Đồ án Kết cấu nhà thép - nhà dân dụng

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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: "Ƴ2 = 35 (daN/m2) - 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 ⇔ h= × 2 = 0.07 = 7 ( )≥ℎ = 5 ( ) 40 Choose:Thickness of slab 8 (cm) = 80 (mm) Student: Thanh Nguyen Ngo - 172216544 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 Factor of Factored Load No. Types of loading (daN/m2) Safety n (daN/m2) Layer of ceramic tile, t = 8 1 16 1.1 17.6 mm Layer of mortar, t = 15 mm 2 30 1.3 39 2000x0.015 The concrete slab, t = 80 mm 3 208 1.1 228.8 2500x0.08 285.4 -> gs (daN/m2) - Determine the Dead Load of roof slab: Table II.2 Dead Load of Slab Load Factor of Factored Load No. Types of loading (daN/m2) Safety n (daN/m2) Layer of the sealing, t =20 1 mm 40 1.3 52 2000x0.02 Layer of the insulation 10 2 20 1.3 26 mm The concrete slab, t = 80 mm 3 208 1.1 228.8 2500x0.08 306.8 -> gs (daN/m2) 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: = + ×2= 1171 (daN/m) = 11.71 (daN/cm) Student: Thanh Nguyen Ngo - 172216544 Page:..
Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham Factored loads: = × + × ×2= 1355.46 (daN/m) = 13.5546 (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 No. Types of load Factord Load (daN) The self-weight of beams have gc = 37.1 (daN/m) 1 219.6 -> (37.1(daN/m)x6/2)x2(m) The self-weight of wall that put on exterior beam : 2 3.3(m) -0.5(m) = 2.8 (m) 3024 -> (180(daN/m2)x2.8(m)x6/2(m))x2 The self-weight of slab with L = 6(m) 3 1712.4 -> (285.4(daN/m)x(6/2)x(2/2))x2(m) GA = 4956 Table IV.1.1a Student: Thanh Nguyen Ngo - 172216544 Page:..
Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham GB = GC No. Types of load Factord Load (daN) The self-weight of beams have gc = 37.1 (daN/m) 1 219.6 -> (37.1(daN/m)x6/2)x2(m) The self-weight of slab with L = 6(m) 2 3681.6 -> gs.(6/2(m))x(2/2)+gs.(6/2)x(2.3/2) GB = 3901.2 Table IV.1.1b GBC > GAB No. Types of load Factord Load (daN) The self-weight of beams have gc = 37.1 (daN/m) 1 219.6 -> (37.1(daN/m)x6/2)x2(m) The self-weight of gypsum partition tile with the hight of girder : 2 3.3(m) -0.5(m) = 2.8 (m) 107.80 -> (35(daN/m2)x2.8(m)x6./2(m))x2 The self-weight of slab with L = 6(m) 3 3938.52 -> gs.(6/2(m))x(2.3/2)+gs.(6/2)x(2.3/2) GBC = 4265.92 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 No. Types of load Factored Load The self-weight of beams have gc = 37.1 (daN/m) 1 219.6 -> (37.1(daN/m)x6/2)x2(m) 2 The self-weight of slab with L = 6(m) 1840.8 GAm = 2060.4 Table IV.1.2a GBm = GCm No. Types of load Factored Load The self-weight of beams have gc = 37.1 (daN/m) 1 219.6 -> (37.1(daN/m)x6/2)x2(m) The self-weight of slab with L = 6(m) 2 3983.52 -> gsm.(6/2(m))x(2/2)+gsm.(6/2)x(2.3/2) GBm = 4203.12 Table IV.1.2b Student: Thanh Nguyen Ngo - 172216544 Page:..
Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham GBCm > GABm Số TT Loại tải trọng Kết quả (daN) The self-weight of beam have gc = 37.1 (daN/m) 1 219.6 -> 37.1(daN/m)x6/2(m) The self-weight of slab with L = 6(m) 2 4261.44 -> gsm.(6/2(m))x(2.3/2)+gsm.(6/2)x(2.3/2) GBCm = 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. Types of load Factored Load 1 P1 = pc x 6x2/2x1.3 1560 P1 = 1560 P2 No. Types of load Factored Load P2 = pc x 6.x1x1x1.3x2 1 3120 -> 200(daN/m)x6(m)x1x1x1.3x2 P2 = 3120 P3 No. Types of load Factored Load P3 = pc x 6x1x2,3/2x1.3x2 1 2484 -> 300(daN/m)x6(m)x1.15x1/2x1.3x2 P3 = 2484 Student: Thanh Nguyen Ngo - 172216544 Page:..
Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham P4 No. Types of load Factored Load P3 = pc x 6x1x2.3x1.3x2 1 4968 -> 300(daN/m)x6(m)x2.3x1.3x2 P4 = 9936 2.2> Roof-Live Load 1 Table IV.2.2 Calculate Roof-Live Load 1 P1m No. Types of load Factored Load P1m = pc x 6x1x1/2x1.3 1 585 -> 75(daN/m)x6(m)x1x1/2x1.3 P1m = 585 P2m No. Types of load Factored Load P2m = pc x 6x1xx1.3x2 1 1170 -> 75(daN/m)x6(m)x1x1x1.3x2 P2m = 1170 Fig IV.2.1 Live Load 1 2.3> Live Load 2 Table IV.2.3 Calculate Live Load 2 P1 No. Types of load Factored Load (daN) P1 = pc x 6x2/2x1.3 1 1560 -> 200(daN/m)x6(m)x1x1/2x1.3 P1 = 1560 Student: Thanh Nguyen Ngo - 172216544 Page:..
Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham P2 No. Types of load Factored Load (daN) 1 P2 = pc x 6.x1x1x1.3x2 3120 P2 = 3120 P3 No. Types of load Factored Load (daN) 1 P3 = pc x 6x1x2,3/2x1.3x2 2484 P3 = 2484 P4 No. Types of load Factored Load (daN) 1 P3 = pc x 6x1x2.3x1.3x2 4968 P4 = 9936 2.4> Roof-Live Load 2 Table IV.2.4 Calculate Roof-Live Load 2 P3m No. Types of Load Factored Load(daN) 1 P3m = pc x 6x2,3/2x1.3x2 672.75 P3m = 672.75 P4m No. Types of Load Factored Load(daN) P3m = pc x 6x2,3x1.3x2 1 1345.5 -> 75(daN/m)x6(m)x2.3x1.3x2 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= × × × đ = × × × + đ= đ × 2+ = × 2 With W0 = 65 (daN/m2) n= 1.2 Cđ = 0.8 Ch = 0.6 + 6+6 = = 2 2 6 (m) 3.2> Calculate Wind Load Table IV.3.1 Calculate Wind Load Ht Z Wđ Wh qh Floors k qđ (daN/m) (m) (m) (daN/m2) (daN/m2) (daN/m) 1 4.2 4.2 0.848 52.92 39.69 317.4912 238.12 2 3.3 7.5 0.94 58.66 43.99 351.936 263.95 3 3.3 10.8 1.013 63.21 47.41 379.2672 284.45 4 3.3 14.1 1.066 66.52 49.89 399.1104 299.33 5 3.3 17.4 1.104 68.89 51.67 413.3376 310.00 Fig IV.3.1 Wind Left Student: Thanh Nguyen Ngo - 172216544 Page:..
Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham Fig IV.3.2 Wind Right ậ 386 ℎ ậly=μ128 lx=μ 1247 ℎ87 17ℎ 15 h ℎ251 ∑ ∑ 66 10 477 5542ả02 128 86 99 87 7 ả 0 25 Student: Thanh Nguyen Ngo - 172216544 Page:..
Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham V. THE COMBINATION OF INTERNAL FORCE TABLE V.1.1 INTERNAL FORCE OF ELEMENTS INTERNAL FORCE OF ELEMENTS Axial Force Shear Force Moment Name No. Position N Kn. V K.n Kn.m 0 Max -626.14 7.58 37.85 4.2 Max -626.14 -4.40 131.52 Column 1 0 Min -1017.55 -62.63 -112.63 4.2 Min -1017.55 -53.63 28.73 0 Max -498.38 -36.20 -76.77 3.3 Max -498.38 -47.82 153.42 Column 2 0 Min -805.19 -101.35 -168.10 3.3 Min -805.19 -93.51 61.86 0 Max -368.90 -32.67 -64.42 3.3 Max -368.90 -45.18 146.57 Column 3 0 Min -576.45 -90.49 -138.13 3.3 Min -576.45 -82.14 64.04 0 Max -237.63 -32.43 -70.59 3.3 Max -237.63 -44.26 134.55 Column 4 0 Min -363.25 -86.35 -136.68 3.3 Min -363.25 -81.01 58.16 0 Max -104.19 -61.87 -88.61 3.3 Max -104.19 -67.08 183.57 Column 5 0 Min -134.24 -103.11 -148.38 3.3 Min -134.24 -101.48 136.40 0 Max -706.23 56.34 106.48 4.2 Max -706.23 56.34 17.98 Column 6 0 Min -1345.34 -16.89 -52.98 4.2 Min -1345.34 -16.89 -130.17 0 Max -574.18 91.70 155.99 3.3 Max -574.18 91.70 -13.38 Column 7 0 Min -1066.05 11.80 25.55 3.3 Min -1066.05 11.80 -151.32 0 Max -439.86 75.86 124.57 3.3 Max -439.86 75.86 -22.06 Column 8 0 Min -771.92 15.55 26.73 3.3 Min -771.92 15.55 -135.05 0 Max -300.10 67.43 114.67 3.3 Max -300.10 67.43 -26.90 Column 9 0 Min -493.45 20.55 37.94 3.3 Min -493.45 20.55 -113.76 0 Max -155.24 74.14 120.17 3.3 Max -155.24 74.14 -90.28 Column 10 0 Min -199.58 47.90 58.18 3.3 Min -199.58 47.90 -135.74 Student: Thanh Nguyen Ngo- 172216544 Page:........
Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham TABLE V.1.2 INTERNAL FORCE OF ELEMENTS INTERNAL FORCE OF ELEMENTS Axial Force Shear Force Moment Name No. Position N Kn. V K.n Kn.m 0 Max -706.23 16.89 52.98 4.2 Max -706.23 16.89 130.16 Column 11 0 Min -1344.81 -56.35 -106.51 4.2 Min -1344.81 -56.35 -17.98 0 Max -574.18 -11.80 -25.55 3.3 Max -574.18 -11.80 151.32 Column 12 0 Min -1065.57 -91.72 -156.04 3.3 Min -1065.57 -91.72 13.38 0 Max -439.86 -15.55 -26.79 3.3 Max -439.86 -15.55 135.08 Column 13 0 Min -771.55 -75.89 -124.57 3.3 Min -771.55 -75.89 22.06 0 Max -300.10 -20.55 -37.94 3.3 Max -300.10 -20.55 113.76 Column 14 0 Min -493.19 -67.46 -114.72 3.3 Min -493.19 -67.46 26.90 0 Max -155.24 -47.90 -58.25 3.3 Max -155.24 -47.90 135.82 Column 15 0 Min -199.46 -74.18 -120.17 3.3 Min -199.46 -74.18 90.28 0 Max -626.14 62.63 112.61 4.2 Max -626.14 53.64 -28.73 Column 16 0 Min -980.61 -7.58 -37.85 4.2 Min -980.61 4.40 -131.56 0 Max -498.38 101.37 168.12 3.3 Max -498.38 93.53 -61.86 Column 17 0 Min -783.84 36.20 76.77 3.3 Min -783.84 47.82 -153.47 0 Max -368.90 90.52 138.16 3.3 Max -368.90 82.17 -64.04 Column 18 0 Min -555.07 32.67 64.42 3.3 Min -555.07 45.18 -146.63 0 Max -237.63 86.38 136.73 3.3 Max -237.63 81.04 -58.16 Column 19 0 Min -357.45 32.43 70.59 3.3 Min -357.45 44.26 -134.60 0 Max -104.19 103.16 148.38 3.3 Max -104.19 101.52 -136.40 Column 20 0 Min -128.41 61.87 88.66 3.3 Min -128.41 67.08 -183.68 Student: Thanh Nguyen Ngo- 172216544 Page:........
Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham TABLE V.1.3 INTERNAL FORCE OF ELEMENTS INTERNAL FORCE OF ELEMENTS Axial Force N Shear Force Moment Name No. Position Kn. V K.n Kn.m 0 Max 56.76 -77.99 -113.43 5 Max 56.76 8.23 168.88 10 Max 56.76 150.53 -127.07 Dầm 21 0 Min 28.25 -147.70 -289.39 5 Min 28.25 -6.43 89.88 10 Min 28.25 79.78 -300.73 0 Max 1.68 -79.72 -126.28 5 Max 1.68 6.50 165.42 10 Max 1.68 149.46 -134.08 Dầm 22 0 Min -18.16 -147.81 -291.56 5 Min -18.16 -5.49 85.60 10 Min -18.16 80.72 -297.13 0 Max 4.30 -81.50 -134.64 5 Max 4.30 4.72 166.56 10 Max 4.30 149.40 -141.58 Dầm 23 0 Min -12.75 -147.85 -282.33 5 Min -12.75 -3.73 86.93 10 Min -12.75 82.49 -288.10 0 Max 33.27 -83.67 -148.41 5 Max 33.27 2.65 164.09 10 Max 33.27 149.19 -151.32 Dầm 24 0 Min 9.66 -148.08 -276.99 5 Min 9.66 -2.16 85.57 10 Min 9.66 84.05 -282.72 0 Max -67.08 -83.37 -136.40 5 Max -67.08 3.14 135.18 10 Max -67.08 112.13 -157.05 Dầm 25 0 Min -101.48 -107.57 -183.57 5 Min -101.48 1.06 99.72 10 Min -101.48 87.37 -205.02 0 Max 11.71 1.63 12.85 2.3 Max 11.71 5.25 42.45 2.3 Max 11.71 66.79 42.45 4.6 Max 11.71 70.41 12.85 Dầm 26 0 Min 4.30 -70.43 -139.77 2.3 Min 4.30 -66.81 -18.02 2.3 Min 4.30 -5.25 -18.02 4.6 Min 4.30 -1.63 -139.67 0 Max -0.70 -2.38 9.45 2.3 Max -0.70 1.24 52.00 2.3 Max -0.70 63.14 52.00 4.6 Max -0.70 66.76 9.45 Dầm 27 0 Min -2.32 -66.83 -122.15 2.3 Min -2.32 -63.21 -11.99 2.3 Min -2.32 -1.24 -11.99 4.6 Min -2.32 2.38 -121.99 Student: Thanh Nguyen Ngo- 172216544 Page:........
Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham 0 Max 0.06 -9.59 -9.00 2.3 Max 0.06 -5.97 50.04 2.3 Max 0.06 56.68 50.04 4.6 Max 0.06 60.30 -9.00 Dầm 28 0 Min -2.02 -60.35 -109.54 2.3 Min -2.02 -56.73 -14.25 2.3 Min -2.02 5.97 -14.25 4.6 Min -2.02 9.59 -109.34 0 Max 8.35 -16.88 -21.49 2.3 Max 8.35 -13.25 52.99 2.3 Max 8.35 50.08 52.99 4.6 Max 8.35 53.70 -21.49 Dầm 29 0 Min 3.05 -53.80 -91.05 2.3 Min 3.05 -50.18 -9.59 2.3 Min 3.05 13.25 -9.59 4.6 Min 3.05 16.88 -90.82 0 Max -19.19 -23.74 -55.23 2.3 Max -19.19 -20.12 11.33 2.3 Max -19.19 30.70 11.33 4.6 Max -19.19 34.33 -55.23 Dầm 30 0 Min -27.57 -34.37 -79.81 2.3 Min -27.57 -30.75 -20.36 2.3 Min -27.57 20.12 -20.36 4.6 Min -27.57 23.74 -79.61 0 Max 56.76 -79.78 -127.07 5 Max 56.76 6.43 168.88 10 Max 56.76 147.71 -113.43 Dầm 31 0 Min 28.27 -150.51 -300.67 5 Min 28.27 -8.23 89.88 10 Min 28.27 77.99 -289.44 0 Max 1.68 -80.72 -134.08 5 Max 1.68 5.49 165.42 10 Max 1.68 147.81 -126.28 Dầm 32 0 Min -18.15 -149.44 -297.05 5 Min -18.15 -6.50 85.61 10 Min -18.15 79.72 -291.63 0 Max 4.30 -82.49 -141.58 5 Max 4.30 3.73 166.57 10 Max 4.30 147.87 -134.64 Dầm 33 0 Min -12.75 -149.38 -287.98 5 Min -12.75 -4.72 86.93 10 Min -12.75 81.50 -282.43 0 Max 33.27 -84.05 -151.32 5 Max 33.27 2.16 164.09 10 Max 33.27 148.08 -148.41 Dầm 34 0 Min 9.67 -149.16 -282.59 5 Min 9.67 -2.63 85.58 10 Min 9.67 83.67 -277.11 Student: Thanh Nguyen Ngo- 172216544 Page:........
Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham TABLE V.1.3 INTERNAL FORCE OF ELEMENTS INTERNAL FORCE OF ELEMENTS Axial Force N Shear Force Moment Name No. Position Kn. V K.n Kn.m 0 Max -67.08 -87.37 -157.05 5 Max -67.08 -1.06 135.20 10 Max -67.08 107.59 -136.40 Dầm 35 0 Min -101.52 -112.10 -204.88 5 Min -101.52 -3.12 99.72 10 Min -101.52 83.37 -183.68 Student: Thanh Nguyen Ngo- 172216544 Page:........
SAP2000 9/24/15 0:16:14 SAP2000 v16.0.0 - File:DATHEPNEW - Moment 3-3 Diagram (BAO) - KN, m, C Units
SAP2000 9/24/15 0:17:08 SAP2000 v16.0.0 - File:DATHEPNEW - Axial Force Diagram (BAO) - KN, m, C Units
SAP2000 9/24/15 0:16:49 SAP2000 v16.0.0 - File:DATHEPNEW - Shear Force 2-2 Diagram (BAO) - KN, m, C Units
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 : lx=μ×H=1×4.2= 4.2 (m) = 420 (cm) *The effective length with Minor Axis: ly=μ×H=0.7×4.2= 2.94 (m) = 294 (cm) * 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 = × 1.25 + 2.8 × = 91.85 (cm2) 21 × 1 62.3 *Determine bf, tf and tw: 1 1 Required: b= ÷ = 24 (cm) 20 30 *The thickness of the web be choose: 1 1 tw = ÷ ℎ ≥ 0.6 = 1.2 (cm) 60 120 *The thickness of the flange be choose: 21 tf ≥ × = 21 × = 0.66 (cm) 21000 tf ≥ = 1.2 (cm) => Choose tf = 1.4 (cm) *The dimension of column be choose: The flange: (1.4x24) cm The web : (1.2x45.2) 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 Instructor: Msc. Viet Hieu Pham A = 121.44 cm2 − = = 11.4 (cm) 2 ×ℎ ×ℎ = −2 = 45727.7248 (cm4) 12 12 ℎ × × = +2 = 3232.1088 (cm4) 12 12 = / = 19.4048 (cm) = / = 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 . 2 ℎụ ụ ó The value of interpolation = 0.607 Check left-side of expression: = = 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 = 56.988 < = 3.14 × √ = 99 = 1 = 0.7 = = 0.66941 1+ With: = 1.8021 < 2.5, we have equation: = 1 − 0.073 − 5.53 × × ̅× ̅= 0.83677 SVTH: Ngô Thanh Nguyên -172216544 Page: