1.1 �16.5 2.72 (kN / m) Slant segment : GRADUATION THESIS PAGE 10 INSTRUCTOR �h h t2 � �2.35 0.75 � g t2 nb t � t1 t 1.1�0.2 �� �16.5 5.62 (kN / m) � � � � � � With : LL1, LL2 – wall’s heigLLs of staircase breaks 1,2 Stairs transmiDEADe to: Staircase break 1, has RC = RD = 12.30 (kN/m) Load of slant slab by staircase : RG = qs2A2 = 12.34 �1.3 = 16.042 (kN/m) Total load on the beam is: Staircase break 1: g d1 gd1 g t1 R C 1.485 8.53 12.30 22.32 kN / m Staircase break 2: g d3 g d1 g t3 R D 1.485 2.72 12.30 16.505 kN / m Slant slab : g d2 g d2 g t R G 1.66 5.62 16.042 23.32 kN / m We have calculation chart of load on staircase breaks are shown in Table 3.10, Table 3.11 and Table 3.12: Table 3.10 - Load chart on staircase break GRADUATION THESIS PAGE 26 Fig 4.9 Dead load on secondary beam axis B,C Fig 4.10 Dead load on main beam axis 3,4 Fig 4.11 Dead load on secondary beam axis 3,4 INSTRUCTOR GRADUATION THESIS PAGE 27 Fig 4.12 Live load on main beam axis B,C Fig 4.13 Live load on secondary beam axis B,C Fig 4.14 Live load on main beam axis 3,4 INSTRUCTOR GRADUATION THESIS PAGE 28 INSTRUCTOR Fig 4.15 Live load on secondary beam axis 3,4 Fig 4.16 Live load by water pressure on side slab transmis to top slab GRADUATION THESIS PAGE 29 INSTRUCTOR Fig 4.17 Live load by water pressure on side slab transmis to boDEADom slab Fig 4.18 Wind X GRADUATION THESIS PAGE 30 Fig 4.19 Wind XX Fig 4.20 Wind Y INSTRUCTOR GRADUATION THESIS PAGE 31 INSTRUCTOR Fig 4.21 Wind YY 2.8.2 Load cases We have load cases is shown in Table 4.7: Case Load cases Combo Case Combo DL + LL DL + 0.9(LL + WIND X) DL + WIND X DL + 0.9(LL + WIND XX) DL + WIND XX DL + 0.9(LL + WIND Y) DL + WIND Y DL + 0.9(LL + WIND YY) DL + WIND YY 10 ENVELOP Table 4.7 Load cases 2.8.3 Results of “SAP 2000” We have all the moments, shears and axial forces is shown from Fig 4.22 to Fig 4.30 GRADUATION THESIS PAGE 32 Fig 4.22 Moment main beam axis B,C Fig 4.23 Moment main secondary beam axis B,C Fig 4.24 Moment main beam axis 3,4 INSTRUCTOR GRADUATION THESIS PAGE 33 Fig 4.25 Moment secondary beam axis 3,4 Fig 4.26 Shear main beam axis B,C Fig 4.27 Shear secondary beam axis B,C INSTRUCTOR GRADUATION THESIS PAGE 34 “ ” Fig 4.28 Shear main beam axis 3,4 Fig 4.29 Shear secondary beam axis 3,4 Fig 4.30 Axial force for calculate column reinforcement INSTRUCTOR GRADUATION THESIS PAGE 35 INSTRUCTOR 2.9 REINFORCEMEN T CALCULATION FOR BEAMS OF WATER TANK 2.9.1 Horizontal reinforcement Top slab main beam (200 x 500): We assume that protective concrete is a = 50 mm Ratio of reinforcement: 0.1% max R b R b 0.604 �0.9 �14.5 2.8% Rs 280 for reinforcement group AII Moment in mid-span M = 42.38 (kNm) Working heigLL: a 50mm � h o h a 500 50 450mm We have m : M 42.38 0.082 R 0.421 b R b bh o 0.9 �14.5 �103 �0.2 �0.452 We have 2 �0.082 0.09 Reinforcement : Astt b R b bh o 0.09 �0.9 �14.5 �200 �450 377 (mm ) Rs 280 c Choose 216, As 402 (mm ) Ratio : Astt 402 �100 0.45% bh o 200 �450 0.1% 0.45% max 2.16% Eligiable Moment at suport M = 27.74 (kNm) Working heigLL : a 50mm � h o h a 500 50 450mm We have m : M 27.74 0.053 R 0.421 b R b bh o2 0.9 �14.5 �103 �0.2 �0.452 GRADUATION THESIS PAGE 36 INSTRUCTOR We have 2 �0.053 0.06 Choose : Astt b R b bh o 0.06 �0.9 �14.5 �200 �350 252(mm ) Rs 280 c Choose 214, A s 308 (mm ) Ratio : A stt 308 �100 0.34% bh o 200 �450 0.1% 0.34% max 2.16% Eligible Use the same method for all beams we have horizontal reinforement of all beams is shown in Table 4.7: tt ch A A Beam M s s αm ξ Choose Ratio (mm) (kN.m) (mm ) (mm2) 314 Msp 461 40.7 0.127 0.137 446 0.66 TB 320 (200 x 450) Mspan 941 91.7 0.287 0.347 869 1.34 314 Msp 461 40.7 0.127 0.137 446 0.66 TB 416 (200 x 450) Mspan 804 83.6 0.261 0.309 774 1.15 14 Msp 308 29.1 0.055 0.057 238 0.34 TB + 216 (200 x 500) Mspan 402 44.9 0.085 0.089 373 0.45 416 Msp 804 61.2 0.116 0.123 776 0.89 BB + 220 (200 x 500) Mspan 628 49.7 0.094 0.099 622 0.70 216 Msp 402 40.7 0.098 0.103 383 0.57 BB 332 (200 x 450) Mspan 199.4 0.478 0.788 2254 2413 1.91 216 Msp 402 40.7 0.097 0.103 383 0.50 BB 28 (200 x 450) Mspan 169.2 0.405 0.565 1615 1847 1.58 320 Msp 941 69.7 0.167 0.184 887 0.70 BB + 320 (200 x 500) Mspan 941 55.1 0.132 0.142 894 0.70 Table 4.7 Horizontal reinforcement of water tank beams 2.9.2 Vertical reinforcement Beam TB1,3 (200 x 500) mm has Qmax = 42.98 kN �62 d sw mm A w 28.3 mm Choose stirrup has b2 2, f 0, n 2, b 0.9 GRADUATION THESIS PAGE 37 INSTRUCTOR Calcutation of design stirrup distance: R sw n..dsw b2 n b R bt bh 02 s tt Q 2max 175 �2 � �62 �2 �0.9 �1.05 �200 �4502 42.98 �10 1640 mm Maximun stirrup: s max b4 n b R bt b.h 02 1.5 �0.9 �1.05 �200 �450 1335 mm Q max 42.98 �103 Structural stirrup for h = 500 mm > 450 mm According to “8.7.6 TCVN 5574-2012”, We have : �h � s ct � ;500mm � 170mm �3 for ¼ span Distance of stirrup : s �min sct ,s tt ,s max 170;1640;1335 Choose for ¼ span Result : 6a170 in ¼ span Olique reinforcement for beam ( destroyed 450 degree): Q Q b Qsw b2 (1 f n )R bt bh h0 Asw R sw s With retangle shape f , the beam has axial force n qsw R sw A sw 175 �2 � �6 65.96(N / mm) s �170 h0 A sw R sw s �1.05 �200 �450 450 �65.96 �103 181kN Q b2 (1 f n )R bt bh � Q max 42.98(kN) Q 232(kN) => Enough to resist axial force no need arrange olique reinforcement Stirrup calculation for ½ span Q = 33.22 (kN.m) Axial force resistance of beam GRADUATION THESIS PAGE 38 INSTRUCTOR Q0 b3 (1 n )R bt �b �h 0.6 �1.2 �200 �450 �103 64.8(kN) Q max 33.22kN � Enough to resist axial force so we arrange structural stirrup �3h � sct � ;500mm � 375mm �4 for ½ span 6a300 Result: for ẵ span We have stirrup for ẳ span and ½ is shown in Table 4.8 and Table 4.9: L/4 Beam TB TB TB 4,6 BB 1,3 BB BB BB 4,6 Beam TB TB TB 4,6 BB 1,3 BB BB BB 4,6 Qmax (kN) 70.76 69.53 44.64 85.28 142.36 137.19 85.79 sDEAD (mm) 366 379 1520 624 118 127 621 sct (mm) “300” 300 375 375 337.5 337.5 375 smax (mm) 490 499 1286 1009 318 330 1003 sct (mm) 150 150 170 170 150 150 170 Choose 6a150 6a150 6a170 6a170 8a150 8a150 6a170 Table 4.8 Stirrup for ¼ span Sc (mm) “300” “300” “300” “300” “300” “300” “300” Table 4.9 Stirrup for ½ span [Q] (kN) 170 170 181 313 313 194 194 Choose 6a300 6a300 6a300 6a300 8a300 8a300 6a300 Intersection reinforcement Intersection of beam TB1 (3) and TB5 We have shear force between TB1 (3) and TB5 is shown in Fig 4.31: Qmax ( too much for arrangement space) so we need calculate bentup bar : Area bent-up bar in one side : F 2R sw sin We arrange for each one side a stirrup 6a50 Bent-up reinforcement : Asw A sw F 2n d f d R s 69.53 �103 �� 28.3 �175 2.4 cm 2R sw sin 45� �175 �sin 45� so we choose 214 GRADUATION THESIS PAGE 40 INSTRUCTOR We use same method for other intersections ( beam TB2 vs TB4(6), beam BB2 vs BB4(6), beam BB5 vs BB1(3) ) is shown in Table 4.10: Intersection TB2-TB4(6) BB2-BB4(6) BB5-BB1(3) hmb hsb F (mm) 500 (mm) 450 (kN) 70.67 ds asw (mm) (mm ) 28.3 m 500 450 142.36 50.3 500 450 137.19 50.3 Table 4.10 Intersection beam reinforcement Asw c (mm2) 245 (mm) 504 483 214 128 125 ... 0.03 (kN.m/m) M1 All M2 MI MII 1 .29 1. 12 2.98 2. 60 Design reinforcement Asds Ø a (mm2/m) (mm) (mm) 77 75 Asc (mm2/m ) Ratio bt (% ) 20 0 141 0.19 20 0 141 0 .21 181 20 0 25 0 0.33 157 180 157 0 .21 ... Layers s (m) γ (kN/m 3) Standard load n `(kN/m 2) Design load (kN/m 2) Cemaric 0.01 20 0 .20 1.1 0 .22 Plaster 0.03 18 0.54 1.3 0.70 0.10 25 2. 5 1.1 2. 75 Water prooffing - - 0. 02 1.3 0. 026 Mortar... Layers s (m) Dead load γ (kN/m 3) Standard load `(kN/m 2) Design n load (kN/m 2) Cemaric 0.01 20 0 .2 1.1 0 .22 Plaster 0.03 18 0.54 1.3 0.70 Water prooffing - - 0. 02 1.3 0. 026 Reinforcement 0.10 25 2. 5