Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham A) DATA I.> Initial Data Spans L1 = 10 (m) L2 = 4.6 (m) - Column spacing B1 =: 6 (m) B2 = 6 (m) - Hight of floor: Ht = 3.3 (m) - Design frame: Frame 4th - Minh Hóa - Quảng Bình -> Zone of wind I A -> W0 = 65 (daN/m2) - Wall be built by perforated, thickness 100 mm put on exterior beam of construction : "Ƴ1 = 180 (daN/m2) - Assume the Gypsum partition tile put on beam: 35 (daN/m2) "Ƴ2 = - Live load of office: pc = 2 (kN/m2) - Live Load of corridor : pc = 3 (kN/m2) - Live Roof : pc = 0.75 (kN/m2) - Concrete Roof-Slab have sealing and insulation coat . - Grade of steel: CCT34 -> f = 21 (daN/mm2) - Type of Welding stick: N42 - Grade of bolt 5.8 B) CACULATING AND PROCESSING OF DATA I.> Determine the beam gird: - Design frame 4th -> We have the plan of construction Fig I.1 : Fig I.1 : The Plan construction and Beam gird system II.> Determine the thickness, self-weight of slab and loading. - Dimension of slab 2x6 (m) - The thickness of slab be detemined follow fomular: h= × ≥ℎ = 5 ( ) 1.4 × 2 = 0.07 = 7 ( 40 Choose:Thickness of slab 8 (cm) = ⇔ h= Student: Thanh Nguyen Ngo - 172216544 )≥ℎ = 5 ( ) 80 (mm) Page:.. Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham - Determine the Dead Load of slab: Table II.1 Dead Load of Slab Load Types of loading No. (daN/m2) Layer of ceramic tile, t = 8 16 1 mm Layer of mortar, t = 15 mm 30 2 2000x0.015 3 The concrete slab, t = 80 mm 2500x0.08 The concrete slab, t = 80 mm 2500x0.08 Factored Load (daN/m2) 1.1 17.6 1.3 39 1.1 228.8 -> gs 285.4 (daN/m2) Factor of Safety n Factored Load (daN/m2) 1.3 52 1.3 26 1.1 228.8 -> gs 306.8 (daN/m2) 208 - Determine the Dead Load of roof slab: Table II.2 Dead Load of Slab Load Types of loading No. (daN/m2) Layer of the sealing, t =20 mm 40 1 2000x0.02 Layer of the insulation 10 20 2 mm 3 Factor of Safety n 208 III.> Determine the preminary dimensions of beam and girder: 1.> Determine the dimension of beam - Calculating model: Fig III.1.1: Caculating and Internal Force Model - Determine the Loading and Internal force: Factor loads: 1171 (daN/m) = + ×2= = 11.71 (daN/cm) Student: Thanh Nguyen Ngo - 172216544 Page:.. Project: Structural Steel Design Factored loads: = × Instructor: Ms.c Viet Hieu Pham + × ×2= = 1355.46 13.5546 (daN/m) (daN/cm) × = = 715609 (daN.cm) 8 × = = 4403.8 (daN) 2 = = 340.77 (cm3) × From Wx = 340 (cm3) seaching table of I.6 appendix I [2], Use I-Shape , I 27: Fig III.1.2 Dimension of beam Wx = 371 (cm3) A = 40.2 (cm2) Ix = 5010 (cm4) b = 12.5 (cm) h = 27 (cm) d = 0.6 (cm) t = 0.98 (cm) gc = 31.5 (kN/cm) S = 210 (cm3) 2.> Determine the dimension of girder - Choose preminary dimension of girder to calculate load act to frame; h= 50 (cm) Fig III.2.1 The model of transverce frame Student: Thanh Nguyen Ngo - 172216544 Page:.. Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham IV.> Determine the loading act to frame: Fig IV.1: Model of the loading transefer 1.> Determine Dead Load 1.1> The Distribution Dead Load: - The self-weight of gypsum partition tile with the hight of girder h = 50 (cm) Hv = Ht- Hdc = 2.8 m ->gv = 107.8 (daN/m) - The self-weight of girder: Asumme the self-weight of girder is g = 1.5 Kn/m = 150 daN/m -> gdc =157.5 (daN/m) 1.2> Consentated Dead Load Fig IV.1.1 The model of charging Load Table IV.1.1 : Caculate the concentrated Load GA = GD Types of load No. The self-weight of beams have gc = 37.1 (daN/m) 1 -> (37.1(daN/m)x6/2)x2(m) The self-weight of wall that put on exterior beam : 3.3(m) -0.5(m) = 2.8 (m) 2 -> (180(daN/m2)x2.8(m)x6/2(m))x2 The self-weight of slab with L = 6(m) 3 -> (285.4(daN/m)x(6/2)x(2/2))x2(m) GA = Table IV.1.1a Student: Thanh Nguyen Ngo - 172216544 Factord Load (daN) 219.6 3024 1712.4 4956 Page:.. Project: Structural Steel Design No. 1 2 No. 1 2 3 Instructor: Ms.c Viet Hieu Pham GB = GC Types of load Factord Load (daN) The self-weight of beams have gc = 37.1 (daN/m) 219.6 -> (37.1(daN/m)x6/2)x2(m) The self-weight of slab with L = 6(m) 3681.6 -> gs.(6/2(m))x(2/2)+gs.(6/2)x(2.3/2) 3901.2 GB = Table IV.1.1b GBC > GAB Types of load Factord Load (daN) The self-weight of beams have gc = 37.1 (daN/m) 219.6 -> (37.1(daN/m)x6/2)x2(m) The self-weight of gypsum partition tile with the hight of girder : 107.80 3.3(m) -0.5(m) = 2.8 (m) -> (35(daN/m2)x2.8(m)x6./2(m))x2 The self-weight of slab with L = 6(m) 3938.52 -> gs.(6/2(m))x(2.3/2)+gs.(6/2)x(2.3/2) 4265.92 GBC = Table IV.1.1c 1.3> Determine the Roof-Dead Load Fig IV.1.2 The model of charging Roof-Load Table IV.1.2: Calculate the concentrated Load GAm =GDm Types of load No. 1 2 The self-weight of beams have gc = 37.1 (daN/m) -> (37.1(daN/m)x6/2)x2(m) The self-weight of slab with L = 6(m) GAm = Table IV.1.2a GBm = GCm Types of load No. 1 2 The self-weight of beams have gc = 37.1 (daN/m) -> (37.1(daN/m)x6/2)x2(m) The self-weight of slab with L = 6(m) -> gsm.(6/2(m))x(2/2)+gsm.(6/2)x(2.3/2) GBm = Table IV.1.2b Student: Thanh Nguyen Ngo - 172216544 Factored Load 219.6 1840.8 2060.4 Factored Load 219.6 3983.52 4203.12 Page:.. Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham GBCm > GABm Số TT 1 2 Loại tải trọng Kết quả (daN) The self-weight of beam have gc = 37.1 (daN/m) -> 37.1(daN/m)x6/2(m) The self-weight of slab with L = 6(m) -> gsm.(6/2(m))x(2.3/2)+gsm.(6/2)x(2.3/2) GBCm = 219.6 4261.44 4481.04 Fig IV.1.3 The model of Dead Load 2.> Determine Live Load act to Frame 2.1> Live Load 1 Table IV.2.1 Calculate Live Load 1 P1 No. 1 Types of load P1 = pc x 6x2/2x1.3 P1 = Factored Load 1560 1560 P2 No. 1 Types of load P2 = pc x 6.x1x1x1.3x2 -> 200(daN/m)x6(m)x1x1x1.3x2 Factored Load 3120 P2 = 3120 P3 No. 1 Types of load P3 = pc x 6x1x2,3/2x1.3x2 -> 300(daN/m)x6(m)x1.15x1/2x1.3x2 Factored Load 2484 P3 = Student: Thanh Nguyen Ngo - 172216544 2484 Page:.. Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham P4 No. 1 Types of load P3 = pc x 6x1x2.3x1.3x2 -> 300(daN/m)x6(m)x2.3x1.3x2 Factored Load 4968 P4 = 9936 2.2> Roof-Live Load 1 Table IV.2.2 Calculate Roof-Live Load 1 P1m No. 1 Types of load P1m = pc x 6x1x1/2x1.3 -> 75(daN/m)x6(m)x1x1/2x1.3 Factored Load 585 P1m = 585 P2m No. 1 Types of load P2m = pc x 6x1xx1.3x2 -> 75(daN/m)x6(m)x1x1x1.3x2 Factored Load 1170 P2m = 1170 Fig IV.2.1 Live Load 1 2.3> Live Load 2 Table IV.2.3 Calculate Live Load 2 P1 No. 1 Types of load P1 = pc x 6x2/2x1.3 -> 200(daN/m)x6(m)x1x1/2x1.3 Factored Load (daN) 1560 P1 = Student: Thanh Nguyen Ngo - 172216544 1560 Page:.. Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham P2 No. 1 Types of load P2 = pc x 6.x1x1x1.3x2 Factored Load (daN) 3120 P2 = 3120 P3 = Factored Load (daN) 2484 2484 P4 = Factored Load (daN) 4968 9936 P3m = Factored Load(daN) 672.75 672.75 P3 1 Types of load P3 = pc x 6x1x2,3/2x1.3x2 1 P4 Types of load P3 = pc x 6x1x2.3x1.3x2 No. No. 2.4> Roof-Live Load 2 Table IV.2.4 Calculate Roof-Live Load 2 P3m No. 1 Types of Load P3m = pc x 6x2,3/2x1.3x2 P4m No. 1 Types of Load P3m = pc x 6x2,3x1.3x2 -> 75(daN/m)x6(m)x2.3x1.3x2 Factored Load(daN) 1345.5 P4m = 1345.5 Fig IV.2.2 Live Load 2 Student: Thanh Nguyen Ngo - 172216544 Page:.. Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham 3.> Determine the Wind Load act to Frame 3.1> Calculating formulas đ= × × × = × × × + 2+ 2 đ= = With W0 = đ × × đ 65 (daN/m2) n= Cđ = 0.8 Ch = + 6+6 = = 2 2 6 (m) 3.2> Calculate Wind Load Table IV.3.1 Calculate Wind Load Wđ Ht Z Floors k (daN/m2) (m) (m) 52.92 1 4.2 4.2 0.848 58.66 2 3.3 7.5 0.94 63.21 3 3.3 10.8 1.013 66.52 4 3.3 14.1 1.066 68.89 5 3.3 17.4 1.104 1.2 0.6 Wh (daN/m2) 39.69 43.99 47.41 49.89 51.67 qđ (daN/m) 317.4912 351.936 379.2672 399.1104 413.3376 qh (daN/m) 238.12 263.95 284.45 299.33 310.00 Fig IV.3.1 Wind Left Student: Thanh Nguyen Ngo - 172216544 Page:.. Project: Structural Steel Design 128 1247 ℎ ℎ87 17ℎ 15 h ℎ251 ậly=μ lx=μ ậ 386 ∑ ∑ Instructor: Ms.c Viet Hieu Pham 66 Fig IV.3.2 Wind Right 10 477 128 86 87 7 5542ả02 99 0 25 ả Student: Thanh Nguyen Ngo - 172216544 Page:.. Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham V. THE COMBINATION OF INTERNAL FORCE Name Column Column Column Column Column Column Column Column Column Column TABLE V.1.1 INTERNAL FORCE OF ELEMENTS INTERNAL FORCE OF ELEMENTS Axial Force Shear Force No. Position N Kn. V K.n 0 Max -626.14 7.58 4.2 Max -626.14 -4.40 1 0 Min -1017.55 -62.63 4.2 Min -1017.55 -53.63 0 Max -498.38 -36.20 3.3 Max -498.38 -47.82 2 0 Min -805.19 -101.35 3.3 Min -805.19 -93.51 0 Max -368.90 -32.67 3.3 Max -368.90 -45.18 3 0 Min -576.45 -90.49 3.3 Min -576.45 -82.14 0 Max -237.63 -32.43 3.3 Max -237.63 -44.26 4 0 Min -363.25 -86.35 3.3 Min -363.25 -81.01 0 Max -104.19 -61.87 3.3 Max -104.19 -67.08 5 0 Min -134.24 -103.11 3.3 Min -134.24 -101.48 0 Max -706.23 56.34 4.2 Max -706.23 56.34 6 0 Min -1345.34 -16.89 4.2 Min -1345.34 -16.89 0 Max -574.18 91.70 3.3 Max -574.18 91.70 7 0 Min -1066.05 11.80 3.3 Min -1066.05 11.80 0 Max -439.86 75.86 3.3 Max -439.86 75.86 8 0 Min -771.92 15.55 3.3 Min -771.92 15.55 0 Max -300.10 67.43 3.3 Max -300.10 67.43 9 0 Min -493.45 20.55 3.3 Min -493.45 20.55 0 Max -155.24 74.14 3.3 Max -155.24 74.14 10 0 Min -199.58 47.90 3.3 Min -199.58 47.90 Student: Thanh Nguyen Ngo- 172216544 Moment Kn.m 37.85 131.52 -112.63 28.73 -76.77 153.42 -168.10 61.86 -64.42 146.57 -138.13 64.04 -70.59 134.55 -136.68 58.16 -88.61 183.57 -148.38 136.40 106.48 17.98 -52.98 -130.17 155.99 -13.38 25.55 -151.32 124.57 -22.06 26.73 -135.05 114.67 -26.90 37.94 -113.76 120.17 -90.28 58.18 -135.74 Page:........ Project: Structural Steel Design Name Column Column Column Column Column Column Column Column Column Column TABLE V.1.2 INTERNAL FORCE OF ELEMENTS INTERNAL FORCE OF ELEMENTS Axial Force Shear Force No. Position N Kn. V K.n 0 Max -706.23 16.89 4.2 Max -706.23 16.89 11 0 Min -1344.81 -56.35 4.2 Min -1344.81 -56.35 0 Max -574.18 -11.80 3.3 Max -574.18 -11.80 12 0 Min -1065.57 -91.72 3.3 Min -1065.57 -91.72 0 Max -439.86 -15.55 3.3 Max -439.86 -15.55 13 0 Min -771.55 -75.89 3.3 Min -771.55 -75.89 0 Max -300.10 -20.55 3.3 Max -300.10 -20.55 14 0 Min -493.19 -67.46 3.3 Min -493.19 -67.46 0 Max -155.24 -47.90 3.3 Max -155.24 -47.90 15 0 Min -199.46 -74.18 3.3 Min -199.46 -74.18 0 Max -626.14 62.63 4.2 Max -626.14 53.64 16 0 Min -980.61 -7.58 4.2 Min -980.61 4.40 0 Max -498.38 101.37 3.3 Max -498.38 93.53 17 0 Min -783.84 36.20 3.3 Min -783.84 47.82 0 Max -368.90 90.52 3.3 Max -368.90 82.17 18 0 Min -555.07 32.67 3.3 Min -555.07 45.18 0 Max -237.63 86.38 3.3 Max -237.63 81.04 19 0 Min -357.45 32.43 3.3 Min -357.45 44.26 0 Max -104.19 103.16 3.3 Max -104.19 101.52 20 0 Min -128.41 61.87 3.3 Min -128.41 67.08 Student: Thanh Nguyen Ngo- 172216544 Instructor: Msc. Viet Hieu Pham Moment Kn.m 52.98 130.16 -106.51 -17.98 -25.55 151.32 -156.04 13.38 -26.79 135.08 -124.57 22.06 -37.94 113.76 -114.72 26.90 -58.25 135.82 -120.17 90.28 112.61 -28.73 -37.85 -131.56 168.12 -61.86 76.77 -153.47 138.16 -64.04 64.42 -146.63 136.73 -58.16 70.59 -134.60 148.38 -136.40 88.66 -183.68 Page:........ Project: Structural Steel Design Name Dầm Dầm Dầm Dầm Dầm Dầm Dầm TABLE V.1.3 INTERNAL FORCE OF ELEMENTS INTERNAL FORCE OF ELEMENTS Axial Force N Shear Force No. Position Kn. V K.n 0 Max 56.76 -77.99 5 Max 56.76 8.23 10 Max 56.76 150.53 21 0 Min 28.25 -147.70 5 Min 28.25 -6.43 10 Min 28.25 79.78 0 Max 1.68 -79.72 5 Max 1.68 6.50 10 Max 1.68 149.46 22 0 Min -18.16 -147.81 5 Min -18.16 -5.49 10 Min -18.16 80.72 0 Max 4.30 -81.50 5 Max 4.30 4.72 10 Max 4.30 149.40 23 0 Min -12.75 -147.85 5 Min -12.75 -3.73 10 Min -12.75 82.49 0 Max 33.27 -83.67 5 Max 33.27 2.65 10 Max 33.27 149.19 24 0 Min 9.66 -148.08 5 Min 9.66 -2.16 10 Min 9.66 84.05 0 Max -67.08 -83.37 5 Max -67.08 3.14 10 Max -67.08 112.13 25 0 Min -101.48 -107.57 5 Min -101.48 1.06 10 Min -101.48 87.37 0 Max 11.71 1.63 2.3 Max 11.71 5.25 2.3 Max 11.71 66.79 4.6 Max 11.71 70.41 26 0 Min 4.30 -70.43 2.3 Min 4.30 -66.81 2.3 Min 4.30 -5.25 4.6 Min 4.30 -1.63 0 Max -0.70 -2.38 2.3 Max -0.70 1.24 2.3 Max -0.70 63.14 4.6 Max -0.70 66.76 27 0 Min -2.32 -66.83 2.3 Min -2.32 -63.21 2.3 Min -2.32 -1.24 4.6 Min -2.32 2.38 Student: Thanh Nguyen Ngo- 172216544 Instructor: Msc. Viet Hieu Pham Moment Kn.m -113.43 168.88 -127.07 -289.39 89.88 -300.73 -126.28 165.42 -134.08 -291.56 85.60 -297.13 -134.64 166.56 -141.58 -282.33 86.93 -288.10 -148.41 164.09 -151.32 -276.99 85.57 -282.72 -136.40 135.18 -157.05 -183.57 99.72 -205.02 12.85 42.45 42.45 12.85 -139.77 -18.02 -18.02 -139.67 9.45 52.00 52.00 9.45 -122.15 -11.99 -11.99 -121.99 Page:........ Project: Structural Steel Design Dầm 28 Dầm 29 Dầm 30 Dầm 31 Dầm 32 Dầm 33 Dầm 34 0 2.3 2.3 4.6 0 2.3 2.3 4.6 0 2.3 2.3 4.6 0 2.3 2.3 4.6 0 2.3 2.3 4.6 0 2.3 2.3 4.6 0 5 10 0 5 10 0 5 10 0 5 10 0 5 10 0 5 10 0 5 10 0 5 10 Instructor: Msc. Viet Hieu Pham Max Max Max Max Min Min Min Min Max Max Max Max Min Min Min Min Max Max Max Max Min Min Min Min Max Max Max Min Min Min Max Max Max Min Min Min Max Max Max Min Min Min Max Max Max Min Min Min Student: Thanh Nguyen Ngo- 172216544 0.06 0.06 0.06 0.06 -2.02 -2.02 -2.02 -2.02 8.35 8.35 8.35 8.35 3.05 3.05 3.05 3.05 -19.19 -19.19 -19.19 -19.19 -27.57 -27.57 -27.57 -27.57 56.76 56.76 56.76 28.27 28.27 28.27 1.68 1.68 1.68 -18.15 -18.15 -18.15 4.30 4.30 4.30 -12.75 -12.75 -12.75 33.27 33.27 33.27 9.67 9.67 9.67 -9.59 -5.97 56.68 60.30 -60.35 -56.73 5.97 9.59 -16.88 -13.25 50.08 53.70 -53.80 -50.18 13.25 16.88 -23.74 -20.12 30.70 34.33 -34.37 -30.75 20.12 23.74 -79.78 6.43 147.71 -150.51 -8.23 77.99 -80.72 5.49 147.81 -149.44 -6.50 79.72 -82.49 3.73 147.87 -149.38 -4.72 81.50 -84.05 2.16 148.08 -149.16 -2.63 83.67 -9.00 50.04 50.04 -9.00 -109.54 -14.25 -14.25 -109.34 -21.49 52.99 52.99 -21.49 -91.05 -9.59 -9.59 -90.82 -55.23 11.33 11.33 -55.23 -79.81 -20.36 -20.36 -79.61 -127.07 168.88 -113.43 -300.67 89.88 -289.44 -134.08 165.42 -126.28 -297.05 85.61 -291.63 -141.58 166.57 -134.64 -287.98 86.93 -282.43 -151.32 164.09 -148.41 -282.59 85.58 -277.11 Page:........ Project: Structural Steel Design Name Dầm TABLE V.1.3 INTERNAL FORCE OF ELEMENTS INTERNAL FORCE OF ELEMENTS Axial Force N Shear Force No. Position Kn. V K.n 0 Max -67.08 -87.37 5 Max -67.08 -1.06 10 Max -67.08 107.59 35 0 Min -101.52 -112.10 5 Min -101.52 -3.12 10 Min -101.52 83.37 Student: Thanh Nguyen Ngo- 172216544 Instructor: Msc. Viet Hieu Pham Moment Kn.m -157.05 135.20 -136.40 -204.88 99.72 -183.68 Page:........ SAP2000 SAP2000 v16.0.0 - File:DATHEPNEW - Moment 3-3 Diagram (BAO) - KN, m, C Units 9/24/15 0:16:14 SAP2000 SAP2000 v16.0.0 - File:DATHEPNEW - Axial Force Diagram (BAO) - KN, m, C Units 9/24/15 0:17:08 SAP2000 SAP2000 v16.0.0 - File:DATHEPNEW - Shear Force 2-2 Diagram (BAO) - KN, m, C Units 9/24/15 0:16:49 Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham C> DIMENSION AND CONNECTION DESIGN I.> Design No.1 column 1. The dimension of column design( Uniform Cross-Section ): *From diagram of moment envelope we have: M = 112.63 (kN.m) V= 62.63 (kN) N = 1017.6 (kN) * The height of storey : ht= 4.2 (m) = 420 (cm) *The effective length with Major Axis : 4.2 (m) = 420 (cm) lx=μ×H=1×4.2= *The effective length with Minor Axis: 2.94 (m) = 294 (cm) ly=μ×H=0.7×4.2= * The shape of column is H-Shape( Symmetry) 1 h 1 Based on Required: có l = 420 (cm), Choose h = 48 (cm) ≤ ≤ , 15 10 * The eccentricity and required area: The eccentricity e: = = 0.11 (m) = 11.1 (cm) Grade of steel: CCT34 with f = 21 (kN/cm2) E= 21000 (kN/cm2) = × 1.25 + 2.2 ÷ 2.8 × × ×ℎ 1017.6 11.26 91.85 (cm2) = × 1.25 + 2.8 × = 21 × 1 62.3 *Determine bf, tf and tw: 1 1 ÷ 20 30 *The thickness of the web be choose: 1 1 tw = ÷ ℎ ≥ 0.6 = 60 120 *The thickness of the flange be choose: Required: tf ≥ × = 21 × b= 21 21000 tf ≥ => Choose tf = 1.4 (cm) *The dimension of column be choose: The flange: (1.4x24) cm The web : (1.2x45.2) cm = 24 (cm) 1.2 (cm) 0.66 (cm) = = 1.2 (cm) Fig I.1 Dimension of No.1 column * The area of colum is: A = 121.4 cm2 Check: So Act< A therefore : The area of column is satisfy 2> Calculate index property and check in dimension of column: SVTH: Ngô Thanh Nguyên -172216544 Page: Project: Structural Steel Design A = 121.44 cm2 − 11.4 (cm) = = 2 ×ℎ ×ℎ 45727.7248 (cm4) = −2 = 12 12 × ℎ × 3232.1088 (cm4) = +2 = 12 12 = / = 19.4048 (cm) = / = Instructor: Msc. Viet Hieu Pham 5.15896 (cm) = = 21.6441 < = 120 = = 56.9882 < = 120 With λ à < = 120 → The dimension of column is satisfy with slenderness. ̅ = × = 0.684 ̅ = × = 1.802 Wx =2 Ix/h = 1905.32 (cm3) × = = 0.70549 × * Seaching of apependix table IV.5, with the type of No.5 dimension, We have: With Af/Aw = 0.61947 η= 1.9 − 0.1 − 0.02(6 − ) ̅ = 1.639 So: me =η mx= 1.16 < 20 *The checking condition for general stability inside of the flexuaral plane : = ≤ × × Have ̅ = 0.747 à = 1.41 ả = 0.607 The value of interpolation Check left-side of expression: . 2 ℎụ ụ ó = 13.804 (kN/cm2) = × Check right-side of expression × = 21 × 1 = 21 ( ) The dimension is statisfy with the general stability conditon *The checking condition for general stability outside of the flexuaral plane : According to the flexuaral plane, we have: mx = m= 0.71 With: mx = 0.7055 < 1 = 3.14 × √ = 56.988 < 99 = = = With: 1+ = = 1 = 0.7 0.66941 1.8021 < 2.5, we have equation: = 1 − 0.073 − 5.53 × SVTH: Ngô Thanh Nguyên -172216544 × ̅× ̅= 0.83677 Page: Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham = 14.9586 (kN/cm2) < × = 21 × 1 = 21 ( ) × × The dimension is statisfy with the general stability outside of the flexuaral plane condition *The local stability conditon of dimension be calculated: *With the flange of column: We have: ≤ Left-side of expression = Right-side of expression 8.14286 = 0.36 + 0.1 ̅ × 17.083 = The flange is statisfy with the local stability condition *With the web: ℎ ℎ ≤ Left-side expression ℎ Right-side experssion ℎ = 37.6667 = 0.70549 < 1 = 1.3 + 0.15 × ̅ = 0.74 < 2 × = 43.3318 Local stability is not problem II.> Design No.6 Column 1. The dimension of column design( Uniform Cross-Section ): *From diagram of moment envelope we have: M = 130.17 (kN.m) V= 56.34 (kN) N = 1345.3 (kN) The height of storey 4.2 (m) = 420 (cm) *The effective length with Major Axis : 4.2 (m) = 420 (cm) lx=μ×H=1×4.2= *The effective length with Minor Axis: 2.94 (m) = 294 (cm) ly=μ×H=0.7×4.2= * The shape of column is H-Shape( Symmetry) 1 h 1 ≤ ≤ , Based on Required: có l = 420 (cm), chọn h = 15 10 * The eccentricity and required area: The eccentricity e: = = 50 (cm) 0.10 (m) = 9.68 (cm) Grade of steel CCT34 with: f = 21 (kN/cm2) E= 21000 (kN/cm2) = × × 1.25 + 2.2 ÷ 2.8 × ×ℎ 896.03 13.2 114.8 (cm2) × 1.25 + 2.8 × = 21 × 1 42 *Determine bf, tf and tw: 1 1 Based on Required: b= ÷ = 25 (cm) 20 30 *The thickness of the web be choose: = SVTH: Ngô Thanh Nguyên -172216544 Page: Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham 1 1 ÷ ℎ ≥ 0.6 60 120 tw = 1.2 (cm) = *The thickness of the flange be choose: tf ≥ × = 21 × 21 21000 0.66 (cm) = tf ≥ => Choose tf = 1.4 (cm) *The dimension of column be choose: The flange: (1.4x25) cm The web : (1.2x47.2) cm = 1.2 (cm) Fig II.1 Dimension of No.6 column * The area of colum is: A= 126.6 cm2 Check: So Act< A therefore : The area of column is satisfy 2> Calculate index property and check in dimension of column: A = 126.64 cm2 − = = 11.9 (cm) 2 ×ℎ ×ℎ −2 = 12 12 × ℎ × = +2 = 12 12 = / = 20.2365 (cm) = = / = 51861.1381 (cm4) 3652.63013 (cm4) 5.37053 (cm) = = 20.7546 < = 120 = = 54.7432 < = 120 With λ à < = 120 → The dimension of column is satisfy with slenderness. ̅ = × = 0.656 ̅ = × = 1.731 Wx =2 Ix/h = × = = × 2074.45 (cm3) 0.59067 SVTH: Ngô Thanh Nguyên -172216544 Page: Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham * Seaching of apependix table IV.5, with the type of No.5 dimension, We have: Với Af/Aw = 0.61794 >=1 η= 1.9 − 0.1 − 0.02(6 − ) ̅ = 1.654 Vậy me =η mx= 0.977 < 20 *The checking condition for general stability inside of the flexuaral plane : = ≤ × × Have ̅ = 0.651 à = 0.704 ả = The value of interpolation 0.72 Check left-side of expression: . 2 ℎụ ụ ó = = 14.75 (kN/cm2) × Check right-side of expression: × = 21 × 1 = 21 ( ) The dimension is statisfy with the general stability conditon *The checking condition for general stability outside of the flexuaral plane : According to the flexuaral plane, we have: mx = m= 0.59 With: mx = 0.5907 < 1 = 54.743 = 1 = With: 1+ = = = 3.14 × √ < = = 99 0.7 0.70748 < 1.7311 2.5, we have equation: = 1 − 0.073 − 5.53 × × ̅× ̅= 0.84632 = 17.7424 (kN/cm2) < × = 21 × 1 = 21 ( ) × × The dimension is statisfy with the general stability outside of the flexuaral plane condition *The local stability conditon of dimension be calculated: *With the flange: We have: ≤ Left-side of expression = 8.5 Right-side of = 0.36 + 0.1 ̅ × = 16.8585 expression The flange is statisfy with the local stability condition *With the web: ℎ ℎ ≤ Left-side of expression Right-side of expression ℎ ℎ = 39.3333 = 0.59067 < 1 = 1.3 + 0.15 × ̅ = 0.651 < 2 × = 43.1528 Local stability is not problem III. Design No.21 Girder *From diagram of moment envelope we have: M = 300.73 (kN.m) V = 150.53 (kN) SVTH: Ngô Thanh Nguyên -172216544 Page: Project: Structural Steel Design N= 56.76 (kN) 1>.Choose dimension of the girder: *The height of girder: (h): ~ 6 ÷ 22 , Assume: About economic value: ℎ = 1.15 ÷ 1.2 × Instructor: Msc. Viet Hieu Pham Chọn tw = × = × 0.8 (cm) 50.77 (cm) Choose h = 46 (cm) *Check the thickness of web according to the resistance shear condition: 3 ≥ × = 0.43 (cm) 2 ℎ × × The web is statisfied *The thickness of the flange: × ≥ × × ℎ − 2 ×ℎ 12 ℎ = ℎ − 15 ÷ 20 →ℎ × ≥ 29.97 (cm2) *About detailing: = 10 ÷ 24 ≥ = ≤ 30 ≤ = , Choose tf = 1.2 × = 44.8 (cm) 1.00 (cm) 1 1 ÷ ℎ, 2 5 (cm), 2 ℎ ≥ 180; bf = 23 1 ℎ 10 (cm) Fig.III.1 Dimention of No.21 Girder 2>Check the dimesion of section for girder : *Calculate index property and check in dimension of column: − 11.00 (cm) = = 2 ×ℎ ×ℎ 34610.5973 (cm4) = −2 = 12 12 Wx =2 Ix/h = 1504.81 (cm3) A= 98.80 (cm2) ℎ 618.2 (cm3) = × × = 2 *Check strength condition when M and N simultaneously support to girder : + ≤ × Left-side of 20.56 (kN/cm2) + = expression Right-side of × = 21 × 1 = 21( ) expression Strength condition is not problem *Check the equivalent stress condition when M and N simultaneously support to girder: = +3 ≤ 1.15 × Left-side of expression ℎ 18.6813 = × = ℎ = +3 = × = × = 2.69 19.25 Right-side of expression 1.15 × × = 24.15 SVTH: Ngô Thanh Nguyên -172216544 Page: Project: Structural 1 15 ×Steel × Design The equivalen stress condition is not problem * Check the local stability condition of girder: *The flange ≤ 0.5 × = = Instructor: Msc. Viet Hieu Pham 15.81 9.17 The local stability of flange is not problem * Check the local stability condition of the web when supported by normal stress: ℎ ≤ 5.5 × ℎ = = 173.93 54.5 The local stability of web when supported by normmal stress is not problem * Check the local stability condition of the web when supported by shear stress: ℎ ℎ × × ≤ 3.2 1.72 = The local stability when supported by shear stress is not problem *Check the general stability : ≤ 0.41 + 0.0032 = + 0.73 − 0.016 × ℎ × = 21.78 8.6957 The general stability do not check IV. Design No.6 Girder *From diagram of moment envelope we have: M = 139.77 (kN.m) V= 70.43 (kN) N= 11.71 (kN) 1>.Choose dimension of the girder: *The height of girder: (h): ~ 6 ÷ 22 , Assume: Chọn tw = About economic value: ℎ × = 1.15 ÷ 1.2 × × × = 0.8 (cm) 34.61 (cm) Choose h = 32 (cm) *Check the thickness of web according to the resistance shear condition: 3 ≥ × = 0.30 (cm) 2 ℎ × × The web is statisfied *The thickness of the flange: × ≥ × × ℎ − 2 ×ℎ 12 ℎ = ℎ − 15 ÷ 20 →ℎ 2 21.18 (cm ) × ≥ *About detail: = 10 ÷ 24 ≥ = SVTH: Ngô Thanh Nguyên -172216544 × = 2 ℎ 30.8 (cm) 0.80 (cm) Page: Project: Structural Steel 10 Design 24 ≥ ≤ 30 ≤ = , Choose tf = 1.2 Instructor: Msc. Viet Hieu Pham 1 1 ÷ ℎ, 2 5 (cm), ≥ 180; bf = 20 1 ℎ 10 (cm) Fig.IV.1 Dimention of No.21 Girder 2>Check the dimesion of section for girder : *Calculate index property and check in dimension of column: − = = 9.60 (cm) 2 ×ℎ ×ℎ 13118.3957 (cm4) = −2 = 12 12 Wx =2 Ix/h = 819.9 (cm3) A= 71.68 (cm2) ℎ 369.6 (cm3) = × × = 2 *Check strength condition when M and N simultaneously support to girder : + ≤ × Left-side of + = 17.21 (kN/cm2) expression Right-side of × = 21 × 1 = 21( ) expression Strength condition is not problem *Check the equivalent stress condition when M and N simultaneously support to girder: = +3 ≤ 1.15 × Left-side of expression ℎ = × = 15.449 ℎ = +3 = × = × = 2.48 16.035257 Right-side of expression 1.15 × × = 24.15 The equivalen stress condition is not problem * Check the local stability condition of girder: * The flange: ≤ 0.5 × = = 15.81 8.00 The local stability of the flange is not problem * Check the local stability condition of the web when supported by normal stress: ℎ ≤ 5.5 × ℎ = = 173.93 37 The local stability of the web when supported by normmal stress is not problem * Check the local stability condition of the web when supported by shear stress: SVTH: Ngô Thanh Nguyên -172216544 Page: Project: Structural Steel Design ℎ ℎ × × Instructor: Msc. Viet Hieu Pham ≤ 3.2 = 1.17 The local stability when supported by shear stress is not problem *Check the general stability : ≤ 0.41 + 0.0032 = + 0.73 − 0.016 × × ℎ × = 24.17 10 The general stability do not check V> Beam-Column Connections Design (Beam:No.21 and Column: No1 : * From Table IV.1.3 we choose the most dangerous moment of Beam: No.21 M= 289.4 (kN.m) V= 128.53 (kN) N= 44.244 (kN) 1. Determine Bolts: Use Bold have grade of bolts 5.8 , The diameter of bold is Arrangement bolts into 2 line, with distance of bolds accrording to table I.13 appendix I, [2] *Tensile capacity of a bold: = = 163.2 (kN) With ftb is Tensile strength of a bold ( table I.9 appendix I, [2]) ftb = 20 (kN/cm2) With Abn is the net area of a bolt ( Table I.11, appendix I, [2]) Abn = 8.16 (cm2) * The lateral resistance of the bolt : 0.25 = = 0.7 × 95 × 4.59 × 1 × ×1= 1.7 fub - Tensile strength of hight strength bolts in frictional connection , fub = 0.7*fub fub -The break tension strength ( Table I.12 , appendix I, [2]) fub = 95 (kN/cm2), grade of steel 40Cr Ƴb1 = 1 μ,Ƴb2 Friction ratio and safety ratio. = 0.25, d= 36 (mm) 44.89 (kN) = 1.7 nf -The quatity of frictional face, nf =1 * The tension force were applied the bolt in the farthest: SVTH: Ngô Thanh Nguyên -172216544 Page: Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham ℎ1 289.4 × 10 × 47 44.24 161 (kN) = ± = + = , 2 × (12 + 24 + 36 + 47 ) 10 2∑ℎ So all of bolts were satisfacted * Check condition of the bolt be bearing shear force 128.53 = = 12.85 (kN) < 45 (kN) × = 10 2> Determine the collar * Determine the thickness of collar according to bending codition = 1.1 × = 1.1 × , = 1.1 × + ∑ +ℎ = 1.1 × 12.5 × 161 = 23 + 12.5 21 1.76 (cm) 10 × 161(12 + 24 + 36 + 47) = 37 23 + 47 21 2.03 (cm) Chọn t = 2.1 (cm) 3. Determine the weld be connected between the collar and girder *The total length of welds at outside flange = 2 × 23 − 1 + 2 × 11.5 − 1 = 65 (cm) * The tension force in outside plate according to the applied mement and axial force = 651.253 (kN) ± = ℎ 2 We have: Stell CCT34: fu 34 (kN/cm2) Weld N42 18 (kN/cm2) = = 0.7 1 → × = × ; × = → × = 12.6 (kN/cm2) 571.591 ℎ = = = ∑ × 54 × 12.6 × 1 × ; 0.80 × 0.45 × (cm) Choose the height of weld at the web is hf= 1 (cm) The requirement height of weld be connected the web and collar: ℎ = ∑ × = 128.53 = 2 × (31 − 1) × 12.6 × 1 0.17 (cm) Choose the height of weld at the web is: hf= 0.6 (cm) V> Beam-Column Connections Design (Beam:No.21 and Column: No1 : * From Table IV.1.3 we choose the most dangerous moment of Beam: No.21 M= 139.77 (kN.m) V= 70.43 (kN) N= 11.71 (kN) 1. Determine Bolts: Use Bold have grade of bolts 5.8 , The diameter of bold is Arrangement bolts into 2 line, with distance of bolds accrording to table I.13 appendix I, [2] SVTH: Ngô Thanh Nguyên -172216544 d= 24 (mm) Page: Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham *Tensile capacity of a bold: = = 163.2 (kN) With ftb is Tensile strength of a bold ( table I.9 appendix I, [2]) ftb = 20 (kN/cm2) With Abn is the net area of a bolt ( Table I.11, appendix I, [2]) Abn = 8.16 (cm2) * The lateral resistance of the bolt : 0.25 = = 0.7 × 95 × 4.59 × 1 × ×1= 1.7 fub - Tensile strength of hight strength bolts in frictional connection , fub = 0.7*fub fub -The break tension strength ( Table I.12 , appendix I, [2]) fub = 95 (kN/cm2), grade of steel 40Cr Ƴb1 = 1 μ,Ƴb2 Friction ratio and safety ratio. = 0.25, = 1.7 nf -The quatity of frictional face, nf =1 * The tension force were applied the bolt in the farthest: ℎ1 139.77 × 10 × 32.8 11.71 = ± = + = , 2 × (10 + 20 + 32.8 ) 8 2∑ℎ So all of bolts were satisfacted * Check condition of the bolt be bearing shear force 70.43 7.04 (kN) < 45 (kN) = = × = 10 2> Determine the collar * Determine the thickness of collar according to bending codition = 1.1 × , + 44.89 (kN) = 1.1 × 10 × 151 = 20 + 11.2 21 1.70 (cm) ∑ 11.2 × 151(10 + 20 + 32.8) = 1.1 × = +ℎ 20 + 32.8 21 Chọn t = 2 (cm) 3. Determine the weld be connected between the collar and girder *The total length of welds at outside flange = 1.1 × = 2 × 20 − 1 + 2 × 10.2 − 1 152 (kN) 1.87 (cm) 56.4 (cm) * The tension force in outside plate according to the applied mement and axial force = ± = ℎ 2 We have: Stell CCT34: fu Weld N42 204.307 (kN) 34 18 SVTH: Ngô Thanh Nguyên -172216544 (kN/cm2) (kN/cm2) Page: Project: Structural Steel Design = 0.7 → × = × ; → × = 12.6 ℎ = ∑ Instructor: Msc. Viet Hieu Pham = 1 × = (kN/cm2) 204.307 = = 56.4 × 12.6 × 1 × × ; 0.29 × 0.45 × (cm) Choose the height of weld at the web is hf= 0.5 (cm) The requirement height of weld be connected the web and collar: 70.43 0.09 (cm) ∑ × 56.4 × (31 − 1) × 12.6 × 1 Choose the height of weld at the web is hf= 0.5 (cm) VII> Detail of The base colum: No1 1. Determine the base plate of column From table V.1 we have M= 112.63 (kN.m) V= 62.63 (kN) N= 1017.6 (kN) According to the dimension of column No.1, Choose 8 anchor bolts for base column Determin the dimension of base plate: The width of base plate Bbd = b + 2c1 = 36 (cm) The grade of concrete B20 1.15 (kN.m), = × ×∝= 1.15 × 1.2 × 1 = 1.38 (kN/cm2) , The length of base plate be determined follow the partial press of footing concrete ℎ = = = 2× × ≥ × + , 2× × × + , 2× 6× × × , 46.72 (cm) According to the design condition and distance of the anchor bolts, assume c2=12cm and thickness of base palte ts=1cm. So, determine the length of base plate 74 (cm) =h+2 +2 = Determine the stress reaction of footing concrete under the base plate: 6 = + = 0.725 (kN/cm2) × × 6 = − = 0.04 (kN/cm2) × × With ≤ × , = 0.75 × 1.38 = 1.035 2 Fig. VII.1 Detail of base column *According to the divide space of base plate and interpolated value, defind table: Table VII.1 2 M1 = αbσd Quatity αb Space b2 a2 b2/a2 edge (kN.cm) 0.06 20.764 Space 1 2 edge 9.9 21.2 0.467 SVTH: Ngô Thanh Nguyên -172216544 Page: Project: Structural Steel Design 0.092 Space 2 3 edge 17.5 23.5 0.745 The thickness of base plate be calculated follow the formular = 6× = × Instructor: Msc. Viet Hieu Pham 28.576 2.857 (cm) Choose tbd 3 (cm) 2> Determine the reinforced plate at the base of column No.1 Select the dimension of the reinforced beam at the bas of column No.1 Thicknesstdd = 1 (cm) Width: Bdd = 36 (cm) The height: hdd follow the weld be connected the reinforced beam at base column and column No.1 The applied forced to the reinforced beam follow the stress reaction of footing concrete Ndd = (12 + 11.75) x 36.5 x 0.58 = 495.9 (kN) According the design condition, choose height of weld hf =1 Determine the design length of weld: = = 19.68 (cm) 2×ℎ × × × Select hdd= 22 (cm) 3> Determine the stiffener A The model calculation like is the congxon fixed to the column web qs = 0.48 x (23.5)= 11.71 (kN/cm) × 1793.4 (kN.cm) = = 2 V s = qs x l s = 187.392 (kN) Determine the height of stiffener A ℎ ≥ 6× × × = 22.64 (cm) Select hs= 24 (cm) Check follow the equivalent stress = +3 = 6× ×ℎ +3× ×ℎ = 23.1 (kN/cm2) 1.15 × × = 24.15 (kN/cm2) < 1.15 ⟹ Statisfaction Follwing the design condition, select height of weld hf =1cm Area and Moment of weld Aw = 2 x 1 x (24-1) = 46 (cm) 1 × 24 − 1 176 (cm3) =2× = 6 Check the ability force of weld = + = 11 (kN/cm2) × = 12.6 (kN/cm2) ậ < × ⟹ OK 4> Determine the stiffnener B The width of the load tranfer to stiffener is 1.5ls=1.5x12=18cm qs = 0.77 x (18)= SVTH: Ngô Thanh Nguyên -172216544 13.86 (kN/cm) Page: Project: Structural Steel Design × = = 2 V s = qs x l s = Instructor: Msc. Viet Hieu Pham 997.92 (kN.cm) 221.76 (kN) Select ts = 1cm Determine the height of stiffener B 6× × × ℎ ≥ = 15.41 (cm) Select hs= 22 (cm) Check follow the equivalent stress = +3 = 6× ×ℎ +3× ×ℎ = 21.4 (kN/cm2) 1.15 × × = 24.15 (kN/cm2) ậ < 1.15 ⟹ (Ok) Follwing the design condition, select height of weld hf =1cm Area and Moment of weld Aw = 2 x 1 x (22-1) = 42 (cm) 1 × 22 − 1 147 (cm3) =2× = 6 Check the ability force of weld = + = 8.6 (kN/cm2) × = 12.6 (kN/cm2) ậ < × ⟹ (OK) 5> Calculate the anchor bolt From Table V.1 we have M= 28.73 (kN.m) V= -62.63 (kN) N= -1018 (kN) The length of the compression concrete zone under the base plate 74 (cm) Select the margin of the base plate to center of anchor bolt is 6cm, we have: = 2 − 3 = 12.3333 (cm) = 43.3333 (cm) − − 6 = 3 The total tension force of the anchor bolt: − × -223.31 (kN) = = Select the design anchor botl at the base column No,1 Select anchor bolt d22, As= 3.08 cm2, Table II.2, appendix II {3} Check again the total tension force = + = -508.39 (kN) 2 T2 < T1 The diameter of boltl is OK 6> Determine the weld be connected the column and the base plate The weld at the flange be braced moment and axial force The weld at the web be braced shear force The calculated internal force be followed the envelope internal force The tension force at the column web SVTH: Ngô Thanh Nguyên -172216544 Page: Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham = ± = -448.92 (kN) ℎ 2 The total length of the line connect at column flange (include the weld be conneted the reinforced plate at base column and the base plate 33 − 1 21 − 1 33 − 21 =2× −1 +2× −1 +2× − 1 = 2 2 2 The requirement height of the weld at the column flange ℎ = ℎ = 57 (cm) 0.63 (cm) = ∑ × × The requirement height of the weld at the column web ∑ Select hf= 0.6 cm. × × = 0.07 (cm) VIII> Detail of The base colum: No.6 1. Tính toán bản đế 1. Determine the base plate of column M= 106.48 (kN.m) V= 56.34 (kN) N= 1345.3 (kN) According to the dimension of column No.1, Choose 8 anchor bolts for base column Determin the dimension of base plate: Bbd = b + 2c1 = 37 (cm) The grade of concrete B20 1.15 (kN.m), = × ×∝= 1.15 × 1.2 × 1 = 1.38 (kN/cm2) , The length of base plate be determined follow the partial press of footing concrete = 2× ≥ × × + , 2× × × + , 2× 6× × × , 51.369 (cm) According to the design condition and distance of the anchor bolts, assume c2=12cm and thickness of base palte ts=1cm. So, determine the length of base plate 76 (cm) =h+2 +2 = Determine the stress reaction of footing concrete under the base plate: 6 = + = 0.777 (kN/cm2) × × 6 SVTH: Ngô Thanh Nguyên -172216544 Page: Project: Structural Steel Design × = ≤ Với − × × , × 6 × Instructor: Msc. Viet Hieu Pham = = 0.75 × 1.38 = 1.035 0.11 (kN/cm2) 2 Fig VIII.1 Detail of base column *According to the divide space of base plate and interpolated value, defind table: Table VIII.1 M1 = αbσd2 αb Ô sàn Bản kê b2 a2 b2/a2 (kN.cm) 0.06 20.764 Ô2 2 cạnh 9.9 21.2 0.467 0.091 30.589 Ô1 3 cạnh 18 24.5 0.735 The thickness of base plate be calculated follow the formular = 6× × = 2.956 (cm) Choose tbd 3 (cm) 2> Determine the reinforced plate at the base of column No.1 Select the dimension of the reinforced beam at the bas of column No.1 Thickness 1 (cm) Width: 37 (cm) The height: hdd follow the weld be connected the reinforced beam at base column and column No.1 The applied forced to the reinforced beam follow the stress reaction of footing concrete Ndd = (12 + 11.75) x 36.5 x 0.58 = 509.7 (kN) According the design condition, choose height of weld hf =1 Determine the design length of weld: = = 20.23 (cm) 2×ℎ × × × Select hdd= 22 (cm) 3> Determine the stiffener A The model calculation like is the congxon fixed to the column web qs = 0.73 x (24.5)= 17.89 (kN/cm) × 2897.37 (kN.cm) = = 2 V s = qs x l s = 321.93 (kN) Determine the height of stiffener A ℎ ≥ 6× × × = 28.77 (cm) Chọn hs = 30 (cm) Check follow the equivalent stress SVTH: Ngô Thanh Nguyên -172216544 Page: Project: Structural Steel Design = +3 Instructor: Msc. Viet Hieu Pham 6× ×ℎ = +3× ×ℎ = 22.3 (kN/cm2) = 23.2 (kN/cm2) 1.15 × × = 24.15 (kN/cm2) ậ < 1.15 ⟹ OK Following the design condition, select height of weld hf =1cm Area and Moment of weld Aw = 2 x 1 x (30-1) = 69.6 (cm) 1 × 30 − 1 280 (cm3) =2× = 6 Check the ability force of weld = + 11.3 (kN/cm2) = × = 12.6 (kN/cm2) ậ < × ⟹ OK 4> Determine the stiffnener B The width of the load tranfer to stiffener is 1.5ls=1.5x12=18cm qs = 0.78 x (18)= × = = 2 V s = qs x l s = 13.86 (kN/cm) 997.92 (kN.cm) 249.48 (kN) Determine the height of stiffener B ℎ ≥ 6× × × = 15.41 (cm) Select hs= 22 (cm) Check follow the equivalent stress = +3 = 6× ×ℎ +3× ×ℎ 1.15 × × = 24.15 (kN/cm2) ậ < 1.15 ⟹ OK Follwing the design condition, select height of weld hf =1cm Area and Moment of weld Aw = 2 x 1 x (24-1) = 42 (cm) 1 × 24 − 1 176 (cm3) =2× = 6 Check the ability force of weld = + = 8.2 (kN/cm2) × = 12.6 (kN/cm2) ậ < × ⟹ OK 5> Calculate the anchor bolt From Table V.1 we have M= 28.73 (kN.m) SVTH: Ngô Thanh Nguyên -172216544 Page: Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham V= -56.34 (kN) N= -1345 (kN) The length of the compression concrete zone under the base plate 76 (cm) Select the margin of the base plate to center of anchor bolt is 6cm, we have: = 2 − 3 = 11.6667 (cm) = 42.6667 (cm) − − 6 = 3 The total tension force of the anchor bolt: − × -300.53 (kN) = = Select the design anchor botl at the base column No,1 Select anchor bolt d22, As= 3.08 cm2, Table II.2, appendix II {3} Check again the total tension force = + = -672.29 (kN) 2 T2 < T1 The diameter of boltl is OK 6> Determine the weld be connected the column and the base plate The weld at the flange be braced moment and axial force The weld at the web be braced shear force The calculated internal force be followed the envelope internal force The tension force at the column web = ± = -612.82 (kN) ℎ 2 The total length of the line connect at column flange (include the weld be conneted the reinforced plate at base column and the base plate 33 − 1 21 − 1 33 − 21 =2× −1 +2× −1 +2× − 1 = 2 2 2 The requirement height of the weld at the column flange ℎ = ℎ = 57 (cm) 0.85 (cm) = ∑ × × The requirement height of the weld at the column web ∑ Select hf= 0.6 cm. × × SVTH: Ngô Thanh Nguyên -172216544 = 0.06 (cm) Page: [...]... of the weld at the column flange ℎ = ℎ = 57 (cm) 0.63 (cm) = ∑ × × The requirement height of the weld at the column web ∑ Select hf= 0.6 cm × × = 0.07 (cm) VIII> Detail of The base colum: No.6 1 Tính toán bản đế 1 Determine the base plate of column M= 106.48 (kN.m) V= 56.34 (kN) N= 1345.3 (kN) According to the dimension of column No.1, Choose 8 anchor bolts for base column Determin the dimension of base ... Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham GBCm > GABm Số TT Loại tải trọng Kết (daN) The self-weight of beam have gc = 37.1 (daN/m) -> 37.1(daN/m)x6/2(m) The self-weight of... the column web ∑ Select hf= 0.6 cm × × = 0.07 (cm) VIII> Detail of The base colum: No.6 Tính toán đế Determine the base plate of column M= 106.48 (kN.m) V= 56.34 (kN) N= 1345.3 (kN) According