Tài liệu đồ án kết cấu nhà thép - nhà dân dụng

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: