Đồ án Kết cấu nhà thép – nhà dân dụng
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Đồ án Kết cấu nhà thép – nhà dân dụng
Nội dung Text: 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:
“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:..
- 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:..
- 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 LoadFig 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.1.1aStudent: 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.1.1c1.3> 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.1.2bStudent: 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 = 4481.04Fig 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 1P1
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 = 2484Student: 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 6x1x2.3×1.3×2
1 4968
-> 300(daN/m)x6(m)x2.3×1.3×2
P4 = 9936
2.2> Roof-Live Load 1Table 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 = 1170Fig IV.2.1 Live Load 1
2.3> Live Load 2Table 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 = 1560Student: 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.x1x1x1.3×2 3120
P2 = 3120P3
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 = 99362.4> 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 = 1345.5Fig 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= 1.2Cđ = 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 310.00Fig 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 25Student: 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 -135.74Student: 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 -183.68Student: 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 -121.99Student: 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 -277.11Student: 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 -183.68Student: 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: (1.4×24) cm
The web : (1.2×45.2) cmFig 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:
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