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Đồ án Kết cấu nhà thép – nhà dân dụng

Đồ án Kết cấu nhà thép – nhà dân dụng
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Đồ án Kết cấu nhà thép – nhà dân dụng

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Nội dung Text: Đồ án Kết cấu nhà thép – nhà dân dụng

  1. 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 2×6 (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:..

  2. 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
    2000×0.015
    The concrete slab, t = 80 mm
    3 208 1.1 228.8
    2500×0.08
    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
    2000×0.02
    Layer of the insulation 10
    2 20 1.3 26
    mm
    The concrete slab, t = 80 mm
    3 208 1.1 228.8
    2500×0.08
    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:..

  3. 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:..

  4. 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:..

  5. 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:..

  6. 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 6×2/2×1.3 1560
    P1 = 1560
    P2
    No. Types of load Factored Load
    P2 = pc x 6.x1x1x1.3×2
    1 3120
    -> 200(daN/m)x6(m)x1x1x1.3×2
    P2 = 3120
    P3
    No. Types of load Factored Load
    P3 = pc x 6x1x2,3/2×1.3×2
    1 2484
    -> 300(daN/m)x6(m)x1.15×1/2×1.3×2
    P3 = 2484

    Student: Thanh Nguyen Ngo – 172216544 Page:..

  7. Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham

    P4
    No. Types of load Factored Load
    P3 = pc x 6x1x2.3×1.3×2
    1 4968
    -> 300(daN/m)x6(m)x2.3×1.3×2
    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 6x1x1/2×1.3
    1 585
    -> 75(daN/m)x6(m)x1x1/2×1.3
    P1m = 585
    P2m
    No. Types of load Factored Load
    P2m = pc x 6x1xx1.3×2
    1 1170
    -> 75(daN/m)x6(m)x1x1x1.3×2
    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 6×2/2×1.3
    1 1560
    -> 200(daN/m)x6(m)x1x1/2×1.3
    P1 = 1560

    Student: Thanh Nguyen Ngo – 172216544 Page:..

  8. Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham

    P2
    No. Types of load Factored Load (daN)
    1 P2 = pc x 6.x1x1x1.3×2 3120
    P2 = 3120

    P3

    No. Types of load Factored Load (daN)
    1 P3 = pc x 6x1x2,3/2×1.3×2 2484
    P3 = 2484
    P4
    No. Types of load Factored Load (daN)
    1 P3 = pc x 6x1x2.3×1.3×2 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 6×2,3/2×1.3×2 672.75
    P3m = 672.75
    P4m
    No. Types of Load Factored Load(daN)
    P3m = pc x 6×2,3×1.3×2
    1 1345.5
    -> 75(daN/m)x6(m)x2.3×1.3×2
    P4m = 1345.5

    Fig IV.2.2 Live Load 2

    Student: Thanh Nguyen Ngo – 172216544 Page:..

  9. 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:..

  10. Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham

    Fig IV.3.2 Wind Right
    ậ 386

    ậly=μ128
    lx=μ 1247
    ℎ87
    17ℎ 15
    h ℎ251
    ∑ ∑
    66 10
    477
    5542ả02
    128
    86
    99 87
    7

    0 25

    Student: Thanh Nguyen Ngo – 172216544 Page:..

  11. 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:……..

  12. 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:……..

  13. 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:……..

  14. 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:……..

  15. 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:……..

  16. SAP2000 9/24/15 0:16:14

    SAP2000 v16.0.0 – File:DATHEPNEW – Moment 3-3 Diagram (BAO) – KN, m, C Units

  17. SAP2000 9/24/15 0:17:08

    SAP2000 v16.0.0 – File:DATHEPNEW – Axial Force Diagram (BAO) – KN, m, C Units

  18. SAP2000 9/24/15 0:16:49

    SAP2000 v16.0.0 – File:DATHEPNEW – Shear Force 2-2 Diagram (BAO) – KN, m, C Units

  19. 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: (1.4×24) cm
    The web : (1.2×45.2) 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:

  20. 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:

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